Does anyone recognise this serie?

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The discussion revolves around the series Σ_{k=0}^{∞} (a^k / (k-x)!) and its behavior based on the value of x. When x is positive, the series involves factorials of negative numbers, which is considered unusual. If x is not an integer, it leads to factorials of non-integer values, also deemed unconventional. For negative integer values of x, the series can be expressed in terms of the exponential function, with modifications involving polynomial subtraction. The conversation highlights the complexities of factorials in series and their implications in mathematical analysis.
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\Sigma_{k=0}^{\infty}\frac{a^k}{(k-x)!}

Thanks!
 
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_joey said:
\Sigma_{k=0}^{\infty}\frac{a^k}{(k-x)!}

Thanks!

If x is positive, then you have factorials of a negative number, which is a tad unusual.

If x is not an integer, then you have factorials of a non-integer. Also unusual.

If x is a negative integer, you have

<br /> \begin{align*}<br /> \Sigma_{k=0}^{\infty}\frac{a^k}{(k-x)!} &amp; = a^x \Sigma_{k=0}^{\infty}\frac{a^{k-x}}{(k-x)!} \\<br /> &amp; = a^x \Sigma_{k=-x}^{\infty} \frac{a^k}{k!} \\<br /> &amp; = a^x \left( e^a - \Sigma_{k = 0}^{-x-1} \frac{a^k}{k!} \right)<br /> \end{align*}​

This is the exponential function, scaled and translated.

P.S. Added in edit. Bad description there sorry. It is not scaled and translated by a constant. You subtract a polynomial, and then divide by a-x.

Cheers -- sylas
 
Last edited:

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