/sci/ - Science & Math

>>4266870
>It looks like you went to a lot of effort to prove something completely trivial.

It looks like you don't even understand the question.

Apply the following inductively to <span class="math">F(x) = f(x) - 1/(x^2+1), F(x) = f'(x) - [1/(x^2+1)]', etc.[/spoiler]

Lemma: Let <span class="math">F(x)[/spoiler] be infinitely differentiable and suppose there is a strictly decreasing sequence <span class="math">x_1 > x_2 > \cdots > 0[/spoiler] with limit zero such that <span class="math">F(x_i) = 0[/spoiler] for all <span class="math">i[/spoiler], and also <span class="math">F(0) = 0[/spoiler]. Then <span class="math">F'(0) = 0[/spoiler] and there exists another strictly decreasing sequence <span class="math">y_1 > y_2 > \cdots > 0[/spoiler] with limit zero such that <span class="math">F'(y_i) = 0[/spoiler].

Proof: The sequence of difference quotients <span class="math">(F(x_i) - F(0))/x_i[/spoiler] is identically zero, and since <span class="math">x_i \rightarrow 0[/spoiler], <span class="math">F'(0) = \lim_{i \rightarrow \infty} (F(x_i) - F(0))/x_i = \lim_{i \rightarrow \infty} 0 = 0[/spoiler] by continuity of <span class="math">F'[/spoiler]. The sequence <span class="math">y_i[/spoiler] exists by Rolle's theorem, choosing a <span class="math">y_i[/spoiler] between each pair of consecutive <span class="math">x_i[/spoiler]'s.

Therefore <span class="math">f^{(n)}(x) - \frac{d^n}{dx^n} 1/(1+x^2)[/spoiler] is zero at <span class="math">x=0[/spoiler] for all <span class="math">n[/spoiler] and the answers are what >>4266592 said.


Now pony tiem

I AM BEST TRIPFAG :V

>>4253460
I'm the girl with the mid-length brown hair and weird glasses. Sneak up behind me and grab my boobs and I promise to sleep with you.

>>4253399
<span class="math">4x^2-1[/spoiler] should be <span class="math">4x^2-2[/spoiler] those two places.

The solutions of <span class="math">x^2=y^2+1[/spoiler] are given by <span class="math">x=\frac{r+r^{-1}}{2}, y = \frac{r-r^{-1}}{2}, r \in \mathbb{F}_p^*[/spoiler].

Two numbers <span class="math">r,s \in \mathbb{F}_p^{*} [/spoiler] give the same value of <span class="math">x^2[/spoiler] iff <span class="math">\left(\frac{r+r^{-1}}{2}\right)^2 = \left(\frac{s+s^{-1}}{2}\right)^2[/spoiler] iff <span class="math">r^2+r^{-2} = s^2+s^{-2} (=4x^2-1)[/spoiler]. But then <span class="math">\{r^2, r^{-2}\}[/spoiler] and <span class="math">\{s^2, s^{-2}\}[/spoiler] are both the set of roots of <span class="math">t^2-(4x^2-1)t+1[/spoiler], so <span class="math">\{r^2, r^{-2}\} = \{s^2, s^{-2}\}[/spoiler]. The number we want is the number of distinct sets <span class="math">\{r^2, r^{-2}\}, r \in \mathbb{F}_p^*[/spoiler]

<span class="math">1[/spoiler] and <span class="math">-1[/spoiler] are the only numbers which are their own inverses, and <span class="math">1[/spoiler] is always a residue. If <span class="math">-1[/spoiler] is not a residue, i.e. <span class="math">p \equiv 3 \pmod 4[/spoiler], the total is one plus half the number of remaining residues, or <span class="math">1 + \frac{1}{2}\left(\frac{p-1}{2}-1\right) = \frac{p+1}{4}[/spoiler].

OTOH if <span class="math">-1[/spoiler] is a residue so <span class="math">p \equiv 1 \pmod 4[/spoiler], the total is <span class="math">2 + \frac{1}{2}\left(\frac{p-1}{2}-2\right) = \frac{p+3}{4}[/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]