While finding the Taylor Series of a function, when are you allowed to substitute? And why?

For example:

Around $x=0$ for $e^{2x}$ I apparently am allowed to substitute $u=2x$ and then use the known series for $e^u$. But for $e^{x+1}$ I am not allowed to substitute $u=x+1$.

I know the technique for finding the Taylor Series of $e^{x+1}$ around $x=0$ by taking $e^{x+1}=e\times e^x$. However, I am looking for understanding and intuition for when and why it is allowed to apply substitution.

Note: there are several question that are similar to this one, but I have found none that actually answers the question "why"; or that shows a complete proof.

EDIT: Thanks to the answer of Markus Scheuer I should refine the question to cases where the series is finite, for example $n\to3$

  • $\begingroup$ The real questioni is: why on Earth shall you substitute $u = 2x$?? Leave $e^{2x}$ and use the Taylor series for that! $\endgroup$
    – Enrico M.
    Nov 12, 2016 at 10:19
  • $\begingroup$ @AlanTuring The point is that I try to understand why I can do that for $e^{2x}$ but not for $e^{x+1}$ $\endgroup$ Nov 12, 2016 at 10:20

3 Answers 3


A function $f(x)$ analytic at $x=0$ can be represented as power series within an open disc with radius of convergence $R$. \begin{align*} f(x)=\sum_{n=0}^\infty a_nx^n\qquad\qquad \qquad |x|<R \end{align*}

Any substitution $x=g(u)$ is admissible as long as we respect the radius of convergence. \begin{align*} f(g(u))=\sum_{n=0}^\infty a_n \left(g(u)\right)^n\qquad\qquad\quad |g(u)|<R \end{align*}

We know $f(u)=e^u$ can be represented as Taylor series convergent for all $u\in\mathbb{R}$, i.e. the radius of convergence $R=\infty$. \begin{align*} f(u)=e^u=\sum_{n=0}^\infty \frac{u^n}{n!}\qquad\qquad\qquad u\in \mathbb{R} \end{align*}

Substitution $u=2x$

We consider

\begin{align*} f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad 2x\in \mathbb{R} \end{align*}

This substitution is admissible for all $x \in \mathbb{R}$ since $$2x\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(2x)=e^{2x}$ is $\infty$.

We obtain \begin{align*} f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R} \end{align*}

Substitution $u=x+1$

We consider

\begin{align*} f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x+1\in \mathbb{R} \end{align*}

This substitution is admissible for all $x \in \mathbb{R}$ since $$x+1\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(x+1)=e^{x+1}$ is $\infty$.

We obtain \begin{align*} f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R} \end{align*}

We also obtain \begin{align*} e\cdot e^x&=\left(\sum_{k=0}^\infty \frac{1}{k!}\right)\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty \left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{1}{k!}\cdot\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty \left(\sum_{l=0}^n\frac{1}{(n-l)!}\cdot\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty\left(\sum_{l=0}^n\binom{n}{l}x^l\right)\frac{1}{n!}\\ &=\sum_{n=0}^\infty\frac{(x+1)^n}{n!}\\ &=e^{x+1} \end{align*}

Conclusion: We can use any substitution for convenience as long as we respect the radius of convergence.

  • $\begingroup$ I understand now that one is allowed to substitute in this case if $n\to\infty$. Is it correct to say that the series of $e^{x+1}$ that is the result of substitution is centered around $x=-1$, while the series that is calulated directly is centered around $x=0$? And doesn't this make a difference if I want to give an approximation of $e^{x+1}$ around $x=0$? I mean if I want to approximate I usually don't let $n\to\infty$, but rather to a fixed number, say $n\to3$. Would that make the case different? $\endgroup$ Nov 14, 2016 at 15:07
  • $\begingroup$ @ImreVégh: Yes, it is correct that the representation of $e^{x+1}$ above is centered around $x=-1$. Note, that if we have a power series centered around some value $x=x_0$, we can select any value $x_1$ within the disc of convergence $x_1\in d(x_0,R)$ of $f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ as new center by expanding $\sum_{n=0}^\infty a_n ((x-x_1)-(x_0-x_1))^n$ at powers of $x-x_1$. The new radius of convergence $R_1$ is at least as big, that $d(x_1,R_1)\subset d(x_0,R)$. $\endgroup$ Nov 14, 2016 at 15:23
  • $\begingroup$ @ImreVégh: If we want to expand the special case $e^{x+1}$ around $x=0$ we can go the easy way (as you already did) and write $e^{x+1}=e\cdot e^x=\sum_{n=0}^\infty \frac{e}{n!} x^n$ with coefficients $\frac{e}{n!}$ of $x^n$. The somewhat more cumbersome way would be to expand $(x+1)^n$ in $\sum_{n=0}^\infty \frac{(x+1)^n}{n!}$ and arrange the terms according to increasing powers of $x$. $\endgroup$ Nov 14, 2016 at 15:27

If $ f(x)=P_n(x)+x^n\epsilon(x)$ then


with $\lim_{x\to 0}\epsilon(x)=0$.

thus we need that

$\lim_{x\to 0}\epsilon(u(x))=0$

so, we must have

$$\lim_{x\to 0}u(x)=0$$

  • $\begingroup$ Hmm, I understand. And why is substitution allowed in the first place? When $\lim_{x\to 0}\epsilon(u(x))=0$? $\endgroup$ Nov 12, 2016 at 9:31

The quantity $2x$ is a product and as $x\to 0$ it remains a small number.

The quantity $x+n$ for $n\neq 0$ is not a little quantity anymore, and so you are not anymore around zero but you're around $n$.


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