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Since $e$ is a real number I know that $e^0 = 1$, but when I enter $z=0$ into the power series definition of $e^z$ I get an output of $0$. Am I doing something wrong?

$$e^z = \sum_{n=0}^\infty \frac{1}{n!}z^n$$

Setting $z=0$:

$$e^0 = \sum_{n=0}^\infty \frac{1}{n!}0^n = \sum_{n=0}^\infty 0 = 0$$

What have I done wrong?

guywithllama
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  • You need to know the conventions of power series, that is that $0!=1$ and $z^0=1$ (for all $z$) and then the zero-th term is indeed $1$. – ancient mathematician Mar 30 '24 at 12:20
  • Note that $0^0=1$ by definition. So the first summand is $1$. – Mark Mar 30 '24 at 12:21
  • Thank you so much! I understand now. – guywithllama Mar 30 '24 at 12:26
  • @Mark $0^0 = \lim_{x\to 0^+}x^x= \lim_{x\to 0^+}e^{x\ln(x)}=e^0=1$. – hamam_Abdallah Mar 30 '24 at 12:30
  • @hamam_Abdallah That indeed gives more motivation to defining $0^0=1$, to make $x^x$ continuous at $0$. But I usually prefer to tell my students that just like with $0!=1$, this is just a good definition, which works nicely in many formulas, such as the Taylor expansion. (as otherwise we would have to write the first summand separately) – Mark Mar 30 '24 at 12:36
  • See all the other posts here about $0^0$. For example, $0^0 = 1$ if the exponent is the integer zero (as here). But $0^0=1$ is not so clear when the exponent is the real number zero. – GEdgar Mar 30 '24 at 13:34

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