Ossih Posted January 25, 2015 Report Share Posted January 25, 2015 Hey guys I was doing a question on the calculus option on power series and I'm stuck on this question. I got part (a), but I'm not able to reach the conclusion in part (b). Any help would be appreciated Thanks! Reply Link to post Share on other sites More sharing options...
maroctam Posted January 28, 2015 Report Share Posted January 28, 2015 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) Reply Link to post Share on other sites More sharing options...
Ossih Posted January 29, 2015 Author Report Share Posted January 29, 2015 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 = 0Similarly, 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) 1 Reply Link to post Share on other sites More sharing options...
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