need help in math :)

[wesleyanne]

hey ppl :) my friends and i are working together on this problem but we can't seem to solve it. help? gracias~

lim as k--> infinity ( sum from n= 1 to n= k of (1/sqrt(nk)))

[__earth]

shouldnt it be zero?

the sqrt of infinity is infinity.

[__earth]

shouldnt it be zero?

the sqrt of infinity is infinity.

[wesleyanne]

shouldnt it be zero?

the sqrt of infinity is infinity.

oops, my bad :) forgot to include this in my first post :) well, actually the answer is 2, but we need to come up with a way to prove it :) :) :)

[wesleyanne]

shouldnt it be zero?

the sqrt of infinity is infinity.

oops, my bad :) forgot to include this in my first post :) well, actually the answer is 2, but we need to come up with a way to prove it :) :) :)

[__earth]

it has been awhile since i last touched calc but

limit k -> infinity, Sigma n = 1..k of (1/sqrt(nk)

= { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(n) } + { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(k) }

= { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(n) } + 0

= { limit k -> infinity, [ 1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) ] }

= { limit k -> infinity, [ 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) ] }

= 1 + { limit k -> infinity, [ 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) ] }

= 1 + 1 (hah! i presume that the sigma n = 2..k == 1. if you want to calculate it, be my guest)

= 2

I think that would do it though its brute force. if you could remember the closed form of Sigma [1/sqrt(a)], maybe that would make the answer cleaner.

sorry about saying its zero earlier. didnt really put my mind into it at first.

[__earth]

it has been awhile since i last touched calc but

limit k -> infinity, Sigma n = 1..k of (1/sqrt(nk)

= { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(n) } + { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(k) }

= { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(n) } + 0

= { limit k -> infinity, [ 1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) ] }

= { limit k -> infinity, [ 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) ] }

= 1 + { limit k -> infinity, [ 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) ] }

= 1 + 1 (hah! i presume that the sigma n = 2..k == 1. if you want to calculate it, be my guest)

= 2

I think that would do it though its brute force. if you could remember the closed form of Sigma [1/sqrt(a)], maybe that would make the answer cleaner.

sorry about saying its zero earlier. didnt really put my mind into it at first.

[bachok83]

cool ___earth.. i think it's correct...

[topdog]

limit k -> infinity, Sigma n = 1..k of (1/sqrt(nk)

= { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(n) } + { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(k) }

that's not right...can't split it like that.

i can't solve it though (my math sucks). but using matlab we can see that the limit approaches 2.
e.g. n=1:100000;
sum(1./sqrt(n.*100000))

returns 1.9954 which is ~2. heh...

where are the math geniuses when we need them? hello?

[topdog]

limit k -> infinity, Sigma n = 1..k of (1/sqrt(nk)

= { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(n) } + { limit k -> infinity, (Sigma n = 1..k of (1/sqrt(k) }

that's not right...can't split it like that.

i can't solve it though (my math sucks). but using matlab we can see that the limit approaches 2.
e.g. n=1:100000;
sum(1./sqrt(n.*100000))

returns 1.9954 which is ~2. heh...

where are the math geniuses when we need them? hello?

[__earth]

yeah, cant do that LOL!
what on earh was i doing LOL!

maybe i should've done it like this

1/sqrt(nk)

= [1/sqrt(nk)]*[sqrt(nk)/sqrt(nk)]

= sqrt(nk)/nk

= sqrt(n)/nk + sqrt(k)/nk

after that, err...

[__earth]

yeah, cant do that LOL!
what on earh was i doing LOL!

maybe i should've done it like this

1/sqrt(nk)

= [1/sqrt(nk)]*[sqrt(nk)/sqrt(nk)]

= sqrt(nk)/nk

= sqrt(n)/nk + sqrt(k)/nk

after that, err...

Here are is the sketch of the proof.

The main step involves evaluating the sum
1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k)

We can do this by integral approximation (i.e. approximating the sum by integrals). If you draw the function 1/sqrt(x) on the graph and try to visuliaze approximating the area under the graph by rectangles of length 1, you will notice that

\integrate_{1 to k+1} 1/sqrt(x) <= \sum_{i=1 to k} 1/sqrt(i)

and

\sum_{i=1 to k} 1/sqrt(i) <= 1 + \integrate_{1 to k} 1/sqrt(x)

If we let A = \sum_{i=1 to k} 1/sqrt(i) then

2 sqrt(k+1) - 2 <= A <= 2sqrt(k) - 1

So, lim (k -> +oo) A/sqrt(k) = 2

Here are is the sketch of the proof.

The main step involves evaluating the sum
1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k)

We can do this by integral approximation (i.e. approximating the sum by integrals). If you draw the function 1/sqrt(x) on the graph and try to visuliaze approximating the area under the graph by rectangles of length 1, you will notice that

\integrate_{1 to k+1} 1/sqrt(x) <= \sum_{i=1 to k} 1/sqrt(i)

and

\sum_{i=1 to k} 1/sqrt(i) <= 1 + \integrate_{1 to k} 1/sqrt(x)

If we let A = \sum_{i=1 to k} 1/sqrt(i) then

2 sqrt(k+1) - 2 <= A <= 2sqrt(k) - 1

So, lim (k -> +oo) A/sqrt(k) = 2

BTW wesleyanne, I don't suppose this is a homework problem, is it?

BTW wesleyanne, I don't suppose this is a homework problem, is it?

[wesleyanne]

BTW wesleyanne, I don't suppose this is a homework problem, is it?

first of all, thanx for the help :) truly appreciate it....

to answer your question:

A dorm mate of mine has this problem taped on her door, initially thought the series diverges but she also had the answer 2 down. Having taken some math classes before, (obviously not enough :oops: ) i tried to solve it, but obviously couldn't, asked my friends, and they couldn't so I decided to post it up here :)

anyway, thanx :) kinda makes me wish i took real analysis ;) hahaha :) but considering im a bus major, that would be asking for trouble :)

[wesleyanne]

BTW wesleyanne, I don't suppose this is a homework problem, is it?

first of all, thanx for the help :) truly appreciate it....

to answer your question:

A dorm mate of mine has this problem taped on her door, initially thought the series diverges but she also had the answer 2 down. Having taken some math classes before, (obviously not enough :oops: ) i tried to solve it, but obviously couldn't, asked my friends, and they couldn't so I decided to post it up here :)

anyway, thanx :) kinda makes me wish i took real analysis ;) hahaha :) but considering im a bus major, that would be asking for trouble :)

[jun_wolverine]

i have a question

this question is not one of my homework assignments,

Someone asked me to do this but i had no idea how to
here goes

solve for t

the equation is as follow:

At^2 +Be^t + C = 0

where A,B,C are constants

i have no idea how to do this by hand...
the general idea on how to do this will be good enough

P/S: no use of calculators/matlab/mathematica

[jun_wolverine]

i have a question

this question is not one of my homework assignments,

Someone asked me to do this but i had no idea how to
here goes

solve for t

the equation is as follow:

At^2 +Be^t + C = 0

where A,B,C are constants

i have no idea how to do this by hand...
the general idea on how to do this will be good enough

P/S: no use of calculators/matlab/mathematica

[Randomphantom]

Saw this question, seen a similar one so i gave it a try...

sqrt(N) + sqrt(N-1) < 2sqrt(N) < sqrt(N+1) + sqrt(N)
therefore 1/(sqrt(N+1) + sqrt(N)) < 1/(2sqrt(N)) < 1/(sqrt(N) + sqrt(N-1))

now 1 / (sqrt(N+1) + sqrt(N)) = sqrt(N+1) - sqrt(N)
and 1 / (sqrt(N) + sqrt(N-1)) = sqrt(N) - sqrt(N-1)

we get

2(sqrt(N+1) - sqrt(N)) < 1/sqrt(N) < 2(sqrt(N) - sqrt(N-1))

sigma (N=1..k) of 2(sqrt(N+1) - sqrt(N)) < sigma (N=1..k) of 1/sqrt(N) **main step** < sigma (N=1..k) of 2(sqrt(N) - sqrt(N-1))

after cancelling terms you get

2(sqrt(k+1) - 1) < sigma (N=1..k) of 1/sqrt(N) < 2sqrt(k)

divide by sqrt(k)

2(sqrt(k+1) - 1)/sqrt(k) < sigma (N=1..k) of 1/sqrt(Nk) < 2

To evaluate use squeezing
if Im not wrong then

lim k-->infinity of 2(sqrt(k+1) - 1)/sqrt(k)
=lim k-->infinity of 2(sqrt(1 + 1/k) - 1/sqrt(k))
= 2 (squeezed!)

[luke]

Reusing this thread ... I need a bit of help for this modular arithmetic problem (well not really a problem but read on please)

Suppose we have n integers {A1, A2, ... An} that do not have any common factors other than 1. What is the term for M, the largest integer that fulfills the following congruence equation?

A1 ≡ A2 ≡ .... ≡ An (mod M)

For example: 1, 4 and 10 have no common factor other than 1.
The largest M that fulfills the equation 1 ≡ 4 ≡ 10 (mod M) is 3.
Thus, 3 is <term> of 1, 4 and 10. Is there an existing term that can be put in the statement?

[youngyew]

Don't think so. Just call it the biggest natural number n for which A_i are congruent in modulo N. Or co-mo-do if you feel like it. :P

[luke]

I think I'd coin the term "the greatest modulus producing common residue for" as in "3 is the greatest modulus producing common residue for 1, 4 and 10" ...

shorthand: gmpcr(1,4,10) = 3 or gmcr(1,4,10) = 3

umm .. where should I go to register the term? :nod

[bluez_aspic]

Haha start your own wikipedia entry