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Solving inhomogenous wave equation |
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| Sep21-08, 07:11 PM | #52 |
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Solving inhomogenous wave equation
Yep, and
[tex]\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}} =\frac{\pi^{4}}{96}[/tex] can be used to answer your question by substituting n=m+1 : [tex]\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}} = \sum_{m+1=1}^{\infty} \frac{1}{(2(m+1)-1)^{4}} = \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{4}}=\frac{\pi^{4}}{96}[/tex] |
| Sep21-08, 07:14 PM | #53 |
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Thank you SO much for your patience. Tomorrow I am going through this problem from the start again to make sure I get each step.
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| Sep21-08, 07:25 PM | #54 |
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