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Calculus Option Question (Power Series)


Ossih

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haven't tried this (and you probably have), but this may help: use ak=1 in part a)

 

would you care to show what you did for the first part? (i'm feeling a bit lazy to try myself haha)

 

Hiya!

 

I did manage to get part (b) too, so here goes.

 

Part (a)

 

(1 + Ax + Bx2) ∑ akxk

 

         a0 + a1x + a2x2 + a3x3 + a4x4 + ...

           + Aa0x + Aa1x2+ ...

                      + Ba0x2 + ...

 

Now, ak + Aak-1 + Bk-2 = 0 

So a2x2 + Aa1x2 + Ba0x2 = 0

Similarly, the next 'column' will cancel out and so on.

 

So the sum = a0 + a1x + Aa0x = a0 + (a1 + Aa0)x

 

Part (b)

 

∑ xk is ∑akxk with ak = 1

 

If I look closely at the right hand side of what I want, I see its 1/(1-x)

 

I also know that ∑akxk = [a0 + (a1 + Aa0)x] / (1 + Ax + Bx2) [if the condition that ak + Aak-1 + Bk-2 = 0  is satisfied]

 

To get 1-x in the denominator, I'd ideally want an A = -1 and B = 0

 

For the sequence where ak = 1, A = -1, B = 0 

 

ak + Aak-1 + Bk-2 = 1 + (-1)(1) + (0)(1) = 0. Therefore, this condition is satisfied.

 

So, ∑xk = [1 + (1-1)x] / [1 + (-1)x + 0x2] = 1/(1-x)

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