Help with linear homogeneous recurrence relations

Bucs44
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Homework Statement


Here's my problem - Give the order of linear homogeneous recurrence relations with constant coefficients for: An = 2na(n-1)



The Attempt at a Solution

I have no idea on how to start this problem - Any help would be greatly appreciated.
 
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First, I think you mean An= 2An-1. Be careful to distinguish between "A" and "a"!

Notice that this problem does not (yet) ask you to solve the equation! It just asks that you state its order. Do you know the definition of "order" of a recurrence relation? I suspect the way to "start this problem" is to look up "order"!
 
=2(2An-1 + 1) + 1
=2^2An-1 + 2 + 1

Is this right?
 
The order is just the number of "previous" terms, in which case the order is 2
 
No, in the recursion An= 2An-1, An depends on the value of A one place before it. The order is 1.

As for
=2(2An-1 + 1) + 1
=2^2An-1 + 2 + 1

I can't tell whether it is correct or not because you haven't told me what it is supposed to equal!

Once again, is this intended to be An= 2Sn-1? If so, I cannot see where you are getting the "+1" terms from.

Suppose A0= 1. What is A1? A2?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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