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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] is given by:
x^3 = a0/2 + Σ (an*cos(nx) + bn*sin(nx))
where the coefficients are calculated as follows:
a0 = (1/π) ∫[-π, π] x^3 dx = 0 (since x^3 is an odd function)
an = (1/π) ∫[-π, π] x^3*cos(nx) dx = 0 (since the integrand is an odd function)
bn = (1/π) ∫[-π, π] x^3*sin(nx) dx
To calculate bn, we can use integration by parts:
bn = (1/π) ∫[-π, π] x^3*sin(nx) dx
= (1/π) * [-x^3*cos(nx)/(n) + ∫[π, -π] 3x^2*cos(nx)/(n) dx]
= (1/π) * [-x^3*cos(nx)/(n) + 3/n * ∫[π, -π] x^2*cos(nx) dx]
Now, we can use integration by parts again to evaluate the integral:
bn = (1/π) * [-x^3*cos(nx)/(n) + 3/n * (-x^2*sin(nx)/(n) + ∫[π, -π] 2x*sin(nx) dx)]
= (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^2 * ∫[π, -π] x*cos(nx) dx]
= (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^2 * (-x*sin(nx)/(n) + ∫[π, -π] sin(nx) dx)]
= (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^3 * x*sin(nx) - 6/n^3 * cos(nx)]
Therefore, the Fourier series expansion of x^3 in the interval [-π, π] is:
x^3 = Σ (bn*sin(nx))
where bn = (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^3 * x*sin(nx) - 6/n^3 * cos(nx)]
x^3 = a0/2 + Σ (an*cos(nx) + bn*sin(nx))
where the coefficients are calculated as follows:
a0 = (1/π) ∫[-π, π] x^3 dx = 0 (since x^3 is an odd function)
an = (1/π) ∫[-π, π] x^3*cos(nx) dx = 0 (since the integrand is an odd function)
bn = (1/π) ∫[-π, π] x^3*sin(nx) dx
To calculate bn, we can use integration by parts:
bn = (1/π) ∫[-π, π] x^3*sin(nx) dx
= (1/π) * [-x^3*cos(nx)/(n) + ∫[π, -π] 3x^2*cos(nx)/(n) dx]
= (1/π) * [-x^3*cos(nx)/(n) + 3/n * ∫[π, -π] x^2*cos(nx) dx]
Now, we can use integration by parts again to evaluate the integral:
bn = (1/π) * [-x^3*cos(nx)/(n) + 3/n * (-x^2*sin(nx)/(n) + ∫[π, -π] 2x*sin(nx) dx)]
= (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^2 * ∫[π, -π] x*cos(nx) dx]
= (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^2 * (-x*sin(nx)/(n) + ∫[π, -π] sin(nx) dx)]
= (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^3 * x*sin(nx) - 6/n^3 * cos(nx)]
Therefore, the Fourier series expansion of x^3 in the interval [-π, π] is:
x^3 = Σ (bn*sin(nx))
where bn = (1/π) * [-x^3*cos(nx)/(n) - 3/n^2 * x^2*sin(nx) + 6/n^3 * x*sin(nx) - 6/n^3 * cos(nx)]
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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