Math Is Fun Forum

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Absolutely "don't knowing what's going on"! Please have a look...

OK. Something pretty strange is going on here...
It's FACT that for positive integer k, we have:

OK?
Noww..., we can write this in this way: (k is positive)

Now, because k is positive integer, such is:

So we can write it as we can write k:

and...

Plugging this into the expression for k:

Now we do the same thing, but for

:

Plugging again:

As we continue infinitely many times, wa have:

So we got nice representation of k as nested radicals.
For example, put k=1:

BUT 1 IS POSITIVE, AND ALL THE NUMBERS IN THE RADICALS (EXCEPT THE LAST, WHICH IS INFINITLY DEEP) ARE NEGATIVE!!!!

Last edited by krassi_holmz (2007-08-06 01:33:23)

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Absolutely "don't knowing what's going on"! Please have a look...

But the last is infinity if you allow infinite radicals.

If you don't, stop it at some finite step and you can just varify it.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Absolutely "don't knowing what's going on"! Please have a look...

BUT 1 IS POSITIVE, AND ALL THE NUMBERS IN THE RADICALS (EXCEPT THE LAST, WHICH IS INFINITLY DEEP) ARE NEGATIVE!!!!

I have yet to check over your work.  But I don't see why you have a problem with the above.  Infinity acts in weird ways, you know that krassi.  I don't see why it isn't possible for the infinite expression on the right to be 1.

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"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..."

HallsofIvy

Guest

Re: Absolutely "don't knowing what's going on"! Please have a look...

OK. Something pretty strange is going on here...
It's FACT that for positive integer k, we have:

OK?
Noww..., we can write this in this way: (k is positive)

Now, because k is positive integer, such is:

So we can write it as we can write k:

and...

Plugging this into the expression for k:

Now we do the same thing, but for

:

Plugging again:

As we continue infinitely many times, wa have:

So we got nice representation of k as nested radicals.
For example, put k=1:

BUT 1 IS POSITIVE, AND ALL THE NUMBERS IN THE RADICALS (EXCEPT THE LAST, WHICH IS INFINITLY DEEP) ARE NEGATIVE!!!!

If you "continue infinitely many times" you have a divergent sequence so your final conclusion is not valid.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Absolutely "don't knowing what's going on"! Please have a look...

If you "continue infinitely many times" you have a divergent sequence so your final conclusion is not valid.

Do you have proof of this?

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"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..."

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Absolutely "don't knowing what's going on"! Please have a look...

I posted this, because there are identities, extracted by the exactly same way:
A simular nested radical is discovered by Ramanujan:

And this is true! But the last term here is infinity too.
Why this method works in some cases, in others - not?
And indeed, infinity is weird, so this may be true.
I'm maybe not sure the result will be one, but I'm pretty sure that this is not a divergent process!

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krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Absolutely "don't knowing what's going on"! Please have a look...

But the last is infinity if you allow infinite radicals.

Yes, it will be infinity (informal)
But the point is that (I hope you'll understand) the last term doesn't matter, because it's infinitely nested, so you "will never get up with it". So when we go to infinity, the last term "dissapears" and leaves only this "odd" negative numbers!

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Absolutely "don't knowing what's going on"! Please have a look...

Knowing your knowledge, krassi, you probably aren't making this mistake, but from what you wrote it sure sounds like you are.  In other words, I don't mean for this to be offending if it is. 

Negative numbers under radicals aren't negative, they are complex.  And when you have nested radicals, the complex parts can cancel each other out leaving a real number.

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"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..."

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Absolutely "don't knowing what's going on"! Please have a look...

Negative numbers under radicals aren't negative, they are complex.  And when you have nested radicals, the complex parts can cancel each other out leaving a real number.

You're right, but it's tricky that no complex numbers will occour in the nested radicals!
If you look at the first convergent:

,
you'll see that under the radical there is preciesly positive number, so we'll not have complex numbers at this stage.
But this happens with the second convergent too and with the others... - the last term always eliminates the root-of-negative thing!
So, even paradoxal at first sight, we can't even speak of complex numbers inside the radicals!
Or, in other words, every "sub-radical" is real, and it's bigger enough not to give negative answer when we substract its negative-fellow.

Last edited by krassi_holmz (2007-08-06 19:23:32)

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