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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?

nofe
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    What is the radius of convergence of your starting series? – David Mitra Feb 24 '12 at 18:36
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    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
  • why must it be close to $0$ ? – nofe Feb 24 '12 at 18:46
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    @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
  • @nofe By definition, the Mc Laurin series expands the function at $x=0$. – Pedro Feb 24 '12 at 18:58
  • @AndréNicolas I got it, thanks a lot. – nofe Feb 24 '12 at 18:59

2 Answers2

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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 $a_n x^n$).

David Mitra
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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 $0$.

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Pedro
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