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26 
Posts: 179 Joined: 15 May 2007 Last Visit: 29 Apr 2015
LD count: 6
 


Posted: Sun 17 Jun, 2007 


what does that big line mean?
Current LD goal(s): One more.



 
 
   




31 
Posts: 210 Joined: 23 Dec 2005 Last 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.



 
 
   





Posts: 523 Joined: 25 Sep 2006 Last 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



 
 
   




31 
Posts: 3 Joined: 07 Nov 2007 Last 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



 
 
   




26 
Posts: 1033 Joined: 10 Sep 2006 Last 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



 
 
   





Posts: 5954 Joined: 03 Dec 2005 Last 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)?



 
 
   




32 
Posts: 175 Joined: 06 Oct 2005 Last 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<az+zb+2z^2/(z+az+zb+z^2)<9/6
we can add z and z cause zz=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^2z)/(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)^2ab]/(ab+a^2b+b^2a+(ab)^2)<3/6
2/6< [ab(ab1)]/ab(1+b+a+ab)<3/6
2/6<(ab1)/(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...



 
 
   




26 
Posts: 1033 Joined: 10 Sep 2006 Last Visit: 22 Sep 2015
LD count: Never enough
Location: Greece  making ebil plans  


Posted: Tue 20 Nov, 2007 





 
 
   





Posts: 523 Joined: 25 Sep 2006 Last 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 xpi = +(x3).
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.



 
 
   




26 
Posts: 493 Joined: 03 Feb 2007 Last 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?



 
 
   




28 
Posts: 836 Joined: 09 Apr 2006 Last 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. =/



 
 
   




31 
Posts: 210 Joined: 23 Dec 2005 Last 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.



 
 
   




151

Posts: 2490 Joined: 21 Jul 2008 Last Visit: 19 Apr 2019
LD count: %d
Location: Universe............. Favorite Animal: Mew....... Nightmare count: 4  


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



 
 
   




31

Posts: 2778 Joined: 30 Jul 2005 Last 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



 
 
   





Posts: 5954 Joined: 03 Dec 2005 Last Visit: 29 Mar 2018
LD count: a bunch.
Location: fleeting.  


Posted: Wed 24 Sep, 2008 


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



 
 
   
 
 