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This is a fairly basic result, but I could not find anything about it here.

How do you prove that the following relationship exists, and where does the basis for it originate from:

$$(1+r/∞)^∞=\left( \left( 1+\frac{1}{\infty}\right) ^{\infty}\right)^r$$


Improved by hjhjhj57:

$$\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n=\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^r$$


That is to say, a constantly compounded interest at the period's interest rate $r$ is equivalent to a constantly compounded interest at a period's rate of $100$ % to the power of $r$.

The second formula is equivalent to the exponent function $e^r$. Can other functions with the same features (namely that the $f(dx) = f(x)$), be derived from the formula by changing the $r$ and maybe with some other alterations? I am asking this since using the $100$ % interest rate as a definition for $e$ seems somewhat arbitrary and I am wondering if other functions with the same features can be derived from the formula.

EDIT: The idea of this question is not necessarily to prove the equation above holds, but to explain where the relationship or the intuition comes from in the first place or how it was first discovered. In other words why does $e^r$ represent growth factor when the period's rate is $r$ and added constantly, given that $e^1$ equals period's rate of $100$%.

The second question was already answered here: Function is equal to its own derivative.

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  • $\begingroup$ I'm not sure this notation helps at all. You should try using the binomial theorem. $\endgroup$ – hjhjhj57 Jun 11 '15 at 19:33
  • $\begingroup$ @calculus I edited the original equation the OP posted. You can edit the post to correct it. $\endgroup$ – hjhjhj57 Jun 11 '15 at 22:11
  • $\begingroup$ @hjhjhj57 Thanks for the information. A have made an edit. I have posted both notations. $\endgroup$ – callculus Jun 11 '15 at 22:28
  • $\begingroup$ @calculus not sure that I am understanding you. The RHS of my equation without the power of r is simply definition for e, so that equation ^r = e^r. In addition, I am not getting the correct result using your formula. $\endgroup$ – Dole Jun 12 '15 at 2:18
  • $\begingroup$ @Dole You are right. I missed, that there is no r at the numerator anymore on the RHS. Sorry for the confusion. $\endgroup$ – callculus Jun 12 '15 at 3:59
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Consider the derivative of the differences with respect to r.

$\frac{1}{n}(1+\frac{r}{n})^{n-1}-nlog((1+\frac{1}{n})^{nr})(1+\frac{1}{n})^{nr-1}$

This expression tends towards 0 as n tends towards infinity so they are equal to each other.

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  • $\begingroup$ I think it's clear based on the reference to exponential functions that he means the limit definition. Better to leave a constructive comment and actually help OP. $\endgroup$ – Alfred Yerger Jun 11 '15 at 19:35

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