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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

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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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

(i think the above is right)

The Beginning Of All Things To End.

The End Of All Things To Come.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

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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**shocamefromebay****Member**- Registered: 2007-05-30
- Posts: 103

hmmm

what are the possiblities that this does not get infinitly close to one number

and just constantly increases?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**shocamefromebay****Member**- Registered: 2007-05-30
- Posts: 103

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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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

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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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

The individual part sums are all divergent?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

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**shocamefromebay****Member**- Registered: 2007-05-30
- Posts: 103

luca-deltodesco wrote:

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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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Funny luca, I would probably say the same thing about 1/n. Your eyes may deceive you. Don't trust them.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

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**shocamefromebay****Member**- Registered: 2007-05-30
- Posts: 103

teh solution for this one is e

but before theinvention/discovery/creation of e

no one could have figured it out

what are the possibilities taht this infinite series is like the one shown above

that there really is no rational answer taht we can put in fractional form?

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