Showing convergence with complex numbers

Question

I would like to show that if $|a-b|<\delta$ then $|e^{a}-e^b|<\epsilon$. Where $a$ is complex and $b$ is real.

In essence if the difference between a and b is small then the difference between e^a and e^b is small

I was hoping to use The Mean value theorem and say the following:

Take $f(y)=e^y$ then $|f(a)-f(b)|=|a-b||f'(u)|$ and so as $|a-b|<\delta$ then $|f(a)-f(b)|<\epsilon$

unfortunately the mean value theorem follows a different structur when dealing with complex values (according to https://en.wikipedia.org/wiki/Mean_value_theorem)

I have a feeling that my statement is not necessarily true because if i write the $|e^a-e^b|$ as a sum I get

$$ \left|\sum_{n=0}^{\infty} \frac{(a^n-b^n)}{n!}\right| \leq \sum_{n=0}^{\infty} \frac{(|a^n-b^n|)}{n!} $$

and just because $|a-b|<\epsilon$ we don't necessarily have $|a^n-b^n|<\epsilon$ as $n$ goes to $\infty$

Any help would be much appreciated

edit:

in original i wrote "I would like to show that if $|a-b|<\epsilon$ then $|e^{a}-e^b|<\epsilon$. Where $a$ is complex and $b$ is real."

Yes, the statement is false. We have, for example, $|2 - 1| < 2$ but $|e^2 - e| = e(e - 1) > 2$. — Travis Willse, Mar 31 '16 at 14:14
This is not true. For a counterexample, take for example $a=1$, $b=2$, $\epsilon=3$. — hmakholm left over Monica, Mar 31 '16 at 14:15

score 2 · Accepted Answer · answered Mar 31 '16 at 14:48

2

Another approach: $$ \begin{align} \left|e^a-e^b\right| &=\left|b-a\right|\left|\int_0^1e^{bt+a(1-t)}\,\mathrm{d}t\,\right|\\ &\le\left|b-a\right|\int_0^1e^{\mathrm{Re}(bt+a(1-t))}\,\mathrm{d}t\\[6pt] &\le\left|b-a\right|e^{\max(\mathrm{Re}(a),\mathrm{Re}(b))} \end{align} $$

answered Mar 31 '16 at 14:48

robjohn

345,667

If Socrates asks "what is beauty?" i will show him this. Thank you. – Oliver Brace Mar 31 '16 at 16:35

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

Here is a brute force approach. Let $z=a-b$ with $\text{Re}(z)=x$ and $\text{Im}(z)=y$.

NOTE:

In the ensuing development, we will use the inequalities

$$\begin{align} (1-e^{-|x|})^2&\le x^2 \tag 1\\\\ \sin^2(y)&\le y^2 \tag 2 \end{align}$$

which can be easily established a number of ways. For example, $(1)$ can be shown from the inequality established in This Answer using the limit definition of the exponential function and Bernoulli's Inequality. And $(2)$ can be established from elementary geometry. Alternatively, both can be shown using the mean value theorem.

We begin with the term $|e^a-e^b|$ and write

$$\begin{align} |e^a-e^b|&=|e^b||e^z-1|\\\\ &=|e^b||(e^x\cos(y)-1)+ie^x\sin(y)|\\\\ &=|e^b|\sqrt{e^{2x}-2e^x\cos(y)+1}\\\\ &=|e^b|\sqrt{(e^x-1)^2+2e^x(1-\cos(y))}\\\\ &=|e^b|\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)} \tag 3\\\\ &=|e^b|e^x\sqrt{(1-e^{-x})^2+4e^{-x}\sin^2(y/2)} \tag 4 \end{align}$$

For $x\ge 0$, we have using $(4)$

$$\begin{align} |e^b|e^x\sqrt{(1-e^{-x})^2+4e^{-x}\sin^2(y/2)}&\le |e^b|\,e^x\,\sqrt{x^2+y^2}\\\\ &=e^{x+\text{Re}(b)}|z|\\\\ &=e^{\text{Re}(a)}|a-b| \end{align}$$

For $x<0$, we have using $(3)$

$$\begin{align} |e^b|\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)}&=|e^b|\sqrt{(1-e^{-|x|})^2+4e^{-|x|}\sin^2(y/2)}\\\\ &\le |e^b|\sqrt{x^2+y^2}\\\\ &= e^{\text{Re}(b)}|z|\\\\ &=e^{\text{Re}(b)}|a-b| \end{align}$$

And now one can easily construct a $\delta-\epsilon$ proof.

Showing convergence with complex numbers

2 Answers2