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I'm trying learn to apply Neuton rules directly, and I've solved it this way:
As with any finite addition sequence, all addition is associative:
Absolutely right! That reminds me of this:
as your sum, but goes to infinity:
Why do we get different answers?
Because, as ricky pointed, addition is not nessesarly associative with infinite sequences.
Another counting system without null:
1-1
2-11
3-111
4-1111
ect.
Actually your question is very famous and hard. Such questins are known as spere-packing problems.
Here's a start:
http://en.wikipedia.org/wiki/Sphere_packing
If you want more information, i'm sure google will help.
If you use it very fraquently, you can write in the beginning "S(x) stands for..." and then use it.
I've seen a notebook with properties of the arctangens function, written 'A' for short.
We don't need a special function, because it's
when convergant.What is "set variable"?
For the series to be convergent, is't nessesary x to be in the unary complex circle: Abs(x)<1.
I tried the brackets, no solution!
You can't do it without brackets:
Why are recurrence relationships written in the form, for example:
?Thanks.
That's not true. You can write it as you like. The difference is only in the domain for n-s. For example, if you're specifying the Fibonnaci sequence, you may write:
oh!
That's fascinating!
after you go around it, try going inside!
It's like moving in cheese!
Watch this:
http://www.youtube.com/watch?v=d7eGypGOlOc
It's the only solution. Look at a plot of the above function to see it's strictly increasing for x>1 and x<-1.
x^3-3x^2 = x(x^2-3x) etc.
And here is the plot - all points lie on a same plane!
OK.
First, assuming you want only integer answers...
There are many ways, but I use this:
(it's long...)
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:
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!
I posted this, because there are identities, extracted by the exactly same way:
A simular nested radical is discovered by Ramanujan:
OK. Something pretty strange is going on here...
It's FACT that for positive integer k, we have:
So we got nice representation of k as nested radicals.
For example, put k=1:
Calculating on a mental abacus? Cool!
So close...but so far away...
What do you mean?
Exaactly!