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Calculate the first three coefficients of the reciprocal of the power series of the functions: i. cosx (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} $

Maclaurin expansion of cosx: cosx =$ 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!} + ...$

My attempt: let $R_c$ denote the reciprocal of cosx ( $1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!} + ...$ ) $R_c$ =$ 1$ From here I don't know how to apply Cauchy product, can anyone help?

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  • $\begingroup$ Here $c_n$ are the coefficients of $F(x)\frac{1}{F(x)} = 1$. $\endgroup$
    – reuns
    Commented Jul 14, 2017 at 15:20

1 Answer 1

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What you call $R_c$ is usually called $\sec x$. Let $\sec x=b_0+b_1x+b_2x^2+\cdots$. Then $\cos x=\sum a_nx^n$ where $a_{2m}=(-1)^m/(2m)!$ and $a_{2m+1}=0$. Then $\cos x\sec x=1=\sum c_nx^n$ where $c_0=1$ and the remaining $c_n=0$.

Now $$1=c_0=a_0b_0=b_0.$$ So $b_0=1$. Also $$0=c_1=a_0b_1+a_1b_0=b_1+0.$$ So $b_1=0$. Anyway it's obvious (why?) that $b_n=0$ whenever $n$ is odd. Next $$0=c_2=a_0b_2+a_1b_1+a_2b_0=b_2+0-\frac1{2}.$$ So $b_2=1/2$. Skip the next stage as we know $b_3=0$. Then $$0=c_4=a_0b_4+a_2b_2+a_4b_0=b_4-\frac14+\frac1{24}$$ giving $b_4=5/24$. Keep going, until you get bored...

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