250+ TOP MCQs on Fourier Series Expansions and Answers

Fourier Analysis Multiple Choice Questions on “Fourier Series Expansions”.

1. Which of the following is not Dirichlet’s condition for the Fourier series expansion?
a) f(x) is periodic, single valued, finite
b) f(x) has finite number of discontinuities in only one period
c) f(x) has finite number of maxima and minima
d) f(x) is a periodic, single valued, finite
View Answer

Answer: d
Explanation: Dirichlet’s condition for Fourier series expansion is f(x) should be periodic, single valued and finite; f(x) should have finite number of discontinuities in one period and f(x) should have finite number of maxima and minima in a period.

2. At the point of discontinuity, sum of the series is equal to ___________
a) (frac{1}{2} [f(x+0) – f(x-0)] )
b) (frac{1}{2} [f(x+0) + f(x-0)] )
c) (frac{1}{4} [f(x+0) – f(x-0)] )
d) (frac{1}{4} [f(x+0) + f(x-0)] )
View Answer

Answer: b
Explanation: When there is a point of discontinuity, the value of the function at that point is found by taking the average of the limit of the function in the left hand side of the discontinuous point and right hand side of the discontinuous point. Hence the value of the function at that point of discontinuity is (frac{1}{2} [f(x+0) + f(x-0)] ).

3. What is the Fourier series expansion of the function f(x) in the interval (c, c+2π)?
a) (frac{a_0}{2}+∑_{n=1}^∞ a_n cos(nx) +∑_{n=1}^∞ b_n sin(nx) )
b) (a_0+∑_{n=1}^∞ a_n cos(nx) +∑_{n=1}^∞ b_n sin(nx) )
c) (frac{a_0}{2}+∑_{n=0}^∞ a_n cos(nx) +∑_{n=0}^∞ b_n sin(nx) )
d) (a_0+∑_{n=0}^∞ a_n cos(nx) + ∑_{n=0}^∞ b_n sin(nx) )
View Answer

Answer: a
Explanation: Fourier series expantion of the function f(x) in the interval (c, c+2π) is given by (frac{a_0}{2}+∑_{n=1}^∞ a_n cos(nx) +∑_{n=1}^∞ b_n sin(nx) ) where, a0 is found by using n=0, in the formula for finding an. bn is found by using sin(nx) instead of cos(nx) in the formula to find an.

4. If the function f(x) is even, then which of the following is zero?
a) an
b) bn
c) a0
d) nothing is zero
View Answer

Answer: b
Explanation: Since bn includes sin(nx) term which is an odd function, odd times even function is always odd. So, the integral gives zero as the result.

5. If the function f(x) is odd, then which of the only coefficient is present?
a) an
b) bn
c) a0
d) everything is present
View Answer

Answer: b
Explanation: Since to find bn we have sin(nx) and the function we have is also odd function, the product of odd function and another odd function yields even function, the only coefficient which exists is bn.

6. Find a0 of the function (f(x) = sqrt{frac{1-cosx}{2}}.)
a) (frac{4}{π} )
b) (frac{2}{π} )
c) (frac{π}{4} )
d) (frac{π}{2} )
View Answer

Answer: a
Explanation: (f(x) = sqrt{frac{1-cosx}{2}} = sin(frac{x}{2}) )
(a_0 = frac{1}{π} int_0^{2π} sin(frac{x}{2})dx )
( = frac{1}{π} (-cos(frac{x}{2}))^2 )
( = frac{(-2)}{ π (-2)} )
( = frac{4}{π} )

7. Find a0 of the function (f(x) = frac{1}{4} (π-x)^2.)
a) (frac{π^2}{6} )
b) (frac{π^2}{12} )
c) (5frac{π^2}{6} )
d) (5frac{π^2}{12} )
View Answer

Answer: a
Explanation: (f(x) = frac{1}{4} (π-x)^2 )
(a_0 = frac{1}{π} int_0^{2π} frac{1}{4} (π-x)^2 )
= from 0 to 2 π
( = frac{1}{2} (frac{1}{3} π^2 ) )
( = frac{π^2}{6} ).

8. Find the sum of (frac{1}{1^2} + frac{1}{3^2} + frac{1}{5^2} ) +……… using Fourier series expansion if f(x) = a when [0,π] and 2 π – x when [ π, 2 π].
a) (frac{π^2}{8} )
b) (frac{π^2}{4} )
c) (frac{π^2}{16} )
d) (frac{π^2}{2} )
View Answer

Answer: a
Explanation: (a_0 = frac{1}{π} (int_0^π x dx)(int_π^{2π} (2π-x) dx) )
( = frac{1}{π} [π^2+4π^2-4 frac{π^2}{2}-2π^2+frac{π^2}{2}] )
( = π.)
(a_n = frac{1}{π} (int_0^π (xcos(nx)dx) )(int_π^{2π}((2π-x)cos(nx))dx) )
( = frac{-4}{πn^2} ) when n is odd and 0 when n is even
( b_n = 0 )
now, substituting x=0 in the given function and the Fourier series expansion, we get,
(0 = frac{π}{2}-frac{4}{π} (frac{1}{1^2} + frac{1}{2^2} + frac{1}{3^2} +….) )
Therefore, (frac{1}{1^2} + frac{1}{3^2} + frac{1}{5^2} +………= frac{π^2}{8} ).

9. Find an if the function f(x) = x – x3.
a) finite value
b) infinite value
c) zero
d) can’t be found
View Answer

Answer: c
Explanation: Since the function is odd function, odd times even function (cos(nx)) is odd function. So in the given interval, the coefficient is zero.

10. Find bn if the function f(x) = x2.
a) finite value
b) infinite value
c) zero
d) can’t be found
View Answer

Answer: c
Explanation: Since the given function is even function, even times odd function(sin(nx)) gives odd function. Hence the coefficient in the given interval is zero.

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250+ TOP MCQs on Expansion of Trigonometric Functions and Answers

Complex Analysis Multiple Choice Questions on “Expansion of Trigonometric Functions”.

1. The Taylor series for f(x)=7x2-6x+1 at x=2 is given by a+b(x-2)+c(x-2)2. Find the value of a+b+c.
a) -1
b) 0
c) 17
d) 46
Answer: d
Explanation: We know
(f(x)=7x^2-6x+1)
(f'(x)=14x-6)
(f”(x)=14)
(f”'(x)=0)
Thus for n>=3, the derivative of the function is 0.
As per the Taylor Series,
(7x^2-6x+1=sum_{n=0}^{infty} frac{f^n (2)(x-2)^n}{n!})
(7x^2-6x+1=f(2)+f'(2)(x-2)+frac{1}{2} f”(2) (x-2)^2+0)
(7x^2-6x+1=17+22(x-2)+7(x-2)^2)
Thus, a=17, b=22, c=7
a+b+c=46
Thus the answer is 46.

2. Find the Taylor Series for the function (f(x)=e^{-6x}) about x=-4.
a) (sum_{n=0}^{infty} frac{(-6)^n}{n!} e^{12} (x+4)^n)
b) (sum_{n=0}^{infty} frac{(-6)^n}{n!} e^{24} (x-4)^n)
c) (sum_{n=0}^{infty} frac{(-6)^n}{n!} e^{24} (x+4)^n)
d) (sum_{n=0}^{infty} frac{(-4)^n}{n!} e^{24} (x+4)^n)
Answer: c
Explanation: We start by finding the derivative of the given function,
(f(x)=e^{-6x})
(f'(x) = -6e^{-6x})
(f”(x) = 36e^{-6x})
(f”'(x) = -216e^{-6x})
(f””(x) = 1296e^{-6x})
Thus we take derivative of maximum to the fourth order.
Thus according to formula of Taylor series about x=-4
(e^{-6x}=sum_{n=0}^{infty} frac{f^n (-4)}{n!} (x+4)^n)
(e^{-6x}=sum_{n=0}^{infty} frac{(-6)^n}{n!} e^{24} (x+4)^n)
Thus the Taylor Series is given by
(e^{-6x}=sum_{n=0}^{infty} frac{(-6)^n}{n!} e^{24} (x+4)^n).

Global Education & Learning Series – Complex Analysis.

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