Use the power series definition of $e^z$ to prove that for any $h\in \mathbb{C}$ and $t>0$, we have

$$\left|\frac{t^h-1}{h} \right|\leq t^{|h|}|\ln t|$$

and furthermore,

$$\left|\frac{t^h-1}{h}-\ln t \right|\leq |h|t^{|h|}\ln^{2} t$$

I first tried to approach it first by using the fact that $t^h=e^{h\ln t}$ to get

$$\left|\frac{t^h-1}{h} \right| = \left|\frac{e^{h\ln t}-1}{h} \right|$$

Since the power series expansion of $e^z$ is $\sum_{0}^{\infty}\frac{z^n}{n!}$, I then substituted that to get

$$\left|\frac{e^{h\ln t}-1}{h} \right| = \left|\frac{(\sum_{0}^{\infty}\frac{z^n}{n!})^{\ln t}-1}{h} \right|$$

But then, I don't know how I should be approaching this problem. I feel like I'm no where near the right track.

  • 1
    $\begingroup$ Your substitution is wrong $$e^{z} = \sum_{k \ge 0} \frac{z^{k}}{k!} \implies e^{h \ln t} = \sum_{k \ge 0} \frac{(h \ln t)^{k}}{k!}$$ $\endgroup$ – Mattos Sep 27 '18 at 2:34
  • $\begingroup$ @Mattos That was stupid of me to not think of that. Then, what do you do with that $-1$? Should I change that into $e^{i\pi}$ and do the same expansion? I was trying to do that, but it comes out pretty messy $\endgroup$ – Ya G Sep 27 '18 at 2:45
  • $\begingroup$ $$e^{h \ln t} - 1 = \sum_{k \ge 0} \frac{(h \ln t)^{k}}{k!} - 1 = \sum_{k \ge 1} \frac{(h \ln t)^{k}}{k!}$$ $\endgroup$ – Mattos Sep 27 '18 at 2:46
  • $\begingroup$ @Mattos $$\begin{cases}|\frac{\sum_{k=1}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}|& =|\frac{h\ln t+\sum_{k=2}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}|\\ & \leq|\frac{h\ln t}{h}|+|\frac{\sum_{k=0}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}|\\&=|\ln t|+|\frac{t^{h}}{h}|\end{cases}$$ $\endgroup$ – Ya G Sep 27 '18 at 3:01
  • $\begingroup$ @Mattos Can you give me a little more hint please? $\endgroup$ – Ya G Sep 27 '18 at 3:17

Starting from the comment : $$\left|\frac{\sum_{k=1}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}\right| \leq \frac{\sum_{k=2}^{\infty}|\frac{(h\ln t)^{k}}{k!}|}{|h|}\\ \leq \frac{\sum_{k=2}^{\infty}\frac{|h|^k|\ln t|^{k}}{k!}}{|h|}\\ \leq \sum_{k=2}^{\infty}\frac{|h|^{k-1}|\ln t|^{k}}{k!}\\ \leq |\ln t|\sum_{k=2}^{\infty}\frac{|h|^{k-1}|\ln t|^{k-1}}{k!}$$ Then you notice that : $$\frac{|h|^{k-1}|\ln t|^{k-1}}{k!}\leq \frac{|h|^{k-1}|\ln t|^{k-1}}{(k-1)!}$$ So we have : $$\left|\frac{\sum_{k=1}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}\right| \leq |\ln t|\sum_{k=2}^{\infty}\frac{|h|^{k-1}|\ln t|^{k-1}}{(k-1)!}\\ \leq |\ln t|\sum_{k=1}^{\infty}\frac{|h|^{k}|\ln t|^{k}}{k!}$$ You notice that : $$\sum_{k=1}^{\infty}\frac{|h|^{k}|\ln t|^{k}}{k!}=\sum_{k=1}^{\infty}\frac{|h\ln t|^{k}}{k!}=\sum_{k=0}^{\infty}\frac{|h\ln t|^{k}}{k!}-1=|t|^{|h|}-1=t^{|h|}-1$$ The last equality is true because $t>0$. Finally : $$\left|\frac{\sum_{k=1}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}\right| \leq |\ln t|(t^{|h|}-1)\leq |\ln t|t^{|h|}$$


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