# Why the development of continuous compound growth function divide the rate equally among intervals?

I just read from this article about the relations between continuous compound growth and the mysterious constant $e=2.71818...$.

In that article, it developed that for a growth rate of 100% in a time period, the final continuous growth is:

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

I am wondering about the $\frac{1}{n}$ part. It seems the growth rate is evenly divided into $n$ parts.

Why do we have to/can divide it that way? What's the rationale?

## ADD 1

There can be many ways to divide the interval of increase. But for a specific number of intervals of $n$, the even approach seems to give the biggest product.

For example, suppose $n=4$. If we divide the increase into 4 separate ones like below:

$$0.2, 0.2, 0.3, 0.3$$

no matter how long each interval is, the final outcome can be calculated as: $$1.2*1.2*1.3*1.3=2.4336$$

While the evenly approach gives: $$1.25*1.25*1.25*1.25 = 2.44140625$$

So the even approach is bigger. That is to say, if you want to get the maximum product while keeping the same sum, the evenly division is the way to go. (I remember there's a theorem for this.)

But I still don't know why we have to divide the total increase into each intervals, i.e. why we have to follow $0.2+0.2+0.3+0.3=1$.

• "It seems the growth rate is evenly divided into nn parts." Rather, the author of the post decides to consider the combined effect of a growth of $100\%/n$ over $n$ time periods and they realize that it roughly yields a growth of $e-1$ when $n$ is large. I fail to see much "rationale" in this... (Additionally, their post confuses the growth and the growth plus 1.) – Did Oct 31 '16 at 8:55
• We do not have to divide it this way, but this seems the most straightforward way to divide it. So if we are to divide our return of 100% in 2 parts (one to be given halfway and another one at the end) then 50% and 50% seems the most straightforward thing to do. It's just the simplest linear relationship between time and return: 1 unit of time returns 100%, 0.5 units of time return 50%. Maybe I did not understand your question. What other partitions do you see as more obvious? – Thanassis Oct 31 '16 at 9:15
• That article really doesn't help explain things; math.stackexchange.com/questions/1831666/…. – user301988 Oct 31 '16 at 10:05
• @selfawareuser The question you link is concerned with compounding interest (and I think it implicitly confounds compounded and simple interests but that's a different discussion). When the article does the division 1/n, it does not divide compounded interest. This is the nominal non-compounded interest of 100%. The article shows what happens when we take this non-compounding interest and we compound it continuously. – Thanassis Nov 1 '16 at 0:30

## 1 Answer

I am still not sure where your question comes from (what is it that confuses you more) but I'll try an answer based on your added information.

You do not have to divide the time interval in equal intervals. It's just one straightforward way to divide it. If you divide it a different way, you will get different answers for finite steps of compounding interest (as you discovered), but you will still get the same answer for the continuously compounded case: $e$. So as long as you take the limit of your process to infinite intervals, the answer will be $e$.

Let's take the limit of your example to infinite intervals. One easy way to do it, is just separate each of your initial intervals $0.2, 0.2, 0.3, 0.3$ in $n$ equal intervals. Notice that the intervals we end up with might have different sizes. For example one interval can be $0.2/n$ while another is $0.3/n$ (the size depends on which was the initial interval we started dividing). Notice what happens when we take the limit for $n \rightarrow \infty$. We can bunch together all infinitesimal intervals that came from the first initial interval of $0.2$. The return of these is $e^{0.2}$. We do this for all of your initial intervals. The total return is: $$e^{0.2}\times e^{0.2}\times e^{0.3}\times e^{0.3} = e^{0.2+0.2+0.3+0.3} = e$$

The other interesting thing is that you can view this arbitrary division as time intervals that all have proportional rates to the whole time unit rate or you can view it as arbitrary rates for arbitrary intervals. To ease the discussion, let's make the whole time unit to be 1 year and the whole time unit return rate to be 1 (100%). Note also that the rate mentioned here is not compounded. It is the return from your capital only, after one year without compounding (without gaining interest on partial interest). Now consider the following irregular example where for the first month the rate is $0.5$, while for the rest $5$ months it is $0.1$, and the last $6$ months it is $0$. The only limitation is that the time intervals have to add up to 1 year, and the rates have to add up to the total yearly rate of 1. If we calculate the total return assuming continuously compounded interest, it still is $e$.

Now about your last question on why the intervals (or equivalently the rates) have to add up to 1. This is the question that confuses me the most and I do not understand where it comes from. If you have a simple return rate of 100% and you want to break it up, you can do it in any way you want. But adding all the partitions together must total 100% (or whatever your rate is). Otherwise you would have a different return rate.

Say you have one bacterium, and you know that from this bacterium (not its children) you get 4 more bacteria after 4 hours. It does not matter how you get them. It might be that you get 1 each hour, or 3 the first hour and 1 the next 3 hours, or vice versa, but the end result has to be 4 bacteria in 4 hours. If the partial numbers(rates) summed up to something else then you would have a different rate :)

Now let's say that we want to take into account compounded returns (taking into account what the children bacteria produce too) and find the total return after $400$ hours. Since the steps are many we can treat this as a continuously compounded process* and simply say the result is $e^{400}$. It does not matter whether the simple rate of +4 bacteria every 4 hours was uniform within these 4 hours or not. The result is the same.

(*Of course bacteria multiplication is a discrete process, so the steps are finite, and hence the results will differ with different rate profiles, but the more the steps the smaller the difference will be.)

• @smwikipedia I am wondering whether my answer helped with some of your questions around $e$ and continual growth? If you still have questions it might be beneficial to chat in a chatroom so I get a better perspective of what confuses you. – Thanassis Nov 11 '16 at 21:24