Question

Question on calculus

Answer 1

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

Answer 2

find the value

Answer 3

PI e^(r/n^2) correct

Answer 4

where PI is multiplication i assume

Answer 5

you want the limit as n goes to infinity

Answer 6

yes....

Answer 7

also its not defined for n = 0

Answer 8

that should be from 1 to n ?

Answer 9

wait, you probably meant sum, not PRODUCT

Answer 10

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

Answer 11

it is defined for r=0

Answer 12

then your question is ambiguous. theres something wrong with it

Answer 13

and it is product

Answer 14

youre messing up your indices

Answer 15

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

Answer 16

is this online, can i see it

Answer 17

can you draw it exactly as it appears

Answer 18

no we dont work online in india

Answer 19

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

Answer 20

please click on that, and draw the mother trucking expression

Answer 21

im in usa , here we do everything online.

Answer 22

oh wait..i got it..thnx neway

Answer 23

wait, can you show me the expression

Answer 24

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

Answer 25

you can draw the expression there

Answer 26

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

Answer 27

ohhh

Answer 28

i got it neway..thnx

Answer 29

how did you get solution

Answer 30

e^(1/2)

Answer 31

ok change the exponent product to sum of exponents

Answer 32

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

Answer 33

then exponentiate on both sides

Answer 34

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

Answer 35

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

Answer 36

oh integral e^(1/x) dx ?

Answer 37

no man

Answer 38

right, e^x

Answer 39

ln (e^x)

Answer 40

theres a direct approach though that works

Answer 41

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

Answer 42

right

Answer 43

and then you can take limits directly

Answer 44

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

Answer 45

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

Answer 46

yes....thanks a lot guys

Answer 47

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

Answer 48

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)

Answer 49

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)

Answer 50

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

Answer 51

log of a product is a sum

Answer 52

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

Answer 53

ln y = lim n-> oo sum k/n *1/n for k= 0..n , so ln y = integral 1/x from 0 to 1 ?

Answer 54

my bad,

Answer 55

ln y = integral x from 0 to 1

Answer 56

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)

Answer 57

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

Answer 58

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

Answer 59

i know im splitting hairs here but

Answer 60

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

Answer 61

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

Answer 62

hmmm, then its constant, then ...

Answer 63

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

Answer 64

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

Answer 65

oh so that 1/n vanishes in the limit

Answer 66

\(y=\left(\prod_{r=0}^n e^{r/n}\right)^{1/n}=\prod_{r=0}^n e^{r/n^2}\)
\(\ln y=\sum_{r=0}^n \frac{r}{n^2}=\sum_{r=1}^n \frac{r}{n}\cdot \frac{1}{n}\)
If we take the limit as \(n\to \infty\) then we have
\(\int_0^1 x\, dx=1/2.\)
So \(\lim y= \lim e^{\ln y}=e^{1/2}\)

Answer 67

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 ]

Answer 68

oh woops

Answer 69

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 ]

Answer 70

err, that should be zero the first term

Answer 71

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

Answer 72

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

Answer 73

yes

Answer 74

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

Answer 75

hehe, i like my discrete proof though

Answer 76

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

Answer 77

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

Answer 78

the discrete approach

Answer 79

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 :).

Answer 80

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

Answer 81

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

Answer 82

oh the torus, what was it again

Answer 83

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