Q. 78 - Calculus

Question

In Exercises 77–82,

(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

${10}^{-6}$

$e^{0.9}$

Step-by-Step Solution

Answer

(a) The proper Maclaurin series is $e^{0.9}=\sum _{k=0}^{\infty }\frac{1}{k!}(0.9{)}^k$

(b) The Lagrange's form for the remainder is $\left|R_5\left(x\right)\right|\le \frac{e^{0.9}(0.9{)}^6}{720}$

(c) The required smallest value is $9$

Step 1 Given Information

The quantity is $e^{0.9}$

Part (a) Step 1 Finding the Maclaurin series

The objective is to express the given quantity in series form using appropriate Maclaurin series. The Maclaurin series for the function $f\left(x\right)=e^x$ is,
$e^x=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$
Substitute $0.9$ for $x$ in the above series.
Therefore, the required series form of the given series is,
$e^{0.9}=\sum _{k=0}^{\infty }\frac{1}{k!}(0.9{)}^k$

Part (b) Step 1 Lagrange Remainder

The Lagrange's form for the remainder is
$R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$
So, the Lagrange's form for the remainder to bound the error in using the $5^{\text{th }}$ degree Maclaurin polynomial to approximate the given quantity is,
$\left|R_5\left(x\right)\right|=\left|\frac{f^{\left(6\right)}\left(c\right)}{6!}{x}^6\right|$
The value of $f^{\left(6\right)}\left(x\right)=e^x$ since $f\left(x\right)=e^x$.
Thus,
$\left|R_5\left(x\right)\right|\le \left|\frac{f^{\left(6\right)}\left(c\right)}{6!}{x}^6\right|$

Substitute $0.9$ for $c$ and $x$ in the above expression. Thus, the value of $\left|R_5\left(x\right)\right|$ is,
$\left|R_5\left(x\right)\right|\le \frac{e^{0.9}(0.9{)}^6}{720}$

Part (c) Step 1 Calculation

Since $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$
Implies that,

$e^{0.9}=1+0.9+\frac{(0.9{)}^2}{2!}+\frac{(0.9{)}^3}{3!}+\frac{(0.9{)}^4}{4!}+\frac{(0.9{)}^5}{5!}+\frac{(0.9{)}^6}{6!}$

$+\frac{(0.9{)}^7}{7!}+\frac{(0.9{)}^8}{8!}+\frac{(0.9{)}^9}{9!}\cdots$
Hence, by trial and error, the smallest value of $n$ so that the $n$th degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within ${10}^{-6}$  is $9$ .