/sci/ - Science & Math

now this I like

alright:
I want to show that <span class="math"><n>=p [/spoiler] if and only if <span class="math">p(p-1)<n \leq p(p+1)[/spoiler].

>>5888099
I'll show later that the square root of n is a half integer if and only if it is an integer.
Assuming that for the moment, we have <span class="math"><n>=p [/spoiler] if and only if:
<span class="math">p-1/2 < \sqrt{n} < p+1/2 [/spoiler], ie <span class="math">p^2-p+1/4 < n <p^2+p+1/4 [/spoiler].
The left side and right side can never be integers, so if n is a half integer, it is necessarily an integer anyway.
The left side can be replaced by the first bigger integer, and the right side can be replaced by the first smaller integer.
We obtain: <span class="math"> p^2-p=p(p-1) < n < p^2 + p=p(p+1) [/spoiler], which is what we wanted to show.

>>5888127
sorry the end should be: [/math]p(p-1)<n\leq p(p+1)[/math] as I replaced the left side with the first smaller integer.
now we can calculate the sum
<span class="math"> \displaystyle \sum_{n=1}^{infty} \frac{2^{<n>}+2^{-<n>}}{2^n}= \sum_{n=1}^{infty} u_n [/spoiler].

>>5888127
sorry the end should be: <span class="math">p(p-1)<n\leq p(p+1)[/spoiler] as I replaced the left side with the first smaller integer.
now we can calculate the sum <span class="math">\displaystyle \sum_{n=1}^{\infty} \frac{2^{<n>}+2^{-<n>}}{2^n}= \sum_{n=1}^{\infty} u_n[/spoiler].

>>5888139
forget u_n.

anyway the sum is also equal to
<span class="math">\displaystyle S= \sum_{p=1}^{\infty} \sum_{n=p(p-1)+1}^{p(p+1)} \frac{2^p + 2^{-p}}{2^n}[/spoiler].

modulo my mistakes, I find <span class="math"> \displaystyle S=2. \sum_{p=1}^{\infty} \frac{1}{2^{(p-1)^2}} - \frac{1}{2^{(p+1)^2}}[/spoiler].

>>5888161
which leads simply to <span class="math">S=2.(1 + \frac{1}{2})=3 [/spoiler] (S is a telesoping series)

>>5888161
very nice!

The answer's three. I know this because I am a math wizard.

>>5888170
Yes, but does .999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 = 1?

>>5888630
no

>>5888170
>'s three. I know this because I am a math wizard.
Nice :)

I tried this one >>5889840 but no answer.
Again, my english is terrible, soprry

I think for the sum we could maybe argue more "directly" :
In fact, <n> = q (in N) means n is in {q^2-q+1; .... ; q^2+q }

When we "change" q :
n+q cover N except integers like A^2 (squared integer ?)
And n-q cover all integers, and numbers like A^2 twice !

So n + or - <n> is {0;0;1;1;2;2;....;p;p;...}
(I know you can't write set like this but you understand, all intergers are covered twice)

And finally, we can compute the sum easiliy : S = 2^0 + blabla
where blabla=2 (trivial sum) and so S=3

Is it more simple ? Not sure, actually :)