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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:)

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