# Thread: Standard Taylor series

1. ## 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
Could someone clarify this, please.

I always assume I'm wrong, but rarely see why.

2. ## Re: Standard Taylor series

Can you simplify $\displaystyle \frac{1}{2}(-x)^2$?

-(x^2)/2
Yes!

4. ## Re: Standard Taylor series

You need to review arithmetic with negative numbers, e.g., this page: "Why is the Product of Negative Numbers Positive?"

5. ## Re: Standard Taylor series

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
Could someone clarify this, please.

I always assume I'm wrong, but rarely see why.
you have made a mistake. Note that in log(1+x) all the even powers of x are - terms and all odd powers of x are + terms So when you replace x by -x even powers will stay - and odd powers will become -