Pairs upstairs

Calculus Level 5

Calculate

$\displaystyle\ln\left(\sum\limits_{\gamma_1\in\{-1,1\}}\ldots\sum\limits_{\gamma_{2014}\in\{-1,1\}}e^{\gamma_1\gamma_2}e^{\gamma_2\gamma_3}\ldots e^{\gamma_{2013}\gamma_{2014}}\right).$

Note: This kind of summation has applications in the statistical mechanics of magnets and simple models of neural networks.

The answer is 2269.2.

1 solution

Patrick Corn
Mar 6, 2014

Letting $S_n$ be the sum, we note that $S_n = e S_{n-1} + e^{-1} S_{n-1}$ by separating it into the cases where $\gamma_n = \gamma_{n-1}$ and $\gamma_n = -\gamma_{n-1}$ .

So $S_n = (e+e^{-1})S_{n-1}$ .

Applying this inductively, we get $S_n = (e+e^{-1})^{n-2}S_2$ .

Since $S_2 = 2(e+e^{-1})$ , we get $S_n = 2(e+e^{-1})^{n-1}$ .

So $\boxed{\ln S_n = \ln 2 + \left(n-1\right)\ln \left(e + e^{-1}\right)}$ .

Plug in $n = 2014$ to get $\ln(S_{2014}) \approx 2269.2$ .

Why not $\ln S_n=2014\ln(e+e^{-1})$ by the method of generating functions?

Cody Johnson - 6 years, 11 months ago

Pairs upstairs

The answer is 2269.2.

1 solution

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