A86: Re: Solving Compound Intrest for N
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A86: Re: Solving Compound Intrest for N
What I meant was plugging in consecutive values for N and solving for p
then checking to see if the p's match up...
I know that just incrementing the n trials would take for ever but what
about some sort of system where the closer the returned p gets to the
supplied p then the smaller the incrementation gets..
Just an idea...
How does solver do it?
Thanks for you help,
Chris
On Mon, 19 Apr 1999 14:15:29 -0500 "David Phillips" <david@acz.org>
writes:
>
>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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>
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