Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,982

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.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,982

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.

Offline

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

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.

Offline

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

-----------------------------------------

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

Offline

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

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!

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

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

Offline

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

thanks MathsIsFun! : )

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,982

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.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,982

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.

Offline

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

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.

Offline

Pages: **1**