# Show that the given function is decreasing

• idobido
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.

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!

Last edited:
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}
## ?

idobido said:
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.

fresh_42 said:
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
##

idobido said:
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.

did you notice that ##
fresh_42 said:
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?

idobido said:
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.

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.

Last edited:
fresh_42
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!

fresh_42

## What does it mean for a function to be decreasing?

A function is said to be decreasing on an interval if, for any two points $$x_1$$ and $$x_2$$ within that interval, where $$x_1 < x_2$$, the function values satisfy $$f(x_1) \geq f(x_2)$$. This means that as the input values increase, the output values either stay the same or decrease.

## How do I determine if a function is decreasing?

To determine if a function is decreasing, you can use the first derivative test. If the derivative $$f'(x)$$ is less than or equal to zero for all $$x$$ in an interval, then the function is decreasing on that interval. Specifically, if $$f'(x) < 0$$ for all $$x$$ in the interval, the function is strictly decreasing.

## What role does the first derivative play in showing that a function is decreasing?

The first derivative of a function, $$f'(x)$$, represents the rate of change of the function. If $$f'(x)$$ is negative over an interval, it indicates that the function is decreasing on that interval because the slope of the tangent line to the curve is negative.

## Can a function be decreasing if its derivative is zero?

If the derivative $$f'(x) = 0$$ over an interval, the function is constant on that interval, meaning it does not increase or decrease. For a function to be strictly decreasing, the derivative must be less than zero. However, if $$f'(x) \leq 0$$, the function is non-increasing, which includes both decreasing and constant behaviors.

## What are some examples of decreasing functions?

Examples of decreasing functions include $$f(x) = -x$$, $$f(x) = e^{-x}$$, and $$f(x) = \frac{1}{x}$$ for $$x > 0$$. In these cases, the derivatives are $$f'(x) = -1$$, $$f'(x) = -e^{-x}$$, and $$f'(x) = -\frac{1}{x^2}$$, respectively, all of which are less than zero over their respective domains.

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