first order linear difference equation

Scubasteve123

just studying for a maths method test i have on tuesday and would love sum help, this is the thing thats really annoying me at the moment

a) consider the first order linear difference equation

xn+1 = Axn + B, x0 = C

where A = 0, B and C are constants. show that the solution may be expressed in the form

xn = A^nC + B((A^n - 1)/(A - 1)), A does not = 1

xn = C + Bn, A = 1
for n= 0,1,2,3,......

ok i get part a) its part b i struggle with a small bit!

b) use the result in (a), for C > 0, and determine the long term behaviour of xn when
1) A = 1/2
2) A= 1, B = 0
3) A = 2, B = 2

after this theres this small bit
c) use the result in (a) to calculate the monthly repayment on a loan of 10,000 to be paid back over 5 years at a yearly interest rate of 8% compunded every 3 months. calculate also total amount paid back (this part isnt as hard for me, i understand it a little better i think)

Michael Collins

For part (b) with A = 1/2, take the limit of the first formula for x(n), as n tends to infinity. Your answer will be in terms of B.

For part (b) with A = 1, B = 0, use the second formula for x(n), with B = 0 this formula is indepent of 'n', so that will be its long term final value (in terms of C).

for part (b) with A = 2, B = 2, take the limit of the first formula for x(n), as n tends to infinity. Hint: it may not exist! In which case, what does this say about the long term value for x?

Scubasteve123

so do u mean for A=1/2 to input it into xn = A^nC + B((1-A^n)/(1-A)) so it will look like xn = (1/2)^nC + B((1-(1/2)^n)/(1-(1/2))

i know im very behind and not much good at anything :L
and then have xn = e.g. x1 and carry on e.g. x1 = (1/2)^1C + B((1-(1/2)^1)/(1/2)) = x1 = (1/2)C +B

so wat do u do with the C then? thats were i think im having the most problem?
and after that when i have to determine the long term behaviour of xn..do i have to talk about divergent or convergent?