The Taylor series expansion of $f(x)=e^x$ in $a=1$ is:

$T(x,a=1)=\sum_{n=0}^{\infty}\frac{e\left(x-1\right)^{\ n}}{n!}$

The interval of convergence using the absolute ratio test is $x\in(-\infty,\infty)$

However, if we evaluate the series in $x=1$ we obtain:

$T(x=1,a=1)=\sum_{n=0}^{\infty}\frac{e·0^{\ n}}{n!}=0$

Instead of $f(x=1)=e^1=1$

What is wrong with this calculation?

  • $\begingroup$ In $\sum_{n=0}^{\infty}\frac{e\left(x-1\right)^{\ n}}{n!}$ the starting term is always interpreted as $e$ even when $x=1$. $\endgroup$ Commented Dec 3, 2023 at 12:29
  • 1
    $\begingroup$ ... i.e $0^0=1$. You use the same value for example in the binomial $(a+b)^n =\sum\limits_{i=0}^n {n \choose i}a^ib^{n-i}$ when $a$ or $b=0$, or in the exponential series for $e^x=\sum\limits_{i=0}^\infty \frac{x^i}{i!}$ when $x=0$ to get $e^0=1$. $\endgroup$
    – Henry
    Commented Dec 3, 2023 at 12:38


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