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Topic: MEDIUM: E vs. PI (Read 1956 times) |
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S. Owen
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MEDIUM: E vs. PI
« on: Aug 13th, 2002, 12:43pm » |
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Not that I've got something here that many haven't already figured out, but I like the "hints" system of writing up solutions on these boards. So, here goes.
Hint 1:
Let pi = e^k for some k; k is a little bigger than 1.
Hint 2:
Therefore you are comparing
e^(e^k) and (e^k)^e
or...
e^(e^k) and e^(ke)
Answer:
Well, you know that e^0 > 0e, and that e^1 = 1e, and that e^2 > 2e, and could probably correctly intuit that e^k > ke whenever k != 1. Therefore you can correctly figure that e^pi > pi^e.
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| « Last Edit: Aug 14th, 2002, 1:48pm by S. Owen » |
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anshil
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Re: MEDIUM: E vs. PI
« Reply #1 on: Aug 14th, 2002, 12:21pm » |
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Honestly a really stupid puzzle, I took just my calculator hit some keys, and tata! solved.
puzzle score: 1
fun/story score: 0
surprise score: 0
stupidy: 10
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william wu
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Re: MEDIUM: E vs. PI
« Reply #2 on: Aug 14th, 2002, 1:06pm » |
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"Without computing their actual values, which is greater, e^(pi) or (pi)^e?"
S.Owen: Good hints
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| « Last Edit: Aug 14th, 2002, 1:44pm by william wu » |
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[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
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Eric Yeh
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Re: MEDIUM: E vs. PI
« Reply #3 on: Aug 14th, 2002, 4:10pm » |
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Just to formalize, let's add that taking the derivative of what S.Owen wrote gives you a quick proof of the minimum. Sorry, too many years of gradnig papers makes phrases like "could probably correctly intuit" just scream red flag to me!!!
Also, note that the soln generalizes to any x^y vs y^x fairly easily.
Best,
Eric
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| « Last Edit: Aug 14th, 2002, 4:11pm by Eric Yeh » |
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"It is better to have puzzled and failed than never to have puzzled at all."
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anshil
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Re: MEDIUM: E vs. PI
« Reply #4 on: Aug 19th, 2002, 12:36am » |
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I give following postulate:
for all x > 0 and y > 0 is true:
if x > y then x ^ y > y ^ x
if x = y then x ^ y = y ^ x
if x < y then x ^ y < y ^ x
Can you proove me wrong?
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AlexH
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Re: MEDIUM: E vs. PI
« Reply #5 on: Aug 19th, 2002, 2:31am » |
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Rethink that anshil. 
x = 5, y=2
5 > 2 but 52 < 25
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| « Last Edit: Aug 19th, 2002, 2:31am by AlexH » |
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anshil
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Re: MEDIUM: E vs. PI
« Reply #6 on: Aug 19th, 2002, 3:50am » |
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Very interresting, if they change in size at a certain point, do you think there exists a x ^ y = y ^ x for a x != y ?
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anshil
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Re: MEDIUM: E vs. PI
« Reply #7 on: Aug 19th, 2002, 4:09am » |
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Okay 4 ^ 2 = 2 ^ 4 =
So if x = 2, there exists a y that is not x, but x ^ y = y ^ x.
Does such value exist for every x? How could a formula look like that calculates the matching pair for symetric potentation.
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S. Owen
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Re: MEDIUM: E vs. PI
« Reply #8 on: Aug 19th, 2002, 1:28pm » |
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Assuming x,y > 0, you can do this...
x^y = y^x
y(log_y(x)) = x
y(ln(x)/ln(y)) = x
y/ln(y) = x/ln(x)
So the question is whether for a given x there is a distinct other value of y that satisfies this equation. So I cheated and graphed x/ln(x), and it shows that there will be another such a value for x>1.
I can't figure out how to express the other value in terms of x though.
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anshil
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Re: MEDIUM: E vs. PI
« Reply #9 on: Aug 20th, 2002, 7:22am » |
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y/ln(y) = x/ln(x)
ln(x) / x = ln(y) / y
ln(x) * x ^(-1) = ln(y) * y ^ (-1)
ln(x - 1) = ln(y - 1)
x = y
Hey thats true, but damm I lost the other solution somewhere.
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AlexH
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Re: MEDIUM: E vs. PI
« Reply #10 on: Aug 20th, 2002, 11:43am » |
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The line where you get ln(x-1) = ln(y-1) doesn't follow from the line before it. You'd actually get
ln(x1/x) = ln(y1/y)
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NickH
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Re: MEDIUM: E vs. PI
« Reply #11 on: Aug 23rd, 2002, 1:12pm » |
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Another solution to the original puzzle is to note that, for real x > 0, exp(x) > 1 + x.
Substitute x = pi/e - 1.
exp(pi/e - 1) > pi/e.
exp(pi/e) > pi.
exp(pi) > pi^e.
Nick
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NickH
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Re: MEDIUM: E vs. PI
« Reply #12 on: Aug 31st, 2002, 2:33pm » |
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Pi goes on and on and on...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed?
-- Martin Gardner
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That Don Guy
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Well, there's a general solution for e^x and x^e (x > 0 and x <> e):
Let f(x) = ln x / x.
f '(x) = (x * 1/x - ln x * 1) / x^2 = (1 - ln x)/x2, which is zero only at x = e.
f ''(x) = (x2 * (-1/x) - (1 - ln x) * 2x) / x4; for x = e, this is -e, so f(x) is a maximum by x = e.
Therefore ln e / e > ln x / x for x > 0 and x <> e.
Since ex is strictly increasing, a > b -> ea > eb, so e(ln e / e) > e(ln x / x) ->
(eln e)1/e > (xln x)1/x ->
e1/e > x1/x ->
(e1/e)ex > (x1/x)ex ->
ex > xe.
I've been trying to work on a general ab > ba for a,b > 0, but I get as far as: if a and b < e or a and b > e, the base that is closest to e is the larger number. For a > e and b < e, I've gotten the numbers equal if a = kk/(k-1) and b = k1/(k-1) for k > 0, but can't reduce that to a in terms of b.
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rugga
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Re: MEDIUM: E vs. PI
« Reply #14 on: Sep 14th, 2002, 4:43am » |
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Nice analysis, Don guy.
The bit about kk/k-1 and k1/k-1 has a certain beauty. Trying to express a in terms of b is an interesting challenge. I also got stuck. But absent that, you can always fall back on what you already proved:
Whichever of (ln a) / a or (ln b) / b is larger corresponds to the base that gives the larger result.
BTW, trying not to give too much away, there's also a connection with the 271 problem.
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ThudnBlunder
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Re: MEDIUM: E vs. PI
« Reply #15 on: May 14th, 2003, 11:53am » |
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Let x = (1 + 1/t)t
Let y = (1 + 1/t)1 + t
t not in [-1, 0],
Then xy = yx
Setting t = any integer other than -1 and 0 yields all
solutions in distinct rationals (x,y).
eg, t = 2 gives {9/4, 27/8}
If t in Q, we can also have (eg)
x = 31/2, y = 271/2
or, if t in R, we can have (eg)
x = pi, y =2.382179088...
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| « Last Edit: May 14th, 2003, 12:45pm by ThudnBlunder » |
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