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H: Integer Part of sequence convergence I was trying to solve the following exercise. If $(a_n) \in \mathbb{R}$ and $(a_n)\rightarrow {1}/{2}$ show that $[a_n] \rightarrow 0$ , where $[~]$ the integer part. I was trying to solve it using the ε-definition. $\forall ε>0 ~ \exists n_0 \in \mathbb{N} : \|a_n - 1/2\|< ε ~,~~ \forall n \geq n_0$. I want $\forall ε>0 ~ \exists n_1 \in \mathbb{N} : \|[a_n] \|< ε ~,~~ \forall n \geq n_1$. Also I know that $[a_n]\leq a_n < [a_n]+1 $ I tried some inequalities with the above absolute values and the integer part definition but I'm really stuck... Any help appreciated. AI: Since the sequence converges to 1/2, you have that there exists an $n_0$ such that, for all $n>n_0$, $\|a_n-1/2\|<1/4$, say. So, for $n>n_0$, $\frac{1}{4} < a_n < \frac{3}{4}$, and so, for those $n$, $[a_n]=0$. You can probably finish the proof from there.
H: Direct product of Sylow subgroups Proposition II.7.5 of Hungerford's Algebra goes Proposition 7.5. A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups. Let $G = (\mathbb{Z}_6, +)$. $G$ is abelian, so it is nilpotent and thus by the proposition is the direct product of its Sylow subgroups. The Sylow subgroups of $G$ are $H = (\mathbb{Z}_3, +)$ and $K = (\mathbb{Z}_2, +)$, so the proposition says $G = H \times K$. In particular, $0 \in H$ and $0 \in K$, so $(0,0) \in H \times K = G$... but that's not true, since $G = \{0,1,2,3,4,5\}$. Where is/are my misunderstanding(s)? AI: When we say a group "is a direct product," it suffices to simply interpret this as saying the group is isomorphic to said direct product. However it is important to understand the difference between thinking about direct products internally as opposed to externally. There is a psychological difference between an internal and an external direct product. In an external direct product $H\times K$, the elements look like tuples $(h,k)$ with $h\in H$ and $k\in K$. Thus one may say that anything that is not a tuple of this form is not in $H\times K$. An internal direct product is structurally a direct product, but the elements may or may not literally be in the form of tuples. More on this now. Let's look at how the internal direct product is motivated. We see that there is a "copy" of $H$ sitting inside the direct product $H\times K$ as $H\times 1=\{(h,1):h\in H\}$. Similarly there is a "copy" of $K$ sitting inside as $1\times K$. (If the groups are not written multiplicatively then the numeral "$1$" might be misleading.) Everything in $H\times K$ can be written as a product of something in $H\times1$ times something in $1\times K$ (in particular $(h,k)=(h,1)(1,k)$). The two subgroups $H\times1$ and $1\times K$ intersect trivially as $(1,1)$. And everything in $H\times1$ commutes with everything in $K\times1$. And thus we have our definition. We say $G$ is a direct product of subgroups $H$ and $K$ if: $G=HK$, i.e. every element $g\in G$ can be written as $g=hk$ with $h\in H$ and $k\in K$; $H\cap K=1$, i.e. $H$ and $K$ intersect trivially; $[H,K]=1$, i.e. everything in $H$ commutes with everything in $K$. Fact: If $H$ and $K$ are two groups, then the external direct product $H\times K$ is an internal direct product of its subgroups $H\times1$ and $1\times K$ (which are canonincal copies of $H$ and $K$ sitting inside $H\times K$). Conversely, if $G$ is an internal direct product of subgroups $H$ and $K$, then there is an isomorphism $H\times K\cong G$ given by $(h,k)\mapsto hk$ between the external direct product $H\times K$ and the original group $G$ (which, remember, is an internal direct product of $H$ and $K$). This tells us that internal and external direct products are actually the same (the are isomorphic), but the elements inside might look different. In my opinion the best way to think about the difference is as follows: a direct product is external if it was constructed that way to begin with, and it is internal if it was only discovered that way after analyzing it. If I were to, say, merely relabel the elements of an external direct product while keeping the whole group structure the exact same, the concept of internal direct product allows us to still recognize the group as a direct product, even if the elements no longer look like tuples. In particular, $\Bbb Z/6\Bbb Z$ is an internal direct product of its subgroups $2\Bbb Z/6\Bbb Z$ and $3\Bbb Z/6\Bbb Z$ (which are isomorphic to $\Bbb Z/3\Bbb Z$ and $\Bbb Z/2\Bbb Z$ respectively). More generally we have an isomorphism $$\frac{\Bbb Z}{p_1^{e_1}\cdots p_r^{e_r}\Bbb Z}\cong\frac{\Bbb Z}{p_1^{e_1}\Bbb Z}\times\cdots\times\frac{\Bbb Z}{p_1^{e_1}\Bbb Z}.$$ This is the content of the abstract version of Sun-Ze (better known as the Chinese remainder theorem). It is recommended that students become familiar with the structure theory of abelian groups and cyclic groups in particular as well as the basic construction of direct products before dwelling on more advanced topics like subgroup series, Sylow subgroups and nilpotence.
H: Sum of the Series , Calculus Homework I put this into WolfRam and got e^(3/5), but I am trying to figure out how to arrive to that answer? AI: We know, $$e^x=\sum_{0\le n<\infty}\frac{x^n}{n!}$$ Here $$\frac{3^n}{5^n\cdot n!}=\frac{\left(\frac35\right)^n}{n!}$$
H: Looking into mappings I'm interested in looking at mappings of functions. For example, how would I come up with a function $f:x\in[0,\infty)\|\longmapsto\ (-\infty, 0]$ where $f(x)=x^2.$ Basically I want to glue every point to the right of this function to the left. AI: HINT: Construct a function mapping positive elements to their additive inverse. This is impossible with $f(x) = x^2$.
H: Is there a possibility for two different primes to have any of its powers to be the same? Say there are two primes $P_1$ and $P_2$ where $P_1 \neq P_2$. Is there a possibility for some $m$, $n$ ($m \neq 0, n \neq 0$) such that $P_1^m = P_2^n$. AI: No. Because then $P_1$ divides both sides. And if a prime divides a product of terms, it must divide at least one of the terms. In particular, it must divide $P_2$, contradicting that $P_2$ is distinct.
H: Example of two series with certain properties? Find 2 series $\sum a_k$ and $\sum b_k$ such that $\sum b_k$ converges conditionally, $\dfrac{a_k}{b_k} \rightarrow 1$ as $k \rightarrow \infty$, and $\sum a_k$ diverges. Can someone give me a hint with this? Thanks. AI: Let$$b_n=\dfrac{(-1)^n \sqrt{2}}{\sqrt{n}}, \;\; a_n=(-1)^n u_n,\;\;\; n\geqslant {2},$$ where $$u_n=\begin{cases}\dfrac{1}{\sqrt{k+1}-1},&n=2k, \\ \dfrac{1}{\sqrt{k+1}+1},&n=2k+1, \;\;k\in\mathbb{N}. \end{cases}$$ Partial sums for $\sum{a_n}$ $$S_{2k+1}=\sum\limits_{i=1}^{2k+1}{\left(\dfrac{1}{\sqrt{i+1}-1}-\dfrac{1}{\sqrt{i+1}+1} \right)}=\sum\limits_{i=1}^{2k+1}\dfrac{2}{i}\;\underset{k\to\infty}{\;\longrightarrow{\infty}}.$$
H: Given $A$ and $B$ positive-definite matrices and $Q$ unitary matrix, prove that if $A = BQ$, then $A=B$. Given $A$ and $B$ positive-definite matrices and $Q$ unitary matrix, prove that if $A = BQ$, then $A=B$. $Q$ is unitary, so $QQ^=I$ If $A$ and $B$ are positive-definite, than $A=A^$ and $B=B^$. $A^=(BQ)^=Q^B^$ $A^2=AA=AA^=(BQ)(Q^B^)=B(QQ^)B^=BIB^*=B^2$ $A^2=B^2$. I don't know how to use the fact that $A$ and $B$ are positive-definite to finish the proof. AI: Hint: By considering $x^\ast Ax$ for each eigenvector $x$ of $Q$, show that $Q=I$.
H: Prove if $\{t_n\}_{n\in\mathbb{N}}\to t$ and $t_n\geq 0\forall n\in\mathbb{N}$, then $t\geq 0$ I've been working on some sequence practice problems in Steven Lay's Introduction to Analysis With an Introduction to Proof for my introductory real analysis course, as we are starting our unit on sequences next week. I encountered this problem yesterday and wasn't sure how to start it. Any hints would be welcome- thanks. Prove if $\{t_n\}_{n\in\mathbb{N}}\to t$ and $t_n\geq 0\forall n\in\mathbb{N}$, then $t\geq 0$. AI: $t<0\Rightarrow t\in (-\infty,0)\Rightarrow \exists \delta>0\ni (t-\delta,t+\delta)\subseteq (-\infty,0)\Rightarrow (t-\delta,t+\delta)\cap\{t_n:t_n\ge 0\forall n\in\mathbb{N}\}=\emptyset$ which means $t$ is not limit point of your sequence!
H: The order of equalities First note that I am not a mathematician. I do use it for my studies, but I am not reading anything remotely complicated in regards to maths. That said, here is my question: Today I found myself wanting to write a probability, first in terms of a fraction and then in terms of a decimal value. I wrote down this: $$ P \geq \left( \frac{1}{2} \right)^3 = 0.125 $$ That is to say, the probability is at least a half to the power of three which equals $0.125$, but I am afraid that I am in fact writing the probability is at least a half to the power of three, which means that the probability equals $0.125$ which would be incorrect. Does the equals sign bind stronger than greater-than-equals, meaning the first interpretation is correct, or do the have the same strength so that the second interpretation is the correct one? If the latter is true, is there a different way to express the original meaning without using biimplication? I know I can write something like $$ P \geq \left( \frac{1}{2} \right)^3 \Leftrightarrow P \geq 0.125 $$ but I like the other format better. If it had to do with multiplication instead of equalities I could use paranthesis to express the order of operations, but I do not think I have ever seen this used for clarifying the "order of equalitites"? Thank you. AI: You may be overthinking this. If your probability $P \geq \left( \frac{1}{2} \right)^3$, then $P \geq 0.125$, and there is nothing mathematically wrong with saying that $P \geq \left( \frac{1}{2} \right)^3 = 0.125$. You can leave that first statement as is with no risk of having someone pick on it. Anyone understanding what you are doing will understand your inference. People write similar phrases in published papers all the time.
H: prove that $T''$ is not injective (difficult computation) Let $T:c_0 \to c_0$ defined by $T(\{s_j\}_j)=\{s_{j+1}-s_j\}_j$. Prove that $T''$ is not injective. I tried even knowing that $(c_0)' \sim l^1$,in fact if $F\in (c_0)'$ then there exist $s=(s_j)\in l^1$ such that $F=F_s$ and $F_s(t_j)=\sum s_jt_j$ , and similar for $(l^1)' \sim l^{\infty}$. It's easy to prove that $T''$ is injective if and only iff the image of $T'$ is dense on $X'$. Maybe this is easy to prove. AI: Hint: for $s \in \ell^\infty$, $(T'' s)_j = s_{j+1} - s_j$ as well. So what $s$ would make those all $0$?
H: Why is the Uniform Boundedness Theorem not true for all normed vector spaces? A tentative statement of the Uniform Boundedness Theorem for any normed vector space would be Let $(T_n)$ be a sequence of bounded linear operators $T_n:X\to Y$ such that $(\\|T_n x\\|)$ is bounded for every $x\in X$. Then the sequence of the norms $\\|T_n\\|$ is bounded, that is, there is a $c$ such that $\\|T_n\\|\leq c$. Let us assume that $\\|T_n\\|\not \leq c$ for some $c\in \Bbb{R}$. This would directly imply that for some $x\in X$, the sequence $(\\|T_nx\\|)$ would also not be bounded. QED. How is the argument wrong? Thanks in advance! AI: For example, consider the normed vector space $V$ of sequences $s = (s_1,s_2,\ldots)$ that have only finitely many nonzero elements, with $\\|s\\| = \max_n \|s_n\|$. Define $T_n : V \to \mathbb R$ by $T_n s = n s_n$. For every $s \in V$, all but finitely many $T_n s$ are $0$, so $\{\\|T_n s\\|: n \in {\mathbb N}\}$ is a finite set and thus bounded. But $\\|T_n\\| = n$, so $\{\\|T_n\\|: n \in {\mathbb N}\} = \mathbb N$ is unbounded.
H: Proof of natural log identities I need to prove a few of the following identities from a real analysis perspective- this means I do not have access the $\ln e^2 = 2$ type definition of the log function. I am developing the log function from the definition $log x = \int_1^x \frac1t \mathrm dt$ for $0 < x$. I need to prove that: $\ln(ab) = \ln(a) + \ln(b)$ first. I imagine the proof for that will apply itself pretty directly to $\ln(x^n) = n\ln(x)$. I've been given the hint that the idea here is to define $f(x) = \ln(ax)$ and show that $f'(x) = \ln(x)'$, implying $f(x) = L(ax) = L(x) + k$, where I can show that $k = L(a)$. Any help with starting this? AI: Let $f(x) = \ln(x)$ and let $g(x) = \ln(ax)$, where $a$ is some constant. We claim that $f$ and $g$ only differ by a constant. To see this, it suffices to prove that they have the same derivative. Indeed, by the fundamental theorem of calculus and chain rule, we have that: \begin{align} g'(x) &= \frac{d}{dx} \ln(ax) = \frac{d}{dx} \int_1^{ax} \frac{dt}{t} = \frac{1}{ax} \cdot a = \frac{1}{x} = \frac{d}{dx} \int_1^{x} \frac{dt}{t} = \frac{d}{dx} \ln(x) = f'(x) \end{align} Hence, we have that $g(x) = f(x) + k$ for some constant $k$. Now substitute $x=1$. This yields: \begin{align} g(1) &= f(1) + k \\ \ln(a \cdot 1) &= \ln(1) + k \\ \int_1^{a \cdot 1}\frac{dt}{t} &= \int_1^1\frac{dt}{t} + k \\ \int_1^{a}\frac{dt}{t} &= 0 + k \\ \ln(a) &= k \\ \end{align} So we have that $\ln(ax) = \ln(a) + \ln(x)$, as desired.
H: Calculating 6 decimal digits of $3^{\sqrt2}$ using a calculator. How can we calculate $3^{\sqrt2}$ to 6 decimal digits, using only a semi-basic calculator (Which has the square root too) and a pen and paper? I asked this question from my teacher and he gave me a hint: "Compute the binary digits of $\sqrt2$." So what i did was this: Suppose $A=\sqrt2$ can be written in binary as: $A=a_0/a_1a_2a_3...$ and $a_i\in 0,1$. We have: $2^kA=a_0a_1a_2...a_k/a_{k+1}a_{k+2}...$ so $a_k$ is the first digit of the binary expansion of $[2^kA]$ and thus we can calculate it using the calculator. But i don't know how to go further from here. AI: Basically you're approximating $\sqrt{2}$ by a sum $1/2^{n_1} + 1/2^{n_2} + \ldots + 1/2^{n_k}$. So $$ 3^\sqrt{2} \approx 3^{1/2^{n_1}} 3^{1/2^{n_2}} \ldots 3^{1/2^{n_k}}$$ and each $3^{1/2^{m}}$ can be found by starting with $3$ and hitting the square root button $m$ times.
H: Find the Bounding Rectangle of Rotated Rectangle I have rectangle with co-ordinates(x1,y1) and (x2,y2) and I have to rotate the rectangle an amount of θ about it centre using Rotation Matrix \| cosθ sinθ \| \| -sinθ cosθ \| I need to find the co-ordinates of bounding rectangle after rotation. Before rotation 0,0 \|"""""""""""""""""""""""""""""""""""""""""""\| \| \| \| x1,y1 \| \| \|"""""""""""""\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| """""""""""""" x2,y2 \| \| \| \| \| """"""""""""""""""""""""""""""""""""""""""" W,H After rotation 0,0 \|"""""""""""""""""""""""""""""""""""""""""""\| \| ?,? \| \| \|""""/\"""""\| \| \| \| / \ \| \| \| \| / \ \| \| \| \| / /\| \| \| \|/ / \| \| \| \|\ / \| \| \| \| \ / \| \| \| \| \ / \| \| \| """"""""""" ?,? \| \| \| \| \| """"""""""""""""""""""""""""""""""""""""""" W,H Is there any general equation for finding the co-ordinates of bounding rectangle?. Thanks.... Haris. AI: For each of the four original corners: $$ (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) $$ use the rotation matrix to obtain four new corners. Then to obtain the two ordered pairs that define your bounding rectangle, let one ordered pair have the minimum $x$ and $y$ values out of all the new corners and let the other ordered pair have the maximum $x$ and $y$ values.
H: transpose of the exponential operator Let $X$ be a Banach space and $T:X\to X$ be a continuous and linear operator. What is the transpose operator of $e^T?$ I would like to prove that $e^{T'}=(e^T)'$. At least that equality make perfect sense because they are defined in the same domain. But if $f\in X'$ $$ (e^T)'f=f(e^T)=f(\sum \frac{T^k}{k!})=\sum\frac{f(T^k)}{k!}$$ $$ (e^{T'})(f)=e^{T'f}=e^{f(T)}=\sum \frac{(f(T))^k}{k!} $$ I'm not sure if they are the same. If not I'm still looking for some possible expression for $(e^T)'$. Please help me! AI: For $A,B\in\mathcal{B}(X)$ we have $(AB)'=B'A'$. As the consequence $(T^k)'=(T')^k$. Therefore $$ e^{T'} =\sum\limits_{k=0}^\infty\frac{1}{k!}(T')^k =\sum\limits_{k=0}^\infty\frac{1}{k!}(T^k)' =\left(\sum\limits_{k=0}^\infty\frac{1}{k!}T^k\right)' =(e^T)' $$
H: Why is normalization of inequalities possible? I have seen, in many proofs for inequalities, the author does something called normalization. I believe this is only possible for homogeneous inequalities. I saw this in a proof of Nesbitt's inequality: $$\frac{a}{b+c} + \frac{b}{a +c} + \frac{c}{a+b} \geq \frac32$$ The above can be transformed as: $$\frac{a+b+c}{b+c} + \frac{a+b+c}{a+c} + \frac{a+b+c}{a+b} - 3$$ $$=(a+b+c)\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3$$ $$=\frac{1}{2}[(a+b) + (b+c) + (c+a)]\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3$$ Now, the author says, since the inequality is homogeneous, we may normalize $a+b+c = 1$ From the $AM-HM$ inequality: $$\frac{\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}}{3} \geq \frac{3}{2(a+b+c)}$$ $$\implies \frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b} \geq \frac92$$ Since $a+b+c = 1$, we may multiply the LHS with $\frac12(a+b+b+c+c+a)$, without changing its value $$\implies \frac{1}{2}[(a+b) + (b+c) + (c+a)]\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) \geq \frac92$$ Subtracting $3$ from both sides gives us the result. I don't understand the normalization procedure at all. I can't see why it is possible to do this. It also seems a bit too good to be true. AI: Say $a + b+ c = k$. Let $a = k\alpha$, $b = k\beta$, $c = k\gamma$. Express the inequality in terms of $\alpha, \beta, \gamma$. What happens to $k$?
H: How do I draw a diagram for a function space? If one considers a single function, then one can just draw its diagram as a Cartesian product. So it's relatively easy to contribute one's intuition to an argument. However, when it is a function space, I completely lose my intuition. And I am sure this way of studying is bad since I cannot confirm myself why such theorems have to be true. Ascoli's (classical) theorem is one example. Firstly, I learned this theorem when I studied Rudin-Analysis. There he attacks the theorem directly. And I had no idea why it has to be. Then, I have encountered this theorem again in Munkres-Topology. Here, he attacks the theorem via the uniform topology on ${\mathbb{R}^n}^X$. Still, I have no idea why it has to be. How do I literally draw a diagram for, or imagine a function space? AI: I am not sure it is possible to visualize function spaces in the way that you can visualize $\mathbb{R}^2$, but here is one attempt to see Arzela-Ascoli : Consider the usual $\epsilon-\delta$ picture of a limit (See this, for instance).Now, equicontinuity of a family $S = \{f_{\alpha}\}$ says that, for any such horizonal $\epsilon-$strip, there is a vertical $\delta$-strip such that, for any $x$ within that $\delta$-strip, the corresponding values $\{f_{\alpha}(x)\}$ are all in that horizontal $\epsilon$-strip. Suppose $f:[0,1] \to [0,1]$ is continuous. Visualize an open ball $B(f,r)$ around a function $f\in C[0,1]$ as a band of width $r/2$ on either side of the curve (ie. It would be bounded by the curves $x \mapsto f(x) + r/2$ and $x \mapsto f(x) - r/2$) (See this, for instance) Now suppose $S := \{f_{\alpha} :[0,1]\to [0,1]\}$ is equicontinuous. You want to know why it is compact. So start with infinitely many bands $B(f_{\alpha}, \epsilon_{\alpha})$. Now fix $x \in [0,1]$ and look at all the bands "above" it. Since the bands above it do not go beyond the largest $Y$-value (ie. 1), there are only finitely many bands that are needed to cover $\{f_{\alpha}(x)\}$. Each of these finitely many bands contribute an $\epsilon_{\alpha}$, of which you can take the minimum - call that $\epsilon_x$. Now there are finitely many horizontal strips of radius $\epsilon_x$ that together cover all the $Y-$values, $\{f_{\alpha}(x)\}$. Choose a vertical $\delta_x$-strip as in (1). And note that, for any $z \in (x-\delta_x,x+\delta_x)$, the corresponding $Y-$values $\{f_{\alpha}(z)\}$ are all in one of those finitely many horizontal $\epsilon_x$-strips. Now is where the compactness of the domain comes in. There are only finitely many such $\delta_x$ that are needed to cover all of $[0,1]$. Each $\delta_x$ needs only finitely many $B(f_{\alpha},\epsilon_{\alpha})$; which gives compactness. I have a pretty good picture in my mind right now, and I hope this explanation was able to translate that into words. Unfortunately, I am not sure how to represent this graphically. If someone can suggest a nice way to do this, then that would help greatly.
H: Books about manifolds? I would like to learn about manifolds. Please can someone recommend me a good book to learn about manifolds? AI: I like Introduction to Smooth Manifolds by John M. Lee. The wikipedia article on manifolds is also quite nice and contains a number of references.
H: Isomorphism of quotient ring of polynomial ring Are $F_3[x]/(x^2-2)$ and $F_3[x]/(x^2-2x-1)$ isomorphic? I know that $x^2-2$ and $x^2-2x-1$ are irreducible but how to determine if they are isomorphic or not? Here, $F_3$ means finite field of order 3. AI: One of the most important facts about finite fields is that all finite fields of the same size are isomorphic! Finding the isomorphism can be a little trickier. For a problem expressed like yours, the most direct approach would be something like finding a root of the polynomial $t^2 - 2$ in the field $\mathbf{F}_3[x] / (x^2 - 2x - 1)$. Aside: it sometimes helps to use different indeterminate variables to keep everything straight: e.g. let your two fields be $\mathbf{F}_3[x]/(x^2 - 2)$ and $\mathbf{F}_3[y]/(y^2 - 2y - 1)$.
H: 1-F(x) as F(x) goes to 1 I stumbled into the following statement and I am not sure how to prove it. The statement is : Given a distribution function $F(x)$ we can represent the survival function $1-F(x)$ as $-logF(x)$ for $F(x)\rightarrow 1$. More explicitly $1-F(x)\sim{-logF(x)}$ as $F(x)\rightarrow 1.$ Any references would be appreciated.Thank you. AI: We cannot. What you are referring to is probably the fact that $\frac{-\log u}{1-u}\to1$ when $u\to1$.
H: How to really understand the tensor algebra? If $V$ is a vector space over $F$, then we define $T^r_0(V)=V^{\otimes r}$, then we define the algebra of contravariant tensors to be $$T(V)=\bigoplus_{r=0}^\infty T^r_0(V)$$ together with the tensor product. But I'm really confused with that. The set $T(V)$ by construction, is the set of all sequences $T=(T_i)$ with $T_i=0$ for all but finitely many indexes $i\in \Bbb N$ with addition defined by $(T+S)_i = T_i+S_i$ and $(kT)_i=kT_i$. The book I'm working with doesn't tell how the tensor product is defined in $T(V)$, but I think it is defined by $(T\otimes S)_i = T_i\otimes S_i$ which is the most natural thing to do. But I can't really understand how we should think about this space, it seems like to be putting together all kinds of contravariant tensors, but I can't really see why do we need this space, and how should we work with it. I tried showing that this is really an algebra, but I'm confused if I did it right. I did the following: let $T,S\in T(V)$ and $a,b\in F$, then $(T\otimes (S+R))_i=T_i\otimes (S+R)_i,$ then from multilinearity of the tensor roduct on each factor $(T\otimes(S+R))_i=T_i\otimes S_i+T_i\otimes R_i$ wich means that $(T\otimes (S+R))_i = (T\otimes S)_i + (T\otimes R)_i$ and finally $$(T\otimes (S+R))_i = ((T\otimes S)+(T\otimes R))_i\Longrightarrow T\otimes(S+R)=T\otimes S + T\otimes R,$$ the same for distribuitivity on the other side and for multiplication by scalars. So, is this proof right? And in general, how should we really understand and think about the tensor algebra of a vector space? Thanks very much in advance! AI: Your proof looks correct to me, and I don't think you said anything else wrong either. I prefer not to think of $T(V)$ in terms of sequences like this. For me, an element of $T(V)$ is a linear combination of tensors of possibly mixed degree; to convert a sequence into this, just add up the terms, which you can do because all but finitely many are zero. Then the tensor product can essentially be defined to be distributive. You already know how to take the tensor product of two pure tensors, so to take the product of two combinations, just expand by the distributive law and then use the definition for pure tensors. As an analogy, this is kind of like thinking of the polynomial ring $\mathbb{C}[x]$ as: $$\mathbb{C}[x]=\bigoplus_{i=0}^\infty\mathbb{C}\langle x^i\rangle$$ where $\mathbb{C}\langle x^i\rangle$ is the one-dimensional vector space spanned by $x^i$. Then we can define $x^i\cdot x^j=x^{i+j}$ and extend this product to $\mathbb{C}[x]$ uniquely by insisting it is multilinear.
H: if $(ab)^{3}=e\Rightarrow(ba)^{3}=e$, this is true? for $a,b\in G$ (G is a group) I have to prove that $if (ab)^{3}=e\Rightarrow(ba)^{3}=e$ or to give an example that this is false... I belive it false and I try to find an exaple - please help and tell if I'm right (and I'd like to get an exaple) or I wrong... Thank you! AI: $$a(babab)=e$$ so $$babab=a^{-1}$$ and so $$(babab)a=a^{-1}a=e$$
H: What's wrong with this argument that $[0,1]$ is countable? Every real number in $[0,1]$ has a decimal expansion $0.d_1d_2d_3...$, so construct an infinite tree rooted at 0 where each node has branches leading to $\{0,1,2,3,4,5,6,7,8,9\} $, and let each path through the tree represent the successive decimal digits of a real number. To enumerate them, go through the tree breadth-first i.e. layer-by-layer. AI: While it’s true that the tree has only countably many nodes, it has uncountably many branches, and the real numbers in $[0,1]$ correspond to the branches, not to the nodes. In fact the nodes correspond only to the real numbers that have terminating decimal expansions, which are a proper subset of the rational numbers.
H: Is $ord(a)=1$ equivalent to $a=e$? I write at my notebook: $ord(a)=?? \Leftrightarrow a=e$ and I forgor to write the number after $ord(a)$, I guess that it was "1", I'm right? Thank you!! AI: yes. By definition of order you can "feel" that.
H: Show that $\{ 1, 1-x , 1-2x + {1 \over 2} x^2\}$ a orthogonal system I have to show that the following set: $$A = \left\{ 1, 1-x , 1-2x + {1 \over 2} x^2\right\}$$ is orthogonal system in relative to the inner product $$\langle f, g\rangle = \int ^\infty_0 f(x)g(x)e^{-x}dx$$. as far as I know, in order that $A$ will be a orthogonal system, for any $f,g \in A$, $$\langle f, g\rangle = \begin{cases}0 & f \ne g \\ 1 & f = g \end{cases}$$ Lets take $f = 1, g = (1 - 2x + \frac{1}{2}x^2)$, \begin{align} \langle f, g\rangle = &\int^\infty_0 1\cdot(1 - 2x + \frac{x^2}{2})dx\\ = & \int^\infty_01-2x + \frac{x^2}{2} dx \\ = & \int^\infty_01 - 2x + \frac{1}{2}x^2 \\ = & \int^\infty_01 - \int^\infty_02x + \int^\infty_0 \frac {1}{2} x^2\\ = & x - \frac{x^2}{2} + \frac{x^3}{6}\bigg\|^\infty_0 \\ = & 0 - (\infty - \frac{\infty^2}{2} + \frac{\infty^3}{6}) = \infty. \end{align} So how that set is orthogonal? thanks in advance. AI: You've forgotten the $e^{-x}$ term in $\langle f,g\rangle$. $$\langle f,g\rangle =\int_0^\infty 1\cdot(1-2x+\frac{x^2}{2})\cdot e^{-x} \mathrm{d}x=0$$ etc.
H: Let $\,f$ be a real differentiable function defined on $\,[a,b]$,where the derivative is an increasing function I am stuck on the following problem: Let $\,f$ be a real differentiable function defined on $\,[a,b]$,where the derivative is an increasing function and $x_0 \in [a,b]$. Then which of the following statements is correct? $f(x) \le f(x_0)+(x-x_0)f'(x_0), \forall x \in [a,b]$ $f(x) \ge f(x_0)+(x-x_0)f'(x_0), \forall x \in [a,b]$ My Attempt: Since $f'(x)$ is an increasing function $\forall x \in [a,b]$, $\frac{d}{dx}f'(x) \ge 0$ and so at $x_0, \frac{d}{dx}f'(x_0) \ge 0 \implies \lim_{x \to x_0} {{f'(x)-f'(x_0)}\over {x-x_0}} \ge 0$ and I am stuck and could not get rid of limit. Can someone explain? Thanks and regards to all. AI: You don't know that you can differentiate $f$ twice. But, you do know that $f(x) = f(x_0)+(x-x_0) f'(c)$ for some $c \in (x_0,x)$ (why!?). So, can you say which inequality is correct for $f(x_0) + (x-x_0)f'(c)$ based on the increasing nature of $f'$?
H: Solution of the differential equation $\ddot{x}(t)=\alpha\dot{x}(t)x(t)+\beta x(t)^3$ I have to solve the following nonlinear differential equation: $$\ddot{x}(t)=\alpha\dot{x}(t)x(t)+\beta x(t)^3$$ with initial conditions: $x(0)=x_0$ and $\dot{x}(0)=x_1$ Is it possible to solve it without the use of numerical techniques? Thanks. AI: It doesn't involve $t$ directly, so you can turn it into a first-order equation by letting $y=\dot{x}$ and then $$\ddot{x}=\frac{dy}{dx}\frac{dx}{dt}=y\frac{dy}{dx}$$. Then let $z=y/x^2$ and the equation becomes separable. I get $$z^{\beta}e^{\alpha z}=Ax$$ for any $A$. That can be solved using the Lambert W function, so $t=\int dx/[x^2W(...)]$
H: is the number algebraic? Is the number $\alpha=1+\sqrt{2}+\sqrt{3}$ algebraic? My first attempt was to try a polynomial for which $p(\alpha)=0$ for some $p(x)=a_{0}+a_{1}b_{1}+\cdots +b_{n-1}x^{n-1}$ i. e $x=1+\sqrt{2}+\sqrt{3}$ and then square it many times to get rid of the irrationals. This procedure was futile. Second attempt: I remember from the lecture we had that if $L=K(\alpha, \beta)$ with $\alpha,\beta$ algebraic over $K$ then $[L:K]<\infty$ moreover $[K(\gamma):K]<\infty$ for $\gamma=\alpha\pm \beta$ and $\gamma=\alpha\beta$ and $\gamma=\frac{\alpha}{\beta},\beta\neq 0$ as $K(\gamma)\subseteq L$ hence $\gamma$ is algebraic over $K$. So applying the above to the problem then $\alpha=1+\sqrt{2}+\sqrt{3}$ is algebraic then over $K$ Right? Or is there any other way to show it? But I would like to know if it is possible to find a minimal polynomial having $\alpha$ as a zero? How to do that? AI: $$\alpha=1+\sqrt2+\sqrt3$$ $$\alpha-1=\sqrt2+\sqrt3$$ $$(\alpha-1)^2=5+2\sqrt6$$ $$(\alpha-1)^2-5=2\sqrt6$$ $$((\alpha-1)^2-5)^2=24$$
H: Necessary and sufficient conditions for the embeddability of a semigroup in a group According to wikipedia, The first set of necessary and sufficient conditions for the embeddability of a semigroup in a group were given in (Malcev 1939).[5] Though theoretically important, the conditions are countably infinite in number and no finite subset will suffice, as shown in (Malcev 1940).[6] Although very important, Malcev's two papers share a fundamental flaw; they're both written in Russian, and alas, I do not speak Russian. Question 1. What was Malcev's countably infinite set of (necessary and sufficient) conditions? Question 2. And what other characterizations have been discovered since 1940? AI: You can read the Malcev's theory in 2nd vol. of Algebraic theory of semigroups by Clifford and Preston. You will also find there the more late results of Lambeck. Addendum: N. Bouleau, Classes de semigroupes immersible dans groupes, Semigroup Forum,6(1973), N.2, pp.100-112. S.I. Adyan, Defining relations and algorithmic problems for groups and semigroups. Proc. Steklov Inst. Math. 85(1966), 152 p.; translation from Tr. Mat. Inst. Steklov 85(1966), 123 p.
H: Show that $1<\sqrt{1+x^3}<1+x^3$ for $x>0$ The problem: Show that $1<\sqrt{1+x^3}<1+x^3$ for $x>0$ How do I solve this using the Fundamental Theorem of Calculus AI: As other people have pointed out, you certainly don't need the Fundamental Theorem of Calculus to prove these inequalities, but if you do want to use it, here's a way to get one of them at least: If $F(t)=\sqrt{1+t^3}$, then $F'(t)=3t^2/2\sqrt{1+t^3}$, which is clearly postive for any $t\not=0$ in its domain, which means that it has a positive integral over any interval of positive length. Thus, by FTC, $$0\lt\int_0^x F'(t)dt = F(x)-F(0)=\sqrt{1+x^3}-1$$ for $x\gt0$, which gives the first inequality. You can get the other inequality by using $F(t)=t^3-\sqrt{1+t^3}$, but showing $F'(t)\gt0$ requires showing $1\lt2\sqrt{1+t^3}$ for $t\gt0$, which I suppose you can justify because you've already shown that $1\lt\sqrt{1+x^3}$ for $x\gt0$.
H: Continuous function: $ Z=${$x \in \mathbb{R} : f(x) = 0$} is closed Consider $f: \mathbb{R} \to \mathbb{R}$ a continuous function. Show that the set $Z=${$x \in \mathbb{R}: f(x) = 0$} is closed My attempt: My idea is to show that if $a \in \bar{Z}$ (the set of adherent points), then $f(a) = 0$. Considering only the case that $ \bar{Z} \neq ${} (since $\bar{Z} =${} implies $Z=${}, which is cleary closed), let $a \in \bar{Z}$. Since $f$ is continuous, $\forall \varepsilon > 0, \exists \delta > 0 $ such that $ \|x-a\| < \delta \Rightarrow \|f(x) - f(a) \| < \varepsilon $. As $a$ is adherent, we can take $x \in Z$ such that $x \in B_{\delta}(a)$. In this case, we will have $\|0 - f(a) \| < \varepsilon $ Now, I was thinking: Since it is true for really small values of $ \varepsilon$, does it imply that $f(a) = 0$? I don't know if it is rigorous. Thanks you very much! Edit: Clearly I do not want to consider that $a \in Z$ previously. Maybe it is better suppose, by contradiction, that $a \not \in Z$ and, then, I can use the existance of $x \in B_{\delta}(a) \cap Z$ AI: If you’ve already proved that a function $f$ is continuous if and only if $f^{-1}[F]$ is closed whenever $F$ is closed, then you can simply observe that $\{0\}$ is a closed set in $\Bbb R$, so $Z=f^{-1}[\{0\}]$ must be closed as well. If not, I think that the simplest argument is to show that if $a\in\Bbb R\setminus Z$, then there is a $\delta>0$ such that $f(x)\ne 0$ whenever $\|x-a\|<\delta$: this in turn shows that $a\notin\operatorname{cl}Z$. If $a\in\Bbb R\setminus Z$, then $f(a)\ne 0$. Let $\epsilon=\|f(x)\|>0$; $f$ is continuous, so there is a $\delta>0$ such that $$\|f(x)-f(a)\|<\epsilon=\|f(a)\|$$ whenever $\|x-a\|<\delta$. But if $\|f(x)-f(a)\|<\|f(a)\|$, then clearly $f(x)\ne 0$, and we’ve found the desired $\delta$. An argument closer to what you attempted uses sequences, but it’s available only if you’ve already proved that if $f$ is continuous, and $\langle x_n:n\in\Bbb N\rangle$ converges to $a$, then $\langle f(x_n):n\in\Bbb N\rangle$ converges to $f(a)$. If $a\in\operatorname{cl}Z$, then there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $Z$ that converges to $a$. But $f(x_n)=0$ for each $n\in\Bbb N$, so $\langle 0,0,0,\ldots\rangle$ converges to $f(a)$, so $f(a)=0$, and therefore $a\in Z$.
H: Ring-Homomorphism from $\mathbb{Z}_{2}$ to $\mathbb{Z}_{2n}$ Let $n$ be a positive integer. Then the problem is to show that there is a ring-homomorphism from $\mathbb{Z}_{2}$ to $\mathbb{Z}_{2n}$ if and only if $n$ is odd. My effort : let $\phi$ be such a ring homomorphism (apart from the zero-map). then if $\phi(1)\ =\ a$ and $1.1\ =\ 1$ in $\mathbb{Z}_{2}$ so in $\mathbb{Z}_{2n}$ we must have $a^{2}\ =\ a$. Moreover $\phi(1+1)\ =\ \phi(0)\ =\ 2a$ and so $\|2a\|$ must divide $2n$. But I can't deduce anything from these. Thanks AI: Consider $\phi : \mathbb{Z}_2 \rightarrow \mathbb{Z}_4$ you should be able to see that $\phi(\bar{0})=\bar{0}$ i.e., $\phi(2.\bar{1})=\bar{0}$ i.e., $2.\phi(\bar{1})=\bar{0}$ Now only possibility for $\phi(\bar{1})$ excluding $\bar{0}$ is $\phi(\bar{1})=\bar{2}$. But then, your map should preserve multiplication property also... I mean you should have $\phi(\bar{1})=\phi(\bar{1}.\bar{1})=\phi(\bar{1}).\phi(\bar{1})=\bar{2}.\bar{2}=\bar{0}$ so the only possibility which we have got is not even well defined as $\phi(\bar{1})$ is taking two distinct values. More generally $\bar{1}$ should be send to $\bar{n}$ and in that case, for similar reasn as above, $\phi(\bar{1})=\phi(\bar{1}.\bar{1})=\phi(\bar{1}).\phi(\bar{1})=\bar{n}.\bar{n}=\bar{n}^2$ the worst case we should not have is $n^2$ being divided by $2n$ i.e., $n$ being divided by $2$.. which means $n$ has to be odd.... I hope you can do the other way....
H: select the min of a set of sums I want to the selection of the minimum element in a set of sums; I have a cluster with n elements. In this set, we want to select whats called the clustroid. The clustroid is defined to be the element whos sum of the distance to every other element in the cluster is the minimum. The distance is calculated using the function d(element1, element2). How is this definition of the clustroid expressed mathematically? AI: Let your set be $\{c_1, c_2, ....., c_n\}$ and let $c_j$ be the clustroid element. You can define $c_j$ as the element satisfying the following: $$\exists ! c_j ,\forall k \in \mathbb{Z}^+, k \leq n, \sum_{i=1}^n d(c_j,c_i) < \sum_{i=1}^n d(c_k,c_i) \iff c_j \text{is clustroid.}$$ As I don't know the exact definition, I'm not sure if this element should be unique or not. What I mean is, there are two elements with the minimum sum, then you can change your definition as: $$\,\forall k \in \mathbb{Z}^+, k \leq n, \sum_{i=1}^n d(c_j,c_i) \leq \sum_{i=1}^n d(c_k,c_i) \iff c_j \text{is clustroid.}$$
H: Is $\int\frac{\cos^5x\sin^3x}{1+\cos2x}dx = \frac{\sin^4x}{8}-\frac{\sin^6x}{12} +C$ or $\frac{\cos^6x}{12}-\frac{\cos^4x}{8} +C$? $\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx = \dfrac{\sin ^4x}{8}-\dfrac{\sin ^6x}{12} +C$ or $\dfrac{\cos ^6x}{12}-\dfrac{\sin ^4x}{8} +C$? $\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx$ = $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$ 1) $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$ = $\dfrac{1}{2}\int{\cos ^2x\sin ^3x\cos x} dx$ = $\dfrac{1}{2}\int{(1-\sin ^2x)\sin ^3x} d\sin x$ = $\dfrac{1}{2}\int{\sin ^3x-\sin ^5x} d\sin x$ = $\dfrac{\sin ^4x}{8}-\dfrac{\sin ^6x}{12} +C$ 2) $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$ = $\dfrac{1}{2}\int{\cos ^3x\sin ^2x\sin x} dx$ = $-\dfrac{1}{2}\int{(1-\cos ^2x)\cos ^3x} d\cos x$ = $-\dfrac{1}{2}\int{\cos ^3x-\cos ^5x} d\cos x$ = $\dfrac{\cos ^6x}{12}-\dfrac{\cos ^4x}{8} +C$ Which one is correct? or both are correct? AI: You can differentiate and simplify your two answers to see that they are both correct. If you take the first answer, expand it using the identity $\sin^2 x = 1-\cos^2 x$, and simplify, you will get the second answer plus an additional constant. Thus the two expressions (ignoring the $C$'s) differ by a constant.
H: Sum of real roots of the equation $x^2 + 5\|x\| + 6 = 0$? Sum of real roots of the equation $x^2 + 5\|x\| +6 = 0$ AI: Hint: It $r$ is a solution then so is $-r$ Note: After seeing Ryan's answer, I realized that the solution set is empty. Thus, Ryan's answer is the correct answer. Now if we are looking for solutions inside $\mathbb{C}$, my hint can be used to deduce that either the solution set is infinite (hence the sum is undefined) or the sum is zero
H: Calculating derivative In order to solve a mathematical problem I have to calculate the following derivative: $\frac{\delta}{\delta k}\frac{11 + \sum_{i = 0}^{k-1}i}{k}$ Does anyone know this derivative? AI: $$f(k):=\frac{11+\sum_{i=0}^{k-1}i}h=\frac{11+\frac{(k-1)k}2}k=\frac{22+k^2-k}{2k}=\frac k2-\frac12+\frac{11}k\implies$$ $$f'(k)=\frac12-\frac{11}{k^2}$$
H: If $\lim_{x \to p} f(t)=L$ how do I prove that $\lim_{x \to p} \frac{1}{f(t)}=1/L?$ If $\lim_{x \to p} f(t)=L$ how do I prove that $\lim_{x \to p} \frac{1}{f(t)}=1/L?$ My attempt: $\|\frac{1}{f(x)}-\frac{1}{L}\|=\frac{\|f(x)-L\|}{\|L\|\|f(x)\|}$ So the only thing stopping me from going on with the proof is the $\|f(x)\|$ term in the denominator. Somehow, I need to show that $f(x) \geq M$ for some $ \delta>0$ which implies $\frac{1}{f(x)} \leq \frac{1}{M}$ Can someone please give me a hint? Thanks! AI: Hint: Given any $\epsilon > 0, \exists \delta > 0:$ $$\|x-p\|<\delta \Rightarrow \|f(x)-L\| < \epsilon$$ You then get $$\|f(x)\| \geq \|L\| - \|L-f(x)\| > \|L\| - \epsilon $$ Pick $\epsilon < \|L\|$. Can you proceed from here?
H: Help with the chain rule $h(t)=f(t, X(t))$ Assume we have the function $h(t)=f(t, X(t)): \mathbb{R}\rightarrow \mathbb{R}$. How to I calculate $h'$? I thought of letting $g:t \rightarrow(t,X(t))$ and then $h' = g'(t)f'(g(t)) = (1, X_{t})f'(t,X(t))$ but it is not a scalar... so where is the problem? AI: The problem is that you did not understand the chain rule. Since you are working with functions of several variables, the order in which you "multiply" the derivatives matters. Since $h = f \circ g$, you should "multiply" $f'(t,X(y))(1,X'(t))$, where $f'$ is the Jacobian matrix of $f$. More explicitly, $$ h'(t) = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial X}X'(t). $$
H: Is it true that if $X$ is connected, then for every nonempty proper subset $A$ of $X$, we have $\mathbf{Bd} \ne \emptyset$ Is it true that if $X$ is connected, then for every nonempty proper subset $A$ of $X$, we have $\mathbf{Bd} \ne \emptyset$? Does the converse hold? I start by trying to understand what $\mathbf{Bd}$ means in "formal terms" (i.e. $\bar{A} \cap \overline{X-A}$ is the 'boundary' of $A \in X$)? This means I'm supposed to figure out if the boundary of subset of a connected space is empty or not, right? I think that if a set is connected then it contains no partition into two sets, so there would exist points that are both in $A$ and in the complement of $A$ in $X$. This would be an informal definition of a boundary. (A boundary contains points in a set $A$ and points outside of it, right?) What I mean is, if $X$ is connected, and $A \subset X$ is connected, then $A$ could have a boundary which means that technically, $X$ cannot be partitioned into $A$ and $A$-complement without making the boundary of $A$ an empty set. Right? AI: Be careful: any set with more than one element can be partitioned into two non-empty sets; a connected set cannot be partitioned into two non-empty relatively open sets. And in your last paragraph it’s the closures of $A$ and $X\setminus A$ whose disjointness makes the boundary of $A$ empty. The easiest way to prove the result is to prove the contrapositive. Suppose that $A$ is a non-empty proper subset of a space $X$, and suppose further that $\operatorname{bdry}A=\varnothing$. As you say, $\operatorname{bdry}A=(\operatorname{cl}A)\cap\operatorname{cl}(X\setminus A)$, so $\operatorname{cl}A$ and $\operatorname{cl}(X\setminus A)$ are disjoint closed sets whose union is $X$. Each of them is therefore the complement of a closed set, so each is open. Moreover, $\operatorname{cl}A\supseteq A\ne\varnothing$ and $\operatorname{cl}(X\setminus A)\supseteq X\setminus A\ne\varnothing$, so $\{\operatorname{cl}A,\operatorname{cl}(X\setminus A)\}$ is a partition of $X$ into non-empty open sets, and $X$ therefore is not connected.
H: How to get the Galois group for $\mathbb{R}/\mathbb{Q}$ For the real number field $\mathbb{R}$ and the rational number field $\mathbb{Q}$, how to get the Galois group ${\rm{Gal}}(\mathbb{R/Q})$? AI: Let $f\in Aut_{\mathbb{Q}}(\mathbb{R})$. Hints: 1) $\forall a,b\in\mathbb{R} [a\leq b \rightarrow f(a)\leq f(b) ]$. To show this, note that $f(b)-f(a)=f(b-a)=f((\sqrt{b-a})^2)=f(\sqrt{b-a})^2\geq 0$ 2) Let $x$ be any real number. Let $r_n$ be an increasing sequence of rationals that converge to $x$. we find that $\forall n\in \mathbb{N} [r_n=f(r_n)\leq f(x)]$. By taking limits, we find that $x\leq f(x) $ 3) Let $s_n$ be any decreasing sequence that converges to $x$. imitate the argument of step 2 to deduce that $x\geq f(x)$ for every real $x$ Thus, $f$ is the identity. Hence $Aut_{\mathbb{Q}}(\mathbb{R})$ is trivial.
H: Two graphs with the same number of edges and vertices but not isomorphic? Can anyone give me an example of two graphs that have the same number of edges and vertices but is not isomorphic? AI: HINT: Any tree with $n$ vertices has exactly $n-1$ edges. Find two non-isomorphic trees with the same number of vertices, and you’re done.
H: Applying prices to augmented matrices The question is as follows (translated): A company wants to rent 20 buses. These 2 buses are to hold 1000 people. They can choose between 3 types, 30, 40 and 60 man buses. How many of each kind can the company rent to satisfy the constraints? Solve with Gauss-Jordan. So I solve this problem and get the solutions: x1 (30 man)=2t-20 x2 (40 man)=40-3t 10$\leq$t$\leq$13 where t denotes the 60 man bus. Then comes the second part of the question: The rent is 100.000 for the 30 man bus, 130.000 for the 40 man bus and 150.000 for the 60 man bus. Which of the solutions to question 1 will give the company the best price? So I'm wondering if I should solve this with Gauss-Jordan (or if it is possible) or whether I can simply do like I usually do and just CHECK each solution. That doesn't seem very professional, though! Thanks! AI: Essentially, as long as your working is sound, then it doesn't matter what you do to arrive at the correct solution and so you'll generally want to use whichever method gets you there the fastest. In this case, pretty much no other method will require less effort than simply substituting the four values of $t$ you have. For other questions though you may have many more cases to check and so this may become unfavourable, in which case you may want to use a calculus approach.
H: Galois extensions of Local fields Let $ L / K $ be a Galois extension of local fields. My question: why $L / K$ is necessarily of finite degree ?? thanks. AI: Assuming the inclusion $K\hookrightarrow L$ is continuous, so that $L$ is a topological vector space over $K$, this is true, though the condition that $L/K$ is Galois is not needed. If $V$ is a Hausdorff topological vector space over a local field $K$, then $V$ is locally compact if and only if $V$ is finite-dimensional over $K$. This is a special case of Theorem 3, No. 4, $\S 2$ from Bourbaki's Topological Vector Spaces.
H: Analytical solution of nonlinear ordinary differential equation I have following first order nonlinear ordinary differential and i was wondering if you can suggest some method by which either i can get an exact solution or approaximate and converging perturbative solution. $$ \frac{dx}{dt} = 2Wx + 2xy - 4x^{3} $$ $$ \frac{dy}{dt} = \gamma (x^{2} - y) $$ Kindly help me with any methods you that might work and it will be great if you can provide few references where i can read about those methods. Also If somebody can help me about how I can use fixed point analytic method to solve this differential equations and some references on it, will be very useful too. Any help will be highly helpful. Thanks a lot in advance. PS. I tried homotopy perturbation analysis and simple iteration procedure to try to solve it and it diverges after some time(good only for early short times). AI: If you can assume that $\gamma$ is a small parameter, then write $$x(t) = x_0(t) + \gamma x_1(t) + \gamma^2 x_2(t) + \cdots$$ $$y(t) = y_0(t) + \gamma y_1(t) + \gamma^2 y_2(t) + \cdots$$ i.e., assume such convergent series exist. A zeroth order solution is $y_0(t)=y(0)=y_0$, a constant, and the equation for $x_0(t)$ becomes $$\frac{dx_0}{dt} = 2 (W+y_0) x_0-4 x_0^3$$ This equation is integrable: $$\int \frac{dx_0}{2 (W+y_0) x_0-4 x_0^3} = t \implies \frac{1}{4 (W + y_0)} \log{\left (\frac{x_0(t)}{2 x_0(t)^2-W-y_0}\right)} = t+C$$ Solve for $x_0(t)$, then plug into the first order equation for $y_1(t)$: $$\gamma \frac{dy_1}{dt} = \gamma(x_0(t)^2-y_0) \implies \frac{dy_1}{dt} = x_0(t)^2-y_0$$ Integrate with respect to $t$ to get $y_1(t)$, then plug into $x$ equation: $$\frac{d}{dt} (x_0+\gamma x_1) = 2 (W + y_0+\gamma y_1) (x_0+\gamma x_1) - 4 (x_0+\gamma x_1)^3$$ Note that $(x_0+\gamma x_1)^3 = x_0^3 + 3 \gamma x_0^2 x_1 + O(\gamma^2)$. Coefficient of $\gamma^0$ is zero because of above equation. Equating coefficients of $\gamma^1$, we get $$\frac{d x_1}{dt} = 2 (W+y_0) x_1 + x_0 y_1 - 12 x_0^2 x_1$$ This is an inhomogeneous 1st order equation for $x_1$, which may be solved using known techiques (e.g., integration factor). At this point, you may repeat this process to get higher powers of $\gamma$. I do not have a proof that the resulting series converges.
H: tautologies and contradictions with $r$ I'm really struggling to understand tautologies and contradictions. I've been able to do $(p \rightarrow q) \leftrightarrow (\lnot q \rightarrow \lnot p)$ and I understand why it is a tautology, however I don't understand when they involve the character r. For example, $$p \rightarrow (q \rightarrow r)$$ How do I construct a truth table from this? What even is r? I'm so confused. If someone could actually explain the concept or knows of a website that explains it well I'm open to anything! AI: $r$ is simply another variable, just like $p$ and $q$ are variables. That means the truth value of the statement is a function of the truth values of $p, q, r$. So you need $2^3 = 8$ rows in your truth table in order to consider all $8$ possible distinct truth-value assignments to the variables $p, q, r$. The number of rows you need for a truth-table for any statement is a function of the number of distinct variables in the statement: if there are $n$ variables, then you need to consider $\,2^n\,$ possible distinct truth-value assignments to those variables, since each of the $n$ variables can take on two possible truth-values: true or false. Wolfram Alpha does a nice job with truth-tables: Note the $8$ rows needed. Note that the only possible truth value assignment that makes the entire statement false is when $p$ and $q$ are both true, but $r$ is false. Then, and only then, we have that $$\;\underbrace{\underbrace{p}_{T} \rightarrow (\underbrace{\underbrace{q}_{T} \rightarrow \underbrace{r}_{F}}_{F}}_{F})\;\text{ is false.}$$ The statement, however, is not a tautology nor a contradiction, as your title suggests. The truth value of the statement is contingent: it depends on the truth-value assignments of the variables. In a tautology, a statement is true under every possible truth-value assignments of its variables, and In a contradiction, the statement is false under every possible truth-value assignments of its variables.
H: Simplifying square root with fraction I'm not sure about this equality $$4(-3+\sqrt {15})/4)^2 = (9-6 \sqrt{15} +15)/4$$ Hope some one can enlighten me. I will be facing more of such fractions, please guide me on how to solve/simplify in easy method. Thanks :) AI: I’ll even finish the simplification: $$\begin{align} 4\left(\frac{-3+\sqrt{15}}4\right)^2&=4\cdot\frac{(-3+\sqrt{15})(-3+\sqrt{15}}{4\cdot4}\\ &=\frac{(-3)^2+2(-3)\sqrt{15}+(\sqrt{15})^2}4\\ &=\frac{9-6\sqrt{15}+15}4\\ &=\frac{24-6\sqrt{15}}4\\ &=6-\frac32\sqrt{15}\;. \end{align}$$ In short: cancel a factor of $4$, and multiply out the numerator.
H: Are critical points fixed? Let $M$ be a smooth manifold (compact, connected, without boundary and oriented if you wish) with a smooth action of $S^1$. Let $f:M\rightarrow\mathbb{R}$ be an invariant function $f$. I know how to prove that a fixed point of the action is a critical point of $f$. What I don't know is if the converse is true: I can prove that if a point is critical, then all the points in its orbit are critical as well, but that's all. So the questions are: 1) Are the critical points of $f$ necessarily fixed? 2) If the answer to 1) is "yes": any hint on the proof? 3) If the answer to 1) is "no": any counterexample? Is there any extra condition that makes the answer of 1) to be "yes"? AI: No; take the unit sphere, with $S^1$ acting by rotating the sphere about the $z$ axis, and let $f(x,y,z) = x^2+y^2.$ The two fixed points (poles) are critical points of $f$, but so is every point along the equator.
H: Monoids as categories; does this construction have a name? We can view a monoid $M$ as a category with a single object. However, there is another way to make $M$ into a category. Take the elements of $M$ as objects, and define $\mathrm{Hom}(x,y)$ to be set of all triples $(x,y,a)$ such that $ax=y$. Define composition such that $(y,z,b) \circ (x,y,a) = (x,z,ba)$ and identity arrows by $\mathrm{id}_x = (x,x,1).$ I'd like to learn more about this construction. Does it have a name? AI: What's going on here is that you've allowed the monoid $M$ to operate on the set $M$, and you've successfully expressed this monoid action as a category. For a general monoid $M$ acting on a set $X$, you can view the elements of $X$ as objects of a category and use the elements of $M$ as arrows between the objects.
H: $\mu(A_i)>0$ for countably many $i$ Let $\mu$ be a $\sigma$-finite measure on a $\sigma$-Algebra $\mathcal A$ and $A_i\in\mathcal A$ ($i\in I$) subsets with $A_i\cap A_j=\emptyset$ for $i\neq j$. Then $\mu(A_i)>0$ for at most countably many $i\in I$. I know $\sigma$-finite means that there exists a sequence $(A_n)\subset\mathcal A$ with $\Omega=\sum_{i=1}^\infty A_i$ such that $A_i\cap A_j=\emptyset$ for $i\neq j$ and $\mu(A_n)<\infty$ for all $n$. But I don't see how I can show the above using this. AI: Let $(B_n)_{n\geqslant 1}$ be an increasing sequence sets of finite measure which covers $\Omega$. Fix integers $n$ and $p\geqslant 1$. The set $J_{n,p}:=\{i\in I\mid\mu(A_i\cap B_n)\geqslant p^{-1}\}$ is finite. Hence the set $\bigcup\limits_{n,p\geqslant 1}J_{n,p}$ is at most countable.
H: Limit of subtracting fractions from 1 Suppose you have the sequence of fractions $\left\{\frac{1}{a} : a \in \mathbb{N}\right\}$ ($\frac{1}{2},\frac{1}{3}$ and so on). Now you start with $1$ and subtract every item of the sequence as long as the result is larger than $0$. You would start with subtracting $\frac{1}{2}$ and $\frac{1}{3}$, but then skip $4-6$ because the result would be 0 or smaller. You continue with $\frac{1}{7}$ and $\frac{1}{43}$. Is there any lower limit to how small a number you can get or can you get as close to $0$ as you like? AI: The limit is $0$, and you approach it fast (with few subtractions). Say after $k$ subtractions the remainder is $\dfrac{1}{n_k}$. Then the next number you subtract is $\dfrac{1}{n_k+1}$, and $$\frac{1}{n_{k+1}} = \frac{1}{n_k} - \frac{1}{n_k+1} = \frac{1}{n_k(n_k+1)}.$$ The denominator satisfies $n_k \geqslant 2^{2^{k-1}}$ for $k \geqslant 1$, that is an extremely fast growing sequence.
H: Number of elements in a group and its subgroups (GS 2013) Every countable group has only countably many distinct subgroups. The above statement is false. How to show it? One counterexample may be sufficient, but I am blind to find it out. I have considered some counterexample only like $(\mathbb{Z}, +)$, $(\mathbb{Q}, +)$ and $(\mathbb{R}, +)$. Is there any relationship between number of elements in a group and number of its subgroup? I do not know. Please discuss a little. What will be if the group be uncountable? Thank you for your help. AI: The countably infinite sum $S$ of copies of the group $G=\mathbb Z/2\mathbb Z$ (cyclic group of order two) indexed by $I$ is countable. Every subset $A$ of the index set $I$ corresponds to a subgroup of $S$ consisting of elements with nonzero components only in the copies of $G$ corresponding to the subset $A\subset I$. This gives uncountably many subgroups because the set of subsets of $I$ is uncountable.
H: Urn problem - black and white balls I have a problem with the following exercise: We have an urn with one black and one white ball. At time 1 you take one of the balls from the urn randomly. Then you take this ball and replace it by two balls of the same colour. For example you take one white ball, then you replace it by two white balls and you put two white balls back in the urn. Then you continue doing this procedure. $W_n$ denotes the number of white balls at time $n$, then define $X_n=\frac{W_n}{n+1}$. This $X_n$ should converge to something , lets call it $S$. I want to compute the expectation and the variance from $S$. My first thought was to get the expectation of $W_n$. Therefore I need some formula for $W_n$. $W_2=1$ with prob. $1-p$ and $W_2=2$ with prob $p$. Probability $p$ is the probability that you take a white ball. Continuing leads me to $W_n=1$ with prob. $1-p$, $W_n=2$ with prob $p$....$W_n=n$ with prob $p^{n-1}$, therefore $E[W_n]=1-p+2p+3p^2+...+np^{n-1}$ How can I proceed? AI: Since the problem is symmetric with respect to white balls and black balls, the $E[W_n]= \frac{n+1}{2}$
H: Infinite line is closed in $\mathbb{R}^n$ I have been reading the book "Elements of the functional analisys", by Kolmogorov and Fomin. At the chapter of Normed Linear Spaces, page 73 to be precise, the author makes the following definitions: A linear mainfold $L$ in a normed linear space $\mathbb{R}$ is any set of elements satisfying the following condition: if $x,y$ belong to $L$, then for any two arbitary numbers $a$ and $b$, $a\cdot x + b\cdot y$ does also belong to $L$. A subspace of a space $\mathbb{R}$ is a closed linear mainfold in $\mathbb{R}$. After defining linear mainfold and subspace the author proposes the reading to prove that given any euclidean space $\mathbb{R}^n$, evey linear mainfold is a subspace; i.e every linear mainfold is closed. I would greatly appreciate any guidance upon this proof.( Especially if you could first help me prove that, since any straight line in $\mathbb{R}^n$ is a linear mainfold, that every straight line is closed in $\mathbb{R}^n$, in the sense that it contains its limit points) AI: Hint A straight line passing through origin is one dimensional subspace of $\mathbb{R}^n$ Every finite-dimensional subspace of a normed space is closed and complete. A straight line through the points $a,b\in\mathbb{R}^n$ is subset $L=\{a(1-t)+bt:t\in\mathbb{R}\}$ of $\mathbb{R}^n$ $L$ is closed as it contains all its limit points. Let $x_n\in L\ni x_n\to x$ enough to show $x\in L$ $x_n= a(1-t_n)+bt_n\in L$, as $a(1-t)+bt$ is continuous so $a(1-t_n)+bt_n\to a(1-k)+bk=x$ for some $k\in\mathbb{R}$, so $x\in L$ too. Done! In other way, a straight line $\{ax+by+c=0\}$ in $\mathbb{R}^2$ say, is just inverse image of $\{0\}$ under continuous map $\mathbb{R}^2\to\mathbb{R}:(x,y)\to ax+by+c$ Can you generalize for $\mathbb{R}^n$?
H: Generalizations of the quadratic formula The quadratic formula can be used to find the roots of any quadratic polynomial of the form $ax^2 + bx + c$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The derivation is simple enough and uses a technique called completing the square. Is there a formula to solve cubic equations of the form: $$ax^3 + bx^2 + cx + d$$ Or a general formula to solve an $n$th degree polynomial? EDIT: According to the answers below, and some mathematicians, such a formula isn't possible. But why is that? I mean, consider an $n$th degree polynomial $p(x)$ with roots $r_1, r_2, \dots, r_n$: $$p(x) = ax^n - \left(\sum r_i\right)ax^{n-1} + \left(\sum_{i, j} r_ir_j\right)ax^{n-2} + \dots+(-1)^n(r_1r_2\dots r_n)$$ We have $n$ equations with $n$ unknowns, surely this can be solved using a general algorithm? AI: To your first question, yes. There exist explicit forms for the solutions for cubic equations as well as quartic equations. However, it was proven by Abel and Ruffini that such explicit solutions don't exist for 5th degree or higher polynomials. There is no general form for the solutions.
H: Expected Value Word Problem I have another problem, that is so: The probability that a roulette wheel stops on a red number is 18/37. For each bet on “red” you are returned twice your bet (including your bet) if the wheel stops on a red number, and lose your money if it does not. If you bet $1 on each of the 10 consecutive plays, what is your expected winnings? I started the question by declaring my random variables. I also know that each spin of the roulette wheel is independent of each other. let X be the number of wins in the 10 consecutive plays X ~ Binomial(10, 18/37), for any x such that $1 \leq x\leq 10$ By definition, my expected value function will look: $E(x)=\displaystyle\sum\limits_{x=1}^{10} 3xf(x)$ What I am confused about is, how I determine my upper bound for my wins (how many wins can I get in 10 rolls), so that I can split my summation into two summations. Since I do not know my probability distribution function, is my assumption that X follows a Binomial distribution correct? AI: You are correct $($assuming you are talking about French Roulette, which has no $00)$ that $X\sim\operatorname{Binomial}(10,\frac{18}{37})$. However, you've miscalculated. If you win, then you are given twice your bet including your bet, not in addition to your bet. That is, if you bet on red and win, then you only double your bet, not triple it. Your net gain for each win, then, is only a dollar, since you're betting a dollar each time. You'd be better off starting by computing your expected number of wins, which is just $10\cdot\frac{18}{37}$. At that point, the answer will depend on what is meant by "expected winnings." If it means "expected net gain," then note that you will gain a dollar for each win and lose a dollar for each loss, so since your expected losses are $10\cdot\frac{19}{37},$ then your expected net gain is $10\cdot\frac{18}{37}-10\cdot\frac{19}{37}=-\frac{10}{37}$ dollars. If it means "expected gain," then the answer is simply $10\cdot\frac{18}{37}$ dollars. If it means "expected total amount that you will be given by the croupier," then the answer will be $2\cdot10\cdot\frac{18}{37}$ dollars. (Note: If you simply subtract the $10$ dollars that you bet from this, then you will find the expected net gain in another way.) Addendum: More generally, suppose that you play $n$ independent games with win probability $\theta.$ Let $X$ be the number of wins, so that $X\sim\operatorname{Binomial}(n,\theta).$ Then your expected number of wins is $$\Bbb E[X] = \sum_{x=0}^n\binom{n}{x}\theta^x(1-\theta)^{n-x},$$ but that's a bear to actually calculate. Instead, we will let $X_k$ be the number of wins on the $k$th trial for $1\le k\le n,$ so that $X\sim\operatorname{Bernoulli}(\theta)$ for each $k,$ and $$X=\sum_{k=1}^nX_k.$$ Since expectation is linear over sums of random variables, then $$\Bbb E[X]=\Bbb E\left[\sum_{k=1}^nX_k\right]=\sum_{k=1}^n\Bbb E[X_k]=\sum_{k=1}^n\sum_{x=0}^1x\theta^x(1-\theta)^{1-x}=\sum_{k=1}^n1\cdot\theta^1(1-\theta)^{1-1}=\sum_{k=1}^n\theta=n\theta.$$ So, our expected number of wins is $n\theta,$ which in your particular example is $10\cdot\frac{18}{37}.$
H: Proof by induction Involving Factorials My "factorial" abilities are a slightly rusty and although I know of a few simplifications such as: $(n+1)\,n! = (n+1)!$, I'm stuck I have to prove by induction that: $$\sum_{i=1}^n\frac{i-1}{i!} = \frac{n!-1}{n!}$$ I get so far as: $$\frac{k!-1}{k!} + \frac{(k+1)-1}{(k+1)!} = \frac{(k+1)!(k!-1) + k\cdot k!}{k!(k+1)!}$$ and I know I should get: $$\frac{(k+1)! -1}{(k+1)!}.$$ I don't see how, though. Any help would be appreciated, thanks! AI: Hint: Instead of taking $k!(k+1)!$ as the common demoninator, simply take $(k+1)!$ as the common denominator. Then $$\frac{k!-1}{k!}+\frac{(k+1)-1}{(k+1)!}=\frac{k!-1}{k!}+\frac{k}{(k+1)!}=\frac{(k!-1)(k+1)}{(k+1)!}+\frac{k}{(k+1)!}.$$ Can you take it from there?
H: Transitive closure Given $M=\{n\in\Bbb Z: 0\le n\le 30\}$ find the transitive closure of the relation $R\subset M\times M$ defined by $R=\{(n,m): m=3n+1\}\cup\{(8,16)\}$ So, I know that a transitive closure is the least possible subset that $R$ takes to be transitive. But, what can I do to know the pairs that are missing in this relation. Or better, how can I describe all the pairs to recognize the missing pairs? AI: Start by writing out $R$ explicitly: $$R=\{\langle 0,1\rangle,\langle 1,4\rangle,\langle 2,7\rangle,\langle 3,10\rangle,\langle 4,13\rangle,\langle 5,16\rangle,\langle 6,19\rangle,\langle 7,22\rangle,\langle 8,16\rangle,\langle 8,25\rangle,\langle 9,28\rangle\}$$ Now look for the ‘linked’ pairs, like $\langle 0,\color{brown}1\rangle$ and $\langle\color{brown}1,4\rangle$: transitivity says that when you have linked pairs like that in the relation, you must also have corresponding the ‘shortcut’ pair, in this case $\langle 1,4\rangle$. Here the linked pairs are: $$\begin{align} &\langle 0,1\rangle\quad\text{and}\quad\langle 1,4\rangle\;,\\ &\langle 1,4\rangle\quad\text{and}\quad\langle 4,13\rangle\;,\text{ and}\\ &\langle 2,7\rangle\quad\text{and}\quad\langle 7,22\rangle\;, \end{align}$$ so we have to add the shortcut pairs $\langle 0,4\rangle$, $\langle 1,13\rangle$, and $\langle 2,22\rangle$. Now repeat the process: for example, we now have the linked pairs $\langle 0,4\rangle$ and $\langle 4,13\rangle$, so we need to add $\langle 0,13\rangle$. When you finish a second pass, repeat the process again, if necessary, and keep repeating it until you have no linked pairs without their corresponding shortcut.
H: $(a+b)^p = a^p+b^p$ if $p$ prime and $a,b \in \mathbb{F}_p$ Can someone please explain why \begin{align} (a+b)^p = a^p+b^p \end{align} if $p$ is prime number and $a,b \in \mathbb{F}_p$ I tried to proof it that way \begin{align} (a+b)^p = \sum_{j=0}^{p}{p \choose j}a^{p-j}b^j = a^p+b^p + \sum_{j=1}^{p-1}{p \choose j}a^{p-j}b^j \end{align} but I can't figure out why $\sum_{j=1}^{p-1}{p \choose j}a^{p-j}b^j = 0$ Thanks in advance! AI: $${p\choose k} = \frac{p(p-1) \ldots (p-k+1)}{k!}$$ If $0<k<p$ all numbers in the denominator are less than $p$, so $p$ cant be reduces. Therefore ${p\choose k}\equiv 0\pmod p$ for $0<k<p$. So $$(a+b)^p = \sum_{k=0}^{p}a^kb^{p-k}{p\choose k} = a^p + b^p$$
H: On finding polynomials that approximate a function and its derivative (extensions of Stone-Weierstrass?) The Stone-Weierstrass Theorem tells us that we can approximate any continuous $f:\mathbb{R}^n\to\mathbb{R}$ arbitrary well on a compact subset of $\mathbb{R}^n$ by some polynomial. Suppose that $f$ is continuously differentiable, under what conditions can we guarantee we can find a polynomial $p$ such that $p$ approximates arbitrarily well $f$ and the partial derivatives approximate of $p$ arbitrarily well those of $f$ (on a compact subset of $\mathbb{R}^n$)? References are welcome. See below for my (probably naive) approach to the question. It is straightforward to find conditions if $n=1$: Let $K:=[a,b]\subseteq\mathbb{R}$ and let $K_+$ denote some open cover of $K$. Suppose that $f:K_+\to\mathbb{R}$ is continuously differentiable and let $f'$ denote its derivative. Then, by Weierstrass's approximation Theorem we have that there exists a polynomial $q$ such that $$\|\|f'(x)-q(x)\|\|\leq \varepsilon$$ for all $x\in K$. Choose some $y\in K$. For any $x\in K$ define $$p(x):=f(y)+\int_y^xq(t)dt.\quad\quad()$$ By the Fundamental Theorem of calculus we have that $p'=q$. So for any $x\in K$ $$\|\|f(x)-p(x)\|\|=\left\|\left\|\int_y^x (f'(t)-q(t))dt\right\|\right\|\leq \int_y^x \|\|f'(t)-q(t)\|\|dt\leq \varepsilon\|\|x-y\|\|\leq \varepsilon (a-b).$$ However, the above argument doesn't work in higher dimensions because the analogue of $()$ does not imply that $\nabla p = q$ (since not every vector of polynomials spans a gradient field). Can anyone suggest a way around this? That is, a way to to show that for continuously differentiable $f:\mathbb{R}^n\to\mathbb{R}$ and for any $\varepsilon>0$, we can always find a polynomial $p$ such that $$\|\|f(x)-p(x)\|\|+\|\|\nabla f(x)-\nabla p(x)\|\|\leq \varepsilon$$ for all $x$ in some compact set $K$? Feel free to add extra conditions on $K$, if it makes life easier. AI: If we assume a little more smoothness, we can use a trick similar to the one in this question. Let me take $n=2$ for concreteness. Suppose $f$ is $C^2$ on a neighborhood $U$ of $K$. With an appropriate cutoff function we can find a $C^2$, compactly supported $g$ which agrees with $f$ on a (possibly smaller) neighborhood of $K$. We can also find a large enough square $Q = (-M,M)$ which contains the support of $g$, so that $g$ and all its derivatives vanish on the boundary $\partial Q$. Fix $\epsilon > 0$. The second partial $\partial_x \partial_y g$ is continuous, so we can find a polynomial $p_0$ with $\|p_0 - \partial_x \partial_y g\| < \epsilon$ on $\bar{Q}$. Set $$p(x,y) = \int_{-M}^x \int_{-M}^y p_0(s,t)\,dt\,ds.$$ Then by the fundamental theorem of calculus and Fubini's theorem we have, for any $(x,y) \in Q$, $$\begin{align}\|p(x,y) - g(x,y)\| &= \left\|\int_{-M}^x \int_{-M}^y p_0(s,t) - \partial_x \partial_y g(s,t)\,ds\,dt\right\|\\ & \le \int_{-M}^x \int_{-M}^y \|p_0(s,t) - \partial_x \partial_y g(s,t)\|\,ds\,dt \\&\le (2M)^2 \epsilon \end{align}$$ and $$\begin{align}\|\partial_x p(x,y) - \partial_x g(x,y)\| &= \left\| \int_{-M}^y p_0(s,t) - \partial_x \partial_y g(s,t)\,dt\right\|\\ & \le \int_{-M}^y \|p_0(s,t) - \partial_x \partial_y g(s,t)\|\,dt \\&\le 2M \epsilon. \end{align}$$ A similar argument works for $\partial_y$. Since $f=g$ on a neighborhood of $K$, we have that $p$ and $f$, as well as their derivatives, are uniformly close on a neighborhood of $K$. In $\mathbb{R}^n$, this approach requires us to assume that $f$ is $C^n$.
H: Show that $\sqrt{2+\sqrt{2+\sqrt{2...}}}$ converges to 2 Consider the sequence defined by $a_1 = \sqrt{2}$, $a_2 = \sqrt{2 + \sqrt{2}}$, so that in general, $a_n = \sqrt{2 + a_{n - 1}}$ for $n > 1$. I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least upper bound of this sequence. But how? AI: Let $ x = \sqrt {2 + \sqrt {2 + \sqrt {2 + \cdots}}} $. Then, note that $$ x^2 = 2 + \sqrt {2 + \sqrt {2 + \cdots}} = 2 + x \implies x^2 - x - 2 = 0. $$Note that the two solutions to this equation are $x=2$ and $x=-1$, but since this square root cannot be negative, it must be $2$.
H: Simple question about parametric equations of a plane in 3D I'm quite rusty in Linear Algebra. If you have a plane in 3D with the equation $z=2$, what does $x$ and $y$ equal? Does $x=t$ and $y=t$? Because if I graph that in Wolfram Alpha, I don't get a horizontal plane in 3D at $z=2$: http://www.wolframalpha.com/input/?i=graph+z%3D2%2Cx%3Dt%2Cy%3Dt AI: $x, y$ can take on any value whatsoever, but they need not be equal. So if you insist on parameters for $x$ and $y$, you need separate parameters for each: $x = t, y = s$, with $t, s \in \mathbb R$. So the points on the plane are simply of the form $(x, y, 2)$. Note that it suffices to express your plane as simply $z = 2:$
H: Equivalence relation class $\bar{0}$ In the set $\mathbb{Z}$ we define the following relation: $$a\Re b \iff a\equiv \bmod2\text{ and }a\equiv \bmod3$$ 1)Prove that $\Re$ is an equivalence relation. (Done) 2) Describe the equivalence class $\bar{0}$. How many different equivalence classes exist? My thought on $\bar{0}$ is : $$\bar{0} = a \in \mathbb{Z} / a\Re b \implies a\equiv 0\bmod2\text{ and }a\equiv 0\bmod3$$ AI: Your thoughts?: exactly. Now, $a \equiv 0 \pmod 2 $ and $a \equiv 0 \pmod 3 \implies a \equiv 0 \pmod 6$. So, the equivalence class of $\bar{0}$ is equal to the set of all integer multiples of $6$: $$\bar{0} = \{6k\mid k\in \mathbb Z\}$$ Can you see that the equivalence classes of $\Re$ are the residue classes, modulo $6$?
H: Independence of random variables measure theory I wish to show that for two random variables $X$ and $Y$, the condition $P(X\leq x, Y\leq y ) = P(X\leq x)P(Y\leq y)$ implies that X and Y are independent. I am approaching this problem from a measure theoretic perspective. So in particular I can write that $P(X\leq x) = \mu_X ((-\infty, x])$ and $P(Y\leq y) = \mu_Y ((-\infty, y])$. Also the independence condition here is that $\sigma(X)$ and $\sigma(Y)$ are independent, where these are the sigma algebras generated by the random variables. AI: Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $\mathbb{R}$. Show that $\sigma(X) = \{X^{-1}(B) : B \in \mathcal{B}\}$. ($\supset$ is clear. For $\subset$, show the right side is a $\sigma$-algebra.) Fix $y \in \mathbb{R}$. Let $\mathcal{L} = \{ B \in \mathcal{B} : P(X \in B, Y \le y) = P(X \in B) P(Y \le y)\}$. Show that $\mathcal{L}$ is a $\lambda$-system. Let $\mathcal{P} = \{ (-\infty, x] : x \in \mathbb{R} \}$. Show that $\mathcal{P}$ is a $\pi$-system, $\mathcal{P} \subset \mathcal{L}$ and that $\sigma(\mathcal{P}) = \mathcal{B}$. By the $\pi$-$\lambda$ theorem, $\mathcal{B} \subset \mathcal{L}$. Conclude that for every $B \in \mathcal{B}$ and $y \in Y$, $P(X \in B, Y \le y) = P(X \in B) P(Y \le y)$. Fix $B \in \mathcal{B}$. Let $\mathcal{L}' = \{C \in \mathcal{B} : P(X \in B, Y \in C) = P(X \in B) P(Y \in C)\}$. Show that $\mathcal{L}'$ is a $\lambda$-system, and proceed as before to show $\mathcal{B} \subset \mathcal{L}'$. We have now shown that $P(X \in B, Y \in C) = P(X \in B) P(Y \in C)$ for all $B, C \in \mathcal{B}$. By (1), this says that $\sigma(X)$ and $\sigma(Y)$ are independent.
H: Prove $\lim\{s_n\}=+\infty\iff\lim\left(\frac{1}{\{s_n\}}\right)=0$ Prove the following: Given that $\{s_n\}$is a sequence of positive numbers. Then $$\lim s_n=+\infty\iff\lim\left(\frac{1}{s_n}\right)=0$$ My attempt at proving this: For $\lim s_n=+\infty\implies \lim\left(\dfrac{1}{s_n}\right)=0$, suppose $\lim s_n=+\infty.$ Given any $\epsilon>0$, let $K=\frac{1}{\epsilon}$. Then there is a $N\in\mathbb{N}\ni n \geq N\implies s_n>K=\frac{1}{\epsilon}$. Since each of the $s_n$'s are positive, $\mid \dfrac{1}{s_n}\mid<\epsilon$, for $n\geq N.$ Therefore $\lim(\dfrac{1}{s_n})=0.$ Regarding the converse I'm having a bit more trouble; for $\lim\left(\dfrac{1}{s_n}\right)=0\implies\lim s_n=+\infty$. Suppose that $\lim \left(\dfrac{1}{s_n}\right)=0$. Given any $\epsilon>0$, let $M=\epsilon$. Then there is $N\in\mathbb{N}\ni n\geq N \implies \|s_n-\infty\|<\epsilon$. Of course this makes no sense, as you can't subtract $\infty$, so I'm not sure how to proceed. Any hints/tips would be appreciated. Thanks in advance. AI: What you have to show is the following: For every $K\in\mathbb{N}$ there is a $N\in\mathbb{N}$ such that $s_n\ge K$ for all $n\ge N$. This means that your sequence $\{s_n\}$ is converging to $+\infty$. So let $K\in\mathbb{N}$ given. By assumption there is a $N\in\mathbb{N}$ such that for all $n\ge N$: $\frac{1}{s_n}\le\frac{1}{K}\iff K\le s_n$ for all $n\ge N$ and you are done. Note, the last equivalence $\frac{1}{s_n}\le\frac{1}{K}\iff K\le s_n$ is the reason why your claim is true.
H: Dimension of a vector space-regarding I want to find the dimension of the vector space $V=\{u\in \mathbb{R}^3:Mu^{t}=u^{t}\}$, where $M=\begin{pmatrix} 1&0&0\\ 0&\cos \theta& -\sin \theta\\ 0& \sin \theta& \cos \theta\end{pmatrix}$, $0<\theta<\dfrac{\pi}{2}$. I feel no vector $u$ except $(0,0,0)$ will satisfy this. Thus the dimension of $V$ is $0$. Am I right? AI: No, you forgot the vector $(1,0,0)$. Notice that $M$ is a rotation of $\theta$ radians around the $x$-axis, thus that is the unique eigenvector (up to scalar multiples).
H: cardinality of $S_{\mathbb{N}}$ When I am proving something, I got a doubt. what is the cardinality of $S_{\mathbb{N}}$, the set of all bijections from $\mathbb{N}$ to $\mathbb{N}$? I hope it is countable, because that will make my life easier. Thanks in Advance. AI: It is not countable: it is $2^{\aleph_0}=\mathfrak{c}$, the cardinality of $\Bbb R$, the real line, and of $\wp(\Bbb N)$. To see that it is at least this big, let $A$ be any subset of $\Bbb N$. We can define a bijection $f_A\in S_{\Bbb N}$ as follows: $$f_A:\Bbb N\to\Bbb N:n\mapsto\begin{cases} n+1,&\text{if }n\in 2A\\ n-1,&\text{if }n\in 2A+1\\ n,&\text{otherwise}\;. \end{cases}$$ Here $2A=\{2n:n\in A\}$, and $2A+1=\{2n+1:n\in A\}$. As an example, if $A=\{1,4\}$, then $2A=\{2,8\}$, $2A+1=\{3,9\}$, and $f_A$ is the bijection that interchanges $2$ and $3$ and interchanges $8$ and $9$ while leaving every other natural number alone. It’s not hard to check that if $A,B\subseteq\Bbb N$, and $A\ne B$, then $f_A\ne f_B$.
H: How to convert parametric form to a single algebraic equation? I'm pretty sure this is impossible to do but here is my attempt. Parametric form: $$x=1+t\\y=2+2t\\z=3+3t$$ Attempt: $$(x,y,z)=(1+t,2+2t,3+3t)$$ That didn't really get me anywhere, so here I tried to put it into symmetric form: $$x-1=\frac{y-2}2=\frac{z-3}3$$ Basically, I'm trying to get it into something like this: $$z^2+xy-2x-y^2=1$$ I am aware that represents a plane but I'm trying to get an equation to represent a line. If this is not possible, what can an equation represent? Only planes? Is there any easy way to tell what an equation represents from just looking at it if it can represent more than one thing? Also, bonus question: I am trying to put the equation $z=2$ into a function. Is $F(x,y,z)=z-2$ correct? My graph of F doesn't match z=2: http://www.wolframalpha.com/input/?i=graph+F%28x%2Cy%2Cz%29%3Dz-2 AI: An equation $L=R$ can be rewritten as $L-R=0$. Suppose now that you have equations $L_1=R_1$, $L_2=R_2$, etc. Well, you can rewrite them all and then pull a rabbit out of your hat: $$(L_1-R_1)^2+(L_2-R_2)^2+\cdots+(L_n-R_n)^2=0.$$ This sort of thing is not done terribly often, I don't think, because it is usually much easier to manipulate a system of simple equations than one somewhat monstrous one, but there are probably some applications for it I don't know about. Note: this technique does not work for complex numbers. It's a totally ordered ring only sort of thing.
H: Combinatorial Proof Of A Number Theory Theorem--Confusion I came across a combinatorial proof of the Fermat's Little Theorem which states that If $p$ is a prime number then the number ($a$$p$-$a$) is a multiple of $p$ for any natural number $a$. Let me write down the proof. PROOF We have pearls of $a$ colours . From these we make necklaces of exactly $p$ pearls . First,we make a string of pearls . There are $a$$p$ different strings.If we throw away the $a$ one one -coloured pearls ($a$$p$-$a$) strings will remain.We connect the ends of each string to get necklaces.We find that two strings that differ only by a cyclic permutation of its pearls result in indistinguishable necklaces.But there are $p$ cyclic permutations of $p$ pearls on a string . Hence the number of distinct necklaces is [($a$$p$-$a$)/$p$]. Because of this interpretation this is an integer.Thus ($a$$p$-$a$) is a multiple of $p$ for any natural number $a$. [HENCE PROVED] MY CONFUSIONS What is the use of the fact that $p$ is a prime in this proof? If this proof is true then why is this not true for every positve integer $p$? AI: The crux lies in the statement But there are p cyclic permutations of p pearls on a string. We require the fact that $p$ is prime to establish that all $p$ cyclic permutations give distinct strings. Consider a string with colours $a_0a_1 \ldots a_{p-1}$. The cyclic permutations are $a_ia_{i+1} \ldots a_{i-1}$ where $0 \leq i \leq p-1$. (Note: Indices taken $\pmod{p}$) If any $2$ cyclic permutations $a_ia_{i+1} \ldots a_{i-1}$ and $a_ja_{j+1} \ldots a_{j-1}$ give the same string, where $i \not \equiv j \pmod{p}$, then $a_i=a_j, a_{i+1}=a_{j+1}, \ldots , a_{i-1}=a_{j-1}$, i.e. $a_{i+k}=a_{j+k}$ for $0 \leq k \leq p-1$. So $a_n=a_{n+(j-i)}$ for all $n$. Thus $a_0=a_{j-i}=a_{2(j-i)}= \ldots =a_{(p-1)(j-i)}$. Now, since $p \nmid j-i$ and $p$ is prime, $(j-i, p)=1$. Thus $0, (j-i), 2(j-i), \ldots, (p-1)(j-i)$ form a complete residue system $\pmod{p}$, so $a_0=a_1= \ldots =a_{p-1}$. But we have excluded all strings with one colour, so we get a contradiction. Thus when $p$ is prime, we are able to establish that all $p$ cyclic permutations are distinct. Example for how this fails when $p$ is not prime: On the other hand, let us consider $p=4$ as an example of when $p$ is not prime. We could have a string $1212$, and the cyclic permutations are only $1212$ and $2121$. If we rotate it again we get $1212$ once again. Thus there are only 2 cyclic permutations of this string and not $p=4$ permutations.
H: Taking "Absolute Value Operator" as a common factor? If I have an equation like this and Im trying to solve for X \|x\| + 4\|x\| = 40 Can I take the absolute Value (Modulus) as a common factor? \|x + 4x\| = 40 and the proceed to solve for X? AI: In general $\|a\|+\|b\|\ne\|a+b\|$ See here, for the general relation We can write $$\|x\|(1+4)=40\implies 5\|x\|=40$$ Things become more clear if we substitute $\|x\|$ with $y$
H: Why does $e$ seem to be an intuitive number? I often find two numbers roughly "in the same ballpark" if they are within a factor of about $e$ of each other. For example, if I know computers generally cost upward of $\$1000$, then $\$2700$ would probably be the most I would a priori feel is a reasonable upper limit for computer prices. Is this just a coincidence, or is there some mathematical deep reason why $e$ is inherently the "best" order-of-magnitude exponent? AI: I think that we subconsciously think in a logarithmic way and since the root of 10 is around 3.162, we think of 3 as half way the order of magnitude. So anything under 3,000 we perceive in the order of magnitude of 1,000 and anything over that we perceive as being in the 10,000 order of magnitude. In other words if 1,000 is 3 in 10 based logarithm and 10,000 is 4 in 10 based logarithm then 3,162 is the 3.5 on this scale and anything under 3,000 is closer to 3 and almost everything over 3,000 is closer to 4.
H: Normal Subgroups and their Qualities Hope all you are healthy and in peace Suppose N is a normal subgroup of G. If every subgroup of N is normal also in G, could we deduce that centralizer of N in N is N itself ? AI: Short answer: no. Suppose $G$ is the Quaternion group and $N = G$. Then $N$ is normal in $G$. Every subgroup of $G$ is also normal in $G$. But $G$ is not abelian: its centralizer in itself (i.e. its center) is $\{-1, 1\}$.
H: Convergence of a function and interval topology Let $(X,\leq)$ be a well-ordered set that contains exactly one element $x$ such that $y<x$ for uncountably many values of $y\in X$. Let $f(x)=1$ and $f(y)=0$ for all other values of $y\in X$. For the interval topology $\tau$ on $X$, show that for every sequence $u_n\to u$ in $X$, $f(u_n)\to f(u)$. This is what I've tried. Let $u=x$ and $u_n\to u$. Let $B$ be a neighborhood of $f(u)=1$. It is enough to consider $B=\{1\}$. We need to show that there exists an $m$ such that $u_n=x$ for all $n\geq m$. Suppose there exists a subsequence $(u_{n_k})$ such that $u_{n_k}<x$ for all $k$. Then for each $k$ there exist uncountably many $w_k$ such that $u_{n_k}< w_k<x$, but I don't know how to proceed from here. AI: HINT: Suppose that $\langle y_n:n\in\Bbb N\rangle$ is a sequence such that $y_n<x$ for all $n\in\Bbb N$. For each $n\in\Bbb N$ let $A_n=\{y\in X:y<y_n\}$. Each $A_n$ is countable, so $A=\bigcup_{n\in\Bbb N}A_n$ is countable, and $X\setminus A\ne\varnothing$. Let $y=\min(X\setminus A)$. Show that $y<x$, and use this to show that $\langle y_n:n\in\Bbb N\rangle$ does not converge to $x$.
H: Meaninig of a symbol at the Circle Group I have $U_{20}$ (at the meainig of the Circle Group), What is the meaning of $W_{20}^{8}$?? What is the 20 and what is the 8? Thank you! AI: Assuming you mean $W = \omega$, then I suspect $\;\omega^8_{20}\;$ denotes the $8$th root of unity in $U_{20}= \left\{\omega^k_{20} = e^{\large\frac{ik2\pi}{20}}\mid\in \mathbb Z, 1 \leq k \leq 20\right\}$, where $\omega$ denotes the element $e^{\large\frac{2i\pi}{20}} = e^{\large \frac{i\pi}{10}}\in U_{20}$ that generates all of $U_{20}$. So $\omega^8_{20}$ would the element $e^{\large\frac{8i\pi}{10}} = e^{4i\pi/5}$.
H: Rate of change of radius and volume They put a gas bubble in someone's eye. The volume of a gas bubble changes from $0.4$ $cc$ to $1.6$ $cc$ in $74$ hours. Assuming that the rate of change of the radius is constant, find (a) The rate at which the radius changes; (b) The rate at which the volume of the bubble is increasing at any volume $V$; (c) The rate at which the volume is increasing when the volume is $1$ $cc$. (Note: The volume of a ball of radius $r$ is $\frac{4}{3}\pi r^3$. Assume the bubble is spherical.) Explanation would be appreciated. I did differentiate the $\frac{4}{3}\pi r^3$ with respect to r that is $4\pi r^2$ and then made that equal to $1.66-06$ which is the rate change of $V$. But I Don't know if I am doing it right. AI: $v=\frac{4\pi r^3}{3}$ or $r=(3v/4\pi)^{1/3}$ $\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$ substitute for $r$ we get and letting $dr/dt=K$ (constant) then $v^{-2/3}dv=4\pi K(3/(4\pi))^{2/3} dt$ $3v^{1/3} = 4\pi K(3/(4\pi))^{2/3} dt$ Integrating between limits $0.4$ and $1.6$ for V and $t=0$ to $74$, we get $K= \frac{1.6^{1/3}-.4^{1/3}}{74 (4\pi/3)^{1/3}}$ which works out to $3.628E-3 cm/hr$ as the rate of change of radius. b) $dv/dt=(36\pi)^{1/3} KV^{2/3}$ c) substitute $v=1$ in above we get $(36\pi)^{1/3} K$
H: Balls and bins with 2 balls and 2 bins I'm trying to understand a proof in a paper I'm reading. It relies on a balls and bins problem. Here is what I'm trying to figure out: We want the maximum number of balls in a bin. We have 2 balls and 2 bins. We have a lower bound of 3/2. How do I see that the lower bound is 3/2? Here is the context: To compute the social cost of the equilibrium we see this as the problem of throwing m balls into m bins. The social cost of the equilibrium is equal to the expected maximum number of balls in a bin which is well known to be (log m/ log log m) Given that the optimal solution has cost 1, the lower bound follows. For m = 2, this gives a lower bound of 3/2. (taken from Section 3 of http://cgi.di.uoa.gr/~elias/publications/paper-kp09.pdf) AI: There is a 50% chance that there are two balls in one bin and 50% there are one ball in each bin. The maximum of the first case is 2; the maximum of the second is 1. $$.52+.51 = \frac{3}{2}$$ The above explains why the maximum for a 2x2 bin is what it is if the likely hood is 50:50. If it is p: $$p^22+2p(1-p)1+2*(1-p)^2 = 2(p+(1-p))^2-2p(1-p) = 2-2p(1-p)$$ $p(1-p)$ between 0 and 1 has a maximum at $p=.5$ of $.5(1-.5)=.25$ Ergo: the minimum of the problem is 1.5
H: The set $F_1\subset X_1$ is closed set in $X_1$ if and only if there is an closed set $F$ in $X$ such that $F\cap X_1=F_1$ I need th proving this theorem: Let $(X, d)$ is metric space and let $(X_1, d_1)$ is its subspace. The set $G_1\subset X_1$ is open set in $X_1$ if and only if there is an open set $G$ in $X$ such that $G\cap X_1=G_1$. The set $F_1\subset X_1$ is closed set in $X_1$ if and only if there is an closed set $F$ in $X$ such that $F\cap X_1=F_1$. Prove: First part of the theorem I've proved i.e. (The set $G_1\subset X_1$ is open set in $X_1$ if and only if there is an open set $G$ in $X$ such that $G\cap X_1=G_1$.) But I do not know how to prove the second part of the theorem (The set $F_1\subset X_1$ is closed set in $X_1$ if and only if there is an closed set $F$ in $X$ such that $F\cap X_1=F_1$). Please if anyone can help me to prove the second part of the theorem, thank you, for your help and your attention AI: So $\;F_1\subset X\;$ is closed in $\;X_1\;$ iff $\;X\setminus F_1\;$ is open iff there exists open $\;G\subset X\;$ s.t. $$(X\setminus F_1)\cap G\;\;\text{is open}\iff X\setminus\left((X\setminus F_1)\cap G\right)\stackrel{\text{de Morgan}}=\left(F_1\right)\cup\left(X\setminus G\right)\;\;\text{is closed}$$ (since $\;X\setminus(X\setminus F_1)=F_1\;$ ) , and we're done since trivially $$F_1=F_1\cap\left[\left(F_1\right)\cup\left(X\setminus G\right)\right] $$
H: Proving A Trigonometric Identity- Double Angles $(\cos(2x)-\sin(2x))(\sin(2x)+\cos(2x)) = \cos(4x)$ I'm trying to prove that the left side equals the right side. I'm just stuck on which double angle formula of cosine to use. AI: $$(\cos(2x)-\sin(2x))(\sin(2x)+\cos(2x))=(\cos^2(2x)-\sin^2(2x)) = \cos(4x)$$ From $$\cos(a+b)=\cos a \cos b-\sin a\sin b$$ if $a=2x,b=2x$ then $$\cos(4x)=\cos^22x-\sin^22x$$
H: Does a point lie on a line with a parametric equation Does the point $(0, 5, 5)$ line on the line with the parametric equations: $x = 3 - t\\y = 2 + t\\z = 2 + 2t$ This is the first time I see one of these, right now I assume it is as simple as solving $t$ and plugging it into the equations as such if $x = 0$ then $t = 3$ then I plug this and get $y = 2 + 3 = 5\\z = 2+2(3) = 8$ hence $(0, 5, 5)$ is not on the line. Is this the correct way to solve such a problem? AI: That's correct! The corresponding system (using the values for $x, y, z$ in your "test point") is inconsistent, so the "test point" fails to lie on given line. In particular, note that there is no value for $t$ that satisfies both $2 + t = 5$ and $2 + 2t = 5$.
H: Easy Probability Question (Independent Events) Suppose I have two doors. One of them has a probability of $1/9$ to contain X, the other has a probability of $2/3$ to contain X. Then, supposing I pick randomly one of the two doors, what is the probability that it contains X? (If one contains X, the other can also contain X. They are independent but not mutually exclusive.) I'm not sure what the solution is - is it just the average of the probabilities? I need this as a stepping stone in a larger argument. Thanks. AI: Hint: Let $E$ be the event that the door you open contains $X$. Assuming that you must choose either door $1$ or door $2$, but not both: $$ P(E) = P(E~\|~\text{choose door } 1)P(\text{choose door} 1) + P(E~\|~\text{choose door } 2)P(\text{choose door} 2) $$
H: Proving that $\sum_p\frac{1}{p+1}$ diverges How does one prove $$\sum_{p\in\Bbb P}\frac1{p+1}=\infty.$$ Where $\Bbb P$ denotes the set of prime numbers. I have attempted forming an inequality by playing around with Euler's work on the reciprocals of primes. Robjohn showed me an inequality in chat that I do not understand and I was wondering if there was another way to do this. The work I have done is inconclusive and is nothing more than stating $$\sum_{p\in\Bbb P}\frac1p\ge\sum_{p\in\Bbb P}\frac1{p+1}$$ AI: $$ \sum \frac{1}{p+p} < \sum \frac{1}{p+1} \implies \frac{1}{2} \sum \frac{1}{p} < \sum \frac{1}{p+1}. $$
H: Trigonometry- Double Angle Trigonometric Equations $$\tan(3x)=((\tan x(3-\tan^2x))/(1-3\tan^2x))$$ I'm just having a hard time seeing where the double angle formula fits in with verifying this equation. AI: Using a Adicional formula: $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y},$$ we have: $$\tan(2x+x)=\frac{\tan 2x+\tan x}{1-\tan 2x\tan x}=\frac{\tan (x+x)+\tan x}{1-\tan (x+x) \tan x}=\frac{\frac{\tan x + \tan x}{1-\tan x \tan x } +\tan x}{1-\frac{\tan x + \tan x}{1-\tan x \tan x } \tan x}$$ $$=\frac{\tan x +\frac{2\tan x}{1-\tan^2 x}}{1-\tan x \frac{2\tan x}{1-\tan^2 x}}=\frac{\frac{\tan x -\tan^3 x + 2\tan x}{1-\tan^2 x}}{\frac{1-\tan^2 x-2\tan^2 x}{1-\tan^2 x}}$$ $$=\frac{{\tan x -\tan^3 x + 2\tan x}}{{1-\tan^2 x-2\tan^2 x}}=\frac{{3\tan x -\tan^3 x }}{{1-3\tan^2 x}}=\frac{\tan x (3-\tan^2 x)}{1-3\tan^2 x}$$
H: The most complete reference for identities and special values for polylogarithm and polygamma functions I am looking for a book, paper, web site, etc. (or several ones) containing the most complete list of identities and special values for the polylogarithm $\operatorname{Li}_s(z)$ and polygamma $\psi^{(a)}(z)$ functions, including generalizations to negative, non-integer and complex orders and arguments. As far as I know, there is an ongoing research in this area, and many new and curious identities occur from time to time. Here is the list of resources I found so far: Dilogarithm at Wikipedia Dilogarithm at MathWorld PolyLog at The Wolfram Functions Site Trilogarithm at MathWorld Polylogarithm at Wikipedia Digamma Function at Wikipedia Digamma Function at MathWorld Trigamma Function at MathWorld PolyGamma at The Wolfram Functions Site AI: DLMF cites Lewin (1981), Kölbig (1986), Maximon (2003), Prudnikov et al. which can be found in the bibliography of Dlmf.
H: Probability that $a^2 \equiv 1 \pmod{10}$ when $a$ is chosen randomly from a set Out of the set $\{1,2,...,n\}$ we choose randomly a number $a$. Find the probability $p_n$ that $a^2 \equiv 1 \pmod{10}$, and find $\lim_{n \to \infty} {p_n}$ Ideas anyone? AI: All squares mod $10$ end with a $0,1,4,5,6,9$. We know that the only numbers that when squared end with a $1$ are numbers that end in $1$ and $9$. So this question is basically asking for "What is the probability that a number in a given set ends with a $1$ or a $9$?"
H: Is this equation with two unknowns solvable? Can this equation be solved? If so how? I would like to find both $X$ and $Z$ . $4.33=\dfrac{0.4397-Z}{X-0.4397}$ where $Z$ is known to be in the range of $0.1931$ to $0.2352$ and $X$ is known to be in the range of $0.3549$ to $0.5576$ Ps. I am new here so if I could phrase my title better or format my question better please tell me. Also I don't know what type of equation this would be, that is if it is a legitimate equation, so I may have tagged it incorrectly. Kind Regards AI: In general, it is not possible to solve for both variables when there are two variables in a single equation (certain rare cases do exist though, such as in the equation, $x^2+y^2=0$, where $x$ and $y$ are both real). In more generality, it is typically not possible to solve for all the variables in a system of $m$ equations in $n$ variables if $n > m$. Hope this helps!
H: Prove that $\{z\in \mathbb{C}, \|z\|=1\}\cong\mathbb{R}/\mathbb{Z}$ How to prove that $T=\{z\in \mathbb{C}, \|z\|=1\}$, $T\cong\mathbb{R}/\mathbb{Z}$? AI: Let $\phi: \mathbb{R} \to \mathbb{C}$ be given by $\phi(t) = e^{2 \pi i t}$. Suppose $\phi(s)=\phi(t)$, what does that say about the relationship between $s$ and $t$?
H: How can the derivative of arc length be anything except zero? If a and b are constants, then the above definite integral (arc length) has to be some constant. How can the derivative of s be anything except zero (this is contradicted in the blue box)? This method of evaluation is for line integrals by the way. AI: Pick a fixed starting time $t_0$, then you may define the arc length travelled after a time $t$ and denote it by $s(t)$. This is not constant in $t$, as you travel along the curve, the travelled distance increases. So $\frac{\mathrm ds}{\mathrm dt}$ is the velocity at a certain time.
H: Showing that $\mathcal{M} \preccurlyeq \mathcal{N} \implies \mathcal{M} \equiv \mathcal{N}$. Suppose that $\mathcal{M} \preccurlyeq \mathcal{N}$. Then by definition we have that $\mathcal{M}$ is a substructure of $\mathcal{N}$ s.t. for any (possibly empty) tuple $\overline{a}$ from $M^n$ and for any $\phi \in L$, we have that $\mathcal{N} \models \phi(\overline{a})$ implies that $\mathcal{M} \models \phi(\overline{a})$. Now we aim to show that $\mathcal{M} \equiv \mathcal{N}$. This is true iff $Th(\mathcal{M}) = Th(\mathcal{N})$. I can only think of how to show that $Th(\mathcal{M}) \supseteq Th(\mathcal{N})$. This is done as follows: let $\phi \in Th(\mathcal{N})$. Then for any tuple $\overline{a} \in M^n$, we have that $\mathcal{N} \models \phi(\overline{a})$. Among these tuples are the empty tuple, so that we can say $$ \mathcal{N} \models \phi() $$ so that of course $$ \mathcal{N} \models \phi $$ and hence applying the definition of $\preccurlyeq$ we get: $$ \mathcal{M} \models \phi() \implies \mathcal{M} \models \phi $$ Then we've established that $Th(\mathcal{N}) \subseteq Th(\mathcal{M})$. But how do we show that $Th(\mathcal{M}) \subseteq Th(\mathcal{N})$? AI: Suppose $\phi\in\text{Th}(\mathcal{M})$. Then $\lnot\phi\not\in\text{Th}(\mathcal{M})$. So by your argument, $\lnot\phi\not\in\text{Th}(\mathcal{N})$, and hence $\phi\in\text{Th}(\mathcal{N})$. More generally, if $T$ and $T'$ are complete theories, then $T\subseteq T'$ implies $T = T'$. Comment: The definition of $\preccurlyeq$ is often taken to be $\mathcal{N}\models \phi(\overline{a})$ if and only if $\mathcal{M}\models\phi(\overline{a})$, for $\overline{a}$ from $\mathcal{M}$. This is equivalent, of course.
H: When does $\mathbb E_\mathbb P[X]=0$ imply $\mathbb E_\mathbb P[X\mid\mathcal E]= 0$ $\mathbb P$-a.s. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, $\mathcal E\subseteq\mathcal F$ a sub-$\sigma$-algebra, $X:\Omega\rightarrow\mathbb R^d$ a $\mathcal F$-measurable map. My question is, when does $\mathbb E_\mathbb P[X]=0$ imply that $\mathbb E_\mathbb P[X\mid\mathcal E]= 0$ $\mathbb P$-a.s. ? Clearly, if $\mathcal E$ is trivial, then this implication holds. But what are other necessary and sufficient conditions on $\mathbb P$ and/or $\mathcal E$ in order that this implication holds? For example, does the implication hold if we assume that $\mathbb P$ is of the form $\mathbb P=\sum_{i=1}^Na_i\delta_{\omega_i}$ for $a_i\in (0,1]$, $N\in\mathbb N\cup\{\infty\}$? AI: Assume that $\mathbb P$ and $\cal E$ are such that the implication holds. If $X$ is $\cal E$-measurable, define $X':=X-\mathbb E(X)$. Then we should ahve $$\mathbb E_{\mathbb P}[X'\mid \mathcal E]=0,$$ hence $X$ is $\mathbb P$-almost surely constant. Therefore, considering characteristic functions, $\mathbb P(E)\in\{0,1\}$ for each $E\in\mathcal E$.
H: open cover and boundedness Use the open cover characterization of compactness to prove that if $f:[a,b]→X$ is a continuous function and $X$ is a metric space, the $f$ is bounded. Proof: To show $f$ is bounded we must show that $f([a,b])$ is a bounded subset of $X$. So if $f([a,b]) \subseteq X$ then $[a,b]$ is compact if and only if $[a,b]$ is a closed and bounded subset of $\mathbb{R}$. Let $\{x_{n}\}$ be a sequence of elements of $[a,b]$. By the Bolzano-Weierstrass Theorem $\{x_n\}$ contains a convergent sub-sequence with limit $x \in \mathbb{R}$. Since $[a,b]$ is closed, $x \in [a,b]$. Thus, $[a,b]$ is compact. Now we must show $f([a,b])$ is compact in $X$. Suppose $f$ is continuous and $X$ is a metric space. Let $\{U_{x}:x \in \mathscr{U} \}$ be an open cover of $f(Y)$ where $Y = [a,b]$. For each point $y \in Y$, $f(y) \in f(Y)$ so $f(y) \in U_{x_{y}}$ for some $x_{y} \in \mathscr{U}$, that is $y \in f^{-1}(U_{x_{y}})$. Thus, the collection $\{f^{-1}(U_{x}): x \in \mathscr{U}$ is a cover of $Y$. Since $f$ is continuous each set $f^{-1}(U_{x})$ is open in $Y$, so $\{ f^{-1}(U_{x}):x \in \mathscr{U} \}$ is an open cover of $Y$. Since $Y$ is compact, it has a finite subcover of $\{f^{-1}(U_{x_{}}),...f^{-1}(U_{x_{n}})\}$. Given $x \in f(Y)$, there exists $y\in Y$ such that $f(y)=x$, and $y \in f^{-1}(U_{x_{i}})$ for some $i$, so $x \in U_{x_{i}}$. Thus, $\{U_{x},....,U_{x_{n}}\}$ is a finite subcover of $f(Y)$, so $f(Y)$ is compact. So by definition, since $f(Y)$ is compact then $f$ is bounded. I was told this was right but I was missing how to show why $f(Y)$ is compact. Any ideas how to clear up what I'm missing? AI: This seems very complicated. Let $y \in X$. The collection of balls $B(y,n)$ is an open cover of $X$. Hence $f^{-1}(B(y,n)\}$ forms an open cover of $[a,b]$ since $f$ is continuous. Since there must be a finite subcover, choose $n^$ to be the largest $n$ in all the balls. Then $f(x) \in B(y,n^)$ for all $x \in X$. Hence $f$ is bounded. Addendum: To show why $f(Y)$ is compact: Suppose $\{ U_\alpha \}$ is an open cover of $f(Y)$. Then $\{ f^{-1}(U_\alpha) \}$ is an open cover of $Y$ because $f$ is continuous (this is the key step here). Since $Y$ is compact, there is a finite subcover $f^{-1}(U_{\alpha_1}),...,f^{-1}(U_{\alpha_n})$. That is, $Y = f^{-1}(U_{\alpha_1}) \cup \cdots \cup f^{-1}(U_{\alpha_n})$. Hence $f(Y) \subset U_{\alpha_1} \cup \cdots \cup U_{\alpha_n}$, that is, $U_{\alpha_1}, ..., U_{\alpha_n}$ is a finite cover of $f(Y)$. Hence every open cover of $f(Y)$ has a finite subcover.
H: Expectation throwing balls into boxes I was thinking about this situation: Suppose there are $n$ boxes. In each box we randomly throw one of the balls numbered $1,2,\ldots,k$, independently of other boxes. Let $X$ be the number of boxes with ball number $1$. What is $E[X]$, and what is $E[\dfrac{1}{X}\mid X>0]$? It seems that there should be "roughly" $n/k$ boxes with ball number $1$, so $E[X]=n/k$, and $E[\dfrac{1}{X}\mid X>0]=k/n$. But is it right? Edit: I've asked a new question here to make it direct to the point. AI: You're right about $E[X]$ but wrong about $E[1/X]$. Since $X$ can take the value $0$ with nonzero probability, $E[1/X]$ is undefined. Added after OP's edit: Even if you condition on $X$ being positive, $k/n$ can't be the right answer in general. Since $X$ is the number of boxes with ball number 1, it's a whole number and consequently $1/X\le1$ if $X\gt0$, which means the expected value can't exceed $1$. But $k/n\gt1$ if $k\gt n$.
H: Random variables $X$ and $Y$ that are not independent I am trying to solve a stat question, here is the question: Give an example of two random variables $X$ and $Y$, each taking values in the set $\{1,2,3\}$ such that $P(X = 1; Y = 1) = P(X = 1)P(Y = 1)$, but $X$ and $Y$ are not independent. A friend says that the answer is: $$P(X=1,Y=1)=P(X=1,Y=2)=P(X=2,Y=1)=P(X=3,Y=3)=\frac{1}{4}$$ I don't quite get why that is the case. Where did $1/4$ come from? Thanks AI: Observe that $$P(X=1)=P(X=1,Y=1)+P(X=1,Y=2)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$ and $$P(Y=1)=P(X=1,Y=1)+P(X=2,Y=1)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$$ Furthermore, $$P(X=1;Y=1)=\frac{1}{4},$$ so the first criterion is satisfied. Also, since $X=3$ implies $Y=3$, it is obvious that $X$ and $Y$ are not independent in the given answer. The value $1/4$ was chosen so that the sum of all probabilities is $1$.
H: Compute $\langle 5 \rangle$ in integers Compute $\langle 5 \rangle$ in integers. I thought the answer would have been $$\{5^n\| \text{ for $n$ in integers }\}$$ However my teacher has marked it $$\{5*n\|\text{ for $n$ in integers }\}$$ What have I done wrong? AI: The integers do not form a group under multiplication, for almost no element has an inverse under multiplication. The integers do form a group under addition, however. So if we're viewing $\Bbb{Z}$ as a cyclic group in the usual sense, the operation is addition and $$\langle 5 \rangle = \{..., -5-5, -5, 0, 5, 5 + 5, 5 + 5 + 5 , ...\} = \{5n : n \in \Bbb{Z}\}$$
H: Computing $E[1/X]$ for binomial $X$ Let $X$ be a binomial distributed with parameters $n$ and $p=1/k$. Then we have $E[X]=n/k$. But what is $E[\dfrac{1}{X}\mid X>0]$? Is there a nice closed formula getting the exact value or approximation? AI: I don't know a closed formula for this, but you could simply calculate it as $E[\frac{1}{X}\|X>0]=\sum\limits_{i=1}^n i^{-1}\frac{B(i;n,p)}{1-B(0;n,p)}$
H: eigenvalues of $AB$ are eigenvalues of $\sqrt{B} A \sqrt{B}$ Suppose $A,B$ are symmetric positive definite matrices. An author claims that the spectrum of $AB$ is the spectrum of $\sqrt{B}A\sqrt{B}$. Why? Certainly they have the same trace by cyclic permutation. And they have the same determinant by multiplicativity. Ideas: Maybe use the minimax principle? AI: Notice that $$AB = B^{-1/2} (B^{1/2} A B^{1/2}) B^{1/2},$$ so these two are similar.
H: Line integrals giving different values depending on what it is integrated against? For example, take a look at the solution to 4: Considering the interval is always the same (from x=1 to x=8), how can it give different values depending on what you integrate against? AI: It's probably easiest to explain it starting with $ds$: Given a curve, $C$, and point function $G(x,y)$, $ds$ is a differential distance or "line element" along the curve, as I'm sure you're aware, whereas $dx$ and $dy$ are projections of that line element onto the $x$ and $y$ axis, respectively. In vector form, if $d\vec{s}$ is the tangent vector at the points $(x,y)$, then $dx=d\vec{s}\cdot \vec{i}, dy=d\vec{s}\cdot \vec{j}$. You are basically only adding up one component of the motion in the $dx,dy$ cases. Applying this to the point function $G(x,y)$, you are projecting the derivative $\frac{d}{ds}G(x,y)$ onto the x and y axes: $\frac{dG_C}{dx}\|_{(x,y)} = [\frac{dG}{ds}\frac{ds}{dx}]_{(x,y)}$
H: Does a differentiable $f[g(x)]$ imply a differentiable $g(x)$ or the reverse? Does a differentiable $f[g(x)]$ imply a differentiable $g(x)$? Does a differentiable $g(x)$ imply a differentiable $f[g(x)]$? Thanks in advance AI: Not at all. Take $f$ to be the $0$ function, so that $f\circ g$ is readily differentiable, regardless of how nasty $g$ is. By the same token, let $f$ be any non-differentiable function, and let $g$ be the identity function, so that $f\circ g=f$ is non-differentiable, even though $g$ is differentiable.
H: What is the closure of these sets? I am working through the problems in Topology by Munkres. This comes from Section 17, #17 on page 101. Consider the lower limit topology on $\mathbb{R}$ and the topology given by the basis $\mathcal{C}$, where $\mathcal{C} = \{[a,b)\text{ such that }a \lt b, a\text{ and }b \text{ rational}\}$. Determine the closure of the intervals $A = (0, \sqrt{2})$ and $B = (\sqrt{2}, 3)$ in these two topologies." Here is my answer. Closure of $A = (0, \sqrt{2})$ in the lower limit topology is $[0, \sqrt {2})$. Every neighborhood around $0$ must intersect $A$, so $0$ is an element of $\overline{A}$. Also, $[\sqrt 2, 4)$ is an open set which doesn't intersect $A$, thus $\sqrt{2}$ is not an element of $\overline{A}$. Closure of $B = (\sqrt{2}, 3)$ in the lower limit topology is $[\sqrt{2}, 3)$. Every neighborhood around $\sqrt{2}$ must intersect $B$, so $\sqrt{2}$ is an element of $B$-closure. Also, $[3, 4)$ is an open set which doesn't intersect $B$, thus $3$ is not an element of $B$-closure. Closure of $A = (0, \sqrt{2})$ in the topology generated by the basis $\mathcal{C}$ is $[0,\sqrt{2}]$. Every neighborhood around $0$ must intersect $A$, so $0$ is an element of $A$-closure. There do not exist open sets like $[\sqrt{2}, 4)$, where $\sqrt{2}$ is the lowest element in the set. Thus, any open set containing $\sqrt{2}$ must contain an element lower than $\sqrt{2}$. Thus $\sqrt{2}$ is an element of $A$-closure. Closure of $B = (\sqrt{2}, 3)$ in the topology generated by the basis $\mathcal{C}$ is $[\sqrt{2}, 3)$. Every neighborhood around $\sqrt{2}$ must intersect $B$, so $\sqrt{2}$ is an element of $B$-closure. Also, $[3, 4)$ is an open set which doesn't intersect $B$, thus $3$ is not an element of $B$-closure. Are these correct? Any feedback would be most appreciated. Thanks. AI: These are all correct. It's worth noting, though, that you should justify (for example) why there are no other points in the closure of $(0,\sqrt2)$ in the lower limit topology. You've demonstrated why $\sqrt2$ isn't in the closure, but why isn't there any element greater than $\sqrt2$? Why isn't there any negative element?
H: Increasing in each point implies increasing. $\newcommand{\R}{\mathbb R}$ Let $f: \R \to \R$ be a function. We say $f$ is increasing in $p \in \R$ iff $\exists \delta > 0 : x \in (p-\delta,p), y \in (p,p+\delta) \Rightarrow f(x) \leq f(p) \leq f(y)$. We say $f$ is increasing if $f(a) \leq f(b)$ if $a \leq b$. If $f$ is increasing in each point $p \in \R$ then $f$ is increasing. I tried to do this as follows: Assume $f(a) > f(b)$ for some $a < b$ in $\R$. Let $c := \frac{a+b} 2$. Then $f(a) > f(c)$ of $f(c) > f(b)$. In the first case let $a_1 := a$ and $b_1 := c$ and in the second case let $a_1 := c$ and $b_1 := b$. Now proceed like this. This gives intervals $[a_n,b_n]$ s.t. $f(a_n) > f(b_n)$ and $[a_{n+1},b_{n+1}]\subseteq [a_n,b_n]$. Since all intervals are non empty and compact and decreasing the intersection is non empty. So let $x \in \cap_n [a_n,b_n]$. I think that $f$ is not increasing in $x$ but I can't show it. Please help ! AI: If $f$ is locally increasing at $x$, then there is a $δ$ such that $f(u)\le f(x)\le f(v)$ for all $x-δ<u\le x\le v<x+δ$. On the other hand, there is a $k\in N$ such that $[a_k,b_k]\subset(x-δ,x+δ)$. We have $f(a_k)>f(b_k)$. But $x-δ<a_k\le x\le b_k<x+δ$, so $f(a_k)\le f(b_k)$ which is a contradiction.
H: Proving the interior of a set [Homework] I have to find the interior of the following set: $E = [0,5] \cup (5,7)$, and prove it. I found the union of the set to be $[0,7)$, and thus found the interior to be $(0,7)$. I'm not sure how to prove this is true using the definition of a closed set i.e. there exists a delta such that for an $x$ element of $E$, $(x - \delta, x + \delta)$ is contained in $E$? Sorry about the lack of quantifiers, I haven't learned how to write them yet. Any step in the right direction would be greatly appreciated. Thank you in advance! AI: Hint: For $x \in (0,7)$ let $\delta := \min\{7-x, x\} > 0$.
H: How much math does one need to know to do philosophy of math? I'm looking for advice from mathematicians who also study philosophy of math (PoM). Due to interest I'd like to study PoM as a hobby, but I'm worried if I don't understand math well enough from a pure math perspective I will make errors in reasoning about the nature of math. I am enrolled in a STEM field so I have some applied math/calculus background and am willing to invest time into studying pure math. I just think that if I don't learn the better details of how proofs are created or have a broader knowledge of math than calculus I may have false ideas of how math is done and this will affect my thinking in studying PoM. Thank you for your time. AI: It depends which areas of the philosophy of mathematics you want to study. We can usefully divide the field into (A) very general Big Picture questions about how mathematics fits into our general views about the world and our knowledge of it; and then there are (B) more specific questions that arise from reflecting on some of the details of mathematical practice. You don't need a lot of mathematical knowledge to tackle the first sort of question (though you need a lot of other philosophical knowledge); you need more, perhaps a great deal more, mathematical knowledge (but less general philosophy) to tackle the second sort of question. Let me spell that out a bit (with the preliminary remark that I wouldn't want to say that there is a really sharp division here: still, I suggest that it is a very useful first approximation to think in terms of there being two different sorts of question here.) (A) There’s a lovely quote from the great philosopher Wilfrid Sellars that many modern philosophers in the Anglo-American tradition [apologies to those Down Under and in Scandinavia ...] would also take as their motto: The aim of philosophy, abstractly formulated, is to understand how things in the broadest possible sense of the term hang together in the broadest possible sense of the term. Concerning mathematics, then, we might wonder: how do the abstract entities that maths seems to talk about fit into our predominantly naturalistic world view (in which empirical science, in the end, gets to call the shots about what is real and what is not)? How do we get to know about these supposed abstract entities (gathering knowledge seems normally to involve some sort of causal interactions with the things we are trying to find out about, but we can’t get a causal grip on the abstract entities of mathematics)? Hmmmm: what maths is about and how we get to know about it — or if you prefer than in Greek, the ontology and epistemology of maths — seems very puzzlingly disconnected from the world, and from our cognitive capacities in getting a grip on the world, as revealed by our best going science. And yet, … And yet maths is intrinsically bound up with, seems to be positively indispensable to, our best going science. That’s odd! How is it that enquiry into the abstract realms of mathematics gets to be empirically so damned useful? A puzzle that prompted the physicist Eugene Wigner to write a famous paper called “The Unreasonable Effectiveness of Mathematics in the Natural Sciences”. Well, perhaps it’s the very idea of mathematics describing an abstract realm sharply marked off from the rest of the universe — roughly, Platonism (for a short-hand label) — that gets us into trouble. But in that case, what else is mathematics about? Structures in some sense (where structures can be exemplified in the non-mathematical world too, which is how maths gets applied)? — so, ahah!, maybe we should go for some kind of Structuralism about maths? But then, on second thoughts, what are structures if not very abstract entities, after all? Hmmmm. Maybe mathematics is really best thought of as not being about anything “out there” at all, and we should go for some kind of sophisticated version of Formalism -- perhaps it is all just symbol shuffling, that doesn't have to hook up to some abstract Platonic reality). And so we get swept away into esoteric philosophical fights, as the big Isms slug it out. Well, I caricature of course! -- but the key point is that you don't need to know a great deal of advanced maths to follow these fights, they arise from quite elementary reflections on the school-room beginnings of maths and on the applications of elementary mathematics. But to get anywhere, you do need to be able to follow arguments in general metaphysics and epistemology, i.e. follow arguments about what there is and about how we can know about it. (B) However, philosophers of mathematics talk about much more than this Big Picture stuff. To be sure, the beginning undergraduate curriculum in the philosophy of mathematics tends to concentrate in that region: e.g. for an excellent textbook see Stewart Shapiro’s very readable Thinking about Mathematics (OUP, 2000). But the philosophers also worry about more specific questions like this: Have we any reason to suppose that the Continuum Hypothesis has a determinate truth-value? How do we decide on new axioms for set theory as we beef up ZFC trying to decide the likes of the Continuum Hypothesis? Anyway, what’s so great about ZFC as against other set theories (does it have a privileged motivation)? In what sense if any does set theory serve as a foundation for mathematics? Is there some sense in which topos theory, say, is a rival foundation? What kind of explanations/insights do very abstract theories like category theory give us? What makes for an explanatory proof in mathematics anyway? Is the phenomenon of mathematical depth just in the eye of the beholder, or is there something objective there? What are we to make of the reverse mathematics project (which shows that applicable mathematics can be founded in a very weak system of so-called predicative second-order arithmetic)? Must every genuine proof be formalisable, and if so, using what grade of logical apparatus? Are there irreducibly diagrammatic proofs? … That's only the beginnings of a list which could go on. And on. But the point is already made. These questions, standing-back-a-bit and reflecting on our mathematical practice, can still reasonably enough be called philosophical questions (even if they don’t quite fit Sellars’s motto). They are more local than the Big Picture questions which arise from looking over our shoulders and comparing mathematics with some other form of enquiry and wondering how they fit together. These are questions are internal to the mathematical enterprise, discussed by mathematically informed philosophers, as well as by philosophically minded mathematicians -- and you do need varying amounts of serious mathematics to tackle them. You can't, for an obvious example, discuss the foundational significance of category theory if you know no category theory. So, in summary, you don't need a lot of mathematics to follow debates on some central Big Picture ontological and epistemological questions in the philosophy of mathematics. But other areas of the philosophy of mathematics focus in on specific areas of mathematical practice, and then you do have to know quite a bit of maths to follow them.

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