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 • 71

  • cookiebuster
    cookiebuster 13 days ago

    when reality just breaks...3:26

  • Chaw Den
    Chaw Den 27 days ago

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

  • Sagnik Banerjee
    Sagnik Banerjee 2 months ago

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

  • Zυвi
    Zυвi 2 months ago

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

  • Tristan Cole
    Tristan Cole 3 months ago

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

  • Serengeti Ghasa
    Serengeti Ghasa 4 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
    spicca 4 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
    Angel Mendez-Rivera 4 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
    Shashank S I 4 months ago

    Awesome solution 😀

  • Shinigami Steve
    Shinigami Steve 4 months ago

    Awesome proof

  • Mark Levin
    Mark Levin 4 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
    Joe Potillor 4 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
    Lily Lopez 5 months ago

    okay but i love your accent :(

  • Ding Hui Ng
    Ding Hui Ng 5 months ago

    I perfer this proof

  • Radio TV
    Radio TV 5 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
    James Wilson 6 months ago

    Very clever trick!

  • sakib shahriar
    sakib shahriar 7 months ago

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

  • Rachid Iksi
    Rachid Iksi 8 months ago

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

  • Gregory Fenn
    Gregory Fenn 9 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 Empeigne
    Michael Empeigne 9 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 ).

  • MrOligi3003
    MrOligi3003 10 months ago

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

  • LiZichong
    LiZichong 10 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_bv
      jose_bv 6 months ago

      r/iamverysmart

    • James Liu
      James Liu 10 months ago

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

    • LiZichong
      LiZichong 10 months 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
      LetsSolveMathProblems  10 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 Math
    About Math 10 months 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
    Kyx 10 months ago +67

    Engineer: they both = 9

    • August Liu
      August Liu Month ago

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

    • Sahil Baori
      Sahil Baori 2 months ago

      @Rohan LOL!

    • Rohan
      Rohan 2 months ago

      can't even do a shit joke correctly

    • ShouTZ
      ShouTZ 4 months ago

      i was about to type this lmaoooo

    • Evan Murphy
      Evan Murphy 5 months ago +9

      27?

  • Debajyoti Guha
    Debajyoti Guha 10 months ago +2

    You're really good. You deserve more subscribers.

  • Tasty Rainbro
    Tasty Rainbro 10 months 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
    HPP 10 months 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.

    • }{
      }{ 10 months ago

      For any real 1

  • Stephen H
    Stephen H 10 months ago +4

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

  • Math Life
    Math Life 10 months ago

    y=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 vídeo

    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
      Angel Mendez-Rivera 4 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
    Metalhammer1993 10 months 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
    pandascarpo 10 months ago

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

  • Mohammad Rehmaan
    Mohammad Rehmaan 10 months 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
    MCMage 10 months 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
      Angel Mendez-Rivera 4 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
    gillian2611 10 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 Temesvári
    Csanád Temesvári 10 months ago

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

  • 王万网
    王万网 10 months 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
    Kazi Abu Rousan 10 months ago

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

  • Smiley1000
    Smiley1000 10 months ago +4

    What about generalizing the e to some variable?

    • Angel Mendez-Rivera
      Angel Mendez-Rivera 4 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.

  • Mathedidasko
    Mathedidasko 10 months ago

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

  • khaled qaraman
    khaled qaraman 10 months ago +2

    great video

  • HelloItsMe
    HelloItsMe 10 months ago +4

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

    • HelloItsMe
      HelloItsMe 10 months ago +2

      jkid1134 oh 😂

    • jkid1134
      jkid1134 10 months ago +5

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

  • Dr Din
    Dr Din 10 months ago +2

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

  • Adlet Irlanuly
    Adlet Irlanuly 10 months ago +5

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

  • E Rock
    E Rock 10 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 Pie
      xXappleXx Pie 10 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 Rock
      E Rock 10 months ago

      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
      LetsSolveMathProblems  10 months ago +5

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

  • Deepak Maheshwari
    Deepak Maheshwari 10 months ago +4

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

  • Joe
    Joe 10 months ago +2

    AAAAAAAAAAAAAAAAAAAAAA COOL

  • Krosis
    Krosis 10 months ago +16

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

  • Jonathan Liu
    Jonathan Liu 10 months ago +4

    Great method!

  • end my misery
    end my misery 10 months ago +3

    great video! :) really liked this method

  • Matt GSM
    Matt GSM 10 months ago +51

    BlackPenRedPen did a video about this! I love this solution

    • space time
      space time 6 months ago +3

      This solution is a lot better