A86: Re: Re: Solving Compound Intrest for N
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A86: Re: Re: Solving Compound Intrest for N
You can't. There is no way to solve that in terms of n, without
approximating. Some people I talked to suggested turning it into a power
series:
A = P + (tn/n) + ((t^2-1)(r^2)/(2!)(n^2)) + (((t^2-1)(t-2)(r^3))/(3!)(n^3))
+ ...
I don't know what you could do past that. (looks a lot better when written
out) Try approximating it to the first four or five terms and using an
89/92 to solve for n.
> How would I get the value of n then...
>
> Would I have to create some sort of loop pluging in numbers until one
> checks out...
>
> On Sun, 18 Apr 1999 23:55:30 -0500 "David Phillips" <david@acz.org>
> writes:
> >
> >Hmm, my 89 and 92 say that the solution is
> >
> >ln(p(r/n + 1)) * n = ln(a) / t AND a >= 0
> >
> >It isn't possible to solve it past that. Even Mathematica can't do
> >it!
> >
> >> I managed to solve the equation A = P(1+r/n)^(nt) for everything but
> >n...
> >>
> >> Is it even possible to solve this equation for n by some theorem or
> >> property that I have not yet learnt.
> >
> >
> >
> >
>
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