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.

  • 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 $\exp(z+w) = \exp(z)\exp(w)$? – Bungo May 11 at 22:41
  • @Bungo Not much. I know also that $\exp(z+w)=\exp(z) \exp(w)$. – Hendrra May 11 at 22:44
  • 2
    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. – Bungo May 11 at 22:47
  • @Bungo thank you! I think that makes the point :) – Hendrra May 11 at 22:49
  • @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. – Mark Viola May 11 at 23:25
up vote 1 down vote accepted

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.

  • 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. – Michael Hardy May 12 at 0:19
  • You say IF the function is increasing THEN blahblahblah, and blahblahblah is clearly true. You need to interchange the "if" and the "then". $\qquad$ – Michael Hardy May 12 at 0:20

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.