taylor series f(x) = ln(x) c = 1

Question

experimentx · Answer

konradzuse · Answer

Yeah, that's what I got too...  But apparently not correct.....

experimentx · Answer

let x = u + 1, expand it at u=0

experimentx · Answer

konradzuse · Answer

well actually the sumation starts at 0.

konradzuse · Answer

SO maybe ^n-1?

experimentx · Answer

f(x) = ln (x), x=1 => f(x) = 0
f'(x) = 1/x, x=1 => f(x) = 1
f''(x) = -1/x^2 => f(x) = -1
.. so on and on

konradzuse · Answer

hmm...

experimentx · Answer

use the regular formula ..
f^n(1)(x-1)^n/n!

konradzuse · Answer

f^n(1)?

experimentx · Answer

that's nth derivative of log(x) at x=1

konradzuse · Answer

Ok, so the same as what wolfram said, except n! instead of n?

experimentx · Answer

well, you will get some numbers from the derivative that would help cancelling some parts of n!

experimentx · Answer

it is essentially same as expansion of ln(1+u) at u=0 http://home.scarlet.be/math/taylor.htm#ln(1+x)-and-its-Macl then put back u = x-1

turingtest · Answer

where are you stuck?

turingtest · Answer

the Taylor expansion for a function around x=a is$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)$$

turingtest · Answer

in this case a=1 so we have$$\ln x=\sum_{n=0}^\infty\frac{f^{(n)}(1)}{n!}(x-1)$$so for the first term, what is f(1) (that is when n=0) ??