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find
the value of this infinite series
Last edited by RauLiTo (2007-06-21 10:53:12)
ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
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(i think the above is right)
The Beginning Of All Things To End.
The End Of All Things To Come.
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However this formula almost only makes it harder to solve.
I prefer the original one.:/
Last edited by George,Y (2007-06-21 13:09:59)
X'(y-Xβ)=0
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hmmm
what are the possiblities that this does not get infinitly close to one number
and just constantly increases?
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Whoops, forgot to come back to this problem.
what are the possiblities that this does not get infinitly close to one number
and just constantly increases?
None. It can be easily shown that each term is less than 1/n^2, and the sum of 1/n^2 converges. Thus, this must converge.
So we may say:
The rest is just algebra.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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No, it definitely converges to something. The series is equivalent to this:
Each term in that sum is smaller than the equivalent term in
, but that sum is finite.Edit: Heh, post collision and we say almost exactly the same thing. Weird.
Why did the vector cross the road?
It wanted to be normal.
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Edit: Heh, post collision and we say almost exactly the same thing. Weird.
I wouldn't say weird. 1/n^2 is a very popular series and the original series is 1 / something*something else*something else. It only makes sense to compare this to 1/n^2.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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well
i made a prog on my calc about 5 mins ago to do sequences for me
and usually if the number converges
it jsut rounds and give me that number that it converges to
and i put this one into it
and it just kept increasing
for like 45 mins
then i just turned it off
and the calc couldnt even turn the number into fractions form
so i think if its byond the power of my calcy
its beyond the power of me
but this shows that this problem
is pretty difficult
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shocamefromebay, no trust me, it DOES converge, you program is wrong:
simple AS2 script:
var s:Number = 0;
var i:Number = 1;
function onEnterFrame():Void
{
for(var j:Number = 0; j<5000; j++)
{
s+= 1/(3*i*(3*i-1)*(3*i-2));
i++;
}
trace(i+" : "+s);
}
printing a value every 5000 times, you get to:
140001 : 0.178796768890753
from which point it doesn't change
The Beginning Of All Things To End.
The End Of All Things To Come.
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Probably too late to say this as we now have an answer, but it simplifies a bit to give:
I'm not sure if that helps any, but it would be nice to find an exact answer if we can.
Why did the vector cross the road?
It wanted to be normal.
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Anyone know why what mathsyperson did is wrong?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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The individual part sums are all divergent?
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Well, that's part of it.
Mathsyperson rewrote it as an alternating series, which is fine. But an alternating series can only have its terms reordered if it is absolutely convergent. This one isn't.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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shocamefromebay, no trust me, it DOES converge, you program is wrong:
this doesnt necessarily mean that its wrong
it just means that it has reached the point where it converged yet
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it quickly converges in the small number of decimal places, after that it takes some time for the rest of the decimal places to follow, but even after say 10 iterations, its clear its converging;
The Beginning Of All Things To End.
The End Of All Things To Come.
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Funny luca, I would probably say the same thing about 1/n. Your eyes may deceive you. Don't trust them.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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but 1/n doesn't do that, it constantly changes, maybe a few decimal places remain consistent, but then they change a few iterations later, well, maybe i exxagerated a bit on 10 iterations , but the first 100 iterations of the series printing every 5 iterations: after 5 iterations the first 3 decimal places never change, after 15 iterations it has the fourth decimal place, then it starts taking longer and longer to converge.
5 0.1781080031080031
10 0.17861794090836494
15 0.1787163312115937
20 0.17875125640305384
25 0.1787675393766264
30 0.17877642383782857
35 0.17878179703796668
40 0.17878529205149957
45 0.17878769217497648
50 0.17878941118940814
55 0.1787906843899262
60 0.17879165359266566
65 0.1787924084010432
70 0.17879300768311585
75 0.17879349140653122
80 0.17879388747893882
85 0.17879421586516644
90 0.17879449115322782
95 0.17879472420280887
after 140,000 iterations: 0.17879676889075347
after 1,000,000 iterations: 0.17879676889075347
after 100,000,000 iterations: 0.17879676889075347
pretty safe to say it converges.
----
actually, why doesn't 1/n converge? because 1/n gets smaller and smaller towards 0, i would have thought it would converge?
Last edited by luca-deltodesco (2007-06-25 17:56:56)
The Beginning Of All Things To End.
The End Of All Things To Come.
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actually, why doesn't 1/n converge? because 1/n gets smaller and smaller towards 0, i would have thought it would converge?
That's a good question and I wish I had a sufficient answer. Basically, it just doesn't converge fast enough. But I still find series like this mysterious.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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