$f(x)=e^x$ be approximated by Taylor's polynomial of degree $n$ at the point ...

Question

I am stuck on the following problem:

Let $f(x)=e^x$ be approximated by Taylor's polynomial of degree $n$ at the point $x=\frac12$ and on the entire interval $[0,1]$. If the absolute error in this approximation does not exceed $10^{-2},$ then the value of $n$ should be taken as

1. $0$

2. $1$

3. $2$

4. $3$

I am not sure how to tackle it. Can someone explain? Thanks in advance for your time.

What I know is given below:

Here I have to determine the value of $n$ so that $R_n(x)<10^{-2}$.That is where I am stuck and could not do my calculations right.

Given that $f(x) =e^x$. Here, we obtain: \begin{align*} f(x) &\approx p_3(x) \\ &= f(0) + f'(0)(x - 0) + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 \\ \end{align*} Thus, substituting $x = 0.5$ and taking $n=3$ yields: \begin{align*} f(0.5) &\approx p_3(0.5) \\ &= 1 + 0.5+\frac{1}{2}(0.5)^2 + \frac{1}{6}(0.5)^3 \\ &\approx 1+0.5+0.125+0.0208 \\ &=1.6458 \end{align*}

Answer 1

Observe, you can add zero and use the known MacClaurin series for the exponential: $$e^x = e^{x-1/2+1/2} =e^{x-1/2}e^{1/2}= \sqrt{e}(e^{x-1/2})$$ hence $$ e^x = \sqrt{e}\left(1+(x-1/2)+\frac{1}{2}(x-1/2)^2+ \cdots +\frac{1}{k!}(x-1/2)^k+\cdots \right)$$ Now, you just need to determine which remainder theorem you wish to apply given the interval $[0,1]$ is the desired interval of application. The coefficient of the $k$-th term in the Taylor expansion is seen to be $\sqrt{e}/k!$

A Theorem (from Ash's Real Analysis) which I'll use: If $f$ has $k$ derivatives on a closed interval $I$ with $\partial I = \{a,b \}$ then $$ f(b) = T_k(b) +R_{k}(b) = \sum_{j=0}^k \frac{f^{(j)}(a)}{j!}(b-a)^j +R_{k}(b) $$ where $R_{k}(b) = f(b)-T_k(b)$ is the $k$-th remainder. Moreover, there exists $c \in int(I)$ such that $$ R_k(b) = \frac{f^{(k+1)}(c)}{(k+1)!}(b-a)^{k+1}. $$

Ok, so think of $a=1/2$ and $b=x \in [0,1]$ you see that $(b-a) \leq 1/2$. Moreover, we also observe $$\frac{f^{(k+1)}(c)}{(k+1)!} = \frac{\sqrt{e}}{(k+1)!}$$ Therefore, $$ |R_k(x)| \leq \frac{\sqrt{e}}{(k+1)!}\left( \frac{1}{2} \right)^{k+1} $$ You wish to find $k$ for which $|R_k(x)| < 0.01$. I found with wolfram alpha that $k=3$ gives $|R_k(x)| < 0.0043$.