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I've been struggling for a while with the following problem:

Which is bigger: $(\pi+1)^{\pi+1}$ or $\pi^{\pi+2}$?

Needless to say software aid is not allowed. All manual calculations should be possible to be performed in reasonable time (for example, I did read some solutions to similar problems that to my understanding include calculating 113th power of a decimal number, and this does not count as "reasonable" manual calculation).


Probably useless for the solution, but nevertheless interesting graphs of $(x+1)^{x+1}-x^{x+2}$:

enter image description here

enter image description here

enter image description here


Using computer:

$(\pi+1)^{\pi+1} = 359.796$

$\pi^{\pi+2} = 359.867$

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  • $\begingroup$ Calculation verifies that the first is smaller by about $0.01\%$. $\endgroup$ – Omnomnomnom May 11 '15 at 23:59
  • $\begingroup$ Correct, and such small difference makes it hard to prove with rational approximations of $\pi$, @Omnomnomnom $\endgroup$ – VividD May 12 '15 at 0:02
  • $\begingroup$ since the inequality is very sharp, I think you might need a very close approx for $\pi$ $\endgroup$ – Yimin May 12 '15 at 0:05
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    $\begingroup$ I really hope needles aren't actually needed to solve this. $\endgroup$ – alex.jordan May 12 '15 at 0:14
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    $\begingroup$ this Q is exactly the same as the one you reference: $\left(1+\frac{1}{\pi}\right)^{\pi+1}<\pi \to \left(\frac{\pi+1}{\pi}\right)^{\pi+1}<\pi \to \frac{(\pi+1)^{\pi+1}}{(\pi)^{\pi+2}}<1$ $\endgroup$ – JonMark Perry May 20 '15 at 19:17
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Just to be clear on what it would take to decide this based on rational approximation: we want to compare the ratio $$ f(x)=\frac{x^x}{(x-1)^{x+1}} $$ to unity at $x=(\pi + 1)$. Its logarithm is $$ \begin{eqnarray} g(x)=x\log x - (x+1)\log(x-1) &=& x\log x - (x+1)\log x - (x+1)\log(1-1/x) \\ &=& -\log x - (x+1)\log(1-1/x) \\ &=& -\log x + (x+1)\sum_{k=1}^{\infty}\frac{1}{k}x^{-k} \\ &=& -\log x + 1 +\sum_{k=1}^{\infty}\left(\frac{1}{k}+\frac{1}{k+1}\right)x^{-k}, \end{eqnarray} $$ valid for $x>1$, the derivative of which is $$ g'(x)=-\frac{1}{x}-\sum_{k=2}^{\infty}\left(1+\frac{k-1}{k}\right)x^{-k}, $$ which is clearly negative. So $f(x)$ is decreasing, and is less than one at $x=(\pi + 1)$ if it less than one for some rational $(p/q) < \pi+1$. A good rational underestimate for $\pi$ is given by $\frac{333}{106}\approx 3.14151 < \pi$. Indeed, one finds numerically that $f(333/106 + 1) =f(439/106) < 1$ by a few percent: $$ f\left(\frac{439}{106}\right)^{106}=\frac{(439/106)^{439}}{(333/106)^{545}}=\frac{439^{439} 106^{106}}{333^{545}} \approx 0.98. $$ Despite being integers, though, the numbers involved are too large for manual calculation. We can do better by looking at the second derivative of $f(x)$: it is $f''(x) = (g'(x)^2 + g''(x))f(x)$, which is clearly positive, since $g''(x)$ is. Therefore $f(x)$ always lies below any line segment that connects two points on it. So for any $x < y < (\pi+1) < z$, we have $$ f(\pi + 1) < f(y) < \lambda f(x) + (1-\lambda) f(z), $$ where $\lambda = (z-y)/(z-x)$. This estimate is good enough for $(x,y,z)=\left(4,\frac{439}{106},\frac{29}{7}\right)$, for which $\lambda = 1/106$, $f(x)=256/243$, and $$f(z)=7\sqrt[7]{\frac{29^{29}}{22^{36}}}=\frac{7\cdot 29^{4}}{22^{5}}\sqrt[7]{\frac{29}{22}}.$$ Putting it together, we need to show that $$ \frac{128}{12879} + \frac{105\cdot 7 \cdot 29^{4}}{106\cdot 22^{5}}\sqrt[7]{\frac{29}{22}} < 1, $$ or $$ \sqrt[7]{1+\frac{7}{22}} < \frac{12751\cdot 106 \cdot 22^5}{12879 \cdot 105 \cdot 7 \cdot 29^4}. $$ Now, the root is less than its fifth-order Taylor series approximation: $$ \sqrt[7]{1+7x} < 1 + x - 3x^2 + 13x^3 - 65x^4 + 351x^5 = \frac{5361157}{5153632}, $$ evaluated at $x=1/22$. So in the end it will suffice to show that $$ {5361157 \cdot 12879 \cdot 105 \cdot 7 \cdot 29^4} < {5153632 \cdot 12751\cdot 106 \cdot 22^5}. $$ While these are certainly big numbers (about $20$ digits), they're much smaller than the thousand-digit monsters involved in the approach relying only on the first derivative, and I think you could argue that even if you didn't want to, you could carry out this calculation by hand to prove that $(\pi+1)^{\pi+1} < \pi^{\pi+2}$.

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  • $\begingroup$ Thanks a lot, I would say this is the cleanest and the shortest solution, but the last step bothers me, I don't see the way to calculate it manually in reasonable time. :( $\endgroup$ – VividD May 21 '15 at 21:09
  • $\begingroup$ OK, you surprised me with the new addition to the answer, let me take a look... $\endgroup$ – VividD May 21 '15 at 21:56
  • $\begingroup$ I agree that establishing ${5361157 \cdot 12879 \cdot 105 \cdot 7 \cdot 29^4} < {5153632 \cdot 12751\cdot 106 \cdot 22^5}$ manually is reasonable. $\endgroup$ – VividD May 21 '15 at 22:03
  • $\begingroup$ Mind-boggling! Thanks again! $\endgroup$ – VividD May 21 '15 at 22:07
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In my attempts to solve the problem of the OP, I found it useful to choose a slightly broader perspective. I started by defining a function $f(x)$ in which an adjustable parameter $a$ appears:

$$f(x) = (x + a)log(x+a) - (x+2a)log(x)$$

In terms of parameter $a$, we seek the value of $x$ for which $f(x)$ equals zero. Later on we will focus on the case $a=1$.

There exist rational solutions for $x$ and $a$ whenever $x = n a$, where $n$ is a natural number. Here are the first four of these solutions: $ \{n = 1, x=4, a=4 \}$; $ \{n = 2, x= 27/8, a = 27/16 \}$; $ \{n=3, x = 256/81, a=256/243 \}$; $ \{n=4, x=3125/1024, a = 3125/4096 \}$. Note that the solution for $n=3$ is close to the values occurring in the OP's problem: $x= \pi$, $a=1$.

We now seek the general solution for the case that $a << x$. Expanding the $ log(x+a)$ term in $f(x)$, dividing by $a$ and rearranging terms leads to:

$$log(x) = 1 + \frac {1}{2}(a/x) - \frac {1}{6}(a/x)^2 + \frac {1}{12}(a/x)^3 - \frac{1}{20}(a/x)^4 + \frac {1}{30}(a/x)^5 - \cdots$$

To get rid of the remaining logarithmic term, we substitute both sides of the equation in the exponential function. Next we divide by $e$. Then we take the reciprocal of both sides. This yields:

$$z = exp \{- \frac {1}{2}pz + \frac {1}{6}(pz)^2 - \frac {1}{12}(pz)^3 + \frac {1}{20}(pz)^4 - \frac {1}{30}(pz)^5 + \cdots \}$$

Where we have introduced the new variable $z = e/x$ and the new parameter $p = a/e$. Expanding the RHS in a Taylor series yields:

$$z = 1 - \frac {1}{2}(pz) + \frac {7}{24} (pz)^2 - \frac {3}{16} (pz)^3 + \frac {743}{5760}(pz)^4 - \frac {215}{2304}(pz)^5 + \cdots$$

From this result we obtain the power series for $z$ in terms of $p$:

$$z = 1 - \frac {1}{2}p + \frac {13}{24}p^2 - \frac {3}{4}p^3 + \frac {6763}{5760}p^4 - \frac {285}{144}p^5 + \cdots$$

Converting this result for $z$ into a series for $x$ gives:

$$x/e = 1 + \frac {1}{2}(a/e) - \frac {7}{24}(a/e)^2 + \frac {1}{3}(a/e)^3 - \frac {911}{1920}(a/e)^4 + \frac {34}{45}(a/e)^5 - \frac {748045}{580608}(a/e)^6 + \cdots$$

For small values of $(a/e)$ the series converges well, since consecutive terms alternate in sign while the pre-factors are roughly similar in magnitude. Unfortunately for the case $a=1$ we do not get the high accuracy that is required to answer the OP's question with certainty.

I chose a simple method to improve the convergence. As pointed out above, there is an rational solution to $x(a)$ quite near $a=1$, namely $x =256/81, a = 256/243$. We can simply add an extra term of order $(n+1)$ to an approximation of order $n$, in order to force the solution through this exact value. If I do this, the results are: $x0 = 3.1380$, $x1 = 3.1421$, $x2 = 3.1405$, $x3 = 3.1413$, $x4 = 3.1409$, $x5 = 3.1411$, $x6 = 3.1410$. From this set of values we can conclude that the zero of $f(x)$ occurs very near the value $x = 3.1410$. The uncertainty is smaller than $0.0001$.

A different approach is to express the series in a Padé form. This is an elegant and effective method to improve convergence of a series. The $4$-parameter Padé representation for $x$ is:

$$x/e = \frac {1 + (A+B+C+D)(a/e) + (AC+AD+BD)(a/e)^2}{1 + (B+C+D)(a/e) + BD(a/e)^2} $$

The parameter values are found to be: $A = 1/2$, $B = 7/12$, $C = 47/84$ , $D = 11313/19740$. To enhance the formula further, we adjust the least significant parameter (which is $D$) so that the formula satisfies the exact solution $x = 256/81$ when $a = 256/243$. The new value for $D$ is $0.48387$, which is somewhat smaller than the original value $0.57310$. The formula works very well for all $a$ in the range $0$ to $1.05$. For $a=1$ we obtain the result $x = 3.14105$ with an accuracy of $0.00001$.

In conclusion: for $a=1$ the solution to $f(x) = 0$ occurs for a value of $x$ which is smaller than $\pi = 3.14159$. Hence the function $f(x)$ is negative for $a=1$ and $x= \pi$.

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    $\begingroup$ I liked this answer very much. A lot of ideas, a lot of work, a lot of results. I am going to reward it with 100 points (current bounty). This is going to happen in 5 or 7 days, in accordance with site rules. $\endgroup$ – VividD May 15 '15 at 22:34
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Unfortunately, needles are needed. This work is for $(\pi + 1)/\pi$, not $(\pi+2)/(\pi + 1)$, if you are going to make it more precise, third order approximation is needed.

I think needles are not necessary. $a = \pi$, We try to prove

$\ln(a + 1)/\ln(a) < (\frac{a+1}{a})$.

well, $a \sim 355/113\sim 3$, the error on estimate of $\pi$ is very tiny, will not harm our result, we do $a = 3 + b$, thus $b = 16/113\sim 1/7 \sim 0.14$

we just show it for

$\frac{\ln(4) + \ln(1+ b/4)}{\ln(3) + \ln(1 + b/3)}$

we use Taylor expansion on $\ln(1 + b/4)$ and $\ln(1 + b/3)$ to the first order term.

The following is estimate of error from truncation.

$$\vert\frac{u}{v} - \frac{u + s}{v + t}\vert \le \frac{|ut| + |vs|}{v(v+t)}$$

since $\frac{u}{v}$ is around $4/3 < 2$, $v$ is bounded below by $1$, we know that the error will be bounded by

$2(|t| + |s|) \le 2(b/3)^2 \le 1/100$.

$\ln(1 + b/4) \sim b/4 \sim 0.0354$

$\ln(1 + b/3) \sim b/3 \sim 0.0472$

$\ln(4) = 2\ln(2) \sim 1.386$

$\ln(3) \sim 1.099$

result is $\frac{1.386 + 0.0354}{1.099 + 0.0472} = 1.2400 < (\pi + 1)/(\pi) = 1.32$

By the way,

$\log(\pi + 1)/(\log(\pi)) \sim 1.2414$

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    $\begingroup$ The form of the question means that you should replace $\frac{\pi+1}{\pi}$ with $\frac{\pi+2}{\pi+1}$. Otherwise you are proving $\pi^{\pi+1} > (\pi+1)^{\pi}$ which is much less sharp. $\endgroup$ – Rolf Hoyer May 12 '15 at 1:02
  • $\begingroup$ @RolfHoyer you are right, I just notice that, then using first order approx is far from enough, in this case, 3rd order is necessary. $\endgroup$ – Yimin May 12 '15 at 1:51
  • $\begingroup$ What does "needles" refer to, in this instance? $\endgroup$ – Charles May 20 '15 at 15:15
  • $\begingroup$ @Charles see the comments which follows the question. $\endgroup$ – Yimin May 20 '15 at 15:31
  • $\begingroup$ @Charles In original version of the question, I wrote "needles" instead of "needless", and this became a sort of internal joke. $\endgroup$ – VividD May 22 '15 at 5:54
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Consider the function f(x) =($\frac{x+1}{x}$)^(x+1); the sign of the derivative on [3, 4] depends of the expression xln($\frac{x+1}{x})$ - 1 and both f ‘(3) and f ‘(4) are negative and not null on the interval. Hence f is decreasing over [3, 4]. We have f(3) = 3.160493827 > $\pi$ and f(4) = 3.051757813 < $\pi$ . We note that f(3) is nearer of $\pi$ than f(4); calculate f(3.14) = ($\frac{414}{314})^{4.14}$ = 3.141178 < $\pi$. Therefore f($\pi$) < $\pi$. Consequently the second expression is bigger.

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  • $\begingroup$ ($\frac{414}{314})^{4.14}$ - how to calculate manually? $\endgroup$ – VividD May 12 '15 at 8:57
  • $\begingroup$ I thought he was allowed a single calculation at most; however an approach to $10^{-4}$ is enough to have the inequality less than $\pi$ $\endgroup$ – Piquito May 12 '15 at 12:58
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Let $x=(\pi+1)^{\pi+1}$ and $y=\pi^{\pi+2}$

Since $\ln x=(\pi+1)ln(\pi+1)$ and $\ln y=(\pi+2)\ln\pi$ and $\ln x$ is increasing , compare $\ln x$ and $\ln y$

So I depend on a internet calculating, $\ln x-\ln y<0$ and so $x<y$

But $\ln x-\ln y=-0.00019...$ tell us that it is difficult to clear for manually.

(I tried to prove for " $f(x)=(x+1)\ln(x+1)-(x+2)\ln x$ is decreasing and $f(\pi)<0$"

But $f(3.14)>0$ ,according to calculating)

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