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anonymous

  • one year ago

Show that \[\prod_{n=2}^N\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}\]

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  1. ybarrap
    • one year ago
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    Simplifying the product a bit: $$ \prod_{n=2}^N\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}\\ =\prod_{n=2}^N\frac{\left((n-1)(n+1)\right)^{n^2}}{n^{(n+1)^2+(n-1)^2}}\\ =\prod_{n=2}^N\cfrac{(n^2-1)^{n^2}}{n^{2(n^2+1)}}\\ =\prod_{n=2}^N\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2} $$ Using this simplified form, we use Induction to prove: $$ =\prod_{n=2}^N\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2}=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}} $$ Basis: \(n=2\) $$ \cfrac{1}{2^2}\left(\frac{2^2-1}{2^2}\right)^{2^2}=\frac{3^4}{2^{10}}\\ =\frac{(2+1)^{2^2}}{2\cdot 2^{(2+1)^2}}=\frac{3^4}{2^{10}}\\ $$ Inductive Step: We assume equation holds for \(k\), then for \(k+1\) $$ =\prod_{n=2}^{k+1}\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2}\\ =\frac{1}{(k+1)^2}\left (\frac{(k+1)^2-1}{(k+1)^2}\right )^{(k+1)^2}\prod_{n=2}^{k}\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2}\\ =\frac{1}{(k+1)^2}\left (\frac{(k+1)^2-1}{(k+1)^2}\right )^{(k+1)^2}\frac{(k+1)^{k^2}}{2k^{(k+1)^2}}\text{ by Induction Hypothesis}\\ =\frac{1}{(k+1)^2}\frac{(k+1)^{k^2}}{(k+1)^{2(k+1)^2}}\times \frac{ \left((k+1)^2-1\right )^{(k+1)^2}}{2k^{k+1)^2}}\\ =\frac{1}{(k+1)^{(2-k^2+2(k+1)^2)}}\times \frac{\left(k^2+2k\right)^{(k+1)^2}}{2k^{(k+1)^2}} \\ =\frac{1}{(k+1)^{((k+1)+1)^2}}\frac{\left( (k+1)+1\right )^{(k+1)^2 }}{2}\\ =\frac{\left( (k+1)+1\right )^{(k+1)^2}}{2(k+1)^{((k+1)+1)^2}}\\ $$ QED Does this make sense?

  2. anonymous
    • one year ago
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    actually those 'simplifications' were a bad idea: $$\prod_{n=2}^N(n-1)^{n^2}\cdot\prod_{n=2}^N(n+1)^{n^2}\cdot\prod_{n=2}^Nn^{-(n+1)^2}\cdot\prod_{n=2}^Nn^{-(n-1)^2}\\\begin{align*}\quad&=\prod_{n=1}^{N-1}n^{(n+1)^2}\cdot\prod_{n=3}^{N+1} n^{(n-1)^2}\cdot\prod_{n=2}^Nn^{-(n+1)^2}\cdot\prod_{n=2}^Nn^{-(n-1)^2}\\&=1^{2^2}\cdot\prod_{n=2}^{N-1}n^{(n+1)^2}\cdot\prod_{n=3}^{N+1} n^{(n-1)^2}\cdot\prod_{n=2}^{N}n^{-(n+1)^2}\cdot\prod_{n=3}^Nn^{-(n-1)^2}\cdot2^{-1^2}\\&=\frac12\cdot\prod_{n=2}^{N-1}n^{(n+1)^2}\cdot(N+1)^{N^2}\cdot\prod_{n=3}^{N} n^{(n-1)^2}\cdot N^{-(N+1)^2}\cdot\prod_{n=2}^{N-1}n^{-(n+1)^2}\cdot\prod_{n=3}^Nn^{-(n-1)^2}\\&=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}\end{align*}$$

  3. anonymous
    • one year ago
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    this is a telescoping product and it works best factored

  4. ybarrap
    • one year ago
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    Brilliant!

  5. anonymous
    • one year ago
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    Nicely done! This problem came from an interesting question about an infinite product representation of \(\pi\): \[\pi=e^{3/2}\prod_{n=2}^\infty e\left(1-\frac{1}{n^2}\right)^{n^2}\] For those interested, here's the link for the rest of the derivation: http://math.stackexchange.com/a/1346822/170231

  6. ybarrap
    • one year ago
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    Ahhh... even simpler! $$ \prod_{n=2}^N\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}\\ =\prod_{n=2}^N\frac{(n-1)^{n^2}}{n^{(n+1)^2}}\prod_{n=2}^N\frac{(n+1)^{n^2}}{n^{(n-1)^2}}\\ \text{But, }\\ \prod_{n=2}^N\frac{(n-1)^{n^2}}{n^{(n+1)^2}}=\frac{\cancel{(N-1)^{N^2}}}{N^{(N+1)^2} }\frac{(N-2)^{(N-1)^2}}{\cancel{(N-1)^{N^2} }}\cdots \frac{\cancel{2^{3^2}}}{3^{4^2}}\frac{(2-1)^{2^2}}{\cancel{2^{3^2 }}}=\frac{1}{N^{(N+1)^2}}\\ \text{and }\\ \prod_{n=2}^N\frac{(n+1)^{n^2}}{n^{(n-1)^2}}=\frac{(N+1)^{N^2}}{\cancel{N^{(N-1)^2}}}\frac{\cancel{N^{(N-1)^2}}}{(N-1)^{(N-2)^2}}\cdots \frac{4^{2^2}}{\cancel{3^{2^2}}}\frac{\cancel{3^{2^2}}}{2^{1^2}}=\frac{(N+1)^{N^2}}{2}\\ \text{then}\\ =\prod_{n=2}^N\frac{(n-1)^{n^2}}{n^{(n+1)^2}}\prod_{n=2}^N\frac{(n+1)^{n^2}}{n^{(n-1)^2}}=\frac{1}{N^{(N+1)^2}}\frac{(N+1)^{N^2}}{2}\\ $$ Thanks @SithsAndGiggles for this problem!

  7. anonymous
    • one year ago
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    eh that's more or less exactly the thing I did, I just separated into four rather than two products so you could see the cancellation more clearly without the risk of losing track of terms

  8. anonymous
    • one year ago
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    but yeah, that illustrates the telescoping more clearly :p

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