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FP2 maclurin series help

http://www.ocr.org.uk/Images/65004-question-paper-unit-4726-01-further-pure-mathematics-2.pdf

I'm a bit stuck on question 2. So I've found the macluren series for e2xe^{2x} and for sin(x)sin(x). But then I'm not to sure what to do, I expanded out the binomial to get the first 3 terms not sure if it's right...
Post your working.

Work out the expansion of e2x(1+sin⁡x)e^{2x} (1+\sin x) first though (multiply your previous results).
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Music99
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Original post by Mr M
Post your working.

Work out the expansion of e2x(1+sin⁡x)e^{2x} (1+\sin x) first though (multiply your previous results).


So we have e2x=1+2x+2x22!+... e^{2x}=1+2x+\frac{2x^2}{2!}+...
sin(x)=x−x33!+x55!+...sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...

So (1+sinx)e2x=e2x+sinxe2x=1+2x+2x22!+...+(x−x33!+x55!)(1+2x+2x22!)(1+sinx)e^{2x}=e^{2x}+sinxe^{2x}=1+2x+\frac{2x^2}{2!}+...+(x-\frac{x^3}{3!}+\frac{x^5}{5!})(1+2x+\frac{2x^2}{2!})
Original post by Music99
So we have e2x=1+2x+2x22!+... e^{2x}=1+2x+\frac{2x^2}{2!}+...
sin(x)=x−x33!+x55!+...sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...

So (1+sinx)e2x=e2x+sinxe2x=1+2x+2x22!+...+(x−x33!+x55!)(1+2x+2x22!)(1+sinx)e^{2x}=e^{2x}+sinxe^{2x}=1+2x+\frac{2x^2}{2!}+...+(x-\frac{x^3}{3!}+\frac{x^5}{5!})(1+2x+\frac{2x^2}{2!})


Blimey you are making that too complicated. You don't need so many terms. Read the question.

Add 1 to your expansion to sin x to find the first bracket.

Multiply the terms and simplify.
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Music99
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Original post by Mr M
Blimey you are making that too complicated. You don't need so many terms. Read the question.

Add 1 to your expansion to sin x to find the first bracket.

Multiply the terms and simplify.


Yeah I was thinking that as I was typing it out. So then for the first bracket it's just 1+x−x33!1+x-\frac{x^3}{3!}?
Original post by Music99
Yeah I was thinking that as I was typing it out. So then for the first bracket it's just 1+x−x33!1+x-\frac{x^3}{3!}?


Yes. Simplify 3! to 6. You won't need that x^3 term in the end.
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Music99
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Original post by Mr M
Yes. Simplify 3! to 6. You won't need that x^3 term in the end.


Okay, I'll play around with it and post back If I have any questions or get stuck. Thanks for the help. I really need to get the textbook it sucks not having it aha.
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Music99
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Original post by Mr M
Yes. Simplify 3! to 6. You won't need that x^3 term in the end.


I got it right :smile:. Thank you. Would you mind helping me on question 5 also?

I'm on the first part. I've integrated by parts to get

(1−2x)n(ex)+2n∫(1−2x)n−1(ex)dx(1-2x)^n(e^x)+2n\int(1-2x)^{n-1}(e^x)dx Limits are 1/2 and 0,but I don't know how to latex them. I simplified it down to by saying the integral is just 2nIn−12nI_{n-1} , although I'm not sure if that's right, or do I just integrated again?

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