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# Maclaurins Series Expansions, FP3 watch

1. I've been doing the miscallaneous exercises from the AQA FP3 textbook, and I'm really stuck on the last question about Maclaurin's series. It asks you to show that the series for 1/(sin 2x) is 1/(2x) + (1/3) x + (7/45) x^3, having already found sin (2x). I tried rewriting it as (1+(sin(2x) -1))^(-1) and using binomial series, but it was impossible to calculate as each term in x changed each time you added a new term, so it would change infinitely. Also, isn't f(0) for this function undefined? So you couldn't get a series anyway?
Can somebody please explain this to me!
Thanks
p.s. It then asks for 1/(1-e^(-x)), having already found 1-e^(-x), which I can't do either.
2. (Original post by lambda98)
I've been doing the miscallaneous exercises from the AQA FP3 textbook, and I'm really stuck on the last question about Maclaurin's series. It asks you to show that the series for 1/(sin 2x) is 1/(2x) + (1/3) x + (7/45) x^3, having already found sin (2x). I tried rewriting it as (1+(sin(2x) -1))^(-1) and using binomial series, but it was impossible to calculate as each term in x changed each time you added a new term, so it would change infinitely. Also, isn't f(0) for this function undefined? So you couldn't get a series anyway?
Can somebody please explain this to me!
Thanks
p.s. It then asks for 1/(1-e^(-x)), having already found 1-e^(-x), which I can't do either.
Perhaps this would be better off in the Maths forum.
3. (Original post by lambda98)
I've been doing the miscallaneous exercises from the AQA FP3 textbook, and I'm really stuck on the last question about Maclaurin's series. It asks you to show that the series for 1/(sin 2x) is 1/(2x) + (1/3) x + (7/45) x^3, having already found sin (2x). I tried rewriting it as (1+(sin(2x) -1))^(-1) and using binomial series, but it was impossible to calculate as each term in x changed each time you added a new term, so it would change infinitely. Also, isn't f(0) for this function undefined? So you couldn't get a series anyway?
Can somebody please explain this to me!
Thanks
p.s. It then asks for 1/(1-e^(-x)), having already found 1-e^(-x), which I can't do either.

I feel like there is a trick here, finding the it normally I think is impossible since as you said f(0) would be undefined. The term that has gotten my interest is 1/2x which I think carries a lot of weight here since it is abnormal.
4. I think I have it, the 1/2x term should be a major clue however I could see why you don't immediately see it.

Maclaurin expansions only exist for functions which are defined at f(0) and since 1/sin(2x) isn't defined at 0 a standard Maclaurin expansion doesn't exist for it, another approach would be finding the series of a similar function which does exist at x = 0 and then manipulating every term in the series to get the original function.

Spoiler:
Show

using that for small x.

5. (Original post by SeanFM)
Perhaps this would be better off in the Maths forum.
Oh right yeah thanks!
6. i used the McLauren expansion for sin(2x)... then applied the binomial expansion to the first few terms of the reciprocal , having factorised out 1/2x to give 1 as the first term.
7. Moved to maths.
8. this expansion is nonsense about 0
9. This question is far simpler than you are thinking it is.

Calculating the series directly is beyond A level - you need to calculate the Laurent series for that.

BUT you have all the information you need from the series for Sin(2x) itself.

Just take the reciprocal of that series :

factor out 2x

and then expand the term in brackets as a binomial series. Then divide each term from that by the 2x you factored out at the start.
10. (Original post by kc_chiefs)
This question is far simpler than you are thinking it is.

Calculating the series directly is beyond A level - you need to calculate the Laurent series for that.

BUT you have all the information you need from the series for Sin(2x) itself.

Just take the reciprocal of that series :

factor out 2x

and then expand the term in brackets as a binomial series. Then divide each term from that by the 2x you factored out at the start.
Thank you!

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