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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?

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  • $\begingroup$ 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
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Hint:

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

$$(1+x+x^2)^{-1/2}=(1-x)^{1/2}(1-x^3)^{-1/2}$$

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

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  • $\begingroup$ 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

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