0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ You could just substitute $\sqrt{x}$ in for $x$ in the original expansion.. no? $\endgroup$
    – homosapien
    Commented Jun 11, 2023 at 23:29

2 Answers 2

2
$\begingroup$

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, $$\lim_{x\to 0}\frac{e^{\sqrt{x}}}{2\sqrt{x}}=\frac{1}{0}=\infty.$$

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – keepfrog
    Commented Jun 11, 2023 at 23:47
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jun 11, 2023 at 23:50
0
$\begingroup$

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 $x$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .