So far I have been able to expand most rational and transcendental functions into the Maclaurin series. But irrational functions such as this one $\dfrac{1}{\sqrt{1+x+x^2}}$ confuse me. I try to use the binomial series formula but the general formula only gives me $2$ terms. What I have tried to do is to set up a dummy variable $X=x+x^2$ so that my function has the form $(1+X)^\frac{1}{2}$ and expand them, then reinsert the variable into the expression.

Is this method correct or not? Is there a more general binomial series formula to deal with non-factorable irrational expression?

  • $\begingroup$ That should work. The key point is that power series about a point are unique if they exist. So if you find one power series that converges to the desired expression, it is the power series. $\endgroup$ – Charles Hudgins Nov 21 '19 at 2:40


As $1-x^3=(1-x)(1+x+x^2),$


Use https://en.m.wikipedia.org/wiki/Binomial_series assuming $|x|<1$

  • $\begingroup$ How do you factor so fast that expression? $\endgroup$ – James Warthington Nov 21 '19 at 2:41
  • $\begingroup$ @James, Please find the updated version $\endgroup$ – lab bhattacharjee Nov 21 '19 at 2:42
  • $\begingroup$ What will happen if the generating function is irrational and is not factorable? $\endgroup$ – James Warthington Nov 21 '19 at 2:47
  • $\begingroup$ @JamesWarthington (1) People have the factorization of $A^3-B^3$ memorized. It's also a special case of the (homogeneous version) of the geometric sum formula, which gives a factorization of $A^n-B^n$ for general $n$, as well as $A^n+B^n$ for odd $n$. (2) All polynomials are factorizable with complex numbers. What kind of functions are you talking about otherwise? Be specific what you mean by "irrational" function - is it just an expression built from rational expressions and taking radicals? There may be no general answer better than "do it piece by piece" it's so broad. $\endgroup$ – runway44 Nov 23 '19 at 0:33

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.