Thread: multipication of power series

1. multipication of power series

I was wondering if someone could please check my work because i keep getting this one wrong:

so the function is:
$e^8x ln\left(1 - \frac{x}{3}\right)$

and so far i have
$e^8x = 1 + 8x + \frac{64x^2}{2!} + ...$
$ln\left(1 - \frac{x}{3}\right) = -x - \frac{x^2}{6}
- \frac{x^3}{27}$

and then when i multiple (i only need first 3 nonzero terms) and collect like terms i get:
$= -x - \frac{49x^2}{6} - \frac{37x^3}{27}$

2. Hello,
Originally Posted by Tascja
I was wondering if someone could please check my work because i keep getting this one wrong:

so the function is:
$e^8x ln\left(1 - \frac{x}{3}\right)$

and so far i have
$e^8x = 1 + 8x + \frac{64x^2}{2!} + ...$
$ln\left(1 - \frac{x}{3}\right) = -x - \frac{x^2}{6}
- \frac{x^3}{27}$

and then when i multiple (i only need first 3 nonzero terms) and collect like terms i get:
$= -x - \frac{49x^2}{6} - \frac{37x^3}{27}$
that's because you got wrong in the logarithm power series :
$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots$

so for $\ln(1-x/3)$, it should be $-{\color{red}\frac x3}-\frac{({\color{red}x/3})^2}{2}-\frac{({\color{red}x/3})^3}{3}-\dots=-\frac x3-\frac{x^2}{18}-\frac{x^3}{81}-\dots$

3. oh okay thanks that makes a lot more sense!