# e^pi or pi^e: Which One is Greater?

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• Published on Aug 8, 2018
• We use an elegant proof to compare e^pi and pi^e without a calculator and generalize the result to compare e^k and k^e for any nonnegative k.
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## Comments • 74

• tktsom Month ago

oh nice
I used f(x)=ln(e^x/x^e)=x-elnx
df/dx>0 in x>e and f(e)=0
so f(π)>0 and then e^π/π^e>1

• Anupam Pandey 2 months ago

Superb

• cookiebuster 3 months ago

when reality just breaks...3:26

• Chaw Den 3 months ago

Compare the order of growth is the fastest way
O(x^n) >> O(n^x)
pi=3.14, e=2.78

• Sagnik Banerjee 4 months ago

I have generalised this result for x^y>y^x where (x,y>0)

• Zυвi 4 months ago

well pi = e = 2 so inequality isn't actually strict

• Tristan Cole 6 months ago

Is there some way to find a number x where x^k=k^x for all values k?

• Serengeti Ghasa 6 months ago

So simple but awesome method without using calculus. I loved this way for a classic question consist in almost all calculus book (maxima and minima ) .
With love and regards

• spicca 6 months ago

i tried to solve as easier way by ln in both sides and compare that so
if e^pi>pi^e
pi>eln(pi)
pi/e>ln(pi)
but i stucked here plz anyone to help (if i can found a good number between them it will be solved..)

• Angel Mendez-Rivera 7 months ago

I think that the problem with this solution is that you asserted that e^x > x + 1 for nonzero x without proof, and the proof requires calculus, but if you are going to involve calculus in the proof, then it simply is easier and quicker to find the critical points of the function e ln(x) - x.

• Shashank S I 7 months ago

Awesome solution 😀

• Shinigami Steve 7 months ago

Awesome proof

• Mark Levin 7 months ago

There is a constant c so whenever a>b>c b to the power of a is always greater than a to the power of b. That constant c is less than e. So this is a particular case.

• Joe Potillor 7 months ago

Change both to base e, so (pi)^e = e^(e ln (pi)) Vs e^(pi). From there comparison tells us which one is bigger.

• Lily Lopez 7 months ago

okay but i love your accent :(

• Ding Hui Ng 8 months ago

I perfer this proof

• Radio TV 8 months ago +16

Engineers be like
Yeah guys so that about wraps up this problem so just let π = e = 3 and that should be good enough

• James Wilson 9 months ago

Very clever trick!

• sakib shahriar 10 months ago

another solve plz check this
usclip.net/video/PBVbyfqLuvk/video.html

• Rachid Iksi 10 months ago

By studiyng logx/x we get easily the result this function decrease over e

• Gregory Fenn Year ago +1

I don't know how you knew to use the inequality e^x > 1+x, but I did it a different way using only the following assumptions:
(1) if f(x) = x - e*log(x), then f'(x) = 1 - e/x
(2) pi > 0
(3) log(e) = 1
(4) log(a*b) = log(a) + log(b)
(5) log(0+) = -infinity < 0
Proof that e^pi > pi^e:
let X = e^pi, Y = pi^e, Z = X/Y, L = log(Z) = pi - e*log(e). Clearly, X>Y iff Z > 1 iff L > 0,
define f(x) = x - e*log(x) for any x>0. Notice that L = f(pi). We claim that f(x) >0 for all x, except f(e) = 0. Thus, f(pi) > 0, so L > 0.
(A) f(0+) infinity - -infinity > 0.
(B) f(1) = 1
(C) f(e) = 0
(D) f(infinity) = infinity > 0
f'(x) = 1 - e/x
(E) So f'(x) = 0 IFF x = e.
At this point, I suggest you draw graph f(x), using the observations (A-E).
So the only turning point for f(x) is at x=e. Hence this is the unique minimum, thus f(x) > 0 for all x != e.

(You can be even more rigorous and show f''(x) > 0, so that all turning points are local minima, which verifies that f(e) IS a minima, although that should be obvious by now).

• Michael Empeigne Year ago

(e^pi)^(1 / e*pi ) or (pi^e)^(1 / e*pi ).......... e^(1 / e ) or pi^( 1 / pi ) both of these are of the form x^( 1 / x )......Now let's do some differentiation of x^( 1 / x ) Let y = x^( 1 / x )........ ln y = ( 1 / x ) ln x.......( 1 / y )*dy / dx = ( 1 / x ) * ( 1 / x ) + ( - 1 / x^2 ) ln x dy / dx = y * [ 1 - ln x ] ( 1 / x^2 )...... let dy / dx = 0.......... 0 = x^( 1 / x ) * ( 1 - ln x ) ( 1 / x^2 ) The only part of this derivative that can equal 0 is ( 1 - ln x ). Therefore, 1 - ln x = 0 That gives x = e...... Now to check if this is a maximum or minimum using the first derivative test......test value of x = 1 produces a positive........test value of x = 3 produces a negative........This means that x = e is a maximum. As a result, e^(1 / e ) is greater than pi^(1 / pi ) which in turn means that e^(pi) is greater than pi^(e ).

• MrOligi3003 Year ago

But floor(e^pi) = ceiling(pi^e), so the values are pretty close

• LiZichong Year ago +1

Nice video. I have a story on this topic too. Back in 3rd grade, a few times before asleep I was thinking about "With fixed sum, how to get highest product". I immediately realized that compared to other numbers, 2 is a good choice and 3 is even better. Then I try to generalize this problem into first decimal and eventually concluded 2.7 was the best. Only later in middle school did I realize this process would converge to e.

• jose_bv 9 months ago

r/iamverysmart

• James Liu Year ago

李子翀 the Chinese imo team? You’re insane wow

• LiZichong Year ago

I'm probably an outlier. I was trained well in math very early. I was even fortunate enough to be part of Chinese math team before coming to the States.

• LetsSolveMathProblems  Year ago +1

I am amazed that you were creative and knowledgeable enough to generalize such a problem in 3rd grade! I personally was not even aware of the number e until seventh or eighth grade, if I remember correctly.

• About Math Year ago

Ill just suggest my solution.. Lets suppose that e^π>π^e then πln (e)>eln (π) and ln (e)/e>ln (π)/π..Consider f (x)=lnx/x x>=e you can see that f is decreasing on so for e f (e)>f (π) and my supposal is correct.

• Kyx Year ago +75

Engineer: they both = 9

• JS K 2 months ago

no engineer in the world regard pi or e as 3..

• August Liu 3 months ago

I laughed so hard when my engineer friend told me they use pi=3=10

• Sahil Baori 4 months ago

@Rohan LOL!

• Rohan 5 months ago

can't even do a shit joke correctly

• ShouTZ 7 months ago

i was about to type this lmaoooo

• Debajyoti Guha Year ago +2

You're really good. You deserve more subscribers.

• Tasty Rainbro Year ago

2.7^3.1 > 3.1^2.7
Aproximated by simply count for
3^4 > 4^3.
I was right.
If a>b for a^b ~ b^a
Then the expression with the bigger base wins.
Btw i know 20 decimals of pi.

• HPP Year ago

Nice problem with a nice solution! This problem got me pondering though, whether if x and y are irrational numbers, and x is not equal to y, would the equality x^y = y^x hold true for any irrational numbers x and y? I conjecture that x^y would never equal y^x, but I cannot come up with a proof yet.

• }{ Year ago

For any real 1

• MC Urist Year ago +4

It's trivial to prove x^y>y^x when e

• Math Life Year ago

y=lnx/x, y'e ; ln pi/pi < ln e/e ; e*lnpi < pi*lne ; ln(pi^e)< ln(e^pi); pi^e

• I thought that by "generalizing the result" you meant proving the relation for any arbitrary a^b and b^a, with a not necessarily being e. Very nice video though.

• Angel Mendez-Rivera 7 months ago +1

Void of Kathaaria Well, doing that is equivalent to this, since a = e^ln(a), and we can define c = b ln(a)

• Metalhammer1993 Year ago

yay i got something right for a change. well not really i didn´t come u pwith a proof but at least my hunch was correct. that counts as something right? my though process was pretty simplistic. e and pi are both around 3. one a birt more one a bit less so a change in the base probably wouldn´t make that much of a differenz. e^2 and pi^2 are probably not too far off from each other (no calculator at hand rn) e^3 and pi^3 prob a bit further. so i´d say round about the same base just to a bigger power will obviously be bigger. i know that argument is pretty shrewd to weigh the exponent heavier than the base, but it for once worked. almost. it gave me the right answer. without proof tough^^

• pandascarpo Year ago

...or you could just take the logs of both the terms...

• Mohammad Rehmaan Year ago

You can take log of both the numbers ln(e^π) is π equal 3.14 approx. If we take log of π^e it becomes ln(π^e) which is elnπ that is surely less than 3.14 hence e^π is greater

• MCMage Year ago

By the way, who said BPRP did it first, YOURE WRONG, i am Bprps fan but both showed us Really REALLY different answers. I like this one because it is less confusing and uses some kind of number theory and graphs

• Angel Mendez-Rivera 7 months ago

MCMage Number theory is defined as the study of the arithmetic of natural numbers, so I do not know what you are talking about. Also, BPRP objectively DID do it first, and the fact that his approach was different does not change this. Perhaps you need to learn some formal logic to understand how to properly make logical conclusions. And using calculus is objectively the most straightforward way to do this. This solution also used calculus, actually, just less rigorously.

• gillian2611 Year ago +5

Around the same idea, have you already made a video of "Which is greater : n^(n+1) or (n+1)^n ?"

• BlackPenRedPen madethe same video with the function x^1/x and the maximum values

• 王万网 Year ago

MORE difficult,can you compear to 3^e, e^3, pi^e, e^pi, pi^3, pi^3? (no calculator,using the
function f(x)=(lnx)/x pi=3.1415926... e=2.71828...）

• Kazi Abu Rousan Year ago

This can also be done using y=x^(1/x) function

• Smiley1000 Year ago +4

What about generalizing the e to some variable?

• Angel Mendez-Rivera 7 months ago

Smiley1000 Not necessary since if the base is a, then a = e^ln(a) for all a, and then you can work with c = b ln(a) instead of b.

• Otium Abscondita Year ago

usclip.net/video/OqWbnM54qMw/video.html look at this and tell me if you like it.

• khaled qaraman Year ago +2

great video

• HelloItsMe Year ago +4

Haha I can just guess this by feeling that powers have more power than bases so e^pi is greater 😂

• HelloItsMe Year ago +2

jkid1134 oh 😂

• jkid1134 Year ago +5

HelloItsMe BUT
2^3 = 8 < 3^2 = 9
Really makes you think 🤔

• Dr Din Year ago +2

I like proving it with x^(1/x)

• Adlet Irlanuly Year ago +5

I wanted to go to the restroom but I lasted to keep watching

• E Rock Year ago +17

I gave a solution just like this on Stack although I just mentioned that the inequality can be easily seen from the Taylor expansion for e^x. The problem is simple but very fun!

• xXappleXx Pie Year ago +3

LetsSolveMathProblems Even better, you can derive it from the fact (1+x/n)^n is an increasing sequence which converges to e^x

• E Rock Year ago +1

LetsSolveMathProblems it is a little more work for that one particular case, admittedly. It’s just a matter of realizing that if -1 < x < 0 then x^(2k) > |x^(2k+1)| and so x^(2k)/((2k)!) - x^(2k+1)/((2k+1)!) >0. This tells us that if we consider the Taylor expansion from the x^2 term and on then we can just consider the terms in pairs and realize that everything is positive. Perhaps it isn’t as quick to see as the con cavity argument though.

• LetsSolveMathProblems  Year ago +5

The only downside of Taylor expansion argument is that we have to do a little bit more work for -1

• Deepak Maheshwari Year ago +4

Its my class ilustration of iit exam of india but i have two methods of solving but its cool though

• Joe Year ago +2

AAAAAAAAAAAAAAAAAAAAAA COOL

• Krosis Year ago +16

I just watcher blackpenredpen's video about this problem 5 minutes ago...

• Mihai Ciorobitca Year ago

Krosis wow awesome !

• Jonathan Liu Year ago +4

Great method!

• end my misery Year ago +3

great video! :) really liked this method

• Matt GSM Year ago +51

BlackPenRedPen did a video about this! I love this solution

• space time 9 months ago +3

This solution is a lot better