# Thread: taylor series

1. ## taylor series

how would you do this?
Find the Taylor series for ln(x) about 2.

2. Originally Posted by davecs77
how would you do this?
Find the Taylor series for ln(x) about 2.
see here

if you need further explanation, just say so

topsquark showed you the main process of deriving something like this, but again, memorization could be used here, the page i gave you has the general formulas

3. Originally Posted by davecs77
how would you do this?
Find the Taylor series for ln(x) about 2.
Just for the record, let $f(x) = ln(x)$

$f(x) = ln(x)$ ==> $f(2) = ln(2)$

$f^{\prime}(x) = \frac{1}{x}$ ==> $f^{\prime}(2) = \frac{1}{2}$

$f^{\prime \prime}(x) = -\frac{1}{x^2}$ ==> $f^{\prime \prime}(2) = -\frac{1}{4}$

etc.

So
$f(x) \approx ln(2) + \frac{1}{1!}\frac{1}{2}(x - 2) - \frac{1}{2!}\frac{1}{4}(x - 2)^2 + ~ ...$

$f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...$

-Dan

4. Originally Posted by topsquark
Just for the record, let $f(x) = ln(x)$

$f(x) = ln(x)$ ==> $f(2) = ln(2)$

$f^{\prime}(x) = \frac{1}{x}$ ==> $f^{\prime}(2) = \frac{1}{2}$

$f^{\prime \prime}(x) = -\frac{1}{x^2}$ ==> $f^{\prime \prime}(2) = -\frac{1}{4}$

etc.

So
$f(x) \approx ln(2) + \frac{1}{1!}\frac{1}{2}x - \frac{1}{2!}\frac{1}{4}x^2 + ~ ...$

$f(x) \approx ln(2) + \frac{1}{2}x - \frac{1}{8}x^2 + ~ ...$

-Dan
How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2)

5. Originally Posted by davecs77
How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2)

The answer in my book is ln2 + summation of n = 1 going to infinity of
(-1)^(n - 1) times ((x - 2)^n)/(n2^n)

I get the first part..ln2 but dont understand the rest. Thank you.

6. Originally Posted by davecs77
How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2)
Crap! I goofed!

The general form for the Taylor series of f(x) about x = a is:
$f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + ~ ... ~ + \frac{1}{n!}f^{(n)}(a)(x - a)^n + ~ ...$

So for a = 2 the first coefficient is
$\frac{1}{1!}f^{\prime}(2) = 1 \cdot \frac{1}{2} = \frac{1}{2}$

The second coefficient is
$\frac{1}{2!}f^{\prime \prime}(2) = \frac{1}{2} \cdot -\frac{1}{4} = -\frac{1}{8}$

So the series should be:
$f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...$

You can work the general formula for the nth term. It's easy enough.

-Dan

7. Originally Posted by topsquark
Crap! I goofed!
can a Mormon say that? you know, the C-word, i only quoted "goofed" because it was funny as h

8. Originally Posted by topsquark
Crap! I goofed!

The general form for the Taylor series of f(x) about x = a is:
$f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + ~ ... ~ + \frac{1}{n!}f^{(n)}(a)(x - a)^n + ~ ...$

So for a = 2 the first coefficient is
$\frac{1}{1!}f^{\prime}(2) = 1 \cdot \frac{1}{2} = \frac{1}{2}$

The second coefficient is
$\frac{1}{2!}f^{\prime \prime}(2) = \frac{1}{2} \cdot -\frac{1}{4} = -\frac{1}{8}$

So the series should be:
$f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...$

You can work the general formula for the nth term. It's easy enough.

-Dan
You are only doing 2 coefficients since a = 2?

9. Originally Posted by davecs77
You are only doing 2 coefficients since a = 2?
No, I only did two terms since I'm lazy. The series is infinite with each term being smaller than the last.

-Dan

10. Originally Posted by davecs77
how would you do this?
Find the Taylor series for ln(x) about 2.
An indirect way to find the taylor series is to use the following known Maclaurin series

11. Originally Posted by curvature
An indirect way to find the taylor series is to use the following known Maclaurin series
Then