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AsianSensation




PostPosted: Sun Jun 29, 2003 4:46 pm   Post subject: (No subject)

SilverSprite wrote:
asian. your way cooler solution is the same solution as mine..just presented differently:P


um..have you read the entire thing completely? Yours compare parity, and the other solution compares answers in modulus form. You used a bit of modulus math, in fact, that modulus math you used wasn't even needed, you could have simplified and got the same result, and it comes down to parity.

SilverSprite wrote:
and 4 in modulus 4 is the same as 0


um.....what? I know that's true, but what does that have to do with the other solution?
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AsianSensation




PostPosted: Sun Jun 29, 2003 5:31 pm   Post subject: (No subject)

Solutions to Question #2:

1 + 1/1! = 2

now we have to prove
1/2! + 1/3! + ... + 1/n! < 1

First we look at the sum of a geometric series, with r term 1/2.
1/2 + 1/(2^2) + 1/(2^3) + ... + 1/(2^n)

compare each one of these terms with the series
1/2! + 1/3! + ... + 1/n!

1) 1/2, 1/4, 1/8, ... 1/(2^n)
2) 1/2, 1/6, 1/24, ...1/n!

it is clearly seen that 1) > 2)

the sum of 1) is (1/2(1 - (1/2)^n))/(1/2)
and that is always less than 1.

therefore, 2 + something less than 1 is obviously less than 3.

therefore, 1 + 1/1! + 1/2! + 1/3! + ... + 1/n! < 3

*note: post any other alternate solutions, even if it is not correct, but if you think have a good approach, then I will still award bits.
SilverSprite




PostPosted: Sun Jun 29, 2003 5:58 pm   Post subject: (No subject)

okkk that solution is wrong.. i think you remembered something else and tried to copy it but you did it wrong..
Quote:
1/2! + 1/3! + ... + 1/n!

after taking out a factor of 1/2 does not become
Quote:
1/2(1/1! + 1/2! + 1/3! + ... + 1/(n - 1)!)

or am i missing something here?
i am completely lost with your solution.. you dont have any equations just expressions.. dont be too quick to award yourself bits Razz
AsianSensation




PostPosted: Sun Jun 29, 2003 6:21 pm   Post subject: (No subject)

lol, I don't actually award myself bits, but anyways, I see what I did wrong, so I am going to edit that post and put the correct solutions up instead.

btw, is it only us two that's participating in this? come on peoples, put in all your solutions.

anyways, +5 bits to SilverSprite for pointing out my mistake.
SilverSprite




PostPosted: Sun Jun 29, 2003 6:33 pm   Post subject: (No subject)

that is bugz's solution...
SilverSprite




PostPosted: Sun Jun 29, 2003 6:33 pm   Post subject: (No subject)

erase your first solution eh:P save your own skin
SilverSprite




PostPosted: Sun Jun 29, 2003 6:36 pm   Post subject: (No subject)

btw.. your equation 1 is not greater than equation 2 for n=1 lol
AsianSensation




PostPosted: Sun Jun 29, 2003 6:38 pm   Post subject: (No subject)

what is, the one I edited?

well, I didn't ask him or anyone else, I came up with that by myself, there is another solution to that, one kinda similar, it uses the same ideas, so I'm not going to post it up here.

Anyways...

Question #3:

let Sn = 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n)

show that 2*sqrt(n + 1) - 2 < Sn < 2*sqrt(n) - 1

for n > 1
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bugzpodder




PostPosted: Wed Jul 02, 2003 10:18 pm   Post subject: (No subject)

i solved Q#3 so i get bits Very HappyVery HappyVery Happy
bugzpodder




PostPosted: Wed Jul 02, 2003 11:34 pm   Post subject: (No subject)

here is my question (i am too cheap to offer any bits so dont ask for them)

prove if p+2 and p^2+2 are prime,then p^3+2 is also prime, with p being a prime itself. SilverSprite isnt allowed to answer. he should know why.
AsianSensation




PostPosted: Thu Jul 03, 2003 9:04 am   Post subject: (No subject)

yeah, bugz gets the bits for #3, but you still have to post the solutions up, I want to see a "real" solution, instead of my solution (if you call random pen scratches on scrap paper solutions)
bugzpodder




PostPosted: Thu Jul 03, 2003 10:56 am   Post subject: (No subject)

proof is located here: http://www.sosmath.com/CBB/viewtopic.php?p=3296#3296
bugzpodder




PostPosted: Thu Jul 03, 2003 10:59 am   Post subject: (No subject)

My proof for question #2:
since lim (k-> inf) 1/1!+1/2!+1/3!+...+1/k! = e ~ 2.71828...
therefore 1/1!+1/2!+1/3!+...+1/n! < e < 3
another 30 bits? Very Happy
AsianSensation




PostPosted: Thu Jul 03, 2003 1:21 pm   Post subject: (No subject)

that's cool, yeah, I'll give you another 30 bits, just keep pumping out those good solutions.

and since bugz already gave a question, let that question be question #4 then.
bugzpodder




PostPosted: Thu Jul 03, 2003 1:25 pm   Post subject: (No subject)

i changed my mind if anyone solves 4, i'll give the 30 bits AS gave me to give you (Excluding SS of course)
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