Q. 77 - Calculus

Question

(a) Use appropriate Maclaurin series to express the quantities in series form. 

(b) Use Lagrange’s form for the remainder to bound the error in using the 5th degree Maclaurin polynomial to approximate the given quantity.

 (c) Find the smallest value of n so that the nth degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within ${10}^{-6}$.

 

$e^{0.3}$

Step-by-Step Solution

Answer

Part (a) The required series form of the given series is $e^{0.3}=\sum _{k=0}^{\infty }\frac{1}{k!}(0.3{)}^k$

Part (b) The required value of $\left|R_5\left(x\right)\right|$ is $\left|R_5\left(x\right)\right|\le \frac{e^{0.3}(0.3{)}^6}{720}$

Part (c) The minimum value of n for which the nth degree Maclaurin polynomial approximation to the provided quantity is guaranteed to be accurate to within ${10}^{-6}$ is 7 determined through trial and error.

Part (a) Step 1: Given information

The quantity is $e^{0.3}$.

Part (a) Step 2: Find the required series form

Let us consider the quantity, $e^{0.3}$

The goal is to use appropriate Maclaurin series to express the given quantity in series form.

The Maclaurin series for the function $f\left(x\right)=e^x$ is,

$e^x=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$

Insert $0.3$ for $x$ in the above series.

Thus, the given series' needed series form is,

$\begin{array}{rcl}{e}^{0.3}& =& \sum _{k=0}^{\infty }\frac{1}{k!}(0.3{)}^k\end{array}$

Part (b) Step 1: Given information

The given value is $\left|{\mathrm{R}}_5\left(\mathrm{x}\right)\right|$

Part (b) Step 2: Find the value for R 5 ( x )

For the remainder, the Lagrange form is ${\mathrm{R}}_{\mathrm{n}}\left(\mathrm{x}\right)=\frac{{\mathrm{f}}^{(\mathrm{n}+1)}\left(\mathrm{c}\right)}{(\mathrm{n}+1)!}{\mathrm{x}}^{\mathrm{n}+1}$

So, using the  $5^{\mathrm{th}}$degree Maclaurin polynomial to approximate the provided quantity, the Lagrange's form for the residual to bound the error is, $\left|{\mathrm{R}}_{\mathrm{s}}\left(\mathrm{x}\right)\right|=\left|\frac{{\mathrm{f}}^{\left(6\right)}\left(\mathrm{c}\right)}{6!}{\mathrm{x}}^6\right|$

The value of ${\mathrm{f}}^{\left(6\right)}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$ since $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$

Thus, $\left|{\mathrm{R}}_5\left(\mathrm{x}\right)\right|\le \left|\frac{{\mathrm{f}}^{\left(6\right)}(0.3)}{6!}{\mathrm{x}}^6\right|$

insert $0.3$ for c and x in the above expression

Thus, the value of $\left|{\mathrm{R}}_5\left(\mathrm{x}\right)\right|$ is, $\left|{\mathrm{R}}_5\left(\mathrm{x}\right)\right|\le \frac{{\mathrm{e}}^{0.3}(0.3{)}^6}{720}$

Part (c) Step 1 : Given information

The given quantity is ${10}^{-6}$

Part (c) Step 2: Find the value of n which is smallest

As , $e^x=1+x+\frac{x^2}{21}+\frac{x^3}{31}+\cdots$

It implies that ,

$e^{0.3}=1+0.3+\frac{(0.3{)}^2}{2!}+\frac{(0.3{)}^3}{3!}+\cdots$

Thus, the minimum value of n for which the ${\mathrm{n}}^{\mathrm{th}}$ degree Maclaurin polynomial approximation to the provided quantity is guaranteed to be accurate to within  ${10}^{-6}$is 7 determined through trial and error.