# Taylor expansion of $\exp(x^{1/2})$

I want to compute the first-order Taylor expansion of the function $$g(x)=\exp(x^{1/2})$$ at $$x=0$$. Since $$g'(x)=\frac{e^{\sqrt{x}}}{2\sqrt{x}}$$, I have $$g(x)=g(0)+\left(\lim_{x'\to0} g'(x') \right)x+o(x^{-1}).$$

Using the computed first derivative, I have $$\lim_{x'\to0} g'(x')=\lim_{x'\to0}\frac{e^{\sqrt{x'}}}{2\sqrt{x'}}=\infty.$$

Thus, I obtain $$g(x)\approx 1+\infty x$$. However, when I use Mathematica, it gives $$g(x)\approx 1 - \sqrt{x} + \frac{x}{2},$$ which looks correct by plotting this function. What am I missing?

Edit: I corrected the mistake in the limit.

• You could just substitute $\sqrt{x}$ in for $x$ in the original expansion.. no? Commented Jun 11, 2023 at 23:29

Since $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots=\sum_{n=0}^\infty\frac{x^n}{n!},$$ then just subtituting $$x$$ by $$x^{1/2}$$ we have $$e^{x^{1/2}}=1+x^{1/2}+\frac{x}{2}+\cdots=\sum_{n=0}^\infty \frac{x^{n/2}}{n!}.$$ Note that this is not a Taylor series, since the exponents are fractional. That's because $$\sqrt{x}$$ is not analytic in $$x=0$$. The given is the Puiseux series of the function. Also you computed that limit terribly wrong, $$\lim_{x\to 0}\frac{e^{\sqrt{x}}}{2\sqrt{x}}=\frac{1}{0}=\infty.$$
• How can I reconcile the fact that (i) $e^{\sqrt{x}}$ is not analytic at $x=0$ thus the Taylor expansion does not exist at $x=0$ and (ii) for the Taylor expansion of $e^x$ one can replace $x$ by $x^{1/2}$.? Do you mean the thing that I get from substituting is not a Taylor series and called the Puiseux series? Commented Jun 11, 2023 at 23:47
• Exactly, from the Taylor series of $e^x$ we can obtain what is called a formal Laurent series (Puiseux series) by substituting $x$ by $\sqrt{x}$, but the expansion you obtain is not a Taylor series. The reason is that is not analytic, since the limit $\lim_{x\to 0}f'(x)$ does not exist. Commented Jun 11, 2023 at 23:50
If $$f(x)=e^x= \sum_{n=0}^\infty \frac{x^n}{n!}$$, then
$$g(x)=e^{x^{\frac{1}{2}}}=\sum_{n=0}^\infty \frac{x^{\frac{n}{2}}}{n!}$$
As needed. You just substitute $$x^\frac{1}{2}$$ for $$x$$.