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Binomial expansion with negative power

A question in the book seems to be incorrect. My expansion answer didn’t work out as expected so I turned to the solution bank that did agree. However, when I came to try to use th q(0.1) part the values of actual and approximation were miles apart. So have I done something wrong or does the binomial expansion sometimes not work.
Thanks
Reply 1
The question and working refer to two different questions? The series expansion for the posted question is correct (well the first two given terms are)
https://www.wolframalpha.com/input?i=series+at+x%3D0+%289x%5E2%2B26x%2B20%29%2F%28%281%2Bx%29%282%2Bx%29%29

Edit - the wolfram link is losing "+" terms again, just put them in the orange box and hit enter and Ill edit the link later when I remember how to correct the plus symbols.
(edited 1 month ago)
Hi I'm losing it aren't I. Yes, I've sent the photo of one question and the answer to a different one. And now, although the % error was not required in the actual question, I've tried it and it's much nearer. Thank you so much and very well spotted.
Reply 3
Original post by maggiehodgson
Hi I'm losing it aren't I. Yes, I've sent the photo of one question and the answer to a different one. And now, although the % error was not required in the actual question, I've tried it and it's much nearer. Thank you so much and very well spotted.

Just post your new working if youre still unsure.
Original post by mqb2766
Just post your new working if youre still unsure.

Thanks, but I'm fully with it now. I was comparing my answer to question 15 in Mixed Exercises C whilst actually attempting the Paper 2 question 9. No wonder they didn't match. All is sorted now. Silly me.
Reply 5
Original post by maggiehodgson
Thanks, but I'm fully with it now. I was comparing my answer to question 15 in Mixed Exercises C whilst actually attempting the Paper 2 question 9. No wonder they didn't match. All is sorted now. Silly me.

Good. Im not too sure of how youre taught it, but another way (for the wrong question) is
1/2 * (9x^2+26x+20) * (1 + (3x+x^2)/2)^(-1)
and you can do the binomial series for (1 + (3x+x^2)/2)^(-1) where the usual x is replaced by (3x+x^2)/2 and then multiply that by (9x^2+26x+20) (and 1/2) to get the result

Id guess they want you to do it your way as theyve given you the denominator factorised. Its a bit of an algebra slog whichever way you do it.
(edited 1 month ago)
Original post by mqb2766
Good. Im not too sure of how youre taught it, but another way (for the wrong question) is
1/2 * (9x^2+26x+20) * (1 + (3x+x^2)/2)^(-1)
and you can do the binomial series for (1 + (3x+x^2)/2)^(-1) where the usual x is replaced by (3x+x^2)/2 and then multiply that by (9x^2+26x+20) (and 1/2) to get the result
Id guess they want you to do it your way as theyve given you the denominator factorised. Its a bit of an algebra slog whichever way you do it.

There is a slight danger to expand that way, because of radius of convergence issues.

But all is good here - we just need to justify that |(3x+x^2)/2| < 1 indeed implies |x|<1, which is probably not required in A-Levels anyway.
Reply 7
If you like differentiation, this can (have to make a check though, at 0 can be differentiated) be attempted by McLaurin's series. Maybe for further maths.
Original post by tonyiptony
There is a slight danger to expand that way, because of radius of convergence issues.
But all is good here - we just need to justify that |(3x+x^2)/2| < 1 indeed implies |x|<1, which is probably not required in A-Levels anyway.

It doesn't really matter; regardless of what approach you take - as long as you get a series with a non-zero radius of convergence, it has to match the Taylor series (and therefore have the same radius of convergence).

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