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Transcendental Equation x = a^x

Date: 03/17/2004 at 15:34:48
From: Matt
Subject: Find intersection of y = x and y = a^x

Given y1(x) = x and y2(x) = A^x, where A is a constant real number
between 0 and 1, what is the analytic expression for the intersection
of y1(x) and y2(x) in terms of A?
The difficulty is in isolating A on one side of the equation.  Might 
be an infinite sum or product?

Date: 03/18/2004 at 08:54:33
From: Doctor Vogler
Subject: Re: Find intersection of y = x and y = a^x

Hi Matt,

What you have is called a transcendental equation,

  x = A^x,

and it has no "closed-form" solution.  That means that you can't write 
out a formula for the solution in terms of A by using functions that
you are familiar with.

What most people do with an equation like this is pick an A, graph 
both functions on a graphing calculator, and find the intersection. 
That's called "numerical methods" by mathematicians and it gives you
an approximate answer.

What a mathematician does is first try to prove that there is always 
exactly one solution.  In this case, there is not.  So then he would
find for which values of A there is no solution, for which values of A
there is exactly one solution, and for the other values of A how many
solutions there are.  This can be done with a little calculus and
isn't very hard.  

Then, if he so desires, he can define a function f(A) (or whatever he
wants to call it), and then describe several of its properties (like
where it's defined, when it's increasing or decreasing, how to find
its derivative, and a few known exact values).  This is the same sort
of thinking that was used to create functions like arctan, for instance.

If you have any questions or need more help, please write back and 
show me what you have been able to do, and I will try to offer further 

- Doctor Vogler, The Math Forum 

Date: 03/18/2004 at 11:27:58
From: Matt
Subject: Thank you (Find intersection of y = x and y = a^x)

Dr. Vogler - 

Thanks for taking the time to answer my question.  I had a feeling it
might be something like that.  I'll continue to play with this some
more and will follow your suggestion on defining it.  At this point it
seems that there's one solution for all 0 < A <= 1 and no solution for
A = 0 (with 0^0 = 1) or A > 1, though I'd say there's a right handed
limit of 0 at 0.  The A < 0 case will take some brushing up on complex 
numbers on my part.  It'll take me a bit to get up to trying to find a
derivative for it, but I'll give it a shot.

Thanks again for your help.  I may respond again if I get stuck!
Associated Topics:
College Analysis

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