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myininaya

  • 5 years ago

calculus dream problem? (fg)'=f'g' and (fg)''=f''g'' I don't think this what you guys are looking for but here what i did

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  1. watchmath
    • 5 years ago
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    again? :D

  2. anonymous
    • 5 years ago
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    watchmath: ur problem has become famous..:D

  3. myininaya
    • 5 years ago
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    1 Attachment
  4. myininaya
    • 5 years ago
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    so this is your problem?

  5. myininaya
    • 5 years ago
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    have you done it?

  6. watchmath
    • 5 years ago
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    Check it here. It is solution for f. The g will be of the same form, but then the coefficients need to be adjusted.

  7. myininaya
    • 5 years ago
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    i had a dream about this problem last night

  8. watchmath
    • 5 years ago
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    :D :D

  9. myininaya
    • 5 years ago
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    i feel like i really dont sleep when i dream about math problems

  10. myininaya
    • 5 years ago
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    i have to look at this later but im very interested thanks for the link

  11. myininaya
    • 5 years ago
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    pdf*

  12. watchmath
    • 5 years ago
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    you are welcome :)

  13. anonymous
    • 5 years ago
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    i got as far as g = C e^ int ( f' / ( f' - f)

  14. anonymous
    • 5 years ago
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    i think, why did someone say , let F = e ^f, and G = e^g

  15. anonymous
    • 5 years ago
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    then ln F = f , and ln G = f ,

  16. watchmath
    • 5 years ago
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    Yes, that's good laplace. Then basically the idea is that from that g we can express, g' and g'' in terms of g. By doing that, then we have an equation in terms of f,f' and f'' only.

  17. watchmath
    • 5 years ago
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    I mean after you find \(g=Ce^{\int f'/(f'-f)}\)

  18. watchmath
    • 5 years ago
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    Laplace, are you cantorset?

  19. anonymous
    • 5 years ago
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    yes, but dont tell anyone

  20. watchmath
    • 5 years ago
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    :D :D

  21. anonymous
    • 5 years ago
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    my account has been suspended,

  22. anonymous
    • 5 years ago
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    im here to do math, not to debate with admin :)

  23. anonymous
    • 5 years ago
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    so from that g , whats g ' , its ...

  24. watchmath
    • 5 years ago
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    do you see the pdf file that I attached above?

  25. anonymous
    • 5 years ago
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    g ' = f' ( f' - f) C e ^ ( int f' / f' - f ) ?

  26. watchmath
    • 5 years ago
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    correct, which is f'g/(f'-f)

  27. anonymous
    • 5 years ago
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    oh im going backwards,

  28. anonymous
    • 5 years ago
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    this problem has many twists and turns

  29. anonymous
    • 5 years ago
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    one sec , im going to read

  30. watchmath
    • 5 years ago
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    there is some little mistake whenn I said that if f'=f then f'g=0 it should be then fg'=0. But similar to what I wrote we will arrive a t contradiction.

  31. anonymous
    • 5 years ago
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    you can fix it and send me new pdf?

  32. anonymous
    • 5 years ago
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    or you already updated? i dont know how to write pdf's, thats pretty cool

  33. watchmath
    • 5 years ago
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    hold on

  34. anonymous
    • 5 years ago
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    i also want to solve that torus problem,

  35. watchmath
    • 5 years ago
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    Here is the updated version:

  36. watchmath
    • 5 years ago
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    Are you teaching as a TA laplace?

  37. anonymous
    • 5 years ago
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    no, i tutor at the moment online

  38. anonymous
    • 5 years ago
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    a TA is what? a teacher assistant?

  39. watchmath
    • 5 years ago
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    Yes, usually if you take a graduate program in math you are a TA as well.

  40. anonymous
    • 5 years ago
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    i should,

  41. anonymous
    • 5 years ago
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    sorry was away, i enjoy math. but sometimes i cant remember the last problem i was working on. some people say its ok to forget all the loose ends problems, and start fresh. dunno. i have various interests in math, linear algebra, diff eq. they always end up in an interesting problem

  42. watchmath
    • 5 years ago
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    If you have an interesting problem please post it here: http://www.ask.watchmath.com

  43. anonymous
    • 5 years ago
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    thanks

  44. anonymous
    • 5 years ago
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    whats an easy way to draw a graph and upload it? microsoft paint?

  45. watchmath
    • 5 years ago
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    is it a graph of a well defined function? if it is you can plot it at http://fooplot.com and save it as an image

  46. anonymous
    • 5 years ago
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    and how would you upload it, whats a good site

  47. anonymous
    • 5 years ago
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    i use sendspace.com, not sure how long they keep it up

  48. watchmath
    • 5 years ago
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    where do you want to upload the picture?

  49. anonymous
    • 5 years ago
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    well i can draw problems and solve them, and you can look over

  50. anonymous
    • 5 years ago
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    www.sendspace.com , is a website that you can upload anything

  51. anonymous
    • 5 years ago
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    ok heres a question i was working on, i got no solution 2/ln 3 * ln ( x - 5) = log base 3 ( x-2) + log base 3 (5x) , i made the right side a product and then changed base

  52. watchmath
    • 5 years ago
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    so the left hand side is \(\frac{2}{\ln 3\ln(x-5)}\)

  53. anonymous
    • 5 years ago
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    nope, ln (x-5) in numerator

  54. anonymous
    • 5 years ago
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    i know, its ambiguous when you have a bad student :)

  55. anonymous
    • 5 years ago
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    im careful, i use CAS (computer algebra software)

  56. anonymous
    • 5 years ago
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    maple, did you know that maple cant find tan (105 degrees) ,

  57. anonymous
    • 5 years ago
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    tan ( 105* Pi/180) , its an exact angle, i want to complain to the makers

  58. watchmath
    • 5 years ago
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    Oh ok, then by the change of base formula we have \(2\log_3(x-5)=\log_3((x-2)(5x))\) So \((x-5)^2=(x-2)(5x)\) \(x^2-10x+255=5x^2-10x\) \(4x^2=225\) \(x=\pm\frac{15}{2}\) only the postive one works

  59. watchmath
    • 5 years ago
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    sorry ..... It should be 4x^2=25 So x=5/2 then no solution as well

  60. anonymous
    • 5 years ago
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    hah, yeah

  61. anonymous
    • 5 years ago
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    weird to give a student

  62. anonymous
    • 5 years ago
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    you changed the left side to log base 3

  63. anonymous
    • 5 years ago
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    2/ln 3 * ln ( x - 5) = log base 3 ( x-2) + log base 3 (5x) , what happened to your left side?

  64. anonymous
    • 5 years ago
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    your left side is log _3 (x-5) / log _3 ln 3 * 2/ln 3

  65. watchmath
    • 5 years ago
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    remember that \[\log_ba=\frac{\log_c a}{\log_cb}\] Where you can choose your c as you like. Choosince c to be the natural number e, then the left side is exactly \(\log_3((x-5))\)

  66. anonymous
    • 5 years ago
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    right, i followed the rule

  67. anonymous
    • 5 years ago
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    but you have 2/ln 3 *log _3 (x-5) / log _3 (ln 3)

  68. anonymous
    • 5 years ago
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    and whats log _3 of ln 3 ? log base 3 of ln 3

  69. watchmath
    • 5 years ago
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    In that case I thin we can't simplify it nicely.

  70. anonymous
    • 5 years ago
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    its funny you did that and still got right answer. my instinct would be to change the right side to ln instead of log base 3

  71. anonymous
    • 5 years ago
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    so it must be the case that ln 3 * log_3 ln 3 = 1 ?

  72. watchmath
    • 5 years ago
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    no

  73. watchmath
    • 5 years ago
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    it is equal to ln(ln 3)

  74. anonymous
    • 5 years ago
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    oh i think i see it now

  75. anonymous
    • 5 years ago
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    for the left side we have 2/( ln 3) * ln (x-5) = 2/ln 3 * log_3 (x-5) / log _3 (e) = 2 *log_3 (x-5) * 1/ ( ln 3 * log _3 (e) = log _3 (x-5)^2 * 1/ ( log _3 e^(ln 3) ) = log _3 (x-5)^2* 1 / ( log _3 3 ) = log_3 (x-5)^2 * 1

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