Problem

By expanding x⁴ by a Fourier series, obtain

∞ ∑ n=1  1 n4

Solution

a0 = 1 π ⌠ ⌡ π −π  x4 dx = 2 π4 5 ,

am = 1 π ⌠ ⌡ π −π  x4 cos m x dx = − 48 (−1)m m4 + 8 π2 (−1)m m2

so

f(x) = π4 5 + ∞ ∑ m=1  ⎛ ⎝ − 48 (−1)m m4 + 8 π2 (−1)m m2 ⎞ ⎠ cos m x.

Substitute x = π in the both sides gives

π4 = π4 5 + ∞ ∑ m=1  ⎛ ⎝ − 48 m4 + 8 π2 m2 ⎞ ⎠ (1) = π4 5 − 48 ∞ ∑ m=1  1 m4 + 8 π2 ∞ ∑ m=1  1 m2 (2)

Noting that

∞ ∑ m=1  1 m2 = π2 6 ,

the above can be solved as

∞ ∑ m=1  1 m4 = π4 90 .

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File translated from T_EX by T_TH, version 4.03.
On 22 Oct 2023, 22:20.