45 lines
1.2 KiB
Markdown
45 lines
1.2 KiB
Markdown
#Math #NT
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# Basel Problem Solution
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## Base Sum
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$$
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\frac {\pi^2} 4 \\
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= \frac {\pi^2} 4 \csc^2 (\frac \pi 2) \\
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= \frac {\pi^2} {4^2} (\csc^2 (\frac \pi 4) + \csc^2 (\frac \pi 4 + \pi))
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$$
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Do this operation $a$ times, with the above equation being the second time:
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$$
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= \frac {\pi^2} {4^{a + 1}}\sum _{n = 1}^{2^{a}} \csc^2(\frac \pi {2^{a+1}} + \frac \pi {2^a}) \\
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= \sum _{n = 1}^{2^{a}} \frac {\pi^2} {4^{a + 1}} \csc^2(\frac \pi {2^{a+1}} + \frac \pi {2^a}) \\
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= \sum _{n = 1}^{2^{a}} \frac {\pi ^2}{4^{a + 1}} \csc^2(\frac \pi {2^{a+1}} + \frac \pi {2^a}) \\
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= \sum _{n = 1}^{2^{a}} (\frac {2^{a + 1}} \pi \sin(\frac \pi {2^{a+1}} + \frac \pi {2^a}))^{-2} \\
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$$
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As $a$ approaches $\infty$:
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$$
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= 2\sum _{n=1}^{\infty} (2n - 1)^{-2}
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$$
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Therefore:
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$$
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\sum _{n = 1}^{\infty} (2n - 1)^{-2} = \frac {\pi^2} {8}
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$$
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## Manipulating this Sum
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$$
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\sum _{n = 1}^{\infty} (2n)^{-2} = \frac 1 4 \sum _{n = 1}^{\infty} n^{-2} \\\sum _{n = 1}^{\infty} (2n - 1)^{-2} = \frac 3 4 \sum _{n = 1}^{\infty} n^{-2} \\\frac {\pi ^2} 8 = \frac 3 4 \sum _{n = 1}^{\infty} n^{-2} \\
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\frac {\pi ^2} 6 = \sum _{n = 1}^{\infty} n^{-2} \\
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$$
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Therefore
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$$
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\frac {\pi ^2} 6 = \sum _{n = 1}^{\infty} n^{-2} \\
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$$ |