Find $\mathcal{O}(x)$ of $\frac{x^2+x^3+x^4}{(x+x^2+x^3)^{3/2}}$

Question

Consider $$f(x)=\frac{x^2+x^3+x^4}{(x+x^2+x^3)^{3/2}}.$$ Aim: Write $f(x) = g(x) + \mathcal{O}(x)$ as $x\downarrow0$.

I can change the fraction to $$f(x)=\frac{x^{1/2}+x^{3/2}+x^{5/2}}{(1+x+x^2)^{3/2}}.$$ I can now use the Taylor approximation $\frac{1}{(1+y)^{3/2}}=1-ay+by^2\mp...$ and set $y=x+x^2$. Multiplying the numerator with the Taylor approximation gives \begin{align} f(x)&=\left(x^{1/2}+x^{3/2}+x^{5/2}\right)\left(1-a(x+x^2)+b(x+x^2)^2\mp...\right) \\ &= \sqrt{x}+\mathcal{O}(x). \end{align}

Question: Is my approach correct and can I get there without a Taylor approximation?

Answer 1

To show $f(x) = x^{1/2} + O(x)$, it suffices to show $\left|\frac{f(x)-x^{1/2}}{x}\right| \le C$ for sufficiently small $x$.

$$\frac{f(x)-x^{1/2}}{x} = \frac{x^{1/2} + x^{3/2} + x^{5/2} - x^{1/2} - x^{4/2} - x^{7/2} }{x(1+x+x^2)^{3/2}} = \frac{x^{1/2} + x^{3/2} - x - x^{5/2}}{(1+x+x^2)^{3/2}} \overset{x \downarrow 0}{\longrightarrow} 0.$$

I think you can actually show the stronger statement that $f(x) = x^{1/2} + O(x^{3/2})$, which would imply $f(x) = x^{1/2} + O(x)$.

Answer 2

Your method is alright, though it would have been less error prone to just take the Taylor expansion of $\frac 1{(1+x+x^2)^{3/2}}$ directly. In your case, you have to be careful that the substitution doesn't change the relevant point for which you developed the Taylor series. Here it works out since we care about $x\to 0$, for which we have $x+x^2\to 0$, which converges to the same point around which we developed the Taylor-series for $\frac1{1+y}$.

One specific error you made however is that you forgot about the $(\cdot)^{3/2}$ in $$ f(x)= \left(x^{1/2}+x^{3/2}+x^{5/2}\right)\left(1-a(x+x^2)+b(x+x^2)^2\mp...\right)^{3/2} $$ because of which the following calculation which gives you $f(x)=\sqrt x+\mathcal{O}(x^{3/2})$ isn't as nice.

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In general, given a rational polynomial $p(x)/q(x)$ I think the most comfortable method is developing $\frac1{q(x)}$ into a Taylor series.