# Standard Taylor series

• Jun 12th 2012, 02:27 AM
froodles01
Standard Taylor series
Question is: Using a standard Taylor series, write down the Taylor series about 0 for the function

f(x) = ln(1 x),
giving all terms up to and including that include x^4
Now, Standard Taylor series of ln(1+x), I understand, is;

ln(1+x) = x - (1/2 x^2) +(1/3 x^3) - (1/4 x^4) + . . . . .
so substitute -x for +x I get

ln(1-x) = -x -(-1/2 x^2) + (-1/3 x^3) - (-1/4 x^4) + . . . .
Ln(1-x) = -x + (x^2)/2 - (x^3)/3 + (x^4)/4 - . . . . . .

But the answer I have been given as a specimen makes all of the signs -ve

I always assume I'm wrong, but rarely see why.
• Jun 12th 2012, 04:57 AM
emakarov
Re: Standard Taylor series
Can you simplify $\frac{1}{2}(-x)^2$?
• Jun 12th 2012, 05:30 AM
froodles01
Re: Standard Taylor series
-(x^2)/2
Yes!
• Jun 12th 2012, 05:37 AM
emakarov
Re: Standard Taylor series
You need to review arithmetic with negative numbers, e.g., this page: "Why is the Product of Negative Numbers Positive?"
• Jun 12th 2012, 06:08 AM
biffboy
Re: Standard Taylor series
Quote:

Originally Posted by froodles01
Question is: Using a standard Taylor series, write down the Taylor series about 0 for the function

f(x) = ln(1 x),
giving all terms up to and including that include x^4
Now, Standard Taylor series of ln(1+x), I understand, is;

ln(1+x) = x - (1/2 x^2) +(1/3 x^3) - (1/4 x^4) + . . . . .
so substitute -x for +x I get

ln(1-x) = -x -(-1/2 x^2) + (-1/3 x^3) - (-1/4 x^4) + . . . .
Ln(1-x) = -x + (x^2)/2 - (x^3)/3 + (x^4)/4 - . . . . . .

But the answer I have been given as a specimen makes all of the signs -ve