e^Pi vs Pi^e : which is larger? (pi day battle)

This is a short, animated visual proof demonstrating which of the values e^pi or pi^e is larger (where pi is the circle constant and e is Euler's constant) .

For two visual proofs of a more general version of this inequality, see
and

#math​ #inequality #manim​ #animation​ #theorem​ #pww​ #proofwithoutwords​ #visualproof​ #proof​ #iteachmath #calculus #mathshorts​ #mathvideo​ #mtbos​ #euler #pi #piday

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[liambohl]

I understood all the steps as you went through them, but I would never think to put those steps together to answer this question.

[lilylovecraft]

Without a calculator...'First plot...' *grabs calculator*

[lavieag]

e^x >= x^e in every case while the only way it is equal is x = e

[yusufdenli]

This is one of the most beautiful proofs I've ever seen.

[monkeyhook]

You know I actually hadn’t wondered that

[awesomelink]

Fun fact: For any positive real number x, e^x > x^e. Except if x = e, then they're equal. Also, the only time the difference decreases is between 1 and e.

[timwildauer]

How to draw an owl:
1. Draw a circle for the head
2. Draw the rest of the owl

[三点一四]

Another way :
Take the function f(x) = x^(1/x). Its derivative is x^(1/x) ((1-ln(x))/x²) so the function has a maximum on x = e
Therefore, e^(1/e) ＞ π^(1/π)
Raising both sides to the πe power we get e^π ＞ π^e

[seppeswart]

Another way would be to take the Ln of both sides, take the exponent out and use that if x<y then ln(x) < ln(y).

[ericpmoss]

Very elegant, BUT... what is the intuition that leads one to examine y = 1/x in the first place? And then why the area under it, and I suppose, why from e to pi? THAT is the spark many of us are missing.

[quintopia]

Seems overcomplicated. Just start by proving x^y>y^x for all e<=x<y and then this falls out as a special case. The former follows from the fact that ln(x)/x is monotonic decreasing on x>e, which is obvious from its derivative. No need for integrals or looking at areas.

[ozargaman]

So for every number n>e, eⁿ>n^e

[divyansh]

Raise both to the power of 1/eπ it becomes a function of x^1/x which has its max value at e (can be found by finding the derivative of the function) so yes e^π is greater than π^e.

Looks like someone else in the comments did this too.

[smitshah]

One other way to do it is to find derivative of x^(1/x) which will result the function been decreasing for x>e so, π^(1/π)<e^(1/e) and after interchanging powers, π^(e) <e^π...

[naitikjain]

Do not need to go so complex
Just draw graph of y=x to power 1/x, it is also called gaussian curve, its peak reaches as x=e, so since pi>e so e to power 1/e >pi to power 1/pi then after raising power e×pi both sides, we will get e to power pi> pi to power e.🎉

[TheOneMaddin]

That's cool and all, but you could also use your calculus skills to show that x^y < y^x whenever e <= y < x. So there is absolutely no reason to use pi here.

[PC_Simo]

I just threw a wild guess that; since 2 < e < π (that is, π is greater, than e, and *_BOTH_* are greater, than 2); then, e^π should be larger, than π^e 😅.

[rizwannazir]

I am so relieved i understood every bit of it. Thanks a billion to my calculus professor. Respect

[yurenchu]

Nice proof; but it still requires knowing that e < π .

By the way, for _any_ positive real number a ≠ e,
e^a > a^e

[dragoscalin]

Foarte elegantă problema și rezolvarea ei. Felicitări.