Chapter 11, Problem 1CRQ

a.

To determine

To fill: The blank in the statement “The nth Taylor polynomial of f at a is the polynomial $P_n\left(x\right)=\_\_\_\_\_\_\_\_\_\_\_$”.

Answer to Problem 1CRQ

The nth Taylor polynomial of f at a is the polynomial $\underset{\_}{P_n\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)+\cdots +\frac{f^{\left(n\right)}\left(a\right)}{n!}{\left(x-a\right)}^n}$.

Explanation of Solution

Let f be a function and its first n derivatives are defined at a.

Then the nth Taylor polynomial of f at a is the polynomial, $P_n\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)+\frac{f^{″}\left(a\right)}{2!}{\left(x-a\right)}^2+\cdots +\frac{f^{\left(n\right)}\left(a\right)}{n!}{\left(x-a\right)}^n$.

Therefore, the nth Taylor polynomial of f at a is the polynomial $\underset{\_}{P_n\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)+\cdots +\frac{f^{\left(n\right)}\left(a\right)}{n!}{\left(x-a\right)}^n}$.

b.

To determine

To fill: The blank in the statement “In terms of f and its derivatives, we have $P_n\left(a\right)=\_\_\_\_\_\_,{P^{\prime }}_n\left(a\right)=\_\_\_\_\_\_,\dots ,P_n{}^{\left(n\right)}\left(a\right)=\_\_\_\_\_\_\_$”.

Answer to Problem 1CRQ

In terms of f and its derivatives, we have $P_n\left(a\right)=\underset{\_}{f\left(a\right)},{P^{\prime }}_n\left(a\right)=\underset{\_}{f^{\prime }\left(a\right)},\dots ,P_n{}^{\left(n\right)}\left(a\right)=\underset{\_}{\frac{f^{\left(n\right)}\left(a\right)}{n!}}$.

Explanation of Solution

Let f be a function and its first n derivatives are defined at a.

Then the nth Taylor polynomial of f at a is the polynomial, $P_n\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)+\frac{f^{″}\left(a\right)}{2!}{\left(x-a\right)}^2+\cdots +\frac{f^{\left(n\right)}\left(a\right)}{n!}{\left(x-a\right)}^n$.

So it coincides with $f\left(x\right),f^{\prime }\left(x\right),\dots ,f^{\left(n\right)}\left(x\right)$ at a, that is, $P_n\left(a\right)=f\left(a\right),{P^{\prime }}_n\left(a\right)=f^{\prime }\left(a\right),\dots ,P_n{}^{\left(n\right)}\left(a\right)=\frac{f^{\left(n\right)}\left(a\right)}{n!}$.

Therefore, in terms of f and its derivatives, we have $P_n\left(a\right)=\underset{\_}{f\left(a\right)},{P^{\prime }}_n\left(a\right)=\underset{\_}{f^{\prime }\left(a\right)},\dots ,P_n{}^{\left(n\right)}\left(a\right)=\underset{\_}{\frac{f^{\left(n\right)}\left(a\right)}{n!}}$.