Let's define $\exp(z)$ in the following form: $$\exp(z) = \sum \limits_{n=0}^{\infty} \frac{z^n}{n!}, z \in \mathbb{C}.$$ We are to show that the real part of the function defined above is increasing and positive $\forall_{x \in \mathbb{R}}$.
I found some information linked to $\exp$ and my problem here but I don't understand the proof and I don't know if it's enough.

  • $\begingroup$ Without context, this is hard to answer. What facts about $\exp$ do you already have available? In particular, have you established that $\exp' = \exp$ and that $\exp(z+w) = \exp(z)\exp(w)$? $\endgroup$ – Bungo May 11 '18 at 22:41
  • $\begingroup$ @Bungo Not much. I know also that $\exp(z+w)=\exp(z) \exp(w)$. $\endgroup$ – Hendrra May 11 '18 at 22:44
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    $\begingroup$ From the series definition it's clear that $\exp(x) > 0$ for all $x > 0$ (the terms are all positive) and that $\exp(0) = 1$. Then from $\exp(x)\exp(-x) = \exp(x-x) = \exp(0) = 1$, it follows that $\exp(-x) > 0$, so we can conclude that $\exp(x) > 0$ for all $x \in \mathbb R$. If you also have (or can establish) the fact that $\exp' = \exp$, then $\exp' > 0$, so the function is increasing. $\endgroup$ – Bungo May 11 '18 at 22:47
  • $\begingroup$ @Bungo thank you! I think that makes the point :) $\endgroup$ – Hendrra May 11 '18 at 22:49
  • $\begingroup$ @bungo The real part of $e^z$ is $\text{Re}(e^z)=e^{\text{Re}(z)}\cos(\text{Im}(z))=e^x\cos(y)$. So, what does it mean to be increasing here? I believe that the OP wants to show that $|e^z|=e^x$ is increasing. $\endgroup$ – Mark Viola May 11 '18 at 23:25

We have a function $f: \mathbb R \to \mathbb R$ such that $f(0)=1$ and $f>1$ on $(0,\infty).$ We also know $f(x+y)= f(x)f(y)$ for all $x,y\in \mathbb R.$ Thus for any real $x,$ $f(x-x)=f(0)=1=f(x)f(-x),$ which implies $f(-x)=1/f(x).$ From this it follows that $f>0$ on $(-\infty,0).$ Hence $f>0$ everywhere.

Finally, suppose $x<y.$ Then $0<y-x,$ hence

$$1=f(0) < f(y-x) = f(y)f(-x)= f(y)/f(x).$$

This implies $f(x) < f(y)$ and we're done.


$\exp(0) = 1$

If $x>0$ then $\exp(x)>1$

$\exp(x) - 1 = \sum_\limits{n=1}^{\infty} \frac {x^n}{n!}$

every term on the right is positive, so the sum must be positive.

$\exp(x+h) = \exp(x)\exp(h)$

$\exp(x-x) = \exp(x)\exp(-x) = exp(0) = 1\\ \exp(-x) = \frac {1}{\exp(x)}$

For all real $x, \exp(x)>0$

$\exp(x)$ is increasing if for all $h>0, \exp(x+h) > \exp(x)$

$\exp(x)\exp(h)>\exp(x)\\ \exp(x)(\exp(h)-1)>0$

Since both factors are positive, it is clearly true.

  • $\begingroup$ Proving that $\exp(x+h) = \exp(x)\exp(h)$ based on the power series alone may be more work that what has been done here. $\endgroup$ – Michael Hardy May 12 '18 at 0:19
  • $\begingroup$ You say IF the function is increasing THEN blahblahblah, and blahblahblah is clearly true. You need to interchange the "if" and the "then". $\qquad$ $\endgroup$ – Michael Hardy May 12 '18 at 0:20

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