Difference between revisions of "Euler E generating function"

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==Theorem==
 
==Theorem==
The following formula holds:
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The following formula holds for $|z|<\pi$:
 
$$\dfrac{2e^{xt}}{e^t+1} = \sum_{k=0}^{\infty} \dfrac{E_n(x)t^n}{n!},$$
 
$$\dfrac{2e^{xt}}{e^t+1} = \sum_{k=0}^{\infty} \dfrac{E_n(x)t^n}{n!},$$
 
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.
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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Euler numbers|next=findme}}: $\S 1.14 (2)$
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Euler numbers|next=Euler E n'(x)=nE n-1(x)}}: $\S 1.14 (2)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 01:05, 4 March 2018

Theorem

The following formula holds for $|z|<\pi$: $$\dfrac{2e^{xt}}{e^t+1} = \sum_{k=0}^{\infty} \dfrac{E_n(x)t^n}{n!},$$ where $e^{xt}$ denotes the exponential function and $E_n$ denotes an Euler E polynomial.

Proof

References