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Suppose that \$1000 is invested at 5% interest compounded continuously. At 20 years, what is the rate of increase? After how many years will the account reach \$2000?

Could you explain how solve it? Please explain.

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Compounding daily for a yearly interest rate means that you must split the yearly rate into daily terms and compound by that rate once for each day. Compounding continuously means that you need to split up the interest rate into an increasingly small number of pieces and compound for each piece. Can you come up with a formula that expresses this compounding? –  abiessu Dec 4 '13 at 15:01
    
Why once per day? Unless otherwise stated, I think we must assume the time period is as given, i.e. a year. Banks won't ever agree to pay you daily if you've invested under yearly conditions: it will cost them much more . –  DonAntonio Dec 4 '13 at 15:02
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@DonAntonio: I was offering an example for how to approach the problem; it seems to me that the intent is to get the student to arrive at one of the limit formulas for $e^{rx}$... –  abiessu Dec 4 '13 at 15:07
    
And of course I said "increasingly small number of pieces" when I meant "increasing number of pieces, each of which is decreasing in size". –  abiessu Dec 4 '13 at 15:10
    
Perhaps you're right, @abiessu:I am noting now that this is a calculus problem so a continuous variable would likely kick in. –  DonAntonio Dec 4 '13 at 15:10

3 Answers 3

Taking the usual compounding-interest formula to be $A=P\left(1+\frac r{100}\right)^n$ for compounding the interest $n$ times at interest rate $r$, what happens if we have to compound more often? Then we must break up the percentage into the number of subdivisions of the primary time scale (in this case, years), and then multiply the number of compounding events by the number of subdivisions.

Taking a yearly rate of $5$% as an example, and doing daily compounding, we would change the formula to

$$A=P\left(1+\frac 5{365\cdot 100}\right)^{n\cdot 365}$$

We can rewrite this formula as a limit to approach the continuous case:

$$A=P\cdot\left( \lim_{n\to\infty}\left(1+\frac {\frac 5{100}}{n}\right)^n\right)^t$$

Here we have exchanged the $365$ from the example with $n$ as the number of subdivisions, and used $t$ as the measurement of the number of years during which the compounding process has occurred.

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Hints:

The formula is

$$F=Ba^t\;\;,\;\;X=\text{Final Ammount}\;,\;\;B=\text{Initial Ammount}\;,\;\;a=\text{the growth/decay factor}$$

$$a=\begin{cases}1+\frac r{100}&,\;\;\text{if it is a growth problem}\\{}\\1-\frac r{100}&,\;\;\text{if it is a decay problem}\end{cases}$$

$$r=\text{percentage}\;\;,\;\;t=\text{ ammount of time periods elapsed}$$

In our particular case, we have

$$B=1000\;\;,\;\;r=5\;\;,\;t=\text{time in years (apparently)}$$

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Formula for continuous compounding is

Future Value = Present Value * ($e^{rt}$)

In 20 years, FV = 1000*($e^{.05*20}$)

FV = 1000*e = $2718.28

Rate of Increase could be defined as (FV-PV)/PV = 1718.28/1000 = 171.8%

Second part of the answer is:

FV = 2000, PV = 1000, r = 0.05, Find T

T = $\frac{1}{r}$*$\ln\frac{FV}{PV}$

If you plug in numbers, then

T = 13.862 Years.

Thanks

Satish

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