...... ﻗﺎﻋﺪﺓ ﺍﻷﺳﻄﻮﺍﻧﺔ . ﻭﻟﺬﻟﻚ ﻓﻌﻤﻠﻴﺔ ﺍﳌﻜﺎﻣﻠﺔ ﺍﳌﺬﻛﻮﺭﺓ ﺗﻌﻄﻲ. ﻣﺒﺎﺷﺮﺓ. : . 2. 2. 0
h. V r dz. r h π π. = = ∫ ...... Querre J. : Cours d'Algèbre, Masson, 1976. 9.
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و ارس ﺍﻟﻔﺼﻞ :1ﺍﳌﺘﺘﺎﻟﻴﺎﺕ
09
-1ﻣﻘﺪﻣﺔ
14
-2ﺗﻌﺎﺭﻳﻒ
20
-3ﻧﺘﺎﺋﺞ ﻭﺧﻮﺍﺹ
30
-4ﺗﻄﺒﻴﻖ
39
-5ﺧﻮﺍﺹ ﺃﺧﺮﻯ ﻟﻠﺪﺍﻟﺔ ﺍﻷﺳﻴﺔ
41
-6ﻣﻮﺿﻮﻉ ﻟﻠﺪﺭﺍﺳﺔ
42
-7ﲤﺎﺭﻳﻦ
53
ﺍﻟﻔﺼﻞ :2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
55
-1ﻣﻘﺪﻣﺔ
56
-2ﻋﻤﻮﻣﻴﺎﺕ ﻋﻠﻰ ﺍﻟﺪﻭﺍﻝ
64
-3ﺍﻟﻨﻬﺎﻳﺎﺕ
75
-4ﺍﻻﺳﺘﻤﺮﺍﺭ
75
-5ﺍﻻﺷﺘﻘﺎﻕ
86
ﺍﻟﻔﺼﻞ :3ﺍﳊﺴﺎﺏ ﺍﻟﺘﻜﺎﻣﻠﻲ
97
-1ﻣﻘﺪﻣﺔ
99
-2ﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺮﳝﺎﱐ
100
-3ﺍﻟﺘﻜﺎﻣﻞ ﻏﲑ ﺍﶈﺪﺩ
114
-4ﻣﻦ ﻃﺮﻕ ﺍﳌﻜﺎﻣﻠﺔ
119
-5ﻃﻮﻝ ﻗﻮﺱ ﻣﻨﺤﲎ
128
-6ﺣﺴﺎﺏ ﺑﻌﺾ ﺍﳌﺴﺎﺣﺎﺕ ﻭﺍﳊﺠﻮﻡ
138
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 ﻓﺤﺴﺎﺎ 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 ﻭﺍﺴﻤﺎﺕ 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 ﺃﺎ ﺗﺘﻤﻴﺰ 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 ...ﺇﻢ 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(؟ ﺇﺎ 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
ا : 1ا ت
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؟ ﺇﺎ 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 ﺍﻤﻮﻋﺔ 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 ﺃﺎ 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 ﻳﺘﺄﺧﺮ 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
ﺇﺎ ﻣﺘﺰﺍﻳﺪﺓ )ﻣﺘﻨﺎﻗﺼﺔ ،ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ(
،ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ( ﻣﻦ ﺃﺟﻞ ﻛﻞ
n
ﰲ
ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻄﺒﻴﻌﻴﺔ. -ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ
ﺣﻘﻴﻘﻴﺔ (un ) n∈ℕ
ﺇﺎ ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﻋﻠﻰ ﺇﺫﺍ ﻭﺟﺪ
ﺛﺎﺑﺖ Mﻣﻮﺟﺐ ﲝﻴﺚ ∀n ∈ ℕ, un ≤ M . ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﺣﻘﻴﻘﻴﺔ (un )n∈ℕﺇﺎ ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﺩﱏ ﺇﺫﺍ ﻭﺟﺪ ﺛﺎﺑﺖM
ﲝﻴﺚ
. ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﻋﺪﺩﻳﺔ (un )n∈ℕﺇﺎ 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 ﺍﻤﻮﻋﺔ
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
-ﺍﳌﺘﺘﺎﻟﻴﺔ
un = n
-ﺍﳌﺘﺘﺎﻟﻴﺔ
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 →+
ﺇﺎ ﻣﺘﻘﺎﺭﺑﺔ ﺇﺫﺍ ﻭﺟﺪ ﻋﺪﺩ
)(un
u
ﲝﻴﺚ
ﺃﻱ
un − u < ε .
∀ε > 0,
⇒ ∃n0 ∈ ℕ : n ≥ n0
ﻧﻘﻮﻝ ﻋﻦ ﻣﺘﺘﺎﻟﻴﺔ ﺇﺎ 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 ﺇﺎﻣﺘﺒﺎﻋﺪﺓ .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
ﺍﳌﻌﺮﻓﺔ ﺑـ
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 ﺇﺎ ﺟﺰﺋﻴﺔ ﻣﻦ ( (un )n∈ℕﺇﺫﺍ ﻭﺟﺪ ﺗﻄﺒﻴﻖ
f :ℕ → ℕ
ﻣﺘﺰﺍﻳﺪ ﲤﺎﻣﺎ ﲝﻴﺚ vn = u f ( n ) .
) (vn
ﺇﺎ
∀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
ﺇﺎ 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 ﺇﺛﺒﺎﺎ : 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 ﺃﺎ ﺗﺆﻭﻝ ﺇﱃ
∞+
)ﻭﻫﺬﺍ ﻻ ﻳﻌﲏ ﺃﺎ ﻣﺘﻘﺎﺭﺑﺔ ﻷﻥ ﺍﻟﺘﻘﺎﺭﺏ ﻳﺴﺘﻮﺟﺐ ﺃﻥ ﺗﻜﻮﻥ ﺍﻟﻨﻬﺎﻳﺔ ﰲ ℝ
...ﻭ
∞+
ﻻ ﻳﻨﺘﻤﻲ ﺇﱃ .( ℝﻛﻴﻒ ﻧﱪﺭ ﺃﺎ 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 ﺃﺎ ﺗﺴﻤﺢ ﺑﺪﺭﺍﺳﺔ ﻃﺒﻴﻌﺔ ﻣﺘﺘﺎﻟﻴﺔ )ﺃﻱ ﻣﻌﺮﻓﺔ ﻣﺎ ﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺘﻘﺎﺭﺑﺔ ﺃﻡ ﻣﺘﺒﺎﻋﺪﺓ( ﺩﻭﻥ ﻣﻌﺮﻓﺔ ﺎﻳﺘﻬﺎ )ﰲ ﺣﺎﻟﺔ ﺗﻘﺎﺭﺎ(.
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
⇒
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
ﺛﺎﻧﻴﺎ : :ﻧﻔﺮﺽ ﺃﻧﻪ ﻣﻬﻤﺎ ﻛﺎﻧﺖ f ( xn ) = c
، nlimﻭﺃﻥ ∞→+
ﺍﳌﺘﺘﺎﻟﻴﺔ ) ( xn
f ( x) ≠ c
ﻣﻦ
I
ﺍﳌﺘﻘﺎﺭﺑﺔ ﳓﻮ
a
ﻓﺈﻥ
. limﻧﻌﺒﺮ 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
∞+
ﺇﺎ 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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﺑﺪﻻﻟﺔ ﻗﻴﻢ
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
∞−
∞−
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
. 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
ﺑﺪﻻﻟﺔ ﻗﻴﻢ
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 . ﻭﻧﻌﺮﻑ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻋﻠﻰ ﺍﺎﻝ Iﺇﻥ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﻋﻨﺪ ﻛﻞ ﻧﻘﻄﺔ ﻣﻦ . Iﻭﻧﻌﺮﻑ 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 ﻧﻔﺴﺮ 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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﺇﺎ ﺩﺍﻟﺔ ﻏﲑ ﻣﺴﺘﻤﺮﺓ ﻋﻨﺪ ﻋﺪﺩ ﻏﲑ ﻣﻨﺘﻪ ﻣﻦ ﺍﻟﻨﻘﺎﻁ .ﳎﻤﻮﻋﺔ ﻫﺬﻩ ﺍﻟﻨﻘﺎﻁ ﻫﻲ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ.
ﺑﻴﺎﻥ ا ا g
79
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
ﻣﺜﺎﻝ 3 ﳝﻜﻦ ﺃﻳﻀﺎ ﺍﻟﺘﻔﻜﲑ ﰲ ﺍﻟﺪﺍﻟﺔ ﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ﺍﺎﻝ ] [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 ﺃﺎ ﺩﺍﻟﺔ ﻻ ﳝﻜﻦ ﺭﺳﻢ ﺑﻴﺎﺎ ﻭﻫﻲ ﻏﲑ ﻣﺴﺘﻤﺮﺓ ﰲ ﺃﻳﺔ ﻧﻘﻄﺔ ﻣﻦ . ℝ ﺩﻋﻨﺎ ﻧﻨﻬﻲ ﻫﺬﺍ ﺍﳌﻘﻄﻊ ﺑﺘﻘﺪﱘ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺗﺎﺭﻛﲔ ﺑﺮﻫﺎﺎ ﻟﻠﻘﺎﺭﺉ.
80
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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 ﺃﺎ 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 ﻭﺎﻳﺘﻬﺎ 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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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 .ﻗﺪﻡ 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
ﻣﻦ ] [a, bﲝﻴﺚ
] 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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﻣﻦ ] [a, bﲝﻴﺚ
= ) . f (ξ
)sup f ( x
] x∈[a ,b
ﻳﺪﺭﻙ ﺣﺪﻩ ﺍﻷﻋﻠﻰ.
* ﻗﺪﻡ ﺗﻔﺎﺻﻴﻞ ﺇﺩﺭﺍﻙ
f
ﳊﺪﻩ ﺍﻷﺩﱏ.
ﻣﻼﺣﻈﺔ ﻻﺣﻆ ﺃﻥ ﳏﺪﻭﺩﻳﺔ ﺍﺎﻝ ﻣﻬﻤﺔ ﰲ ﻫﺬﻩ ﺍﻟﻨﻈﺮﻳﺔ .ﻟﻠﺘﺄﻛﺪ ﻣﻦ ﺫﻟﻚ ﺧﺬ ﻣﺜﻼ ﺇﺣﺪﻯ ﺍﻟﺪﺍﻟﺘﲔ :ﺩﺍﻟﺔ ﺍﻟﻈﻞ ﻋﻠﻰ ﺍﺎﻝ
π π − 2 , 2
ﺃﻭ
ﺍﻟﺪﺍﻟﺔ g
ﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ] ]0,1ﺑـ . g ( x) = 1 x
ﻻﺣﻆ ﺃﻳﻀﺎ ﺃﻥ ﻏﻠﻖ ﺍﺎﻝ ﻣﻬﻢ ﰲ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ .ﻟﻠﺘﺄﻛﺪ ﻣﻦ ﺫﻟﻚ ﺍﻋﺘﱪ ﻛﻤﺜﺎﻝ
ﺍﻟﺪﺍﻟﺔ g
ﺃﻭ ﺍﻟﺪﺍﻟﺔ hﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ﺍﺎﻝ [ [1,2004ﺑـ
h( x) = x
ﻓﻬﻲ ﻻ ﺗﺪﺭﻙ ﺣﺪﻫﺎ ﺍﻷﻋﻠﻰ ﰲ ﺍﺎﻝ 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 .ﻭﻧﻌﺒﺮ ﻋﻦ ﺫﻟﻚ ﺭﻳﺎﺿﻴﺎ ﺑﺎﻟﻘﻮﻝ :ﺇﺫﺍ 83
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﻣﺮﻛﺰﻩ
c
ﲢﺘﻔﻆ ﻓﻴﻪ
f (b) < 0
ﻭ . c = sup Xﻭﻟﻨﺒﻴﻦ ﺃﻥ . f (c) = 0
ﻣﻦ ﺍﻟﻮﺍﺿﺢ ﺃﻥ . a < c < bﻓﻠﻮ ﻛﺎﻥ f
( .ﻭﻟﻨﻀﻊ
f (c ) ≠ 0
ﻟﻮﺟﺪ ﳎﺎﻝ
ﺑﺈﺷﺎﺭﺎ ﺍﳌﻮﺟﺒﺔ ،ﻭﺫﻟﻚ ﺑﻔﻀﻞ ﺍﺳﺘﻤﺮﺍﺭ
ﺫﻟﻚ( .ﻭﻫﺬﺍ ﻳﻨﺎﻗﺾ ﺍﻟﻘﻮﻝ
c = sup X
f
)ﻭﺿﺢ
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 ﺍﺎﻝ [ ]x1 , x2ﳛﻘﻖ . f ( x0 ) = c 84
و
) f ( x2
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
ﻣﻼﺣﻈﺔ ﻳﻨﺘﺞ ﻣﻦ ﺫﻟﻚ ﺃﻥ ﺻﻮﺭﺓ ﳎﺎﻝ ﻋﱪ ﺩﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﻫﻲ ﺃﻳﻀﺎ ﳎﺎﻝ .ﻛﻤﺎ ﻳﻨﺘﺞ ﻣﻦ ﻫﺬﻩ ﺍﻟﻨﻈﺮﻳﺔ ﻭﺍﻟﱵ ﺳﺒﻘﺘﻬﺎ ﰲ ﺣﺎﻟﺔ ﺗﺮﺍﺹ ﺍﺎﻝ ] [a, bﺃﻥ ﺻﻮﺭﺓ ﻫﺬﺍ ﺍﺎﻝ ﻫﻲ ﺍﺎﻝ
. sup
f ( x), inf f ( x) ] x∈[a ,b ] x∈[a ,b
85
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
.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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
ﻗﺪ ﻳﻜﻮﻥ
f
ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ ﻣﻦ ﺍﻟﻴﻤﲔ ﻭﻣﻦ ﺍﻟﻴﺴﺎﺭ
ﻋﻨﺪ x0
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
ﻋﻨﺪ x0
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﻋﻨﺪ x0
ﻓﺈﻧﻪ ﻣﺴﺘﻤﺮ ﻋﻨﺪ ﻫﺬﻩ ﺍﻟﻨﻘﻄﺔ.
ﺍﻟﱪﻫﺎﻥ : ﻟﻴﻜﻦ
f
ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ
ﺃﻱ
ﻋﻨﺪ x0
)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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﻋﻨﺪ x0
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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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
ﻋﻨﺪ x0
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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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 ﺍﺎﻝ 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 ﺍﺎﻝ ﺍﳌﻔﺘﻮﺡ []−1,1
ﻓﻨﺠﺪ ∀x∈]−1,1[, arccos'x= −1 1− x 2
92
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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 : ﻧﻌﺮﻑ 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
ﻗﺎﺑﻼ ﻟﻼﺷﺘﻘﺎﻕ
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.
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
ﳝﻜﻦ ﻟﺘﺎﺑﻊ ﺃﻥ ﻳﻘﺒﻞ ﻗﻴﻤﺔ ﻗﺼﻮﻯ ﻋﻨﺪ . x0ﻣﺜﻼ ﺍﻟﺘﺎﺑﻊ
x֏ x
ﻋﻨﺪ x0
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 ﺍﺎﻝ ] [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
ﺍﻟﻔﺼﻞ : 2ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻻﺷﺘﻘﺎﻕ )ﻣﺮﺍﺟﻌﺔ(
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 ﺍﺎﻝ ] [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 ﺍﳌﺒﻴﻨﺔ ﺃﺩﻧﺎﻩ ﳝﻜﻦ ﲤﺜﻴﻠﻬﺎ ﺑﺘﻜﺎﻣﻞ ﺍﻟﺪﺍﻟﺔ
f
ﻋﻠﻰ ﺍﺎﻝ ] : [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 .
ﺇﺎ 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 :ﺇﺎ 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 ﺍﺎﻝ ] [0,1ﺑـ
1 f ( ) =1 n
ﻣﻦ ﺃﺟﻞ
*n ∈ ℕ
ﻭﺍﳌﻨﻌﺪﻣﺔ ﰲ ﺑﺎﻗﻲ ﻧﻘﺎﻁ ﺍﺎﻝ ] . [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" ﺍﻤﻮﻋﺎﺕ 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 ﺍﺎﻝ
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 ﺍﺎﻝ 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
ﺍﺎﻟﲔ ] [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 ﻧﻌﺮﻑ 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 ﺍﺎﻝ . 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 ﺍﺎﻝ Iﻓﺈﻥ
f
ﻳﻘﺒﻞ ﻋﺪﺩﺍ ﻏﲑ ﻣﻨﺘﻪ ﻣﻦ ﺍﻟﺘﻮﺍﺑﻊ ﺍﻷﺻﻠﻴﺔ ﻋﻠﻰ ﺍﺎﻝ . 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 ﺍﺎﻝ ]. [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 ﺍﺎﻝ ][ 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
ﺍﺎﻝ ، 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 ﺍﺎﻝ . 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 ﺍﺎﻝ . 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 ﺍﺎﻝ ، 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 ﺍﻤﻮﻋﺔ }
) ] 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 ﺍﻤﻮﻋﺔ 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
ﺇﻤﺎ 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
ﺍﳌﻌﺮﻑ ﺑـ
) ) 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
ﺍﳌﻌﺮﻑ ﺑـ
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
ﺍﳌﻌﺮﻑ ﺑـ
)(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 ﺍﺴﻤﺎﺕ 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 ﻭﺍﺴﻤﺎﺕ 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 ﻋﺮﻓﻨﺎﻩ 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 ﻣﺒﻴﻦ ﰲ ﺍﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ :
140
ا : 3ا ب ا
ﻭﻫﻨﺎﻙ ﺃﻧﻮﺍﻉ ﻛﺜﲑﺓ ﻣﻦ ﺍﳌﺨﺮﻭﻃﺎﺕ ﻣﻨﻬﺎ ﺗﻠﻚ ﺍﳌﺴﻤﺎﺓ ﺑﺎﳌﺨﺮﻭﻃﺎﺕ ﺍﻟﻜﺮﻭﻳﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﺭﺃﺳﻬﺎ ﰲ ﻣﺮﻛﺰ ﻛﺮﺓ ﻭﻗﺎﻋﺪﺎ ﺟﺰﺀﺍ ﻣﻦ ﺳﻄﺢ ﺍﻟﻜﺮﺓ ﻛﻤﺎ ﻫﻮ ﻣﺒﻴﻦ 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
ﻣﺮﻛﺰﻩ
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 ﺃﺎ ﺗﺴﺎﻭﻱ ﻣﺴﺎﺣﺔ ﺟﺰﺀ ﺍﻟﻘﺮﺹ ﺍﻟﺬﻱ
ﻣﺮﻛﺰﻩ
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 ﻣﺒﻴﻦ 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 ﲤﻴﺰﻫﺎ 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
=
ﻭﺍﺭﺗﻔﺎﻋﻪ
h
ﻫﻮ
= .V
.4ﺣﺠﻢ ﺍﻷﺳﻄﻮﺍﻧﺔ ﺇﺫﺍ ﻛﺎﻥ
h
ﻫﻮ ﺍﺭﺗﻔﺎﻉ ﻫﺬﻩ ﺍﻷﺳﻄﻮﺍﻧﺔ ﺍﻟﺪﻭﺭﺍﻧﻴﺔ ،ﻭﻛﺎﻥ
ﻗﻄﺮ ﻗﺎﻋﺪﺎ 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 ﻣﺘﻐﻴﺮ 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 ﻗﺎﻋﺪﺎ
R
ﻭﺍﺭﺗﻔﺎﻋﻬﺎ
h
R
ﻭﺍﺭﺗﻔﺎﻋﻪ
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 ﺑﻴﻦ ﺃﻥ ﻣﺴﺎﺣﺔ ﺳﻄﺢ ﻛﺮﺓ ﺗﺴﺎﻭﻱ ﺍﳌﺴﺎﺣﺔ ﺍﳉﺎﻧﺒﻴﺔ ﻟﻸﺳﻄﻮﺍﻧﺔ ﺍﻟﱵ ﻗﻄﺮ ﻗﺎﻋﺪﺎ 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 ﻧﻘﺪﻡ 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 ﻗﺎﻋﺪﺎ ﻭﺍﺭﺗﻔﺎﻋﻬﺎ
2ab + 2bc + 2ac
ﺍﳌﺴﺎﺣﺔ ﺍﳉﺎﻧﺒﻴﺔ :
R
2π R.h
h
ﺍﳌﺴﺎﺣﺔ ﺍﻟﻜﻠﻴﺔ : 2π R.h + 2π R 2
ﳐﺮﻭﻁ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪﺎ
R
ﻭﺍﺭﺗﻔﺎﻋﻬﺎ
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 ﻗﺎﻋﺪﺎ ﻭﺍﺭﺗﻔﺎﻋﻬﺎ
R
π R 2 .h
h
ﻣﺴﺎﺣﺔ ﻗﺎﻋﺪﺎ ﰲ
ﺃﺳﻄﻮﺍﻧﺔ
ﺍﺭﺗﻔﺎﻋﻬﺎ ﳐﺮﻭﻁ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪﺎ
R
ﻭﺍﺭﺗﻔﺎﻋﻬﺎ
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 ﺍﺴﻢ 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 ِﳌﻌﺪﺍﺩ
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 ﻧﺘﻮﺳﻊ ﰲ ﻣﻮﺿﻮﻉ ﺍﻟﻌﻼﻗﺎﺕ ﺣﻴﺚ ﻗﺪﻣﻨﺎ ﺑﺈﳚﺎﺯ ،ﻣﻊ ﺑﻌﺾ ﺍﻷﻣﺜﻠﺔ ،ﻋﻼﻗﱵ ﺍﻟﺘﻜﺎﻓﺆ ﻭﺍﻟﺘﺮﺗﻴﺐ .ﻭﻻ ﺷﻚ ﺃﻥ ﺍﳌﺪﺭﺱ ﺳﻴﺘﻤﻜﹼﻦ ﻣﻦ ﺗﻘﺪﱘ ﻋﺪﺓ ﻣﻮﺍﺿﻴﻊ ﱂ ﻧﺘﻄﺮﻕ ﺇﻟﻴﻬﺎ ﻫﻨﺎ ﻛﻤﻮﺍﺿﻴﻊ ﻟﻠﺪﺭﺍﺳﺔ ﺧﻼﻝ ﺍﻟﺴﻨﺔ ﺍﻟﺪﺭﺍﺳﻴﺔ .ﻭﻣﻦ ﺗﻠﻚ ﺍﳌﻮﺍﺿﻴﻊ ﳝﻜﻦ ﺍﻗﺘﺮﺍﺡ ﻧﻈﺮﻳﺔ ﺍﻤﻮﻋﺎﺕ ،ﻭﺧﻮﺍﺹ ﺍﻟﺘﻄﺒﻴﻘﺎﺕ ،ﻭﺍﻟﺘﻌﻤﻖ ﰲ ﻣﻮﺿﻮﻉ ﺍﳌﻨﻄﻖ ﺍﻟﺮﻳﺎﺿﻲ ،ﻭﻣﻮﺍﺿﻴﻊ ﺃﺧﺮﻯ ﻛﺜﲑﺓ.
165
ﺍﻟﻔﺼﻞ : 1ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ
.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 ﺰﻭﺩ 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
ﺍﻟﻔﺼﻞ : 1ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ
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
ﺍﻟﻔﺼﻞ : 1ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ
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
ﺍﻟﻔﺼﻞ : 1ﺍﳌﻨﻄﻖ ﻭﺍﻟﻌﻼﻗﺎﺕ
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
ﺍﻟﻔﺼﻞ : 2ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ
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
ﺍﳌﻌﺮﻓﺔ ﺑـ
ﺍﳌﻨﺘﻤﻴﺔ ﺇﱃ G؟ ﺇﺫﺍ ﺭﻣﺰﻧﺎ
ﻓﺈﻧﻨﺎ ﻧﻼﺣﻆ ﺑﺄﺎ 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
ﺍﻟﻔﺼﻞ : 2ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ
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 ﺇﺎ - 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 ﺇﺎ
- pﺯﻣﺮﺓ ﺳﻴﻠﻮ )ﺃﻭ - pﺯﻣﺮﺓ ﺳﻴﻠﻮﻳﺔ ﺃﻭ - pﺳﻴﻠﻮ( ﺇﻥ ﻛﺎﻧﺖ ﺭﺗﺒﺘﻬﺎ ﺗﺴﺎﻭﻱ . pα ﺍﻟﻨﻈﺮﻳﺔ ﺍﻷﻭﱃ ﻟﺴﻴﻠﻮ ﺇﺫﺍ ﻛﺎﻧﺖ n = pα .m
ﺣﻴﺚ
G
ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﺭﺗﺒﺘﻬﺎ
n
ﻭﻛﺎﻥ
p
ﻋﺪﺩ ﺃﻭﱄ ﻻ ﻳﻘﺴﻢ
m
ﻓﺈﻥ
ﺳﻴﻠﻮﻳﺔ.
221
n
G
ﳛﻘﻖ ﺍﻟﻌﻼﻗﺔ ﲤﻠﻚ ﺯﻣﺮﺓ - p
ﺍﻟﻔﺼﻞ : 2ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ
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,.
ﺇﻤﺎ ﻣﺘﺮﺍﻓﻘﺎﻥ ﺇﻥ ﻭﺟﺪ
ﺍﻟﻔﺼﻞ : 2ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ
ﲤﺮﻳﻦ 3 ﺇﺫﺍ ﻛﺎﻧﺖ ﺯﻣﺮﺗﺎﻥ ﺟﺰﺋﻴﺘﺎﻥ ﻣﻦ ﺯﻣﺮﺓ
ﻣﺘﺮﺍﻓﻘﺘﲔ ﻓﺈﻤﺎ
)(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
ﺯﻣﺮﺓ ﻣﻨﺘﻬﻴﺔ ﺭﺗﺒﺘﻬﺎ
n
p
ﻋﺪﺩ ﺃﻭﱄ ﻻ ﻳﻘﺴﻢ
m
ﻭﻛﺎﻥ
n
ﳛﻘﻖ ﺍﻟﻌﻼﻗﺔ
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
ﺍﻟﻔﺼﻞ : 2ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ
ﺗﻄﺒﻴﻖ ﳝﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ ﻧﻈﺮﻳﺎﺕ ﺳﻴﻠﻮ ﻹﺛﺒﺎﺕ ﺃﻥ ﻛﻞ ﺯﻣﺮﺓ ﺭﺗﺒﺘﻬﺎ ﺗﺴﺎﻭﻱ 63
ﻻ ﳝﻜﻦ ﺃﻥ ﺗﻜﻮﻥ ﺑﺴﻴﻄﺔ .ﻟﻨﺮ ﺫﻟﻚ : ﻧﺬﻛﺮ ﺃﻧﻨﺎ ﻧﻘﻮﻝ ﻋﻦ ﺯﻣﺮﺓ ﺇﺎ 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
ﻋﺪﺩ ﺃﻭﱄ ﻻ ﻳﻘﺴﻢ
ﺍﻟـ - pﺳﻴﻠﻮﻳﺔ. ﺃﺛﺒﺖ ﺃﻥ
k
ﺃﻭﱄ ﻣﻊ .. p
224
ﻭﻛﺎﻥ
n m
n
ﳛﻘﻖ ﺍﻟﻌﻼﻗﺔ
.ﻭﻟﻴﻜﻦ
ﻋﺪﺩ ﺍﻟﺰﻣﺮ
k
ﺍﻟﻔﺼﻞ : 2ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ
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 .
(2ﺑﻴﻦ 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
ﺍﻟﻔﺼﻞ : 2ﺍﻟﺒﲎ ﺍﳉﱪﻳﺔ
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