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 suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
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!
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum