Monster Limit 3


\(\displaystyle f(x) = \sum_{n=0}^{\infty} f_n x^n\), \(\displaystyle F(n) = \sum_{r=0}^{n} f_{r}\) and \(\displaystyle \lim_{n \to \infty} \dfrac{F(n)}{n^{r_f}} = \text{d}_f \)


g(x)=n=0gnxn\displaystyle g(x) = \sum_{n=0}^{\infty} g_n x^n, G(n)=r=0ngr\displaystyle G(n) = \sum_{r=0}^n g_r and limnG(n)nrg=dg\displaystyle \lim_{n \to \infty} \dfrac{G(n)}{n^{r_g}} = \text{d}_g

From Monster Limit 2, we have,

{limx1(1x)rf f(x)=df Γ(1+rf)limx1(1x)rg g(x)=dg Γ(1+rg)\begin{cases} \lim_{x \to 1^{-}} (1-x)^{r_f} \ f(x) = \text{d}_f \ \Gamma(1+r_f)\\ \lim_{x \to 1^{-}} (1-x)^{r_g} \ g(x) = \text{d}_g \ \Gamma(1+r_g) \end{cases}

Now, consider the same quantities for fg(x)=f(x)g(x)fg(x) = f(x)g(x), i.e,

fg(x)=n=0fgnxn\displaystyle fg(x) = \sum_{n=0}^{\infty} {fg}_n x^n, FG(n)=r=0nfgr\displaystyle FG(n) = \sum_{r=0}^{n} {fg}_{r} and limnFG(n)nrfg=dfg\displaystyle \lim_{n \to \infty} \dfrac{FG(n)}{n^{r_{fg}}} = \text{d}_{fg}


limx1(1x)rfg fg(x)=dfg Γ(1+rfg)\displaystyle \lim_{x \to 1^{-}} (1-x)^{r_{fg}} \ fg(x) = \text{d}_{fg} \ \Gamma(1+r_{fg})

Prove That
dfg=dfdgΓ(1+rf)Γ(1+rg)Γ(1+rf+rg)\displaystyle \text{d}_{fg} = \text{d}_{f} \text{d}_{g} \dfrac{\Gamma(1+r_f) \Gamma(1+r_g)}{\Gamma(1+r_f + r_g)}

Note by Ishan Singh
2 years, 9 months ago

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