For ln(2+x) can this be written as 2ln(1+x) or is it ln2 + ln(1+x)?
Neither! Your second attempt is closer to the target. Start with ln(2+x)=ln[2(1+x/2)].
Can i write ln(2+x) as ln[1+(1+x)] ?? would this work the same as ln[2(1+x/2)] ??
You can certainly write ln(2+x) as ln[1+(1+x)], but then, what will you do with that? That would be only useful to get Taylor representation around a=-1, not so for Maclaurin representation.
Ok but if I have ln(2+x)=ln[2(1+x/2)] does this equal ln2 + ln(1+x/2)?
So then for ln(2+x)*Tan-(x^2) I get [ln2 + (x + x^2/2 +x^3/3 ...)]*(x^2 + x^6/3 + x^10/5...)
DO I multiply ln2 with (x^2 +...) and end with
x + (x^2/2 + ln2x^2) + x^3/3
or am I missing something?
You multiply series like polynomials: each term with each term. So, you should not end up with you wrote above.
Candice, I would check your numbers for expanding ln(2+x)*tan-(x^2) because they are off. ln (2+x) is alternating too.
In my notes I have ln(1+x)=sum of x^n/n
is this not right?
ln(1+x)= Sigma (-1)^(n+1) x^n/n (sigma is from 1 to infinity)
also tan-1(x) = (-1)^n x^(2n+1)/ (2n+1)
so they both will be alternating in sign.
I had the same numbers as you for tan-(x^2)but with alternating signs
but for ln(2+x) i had (ln 2 + x/2 - x^2/8 + x^3/24 - ...)
okay thanks a lot, that helps.