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Math Help - Comparison of simple and compounded interest function

  1. #1
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    Comparison of simple and compounded interest function

    Hello everyone,

    I had an assignment that one of my actuarial math teacher gave me and there is a question that need to be proved.

    6. Assume that 0<i<1. Show that

    a) (1+i)^t < 1+it for 0<t<1

    b) (1+i)^t > 1+it for 1<t

    HINT: if f(a)=0 and f'(x)>0 for a<x<b then f(b)>0


    The problem must be solved with the HINT he gave us.

    Thank for your help.
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  2. #2
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    Quote Originally Posted by solvj View Post
    Just to see where you're at, do you agree with following:


    > 6. Assume that 0<i<1.


    12% annual compounded monthly means i = .01,
    and 1.01^12 is the future value of $1 after 1 year,
    and the effective annual rate becomes 12.6825%

    >a) (1+i)^t < 1+it for 0<t<1


    If rate = 12% compounded annually, then i = .12, and above means
    the future value of $1 after a period of time less than 1 year.

    > b) (1+i)^t > 1+it for 1<t


    If rate = 12% compounded annually, then i = .12, and above means
    the future value of $1 after a period of time greater than 1 year, and
    if the future value is to be after 3 years and 7 months, then the
    value of t would be 3 + 7/12 = 43/12.


    .
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  3. #3
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    This can be an exemple but it doesn't prove that

    a) (1+i)^t < 1+it for 0<t<1
    or
    b) (1+i)^t > 1+it for 1<t

    I need a proof for "t" in the real number

    Thank
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  4. #4
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    What I said has NOTHING to do with a proof.
    I asked if you agreed with (thus understood) what I said.
    If not, not much use working on a proof.

    Also, I'm a little perplexed with your subject title:
    "Comparison of simple and compounded interest function".
    There is nothing in your post that refers to simple interest.
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  5. #5
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    Sorry for the misunderstood, but yes I agree with what you said.
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