# Understanding e when there is continuous compounding at less than 100%

I am not a math professional but I know the formula for continuous compounding and also (finally) just studied its derivation using the limit when n --> infinity; how r, n, t all play together in the formula when n/r = m and all of that which accounts for situations when r (growth rate) is not equal to 100%. I understand it but don't understand the intuition behind the substitution n=mr. How in the world one would intuitively understand and get an idea like that -i.e. by substituting n=mr, we can pull out r from inside the limit to outside?

What I fail to understand is the intuitive part behind the merging of the rate and time. Using a simple example as a case in point. Assuming a 100% continuous compounding rate, a 1 dollar grows to e at the end of year 1. At the end of year 2, it grows to e.e = e^2. This is fine so far.

Now I say that I cannot wait two years, but I want the same e^2 amount at the end of year 1 instead of year 2 end. This implies that the rate of growth has to be naturally higher since there is less time. The formula says because (e^2)^1 = (e^1)^2 the rate and time can be interchanged. So a 200% rate is required.

I understand the formula, but the intuition is lacking. If the amount at the end of two years is e^2, the amount after half the time (i.e year 1 end) is square root of (e^2) which is e. I understand this part. While 200% fits the formula, I don't intuitively understand how the 200% growth rate (continuous compounding, that is) is required during year 1 to make the amount e^2 at the end of year 1.

Thanks

• Your question refers to the variables $r$, $n$, and $t$, but doesn't give the formula you intend for them to appear in, which is necessary to understand what you mean. In fact, this seems to be a problem more generally. Please explain the context of your question so we can understand the question. – dfeuer Jan 23 '14 at 7:28
• what to you mean by a "100% continuous compounding rate" surely that after a year it has grown to 200% , not to 272%? – Willemien Jan 23 '14 at 8:32
• Sorry I should have been more clearer. r is the continuously compounded rate of growth. n is the number of years or periods for which the continuous growth happens. time t is "merging" of the rate and # of periods, ie. if growth rate doubles, the same amount (Final Value) can be reached in half the number of periods or vice-versa. so t is rn. The intuition of t=rn in the context of continuous compounding is where I have difficulty in understanding. – kumar k Jan 25 '14 at 6:09
• Willemein - what I meant to say is that a 100% rate of interest or growth is compounded continuously to get e at the end of year 1 and e^2 at the end of year 2. I didn't mean to say that the final value is 100% of the initial value. I should have been more clearer. – kumar k Jan 25 '14 at 6:46

In the previous case where continuous interest rate is $100\%$, it takes two year for your amount to grow to $e^2$ times.
Now, if instead, you wish your amount to grow to $e^2$ times in one year. Then for each of the half years, your amount has to grow to $e$ times. This corresponds to $100\%$ interest rate ever half year, or $200\%$ interest rate per annum in annual percentage rate.