What is the coefficient of $\frac{y^3}{x^8}$ in $(x+y)^{-5}$? when $|\frac{y}{x}|<1$ (EAMCET 2020)
I am using the generalized binomial theorem $$(x+y)^r=\sum_{k=0}^{\infty}{r \choose k} x^{r-k}y^k$$
Substituting $n=-5$ and $r=3$ in the above gives: $${5+3-1 \choose 3}(-1)^3=-{7 \choose 3}=-35$$
I did not use the condition $|\frac{y}{x}|<1$ anywhere. Was it required as an assumption for what I did? What would removing that condition do?
Note:
- I do not know Laurent expansions (though I do know Taylor series), but neither are included in the syllabus of this exam. So answers that refer to the binomial theorem, and how to use it would be appreciated.
- I am looking at the question from a generating function point of view. So convergence issues are not primary (unless they relate to generating functions).
Yes.
When $|\frac{y}{x}|<1$, the Laurent expansion of $(x+y)^{-5}$ is obtained by taking the Taylor expansion of $(1+\theta)^{-5}$, substituting $\theta=\frac{y}{x}$, and multiplying by $x^{-5}$.
When $|\frac{x}{y}|<1$, the Laurent expansion of $(x+y)^{-5}$ is obtained by taking the Taylor expansion of $(1+\beta)^{-5}$, substituting $\beta=\frac{x}{y}$, and multiplying by $y^{-5}$.
The former series does not converge when $|\frac{x}{y}|<1$, and vice versa.
Basically, yes it does matter, the coefficients are NOT the same in the other case.