2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! I hope that will help me! :) $\endgroup$ – Finn Eggers Feb 23 '18 at 8:13
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.