y = f(x) = a^x y = g(x)= x^a Does there exist an 'a'

>>8691757
x, 0

>>8691762
I should've specified that I wanted a to be a number, either real or complex (excluding a = x) and a = 0 is trivial.

>>8691757
Well with a=0 g(x)=1 and f(x) = 0 so f'x and g'x are both 0 and growing at the same speed.

>>8691955
as I just wrote a = 0 is trivial

>>8691955
If you ignore the solutions then there are none.

"At all points (x,y)" implies that y=f(x)=g(x).

There are no constant solutions for a.

Not sure about complex values but
[math]
f(x,y) = x^{y} \\
g(x,y) = y^{x} \\
[/math]

[math]
x^{y} = y^{x} \\
y\ln{x} = x\ln{y} \\
\frac{1}{x}\ln{x} = \frac{1}{y}\ln{y} \\
x^{\frac{1}{x}} = y^{\frac{1}{y}} \\
[/math]

which clearly only is equal when x = y

>>8691757
These functions will have at most only two common points, so it's not clear what you mean by having the same rate at all points.

NP != P

>>8691994
I assume anon means that the derivatives are equal at all points.

File: Screen Shot 2017-02-21 at 17.46.11.png (43KB, 1504x296px) Image search: [Google]

43KB, 1504x296px

>>8692040
correct

Here's a picture with a more refined statement of the problem

Set them equal to each other and solve for x. There's your answer.

>>8691757
If the derivatives are equal their anti-derivatives can only vary by constant (fundamental theorem of calculus).

So it is clear that g(x)=f(x)+c which leadsto the conclusion c has to be 0 (c is not depended on a).

We now know that this is only possible if f(x)=g(x) so x^a=a^x you can do the rest yourself.

>>8691959
It's useful to appreciate the trivial case. It's the basepoint. Whenever a modulii space or something like it comes with a basepoint, you know things are tied into homotopy theory, category theory, and the depths of mathematics as a whole.

>>8692229
ah yes of course, I feel stupid now for not realising that lol

>>8691767
zero is not trivial it is the most important number.