I was bored in maths (we were doing some generic fractional powers and negative powers)...so my mind wandered a bit, and I came up with:
A^B > B^A when A < B.
With the exception of A being 2 and B being 4, which both answers result in 16.
(So in reality it'd be A^B >= B^A when A < B).

I was wondering if there were any other numbers which this happens for, I've tried 1&2, 3&9 and those didn't work. Anyone know any which do?

For natural numbers? Well, if A = 1, A^B = 1 for any B, so if A = 1, the relation is trivially true.

If A = 2 and B = 3, A^B = 8, and B^A = 9, so your conjecture is not generally true by this very simple case. It's also not true for A =3 and B = 4, or for A = 4 and B = 5. Maybe you should conjecture the inequality in a different way.

Here's how the pattern is progressing for positive integers. The table is colour coded and shows AB - BA.



The exceptions are clearly shown:

-A=1
-A=2 and B=3
-A=2 and B=4

(You can prove the pattern for most cases easily enough by assuming A>2 and B>A and then taking the differential with respect to B of AB - BA, keeping A constant, which is ln(A)AB - ABA-1 and noting that since B>A>1 we have ABA-1<BA and also noting that since A>2 and a natural number ln(A)AB>AB, hence the gradient is strictly positive and the function is monotonic on that region)

Nice.

Some bugger had moved the series->fill and conditional formatting options when I wasn't looking. Took me several minutes to locate them.

It was the last paragraph that I liked.

Nice job, Thommo.

You can picture it too. If you think about the graphs, you know that A^x and x^A are going to intercept twice: once when A = x, and again since A^x hits x^A from below but then comes to completely dominate x^A. When you restrict yourself to A being natural, you can get a stronger conclusion.

These intuitive observations can and must be cashed out in the exact sort of calculus that Thommo went through.

There is a third intersection when A is even and x is negative.

scott1328 wrote:There is a third intersection when A is even and x is negative.

Yes, you're right. Forgot to mention I was just thinking about the positive quadrant.

The other numbers don't really exist anyway.

VazScep wrote:

scott1328 wrote:There is a third intersection when A is even and x is negative.

Yes, you're right. Forgot to mention I was just thinking about the positive quadrant.

Which leads me to ask:

Is there a general solution for the roots of the equation

A^X - X^A = 0 when A is a positive even integer?

for A=2, the roots are 2,4, and ~-0.766614...

Evolving wrote:The other numbers don't really exist anyway.

"Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk"

Wie wahr, wie wahr.

LjSpike wrote:I was bored in maths (we were doing some generic fractional powers and negative powers)...so my mind wandered a bit, and I came up with:
A^B > B^A when A < B.
With the exception of A being 2 and B being 4, which both answers result in 16.
(So in reality it'd be A^B >= B^A when A < B).

I was wondering if there were any other numbers which this happens for, I've tried 1&2, 3&9 and those didn't work. Anyone know any which do?

Well, if A=0 then the inequality does not hold.

Veida wrote:

LjSpike wrote:I was bored in maths (we were doing some generic fractional powers and negative powers)...so my mind wandered a bit, and I came up with:
A^B > B^A when A < B.
With the exception of A being 2 and B being 4, which both answers result in 16.
(So in reality it'd be A^B >= B^A when A < B).

I was wondering if there were any other numbers which this happens for, I've tried 1&2, 3&9 and those didn't work. Anyone know any which do?

Well, if A=0 then the inequality does not hold.

Maybe I should limit it to positive integers? Still, I like the idea of using a table, I'll go sprint over to excel and extrapolate ridiculously.

As already pointed out loads of exceptions as soon as you allow negatives. Quite obvious when you think about the sign flipping of multiplying negative numbers by themselves repeatedly.

Thommo wrote:As already pointed out loads of exceptions as soon as you allow negatives. Quite obvious when you think about the sign flipping of multiplying negative numbers by themselves repeatedly.

What conditional formatting and equations did you use to do the excel sheet, I've not used excel in forever...(about 3, 4 years?)

I made three sheets:

-first sheet I've just used hard numbers (series fill) in row 1 and column A, and then used the formula =$A2^B$1 for the cells, then copy and paste into all other cells in rows 2-31 and columns B-AE
-I've then conditionally formatted (using excel 2010, as I whinged about above conditional formatting commands have changed with version) using the command conditional formatting->New rule->Use a formula to determine which cells to format three times to input the rules =B2<0 (note the relative cell reference allows this to adjust to each cell in the range) for colour red, then the second new rule (so ok the first, choose the options from the menu for a new rule again) =B2=0 for colour orange, ok that and make a third rule =B2>0 for colour green.
-Copy the whole sheet (click the square at the very top left of the row and column headers that has no text and a small triangle in it, to the left of "A" and above "1") and paste it onto sheet 2 and onto sheet 3
-Go into sheet 2 and modify the formula to be =B$1^$A2 in cell B2 and then copy that formula and again paste into all relevant cells.
-Go into sheet 3 and modify the formula in B2 to be =Sheet2!B2-Sheet1!B2 and copy and paste that into all the cells in the range.

Voila!

(To be fair you could just do it with one sheet by using the formula =B$1^$A2-$A2^B$1 instead I suppose)

Oh, and happy birthday!

Thommo wrote:Oh, and happy birthday!

Oooh thanks!