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agentx5

Topic: \(\textbf{Calculus 2}\) Evaluate the indefinite integral as an infinite series. \[\int \frac{\cos x − 1}{x} \ \ \ dx \] What the heck? I mean putting these in Taylor series forms are hard enough, but this? Not even sure where to start here >_< It worries me because the final is coming up and this is a hole in my knowledge... So, um, help?

  • one year ago
  • one year ago

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  1. telliott99
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    Well, could you start with the infinite series for cosine x?

    • one year ago
  2. telliott99
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    You'd have (1/x) (-x^2/2! + x^4/4! .....)

    • one year ago
  3. Spacelimbus
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    \[ \int \frac{\cos x}{x}dx - \int \frac{1}{x}dx \]

    • one year ago
  4. agentx5
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    Let's see... \[\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \text{for all x = } \mathbb{R}\]

    • one year ago
  5. agentx5
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    That's the MacLaurin equivalent for cosine

    • one year ago
  6. Spacelimbus
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    \[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \]

    • one year ago
  7. Spacelimbus
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    Is it? Did I look at the wrong one? (-:

    • one year ago
  8. Spacelimbus
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    oh yes that's the hyperbolic one! SOrry!

    • one year ago
  9. agentx5
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    Oh no wait I have the wrong one, I wrote sine by mistake

    • one year ago
  10. agentx5
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    Arg soo confusing lol

    • one year ago
  11. Spacelimbus
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    lets both check again then hehe.

    • one year ago
  12. Spacelimbus
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    yes I believe I have the right one

    • one year ago
  13. agentx5
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    I thought you couldn't just do 1/x , you'd have to convert that to a series as well, yes?

    • one year ago
  14. experimentX
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    \[ \cos x = \sum_{n = 0}^\infty {(-1)^nx^{2n}\over (2n)!}\] which makes \[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = {(-1)^nx^{2n}\over 2n(2n)!}\]

    • one year ago
  15. experimentX
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    +c

    • one year ago
  16. agentx5
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    Anti-derivative power rule?

    • one year ago
  17. Spacelimbus
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    \[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \] Distribute it out \[ \frac{\cos x}{x} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n-1}}{(2n)!} \]

    • one year ago
  18. Spacelimbus
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    yes that's it @agentx5

    • one year ago
  19. mahmit2012
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    all has mistake!

    • one year ago
  20. Spacelimbus
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    where @mahmit2012 ?

    • one year ago
  21. mahmit2012
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    all !but it is time to pray i wll come back and show you

    • one year ago
  22. experimentX
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    \[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = \sum_{n=1}^\infty {(-1)^nx^{2n}\over 2n(2n)!} \]

    • one year ago
  23. mahmit2012
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    after pray !

    • one year ago
  24. experimentX
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    lol ... where did i make mistake??

    • one year ago
  25. Spacelimbus
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    Did we do the same @experimentX ? besides that you completed it already.

    • one year ago
  26. Spacelimbus
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    I think we did.

    • one year ago
  27. experimentX
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    n=1, n=0

    • one year ago
  28. Spacelimbus
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    yes but obviously "ALL" is wrong :D

    • one year ago
  29. mahmit2012
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    first of all the original question has interval [0,inf) and it has many different solution to get the answer. and if you are looking for undefined integral so just a matching series is the answer ! but with a log function.

    • one year ago
  30. mahmit2012
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    |dw:1343670857318:dw|

    • one year ago
  31. agentx5
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    *tilting head sideways to read* :-D

    • one year ago
  32. mahmit2012
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    |dw:1343670990609:dw|

    • one year ago
  33. mahmit2012
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    |dw:1343671203407:dw|

    • one year ago
  34. agentx5
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    In that first step you just used the cosine equivalent from MacLaurin right? Then you're dividing that polynomial by x, then... Pattern?

    • one year ago
  35. agentx5
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    Well the final answer is probably going to be ln|x| + something, I think...

    • one year ago
  36. mahmit2012
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    yes but you didn't allow to separate 1 and cos because of in cosx/x you have a log function.

    • one year ago
  37. experimentX
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    can't we cancel 1 and start from n=1 instead?

    • one year ago
  38. mahmit2012
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    yes but if you start with 1-cosx in num so you wont confront the mistake.

    • one year ago
  39. mahmit2012
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    take a look my solution and figure out the mistake.

    • one year ago
  40. Spacelimbus
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    @mahmit2012, I think I understand the main problem, it was the 'splitting' up of terms right? Separating the -1 or -cos(x) from the remaining term.

    • one year ago
  41. Spacelimbus
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    Better: Substitute it directly and derive the formula in form of a series representation?

    • one year ago
  42. experimentX
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    yeah ... my case (-1)^n was mistake ..

    • one year ago
  43. mahmit2012
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    and for the last this problem is originally has defined to 0 to inf.

    • one year ago
  44. mahmit2012
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    |dw:1343671601188:dw|

    • one year ago
  45. Spacelimbus
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    I understand, thanks @mahmit2012

    • one year ago
  46. experimentX
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    perhaps this is related to http://en.wikipedia.org/wiki/Dirichlet_integral

    • one year ago
  47. mahmit2012
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    Not exactly but a little bit !

    • one year ago
  48. mukushla
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    *

    • one year ago
  49. agentx5
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    Um, how did you do this step? \[\frac{x^2}{2\cdot 2!}-\frac{x^4}{4\cdot 4!}+\frac{x^4}{6\cdot 6!}-... = \sum_{n=1}^{\infty} (-1)^n \cdot \frac{1}{2n(2n!)} \cdot x^{2n}\]

    • one year ago
  50. experimentX
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    in the series of cosine ... there is 1 ... cancel out 1 ... it's simply ... shifting n=0 to n=1

    • one year ago
  51. agentx5
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    actually that needs to be n \(\pm\) 1on the alternating part

    • one year ago
  52. experimentX
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    and note that you are making same mistake as I previously did, (-1)^n+1 or n-1 ... note that this also shifts.

    • one year ago
  53. agentx5
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    @experimentX you mean shifting the index eh?

    • one year ago
  54. experimentX
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    yep

    • one year ago
  55. experimentX
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    since the one gets cancelled ... the after that x is divided.

    • one year ago
  56. agentx5
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    Odd though, because it's showing @mahmit2012's answer, put into the index (by increasing all n's by n+2) they want, as incorrect. They have the sigma upper and lower bounds fixed at n=2 to n=\(\infty\) ________ + C

    • one year ago
  57. agentx5
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    Then of course I'll need to use a calculator to find a certain number of significant figures blah blah but it's this part that's costing me 15 points on this homework (everything else is correct but this one problem, which of course would have to be the most points)

    • one year ago
  58. agentx5
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    I read that link you posted @experimentX , I didn't quite understand it. New concepts I've not encountered before so a little overwhelming lol

    • one year ago
  59. agentx5
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    The most I got out of it was that some things like that (sin(x))/x function are equal to 90 degrees, or \(\pi\)/2 radians

    • one year ago
  60. experimentX
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    i don't understand it too well ... it's the second time i'm seeing the the definite integral with Laplace transform giving it's value.

    • one year ago
  61. experimentX
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    this gives the value of |dw:1343673440350:dw|

    • one year ago
  62. agentx5
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    I've never even heard of Laplace Transform until now :3

    • one year ago
  63. telliott99
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    Can one of you guys summarize this for me, because I'm not getting it? I understand where the series went wrong I don't undertand Laplace Transform---is that where I need to go? I didn't find @mahmit2012 answer, just some demos of incorrect proofs. Are you, in the end, converting integral ( (cos x -1)/x) dx into integral (sin x / x) dx How can that be?

    • one year ago
  64. agentx5
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    I agree with @telliott99 , I'm confused and don't understand what's going on with that, or how it answers my original question (or if it even does)

    • one year ago
  65. telliott99
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    OK. I think the thing to do is to learn the Laplace Transform. I have these notes: http://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx But it's going to take a while. I know calculus pretty well and some LA but no DEs.

    • one year ago
  66. agentx5
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    I haven't learned either of those yet, the section from a chapter of the text on infinite series.

    • one year ago
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