Test: Fourier Series- 1 - Civil Engineering (CE) MCQ

f(x) = x³
find f(x) is even or odd
put x = -x
f(-x) = - x³
f(x) = -f(-x) hence it is odd function
for odd function, a_o = a_n = 0 
Fourier Series for odd function has only b_n term

Hence only sine terms are left in Fourier expansion of x³
Additional Information
Fourier Series

Concept:
Consider the infinite series ∑u_n = u₁ + u₂ + u₃ + … ∞

• Convergence test:
• If u_n tends to a finite or unique limit as n → ∞, the series is said to be convergent.
• If u_n tends to ± ∞ as n → ∞, the series is said to be divergent.
• If u_n does not tend to a unique limit as n → ∞, the series is said to be oscillatory or non-convergent.

Calculation:
Given series is 

n^th term of series is given by

Un will provide definite and unique value if
p - 1 > 1
⇒ p > 2 

Concept:
Fourier Series is defined as


Calculation:
Given:
f(t) is an even periodic function, Since
f(-t) = f(t)
The constant term in the Fourier series is: a_o/2

Since the function is not continious, we need to break the integral according to interval
from -1 to 0, f(t) = 0,
from 0 to 1, f(t) = 4

The constant term is:
a₀/2 = 4/2 = 2
Mistake Points
Avoid using the integral property of even function because the given function f(t) is not continuous. Though the function given is an even periodic function but it is not continuous

Concept:
Fourier series:
The Fourier series for the function f(x) in the interval α < x < α + 2π is given by:

where

Even and odd function:

Calculation:
Given:

For f(x) = xcos⁡x
f(x) is an odd function
 [Option 1 is correct].

For f(x) = xcos⁡x is an odd function whereas cos⁡nx is an even function. 
The product of odd and even function is an odd function.
 [Option 2 is correct].
For f(x) = xsin⁡x is an even function whereas cos⁡nx is an even function.
The product of two even function is an even function.
 [Option 3 is incorrect].
For f(x) = xsin⁡x is an even function whereas sin⁡nx is an odd function.  
The product of even and odd function is an odd function.
 [Option 4 is correct].

Concept:
If the given is infinite series, find the partial sum (i.e. sum of first nth term )
If the sequence of partial sums is a convergent sequence (i.e. its limit exists and is finite) then the series is also called convergent 
Calculations:
Consider the  given series is 
The n^th term of the series is a_n = 
Given is infinite series, find the partial sum (i.e. sum of first n^th term)
If the sequence of partial sums is a convergent sequence (i.e. its limit exists and is finite) then the series is also called convergent 

⇒ lim_{n → ∞}a_n = 0, limit exists and is finite
The series   is convergent.

Concept:
If f(x) is periodic function of period ‘’T’’ then f(x) can be expressed as below:

If f(x) is odd function, then only the coefficients of sin nx exists (i.e. a_n = 0 & a₀ = 0).
Calculation:
We are given f(x) = x for x ∈ (-π, π)
sin a f(x) = x is odd So, an = 0, a₀ = 0
So, now we have to find b_n,



Hence required sum of series is 
Alternate Method:
Taylor expansion of tan^-1x is expressed in below:

Put x = 9, above expansion series

Hence π/4 is required sum of series.

f(x) = x³
find f(x) is even or odd
put x = -x
f(-x) = - x³
f(x) = -f(-x) hence it is odd function
for odd function, a_o = a_n = 0 
Fourier Series for odd function has only bn term

Hence only sine terms are left in Fourier expansion of x³
Additional Information
Fourier Series

Concept:
Let f(x) is a periodic function defined in (C, C + 2L) with period 2L, then the Fourier series of f(x) is

Where the Fourier series coefficients a₀, a_n, and b_n are given by

• If f(x) is an odd function, then only bn exists where a0 and bn are zero.
• If f(x) is an even function, then both a0 and an exists where bn is zero.

Calculation:
\(f\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} { - π }&{if\;}&{ - π
Fundamental period = 2π
The given function is an odd function and hence only b_n exists.

Here L = 2π

Now, the Fourier series of f(x) is,

The coefficient of sin 5x = b₅

Dirichlet Conditions in Fourier Transformation are as follows:

• f(x) must absolutely integrable over a period, i.e., 
• f(x) must have a finite number of extrema in any given interval, i.e. there must be a finite number of maxima and minima in the interval.
• f(x) must have a finite number of discontinues in any given interval, however, the discontinuity cannot be infinite.
• f(x) must be bounded.

Concept:
Improper Integral:  If a function f on [a, b] have infinite value then it is called is improper integral

• Improper Integral of the First kind:  is said to be the improper integral of the first kind if a = -∞ or b = ∞ or both.
• Improper Integral of the second kind:  is said to be the improper integral of the second kind if a or b is finite but f(x) is infinite for some x ∈ [a, b].

​​If the integration of the improper integral exists, then it is called as Converges, But if the limit of integration fails to exist, then the improper integral is said to be Diverge.
​Calculation: 
Given:

By using the Leibniz rule
Let, a is any parameter then,
 ...(1)
Differentiating w.r.t.a



Integrating both side w.r.t a

put x = ∞ in both (1) and (2)
From equation (1),

I(∞) = 0
From equation (2),

put a = 0, we get

Therefore we can conclude that, 
Hence,It is improper integral but having finite value so, It is Converges in [0, ∞].