maclaurin series question

**pb6883** · August 8th 2009, 08:17 AM

hello, if possible hep with the following:

Q: By expanding (1-4x)^1/2 and integrating term by term, or otherwise, find the series expansion for arcsin(2x), when |x|<1/2, as far as the term in x^7.

A: I have done it by consequatively differentiating until the seventh order and then evaluating at x = 0 to produce the desired series.

However I am at a loss as to how to do it by "expanding (1-4x)^1/2 and integrating term by term"..

...I know its basic stuff but I would much appreciate any help!

**Random Variable** · August 8th 2009, 08:54 AM

Do you mean $\frac{1}{\sqrt{1-4x^{2}}}$ ?

**pb6883** · August 8th 2009, 09:00 AM

I do! I am ever so sorry... Thanks for spotting it

**Failure** · August 8th 2009, 09:02 AM

Originally Posted by pb6883

hello, if possible hep with the following:

Q: By expanding (1-4x)^1/2 and integrating term by term, or otherwise, find the series expansion for arcsin(2x), when |x|<1/2, as far as the term in x^7.

A: I have done it by consequatively differentiating until the seventh order and then evaluating at x = 0 to produce the desired series.

However I am at a loss as to how to do it by "expanding (1-4x)^1/2 and integrating term by term"..

...I know its basic stuff but I would much appreciate any help!

Are you sure that you were asked to expand $(1-4x)^{1/2}$ and integrate the resulting series to get the expansion for $\mathrm{arcsin}(2x)$ ? - Because it is not at all clear what $\int (1-4x)^{1/2}$ has got to do with $\mathrm{arcsin}(2x)$ .
Maybe you were asked to expand $(1-4x^2)^{-1/2}$ instead? Because in this case, it is quite obvious how the integral of this term is related to $\mathrm{arcsin}(2x)$ .

**Failure** · August 8th 2009, 09:07 AM

Originally Posted by pb6883

I do! I am ever so sorry... Thanks for spotting it

Ok, in that case you have that $\int (1-4x^2)^{-1/2}\, dx=\frac{\mathrm{arcsin}(x)}{2}+C$ . So multiply the integrated series by 2 and check for a possibly necessary adjustment of the C. (Actually, C=0 will do - I think.)

**Random Variable** · August 8th 2009, 09:08 AM

You could use the generalized binomial theorem to expand $\frac{1}{\sqrt{1-4x^{2}}} = (1-4x^2)^{-1/2}$

$(1+x)^{n} = 1 +nx +\frac{n(n-1)x^{2}}{2!} + \frac{n(n-1)(n-2)x^{3}}{3!} + ...$

**Bruno J.** · August 8th 2009, 09:09 AM

You can show easily by induction that

$\frac{1}{\sqrt{1-t}} = 1+\sum_{j=1}^\infty\frac{1\cdot 3 \cdot ... \cdot (2j-1)}{2^jj!}t^j$

Now if you use

$\int_0^x\frac{1}{\sqrt{1-t^2}}\: dt = \arcsin x$

you can get the series for arcsin by replacing $t$ by $t^2$ in the first identity and integrating the resulting series term by term.

**pb6883** · August 8th 2009, 09:25 AM

thanks all, your comments have really helped (p.s. i told you it was basic

)

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