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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.

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  • male robot hal style avatar for user Mitchell McGill
    Try as I might, I cannot understand why this formula is correct

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

    And this one isn't


    What is the significance of the n vs. the t? Why aren't they used as a single variable?
    (21 votes)
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    • aqualine ultimate style avatar for user Euler
      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)
      (15 votes)
  • blobby green style avatar for user Marco Birnkammer
    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?
    (10 votes)
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  • male robot donald style avatar for user Doug
    I want to know why the rate is divided by time (r/n)? If somebody could explain how that is derived?
    (7 votes)
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    • leaf green style avatar for user 20Kor
      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.
      (14 votes)
  • duskpin ultimate style avatar for user Adis Music
    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 :)
    (8 votes)
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    • blobby green style avatar for user Michael Primavera
      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.
      (7 votes)
  • orange juice squid orange style avatar for user Boston Abrams
    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)
    (4 votes)
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  • piceratops seed style avatar for user braveheart
    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?
    (3 votes)
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    • ohnoes default style avatar for user Tejas
      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).
      (6 votes)
  • blobby green style avatar for user Abrahan Moreno
    Do you really use this in your live?
    (5 votes)
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  • purple pi teal style avatar for user Gustavo Delazeri
    why continuously compounding interest is useful?
    (3 votes)
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  • leafers tree style avatar for user Vishwa Patel
    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) ?
    (3 votes)
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    • sneak peak green style avatar for user LevRoz
      probably too late, but should be done with natural log, so ln(2)/0.0545, that comes from inserting values into formula whilst assuming that Principal is 1 for the sake of simplicity, 1*e^(0.0545*t)=2, hence ln(2)= 0.0545*t, coming to the answer above
      (1 vote)
  • piceratops seed style avatar for user Jess Orellanes
    Will I survive without understanding this?
    (5 votes)
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Video transcript

Let's say that we're looking to borrow $50. We can say that our principal is $50. We're going to borrow it for 3 years. Our time, let's say T in years is 3. Let's say, we're not going to just compound per year. We're going to compound 4 times a year, or every 3 months. Let's say that our interest rate ... if we were to only compound once per year, it would be 10%. Since we're going to compound 4 times a year, we're going to see ... We're going to divide this by 4 to see how much we compound each period. 10% is the same thing as 0.10. Let's write an expression. I encourage you actually to pause this video and try to write an expression for the amount that you would have to pay back if you were to do this. If you were to borrow $50 over 3 years, compounding 4 times a year, each period you would be compounding 10% divided 4%. How much would you have to pay back in 3 years? Let's write it out. $50, that's your principal. You're going to multiply that, so you could compound it. Each time, each period, each of these 3 x 4 periods. You have 3 years, each of them divide into 4 sections, so you're going to have 12 periods. Each of them you're going to compound by 1 plus this R. I'll write that as a decimal. 0.10 divided by the number of times you're compounding per year to the ... Well, you would be raising it to the nth power, if this was only over a year. There's 4 periods and you would raise it to the 4th power if it was only a year, but this is 3 years. You're going to be doing this 3 x 4. You're going to have 4 periods, 3 times. Let me write this. It's going to be 4 ... Actually, instead of N right over here let me write the 4, so you can see all the numbers. You're going to do this 4 x 3, to the 4 x 3 power. I encourage you, if you want, you could pause the video and you can use your calculator to actually calculate what that is. The whole point of this is just to use real numbers to see why this actually makes sense. This is your principal. Each time you're going to be multiplying that times 1.025. You're going to be growing it by 2 1/2% and you're going to do this 12 times, because there's 12 periods. 4 periods per year times 3 years. This is going to be how much you have to pay back. If we wanted to write this in a little bit more abstract terms, we could write this as P(1 +). I'll do this a close parentheses, since it's the same color. R over N to the N x T power. You could pick your P, your Ts, your Ns and your R and you could put it here and that's essentially how much you're going to have to pay back. An interesting thing, and you saw that we had this up here from a previous video, where we took a limit as N approaches infinity. Let's do the same thing here. Let's think about what that would mean. If we took the limit as N approaches infinity, if we took the limit of this as N approaches infinity, what is this conceptually? We're dividing our year into more and more and more chunks, an infinite number of chunks. You could really say, "This would be the case where we're doing continuous compound interest. Which is a fascinating concept to me. You're dividing your time period in an infinite number of chunks and then compounding just an infinitely small extra amount every one of those periods. You can actually come up with an expression for that. As we see, that this actually doesn't just go unbounded and give us crazy things, that we can actually use this to come up with a formula for continuously compounding interest. Which is used heavily in finance and banking and, as you can imagine, a bunch of things, actually many things outside of finance and banking, exponential growth, etc., etc. Let's see if we can actually try to evaluate this thing right over here. The one thing I am going to do to simplify this, is to do a substitution. I'm going to define a variable. The whole goal is so that I can get it into a form that looks something like this. I'm going to define a variable X. I'm going to say that X is the reciprocal of R over N, so that I can get a 1 over X right over here. I'll write that as N over R. X is equal to N over R, or we could write this as N is equal to X x R. If we make that substitution the limit is N approaches infinite. If we make the limit as X approaches infinite, then N is going to go to infinite as well. If N goes to infinite, then X is going to go to infinite as well. R, right over here, is just a constant. We're just assuming that that's a given, that N is what we're really seeing what happens as we change it. We could rewrite this thing right over here. I'm doing it. I'm not being as super rigorous, but it's really to give you an intuition for where the formula we're about to see comes from. Let's rewrite this as the limit is X approaches infinite. The limit of constant times some expression. We could take the constant out. We could say that's going to be P times the limit as X approaches infinite of 1 plus. R over N is 1 over X. 1+1 over X to the ... N is X x R. N is X x R, so let me write that, to the X x R, R x T power. All of this business is the exact same thing. Let me rewrite this. Let me copy and paste this part right over here. Copy. This is the same thing. This is equal to P times (let me put some parenthesis here) times (maybe that's too big) times the limit. This limit right over here. If I raise something to the product of these, I'm taking X x R x T, that's the same thing as doing this whole thing to the X and then raising that to the RT power. This comes from exponent properties, that you might have learned before. These 2 things are equivalent. I'm doing a couple of steps in the process here, but hopefully this seems reasonably intuitive for you. I'm really just using the property. The limit as, let's say, X approaches C of F of X to the, let's call it, to the XRT power. This is the same thing as the limit as X approaches C of F of X to the X and then all of that raised to the RT power. What is this stuff right over here? What is all of this business that's inside the parentheses? We've seen that before. All of this, all of that is equal to E. We can write this. This is exciting. This is formula for continuous compounding interest. If we continuously compound, we're going to have to pay back our principal times E, to the RT power. Let's do a concrete example here. If you were to borrow $50, over 3 years, 10% interest, but you're not compounding just 4 times a year, you're going to compound an infinite times per year. You're going to be continuous compounding. We can see how much you would actually have to pay back. It is going to be 50 x E to the ... Our rate is .1. Just let me put some parentheses here. 0.1 x time, so times 3 years. T as in years. We assumed it was in years. We get ... You would have to pay back $67. If we're to round ... $67.49 if you were to round.