Re: exponential equation with constant
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 2 Jun 2005 21:26:27 GMT
In article <1117738290.754362.276360@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<quiasmox@xxxxxxxxx> wrote:
> I had a problem in an electrical circuits class, which was to find the
>point at which the voltage on two exponential curves is equal. These
>were the decaying voltage on an inductor and the rising voltage on a
>capacitor, the curves starting from a fixed voltage and from zero
>(respectively) at the same instant. The equated formulas were
>VL = 2* exp(-2000t) = 1 - exp(-1000t) = VC
>By luck, I found I could solve it for t by turning it into a quadratic-
>let y = exp(-1000t)
>2y^2 + y - 1 = 0 = (2y-1)(y+1)
>Then, using (2y-1),
>2 exp(-1000t) = 1,
>ln(2) = 1000t
>t = 693.2 us.
>This solution depends on the coincidence that one exponent is exactly
>twice the other.
>This brings me to the question -- is there a way, other than computer
>simulation, to solve a*exp(-ct) = 1 - b*exp(-dt) for t?
Similar to your solution: let x = exp(-ct). Then you want to solve
a*x + b*x^e = 1 where e = d/c. I'll assume e > 1 (if e < 1 you can
interchange the terms a*exp(-ct) and b*exp(-dt)). In only a few cases
you have a polynomial equation that can be solved using radicals.
However, at least if b/(a^e) is small there is a series solution
x =
1/a sum_{n=0}^infty (-1)^n GAMMA(ne+1)/(n! GAMMA(n(e-1)+2)) (b/a^e)^n
If e is rational, this can be expressed in terms of hypergeometic
functions.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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