Calculate the first three coefficients of the reciprocal of the power series of the functions: i.$ (1+x)^m$ (Hint: Use Cauchy product)

I am working on formal power series now and I have this question. Here are some definitions that must known:

Given two formal series $F$(x) and $G$(x), if $F$(x)$G$(x) = 1, then we say that $F$ is the reciprocal of $G$.

Cauchy Product: $ \sum_{n=0}^\infty a_n x^n \sum_{n=0}^\infty b_n x^n = \sum_{n=0}^\infty c_n x^n\;where\;c_n = \sum_{k=0}^n a_k b_{n-k} $

My attempt: let $R$ denote the reciprocal of $(1+x)^m$ Now, we have

($1 + mx + \frac {m(m-1)x^2}{2!} + \frac {m(m-1)(m-2)}{3!} + ... ) R=1$

From here I don't know how to apply Cauchy product, can anyone help?


1 Answer 1



As $(1+x)^m(1-x)^{-m}=1$

if $((1-x)^{-m}=\sum_{r=0}^\infty a_rx^r,$

we have $$1=\left(1+mx+\dfrac{m(m-1)}2x^2+\cdots+x^m\right)(\sum_{r=0}^\infty a_rx^r)$$

Comparing the constants $$1=1\cdot a_0$$

Comparing the coefficients of $x$ $$0=m+a_1$$

Comparing the coefficients of $x^2$ $$0=a_2+ma_1+\dfrac{m(m-1)}2a_0$$

  • $\begingroup$ You meant $(1+x)^{-m}$ as the reciprocal of $(1+x)^m $ right? $\endgroup$ Commented Jul 15, 2017 at 10:42
  • $\begingroup$ @LeylaAlkan, True $\endgroup$ Commented Jul 15, 2017 at 10:48

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