## anonymous 5 years ago Question on calculus

1. anonymous

$\prod_{0}^{n}e ^{r/n})]^{1/n}$

2. anonymous

find the value

3. anonymous

PI e^(r/n^2) correct

4. anonymous

where PI is multiplication i assume

5. anonymous

you want the limit as n goes to infinity

6. anonymous

yes....

7. anonymous

also its not defined for n = 0

8. anonymous

that should be from 1 to n ?

9. anonymous

wait, you probably meant sum, not PRODUCT

10. anonymous

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

11. anonymous

it is defined for r=0

12. anonymous

then your question is ambiguous. theres something wrong with it

13. anonymous

and it is product

14. anonymous

15. anonymous

it says the answer is e^(1/2)

16. anonymous

is this online, can i see it

17. anonymous

can you draw it exactly as it appears

18. anonymous

no we dont work online in india

19. anonymous
20. anonymous

please click on that, and draw the mother trucking expression

21. anonymous

im in usa , here we do everything online.

22. anonymous

oh wait..i got it..thnx neway

23. anonymous

wait, can you show me the expression

24. anonymous
25. anonymous

you can draw the expression there

26. anonymous

$\lim_{n \rightarrow \infty}[(\prod_{r=0}^{n} (e ^{r/n})]^{1/n}$

27. anonymous

ohhh

28. anonymous

i got it neway..thnx

29. anonymous

how did you get solution

30. watchmath

e^(1/2)

31. anonymous

ok change the exponent product to sum of exponents

32. anonymous

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

33. anonymous

then exponentiate on both sides

34. anonymous

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

lim [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)

36. anonymous

oh integral e^(1/x) dx ?

37. anonymous

no man

38. anonymous

right, e^x

39. anonymous

ln (e^x)

40. anonymous

theres a direct approach though that works

41. watchmath

cantorset, your expression is simpliefied to $$e^{n(n+1)/(2n^2)}$$

42. anonymous

right

43. anonymous

and then you can take limits directly

44. anonymous

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

45. anonymous

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

46. anonymous

yes....thanks a lot guys

47. anonymous

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

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

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

you get $\log (I) = \int\limits_{0}^{1}xdx$

51. anonymous

log of a product is a sum

52. anonymous

let y = lim product (e^(k/n))*1/n for k = 0..n , take ln of both sides

53. anonymous

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

55. anonymous

ln y = integral x from 0 to 1

56. anonymous

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

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

a riemann sum should have n terms, this has n+1 terms

59. anonymous

i know im splitting hairs here but

60. watchmath

well, you can ignore the e^(0/n) since it is equal to 1.

61. anonymous

delta x = 1 - 0 / n , so delta x = 1/n

62. anonymous

hmmm, then its constant, then ...

63. anonymous

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

64. anonymous

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

oh so that 1/n vanishes in the limit

66. watchmath

$$y=\left(\prod_{r=0}^n e^{r/n}\right)^{1/n}=\prod_{r=0}^n e^{r/n^2}$$ $$\ln y=\sum_{r=0}^n \frac{r}{n^2}=\sum_{r=1}^n \frac{r}{n}\cdot \frac{1}{n}$$ 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

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

oh woops

69. anonymous

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

err, that should be zero the first term

71. watchmath

I see ... so you mean 1/n*0 right?

72. anonymous

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

yes

74. watchmath

Yes, that's right it is the same as what I wrote above :)

75. anonymous

hehe, i like my discrete proof though

76. anonymous

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

my proof does not use fundamental theorem of calculus, so it may have an advantage

78. anonymous

the discrete approach

79. watchmath

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

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

its just weird doing the whole 6 - x , when it revolves around say x = 6

82. anonymous

oh the torus, what was it again

83. anonymous

( x - R)^2 + y^2) ,