update

Author: dirichy

AlgebraicGeometry/1/solution/AlgebraicGeometry1.tex

@@ -21,6 +21,7 @@
 \begin{problem}
   $P$ is an ideal of a unitary commutative ring $A$, then $P$ is prime ideal of $A\iff A/P$ is integral domain.
 \end{problem}
+\(\)
 \begin{solution}
   $\Rightarrow $:
 
@@ -85,7 +86,7 @@
 \begin{solution}
   \begin{lemma}
     \label{lem:2}
-    $K$ is a infinite field, $f\in K[x_1,x_2,\cdots x_n]\setminus \{0\}$, then exists $p\in \A_K^n,f(p)\neq 0$.
+    $K$ is an infinite field, $f\in K[x_1,x_2,\cdots x_n]\setminus \{0\}$, then exists $p\in \A_K^n,f(p)\neq 0$.
   \end{lemma}
   \begin{proof}
     Use MI. When $n=0,K[x_1,x_2,\cdots x_n]=K$, so it's obvious. Assume for $n=k$ it's right, when goes to $k+1$:
@@ -102,5 +103,11 @@
   \(\phi\)
   So there is not a pair of points can be seperated by two disjoint open set.
 \end{solution}
+\begin{figure}
+  \centering
+  \includegraphics{~/Pictures/2-3.jpg}%![aaa](~/Pictures/2-3.jpg)
+
+\end{figure}
 
 \end{document}
+

AlgebraicGeometry/10/solution/AlgebraicGeometry10.tex

@@ -9,65 +9,66 @@
 \large
 \setlength{\baselineskip}{1.2em}
 \ifpreface
-    \input{../../../global/preface}
-\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+  \input{../../../global/preface}
+  \newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
 \else
-\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
-\maketitle
+  \newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+  \maketitle
 \fi
 %from_here_to_type
 \begin{problem}
-  Assume \(V \subset \mathbbm{A}_k^{n}\) is irreducible, and \(p \in V\). 
-  Let \(m_p := \{f \in k[V] : f(p) = 0\}, M_p := \{f \in k[x_1,\cdots,x_n]: f(p) = 0\}\). 
-  Prove that \(m_p \cong M_p/\ide(V)\). 
+  Assume \(V \subset \mathbbm{A}_k^{n}\) is irreducible, and \(p \in V\).
+  Let \(m_p := \{f \in k[V] : f(p) = 0\}, M_p := \{f \in k[x_1,\cdots,x_n]: f(p) = 0\}\).
+  Prove that \(m_p \cong M_p/\ide(V)\).
 \end{problem}
-
 \begin{solution}
   Consider \( \theta : M_p \to m_p ,\theta(f) := f + \ide(V)\).
-  Since \( p \in V\) we get \(\theta\) is well-defined. 
-  And easily we get \(\theta\) is homomorphism. Now consider \(\ker \theta\). 
-  Obviously \(\ide(V) \subset \ker \theta\), now we prove \(\ker \theta \subset \ide(V)\). 
-  Assume \( f \in \ker\theta\), to prove \(f \in \ide(V)\). 
-  Since \(\theta(f)=\ide(V)\), we get \(f+\ide(V)=\ide(V)\), so \(f \in \ide(V)\). 
-  So \(\ker \theta = \ide(V)\). And easily \(\theta\) is surjective, so we get 
-  \(m_p=M_p/\ker \theta=M_p / \ide(V)\). 
+  Since \( p \in V\) we get \(\theta\) is well-defined.
+  And easily we get \(\theta\) is homomorphism. Now consider \(\ker \theta\).
+  Obviously \(\ide(V) \subset \ker \theta\), now we prove \(\ker \theta \subset \ide(V)\).
+  Assume \( f \in \ker\theta\), to prove \(f \in \ide(V)\).
+  Since \(\theta(f)=\ide(V)\), we get \(f+\ide(V)=\ide(V)\), so \(f \in \ide(V)\).
+  So \(\ker \theta = \ide(V)\). And easily \(\theta\) is surjective, so we get
+  \(m_p=M_p/\ker \theta=M_p / \ide(V)\).
 \end{solution}
 \begin{problem}
-  Prove that \(M_p/M_p^2+\ide(V) \cong m_p/m_p^2\). 
+  Prove that \(M_p/M_p^2+\ide(V) \cong m_p/m_p^2\).
 \end{problem}
-
 \begin{solution}
-  Consider \(\theta : M_p \to m_p/m_p^2, f \mapsto f|_V+m_p^2\). Easily \(\theta\) is homomorphism and surjective. 
-  Now we prove \(\ker \theta = M_p^2+\ide(V)\). 
+  Consider \(\theta : M_p \to m_p/m_p^2, f \mapsto f|_V+m_p^2\). Easily \(\theta\) is homomorphism and surjective.
+  Now we prove \(\ker \theta = M_p^2+\ide(V)\).
 
   On one hand, assume \(f \in \ker \theta\), i.e., \(f|_V \in m_p^2\).
-  Then \(\exists g_1,\cdots,g_n,h_1,\cdots,h_n \in m_p\) such that \(f|_V=\sum_{k=1}^{n} g_k h_k\). 
-  Assume \(g_k=G_k|_V,h_k=H_k|_V\). Then \(G_k,H_k \in M_p\). 
-  Consider \(f-\sum_{k=1}^{n} G_k H_k=:h \in k[x_1,\cdots,x_n]\), easily to know \(h(x)=0,\forall x \in V\). 
-  So \(h \in \ide(V)\), thus \(f \in M_p^2+\ide(V)\). 
+  Then \(\exists g_1,\cdots,g_n,h_1,\cdots,h_n \in m_p\) such that \(f|_V=\sum_{k=1}^{n} g_k h_k\).
+  Assume \(g_k=G_k|_V,h_k=H_k|_V\). Then \(G_k,H_k \in M_p\).
+  Consider \(f-\sum_{k=1}^{n} G_k H_k=:h \in k[x_1,\cdots,x_n]\), easily to know \(h(x)=0,\forall x \in V\).
+  So \(h \in \ide(V)\), thus \(f \in M_p^2+\ide(V)\).
 
-  On the other hand, assume \(f \in M_p^2+\ide(V)\), to prove \(\theta f =0\). 
-  Assume \(f=\sum_{k=1}^{n} G_k H_k+h\), where \(G_k,H_k \in M_p\) and \(h \in \ide(V)\). 
-  Then \(\theta f = \sum_{k=1}^{n} g_k h_k +m_p^2\), where \(g_k=G_k|_V,h_k=H_k|_V\). 
-  So we get \(\theta f = m_p^2=0\). 
+  On the other hand, assume \(f \in M_p^2+\ide(V)\), to prove \(\theta f =0\).
+  Assume \(f=\sum_{k=1}^{n} G_k H_k+h\), where \(G_k,H_k \in M_p\) and \(h \in \ide(V)\).
+  Then \(\theta f = \sum_{k=1}^{n} g_k h_k +m_p^2\), where \(g_k=G_k|_V,h_k=H_k|_V\).
+  So we get \(\theta f = m_p^2=0\).
 
-  Finally we get \(m_p/m_p^2 = M_p/ \ker \theta = M_p / M_p^2+\ide(V)\). 
+  Finally we get \(m_p/m_p^2 = M_p/ \ker \theta = M_p / M_p^2+\ide(V)\).
 \end{solution}
 \begin{problem}
-  Assume \(V \subset \mathbbm{A}_k^n\) is irreducible, and \(p \in V,f \in k[V], f(p)\neq 0\). 
-  Consider \(V_f:= \{x \in V: f(x)\neq 0 \}\). Let \(\theta: V_f \to \mathbbm{A}_k^{n+1},x \mapsto (x,\frac{1}{f(x)})\). 
-  Let \(U = \theta(V_f)\), prove that \(T_p V \cong T_{\theta(p)} U\). 
+  Assume \(V \subset \mathbbm{A}_k^n\) is irreducible, and \(p \in V,f \in k[V], f(p)\neq 0\).
+  Consider \(V_f:= \{x \in V: f(x)\neq 0 \}\). Let \(\theta: V_f \to \mathbbm{A}_k^{n+1},x \mapsto (x,\frac{1}{f(x)})\).
+  Let \(U = \theta(V_f)\), prove that \(T_p V \cong T_{\theta(p)} U\).
 \end{problem}
 
 \begin{solution}
-  Write \(k[\mathbbm{A}_k^n]=k[x_1,\cdots , x_n ,y]\). Assume \(V = \van(I)= \van(f_1,\cdots, f_m)\), where \(f_i \in k[x_1,\cdots,x_n ]\). 
-  Then \(U = \van(f_1,\cdots,f_n, yf-1)\). Now consider \(\tau: T_p(V) \to \mathbbm{A}_k^{n+1}, \tau(x):=(x,\frac{1}{f(p)}-\frac{f_p^{(1)}(x)}{f^2(p)})\). 
-  Now we prove \(\tau(T_p V)=T_{\theta(p)} U\). 
-  Only need to prove \((yf-1)_{\theta(p)}^{(1)}(\tau(x))=0\). 
-  i.e., \(\frac{f_p^{(1)}(x)}{f(p)}+f(p)(y-\frac{1}{f(p)})=0\), where \(y = \frac{1}{f(p)}-\frac{f_p^{(1)}}{f^2(p)}\). 
-  Substitute \(y\) into the equation, we get is't obvious. 
+  Write \(k[\mathbbm{A}_k^n]=k[x_1,\cdots , x_n ,y]\). Assume \(V = \van(I)= \van(f_1,\cdots, f_m)\), where \(f_i \in k[x_1,\cdots,x_n ]\).
+  Then \(U = \van(f_1,\cdots,f_n, yf-1)\). Now consider \(\tau: T_p(V) \to \mathbbm{A}_k^{n+1}, \tau(x):=(x,\frac{1}{f(p)}-\frac{f_p^{(1)}(x)}{f^2(p)})\).
+  Now we prove \(\tau(T_p V)=T_{\theta(p)} U\).
+  Only need to prove \((yf-1)_{\theta(p)}^{(1)}(\tau(x))=0\).
+  i.e., \(\frac{f_p^{(1)}(x)}{f(p)}+f(p)(y-\frac{1}{f(p)})=0\), where \(y = \frac{1}{f(p)}-\frac{f_p^{(1)}}{f^2(p)}\).
+  Substitute \(y\) into the equation, we get is't obvious.
 
-  Obviously \(\tau\) is injective, so it's isomorphic. 
+  Obviously \(\tau\) is injective, so it's isomorphic.
 \end{solution}
-
+\begin{figure}
+  \centering
+  \includegraphics{~/Pictures/pic2/365/IMG_3135.JPG}
+\end{figure}
 \end{document}

MarkovProcess/14/solution/MarkovProcess14.pdf

@@ -28,7 +28,7 @@
   Easily \(N(t)+1 \geq T \geq N(t)\).
   So \(\int_{0}^{T}R(t) \d t \leq \sum_{i=1}^{N(T)+1} \int_{S_{i-1}}^{S_i}(S_i-t) \d t=\frac{1}{2}\sum_{i=1}^{N(t)+1} (S_i-S_{i-1})^2=\frac{1}{2} \sum_{i=1}^{N(T)+1} \xi_i^2\).
   For the same reason, we get that \(\int_{0}^{T}R(t) \d t \geq \frac{1}{2} \sum_{i=1}^{N(T)} \xi_i^2\).
-
+  \(\int_{0}^T R(t)\d t \geq \frac{1}{2}\sum_{i=1}^{N} \xi_i^2\)
   Easy to know that \(\lim_{T \to \infty} \frac{1}{T} \sum_{i=1}^{N(T)} \xi_i^2=\lim_{T \to \infty} \frac{1}{T} \sum_{i=1}^{N(T)+1} \xi_i^2=\frac{\mathbb{E}(X^2)}{\mathbb{E}(X)^2}\).
   So finally we get that
   \[
@@ -76,4 +76,27 @@
       So \(T=9.76\) can make the fee get minimum.
   \end{enumerate}
 \end{solution}
+\begin{problem}\label{pro:4 }
+  A kind of product is qualified with probability \(p(0 < p < 1)\).
+  We sample these product by the following way: we check all the product at first until there appears
+  \(k\) qualified product sequently. Then we check the rest of product by probability \(\alpha(0 < \alpha <1)\)
+  until there appears one unqualified product, then one circle ends. Next we restart another checking
+  circle. Please find out the proportion of checked product after a long time.
+\end{problem}
+\begin{solution}
+  For sake of convenience, we call the \(k\) qualified products appearing sequently as \(k\) qualified sequence.
+  Assume the proportion of checked product after a long time is \(\beta\).
+  Let \(N_k\) be the amount of product when the first \(k\) qualified sequence ends. \(M_k=\mathbb{E}(N_k)\). \(G_k\) is the
+  event that the next one is qualified after the first \(k-1\) qualified sequence ends. Obviously, \(\mathbb{E}(N_k-N_{k-1} \mid G_k)=1\).
+  And \(\mathbb{E}(N_k-N_{k-1} \mid G_k)=\mathbb{E}(N_k) + 1\).
+  Therefore, \(\mathbb{E}(N_k-N_{k-1})=p + (1-p)(\mathbb{E}(N_k) + 1)\). So \(M_{k-} M_{k-1}=p + (1-p)( 1 + M_k)\).
+  Then \(pM_k=M_{k-1} + 1\). Thus, \(M_k=\frac{\frac{1}{p^k}-1}{1-p}\).
+  Let \(A=\{\)The amount of product checked in one circle\(\}\), \(B=\{\)The amount of product in one circle\(\}\).
+  So finding one unqualified product need average checking time \(\frac{1}{1-p}\) according to geometric distribution.
+  Then we averagely need \(\frac{1}{\alpha(1-p)}\) product to find out the unqualified one.
+  Then \(\mathbb{E}(A)=M_k + \frac{1}{1-p}, \mathbb{E}(B)=M_k + \frac{1}{\alpha(1-p)}\).
+  So \[
+    \beta = \frac{\mathbb{E}(A)}{\mathbb{E}(B)}=\frac{\frac{\frac{1}{p^k}-1}{1-p} + \frac{1}{1-p}}{\frac{\frac{1}{p^k}-1}{1-p} + \frac{1}{\alpha(1-p)}}=\frac{\alpha}{\alpha + p^k-\alpha p^k}
+  \]
+\end{solution}
 \end{document}

MarkovProcess/14/solution/MarkovProcess14.tex

@@ -1,3 +1,11 @@
+local function isprime(n)
+	for i = 2, n - 1 do
+		if n % i == 0 then
+			return false
+		end
+	end
+	return true
+end
 local function legendre(p, q)
 	if p % q == 0 then
 		return 0
@@ -16,9 +24,12 @@ local function legendre(p, q)
 	return -1
 end
 local function defactor(n)
-	local result = {}
+	local result = { isprime = true }
 	for i = 2, n, 1 do
 		if n % i == 0 then
+			if i < n then
+				result.isprime = false
+			end
 			result[i] = 0
 			while n % i == 0 do
 				result[i] = result[i] + 1
@@ -42,17 +53,18 @@ local function termnext(term)
 			return { { type = "number", value = (term.lower * term.lower - 1) % 16 == 0 and 1 or -1 } }
 		end
 		local factors = defactor(term.upper)
-		if #factors == 1 then
+		if factors.isprime then
 			state = {
 				{ type = "number", value = (term.upper % 4 == 1 or term.lower % 4 == 1) and 1 or -1 },
 				{ type = "legendre", upper = term.lower, lower = term.upper },
 			}
 			return state
 		end
 		for k, v in pairs(factors) do
-			print(k .. "," .. v)
-			if v % 2 == 1 then
-				state[#state + 1] = { type = legendre, upper = k, lower = term.lower }
+			if type(v) == "number" then
+				if v % 2 == 1 then
+					state[#state + 1] = { type = "legendre", upper = k, lower = term.lower }
+				end
 			end
 		end
 		return state
@@ -67,16 +79,69 @@ local function term2str(term)
 		str = [[\legendre{]] .. term.upper .. [[}{]] .. term.lower .. [[}]]
 	end
 	if term.type == "number" then
-		print(term.value)
+		-- print(term.value)
 		str = tostring(term.value)
 	end
 	return str
 end
 local function state2str(state)
 	local str = ""
+	local number = 1
 	for k, v in ipairs(state) do
-		str = str .. term2str(v)
+		if v.type == "legendre" then
+			str = str .. term2str(v)
+		end
+		if v.type == "number" then
+			number = number * v.value
+		end
+	end
+	if str == "" then
+		str = tostring(number)
+	elseif number == -1 then
+		str = "-" .. str
 	end
 	return str
 end
-print(state2str(termnext({ type = "legendre", upper = 6, lower = 13 })))
+local function statenext(oldstate)
+	local newstate = { { type = "number", value = 1 } }
+	local number = 1
+	for _, v in ipairs(oldstate) do
+		local tempstate = termnext(v)
+		for _, y in ipairs(tempstate) do
+			if y.type == "legendre" then
+				newstate[#newstate + 1] = y
+			elseif y.type == "number" then
+				number = number * y.value
+			end
+		end
+	end
+	newstate[1]["value"] = number
+	return newstate
+end
+local function legendre_with_calculation(p, q)
+	if p % q == 0 then
+		return 0
+	end
+	if q == 0 then
+		error("the second argument of legendre should be non-zero!")
+	end
+	if q < 0 then
+		q = -q
+	end
+	local state = { { type = "legendre", upper = p, lower = q } }
+	local calculation = "\\[\n" .. state2str(state)
+	while #state > 1 or state[1].type == "legendre" do
+		state = statenext(state)
+		calculation = calculation .. "\\\\\n=" .. state2str(state)
+	end
+	calculation = calculation .. "\n\\]"
+	return calculation
+end
+for i = 1, 100, 1 do
+	local p = math.random(1, 1000)
+	local q = math.random(1001, 10000)
+	while not isprime(q) do
+		q = math.random(1001, 10000)
+	end
+	print(legendre_with_calculation(p, q))
+end

NumberTheory/8/solution/legendre.lua

@@ -0,0 +1 @@
+3