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I've managed to show it is zero stable and not consistent, and that it is not convergent because zero-stable + consistent = convergent. Now I'm stuck with the part bracketed in red. How would I choose tau and find a general solution for U^n?

Do I just take tau=1/N and so then N=1/tau?

For the general solution for U^n I tried doing:

U^0 = 0

U^1 = tau/3

U^2 = U^1 (from the 1st equation, left hand side) = tau/3

U^3 = U^2 = tau/3

...

U^n = (n/n) tau/3. Is this correct?

Many thanks!

By the way, how would I go about showing the limit as required in the question?

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You can write formulas here by including $\TeX$ code in dollar signs, for instance $\lim_{\tau\to0}U^N\ne u(1)$. You may find people are more willing to answer your question if you make the formulas more easily readable. If you don't know how to write something in $\TeX$, you can look around the site; you can right-click on any formula you see and select "Show Source" to see the $\TeX$ code for it. –  joriki Nov 7 '11 at 9:02
    
Alright thanks, i will try to look at it soon! –  John Southall Nov 7 '11 at 21:25

1 Answer 1

up vote 1 down vote accepted

Everything seems fine until where you write "U^2 = U^1 (from the 1st equation, left hand side) = tau/3". Perhaps this is a typo, but when you put n = 0 in the first equation, you should get "U^2 - U^1 = tau/3". From this you can determine U^2. Continue to get U^n.

For the limit, fill in the general formula you derived for U^N. This will be something that depends on both N and tau. However, N and tau are related by N * tau = 1. Use this to get rid of N so that you have an expression that only depends on tau. And then take the limit as tau goes to zero.

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Thanks, I've solved it now! Yeah.. i think I made a mistake someone in my handwritten workings, thanks again. Cheers! –  John Southall Nov 7 '11 at 21:24

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