Difference between revisions of "Euler E generating function"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\dfrac{2e^{xt}}{e^t+1} = \sum_{k=...") |
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\dfrac{2e^{xt}}{e^t+1} = \sum_{k=0}^{\infty} \dfrac{E_n(x)t^n}{n!},$$ | $$\dfrac{2e^{xt}}{e^t+1} = \sum_{k=0}^{\infty} \dfrac{E_n(x)t^n}{n!},$$ | ||
where $e^{xt}$ denotes the [[exponential function]] and $E_n$ denotes an [[Euler E]] polynomial. | where $e^{xt}$ denotes the [[exponential function]] and $E_n$ denotes an [[Euler E]] polynomial. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
Revision as of 03:51, 6 June 2016
Theorem
The following formula holds: $$\dfrac{2e^{xt}}{e^t+1} = \sum_{k=0}^{\infty} \dfrac{E_n(x)t^n}{n!},$$ where $e^{xt}$ denotes the exponential function and $E_n$ denotes an Euler E polynomial.
