Binomial expansion on $\sqrt{1+\frac{4}{x^2}+\frac{1}{x^3}}$

Question

Yesterday, I've posted a question where this term was being used:

$$\sqrt{1+\dfrac{4}{x^2}+\dfrac{1}{x^3}}{}$$

One of the solutions stated the following (For x getting really big).

$$\sqrt{1+\frac{4}{x^2}+\frac{1}{x^3}}{} \approx 1 + \dfrac{1}{2}\left(\frac{4}{x^2}+\frac{1}{x^3}\right)$$

Also the solution stated that he used to the binomial expansion. I tried to get to this solution by myself but did not really get to a solution. It would be great if someone could explain why the squareroot can be written like that for big x.

Tomorrow, I am going to have a exam for the university and I think that this could help me in a lot of cases but I am not allowed to use it without an explanation.

Greetings, Finn

Answer 1

In general we have that for $x\to 0$

$$(1+x)^a=1+ax+o(x)$$

where $o(x)$ represent a term of order great the $x$ and thus negligeble with respect to $x$ when $x\to 0$, thus in this case we can also write

$$(1+x)^a\sim 1+ax$$

as infinitesimal approximation.

Notably in this case we are taking just a first order approximation

$$\sqrt{1+\frac{4}{x^2}+\frac{1}{x^3}}{} =1 + \dfrac{1}{2}\left(\frac{4}{x^2}+\frac{1}{x^3}\right)+o\left(\frac1{x^3}\right)$$

$$\sqrt{1+\frac{4}{x^2}+\frac{1}{x^3}}{} \sim 1 + \dfrac{1}{2}\left(\frac{4}{x^2}+\frac{1}{x^3}\right)$$ Be aware that this kind of approximation must be handled carefully since in some cases could lead in error. Indeed in some cases we could have to expand at a order greater than the first. See here for the general binomial expansion.

Answer 2

What is the expansion of $f(x)=\sqrt{1+x}$. Lets say the following $$f(x)=\sqrt{1+x}=a_0+a_1x+a_2x^2\cdots$$ Now if we calculate the coefficients then we should have the maclurian series for $f(x)$.

We can calculate $a_0$ by putting $x=0$. This gives $a_0=1$.

Next we take the first derivative of both sides. $$\frac d{dx}(\sqrt{1+x})=\frac{d}{dx}(a_0+a_1x+a_2x^2\cdots) \\ \implies \frac{1}{2\sqrt{1+x}}=a_1+2a_2x+\cdots$$ We again substitute $x=0$, we get $a_1=\frac12$. We can keep taking derivatives and putting $x=0$ to get however many terms we want. In your case you just replace $x$ by $\dfrac{4}{x^2}+\dfrac1{x^3}$