Prove of inequality (exponential function)

[Galenus]

Problem: For [MATH] {z} \in \mathbb{C}, |{z}|<1,[/MATH] show that [MATH] |\exp(z)| \leq \frac{1}{1-\Re(z) } [/MATH]
My approach: First I noticed that [MATH] |\exp(z)| = \exp(\Re(z)) [/MATH].
Saying that [MATH] a = \Re(z) [/MATH] it's true by definition that [MATH] \exp(a) = \lim_{n \rightarrow \infty} (1 + \frac{a}{n})^{n} [/MATH],
so I tried to use [MATH] \exp(a) \geq (1 + a) [/MATH] and [MATH] a < 1 [/MATH] to arrive at [MATH] \exp(a) \leq (1 - a)^{-1} [/MATH].

I don't know if this is the right approach and I somehow really confused myself ?

 

[Cubist]

First I noticed that [MATH] |\exp(z)| = \exp(\Re(z)) [/MATH].

I think you'll have to show this, which can be done because your observation is correct!

Saying that [MATH] a = \Re(z) [/MATH] it's true by definition that [MATH] \exp(a) = \lim_{n \rightarrow \infty} (1 + \frac{a}{n})^{n} [/MATH],
so I tried to use [MATH] \exp(a) \geq (1 + a) [/MATH] and [MATH] a < 1 [/MATH] to arrive at [MATH] \exp(a) \leq (1 - a)^{-1} [/MATH].

I don't know if this is the right approach and I somehow really confused myself ?

I'd do this part of the proof by considering the Maclaurin series of:-

• exp(a) when a≥0 ....... EDIT: better to just extend the range on the next bullet point from 0<b<1 to -1<b<1
• exp(-b) when -1<b<1

 

[Galenus]

I think you'll have to show this, which can be done because your observation is correct!

This can be shown as follows:
[MATH] \left | \exp(z) \right |^2 = \exp(z)\overline{\exp(z)} = \exp(z)\exp(\overline{z}) = \exp(z + \overline{z}) = \exp(2\Re(z))= \exp(\Re(z))^2 [/MATH]
[MATH] \Rightarrow \left | \exp(z) \right | = \exp(\Re(z)) [/MATH]

I'd do this part of the proof by considering the Maclaurin series of:-

• exp(a) when a≥0 ....... EDIT: better to just extend the range on the next bullet point from 0<b<1 to -1<b<1
• exp(-b) when -1<b<1

Thanks for the hint! This is what I get:

[MATH] \exp(-a) = \sum_{n=0}^{\infty}\frac{(-1)^na^n}{n!} = 1 - a + \frac{a^2}{2} - \frac{a^3}{6} + \ldots [/MATH]
But since the sequence [MATH]\frac{a^n}{n!}[/MATH] is monotonically decreasing for [MATH]-1<a<1[/MATH] it follows that

[MATH] \exp(-a) \geq 1-a [/MATH]
[MATH] \Rightarrow \exp(a) \leq \frac{1}{1-a} [/MATH]

And together [MATH]\left | \exp(z) \right | \leq \frac{1}{1-\Re(z)}[/MATH]
Does this look good ?

 

[Cubist]

[MATH] \left | \exp(z) \right |^2 = \exp(z)\overline{\exp(z)} = \exp(z)\exp(\overline{z}) = \exp(z + \overline{z}) = \exp(2\Re(z))= \exp(\Re(z))^2 [/MATH]

Nice, but it might be good to avoid the square (which can introduce extra solutions, after all \( (-2)^2=2^2\). I'd recommend you approach this by writing z as x+i*y and then split the exponent \( |e^{x+iy}|=|e^xe^{iy}|=|e^x\left( cos+...\right)| \) and continue

But since the sequence [MATH]\frac{a^n}{n!}[/MATH] is monotonically decreasing for [MATH]-1<a<1[/MATH] it follows that

I'm not sure about your argument. It's a bit of a problem that the sign flips every term. Therefore I was thinking of pairing up term 3&4, 5&6 such that the pairs are positive when -1<b<1. Can you show this?

 

[Galenus]

Nice, but it might be good to avoid the square (which can introduce extra solutions, after all \( (-2)^2=2^2\). I'd recommend you approach this by writing z as x+i*y and then split the exponent \( |e^{x+iy}|=|e^xe^{iy}|=|e^x\left( cos+...\right)| \) and continue

You mean like this?

\( |e^{x+iy}|=|e^xe^{iy}|=|e^x\left( cos(y)+isin(y)\right)| = |e^x| |cos(y)+isin(y)| = e^x \)

where in the last equality I used that \( e^x \) is strictly positive and \( cos^2(y) + sin^2(y) = 1 \).

I'm not sure about your argument. It's a bit of a problem that the sign flips every term. Therefore I was thinking of pairing up term 3&4, 5&6 such that the pairs are positive when -1<b<1. Can you show this?

OK let me try to be rigorous here:

Let \(b_n = \frac{(-1)^na^n}{n!} \) be a sequence starting at n=0.

Then \( c_n = b_{2n} + b_{2n+1} = \frac{a^{2n}}{(2n)!} - \frac{a^{2n+1}}{(2n+1)!} = \frac{a^{2n}}{(2n)!}(1 - \frac{a}{2n+1}) > 0\)
because \(a\) raised to the even number 2n is certainly >0, and \( a < 1 \Rightarrow \frac{a}{2n+1} < 1 \Rightarrow 1 - \frac{a}{2n+1} > 0\).


Now \( \sum_{n=0}^{\infty}\frac{(-1)^na^n}{n!} = \sum_{n=0}^{\infty}b_{n} = \sum_{n=0}^{\infty}c_{n} \geq c_0 = 1-a \).



btw: being really unexperienced in using latex, this is some good training for me ^^

 

[Cubist]

Excellent! It looks cast-iron to me. Now just tie the various parts together...

[math] |e^{x+iy}|=|e^xe^{iy}|=|e^x\left( \cos(y)+i\sin(y)\right)| = |e^x| |\cos(y)+i\sin(y)| = e^x [/math]
where in the last equality I used that \( e^x \) is strictly positive and \( \cos^2(y) + \sin^2(y) = 1 \)

Therefore...

[MATH] |\exp(z)| = \exp(\Re(z)) \quad\enclose{box}{A} [/MATH]

Using the Maclaurin series...

[MATH] \exp(-a) = \sum_{n=0}^{\infty}\frac{(-1)^na^n}{n!} = 1 - a + \frac{a^2}{2} - \frac{a^3}{6} + \ldots [/MATH]

Let \(b_n = \frac{(-1)^na^n}{n!} \) be a sequence starting at n=0.

Then \( c_n = b_{2n} + b_{2n+1} = \frac{a^{2n}}{(2n)!} - \frac{a^{2n+1}}{(2n+1)!} = \frac{a^{2n}}{(2n)!}(1 - \frac{a}{2n+1}) > 0\)
because \(a\) raised to the even number 2n is certainly >0, and \( a < 1 \Rightarrow \frac{a}{2n+1} < 1 \Rightarrow 1 - \frac{a}{2n+1} > 0\).

Now [math] \sum_{n=0}^{\infty}b_{n} = \sum_{n=0}^{\infty}c_{n} \geq c_0 \) where \( c_0 = 1-a [/math]

... it follows that

[MATH] \exp(-a) \geq 1-a [/MATH]
[MATH] \Rightarrow \exp(a) \leq \frac{1}{1-a} [/MATH]
And together with eqn A gives [MATH]\left | \exp(z) \right | \leq \frac{1}{1-\Re(z)}[/MATH]

--

btw: being really unexperienced in using latex, this is some good training for me ^^

I'll bet! Well done. An extra tip - you can precede sin and cos with \ so that they don't appear in italics (helps to distinguish from variables)

 

[Cubist]

FYI: I'd present the inequality proof like this (to give you another option)...

[math]\exp(-a) = 1 - a + \frac{a^2}{2!} - \frac{a^3}{3!} + \frac{a^4}{4!} - \frac{a^5}{5!} + \cdots[/math]
[math]= 1 - a + \left(\frac{a^2}{2!} - \frac{a^3}{3!}\right) + \left(\frac{a^4}{4!} - \frac{a^5}{5!} \right) + \cdots[/math]
[math] = 1 - a + \left(\frac{3a^2-a^3}{3!} \right) + \left(\frac{5a^4-a^5}{5!}\right) + \cdots [/math]
[math] = 1 - a + \left(\frac{a^2(3-a)}{3!} \right) + \left(\frac{a^4(5-a)}{5!}\right) + \cdots [/math]
Therefore [math]\exp(-a) \geq 1-a\, [/math] since clearly the terms in brackets are +ve when -1<a<1

EDIT: This seems to actually hold for all a<3. But when taking the reciprocal of the inequality the direction no longer reverses if a>1, and let's also avoid a=1 since 1/0 isn't good!