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The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation has
- a)only sine terms
- b)only cosine terms
- c)both sine and cosine terms
- d)only sine terms and a non-zero constant
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The Fourier series expansion of x3 in the interval −1 ≤ x <...
The Fourier series expansion of x^3 in the interval [-π, π] can be obtained by finding the coefficients of the sine and cosine terms in the expansion.
The Fourier series expansion of a function f(x) in the interval [-π, π] is given by:
f(x) = a0/2 + Σ(an*cos(nx) + bn*sin(nx))
where a0, an, and bn are the coefficients to be determined.
To find the coefficients, we need to calculate the following integrals:
a0 = (1/π) * ∫(x^3)dx from -π to π
an = (1/π) * ∫(x^3*cos(nx))dx from -π to π
bn = (1/π) * ∫(x^3*sin(nx))dx from -π to π
Evaluating these integrals, we get:
a0 = (1/π) * ∫(x^3)dx from -π to π
= (1/π) * [(1/4)x^4] from -π to π
= (1/π) * [(1/4)(π^4 - (-π)^4)]
= (1/π) * [(1/4)(π^4 - π^4)]
= 0
an = (1/π) * ∫(x^3*cos(nx))dx from -π to π
= (1/π) * [(x^3/n)*sin(nx) - (3/n)∫(x^2*sin(nx))dx] from -π to π
= 0
bn = (1/π) * ∫(x^3*sin(nx))dx from -π to π
= (1/π) * [(-x^3/n)*cos(nx) + (3/n)∫(x^2*cos(nx))dx] from -π to π
= 0
Therefore, the Fourier series expansion of x^3 in the interval [-π, π] is:
x^3 ≈ 0
This means that the function x^3 can be approximated by a constant value of 0 in the Fourier series expansion.
The Fourier series expansion of a function f(x) in the interval [-π, π] is given by:
f(x) = a0/2 + Σ(an*cos(nx) + bn*sin(nx))
where a0, an, and bn are the coefficients to be determined.
To find the coefficients, we need to calculate the following integrals:
a0 = (1/π) * ∫(x^3)dx from -π to π
an = (1/π) * ∫(x^3*cos(nx))dx from -π to π
bn = (1/π) * ∫(x^3*sin(nx))dx from -π to π
Evaluating these integrals, we get:
a0 = (1/π) * ∫(x^3)dx from -π to π
= (1/π) * [(1/4)x^4] from -π to π
= (1/π) * [(1/4)(π^4 - (-π)^4)]
= (1/π) * [(1/4)(π^4 - π^4)]
= 0
an = (1/π) * ∫(x^3*cos(nx))dx from -π to π
= (1/π) * [(x^3/n)*sin(nx) - (3/n)∫(x^2*sin(nx))dx] from -π to π
= 0
bn = (1/π) * ∫(x^3*sin(nx))dx from -π to π
= (1/π) * [(-x^3/n)*cos(nx) + (3/n)∫(x^2*cos(nx))dx] from -π to π
= 0
Therefore, the Fourier series expansion of x^3 in the interval [-π, π] is:
x^3 ≈ 0
This means that the function x^3 can be approximated by a constant value of 0 in the Fourier series expansion.
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Community Answer
The Fourier series expansion of x3 in the interval −1 ≤ x <...
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, ao = an = 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
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, ao = an = 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
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The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation hasa)only sine termsb)only cosine termsc)both sine and cosine termsd)only sine terms and a non-zero constantCorrect answer is option 'A'. Can you explain this answer?
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