# Taylor's theorem

• May 21st 2007, 11:53 AM
harry
Taylor's theorem
multiple choice
• May 22nd 2007, 01:44 PM
CaptainBlack
(1) Find the sum:

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

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

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

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

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

RonL
• May 22nd 2007, 02:28 PM
ThePerfectHacker
There are many verions, I use the the following one.

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

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

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

So it is 2 decimal points.