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polivijuye

polivijuye

Answered question

2022-06-25

Is there a way to prove this relation?:
n = 0 1 s 2 + ( 1 + 2 n ) 2 ω 2 = π tanh ( π s 2 ω ) 4 s ω
And find the conditions for which this equality hold?

Answer & Explanation

Zayden Andrade

Zayden Andrade

Beginner2022-06-26Added 22 answers

n = 0 1 s 2 + ( 1 + 2 n ) 2 ω 2 = 1 4 ω 2 n = 0 1 ( n + 1 / 2 ) 2 + s 2 / ( 4 ω 2 ) =   1 4 ω 2 n = 0 1 [ n + 1 / 2 + s i / ( 2 ω ) ] [ n + 1 / 2 s i / ( 2 ω ) ] =   1 4 ω 2 Ψ ( 1 / 2 + s i / [ 2 ω ] ) Ψ ( 1 / 2 s i / [ 2 ω ] ) s i / ω =   i 4 s ω [ π cot ( π [ 1 2 s 2 ω i ] ) ] = π i 4 s ω tan ( π s 2 ω i ) = π i 4 s ω [ i tanh ( π s 2 ω ) ] =   π 4 s ω tanh ( π s 2 ω )

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