Fourier series questions with solutions are provided here for students to practice. Fourier series questions are often asked in many competitive exams, and these questions will help and amp up your preparations. Fourier series is a summation that represents periodic functions as waves of simple periodic functions – sine and cosine functions. The main requirement of the Fourier series is to generate infinite periodic sinusoidal waves.

Periodic Functions: For any positive number T, if for a given function F,

F(x + T) = F(x) ∀ x ∈ R

Then F is called a periodic function, and T is called the period of the function. The function repeats its value T, 2T, 3T, … intervals.

For example: Let f = sin x, then sin (x + 2𝜋) = sin x, hence period of sine function is 2n𝜋 for n = 1, 2, 3, …

Fourier Series: For a given function f, we can find the real numbers a₀, a₁, a₂, …a_n,… , b₁, b₂, …, b_n, … by the Euler-Fourier formulae:

\begin{array}{l} a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s n x d x (n = 0, 1, 2, 3, \dots) \end{array}

\begin{array}{l} b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) s i n n x d x (n = 1, 2, 3, \dots) \end{array}

Then the Fourier series of f is given by:

\begin{array}{l} \frac{1}{2} a_{0} + \sum_{n = 1}^{\infty} (a_{n} c o s n x + b_{n} s i n n x) \end{array}

For a function f(x) defined in the interval (–L, L) such that f(x + 2L) = f(x), that is, 2L is the period of f, then the Fourier series of f is given by:

\begin{array}{l} \frac{1}{2} a_{0} + \sum_{n = 1}^{\infty} (a_{n} c o s \frac{n π x}{L} + b_{n} s i n \frac{n π x}{L}) \end{array}

Where,

\begin{array}{l} a_{n} = \frac{1}{L} \int_{- L}^{L} f (x) c o s \frac{n π x}{L} d x (n = 0, 1, 2, 3, \dots) \end{array}

\begin{array}{l} b_{n} = \frac{1}{L} \int_{- L}^{L} f (x) s i n \frac{n π x}{L} d x (n = 1, 2, 3, \dots) \end{array}

Learn more about Fourier Series.

Fourier Series Questions with Solutions

Now let us solve questions on the Fourier series.

Question 1:

Find the Fourier series of the function f(x) = x², –𝜋 < x < 𝜋.

Solution:

Let us find the values of the real numbers a₀, a_n, and b_n. The period of the given function is 2𝜋, then,

\begin{array}{l} a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \int_{- π}^{π} x^{2} d x \end{array}

\begin{array}{l} = \frac{1}{π} {[\frac{x^{3}}{3}]}_{- π}^{π} = \frac{2 π^{2}}{3} \end{array}

⇒ a₀ = 2𝜋²/3.

\begin{array}{l} a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s n x d x = \frac{2}{π} \int_{0}^{π} x^{2} c o s n x d x [\int_{- a}^{a} f (x) d x = {\begin{array}{c} 2 \int_{0}^{a} f (x) d x & , w h e n f i s a n e v e n f u n c t i o n \\ 0 & , w h e n f i s a n o d d f u n c t i o n \end{array}] \end{array}

\begin{array}{l} = \frac{2}{π} {[x^{2} \frac{s i n n x}{n} + 2 x (\frac{c o s n x}{n^{2}}) + 2 (- \frac{s i n n x}{n^{2}})]}_{0}^{π} \end{array}

\begin{array}{l} = \frac{4 (- 1)^{n}}{n^{2}} [∵ s i n n π = 0 a n d c o s n π = (- 1)^{n}] \end{array}

\begin{array}{l} \Rightarrow a_{n} = {\begin{array}{c} - \frac{4}{n^{2}} & , w h e n n i s o d d \\ \frac{4}{n^{2}} & , w h e n n i s e v e n \end{array} \end{array}

Now x²sin nx is an odd function

∴ b_n= 0.

The Fourier series of x² is

\begin{array}{l} x^{2} = \frac{π^{2}}{3} + \sum_{n = 1}^{\infty} \frac{4 (- 1)^{n}}{n^{2}} c o s n x + \sum_{n = 1}^{\infty} 0. s i n n x \end{array}

\begin{array}{l} x^{2} = \frac{π^{2}}{3} + 4 [- c o s x + \frac{1}{4} c o s 2 x - \frac{1}{9} c o s 3 x + \dots] \end{array}

Question 2:

Find the Fourier series to represent e^ax for x ∈ (–𝜋, 𝜋).

Solution:

Since x ∈ (–𝜋, 𝜋), therefore, the period of the given function is 2𝜋. Then,

\begin{array}{l} a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \int_{- π}^{π} e^{a x} d x \end{array}

\begin{array}{l} = \frac{1}{π} {[\frac{e^{a x}}{a}]}_{- π}^{π} = \frac{1}{π} [\frac{e^{a π} - e^{- a π}}{a}] = \frac{2 s i n h a π}{a π} \end{array}

\begin{array}{l} a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s n x d x = \frac{1}{π} \int_{- π}^{π} e^{a x} c o s n x d x \end{array}

\begin{array}{l} = \frac{1}{π} {[\frac{e^{a x}}{(a^{2} + n^{2})} (a c o s n x + n s i n n x)]}_{- π}^{π} = \frac{1}{π} \frac{a c o s n π}{(a^{2} + n^{2})} (e^{a π} - e^{- a π}) = \frac{1}{π} \frac{2 a (- 1)^{n} . s i n h a π}{(a^{2} + n^{2})} \end{array}

Similarly,

\begin{array}{l} b_{n} = \frac{1}{π} \frac{2 n (- 1)^{n} . s i n h a π}{(a^{2} + n^{2})} \end{array}

\begin{array}{l} ∴ e^{a x} = \frac{s i n h a π}{a π} + \sum_{n = 1}^{\infty} \frac{2 a (- 1)^{n} s i n h a π}{π (a^{2} + n^{2})} c o s n x + \sum_{n = 1}^{\infty} \frac{2 n (- 1)^{n} s i n h a π}{π (a^{2} + n^{2})} s i n n x \end{array}

Question 3:

If f(x) = cos x for 0 < x < 𝜋, f(x) = 50 for 𝜋 ≤ x < 2𝜋 and f(x + 2𝜋) = f(x) ∀ x. Find the sum of the Fourier series of f at x = 𝜋.

Solution:

Sum of the Fourier series at x = 𝜋 is given by

½ [f(𝜋–) + f(𝜋+)] = ½ (cos 𝜋 + 50) = ½ ( –1 + 50) = 24.5.

Question 4:

Find the Fourier series of the periodic function f(x), such that

\begin{array}{l} f (x) = {\begin{array}{c} - π & , w h e n - π < x < 0 \\ x & w h e n - 0 < x < π \end{array} \end{array}

Solution:

Clearly, the period of the function of 2𝜋, then,

\begin{array}{l} a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \int_{- π}^{0} - π d x + \frac{1}{π} \int_{0}^{π} x d x = - [x]_{- π}^{0} + \frac{1}{π} {[\frac{x^{2}}{2}]}_{0}^{π} \end{array}

\begin{array}{l} = - π + \frac{π}{2} = - \frac{π}{2} \end{array}

\begin{array}{l} a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s n x d x = \frac{1}{π} \int_{- π}^{0} (- π) c o s n x d x + \frac{1}{π} \int_{0}^{π} x c o s n x d x \end{array}

\begin{array}{l} = - {[\frac{s i n n x}{n}]}_{- π}^{0} + \frac{1}{π} {[x \frac{s i n n x}{n} + \frac{c o s n x}{n^{2}}]}_{0}^{π} = \frac{c o s n π - 1}{π n^{2}} \end{array}

\begin{array}{l} \Rightarrow a_{n} = {\begin{array}{c} 0 & , w h e n n i s e v e n \\ - \frac{2}{π n^{2}} & , w h e n n i s o d d \end{array} \end{array}

\begin{array}{l} b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) s i n n x d x = \frac{1}{π} \int_{- π}^{0} (- π) s i n n x d x + \frac{1}{π} \int_{0}^{π} x s i n n x d x \end{array}

\begin{array}{l} = \frac{1}{π} {[- π \frac{c o s n x}{n}]}_{- π}^{0} + \frac{1}{π} {[\frac{- x c o s n x}{n} + (\frac{s i n n x}{n^{2}})]}_{0}^{π} = \frac{1 - 2 c o s n π}{n} \end{array}

\begin{array}{l} \Rightarrow b_{n} = {\begin{array}{c} - \frac{1}{n} & , w h e n n i s e v e n \\ \frac{3}{n} & , w h e n n i s o d d \end{array} \end{array}

Putting the values of a’s and b’s we get the required fourier series:

f(x) = – 𝜋/4 – 2/𝜋[cos x + 1/3² cos 3x + 1/5² cos 5x + …] + [3 sin x – ½ sin 2x + 3/3 sin 3x – …]

Dirichlet’s Condition

Dirichlet’s condition for Fourier expansion of a given function f are:

The function is well-defined and single-valued except at a finite number of points in (–L, L).
The function is with period 2L
f(x) and f’(x) are piecewise continuous in (–L, L)
The function has finite number of maximas and minimas.

Then the Fourier expansion of the function converges to

f(x) if x is a point of continuity
½ f(x + 0) + f(x – 0) if x is a point of discontinuity.

Half-range Fourier Sine or Cosine series

A Fourier series with only sine or cosine terms is called half-range series. This series is defined in (0, L) or (–L, 0). In such case we have,

\begin{array}{l} a_{n} = 0, b_{n} = \frac{2}{L} \int_{0}^{L} f (x) s i n \frac{n π x}{L} d x f o r h a l f r a n g e s i n e s e r i e s \end{array}

\begin{array}{l} b_{n} = 0, a_{n} = \frac{2}{L} \int_{0}^{L} f (x) c o s \frac{n π x}{L} d x f o r h a l f r a n g e c o s i n e s e r i e s \end{array}

Question 5:

Prove the following for k = 1, 2, 3, …

\begin{array}{l} \int_{- L}^{L} s i n \frac{k π x}{L} d x = \int_{- L}^{L} c o s \frac{k π x}{L} = 0 \end{array}

Solution:

Now, sin (–𝜃) = – sin 𝜃

And

\begin{array}{l} \int_{- a}^{a} f (x) d x = {\begin{array}{c} 2 \int_{0}^{a} f (x) d x & w h e n f i s a n e v e n f u n c t i o n \\ 0 & w h e n f i s a n o d d f u n c t i o n \end{array} \end{array}

\begin{array}{l} ∴ \int_{- L}^{L} s i n \frac{k π x}{L} d x = 0 f o r k = 1, 2, 3, \dots \end{array}

\begin{array}{l} \int_{- L}^{L} c o s \frac{k π x}{L} d x = \frac{L}{k π x} {[s i n \frac{k π x}{L}]}_{- L}^{L} = \frac{L}{k π x} (s i n k π + s i n k π) = 0 (∵ s i n n π = 0) \end{array}

Also Read:

Question 6:

Find the Fourier series of |x| where –𝜋 < x < 𝜋.

Solution:

Let f(x) = |x| and period of f is 2𝜋.

\begin{array}{l} a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \int_{- π}^{π} | x | d x = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} {[\frac{x^{2}}{2}]}_{0}^{π} d x = π \end{array}

\begin{array}{l} a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s n x d x = \frac{2}{π} \int_{0}^{π} | x | c o s n x d x \end{array}

\begin{array}{l} = \frac{2}{π} {[\frac{x s i n n x}{n} + \frac{c o s n x}{n^{2}}]}_{0}^{π} = \frac{2}{π} [\frac{c o s n π}{n^{2}} - \frac{1}{n^{2}}] \end{array}

= 0 if n is even

= –4/𝜋n² if n is odd.

And b_n = 0, since |x| is an even function.

Therefore, the fourier series of |x| is

𝜋/2 + 4/𝜋 [cos x + (cos 3x)/3² + (cos 5x)/5² + … ]

Question 7:

Find the value of the real number a₀ of the Fourier series if f(x) = x² for 0 ≤ x ≤ 2𝜋.

Solution:

Given f(x) = x², for this also the period of the function is 2𝜋

\begin{array}{l} a_{0} = \frac{1}{π} \int_{0}^{2 π} f (x) d x = \frac{1}{π} \int_{0}^{2 π} x^{2} d x = \frac{1}{π} {[\frac{x^{3}}{3}]}_{0}^{2 π} = \frac{8 π^{2}}{3} \end{array}

∴ a₀ = 8𝜋²/3.

Question 8:

Express f(x) = sin(𝜋x/l) as half range cosine series for 0 ≤ x ≤ l.

Solution:

Let sin(𝜋x/l) = a₀/2 + ∑ a_n cos(n𝜋x/l) where

\begin{array}{l} a_{0} = \frac{2}{l} \int_{0}^{1} f (x) d x = \frac{2}{l} \int_{0}^{1} s i n (\frac{π x}{l}) d x = \frac{2}{l} {[\frac{c o s \frac{π x}{l}}{π / l}]}_{0}^{l} \end{array}

\begin{array}{l} = - \frac{2}{l} c o s (π - 1) = \frac{4}{π} \end{array}

\begin{array}{l} a_{n} = \frac{2}{l} \int_{0}^{l} s i n \frac{π x}{l} c o s \frac{n π x}{l} d x = \frac{1}{l} \int_{0}^{l} [s i n (n + 1) \frac{π x}{l} - s i n (n - 1) \frac{π x}{l}] d x \end{array}

\begin{array}{l} = \frac{1}{l} {[- \frac{c o s (n + 1) \frac{π x}{l}}{(n + 1) \frac{π}{l}} + \frac{c o s (n - 1) \frac{π x}{l}}{(n - 1) \frac{π}{l}}]}_{0}^{l} = \frac{1}{π} [{- \frac{c o s (n + 1) π}{(n + 1)}} - {- \frac{1}{n + 1} + \frac{1}{n - 1}}] \end{array}

When n is odd, a_n = 0

and when n is even, a_n = –4/[𝜋(n + 1)(n – 1)]

∴ the half-range fourier series for sin(𝜋x/l) is given by

\begin{array}{l} s i n (\frac{π x}{l}) = \frac{2}{π} - \frac{4}{π} [\frac{c o s \frac{2 π x}{l}}{1.3} + \frac{c o s \frac{4 π x}{l}}{3.5} + \frac{c o s \frac{6 π x}{l}}{5.7} + \dots] \end{array}

Question 9:

Express the given function f(x) as Fourier sine half range series.

\begin{array}{l} f (x) = {\begin{array}{c} x & , 0 \leq x \leq \frac{π}{2} \\ π - x & , \frac{π}{2} < x \leq π \end{array} \end{array}

Solution:

Let the Fourier sine hailf-range series is given by

f(x) = ∑_nb_n sin nx where

\begin{array}{l} b_{n} = \frac{2}{π} \int_{0}^{π} f (x) s i n n x d x = \frac{2}{π} \int_{0}^{π / 2} x s i n n x d x + \frac{2}{π} \int_{π / 2}^{π} (π - x) s i n n x d x \end{array}

\begin{array}{l} = \frac{2}{π} {[x . (- \frac{c o s n x}{n}) + (\frac{s i n n x}{n^{2}})]}_{0}^{π / 2} + \frac{2}{π} {[(π - x) . (- \frac{c o s n x}{n}) - (\frac{s i n n x}{n^{2}})]}_{π / 2}^{π} \end{array}

\begin{array}{l} = \frac{2}{π} [- \frac{π}{2 n} c o s \frac{n π}{2} + \frac{1}{n^{2}} s i n \frac{n π}{2}] + \frac{2}{π} [\frac{π}{2 n} c o s \frac{n π}{2} + \frac{1}{n^{2}} s i n \frac{n π}{2}] = \frac{4}{π n^{2}} s i n \frac{n π}{2} \end{array}

When n is even, b_n = 0

When n is odd, b_n = 4/n²𝜋

Hence, the Fourier sine half-range series is given by

\begin{array}{l} f (x) = \frac{4}{π} [\frac{1}{1^{2}} s i n x - \frac{1}{3^{2}} s i n 3 x + \frac{1}{5^{2}} s i n 5 x - \dots] \end{array}

Question 10:

Deduce the half-range cosine series for f(x) = kx when 0 ≤ x ≤ l/2 and f(x) = k(l – x) when l/2 < x ≤ l.

Solution:

Let the Fourier cosine half-range series of f is

f(x) = a₀/2 + ∑_n a_ncos (n𝜋x/l)

\begin{array}{l} a_{0} = \frac{2}{l} \int_{0}^{l} f (x) d x = \frac{2}{l} \int_{0}^{l / 2} k x d x + \frac{2}{l} \int_{l / 2}^{l} k (l - x) d x \end{array}

\begin{array}{l} = \frac{2}{l} {[\frac{k x^{2}}{2}]}_{0}^{l / 2} + \frac{2}{l} {[k l x - \frac{k l x^{2}}{2}]}_{l / 2}^{l} = \frac{k l}{2} \end{array}

And for n = 1, 2, 3, …

\begin{array}{l} a_{n} = \frac{2}{l} \int_{0}^{l} f (x) c o s \frac{n π x}{l} d x = \frac{2}{l} \int_{0}^{l 2} k x c o s \frac{n π x}{l} d x + \frac{2}{l} \int_{l / 2}^{l} k (l - x) c o s \frac{n π x}{l} d x \end{array}

\begin{array}{l} = \frac{2}{l} [\frac{2 k l^{2}}{n^{2} π^{2}} c o s \frac{n π}{2} - \frac{k l^{2}}{n^{2} π^{2}} - \frac{k l^{2}}{n^{2} π^{2}} c o s n π] = \frac{2 k l}{n^{2} π^{2}} [2 c o s \frac{n π}{2} - 1 - c o s n π] \end{array}

When n is odd cos (n𝜋/2) = 0 and cos n𝜋 = –1 ∴ a_n = 0

When n is even,

a₂ = 8kl/2²𝜋², a₄ = 0, a₆ = 8kl/6²𝜋² and so on.

Hence, the Fourier half-range cosine series of f is

\begin{array}{l} f (x) = \frac{k l}{4} - \frac{8 k}{π^{2}} (\frac{1}{2^{2}} c o s \frac{2 π x}{l} + \frac{1}{6^{2}} c o s \frac{6 π x}{l} + \dots) \end{array}

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Practice Questions on Fourier Series

1. Find the Fourier series of the following:

(i) x – x² for –𝜋 ≤ x ≤ 𝜋

(ii) f(x) = –1 for –𝜋 < x < 0 and f(x) = 1 for 0 ≤ x ≤ 𝜋.

2. For f(x) = x²(1 – x⁴) find the b_n of the Fourier series x ∈ (–l,l).

3. Write the Fourier sine series for f(x) = k in (0, 𝜋).

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Fourier Series Questions with Solutions

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Practice Questions on Fourier Series

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