How to prove that $f(x)=\frac{ \cot(\frac{\pi}{x+1}) }{ \cot(\frac{\pi}{x}) }\cdot\frac{x}{x+1}$ is strictly deceasing for $x>2$?

Question

I read an article that claimed that among two regular polygons with equal perimeters, the one with more sides has a larger area, in other words $\frac{ \cot(\frac{\pi}{n+1}) }{ \cot(\frac{\pi}{n}) }\cdot\frac{n}{n+1} >1$ for $n \ge 3$.

I became curious about $f(x):=\frac{ \cot(\frac{\pi}{x+1}) }{ \cot(\frac{\pi}{x}) }\cdot\frac{x}{x+1}$ and decided to graph it. It seemed to be strictly decreasing on $(2,\infty)$

$$f'(x)=\frac{\cot(\frac{\pi}{x}) \cdot \csc(\frac{\pi}{x+1})\cdot \frac{\pi}{(x+1)^2}- \cot(\frac{\pi}{x+1}) \cdot \csc(\frac{\pi}{x})\cdot \frac{\pi}{x^2}}{\cot^2(\frac{\pi}{x}) }\bigg(1 -\frac{1}{x+1} \bigg) +\frac{ \cot(\frac{\pi}{x+1}) }{ \cot(\frac{\pi}{x}) }\cdot\frac{1}{(x+1)^2} $$

I tried to prove this observation, but I failed. Finding the derivative of the function and then showing that $f′(x)<0$ for $x>2$ seemed impossible and dreadful, as the derivative involved trigonometric and rational functions.

I also want to know how to prove that $\frac{\cot(\frac{\pi}{x})}{x}$ is strictly increasing for $x\ge 3$ or in other words among two regular polygons with equal perimeters, the one with more sides has a larger area

Answer 1

It suffices to prove that, for all $x > 2$, $$(\ln f(x))' < 0, $$ or $$\left(\ln \cot \frac{\pi}{x + 1} - \ln\cot \frac{\pi}{x} + \ln x - \ln(x + 1)\right)' < 0,$$ or $$\frac{2\pi}{(x+1)^2\sin\frac{2\pi}{x+1}} - \frac{2\pi}{x^2\sin\frac{2\pi}{x}} + \frac{1}{x} - \frac{1}{x + 1} < 0. \tag{1} $$

Note that $g(u) := \frac{u}{\sin u}$ is non-decreasing on $(0, \pi)$. We have $$\frac{2\pi}{(x + 1)\sin \frac{2\pi}{x+ 1}} \le \frac{2\pi}{x\sin \frac{2\pi}{x}}. \tag{2}$$ (Note: We have $g'(u) = \frac{\sin u - u\cos u}{\sin^2 u} \ge 0$ on $(0, \pi)$.)

From (1) and (2), it suffices to prove that, for all $x > 2$, $$\frac{1}{x + 1} \cdot \frac{2\pi}{x\sin \frac{2\pi}{x}} - \frac{2\pi}{x^2\sin\frac{2\pi}{x}} + \frac{1}{x} - \frac{1}{x + 1} < 0,$$ or $$\sin \frac{2\pi}{x} < \frac{2\pi}{x}$$ which is true.

We are done.

Answer 2

Added an Addendum to the end of my answer to respond to an additional question from pie.

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Preliminary Results:
PR-1
For $~0 < x < \pi/2, ~~x > \sin(x).~$
Proof
$~f(x) = x - \sin(x) \implies f(0) = 0 ~~\text{and} ~~f'(x) = 1 - \cos(x). $
Therefore, $~f(x)~$ is strictly increasing for $~0 < x < \pi/2.~$

PR-2
$\cos^2(\theta) \times \tan(\theta) = \dfrac{1}{2}\sin(2\theta).$
Proof
$\cos^2(\theta) \times \dfrac{\sin(\theta)}{\cos(\theta)} = \sin(\theta)\cos(\theta).$

PR-3
$\displaystyle \frac{d\left[ ~\frac{n-2}{n} ~\right]}{dn} = \frac{2}{n^2}.$
Proof
By the quotient rule for derivatives, you have:
$\displaystyle \frac{1}{n^2} \times \left[ ~n - (n-2) ~\right].$

PR-4
$\displaystyle \frac{d\left[ ~\tan(\theta) ~\right]}{d\theta} = \cos^{-2}(\theta).$
Proof
By the quotient rule for derivatives: $\displaystyle \frac{d\left[ ~\frac{\sin(\theta)}{\cos(\theta)} ~\right]}{d\theta} $
$\displaystyle = \cos^{-2}(\theta) \times \left\{ ~\left[ ~ \cos(\theta)\cos(\theta)~\right] - \left[ ~\sin(\theta) \times (-1) \times \sin(\theta) ~\right] ~\right\}$

$\displaystyle = \cos^{-2}(\theta) \times 1.$

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The goal is to show that if the perimeter of a regular polygon is fixed, that the area is strictly increasing when the number of sides increases.

I am unsure how you derived your formula.

In the diagram above, you have a regular polygon with $~n~$ sides, with a total perimeter of $~P,~$ a fixed constant.

Each side has a length of $~\dfrac{P}{n},~$ and an altitude of $~\dfrac{P}{2n} \times \tan(\theta).$

Each angle $~\theta~$ equals $~\dfrac{\pi(n-2)}{2n}.~$

This is because there are $~n~$ triangles, so the sum of all of the angles is $~n\pi.~$ Then, in the $~n~$ triangles, the sum of the angles at the center of the polygon is $~2\pi.~$ Finally, the remainder, $~\pi(n-2)~$ is distributed equally among the $~2n~$ angles represented by $~\theta.$

Therefore, the total area of the polygon is

$$n \times \frac{1}{2} \times \frac{P}{n} \times \frac{P}{2n} \times \tan\left[ ~\frac{\pi(n-2)}{2n} ~\right]. \tag1 $$

The formula in (1) above equals

$$\text{constant} \times \frac{1}{n} \times \tan\left[ ~\frac{\pi(n-2)}{2n} ~\right] ~: ~~~\text{constant} = \frac{P^2}{4}. \tag2 $$

So, the goal is to show that

$$f(n) = \frac{1}{n} \times \tan\left[ ~\frac{\pi(n-2)}{2n} ~\right]$$

is strictly increasing.

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Using the PR-3, PR-4, and the quotient rule for derivatives:

$$f'(n) = \left\{ ~\frac{1}{n} \times \cos^{-2}\left[ ~\frac{\pi(n-2)}{2n} ~\right] \times \frac{\pi}{n^2} ~\right\} + \left\{ ~\tan\left[ ~\frac{\pi(n-2)}{2n} ~\right] \times \frac{-1}{n^2}. ~\right\} $$

Let $\displaystyle ~g(n) = \frac{\pi}{n} - \left\{ ~\frac{1}{2} \times \sin\left[ ~\frac{\pi(n-2)}{n} ~\right] ~\right\}.$

Then, using PR-2,

$$f'(n) = ~\frac{1}{n^2} \times \cos^{-2}\left[ ~\frac{\pi(n-2)}{2n} ~\right] \times g(n). \tag3 $$

In (3) above, the first two factors are always positive, so the problem reduces to showing that $~g(n)~$ is always positive for $~n \in \Bbb{Z_{\geq 3}}.$

Since $~\displaystyle \left[ ~\frac{\pi(n-2)}{n} ~\right] = \pi - \frac{2\pi}{n},~$ and since $~\sin(\theta) = \sin(\pi - \theta),~ g(n)~$ may be re-expressed as

$$g(n) = \frac{1}{2} \times \left\{ ~\frac{2\pi}{n} - \sin\left[ ~\frac{2\pi}{n} ~\right] ~\right\}.$$

Now, let $~x = \dfrac{2\pi}{n}.$

By PR-1, $~g(n)~$ is always positive.

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$\underline{\text{Addendum}}$

Responding to an additional question from pie.

I also want to know how to prove that $\frac{\cot(\frac{\pi}{x})}{x}$ is strictly increasing for $x\ge 3$

Virtually the same approach taken in the first part of my answer will work here.

Let $~\displaystyle f(x) = \frac{\cot\left(\frac{\pi}{x}\right)}{x} \implies $

$$f'(x) = \frac{1}{x^2} \times \left\{ ~ \left[ ~x \times (-1) \times \sin^{-2}\left(\frac{\pi}{x}\right) \times \frac{-\pi}{x^2} ~\right] - \left[ ~\cot\left(\frac{\pi}{x}\right) ~\right] ~\right\}$$

$$= \frac{1}{x^2} \times \left\{ ~ \left[ ~ \frac{\pi}{x} \times \sin^{-2}\left(\frac{\pi}{x}\right) ~\right] - \left[ ~\cot\left(\frac{\pi}{x}\right) ~\right] ~\right\}.$$

Now, similar to PR-2, you have that

$$\sin^2(\theta) \times \cot(\theta) = \dfrac{1}{2}\sin(2\theta).$$

Therefore,

$$f'(x) = \frac{1}{x^2} \times \sin^{-2}\left(\frac{\pi}{x}\right) \times \left\{ ~ \left[ ~ \frac{\pi}{x} ~\right] - \left[ \frac{1}{2} ~\sin\left(\frac{2\pi}{x}\right) ~\right] ~\right\}$$

$$= \frac{1}{2x^2} \times \sin^{-2}\left(\frac{\pi}{x}\right) \times \left\{ ~ \left[ ~ \frac{2\pi}{x} ~\right] - \left[ ~\sin\left(\frac{2\pi}{x}\right) ~\right] ~\right\}.$$

Here, the factors $~\displaystyle \frac{1}{2x^2} ~$ and $~\displaystyle \sin^{-2}\left(\frac{\pi}{x}\right) ~$ are always positive for $~x \geq 3.~$

Further, by PR-1, you have that

$\displaystyle \left\{ ~ \left[ ~ \frac{2\pi}{x} ~\right] - \left[ ~\sin\left(\frac{2\pi}{x}\right) ~\right] ~\right\} ~$

is always positive for $~x \geq 3.$

Answer 3

Expansion:

Recognizing the zeros of $\sin x$ and expressing it as a product:

$$ \sin(x) = x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\dots $$

Regrouping the terms yields:

$$ \sin(x) = x(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\dots $$

$$ = x \prod_{n\in\mathbb{N}^*}(x^2-n^2\pi^2) $$

$$ \Leftrightarrow \sin(x) = x \prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right) $$

This is a well-defined factorized series for $\sin(x)$.

The product expansion for the sine function is given by:

$$ \sin x = x \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right) $$

This expression represents the infinite product expansion of the sine function in terms of its zeros, where $n$ ranges over all positive integers.

Now, applying the natural logarithm ($\log$) to both sides of the equation:

$$ \log(\sin x) = \log\left(x \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)\right) $$

Using the properties of logarithms, the product turns into a sum:

$$ \log(\sin x) = \log x + \sum_{n=1}^\infty \log\left(1-\frac{x^2}{n^2\pi^2}\right) $$

Now, differentiating both sides with respect to $x$:

$$ \frac{1}{\sin x} \cdot \cos x = \frac{1}{x} + \sum_{n=1}^\infty \frac{-2x}{n^2\pi^2 \left(1-\frac{x^2}{n^2\pi^2}\right)} $$

Simplifying, we get the expression for the cotangent function:

$$ \cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \frac{1}{1-\frac{x^2}{n^2\pi^2}}\right] $$

This formula represents the cotangent function as an infinite sum involving terms that account for the zeros of the sine function. Each term in the sum corresponds to a specific zero of the sine function, and the series converges for all valid values of $x$.

The cotangent function ($\cot(z)$) can be expressed as:

$$ \cot(z) = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{z-n\pi} = \frac{1}{z} + 2z\sum_{k=1}^\infty \frac{1}{z^2 - (k\pi)^2} $$

Similar steps can be done for proving $\tan x$ expansion with the help of expressing $\cos x$ as the product of its zeros. Alternatively, similar results can be obtained using the residue theorem in complex analysis or by considering the asymptotes of the functions. Feel free to ask for assistance in proving with any of these methods.

The tangent function ($\tan(z)$) can be expressed as an infinite series:

$$ \tan(z) = \sum_{n=0}^\infty \frac{8z}{(2n+1)^2\pi^2-4z^2} $$

Now, coming back to your problem:

$$ f(x) = \frac{\cot\left(\frac{\pi}{x+1}\right)}{(x+1)} \cdot x \tan\left(\frac{\pi}{x}\right) $$

now $$ x \tan\left(\frac{\pi}{x}\right) = \sum_{n=0}^\infty \frac{8x\frac{\pi}{x}}{(2n+1)^2\pi^2-4\left(\frac{\pi}{x}\right)^2} $$ we can see here as x increases the denominator term will become larger because 4/x becomes smaller so it is a decreasing function $$ x \tan\left(\frac{\pi}{x}\right) = \sum_{n=0}^\infty \frac{8\pi}{(2n+1)^2\pi^2-4\left(\frac{\pi}{x}\right)^2} $$ for first $$ \cot\left(\frac{\pi}{x+1}\right) = \frac{1}{\frac{\pi}{x+1}} + 2\left(\frac{\pi}{x+1}\right)\sum_{k=1}^\infty \frac{1}{\left(\frac{\pi}{x+1}\right)^2 - (k\pi)^2} $$ as we increase x the function will be decreasing because as we increase x it will the overall quantity or sum which is being added to 1/$\pi$ $$ \frac{1}{x+1} \cot\left(\frac{\pi}{x+1}\right) = \frac{1}{\pi} + 2\pi \sum_{k=1}^\infty \frac{1}{\left(\left(\frac{\pi}{x+1}\right)^2 - (k\pi)^2\right)} $$

conclusion

so this is an decreasing funtion $$ x \tan\left(\frac{\pi}{x}\right) \cdot \left(\frac{1}{x+1} \cot\left(\frac{\pi}{x+1}\right)\right) = \left(\sum_{n=0}^\infty \frac{8\pi}{(2n+1)^2\pi^2-4\left(\frac{\pi}{x}\right)^2}\right) \cdot \left(\frac{1}{\pi} + 2\pi \sum_{k=1}^\infty \frac{1}{\left(\left(\frac{\pi}{x+1}\right)^2 - (k\pi)^2\right)}\right) $$

this might not be prettiest but its very easy to follow and intutive

i hope it helps

Answer 4

here is an elegant approach i came up with after thinking a bit

Let $ g(x) = x \tan\left(\frac{\pi}{x}\right) $.

Then, the given function is $ f(x) = \frac{g(x)}{g(x+1)} $.

Now, if $ f(x) $ is decreasing for $ x > 2 $, then $ f(x+1) - f(x) < 0 $ for any $ x > 2 $. the difference between the fractions (denominator - numerator) in $ f(x) $ must be increasing as $ x $ increases.

Therefore, $ g(x+2) - g(x+1) > g(x+1) - g(x) $, suggesting that $ g'(x+1) - g'(x) > 0 $.

Consequently, $ g''(x) > 0 $ for all values greater than 2. Hence, the function is strictly decreasing.

lets check

The first derivative of $ g(x) $ is given by $$ g'(x) = \frac{d}{dx} [g(x)] = \tan\left(\frac{\pi}{x}\right) - \frac{\pi \sec^2\left(\frac{\pi}{x}\right)}{x}. $$

The second derivative of $ g(x) $ is $$ g''(x) = \frac{d^2}{dx^2} [g(x)] = \frac{2\pi^2 \sec^2\left(\frac{\pi}{x}\right) \tan\left(\frac{\pi}{x}\right)}{x^3}. $$

its clear that g''(x) is positive for all x>2

Idea

the rate at which the denominator is increasing exceeds the rate at which the numerator is increasing as $ x $ increases. This causes the overall fraction to decrease, indicating that $ f(x) $ is decreasing.