SOLUTION: an extra credit for my honors math class. i cant figure it out for the life of me... derive the formula A=Pe^rt from the general formula for compounding interestL A=P(1+r/n)^n*t

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: an extra credit for my honors math class. i cant figure it out for the life of me... derive the formula A=Pe^rt from the general formula for compounding interestL A=P(1+r/n)^n*t      Log On

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Question 30099: an extra credit for my honors math class. i cant figure it out for the life of me...
derive the formula A=Pe^rt from the general formula for compounding interestL A=P(1+r/n)^n*t

Answer by Fermat(127) About Me  (Show Source):
You can put this solution on YOUR website!
A+=+P%281%2Br%2Fn%29%5Ent - interest compounded periodically
A+=+Pe%5Ert ------- interest compounded continuously
When interest is compounded continously, this means letting the number of periods in a year increase without bound.
In other words, we find the limit, as n tends to infinity of A+=+P%281%2Br%2Fn%29%5Ent
Lim+%281%2Br%2Fn%29%5Ent
n+-%3E+infty
Binomial theorem
%281%2Bx%29%5En+=+1+%2B+nx+%2B+n%28n-1%29x%5E2%2F2+%2B+n%28n-1%29%28n-2%29x%5E3%2F3%21+%2B+n%28n-1%29%28n-2%29%28n-3%29x%5E4%2F4%21+...
Applying the Binomial theorem to %281%2Br%2Fn%29%5Ent, we get
1+%2B+%28nt%29%28r%2Fn%29+%2B+%28nt%29%28nt-1%29%28r%2Fn%29%5E2%2F2+%2B+%28nt%29%28nt-1%29%28nt-2%29%28r%2Fn%29%5E3%2F3%21+%2B+%28nt%29%28nt-1%29%28nt-2%29%28nt-3%29%28r%2Fn%29%5E4%2F4%21+...
1+%2B+%28tr%29+%2B+%28t%29%28t-1%2Fn%29r%5E2%2F2+%2B+%28t%29%28t-1%2Fn%29%28t-2%2Fn%29r%5E3%2F3%21+%2B+%28t%29%28t-1%2Fn%29%28t-2%2Fn%29%28t-3%2Fn%29r%5E4%2F4%21+...
as n-%3E+infty, 1%2Fn+-%3E+0, giving
1+%2B+tr+%2B+t%5E2r%5E2%2F2+%2B+t%5E3r%5E3%2F3%21+%2B+t%5E4r%5E4%2F4%21+...
1+%2B+%28rt%29+%2B+%28rt%29%5E2%2F2+%2B+%28rt%29%5E3%2F3%21+%2B+%28rt%29%5E4%2F4%21+...
and the above is the Maclaurin series for e%5E%28rt%29
i.e. %281%2Br%2Fn%29%5Ent+-%3E+e%5E%28rt%29 as n+-%3E+infty
We wanted to find the limit, as n tends to infinity of A+=+P%281%2Br%2Fn%29%5Ent and we have the limit, as n tends to infinity, of %281%2Br%2Fn%29%5Ent which is e%5E%28rt%29. We substitute e%5E%28rt%29 for A+=+P%281%2Br%2Fn%29%5Ent.
So for continuous compounding we can write,
A+=+Pe%5E%28rt%29
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