# Non-calculator proof that $\pi^\pi -\pi \lt \frac{100}{3}$

I am looking for a few non-computational, non-calculator proof of the following inequality:

$$\pi^\pi -\pi \lt \frac{100}{3}$$

I can't really seem to come up with a proof because of that killer $\pi^\pi$ term.

• What would constitue a "non-computational" proof? Especially considering that this looks like a purely numeric problem. (And the bound is pretty sharp.)
– mrf
Commented Jan 18, 2016 at 8:51
• @mjl A non-computational proof is one that does not just approximate pi and calculate from that.
– user266519
Commented Jan 18, 2016 at 8:56
• @John: It will need a pretty clever argument, since the lefthand side is about $33.320567$. Commented Jan 18, 2016 at 9:02
• The eternal question... :-$)$ Commented Jan 18, 2016 at 9:50
• @JohnVine This is not for MathOverflow Commented Jan 25, 2016 at 18:36

The inequality $x^x-x<100/3$ is true precisely when $0<x<3.14175798627357...$, so intuitively a solution to this problem must in some fashion make use of the fact that $$\pi<3.14175798627357\ldots.$$ The upper bound $\pi<22/7$ is insufficient for this, but the upper bound $355/113$ suffices. (The bound of $355/113$ dates to the fifth century CE, which I personally would be prepared to admit as "non-calculator".) One could therefore proceed as follows: since $x \mapsto x^x-x$ is increasing for $x>1$ (an easy calculus exercise), it is sufficient to show $$\left(\frac{355}{113}\right)^{\frac{355}{113}}-\frac{355}{113}<\frac{100}{3}$$ which is to say $$\left(\frac{355}{113}\right)^{\frac{355}{113}}<\frac{12365}{339}$$ or $$3\times 355^{\frac{355}{113}}<12365\times 113^{\frac{242}{113}}$$ or $$3^{113}\times 355^{355}<12365^{113}\times 113^{242}.$$ It is not immediately obvious to me that this inequality cannot be obtained by hand within a broadly viable timeframe, but before attempting it I would probably start reading up on things like Karatsuba multiplication, and work out carefully beforehand how many pages I might need to fill with long multiplication calculations and endless repeated squaring. The gap between the two numbers is very small, one being about $1.7\times 10^{959}$ and the other around $1.8\times 10^{959}$. In any event, I hope that the sharpness of the inequality which needs to be proved illustrates the difficulty of the problem.
• Yes, you are right; I made an arithmetical error. But I notice you have swapped $113$ for $133$ up there. That's a more serious problem. Commented Jan 25, 2016 at 18:16
• The fact that both numbers have the same number of digits (ETA: when $113$ is properly substituted for $133$) indicates that the razor must be sharper than a few powers of $2$. Commented Jan 25, 2016 at 18:18
• For the benefit of voters: The current version of the answer (which references Karatsuba multiplication) requires $22$ multiplications (float-level precision should be enough) on the LHS and $21$ multiplications on the RHS. That's tedious but doable. Commented Jan 25, 2016 at 18:43