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

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,344

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


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,344

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! smile


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

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.

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#4 2005-07-24 13:01:52

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Which of the two is greater?

-----------------------------------------
I think that finding these crit. pts. should involve logarythms. just a supposition=P

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#5 2005-07-24 14:22:22

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

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!

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#6 2005-07-24 18:08:09

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,626

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

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#7 2005-07-25 01:23:56

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Which of the two is greater?

thanks MathsIsFun! : )

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#8 2005-07-25 16:20:06

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,344

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 + ∞


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#9 2005-07-31 22:21:46

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,344

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! roll


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#10 2005-08-07 11:05:54

NIH
Member
Registered: 2005-06-14
Posts: 33

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/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


2 + 2 = 5, for large values of 2.

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