Variable Within and Outside an Exponent

Date: 07/29/97 at 12:21:23
From: John Shannonhouse
Subject: Variable Within and Outside an Exponent

Hello,

I am trying to figure out how to solve for t in the equation:

	d = a*t + b*e^-(c*t)

where a, b and c are constants and e is the exponential.  I am pretty 
sure that the solution can only be found by an algorithm because t is 
found both in an exponent and outside an exponent.  What is the 
algorithm?  Or is there a solution that I have not seen?

Thanks,
John Shannonhouse

Date: 07/31/97 at 15:58:02
From: Doctor Rob
Subject: Re: Variable Within and Outside an Exponent

In general there is no closed form solution for this equation.

For a particular choice of a, b, c, and d, you can solve for t using
Newton's Method (yes, Sir Isaac Newton!).  It goes like this.

Let F be a function of t. To find a root of F(t) = 0, pick a starting
place t[0], which is a guess at the root. Let F'(t) be the derivative
of F(t) with respect to t. Use the recursion

   t[n+1] = t[n] - F(t[n])/F'(t[n])

until you have enough accuracy, or until it is obvious that the 
sequence {t[n]} will not converge. There is a theory of convergence, 
well setout in any book on Numerical Analysis. If any t[i] is such 
that |F'(t[i])| < 1, then the sequence converges, and from some point 
on, each iterate agrees with the limit in about double the number of
significant figures as its predecessor does.

If convergence fails, then you may be able to rewrite the equation
in a different form.  For example, in your case, you could say that
since 

   d = a*t + b*e^-(c*t),
   (d - a*t)/b = e^-(c*t),

so

   ln[(d - a*t)/b] + c*t = 0.

This latter form can serve as F(t) = 0, and may (or may not) have
different, better convergence properties at the root you seek.  Other
transformations of the equation may also be useful, so long as a root
of the transformed equation leads you to a root of the original.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/