munchkin1212
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Hi!

The question is by expanding (1-4x^2)-(1/2) and integrating term by term, find the series expansion for arcsin(2x) as far as the term x^7.

I know how to do normal Maclaurin expansion but I keep getting stuck here. Each time I differentiate, I just end up with all of the f'(0)= 0 which means there isn't a proper expansion.

Any help??
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RDKGames
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(Original post by munchkin1212)
Hi!

The question is by expanding (1-4x^2)-(1/2) and integrating term by term, find the series expansion for arcsin(2x) as far as the term x^7.

I know how to do normal Maclaurin expansion but I keep getting stuck here. Each time I differentiate, I just end up with all of the f'(0)= 0 which means there isn't a proper expansion.

Any help??
So you have \displaystyle \frac{1}{\sqrt{1-(2x)^2}}, what is your expansion? You're told to integrate each term, not differentiate them. Also, might help to write down the derivative of \arcsin(2x)
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munchkin1212
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(Original post by RDKGames)
So you have \displaystyle \frac{1}{\sqrt{1-(2x)^2}}, what is your expansion? You're told to integrate each term, not differentiate them.
I don't know how to expand it - Binomial expansion?
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RDKGames
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(Original post by munchkin1212)
I don't know how to expand it - Binomial expansion?
Yes

EDIT: If you MUST use MacLaurin's and you've failed to apply it, post your working because it should work.
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munchkin1212
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(Original post by RDKGames)
Yes

EDIT: If you MUST use MacLaurin's and you've failed to apply it, post your working because it should work.
It's much simpler now I've used Binomial, thanks.
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munchkin1212
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(Original post by RDKGames)
Yes

EDIT: If you MUST use MacLaurin's and you've failed to apply it, post your working because it should work.
In another question, it says:

By integrating 1/(1+x^2) and its Maclaurin expansion, find the Maclaurin series for arctanx.

I'm struggling to find the Maclaurin series for (1+x^2)^-1

Each time I differentiate I just get f'(0) and f''(0) to equal 0. Where am I going wrong?
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RDKGames
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(Original post by munchkin1212)
In another question, it says:

By integrating 1/(1+x^2) and its Maclaurin expansion, find the Maclaurin series for arctanx.

I'm struggling to find the Maclaurin series for (1+x^2)^-1

Each time I differentiate I just get f'(0) and f''(0) to equal 0. Where am I going wrong?
You're probably doing the same mistake as before.

y=(1+x^2)^{-1}

y'=-2x(1+x^2)^{-2}

y''=-2x\cdot (-4x)(1+x^2)^{-3}-2(1+x^2)^{-2} and clearly y''(0)\neq 0
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munchkin1212
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(Original post by RDKGames)
You're probably doing the same mistake as before.

y=(1+x^2)^{-1}

y'=-2x(1+x^2)^{-2}

y''=-2x\cdot (-4x)(1+x^2)^{-3}-2(1+x^2)^{-2} and clearly y''(0)\neq 0
Eugh I completely forgot I had to use the product rule! Sorry to have bothered you with my silly mistakes - thanks again!
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Zacken
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(Original post by munchkin1212)
Eugh I completely forgot I had to use the product rule! Sorry to have bothered you with my silly mistakes - thanks again!
By the way, this is yet another problem where you shouldn't be differentiating over and over again. Instead, you should realise that \frac{1}{1-a} is a geometric series 1 + a  +a^2 + \cdots so \frac{1}{1+(-x^2)} = 1 - x^2 + x^4 - \cdots

This is only valid for |x| < 1 though, but using \arctan \frac{1}{x} = \frac{\pi}{2} - \arctan x you can recover a series for |x| > 1 although I suspect you don't need that in this question.
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