The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$

I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?

  • $\begingroup$ Are you supposed to know about Taylor series ? $\endgroup$ – Claude Leibovici Dec 7 '17 at 12:16
  • $\begingroup$ For the binomial case, the answer is like this. It doesn't seem too difficult to adapt to the multinomial case. $\endgroup$ – Arthur Dec 7 '17 at 12:20


The coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$


the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3+\cdots)^{1/2}$

Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,



  • 1
    $\begingroup$ Really nice and simple +1 $\endgroup$ – Paramanand Singh Dec 7 '17 at 12:43

Say $$(1-2x+3x^2-4x^3)^{1/2} =a+bx+cx^2+dx^3...$$ then

$$1-2x+3x^2-4x^3 =(a+bx+cx^2+dx^3...)^2$$ but $$(a+bx+cx^2+dx^3...)^2 = a^2+2abx+(2ac+b^2)x^2+2(ad+bc)x^3+...$$

So $a=\pm 1$.

If $a=1$ then $b=-1$ and $c=1$ and $d=-1$


Expand $(1+u)^{\tfrac12}$ up to order $3$: $$(1+u)^{\tfrac12}=1+\frac12 u-\frac18u^2+\frac1{16}u^3+o(u^3),$$ and compose with $u=-2x+3x^2-4x^3$:

  • $u^2=4x^2-12x^3+o(x^3)$,

  • $u^3=u^2\cdot u=-8x^3+o(x^3)$.

One finally obtains$$1-x+x^2-x^3+o(x^3).$$


It is quite practical to exploit the identity $$ (1+x)^2 (1-2x+3x^2-4x^3) = 1-5x^4-4x^5 \tag{A}$$ from which $$\begin{eqnarray*} \sqrt{1-2x+3x^2-4x^3} &=& \frac{\sqrt{1-5x^4-4x^3}}{1+x} \\&=&\left(1+O(x^4)\right)(1-x+x^2-x^3+O(x^4))\tag{B}\end{eqnarray*}$$ and the coefficient of $x^3$ in the RHS of $(B)$ is trivially $\color{red}{-1}$.


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