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I have the following summation:

$$\sum_{k=0}^\infty(1-e^x)^k=\sum_{k=0}^\infty\sum_{j=0}^k\binom{k}{j}(-1)^je^{jx}$$ Then $$e^{jx}=\sum_{i=0}^\infty j^i\frac{x^i}{i!}$$ So, $$\sum_{k=0}^\infty\sum_{j=0}^k\binom{k}{j}(-1)^j\sum_{i=0}^\infty j^i\frac{x^i}{i!}=\sum_{k=0}^\infty\sum_{j=0}^k\sum_{i=0}^\infty\binom{k}{j}(-1)^j j^i\frac{x^i}{i!}$$ However, when I run the program, my first term with have $j^i=0^0$ I don't feel like any of the math is wrong above, so what is happening?

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  • $\begingroup$ What values of $x$ are you using? If $-1<1-e^x<1$ then you've got a geometric series $\sum_{k=0}^\infty r^k = 1/(1-r)$ $\endgroup$
    – grand_chat
    Jun 5 '15 at 22:54
  • $\begingroup$ I was more interested in finding coefficients. The work I'm doing is on formal power series and I'm trying to extract a coefficient. But in my attempt to obtain an explicit formula for the coefficients, I get this. $\endgroup$
    – Iceman
    Jun 5 '15 at 23:16
  • $\begingroup$ Got it. In the power series expansion for $e^x$ the $i=0$ term is always 1. You can finesse the indeterminate form by representing $e^x$ as $e^x=1+\sum_{i=1}^\infty (\cdots)$ $\endgroup$
    – grand_chat
    Jun 6 '15 at 1:05
  • $\begingroup$ the problem arose in my mathematica formula. That is why. I will try and amend my calculation and formula to see if that removes the "indeterminant" error. Thanks. $\endgroup$
    – Iceman
    Jun 6 '15 at 1:07
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In this case, $0^0=1$ for it is what happens for $n=0$ term when $x=0$ in $e^x=\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n!}$.

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  • $\begingroup$ THat is what I thought was happening. I'm doing calculations in mathematica and mathematica tells me it is indeterminant. therefore I can't move on. $\endgroup$
    – Iceman
    Jun 5 '15 at 23:17
  • $\begingroup$ Again, that is what I thought. I suppose then my question is best asked in the mathematica.stackexchange forum. thanks $\endgroup$
    – Iceman
    Jun 6 '15 at 1:06

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