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Let $f(x) = \frac {x^3}{(x+1)^2}$. Find constants a, b, c, so that $f(x) = ax + b + \frac cx + o(\frac 1x)$ as $x$ goes to $\pm \infty$. So i know that i can't take Taylor series as $x$ goes to infinity. So i am assuming i have to make some kind of substitution. I tried making $x = \frac 1u$ but i get nowhere.

$$f(x) = \frac {x^3}{(x+1)^2} = x^3(x+1)^{-2}=\frac{1}{u^3}(1 + \frac 1u)$$
$$\lim_{u\to 0}\frac{1}{u^3}(1 + \frac 1u) = \lim_{u\to 0}\frac{1}{u^3}(1 - \frac 2u + \frac{3}{u^2} + o({u^3})) =\lim_{u\to 0} u^3 - \frac {2}{u^4} + \frac{3}{u^5} + o(\frac 1u) = \lim_{u\to 0} \frac{1}{x^3} - 2x^4 + 3x^5 + o(x^5)$$

Thanks.

Siminore
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GovernmentFX
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  • Technically this is not a Taylor's series, as it contains terms like $\frac{1}{x}$, when Taylor's series all terms must be polynomial. – lsoranco Jan 09 '14 at 14:40
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    You lost somewhere power -2. You can do this using long division too (dividing x^3 by x^2 + 2 x + 1). – Claude Leibovici Jan 09 '14 at 14:44

2 Answers2

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Here is how you approach the problem. Put $x=\frac{1}{t}$ and then find the Laurent series at the point $t=0$

$$ f(x) = \frac {(1/t)^3}{(1/t+1)^2}=\frac{1}{t(1+t)^2} ={t}^{-1}-2+3\,t-4\,{t}^{2}+5\,{t}^{3}+ ( {t}^{4}) .$$

Now, substitute $t=1/x$ in the above expansion. See related problem.

Note: Note that, you only need to find the Taylor series of $\frac{1}{(1+t)^2}$ at $t=0$, since

$$ \frac{1}{t(1+t)^2}= \frac{1}{t} \frac{1}{(1+t)^2} .$$

Community
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Mhenni Benghorbal
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I would start off here by using long division to simplify things a bit: $$ \frac{x^3}{(x+1)^2}=\frac{x^3}{x^2+2x+1}=\cdots=x-2+\frac{3x+2}{x^2+2x+1}. $$

Heuristically speaking, it is clear that as $x\rightarrow\infty$, that last bit is $\frac{3}{x}+o(\frac{1}{x})$. So, this would suggested that $$ \frac{x^3}{(x+1)^2}=x-2+\frac{3}{x}+o\left(\frac{1}{x}\right)\text{ as }x\to\infty. $$ Having used long division / heuristics to find the answer, now it is only a matter of justifying it; and, that shouldn't be too bad.

Nick Peterson
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