Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

For (a) I tried using a shortcut, and rewrote $1/\sqrt{x^2+9}$ as follows:

$(x^2+9)^{-1/2} = [1+(8+x^2)]^{-1/2}$ Then I used the known Maclaurin expansion for $(1+x)^{\alpha} = 1 + \alpha x + (\alpha(\alpha-1)x^2) 1/2! + ... + $ using $\alpha = -0.5$ and $x = 8+x^2$ .

I got completely different results from the answer. What am I doing wrong here?

share|cite|improve this question
What is the radius of convergence of your starting series? – David Mitra Feb 24 '12 at 18:36
You are not expanding about $x=0$. What's worse, your series will not converge. Try $(1/3)(1+(x/3)^2)^{-1/2}$, it will work out fine. – André Nicolas Feb 24 '12 at 18:37
what do you mean? isn't a maclaurin expansion always expanded about $x=0$ ? – nofe Feb 24 '12 at 18:39
@nofe: Yes, but $8+x^2$ is nowhere near $0$. If you use the expansion of $(1+u)^{-1/2}$, as you did, you are taking $u=8+x^2$! Use the alternate rewrite I suggested above. – André Nicolas Feb 24 '12 at 18:43
@nofe: The MacLaurin series for $(1+u)^{-1/2}$ that you substituted into only converges when $|u|<1$. By the way, if you want to send a message to me, it is best to begin it as I did this message. It is only by accident that I looked at this question again, and was not notified of a message. – André Nicolas Feb 24 '12 at 18:51
up vote 3 down vote accepted

The radius of convergence for the Maclauren series for $(1+x)^{-1/2}$ is $1$. This means that the series will diverge if you evaluate it at a number whose absolute value is larger than 1, such as $8+x^2$.

Generally, if the power series $f(x)=\sum a_n x^n$ has radius of convergence $0<r<\infty$ then you could substitute say $$ f(x^2) =\sum a_n x^{2n} $$ and this would be valid for $x^2<r$. It would not be valid for $x^2>r$.

A substitution $x=y$ into the Maclaurin series for $(1+x)^{-1/2}$ would only be valid when $y<1$. But, you tried to substitute $y=8+x^2$, which never satisfies this.

Moreover, as André points out in the comments, plugging in $8+x^2$, even if it resulted in a convergent series, would not directly result in the Maclaurin series for your function (it wouldn't be a sum of terms of the form $a_n x^n$).

share|cite|improve this answer

The coefficients you have found are correct. The problem is that the second approach is very complicated in the sense that it will give a series with each term containing a binomial series, so you would be handling a very complicated expression.

You could consider the general solution via the binomial theorem:

$${\left( {{x^2} + 9} \right)^{ - 1/2}} = \frac{1}{3}{\left( {1 + {{\left( {\frac{x}{3}} \right)}^2}} \right)^{ - 1/2}}$$

$${\left( {1 + {{\left( {\frac{x}{3}} \right)}^2}} \right)^{ - 1/2}} = \sum\limits_{k = 0}^\infty {-1/2 \choose k} {\frac{{{x^{2k}}}}{{{9^k}}}} $$

You can find many places where you'll get the closed for of that binomial coefficient:

$${-1/2 \choose k} = (-1)^k \frac{(2k-1)!!}{(2k)!!}= \frac{1}{(-4)^k}{2k \choose k}$$

so you finally have

$$\frac{1}{3}{\left( {1 + {{\left( {\frac{x}{3}} \right)}^2}} \right)^{ - 1/2}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}} \frac{{{x^{2k}}}}{{{3^{2k + 1}}}}$$

$$\frac{1}{{\sqrt {{x^2} + 9} }} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}} \frac{{{x^{2k}}}}{{{3^{2k + 1}}}}$$

Note that all terms are of even powers, so that as you showed all odd terms are $0$.

share|cite|improve this answer
what does $!!$ double factorial mean? – nofe Feb 24 '12 at 18:56
Whoops! Thought you knew. It means multiplying by the precending numbers of the same parity. This is $$8!! = 8\cdot 6 \cdot4\cdot 2 $$ $$9!! = 9\cdot 7 \cdot5\cdot 3\cdot 1 $$ – Pedro Tamaroff Feb 24 '12 at 19:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.