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In showing that $\log^\alpha{(1+x)}$ is $O((x)^\alpha)$ at $1$, for $\alpha>0$, one can note that

$$\left ( \frac{\log{(1+x)}}{x} \right )^\alpha \overset{x\to 0}{\longrightarrow} \left ( 1\right )^\alpha = 1.$$

So we know that $$\log^\alpha{(1+x)}= (x)^\alpha + o((x)^\alpha).$$

But how would I find $\beta > \alpha$ and $c\in \mathbb{R}$ such that

$$\log^\alpha{(1+x)}= (x)^\alpha + c(x)^\beta + o((x)^\beta)?$$

I'd like to raise the power series to an exponent:

$$\log^\alpha(1+x) = \left ( x - x^2/2 + x^3/3 + \dotsb \right) ^ \alpha = \,\,\,??$$

Is there a version of the multinomial theorem I need to use here? This page I found gives me a formula for a multinomial when the first term is larger than the sums of the rest of the terms, but I don't think I want to use that...

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    $\begingroup$ You may write $(x+O(x^2))^\alpha$ as $x^\alpha(1+O(x))^\alpha$ and apply binomial expansion to $(1+t)^\alpha$ for $t$ small. $\endgroup$
    – 23rd
    Commented Aug 1, 2013 at 19:20
  • $\begingroup$ Definitely the latter...let me fix that $\endgroup$
    – Eric Auld
    Commented Aug 1, 2013 at 19:46
  • $\begingroup$ Is there a general method when one has more than two terms? $\endgroup$
    – Eric Auld
    Commented Aug 1, 2013 at 19:47
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    $\begingroup$ I am not clear how general for what expansion you need. Could you be more specific? $\endgroup$
    – 23rd
    Commented Aug 1, 2013 at 19:51
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    $\begingroup$ Binomial expansion can give you as precise as you wish. For example, up to $t^2$ term, $(1+t)^\alpha=1+\alpha t+\frac{\alpha(\alpha-1)}{2}t^2+O(t^3)$. For $t=O(x)$, you will get $1+\alpha O(x)$; for $t=ax+o(x)$, you will get $1+a\alpha x+o(x)$; for $t=ax+bx^2+o(x^2)$, you will get $1+a\alpha x+(b\alpha+\frac{a^2\alpha(\alpha-1)}{2})x^2+o(x^2)$. $\endgroup$
    – 23rd
    Commented Aug 1, 2013 at 20:07

1 Answer 1

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Landscape has provided a very nice answer in the comments. I will summarize in this community wiki, hoping I am getting it right:

Given a power series to a (possibly fractional) exponent $\alpha > 0$, say the one in the original question:

$$ \left ( x - x^2/2 + x^3/3 + \dotsb \right) ^ \alpha,$$

one can factor out the $x$ term, and truncate at whatever term one desires

$$ x^\alpha \left ( 1 - x/2 + x^2/3 + o(x^2) \right) ^ \alpha$$

and then apply the generalized binomial theorem

$$= x^\alpha \left ( 1 + \alpha\cdot[- x/2 + x^2/3 + o(x^2) ] + \frac {\alpha (\alpha - 1) } {2} [ - x/2 + x^2/3 + o(x^2) ]^2 \\ + o( [ - x/2 + x^2/3 + o(x^2) ]^2) \right) \\ = x^\alpha - \frac {\alpha} {2} x^{\alpha + 1} + \left ( \frac {\alpha} {3} + \frac {\alpha(\alpha-1)} {8} \right )x^{\alpha + 2} + o(x^{\alpha + 2}).$$

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