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H: Integral of derivatives and conjugate For function $f,g$ in the Schwartz class, I want to show that $$\int_{\mathbb{R}}\left(\frac{d}{dx}f(x)+xf(x)\right)\overline{g(x)}dx=\int_{\mathbb{R}}f(x)\overline{\left(-\frac{d}{dx}g(x)+xg(x)\right)}dx$$ But I don't see how to start, since there isn't any easy way to manipulate the integrals AI: Notice that the term with $xf(x)\overline{g(x)}$ doesn't change on either side. Eliminating that, you're left with $$\int_{\mathbb{R}} \frac{d}{dx}f(x)\overline{g(x)}dx = -\int_{\mathbb{R}} f(x)\frac{d}{dx}\overline{g(x)}dx.$$ If you make use of integration by parts and the fact that conjugation commutes with differentiation, you will have the result. This kind of result is actually a pretty nifty one that gets use in quantum mechanics when dealing with quantum mechanical charge operators. The charge operators for the harmonic oscillator Hamiltonian are (up to normalization) $-\frac{d}{dx} + x$ and $\frac{d}{dx} + x$, which are adjoints of each other. You can think of your result being $\langle (-\frac{d}{dx}+x)f,g\rangle = \langle f,(\frac{d}{dx}+x)g\rangle$ (if you embed the Schwartz class into $L^2$ with the usual inner product).
H: Mean value theorem and differentiability Let $a,b\in \mathbb{R}$, $a<b$ and let $f$ be differentiable real-valued function on an open subset of $\mathbb{R}$ that contains $[a,b].$ Show that if $\lambda$ is any real number between $f'(a)$ and $f'(b)$ then there exists a number $c\in(a,b)$ such that $\lambda = f'(c)$. I know I have to combine the mean value theorem with the intermediate value theorem for the function $\frac{f(x_1)-f(x_2)}{x_1-x_2}$ on the set $\{ (x_1,x_2)\in E^2:a\le x_1<x_2 \le b\}.$ How can I do that? AI: I think we should add the condition that $f'(a)\ne f'(b)$, otherwise $f(x)=\sin x$, $a=-\dfrac{\pi}{2}$ and $b=\dfrac{\pi}{2}$ would be a counterexample. Furthermore, $\lambda\ne f'(a)$ or $f'(b)$. If we donote $g(x)=f(x)-\lambda x$, then the problem is equivalent to "If $f'(a)f'(b)<0$ then exists a $\nu\in(a,b)$, s.t. $f'(\nu)=0$". Proof: Assume $f'(a)<0,f'(b)>0$, then $f(x)$ must reach its minimum value in $(a,b)$. This point $x=\nu$ has the property $f'(\nu)=0$. If $f'(a)>0,f'(b)<0$, then $f(x)$ must reach its maximum value in $(a,b)$. This point $x=\mu$ has the property $f'(\mu)=0$.
H: Finding the Nth number in a generated list I am generating numbers as follows: Let the first digit range from 1 to 2 inclusive. Let the second digit range from 1 to 3 inclusive. Let the last digit range from 1 to 2 inclusive. I am then generating all the possible numbers by cycling the last digit, then the middle, then the first to get this: 1 1 1 1 1 2 1 2 1 1 2 2 1 3 1 1 3 2 2 1 1 2 1 2 and so on. How would you find the Nth number in the list generated using this method. For example, querying for the 2nd number would return 111, while querying for the 3rd would return 121. AI: The first digit is $\lceil \frac{N}{6}\rceil$, where $\lceil \cdot \rceil$ denotes the ceiling function. The second digit is $\lceil \frac{N \pmod 6}{2}\rceil$, where $\pmod{6} $ denotes modular arithmetic. The third digit is $2-(N \pmod{2})$.
H: I need a proof of a theorem about generated group how to use definition 2.7 to prove theorem 2.8? AI: Let $H$ be the collection of all finite products $a_1^{n_1} a_2^{n_2} \ldots a_t^{n_t} (a_i \in X, n_i \in \mathbb{Z})$. Check that : a) $H$ is a subgroup of $G$ (Use the subgroup test : if $x, y\in H$, conclude that $xy^{-1} \in H$) b) If $K$ is any subgroup of $G$ containing $X$, then $H\subset K$. Proof : If $X \subset K$, then for any $a_i \in X, n_i \in \mathbb{Z}$, $a_i^{n_i} \in K$, and hence by closure, $a_1^{n_1} a_2^{n_2}\ldots a_t^{n_t} \in K$. Hence, $H\subset K$
H: What would be the simplified form of this expression? I'm working on a Homework problem involving Convergence of Random variables and I've arrived at an expression which looks like follows: $$ M_{X_n}(ju)= \prod_{k=1}^{n}0.5\frac{1-e^{\frac{ju}{2^{k-1}}}}{1-e^\frac{ju}{2^k}}$$ ,where $M_X(ju)$ denotes the characteristic function of a random variable $X$. It'd be great to have an hint to proceed as I've not seen(or remember) a product like this before and I'd love to show my work here but this is just a single step of a big homework problem and I'm just stuck at this part. Unfortunately, I don't seem to have too many thoughts about this particular step. (I know some work has to be shown, but really I don't seem to find a way through this step and that's the reason why I'm posting this; Thank You !!) Edit: I made an error in the initial equation and I've rectified it. $0.5$ should have been inside the product. AI: Hint: if $e^{ju/2^{k}} = t$, then $t^2 = e^{ju/2^{k-1}}$
H: What is the first $w$ such that a rectangle, $R_{w\times w-1}$ is minimally-square-partitioned by less than $w$ squares. Motivated by: Tiling an orthogonal polygon with squares, How to prove that the minimum square partition of a 3X2 rectangle has 3 squares, Minimum square partitions for 4x3 and 5x4 rectangles, What is the minimum square partition of an almost-square rectangle?. Let: An almost-square-rectangle be a rectangle that has a width $w$ and height $h=w-1$. A square partitioning be a covering by non-overlapping squares; the entire rectangle must be covered, all the squares must be disjoint. A minimum-square-partitioning be a square partitioning, for which is no square partitioning that is made of a lesser number of squares. The $R_{w\times w-1}$ rectangles illustrated below are minimally-square-partitioned by $w$ squares. Proofs here: How to prove that the minimum square partition of a 3X2 rectangle has 3 squares and Minimum square partitions for 4x3 and 5x4 rectangles). However, in general, a $R_{w\times w-1}$ rectangle can be partitioned into $\mathcal O(\log w)$ squares (see What is the minimum square partition of an almost-square rectangle?, and Minimum tiling of a rectangle by squares, which refer to a paper called Tiling a Rectangle with the Fewest Squares by Richard Kenyon. My question: What is the minimum $w$, such that a $w\times w-1$ rectangle, $R_{w\times w-1}$, can be minimally-square-partitioned into $f(w)$ squares, and $f(w)<w$. AI: By the proofs cited above, $w\geq 6$, and by Yuval's answer cited above, $w\leq 7$. So the only question is now whether a $6 \times 5$ rectangle can be covered with 5 squares. I think the answer is yes: it can be covered by three $2 \times 2$ squares and two $3 \times 3$ squares. So $w=6$.
H: Convolution convergent in $L^\infty$ Suppose $f\in L^\infty(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_\epsilon(x)=\dfrac{1}{\epsilon}K\left(\dfrac{x}{\epsilon}\right)$$ Is it always true that $\lim_{\epsilon\rightarrow 0}\\|f\ast K_\epsilon-f\\|_\infty=0$? I think it's always true for $L^p$, where $1\leq p<\infty$, but what about for $L^\infty$? AI: It is a well-known fact that smooth functions are not dense in $L_\infty$. Take $f(x)=\mathrm{sign}\,x$ and $K$ be a smooth non-negative even function with a compact support. Then $f*K_\epsilon(0)=0$ and $\\|f\ast K_\epsilon-f\\|_\infty=1$.
H: Show that $\lim_{h\to 0}[1/(f(a+h)-f(a))-1/hf'(a)]=-f''(a)/2f'(a)^2$. Given that $f'(x)\ne0$ show that $\lim_{h\to 0}[1/(f(a+h)-f(a))-1/hf'(a)]=-f''(a)/2f'(a)^2$. By wrriting $f'(a)$ into its limit definitions, LHS seems to be $0$, so how to do this problem? Thanks. AI: Let $$ L(h)=\frac{1}{f(a+h)-f(a)}-\frac{1}{hf'(a)}. $$ Then $$ L(h)=-\frac{f(a+h)-f(a)-hf'(a)}{hf'(a)(f(a+h)-f(a))} $$ By Taylor’s formula we have $\frac{f(a+h)-f(a)-hf'(a)}{h^2} \to \frac{f''(a)}{2}$ when $h\to 0$, so $$ L(h)=-\frac{f(a+h)-f(a)-hf'(a)}{h^2} \frac{1}{f'(a)}\frac{1}{\frac{f(a+h)-f(a)}{h}} \to \frac{-f''(a)}{2} \frac{1}{f'(a)^2} \ \text{when} \ h\to 0 $$
H: How prove this $d(\alpha,W)$ is Gram determinant? if $W$ is subspace of $V$.and $V$ is $n$-dimensional Euclidean space. and $\alpha\in V$, Define the distance from $\alpha$ to $W$ $$d(\alpha,W)=\|\alpha-\alpha'\|$$ where $\alpha'$ is $\alpha$ orthogonal projection on the subspace $W$. such $\alpha_{1},\alpha_{2},\alpha_{3},\cdots,\alpha_{m}$ is $W$ one of basis. show that $$d(\alpha,W)=\sqrt{\dfrac{\|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha)\|}{\|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})\|}}$$ where $G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$ is Gram matrix. my try: since this Gram determinant define:http://en.wikipedia.org/wiki/Gramian_matrix But I can't prove this problem,I hope someone can help me.Thank you AI: If you have covered the relation of the determinant of a Gram matrix and the (hyper)volume of the $m$-dimensional parallelotope $V_m(\alpha_1,\ldots,\alpha_m)$ determined by having vectors $\alpha_1,\ldots,\alpha_m$ as its edges, namely $$ V_m(\alpha_1,\ldots,\alpha_m)=\sqrt{\|G(\alpha_1,\ldots,\alpha_m)\|}, $$ then your claim follows easily. It is the $(m+1)$-dimensional generalization of the fact that the area of a parallelogram (in the plane) is its base times the height, where height equals the distance of the endpoint of the vector $\alpha_2$ from the baseline generated by $\alpha_1$. In the high-dimensional analogue the $m$-dimensional parallelotope with edges $\alpha_1,\ldots,\alpha_m$ takes the role of the "base", and the distance of the endpoint of the remaining edge vector $\alpha_{m+1}$ from the base takes the role of "height" $h$. So from the equation $$ V_{m+1}(\alpha_1,\ldots,\alpha_m,\alpha_{m+1})=V_m(\alpha_1,\ldots,\alpha_m)\cdot h =V_m(\alpha_1,\ldots,\alpha_m)\cdot d(\alpha_{m+1},W) $$ we get $$ d(\alpha_{m+1},W)=h=\frac{V_{m+1}(\alpha_1,\ldots,\alpha_m,\alpha_{m+1})}{V_m(\alpha_1,\ldots,\alpha_m)}= \sqrt{\frac{\|G(\alpha_1,\ldots,\alpha_m,\alpha_{m+1})\|}{\|G(\alpha_1,\ldots,\alpha_m)\|}}. $$
H: Two questions on torsion set Let $R$ be a ring and $T$ be the set of torsion elements of $R$. Prove that the only torsion element of $R/T$ is $\{0\}$. I can show that $T$ is an ideal of $R$. Or in fact, $T$ is a normal subgroup of $R$ if we treat $R$ simply as an additive group. If I take this latter view, then I guess I can use the first isomorphism theorem to prove this statement. Define $\psi: R\rightarrow\{0, \infty\}$ by sending element of $R$ to 0 if its order is finite and $\infty$ otherwise. This is an onto homomorphism if we assume that $0+\infty=\infty$. Therefore, $R/T\cong\{0,\infty\}$ since the kernel of $\psi$ is $T$. Then I can conclude that $R/T$ is torsion free since $0$ is the only element of $\{0, \infty\}$ which is of finite order. Is this argument right, please? Thank you! Give an example of a ring $R$ for which $T\neq \{0\}$ and $T\neq R$. I tried $2\times 2$ matrices with entries in $\mathbb{Z}/4\mathbb{Z}$. It seems to work. I am wondering whether there is a general rule for this, please? Thank you! BTW, a torsion element is defined as an element with finite order. AI: For the proof, check here: Prove that $M/Tor(M) $ is torsion-free. For an example of a ring with some elements but not all , torsion, take $\mathbb Z\oplus \mathbb Z_n$ , where the only torsion elements are $\{ (0,z_n): z_n \in \mathbb Z_n\}$
H: Coordinates of point that is on the end of perpendicular line Lines AB and CD are perpendicular. Points A, B, D can have any random coordinates and we know there value. Line CD can have any random length and we know its value. How can we calculate coordinates of point C ? AI: Suppose $A(a_1,a_2),B(b_1,b_2)$ and $D(d_1,d_2)$. The line $(\epsilon_1)$ passing from these points has an equation of the form $Ax+By+C=0$. Since $DC$ (lets call her $\epsilon_2$ ) is perpendicular to $\epsilon_1$ lets distinquish the following cases: A. If $\epsilon_1$ has an equation of the form $x=x_0,x_0\in\mathbb{R}$. Then $\epsilon_2$ has an equation of the form $y=y_0,y_0\in\mathbb{R}$. You can easily find $y_0$ from $D$ and then find the coordinates of $C$ from the given length. B. If $\epsilon_1$ has an equation of the form $y=y_0,y_0\in\mathbb{R}$. Then $\epsilon_2$ has an equation of the form $x=x_0,x_0\in\mathbb{R}$. You can easily find $x_0$ from $D$ and then find the coordinates of $C$ from the given length. C. If $\epsilon_1$ isn't perpendicular to any of the axes $xx',yy'$ then $\epsilon_1$ has an equation of the form $y=ax+b$ (find $a,b$).Then, $\epsilon_2$ has an equation of the form $y-d_2=\lambda(x-d_1)$ where $\lambda\cdot a=-1$. The circle $C$ with center $D(d_1,d_2)$ and radius the given length $r$ has an equation of the form $(x-d_1)^2+(y-d_2)^2=r^2$. Solving the system of equations for $C$ and $\epsilon_2$ will give you the desired coordinates. (Notice: you'll find 2 points in each case!)
H: If $G$ is a group and $N$ is a nontrivial normal subgroup, can $G/N \cong G$? I know $G/N$ is isomorphic to a proper subgroup of $G$ in this case, so the gut instinct I had was 'no'. But there are examples of groups that are isomorphic to proper subgroups, such as the integers being isomorphic to the even integers, so that reasoning doesn't work. However in this case the even integers are not a quotient of the integers. edit: I realize now that $G/N$ is not necessarily isomorphic to a proper subgroup of $G$, just a subgroup of $G$. AI: Take $G = \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus \ldots $ and $N = \mathbb{Z}\oplus \{0\}\oplus \{0\} \ldots$
H: Minimize distance from a point to a parabola Find the point on the parabola $$y^2=3x$$ that is closest to the point $(1,7)$. AI: A point on that parabola can be parametrized as $(\frac{y^2}{3},y)$. Now find the distance between $(\frac{y^2}{3},y)$ and $(1,7)$ and minimize it.
H: Strange mistakes when calculate limits I have difficulties with calculating the following limits. W\|A gives the correct answers for both of them: $$ \lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \\ = \lim_{x \to +\infty} x\left(\sqrt{1+ \frac {\sqrt x}{x}} + \sqrt{1 - \frac{\sqrt x}{x}} - 2\right) = \left[t = \frac 1x\right] = \lim_{t \to 0+} \frac{\sqrt{1+\sqrt t} - 1 +\sqrt{1-\sqrt t} - 1}{t} = \left[c = \sqrt t\right] = \lim_{c \to 0+} \frac{\sqrt{1+c} - 1 +\sqrt{1-c} - 1}{c^2} = \frac 12c + \left(-\frac 12 c\right) = 0 $$ It's because: $$ \lim_{c \to 0+} \frac{\sqrt{1+c}-1}{c} = \left\|k = \sqrt{1+c}\right\| = \lim_{k \to 1+} \frac{k-1}{k^2-1} = \lim_{k \to 1+} \frac{1}{k+1} = \frac 12 $$ The second part is similar to it. But the answer is -$\frac 14$. And the second limit is: $$ \lim_{x \to 0} \left(\frac{e^x - \sin x}{\sqrt{1 - 2x} + \log(x+1)}\right)^\cfrac{1}{x^2} = \exp\left(\lim_{x \to 0} \frac{1}{x^2}\left(\frac{e^x-\sin x}{\sqrt{1-2x} + \log(x+1)}-1\right)\right) = \exp\left(\lim_{x \to 0}\frac{e^x - \sin x - \sqrt{1-2x} - \log(x+1)}{x^2} \cdot \frac{1}{\sqrt{1-2x}+\log(x+1)}\right) = \exp\left(\lim_{x \to 0} \left(\frac{e^x-1}{x \cdot x} - \frac{\sin x}{x \cdot x} - \frac{\sqrt{1-2x}-1}{x \cdot x} - \frac{\log(x+1)}{x \cdot x}\right)\right) = \exp\left(\lim_{x \to 0} \left(\frac 1x - \frac 1x + \frac 1x - \frac 1x\right)\right) = e^0 = 1 $$ The answer is $e^\frac 32$. Where am I wrong? AI: How do you get from $$\lim\limits_{c \rightarrow 0^+} \frac{\sqrt{1+c} - 1 + \sqrt{1-c} - 1}{c^2}$$ to $$\frac{1}{2}c - \frac{1}{2}c?$$ You cannot just evaluate part of the limit and leave another part. Compare your mistake with "evaluating" $\lim_{x \rightarrow 0} \frac{x^2}{x^2}$ by noting that $\lim_{x \rightarrow 0} \frac{x^2}{x}$ is zero and then concluding that $$\lim_{x \rightarrow 0} \frac{x^2}{x^2} = \lim_{x \rightarrow 0} \frac{0}{x} = 0$$ In terms of Taylor series, $\frac{\sqrt{1+x} - 1}{x}$ is approximated by $\frac{1}{2} - \frac{x}{8} + O(x^2)$. So the limit as $x \rightarrow 0$ is equal to $\frac{1}{2}$, but $\frac{\sqrt{1+x} - 1}{x^2}$ is approximated by the Taylor series $\frac{1}{2x} - \frac{1}{8} + O(x)$. Notice that this is not equal to $\frac{1}{2x}$, which you substitute it with! In a similar way, $\frac{\sqrt{1-x} - 1}{x}$ is approximated by the Taylor series $-\frac{1}{2x} - \frac{1}{8} + O(x)$, and adding these up gives $-\frac{1}{4} + O(x)$, and thus the limit of $\frac{\sqrt{1+x} - 1 + \sqrt{1 - x} - 1}{x^2}$ as $x \rightarrow 0$ is equal to -$\frac{1}{4}$.
H: infinite intersection of 2 uncountable sets Is it true that if the intersection of 2 uncountable sets is infinite, then the intersection is definitely uncountable? How do I start disproving/ proving this statement? AI: False: Construct 2 sets: $A=\mathbb{N}\cup(0,1), B=\mathbb{N}\cup(2,3)$.
H: Application of Sylow's theorem Let $p>q$ be primes. $ (1): \exists $ non-abelian group of order $pq$ $\Longleftrightarrow$ $p \equiv 1 (mod \ q)$ $(2):$ Any $2$ non-abelian groups of order $pq$ are isomorphic to each other. Proof of claim $(1):$ Suppose $\exists$non-abelian $G$ of order $pq.$ Let $P$ be the $p-$sylow subgroup of $G$ and $N(P)$ be the normalizer of $P$ in $G.$ Then, $P \leq N(P) \leq G$ and so $P = N(P)$ or $P=G.$ If $P=N(P).$ Then, $\|G:N(P)\|= q = 1 + kp$ $(k > 0),$ by Sylow's theorem. This contradicts that $p>q.$ So $G = N(P)$ and $\|G:N(P)\|=1.$ How do I proceed to show $p \equiv 1 (mod \ q) ? $ Conversely, suppose $p \not \equiv 1 (mod \ q).$ How do I proceed from here? $(2):$ Could anyone advise on the proof ? Thank you very much. AI: Suppose $p > q$ and $P$ is a $p-$Sylow subgroup, $Q$ is the q-Sylow subgroup. You know that the p-Sylow subgroup $P$ is normal in $G$. Now suppose $p\neq 1\pmod{q}$, then the number $n_q$ of q-Sylow subgroups must also be 1, and hence $Q\triangleleft G$. Now check that $G = PQ$ and since $P\cap Q =\{e\}$, $G$ is abelian; and in fact $$ G \cong P\times Q \cong \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z} \cong \mathbb{Z}/pq\mathbb{Z} $$ Now if $p\equiv 1\pmod{q}$, then it is possible that $n_q = p$; and so $Q$ may not be normal in $G$. Also, there is a non-trivial homomorphism $\alpha : Q \to Aut(P) \cong (\mathbb{Z}/p\mathbb{Z})^{\times}$. This will give a non-trivial semi-direct product. Furthermore, any two non-zero homomorphism $\alpha, \beta : Q\to Aut(P)$ will have the same image in $Aut(P)$ (since $Aut(P)$ is cyclic and so has a unique subgroup of order $q = \|Q\|$). Hence, the two corresponding semi-direct products will be isomorphic.
H: How to simplify this summation equation? If I have $$ \sum_{n=1}^K nx^{n-1} = \frac{d}{dx}\frac{1-x^{K+1}}{1-x}. $$ How to calculate this formula based on above equation ? $$a = \sum_{n=2}^{K}n\left(\frac{\gamma}{2}\right)^{n-1}\rho^n$$ AI: Hint: Factor out $\rho$ and put it outside the summation. $$a = \sum_{n=2}^{K}n\left(\frac{\gamma}{2}\right)^{n-1}\rho^n = \rho\sum_{n=2}^{K}n\left(\frac{\gamma}{2}\right)^{n-1}\rho^{n-1}=\rho\left(\left(\sum_{n=1}^{K}n\left(\frac{\gamma}{2}\right)^{n-1}\rho^{n-1}\right)-1\right)$$
H: Isomorphism of R-modules Does somebody has an example where the left $R$-modules $R^m$ and $R^n$ are isomorphic for all positive integers $m$, and $n$? AI: Let $\mathrm{CFM}_\mathbb{N}(R)$ denote the ring of "column finite matrices", where $R$ is some ring.. Then, one can show that $\mathrm{CFM}_\mathbb{N}(R)\to\mathrm{CFM}_\mathbb{N}(R)^2$ defined by $$M\mapsto (\text{odd indexed columns of }M,\text{even indexed columns of }M)$$ is an isomorphism of $\mathrm{CFM}_\mathbb{N}(R)$-modules. It clearly then follows that $\mathrm{CFM}_\mathbb{N}(R)^m\cong\mathrm{CFM}_\mathbb{N}(R)^n$ for all $n$ and $m$. I believe this example is in Dummit and Foote.
H: How find this continued fraction Question: let $x$ such this Continued fraction $$x=[0;1,3,5,7,9,11,13,\cdots]$$ How find the vaule of $x$. (can see:http://en.wikipedia.org/wiki/Continued_fraction) My try: I know this $$e=[2;1,2,1,1,4,1,1,6,1,1,8,\cdots] $$and $$\pi=[3;7,15,1,292,1,1,1,2,1,3,1,\cdots]$$ But for my problem,I can't find solve it.Thank you AI: See Gauss's continued fraction: This particular expansion is known as Lambert's continued fraction and dates back to 1768.
H: Find an value of real number M such that $\frac{2+\sin(n+1)}{2+\sin(n)}\leq M<3$ This sequence is certainly limited \begin{equation} a_n=\frac{2+\sin(n+1)}{2+\sin(n)} \end{equation} For example, if $\sin(n+1)=1$ and $\sin(n)=-1$, we have: $a_n\leq 3$. But I believe that there is $0<M<3$ such that $a_n\leq M$. I would like to find this value of M. Who has any suggestions for me? Thank you very much. AI: Fix some $\alpha$ and, for every real number $x$, let $$a(x)=\dfrac{2+\sin(x+\alpha)}{2+\sin(x)}. $$ Using the addition formula $\sin(x+\alpha)=\cos(\alpha)\sin(x)+\sin(\alpha)\cos(x)$, one sees that $a(x)\leqslant M$ if and only if $$ \sin(\alpha)\cos(x)+(\cos(\alpha)-M)\sin(x)\leqslant2M-2. $$ There exists $(r,t)$, $r\gt0$, such that $\sin(\alpha)=r\sin(t)$ and $\cos(\alpha)-M=r\cos(t)$, thus the LHS is $r\sin(x+t)\leqslant r$. This shows that if $2M-2\geqslant r$ then $a(x)\leqslant M$ for every real number $x$. Assume that $M\geqslant1$. Then $r^2=\sin(\alpha)^2+(\cos(\alpha)-M)^2$ hence the condition is $$ 4(M-1)^2\geqslant1-2\cos(\alpha)M+M^2,\tag{$\ast$} $$ that is, $$ 3M^2-2(4-\cos(\alpha))M+3\geqslant0,\qquad M\geqslant1. $$ Numerically, this means one can choose $$ M_\alpha=\frac13\left(4-\cos(\alpha)+\sqrt{7-8\cos(\alpha)+\cos(\alpha)^2}\right). $$ Thus, $1\leqslant M_\alpha\leqslant3$ for every $\alpha$ and $M_\alpha\lt3$ for every $\alpha$ not in $\pi+2\pi\mathbb Z$. In particular, $M_1\lt1.73$. To show more modestly that $M=2$ works when $\alpha=1$, it suffices to check that, for $M=2$, $(\ast)$ reads $\cos(1)\geqslant\frac14$, and to note that this last inequality is obvious since $1\leqslant\frac\pi3$ and $\cos(\frac\pi3)=\frac12$ hence $\cos(1)\geqslant\frac12$.
H: Question about $f :\mathbb{R}\rightarrow \mathbb{R}$ defined as $f(x)=\|x\|^{\frac{3}{2}}$ (TIFR GS $2010$) Question is : I am not sure how to check for differentiability, only thing i know is how to see for differentiability of $f(x)=\|x\|$ $\lim_ {x\rightarrow 0} \frac{f(x)}{x}=\lim_ {x\rightarrow 0} \frac{\|x\|}{x}$ But then, when $x$ approach $0$ from positive line we have $\lim_ {x\rightarrow 0^+} \frac{\|x\|}{x}=1$ and when $x$ approach $0$ from negativeline we have $\lim_ {x\rightarrow 0^-} \frac{\|x\|}{x}=-1$ So, limit does not exists so the function $f(x)=\|x\|$ is not differentiable at $0$ But, I am not getting any idea how to proceed in case of $f(x)=\|x\|^{\frac{3}{2}}$ $lim_ {x\rightarrow 0} \frac{f(x)}{x}=\lim_ {x\rightarrow 0} \frac{\|x\|^{\frac{3}{2}}}{x}$ I was thinking to write this as $$\lim_ {x\rightarrow 0} \frac{\|x\|^{\frac{3}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{\|x\|^{2-\frac{1}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{\|x\|^2 .\|x\|^{\frac{-1}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{x}{\|x\|^{\frac{1}{2}}}$$ as $x$ is in square rot in denominator i am unable to see what are its left and right limits for $x=0$ Please help me to proceed further.. Thank you AI: It might be easier to denote $\|x\|^{3/2} = x\cdot\sigma(x)\sqrt{\|x\|}$, where $\sigma(x)$ is the sign of $x$. Then $$\left\|\frac{\|x\|^{3/2}}{x}-0\right\|=\sqrt{\|x\|}\to 0.$$
H: calculus the following functions could you find the minimum of? In finding the point on the line y=x closest to the point (0,1) which of the following functions could you find the minimum of? a. $2x^2+2x+2$ b. $x^2-2x+2$ c. $2x^2-2x+1$ d. $2x^2-x+2$ e. $x^2-1$ Anyone can explain? AI: If the $d$ is the distance, $$d^2=(0-x)^2+(1-y)^2=(0-y)^2+(1-y)^2=2y^2-2y+1=2x^2-2x+1$$ As $d>0,$ $d^2$ will be minimum $\iff d$ is minimum
H: How to solve this system of differential equations? $\begin{cases} \dot{p}=-p+2x\\ \dot{x}=\frac{1}{2}p + x\\ \end{cases}$ I don't know how I can express $p$ only in terms of $p$'s or $x$ only in terms of $x$'s. Can anyone please help? AI: $x=(1/2)(p'+p)$, $x'=(1/2)(p''+p')$, $(1/2)(p''+p')=(1/2)p+(1/2)(p'+p)$. (edited to correct a typo noted in the comments)
H: I need some help with the derivative of this function. Hey guys i was wondering , what is the derivative function of this function. f(x) = $\sqrt{x} - e^{-x}$ Any advise will be greated. AI: In general, $$\frac{d}{dx}(x^n) = nx^{n-1}$$ and $$\frac{d}{dx}(e^{ax}) = ae^{ax}.$$ Observing that $$f(x) = x^{\frac{1}{2}} - e^{-x}$$ we apply these general results to find that $$f'(x) = \frac {1}{2}x^{-\frac{1}{2}}+e^{-x}$$
H: Prove that there are rational polynomials $p,q$ such that $p(x)(x^4+2x^2+1)+ q(x)(x^4-3x^2-4) = x^2+1$ Prove that there are polynomials $p(x), q(x)$ in $\mathbb{Q}[X]$ such that: $$p(x)(x^4+2x^2+1)+ q(x)(x^4-3x^2-4) = x^2+1$$ Is it still true if we replace $x^2+1$ with $x+5 $? So: I know how to prove this: it's by the Polynomial remainder theorem. Someone told me that I need to move to the complex field, and put in the value $x =i $ to get $0 = 0 $, but if we replace $x^2+1$ in $x+5$ we will get $0 =i+5 $, which is false; therefore it is not true for $x+5$, but I don't understand this explanation because our field is $\mathbb{Q}[x]$ and not complex. AI: You know that $(x^{2} + 1) \| (x^{4} + 2x^{2} + 1)$ and $(x^{2} + 1) \| (x^{4} - 3x^{2} - 4)$ and hence it divides $p(x)(x^{4} + 2x^{2} + 1) + q(x)(x^{4} - 3x^{2} - 4)$. If this expression was $(x + 5)$ then we would have $x^{2} + 1$ divide $x + 5$ which is non-sense. Hence the $x^{2} + 1$ in part 1 can't be replaced by $x + 5$. The reason part 1 holds is because of the Euclid's algo for GCD of $(x^{4} + 2x^{2} + 1)$ and $(x^{4} - 3x^{2} - 4)$
H: Multiplicative function, $\mathbb{Z}[a+b\frac{1+\sqrt{-19}}{2}]$ Could you tell me how to prove that this function is multiplicative? $R = \mathbb{Z}[a+b\frac{1+\sqrt{-19}}{2}]$ $f: R \ni a+b\frac{1+\sqrt{-19}}{2} \rightarrow a^2 + ab + 5b^2 \in \mathbb{Z}$ Here's my attempt to solve it: $f(1) = 1$ I multiply $(a+b\frac{1+\sqrt{-19}}{2})(c+d\frac{1+\sqrt{-19}}{2})$ and write it as $ac-5bd + (ad+bd+cb)(\frac{1+\sqrt{-19}}{2})$ and then I apply to it. Then I multiply $(a^2 + ab + 5b^2)(c^2 + cd + 5d^2)$ and Wolphram says that the difference is 0. I know this isn't a correct solution, because it's very time-consuming, and a bit naive. Could you tell me how to solve it in a more elegant manner? Thank you. AI: Hint: If $$z=a+b\frac{1+\sqrt{-19}}2,$$ then what is $z\overline{z}$?
H: Is my proof for $\lim_{n\to\infty}\frac3{n+1}$ wrong? To calculate $\lim_{n\to\infty}\frac3{n+1}$ I do: $$\lim_{n\to\infty}\frac3{n+1}=\lim_{n\to\infty}\frac3\infty=0$$ but my teacher is not convinced so how can I do?? AI: O.K. A proof without $\epsilon$ and $\delta$ can it be as follow. Assuming that $\lim_{n\to\infty}\frac{1}{n}=0$ we have that $0<\frac{3}{n+1}\leq \frac{1}{n}$ for all $n>50$ for example (since we use n large this no matter) so $\lim \frac{3}{n+1}\leq \lim\frac{1}{n}=0$ but $\frac{3}{n+1}>0$ so $\lim\frac{3}{n+1}\geq 0$ i.e. $\lim\frac{3}{n+1}=0$.
H: combinations in a circuit in a circuit there are 3 switches, of which 1 & 2 are in series., 3 is in parralel with 1&2. all three switches can either be open or closed. how many ways of opening/closing all three different switches are there where there is electricity flowing through the circuit? i would like to say $2!2!2!$ but that is for all combinations. What is the best way to think about this? AI: Let's call $C_1$ the bit with switches 1 and 2 and $C_2$ the bit with just switch 3. In total you have $2^3$ ways of combining you close/open switches as each has only 2 possible positions. Now you need to count how many ways there are so that it no electricity flows through and subtract that from $2^3 = 8$. For no electricity to flow through, you need both $C_1$ and $C_2$ to be open. $C_2$ has only one way to be open that is to open switch 3. $C_1$ on the other hand has three different ways of being open. Both switches open, 1 only or 2 only. So you have $1\times3 = 3$ ways of having an open circuit so $8-3 = 5$ ways of having electricity flowing through. Another approach is to directly count all possibilities for the electricity to flow through. Because you only have three switches, I suggest you do a tree with each switch begin either open or close and for each combination of the three, deduce whether the electricity flows through.
H: why for every $ f\in C(\sigma(x))$ we have $ \Phi (f(x))= f(\Phi(x))$? In a book about $ C^* $-algebra, in the section of continuous functional calculus says that: Suppose $ x $ is a normal element of $ C^$-algebra $ A $, then the continuous functional calculus has this property that If $ \Phi: A \to B $ is a $ C^$-homorphism ($B$ is an arbitrary $C^*$-algebra) then for every $ f\in C(\sigma(x))$ we have $ \Phi (f(x))= f(\Phi(x))$. Please help me about the proof of this statement. thanks AI: This is clear if $f$ is a polynomial in $t,\overline{t}$. But these polynomials are dense in $C(\sigma(x))$ (and actually this is how one constructs the functional calculus) and both sides of the equation are continuous in $f$. This proves the equation in general.
H: Complex number ( equation) Find the number of solutions of $z^3+\overline{z}=0$ ... Problem : Find the number of solutions of $z^3+\overline{z}=0$ Solution : $z^3 =-\overline{z} \Rightarrow \|z\|^3 = \|(-\overline{z})\|$ $\Rightarrow \|z\|^3 = \|z\| \Rightarrow \|z\|^3-\|z\| =0$ $\Rightarrow \|z\|(\|z\|^2-1)=0 \Rightarrow \|z\| =0 ; \|z\|^2=1 \Rightarrow \|z\| = \pm 1$ $\|z\|^2=1 \Rightarrow z.\overline{z} =1 \Rightarrow z.\overline{z} = 1$ Now the equation can be written as $z^3+\frac{1}{z}=0$ $\Rightarrow z^4+1 =0$ This equation has four roots. But answer is five roots. Please suggest AI: As @Daniel Fischer already commented: From what you did in your calculation $$\|z\|(\|z\|^2-1)=0 \Rightarrow \color{orange}{\|z\| =0} ; \|z\|^2=1 \Rightarrow \|z\| = \pm 1$$ and from $\|z\|^2=1 \implies \ldots \implies z^4+1=0$ which gives $4$ roots plus $z=0$ gives one root.
H: Finding the Expected value You are in a room; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors. If you choose the ath door, it can either take you back to the same position where you begun in ti minutes, or can take you out of the maze after ti minutes. If you come back to the same position, you can't remember anything. So, every time you come to the beginning position, you have no past experience. Now you want to find the expected time to get out of the room. In the case if the particular door will get you back to same position.You'll be given the minutes in negative. Suppose if the input is of the form For each door, -6 -9 ....-value Here everything is negative,so we cannot come out of the door.We have the probability here as "infinity" Suppose if we have: 3 doors with values 3 -6 -9 The expected value is 18 ? but i don't understand how the answer is 18.Can someone please explain in simple terms ? Source : This is a programming contest question. AI: Let $E$ be the expected number of minutes to escape. Then anytime you are in the room about to choose a door among the three mentioned, $$E = \begin{cases} 3 & \text{ probability} \frac13 \\ 6+E & \text{ probability} \frac13 \\ 9+E & \text{ probability} \frac13 \end{cases}$$ Thus we have $E = \frac13 \times 3 + \frac13 \times (6+E) + \frac13 \times (9+E) \implies 3E = 18 + 2E \implies E = 18$. You can extend the logic to more doors / other times...
H: Need a counter example for series convergence I need some advice for constructing a counter example for $\sum\limits_{i=1}^\infty a_i$ converge but $\sum\limits_{i=1}^\infty \frac{a_i}{i}$ diverges. AI: There is no counterexample. Take $t_n=\frac{1}{n}$ then $t_n$ is decreasing & positive with limit $0$. It is given that $\displaystyle\sum_{n\geq 1} a_n$ converges. By Dirichlet's test it follows that $\displaystyle \sum_{n\geq 1} a_n t_n$ converges.
H: Laplace's Equation for a Radial Function (cylindrical co-ord) I'm working on a question which has lead me to the Laplace equation in cylindrical coordinates. I've looked it up and found that, for the radial component, this is equivalent to $$\nabla^2\ f = \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right) = 0.$$ I think I should be able to separate this out somehow to get something like the form $$a\frac{\partial f}{\partial r}+b\frac{\partial^2 f}{\partial r^2} =0.$$ Any ideas? I hope I've made this clear enough! AI: If you know how to differentiate a product, then $$ \frac{\partial}{\partial r} \left( r \frac{\partial f}{\partial r} \right) = \frac{\partial f}{\partial r} + r \frac{\partial^2 f}{\partial r^2}. $$
H: $\|a_n - a_m\| \le \sum_{k=m}^{n-1} \|a_{k+1} - a_k\|$ How to show, that every for sequence $(a_n)_{n\ge_1}$ with $n \gt m$ following holds ? $$\|a_n - a_m\| \le \sum_{k=m}^{n-1} \|a_{k+1} - a_k\|$$ AI: Let $d=(n-m)$. We have $n-(d-1)=m+1 $ and \begin{align} \|a_n-a_m\| =& \|a_n\color{red}{-a_{n-1}+a_{n-1}}-a_m\|\\ =& \|(a_n-a_{n-1})+a_{n-1}\color{red}{-a_{n-2}+a_{n-2}}-a_m\|\\ =& \|(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+a_{n-2}\color{red}{-a_{n-3}+a_{n-3}}-a_m\|\\ \vdots\;& \hspace{4cm}\vdots\\ =& \|(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+(a_{n-2}-a_{n-3})+\cdots\color{red}{-a_{n-(d-1)}+a_{n-(d-1)}}-a_m\|\\ =& \|(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+(a_{n-2}-a_{n-3})+\cdots+(a_{m-1}-a_m)\|\\ \end{align}
H: On irreducible polynomial Prove that the polynomial $f(X)=X^5-9X^3+15X+6$ is irreducible over $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ Apply Eisenstein's Irreducibility Criterion with $p=3$ we see that $f$ is irreducible over $\mathbb{Q}$. Can conclude that $f$ is irreducible over $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$? AI: You can't directly conclude that $f$ is irreducible over $\mathbb{Q}(\sqrt{2},\sqrt{3})$ from its irreducibility over $\mathbb{Q}$. But since the degree of $f$ is $5$, you can conclude that for any zero $\alpha$ of $f$, $$[\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]$$ is a multiple of $5$. On the other hand, $$\begin{align} [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}] &= [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})]\cdot [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\\ &= 4[\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})], \end{align}$$ so $$5 \mid [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})].$$
H: Calculate the area between $f(t) = \cos(t)$ and the $t$ axis as $t$ varies from $0$ to $\pi/4$ The problem below is part of a multiple section question that I'm hoping to solve. Calculate the area between $f(t) = \cos(t)$ and the $t$ axis as $t$ varies from $0$ to $\pi/4$. AI: $$\text{Area}=\int_0^{\pi/4} \cos(t) \space\text{d}t=[\sin(t)]_0^{\pi/4}=\sin(\pi/4)-\sin(0)=\dfrac{1}{\sqrt{2}}-0=0.5\sqrt{2}$$
H: $f(x)=\sin x^3$ for $x\in \mathbb{R}$ is not uniformly continuous Question is to prove that : $f(x)=\sin x^3$ for $x\in \mathbb{R}$ is not uniformly continuous. What would my first observation in checking uniform continuity is to check if its derivative is bounded. In this case its derivative $f'(x)=3x^2\sin x^3$ which is unbounded. So, I can not rely on this. I tried with definition : $\|f(x)-f(y)\|=\|\sin x^3-\sin y^3\|=\|2\cos (\frac{x^3+y^3}{2})\sin (\frac{x^3-y^3}{2})\|\leq 2 \|\sin (\frac{x^3-y^3}{2})\|$ I see that $\sin x \leq x$.. This may not imply $\|\sin x\|\leq \|x\|$ but i am assuming it... So, $\|f(x)-f(y)\|\leq 2 \|\sin (\frac{x^3-y^3}{2})\|\leq 2\frac{\|x^3-y^3\|}{2}=\|x^3-y^3\|=\|x-y\|\|x^2+xy+y^2\|$. Though $\|x-y\|$ is small i should make $\|f(x)-f(y)\|$ considerably large. I now have $\|f(x)-f(y)\|\leq 2 \|x-y\|\|x^2+xy+y^2\|$ I some how can sense that I can make $\|x^2+xy+y^2\|$ large enough though $\|x-y\|$ is small but not so sure how to make this in $\epsilon-\delta$ case. Please help me to solve this. Thank you. EDIT : I know that uniform continuous functions takes cauchy sequence to cauchy sequence and tried to use it in this case. But, I could not find correct cauchy sequence that would help me to go through this. I would be thankful if some one can help me in this way too. AI: Let $x_n=(n\pi+\pi/2)^{1/3}$, then $x_{n+1}-x_n\to0$ but $f(x_{n+1})-f(x_n)=\pm2$ hence $f(x_{n+1})-f(x_n)$ does not converge to $0$. Therefore, no uniform continuity.
H: Find the integral of $\tan^4(x) \sec(x)$ I want to know if there is a shorter way to find integral of $\tan^4(x) \sec(x)$ without using reduction formula of $\sec(x)$ because it's really takes a long time. Thanks all AI: Write it as $$ \int\frac{\sin^4x}{\cos^5x}\,dx= \int\frac{\sin^4{x}}{\cos^6x}\cos x\,dx= \int\frac{\sin^4{x}}{(1-\sin^2x)^3}\cos x\,dx $$ Can you see an obvious substitution?
H: Combination and Permutation- examples I am not sure how to answer this question. I have a word TOKYO. a) how many different can I arrange the letters in a row? b) in how many of these will O's be together? c) how many will the two O's not together? My thoughts: a) I figure out that this is a permutation because order is important hence. I guess "different" also means "unique"?. Also am assuming the question says we should use all the 5 letters. unique words $= \dfrac{5!}{2!\times1!\times1!\times1!} = 60 $ b)for this question, I got thoroughly confused. When I did the words manually, I managed to get 24 words (e.g. OOTKY OOTYK OOKTY .....YTKOO YKTOO.) But I cant seem to figure out the formula for this. I thought it is also a permutation: $5P2 which would be \dfrac{5!}{(5-2)!} = \dfrac{5!}{3!} = 20 $ c) for this question, I subtract answers in $ a-b =60-20 = 30 $ AI: a) Correct! b) Treat the 2 Os as one group, so there are 4 groups in total, hence 4! c) a)-b)
H: Help With Eigenvectors and Dynamical Systems I have the following system of differential equations: $ \frac{d}{dt} \left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c c} -(a+b) & 0\\ a & -(a+b)\\ \end{array} \right] \left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] $ I believe I have correctly obtained the eigenvalue to be $\lambda = -(a+b)$ and associated eigenvector to be $\nu = \left[ \begin{array}{c} 0\\ 1\\\end{array} \right]$. I am trying to solve the system by the following: $\left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c} \nu & \nu\end{array} \right] \left[ \begin{array}{c c} \exp^{\lambda t} & 0\\ 0 & \exp^{\lambda t}\\ \end{array} \right]\left[ \begin{array}{c} A(0)\\ N(0)\\\end{array} \right]$ $\left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c c} 0 & 0\\ 1 & 1\\ \end{array} \right] \left[ \begin{array}{c c} \exp^{-(a+b)t} & 0\\ 0 & \exp^{-(a+b)t}\\ \end{array} \right]\left[ \begin{array}{c} A(0)\\ N(0)\\\end{array} \right]$ $\left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c} 0\\ A(0)\exp^{-(a+b)t}+N(0)\exp^{-(a+b)t}\\\end{array} \right]$ Clearly I am doing something wrong. I suspect it has something to do with the fact that there is only one eigenvalue for this system. I was hoping Someone could help me figure this out? Thanks Mark AI: Hints: You found the correct eigenvalue and it is a repeated eigenvalue $\lambda_{1,2} = -(a+b)$. You need to find two linearly independent eigenvectors. $[A-\lambda I]v = 0$ only produces one linearly independent eigenvector, so you must find a generalized one. The first eigenvector you found is correct. Find a linearly independent generalized second eigenvector. For the second eigenvector, try $[A-\lambda I]v_2 = v_1$. Once you have found that eigenvector, you can write the solution as (because of the repeated eigenvalue): $$X(t) = \left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = c_1~e^{\lambda~t}~(t~v_1+v_2) + c_2~e^{\lambda~t}~v_1$$
H: Extend a function as odd/even periodic function Let $f$ be the function $f(x) = x^2 + 2 $, where $ 0<x<1 $. Extend the function $f(x)$ (1) As an odd periodic function with period $2$ (2) As an even periodic function with period $2$ (3) As a periodic function with period $1$ I know what exactly are odd and even functions. But have no idea how to extend as such. Please explain with a correct method. AI: Even case Due to symmetry we need to define $f(x)=x^2+2$ on $(-1,0)$. It remains to discuss the $2$-periodicity. One starts with "special points", for example $x=0$. Then $$f(0)=0^2+2=\text{(2-periodicity)}=f(0+2)=f(2),$$ i.e. $f(2):=2$. To find the extension of $f$ on $(2,3)$ one continues with all $\delta\in (0,1)$, i.e. $$f(\delta)=\delta^2+2=\text{(2-periodicity)}=f(\delta+2),$$ where $\delta+2\in (2,3)$. Similarly, considering all points $-\delta\in(-1,0)$, using $2$-periodicity one finds the extension of $f$ on the interval $(1,2)$ via $-\delta+2\in (1,2)$, as we did above. The extension of $f$ for all other points easily follow by drawing. Odd case Due to symmetry we need to define $f(x)=-x^2-2$ on $(-1,0)$. One selects once again the "special points", for example $x=0$ as in the even case. The analysis is completely similar, with due changes.
H: Simple Modular Arithmetic We know that, for example, $2x \equiv 3 \mod 4$ has no solutions since $2\mid 2x, 2\mid 4,$ however 2 does not divide 3. So my question is, how does one get from $2x \equiv 3 \mod 5$ to $x \equiv 4 \mod 5$? In other words, could someone please explain the steps involved going from $2x = 3 + 5k$ to $x = 4 + 5k$, where $k\ \epsilon \ \mathbb{Z}$. AI: Well, there can be many ways. For example, double the congruence $2x\equiv 3\pmod 5$: $\ 4x\equiv 6\pmod 5$, but because $4\equiv -1$ and $6\equiv 1$, this yields $$-x\equiv 1\pmod 5$$ which gives $x\equiv -1\ (\equiv 4) \pmod 5$.
H: Suppose $f$ is entire and for all $z \in \mathbb{C}$, $f(z)=f(\frac{1}{z})$. Prove that $f$ is constant. Suppose $f$ is entire and for all $z \in \mathbb{C}$, $f(z)=f(\frac{1}{z})$. Prove that $f$ is constant. I want to prove $f$ is bounded. Then by using Liouville theorem, $f$ is constant. But I can't proceed. Can anyone give some hint? AI: Note : If $f$ is entire, then, in particular it is bounded on closed balls around the origin. Hence, if $D = \{z : \|z\| \leq 1\}$, then $f$ is bounded on $D$ by some $M > 0$. Hence, if $\|z\| \geq 1$, then $1/z \in D$, and hence $$ \|f(z)\| = \|f(1/z)\| \leq M $$ And hence, $\|f(z)\| \leq M$ for all $z\in \mathbb{C}$. Hence, Louiville applies.
H: Any angle divides a plane into two regions I am studying Euclidian geometry and I noticed that any angle divides a plane into two regions: an inside and an outside. Is there a need for a proof of this (something along the lines of Jordan theorem), or is it just "obvious"? Browsing the internet, I came across a following simpler version: any line divides a plane into two regions. Maybe someone will find it relevant. My current understanding is that the PSA plays a key role in this sort of thing. Unless you are doing it the analytic geometry way, in which case the PSA must be somehow already "coded in" ...I think. AI: Let $\vec v$ (the vertex), $\vec a$, $\vec b$ (the rays) of an angle in $\boldsymbol R^2$. The interior $I$ of the angle may be defined as $I:=\{\vec v+t\cdot\vec a+s\cdot\vec b\mid t,s\in\boldsymbol R_+\cup\{0\}\}$. Can you go from here?
H: Did I do this assignment right? Antisymmetric relation. I need to prove or disprove, that R is antisymmetric. This is my set: $$ R=\{(1,1),(1,2), (1,4), (2,1), (2,2), (3,2), (3,3), (4,4)\} $$ I proved that it is not antisymmetric in the following manner: $$ \text{Definition of antisymmetric relation}\\ (x,y)\in R \land (y,x)\in R \implies x=y\\ (x,y)\in R \land x\not = y \implies (y,x) \notin R\\ \text{I found a example in set where this does not apply}\\ (2,1)\in R \land 2\not = 1 \implies (1,2) \notin R\\ \text{This is false, so that means that the relation is not antisymmetric.} $$ This means thet this set is not antisymmetric?? Right?? Thanks!!! AI: Your reasoning is right but you need to write it a bit more clearly. Something like the following should be fine. Suppose $R$ is anti-symmetric. We see that $(2,1)\in R$ and $2\neq 1$ and so by anti-symmetry we should have that $(1,2)\notin R$. However, this is not the case because $(1,2)\in R$ and so we have reached a contradiction. It follows that the assumption that $R$ is anti-symmetric is false. I should probably add, getting used to using symbols when writing proofs is good... but honestly, most people prefer reading proper English sentences in proofs. You should get into the habit of writing your math out in prose, sure with the occasional symbol where needed, so that you can actually read it out loud and it makes sense. It also shows that you actually understand what you're writing, and you're not just trying to remember a lot of rules about symbol manipulation.
H: Steps to find gradient of this? I don't understand how to get from .54 to .55, any help? Edit: Wolfram alpha gives the answer as this http://www.wolframalpha.com/input/?i=grad+Ucos%28theta%29%28r%2Ba%5E2%2Fr%29 I can't seem to convert their answer to the answer needed even though it is similar. AI: Hint: Remember that $x = r\cos(\theta)$ in cylindrical coordinates. You have $$ \phi = U \, r\cos(\theta) + U \, \frac{a^2}{r} \cos(\theta) = U\,x+U\,\frac{a^2}{r}\cos(\theta) $$ Take this gradient in cylindrical coordinates (with $x$ the rectangular coordinate) $$ \nabla \phi = \frac{\partial \phi}{\partial x}\mathbf{\hat{e}_x} + \frac{1}{r} \frac{\partial \phi}{\partial \theta} \mathbf{\hat{e}_{\theta}} + \frac{\partial \phi}{\partial r} \mathbf{\hat{e}_{r}} \\ = U\mathbf{\hat{e}_{x}} + U \frac{1}{r}\left( - \frac{a^2}{r} \sin(\theta) \right)\mathbf{\hat{e}_{\theta}} + U \left( - \frac{a^2}{r^2} \cos(\theta) \right) \mathbf{\hat{e}_{r}}$$ Now, after minor algebraic manipulation you will see that the equation you are curious about is exactly $- \nabla \phi$.
H: Convolution is uniformly continuous and bounded Suppose $f\in L^\infty(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Show that the convolution $f\ast K$ is a uniformly continuous and bounded function. The definition of the convolution is $(f\ast K)(x)=\int_\mathbb{R}f(x-y)K(y)dy$. There is the inequality $\\|f\ast K\\|_\infty\leq\\|f\\|_\infty\\|K\\|_1$, which yields that $f\ast K$ is bounded. But what about uniform continuity? AI: First, assume that $K$ is continuous with compact support. Then using $$\|(f\star K)(x)-(f\star K)(x')\|\leqslant\int_\mathbb R\|f(y)\|\cdot \|K(x-y)-K(x'-y)\|\mathrm dy,$$ boundedness of $f$, and uniform continuity of $K$, we get what we want. Then we conclude by a density argument: if $K_n\to K$ in $L^1$, then $(f\star K_n)_n$ converges uniformly on the real line to $f\star K$. Indeed, we have $$\|f\star K_n(x)-f\star K(x)\|\leqslant \left\|\int_\mathbb Rf(x-t)K_n(t)\mathrm dt-\int_\mathbb Rf(x-t)K(t)\mathrm dt\right\|\leqslant\lVert f\rVert_\infty\lVert K_n-K\rVert_1.$$
H: transformation of DFT matrix $\mathbf{F}$ is a unitary DFT matrix where the $(m,n)$-th entry of $\mathbf{F}$ is given by $\frac{1}{\sqrt{M}}e^{-\imath2\pi(m-1)(n-1)/M}$. Note that $\imath=\sqrt{-1}$. Let $\mathbf{A}$ be a matrix where the the $(m,n)$-th entry of $\mathbf{A}$ is given by $\frac{1}{\sqrt{M}}(-1)^{m-1}e^{-\imath2\pi(m-1)(n-1)/M}$. Then how can $\mathbf{A}$ be transformed to $\mathbf{F}$ via some row or column pemutations? AI: $$(-1)^{m-1} = e^{\pi i (m-1)}$$ To see what's going on, it may help to write $$\zeta = e^{2 \pi i / M}$$ and then write all of the exponentials as powers of $\zeta$, noting that the exponents only matter modulo $M$.
H: choosing odd number of balls from $n$ balls How many ways can an odd number of balls be chosen from $n$ balls? I tried enumerating, but it's really too tedious. :/ Any help is appreciated. AI: Hint: #ways of choosing even balls = #ways of choosing odd balls. and total number of ways of choosing balls = $2^n$ (Why?)
H: Set problem where $f=A\times B$ The problem states: If there is function $f:A\to B$ and $$\left(f=A \times B \right) \iff \left( A= \varnothing \text{ or } \|B\| = 1\right)$$ (where $A \times B$ is the cartesian product and $\|B\|$ is the cardinality of $B$) So I must demonstrate that equivalence. I've tried to slove it, but i can't wrap my head around what $f=A\times B$ means. Please help me! AI: Recall that $A\times B$ is the set of all possible ordered pairs from $A$ and $B$, a function is just a subset of these pairs. For example if $A=\{x,y\}$ then a function whose domain is $A$ will only have two elements $\langle x,f(x)\rangle$ and $\langle y,f(y)\rangle$. You are asked to show that if $f$ has all the possible ordered pairs, then either $A$ is the empty set or $B$ is a singleton. To do so, first you have the show that if $A$ is the empty set, or $B=\{b\}$, then $A\times B$ is a function; and in the other direction show either that from the assumption $f=A\times B$ you can conclude at least one of the things hold ($A=\varnothing$ or $B=\{b\}$), or work towards contradiction or contrapositive and assume that $A$ is not empty, $B$ is either empty, or have two elements, and conclude that $A\times B$ cannot be a function whose domain is $A$.
H: Simple Question about the norm. Suppose we have a Banach space $V$ with a norm $\\|\cdot\\|:V\to \mathbb{R}$. Is the following true for all linearly independent vectors $x,y\in V$?: $$\\|x+y\\|\geq \\|y\\|,~~~~\\|x+y\\|\geq \\|x\\|$$ AI: No, take $V=\mathbb{R}^2$, $x=(1,0)$ and $y = (-1,0.1)$.
H: Prove $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ Prove that $$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ for $a,b,c>0$ Any hints/solutions? AI: Expanding the terms, we want to show that $$ \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq 1 + 1 + 1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} $$ This is true because $ \frac{a^2}{b^2} + \frac{b}{a} + \frac{b}{a} \geq 3$ and $ \frac{a^2}{b^2} + 1 \geq 2 \frac{a}{b}$ by direct AM-GM. Add up the cyclic versions of the above inequalities.
H: A plane is a surface. I am trying to show that a plane is a surface. I posted what I did. I find $σ$ But I cannot find $σ^{-1}$. Also as I said, I need to verify $σ$ is 1-1 continuous and continuous inverse. How can I do all? Please help me to end my proof? Thanks. AI: You are asking, how to invert the map $(x, y) \mapsto (x, y, \text{whatever})$. Note that you can assume $\text{whatever}$ to be correct. That is, you don't have to worry whether the given point belongs to the plane. This is more or less by convention. For continuity you need to show that if distance between two points in $\mathbb{R}^2$ is $d(a, b)$, then the distance between their images is less than $kd(a, b)$, for some fixed constant $k$, depending on $a, b, c, d$. That is, small distances are mapped to small distances. Note that it doesn't matter how large $k$ is - you may save yourself some work if you take it to be larger than it needs to be. I'd suggest writing an affine transformation from $z=0$ plane to $ax+by+cz=d$ plane. Continuity and bijectivity of this map are trivial results in linear algebra. You still have to show the same for $f(x, y) = (x, y, 0)$, but it shouldn't be a problem. Note that now you don't have to assume $c \neq 0$, as you did.
H: Proof that the derivative of a linear function is $0$. We have a function $f(x)=x$ which is defined and is continuous on the set $S$ of all real numbers. The derivative at point $x$, $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}h$$ Using the theorems of limits we can replace the above limit equation by:$$f'(x)=\frac{\lim_{h\to 0}{f(x+h)}-f(x)} {\lim_{h\to 0}h}$$ Since $f$ is continuous at all points $f(x+h) \to f(x)$ as $h\to 0$ so $$f'(x)=\frac{0} {\lim_{h\to 0}h}$$ so $f'(x)=0$. What is wrong here ??? This is a continuation of my other question Proving the derivative is $0$ at the extremum and all derivatives are $0$. AI: that part is wrong: "Using the theorems of limits we can replace the above limit equation by:" you need to know both limits (of the denominator and enumarator) exist, are finite, denominator's limit $\not= 0$. in your case $\lim_{h \rightarrow 0} = 0$ so you can't do that.
H: Image of koebe map Let $f(z)=\frac {z}{(1-z)^2}$ how can I find the image of unit disk under this transformation? Also is it possible to find an explicit expression for the inverse of this transformation? AI: Note that $$\frac{z}{(1-z)^2} = \frac{1}{4} \left( \frac{1+z}{1-z} \right)^2 - \frac{1}{4}$$ and that $$z \mapsto \frac{1+z}{1-z}$$ maps the unit disc to the right half plane. The image of the Koebe map is therefore $\mathbb{C} \setminus (-\infty, -\frac{1}{4}]$.
H: How to master integration and differentiation? We have learnt in school about differentiation and integration, however I find my knowledge fairly poor. I mean I have problems with taking the derivative/integral even simple functions. So I would like to get some links to articles and guides where I could learn the derivatives and primitive functions of elementary functions and also familiarize with basic techniques to evaluate these. AI: First of all, if you're struggling with derivatives, you've got almost three strikes against you when it comes to knowing how to integrate! So focus first on getting very familiar with derivatives. Then integration follows much more easily. For both taking derivatives and integrating, Paul's Online Math Notes are an excellent source for tutorials, practice problems, and "cheat sheets". I've linked you to the notes involving derivatives. You can review just about everything covered in Calculus I, II, and III, all at the same site, by navigating the tabs and the menu on the left of the webpages. For quick guides ("cheat sheets"), Paul's Notes provides links here to pdf documents that you can download to your own computer for reviewing "off-line." For your convenience, I'll include links to the study sheet on derivatives, and the study sheet on integrals. IMPORTANT: Apart from that, but more importantly, if you want to master taking derivatives of functions, and integration, you'll need to devote yourself to practice, and lots of it. Mathematics is not a spectator sport. To develop competence and mastery, you need to do math, and not just "read about it."
H: Why is every continuously differentiable function with a uniform bounded derivative lipschitz continuous I only know how to prove this for functions on a convex set by using the mean value theorem, but is this also true for this general case when nothing is said about the domain of the function besides the fact that it is a subset of $\mathbb{R}^n$? AI: The function $f(x) = \operatorname{sgn} x$ defined on the domain $D = \mathbb{R} \setminus \{0\}$ is differentiable at every point of $D$ and the derivative is continuous and bounded (it's identically zero), but it is not Lipschitz.
H: Convergent series: am I doing it right? I have $$ \sum_{k=1}^\infty \frac{5k-2}{(3{k}^{2}-2)\sqrt[3]{k+6}} $$ I get (i did not write all solution as it is quite hard to me to put this in LaTeX by myself) $$ \lim_{k\rightarrow \infty} \frac{{a}_{k+1}}{{a}_{k}}=...=\lim_{k\rightarrow \infty}\frac{15{k}^{3\frac{1}{3}}}{30{k}^{3\frac{1}{3}}}=\frac{3}{10} $$ As $$\frac{3}{10}<1$$ series is convergent. Is that right? AI: It is better to use the comparison test. For the comparison test to work we need to be aware of a well known series with which to compare the given series. In this direction we will use the following standard result: The series $$\sum_{k = 1}^{\infty}\dfrac{1}{k^{p}}$$ is convergent if $p > 1$ and divergent if $p \leq 1$. This result itself can be proved using Cauchy-Maclaurin Integral test or Cauchy Condensation test. Now we can informally check that for given series $\sum a_{k}$, $$a_{k} = \frac{5k-2}{(3{k}^{2}-2)\sqrt[3]{k+6}} \approx \dfrac{5k}{3k^{2}\cdot k^{1/3}} = \frac{5}{3k^{4/3}}$$ Let $b_{k} = 1/k^{4/3}$ so that $\sum b_{k}$ is convergent (from the result in italics above with $p = 4/3$). Clearly $$\lim_{k \to \infty}\frac{a_{k}}{b_{k}} = \frac{5}{3} \neq 0$$ (this limit needs to be evaluated properly in an exam, the above informal approximation shows why it should be $5/3$). Hence by comparison test the behavior of both series $\sum a_{k}$ and $\sum b_{k}$ is same with regard to convergence. It follows now that the given series $\sum a_{k}$ is convergent.
H: What is the steps of solution for limit of division of trigonometric functions? Given the $\lim_{x\to \pi} \dfrac{\sin(3x)}{\sin(5x)}$. The answer is $\dfrac 3 5$ but what are the solution steps? AI: Use L'Hopital's rule (L.H.): $$\lim_{x\to \pi} \dfrac{\sin(3x)}{\sin(5x)} \quad \overset{L.H.}{=} \quad \lim_{x \to \pi}\dfrac{\Big(\sin(3x)\Big)'}{\Big(\sin(5x)\Big)'} = \lim_{x\to \pi} \dfrac {3\cos 3x}{5 \cos 5x} = \dfrac {3\cdot 1}{5 \cdot 1} = \dfrac 35$$
H: Integrating Real Function in the Complex Plane Question: Evaluate the integral $$\int_{-\infty}^{\infty}\frac{\sin(x)}{x(x^2+a^2)}=Im\left ( \frac{e^{ix}}{x(x^2+a^2)} \right)$$ Attempt: To evaluate this integral, I use the residue theorem (noting that we have poles $z=0$ and $z=ia$ within our contour): $$2\pi i Res(f(z),z_0)=2\pi i \left (\lim_{z\rightarrow 0}\frac{e^{iz}}{z(z^2+a^2)}+\lim_{z\rightarrow ia}\frac{e^{iz}}{z(z^2+a^2)} \right )$$ $$=2\pi i \left(\frac{i}{a^2}-\frac{ie^{-a}}{2a^2} \right)$$ Therefore, our integral above should be $$\frac{\pi(2-e^{-a})}{a^2}$$ However, in Mathematica, it says the answer should be $$\frac{\pi(1-e^{-a})}{a^2}$$ What did I do incorrectly; i.e., where did a factor of 2 come from? Thank you in advance. AI: noting that we have poles $z=0$ and $z=ia$ within our contour Not quite. The pole in $z = 0$ lies on the contour, it is not enclosed by the contour. The informal short-cut says that the pole lies half in the upper and half in the lower half plane, so it contributes only half its residue to the sum, and $$\begin{align} \int_{-\infty}^\infty \frac{\sin x}{x(x^2+a^2)}\,dx &= \operatorname{Im} 2\pi i\left(\frac12\operatorname{Res}\left(\frac{e^{iz}}{z(z^2+a^2)};0\right) + \operatorname{Res}\left(\frac{e^{iz}}{z(z^2+a^2)};ia\right)\right)\\ &= \operatorname{Im} 2\pi i \left(\frac{1}{2a^2} + \frac{e^{-a}}{(ia)(2ia)}\right)\\ &= \frac{\pi(1-e^{-a})}{a^2}. \end{align}$$ The formal way modifies the contour of integration by using a small semicircle around the pole in $z=0$ to get a closed contour that doesn't pass through any pole of the integrand, and then subtracts the integral over the small semicircle, and lets the radius of that semicircle tend to $0$. In the limit, you add or subtract $\pi i$ times the residue in $0$, depending on whether you chose the semicircle in the upper or lower half plane to avoid the pole in $0$.
H: recurrence relation It was some time ago I studied recurrence relations and I came across this one that I cannot solve: $a_{n+3}=-3a_{n+2}+4a_{n}$ with $a_{0}=2$ and $a_{1}=-5$ Ansatz: $a_{n}=r^{0}$ then I get $r^{3}+3r^{2}-4=0$ with roots $r=1$ and $r=-2$ of double multiplicity so the solution is $a_{n}=A+B(-2)^{n}+Cn(-2)^{n}$ is this correct? But I cannot find constants as I have 2 equations with 3 indeterminants... AI: What you’ve done is correct. There’s no way to pin down $A,B$, and $C$ with the available information: it’s a third-order recurrence, so you need three initial values. The best you can do is solve for $A,B$, and $C$ in terms of the missing value $a_2$.
H: how to prove $\int_{0}^{a}B(t)dt\sim N(0,\frac{a^3}{3})$ Let $B(t)$ is Brownian Motion. I want to prove the integral $\int_{0}^{a}B(t)dt$ has normal distribution , $N(0,\frac{a^3}{3})$. means $\int_{0}^{a}B(t)dt\sim N(0,\frac{a^3}{3})$ AI: This is a normal random variable as a barycenter of normal random variables. Its mean is $\int\limits_0^aE[B_t]\mathrm dt=$ $___$. Its variance is $\int\limits_0^a\int\limits_0^aE[B_tB_s]\mathrm dt\mathrm ds=\int\limits_0^a\int\limits_0^a\min(t,s)\mathrm dt\mathrm ds=$ $___$.
H: Using the ratio Test to see if a series converges or diverges I need to find if $\sum_{n=1}^\infty\dfrac{3^n}{2^n+1}$ converges or diverges. I am trying to use the ratio test, which gives $\dfrac{3(2^n)+1}{2(2^n)+1}$. I am struggling to rearrange this into a form where it is clear what it tends to as n tends to infinity, or is it just an inconclusive test? AI: Sorry but the ratio test seems slightly displaced here. Note that, since $\frac{3^n}{2^n+1}\geqslant1$ for every $n\geqslant1$, $\frac{3^n}{2^n+1}$ does not converge to zero, hence the series $\sum\limits_n\frac{3^n}{2^n+1}$ diverges.
H: Does $\pi \ \| \ 2 \pi$ Does $\pi$ divide $2 \pi?$ Clearly $\frac{2 \pi}{\pi}=2$ and 2 is an integer, so it would seem to make sense to say that $\pi \ \| \ 2 \pi$. Does it make sense to write, for example, $$\pi \ \| \ x \implies \sin(x)=0?$$ AI: This isn't completely formal, as usually divisibility is only defined on the integers. However, if we define divisibility on the reals, or say a ring extension of $\mathbb{Z}$, we can use the generalized definition $$ a\mid b \Longleftrightarrow \exists c\in \mathbb{Z}\text{ such that } b=ac $$ In which case your assertion would be correct, taking $c=2$.
H: Irreducible polynomial $f$ as quotient - effect on $\mathbb{Z}_5[x]$ I want to get a better understanding of quotient rings so I have two questions. Let $f(x) = x^2 + 2$ Let $R = \mathbb{Z}_{5}/(f(x))$ Now as $f$ is irreducible in $\mathbb{Z}_{5}$ we have that $R$ is a field with elements being all polynomials in $\mathbb{Z}_{5}$ with degree less than $2$, i.e. $\{0, 1, 2, 3, 4,$ $x, x + 1, \dots, x + 4,$ $,\dots,$ $4x, 4x + 1, ..., 4x + 4 \}$ First question - But how can, say, the element $x \in R$ be a unit? Also now if we let the quotient be a reducible polynomial, the ring is not supposed to be a field? I.e. Let $g(x) = x^2 + 1$ Let $S = \mathbb{Z}_{5}/(g(x))$ Second question - It seems to me that $S$ will have exactly the same elements as $R$ and hence it will also be a field? AI: In $R$: $x^2 + 2 = 0$, so $(2x)x = 2 x^2 = -4 = 1$, i.e., $x^{-1} = 2x$. For the second question, $R$ does not have the same elements as $S$; $R$ and $S$ only happen to have the same number of elements and they can be represented by the same elements of ${\mathbb Z}_5[x]$, but that's it. Now because $x^2 + 1$ is reducible over ${\mathbb Z}_5$ ($x^2 + 1 = (x + 2)(x - 2)$), the ring $S$ is not an integral domain ($(x + 2)(x-2) = 0$ in $S$) and therefore not a field. It may be worthwhile to stress that elements of $R$ (and of $S$) are not polynomials over ${\mathbb Z}_5$ with degree less than 2, they can merely be represented by those polynomials. The elements are residue classes; the residue class of $h(x) \in {\mathbb{Z}_5[x]}$ is $\{ h(x) + a(x) f(x) \;\mid\; a(x) \in {\mathbb Z}_5[x] \}$ (and letting $h(x)$ range over the polynomials of degree less than 2, you get every residue class exactly once). Writing down these residue classes all the time is not really enlightening, which is why most of the time you just write $h(x)$ (saying that computations are "in $R$" or "modulo $(f(x))$") or maybe $\overline{h(x)}$.
H: Evaluate the limit $\lim_{x\rightarrow 0} \frac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}$ Trying to find $$\lim_{x\rightarrow 0} \dfrac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}=\lim_{x\rightarrow 0} \dfrac{(1-\sin(5x))-(1+\sin(5x))}{(x^2+x)(\sqrt{1-\sin(5x)}+\sqrt{1+\sin(5x)})}=\lim_{x\rightarrow 0} \dfrac{-2\sin(5x)}{(x^2+x)(\sqrt{1-\sin(5x)}+\sqrt{1+\sin(5x)})}$$ How to solve it? AI: $$\lim_{x\to 0}\dfrac{(1-\sin{(5x)})-(1+\sin{(5x)})}{x(x+1)(\sqrt{1-\sin{(5x)}}+\sqrt{1+\sin{(5x)}})}=\lim_{x\to 0}\dfrac{-2\sin{(5x)}}{2x}=-5$$
H: How to find $S$ such that $S = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)3^n}$? How to find $S$ such that $$S = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)3^n}$$ I know I have somehow to manipulate a integrals and derivative but I can't see how. I'd love if anyone can solve me this question with full solution. Thanks in advance! AI: Use the fact that $\sum_{n=0}^{\infty}a^n=\frac{1}{1-a}$ $$f(x)=\sum_{n=0}^{\infty}\left(\frac{-x}{3}\right)^n=\frac{1}{1+\frac{x}{3}}=\frac{3}{3+x}$$ $$g(x)=f(x^2)=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{3^n}$$ Then use the fact that derivative of $\frac{x^{2n+1}}{2n+1}$ is $x^{2n}$ $$h(x)=\int_0^x g(t)dt=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)3^n}$$ Now if we take $x=1$ : $$S=h(1)=\int_0^1\frac{3}{3+t^2}dt$$ Note that derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$.
H: The continuity of the expectation of a continuous stochastic procees Let $X_t$ be a continuous stochastic process on a filtered space $(\Omega, \mathcal F, \mathcal F_t, \mathbb P)$. Is $\mathbb E[X_t]$ necessarily a continuous function? My first answer would be no. For example if $X_t$ admits densities $f(t,x)$, the first equality in: $$\lim_{t \rightarrow t_0} \int_{\mathbb R} x f(t, x) dx=\int_{\mathbb R} \lim_{t \rightarrow t_0} x f(t,x)=\int_{\mathbb R} x f(t_0,x) $$ requires $f$ to be continuous in $t$ uniformly in $x$ to hold. Examples where $\mathbb E[X_t]$ is indeed continuous are abundant. Counterexamples where it is not? Thanks. AI: One counterexample is the following: Let $B_s$ be a standard Brownian motion, and let $S = \inf\{s : B_s = 1\}$ be the first time it hits 1. Since Brownian motion is recurrent, $S < \infty$ almost surely. Let $Y_s = B_{s \wedge S}$; then $Y_s$ is a continuous martingale and $\lim_{s \to \infty} Y_s = 1$ almost surely. Finally, let $$X_t = \begin{cases} Y_{t/(1-t)}, & 0 \le t < 1 \\ 1, & t \ge 1.\end{cases}$$ $X_t$ is continuous since $\lim_{t \to 1^-} X_t = \lim_{s \to +\infty} Y_s = 1$. But for $t < 1$, we have $E X_t = E Y_{t/(1-t)} = 0$ since $Y_s$ is a martingale, and for $t \ge 1$, $E X_t = 1$. So $E X_t$ is discontinuous. $X_t$ is a useful example to keep in mind; for instance, it is a local martingale with respect to its natural filtration, but not a martingale. Edit: Hans asks in comments whether we can have an example where the second moments are finite but discontinuous. The answer is yes. Let $X_t$ be as above and set $Z_t = \sqrt{1-X_t}$. Note that $X_t \le 1$ everywhere so the square root makes sense, and $Z_t$ is a continuous process. Then we have $$E[Z_t^2] = \begin{cases} 1, & t < 1 \\ 0, & t \ge 1 \end{cases}$$ which is discontinuous. The second moments $E[Z_t^2]$ are not only finite but uniformly bounded. Moreover, if we wish to consider variance (centered second moment) instead, notice that by continuity of $Z_t$, as $t \uparrow 1$ we have $Z_t \to Z_1 = 0$ almost surely. Since we showed $E[Z_t^2]$ is uniformly bounded, we have that $\{Z_t\}$ is uniformly integrable (see https://math.stackexchange.com/a/184484/822), hence $E Z_t \to E Z_1 = 0$. So as $t \uparrow 1$, $\operatorname{Var}(Z_t) = E[Z_t^2] - (E Z_t)^2 \to 1$, whereas $\operatorname{Var}(Z_1) = 0$.
H: If $\|X_{n}\| \leq Y$ almost surely, show that $\sup_{n}\|X_{n}\|\leq Y$ almost surely as well. Suppose $\|X_{n}\|\leq Y$ a.s., each $n$, $n=1,2,3,\cdots$. Show that $\sup_{n}\|X_{n}\|\leq Y$ a.s. also. This seems pretty intuitive to me, since if $\|X_{n}\|\leq Y$ a.s., it is bounded above by $Y$, and the sup is the least of these upper bounds. But, somehow I feel like there is more to it than that (especially since this is a starred exercise in my textbook, indicating that it is more challenging than the others?) What exactly is the trick to showing this mathematically rigorously, thereby showing what the problem is actually asking me to show? AI: Let $N_n $ be the set for which $x_n \not \le Y$. Then $\mu N_n = 0$. Let $N = \cup_n N_n$, then we have $\mu N = 0$. Then we have $X_n(\omega) \le Y(\omega) $ for all $\omega \in N^c$, hence $\sup_n X_n(\omega) \le Y(\omega) $ for all $\omega \in N^c$.
H: complex numbers and 2x2 matrices Is it correct that set ${\mathbb C}$ is isomorphic to the set of following 2x2 matrices: $$\left( \begin{array}{cc} a &-b\\ b &a \end{array}\right) $$ $a \in {\mathbb R}$ and $b \in {\mathbb R} $? In other words: are these two sets identical? AI: Yes. Provided that $a^2+b^2 \neq 0$, that means the determinant is non-zero. the map $\varphi: (\mathbb{C}-\{0\},.) \to (M, \times)$ given by $(a+bi) \mapsto \left( \begin{array}{cc} a &-b\\ b &a \end{array}\right)$ is an isomorphism. You can check it easily ($M$ is the set of all such matrices and $\times$ is matrix multiplication)
H: Probability Revision An urn contains $2$ black balls and $3$ white balls. Two balls are drawn at random from the urn. Find the probability that both balls are black. My approach towards this problem:- No. of black balls = 2 No. of white balls = 3 Total no. of balls = 5 2 balls are drawn at random P(E) = ($2$C2)/$5$C2 = $1/10$ is this the right answer? If not, please help. My textbook says the correct answer is $3/10$. I can't figure it how.. AI: Your textbook's answer appears to be wrong.
H: permuting digits of a number Given a number $N$ with upto 18 digits, We need to find how many numbers smaller than $N$ can be formed using the same digits. Eg: for 725 we can form 527,572,257,275 hence answer=4 AI: It's not going to have a nice formula, but you can formulate a recursive algorithm to compute it. The basic idea is that the numbers of length $n$ are all those formed from: a) a smaller digit of $N$ that the lead digit and any permutation of the rest b) the same lead digit as $N$ and a permutation of the rest that is smaller than the number formed after removing the first digit of $N$ a) yields to the calculation of permutations, accounting for the multiplicity of each digit. You have to decide whether a leading zero is permitted. It clearly is after the first step b) is a recursive call, with a number one digit shorter.
H: Tempered fundamental solutions According to the Malgrange–Ehrenpreis theorem every nontrivial linear constant coefficient PDO $P(\partial)$ admits a fundamental solution $E\in\mathscr{D}'$; I wonder whether $P(\partial)$ admits a tempered fundamental solution, namely a $E\in\mathscr{S}'$ with $P(\partial) E=\delta_0$. AI: Indeed, this has been demonstrated by L. Hörmander in 1958 ('On the division of distributions by polynomials', Arkiv för Matematik Band 3, nr 53; Theorem 5). In fact he showed the division problem tempered distributions; and proved that if $P$ is a polynomial that does not vanish identically, the multiplication mapping $\mathscr{S}\rightarrow\mathscr{S}$ defined by $f\mapsto Pf$ admits a continuous inverse. From this result the rest easily follows.
H: Probability of baby be right handed I'm trying to resolv this problem but I think i'm not getting the correct answer. The anwser says that there is 50% probability that the baby is right handed if the test accuracy is 90%, and 69% probability the baby is right handed if the accuracy of the test is 80%. The problem is : A doctor says that an infant who predominantly turns her head to the right while lying on her back will be right-handed and one who predominantly turns her head to the left will be left-handed. Given that 90% of the population is right-handed, what is my daughter Clara's probability of being right-handed, given that she predominantly turns her head to the left assuming that the test is 90% accurate? What if the test is 80% accurate? What I have done is the normalisation using this formula : P(M\|s) = P(s\|m)P(m) / (P(s\|m)P(m) + P(s\|not m)P(not m)) And my variables are : P(right-handed) = 0.9 P(not right-handed) = 0.1 P(test \| right-handed) = 0.9 P(not test \| not right-handed) = 0.9 My equation is : P(not right-handed \| test) = P(test \| not right-handed) * P(not right-handed) / P(test \| not right-handed) * P(not right-handed) + P(test \| right-handed) * P(right-handed) What I am doing wrong? AI: Simple problems like this are sometimes easier to work out in tabular form. If the test is $80$% accurate, you have the following distribution of percentages: $$\begin{array}{r\|cc\|c} &\text{accurate}&\text{not accurate}&\text{total}\\ \hline \text{right-handed}&72&\color{brown}{18}&90\\\ \text{left-handed}&\color{brown}8&2&10\\ \hline \text{total}&80&20&100 \end{array}$$ The two brown figures represent the percentage of the population that as infants predominantly turn their heads to the left: they’re the right-handers for whom the test is not accurate and the left-handers for whom it is accurate. Within this group right-handers are $$\frac{18}{18+8}=\frac{18}{26}\approx0.6923$$ of the total, or about $69$%. You can also use this table and calculation to check your own, by identifying the component probabilities in yours with entries in the table.
H: Help with intro to number theory question Let $a$, $b$, $m$ and $n$ be integers with $m > 0$ and $n > 0$. If $(n,m)\mid(a−b)$, then the system $$\begin{cases} x\equiv a\pmod m \\ x\equiv b\pmod m \end{cases} $$ has a solution. Describe how to find a solution. Can anyone explain how to solve this for me? I have been looking at it, but I do not know where to begin or what to look for when solving it. AI: The solutions of $x\equiv a\pmod n$ are $x=a+kn$ for any $k\in\Bbb Z$. Similarly, that of $x\equiv b\pmod m$ are $x=b+lm$ with $l\in\Bbb Z$. By Bezout's identity, we have $u_0,v_0\in\Bbb Z$ such that $(n,m)=u_0n+v_0m$. By the hipothesis, this implies that $$b-a=un+vm$$ for $u=\frac{b-a}{(n,m)}\cdot u_0,\ v=\frac{b-a}{(n,m)}\cdot v_0$. But then consider $$x:=a+un=b-vm\,.$$
H: $\gcd$ of polynomials over a field I have the polynomials $f,g\neq 0 $ over a field $F$. We know also that $\gcd(f,g)=1$ and $$ \det \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}\neq 0. $$ I need to prove that $\gcd(af+bg,cf+dg) = 1 $ for every $a,b,c,d \in F$. I really do not know how to start answer the question. Thanks for helpers! AI: Just work through the equations of gcd (bearing in mind that you're working in $f(x)$, hence constants do not matter): $ \gcd(af+bg, cf+dg) = \gcd(adf+bdg, cbf + bdg) = \gcd( (ad-bc)f, cbf + bdg) = \gcd( f, cbf+bdg) = \gcd(f, bdg) = \gcd(f, g) = 1$
H: Boolean Algebra - Why is the result 1? Given: = !(A * (!B + C)) + !(!B * !C) = !A + (B * !C) + !B + C Where: ! = NOT + = OR * = AND I'm having some trouble to why !A + (B * !C) + !B + C simplifies to 1? Can someone shed some light on this please? It would be very much appreciated. AI: Use the distributive property to obtain an "or clause" consisting of $(B\; + \;!B + \cdots = 1)(!C\;+ \;C + \cdots = 1)$ $$\begin{align} \lnot A + (B \lnot C) + (\lnot B + C) & = \lnot A + (B + \lnot B + C)(\lnot C + \lnot B + C)\\ \\ & = \lnot A + (1 + C)(1+ \lnot B) \\ \\ &= \lnot A + 1*1 \\ \\ & = \lnot A + 1 = 1\end{align}$$ That is, $B \lor \lnot B$ is always true, as is $C \lor \lnot C$, and since the clause in which they occur one clause in a disjunction of clauses, the entire statement is there by true, since $P + (1)(1) = P + 1 = 1$, where $P$ can be any statements whatsoever.
H: Algebraic structures with one binary operation I need your help.I am studying about algebraic structures.I have a question that I can't solve .. Let $G=\langle A,*\rangle$ be a monoid with the identity $e$. Let $a\in A$. İs it possible that $a$ has two different left-inverses, $L_1$ and $L_2$? Thank you very much for your answers. AI: Hint: For an example of such a thing consider the monoid of set maps $\mathbb{N \to N}$ with composition as the operation. A function has a left-inverse if and only if it's injective, so pick an injective map (for example $x \mapsto 2x$) and try and find two different left-inverses.
H: Cardinalities of Sylow subgroups I must be missing something very simple here. Problem: Let $\|G\|=56$. Let H be a Sylow 7-subgroup of G, and suppose H is not a normal subgroup of G. What are the cardinalities of the Sylow 7-subgroups of G and the Sylow 2-subgroups of G? My attempt: The cardinalities of the Sylow 7-subgroups in G can be either 56 or 8, and the cardinalities of the Sylow 2-subgroups in G can be either 28, 14, or 7. Is this correct? Thanks. AI: The cardinality of a Sylow $p$-subgroup is always a power of $p$. In particular, it is the largest power of $p$ that divides the order of $G$. Based on your guesses, you might be thinking of the number of Sylow $p$-subgroups. In fact, the question sounds like it's asking for the number of Sylow $p$-subgroups and not their cardinalities. The fact that $H$ is not normal in $G$ has no bearing on the cardinality of any Sylow subgroup.
H: Elementary number theory problem for homological algebra Hello to everybody: I'd like to know if the following statement is true or not, since if it's false it will help me solving a problem for exact sequences of modules. $``$Given $(a,b,m) \in \mathbb{Z}^3$ if $[a,m]= [b,m]$ and $a \equiv b (\textrm{mod} \ m)$ then $a\in \{b,-b\}"$. Where I denote by $[a,m]$ the least common multiple of $a,m$. I tried to find a counterexample but I didn't succeed. Thanks in advance! AI: For $m = 0$, we have $\lvert a\rvert = [a,m] = [b,m] = \lvert b\rvert$, if the least common multiple is defined to be non-negative, so it's okay. If $m \neq 0$, then we have $$\frac{\lvert am\rvert}{\gcd(a,m)} = [a,m] = [b,m] = \frac{\lvert bm\rvert}{\gcd(b,m)},$$ and $a \equiv b \pmod{m}$ ensures $\gcd(a,m) = \gcd(b,m)$, hence $\lvert a \rvert = \lvert b\rvert$ follows.
H: Convolution convergent in $L^p$ Suppose $f\in L^p(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_t(x)=\dfrac{1}{t}K\left(\dfrac{x}{t}\right)$$ I'm trying to prove that $\lim_{t\rightarrow 0}\\|f\ast K_t-f\\|_p=0$. I choose a compactly supported function $g\in C^\infty$ such that $\\|f-g\\|_p<\epsilon$ (possible because this class is dense in $L^p(\mathbb{R})$). Then I want to bound $$\\|f\ast K_t-f\\|_p\leq \\|f\ast K_t-g\ast K_t\\|_p+\\|g\ast K_t-g\\|_p+\\|g-f\\|_p$$ We have of course $\\|g-f\\|_p<\epsilon$. We have $\\|(f-g)\ast K_t\\|_p\leq \\|f-g\\|_p\\|K_t\\|_1<\epsilon\\|K_t\\|_1$. But how can we bound the term $\\|K_t\\|_1$? AI: Okay, I found that $\\|K_t\\|_1=\\|K\\|_1$ for all $t$, so that gives the necessary bound.
H: Prove that a Language is Non-Regular Using Closure Properties Use the closure properties of regular languages and a language $B$ known to be non-regular to prove that a language $A$ is not regular. My understanding is that the closure properties only apply when both languages are regular. So, I'm not sure what such a proof would look like and I'm looking for an outline of what the proof would look like. AI: The class of regular languages is closed under intersection. Suppose that you can find a regular language $L$ such that $B=L\cap A$; if $A$ were regular, then $B$, being the intersection of two regular languages, would also be regular. However, we know that $B$ is not regular, so $A$ cannot be regular. Other closure properties can be used in the same way.
H: Is it possible to formulate any line or curve in 2d space ?! I wonder how we can explain any kind of lines or curves as a formula! more clear,can we say that "any kind of lines or curves have a formula but we cannot find their formula"?! AI: It makes no sense just to say if you have a line or curve, can you find a function or relation to represent it. Most people would take a line or curve to mean you already have a function or relation you are using to describe them. If you mean if you were to draw some kind of 'curve' or sketch on a piece of graph paper, could you find a function or relation to represent it. Then the answer is a definite maybe. Let me explain. The naive answer is that you could merely write down a rule $f(x)=y$ for every point $(x,y)$ on the graph. But in practice you'd have to write an uncountable infinite amount of points--impractical. So on to the more practical answer: Most shapes you could draw on a sheet of paper could be easily put into the form of $f(x)=$ something or $f(y)=$ something or maybe something in a parametric form $x(t)=$ something $y(t)=$ something. Examples would be lines, polynomial-like curves, trig-like curves, circles, et cetera. Or you could use equations similar to how we define elliptic curves. For functions which are 'smooth', we can approximate them using polynomials (this is Taylor's Theorem). Then we have uglier functions like $$ f(x)= \begin{cases} x, \text{ if } x \in \mathbb{Q} \\ 0, \text{ if }x \in \mathbb{Q}^C \end{cases} $$ Then again, one might say these aren't included in your answer as no human could draw it. In any case, even for this nasty, impossible to graph function, we have a function above to describe it. Think of the graph as a set of numbers, then using the same type of constructions used in constructing Lebesgue measurable functions, we should be able to find some type of approximation that will fit the function just fine (though it won't be pretty) in the sense it converges to the graph you would have drawn. However, this just covers 'most' things you would draw in the plane. There are an infinite amount of things that one could draw. So I certainly don't know if there exists things you could at least find a relation for. If there were to be such things, they would be very odd indeed and I doubt they would fit your criterion of being able to be drawn by a human. In any case, even the oddest functions you could imagine can be given piecewise relations that construct them. A absolutely stunning example of this was given using Mathematica on the Mathematica blog. I suggest you check these out as they are probably the perfect example of what you were looking for: Making Formulas for Everything — From Pi to the Pink Panther to Sir Isaac Newton Even More Formulas for Everything — From Filled Algebraic Curves to the Twitter Bird, the American Flag, Chocolate Easter Bunnies, and the Superman Solid
H: Prove the following combination? I need a quick proof of the following combination $$\binom{n+1}1+\binom{n+1}2+\binom{n+1}3+\dots++\binom{n+1}{n+1}=2^{n+1}-1$$ AI: Daniel Fischer has already given you a very nice answer. However, it might be nice to see an alternative more combinatorial approach. Imagine you are trying to compute the ways you can form a subset from $n+1$ objects, without that subset being empty. It should be the ways you can choose $1$ object from among those $n+1$ objects, plus the ways you can choose $2$ objects from among those $n+1$ objects, and so on up until you are choosing all $n+1$ objects to form your subset. We can compute this sum (that is, the number of subsets we can form) by imagining a subset as a bit string of length $n+1$, where each entry is $0$ if the object is not included in the subset, and each entry is a $1$ if that object IS included in the subset. There are clearly $2^{n+1}$ such bit strings, but we must subtract $1$, as we cannot include the bit string where all entries are $0$, i.e. the empty set. Hence, we arrive exactly at your identity.
H: Two general questions regarding intervals in $\mathbf{R}$ Does an open interval contain infinitely many closed intervals? Is the interval $(-1,1)$ equal to $[-a,a]$ where $a<1$? Why? AI: HINT for the second question: If $a<1$, then $a<\frac12(a+1)<1$. Added: For the first question, let $(a,b)$ be a non-empty closed interval. Pick any $x\in(a,b)$. Consider the intervals $[c,x]$ for $c\le x<b$.
H: Taylor/Maclaurin Series Exam Question. Show that if x is small compared with unity, then $$f(x)=\frac{(1-x)^\frac{-2}{3}+(1-4x)^\frac{-1}{3}}{(1-3x)^\frac{-1}{3}+(1-4x)^\frac{-1}{4}}=1-\frac{7x^2}{36}.$$ I've expanded all the brackets of f(x) up to the 2nd order and I'v ended up with $$f(x)=\frac{2+2x+\frac{37x^2}{9}}{2+2x+\frac{9x^2}{2}}$$ I can't seem to be able to simplify this to get $1-\frac{7x^2}{36}$. Can someone tell me what to do from here? I'm certain that the question does not require me to find $f'(o)$ and $f''(o)$ as this would be to time-consuming and this is an exam question. Thank you. AI: You can convert $(37/9)x^2$ to $(74/18)x^2$ and $(9/2)x^2$ to $(81/18)x^2$. Using this you can get: $$ f(x) = \frac{2 + 2x + \frac{81x^2}{18} - \frac{7x^2}{18}}{2 + 2x + \frac{81x^2}{18}}$$ This simplifies to $$ f(x) = 1 - \frac{\frac{7x^2}{18}}{2 + 2x + \frac{81x^2}{18}}$$ For small $x$ or $x$ close to zero, you can simplify the denominator to $2$ because $2x + \frac{81x^2}{18} \approx 0 $. This gives you $$f(x) = 1 - \frac{7x^2}{36}$$
H: Set logic to propositional logic How would you convert set logic to propositional logic? In particular, I'm not sure how to handle converting $\subseteq$ For example: $$A-(\bar{B} \cup \bar{C}) \subseteq B \cap C$$ My attempt at converting to propositional logic: $$ x \in A \land (x \in B \lor x \in C) \implies x \in B \land x \in B$$ Is this the right way? AI: Yes, you're correct that $\subseteq$ corresponds to the conditional connective $\rightarrow$. But you've made an error: The antecedent in your translation corresponds to $A - (B \cup C)$, whereas we need to translate $A - (\overline B \cup \overline C)$. So we need DeMorgan's: In particular, note how the side to the left of the conditional is translated: $$A-(\overline{B} \cup \overline{C}) \subseteq B \cap C$$ $$x\in A \land \lnot [x \in (\overline B \cup \overline C)] \rightarrow x \in B \land x \in C$$ $$\iff A \land \lnot[\lnot (x \in B\cap C)] \rightarrow x \in B \land x \in C\tag{DeMorgan's}$$ $$\iff x \in A \land (x\in B \land x \in C) \rightarrow x \in B \land x\in C$$
H: Sum of the series $\sum_{n \ge 0}{\frac{x^{4n+1}}{(4n+1)!}}$ I want to determine the sum of the series $$\sum_{n \ge 0}{\frac{x^{4n+1}}{(4n+1)!}}$$ I know this has to do with the sum $$\sum_{n \ge 0}{\frac{x^{n}}{(n)!}}=e^x\;\; \forall x\in \mathbb R$$ But i can't see how to start. Thank you for your help!! AI: Recall that $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$ and $$\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots.$$ Add, divide by $2$. If the $\sinh$ function is unfamiliar, replace it by the equivalent $\frac{e^x-e^{-x}}{2}$. Use the power series for $e^t$ to find the power series for $\frac{e^{x}-e^{-x}}{2}$.
H: Bounding for convolution convergence Suppose $f\in L^p(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_t(x)=\dfrac{1}{t}K\left(\dfrac{x}{t}\right)$$ I'm trying to prove that $\lim_{t\rightarrow 0}\\|f\ast K_t-f\\|_p=0$. I choose a compactly supported function $g\in C^\infty$ such that $\\|f-g\\|_p<\epsilon$ (possible because this class is dense in $L^p(\mathbb{R})$). Then I want to bound $$\\|f\ast K_t-f\\|_p\leq \\|f\ast K_t-g\ast K_t\\|_p+\\|g\ast K_t-g\\|_p+\\|g-f\\|_p$$ We have of course $\\|g-f\\|_p<\epsilon$. We have $\\|(f-g)\ast K_t\\|_p\leq \\|f-g\\|_p\\|K_t\\|_1<\epsilon\\|K_t\\|_1=\epsilon\\|K\\|_1$. Now I need to bound $\\|g\ast K_t-g\\|_p$. I have $$\left\|(g\ast K_t)(x)-g(x)\right\|_p=\left\|\int_\mathbb{R}(g(x-y)-g(x))K_t(y)dy\right\|$$ Since $g$ is continuous and compactly supported, it is uniformly continuous. But what can I do with the $K_t(y)$? AI: In the last displayed integral, use the substitution $s:=y/t$. Then $$\lVert g\star K_t-g\rVert_p^p=\int_{\mathbb R}\left\|\int_{\mathbb R}(g(x-ts)-g(x))K(s)\mathrm ds\right\|^p\mathrm dx.$$ I assume $K\geqslant 0$ (not restrictive). Then by Jensen's inequality, it's enough to prove that $$\lim_{t\to 0}\iint_{\mathbb R^2}\|g(x-ts)-g(x)\|^pK(s)\mathrm ds\mathrm dx=0.$$ This can be done by an approximation argument: take $\varepsilon\gt 0$: then there is $R$ such that $\int_{\mathbb R\setminus [-R,R]}K(s)\mathrm ds\lt \varepsilon$. Then we have $$\int_{\mathbb R}\int_{\mathbb R\setminus [-R,R]}\|g(x-ts)-g(x)\|^pK(s)\mathrm ds\mathrm dx\leqslant 2\varepsilon\lVert g\rVert_p.$$ We have $$\int_{\mathbb R}\int_{[-R,R]}\|g(x-ts)-g(x)\|^pK(s)\mathrm ds\mathrm dx\leqslant 2R\sup\|g'\|t^p\lVert K\rVert_1.$$
H: power series help for real this time I'm given $$f(x)= \frac{1}{x+2} $$ I know i have to make it look like: $$f(x) = \frac{1}{1-x} $$ but I have no clue how to go about that. Any suggestions? or maybe I could divide by 2 throughout? AI: Hints: $$\frac1{2+x}=\frac12\frac1{1+\frac x2}\;,\;\;\left\|\frac x2\right\|<1\iff\ldots$$
H: Prove $(3x^2+3) \geq (x+1)^2+1$ $(3x^2+3) \geq (x+1)^2+1$ I tried using a direct proof but I think I got stumped along the way. $3x^2+3 \geq x^2+2x+2$ $2x^2+1 \geq 2x$ $2(x^2) +1 \geq 2x$ $x^2 + (1/2) \geq x$ How can I make this appear more clear? I don't think this is evident that it is true. AI: Try completing the square: \begin{align} x^2 + 1/2 \geq x &\iff x^2 - x + 1/2 \geq 0 \\ &\iff (x^2 - x + 1/4) - 1/4 + 1/2 \geq 0 \\ &\iff (x - 1/2)^2 + 1/4 \geq 0 \\ \end{align} which is always true for any $x \in \mathbb R$. So the original inequality must also always be true for all $x \in \mathbb R$.
H: $\dim(\mbox{im}(f)) = \dim(U)$ and $\dim(\ker(g)) = \dim(V)-\dim(U)$ Let $U, V, W$ be vector-spaces over the same field $K$ with linear transformations $$f:U\to V , g:V\to W $$ so that $$g\circ f$$ is an Isomorphism. Show that $$\dim(\mbox{im}(f)) = \dim(U)$$ and $$\dim(\ker(g)) = \dim(V)-\dim(U)$$$$$$My idea is to say that because of the Isomorphism between the linear transformations the vector spaces equal each other because of the bijection. Is this sufficient enough (or even right)? AI: Hint: since $g\circ f$ is an isomorphism, you can prove that $f$ is injective and that $g$ is surjective. Apply rank-nullity. Why is $f$ injective? Suppose $f(u_1)=f(u_2)$; then $g(f(u_1))=g(f(u_2))$ or $g\circ f(u_1)=g\circ f(u_2)$. By hypothesis $g\circ f$ is injective, so … Why is $g$ surjective? Suppose $w\in W$; then, since $g\circ f$ is surjective by hypothesis, $w=g\circ f(u)$ for some $u\in U$. Can you find $v\in V$ such that $w=g(v)$?
H: Normal spaces in box and uniform topology Is $\mathbb{R^\omega}$ normal in product topology? In the uniform topology? and In box topology? My attempt:I know it is normal in uniform topology because it is metrizable. I would gues that it is normal in product topology. But do not know how to go about proving it. AI: It’s a standard theorem that the ordinary product of at most countably many metric spaces is metrizable, so $\Bbb R^\omega$ is normal in both the uniform and the usual product topologies. The continuum hypothesis implies that it is paracompact and therefore normal in the box topology; I’m not sure whether there are known models of set theory in which it is definitely not normal in the box topology.
H: What are the angle brackets in Linear Algebra? In my linear algebra book, they have angle brackets around two different vectors, so it looks like this: $\langle\mathbf{u_2},\mathbf{v}_1\rangle$. They don't use angle brackets to define vectors, but use regular parenthesis instead. For the Gram-Schmidt process, they define $\mathbf{v}_1 = \mathbf{u}_1 = (1,1,1)$ and $\mathbf{v}_2 = \mathbf{u}_2 = \mathbf{u}_2 - \dfrac{\langle\mathbf{u}_2, \mathbf{v}_1\rangle}{\\|\mathbf{v}_1\\|^2} \mathbf{v}_1$ where $\mathbf{u}_2 = (0,1,1)$ They conclude that that formula is equal to $(0,1,1) - \dfrac{2}{3}(1,1,1)$. What operation is the angle brackets to have that result? AI: The angled brackets represent an inner product. The best known one is the scalar product or the dot product. If ${\bf u} = (u_1,u_2,u_3)$ and ${\bf v} = (v_1,v_2,v_3)$, then the dot product is given by $$\langle {\bf u},{\bf v} \rangle = u_1v_1 + u_2v_2+u_3v_3$$ It has many useful properties. First $\langle {\bf u},{\bf u} \rangle = \\|{\bf u}\\|^2$, and second if ${\bf u}$ and ${\bf v}$ are both non-zero then $\langle {\bf u},{\bf v}\rangle = 0$ if and only if ${\bf u}$ and ${\bf v}$ are orthogonal. In general: $$\langle {\bf u}, {\bf v} \rangle = \\|{\bf u}\\| \\|{\bf v}\\| \cos\theta$$ where $\theta$ is the angle between ${\bf u}$ and ${\bf v}$. This idea can be generalised. Notice that $\langle {\bf u},{\bf v} \rangle = {\bf u}E{\bf v}^{\top}$, where $E$ is the 3-by-3 identity matrix and ${\bf u}$ and ${\bf v}$ are being thought of as 1-by-3 matrices. For any 3-by-3 matrix, say $M$, we can define $\langle {\bf u},{\bf v}\rangle_M := {\bf u}M{\bf v}^{\top}$. Different matrices gives rise to different $\langle {\bf u},{\bf v}\rangle_M$. We usually assume that $M$ is a positive definite matrix.
H: Tiling a $23 \times 23$ square by $1 \times 1$, $2 \times 2$, and $3 \times 3$ tiles A $23 \times 23$ square is tiled by $1 \times 1$, $2 \times 2$, and $3 \times 3$ tiles. Prove that at least one 1 x 1 tile must be used. Find such a tiling with exactly one $1 \times 1$ tile. Hint: put a number in each $1 \times 1$ square of the big square so that $2 \times 2$ and $3 \times 3$ tiles cover a total divisible by $3$. I'm not sure how to utilize the hint in solving this problem. AI: Further hint: $$ \matrix{1 & 0 & 1 & 0 & \ldots\\ 0 & 2 & 0 & 2 & \ldots\\ 1 & 0 & 1 & 0 & \ldots\\ 0 & 2 & 0 & 2 & \ldots\\ \ldots & \ldots & \ldots & \ldots & \ldots &\cr}$$ What is the total in the $23 \times 23$ square?
H: A Noetherian module annihilated by a power of maximal ideal must has finite length. Let $M$ be a Noetherian $R$-module and $P^kM=0$ from some maximal ideal $P$ of $R$ and some integer $k$. How to show that $M$ has finite length? The length of a module is defined to be the maximum length of the chain of submodule: $$ 0=M_0<M_1<\cdots<M_{n-1}<M_n=M $$ I have tried following. We can assume there is no strict submodule between $M_i$ and $M_{i+1}$, and try to prove such $n$ is bounded. Then we have $$ M_i/M_{i+1}\cong R/Q_i $$ where $Q_i=\operatorname{ann}_R(M_i/M_{i+1})$ is maximal. Then I cannot move on. I tried to look at the localization of the chain at $P$, but it seems to provide nothing. Could anyone help? AI: WLOG, choose $k$ to be the smallest possible. Consider the sequence of submodules $$()\qquad0=MP^{k} \subsetneq MP^{k-1} \subsetneq \dots \subsetneq MP \subsetneq M \,.$$ The goal is to show that every consequtive factor of this sequence is of finite length. For arbitrary $l \in \{0,1, \dots, k-1\},$ consider the factor $MP^l/MP^{l+1}$. Since $M$ is noetherian, this clearly is a finitely generated module. The annihilator $\mathrm{Ann}_R(MP^l/MP^{l+1})$ clearly contains the maximal ideal $P$. On the other hand, from the strictness of the inclusion $MP^{l+1} \subsetneq MP^{l}$ it follows that $1 \notin \mathrm{Ann}_R(MP^l/MP^{l+1})$, hence (since $P$ is maximal) $\mathrm{Ann}_R(MP^l/MP^{l+1})=P,$ a maximal ideal. Now, since any module $N$ can be considered as $R/\mathrm{Ann}_R(N)$-module (with the multiplication defined by $n \cdot (r+\mathrm{Ann}_R(N)):=nr$ and the important property that the lattice of submodules does not change by this shift of perspective), we can see that $MP^l/MP^{l+1}$ is actually finitely-generated $R/P$-module, i.e. a vector space of finite dimension. Hence, it is of finite length. Adding more details: It is a well-known fact that a module $M$ is of finite length iff it has a finite composition series, i.e. a finite chain of submodules from $0$ to $M$ with the consecutive factors simple. Now, we have shown (for arbitrary $l$) that $MP^l/MP^{l+1}$ are of finite length, hence there exist a composition series $$0=MP^{l+1}/MP^{l+1}=N_0^{l} \subseteq N_{1}^{l} \subseteq \dots \subseteq N_{k_l}^{l}=MP^{l}/MP^{l+1}$$ with simple consecutive factors. After applying the correspondence theorem, we obtain a chain of submodules of $M$ $$MP^{l+1}=\overline{N_0^{l}}\subseteq \overline{N_1^{l}} \subseteq \dots \subseteq \overline{N_{k_l}^{l}}=MP^{l}$$ again with simple consecutive factors. Doing this for every $l$, we obtain a refinement of the series $()$ with simple consecutive factors of finite length, i.e. a composition series of module $M$ of finite length.
H: Convergence of formal power series substitution Prove that the substitution of formal power series $F(G(x))=\sum_{k\geq0}f_k \frac{G(x)^k}{n!}$ converges for every $F$ if and only if $G(0)=0$ AI: Let's call the degree (the $x$-adic valuation really) of a formal power series the lowest exponent appearing in a non-zero term. Assume first that $G(0)=0$. That means that $\deg G(x)\ge1$. Therefore $\deg G(x)^k\ge k$ for all $k$. So if we look at terms of degree $\ell$ for some fixed natural number $\ell$, only the terms of $f_k G(x)^k/k!$ with $k\le \ell$ will contribute. But these are only finitely many formal power series, so there is no problem. Thus the sum converges. On the other hand, if $G(0)\neq0$, then each and every one of the series $G(x)^k/k!$ has a non-zero constant term. So if $F(x)$ has infinitely many terms, then there are infinitely many non-zero constant terms in the sum, and thus the sum won't converge in the ring of formal power series.
H: Derivative of $e^{-x} - xe^{-x}$ Find $f'(x)$: $$f(x) = e^{-x} - xe^{-x}$$ $$f'(x) = -e^{-x}-(x)'(e^{-x})+(x)(e^{-x})'$$ $$f'(x) = -e^{-x}-e^{-x}-xe^{-x}$$ $$f'(x) = -2e^{-x}-xe^{-x}$$ $$f'(x) = -e^{-x}(2+x)$$ However the answer states: $$f'(x) = e^{-x}(x-2)$$ I am not sure where I made a mistake. AI: $$f'(x) = -e^{-x}-(x)'(e^{-x})+(x)(e^{-x})'$$ Should be $$f'(x) = -e^{-x}-[(x)'(e^{-x})+(x)(e^{-x})']$$ $$f'(x) = -e^{-x}-(x)'(e^{-x})-(x)(e^{-x})'$$
H: Number invariant problem: replacing any two numbers $a$ and $b$ with $a - 1$ and $b + 3$ Numbers 1, 2, 3, ..., 2014 are written on a blackboard. Every now and then somebody picks two numbers $a$ and $b$ and replaces them by $a - 1$, $b + 3$. Is it possible that at some point all numbers on the blackboard are even? Can they all be odd? I cannot figure out how to use the terms $a - 1$ and $b + 3$ to form a invariant. AI: It is not possible, since with every replacing, the number of odd numbers either stays the same, increases by 2 or decreases by 2. Since at the start there is an odd number of odd numbers, it will always stay odd, especially it will never be 2014. The same argument applies for the number of even numbers.
H: When should I use the Central Limit Theorem? I am facing the following question: Assume you have invited a $100$ people to a party. The probability that one would decide to come to the party is $0.75$. What is the probability that more 70 but not more than 80 people will decide to come to your party? I can tell that if $X$ is the number of people that have decided to come to the party, I can easily say that $X \sim Bin(100, 0.75)$. I can use that to tell what is $P(70 < k=x <80)$ Why would I want to use the Central Limit Theorem and move to $N \dot{\sim}(100\cdot0.75, 100\cdot0.25)$? AI: If you have a good calculator that will be able to handle binomial coefficients with $n= 100$, it's feasible, if a bit tedious, to compute and add $10$ probabilities. But what if you invited $10$ million people and wanted to know the probability that between $7,490,000$ and $7,500,000$ would come? (I guess if you can afford to throw that big a party, you can afford to hire someone to compute that for you)

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