CALCULUS

y x-axis

Calculus

a

Integration-x-axis-

Fundamental Theorem

Integration is along x-axis

Relates integration and differentiation

b

Line integral can be given as sum of integration along x-axis and y-axis

Line Integral y

b a

Integration is on a curve in a planecan be broken as an integration along x-axis and an integration along y -axis.

x-axis

Green’s Theorem Relates line integral with double integral.

similar

y

Region

Double Integral Integral is on a region in x-y plane.

Analogous to line integrals Surface integral can be given as sum of double integrations along x-y-plane , x-z-plane and y-z plane.

x-axis

Surface in space

Stokes’s Theorem Relates line integral to surface integral. Can be seen as generalization of Green’s theorem. Becomes Green’s theorem in plane region.

Surface Integral Can be broken up into double integrals in x-y, x-z and y-z planes (analogous to line integral).

Triple Integral Integration is in a volume in xyz space.

Divergence Theorem Relates surface integral with triple integralAnalogous to Green’s theorem.

2

Scalar b

y-axis

(F dx F dy F dz) 1

b

or y-axis

2

a

b

Representation

x-axis

(F1x'F2 y'F3 z' )dt (in parametric form).

a

a

b

x-axis

Line Integration

Integration along x-axis

Vector

Line integral is integral on a plane and can be given as a sum of integrals both along x and y axis.

b

f ( x)dx

a

b

b

a

a

Using ‘t’ as a parameter

1

A

Ft

D

F

A D

Fn

B E

Q

P

x

0

a)

Ft

dy

dx x

0

b)

Consider the case of a force acting on a particle in the x-y plane. If Ft is the tangential force and Fn is the normal force at a point D then to move a small distance ds, Ftds is the work done .

Integration can give the work done by a variable force along the x-axis in the case of integration along x-axis. Here the work done is found by integration along curve AB. The total work done is same as sum of work (work is scalar) along both x and y-axis.

F ds Pdx Qdy t

AB

AB

b

a

a

Path independence, Theorems ds

y

b

( r )' dr / dt

2

Understanding Line Integral using work done B

The value of the integral can be given as a vector rather than a scalar.

F(r) dr F(r(t)) r' (t) dt

Fds (F dx F dy) y

3

a

y-axis

y-axis

b a

a

b

x-axis

a

a

b

b

b

a

a

Pu ( x ) dx Pv ( x ) dx

Line Integral b

f ( x)dx

a

Pdx Pu ( x ) dx Pv ( x ) dx b

Integration -x-axis b

b

x-axis

a

P

u ( x)

Pv ( x ) dx

a

b

Fds ( F1dx F2 dy) a

Green’s Theorem P R y dydx Pdx, Also we find

Q

Double Integral

R

R

d b f ( x, y )dxdy f ( x, y )dx dy c a

b

Double integral is carried out by using the fundamental theory of computation, by which (see Fig.)

b v( x) P P dydx dy dx y y a u( x)

P

v( x) u( x)

a

b

dx

P

u ( x)

a

Pv ( x ) dx

x dydx Qdy

Hence we get

R

Q P R x y dydx Pdx Qdy

Fourier Series, Transform

Fourier Fourier Series, 2p

Fourier Series, 2p, Complex form

Fourier-Series, Integrals, Transforms Fourier Series of any Period, 2L

Even Function, Fourier Cosine Series Odd Function, Fourier Sine Series

Application-ODE

Approximation

Half Range Expans ions

Fourier Integral

Fourier Cosine Integral

Fourier Sine Integral

Fourier Integral, Complex Form

Fourier Transform

Fourier Cosine Transform

Fourier Sine Transform

Discrete/Fast Fourier Transform,

Fourier Series of any Period, 2L

Fourier Series, 2p

Fourier Integral, L→∞

Fourier Integral, Complex Form

Fourier Transform Separate exponential And re-arrange

1) Write in Cos (wx-wv) terms 2) Eqn. in Sin (wx-wv) terms is 0. 3) Change to complex

To change scale first Write equation of Fourier series using ‘v’.

Euler formulas To derive Euler’s formulas use orthogonality condition of Trigonometric system.

Writing v=x For even function

Substituting

For sin function

We get

To derive substitute above value of f(x) and find

Similarly to derive substitute above value of f(x) and find

g(v)=f(x) gives

Fourier Cosine Fourier Sine Series For even function, Cosine Series For odd function, Sine Series

Fourier Series, 2p, Complex form

Fourier Cosine Fourier Sine Integral For even function, Cosine Integral For odd function, Sine Integral

Fourier Cosine Fourier Sine Transform 1) Re-arranging the eqn. for Cosine Integral 2) Writing v=x

A(w), B(w) slightly diff.

Substituting

Writing v=x

Discrete/Fast Fourier Transform,

LAPLACE TRANSFORM

Existence • Growth restriction • Piecewise continuous

Kernel of Transform

Laplace Transforms

Transform Inverse of Transform

Transforms of Diff/Integration

Diff/Integration of Transforms

Operation on Functions Operation on Transforms

Transform of Functions

Addition of Transforms / S-shifting(Associated with multiplicati on of function by e..)

Differentiation of Functions (Associated with multiplication)

Integration of Functions (associated with Division)

Differential Equations Given problem

S-space Subsidiary Eqn.

Solve for y(t)

Solve for Y in S-space

t-space

Unit step Function/tshifting (Associated with multiplication of Transform by e..)

Dirac Delta function

Multiplication of Transforms/ Convolution

Differentiation of Transforms (Associated with multiplication)

Integration of Transforms (Associated with Division)

Kernel of Transform

Transform

Laplace Transforms

Inverse of Transform Transforms of Diff/Integration

Diff/Integration of Transforms

Operation on Functions (Trans of Funcs, Diff, Integration, Diff Eqn., Unit Step/Dirac Delta) Operation on Transforms, Add, Multiply, Diff., Integrate Addition of Transforms/ S-shifting-

Transform of Functions

(Associated with multiplication of function by e..)

Differentiation of Functions

Unit step Function/t-shifting

Integration of Functions

(Associated with multiplication)

(Associated with multiplication of Transform by e..)

(associated with Division)

Dirac Delta function

Multiplication of Transforms/ Convolution

Differentiation of Transforms (Associated with multiplication)

Integration of Transforms (Associated with Division)

Unit Step

For f(t)=1

Area=1

1

Transform

Cos wt

Sin wt

Cosh at (rem---with s)

u(t-a)

1

s-shifting

Function

1/k

u(t-0)=u(t)

Differential Equations

Sinh at

Input is r(t), Output is y Substituting above equations For Differentiation we get

m

t-space

S-space

Given problem

Subsidiary Eqn.

Solve for y(t)

Solve for Y in Sspace

k

m

m

a

f(t) =msin t

a+k

a

a

Then Y=RQ a

Special linear ODE’s

f(t-a)u(t-a)= f(t)u(t-a)= msin t u(t-a) msin(t-a) u(t-a)

t-shifting

For Non-homogenous Linear ODE’s If

Dirac-Delta

Kernel of Transform

Laplace Transform

(Examples from Kreyszig)

Transform Inverse of Transform

On Functions

Transform of Functions

Addition/ S-shifting

Transform of Functions

Addition

Differentiation of Functions

On Transforms

Integration of Functions

Differentiation

Dirac Delta/Impulse Differential Eqn

From Functions Find Transform

Indicates Integration

By differentiation formula

Unit step/tshifting

Indicates Shifting

By Integration formula

From (1) and (2)

Write f(t) in terms of unit step Use

Solving Inverses

Inverse of g(t)=1

Differentiation/ Integration of Transforms Since

Finding Transforms

Hence

Inverse Transforms ‘s’ and ‘t’ shift

t-shift

S-shifting

Multiplication of Transforms/ Convolution

Finding Inverses a) Given F(s) we find F’(s). SinceL tf(t)=-F’(s) we can find f(t)

s-shift a=3

Differential Eqn.

From s-shifting of Cos and Sin formulas,

a=-1, w=20

a1,w=20

Differential Equations

Differential Eqn. Same as normal Diff. eqn. except Input is unit step Function, f(t) obtained in steps as

For y(0)=y’(0)=0 Y=RQ

Knowing ‘q’ and ‘r’ we can solve in an alternate manner.

Special linear ODE’s With variable Coefficient

Example-Laguerre’s ODE

T-space

ProblemGiven-Diff. Eqn., Initial Values y”-y=t, y(0)=1, y’(0)=1

S-space Write Eqn. in ‘s’ terms

Re-arrange

Solution - S-space

Solution - t-space Integral Eqn.

Certain Integral Equations Can be solved by convolution

Can be written as

Since we know Transform of ty” and ty’ we can find Eqn. in s-space

COMPLEX ANALYSIS

Harmonic Conjugates From u one can get v and vice versa

Complex Laplace’s Eqn. Differentiating 1st Cauchy Riemann eqn. in x and 2nd in y we get one laplace eqn.

Limits, Diff. in Complex plane. Function

Hyp.

In real we approach along the axis-in complex we approach in a plane

Definition of Complex integral Integral is on complex plane (has some similarity with line integral). The integral can be written in terms of u and v.

Cauchy-Riemann

Cauchy-Integral Theorem

Path-Independence

Approaching limit Dx and Dy →0 and Dy and Dx →0 way and equating gives CauchyRiemann.

For analytic, simply connectedwriting the integral in terms of u and v, applying Green’s theorem and then Cauchy Riemann the integral is shown to be 0.

For analytic, simply connected integration from A to B and B to A is 0 (by Cauchy Integral Theorem ) –hence it can be shown that integration from A to B along two diff. paths give same result.

Derivatives Exp.

Trig.

Path-independence Dependence If function is not analytic integral it is path dependent. Thus integration of 1/(z-z0 ) along a unit circle |z|=1 is 2pi.

Log.

We can start from Exp. From exp. Eqns.we can get Trig. Eqns Trig. Eqns. are inter-related with. Hyp. Exp. are also related to Log.

Cauchy-Integral Formula The integral of f(z)/(z-z0 ) can be written as integral of f(z0)/(z-z0) and (f(z)f(z0)) /(z-z0 ) . The second integral is shown to be 0. Hence the answer to the integral of f(z)/(z-z0) is 2pif(z0) (see path depen-)

Power Series Representation

Liouville’s Theorem

Integration is carried about a circle C of radius ‘r’. M is greater than absolute value of f(z). Substituting values we get n!M/rn greater than the absolute value of fn (z0 ).

For an entire function if |f(z0 )|

y x-axis

Calculus

a

Integration-x-axis-

Fundamental Theorem

Integration is along x-axis

Relates integration and differentiation

b

Line integral can be given as sum of integration along x-axis and y-axis

Line Integral y

b a

Integration is on a curve in a planecan be broken as an integration along x-axis and an integration along y -axis.

x-axis

Green’s Theorem Relates line integral with double integral.

similar

y

Region

Double Integral Integral is on a region in x-y plane.

Analogous to line integrals Surface integral can be given as sum of double integrations along x-y-plane , x-z-plane and y-z plane.

x-axis

Surface in space

Stokes’s Theorem Relates line integral to surface integral. Can be seen as generalization of Green’s theorem. Becomes Green’s theorem in plane region.

Surface Integral Can be broken up into double integrals in x-y, x-z and y-z planes (analogous to line integral).

Triple Integral Integration is in a volume in xyz space.

Divergence Theorem Relates surface integral with triple integralAnalogous to Green’s theorem.

2

Scalar b

y-axis

(F dx F dy F dz) 1

b

or y-axis

2

a

b

Representation

x-axis

(F1x'F2 y'F3 z' )dt (in parametric form).

a

a

b

x-axis

Line Integration

Integration along x-axis

Vector

Line integral is integral on a plane and can be given as a sum of integrals both along x and y axis.

b

f ( x)dx

a

b

b

a

a

Using ‘t’ as a parameter

1

A

Ft

D

F

A D

Fn

B E

Q

P

x

0

a)

Ft

dy

dx x

0

b)

Consider the case of a force acting on a particle in the x-y plane. If Ft is the tangential force and Fn is the normal force at a point D then to move a small distance ds, Ftds is the work done .

Integration can give the work done by a variable force along the x-axis in the case of integration along x-axis. Here the work done is found by integration along curve AB. The total work done is same as sum of work (work is scalar) along both x and y-axis.

F ds Pdx Qdy t

AB

AB

b

a

a

Path independence, Theorems ds

y

b

( r )' dr / dt

2

Understanding Line Integral using work done B

The value of the integral can be given as a vector rather than a scalar.

F(r) dr F(r(t)) r' (t) dt

Fds (F dx F dy) y

3

a

y-axis

y-axis

b a

a

b

x-axis

a

a

b

b

b

a

a

Pu ( x ) dx Pv ( x ) dx

Line Integral b

f ( x)dx

a

Pdx Pu ( x ) dx Pv ( x ) dx b

Integration -x-axis b

b

x-axis

a

P

u ( x)

Pv ( x ) dx

a

b

Fds ( F1dx F2 dy) a

Green’s Theorem P R y dydx Pdx, Also we find

Q

Double Integral

R

R

d b f ( x, y )dxdy f ( x, y )dx dy c a

b

Double integral is carried out by using the fundamental theory of computation, by which (see Fig.)

b v( x) P P dydx dy dx y y a u( x)

P

v( x) u( x)

a

b

dx

P

u ( x)

a

Pv ( x ) dx

x dydx Qdy

Hence we get

R

Q P R x y dydx Pdx Qdy

Fourier Series, Transform

Fourier Fourier Series, 2p

Fourier Series, 2p, Complex form

Fourier-Series, Integrals, Transforms Fourier Series of any Period, 2L

Even Function, Fourier Cosine Series Odd Function, Fourier Sine Series

Application-ODE

Approximation

Half Range Expans ions

Fourier Integral

Fourier Cosine Integral

Fourier Sine Integral

Fourier Integral, Complex Form

Fourier Transform

Fourier Cosine Transform

Fourier Sine Transform

Discrete/Fast Fourier Transform,

Fourier Series of any Period, 2L

Fourier Series, 2p

Fourier Integral, L→∞

Fourier Integral, Complex Form

Fourier Transform Separate exponential And re-arrange

1) Write in Cos (wx-wv) terms 2) Eqn. in Sin (wx-wv) terms is 0. 3) Change to complex

To change scale first Write equation of Fourier series using ‘v’.

Euler formulas To derive Euler’s formulas use orthogonality condition of Trigonometric system.

Writing v=x For even function

Substituting

For sin function

We get

To derive substitute above value of f(x) and find

Similarly to derive substitute above value of f(x) and find

g(v)=f(x) gives

Fourier Cosine Fourier Sine Series For even function, Cosine Series For odd function, Sine Series

Fourier Series, 2p, Complex form

Fourier Cosine Fourier Sine Integral For even function, Cosine Integral For odd function, Sine Integral

Fourier Cosine Fourier Sine Transform 1) Re-arranging the eqn. for Cosine Integral 2) Writing v=x

A(w), B(w) slightly diff.

Substituting

Writing v=x

Discrete/Fast Fourier Transform,

LAPLACE TRANSFORM

Existence • Growth restriction • Piecewise continuous

Kernel of Transform

Laplace Transforms

Transform Inverse of Transform

Transforms of Diff/Integration

Diff/Integration of Transforms

Operation on Functions Operation on Transforms

Transform of Functions

Addition of Transforms / S-shifting(Associated with multiplicati on of function by e..)

Differentiation of Functions (Associated with multiplication)

Integration of Functions (associated with Division)

Differential Equations Given problem

S-space Subsidiary Eqn.

Solve for y(t)

Solve for Y in S-space

t-space

Unit step Function/tshifting (Associated with multiplication of Transform by e..)

Dirac Delta function

Multiplication of Transforms/ Convolution

Differentiation of Transforms (Associated with multiplication)

Integration of Transforms (Associated with Division)

Kernel of Transform

Transform

Laplace Transforms

Inverse of Transform Transforms of Diff/Integration

Diff/Integration of Transforms

Operation on Functions (Trans of Funcs, Diff, Integration, Diff Eqn., Unit Step/Dirac Delta) Operation on Transforms, Add, Multiply, Diff., Integrate Addition of Transforms/ S-shifting-

Transform of Functions

(Associated with multiplication of function by e..)

Differentiation of Functions

Unit step Function/t-shifting

Integration of Functions

(Associated with multiplication)

(Associated with multiplication of Transform by e..)

(associated with Division)

Dirac Delta function

Multiplication of Transforms/ Convolution

Differentiation of Transforms (Associated with multiplication)

Integration of Transforms (Associated with Division)

Unit Step

For f(t)=1

Area=1

1

Transform

Cos wt

Sin wt

Cosh at (rem---with s)

u(t-a)

1

s-shifting

Function

1/k

u(t-0)=u(t)

Differential Equations

Sinh at

Input is r(t), Output is y Substituting above equations For Differentiation we get

m

t-space

S-space

Given problem

Subsidiary Eqn.

Solve for y(t)

Solve for Y in Sspace

k

m

m

a

f(t) =msin t

a+k

a

a

Then Y=RQ a

Special linear ODE’s

f(t-a)u(t-a)= f(t)u(t-a)= msin t u(t-a) msin(t-a) u(t-a)

t-shifting

For Non-homogenous Linear ODE’s If

Dirac-Delta

Kernel of Transform

Laplace Transform

(Examples from Kreyszig)

Transform Inverse of Transform

On Functions

Transform of Functions

Addition/ S-shifting

Transform of Functions

Addition

Differentiation of Functions

On Transforms

Integration of Functions

Differentiation

Dirac Delta/Impulse Differential Eqn

From Functions Find Transform

Indicates Integration

By differentiation formula

Unit step/tshifting

Indicates Shifting

By Integration formula

From (1) and (2)

Write f(t) in terms of unit step Use

Solving Inverses

Inverse of g(t)=1

Differentiation/ Integration of Transforms Since

Finding Transforms

Hence

Inverse Transforms ‘s’ and ‘t’ shift

t-shift

S-shifting

Multiplication of Transforms/ Convolution

Finding Inverses a) Given F(s) we find F’(s). SinceL tf(t)=-F’(s) we can find f(t)

s-shift a=3

Differential Eqn.

From s-shifting of Cos and Sin formulas,

a=-1, w=20

a1,w=20

Differential Equations

Differential Eqn. Same as normal Diff. eqn. except Input is unit step Function, f(t) obtained in steps as

For y(0)=y’(0)=0 Y=RQ

Knowing ‘q’ and ‘r’ we can solve in an alternate manner.

Special linear ODE’s With variable Coefficient

Example-Laguerre’s ODE

T-space

ProblemGiven-Diff. Eqn., Initial Values y”-y=t, y(0)=1, y’(0)=1

S-space Write Eqn. in ‘s’ terms

Re-arrange

Solution - S-space

Solution - t-space Integral Eqn.

Certain Integral Equations Can be solved by convolution

Can be written as

Since we know Transform of ty” and ty’ we can find Eqn. in s-space

COMPLEX ANALYSIS

Harmonic Conjugates From u one can get v and vice versa

Complex Laplace’s Eqn. Differentiating 1st Cauchy Riemann eqn. in x and 2nd in y we get one laplace eqn.

Limits, Diff. in Complex plane. Function

Hyp.

In real we approach along the axis-in complex we approach in a plane

Definition of Complex integral Integral is on complex plane (has some similarity with line integral). The integral can be written in terms of u and v.

Cauchy-Riemann

Cauchy-Integral Theorem

Path-Independence

Approaching limit Dx and Dy →0 and Dy and Dx →0 way and equating gives CauchyRiemann.

For analytic, simply connectedwriting the integral in terms of u and v, applying Green’s theorem and then Cauchy Riemann the integral is shown to be 0.

For analytic, simply connected integration from A to B and B to A is 0 (by Cauchy Integral Theorem ) –hence it can be shown that integration from A to B along two diff. paths give same result.

Derivatives Exp.

Trig.

Path-independence Dependence If function is not analytic integral it is path dependent. Thus integration of 1/(z-z0 ) along a unit circle |z|=1 is 2pi.

Log.

We can start from Exp. From exp. Eqns.we can get Trig. Eqns Trig. Eqns. are inter-related with. Hyp. Exp. are also related to Log.

Cauchy-Integral Formula The integral of f(z)/(z-z0 ) can be written as integral of f(z0)/(z-z0) and (f(z)f(z0)) /(z-z0 ) . The second integral is shown to be 0. Hence the answer to the integral of f(z)/(z-z0) is 2pif(z0) (see path depen-)

Power Series Representation

Liouville’s Theorem

Integration is carried about a circle C of radius ‘r’. M is greater than absolute value of f(z). Substituting values we get n!M/rn greater than the absolute value of fn (z0 ).

For an entire function if |f(z0 )|