Hi Krishna, thanks for the comment! I just fixed up the two typos, appreciate the pointers :).
I’m really happy you’re finding the site useful, it’s nice to interact with people who have a similar take on learning (that it can be about understanding and not just facts).
Again, thanks for the comment!
I’m a college finite math student and I was wondering if you could help me figure out how to convert a continuous rate into APY. In the way that something like 7.23% at a continuous rate is equal to APY. I’m trying to pick this up through my math book and can not figure it out at all.
Hi Andrew, great question. When you have a 7.23% continuous rate, if you start with $1 at the end of the year you have
$1 * e^(.0723) = 1.07497778903 ~ 1.075
So 7.5% is your effective “APY”, that is your increase from the beginning of the year. Hope this helps!
Thank you for another great article that does as intended. I hope mathematicians will get the idea and follow suit.
There are two things I’d like to suggest, as they occurerd to me:
You show what happens if you compound twice in one year, how each additional compounded total itself adds to the next total to be compounded. The question I had, and I felt some students might have, is, what do you do if you want to make sure that at the end of the year the total interest does not exceed, say 50%, no matter how often it is compounded during the year. If at six months the interest of 50% makes the end result higher than a total of 50%, to make $131.25 (from the above examples), what percent would the halfway mark need to be to make the end result equal $150, i.e. 50% interest on P=$100?
What would the explanation and especially the equation be for a savings account that has an additional constant amount of money added per month, e.g. Start with $100, know the interest rate is 3%, but each month, besides the compounded interest, another $25 is added from each additional paycheck?
@Armin: Glad you enjoyed it, those are great questions.
- This is where the natural log comes into play. If you want a “final” return of 50% (that is, 1.00 becomes 1.50) you can do ln(1.50) = .405.
That means an interest rate of 40.5% will get you 50% return if it’s compounded as fast as possible. So, 40.5% is the “safe” rate you can use.
If you know you’ll only compound twice, then you can solve the equation:
(1 + r)^2 = 1.5
(1 + r) = sqrt(1.5) [take square root]
r = sqrt(1.5) - 1 [subtract 1]
r ~ .22
So, if you only compound twice (halfway & end of year) then you are safe with a rate of 22%.
- That’s a really good question. The formula is a bit more complicated because you have to account for each deposit, which come in at different times. There’s more info here:
I think that would make a good follow-up article, as a similar formula is used to calculate loan payments (you pay the same amount each month for the loan, but how to do they work out that number given the interest?).
I have this question:
How does ‘e’ relate to compound interest?
@George: Hi, e is the result of 100%, continuously compounded interest. You can try the article on e (see first paragraph) for more details.
Your explanation is fantastic.
I have maintained a blog which is basically open course ware for Masters in Renewable Energy www.ioemsre.wordpress.com . I would like to take your permission to put this article there. Please let me know.
@Ram: Hi, you can excerpt/link to the article as long as there’s attribution. Thanks!
I always wanted to learn this calculus stuff. Though I seemed to have survived the last 40 years of electronics and computer theory without it, I’ve always had a curiosity about just what all those squiggly lines were on the old chalk boards. I think you have succeeded in clearing up some of the fog (which I was unable to clear no matter how many books downloaded by http://rapid4me.com and read without understanding). so far so good anyway:) Please keep up the good work you have been doing on this web site. I really have enjoyed all of your articles.
@Helen: Thanks for the comment! I’m really happy the site was useful for you, there are so many things in my mathematical past that I want to revisit to understand better :).
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If r=12%, p=100, n=1.5 (one and half year), and annually compound, what is the payment after one and half year?
Is that 100*(1+12%)^1.5 or 100*(1+12%)(1+12%*0.5)
Could answer the above question?
Excellent communication and visual presentation. Bravo. My a-ha moment was how to read JX = X% (eg: J12 = 10%) Some how in mind’s eye, I was incorrectly visualizing when reading J12 = 10% to mean 10% was being earned every one of those 12 periodic periods.
After reading your excellent website, I now understand (continuing using J12 as the example) that J12 really means that interest is being paid 12 times through out the year and each time it is paid, it is being added to the principle to be put to work to earn additional interest as time marches to reach the one year milestone. Furthermore, it also explains my second mystery as to why the nominal rate is divided by the number of periodic periods. Why? Because the J12 is focusing on the frequency of interest being paid in the year and is independent of the nominal rate. Therefore if there are 12 periodic periods, this means the interest of 10% is divided by 12 to equal .8333333 per periodic period etc. Thank you. Thank you. One last question…What does “J” stand for??? What is its history? I am very curious to knowing the history of “J” Thank you very much. You have been a great help. Jan
no one has explained it to me so clearly…thanks a lot
Great explanation, you have the knack for getting to the intuitive core of concepts!
One small quibble with the “cheatsheet” table at the beginning: t is being used in two different senses, which could confuse some (it did me). The compound formula uses t as the number of periods per year (n), whereas the continuous growth formula uses t as the total number of periods.
When you take the limit of (1+r/t)^tn, you get e^rn (not e^rt). I realize you’re sticking to the conventional lettering, but I think it’s confusing.
@Jan: Thanks for the comment! Just summarizing from email, but I think the J notation may have evolved due to the need for another letter (not i) to represent an item. In math it’s very common to say the “ith” element, but since that was taken “j” may have been a fit. Not sure though!
@Toban: Great feedback – we should compound “n” times for year and do that for “t” periods of time to keep it consistent. Thanks!
in the formula for compound interest in the brackets there is the number 1, could you tell me what that represents or what is it’s significance
@ken: Hi, the formula for compound interest is
P * (1 + r)^n
where P is your initial amount, r is your interest rate (10% = .10), and n is the number of years. The 1 is needed because your interest rate is a percentage increase from 1.0 (i.e, 1.0 + .10 = 1.1, which is a 10% increase over 1.0).
You could write
P * (T)^n
where T is the “total” amount after 1 year (1.1). But, it’s usually easier to see the interest rate separated out in each formula.
[…] Examples make everything more fun. A quick note: We’re so used to formulas like 2^x and regular, compound interest that it’s easy to get confused (myself included). Read more about simple, compound and continuous growth. […]