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‫ و ارس‬ ‫ﺍﻟﻔﺼﻞ ‪ :1‬ﺍﳌﺘﺘﺎﻟﻴﺎﺕ‬

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‫‪ -1‬ﻣﻘﺪﻣﺔ‬

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‫‪ -2‬ﺗﻌﺎﺭﻳﻒ‬

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‫‪ -3‬ﻧﺘﺎﺋﺞ ﻭﺧﻮﺍﺹ‬

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‫‪ -4‬ﺗﻄﺒﻴﻖ‬

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‫‪ -5‬ﺧﻮﺍﺹ ﺃﺧﺮﻯ ﻟﻠﺪﺍﻟﺔ ﺍﻷﺳﻴﺔ‬

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‫‪ -6‬ﻣﻮﺿﻮﻉ ﻟﻠﺪﺭﺍﺳﺔ‬

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‫‪ -7‬ﲤﺎﺭﻳﻦ‬

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‫ﺍﻟﻔﺼﻞ ‪ :2‬ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(‬

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‫‪ -2‬ﻋﻤﻮﻣﻴﺎﺕ ﻋﻠﻰ ﺍﻟﺪﻭﺍﻝ‬

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‫‪ -3‬ﺍﻟﻨﻬﺎﻳﺎﺕ‬

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‫‪ -4‬ﺍﻻﺳﺘﻤﺮﺍﺭ‬

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‫‪ -5‬ﺍﻻﺷﺘﻘﺎﻕ‬

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‫ﺍﻟﻔﺼﻞ ‪:3‬ﺍﳊﺴﺎﺏ ﺍﻟﺘﻜﺎﻣﻠﻲ‬

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‫‪ -2‬ﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺮﳝﺎﱐ‬

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‫‪ -3‬ﺍﻟﺘﻜﺎﻣﻞ ﻏﲑ ﺍﶈﺪﺩ‬

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‫‪ -4‬ﻣﻦ ﻃﺮﻕ ﺍﳌﻜﺎﻣﻠﺔ‬

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‫‪ -5‬ﻃﻮﻝ ﻗﻮﺱ ﻣﻨﺤﲎ‬

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‫‪ -6‬ﺣﺴﺎﺏ ﺑﻌﺾ ﺍﳌﺴﺎﺣﺎﺕ ﻭﺍﳊﺠﻮﻡ‬

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‫ﺗﻘﺪﱘ ‪:‬‬ ‫ﹸﻛﺘِﺐ ﻫﺬﺍ ﺍﻟﺪﺭﺱ ﰲ ﺍﻟﺘﺤﻠﻴﻞ ﻭﻓﻖ ﺍﻟﱪﻧﺎﻣﺞ ﺍﳌﺴﻄﺮ ﻣﻦ ﻃﺮﻑ‬ ‫ﻭﺯﺍﺭﺓ ﺍﻟﺘﺮﺑﻴﺔ ﺍﻟﻮﻃﻨﻴﺔ ﺍﳍﺎﺩﻑ ﺇﱃ ﺗﻜﻮﻳﻦ ﻣﻔﺘﺸﻲ ﺍﻟﺘﻌﻠﻴﻢ ﺍﳌﺘﻮﺳﻂ ﰲ ﻣﺎﺩﺓ‬ ‫ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‪ .‬ﻭﻣﻦ ﺍﳌﻬﻢ ﺃﻥ ﻧﻼﺣﻆ ﻫﻨﺎ ﺑﺄﻥ ﺍﻟﱪﻧﺎﻣﺞ ﺍﻟﺮﲰﻲ ﻳﺸﻤﻞ ﰲ‬ ‫ﺍﻟﻮﺍﻗﻊ ﻭﺣﺪﺗﲔ‪ ،‬ﳘﺎ ‪ (1 :‬ﺍﳌﺘﺘﺎﻟﻴﺎﺕ‪ (2 ،‬ﺍﳊﺴﺎﺏ ﺍﻟﺘﻜﺎﻣﻠﻲ‪ .‬ﻏﲑ ﺃﻧﻪ ﻣﻦ‬ ‫ﺍﳌﻌﻠﻮﻡ ﺑﺄﻥ ﺍﻹﳌﺎﻡ ﲟﻔﻬﻮﻡ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﺳﻴﻤﺎ ﺃﻥ ﺍﻟﱪﻧﺎﻣﺞ ﻳﻄﻠﺐ‬ ‫ﻣﺜﻼ ﺗﻘﺪﱘ ﺍﳌﻜﺎﻣﻠﺔ ﺑﺎﻟﺘﺠﺰﺋﺔ( ﺿﺮﻭﺭﻱ ﻟﺘﻘﺪﱘ ﻭﺩﺭﺍﺳﺔ ﺍﻟﺘﻜﺎﻣﻼﺕ‪ .‬ﻭﻋﻠﻴﻪ‬ ‫ﺍﺭﺗﺄﻳﻨﺎ ﺗﻨﺎﻭﻝ ﻣﻮﺿﻮﻉ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ ﻗﺒﻞ ﺍﻟﺘﻜﺎﻣﻞ‪ .‬ﺫﻟﻚ ﻣﺎ ﻳﻌﺎﳉﻪ‬ ‫ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﱐ‪ .‬ﻭﺑﻄﺒﻴﻌﺔ ﺍﳊﺎﻝ ﻓﺎﻻﺳﺘﻤﺮﺍﺭ ﻻ ﳝﻜﻦ ﺗﻘﺪﳝﻪ ﻗﺒﻞ ﺗﻘﺪﱘ ﻣﻔﻬﻮﻡ‬ ‫ﺍﻟﻨﻬﺎﻳﺔ‪ .‬ﻭﻟﺬﺍ ﳚﺪ ﺍﻟﻘﺎﺭﺉ ﺣﺪﻳﺜﺎ ﻋﻦ ﺍﻟﺪﻭﺍﻝ ﻭﺍﻟﻨﻬﺎﻳﺎﺕ ﰲ ﺑﺪﺍﻳﺔ ﺍﻟﻔﺼﻞ‬ ‫ﺍﻟﺜﺎﱐ‪ .‬ﻭﻗﺪ ﺍﻋﺘﱪﻧﺎ ﻫﺬﺍ ﺍﻟﻔﺼﻞ ﻓﺼﻞ "ﻣﺮﺍﺟﻌﺔ"‪.‬‬ ‫ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‪ ،‬ﻳﺸﲑ ﺍﻟﱪﻧﺎﻣﺞ ﰲ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺜﺎﻧﻴﺔ ﺇﱃ ﺣﺴﺎﺏ‬ ‫ﻃﻮﻝ ﻗﻮﺱ ﻭﺣﺴﺎﺏ "ﺍﳌﺴﺎﺣﺎﺕ ﻭﺍﳊﺠﻮﻡ"‪ .‬ﻓﺄﻣﺎ ﺣﺴﺎﺏ ﻃﻮﻝ ﻗﻮﺱ‬ ‫ﻣﻨﺤﲎ ﻓﻬﺬﺍ ﻣﺎ ﻗﺪﻣﻨﺎﻩ ﰲ ﺍﳉﺰﺀ ﺍﻷﺧﲑ ﻣﻦ ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻟﺚ‪ .‬ﻭﻟﻌﻞ ﺍﻟﻘﺎﺭﺉ‬ ‫ﻳﻼﺣﻆ ﺃﻧﻨﺎ ﱂ ﻧﺘﻄﺮﻕ ﻟﻠﺪﺍﻟﺔ ﺍﻟﺸﻌﺎﻋﻴﺔ‪ ،‬ﻭﻫﻮ ﻣﻮﺿﻮﻉ ﻣﻬﻢ ﻻﺳﺘﻴﻌﺎﺏ‬ ‫ﻃﺮﻳﻘﺔ ﺗﻌﺮﻳﻒ ﺍﻟﻘﻮﺱ ﻭﺍﳌﻨﺤﲏ‪ ،‬ﺇﻻ ﺃﻥ ﺍﻟﱪﻧﺎﻣﺞ ﱂ ﻳﺸﺮ ﳍﺬﺍ ﺍﻟﻨﻮﻉ ﻣﻦ‬ ‫ﺍﻟﺪﻭﺍﻝ‪.‬‬ ‫ﻭﰲ ﻣﺎ ﳜﺺ ﺣﺴﺎﺏ ﺍﳌﺴﺎﺣﺎﺕ‪ ،‬ﻓﺈﺫﺍ ﻛﺎﻧﺖ ﺍﳌﺴﺎﺣﺔ ﺍﳌﺸﺎﺭ ﺇﻟﻴﻬﺎ‬ ‫ﻣﺴﺎﺣﺔ ﺳﻄﺢ ﻣﺴﺘﻮ ﳏﺪﻭﺩ ﺑﺒﻴﺎﻥ ﺩﺍﻟﺔ ﻓﺤﺴﺎ‪‬ﺎ ﻳﺄﰐ ﻣﺒﺎﺷﺮﺓ ﻣﻦ ﺗﻜﺎﻣﻞ‬ ‫‪7‬‬

‫ﺭﳝﺎﻥ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺿﻤﻦ ﺍﻟﺪﺭﺱ‪ .‬ﻟﻜﻦ ﺣﺴﺎﺏ ﻣﺴﺎﺣﺔ ﺳﻄﺢ ﻏﲑ‬ ‫ﻣﺴﺘﻮ )ﰲ ﺍﻟﻔﻀﺎﺀ( ﻭﺣﺴﺎﺏ ﺍﳊﺠﻮﻡ ﻳﺘﻄﻠﺐ ﺇﺩﺧﺎﻝ ﺍﻟﺘﻜﺎﻣﻼﺕ ﺍﳌﻀﺎﻋﻔﺔ‬ ‫)ﺍﻟﺜﻨﺎﺋﻴﺔ ﻭﺍﻟﺜﻼﺛﻴﺔ‪ (... ،‬ﺑﻜﺜﲑ ﻣﻦ ﺍﻟﺘﻔﺎﺻﻴﻞ‪ .‬ﺇﻻﹼ ﺃﻧﻨﺎ ﱂ ﳒﺪ ﺇﺷﺎﺭﺓ ﰲ‬ ‫ﺍﻟﱪﻧﺎﻣﺞ ﺇﱃ ﻫﺬﺍ ﺍﻟﻨﻮﻉ ﻣﻦ ﺍﻟﺘﻜﺎﻣﻼﺕ‪ .‬ﻓﻜﻴﻒ ﺑﻨﺎ ﻧﻘﺪﻡ ﻫﺬﺍ ﺍﳌﻮﺿﻮﻉ‬ ‫ﺑﺪﻭﻥ ﺍﻷﺩﻭﺍﺕ ﺍﻟﻼﺯﻣﺔ؟‬ ‫ﰒ ﺇﻥ ﺍﻟﱪﻧﺎﻣﺞ ﻳﺸﲑ ﺇﱃ ﺍﳌﺪﺓ ﺍﳌﺨﺼﺼﺔ ﳌﺎﺩﺓ ﺍﻟﺘﺤﻠﻴﻞ ﻓﻴﺤﺪﺩﻫﺎ‬ ‫ﺑـ ‪ 48‬ﺳﺎﻋﺔ ! ﻭﻫﻲ ﻣﺪﺓ ﻻ ﺗﺴﻤﺢ ﺑﺎﻟﺘﻄﺮﻕ ﻟﺘﻠﻚ ﺍﻷﺩﻭﺍﺕ‪ .‬ﻟﺬﻟﻚ ﻛﻠﻪ‬ ‫ﺍﻛﺘﻔﻴﻨﺎ ﻫﻨﺎ ﺑﺘﻘﺪﱘ ﻣﺴﺎﺣﺎﺕ ﻭﺣﺠﻮﻡ ﺍﻷﺷﻜﺎﻝ ﻭﺍ‪‬ﺴﻤﺎﺕ ﺍﳌﺄﻟﻮﻓﺔ‪ ،‬ﻣﺜﻞ‬ ‫ﺍﳌﺨﺮﻭﻁ ﻭﺍﻻﺳﻄﻮﺍﻧﺔ ﻭﺍﻟﻜﺮﺓ‪ ،‬ﺩﻭﻥ ﺗﻔﺼﻴﻞ ﺍﻟﻄﺮﻕ ﺍﳊﺴﺎﺑﻴﺔ ﺍﳌﺴﺘﺨﺪﻣﺔ‬ ‫ﻟﻠﺘﻜﺎﻣﻼﺕ‪.‬‬ ‫ﻧﻮﺻﻲ ﺃﻥ ﺗﻘﺪﻡ "ﺍﻟﺜﻐﺮﺍﺕ" ﺍﳌﻮﺟﻮﺩﺓ ﰲ ﺍﻟﱪﻧﺎﻣﺞ ﻛﻤﻮﺍﺿﻴﻊ‬ ‫ﺩﺭﺍﺳﺔ ﻟﻠﻄﻠﺒﺔ ﺍﳌﻔﺘﺸﲔ‪ .‬ﻭﻣﻦ ﺗﻠﻚ ﺍﳌﻮﺍﺿﻴﻊ ﻧﻘﺘﺮﺡ ‪ (1 :‬ﺧﻮﺍﺹ ﺍﻟﺪﻭﺍﻝ‬ ‫ﺍﻟﺸﻌﺎﻋﻴﺔ‪ (2 ،‬ﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺜﻨﺎﺋﻲ )ﺗﻌﺮﻳﻔﻪ ﻭﺧﻮﺍﺻﻪ(‪ (3 ،‬ﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺜﻼﺛﻲ‬ ‫)ﺗﻌﺮﻳﻔﻪ ﻭﺧﻮﺍﺻﻪ(‪ (4 ،‬ﺍﺳﺘﻌﻤﺎﻝ ﺍﻟﺘﻜﺎﻣﻼﺕ ﳊﺴﺎﺏ ﻣﺴﺎﺣﺎﺕ ﺍﻟﺴﻄﻮﺡ‬ ‫ﻏﲑ ﺍﳌﺴﺘﻮﻳﺔ‪ (5 ،‬ﺍﺳﺘﻌﻤﺎﻝ ﺍﻟﺘﻜﺎﻣﻼﺕ ﺍﳌﻀﺎﻋﻔﺔ ﳊﺴﺎﺏ ﺍﳊﺠﻮﻡ‪ ،‬ﺍﱁ‪.‬‬ ‫ﻧﺘﻤﲎ ﺃﻥ ﻳﻜﻮﻥ ﻫﺬﺍ ﺍﻟﺪﺭﺱ ﻣﻔﻴﺪﺍ ﻟﻠﻄﺎﻟﺐ ﺍﳌﻔﺘﺶ‪.‬‬ ‫ﺃﺑﻮ ﺑﻜﺮ ﺧﺎﻟﺪ ﺳﻌﺪ ﺍﷲ‬ ‫ﻗﺴﻢ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‬ ‫ا ر ا ة‪ ،‬ا ‪ ،‬ا ا ‬ ‫‪8‬‬

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‫‪ .1‬ﻣﻘﺪﻣﺔ ‪:‬‬ ‫ﻻ ﺷﻚ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﺗﻌﺘﱪ ﻣﻦ ﺃﻫﻢ‪ ‬ﺍﻷﺩﻭﺍﺕ ﺍﳌﺴﺘﺨﺪﻣﺔ ﰲ ﻛﺎﻓﺔ‬ ‫ﺍﻟﱪﺍﻫﲔ ﺍﻟﻮﺍﺭﺩﺓ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﲜﻤﻴﻊ ﻓﺮﻭﻋﻬﺎ‪ .‬ﺫﻟﻚ ﺃ‪‬ﺎ ﺗﺘﻤﻴ‪‬ﺰ ﺑﺼﻔﺔ‬ ‫"ﺍﻟﺘﻘﻄﻊ" ﺍﻟﻨﺎﲨﺔ ﻋﻦ ﺍﺭﺗﺒﺎﻁ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﺑﺎﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ ﺍﻟﱵ ﻳﺪﺭﻛﻬﺎ‬ ‫ﻓﻜﺮﻧﺎ ﺃﻛﺜﺮ ﳑﺎ ﻳﺪﺭﻙ ﺍﻷﻋﺪﺍﺩ ﺍﻷﺧﺮﻯ ﻛﺎﻷﻋﺪﺍﺩ ﺍﳉﺬﺭﻳﺔ ﺃﻭ ﺍﳊﻘﻴﻘﻴﺔ ﺃﻭ‬ ‫ﺍﳌﺮﻛﺒﺔ )ﺍﻟﻌﻘﺪﻳﺔ(‪ .‬ﻭﻟﻌﻞ ﺃﻓﻀﻞ ﺩﻟﻴﻞ ﻋﻠﻰ ﺫﻟﻚ ﻇﻬﻮﺭ ﻭﺍﺳﺘﺨﺪﺍﻡ ﺍﻷﻋﺪﺍﺩ‬ ‫ﺍﻟﻄﺒﻴﻌﻴﺔ ﻗﺒﻞ ﺳﺎﺋﺮ ﺃﳕﺎﻁ ﺍﻷﻋﺪﺍﺩ ﺍﻷﺧﺮﻯ‪.‬‬ ‫ﻭﺍﳌﻼﺣﻆ ﺃﻥ ﺍﻟﺮﻳﺎﺿﻴﲔ ﻟﻴﺴﻮﺍ ﺍﻟﻮﺣﻴﺪﻳﻦ ﺍﻟﺬﻳﻦ ﻳﻔﻀﻠﻮﻥ ﺍﻟﻌﻤﻞ‬ ‫ﺑﺎﳌﺘﺘﺎﻟﻴﺎﺕ ﺑﺪﻝ ﺍﻷﺩﻭﺍﺕ ﺍﻷﺧﺮﻯ )ﻛﺎﻟﺪﻭﺍﻝ ﻣﺜﻼ(‪ .‬ﺃﻧﻈﺮ ﺇﱃ ﺍﳉﻐﺮﺍﻓﻴﲔ‬ ‫ﻭﺍﻹﺣﺼﺎﺋﻴﲔ ﻭﺍﻟﻔﻴﺰﻳﺎﺋﻴﲔ ﻭﻏﲑﻫﻢ ﻣﻦ ﺍﻟﻌﺎﻣﻠﲔ ﰲ ﺣﻘﻮﻝ ﺍﳌﻌﺮﻓﺔ ﺍﳌﺨﺘﻠﻔﺔ‬ ‫‪ ...‬ﺇ‪‬ﻢ ﲨﻴﻌﺎ ﻳﺴﺘﺨﺪﻣﻮﻥ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﻭﻻ ﻳﻠﺠﺆﻭﻥ ﺇﱃ ﺍﻟﺪﻭﺍﻝ ﺇﻻ ﻋﻨﺪ‬ ‫ﺍﻟﻀﺮﻭﺭﺓ‪ .‬ﻭﻣﻦ ﱂ ﻳﺴﻤﻊ ﻣﺜﻼ ﲟﺘﺘﺎﻟﻴﺔ ﻓﻴﺒﻮﻧﺎﺗﺸﻲ ‪Fibonacci‬‬ ‫)‪ ،(1250-1170‬ﺃﻱ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﺃﻱ ﻋﻨﺼﺮ ﻣﻨﻬﺎ ﻳﺴﺎﻭﻱ‬ ‫ﳎﻤﻮﻉ ﺍﻟﻌﻨﺼﺮﻳﻦ ﺍﻟﺴﺎﺑﻘﲔ ﻟﻪ‪ ،‬ﻣﻊ ﺍﻟﻌﻠﻢ ﺃﻥ ﺍﻟﻌﻨﺼﺮﻳﻦ ﺍﻷﻭﻝ ﻭﺍﻟﺜﺎﱐ‬ ‫ﻣﻌﻠﻮﻣﺎﻥ(؟ ﺇ‪‬ﺎ ﻣﺘﺘﺎﻟﻴﺔ ﺗﺪﺧﻞ ﰲ ﺗﻮﺯﻳﻊ ﻭﺗﻨﻈﻴﻢ ﻣﻮﺍﻗﻊ ﻭﺭﻕ ﺑﻌﺾ ﺍﻟﻨﺒﺎﺗﺎﺕ‬ ‫ﺣﻮﻝ ﺍﻷﻏﺼﺎﻥ‪ .‬ﻭﺍﻷﻏﺮﺏ ﻣﻦ ﺫﻟﻚ ﺃﻥ ﻫﺬﺍ ﺍﻟﺘﻮﺯﻳﻊ ﻳﻀﻤﻦ ﻭﺻﻮﻝ ﺃﺷﻌﺔ‬ ‫ﺍﻟﺸﻤﺲ ﺑﺄﻛﱪ ﻗﺪﺭ ﳑﻜﻦ ﺇﱃ ﺃﻭﺭﺍﻕ ﻫﺬﻩ ﺍﻟﻨﺒﺎﺗﺎﺕ‪ .‬ﻭﻗﺪ ﺃﺛﺒﺖ ‪R.‬‬ ‫‪ Jones‬ﻋﺎﻡ ‪ 1975‬ﺑﺄﻥ ﻋﻨﺎﺻﺮ ﻫﺬﻩ ﺍﳌﺘﺘﺎﻟﻴﺔ ﲤﺜﻞ ﺟﺬﻭﺭﺍ ﻟﻜﺜﲑﺍﺕ‬ ‫ﺣﺪﻭﺩ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ﺍﳋﺎﻣﺴﺔ‪.‬‬ ‫‪11‬‬

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‫ﻛﻤﺎ ﺃﻥ ﳍﺬﻩ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﺻﻠﺔ ﺑﻘﺎﻧﻮﻥ ﺗﻮﺍﻟﺪ ﺑﻌﺾ ﺍﳊﻴﻮﺍﻧﺎﺕ‬ ‫ﻛﺎﻷﺭﺍﻧﺐ‪ .‬ﻭﻣﻦ ﺍﳌﻌﻠﻮﻡ ﺃﻥ ﻓﻴﺒﻮﻧﺎﺗﺸﻲ ﺃﺛﺒﺖ ﺃﻥ ﻣﺘﺘﺎﻟﻴﺘﻪ ﲤﺜﻞ ﺣﻼ ﻟﻠﻤﺴﺄﻟﺔ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ ‪ :‬ﻛﻢ ﺯﻭﺟﺎ ﻣﻦ ﺍﻷﺭﺍﻧﺐ ﳝﻜﻦ ﺍﳊﺼﻮﻝ ﻋﻠﻴﻬﺎ ﺧﻼﻝ ﺳﻨﺔ ﻋﻨﺪﻣﺎ‬ ‫ﻳﻜﻮﻥ ﻟﻨﺎ ﰲ ﺍﻟﺒﺪﺍﻳﺔ ﺯﻭﺝ ﻭﺍﺣﺪ ﻭﺇﺫﺍ ﻋﻠﻤﻨﺎ ﺃﻥ ﻛﻞ ﺯﻭﺝ ﻳﻠﺪ ﺯﻭﺟﺎ ﺁﺧﺮ‬ ‫ﻛﻞ ﺷﻬﺮ؟‬

‫ﺃﻳﻦ ﳒﺪ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ؟ ﺇ‪‬ﺎ ﻣﻮﺟﻮﺩﺓ ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﳌﺜﺎﻝ‬ ‫ﰲ‪:‬‬ ‫ ﻣﻔﻬﻮﻡ ﺍﻟﻜﺜﺎﻓﺔ ‪ :‬ﻛﺜﺎﻓﺔ ﳎﻤﻮﻋﺔ ﺟﺰﺋﻴﺔ ﻣﻦ ﻓﻀﺎﺀ ﻃﺒﻮﻟﻮﺟﻲ ﰲ‬‫ﻧﻔﺲ ﺍﻟﻔﻀﺎﺀ ﺃﻭ ﻓﻀﺎﺀ ﺁﺧﺮ‪ .‬ﻓﺄﻧﺖ ﺇﺫﺍ ﺃﺭﺩﺕ ﻣﺜﻼ ﺇﺛﺒﺎﺕ ﻣﺴﺎﻭﺍﺓ ﺃﻭ‬ ‫ﻣﺘﺒﺎﻳﻨﺔ ﰲ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﳊﻘﻴﻘﻴﺔ ﻳﻜﻔﻴﻚ ﰲ ﺃﻏﻠﺐ ﺍﻷﺣﻴﺎﻥ ﺃﻥ ﺗﺜﺒﺘﻬﺎ ﰲ‬ ‫ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻨﺎﻃﻘﺔ‪ ،‬ﻭﻫﺬﺍ ﺑﻔﻀﻞ ﻛﺜﺎﻓﺔ ﻫﺬﻩ ﺍ‪‬ﻤﻮﻋﺔ ﺍﻷﺧﲑﺓ ﰲ‬ ‫ﳎﻤﻮﻋﺔ ﺍﻟﻌﺪﺍﺩ ﺍﳊﻘﻴﻘﻴﺔ‪.‬‬ ‫ ﺩﺭﺍﺳﺔ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ‪ :‬ﳓﺼﻞ ﻋﻠﻰ ﺣﻠﻮﻝ ﻫﺬﻩ‬‫ﺍﳌﻌﺎﺩﻻﺕ ﰲ ﺍﻟﻜﺜﲑ ﻣﻦ ﺍﻷﺣﻴﺎﻥ ﻛﻨﻬﺎﻳﺎﺕ ﻣﺘﺘﺎﻟﻴﺎﺕ ﺗﻘﺮﺑﻨﺎ ﺷﻴﺌﺎ ﻓﺸﻴﺌﺎ ﻣﻦ‬ ‫ﺍﳊﻞ ﺍﻟﺪﻗﻴﻖ‪.‬‬ ‫ ﺍﳊﺴﺎﺏ )ﺃﻭ ﺍﻟﺘﺤﻠﻴﻞ( ﺍﻟﻌﺪﺩﻱ ‪ :‬ﺍﻟﺘﻘﺮﻳﺒﺎﺕ ﻭﺗﻘﺪﻳﺮﺍﺕ‬‫ﺍﻷﺧﻄﺎﺀ ﺗﺘﻢ ﻋﻤﻮﻣﺎ ﻋﱪ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ‪.‬‬ ‫ ﺗﻌﺮﻳﻒ ﻣﻔﺎﻫﻴﻢ ﺭﻳﺎﺿﻴﺔ ﺃﺧﺮﻯ ‪ :‬ﺍﻻﻧﺘﻘﺎﻝ ﻣﺜﻼ ﻣﻦ ﺗﻌﺮﻳﻒ‬‫ﻣﻔﻬﻮﻡ ﺍﳌﻜﺎﻣﻠﺔ ﻟﻠﺪﺍﻟﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﳎﺎﻝ ﺣﻘﻴﻘﻲ ﻭﺗﺄﺧﺬ ﻗﻴﻤﻬﺎ ﰲ ﻓﻀﺎﺀ ﳎﺮﺩ‬ ‫‪12‬‬

‫ا ‪ : 1‬ا ت‬

‫ ﻓﻀﺎﺀ ﺑﺎﻧﺎﺧﻲ ‪ (1945-1892) Banach‬ﻣﺜﻞ ‪ - ℝ n‬ﳝﺮ ﻋﱪ‬‫ﺍﳌﺘﺘﺎﻟﻴﺎﺕ‪.‬‬ ‫ﻭﻣﻦ ﺍﻟﺘﻄﺒﻴﻘﺎﺕ ﺍﻟﱵ ﳒﺪﻫﺎ ﰲ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﺃ‪‬ﺎ ﲤﻜﻦ ﻣﻦ ﺗﻌﺮﻳﻒ‬ ‫ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺄﻟﻮﻓﺔ ﻣﺜﻞ‬ ‫ ﺍﻟﺪﺍﻟﺔ ﺍﻷﺳﻴﺔ‪،‬‬‫ ﺍﻟﺪﺍﻟﺔ ﺍﳌﺜﻠﺜﻴﺔ ﺟﺐ‪،‬‬‫ ﺍﻟﺪﺍﻟﺔ ﺍﳌﺜﻠﺜﻴﺔ ﲡﺐ‪،‬‬‫ ﺍﻟﺪﺍﻟﺔ ﺍﻟﻠﻮﻏﺎﺭﻳﺘﻤﻴﺔ )ﺑﻮﺻﻔﻬﺎ ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ ﺍﻷﺳﻴﺔ(‪،‬‬‫ ﺍﻟﺪﺍﻟﺔ ﺍﳌﺜﻠﺜﻴﺔ ﻇﻞ )ﺑﻮﺻﻔﻬﺎ ﻧﺴﺒﺔ ﻟﻠﺪﺍﻟﺘﲔ ﺍﳌﺜﻠﺜﻴﺘﲔ ﺟﺐ ﻭ‬‫ﲡﺐ(‪.‬‬ ‫ﻭﻟﻌﻠﻪ ﻣﻦ ﺍﳌﻔﻴﺪ ﺃﻥ ﻧﺘﺴﺎﺀﻝ ﻫﻨﺎ ‪ :‬ﻫﻞ ﻣﻦ ﺍﳌﺆﻛﺪ ﺃﻥ ﻣﻔﻬﻮﻡ ﺍﻟﺪﺍﻟﺔ‬ ‫)ﺃﻭ ﺍﻟﺘﻄﺒﻴﻖ( ﺃﻗﺮﺏ ﻟﺬﻫﻦ ﺍﻟﺘﻠﻤﻴﺬ ﻣﻦ ﻣﻔﻬﻮﻡ ﺍﳌﺘﺘﺎﻟﻴﺔ؟ ﺇﺫﺍ ﻛﺎﻥ ﺍﳉﻮﺍﺏ‬ ‫ﺑﺎﻟﻨﻔﻲ‪ ،‬ﻓﻠﻤﺎﺫﺍ ﻳﺘﺄﺧ‪‬ﺮ ﺗﻘﺪﱘ ﻫﺬﺍ ﺍﳌﻔﻬﻮﻡ ﻟﻠﺘﻠﻤﻴﺬ ﻣﻘﺎﺭﻧﺔ ﲟﻔﻬﻮﻡ ﺍﻟﺪﺍﻟﺔ؟‬

‫‪13‬‬

‫ا ‪ : 1‬ا ت‬

‫‪ .2‬ﺗﻌﺎﺭﻳﻒ ‪:‬‬ ‫ﺗﻌﺮﻳﻒ )ﺍﳌﺘﺘﺎﻟﻴﺔ(‬ ‫ ﻳﺴﻤﻰ ﻛﻞ ﺗﻄﺒﻴﻖ ﻣﻦ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ ﰲ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ‬‫ﺍﳊﻘﻴﻘﻴﺔ ﻣﺘﺘﺎﻟﻴﺔ ﺣﻘﻴﻘﻴﺔ‪.‬‬ ‫ ﻳﺴﻤﻰ ﻛﻞ ﺗﻄﺒﻴﻖ ﻣﻦ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ ﰲ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ‬‫ﺍﳌﺮﻛﺒﺔ ﻣﺘﺘﺎﻟﻴﺔ ﻣﺮﻛﺒﺔ‪.‬‬ ‫ ﻳﺴﻤﻰ ﻛﻞ ﺗﻄﺒﻴﻖ ﻣﻦ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ ﰲ ﳎﻤﻮﻋﺔ ﺃﻋﺪﺍﺩ‬‫)ﻃﺒﻴﻌﻴﺔ ﺃﻭ ﺻﺤﻴﺤﺔ ﺃﻭ ﻧﺎﻃﻘﺔ ﺃﻭ ﺣﻘﻴﻘﻴﺔ ﺃﻭ ﻣﺮﻛﺒﺔ( ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ‪.‬‬ ‫ ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ‬‫ﺇﺫﺍ ﻛﺎﻥ‬

‫ﺣﻘﻴﻘﻴﺔ )‪(un‬‬

‫‪un ≥un +1 ) un ≤un +1‬‬

‫ﺇ‪‬ﺎ ﻣﺘﺰﺍﻳﺪﺓ )ﻣﺘﻨﺎﻗﺼﺔ‪ ،‬ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ(‬

‫‪ ،‬ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ( ﻣﻦ ﺃﺟﻞ ﻛﻞ‬

‫‪n‬‬

‫ﰲ‬

‫ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ‪.‬‬ ‫‪ -‬ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ‬

‫ﺣﻘﻴﻘﻴﺔ ‪(un ) n∈ℕ‬‬

‫ﺇ‪‬ﺎ ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﻋﻠﻰ ﺇﺫﺍ ﻭﺟﺪ‬

‫ﺛﺎﺑﺖ ‪ M‬ﻣﻮﺟﺐ ﲝﻴﺚ‬ ‫‪∀n ∈ ℕ, un ≤ M .‬‬ ‫ ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﺣﻘﻴﻘﻴﺔ ‪ (un )n∈ℕ‬ﺇ‪‬ﺎ ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﺩﱏ ﺇﺫﺍ ﻭﺟﺪ ﺛﺎﺑﺖ‬‫‪M‬‬

‫ﲝﻴﺚ‬

‫‪.‬‬ ‫ ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ ‪ (un )n∈ℕ‬ﺇ‪‬ﺎ ﳏﺪﻭﺩﺓ ﺇﺫﺍ ﻭﺟﺪ ﺛﺎﺑﺖ‬‫‪∀n ∈ ℕ, un ≥ M‬‬

‫ﻣﻮﺟﺐ ﲝﻴﺚ‬ ‫‪.‬‬

‫‪∀n ∈ ℕ, un ≤ M‬‬ ‫‪14‬‬

‫‪M‬‬

‫ا ‪ : 1‬ا ت‬

‫ﻣﻼﺣﻈﺎﺕ ‪:‬‬ ‫‪ .1‬ﳓﺎﻓﻆ ﻋﻠﻰ ﻣﺼﻄﻠﺢ ﻣﺘﺘﺎﻟﻴﺔ ﺣﱴ ﻟﻮ ﺍﻧﻄﻠﻖ ﺍﻟﺘﻄﺒﻴﻖ ﻣﻦ ﺟﺰﺀ‬ ‫ﻣﻦ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ‪ .‬ﻭﻧﺸﲑ ﻋﺎﺩﺓ ﺍﱃ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﺑﺮﻣﺰ‪ ،‬ﻣﺜﻞ‬ ‫)‪ ، (un‬ﺩﻭﻥ ﺍﻹﺷﺎﺭﺓ ﺇﱃ ﳎﻤﻮﻋﺔ ﺗﻌﺮﻳﻔﻬﺎ ﺇﻻ ﺇﺫﺍ ﺩﻋﺖ ﺍﻟﻀﺮﻭﺭﺓ ﺇﱃ‬ ‫ﺫﻟﻚ‪.‬‬ ‫‪ .2‬ﳝﻜﻦ ﺇﺛﺒﺎﺕ ﺃﻥ ﻣﺘﺘﺎﻟﻴﺔ‬

‫ﺗﻜﻮﻥ ﳏﺪﻭﺩﺓ ﺇﺫﺍ‬

‫ﺣﻘﻴﻘﻴﺔ ‪(un )n∈ℕ‬‬

‫ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ ﳏﺪﻭﺩﺓ ﰲ ﺁﻥ ﻭﺍﺣﺪ ﻣﻦ ﺍﻷﻋﻠﻰ ﻭﻣﻦ ﺍﻷﺩﱏ‪.‬‬ ‫ﺳﻮﻑ ﻧﻌﺘﱪ‪ ،‬ﰲ ﻣﺎ ﻳﻠﻲ‪ ،‬ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﻛﺎﻣﻞ‬

‫‪.3‬‬

‫ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ ﺇﻻ ﺇﺫﺍ ﺃﺷﺮﻧﺎ ﺇﱃ ﻋﻜﺲ ﺫﻟﻚ‪.‬‬ ‫ار ‪ :‬ﻻ ﳚﻮﺯ ﺃﻥ ﻧﺘﺤﺪﺙ ﻋﻦ ﺭﺗﺎﺑﺔ ﻣﺘﺘﺎﻟﻴﺔ ﺇﺫﺍ ﺃﺧﺬﺕ ﻗﻴﻤﺎ ﻋﻘﺪﻳﺔ‬ ‫)ﻣﺮﻛﺒﺔ( ﻏﲑ ﺣﻘﻴﻘﻴﺔ ‪ ...‬ﺫﻟﻚ ﺃﻥ ﺍ‪‬ﻤﻮﻋﺔ‬

‫‪ℂ‬‬

‫ﻏﲑ ﻣﺮﺗﺒﺔ ﺗﺮﺗﻴﺒﺎ‬

‫"ﻃﺒﻴﻌﻴﺎ" !‬ ‫ﺃﻣﺜﻠﺔ ‪:‬‬ ‫‪ (1‬ﺇﻟﻴﻚ ﺑﻌﺾ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﱵ ﺗﻌﺮﻑ ﻣﺘﺘﺎﻟﻴﺎﺕ‬ ‫‪un = n‬‬

‫‪،‬‬

‫‪1‬‬ ‫‪n+2‬‬

‫= ‪un‬‬

‫‪،‬‬

‫‪n2‬‬ ‫‪، un = − n‬‬ ‫‪3‬‬ ‫‪n +1‬‬ ‫‪n+2‬‬

‫‪15‬‬

‫‪un = (−1) n :‬‬

‫‪. un = (−1)n i +‬‬

‫‪،‬‬

‫ا ‪ : 1‬ا ت‬

‫‪ (2‬ﺣﻮﻝ ﺍﳌﺘﺘﺎﻟﻴﺘﲔ ﺍﳊﺴﺎﺑﻴﺔ ﻭﺍﳍﻨﺪﺳﻴﺔ ‪ :‬ﺗﻌﺮﻑ ﺍﳌﺘﺘﺎﻟﻴﺘﺎﻥ‬ ‫ﺍﳊﺴﺎﺑﻴﺔ ﻭﺍﳍﻨﺪﺳﻴﺔ ﺑﺎﻟﻌﻼﻗﺘﲔ ‪:‬‬ ‫*‪n ∈ ℕ‬‬

‫)ﻣﺘﺘﺎﻟﻴﺔ ﺣﺴﺎﺑﻴﺔ ﺃﺳﺎﺳﻬﺎ ‪ ،( r‬ﻭ‬

‫ﻏﲑ ﻣﻨﻌﺪﻡ ﻭ‬

‫*‪n ∈ ℕ‬‬

‫ﺣﻴﺚ‬

‫‪un = un −1 + r‬‬

‫‪un = aun −1‬‬

‫‪r‬‬

‫ﺛﺎﺑﺖ ﻭ‬

‫ﺣﻴﺚ‬

‫ﻋﺪﺩ‬

‫‪a‬‬

‫)ﻣﺘﺘﺎﻟﻴﺔ ﻫﻨﺪﺳﻴﺔ ﺃﺳﺎﺳﻬﺎ ‪ .( a‬ﻻﺣﻆ ﺃﻧﻪ ﳝﻜﻦ‬

‫ﻛﺘﺎﺑﺔ ﺍﳊﺪ ﺍﻟﻌﺎﻡ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ ﺍﳊﺴﺎﺑﻴﺔ ﻋﻠﻰ ﺍﻟﺸﻜﻞ‬ ‫ﺍﻟﻌﺎﻡ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ ﺍﳍﻨﺪﺳﻴﺔ ﻋﻠﻰ ﺍﻟﺸﻜﻞ ‪:‬‬

‫‪= an‬‬

‫‪un = un −1 + r‬‬

‫ﺗﻜﺎﻓﺊ‬

‫‪un = aun −1‬‬

‫ﺗﻜﺎﻓﺊ‬

‫‪un = nr‬‬

‫ﻭﻛﺘﺎﺑﺔ ﺍﳊﺪ‬

‫‪ ، un‬ﺃﻱ ﺃﻥ ‪:‬‬ ‫‪= nr‬‬ ‫‪= an‬‬

‫‪، un‬‬

‫‪. un‬‬

‫‪ (3‬ﺣﻮﻝ ﺍﻟﺮﺗﺎﺑﺔ ‪:‬‬ ‫ﻏﲑ ﺭﺗﻴﺒﺔ‪.‬‬

‫‪ -‬ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪un = (−1) n‬‬


‫‪un = n‬‬


‫‪1‬‬ ‫‪n+2‬‬ ‫‪ un = − n‬ﻣﺘﺰﺍﻳﺪﺓ‪.‬‬ ‫‪n+2‬‬ ‫‪2‬‬ ‫‪ un = (−1)n i + 3n‬ﻏﲑ‬ ‫‪n +1‬‬

‫ ﺍﳌﺘﺘﺎﻟﻴﺔ‬‫‪ -‬ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫= ‪un‬‬

‫ﻣﺘﺰﺍﻳﺪﺓ‪.‬‬ ‫ﻣﺘﻨﺎﻗﺼﺔ‪.‬‬

‫ﺭﺗﻴﺒﺔ‪.‬‬

‫ﺳﻮﻑ ﻧﻌﺘﱪ‪ ،‬ﰲ ﻣﺎ ﻳﻠﻲ‪ ،‬ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﻛﺎﻣﻞ ﳎﻤﻮﻋﺔ‬ ‫ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ ﺇﻻ ﺇﺫﺍ ﺃﺷﺮﻧﺎ ﺇﱃ ﻋﻜﺲ ﺫﻟﻚ‪.‬‬

‫‪16‬‬

‫ا ‪ : 1‬ا ت‬

‫ﺗﻌﺮﻳﻒ )ﺍﻟﺘﻘﺎﺭﺏ(‬ ‫ ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ‬‫‪lim un − u = 0‬‬

‫∞‪n →+‬‬

‫ﺇ‪‬ﺎ ﻣﺘﻘﺎﺭﺑﺔ ﺇﺫﺍ ﻭﺟﺪ ﻋﺪﺩ‬

‫)‪(un‬‬

‫‪u‬‬

‫ﲝﻴﺚ‬

‫ﺃﻱ‬

‫‪un − u < ε .‬‬

‫‪∀ε > 0,‬‬

‫⇒ ‪∃n0 ∈ ℕ : n ≥ n0‬‬

‫ ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﺇ‪‬ﺎ ﻣﺘﺒﺎﻋﺪﺓ ﻋﻨﺪﻣﺎ ﻻ ﺗﻜﻮﻥ ﻣﺘﻘﺎﺭﺑﺔ‪.‬‬‫ﺗﻌﻤﻴﻢ ‪:‬‬ ‫‪ -‬ﻧﻘﻮﻝ ﺇﻥ ﻣﺘﺘﺎﻟﻴﺔ‬

‫)‪(un‬‬

‫ﺗﺆﻭﻝ ﺇﱃ‬

‫∞‪+‬‬

‫ﺇﺫﺍ ﻛﺎﻥ ‪:‬‬

‫‪∀A > 0, ∃n0 ∈ ℕ : n ≥ n0 ⇒ un > A.‬‬

‫‪ -‬ﻧﻘﻮﻝ ﺇﻥ ﻣﺘﺘﺎﻟﻴﺔ‬

‫)‪(un‬‬

‫ﺗﺆﻭﻝ ﺇﱃ‬

‫∞‪−‬‬

‫ﺇﺫﺍ ﻛﺎﻥ ‪:‬‬

‫‪∀A < 0, ∃n0 ∈ ℕ : n ≥ n0 ⇒ un < A.‬‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ ﻻ ﻧﻘﻮﻝ ﰲ ﺍﳊﺎﻟﺘﲔ ﺍﻷﺧﲑﺗﲔ ﺇﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ ﻣﺘﻘﺎﺭﺑﺔ‪ ،‬ﺑﻞ ﻧﻘﻮﻝ ﺇ‪‬ﺎ‬‫ﻣﺘﺒﺎﻋﺪﺓ‪ .‬ﻓﻌﻠﻰ ﺳﺒﻴﻞ ﺍﳌﺜﺎﻝ ﻧﻼﺣﻆ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺍﳊﺴﺎﺑﻴﺔ ﺍﳌﻌﺮﻓﺔ ﺑـ‬ ‫‪un = un −1 + r‬‬

‫ﺣﻴﺚ‬

‫‪r‬‬

‫ﺛﺎﺑﺖ و *‪ ، n ∈ ℕ‬ﺃﻱ‬

‫ﰲ ﺣﺎﻟﺔ ‪ . r = 0‬ﺃﻣﺎ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺍﳍﻨﺪﺳﻴﺔ‬ ‫ﻣﻨﻌﺪﻡ ﻭ *‪ ( n ∈ ℕ‬ﺃﻱ‬

‫‪un = a n‬‬

‫‪un = aun −1‬‬

‫‪un = nr‬‬

‫ﻣﺘﺒﺎﻋﺪﺓ ﺇﻻ‬

‫)ﺣﻴﺚ‬

‫ﻋﺪﺩ ﻏﲑ‬

‫‪a‬‬

‫ﻓﻬﻲ ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ‪ 0‬ﻣﻦ ﺃﺟﻞ‬

‫‪a 1‬ﺇﺫﻥ ﻳﻜﻔﻲ‬ ‫‪ε‬‬

‫ﺍﳌﻌﺮﻓﺔ ﺑـ‬

‫‪1‬‬ ‫‪n‬‬

‫‪1‬‬ ‫‪1‬‬ ‫‪−0 = ‪n 0‬‬

‫=‬

‫) ‪(u n‬‬

‫‪ε‬‬

‫ﺃﺷﺮﻧﺎ ﻓﺈﻥ ﺃﻱ ﻋﺪﺩ ﻃﺒﻴﻌﻲ ﺃﻛﱪ ﻣﻦ‬ ‫ﻟـ‬

‫‪n0‬‬

‫=‬

‫‪ u n‬ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ‪.0‬‬

‫‪ . u n − u‬ﻭﻫﺬﺍ ﻳﺆﺩﻱ ﺇﱃ‬

‫‪ ،‬ﻣﺜﻼ ‪ . n 0 =  1  + 1 :‬ﻭﻛﻤﺎ‬ ‫‪ε ‬‬

‫‪1‬‬ ‫‪ ε  + 1‬‬

‫ﺻﺎﱀ ﺃﻥ ﻳﻜﻮﻥ ﻗﻴﻤﺔ‬

‫ﻷﻧﻪ ﳛﻘﻖ ﺑﺎﻟﻀﺮﻭﺭﺓ ﺍﻟﺘﻌﺮﻳﻒ‪.‬‬

‫‪ (2‬ﺃﺛﺒﺖ ﺃﻥ‬

‫ﺍﳌﺘﺘﺎﻟﻴﺔ ) ‪(u n‬‬


‫‪3n − 1‬‬ ‫‪n +1‬‬

‫=‬

‫‪ u n‬ﻣﺘﻘﺎﺭﺑﺔ‬

‫ﳓﻮ‪.3‬‬ ‫ﺍﻹﺛﺒﺎﺕ ‪ :‬ﻧﻜﺘﺐ‬

‫ﻭﻣﻨﻪ ﻳﺄﰐ‬

‫‪3n − 1‬‬ ‫‪−1 − 3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫= ‪−3‬‬ ‫=‬ ‫‪< 4‬ﺇﺫﻥ ﻳﻜﻔﻲ ﺃﻥ ﳔﺘﺎﺭ ‪، n 0 > 4‬‬

‫‪ε‬‬

‫‪ε‬‬

‫‪. n 0 =  4  + 1‬‬ ‫‪ε ‬‬

‫‪18‬‬

‫= ‪un −u‬‬

‫ﻣﺜﻼ‬

‫ا ‪ : 1‬ا ت‬

‫ﻣﻼﺣﻈﺔ‬ ‫ﺇﺫﺍ ﺍﻫﺘﻤﻤﻨﺎ ﺑﺎﳊﺴﺎﺏ ﺍﻟﺘﻘﺮﻳﱯ ﻟﻠﻨﻬﺎﻳﺔ ‪ u‬ﳌﺘﺘﺎﻟﻴﺔ ﻣﺘﻘﺎﺭﺑﺔ ) ‪، (un‬‬ ‫ﻭﻛﺎﻧﺖ ﻫﺬﻩ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺭﺗﻴﺒﺔ ﻓﺒﻘﺪﺭ ﻣﺎ ﻳﻜﱪ ‪ n‬ﺑﻘﺪﺭ ﻣﺎ ﻳﻜﻮﻥ ‪ un‬ﻗﺮﻳﺒﺎ‬ ‫ﻣﻦ ﺍﻟﻨﻬﺎﻳﺔ ‪ . u‬ﻓﻌﻠﻰ ﺳﺒﻴﻞ ﺍﳌﺜﺎﻝ ﳒﺪ ﺃﻥ ‪ u100‬ﺃﻗﺮﺏ ﺇﱃ ﺍﻟﻨﻬﺎﻳﺔ ﻣﻦ‬ ‫‪ . u99‬ﺇﻣﺎ ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﳌﺘﺘﺎﻟﻴﺔ ﻏﲑ ﺭﺗﻴﺒﺔ ﻓﻬﺬﻩ ﺍﳋﺎﺻﻴﺔ ﺧﺎﻃﺌﺔ ﺣﻴﺚ ﳚﻮﺯ‬ ‫ﺃﻥ ﻳﻜﻮﻥ ‪ u99‬ﺃﻗﺮﺏ ﺇﱃ ﺍﻟﻨﻬﺎﻳﺔ ﻣﻦ ‪ . u100‬ﻏﲑ ﺃﻥ ﻫﺬﺍ ﻻ ﳝﻨﻊ ﺃﻧﻪ‬ ‫ﻋﻨﺪﻣﺎ ﺗﻜﻮﻥ ‪ n‬ﻛﺒﲑﺓ ﻓﺈﻥ ‪ un‬ﺳﻴﻜﻮﻥ ﻗﺮﻳﺒﺎ ﻣﻦ ﺍﻟﻨﻬﺎﻳﺔ ‪ ، u‬ﲟﻌﲎ ﺃﻥ‬ ‫‪ u − un‬ﺳﻴﻜﻮﻥ ﺻﻐﲑﺍ ﺟﺪﺍ ﺣﱴ ﻟﻮ ﻛﺎﻧﺖ ﺍﳌﺘﺘﺎﻟﻴﺔ ﻋﻘﺪﻳﺔ‪.‬‬ ‫ﺗﻌﺮﻳﻒ )ﺍﳌﺘﺘﺎﻟﻴﺔ ﺍﳉﺰﺋﻴﺔ(‬ ‫ﻟﺘﻜﻦ‬

‫‪(un ) n∈ℕ‬‬

‫ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ‪ .‬ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ‬

‫ﻣﺴﺘﺨﺮﺟﺔ ﻣﻦ‬

‫‪(un ) n∈ℕ‬‬

‫)ﺃﻭ ﺇ‪‬ﺎ ﺟﺰﺋﻴﺔ ﻣﻦ ‪ ( (un )n∈ℕ‬ﺇﺫﺍ ﻭﺟﺪ ﺗﻄﺒﻴﻖ‬

‫‪f :ℕ → ℕ‬‬

‫ﻣﺘﺰﺍﻳﺪ ﲤﺎﻣﺎ ﲝﻴﺚ‬ ‫‪vn = u f ( n ) .‬‬

‫) ‪(vn‬‬

‫ﺇ‪‬ﺎ‬

‫‪∀n ∈ ℕ :‬‬

‫ﻣﺜﺎﻝ‬ ‫ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪u2 n = 1‬‬

‫ﻫﻲ ﻣﺘﺘﺎﻟﻴﺔ ﺟﺰﺋﻴﺔ ﻣﻦ ﺍﳌﺘﺘﺎﻟﻴﺔ ‪. un = (−1)n‬‬

‫ﻭﻛﺬﻟﻚ ﺷﺄﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ ‪. u2 n +1 = −1‬‬ ‫ﺗﻌﺮﻳﻒ )ﺍﳌﺘﺘﺎﻟﻴﺔ ﺍﻟﻜﻮﺷﻴﺔ(‬ ‫ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ‬

‫) ‪(un‬‬

‫ﺇ‪‬ﺎ ﻛﻮﺷﻴﺔ‪ ،‬ﺃﻭ ﻟﻜﻮﺷﻲ‬

‫‪ ، (1857-1789) Cauchy‬ﺇﺫﺍ ﻛﺎﻥ ‪:‬‬ ‫‪un − um < ε‬‬

‫⇒ ‪∃n0 ∈ ℕ : n ≥ n0 ∧ m ≥ n0‬‬

‫‪∀ε > 0,‬‬

‫ﳝﻜﻦ ﺃﻳﻀﺎ ﺍﻟﺘﻌﺒﲑ ﻋﻦ ﻫﺬﻩ ﺍﻟﻌﻼﻗﺔ ﺑﺎﻟﻜﺘﺎﺑﺔ ‪:‬‬ ‫‪un + p − u n < ε‬‬

‫⇒ ‪∃n0 ∈ ℕ : n ≥ n0 ∧ p ∈ ℕ‬‬ ‫‪19‬‬

‫‪∀ε > 0,‬‬

‫ا ‪ : 1‬ا ت‬

‫‪ .3‬ﻧﺘﺎﺋﺞ ﻭﺧﻮﺍﺹ ‪:‬‬ ‫ﻧﻈﺮﻳﺔ )ﻭﺣﺪﺍﻧﻴﺔ ﺍﻟﻨﻬﺎﻳﺔ(‬ ‫ﺇﺫﺍ ﺗﻘﺎﺭﺑﺖ ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ ﻓﺈﻥ ‪‬ﺎﻳﺘﻬﺎ ﻭﺣﻴﺪﺓ‪.‬‬ ‫ﺍﻟﱪﻫﺎﻥ‬ ‫ﻟﻨﺘﺄﻛﺪ ﻣﻦ ﺫﻟﻚ ﺑﺎﳋﻠﻒ )ﻧﻜﺘﻔﻲ ﻫﻨﺎ ﺑﺎﻋﺘﺒﺎﺭ ﺣﺎﻟﺔ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ‬ ‫ﺍﳊﻘﻴﻘﻴﺔ( ‪ :‬ﻧﻔﺮﺽ ﻭﺟﻮﺩ ‪‬ﺎﻳﺘﲔ ﳐﺘﻠﻔﺘﲔ ‪ u‬ﻭ‬

‫' ‪ u‬ﳌﺘﺘﺎﻟﻴﺔ )‪(un‬‬

‫ﺑﺎﻋﺘﺒﺎﺭ‬

‫ﻣﺜﻼ ﺃﻥ ' ‪ . u < u‬ﻭﻟﻨﺨﺘﺮ ﰲ ﺍﻟﺘﻌﺮﻳﻒ ﺍﻟﺴﺎﺑﻖ ‪ . ε = u '− u‬ﻭﻣﻦ ﰒﹼ‬ ‫‪2‬‬

‫ﻓﺈﻥ‬ ‫‪un − u < ε‬‬ ‫‪un − u ' < ε‬‬

‫ﻭﻋﻨﺪﻣﺎ ﻧﻀﻊ‬

‫⇒ ‪∃n0 ∈ ℕ : n ≥ n0‬‬ ‫⇒ ‪∃n0' ∈ ℕ : n ≥ n0‬‬

‫) ‪N = max(n0 , n '0‬‬

‫ﻧﺴﺘﻨﺘﺞ‬

‫‪u '− u‬‬ ‫‪u '− u‬‬ ‫‪< un < u +‬‬ ‫‪2‬‬ ‫‪2‬‬

‫‪n ≥ N ⇒ u '−‬‬

‫ﺃﻱ ‪:‬‬ ‫‪u '+ u‬‬ ‫‪u '+ u‬‬ ‫< ‪< un‬‬ ‫‪2‬‬ ‫‪2‬‬

‫⇒ ‪n≥N‬‬

‫ﻭﰲ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗﻨﺎﻗﺾ ﻭﺍﺿﺢ‪ .‬ﻭﻣﻨﻪ ﺍﳌﻄﻠﻮﺏ‪.‬‬ ‫‪20‬‬

‫ا ‪ : 1‬ا ت‬

‫ﺇﻟﻴﻚ ﻫﺬﻩ ﺍﳋﻮﺍﺹ ﺍﳌﺘﻌﻠﻘﺔ ﺑﺎﻟﺘﻘﺎﺭﺏ‪ ،‬ﻭﻫﻲ ﺧﻮﺍﺹ ﻧﺴﺘﻌﻤﻠﻬﺎ‬ ‫ﺑﻜﺜﺮﺓ ﰲ ﺍﻟﱪﺍﻫﲔ ﻭﺩﺭﺍﺳﺔ ﻃﺒﻴﻌﺔ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ‪ ،‬ﻭﺃﺣﻴﺎﻧﺎ ﺩﻭﻥ ﺃﻥ ﻧﺪﺭﻱ‪.‬‬ ‫ﻧﻄﻠﺐ ﻣﻦ ﺍﻟﻘﺎﺭﺉ ﺍﻟﻌﻤﻞ ﻋﻠﻰ ﺇﺛﺒﺎ‪‬ﺎ ‪:‬‬ ‫ﻋﻨﺪﻣﺎ ﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺘﺎﻥ‬

‫ﻋﺪﺩﻳﺘﺎﻥ ) ‪(un‬‬

‫ﻭ‬

‫) ‪(vn‬‬

‫ﻣﺘﻘﺎﺭﺑﺘﲔ )ﻧﺆﻛﺪ‬

‫ﻫﻨﺎ ﻋﻠﻰ ﺿﺮﻭﺭﺓ ﺗﻘﺎﺭﺏ ﺍﳌﺘﺘﺎﻟﻴﺘﲔ( ﻓﺈﻥ ‪:‬‬ ‫‪، nlim‬‬ ‫‪(un + vn ) = lim un + lim vn (1‬‬ ‫∞‪→+‬‬ ‫∞‪n →+‬‬ ‫∞‪n →+‬‬ ‫‪، nlim‬‬ ‫‪(un − vn ) = lim un − lim vn (2‬‬ ‫∞‪→+‬‬ ‫∞‪n →+‬‬ ‫∞‪n →+‬‬ ‫‪، nlim‬‬ ‫‪(un × vn ) = lim un × lim vn (3‬‬ ‫∞‪→+‬‬ ‫∞‪n →+‬‬ ‫∞‪n →+‬‬ ‫‪ (4‬ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻋﺪﺩ‬ ‫‪، nlim‬‬ ‫‪(λ vn ) = λ lim vn :‬‬ ‫∞‪→+‬‬ ‫∞‪n →+‬‬ ‫‪λ‬‬

‫‪ (5‬ﻋﻨﺪﻣﺎ ﻳﻜﻮﻥ‬

‫)ﺣﻘﻴﻘﻲ ﺃﻭ ﻋﻘﺪﻱ(‬

‫‪lim vn ≠ 0‬‬

‫∞‪n →+‬‬

‫‪:‬‬

‫‪lim un‬‬

‫∞‪n →+‬‬

‫‪lim vn‬‬

‫∞‪n →+‬‬

‫‪(6‬‬

‫‪= lim un‬‬ ‫∞‪n →+‬‬

‫‪ (7‬ﺇﺫﺍ ﻛﺎﻥ‬

‫=‬

‫‪un‬‬ ‫‪، nlim‬‬ ‫∞‪→+‬‬ ‫‪vn‬‬

‫‪، nlim‬‬ ‫‪un‬‬ ‫∞‪→+‬‬

‫‪un ≤ vn‬‬

‫‪. nlim‬‬ ‫ﺍﺑﺘﺪﺍﺀ ﻣﻦ ﺭﺗﺒﺔ ﻣﻌﻴﻨﺔ ﻓﺈﻥ ‪un ≤ lim vn‬‬ ‫∞‪→+‬‬ ‫∞‪n →+‬‬

‫ﻣﻼﺣﻈﺔ‬ ‫ﻧﺴﺘﺨﻠﺺ ﻣﻦ ﺍﳋﺎﺻﻴﺔ ‪ (7‬ﺃﻥ ‪:‬‬ ‫‪un ≤ 0 ⇒ lim un ≤ 0‬‬ ‫∞‪n →+‬‬

‫‪un ≥ 0 ⇒ lim un ≥ 0,‬‬ ‫∞‪n →+‬‬

‫ﻭﺗﺼﺪﻕ ﻫﺎﺗﺎﻥ ﺍﻟﻌﻼﻗﺘﺎﻥ ﺣﱴ ﺇﻥ ﲢﻘﻖ ﻃﺮﻓﺎﻫﺎ ﺍﻷﻭﻻﻥ ﺍﺑﺘﺪﺍﺀ ﻣﻦ ﺭﺗﺒﺔ‬ ‫ﻣﻌﻴﻨﺔ ﻟﻴﺴﺖ ﺑﺎﻟﻀﺮﻭﺭﺓ ﺍﻟﺮﺗﺒﺔ ‪. 0‬‬ ‫‪21‬‬

‫ا ‪ : 1‬ا ت‬

‫ﺍﺣﺬﺭ ‪ :‬ﺇﺫﺍ ﻛﺎﻥ‬

‫‪un > 0‬‬

‫ﻓﻬﺬﺍ ﻳﺆﺩﻱ ﺇﱃ‬

‫ﺍﺑﺘﺪﺍﺀ ﻣﻦ ﺃﻭﻝ ﺭﺗﺒﺔ ﺃﻭ ﺍﺑﺘﺪﺍﺀ ﻣﻦ ﺭﺗﺒﺔ ﻣﻌﻴﻨﺔ‬

‫‪lim un ≥ 0‬‬

‫∞‪n →+‬‬

‫‪ . nlim‬ﻟﻠﺘﺘﺄﻛﺪ ﻣﻦ ﺫﻟﻚ ﺧﺬ‬ ‫‪un > 0‬‬ ‫∞‪→+‬‬

‫‪ ...‬ﻭﻻ ﻳﺆﺩﻱ ﺑﺎﻟﻀﺮﻭﺭﺓ ﺇﱃ‬ ‫‪1‬‬ ‫‪n‬‬

‫= ‪un‬‬

‫ﺍﳌﻮﺟﺒﺔ ﲤﺎﻣﺎ ﻟﻜﻦ ﺫﻟﻚ‬

‫ﱂ ﳝﻨﻊ ﺍﻧﻌﺪﺍﻡ ‪‬ﺎﻳﺘﻬﺎ‪.‬‬ ‫ﻧﻈﺮﻳﺔ )ﺍﻟﺘﻘﺎﺭﺏ ﻭﺍﶈﺪﻭﺩﻳﺔ(‬ ‫ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ ﻣﺘﻘﺎﺭﺑﺔ ﻣﺘﺘﺎﻟﻴﺔ ﳏﺪﻭﺩﺓ‪ .‬ﻭﺍﻟﻌﻜﺲ ﻏﲑ ﺻﺤﻴﺢ‪.‬‬ ‫ﺍﻟﱪﻫﺎﻥ‬ ‫ﻟﻠﺘﺄﻛﺪ ﻣﻦ ﺫﻟﻚ ﻧﻜﺘﺐ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪(un ) n∈ℕ‬‬

‫ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﻋﺪﺩ ‪، u‬‬

‫ﺃﻱ ﺃﻥ‬ ‫‪un − u < ε‬‬

‫⇒ ‪∃n0 ∈ ℕ : n ≥ n0‬‬

‫ﰒ ﳔﺘﺎﺭ ﰲ ﻫﺬﻩ ﺍﻟﻌﻼﻗﺔ‬

‫‪ε =1‬‬

‫‪un − u < 1‬‬

‫‪∀ε > 0,‬‬

‫ﻣﺜﻼ‪ ،‬ﻓﻴﻜﻮﻥ‬

‫⇒ ‪∃n0 ∈ ℕ : n ≥ n0‬‬

‫ﻭﻣﻨﻪ ‪:‬‬ ‫‪.‬‬ ‫ﻟﻴﻜﻦ‬

‫‪K‬‬

‫‪un < 1 + u‬‬

‫⇒ ‪n ≥ n0‬‬

‫ﺣﺎﺩﺍ ﻣﻦ ﺍﻷﻋﻠﻰ ﻟﻠﻤﺠﻤﻮﻋﺔ‬

‫}‬

‫‪, u1 ,..., un0 −1‬‬

‫‪0‬‬

‫‪{u‬‬

‫و‬

‫} ‪ . M = max {1 + u , K‬ﻻﺣﻆ ﻋﻨﺪﺋﺬ ﺃﻥ‪:‬‬ ‫‪∀n ∈ ℕ, un ≤ M .‬‬ ‫وه اب‪.‬‬ ‫ﺍﻟﻌﻜﺲ ﻏﲑ ﺻﺤﻴﺢ ‪ :‬ﺍﳌﺜﺎﻝ ﺍﳌﻀﺎﺩ ﺍﻟﺒﺴﻴﻂ ﻭﺍﻟﺸﻬﲑ ﻳﺘﻤﺜﻞ ﰲ‬ ‫ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪un = (−1) n‬‬

‫ﺍﻟﱵ ﺳﻴﺘﻨﺎﻭﳍﺎ ﺍﳌﺜﺎﻝ ‪ (1‬ﺍﳌﻮﺍﱄ‪.‬‬ ‫‪22‬‬

‫ا ‪ : 1‬ا ت‬

‫ﺃﻣﺜﻠﺔ ‪:‬‬ ‫‪ (1‬ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪un = (−1) n‬‬

‫ﻏﲑ ﻣﺘﻘﺎﺭﺑﺔ )ﻭﻫﻲ ﳏﺪﻭﺩﺓ(‪ .‬ﳝﻜﻦ ﺗﱪﻳﺮ ﺫﻟﻚ‬

‫ﺑﺎﳋﻠﻒ‪ ،‬ﻛﻤﺎ ﻳﻠﻲ ‪ :‬ﻧﻔﺮﺽ ﻭﺟﻮﺩ ‪‬ﺎﻳﺔ‬

‫‪u‬‬

‫ﳍﺬﻩ ﺍﳌﺘﺘﺎﻟﻴﺔ‪ .‬ﺗﺴﺘﻔﻴﺪ ﻣﻦ‬

‫ﺍﻟﻌﻼﻗﺔ‬ ‫‪un − u < ε‬‬

‫ﺑﺎﻋﺘﺒﺎﺭ ﺃﻭﻻ‬

‫⇒ ‪∃n0 ∈ IN : n ≥ n0‬‬

‫‪∀ε > 0,‬‬

‫)‪( 0‬‬

‫ﺯﻭﺟﻴﺎ ﻓﻴﺄﰐ‬

‫‪n‬‬

‫‪1− u < ε‬‬

‫⇒ ‪∀ε > 0, ∃n0 ∈ IN : n ≥ n0‬‬

‫)‪( 1‬‬

‫ﻋﻨﺪﺋﺬ ﻧﻼﺣﻆ ﺃﻥ )‪ (1‬ﺗﺴﺘﻠﺰﻡ ‪. u = 1‬‬ ‫ﰒ ﺑﺎﻋﺘﺒﺎﺭ‬

‫‪n‬‬

‫ﻓﺮﺩﻳﺎ ﻳﺄﰐ ‪:‬‬

‫‪−1 − u < ε‬‬

‫⇒ ‪∀ε > 0, ∃n0 ∈ IN : n ≥ n0‬‬

‫)‪( 2‬‬

‫ﻋﻨﺪﺋﺬ ﻧﻼﺣﻆ ﺃﻥ )‪ (2‬ﺗﺴﺘﻠﺰﻡ ‪ . u = −1‬ﻭﲟﺎ ﺇﻧﻨﺎ ﻻ ﻧﺴﺘﻄﻴﻊ ﺍﳊﺼﻮﻝ‬ ‫ﻋﻠﻰ‬

‫‪u =1‬‬

‫ﻭ‬

‫‪u = −1‬‬

‫ﰲ ﺁﻥ ﻭﺍﺣﺪ ﻧﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻌﻼﻗﺔ )‪(0‬‬

‫ﻣﺴﺘﺤﻴﻠﺔ‪ .‬ﻭﺑﺎﻟﺘﺎﱄ ﻓﺎﳌﺘﺘﺎﻟﻴﺔ ﻣﺘﺒﺎﻋﺪﺓ‪.‬‬ ‫‪ (2‬ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪un = n‬‬

‫ﻣﺘﺒﺎﻋﺪﺓ‪ .‬ﳝﻜﻦ ﻣﻼﺣﻈﺔ ﺃ‪‬ﺎ ﺗﺆﻭﻝ ﺇﱃ‬

‫∞‪+‬‬

‫)ﻭﻫﺬﺍ ﻻ ﻳﻌﲏ ﺃ‪‬ﺎ ﻣﺘﻘﺎﺭﺑﺔ ﻷﻥ ﺍﻟﺘﻘﺎﺭﺏ ﻳﺴﺘﻮﺟﺐ ﺃﻥ ﺗﻜﻮﻥ ﺍﻟﻨﻬﺎﻳﺔ ﰲ‬ ‫‪ℝ‬‬

‫‪ ...‬ﻭ‬

‫∞‪+‬‬

‫ﻻ ﻳﻨﺘﻤﻲ ﺇﱃ ‪ .( ℝ‬ﻛﻴﻒ ﻧﱪﺭ ﺃ‪‬ﺎ ﺗﺆﻭﻝ ﺇﱃ ∞‪ +‬؟‬

‫‪ε‬‬

‫ﻣﻮﺟﺒﺎ )ﻣﻬﻤﺎ ﻛﺎﻥ ﻛﺒﲑﺍ(‪ .‬ﻳﻮﺟﺪ ﺩﻭﻣﺎ ﻋﺪﺩ ﻃﺒﻴﻌﻲ‬

‫ﺧﺬ ﺃﻱ ﻋﺪﺩ‬ ‫‪nε‬‬ ‫‪2‬‬

‫ﲝﻴﺚ ‪:‬‬

‫‪n ≥ nε‬‬

‫⇐‬

‫) ‪ . nε = ([ε ] + 2‬ﺗﺄﻛﺪ ﻣﻦ ﺫﻟﻚ‪.‬‬ ‫‪23‬‬

‫‪n >ε‬‬

‫‪ .‬ﻳﻜﻔﻲ ﺍﺧﺘﻴﺎﺭ ﻣﺜﻼ‬

‫ا ‪ : 1‬ا ت‬

‫‪ (3‬ﺍﳌﺘﺘﺎﻟﻴﺔ‬ ‫ﺍﻟﻌﻼﻗﺔ )‪(0‬‬ ‫‪ (4‬ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪1‬‬ ‫‪n+2‬‬ ‫‪. n0 > 1 :‬‬

‫= ‪un‬‬

‫‪ε‬‬

‫‪3n‬‬ ‫‪n+2‬‬ ‫‪2‬‬

‫ﺍﻟﻌﻼﻗﺔ )‪: (0‬‬ ‫‪ (5‬ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ‪ .0‬ﺫﻟﻚ ﺃﻧﻪ ﻳﻜﻔﻲ ﺃﻥ ﳔﺘﺎﺭ ﰲ‬

‫‪ε‬‬

‫‪un = −‬‬

‫ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ‪ . 3‬ﺫﻟﻚ ﺃﻧﻪ ﻳﻜﻔﻲ ﺃﻥ ﳔﺘﺎﺭ ﰲ‬

‫> ‪. n0‬‬

‫‪n2‬‬ ‫‪n3 + 1‬‬

‫‪un = (−1)n i +‬‬

‫ﻏﲑ ﻣﺘﻘﺎﺭﺑﺔ‪ .‬ﺗﺄﻛﺪ ﻣﻦ ﺫﻟﻚ‬

‫ﺑﺎﻻﺳﺘﻔﺎﺩﺓ ﻣﻦ ﺗﺒﺎﻋﺪ ﺍﳌﺘﺘﺎﻟﻴﺔ ‪. un = (−1)n i‬‬ ‫ﻧﻈﺮﻳﺔ )ﻛﻮﺷﻰ ﻭﺍﻟﺘﻘﺎﺭﺏ(‬ ‫ﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ ﻣﺘﻘﺎﺭﺑﺔ ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ ﻛﻮﺷﻴﺔ‪.‬‬ ‫ﻣﻼﺣﻈﺔ‬ ‫ﺗﻌﻮﺩ ﺃﳘﻴﺔ ﻧﻈﺮﻳﺔ ﻛﻮﺷﻲ ﺇﱃ ﺃ‪‬ﺎ ﺗﺴﻤﺢ ﺑﺪﺭﺍﺳﺔ ﻃﺒﻴﻌﺔ ﻣﺘﺘﺎﻟﻴﺔ )ﺃﻱ‬ ‫ﻣﻌﺮﻓﺔ ﻣﺎ ﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺘﻘﺎﺭﺑﺔ ﺃﻡ ﻣﺘﺒﺎﻋﺪﺓ( ﺩﻭﻥ ﻣﻌﺮﻓﺔ ‪‬ﺎﻳﺘﻬﺎ )ﰲ ﺣﺎﻟﺔ‬ ‫ﺗﻘﺎﺭ‪‬ﺎ(‪.‬‬

‫‪24‬‬

‫ا ‪ : 1‬ا ت‬

‫ﺍﻟﱪﻫﺎﻥ‬ ‫* ﺃﻭﻻ ‪ :‬ﺇﺫﺍ ﻛﺎﻧﺖ‬

‫) ‪(u n‬‬

‫ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ‪ u‬ﻓﺈﻧﻨﺎ ﻧﺴﺘﻄﻴﻊ ﻛﺘﺎﺑﺔ‬

‫) ‪u p − u q = (u p − u ) + (u − u q‬‬ ‫‪≤ u p − u + u − uq .‬‬

‫ﻭﺑﺎﳌﺮﻭﺭ ﺇﱃ ﺍﻟﻨﻬﺎﻳﺔ ﰲ ﺍﻟﻄﺮﻓﲔ ﳒﺪ ‪:‬‬ ‫‪0 ≤ lim u p − u q ≤ lim u p − u + lim u − u q = 0 + 0,‬‬ ‫∞‪q →+‬‬

‫ﻷﻥ‬

‫‪lim u p − u = 0‬‬

‫∞‪p →+‬‬

‫ﺍﳌﺘﺘﺎﻟﻴﺔ‪ .‬ﻭﻣﻨﻪ‬

‫∞‪p →+‬‬

‫و‬

‫∞‪p →+‬‬ ‫∞‪q →+‬‬

‫‪lim u − u q = 0‬‬

‫∞‪q →+‬‬

‫ﺑﻔﻀﻞ ﺗﻘﺎﺭﺏ‬

‫‪. plim‬‬ ‫‪u p − uq‬‬ ‫∞‪→+‬‬

‫‪=0‬‬

‫∞‪q →+‬‬

‫* ﺛﺎﻧﻴﺎ ‪ :‬ﻳﺘﻄﻠﺐ ﻫﺬﺍ ﺍﳉﺰﺀ ﻣﻦ ﺍﻟﱪﻫﺎﻥ ﺍﻹﳌﺎﻡ ﲟﻔﻬﻮﻡ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ‬ ‫ﺍﳌﺘﺠﺎﻭﺭﺓ ﻭﺑﻌﺾ ﺧﻮﺍﺻﻬﺎ )ﻧﻄﻠﺐ ﻣﻦ ﺍﻟﻘﺎﺭﺉ ﺍﻟﺒﺤﺚ ﻋﻦ ﻫﺬﺍ ﺍﳌﻔﻬﻮﻡ‬ ‫ﰲ ﺍﳌﺮﺍﺟﻊ ﺍﻟﻮﺍﺭﺩﺓ ﰲ ﺫﻳﻞ ﺍﻟﺪﺭﺱ(‪ .‬ﻧﻔﺮﺽ ﺍﻵﻥ ﺃﻥ ﺷﺮﻁ ﻛﻮﺷﻲ‬ ‫ﳏﻘﻖ‪ ،‬ﺃﻱ ﺃﻥ‬

‫‪=0‬‬

‫‪ . plim‬ﻭﺍﳌﻄﻠﻮﺏ ﺇﺛﺒﺎﺕ ﺗﻘﺎﺭﺏ ﺍﳌﺘﺘﺎﻟﻴﺔ‪.‬‬ ‫‪u p − uq‬‬ ‫∞‪→+‬‬ ‫∞‪q →+‬‬

‫ﻧﻔﺘﺮﺽ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺣﻘﻴﻘﻴﺔ )ﻳﻜﻔﻲ ﺍﻋﺘﺒﺎﺭ ﻣﺘﺘﺎﻟﻴﱵ ﺍﳉﺰﺀ ﺍﳊﻘﻴﻘﻲ‬ ‫ﻭﺍﳉﺰﺀ ﺍﻟﺘﺨﻴﻠﻲ ﰲ ﺣﺎﻟﺔ ﻣﺘﺘﺎﻟﻴﺔ ﻋﻘﺪﻳﺔ(‪ .‬ﻧﺜﺒﺖ ﰲ ﺍﻟﺒﺪﺍﻳﺔ ﺃﻥ ﺷﺮﻁ‬ ‫ﻛﻮﺷﻲ ﻳﺆﺩﻱ ﺇﱃ ﳏﺪﻭﺩﻳﺔ ﺍﳌﺘﺘﺎﻟﻴﺔ‪.‬‬ ‫ﻟﻴﻜﻦ‬

‫‪0 0 :‬‬

‫ﻣﻼﺣﻈﺔ‬ ‫ﻣﻦ ﺣﻖ ‪ ، α‬ﰲ ﺍﻟﺘﻌﺮﻳﻒ ﺍﻟﺴﺎﺑﻖ‪ ،‬أن ﻳﺘﻌﻠﻖ ﺑـ‬ ‫ﺫﻛﺮﻧﺎ ﰲ ﺣﺎﻟﺔ ﺍﳌﺘﺘﺎﻟﻴﺎﺕ ﺍﳌﺘﻘﺎﺭﺑﺔ ﻓﺈﻥ ﺇﺛﺒﺎﺕ‬ ‫ﺍﻟﺘﻌﺮﻳﻒ ﻳﺘﻤﺜﻞ ﰲ ﲢﺪﻳﺪ ‪ α‬ﺑﺪﻻﻟﺔ‬

‫‪ε‬‬

‫‪ε‬‬

‫و ‪ . a‬ﻭﻛﻤﺎ‬

‫ﺃﻥ ‪lim f ( x) = c‬‬ ‫‪x →a‬‬

‫ﺑﺎﺳﺘﺨﺪﺍﻡ‬

‫ﻭ ‪.a‬‬

‫ﻣﺜﺎﻝ‬ ‫ﻹﺛﺒﺎﺕ‬

‫ﺃﻥ ‪lim x 2 = 9‬‬ ‫‪x →3‬‬

‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺘﻌﺮﻳﻒ ﻧﻼﺣﻆ ﺑﻌﺪ ﺍﺧﺘﻴﺎﺭ‬

‫ﺃﻧﻪ ﻳﻜﻔﻲ ﺃﻥ ﻧﻜﺘﺐ‬

‫‪64‬‬

‫‪ε‬‬

‫ﻛﻴﻔﻴﺎ‬

‫ﺍﻟﻔﺼﻞ ‪ : 2‬ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(‬ ‫‪x2 − 9 ≤ x + 3 x − 3‬‬ ‫‪ 0, ∃α > 0 :‬‬

‫ﺇﻥ ﺍﻟﺘﻌﺮﻳﻒ ﺍﳌﻮﺍﱄ ﻳﻜﺎﻓﺊ ﺍﻟﺴﺎﺑﻖ‪.‬‬ ‫ﺗﻌﺮﻳﻒ )ﺍﻟﻨﻬﺎﻳﺔ "ﻃﺒﻮﻟﻮﺟﻴﺎ"(‬ ‫ﻟﺘﻜﻦ‬

‫‪f : I ⊂ ℝ ‬‬ ‫‪→ℝ‬‬

‫ﺇﻟﻴﻪ ﻧﻘﻄﺔ ‪ . a‬ﻧﻘﻮﻝ ﻋﻦ‬

‫‪f‬‬

‫ﺩﺍﻟﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﳎﺎﻝ ﻣﻔﺘﻮﺡ ‪ I‬ﺗﻨﺘﻤﻲ‬

‫إ‪ /‬ﲤﻠﻚ ‪‬ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ‪ c‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ‬

‫‪a‬‬

‫ﺇﺫﺍ‬

‫ﲢﻘﻖ ﺍﻟﺸﺮﻁ ‪:‬‬ ‫ﻣﻦ ﺃﺟﻞ ﻛﻞ ﳎﺎﻝ‬ ‫‪. f (A∩ I) ⊂ C‬‬

‫‪C‬‬

‫ﻣﺮﻛﺰﻩ‬

‫‪65‬‬

‫‪c‬‬

‫ﻳﻮﺟﺪ ﳎﺎﻝ‬

‫‪A‬‬

‫ﻣﺮﻛﺰﻩ ‪ a‬ﲝﻴﺚ‬

‫ﺍﻟﻔﺼﻞ ‪ : 2‬ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(‬

‫ﻧﻈﺮﻳﺔ )ﺍﻟﻨﻬﺎﻳﺔ ﺑﺎﳌﺘﺘﺎﻟﻴﺎﺕ(‬ ‫ﻟﺘﻜﻦ‬

‫‪f : I ⊂ ℝ ‬‬ ‫‪→ℝ‬‬

‫ﺇﻟﻴﻪ ﻧﻘﻄﺔ ‪ . a‬ﺗﻜﻮﻥ ﻟﻠﺪﺍﻟﺔ‬

‫‪f‬‬

‫ﻛﺎﻥ ‪ :‬ﻣﻬﻤﺎ ﻛﺎﻧﺖ ﺍﳌﺘﺘﺎﻟﻴﺔ‬ ‫‪. nlim‬‬ ‫‪f ( xn ) = c‬‬ ‫∞‪→+‬‬

‫ﺩﺍﻟﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﳎﺎﻝ ﻣﻔﺘﻮﺡ ‪ I‬ﺗﻨﺘﻤﻲ‬

‫‪‬ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ‪ c‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ‬

‫) ‪( xn‬‬

‫ﻣﻦ‬

‫‪I‬‬

‫ﺍﳌﺘﻘﺎﺭﺑﺔ ﳓﻮ‬

‫‪a‬‬

‫‪a‬‬

‫ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ‬

‫ﻓﺈﻥ‬

‫ﺍﻟﱪﻫﺎﻥ‬ ‫ﺃﻭﻻ ‪ :‬ﻧﻔﺮﺽ ﺃﻥ ﻟﻠﺪﺍﻟﺔ ‪ f‬ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ‪ c‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ ‪ . a‬ﻭﻟﺘﻜﻦ‬ ‫ﻣﺘﺘﺎﻟﻴﺔ ) ‪ ( xn‬ﻣﻦ‬ ‫‪(1) ∀ε > 0, ∃n0 > 0 : n ≥ n 0 ⇒ xn − a < ε .‬‬ ‫ﻧﻜﺘﺐ ﺍﻵﻥ ﺃﻥ ﻟﻠﺪﺍﻟﺔ ‪ f‬ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ‪ c‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ ‪: a‬‬ ‫‪(2) ∀ε > 0, ∃α > 0 : x − a < α ⇒ f ( x) − c < ε .‬‬ ‫ﻧﻌﺘﱪ ﺑﻌﺪ ﺫﻟﻚ ‪ . ε > 0‬ﳔﺘﺎﺭ ‪ α > 0‬ﲢﻘﻖ )‪ (2‬ﻭﳔﺘﺎﺭ ﺍﻟﻌﺪﺩ ‪ ε‬ﰲ )‪(1‬‬ ‫‪I‬‬

‫ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ‪ ، a‬ﺃﻱ ﺃﻥ ‪:‬‬

‫ﲝﻴﺚ‪ . ε = α :‬ﻭﻣﻦ ﰒ ﻳﺄﰐ ﻭﺟﻮﺩ ‪ n0‬ﲝﻴﺚ ﺗﺘﺤﻘﻖ )‪ ،(1‬ﺃﻱ‬ ‫‪(3) n ≥ n 0 ⇒ xn − a < α .‬‬ ‫ﺑﺎﻟﺮﺟﻮﻉ ﺇﱃ )‪ (2‬ﻭﺑﺎﻻﺳﺘﻨﺎﺩ ﺇﱃ )‪ (3‬ﻳﺘﻀﺢ ﺃﻥ‬ ‫‪n ≥ n 0 ⇒ xn − a < α‬‬ ‫‪f ( xn ) − c < ε .‬‬

‫ﻭﺑﺎﻟﺘﺎﱄ‬

‫‪f ( xn ) = c‬‬

‫‪. nlim‬‬ ‫∞‪→+‬‬

‫‪66‬‬

‫⇒‬


‫ﺛﺎﻧﻴﺎ ‪ : :‬ﻧﻔﺮﺽ ﺃﻧﻪ ﻣﻬﻤﺎ ﻛﺎﻧﺖ‬ ‫‪f ( xn ) = c‬‬

‫‪ ، nlim‬ﻭﺃﻥ‬ ‫∞‪→+‬‬

‫ﺍﳌﺘﺘﺎﻟﻴﺔ ) ‪( xn‬‬

‫‪f ( x) ≠ c‬‬

‫ﻣﻦ‬

‫‪I‬‬

‫ﺍﳌﺘﻘﺎﺭﺑﺔ ﳓﻮ‬

‫‪a‬‬

‫ﻓﺈﻥ‬

‫‪ . lim‬ﻧﻌﺒ‪‬ﺮ ﻋﻦ ﺍﻟﻌﻼﻗﺔ ﺍﻷﺧﲑﺓ ﺑـ‬ ‫‪x →a‬‬

‫‪∃ε 0 > 0, ∀α > 0, ∃x ∈ I : x − a < α ∧ f ( x) − c ≥ ε 0 .‬‬

‫ﻟﻨﺨﺘﺮ ‪ α = 1‬ﺣﻴﺚ‬ ‫‪n‬‬

‫‪n‬‬

‫ﻋﺪﺩ ﻃﺒﻴﻌﻲ ﻏﲑ ﻣﻨﻌﺪﻡ‪ .‬ﻳﻮﺟﺪ‬

‫ﻣﻦ‬

‫ﻋﻨﺼﺮ ‪xn‬‬

‫‪I‬‬

‫ﳛﻘﻖ‬ ‫‪1‬‬ ‫‪∧ f ( xn ) − c ≥ ε 0 .‬‬ ‫‪n‬‬

‫ﻭﺑﺬﻟﻚ ﻧﻨﺸﺊ‬

‫ﻣﺘﺘﺎﻟﻴﺔ ) ‪( xn‬‬

‫ﻟﻜﻨﻬﺎ ﻻ ﲢﻘﻖ ﺍﻟﻌﻼﻗﺔ‬ ‫ﺍﻟﺘﻨﺎﻗﺾ‬

‫ﻳﺆﺩﻱ‬

‫ﻭﻟﺬﺍ ‪f ( x) = c‬‬

‫ﺗﺘﻘﺎﺭﺏ ﳓﻮ‬

‫‪lim f ( xn ) = c‬‬

‫∞‪n →+‬‬

‫ﺇﱃ‬

‫ﺃﻥ‬

‫‪a‬‬

‫< ‪xn − a‬‬

‫)ﺑﻔﻀﻞ ﺍﻟﻌﻼﻗﺔ‬

‫)ﺑﺴﺒﺐ‬

‫ﻓﺮﺿﻨﺎ‬

‫‪1‬‬ ‫‪n‬‬

‫a‬‬

‫‪x ‬‬ ‫‪→a‬‬

‫‪x‬‬

‫ﻳﻘﺘﺮﺏ ﻣﻦ‬

‫‪a‬‬

‫ﻣﻦ ﺟﻬﺔ‬

‫ﺍﻟﻴﻤﲔ ﻋﻠﻰ ﺍﶈﻮﺭ ﺍﳊﻘﻴﻘﻲ(‪ ،‬ﻭﻧﺘﺤﺪﺙ ﻋﻦ ﺍﻟﻨﻬﺎﻳﺔ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﺇﺫﺍ ﺍﺳﺘﺒﺪﻟﻨﺎ‬ ‫ﺍﻟﻜﺘﺎﺑﺔ )‪lim f ( x‬‬ ‫‪x →a‬‬

‫ﺑﺎﻟﻜﺘﺎﺑﺔ‬

‫‪lim‬‬ ‫‪x A .‬‬

‫ﻭﻧﻘﻮﻝ ﻋﻦ‬ ‫)ﻭﻧﻜﺘﺐ ∞‪f ( x) = −‬‬

‫‪f‬‬

‫∞‪+‬‬

‫ﺇ‪‬ﺎ ﺗﺆﻭﻝ ﺇﱃ‬

‫∞‪−‬‬

‫‪∀A > 0, ∃α > 0 :‬‬

‫ﻋﻨﺪﻣﺎ ﻳﺆﻭﻝ‬

‫‪x‬‬

‫ﺇﱃ‬

‫‪a‬‬

‫‪ ( lim‬ﺇﺫﺍ ﲢﻘﻖ ﺍﻟﺸﺮﻁ ‪:‬‬ ‫‪x →a‬‬

‫‪x − a < α ⇒ f ( x) < − A .‬‬ ‫‪68‬‬

‫‪∀A > 0, ∃α > 0 :‬‬

‫ﺇﱃ‬


‫ﻣﻼﺣﻈﺔ‬ ‫ﺃ( ﻣﻦ ﺍﳋﻮﺍﺹ ﺍﻟﺸﻬﲑﺓ ﻟﻠﻨﻬﺎﻳﺎﺕ ﻧﺬﻛﺮ ‪:‬‬ ‫‪ .1‬ﺎﻳﺔ ﳎﻤﻮﻉ ﺩﺍﻟﺘﲔ ‪ : f + g‬ﻣﻦ ﺍﻟﺴﻬﻞ ﺍﻟﺘﺄﻛﺪ ﻣﻦ ﺻﺤﺔ ﻧﺘﺎﺋﺞ‬ ‫ﻗﻴﻤﺔ ) )‪lim ( f ( x) + g ( x‬‬

‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﺍﻟﺬﻱ ﻳﻮﺿﺢ‬ ‫ﺍﻟﻨﻬﺎﻳﺘﲔ )‪ lim f ( x‬ﻭ )‪lim g ( x‬‬ ‫‪x →a‬‬

‫‪x →a‬‬

‫ﺑﺪﻻﻟﺔ ﻗﻴﻢ‬

‫‪x →a‬‬

‫‪:‬‬ ‫→ = )‪lim f ( x‬‬

‫∞‪−‬‬

‫∞‪+‬‬

‫‪c‬‬

‫∞‪−‬‬

‫∞‪+‬‬

‫'‪c + c‬‬

‫'‪c‬‬

‫?‬

‫∞‪+‬‬

‫∞‪+‬‬

‫∞‪+‬‬

‫∞‪−‬‬

‫?‬

‫∞‪−‬‬

‫∞‪−‬‬

‫‪x →a‬‬

‫↓= )‪lim g ( x‬‬ ‫‪x →a‬‬

‫‪ .2‬ﺎﻳﺔ ﺟﺪﺍﺀ ﺩﺍﻟﺘﲔ‬ ‫ﺍﻟﺘﺎﱄ ﺍﻟﺬﻱ ﻳﻮﺿﺢ‬ ‫ﻭ )‪lim g ( x‬‬ ‫‪x →a‬‬

‫‪f ×g‬‬

‫‪ :‬ﻣﻦ ﺍﻟﺴﻬﻞ ﺍﻟﺘﺄﻛﺪ ﻣﻦ ﺻﺤﺔ ﻧﺘﺎﺋﺞ ﺍﳉﺪﻭﻝ‬

‫ﻗﻴﻤﺔ )‪lim f ( x) × g ( x‬‬ ‫‪x →a‬‬


‫ﺍﻟﻨﻬﺎﻳﺘﲔ )‪lim f ( x‬‬ ‫‪x →a‬‬

‫‪:‬‬ ‫→ = )‪lim f ( x‬‬

‫∞‪−‬‬

‫∞‪+‬‬

‫‪c=0‬‬

‫‪c0‬‬

‫∞‪−‬‬

‫∞‪+‬‬

‫‪0‬‬

‫'‪c×c‬‬

‫'‪c×c‬‬

‫‪c' > 0‬‬

‫∞‪+‬‬

‫∞‪−‬‬

‫‪0‬‬

‫'‪c×c‬‬

‫'‪c×c‬‬

‫‪c' < 0‬‬

‫?‬

‫?‬

‫‪0‬‬

‫‪0‬‬

‫‪0‬‬

‫‪c' = 0‬‬

‫∞‪−‬‬

‫∞‪+‬‬

‫∞‪+‬‬

‫∞‪−‬‬

‫?‬ ‫?‬

‫∞‪−‬‬

‫∞‪+‬‬

‫∞‪+‬‬

‫‪x →a‬‬

‫↓= )‪lim g ( x‬‬ ‫‪x →a‬‬

‫∞‪+‬‬ ‫‪69‬‬

‫∞‪−‬‬

‫∞‪−‬‬


‫‪ . 3‬ﺎﻳﺔ ﺟﺪﺍﺀ ﺩﺍﻟﺔ ﻭﻋﺪﺩ‬ ‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﺍﻟﺬﻱ ﻳﻮﺿﺢ‬ ‫ﻭﺍﻟﻌﺪﺩ‬

‫‪λ‬‬

‫‪λ. f‬‬

‫‪ :‬ﻣﻦ ﺍﻟﺴﻬﻞ ﺍﻟﺘﺄﻛﺪ ﻣﻦ ﺻﺤﺔ ﻧﺘﺎﺋﺞ‬

‫ﻗﻴﻤﺔ )‪lim λ . f ( x‬‬ ‫‪x →a‬‬


‫ﺍﻟﻨﻬﺎﻳﺔ )‪lim f ( x‬‬ ‫‪x →a‬‬

‫‪:‬‬ ‫→ = )‪lim f ( x‬‬

‫∞‪−‬‬

‫∞‪+‬‬

‫‪c‬‬

‫∞‪−‬‬

‫∞‪+‬‬

‫∞‪+‬‬

‫∞‪−‬‬

‫‪λ ×c‬‬ ‫‪λ ×c‬‬

‫‪x →a‬‬

‫↓= ‪λ‬‬

‫‪ . 4‬ﺎﻳﺔ ﻣﻘﻠﻮﺏ ﺩﺍﻟﺔ‬

‫‪1‬‬ ‫‪f‬‬

‫‪ :‬ﻣﻦ ﺍﻟﺴﻬﻞ ﺍﻟﺘﺄﻛﺪ ﻣﻦ ﺻﺤﺔ ﻧﺘﺎﺋﺞ‬ ‫‪1‬‬ ‫ﻗﻴﻤﺔ‬ ‫)‪f ( x‬‬

‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﺍﻟﺬﻱ ﻳﻮﺿﺢ‬ ‫ﺍﻟﻨﻬﺎﻳﺔ )‪lim f ( x‬‬ ‫‪x →a‬‬

‫‪λ >0‬‬ ‫‪λ ‬

‫‪x ‬‬ ‫‪→a‬‬

‫ﻭﺇﺫﺍ ﺍﺳﺘﺒﺪﻟﻨﺎ ﺗﻠﻚ ﺍﻟﻌﻼﻗﺔ‬

‫ﺍﻟﻌﻼﻗﺔ ) ‪lim f ( x) = f (a‬‬ ‫‪x →a‬‬

‫ﻧﻘﻮﻝ ﺇﻥ ‪ f‬ﻣﺴﺘﻤﺮ ﻣﻦ ﺍﻟﻴﻤﲔ ﻋﻨﺪ ‪. a‬‬

‫ﺑﺎﻟﻌﻼﻗﺔ ) ‪f ( x) = f (a‬‬

‫ﻣﺴﺘﻤﺮ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﻋﻨﺪ ‪. a‬‬ ‫‪75‬‬

‫‪lim‬‬ ‫ 0, ∃α > 0 : x − a < α ⇒ f ( x) − f (a) < ε .‬‬ ‫ﻭﻧﻌﺮ‪‬ﻑ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻋﻠﻰ ﺍ‪‬ﺎﻝ ‪ I‬ﺇﻥ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﻋﻨﺪ ﻛﻞ ﻧﻘﻄﺔ‬ ‫ﻣﻦ ‪ . I‬ﻭﻧﻌﺮ‪‬ﻑ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺟﻬﺔ ﻭﺍﺣﺪﺓ )ﻣﻦ ﺍﻟﻴﻤﲔ ﺃﻭ ﻣﻦ ﺍﻟﻴﺴﺎﺭ(‬ ‫ﺑﺘﻘﻴﻴﺪ ﻣﺂﻝ‬

‫‪x‬‬

‫ﳓﻮ‬

‫ﺑﺎﻟﻘﻴﺪ‬

‫‪a‬‬

‫‪x>a‬‬

‫ﺃﻭ ‪ . x < a‬ﻭﺑﻄﺒﻴﻌﺔ ﺍﳊﺎﻝ ﻓﺈﻥ‬

‫ﺍﻻﺳﺘﻤﺮﺍﺭ ﻋﻨﺪ ﻧﻘﻄﺔ ﻳﻌﲏ ﺃﻥ ﻫﻨﺎﻙ ﺍﺳﺘﻤﺮﺍﺭﺍ ﻣﻦ ﺟﻬﱵ ﺗﻠﻚ ﺍﻟﻨﻘﻄﺔ‪.‬‬ ‫ﻛﻴﻒ ﻧﻔﺴ‪‬ﺮ ﺍﻟﻌﻼﻗﺔ ‪: ε − α‬‬ ‫‪f ( x) − f (a) < ε‬‬

‫ﺍﻟﱵ ﺗﻌﱪ ﻋﻦ ﺍﺳﺘﻤﺮﺍﺭ‬ ‫ﻧﻼﺣﻆ ﺃﻭﻻ ﺃﻥ‬

‫‪∀ε > 0, ∃α > 0 : x − a < α‬‬

‫⇒‬

‫ﻋﻨﺪ ‪ a‬؟‬

‫‪f‬‬

‫‪f ( x) − f (a) < ε‬‬

‫ﺗﻌﲏ ﺑﺄﻥ‬

‫)‪f (x‬‬

‫ﻣﺮﻛﺰﻩ ) ‪ ، f (a‬ﻭﻫﻮ [ ‪ . ] f (a) − ε , f (a) + ε‬ﻛﻤﺎ ﺃﻥ‬ ‫‪x‬‬

‫ﺗﻨﺘﻤﻲ ﺇﱃ ﳎﺎﻝ‬

‫‪x −a 0‬‬ ‫‪: x ≤ 0.‬‬

‫‪0‬‬ ‫‪f ( x) = ‬‬ ‫‪1‬‬

‫‪1‬‬ ‫‪0‬‬

‫‪34‬ن ا ا ‪f‬‬

‫ﻣﺴﺘﻤﺮﺓ ﰲ ﻛﻞ ﻣﻜﺎﻥ ﻣﺎ ﻋﺪﺍ ﰲ ﺍﻟﻨﻘﻄﺔ ‪. 0‬‬

‫‪78‬‬


‫ﻣﺜﺎﻝ ‪2‬‬ ‫ﺗﺼﻮﺭ ﺍﻵﻥ ﺃﻧﻨﺎ ﻧﻌﻴﺪ ﺍﻟﻨﻈﺮ ﰲ ﺍﳌﺜﺎﻝ ﺍﻟﺴﺎﺑﻖ ﻭﻧﻜﺮﺭ ﻣﺎ ﺣﺪﺙ ﰲ‬ ‫‪0‬‬

‫ﻋﻨﺪ ﻛﻞ ﻗﻴﻤﺔ ﻟـ‬

‫‪x‬‬

‫ﺗﺴﺎﻭﻱ ﻋﺪﺩﺍ ﺻﺤﻴﺤﺎ‪ ،‬ﺃﻱ ﺃﻧﻨﺎ ﻧﻌﺘﱪ ﺍﻟﺪﺍﻟﺔ‬

‫ﺍﳌﻌﺮﻓﺔ ﻣﺜﻼ ﻛﻤﺎ ﻳﻠﻲ )ﺣﻴﺚ ﻳﺸﲑ‬

‫‪n‬‬

‫ﻟﻌﻨﺼﺮ ﻛﻴﻔﻲ ﻣﻦ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ‬

‫ﺍﻟﺼﺤﻴﺤﺔ(‪:‬‬ ‫‪g ( x) = 1‬‬ ‫‪g ( x) = 0‬‬

‫ﻣﻦ ﺃﺟﻞ‬ ‫ﻣﻦ ﺃﺟﻞ‬

‫[‪x ∈ [2n,2n + 1‬‬

‫[‪x ∈ [2n + 1,2n + 2‬‬

‫ﺇ‪‬ﺎ ﺩﺍﻟﺔ ﻏﲑ ﻣﺴﺘﻤﺮﺓ ﻋﻨﺪ ﻋﺪﺩ ﻏﲑ ﻣﻨﺘﻪ ﻣﻦ ﺍﻟﻨﻘﺎﻁ‪ .‬ﳎﻤﻮﻋﺔ ﻫﺬﻩ ﺍﻟﻨﻘﺎﻁ‬ ‫ﻫﻲ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ‪.‬‬

‫ﺑﻴﺎﻥ ا ا ‪g‬‬

‫‪79‬‬


‫ﻣﺜﺎﻝ ‪3‬‬ ‫ﳝﻜﻦ ﺃﻳﻀﺎ ﺍﻟﺘﻔﻜﲑ ﰲ ﺍﻟﺪﺍﻟﺔ ﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ﺍ‪‬ﺎﻝ‬ ‫]‪ [x‬ﺇﱃ ﺍﳉﺰﺀ ﺍﻟﺼﺤﻴﺢ ﻟـ‬ ‫ﺇﻥ ﺍﻟﺪﺍﻟﺔ‬

‫‪u‬‬

‫‪x‬‬

‫‪:‬‬

‫‪ℝ‬‬

‫ﻛﺎﻟﺘﺎﱄ ﺣﻴﺚ ﻳﺸﲑ‬

‫] ‪u ( x ) = [x‬‬

‫ﻣﺴﺘﻤﺮﺓ ﻣﺎ ﻋﺪﺍ ﻋﻨﺪ ﻗﻴﻢ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ‪.‬‬

‫ﻣﺜﺎﻝ ‪4‬‬

‫ﺑﻴﺎﻥ ا ا ‪u‬‬

‫ﻧﻄﺮﺡ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺘﺎﱄ ‪ :‬ﻫﻞ ﺗﻮﺟﺪ ﺩﺍﻟﺔ ﻟﻴﺴﺖ ﻣﺴﺘﻤﺮﺓ ﰲ ﺃﻳﺔ ﻧﻘﻄﺔ‬ ‫ﻋﻠﻰ ‪ ℝ‬؟ ﻫﺬﺍ ﺍﻟﺴﺆﺍﻝ ﻟﻴﺲ ﻭﻟﻴﺪ ﺍﻟﻴﻮﻡ ﻭﻟﺬﺍ ﲝﺚ ﻓﻴﻪ ﺃﺳﻼﻓﻨﺎ ﻭﺍﻫﺘﺪﻭﺍ ﺇﱃ‬ ‫ﺇﻧﺸﺎﺀ ﺩﺍﻟﺔ ﻣﻦ ﻫﺬﺍ ﺍﻟﻘﺒﻴﻞ‪ .‬ﺧﺬ ﻣﺜﻼ ﺍﻟﺪﺍﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺣﻴﺚ ﻳﺮﻣﺰ‬

‫‪ℚ‬‬

‫‪‬ﻤﻮﻋﺔ‬

‫ﺍﻷﻋﺪﺍﺩ ﺍﻟﻨﺎﻃﻘﺔ ‪:‬‬ ‫‪0 : x ∈ ℚ‬‬ ‫‪v( x ) = ‬‬ ‫‪1 : x ∉ ℚ.‬‬

‫ﻻﺣﻆ ﺃ‪‬ﺎ ﺩﺍﻟﺔ ﻻ ﳝﻜﻦ ﺭﺳﻢ ﺑﻴﺎ‪‬ﺎ ﻭﻫﻲ ﻏﲑ ﻣﺴﺘﻤﺮﺓ ﰲ ﺃﻳﺔ ﻧﻘﻄﺔ ﻣﻦ ‪. ℝ‬‬ ‫ﺩﻋﻨﺎ ﻧﻨﻬﻲ ﻫﺬﺍ ﺍﳌﻘﻄﻊ ﺑﺘﻘﺪﱘ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺗﺎﺭﻛﲔ ﺑﺮﻫﺎ‪‬ﺎ ﻟﻠﻘﺎﺭﺉ‪.‬‬

‫‪80‬‬


‫ﻧﻈﺮﻳﺔ )ﺍﺳﺘﻤﺮﺍﺭ ﺗﺮﻛﻴﺐ ﺍﻟﺪﻭﺍﻝ(‬ ‫ﻟﻴﻜﻦ‬ ‫ﺩﺍﻟﺘﲔ ﻭ‬

‫‪a‬‬

‫ﺟﺰﺀﻳﻦ ﻣﻦ‬

‫‪ A‬ﻭ‪B‬‬

‫‪ ℝ‬ﻭ‪f :A→B‬‬

‫ﻭ‬

‫‪g:B →ℝ‬‬

‫ﻧﻘﻄﺔ ﻣﻦ ‪. A‬‬

‫ﻧﻔﺮﺽ ﻗﻴﺎﻡ ﺍﻟﺸﺮﻃﲔ ‪:‬‬ ‫‪(1‬‬

‫‪f‬‬

‫ﻣﺴﺘﻤﺮ ﻋﻨﺪ ‪، a‬‬

‫‪(2‬‬

‫‪g‬‬

‫ﻣﺴﺘﻤﺮ ﻋﻨﺪ ) ‪. f (a‬‬

‫ﻋﻨﺪﺋﺬ ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺔ‬

‫‪g f : A→ℝ‬‬

‫ﻣﺴﺘﻤﺮﺓ ﻋﻨﺪ ‪. a‬‬

‫ﻧﻈﺮﻳﺔ )ﺍﻻﺳﺘﻤﺮﺍﺭ ﻋﻠﻰ ﻣﺘﺮﺍﺹ(‬ ‫ﻛﻞ ﺩﺍﻟﺔ‬

‫‪f‬‬

‫ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﻣﺘﺮﺍﺹ ]‪ [a, b‬ﺩﺍﻟﺔ ﳏﺪﻭﺩﺓ ﻭﺗﺪﺭﻙ‬

‫ﺣﺪﻳﻬﺎ ﺍﻷﻋﻠﻰ ﻭﺍﻷﺩﱏ‪.‬‬ ‫ﺍﻟﱪﻫﺎﻥ‬ ‫‪ (1‬ﻧﻔﺮﺽ ﺃﻥ‬

‫‪f‬‬

‫ﻏﲑ ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﻋﻠﻰ ﻭﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ]‪. [a, b‬‬

‫ﻻﺣﻆ ﺃ‪‬ﺎ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ‪ .‬ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻋﺪﺩ‬ ‫ﻃﺒﻴﻌﻲ‬

‫‪n‬‬

‫ﺍﳌﺘﺘﺎﻟﻴﺔ‬

‫‪xn‬‬

‫ﻳﻮﺟﺪ ﻋﻨﺼﺮ‬

‫‪xn‬‬

‫ﻣﻦ ]‪ [a, b‬ﲝﻴﺚ‬

‫‪f ( xn ) > n‬‬

‫‪ .‬ﻭﳌﺎ ﻛﺎﻧﺖ‬

‫ﳏﺪﻭﺩﺓ )ﲝﻜﻢ ﺍﻧﻨﺘﻤﺎﺋﻬﺎ ﺇﱃ ﳎﺎﻝ ﳏﺪﻭﺩ( ﻓﺈﻧﻨﺎ ﻧﺴﺘﻄﻴﻊ ﺃﻥ‬

‫ﻧﺴﺘﺨﺮﺝ ﻣﻨﻬﺎ ﻣﺘﺘﺎﻟﻴﺔ ﺟﺰﺋﻴﺔ ‪ xn‬ﻣﺘﻘﺎﺭﺑﺔ ﻭ‪‬ﺎﻳﺘﻬﺎ ‪ x‬ﺗﻨﺘﻤﻲ ﺇﱃ ]‪[a, b‬‬ ‫ﻷﻥ ‪ . a ≤ xn ≤ b‬ﻭﻣﻦ ﰒ ﳓﺼﻞ ﻋﻠﻰ ﺗﻨﺎﻗﺾ ﻳﺘﻤﺜﹼﻞ ﰲ ‪ :‬ﻣﻦ ﺟﻬﺔ ﻟﺪﻳﻨﺎ‬ ‫‪lim f ( x n ) = f ( x) < +∞ .‬‬ ‫∞‪n → +‬‬ ‫‪k‬‬

‫‪k‬‬

‫‪k‬‬

‫‪k‬‬

‫‪81‬‬


‫ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ ﺗﺆﺩﻱ ﺍﻟﻌﻼﻗﺔ‬ ‫‪) = +∞ .‬‬

‫ﺇﱃ‬

‫‪f ( x nk ) > n k‬‬

‫ﻭﺑﺎﻟﺘﺎﱄ ﻓﺎﻟﺪﺍﻟﺔ‬

‫‪f ( x nk‬‬

‫‪lim‬‬

‫∞‪nk → +‬‬

‫ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﻋﻠﻰ‪.‬‬

‫‪f‬‬

‫* ﺍﻟﱪﻫﺎﻥ ﻋﻠﻰ ﺃﻥ‬

‫‪f‬‬

‫ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﺩﱏ ﺷﺒﻴﻪ ﺑﺎﻟﱪﻫﺎﻥ ﻋﻠﻰ ﺍﶈﺪﻭﺩﻳﺔ ﻣﻦ‬

‫ﺍﻷﻋﻠﻰ‪ .‬ﻗﺪ‪‬ﻡ ﺗﻔﺎﺻﻴﻞ ﺍﻟﱪﻫﺎﻥ ﻋﻠﻰ ﺍﶈﺪﻭﺩﻳﺔ ﻣﻦ ﺍﻷﺩﱏ‪.‬‬ ‫ﻧﻮﺍﺻﻞ ا ‪+6‬هن ﺑﺎﻟﺘﺄﻛﺪ ﺍﻵﻥ ﻣﻦ ﺃﻥ ﺍﻟﺪﺍﻟﺔ‬ ‫]‪ [a, b‬ﺗﺪﺭﻙ ﺣﺪﻫﺎ ﺍﻷﻋﻠﻰ‪ ،‬ﺃﻱ ﺃﻧﻪ ﺗﻮﺟﺪ ﻧﻘﻄﺔ‬ ‫)‪. f (ξ ) = sup f ( x‬‬

‫‪ξ‬‬

‫‪f‬‬

‫ﺍﳌﺴﺘﻤﺮﺓ ﻋﻠﻰ‬


‫] ‪x∈[a ,b‬‬

‫ﻧﻘﺪﻡ ﺑﺮﻫﺎﻧﺎ ﺑﺎﳋﻠﻒ ‪ :‬ﻧﻌﻠﻢ ﳑﺎ ﺳﺒﻖ ﺃﻥ‬ ‫‪f ( x) = s‬‬

‫)‪sup f ( x‬‬

‫] ‪x∈[a ,b‬‬

‫ﻣﻮﺟﻮﺩ ﰲ ‪ . ℝ‬ﻟﻨﻀﻊ‬

‫‪ sup‬ﻭﻟﻨﻔﺮﺽ ﺃﻥ‬

‫] ‪x∈[a ,b‬‬

‫‪f ( x) < s .‬‬

‫ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﺪﺍﻟﺔ‬

‫‪u‬‬

‫‪∀x ∈ [a, b] :‬‬

‫ﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ]‪ [a, b‬ﺑـ‬ ‫‪1‬‬ ‫)‪s − f ( x‬‬

‫= )‪u ( x‬‬

‫ﺩﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ]‪ . [a, b‬ﻭﻣﻦ ﰒﹼ ﻓﻬﻲ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻭﳏﺪﻭﺩﺓ ﻋﻠﻰ‬ ‫]‪ . [a, b‬ﻧﻀﻊ ‪ . sup u ( x) = t‬ﻣﻦ ﺍﳌﺆﻛﺪ ﺃﻥ‬ ‫] ‪x∈[a ,b‬‬

‫‪1‬‬ ‫‪> 0.‬‬ ‫)‪s − f ( x‬‬

‫ﻭﻋﻠﻴﻪ‬

‫‪t>0‬‬

‫= )‪u ( x‬‬

‫‪∀x ∈ [a, b ] :‬‬

‫ﻭ‬ ‫‪∀x ∈ [a, b] : 0 < u ( x) ≤ t .‬‬ ‫‪82‬‬


‫ﻭﻣﻨﻪ ﻳﻨﺘﺞ‬ ‫‪1‬‬ ‫‪f (x) ≤ s − < s .‬‬ ‫‪t‬‬

‫ﻭﻫﺬﺍ ﻳﺘﻨﺎﰱ ﻣﻊ ﺍﻟﻘﻮﻝ ﺇﻥ‬ ‫ﺍﻓﺘﺮﺍﺿﻨﺎ‬

‫‪f ( x) < s‬‬

‫‪f ( x) = s‬‬

‫ﺇﺫﻥ ﻓﺈﻥ‬

‫‪ . sup‬ﻫﺬﺍ ﺍﻟﺘﻨﺎﻗﺾ ﻧﺎﺟﻢ ﻣﻦ‬

‫] ‪x∈[a ,b‬‬

‫‪∀x ∈ [a, b] :‬‬

‫ﺍﻟﺬﻱ ﻳﻌﲏ ﺃﻧﻪ ﻻ ﻭﺟﻮﺩ ﻟﻨﻘﻄﺔ‬ ‫‪f‬‬

‫‪∀x ∈ [a, b ] :‬‬

‫‪ξ‬‬


‫= ) ‪. f (ξ‬‬

‫)‪sup f ( x‬‬

‫] ‪x∈[a ,b‬‬

‫ﻳﺪﺭﻙ ﺣﺪﻩ ﺍﻷﻋﻠﻰ‪.‬‬

‫* ﻗﺪ‪‬ﻡ ﺗﻔﺎﺻﻴﻞ ﺇﺩﺭﺍﻙ‬

‫‪f‬‬

‫ﳊﺪﻩ ﺍﻷﺩﱏ‪.‬‬

‫ﻣﻼﺣﻈﺔ‬ ‫ﻻﺣﻆ ﺃﻥ ﳏﺪﻭﺩﻳﺔ ﺍ‪‬ﺎﻝ ﻣﻬﻤﺔ ﰲ ﻫﺬﻩ ﺍﻟﻨﻈﺮﻳﺔ‪ .‬ﻟﻠﺘﺄﻛﺪ ﻣﻦ ﺫﻟﻚ‬ ‫ﺧﺬ ﻣﺜﻼ ﺇﺣﺪﻯ ﺍﻟﺪﺍﻟﺘﲔ ‪ :‬ﺩﺍﻟﺔ ﺍﻟﻈﻞ ﻋﻠﻰ ﺍ‪‬ﺎﻝ‬

‫‪ π π‬‬ ‫‪ − 2 , 2 ‬‬

‫ﺃﻭ‬

‫ﺍﻟﺪﺍﻟﺔ ‪g‬‬

‫ﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ]‪ ]0,1‬ﺑـ ‪. g ( x) = 1‬‬ ‫‪x‬‬

‫ﻻﺣﻆ ﺃﻳﻀﺎ ﺃﻥ ﻏﻠﻖ ﺍ‪‬ﺎﻝ ﻣﻬﻢ ﰲ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ‪ .‬ﻟﻠﺘﺄﻛﺪ ﻣﻦ ﺫﻟﻚ ﺍﻋﺘﱪ‬ ‫ﻛﻤﺜﺎﻝ‬

‫ﺍﻟﺪﺍﻟﺔ ‪g‬‬

‫ﺃﻭ ﺍﻟﺪﺍﻟﺔ ‪ h‬ﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ﺍ‪‬ﺎﻝ [‪ [1,2004‬ﺑـ‬

‫‪h( x) = x‬‬

‫ﻓﻬﻲ ﻻ ﺗﺪﺭﻙ ﺣﺪﻫﺎ ﺍﻷﻋﻠﻰ ﰲ ﺍ‪‬ﺎﻝ ﺍﳌﻌﺘﱪ‪.‬‬ ‫ﻣﻦ ﺍﻟﻨﻈﺮﻳﺎﺕ ﺍﳌﻬﻤﺔ ﺃﻳﻀﺎ ﰲ ﻣﻮﺿﻮﻉ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺴﺘﻤﺮﺓ ﺍﻟﻨﺘﻴﺠﺔ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ ﻧﻈﺮﻳﺔ ﺍﻟﻘﻴﻢ ﺍﻟﻮﺳﻄﻰ ﺍﻟﱵ ﺑﺮﻫﻦ ﻋﻠﻴﻬﺎ ﻷﻭﻝ ﻣﺮﺓ ﺧﻼﻝ ﺍﻟﺮﺑﻊ ﺍﻷﻭﻝ‬ ‫ﻣﻦ ﺍﻟﻘﺮﻥ ﺍﻟﺘﺎﺳﻊ ﻋﺸﺮ ﺍﻟﺘﺸﻴﻜﻲ ﺑﻮﻟﺰﺍﻧﻮ ﻭﺍﻟﻔﺮﻧﺴﻲ ﻛﻮﺷﻲ‪ .‬ﺗﻘﻮﻝ ﻫﺬﻩ‬ ‫ﺍﻟﻨﻈﺮﻳﺔ – ﺑﺘﻌﺒﲑ ﺑﺴﻴﻂ ‪ -‬ﺇﻧﻨﺎ ﻻ ﻧﺴﺘﻄﻴﻊ ﺍﳌﺮﻭﺭ ﻣﻦ ﺿﻔﺔ ﺇﱃ ﺃﺧﺮﻯ ﻋﱪ‬ ‫‪‬ﺮ ﺑﺪﻭﻥ ﻗﻔﺰ ﻭﺩﻭﻥ ﺃﻥ ﺗﺒﺘﻞﹼ ﺃﻗﺪﺍﻣﻨﺎ‪ .‬ﻭﻧﻌﺒ‪‬ﺮ ﻋﻦ ﺫﻟﻚ ﺭﻳﺎﺿﻴﺎ ﺑﺎﻟﻘﻮﻝ ‪ :‬ﺇﺫﺍ‬ ‫‪83‬‬


‫ﺃﺧﺬﺕ ﺩﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﳌﺘﻐﲑ ﻭﺍﺣﺪ ﺇﺷﺎﺭﺗﲔ ﳐﺘﻠﻔﺘﲔ ﻋﻨﺪ‬

‫ﻗﻴﻤﺘﲔ ‪ a‬ﻭ ‪b‬‬

‫ﺑﲔ ‪a‬‬

‫ﻭ ‪ . b‬ﻭﻫﻮ ﻣﺎ‬

‫فﺇن ﻫﺬﻩ ﺍﻟﺪﺍﻟﺔ ﺗﻨﻌﺪﻡ‪ ،‬ﻋﻠﻰ ﺍﻷﻗﻞ ﻣﺮﺓ ﻭﺍﺣﺪﺓ‪،‬‬ ‫ﻳﻘﻮﻝ ﺍﻟﻨﺺ ﺍﳌﺄﻟﻮﻑ ﺍﻟﺘﺎﱄ ‪:‬‬ ‫ﻧﻈﺮﻳﺔ )ﻧﻘﻄﺔ ﺍﻧﻌﺪﺍﻡ(‬ ‫ﻟﺘﻜﻦ‬

‫‪f‬‬

‫ﺩﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﳎﺎﻝ ﻣﺘﺮﺍﺹ ]‪ . [a, b‬ﺇﻥ ﻛﺎﻥ‬ ‫ﻓﺈﻧﻪ ﺗﻮﺟﺪ ﻧﻘﻄﺔ‬

‫‪f (a ). f (b) < 0‬‬

‫ﻣﻦ ]‪ [a, b‬ﲝﻴﺚ ‪. f (c) = 0‬‬

‫‪c‬‬

‫ﺍﻟﱪﻫﺎﻥ‬ ‫ﻟﻨﻔﺮﺽ ﻣﺜﻼ ﺑﺄﻥ‬

‫)ﻭﻣﻨﻪ ﺳﻴﻜﻮﻥ‬

‫‪f (a) > 0‬‬

‫}‪X = {x ∈ [a, b] : f ( x) > 0‬‬


‫‪c‬‬

‫ﲢﺘﻔﻆ ﻓﻴﻪ‬

‫‪f (b) < 0‬‬

‫ﻭ ‪ . c = sup X‬ﻭﻟﻨﺒﻴ‪‬ﻦ ﺃﻥ ‪. f (c) = 0‬‬

‫ﻣﻦ ﺍﻟﻮﺍﺿﺢ ﺃﻥ ‪ . a < c < b‬ﻓﻠﻮ ﻛﺎﻥ‬ ‫‪f‬‬

‫(‪ .‬ﻭﻟﻨﻀﻊ‬

‫‪f (c ) ≠ 0‬‬

‫ﻟﻮﺟﺪ ﳎﺎﻝ‬

‫ﺑﺈﺷﺎﺭ‪‬ﺎ ﺍﳌﻮﺟﺒﺔ‪ ،‬ﻭﺫﻟﻚ ﺑﻔﻀﻞ ﺍﺳﺘﻤﺮﺍﺭ‬

‫ﺫﻟﻚ(‪ .‬ﻭﻫﺬﺍ ﻳﻨﺎﻗﺾ ﺍﻟﻘﻮﻝ‬

‫‪c = sup X‬‬

‫‪f‬‬

‫)ﻭﺿ‪‬ﺢ‬

‫ﺍﻟﺬﻱ ﻳﻌﺮﻑ ﺍﻟﻨﻘﻄﺔ ‪ . c‬ﻭﺑﺎﻟﺘﺎﱄ‬

‫‪ . f (c) = 0‬ﻧﺴﺘﻨﺘﺞ ﻣﻦ ﺫﻟﻚ ﻫﺬﺍ ﺍﻟﺘﻌﻤﻴﻢ ‪:‬‬ ‫ﻧﻈﺮﻳﺔ )ﻧﻈﺮﻳﺔ ﺍﻟﻨﻘﻄﺔ ﺍﳌﺘﻮﺳﻄﺔ(‬ ‫ﻟﺘﻜﻦ‬ ‫ﻭﻟﺘﻜﻦ‬

‫) ‪f ( x1‬‬

‫‪f : I ⊂ ℝ ‬‬ ‫‪→ℝ‬‬

‫ﻭ‬

‫) ‪f ( x2‬‬

‫ﻗﻴﻤﺘﲔ ﻟـ‬

‫ﻋﻨﺪﺋﺬ ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻋﻨﺼﺮ‬ ‫ﻳﻮﺟﺪ ﻋﻨﺼﺮ‬

‫‪x0‬‬

‫ﺩﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﳎﺎﻝ ﻛﻴﻔﻲ ‪. I‬‬ ‫‪c‬‬

‫‪f‬‬

‫ﺣﻴﺚ ‪. x1 < x2‬‬

‫‪"89%‬ر ﺑﲔ‬

‫) ‪f ( x1‬‬

‫ﻣﻦ ﺍ‪‬ﺎﻝ [ ‪ ]x1 , x2‬ﳛﻘﻖ ‪. f ( x0 ) = c‬‬ ‫‪84‬‬

‫و‬

‫) ‪f ( x2‬‬


‫ﻣﻼﺣﻈﺔ‬ ‫ﻳﻨﺘﺞ ﻣﻦ ﺫﻟﻚ ﺃﻥ ﺻﻮﺭﺓ ﳎﺎﻝ ﻋﱪ ﺩﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﻫﻲ ﺃﻳﻀﺎ ﳎﺎﻝ‪ .‬ﻛﻤﺎ‬ ‫ﻳﻨﺘﺞ ﻣﻦ ﻫﺬﻩ ﺍﻟﻨﻈﺮﻳﺔ ﻭﺍﻟﱵ ﺳﺒﻘﺘﻬﺎ ﰲ ﺣﺎﻟﺔ ﺗﺮﺍﺹ ﺍ‪‬ﺎﻝ ]‪ [a, b‬ﺃﻥ ﺻﻮﺭﺓ‬ ‫ﻫﺬﺍ ﺍ‪‬ﺎﻝ ﻫﻲ ﺍ‪‬ﺎﻝ‬

‫‪.  sup‬‬

‫‪‬‬ ‫‪f ( x), inf f ( x)‬‬ ‫] ‪x∈[a ,b‬‬ ‫] ‪ x∈[a ,b‬‬ ‫‪‬‬

‫‪85‬‬


‫‪ .5‬ﺍﻻﺷﺘﻘﺎﻕ ‪:‬‬ ‫ﺗﻌﺮﻳﻒ )ﻣﺸﺘﻖ ﺗﺎﺑﻊ ﻋﻨﺪ ﻧﻘﻄﺔ(‬ ‫ﻟﻴﻜﻦ ‪ I‬ﳎﺎﻻ ﻣﻔﺘﻮﺣﺎ‬ ‫‪f :I →ℝ‬‬

‫ﻣﻦ ‪ ℝ‬ﻭ ‪x0‬‬

‫ﻧﻘﻄﺔ ﻣﻦ ‪ . I‬ﻭ ﻟﻴﻜﻦ‬

‫ﺗﺎﺑﻌﺎ ﺣﻘﻴﻘﻴﺎ‪.‬‬

‫ﻧﻘﻮﻝ ﻋﻦ ‪ f‬ﺇﻧﻪ ﻗﺎﺑﻞ ﻟﻼﺷﺘﻘﺎﻕ ﺍﻟﻨﻘﻄﺔ‬ ‫)‪f(x)− f(x0‬‬ ‫‪x − x0‬‬

‫‪x0‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻨﻬﺎﻳﺔ‬

‫‪lim‬‬

‫‪x → x0‬‬

‫ﻣﻮﺟﻮﺩﺓ ﻭ ﻣﻨﺘﻬﻴﺔ‪ .‬ﺗﺴﻤﻰ ﻫﺬﻩ ﺍﻟﻨﻬﺎﻳﺔ ﺍﻟﻌﺪﺩ ﺍﳌﺸﺘﻖ )ﺃﻭ ﺍﳌﺸﺘﻖ( ﻟﻠﺘﺎﺑﻊ ‪f‬‬ ‫ﻋﻨﺪ‬

‫ﺍﻟﻨﻘﻄﺔ ‪x0‬‬

‫و ﻧﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ )‪ ، f'(x0‬ﺃﻱ‬

‫)‪f(x)− f(x0‬‬ ‫)‪f(x0 +h)− f(x0‬‬ ‫‪= lim‬‬ ‫‪.‬‬ ‫‪h →0‬‬ ‫‪x− x0‬‬ ‫‪h‬‬

‫‪f'(x0)= lim‬‬ ‫‪x → x0‬‬

‫ﻣﻼﺣﻈﺔ‬ ‫ﻧﺴﺘﻨﺘﺞ ﻣﻦ ﻭﺣﺪﺍﻧﻴﺔ ﺍﻟﻨﻬﺎﻳﺔ ﺃﻥ ﻣﺸﺘﻖ ﺗﺎﺑﻊ ‪ f‬ﻋﻨﺪ ﻧﻘﻄﺔ ‪، x0‬‬ ‫ﻭﺣﻴﺪ )ﺇﻥ ﻭﺟﺪ(‪.‬‬ ‫ﻳﺘﻀﺢ ﻣﻦ ﺗﻌﺮﻳﻒ ﺍﳌﺸﺘﻖ ﺑﻜﺘﺎﺑﺔ‬ ‫)‪f(x)− f(x0‬‬ ‫‪= f'(x0)+ε(x),‬‬ ‫‪x− x0‬‬

‫‪. xlim‬‬ ‫ﺣﻴﺚ ‪ε(x)=0‬‬ ‫‪→x‬‬ ‫‪0‬‬

‫‪86‬‬


‫ﺃﻣﺜﻠﺔ‬ ‫ ﺇﺫﺍ ﻛﺎﻥ ‪ f‬ﺛﺎﺑﺘﺎ ﻓﺈﻥ‬‫‪ -‬ﺇﺫﺍ‬

‫ﻛﺎﻥ ‪f(x)= x‬‬

‫ﻟﻴﻜﻦ ‪x‬‬

‫‪f'(x)=0‬‬

‫ﻓﺈﻥ‬

‫ﻣﻬﻤﺎ ﻛﺎﻥ ‪ x‬ﻣﻦ ‪. ℝ‬‬

‫‪f'(x)=1‬‬

‫ﻣﻬﻤﺎ ﻛﺎﻥ ‪ x‬ﻣﻦ ‪. ℝ‬‬

‫=)‪ ، f'(1)= 12 . f(x‬ﻷﻥ‬

‫‪f (x ) − f (1) 1‬‬ ‫=‬ ‫‪x →1‬‬ ‫‪x −1‬‬ ‫‪2‬‬

‫‪lim‬‬

‫)‪f (x ) − f (1‬‬ ‫‪x −1‬‬ ‫=‬ ‫=‬ ‫‪x −1‬‬ ‫‪x −1‬‬

‫‪1‬‬

‫⇒‬

‫‪x +1‬‬

‫ﺗﻌﺮﻳﻒ )ﺍﳌﺸﺘﻖ ﻣﻦ ﺍﻟﻴﻤﲔ ﻭ ﺍﳌﺸﺘﻖ ﻣﻦ ﺍﻟﻴﺴﺎﺭ(‬ ‫ﻟﻴﻜﻦ ‪ I‬ﳎﺎﻻ ﻣﻔﺘﻮﺣﺎ ﻣﻦ ﺍﳌﺸﺘﻖ ﻣﻦ ﺍﻟﻴﻤﲔ ﻭ ﺍﳌﺸﺘﻖ ﻣﻦ ﺍﻟﻴﺴﺎﺭ‬ ‫‪ ℝ‬ﻭ ‪x0‬‬

‫ﻧﻘﻄﺔ ﻣﻦ ‪. I‬‬ ‫ﻭ ﻟﻴﻜﻦ‬

‫‪f :I →ℝ‬‬

‫‪ .1‬ﺇﺫﺍ ﻗﺒﻠﺖ ﺍﻟﻨﺴﺒﺔ‬ ‫ﻧﻘﻮﻝ ﺇﻥ‬

‫‪f‬‬

‫ﺗﺎﺑﻌﺎ ﺣﻘﻴﻘﻴﺎ‪.‬‬ ‫)‪f(x)− f(x0‬‬ ‫‪x− x0‬‬

‫‪‬ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﻣﻦ ﺍﻟﻴﻤﲔ‬

‫ﻋﻨﺪ ‪x0‬‬

‫ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻣﻦ ﺍﻟﻴﻤﲔ ﻋﻨﺪ ‪ . x0‬ﺗﺴﻤﻰ ﻫﺬﻩ ﺍﻟﻨﻬﺎﻳﺔ ﺍﳌﺸﺘﻖ‬

‫ﻣﻦ ﺍﻟﻴﻤﲔ‪ ،‬ﻭﻧﺮﻣﺰ ﳍﺎ ﺑـ )‪. f'd (x0‬‬ ‫‪ .2‬ﺇﺫﺍ ﻗﺒﻠﺖ ﺍﻟﻨﺴﺒﺔ‬ ‫‪x0‬‬

‫ﻧﻘﻮﻝ ﺇﻥ‬

‫‪f‬‬

‫)‪f(x)− f(x0‬‬ ‫‪x− x0‬‬

‫‪‬ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﻋﻨﺪ‬

‫ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﻋﻨﺪ ‪. x0‬‬

‫ﺗﺴﻤﻰ ﻫﺬﻩ ﺍﻟﻨﻬﺎﻳﺔ ﺍﳌﺸﺘﻖ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﻭ ﻧﺮﻣﺰ ﳍﺎ ﺑـ )‪. f'g (x0‬‬ ‫ﻣﻼﺣﻈﺎﺕ ‪:‬‬ ‫ﺣﱴ ﻳﻜﻮﻥ‬

‫‪f‬‬

‫ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻋﻨﺪ‬

‫‪x0‬‬

‫ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻣﻦ ﺍﻟﻴﻤﲔ ﻭﻣﻦ ﺍﻟﻴﺴﺎﺭ ﻋﻨﺪ ‪. x0‬‬ ‫‪87‬‬

‫ﻳﻠﺰﻡ ﻭﻳﻜﻔﻲ ﺃﻥ ﻳﻜﻮﻥ‬

‫‪f‬‬


‫ﻗﺪ ﻳﻜﻮﻥ‬

‫‪f‬‬

‫ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻣﻦ ﺍﻟﻴﻤﲔ ﻭﻣﻦ ﺍﻟﻴﺴﺎﺭ‬


‫ﺩﻭﻥ‬

‫ﺃﻥ ﻳﻜﻮﻥ ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻋﻨﺪ ‪. x0‬‬ ‫ﻣﺜﺎﻝ ﻟﺘﺎﺑﻊ ﻻ ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻣﻊ ﻗﺒﻮﻟﻪ ﻟﻠﻤﺸﺘﻘﲔ ﳝﻴﻨﺎ ﻳﺴﺎﺭﺍ ‪ :‬ﻟﻴﻜﻦ‬ ‫‪ . f(x)= x‬ﻟﻨﺪﺭﺱ ﻗﺎﺑﻠﻴﺔ‬

‫‪f‬‬

‫ﻟﻼﺷﺘﻘﺎﻕ ﻋﻨﺪ ‪.0‬‬

‫)‪f(x)− f(0‬‬ ‫)‪f(x)− f(0‬‬ ‫‪=1 , lim‬‬ ‫‪= −1‬‬ ‫‬ ‫‪x→ 0‬‬

‫ﻟﻜﻨﻪ ﻏﲑ ﻗﺎﺑﻞ‬

‫ﻟﻼﺷﺘﻘﺎﻕ ﻋﻨﺪ ‪. 0‬‬ ‫ﺍﻟﺘﻔﺴﲑ ﺍﳍﻨﺪﺳﻲ ﻟﻠﻤﺸﺘﻖ ‪:‬‬ ‫ﻟﻴﻜﻦ‬ ‫)‪M 0 ، (o,i, j‬‬

‫)‪(C‬‬

‫ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ ﻟﻠﺘﺎﺑﻊ‬

‫ﻧﻘﻄﺔ ﻣﻦ ﺍﳌﻨﺤﲏ‬

‫)‪(C‬‬

‫‪f‬‬

‫ﰲ ﻣﺴﺘﻮ ﻣﻨﺴﻮﺏ ﺇﱃ ﻣﻌﻠﻢ‬

‫ﻓﺎﺻﻠﺘﻬﺎ ‪. x0‬‬

‫ﺇﺫﺍ ﻛﺎﻥ ‪ f‬ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻋﻨﺪ ‪ ، x0‬ﻓﺈﻥ‬

‫)‪(C‬‬

‫ﻳﻘﺒﻞ ﳑﺎﺳﺎ‪ ،‬ﻋﻨﺪ‬

‫ﺍﻟﻨﻘﻄﺔ ‪ ، M 0‬ﻣﻌﺎﺩﻟﺘﻪ‬ ‫‪(x0)(x −x0).‬‬

‫' ‪y = f (x0) +f‬‬

‫ﻣﻴﻞ ﳑﺎﺱ ﻣﻨﺤﲏ ﺍﻟﺘﺎﺑﻊ‬ ‫ﻣﺸﺘﻖ اﻟﺘﺎﺑﻊ‬

‫‪f‬‬

‫‪f‬‬

‫ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ ذات‬

‫ﺍﻟﻔﺎﺻﻠﺔ ‪x0‬‬

‫ﻫﻮ‬

‫ﻋﻨﺪ ‪. x0‬‬

‫ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻣﻦ ﺍﻟﻴﻤﲔ )ﻣﻦ ﺍﻟﻴﺴﺎﺭ‪ ،‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ(‬

‫ﺇﺫﺍ ﻛﺎﻥ‬

‫‪f‬‬

‫ﻓﺈﻥ )‪(C‬‬

‫ﻳﻘﺒﻞ‪ ،‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ ‪ ، M 0‬ﻧﺼﻒ ﳑﺎﺱ ﻣﻦ ﺍﻟﻴﻤﲔ )ﻣﻦ ﺍﻟﻴﺴﺎﺭ‪ ،‬ﻋﻠﻰ‬

‫ﺍﻟﺘﻮﺍﱄ( ﻣﻌﺎﻣﻞ‬

‫ﺗﻮﺟﻴﻬﻪ )‪f'd (x0‬‬

‫) )‪ ، f'g (x0‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ(‪.‬‬

‫‪88‬‬



‫ﺇﺫﺍ ﻛﺎﻧﺖ ‪‬ﺎﻳﺔ ﺍﻟﻨﺴﺒﺔ‬ ‫)‪(C‬‬

‫)‪f(x)− f(x0‬‬ ‫‪x− x0‬‬

‫ﻏﲑ ﻣﻨﺘﻬﻴﺔ ﻋﻨﺪ‬

‫‪x0‬‬

‫ﻓﺈ ﻥ‬

‫ﻳﻘﺒﻞ‪ ،‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ ‪ ، M 0‬ﳑﺎﺳﺎ ‪"%‬ﺍز ﻟـ )‪. (y'oy‬‬

‫ﺃﻣﺜﻠﺔ ‪:‬‬ ‫ﺍﻟﺘﺎﺑﻊ‬

‫‪x֏ x‬‬

‫ﻳﻘﺒﻞ ﻧﺼﻔﻲ ﳑﺎﺱ‪ ،‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ )‪ ، o(0,0‬ﻣﻴﻞ‬

‫ﺃﺣﺪﳘﺎ ‪ 1‬ﻭ ﻣﻴﻞ ﺍﻵﺧﺮ –‪.1‬‬ ‫ﺍﻟﺘﺎﺑﻊ ‪x֏ x‬‬

‫ﻻ ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻣﻦ ﺍﻟﻴﻤﲔ ﻋﻨﺪ ‪ 0‬ﻷﻥ‬ ‫∞‪lim x = lim+ 1 =+‬‬ ‫‪x x →0 x‬‬

‫‪x→0 +‬‬

‫ﺇﺫﻥ ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ‬

‫ﻟﻠﺘﺎﺑﻊ ‪x֏ x‬‬

‫ﻳﻘﺒﻞ ﻧﺼﻒ ﳑﺎﺱ ﻣﻮﺍﺯ ﻟـ )‪. (y'oy‬‬

‫ﻗﻀﻴﺔ )ﻗﺎﺑﻠﻴﺔ ﺍﻻﺷﺘﻘﺎﻕ ﻭﺍﻻﺳﺘﻤﺮﺍﺭ(‬ ‫ﺇﺫﺍ ﻛﺎﻥ‬

‫‪f‬‬

‫ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ‬


‫ﻓﺈﻧﻪ ﻣﺴﺘﻤﺮ ﻋﻨﺪ ﻫﺬﻩ ﺍﻟﻨﻘﻄﺔ‪.‬‬

‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫ﻟﻴﻜﻦ‬

‫‪f‬‬


‫ﺃﻱ‬


‫)‪f(x)− f(x0‬‬ ‫‪= f '(x0).‬‬ ‫‪x− x0‬‬

‫‪lim‬‬

‫‪x→ x0‬‬

‫ﺇﺫﻥ‬ ‫)‪f(x)− f(x0‬‬ ‫‪= f '(x0)+ε(x),‬‬ ‫‪x− x0‬‬

‫‪ . xlim‬ﻧﺴﺘﻨﺘﺞ ﺃﻥ‬ ‫ﺣﻴﺚ ‪ε(x)=0‬‬ ‫‪→x‬‬ ‫‪0‬‬

‫‪f(x)= f(x0)+(x − x0)f '(x0)+ε(x).‬‬

‫ﺑﺄﺧﺬ ‪‬ﺎﻳﺔ ﺍﻟﻄﺮﻓﲔ ﳌﺎ ‪ x‬ﻳﺆﻭﻝ ﺇﱃ‬

‫‪x0‬‬

‫ﳒﺪ‬

‫‪lim f(x)= f(x0).‬‬

‫‪x → x0‬‬

‫ﺃﻱ ﺃﻥ‬

‫‪f‬‬

‫ﻣﺴﺘﻤﺮ ﻋﻨﺪ ‪. x0‬‬ ‫‪89‬‬


‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﺍﻟﻘﻀﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﻘﻀﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺧﺎﻃﺌﺔ‪ .‬ﳝﻜﻦ ﺃﻥ ﻳﻜﻮﻥ ﺗﺎﺑﻊ‬ ‫ﻣﺴﺘﻤﺮﺍ ﻋﻨﺪ ﻧﻘﻄﺔ ﺩﻭﻥ ﺃﻥ ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻋﻨﺪ ﺗﻠﻚ ﺍﻟﻨﻘﻄﺔ‪ .‬ﻣﺜﻞ ﺫﻟﻚ ‪:‬‬ ‫ﺍﻟﺘﺎﺑﻊ‬

‫‪x֏ x‬‬

‫ﻣﺴﺘﻤﺮ ﻋﻨﺪ ‪ ،0‬ﻟﻜﻨﻪ ﻻ ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻋﻨﺪ ‪.0‬‬

‫ﻧﻈﺮﻳﺔ )ﻣﺸﺘﻖ ﺗﺮﻛﻴﺐ ﺗﺎﺑﻌﲔ(‬ ‫ﻟﻴﻜﻦ‬

‫‪f‬‬

‫ﺗﺎﺑﻌﺎ ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ‬


‫ﻭ ‪ g‬ﺗﺎﺑﻌﺎ ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ‬

‫ﻋﻨﺪ )‪. f(x0‬‬ ‫ﻋﻨﺪﺋﺬ ﻳﻘﺒﻞ ﺍﻟﺘﺎﺑﻊ ‪ g f‬ﺍﻻﺷﺘﻘﺎﻕ ﻋﻨﺪ ‪ ، x0‬ﻭﻟﺪﻳﻨﺎ‬ ‫‪(g f)'(x0)= g'[ f(x0)]⋅ f '(x0).‬‬ ‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫ﲟﺎ ﺃﻥ ﺍﻟﺘﺎﺑﻊ ‪ g‬ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ‬

‫ﻋﻨﺪ )‪f(x0‬‬

‫ﻓﺈﻥ‬

‫))‪g(y)− g(f(x0‬‬ ‫‪= g'(f(x0))+ε(y),‬‬ ‫)‪y − f(x0‬‬

‫ﺣﻴﺚ ‪. y →limf(x )ε(y)=0‬‬ ‫‪0‬‬

‫ﺍﻟﺘﺎﺑﻊ‬

‫‪f‬‬

‫‪ . xlim‬ﺑﺄﺧﺬ )‪ y = f(x‬ﳒﺪ‬ ‫ﻣﺴﺘﻤﺮ‪ ،‬ﺇﺫﻥ )‪f(x)= f(x0‬‬ ‫‪→x‬‬ ‫‪0‬‬

‫))‪g(f(x))− g(f(x0‬‬ ‫‪= g'(f(x0))+ε(f(x)).‬‬ ‫)‪f(x)− f(x0‬‬

‫ﺃﻱ‬ ‫)‪g(f(x))− g(f(x0)) f(x)− f(x0‬‬ ‫‪[g'(f(x0))+ε(f(x))] .‬‬ ‫=‬ ‫‪x− x0‬‬ ‫‪x− x0‬‬

‫ﲜﻌﻞ ‪ x‬ﻳﺆﻭﻝ‬

‫ﺇﱃ ‪x0‬‬

‫ﳓﺼﻞ ﻋﻠﻰ‬ ‫‪(g f)'(x0)= g'[ f(x0)]⋅ f'(x0) .‬‬ ‫‪90‬‬


‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺣﺴﺎﺏ ﻣﺸﺘﻖ ﺍﻟﺘﺎﺑﻊ‬

‫‪ln x‬‬

‫‪ . f(x)=sin‬ﻣﻦ ﺃﺟﻞ‬

‫[∞‪x∈]1,+‬‬

‫ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪f ' ( x) = ( ln x ) 'cos ln x‬‬ ‫')‪(ln x‬‬ ‫=‬ ‫‪cos ln x‬‬ ‫‪2 ln x‬‬ ‫‪= 1 cos ln x .‬‬ ‫‪2x ln x‬‬

‫ﻧﻈﺮﻳﺔ )ﻣﺸﺘﻖ ﺗﺎﺑﻊ ﻋﻜﺴﻲ(‬ ‫ﻟﻴﻜﻦ‬ ‫ﻟﻼﺷﺘﻘﺎﻕ‬

‫‪f‬‬


‫ﺗﻄﺒﻴﻘﺎ ﺗﻘﺎﺑﻠﻴﺎ ﻭﻣﺴﺘﻤﺮﺍ ﻋﻠﻰ ﳎﺎﻝ ‪ I‬ﰲ ﳎﺎﻝ ‪ ، J‬ﻭﻗﺎﺑﻼ‬ ‫ﻣﻦ ‪ . I‬ﺇﺫﺍ ﻛﺎﻥ‬

‫‪f '(x0)≠0‬‬

‫ﻓﺈﻥ ﺍﻟﺘﺎﺑﻊ‬

‫ﻳﻘﺒﻞ‬

‫ﺍﻟﻌﻜﺴﻲ ‪f −1‬‬

‫اﻻﺷﺘﻘﺎﻕ ﻋﻨﺪ )‪ ، f(x0‬ﻭﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪(f −1)'(f(x0))= 1 .‬‬ ‫)‪f '(x0‬‬

‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫ﻣﻦ‬

‫ﺃﺟﻞ ‪t∈J‬‬

‫ﻧﻀﻊ )‪ ، x= f −1(t‬ﺇﺫﻥ )‪ . t = f(x‬ﻟﺪﻳﻨﺎ ‪:‬‬ ‫)‪f −1(t)− f −1(t0‬‬ ‫‪= x− x0‬‬ ‫‪t −t0‬‬ ‫)‪f(x)− f(x0‬‬ ‫‪1‬‬ ‫=‬ ‫)‪f(x)− f(x0‬‬ ‫‪x− x0‬‬

‫ﲟﺎ ﺃﻥ‬

‫ﺍﻟﺘﺎﺑﻊ ‪f −1‬‬

‫ﻛﻤﺎ ﺃﻥ‬

‫‪f‬‬

‫ﻣﺴﺘﻤﺮ ﻋﻨﺪ‬

‫‪t0‬‬

‫ﻳﻘﺒﻞ اﻻﺷﺘﻘﺎﻕ ﻋﻨﺪ‬

‫ﻓﺈﻥ‬ ‫‪x0‬‬

‫)‪lim f −1(t)= f −1(t0‬‬ ‫‪t →t 0‬‬

‫و ‪ . f '(x0)≠0‬ﺇﺫﻥ‬

‫‪f −1(t)− f −1(t0) 1‬‬ ‫=‬ ‫‪.‬‬ ‫‪t −t0‬‬ ‫)‪f'(x0‬‬

‫ﻭ ﻫﻮ ﺍﳌﻄﻠﻮﺏ‪.‬‬ ‫‪91‬‬

‫‪lim‬‬ ‫‪t →t 0‬‬

‫‪. lim‬‬ ‫ﺃﻱ ‪x= x0‬‬ ‫‪t →t‬‬ ‫‪0‬‬


‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫إﺫ ا‬

‫ﻛﺎﻥ ‪f'(x0)=0‬‬

‫ﻓﺈﻥ ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ‬

‫ﻟﻠﺘﺎﺑﻊ ‪f −1‬‬

‫ﻳﻘﺒﻞ‪ ،‬ﻋﻨﺪ‬

‫ﺍﻟﻨﻘﻄﺔ )‪ ، t 0= f(x0‬ﳑﺎﺳﺎ ﻣﻮﺍﺯﻳﺎ ﻟـ ‪. y'oy‬‬ ‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺣﺴﺎﺏ ﻣﺸﺘﻖ ﺍﻟﺘﺎﺑﻊ‬

‫‪arcsin‬‬

‫ﻋﻠﻰ ﺍ‪‬ﺎﻝ ﺍﳌﻔﺘﻮﺡ [‪ . ]−1,1‬ﻟﺪﻳﻨﺎ‬

‫‪1‬‬ ‫)‪sin'(arcsin x‬‬

‫=‪arcsin'x‬‬ ‫‪1‬‬ ‫)‪cos(arcsin x‬‬

‫=‬

‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻌﻼﻗﺔ ﺍﳌﺜﻠﺜﻴﺔ ﺍﻟﺸﻬﲑﺓ ‪ ، sin 2α +cos2α =1‬ﳒﺪ‬ ‫‪cos(arcsin x)= 1−sin 2(arcsin x) = 1− x 2‬‬

‫و ﻣﻨﻪ‬ ‫‪1‬‬ ‫‪1− x 2‬‬

‫ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﳓﺴﺐ ﻣﺸﺘﻖ‬

‫ا ‪arccos >4#‬‬

‫= ‪∀x∈]−1,1[, arcsin'x‬‬

‫ﻋﻠﻰ ﺍ‪‬ﺎﻝ ﺍﳌﻔﺘﻮﺡ [‪]−1,1‬‬

‫ﻓﻨﺠﺪ‬ ‫‪∀x∈]−1,1[, arccos'x= −1‬‬ ‫‪1− x 2‬‬

‫‪92‬‬


‫ﺗﻌﺮﻳﻒ )ﺍﳌﺸﺘﻘﺎﺕ ﺫﺍﺕ ﺍﻟﺮﺗﺐ ﺍﻟﻌﻠﻴﺎ(‬ ‫ﻟﻴﻜﻦ‬ ‫ﺇﺫﺍ‬

‫ﺗﺎﺑﻌﺎ ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻋﻠﻰ ﳎﺎﻝ ‪. I‬‬

‫‪f‬‬

‫ﺍﻻﺷﺘﻘﺎﻕ ﻋﻠﻰ ‪ I‬ﻧﻘﻮﻝ ﺇﻥ‬

‫ﻗﺒﻞ '‪f‬‬

‫‪f‬‬

‫ﻳﻘﺒﻞ ﻣﺸﺘﻘﺎ ﻣﻦ ﺍﻟﺮﺗﺒﺔ ﺍﻟﺜﺎﻧﻴﺔ‪،‬‬

‫ﻭﻧﺮﻣﺰ ﻟﻠﻤﺸﺘﻖ ﺍﻟﺜﺎﱐ ﺑﺎﻟﺮﻣﺰ ''‪. f‬‬ ‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﻧﻌﺮ‪‬ﻑ ﺑﺎﻟﺘﺮﺍﺟﻊ ا ‪ % @#AB‬ا ‪ n 6?+‬ﻟـ ‪ ، f‬ﻭﻧﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ‬ ‫)‪ ، f (n‬وه" ?‪ ً$+,‬ﻣﺸﺘﻖ‬

‫ﺍﻟﺘﺎﺑﻊ )‪f (n −1‬‬

‫ﺃﻱ‬

‫' ) )‪f ( n ) = ( f ( n −1‬‬

‫ﺇﺫﺍ آن‬ ‫ﺍﻟﺘﺎﺑﻊ‬ ‫ﻋﻦ‬

‫‪f‬‬

‫‪f‬‬


‫‪n‬‬

‫ﻣﺮﺓ ﻭﻛﺎﻥ‬

‫‪∀n ∈ ℕ∗ ,‬‬

‫ﺍﳌﺸﺘﻖ )‪f (n‬‬

‫ﻣﺴﺘﻤﺮﺍ ﻧﻘﻮﻝ ﺇﻥ‬

‫ﻣﻦ ﺍﻟﺼﻨﻒ ‪ ، C n‬ﺃﻭ ﺇﻧﻪ ﻗﺎﺑﻞ ﻟﻼﺷﺘﻘﺎﻕ ‪ n‬ﻣﺮﺓ ﺑﺎﺳﺘﻤﺮﺍﺭ‪ .‬ﻭﻧﻘﻮﻝ‬

‫‪f‬‬

‫ﺇﻧﻪ ﻣﻦ‬

‫ﺍﻟﺼﻨﻒ ‪C 0‬‬

‫ﺇﺫﺍ ﻛﺎﻥ ﻣﺴﺘﻤﺮﺍ‪.‬‬

‫ﻧﻈﺮﻳﺔ )ﺷﺮﻁ ﻻﺯﻡ ﻟﻮﺟﻮﺩ ﻗﻴﻤﺔ ﻗﺼﻮﻯ(‬ ‫ﺇﺫﺍ ﻛﺎﻥ ﻟﻠﺘﺎﺑﻊ‬

‫‪f‬‬

‫ﻗﻴﻤﺔ ﻗﺼﻮﻯ ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ‬

‫‪ x0‬ﻭﻛﺎﻥ )‪f'(x0‬‬

‫ﻣﻮﺟﻮﺩﺍ ﻓﺈﻥ ‪. f'(x0)=0‬‬ ‫ﻣﻼﺣﻈﺎﺕ ‪:‬‬ ‫ﺇﻥ ﺍﻟﻘﻀﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ ﺧﺎﻃﺌﺔ ‪ :‬ﺇﺫﺍ‬

‫ﻛﺎﻥ ‪f'(x0)=0‬‬

‫ﻟﻘﻴﻤﺔ ﻗﺼﻮﻯ ﻟﻴﺲ ﺃﻣﺮﺍ ﻣﺆﻛﺪﺍ‪ .‬ﻣﺜﺎﻝ ﺫﻟﻚ ﺍﻟﺘﺎﺑﻊ‬ ‫‪x0 =0‬‬

‫‪:‬‬

‫‪f'(0)=0‬‬

‫ﰲ ﺣﲔ ﺃﻥ‬

‫ﺍﻟﺘﺎﺑﻊ ‪x֏ x3‬‬

‫‪93‬‬

‫‪x֏ x3‬‬

‫ﻓﺈﻥ ﻗﺒﻮﻝ‬

‫‪f‬‬

‫ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ‬

‫ﻻ ﻳﻘﺒﻞ ﺃﻳﺔ ﻗﻴﻤﺔ ﻗﺼﻮﻯ‪.‬‬


‫ﳝﻜﻦ ﻟﺘﺎﺑﻊ ﺃﻥ ﻳﻘﺒﻞ ﻗﻴﻤﺔ ﻗﺼﻮﻯ‬ ‫ﻋﻨﺪ ‪ . x0‬ﻣﺜﻼ ﺍﻟﺘﺎﺑﻊ‬

‫‪x֏ x‬‬


‫ﺩﻭﻥ ﺃﻥ ﻳﻜﻮﻥ ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ‬

‫ﻳﻘﺒﻞ ﻗﻴﻤﺔ ﺻﻐﺮﻯ‬

‫ﻋﻨﺪ ‪x0 =0‬‬

‫ﰲ ﺍﻟﻮﻗﺖ ﺍﻟﺬﻱ‬

‫ﻻ ﻳﻘﺒﻞ ﻓﻴﻪ ﺍﻻﺷﺘﻘﺎﻕ ﻋﻨﺪ ﻫﺬﻩ ﺍﻟﻨﻘﻄﺔ‪.‬‬ ‫ﻧﻈﺮﻳﺔ )ﺭﻭﻝ ‪((1719-1652) Rolle‬‬ ‫ﻟﻴﻜﻦ‬

‫‪f : [ a, b ] → ℝ‬‬

‫ﺗﺎﺑﻌﺎ ﻣﺴﺘﻤﺮﺍ ﻭﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻋﻠﻰ [‪]a,b‬‬

‫ﲝﻴﺚ )‪. f(a)= f(b‬‬ ‫ﻋﻨﺪﺋﺬ ﺗﻮﺟﺪ ﻧﻘﻄﺔ‬

‫[‪c∈]a,b‬‬

‫ﲢﻘﻖ ‪. f'(c)=0‬‬

‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫ﺍﻟﺘﺎﺑﻊ ‪ f‬ﻣﺴﺘﻤﺮ ﻋﻠﻰ ﺍ‪‬ﺎﻝ ]‪ [a,b‬ﻭﻟﺬﺍ ﻓﻬﻮ ﳏﺪﻭﺩ‬

‫ﻭﺣﺪﺍﻩ ‪m‬‬

‫ﻭ ‪.M‬‬ ‫ﺇﺫﺍ ﻛﺎﻥ‬

‫‪m= M‬‬

‫ﺇﺫﺍ ﻛﺎﻥ‬

‫‪m≠ M‬‬

‫ﻓﺈﻥ ﺍﻟﺘﺎﺑﻊ ‪ f‬ﺛﺎﺑﺖ‪.‬‬ ‫ﻓﺈﻥ ﺍﻟﺘﺎﺑﻊ‬

‫‪f‬‬

‫ﻭﻣﻨﻪ ‪f'(c)=0‬‬

‫‪. ∀c∈]a,b[,‬‬

‫ﻳﺪﺭﻙ ﻋﻠﻰ ﺍﻷﻗﻞ ﺃﺣﺪ ﺣﺪﻳﻪ ﻋﻨﺪ‬

‫ﻧﻘﻄﺔ ‪ c‬ﳐﺘﻠﻔﺔ ﻋﻦ ‪ a‬ﻭ ‪ ، b‬ﺃﻱ‬ ‫‪∃c∈]a,b[, f'(c)=0 .‬‬ ‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﺇﺫﺍ‬

‫ﻛﺎﻥ ‪f(a)= f(b)=0‬‬

‫ﳝﻜﻨﻨﺎ ﺻﻴﺎﻏﺔ ﻧﻈﺮﻳﺔ ﺭﻭﻝ ﻋﻠﻰ ﺍﻟﺸﻜﻞ ‪:‬‬

‫ﺑﲔ ﻛﻞ ﺻﻔﺮﻳﻦ ﻟﻠﺘﺎﺑﻊ ﺍﻟﻘﺎﺑﻞ ﻟﻼﺷﺘﻘﺎﻕ‬ ‫'‪. f‬‬

‫‪94‬‬

‫‪f‬‬

‫ﻳﻮﺟﺪ ﺻﻔﺮ ﻋﻠﻰ ﺍﻷﻗﻞ ﻟﻠﺘﺎﺑﻊ‬


‫ﻧﻈﺮﻳﺔ )ﺍﻟﺘﺰﺍﻳﺪﺍﺕ ﺍﳌﻨﺘﻬﻴﺔ(‬ ‫ﻟﻴﻜﻦ‬

‫‪f : [ a, b ] → ℝ‬‬

‫ﻋﻨﺪﺋﺬ ﺗﻮﺟﺪ‬

‫ﺗﺎﺑﻌﺎ ﻣﺴﺘﻤﺮﺍ ﻭﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻋﻠﻰ [‪. ]a,b‬‬

‫ﻧﻘﻄﺔ [‪c∈]a,b‬‬

‫ﲢﻘﻖ‬

‫‪f(b)− f(a)=(b−a)f'(c).‬‬

‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫?‪ 1?E‬ه‪ FG‬ﺍﻟﻨﻈﺮﻳﺔ ﻣﻦ ﻧﻈﺮﻳﺔ ﺭﻭﻝ‪ .‬ﻧﻀﻊ‬ ‫)‪f(b)− f(a‬‬ ‫‪(x−a).‬‬ ‫‪b−a‬‬

‫‪φ(x)= f(x)− f(a)−‬‬

‫ﺍﻟﺘﺎﺑﻊ ‪ φ‬ﻣﺴﺘﻤﺮ ﻋﻠﻰ ]‪ [a,b‬ﻭﻗﺎﺑﻞ ﻟﻼﺷﺘﻘﺎﻕ ﻋﻠﻰ [‪ . ]a,b‬ﻭﻟﺪﻳﻨﺎ‬ ‫‪. φ(a)=φ(b)=0‬‬ ‫ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺭﻭﻝ ﻳﻮﺟﺪ [‪ c∈]a,b‬ﳛﻘﻖ ‪ φ'(c)=0‬ﺃﻱ ‪:‬‬ ‫)‪f(b)− f(a‬‬ ‫‪=0.‬‬ ‫‪b−a‬‬

‫‪f'(c)−‬‬

‫ﻭ ﻣﻨﻪ ‪:‬‬ ‫‪f(b)− f(a)=(b−a)f'(c).‬‬

‫ﻧﻈﺮﻳﺔ )ﻗﺎﻋﺪﺓ ﻟﻮﺑﻴﺘﺎﻝ )‪((1704-1661‬‬ ‫ﻟﻴﻜﻦ ‪ f‬ﻭ ‪ g‬ﺗﺎﺑﻌﲔ ﻣﺴﺘﻤﺮﻳﻦ ﻋﻠﻰ ﺍ‪‬ﺎﻝ ]‪ [a,b‬ﻭﻗﺎﺑﻠﲔ ﻟﻼﺷﺘﻘﺎﻕ‬ ‫ﻋﻠﻰ [‪ . ]a,b‬ﻭﻟﻴﻜﻦ [‪. x0∈]a,b‬‬ ‫ ‪GH‬‬ ‫)‪f(x)− f(x0‬‬ ‫‪lim‬‬ ‫‪=l .‬‬ ‫)‪x → x g(x) − g(x0‬‬ ‫‪0‬‬

‫)‪f'(x‬‬ ‫⇒ ‪=l‬‬ ‫)‪g'(x‬‬

‫‪lim‬‬

‫‪x → x0‬‬

‫ﺍﻟﱪﻫﺎﻥ ‪ :‬ﻳﺄﰐ ﻣﻦ ﻧﻈﺮﻳﺔ ﺍﻟﺘﺰﺍﻳﺪﺍﺕ ﺍﳌﻨﺘﻬﻴﺔ )ﺍﳌﻌﻤﻤﺔ(‪.‬‬ ‫‪95‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .1‬ﻣﻘﺪﻣﺔ ‪:‬‬ ‫ﻳﻌﺘﱪ ﺍﳊﺴﺎﺏ ﺍﻟﺘﻜﺎﻣﻠﻲ ﺃﺩﺍﺓ ﻓﻌﺎﻟﺔ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﻭﺑﺎﻗﻲ ﺍﻟﻔﺮﻭﻉ‬ ‫ﺍﻟﻌﻠﻤﻴﺔ ﺇﺫ ﻳﺴﻤﺢ ﰲ ﺍﻟﻜﺜﲑ ﻣﻦ ﺍﳊﺎﻻﺕ ﺑﺎﳊﺼﻮﻝ ﻋﻠﻰ ﻧﺘﺎﺋﺞ ﻫﺎﻣﺔ ﺗﺘﻌﻠﻖ‬ ‫ﲝﺴﺎﺏ ﺍﻷﻃﻮﺍﻝ ﻭﺍﳌﺴﺎﺣﺎﺕ ﻭﺍﳊﺠﻮﻡ ﻭﻗﻴﻢ ﺃﺧﺮﻯ ﺫﺍﺕ ﻃﺎﺑﻊ ﻓﻴﺰﻳﺎﺋﻲ ﺃﻭ‬ ‫ﺍﻗﺘﺼﺎﺩﻱ‪ ،‬ﺍﱁ‪ .‬ﻭﺍﳌﻼﺣﻆ ﺃﻥ ﻣﺴﺎﺋﻞ ﺣﺴﺎﺏ ﺍﳌﺴﺎﺣﺎﺕ ﻟﻴﺴﺖ ﻭﻟﻴﺪﺓ ﻫﺬﺍ‬ ‫ﺍﻟﻌﺼﺮ ﺑﻞ ﻳﻌﻮﺩ ﻃﺮﺣﻬﺎ ﺇﱃ ﺍﻟﻌﺼﻮﺭ ﺍﻟﻘﺪﳝﺔ‪ ،‬ﺃﻣﺎ ﺍﳊﺴﺎﺏ ﺍﻟﺘﻜﺎﻣﻠﻲ‬ ‫ﲟﻔﻬﻮﻣﻪ ﺍﳊﺪﻳﺚ ﻓﻬﻮ ﻣﻦ ﺇﻧﺘﺎﺝ ﺍﻟﻘﺮﻭﻥ ﺍﻷﺧﲑﺓ ﺑﺪﺀﺍ ﻣﻦ ﺍﻟﻘﺮﻥ ﺍﻟﺴﺎﺑﻊ‬ ‫ﻋﺸﺮ )ﻧﻴﻮﺗﻦ ‪ Newton‬ﻭﻻﻳﺒﻨﺘﺰ ‪ (Leibniz‬ﺇﱃ ﻳﻮﻣﻨﺎ ﻫﺬﺍ )ﻟﻮﺑﻴﻎ‬ ‫‪ ،Lebesgue‬ﺑﻮﺧﻨﲑ‪ ،Bochner‬ﺑﻴﺘﻴﺲ ‪.(...Pettis‬‬

‫‪99‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .2‬ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ ‪:‬‬ ‫ﳝﻜﻦ ﻣﻦ ﺍﻟﻨﺎﺣﻴﺔ ﺍﻟﻌﻤﻠﻴﺔ ﺗﻘﺪﱘ ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ ﻋﻠﻰ ﺃﻧﻪ ﻳﻨﻄﻠﻖ ﻣﻦ‬ ‫ﺍﳊﺎﺟﺔ ﺇﱃ ﺍﻟﺘﻌﺒﲑ ﻋﻦ ﻣﺴﺎﺣﺔ ﻣﻨﺤﺼﺮﺓ ﺑﲔ ﺑﻴﺎﻥ ﺗﺎﺑﻊ ﻭﳏﻮﺭ ﺍﻟﻔﻮﺍﺻﻞ‬ ‫ﻭﻣﺴﺘﻘﻴﻤﲔ ﺷﺎﻗﻮﻟﻴﲔ‪ .‬ﻓﺎﳌﺴﺎﺣﺔ ﺍﻟﺪﺍﻛﻨﺔ ﺍﳌﺒﻴ‪‬ﻨﺔ ﺃﺩﻧﺎﻩ ﳝﻜﻦ ﲤﺜﻴﻠﻬﺎ ﺑﺘﻜﺎﻣﻞ‬ ‫ﺍﻟﺪﺍﻟﺔ‬

‫‪f‬‬

‫ﻋﻠﻰ ﺍ‪‬ﺎﻝ ] ‪: [a, b‬‬

‫ﻟﻜﻦ ﺍﻟﺘﻌﺮﻳﻒ ﺍﻟﺪﻗﻴﻖ ﳍﺬﺍ ﺍﻟﺘﻜﺎﻣﻞ ﻳﺘﻄﻠﺐ ﻣﻨﺎ ﺍﻻﻧﻄﻼﻕ ﻣﻦ ﺩﻭﺍﻝ‬ ‫ﺑﺴﻴﻄﺔ ﺗﺴﻤﻰ ﺍﻟﺪﻭﺍﻝ ﺍﻟﺪﺭﺟﻴﺔ‪ .‬ﻓﻤﺎ ﻫﻲ ﻫﺬﻩ ﺍﻟﺪﻭﺍﻝ؟‬ ‫ﺗﻌﺮﻳﻒ )ﺍﻟﺪﺍﻟﺔ ﺍﻟﺪﺭﺟﻴﺔ(‬ ‫ﻟﻴﻜﻦ‬

‫] ‪I = [a , b‬‬

‫ﻧﻘﻮﻝ ﻋﻦ ﺩﺍﻟﺔ‬

‫ﳎﺎﻻ ﻣﻦ ‪. ℝ‬‬

‫‪f :I →ℝ‬‬

‫‪a = 0 < a1 < a2 < .... < an = b‬‬

‫‪f (x ) = c i .‬‬

‫ﺇ‪‬ﺎ ﺩﺭﺟﻴﺔ ﺇﺫﺍ ﻭﺟﺪﺕ ﺗﻘﺴﻴﻤﺔ‬

‫ﻟﻠﻤﺠﺎﻝ‬

‫‪∀x ∈ [ai , ai +1 [ ,‬‬

‫‪100‬‬

‫‪I‬‬

‫ﲝﻴﺚ ﻳﻜﻮﻥ‬

‫‪∀i = 0,...n − 1, ∃c i ∈ ℝ :‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬

‫ﳝﻜﻦ ﰲ ﺍﻟﻜﺘﺎﺑﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗﻌﻮﻳﺾ [ ‪ [ai , ai +1‬ﺑـ [ ‪]ai , ai +1‬‬ ‫ﻭﺍﻋﺘﺒﺎﺭ ﺃﻳﺔ ﻗﻴﻢ ﻟـ ‪ f‬ﻋﻨﺪ ﺍﻟﻨﻘﺎﻁ ‪. (ai )i =0,...,n −1‬‬ ‫ﻻﺣﻆ ﺃﻧﻨﺎ ﻧﺴﺘﻄﻴﻊ ﺍﻟﺘﻌﺒﲑ ﻋﻦ ﺍﳌﺴﺎﺣﺔ ﺍﳌﻨﺤﺼﺮﺓ ﺑﲔ ﺑﻴﺎﻥ ﺗﺎﺑﻊ ‪f‬‬ ‫ﻭﳏﻮﺭ ﺍﻟﻔﻮﺍﺻﻞ ﻭﺍﳌﺴﺘﻘﻴﻤﲔ ﺍﻟﺸﺎﻗﻮﻟﻴﲔ ﺍﳌﻌﺮﻓﲔ ﺑﺎﳌﻌﺎﺩﻟﺘﲔ ‪ x = a‬ﻭ‬ ‫‪ x = b‬ﻛﻤﺎ ﻳﻠﻲ ‪ :‬ﺇ‪‬ﺎ ﺗﺴﺎﻭﻱ‬ ‫‪(ai +1 − ai ) .‬‬

‫ﻭﻧﻜﺘﺐ ‪:‬‬

‫‪n −1‬‬

‫‪n −1‬‬

‫‪i‬‬

‫‪∑c‬‬

‫) ‪(x )dx = ∑ c i (ai +1 − ai‬‬ ‫‪i =0‬‬

‫‪i =0‬‬

‫‪ . ∫a f‬ﻣﻦ ﺍﳌﻬﻢ ﺃﻥ ﺗﻼﺣﻆ‬ ‫‪b‬‬

‫ﺃﻳﻀﺎ ﺃﻥ ﻫﺬﺍ ﺍﻟﺘﻜﺎﻣﻞ )ﺃﻱ ﺍﳌﺴﺎﺣﺔ( ﻻ ﺗﺘﻌﻠﻖ ﺑﻘﻴﻢ‬

‫‪f‬‬

‫ﻋﻨﺪ ﺍﻟﻨﻘﺎﻁ‬

‫‪. (ai )i =0,...,n −1‬‬ ‫ﺗﻠﻚ ﻫﻲ ﺍﳌﺮﺣﻠﺔ ﺍﻷﻭﱃ ﺍﳌﺆﺩﻳﺔ ﺇﱃ ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ‪ .‬ﻭﻫﻨﺎﻙ ﻣﺮﺣﻠﺔ‬ ‫ﺛﺎﻧﻴﺔ ﻧﻮﺩ ﺍﳌﺮﻭﺭ ﻋﻠﻴﻬﺎ ﻣﺮ ﺍﻟﻜﺮﺍﻡ‪ ،‬ﻭﻫﻲ ﺗﺘﻌﻠﻖ ﺑﺘﻜﺎﻣﻞ ﻓﺌﺔ ﻣﻦ ﺍﻟﺪﻭﺍﻝ‬ ‫ﺗﺴﻤﻰ ﺍﻟﺪﻭﺍﻝ "ﺍﳌﺴﻮﺍﺓ"‪ .‬ﻭﺗﻄﻠﻖ ﻫﺬﻩ ﺍﻟﺼﻔﺔ ﻋﻠﻰ ﻛﻞ ﺩﺍﻟﺔ ﳝﻜﻦ ﺍﳊﺼﻮﻝ‬ ‫ﻋﻠﻴﻬﺎ ﻛﻨﻬﺎﻳﺔ )ﻣﻨﺘﻈﻤﺔ( ﳌﺘﺘﺎﻟﻴﺔ ﺩﻭﺍﻝ ﺩﺭﺟﻴﺔ‪ .‬ﻟﻦ ﻧﺘﻮﻗﻒ ﻋﻨﺪ ﻫﺬﻩ ﺍﻟﻔﺌﺔ ﻣﻦ‬ ‫ﺍﻟﺪﻭﺍﻝ ﻷﻥ ﺗﻔﺎﺻﻴﻠﻬﺎ ﺗﺘﻄﻠﺐ ﺇﺩﺧﺎﻝ ﻣﻔﺎﻫﻴﻢ ﱂ ﻧﺘﻄﺮﻕ ﺇﻟﻴﻬﺎ )ﻭﱂ ﺗﺬﻛﺮ ﰲ‬ ‫ﺍﻟﱪﻧﺎﻣﺞ(‪.‬‬ ‫ﻭﺗﻌﺮﻳﻒ ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ ﺍﻟﺬﻱ ﺳﻨﻘﺪﻣﻪ ﺑﻌﺪ ﺣﲔ ﻣﻦ ﺃﺟﻞ ﺗﺎﺑﻊ‬ ‫ﻣﺴﺘﻤﺮ ﺃﻭ ﻣﺴﺘﻤﺮ ﺑﺘﻘﻄﻊ ﻳﺼﺪﻕ ﻋﻠﻰ ﻫﺬﻩ ﺍﻟﻔﺌﺔ ﻣﻦ ﺍﻟﺪﻭﺍﻝ ﺃﻳﻀﺎ‪ .‬ﺫﻟﻚ‬

‫‪101‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﺃﻥ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺴﺘﻤﺮﺓ ﻭﺍﳌﺴﺘﻤﺮﺓ ﺑﺘﻘﻄﻊ ﺩﻭﺍﻝ "ﻣﺴﻮﺍﺓ" )ﺍﻟﻌﻜﺲ ﻏﲑ‬ ‫ﺻﺤﻴﺢ(‪.‬‬ ‫ﻭﻳﺒﺤﺚ ﺍﻟﺮﻳﺎﺿﻴﻮﻥ ﰲ ﺗﻌﻤﻴﻢ ﻣﻔﻬﻮﻡ ﺍﳌﻜﺎﻣﻠﺔ‪ .‬ﻭﻣﻦ ﺑﲔ ﺍﻷﺳﺌﻠﺔ‬ ‫ﺍﳌﻄﺮﻭﺣﺔ ﻫﻲ ﺇﳚﺎﺩ ﺃﻛﱪ ﳎﻤﻮﻋﺔ ﺗﻮﺍﺑﻊ ﺗﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ‪ .‬ﻭﺇﺫﺍ ﻛﺎﻥ ﺇﺩﺧﺎﻝ‬ ‫ﻣﻔﻬﻮﻡ ﺍﻟﺘﻮﺍﺑﻊ ﺍﳌﺴﻮﺍﺓ ﻗﺪ ﺃﺟﺎﺏ ﻋﻠﻰ ﺟﺰﺀ ﻣﻦ ﺍﻟﺴﺆﺍﻝ ﻓﺈﻧﻨﺎ ﻧﻼﺣﻆ ﺃﻥ‬ ‫ﻫﻨﺎﻙ ﺗﻮﺍﺑﻊ ﻣﺴﻮﺍﺓ ﻭﺭﻏﻢ ﺫﻟﻚ ﻓﻬﻲ ﻻ ﺗﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ ﲟﻔﻬﻮﻡ ﺭﳝﺎﻥ‪ .‬ﻣﺜﺎﻝ‬ ‫ﺫﻟﻚ ﺍﻟﺘﺎﺑﻊ ﺍﳌﻌﺮﻑ ﻋﻠﻰ ﺍ‪‬ﺎﻝ ]‪ [0,1‬ﺑـ‬

‫‪1‬‬ ‫‪f ( ) =1‬‬ ‫‪n‬‬

‫ﻣﻦ ﺃﺟﻞ‬

‫*‪n ∈ ℕ‬‬

‫ﻭﺍﳌﻨﻌﺪﻣﺔ ﰲ ﺑﺎﻗﻲ ﻧﻘﺎﻁ ﺍ‪‬ﺎﻝ ]‪ . [0,1‬ﻫﺬﺍ ﺍﻟﺘﺎﺑﻊ ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ ﲟﻔﻬﻮﻡ‬ ‫ﺭﳝﺎﻥ ﺭﻏﻢ ﺃﻧﻪ ﻏﲑ ﻣﺴﻮﻯ‪.‬‬ ‫ﻭﻟﻌﻞ ﺍﻟﻘﺎﺭﺉ ﻳﺘﺴﺎﺀﻝ ﻋﻤﺎ ﺇﺫﺍ ﻛﺎﻧﺖ ﻫﻨﺎﻙ ﻓﺎﺋﺪﺓ ﺟﺎﺩﺓ ﻣﻦ ﻭﺭﺍﺀ‬ ‫ﺍﻟﺒﺤﺚ ﻋﻦ ﻣﻜﺎﻣﻠﺔ ﻣﺜﻞ ﺗﻠﻚ ﺍﻟﺪﻭﺍﻝ ﻏﲑ ﺍﳌﺄﻟﻮﻓﺔ‪ .‬ﺗﺜﺒﺖ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‬ ‫ﺍﳌﺘﻘﺪﻣﺔ ﺃﻥ ﺍﻷﻣﺮ ﻻ ﻳﺘﻌﻠﻖ ﺑﻌﺒﺚ ﺭﻳﺎﺿﻲ‪ .‬ﲟﻌﲎ ﺃﻧﻨﺎ ﻣﻄﺎﻟﺒﲔ ﺑﺎﻟﺒﺤﺚ ﻋﻦ‬ ‫ﻣﻜﺎﻣﻠﺔ ﺗﻮﺍﺑﻊ ﺃﻋﻢ ﻣﻦ ﺍﻟﺘﻮﺍﺑﻊ ﺍﳌﺴﻮﺍﺓ‪ .‬ﻭﻟﺬﻟﻚ ﺃﺩﺧﻠﺖ ﻋﺪﺓ ﻣﻔﺎﻫﻴﻢ‬ ‫ﻟﻠﻤﻜﺎﻣﻠﺔ ﺃﳘﻬﺎ ﺗﻜﺎﻣﻞ ﻟﻮﺑﻴﻎ ‪ Lebesgue‬ﺍﻟﺬﻱ ﻳﺴﻤﺢ ﲟﻜﺎﻣﻠﺔ ﻓﺌﺔ‬ ‫ﺃﻭﺳﻊ ﻣﻦ ﺍﻟﺘﻮﺍﺑﻊ‪ .‬ﻳﻨﺒﻐﻲ ﺃﻥ ﻧﻼﺣﻆ ﺑﺄﻥ ﺗﻜﺎﻣﻠﻲ ﺭﳝﺎﻥ ﻭﻟﻮﺑﻴﻎ ﻳﺘﻄﻠﺒﺎﻥ ﺃﻥ‬ ‫ﻳﻜﻮﻥ ﻓﻀﺎﺀ ﻭﺻﻮﻝ ﺍﻟﺪﺍﻟﺔ ﻣﺮﺗﺒﺎ‪ .‬ﻭﺑﻄﺒﻴﻌﺔ ﺍﳊﺎﻝ ﻓﻘﺪ ﺳﻌﻰ ﺍﻟﺮﻳﺎﺿﻴﻮﻥ ﺇﱃ‬ ‫ﺗﻌﻤﻴﻢ ﻣﻔﻬﻮﻡ ﺍﳌﻜﺎﻣﻠﺔ ﺇﱃ ﺍﳊﺎﻟﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﻓﻴﻬﺎ ﻓﻀﺎﺀ ﺍﻟﻮﺻﻮﻝ ﻏﲑ‬ ‫ﻣﺮﺗﺐ‪ ،‬ﺫﻟﻚ ﻣﺎ ﻗﺎﻡ ﺑﻪ ﻣﺜﻼ ﺑﻮﺧﻨﺮ ‪.Bochner‬‬

‫‪102‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻭﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ﺃﻥ ﺗﻜﺎﻣﻞ ﻟﻮﺑﻴﻎ ﻟﻴﺲ ﰲ ﺍﻟﱪﻧﺎﻣﺞ ﺍﳌﺴﻄﺮ‬ ‫ﻟﻠﻤﻔﺘﺸﲔ ﺇﻻ ﺃﻥ ﺃﳘﻴﺘﻪ ﰲ ﺍﻟﺘﺤﻠﻴﻞ ﺍﻟﺮﻳﺎﺿﻲ ﻛﺒﲑﺓ ﺟﺪﺍ‪ ،‬ﻭﻫﻮ ﺍﻷﻛﺜﺮ‬ ‫ﺗﺪﺍﻭﻻ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺍﳊﺪﻳﺜﺔ ﻷﻧﻪ ﻳﺴﻤﺢ ﺑﺈﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ "ﺍﳌﺮﻭﺭ ﺇﱃ ﺍﻟﻨﻬﺎﻳﺔ"‬ ‫)ﻛﺎﳌﺒﺎﺩﻟﺔ ﺑﲔ ﺭﻣﺰ ﺍﻟﺘﻜﺎﻣﻞ ﻭﺍﻟﻨﻬﺎﻳﺔ ﰲ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﺮﻳﺎﺿﻴﺔ( ﺑﺪﻭﻥ ﺗﻌﻘﻴﺪﺍﺕ‬ ‫ﻣﻘﺎﺭﻧﺔ ﺑﺘﻜﺎﻣﻞ ﺭﳝﺎﻥ‪.‬‬ ‫ﻭﻣﻦ ﺍﻟﻨﺎﺣﻴﺔ ﺍﻟﺘﺎﺭﳜﻴﺔ ﻓﻘﺪ ﺍﻧﻄﻠﻖ ﻟﻮﺑﻴﻎ ﻣﻦ ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ ﻹﻧﺸﺎﺀ‬ ‫ﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺬﻱ ﳛﻤﻞ ﺍﻟﻴﻮﻡ ﺍﲰﻪ‪ .‬ﻳﻘﻮﻝ ﻟﻮﺑﻴﻎ ﰲ ﻣﻘﺎﺭﻧﺔ ﻣﻔﻬﻮﻣﻲ ﺗﻜﺎﻣﻞ‬ ‫ﺭﳝﺎﻥ ﻣﻊ ﺗﻜﺎﻣﻠﻪ ‪ " :‬ﺗﺼﻮﺭ ﺃﻥ ﻋﻠﻲ ﺩﻓﻊ ﻣﺒﻠﻎ ﻣﻌﲔ‪ .‬ﳝﻜﻨﲏ ﺃﻥ ﺃﺧﺮﺝ ﻣﻦ‬ ‫ﳏﻔﻈﺔ ﻧﻘﻮﺩﻱ ﻗﻄﻌﺎ ﺍﻟﻮﺍﺣﺪﺓ ﺗﻠﻮ ﺍﻷﺧﺮﻯ‪ ،‬ﺑﺎﻟﺼﺪﻓﺔ‪ ،‬ﺣﱴ ﻳﺼﺒﺢ ﳎﻤﻮﻋﻬﺎ‬ ‫ﻣﺴﺎﻭﻳﺎ ﻟﻠﻤﺒﻠﻎ ﺍﳌﻄﻠﻮﺏ‪ .‬ﻛﻤﺎ ﳝﻜﻨﲏ ﺃﻥ ﺃﺧﺮﺝ ﻛﻞ ﻧﻘﻮﺩﻱ ﺩﻓﻌﺔ ﻭﺍﺣﺪﺓ‪،‬‬ ‫ﻭﺍﺧﺘﺎﺭ ﻣﻨﻬﺎ ﺍﻟﻘﻄﻊ ﺣﺴﺐ ﻗﻴﻤﻬﺎ ﻟﻴﻜﻮﻥ ﳎﻤﻮﻋﻬﺎ ﻣﺴﺎﻭﻳﺎ ﻟﻠﻤﺒﻠﻎ‬ ‫ﺍﳌﻄﻠﻮﺏ"‪.‬‬ ‫ﰒ ﻳﻘﻮﻝ ﻟﻮﺑﻴﻎ ﺇﻥ ﺍﻟﻄﺮﻳﻘﺔ ﺍﻷﻭﱃ ﻫﻲ ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ‪ ،‬ﺃﻣﺎ ﺍﻟﻄﺮﻳﻘﺔ‬ ‫ﺍﻟﺜﺎﻧﻴﺔ ﻓﻬﻲ ﺗﻜﺎﻣﻞ ﻟﻮﺑﻴﻎ‪ .‬ﲟﻌﲎ ﺃﻥ ﺗﻜﺎﻣﻞ ﻟﻮﺑﻴﻎ ﳝﺴﺢ ﺃﻓﻘﻴﺎ ﺍﻟﻘﻄﻌﺔ‬ ‫ﺍﳌﺴﺘﻘﻴﻤﺔ ﺍﻟﱵ ﳒﺮﻱ ﻋﻠﻴﻬﺎ ﺍﳌﻜﺎﻣﻠﺔ )ﺍﳌﻌﺮﻓﺔ ﻋﻠﻴﻬﺎ ﺍﻟﺪﺍﻟﺔ ‪ ( f‬ﻭﻳﻘﻴﺲ‬ ‫"ﺍﻻﺭﺗﻔﺎﻋﺎﺕ" ﺍﻟﻮﺍﺣﺪ ﺗﻠﻮ ﺍﻵﺧﺮ‪ .‬ﺃﻣﺎ ﺗﻜﺎﻣﻞ ﻟﻮﺑﻴﻎ ﻓﻴﻌﺘﱪ "ﺣﺠﻢ"‬ ‫ﺍ‪‬ﻤﻮﻋﺎﺕ ﺍﶈﺼﻮﺭﺓ ﺑﲔ‬

‫‪f =y‬‬

‫ﻭﳏﻮﺭ ﺍﻟﻔﻮﺍﺻﻞ‪ .‬ﻭﻗﺪ ﲰﺢ ﺗﻜﺎﻣﻞ ﻟﻮﺑﻴﻎ‬

‫ﺑﻈﻬﻮﺭ ﻓﺮﻉ ﺟﺪﻳﺪ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﻭﻫﻮ ﻧﻈﺮﻳﺔ ﺍﻟﻘﻴﺎﺱ‪.‬‬

‫‪103‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﺗﻌﺮﻳﻒ )ﳎﻤﻮﻉ ﺭﳝﺎﻥ(‬ ‫] ‪I = [a , b‬‬

‫ﻟﻴﻜﻦ‬

‫ﳎﺎﻻ‬

‫}‪d n = {a =0 < a1 < a2 < .... < an = b‬‬

‫ﻭﻟﺘﻜﻦ ﺩﺍﻟﺔ‬

‫ﻣﻦ‬

‫‪.ℝ‬‬

‫ﻭﻟﺘﻜﻦ‬

‫ﺗﻘﺴﻴﻤﺔ ﻟﻠﻤﺠﺎﻝ ‪. I‬‬

‫ﳏﺪﻭﺩﺓ‪ .‬ﻧﺴﻤﻲ‬

‫‪f :I →ℝ‬‬

‫‪n −1‬‬

‫) ‪R ( f , d n , c) = ∑ f (ci )(ai +1 − ai‬‬ ‫‪i =0‬‬

‫ ] ‪ ci ∈ [ ai , ai +1‬أ آ *‪ ، n ∈ ℕ‬ع رن ‪f‬‬

‫و‪ $%‬ا"! ‪ d n‬و ا("ط ‪. c = (ci )i =0,...,n −1‬‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﳝﺜﹼﻞ‬

‫‪n −1‬‬

‫) ‪R ( f , d n , c) = ∑ f (ci )(ai +1 − ai‬‬

‫ﺍﳌﺴﺘﻄﻴﻼﺕ ﺍﻟﱵ ﺑﻌﺪﺍﻫﺎ‬

‫‪i =0‬‬

‫‪ai +1 − ai‬‬

‫ﻭ‬

‫) ‪f (ci‬‬

‫ﺃﺩﻧﺎﻩ ‪:‬‬

‫‪104‬‬

‫ﳎﻤﻮﻉ ﻣﺴﺎﺣﺎﺕ‬

‫ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﰲ ﺍﻟﺸﻜﻞ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﺗﻌﺮﻳﻒ )ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ(‬ ‫ﻟﻴﻜﻦ‬

‫] ‪I = [a , b‬‬

‫ﳎﺎﻻ ﻣﻦ‬

‫ﻭ‬

‫‪ℝ‬‬

‫‪f :I →ℝ‬‬ ‫‪n −1‬‬

‫ﻛﺎﻧﺖ ‪‬ﺎﻳﺔ ﳎﻤﻮﻉ ﺭﳝﺎﻥ‬

‫ﻣﻮﺟﻮﺩﺓ‬

‫) ‪R ( f , d n , c) = ∑ f (ci )(ai +1 − ai‬‬ ‫‪i =0‬‬

‫ﻭﻻ ﺗﺘﻌﻠﻖ ﺑﺎﺧﺘﻴﺎﺭ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺘﻘﺴﻴﻤﺎﺕ‬ ‫ﻓﺈﻧﻨﺎ ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺘﺎﺑﻊ‬

‫‪c = (ci )i =0,...,n −1‬‬

‫ﺩﺍﻟﺔ ﳏﺪﻭﺩﺓ‪ .‬ﺇﺫﺍ‬

‫*‪(d n ) n∈ℕ‬‬ ‫‪f‬‬

‫ﻭﺍﺧﺘﻴﺎﺭ ﲨﻠﺔ ﺍﻟﻨﻘﺎﻁ‬

‫ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ ﻋﻠﻰ ﺍ‪‬ﺎﻝ‬

‫‪I‬‬

‫ﲟﻔﻬﻮﻡ ﺭﳝﺎﻥ‪ .‬ﻭﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ ﳍﺬﺍ ﺍﻟﺘﺎﺑﻊ ﻫﻮ )ﺗﻌﺮﻳﻔﹰﺎ( ‪:‬‬ ‫‪lim R ( f , d n , c) = ∫ f ( x)dx .‬‬ ‫‪b‬‬

‫∞‪n →+‬‬

‫‪a‬‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﺇﺫﺍ ﻛﻨﺎ ﻧﻌﻠﻢ ﺃﻥ ﺍﻟﺘﺎﺑﻊ‬ ‫ﺭﳝﺎﻥ‬

‫ﻓﺈﻧﻨﺎ‬

‫‪f‬‬

‫ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ ﻋﻠﻰ ﺍ‪‬ﺎﻝ‬ ‫ﺍﺧﺘﻴﺎﺭ‬

‫ﻧﺴﺘﻄﻴﻊ‬

‫}‪d n = {a =0 < a1 < a2 < .... < an = b‬‬ ‫‪b−a‬‬ ‫‪n‬‬

‫= ‪ai +1 − ai‬‬

‫] ‪I = [a , b‬‬

‫ﻣﺘﺘﺎﻟﻴﺔ‬

‫ﲟﻔﻬﻮﻡ‬

‫ﺗﻘﺴﻴﻤﺎﺕ‬

‫ﻣﺘﺠﺎﻧﺴﺔ‪ ،‬ﺃﻱ ﻣﺘﺴﺎﻭﻳﺔ ﺍﳋﻄﻮﺓ ‪:‬‬

‫ﻣﻦ ﺃﺟﻞ ﻛﻞ ‪ . i = 0,..., n‬ﻭﻋﻨﺪﺋﺬ ﻳﻜﻮﻥ )ﰲ ﺣﺎﻟﺔ‬

‫ﻭﺟﻮﺩ ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ( ‪:‬‬ ‫) ‪f (x )dx = lim R (f , d n , c‬‬ ‫∞‪n →+‬‬

‫) ‪)(ai +1 − ai‬‬

‫‪n −1‬‬

‫‪i‬‬

‫‪∑ f (c‬‬ ‫‪i =0‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫‪= lim‬‬

‫∞‪n →+‬‬

‫‪b − a n −1‬‬ ‫‪∑ f (c i ).‬‬ ‫∞‪n →+‬‬ ‫‪n i =0‬‬

‫‪= lim‬‬

‫ﻭﺇﺫﺍ ﻣﺎ ﺍﺧﺘﺮﻧﺎ ﰲ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺑﻘﺔ‬

‫‪b−a‬‬ ‫‪n‬‬

‫‪ci = ai = a + i‬‬

‫ﺃﺟﻞ ﻛﻞ ‪ i‬ﻓﺈﻧﻨﺎ ﳓﺼﻞ ﻋﻠﻰ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ )ﰲ ﺣﺎﻟﺔ ﻭﺟﻮﺩ ﺍﻟﺘﻜﺎﻣﻞ( ‪:‬‬ ‫‪1 n −1‬‬ ‫‪b−a‬‬ ‫‪.∑ f (a + i‬‬ ‫‪).‬‬ ‫‪n →+∞ n‬‬ ‫‪n‬‬ ‫‪i=0‬‬

‫‪f ( x)dx = (b − a ) lim‬‬

‫‪105‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫ﻣﻦ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻧﺆﻛﺪ ﻋﻠﻰ ﺃﻥ ﻗﻴﺎﻡ ﻫﺬﻩ ﺍﻟﻌﻼﻗﺔ ﻻ ﺗﺆﺩﻱ ﺣﺘﻤﺎ ﺇﱃ ﻭﺟﻮﺩ ﺗﻜﺎﻣﻞ‬ ‫ﺭﳝﺎﻥ‪.‬‬ ‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﻧﻌﺘﱪ ﺗﻘﺴﻴﻤﺔ ﻣﺘﺠﺎﻧﺴﺔ ﻭﻧﻀﻊ ‪:‬‬ ‫‪b −a‬‬ ‫) ) ‪( f (a0 ) + f (x 1 ) + ... + f (x n −1‬‬ ‫‪n‬‬ ‫‪b −a‬‬ ‫= ‪Jn‬‬ ‫‪( f (x 1 ) + ... + f (x n ) ) .‬‬ ‫‪n‬‬

‫= ‪In‬‬

‫ﻧﻼﺣﻆ ﺃﻥ‬ ‫‪b −a‬‬ ‫) ) ‪( f (an ) − f (a0‬‬ ‫‪n‬‬ ‫‪b −a‬‬ ‫=‬ ‫‪( f (b ) − f (a ) ) .‬‬ ‫‪n‬‬

‫= ‪Jn − In‬‬

‫ﻭﺃﻥ‬

‫‪b −a‬‬ ‫‪( f (b ) − f (a ) ) = 0‬‬ ‫∞‪n →+‬‬ ‫‪n‬‬ ‫‪lim‬‬

‫ﻓﺈﺫﺍ ﺍﻓﺘﺮﺿﺎ ﺃﻥ‬

‫ﺍﳌﺘﺘﺎﻟﻴﺔ ‪I n‬‬

‫ﻣﺘﻘﺎﺭﺑﺔ ﻓﺈﻥ ﺍﻷﻣﺮ ﻛﺬﻟﻚ ﺑﺎﻟﻨﺴﺒﺔ‬

‫ﻟﻠﻤﺘﺘﺎﻟﻴﺔ ‪ ، J n‬ﻭﺍﻟﻌﻜﺲ ﺑﺎﻟﻌﻜﺲ‪ .‬ﻭﰲ ﺣﺎﻟﺔ ﺍﻟﺘﻘﺎﺭﺏ ﳓﺼﻞ ﻋﻠﻰ ﺍﳌﺴﺎﻭﺍﺓ‬ ‫‪(x )dx = lim I n = lim J n‬‬ ‫∞‪n →+‬‬

‫∞‪n →+‬‬

‫‪. ∫a f‬‬ ‫‪b‬‬

‫‪106‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺍﺣﺴﺐ ﺍﻟﺘﻜﺎﻣﻞ ‪ ∫0 x 2dx‬ﺑﺎﻋﺘﺒﺎﺭ ﺃﻧﻪ ﻣﻮﺟﻮﺩ‪.‬‬ ‫ﻧﺴﺘﻌﻤﻞ ﺗﻘﺴﻴﻤﺔ ﻣﺘﺠﺎﻧﺴﺔ ﻓﻴﺄﰐ ) ﻋﻠﻤﺎ ﺃﻥ‬ ‫‪1‬‬

‫)‪( ∑ i 2 = n (n − 1)(2n − 1‬‬ ‫‪n −1‬‬

‫‪6‬‬

‫‪i =0‬‬

‫‪(1 − 0) n −1 ‬‬ ‫‪(1 − 0) ‬‬ ‫‪f 0 + i‬‬ ‫∑‬ ‫‪‬‬ ‫∞‪n →+‬‬ ‫‪n i =0 ‬‬ ‫‪n ‬‬ ‫‪(1 − 0) n −1 ‬‬ ‫‪(1 − 0) ‬‬ ‫‪f 0 + i‬‬ ‫‪= lim‬‬ ‫∑‬ ‫‪‬‬ ‫∞‪n →+‬‬ ‫‪n i =0 ‬‬ ‫‪n ‬‬

‫‪x 2dx = lim‬‬

‫‪1‬‬

‫∫‬

‫‪0‬‬

‫‪1 n −1  i ‬‬ ‫‪f  ‬‬ ‫∑‬ ‫‪n →+∞ n‬‬ ‫‪n‬‬ ‫‪i =0‬‬ ‫‪n −1 2‬‬ ‫‪i‬‬ ‫‪1‬‬ ‫‪= lim ∑ 2‬‬ ‫‪n →+∞ n‬‬ ‫‪i =0 n‬‬ ‫‪= lim‬‬

‫ﺃﻱ‬ ‫‪1 n −1 2‬‬ ‫‪i‬‬ ‫∑‬ ‫‪n →+∞ n 3‬‬ ‫‪i =0‬‬ ‫)‪n (n − 1)(2n − 1‬‬ ‫‪= lim‬‬ ‫∞‪n →+‬‬ ‫‪6n 3‬‬ ‫‪2‬‬ ‫=‬ ‫‪6‬‬ ‫‪1‬‬ ‫‪= .‬‬ ‫‪3‬‬

‫‪x 2dx = lim‬‬

‫‪1‬‬

‫∫‬

‫‪0‬‬

‫ﺗﻘﺪﻡ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺷﺮﻃﺎ ﻻﺯﻣﺎ ﻭﻛﺎﻑ ﻟﻜﻲ ﻳﻘﺒﻞ ﺗﺎﺑﻊ ﳏﺪﻭﺩ‬ ‫ﺍﳌﻜﺎﻣﻠﺔ ﻋﻠﻰ ﳎﺎﻝ ﻣﺘﺮﺍﺹ‪.‬‬

‫‪107‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻧﻈﺮﻳﺔ )ﺩﺍﺭﺑﻮ ‪(Darboux‬‬ ‫ﳓﺎﻓﻆ ﻋﻠﻰ ﺍﻟﺮﻣﻮﺯ ﺍﻟﺴﺎﺑﻘﺔ‪.‬‬ ‫ﻟﻴﻜﻦ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﺗﺎﺑﻌﺎ ﳏﺪﻭﺩﺍ‪ .‬ﻧﻀﻊ‬ ‫‪n −1‬‬

‫‪S (f , d n ) = ∑ (ai +1 − ai ). sup f (x i ) ,‬‬ ‫] ‪x i ∈[ai ,ai +1‬‬

‫‪i =0‬‬

‫‪n −1‬‬

‫‪s (f , d n ) = ∑ (ai +1 − ai ). inf f (x i ) .‬‬ ‫] ‪x i ∈[ai ,ai +1‬‬

‫ﻳﺴﻤﻰ‬

‫) ‪ S ( f , d n‬ﻭ ) ‪s (f , d n‬‬

‫ﻳﻜﻮﻥ‬

‫‪f : [a , b ] → ℝ‬‬

‫‪.‬‬

‫‪i =0‬‬

‫ﳎﻤﻮﻋﻲ ﺩﺍﺭﺑﻮ ﺍﻟﻌﻠﻮﻱ ﻭﺍﻟﺴﻔﻠﻲ‪.‬‬

‫ﻗﺎﺑﻼ ﻟﻠﻤﻜﺎﻣﻠﺔ ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻥ‬

‫‪∀ε > 0, ∃d n : S (f , d n ) − s (f , d n ) < ε‬‬

‫ﺗﻄﺒﻴﻘﺎ ﻟﻨﻈﺮﻳﺔ ﺩﺍﺭﺑﻮ ﻧﻘﺪﻡ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫ﻧﻈﺮﻳﺔ‬ ‫ﻛﻞ ﺗﺎﺑﻊ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﺭﺗﻴﺐ ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ‪.‬‬

‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫ﻧﻄﺒﻖ ﻧﻈﺮﻳﺔ ﺩﺍﺭﺑﻮ‪ .‬ﻧﻔﺮﺽ ﻣﺜﻼ ﺃﻥ ‪ f‬ﻣﺘﺰﺍﻳﺪ ﲤﺎﻣﺎ‪ .‬ﻓﻨﻼﺣﻆ ﺃﻥ‬ ‫‪n −1‬‬

‫‪n −1‬‬

‫) ‪S (f ,d n ) = ∑ (ai +1 − ai ). sup f (x i ) = ∑ (ai +1 − ai ).f (ai +1‬‬ ‫] ‪x i ∈[ai ,ai +1‬‬

‫‪i =0‬‬

‫‪n −1‬‬

‫‪i =0‬‬

‫‪n −1‬‬

‫) ‪s (f , d n ) = ∑ (ai +1 − ai ). inf f (x i ) = ∑ (ai +1 − ai ).f (ai‬‬ ‫‪i =0‬‬

‫] ‪x i ∈[ai ,ai +1‬‬

‫‪108‬‬

‫‪i =0‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻭﻣﻨﻪ ‪:‬‬ ‫‪n −1‬‬

‫) ) ‪S (f , d n ) − s (f , d n ) = ∑ (ai +1 − ai ). ( f (ai +1 ) − f (ai‬‬ ‫‪i =0‬‬

‫‪n −1‬‬

‫) ) ‪≤ max(ai +1 − ai )∑ ( f (ai +1 ) − f (ai‬‬ ‫‪i‬‬

‫‪i =0‬‬

‫‪= (f (b ) − f (a )).max(ai +1 − ai ).‬‬ ‫‪i‬‬

‫ﻟﻴﻜﻦ ‪ . 0 < ε‬ﳔﺘﺎﺭ ﺍﻟﺘﻘﺴﻴﻤﺔ‬

‫ﺍﳌﺘﺒﺎﻳﻨﺔ‬

‫) ‪(f (b ) − f (a )).max(ai +1 − ai‬‬ ‫‪i‬‬

‫‪(f (b ) − f (a )).max(ai +1 − ai ) < ε‬‬ ‫‪i‬‬

‫‪ai +1 − ai‬‬

‫‪dn‬‬

‫ﲝﻴﺚ ﲢﻘﻖ ﺍﻟﻌﺒﺎﺭﺓ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ‬

‫‪ ..‬ﻫﺬﺍ ﻳﻌﲏ ﺃﻧﻨﺎ ﳔﺘﺎﺭ ﺍﻟﻔﺮﻭﻕ‬

‫ﺻﻐﲑﺓ ﺑﻜﻔﺎﻳﺔ‪ .‬ﻭﺑﺬﻟﻚ ﻳﺘﺄﻛﺪ ﺍﻟﺸﺮﻁ ﺍﻟﻼﺯﻡ ﻭﺍﻟﻜﺎﰲ ﺍﻟﻮﺍﺭﺩ ﰲ‬

‫ﻧﻈﺮﻳﺔ ﺩﺍﺭﺑﻮ‪.‬‬ ‫ﻧﻼﺣﻆ ﺃﻧﻪ ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﺎﺑﻊ ﺛﺎﺑﺘﺎ ﻓﺈﻥ ﳎﻤﻮﻋﻲ ﺩﺍﺭﺑﻮ ﺍﻟﻌﻠﻮﻱ‬ ‫ﻭﺍﻟﺴﻔﻠﻲ ﻣﺘﺴﺎﻭﻳﺎﻥ‪ ،‬ﻭﻣﻦ ﰒ ﻓﻔﺮﻗﻬﻤﺎ ﻣﻨﻌﺪﻡ ﻭﳛﻘﻖ ﺑﺪﺍﻫ ﹰﺔ ﺷﺮﻁ ﺩﺍﺭﺑﻮ‪.‬‬ ‫ﻭﻫﺬﺍ ﺗﻄﺒﻴﻖ ﺛﺎﻥ ﻟﻨﻈﺮﻳﺔ ﺩﺍﺭﺑﻮ ‪ :‬ﻣﻦ ﺑﲔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻟﻘﺎﺑﻠﺔ ﻟﻠﻤﻜﺎﻣﻠﺔ‬ ‫ﺍﻟﺘﻮﺍﺑﻊ ﺍﳌﺴﺘﻤﺮﺓ‪ .‬ﺫﻟﻚ ﻣﺎ ﺗﻨﺺ‪ ‬ﻋﻠﻴﻪ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﺎﻟﻴﺔ‪.‬‬ ‫ﻧﻈﺮﻳﺔ‬ ‫ﻛﻞ ﺗﺎﺑﻊ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﻣﺴﺘﻤﺮ ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ‪.‬‬

‫‪109‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫ﻧﺴﺘﻔﻴﺪ ﻫﻨﺎ ﻣﻦ ﻧﻈﺮﻳﺔ ﲤﻴﺰ ﺍﻟﺘﻮﺍﺑﻊ ﺍﳌﺴﺘﻤﺮﺓ ﻋﻠﻰ ﳎﺎﻝ ﻣﺘﺮﺍﺹ‪،‬‬ ‫ﻭﻫﻲ ﺍﻟﻘﺎﺋﻠﺔ ﺇﻥ ﻛﻞ ﺗﺎﺑﻊ ﻣﺴﺘﻤﺮ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﻋﻠﻰ ﺍﳌﺘﺮﺍﺹ ] ‪[a, b‬‬

‫ﻣﺴﺘﻤﺮ ﺑﺎﻧﺘﻈﺎﻡ‪ ،‬ﺃﻱ‬ ‫‪∀ε > 0, ∃α > 0 : x '− x " < α ⇒ f (x ') − f (x ") < ε‬‬

‫ﻭﻻ ﻣﺎﻧﻊ ﺃﻥ ﻧﻜﺘﺐ ﻫﺬﻩ ﺍﻟﻌﻼﻗﺔ ﻋﻠﻰ ﺍﻟﺸﻜﻞ ﺇﻥ ﺳﻬﻠﺖ ﻋﻠﻴﻨﺎ‬ ‫ﺍﳊﺴﺎﺑﺎﺕ ‪:‬‬ ‫‪ε‬‬ ‫‪b −a‬‬

‫< )" ‪∀ε > 0, ∃α > 0 : x '− x " < α ⇒ f (x ') − f (x‬‬

‫(‪ . max‬ﻻﺣﻆ‬ ‫ﻟﺘﻜﻦ ‪ . 0 < ε‬ﳔﺘﺎﺭ ﺗﻘﺴﻴﻤﺔ ﲝﻴﺚ ‪ai +1 − ai ) < α‬‬ ‫‪i‬‬ ‫ﺃﻥ ﺫﻟﻚ ﻳﺴﺘﻠﺰﻡ‬ ‫‪.‬‬

‫‪ε‬‬ ‫‪b −a‬‬

‫< ) ‪f (x i‬‬

‫‪inf‬‬

‫] ‪x i ∈[ai ,ai +1‬‬

‫‪sup f (x i ) −‬‬

‫] ‪x i ∈[ai ,ai +1‬‬

‫ﻭﻋﻨﺪﺋﺬ ﻓﺈﻥ‬ ‫‪n −1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪S (f ,dn ) − s (f ,dn ) = ∑ (ai +1 − ai ). sup f (x i ) − inf f (x i ) ‬‬ ‫]‪x i ∈[ai ,ai +1‬‬ ‫‪i =0‬‬ ‫]‪ x i ∈[ai ,ai +1‬‬ ‫‪‬‬

‫) ‪− ai‬‬

‫‪110‬‬

‫‪n −1‬‬

‫‪∑ (a‬‬

‫‪i +1‬‬

‫‪ε‬‬

‫≤‬

‫‪b − a i =0‬‬ ‫‪= ε.‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﻧﻮﺟﺰ ﰲ ﻣﺎ ﻳﻠﻲ ﺑﻌﺾ ﺧﻮﺍﺹ ﺗﻜﺎﻣﻞ ﺭﳝﺎﻥ‪ ،‬ﻭﳝﻜﻦ ﻟﻠﻘﺎﺭﺉ‬ ‫ﺍﻟﺘﺄﻛﺪ ﻣﻦ ﺻﺤﺘﻬﺎ )ﻋﻠﻤﺎ ﺃﻧﻨﺎ ﻧﻀﻊ ﺍﺗﻔﺎﻗﺎ ‪:‬‬ ‫‪ .1‬ﺇﺫﺍ ﻛﺎﻥ‬

‫‪f : [a , b ] → ℝ‬‬

‫‪(x )dx = 0‬‬

‫‪: ( ∫a f‬‬ ‫‪a‬‬

‫ﻗﺎﺑﻼ ﻟﻠﻤﻜﺎﻣﻠﺔ ﻓﺈﻧﻪ ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ‬

‫ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ﺟﺰﺋﻲ ﻣﻦ ] ‪. [a, b‬‬ ‫‪ .2‬ﻋﻼﻗﺔ ﺷﺎﻝ ‪ :Chasles‬ﺇﺫﺍ ﻛﺎﻥ‬ ‫ﻟﻠﻤﻜﺎﻣﻠﺔ ﻭﻛﺎﻥ‬

‫[ ‪c ∈ ]a, b‬‬

‫ﻓﺈﻥ‬

‫‪f‬‬

‫‪f : [a , b ] → ℝ‬‬

‫ﻗﺎﺑﻼ‬

‫ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ ﻋﻠﻰ ﻛﻞ ﻣﻦ‬

‫ﺍ‪‬ﺎﻟﲔ ] ‪ [a, c‬ﻭ ] ‪ ، [c ,b‬ﻭﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪.‬‬

‫‪b‬‬

‫‪c‬‬

‫‪c‬‬

‫‪a‬‬

‫‪f (x )dx = ∫ f (x )dx + ∫ f (x )dx‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫ﳝﻜﻦ ﺗﻌﻤﻴﻢ ﻫﺬﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺇﱃ ﺃﻛﺜﺮ ﻣﻦ ﻧﻘﻄﺔ ‪. c‬‬ ‫‪ .3‬ﻟﺪﻳﻨﺎ ‪:‬‬

‫‪a‬‬

‫‪(x )dx = − ∫ f (x )dx‬‬

‫‪ .4‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﺎﺑﻊ‬

‫‪b‬‬

‫‪f : [a , b ] → ℝ‬‬

‫‪. ∫a f‬‬ ‫‪b‬‬

‫ﳏﺪﻭﺩﺍ ﻭﻗﺎﺑﻼ ﻟﻠﻤﻜﺎﻣﻠﺔ ﻋﻠﻰ‬

‫ﻛﻞ ﳎﺎﻝ ]' ‪ ]a, b [ ⊃ [a ',b‬ﻓﺈﻧﻪ ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ ﻋﻠﻰ ] ‪. [a, b‬‬

‫‪ .5‬ﺗﺸﻜﻞ ﳎﻤﻮﻋﺔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻟﻘﺎﺑﻠﺔ ﻟﻠﻤﻜﺎﻣﻠﺔ ﻋﻠﻰ ﳎﺎﻝ ] ‪[a, b‬‬ ‫ﻓﻀﺎﺀ ﺷﻌﺎﻋﻴﺎ ﻋﻠﻰ ‪ : ℝ‬ﺇﺫﺍ ﻛﺎﻥ ‪ f‬ﻭ ‪ g‬ﻗﺎﺑﻠﲔ ﻟﻠﻤﻜﺎﻣﻠﺔ ﻋﻠﻰ ] ‪[a, b‬‬

‫‪111‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻓﺈﻥ‬ ‫‪(x ) + g (x ) ) dx = ∫ f (x )dx + ∫ g (x )dx ,‬‬ ‫‪b‬‬

‫‪b‬‬

‫‪a‬‬

‫‪a‬‬

‫‪∫a ( f‬‬ ‫‪b‬‬

‫‪b‬‬

‫‪b‬‬

‫‪a‬‬

‫‪a‬‬

‫‪λ ∫ f (x )dx = ∫ λ.f (x )dx ; λ ∈ ℝ.‬‬

‫ﻭﺑﺎﻟﺘﺎﱄ ﻓﺈﻥ ﺗﻄﺒﻴﻖ ﺍﳌﻜﺎﻣﻠﺔ ‪ ∫a‬ﻣﻦ ﺍﻟﻔﻀﺎﺀ ﺍﻟﺸﻌﺎﻋﻲ ﺍﳌﺆﻟﻒ ﻣﻦ‬ ‫ﺍﻟﺘﻮﺍﺑﻊ ﺍﻟﻘﺎﺑﻠﺔ ﻟﻠﻤﻜﺎﻣﻠﺔ‬ ‫‪b‬‬

‫ﳓﻮ‬

‫‪ℝ‬‬

‫ﺗﻄﺒﻖ ﺧﻄﻲ‪ .‬ﳝﻜﻦ ﺗﻌﻤﻴﻢ ﺫﻟﻚ ﺇﱃ ﺍﻟﺘﻮﺍﺑﻊ ﺫﺍﺕ ﺍﻟﻘﻴﻢ ﺍﻟﻌﻘﺪﻳﺔ‪.‬‬ ‫ﻛﻤﺎ ﺃﻥ ﺟﺪﺍﺀ ﺗﺎﺑﻌﲔ ﻗﺎﺑﻠﲔ ﻟﻠﻤﻜﺎﻣﻠﺔ ﺗﺎﺑﻊ ﻗﺎﺑﻞ ﻟﻠﻤﻜﺎﻣﻠﺔ‪.‬‬ ‫‪ .6‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﺎﺑﻊ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﻗﺎﺑﻼ ﻟﻠﻤﻜﺎﻣﻠﺔ ﻓﺈﻥ ﺍﻻﺳﺘﻠﺰﺍﻡ‬

‫ﺍﻟﺘﺎﱄ ﻗﺎﺋﻢ ‪:‬‬ ‫‪f ( x)dx ≥0 .‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫⇒ ‪∀x ∈ [ a, b ] , f ( x) ≥ 0‬‬

‫ﻭﻣﻨﻪ ﻳﺄﰐ ‪ :‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﺎﺑﻌﺎﻥ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﻭ‬

‫‪g : [ a, b ] → ℝ‬‬

‫ﻗﺎﺑﻼﻥ ﻟﻠﻤﻜﺎﻣﻠﺔ ﻓﺈﻥ ﺍﻻﺳﺘﻠﺰﺍﻡ ﺍﻟﺘﺎﱄ ﻗﺎﺋﻢ ‪:‬‬ ‫‪g ( x)dx .‬‬

‫ﻭﳌﺎ ﻛﺎﻥ‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫≥ ‪f ( x)dx‬‬

‫‪b‬‬

‫⇒ )‪∀x ∈ [ a, b ] , f ( x) ≥ g ( x‬‬

‫∫‬

‫‪a‬‬

‫ﻓﺈﻥ ﻫﺬﻩ ﺍﳋﺎﺻﻴﺔ ﺗﺴﺘﻠﺰﻡ ﺃﻥ ﻟﺪﻳﻨﺎ ﺩﻭﻣﺎ ‪:‬‬

‫‪f ≤ f‬‬

‫‪f ( x) dx .‬‬

‫‪ .7‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﺎﺑﻊ‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫≤ ‪f ( x)dx‬‬

‫‪f : [a , b ] → ℝ‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫ﻣﺴﺘﻤﺮﺍ ﻓﺈﻥ ﺍﻻﺳﺘﻠﺰﺍﻡ ﺍﻟﺘﺎﱄ‬

‫ﻗﺎﺋﻢ ‪:‬‬ ‫‪f ( x)dx = 0 ⇒ ∀x ∈ [ a, b ] , f ( x) = 0 .‬‬

‫ﺍﻟﱪﻫﺎﻥ ﻳﺘﻢ ﺑﺎﳋﻠﻒ‪.‬‬ ‫‪112‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .8‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﺎﺑﻌﺎﻥ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﻭ‬

‫‪g : [ a, b ] → ℝ‬‬

‫ﻗﺎﺑﻼﻥ‬

‫ﻟﻠﻤﻜﺎﻣﻠﺔ ﻓﺈﻥ )ﻣﺘﺒﺎﻳﻨﺔ ﻛﻮﺷﻲ – ﺷﻔﺎﺭﺗﺰ ‪Cauchy-‬‬ ‫‪: (Schwarz‬‬ ‫‪g 2 ( x)dx .‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫‪b‬‬

‫× ‪f 2 ( x)dx‬‬

‫∫‬

‫‪b‬‬

‫≤ ‪f ( x) g ( x)dx‬‬

‫‪a‬‬

‫∫‬

‫‪a‬‬

‫ﻟﻨﱪﻫﻦ ﻋﻠﻰ ﻫﺬﻩ ﺍﳋﺎﺻﻴﺔ‪ .‬ﻧﻌﻠﻢ ﺃﻥ ﳎﻤﻮﻋﺔ ﺍﻟﺘﻮﺍﺑﻊ‬ ‫ﺍﻟﻘﺎﺑﻠﺔ ﻟﻠﻤﻜﺎﻣﻠﺔ ﺗﺸﻜﻞ ﻓﻀﺎﺀ ﺷﻌﺎﻋﻴﺎ‪ ،‬ﻭﺑﺎﻟﺘﺎﱄ ﻓﺎﻟﺘﺎﺑﻊ‬

‫‪g+λf‬‬

‫ﻳﻘﺒﻞ ﺍﳌﻜﺎﻣﻠﺔ ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻋﺪﺩ ﺣﻘﻴﻘﻲ ‪ . λ‬ﻧﻌﺘﱪ ﺛﻼﺛﻲ ﺍﳊﺪﻭﺩ‬ ‫ﺑﺎﻟﻨﺴﺒﺔ ﻟـ ‪ ، λ‬ﻣﻊ ﺍﳌﻼﺣﻈﺔ ﺃﻧﻪ ﻣﻮﺟﺐ )ﻭﺑﺎﻟﺘﺎﱄ ﻓﻬﻮ ﳛﺎﻓﻆ ﻋﻠﻰ‬ ‫ﺇﺷﺎﺭﺗﻪ(‪:‬‬ ‫‪b‬‬

‫‪b‬‬

‫‪a‬‬

‫‪a‬‬

‫‪f 2 ( x)dx + 2λ ∫ f ( x) g ( x)dx + ∫ g 2 ( x)dx‬‬

‫‪2‬‬

‫‪( f ( x) + λ g ( x) ) dx = ∫a‬‬

‫‪b‬‬

‫‪b‬‬

‫∫ = ) ‪P(λ‬‬

‫‪a‬‬

‫ﻭﻣﻦ ﰒ ﻓﻤﻤﻴﺰﻩ )ﺍﳌﺨﺘﺼﺮ( ﺳﺎﻟﺐ‪ .‬ﻭﻫﺬﺍ ﻳﻌﲏ ‪:‬‬ ‫‪f 2 ( x)dx × ∫ g 2 ( x)dx .‬‬ ‫‪b‬‬

‫‪a‬‬

‫‪b‬‬

‫∫‬

‫‪a‬‬

‫ﻭﻣﻨﻪ ﺗﺄﰐ ﻋﻼﻗﺔ ﻛﻮﺷﻲ – ﺷﻔﺎﺭﺗﺰ‪.‬‬

‫‪113‬‬

‫≤‬

‫‪2‬‬

‫)‬

‫‪f ( x) g ( x)dx‬‬

‫‪b‬‬

‫∫(‬

‫‪a‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .3‬ﺍﻟﺘﻜﺎﻣﻞ ﻏﲑ ﺍﶈﺪﺩ‬ ‫ﳝﻜﻦ ﺃﻥ ﻧﻌﺮ‪‬ﻑ ﺍﻟﺘﻜﺎﻣﻞ ﻏﲑ ﺍﶈﺪﺩ ﺑﺎﻟﻄﺮﻳﻘﺔ ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﺗﻌﺮﻳﻒ )ﺍﻟﺘﻜﺎﻣﻞ ﻏﲑ ﺍﶈﺪﺩ‪ /‬ﺍﻟﺘﺎﺑﻊ ﺍﻷﺻﻠﻲ(‬ ‫‪f :I →ℝ‬‬

‫ﺗﺎﺑﻌﺎ ﻣﻌﺮﻓﺎ ﻋﻠﻰ ﳎﺎﻝ‬

‫ﻏﲑ ﻣﻐﻠﻖ(‪ .‬ﻳﺴﻤﻰ ﻛﻞ ﺗﺎﺑﻊ‬

‫)ﻣﻌﺮﻑ ﻋﻠﻰ ﳎﺎﻝ‬

‫ﻟﻴﻜﻦ‬

‫‪F‬‬

‫) ‪I = (a , b‬‬ ‫‪⊇J‬‬

‫)ﻣﻐﻠﻖ ﺃﻭ‬

‫‪ ( I‬ﳛﻘﻖ ﺍﳌﻌﺎﺩﻟﺔ‬

‫)ﺍﻟﺘﻔﺎﺿﻠﻴﺔ(‬ ‫‪∀x ∈ J ,‬‬

‫) ‪F '(x ) = f (x‬‬

‫ﺗﻜﺎﻣﻞ ﻏﲑ ﳏﺪﺩ ﻟﻠﺘﺎﺑﻊ ‪ . f‬ﻭﻧﺮﻣﺰ ﳍﺬﺍ ﺍﻟﺘﻜﺎﻣﻞ ﺑـ‬ ‫ﻳﺴﻤﻰ ﺍﻟﺘﺎﺑﻊ‬

‫‪F‬‬

‫ﺗﺎﺑﻌﺎ ﺃﺻﻠﻴﺎ ﻟـ‬

‫‪f‬‬

‫‪(x )dx‬‬

‫‪.F = ∫f‬‬

‫ﻋﻠﻰ ﺍ‪‬ﺎﻝ ‪. J‬‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﻻﺣﻆ ﺃﻥ ﻛﻞ ﺗﺎﺑﻊ ﺃﺻﻠﻲ ﻣﺴﺘﻤﺮ ﺇﺫ ﺃﻧﻪ ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ‪ .‬ﺗﺄﻣﻞ‬ ‫ﰲ ﺍﳌﺜﺎﻝ ﺍﻟﺘﺎﱄ‪ :‬ﻫﻞ ﻳﻘﺒﻞ ﺍﻟﺘﺎﺑﻊ‬

‫‪f :ℝ → ℝ‬‬

‫ﻏﲑ ﺍﳌﺴﺘﻤﺮ ﺍﳌﻌﺮﻑ ﺑـ‬

‫‪ 2, x > 0‬‬ ‫‪f (x ) = ‬‬ ‫‪0, x ≤ 0‬‬

‫ﺗﺎﺑﻌﺎ ﺃﺻﻠﻴﺎ؟ ﺍﳉﻮﺍﺏ ‪ :‬ﻧﻌﻢ‪ ،‬ﻭﻫﺬﺍ ﺗﺎﺑﻊ ﺃﺻﻠﻲ ﻟﻪ ‪:‬‬ ‫‪ 2x , x > 0‬‬ ‫‪F (x ) = ‬‬ ‫‪0, x ≤ 0.‬‬

‫ﺇﻥ‬

‫‪f‬‬

‫ﻏﲑ ﻣﺴﺘﻤﺮ )ﻋﻨﺪ ﺍﻟﺼﻔﺮ( ﺑﻴﻨﻤﺎ ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﺘﺎﺑﻊ ﺍﻷﺻﻠﻲ‬

‫ﻣﺴﺘﻤﺮ‪.‬‬ ‫‪114‬‬

‫‪F‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻧﻈﺮﻳﺔ )ﳎﻤﻮﻋﺔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ(‬ ‫ﺇﺫﺍ ﻗﺒﻞ ﺍﻟﺘﺎﺑﻊ‬

‫‪f :I →ℝ‬‬

‫ﺗﺎﺑﻌﺎ ﺃﺻﻠﻴﺎ ‪ F‬ﻋﻠﻰ ﺍ‪‬ﺎﻝ ‪ I‬ﻓﺈﻥ‬

‫‪f‬‬

‫ﻳﻘﺒﻞ ﻋﺪﺩﺍ ﻏﲑ ﻣﻨﺘﻪ ﻣﻦ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻋﻠﻰ ﺍ‪‬ﺎﻝ ‪. I‬‬ ‫ﻛﻞ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ‬ ‫ﺣﻴﺚ‬

‫‪λ‬‬

‫‪G‬‬

‫ﺗﻜﺘﺐ ﻋﻨﺪﺋﺬ ﻋﻠﻰ ﺍﻟﺸﻜﻞ‬

‫‪G = F +λ‬‬

‫ﺛﺎﺑﺖ ﺣﻘﻴﻘﻲ‪ ،‬ﺃﻱ ﺃﻥ ‪:‬‬ ‫‪G (x ) = F ( x ) + λ .‬‬

‫‪∀x ∈ I ,‬‬

‫ﻫﻨﺎﻙ ﺗﻮﺍﺑﻊ ﻻ ﺗﻘﺒﻞ ﺗﻮﺍﺑﻊ ﺃﺻﻠﻴﺔ‪ ،‬ﻭﺃﺧﺮﻯ ﺗﻘﺒﻞ ﻣﺜﻞ ﺗﻠﻚ ﺍﻟﺘﻮﺍﺑﻊ‪.‬‬ ‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺗﻘﺪﻡ ﻓﺌﺔ ﺷﻬﲑﺓ ﻣﻦ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻟﱵ ﺗﻘﺒﻞ ﺗﻮﺍﺑﻊ ﺃﺻﻠﻴﺔ‪.‬‬ ‫ﻧﻈﺮﻳﺔ )ﻭﺟﻮﺩ ﺗﺎﺑﻊ ﺃﺻﻠﻲ(‬ ‫ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﺎﺑﻊ‬

‫‪f :I →ℝ‬‬

‫ﻣﺴﺘﻤﺮﺍ ﻓﺈﻧﻪ ﻳﻘﺒﻞ ﻋﺪﺩﺍ ﻏﲑ ﻣﻨﺘﻪ‬

‫ﻣﻦ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ‪.‬‬ ‫ﻟﻴﻜﻦ‬

‫‪f :I →ℝ‬‬

‫ﺗﺎﺑﻌﺎ ﻣﺴﺘﻤﺮﺍ‪ .‬ﺇﻧﻪ ﻳﻘﺒﻞ ﻋﺪﺩﺍ ﻏﲑ ﻣﻨﺘﻪ ﻣﻦ‬

‫ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻋﻠﻰ ‪ . I‬وﺇﺫﺍ ﻛﺎﻥ‬

‫‪F‬‬

‫ﻭ‬

‫ﻋﻠﻰ ‪ I‬ﻓﺈﻧﻪ ﻳﻮﺟﺪ ﻋﺪﺩ ﺣﻘﻴﻘﻲ‬ ‫‪G (x ) = F (x ) + λ .‬‬ ‫ﻧﻼﺣﻆ ﻋﻨﺪﺋﺬ‬ ‫‪λ‬‬

‫‪G‬‬

‫ﺗﺎﺑﻌﲔ ﺃﺻﻠﻴﲔ ﻟـ‬

‫ﲝﻴﺚ‬

‫‪115‬‬

‫‪∀x ∈ I ,‬‬

‫‪f‬‬

‫ا ‪ : 3‬ا ب ا ‬ ‫‪∀(a,b) ∈I 2 ,‬‬

‫‪G(b) −G(a) = F(b) − F(a).‬‬

‫ﺗﻌﺮﻳﻒ )ﺗﻜﺎﻣﻞ ﺗﺎﺑﻊ(‬ ‫ﻟﻴﻜﻦ‬ ‫ﺗﻜﺎﻣﻞ‬

‫‪f‬‬

‫‪f :I →ℝ‬‬

‫ﺗﺎﺑﻌﺎ ﻣﺴﺘﻤﺮﺍ‪ .‬ﻳﺴﻤﻰ ﺍﻟﻌﺪﺩ‬

‫)‪F(b)− F(a‬‬

‫‪b‬‬

‫ﻣﻦ‬

‫‪a‬‬

‫ﺇﱃ ‪ b‬ﻭ ﻧﻜﺘﺐ ‪ [F(x)]ba‬ﺃﻭ ‪ ، ∫ f(x)dx‬ﺃﻱ‬ ‫‪a‬‬ ‫‪b‬‬

‫‪= F (b ) − F (a ).‬‬

‫]) ‪∫ f (x )dx = [ F (x‬‬

‫‪b‬‬ ‫‪a‬‬

‫‪a‬‬

‫ﻭﻧﻘﻮﻝ ﺇﻥ ‪ f‬ﻗﺎﺑﻞ ﻟﻠﻤﻜﺎﻣﻠﺔ ﻋﻠﻰ ﺍ‪‬ﺎﻝ ]‪. [a,b‬‬ ‫ﻣﺜﺎﻻﻥ‬ ‫‪(1‬‬ ‫) ‪= λ (b − a‬‬

‫‪b‬‬

‫] ‪∫ λ dx = [λ x‬‬

‫‪∀λ ∈ ℝ,‬‬

‫‪∀(a, b) ∈ ℝ 2 ,‬‬

‫‪b‬‬ ‫‪a‬‬

‫‪a‬‬

‫‪(2‬‬

‫‪1‬‬ ‫‪π π π‬‬ ‫‪1‬‬ ‫= ‪dx = [ arctan ] 1 = −‬‬ ‫‪2‬‬ ‫‪4 6 12‬‬ ‫‪1+ x‬‬ ‫‪3‬‬

‫‪1‬‬

‫∫‪.‬‬ ‫‪1‬‬ ‫‪3‬‬

‫ﻣﻼﺣﻈﺎﺕ ‪:‬‬ ‫‪ (1‬ﳝﻜﻦ ﺍﻟﱪﻫﺎﻥ ﻋﻠﻰ ﺃﻥ ﻗﺎﺑﻠﻴﺔ ﺍﳌﻜﺎﻣﻠﺔ ﻫﻨﺎ ﻫﻲ ﲟﻔﻬﻮﻡ ﺭﳝﺎﻥ‪.‬‬ ‫‪ (2‬ﳝﻜﻦ ﺇﺛﺒﺎﺕ ﻋﻼﻗﺔ ﺷﺎﻝ ﺍﻟﺴﺎﻟﻔﺔ ﺍﻟﺬﻛﺮ ﺍﺳﺘﻨﺎﺩﺍ ﺇﱃ ﺧﻮﺍﺹ‬ ‫ﺍﻟﺪﻭﺍﻝ ﺍﻷﺻﻠﻴﺔ‪ :‬ﺇﺫﺍ ﻛﺎﻥ‬

‫‪F‬‬

‫ﺗﺎﺑﻌﺎ ﺃﺻﻠﻴﺎ ﻟﻠﺘﺎﺑﻊ‬

‫ﻓﺈﻥ‬ ‫‪b‬‬

‫)‪∫ f(x)dx=F(b)− F(a‬‬ ‫‪a‬‬

‫‪116‬‬

‫‪f‬‬

‫ﻋﻠﻰ ﺍ‪‬ﺎﻝ ]‪[ a; b‬‬

‫ا ‪ : 3‬ا ب ا ‬ ‫)‪= F(b)− F(c)+ F(c) − F(a‬‬ ‫‪c‬‬

‫‪b‬‬

‫‪a‬‬

‫‪c‬‬

‫‪= ∫ f(x)dx + ∫ f(x)dx.‬‬

‫‪ (3‬ﳝﻜﻦ ﺃﻳﻀﺎ ﺇﺛﺒﺎﺕ ﺧﻄﻴﺔ ﻋﻤﻠﻴﺔ ﺍﳌﻜﺎﻣﻠﺔ ﺍﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺍﻟﺪﻭﺍﻝ‬ ‫ﺍﻷﺻﻠﻴﺔ ‪:‬‬ ‫ﻟﻴﻜﻦ‬ ‫ﻭ‬

‫‪β‬‬

‫‪f‬‬

‫ﻭ‬

‫‪g‬‬

‫ﺗﺎﺑﻌﲔ ﻣﺴﺘﻤﺮﻳﻦ ﻋﻠﻰ ﳎﺎﻝ ]‪ ، I = [ a, b‬ﻭﻟﻴﻜﻦ‬

‫‪α‬‬

‫ﻋﺪﺩﻳﻦ ﺣﻘﻴﻘﻴﲔ‪ .‬ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪b‬‬

‫‪b‬‬

‫‪b‬‬

‫‪a‬‬

‫‪a‬‬

‫‪a‬‬

‫‪∫ (α f + β g )( x)dx = α ∫ f ( x)dx + β ∫ g ( x)dx.‬‬ ‫ﻟﺮﺅﻳﺔ ﺫﻟﻚ ﻧﻌﺘﱪ ﺩﺍﻟﺘﲔ ﺃﺻﻠﻴﺘﲔ‬

‫‪F‬‬

‫ﻭ‬

‫‪G‬‬

‫ﻟﻠﺘﺎﺑﻌﲔ‬

‫‪f‬‬

‫ﻭ‬

‫ﻋﻠﻰ‬

‫‪g‬‬

‫ﺍ‪‬ﺎﻝ ‪ ، I‬ﻭﻧﻼﺣﻆ ‪:‬‬ ‫‪(αF + βG)'=αf + βg.‬‬

‫ﻭﻣﻨﻪ ﻳﺘﻀﺢ ﺃﻥ ‪ αF + βG‬ﺩﺍﻟﺔ ﺃﺻﻠﻴﺔ ﻟﻠﺪﺍﻟﺔ ‪ αf + βg‬ﻋﻠﻰ ‪ . I‬ﺇﺫﻥ‬ ‫‪b‬‬

‫])‪∫ (α f + β g )( x)dx = [(α F + β G )( x‬‬

‫‪b‬‬ ‫‪a‬‬

‫‪a‬‬

‫)) ‪= (α F (b) + β G (b)) − (α F ( a ) + β G ( a‬‬ ‫) ) ‪= α ( F (b) − F ( a ) ) + β ( G (b) − G ( a‬‬ ‫‪b‬‬

‫‪b‬‬

‫‪a‬‬

‫‪a‬‬

‫‪= α ∫ f ( x ) dx + β ∫ g ( x ) dx.‬‬

‫‪117‬‬

‫ ا ب ا‬: 3 ‫ا‬

‫ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﺍﳌﺘﺪﺍﻭﻟﺔ‬ ‫ت‬42 α ∈ ℝ,

/0‫وال ا‬.‫ا‬ α +1

α ≠ −1

x

α +1

+k

‫ا‬.‫ا‬ xα

ln x + k

1 x

−cos x+ k

sin x

sin x+ k −cot anx+ k

cos x 1 sin 2 x

tan x+ k

1 cos 2 x

−ln cos x + k

tan x

ln sin x + k

λ∈IR*

1 eλx + k

λ

−arccos x + k'=

118

ctnx eλx

arctan x+ k

1 x 2 +1

arcsin x+ k

1 1− x 2

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .4‬ﻣﻦ ﻃﺮﻕ ﺍﳌﻜﺎﻣﻠﺔ ‪:‬‬ ‫‪ .1‬ﺗﺒﺪﻳﻞ ﺍﳌﺘﻐﲑ ‪:‬‬ ‫ﻧﻈﺮﻳﺔ )ﺣﺎﻟﺔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ(‬ ‫ﻟﻴﻜﻦ‬

‫‪u:I → J‬‬

‫ﺍﳌﺴﺘﻖ ﻣﺴﺘﻤﺮ(‪ .‬ﻭﻟﻴﻜﻦ‬ ‫ﺃﺻﻠﻴﺎ ﻟـ‬

‫‪f‬‬

‫ﻋﻠﻰ ‪ J‬ﻓﺈﻥ‬

‫ﺗﺎﺑﻌﺎ ﻣﻦ‬ ‫‪f‬‬

‫ﺍﻟﺼﻨﻒ ‪C1‬‬

‫)ﺃﻱ ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻭﺍﻟﺘﺎﺑﻊ‬

‫ﺗﺎﺑﻌﺎ ﻣﺴﺘﻤﺮﺍ ﻋﻠﻰ ﺍ‪‬ﺎﻝ ‪ . J‬ﺇﺫﺍ ﻛﺎﻥ‬

‫‪F u‬‬

‫ﺗﺎﺑﻊ ﺃﺻﻠﻲ ﻟﻠﺘﺎﺑﻊ‬

‫' ‪( f u )u‬‬

‫‪∫ f(u(x))u'(x)dx= F u(x)+c.‬‬ ‫ﺣﻴﺚ‬

‫ﺛﺎﺑﺖ ﺣﻘﻴﻘﻲ‪.‬‬

‫‪c‬‬

‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬ ‫ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﻣﺸﺘﻖ ﺗﺮﻛﻴﺐ ﺗﺎﺑﻌﲔ ﻟﺪﻳﻨﺎ‬ ‫'‪(F u)'=(F'u)u‬‬ ‫‪=(f u)u' .‬‬

‫ﻭﻋﻠﻴﻪ‬

‫ﻓﺈﻥ ‪F u‬‬

‫ﺗﺎﺑﻊ ﺃﺻﻠﻲ ﻟﻠﺘﺎﺑﻊ '‪. (f u)u‬‬

‫ﻣﺜﺎﻻﻥ‬ ‫‪ .1‬ﻟﺪﻳﻨﺎ ﻣﻦ ﺃﺟﻞ ﻛﻞ‬

‫‪c∈ℝ‬‬

‫‪∫ ln xdx = ∫ 1.ln xdx‬‬ ‫‪1‬‬ ‫‪= x ln x − ∫ x ⋅ dx‬‬ ‫‪x‬‬ ‫‪= x ln x − x + c.‬‬

‫‪119‬‬

‫ﻭ‬

‫‪F‬‬

‫ﺗﺎﺑﻌﺎ‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .2‬ﻟﺪﻳﻨﺎ ﻣﻦ ﺃﺟﻞ ﻛﻞ‬ ‫‪dx‬‬

‫‪c∈ℝ‬‬ ‫‪x‬‬

‫‪2‬‬

‫‪∫ arctan xdx =x arctan x − ∫ 1 + x‬‬

‫‪1‬‬ ‫‪= x arctan x − ln(1 + x 2 ) + c.‬‬ ‫‪2‬‬

‫ﻭﻗﺪ ﺍﺧﺘﺮﻧﺎ ﻫﻨﺎ‬

‫‪f ( x) = x‬‬

‫ﻭ ‪. g ( x) = arctan x‬‬

‫ﻧﻈﺮﻳﺔ )ﺗﺒﺪﻳﻞ ﺍﳌﺘﻐﲑ(‬ ‫ﻟﻴﻜﻦ‬

‫‪f‬‬

‫ﻭﻟﻴﻜﻦ‬ ‫ﲝﻴﺚ ‪u(α)=a‬‬ ‫ﻭ ‪u(x) ، β‬‬

‫ﺗﺎﺑﻌﺎ ﻣﺴﺘﻤﺮﺍ ﻋﻠﻰ ﳎﺎﻝ‬ ‫‪u‬‬

‫‪J‬‬

‫ﻳﺸﻤﻞ ﺍﻟﻌﺪﺩﻳﲔ ‪ a‬ﻭ ‪.b‬‬

‫ﺗﺎﺑﻌﺎ ﻣﻦ ﺍﻟﺼﻨﻒ ‪ C1‬ﻋﻠﻰ ﳎﺎﻝ‬

‫ﻭ ‪ . u(β)=b‬ﻧﻔﺮﺽ ﻣﻦ ﺃﺟﻞ ﻛﻞ‬

‫ﻳﻨﺘﻤﻲ ﺇﱃ ﺍ‪‬ﺎﻝ ‪ . J‬ﻋﻨﺪﺋﺬ‬ ‫‪β‬‬

‫‪b‬‬

‫‪∫ f(t)dt =α∫ f(u(x))u'(x)dx.‬‬ ‫‪a‬‬

‫‪120‬‬

‫‪I‬‬

‫‪x‬‬

‫ﻳﺸﻤﻞ‬

‫‪ α‬و‪β‬‬

‫ﳏﺼﻮﺭ ﺑﲔ‬

‫‪α‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﺍﻟﱪﻫﺎﻥ‬ ‫ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ )ﺣﺎﻟﺔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ(‪ ،‬ﺇﺫﺍ‬ ‫ﺃﺻﻠﻴﺎ ﻟـ‬

‫‪f‬‬

‫ﻛﺎﻥ ‪F‬‬

‫ﺗﺎﺑﻌﺎ‬

‫ﻓﺈﻧﻪ‬ ‫‪β‬‬

‫‪∫ f (u ( x))u '( x)dx = [ F u ( x)]α‬‬ ‫‪α‬‬

‫‪β‬‬

‫)) ‪= F (u ( β )) − F (u (α‬‬ ‫) ‪= F (b) − F (a‬‬ ‫‪b‬‬

‫‪= ∫ f (t )dt.‬‬ ‫‪a‬‬

‫‪b‬‬

‫ﻣﻦ ﺍﻟﻨﺎﺣﻴﺔ ﺍﻟﻌﻤﻠﻴﺔ ﻧﻘﻮﻡ ﲟﺎ ﻳﻠﻲ ‪ :‬ﳊﺴﺎﺏ ﺍﻟﺘﻜﺎﻣﻞ ‪ ∫ f(t)dt‬ﻧﻀﻊ‬ ‫‪a‬‬ ‫)‪t=u(x‬‬

‫و ‪ dt‬ﺗﺼﺒﺢ ‪ ، u'(x)dx‬ﻭﺣﺪﺍ ﺍﻟﺘﻜﺎﻣﻞ ‪ a‬ﻭ ‪ b‬ﻳﺼﺒﺤﺎﻥ‬

‫و ‪ ، β‬ﺣﻴﺚ ‪u(α)=a‬‬ ‫‪f‬‬

‫ﻭ‬

‫‪u‬‬

‫‪α‬‬

‫ﻭ ‪ . u(β)=b‬ﻭﻳﻨﺒﻐﻲ ﺃﻻ ﻧﻨﺴﻰ ﺍﻟﺘﺄﻛﺪ ﻣﻦ ﺃﻥ ﺍﻟﺘﺎﺑﻌﲔ‬

‫ﳛﻘﻘﺎﻥ ﺷﺮﻭﻁ ﺍﻟﻨﻈﺮﻳﺔ‪.‬‬

‫ﻣﺜﺎﻝ‬ ‫ﺃﺣﺴﺐ ﺍﻟﺘﻜﺎﻣﻞ‬

‫‪1−t 2 dt‬‬

‫‪1‬‬

‫∫= ‪. I‬‬ ‫‪0‬‬

‫ﻧﻀﻊ ‪dt ، t =sin x‬‬

‫ﺗﺼﺒﺢ ‪cos xdx‬‬

‫ﻭﺣﺪﺍ ﺍﻟﺘﻜﺎﻣﻞ ﺍﳉﺪﻳﺪ ﳘﺎ ‪0‬‬

‫و ‪ . π2‬ﺇﺫﻥ‬ ‫‪π‬‬ ‫‪2‬‬

‫‪I = ∫ 1−sin 2 x cos xdx‬‬ ‫‪0‬‬

‫‪π‬‬ ‫‪2‬‬

‫‪= ∫cos 2 xdx‬‬ ‫‪0‬‬

‫‪121‬‬

‫ا ‪ : 3‬ا ب ا ‬ ‫‪π‬‬ ‫‪2‬‬

‫‪= ∫1+cos2x dx‬‬ ‫‪2‬‬ ‫‪0‬‬

‫[‬

‫]‬

‫‪π‬‬

‫‪= x + sin 2x 2‬‬ ‫‪2‬‬ ‫‪4 0‬‬

‫‪=π‬‬ ‫‪4‬‬

‫ﻋﻠﻤﺎ ﺃﻥ ﺷﺮﻭﻁ ﺍﻟﻨﻈﺮﻳﺔ ﳏﻘﻘﺔ ﰲ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻟﻮﺍﺭﺩ ﰲ ﺍﳌﺜﺎﻝ‪.‬‬ ‫‪ .2‬ﺍﳌﻜﺎﻣﻠﺔ ﺑﺎﻟﺘﺠﺰﺋﺔ‬ ‫ﻧﻌﻠﻢ‬

‫ﺗﺎﺑﻊ ﺃﺻﻠﻲ ـ ‪ . f ⋅g‬ﻭﻣﻦ ﰒ ﺗﺄﰐ ﺍﻟﻨﻈﺮﻳﺔ‬

‫ﺃﻥ '‪f'⋅g + f ⋅g‬‬

‫ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﻧﻈﺮﻳﺔ )ﺍﳌﻜﺎﻣﻠﺔ ﺑﺎﻟﺘﺠﺰﺋﺔ(‬ ‫ﺇﺫﺍ ﻛﺎﻥ‬

‫ﺗﺎﺑﻌﲔ ﻣﻦ‬

‫‪ f‬ﻭ‪g‬‬

‫ﺍﻟﺼﻨﻒ ‪C1‬‬

‫ﻋﻠﻰ ﺍ‪‬ﺎﻝ ‪ ، I‬ﻭﻛﺎﻧﺖ ‪a‬‬

‫ﻭ ‪ b‬ﻧﻘﻄﺘﲔ ﻣﻦ ‪ I‬ﻓﺈﻥ‬ ‫‪b‬‬

‫‪b‬‬

‫‪∫ f'(x)g(x)dx=[ f(x)g(x)] −∫ f(x)g'(x)dx.‬‬ ‫‪b‬‬ ‫‪a‬‬

‫‪a‬‬

‫‪a‬‬

‫ﻣﺜﺎﻝ‬ ‫ﺍﺣﺴﺐ ‪. ∫0 xe− x dx‬‬ ‫‪2‬‬

‫ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪xe − x dx =  − xe − x  − ∫ −e − x dx‬‬ ‫‪0‬‬

‫‪2‬‬

‫‪−x‬‬

‫‪0‬‬

‫‪−2‬‬

‫‪= −2e +  −e ‬‬ ‫‪0‬‬ ‫‪= −2e−2 − e −2 + 1‬‬ ‫‪= −3e−2 + 1.‬‬ ‫‪122‬‬

‫‪2‬‬

‫∫‬

‫‪0‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .3‬ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻟﺘﺎﺑﻊ ﻛﺴﺮﻱ‬ ‫ﻳﻜﺘﺐ ﻛﻞ ﻛﺴﺮ ﻧﺎﻃﻖ ﺑﻜﻴﻔﻴﺔ ﻭﺣﻴﺪﺓ ﻛﻤﺠﻤﻮﻉ ﻟﻜﺜﲑ ﺣﺪﻭﺩ ﻭﻟﻌﺪﺩ ﻣﻨﺘﻪ‬ ‫ﻣﻦ ﺍﻟﻜﺴﻮﺭ ﺍﻟﻨﺎﻃﻘﺔ ﺫﺍﺕ ﺍﻟﺸﻜﻞ )ﺣﻴﺚ‬

‫‪m‬‬

‫ﻭ ‪ n‬ﻋﺪﺩﺍﻥ ﻃﺒﻴﻌﻴﺎﻥ( ‪:‬‬

‫‪A‬‬ ‫‪,‬‬ ‫‪( x − a)n‬‬ ‫‪Bx + C‬‬ ‫‪.‬‬ ‫‪(( x − b) 2 + c 2 ) m‬‬

‫ﻭﻋﻠﻴﻪ ﻳﻜﻔﻲ ﺍﻟﺒﺤﺚ ﻋﻦ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻟﻠﻜﺴﺮﻳﻦ ﺍﻟﺴﺎﺑﻘﲔ‪.‬‬ ‫ﺣﺎﻟﺔ ‪: 1‬‬ ‫ﺗﻌﻴﲔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻟﻠﺘﺎﺑﻊ ‪. x֏ (x−1a)n‬‬ ‫‪ .1‬ﺇﺫﺍ ﻛﺎﻥ‬

‫‪n=1‬‬

‫ﻓﺈﻥ‬

‫‪ .2‬ﺇﺫﺍ ﻛﺎﻥ‬

‫‪n>1‬‬

‫ﻓﺈﻥ‬

‫‪∫ x−1a dx=ln x−a +c‬‬ ‫‪−1‬‬ ‫‪+c‬‬ ‫‪(n−1)(x−a)n −1‬‬

‫= ‪dx‬‬

‫‪n‬‬

‫)‪∫ (x−1a‬‬

‫ﺣﺎﻟﺔ ‪: 2‬‬ ‫ﺗﻌﻴﲔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻟﻠﺘﺎﺑﻊ‬

‫‪1‬‬ ‫‪x 2 + px+ q‬‬

‫ﺟﺬﻭﺭﺍ ﺣﻘﻴﻘﻴﺔ‪ .‬ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻳﻜﻮﻥ‬

‫֏‪x‬‬

‫‪x 2 + px+ q‬‬

‫إذا ﱂ‬

‫ﻳﻘﺒﻞ ‪x 2 + px+ q‬‬

‫ﻣﻮﺟﺒﺎ ﲤﺎﻣﺎ‪.‬‬

‫ﻧﻀﻊ ‪ . x2 + px+ q =u 2 +a 2‬ﻋﻨﺪﺋﺬ‬ ‫‪1 dx= 1 arctan u +c‬‬ ‫‪+a2‬‬ ‫‪a‬‬ ‫‪a‬‬

‫‪123‬‬

‫‪2‬‬

‫‪∫u‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻣﺜﺎﻝ‬ ‫ﺣﺴﺎﺏ ‪. ∫ x2 −1x+1dx‬‬ ‫ﻟﺪﻳﻨﺎ ‪ ، x2 − x+1=(x− 12)2 + 43‬ﺇﺫﻥ‬

‫‪∫ x −1x+1dx=∫ (x− 11) + 3 dx‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪4‬‬

‫‪2‬‬

‫‪x− 1‬‬ ‫‪1‬‬ ‫=‬ ‫‪arctan 2 +c‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪= 2 arctan 2x −1+c‬‬ ‫‪3‬‬ ‫‪3‬‬

‫ﺣﺎﻟﺔ ‪: 3‬‬ ‫ﺗﻌﻴﲔ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻟﻠﺘﺎﺑﻊ‬ ‫ﻳﻘﺒﻞ ‪x 2 + px+ q‬‬

‫‪x֏ 2ax +b‬‬ ‫‪x + px+ q‬‬

‫ﺇﺫﺍ ﱂ‬

‫ﺟﺬﻭﺭﺍ ﺣﻘﻴﻘﻴﺔ‪.‬‬

‫ﳊﺴﺎﺏ ﻫﺬﺍ ﺍﻟﺘﻜﺎﻣﻞ ﻧﻈﻬﺮ ﻣﺸﺘﻖ ﺍﳌﻘﺎﻡ ﰲ ﺍﻟﺒﺴﻂ ﻓﻨﺤﺼﻞ ﻋﻠﻰ ﺗﻜﺎﻣﻠﲔ‬ ‫ﺃﺣﺪﳘﺎ ﻣﻦ ﺍﻟﺸﻜﻞ ﺍﳌﺪﺭﻭﺱ ﺳﺎﺑﻘﺎ‪.‬‬ ‫ﻣﺜﺎﻝ‬ ‫ﺣﺴﺎﺏ ‪. ∫ x−23−xx++21dx‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪− (2x − 1) +‬‬ ‫‪−3x + 2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪∫ x 2 − x + 1 dx = ∫ x 2 − x + 1 dx‬‬ ‫‪=− 3 ∫ 22x−1 dx+ 1 ∫ 2 1 dx‬‬ ‫‪2 x − x+1‬‬ ‫‪2 x − x+1‬‬ ‫‪=− 3 ln(x 2 − x +1)+ 1 ⋅ 2 arctan 2x−1+c‬‬ ‫‪2‬‬ ‫‪2 3‬‬ ‫‪3‬‬

‫‪124‬‬

‫ ا ب ا‬: 3 ‫ا‬

1 ‫ﲤﺮﻳﻦ‬ ‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﻃﺮﻳﻘﺔ ﺗﺒﺪﻳﻞ ﺍﳌﺘﻐﲑ ﺃﺛﺒﺖ ﻣﺎ ﻳﻠﻲ‬ x 1 dx = ln(a 2 + x 2 ) + c , 2 2 a +x . ∫ 2 1 2 dx = 1 arctan x + c , a a a +x

.∫

2

(a ∈ ℝ)

(1

(a ∈ ℝ* )

(2 ‫ﺍﳊﻞ‬

‫ ﺇﺫﻥ‬. dt =2xdx ‫ﳒﺪ‬

∫ a +x x 2

2

t =a 2 + x 2

‫( ﺑﻮﺿﻊ‬1

dx= 1 ∫1dt 2 t

= 1 lnt+c 2 = 1 ln(a 2 + x 2)+c 2

2 ‫( ﻟﺪﻳﻨﺎ‬

∫ a +1 x 2

2

dx = 12 ∫ 1 dx . a 1+( x )2 a

‫ ﺇﺫﻥ‬. dt = 1a dx ‫ﻓﻨﺠﺪ‬

∫ a +1 x 2

2

dx = 12 ∫ a 2 dt a 1+t

= 1 arctant +c a = 1 arctan x +c a a

125

t= x a

‫ﻧﻀﻊ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﲤﺮﻳﻦ ‪2‬‬ ‫ﺍﺣﺴﺐ ﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺘﺎﱄ‬

‫‪∫ (x+1)2(xx−1+ x+1) dx.‬‬ ‫‪2‬‬

‫‪2‬‬

‫ﻧﻘﻮﻡ ﻗﺲ ﺍﻟﺒﺪﺍﻳﺔ ﺑﺘﻔﻜﻴﻚ ﺍﻟﻜﺴﺮ ﻋﻠﻰ ﺍﻟﺸﻜﻞ‬ ‫‪2x−1‬‬ ‫‪= a + b + cx + d .‬‬ ‫‪(x +1)2(x 2 + x +1) (x+1)2 x+1 x 2 + x+1‬‬

‫ﻭﺑﺎﳌﻄﺎﺑﻘﺔ ﳒﺪ‬

‫‪c = 1 ، b = −1 ، a = −3‬‬

‫‪dx− ∫ 1 dx + ∫ 2x+3 dx .‬‬ ‫‪x +1‬‬ ‫‪x + x +1‬‬

‫‪2‬‬

‫‪ . d =3 ،‬ﻭﻣﻨﻪ‬

‫)‪∫ (x+1)2(xx−1+ x+1) dx=3∫ (x+−11‬‬ ‫‪2‬‬

‫‪2‬‬

‫ﻧﻼﺣﻆ ﺑﻌﺪ ﺫﻟﻚ ‪:‬‬ ‫‪dx = 1 +c1 ,‬‬ ‫‪x +1‬‬

‫‪2‬‬

‫)‪∫ (x+−11‬‬

‫ﻭ‬

‫‪∫ x1+1dx=ln x+1 +c ,‬‬ ‫‪2‬‬

‫ﻭ‬ ‫‪x +3‬‬ ‫‪1/ 2(2x + 1) + 5 / 2‬‬ ‫∫ = ‪dx‬‬ ‫‪dx‬‬ ‫‪+ x +1‬‬ ‫‪x 2 + x +1‬‬

‫‪2‬‬

‫‪1‬‬ ‫‪2x + 1‬‬ ‫‪5‬‬ ‫‪1‬‬ ‫‪dx + ∫ 2‬‬ ‫‪dx‬‬ ‫‪2‬‬ ‫∫‬ ‫‪2 x + x +1‬‬ ‫‪2 x + x +1‬‬

‫=‬

‫‪1‬‬ ‫‪5‬‬ ‫∫ ‪ln(x 2 + x + 1) +‬‬ ‫‪2‬‬ ‫‪2‬‬

‫=‬

‫‪1‬‬ ‫‪dx‬‬ ‫‪1 2 3‬‬ ‫‪(x + ) +‬‬ ‫‪2‬‬ ‫‪4‬‬

‫‪126‬‬

‫‪∫x‬‬

‫ ا ب ا‬: 3 ‫ا‬

1 x + 1 5 1 2 +c = ln(x 2 + x + 1) + ⋅ arctan 3 2 2 3 3 4 4 1 5 x +1 = ln(x 2 + x + 1) + ⋅ arctan + c3 . 2 3 3

‫ﻭﻋﻠﻴﻪ‬ I =

3 1 5 x +2 − ln x + 1 + ln(x 2 + x + 1) + arctan + c. x +1 2 3 3

127

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .5‬ﻃﻮﻝ ﻗﻮﺱ ﻣﻦ ﻣﻨﺤﲎ ‪:‬‬ ‫ﻳﺘﻄﻠﺐ ﺣﺴﺎﺏ ﺃﻃﻮﺍﻝ ﺍﻷﻗﻮﺍﺱ ﺗﻌﺮﻳﻒ ﺍﻟﻘﻮﺱ ﻭﺍﳊﺪﻳﺚ ﻋﻦ‬ ‫ﺍﻟﺘﻮﺍﺑﻊ ﺍﻟﱵ ﺗﻜﻮﻥ ﳎﻤﻮﻋﺔ ﺗﻌﺮﻳﻔﻬﺎ ‪ . ℝ n‬ﺳﻮﻑ ﻳﻜﻮﻥ ﻫﺬﺍ ﺍﳊﺪﻳﺚ‬ ‫ﻣﻘﺘﻀﺒﺎ‪ ،‬ﻭﻧﻜﺘﻔﻲ ﺑﺄﻗﺼﺮ ﺍﻟﻄﺮﻕ ﻟﻠﻮﺻﻮﻝ ﺇﱃ ﺗﻌﺮﻳﻒ ﻃﻮﻝ ﻗﻮﺱ‪.‬‬ ‫ﺗﻌﺮﻳﻒ )ﺍﳌﻨﺤﲎ(‬ ‫ﺍﳌﻨﺤﲎ ﰲ ‪ ℝ n‬ﻫﻮ ﺗﺎﺑﻊ ‪. f : [a, b ] → ℝ n‬‬ ‫ﻧﺴﻤﻲ ﺍﻟﻨﻘﻄﺘﲔ ) ‪ f (a‬ﻭ ) ‪ f (b‬ﻃﺮﰲ ﺍﳌﻨﺤﲎ‪.‬‬ ‫ﻧﻘﻮﻝ ﻋﻦ ﺍﳌﻨﺤﲎ ﺇﻧﻪ ﻣﻐﻠﻖ ﺇﺫﺍ ﻛﺎﻥ‬

‫) ‪(a ) = f (b‬‬

‫ﻧﻘﻮﻝ ﻋﻦ ﺍﳌﻨﺤﲎ ﺇﻧﻪ ﺑﺴﻴﻂ ﺇﺫﺍ ﻛﺎﻥ‬

‫‪f‬‬

‫‪.f‬‬

‫ﻣﺘﺒﺎﻳﻨﺎ ﻋﻠﻰ ] ‪ ، [a, b‬ﺃﻱ‬

‫ﺇﺫﺍ ﻛﺎﻥ ﺍﳌﻨﺤﲎ ﻻ ﻳﺘﻘﺎﻃﻊ ﻣﻊ ﺫﺍﺗﻪ‪.‬‬ ‫ﻧﻘﻮﻝ ﻋﻦ ﺍﳌﻨﺤﲎ ﺇﻧﻪ ﻣﻨﺤﲎ ﺟﻮﺭﺩﺍﻥ ‪ Jordan‬ﺇﺫﺍ ﻛﺎﻥ ﻣﻐﻠﻘﺎ‬ ‫ﻭﻣﺘﺒﺎﻳﻨﺎ ﻋﻠﻰ [ ‪. ]a, b‬‬ ‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺻﻮﺭﺓ‬

‫ﺍﳌﻨﺤﲎ‬

‫) ‪f (t ) = ( sin t , cos t‬‬

‫‪f : [ 0, 2π ] → ℝ 2‬‬

‫ﺍﳌﻌﺮﻑ‬

‫ﻫﻲ ﺩﺍﺋﺮﺓ ﺍﻟﻮﺣﺪﺓ ﰲ ﺍﳌﺴﺘﻮﻱ‪ .‬ﻭﻫﻮ ﻣﻨﺤﲎ ﻣﻐﻠﻖ‬

‫ﻭﺟﻮﺭﺩﺍﱐ‪ .‬ﻭﻫﻮ ﻏﲑ ﺑﺴﻴﻂ ﻟﻮ ﺍﺳﺘﺒﺪﻟﻨﺎ ] ‪ [0, 2π‬ﺑـ ‪. ℝ‬‬ ‫ﻧﺘﻨﺎﻭﻝ ﺍﻵﻥ ﺗﻌﺮﻳﻒ ﻃﻮﻝ ﻗﻮﺱ )ﺃﻭ ﻣﻨﺤﲎ( ‪: [a, b ] → ℝ n‬‬ ‫ﻟﺘﻜﻦ‬

‫ﺑـ‬

‫} ‪d n = {a = a0 < a1 < ... < an = b‬‬ ‫‪128‬‬

‫‪f‬‬

‫‪:‬‬

‫ﺗﻘﺴﻴﻤﺔ ﻟﻠﻤﺠﺎﻝ ] ‪ . [a, b‬ﻧﻀﻊ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻣﻦ ﺃﺟﻞ ﻛﻞ‬

‫) ‪x i = f (ai‬‬

‫‪= 0,1,..., n‬‬

‫ﺍﻟﻘﻄﻊ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ai , ai +1‬ﻣﻦ ﺃﺟﻞ ﻛﻞ‬

‫‪ . i‬ﻧﺮﻣﺰ ﺑـ‬

‫‪Ld n‬‬

‫‪= 0,1,..., n‬‬

‫‪ ، i‬ﺃﻱ‬

‫‪n −1‬‬

‫‪n −1‬‬

‫‪i =0‬‬

‫‪i =0‬‬

‫‪‬ﻤﻮﻉ ﺃﻃﻮﺍﻝ‬

‫) ‪Ld n = ∑ x i +1 − x i = ∑ f (ai +1 ) − f (ai‬‬

‫ﺣﻴﺚ ﻳﺮﻣﺰ‬

‫‪.‬‬

‫ﻟﻠﻨﻈﻴﻢ ﺍﻷﻗﻠﻴﺪﻱ ﰲ ‪. ℝ n‬‬

‫ﻣﻦ ﺍﻟﻮﺍﺿﺢ ﺃﻧﻪ ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺍﻟﺘﻘﺴﻴﻤﺔ‬

‫ﺃﺩﻕ ﻛﻠﻤﺎ ﺍﻗﺘﺮﺏ ﺍﻟﻄﻮﻝ‬

‫‪dn‬‬

‫‪Ld n‬‬

‫ﻣﻦ ﻃﻮﻝ ﺍﻟﻘﻮﺱ ‪. f‬‬ ‫ﺗﻌﺮﻳﻒ )ﻃﻮﻝ ﻗﻮﺱ(‬ ‫)] ‪P ([a, b‬‬

‫ﻧﺮﻣﺰ ﺑـ‬

‫‪‬ﻤﻮﻋﺔ ﺍﻟﺘﻘﺴﻴﻤﺎﺕ ﺍﳌﻤﻜﻨﺔ ﻟﻠﻤﺠﺎﻝ‬

‫] ‪. [a , b‬‬ ‫ﻧﻘﻮﻝ ﺇﻥ ﺍﳌﻨﺤﲎ‬

‫‪f : [a , b ] → ℝ n‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ ﺍ‪‬ﻤﻮﻋﺔ }‬

‫) ] ‪d n ∈ P ( [a , b‬‬

‫‪,‬‬

‫‪n‬‬

‫ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻧﺴﻤﻲ ﻃﻮﻝ ﺍﻟﻘﻮﺱ‬ ‫) ‪L (f‬‬

‫ﻗﺎﺑﻞ ﻟﻠﺘﻘﻮﱘ ‪rectifiable‬‬

‫‪ {Ld‬ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﻋﻠﻰ‪.‬‬ ‫‪f : [a , b ] → ℝ n‬‬

‫ﺍﳊﺪ ﺍﻷﻋﻠﻰ‬

‫ﻟﻠﻤﺠﻤﻮﻋﺔ ﺍﻟﺴﺎﺑﻘﺔ‪ ،‬ﺃﻱ‬

‫‪}.‬‬

‫) ] ‪d n ∈ P ( [a , b‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ ﺍ‪‬ﻤﻮﻋﺔ‬ ‫ﺍﻷﻋﻠﻰ ﻓﻬﺬﺍ ﻳﻌﲏ ﺃﻥ‬

‫∞=)‬

‫{‬

‫‪L (f ) = sup Ld n ,‬‬

‫}‬

‫) ] ‪d n ∈ P ( [a , b‬‬

‫‪. L (f‬‬

‫‪129‬‬

‫‪,‬‬

‫‪dn‬‬

‫‪{L‬‬

‫ﻏﲑ ﳏﺪﻭﺩﺓ ﻣﻦ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﳝﻜﻦ ﺇﺛﺒﺎﺕ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﻧﻈﺮﻳﺔ )ﺣﺴﺎﺏ ﻃﻮﻝ ﻗﻮﺱ(‬ ‫‪f : [a , b ] → ℝ n‬‬

‫ﻣﻨﺤﻨﻴﺎ ﻳﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻭﻣﺸﺘﻘﻪ ﻣﺴﺘﻤﺮﺍ‬

‫)ﺃﻱ ﺃﻥ ﻛﻞ ﻣﺮﻛﺒﺔ ﻣﻦ ﻣﺮﻛﺒﺎﺕ‬

‫ﺗﻘﺒﻞ ﺍﻻﺷﺘﻘﺎﻕ ﻭﻣﺸﺘﻘﻬﺎ‬

‫ﻟﻴﻜﻦ‬

‫ﻣﺴﺘﻤﺮ(‪ .‬ﻋﻨﺪﺋﺬ ﻳﻜﻮﻥ‬

‫) ‪f = (f 1 , f 2‬‬

‫ﻗﺎﺑﻼ ﻟﻠﺘﻘﻮﱘ ﻭﻃﻮﻟﻪ ﻫﻮ‬

‫‪f‬‬

‫‪b‬‬

‫‪L (f ) = ∫ f '(t ) dt‬‬ ‫‪a‬‬

‫‪2‬‬

‫)‬

‫‪(t ) dt .‬‬

‫'‬ ‫‪2‬‬

‫‪) + (f‬‬ ‫‪2‬‬

‫) ‪(t‬‬

‫'‬ ‫‪1‬‬

‫‪(f‬‬

‫‪b‬‬

‫∫=‬

‫‪a‬‬

‫ﻣﺜﺎﻝ ‪1‬‬ ‫ﺍﺣﺴﺐ ﳏﻴﻂ ﺍﻟﺪﺍﺋﺮﺓ ﺫﺍﺕ ﻧﺼﻒ ﺍﻟﻘﻄﺮ ‪. R‬‬ ‫ﻧﺼﻒ‬

‫ﺍﳌﻨﺤﲏ‬

‫‪f : [ 0, 2π ] → ℝ 2‬‬

‫ﺍﻟﺬﻱ‬

‫ﺍﳌﻌﺮﻑ ﺑـ‬

‫) ‪'(t ) = (R cos t , − R sin t‬‬

‫‪.‬‬

‫ﳝﺜﻞ‬

‫ﺍﳌﻌﻄﺎﺓ‬

‫ﺍﻟﺪﺍﺋﺮﺓ‬

‫) ‪(t ) = (R sin t , R cos t‬‬

‫‪ . f‬ﻟﺪﻳﻨﺎ ﻋﻨﺪﺋﺬ‬

‫‪ . f‬ﻭﻣﻨﻪ ‪:‬‬

‫‪f '(t ) = R 2 cos 2 t + R 2 sin 2 t = R‬‬

‫ﺇﺫﻥ ‪:‬‬ ‫‪f '(t ) dt‬‬ ‫‪2‬‬

‫)‬

‫‪(t ) dt‬‬

‫'‬ ‫‪2‬‬

‫‪) + (f‬‬ ‫‪2‬‬

‫) ‪(t‬‬

‫'‬ ‫‪1‬‬

‫‪2π‬‬

‫∫ = ) ‪L (f‬‬

‫‪0‬‬

‫‪(f‬‬

‫‪2π‬‬

‫∫=‬

‫‪0‬‬ ‫‪2π‬‬

‫‪= ∫ Rdt‬‬ ‫‪0‬‬

‫‪2π‬‬

‫‪= R ∫ dt‬‬ ‫‪0‬‬

‫‪= 2π R .‬‬

‫ﻭﻫﺬﺍ ﻫﻮ ﺑﺎﻟﻀﺒﻂ ﳏﻴﻂ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻧﻌﺮﻓﻪ ﻣﻨﺬ ﺯﻣﺎﻥ !‬ ‫‪130‬‬

‫ﺑﺎﻟﺘﺎﺑﻊ‬

‫ ا ب ا‬: 3 ‫ا‬

: 2 ‫ﻣﺜﺎﻝ‬ ‫ﺍﳌﻌﺮﻑ ﺑـ‬

f : [ 0,T ] → ℝ3

‫ﺃﻋﺪﺍﺩ ﺣﻘﻴﻘﻴﺔ ﻣﻮﺟﺒﺔ‬

R

: ‫ ﻭﻣﻨﻪ‬. f

‫ﻭ‬

c

‫ﺍﺣﺴﺐ ﻃﻮﻝ ﺍﻟﻘﻮﺱ ﺍﳊﻠﺰﻭﱐ‬ ‫ﻭ‬

T

‫ﺣﻴﺚ‬

f (t ))(R cos t , R sin t , ct )

'(t ) = (− R sin t , R cos t , c )

‫ ﻟﺪﻳﻨﺎ ﻋﻨﺪﺋﺬ‬.‫ﲤﺎﻣﺎ‬

f '(t ) = R 2 sin 2 t + R 2 cos 2 t + c 2 = R 2 + c 2

: ‫ﺇﺫﻥ‬ T

L (f ) = ∫ f '(t ) dt 0

=∫

T

0

=∫

T

0

(f

' 1

(t )

) + (f 2

' 2

(t )

) + (f 2

' 2

)

2

(t ) dt

R 2 + c 2 dt T

= R 2 + c 2 ∫ dt 0

=T

R 2 +c2.

: 3 ‫ﻣﺜﺎﻝ‬ ‫ﺍﳌﻌﺮﻑ ﺑـ‬

f : [ 0, 2] → ℝ 3

‫ﺍﺣﺴﺐ ﻃﻮﻝ ﻗﻮﺱ ﺍﳌﻨﺤﲏ‬ .f

.f

(t ) = (e t cos t , e t sin t , 0)

'(t ) = (e t (cos t − sin t ), e t (sin t + cos t ), 0 )

‫ﻟﺪﻳﻨﺎ ﻋﻨﺪﺋﺬ‬ :‫ﻭﻣﻨﻪ‬

f '(t ) = e 2t (cos t − sin t )2 + e 2t (sin t + cost )2 + 0 = e t 2. 131

‫ا ‪ : 3‬ا ب ا ‬

‫ﺇﺫﻥ ‪:‬‬ ‫‪T‬‬

‫‪L (f ) = ∫ f '(t ) dt‬‬ ‫‪0‬‬

‫‪2‬‬

‫‪= 2 ∫ e t dt‬‬ ‫‪0‬‬

‫‪= 2(e 2 − 1).‬‬

‫ﺗﻌﺮﻳﻒ )ﺗﻜﺎﻓﺆ ﻣﻨﺤﻨﻴﲔ(‬ ‫ﻧﻘﻮﻝ ﻋﻦ ﺍﳌﻨﺤﻨﻴﲔ‬

‫‪f : [a , b ] → ℝ n‬‬

‫ﻭ‬

‫] ‪h : [c , d ] → [a, b‬‬

‫ﻣﺘﻜﺎﻓﺌﺎﻥ ﺇﺫﺍ ﻭﺟﺪ ﺗﺎﺑﻊ ﺗﻘﺎﺑﻠﻲ ﻭﻣﺴﺘﻤﺮ‬ ‫‪= f h‬‬

‫‪g : [c , d ] → ℝ n‬‬

‫ﺇ‪‬ﻤﺎ‬ ‫ﲝﻴﺚ‬

‫‪.g‬‬

‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺍﳌﻨﺤﻨﻴﺎﻥ‬ ‫)‪f (t ) = (t , 0‬‬

‫‪f : [ 0, 2] → ℝ 2‬‬

‫ﻭ‬

‫ﻭ‬

‫‪g : [ −1,1] → ℝ 2‬‬

‫)‪g (t ) = (t 2 + t , 0‬‬

‫ﺍﳌﻌﺮﻓﺎﻥ ﺑـ‬

‫ﻣﺘﻜﺎﻓﺌﺎﻥ‪ .‬ﻳﻜﻔﻲ ﻭﺿﻊ‬

‫‪. h (t ) = t 2 + t‬‬ ‫ﲤﺮﻳﻦ ‪3‬‬ ‫ﺃﺛﺒﺖ ﺃﻥ ﺍﳌﻨﺤﻨﻴﲔ‬ ‫ﺑـ‬

‫‪2 − 2t 2 4t‬‬ ‫‪,‬‬ ‫)‬ ‫‪1+t 2 1+t 2‬‬

‫‪f : [ 0,1] → ℝ 2‬‬

‫( = ) ‪f (t‬‬

‫ﻭ‬

‫ﻭ‬

‫‪g : [ 0, π ] → ℝ 2‬‬

‫‪s‬‬ ‫‪s‬‬ ‫) ‪g (s ) = (2 cos , 2sin‬‬ ‫‪2‬‬ ‫‪2‬‬

‫ﺍﳌﻌﺮﻓﲔ‬

‫ﻣﺘﻜﺎﻓﺌﺎﻥ‪.‬‬

‫ﺍﳊﻞ‬ ‫ﻟﻨﻜﺘﺐ ﺷﻜﻠﻴﺎ ﺃﻥ ﺍﻟﺘﺎﺑﻊ‬

‫]‪h : [ 0, π ] → [ 0,1‬‬

‫ﳛﻘﻖ‬

‫ﺃﻱ‬ ‫) ) ‪g (s ) = f ( h (s‬‬ ‫‪132‬‬

‫‪∀s ∈ [ 0, π ] ,‬‬

‫‪= f h‬‬

‫‪،g‬‬

‫ ا ب ا‬: 3 ‫ا‬

‫ﻭﻫﺬﺍ ﻳﻌﲏ‬ ∀s ∈[ 0,π] ,

 2−2h2 (s) 4h(s)  s s g(s) = f ( h(s)) =  ,  = (2cos ,sin ) 2 2 2 2  1+h (s) 1+h (s) 

: ‫ﻭﻣﻨﻪ‬ 1 − h 2 (s ) s = cos  2 2 1 + h (s ) ∀s ∈ [ 0, π ] ,   2h (s ) = sin s 1 + h 2 (s ) 2

‫ﻭﺑﺎﻟﺘﺎﱄ‬ ∀s ∈ [ 0, π ] , cos

s 1 + h 2 (s ) − 2h 2 (s ) = 2 1 + h 2 (s ) 2h 2 (s ) = 1− 1 + h 2 (s )

‫ ﻭﻣﻦ ﰒﹼ‬. h (0) = 0 ‫ﺗﺴﺘﻠﺰﻡ‬

s = 1 − h (s ).sin . 2 2h (s ) s = sin 2 2 1 + h (s )

‫ﻣﻊ ﺍﻟﻌﻠﻢ ﺃﻥ‬

‫ﻧﺴﺘﺨﻠﺺ ﺑﻨﺎﺀ ﻋﻤﺎ ﺳﺒﻖ ﺃﻥ‬ s = 0, 0,  s  h (s ) = 1 − cos 2 , s ∈ ]0, π ] .  s  sin 2 

133

‫ا ‪ : 3‬ا ب ا ‬

‫ﺇﻥ ﺍﻟﺘﺎﺑﻊ ‪ h‬ﻣﺴﺘﻤﺮ ﻭﺗﻘﺎﺑﻠﻲ ﻣﻦ ] ‪ [0, π‬ﺇﱃ ]‪ [0,1‬ﺣﻴﺚ ﺃﻧﻪ ﻳﻘﺒﻞ‬ ‫ﺍﻻﺷﺘﻘﺎﻕ ﻭﻣﺸﺘﻘﻪ ﻣﻮﺟﺐ ﲤﺎﻣﺎ )ﻓﻬﻮ ﺇﺫﻥ ﻣﺘﺰﺍﻳﺪ ﲤﺎﻣﺎ( ‪:‬‬ ‫‪,s = 0‬‬ ‫‪, s ∈ ]0, π ] .‬‬

‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪h '(s ) =  1 − cos s‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪s‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪ sin 2‬‬

‫‪134‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻧﻈﺮﻳﺔ )ﻃﻮﹶﻟﺎ ﻣﻨﺤﻨﻴﲔ ﻣﺘﻜﺎﻓﺌﲔ(‬ ‫‪f : [a , b ] → ℝ n‬‬

‫ﺇﺫﺍ ﻛﺎﻥ ﻣﻨﺤﻨﻴﺎﻥ‬

‫ﻭ‬

‫‪g : [c , d ] → ℝ n‬‬

‫ﻣﺘﻜﺎﻓﺌﻲ‪ ،‬ﻭﻛﺎﻥ ﺃﺣﺪﳘﺎ ﻗﺎﺑﻼ ﻟﻠﺘﻘﻮﱘ ﻓﺈﻥ ﺍﻵﺧﺮ ﻳﻘﺒﻞ ﺃﻳﻀﺎ ﺍﻟﺘﻘﻮﱘ‪ ،‬ﻭﳍﻤﺎ‬ ‫ﻧﻔﺲ ﺍﻟﻄﻮﻝ‪.‬‬ ‫ﻣﺜﺎﻝ‬ ‫‪f : [ 0,1] → ℝ 2‬‬

‫ﻧﻌﻠﻢ ﺃﻥ ﺍﳌﻨﺤﻨﻴﲔ‬ ‫ﺑـ‬

‫‪2 − 2t 2 4t‬‬ ‫‪,‬‬ ‫)‬ ‫‪1+t 2 1+t 2‬‬

‫ﻭ‬

‫( = ) ‪f (t‬‬

‫‪g : [ 0, π ] → ℝ 2‬‬

‫ﻭ‬

‫‪s‬‬ ‫‪s‬‬ ‫) ‪g (s ) = (2 cos , 2sin‬‬ ‫‪2‬‬ ‫‪2‬‬

‫ﺍﳌﻌﺮﻓﲔ‬

‫ﻣﺘﻜﺎﻓﺌﺎﻥ‪.‬‬

‫ﻧﻼﺣﻆ ﺃﻥ‬ ‫‪2‬‬

‫)‬

‫‪(s ) ds‬‬

‫'‬ ‫‪2‬‬

‫‪( g (s ) ) + ( g‬‬ ‫‪2‬‬

‫‪s‬‬ ‫‪s‬‬ ‫‪+ cos 2 ds‬‬ ‫‪2‬‬ ‫‪2‬‬

‫‪π‬‬

‫'‬ ‫‪1‬‬

‫∫ = ) ‪L (g‬‬

‫‪0‬‬

‫‪sin 2‬‬

‫‪π‬‬

‫∫=‬

‫‪0‬‬

‫‪π‬‬

‫‪= ∫ ds‬‬ ‫‪0‬‬

‫‪= π.‬‬

‫ﻭﺑﺎﻟﺘﺎﱄ ﻓﺈﻥ‬ ‫‪) =π‬‬

‫‪f : [ 0,1] → ℝ 2‬‬

‫‪. L (f‬‬

‫‪135‬‬

‫ﻳﻘﺒﻞ ﺍﻟﺘﻘﻮﱘ ﻭﻃﻮﻟﻪ ﻳﺴﺎﻭﻱ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﲤﺮﻳﻦ ‪: 4‬‬ ‫ﻟﻴﻜﻦ‬

‫‪f : [a , b ] → ℝ‬‬

‫ﺗﺎﺑﻌﺎ ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ‪ .‬ﺃﺛﺒﺖ ﺃﻥ‬

‫ﻃﻮﻝ ﺑﻴﺎﻧﻪ ﻳﺴﺎﻭﻱ‬ ‫‪1 + ( f '(s ) ) ds .‬‬

‫‪b‬‬

‫‪2‬‬

‫∫= ‪l‬‬

‫‪a‬‬

‫ﺍﳊﻞ‬ ‫ﺍﻟﺘﺎﺑﻊ ‪g : [a, b ] → ℝ 2‬‬

‫ﻳﻜﻔﻲ ﺍﻋﺘﺒﺎﺭ‬


‫) ) ‪ g (t ) = (t , f (t‬ﺍﻟﺬﻱ ﳝﺜﻞ ﺑﻴﺎﻥ ﺍﻟﺘﺎﺑﻊ ‪. f : [a, b ] → ℝ‬‬ ‫ﻭﻋﻠﻴﻪ ﻓﻄﻮﻝ ﺍﻟﻘﻮﺱ ‪ g : [a,b ] → ℝ 2‬ﻫﻮ ﻃﻮﻝ ﺑﻴﺎﻥ‬ ‫‪ . f : [a, b ] → ℝ‬ﺇﺫﻥ‬ ‫‪2‬‬

‫)‬

‫‪(s ) ds‬‬

‫'‬ ‫‪2‬‬

‫‪( g (s ) ) + ( g‬‬ ‫‪2‬‬

‫‪2‬‬

‫)‬

‫'‬ ‫‪1‬‬

‫‪π‬‬

‫‪0‬‬

‫(‬

‫‪π‬‬

‫‪1 + f ' (s ) ds .‬‬

‫ﻣﺜﺎﻝ ﺫﻟﻚ ‪ :‬ﺍﻟﺘﺎﺑﻊ‬

‫‪f : [ 0, a ] → ℝ‬‬

‫∫ = ) ‪l = L (g‬‬ ‫∫=‬

‫‪0‬‬


‫‪2‬‬

‫‪(x ) = x‬‬

‫‪ . f‬ﺇﻥ‬

‫ﻃﻮﻝ ﺑﻴﺎﻧﻪ ﻳﺴﺎﻭﻱ ‪:‬‬ ‫‪3‬‬

‫‪1‬‬ ‫‪1‬‬ ‫‪1+2xdx = (1 + 2a ) 2 − .‬‬ ‫‪3‬‬ ‫‪3‬‬

‫‪a‬‬

‫∫= ‪l‬‬

‫‪0‬‬

‫ﲤﺮﻳﻦ ‪: 5‬‬ ‫ﻋﲔ ﻃﻮﻝ ﺍﻟﻘﻮﺱ ﺍﳌﻐﻠﻖ ﺍﳌﻌﺮﻑ ﺑـ‬

‫‪3‬‬

‫‪3‬‬

‫‪3‬‬

‫‪x 2 + y 2 = a2‬‬

‫ﺣﻴﺚ‬

‫‪.0 < a‬‬ ‫ﺇﺭﺷﺎﺩ ‪ :‬ﳝﻜﻨﻚ ﻭﺿﻊ‬

‫) ‪(x , y ) = ( a cos3 t , a sin 3 t‬‬

‫ﻭﺳﻴﻂ‪.‬‬ ‫‪136‬‬

‫ﺣﻴﺚ‬

‫‪t‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﺍﳊﻞ‬ ‫‪3‬‬ ‫‪2‬‬

‫ﻧﻼﺣﻆ ﺃﻥ‬ ‫‪≤a‬‬

‫‪=a‬‬

‫‪3‬‬ ‫‪2‬‬

‫‪3‬‬ ‫‪2‬‬

‫‪x +y‬‬

‫‪ . 0 ≤ y‬و ;‪ 9 :‬ﻭﺿﻊ ) ‪. (x , y ) = ( a cos3 t , a sin 3 t‬‬

‫ﻭﺑﺬﻟﻚ ﻧﺘﺄﻛﺪ ﻓﻌﻼ ﺃﻥ‬

‫‪3‬‬ ‫‪2‬‬

‫‪=a‬‬

‫‪3‬‬ ‫‪2‬‬

‫‪f : [ 0, 2π ] → ℝ 2‬‬


‫)‪(0) = f (π ) = (a , 0‬‬

‫‪3‬‬ ‫‪2‬‬

‫‪+y‬‬

‫ﺍﳌﻄﻠﻮﺏ ﺗﻌﻴﲔ ﺍﻟﻄﻮﻝ‬ ‫ﺃﻥ‬

‫ﺗﺴﺘﻠﺰﻡ ﺃﻥ‬

‫‪0≤x ≤a‬‬

‫ﻭ‬

‫‪l‬‬

‫‪ . x‬ﺍﻧﻈﺮ ﺍﻟﺸﻜﻞ ﺍﳌﻤﺜﻞ ﻟﻠﻤﻨﺤﲎ‬

‫ﻟﻠﻘﻮﺱ ﺍﳌﻐﻠﻖ‪ .‬ﻧﻼﺣﻆ ﺃﻥ ﺍﳌﻨﺤﲏ‬

‫) ‪f (t ) = ( a cos 3 t , a sin 3 t‬‬

‫‪ . f‬ﻭﻣﻦ ﰒ ﻓﺈﻥ‬ ‫‪2‬‬

‫)‬

‫) ‪= L (f‬‬

‫( )‬ ‫‪2‬‬

‫‪.l‬‬

‫‪(t ) + f 2' (t ) dt‬‬

‫'‬ ‫‪1‬‬

‫‪(f‬‬

‫‪(3a cos t  [−sint ]) + (3a sin t  [cost ]) dt‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫ﻣﻐﻠﻖ ﺣﻴﺚ‬

‫‪2π‬‬

‫∫ = ) ‪l = L (f‬‬

‫‪0‬‬

‫‪2π‬‬

‫∫=‬

‫‪0‬‬ ‫‪2π‬‬

‫‪= ∫ 3a cost .sin t dt‬‬ ‫‪0‬‬

‫‪= 6a.‬‬

‫‪137‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .6‬ﺣﺴﺎﺏ ﺑﻌﺾ ﺍﳌﺴﺎﺣﺎﺕ ﻭﺍﳊﺠﻮﻡ‬ ‫ﻳﺘﻄﻠﺐ ﺣﺴﺎﺏ ﻣﺴﺎﺣﺎﺕ ﺍﻟﺴﻄﻮﺡ ﺍﳊﺠﻮﻡ ﺍ‪‬ﺴﻤﺎﺕ ﺇﺩﺧﺎﻝ‬ ‫ﺍﻟﺘﻜﺎﻣﻼﺕ ﺍﳌﻀﺎﻋﻔﺔ )ﺍﻟﺜﻨﺎﺋﻲ‪ ،‬ﺍﻟﺜﻼﺛﻲ‪ ،(... ،‬ﻭﻫﻮ ﺗﻌﻤﻴﻢ ﻟﻠﺘﻜﺎﻣﻞ‬ ‫ﺍﻷﺣﺎﺩﻱ‪ ،‬ﻟﻜﻨﻪ ﻳﺘﻄﻠﺐ ﲤﻬﻴﺪﺍﺕ ﻣﻄﻮﻟﺔ‪ .‬ﻭﻋﻠﻴﻪ ﺳﻨﻜﺘﻔﻲ ﰲ ﻫﺬﺍ ﺍﳌﻮﺿﻮﻉ‬ ‫ﺑﺘﻘﺪﱘ ﻣﺴﺎﺣﺎﺕ ﻭﺣﺠﻮﻡ ﻋﺪﺩ ﻗﻠﻴﻞ ﻣﻦ ﺍﻟﺴﻄﻮﺡ ﻭﺍ‪‬ﺴﻤﺎﺕ ﺩﻭﻥ‬ ‫ﺍﻟﻐﻮﺹ ﰲ ﺍﳉﺎﻧﺐ ﺍﻟﺘﻘﲏ ﺍﳌﺘﻌﻠﻖ ﺑﺎﻟﺘﻜﺎﻣﻼﺕ‪.‬‬ ‫‪ .1‬ﺗﻌﺮﻳﻒ ﺍﳌﺨﺮﻭﻁ ‪:‬‬ ‫ﻳﺴﺘﺤﺴﻦ ﺍﳊﺪﻳﺚ ﻋﻦ ﺗﻌﺮﻳﻒ ﺍﳌﺨﺮﻭﻁ ﻗﺒﻞ ﺗﻘﺪﱘ ﻣﺴﺎﺣﺘﻪ‬ ‫ﻭﺣﺠﻤﻪ‪ .‬ﰲ ﺍﳍﻨﺪﺳﺔ ﺍﻷﻭﻟﻴﺔ ﺍﳌﺨﺮﻭﻁ ﻫﻮ ﺍﻟﺴﻄﺢ ﺍﻟﺬﻱ ﳓﺼﻞ ﻋﻠﻴﻪ ﲜﻌﻞ‬ ‫ﻣﺜﻠﺚ ﻗﺎﺋﻢ ﻳﺪﻭﺭ ﺣﻮﻝ ﺃﺣﺪ ﺿﻠﻌﻴﻪ ﺍﻟﻘﺎﺋﻤﲔ‪ .‬ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺗﺴﻤﻰ ﺍﳌﺴﺎﺣﺔ‬ ‫ﺍﻟﱵ ﳝﺴﺤﻬﺎ )ﻭﻫﻲ ﻗﺮﺹ( ﺍﻟﻀﻠﻊ ﺍﻟﻘﺎﺋﻢ ﺍﻵﺧﺮ ﻗﺎﻋﺪﺓ ﺍﳌﺨﺮﻭﻁ‪.‬‬ ‫ﺃﻣﺎ ﺍﺭﺗﻔﺎﻉ ﺍﳌﺨﺮﻭﻁ ﻓﻬﻮ ﻃﻮﻝ ﺍﻟﻀﻠﻊ ﺍﻟﺬﻱ ﻳﺪﻭﺭ ﺣﻮﻟﻪ ﺍﳌﺜﻠﺚ‪.‬‬ ‫ﻛﻤﺎ ﻳﺴﻤﻰ ﻃﺮﻑ ﻫﺬﺍ ﺍﻟﻀﻠﻊ ﺍﻟﺬﻱ ﻻ ﳝﺲ ﺍﻟﻘﺎﻋﺪﺓ ﺭﺃﺱ ﺍﳌﺨﺮﻭﻁ‪.‬‬ ‫ﻧﺼﻒ ﺯﺍﻭﻳﺔ ﺍﻟﺮﺃﺱ ﻟﻠﻤﺨﺮﻭﻁ ﻫﻲ ﺍﻟﺰﺍﻭﻳﺔ ﺍﳌﺜﻠﺚ ﺍﻟﻘﺎﺋﻢ ﺍﻟﱵ‬ ‫ﺭﺃﺳﻬﺎ ﺭﺃﺱ ﺍﳌﺨﺮﻭﻁ‪.‬‬

‫‪138‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻭﺑﻌﺒﺎﺭﺓ ﺃﻭﺿﺢ ‪:‬‬ ‫ﺇﺫﺍ ﻛﺎﻥ‬ ‫‪AB‬‬

‫‪ABC‬‬

‫ﻣﺜﻠﺜﺎ ﻗﺎﺋﻤﺎ ﰲ‬

‫‪B‬‬

‫ﻭﺟﻌﻠﻨﺎﻩ ﻳﺪﻭﺭ ﺣﻮﻝ ﺍﻟﻀﻠﻊ‬

‫ﻓﺈﻥ‬ ‫*‬

‫‪A‬‬

‫ﻫﻮ ﺭﺃﺱ ﺍﳌﺨﺮﻭﻁ‪،‬‬

‫* ﺍﻟﻄﻮﻝ‬

‫‪BC‬‬

‫ﻫﻮ ﺍﺭﺗﻔﺎﻋﻪ‪،‬‬

‫* ﻗﺎﻋﺪﺗﻪ ﻫﻲ ﺍﻟﻘﺮﺹ ﺍﻟﺬﻱ ﻣﺮﻛﺰﻩ‬ ‫‪BC‬‬

‫‪B‬‬

‫ﻭﻧﺼﻒ ﻗﻄﺮﻩ‬

‫ﺍﻟﻮﺍﻗﻊ ﰲ ﺍﳌﺴﺘﻮﻱ ﺍﻟﻌﻤﻮﺩﻱ ﻋﻠﻰ ﺍﳌﺴﺘﻘﻴﻢ )‪ ، ( AB‬ﻭﻧﺼﻒ‬

‫ﺯﺍﻭﻳﺘﻪ ﻫﻲ ‬ ‫‪. BAC‬‬

‫ﻣﻼﺣﻈﺔ‪:‬‬ ‫‪ (1‬ﻳﺴﻤﻰ ﺍﳌﺨﺮﻭﻁ ﺍﻟﺬﻱ ﻋﺮ‪‬ﻓﻨﺎﻩ ﺁﻧﻔﺎ ﳐﺮﻭﻃﺎ ﺩﻭﺭﺍﻧﻴﺎ ﺃﻭ ﻗﺎﺋﻤﺎ‪،‬‬ ‫ﻭﻫﻮ ﺍﻟﺬﻱ ﻧﻌﻨﻴﻪ ﻋﻤﻮﻣﺎ‪ .‬ﻟﻜﻦ ﺍﳌﻔﻬﻮﻡ ﺍﻟﻌﺎﻡ )ﺍﻟﺘﻈﺮﻱ( ﻟﻠﻤﺨﺮﻭﻁ ﰲ‬ ‫ﺍﳍﻨﺪﺳﺔ ﻳﺸﻤﻞ ﺳﻄﻮﺣﺎ ﺃﺧﺮﻯ‪:‬‬ ‫ﺧﺬ‬

‫ﻣﺴﺘﻘﻴﻤﺎ )‪( D‬‬

‫ﻭﺍﻋﺘﱪ ﻋﻠﻴﻪ ﻧﻘﻄﺔ ﻣﺜﺒﺘﺔ ‪ ، M‬ﻭﺧﺬ ﻣﻨﺤﻨﻴﺎ ﻣﻐﻠﻘﺎ‬

‫) ‪ . (C‬ﰒ ﺍﺟﻌﻞ ﺍﳌﺴﺘﻘﻴﻢ‬ ‫)‪( D‬‬

‫ﻳﺪﻭﺭ ﺣﻮﻝ ﻧﻘﻄﺔ ﻣﺜﺒﺘﺔ ﻣﻨﻪ‬

‫‪A‬‬

‫ﲝﻴﺚ ﲤﺴﺢ ﺍﻟﻨﻘﻄﺔ‬

‫‪M‬‬

‫ﺍﳌﻨﺤﲏ‬

‫) ‪ . (C‬ﺇﻥ ﺍﻟﺴﻄﺢ ﺍﶈﺼﻞ ﻋﻠﻴﻪ ‪‬ﺬﻩ ﺍﻟﻄﺮﻳﻘﺔ ﻳﺴﻤﻰ ﳐﺮﻭﻃﺎ ﻣﻮﻟﺪﺍ‬ ‫ﺑﺎﳌﺴﺘﻘﻴﻢ )‪ ، ( D‬ﺭﺃﺳﻪ ‪ . A‬ﺍﳌﺨﺮﻭﻁ ﺍﶈﺼﻞ ﻋﻠﻴﻪ ﻋﻨﺪﺋﺬ ﻳﻜﻮﻥ ﻣﻦ‬ ‫ﺍﻟﺸﻜﻞ‬

‫‪139‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﺫﻟﻚ ﻫﻮ ﺍﳌﺨﺮﻭﻁ ﺍﻟﺬﻱ ﻧﻌﺘﱪﻩ ﻣﺜﻼ ﻋﻨﺪﻣﺎ ﻳﺘﻌﻠﻖ ﺍﻷﻣﺮ ﺑﺎﻟﻘﻄﻮﻉ‬ ‫ﺍﳌﺨﺮﻭﻃﻴﺔ ﻟﻠﺤﺼﻮﻝ ﺑﻮﺟﻪ ﺧﺎﺹ ﻋﻠﻰ ﺍﻟﻘﻄﻊ ﺍﻟﺰﺍﺋﺪ‪ .‬ﻟﻜﻨﻨﺎ ﻏﺎﻟﺒﺎ ﻣﺎ ﻧﻌﲏ‬ ‫ﺑﺎﳌﺨﺮﻭﻁ ﻧﺼﻒ ﻫﺬﺍ ﺍﻟﺴﻄﺢ ﻏﲑ ﺍﶈﺪﻭﺩ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ ‪:‬‬

‫ﺑﻞ ﻧﻜﺘﻔﻲ ﰲ ﺃﻏﻠﺐ ﺍﻷﺣﻴﺎﻥ )ﻛﻤﺎ ﻗﺪﻣﻨﺎ ﰲ ﺗﻌﺮﻳﻒ ﺍﳌﺨﺮﻭﻁ ﺍﻟﺪﻭﺭﺍﱐ(‬ ‫ﲟﺨﺮﻭﻁ ﳏﺪﻭﺩ ﻣﻦ ﺟﻬﺔ ﻗﺎﻋﺪﺗﻪ ﻛﻤﺎ ﻫﻮ ﻣﺒﻴ‪‬ﻦ ﰲ ﺍﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ ‪:‬‬

‫‪140‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻭﻫﻨﺎﻙ ﺃﻧﻮﺍﻉ ﻛﺜﲑﺓ ﻣﻦ ﺍﳌﺨﺮﻭﻃﺎﺕ ﻣﻨﻬﺎ ﺗﻠﻚ ﺍﳌﺴﻤﺎﺓ‬ ‫ﺑﺎﳌﺨﺮﻭﻃﺎﺕ ﺍﻟﻜﺮﻭﻳﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﺭﺃﺳﻬﺎ ﰲ ﻣﺮﻛﺰ ﻛﺮﺓ ﻭﻗﺎﻋﺪ‪‬ﺎ ﺟﺰﺀﺍ ﻣﻦ‬ ‫ﺳﻄﺢ ﺍﻟﻜﺮﺓ ﻛﻤﺎ ﻫﻮ ﻣﺒﻴ‪‬ﻦ ﰲ ﺍﻟﺸﻜﻞ ﺃﺩﻧﺎﻩ ‪:‬‬

‫‪ (2‬ﳝﻜﻦ ‪ -‬ﺧﻼﻓﺎ ﳌﺎ ﺍﻋﺘﱪﻧﺎﻩ ﰲ ﺍﻟﺘﻌﺮﻳﻒ ‪ -‬ﺃﻥ ﻳﻜﻮﻥ ﺍﳌﺨﺮﻭﻁ‬ ‫ﻏﲑ ﺩﻭﺭﺍﱐ‪ ،‬ﻛﺄﻥ ﻳﻜﻮﻥ ﺍﳌﻨﺤﲎ‬ ‫ﻧﺼﻒ ﻗﻄﺮﻫﺎ‬

‫‪r‬‬

‫) ‪(C‬‬

‫ﺍﳌﻌﺘﱪ ﰲ ﺍﳌﻼﺣﻈﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺩﺍﺋﺮﺓ‬

‫‪ ...‬ﻟﻜﻦ ﻣﺴﻘﻂ ﺭﺃﺱ ﺍﳌﺨﺮﻭﻁ‬

‫ﻟﻴﺲ ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ‬

‫) ‪(C‬‬

‫‪:‬‬

‫‪141‬‬

‫‪A‬‬

‫ﻋﻠﻰ ﻣﺴﺘﻮﻱ ﺍﻟﻘﺎﻋﺪﺓ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻧﻼﺣﻆ ﺃﻥ ﺍﻻﺭﺗﻔﺎﻉ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻫﻮ ﺍﳌﺴﺎﻓﺔ‬ ‫ﺍﻟﺮﺃﺱ ﻋﻦ ﻣﺴﺘﻮﻱ ﺍﻟﻘﺎﻋﺪﺓ ‪:‬‬

‫‪142‬‬

‫‪h‬‬

‫ﺍﻟﱵ ﺗﻔﺼﻞ‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .2‬ﻣﺴﺎﺣﺔ ﺍﳌﺨﺮﻭﻁ‬ ‫ﻧﻌﺘﱪ ﳐﺮﻭﻃﺎ ﺩﻭﺭﺍﻧﻴﺎ ﺗﺼﻤﻴﻤﻪ ﻣﻦ ﺍﻟﺸﻜﻞ‬

‫ﺭﺃﺳﻪ‬

‫‪A‬‬

‫ﻭﻗﺎﻋﺪﺗﻪ ﻗﺮﺹ ﻧﺼﻒ ﻗﻄﺮﻩ‬

‫ﻭﻃﻮﻝ ﺍﳌﺴﺎﻓﺔ ﺍﻟﻔﺎﺻﻠﺔ‬

‫‪r‬‬

‫ﺑﲔ ﺍﻟﺮﺃﺱ ﻭﻧﻘﻄﺔ ﻣﻦ ﻧﻘﺎﻁ ﺣﺎﻓﺔ ﺍﻟﻘﺎﻋﺪﺓ ﻳﺴﺎﻭﻱ ‪ . a‬ﺇﻥ ﺍﳌﺴﺎﺣﺔ ﺍﳉﺎﻧﺒﻴﺔ‬ ‫‪S‬‬

‫ ‪ ،‬ﺃﻱ ﺟﺰﺀ ﺍﻟﻘﺮﺹ ﺍﻟﺬﻱ‬ ‫ﻟﻠﻤﺨﺮﻭﻁ ﻫﻮ ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﺚ ﺍﳌﻨﺤﲏ ‪ABC‬‬


‫‪A‬‬

‫ﻭﻧﺼﻒ ﻗﻄﺮﻩ‬

‫‪a‬‬

‫ﻭﺯﺍﻭﻳﺘﻪ ‪ . α‬ﻟﻨﺤﺴﺐ‬

‫ﻣﻦ ﺃﺟﻞ ﺫﻟﻚ ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﻄﻮﻝ‬

‫‪L‬‬

‫‪S‬‬

‫‪:‬‬

‫ﻟﻠﻘﻮﺱ‬

‫‪BC‬‬

‫ﻫﻮ ﳏﻴﻂ‬

‫ﺍﻟﺪﺍﺋﺮﺓ ﺫﺍﺕ ﻧﺼﻒ ﺍﻟﻘﻄﺮ ‪ . r‬ﻭﻟﺬﺍ ﳝﻜﻦ ﺣﺴﺎﺏ ﻃﻮﻝ ﻫﺬﺍ ﺍﻟﻘﻮﺱ‬ ‫ﺑﻄﺮﻳﻘﺘﲔ‪ ،‬ﻓﻬﻮ ﻳﺴﺎﻭﻱ ‪:‬‬ ‫‪L = 2π r‬‬

‫)ﳏﻴﻂ ﺍﻟﺪﺍﺋﺮﺓ ﺫﺍﺕ ﻧﺼﻒ ﺍﻟﻘﻄﺮ ‪( r‬‬

‫ﻭﻳﺴﺎﻭﻱ ﺃﻳﻀﺎ ‪:‬‬ ‫‪143‬‬

‫‪α‬‬ ‫‪= a.α‬‬ ‫‪2π‬‬

‫‪. L = 2π a‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻭﻣﻨﻪ ﻳﻨﺘﺞ ‪. a.α = 2π r :‬‬ ‫ﻭﺑﺎﻟﺘﺎﱄ ﻓﺈﻥ ﻗﻴﻤﺔ ‪ α‬ﺑﺪﻻﻟﺔ‬

‫ﻭ‬

‫‪r‬‬

‫‪a‬‬

‫ﻫﻲ ‪. α = 2π r :‬‬ ‫‪a‬‬

‫ﻣﺎﺫﺍ ﺗﺴﺎﻭﻱ ﺍﳌﺴﺎﺣﺔ ‪ S‬؟ ﲟﺎ ﺃ‪‬ﺎ ﺗﺴﺎﻭﻱ ﻣﺴﺎﺣﺔ ﺟﺰﺀ ﺍﻟﻘﺮﺹ‬ ‫ﺍﻟﺬﻱ‬


‫‪α a 2α‬‬ ‫=‬ ‫‪2π‬‬ ‫‪2‬‬

‫‪A‬‬

‫ﻗﻄﺮﻩ‬

‫ﻭﻧﺼﻒ‬

‫ﻭﺯﺍﻭﻳﺘﻪ‬

‫‪a‬‬

‫‪α‬‬

‫ﻓﺈﻥ‬

‫‪. S = π a2‬‬

‫ﻭﲟﺎ ﺃﻥ ‪ α = 2π r‬ﻓﺈﻥ ‪:‬‬ ‫‪a‬‬

‫‪a α a 2π r‬‬ ‫=‬ ‫‪= π ar .‬‬ ‫‪2‬‬ ‫‪2 a‬‬ ‫‪2‬‬

‫‪2‬‬

‫=‪S‬‬

‫ﻭﺇﻥ ﲝﺜﻨﺎ ﻋﻦ ﺍﳌﺴﺎﺣﺔ ﺍﻟﻜﻠﻴﺔ ﻓﻌﻠﻴﻨﺎ ﺃﻥ ﻧﻀﻴﻒ ﺇﱃ ﺍﳌﺴﺎﺣﺔ ﺍﻟﺴﺎﺑﻘﺔ ﻣﺴﺎﺣﺔ‬ ‫ﺍﻟﻘﺎﻋﺪﺓ‪ ،‬ﺃﻱ ﺃﻥ ﺍﳌﺴﺎﺣﺔ ﺍﻟﻜﻠﻴﺔ ﻫﻲ ‪. π ar + π r 2 :‬‬ ‫‪ .3‬ﺣﺠﻢ ﺍﳌﺨﺮﻭﻁ‬ ‫ﻧﻌﺘﱪ ﳐﺮﻭﻃﺎ ﺩﻭﺭﺍﻧﻴﺎ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪﺗﻪ‬ ‫ﻣﺒﻴ‪‬ﻦ ﰲ ﺍﻟﺸﻜﻞ ﺍﳌﻮﺍﱄ )ﺍﳌﻤﺜﻞ ﳌﻘﻄﻊ ﳐﺮﻭﻁ( ‪:‬‬ ‫‪z‬‬ ‫‪h‬‬

‫‪r‬‬ ‫‪t‬‬

‫‪R‬‬

‫‪0‬‬

‫‪144‬‬

‫‪R‬‬

‫ﻭﺍﺭﺗﻔﺎﻋﻪ‬

‫‪h‬‬

‫ﻛﻤﺎ ﻫﻮ‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻟﻨﺤﺴﺐ ﺃﻭﻻ‬ ‫‪r‬‬ ‫‪R‬‬ ‫=‬ ‫‪h−t h‬‬

‫‪r‬‬

‫ﺑﺪﻻﻟﺔ‬

‫‪،‬‬

‫‪h‬‬

‫‪ . R ،‬ﻟﺪﻳﻨﺎ ﺍﻟﻌﻼﻗﺔ )ﻧﻈﺮﻳﺔ ﻃﺎﻟﺲ( ‪:‬‬

‫‪t‬‬

‫‪ .‬ﻭﻣﻨﻪ ) ‪. r = R (h − t‬‬ ‫‪h‬‬

‫ﳝﻜﻦ ﺣﺴﺎﺏ ﺍﳊﺠﻢ ‪ V‬ﻟﻠﻤﺨﺮﻭﻁ ﺣﺴﺐ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪h‬‬

‫‪V = ∫ Vt dt‬‬ ‫‪0‬‬

‫ﺣﻴﺚ ﳝﺜﻞ‬

‫‪Vt‬‬

‫ﻣﺴﺎﺣﺔ ﺍﻟﻘﺮﺹ ﺍﶈﺼﻞ ﻋﻠﻴﻪ ﰲ ﺗﻘﺎﻃﻊ ﺍﳌﺨﺮﻭﻁ ﻣﻊ ﻣﺴﺘﻮ‬

‫ﻳﻮﺍﺯﻱ ﺍﻟﻘﺎﻋﺪﺓ ﻭﻳﻘﻄﻊ ﺍﶈﻮﺭ ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ ﺍﻟﱵ ﲤﻴ‪‬ﺰﻫﺎ ﺍﻹﺣﺪﺍﺛﻴﺔ ‪ . t‬ﻣﺴﺎﺣﺔ‬ ‫ﻫﺬﺍ ﺍﻟﻘﺮﺹ ﻫﻲ ‪:‬‬ ‫‪Vt = π r 2‬‬ ‫‪2‬‬

‫‪t‬‬ ‫‪R2t 2‬‬ ‫‪+π 2 .‬‬ ‫‪h‬‬ ‫‪h‬‬

‫‪R‬‬ ‫‪‬‬ ‫‪= π  (h − t ) ‬‬ ‫‪h‬‬ ‫‪‬‬ ‫‪= π R 2 − 2π R 2‬‬

‫ﻭﻟﺬﻟﻚ ﺗﻜﺘﺐ ﺍﻟﻌﻼﻗﺔ ‪ V = ∫0 Vt dt‬ﻋﻠﻰ ﺍﻟﺸﻜﻞ ‪:‬‬ ‫‪h‬‬

‫‪145‬‬

‫ا ‪ : 3‬ا ب ا ‬ ‫‪h‬‬

‫‪Vt dt‬‬

‫∫‬

‫‪0‬‬

‫= ‪V‬‬

‫‪‬‬ ‫‪R 2t 2 ‬‬ ‫‪2‬‬ ‫‪2 t‬‬ ‫‪π‬‬ ‫‪R‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪π‬‬ ‫‪R‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫‪dt‬‬ ‫‪∫0 ‬‬ ‫‪h‬‬ ‫‪h2 ‬‬ ‫‪2‬‬ ‫‪h‬‬ ‫‪h 2π R‬‬ ‫‪h‬‬ ‫‪R 2t 2‬‬ ‫∫ ‪= ∫ π R 2 dt −‬‬ ‫‪tdt + ∫ π‬‬ ‫‪dt‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪h‬‬ ‫‪h2‬‬ ‫‪2π R 2 h‬‬ ‫‪R2 h 2‬‬ ‫‪π‬‬ ‫‪tdt‬‬ ‫‪t dt‬‬ ‫‪= π R2h −‬‬ ‫‪+‬‬ ‫‪h ∫0‬‬ ‫‪h 2 ∫0‬‬ ‫‪R 2 h3‬‬ ‫× ‪= π R2h − π R2h + π 2‬‬ ‫‪3‬‬ ‫‪h‬‬ ‫‪2‬‬ ‫‪πR h‬‬ ‫‪.‬‬ ‫=‬ ‫‪3‬‬ ‫‪h‬‬

‫ﻭﺑﺎﻟﺘﺎﱄ ﻓﺤﺠﻢ ﳐﺮﻭﻁ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪﺗﻪ‬ ‫‪π R2 h‬‬ ‫‪3‬‬

‫‪R‬‬

‫=‬


‫‪h‬‬

‫ﻫﻮ‬

‫= ‪.V‬‬

‫‪ .4‬ﺣﺠﻢ ﺍﻷﺳﻄﻮﺍﻧﺔ‬ ‫ﺇﺫﺍ ﻛﺎﻥ‬

‫‪h‬‬

‫ﻫﻮ ﺍﺭﺗﻔﺎﻉ ﻫﺬﻩ ﺍﻷﺳﻄﻮﺍﻧﺔ ﺍﻟﺪﻭﺭﺍﻧﻴﺔ‪ ،‬ﻭﻛﺎﻥ‬

‫ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ ﻓﺈﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﱵ ﺗﻌﻄﻲ ﺍﳊﺠﻢ‬

‫‪V‬‬

‫‪r‬‬

‫ﻧﺼﻒ‬

‫)ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺘﻜﺎﻣﻞ( ﺗﺘﻤﺜﻞ‬

‫ﰲ ﻣﻜﺎﻣﻠﺔ ﻣﺴﺎﺣﺔ ﻣﻘﻄﻊ ﻣﻦ ﺍﻷﺳﻄﻮﺍﻧﺔ ‪ -‬ﳓﺼﻞ ﻋﻠﻴﻪ ﻛﺘﻘﺎﻃﻊ ﻣﺴﺘﻮ‬ ‫ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﳏﻮﺭ ﺍﻷﺳﻄﻮﺍﻧﺔ ﻣﻊ ﳎﺴﻢ ﺍﻷﺳﻄﻮﺍﻧﺔ ‪ -‬ﻭﳒﻌﻞ ﻣﺘﻐﻴ‪‬ﺮ‬ ‫ﺍﳌﻜﺎﻣﻠﺔ ﳝﺴﺢ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ]‪. [ 0, h‬‬ ‫ﻻﺣﻆ ﺃﻥ ﻫﺬﺍ ﺍﳌﻘﻄﻊ ﻫﻮ ﻗﺮﺹ ﺛﺎﺑﺖ ﺍﳌﺴﺎﺣﺔ ﻭﻣﺴﺎﺣﺘﻪ ﻫﻲ ﻣﺴﺎﺣﺔ‬ ‫ﻗﺎﻋﺪﺓ ﺍﻷﺳﻄﻮﺍﻧﺔ‪ .‬ﻭﻟﺬﻟﻚ ﻓﻌﻤﻠﻴﺔ ﺍﳌﻜﺎﻣﻠﺔ ﺍﳌﺬﻛﻮﺭﺓ ﺗﻌﻄﻲ ﻣﺒﺎﺷﺮﺓ ‪:‬‬ ‫‪V = ∫ π r 2 dz = π r 2 h .‬‬ ‫‪h‬‬

‫‪0‬‬

‫‪146‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﻧﻼﺣﻆ ﻋﻨﺪ ﻣﻘﺎﺭﻧﺔ ﺣﺠﻢ ﳐﺮﻭﻁ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪﺗﻪ‬ ‫‪h‬‬

‫ﲝﺠﻢ ﺃﺳﻄﻮﺍﻧﺔ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ‬

‫‪R‬‬

‫ﻭﺍﺭﺗﻔﺎﻋﻬﺎ‬

‫‪h‬‬

‫‪R‬‬


‫ﺃﻥ ﺣﺠﻢ ﺍﳌﺨﺮﻭﻁ‬

‫ﻳﺴﺎﻭﻱ ﺛﻠﺚ ﺣﺠﻢ ﺍﻹﺳﻄﻮﺍﻧﺔ‪.‬‬ ‫ﻳﻌﲏ ﺫﻟﻚ ﺃﻧﻨﺎ ﻧﺴﺘﻄﻴﻊ "ﻭﺿﻊ" ‪ 3‬ﳐﺮﻭﻃﺎﺕ ﺩﺍﺧﻞ ﺇﺳﻄﻮﺍﻧﺔ ﺇﻥ‬ ‫ﻛﺎﻥ ﳍﺎ ﻧﻔﺲ ﺍﻻﺭﺗﻔﺎﻉ ﻭﻧﻔﺲ ﻧﺼﻒ ﺍﻟﻘﻄﺮ‪ ...‬ﰲ ﺣﲔ ﺃﻥ ﻣﺴﺎﺣﺔ‬ ‫ﻣﺴﺘﻄﻴﻞ ﺗﺴﺎﻭﻱ ﻧﺼﻒ ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﺚ ﺍﻟﺬﻱ ﻳﻜﻮﻥ ﺍﺭﺗﻔﺎﻋﻪ ﻋﺮﺽ‬ ‫ﺍﳌﺴﺘﻄﻴﻞ ﻭﻗﺎﻋﺪﺗﻪ ﻃﻮﻝ ﺫﻟﻚ ﺍﳌﺴﺘﻄﻴﻞ ! ﺃﻻ ﻳﺮﺟﻊ ﺫﻟﻚ ﺇﱃ ﺍﻻﻧﺘﻘﺎﻝ ﻣﻦ‬ ‫ﺍﳌﺴﺘﻮﻱ ﺇﱃ ﺍﻟﻔﻀﺎﺀ ‪ ...‬ﺍﻻﻧﺘﻘﺎﻝ ﻣﻦ ﺍﻟﺒﻌﺪ ‪ 2‬ﺇﱃ ﺍﻟﺒﻌﺪ ‪3‬؟‬ ‫‪ .5‬ﺣﺠﻢ ﺍﻟﻜﺮﺓ ‪:‬‬ ‫ﻫﻨﺎﻙ ﺧﺎﺻﻴﺔ ﻣﻬﻤﺔ ﻟﻠﻜﺮﺓ ‪ :‬ﳝﺜﻞ ﺳﻄﺢ ﺍﻟﻜﺮﺓ ﺃﺻﻐﺮ ﻣﺴﺎﺣﺔ ﳑﻜﻨﺔ‬ ‫ﻣﻦ ﺑﲔ ﺍﻟﺴﻄﻮﺡ ﺍﻟﱵ ﲢﻴﻂ ﲝﺠﻢ ﻣﻌﻄﻰ‪.‬‬ ‫ﲟﻌﲎ ﺃﻧﻪ ﺇﺫﺍ ﺃﻋﻄﻲ ﺣﺠﻢ ﻭﻃﻠﺐ ﻭﺿﻌﻪ ﺩﺍﺧﻞ ﺇﻧﺎﺀ ﻭﺃﺭﺩﻧﺎ ﺃﻥ‬ ‫ﻳﻜﻮﻥ ﺳﻄﺢ ﺍﻹﻧﺎﺀ ﺃﺻﻐﺮﻳﺎ ﻓﻼ ﺑﺪ ﺃﻥ ﳔﺘﺎﺭ ﺍﻹﻧﺎﺀ ﻛﺮﻭﻱ ﺍﻟﺸﻜﻞ‪.‬‬ ‫ﻛﻤﺎ ﺃﻥ ﺍﻟﻜﺮﺓ ﲢﺘﻮﻱ ﻋﻠﻰ ﺃﻛﱪ ﺣﺠﻢ ﳑﻜﻦ ﻣﻦ ﺑﲔ ﺍﻟﺴﻄﻮﺡ ﺍﻟﱵ‬ ‫ﳍﺎ ﻣﺴﺎﺣﺔ ﻣﻌﻄﺎﺓ‪ .‬ﲟﻌﲎ ﺃﻧﻪ ﺇﺫﺍ ﺃﻋﻄﻲ ﺳﻄﺢ ﻣﺴﺎﺣﺘﻪ ﻣﻌﻠﻮﻣﺔ ﻭﺃﺭﺩﻧﺎ ﺃﻥ‬ ‫ﻧﻌﻄﻲ ﻟﻪ ﺷﻜﻼ ﳚﻌﻠﻪ ﳛﺘﻮﻱ ﻋﻠﻰ ﺃﻛﱪ ﺣﺠﻢ ﳑﻜﻦ ﻓﻼ ﺑﺪ ﺃﻥ ﳒﻌﻠﻪ ﻳﺄﺧﺬ‬ ‫ﺷﻜﻞ ﻛﺮﺓ‪.‬‬ ‫‪147‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻫﺬﻩ ﺍﳋﺎﺻﻴﺔ ﻫﻲ ﺍﻟﱵ ﲡﻌﻞ ﻓﻘﻌﺎﺕ ﺍﻟﺼﺎﺑﻮﻥ ﻭﻗﻄﺮﺍﺕ ﺍﳌﺎﺀ –‬ ‫ﻋﻨﺪﻣﺎ ‪‬ﻤﻞ ﺗﺄﺛﲑ ﺍﳉﺎﺫﺑﻴﺔ – ﺗﺄﺧﺬ ﺃﺷﻜﺎﻻ ﻛﺮﻭﻳﺔ ﺫﻟﻚ ﺃﻥ ﺍﻟﻀﻐﻂ‬ ‫ﺍﻟﺴﻄﺤﻲ ﻳﺴﻌﻰ ﺩﻭﻣﺎ ﺇﱃ ﺗﺼﻐﲑ ﺍﳌﺴﺎﺣﺔ‪.‬‬ ‫ﻧﻌﺘﱪ ﻛﺮﺓ ﻧﺼﻒ ﻗﻄﺮﻫﺎ‬ ‫ﺍﳊﺠﻢ‬

‫‪V‬‬

‫‪R‬‬

‫ﻭﻧﺮﺳﻢ ﺍﻟﺸﻜﻞ ﺍﳌﻮﺍﱄ‪ ،‬ﻭﳓﺴﺐ‬

‫ﻟﻠﻜﺮﺓ ﺑﻄﺮﻳﻘﺔ ﳑﺎﺛﻠﺔ ﻟﺘﻠﻚ ﺍﻟﱵ ﺍﺗﺒﻌﻨﺎﻫﺎ ﰲ ﺣﺴﺎﺏ ﺣﺠﻢ‬

‫ﺍﳌﺨﺮﻭﻁ‪ .‬ﻣﻦ ﺃﺟﻞ ﺫﻟﻚ ﻧﻘﻄﻊ ﺍﻟﻜﺮﺓ ﲟﺴﺘﻮ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺍﶈﻮﺭ‬

‫) ‪(Oz‬‬

‫ﻓﻴﻘﻄﻊ ﻫﺬﺍ ﺍﶈﻮﺭ ﻋﻨﺪ ﺍﻟﻨﻘﻄﺔ ‪ . t‬ﻭﻳﻜﻮﻥ ﺗﻘﺎﻃﻊ ﺍﻟﻜﺮﺓ ﻣﻊ ﺍﳌﺴﺘﻮﻱ ﻗﺮﺻﺎ‬ ‫ﻣﺮﻛﺰﻩ ‪ t‬ﻭﻧﺼﻒ ﻗﻄﺮﻩ ‪. r‬‬ ‫‪z‬‬ ‫‪R‬‬ ‫‪r‬‬ ‫‪t‬‬

‫‪0‬‬

‫‪R‬‬

‫ﻛﻢ ﻳﺴﺎﻭﻱ‬

‫‪r‬‬

‫ﺑﺪﻻﻟﺔ‬

‫‪R‬‬

‫ﻭ ‪ t‬؟ ﺑﺘﻄﺒﻴﻖ ﻧﻈﺮﻳﺔ ﻓﻴﺜﺎﻏﻮﺭﺱ ﳓﺼﻞ‬

‫ﻋﻠﻰ ﺍﻟﻌﻼﻗﺔ ‪ . r 2 + t 2 = R 2 :‬ﻭﻣﻨﻪ ‪ . r 2 = R 2 − t 2‬ﻳﻌﱪ ﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺘﺎﱄ‬ ‫ﻋﻦ ﺣﺠﻢ ﻧﺼﻒ ﺍﻟﻜﺮﺓ‬ ‫‪π r 2 dt = ∫ π ( R 2 − t 2 ) dt‬‬ ‫‪R‬‬

‫‪0‬‬

‫‪148‬‬

‫‪R‬‬

‫∫‬

‫‪0‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻭﻣﻨﻪ ﻓﺈﻥ ﺣﺠﻢ ﺍﻟﻜﺮﺓ ﻫﻮ ‪:‬‬ ‫‪V = 2 ∫ π ( R 2 − t 2 ) dt‬‬ ‫‪R‬‬

‫‪0‬‬

‫‪2π R 3‬‬ ‫‪3‬‬

‫‪= 2π R 3 −‬‬ ‫‪4π R 3‬‬ ‫‪.‬‬ ‫‪3‬‬

‫=‬

‫ﻣﻼﺣﻈﺔ ‪:‬‬ ‫ﻛﺎﻥ ﺃﺭﲬﻴﺪﺱ ﻗﺪ ﺑﻴ‪‬ﻦ ﺃﻥ ﻣﺴﺎﺣﺔ ﺳﻄﺢ ﻛﺮﺓ ﺗﺴﺎﻭﻱ ﺍﳌﺴﺎﺣﺔ‬ ‫ﺍﳉﺎﻧﺒﻴﺔ ﻟﻸﺳﻄﻮﺍﻧﺔ ﺍﻟﱵ ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ ﻳﺴﺎﻭﻱ ﻗﻄﺮ ﺍﻟﻜﺮﺓ ﻭﺍﺭﺗﻔﺎﻋﻬﺎ ﻳﺴﺎﻭﻱ‬ ‫ﺃﻳﻀﺎ ﻗﻄﺮ ﺍﻟﻜﺮﺓ‪.‬‬

‫ﻭﻟﺬﻟﻚ ﻧﺴﺘﻨﺘﺞ ﳑﺎ ﻗﺎﻟﻪ ﺃﺭﲬﻴﺪﺱ ﺃﻥ ﻣﺴﺎﺣﺔ ﺳﻄﺢ ﻛﺮﺓ ﻧﺼﻒ ﻗﻄﺮﻫﺎ‬ ‫‪R‬‬

‫ﻫﻮ ‪. 4π R 2‬‬

‫‪149‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .6‬ﺟﺪﻭﻝ ﺍﳌﺴﺎﺣﺎﺕ‬ ‫ﻧﻘﺪ‪‬ﻡ ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻣﺴﺎﺣﺎﺕ ﺑﻌﺾ ﺍﻷﺷﻜﺎﻝ ﺍﳍﻨﺪﺳﻴﺔ ﰲ‬ ‫ﺍﻟﻔﻀﺎﺀ‪:‬‬ ‫ﺍﳌﺴﺎﺣﺔ‬

‫ﺍﻟﺸﻜﻞ‬ ‫ﻣﻜﻌﺐ ﻃﻮﻝ ﺿﻠﻌﻪ‬

‫‪6a 2‬‬

‫‪a‬‬

‫ﻣﺘﻮﺍﺯﻱ ﺍﳌﺴﺘﻄﻴﻼﺕ ﺃﺑﻌﺎﺩﻩ‬

‫‪a‬‬

‫‪،‬‬

‫‪c ،b‬‬

‫ﺃﺳﻄﻮﺍﻧﺔ ﺩﻭﺭﺍﻧﻴﺔ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ‬ ‫ﻭﺍﺭﺗﻔﺎﻋﻬﺎ‬

‫‪2ab + 2bc + 2ac‬‬

‫ﺍﳌﺴﺎﺣﺔ ﺍﳉﺎﻧﺒﻴﺔ ‪:‬‬

‫‪R‬‬

‫‪2π R.h‬‬

‫‪h‬‬

‫ﺍﳌﺴﺎﺣﺔ ﺍﻟﻜﻠﻴﺔ ‪:‬‬ ‫‪2π R.h + 2π R 2‬‬

‫ﳐﺮﻭﻁ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ‬

‫‪R‬‬


‫‪h‬‬

‫ﺍﳌﺴﺎﺣﺔ ﺍﳉﺎﻧﺒﻴﺔ ‪:‬‬ ‫‪π r r 2 + h2‬‬

‫ﺍﳌﺴﺎﺣﺔ ﺍﻟﻜﻠﻴﺔ ‪:‬‬ ‫‪π r r 2 + h2 + π r 2‬‬

‫ﻛﺮﺓ ﻧﺼﻒ ﻗﻄﺮﻫﺎ‬

‫‪R‬‬

‫‪4π R 2‬‬

‫‪150‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫‪ .7‬ﺟﺪﻭﻝ ﺍﳊﺠﻮﻡ‬ ‫ﻧﻘﺪﻡ ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻗﺎﺋﻤﺔ ﺗﻮﺿﺢ ﺣﺠﻮﻡ ﺃﻫﻢ ﺍﻷﺷﻜﺎﻝ‬ ‫ﺍﳍﻨﺪﺳﻴﺔ ﰲ ﺍﻟﻔﻀﺎﺀ ‪:‬‬ ‫ﺍﳊﺠﻢ‬

‫ﺍﻟﺸﻜﻞ‬ ‫ﻣﻜﻌﺐ ﻃﻮﻝ ﺿﻠﻌﻪ‬

‫‪a3‬‬

‫‪a‬‬

‫ﻣﺘﻮﺍﺯﻱ ﺍﳌﺴﺘﻄﻴﻼﺕ ﺃﺑﻌﺎﺩﻩ‬

‫‪a‬‬

‫‪،‬‬

‫‪b‬‬

‫‪،‬‬

‫‪c‬‬

‫‪a.b.c‬‬

‫ﻣﺴﺎﺣﺔ ﻗﺎﻋﺪﺗﻪ ﰲ ﺍﺭﺗﻔﺎﻋﻪ‬

‫ﻣﺘﻮﺍﺯﻱ ﻭﺟﻮﻩ‬ ‫ﺃﺳﻄﻮﺍﻧﺔ ﺩﻭﺭﺍﻧﻴﺔ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ‬ ‫ﻭﺍﺭﺗﻔﺎﻋﻬﺎ‬

‫‪R‬‬

‫‪π R 2 .h‬‬

‫‪h‬‬

‫ﻣﺴﺎﺣﺔ ﻗﺎﻋﺪ‪‬ﺎ ﰲ‬

‫ﺃﺳﻄﻮﺍﻧﺔ‬

‫ﺍﺭﺗﻔﺎﻋﻬﺎ‬ ‫ﳐﺮﻭﻁ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ‬

‫‪R‬‬


‫‪h‬‬

‫‪π R 2 .h‬‬ ‫‪3‬‬

‫ﺛﻠﺚ ﺟﺪﺍﺀ ﻣﺴﺎﺣﺔ‬

‫ﺍﳍﺮﻡ‬

‫ﻗﺎﻋﺪﺗﻪ ﰲ ﺍﺭﺗﻔﺎﻋﻪ‬ ‫ﻛﺮﺓ ﻧﺼﻒ ﻗﻄﺮﻫﺎ‬

‫‪4π R 3‬‬ ‫‪3‬‬

‫‪R‬‬

‫ﳎﺴﻢ ﻧﺎﻗﺼﻲ ﺃﻧﺼﺎﻑ ﳏﺎﻭﺭﻩ‬ ‫ﺟﺬﻉ ﳐﺮﻭﻁ ﺍﺭﺗﻔﺎﻋﻪ‬ ‫ﻗﺎﻋﺪﺗﺎﻩ‬

‫‪R‬‬

‫ﻭ‬

‫‪h‬‬

‫‪a‬‬

‫‪،b ،‬‬

‫‪c‬‬

‫‪4π‬‬ ‫‪a.b .c‬‬ ‫‪3‬‬

‫ﻭﻧﺼﻔﺎ ﻗﻄﺮﻱ )‬

‫‪R‬‬

‫‪3‬‬

‫'‪R‬‬

‫ﺟﺬﻉ ﻛﺮﺓ ﺍﺭﺗﻔﺎﻋﻪ ‪ h‬ﻭﻧﺼﻔﺎ ﻗﻄﺮﻱ ﻗﺎﻋﺪﺗﺎﻩ )‬ ‫ﻭ‬

‫' ‪+ R '2 + RR‬‬

‫‪2‬‬

‫‪(R‬‬

‫‪πh‬‬

‫'‪R‬‬

‫‪151‬‬

‫‪+ 3R '2 + h 2‬‬

‫‪2‬‬

‫‪( 3R‬‬

‫‪πh‬‬ ‫‪6‬‬

‫ا ‪ : 3‬ا ب ا ‬

‫ﻗﺒﺔ ﻛﺮﺓ ﺍﺭﺗﻔﺎﻋﻬﺎ ‪ h‬ﻭﻧﺼﻒ ﻗﻄﺮ ﺍﻟﻜﺮﺓ‬

‫‪h‬‬

‫)‪(3r − h‬‬

‫‪π h2‬‬ ‫‪3‬‬

‫ﺍﻟﻘﺎﻋﺪﺓ ﺍﻟﻌﺎﻣﺔ ﳊﺴﺎﺏ ﺣﺠﻢ ﳎﺴﻢ‬ ‫ﳊﺴﺎﺏ ﺣﺠﻢ ﺃﻱ ﳎﺴﻢ ﻧﺴﺘﺨﺪﻡ ﺍﻟﻌﻼﻗﺔ‬

‫‪A (t )dt‬‬

‫∫ ﺍﻟﱵ ﺗﻌﱪ‬

‫ﻋﻦ ﺍﳊﺠﻢ ﺍﳌﻄﻠﻮﺏ ﺣﻴﺚ ‪:‬‬ ‫‪(1‬‬

‫) ‪A (t‬‬

‫ﻫﻲ ﻣﺴﺎﺣﺔ ﻣﻘﻄﻊ ﺍ‪‬ﺴﻢ ﺍﻟﻌﻤﻮﺩﻱ ﻋﻠﻰ "ﺍﶈﻮﺭ"‬

‫ﺍﻟﺬﻱ ﻳﻨﺘﻤﻲ ﺇﻟﻴﻪ ‪، t‬‬ ‫‪ t (2‬ﲤﺴﺢ "ﺍﻻﺭﺗﻔﺎﻉ"‪.‬‬

‫‪152‬‬

‫ﺑﻌﺾ ﺍﳌﺮﺍﺟﻊ ‪:‬‬ ‫‪ .1‬ﺯﻳﺘﻮﱐ ﻝ‪ : .‬ﺩﺭﻭﺱ ﻣﻮﺟﻬﺔ ﻟﺘﻜﻮﻳﻦ ﺃﺳﺎﺗﺬﺓ ﺍﻟﺘﻌﻠﻴﻢ ﺍﳌﺘﻮﺳﻂ‪،‬‬ ‫ﺍﻟﺮﻳﺎﺿﻴﺎﺕ )ﺑﻴﻮﻟﻮﺟﻴﺎ(‪ ،‬ﺍﻹﺭﺳﺎﻝ ﺍﻷﻭﻝ‪ ،‬ﻭﺯﺍﺭﺓ ﺍﻟﺘﻌﻠﻴﻢ ﺍﻟﻌﺎﱄ ﻭﺍﻟﺒﺤﺚ‬ ‫ﺍﻟﻌﻠﻤﻲ‪.2006 ،‬‬ ‫‪ .2‬ﺳﻌﺪ ﺍﷲ ﺃ‪.‬ﺥ‪ : .‬ﺩﺭﻭﺱ ﻣﻮﺟﻬﺔ ﻟﺘﻜﻮﻳﻦ ﺃﺳﺎﺗﺬﺓ ﺍﻟﺘﻌﻠﻴﻢ ﺍﳌﺘﻮﺳﻂ‪،‬‬ ‫ﺍﻟﺘﺤﻠﻴﻞ ‪) 1‬ﺭﻳﺎﺿﻴﺎﺕ ﻭﺗﻜﻨﻮﻟﻮﺟﻴﺎ(‪ ،‬ﺍﻹﺭﺳﺎﻝ ﺍﻷﻭﻝ‪ ،‬ﻭﺯﺍﺭﺓ ﺍﻟﺘﻌﻠﻴﻢ ﺍﻟﻌﺎﱄ‬ ‫ﻭﺍﻟﺒﺤﺚ ﺍﻟﻌﻠﻤﻲ‪.2006 ،‬‬ ‫‪ .3‬ﺳﻌﺪ ﺍﷲ ﺃ‪.‬ﺥ‪ : .‬ﺩﺭﻭﺱ ﻣﻮﺟﻬﺔ ﻟﺘﺤﺴﲔ ﻣﺴﺘﻮﻯ ﺃﺳﺎﺗﺬﺓ ﺍﻟﺘﻌﻠﻴﻢ‬ ‫ﺍﻟﺜﺎﻧﻮﻱ‪ ،‬ﺍﳌﻌﻬﺪ ﺍﻟﻮﻃﲏ ﻟﺘﻜﻮﻳﻦ ﻣﺴﺘﺨﺪﻣﻲ ﺍﻟﺘﺮﺑﻴﺔ ﻭﲢﺴﲔ ﻣﺴﺘﻮﺍﻫﻢ‪،‬‬ ‫ﺍﳊﺮﺍﺵ‪ ،‬ﺍﳉﺰﺍﺋﺮ‪.2004 ،‬‬ ‫‪4. Allab K. : Eléments d'analyse, O.P.U., Alger, 1980.‬‬ ‫) إ ا أب‪ .‬خ‪ %& .‬ا‪ # $‬د! ان‬ ‫ا ‪+# -.‬ت ا )‪ ،+‬ا )(ا'(‬ ‫‪5. Couty R. Ezra J. : Analyse, Armand Colin, Paris,‬‬ ‫‪1967.‬‬

‫) إ ا ! &‪ # 0# 1‬د! ان‬ ‫ا ‪+# -.‬ت ا )‪ ،+‬ا )(ا'(‬

‫‪153‬‬

6. Medjadi D.E., Boukra M., Djadane A., Sadallah B.-K. : Analyse mathématique , Tome 1, O.P.U., Alger 1994.

7. Dieudonné J. : Calcul infinitésimal , Hermann, Paris, 1980.

8. Kolmogorov, Fomine : Éléments de la théorie des fonctions et de l’analyse fonctionnelle, Ed. Mir, 1976

‫ د! ان‬# $‫ ا‬%& .‫ خ‬.‫) إ ا أ‬ ('‫ ا )(ا‬،+) ‫ت ا‬+# -. ‫ا‬

154

‫ و ارس‬ ‫ﺗﻘﺪﱘ‬

‫‪161‬‬ ‫ﺍﻟﻔﺼﻞ ‪ :1‬ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ‬

‫‪ .1‬ﺍﳌﻨﻄﻖ‬

‫‪163‬‬ ‫‪166‬‬

‫‪ 1.1‬ﺗﻌﺎﺭﻳﻒ‬

‫‪166‬‬

‫‪ 2.1‬ﻫﺸﺎﺷﺔ ﺍﳌﻨﻄﻖ ﺍﻟﺮﻳﺎﺿﻲ‬

‫‪174‬‬

‫‪ 3.1‬ﺃﳕﺎﻁ ﺍﻟﱪﻫﺎﻥ‬

‫‪185‬‬

‫‪ .2‬ﺍﻟﻌﻼﻗﺎﺕ‬

‫‪190‬‬ ‫ﺍﻟﻔﺼﻞ ‪ :2‬ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ‬

‫‪ .1‬ﺍﻟﺰﻣــﺮﺓ‬

‫‪199‬‬ ‫‪201‬‬

‫‪ 1.1‬ﻣﻘﺪﻣﺔ‬

‫‪201‬‬

‫‪ 2.1‬ﺗﻌﺎﺭﻳﻒ ﻭﺃﻣﺜﻠﺔ‬

‫‪204‬‬

‫‪ 3.1‬ﳌﺎﺫﺍ ﻳﻬﺘﻢ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﻮﻥ ﺑﺎﻟﺮﻣﺰ؟‬

‫‪213‬‬

‫‪ 4.1‬ﺗﻌﺎﺭﻳﻒ ﻭﻧﺘﺎﺋﺞ‬

‫‪215‬‬

‫‪ 5.1‬ﲤﺎﺭﻳﻦ ﺃﺳﺎﺳﻴﺔ‬

‫‪225‬‬

‫‪ .2‬ﺍﳊﻠﻘــﺔ‬

‫‪228‬‬ ‫‪ 1.2‬ﺗﻌﺮﻳﻒ ﺍﳊﻠﻘﺔ ﻭﺃﻣﺜﻠﺔ ﺃﻭﻟﻴﺔ‬

‫‪161‬‬

‫‪ 2.2‬ﺗﻌﺎﺭﻳﻒ ﻋﻨﺎﺻﺮ ﺧﺎﺻﺔ ﰲ ﺍﳊﻠﻘﺎﺕ‬

‫‪236‬‬

‫‪ 3.2‬ﺃﺟﺰﺍﺀ ﺧﺎﺻﺔ ﻣﻦ ﺍﳊﻠﻘﺎﺕ‬

‫‪243‬‬

‫‪ 4.2‬ﺣﻠﻘﺎﺕ ﺃﺧﺮﻯ ﻭﲤﺎﺛﻞ ﺍﳊﻠﻘﺎﺕ‬

‫‪248‬‬

‫‪157‬‬

‫‪ .3‬ﺍﳊﻘﻞ )ﺍﳉﺴﻢ(‬

‫‪254‬‬ ‫‪ 1.3‬ﺗﻌﺎﺭﻳﻒ ﻭﻧﺘﺎﺋﺞ‬

‫‪ .4‬ﶈﺔ ﺗﺎﺭﳜﻴﺔ‬

‫‪254‬‬ ‫‪263‬‬

‫ﺍﻟﻔﺼﻞ ‪ :3‬ﳎﻤﻮﻋﺎﺕ ﺍﻷﻋﺪﺍﺩ‬ ‫‪ .1‬ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ‬

‫‪277‬‬ ‫‪280‬‬

‫‪ 1.1‬ﺇﻧﺸﺎﺀ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ‬

‫‪280‬‬

‫‪ 2.1‬ﺍﻷﻋﺪﺍﺩ ﺍﻷﻭﻟﻴﺔ‬

‫‪285‬‬

‫‪ 3.1‬ﺍﻷﻋﺪﺍﺩ ﺍﻷﻭﻟﻴﺔ ﺍﻟﻮﺍﺣﺪﻳﺔ ﺍﻟﺘﻜﺮﺍﺭﻳﺔ‬

‫‪291‬‬

‫‪ 4.1‬ﺍﻷﻋﺪﺍﺩ ﺍﻷﻭﻟﻴﺔ ﺍﳌﺘﻨﺎﻇﺮﺓ‬

‫‪293‬‬

‫‪ 5.1‬ﺍﻷﻋﺪﺍﺩ ﺍﳌﻘﺘﺼﺪﺓ ﻭﺍﻷﻋﺪﺍﺩ ﺍﳌﺒﺬﺭﺓ‬

‫‪295‬‬

‫‪ .2‬ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ‬

‫‪296‬‬

‫‪ 1.2‬ﺇﻧﺸﺎﺀ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ‬

‫‪296‬‬

‫‪ 2.2‬ﻣﻦ ﺧﻮﺍﺹ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ‬

‫‪299‬‬

‫‪ .3‬ﺍﻷﻋﺪﺍﺩ ﺍﻟﻨﺎﻃﻘﺔ‬

‫‪303‬‬ ‫‪ 1.3‬ﺇﻧﺸﺎﺀ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻨﺎﻃﻘﺔ‬

‫‪ .4‬ﺍﻷﻋﺪﺍﺩ ﺍﳊﻘﻴﻘﻴﺔ‬

‫‪303‬‬ ‫‪307‬‬

‫‪ 1.4‬ﺇﻧﺸﺎﺀ ﲟﺘﺘﺎﻟﻴﺎﺕ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻨﺎﻃﻘﺔ‬

‫‪307‬‬

‫‪ 2.4‬ﺍﻹﻧﺸﺎﺀ ﲟﻘﺎﻃﻊ ﺩﻳﺪﻛﻴﻨﺪ ‪Dedekind‬‬

‫‪309‬‬

‫‪ .5‬ﺍﻷﻋﺪﺍﺩ ﺍﳌﺮﻛﺒﺔ‬

‫‪312‬‬ ‫‪ 1.5‬ﺇﻧﺸﺎﺀ ﺑﺈﺳﺘﺨﺪﺍﻡ ﺍﳉﺪﺍﺀ ﺍﻟﺪﻳﻜﺎﺭﰐ ‪R²‬‬

‫‪313‬‬

‫‪ 2.5‬ﺇﻧﺸﺎﺀ ﺑﺈﺳﺘﺨﺪﺍﻡ ﻛﺜﲑﺍﺕ ﺍﳊﺪﻭﺩ‬

‫‪314‬‬

‫‪158‬‬

‫‪ 3.5‬ﺇﻧﺸﺎﺀ ﺑﺈﺳﺘﺨﺪﺍﻡ ﺍﳌﺼﻔﻮﻓﺎﺕ‬

‫‪315‬‬

‫‪ 4.5‬ﻛﺘﺎﺑﺔ ﺍﻷﻋﺪﺍﺩ ﺍﳌﺮﻛﺒﺔ‬

‫‪316‬‬

‫‪ 5.5‬ﺍﻟﺮﺅﻳﺔ ﺍﳍﻨﺪﺳﻴﺔ ﻟﻸﻋﺪﺍﺩ ﺍﳌﺮﻛﺒﺔ‬

‫‪317‬‬

‫‪ .6‬ﺗﻄﻮﺭ ﺍﻟﻌﺪﺩ‬

‫‪318‬‬ ‫‪ 1.6‬ﺍﻟﻌﺪﺩ ﻗﺒﻞ ﻣﺌﺎﺕ ﺁﻻﻑ ﺍﻟﺴﻨﲔ‬

‫‪319‬‬

‫‪ 2.6‬ﺍﳊﺴﺎﺏ ﻋﻨﺪ ﻗﺪﻣﺎﺀ ﺍﳌﺼﺮﻳﲔ‬

‫‪321‬‬

‫‪ 3.6‬ﻭﻋﻨﺪ ﺍﻟﺒﺎﺑﻠﻴﲔ ﻭﻣﻦ ﺧﻠﹶﻔﻬﻢ‬

‫‪322‬‬

‫‪ 4.6‬ﺗﻨﻮﻉ ﺃﻧﻈﻤﺔ ﺍﻟﻌﺪ‪ ‬ﻭﻛﺘﺎﺑﺔ ﺍﻷﺭﻗﺎﻡ‬

‫‪323‬‬

‫‪ 5.6‬ﺍﳍﻨﻮﺩ ﻭﺍﻟﺘﺮﻗﻴﻢ‬

‫‪325‬‬

‫‪ 6.6‬ﺍﻷﺭﻗﺎﻡ ﻭﺍﳊﺴﺎﺏ ﻋﻨﺪ ﺍﻟﻌﺮﺏ‬

‫‪327‬‬

‫‪ 7.6‬ﺇﻫﺘﻤﺎﻡ ﻋﻠﻤﺎﺋﻨﺎ ﺑﺎﻟﻌﺪﺩ ﻭﺍﳊﺴﺎﺏ‬

‫‪330‬‬

‫‪ 8.6‬ﻗﺒﻞ ﺍﳌﻴﻼﺩ ‪ ...‬ﻛﺎﻥ ﺍ ِﳌﻌ‪‬ﺪﺍﺩ‬

‫‪331‬‬

‫‪ 9.6‬ﺍﳊﺎﺟﺔ ﺃﻡ ﺍﻻﺧﺘﺮﺍﻉ‬

‫‪333‬‬

‫‪ 10.6‬ﺑﺪﺍﻳﺔ ﻣﺸﻮﺍﺭ ﺍﻵﻟﺔ‬

‫‪334‬‬

‫‪ 11.6‬ﻭﻳﺘﻮﺍﺻﻞ ﻣﺸﻮﺍﺭ ﺍﻻﺑﺘﻜﺎﺭﺍﺕ‬

‫‪336‬‬

‫ﺍﳌﺮﺍﺟﻊ‬

‫‪341‬‬

‫‪159‬‬

‫ﺗﻘﺪﱘ‪:‬‬ ‫ﹸﻛﺘِﺐ ﻫﺬﺍ ﺍﻟﺪﺭﺱ ﰲ ﺍﳉﱪ ﻭﻓﻖ ﺍﻟﱪﻧﺎﻣﺞ ﺍﳌﺴﻄﺮ ﻣﻦ ﻃﺮﻑ ﻭﺯﺍﺭﺓ‬ ‫ﺍﻟﺘﺮﺑﻴﺔ ﺍﻟﻮﻃﻨﻴﺔ ﺍﳍﺎﺩﻑ ﺇﱃ ﺗﻜﻮﻳﻦ ﻣﻔﺘﺸﻲ ﺍﻟﺘﻌﻠﻴﻢ ﺍﳌﺘﻮﺳﻂ ﰲ ﻣﺎﺩﺓ‬ ‫ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‪ .‬ﻭﻗﺪ ﺍﺭﺗﺄﻳﻨﺎ ﺗﻘﺴﻴﻢ ﻫﺬﺍ ﺍﻟﱪﻧﺎﻣﺞ ﺇﱃ ﺛﻼﺙ ﻭﺣﺪﺍﺕ‪ ،‬ﻫﻲ ‪:‬‬ ‫‪ (1‬ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ‪،‬‬ ‫‪ (2‬ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ‪،‬‬ ‫‪ (3‬ﳎﻤﻮﻋﺎﺕ ﺍﻷﻋﺪﺍﺩ‪.‬‬ ‫ﻭﱂ ‪‬ﺘﻢ ﺑﺎﳉﺎﻧﺐ ﺍﻟﺘﺪﺭﻳﱯ )ﺍﻟﺘﻤﺎﺭﻳﻦ( ﺇﻻ ﰲ ﻣﻮﺍﻗﻊ ﻗﻠﻴﻠﺔ‪ ،‬ﻭﺃﻏﻔﻠﻨﺎ‬ ‫ﻫﺬﺍ ﺍﳌﻮﺿﻮﻉ‪ ،‬ﺳﻴﻤﺎ ﰲ ﺩﺭﺍﺳﺔ ﳎﻤﻮﻋﺎﺕ ﺍﻷﻋﺪﺍﺩ ﺣﻴﺚ ﺭﻛﺰﻧﺎ ﻋﻠﻰ‬ ‫ﻣﻔﻬﻮﻡ ﺍﻹﻧﺸﺎﺀ‪.‬‬ ‫ﰒ ﺇﻥ ﺍﻟﱪﻧﺎﻣﺞ ﻳﺸﲑ ﺇﱃ ﺍﳌﺪﺓ ﺍﳌﺨﺼﺼﺔ ﳌﺎﺩﺓ ﺍﳉﱪ ﻓﻴﺤﺪﺩﻫﺎ ﺑـ‬ ‫‪ 48‬ﺳﺎﻋﺔ ! ﻭﻫﻲ ﻣﺪﺓ ﻻ ﺗﺴﻤﺢ ﺑﺎﻟﺘﻄﺮﻕ ﻟﻜﻞ ﺍﳌﻮﺍﺿﻴﻊ ﺍﳌﻘﺮﺭﺓ‬ ‫ﺑﺎﻟﺘﻔﺼﻴﻞ‪ ،‬ﻭﻟﺬﻟﻚ ﻻ ﺑﺪ ﻣﻦ ﲣﺼﻴﺺ ﺍﻟﺒﻌﺾ ﻣﻨﻬﺎ )ﺳﻴﻤﺎ ﰲ ﻓﺼﻞ‬ ‫ﳎﻤﻮﻋﺎﺕ ﺍﻷﻋﺪﺍﺩ( ﻛﻤﺤﺎﻭﺭ ﻟﻠﺒﺤﻮﺙ ﻳﻘﺪﻣﻬﺎ ﺍﻟﻄﻠﺒﺔ ﺍﳌﻔﺘﺸﻮﻥ ﺧﻼﻝ‬ ‫ﺍﻟﺴﻨﺔ ﺍﻟﺪﺭﺍﺳﻴﺔ‪.‬‬ ‫ﻧﺘﻤﲎ ﺃﻥ ﻳﻜﻮﻥ ﻫﺬﺍ ﺍﻟﺪﺭﺱ ﻣﻔﻴﺪﺍ ﻟﻠﻄﺎﻟﺐ ﺍﳌﻔﺘﺶ‪.‬‬ ‫ﺃﺑﻮ ﺑﻜﺮ ﺧﺎﻟﺪ ﺳﻌﺪ ﺍﷲ‬ ‫ﻗﺴﻢ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‬ ‫ﺍﳌﺪﺭﺳﺔ ﺍﻟﻌﻠﻴﺎ ﻟﻸﺳﺎﺗﺬﺓ‪ ،‬ﺍﻟﻘﺒﺔ‪ ،‬ﺍﳉﺰﺍﺋﺮ‬

‫ﺍﻟﻔﺼﻞ ‪ : 1‬ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ‬

‫ﻣﻘﺪﻣﺔ‪:‬‬ ‫ﻳﺸﲑ ﺍﻟﱪﻧﺎﻣﺞ ﺍﳌﺴﻄﺮ ﰲ ﻓﺼﻞ ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ ﺇﱃ ﺃﻥ‬ ‫ﺍﻷﻣﺮ ﻳﺘﻌﻠﻖ ﲟﻮﺿﻮﻉ ﻳﻠﹼﻢ ﺑﻪ ﺍﻟﻄﺎﻟﺐ ﺍﳌﻔﺘﺶ‪ .‬ﻭﻋﻠﻴﻪ ﻓﺎﳌﻄﻠﻮﺏ ﻫﻮ‬ ‫ﺯﻳﺎﺩﺓ ﺗﺮﺳﻴﺦ ﺍﳌﻔﺎﻫﻴﻢ‪ .‬ﻭﻣﻦ ﻫﺬﺍ ﺍﳌﻨﻈﻮﺭ ﻓﻘﺪ ﺍﻛﺘﻔﻴﻨﺎ ﺑﺎﻟﺘﺬﻛﲑ ﺑﺄﺑﺮﺯ‬ ‫ﺍﻟﺘﻌﺎﺭﻳﻒ ﰲ ﺍﳌﻨﻄﻖ )ﺍﻟﻘﻀﻴﺔ‪ ،‬ﺍﻟﺮﻭﺍﺑﻂ‪ (... ،‬ﻭﺃﺩﳎﻨﺎ ﻧﺼﺎ ﻧﻘﺪﻳﺎ‬ ‫ﻣﻄﻮﻻ ﳋﹼﺺ ﺁﺭﺍﺀ ﺭﻳﺎﺿﻴﲔ ﻏﲑ ﻣﺄﻟﻮﻓﺔ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‪ ،‬ﻻ ﺷﻚ ﺃﻥ‬ ‫ﺍﻟﻘﺎﺭﺉ ﺳﻴﺠﺪ ﻓﻴﻪ ﺑﻌﺾ ﺍﻟﻐﺮﺍﺑﺔ‪ .‬ﻭﻣﻊ ﺫﻟﻚ ﻧﺮﻯ ﺃﻧﻪ ﻣﻦ ﺍﳌﻔﻴﺪ ﺃﻥ‬ ‫ﻳﺘﺄﻣﻞ ﻓﻴﻪ ﺍﻟﻄﺎﻟﺐ ﺍﳌﻔﺘﺶ ﻟﻴﺘﻌﺮﻑ ﻋﻠﻰ ﺣﺪﻭﺩ ﺍﳌﻨﻄﻖ ﻭﻣﺪﻯ ﺩﻗﺔ‬ ‫ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‪ .‬ﻛﻤﺎ ﺃﺷﺮﻧﺎ ﻟﺒﻌﺾ ﺃﳕﺎﻁ ﺍﻟﱪﻫﺎﻥ‪ ،‬ﻭﻫﻲ ﺃﻳﻀﺎ ﻣﻌﺮﻭﻓﺔ‬ ‫ﻟﺪﻯ ﺍﻟﻘﺎﺭﺉ‪.‬‬ ‫ﻭﱂ ﻧﺘﻮﺳ‪‬ﻊ ﰲ ﻣﻮﺿﻮﻉ ﺍﻟﻌﻼﻗﺎﺕ ﺣﻴﺚ ﻗﺪﻣﻨﺎ ﺑﺈﳚﺎﺯ‪ ،‬ﻣﻊ ﺑﻌﺾ‬ ‫ﺍﻷﻣﺜﻠﺔ‪ ،‬ﻋﻼﻗﱵ ﺍﻟﺘﻜﺎﻓﺆ ﻭﺍﻟﺘﺮﺗﻴﺐ‪ .‬ﻭﻻ ﺷﻚ ﺃﻥ ﺍﳌﺪﺭ‪‬ﺱ ﺳﻴﺘﻤﻜﹼﻦ‬ ‫ﻣﻦ ﺗﻘﺪﱘ ﻋﺪﺓ ﻣﻮﺍﺿﻴﻊ ﱂ ﻧﺘﻄﺮﻕ ﺇﻟﻴﻬﺎ ﻫﻨﺎ ﻛﻤﻮﺍﺿﻴﻊ ﻟﻠﺪﺭﺍﺳﺔ‬ ‫ﺧﻼﻝ ﺍﻟﺴﻨﺔ ﺍﻟﺪﺭﺍﺳﻴﺔ‪ .‬ﻭﻣﻦ ﺗﻠﻚ ﺍﳌﻮﺍﺿﻴﻊ ﳝﻜﻦ ﺍﻗﺘﺮﺍﺡ ﻧﻈﺮﻳﺔ‬ ‫ﺍ‪‬ﻤﻮﻋﺎﺕ‪ ،‬ﻭﺧﻮﺍﺹ ﺍﻟﺘﻄﺒﻴﻘﺎﺕ‪ ،‬ﻭﺍﻟﺘﻌﻤﻖ ﰲ ﻣﻮﺿﻮﻉ ﺍﳌﻨﻄﻖ‬ ‫ﺍﻟﺮﻳﺎﺿﻲ‪ ،‬ﻭﻣﻮﺍﺿﻴﻊ ﺃﺧﺮﻯ ﻛﺜﲑﺓ‪.‬‬

‫‪165‬‬


‫‪ .1‬ﺍﳌﻨﻄﻖ ‪:‬‬ ‫‪1.1‬‬

‫ﺗﻌﺎﺭﻳﻒ ‪:‬‬

‫ﺗﻌﺮﻳﻒ )ﺍﻟﻘﻀﻴﺔ(‬ ‫ﻧﺴﻤﻲ ﻗﻀﻴﺔ ﻛﻞ ﻋﺒﺎﺭﺓ ﲢﺘﻤﻞ ﺍﻟﺼﺤﺔ ﺃﻭ ﺍﳋﻄﺄ‪.‬‬ ‫ﺃﻣﺜﻠﺔ‬ ‫‪" (1‬ﺍﻟﻌﺒﺎﺭﺓ ‪ " 8 < 5‬ﻗﻀﻴﺔ‪ ،‬ﻭﻫﻲ ﻗﻀﻴﺔ ﺧﺎﻃﺌﺔ‪.‬‬ ‫‪" (2‬ﺍﻟﺮﺑﺎﻁ ﻋﺎﺻﻤﺔ ﺍﳌﻐﺮﺏ" ﻗﻀﻴﺔ‪ ،‬ﻭﻫﻲ ﻗﻀﻴﺔ ﺻﺤﻴﺤﺔ‪.‬‬ ‫‪ " ∃x : x 2 = y " (3‬ﻋﺒﺎﺭﺓ ﻻ ﲤﺜﹼﻞ ﻗﻀﻴﺔ ﻷﻧﻨﺎ ﻻ ﻧﺴﺘﻄﻴﻊ‬ ‫ﺍﻟﺒﺖ ﰲ ﺻﺤﺘﻬﺎ ﺃﻭ ﺧﻄﺌﻬﺎ ﻣﺎ ﱂ ‪‬ﻧ ‪‬ﺰﻭ‪‬ﺩ ﺑﺘﻮﺿﻴﺤﺎﺕ ﺇﺿﺎﻓﻴﺔ ﺣﻮﻝ‬ ‫ﺍﻟﻌﻨﺼﺮﻳﻦ‬

‫‪x‬‬

‫ﻭ ‪.y‬‬

‫ﻣﻼﺣﻈﺔ‬ ‫ﻻﺣﻆ ﺃﻥ ﺑﻌﺾ ﺍﻟﻘﻀﺎﻳﺎ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺗﻀﻢ ﻣﺘﻐﲑﺍﺕ‪،‬‬ ‫ﻭﻳﻨﺒﻐﻲ ﺍﳊﺼﻮﻝ ﻋﻠﻰ ﻣﻌﻠﻮﻣﺎﺕ ﺇﺿﺎﻓﻴﺔ ﻟﻠﺒﺖ ﰲ ﺻﺤﺘﻬﺎ‪.‬‬ ‫ﺗﻌﺮﻳﻒ )ﻧﻔﻲ ﺍﻟﻘﻀﻴﺔ(‬

‫‪166‬‬


‫ﻧﻔﻲ ﻗﻀﻴﺔ‬ ‫ﺗﻜﻮﻥ‬

‫‪P‬‬

‫‪P‬‬

‫ﻫﻲ ﺍﻟﻘﻀﻴﺔ ﺍﻟﱵ ﺗﻜﻮﻥ ﺻﺤﻴﺤﺔ ﻋﻨﺪﻣﺎ‬

‫ﺧﺎﻃﺌﺔ‪ ،‬ﻭﺗﻜﻮﻥ ﺧﺎﻃﺌﺔ ﻋﻨﺪﻣﺎ ﺗﻜﻮﻥ‬

‫ﻧﺮﻣﺰ ﻏﺎﻟﺒﺎ ﻟﻨﻔﻲ ﺍﻟﻘﻀﻴﺔ‬

‫‪P‬‬

‫ﺻﺤﻴﺤﺔ‪.‬‬

‫ﺑـ ‪. P‬‬

‫‪P‬‬

‫ﻣﺜﺎﻝ‪:‬‬ ‫‪ (1‬ﻧﻔﻲ ﺍﻟﻘﻀﻴﺔ "ﺍﻟﺮﺑﺎﻁ ﻋﺎﺻﻤﺔ ﺍﳌﻐﺮﺏ" ﻫﻲ " ﺍﻟﺮﺑﺎﻁ‬ ‫ﻟﻴﺴﺖ ﻋﺎﺻﻤﺔ ﺍﳌﻐﺮﺏ"‪.‬‬ ‫‪ (2‬ﻧﻌﺘﱪ ﰲ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ‬

‫ﺍﳊﻘﻴﻘﻴﺔ ‪ℝ‬‬

‫ﺍﻟﻘﻀﻴﺔ‬

‫‪P‬‬

‫ﺍﳌﺘﻤﺜﻠﺔ ﰲ ﺍﳋﺎﺻﻴﺔ "‪ . " x ∈ ℝ, x ≤ 2‬ﻋﻨﺪﺋﺬ ﻳﻜﻮﻥ ﻧﻔﻲ ﻫﺬﻩ‬ ‫ﺍﻟﻘﻀﻴﺔ "‪. " x ∈ ℝ, x > 2‬‬ ‫‪ (3‬ﳕﺜﹼﻞ ﻋﺎﺩﺓ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ﻗﻀﻴﺔ ﻭﻧﻔﻴﻬﺎ ﲜﺪﻭﻝ ﻛﻤﺎ ﻳﻠﻲ‪،‬‬ ‫ﺣﻴﺚ ﻳﺮﻣﺰ ‪ 1‬ﻟﺼﺤﺔ ﺍﻟﻘﻀﺎﻳﺎ ﻭ‬

‫‪0‬‬

‫ﳋﻄﺌﻬﺎ ‪:‬‬

‫‪P‬‬

‫‪P‬‬

‫‪0‬‬ ‫‪1‬‬

‫‪1‬‬ ‫‪0‬‬

‫ﺗﻌﺮﻳﻒ )ﺍﻟﻮﺻﻞ(‬ ‫ﻟﺘﻜﻦ‬

‫‪P‬‬

‫ﻭ‬

‫‪Q‬‬

‫ﻗﻀﻴﺘﲔ‪ .‬ﻭﺻﻞ ﺍﻟﻘﻀﻴﺘﲔ‬

‫‪P‬‬

‫ﻭ‬

‫‪Q‬‬

‫ﻫﻮ‬

‫ﺍﻟﻘﻀﻴﺔ‪ ،‬ﺫﺍﺕ ﺍﻟﺮﻣﺰ ‪ ، P ∧ Q‬ﺍﻟﱵ ﺗﻜﻮﻥ ﺻﺤﻴﺤﺔ ﺇﺫﺍ ﻭﻓﻘﻂ ﻛﺎﻧﺖ‬ ‫ﻛﻞ ﻣﻦ‬

‫‪P‬‬

‫ﻭ‬

‫‪Q‬‬

‫ﺻﺤﻴﺤﺔ‪.‬‬

‫ﺇﺫﺍ ﻛﺎﻥ ﻭﺻﻞ ﺍﻟﻘﻀﻴﺘﲔ‬

‫‪P‬‬

‫ﻣﺘﻨﺎﻗﻀﺘﺎﻥ )ﺃﻭ ﻏﲑ ﻣﻨﺴﺠﻤﺘﲔ(‪.‬‬ ‫‪167‬‬

‫ﻭ‬

‫‪Q‬‬

‫ﺧﺎﻃﺌﺎ ﻗﻠﻨﺎ ﺇﻥ‬

‫‪P‬‬

‫ﻭ‬

‫‪Q‬‬


‫ﻣﻼﺣﻈﺔ‬ ‫ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﺘﻌﺮﻳﻒ ﺍﻟﺴﺎﺑﻘﺔ ﻳﺆﺩﻱ ﺇﱃ ﺃﻥ ﻗﻀﻴﺔ ﺍﻟﻮﺻﻞ‬ ‫‪P∧Q‬‬

‫ﺧﺎﻃﺌﺔ ﰲ ﻛﻞ ﺣﺎﻟﺔ ﻣﻦ ﺍﳊﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪(1‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ‬

‫‪P‬‬

‫ﺧﺎﻃﺌﺔ‪.‬‬

‫‪(2‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ‬

‫‪Q‬‬

‫ﺧﺎﻃﺌﺔ‪.‬‬

‫‪(3‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ ﻛﻞﹼ ﻣﻦ ‪ P‬ﻭ ‪ Q‬ﺧﺎﻃﺌﺔ‪.‬‬

‫ﻣﺜﺎﻝ‬ ‫ﻧﻌﺘﱪ ﺍﻟﻘﻀﻴﺔ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ‬ ‫‪1‬‬ ‫‪2‬‬

‫‪1‬‬ ‫‪2‬‬

‫≤‪x‬‬

‫‪P‬‬

‫ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬

‫‪1‬‬ ‫‪2‬‬

‫≥‪x‬‬

‫‪ ، x ∈ ℚ,‬ﻭﺍﻟﻘﻀﻴﺔ‬

‫‪ . x ∈ ℚ,‬ﻋﻨﺪﺋﺬ ﺗﻜﻮﻥ ﻗﻀﻴﺔ ﺍﻟﻮﺻﻞ‬

‫‪P∧Q‬‬

‫‪Q‬‬

‫ﻫﻲ‬

‫‪. x ∈ ℚ,‬‬

‫=‪x‬‬

‫ﺗﻌﺮﻳﻒ )ﺍﻟﻔﺼﻞ(‬ ‫ﻟﺘﻜﻦ‬

‫‪P‬‬

‫ﻭ‬

‫‪Q‬‬

‫ﻗﻀﻴﺘﲔ‪ .‬ﻓﺼﻞ ﺍﻟﻘﻀﻴﺘﲔ‬

‫‪P‬‬

‫ﻭ‬

‫‪Q‬‬

‫ﻫﻮ‬

‫ﺍﻟﻘﻀﻴﺔ‪ ،‬ﺫﺍﺕ ﺍﻟﺮﻣﺰ ‪ ، P ∨ Q‬ﺍﻟﱵ ﺗﻜﻮﻥ ﺻﺤﻴﺤﺔ ﺇﺫﺍ ﺇﺣﺪﻯ‬ ‫ﺍﻟﻘﻀﻴﺘﲔ‬

‫‪P‬‬

‫ﻭ‬

‫‪Q‬‬

‫ﺃﻭ ﻛﻼﳘﺎ‪.‬‬

‫‪168‬‬


‫ﻣﻼﺣﻈﺔ‬ ‫ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﺘﻌﺮﻳﻒ ﺍﻟﺴﺎﺑﻖ ﻳﺆﺩﻱ ﺇﱃ ﺃﻥ ﻗﻀﻴﺔ ﺍﻟﻔﺼﻞ‬ ‫‪P∨Q‬‬

‫ﺧﺎﻃﺌﺔ ﺇﺫﺍ ﻛﺎﻧﺖ ﻛﻞﹼ ﻣﻦ ﺍﻟﻘﻀﻴﺘﲔ‬

‫‪ P‬ﻭ‪Q‬‬

‫ﺧﺎﻃﺌﺔ‪.‬‬ ‫ﻣﺜﺎﻝ‬ ‫ﻧﻌﺘﱪ ﺍﻟﻘﻀﻴﺔ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ‬ ‫‪1‬‬ ‫‪2‬‬

‫‪1‬‬ ‫"‬ ‫‪2‬‬

‫≠‪x‬‬

‫n‬‬

‫ﺗﺆﺩﻱ ﺇﱃ ‪ . a m− n = 1‬ﻭﻫﺬﺍ ﻳﺴﺘﻠﺰﻡ ﺃﻥ ﺍﻟﺰﻣﺮﺓ‬

‫)ﺭﺗﺒﺘﻬﺎ ﺃﺻﻐﺮ ﻣﻦ ﺍﻟﻌﺪﺩ‬

‫‪m−n‬‬

‫‪am = an‬‬

‫‪a‬‬

‫ﻣﻨﺘﻬﻴﺔ‬

‫ﺃﻭ ﺗﺴﺎﻭﻳﻪ( ‪ ...‬ﻭﳓﻦ ﺍﻓﺘﺮﺿﻨﺎ ﺃﻥ‬

‫ﻏﲑ ﻣﻨﺘﻬﻴﺔ‪.‬‬

‫‪218‬‬

‫ﻣﻊ‬ ‫‪a‬‬

‫ﺍﻟﻔﺼﻞ ‪: 2‬ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ‬

‫ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‪ ،‬ﻧﻼﺣﻆ ﺃﻥ ﻟﺪﻳﻨﺎ ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻋﺪﺩﻳﻦ ﺻﺤﻴﺤﲔ‬ ‫ﻭ‬

‫‪n‬‬

‫‪m‬‬

‫‪:‬‬ ‫‪ϕ ( n + m) = a n + m‬‬ ‫‪= an × am‬‬ ‫)‪= ϕ (n).ϕ (m‬‬

‫ﻭﺑﺎﻟﺘﺎﱄ ﻓﺎﻟﺘﻄﺒﻴﻖ‬ ‫)‪,.‬‬

‫ﺗﺸﺎﻛﻞ )ﺃﻱ ﲤﺎﺛﻞ ﺗﻘﺎﺑﻠﻲ( ﺑﲔ ﺍﻟﺰﻣﺮﺗﲔ‬

‫‪ϕ‬‬

‫ﻭ‬

‫) ‪( ℤ, +‬‬

‫‪.( a‬‬

‫‪ (2‬ﻟﺘﻜﻦ‬ ‫ﺍﻟﺰﻣﺮﺓ‬

‫‪a‬‬

‫ﺯﻣﺮﺓ ﻣﻮﻟﺪﺓ ﺑﻌﻨﺼﺮ ﻭﺍﺣﺪ ﻭﻣﻨﺘﻬﻴﺔ ﺭﺗﺒﺘﻬﺎ ‪ . n‬ﻧﻌﻠﻢ ﺃﻥ‬

‫) ‪(ℤ / nℤ, +‬‬

‫}‬

‫ﻣﻮﻟﺪﺓ ﺑﻌﻨﺼﺮ ﻭﺍﺣﺪ‪ ،‬ﺭﺗﺒﺘﻬﺎ ‪ ، n‬ﻭﳝﻜﻦ ﻛﺘﺎﺑﺔ‬

‫{‬

‫‪(ℤ / nℤ, + ) = 0,1, 2,..., n − 1‬‬

‫ﻭﺗﻌﺮﻳﻒ ﺍﻟﺘﻄﺒﻴﻖ ‪:‬‬ ‫‪.‬‬

‫)‪ϕ : (ℤ / nℤ, + ) → ( a ,.‬‬ ‫‪p ֏ ap‬‬

‫ﺗﺄﻛﺪ‪ -‬ﻛﻤﺎ ﺟﺎﺀ ﰲ ﺍﻟﻘﻀﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ‪ -‬ﻣﻦ ﺃﻥ‬ ‫ﺗﻘﺎﺑﻠﻲ( ﺑﲔ ﺍﻟﺰﻣﺮﺗﲔ‬

‫) ‪(ℤ / nℤ, +‬‬

‫ﻭ‬

‫)‪,.‬‬

‫‪ϕ‬‬

‫ﺗﺸﺎﻛﻞ )ﺃﻱ ﲤﺎﺛﻞ‬

‫‪.( a‬‬

‫ﺗﻘﻮﻝ ﻧﻈﺮﻳﺔ ﻻﻏﺮﺍﻧﺞ ﺇﻥ ﺭﺗﺒﺔ ﻋﻨﺼﺮ ﰲ ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﺗﻘﺴﻢ ﺭﺗﺒﺔ‬ ‫ﺍﻟﺰﻣﺮﺓ‪ .‬ﳝﻜﻦ ﺃﻥ ﻧﺼﻴﻎ ﻫﺬﻩ ﻧﻈﺮﻳﺔ ﺃﻳﻀﺎ ﻛﻤﺎ ﻳﻠﻲ ‪ :‬ﺇﺫﺍ ﻛﺎﻧﺖ‬ ‫ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ )ﻣﻦ ﺯﻣﺮﺓ‬

‫‪G‬‬

‫‪m‬‬

‫ﻣﻌﻄﺎﺓ ﺭﺗﺒﺘﻬﺎ ‪ ( n‬ﻣﻮﻟﺪﺓ ﻋﻦ ﻋﻨﺼﺮ ﻓﺈﻥ‬

‫ﻳﻘﺴﻢ ‪. n‬‬ ‫‪219‬‬

‫ﺭﺗﺒﺔ‬ ‫‪m‬‬


‫ﻛﻴﻒ ﻧـﺜﺒﺖ ﺫﻟﻚ؟ ﻟﺘﻜﻦ‬

‫‪H‬‬

‫ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ ﺭﺗﺒﺘﻬﺎ‬

‫ﺭﺗﺒﺘﻬﺎ ‪ . n‬ﻣﺎﺫﺍ ﳝﻜﻦ ﺍﻟﻘﻮﻝ ﺣﻮﻝ ﳎﻤﻮﻋﺔ ﺍﻟﻌﻨﺎﺻﺮ‬ ‫} ‪x = { y ∈ G, y ∈ xH‬‬

‫ﺑـ‬

‫‪A‬‬

‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬

‫ﻣﻦ ﺃﺟﻞ ﺍﻟﻌﻨﺎﺻﺮ‬

‫}‬

‫{‬

‫‪A = x : x ∈G‬‬

‫‪x‬‬

‫‪x‬‬

‫‪m‬‬

‫ﻣﻦ ﺯﻣﺮﺓ‬


‫ﺍﳌﻨﺘﻤﻴﺔ ﺇﱃ ‪ G‬؟ ﺇﺫﺍ ﺭﻣﺰﻧﺎ‬

‫ﻓﺈﻧﻨﺎ ﻧﻼﺣﻆ ﺑﺄ‪‬ﺎ ﺗﺘﺸﻜﻞ ﻣﻦ‬

‫ﺃﺻﻨﺎﻑ ﺗﻜﺎﻓﺆ ﻭﺃﻥ ﻋﺪﺩ ﻋﻨﺎﺻﺮ ﻛﻞ ﺻﻨﻒ ﺗﻜﺎﻓﺆ ﻳﺴﺎﻭﻱ‬

‫‪m‬‬

‫ﻷﻥ ﺍﻟﺘﻄﺒﻴﻖ‬

‫‪f x : x → xH‬‬ ‫‪y ֏ xy‬‬

‫ﺗﻘﺎﺑﻞ )ﺗﺄﻛﺪ ﻣﻦ ﺫﻟﻚ(‪ .‬ﻭﻛﺬﻟﻚ ﺍﻷﻣﺮ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﻄﺒﻴﻖ‬ ‫‪g x : xH → H‬‬ ‫‪xy ֏ y.‬‬

‫ﻡ ﺇﻥ ﺍﻟﻌﻨﺎﺻﺮ‬

‫‪x‬‬

‫ﺗﺸﻜﻞ ﲡﺰﺋﺔ ﻟـ ‪ . G‬ﺃﻛﻤﻞ ﺍﻟﱪﻫﺎﻥ‪.‬‬

‫ﻫﻨﺎﻙ ﻣﻦ ﻳﻨﺺ ﻋﻠﻰ ﻧﻈﺮﻳﺔ ﻻﻏﺮﺍﻧﺞ ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﱄ ‪ :‬ﻟﺘﻜﻦ‬ ‫ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﻭ‬

‫‪H‬‬

‫ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ ﻣﻨﻬﺎ‪ .‬ﺇﻥ ﺭﺗﺒﺔ‬

‫‪H‬‬

‫‪G‬‬

‫ﺗﻘﺴﻢ ﺭﺗﺒﺔ ‪. G‬‬

‫ﲤﺮﻳﻦ ‪2‬‬ ‫ﻟﺘﻜﻦ‬

‫)‪(G,.‬‬

‫ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﺭﺗﺒﺘﻬﺎ‬

‫‪n‬‬

‫ﻟﻌﻨﺼﺮﻫﺎ ﺍﳊﻴﺎﺩﻱ ﺑـ ‪ .1‬ﺃﺛﺒﺖ ﺃﻥ ‪. a n = 1‬‬

‫‪220‬‬

‫ﻭ‬

‫‪a‬‬

‫ﻋﻨﺼﺮﺍ ﻛﻴﻔﻴﺎ ﻣﻨﻬﺎ‪ ،‬ﻧﺮﻣﺰ‬


‫ﺇﻟﻴﻚ ﰲ ﻧﻔﺲ ﺍﻟﺴﻴﺎﻕ ﻫﺬﻩ ﺍﻟﻨﻈﺮﻳﺔ ‪:‬‬ ‫ﻧﻈﺮﻳﺔ ﻛﻮﺷﻲ‬ ‫ﻟﻴﻜﻦ‬

‫‪p‬‬

‫ﻋﺪﺩﺍ ﺃﻭﻟﻴﺎ ﻳﻘﺴﻢ ﺭﺗﺒﺔ ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ‪. G‬‬

‫ﻋﻨﺪﺋﺬ ﻳﻮﺟﺪ ﻋﻠﻰ ﺍﻷﻗﻞ ﻋﻨﺼﺮ ﻣﻦ‬

‫‪G‬‬

‫ﺭﺗﺒﺘﻪ ‪. p‬‬

‫ﺗﻌﺮﻳﻒ )ﺯﻣﺮ ﺳﻴﻠﻮ(‬ ‫ﻟﻴﻜﻦ‬

‫‪p‬‬

‫ﻋﺪﺩﺍ ﺃﻭﻟﻴﺎ‪.‬‬

‫‪ (1‬ﻧﻘﻮﻝ ﻋﻦ ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﺇ‪‬ﺎ ‪ - p‬ﺯﻣﺮﺓ ﺇﻥ ﻛﺎﻧﺖ ﺭﺗﺒﺘﻬﺎ‬ ‫ﺗﺴﺎﻭﻱ ﻗﻮﺓ ﻟﻠﻌﺪﺩ‬

‫‪p‬‬

‫)ﺃﻱ ﺇﻥ ﻛﺘﺒﺖ ﺭﺗﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺸﻜﻞ ‪ ، p k‬ﺣﻴﺚ‬

‫‪k‬‬

‫ﻋﺪﺩ ﻃﺒﻴﻌﻲ(‪.‬‬ ‫‪ .2‬ﺇﺫﺍ ﻛﺎﻧﺖ ﺭﺗﺒﺔ ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ‬ ‫‪n = pα .m‬‬

‫ﺣﻴﺚ‬

‫‪p‬‬

‫ﻻ ﻳﻘﺴﻢ‬

‫‪m‬‬

‫‪G‬‬

‫ﺗﻜﺘﺐ ﻋﻠﻰ ﺍﻟﺸﻜﻞ‬

‫ﻓﺈﻧﻨﺎ ﻧﻘﻮﻝ ﻋﻦ ﻛﻞ ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ ﺇ‪‬ﺎ‬

‫‪ - p‬ﺯﻣﺮﺓ ﺳﻴﻠﻮ )ﺃﻭ ‪ - p‬ﺯﻣﺮﺓ ﺳﻴﻠﻮﻳﺔ ﺃﻭ ‪ - p‬ﺳﻴﻠﻮ( ﺇﻥ ﻛﺎﻧﺖ ﺭﺗﺒﺘﻬﺎ‬ ‫ﺗﺴﺎﻭﻱ ‪. pα‬‬ ‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻷﻭﱃ ﻟﺴﻴﻠﻮ‬ ‫ﺇﺫﺍ ﻛﺎﻧﺖ‬ ‫‪n = pα .m‬‬

‫ﺣﻴﺚ‬

‫‪G‬‬


‫‪n‬‬

‫ﻭﻛﺎﻥ‬

‫‪p‬‬

‫ﻋﺪﺩ ﺃﻭﱄ ﻻ ﻳﻘﺴﻢ‬

‫‪m‬‬

‫ﻓﺈﻥ‬

‫ﺳﻴﻠﻮﻳﺔ‪.‬‬

‫‪221‬‬

‫‪n‬‬

‫‪G‬‬

‫ﳛﻘﻖ ﺍﻟﻌﻼﻗﺔ‬ ‫ﲤﻠﻚ ﺯﻣﺮﺓ ‪- p‬‬


‫ﻣﻼﺣﻈﺔ‬ ‫ﻻﺣﻆ ﺃﻥ ﻫﺬﻩ ﺍﻟﻨﻈﺮﻳﺔ ﺗﺆﺩﻱ ﺇﱃ ﻧﻈﺮﻳﺔ ﻛﻮﺷﻲ‪ .‬ﻟﻨﻮﺿﺢ ﺫﻟﻚ ‪:‬‬ ‫ﻟﻴﻜﻦ‬ ‫ﻋﺪﺩﺍﻥ‬ ‫‪m‬‬

‫ﻋﺪﺩﺍ ﺃﻭﻟﻴﺎ ﻳﻘﺴﻢ ﺍﻟﺮﺗﺒﺔ‬

‫‪p‬‬

‫ﻃﺒﻴﻌﻴﺎﻥ ‪m‬‬

‫ﻭ ‪ α‬ﳛﻘﻘﺎﻥ ﺍﻟﻌﻼﻗﺔ‬

‫‪n‬‬

‫ﻟﺰﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ‪ . G‬ﺇﺫﻥ ﻳﻮﺟﺪ‬

‫‪n = pα .m‬‬

‫ﻣﻊ ﺍﻟﻘﻴﺪ‬

‫‪p‬‬

‫ﻻ ﻳﻘﺴﻢ‬

‫)ﺫﻟﻚ ﺃﻧﻪ ﺇﺫﺍ ﺣﺪﺙ ﺍﻟﻌﻜﺲ ﻓﻤﺎ ﻋﻠﻴﻨﺎ ﺳﻮﻯ ﺗﻐﻴﲑ ﺍﻷﺱ ‪ .( α‬ﻭﻣﻦ ﰒ‬

‫ﺗﺘﺤﻘﻖ ﺷﺮﻭﻁ ﻧﻈﺮﻳﺔ ﺳﻴﻠﻮ‪ .‬ﺇﺫﻥ‬

‫‪G‬‬

‫ﲤﻠﻚ ﺯﻣﺮﺓ ‪ - p‬ﺳﻴﻠﻮﻳﺔ‪ ،‬ﺃﻱ ﺯﻣﺮﺓ‬

‫ﺭﺗﺒﺘﻬﺎ ﻣﻦ ﺍﻟﺸﻜﻞ ‪. p β‬‬ ‫ﻟﻜﻦ‬

‫‪p‬‬

‫ﻋﺪﺩ ﺃﻭﱄ ﻳﻘﺴﻢ ﺗﻠﻚ ﺍﻟﺮﺗﺒﺔ‪ .‬ﻭﺣﺴﺐ ﻧﻈﺮﻳﺔ ﻻﻏﺮﺍﻧﺞ‬

‫ﻓﺈﻧﻪ ﺗﻮﺟﺪ ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ‬ ‫‪H‬‬

‫‪H‬‬

‫ﺭﺗﺒﺘﻬﺎ ﺍﻟﻘﺎﺳﻢ ‪ . p‬ﻭﻣﻦ ﰒ ﻳﻮﺟﺪ ﻋﻨﺼﺮ‬

‫ﺭﺗﺒﺘﻪ ‪ . p‬ﳌﺎﺫﺍ؟ ﻧﻌﺘﱪ ﻋﻨﺼﺮﺍ‬

‫‪a‬‬

‫ﻣﻦ‬

‫‪H‬‬

‫ﳜﺘﻠﻒ ﻋﻦ ﺍﻟﻌﻨﺼﺮ ﺍﳊﻴﺎﺩﻱ‪.‬‬

‫ﻣﺎ ﻫﻲ ﺭﺗﺒﺘﻪ؟ ﻫﻞ ﳝﻜﻦ ﺃﻥ ﺗﻜﻮﻥ ﺃﺻﻐﺮ ﲤﺎﻣﺎ ﻣﻦ ‪ p‬؟ ﻻ ! ﻷﻥ‬ ‫ﻭ ‪ a‬ﻣﻦ‬

‫‪H‬‬

‫‪a‬‬

‫‪p‬‬

‫ﳜﺘﻠﻒ ﻋﻦ ﺍﻟﻌﻨﺼﺮ ﺍﳊﻴﺎﺩﻱ‪ .‬ﻫﻞ ﳝﻜﻦ ﺃﻥ ﺗﻜﻮﻥ ﺭﺗﺒﺔ‬

‫ﺃﻭﱄ‬ ‫‪a‬‬

‫ﺃﻛﱪ‬

‫ﲤﺎﻣﺎ ﻣﻦ ‪ p‬؟ ﻻ ! ﻷﻧﻪ ﻻ ﳝﻜﻦ ﺃﻥ ﺗﺘﺠﺎﻭﺯ ﺭﺗﺒﺔ ‪. H‬‬ ‫ﺗﻌﺮﻳﻒ )ﺍﻟﺘﺮﺍﻓﻖ(‬ ‫ﻧﻘﻮﻝ ﻋﻦ ﺯﻣﺮﺗﲔ ﺟﺰﺋﻴﺘﲔ‬ ‫ﻋﻨﺼﺮ‬

‫‪a‬‬

‫ﻣﻦ‬

‫‪G‬‬

‫‪H‬‬

‫ﻭ‬

‫‪K‬‬

‫ﻣﻦ‬

‫ﲝﻴﺚ ‪. aHa −1 = K‬‬

‫‪222‬‬

‫ﻣﻦ‬

‫)‪(G,.‬‬

‫ﺇ‪‬ﻤﺎ ﻣﺘﺮﺍﻓﻘﺎﻥ ﺇﻥ ﻭﺟﺪ‬


‫ﲤﺮﻳﻦ ‪3‬‬ ‫ﺇﺫﺍ ﻛﺎﻧﺖ ﺯﻣﺮﺗﺎﻥ ﺟﺰﺋﻴﺘﺎﻥ ﻣﻦ ﺯﻣﺮﺓ‬

‫ﻣﺘﺮﺍﻓﻘﺘﲔ ﻓﺈ‪‬ﻤﺎ‬

‫)‪(G,.‬‬

‫ﻣﺘﻘﺎﺑﻼﻥ )ﻭﻣﻦ ﰒ ﻓﻠﻬﻤﺎ ﻧﻔﺲ ﺍﻟﺮﺗﺒﺔ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﱵ ﺗﻜﻮﻥ ﻓﻴﻬﺎ ﺍﻟﺰﻣﺮﺓ‬ ‫)‪(G,.‬‬

‫ﻣﻨﺘﻬﻴﺔ(‪.‬‬

‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺜﺎﻧﻴﺔ ﻟﺴﻴﻠﻮ‬ ‫ﺇﺫﺍ ﻛﺎﻧﺖ‬ ‫‪n = pα .m‬‬

‫ﺣﻴﺚ‬

‫‪G‬‬

‫ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﺭﺗﺒﺘﻬﺎ ‪ ، n‬ﻭﻛﺎﻥ‬ ‫ﻋﺪﺩ ﺃﻭﱄ ﻻ ﻳﻘﺴﻢ‬

‫‪p‬‬

‫‪m‬‬

‫ﻓﺈﻥ‬

‫‪n‬‬

‫‪G‬‬

‫ﳛﻘﻖ ﺍﻟﻌﻼﻗﺔ‬

‫ﳝﺘﻠﻚ )ﺣﺴﺐ‬

‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ( ﺯﻣﺮﺓ ‪ - p‬ﺳﻴﻠﻮﻳﺔ ﻋﻠﻰ ﺍﻷﻗﻞ‪ .‬ﺇﻥ ﺍﻟﺰﻣﺮ ‪ - p‬ﺍﻟﺴﻴﻠﻮﻳﺔ‬ ‫ﻛﻠﻬﺎ ﻣﺘﺮﺍﻓﻘﺔ ﻭﻋﺪﺩﻫﺎ ﻳﻘﺴﻢ ‪. n‬‬ ‫ﲤﺮﻳﻦ ‪4‬‬ ‫ﺃﺛﺒﺖ ﺃﻥ ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ‬ ‫ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﻥ ﻛﺎﻧﺖ‬

‫‪H‬‬

‫‪H‬‬

‫‪ - p‬ﺳﻴﻠﻮﻳﺔ ﻣﻦ ﺯﻣﺮﺓ‬

‫‪G‬‬

‫ﺗﻜﻮﻥ ﳑﻴﺰﺓ‬

‫ﺍﻟﺰﻣﺮﺓ ﺍﻟـ ‪ - p‬ﺳﻴﻠﻮﻳﺔ ﺍﻟﻮﺣﻴﺪﺓ‪.‬‬

‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺜﺎﻟﺜﺔ ﻟﺴﻴﻠﻮ‬ ‫ﺇﺫﺍ ﻛﺎﻧﺖ‬ ‫‪n = pα .m‬‬

‫ﺣﻴﺚ‬

‫‪G‬‬


‫‪n‬‬

‫‪p‬‬


‫‪m‬‬

‫ﻭﻛﺎﻥ‬

‫‪n‬‬


‫ﻓﺈﻥ‬

‫‪G‬‬

‫ﳝﺘﻠﻚ )ﺣﺴﺐ‬

‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ( ﺯﻣﺮﺓ ‪ - p‬ﺳﻴﻠﻮﻳﺔ ﻋﻠﻰ ﺍﻷﻗﻞ‪ .‬ﺇﻥ ﺍﻟﻌﺪﺩ‬ ‫ﺍﻟﺴﻴﻠﻮﻳﺔ ﳛﻘﻖ ﺍﻟﻌﻼﻗﺔ )‪) ، k ≡ 1(mod p‬ﺃﻱ ‪.( k = l. p + 1‬‬

‫‪223‬‬

‫‪k‬‬

‫ﻟﻠﺰﻣﺮ ‪- p‬‬


‫ﺗﻄﺒﻴﻖ‬ ‫ﳝﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ ﻧﻈﺮﻳﺎﺕ ﺳﻴﻠﻮ ﻹﺛﺒﺎﺕ ﺃﻥ ﻛﻞ ﺯﻣﺮﺓ ﺭﺗﺒﺘﻬﺎ ﺗﺴﺎﻭﻱ‬ ‫‪63‬‬

‫ﻻ ﳝﻜﻦ ﺃﻥ ﺗﻜﻮﻥ ﺑﺴﻴﻄﺔ‪ .‬ﻟﻨﺮ ﺫﻟﻚ ‪:‬‬ ‫ﻧﺬﻛﺮ ﺃﻧﻨﺎ ﻧﻘﻮﻝ ﻋﻦ ﺯﻣﺮﺓ ﺇ‪‬ﺎ ﺑﺴﻴﻄﺔ ﺇﻥ ﻛﺎﻧﺖ ﺍﻟﺰﻣﺮ ﺍﳉﺰﺋﻴﺔ‬

‫ﺍﳌﻤﻴﺰﺓ ﺍﻟﻮﺣﻴﺪﺓ ﻫﻲ ﺍﻟﺰﻣﺮﺓ ﻧﻔﺴﻬﺎ ﻭﺯﻣﺮﺓ ﺍﻟﻌﻨﺼﺮ ﺍﳊﻴﺎﺩﻱ‪ .‬ﻧﻌﺘﱪ ﻓﻴﻤﺎ ﺳﺒﻖ‬ ‫‪ . p = 7‬ﻛﻢ ﻳﺒﻠﻎ ﻋﺪﺩ ﺍﻟﺰﻣﺮ ﺍﳉﺰﺋﻴﺔ ‪- 7‬ﺳﻴﻠﻮﻳﺔ؟ ﺇﻧﻪ ﻋﺪﺩ‬

‫‪k‬‬

‫ﻳﻘﺴﻢ‬

‫‪63‬‬

‫ﻭﻳﻜﺘﺐ ﻋﻠﻰ ﺍﻟﺸﻜﻞ ‪ . k = l.7 + 1‬ﺍﻟﻌﺪﺩ ﺍﻟﻮﺣﻴﺪ ﺍﻟﺬﻱ ﳛﻘﻖ ﺫﻟﻚ ﻫﻮ‬ ‫‪ . k = 1‬ﺇﺫ ﻫﻨﺎﻙ ﺯﻣﺮﺓ ‪ - 7‬ﺳﻴﻠﻮﻳﺔ ﻭﺍﺣﺪﺓ‪ .‬ﻭﺣﺴﺐ ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺴﺎﺑﻖ ﻓﻬﺬﺍ‬ ‫ﻳﺴﺘﻠﺰﻡ ﺃﻥ ﻫﺬﻩ ﺍﻟﺰﻣﺮﺓ ﳑﻴﺰﺓ‪ .‬ﻭﻣﻦ ﰒ ﻓﺎﻟﺰﻣﺮﺓ ﺍﻟﱵ ﺭﺗﺒﺘﻬﺎ‬

‫‪63‬‬

‫ﻟﻴﺴﺖ‬

‫ﺑﺴﻴﻄﺔ‪.‬‬ ‫ﲤﺮﻳﻦ ‪5‬‬ ‫ﻟﺘﻜﻦ‬ ‫‪n = pα .m‬‬

‫‪G‬‬

‫ﺣﻴﺚ‬

‫ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﺭﺗﺒﺘﻬﺎ‬ ‫‪p‬‬


‫ﺍﻟـ ‪ - p‬ﺳﻴﻠﻮﻳﺔ‪.‬‬ ‫ﺃﺛﺒﺖ ﺃﻥ‬

‫‪k‬‬

‫ﺃﻭﱄ ﻣﻊ ‪.. p‬‬

‫‪224‬‬

‫ﻭﻛﺎﻥ‬

‫‪n‬‬ ‫‪m‬‬

‫‪n‬‬


‫‪ .‬ﻭﻟﻴﻜﻦ‬

‫ﻋﺪﺩ ﺍﻟﺰﻣﺮ‬

‫‪k‬‬


‫‪ 5.1‬ﲤﺎﺭﻳﻦ ﺃﺳﺎﺳﻴﺔ ‪:‬‬ ‫ﲤﺮﻳﻦ ‪1‬‬ ‫ﺃﺛﺒﺖ ﺃﻥ‬

‫)‪(G,.‬‬

‫ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ ﻣﻦ‬

‫)‪(G,.‬‬

‫ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻥ ‪:‬‬

‫‪∀x ∈ H , ∀y ∈ H : x. y ∈ H ,‬‬ ‫‪‬‬ ‫‪−1‬‬ ‫‪x ∈ H ⇒ x ∈ H.‬‬

‫ﲤﺮﻳﻦ ‪2‬‬ ‫ﺃﺛﺒﺖ ﺃﻥ ﺗﻘﺎﻃﻊ ﺯﻣﺮﺗﲔ ﺟﺰﺋﻴﺘﲔ ﻫﻮ ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ‪.‬‬ ‫ﲤﺮﻳﻦ ‪3‬‬ ‫ﻟﺘﻜﻦ‬

‫)‪(G,.‬‬

‫ﺯﻣﺮﺓ ﻭ‬

‫ﻫﻲ ﳎﻤﻮﻋﺔ ﻋﻨﺎﺻﺮ‬

‫‪G‬‬

‫ﺟﺰﺀﺍ ﻣﻦ ‪ . G‬ﺃﺛﺒﺖ ﺃﻥ ﺍﻟﺰﻣﺮﺓ ﺍﳌ ‪‬ﻮﻟﱠﺪﺓ‬

‫‪A‬‬

‫>‪< A‬‬

‫ﺍﻟﱵ ﺗﻜﺘﺐ ﻋﻠﻰ ﺷﻜﻞ ﺟﺪﺍﺀ ﺃﺳﺲ ﻟﻌﻨﺎﺻﺮ ‪. A‬‬

‫ﲤﺮﻳﻦ ‪4‬‬ ‫ﻟﺘﻜﻦ‬

‫)‪(G,.‬‬

‫‪ (1‬ﺃﺛﺒﺖ ﺃﻥ‬

‫ﺯﻣﺮﺓ ﻭ‬ ‫)‪( H ,.‬‬

‫)‪( H ,.‬‬

‫ﺯﻣﺮﺓ ﺟﺰﺋﻴﺔ ﻣﻨﻪ‪.‬‬

‫ﺗﻜﻮﻥ ﳑﻴﺰﺓ ﺇﺫﺍ ﻛﺎﻥ ‪:‬‬ ‫‪∀x ∈ G, xHx −1 = H .‬‬

‫‪ (2‬ﺑﻴ‪‬ﻦ ﺃﻥ ﺍﻟﺸﺮﻁ ﺍﻟﺴﺎﺑﻖ ﻳﻜﺎﻓﺊ ‪:‬‬ ‫‪∀x ∈ G , xH = Hx.‬‬

‫ﲤﺮﻳﻦ ‪5‬‬ ‫‪ (1‬ﺗﺄﻛﺪ ﻣﻦ ﺃﻥ ﺍﳉﺪﺍﺀ ﺯﻣﺮﺗﲔ ﳝﺜﻞ ﺯﻣﺮﺓ‪.‬‬ ‫‪ (2‬ﻟﺘﻜﻦ‬

‫‪n (Gi ,.i )i =1,...,n‬‬

‫ﺯﻣﺮﺓ‪ .‬ﻭﺿﺢ ﻛﻴﻒ ﳝﻜﻦ ﺗﻌﺮﻳﻒ ﺯﻣﺮﺓ ﺍﳉﺪﺍﺀ‬

‫ﺍﻟﺪﻳﻜﺎﺭﰐ ‪. G1 × G2 × ... × Gn‬‬ ‫‪225‬‬


‫ﲤﺮﻳﻦ ‪6‬‬ ‫ﺃﺛﺒﺖ ﺃﻧﻪ ﺇﺫﺍ ﻛﺎﻧﺖ‬

‫)‪(G,.‬‬

‫ﺯﻣﺮﺓ ﻭ‬

‫‪a ∈G‬‬

‫ﻓﺈﻥ ‪:‬‬

‫‪∀m, ∀n ∈ ℤ, a m .a n = a m + n‬‬ ‫‪( a m ) n = a m .n .‬‬

‫ﻣﻊ ﺍﻟﻌﻠﻢ ﺃﻥ ‪ 1‬ﻫﻮ ﺍﻟﻌﻨﺼﺮ ﺍﳊﻴﺎﺩﻱ ﰲ ﺍﻟﺰﻣﺮﺓ ﻭﺃﻥ ‪:‬‬ ‫‪a0 = 1‬‬ ‫‪∀p ∈ ℤ − , a p = (a −1 ) − p = (a −1 ).(a −1 )...(a −1 ) .‬‬ ‫ ‬ ‫) ‪ ( − p‬ة‬

‫ﲤﺮﻳﻦ ‪7‬‬ ‫ﺇﺫﺍ ﻛﺎﻧﺖ‬ ‫ﻋﺪﺩ ﻃﺒﻴﻌﻲ‬

‫)‪(G,.‬‬

‫‪0