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The growth in an exponential growth system can be described by: $y = i\cdot(1+\frac{r}{n})^{tn}$ . I want to prove that the overall rate of growth $r$ despite being divided by the number of times compounded ($n$), is still the same ($r$) when applied the power $n$. It doesn't seem right that NET exponential growth rate should change regardless of how many times we compound the system. For example, if I have a 100 percent interest rate and it is compounded once annually: $y = i\cdot(1+(1/1))^1 \cdot 1 = i \cdot 2$. I want to show that if we do 50 percent interest semi-annually, the rate of growth over a full year is still 100 percent. If its true of course.

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    $\begingroup$ It's not true; that is the whole point of compound interest. $1.5^2=2.25$, which is better than 100% annually. $\endgroup$ – symplectomorphic Nov 2 '17 at 6:25
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Your guess is incorrect: you are looking at what people call compound growth.

If you had \$1 and received 100% interest in a year, you would expect to have \$2 by the end of the year. However, if you had 50% compounded twice a year, you would earn an additional 50% interest on the \$0.50 that you received at mid year (hence the term 'compound interest', where you earn interest on interest), so you would finish with \$2.25.

Thus compounding the system indeed increases the growth rate, but there's a limit to this: as you compound more frequently (say every second), the most you can have by the end of the year is about \$2.72. This number is called $e$, and one way we define this in mathematics is the limit as $n$ goes to infinity of $\left(1+\frac{1}{n}\right)^n$.

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  • $\begingroup$ Is there any way to model a system where you can have 50 percent growth rate twice and end up with the original amount using the exponential growth formula? $\endgroup$ – A.AK Nov 2 '17 at 6:29
  • $\begingroup$ What do you mean by the 'original amount'? If you're saying 100%, and using compound interest of 50%, then the answer is of course not: the formula gives you what it gives. To get a 100% return through paying a 50% interest rate twice, you would have to be doing a principal-only interest, where you pay interest on the initial \$1 only. This is equivalent to calculating the interest you would pay in a year (\$1) and distributing it evenly over the number of payments to make (e.g. $\frac{\$1}{2} = \$0.50$ twice in a year). $\endgroup$ – Ken Wei Nov 2 '17 at 6:34
  • $\begingroup$ I meant original growth rate of 100 percent. Ok, that makes more sense. Also is there a reason why r is divided by the number of times compounded, it could just as easily be divided any other amount of times. Is there any significance in dividing r evenly over all the time intervals? $\endgroup$ – A.AK Nov 2 '17 at 6:37
  • $\begingroup$ @A.AK The reason to take $\frac rn$ is in order to have the same growth rate during the first $\frac1n$ as you would have without compounding for the same $\frac1n$ of a year. Also, this way you will end up with the same interest $r$ on the principal - everything else will be compound and compound on compound and ... $\endgroup$ – Hagen von Eitzen Nov 2 '17 at 6:45
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You are looking at the difference of $$1+na$$ and $$(1+a)^n=1+na+\frac{n(n-1)}2a^2+\dots+\binom nk a^k+\dots+a^n.$$ As $a=\frac rn$ is positive, the second binomial sum has more positive terms than the first, and is thus larger. The only way to get equality is for $a=0$, that is with no growth rate at all.

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