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\[\ln(-1) + \sum_{n=1}^{\infty}(-1)^{n+1}((x-2)^{n})/n\]

I am not sure if you're taylor series is right.

It shouldn't be alternating, the terms should all be negative

Yes, I know ln(1+x)
=> x - x^2/2 + x^3/3 - x^4/4

Replace all the x's with -x and you get -x -x^2/2 -x^3/3 -x^4/4

compute a general formula for the derivative of ln(1-x) <= what do you mean by this.

I think it would be \[\sum_{k=0}^{\infty} (-1)^k \frac{(-x)^{k+1}}{k+1}\] rather than what you have.

If I can do it for ln(1+x) its definetly \[\sum_{k=0}^{inftny} (-1)^k \frac{(x)^{k+1}}{k+1} \]

and then evaluated at a=2 we get.

i pi+(x-2)-1/2 (x-2)^2+1/3 (x-2)^3-1/4 (x-2)^4+1/5 (x-2)^5

ipi = ln(-1) just as you got.

i apologize for not regularly checking this, please wait as i read your replies!

my formula for ln(1-x) is wrong, i just evaluate at a=2 by hand. and that's what i got.

well that should fix ln(-1)

haha, yeah. i will now reattempt the problem, which will be similar to ln(3). thank you very much!!

but I just took the series for ln(1+x) and replaced it with -x, and found them to be alternating.

I am still getting all terms negative.

in that case its correct.

You can run it on maple to be sure.

yea get rid of the n on top of (-1)

I'll run it for you one sec.

so it would be..
\[\ln(3) + \sum_{n=1}^{\infty}(-1)(x+2)^n/(-3)^nn\]

without the (-3)

so simply (3)^n*n?

yes.

last confirm..
\[\ln(3) + \sum_{n=1}^{\infty}(-1)(x+2)^n/3^nn\]

Yes =] you can take the negative out too if you want, by the linearity property of the sum.

as in make it ln(3) - sum?

yes

i think i am going braindead.. hahaha. okay!! thank you!!

is this calc ii or i

ii at ucb

what's ucb

berkeley?

yes

Awesome

np.