Unit 4

Fourier series and Vector Calculus

Full range Fourier series

Definition:

Fourier Series

Let f(x) be a periodic function of period 2L. Defined in the internal and satisfied Dirichlet's conditions, then f(x) can be expressed as,

.

Where ao, an, bn are called Fourier constant’s or Fourier coefficients and are given by,

Note:

That there are only 4 intervals as below. i.e. is divided into following four intervals.

Note that for [0, 2L] we put c = 0

Hence Fourier series in this interval will be,

Where

Simillarly, for the interval we put c = 0,

Hence Fourier series in this interval will be

Where

Note that for the interval [-L, L] i.e. put C = -L,

First we check whether f(x) is even function or odd.

Case I:-

If f(x) is even function. Then we get half range cosine series as,

Where

Case II:-

If f(x) is odd function. Then we get half range sine series as,

Where

Simillarly

Note that for that interval i.e. put ,

First we check wheatear f(x) is even or odd function.

Case I:-

If f(x) is even function then we get half range cosine series as

Where

Case II:-

If f(x) is odd function then we get half range sine series as,

Where

Note that

- For half range cosine series i.e. f(x) is even function bn = 0
- For half range sine series i.e. f(x) is odd function ao = an = 0

Exercise

Find the Fourier series of f(x) = x in the interval

Solution:

Here ;

It’s Fourier series is given by

… (1)

Where

&

Hence the required Fourier series is

- Find the Fourier series for

in the interval

Hence deduce that

Solution:

Here ;

Hence it’s Fourier series is,

… (1)

Where

&

Hence equation (1) becomes

Put we get

i.e.

2. Find a Fourier series expansion in the interval for

;

;

Solution:

Here

;

;

Hence it’s Fourier series expansion is,

… (1)

Where

And

Hence equation (1) becomes

3. Find a Fourier series of

;

;

Solution:

Here

;

;

Here f(x) is odd function Hence we get half range sine series i.e.

… (1)

Where

Hence equation (1) becomes,

Half range Fourier series

4. Find a Fourier series for

;

Solution:

Here

;

Since f(x) is even function hence

It’s Fourier series is

… (1)

Where

Hence equation (1) becomes,

5. Find half range cosine series of in the interval and hence deduce that

a)

b)

Solution:

Here

;

Hence it’s half range cosine series is,

… (1)

Where

Hence equation (1) becomes,

… (2)

Put x = 0, we get

Hence the result

Put we get,

i.e.

Cosine Series:

Example 1:

Using complex form,find the Fourier series of the function

f(x) = sinnx =

Solution:

We calculate the coefficients

=

=

Hence the Fourier series of the function in complex form is

We can transform the series and write it in the real form by renaming as

n=2k-1,n=

=

Example 2:

Using complex form find the Fourier series of the function f(x) = x2, defined on te interval [-1,1]

Solution:

Here the half-period is L=1.Therefore,the co-efficient c0 is,

For n

Integrating by parts twice,we obtain

=

=

= .

= .

Example 1: Find the gradient of the following:

Solution:

y=

= .

= 2x+

Example 2:

Find the curl of F(x,y,z) = 3i+2zj-xk

Curl F =

=

= i -

= (0-2)i-(-1-0)j+(0-0)k

= -2i+j

Example 3:

What is the curl of the vector field F= ( x +y +z ,x-y-z,)?

Solution:

Curl F =

=

=

= (2y+1)i-(2x-1)j+(1-1)k

= (2y+1)i+(1-2x)j+0k

= (2y+1, 1-2x,0)

Example 4: Compute where F= (3x+ and s is the surface of the box such that 0 use outward normal n

Solution: Writing the given vector fields in a suitable manner for finding divergence

Div F =3+2y+x

We use the divergence theorem to convert the surface integral into a triple integral

Where B is the box 0 , 0

We compute the triple integral of div F=3+2y+x over the box B

=

=

= 36+3=39

Example 5: For F= ( use divergence theorem to evaluate where s is the dphere of radius 3 centred at origin.

Solution: Since div F= , the surface integral is equal to the triple integral.

To evaluate the triple integral we can change value of variables to spherical co-ordinates,

The integral is = .For spherical co-ordinates, we know that the jacobian determinant is dV = .therefore, the integral is

=

=

=

Reference Book:

1. G.B. Thomas and R.L. Finney, Calculus and Analytic geometry, 9th Edition, Pearson, Reprint, 2002.

2. Erwin kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons,2006.

3. Veerarajan T., Engineering Mathematics for the first year, Tata McGraw-Hill, New Delhi,2008.

4. Ramana B.V., Higher Engineering Mathematics, Tata McGraw Hill New Delhi, 11th Reprint, 2010.

5. D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.

6. B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010