Which of the two is greater?

Jai Ganesh · 2005-07-13 21:39:52

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

Jai Ganesh · 2005-07-22 22:25:51

Mathsy, I think you missed this post!
With all the available technology, you could have well improved upon those digits!

mathsyperson · 2005-07-22 23:39:03

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.

MathsIsFun · 2005-07-24 18:08:09

Hello, and welcome to the forum kylekatarn !

I will let Ganesh reply to this, but just thought I would say hi.

Jai Ganesh · 2005-07-25 16:20:06

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 + ∞

Jai Ganesh · 2005-07-31 22:21:46

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!

NIH · 2005-08-07 11:05:54

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/LambertW-Function.html
http://www.americanscientist.org/template/AssetDetail/assetid/40804;_voi8-8bIm
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

Math Is Fun Forum

#1 2005-07-13 21:39:52

Which of the two is greater?

#2 2005-07-22 22:25:51

Re: Which of the two is greater?

#3 2005-07-22 23:39:03

Re: Which of the two is greater?

#4 2005-07-24 18:08:09

Re: Which of the two is greater?

#5 2005-07-25 16:20:06

Re: Which of the two is greater?

#6 2005-07-31 22:21:46

Re: Which of the two is greater?

#7 2005-08-07 11:05:54

Re: Which of the two is greater?

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