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I have been introduced to the Substitution Rule of Taylor Polynomials. I have not found a proof as to why it holds and there are some aspects that are still not clear to me. $$\\$$ For example: Let $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$$ We can find the Taylor expansion (at $0$) of $e^{-x^2}$ by plugging in $-x^2$ to the previous series: $$e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}=1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\cdots$$ But what if I want to find the expansion of $e^{\sin(x)}$ at $0$? Would it be correct to plug in $\sin(x)$? Specifically: $$e^{\sin(x)}=\sum_{n=0}^{\infty}\frac{\sin^n(x)}{n!}=1+\sin(x)+\frac{\sin^2(x)}{2}+\frac{\sin^3(x)}{6}+\cdots$$ However the above expression is not a polynomial, so that's were my doubt arises. Also, does the substitution rule apply only for Maclaurin Polynomials?

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  • $\begingroup$ You could plug in the series for $\sin x$. $\endgroup$ – quasi Mar 28 '17 at 4:42
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Making it more general, when you want the Taylor expansion of $e^{f(x)}$ around $x=0$, you have two ways for the series composition.

The first one, as you did, is to set $f(x)=t$, to use the classical expansion of $e^t$ and later, just as quasi commented, replace $t$ by its series expansion. This would give $$e^{f(x)}=e^{f(0)}+e^{f(0)} f'(0)\,x+\frac{1}{2} e^{f(0)} \left(f''(0)+f'(0)^2\right)\,x^2+\frac{1}{6} e^{f(0)} \left(f'''(0)+f'(0)^3+3 f'(0) f''(0)\right)\,x^3+O\left(x^4\right)$$ The other way is to use the expansion $$f(x)=f(0)+ f'(0)\,x+\frac{1}{2} f''(0)\,x^2+\frac{1}{6} f'''(0) \,x^3+O\left(x^4\right)$$ Name the above $t$ and use the expansion of $e^t$ for the same result.

Applying it to $f(x)=\sin(x)$ both would lead to $$e^{\sin(x)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+O\left(x^7\right)$$ Just for the fun, use it for $x=\frac \pi 6$. This would give $$\sqrt e=1+\frac{\pi }{6}+\frac{\pi ^2}{72}-\frac{\pi ^4}{10368}-\frac{\pi ^5}{116640}-\frac{\pi ^6}{11197440}+\cdots$$ The rhs is $\approx 1.64857$ while $\sqrt e\approx 1.64872$.

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