Show that the given function is decreasing

[idobido]

Homework Statement

show that the given function is decreasing.

Relevant Equations

derivative
induction

As a follow up for : https://www.physicsforums.com/threa...there-is-i-n-s-t-1-1-k-i-1-2-k-i-1-4.1054669/

show that $\alpha\left(k\right)\ :=\ \left(1-\tfrac{1}{k}\right)^{\ln\left(2\right)k}-\left(1-\tfrac{2}{k}\right)^{\ln\left(2\right)k}$ is decreasing for $k\in\left[3,\infty\right)$

thing I've tried:
showing that first derivative is non-positive, but I've got complicated expression i cannot handle with ($k = x$):

$\dfrac{\left(\frac{x-1}{x}\right)^{\ln\left(2\right)\,x}\,\left(\ln\left(2\right)\ln\left(\frac{x-1}{x}\right)\left(x-1\right)+\ln\left(2\right)\right)}{x-1}-\dfrac{\left(\frac{x-2}{x}\right)^{\ln\left(2\right)\,x}\,\left(\ln\left(2\right)\ln\left(\frac{x-2}{x}\right)\left(x-2\right)+2\ln\left(2\right)\right)}{x-2}$

trying to compare
$\alpha\left(k\right)\$ with $\alpha\left(k+1\right)\$
also got complicated.

 

[fresh_42]

Set $f(k) =\left(1-\dfrac{1}{k}\right)^{\log 2},$ $\varepsilon(k) =\dfrac{1}{k}$ and $g(k)=(f(k)-\varepsilon(k) )^{\log 2}$ and show that
\begin{align*}
\dfrac{d}{dk}\left(f(k)^k-g(k)^k\right)&<0
\end{align*}
... and don't forget to write down your final calculation backward!

 

[idobido]

did you notice that $g(k) = \left( \left(1 - \frac{1}{k}\right)^{\log 2} - \frac{1}{k} \right)^{\log 2} \neq \left(1 - \frac{2}{k}\right)^{\log 2}$ ?

 

[fresh_42]

did you notice that $g(k) = \left( \left(1 - \frac{1}{k}\right)^{\log 2} - \frac{1}{k} \right)^{\log 2} \neq \left(1 - \frac{2}{k}\right)^{\log 2}$

You are right, I overlooked that. So set $f(k)=1-\dfrac{1}{k}$ and $g(k)=f(k)-\dfrac{1}{k}.$ Shouldn't make a difference since we cancel $k\log(2)$ anyway.

 

[idobido]

You are right, I overlooked that. So set $f(k)=1-\dfrac{1}{k}$ and $g(k)=f(k)-\dfrac{1}{k}.$ Shouldn't make a difference since we cancel $k\log(2)$ anyway.

how does it cancelled i've got
$\frac{d}{dk}\left( f(k)^k - g(k)^k \right) = \left( \ln f(k) + k \cdot \frac{1}{f(k)} \cdot f'(k) \right) \cdot f(k)^k - \left( \ln g(k) + k \cdot \frac{1}{g(k)} \cdot g'(k) \right) \cdot g(k)^k$

 

[fresh_42]

how does it cancelled i've got
$\frac{d}{dk}\left( f(k)^k - g(k)^k \right) = \left( \ln f(k) + k \cdot \frac{1}{f(k)} \cdot f'(k) \right) \cdot f(k)^k - \left( \ln g(k) + k \cdot \frac{1}{g(k)} \cdot g'(k) \right) \cdot g(k)^k$

You are thinking way too complicated. We want to show that
\begin{align*}
0&>\dfrac{d}{dk}\alpha(k) \\
0&>\dfrac{d}{dk}\left[\left(1-\dfrac{1}{k}\right)^{k\log 2}-\left(1-\dfrac{2}{k}\right)^{k\log 2}\right]\\
0&>\dfrac{d}{dk}\left(1-\dfrac{1}{k}\right)^{k\log 2}-\dfrac{d}{dk}\left(1-\dfrac{2}{k}\right)^{k\log 2}\\
0&>{k\log 2}\dfrac{d}{dk}\left(1-\dfrac{1}{k}\right)-{k\log 2}\dfrac{d}{dk}\left(1-\dfrac{2}{k}\right)\\
&\phantom{>}\ldots
\end{align*}
... and at the end of it: rewrite it backward.

 

[idobido]

did you notice that ##

You are thinking way too complicated. We want to show that
\begin{align*}
0&>\dfrac{d}{dk}\alpha(k) \\
0&>\dfrac{d}{dk}\left[\left(1-\dfrac{1}{k}\right)^{k\log 2}-\left(1-\dfrac{2}{k}\right)^{k\log 2}\right]\\
0&>\dfrac{d}{dk}\left(1-\dfrac{1}{k}\right)^{k\log 2}-\dfrac{d}{dk}\left(1-\dfrac{2}{k}\right)^{k\log 2}\\
0&>{k\log 2}\dfrac{d}{dk}\left(1-\dfrac{1}{k}\right)-{k\log 2}\dfrac{d}{dk}\left(1-\dfrac{2}{k}\right)\\
&\phantom{>}\ldots
\end{align*}
... and at the end of it: rewrite it backward.

i really can't understand why the last result is true, you are applying $\sqrt{k \cdot \log 2}$
on both expressions inside the derivative brackets i.e. (before the derivative)? why is that true that the inequality will hold after the derivative?

 

[fresh_42]

did you notice that $i really can't understand why the last result is true, you are applying$
\sqrt{k \cdot \log 2}
##
on both expressions inside the derivative brackets i.e. (before the derivative)? why is that true that the inequality will hold after the derivative?

You are right. I made a mistake with the chain rule. Sorry.

 

[PeroK]

Note: I took a wrong shortcut with the derivative, so this is wrong. Corrected solution is on the way!

I think you can crank it out using calculus. Let:
$$f(x) = \big (\frac {x-1} x\big)^{ax} - \big (\frac {x-2} x\big)^{ax}$$Then$$f'(x) = a\ln\big (\frac {x-1} x\big)\big (\frac {x-1} x\big)^{ax}\big(\frac 1 {x^2}\big) - a\ln\big (\frac {x-2} x\big)\big (\frac {x-2} x\big)^{ax}\big(\frac 2 {x^2}\big)$$Looking for $f'(x) = 0$ gives$$\frac{\ln\big (\frac x {x-1} \big)}{\ln\big (\frac x {x-2} \big)} = 2\big (\frac{x-2}{x-1})^{ax}$$The left-hand side (for $x \ge 3$) is an increasing function with a limit of $\frac 1 2$. The right-hand side is increasing (for $x \ge 3$) with a limit of $1$. And, in any case, for $a = \ln 2$, it is greater than $\frac 1 2$.

This means that for $a = \ln 2$ the function has no turning points for $x \ge 3$, hence is decreasing.

 

[PeroK]

I think I have managed to crank out a solution this time. The derivative in post #1 is correct. Using $a = \ln(2)$:
$$f(x) =\big (\frac {x-1} x\big)^{ax} - \big (\frac {x-2} x\big)^{ax}$$Then$$f'(x) = a\big (\frac {x-1} x\big)^{ax}\bigg [\ln\big (\frac {x-1} x\big) + \frac 1 {x-1} \bigg ] - a\big (\frac {x-2} x\big)^{ax}\bigg [\ln\big (\frac {x-2} x\big) + \frac 2 {x-2} \bigg ]$$By estimating $f(3)$ and $f(4)$ it is enough to show that $f'(x) \ne 0$ for $x > 3$. Then we can conclude that $f$ is decreasing for $x > 3$. We have $f'(x) = 0$ iff:
$$\frac{\ln(x-1) - ln(x) + \frac 1 {x-1}}{\ln(x-2) - ln(x) + \frac 2 {x-2}} = \big (\frac{x-2}{x-1})^{ax}$$The RHS is increasing from greater than $\frac 1 4$ to $1$. It is enough to show that the LHS is less than $\frac 1 4$ for $x \ge 3$. Note that $p(3) \approx 0.1$ and $\lim_{x \to \infty} LHS = \frac 1 4$. In other words, for $x \ge 3$ it is enough to show that:
$$\ln(x-2) - ln(x) + \frac 2 {x-2} > 4 \big [\ln(x-1) - ln(x) + \frac 1 {x-1}\big]$$The trick I used is to set $y = \frac 1 x$ and then we need to show that for $0 < y < \frac 1 3$ we have:
$$\ln(1 - 2y) + \frac{2y}{1-2y} > 4\big [ \ln(1-y) + \frac{y}{1-y} \big ]$$Using Taylor series we have:
$$\ln(1 - 2y) + \frac{2y}{1-2y} = \frac 1 2 (2y)^2 + \frac 2 3 (2y)^3 + \frac 3 4 (2y)^4 \dots$$$$4\big [\ln(1 - y) + \frac{y}{1-y} \big ] = \frac 1 2 (2y)^2 + \frac 2 3 (\frac 1 2)(2y)^3 + \frac 3 4 (\frac 1 4) (2y)^4 \dots$$And that's it, I hope!