Fourier Series: Assignment | Mathematical Methods - Physics PDF Download

Q.1. Expand  in trigonometric Fourier series. Draw three periods and use it to derive the following relations.

Thus the trigonometric Fourier series of the given function is

The graph of the function is shown in the figure

(i) At x = 0 , the Fourier series converges to

(ii) At x = π we have

(iii) we have at x = π/2

Thus

Hence

Q.2. Expand the given function in Fourier Series  and hence find the value of

Given Function

Here, 2l = 2 ⇒ l = 1



Thus, b_n = 4/nπ for odd n.

= 0 for even n.
Hence the Fourier Series of the given function is

Q.3.
The function shown in the graph is plotted for one period. Sketch it over three periods and use the graph to find the Fourier series of the function shown.

f (x) = -(x + π), -π < x < 0 

x, 0 < x < π
Hence





= 0 for even n


Thus, b_n = 2/n for odd n
= 0 for even n
Thus the Fourier series of the given function is

Q.4. Expand the given function in Fourier Series  and hence find the value of

Thus Fourier series is

At x = 1, f (x) = 2

Q.5. Find the Fourier series of

Now,  for n = 1, 5, 9, 13
 for 3, 7,11,15
and  for even n.
Thus, a_n = 1/nπ  for n = 1, 5, 9, 13
 for n 3, 7,11,15
We can write the general formula for  for even n and 

 for odd n.

Now,  for 2, 6,10,14....
= +1 for n = 4, 8,12,16 ......
= 0 for odd n
Hence b_n = 2/nπ  for  n = 2, 6,10,14....
= 0 for n = 4, 8, 12, 16 ....

= 1/nπ for odd n.
We can write the general formula for even n

Thus b_n = 1/nπ , for odd n.

Thus the Fourier series of the given function is

Q.6. Expand the given function in Fourier Series  and hence find the value of

The given function is

Here, 2l = 2 ⇒ l = 1

 for odd n and a_n a = for even n.

Thus the Fourier Series of the given function is

Putting x = 0 gives

Q.7. Find the Fourier series of f (x) = x² , 0 < x < 2π

The given function is f (x) = x² , 0 < x < 2π

Thus, b_n = -4π/n for all n.
Thus the Fourier series of the given function is

Q.8. Expand the given function in Fourier Series f(x) = π sinπx, 0 < x < 1 and hence find the value of

Here, 2l = 1 ⇒ l = 1/2

For all n

Thus the Fourier Series of the given function is given by

At x = 0 , the function is discontinuous and the fourier series converges to

 

Q.9. Find the Fourier series of 

The given function is 

We know that

We see that this formula holds for n ≠ 1.

Hence we directly calculate

Now let us evaluate a_n using equation (I)


when n is odd, (n + 1) is even and (n - 1) is even. Thus for odd n,

When n is even (n + 1) and (n - 1) are odd, hence


Thus a_n = 0 for odd n

Now we have

We see that we cannot calculate b₁ using this formula.
Let us evaluate b₁ directly

Let us evaluate other b_n ’s using equation (ii)

Thus b_n = 0 for all n.

Hence the Fourier Series of the given function is

In the expanded form

Q.10. Expand the given functions in Fourier Series f(x) = e^x ,  0 < x < 2 and hence find the value of

f(x) = e^x ,  0 < x < 2

Hence,






Thus the Fourier Series of the given function is

In the expanded form we can write

At x = 0, the function is discontinuous and the Fourier series converges to the average of its left hand and right hand limit.
Thus , at x = 0