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Muzzius
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PostPosted: Sun 17 Jun, 2007  Reply with quote

what does that big line mean?


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Kenneth
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PostPosted: Mon 18 Jun, 2007  Reply with quote

The one on the far left? That's an integral sign. You'll learn about it if you take a math class on calculus. It can mean either finding the antiderivative of a function, or, as in this case, finding the area under the graph of a function. For more info, and probably a better explanation on what an integral is, see Wikipidia's entry on integrals. :D

For anyone who is interested, this integral is so important that it has it's own name, the Gaussian Integral, and a proof of it can also be found on Wikipedia.


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Bombax
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PostPosted: Mon 18 Jun, 2007  Reply with quote

check wrote:
I have pi memorised to 362 places. smile Pi is infinite (I've read books on it!), it can't end. You can't say 'x' = pi, because we don't know what pi fully is. Sure, you can say x = 3.14, but that's nothing spectacular - it proves nothing. You can't say that x = pi, because pi never ends. Ever.

Anyone know of the Riemann hypothesis? Prime numbers are more intricate than you may think!
Eh, you're wrong. You can give any variable any value, even irrational ones. You may protest I give x = pi, but do you protest against me solving an equation and finding x = sqrt(2), which is also an irrational number? No, I don't think so ^^


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stopusingoil
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Re: The Maths Topic
PostPosted: Thu 15 Nov, 2007  Reply with quote

Python Angel wrote:
Whatever π is, I can always base it of another value right? A value we call x? So now I have this little starter equation:
x = (π + 3) / 2
Shouldn't be too hard for any of you to understand.


This is the major flaw in your proposal.


Essentially, you are letting x = π.

That means that the equation x = (π + 3) / 2 is incomplete. If π=x, then you should not have both x and π in your equation. You need to finish your substitution of x for π.

If you do this, your equation will then look like:
x = (x + 3) / 2
2x = x + 3
x = 3

Now it would seem that you have proved π=3. But no. The equation you began with was completely made up. What you have found is that x=3 in your equation. You need to prove that x=π. This is impossible. An attempt looks like this:

x = π
3 = π
3 = 3.14...

So really, I think that cookie belongs to me smile


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Queen SD
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PostPosted: Mon 19 Nov, 2007  Reply with quote

Can anyone solve this?

a,b > 0 and a+b=z
prove that:
(4/3z)<(1/a+z) + (1/b+z)<3/2z


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Bruno
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PostPosted: Tue 20 Nov, 2007  Reply with quote

Αλεξάνδρα wrote:
Can anyone solve this?

a,b > 0 and a+b=z
prove that:
(4/3z)<(1/a+z) + (1/b+z)<3/2z

Your notation isn't helping much

(4/3z) is supposed to mean 4/(3z) or (4/3)z ?

(1/a+z) is (1/a)+z or 1/(a+z)?


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PostPosted: Tue 20 Nov, 2007  Reply with quote

Is this some kind of an exersize for school??
Let's see!

a,b>0 and a*b=z => z>0
so we can multiply z!

4/3< [(b+z)z+(a+z)z]/(a+z)(b+z)< 3/2
4/3 < z[(b+z)+(a+z)]/(ab+az+zb+z^2)/3/2
8/6<[z(a+b+2z)]/(z+az+zb+z^2)<9/6
8/6<az+zb+2z^2/(z+az+zb+z^2)<9/6
we can add z and -z cause z-z=0
8/6<[(az+zb+z^2+z)-z+z^2]/(z+az+bz+z^2)<9/6
8/6<1+ (-z+z^2)/(z+az+bz+z^2)<9/6

1=6/6 and z^2>-z for every z in R

so 8/6<1+(something >0)<9/6

2/6<(z^2-z)/(z+az+bz+z^2)<3/6

I think it goes further but i'm a bit bored right now:Ptounge2
so this is for you:P

Ok i'll take it a bit further!
2/6<[(ab)^2-ab]/(ab+a^2b+b^2a+(ab)^2)<3/6
2/6< [ab(ab-1)]/ab(1+b+a+ab)<3/6
2/6<(ab-1)/(1+b+a+ab)<3/6

(thank god a,b>0 cause if any one of them was 0 then,this one could not be solved)
If ab=1 then 2/6<0<3/6 which is wrong!
So i might have done a mistake somewhere!If someone else doesn't solve this,i'll look at it later:P

a second edit! ab=a*b and so on...


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Queen SD
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PostPosted: Tue 20 Nov, 2007  Reply with quote

Bruno wrote:
Αλεξάνδρ& #945; wrote:
Can anyone solve this?

a,b > 0 and a+b=z
prove that:
(4/3z)<(1/a+z) + (1/b+z)<3/2z

Your notation isn't helping much

(4/3z) is supposed to mean 4/(3z) or (4/3)z ?

(1/a+z) is (1/a)+z or 1/(a+z)?

oh heh it supposed to mean 4/(3z) and 1/(a+z) shy2
Its just an excersise i found a few days ago, tried solving it and failed miserably .

EDIT: hold on... i solved the first part --> 4/(3z)<1/(a+z) + 1/(b+z)
(i wrote the answer on paint cuz it would take me too long to type it and id probably get things wrong again - not used to typing maths tounge2 . Sorry for the letters but they were the best i could do )

Click here to see the hidden message (It might contain spoilers)


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Bombax
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Re: The Maths Topic
PostPosted: Wed 28 Nov, 2007  Reply with quote

stopusingoil wrote:
Python Angel wrote:
Whatever π is, I can always base it of another value right? A value we call x? So now I have this little starter equation:
x = (π + 3) / 2
Shouldn't be too hard for any of you to understand.


This is the major flaw in your proposal.


Essentially, you are letting x = π.

That means that the equation x = (π + 3) / 2 is incomplete. If π=x, then you should not have both x and π in your equation. You need to finish your substitution of x for π.

If you do this, your equation will then look like:
x = (x + 3) / 2
2x = x + 3
x = 3

Now it would seem that you have proved π=3. But no. The equation you began with was completely made up. What you have found is that x=3 in your equation. You need to prove that x=π. This is impossible. An attempt looks like this:

x = π
3 = π
3 = 3.14...

So really, I think that cookie belongs to me smile


Eh, no. Leave it to the math freak. lachgroen I have not found the value of x, I have "found" the value of pi. Essentially I haven't, because the first equation is completely correct; I simply made incorrect transformations in the process to "prove" that pi was 3. Basically, I have incorrectly elmininated our unknown variable x. You see, I can make whatever statement mathematically to assign (or in this case, correlate) the variable x. pi in this case is some constant.

The incorrect part lies when I take the square root out of both sides of the equation; the value within the paranthesis may as well be it's negative correspondance; and when you actually add that, you don't get pi = 3 anymore. It may as well be x-pi = +-(x-3).

Alright, if you still want to argue about it, try replacing pi with "4" everywhere in the "proof". The first equation will be correct, yet the end result (that 4=3) will trivially be incorrect.


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Phi_guy
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Re: The Maths Topic
PostPosted: Thu 27 Dec, 2007  Reply with quote

stopusingoil wrote:
Python Angel wrote:
Whatever π is, I can always base it of another value right? A value we call x? So now I have this little starter equation:
x = (π + 3) / 2
Shouldn't be too hard for any of you to understand.


This is the major flaw in your proposal.


Essentially, you are letting x = π.

That means that the equation x = (π + 3) / 2 is incomplete. If π=x, then you should not have both x and π in your equation. You need to finish your substitution of x for π.

If you do this, your equation will then look like:
x = (x + 3) / 2
2x = x + 3
x = 3

Now it would seem that you have proved π=3. But no. The equation you began with was completely made up. What you have found is that x=3 in your equation. You need to prove that x=π. This is impossible. An attempt looks like this:

x = π
3 = π
3 = 3.14...

So really, I think that cookie belongs to me smile


Where does he ever say that π=x?


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Wissam
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PostPosted: Sat 29 Dec, 2007  Reply with quote

Hi, I'm studying Statistics 1 and I'm stuck on a question on Discrete Random Variables. I have the question and the answer from the mark scheme but I don't understand how they answered it. If anyone can please explain what I have to do, to answer it would be greatly appreciated.


It's supposed to be an easy question but I don't get it. =/


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Kenneth
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PostPosted: Sat 29 Dec, 2007  Reply with quote

The first table shows the probablity that x is smaller than or equal to a certain value.
The second table will show the probablity that x is precicely equal to a certain value.

Starting with the first entry, from the first table we see that the probablity that x is smaller than or equal to 1 is 0.1. The question then is, what is the probablity that x is precicely 1? Since the probablity that x is smaller than 1 is 0 (1 is the lowest value it can take) the probability that x is precicely 1 must also equal 0.1.

For the second entry, we see from table 1 that the probability that x is smaller than or equal to 2 is 0.2. This means that we have
(probability that x is smaller than 2) + (probability that x is precicely 2) = 0.2
The unknown we are looking for here is the probability that x is precicely 2. The probability that x is smaller than 2 must be equal to the probability that x is precicely 1, since the only smaller value than 2 x can take is 1. Thus we have
0.1 + (probability that x is precicely 2) = 0.2
and we see that the probability that x is precicely 2, and thus the second entry in the solution table, is 0.1.

For the third entry, we carry on the same way. We have
(probability that x is smaller than 3) + (probability that x is precicely 3) = 0.25
The probability that x is smaller than 3 must be the probability that x is precicely 1 + the probability that x is precicely 2, since 1 and 2 are the only lower values x can take. Thus we have
(probability that x is smaller than 3) = 0.1 + 0.1 = 0.2 and
0.2 + (probability that x is precicely 3) = 0.25
and thus (probability that x is precicely 3) = 0.05 - this is the third entry in the solution table.

For the rest, the same general pattern (which I've hopefully made clear by now) is followed. Hope this helps. ^^


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Mew151
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PostPosted: Mon 15 Sep, 2008  Reply with quote

Proof that .999999... equals 1:

.(9)/3=.(3)

.(3) = 1/3
1/3 · 3 = 3/3
.(3) · 3 = .(9)
So...
.(9) = 3/3 = 1



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Magnus
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PostPosted: Wed 24 Sep, 2008  Reply with quote

well, the error lies in that .(3)=1/3 that's not true wink

1/3=.333333333333333...

so therefore it's not a proof of .9999999... equals 1



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Bruno
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PostPosted: Wed 24 Sep, 2008  Reply with quote

Mag, .(3) = .33333333..., it's just a different notation.

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