Question

Power Series Representation
Please Check if my work is correct.
Thanks

Answer 1

Everything is right up to your approximation.
\[\ln(1-x)=-\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}=-\sum_{n=1}^\infty \frac{x^n}{n}\approx-x-\frac{x^2}{2}-\frac{x^3}{3}\]
You want to approximate \(\ln.75\), which means you want to find
\[\ln(.75)=\ln(1-.25)~~\Rightarrow~~\text{substitute .25 for }x\]
\[\ln(.75)\approx-\frac{1}{4}-\frac{1}{2}\left(\frac{1}{4}\right)^2-\frac{1}{3}\left(\frac{1}{4}\right)^3\]

Answer 2

wait, why do i need to substitute .25 instead?

Answer 3

You've found the power series rep for \(\ln(1-x)\), not \(\ln x\).

You want to find \(\ln(.75)\), but in order to use the power series you've found, you must write it as \(\ln(1-.25)\), since \(1-.25=.75\). Then, according to the power series' first three terms, plugging in \(.25\) will give you the desired approximation.

Answer 4

ooh i get it now.
Thanks a lot for the correction

Answer 5

yw

Answer 6

Oh, one red flag for telling if you made a mistake... If you're trying to approximate \(\ln(.75)\), you probably won't have any expression containing \(\ln(0.75)\).

Answer 7

... as your answer, I mean.

Answer 8

Aright