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Math Help - Showing LHS = RHS

  1. #1
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    Showing LHS = RHS

    Ok, i have go to a point working through a question where i am having to show that the LHS = RHS but cant seem to.... is the following possible or have i gone wrong somewhere earlier in my calculations?

    If P(Ak) = a + b* ({q/p})^k

    Then use the equivalent formula for P(Ak+1) and P(Ak-1) and show that....

    P(Ak+1) =
    \frac{P(Ak) - q*P(Ak-1)}{p}
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by sirellwood View Post
    Ok, i have go to a point working through a question where i am having to show that the LHS = RHS but cant seem to.... is the following possible or have i gone wrong somewhere earlier in my calculations?

    If P(Ak) = a + b* ({q/p})^k

    Then use the equivalent formula for P(Ak+1) and P(Ak-1) and show that....

    P(Ak+1) =
    \frac{P(Ak) - q*P(Ak-1)}{p}
    Something isn't right about this "equality" as I work it out. What is the original question?
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    Iv uploaded the relevant document for the question. Sorry, i probably didnt include all relevent information of the part i was asking.... i have however managed to do that proof, part (iii), but i am now stuck on part (iv)!

    Iv got those boundary conditions A0 = 1 and AN = 1, but im not compeltely sure how to set up the proof.... once iv got it set up, the algebra part should be ok....
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