# A shortcut to the Maclaurin series expansion?

• May 6th 2011, 10:32 AM
mtpastille
A shortcut to the Maclaurin series expansion?
(Not sure if this should go into the calculus section; feel free to move it.)

Question: Find the Maclaurin series for $(p+qx^2)^r$ up to and including the term in x^4 where $p, q, r \in \mathbb{R}$

The key gives the following answer:
$p^r(1+\frac{p}{q}x^2)=p^r(1+r\frac{q}{p}x^2+\frac{ r(r-1)}{2}\frac{q^2}{p^2}x^4)$

I understand how they can move out $p^r$ and expand the parentheses binomially to get the right hand side of the equation, but there are two things that puzzle me:

1. The question asks for me to find a Maclaurin series for the expression. To me, that means finding the first derivatives and of the expression, dividing them by a factorial and multiplying by an $x^n$ term. This is not done here. Why does the answer given correspond to a Maclaurin series without performing these steps?
2. In which cases is it valid to do the above trick? When should one think of using it?
• May 6th 2011, 11:30 AM
Opalg
Quote:

Originally Posted by mtpastille
Question: Find the Maclaurin series for $(p+qx^2)^r$ up to and including the term in x^4 where $p, q, r \in \mathbb{R}$

The key gives the following answer:
$p^r(1+\frac{p}{q}x^2)=p^r(1+r\frac{q}{p}x^2+\frac{ r(r-1)}{2}\frac{q^2}{p^2}x^4)$

I understand how they can move out $p^r$ and expand the parentheses binomially to get the right hand side of the equation, but there are two things that puzzle me:

1. The question asks for me to find a Maclaurin series for the expression. To me, that means finding the first derivatives and of the expression, dividing them by a factorial and multiplying by an $x^n$ term. This is not done here. Why does the answer given correspond to a Maclaurin series without performing these steps?

Bad terminology. I would call it a binomial series, not a Maclaurin series.

Quote:

Originally Posted by mtpastille
2. In which cases is it valid to do the above trick? When should one think of using it?

Except in the elementary case where r is a positive integer, this binomial expansion will only converge when $\bigl|\tfrac qpx^2\bigr|<1$. That is probably the only situation when you would want to use this result.
• May 6th 2011, 11:31 AM
FernandoRevilla
Quote:

Originally Posted by mtpastille
2. In which cases is it valid to do the above trick? When should one think of using it?

Always

$(1+u)^r=\displaystyle\sum_{n=0}^{+\infty}\binom{r} {n}u^n \quad (|u|<1,\;r\in \mathbb{R})$

Edited: Sorry, I didn't see Opalg's post.