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H: What is a set function that returns another set of points called? I have a set of points $S = \{x_i\}_{i = 1}^m, x_i \in \mathbb{R}^n \forall i$. Now, I have a set function $f$ which operates as follows: $$f(S) = GX^T$$ where $G \in \{0,1\}^{m\times m}$ and $X = [x_1 x_2 \ldots x_m]$. Is there a specific name for such $f$'s? AI: The function $f$ does not return a set of points: it returns an $m\times 1$ matrix of real $n$-tuples. And the input is not a set $S=\{x_k:k=1,\dots,m\}\subseteq\Bbb R^n$: you’re using the indexing of the $x_k$ in an essential way, so the input to $f$ is actually an $m$-tuple of vectors in $\Bbb R^n$. Specifically, $$f:\left(\Bbb R^n\right)^m\to\left(\Bbb R^n\right)^{m\times 1}:\langle x_1,\dots,x_m\rangle\mapsto G\begin{bmatrix}x_1\\x_2\\\vdots\\x_m\end{bmatrix}\;.$$ It’s simply a linear transformation between two finite-dimensional real vector spaces.
H: Definite integral of a periodic funtion, offset by the period, equals the original definite integral Suppose $f: \mathbb R \to \mathbb R$ is Riemann integrable on every finite interval and periodic with period $T>0$. Then for every interval $[a,b]$: $$ \int_a^b f = \int_c^d f,$$ where $c = a+T$ and $d = b+T$. I don't understand why, if so, this is true. How van I explain it? I could do it if I knew that the antiderivative is periodic, but do we even know that there is an antiderivative? AI: By the change variable $x=u-T$ so $du=dx$ we have $$\int_a^b f(x)dx=\int_{a+T}^{b+T}f(u-T)du=\int_c^df(u)du$$
H: If derivative of $e^{ax} \cos{bx}$ with respect to $x$ is $re^{ax}\cos(bx + \tan^{-1} \frac {b} {a})$ Problem: If derivative of $e^{ax} \cos{bx}$ with respect to $x$ is $re^{ax}\cos(bx + \tan^{-1} \frac {b} {a})$. Then find $r$ when $a>0,b>0$ Solution: Differentiating $e^{ax} \cos{bx}$ w.r.t $x$ we get $ ae^{ax} \cos{bx} -be^{ax} \sin{bx} $ But I don't know,how to convert it into $re^{ax}\cos(bx + \tan^{-1} \frac {b} {a})$ AI: Hint: Use the identity $$ \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B). $$ to expand $$ \cos(bx + \tan^{-1} \frac {b} {a}) $$ and then use the facts $$ \cos(\arctan(t))=\frac{1}{\sqrt{1+t^2}},\quad \sin(\arctan(t))=\frac{t}{\sqrt{1+t^2}}. $$
H: A question on generators Suppose $I$ is an ideal in a ring $R$ which is finitely generated. Suppose on the other hand that there is some (possibly other) set of generators $\{g_t\colon t\in T\}\subset I$ which also generates $I$ as an ideal. Can we find a finite subset $T_0\subset T$ such that $\{g_t\colon t\in T_0\}$ generates $I$? AI: $I$ is generated by $(f_1, \ldots, f_n)$ as per your first assumption. Since the family of $(g_t)_{t\in T}$ generates $I$, for all $i$ we have $f_i = \sum a_{i,t} g_t$, the sum being finite. Take $T_0$ to be the set of all $g_t$'s that appear in those $n$ summations, $T_0$ is finite and still generates $(f_1,\ldots, f_n)$ so it generates $I$.
H: Relationship between positive definite matrix eigenvalues and the principal axes lengths of ellipsoid In many textbooks of pattern recognition I have seen the following statement: If a matrix $A$ in the quadratic form $F(\text{x}) = \text{x}^{T}A\text{x}$ is positive definite, then it follows that the surfaces of constant $F(\text{x})$ are hyperellipsoids, with principal axes having lengths proportional to $\lambda_{k}^{-1/2}$. $A$ is a $n \times n$ matrix, $\lambda_k$ is an eigenvalue of $A$, $\text{x}$ is a non-zero vector and $k = 1, ..., n$. My question is: "Why are the principal axes of hyperellipsoids proportional to the eigenvalues of $A$? Why are they proportional to $\lambda_{k}^{-1/2}$? And why are the surfaces of constant $F(\text{x})$ hyperellipsoids?" Can someone give me a proof? Visualization perhaps? etc. Anything that would make me see that this indeed is the case :) Hope my question is clear :) Thank you for any help! AI: If $A \in \def\R{\mathbb R}\R^{n\times n}$ is positive definite, there is an orthogonal $Q \in O(n)$ (that is a matrix whose columns form an orthonormal basis of $\R^{n}$) such that $QAQ^t = \mathrm{diag}(\lambda_1, \ldots, \lambda_n) =: \Lambda$ with the $\lambda_k$ positive. For $c > 0$ and $x \in \R^n$ we have \begin{align} F(x) &= c\\ \iff x^t Ax &= c \\ \iff x^t Q^t QAQ^t Qx &= c\\ \iff (Qx)^t \Lambda Qx &= c\\ \iff \sum_k \lambda_k (Qx)_k^2 &= c\\ \iff \sum_k \frac{(Qx)_k^2}{(c^{1/2}\lambda_k^{-1/2})^2} &= 1 \end{align} So in the transformed coordinates $y = Qx$ given by $Q$ we see that the hypersurface $\{F = c\}$ is given by $$ \frac{y_1^2}{(c^{1/2}\lambda_1^{-1/2})^2} + \cdots + \frac{y_n^2}{(c^{1/2}\lambda_n^{-1/2})^2} = 1 $$ that is it forms a hyperellipsoid with principal axes in direction $Qe_k$ (where $e_k$ denotes the $k$-th standard unit vecotr) having lengths $c^{1/2}\lambda_k^{-1/2}$.
H: Commuting Exponential Matrices Let $x(t)=\exp(tA)\exp(tB)$ and $y(t)=\exp(t(A+B))$. Show that if $AB=BA$ then $x(t)$ and $y(t)$ satisfy the same initial value problem for ODEs and therefore must be equal. $A, B$ square matrices. AI: Hint: (1) What ODEs do $x$ and $y$ fulfill? Compute $x'$ and $y'$, try to find the given terms for $x$ and $y$ in your computed expressions. Write down a differential equation for $x$ and $y$? Are they equal? (2) Compute $x(0)$ and $y(0)$. Are these values equal?
H: ball on the moon part 3 This the final parts to my previous question ( On the surface of the moon ) The original question reads "On the surface of the moon, acceleration due to gravity is approximately 5.3 feet per second squared. Suppose a baseball is thrown upward from a height of 6 feet with an initial velocity of 15 feet per second. A)Determine the maximum height attained by the baseball B)Determine how long it takes the ball to hit the surface of the moon. C) Find the average velocity of the ball over the time it's in flight D) Find the moment when the ball's instantaneous velocity is the same as your answer in the previous part. E) What assurance do we have that there is an answer to part D? Part A is answered AI: (B) You now know that $S(t)=-2.65t^2+15t+6$, where $S(t)$ is the height above the moon’s surface at time $t$. The ball leaves your hand at time $t=0$, and it hits the moon when $S(t)=0$. Thus, all you have to do is solve $S(t)=0$ for $t$. (C) The ball started at a height of $6$ ft and ended at a height of $0$ ft, so it travelled a net distance of how much? Be careful: this is the net distance, and the algebraic sign matters. From (B) you know how long it took. Divide the distance travelled by the time, and of course you get the average velocity $v_{\text{avg}}$. (D) You know that the velocity at time $t$ is $v(t)=-5.3t+15$, and you know $v_{\text{avg}}$ from (C), so you just have to solve $v(t)=v_{\text{avg}}$ for $t$. (E) HINT: Mean Value Theorem.
H: True Or not: Compact iff every continuous function is bounded Let $X$ be a topological space. My question is: If $f:X\to \mathbb{R}$ is bounded for all such continuous $f$, then is $X$ compact. Is is really? If $X$ is the subset of $\mathbb{R}^d$, then it is clear, beacause with Heine-Borel we get what we want (closed and bounded (with the help of the norm)), but is it true in general? I really hope there exists a non-compact space with the property above. AI: Let $X = [0,\omega_1)$. Then $X$ is non-compact as the $[0,\alpha)$, $\alpha < \omega_1$ form an open cover without even a countable subcover. On the other hand, let $f \colon X \to \mathbb R$ continuous. Suppose $f%$ were unbounded. Then there is a sequence $\alpha_n < \omega_1$ such that $f(\alpha_n) \ge n$. As $[0,\alpha]$ is compact for $\alpha < \omega_1$ and $f$ is continuous, $f([0,\alpha])$ is bounded. So we may arrange that $\alpha_{n+1} > \alpha_n$ for $n < \omega$. Let $\alpha^* = \sup_n \alpha_n$. By continuity $f(\alpha^*) = \lim_n f(\alpha_n)$. But $f(\alpha_n) \ge n$ for each $n$. Contradiction, hence $f$ is bounded.
H: Conjecture on eigenvalue property of a matrix subspace Suppose we have a full rank positive definite Hermitian matrix $A\in \mathbb{C}^{n \times n}$ with eigenvalues $\lambda_1>\lambda_2> \dots >\lambda_n$. Consider a semi-orthogonal matrix $X \in \mathbb{C}^{n \times p}$ (i.e., $X^* X = I_p$) spanning a subspace of $A$. Denote the eigenvalues of matrix $B = X^* A X$ to be $s_1 > s_2 > \dots > s_p$. Is it possible to show that $\lambda_j \geq s_j, \forall 1\leq j \leq p$? I ran through a simulation for $10^5$ trials over $n=10$ and $p=4$, and the results suggested so. But I cannot prove it or raise a counter example. AI: Let $X = U {\rm I}_{n \times p} V^$ be an SVD of $X$: $U$ is unitary of order $n$, $V$ is unitary of order $p$, and ${\rm I}_{n \times p}$ is diagonal of order $n \times p$ with all diagonal elements equal to $1$. Then $$B = X^ A X = V {\rm I}_{p \times n} U^* A U {\rm I}_{n \times p} V^,$$ so $B$ is (unitarily) similar to $B' := {\rm I}_{p \times n} U^ A U {\rm I}_{n \times p}$, which means that $B$ and $B'$ have the same eigenvalues. However, $A' := U^* A U$ is (unitarily) similar to $A$, so they have the same eigenvalues as well. Notice that ${\rm I}_{p \times n} A' {\rm I}_{n \times p}$ is the top left principal submatrix of $A'$, so you can apply Cauchy interlacing theorem, which will give you a bit more than you asked for: $$\lambda_j \ge s_j \ge \lambda_{n-p+j}, \quad \text{for $1 \le j \le p$}.$$
H: The national lottery When playing the lottery you have to pick 6 numbers out of 45 possibilities. Since the order of the numbers don't matter, the number of possible combinations for the jackpot (and hence the 6 correct numbers) is given by: $ \dbinom{45}{6} = C^6_{45} = \frac{45!}{(45-6)!6!} = 8145060 $ Assuming the numbers are picked at random out of a uniform distribution the chance of winning the jackpot is given by: $P(win) = \frac{1}{8145060}\approx 0.0000123 \%$ Now I can figure two possible scenarios: You play every week with the same numbers hoping that one day you'll get the jackpot. You play every week with a different set of numbers hoping that one day you'll get the jackpot. Now I was wondering if there is a difference in the chance of winning between the two methods. I was thinking that in the first case the chances of winning are larger because if you stick to your number the alternatives might run out. While if you switch you don't have the other options that become eliminated. But then I started wondering, because in the Monty Hall paradox the chances become larger if you switch. For the first case I believe the chance of winning in the $n^{th}$ drawing is simply calculated as: $1-(1-P(win))^n$ But in the second case I wouldn't know how to succeed. AI: Assuming the outcome of each lottery raffle is completely random (and thus assuming every result is independent from all the preceeding ones), the chance to win writing down a random number each time is the same, i.e. $\,\frac1{8145060}\,$
H: Powers of a unitary matrix I'm trying to find the minimum exponent $M\in \mathbb{N}$, such that for a certain unitary matrix $F\in \mathbb{C}^{N}$, $$F^M = 1_{\mathbb{C}^{N}}.$$ I don't think it matters, but it's the DFT matrix. Now, I noticed that if $F^M = 1$, then $F^{M/2} = ({F^{M/2}})^\dagger$. However, I can't find a justification as to why M must be even. I'm not even sure if it has to be. (for the DFT matrix, M=4, though I'm still trying to figure out why it isn't 2) Any ideas? AI: It seems to me that this comes from the eigenvalues of a DFT matrix. Raising a matrix to the power $k$ raises its eigenvalues to the power $k$. So, if the order of your matrix is $2$, you have $1$ eigenvalue equal to $-1$; when you have order greater than $2$, you have at least one eigenvalue equal to $-i$ (and, for order at least $4$, you have $i$ as well, but this doesn't change anything). See the table on the above link. Notice that this does not hold for a general unitary matrix. For a counterexample, just take a 2D rotation by an angle $\varphi$ which is not of a form $\varphi = q\pi$ for $q \in \mathbb{Q}$ and no power of your matrix will ever become identity. Also, rotation for $\varphi = 2\pi/3$ will become identity when raised to the third (so, odd) power.
H: Show that ($\ell^1$, $\\|\cdot\\|_1$) is complete Show that the vector space $\ell^1 : = \{(a_n) : \sum_n\|a_n\| < \infty\}$ with the norm $\\|(a_n)\\|_1 : = \sum_n\|a_n\|$ where $(a_n)$ are sequences in $\mathbb C$ is complete. Thanks in advance. AI: Consider a Cauchy sequence $\{x^n\}$in $l_1$. Where $x^n = (x_1^n, x_2^n, \dots)$. For any $\epsilon >0$ there exist $k_1 \in \mathbb{N} $ s.t. $\\|x^p - x^q\\|_1 < \epsilon$ when $p, q > k_1$. So $\sum_{i=1}^{\infty} \|x_i^p -x_i^q\| < \epsilon$. Thus for any fixed subscript $i$ the sequence $\{x_i^n\}$ is Cauchy on $\mathbb{C}$. As $\mathbb{C}$ is complete the sequence is convergent and let it converges to $x_i$ . Consider the element $x = (x_1, x_2, \dots)$. Now two points to be shown $x \in l_1$. $\{x^n\}$ converges to $x$. The sequence $\{x^n\}$ is Cauchy and hence it is bounded. Thus $\sum_{i=1}^{\infty} \|x_i^n\| < C$ for some $C \in \mathbb{R}$. Thus $\sum_{i=1}^k\|x_i^n\| < C$ for any arbitrary $k$. Taking $n \rightarrow \infty$ we shall get $\sum_{i=1}^{k}\|x_i\| < C$. Now taking $k \rightarrow \infty$ we shall get $\sum_{i=1}^{\infty}\|x_i\|$ is bounded. Now form the inequality from the condition of Cauchy Sequence taking one supercsript $\rightarrow \infty$ we can show the sequence $\{x^n\}$ converge to $x$. That is, $$\lim_{q \to \infty}\sum_i^k\|x^p_i-x^q_i\| = \sum_i^k\|x^p_i-x_i\| \leq \epsilon$$ for all $k$. Hence the proof.
H: Is $\frac{n!}{\left(n/2\right)!\left(n/2\right)!}$ a natural number for $n$ even? Probably this is a easy question, but I was unable to solve it. Let $n$ be a even natural number. Is true that the following number is natural for all $n$? $$\frac{n!}{\left(n/2\right)!\left(n/2\right)!}$$ I can see, for example, that $(n/2)!$ divides $n!$, but then I can't conclude by using this arguing. Thank you AI: Yes, it is. Suppose that $n=2m$; then $$\frac{n!}{(n/2)!(n/2)!}=\frac{(2m)!}{m!m!}=\binom{2m}m\;,$$ which is the number of $m$-element subsets of a set of $2m$ things. This clearly must be an integer. For more information, see the Wikipedia article on binomial coefficients.
H: Trigonometry question: Find in simplest surd form: $\cos 195^{\circ}$ Find in simplest surd form: $\cos 195^{\circ}$. Ive recently been doing the trigonometry topic form textbook and have oftenly come across these questions. Can someone please justify how you do this question? Ive tried many times but no luck. AI: Assume that we are working with angles in degrees. Hint: $$\cos(195) = \cos(180 + 15)$$ Another hint: Now split $\cos(180 + 15)$ using $\cos(A+B) = \cos(A)\cos(B)-\sin(A)\sin(B)$. And another: Substitute $\sin(180) = 0$ and $\cos(180) = -1$. Nearly done: Notice that $\cos(15) = \cos(30/2)$ and remember that $\cos(2A) = 2\cos^{2}(A) - 1$. Now draw the 30/60/90 triangle and evaluate.
H: Understanding Power Series Multiplication Step Working on Spivak's Calculus problems, I searched online, trying to understand the solution provided for Problem 4a of Chapter 2. I found the question I needed: Spivak's Calculus - Exercise 4.a of 2nd chapter. However, the answer provided there started with the following equation, and then went on from it to explain other things, which I could understand. But this part, which lies at the foundation of the argument, I don't understand. Could I get an explanation for why it is that: $$ \begin{align} \left(\sum_{k=0}^\infty a_kx^k\right)\left(\sum_{k=0}^\infty b_kx^k\right) &=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_j\color{#C00000}{x^j}b_{k-j}\color{#C00000}{x^{k-j}}\right) \end{align} $$ Keep in mind I'm a beginner, working on an introductory Calculus book as my first exposure to the subject. I'd appreciate both intuitive and rigorous answers. AI: Write the product out longhand: $$\left(a_0+a_1x+a_2x^2+a_3x^3+\ldots\right)\left(b_0+b_1x+b_2x^2+b_3x^3+\ldots\right)\;.$$ This is the sum of all possible products of the form $(a_kx^k)(b_\ell x^\ell)=a_kb_\ell x^{k+\ell}$. The $x^n$ term in the product will therefore be the sum of all of these term $a_kb_\ell x^{k+\ell}$ for which $k+\ell=n$. Let’s look specifically at $n=3$, just to get a clearer picture of what’s going on. If $k+\ell=3$, where $k$ and $\ell$ are exponents in the two factors, then clearly $\langle k,\ell\rangle$ must be one of the pairs $\langle 0,3\rangle,\langle 1,2\rangle,\langle 2,1\rangle$, or $\langle 3,0\rangle$. Thus, the $x^3$ term in the product is $$(a_0x^0)(b_3x^3)+(a_1x^1)(b_2x^2)+(a_2x^2)(b_1x^1)+(a_3x^3)(b_0x^0)=\sum_{k=0}^3(a_kx^k)(b_{3-k}x^{3-k})\;,$$ or $$\left(a_0b_3+a_1b_2+a_2b_1+a_3b_0\right)x^3=\left(\sum_{k=0}^3a_kb_{3-k}\right)x^3\;.$$ If you generalize this to an arbitrary power $n$ of $x$, you get $$(a_0x^0)(b_nx^n)+(a_1x^1)(b_{n-1}x^{n-1})+\ldots+(a_{n-1}x^{n-1})(b_1x^1)+(a_nx^n)(b_0x^0)=\sum_{k=0}^n(a_kx^k)(b_{n-k}x^{n-k})\;,$$ or $$(a_0b_n+a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0)x^n=\left(\sum_{k=0}^na_kb_{n-k}\right)x^n\;.$$ In other words, the coefficient of $x^n$ in the product series is $$\sum_{k=0}^na_kb_{n-k}\;.$$
H: Group Theory Normal Subgroups Let $G$ be a group of order $8$ with $x\in G$ such that $o(x)=4$. Prove that $x^2\in Z(G)$, where $$Z(G)=\{ x \in G \mid xg=gx\text{ for all }g\in G\}.$$ AI: HINT: Any subgroup of index $2$ is normal in $G$.
H: Is an automorphism of the field of real numbers the identity map? Is an automorphism of the field of real numbers $\mathbb{R}$ the identity map? If yes, how can we prove it? Remark An automorphism of $\mathbb{R}$ may not be continuous. AI: Hint: Let $\phi$ be a field automorphism of $\mathbb R$. Then prove: $\phi$ sends positive numbers to positive numbers $\phi$ is increasing $\phi$ is continuous $\phi$ is the identity on $\mathbb Q$ $\phi$ is the identity on $\mathbb R$.
H: A function in a Real Vector Space V (where V is the set of all complex-valued functions f on the real line ), which is NOT real-valued. Question: Let $V$ be the set of all complex-valued functions $f$ on the real line such that (for all $t \in \Bbb R) \ \ f(-t) = \overline {f(t)} $. The bar denotes complex conjugation. (i) Show that $V$, with the operations $(f + g)(t) = f(t) + g(t)$ and $(cf)(t) = cf(t)$ is a vector space over the field of real numbers but NOT a vector space over the filed of complex numbers. (ii) Give an example of a function in $ V$ which is NOT real-valued. My Try: (i) $ (f + g)(-t) = f(-t) + g(-t) = \overline {f(t)} + \overline {g(t)} = \overline {f(t) + g(t)} = \overline {(f + g)(t)}$ . $ (cf)(-t)=cf(-t)=c \overline{f(t)}= \overline{c f(t)} $ . Hence, a subset $V$ of the real vector space of all functions from $\Bbb R $ to $\Bbb C $ is closed under addition and multiplication by real numbers. This means that $V$ is a subspace and satisﬁes all properties of a vector space. Hence $V$ is a vector space over the field of real numbers. BUT I DON'T KNOW HOW TO PROVE that $V$ is NOT a Vector Space over the filed of complex numbers. Please help me solve this. (ii) An example of a non-real-valued function in $V$ is $f(t) = it$ Now, how will I prove that this function is in $V$ ? My attempt is given below: $f(t) = it$ $\Rightarrow f(-t)= i(-t)=-it$ or $f(t) = it$ $\Rightarrow f[-(-t)]=it$ $\Rightarrow \overline {f(-t)}=it$ $f(t)+f(-t)=0 \in V$ This is University exam question and I'm preparing for the same exam. Part (i) is for 10 marks and Part (ii) is for 5 marks. If there is any mistake, please correct the solution given by me and help me complete the solution. Thanks. AI: If $\,V\,$ were a complex linear space then it'd fulfill $$(if)(-x)=\overline{(if)(x)}\;,\;\;\forall\,f\in V$$ Yet for $\,f(x)=1\;,\;\;\forall\,x\in\Bbb R\;$ , we get: $$\begin{align}(if)(-x)&=i(f(-x))=i(1)=i\\ \overline{(if)x}&=\overline{i(f(x)}=\overline i\overline{f(x)}=-i(1)=-i\end{align}$$ For part two I think it's easier and simpler: $$\forall\,t\in\Bbb R\;,\;\;f(t):=it\implies f(-t)=-it=\overline{it}=\overline{f(t)}$$
H: Rational number to the power of irrational number = irrational number. True? I suggested the following problem to my friend: prove that there exist irrational numbers $a$ and $b$ such that $a^b$ is rational. The problem seems to have been discussed in this question. Now, his inital solution was like this: let's take a rational number $r$ and an irrational number $i$. Let's assume $$a = r^i$$ $$b = \frac{1}{i}$$ So we have $$a^b = (r^i)^\frac{1}{i} = r$$ which is rational per initial supposition. $b$ is obviously irrational if $i$ is. My friend says that it is also obvious that if $r$ is rational and $i$ is irrational, then $r^i$ is irrational. I quickly objected saying that $r = 1$ is an easy counterexample. To which my friend said, OK, for any positive rational number $r$, other than 1 and for any irrational number $i$ $r^i$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample? Let's stick to real numbers only (i.e. let's forget about complex numbers for now). AI: Consider $2^{\log_2 3}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
H: Predicting the number of decimal digits needed to express a rational number The number $1/6$ can be expressed with only two digits (and a repeat sign denoted as $^\overline{}$), $$ \frac{1}{6} = \,.1\overline{6}$$ Meanwhile, it takes 49 digits to express the number $1/221$, since a string of 49 digits repeats: $$\frac{1}{221} = .\overline{004524886877828054298642533936651583710407239819}$$ Yet for $1/223$, 222 digits repeat, giving a total of 224 digits needed to express the number. If $f:\mathbb{Q}\rightarrow\mathbb{N}$ is a function that gives the smallest number of digits needed to express a rational number in decimal notation, what can we say about $f$? For example, if we do not consider the negative sign to be a digit, then $f$ is an odd function. Other than that, is there any pattern to it at all? AI: Consider the fraction $1/m$. Write $m=2^a 5^b v$ with $\gcd(v,10)=1$. Then the periodic part of $1/m$ has length $e$, where $e$ is the smallest positive number such that $v$ divides $10^e-1$. The non-repeating part has length $f=\max(a,b)$. There are no easy formulas for either $e$ or $f$ in terms of $m$.
H: Is a differentiable probability density function bounded? Let $g$ be a lebesgue probability density function, differentiable, such that $\sup_x\|g'(x)\| < \infty.$ Is $g$ bounded? AI: Let $M := \sup_x \lvert g'(x)\rvert$. Let $n \geqslant 2$. Suppose there exists an $x_n$ with $g(x_n) = n$. Let $x_{n-1}$ be a closest point to $x_n$ with $g(x_{n-1}) = n-1$ (since $g$ is continuous, and $\int g(x)\,dx = 1$, there exist points with $g(x) = n-1$, as $\liminf\limits_{x\to\pm\infty} g(x) = 0$). Without loss of generality, $x_{n-1} < x_n$. Let $$ y_n = \min \{ x : x > x_{n-1}, g(x) = n\}. $$ Then $n-1 \leqslant f(x) \leqslant n$ for $x_{n-1} \leqslant x \leqslant y_n$, and $$1 = g(y_n) - g(x_{n-1}) = \int_{x_{n-1}}^{y_n} g'(x) \,dx \leqslant (y_n - x_{n-1})\cdot M,$$ hence $\lambda(\{x : n-1\leqslant g(x) \leqslant n\}) \geqslant \frac{1}{M}$, where $\lambda$ is the Lebesgue measure. But since $\int g(x)\,dx = 1$ we have $$ (n-1)\cdot \lambda(\{x : n-1\leqslant g(x) \leqslant n\}) \leqslant \int g(x)\,dx = 1, $$ hence $\frac{1}{M} \leqslant \frac{1}{n-1}$ or $n \leqslant M + 1$.
H: A not smooth distribution Is exist the not smooth distribution which satisfying: $\left ( D_{t}^{2}-D_{x}^{2} \right )u(x,t)=0$ I can't find at least one not smooth distribution like this... Thanks for the help! AI: A classic! This example upset a lot of people c. 1800 and even before. It it the one-dimensional wave equation. For any distribution $v$, $u_{\pm} (x,t)=v(x\pm t)$ (suitably interpreted) solves that equation. "Suitably interpreted" is just a generalization of what it should mean for classical functions, but can be made precise in several (provably equivalent) ways: let $T_x$ be translation of functions or distributions, then "$v(x+t)$" can be defined as $T_x v$, where the dummy variable is $t$.
H: Statistics I - Bayes's Theorem A desk has three drawers. The first contains two gold coins, the second has two silver coins and the third has one gold and one silver coin. A drawer is selected at random and a coin is drawn at random from the drawer. Suppose that the coin selected was silver. Use Bayes's Theorem to find the probability that the other coin in that drawer is gold. AI: You select a drawer and select a coin from that drawer. Let $A$ denote the event that the coin that you select is silver. Let $B$ denote the event that the other coin in the drawer is gold. You are equally likely to pick any coin, so $$P(A) = \frac{3\text{ silver coins}}{6 \text{ coins}} = \frac 12$$ There are 6 different ways you could have selected a coin. Only one of them results in you selecting a silver coin AND the other coin in the drawer being gold, so $$P(A \cap B) = \frac{1\text{ desired outcome}}{6 \text{ total outcomes}} = \frac 16$$ By the definition of conditional probability, $$P(B \mid A) = \frac{P(B \cap A)}{P(A)} = \frac{1/6}{1/2} = \frac 13$$ As I mentioned in my comment, Bayes's Theorem is useful if you want to know $P(B\mid A)$ but you only know $P(A\mid B)$. Here, we don't know $P(A \mid B)$ either, so Bayes's Theorem doesn't really help.
H: Parametric Equations rotation of axes For old coordinates $(x,y)$ the new coordinates $(u,v)$ are related like this: $x = u\cos(\theta) - v\sin(\theta)$ $y = u\sin(\theta) + v\cos(\theta)$ So would it be correct to say that to rotate the axis for a parametric equation defined by $x = f(u)$ and $y = g(u)$ I need to multiply $f(u)$ with the $x$ rotation equation and $g(u)$ with $y$ rotation equation? I can't find any resources online for this topic. AI: If you have a parametric equation $$\begin{cases}x = f(u) \\ y = g (u)\end{cases}$$ and you want to rotate the image by $\theta$, you can just take $$\begin{cases}x' = x\cos \theta - y \sin \theta = (\cos\theta) f(u) - (\sin\theta)g(u) \\ y' = x \sin\theta + y\cos\theta = (\sin\theta) f(u) + (\cos\theta)g(u). \end{cases}$$
H: On linear transformations Do there exist continuous linear maps $T\colon V \to V$ such that $T^{-1}\colon V\to V$ exists but is not continuous? Clearly if $V$ has finite dimension the answer is no. AI: By the open mapping theorem, that can't happen when $V$ is a complete metrizable TVS (over $\mathbb{R}$ or $\mathbb{C}$), but it is easy to construct examples for (some) other situations, e.g. let $V = \mathbb{C}^\infty$ the space of all complex sequences with only finitely many nonzero terms. Endow $V$ with any $\ell^p$ norm you particularly like, and let $$T(x) = \left(\frac{x_k}{2^k}\right)_{k\in \mathbb{N}}.$$ Since every sequence in $V$ has only finitely many nonzero terms, $T$ is bijective, but $T^{-1}$ is not bounded.
H: Solution of pendulum linked to Weierstrass $\wp$-function I've been working through a question about the equation of motion of a pendulum. I have to now solve the equation of the form: $$u'^2=u^3+au+b,$$ where $a=(\frac{g^2}{l^2}-\frac{c^2}{3})$ and $b=(\frac{g^2c}{3l^2}-\frac{2c^3}{27})$, using separation of variables. So, I have separated the equation to get: $$\int\frac{1}{\sqrt{(u^3+au+b)}}\,\mathrm du=\int1\,\mathrm dt$$ Now, I could just integrate this directly and attempt a solution, but the next part of the question asks what the solution has to do with the Weierstrass $\wp$-function. Is there a trick we can use to find the solution? AI: Your answer is straight because your $u$ corresponds to a Weierstrass P-function. http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions#Differential_equation
H: Ideas about a transformation matrix I have a problem in 3D where I have three vectors: $\boldsymbol{\omega}$, $\boldsymbol{m}$ and $\boldsymbol{T}$. The first $\boldsymbol{\omega}$ is the variable I solve for, $\boldsymbol{m}$ is fixed and the last is related to the others by $$ \boldsymbol{T} \sim (\boldsymbol{m}^{T} \boldsymbol{m} - I) \boldsymbol{\omega} $$ Hopefully I have this correct, the $(\boldsymbol{m}^{T} \boldsymbol{m} )$ term creates a $3 \times 3$ array and $I$ is the identity matrix. I think this should be some kind of orthogonal projection of $\boldsymbol{\omega}$ onto a plane orthogonal to $\boldsymbol{m}$ but I can't seem to picture it. So does anyone have an idea or references for what kind of transformation this is? AI: Given that you're multiplying $\omega$ from the left to get another vector, I can only presume that $T$, $m$, and $\omega$ are column vectors, in which case you should have $mm^T$ instead of $m^T m$, which would just be a scalar. Now, if $m$ is a unit vector, then $m m^T$ is precisely the orthogonal projection onto the line $\mathbb{R}m$ spanned by $m$ (otherwise, you need to take $\tfrac{1}{\\|m\\|^2}mm^T$). Hence, $I - mm^T$ (or more generally, $I - \tfrac{1}{\\|m\\|^2}mm^T$ if $m$ isn't a unit vector) is, for completely general reasons, the orthogonal projection onto the orthogonal complement $(\mathbb{R}m)^\perp$ of $\mathbb{R}m$, that is, onto the plane orthogonal to $m$. Thus, if $m$ is a unit vector, then your relation implies that $T$ is proportional to the orthogonal projection of $\omega$ onto the plane orthogonal to $m$, as you suspected.
H: How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$ Problem : How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$ I am unable to find out how to proceed in this problem.. this is a problem of arithmetic progression... Please suggest how to proceed...Thanks.. AI: HINT: As $1+r^2+r^4=(1+r^2)^2-(r)^2=(1+r+r^2)(1-r+r^2)$ and $(1+r+r^2)-(1-r+r^2)=2r,$ $$\frac r{1+r^2+r^4}$$ $$=\frac12\cdot\frac{2r}{(1+r+r^2)(1-r+r^2)}$$ $$=\frac12\cdot\frac{(1+r+r^2)-(1-r+r^2)}{(1+r+r^2)(1-r+r^2)}$$ $$=\frac12\left(\frac1{1-r+r^2}-\frac1{1+r+r^2}\right)$$ Put the values of $r=1,2,\cdots.. n-1,n$ to find the partial sum and recognize the Telescoping Sum/Series which is evident as $1-(r+1)+(r+1)^2=1+r+r^2$ Then set $n\to\infty$
H: Sums and products involving Fibonacci In summary, if $\phi$ is the golden ratio, I want to show: \begin{align} \sum_{n=1}^\infty \frac1{F_n} &= 4-\phi \\ \sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{F_nF_{n+1}} &= \phi-1 \\ \prod_{n=2}^\infty \bigg( 1 + \frac{(-1)^n}{F_n^2} \bigg) &= \phi. \end{align} AI: The first equation is obviously wrong, since $$\frac{1}{1} + \frac{1}{1} + \frac{1}{2} = \frac{5}{2} > 4 - \phi = 4 - \frac{1+\sqrt{5}}{2} \approx 2.381966.$$ The other two equations can be easily derived by considering $$F_n^2 - F_{n-1}F_{n+1}.$$
H: Combination and Permutation In how many ways can the letters of the word VANESSA be arranged so that no two vowels are together? The answer is $900$ but I get: $2!/2!\cdot\binom{3}{2}\cdot 2!/2! = 3$. AI: Let's lay down the consonants first. There are $4!/2!=12$ ways to arrange the letters $V,N,S,S$. Let's take $SNVS$ for example. How can we insert the vowels? $$SNVS$$ Each $$ represents either $1$ or $0$ vowels. We need to pick three of these stars to place our vowels, and the other two will just be "empty." There are $\binom 5 3 = 10$ ways to choose three stars. Let's take $SNV*S$ for example. Now we need to place the vowels $A, A, E$ into the 3 stars. There are $3!/2!=3$ ways to do that. So, $12 \cdot 10 \cdot 3 = 360$ ...can someone find my mistake? So it seems that the question is asking for the complement (thanks Douglas!): how many arrangements have [at least] two adjacent vowels. Simply take the total number of arrangements , which is $\frac{7!}{2!2!}=1260$ and subtract our former answer: $1260-360=\boxed{900}$.
H: $\nabla \varphi \overset{?}{=} \nabla \cdot \varphi \bar{\bar{I}}$ where $\varphi$ is scalar, $\bar{\bar{I}}$ is identity tensor I am trying to determine if these two are equivalent. I have a function written with both terms, and this is the only discrepancy. The gradient increases the rank of the scalar to a vector, while the divergence reduces the rank of a tensor to a vector, but is this statement correct? AI: Yes. Maybe this is easiest to see in coordinates: $$g^{ij}\nabla_i(\varphi\delta)_{jk}=g^{ij}\nabla_i\varphi\delta_{jk}+g^{ij}\varphi\nabla_i\delta_{jk}=g^{ij}\nabla_i\varphi\delta_{jk}=\nabla_k\varphi.$$ Without coordinates, this would be $$\nabla\cdot\varphi\overline{I}=\operatorname{tr}_{12}\nabla\varphi\otimes \overline{I}+\varphi\nabla\overline{I}=\operatorname{tr}_{12}\nabla\varphi\otimes\overline{I}=\nabla\varphi.$$
H: Create function F() from Points I would like to recreate a function only by knowing points on the graph. So I would have the points A(x/y) B(x/y) C(x/y) and would like to create its f() Is this possible? I heard this should be possible with a Taylor Series but to do a Taylor Series wouldn't I need a f() and its derivative in the first place? AI: A common method for determining a function which runs through some number of points is Lagrange interpolation. There are explicit formulas you can look up (Hoffman and Kunz linear algebra textbook, or look it up on Wikipedia/MathWorld) which help you construct a polynomial function which has intersection points with the desired ones you plugged into the formula. Keep in mind that there need not be a unique continuous function which runs through a given set of points in the plane and that there are a variety of interpolation techniques. You can also use the Stone-Weierstrass theorem to construct a polynomial function in a similar way, but this is a bit harder to understand, Lagrange interpolation is probably what you are looking for. Reference: Lagrange Interpolation on Wolfram MathWorld
H: Complex Exponent of Complex Numbers How does one find the algebraic solution of a Complex number raised to the power of another Complex number? Here is the work I have done so far, if there are any mistakes please inform me. A real number with a Complex Exponent: $$A \in \Bbb R, \space z \in \Bbb C,\space z = x + iy$$ $$A^z = A^{x+iy} = A^xA^{iy}$$ $$\implies e^{iy\ln A} = A^{iy}$$ $$\exp(iy\ln A) = \cos (y\ln A)+i\sin(y\ln A)$$ $$A^z = A^x[\cos (y\ln A)+i\sin(y\ln A)]$$ A Complex number with a Complex Exponent: [Using previous variables] $$C \in \Bbb C,\space C = a +bi\space\|\space re^{i\theta}, \theta =\arg C$$ $$C^z = (a+ib)^z$$ [After previous mistake the following notes are inaccurate] $$C^z = a^x[\cos (y\ln a)+i\sin(y\ln a)] + ib^xib^{iy}$$ And this is as far as I could go as the expansion of $ib^xib^{iy}$ is being problematic. Could I have some help with this expansion? AI: There is a mistake in $C^z=a^z+ib^z$, it should read $C^z=(a+ib)^z$ as I have noted in the comments. Anyway, it is much better to do the computation in polar form. In general, whenever powers are involed, polar form is more appropiate. Bearing that in mind, notice that if $C=re^{i\theta}$, we have $$C^z=\bigl(re^{i\theta}\bigr)^z=r^ze^{iz\,\theta}=r^ze^{i(x+iy)\,\theta}=r^ze^{-y\,\theta}e^{ix\,\theta}$$ Since you have worked out $r^z$ before, we are done!
H: proof of $ A - \left (B \cap C \right)= \left (A - B \right) \cup \left (A - C \right)$ I am trying to prove $ A - \left (B \cap C \right)= \left (A - B \right) \cup \left (A - C \right)$ I came up with this proof: Let $ S \{ x \| x \in A - \left(B \cap C \right)\}$ Let $ Q \{y \| y \in \left (A - B \right) \cup \left (A - C \right)\} $ All $x \in A $ and $ x \notin B,C$. All $y \in A $ and $ y \notin B,C$. Because all $x$ fit the definition of $Q$ then we say $S \subseteq Q$ Because all $y$ fit the definition of $S$ then we say $Q \subseteq S$ Since $ S \subseteq Q$ and $Q \subseteq S$ then $ S = Q \implies A - \left (B \cap C \right)= \left (A - B \right) \cup \left (A - C \right)$ $ \blacksquare $ But then quickly realized it was wrong because the $x$ from set $S$ must meet the following criteria: $x \in A $ and $ x \notin B \wedge C $ whereas the $y$ from set $Q$ must meet different criteria: $y \in A $ and $ y \notin B \vee C$ If you make a Venn diagram for the three sets and let it represent $A$, $B$, and $C$ then you see that when you take away $\left (B \cap C \right)$ from $A$ what you are taking away is the center where all three sets meet(i.e $A \cap B \cap C$) If you evaluate $\left (A - B \right) \cup \left (A - C \right)$ you take away from $A$ the center ($A \cap B \cap C$) and both $A \cap B$ and $A \cap C$, which would render the theorem I am trying to prove wrong. I don't know if I am thinking about it incorrectly, but the theorem is out of Calculus Vol. 1 By Tom M. Apostol. AI: One can produce a watertight proof by showing a double inclusion. First $A\setminus (B\cap C)\subseteq (A\setminus C)\cup (A\setminus B)$ P Pick $x\in A\setminus (B\cap C)$. Then $x\in A$ and $x\notin B\cap C$. This means that $x\in A$ and $x\notin B$ or $x\notin C$. In any case, $x\in A\setminus B$ or $x\in A\setminus C$ (why?), so $x\in (A\setminus B)\cup (A\setminus C)$. Second, $A\setminus (B\cap C)\supseteq (A\setminus C)\cup (A\setminus B)$ P Suppose $x\in (A\setminus C)\cup (A\setminus B)$. Then $x\in A\setminus C$ or $x\in A\setminus B$. Thus $x\in A$ and $x\notin C$; or $x\in A$ and $x\notin B$. But this is the same as saying $x\in A$ and, $x\notin C$ or $x\notin B$. And $x\notin C$ or $x\notin B$ is the same as $x\notin B\cap C$. Thus $x\in A,x\notin B\cap C$, i.e. $x\in A\setminus (B\cap C)$ Note we use the De Morgan Laws $$\mathscr C\left(\bigcup A\right)=\bigcap \mathscr CA$$ $$\mathscr C\left(\bigcap A\right)=\bigcup \mathscr CA$$
H: Difference between basis and subbasis in a topology? I was reading Topology from Munkres and got confused by the definition of a subbasis. What is/are the difference between basis and subbasis in a topology? AI: Bases and subbases "generate" a topology in different ways. Every open set is a union of basis elements. Every open set is a union of finite intersections of subbasis elements. For this reason, we can take a smaller set as our subbasis, and that sometimes makes proving things about the topology easier. We get to use a smaller set for our proof, but we pay for it; with a subbasis we need to worry about finite intersections, whereas we did not have to worry about that in the case of a basis.
H: Proof of Hilbert's Nullstellensatz, weak form. The statement of Hilbert's Nullstellensatz, weak form, as given here is "Let $f_1,f_2,\dots,f_n$ be polynomials in $K[x_1,x_2,\dots,x_n]$, where $K$ is an algebraically closed field. Then $1=\sum{g_t f_t}$ for suitable $g_t\in K[x_1,x_2,\dots,x_n]$ if and ony if the algebraic variety $V(f_1,f_2,\dots,f_n)\in \Bbb{K}^n =\emptyset$" I'd like to start off by proving that if $K$ is an algebraically closed field, then any polynomial $f\in K[x_1,x_2,x_3,\dots x_n]$ has a solution in $\Bbb{K}^n$. We know that any polynomial in $K[x]$ has a solution in $K$. Let us take $f\in K[x_1,x_2,x_3,\dots x_n]$. If we have $x_1=x_2=\dots x_n$, then this becomes a polynomial in $K[x_1]$. This has a solution in $K$. Hence, the original $f$ has a solution in $\Bbb{K}^n$, where all the elements in the n-tuple are equal. Proof of Hilbert's Weak Nullstellensatz: If $V(f_1,f_2,\dots f_n)=\emptyset$, then not all the polynomials can have the same factor in common. If all polynomials have a factor in common, then that polynomial will have a solution in $\Bbb{K}^n$. This contradicts the fact that $V(f_1,f_2,\dots f_n)=\emptyset$. Hence, $1=\sum{g_t f_t}$. If $1=\sum{g_t f_t}$, then all the polynomials do not have any factor in common. Hence, $V(f_1,f_2,\dots f_n)=\emptyset$. Is this proof sound? The proof that I'm currently reading is vastly different from this. Thanks in advance! AI: This proof is not sound as written, because certain polynomials (namely non-zero constant ones) don't have zeroes in $K$. E.g. if $n = 2$ and $f = 1 + x_1 - x_2$, and you set $x = x_1 = x_2$, this reduces to the constant polynomial $1$, which has no zero in $K$. The second step is also incorrect. E.g. in $K[x_1,x_2],$ the polynomials $x_1$ and $x_2$ have no factor in common, but they don't generate the unit ideal.
H: Does an analytic function maps a simple connected region into a simple connected region? Suppose $f$ is analytic, say, in $\mathbb{C}$, and suppose $\Omega$ is a bounded simple connected open domain whose boundary we denote as $\Gamma$, then is $f(\Omega)$ also a simple connected domain whose boundary is $f(\Gamma)$? I think $f(\Omega)$ is also connected becasue the continuity of $f$ suffices, but I'm not sure whether $f(\Omega)$ is simple connected and whether $f(\Gamma)$ will be the boundary of the domain. Sorry for the above too simple question... Now I put an additional condition on $f$, assuming that $f$ maps $\Gamma$ injectively into $f(\Gamma)$, then what can we say about $f(\Omega)$ ? Or, what if $f$ is injective on $\overline{\Omega}$ ? AI: As @nsrt indicates, the set $\{ z \mid \Im(z) \gt 0, 1 \lt \|z\| \lt 2 \}$ maps under $z\mapsto z^3$ to $\{\|z\| \lt 8\}\setminus \{\|z\|\le 1\}$, which is not simply connected.
H: real and imaginary part in $\sin z$ where z is complex I wanted to know, how can I determine the real and imaginary part in $\sin z$ where $z \in \Bbb{C}$? Well, this is a part of a series of questions comprising the same in $\log z$ and $\tan^{-1} z$ I was able to solve this but no idea on how to solve for $\sin z$. Any help appreciated. Thanks. AI: If $z=x+iy, \sin(x+iy)=\sin x\cos (iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$ using the relationship between Hyperbolic & Trigonometric ratios
H: About Regulated and Bounded Variations in Banach Spaces In the following definitions, we assumed that $(X,\left\\|\cdot\right\\|)$ is a Banach space. Definition 1. $f:[a,b]\to X$ is of bounded variation on $[a,b]$ if $$\operatorname{Var}(f;[a,b])=\operatorname{sup}(D)\sum \left\\|f(v)-f(u)\right\\|<+\infty$$ where the supremum is taken over all divisions $D=\{[u,v]\}$ of $[a,b]$. Definition 2. $f:[a,b]\to X$ is regulated if it has one-sided limits at every point of $[a,b]$, i.e. for every $c\in [a,b)$ there is a value $f(c+)\in X$ such that $$\lim_{t\to c^+}\left\\|f(t)-f(c+)\right\\|=0$$ and if for every $c\in (a,b]$ there is a value $f(c-)\in X$ such that $$\lim_{t\to c^-}\left\\|f(t)-f(c-)\right\\|=0.$$ Question. Using the preceding definitions, how do we show that every function of bounded variation is regulated. I need some help on this and many thanks in advance... AI: Proceed by contradiction. Assume for example that $f(c^-)$ does not exist for some $c\in (a,b]$. Then $f$ does not satisfies Cauchy's criterion at $c^-$. So one can find some $\varepsilon >0$ and two sequences $(u_n)$, $(v_n)$ such that $u_n<v_n<u_{n+1}<v_{n+1}<c$ for all $n$ and $\lim_n{u_n}=c=\lim_n v_n$ yet $\Vert f(v_n)-v(u_n)\Vert\geq \varepsilon$. For any $N\in\mathbb N$ we then have $\sum_{n=1}^N \Vert f(v_n)-f(u_n)\Vert\geq N\varepsilon$, which can be made arbitrarily large. Since $\{[u_1,v_1], \dots ,[u_N,v_N]\}$ can be "extended" into a division of $[a,b]$ (the intervals are pairwise disjoint), this shows that $f$ does not have bounded variation.
H: Zeroes of a holomorphic map $f:\mathbb{C}^n\to \mathbb{C}$ for $n\geq 2$ Why can the zeros of a holomorphic map $f:\mathbb{C}^n\to \mathbb{C}$ with $n\geq 2$ have no isolated zeros (or poles if we write it as meromorphic)? Someone says the $n$-times Cauchy Integral formula is enough, but how does it work? AI: This is a consequence of the Weierstrass Preparation Theorem. See http://en.wikipedia.org/wiki/Weierstrass_preparation_theorem. If you still believe in books, any book on several complex variables will address this; it's also discussed on pp. 7-9 of Griffiths and Harris.
H: Expansion of $\sin x$ I wanted to know, how can I derive: $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots$$ AI: Use Mclaren series : $f(x) = f(0)+f^{'}(0)\frac{x}{1!}+f^{''}(0)\frac{x^2}{2!}+f^{'''}(0)\frac{x^3}{3!}+\cdots$ for $f(x)=\sin x$
H: Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Problem The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation: $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Evaluate the limit $\lim_{n\to\infty} a_n$. Solution Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n < a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 < \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n < a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} < \sqrt{2+a_{n+1}},$$ that is $a_{n+1} < a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 < a_n < 2$$ holds for $n=1$, that is $0 < \sqrt2 < 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 < a_n < 2.$$ Adding two and taking square roots, we have that $$0 < \sqrt2 < \sqrt{2+a_n} < \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 > 0$. Is my solution correct? AI: Looks great. Here is a fun trick I've seen to answer this question. Using the half angle formula, notice the following: $$\cos\left(\frac{\pi}{4}\right)=\frac{1}{2}\sqrt 2\\\cos\left(\frac{\pi}{8}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt 2)}=\frac{1}{2}\sqrt{2+\sqrt 2}\\\cos\left(\frac{\pi}{16}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt{ 2+\sqrt2})}=\frac{1}{2}\sqrt{2+\sqrt {2+\sqrt 2}}\\\vdots\\\cos\left(\frac{\pi}{2^{n+1}}\right)=\underbrace{\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+{\ldots}}}}}_\text{n times}=\frac{1}{2}a_{n}$$ Now let $n$ approach infinity.
H: Properties of Triangle - Trigo Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +c\cos(A-\theta) = b\cos\theta$ Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +\cos(A-\theta) = b\cos\theta$ My approach : Using $\cos(A+B) =\cos A\cos B -\sin A\sin B and \cos(A-B) = \cos A\cos B +\sin A\sin B$, we get: \begin{equation} a(\cos C\cos\theta -\sin C\sin\theta) +c(\cos A\cos\theta +\sin A\sin\theta)\quad\\=\cos\theta(a\cos C+c\cos A) +\sin\theta(c\sin A -a\sin C)\qquad(\text{i}) \end{equation} By using projection formula which states: $b=a\cos C +c \cos A $ (i) will become : $$b\cos\theta +\sin\theta (c\sin A -a\sin C)$$ How do I proceed from here? AI: Law of sines states that $$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$$ which means that $c \sin{A} - a \sin{C} = 0$. The desired result follows.
H: What is $\mathfrak a$? I'm currently reading Mendelson's Introduction to topology and have came across this theorem: Theorem 3.8: Let a neighborhood in a topological space be defined by Definition 2.2 and an open set in a neighborhood space be defined by Definition 3.5. Then the neighborhoods of a topological space $( X,\mathfrak T)$ give rise to a neighborhood space $ (X,\mathfrak N) = \mathfrak a(X,\mathfrak T)$ and the open sets of a neighborhood space $ (Y,\mathfrak N')$ give rise to a topological space $ (Y,\mathfrak T')=\mathfrak a' (Y,\mathfrak N')$. Furthermore, for each topological space $(X,\mathfrak T)$, $$(X,\mathfrak T)=\mathfrak a'(\mathfrak a(X,\mathfrak T))$$ and for each neighborhood space $(X,\mathfrak N)$, $$ (X,\mathfrak N)=\mathfrak a(\mathfrak a'(X,\mathfrak N)),$$ thus establishing a one-one correspondence between the collection of all topological spaces and the collection of all neighbourhood spaces. Definition 2.2: Given a topological space $(X,\mathfrak T)$, a subset $\mathrm N$ of $X$ is called a neighborhood of a point $a \in X$ if $N$ contains an open set that contains $a$. Definition 3.5: In a neighborhood space, a subset $O$ is said to be open if it's a neighborhood of each of it's points. So, my question is: What are $\mathfrak a$ and $\mathfrak a'$? AI: The mappings $\mathfrak{a}$ and $\mathfrak{a}'$ are implicitly defined in the statement of the theorem. $\mathfrak{a}$ maps a topological space $(X,\mathfrak{T})$ to the neighbourhood space $(X,\mathfrak{N})$, where $\mathfrak{N}$ is the family of neighbourhoods of all $x\in X$ in the topology $\mathfrak{T}$, and $\mathfrak{a}'$ is the map in the other direction, that constructs a topological space $(X,\mathfrak{T})$ from a neighbourhood space $(X,\mathfrak{N})$ by letting $\mathfrak{T}$ be the family of sets that are neighbourhoods of all their elements. The theorem asserts that the family of neighbourhoods in a topological space satisfy the axioms for a neighbourhood space, the family of sets in a neighbourhood space that are neighbourhoods of all their elements satisfy the axioms of a topology, going back and forth with these constructions takes you back where you started (i.e. $\mathfrak{a}'$ is the inverse of $\mathfrak{a}$). Put in another way, neighbourhood spaces and topological spaces are equivalent, and you can choose whichever viewpoint happens to be more convenient.
H: How to resolve this probability question? today I was answering a exam and I get a problem which I have no idea how to resolve it. Here is the announcement $500$ people attend a nightclub. Those who are members of the club pay 14 dlls, and those who are not members paid 20 dlls All ($100\%$) of those who are members attend, and $70\%$ of non-members attend. How much money did the club make from these $500$ people? It does not give more details. Can you give an explanation about how to resolve it? AI: Let's start with what we know. The probability that someone attended (denoted by $A$) if they are a member (denoted by $M$) is $P(A \| M) = 1$. The probability that someone attended if they are not a member (denoted by $N$) is $P(A \| N) = 0.7$ #M + 0.7#N = 500, where $\#M$ denotes the number of members. From the latter, we have that $\#N = \frac{500 - \#M}{0.7}$. From Bayes' Theorem, we have $$P(M \| A) = \frac{P(A \| M)P(M)}{P(A \| M)P(M) + P(A \| N)P(N)}.$$ This gives us the probability that someone who did attend the party is a member. Now, we don't know the probability that someone is a member ($P(M)$), or that probability that someone who attended the party is a member ($P(M\|A)$). But we can fill some things out and maybe see what we can do. $$P(M\|A) = \frac{1 \cdot P(M)}{1\cdot P(M) + 0.7\cdot P(N)}$$ We can write a similar equation for $P(N\|A)$. Now, the amount of money made is $$500\left[ 20 P(N\|A) + 14 P(M\|A)\right].$$ Remember, the probabilities lie in the interval $[0,1]$, which is why we multiply by 500. So, we have $$\textrm{Income} = 500 \left[\frac{20 \cdot 0.7 P(N) + 14 P(M)}{P(M)+0.7 P(N)}\right] \\ = 500\cdot 14\left[\frac{P(N)+P(M)}{P(M)+0.7P(N)}\right].$$ But, someone is either a member, or not. So $P(M)+P(N) = 1$. Now, for the coup de grace. Remember that $\#N = \frac{500-\#M}{0.7}$. This means that $$P(N) = \frac{\frac{500-\#M}{0.7}}{\#M+\frac{500-\#M}{0.7}}, \\ P(M) = \frac{\#M}{\#M + \frac{500-\#M}{0.7}}. $$ So $$P(M) + 0.7P(N) = \frac{\#M + 500 - \#M}{\#M+\frac{500-\#M}{0.7}} \\ = \frac{500}{\left(1-\frac{1}{0.7}\right)\#M + \frac{500}{0.7}} $$ Therefore, the total money obtained $$\textrm{Income} = 500 \cdot 14 \left[\frac{\left(1-\frac{1}{0.7}\right)\#M + \frac{500}{0.7}}{500}\right]$$ is a function of the total number of members. If there are no members, $\#M = 0$, then the income is easily found to be $500 \cdot \frac{14}{0.7} = 500 \cdot 20 = 10000$, which matches with the obvious definition. If there are 500 members, then we have $500\cdot 14\left(\frac{500 - \frac{500}{0.7}+\frac{500}{0.7}}{500}\right) = 500\cdot 14$ pesos income. As a result, we find that the income is a linear function of the number of total members of the club.
H: Prove that in an obtuse triangle $\angle HAO = \angle B - \angle C$ Consider the following triangle with orthocentre $H$ and circumcentre $O$. Prove that $\angle HAO = \angle B - \angle C$. I am familiar with the proof for this when $ABC$ is acute, I wanted to prove it when it is obtuse. $\angle HAO = \angle HAC + \angle CAO$ Consider $\Delta DAC$, clearly $\angle HAC = 90 - \angle C$. It remains to prove that $\angle CAO = \angle B - 90$. Or, $\angle AOI = 180 - \angle B = \angle A + \angle C$. This is where I'm stuck. EDIT: I realized that in quadrilateral $AOB'B$, $\angle AOB' = 180 - \angle B$, so the proof reduces to proving that $\angle AOB' = \angle AOI$ AI: I got the answer! Let $OG$ intersect $BC$ at $L$ (I forgot to name the point). Let $\angle B = b, \angle EBB' = a$. $\angle EBB' = \angle OLB' = a \implies \angle LOB' = 90 - a$ Similarly, $\angle AOG = 90 - (b - a)$ $\implies \angle AOB' = \angle AOG + \angle OLB' = 90 - a + 90 - (b - a) = 180 - b$ as desired.
H: $PGL(n + 1)$ acts on sets of $n + 2$ points in $\mathbb{P}^n$ transitively: proof without determinants? It is "well known" that if $p_1, \dots, p_{n + 1}$ are points in $\mathbb{P}^{n -1}$ (over $\mathbb{C}$, say) in general position, and $q_1, \dots, q_{n + 1}$ are another set of such points, then there is a unique matrix in $G \in PGL(n - 1)$ sending the first set to the second. Not wishing to take the author's assertion for granted, I came up with the following proof. Take $p_i$ to be the standard basis vector in $\mathbb{C}^n$ for $1 \leq i \leq n$, and $p_{n + 1} = (1, 1, \cdots, 1)$. Let $A$ be the $n \times n$ matrix constructed by taking the first $n$ of the $q$'s as column vectors, then we must have $G = A D$ for some diagonal matrix $D$, none of whose entries is $0$. Let $d$ be the vector consisting of the diagonal elements of $D$, and $v$ be the vector consisting of the coordinates of $q_{n + 1}$. Then the requirement that $G$ send $p_{n + 1}$ to $q_{n + 1}$ amounts to solving the equation $A d = v$. We see that $A$ must be invertible, and writing $d = A^{-1} v$ and using the cofactor expansion for $A^{-1}$, the condition that each $d_i$ be nonzero can be seen to be equivalent to the condition that the points other than $p_i$ are in general position: $A^{-1} v$ is equal, up to a scalar, to the vector of the $n$ relevant matrix determinants. This proof is fine, and perhaps optimal in some sense: each of the hypotheses gets used exactly once. But it requires an arbitrary choice at the beginning, and relies on messy formulas for matrix determinants. Can this be avoided? AI: I'm not quite sure how you got so involved with determinants. Do we agree that for $n+1$ points in $\mathbb P^{n-1}$ to be in general position means that any $n$ of them span $\mathbb P^{n-1}$? First, your "arbitrary choice" at the beginning is just fine. If you show that there's a group element carrying the $n+1$ "standard" points $P_0=[1,0,\dots,0]$, $P_1=[0,1,\dots,0]$, $\dots$, $P_{n-1}=[0,0,\dots,1]$, and $P_n=[1,1,\dots,1]$ to $n+1$ arbitrary points in general position, then you get the general case by composition (using the inverse in the first case): $Q_0,\dots,Q_n\rightsquigarrow P_0,\dots,P_n \rightsquigarrow R_0,\dots,R_n$. A more conceptual way to understand what you're doing is this. Given $Q_0,\dots,Q_n\in\mathbb P^{n-1}$ in general position, we can choose homogeneous coordinate vectors $v_0,\dots,v_n\in\mathbb C^n$ so that $v_n = \sum\limits_{j=0}^{n-1} v_j$. Namely, for any choice, by general position, we have $v_n = \sum\limits_{j=0}^{n-1}\lambda_j v_j$ for some $\lambda_j\ne 0$. Now rescale. So now you take your matrix $A$ to have columns $v_0,\dots,v_{n-1}$ and clearly $AP_j = Q_j$ for all $j=0,\dots,n$.
H: Solving Summation Expressions I would like to know how do you solve summation expressions in an easy way (from my understanding). I am computer science student analyzing for loops and finding it's time complexity. e.g Code: for i=1 to n x++ end for Summation: n ∑ 1 i=1 Solving: = ∑ [n-1+1] (topLimit - bottomLimit + 1) = n (summation formula said ∑ 1 = 1+1+1+1+ ... + 1 = n) The time complexity of the for loop is: O(n) Code for(i=0; i<=n i++) for(j=i; j<=n; j++) x++; Question: How do you solve: n n ∑ [∑ 1] i=1 j=i My Solution: n = ∑ [n-i+1] i=1 = not sure how to progress from here (should i do another topLimit - bottomLimit + [n-i+1]?) The problem i am having is simplifying so i can get to say i, 1/i, i^2, .. i.e something i can use a summation formula on. I know the answer supposed to be: (n(n+1))/2. AI: You correctly got down to this sum: $$ \sum_{i=1}^n (n-i+1) $$ Now use the fact that finite sums are just additions, so you can use all the algebra rules you're used to and write it as $$ \begin{align} \sum_{i=1}^n ((n+1)-i) &= \sum_{i=1}^n(n+1)-\sum_{i=1}^n i\\ &=(n+1)\sum_{i=1}^n 1-\sum_{i=1}^n i \end{align} $$ Can you take it from there?
H: Every infinite subset of a countable set is countable. Here is the proof I tried to weave while trying to prove this theorem: Theorem. Every infinite subset of a countable set is countable. Proof. Let $A$ be a countable set and $E\subset A$ be infinite. Then $A\thicksim\mathbb{N}$. This implies that there is a sequence $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ where $x_{n}\in A$. Construct a sequence $\left\{n_{k}\right\}_{k\in\mathbb{N}}$ where $x_{n_{k}}\in A$, $n_{1}$ is the smallest positive integer such that $x_{n_{1}}\in A$, and $n_{k+1}>n_{k}$. Then $$ E=\bigcup_{k=1}^{\infty}x_{n_{k}}, $$ and $E$ is countable, because $E\thicksim\mathbb{N}$. $\blacksquare$ Is it convincing? AI: I think some things can be written in a clearer manner. First, I would change This implies that there is a sequence $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ where $x_{n}\in A$. For This implies we can write $A=\{x_n:n\geqslant 1\}$ or Let $\{x_n:n\geqslant 1\}$ be an enumeration of $A$. Then, I would say If $S$ is a subset of the natural numbers, let $\min S$ denote least element of $S$. Define $S_1=\{k:x_k\in E\}$. $S_2=S_1\setminus \{\min S_1\}$ and in general $$S_{n+1}=S_{n}\setminus\{\min S_1,\ldots,\min S_{n}\}$$ Then define $n_k=\mathscr \min S_k$ I guess the idea is clear: consider the set of subscripts such that $x_k\in E$. By the well ordering of the natural numbers, we can extract a sequence $n_k$ such that $$n_1<n_2<\cdots\\E=\{x_{n_k}:k\geqslant 1\}$$ by considering the first subscript with $x_k\in E$ removing this one from the list and looking at the new first subscript (our second in the list) and so on. Some details should be addressed $(1)$ The set $S_{n}$ is never empty. Reason: Since $E\subseteq A$; $S_1$ is not empty. Moreover, $E$ is by assumption infinite, thus removing one element every time cannot exhaust it. $(2)$ The construction exhausts the elements of $E$ -- that is, it is a surjection. Reason: Pick $m$ such that $x_m\in E$. We need to find $k$ such that $n_k=m$. Consider the finite set $\{x_1,\dots,x_m\}$. Keeping the order, remove all elements such that $x_i\notin E$. We're left with a finite set, and it must be the case $\{x_{n_1},\dots,x_{n_k}\}$ for some $k$, and $n_k=m$, by definition of the $n_k$. $(3)$ The construction is an injection. Reason: By construction, $n_k\neq n_j$ if $j\neq k$ for if $j>k$ then $n_k\notin S_{j}$. Conclusion We obtain an bijection of $E$ with an infinite subset $F$ of $\Bbb N$. Thus $E\simeq F\simeq \Bbb N$, that is $E\simeq \Bbb N$.
H: What is the minimum of $\left\|z-2(1+i)\right\|+\left\|z+1-5i\right\|+\left\|z-6+2i\right\|$ over all complex numbers? Find the Least value of $\left\|z-2(1+i)\right\|+\left\|z+1-5i\right\|+\left\|z-6+2i\right\|$ My try:: Let $A(2,2)$ and $B(-1,5)$ and $C(6,-2)$ and $P(x,y)$ be a point Here $A,B$ and $C$ are the point of a $\triangle$ and $P(x,y)$ be any point inside $\triangle ABC$ So We have to minimize $PA+PB+PC$ How Can I Minimize it, plz explain me in Detail and also what should be the answer of that Question, Thanks AI: Hint: Notice that your three points are collinear, so that the point you're looking for lies on the line defined by the three points. By drawing a picture, and testing out some different points on the line, see if you can figure out which point on the line is your desired point.
H: Evaluate $\int_0^R \frac{r^2}{(1+r^2)^2}dr.$ I am trying to evaluate the following integral: $$\int_0^R \frac{r^2}{(1+r^2)^2}dr.$$ I might substitute $u=r^2$, but I don't find $du$ anywhere. Obviously the integral should be bounded on $R\in [0,\infty)$. Any ideas? AI: Substitute $r=\tan{t}$; $dr = \sec^2{t} \, dt$. Then the integral is equal to $$\int_0^{\arctan{R}} dt \, \sec^2{t} \, \frac{\tan^2{t}}{ \sec^4{t}} = \int_0^{\arctan{R}} dt \,\sin^2{t}$$ which is then straightforward to evaluate: $$\frac12 \left [t - \sin{t} \cos{t} \right ]_{0}^{\arctan{R}} = \frac12 \left [\arctan{R} - \frac{R}{1+R^2}\right ]$$
H: Is it really true that $A^2 = -A \Leftrightarrow (I + A)^2 = A$? $A$ is a generic square matrix and $I$ is the identity matrix. I failed to prove that, but I managed to disprove it: \begin{align} A &= [-1] \\ A^2 &= ([-1])^2 = [1] = -A \\ (I + A)^2 &= ([1] + [-1])^2 = [0] \neq A \end{align} So... am I missing something? Or is the text wrong? AI: $$(I+A)^2$$ $$I^2+2A+A^2$$ $$I+2A+A^2$$ Now use $A^2=-A$ $$I+2A-A=I+A$$ Text is wrong
H: Exterior Product $d\Phi_1\wedge d\Phi_2$ and spherical coordinates One short question: If $\Phi\colon\mathbb{R}^3\to\mathbb{R}^3$, defined by $$ \begin{pmatrix}r\\\vartheta\\\phi\end{pmatrix}\mapsto\begin{pmatrix}r\sin \vartheta\cos \phi\\r\sin \vartheta \sin\phi\\r\cos\vartheta\end{pmatrix}, $$ what are then the exterior products $$ d\Phi_1\wedge d\Phi_2,~~~~~~~~~~d\Phi_1\wedge d\Phi_3,~~~~~~~~~~d\Phi_2\wedge d\Phi_3? $$ AI: Hint: $$d\Phi_1:=\frac{\partial \Phi_1}{\partial r}dr+\frac{\partial \Phi_1}{\partial \theta}d\theta+\frac{\partial \Phi_1}{\partial \phi}d\phi= \sin\theta\cos\phi dr+r\cos\theta\cos\phi d\theta-r\sin\theta \sin\phi d\phi,$$ and similarly for the other components $\Phi_2$, $\Phi_3$ of $\Phi$, i.e. $$d\Phi_2:=\frac{\partial \Phi_2}{\partial r}dr+\frac{\partial \Phi_2}{\partial \theta}d\theta+\frac{\partial \Phi_2}{\partial \phi}d\phi= \sin\theta\sin\phi dr+r\cos\theta\sin\phi d\theta+r\sin\theta\cos\phi d\phi,$$ $$d\Phi_3:=\frac{\partial \Phi_3}{\partial r}dr+\frac{\partial \Phi_3}{\partial \theta}d\theta+\frac{\partial \Phi_3}{\partial \phi}d\phi= \cos\theta dr-r\sin\theta d\theta,$$ as $\frac{\partial \Phi_3}{\partial \phi}=\frac{\partial r\cos\theta}{\partial \phi}=0.$ To arrive at the answer you need to remember that, by definition of wedge product $$dx_i\wedge dx_j=-dx_j\wedge dx_i,$$ for all $i,j=1,2,3$ with $x_1=r$, $x_2=\theta$ and $x_3=\phi$. In particular $$dr\wedge dr=d\theta\wedge d\theta=d\phi\wedge d\phi=0.~()$$ Moreover $f(r,\theta,\phi)dx_i\wedge g(r,\theta,\phi)dx_j:= f(r,\theta,\phi)g(r,\theta,\phi)dx_i\wedge dx_j$, for all functions $f$ and $g$. Let us compute $d\Phi_1\wedge d\Phi_2$: using the above formulae we have $$d\Phi_1\wedge d\Phi_2=(\sin\theta\sin\phi dr+r\cos\theta\sin\phi d\theta+r\sin\theta\cos\phi d\phi)\wedge(\sin\theta\sin\phi dr+r\cos\theta\sin\phi d\theta+r\sin\theta\cos\phi d\phi); $$ the wedge product $\wedge$ is linear: using the property $()$ we collect only the non trivial contributions arriving at $$d\Phi_1\wedge d\Phi_2=r\sin\theta\sin\phi \cos\theta\sin\phi dr\wedge d\theta+ r \sin^2\theta\sin\phi \cos\phi dr\wedge d\phi+\\ r\cos\theta \sin\theta \sin^2\phi d\theta\wedge dr+ r^2\cos\theta r\sin\theta \sin\phi \cos\phi d\theta\wedge d\phi+\\ r\sin^2\theta\cos\phi\sin\phi d\phi\wedge dr + r^2\sin\theta\cos\theta\cos\phi\sin\phi d\phi \wedge d\theta; $$ using $dr\wedge d\theta=-d\theta\wedge dr$ etc...we arrive at the final result, i.e. $$d\Phi_1\wedge d\Phi_2=(r\sin\theta\sin\phi \cos\theta\sin\phi-r\cos\theta \sin\theta \sin^2\phi) dr\wedge d\theta+ (r \sin^2\theta\sin\phi \cos\phi-r\sin^2\theta\cos\phi\sin\phi ) dr\wedge d\phi+\\ (r^2\cos\theta r\sin\theta \sin\phi \cos\phi-r^2\sin\theta\cos\theta\cos\phi\sin\phi ) d\theta\wedge d\phi. $$
H: Express $f'''_{xxx} and f'''_{yyy}$ in terms of $f'''_{uuu} and f'''_{vvv}$. Let $f(x,y)\in C^3(\mathbf{R}^2)$ and let $u=x+y$ and $v=y$. Express $f'''_{xxx} and f'''_{yyy}$ in terms of $f'''_{uuu} and f'''_{vvv}$. I'm supposed to use the chain rule, how do I go about? Thanks! Alexander AI: If $f$ is of the form $f(x_1,\ldots,x_n)$ I will use the notation $D_if$ for the partial derivativeof $f$ w.r.t. the $i$-th variable. Similarly, $D_{k,i}f=D_kD_if$ denotes the partial derivative of $D_if$ w.r.t. the $k$-th derivative. Then $$\begin{align}D_1\left(f(x+y,y)\right)&=D_1f(x+y,y)D_1(x+y)+D_2f(x+y,y)D_1(y)\\&=D_1f(x+y,y)\end{align}$$ $$\begin{align} D_2 (f(x+y,y))&=D_1f(x+y,y)D_2(x+y)+D_2f(x+y,y)D_2(y)\\&=D_1f(x+y,y)+D_2f(x+y,y)\end{align}$$ Continuing, $$\begin{align} D_{1,1}(f(x+y,y))&=D_1(D_1(f(x+y,y))\\&=D_1(D_1f(x+y,y))\\&=D_{1,1}f(x+y,y)D_1(x+y)+D_{2,1}f(x+y,y)D_1(y)\\&=D_{1,1}f(x+y,y)\end{align}$$ $$\begin{align} D_{2,2}(f(x+y,y))&=D_2(D_2(f(x+y,y))\\&=D_2(D_1f(x+y,y))+D_2(D_1f(x+y,y))\\ &={{D_{1,1}}f\left( {x + y,y} \right){D_2}\left( {x + y} \right) + {D_{2,1}}f\left( {x + y,y} \right){D_2}\left( {x + y} \right)}\\ &+{D_{1,2}}f(x + y,y){D_2}\left( {x + y} \right) + {D_{2,2}}f(x + y,y){D_2}\left( y \right)\\&={D_{1,1}}f\left( {x + y,y} \right) + {D_{2,1}}f\left( {x + y,y} \right) + {D_{1,2}}f(x + y,y) + {D_{2,2}}f(x + y,y)\\&= {D_{1,1}}f\left( {x + y,y} \right) + 2{D_{1,2}}f(x + y,y) + {D_{2,2}}f(x + y,y)\end{align}$$ ...and so on. Note that $D_2(f(x+y,y))$ here denotes the partial derivative of the function composite function $f\circ g$, where $g(x,y)=(x+y,y)$, in parenthesis, while $D_2f(x+y,y)$ denotes the partial derivative $D_2f$ evaluated at $(x+y,y)$.
H: Prove that $ n < 2^{n}$ for all natural numbers $n$. Prove that $ n < 2^{n} $ for all natural numbers $n$. I tried this with induction: Inequality clearly holds when $n=1$. Supposing that when $n=k$, $k<2^{k}$. Considering $k+1 <2^{k}+1$, but where do I go from here? Any other methods maybe? AI: Proof by induction. Let $n \in \mathbb{N}$. Step $1.$: Let $n=1$ $\Rightarrow$ $n\lt2^{n}$ holds, since $ 1\lt 2$. Step $2.$: Assume $ n \lt 2^{n}$ holds where $n=k$ and $k \geq 1$. Step $3.$: Prove $n \lt 2^{n}$ holds for $n = k+ 1$ and $ k\geq 1$ to complete the proof. $k \lt 2^{k}$, using step $2$. $2\times k \lt 2\times2^{k}$ $ 2k \lt 2^{k+1}\quad(1)$ On the other hand, $k \gt 1 \Rightarrow k + 1 \lt k+k = 2k$. Hence $k+1\lt2k\quad(2)$ By merging results (1) and (2). $k + 1 \lt 2k \lt 2^{k+1}$ $k + 1 \lt 2^{k+1}$ Hence, $ n \lt 2^{n}$ holds for all $ n \in \mathbb{N}$
H: Geometric description of points in the complex plane What does the following inequality look like if sketched in the complex plane: $$\operatorname{Re}(az+b)>0$$ The $a$ and $b$ are both complex numbers (function parameters). I understand that $\operatorname{Re}(z)>0$ will yield all of the complex numbers to the right of the imaginary axis (quadrants 1 and 4). I also understand that $\operatorname{Re}(z+b)>0$ will shift the shaded region $\operatorname{Re}(b)$ to the left of the imaginary axis. What does multiplying the $z$ by $a$ do to the inequality ($\operatorname{Re}(az)>0$)? Complex multiplication involves dilation and rotation, so would the shaded (included) region rotate? AI: If $z=x+iy$, $$Re(az+b)=Re(a)x-Im(a)y+Re(b)$$ So, $$Re(az+b)>0\Rightarrow mx-ny+c>0$$ where $m=Re(a),\ n=Im(a),\ c=Re(b)$ So it is the one half of $\mathbb{R}^2$ partitioned by the line $\displaystyle mx-ny+c=0$ So, basically, it is $Re(z)>0$ rotated by $a$ and translated by $b$.
H: The Schwarz Reflection Principle for a circle I'm working on the following exercise (not homework) from Ahlfors' text: " If $f(z)$ is analytic in $\|z\| \leq 1$ and satisfies $\|f\| = 1$ on $\|z\| = 1$, show that $f(z)$ is rational." I already know about the reflection principle for the case of a half plane, so I tried using the "Cayley transform" $$T (\zeta)=\frac{\zeta-i}{\zeta+i}$$ Which maps the closed upper half plane onto the closed unit disk with $1$ removed. I defined $$g(\zeta)=(T^{-1} \circ f \circ T)(\zeta)=i\frac{1+f \left( \frac{\zeta-i}{\zeta+i} \right)}{1-f \left( \frac{\zeta-i}{\zeta+i} \right)},$$ And tried to apply the reflection principle in the book. $g$ is indeed analytic in the upper half plane, but for $\zeta \in \mathbb R$, I'm afraid that $g$ might get infinite (because on the boundary, $f$ takes values on the unit circle). If so, it will not be continuous and not even real, and the reflection principle is not applicable. Am I missing something here? After all Ahlfors does mention in the text a generalized reflection principle for arbitrary circles $C,C'$. Thanks AI: In the given situation, we can proceed directly. The reflection in the unit circle is given by $$\rho(z) = \overline{z}^{-1},$$ so by setting $$g(z) = \frac{1}{\overline{f(\overline{z}^{-1})}},$$ we obtain a function $g$ that is meromorphic in the outside of the unit disk. Since $f$ can have only finitely many zeros in $\mathbb{D}$, $g$ has only finitely man poles in $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$, and since $\lvert f(z)\rvert = 1$ for $\lvert z\rvert = 1$, the function $$h(z) = \begin{cases}f(z) &, \lvert z\rvert \leqslant 1\\ g(z) &, \lvert z\rvert > 1\end{cases}$$ is continuous (outside the poles, none of which lies on $\partial \mathbb{D}$), and holomorphic outside $\partial \mathbb{D} \cup \{\text{poles}\}$. By a small modification of Morera's theorem (you can map each arc on the circle to the real axis by a Möbius transformation), it is meromorphic on all of $\hat{\mathbb{C}}$, hence rational. You can also use the Cayley transform as you started with, if $f$ is not constant, then $f$ can take the value $1$ only finitely often on $\partial\mathbb{D}$, and $g = T^{-1}\circ f \circ T$ has only finitely many poles on $\mathbb{R}$, and either a pole or a removable singularity in $\infty$, on each interval between two poles, you can apply the ultra-classic reflection principle to see that $g$ can be extended by reflection to a meromorphic function on $\hat{\mathbb{C}} \setminus \{\text{poles}\}$, hence is rational.
H: Gaps between primes I recently watched a video about the recent breakthrough involving the gaps between primes. I have an idea that I'm sure is wrong, but I don't know why. If you take the product of all prime numbers up to a certain number and call it x, won't x-1 and x+1 always be primes? And since they always differ by 2, doesn't that make there an infinite number of primes that differ by 2? Once again, I know that I'm wrong, but I would like to know why. AI: Say $p_k$ is the largest prime $\le$ some number $n$. Then take $\displaystyle x=p_1p_2\cdots p_k$ Now, of course $x-1$ and $x+1$ are not divisible by any of $p_i,\ 1\le i\le k$. But there may be $p_k<p_i< x-1$ and $p_k<p_j<x+1$ such that $p_i\|x-1$ and $p_j\|x+1$. Just look at the example given by @DanielFischer to appreciate this fact.
H: Does $2^{\mathfrak m}=2^{\mathfrak n}$ imply $\mathfrak m=\mathfrak n$? Suppose $\mathfrak m$ and $\mathfrak n$ are infinite cardinals. Does $2^{\mathfrak m}=2^{\mathfrak n}$ imply $\mathfrak m=\mathfrak n$? AI: This is independent of ZFC. It is implied by GCH for example, but there exist models where (say) $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$.
H: Get the position by only knowing a landmark and the relative position to it I have a little problem finding a solution for this problem: My position is unknown. I can see a landmark (angle, range) and so I know the position $LM_{local}$ in my local coordinate frame. I recognized this landmark, and so I also know the global position of this landmark. LM(x,y, $\theta$) But I have a problem finding my global position. I only get a correct position at a specific orientation of myself. When I turn myself in place, the calculated position goes in a circle around the landmark... I think my problem could be, not knowing the orientation of the landmark in my local coordinate system... But I am not sure... Is this problem solvable with one landmark? The green elements are known, and the red X is the position to be calculated. AI: In vector notation, you know the vector of the landmark $\vec{L}$ with respect to some origin $O$, and the position of the landmark relative to unknown position $\vec{R}$. The unknown position $\vec{P}$ with respect to $O$ is simply $$\vec{P}=\vec{L}-\vec{R}$$ The vector subtraction means that, when a vector is expressed in terms of Cartesian coordinates, you subtract each component. EDIT I may have oversimplified this. Basically, I think you know a position of a landmark with respect to both a local and global origin, and you know how you are oriented. You just do not know where you are with respect to the global origin. The relationship between the local and global coordinate system is as follows: $$x'=x_0+x \cos{\theta}-y \sin{\theta}$$ $$y'=y_0+x \sin{\theta}+y \cos{\theta}$$ where $(x_0,y_0)$ is the unknown location of the local origin in the global coordinate system, $(x,y)$ is the location of the landmark in the global coordinate system, $(x',y')$ is the position of the landmark in the local coordinate system, and $\theta$ is your orientation angle. Knowing your data, it is straightforward to find your position.
H: Finding the angles of the start and endpoint of an arc I have a line $100$ $mm$ long and I want to draw an arc from endpoint to endpoint with a height of 3mm. I use this formula to find the radius of the arc $$\frac{W . W}{ 8 H} + \frac{H} {2}= 418.1667.$$ But the CAD software I'm using, FreeCAD, uses radius and the angles for the start and endpoints to draw the arc. How do I find the start and end angles that will give me a 100mm straight line between them? AI: See the picture below. From Pythagoras, $R^2=50^2+(R-3)^2,0=2500-6R+9,R=\frac {2509}6 $ confirming your value. Then $\sin \theta=\frac {300}{2509}, \theta = \arcsin \frac{300}{2509}\approx 6.867^\circ$. Since the tangent is perpendicular to the radius at the point of tangency, this is the angle between the tangent and the chord, which I suspect is the angle you are looking for.
H: Euler's identity: why is the $e$ in $e^{ix}$? What if it were some other constant like $2^{ix}$? $e^{ix}$ describes a unit circle in polar coordinates on the complex plane, where x is the angle (in radians) counterclockwise of the positive real axis. My intuition behind this is that $\frac{d}{dx}e^{ix}=i\cdot e^{ix}$. Since multiplication by i is a 90-degree rotation, we could think of $e^{ix}$ as the position vector of a particle and $\frac{d}{dx}e^{ix} = i\cdot e^{ix} $ as its velocity (x could be time). The velocity is always perpendicular to the position vector, so we have circular motion. Hopefully I've described this well, see also http://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/ if you don't understand where I'm coming from. What I don't understand is why you need the "e" in Euler's identity. What if it were some other constant: for example, 2? You wouldn't get a circle, but how can I visualize what it is that you would get? For example, what would $2^{ix}$ look like on the complex plane? I note that $2^{ix} = e^{ix\cdot ln(2)}$, and we could substitute that into Euler's identity and get $e^{ix\cdot ln(2)}=cos(x\cdot ln2) + i\cdot sin(x\cdot ln2)$. So my question really has two related parts: 1) Why do we take e (and not some other number) to the power of ix to get a circle? 2) What would it look like if we took some other number to the power of ix? $e^{ix}$ really gives us a constant-radius spiral in three dimensions (e.g. http://www.songho.ca/math/euler/euler.html), what would $2^{ix}$ look like in complex 3d space? How could I have figured that out? Thank you for your help. AI: We in general define (ignoring $a=0$) $$a^{ix}\equiv e^{i x\ln a}=e^{ixR -xM}$$ where we take the principal value of the logarithm and let $\ln a= R+iM$ be split into real and imaginary parts. If your only aim is to have the locus being a circle then any $a$ such that $M=0$ will do - equivalently, you need $a>0$. The only difference from the $e$ case is that the speed at which you go round the circle is rescaled by $R$. (Your nice idea of differentiation and noticing orthogonality still works here.) If you want the circle to be traversed at a speed such that $x$ is $2\pi$ periodic then you need $R=1\iff a=e$. If you consider negative $a$ or general imaginary $a$ then you can see from the above formula that you get a circle multiplied by a new term $e^{-xM}$ which stretches it as it's drawn. This makes a (logarithmic) spiral. In 3D space, the rescaled circles become helices which are more or less stretched out (like springs are compressed or elongated). The imaginary ones give various spirals stretched out across space. Graph them if you're interested by defining parametric equations $x=t,y=e^{-Mt}\cos (Rt),z=e^{-Mt}\sin (Rt)$.
H: Find the upper and lower limits of the sequence Finding the results (4.1) (4.2) (4.3) was really easy. But it's hard to understand what lines in the red box function or work in this solution. What I understood about lines in the red box is that if there are infinitely many odds and evens, then it is not Cauchy, and if there are either only infinitely many odds or evens, then it is Cauchy. I don't even get what this assumption means. I think there are infinitely many odds and evens in n... How is proving if it is Cauchy or not related with finding the upper and lower limits of the sequence? Thank you! AI: Since the reals are a complete metric space, a sequence converges if and only if it is Cauchy. When finding upper and lower limits of a sequence, we only consider the set of all convergent subsequences. Thus, if a subsequence is not Cauchy, it does not converge, so we ignore it. The solution is saying that there are only two values to which a subsequence can converge, so these must be the lower and upper limits.
H: The preimage of $\triangle$ is a compact zero-dimensional manifold. $f: X \to Y$, $g: Z \to Y$ and $Z$ are appropriate for intersection theory ($X,Y,Z$ are boundaryless oriented manifolds, $X,Z$ is compact, $Z$ is closed submanifold of $Y$, and $\dim X + \dim Z = \dim Y$), $f$ is transversal to $Z$. (1) If $\triangle$ denotes the diagonal of $Y \times Y$, and $f \times g : X \times Z \to Y \times Y$ is the product map, then $f(x) = g(z)$ precisely at pairs $(x,y)$ in $(f \times g)^{-1}(\triangle)$. Prove $\dim (X \times Z) = \operatorname{codim} \triangle$. (2) If $f \times g \pitchfork \triangle$, the preimage of $\triangle$ is a zero-dimensional manifold. (3) The preimage of $\triangle$ is compact. AI: Jellyfish, you need to pay attention to details yourself! Aren't $X$ and $Z$ complementary dimension in $Y$? Do the arithmetic. No, it doesn't assume transversality of the maps.
H: Probability of drawing a pair from a poker hand, unordered with replacement? I am wondering what is the probability with which you can draw a pair in a 5-card hand from a standard 52-card deck, if order does not matter in the context of cards in the hand, and if the cards can be replaced with every draw? AI: To find the probability of a one-pair hand, it will be easier if we take the order of drawing into account. It is true that all we care about is what hand we end up with. But taking order into account will get us the right answer without the need to look at special cases. There are $52^5$ equally likely sequences of $5$ cards. We count the number of $5$-card sequences that satisfy the "one pair" condition. Counting "hands" will not work unless we are very careful, for not all hands are equally likely. The two locations in the draw where the pair will be can be chosen in $\binom{5}{2}$ ways. Once we have chosen the two locations, the kind we have two of can be chosen in $13$ ways. Once we have done that, the leftmost chosen "pair" location can be filled in $4$ ways, and for each such way we can fill the other pair location in $4$ ways, for a total of $4^2$. Now it is time to deal with the remaining $3$ positions. There are $48$ ways to fill the leftmost empty position. For each of these, we can fill the next position in $44$ ways, and then the last position can be filled in $40$ ways. The number of one-pair sequences is therefore $$\binom{5}{2}(13)(4^2)(48)(44)(40).$$ Divide by $52^5$ for the probability. Remark: We interpreted "the cards can be replaced at every draw" to mean that the drawn card is replaced in the deck before the next draw. So the model is that we draw a card, record its value, replace and shuffle, doing this a total of $5$ times. We want the probability that among the $5$ recorded cards, there are exactly two cards (possibly identical) of one kind, and $3$ useless cards all of different kinds.
H: Combinatorics - placing numbered balls (distinguishable) into distinguishable bins Say we have 10 balls, numbered $1$ through $10$, and $30$ distinguishable bins. How many ways are there to distribute the balls among the bins? I think the answer is just $30^{10}$ for this. Is that correct? If not, what am I missing here? AI: You're not missing anything: your conclusion is correct, barring any restrictions on how the balls are to be distributed: there are indeed $30^{10}$ ways of distributing the numbered balls.
H: Problem about intermediate fields in the extension Let $E,K$ be intermediate fields in the extension $L/F$ (a) If $[EK:F]$ is finite, then $$[EK:F] \leq [E:F][K:F] $$ (b) If $E$ and $K$ are algebraic over $F$, then so is $EK$ For (a), I try two ways: (1) I try to prove that $[EK:K]\leq [E:F]$ and $[EK:E]\leq [K:F]$. (2) $x_1,\dotsc,x_m$ are base of $E/F$, $y_1,\dotsc,y_n$ are base of $K/F$, but how to prove that $\{x_iy_j\}$ span $EK$? AI: Hints: A. Pick bases $S$ (resp. $T$) for $E/F$ (resp. $K/F$(. Show that the products $st,s\in S, t\in T$ span all of $EK$. B. Review what you know about products and sums of two algebraic elements. Extending... In both parts, A and B, you know that $E$ and $K$ are algebraic. This implies that the compositum is $$ EK=\{\sum_{i=1}^n e_ik_i\mid n\in\mathbb{N}; e_i\in E, k_i\in K\ \text{for all $i$}\}, $$ i.e. the elements of the compositum are finite sums of products of elements of $E$ and $K$. For part A you can pick bases $S=\{s_1,s_2,\ldots,s_m\}$ and $T=\{t_1,t_2,\ldots,t_n\}$ as in the hint. Take a sum $\sum_i e_ik_i$ as above. For all $i$, you can write $e_i=\sum_j a_{ij}s_j$ and $k_i=\sum_\ell b_{i\ell}t_\ell$ with all the coefficients $a_{ij},b_{i\ell}\in F$. Plugging these in you see that $$ \sum_ie_ik_i=\sum_{j=1}^m\sum_{\ell=1}^n\left(\sum_{p,q}a_{pj}b_{q\ell}\right)s_jt_\ell. $$ So all the elements of $EK$ can be written as $F$-linear combinations of the elements $s_jt_\ell$. Therefore the elements $s_jt_\ell$ span $EK$. There are $mn=[E:F][K:F]$ of them. In part B you are supposed to recall results saying that if all the elements $e_i$ and $k_i$ are algebraic over $F$, then so are the products $e_ik_i$. And also that the sum $$ e_1k_1+e_2k_2+e_3k_3+\cdots+e_nk_n $$ is algebraic as a sum of algebraic elements.
H: Number of non-negative integer solutions for linear equations with constants How do we find the number of non-negative integer solutions for linear equation of the form: $$a \cdot x + b \cdot y = c$$ Where $a, b, c$ are constants and $x,y$ are the variables ? AI: Not a complete answer, but a relatively simple one and approximate one. By Schur's theorem of combinatorics?, the number of solutions is asymptotically ($c \to \infty$): $$ \frac{c}{ab} $$ Schur's theorem of combinatorics states that the number of solutions of (with $a_i$ relatively prime): $$ \sum_{i=1}^M a_i x_i = c $$ is: $$ \frac{c^{M-1}}{(M-1)!\prod a_i} $$ ? This name is used by Wilf's Generatingfunctionology, but I cannot seem to find it elsewhere. It appears that Schur has many theorems.
H: Examine for absolute and conditional convergence Examine the following series for absolute/conditional convergence: $$S_n = \sum_{n=1}^{\infty}{n^2\sin(n\pi/2)\sin(\pi/n^3)}$$ Attempt: I see that $$\sin(k\pi/2)= 0 ,\quad k=2n$$ and $$\sin(k\pi/2)=(-1)^k, \quad k=2n-1$$ In other words, the series terms are zero for all $n$ even, and when $n$ is odd, the terms oscillate in signs. So I was able to reduce the series to the following: $$S_n = \sum_{k=1}^{\infty}{(-1)^k\cdot (2k-1)^2\cdot \sin(\pi/(2k-1)^3)}$$ Using the alternating series test, I see that it satisfies the condition that $$\lim_{n \to \infty}{b_n} = 0$$ But I am unsure of how to proceed and show that the terms $b_n$ are monotonically decreasing. Also I claim that the series is not convergent absolutely,though unclear on how to show it( I was thinking of doing limit comparison test) . AI: As you observe, this series is equivalent to $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\sin(\pi/n^3).$$ Using the series expansion $\sin x = x - \frac{1}{3!} x^3 + \frac{1}{5!}x^5-\cdots$ we get that $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\sin(\pi/n^3)=\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\left(\frac{\pi}{n^3}-\frac{\pi^3}{3!\cdot n^9}+\frac{\pi^5}{5!\cdot n^{15}}-\cdots\right)$$ and observing that $$\left\|(-1)^{(n+1)/2}n^2\left(-\frac{\pi^3}{3!\cdot n^9}+\frac{\pi^5}{5!\cdot n^{15}}-\cdots\right)\right\|\leq \frac{\pi^3}{3!\cdot n^7}$$ we see that the series $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\left(-\frac{\pi^3}{3!\cdot n^9}+\frac{\pi^5}{5!\cdot n^{15}}-\cdots\right)$$ is absolutely convergent, so the absolute/conditional convergence of our original series is determined by the series $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\frac{\pi}{n^3}=\pi\sum\limits_{n\text{ odd}}(-1)^{(n+1)/2}\frac{1}{n}$$ which is easily seen to be conditionally but not absolutely convergent.
H: Why do I keep getting a vertcat error in MATLAB? For some reason the following code gives me the following error: Error using CI2 (line 94) Error using vertcat Dimensions of matrices being concatenated are not consistent. The code is: clear all % Odds Ratio for no clinically significant change, according to original % Study a=[0.48; %Aripiprazole 0.70; %Olanzapine 0.66; %Quetiapine added to ADs 0.57]; %Risperidone CI=[0.37 0.63; 0.48 1.02; 0.51 0.87; 0.36 0.89]'; figure(1) % Setting the figure to figure 1 clf(1) plot(1:length(CI),a,'O','markersize', 6) % plot the mean hold on; plot(1:length(CI),CI(1,:),'v','markersize', 6) % plot lower CI boundary hold on; plot(1:length(CI),CI(2,:),'^','markersize', 6) % plot upper CI boundary hold on; for I = 1:length(CI) % connect upper and lower bound with a line line([I I],[CI(1,I) CI(2,I)]) hold on; end; xlim([0.5 4.5]) my_labels = ['Aripiprazole'; 'Olanzapine '; 'Quetiapine1 '; 'Risperidone ']; title('The Efficacy of SGAs as adjuncts in the Treatment of MDD (OR for non-response)') set(gca,'XTick',[1 2 3 4]); set(gca,'XTickLabel',my_labels); xticklabel_rotate([],45) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Odds Ratio for No Clinically Significant Change (CGI) clear all a=[0.51; %Aripiprazole 0.64]; %Quetiapine added to ADs CI=[0.34 0.78; 0.49 0.84]'; figure(2) % Setting the figure to figure 2 clf plot(1:length(CI),a,'O','markersize', 6) % plot the mean hold on; plot(1:length(CI),CI(1,:),'v','markersize', 6) % plot lower CI boundary hold on; plot(1:length(CI),CI(2,:),'^','markersize', 6) % plot upper CI boundary hold on; for I = 1:length(CI) % connect upper and lower bound with a line line([I I],[CI(1,I) CI(2,I)]) hold on; end xlim([0.75 2.25]) my_labels = ['Aripiprazole '; 'Quetiapine-1 ']; title('The Efficacy of SGAs as adjuncts in the Treatment of MDD (OR for non-response [CGI])') set(gca,'XTick',[1 2]); set(gca,'XTickLabel',my_labels); xticklabel_rotate([],45) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Odds Ratio for number of participants WITHOUT a remission clear all a=[0.48; %Aripiprazole 0.60; %Olanzapine 0.64; %Quetiapine 0.39]; CI=[0.37 0.63; 0.51 1.04 0.49 0.84 0.22 0.69]'; figure(3) % Setting the figure to figure 3 clf plot(1:length(CI),a,'O','markersize', 6) % plot the mean hold on; plot(1:length(CI),CI(1,:),'v','markersize', 6) % plot lower CI boundary hold on; plot(1:length(CI),CI(2,:),'^','markersize', 6) % plot upper CI boundary hold on; for I = 1:length(CI) % connect upper and lower bound with a line line([I I],[CI(1,I) CI(2,I)]) hold on; end xlim([0.5 4.5]) my_labels = ['Aripiprazole '; 'Olanzapine'; 'Quetiapine '; 'Risperidone']; title('The Efficacy of SGAs as adjuncts in the Treatment of MDD (OR for non-response [CGI])') set(gca,'XTick',[1 2 3 4]); set(gca,'XTickLabel',my_labels); xticklabel_rotate([],45) AI: As I mentioned in your other question, the labels are character arrays. You can only concatenate arrays of the same size. So ['dog';.. 'cat'] is ok, but ['banana';... 'house'] is not. You have to add spaces to make sure that all of your lines are as long as the longest entry. ['banana'; 'house ';] As I alluded to in the other question, and as Daryl suggested, you can use a cell array: my_labels = {'Aripiprazole','Olanzapine','Quetiapine','Risperidone'}; Notice how they're commas instead of semi-colons, and note the curly braces instead of square brackets. This is probably easier to do, and it is probably the right way to do it.
H: How many 32-bit strings have fewer 1s than 0s? Now I know the answer is the summation of $32$ choose $0$ all the way to $32$ choose $15$. My class came up with this formula. I thought I understood how we got it but now I don't. If anyone would explain this method of shortening that calculation that would be great. if $n$ is even $\ \ 2^{n-1 } -$ $n-1 \choose n/2$. AI: There's a bijection between the strings with fewer 0s than 1s and vice versa; you only need to count the number of strings with exactly the same number of each. There are $\binom{32}{16}$ of these. Of the remaining strings, half satisfy your needs, so the answer is $$\frac{1}{2}\left(2^{32} - \binom{32}{16}\right).$$
H: Is this a Combination problem? In how many ways can the letters of the word ADRIENNE be arranged so that no two vowels are together? Simplify your answer. My answer is the following: $\dfrac{8!}{2!2!} - \dfrac{4!}{2!} \cdot \large ^5C_4 \cdot \dfrac{4!}{2!} = 9,360$ . But my instructor's answer is $720$. Why? AI: There are five possible setup. You can have one of $$ VCVCVCVC\\ CVCVCVCV\\ VCCVCVCV\\ VCVCCVCV\\ VCVCVCCV\\ $$ Each of them have $\frac{4!}{2!}\frac{4!}{2!}$ ways of occuring, making it $120$ per setup. Five of them makes $720$.
H: How would you interpret this question focusing on problem solving? The first step of problem solving is to understand what the problem is asking, that is where I am stuck. One of the legs of a right triangle has length 4 cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse. This is the picture I first came up with, since a is perpendicular to h. But the answer is $4\sqrt{h^2-16}/h$. This means they do not really want $a$, as that would seem to be half $h$. I need a fourth opinion here, I have asked others and no luck here. AI: Note that $b=\sqrt{h^2-16}$. Also, by similar triangles (or using the sine of the angle at the bottom left corner), $$\frac{a}{4}=\frac{\sqrt{h^2-16}}{h}.$$ Remark: The reason you say that $a$ is half of $h$ is that your picture has $b$ almost the same length as the bottom side, so the whole picture looks like a square. But it need not be. The side $b$ could be a lot bigger or smaller than $4$.
H: The tangent space of $\mathrm{Aut}(T_eG)$ Let $G$ be a Lie group and $e \in G$ be the identity. I want to understand the following sentence. “$\operatorname{Aut}(T_eG)$ being just an open subset of the vector space of endomorphisms of $T_eG$, its tangent space at the identity is naturally identified with $\operatorname{End}(T_eG)$.” I don’t understand why $\operatorname{Aut}(T_eG)$ is an open subset. how to naturally identify $T_e(\operatorname{Aut}(T_eG))$ with $\operatorname{End}(T_e G)$. I think I am lacking basic knowledge about tangent spaces. I appreciate any help. AI: If $G$ is a manifold of dimension $n$, then $T_eG$ is just a $n$-dimensional real vector space, so that after choosing a basis of $T_eG$, we can make an identification $T_eG\cong \mathbb{R}^n$. The endomorphisms of $\mathbb{R}^n$ (i.e., the linear maps from $\mathbb{R}^n$ to itself) can be identified with the $n\times n$ matrices, and hence so can the endomorphisms of $T_eG$. The automorphisms of $\mathbb{R}^n$ are the invertible, a.k.a. “non-singular” matrices. The space of $n\times n$ matrices has its own topology, and in this topology, the invertible matrices form an open set; that is covered in this older answer of mine. As to your second question, for any real vector space $V$ of finite dimension $n$, and any $v\in V$, there is a natural identification of $T_vV$ with $V$ (this is Prop 3.8 in Lee’s Introduction to Smooth Manifolds, 1st ed.). That gives us an identification $$T_\text{id}(\operatorname{End}(T_eG))\cong \operatorname{End}(T_eG)$$ where $\text{id}\in\operatorname{End}(T_eG)$ is the identity map from $T_eG$ to itself (I assume this is what you intended in your problem statement). Additionally, for any manifold $M$, any open set $U\subseteq M$, and any $p\in U$, there is a natural identification of $T_pU$ with $T_pM$ (this is Prop 3.7, ibid.). Thus, we have a sequence of identifications $$T_\text{id}(\operatorname{Aut}(T_eG))\cong T_\text{id}(\operatorname{End}(T_eG))\cong \operatorname{End}(T_eG).$$
H: What is the function to the following graph? What is the function to the following graphs? I am just looking for a rough estimate. It doesn't need to match the exact graph. AI: It depends on what you think is happening to the right. Clearly it goes to infinity for $x=0$ so there is a denominator of $x^n$. If they each approach a constant, I would say each is $a+x^{-n}$. The fact that the corner on the green one is sharper than the others would say it has a greater $n$. I would collect some points from each graph, estimate the asymptote, and see what $n$ fits best. If you think the rise of the red one toward the right is real, you could add a term $+bx$ for a rather small $b$.
H: Searching a function expressing $\sin x$ versus $\cos x$ I'm searching a specific function, I searched everywhere and didn't find anything. Here is my problem with $X=\cos x$ I'm trying to find $f(X)= \sin x$, so that this function shall use only $ \cos x$ as variable. Can someone help me? If you think that ain't achievable let me know it! Thanks in advance. Thanks for your answers but they are incorrect,let me explain. Everyone knows that $\sin^2x+\cos^2x=1$ the issue is that $\sqrt{1-\cos^2x}$ is not equal to $\sin x$ but the absolute value of $\sin x$. By another means I've already found such a result. So the question remains. So the main trouble is now to express using exclusively $\cos x $ as a variable the sign function of $\sin x $ which seems for me very difficult and the reason why I'm asking for assistance. Let me clarify i'm not asking how to change sinus into cosine but if anyone knows a special function that does. Once again Thanks by advance. AI: There is no such function. As Amire Bendjeddou mentioned, about the best you can get is $\sin x = \cos(\tfrac\pi2-x)$. The problem is that the sine and cosine functions can't be calculated purely from each other. For example, $$\sin 0 = \sin \pi$$ but $$\cos 0 \ne \cos \pi.$$ Similarly, $$\cos\frac\pi 2 = \cos\frac{3\pi}2,$$ but $$\sin \frac\pi 2 \ne \sin\frac{3\pi}2.$$
H: Diophantine Equation: $xy+ax+by+c=0$ How to find integer solutions $x,y$ of $xy+ax+by+c=0$ for given $a,b,c \in \mathbb{Z}$? Is there somewhere a treatise on this kind of equations? AI: Remark: We will assume that the equation is $xy+ax+by+c=0$. If it really is $xy+(a+b)x+c=0$, then it can be done in the same way as the problem below, but simpler. The equation then becomes $x(y+a+b)=-c$, and we are looking for factors of $-c$. Hint: Rewrite as $(x+b)(y+a)=ab-c$. We can find the solutions in a straightforward way once we find all divisors of $ab-c$. This includes negative divisors.
H: Isomorphism of polynomial rings in several variables I have been struggling with the following problem: How can one prove that if there is an isomorphism between several variable polynomial rings over a field $K$, $ \varphi : K[X_1, \dots, X_n] \to K[X_1, \dots, X_m]$ such that $\varphi$ restricted to $K$ is $id$, then one must have $m=n$. Any help appreciated, thanks. AI: Extend $\varphi$ to a $K$-isomorphism of fields between $K(X_1, \dots, X_n)$ and $K(X_1, \dots, X_m)$. Since both are field extensions of $K$ and their transcendence degree are $n$, respectively $m$, then $m=n$.
H: Weak flat condition? Let $R$ be a unit ring (not necessarily commutative). Then it is clear that for a right $R$-module $M$ we have: $M$ is flat $R$-module $\Rightarrow$ for any left $R$-module $E$ with $E\otimes_{R}M=0$ and any submodule $E'\leq E$, we have $E'\otimes_{R}M=0$. Can anyone give me a counterexample to show the inverse implication is not correct? Thanks in advance! AI: Let $M=R\oplus T$ where $T$ is not flat. Then $M\otimes E=E\oplus (T\otimes E)$ which is 0, if and only if $E=0$.
H: Why is it necessary $n\geq 2$ in this problem about boundary and isolated point? Let $A\subset\mathbb{R}^n$ be an open set, where $n\geq2$. Prove that given $a\in\mathbb{R}^n-A$, are equivalent: (i) the set $A\cup\{a\}$ is open; (ii) $a$ is an isolated point of the boundary of $A$; (iii) There exists $r>0$ such that $B(a;r)-\{a\}\subset A$. Why is it necessary $n\geq 2$ in this problem? I'm confusing because I belive that I've proved (i)$\Leftrightarrow$(iii) and (iii) $\Rightarrow$(ii) without using this hypothesis. Furthermore, I would like to know how to prove (ii) $\Rightarrow$(i) (or (ii) $\Rightarrow$(iii)). Thanks. AI: You're right, you don't need $n \geqslant 2$ for the equivalence $(i) \iff (iii)$ or the implication $(iii) \Rightarrow (ii)$. What you need $n \geqslant 2$ for is the implication $(ii) \Rightarrow (iii)$. Let's look at the example $A = (0,\,1) \subset \mathbb{R}^1$ to see where it breaks down for $n = 1$. Let's take $a = 0$. $a$ is an isolated boundary point, for example the $\frac12$-neighbourhood of $0$ doesn't contain another boundary point. Now, any $\varepsilon$-neighbourhood of $0$, for $0 < \varepsilon < 1$, consists of three parts $(0,\, \varepsilon) \subset A$, $\{0\} \subset \partial A$, $(-\varepsilon,\,0) \subset (\mathbb{R}\setminus A)^\circ$. The part inside $A$ and the part "outside" $A$ are separated by the part on the boundary of $A$. That is generally so, for each $M \subset X$, where $X$ is a topological space, you have a disjoint union $$X = \overset{\circ}{M} \,\dot{\cup}\, \partial M\, \dot{\cup}\, (X\setminus M)^\circ.$$ Now, if you have an isolated boundary point $p$ of $M$, a small enough punctured open neighbourhood of $p$ (that is, a $V \setminus \{p\}$, where $V$ is an open neighbourhood of $p$), splits into two disjoint open parts, one in the interior of $M$, the other in the exterior of $M$, since $$V = (V \cap \overset{\circ}{M})\, \dot{\cup}\, (V \cap \partial M)\,\dot{\cup}\, \bigl(V \cap (X\setminus M)^\circ\bigr)$$ and $V\cap \partial M = \{p\}$ if $V$ is small enough. For $X = \mathbb{R}^n$, we have that an $\varepsilon$-neighbourhood minus the centre point is connected when $n \geqslant 2$, but not connected when $n = 1$. So in the case $n \geqslant 2$, either $(B_\varepsilon(a) \setminus \{a\}) \cap A$ or $(B_\varepsilon(a)\setminus \{a\}) \cap (\mathbb{R}^n \setminus \overline{A})$ must be empty, since by definition a connected set cannot be written as the disjoint union of two nonempty open parts. $(B_\varepsilon(a) \setminus \{a\}) \cap A$ cannot be empty, since $a \in \partial A$ and $a \notin A$, hence $(B_\varepsilon(a) \setminus \{a\}) \subset A$.
H: Combinations: A Generalization on a Classic Problem I'm asked the following question: How many integers from $1$ to $10,000$, inclusive, are multiples of $5$ or $7$ or both? I've got the answer using this: unsigned a_5 = 0; unsigned a_7 = 0; unsigned a_both = 0; unsigned a_35 = 0; for( unsigned i = 1 ; i <= 10000 ; ++i ) { if( i % 5 == 0 ) { ++a_5; } if( i % 7 == 0 ) { ++a_7; } if( ( i % 5 == 0 ) && (i % 7 == 0 ) ) { ++a_both; } if( i % 35 == 0 ) { ++a_35; } } cout << "a_5 = " << a_5 << endl << endl; cout << "a_7 = " << a_7 << endl << endl; cout << "a_both = " << a_both << endl << endl; cout << "a_35 = " << a_35 << endl << endl; What I'd like to know is in general How many integers from $a_m$ to $a_n$, inclusive, are multiples of $a_m\leq k\leq a_n$? AI: The multiples of $5$ in the given range are $1\cdot5,2\cdot5,\ldots,2000\cdot5$, so there are $2000$ of them. The multiples of $7$ are $1\cdot7,2\cdot7,\ldots,\left\lfloor\frac{10000}7\right\rfloor\cdot7=1428\cdot7$, so there are $1428$ of them. However, every multiple of $35$ is in both lists, so the sum $2000+1428$ counts those multiples of $35$ twice. These multiples are $1\cdot35,2\cdot35,\ldots,\left\lfloor\frac{10000}{35}\right\rfloor\cdot35=285\cdot35$, so there are $285$ of them. Thus, the number of integers from $1$ through $10,000$ that are multiples of $5$, of $7$, or of both is $$2000+1428-285=3143\;.$$ The same ideas work in general: the number of multiples of $d$ between $1$ and $n$, inclusive, is $\left\lfloor\frac{n}d\right\rfloor$. For multiples of more than two numbers, however, you’ll need to use the inclusion-exclusion principle in greater generality to deal properly with the integers that are multiples of more than one of the numbers.
H: Showing that if an equation has a unique solution for one variable, then it has unique solutions for all. I have a problem and a proposed solution. Please tell me if I'm correct. Problem: Let $A$ be a square matrix. Show that if the system $AX=B$ has a unique solution for some particular column vector B, then it has a unique solution for all $B$. Solution: If $AX=B$ has a unique solution for some column vector $B$, then $A$ in reduced row echelon form has a pivot in each column and $A$ can be reduced to $I_n$, for $A$,$\\ n \times n$. Since the number of equations = the number of unknowns, we will have column vector $(n \times 1)$ of $x_i$'s = column vector $n \times 1$ of $b_i$'s. Hence, varying $B$ is equivalent to varying $X$ and will create a new solution for every change made to $B$. Thanks! AI: Your argument is essentially correct, but the end of it is a bit vague. What you’ve shown is that if $AX=B$ has a unique solution for some $B$, then $A$ can be row-reduced to $I_n$. This row-reduction is independent of the augmentation column in the augmented matrix, so for any $n\times 1$ column vector $C$ we can row-reduce the augmented matrix $[A\mid C]$ to some $[I_n\mid Y]$, and $Y$ will be the unique solution to $AX=C$. Thus, $AX=C$ has a unique solution for each $C$. There are many ways to see why this shows that $A$ is invertible (non-singular). If you know that when $[A\mid I_n]$ can be row-reduced to something of the form $[I_n\mid D]$, then $D=A^{-1}$, the result is clear. If you know that row-reduction can be accomplished by premultiplying by elementary matrices, then you can argue that row-reducibility of $A$ to $I_n$ means that there are elementary matrices $E_1,\dots,E_m$ such that $E_mE_{m-1}\dots E_2E_1A=I_n$ and hence $E_mE_{m-1}\dots E_2E_1=A^{-1}$.
H: Planar Graph with Maximum Number of Edges and 3-Colouring in Eulerian Show that a planar graph with $n$ vertices and $3n-6$ edges with $\chi=3$ is Eulerian. $\chi=3$ means there is a optimal vertex colouring with three colours. Eulerian means that the graph admits an Eulerian cycle (a cycle which contains each edge exactly once). My thoughts on this: I know that a planar graph has at most $3n-6$ edges and that a planar graph is maximal iff each face is a triangle. So in the present case each face is a triangle. Moreover it would suffice to prove that all vertex degrees are even since this is equivalent to being Eulerian. I don't see how the information that $\chi=3$ comes in. AI: A graph is Eulerian if and only if each vertex has even degree. So suppose there exists a vertex $v$ of odd degree in your graph. Now look at $v$'s neighbors. Since there are $3n - 6$ edges, the graph is maximally planar. That means that $v$ must have at least $3$ neighbors, and they must be connected in a wheel graph with $v$ at the center. Now what can you say about coloring a wheel graph with an odd length outer cycle?
H: $Ax=b\Leftrightarrow b\in\left(\ker A^\right)^\perp$ Let $A:\mathbb{R}^m\to\mathbb{R}^n$ be a linear map and $A^:\mathbb{R}^n\to\mathbb{R}^m$ be the adjoint of $A$ (that's $\langle Ax,y\rangle=\langle x,A^y\rangle$ for all $x\in\mathbb{R}^m,y\in\mathbb{R}^n$). Given $b\in\mathbb{R}^n$, are equivalent: (i) there exists $x\in\mathbb{R}^n$ such that $Ax=b$; (ii) $\langle b,z\rangle=0$ for all $z\in\ker A^ $. I don't know how to prove (ii)$\Rightarrow$(i). Can someone help me? Thanks. AI: \begin{align} w \in \ker A^ &\iff A^* w = 0 \\ &\iff \langle v, A^* w\rangle = 0 \quad \forall v \in \mathbb R^m \\ &\iff \langle Av, w\rangle = 0 \quad \forall v \in \mathbb R^m \\ &\iff w \in \left(\operatorname{im}A\right)^\perp \end{align} Hence $\ker A^ = \left(\operatorname{im}A\right)^\perp$. Since $\left(V^\perp\right)^\perp = V$ for any subspace $V \subset \mathbb R^n$, it follows that $\left(\ker A^*\right)^\perp = \left(\left(\operatorname{im}A\right)^\perp\right)^\perp = \operatorname{im}A$ as desired.
H: The largest number to break a conjecture There are several conjectures in Mathematics that seem to be true but have not been proved. Of course, as computing power increased, folks have expanded their search for counterexamples ever and ever upwards. Providing a counterexample to a conjecture with a very large number would be interesting, but I cannot think of any non-trivial examples where a really large number has been found to disprove a (non-trivial) conjecture. I've seen plenty of large numbers serving as bounds to some value, but usually this is something known to be bounded (i.e. finite) anyway. Out of curiosity, what's the largest counterexample you've seen to disprove a conjecture? AI: Google about Polya's conjecture. I don't know if it is the largest, though. The smallest number that is a counterexample is $906.150.257$, yet it is quite large.
H: Substring Occurrence in a String of Size $5$ Question: How many strings can be formed by ordering the letters $ABCDE$ so that each string contains the substring $DB$ or the substring $BE$ or both? Attempt: There are four possible ways to order both the $(DB)$-substring and the $(BE)$-substring amongst the other three remaining; that is you have the following cases: $$\overset{\text{Case 1}}{(DB)\star\star\star} ~~~~~~ \overset{\text{Case 2}}{\star(DB)\star\star} ~~~~~~ \overset{\text{Case 3}}{\star\star(DB)\star} ~~~~~~ \overset{\text{Case 4}}{\star\star\star(DB)}$$ The same goes for the $(BE)$-substring: $$\overset{\text{Case 1}}{(BE)\star\star\star} ~~~~~~ \overset{\text{Case 2}}{\star(BE)\star\star} ~~~~~~ \overset{\text{Case 3}}{\star\star(BE)\star} ~~~~~~ \overset{\text{Case 4}}{\star\star\star(BE)}$$ Now, for the situation where both substrings are found we have that the substring $(DBE)$ is to be sought, namely we have the following cases: $$\overset{\text{Case 1}}{(DBE)\star\star} ~~~~~~ \overset{\text{Case 2}}{\star(DBE)\star} ~~~~~~ \overset{\text{Case 3}}{\star\star(DBE)}$$ For both the $(DB)$-substrings and $(BE)$-substrings we have for each case a number of $1\cdot 3!=3\cdot 1 \cdot 2!=3\cdot 2\cdot 1\cdot 1=3!\cdot 1=3!$ outcomes; that is, for the $(DB)$-substring we have $4\cdot 3!$ variations, and for the $(BE)$-substring we have $4\cdot 3!$ variations, so there is a total of $2\cdot 4\cdot 3!$ variations. Although, notice that we have counted too many, namely the $(DBE)$-substring occurences. Using the same logic from above, we see that the number of variations for the $(DBE)$-substring occurrences is $3\cdot 2!$. This needs to be subtracted from the total above: $$2\cdot 4\cdot 3!-3\cdot 2!=42$$ AI: This is also correct, and here again you can also think of it a little differently. There are $4!$ ways to arrange the $4$ items consisting of a two-letter block and $3$ single letters, so there are $4!$ strings with $DB$ and $4!$ strings with $BE$. There are $3!$ ways to arrange the $3$ items consisting of the three-letter string $DBE$ and the remaining two single letters. Then apply the inclusion-exclusion argument that you used to get a total of $2\cdot4!-3!=48-6=42$ strings.
H: Martians and Jovians In how many ways can five distinct Martians and eight distinct Jovians wait in line if no two Martians stand together? AI: Make a lineup of $8$ letters $J$ like this: $$J \qquad J \qquad J \qquad J \qquad J \qquad J \qquad J \qquad J \qquad$$ There are $7$ gaps between $J$'s that we could slip an $M$ into, plus the $2$ "endgaps," for a total of $9$ places. So places for the Martians can be chosen in $\binom{9}{5}$ ways. Multiply by $8!5!$ because these are distinct individuals. So once we have chosen the places for the Jovians, and the places for the Martians, we can insert the Jovians in $8!$ orders, and for each way we can insert the Martians in $5!$ different orders.
H: Eigenvalues of a second derivative I have a function f(r) that describes a Gaussian random field. A second derivative can be formed $\nabla_i \nabla_j f(r)$. I am looking at a paper that claims that in finding the extremum, the eigenvalues of the second derivative operator should be negative definite. I was curious whether anyone knew where this rule came from? I've never seen anything written up about this particular second derivative except in the context of Riemann tensors. Any insight would be appreciated! AI: A matrix $A$ is negative definite if $$\forall x\ne0,\quad\langle x,Ax\rangle<0$$ hence in a critical point of $f$ i.e. $\nabla f(a)=0$ and if the Hessian matrix is definite negative then: \begin{align} f(x) = f(a)+\underbrace{\nabla f(a)}_{=0} \cdot (x-a) + \frac{1}{2}\underbrace{(x-a)^T \mathbb{H}(a) (x-a)}_{<0}+ o(\|\|x-a\|\|^{2}). \end{align} hence we see that $f(x)<f(a)$ in a neighberhood of $a$ so $f(a)$ is a local minimum of $f$.
H: Induction Proof for a series expansion of a function I have done induction proofs of many different types, but trying to prove by induction that a derivative from the Taylor series expansion of a function has me stumped in terms of how to get the final step to work. In most induction proofs we start with a base case of $n=1$, then assume $N=n$ works, then apply $N+1$, and usually make use of the assumption to get a statement that is TRUE to prove it. Here is the function: $$f(x)= \ln (1+x).$$ And here is the equation we wish to apply Induction to: $$f^{(n)}(x)= (-1)^{n-1}\frac{(n-1)!}{(1+x)^{n}}.$$ The approach I took was to apply $N+1$ where ever I see $n$, and the other thing that I did was to just differentiate the equation which was straight forward, and then both were the same. BUT i don't feel that this was a good proof because in Induction you usually make use of step 2 the equation itself. Hope someone can guide me or explain to me how one goes about doing an induction proof that has derivatives involved of this sort. AI: I suppose you had no problem to check that the formula wich we have to proof is true for $n=1$. Wich is step 1 of our induction proof. So we go to step 2 : We show : if $ \ \ \ \ f^{(n)}(x)= (-1)^{n-1}\dfrac{(n-1)!}{(1+x)^{n}}$ $\ \ \ \ \ \ \ \ \ \ \ \ (1)$ is valid for $n$ then it is valid for $n+1$. Wich is step 2 of te induction proof. Now knowing that $(1)$ holds we defferintiate both sides of the equation : $f^{(n+1)}(x)= -n \cdot (-1)^{n-1}\dfrac{(n-1)!}{(1+x)^{n+1}}= (-1)^{n}\dfrac{(n)!}{(1+x)^{n+1}}$ As you can see $(1)$ is valid for $n+1$. Indeed as you claimed "in induction we make use of step 2 the equation itself" we actually used $(1)$ wich is "the equation" and manipulated it to show that it also holds for $n+1$. As far as i understood you made the same steps. I hope this helps to convince you that you were right.
H: $\frac{\sin x}{x^5} - \frac{1}{x^4} \underset{x\to 0}{\approx} \frac{-1}{6} \cdot \frac{1}{x^2}$, right? I was reading an set of notes about Taylor series, and I came across a part I think is a typo. I want to make sure, because I want to understand this stuff correctly. Here is the relevant page of the article. You can see where I've indicated the typo in the margin. In the space I've marked, do you think it should say $\frac{1}{x^2}\cdot \frac{-1}{6}$? AI: Yes you're right : $$\frac{\sin x}{x^5}-\frac{1}{x^4}=\frac{\sin x-x}{x^5}\sim_0 \frac{-1/6 x^3}{x^5}=\frac{-1}{6x^2}$$
H: The evaluation of the infinite product $\prod_{k=2}^{\infty} \frac{k^{2}-1}{k^{2}+1}$ How does one show that$$ \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} =\frac{\pi}{\sinh \pi} ?$$ My attempt: $$ \begin{align} \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} &= \lim_{n \to \infty} \prod_{k=2}^{n}\frac{(k-1)(k+1)}{(k-i)(k+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n) \Gamma(n+2) \Gamma(2-i) \Gamma(2+i)} {2 \Gamma(n+1-i) \Gamma(n+1+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2) (1-i) \Gamma(1-i) (1+i)i \Gamma(i)}{2 \Gamma(n+1-i)\Gamma(n+1+i)} \\ &= \frac{\pi}{\sinh \pi}\lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} \end{align}$$ I'm not sure how to go about showing that the limit evaluates to $1$. EDIT: To evaluate that limit we can use the fact that $ \displaystyle \frac{\Gamma(n)}{\Gamma(n+z)} \sim n^{-z}$ as $ n \to \infty$. $$ \begin{align} \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} &= \lim_{n \to \infty} \frac{\Gamma(n) (n+1)n \Gamma(n)}{(n-i)\Gamma(n-i) (n+i)\Gamma(n+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)n^{-i}}{\Gamma(n-i)} \frac{\Gamma(n) n^{i}}{\Gamma(n+i)} \frac{n^2+n}{n^{2}+1} \\ &= (1)(1)(1) \\ &= 1 \end{align}$$ AI: If you know the product representation $$\sin (\pi z) = \pi z \prod_{k = 1}^\infty \left(1 - \frac{z^2}{k^2}\right),\tag{1}$$ it is rather easy. Setting $z = i$, we obtain $$\frac{\sin (\pi i)}{\pi i} = \frac{\sinh \pi}{\pi} = \prod_{k=1}^\infty \left(1 - \frac{i^2}{k^2}\right) = \prod_{k=1}^\infty \left(1 + \frac{1}{k^2}\right) = 2 \prod_{k=2}^\infty \left(\frac{k^2+1}{k^2}\right).$$ On the other hand, $$\prod_{k=2}^n \left(\frac{k^2-1}{k^2}\right) = \frac12\cdot \frac32\cdot \frac23\cdot \frac43 \dotsb \frac{n-1}{n}\cdot \frac{n+1}{n} = \frac12\cdot\frac{n+1}{n},$$ so $$\prod_{k=2}^\infty \left(\frac{k^2-1}{k^2}\right) = \frac12.$$ Now divide.
H: Showing that two topologies on the unit circle are the same Consider the unit circle, described two ways. The first is as a quotient space, as in What does it mean to "identify" points of a topological space?. (I'm using the first definition of its topology from http://en.wikipedia.org/wiki/Quotient_space) Let's call this version [0,1]/~. The second is as a set of points in the complex plane, as in http://www.maths.bristol.ac.uk/~maxcu/Rotations.pdf. Let's call this version $S^{1}$. This second link defines a metric distance based on arc length. How can I show that the topology generated by this metric is the same as the quotient space topology? What I'm trying: I'm trying to show that the map $\phi: [0,1]/$~ $\rightarrow S^{1}$ defined by $\phi([x]) = e^{2\pi i x}$ is a homeomorphism. Note that it's a bijection. To show that the map is open: Suppose $U$ is an open set in [0,1]/~. Then $\cup_{[a] \in U}[a]$ is open in [0,1] (using the subspace topology from $\mathbb{R}$). So $\cup_{[a] \in U}[a]$ = $[0,1] \cap \cup_{n=1}^{\infty}\mathring{I_{n}} = \cup_{n=1}^{\infty}([0,1] \cap \mathring{I_{n}})$ where $\cup_{n=1}^{\infty}\mathring{I}$ is a countable disjoint union of open intervals. Then $\cup_{n=1}^{\infty}([0,1] \cap \mathring{I_{n}})$ is also a countable disjoint union of open intervals, viewing [0,1] as a topological space. It suffices to show that each $\phi([0,1] \cap \mathring{I_{n}})$ is open in the metric of $S^{1}$. If $[0,1] \cap \mathring{I_{n}}$ is the empty set, then we're o.k. Say $\mathring{I_{n}} = (a_{n},b_{n})$ is contained in (0,1). Let $e^{2\pi i x} \in \mathbb{C}$ be in $\phi(\mathring{I_{n}})$. Let $\delta = min \{d(e^{2\pi i a_{n}},e^{2\pi i x}), d(e^{2\pi i b_{n}},e^{2\pi i x})\}$. We claim $B_{\delta}(e^{2\pi i x})\subseteq \phi(\mathring{I_{n}})$. Let $ e^{2\pi i y} \in B_{\delta}(e^{2\pi i x})$. Then $d(e^{2\pi i y},e^{2\pi i x}) < \delta$. i.e. $min(1-\|x-y\|,\|x-y\|) < min(1-\|x-a_{n}\|,\|x-a_{n}\|,1-\|x-b_{n}\|,\|x-b_{n}\|)$. We'd be done if we could should that $\|a_{n}\| < \|y\| < \|b_{n}\|$. I've played around with cases a bit, but have gotten stuck. Any hints or suggestions? Am I making this too messy? Are the two topologies even the same? Thanks, and regards. AI: Since both topologies make the unit circle a compact Hausdorff space, it is sufficient to show that $\phi \colon [0,\,1]/\sim \to S^1,\; \phi([x]) = e^{2\pi i x}$ is a continuous bijection. That it is a bijection follows from elementary properties of the exponential functions resp $\sin$ and $\cos$, and the continuity follows from general properties of quotient topologies, since the lift to $[0,\,1]$ is continuous. Since $[0,\,1]/\sim$ is compact, every continuous mapping into a Hausdorff space is closed (a closed subspace of $[0,\,1]/\sim$ is compact, hence its image is compact, hence closed). A closed and continuous bijection is a homeomorphism.
H: $X$ is a complete metric space, $Y$ is compact. $X \times Y$ is Baire? Requesting a hint or solution. X is a complete metric space and Y is a compact hausdorff space. Trying to show that $X \times Y$ is a Baire space. AI: $\newcommand{\cl}{\operatorname{cl}}$You need to show that if $G_n$ is a dense open subset of $X\times Y$ for each $n\in\Bbb N$, then $G=\bigcap_{n\in\Bbb N}G_n$ is dense in $X\times Y$. You can do this by imitating the proofs of the Baire category theorem for complete metric spaces and compact Hausdorff spaces simultaneously. Without loss of generality we may assume that $G_0\supseteq G_1\supseteq G_2\supseteq\ldots\;$. (Why?) Let $U$ be a non-empty open set in $X\times Y$. $G_0$ is dense in $X\times Y$, so we can pick a point $p_0\in U\cap G_0$, say $p_0=\langle x_0,y_0\rangle$. Choose open nbhds $V_0$ of $x_0$ and $W_0$ of $y_0$ such that $\cl_XV_0\times\cl_YW_0\subseteq U\cap G_0$ and $\operatorname{diam}(\cl_XV_0)\le 1$, and let $U_1=V_0\times W_0$. Given $U_n$, choose $p_n=\langle x_n,y_n\rangle\in U_n\cap G_n$ and proceed similarly, but make $\operatorname{diam}(\cl_XV_n)\le 2^{-n}$. Can you finish it from here?
H: Exact form of pdf of maximum of normal random variables $$ z = max(x+b,y) $$ where x ~ N(m1,s1) and y~N(m2,s2), b is a contant What's the pdf of z? Or exact form of E(z)? (E is expectation operator) To the best of my guessing from the literature it is related with Weibull (https://en.wikipedia.org/wiki/Weibull_distribution) but I can't derive the exact pdf of z or exact E(z). If there is a formula for arbitrary number of variables and covariance matrix, then that will be even better. ===================================== Should the answer be like the following? $$P(z) = P_2(z) \int_{-\infty}^{z-b} P_1(x)dx\; + P_1(z-b) \int_{-\infty}^z P_2(y)dy\; $$ $$ =\frac{1}{2\pi\sigma_{1}\sigma_{2}}\left[\exp\left(\frac{-(z-\mu_{2})^{2}}{2\sigma_{2}^{2}}\right)\int_{-\infty}^{z-b}\exp\left(\frac{-(x-\mu_{1})^{2}}{2\sigma_{1}^{2}}\right)dx+\exp\left(\frac{-(z-b-\mu_{1})^{2}}{2\sigma_{1}^{2}}\right)\int_{-\infty}^{z}\exp\left(\frac{-(y-\mu_{2})^{2}}{2\sigma_{2}^{2}}\right)dy\right] $$ $$ =\frac{1}{2\pi\sigma_{1}\sigma_{2}}\left[\exp\left(\frac{-(z-\mu_{2})^{2}}{2\sigma_{2}^{2}}\right)\sigma_{1}\sqrt{\frac{\pi}{2}}\mathrm{erfc}\left(\frac{\mu_{1}-z+b}{\sigma_{1}\sqrt{2}}\right)+\exp\left(\frac{-(z-b-\mu_{1})^{2}}{2\sigma_{1}^{2}}\right)\sigma_{2}\sqrt{\frac{\pi}{2}}\mathrm{erfc}\left(\frac{\mu_{2}-z}{\sigma_{2}\sqrt{2}}\right)\right] $$ AI: I denote the pdf of $x$ by $P_1(x)$ and the pdf of $y$ by $P_2(y)$. Then the pdf $P(z)$ of $z=\max(x+b,y)$ is given by $$P(z) = \int_{-\infty}^{\infty} dx \int_{x+b}^\infty dy\; P_1(x) P_2(y) \delta(z-y) + \int_{y-b}^\infty dx \int_{-\infty}^\infty dy\; P_1(x) P_2(y) \delta(z-x-b)$$ $$= P_2(z) \int_{-\infty}^{z-b} P_1(x)\;dx + P_1(z-b) \int_{-\infty}^z P_2(y)\;dy$$ For normal $P_1$ and $P_2$ the integrals evaluate to error functions.
H: Men and Women: Committee Selection There is a club consisting of six distinct men and seven distinct women. How many ways can we select a committee of three men and four women? AI: There are "$6$-choose-$3$" $=\binom{6}{3}$ ways to select the men, and "$7$-choose-$4$" =$\binom{7}{4}$ ways to select the women. That's because $(1)$ We have a group of $6$ men, from which we need to choose $3$ for the committee. $(2)$ We have $7$ women from which we need to choose $4$ to sit in on the committee. Since anyone is either on the committee or not, there is no need to consider order: position or arrangement of those chosen for the committee isn't of concern here, so we don't need to consider permuting the chosen men or women. So we just multiply$\;$ "ways of choosing men" $\times$ "ways of choosing women". [Recall th Rule of the Product, also known as the multiplication principle.] $$\binom 63 \cdot \binom 74 = \dfrac{6!}{3!3!} \cdot \dfrac{7!}{4!3!} = \dfrac{6\cdot 5\cdot 4}{3!}\cdot \dfrac{7\cdot 6\cdot 5}{3!}$$
H: Is this infinite series a Fourier series? I have what looks like a Fourier series but I don't quite understand how (or if) it is possible to recover a function from this. $$e^{3i\pi/2}+2e^{3i\pi/2}+3e^{3i\pi/2}+4e^{3i\pi/2}+5e^{3i\pi/2}+\cdots$$ Any ideas? If changes to this series are necessary to make it a Fourier series from which an equation is recoverable then that also would be really interesting. Thanks for the help, much appreciated. AI: You can factorise $e^{3i\pi/2}$ out of this expression, to get $e^{3i\pi/2}(1+2+3+4+5+\dots)$. So it's divergent (undefined, if you prefer). It's not really a fourier series anyway: a fourier series is a function. Perhaps you're thinking of something like $f(x) = \dots + a_{-1}e^{-ix} + a_0 + a_1e^{ix} + a_2e^{2ix} + \dots$. Whenever you choose values for the $a_i$, though, you still have to make sure the function converges - that is, you can't take them all equal to $1$, because at $x = 0$ you get $f(0) = \dots + 1 + 1 + 1 + 1 + \dots$, which diverges.
H: What does it mean to "compute" a generic formula without values? Very elementary math question here in regards to my discrete math class. I've got a problem here that says... "Compute the following": $$\sum_{j=1}^n \frac{1}{j(j+1)}$$ What on earth does it mean to "compute" this? I mean, without any values, does that just mean to find another way of re-arranging the components to get a formula of equal value, just in a different form? I did some iterations by hand and found that for n=4, we end up with: $$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}$$ So it looks like each iteration just increases the denominator by the previous increase's next even number: $\frac{1}{2} \rightarrow \frac{1}{6}$ (increase by 4) $\frac{1}{6} \rightarrow \frac{1}{12}$ (increase by 6) $\frac{1}{12} \rightarrow \frac{1}{20}$ (increase by 8) So I see the pattern. Does that mean for example that I could re-write the above sigma summation as a "rule" for making the sequence? AI: You’re supposed to find a closed form expressing that sum as a function of $n$, one that does not include a summation. An example that you’ve probably seen is the formula for the sum of the first $n$ positive integers: $$\sum_{k=1}^nk=\frac{n(n+1)}2\;.$$ You’re supposed to do something similar here. Here’s a hint to get you started: $$\frac1{k(k+1)}=\frac1k-\frac1{k+1}\;.$$
H: Why in a space in which compact sets have empty interior, the closure of open sets is not totally bounded? I'm trying to understand the proof of the theorem: The Baire space $\mathcal{N}$ is unique up to homeomorphism, non-empty zero-dimentional Polish space for which all compact subsets have empty interior. It is mentioned in the proof that "since a compact set in $X$ (the space satisfies all above conditions to be homeomorphic to $\mathcal{N}$)have empty interior, it follows that the closure of $U$(a open set in $X$) is not totally bounded." I think I may miss something obvious, and I just don't know how to start. AI: The closure of $U$ is not compact, since it has non-empty interior. It is complete, however, since it’s a closed subset of a complete space. Therefore it cannot be totally bounded, since a set is compact iff it is complete and totally bounded.
H: Generalization of Jensen's inequality for integrals? Jensen's inequality for sums says that for $f$ convex, $$f\left(\sum_1^n \alpha_i x_i\right)\leq \sum_1^n \alpha_i f(x_i), \,\,\,\,\text{for } \sum_1^n \alpha_i = 1.$$ I have read that a generalization of it is that $$f\left(\frac{\sum_1^n \alpha_i x_i}{\sum_1^n \alpha_i}\right)\leq \frac{\sum_1^n \alpha_i f(x_i)}{\sum_1^n \alpha_i}.$$ Now I read Jensen's inequality for integrals: for $f,g$ functions, $f$ convex, $g$ and $f\circ g$ integrable, we have $$f\left(\int_0^1g(x) \, dx\right)\leq \int_0^1 (f\circ g)(x)\, dx.$$ What is the analogous generalization of this to an interval other than $[0,1]$? I don't think it's just dividing by the length of the interval... AI: I believe we can prove that in fact it should be the length of the interval by using Jensen's finite equality on the Riemann sums: $f\left(\frac{\sum_1^n \alpha_i x_i}{\sum_1^n \alpha_i}\right)\leq \frac{\sum_1^n \alpha_i f(x_i)}{\sum_1^n \alpha_i},$ where $x_i = g(t^*_j)$ and $\alpha_i=t_{i+1}-t_i$. @Zach has also mentioned that it can be shown by just choosing another measure $\nu$ such that $\nu([a,b])=1$.
H: The union of a sequence of countable sets is countable. While working on the theorem below, I constructed the following proof: Theorem. If $\left\langle E_{n}\right\rangle_{n\in\mathbb{N}}$ is a sequence of countable sets, then $$ \bigcup_{n\in\mathbb N}E_{n} $$ is countable. Proof. Let $S=\left\langle E_{n}\right\rangle_{n\in\mathbb{N}}$ be a sequence of countable sets. Moreover, let $x_{n,m}$ be the $m$th element of the $n$th set in $S$. Construct a sequence of sequences $$ \mathcal S=\left\langle\left\langle x_{n-m+1,m}\right\rangle_{m\in\mathcal N_n}\right\rangle_{n\in\mathbb N}, $$ where $\mathcal N_{n}=\left\{k\in\mathbb N:k\leqslant n\right\}$. Then $\mathcal S$ contains all the elements of $$ T=\bigcup_{n\in\mathbb N}E_{n}. $$ Observe that $\mathcal S$ is a surjection $\mathbb N\to T$, because for every $n\in\mathbb N$, $\mathcal S$ yields a finite, and thus countable, subsequence. Furthermore, because $\mathcal S$ may contain duplicate elements, an $U\subseteq\mathbb N$ can be found such that $\mathcal S$ is an injection $U\to T$. We have then found a bijection $\mathcal S:U\to T$. Hence, $T$ is countable. $\blacksquare$ Does it sound convincing? Edit: I just realized that I cannot pinpoint duplicates because of how $\mathcal S$ is constructed, can I? :-( AI: You’re trying to list $T$ as $$\langle x_{1,1},x_{2,1},x_{1,2},x_{3,1},x_{2,2},x_{1,3},\ldots\rangle\;,\tag{1}$$ but instead you’ve constructed the sequence $$\Big\langle\langle x_{1,1}\rangle,\langle x_{2,1},x_{1,2}\rangle,\langle x_{3,1},x_{2,2},x_{1,3}\rangle,\ldots\Big\rangle\;,$$ a related but definitely different animal. It’s actually easier to say what $n\in\Bbb N$ corresponds to $x_{k,\ell}$ in the list $(1)$ than to go in the forward direction. Count the elements of $(1)$ that precede $x_{k,\ell}$. They certainly include all $x_{i,j}$ such that $i+j<k+\ell$, and there are $$\sum_{n=2}^{k+\ell-1}n=\frac{(k+\ell-1)(k+\ell)}2-1$$ of those. (The summation starts at $2$ because we always have $i+j\ge 2$.) Let $m=k+\ell$; it also includes all $x_{m-i,i}$ such that $1\le i\le\ell$, and there are $\ell-1$ of those. Thus, $$\frac{(k+\ell-1)(k+\ell)}2-1+\ell-1=\frac{(k+\ell-1)(k+\ell)}2+\ell-2$$ terms of $(1)$ precede $x_{k,\ell}$, and $x_{k,\ell}$ is therefore term number $$\frac{(k+\ell-1)(k+\ell)}2+\ell-1$$ of the sequence $(1)$.

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