Expert: Ahmed Salami - 1/6/2011

Question
QUESTION: Hello:

If I'm not mistaken, the compound interest calculation is p(1 + r)^n. This calculation is a simplified calculation of (p + pr).

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: Hi Kenneth,
P(1 + r)^n is not the compound interest but the new amount after n periods.
A = P(1 + r)^n
Therefore,
profit = A - P
= P(1 + r)^n - P
= P[(1 + r)^n - 1]
And we can see that (1 + r)^n - 1 is the percentage of the principal P that gives the profit.

Regards

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

QUESTION: Hello Ahmed:

I apologize for my lack of mathematical knowledge, but I need some further clarification. I still do not understand the following:

"Profit = A - P
= P(1 + r)^n - P
= P[(1 + r)^n - 1]
And we can see that (1 + r)^n - 1 is the percentage of the principal P that gives the profit."

Why is the "P" missing from (1 + r)^n - 1, = P[(1 + r)^n - 1]
equals (1 + r)^n - 1?

P(1 + r)^n - P was once (P + rP)^n - P. Is this correct?

I thank you for your reply and willingness to help!

Answer
Hi Kenneth,
Yes, P(1 + r)^n - P was once (P + rP)^n - P.
Now for the percentage thing. Suppose you have a principal P that is said to yield a percentage of x%. Isnt it just true then that the yield can be written as x%.P?
So, its just the same thing with the expression
Profit = P[(1 + r)^n - 1]
since in this case [(1 + r)^n - 1] is the coefficient of P that gives the profit.
I have to say though that [(1 + r)^n - 1] is in its simplest form and any expansion would seem rather tedious.

Regards