Question on calculus

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Question on calculus

Mathematics
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\[\prod_{0}^{n}e ^{r/n})]^{1/n}\]
find the value
PI e^(r/n^2) correct

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where PI is multiplication i assume
you want the limit as n goes to infinity
yes....
also its not defined for n = 0
that should be from 1 to n ?
wait, you probably meant sum, not PRODUCT
the limit is for r the iteration is r from 0 to n, n tends to inf
it is defined for r=0
then your question is ambiguous. theres something wrong with it
and it is product
youre messing up your indices
it says the answer is e^(1/2)
is this online, can i see it
can you draw it exactly as it appears
no we dont work online in india
here http://www.twiddla.com/543856
please click on that, and draw the mother trucking expression
im in usa , here we do everything online.
oh wait..i got it..thnx neway
wait, can you show me the expression
here http://www.twiddla.com/543856
you can draw the expression there
\[\lim_{n \rightarrow \infty}[(\prod_{r=0}^{n} (e ^{r/n})]^{1/n}\]
ohhh
i got it neway..thnx
how did you get solution
e^(1/2)
ok change the exponent product to sum of exponents
yes....take log on both sides..it becomes integral (xdx) from 0 to 1
then exponentiate on both sides
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
lim [ e^0/n *e^(1/n) *e^(2/n)*...e^(n/n)] ^(1/n)
oh integral e^(1/x) dx ?
no man
right, e^x
ln (e^x)
theres a direct approach though that works
cantorset, your expression is simpliefied to \(e^{n(n+1)/(2n^2)}\)
right
and then you can take limits directly
something to that effect note you will still have to take a log first
if you take the limit as n goes to infinity, you get e^(1/2)
yes....thanks a lot guys
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 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 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)
you get \[\log (I) = \int\limits_{0}^{1}xdx\]
log of a product is a sum
let y = lim product (e^(k/n))*1/n for k = 0..n , take ln of both sides
ln y = lim n-> oo sum k/n *1/n for k= 0..n , so ln y = integral 1/x from 0 to 1 ?
my bad,
ln y = integral x from 0 to 1
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)
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
a riemann sum should have n terms, this has n+1 terms
i know im splitting hairs here but
well, you can ignore the e^(0/n) since it is equal to 1.
delta x = 1 - 0 / n , so delta x = 1/n
hmmm, then its constant, then ...
riemann sum is Sigma f ( a + r * delta x ) * delta x , for r = 1 to n , for right hand rule
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
oh so that 1/n vanishes in the limit
\(y=\left(\prod_{r=0}^n e^{r/n}\right)^{1/n}=\prod_{r=0}^n e^{r/n^2}\) \(\ln y=\sum_{r=0}^n \frac{r}{n^2}=\sum_{r=1}^n \frac{r}{n}\cdot \frac{1}{n}\) If we take the limit as \(n\to \infty\) then we have \(\int_0^1 x\, dx=1/2.\) So \(\lim y= \lim e^{\ln y}=e^{1/2}\)
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 ]
oh woops
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 ]
err, that should be zero the first term
I see ... so you mean 1/n*0 right?
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
yes
Yes, that's right it is the same as what I wrote above :)
hehe, i like my discrete proof though
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
my proof does not use fundamental theorem of calculus, so it may have an advantage
the discrete approach
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 :).
the shell integration is a bit diferent than washer or disc. since there is no inner and outer radius , like in those cases
its just weird doing the whole 6 - x , when it revolves around say x = 6
oh the torus, what was it again
( x - R)^2 + y^2) ,

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