Prove $x>2y$ Given $e^y=\dfrac{x}{1-e^{-x}}$

[anemone]

Hi MHB,

Problem:

Assume $x>0$, and $y$ satisfy that $e^y=\dfrac{x}{1-e^{-x}}$, prove that $x>2y$.

Attempt:

I first tried to express $y$ in terms of $x$ and get:

$y=x+\ln x- \ln (e^x-1)$

and I am aware that one of the method to prove the intended result is to rewrite the equation above as

$y-\dfrac{x}{2}=\dfrac{x}{2}+\ln x- \ln (e^x-1)$

and if I can prove $\dfrac{x}{2}+\ln x- \ln (e^x-1)<0$, which also implies $y-\dfrac{x}{2}<0$, then the result is proved.

Now, I see it that no inequality theorems that I know of could be applied to the expression $\dfrac{x}{2}+\ln x- \ln (e^x-1)$ and hence I get stuck, so stuck that I wish to drop this problem behind my mind!

But, I hope to solve it nonetheless and that's why I posted it here and hope someone can chime into help me out.

Thanks in advance.

 

[MarkFL]

Re: Prove x>2y

Can you show that the given definition of $y$ and the inequality implies:

$$\sinh\left(\frac{x}{2} \right)>\frac{x}{2}$$ ?

After that, a bit of calculus is all you need to demonstrate this is true. :D

 

[anemone]

Re: Prove x>2y

Can you show that the given definition of $y$ and the inequality implies:

$$\sinh\left(\frac{x}{2} \right)>\frac{x}{2}$$ ?

We can simplify the terms of natural logarithm in $\dfrac{x}{2}+\ln x- \ln (e^x-1)$ and get:

$\dfrac{x}{2}+\ln x- \ln (e^x-1)=\dfrac{x}{2}+\ln \dfrac{x}{e^x-1}$

and to be honest with you, I don't understand how to get what you obtained.

 

[MarkFL]

Re: Prove x>2y

I began with the given inequality:

$$x>2y$$

and this implies:

$$e^x>e^{2y}=\left(e^y \right)^2=\left(\frac{x}{1-e^{-x}} \right)^2=\left(\frac{xe^x}{e^x-1} \right)^2$$

Hence:

$$e^x>\frac{x^2e^{2x}}{\left(e^x-1 \right)^2}$$

Multiplying through by $$\frac{\left(e^x-1 \right)^2}{e^{2x}}>0$$ we obtain:

$$e^{-x}\left(e^x-1 \right)^2>x^2$$

Now this implies:

$$e^{-\frac{x}{2}}\left(e^x-1 \right)>x$$

$$e^{\frac{x}{2}}-e^{-\frac{x}{2}}>x$$

Dividing through by 2:

$$\frac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{2}>\frac{x}{2}$$

Using the definition of the hyperbolic sine function, we may now write:

$$\sinh\left(\frac{x}{2} \right)>\frac{x}{2}$$

 

[anemone]

Re: Prove x>2y

Ah, I see...thanks!

 

[chisigma]

Re: Prove x>2y

Hi MHB,

Problem:

Assume $x>0$, and $y$ satisfy that $e^y=\dfrac{x}{1-e^{-x}}$, prove that $x>2y$.

Attempt:

I first tried to express $y$ in terms of $x$ and get:

$y=x+\ln x- \ln (e^x-1)$

and I am aware that one of the method to prove the intended result is to rewrite the equation above as

$y-\dfrac{x}{2}=\dfrac{x}{2}+\ln x- \ln (e^x-1)$

and if I can prove $\dfrac{x}{2}+\ln x- \ln (e^x-1)<0$, which also implies $y-\dfrac{x}{2}<0$, then the result is proved.

Now, I see it that no inequality theorems that I know of could be applied to the expression $\dfrac{x}{2}+\ln x- \ln (e^x-1)$ and hence I get stuck, so stuck that I wish to drop this problem behind my mind!

But, I hope to solve it nonetheless and that's why I posted it here and hope someone can chime into help me out.

Thanks in advance.

I'm afraid that You have to try the McLaurin expansion around x = 0 of the logarhitm of the function... $\displaystyle y = \ln x - \ln (1 - e^{-x})\ (1)$Starting from f(0) and using l'Hopital rule You have... $\displaystyle \lim_{x \rightarrow 0} \frac{x}{1 - e^{-x}} = \lim_{x \rightarrow 0} e^{x} = 1 \implies y(0) = 0\ (2)$

The procedure for the derivatives however is much more complex and 'Monster Wolfram' can supply a significative aid...

ln x - ln [1 - e^(- x)] - Wolfram|Alpha

The McLaurin expansion we are interersted for is...

$\displaystyle y(x) = \frac{x}{2} - \frac{x^{2}}{24} + \frac{x^{4}}{2880} + \mathcal {O} (x^{6})\ (3)$

As curiosity y(x) is defined for all value of x [positive as well as negatives...] and is $y(x) \le \frac{x}{2}$. May be that a simpler approach to the problem exists and, if yes, it must be found... Kind regards $\chi$ $\sigma$

 

[anemone]

Re: Prove x>2y

I'm afraid that You have to try the McLaurin expansion around x = 0 of the logarhitm of the function... $\displaystyle y = \ln x - \ln (1 - e^{-x})\ (1)$Starting from f(0) and using l'Hopital rule You have... $\displaystyle \lim_{x \rightarrow 0} \frac{x}{1 - e^{-x}} = \lim_{x \rightarrow 0} e^{x} = 1 \implies y(0) = 0\ (2)$

The procedure for the derivatives however is much more complex and 'Monster Wolfram' can supply a significative aid...

ln x - ln [1 - e^(- x)] - Wolfram|Alpha

The McLaurin expansion we are interersted for is...

$\displaystyle y(x) = \frac{x}{2} - \frac{x^{2}}{24} + \frac{x^{4}}{2880} + \mathcal {O} (x^{6})\ (3)$

As curiosity y(x) is defined for all value of x [positive as well as negatives...] and is $y(x) \le \frac{x}{2}$.

Hey chisigma, thank you for your reply...and I am happy to learn these two methods to solve the problem!

May be that a simpler approach to the problem exists and, if yes, it must be found...

I think MarkFL's approach is very straightforward and smart! :)

 

[johng1]

Re: Prove x>2y

Here's another solution which just uses a careful application of the mean value theorem.

View attachment 1617