I am aware that this is the Taylor expansion of $e^x$ but how would people have known that the taylor expansion whose derivative is itself is the exponential function $e^x$.
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1It’s the infinite sum of $\frac{x^{n}}{n!}$ not $\frac{1}{n!}$ – Jack Moody Nov 14 '18 at 02:46
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infinite sum of 1/n! is e not eˣ. – Unknown Nov 14 '18 at 02:53
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3The question is unclear. How do you define $e^x$? – Taladris Nov 14 '18 at 02:58
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Probably use the definition of limit for $e$. Then you would get the answer by simple binomial theorem(expansion). – Akash Roy Nov 14 '18 at 03:40
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There are several characterizations of the exponential function. Given one such characterization of $\exp(x)$, one can show that $\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$. See THIS. – Mark Viola Nov 14 '18 at 04:07
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You can write down a generic power series, differentiate it term by term, set the derivative equal to the original function, and solve for the coefficients in the power series. – littleO Nov 14 '18 at 06:57
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Hope this will help a lot. Here are answers different from using Taylor expansion.
Why does the sum of the reciprocals of factorials converge to $e$?
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1Rather than posting a link to a different MSE question, please just flag as a duplicate. In general, link-only answers should be avoided and are more appropriate as comments. – Nov 14 '18 at 02:51
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ok i'll. I am new here thats why i dont know much but next time i'll do. – Unknown Nov 14 '18 at 02:52
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HINT: let $f(x):=\sum_{k=0}^\infty\frac{x^k}{k!}$, then start showing, using binomial expansion and absolute convergence, that $f(x+y)=f(x)\cdot f(y)$ for all $x,y\in\Bbb R$.
Then show that $f(1)=e$, hence $f(n)=e^n$ for all $n\in\Bbb N$. Do the same for the cases $f(1/n)$ noticing that $f(n)\cdot f(1/n)=f(1)$ and that $f(-n)=\frac1{f(n)}$.
Then you knows now that $f(q)=e^q$ for all $q\in\Bbb Q$. Now use the locally uniform convergence of the series to show that $f$ is continuous, hence $$f(x)=\lim_{n\to\infty}f(q_n)=\lim_{n\to\infty}e^{q_n}$$ for any sequence of rationals $(q_n)$ that converge to $x$. By last you will need that to show that
$$\lim_{n\to\infty}e^{q_n}=e^x$$
Masacroso
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