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Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator.

I did in the following way. Are there other ways?

Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then, $$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$ Now, $$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad f''(x)=\frac{e\pi (2\ln x-3)}{x^3},\quad f'''(x)=\frac{e\pi (11-6\ln x)}{x^4}.$$ Since $f'(x)\lt 0$ for $e\lt x\lt\pi$, one has $f(e)\gt f(\pi)$. By Taylor's theorem, there exists a point $c$ in $(e,\pi)$ such that $$\begin{align}f(\pi)&=f(e)+(\pi-e)f'(e)+\frac{(\pi-e)^2}{2}f''(c)\\&\gt f(e)+(\pi-e)\cdot 0+\frac{(\pi-e)^2}{2}\cdot \frac{e\pi(2\ln e-3)}{e^3}\\&=f(e)-\frac{\pi(\pi-e)^2}{2e^2}\tag2\end{align}$$ because $f'''(x)\gt 0\ (e\lt x\lt \pi)$ implies $f''(c)\gt f''(e)$.

By the mean value theorem and $(2)$, $$e^{f(e)}-e^{f(\pi)}\lt (f(e)-f(\pi))e^{f(e)}\lt \frac{\pi (\pi-e)^2}{2e^2}\cdot e^{\pi}=\frac{e\pi(\pi-e)^2}{2}\cdot e^{\pi-3}\tag3$$

Since $e^x\lt \frac{1}{1-x}\ (0\lt x\lt 1)$ and $0\lt \pi-3\lt 1$, $$e^{\pi-3}\lt\frac{1}{4-\pi}\tag4$$

From $(1)(3)(4)$, $$e^{\pi}-{\pi}^e\lt \frac{e\pi(\pi-e)^2}{2}\cdot\frac{1}{4-\pi}\lt\frac{3\times\frac{22}{7}\left(\frac{22}{7}-2.718\right)^2}{2}\cdot\frac{1}{4-\frac{22}{7}}\lt 0.993\lt 1$$

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  • $\begingroup$ I deleted an answer involving numerical approximations, but thought I should mention something of the difficulty involved there. Since the actual difference between the two numbers is already $ \ \approx \ 0.68 \ $ , each of the two values (about 23) must be estimated to a precision of order 1% so that the estimate of the difference remains less than 1 . This turns out to be extremely difficult to arrange for transcendental powers of transcendental numbers if one is using rational numbers to represent sums of series when one must obtain convincing results without a calculator. $\endgroup$ – colormegone Aug 30 '15 at 2:24
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    $\begingroup$ @mathlove If I may make a suggestion, I have been bothered by the last line, since it uses a decimal approximation for $ \ e \ $ , leading to a difference squared which would seem to require a calculator to demonstrate that the value is less than 0.993 . You also use $ \ 3 \ $ as a bound on $ \ e \ $ which makes this coarser than necessary. Since $ \ \frac{19}{7} \ < \ e \ \frac{20}{7} \ $ , you can use $ \ \frac{20}{7} \ $ as the first factor and $ \ \frac{19}{7} \ $ in the difference-squared. This leads to (continued) $\endgroup$ – colormegone Aug 30 '15 at 2:41
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    $\begingroup$ $$ \frac{1}{2} \ \cdot \ \frac{20}{7} \ \cdot \ \frac{22}{7} \ \cdot \ ( \frac{22}{7} - \frac{19}{7})^2 \ \cdot \left(\frac{1}{4 - \frac{22}{7}} \right) \ = \ \frac{20 \cdot 22 \cdot 9 \cdot 7}{2 \cdot 7 \cdot 7 \cdot 49 \cdot 6} \ = \ \frac{330}{343} \ < \ 1 \ \ . $$ The factors cancel in a way that makes simplification possible without a calculator. [This is an omission-typo in the previous comment: the inequality was meant to read $ \ \frac{19}{7} \ < \ e \ < \ \frac{20}{7} \ $ . ] $\endgroup$ – colormegone Aug 30 '15 at 2:44
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    $\begingroup$ Fantastic method of proof! $\endgroup$ – Chandler Watson Sep 4 '15 at 15:03
  • $\begingroup$ There is a way called ceilings and floor technique where only integral parts are taken into consideration so $2^3-3^2=-1( approx). Hence the inequality is true ie $-1<1$. $\endgroup$ – Archis Welankar Nov 15 '15 at 5:19
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Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$. Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up. In general, without any real knowledge of $f$:

$$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y} $$

where the first inequality is something we are trying to prove for $x = e \approx 2.718$ and $y = \pi \approx 3.141$.


Let's try $x = \frac{11}{4}$ and $y = \frac{13}{4}$. Then we have taken off too much slack:

$$ \left( \frac{11}{4} \right)^{\frac{13}{4}} - \left( \frac{13}{4} \right)^{\frac{11}{4}} \approx 1.214$$

Using the continued fractions of $e$ and $\pi $ does give less than one and check with a calculator:

$$ \left( \frac{8}{3} \right)^{\frac{22}{7}} - \left( \frac{22}{7} \right)^{\frac{8}{3}} \approx 0.621$$

Even your proof requires a calculator in the last step... but we did not use the exact values of $\pi, e$.


How good is this approximation? Somehow we should compute how quickly $f$ changes with $x$ and $y$:

$$ \frac{\partial f}{\partial x} = y \, x^{y-1} - (\ln y ) y^x \approx 3 \times 3^{3-1} - (\ln 3) 3^3 \approx -27(\ln \frac{e}{3}) \approx -2.7 $$

The $y$ partial derivative is similar, except it is positive instead of negative.


It can be proven, the continued fraction error of $e$ and $\pi$ are inversely proportional to the denominators:

$$ \left| e - \frac{8}{3} \right| < \frac{1}{3^2} \text{ and } \left| \pi - \frac{22}{7} \right| < \frac{1}{7^2}$$

These would be our estimates of $\epsilon_1$ and $\epsilon_2$. Then using a tiny bit of multivariable calculus:

$$ \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y} \approx \frac{1}{3^2} \times 2.7 + \frac{1}{7^2} \times (-2.7) < 0.355 $$

This should be enough to establish $|f(\pi,e)| < 1$.


This used a calculator in many places but not the exact value of the constants $\pi$ and $e$.

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I looked at a couple of ways of examining this using series expansions. A difficulty that I noted in a comment to mathlove's answer is that this requires rather careful selection of estimators, since the permissible "error budget" is very limited if we are to show that the difference is less than 1 , when its actual value is about 0.682. Another issue is that we would like to use rational approximations in a way that makes these computations look credibly like ones that can be managed without a calculator. I use some "number facts" here that are either generally familiar or can be found readily enough (while these can be obtained using a calculator, that can be avoided if one is acquainted with approximate values); it proves to be very important to use values that closely approximate the needed numbers.

We can attempt to estimate an upper limit the difference $ \ e^{\pi} \ - \ \pi^e \ $ by setting up an inquality in which $ \ e^{\pi} \ $ is overestimated and $ \ \pi^e \ $ is underestimated. In this first one, we will start from

$$ \left( \frac{68}{25} \right)^{22/7} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \left( \frac{157}{50} \right)^{19/7} \ \ $$

$$ \Rightarrow \ \ \left( 3 \ - \ \frac{7}{25} \right)^{3 + \frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \left( 3 \ + \ \frac{7}{50} \right)^{3 - \frac{2}{7}} \ \ $$

$$ \Rightarrow \ \ 3^3 \ \cdot \ 3^{1/7} \ \cdot \left( 1 \ - \ \frac{7}{75} \right)^3 \ \left( 1 \ - \ \frac{7}{75} \right)^{\frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \quad > \ 3^3 \ \cdot \ 3^{-2/7} \ \cdot \left( 1 \ + \ \frac{7}{150} \right)^3 \ \left( 1 \ + \ \frac{7}{150} \right)^{-\frac{2}{7}} \ \ . $$

We can construct Taylor polynomials for the powers of $ \ 3 \ $ based on the function $ \ 3^x \ $ [using $ \ \ln 3 \ \approx \ 1.1 \ $ ], and can apply the (generalized) binomial series for the other factors:

$$ 3^3 \ \cdot \ 3^{1/7} \ \cdot \left( 1 \ - \ \frac{7}{75} \right)^3 \ \left( 1 \ - \ \frac{7}{75} \right)^{\frac{1}{7}} $$ $$ \approx \ 27 \ \left( 1 \ + \ \ln 3 \cdot \frac{1}{7} \ + \ \frac{(\ln 3)^2}{2} \cdot \frac{1}{7^2} \right) \ \left( 1 \ - \ 3 \cdot \frac{7}{75}\ + \ 3 \cdot \frac{7^2}{75^2} \right) $$ $$ \cdot \left( 1 \ - \ \frac{1}{7} \cdot \frac{7}{75} \ + \ \frac{\frac{1}{7} \cdot \frac{-6}{7}}{2} \cdot \frac{7^2}{75^2} \right) \ $$

$$ \approx \ 27 \ \left( 1 \ + \ \frac{11}{10} \cdot \frac{1}{7} \ + \ \frac{6}{10} \cdot \frac{1}{7^2} \right) \ \left( 1 \ - \ \frac{7}{25}\ + \ 3 \cdot \frac{7^2}{75^2} \right) \ \left( 1 \ - \ \frac{1}{75} \ - \ \frac{3}{75^2} \right) \ \ , $$

and similarly,

$$ 3^3 \ \cdot \ 3^{-2/7} \ \cdot \left( 1 \ + \ \frac{7}{150} \right)^3 \ \left( 1 \ + \ \frac{7}{150} \right)^{-\frac{2}{7}} $$ $$ \approx \ 27 \ \left( 1 \ - \ \ln 3 \cdot \frac{2}{7} \ + \ \frac{(\ln 3)^2}{2} \cdot \frac{4}{7^2} \ - \ \frac{(\ln 3)^3}{6} \cdot \frac{8}{7^3} \right) \ \left( 1 \ + \ 3 \cdot \frac{7}{150}\ + \ 3 \cdot \frac{7^2}{150^2} \right) $$ $$ \cdot \left( 1 \ - \ \frac{2}{7} \cdot \frac{7}{150} \ + \ \frac{\frac{-2}{7} \cdot \frac{-9}{7}}{2} \cdot \frac{7^2}{150^2} \right) \ $$

$$ \approx \ 27 \ \left( 1 \ - \ \frac{11}{10} \cdot \frac{2}{7} \ + \ \frac{6}{10} \cdot \frac{4}{7^2} \ - \ \frac{4 \cdot 4}{3 \cdot 3} \cdot \ \frac{1}{7^3} \right) \ \left( 1 \ + \ \frac{7}{50}\ + \ 3 \cdot \frac{7^2}{150^2} \right) $$ $$ \cdot \ \left( 1 \ - \ \frac{1}{75} \ + \ \frac{9}{150^2} \right) \ \ . $$

In order for the series for $ \ 3^{-2/7} \ $ to remain an underestimate, it is necessary to extend the Taylor polynomial to third-degree. Simple linearizations give far too "coarse" approximations to resolve the inequality we desire clearly enough; we are also unable to obtain sufficient precision by only keeping the "linear" terms in the product. There isn't much for it other than to arrange each factor to have a single denominator, which gives us

$$ 27 \ \left[ \ \left( \frac{573}{490} \right) \ \left( \frac{4197}{5625} \right) \ \left( \frac{5547}{5625} \right) \ - \ \left( \frac{11260}{15435} \right) \ \left( \frac{25797}{22500} \right) \ \left( \frac{22209}{22500} \right) \right] \ > \ e^{\pi} \ - \ \pi^e \ \ . $$

From this point, it becomes a question of how much calculation one has the patience to do without the aid of a device; it would likely be easiest to divide out each ratio, multiply the three factors in each term, take the difference between terms, and multiply the result by $ \ 27 \ $ . In the pre-calculator era, this would not have been considered unreasonable: after all, the Ludolphine number $ ( \pi ) $ was computed to 35 decimal places over 400 years ago using polygon-approximations over the course of some years. If we avail ourselves of a calculator at this point, we see that we have

$$ 27 \ \left( \ 1.16939 \cdot 0.74613 \cdot 0.98613 \ - \ 0.72951 \cdot 1.14653 \cdot 0.98707 \right) \ \approx \ 0.9403 \ > \ e^{\pi} \ - \ \pi^e \ \ . $$

This may be compared with the first inequality we wrote, for which our result is an intended overestimate,

$$ \left( \frac{68}{25} \right)^{22/7} \ - \ \left( \frac{157}{50} \right)^{19/7} \ \approx \ 0.8901 > \ e^{\pi} \ - \ \pi^{e} \ \ . $$

$$ \ \ $$

We might well wish for something more tractable, which can be obtained by making the estimate a bit differently. If we're permitted the "number facts" $ \ e^3 \ \approx \ 20.1 \ \ , \ \ \pi^3 \ \approx \ 31 \ \ , \ \pi \ < \ \frac{22}{7} \ $ and $ \ e \ > \ \frac{19}{7} \ $ , we can proceed from

$$ \frac{201}{10} \ \cdot \ e^{1/7} \ > \ e^{3 \ + \ \frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \pi^{3 \ - \ \frac{7}{25}} \ > \ 31 \ \cdot \ \pi^{-7/25} \ > \ 31 \ \cdot \ \left( \frac{\pi}{e} \right)^{-7/25} \ \cdot \ e^{-7/25} \ \ . $$

We would like the factor $ \ \left( \frac{\pi}{e} \right)^{-7/25} \ $ to be underestimated, so we will use the overestimate

$$ \ \left( \frac{\pi}{e} \right) \ < \ \frac{22/7}{19/7} \ = \ \frac{22}{19} \ \ . $$

As above, we will use Taylor polynomials for the powers of $ \ e \ $ and the generalized binomial series to produce

$$ \frac{201}{10} \ \cdot \ \left( 1 \ + \ \frac{1}{7} \ + \ \frac{1}{2 \cdot 7^2} \right) \ > \ e^{\pi} \ > \ \pi^{e} \ > \ 31 \ \cdot \ \left( 1 \ + \ \frac{3}{19} \right)^{-7/25} \ \cdot \ \left( 1 \ - \ \frac{7}{25} \ + \ \frac{7^2}{2 \cdot 25^2} \right) $$

$$ \Rightarrow \ \ \frac{201}{10} \ \cdot \ \left( 1 \ + \ \frac{1}{7} \ + \ \frac{1}{98} \right) \ > \ e^{\pi} \ > \ \pi^{e} \ > \ 31 \ \cdot \ \left( 1 \ - \ \frac{7}{25} \cdot \frac{3}{19} \right) \ \cdot \ \left( 1 \ - \ \frac{7}{25} \ + \ \frac{49}{1250} \right) $$

$$ \Rightarrow \ \ \frac{201}{10} \ \cdot \ \frac{113}{98} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ 31 \ \cdot \ \frac{454}{475} \ \cdot \ \frac{949}{1250} \ \ , $$

these ratios being obtainable by reasonably simple hand-calculation. We can make the work we face in the right-hand term of this inequality a bit less onerous by using

$$ 31 \ \cdot \ \frac{454}{475} \ \cdot \ \frac{949}{1250} \ > \ 31 \ \cdot \ \frac{450}{475} \ \cdot \ \frac{945}{1250} \ = \ 31 \ \cdot \ \frac{18}{19} \ \cdot \ \frac{189}{250} \ \ . $$

Re-arranging the inequality gives us

$$ \frac{201}{10} \ \cdot \ \frac{113}{98} \ - \ 31 \ \cdot \ \frac{18}{19} \ \cdot \ \frac{189}{250} \ = \ \frac{201 \cdot 113 \cdot 19 \cdot 25 \ - \ 31 \cdot 18 \cdot 189 \cdot 98}{980 \cdot 19 \cdot 25} \ > \ e^{\pi} \ - \ \pi^{e} \ \ . $$

While this looks a bit daunting, it is considerably easier to compute by hand than the earlier estimate: a little work with binomials simplifies some of the multiplicaton and the rest is not too bad to work out "long-hand". Hence,

$$ \frac{10,788,675 \ - \ 10,335,276}{465,500} \ = \ \frac{453,399}{465,500} \ > \ e^{\pi} \ - \ \pi^{e} \ \ . $$

So we again establish the desired inequality, this time with more handiwork. We can make use of the calculator (or divide long-hand if we really want to remain "pure") to find that this ratio is about 0.9740 .

I will admit to a bit of guidance from a calculator to make adequate selections for the rational approximations in the interests of saving some time. It would take rather longer than to refine the numbers used in this investigation if one had to work entirely without "machine assistance".

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    $\begingroup$ But $\approx$ is not rigorous enough, I think. $\endgroup$ – Vincenzo Oliva Aug 29 '15 at 17:28
  • $\begingroup$ I found myself drawn back to this problem more than once in the past months and hope I have now answered certain earlier objections. I offer this answer largely to give some sense of the trouble one must go to in order to obtain an estimate of the difference in question that is less than $ \ 1 \ $ . (It is considerably less work to show that it is less than $ \ 1.5 \ $ .) It also makes one thankful for the availability of high-precision calculating devices in the current era, as they save a great deal of time in looking for ways to refine the estimates used. $\endgroup$ – colormegone Jun 2 '16 at 6:30
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Let $x>0$. Consider $e^{e+x}-(e+x)^e=e^e(e^x-(1+\frac xe)^e)=e^e(A-B)$. Now $z\ge \log(1+z)\ge z-\frac{z^2}2$, so $0\le A-B=A(1-\frac BA)\le A\log(A/B)\le A\frac{x^2}{2e}$. Hence, for $x=\pi-e$, we get $$ 0\le e^{\pi}-\pi^e\le e^{\pi-1}(\pi-e)^2/2\,. $$ Now we use the trivial inequality $e\ge 2+\frac 12+\frac 16+\frac 1{24}=2+\frac{17}{24}>2.7$ ($170>168$ if you want to check it mentally) and the much less trivial inequality $\pi<3.15$, for which I have no paper-free proof, to get $(\pi-e)^2\le 0.45^2=0.2025<\frac 29$ (the two-digit numbers ending in $5$ can be squared without paper). So, all we need to show now is that $\log 9=2\log 3\ge 2.15$ or $-\log \frac 13>1.075$. However, writing the first 6 terms of the Taylor series, we see that this logarithm is at least $$ \frac 23+\frac 29+\frac 8{81}+\frac 4{81}+\frac{32}{5\cdot 3\cdot 81} +\frac{64}{6\cdot 9\cdot 81} \ge 1+\frac 3{81}+\frac 2{81}+\frac{2}{15\cdot 81}+\frac 1{81} \\ >1+\frac 6{81}+\frac 1{8\cdot 81}>1+\frac 6{81}+\frac 6{80\cdot 81}=1+\frac 6{80}=1.075 $$ So, if we define $\pi$ to be a number between $e$ and $3.15$, the whole computation (assuming decent memory) requires no pen and paper, not to mention calculators. However, $\pi$ is not defined that way, so the inequality $\pi<3.15$ is a gaping flaw in this argument :-(.

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    $\begingroup$ $\frac{22}{7}-\pi = \int_{0}^{1} \frac{(x-x^{2})^{4}}{1+x^{2}}dx >0$ $\endgroup$ – Paata Ivanishvili Oct 25 '15 at 5:05
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Take the roots against $\pi e$, to get $e^{1/e}$ v $\pi^{1/\pi}$. These are instances if $x^{1/x}$ for $x=e$ and $x=\pi$.

For integers, we see that 9>8 and 2^4=4^2=16 imply that $\sqrt[3]{3} > \sqrt[4]{4} = \sqrt{2}$, from the sixth and eighth roots respectively.

Put $y=\sqrt[x]{x}$, whence $x = y^x$. We note that for values greater than 1, there are two solutions, the second is $z=y^z$. We now suppose z>x, and consider the case w>z. Where w>z, then $w<y^w$, which means that the value of y for w is less than that for z.

In the region where 1 < w < x we note that $y^x > y^w$, still, and that $\sqrt[w]{w} < \sqrt[x]{x} < L$

If now we note that $\pi > 3 > e$, and if we find a corresponding value for 3, then the range from 3 to 3' will contain either pi or e.

We now seek a value r for which $r^3 = 3^r$. Put r=2.5, gives r^3 = 125/8 = 15.625, and $9\sqrt{3}=1.732*9$, at 15.588. This tells us that r<2.5, and we get $r=3'< 2.5 < e < 3 < \pi$.

Since this is a convergence of $x$ and $z$ converge on some limit $L$, we can use brackets of the form of ( and ) to show this convergance, viz

$ 3' (2.5 ( e )) 3 ) \pi$, and thus from $3' ( e ) 3 ) \pi$, $e^\pi > \pi ^e$.

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