Let $g(x)=x^3-x$,and $f(x)$ be a polynomial of degree $\leq100$.If $f(x)$ and $g(x)$ have no common factor and $\frac{d^{100}}{dx^{100}}\left[\frac{f(x)}{g(x)}\right]=\frac{p(x)}{q(x)}$,then find the degrees of the polynomials $p(x)$ and $q(x)$.
I tried this problem.SInce the degree of $f(x)$ is atmost 100,and degree of $g(x)$ is 3,so the atmost degree of $\frac{f(x)}{g(x)}$ is 97 and after differentiation we get degree of $\frac{p(x)}{q(x)}$ as $-3$ but i cannot exactly pinpoint the degree of $p(x)$ and $q(x)$.
Answer in my book says degree of $p(x)$ is $6\times 2^{99}-101$ and the degree of $q(x)$ is $6\times 2^{99}$.How can i get this answer.Please help me.
Best Answer
Each time you derive $\dfrac{f(x)}{g(x)}$ the degree of the denominator will double, since $$\frac{d}{dx}\frac{N(x)}{D(x)}=\frac{N'(x)D(x)-D'(x)N(x)}{(D(x))^2}$$ So after a hundred iterations, the degree of the denominator in your case is $3\times 2 ^{100}=6\times 2^{99}$ You can easily see that the degree of the numerator satisfies the following recurrence relation: $$\text{dg} (n)=\underbrace{\text{dg}(n-1)-1}_{\text{degree of }N'(x)}+\underbrace{3\times 2^n}_{\text{degree of } D(x)}$$ Or: $$\text{dg} (n)=\underbrace{\text{dg}(n-1)}_{\text{degree of }N(x)}+\underbrace{3\times 2^n -1}_{\text{degree of } D'(x)}$$ at the $n^{th}$ derivative, where $\text{dg}(0)=k$ ($k$ is the degree of $f(x)$). Solving the recurrence relation yields:$$\text{dg}(n)=k-n+3\times 2^n-3$$ But if the degree of $f(x)$ is $100$ then the degree of $p(x)$ is $6\times 2^{99}-3$ which could've been obtained from your finding, namely that the degree of the final quotient is $-3$. To get the textbook's answer the degree of $f(x)$ should be $2$.