Posted by Amy on Saturday, March 27, 2010 at 5:36pm.
Same as your example, but solve for j.
In your example, n=365, and in this case n=4 (quarterly).
f=(1+j/n)^n
f=(1+j/4)^4
1.0325=(1+j/4)^4
You can solve for j by taking the fourth root on both sides (square-root twice) to get
1+j/4 = fourth-root(1.0325)
the rest is standard algebra.
Looking at your procedure, I conclude that you are finding the equivalent annual rate for a given rate compounded daily.
I don't like the way you are writing it up. You are writing down steps connected by equal signs when they are not equal.
e.g.
f=(1+i)^n
=(1+(.0650/365))^365
=1.067152848-1
the second and third lines are NOT equal. This may sound picky to you, but math is exact.
I used to teach it this way:
let the annual rate be i
then
(1 + i = (1 + .065/365)^365
1 + i = 1.067152848
i = .067152848
effective annual rate = 6.72%
for your last question, I will assume that it asked the following,
"What annual rate compounded quarterly is equivalent to an annual rate of 3.25%
Let the quarterly rate be j
then ...
1.0325^1 = (1+j)^4
take the 4th root of both side
1.0325^(1/4) = 1+j
1.008027813 = 1+j
j = .008027813
4j = .032111251
the annual rate compounde quarterly will be 3.211%
Thnx Reiny,it is very clear to me.now i can continue the rest which has the similar questions.thnx mathmate too:)
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