Recent content by swtlilsoni

Homework Statement I am doing a power series solution for: (x^2-1)y" + 8xy' + 12y = 0 I rewrote it in terms of power series and transformed everything into one series and finally ended up with the following recurrence relation: an+2= ((n+3)(n+4)an)/((n+2)(n+1)) I plugged in values for n...

Okay thank you. I did not know I had the freedom to choose any point. I thought the answer varied depending on the chosen point so I thought it needed to be specified.

Right given an equation p(x)y" + q(x)y' + r(x)y=0, an ordinary point is any point at which p(x0) does not equal 0. Thus in this equation, the singular points are -1, and 1. Does that mean I can choose any number that works as an ordinary point and plug it in for x0? I can choose 9? And plug...

Homework Statement Using a power series solution, what is the solution to: (x^2-1)y" + 8xy' + 12y = 0 Homework Equations Normally these questions specify (about x0=0) but this one doesn't specify about which point. So if I use the power series equation, what am I supposed to plug in...

good morning! okay so it seems like these proofs are to prove a given number is fibonacci? so do you mean I should solve for the number in terms of n, then try to use one of those proofs to show it is fibonacci?

Here is what my professor wrote (I don't completely understand what he did): An= c1\alphan + c2\betan c1,c2= 3 \pm \sqrt{}15 Bn= c3\deltan + c4\varpin c3,c4= (5 \pm \sqrt{}61)/2 3+\sqrt{}15=\alpha=6.8 3-\sqrt{}15=.9n = approaching zero (5 + \sqrt{}61)/2 = \delta = 6.5 (5 - \sqrt{}61)/2...

it's because it has to be multiplied. For every rearrangement of five identicals, there are two more rearrangements of the others

Homework Statement An+2=6An+1+6An A1=A2=1 Bn+2=5Bn+1+9Bn B1=B2=10100 a) is it true that Bn>An for every integer n > 0? b) is it true that Bn>An for infinetly many integers n>0? The Attempt at a Solution It just seems like these are increasing functions since they both start with...

If you are given a sequence of integers such as: An=xn+yn where x and y are integers. and n=0,1,2,3... how would one find the recurrence relation? I tried writing An+1 in terms of An but it doesn't come out neatly because it doesn't translate so well. And there are terms raised to the n+1...

Ohh okay so it would be 5!3!

Yes, the fibonacci formula I know. It is this: http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html right? But I tried writing it in terms of the formula, but there was no way of showing whether the end result is a fibonacci or not.

I know how to solve second order ones, but how would you solve third order ones? Because the characteristic polynomial would have a third degree so how can one find the roots? I have looked everywhere online to find out but I can't find anything. Please Please tell me!

Homework Statement An=3An-2-2An-3 When A0=3 A1=1 A2=8 I tried to solve it normally like a normal recurrence relation however since it is not A sub n-1, it turns into a polynomial where the variable is raised to the third power which I couldn't factor and the whole thing turned into a mess.

because there are three sets of five identicals

Homework Statement Is: (Fn+1+Fn-1)Fn always a Fibonacci? The Attempt at a Solution I have no clue! I know I'm supposed to show work and all but I'm so lost, any direction would be appreciated. Even if you just give me a link to read.