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anonymous

  • 5 years ago

Question on calculus

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  1. anonymous
    • 5 years ago
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    \[\prod_{0}^{n}e ^{r/n})]^{1/n}\]

  2. anonymous
    • 5 years ago
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    find the value

  3. anonymous
    • 5 years ago
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    PI e^(r/n^2) correct

  4. anonymous
    • 5 years ago
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    where PI is multiplication i assume

  5. anonymous
    • 5 years ago
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    you want the limit as n goes to infinity

  6. anonymous
    • 5 years ago
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    yes....

  7. anonymous
    • 5 years ago
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    also its not defined for n = 0

  8. anonymous
    • 5 years ago
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    that should be from 1 to n ?

  9. anonymous
    • 5 years ago
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    wait, you probably meant sum, not PRODUCT

  10. anonymous
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    the limit is for r the iteration is r from 0 to n, n tends to inf

  11. anonymous
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    it is defined for r=0

  12. anonymous
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    then your question is ambiguous. theres something wrong with it

  13. anonymous
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    and it is product

  14. anonymous
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    youre messing up your indices

  15. anonymous
    • 5 years ago
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    it says the answer is e^(1/2)

  16. anonymous
    • 5 years ago
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    is this online, can i see it

  17. anonymous
    • 5 years ago
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    can you draw it exactly as it appears

  18. anonymous
    • 5 years ago
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    no we dont work online in india

  19. anonymous
    • 5 years ago
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    here http://www.twiddla.com/543856

  20. anonymous
    • 5 years ago
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    please click on that, and draw the mother trucking expression

  21. anonymous
    • 5 years ago
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    im in usa , here we do everything online.

  22. anonymous
    • 5 years ago
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    oh wait..i got it..thnx neway

  23. anonymous
    • 5 years ago
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    wait, can you show me the expression

  24. anonymous
    • 5 years ago
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    here http://www.twiddla.com/543856

  25. anonymous
    • 5 years ago
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    you can draw the expression there

  26. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty}[(\prod_{r=0}^{n} (e ^{r/n})]^{1/n}\]

  27. anonymous
    • 5 years ago
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    ohhh

  28. anonymous
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    i got it neway..thnx

  29. anonymous
    • 5 years ago
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    how did you get solution

  30. watchmath
    • 5 years ago
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    e^(1/2)

  31. anonymous
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    ok change the exponent product to sum of exponents

  32. anonymous
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    yes....take log on both sides..it becomes integral (xdx) from 0 to 1

  33. anonymous
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    then exponentiate on both sides

  34. anonymous
    • 5 years ago
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    replace 1/n by dx r/n by x this will give you an integral......note this is one of the derivations of the basic integral ....check out reimannian sums

  35. anonymous
    • 5 years ago
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    lim [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)

  36. anonymous
    • 5 years ago
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    oh integral e^(1/x) dx ?

  37. anonymous
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    no man

  38. anonymous
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    right, e^x

  39. anonymous
    • 5 years ago
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    ln (e^x)

  40. anonymous
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    theres a direct approach though that works

  41. watchmath
    • 5 years ago
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    cantorset, your expression is simpliefied to \(e^{n(n+1)/(2n^2)}\)

  42. anonymous
    • 5 years ago
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    right

  43. anonymous
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    and then you can take limits directly

  44. anonymous
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    something to that effect note you will still have to take a log first

  45. anonymous
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    if you take the limit as n goes to infinity, you get e^(1/2)

  46. anonymous
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    yes....thanks a lot guys

  47. anonymous
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    lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)= lim [ e^ (1/n + 2/n + 3/n + ... n/n) ] *1/n

  48. anonymous
    • 5 years ago
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    lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)= lim [ e^ (1/n + 2/n + 3/n + ... n/n) ] *1/n = lim e^ [( 1 + 2 + 3 + ... + n ) / n ] ^(1/n)

  49. anonymous
    • 5 years ago
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    lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)= lim [ e^ (1/n + 2/n + 3/n + ... n/n) ] *1/n = lim e^ [( 1 + 2 + 3 + ... + n ) / n ] ^(1/n) =lim e^ (n^2 + n ) / (2n^2)

  50. anonymous
    • 5 years ago
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    you get \[\log (I) = \int\limits_{0}^{1}xdx\]

  51. anonymous
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    log of a product is a sum

  52. anonymous
    • 5 years ago
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    let y = lim product (e^(k/n))*1/n for k = 0..n , take ln of both sides

  53. anonymous
    • 5 years ago
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    ln y = lim n-> oo sum k/n *1/n for k= 0..n , so ln y = integral 1/x from 0 to 1 ?

  54. anonymous
    • 5 years ago
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    my bad,

  55. anonymous
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    ln y = integral x from 0 to 1

  56. anonymous
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    ln [ lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)] = lim ( 1/n (e^0/n + e^(1/n) + .. e^n/n)

  57. anonymous
    • 5 years ago
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    ln [ lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)] = lim ( 1/n (e^0/n + e^(1/n) + .. e^n/n) , but this is not exactly a riemann sum

  58. anonymous
    • 5 years ago
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    a riemann sum should have n terms, this has n+1 terms

  59. anonymous
    • 5 years ago
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    i know im splitting hairs here but

  60. watchmath
    • 5 years ago
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    well, you can ignore the e^(0/n) since it is equal to 1.

  61. anonymous
    • 5 years ago
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    delta x = 1 - 0 / n , so delta x = 1/n

  62. anonymous
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    hmmm, then its constant, then ...

  63. anonymous
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    riemann sum is Sigma f ( a + r * delta x ) * delta x , for r = 1 to n , for right hand rule

  64. anonymous
    • 5 years ago
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    ln [ lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)] = lim ( 1/n (e^0/n + e^(1/n) + .. e^n/n) = 1/n *1 + integral e^x for a = 0 to b = 1

  65. anonymous
    • 5 years ago
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    oh so that 1/n vanishes in the limit

  66. watchmath
    • 5 years ago
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    \(y=\left(\prod_{r=0}^n e^{r/n}\right)^{1/n}=\prod_{r=0}^n e^{r/n^2}\) \(\ln y=\sum_{r=0}^n \frac{r}{n^2}=\sum_{r=1}^n \frac{r}{n}\cdot \frac{1}{n}\) If we take the limit as \(n\to \infty\) then we have \(\int_0^1 x\, dx=1/2.\) So \(\lim y= \lim e^{\ln y}=e^{1/2}\)

  67. anonymous
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    ln [ lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)] = lim ( 1/n (e^0/n + e^(1/n) + .. e^n/n) = lim [ 1/n *1 + Sigma ( e^(r/n)*1/n , r=1..n ]

  68. anonymous
    • 5 years ago
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    oh woops

  69. anonymous
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    ln [ lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)] = lim ( 1/n ( 0/n + 1/n + 2/n + ... n/n) = lim [ 1/n *1 + Sigma (r/n)*1/n , r=1..n ]

  70. anonymous
    • 5 years ago
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    err, that should be zero the first term

  71. watchmath
    • 5 years ago
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    I see ... so you mean 1/n*0 right?

  72. anonymous
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    ln [ lim n> oo [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)] = lim ( 1/n ( 0/n + 1/n + 2/n + ... n/n) = lim [ 0 + Sigma (r/n)*1/n , r=1..n ] = integral

  73. anonymous
    • 5 years ago
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    yes

  74. watchmath
    • 5 years ago
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    Yes, that's right it is the same as what I wrote above :)

  75. anonymous
    • 5 years ago
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    hehe, i like my discrete proof though

  76. anonymous
    • 5 years ago
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    thanks a lot for your help with integration, i just have to practice a bit. and do lots of various scenarios . it helps to draw out the cylinder i guess

  77. anonymous
    • 5 years ago
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    my proof does not use fundamental theorem of calculus, so it may have an advantage

  78. anonymous
    • 5 years ago
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    the discrete approach

  79. watchmath
    • 5 years ago
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    Yes it helps to draw out the cyllinder. After some practices you won't need to draw the cyllinder explicitly :). So did you try the torus problem :).

  80. anonymous
    • 5 years ago
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    the shell integration is a bit diferent than washer or disc. since there is no inner and outer radius , like in those cases

  81. anonymous
    • 5 years ago
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    its just weird doing the whole 6 - x , when it revolves around say x = 6

  82. anonymous
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    oh the torus, what was it again

  83. anonymous
    • 5 years ago
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    ( x - R)^2 + y^2) ,

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