/sci/ - Science & Math

The region will be symmetrical about <span class="math">x=0[/spoiler], so assume <span class="math">x > 0[/spoiler] and therefore all <span class="math">a_i > 0[/spoiler].

If <span class="math">a_n\rightarrow L[/spoiler], then <span class="math">L = \frac{L^2+y^2}{2}[/spoiler], so <span class="math">L = 1 \pm \sqrt{1-y^2}[/spoiler].

Let <span class="math">f(t) = (t^2+y^2)/2[/spoiler]. Then <span class="math">f'(t) = t > 0[/spoiler] for <span class="math">t>0[/spoiler]. Plotting <span class="math">f(t)[/spoiler] against <span class="math">g(t) = t[/spoiler] we see <span class="math">f(t) = t[/spoiler] for <span class="math">t = 1 \pm \sqrt{1-y^2}[/spoiler] and therefore

1) <span class="math">0 < t < f(t) < 1-\sqrt{1-y^2}[/spoiler] for <span class="math">t \in (0, 1-\sqrt{1-y^2})[/spoiler]

2) <span class="math">1-\sqrt{1-y^2} < t < f(t)< 1+\sqrt{1-y^2}[/spoiler] for <span class="math">t \in (1-\sqrt{1-y^2}, 1+\sqrt{1-y^2})[/spoiler]

3) <span class="math">1+\sqrt{1-y^2} < t < f(t) < \infty[/spoiler] for <span class="math">t \in (1+\sqrt{1-y^2}, \infty)[/spoiler]

In all three cases <span class="math">f(t)[/spoiler] is mapping the specified region into itself

1) If <span class="math">a_0 = x \in (0, 1-\sqrt{1-y^2})[/spoiler] the sequence <span class="math">a_n[/spoiler] will be increasing and bounded above by <span class="math">1-\sqrt{1-y^2}[/spoiler], and so convergent.

2) If <span class="math">x \in (1-\sqrt{1-y^2}, 1+\sqrt{1-y^2})[/spoiler], it's increasing and bounded above by <span class="math">1+\sqrt{1-y^2}[/spoiler], so convergent.

3) If <span class="math">x \in (1+\sqrt{1-y^2}, \infty)[/spoiler], it's increasing and the first term is already higher than either possible limit, so it diverges.

The region <span class="math">0< x < 1-\sqrt{1-y^2}[/spoiler] is half a square of side 2 and half a circle of radius 1, so the total region has area <span class="math">2^2+\pi \cdot 1^2 = 4+\pi[/spoiler]

If <span class="math">j = p[/spoiler], then
<div class="math">{p \choose p}{p+p\choose p} = {2p \choose p} = \frac{2(p+1)(p+2)\cdots(p+(p-1))}{(p-1)!} \equiv \frac{2[p(1+2+\cdots+(p-1)) + (p-1)!]}{(p-1)!} \equiv 2[/moot]

If <span class="math">j < p[/spoiler] then
<div class="math">{p \choose j}{p+j \choose j} = \frac{p!}{(p-j)!j!} \frac{(p+j)!}{p!j!} = \frac{p![(p+1)(p+2)\cdots(p+j)]}{(p-j)!j!j!} \equiv \frac{p!j!}{(p-j)!j!j!} = {p \choose j}[/moot]

So
<div class="math">thing\equiv 2+ \left[\sum_{j=0}^{p-1} {p \choose j} \right] = 2+\left[\sum_{j=0}^{p} {p \choose j}-1\right] = 2+(1+1)^p-1 = 2^p+1[/moot]

If <span class="math">j = p[/spoiler], then
<div class="math">{p \choose p}{p+p\choose p} = {2p \choose p} = \frac{2(p+1)(p+2)\cdots(p+(p-1))}{(p-1)!} \equiv \frac{2[p(1+2+\cdots+(p-1)) + (p-1)!]}{(p-1)!} \equiv 2[/moot]

If <span class="math">j < p[/spoiler] then
<div class="math">{p \choose j}{p+j \choose j} = \frac{p!}{(p-j)!j!} \frac{(p+j)!}{p!j!} = \frac{p![(p+1)(p+2)\cdots(p+j)]}{(p-j)!j!j!} \equiv \frac{p!j!}{(p-j)!j!j!} = {p \choose j}[/moot]

So
<div class="math">thing\equiv 2+ \left[\sum_{j=0}^{p-1} {p \choose j} \right] = 2+\left[\sum_{j=0}^{p} {p \choose j}-1\right] = 2+(1+1)^p-1 = 2^p+1[/moot]

>>4188942
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