1. ## Having trouble with maddening Taylor Polynomials

Here's the question:

Calculate the Taylor polynomial T1,n(x) for f(x) = ln x

a. How many decimal digits are guaranteed in approximation of ln 1.5 by T1,10(1.5)?

b. What degree Taylor polynomial will guarantee the above approximation to four decimal digits?

I have read my professor's notes, but it is hard to understand them so if somebody could clarify to me the steps involved in solving this problem it would be much appreciated.

2. Originally Posted by Undefdisfigure
Here's the question:

Calculate the Taylor polynomial T1,n(x) for f(x) = ln x

a. How many decimal digits are guaranteed in approximation of ln 1.5 by T1,10(1.5)?

b. What degree Taylor polynomial will guarantee the above approximation to four decimal digits?

I have read my professor's notes, but it is hard to understand them so if somebody could clarify to me the steps involved in solving this problem it would be much appreciated.
The Taylor series for $f(x)$ about $a$ is:

$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)(x-a)^{k}}{k!}$

and the $n$-th order Taylor polynomial is obtained by truncating this series after the term where $k=n$.

So, for $f(x)=\ln(x)$:

$
T_{1,n}(x)=\sum_{k=0}^{n} \frac{f^{(k)}(1)(x-1)^{k}}{k!}
$

where $f^{(k)}(1)$ is the $k$-th derivative of $\ln(x)$ evaluated at $x=1$.

$D [\ln(x)] = \frac{1}{x}$

$
D^2 [\ln(x)] = D \left[\frac{1}{x} \right]=(-1)~\frac{1}{x^2}
$

:
:

$
D^n [\ln(x)] = D \left[\frac{1}{x} \right]=(-1)^{n-1} (n-1)!~\frac{1}{x^n}
$

Therefore $f^{(n)}(1) = (-1)^{n-1} (n-1)!$.

and so:

$
T_{1,n}(x)=\sum_{k=1}^{n} \frac{(-1)^{k-1}(x-1)^{k}}{k}
$

(the $k=0$ term disappears as $\ln(1)=0$)

RonL

3. Originally Posted by Undefdisfigure
a. How many decimal digits are guaranteed in approximation of ln 1.5 by T1,10(1.5)?

b. What degree Taylor polynomial will guarantee the above approximation to four decimal digits?
For these you need to look at the remainder term for a truncated Taylor series.

RonL