If $$\frac{1}{(1+x)^\frac{1}{4}} = \sum_{n=0}^\infty c_nx^n$$ then $c_2$ is....

I think I know how to attempt the problem but I'm not sure if I'm on the right track

I started out with the known series

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

I then attempted to manipulate the left side so it would be the same as the equation above. Then depending on how the x term looks I could plug in a value for n to find the second coefficient of the series; however, I am having some trouble manipulating the problem due to the quarter power.

So far I got up to here...

$$\frac{1}{1+x}=\sum_{n=0}^\infty x^n(-1)^n$$

I know that changing the $x$ term on the left side will change the $x$ term on the right side, but I'm unclear if I can change the whole denominator on the left side and how that would effect the right side since the quarter power is over the whole denominator.

It is an old final so I do know the answer is $\frac{5}{32}$ but I don't know how to get that answer.

  • $\begingroup$ Also, this is my first time posting so I'm really struggling with the formatting of equations, I'm trying to figure it out right now. Sorry its hard to understand! $\endgroup$ – I.ling Apr 16 '17 at 22:33
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    $\begingroup$ You can write code inside $$ in the editor. I just helped you with it, you can click "edit" to see how i did. $\endgroup$ – mathreadler Apr 16 '17 at 22:34
  • $\begingroup$ The value of $c_n$ depend on the value of $x$ because the series doesnt have an infinite radius of convergence. Then, to make sense, you must assume (by example) that $|x|<1$. Then the MacLaurin series have the desired coefficients. $\endgroup$ – Masacroso Apr 16 '17 at 23:53

Have a Taylor expansion at $0$

$$f(x)=\frac{1}{(1+x)^\frac{1}{4}} =\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n =\sum_{n=0}^\infty c_nx^n$$ then $c_2=\frac{f''(0)}{2!}$

$f''(x)=((1+x)^{-0.25})''=(-0.25 \times (1+x)^{-1.25})'=\frac{5}{16}(1+x)^{-\frac{9}{4}}$

Thus $c_2 = \frac{5}{32}$


There are two ways to do this. First, there is the formula

$$ (1 + x)^\alpha = \sum_{k = 0}^\infty {\alpha \choose k}x^k $$

where for an arbitrary complex number $\alpha$ and $k \in \mathbb N$

$$ {\alpha \choose k} = \frac{\alpha(\alpha - 1)(\alpha - 2) \cdots (\alpha - k + 1)}{k!}. $$

However, I suspect that the problem is instead asking you to do it by calculating derivatives. Recall that if $f(x)$ admits a power series representation at $0$ then

$$ f(x) = \sum_{k = 0}^\infty \frac{f^{(k)}(0)}{k!} x^k. $$

So in this case, the number you are looking for is

$$ c_2 = \frac12 \cdot \left. \frac{d^2}{dx^2} \frac{1}{(1 + x)^{1/4}} \right|_{x \leftarrow 0}. $$

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    $\begingroup$ Clear and generalizable. I especially like the arrow notation on the last line... I haven't seen it before! The period looks a little out of place though $\endgroup$ – Brevan Ellefsen Apr 17 '17 at 1:12
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    $\begingroup$ "Recall that $f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k$" - Equality is not assured for general infinitely differentiable $f$, even inside the radius of convergence. However, I offer this only as a warning for other uses. In this case, we are given that the function is representable by a power series (and could prove it even if it were not given). So it works here. $\endgroup$ – Paul Sinclair Apr 17 '17 at 3:52
  • $\begingroup$ @PaulSinclair Thanks for reminding me. I didn't have a succinct way of writing "for functions that admit a power series representation in a neighbourhood of the origin" or "for functions which are locally the restriction of a complex differentiable function" so I left it out. Which is of course terrible. I have added a caveat that makes what I've written less wrong. $\endgroup$ – Trevor Gunn Apr 17 '17 at 4:29

Here is a simple and straightforward way which avoids any deep theorems of analysis. We have $$\left(\sum c_{n}x^{n}\right)^{4}(1 + x) = 1$$ or $$\left\{c_{0}^{2} + 2c_{0}c_{1}x + (2c_{0}c_{2} + c_{1}^{2})x^{2} + \cdots\right\}^{2}(1 + x) = 1$$ or $$\left\{c_{0}^{4} + 4c_{0}^{3}c_{1}x + \{2c_{0}^{2}(2c_{0}c_{2} + c_{1}^{2}) + 4c_{0}^{2}c_{1}^{2}\}x^{2} + \cdots\right\}(1 + x) = 1$$ and then we have by comparing coefficients $$c_{0}^{4} = 1, 1 + 4c_{0}^{3}c_{1} = 0, 4c_{0}^{3}c_{1} + 2c_{0}^{2}(2c_{0}c_{2} + c_{1}^{2}) + 4c_{0}^{2}c_{1}^{2} = 0\tag{1}$$ From the original equation $\sum c_{n}x^{n} = (1 + x)^{-1/4}$ we get $c_{0} = 1$ by putting $x = 0$. Then putting this value of $c_{0}$ in $(1)$ we get $$c_{1} = -1/4, -1 + 2(2c_{2} + 1/16) + 1/4 = 0$$ or $c_{2} = (1/2)(3/8 - 1/16) = 5/32$.


$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {1 \over \pars{1 + x}^{1/4}} & = \pars{1 + x}^{-1/4} = 1 + \pars{-\,{1 \over 4}}x\ +\ \overbrace{{1 \over 2!} \pars{-\,{1 \over 4}}\pars{-\,{1 \over 4} - 1}} ^{\ds{-1/4 \choose 2}}\ \,x^{2} + \,\mrm{O}\pars{x^{3}} \\[5mm] & = 1 - {1 \over 4}\,x + \color{#f00}{5 \over 32}\,x^{2} + \,\mrm{O}\pars{x^{3}} \end{align}


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