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How to show that $\left|\sum_{k=0}^\infty\frac{(ix)^k}{(k+1)!}\right|\le \left|\sum_{k=0}^\infty\frac{(ix)^k}{k!}\right|=|e^{ix}|=1$ with restrictions, for $x\in\Bbb R$.

To prove this inequality we cant use any related to derivatives, integrals, geometric statements about sine or cosine, or uniform convergence. We can use limits and basic facts about the convergent properties of these power series.

We already knows that $|e^{ix}|=1$ for $x\in\Bbb R$. The inequality is a slight rewrite of

$$\frac{|e^{ix}-1|}{|x|}=\left|\sum_{k=0}^\infty\frac{(ix)^k}{(k+1)!}\right|\le 1,\quad\forall x\in\Bbb R$$

what need to be proved. I dont know exactly what to do here, Im completely lost. The best I can think is to prove something like

$$\forall\epsilon>0,\exists N\in\Bbb N:\left|\sum_{k=0}^n\frac{(ix)^k}{(k+1)!}-L\right|<\epsilon,\quad\forall n\ge N$$ for some $0\le L<1$.

The exercise leave the hint $\lim_{z\to 0}\frac{\exp(z)-1}{z}=1$ for $z\in\Bbb C\setminus\{0\}$, but I dont see how to relate this to our problem, because we need the result for any $x$, not just for $x=0$. Some hint or solution will be appreciated, thank you.

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  • $\begingroup$ Do you need to use all the inequalities or just the last one: $\;\le1\;$ ? $\endgroup$ – DonAntonio Nov 27 '16 at 18:08
  • $\begingroup$ @DonAntonio any one is fine. $\endgroup$ – Masacroso Nov 27 '16 at 18:09
  • $\begingroup$ The first inequality should be obvious. You're dividing by something smaller (a lower factorial) so the result is larger. If you know that $|e^{ix}| = 1$, , it looks like all you have to prove is the middle equality, which you seem to use in the body. $\endgroup$ – Alfred Yerger Nov 27 '16 at 18:14
  • $\begingroup$ @AlfredYerger I dont see how this is obvious because the series is an alternating series of complex numbers. I can suppose that it is obvious but, how to prove it? $\endgroup$ – Masacroso Nov 27 '16 at 18:15
  • $\begingroup$ @Masacroso Do you have access to the triangle inequality for infinite sums? $\endgroup$ – Alfred Yerger Nov 27 '16 at 19:18
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Note: This answer uses the identity $$e^{i(x+y)} = e^{ix} e^{iy}$$ which can be proved using Cauchy's product.


As the hint suggests, we have

$$\lim_{r \to 0} \frac{e^{ir}-1}{ir}=1. $$

For fixed $\epsilon>0$ we can choose $\delta>0$ such that

$$|e^{ir}-1| \leq (1+\epsilon) |r| \qquad \text{for all $|r| \leq \delta$}. \tag{1}$$

Now pick $x \in \mathbb{R}$ and choose $n \in \mathbb{N}$ sufficiently large such that $r := x/n$ satisfies $|r| \leq \delta$. For $$x_j := j r, \qquad j=0,\ldots,n$$ we have

$$e^{ix}-1 = \sum_{j=1}^n (e^{ix_j}-e^{ix_{j-1}}). \tag{2}$$

Using that

$$e^{ix_j}-e^{ix_{j-1}} = e^{ix_{j-1}} (e^{ir}-1)$$

and $|e^{ix_{j-1}}| \leq 1$, we get

$$|e^{ix}-1| \leq \sum_{j=1}^n |e^{ix_j}-e^{ix_{j-1}}| \leq n |e^{ir-1}| \stackrel{(1)}{\leq} (1+\epsilon) n \cdot |r| = (1+\epsilon) |x|.$$ Since $\epsilon>0$ was arbitrary, this gives

$$|e^{ix}-1| \leq |x| \qquad \text{for all $x \in \mathbb{R}$.}$$

Rewriting this identity using the definition of $e^{ix}$, this proves the assertion.

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  • $\begingroup$ I dont follow the identity of $(1)$. We know that $|r|<\delta$ implies $$\left|\frac{e^{ir}-1}{ir}-1\right|<\epsilon$$ Then to assume $(1)$ we need assume first that $|\frac{e^{ir}-1}{ir}|>1$, right? $\endgroup$ – Masacroso Nov 27 '16 at 19:37
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    $\begingroup$ @Masacroso Note that $$\left| \frac{e^{ir}-1}{ir} \right| \leq \left| \frac{e^{ir}-1}{ir}-1 \right| + 1 \leq \epsilon+1$$ for all $|r| \leq \delta$. Multiplying both sides with $|ir|$ gives the desired inequality. $\endgroup$ – saz Nov 27 '16 at 19:39
  • $\begingroup$ Amazing answer. You can use the fact that $|e^{ix_j}|=1$ for any $x_j\in\Bbb R$ instead of the inequality. $\endgroup$ – Masacroso Nov 27 '16 at 19:54
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    $\begingroup$ @Masacroso Yeah, I know, but here it is enough to have "$\leq$". $\endgroup$ – saz Nov 27 '16 at 19:56
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Just for the record I will add a second proof. We knows that

$$e^{ix}=\cos(x)+i\sin(x)$$

and $|e^{ix}|=1$ for all $x\in\Bbb R$. And we want to prove

$$\frac{|e^{ix}-1|}{|x|}\le 1$$

From the last inequality we have the bound

$$|e^{ix}-1|\le |e^{ix}|+1=2\le|x|$$

then for $|x|\ge 2$ the inequality is clear. Now, we will study the case for $|x|<2$. From the Euler's formula we have that

$$\frac{|e^{ix}-1|}{|x|}=\frac{|\cos(x)+i\sin(x)-1|}{|x|}=\\=\frac{\sqrt{(1-\cos (x))^2+\sin^2(x)}}{|x|}=\frac{\sqrt{2(1-\cos(x))}}{|x|}\le 1$$

Then our inequality can be written as

$$1-\cos(x)=\sum_{k=1}^\infty(-1)^{k+1}\frac{x^{2k}}{(2k)!}\le \frac{|x|^2}2\implies\sum_{k=2}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\ge 0$$

Now observe that for $|x|<2$ and $k\ge 2$ the sequence $(x^{2k}/(2k)!)$ decreases monotonically. Then for these alternating series we have the bound

$$|s-s_n|\le |c_{n+1}|$$

where $s:=\sum_{k=0}^\infty (-1)^k c_k$ is an alternating series where $(c_n)\to 0$ monotonically, $s_n$ is a partial sum of the series, and $c_{n+1}$ is an element of the monotone sequence.

Then

$$\left|\sum_{k=2}^\infty(-1)^k\frac{x^{2k}}{(2k)!}-\frac{x^4}{4!}\right|\le\frac{x^6}{6!}\implies \sum_{k=2}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\ge \frac{x^4}{4!}-\frac{x^6}{6!}=\frac{x^4}{4!}\left(1-\frac{x^2}{30}\right)\ge 0$$

whenever $|x|<2$.$\Box$

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