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agentx5 Group Title

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?

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. telliott99 Group Title
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    You'd have (1/x) (-x^2/2! + x^4/4! .....)

    • 2 years ago
  3. Spacelimbus Group Title
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    \[ \int \frac{\cos x}{x}dx - \int \frac{1}{x}dx \]

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

    • 2 years ago
  5. agentx5 Group Title
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    That's the MacLaurin equivalent for cosine

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

    • 2 years ago
  7. Spacelimbus Group Title
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    Is it? Did I look at the wrong one? (-:

    • 2 years ago
  8. Spacelimbus Group Title
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    oh yes that's the hyperbolic one! SOrry!

    • 2 years ago
  9. agentx5 Group Title
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    Oh no wait I have the wrong one, I wrote sine by mistake

    • 2 years ago
  10. agentx5 Group Title
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    Arg soo confusing lol

    • 2 years ago
  11. Spacelimbus Group Title
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    lets both check again then hehe.

    • 2 years ago
  12. Spacelimbus Group Title
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    yes I believe I have the right one

    • 2 years ago
  13. agentx5 Group Title
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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?

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

    • 2 years ago
  15. experimentX Group Title
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    +c

    • 2 years ago
  16. agentx5 Group Title
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    Anti-derivative power rule?

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

    • 2 years ago
  18. Spacelimbus Group Title
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    yes that's it @agentx5

    • 2 years ago
  19. mahmit2012 Group Title
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    all has mistake!

    • 2 years ago
  20. Spacelimbus Group Title
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    where @mahmit2012 ?

    • 2 years ago
  21. mahmit2012 Group Title
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    all !but it is time to pray i wll come back and show you

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

    • 2 years ago
  23. mahmit2012 Group Title
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    after pray !

    • 2 years ago
  24. experimentX Group Title
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    lol ... where did i make mistake??

    • 2 years ago
  25. Spacelimbus Group Title
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    Did we do the same @experimentX ? besides that you completed it already.

    • 2 years ago
  26. Spacelimbus Group Title
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    I think we did.

    • 2 years ago
  27. experimentX Group Title
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    n=1, n=0

    • 2 years ago
  28. Spacelimbus Group Title
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    yes but obviously "ALL" is wrong :D

    • 2 years ago
  29. mahmit2012 Group Title
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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.

    • 2 years ago
  30. mahmit2012 Group Title
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    |dw:1343670857318:dw|

    • 2 years ago
  31. agentx5 Group Title
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    *tilting head sideways to read* :-D

    • 2 years ago
  32. mahmit2012 Group Title
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    |dw:1343670990609:dw|

    • 2 years ago
  33. mahmit2012 Group Title
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    |dw:1343671203407:dw|

    • 2 years ago
  34. agentx5 Group Title
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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?

    • 2 years ago
  35. agentx5 Group Title
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    Well the final answer is probably going to be ln|x| + something, I think...

    • 2 years ago
  36. mahmit2012 Group Title
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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.

    • 2 years ago
  37. experimentX Group Title
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    can't we cancel 1 and start from n=1 instead?

    • 2 years ago
  38. mahmit2012 Group Title
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    yes but if you start with 1-cosx in num so you wont confront the mistake.

    • 2 years ago
  39. mahmit2012 Group Title
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    take a look my solution and figure out the mistake.

    • 2 years ago
  40. Spacelimbus Group Title
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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.

    • 2 years ago
  41. Spacelimbus Group Title
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    Better: Substitute it directly and derive the formula in form of a series representation?

    • 2 years ago
  42. experimentX Group Title
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    yeah ... my case (-1)^n was mistake ..

    • 2 years ago
  43. mahmit2012 Group Title
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    and for the last this problem is originally has defined to 0 to inf.

    • 2 years ago
  44. mahmit2012 Group Title
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    |dw:1343671601188:dw|

    • 2 years ago
  45. Spacelimbus Group Title
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    I understand, thanks @mahmit2012

    • 2 years ago
  46. experimentX Group Title
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    perhaps this is related to http://en.wikipedia.org/wiki/Dirichlet_integral

    • 2 years ago
  47. mahmit2012 Group Title
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    Not exactly but a little bit !

    • 2 years ago
  48. mukushla Group Title
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    *

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

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

    • 2 years ago
  51. agentx5 Group Title
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    actually that needs to be n \(\pm\) 1on the alternating part

    • 2 years ago
  52. experimentX Group Title
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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.

    • 2 years ago
  53. agentx5 Group Title
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    @experimentX you mean shifting the index eh?

    • 2 years ago
  54. experimentX Group Title
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    yep

    • 2 years ago
  55. experimentX Group Title
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    since the one gets cancelled ... the after that x is divided.

    • 2 years ago
  56. agentx5 Group Title
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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

    • 2 years ago
  57. agentx5 Group Title
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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)

    • 2 years ago
  58. agentx5 Group Title
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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

    • 2 years ago
  59. agentx5 Group Title
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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

    • 2 years ago
  60. experimentX Group Title
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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.

    • 2 years ago
  61. experimentX Group Title
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    this gives the value of |dw:1343673440350:dw|

    • 2 years ago
  62. agentx5 Group Title
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    I've never even heard of Laplace Transform until now :3

    • 2 years ago
  63. telliott99 Group Title
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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?

    • 2 years ago
  64. agentx5 Group Title
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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)

    • 2 years ago
  65. telliott99 Group Title
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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.

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

    • 2 years ago
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