# Series Expansion

• January 17th 2009, 09:19 AM
Haris
Series Expansion
$f(x)=(1+x^2)^\frac{1}{4}$

How would you expand this exactly? Would you use the binomial theorem?
• January 17th 2009, 09:30 AM
Jhevon
Quote:

Originally Posted by Haris
$f(x)=(1+x^2)^\frac{1}{4}$

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

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?
• January 17th 2009, 09:40 AM
Haris
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?
• January 17th 2009, 09:51 AM
Jhevon
Quote:

Originally Posted by Haris
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?

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.
• January 17th 2009, 10:08 AM
HallsofIvy
Quote:

Originally Posted by Jhevon
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?

On the contrary, the extended binomial expansion, $(a+ b)^x$ for non-integer x, does lead to an infinite series.

The binomial expansions says that
$(a+ b)^n= \sum \left(\begin{array}{c}n \\ i\end{arrray}\right)a^{n-i}b^i$
If n is a positive integer then
$\left(\begin{array}{c}n \\ i\end{array}\right)$
is defined as $\frac{n!}{i!}{n-i}!= \frac{n(n-1)(n-2)\cdot\cdot\cdot(n- i+1)}{i!}$ 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, $(1+ x^2)^{1/4}= x^{1/2}+ (1/4)x^{-1/2}+ \cdot\cdot\cdot$.
• January 17th 2009, 10:14 AM
Haris
$f(x)=1+0x+\frac{1}{2!}.0x^2+\frac{1}{3!}.0x^3+\fra c{1}{4!}.0x^4+...$

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

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

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

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

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