
 ganesh
 Moderator
Which of the two is greater?
When there are two numbers x and y, such that both x,y ≥1, does it follow that y^x is always greater than x^y if x is greater than y? No. This is true only if y is greater than a certain CRITICAL Value. Many years back, I tried to find this critical value of y for certain values of x.
Value of x Approximate value of y
10 1.3712886 100 1.04955 1000 1.0069805 10,000 1.000922309 100,000 1.00011514925 1,000,000 1.0000138158 10,000,000 1.00000161283 100,000,000 1.0000001843 1,000,000,000 1.0000000208
Illustration: y^100 can be greater than 100^y only if the value of y ≥1.04955
Character is who you are when no one is looking.
 ganesh
 Moderator
Re: Which of the two is greater?
Mathsy, I think you missed this post! With all the available technology, you could have well improved upon those digits!
Character is who you are when no one is looking.
Re: Which of the two is greater?
Occasionally, the box goes away before I can read all the new posts. Usually when there have been lots of new posts and it takes me a long time to read them all. I think that's what happened here.
There's a strong pattern emerging there, though. Do you think it's possible to rearrange x^y=y^x to find y in terms of x? If you could do that, you could find the critical value for any value of x.
I've got as far as the yth root of y = the xth root of x, but now I'm stuck.
Why did the vector cross the road? It wanted to be normal.
 kylekatarn
 Power Member
Re: Which of the two is greater?
 I think that finding these crit. pts. should involve logarythms. just a supposition=P
 kylekatarn
 Power Member
Re: Which of the two is greater?
this topic is amazing!
I did some 'LN' transformations to the expression and I found the equation:
ln( y^(1/y) ) = ln(x)/x
Solving this equation in a computer program like mathematica, maple or derive, you can find the critical value yCrit; for a given x for example, I was able to compute yCritical for several x's with 50 precision digits(but this can be increased): x = xValue y = yCritical  x=10 y = 1.3712885742386235368613621062996899588428544048422 x=100 y = 1.0495191898071712311474936519440559096925868204045 x=1000 y = 1.0069802219160264731969790392939479509214698343986 x=10000 y = 1.0009223085800102005258019267508413188152496261875 x=100000 y = 1.0001151491408378890243699386042389677717403925799 x=1000000 y = 1.0000138157968674942789013367960898614105090318224 x=10000000 y = 1.0000016118134620021095317020510233099984034273812 x=100000000 y = 1.0000001842068583377621834070851145767978026108316 x=1000000000 y = 1.0000000207232664811270553117793424621783832233279 x=10000000000 y = 1.0000000023025851009468928822904460936293772528616  it seems that yCritical "slowly"(?) approaches the limit of 1 as x approaches +oo
looking forward to hear comments on this topic!
 MathsIsFun
 Administrator
Re: Which of the two is greater?
Hello, and welcome to the forum kylekatarn !
I will let Ganesh reply to this, but just thought I would say hi.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
 kylekatarn
 Power Member
Re: Which of the two is greater?
 ganesh
 Moderator
Re: Which of the two is greater?
kylekatarn wrote:it seems that yCritical "slowly"(?) approaches the limit of 1 as x approaches +oo
looking forward to hear comments on this topic!
Yes, you are correct! ycritical approaches 1, but is certainly greater than 1, as x approaches + ∞
Character is who you are when no one is looking.
 ganesh
 Moderator
Re: Which of the two is greater?
mathsyperson wrote:I've got as far as the yth root of y = the xth root of x, but now I'm stuck.
Interestingly, if yth root of y = xth root of x, it does not automatically follow that x=y For example, if x=4 and y=2, then this is true!
Character is who you are when no one is looking.
 NIH
 Member
Re: Which of the two is greater?
It's not possible to express y explicitly in terms of x using any of the standard elementary functions. However, there is a formal solution using something called the Lambert W function. This function can be evaluated using mathematical packages such as Maple and Mathematica, but no calculator currently has a button for it. See the references below for details.
http://mathworld.wolfram.com/LambertWFunction.html http://www.americanscientist.org/template/AssetDetail/assetid/40804;_voi88bIm http://www.orcca.on.ca/LambertW/
Another approach would be to use the Newton Raphson method. If a is an approximation to a root of f(x) = 10^x  x^10 = 0, then a  f(a)/f'(a) will be a better approximation. In this case, we have f'(x) = ln(10) * 10^x  10x^9.
For example, if a = 1.4 is an approximate solution, then 1.4  (10^1.4  1.4^10)/(ln(10) * 10^1.4  10*1.4^9) ~= 1.3744 is a better approximation.
This converges quite rapidly. The next two convergents are, to 10 decimal places, 1.3713296532 and 1.3712885814.
http://www.sosmath.com/calculus/diff/der07/der07.html
Finally, here's a page which will solve, for example, the equation 10^x = x^10, giving several answers in terms of the Lambert W function (aka the ProductLog function), along with the numeric values.
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic
2 + 2 = 5, for large values of 2.
