# Thread: Multiplication of two Taylor series

1. ## Multiplication of two Taylor series

I'm trying to multiply two Taylor series together to find the quartic Taylor polynomial about 0 of a function

$\displaystyle (1+\frac{1}{2}x^2+\frac{1}{6}x^3+....)(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)$

$\displaystyle = 1(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)+\frac{1}{2}x^2(x-\frac{1}{2}x^2+...)+\frac{1}{6}x^3(x-...)$

$\displaystyle = x+(-\frac{1}{2})x^2+(\frac{1}{3}+\frac{1}{2})x^3+(-\frac{1}{4}+\frac{1}{6})x^4$

$\displaystyle = x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{12}x^4$

The problem is when I enter the function into my computer maths program(mathcad) I get.

$\displaystyle = x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{3}x^4$

where am I going wrong

Oh the function by the way is

$\displaystyle f(x)= (e^x-x)(\ln(1+x))$

2. Originally Posted by macca101
I'm trying to multiply two Taylor series together to find the quartic Taylor polynomial about 0 of a function

$\displaystyle (1+\frac{1}{2}x^2+\frac{1}{6}x^3+....)(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)$

$\displaystyle = 1(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)+\frac{1}{2}x^2(x-\frac{1}{2}x^2+...)+\frac{1}{6}x^3(x-...)$

$\displaystyle = x+(-\frac{1}{2})x^2+(\frac{1}{3}+\frac{1}{2})x^3+(-\frac{1}{4}+\frac{1}{6})x^4$

$\displaystyle = x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{12}x^4$

The problem is when I enter the function into my computer maths program(mathcad) I get.

$\displaystyle = x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{3}x^4$

where am I going wrong

Oh the function by the way is

$\displaystyle f(x)= (e^x-x)(\ln(1+x))$
You're missing a -1/4 term

-1/4-1/4+1/6=-1/3

$\displaystyle x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{3}}{2}-\frac{x^{4}}{4}+\frac{x^{4}}{6}=x-\frac{x^{2}}{2}+\frac{5x^{3}}{6}-\frac{x^{4}}{3}$

3. Originally Posted by galactus
[b]You're missing a -1/4 term
Of course I am

Thanks