the f(n) of MCS

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the f(n) of MCS

Post by i am furfur on July 31st 2009, 8:24 pm

Let f(n) = 1/(2^n)
where n >= 1.
as n approaches infinity f(1) + f(2) + ... + f(n) approaches 1.
f(n) is the percentage number of students in n year level in the department of MCS.
Can you prove that a certain student in a year level will have a 50% chance to proceed to the next year level?

Isn't math fun?

someone prove this... saun ra kaau... math15 + math16 + math35...

edit: f(n) was wrong lol... fixed it... but its so easy now... x.x it looked harder when it was wrong lol u dont need induction anymore... *sadface*


Last edited by i am furfur on August 2nd 2009, 8:07 pm; edited 3 times in total

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Re: the f(n) of MCS

Post by Chujutsu on August 2nd 2009, 5:54 pm

I don't get the f(n). If n = 2, what is f(n) or f(2)?
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Re: the f(n) of MCS

Post by i am furfur on August 2nd 2009, 5:57 pm

Chujutsu wrote:I don't get the f(n). If n = 2, what is f(n) or f(2)?

ooops my bad.
f(n) = 1 - 1/2 - 1/3 - 1/n - ...
f(2) = 1 - 1/2

saup akong gi originally post

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Re: the f(n) of MCS

Post by Chujutsu on August 2nd 2009, 6:01 pm

Uhh... f(3) gives 1 - 1/2 - 1/3 which is 1/6. I don't think that's at least 50% chance.
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Re: the f(n) of MCS

Post by i am furfur on August 2nd 2009, 6:40 pm

Chujutsu wrote:Uhh... f(3) gives 1 - 1/2 - 1/3 which is 1/6. I don't think that's at least 50% chance.

f(n) is the percentage number of students in n year level
get the percentage chance of a student in a n year level going to the n+1 year level.

hmm... you have a point... i was thinking (1/2)/2=1/8 x.x

anyway there's something wrong with the function coz f(4) is already negative lol
i'll fix it after i play dota

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Re: the f(n) of MCS

Post by Chujutsu on August 2nd 2009, 6:47 pm

(1/2)/2 in math is 1/4
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Re: the f(n) of MCS

Post by i am furfur on August 2nd 2009, 7:18 pm

Chujutsu wrote:(1/2)/2 in math is 1/4

i know that LOL... changed the f(n) anyway

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Re: the f(n) of MCS

Post by Chujutsu on August 2nd 2009, 7:43 pm

1/(2^n) = (1/2)^n

Since 1/2 is between zero and one, the more it is multiplied with itself, the smaller its value becomes, so...? Less chance?

Anyway, (1/2)^1 is equal to 1/2, or 50%.

Assuming it's sum of f(1), f(2),... all the way up to f(n) instead. If that was true, then simply adding f(n+1) should increase that. After all, it is impossible for (1/2)^n to become negative. Since the sum of f(1), f(2),... all the way up to f(n) is at least 50%, then that would mean the sum of f(1), f(2),... all the way up to f(n+1) is also at least 50%.
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Re: the f(n) of MCS

Post by i am furfur on August 2nd 2009, 8:06 pm

Chujutsu wrote:1/(2^n) = (1/2)^n

Since 1/2 is between zero and one, the more it is multiplied with itself, the smaller its value becomes, so...? Less chance?

nope.
f(n) is decreasing but you're not looking for the population percentages per year
but the rate of passing a year level into the next.

Chujutsu wrote:
Assuming it's sum of f(1), f(2),... all the way up to f(n) instead. If that was true, then simply adding f(n+1) should increase that. After all, it is impossible for (1/2)^n to become negative. Since the sum of f(1), f(2),... all the way up to f(n) is at least 50%, then that would mean the sum of f(1), f(2),... all the way up to f(n+1) is also at least 50%.

if it's the sum then f(1) = 1/2 and f(2) = 3/4
the percentages per year level would be wrong.
f(1) + f(2) = 5/4 which is greater than 100%
is that what you mean?

whereas the original.
f(1) = 1/2, f(2) = 1/4 so on...
then f(1) + f(2) + ... f(infinity) would approximate to 1 or 100%.

anyway of course f(1) + .. + f(n) would be greater than 50% since f(1) is already 50%

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Re: the f(n) of MCS

Post by Chujutsu on August 3rd 2009, 5:51 pm

Let's just stop. This f(n) is still broken.
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Re: the f(n) of MCS

Post by i am furfur on August 3rd 2009, 5:52 pm

Chujutsu wrote:Let's just stop. This f(n) is still broken.

haha... the exponential function works...
anyway... you just need to prove that the ratio of f(n+1)/f(n) = 50%

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Re: the f(n) of MCS

Post by Chujutsu on August 3rd 2009, 5:59 pm

f(1+1)/f(1) = f(2)/f(1) = (1/2)^2 / (1/2)^1 = (1/4)/(1/2) = (1/4) * 2 = 1/2

f(n+1)/f(n) = (1/2)^(n+1) / (1/2)^n = (1/2)^(n+1 - n) = 1/2
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Re: the f(n) of MCS

Post by i am furfur on August 3rd 2009, 6:00 pm

Chujutsu wrote:f(1+1)/f(1) = f(2)/f(1) = (1/2)^2 / (1/2)^1 = (1/4)/(1/2) = (1/4) * 2 = 1/2

f(n+1)/f(n) = (1/2)^(n+1) / (1/2)^n = (1/2)^(n+1 - n) = 1/2

the end XD lol...
anyway forget the first f(n) that was broken.

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Re: the f(n) of MCS

Post by Chujutsu on August 3rd 2009, 6:02 pm

Yeah sure
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Re: the f(n) of MCS

Post by pbwuzherr on August 3rd 2009, 11:58 pm

I gotta post something here,..

....

oh,.!


Let me just say it:

"Q.E.D.
With this,
the proof is complete."

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Re: the f(n) of MCS

Post by Chujutsu on August 4th 2009, 6:24 pm

Q.E.D.?

What is that?
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Re: the f(n) of MCS

Post by diannee on August 4th 2009, 11:51 pm

I DON'T GET THE WHOLE THING lol Smile
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Re: the f(n) of MCS

Post by Chujutsu on August 5th 2009, 11:52 am

Just breathe in slowly, and breathe out... Make sure that you won't bleed!!!

Anyway, just take it piece by piece. That's how I'd look at math at times. You don't have to take everything in all at once.
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Re: the f(n) of MCS

Post by diannee on August 7th 2009, 10:55 pm

Chujutsu wrote:Just breathe in slowly, and breathe out... Make sure that you won't bleed!!!

Anyway, just take it piece by piece. That's how I'd look at math at times. You don't have to take everything in all at once.

HUHU. salig bryt ai .Smile
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Re: the f(n) of MCS

Post by Chujutsu on August 11th 2009, 8:20 pm

Hmm... Maybe I should give some tips with Math? Maybe not in here though...

Would you want some? Very Happy
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