程序代写代做代考 ### Question 3

### Question 3

#### Part a

If $P_n(x)+Q_m(x)$ is the interpolating polynomial for $f(x)+g(x)$, then for any $x_i \in X_f \cup X_g$ , $P_n(x_i)+Q_m(x_i)-f(x_i)-g(x_i)=0$ .

Without loss of generity, we can assume that $x_i \in X_f$, then we can obtain $g(x_i) = Q_m(x_i)$ for any $x_i \in X_f$.

Similarly, we can obtain that $f(x_i) = P_m(x_i)$ for any $x_i \in X_g$.

#### Part b

The order of $P_n(x)+Q_m(x)$ is the maximum of the orders of $P_n(x)$ and $Q_m(x)$ .

$|P_n(x)+Q_m(x)-f(x)-g(x)| \le $ $ |P_n(x)-f(x)| + |Q_m(x)-g(x)| $

So the error bound is the sum of the error bounds of $P_n(x)$ and $Q_m(x)$

#### Part c

If $P_n(x)*Q_m(x)$ is the interpolating polynomial for $f(x)*g(x)$, then for any $x_i \in X_f \cup X_g$ , $P_n(x_i)*Q_m(x_i)-f(x_i)*g(x_i)=0$ .

Without loss of generity, we can assume that $x_i \in X_f$, then we can obtain $P_n(x_i)=f(x_i)$, so $P_n(x_i)(Q_m(x_i)-g(x_i))=0$ .

So if $P_n(x_i) \neq 0$, then $Q_m(x_i)=g(x_i)$ for any $x_i \in X_f$.

Similarly, we can obtain that if $Q_m(x_i) \neq 0$, then $P_n(x_i)=f(x_i)$ for any $x_i \in X_g$.