  The Maths Topic Author  Message MuzziusSomniologist 26 Posts: 179Joined: 15 May 2007Last Visit: 29 Apr 2015 LD count: 6 Posted: Sun 17 Jun, 2007 what does that big line mean? Current LD goal(s): One more.
 KennethDon't Panic 31 Posts: 210Joined: 23 Dec 2005Last Visit: 24 Aug 2010 Location: Norway Posted: Mon 18 Jun, 2007 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. 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.
 BombaxDream Deity Posts: 523Joined: 25 Sep 2006Last Visit: 24 Mar 2018 LD count: Thrice a week Posted: Mon 18 Jun, 2007 check wrote: I have pi memorised to 362 places. 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 stopusingoilNew member 31 Posts: 3Joined: 07 Nov 2007Last Visit: 15 Nov 2007 Location: European Union
Re: The Maths Topic Posted: Thu 15 Nov, 2007 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 Queen SDStill Dreaming 26 Posts: 1033Joined: 10 Sep 2006Last Visit: 22 Sep 2015 LD count: Never enough Location: Greece - making ebil plans Posted: Mon 19 Nov, 2007 Can anyone solve this? a,b > 0 and a+b=z prove that: (4/3z)<(1/a+z) + (1/b+z)<3/2z
 Brunoa smiling haze  Posts: 5954Joined: 03 Dec 2005Last Visit: 29 Mar 2018 LD count: a bunch. Location: fleeting. Posted: Tue 20 Nov, 2007 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)?

 Dawn'sDeathDead 32 Posts: 175Joined: 06 Oct 2005Last Visit: 07 Feb 2013 Location: Lost in Fields... :P Posted: Tue 20 Nov, 2007 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-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:P 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...
 Queen SDStill Dreaming 26 Posts: 1033Joined: 10 Sep 2006Last Visit: 22 Sep 2015 LD count: Never enough Location: Greece - making ebil plans Posted: Tue 20 Nov, 2007 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) 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 . Sorry for the letters but they were the best i could do )

 BombaxDream Deity Posts: 523Joined: 25 Sep 2006Last Visit: 24 Mar 2018 LD count: Thrice a week
Re: The Maths Topic Posted: Wed 28 Nov, 2007 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 Eh, no. Leave it to the math freak. 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.

 Phi_guyAssistant Pig Keeper 27 Posts: 493Joined: 03 Feb 2007Last Visit: 16 Dec 2009 LD count: 4
Re: The Maths Topic Posted: Thu 27 Dec, 2007 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 Where does he ever say that π=x?

 WissamChange is good 29 Posts: 836Joined: 09 Apr 2006Last Visit: 17 Nov 2018 Location: Middle East Posted: Sat 29 Dec, 2007 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. =/
 KennethDon't Panic 31 Posts: 210Joined: 23 Dec 2005Last Visit: 24 Aug 2010 Location: Norway Posted: Sat 29 Dec, 2007 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. Mew151Ebil Mew Idol 151  Posts: 2499Joined: 21 Jul 2008Last Visit: 21 Jul 2019 LD count: %d Location: Prism Stone Posted: Mon 15 Sep, 2008 Proof that .999999... equals 1: .(9)/3=.(3) .(3) = 1/3 1/3 · 3 = 3/3 .(3) · 3 = .(9) So... .(9) = 3/3 = 1 Current LD goal(s): Get a real LD; Shoot fireworks out of my hands; Become top idol Link to My DJ: www.ld4all.com
 MagnusDo a RC, Now! 32 Posts: 2778Joined: 30 Jul 2005Last Visit: 21 Nov 2018 LD count: 48 Location: Sweden Posted: Wed 24 Sep, 2008 well, the error lies in that .(3)=1/3 that's not true 1/3=.333333333333333... so therefore it's not a proof of .9999999... equals 1 Current LD goal(s): Play chess on the moon :), transform into a guinea pig
 Brunoa smiling haze  Posts: 5954Joined: 03 Dec 2005Last Visit: 29 Mar 2018 LD count: a bunch. Location: fleeting. Posted: Wed 24 Sep, 2008 Mag, .(3) = .33333333..., it's just a different notation.  Display posts from previous:  All times are GMT + 2 Hours