Question on calculus

- anonymous

Question on calculus

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- anonymous

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

- anonymous

find the value

- anonymous

PI e^(r/n^2) correct

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## More answers

- anonymous

where PI is multiplication i assume

- anonymous

you want the limit as n goes to infinity

- anonymous

yes....

- anonymous

also its not defined for n = 0

- anonymous

that should be from 1 to n ?

- anonymous

wait, you probably meant sum, not PRODUCT

- anonymous

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

- anonymous

it is defined for r=0

- anonymous

then your question is ambiguous. theres something wrong with it

- anonymous

and it is product

- anonymous

youre messing up your indices

- anonymous

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

- anonymous

is this online, can i see it

- anonymous

can you draw it exactly as it appears

- anonymous

no we dont work online in india

- anonymous

here http://www.twiddla.com/543856

- anonymous

please click on that, and draw the mother trucking expression

- anonymous

im in usa , here we do everything online.

- anonymous

oh wait..i got it..thnx neway

- anonymous

wait, can you show me the expression

- anonymous

here http://www.twiddla.com/543856

- anonymous

you can draw the expression there

- anonymous

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

- anonymous

ohhh

- anonymous

i got it neway..thnx

- anonymous

how did you get solution

- watchmath

e^(1/2)

- anonymous

ok change the exponent product to sum of exponents

- anonymous

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

- anonymous

then exponentiate on both sides

- 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

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

- anonymous

oh integral e^(1/x) dx ?

- anonymous

no man

- anonymous

right, e^x

- anonymous

ln (e^x)

- anonymous

theres a direct approach though that works

- watchmath

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

- anonymous

right

- anonymous

and then you can take limits directly

- anonymous

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

- anonymous

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

- anonymous

yes....thanks a lot guys

- 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

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

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

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

- anonymous

log of a product is a sum

- anonymous

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

- 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

my bad,

- anonymous

ln y = integral x from 0 to 1

- 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

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

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

- anonymous

i know im splitting hairs here but

- watchmath

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

- anonymous

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

- anonymous

hmmm, then its constant, then ...

- anonymous

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

- 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

oh so that 1/n vanishes in the limit

- 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

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

oh woops

- 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

err, that should be zero the first term

- watchmath

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

- 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

yes

- watchmath

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

- anonymous

hehe, i like my discrete proof though

- 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

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

- anonymous

the discrete approach

- 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

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

- anonymous

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

- anonymous

oh the torus, what was it again

- anonymous

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

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