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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?
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Most Upvoted Answer
The Fourier series expansion of x3 in the interval -1 ≤ x <1 wit...
Given : Function f ( x) = x3 ; -1 ≤ x < 1
The above function f (x) is odd.
We know that, odd function contains only odd terms,
a0 and an will be zero.
Therefore, the given function f (x) contains only sine term.
Hence, the correct option is (A).
The above function f (x) is odd.
We know that, odd function contains only odd terms,
a0 and an will be zero.
Therefore, the given function f (x) contains only sine term.
Hence, the correct option is (A).
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Community Answer
The Fourier series expansion of x3 in the interval -1 ≤ x <1 wit...
The Fourier series expansion of x^3 in the interval -1 < x="" />< 1="" can="" be="" found="" by="" using="" the="" />
f(x) = a0/2 + Σ(an*cos(nπx/L) + bn*sin(nπx/L))
where L is the length of the interval (L = 2 in this case), and the coefficients an and bn can be calculated using the formulas:
an = (2/L) * ∫[f(x)*cos(nπx/L) dx] from -L/2 to L/2
bn = (2/L) * ∫[f(x)*sin(nπx/L) dx] from -L/2 to L/2
For x^3 in the interval -1 < x="" />< 1,="" we="" />
a0 = (2/2) * ∫[x^3 dx] from -1 to 1
= ∫[x^3 dx] from -1 to 1
= [x^4/4] from -1 to 1
= (1^4/4) - (-1^4/4)
= 1/4 - 1/4
= 0
Now, let's calculate the coefficients an and bn:
an = (2/2) * ∫[x^3*cos(nπx/2) dx] from -1 to 1
= ∫[x^3*cos(nπx/2) dx] from -1 to 1
= [x^3*sin(nπx/2)/(nπ/2)] from -1 to 1
= [1^3*sin(nπ/2)/(nπ/2)] - (-1^3*sin(nπ/2)/(nπ/2))
= [sin(nπ/2)/(nπ/2)] - [(-1)*sin(nπ/2)/(nπ/2)]
= [sin(nπ/2)/(nπ/2)] + [sin(nπ/2)/(nπ/2)]
= 2*sin(nπ/2)/(nπ/2)
= 2*sin(nπ/2)/(nπ)
bn = (2/2) * ∫[x^3*sin(nπx/2) dx] from -1 to 1
= ∫[x^3*sin(nπx/2) dx] from -1 to 1
= [-x^3*cos(nπx/2)/(nπ/2)] from -1 to 1
= [-1^3*cos(nπ/2)/(nπ/2)] - (-1^3*cos(nπ/2)/(nπ/2))
= [-cos(nπ/2)/(nπ/2)] - [(-1)*cos(nπ/2)/(nπ/2)]
= [-cos(nπ/2)/(nπ/2)] + [cos(nπ/2)/(nπ/2)]
= 0
Therefore, the Fourier series expansion of x^3 in the interval -1 < x="" />< 1="" />
f(x) = Σ(2*sin(nπ/2)/(nπ) * cos(nπx/2))
f(x) = a0/2 + Σ(an*cos(nπx/L) + bn*sin(nπx/L))
where L is the length of the interval (L = 2 in this case), and the coefficients an and bn can be calculated using the formulas:
an = (2/L) * ∫[f(x)*cos(nπx/L) dx] from -L/2 to L/2
bn = (2/L) * ∫[f(x)*sin(nπx/L) dx] from -L/2 to L/2
For x^3 in the interval -1 < x="" />< 1,="" we="" />
a0 = (2/2) * ∫[x^3 dx] from -1 to 1
= ∫[x^3 dx] from -1 to 1
= [x^4/4] from -1 to 1
= (1^4/4) - (-1^4/4)
= 1/4 - 1/4
= 0
Now, let's calculate the coefficients an and bn:
an = (2/2) * ∫[x^3*cos(nπx/2) dx] from -1 to 1
= ∫[x^3*cos(nπx/2) dx] from -1 to 1
= [x^3*sin(nπx/2)/(nπ/2)] from -1 to 1
= [1^3*sin(nπ/2)/(nπ/2)] - (-1^3*sin(nπ/2)/(nπ/2))
= [sin(nπ/2)/(nπ/2)] - [(-1)*sin(nπ/2)/(nπ/2)]
= [sin(nπ/2)/(nπ/2)] + [sin(nπ/2)/(nπ/2)]
= 2*sin(nπ/2)/(nπ/2)
= 2*sin(nπ/2)/(nπ)
bn = (2/2) * ∫[x^3*sin(nπx/2) dx] from -1 to 1
= ∫[x^3*sin(nπx/2) dx] from -1 to 1
= [-x^3*cos(nπx/2)/(nπ/2)] from -1 to 1
= [-1^3*cos(nπ/2)/(nπ/2)] - (-1^3*cos(nπ/2)/(nπ/2))
= [-cos(nπ/2)/(nπ/2)] - [(-1)*cos(nπ/2)/(nπ/2)]
= [-cos(nπ/2)/(nπ/2)] + [cos(nπ/2)/(nπ/2)]
= 0
Therefore, the Fourier series expansion of x^3 in the interval -1 < x="" />< 1="" />
f(x) = Σ(2*sin(nπ/2)/(nπ) * cos(nπx/2))
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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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