# Fourier Analysis Questions and Answers – Fourier Half Range Series

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This set of Fourier Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Fourier Half Range Series”.

1. In half range Fourier series expansion, we know the nature of the function in its full time period.
a) True
b) False

Explanation: In half range Fourier series expansion, we know the nature of the function only in its half of the period. It can be either odd or even function in its full range. We assume it to be even function when we find half range cosine series and we assume it to be odd function when we find half range sine series.

2. In half range cosine Fourier series, we assume the function to be _________
a) Odd function
b) Even function
c) Can’t be determined
d) Can be anything

Explanation: In half range Fourier series expansion, the nature of the function in half of its period is only known. So when we find half range cosine series, there are only cosine terms which imply that the function is even function. f(x) = f(-x).

3. Find the half range sine series of the function f(x) = x, when 0<x<$$\frac{\pi}{2}$$ and (π-x) when $$\frac{\pi}{2}$$<x< π.
a) $$\frac{8}{\pi}[\frac{sinx}{1^{2}} – sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} – sin \frac{(7x)}{7^{2}} +……]$$
b) $$\frac{4}{\pi}[\frac{sinx}{1^{2}} + sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} + sin \frac{(7x)}{7^{2}} +……]$$
c) $$\frac{8}{\pi}[\frac{sinx}{1^{2}} + sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} + sin \frac{(7x)}{7^{2}} +……]$$
d) $$\frac{4}{\pi}[\frac{sinx}{1^{2}} – sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} – sin \frac{(7x)}{7^{2}} +……]$$

Explanation: bn = $$\frac{2}{\pi}[\int_{0}^{\pi⁄2}xsinnxdx+\int_{\pi⁄2}^{\pi}(\pi-x)sinnxdx$$]
= $$\frac{4}{\pi}\frac{sin\left(n\frac{\pi}{2}\right)}{n^2}$$ [this term is zero whenever n is even and when odd, it gives 1 or -1]
= $$\frac{4}{\pi}[\frac{sinx}{1^{2}} – sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} – sin \frac{(7x)}{7^{2}} +……].$$

4. Find bn when we have to find the half range sine series of the function x2 in the interval 0 to 3.
a) -18 $$\frac{cos(nπ)}{nπ}$$
b) 18 $$\frac{cos(nπ)}{nπ}$$
c) -18 $$\frac{cos(n \pi⁄2)}{nπ}$$
d) 18 $$\frac{cos(n \pi⁄2)}{nπ}$$

Explanation: bn = $$\frac{2}{3} \int_0^3 x^{2} sin\left(n\pi\frac{x}{3}\right)dx$$
= $$\frac{2}{3} \left(-27\frac{cos(nπ)}{nπ}\right)$$
= $$-18\frac{cos(nπ)}{nπ}.$$

5. What is the formula for Parseval’s relation in Fourier series expansion?
a) $$\int_{-l}^l (f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ]$$
b) $$\int_{-l}^l (f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2 ) ]$$
c) $$\int_{-l}^l (f(x))^2 dx=l⁄2 [\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ]$$
d) $$l\int_{-l}^l (f(x))^2 dx=[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ]$$

Explanation: The real life significance of Parseval’s relation is to find the energy of the signal in its time period (1 time period). This can be found using the Fourier series coefficients. $$\int_{-l}^l (f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ]$$ is the realtion between the function and the Fourier series coefficients.

6. In Parseval’s relation of Half range Fourier cosine series expansion, which of the following terms doesn’t appear?
a) a0
b) an
c) bn
d) all terms appear

Explanation: In the expansion of a function in half range Fourier cosine series, only an and a0 appear. As the function is considered to be an even function, bn term becomes a null value as the integral becomes zero. So, only bn doesn’t appear in the parseval’s relation of half range Fourier cosine series.

7. Find the value of $$\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2}$$ +….when finding the Half range Fourier sine series of the function f(x) = 1 in 0<x<π.
a) $$\frac{\pi^2}{4}$$
b) $$\frac{\pi^2}{8}$$
c) $$\frac{\pi^2}{2}$$
d) $$3\frac{\pi^2}{8}$$

Explanation: Using parseval’s relation for half range Fourier sine series,
$$\int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ]$$
L.H.S. = $$\int_0^\pi(1)^2 dx = \pi$$
bn = $$\frac{2}{\pi} \int_0^π1.sin(nx)dx$$
= $$\frac{2}{\pi n} (1-(-1)^n )$$
Using parseval’s formula,
$$\pi = \frac{\pi}{2} * \frac{4}{\pi^2} * 2 (\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +….)$$
Therefore,
$$\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +….= \frac{\pi^2}{8}.$$

8. Find the value of$$\frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +\frac{1}{7^4}$$ +….by finding the half range Fourier cosine series of the function f(x) = x in the interval 0<x<l.
a) $$\frac{\pi^4}{12}$$
b) $$\frac{\pi^4}{48}$$
c) $$\frac{\pi^4}{24}$$
d) $$\frac{\pi^4}{96}$$

Explanation: Using Parseval’s relation,
$$\int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+\sum_{n=1}^{∞}(a_n^2+b_n^2 ) ]$$
L.H.S. = $$\int_0^l x^2 dx= \frac{l^3}{3}$$
a0 = $$\frac{2}{l} \int_0^l xdx= l$$
an = $$\frac{2}{l} \int_0^lxcos(\frac{nπx}{l})dx$$
= $$\frac{-4l}{\pi^2} (\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +…)$$
R.H.S. = $$\frac{l}{2} (\frac{l^2}{2}+\left(-4\frac{l}{\pi}\right)^2 (\frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +…..))$$
Therefore,
$$\frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +\frac{1}{7^4} +….= (\frac{l^3}{3}-\frac{l^3}{4})\left(2\frac{\pi^4}{(16l^3 )}\right)$$
= $$\frac{\pi^4}{96}.$$

9. In Parseval’s formula for half range Fourier series, the formula contains l/2 multiplied with the square of individual coefficients.
a) True
b) False

Explanation: The Parseval’s formula for half range Fourier cosine series is $$\int_0^l(f(x))^2 dx=\frac{l}{2} [\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2)]$$
The Parseval’s formula for half range Fourier sine series is $$\int_0^l(f(x))^2 dx=\frac{l}{2} ∑_{n=1}^∞(b_n^2) .$$

10. What is the value of a0 if the function is f(x) = x3 in the interval 0 to 5?
a) 25/4
b) 125/4
c) 625/4
d) 5/4

Explanation: $$\int_0^5 x^3 dx= \int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2)]$$ (from 0 to 5)
= $$\frac{5^4}{4}= \frac{625}{4}.$$

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