# 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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cookiebuster13 days agowhen reality just breaks...3:26

Chaw Den27 days agoCompare the order of growth is the fastest way

O(x^n) >> O(n^x)

pi=3.14, e=2.78

Sagnik Banerjee2 months agoI have generalised this result for x^y>y^x where (x,y>0)

Zυвi2 months agowell pi = e = 2 so inequality isn't actually strict

Tristan Cole3 months agoIs there some way to find a number x where x^k=k^x for all values k?

Serengeti Ghasa4 months agoSo 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

spicca4 months agoi 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-Rivera4 months agoI 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 I4 months agoAwesome solution 😀

Shinigami Steve4 months agoAwesome proof

Mark Levin4 months agoThere 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 Potillor4 months agoChange both to base e, so (pi)^e = e^(e ln (pi)) Vs e^(pi). From there comparison tells us which one is bigger.

Lily Lopez5 months agookay but i love your accent :(

Ding Hui Ng5 months agoI perfer this proof

Radio TV5 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 Wilson6 months agoVery clever trick!

sakib shahriar7 months agoanother solve plz check this

usclip.net/video/PBVbyfqLuvk/video.html

Rachid Iksi8 months agoBy studiyng logx/x we get easily the result this function decrease over e

Gregory Fenn9 months 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 Empeigne9 months 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 ).

MrOligi300310 months agoBut floor(e^pi) = ceiling(pi^e), so the values are pretty close

LiZichong10 months 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_bv6 months agor/iamverysmart

James Liu10 months ago李子翀 the Chinese imo team? You’re insane wow

LiZichong10 months agoI'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.

LetsSolveMathProblems10 months 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 Math10 months agoIll 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.

Kyx10 months ago^{+67}Engineer: they both = 9

August LiuMonth agoI laughed so hard when my engineer friend told me they use pi=3=10

Sahil Baori2 months ago@Rohan LOL!

Rohan2 months agocan't even do a shit joke correctly

ShouTZ4 months agoi was about to type this lmaoooo

Evan Murphy5 months ago^{+9}27?

Debajyoti Guha10 months ago^{+2}You're really good. You deserve more subscribers.

Tasty Rainbro10 months ago2.7^3.1 > 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.

HPP10 months agoNice 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.

}{10 months agoFor any real 1

Stephen H10 months ago^{+4}It's trivial to prove x^y>y^x when e

Math Life10 months agoy=lnx/x, y'e ; ln pi/pi < ln e/e ; e*lnpi < pi*lne ; ln(pi^e)< ln(e^pi); pi^e

100 000 inscritos sem nenhum vídeo10 months ago^{+32}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-Rivera4 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)

Metalhammer199310 months agoyay 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^^

pandascarpo10 months ago...or you could just take the logs of both the terms...

Mohammad Rehmaan10 months agoYou 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

MCMage10 months agoBy 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-Rivera4 months agoMCMage 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.

gillian261110 months ago^{+5}Around the same idea, have you already made a video of "Which is greater : n^(n+1) or (n+1)^n ?"

Csanád Temesvári10 months agoBlackPenRedPen madethe same video with the function x^1/x and the maximum values

王万网10 months agoMORE 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 Rousan10 months agoThis can also be done using y=x^(1/x) function

Smiley100010 months ago^{+4}What about generalizing the e to some variable?

Angel Mendez-Rivera4 months agoSmiley1000 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.

Mathedidasko10 months agousclip.net/video/OqWbnM54qMw/video.html look at this and tell me if you like it.

khaled qaraman10 months ago^{+2}great video

HelloItsMe10 months ago^{+4}Haha I can just guess this by feeling that powers have more power than bases so e^pi is greater 😂

HelloItsMe10 months ago^{+2}jkid1134 oh 😂

jkid113410 months ago^{+5}HelloItsMe BUT

2^3 = 8 < 3^2 = 9

Really makes you think 🤔

Dr Din10 months ago^{+2}I like proving it with x^(1/x)

Adlet Irlanuly10 months ago^{+5}I wanted to go to the restroom but I lasted to keep watching

E Rock10 months 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 Pie10 months 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 Rock10 months agoLetsSolveMathProblems 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.

LetsSolveMathProblems10 months ago^{+5}The only downside of Taylor expansion argument is that we have to do a little bit more work for -1

Deepak Maheshwari10 months ago^{+4}Its my class ilustration of iit exam of india but i have two methods of solving but its cool though

Joe10 months ago^{+2}AAAAAAAAAAAAAAAAAAAAAA COOL

Krosis10 months ago^{+16}I just watcher blackpenredpen's video about this problem 5 minutes ago...

Mihai Ciorobitca10 months agoKrosis wow awesome !

Jonathan Liu10 months ago^{+4}Great method!

end my misery10 months ago^{+3}great video! :) really liked this method

Matt GSM10 months ago^{+51}BlackPenRedPen did a video about this! I love this solution

space time6 months ago^{+3}This solution is a lot better