# math help plz!

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j=6.50%
m=Daily(365)
f(effective rate)=?

f=(1+i)^n
=(1+(.0650/365))^365
=1.067152848-1
=0.067152848
=6.72%
i know how to find the "f"

but if its

j=?
m=Quarterly(4)
f=3.25%

how do i find the "j"???

• math help plz! -

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.

• math help plz! -

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%

• math help plz! -

Thnx Reiny,it is very clear to me.now i can continue the rest which has the similar questions.thnx mathmate too:)