[腦力大挑戰] Mathematical analysis BB班

之後有D好詭異ge 野:
例如 Q^2 同 Q 一樣多
但係 number of sequence of rational numbers 又多過野過 Q (仲要同 R 一樣多)

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出現了

Admin有睇連登

高質post

我識佢

有冇聽過有人Ug讀applied但pg做pure topic

ug 讀 physics pg做pure math都有啦

Theory of ODE 其實係讀啲咩 同普通solve ode嘅course 係咪好大分別

差好遠

theory of ode 係讀 existence uniqueness stability 的野
唔會直接要你去計個solution

stability of fixed point?當年year 2 ODE都要讀

睇完呢個post 搵左本baby rudin 讀

睇緊頭兩頁已經有野唔明，希望大家教下小弟

睇得明個proof 但係想知(3)果part 其實係點諗出黎？謝謝！

佢係用緊D root finding algorithm, 某程度上係出cheat

想 construct 一個 satisfy 到嗰兩個條件嘅 q，而唯一知道嘅係 p^2-2>0（或者 p^2-2<0）。所以要 set：q=p-(p^2-2)/f(p) (where f(p)>0)，另外：
q^2-2=((pf(p)-(p^2-2))^2-2f(p)^2)/f(p)^2
當然最重要嘅係正負號，所以只需要望：
((pf(p)-(p^2-2))^2-2f(p)^2)
=(p^2-2)(f(p))^2-2pf(p)(p^2-2)+(p^2-2)^2
當然我哋所知嘅 function 唔多，除咗 polynomial 之外都冇乜嘢可以揀，如果 f 嘅 degree 係 n 嘅話上面三個 term degree 分別係 2n+2、n+3、4，最理想嘅情況當然係三個 degree 一樣（或者最大嗰兩個 degree 一樣） cancel 走 degree 最大嘅嘢，所以 set n=1，f(x)=ax+b：
(p^2-2)(ap+b)^2-2p(ap+b)(p^2-2)+(p-2)^2
如果 expand 開嘅話 p^4 嘅 coefficient 就係 a^2-2a+1，想令佢變 0 就唯有 set a=1：
(p^2-2)(p+b)^2-2p(p+b)(p^2-2)+(p^2-2)^2
=p^4+2bp^3+(b^2-2)p^2-4bp-2b^2-2p^4-2bp^3+4p^2+4bp+p^4-4p^2+4
=(b^2-2)p^2+(4-2b^2)
=(b^2-2)(p^2-2)
記唔記得以上呢個 expression 係咩？冇錯就係我哋想佢個 sign 跟住 p^2-2 走，即係話要 set b 令到 b^2-2>0，b=1 就唔得，噉唯有 set b=2。
另外要記得 check f(p)=ap+b=p+2>0 for positive p。
所以我哋要 set q=p-(p^2-2)/(p+2)。
如果唔係對住個答案諗我應該諗唔到

如果你想知點解會砌到嗰舊野出嚟，你可以咁諗嘅：
We would like to find q = p - t s.t. if p \in A then t < 0 and q^2 - 2 < 0, and if p \in B then t > 0 and q^2 - 2 > 0. Note that the defining relations for A and B depend on the sign of p^2 - 2, so we could try t = (p^2 - 2)r and see what r satisfies our needs. Consider q^2 -2 = p^2 -2(p^2 - 2)pr + (p^2 - 2)^2r^2 -2 = (p^2 - 2)((p^2 - 2)r^2 - 2pr + 1). Here we notice that if we could pick r > 0 s.t. the later factor is always positive, then we are done. Also note that the roots of the polynomial f(r) :=(p^2 - 2)r^2 - 2pr + 1 are 1/(p - \sqrt{2}) and 1/(p + \sqrt{2}). There are two cases to consider:
Case 1: p^2 - 2 < 0. Then f(r) > 0 when 1/(p - \sqrt{2}) < r < 1/(p + \sqrt{2}).
Case 2: p^2 - 2 > 0. Then f(r) > 0 when r > 1/(p - \sqrt{2}) or r < 1/(p + \sqrt{2}).
Hence if we pick r = 1/(p + 2) it will serve our purposes.

漏左我地要埋q > 0
不過其實要證A冇max同B冇min唔駛咁做

好詳盡，唔該晒

我唔太明點解rudin要咁做

可能rudin覺得呢個function最易理解
反正唔係唯一一個啱用嘅function
畀我會用continuous fraction嚟玩

不過諗唔到佢點解諗到其實好正常
你諗到就做咗寫書嗰個啦