Question 30099
{{{A = P(1+r/n)^nt}}} - interest compounded periodically
{{{A = Pe^rt}}} ------- 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(1+r/n)^nt}}}
{{{Lim (1+r/n)^nt}}}
{{{n -> infty}}}
Binomial theorem
{{{(1+x)^n = 1 + nx + n(n-1)x^2/2 + n(n-1)(n-2)x^3/3! + n(n-1)(n-2)(n-3)x^4/4! ...}}}
Applying the Binomial theorem to {{{(1+r/n)^nt}}}, we get
{{{1 + (nt)(r/n) + (nt)(nt-1)(r/n)^2/2 + (nt)(nt-1)(nt-2)(r/n)^3/3! + (nt)(nt-1)(nt-2)(nt-3)(r/n)^4/4! ...}}}
{{{1 + (tr) + (t)(t-1/n)r^2/2 + (t)(t-1/n)(t-2/n)r^3/3! + (t)(t-1/n)(t-2/n)(t-3/n)r^4/4! ...}}}
as {{{n-> infty}}}, {{{1/n -> 0}}}, giving
{{{1 + tr + t^2r^2/2 + t^3r^3/3! + t^4r^4/4! ...}}}
{{{1 + (rt) + (rt)^2/2 + (rt)^3/3! + (rt)^4/4! ...}}}
and the above is the Maclaurin series for {{{e^(rt)}}}
i.e. {{{(1+r/n)^nt -> e^(rt)}}} as {{{n -> infty}}}
We wanted to find the limit, as n tends to infinity of {{{A = P(1+r/n)^nt}}} and we have the limit, as n tends to infinity, of {{{(1+r/n)^nt}}} which is {{{e^(rt)}}}. We substitute {{{e^(rt)}}} for {{{A = P(1+r/n)^nt}}}.
So for continuous compounding we can write,
{{{A = Pe^(rt)}}}
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