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Secret
Secret
03:52
Bump
Let $t\in \Bbb{T}$, let $[t]_n$ be nth series that tends to $t$
Then $$\lim_{n\to \infty} t-[t]_n=0$$
For all $n < \infty$, $t-[t]_n$ is transcendental, now to determine what is the transcendence degree of each $t-[t]_n$
Secret
Secret
04:10
No that is not useful, as all we have is the sequence 0.xxxxxxx,0.0xxxxxx,0.00xxxxxx ...
Let s,t transcendental, then P(s),P(t) transcendental. Let s+t algebraic, then P(s+t)=P(s)+P(t)+Q(s,t)=0
Let s transcendental. Then for all 1/q^n in Q, we have s-1/q^n transcendental
Pick P_n such that P_n(1/q^n)=0. Then $P_n(s-1/q^n)=P_n(s)-P_n(1/q^n)+Q_n(s,1/q^n)=P_n(s)+Q_n(s,1/q^n)$
$$\lim_{n\to \infty} P_n(s-1/q^n) = \lim_{n\to \infty} P_n(s) + \lim_{n\to\infty} Q_n(s,1/q^n)=P(s)$$
i.e. $$\lim_{n\to \infty}Q_n(s,1/q^n)=0$$
Let P be deg 2
Ax^2+Bx+C
As^2+Bs+C
A(1/q^2n)+B/q^n+C
As^2-2As(1/q^n)+A(1/q^n)+B(s-1/q^n)+C
P(s)-P(1/q^n)+(-2As(1/q^n)-C)
2As(1/q^n)=C
Hmm...
 
14 hours later…
Secret
Secret
19:01
$$P(x)=Ax^2+Bx+C$$
$$P(x+y) = A(x^2+2xy+y^2)+B(x+y)+C$$
$$P(x+y)=(Ax^2+Bx+C)+(Ay^2+By+C)-C+2Axy=P(x)+P(y)-P(0)+2Axy$$
$$P(s) = \lim_{n\to \infty}P_n(s-\frac{1}{q^n}) = \lim_{n\to \infty} P_n(s) + \lim_{n\to \infty} P_n(\frac{1}{q^n}) -P(0) - 2s\frac{1}{q^n}$$
Pick $P_n$ to vanish the rationals $\frac{1}{q^n}$
$$P(s) = \lim_{n\to \infty}P_n(s-\frac{1}{q^n}) = \lim_{n\to \infty} P_n(s) - \lim_{n\to \infty}P_n(0) - \lim_{n\to \infty} 2s\frac{1}{q^n}$$
$$0 = \lim_{n\to \infty}P_n(s-\frac{1}{q^n}) = - \lim_{n\to \infty}P_n(0) - \lim_{n\to \infty} 2s\frac{1}{q^n}$$
$$\lim_{n\to \infty}P_n(0) = \lim_{n\to \infty} 2s\frac{1}{q^n}$$

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