Expert: Josh - 1/6/2011

Question
QUESTION: Hello:

The compound interest calculation is p(1 + r)^n. This calculation is a simplified calculation of (p + pr)^n.

The annual percentage yield (APY) calculation is (1 + periodic rate)^n - 1, where n equals the number of periods. What is the expanded calculation for this calculation?

For example, p(1 + r)^n equals (p + pr)^n.  What would (1 + periodic rate)^n - 1 be equivalent to, the expanded version?

I thank you for your reply.

ANSWER: Kenneth,

You said "Thanks for the complex reply, but you did not answer my question."
Quite the contrary, I have answered your question fully in the time that I have.

First, I pointed out that your premise "p(1 + r)^n equals (p + pr)^n" is actually incorrect.
Second, I rephrased your question, writing "(1 + periodic rate)^n - 1" as "(1+r)^n -1".
I then ask you to consider the expression (1+r)^n. Writing this explicitly term-by-term requires what is known as "binomial expansion". There is no getting around this.

There is no simple way to explain this. Of course, the answer will involve some algebra which you may not like. But this is what you have asked for. I provided a complete formula for the expansion, which reads
| (1+r)^n = 1 + C(n,1)*r^1 + C(n,2)*r^2 + C(n,3)*r^3 + ... + C(n,n)*r^n
|         = 1 + n*r + [n*(n-1)/2]*r^2 + ... + [n*(n-1)/2]*r^(n-2) + n*r^(n-1) + r^n
This is what "(1+ periodic rate)^n" is equivalent to.

Obviously, the expansion becomes more convoluted as the integer n gets larger.
The only thing in this symbolic expression that might cause confusion are the coefficients C(n,j). It is possible to write this using a summation notation, which would shorten the expression, but I am not sure if you are familiar with it. So, I wrote this long hand, using C(n,j) which is a standard notation.

- Note: C(n,j) correspond to the numbers in the Pascal triangle, if you are interested in number patterns. You can find pictures of this from many websites (just search for Pascal triangle).
- I have also given you a numerical example, for the simple case where n=3. The expansion of (1+r)^3 then becomes 1+3*r+3*r^2+r^3. You may not like the complexity of this, but this is what you asked for. The general answer to your question involves the coefficients C(n,j) which may be defined using factorials, or by looking up the coefficients in a Pascal triangle.

Finally, remember that this may not resemble the thing you were looking for. You may have expected something which looks like (p+pr)^n, but p(1+r)^n does NOT equal (p+pr)^n in the first place.



Kenneth,

What you quoted is actually incorrect. p(1+r)^n is not the same as (p+pr)^n.
In fact, they differ by a factor of p^(n-1), which is enormous for any sizable n.

e.g., If p=1253, r=0.06 and n=7, then p*(1+r)^n = 1884.04, whereas (p+p*r)^n = 7.29 x 10^21.

What you are looking for is the binomial expansion (see http://en.wikipedia.org/wiki/Binomial_theorem for details). In general, the (1+r)^n part will look something like this:
(1+r)^n = 1 + C(n,1)*1^(n-1)*r^1 + C(n,2)*1^(n-2)*r^2 + C(n,3)*1^(n-3)*r^3 + ... + C(n,n)*r^n, where C(n,j) is defined as n!/[(n-j)!*j!] using the factorial notation n!=n*(n-1)*(n-2)*...*2*1.
[e.g., 8!=8*7*6*5*4*3*2*1. Also, 0!=1 by definition.]

Since one raised to the power of anything is 1, the above expression may be simplified as
(1+r)^n = 1 + C(n,1)*r^1 + C(n,2)*r^2 + C(n,3)*r^3 + ... + C(n,n)*r^n
       = 1 + n*r + [n*(n-1)/2]*r^2 + ... + [n*(n-1)/2]*r^(n-2) + n*r^(n-1) + r^n
Notice that the coefficients are symmetrical.

Thus, a direct expansion is not advisable for any integer n of significant size. In the case of n=3, we have a cubic polynomial (1+r)^3 = 1 + 3*r + 3*r^2 + r^n.



---------- FOLLOW-UP ----------

QUESTION: Thanks for your reply.

When I indicated that you did not answer my question, I am referring to how the APY calculation became (1 + periodic rate)^n - 1. Is this not the result of factoring out some common factor from an expanded calculation such as the calculation  (p + pr) becomes p(1 + r)^n. I guess what I am curious to know is what do the 1's represent in the APY calculation?
If you answered my question, I apologize because I do not fully understand your explanation because of the complexity of your answer and the mathematics involved!

I thank you, however, for your efforts and willingness to help!

Answer
Hi Kenneth,

Your original question asks what (1 + periodic rate)^n becomes, not where it comes from. These are different things.

For convenience, once again, let us replace "periodic rate" with r.

Just so that we are absolutely clear on this, let me point out that the expression (1 + r)^n  means multiplying (1 + r) by itself (n-1) times. i.e., (1+r)^n = (1+r)*(1+r)*(1+r)*...*(1+r),  you will perform such multiplication (n-1) times.

I still get the impression that you believe p(1 + r)^n comes from (p + pr)^n. Let me say unequivocally that this is NOT true. If we start with (p+pr)^n, factorizing the content inside the parenthesis, we get (p*(1+r))^n. This simplifies to p^n*(1+r)^n; which is certainly DIFFERENT to p*(1+r)^n.

I am not sure what answer you were expecting, when you asked for the expansion of (1 + periodic rate)^n. If you are interested in the expansion of (1+r)^n, then there is no escaping from the "binomial theorem". I won't go over the binomial expansion again, but you will see a pattern.

When n=1, (1+r)^1 = 1+r.
When n=2, (1+r)^2 = 1+2*r+r^2.
When n=3, (1+r)^3 = 1+3*r+3*r^2+r^3.
When n=4, (1+r)^4 = 1+4*r+6*r^2+4*r^3+r^4.
When n=5, (1+r)^5 = 1+5*r+10*r^2+10*r^3+5*r^4+r^5 and so forth.

If you focus only on the coefficients, you will rediscover the coefficients in the Pascal's triangle (see below). One property is that the coefficients in the expansion are symmetrical. See [http://en.wikipedia.org/wiki/Pascal%27s_triangle] if we want to learn more.

For n=1, {1,1}
For n=2, {1,2,1}
For n=3, {1,3,3,1}
For n=4, {1,4,6,4,1}
For n=5, {1,5,10,10,5,1}

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Now, addressing your second point "curious to know is what do the 1's represent in the APY calculation". Without going into any finance literature, this is my take on this.

First, compound interest is calculated using the formula P*(1+r)^n.
So, after n periods, the total amount is given by P*(1+r)^n.
The net increase is therefore P*(1+r)^n - P.
This may be factorized as P*[(1+r)^n - 1].
So, regardless of the principal, the growth factor (which I guess, this is what you would call the annual percentage yield) is given by (1+r)^n - 1.

Consider an example.
Say, r=0.10 and n=3.
Then, (1+r)^n = (1.1)*(1.1)*(1.1) = 1.331.

If P=10 dollars, then, you end up with 10*1.331 dollars.
Subtracting the principal (P) from this we get 10*1.331 - 10.
This may be factored as 10*[1.331-1].
All that really matters is [1.331-1]=0.331
Whatever P is, it will be increased by a factor of 0.331 under said conditions.