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anonymous
 5 years ago
help me please! pretty urgent..
Compute the Taylor series of ln(1x) at a=2.
I've done some work and the answer I have right now is ln(1) + (
anonymous
 5 years ago
help me please! pretty urgent.. Compute the Taylor series of ln(1x) at a=2. I've done some work and the answer I have right now is ln(1) + (

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\ln(1) + \sum_{n=1}^{\infty}(1)^{n+1}((x2)^{n})/n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what I did was compute a general formula for the derivative of ln(1x) 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
 5 years ago
Best ResponseYou've already chosen the best response.0I am not sure if you're taylor series is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It shouldn't be alternating, the terms should all be negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I know ln(1+x) => x  x^2/2 + x^3/3  x^4/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my general formula for the derivatives is \[\sum_{1}^{\infty}(1)^{n+1}(n1)!\] which I place over n! and multiply by (x2)^n..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Replace all the x's with x and you get x x^2/2 x^3/3 x^4/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0compute a general formula for the derivative of ln(1x) <= what do you mean by this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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(1x) with a=2 is equal to 1, and the second derivative of ln(1x) with a=2 is equal to 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then the third derivative of ln(1x) with a=2 is 2, and the fourth derivative of ln(1x) with a=2 is 6. based off this, i noticed that it is alternating and equal to (n1)!, so i put that back into the top of the taylor series sum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it would be \[\sum_{k=0}^{\infty} (1)^k \frac{(x)^{k+1}}{k+1}\] rather than what you have.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If I can do it for ln(1+x) its definetly \[\sum_{k=0}^{inftny} (1)^k \frac{(x)^{k+1}}{k+1} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then evaluated at a=2 we get.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i pi+(x2)1/2 (x2)^2+1/3 (x2)^31/4 (x2)^4+1/5 (x2)^5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ipi = ln(1) just as you got.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0i apologize for not regularly checking this, please wait as i read your replies!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my formula for ln(1x) is wrong, i just evaluate at a=2 by hand. and that's what i got.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'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
 5 years ago
Best ResponseYou've already chosen the best response.0as in the first derivative of ln(1x) being 1/(x1), and the second derivative being 1/(x1)^2 and etc..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow. 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
 5 years ago
Best ResponseYou've already chosen the best response.0well that should fix ln(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, yeah. i will now reattempt the problem, which will be similar to ln(3). thank you very much!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and they are alternating because f^'(x) = 1/x1 f''(x) = 1/(x1)^2 f'''(x) = 2/(x1)^3 but at a=2, f''''(x) = 6/(x1)^4. All the odd terms will be negative because of 2, and even by design

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but I just took the series for ln(1+x) and replaced it with x, and found them to be alternating.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for ln(1x) 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
 5 years ago
Best ResponseYou've already chosen the best response.0I am still getting all terms negative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0in that case its correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is 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
 5 years ago
Best ResponseYou've already chosen the best response.0You can run it on maple to be sure.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea get rid of the n on top of (1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll run it for you one sec.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0log(3)(x+2)/31/18 (x+2)^21/81 (x+2)^31/324 (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
 5 years ago
Best ResponseYou've already chosen the best response.0so it would be.. \[\ln(3) + \sum_{n=1}^{\infty}(1)(x+2)^n/(3)^nn\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0last confirm.. \[\ln(3) + \sum_{n=1}^{\infty}(1)(x+2)^n/3^nn\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes =] you can take the negative out too if you want, by the linearity property of the sum.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as in make it ln(3)  sum?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think i am going braindead.. hahaha. okay!! thank you!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once again, thank you very much! i think i shall call it a night.. or at least until i post another question here.. thank you!
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