# Showing LHS = RHS

• Feb 26th 2010, 08:42 AM
sirellwood
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) =
$\frac{P(Ak) - q*P(Ak-1)}{p}$
• Feb 26th 2010, 09:00 AM
Chris L T521
Quote:

Originally Posted by sirellwood
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?
• Feb 26th 2010, 10:16 AM
sirellwood
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....