Fourier Series MCQ Quiz - Objective Question with Answer for Fourier Series - Download Free PDF

Concept

The function f(x) is represented by its Fourier series as:

$f(x)=a_o+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cosnx++\sum _{n=1}^{\mathrm{\infty }}{b}_{n}sinnx$

Calculation of a_o, a_n, and b_n

$a_0=\frac{1}{T}{\int }_0^{T}f(x)dx$

$a_n=\frac{2}{T}{\int }_{-T}^{T}f(x)cosnxdx$

$b_n=\frac{2}{T}{\int }_{-T}^{T}f(x)sinnxdx$

Calculation

Given, f(x) = x + π; -π < x < π

$a_0=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }(x+\pi )dx$

$a_o=\frac{1}{2\pi }(\frac{x^2}{2}+\pi x{)}_{-\pi }^{\pi }$

$a_o=\frac{1}{2\pi }(\frac{{\pi }^2}{2}+{\pi }^2-\frac{{\pi }^2}{2}+{\pi }^2)$

$a_o=\pi$

$a_n=\frac{2}{2\pi }{\int }_{-\pi }^{\pi }(x+\pi )cosnxdx$

$a_n=\frac{1}{\pi }((x+\pi )\int cos(nx)dx-\int (\frac{d(x+\pi )}{dx}\int cos(nx)dx){)}_{-\pi }^{\pi }$

$a_n=\frac{1}{\pi }[(x+\pi )\frac{sinnx}{n}+\frac{cosnx}{n^2}{]}_{-\pi }^{\pi }$

$a_n=0$

$b_n=\frac{2}{2\pi }{\int }_{-\pi }^{\pi }(x+\pi )sinnxdx$

$b_n=\frac{1}{\pi }((x+\pi )\int sin(nx)dx-\int (\frac{d(x+\pi )}{dx}\int sin(nx)dx){)}_{-\pi }^{\pi }$

$b_n=\frac{1}{\pi }[(x+\pi )\frac{-cosnx}{n}+\frac{sinnx}{n^2}{]}_{-\pi }^{\pi }$

$b_n=-\frac{2}{n}\cos (n\pi )$

Putting values of  ao, an, and b_n in Fourier expansion of f(x), we get:

$f(x)=a_o+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cosnx+\sum _{n=1}^{\mathrm{\infty }}{b}_{n}sinnx$

$f(x)=\pi +\sum _{i=1}^{\mathrm{\infty }}(0)cosnx+\sum _{i=1}^{\mathrm{\infty }}(-\frac{2}{n}\cos (n\pi ))sinnx$

$f(x)=\pi +\sum _{i=1}^{\mathrm{\infty }}(-\frac{2}{n}\cos (n\pi ))sinnx$

Hence, the correct answer is option 2.

Concept:

The expanded form of the expression (1 - x)-n is,

$(1-x{)}^{-n}=1+nx+\frac{n(n+1)}{2!}\cdot x^2+\frac{(n(n+1)(n+2)}{3!}\cdot x^3+$

Calculation:

Given series is,

$1+\frac{\mathrm{k}}{3}+\frac{\mathrm{k}(\mathrm{k}+1)}{3\times 6}+\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}{3\times 6\times 9}+\dots .$

Suppose its sum is S. Thus,

$\Rightarrow S=1+\frac{\mathrm{k}}{3}+\frac{\mathrm{k}(\mathrm{k}+1)}{3\times 6}+\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}{3\times 6\times 9}+\dots .$

$$\begin{gathered} \Rightarrow S=1+\frac{k}{1}\cdot \frac{1}{3}+\frac{k(k+1)}{1\cdot 2}\cdot \frac{1}{3^2}+\frac{k(k+1)(k+2)}{1\cdot 2\cdot 3}\cdot \frac{1}{3^3}+\cdots \\ \Rightarrow S=1+k\cdot \frac{1}{3}+\frac{k(k+1)}{2!}\cdot (\frac{1}{3}{)}^2+\frac{k(k+1)(k+2)}{3!}\cdot (\frac{1}{3}{)}^3+\cdots \end{gathered}$$

Since, we know that,

$(1-x{)}^{-n}=1+nx+\frac{n(n+1)}{2!}\cdot x^2+\frac{(n(n+1)(n+2)}{3!}\cdot x^3+$

In our case, x = 1/3 and n = k. So, using the above formula, the sum of a given expansion is,

$$\begin{gathered} \Rightarrow S={(1-\frac{1}{3})}^{-k} \\ \Rightarrow S={\left(\frac{2}{3}\right)}^{-k} \\ \Rightarrow S={\left(\frac{3}{2}\right)}^k \end{gathered}$$

So, the sum of the given series is ${\left(\frac{3}{2}\right)}^{\mathrm{k}}$.

Calculation:

From the trigonometric identity:

cos (2x) = 1 - 2sin²(x)

2sin2(x) = 1 - cos (2x)

f(x) = sin2 x = $\frac{1}{2}-\frac{\cos 2x}{2}$

Hence, option 1 is correct.

Concept:

cos 2x = 2 cos²x - 1

cos²x = $\left(\frac{1+\cos 2\mathrm{x}}{2}\right)$

Calculation:

Given that,

$\mathrm{f}(\mathrm{x})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{f}}_{\mathrm{n}}\cos \mathrm{n}\mathrm{x}$

f(x) = cos⁴ x

⇒ cos⁴x = cos²x . cos²x

$=\left(\frac{1+\cos 2\mathrm{x}}{2}\right).\left(\frac{1+\cos 2\mathrm{x}}{2}\right)$

$=\frac{1}{4}[1+2\cos 2\mathrm{x}+{\cos }^{2}2\mathrm{x}]$

$=\frac{1}{4}[1+2\cos 2\mathrm{x}+\frac{(1+\cos 4\mathrm{x})}{2}]$

$=\frac{1}{4}+\frac{\cos 2\mathrm{x}}{2}+\frac{1}{8}+\frac{\cos 4\mathrm{x}}{8}$

$=\frac{3}{8}+\frac{\cos 2\mathrm{x}}{2}+\frac{\cos 4\mathrm{x}}{8}$

when n = 4, F₄ = 1/8, F₅ = 0 [As no term as cos 5x]

so, F₄ + F₅ = 1/8 = 0.125

Explanation:

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

b_n = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)sin\frac{n\pi x}{l}dx$

Hence only sine terms are left in Fourier expansion of x³

Additional Information Fourier Series

f(x) = $\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos\frac{n\pi x}{l}+\sum _{n=1}^{\mathrm{\infty }}{b}_{n}sin\frac{n\pi x}{l}$

a_o = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)dx$

a_n = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)cos\frac{n\pi x}{l}dx$

bn = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)sin\frac{n\pi x}{l}dx$

Concept:

In the interval (-l, l) Fourier series is defined as:-

$f\left(x\right)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos \frac{n\pi x}{l}+\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin \frac{n\pi x}{l}$

Where, $a_0=\frac{1}{l}{\int }_{-l}^{l}f\left(x\right)dx$

$a_n=\frac{1}{l}{\int }_{-l}^{l}f\left(x\right)\cos \frac{n\pi x}{l}dx$

$b_n=\frac{1}{l}{\int }_{-l}^{l}f\left(x\right)\frac{\sin n\pi x}{l}dx$

Euler Definition: In the interval (-π, π)

$f\left(x\right)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos nx+\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin x$

Where,

$a_o=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)dx$

$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\cos nxdx$

$b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin nxdx$

Calculation:

Given,

f(x) = (x – x²)

$\therefore a_0=\frac{1}{l}{\int }_{-l}^{l}f\left(x\right)dx=\frac{1}{\pi }{\int }_{-\pi }^{\pi }\left(x-x^2\right)dx=\frac{1}{\pi }{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_{-\pi }^{\pi }$

Concept:

cos 2x = 2 cos²x - 1

cos²x = $\left(\frac{1+\cos 2\mathrm{x}}{2}\right)$

Calculation:

Given that,

$\mathrm{f}(\mathrm{x})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{f}}_{\mathrm{n}}\cos \mathrm{n}\mathrm{x}$

f(x) = cos⁴ x

⇒ cos⁴x = cos²x . cos²x

$=\left(\frac{1+\cos 2\mathrm{x}}{2}\right).\left(\frac{1+\cos 2\mathrm{x}}{2}\right)$

$=\frac{1}{4}[1+2\cos 2\mathrm{x}+{\cos }^{2}2\mathrm{x}]$

$=\frac{1}{4}[1+2\cos 2\mathrm{x}+\frac{(1+\cos 4\mathrm{x})}{2}]$

$=\frac{1}{4}+\frac{\cos 2\mathrm{x}}{2}+\frac{1}{8}+\frac{\cos 4\mathrm{x}}{8}$

$=\frac{3}{8}+\frac{\cos 2\mathrm{x}}{2}+\frac{\cos 4\mathrm{x}}{8}$

when n = 4, F₄ = 1/8, F₅ = 0 [As no term as cos 5x]

so, F₄ + F₅ = 1/8 = 0.125

Concept:

If the 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 

Calculations:

Consider the  given series is  $\frac{1}{2^2+1}+\frac{\sqrt{2}}{3^2+1}+\frac{\sqrt{3}}{4^2+1}+...$

The n^th term of the series is a_n = ${\displaystyle \frac{\sqrt{\mathrm{n}}}{(\mathrm{n}+1{)}^2+1}}$

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 

⇒ an = ${\displaystyle \frac{\sqrt{\mathrm{n}}}{(\mathrm{n}+1{)}^2+1}}$

⇒$\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{\sqrt{\mathrm{n}}}{(\mathrm{n}+1{)}^2+1}$

 ⇒$\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{\sqrt{\mathrm{n}}}{{\mathrm{n}}^2+2\mathrm{n}+2}$

⇒$\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{\sqrt{\mathrm{n}}}{\sqrt{\mathrm{n}}({\mathrm{n}}^{\frac{3}{2}}+2\sqrt{\mathrm{n}}+\frac{2}{\sqrt{\mathrm{n}}})}$

⇒$\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}=0$, limit exists and is finite

The series $\frac{1}{2^2+1}+\frac{\sqrt{2}}{3^2+1}+\frac{\sqrt{3}}{4^2+1}+...$ is convergent.

Explanation:

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

b_n = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)sin\frac{n\pi x}{l}dx$

Hence only sine terms are left in Fourier expansion of x³

Additional Information Fourier Series

f(x) = $\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos\frac{n\pi x}{l}+\sum _{n=1}^{\mathrm{\infty }}{b}_{n}sin\frac{n\pi x}{l}$

a_o = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)dx$

a_n = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)cos\frac{n\pi x}{l}dx$

bn = $\frac{1}{l}{\int }_{\alpha }^{\alpha +2l}f(x)sin\frac{n\pi x}{l}dx$

${\mathrm{m}}_{\mathrm{t}\mathrm{h}}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m},{\mathrm{X}}_{\mathrm{m}}=\frac{1}{4^{\mathrm{m}}}{\left(\mathrm{x}-1\right)}^{2\mathrm{m}}$

${\left(\mathrm{m}+1\right)}^{\mathrm{t}\mathrm{h}}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m},{\mathrm{X}}_{\mathrm{m}+1}=\frac{1}{4^{\mathrm{m}+1}}{\left(\mathrm{x}-1\right)}^{2\mathrm{m}+2}$

$\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ \mathrm{m}\to \mathrm{\infty }\end{array}\frac{{\mathrm{X}}_{\mathrm{m}+1}}{{\mathrm{X}}_{\mathrm{m}}}=\frac{{\left(\mathrm{x}-1\right)}^2}{4}$

The series converges if $\frac{{\left(\mathrm{x}-1\right)}^2}{4}<1$

⇒ $-1<\mathrm{x}<3$.

Concept:

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

1. Improper Integral of the First kind: $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ is said to be the improper integral of the first kind if a = -∞ or b = ∞ or both.
2. Improper Integral of the second kind: $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ 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:

$f\left(x\right)=\underset{0}{\overset{\infty }{\int }}\frac{sinx}{x}dx$

By using the Leibniz rule

Let, a is any parameter then,

$I\left(a\right)=\underset{0}{\overset{\infty }{\int }}{e}^{-ax}\times \frac{sinx}{x}dx$      _________(1)

Differentiating w.r.t.a

$I^{\prime }\left(a\right)=\underset{0}{\overset{\infty }{\int }}\frac{\delta }{\delta x}\left(e^{-ax}\times \frac{sinx}{x}\right)dx$

$I^{\prime }\left(a\right)=\underset{0}{\overset{\infty }{\int }}-x{e}^{-ax}\left(\frac{sinx}{x}\right)$

$I^{\prime }\left(a\right)=\underset{0}{\overset{\infty }{\int }}{e}^{-ax}sinxdx$

$I^{\prime }\left(a\right)=-{{\left[\frac{e^{-ax}}{a^2+1}\left(=asinx-cosx\right)\right]}^{\infty }}_0$

$I^{\prime }\left(a\right)=-\left[\frac{e^{-\infty }}{a^2+1}\left(-asin\infty -cos\infty \right)-\frac{e^0}{a^2+1}\left(-asin0-cos0\right)\right]$

$I^{\prime }\left(a\right)=-\left[\frac{-1}{a^2+1}\left(-1\right)\right]$

$I^{\prime }\left(a\right)=\frac{-1}{a^2+1}$

Integrating both side w.r.t a

 $I\left(a\right)=-\int \frac{1}{a^2+1}da$

$I\left(a\right)=-{\tan }^{-1}\left(a\right)+c$  _______(2)

put x = ∞ in both (1) and (2)

From equation (1),

$I\left(\infty \right)=\underset{0}{\overset{\infty }{\int }}{e}^{-\infty }\times \frac{sinx}{x}dx$

I(∞) = 0

From equation (2),

$I\left(\infty \right)=-{\tan }^{-1}\left(\infty \right)+c$

$0=-{\tan }^{-1}\left(\infty \right)+c$

$0=-\frac{\pi }{2}+c$

$c=\frac{\pi }{2}$

$I\left(a\right)=-{\tan }^{-1}\left(a\right)+\frac{\pi }{2}$

put a = 0, we get

$I(0)=\frac{\pi }{2}$

Therefore we can conclude that, $\underset{0}{\overset{\infty }{\int }}\frac{sinx}{x}dx=\frac{\pi }{2}$​

Hence,It is improper integral but having finite value so, It is Converges in [0, ∞].

Calculation:

From the trigonometric identity:

cos (2x) = 1 - 2sin²(x)

2sin2(x) = 1 - cos (2x)

f(x) = sin2 x = $\frac{1}{2}-\frac{\cos 2x}{2}$

Hence, option 1 is correct.

Concept:

Fourier Series is defined as

f(x) = $\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos\frac{n\pi x}{l}+\sum _{n=1}^{\mathrm{\infty }}{b}_{n}sin\frac{n\pi x}{l}$

where, $a_o=\frac{1}{l}{\int }_{-l}^{l}f(x)dx$

$a_n=\frac{1}{l}{\int }_{-l}^{l}f(x)cos\frac{n\pi x}{l}dx$

$b_n=\frac{1}{l}{\int }_{-l}^{l}f(x)sin\frac{n\pi x}{l}dx$

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

$a_o=\frac{1}{l}{\int }_{-l}^{l}f(t)dx$

l = 1

$a_o=\frac{1}{1}{\int }_{-1}^{1}f(t)dx$

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

$a_o={\int }_{-1}^{0}0dx+{\int }_0^{1}4dx$ = 4

The constant term is:

$\frac{a_o}{2}$ = 4/2 = 2
Mistake PointsAvoid 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:

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

$f\left(x\right)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos nx+\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin \left(nx\right)$

Where $a_n=\frac{2}{T}{\int }_0^{T}f\left(x\right)\cdot \cos nxdx$

$b_n=\frac{2}{T}{\int }_0^{T}f\left(x\right)\cdot \sin nxdx$

$a_0=\frac{1}{T}{\int }_0^{T}f\left(x\right)dx$

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, a_n = 0, a₀ = 0

So, now we have to find b_n,

$b_n=\frac{2}{2\pi }{\int }_{-\pi }^{\pi }x\cdot \sin \left(nx\right)dx=\frac{1}{\pi }{\int }_{-\pi }^{\pi }x\cdot \mathrm{s}\mathrm{i}\mathrm{n}\left(nx\right)dx$

$\Rightarrow b_n=\frac{2}{\pi }{\int }_0^{\pi }x\cdot \sin \left(nx\right)dx\{\because {\int }_{-a}^{a}f\left(x\right)=2{\int }_0^{a}f\left(x\right)iff\left(x\right)isodd\}$

$\Rightarrow b_n=\frac{2}{n}\cdot (-{(-1)}^n)$

$f\left(x\right)=x=\frac{-2}{n}\cdot {(-1)}^n\cdot \text{sin}\left(nx\right)$

$f\left(x\right)=x=2\sin x-\sin 2x+\frac{2}{3}\sin 3x-\frac{2}{4}\sin 2x+\dots$

Now put $x=\frac{\pi }{2}$ to find $f\left(\frac{\pi }{2}\right)$

$\Rightarrow f\left(\frac{\pi }{4}\right)=\frac{\pi }{2}=\left(2\right)[1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots ]$

$\Rightarrow 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}=\frac{\pi }{4}$

Hence required sum of series is

Alternate Method:

Taylor expansion of tan^-1x is expressed in below:

${\tan }^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots for;\left|x\right|<1$

Put x = 9, above expansion series

${\tan }^{-1}x={\tan }^{-1}1=\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\dots$

Concept:

1. If the value of sequence or the function is less than 1 then it is called convergent

2. If the value of sequence or the function is greater than 1 then it is called divergent

3. If the value of sequence or the function is not fixed, i.e it changes with the value of n, then it is called oscillatory.

Calculation:

Given function is,

$\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ n\to \mathrm{\infty }\end{array}\left[3+{\left(-1\right)}^{\mathrm{n}}\right]$

$\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ n\to \mathrm{\infty }\end{array}\left[3+{\left(-1\right)}^{\mathrm{\infty }}\right]$

For the even function ${\left(-1\right)}^{\mathrm{\infty }}=1$

= 3 + 1

= 4

For the even function ${\left(-1\right)}^{\mathrm{\infty }}=-1$

= 3 – 1

= 2

Hence we can see the function does not have an exact value, it changes with the n value.

∴ The sequence is oscillatory