1. ## Taylor's theorem

multiple choice

2. (1) Find the sum:

$
S = x + \frac{x^2}{2\cdot 2}+ \frac{x^3}{3\cdot 2^2}+ \frac{x^4}{4\cdot 2^3} + ...
$

Consider this as a function of $x$ (note the negative sign in front of the term in $x^4$
I presume is a mistake), the derivative of this is:

$
U(x) = 1 + \frac{x}{ 2}+ \frac{x^2}{ 2^2}+ \frac{x^3}{ 2^3} + ...
$

which is a geometric series and it converges for $|x|<2$.

So write down the sum of this last series, then integrate with the constant
of integration set so that the integral is zero for $x=0$, and that
is your answer (it looks like (d) to me but you will need to check).

RonL

3. There are many verions, I use the the following one.

Taylor's Theorem (Lagrange): Let $f$ be $n$ differenciable on $(a,b)$ (with $a<0) and for $0\not =x\in (a,b)$ the remainder is given by $R_n(x) = \frac{f^{(n)}(y)}{n!}\cdot x^n$ for some $y$ between $0$ and $x$. Where the remainder is the difference between the function and its $n-1$ degree Taylor polynomial $T_{n-1}(x)$, defined as $T_{n-1}(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!}\cdot x^k$ and $R_n(x) = f(x) - T_{n-1}(x)$.

So, if $n=5$ then $T_4(x) = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}$ on $(-.1,.1)$

Then, $R_5(x) = \frac{f^{(5)}(y)}{5!}\cdot x^5 = \frac{e^y}{5!} \leq \frac{e^{.1}}{5!}\approx .003$

So it is 2 decimal points.