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### Course: Finance and capital markets>Unit 1

Lesson 4: Continuous compound interest and e

# Formula for continuously compounding interest

Learn how to calculate interest when interest is compounded continually. We compare the effects of compounding more than annually, building up to interest compounding continually. Created by Sal Khan.

## Want to join the conversation?

• Try as I might, I cannot understand why this formula is correct

P(1 - r/n)^t*n

And this one isn't

P(1-r/n*t)^t*n

What is the significance of the n vs. the t? Why aren't they used as a single variable?
• Good answer....but more simply it's because (1+r/n) represents a single period (ex. one MONTH); (1+r/n)^n represents doing it for a full cycle ('n' times , ex. one YEAR) ; (1+r/n)^tn represents doing it for several cycles (ex. Several YEARS)
• At , Sal explains pretty well that the formula would be 50 (1+0.10/4) raised to the power of 12 (or 4 times 3 to be precise). However, if I compute that in my calculator, I get 67.2444 as a result, and not 67.49 as he indicates at the very end...Any idea why?
• Because at Sal is taking n=4 but in the end is taking n=infinity (continuously compounding interest)
• I want to know why the rate is divided by time (r/n)? If somebody could explain how that is derived?
• Using the video's example, the rate is divided by 4 because it's a yearly rate spread over 4 periods within the year, 3 months each period.

The interest is compounding every period, and once it's finished doing that for a year you will have your annual interest, i.e. 10%. In the example you can see this more-or-less works out:

(1 + 0.10/4)^4

In which 0.10 is your 10% rate, and /4 divides it across the 4 three-month periods. It's then raised to the 4th power because it compounds every period.

If you do the above math you'll find (1+0.10/4)^4 = 1.1038, which we could round to 1.10, which ends up at your 10% rate.

So the example's fancy compounding rate every 3 months effectively amounts to the same thing as a 10% rate for a year's loan.

It's only if somebody borrowed for a longer time period that it would make more of a difference. For example, borrowing at this rate for three years would not mean just paying 3 * 10% on your original amount or something like that. In fact in 3 years the interest would've compounded 12 times, since there's 4 periods every year, and in the end you'd actually be paying 34% on your original amount.

Anyway hopefully that gives some idea of where r/n came from in this case.
• I don't understand how "n" just disappeared from the last formula and still the result was approximately the same. I understood it like "t" in the last formula was n*t in the first and that the "t" represents the period in which the interest is coming. Sal said that it was years but in the first case the period is 3 months not 1 year. Sorry if my English is bad i hope you understood my question :)
• You are right, in that the n "disappeared." It disappeared at when the limit was taken as n goes to infinity. At this point, we are now dealing with a different formula than the original: we are not compounding over n=4 periods, but compounding over n=infinitely many periods. If this seems strange, it's because it is. Check out the previous two videos, if you haven't already; they explain the derivation of e. By taking this limit to compound continuously, you then yield a slightly different answer than if you had just plugged the numbers from the formula at into your calculator. This seems like a small difference, and it can be seen as such given the small percentage difference between the two answers. But if one is dealing with much larger principal and much longer time, the difference will then be exaggerated likewise.
• At, 2 minutes it says that the fraction inside the () is 0.10 / n but it is over 3 years so would't it be n * 3 (years)
• No, n is the number of compounds per period, and r is the interest per period. And t is the number of periods.

So 1 + r/n is the interest per compound (note that "per period" divided out). And n * t is the total number of compounds.
• Is there a practical use of continuously compounding interest in real life? Banks wouldn't want customers to get that kind of interest. Where do we use this in real life?
• Banks actually do use this for demand deposits. They also use it for many loans which they give out, most notably credit card loans. Of course, loans that have a fixed payment schedule, like mortgages, normally won't compound continuously, but instead every payment period (month normally).
• Do you really use this in your live?
• Good question. I was feeling like getting into calculus lesson, rather than simple aritmetic lesson.
• why continuously compounding interest is useful?
• If you are the lender, it's very useful because you earn more interest!
When interest compounds, you are paying interest on interest. The more frequently that happens, the more you end up paying.
• I need help on this homework from school: Mike plans to invest his money at 5.45% interest, compounded continuously. How long will it take for his money to double? (A=Pe^(rt))

Is the answer: log0.0545 (P) ?