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anonymous
 5 years ago
Question on calculus
anonymous
 5 years ago
Question on calculus

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\prod_{0}^{n}e ^{r/n})]^{1/n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where PI is multiplication i assume

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you want the limit as n goes to infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also its not defined for n = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that should be from 1 to n ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, you probably meant sum, not PRODUCT

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the limit is for r the iteration is r from 0 to n, n tends to inf

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is defined for r=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then your question is ambiguous. theres something wrong with it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre messing up your indices

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says the answer is e^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this online, can i see it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you draw it exactly as it appears

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no we dont work online in india

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please click on that, and draw the mother trucking expression

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im in usa , here we do everything online.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait..i got it..thnx neway

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, can you show me the expression

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can draw the expression there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty}[(\prod_{r=0}^{n} (e ^{r/n})]^{1/n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok change the exponent product to sum of exponents

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes....take log on both sides..it becomes integral (xdx) from 0 to 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then exponentiate on both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0replace 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh integral e^(1/x) dx ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0theres a direct approach though that works

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1cantorset, your expression is simpliefied to \(e^{n(n+1)/(2n^2)}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then you can take limits directly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0something to that effect note you will still have to take a log first

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you take the limit as n goes to infinity, you get e^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes....thanks a lot guys

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you get \[\log (I) = \int\limits_{0}^{1}xdx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0log of a product is a sum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let y = lim product (e^(k/n))*1/n for k = 0..n , take ln of both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln y = lim n> oo sum k/n *1/n for k= 0..n , so ln y = integral 1/x from 0 to 1 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln y = integral x from 0 to 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln [ 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln [ 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a riemann sum should have n terms, this has n+1 terms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know im splitting hairs here but

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1well, you can ignore the e^(0/n) since it is equal to 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0delta x = 1  0 / n , so delta x = 1/n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, then its constant, then ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0riemann sum is Sigma f ( a + r * delta x ) * delta x , for r = 1 to n , for right hand rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln [ 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh so that 1/n vanishes in the limit

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln [ 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 ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln [ 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 ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0err, that should be zero the first term

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1I see ... so you mean 1/n*0 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln [ 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

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes, that's right it is the same as what I wrote above :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hehe, i like my discrete proof though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my proof does not use fundamental theorem of calculus, so it may have an advantage

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the discrete approach

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes 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 :).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the shell integration is a bit diferent than washer or disc. since there is no inner and outer radius , like in those cases

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its just weird doing the whole 6  x , when it revolves around say x = 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh the torus, what was it again
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