anonymous
  • anonymous
help me please! pretty urgent.. Compute the Taylor series of ln(1-x) at a=2. I've done some work and the answer I have right now is ln(-1) + (
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\ln(-1) + \sum_{n=1}^{\infty}(-1)^{n+1}((x-2)^{n})/n\]
anonymous
  • anonymous
what I did was compute a general formula for the derivative of ln(1-x) at a=2 and plug that into the taylor series. however, ln(-1) doesn't make much sense.. thank you very much in advance!!
anonymous
  • anonymous
I am not sure if you're taylor series is right.

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anonymous
  • anonymous
It shouldn't be alternating, the terms should all be negative
anonymous
  • anonymous
Yes, I know ln(1+x) => x - x^2/2 + x^3/3 - x^4/4
anonymous
  • anonymous
my general formula for the derivatives is \[\sum_{1}^{\infty}(-1)^{n+1}(n-1)!\] which I place over n! and multiply by (x-2)^n..
anonymous
  • anonymous
Replace all the x's with -x and you get -x -x^2/2 -x^3/3 -x^4/4
anonymous
  • anonymous
compute a general formula for the derivative of ln(1-x) <= what do you mean by this.
anonymous
  • anonymous
i took the first four derivatives and noticed a trend so that I can replace the "top part" of the taylor series, which is f^(n)(a) with some sort of general formula. in this case, i got that the first derivative of ln(1-x) with a=2 is equal to 1, and the second derivative of ln(1-x) with a=2 is equal to -1
anonymous
  • anonymous
and then the third derivative of ln(1-x) with a=2 is 2, and the fourth derivative of ln(1-x) with a=2 is -6. based off this, i noticed that it is alternating and equal to (n-1)!, so i put that back into the top of the taylor series sum
anonymous
  • anonymous
I think it would be \[\sum_{k=0}^{\infty} (-1)^k \frac{(-x)^{k+1}}{k+1}\] rather than what you have.
anonymous
  • anonymous
If I can do it for ln(1+x) its definetly \[\sum_{k=0}^{inftny} (-1)^k \frac{(x)^{k+1}}{k+1} \]
anonymous
  • anonymous
and then evaluated at a=2 we get.
anonymous
  • anonymous
i pi+(x-2)-1/2 (x-2)^2+1/3 (x-2)^3-1/4 (x-2)^4+1/5 (x-2)^5
anonymous
  • anonymous
ipi = ln(-1) just as you got.
anonymous
  • anonymous
The only thing you got messed up on was the negative term alternating, they are all negative. sorry for writing so much i was trying to figure it out for myself.
anonymous
  • anonymous
i apologize for not regularly checking this, please wait as i read your replies!
anonymous
  • anonymous
my formula for ln(1-x) is wrong, i just evaluate at a=2 by hand. and that's what i got.
anonymous
  • anonymous
i'm confused as to why the terms are not alternating. when i plug in 2, i explicitly get negative numbers in an alternating fashion due to the alternating negative sign for the derivative
anonymous
  • anonymous
as in the first derivative of ln(1-x) being 1/(x-1), and the second derivative being -1/(x-1)^2 and etc..
anonymous
  • anonymous
wow. this was actually an assignment for class and i just received notice that there was a typo and that this is centered at a=-2 and not a=2.. i shall reattempt this problem.. thank you very much!!
anonymous
  • anonymous
well that should fix ln(-1)
anonymous
  • anonymous
haha, yeah. i will now reattempt the problem, which will be similar to ln(3). thank you very much!!
anonymous
  • anonymous
and they are alternating because f^'(x) = -1/x-1 f''(x) = -1/(x-1)^2 f'''(x) = 2/(x-1)^3 but at a=-2, f''''(x) = -6/(x-1)^4. All the odd terms will be negative because of -2, and even by design
anonymous
  • anonymous
but I just took the series for ln(1+x) and replaced it with -x, and found them to be alternating.
anonymous
  • anonymous
for ln(1-x) at a=-2, i got the formula.. \[\ln(3) + \sum_{1}^{\infty}(-1)^{n+1}(x+2)^n/3^nn\] does this look correct?
anonymous
  • anonymous
I am still getting all terms negative.
anonymous
  • anonymous
I apologize, the 3^n on the bottom should be (-3)^n, this would multiply with the (-1)^n+1 so that the terms would always be negatie
anonymous
  • anonymous
in that case its correct.
anonymous
  • anonymous
is there a simpler way of stating that it would always be negative? or is this the best way of presenting the answer. also, thank you very much! i am definietly becoming a fan :)
anonymous
  • anonymous
You can run it on maple to be sure.
anonymous
  • anonymous
yea get rid of the n on top of (-1)
anonymous
  • anonymous
I'll run it for you one sec.
anonymous
  • anonymous
log(3)-(x+2)/3-1/18 (x+2)^2-1/81 (x+2)^3-1/324 (x+2)^4-(x+2)^5/1215-(x+2)^6/4374-(x+2)^7/15309-(x+2)^8/52488-(x+2)^9/177147-(x+2)^10
anonymous
  • anonymous
so it would be.. \[\ln(3) + \sum_{n=1}^{\infty}(-1)(x+2)^n/(-3)^nn\]
anonymous
  • anonymous
without the (-3)
anonymous
  • anonymous
so simply (3)^n*n?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
last confirm.. \[\ln(3) + \sum_{n=1}^{\infty}(-1)(x+2)^n/3^nn\]
anonymous
  • anonymous
Yes =] you can take the negative out too if you want, by the linearity property of the sum.
anonymous
  • anonymous
as in make it ln(3) - sum?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i think i am going braindead.. hahaha. okay!! thank you!!
anonymous
  • anonymous
is this calc ii or i
anonymous
  • anonymous
ii at ucb
anonymous
  • anonymous
what's ucb
anonymous
  • anonymous
berkeley?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Awesome
anonymous
  • anonymous
once again, thank you very much! i think i shall call it a night.. or at least until i post another question here.. thank you!
anonymous
  • anonymous
np.

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