Chapter 10.3, Problem 32E

To determine

To find: The quadratic approximation of $f(x)={(1+x)}^k$ (k a constant).

Answer to Problem 32E

That is, $|x{|}^3<0.01,$ , which gives $0\le x<\sqrt[3]{0.01}\approx 0.215$ in the interval $[0,1],$ this occurs when $0\le x<0.215$

Explanation of Solution

Given information:

  $f(x)={(1+x)}^k$ Series of a function $f(x)$ is given by

  $f(x)=f(0)+\left[f^{\prime }(0)\right]x+\frac{\left[f^{″}(0)\right]x^2}{2!}+\frac{\left[f^{‴}(0)\right]x^3}{3!}+\dots \mathrm{..}$

Calculation:

To construct the second order polynomial to evaluate $f(x)$ its first two derivatives at

  $x=0\cdot f(0)=1$

  $f^{\prime }(x)=k{(1+x)}^{k-1}\Rightarrow f^{\prime }(0)=k{f}^{″}(x)=k(k-1){(1+x)}^{k-2}\Rightarrow f^{″}(0)=k(k-1)$

Therefore,

  $P_2(x)=1+kx+\frac{k(k-1)}{2!}{x}^2$

If $k=3,$

  $f(x)={(1+x)}^3$ evaluating the derivatives of $f(x)f^{\prime }(x)=3{(1+x)}^2$

  $f^{″}(x)=6(1+x)f^{‴}(x)=6$

According to the Remainder Estimation Theorem $\left|R_n(x)\right|\le M\frac{r^{n+1}|x-a{|}^{n+1}}{(n+1)!}$ Using Remainder Estimation Theorem with $n=2,M=6$ and $r=1$ Giving $R_2(x)\le \frac{6|x{|}^3}{3!}=|x{|}^3($ (In this particular case it is actually true that $R_2(x)=x^3,$ , since $f(x)$ is a cubic polynomial .) Thus the absolute error is less than $\frac{1}{100}$ whenever $R_2(x)<0.01$ . That is, $|x{|}^3<0.01,$ , which gives $0\le x<\sqrt[3]{0.01}\approx 0.215$ in the interval $[0,1],$ this occurs when $0\le x<0.215$