Substituting function into standard Maclaurin series

Question

I understand that when using standard Maclaurin series (e.g. $e^x = 1+x+\frac{x^2}{2!} + ...$ , we can substitute $x$ with other functions such as $z = x^2$ to transform the standard Maclaurin series into $e^{x^2} = 1+x^2+\frac{x^4}{2!} + ...$ I would like to ask why doesn't this require a $\frac{dz}{dx} = 2x$ component in the expansion since the standard Maclaurin series are derived from differentiation of the original term. Thank you!

Answer 1

The answer is simple:

$$f'(x)=\frac d{dx}e^{x^2}=2xe^{x^2}$$ and $f'(0)=0$ ! [Not a factorial]

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You can try the full computation of the Taylor development of $e^{x^2}$ and that of $e^x$ where you substitute $x$ for $x^2$ and you will see a perfect match. This is because the Taylor development is unique.

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Extra note.

The derivatives of $e^{x^2}$ are in close relation to the Hermite polynomials (https://en.wikipedia.org/wiki/Hermite_polynomials). These are alternatively even and odd, and this is why you will not find odd powers in the Taylor development.

Answer 2

(edit2) As Yves has kindly pointed out, the intuition below is not true for substitution into Taylor series.

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Having a shot at the question generally,

Degree-n taylor polynomials centred at $0$, of $f(x)$ and $f(g(x))$ is given by $f(x) \approx \sum_{k=1}^{n}\frac{f^{(k)}(0)}{k!}x^k $ and $f(g(x)) \approx \sum_{k=1}^{n}\frac{(f\circ{g})^{(k)}(0)}{k!}x^k$

Thus, for a direct substitution of $x=g(x)$ into the degree-n Taylor polynomial of $f(x)$ to be equal to the degree-n Taylor polynomial of $f(g(x))$. We need to be able to express $\sum_{k=1}^{n}\frac{(f\circ{g})^{(k)}(0)}{k!}x^k$ in the form of $\sum_{k=1}^{n}\frac{h^{(k)}(0)}{k!}x^k$, implying $g(0)=0$. Thus, I think direct substitution of $g(x)$ in n-degree Taylor polynomials will yield the expected n-degree Taylor expansion for the composite function, only if $g(0)=0$.

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(edit) For OP's example, let $f(x)=e^x, g(x)=x^2, f(g(x))=e^{x^2}$ and $p(x)$ be n-degree Taylor polynomial of $f(x)$, given by $1+x+\frac{x^2}{2}+\dotsb+\frac{x_n}{n!}$. Considering the first few terms of $f(g(x))$, we have

$f(g(x))\approx f(g(0))+xg'(0)f'(g(0))+\frac{1}{2}\,x^2(g'(0)^2f''(g(0))+g''(0)f'(g(0)))$ $=f(0)+0+\frac{1}{2}\,x^2(2f'(0))$ $=1+x^2=p(x^2)$ where $n=2$

A counterexample where a substitution into a n-degree Taylor polynomial does not yield the expected n-degree Taylor polynomial of the composite function, would be to let $g(x)=x+1, f(g(x))=e^{x+1}$. Considering the same few terms of $f(g(x))$ above, we have $f(1)+x(1)f'(1)=e+ex$. However, a direct substitution for $n=2$ yields $p(x+1) = 1+(x+1)$

The substitution in the first example where $g(0)=0$ reduces $\sum_{k=1}^{n}\frac{(f\circ{g})^{(k)}(0)}{k!}x^k$ to the form $\sum_{k=1}^{n}\frac{h^{(k)}(0)}{k!}x^k$. Thus, there is equality between $p(g(x))$ and similar n-degree Taylor expansion of $f(g(x))$. The second substitution does not do so.