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Unlike the famous question of comparing $e^\pi$ and $\pi^e$, which I solved almost instantly, I am stuck with this problem. My thought was the following.

Since exponential function is order-preserving, we exponentiate both terms and get $\pi$ and $e^{\pi-2}$. Then we study the function $f(x) = e^{x-2} - x$ or the function $g(x) = \frac{e^{x-2}}{x}$, and compare them with zero and one respectively. I tried both. But both involve solving the equation $$e^{x-2} = x.$$

I tried Lagrange error terms and have $$f(x) = -1 + \frac{(x-2)^2}{2!} + R_2(x-2),$$ where $$\frac{(x-2)^3}{3!} \le R_2(x-2) \le \frac{e^{x-2}}{3!} (x-2)^3.$$

It is easy to see that the equation have a root between $3$ and $2 + \sqrt2$. But I don't know how close it is to $\pi$. It is to provide some lower bounds since we can plug in some values and calculate to show that $f(x) > 0$ for such values. But for the upper bound, it is hard to calculate by hands since it has the $e^{x-2}$ factor. At my best attempt by hand, I showed that $f(3.15) > 0$. All it entails is that for all $x \ge 3.15$, $e^{x-2}$ is greater than $x$. But it tells nothing about the other side.

Then I looked at the calculator and find that $e^{\pi-2} < \pi$.

I also tried Newton-Raphson iteration, but it involves a lot of exponentiation which is hard to calculate by hand and also involves approximation by themselves. And I don't know how fast and close the iteration converges to the true root of the equation.

Any other hint for comparing these two number purely by hand?

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    $\begingroup$ Coincidentally enough, the difference is really small, making the common approach of defining a function and root approximating very dicey $\endgroup$ – Dhanvi Sreenivasan Dec 15 '20 at 10:09
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    $\begingroup$ The values are quite close , so I guess a hand-calculation will be somewhat messy. $\endgroup$ – Peter Dec 15 '20 at 10:10
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    $\begingroup$ Here is a suggestion: define the function $f(x) = \text{ln} (\pi x) - \pi x + 2$ for all $x>0$. Since $\lim_{x\to 0^+} f(x) = - \infty$ and computing first derivative one can shows $f$ attains a global maximum at some point $\alpha < 1$ where $f(\alpha)>0$, then Bolzano theorem says there is $c\in (0,1)$ with $f(c)=0$. Now, since $\text{ln}$ is strictly concave, $f$ is also concave. Therefore, given any $t\in (0,1)$, $0<f(t1 + (1-t)c) < t f(1) + (1-t)f(c) = t f(1)$. This leads to $f(1)>0$, which means $\text{ln}(\pi) > \pi -2.$ $\endgroup$ – Senna Dec 15 '20 at 10:10
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    $\begingroup$ Exactly, the difference is like 0.00314 (quite amusing) $\endgroup$ – Dhanvi Sreenivasan Dec 15 '20 at 10:10
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    $\begingroup$ @Senna, I followed your argument until the point where you claimed $$f(t1 + (1-t)c)\stackrel{?}<tf(1)+(1-t)f(c)\qquad\text{for}\qquad t\in(0,1).$$ Is this true only when $f$ is strictly convex? Also note that your argument, if works, should also work with $f(x)=\log(4x)-4x+2$ to show that $0<f(1)=\log(4)-4+2$, which is false. $\endgroup$ – Sangchul Lee Dec 15 '20 at 10:17
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By hand I took $e^{\pi -2}<e\cdot e^{0.1416}<(2.7183)e^{0.1416}.$ Using $B_1=0.1416=0.1+0.04(1+0.04)$ for manual calculation, I computed, to $5$ decimal places, an upper bound $B_2$ for $(B_1)^2/2$ and an upper bound $B_3$ for $B_1B_2/3 $ and an upper bound $B_4$ for $B_1B_3/4,$ etc., until I was sure that the sum of the remaining terms was less than $0.00005,$ to obtain an upper bound $B$ to $4$ decimal places for $e^{0.1416}.$ Then I multiplied $B\times 2.7183$ and got less than $\pi.$

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    $\begingroup$ See the book The Trachtenburg Speed System Of Basic Mathematics for some wonderful ways (that I didn't even use here) to do multiplication. $\endgroup$ – DanielWainfleet Dec 15 '20 at 13:44
  • $\begingroup$ Forgive my ignorance. Could you kindly elaborate on "until the remaining terms was less than $0.00005$"? I believe this is where I am stuck. Is there any theorems about remainder bound I missed? How did you bound the remainder? Thanks. $\endgroup$ – Lei Zhao Dec 16 '20 at 6:30
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    $\begingroup$ Actually $B_4$ was as far as I needed. In the power series for $e^x$, the sum $S$ of the terms of degree $5$ or more in $x$ is $S=(x^5/5!)(1+x/6+x^2/6.7+x^3/6.7.8+...)$ which for $1/6>x>0$ is less than $(x^5/5!)(1+x/6+x^2/6^2+x^3/6^3+...)=(x^5/5!)/(1-x/6).$ And the sum $S$ decreases with decreasing positive $x.$ So with $x=B_1<1/7$ we have $S<((1/7)^5/5!)/(1-1/42)<10^{-5}.$ $\endgroup$ – DanielWainfleet Dec 16 '20 at 20:37
  • $\begingroup$ There are theorems about the remainder of a power series. If $r>0$ and if the series $f(x)=\sum_{j=0}^{\infty}A_jx^j$ converges whenever $|x|<r,$ then for $0<|x|<r$ and $n>0$ we have $f(x)-\sum_{j=0}^{n-1}A_jx^j= \frac {x^n f^{(n)}(y)}{n!}$ for some $y$ strictly between $0$ and $x,$ where $f^{(n)}$ denotes the $n$th derivative of $f$. $\endgroup$ – DanielWainfleet Dec 16 '20 at 21:00
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All the logarithms I know by heart are $$\begin{align}\log_{10}2&=0.30103\\ \ln10&=2.303\\ \ln2&=0.693\end{align}$$ Using the second fact alone we can solve this problem! We know that $$\frac12\ln\left(\frac{1+x}{1-x}\right)=x+R_2(x)$$ Where $$|R_2(x)|\le\frac{|x|^3}{3(1-|x|)^3}$$ for $|x|<1$. Then let $x=\frac{-3}{487}$ so $$\frac12\ln\left(\frac{1+x}{1-x}\right)=\ln\left(\frac{22}{7\sqrt{10}}\right)=\frac{-3}{487}+R_2\left(\frac{-3}{487}\right)$$ So we have $$\begin{align}\ln\left(\frac{22}7\right)&=\frac12\ln10-\frac3{487}+R_2\left(\frac{-3}{487}\right)\gt\frac{2.3025}2-\frac1{160}-\frac1{100^3}\\ &=1.15125-0.00625-10^{-6}=1.145-10^{-6}\gt1.143\gt\frac{22}7-2\end{align}$$ And since $f(x)=\ln x-x+2$ is decreasing for $x\gt1$ and it has been known since the time of Archimedes that $\frac{22}7\gt\pi$ we have established the result.

But if you didn't know that $\ln10=2.303$ to $3$ decimals you might be in for a tougher slog. You could say, for example, that $$\ln10=10\ln\frac54+3\ln\frac{128}{125}\gt20\left(\frac19+\frac1{3\times9^3}\right)+6\left(\frac3{253}\right)$$ So that $$\begin{align}\ln\left(\frac{22}7\right)&\gt\frac{10}9+\frac{10}{2187}+\frac9{253}-\frac3{487}-10^{-6}\\ &\gt1.1111+0.004+0.035-0.007-10^{-6}\gt1.143\\ &\gt\frac{22}7-2\end{align}$$ Where we actually had to carry out one of the long divisions to $2$ significant figures.

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