Taylor's theorem

**harry** · May 21st 2007, 11:53 AM

multiple choice

**CaptainBlack** · May 22nd 2007, 01:44 PM

(1) Find the sum:

$<br /> S = x + \frac{x^2}{2\cdot 2}+ \frac{x^3}{3\cdot 2^2}+ \frac{x^4}{4\cdot 2^3} + ...<br />$

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:

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

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

**ThePerfectHacker** · May 22nd 2007, 02:28 PM

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

Taylor's Theorem (Lagrange): Let $f$ be $n$ differenciable on $(a,b)$ (with $a<0<b$ ) 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.

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