Recent content by stanyeo1984

Oh, so finding the constants and placing them back in the general equation constitutes as solving the equation? - - - Updated - - - Or does question b ask for something else?

Managed to work that one out:D, but what do they mean when they ask me to solve the recurrence relation?

Sorry yes, missed the - sign, but if i place these in the place of the variable in the general solution, won't i get a closed formula? And if that's correct, what do I do when the question asks me to solve?

From here we get, C1=0, C2=0, C3=1 and C4=1

And now I add that to the end of the characteristic solution and solve for C1 and the other constants?

I solve for A=1/3 and therefore, pn= 1/3n3n

I'm only familiar for the method of undetermined coefficients for DE not so much for this, is it possible for you to explain?

I'm thinking that the particular solution pn would be 64.C53n-4

Is the general form C13n+C2(-1)n+C2(-1)n+C2(-1)n Otherwise, I don't know how to deduce the general solution:/

hello, n>=4 just means greater than or equal to 4, I think i got the characteristic roots as (x-3)(x+1)3 But I got stuck after that

Recall that the Fibonacci sequence is defined by the initial conditions F0 = 0 and F1 = 1, and the recurrence relation Fn= Fn-1 + Fn-2 for n >= 2. (a) Let F(z) = F0 +F1z + F2z2 + F3z3 + ··· be the generating function of the Fibonacci numbers. Derive a closed formula for F(z). (b) Consider the...

Recall that the Fibonacci sequence is defined by the initial conditions F0 = 0 and F1 = 1, and the recurrence relation Fn = Fn−1 + Fn−2 for n > 2. (a) Let F(z) = F0 + F1z + F2z 2 + F3z 3 + · · · be the generating function of the Fibonacci numbers. Derive a closed formula for F(z). (b) Consider...

How do I solve this 1. (a) Solve the recurrence relation an =6an−2 +8an−3 +3an−4 +64·3^n−4, n􏰀>=4 where a0 =0,a1 =1,a2 =4 and a3 =33. (b) Write down a closed form of the generating function of the sequence an.