Re: exponential equation with constant

israel@xxxxxxxxxxx (Robert Israel) wrote:
> 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

This sort of problem arises fairly frequently. Having given a series
solution earlier myself, I am interested in your different series solution.

Attempting to use your series to solve the problem original posed here, we
seem to arrive at x + 2*x^2 = 1, so that a = 1, b = 2 and e = 2, and then
b/(a^e) = 2. But b/(a^e) was required to be small (and I presume that 2 is
not adequately small) and so we cannot proceed. Is there perhaps some
reasonably simple way to transform the equation so that your series
solution can then be used?

David Cantrell
.

Relevant Pages

  • Re: exponential equation with constant
    ... > I had a problem in an electrical circuits class, ... >point at which the voltage on two exponential curves is equal. ... I found I could solve it for t by turning it into a quadratic- ...
    (sci.math)
  • Re: exponential equation with constant
    ... >>>point at which the voltage on two exponential curves is equal. ... >> you have a polynomial equation that can be solved using radicals. ... n an odd integer greater than 1 the pole of the GAMMA in the ...
    (sci.math)
  • exponential equation with constant
    ... I had a problem in an electrical circuits class, ... point at which the voltage on two exponential curves is equal. ... I found I could solve it for t by turning it into a quadratic- ...
    (sci.math)
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