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

Embed

- 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.

Your support is truly a huge encouragement.

Please take a second to subscribe in order to send us your valuable support and receive notifications for new videos!

Every subscriber and every like are wholeheartedly appreciated.

tktsomMonth agooh 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 Pandey2 months agoSuperb

cookiebuster3 months agowhen reality just breaks...3:26

Chaw Den3 months agoCompare the order of growth is the fastest way

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

pi=3.14, e=2.78

Sagnik Banerjee4 months agoI have generalised this result for x^y>y^x where (x,y>0)

Zυвi4 months agowell pi = e = 2 so inequality isn't actually strict

Tristan Cole6 months agoIs there some way to find a number x where x^k=k^x for all values k?

Serengeti Ghasa6 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

spicca6 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-Rivera7 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 I7 months agoAwesome solution 😀

Shinigami Steve7 months agoAwesome proof

Mark Levin7 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 Potillor7 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 Lopez7 months agookay but i love your accent :(

Ding Hui Ng8 months agoI perfer this proof

Radio TV8 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 Wilson9 months agoVery clever trick!

sakib shahriar10 months agoanother solve plz check this

usclip.net/video/PBVbyfqLuvk/video.html

Rachid Iksi10 months agoBy studiyng logx/x we get easily the result this function decrease over e

Gregory FennYear 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 EmpeigneYear 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 ).

MrOligi3003Year agoBut floor(e^pi) = ceiling(pi^e), so the values are pretty close

LiZichongYear 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_bv9 months agor/iamverysmart

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

LiZichongYear 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.

LetsSolveMathProblemsYear 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 MathYear 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.

KyxYear ago^{+75}Engineer: they both = 9

JS K2 months agono engineer in the world regard pi or e as 3..

August Liu3 months agoI laughed so hard when my engineer friend told me they use pi=3=10

Sahil Baori4 months ago@Rohan LOL!

Rohan5 months agocan't even do a shit joke correctly

ShouTZ7 months agoi was about to type this lmaoooo

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

Tasty RainbroYear 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.

HPPYear 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.

}{Year agoFor any real 1

MC UristYear ago^{+4}It's trivial to prove x^y>y^x when e

Math LifeYear 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 vídeoYear ago^{+33}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-Rivera7 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)

Metalhammer1993Year 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^^

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

Mohammad RehmaanYear 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

MCMageYear 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-Rivera7 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.

gillian2611Year ago^{+5}Around the same idea, have you already made a video of "Which is greater : n^(n+1) or (n+1)^n ?"

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

王万网Year 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 RousanYear agoThis can also be done using y=x^(1/x) function

Smiley1000Year ago^{+4}What about generalizing the e to some variable?

Angel Mendez-Rivera7 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.

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

khaled qaramanYear ago^{+2}great video

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

HelloItsMeYear ago^{+2}jkid1134 oh 😂

jkid1134Year ago^{+5}HelloItsMe BUT

2^3 = 8 < 3^2 = 9

Really makes you think 🤔

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

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

E RockYear 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 PieYear 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 RockYear 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.

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

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

JoeYear ago^{+2}AAAAAAAAAAAAAAAAAAAAAA COOL

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

Mihai CiorobitcaYear agoKrosis wow awesome !

Jonathan LiuYear ago^{+4}Great method!

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

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

space time9 months ago^{+3}This solution is a lot better