Chapter 8, Problem 10RCC

a.

To determine

To write an expression for nth-degree Taylor polynomial of f centred at a

Answer to Problem 10RCC

Expression for nth-degree Taylor polynomial of f centred at a.

  $\begin{array}{l}{T}_n(x)={\displaystyle \sum _{n=0}^n\frac{f^{(i)}(a)}{i!}}{(}^x\\ \text{ =f(a)+}\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}{(}^x+\cdot \cdot \cdot \cdot \cdot +\frac{f^{(n)}(a)}{n!}{(}^x\end{array}$

Explanation of Solution

Given information:

expression for nth-degree Taylor polynomial of f centred at a, to express.

Expression for nth-degree Taylor polynomial of f centred at a , can be expressed as

  $\begin{array}{l}{T}_n(x)={\displaystyle \sum _{n=0}^n\frac{f^{(i)}(a)}{i!}}{(}^x\\ \text{ =f(a)+}\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}{(}^x+\cdot \cdot \cdot \cdot \cdot +\frac{f^{(n)}(a)}{n!}{(}^x\end{array}$

b.

To determine

To write an expression for the Taylor series of f centred at a.

Answer to Problem 10RCC

  $\begin{array}{l}f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}}{(}^x\\ \text{ =}f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}{(}^x+\frac{f^{‴}(a)}{3!}{(}^x\cdot \cdot \cdot \cdot \end{array}$

Explanation of Solution

Given information:

theexpression for the Taylor series of f centred at a , to express

theexpression for the Taylor series of f centred at a , can be expressed as

  $\begin{array}{l}f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}}{(}^x\\ \text{ =}f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}{(}^x+\frac{f^{‴}(a)}{3!}{(}^x\cdot \cdot \cdot \cdot \end{array}$

c.

To determine

To write an expression for the Maclaurinseries off.

Answer to Problem 10RCC

Expression for the Maclaurin series of f :

  $f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n}=f(0)+\frac{f^{\prime }(0)}{1!}x+\frac{f^{″}(0)}{2!}{x}^2+\cdot \cdot \cdot \cdot$

Explanation of Solution

Given information:

The expression for the Maclaurinseries off , to express

Maclaurin series is the special case of Taylor series when $a=0$

  $f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n}=f(0)+\frac{f^{\prime }(0)}{1!}x+\frac{f^{″}(0)}{2!}{x}^2+\cdot \cdot \cdot \cdot$

d.

To determine

To show that $f(x)$ is equal to sum of its Taylor series.

Explanation of Solution

Given information:

Formula used:.

Proof. :

start by supposing that fis any function that can be represented by a power series

  $f(x)=c_0+c_1(x-a)+c_2{(x-a)}^2+c_3{(x-a)}^3+\cdot \cdot \cdot \cdot \text{ (}i\text{)}$

If we put $x=a$ in equation ( i ) , then $f(a)=c_0$

we can differentiate the series in Equation ( i ) term by term:

  $f^{\prime }(x)=c_1+2c_2(x-a)+3c_3{(x-a)}^2+4c_4{(x-a)}^3+\cdot \cdot \cdot \cdot \text{ (}ii\text{)}$

If we put $x=a$ in equation ( ii ) then $f^{\prime }(a)=c_1$

Differentiate both sides of equation ( ii )

  $f^{″}(x)=2c_{{}_2}+2\cdot 3c_3(x-a)+3\cdot 4c_4{(x-a)}^2+\cdot \cdot \cdot \cdot \text{ (}iii\text{)}$

If we put $x=a$ in equation ( iii ) then ${\text{<msup> f <mo>″</mo></msup> (a)=2c}}_2$

Differentiating equation ( iii )

  $f^{‴}(x)=2\cdot 3c_3+2\cdot 3\cdot 4(x-a)+3\cdot 4\cdot 5{(x-a)}^2+\cdot \cdot \cdot \cdot \text{ (}iiii\text{)}$

If we put $x=a$ in equation ( iiii ) then $\text{<msup> f <mo>‴</mo></msup> (a)=2}\cdot {\text{3c}}_3=3!c_3$

After differentiating n times

  $f^{(n)}(a)=2\cdot 3\cdot 4\cdot \cdot \cdot \cdot n{c}_n=n!c_n$

  $c_n=\frac{f^{(n)}(a)}{n!}$ , $c_n\text{ is the }nth\text{ coefficient}$

Substituting $c_n$ in equation ( i )

  $\begin{array}{l}f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}}{(}^x\\ \text{ =}f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}{(}^x+\frac{f^{‴}(a)}{3!}{(}^x\cdot \cdot \cdot \cdot \end{array}$

Therefore it is shown above that $f(x)$ is sum of its Taylor series.

e.

To determine

To state Taylor’s inequality

Answer to Problem 10RCC

Taylor’s inequality is:

  $\left|f^{(n+1)}(x)\right|\le M\text{ for }\left|x-a\right|\le d,\text{ then the remainder }{R}_n(x)\text{ of the Taylor series satisfies the inequality}$

  $\left|R_n(x)\right|\le \frac{M}{(n-1)!}{\left|x-a\right|}^{n+1}\text{ for}\left|x-a\right|\le d$

Explanation of Solution

Given information:

Taylor’s inequality, to express

Taylor’s inequality is an estimate result for the value of reminder term.

  $\left|f^{(n+1)}(x)\right|\le M\text{ for }\left|x-a\right|\le d,\text{ then the remainder }{R}_n(x)\text{ of the Taylor series satisfies the inequality}$

  $\left|R_n(x)\right|\le \frac{M}{(n-1)!}{\left|x-a\right|}^{n+1}\text{ for}\left|x-a\right|\le d$