Let f(x) = x for-4 sx<4. Because f is odd on [-4, 4), its Fourier cosine coefficients are all zero. Its Fourier sine coefficients are b, =L* sin ("). dx =i * sin ("*) dz=(-1)**128"m² -6 n' The Fourier series of x' on [-4,4] is sin n

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A: We use the formula for solve this

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A: Period  T =3 Frequency  ω=2πT=2π3 a=-2, b=1 The Fourier series is a0  +    ∑K=1∞ ancosKωx+bnsinKωx…

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A: The function is odd so there will be no constant and cosine terms.  The Fourier series will be:

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A: Given,

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A: There is no correct option of first six option, if there is another option then ok.

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A: Consider the provided question, Given function f(x) = sinh(x),   - π &lt; x &lt; π

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Q: 8, Find Fourier series of the function in the Fig. • -2 (XS-1 -1<Xく 1くxく2 fo) -1 T=4 ー3 ー2 子 -2 スナ

A: Fourier series formula for any function: f(x)=ao+∑n=1∞(ancosnπxL+bnsinnπxL)

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Q: Obtain the Fourier series expansion for the function of period 2π as follows f(t)=2-t/π,   -π<x<π.

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Q: Given the piecewise function, what is its fourier series  f(x)={ 0,      -pi ≤x≤0            1,…

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Q: find Fourier series on [0,2n] 0 0<x<7 (7) f(x) =. 1 T<x<2z

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