Variable Within and Outside an ExponentDate: 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/ |
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