anonymous
  • anonymous
Question on calculus
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\prod_{0}^{n}e ^{r/n})]^{1/n}\]
anonymous
  • anonymous
find the value
anonymous
  • anonymous
PI e^(r/n^2) correct

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

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