An almost classic inequality

Question

It is a classical exercise to prove that $e^\pi>\pi^e$. But... is there a way to prove $\sin(e^\pi)<\sin(\pi^e)$ without calculator?

I was trying to prove that $13\pi/2<\pi^e$ and $e^\pi<15\pi/2$ and use monotone decreasing property of $\sin$ in $[13\pi/2,15\pi/3]$, but i couldn't prove the last inequalities.

Answer 1

First, let us prove that $\frac{13\pi}{2}<\pi^e$.

We know that $$e>\frac{19}{7}$$

because the integral

$$\frac{1}{14}\int_0^1 x^2(1-x)^2e^xdx = e-\frac{19}{7}$$

has positive integrand.

Let us use $$\frac{25}{8}<\pi<\frac{22}{7}$$

to transform the inequality

$$\frac{13\pi}{2}<\pi^e$$

into a stricter one with integers only

$$\frac{13}{2}\frac{22}{7}=\frac{143}{7}<\left(\frac{25}{8}\right)^\frac{19}{7}$$

This transforms into $$143^7·8^{19} < 25^{19}·7^7$$ or the lengthy $$176222766583426849287556934139904 < 299603198072873055934906005859375$$

which proves $\frac{13\pi}{2}<\pi^e$.

For $e^\pi<\frac{15\pi}{2}$

Let us apply logarithms to both sides, to obtain the equivalent inequality

$$\pi < \log\frac{15\pi}{2}$$

Now use the inequality $$\frac{25}{8}<\pi < \frac{22}{7}$$

to write the stricter inequality

$$\frac{22}{7} < \log{\frac{3·5^3}{2^4}} = \log(3)+3\log(5)-4\log(2)$$

Now we need lower bounds for $\log(3)$ and $\log(5)$ and an upper bound for $\log(2)$, because of the different signs involved.

Consider the following ternary BBP-type series for $\log(3)$ and $\log(5)$ $$\log(3)=\sum_{k=0}^\infty \frac{1}{9^{k+1}}\left(\frac{9}{2k+1}+\frac{1}{2k+2}\right)$$

$$\log(5)=\frac{4}{27}\sum_{k=0}^\infty \frac{1}{81^k}\left(\frac{9}{4k+1}+\frac{3}{4k+2}+\frac{1}{4k+3}\right)$$

Taking the first two terms yields inequalities

$$log(3)>\frac{355}{324}$$

and

$$log(5)>\frac{13688}{8505}$$

An upper bound for $\log(2)$ may be taken from the Dalzell-type integral $$\int_0^1 \frac{x^3 (1 - x)^3}{8 (1 + x)} dx = \frac{111}{160} - \log(2)$$

so $$log(2)<\frac{111}{160}$$

The resulting stricter inequality

$$\frac{22}{7} < \frac{355}{324} + 3\frac{13688}{8505}-4\frac{111}{160}$$

is easily verified, thus proving

$$e^\pi < \frac{15\pi}{2} $$

Answer 2

For the first,

$$\frac{13}2\pi<\pi^e$$

we merely have to prove that $\frac{13}2<\pi^{e-1}$. Note that $e>\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}$ by cutting off the taylor series at $x=1$; this means

$$\pi^{e-1}>\pi^{1+1/2+1/6}=\pi^{5/3}=\pi^{2/3}\pi>3^{2/3}\pi$$

We are now left to prove $\frac{13}2<3^{2/3}\pi$, which is equivalent to proving $\frac{13^3}{2^3}<3^2\pi$ or

$$\frac{13^3}{2^33^2}=\frac{2197}{72}<\pi^3$$

Now note that $\frac{223}{71}<\pi$ (which is $3+\frac{10}{71}$. According to Wikipedia, "In the 3rd century BCE, Archimedes proved the sharp inequalities  $\frac{223}{71}<\pi<\frac{22}{7}$, by means of regular $96$-gons"), which means

$$\pi^3>\frac{223^3}{71^3}=\frac{11089567}{357911}=\frac{2197\cdot5047+1308}{72\cdot 4971-1}>\frac{2197\cdot5047}{72\cdot 4971}>\frac{2197}{72}$$

So that proves $\frac{13}{2}\pi<\pi^e$.

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Now the second,

$$e^\pi<\frac{15}2\pi$$

Since $\pi<\frac{22}{7}$, and the already mentioned $\pi>\frac{223}{71}$ we prove the original statement by proving

$$e^{22/7}<\frac{15}{2}\frac{223}{71}$$

which we re-write to $e^{22/7}<\frac{3345}{142}$. Now taking the natural logarithm on both sides:

$$\frac{22}7<\log(3345)-\log(142)$$

Now note that

$$\log(n)=\log(n)+H_n-H_n=H_n+\int_1^{n+1}\left(\frac{1}{\lfloor x\rfloor}-\frac1x\right)\text{d}x$$

so that

\begin{align} \log(3345)&=H_{3345}+\int_1^{3346}\left(\frac{1}{\lfloor x\rfloor}-\frac1x\right)\text{d}x\\ &>H_{3345}+\int_1^{143}\left(\frac{1}{\lfloor x\rfloor}-\frac1x\right)\text{d}x\\ &=H_{3345}+\log(142)-H_{142} \end{align}

So that

$$\log(3345)-\log(142)>H_{3345}-H_{142}=\sum_{k=143}^{3345}\frac1k$$

So all we're left with is to prove

$$\frac{22}{7}<\sum_{k=143}^{3345}\frac1k$$

which is very time consuming but can be done by hand. You'd only need to calculate every $\frac1k$ up to $6$ decimal places (and round down from there since it's a lower bound). $5$ decimal places is not enough. Also, you'd only need to go up to about $3307$ instead of $3345$ to pass $\frac{22}7$.