Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
1  
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. –  23rd Aug 1 '13 at 19:20
    
Definitely the latter...let me fix that –  Eric Auld Aug 1 '13 at 19:46
    
Is there a general method when one has more than two terms? –  Eric Auld Aug 1 '13 at 19:47
1  
I am not clear how general for what expansion you need. Could you be more specific? –  23rd Aug 1 '13 at 19:51
1  
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 $1+a\alpha x+(b\alpha+\frac{a^2\alpha(\alpha-1)}{2})x^2+o(x^2)$. –  23rd Aug 1 '13 at 20:07

1 Answer 1

up vote 1 down vote accepted

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

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

share|improve this answer

Your Answer

 
discard

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.