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


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

  • $\begingroup$ Well, what do you get when you form the binomial expansion of $\sqrt{1+u}$? $\endgroup$ – Gerry Myerson Feb 23 '18 at 8:02
  • $\begingroup$ It's just $\sqrt{1+y}=(1+y)^{1/2}=1+y/2+O(y^2)$ as $y\to 0$. $\endgroup$ – Angina Seng Feb 23 '18 at 8:03
  • $\begingroup$ I do not know how to deal with the exponent of $\dfrac{1}{2}$ $\endgroup$ – Finn Eggers Feb 23 '18 at 8:03
  • $\begingroup$ Oh... Okay that seems easy. Thank you!!!! $\endgroup$ – Finn Eggers Feb 23 '18 at 8:03
  • $\begingroup$ @FinnEggers Hi I gave you this hint! $\endgroup$ – user Feb 23 '18 at 8:04

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


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.

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  • $\begingroup$ Thank you! I hope that will help me! :) $\endgroup$ – Finn Eggers Feb 23 '18 at 8:13

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 $\dfrac{4}{x^2}+\dfrac1{x^3}$

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