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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    \[\ln(-1) + \sum_{n=1}^{\infty}(-1)^{n+1}((x-2)^{n})/n\]

  2. anonymous
    • 5 years ago
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    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!!

  3. anonymous
    • 5 years ago
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    I am not sure if you're taylor series is right.

  4. anonymous
    • 5 years ago
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    It shouldn't be alternating, the terms should all be negative

  5. anonymous
    • 5 years ago
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    Yes, I know ln(1+x) => x - x^2/2 + x^3/3 - x^4/4

  6. anonymous
    • 5 years ago
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    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..

  7. anonymous
    • 5 years ago
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    Replace all the x's with -x and you get -x -x^2/2 -x^3/3 -x^4/4

  8. anonymous
    • 5 years ago
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    compute a general formula for the derivative of ln(1-x) <= what do you mean by this.

  9. anonymous
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    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

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

  12. anonymous
    • 5 years ago
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    If I can do it for ln(1+x) its definetly \[\sum_{k=0}^{inftny} (-1)^k \frac{(x)^{k+1}}{k+1} \]

  13. anonymous
    • 5 years ago
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    and then evaluated at a=2 we get.

  14. anonymous
    • 5 years ago
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    i pi+(x-2)-1/2 (x-2)^2+1/3 (x-2)^3-1/4 (x-2)^4+1/5 (x-2)^5

  15. anonymous
    • 5 years ago
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    ipi = ln(-1) just as you got.

  16. anonymous
    • 5 years ago
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    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.

  17. anonymous
    • 5 years ago
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    i apologize for not regularly checking this, please wait as i read your replies!

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

  19. anonymous
    • 5 years ago
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    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

  20. anonymous
    • 5 years ago
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    as in the first derivative of ln(1-x) being 1/(x-1), and the second derivative being -1/(x-1)^2 and etc..

  21. anonymous
    • 5 years ago
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    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!!

  22. anonymous
    • 5 years ago
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    well that should fix ln(-1)

  23. anonymous
    • 5 years ago
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    haha, yeah. i will now reattempt the problem, which will be similar to ln(3). thank you very much!!

  24. anonymous
    • 5 years ago
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    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

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

  26. anonymous
    • 5 years ago
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    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?

  27. anonymous
    • 5 years ago
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    I am still getting all terms negative.

  28. anonymous
    • 5 years ago
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    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

  29. anonymous
    • 5 years ago
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    in that case its correct.

  30. anonymous
    • 5 years ago
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    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 :)

  31. anonymous
    • 5 years ago
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    You can run it on maple to be sure.

  32. anonymous
    • 5 years ago
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    yea get rid of the n on top of (-1)

  33. anonymous
    • 5 years ago
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    I'll run it for you one sec.

  34. anonymous
    • 5 years ago
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    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

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

  36. anonymous
    • 5 years ago
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    without the (-3)

  37. anonymous
    • 5 years ago
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    so simply (3)^n*n?

  38. anonymous
    • 5 years ago
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    yes.

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

  40. anonymous
    • 5 years ago
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    Yes =] you can take the negative out too if you want, by the linearity property of the sum.

  41. anonymous
    • 5 years ago
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    as in make it ln(3) - sum?

  42. anonymous
    • 5 years ago
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    yes

  43. anonymous
    • 5 years ago
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    i think i am going braindead.. hahaha. okay!! thank you!!

  44. anonymous
    • 5 years ago
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    is this calc ii or i

  45. anonymous
    • 5 years ago
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    ii at ucb

  46. anonymous
    • 5 years ago
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    what's ucb

  47. anonymous
    • 5 years ago
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    berkeley?

  48. anonymous
    • 5 years ago
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    yes

  49. anonymous
    • 5 years ago
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    Awesome

  50. anonymous
    • 5 years ago
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    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!

  51. anonymous
    • 5 years ago
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    np.

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