$\displaystyle f(x)=(1+x^2)^\frac{1}{4}$

How would you expand this exactly? Would you use the binomial theorem?

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- Jan 17th 2009, 09:19 AMHarisSeries Expansion
$\displaystyle f(x)=(1+x^2)^\frac{1}{4}$

How would you expand this exactly? Would you use the binomial theorem? - Jan 17th 2009, 09:30 AMJhevon
the binomial expansion doesn't give rise to series, it simply shows you how to expand brackets.

you can write the Taylor series for this. perhaps that's what you want? - Jan 17th 2009, 09:40 AMHaris
Well I thought it was either that or the MacLaurin series. Could someone clarify which one it is and maybe give me a head start?

- Jan 17th 2009, 09:51 AMJhevon
A MacLaurin series

*is*a Taylor series. Specifically one centered around zero. It's usually the one people think about when they think Taylor series. The same link i gave you before talks about them. You can also see here.

So begin by finding a few derivatives of the function, and evaluate them at zero. Then plug them into the form you see. - Jan 17th 2009, 10:08 AMHallsofIvy
On the contrary, the

**extended**binomial expansion, $\displaystyle (a+ b)^x$ for non-integer x, does lead to an infinite series.

The binomial expansions says that

$\displaystyle (a+ b)^n= \sum \left(\begin{array}{c}n \\ i\end{arrray}\right)a^{n-i}b^i$

If n is a positive integer then

$\displaystyle \left(\begin{array}{c}n \\ i\end{array}\right)$

is defined as $\displaystyle \frac{n!}{i!}{n-i}!= \frac{n(n-1)(n-2)\cdot\cdot\cdot(n- i+1)}{i!}$ and is 0 for i> n so the sum terminates.

If n is not a positive integer, then n! is not defined but we can still right out the last of those- but the sum no longer terminates.

In particular, $\displaystyle (1+ x^2)^{1/4}= x^{1/2}+ (1/4)x^{-1/2}+ \cdot\cdot\cdot$. - Jan 17th 2009, 10:14 AMHaris
$\displaystyle f(x)=1+0x+\frac{1}{2!}.0x^2+\frac{1}{3!}.0x^3+\fra c{1}{4!}.0x^4+...$

I got this using the MacLaurin series but somehow it doesnt look as though its correct? - Jan 17th 2009, 10:16 AMo_O
Perhaps the OP is referring to the representation of a binomial as an infinite sum: Binomial Theorem

$\displaystyle (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$$\displaystyle {\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!} + \cdots$

for any $\displaystyle k \in \mathbb{R}$ and if $\displaystyle |y| < 1$. Here, let $\displaystyle y = x^2$ and $\displaystyle k = \tfrac{1}{4}$.

So we have:

$\displaystyle (1 + x^2)^{\frac{1}{4}} = 1 + \left(\tfrac{1}{4}\right)x^2 + \frac{\left(\tfrac{1}{4}\right)\left(-\frac{3}{4}\right)}{2!} \left(x^2\right)^2 + \frac{\left(\tfrac{1}{4}\right)\left(-\frac{3}{4}\right)\left(-\frac{7}{4}\right)}{3!} \left(x^2\right)^3 + \cdots $$\displaystyle {\color{white}.} + \ \frac{ \left( \frac{1}{4}\right) \left(-\frac{3}{4}\right) \left(-\frac{7}{4}\right) \cdots \left(\frac{1}{4} - n + 1\right) }{n!} \left(x^2\right)^n + \cdots$

See where you can go from here. Try to simplify the general term and hopefully you'll be able to get somewhere (Yes)