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H: Show that $\sum_{cyc} J(x,J(y,z))=0$. Let $x,y,z$ be functions of $(u,v)$ and $J$ be the Jacobian matrix. Show that $\sum_{cyc} J(x,J(y,z))=0$. I expanded the thing and realized that the first term in the sum is $x_u(J(y,z_v)+J(y_v,z))-x_v (J(y,z_u)+J(y_u,z))$, but I don't know how to carry on then. Furthermore, I suspect that this has something to do with the jacobi identity, but I don't know are they related. Thanks. AI: I do not fully understand notation (what is the Jacoby matrix of 2 functions?). In any case, if $$J(x,y):=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v} $$ and $x=x(u,v)$, $y=y(u,v)$, then the statement is the Jacobi identity for the Poisson bracket of functions on $\mathbb R^2$, i.e. $$J(x,y):=\{x,y\}. $$
H: Given three vectors, how to find an orthonormal basis closest to them? I know Gram-Schmidt process but that is not what I am looking for. Given three vectors in $\mathbb{R}^3$, $\{v_1,v_2,v_3\}$, I want to find three vectors $\{w_1,w_2,w_3\} \subset \mathbb{R}^3$ such that $$(w_i,w_j) = \delta_{ij}, \quad \forall i,j\in\{1,2,3\},$$ $$D=\sum_{j=1}^3 \\|v_j-w_j\\|^2\ \mbox{is minimal.}$$ The drawback of the GS process is that it assumes a preferred order in the 3-tuple and only changes the length of the first vector, resulting in the change being done to the second and third too large and the distance $D$ suboptimal. I want to deform all three in a "fair" manner. AI: If the norm is the Euclidean $2$-norm, the problem has been studied to death. Put it simply, since $\\|x\\|^2=x^Tx=\operatorname{trace}(xx^T)$, you can rewrite $D$ as a constant plus $-2\operatorname{trace}(VW^T)$. By performing singular value decomposition on $V$, the answer is trivial (but a little care has to be taken if $W$ is not just an orthogonal matrix but a rotation matrix).
H: What can we say about $\dim \operatorname{null}(AB)$ from knowing $p_A$ and $p_B$? Say, there are two matrices $A, B \in \mathbb R^{3,3} $ such that their characteristic polynomials are $p_A(t) = t^3 − t^2 + 2t$ and $p_B(t) = t^3 − 7t^2 + 9t − 3$. What do we know about $\dim \operatorname{null}(AB)$? Clearly, $t=0$ is one of the roots of $p_A$ so $A$ is singular, and therefore $AB$ is singular too. So $\dim \operatorname{null}(AB) \geq 1$. Is there anything else we can conclude? AI: There is a little more we know. $p_B(0) \neq 0$, so $B$ is bijective, and hence $\dim \operatorname{null}(AB) = \dim \operatorname{null} (A)$. Further, $p_A(t) = t(t^2-t+2)$ has three distinct roots (one real, two conjugate complex), so $\dim \operatorname{null} (A) = 1$.
H: proving $\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$ when f is continous on [0,1] $$\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$$ when f is continuous on $[0,1]$ I know it can be proved using bounded convergence theorem but, I wanna know proof using only basic properties of riemann integral and fundamental theorem of calculus and MVT for integrals ... Thank you. AI: take any $\epsilon$, choose $\delta > 0$ such that $\|f(x) - f(0)\| < \epsilon$ on $[0,\delta]$. choose $n$ big enough such that $(1-\epsilon)^n < \delta$ then $$ \|\int_0^1 f(x^n) dx - f(0) \|= \|\int_0^{(1-\epsilon)} [f(x^n) - f(0)]\|dx + \int_{1-\epsilon}^1 [f(x^n) - f(0) ]dx \| \leq $$ $$ \int_0^{(1-\epsilon)} \|f(x^n) - f(0)\|dx + \int_{1-\epsilon}^1 \|f(x^n) - f(0) \|dx $$ first factor is smaller than $\epsilon(1 - \epsilon)$ thanks to the choice of $\delta$ and $n$, second one is smaller than $\epsilon \cdot 2 \sup \|f\|$ because length of your interval of integration is $\epsilon$ so the result follows since $\epsilon$ was arbitrarily small
H: A new way of solving cubics? I found this (from http://www.quora.com/Mathematics/What-are-some-interesting-lesser-known-uses-of-the-quadratic-formula): So my question is: Can this be generalized to solve any depressed cubic [in effect, all the cubics]. Maybe not exactly this method, but some other repeated manipulation of the the quadratic formula? And finally, which is my main aim, can this generalized method give us exact, not-so-messy answers ie. respite from the Cardano's method? AI: What you've discovered is, unfortunately, not a new method, but rather the fact that the cubic up top factors: $$x^3-26x+5 = (x+5)(x^2-5x-1)$$ While at the same time, your equation down below (which, I'll note, you obtained after assuming $x \neq 0$) is either $x(x+5)=0$ or $x^2-5x-1=0$ depending on whether or not you take plus or minus. As an example, you'll find a rough time using that method on $x^3+4x+2$, which is irreducible: $$x^3+4x+2 = x(2)^2+2+x^3$$ $$2 = \frac{-1 \pm \sqrt{1-4x^4}}{2x}$$ $$4x+1 = \pm \sqrt{1-4x^4}$$ Before we finish this note that, when we took the square root, we assumed $1-4x^4$ was positive. This is not true! So already the method has fallen apart; but we'll continue with the algebraic manipulations. (Similarly, we assumed $x \neq 0$.) Now... $$(4x+1)^2 = 1-4x^4$$ $$4x^4+16x^2+8x=0$$ $$x(x^3+4x+2)=0$$ So now we're back where we started (but with an extra solution of $0$ from when we multiplied by $x$ in getting this back into a polynomial)! In addition, we made a bunch of bad assumptions along the way; we did still get back to our original polynomial, but there were quite a few points where our equations were ill-posed. In summary, unfortunately, Cardano's is still the way to go.
H: Fourier and differentiation operators For a function $f:\mathbb{R}\rightarrow\mathbb{R}$ in the Schwartz class, define $$Tf(y)=\dfrac{1}{\sqrt{2\pi}}\int_\mathbb{R}f(x)e^{-ixy}dx$$ We can show that $T^2f(y)=f(-y)$, and $T^4f(y)=f(y)$. Also, define $$Af(y)=yf(y)+\dfrac{d}{dy}f(y)$$ For what value of $a$ is $TA=aAT$? Using $T^4Af(y)=Af(y)$, I can get $$a^4AT^4f(y)=a^4Af(y)=Af(y)$$ so $a^4=1$, so $a=\pm 1,\pm i$. But it doesn't really tell which values of $a$ work. What can I do to identify which values of $a$ work? AI: If you are certain that some value of $a$ must work, then you can plug in a function $f$ that is simple, for example, $f\equiv1$ or something like that, to get information on the value of the constant $a$.
H: Find the point on the parabola Find the point on the parabola $y^{2}=2x$ that is closest to the point $(1,13)$. What is the parabola? and please show the process AI: the distance is: $d^2=(x-x_0)^2+(y-y_0)^2$. In your case: $x_0=1,y_0=13$. So you get: $$d^2=(x-1)^2+(y-13)^2$$ but you know: $y=\sqrt{2x}$. So, you have: $$d^2=(x-1)^2+(\sqrt{2x}-13)^2$$ Now you have to find the minimum of the distance, so you have to calculate: $$\frac{d}{dx}(d^2)=\frac{2x^{3/2}-13\sqrt{2}}{\sqrt{x}}=0$$ From this equation you get: $x=4.388...$ and $y=2.962...$
H: Conditional probability questions? Of a group of children, $0.4$ are boys and $0.6$ are girls. Of the boys, $0.6$ have brown eyes; of the girls, $0.2$ have brown eyes. A child is selected at random from the group. (a) Find the probability that the child is a girl. This is $.6$ (b) Find $P(brown eyes \| boy)$. This is $.6$ (c) Find the probability that the child is a boy with brown eyes. (d) Find the probability that the child is a girl with brown eyes. (e) Find the probability that the child has brown eyes. (f) Find the probability that the child is a girl, given that the child has brown eyes. How can I use $ P(E\|F) = P(E∩F) / P(F)$ for the following questions? AI: Multiply both sides by $P(F)$ to get: $$ P(E\cap F) =P(E\|F)P(F) $$ So for c), for example, you have $$ P(\text{Boy}\cap \text{Brown eyes})=P(\text{Brown eyes}\|\text{Boy})P(\text{Boy}) $$ For d), you do it in the same way. For e), you can use: $$ P(A)=\sum P(A\|B_i)P(B_i) $$ Do you see why? For f), use the original equation (Bayes' theorem) to find it (just like in c), for example).
H: Similar to Poincare inequality on Sobolev spaces The following looks quite similar to Poincare's inequality: Let $\displaystyle{ 1 \leq p < \infty}$ and $\displaystyle{ U \subset \mathbb R^n}$ open and such that $\displaystyle{ U \subset \mathbb R^{n-1} \times (0,L) }$ with $L>0$. Show that for $\displaystyle{ u \in C_c ^ {\infty} (U) \cap W^{1,p} (U) }$ the following inequality hold: $$ \int_U \|u\|^p \leq \frac{L^p}{p} \int_U \| \nabla u\|^p $$. Here it is some thoughts I did, although I didn't manage to prove till the end. Let $ x =(x' ,x_n) $ where $ x' = (x_1, \cdots, x_{n-1} )$. Now integrating with respact to the last variable we get: $\displaystyle{ \|u(x' ,t)\|=\left \| \int_0^{x_n} \partial_n u (x' , t) dt \right \| \leq \int_0^L \| \partial_n u (x' , t) \|dt = \\| 1 \cdot \| \partial_n u (x' , t) \\|_{L^1 ([0,l])} }$ and now form Holder's inequality (where $q$ is the conjugate exponent of $p$) we get that this last is $$ \leq L^{1/q} \left(\int_0^L \| \partial_n u (x' , t) \|^pdt \right)^{1/p} $$ Integrating now with respect to $x_n$ we get that $\displaystyle{ \int_0^L \|u(x' ,t)\|^p dx_n \leq L^{p/q +1 } \left(\int_0^L \| \partial_n u (x' , t) \|^pdt \right) = L^p \int_0^L \|\partial_n u (x' , t) \|^p dt}$ Integrating now with respect to $x'$ we have: $\displaystyle{ \int_U \|u(x)\|^p dx \leq L^p \int_U \| \partial_n u (x' , t) \|^p dx \leq L^p \int_U \|\nabla u(x)\|^p dx }$ I can't see how this $p$ in the denominator of the fraction in the constant appears. Any ideas? AI: We have $$\|u(x',x_n)\|\leqslant \int_0^{x_n}\|\partial_n u(x',t)\|\mathrm dt\leqslant \lVert \partial_n u(x',\cdot)\rVert_px_n^{1-1/p}.$$ The integration with respect to $x_n$ produces the constant $\frac 1p$. Indeed, we have $$\|u(x',x_n)\|^p\leqslant\lVert \partial_n u(x',\cdot)\rVert_p^p\int_0^Lx_n^{p-1}\mathrm dx_n,$$ and the last integral is $[s^p/p]_{s=0}^{s=L}=\frac{L^p}p$. Now we conclude as in the opening post integrating with respect to the other variables.
H: Shortest distance from point and line We want to calculated a shortest distance form point $T(0,1,2)$ and from line of intersection of planes $x+y+z =0$ in $x-z+4=0$ I try this: I have equate both equations $x+y+z =x-z+4$ $y+2z =4$ I get 2 points: $A(0,0,2)$ $B(0,2,1)$ $p= (0,0,2)+t(0,-2,1)$ $T_0=x_0 + \frac{\langle(x_1-x_0)p\rangle}{\langle p,p \rangle} p = (0,0,2)+\frac{ \langle(0,1,0)(0,-2,1)\rangle }{5} \cdot(0,-2,1)$ $= (0,0,2)- \frac{2}{5}(0,-2,1)=(0,\frac{4}{5},\frac{8}{5})$ I use that formula for distance. $d (T,T_0) =\sqrt{ (x_1-x_0)^2 + (y_1-y_0)^2 + (z_1-z_0)^2}$ $=\sqrt{ 0+ (\frac{1}{5})^2+ (\frac{2}{5})^2} = \sqrt{ \frac{1}{25}+ \frac{4}{25}}$ $= \sqrt{ \frac{1}{5} }$ Is this the shortest distance? Thanks AI: Calculate the intersection line: $$\begin{cases}x+y&+&z&=0\\x&-&z&=4\end{cases}\implies y=-4-2z\;,\;\;x=4+z$$ and the line is $\;(4,-4,0)+t(1,-2,1)\;,\;\;t\in\Bbb R\;$ , so the distance is $$\frac{\|\|\;\left[(0,1,2)-(4,-4,0)\right]\times\left[(0,1,2)-(5,-6,1)\right]\;\|\|}{\|\|\;(4,-4,0)-(5,-6,1)\;\|\|}=\frac{\|\|\;(-4,5,2)\times(-5,7,1)\;\|\|}{\|\|\;(-1,2,-1)\;\|\|}=$$ $$\frac{\|\|\;(-9,-6,-3)\;\|\|}{\sqrt6}=\frac{\sqrt{126}}{\sqrt6}=\sqrt{21}$$
H: Composite of two purely inseparable extensions is purely inseparable. Let $L,F$ be extensions of the field $K$ and are contained in a common field. Prove that, if $L$ and $F$ are purely inseparable extensions over $K$ then $LF$ is also a purely inseparable extension over $K$. Is the converse true? How do I prove it? Thanks in advanced. AI: Assume the characteristic of your fields is $p>0$. You want to prove that, for a field extension $L/K$, the set $L^\prime$ of elements of $L$ that are purely inseparable over $K$ is a subfield of $L$. You do this with the following characterization of pure inseparability: an element $x\in L$ is purely inseparable over $K$ if and only if, for some integer $n\geq 0$, $x^{p^n}\in K$. A proof of this fact can be found here: Purely inseparable extension. With this, your result can be proved as follows. Consider the subfield $E$ of $LF$ consisting of elements purely inseparable over $K$. Because $L$ and $F$ are both purely inseparable over $K$, $L,F\subseteq E$, so $LF\subseteq E$, and thus $LF=E$ is purely inseparable over $K$. The converse is also true. More generally, if $K\subseteq L\subseteq E$ and $E/K$ is purely inseparable, then $L/K$ is purely inseparable. This is immediate from the definition of pure inseparability.
H: card expectations Beginning probability question Imagine a standard card deck where cards have the values 2,3,4,...,11(J),12(Q),13(K),14(A). The deck is shuffled so that every permutation is equally likely. I draw cards from the deck, one at a time without replacement, until I draw the queen of spades (Q), at which point I stop drawing. If T represents the number of cards I draw what is the expected value of T. The first card I draw I give it to a member of the audience and from then on, whenever I draw a card with value smaller than the card I gave to the member of the audience, I give it to that member. What are the expected max and minimum values of the cards I give? What is the expected number of cards I give? What is a good way of defining random variables and thinking about this problem? AI: For the first question, each card except the queen of spades has an equal chance to be before or after it. For the second and third, given the rank $r$ of the first card, how many cards do you give to the audience? You probably shouldn't have the word "expected" in the second question. Max will come if you start with an ace, min if you start with a $2$.
H: Determine if these are equivalence relations I would appreciate if someone could go through the task and the answers I've got and check if I've done it correct, if not please correct me. Here is the task: Below we have listed some relationships over The set of $\{a, b, c, d, e\}$. For each of these determine whether it is an equivalence relation, and in that case find $[a]$, which is equivalence class to $a$. $\{\langle a,a\rangle, \langle b,b \rangle, \langle c,c\rangle, \langle d,d\rangle\}$ $\{\langle a,a\rangle, \langle b,b\rangle, \langle c,c\rangle, \langle d,d\rangle, \langle e,e\rangle, \langle a,b\rangle, \langle b,a\rangle\}$ $\{\langle a,a\rangle, \langle b,b\rangle, \langle c,c\rangle, \langle d,d\rangle, \langle e,e\rangle, \langle b,c\rangle, \langle b,d\rangle, \langle c,b\rangle, \langle d,b\rangle\}$ $\{\langle a,a\rangle, \langle b,b\rangle, \langle c,c\rangle, \langle d,d\rangle, \langle e,e\rangle, \langle a,c\rangle, \langle b,c\rangle\}$ $\{\langle a,a\rangle, \langle b,b\rangle, \langle c,c\rangle, \langle d,d\rangle, \langle e,e\rangle, \langle b,d\rangle, \langle d,b\rangle\}$ $\{\langle a,a\rangle, \langle b,b\rangle, \langle c,c\rangle, \langle d,d\rangle, \langle e,e\rangle, \langle b,d\rangle, \langle d,b\rangle, \langle a,c\rangle, \langle c,a\rangle\}$ Here are my answers: $[a] = \{a\}$ $[a]=\{a, b\}$ $[a]=\{a\}$ $[a]=\{a, c\}$ $[a]=\{a\}$ $[a]=\{a, c\}$ If I'm correct $(1)$, $(3)$, $(5)$ and $(4)$ and $(6)$ are equivalence relations. Thanks a lot for your help. AI: A relation $R$ is an equivalence relation if and only if it is reflexive AND symmetric AND transitive. Equivalence classes are determined by an equivalence relation. It makes no sense to speak of equivalence classes when a relation is not an equivalence relation. Equivalence relation does not mean an "equivalent relation". It is a property of a relation, and not a property between different relations. Relation $(1)$ is not reflexive, hence not an equivalence relation. It is lacking the pair $(e, e)$. Relation $(2)$ is an equivalence relation. Now you need to find the equivalence class of $a$: all the elements that are related to $a$. Relation $(3)$ is reflexive and symmetric, but not transitive. $(c, b),$ and $(b, d) \in R$, but $(c, d) \notin R$. Since it is not transitive, $R$ cannot be an equivalence relation. Relation $(4)$ is reflexive and transitive, but not summetric: $(a, c) \in R$, but $(c, a) \notin R$. Relation $(5)$ is an equivalence relation: it is reflexive, symmetric, and transitive. Now you need to find the equivalence class of $a$: [a]. This is the set of all elements in the set which are related to $a$. Relation $(6)$ is also an equivalence relation: it is reflexive, symmetric, and transitive. Now you need to find the equivalence class of $a$: $[a]$. This is the set of all elements which are related to $a$.
H: Evaluate $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } n^{-1/2}$ I know that $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\frac { 1 }{ n } =\ln(2) }$ . How about the series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } \frac { 1 }{ \sqrt { n } }$ To what number does it converge? AI: You are looking for the function $\eta(s)=(1-2^{1-s})\zeta(s)$, where $\eta(s)=\sum (-1)^{n+1}n^{-s}$ and $\zeta(s)=\sum n^{-s}$. Letting $s=1/2$, we can write $\eta(1/2)=(1-\sqrt{2})\zeta(1/2)=0.60489864\ldots$ I do not believe there is a more simple/elementary closed form.
H: Proof of a trigonometric expression Let $f(x) = (\sin \frac{πx}{7})^{-1}$. Prove that $f(3) + f(2) = f(1)$. This is another trig question, which I cannot get how to start with. Sum to product identities also did not work. AI: Let $7\theta=\pi, 4\theta=\pi-3\theta\implies \sin4\theta=\sin(\pi-3\theta)=\sin3\theta$ $$\frac1{\sin3\theta}+\frac1{\sin2\theta}$$ $$=\frac1{\sin4\theta}+\frac1{\sin2\theta}$$ $$=\frac{\sin4\theta+\sin2\theta}{\sin4\theta\sin2\theta}$$ $$=\frac{2\sin3\theta\cos\theta}{\sin4\theta\sin2\theta}\text{ Using } \sin2C+\sin2D=2\sin(C+D)\cos(C-D)$$ $$=\frac{2\cos\theta}{2\sin\theta\cos\theta}$$ $$=\frac1{\sin\theta}$$ All cancellations are legal as $\sin r\theta\ne0$ for $7\not\mid r$
H: Symmetric matrix under orthogonal transformation still symmetric? is there any way to see that a symmetric matrix is still symmetric after applying an orthogonal basis transformation to it? I would say that a proof that refers to the entries of the matrix may be cumbersome, therefore I am asking here for clever ways to show this. AI: The important thing to note here is that the inverse of an orthogonal matrix is its transpose, so the basis change $S \leadsto O^{-1}SO$ is actually (also) $S \leadsto O^TSO$, in which form the symmetry-preserving nature of orthogonal basis changes is near-obvious: $$(O^TSO)^T = O^TS^T(O^T)^T = O^TS^TO = O^TSO.$$
H: Probability questions. The following experiment is used to check a person who claims to have powers of mental telepathy. Six cards are numbered 1 through 6. Seat the person being tested on one side of a screen and a person who will select a card on the other side. The person with the cards shuffles them, selects one at random, and turns it up. The person being tested "receives vibes" and tells which number he thinks is turned up. The card is replaced, and the process is repeated for a total of six times. $a)$ Based on random chance alone, find the probability that the person being tested identifies all six numbers correctly. $b)$ Based on random chance alone, find the probability that the person being tested identifies all six numbers incorrectly. $c)$ Based on random chance alone, find the probability that the person being tested identifies at least one of the six numbers correctly. I know this is a type of conditional probability problem but I have no idea how to start this. Please help me with some pointers. AI: Hint: The probability of guessing a particular one correctly is $\frac 1 6$ The probability of guessing two correctly is $(\frac 1 6)^2$ or $\frac 1 6 \times \frac 1 6$ The probability of guessing at least one correctly (just by chance) is $1-(\frac 5 6)^2$. This is because the subject "did not guess wrong twice". Using the first two rules, you should be able to get a and b. The last one is needed for c.
H: 4 convex sets in a plane have a point in common Let $X_1,X_2,X_3,X_4$ be four sets in the plane such that any three of them have a point in common. Do all four of them have to have a point in common? What if sets are convex? Attempt: I think all of them should not have a point in common, but I dont how them being convex can influence the answer. Thanks for help. AI: It is easy to come up with a counter-example if the sets are not-convex. If the sets are convex, read the technique of Helly's Theorem.
H: Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$ And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}$ is convergent as well. But I find it hard to calculate the sum. can you give me some hints? AI: HINT: As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)$ and $(n^2+1+n)-(n^2+1-n)=2n$ $$\frac n{n^4+n^2+1}=\frac12\left(\frac{2n}{(n^2+1-n)(n^2+1+n)}\right)$$ $$=\frac12\left(\frac{(n^2+1+n)-(n^2+1-n)}{(n^2+1-n)(n^2+1+n)}\right)$$ $$=\frac12\left(\frac1{n^2-n+1}-\frac1{n^2+n+1}\right)$$ Also observe that $: (n+1)^2-(n+1)+1=n^2+n+1$ inviting cancellations
H: find a formula for the sum of combination $5^n {n\choose 0} - 5^{n-1} {n\choose 1} + 5^{n-2}{n\choose2} +...... \pm{n\choose n}$ I was trying to use binomial theorem, but it doesn't seem to work. AI: $$5^n {n\choose 0} - 5^{n-1} {n\choose 1} + 5^{n-2}{n\choose2} +...... \pm{n\choose n}=$$ $$=\sum_{k=0}^n\binom nk(-1)^k5^{n-k}=(-1+5)^n\ldots$$
H: How to solve mechanics problem when acceleration depends on position. I'm curious about how problems such as the following are typically solved analytically, or in computer simulations such as games engines for 2D physics. It seems a bit harder than the typical constant acceleration scenario. Suppose a particle with mass m is free to move inside a smooth 2D basin given by the curve $y=x^2$. Given some initial conditions, such as position $(x_0, x_0^2)$, the problem is to determine the position of the particle at time $t$. The particle is always subject to a gravitational force $(0, -mg)$. The other force acting on it is the reaction from the basin, which has magnitude $mg$, and direction given by the unit normal to the basin at the current position, $\hat{\mathbf{n}}(x)$. Solving for position at time $t$ requires integrating the velocity, and thus the acceleration, up until time $t$. But the acceleration at time $t$ depends on the current position to obtain the normal vector. So it is not clear to me (or I have forgotten) how this can be solved analytically. As well as the question of how this would be solved analytically, I'm also interested in how this may typically be implemented in simulation software. It seems like if you went for a finite element method, however small your time step, you would be moving out of contact with the basin if you took steps along the tangent vector. AI: You can observe that: $$V=mgy=mgx^2$$ and the kinetic energy can be written as: $$T=\frac{1}{2}mv^2=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$$ Bacause you know: $y=x^2$, you get: $\dot{y}=\frac{dy}{dx}\dot{x}=2x\dot{x}$ So you can write the Eulero Lagrange equations putting $L=T-V$ (Lagrangian): $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}=0$$ Writing this equation you get a nonlinear differential equation and if you are luky you can obtain an analytical expression for $x(t)$ and so for $y(t)$
H: Discontinuous function Show that the function, $f:\mathbb{R^2}\rightarrow \mathbb{R}$, $$ f(x,y) = \begin{cases} 1, & \\|(x,y)\\|\geq1\\ 0, & \\|(x,y)\\|<1 \end{cases} $$ is not continuous ($\mathbb{R}$ and $\mathbb{R^2}$ have their usual Euclidean metrics). Could someone check if my proof is correct. Feel like I might have overlooken some fact or left something out. Let d and p be the metrics on $\mathbb{R}$ and $\mathbb{R^2}$ respectively and take $\alpha\in\mathbb{R^2}$ such that $\\|\alpha\\|=1$. So I have to show that there does not exist $\delta>0$ such that if p($(x,y),\alpha)<\delta$, then d($f(x,y),f(\alpha))<\epsilon$. Take $\delta>0$ and let $\epsilon=1$. Let $(x,y)=\alpha-\delta\alpha\implies\\|(x,y)\\|=\\|\alpha-\delta\alpha\\|=\\|\alpha(1-\delta)\\|=\|1-\delta\|\\|\alpha\\|=\|1-\delta\|$ , if $\delta\geq2$ it suffices to show _p_$((x,y),\alpha)<2\leq\delta\implies$_d_$(f(x,y),f(\alpha))<\epsilon$. Hence, for $\delta<2$, we get $\\|(x,y)\\|<1\implies f(x,y)=0\implies d(f(x,y),f(\alpha))=\|0-1\|\geq1.$ Therefore, the function is not continuous. AI: You've got the right idea, but your execution is a bit off. Take $w\in\Bbb R^2$ with $\lVert w\rVert=1,$ take any $\delta>0,$ put $\epsilon=1,$ and take $0<c<\min\{1,\delta\}.$ Then, put $v=(1-c)w,$ instead. You'll see that $\lVert v\rVert=\|1-c\|=1-c<1,$ so that $f(v)=0,$ and so $$d\bigl(f(v),f(w)\bigr)=1\ge\epsilon,$$ but also that $$p(v,w)=\lVert-cw\rVert=\|c\|\lVert w\rVert=\|c\|<\delta,$$ so the function is not continuous.
H: why the Vitali set has size $2^{\aleph_0}$? why the Vitali set has size $2^{\aleph_0}$ and therefore there is a bijection between $\mathbb R$ and the Vitali set? thanks AI: Note that there are $2^{\aleph_0}$ equivalence classes in $\Bbb{R/Q}$. Since a choice function from these classes is injective, this shows that there are exactly $2^{\aleph_0}$ elements in a Vitali set, and certainly there are not more. To see that there are $2^{\aleph_0}$ equivalence classes, note that the vector spaces $\Bbb R$ and $\Bbb{R/Q}$ are isomorphic as vector spaces over $\Bbb Q$. Therefore they have the same dimension and cardinality.
H: Why are group theory and ring theory a part of abstract algebra? I have followed the courses Algebra 1, which was about group theory and Algebra 2, which was about ring theory. I don't think I really understand why those subjects are part of abstract algebra. What does "algebra" has to do with those both subjects ? I tried to google it a little bit, but I get quite confused when I try to google the word "algebra". If I look at this page for example: http://en.wikipedia.org/wiki/Algebra_(ring_theory), then it seems like that "algebra" is a part of ring theory. AI: You are maybe confused by multiple meanings of the word algebra. The WP article you link to is about "an algebra" as the term is used in ring theory. There are also many specific types of algebra (Lie algebra, Boolean algebra) in other domains. Most of these fall within the general area of (abstract) algebra, which is a vast area of mathematics that can be somewhat characterised as dealing with reasoning based on exact equality (rather than approximative equalities and limits). The adjective "abstract" is used to indicate that most of the time reasoning is not for one specific algebraic structure, but for a whole family of structures, with members of the family sharing the same axioms (groups, rings, fields, vector spaces are examples of such families).
H: Notation for translating vectors I'm completely new to vector geometry and recently encountered some new notation (and wholly unfamiliar) for the translation of vectors. $$T:Z \mapsto A + Z$$ The above is described as A translation by vector $A$ Again, I have no idea why they call it a translation by vector $A$ and not of vector $A$. Or maybe I'm misunderstanding the whole thing. Can somebody shed some light on the matter? AI: It is a common shorthand for defining a function. In this case, it is clearly just "translation by $A$", since any $Z$ is mapped to $Z+A$. Quite often, especially when you don't need to name a function, you just use the "$\mapsto$" notation. An argument in a proof of something could be for instance: "Because the function $x\mapsto x^3$ is continuous." Sometimes you even don't 'define' the function for arbitrary $x$ and you let the reader think it out. So for example "the exponential that brings $a\mapsto b$" is the same as "the exponential $x\mapsto (b^{1/a})^x$" and the same as "the function $f:\mathbb R\to\mathbb R$, $f(x)=(b^{1/a})^x$". It is a very convenient notation that avoid too much formalism and too many parentheses.
H: Fibonacci sequences I have the following: $$ f_3+f_6 + \dots+f_{3n} = \frac 12 (f_{3n+2}-1) $$ for $f_0=0$ and $f_1=1$ When I calculate $n\ge2$ and $f_n= f_{n-1}+f_{n-2}$, I get: LHS = 8 while RHS = 10. LHS $$f_6 =f_5+f_4 \\ f_5 = f_4+f_3 \\ f_4 = f_3 + f_2 \\ f_3=f_2+f_1 \\ f_2 = f_1 + f_0 $$ and so : $$f_2=1 \\ f_3=2 \\ f_4 = 3 \\ f_5 = 5 \\ f_6 = 8$$ RHS $$ \frac 12(f_8-1) \\ \equiv \frac 12( (f_7+f_6)-1) \\ \equiv \frac 12( (f_6+f_5)+f_6-1) \\ \equiv \frac 12( (8+5)+8-1) \\ \equiv \frac 12(20) = 10 $$ and so the RHS $\neq$ LHS. So how is it the basis is false when I'm asked to prove for all integers? Or is there something I missed? AI: Don't forget that your left-hand side should be $$f_3+f_6=2+8=10.$$
H: Solving a bijective function task How would you solve this task actually? This is not homework btw, i'm just trying to understand how you can solve tasks like this. Let $S$ be the set $\{$$1,2,3,4,5$$\}$. How many functions from $S$ to $S$ are there and how many of these are bijective? Thanks alot for any help. AI: Suppose that you want to build a function $f$ from a set $A$ of $m$ elements to a set $B$ of $n$ elements. You can list $A$ as $A=\{a_1,a_2,\ldots,a_m\}$. Now go down the line choosing values of $f(a_1)$, $f(a_2)$, and so on. If there is no restriction on your function, there are $n$ possible choices for $f(a_1)$: it can be any element of $B$. There are still $n$ choices for $f(a_2)$, since we imposed no restriction on $f$, and $n$ choices for $f(a_3)$, and so on: for $k=1,\ldots,m$ there are $n$ choices for $f(a_k)$. These choices can be made independently, so by the multiplication principle there are $$\underbrace{n\cdot n\cdot\ldots\cdot n}_{m\text{ copies}}=n^m$$ functions from $A$ to $B$. In your case that’s $5^5=3125$ functions. If $m\le n$, we can talk about injections. There are now $n$ possible choices for $f(a_1)$, since it can be any of the $m$ elements of $B$. For $f(a_2)$, however, there are now only $n-1$ possible choices: if we want a bijection, we can’t re-use $f(a_1)$. Similar reasoning shows that there must be $$\underbrace{n(n-1)(n-2)\ldots(n-m+1)}_{m\text{ factors}}=\frac{n!}{(n-m)!}$$ injections from $A$ to $B$. In your case $m=n=5$, so the injections are precisely the bijections, and their number is $\frac{5!}{0!}=5!=120$.
H: Does sum of random variables remain independent? Suppose we have random variables $X,Y,Z$ such that $X,Z$ are independent, and $Y,Z$ are also independent. Is $X+Y$ independent from $Z$ ? I cannot find a counterexample, and I do not see how to prove it. AI: Toss a fair coin three times. Let $X$ be $1$ if the number of heads on the first and second tosses is odd, $0$ otherwise. Similarly for $Y$ with the first and third tosses, $Z$ with the second and third. Then each pair $(X,Y)$, $(X,Z)$ and $(Y,Z)$ are independent, but $X+Y$ and $Z$ are not independent: in fact $X+Y = 1$ if and only if $Z = 1$.
H: Minimum and maximum values I am not sure if my method for this question is correct: Given the function $$f(x,y)=x^2+y^2+2x+y$$, find it's minimum and maximum values about a closed disc of radius 2 centred at the origin. I took the partial derivatives, $$f_x=2x+2; f_y=2y+1$$ and solved for $x$ and $y$. The minimum value is $f(0,0)$ and the maximum value is $f(-1,-1/2)$. Am I doing the correct thing? The answer I got is wrong according to the key. AI: If you have a boundary which is given by the level curve of a function (here $x^{2} + y^{2}$), you can use the Lagrange multiplier theorem (see here). You have to check the points of the boundary such that the gradient of the function you want to study and the gradient of the function giving the domain are parallel. In this case you have that the only critical point in $\mathbb{R}^{2}$ is $\left(-1 , -1/2 \right)$. $f$ is the sum of two quadratic functions in $x$ adn $y$ separately and with coefficients of $x^{2}$ and $y^{2}$ both positive. Hence $\mathrm{lim}_{\|\|\left(x,y\right)\|\| \to +\infty}f\left(x,y\right)=+\infty$, so the point we have is the absolute minimum, and it lies in the circle. Now to find the maximum in the circle check the points of the circle of radius 2 such that $\left(2x+2,2y+1\right)$ is parallel to $\left(2x,2y\right)$. If you put them in a matrix and want the determinant to be $0$, it gives you the condition $2y=x$.
H: Find upper and lower bounds to the function $f(n)=1\cdot3\cdot5\cdot\ldots\cdot(2n-1)$ where $n\in\Bbb N$ Find upper and lower bounds to the function $$f(n)=1\cdot 3\cdot 5\cdot 7\cdot 9\cdot\ldots\cdot (2n-1)\;,$$ where $n\in\Bbb N$. I got $$\left(\frac{2n-1}e\right)^{(2n-1)/2}$$ as lower bound and $$\left(\frac{2n+1}e\right)^{(2n+1)/2}$$ as upper bound. Just wanted to know if any better bounds were there.. Thanks for the help. AI: Using factorials, we can write $$ f(n) =\frac{(2n)!}{2^n\cdot n!}$$ and then using Stirling's $n!\approx n^ne^{-n}\sqrt{2\pi n}$ find $$\tag1 f(n)\approx \frac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{2^nn^ne^{-n}\sqrt{2\pi n}}=\left(\frac{2n}{e}\right)^n\cdot \sqrt 2$$ so your $f(n)$ grows quite fast (not surprisingly). For an explicit upper/lower bound function, use explicit bounds for Stirling's approximation. Edit: It seems you essentially did that to obtain your bounds with one additional term in the Stirling approximation. Do find something better one might try to look at higher order terms in STirling, but admittedly the expressions turn out ugly (and probably not very helpful) Note that we can also observe that $f(n)={2n\choose n}\cdot \frac{n!}{2^n}$ and then from ${2n\choose n}\le (1+1)^{2n}=4^n$ find $$\tag2f(n)\le 2^nn! $$ as a simple upper bound (and it's only ba a factor $\approx\sqrt {\pi n}$ bigger than the estimate in $(1)$).
H: On polynomial of prime degree. Let $K$ be a field, $f(X)\in K[X]$ be a polynomial of prime degree. Assume that for all extension $L$ of $K$, if $f$ has roots in $L$ then $f$ splits over $L$. Prove that either $f$ is irreducible over $K$ or $f$ splits over $K$. How do I prove it? Thanks a lot. AI: If $f$ has a root in $K$ then $f$ splits over $K$. If not, assume that $f$ is reducible over $K$, take an irreducible factor of $f$ of minimal degree and $a$ a root of this factor. If $L=K(a)$ then any other irreducible factor of $f$ has a root (in fact all roots) in $L$. Now use the degree of these extensions and the prime degree of $f$ to finish the proof.
H: Help required: please evaluate the two given integrals Question: Evaluate the following integral. My answer: $=\displaystyle\cos (6t)+6t . 7t/7 = -\cos(6 . \pi/2) + (6\pi)/2 . (7.(\pi/2)) /7$ Correct answer: $\frac{2}{3}.$ While you are at it could you also answer this one as well (sorry I did not know how to write the equations in proper form here!) AI: Hint: Use the fact that $\frac{d}{dt}\left(\frac{-1}{3}\cos(3t) \right) = \sin(3t)$.
H: Galois group of irreducible polynomial in a field of characteristic zero in which every element is a perfect square This is one of the exercises during my reading of Ian Stewart's Galois Theory. Whether the following statement is true: If $K$ is a field of characteristic zero in which every element is a perfect square, then the Galois group of any irreducible $n$-th degree polynomial over $K$ is isomorphic to $A_n$. I think it might not be true but the examples of such fields in my knowledge are very few. Any thoughts? Thanks! AI: I proffer the following counterexample. First we need to construct such a field $K$. Let $K_0=\Bbb{Q}$, and recursively define fields $K_n, n\in\Bbb{N}$ by the recipe that $K_{\ell+1}$ is gotten from $K_\ell$ by adjoining the square roots of all the elements of $K_{\ell}$ to it. Note that we can do this construction inside $\Bbb{C}$. Then we let $$ K=\bigcup_{n\in\Bbb{N}}K_n\subset \Bbb{C}. $$ Then any element of $K$ belongs to some field $K_\ell$, and thus has a square root in $K_{\ell+1}$, hence also in $K$. If we pick any element $z\in K$, I claim that $[\Bbb{Q}(z):\Bbb{Q}]$ is a power of two. Clearly $z\in K_\ell$ for some $\ell$. W.l.o.g. we can assume that $\ell$ is the smallest natural number with that property. Then there exists a finite set of elements $z_1,z_2,\ldots,z_m\in K_{\ell-1}$ such that $z\in K_{\ell-1}(\sqrt{z_1},\sqrt{z_2},\ldots,\sqrt{z_m})$. This means that we can write $z$ in terms of finitely many element of $K_{\ell-1}$ and their square roots. By induction hypothesis all those elements are algebraic of degree a power of two over $\Bbb{Q}$. Thus so is their compositum. Adjoining the square roots of $z_i, i=1,\ldots,m,$ won't change that fact so we see that $\Bbb{Q}(z)$ is contained in a field of degree a power of two settling the claim. Let's then consider the eleventh cyclotomic polynomial $$ \phi_{11}(x)=x^{10}+x^9+x^8+\cdots+x^2+x+1. $$ By the known theory of cyclotomic fields we know that $\phi_{11}(x)$ factors into a product of two quintic factors over $F=\Bbb{Q}(\sqrt{-11})$, the only quadratic subfield of the eleventh cyclotomic field. Let $f(x)$ be one of those quintic factors. By the above observation $f(x)$ remains irreducible over $K$. If $L=K(e^{2\pi i/11})$ is the splitting field of $f(x)$, we easily see that $Gal(L/K)$ is isomorphic to $Gal(\Bbb{Q}(e^{2\pi i/11})/\Bbb{Q}(\sqrt{-11})\cong C_5$. But $C_5$ is a proper subgroup of $A_5$.
H: Using induction to prove $a_n >2^n$ For the sequence $a_n=2a_{n-1}+1$ where $a_0=1$ Show that $a_n>2^n$ using induction. Use proof by contradiction (minimum counterexample). Attempt: 1. I assume, that $a_n>2^n$ as my induction hypothesys. Now, I try to show that $$a_{n+1}>2^{n+1} \\ 2a_{n}+1>2^{n}*2 \\ a_n+1/2>2^n$$ and I am stuck and dont know how to proceed. 2$.$ Let set $S=(n\in Z_+: a_n\le 2^n)$ be a set of counterexamples. Let $m$ be the smallest element of $S$, then $m-1 \notin S$. Then, I try to produce a contradiction $$a_{m-1}\le2^{m-1} \\ a_{m}\le 2^m+1$$ and again I am stuck. Help apreciated. Thanks. AI: You know that $a_1 = 3 > 2^1$, so assume that $a_n > 2^n$. Then, $$ a_{n+1} = 2a_n + 1 > 2\cdot 2^n + 1 = 2^{n+1} + 1 > 2^{n+1} $$ So, by induction, $a_n > 2^n$ for all $n\in \mathbb{N}$
H: Drawing a hasse diagram of a poset. I have the poset $(\{1,2,3,4,5,6,8,15\}, \|)$, where $\|$ denotes divisibility and was able to come up with the Hasse diagram I feel like I had multiple options to connect the $8$ too as well as the $6$. This is my first hasse diagram and I just wanted to post to make sure I did it correctly. AI: For starters, you have it upside-down: $1$ is the minimum element and should be at the bottom. Apart from that it’s just missing two lines: there should be a line from $2$ to $6$: $2$ is a divisor of $6$, and there is nothing between them (as $4$, for instance, is between $2$ and $8$). Similarly, there should be a line from $3$ to $15$. Everything else is fine: you have no superfluous lines.
H: Max area of triangle -PHP How do i prove that the maximum area that can be obtained among 3 random points in a square is half the area of the square?- I need it to for the following question " Show that among any 9 points inside a triangle of area 1 there are three points which form a triangle of area at most 1/4." Hence there would be at least 3 points in a square of side 1/2. But then how do i proceed? AI: Actually you don't need to prove this to solve your problem, although it is a good question. For your problem, just take the triangle formed by the three midpoints of the sides of the bigger triangle. This triangle divides the bigger in 4 triangles with area $1/4$. Then, given 9 points in the bigger triangle, by the pigeonhole principle, at least 3 of them are in one triangle of area $1/4$. For the different question, here is the proof. Given 3 points in the square, if the points are not at the sides, we can have a different triangle with its vertex at the sides of the square with bigger area than the previous one, e.g. If we now have a triangle with the vertives at the sides, if one of them is not a vertex of the square, we can get a new triangle with bigger area by moving the vertex such that the new side is parallel to the previous one: Now, let $a$ and $b$ be the sizes of the segments in the image: The area $A$ of the triangle is $$A=1-\frac{a}{2}-\frac{b}{2}-\frac{(1-a)(1-b)}{2}=\frac{1-ab}{2}.$$ So, for the maximum area, we must minimize $a$ or $b$ to $0$, which yields $A=1/2$.
H: Show discontinuity of $\frac{xy}{x^2+y^2}$ How to show this function's discontinuity? $ f(n) = \left\{ \begin{array}{l l} \frac{xy}{x^2+y^2} & \quad , \quad(x,y)\neq(0,0)\\ 0 & \quad , \quad(x,y)=(0,0) \end{array} \right.$ AI: $y=mx\Rightarrow f(x,y)=\frac{mx^2}{(1+m^2)x^2}=\frac{m}{1+m^2}$ Does this ring some bell??? I am actually trying to make you more comfortable with this kind of idea. In this case you know limit should be zero and if you consider path $y=x$ then you are done. But, then more generally what would one do if you are not so sure of $\lim$ is you keep changing $m$ in $y=mx$ (some other curve in some other case) and see if the result is same even if i change $m$. In this case we got $f(x,y)=\frac{m}{1+m^2}$. If i approach through $y=x$ keeping $m=1$, I would get the result as $f(x,y)=\frac{1}{2}$ If i approach through $y=2x$ keeping $m=2$, I would get the result as $f(x,y)=\frac{2}{5}$ So, If i approach in two different paths those limits are not equal. So for get about $\lim _{(x,y)\rightarrow (0,0)}f(x,y) $ being equal to $f(0,0)$ (which is what you need for continuity), The limit $\lim _{(x,y)\rightarrow (0,0)}f(x,y) $ does not even exist. So, function is discontinuous....
H: How to prove or disprove finiteness? How to prove or disprove that statement: a group is finite if the set of all its subgroups is finite? AI: A finite group certainly has only finitely many subgroups, so the question really boils down to this: is there an infinite group with only finitely many subgroups? HINT: Let $G$ be an infinite group. Consider two cases: Every element of $G$ has finite order. $G$ contains an infinite cyclic subgroup. What can you say in each case about the number of subgroups of $G$?
H: Help needed with an equivalence relation task on natural numbers I'm having a bit difficulties understanding and solving this task. I would appreciate any help on how you can solve tasks like this. Here is the task: Let ~ be an equivalence relation on the natural numbers, and let $E$ be $[0]$, ie equivalence class to $0$. Prove that E is not equal to the empty set Prove that all natural numbers $x$ and $y$ that are elements of $E$ are such that $x$ ~ $y$. If $[x] = [y]$, then $x$ ~ $y$? If yes, give a proof for it, if no, give a counterexample. Thanks a lot for any help. AI: For the first one, consider what three properties an equivalence relation has, and note that $$E=\{z\in\Bbb N:0\sim z\}.$$ One of the three properties of an equivalence relation will allow you to prove the claim. For the third one, we will use the same property of an equivalence relation. You want to show that if $$\{z\in\Bbb N:x\sim z\}=\{z\in\Bbb N:y\sim z\},$$ then $x\sim y.$ I believe that the second one should read "Prove that all natural numbers $x,y$ which are elements of $E$ are such that $x\sim y.$" That is, we must show that if $x,y\in E,$ then $x\sim y.$ This time, we will be using the other two properties of equivalence relations to prove it. Edit: As you know, a binary relation $R$ is said to be an equivalence relation on a set $A$ if and only if it is reflexive, symmetric, and transitive on $A$. In particular, this means: For all $x\in A,$ we have $x\:R\:x.$ ($R$ is reflexive on $A$.) For all $x,y\in A,$ if $x\:R\:y,$ then $y\:R\:x.$ ($R$ is symmetric on $A$.) For all $x,y,z\in A,$ if $x\:R\:y$ and $y\:R\:z,$ then $x\:R\:z.$ ($R$ is transitive on $A$.) Now, suppose that $R$ is any binary relation, $A$ any set. For each $x\in A,$ define: $$[x]_{A,R}:=\{z\in A:x\:R\:z\}$$ Let's use this notation to rewrite the definitions of reflexivity, symmetry, and transitivity of $R$ on $A$: For all $x\in A,$ we have $x\in[x]_{A,R}.$ ($R$ is reflexive on $A$.) For all $x,y\in A,$ if $y\in[x]_{A,R},$ then $x\in[y]_{A,R}.$ ($R$ is symmetric on $A$.) For all $x,y,z\in A,$ if $y\in[x]_{A,R}$ and $z\in[y]_{A,R},$ then $z\in[x]_{A,R}.$ ($R$ is transitive on $A$.) Observe then that for any $x\in\Bbb N,$ we have $[x]:=[x]_{\Bbb N,\sim}.$ Do you see, then, which of the three properties immediately tells us that $[0]\ne\emptyset$? We will use this property again to show that if $[x]=[y],$ then $y\in[x]$ (why is this true?), and so $x\sim y$ by definition. For $(2),$ we will use the other two properties. In particular, we want to show that if $0\sim x$ and $0\sim y,$ then $x\sim y.$
H: Showing a ring has no non-zero nilpotents via a ring homomorphism Let $\theta\colon R\to S$ be a ring homomorphism. If each of $\ker{\theta}$ and $\operatorname{im}{\theta}$ has the property that its only nilpotent element is $0$, show that the same is true for $R$. AI: If $r \in R$ is nilpotent, then $\phi(r)$ is nilpotent in $\mathrm{im}(\phi)$, hence zero. But then $r$ is in the kernel ... I think you can finish the proof now.
H: Why a non-diagonalizable matrix can be approximated by an infinite sequence of diagonalizable matrices? It is known that any non-diagonalizable matrix, $A$, can be approximated by a set of diagonalizable matrices, e.g. $A \simeq \lim_{k \rightarrow \infty} A_k$. Why this is true? Note: I was faced with it for the first time at a note about a simple proof for Cayley-Hamilton theorem, but I was not able to find it in my books nor in the internet to the extent I've googled. AI: If matrix has distinct eigenvalues, it can be diagonalized. You can perturb a matrix by an arbitrarily small amount so that all eigenvalues are distinct. Give any $A$, let $J$ be the Jordan form. That is, for some $V$, you have $A=U J U^{-1}$. Let $\Delta= \operatorname{diag}(1,2,...,n)$, where $n$ is the dimension of the matrix, and consider the sequence $A_k = A+ \frac{1}{k} U\Delta U^{-1}$ (thanks to p.s. for catching my mistake here). The eigenvalues of $A_k$ are $[J]_{ii} + \frac{1}{k} i$, hence for $k$ large enough, the eigenvalues are distinct, and hence $A_k$ is diagonalizable. Clearly $A_k \to A$, hence the result. Note: If we use the Schur form instead, we can explicitly compute the distance $\\|A-A_k\\|_2 = \frac{n}{k}$.
H: How to prove something about the order of element at a group $G$ is a cyclic group. the order of $G$ is $n$ ($\|G\|=n$). $m\mid n$, I have to prove that there is $b\in G$ that $ord(b)=m$. Know, here is one part of the proof: if $m\mid n$ that means that $n=mk$. Lets mark: $$b=a^k$$ then: $$b^m=(a^k)^m=a^n=e\Rightarrow ord(b)=m$$ Now I need to prove that $m$ is minimal. How I'm prove this? Thank you! AI: Answer to Original Post, Before Edit: The theorem you are trying to prove, as stated, is not true. The converse of Lagrange's Theorem does not, in general, hold. For example, take $A_4$, the alternating group of $S_4$. $\;\|A_4\| = 12$, and clearly, $\;6\mid 12.\;$ However, there is no subgroup of order $6$ in $A_4$, and hence, no element $b\in A_4$ whose order is $6$. So Lagrange holds here, and the corresponding Corollary that for every element $b\in A_4$, the order of $b$ divides the order of $A_4$ is of course true, but it does not follow that if $m$ divides the order of a group, then there is an element of order $m$. With limitations, the theorem is true. Specifically, if $G$ is a finite abelian group, or if the order of $G$ is a power of a prime, or if $G$ is solvable, then the statement is true. After edit: Indeed, this is true for the finite cyclic group $G$. Suggestion about the initial part of your proof: Make sure you make explicit that, since G is cyclic, it is generated by one element, we'll call $a$. So $\langle a\rangle = G$. Then it follows that there is some element $b \in G$ such that $b = a^k$, where $0\leq k \lt n$. To show that $m$ is minimal, we can show that if there exists $0\leq s \lt m$ such that $b^s = e$, then $s = 0$. Or else you can assume, for the sake of contradiction, that there exists an $s$, $0\lt s \lt m$ such that $b^s = e$, and obtain a contradiction.
H: Validity of notation from the aspect of function description I have the following notation that should describe the nature of my function $for \forall a \in A \exists f:A \rightarrow S, A \subset N, S \subset [0,1]^n,\|S\|=n$ Can anyone tell me is the notation correct for the descriptive definition below. Function takes a natural number as an input(from set A) and outputs a vector of probabilities for each state(from set S).The size of the vector is the size of a set S. The probabilities should be all rational numbers from 0 to 1 inclusive. Thank you in advance. M. AI: You are not using $a$ at all. Also, your restriction on $A$ should be quantified. Did you mean to start with $\forall A \subset \mathbb{N} \cdots$ ? You are slightly mixing what you are defining and what you are claiming anyway. As far as I can tell your statement is that: for each such $A$, there is a subset $S \subset [0, 1]^n$ and a function $f: A \to S$ such that $\|S\| = n$. Or is it for each such $A$, and for each subset $S \subset [0, 1]^n$ such that $\|S\| = n$ there is a function $f: A \to S$? Or is it for each such $A$, for each $S \subset [0, 1]^n$ there exists a function such that $f: A \to S$ and $\|S\| = n$? Feel free to pick any of these alternatives, they translate quite literally into $\forall$ and $\exists$ statements the way I have written them.
H: Find the flaw in the attempted proof of the parallel postulate given by J. D. Gergonne The attempted proof: Given $P$ not on line $l$, line $PQ$ perpendicular to $l$ at $Q$, line $m$ perpendicular to $PQ$ at $P$ and point $A \neq P$ on $m$. Then, let $PB$ be the last ray between rays $PA$ and $PQ$ that intersects $l$, $B$ being the point of intersection. There exists a point $C$ on $l$ such that $Q * B * C$ (B is between $Q,C$). It follows that the ray $PB$ is not the last ray between rays $PA$ and $PQ$ that intersects $l$, and hence all rays between $PA$ and $PQ$ meet $l$. Thus $m$ is the only parallel to $l$ through $P$. I can't seem to find it. I'm not sure if it is wrong because he is starting off with right angles, or something else. I feel that because he starts off with right angles, and this clearly only works for right angles, he is basically stating the parallel postulate for all the rays between the two lines as reasons for those not working. By stating and using the parallel postulate, he is being inconsistent. Am I on the right track? AI: "$PB$ is not the last ray between rays $PA$ and $PQ$ that intersects $l$, and hence all rays between $PA$ and $PQ$ meet $l$" The "hence" here is asserted without proof. In fact, it does not follow. All we have shown is that given any point $B$ there is a point $C$ beyond it. For example, given any real number $b$ less than 2, there is another real number less than 2 but greater than $b$. Does it follow that all real numbers are less than 2? Similarly, given any ray through $P$ intersecting $l$ at $B$, we have shown there is another ray through $P$ intersecting $l$ at $C$. Logically, it does not follow that all rays therefore intersect $l$. There's not need for a counterexample; we just need to show the reasoning is flawed. Like many of these attempted proofs that popped up throughout history, it's just a clever way to subtly hide the assumption of the parallel postulate by cloaking it in "geometric intuition". The "hence" is an appeal to the imagination; it is hard to imagine that such a ray between $PA$ and $PQ$ can exist without intersecting $l$, but that's not proof that it doesn't exist. Edit: Ironically, people are downvoting my answer because they fell into the same trap Gergonne fell into. These would probably be the people (many of whom were otherwise skilled mathematicians) who accepted Gergonne's proof back in the day. What Gergonne proves by contradiction is that there is no last ray between $PA$ and $PQ$ that intersects $l$. That is not the same as saying there is no ray between $PA$ and $PQ$ that does not intersect l. They are not the same thing. With a little more structure you can think of it like this: he proves that the set of all the angles between $PQ$ and the rays that intersect $l$ has no greatest element. Logically it's still possible that it's bounded from above by some angle less than a right angle, so he hasn't proved what he set out to prove.
H: Permutations on four consecutive digits yield $n$ such that $2013 < n < 10000$ 2013 is the first year since the Middle Ages that consists of four consecutive digits. How many such years are there still to come after 2013 (and before the year 10000)? AI: HINT: The possible sets of four consecutive digits are $\{0,1,2,3\}$, $\{1,2,3,4\}$, $\{2,3,4,5\}$, and so on up through $\{6,7,8,9\}$. For each of these sets of four digits you should determine how many different years the set can give you after $2013$. The sets whose smallest elements are bigger than $1$ are easiest, because they can’t give you any years at or before $2013$. You’ll have to work a little harder with the first two sets.
H: Floor and ceiling function proof I have the following to prove: $$\lfloor 3x\rfloor = \lfloor x\rfloor + \left\lfloor x+\frac 13 \right\rfloor + \left\lfloor x+\frac 23 \right\rfloor $$ The definition of a floor function is: $ \lfloor x \rfloor = n \le x \lt n+1 $ So my first instinct was to do $ \lfloor 3x\rfloor=3 \lfloor x\rfloor $ and then let $n=\lfloor x\rfloor$ so basically we get $3n$. But if I were to replace both sides of the equation with $n=\lfloor x\rfloor$ I get : $$3n = n + \left\lfloor x + \frac 13 \right\rfloor + \left\lfloor x+\frac 23 \right\rfloor =3n+\frac 33=3n+1$$ But I don't know what to do with this and I'm not sure if this is a formal way of doing these types of proofs. AI: It’s not generally true that $\lfloor 3x\rfloor=3\lfloor x\rfloor$; try $x=\frac13$, for example. One very straightforward approach is to let $n=\lfloor x\rfloor$, so that $n\le x<n+1$, and consider three cases: $n\le x<n+\frac13$; $n+\frac13\le x<n+\frac23$; and $n+\frac23\le x<n+1$.
H: How find this limit $\lim_{x\to 0}\frac{1}{x^4}\left(\frac{1}{x}\left(\frac{1}{\tanh{x}}-\frac{1}{\tan{x}}\right)-\frac{2}{3}\right)=？$ Find this following limit $$\displaystyle \lim_{x\to 0}\dfrac{1}{x^4}\left(\dfrac{1}{x}\left(\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}\right)-\dfrac{2}{3}\right)=？$$ My try: since $$\tanh{x}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}$$ then $$\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}-\dfrac{1}{\tan{x}}$$ then I can't,Thank you AI: $$ \coth x=\frac{1}{x}+\frac{x}{3}-\frac{x^3}{45}+\frac{2x^5}{945}-\frac{x^7}{4725}+... \\ \cot x =\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2x^5}{945}-\frac{x^7}{4725}-... \\[0.9cm] \text{Given limit is}\\ \lim_{x \to 0} \frac{1}{x^4}\left(\frac{1}{x}\left(\frac{2x}{3}+\frac{4x^5}{945} + ...\right)-\frac{2}{3}\right) \\ = \lim_{x \to 0} \frac{1}{x^4}\left(\frac{4x^4}{945} + ...\right) \\ =\frac{4}{945} $$ Don't worry if you can't remember these power series. You can derive them in a jiffy using Bernoulli numbers. But that's for another day.
H: Is my calculation correct about a probability problem? Suppose there are $300$ tickets in the pool, where $7$ of them belong to me. $20$ tickets are randomly taken out of the pool, and are declared as "winning tickets". What is the probability that exactly 4 of the winning tickets are mine? When I tried to solve this I found $$\frac{\binom{20}{4} \left(7 \times 6 \times 5 \times 4 \prod _{j=1}^{16} (-j+293+1)\right)}{\prod _{i=1}^{20} (-i+300+1)} \approx 0.000433665 $$ Is this the correct probability? Thanks. AI: There are $\binom{300}{20}$ ways to make the choice of winning tickets. There are $\binom74\binom{293}{16}$ ways to choose them so that exactly $4$ of the winning tickets are among your $7$ tickets. The desired probability is therefore $$\frac{\binom74\binom{293}{16}}{\binom{300}{20}}\approx0.000433665\;.$$ You did it the hard way, at least from my point of view, but you got there okay.
H: Uncountably many non-homeomorphic compact subsets of the circle As the title says, the question is whether there are uncountably many non-homeomorphic compact subsets of the unit circle. I'm assuming this is true, but I wouldn't mind an elegant proof. AI: For each countable ordinal $\alpha$ let $X_\alpha=\omega^\alpha+1$, where the exponentiation is ordinal exponentiation, and $X_\alpha$ is the space of ordinals less than or equal to $\omega^\alpha$ with the order topology. This is a compact space, and $X_\alpha$ and $X_\beta$ have different Cantor-Bendixson ranks when $\alpha\ne\beta$, so $\{X_\alpha:\alpha<\omega_1\}$ is an uncountable family of pairwise non-homeomorphic countable compact Hausdorff spaces. It’s well-known that every countable ordinal space embeds in $\Bbb R$, hence in $\left[0,\frac12\right)$, and hence in the circle. Added: Let $C$ be the middle-thirds Cantor set, and let $\{p_n:n\in\omega\}$ be an enumeration of the left endpoints of the deleted intervals. If $A\subseteq[\omega_1]^\omega$, let $A=\{\alpha_n:n\in\omega\}$ be the increasing enumeration, and attach a copy of $X_{\alpha_n}$ to $C$ at $p_n$ by identifying $\omega^{\alpha_n}$ with $p_n$; the rest of $X_{\alpha_n}$ should lie in the interval whose left endpoint is $p_n$. Call the resulting space $C_A$. The spaces $C_A$ for $A\in[\omega_1]^\omega$ are pairwise non-homeomorphic and compact, and there are $2^\omega$ of them.
H: What is the group of endomorphisms of $\mathbb{Q}/\mathbb{Z}$ As the question says, I'm trying to work out what $End_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z})$ is. These are just group homomorphisms. But so far all I can see is that its probably enough to see where elements of the form $1/n$ map to, but some hints would be helpful Thank you AI: Decompose $\mathbb{Q}/\mathbb{Z}$ into primary components: define $t_p(G)=\{g\in G: p^ng=0\text{ for some }n>0\}$ where $p$ is a prime and $G$ is an additive abelian group; then $G=\bigoplus_p t_p(G)$. What can you say about $f\colon t_p(G)\to t_q(G)$ when $p\ne q$? What's $\operatorname{End}(t_p(\mathbb{Q}/\mathbb{Z}))$?
H: Showing $\gcd(2^m-1,2^n+1)=1$ A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement If $m$ is odd then $\gcd(2^m-1,2^n+1)=1$. It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory. AI: If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p \neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$. Alternatively, we can use $$\gcd (2^t-1, 2^u-1) = 2^{\gcd (t,u)}-1\tag{1}$$ to conclude $$\gcd (2^m-1, 2^{2n}-1) = 2^{\gcd(m,2n)}-1.$$ But since $m$ is odd, we have $\gcd (m,2n) = \gcd(m,n)$, and hence $$2^{\gcd(m,2n)}-1 \mid 2^n-1,$$ which, since $$\gcd(2^n-1,2^n+1) = \gcd(2^n-1,2) \mid 2$$ and $2^{\gcd(m,2n)}-1$ is odd, implies $\gcd (2^{\gcd(m,2n)}-1,2^n+1) = 1$ and hence $\gcd(2^m-1,2^n+1) = 1$. To see $(1)$, write $u = q\cdot t + r$ with $0 \leqslant r < t$, and $$2^u-1 = 2^r\left(2^{q\cdot t}-1\right) + \left(2^r-1\right),$$ which, since $2^t-1 \mid (2^t)^q-1$, yields $$\gcd(2^t-1,2^u-1) = \gcd(2^t-1,2^r-1),$$ and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
H: $X_{n}$ converges to $X$ in distribution iff $E\{f(X_{n})\} \to E\{f(X)\}$ for all bounded $ f \in C^{\infty}$. Let $(X_{n})_{n\geq 1}$, $X$ be $\mathbb{R}$-valued random variables. Show that $X_{n}$ converges to $X$ in distribution iff $E\{f(X_{n})\}$ converges to $ E\{f(X)\}$ for all bounded $C^{\infty}$ functions $f$. I have sufficiency: if $f$ is in $C^{\infty, b}$, and if $\lim_{n \to \infty} E\{f(X_{n})\}=E\{f(X)\}$, then by a theorem in the text, we have $X_{n}\to^{\mathcal{d}} X$. For necessity, I am thinking that I should use the Monotone Class Theorem (Let $\mathcal{M}$ be a class of bounded functions mapping $\Omega$ into $\mathbb{R}$. Suppose $\mathcal{M}$ is closed under multiplication . Let $\mathcal{A}=\sigma(\mathcal{M})$. Let $\mathcal{H}$ be a vector space of functions with $\mathcal{H}$ containing $\mathcal{M}$. Suppose $\mathcal{H}$ contains the constant functions and is such that whenever $(f_{n})_{n\geq 1}$ is a sequence in $\mathcal{H}$ such that $0\leq f_{1}\leq f_{2} \leq \cdots$, then if $f=\lim_{n\to \infty}f_{n}$ is bounded, then $f$ is in $\mathcal{H}$. Then $\mathcal{H}$ contains all bounded, $\mathcal{A}$-measurable functions.) So, we want $\mathcal{H}$ to be $C^{b}$, and we want $f_{n}$ to be played by $C^{\infty,b}$. I am not sure how to show this mathematically rigorously, however. Also, as a hint, we were told to use the results of the two previous exercises: Let $(X_{n})_{n\geq 1}$, $X$, $Y$ by $\mathbb{R}$-valued random variables, all on the same space, and suppose that $X_{n}+\sigma Y$ converges in distribution to $X+\sigma Y$ for each fixed $\sigma >0$. Show that $X_{n}$ converges to $X$ in distribution. Let $X, Y$ be independent r.v.'s on the same space with values in $\mathbb{R}$ and assume $Y$ is $N(0,1)$. Let $f$ be bounded continuous. Show that $E\{f(X+\sigma Y)\}=E\{f_{\sigma}(X)\}$ where $f_{\sigma}(x)=\frac{1}{\sigma \sqrt{2\pi}}\int_{-\infty}^{\infty}f(z)\exp(-\frac{1}{2}\|z-x\|^{2}/\sigma^{2})dz$. Show that $f_{\sigma}$ is bounded and $C^{\infty}$. I am not sure how these are supposed to help, though. Especially the part about choosing a variable that is $N(0,1)$ and independent to $X$. AI: Using the assumption, we have for each positive $\sigma$ that $$\lim_{n\to \infty}\mathbb E[f_\sigma(X_n)]=\mathbb E[f_\sigma(X)].$$ By the result of 2., the RHS is $\mathbb E[f(X+\sigma Y)]$. If we enlarge the probability space, we can construct $Y$ in order to be independent of the sequence $(X_n)_{n\geqslant 1}$, so still using the result, the LHS is $\lim_{n\to \infty}\mathbb E[f(X_n+\sigma Y)]$. We conclude by 1.
H: Non trivial group homomorphism from $G$ to $H$ Actual Question is to check if there is : Non trivial group homomorphism $\eta : S_3 \rightarrow \mathbb{Z}/3\mathbb{Z}$. What I have tried so far is : I take $(1 2)\in S_3$, this has order $2$. So, order of $\eta(1 2)$ should divide $2$ but then, I have no element of order $2$ in $\mathbb{Z}/3\mathbb{Z}$ which would give me only possibility of $\eta (1 2)$ as $\bar{0}$ with similar reasoning I would see only possibility for images of $(2 3)$ and $(1 3)$ is $\bar{0}$. But then, $\{(1 2), (2 3), (1 3)\}$ generates $S_3$ So, if each generator goes to $\bar{0}$, so should be the images of all elements i.e., all elements must be mapped to $\bar{0}\in \mathbb{Z}/3\mathbb{Z}$. So, I do not have non trivial homomorphism $\eta : S_3 \rightarrow \mathbb{Z}/3\mathbb{Z}$. I like this idea and i would like to see for generalization of this idea. i.e., Can I say something about non existence of a non trivial homomorphism from $G$ to $H$ if i know orders of all elements of $G$ and $H$. I would be thankful if some one can give me an exciting example (though it is subjective) which is not very trivial which follows same idea as above question. Thank you. AI: Since $S_n$ is generated by all transpositions (hence elements of order 2) you can extend your argument to homomorphisms from $S_n$ to any group of odd order.
H: Find the area of triangle There is a square $ABCD$ of side $a$, points $E,F$ lies at centre of respectively $AB,CD$. Line $AE$ intersect with $DF$ at $G$ and $BD$ at $H$. Find area of $DHG$. I don't know why I can't add a comment but thanks for hint, I have already known how to do it AI: Hint: $\text{area of }\triangle DHG=\text{area of }\triangle DAB-\text{area of }\triangle AHB-\text{area of }\triangle ADG$.............
H: Pigeonhole problem - salvaging my solution A student is solving combinatorics problems. Each day he solves at least one problem. He solves no more than 500 problems a year. Prove that there is an interval of days in which he solves 229 problems. This is my approach.Lets consider partial sums. A(x) is a sum of problems after x days. Now what is the remainder of A(x) and 229? A(x) changes as days go by, there are only 229 possible remainders and 365 days - so there must be two days x and y, where A(x) and A(y) have the same remainder. A(x) and A(y) represent sums of problems of two overlapping periods. A(x) is the larger period (you can assume this). A(x)-A(y) is divisible by 229 and is positive. A(x)-A(y) also represents a sum of problems of a certain period (since the two periods are overlapping, just take the larger period and disregard the part of the larger period that is the smaller period. However, I can only show that A(x)-A(y) is divisible by 229, I didnt show that A(x)-A(y) = 229. Is there any way how I can salvage my approach? AI: The only other value for $A(x)-A(y)$ is $458$, because we know it is less than $500$. Repeat your argument for the first $310$ days of the year, as he must have done less than $458$ problems in that span.
H: How are these definite integrals equivalent? $ \int_0^\infty B'(x)S'(t-x) dx = \int_{-\infty}^t B'(t-x)S'(x)dx $ I have been told that the following integrals are equivalent but I cannot figure out how: $ \int_0^\infty B'(x)S'(t-x) dx = \int_{-\infty}^t B'(t-x)S'(x)dx $ Are there some rules on changing limits that I can apply to transform the first one to the second one? AI: when we substitute $ y = t -x \Rightarrow x = t-y $ we get $$ \int_0^{\infty} B'(x)S'(t-x) \ dx = \int_{t}^{-\infty} B'(t-y)S'(t-(t-y)) d(t-y) = -\int_t^{-\infty} B'(t-y)S'(y) \ dy = \int_{-\infty}^t B'(t-y)S'(y) \ dy = \int_{-\infty}^t B'(t-x)S'(x) \ dx $$ Notes : $$ \int_a^b f(x) \ dx = \int_a^b f(y) \ dy $$ $$ -\int_a^b f(x) \ dx = \int_b^a f(y) \ dy $$
H: Defining the Greatest Common Divisor using Symbolic Notation I am trying to write the definition of greatest common divisor using symbolic notation. Here is my current attempt: $d = gcd(m,n) \Leftrightarrow d \in Z \wedge max(d \| m \wedge d \| n)$ Any help or hints are greatly appreciated! Thanks! AI: Let's try to say what you are saying on the right-hand side: $$d = \gcd(m, n) \iff \Big((d\in \mathbb Z \land d\mid m \land d\mid n) \land \forall d' \in \mathbb Z\left((d'\mid m \land d'\mid n) \rightarrow d' \leq d\right)\Big)$$
H: Calculation of $σ_u σ_u$ and $σ_u σ_v$ Accourding to the info which I posted, how can I calculate $σ_u σ_u=\vert\vert σ_u\vert\vert^2$ and $σ_u σ_v$ I am stuck with there. Please show me. Thanks. AI: For example: $$\sigma_u\cdot\sigma_v:=2\cosh u\cosh v\sinh u\sinh v$$ If you really meant the dot product you jsut take the sum of coordinatewise product of the vectors' entries.
H: Prove using mathematical induction that for every positive integer n, $\sum_{i=1}^n ( i * 2^i ) = (n-1) 2^{n+1} + 2 $ Prove using mathematical induction that for every positive integer n, $$\sum_{i=1}^ni2^i=(n-1) 2^{n+1} + 2$$ There is what i did so far : AI: Your solution is good, except the final form should be presented as the $k+1$ step, so it should look like $$\sum_{i=1}^{k+1}i\cdot 2^i=\dots=k\cdot 2^{k+2}+2$$ (where the $\dots$ represent the necessary arithmetic in between.) This will make it clear that the final form is definitely the increment from the $k$th step to the $(k+1)$th step.
H: Finding Laurent series in annulus $1<\|z\|<2$ I want to find the Laurent series for $f(z)=\dfrac{1}{(z-1)(z-2)}$ inside $1<\|z\|<2$. I can apply the formula here to get $$f(z)=\sum_{k=0}^\infty a_kz^k+\sum_{k=1}^\infty b_kz^k$$ where $$a_k=\dfrac{1}{2\pi i}\int_{\|z\|=2}\dfrac{f(z)}{z^{k+1}}dz$$ and $$b_k=\dfrac{1}{2\pi i}\int_{\|z\|=1}f(z){z^{k-1}}dz$$ But how can I calculate those values? AI: Far more simply, observe that $$\frac{1}{(z-1)(z-2)}=\frac{(z-1)-(z-2)}{(z-1)(z-2)}=\frac1{z-2}-\frac1{z-1}.\tag{$\star$}$$ Note that for $z\neq0,2,$ we can write $$\frac1{z-2}=-\frac1{2-z}=-\frac12\cdot\cfrac1{1-\left(\frac{z}{2}\right)}$$ and $$\frac1{z-2}=\frac1{z}\cdot\cfrac1{1-\left(\frac{2}{z}\right)}.$$ Now, one of $\frac1{1-\left(\frac{z}{2}\right)}$ and $\frac1{1-\left(\frac{2}{z}\right)}$ can be expanded as a geometric series in the disk $\|z\|<2,$ and the other can be expanded as a geometric series in the annulus $\|z\|>2$. That is, we will use the fact that $$\frac1{1-w}=\sum_{k=0}^\infty w^k$$ whenever $\|w\|<1$. You should figure out which one works in the disk $\|z\|<2,$ as that is the relevant one. Likewise, we can rewrite $\frac1{z-1}$ in two similar forms, one of which is expandable in $\|z\|<1$ and one of which is expandable in $\|z\|>1.$ You should figure out which one works in the annulus $\|z\|>1,$ as that is the relevant one. Using $(\star)$ with the expansions you found above will give you the desired Laurent series.
H: Markov Chain: starting at $i$ reaching $N$ before $0$ Starting at some state $i$, we have probability of going $P_{i,i+1} = p$ and probability $P_{i,i-1} = 1-p$ what is the probability I reach N before I reach zero? Can I convert this to a gambler's ruin problem where I start with i in bank and reach N before I reach zero? I think the idea is to use the stochastic difference equation: $f_{i+1} - f_i = \frac{q}{p}(f_i - f_{i-1})$, where $f_0 = 0$ and $f_N = 1$ is the absorbing state and this gives rise to $f_i = \frac{1-(q/p)^i}{1-(q/p)^N}$ This gives me the probability of reaching N and not reaching zero, but does it also mean the probability of reaching N before 0? If not, how can I find that? AI: Your formula is correct, assuming $q=1-p$; here's argument in biological terms. This Markov chain is a birth-death process and the concept you are looking for is the fixation probability (called the absorption probability in general). Formula (1) on this page gives the probability of absorbing at $N$ starting at state $i$, where $\gamma_k = \frac{1-p}{p}$ for your process. Formula (1) simplifies to your formula by using also the formula for a finite geometric series -- the ones cancel because the sums start at 1. The fixation probability gives the probability of the process absorbing in the state $N$ when starting from state $i$; one minus this probability gives the fixation probability for state $i=0$ (since there are no other absorbing states). Your second question, the probability of reaching $N$ before $0$, is the same since $0$ and $N$ are both (and the only) absorbing states.
H: How many 5 digits numbers are there, whose digits sum to 22? How many 5 digits numbers are there, whose digits sum to 22? Of course the first digit has to be larger than 0. AI: The objects you are counting may be placed into bijection with solutions to $$x_1+x_2+x_3+x_4+x_5=22$$ such that $0\le x_i\le 9$ and also $x_1\ge 1$. Via the substitution $x_1'=x_1-1$, we may instead solve $$x_1'+x_2+x_3+x_4+x_5=21$$ such that the variables are nonnegative integers, and also $x_1\le 8, x_i\le 9$ ($2\le i\le 5$). Without the upper bound restriction, using stars and bars, there are ${26 \choose 22}$ solutions. Now we must consider the various upper bounds, using inclusion-exclusion. Let $A_1$ denote the set of solutions where $x_1'>8$, and $A_i$ denote the set of solutions where $x_i\ge 10$. By considering the substitution $x_1''=x_1-9\ge 0$, we have $$x_1''+x_2+x_3+x_4+x_5=12$$ Again using the stars-and-bars approach, we have $\|A_1\|={17\choose 13}$. If instead we want $x_2\ge 10$, we use the substitution $x_2'=x_2-10\ge 0$ and $$x_1'+x_2'+x_3+x_4+x_5=11$$ and so $\|A_2\|={16\choose 12}$. By symmetry, $\|A_3\|=\|A_4\|=\|A_5\|$. To complete the problem, you also need to compute all the various $\|A_1\cap A_2\|$ and $\|A_2\cap A_3\|$, and then combine all the data using the inclusion-exclusion principle, which I leave for you as an exercise.
H: Root of $f(z)$ inside $\|z\|<1$ Let $c\in\mathbb{R}$. A non-constant function $f(z)$ is holomorphic in $\|z\|<2$. Suppose $\|f(z)\|=c$ for all $\|z\|=1$. Show that $f(z)$ must have a root in $\|z\|<1$. I'm thinking about the maximum principle, which says $f(z)$ cannot attain a maximum inside $\|z\|<1$. But that still doesn't yield a root. Also, Rouche's theorem might be applicable if there's another function $g(z)$ to be used. AI: $f$ is non-constant. So you must have $\lvert f(z)\rvert < c$ for all $z$ in the unit disk. If $f$ had no zero in the unit disk, what would the maximum principle have to say about $$g(z) = \frac{1}{f(z)}\; ?$$
H: How many possibilities in tinyurl Looking at tinyurl, there is anywhere from 1 digit to 7 digits of I believe 36 choices (lowercase letters a to z and digits 0 to 9) How do I calculate mathmatically the number of permutations of the string with 1 to 7 digits and 36 characters? thanks, Dean AI: $$36^1+36^2+36^3+36^4+36^5+36^6+36^7=80,603,140,212$$
H: What is the expectation of the following random variable Let $X_1,X_2,X_3,\ldots $ be an i.i.d. sequence of uniform random variables over $[0,1]$. Define $$N= \min \{n \geq 1:X_1+\cdots+X_n>1\}$$ Find $P\{N>n\}$ and compute $E[N]$. AI: $S_n=X_1+...+X_n \\ P(S_n \leq t),t<n=\text{The volume between the axises and the hyperplane X_1+X_2+...+X_n=t which is } \frac{t^n}{n!} \\ P(S_n \leq t)=\frac{t^n}{n!},t<n\\ f_{S_n}(t)=\frac{t^{n-1}}{(n-1)!},t<n \\ P(N=n)=\int\limits_0^1f_{S_{n-1}}(t)P(X_n>1-t)dt=\int\limits_0^1\frac{t^{n-2}}{(n-2)!}t dt=\frac{1}{n(n-2)!},n\geq2$
H: Meaning of $x^m + y^n = z^r$ (mod $p_1$) I'm trying to understand how to find a counter example to the Beal Conjecture. One site (here) says that, According to The Prime Pages, the largest primes less than $2^{32}$ are $p_1 = 2^{32}-5$ and $p_2 = 2^{32}-17$. If $x^m + y^n = z^r$ (mod $p_1$) and $x^m + y^n = z^r$ (mod $p_2$). What does modulus $p_1$ and $p_2$ mean? What is p? And what would the modulus be take of? Would it be $x^m + y^n$? Thank you for your help. AI: Programmers think about mod as a binary operation, where you have inputs a and n and the output is Mod(a,n), which is the remainder of a upon division by n. This is not how mathematicians think about modular arithmetic. (The programmer's mod operation is just called the "division algorithm" in number theory.) We think about 'mod' as something called a congruence relation. Numbers $a$ and $b$ are "congruent modulo $n$" if $a$ and $b$ differ by a multiple of $n$. For example, "$5\equiv11$ mod $3$" because the difference between $5$ and $11$ is $6$ which is divisible by $3$. See http://en.wikipedia.org/wiki/Modular_arithmetic for more detailed information.
H: sequences-show that $\inf\{x_{n}: n\in \mathbf{ N}\}>0$ Knowing that $x_{n}>0$ for each $n \in\mathbf{N}$ and $x_{n}\rightarrow x>0$. Let $B$={$x_{n}:n \in\mathbf{N}$}. How could I show that $\inf B>0$?? AI: Look at the definition of limit and fix $\varepsilon = x/4$. Then there is $n$ such that for all $m \geq n$ $\|x_{m}-x\|\leq x/4$. For $m\geq n$ $x_{m} \geq 3/4x$. So the inf restricted to $m \geq n$ is at least $3/4x$. Now look at $m<n$. Since it is a finite set of real numbers, it has a minimum $y$, which is >0 by your assumptions. So the global inf is $\geq \mathrm{min}\left( 3/4x,y\right)>0$.
H: Solving one equation for two unknowns Is there a theorem that states that you need $N$ equations to solve for $N$ unknowns? If I had the following equality, $$x^25^y = 10125$$ isn't it possible to deduce that $x = 9$ and $y = 3$ simply by looking at prime factors? Are there multiple solutions when there are $N$ unknowns and less than $N$ equations? AI: If your system of linear equations is regular (which is a special case of having $n$ equations with $n$ unknown), it is guaranteed to have a unique solution. Your equation above is non-linear, and has an additional constraint that you want only non-negative integer solutions. Of course, you cannot deduce much about such a case only from the number of equations and unknowns. For example $x^2y^7 = 2$ will have no integer solutions, while $x^2y^2 = 2^{32}$ will have lots of integer solutions.
H: Let $\varphi $:$G\to H$ be an onto group homomorphism. Show that if $K \unlhd G$, then $\varphi(K)\unlhd H$. Let $\varphi$: $G\to H$ be an onto group homomorphism. Show that if $K \unlhd G$, then $\varphi(K)\unlhd H$. This problem was in my abstract book, but does not explain why it is true. It is provided as an example and I was wondering if someone could help by providing the subsequent proof of this statement. AI: $\varphi(K)$ is normal in $H$ if and only if $h\varphi(K)h^{-1}=\varphi(K)$ for all $h\in H$. Note that $\varphi(g)\varphi(K)\varphi(g)^{-1} = \varphi(gKg^{-1})=\varphi(K)$ for all $g\in G$. So how can you use the remaining assumptions to "pull" the $h$ inside of $\varphi$?
H: Stability of Analytic Continuation Let $f(z)$ be an analytic function in an open set $U\subset\Bbb{C}$. Recall that an analytic continuation of $f$ is a pair $(F,V)$ such that $U\subset V\subset\Bbb{C}$, $F$ is analytic on $V$, and $F(z)=f(z)$ for all $z\in U$. My question is, how stable is this process? If $\\|f-g\\|$ is small, are we guaranteed $\\|F-G\\|$ small in any reasonable sense? If not, are there easy counterexamples? If the answer depends on the choice of norm, I would find that interesting as well. References gladly accepted in lieu of obvious arguments. Thanks! AI: Not a rigorous answer, but I believe the answer is probably no. Look at the functions $$f(z) = z + z^2 + z^4 + z^8 + \cdots + z^{32}$$ and $$g(z) = z + z^2 + z^4 + z^8 + z^{16} + z^{32} + \cdots$$ inside the open set $\|z\| < 1/2$. Depending on your definition of norm, I suspect that these would be "close", but if you analytically continue them to something like the unit disc, they will not be close by any means, since $f$ is a well-behaved polynomial, but $g$ will be very badly behaved indeed on the unit circle as seen by the results quoted in the second part of my answer here.
H: $C_c^{\infty}(\mathbb R^n)$ is dense in $W^{k,p}(\mathbb R^n)$ As the title say, I want to prove that $C_c^{\infty}(\mathbb R^n)$ is dense in $W^{k,p}(\mathbb R^n)$ i.e. $\displaystyle{ W^{k,p} (\mathbb R^n) = W_0^{k,p}(\mathbb R^n) \quad (\star)}$. In a book I found that in order to prove it, we need the following claim: Claim: Let $\displaystyle{\zeta \in C^{\infty} ([0, \infty)) }$ such that $\zeta (t) =1 $ for $ t \leq 1 $ and $\displaystyle{ \zeta (t) =0 }$ for $ t \geq 2$. For $R>0$ and $x \in \mathbb R^n$ define $\displaystyle{ \eta_R (x) = \zeta \left( \frac{\|x\|}{R} \right)}$. If $ \displaystyle{ u \in W^{k,p} ( \mathbb R^n) }$ with $1 \leq p < \infty $ then, $\displaystyle{ \eta_R u \to u }$ in $W^{k,p}(\mathbb R^n) $ as $ R \to \infty$. If I prove the claim then the $(\star)$ follows using mollifiers. But unfortuently I have stuck, with the proof of the claim. Any ideas? AI: It may help to consider first the case $n=1$ and $k=2$. The fact that $\eta_Ru\to u$ in $L^p$ follows from a monotone convergence argument. Indeed, since $\eta_R(x)=1$ if $\|x\|\leqslant R$, $$\int_\mathbb R\|\eta_R(x)u(x)-u(x)\|^p\mathrm dx=\int_{\{\|x\|\gt R\}}\|u(x)\|^p\mathrm dx.$$ We can use the product rule for derivatives: we get $(\eta_R u)'=\eta_R u'+\eta'_Ru$. The first terms goes to $u'$ by a similar argument as before. For the second one, there is a term $\frac 1R$ which appears when we take the derivative. Since the derivative of $\eta$ is bounded, the light version of the claim follows. For the general case, we have to consider Leibnitz rule. It's the same idea, altought it's more technical. We shall need that each derivative of $\eta$ is bounded.
H: The range of continuous functions and the Intermediate value theorem Let $a,b\in \mathbb{R}$, $a<b$ and let $f$ be a continuous real valued function on $[a,b].$ Prove that if $f$ is one-to-one then $f([a,b])$ is either $[f(a),f(b)]$ or $[f(b),f(a)].$ Suppose $f(a)<f(b)$ then by the Intermediate Value Theorem (IVT) we have if $x\in(a,b)$ then $f(x)\in[f(a),f(b)].$ Arguing by contradiction suppose $f(x)<f(a)$ or $f(x)>f(b).$ If $f(x)<f(a)$ then by IVT $f([x,b])$ contains all points between $f(x)$ and $f(b)$ and contains $f(a)$ contradiction that $f$ is one-to-one. Why does that contradict the assumption that $f$ is one-to-one? AI: If $f(a) \in f([x,b])$, that means there is $y \in [x,b]$ so that $f(y) = f(a)$. Do you see why this cannot happen (if $f$ is one-to-one)?
H: If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue? If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue? I don't really know where to start with this one. I know that $p(0)=det(0I_{n}-A)=det(-A)=(-1)^{n}det(A)$, thus if both $p(0)$ and $det(0) = 0$ then $0$ is an eigenvalue of A and A is not invertible. If neither are $0$, then $0$ is not an eigenvalue of A and thus A is invertible. I'm unsure of how to use this information to prove $\vec{v}$ is also an eigenvector for $A^{-1}$ and how to find a corresponding eigenvalue. AI: If you have $v\neq 0$ such that $A v = \lambda v$, and $A$ is invertible, then $v = \lambda A^{-1} v$. What does that tell you about $\lambda$? $\lambda$ must be non-zero, otherwise this would mean $v=0$, a contradiction. How do you use the above to find an eigenvalue corresponding to $v$? Since $\lambda \neq 0$, we have $A^{-1} v = \frac{1}{\lambda} v$, hence the eigenvalue is $\frac{1}{\lambda}$.
H: Need a hint for sequence convergence homework Hello and thank you for spending time to help me in advance! The following exercise/homework problem has been giving me a difficult time and I am wondering if there is something simple that I am just not seeing at the moment. I thought I could find the maximum sequence by just assuming that none of the $d$s are equal to $0$. Thus $a_n$ becomes $a_n= 2^{-1}+...+2^{-n}$. Then I rewrote this as the sum from $k=1$ to $k=n$ of $2^{-k}$. I wasn't able to figure out how to continue, though. AI: You should probably note that $0 \leq a_{n} = d_{1}\cdot 2^{-1} + ... + d_{n}\cdot 2^{-n} \leq 1\cdot2^{-1} + ... + 1\cdot 2^{-n} \leq \sum_{i=1}^{\infty} \frac{1}{2^{i}} \leq 1$. Note that since $[0,1]$ is closed and $0 \leq a_{n} \leq 1\; \forall\; n \in \mathbb{Z}$, it follows that the sequence converges to some point in $[0,1]$.
H: Prove that $\vec{v}$ is also an eigenvector for $A^{k}$(k = a positive integer). What is the corresponding eigenvalue? Prove that $\vec{v}$ is also an eigenvector for $A^{k}$(k = a positive integer). What is the corresponding eigenvalue? What I have started with is, $A=(CDC^{-1})$ which can be used to prove $A^{2}=(CDC^{-1})(CDC^{-1}) =C(DC^{-1}C)DC^{-1} =CD^{2}C^{-1}$ Thus, $A^{k}=CD^{k}C^{-1}$ I think this will help me find my proof, but I am not sure. AI: Try looking at $A^{k-1}(Av)$. Do this process inductively.
H: Proving if a function $f$ is differentiable and $f'(x)\ne0$ at all $x$, then it is one-to-one Here's what the problem reads: Suppose that the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $f'(x)\neq 0$ for any $x \in (a,b)$. Prove that $f$ must be one-to-one. This looked easy at first, but I'm having a lot of trouble cracking the proof just by examining the definitions. If someone could nudge me in the right direction, that would be extremely helpful. Thanks! AI: Hint: Assume that $f$ takes the same value in two points, and apply the mean value theorem.
H: Expected value - random sets cardinality I've got a set $\|A\|=n$ and two random subsets $B,C\subseteq A$ and $\|B\|=\|C\|=k$. What is the expected value of $\|B\cap C\|$? My approach: I consider $B$ given and try to fill $C$. The probability of picking no common terms ($l=0$) is $\dfrac{n-k\choose k}{n\choose k}$, because there are $n-k\choose k$ ways to pick one of the terms that hasn't been picked already and $n\choose k$ possible scenarios. For $l=1$, I pick one term from $B$ into $C$ and calculate the probability of the remaining $C$-terms not being in $B$. There are $k\choose1$ ways for choosing the common term, and $k-1$ remaining terms: ${k\choose1}\dfrac{n-k\choose k-1}{n\choose k-1}$. So, the expected value is expected to be $$0\cdot\dfrac{{{n-k}\choose k}}{{n\choose k}}+1\cdot {k\choose1}\dfrac{{{n-k}\choose {k-1}}}{{n\choose {k-1}}}+\ldots+k{k\choose k}\dfrac{{{n-k}\choose 0}}{{n\choose0}}$$But the last term is apparently equal to $k$ and other terms are non-negative, so the expected value is at least $k$, and that seems kind of unbelievable. Where did I make a mistake? AI: There’s no harm in taking $B$ to be a fixed $k$-element subset of $A$ and picking $C$ at random. To choose $C$ so that $\|B\cap C\|=j$, we must choose $j$ elements of $B$ and $k-j$ of $A\setminus B$; this can be done in $\binom{k}j\binom{n-k}{k-j}$ ways. There are altogether $\binom{n}k$ ways to choose $C$, so the desired expectation is $$\begin{align} \binom{n}k^{-1}\sum_{j=0}^kj\binom{k}j\binom{n-k}{k-j}&=k\binom{n}k^{-1}\sum_{j=0}^k\binom{k-1}{j-1}\binom{n-k}{k-j}\\\\ &=k\binom{n}k^{-1}\sum_{j=1}^k\binom{k-1}{j-1}\binom{n-k}{k-j}\\\\ &=k\binom{n}k^{-1}\binom{n-1}{k-1}\\\\ &=\frac{k\binom{n-1}{k-1}}{\frac{n}k\cdot\binom{n-1}{k-1}}\\\\ &=\frac{k^2}n\;. \end{align}$$ The step that gets rid of the summation uses Vandermonde’s identity. As a quick sanity check, note that if $k=n$ this gives an expected value of $n$, which is certainly correct, as in that case $B=C=A$, and if $k=1$, it give an expected value of $\frac1n$, which can also be verified directly, since in that case the expected value is evidently $$\frac1n\cdot1+\frac{n-1}n\cdot0=\frac1n\;.$$
H: Question about flipping terms in matrix multiplication in proving that $h(N_n(\mu , K))=\frac{1}{2}\log(2 \pi n)^n \|K\|$ So in my book, it is written: Let $X_1,X_2,...,X_n$ have a multivariate normal distribution with mean $\mu$ and covariance matrix $K$ and $\textbf{X}=(X_1,X_2,...,X_n)$ The above isn't really relevant to my question all that needs to be known is that we're multiplying matrices and we switch the terms of the multiplication. $A=(\textbf{x}-\mu )^{T}K^{-1}(\textbf{x}- \mu)$ $\space \space \space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space $$(1)$ $A$$=\sum_{i,j}(X_i-\mu_i)(K^{-1}_{ij})(X_j-\mu_j)$ $\space \space \space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space $$(2)$ $A$$=\sum_{i,j}(X_i-\mu_i)(X_j-\mu_j)(K^{-1}_{ij})$$\space \space \space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$$(3)$ I don't understand how we went from $(2)$ to $(3)$. Last time I checked, when we multiplied two matrices the number of columns of the left term had to be equal to the number of rows of the right term. This rule doesn't seem to be respected when going from $(2)$ to $(3)$ how is this possible? Thanks for the help! This is from the book by Cover and Thomas on Information theory. It corresponds to equations $(8.37)$ and $(8.38)$ in the book. AI: When you go from 1 to 2 you sort of leave the matrix world. Note the subscripts on $K$. It's the element in row $i$ column $j$ of the inverted covariance matrix. So $K_{ij}$ is simply a scalar.
H: Is it true that $E[X^2]-E[Y^2] = 0?$ Suppose that $X$ and $Y$ are identically distributed and not necessarily independent. Then clearly $E[X]=E[Y]$. But is it also the case that $E[X^2]=E[Y^2]$? Can you think of a counter-example? If not, can you give a quick proof? Thanks AI: If $X$ and $Y$ have the same distribution, they have the same density functions $f_X$ and $f_Y$. So $$\int_{-\infty}^{\infty} t^2 f_X(t)\ dt=\int_{-\infty}^{\infty} t^2 f_Y(t)\ dt$$ (If your probability measures are't absolutely continuous with respect to Lebesgue measure, just replace $f_X(t)\ dt$ with the appropriate measure.)
H: Uniqueness of minimal elements in a totally ordered set Let $R$ be a total order on set $S$. Prove that if $S$ has a minimal element, than the minimum element is unique. I have difficulties with proofs. I know any graph of a total order is a straight line, which clearly has a minimal element. How doI tell when something is mathematically proven? EDIT: this is how I would prove it Let $x, y \in S$ such that x and y are minimal. Since S is a total order $x$ and $y$ are comparable and either case 1: one comes after the other and therefore one isn't the minimal case 2: both are the same and $x=y$ I know this isn't a mathy argument and could use help making it more presentable. AI: If $S$ is a partially ordered set in which is $s$ is a minimal element, then $\forall x\in S(x\leq s\implies x=s)$. Since $R$ is a total order and $s$ is minimal, it follows that $s$ is in fact a minimum. Now to answer your question. Hint: Suppose $s'\in S$ is also a minimum. Relate it do $s$ and try to conclude that $s=s'$. What you wrote is good enough for an answer, but you're asking for a 'more symbolic' answer, so here it is. Full answer: As stated above, since $R$ is a partial order with a minimum element $s\in S$, then $\forall x\in S(x\leq s\implies x=s)$. But $R$ is a total order, thus $\forall x\in S(x\leq s\lor s\leq x)$, therefore, since $s$ is minimal $\forall x\in S(x=s\lor s\leq x)$, i.e., $\forall x\in S(s\leq x)$, that is, $s$ is a minimum of $S$ with respect to the order $R$. Now suppose $s'$ is also a minimum of $S$ with respect to $R$, then, by definition of $s'$ being a minimum, it holds that $s'\leq s$. On the other hand, $s$ is a minimum too, so $s\leq s'$. From $s'\leq s\land s\leq s'$ it follows that $s=s'$. This proves that all minimum elements are the same, that is, there is a unique minimum element. It's probably good to note that the fact that $R$ is a total order is only necessary to prove that a minimum exists. On any poset, if there is a minimum, the reasoning above proves it is unique.
H: Differential Equation Separation I having trouble solving this diffeq. $$\frac{\text{d}P}{\text{d}t} = \frac{(r(t) - B)}{z} \cdot P(t) + c\cdot w$$ ,where $c\cdot w$ is a constant. Normally I would just separate but I do not think I can do that here. Any ideas? AI: The normal method for solving $$\frac{dP}{dt} = g(t)P(t) + k$$ where $k$ is a constant and $g(t)$ is any function of $t$, is to put all $P(t)$ on the same side and multiply as such: \begin{eqnarray} \frac{dP}{dt} - g(t)P(t) &=& K\\ e^{\int^t-g(s)ds}\frac{dP}{dt} - g(t)e^{\int^t-g(s)ds}P(t) &=& Ke^{\int^t-g(s)ds}\\ \frac{d}{dt}\Big(e^{-\int^t g(s)ds}P(t)\Big) &=& Ke^{\int^t-g(s)ds} \end{eqnarray} Then integrate both sides and divide by the exponential to isolate $P(t)$.
H: Prove for a $7\times7$ matrix that the set of all eigenvectors is linearly independent. Suppose $A$ is a $7\times7$ matrix, $\left\{\vec{v_{1}},\vec{v_{2}}\right\}$ is a basis for $\operatorname{Eig}(A,3)$, $\left\{\vec{v_{3}},\vec{v_{4}},\vec{v_{5}}\right\}$ is a basis for $\operatorname{Eig}(A,7)$ and $\left\{\vec{v_{6}},\vec{v_{7}}\right\}$ is a basis for $\operatorname{Eig}(A,-2)$. Prove that the set $\left\{\vec{v_{1}},\vec{v_{2}},\vec{v_{3}},\vec{v_{4}},\vec{v_{5}}\vec{v_{6}},\vec{v_{7}}\right\}$ is linearly independent. What I think I am supposed to do is start with the dependence test equation: $c_{1}\vec{v_{1}}+c_{2}\vec{v_{2}}+c_{3}\vec{v_{3}}+c_{4}\vec{v_{4}}+c_{5}\vec{v_{5}}+c_{6}\vec{v_{7}}+c_{7}\vec{v_{7}} =0$ Edit: I looked up online and found multiply both sides by $A-\lambda_{m}I$ to get $c_{1}(\lambda_{1}-\lambda_{m})\vec{v_{1}}+c_{2}(\lambda_{2}-\lambda_{m})\vec{v_{2}}+c_{3}(\lambda_{3}-\lambda_{m})\vec{v_{3}}+c_{4}(\lambda_{4}-\lambda_{m})\vec{v_{4}}+c_{5}(\lambda_{5}-\lambda_{m})\vec{v_{5}}+c_{6}(\lambda_{6}-\lambda_{m})\vec{v_{6}}+c_{7}(\lambda_{7}-\lambda_{m})\vec{v_{7}} =0$ I'm guessing in this case $\lambda_{m}$ would be $\lambda_{7}$ AI: Hint: Consider what happens when you multiply by $A-\lambda I_7,$ where $I_7$ is the $7\times 7$ identity matrix and $\lambda$ is an eigenvalue of $A$. What happens when you do this for all but one such $\lambda$?
H: How to get the garage to work. (parking functions) At McGeorge's garage every driver has a favourite parking spot. Parking spots are arranged in a line and are numbered 1 through n. A driver always goes to his favourite spot, if it's free he takes it. If it's not he goes to the lowest unoccupied spot after his favourite spot. So if his favourite spot is f he tries at $f+1,f+2...$ until he finds an empty one. If he can't find an empty spot he leaves. let $(a_1,a_2...a_n)$ be the favourite spots of drivers $(a_1,a_2...a_n)$ if driver $a_1$ parks first, then $a_2$... how many lists $(a_1,a_2....a_n)$ are there such that everyone manages to park? AI: There are $(n+1)^{n-1}$ such lists. They are called parking functions, and there is quite a literature on them. The first few pages of this PDF give the most accessible proof that I’ve seen of the result that there are $(n+1)^{n-1}$ of them; it’s due to H.O. Pollak. It’s written rather concisely, but everything is there.
H: Suppose a, b and n are positive integers. Prove that (a^n) \| (b^n) if and only if a \| b. Suppose $a, b$ and $n$ are positive integers. Prove that $a^n\mid b^n$ if and only if $a \mid b$. I have: $$a^n\mid b^n$$ $$\implies b^n = a^n \cdot k$$ $$\implies \sqrt[n]{b^n}=\sqrt[n]{a^n}\cdot k$$ $$\implies a=b\cdot k$$ $$\implies a\mid b$$ Is it really this simple? AI: Factor $a$ and $b$ to prime factors: $$ a = 2^{a_1} 3^{a_2} 5^{a_3} 7^{a_4} \dots $$ $$ b = 2^{b_1} 3^{b_2} 5^{b_3} 7^{b_4} \dots $$ Now $$ a^n = 2^{na_1} 3^{na_2} 5^{na_3} 7^{na_4} \dots $$ $$ b^n = 2^{nb_1} 3^{nb_2} 5^{nb_3} 7^{nb_4} \dots $$ We know that $$ a \| b \Leftrightarrow \forall i \in \mathbb N : a_i \le b_i $$ And finally $$ \forall x, y, n \in \mathbb N : x \le y \Leftrightarrow x n \le y n $$
H: subring of rational numbers and its ideal Let $p$ be a prime number. For any $p$ the subring $\mathbb{Q}_p$ of of the field of rational numbers is defined: $\mathbb{Q}_p=\{\frac{a}{b}\|a,b\mbox{ are integers, $p$ does not divide $b$}\}$ Let $P$ be a subring of $\mathbb{Q}_p$ that is the set of all elements that are NOT invertible in $\mathbb{Q}_p$. i) Show that $P$ is an ideal in $\mathbb{Q}_p$. ii) Show that $P$ is a unique maximal ideal in $\mathbb{Q}_p$. iii) Prove that the quotient ring $\mathbb{Q}_p/P$ is isomorphic to the field $\mathbb{Z}_p$. There was given a hint that one should examine a map $\phi$ s.t. $\phi\left(\frac{a}{b}\right)=\bar{a}\cdot\left(\bar{b}\right)^{-1}$; the last is mod $p$. AI: Change of notation: let me call $\mathbb Z_{(p)}$ what you call $\mathbb Q_p$. It is the localization of $\mathbb Z$ at the multiplicative subset $S=\mathbb Z\setminus p\mathbb Z$. All that lies in $S$ becomes (universally) invertible in $\mathbb Z_{(p)}$ (and we formally write these elements as denominators...). So we have a chain of ring homomorphisms $$\mathbb Z\hookrightarrow \mathbb Z_{(p)}\hookrightarrow\mathbb Q.$$ That said, we have $$ P=\Bigl\{\frac{a}{b}\,:\,a\notin S\Bigr\}=\Bigl\{\frac{a}{b}\,:\,a\in p\mathbb Z\Bigr\}=p\mathbb Z_{(p)}. $$ In other words, $P$ is the ideal generated by $p$, or by $\frac{p}{1}$ if we want to be precise. As $P$ was defined to be $\mathbb Z_{(p)}\setminus \{\textrm{units}\}$, we have that $$\{\textrm{units}\}=\mathbb Z_{(p)}^\times=\mathbb Z_{(p)}\setminus p\mathbb Z_{(p)}.$$ This implies that $p\mathbb Z_{(p)}$ is the unique maximal ideal. Finally, let us follow the hint and define a map $$\phi:\mathbb Z_{(p)}\to \mathbb Z/p\mathbb Z$$ by $\phi(a/b)=\overline a.\overline b^{-1}$. Then $$\ker\phi=\Bigl\{\frac{a}{b}:\overline a.\overline b^{-1}=0_{Z/p\mathbb Z}\Bigr\}=\Bigl\{\frac{a}{b}:\overline a=0_{Z/p\mathbb Z}\Bigr\}=\Bigl\{\frac{a}{b}:p\,\textrm{divides}\,a\Bigr\}=P,$$ where we just used that the inverse of $\overline b$ cannot be zero in $Z/p\mathbb Z$. Aside: in your question you say that $P$ is a subring. It cannot be a subring, as a proper ideal is never a subring.
H: Everything in the Power Set is measurable? Im taking a class in graduate probability. My background is in engineering (very used to math in an applied sense). I am also taking an undergraduate class in real analysis along side (should have taken it before, but I couldn't) I have a couple of questions: We're spending time looking at measurable functions on measurable sets. The definition of a "measurable set" is one who lies in a sigma algebra. My conceptual understanding of a sigma algebra (I know the technical def: countable additivity, etc.) is the resolution with which we understand a certain space - the sets that can be measured - even more simply: the sets we can actually use. We say a sigma algebra is the "domain" of our measure. In other words, a (prob) measure can't measure just any old arbitrary set/sets of set. We define a sigma algebra to handle this, and say our measure operates over this sigma algebra. However, the power set is actually a sigma algebra (the largest one, according to our definition), and yet not every element of the power set is measurable? I'm having a little trouble reconciling my conceptual understanding of a sigma algebra (the behave good-measurable sets) with its actual def (which gives us the power set dilemma). How does the Borel Sigma Algebra fit into this conceptual understanding? How about non measurable sets? Is there a concept of the largest sigma algebra of only measurable sets, which is a subset of the power set? AI: Formally, measure (resp. probability) theory requires us to works with a triple $(\Omega, \mathcal{F}, P)$ where $\Omega$ is the space we are working on, $\mathcal{F}$ is a $\sigma-$algebra and $P$ is a (probability) measure which maps elements of $\mathcal{F}$ to numbers (between $0$ and $1$). We call the elements of $\mathcal{F}$ the "$\mathcal{F}$ measurable sets". For any non-trivial $\Omega$, you will have many potential $\sigma-$algebras that you can use in the place of $\mathcal{F}$. As you say, one option is to take $\mathcal{F} = 2^\Omega$ (the power set of $\Omega$) to be our $\sigma-$algebra. The problem with this choice is that every subset of $\Omega$ is in the power set--everything is measurable here. Why is that an issue? Among other things, it's often too big for $P$ to have nice properties. In many (most, honestly) cases, finding nice properties we want $P$ to have is what really drives the probability, not the particulars of $\mathcal{F}$. Stefan gives the standard non-probabilistic example of this in the comments. The Lebesgue measure, which is the natural notion of volume on the real line, is not compatible with the power set as the $\sigma-$algebra in our triple, so we need to pick a new one. The definition of the Borel $\sigma-$algebra is that it is the smallest $\sigma-$algebra containing the open intervals (which had better be measurable if we are going to define volume). Since this $\sigma-$algebra is compatible with the intuitive notion of volume, it is therefore the smallest $\sigma-$algebra we can choose with the property that $\mu\{(a,b)\} = b-a$ for all open intervals $(a,b)$. Why not stop here? Not all subsets of Borel sets of measure $0$ are measurable and it is often nice for the sake of theory to not have to worry about those sets. The Lebesgue $\sigma-$algebra is what you get if you insist that all subsets of sets of measure zero are measurable. In this case, as in many cases, because this $\sigma-$algebra is so natural we often drop the formalism and just say that a set is "measurable" or "not measurable" on the real line, when what we really mean is that it is measurable with respect to the Lebesgue $\sigma-$algebra or not measurable with respect to the Lebesgue $\sigma-$algebra. I believe that the last issue is the source of your confusion. Whatever you were reading dropped that they were referring to the Lebesgue $\sigma-$algebra. It is not too difficult and not too trivial to construct a set which is Lebesgue measurable but not Borel measurable. In general, most sets you can write down will end up being Borel. By contrast, constructing sets which are not Lebesgue measurable requires using something like the axiom of choice. Analysts are fond of saying that if you can write it down explicitly, it is Lebesgue measurable. Let me make one quick comment about why probabilists like to use the Borel $\sigma-$algebra rather than the Lebesgue $\sigma-$algebra. For an analyst, the definition of a function being measurable is that the inverse image of open sets is measurable. Since probabilists don't require our spaces to have topologies, this really doesn't work for us. For a probabilist, the definition of a function being measurable is that the inverse image of a measurable set is measurable. The Borel $\sigma-$algebra has the nice property that if you compose two Borel measurable functions, you get another Borel measurable function in either definition. This property fails badly for Lebesgue measurable functions with the analysts' definition of measurable.
H: Prove that set of all lines in the plane is uncountable. Let $L$ be the set of all lines in the plane. Prove that $L$ is uncountable, but only countably many of the lines in $L$ contain more than one rational point. Attempt: Well, I was trying to define $L$ using linear combinations of points since a line is a linear combo of two points. So, I wanted to define $L=\{ax+yb:x+y+z=k,k∈Z_+\}$. But, this does not seem right. Anyway, once I define $L$ I would try to find a function from $L$ to some uncountable set D that is onto or a function from D to L that is 1-1. Help appreciated thank you. AI: You can List all lines in the plane this way: Use the fact that a line is described uniquely once you know its slope, and one of its intercepts with the axes. The slope is indexed by the Reals, and so is the intercept, say the x-intercept. So we count all the possible pairs ( slope, x-intercept), and show it is equal to $\|\mathbb R\|$ 1) Consider all lines thru the origin $(0,0)$. These are described uniquely by their slope, and there are $\|\mathbb R\|$ of them, since the slope is parametrized by the Reals. 2) From 1) , we can cover all other cases of lines not going thru $(0,0)$ , by considering all possible ( say x-) intercepts of a line thru any point, with fixed slope$m$. For every line in 1), there are $\mathbb R$ lines not going thru the origin, but with the same slope. This means there are $\|\mathbb R\|\times \|\mathbb R\|=\|\mathbb R\|$ total lines in the plane.
H: isomorphisms- subspaces in topology Consider the following topological spaces: $(X_1,\tau_1)=(\Bbb R,\tau_u)$ and $(X_1,\tau_2)=(\Bbb R, \tau_{kol})$ So the product topology is the following: $(\Bbb R^2, \tau_u \times \tau_{kol})$ I have to describe the subspace topology defined by each subsets. $A=\{(x,y)\in \Bbb R^2: x+y=0\}$ $B=\{(0,0)\} \cup \{(x,y)\in \Bbb R^2: xy=1\}$ In other words, I have to say to which topological space is isomorphic $(A, \tau_A)$ and $(B,\tau_B)$, respectively. Could you help me please? I have defined the base of each topology but I can't see any isomorphism. *Note: $\tau_{kol} = \{ (a,\infty): a \in \Bbb R \} \cup \{\emptyset,\Bbb R\}$ (Topology of Kolmorogov) and $\tau_u$ is the usual topology Thank you for your time AI: The product has a base of open sets of the form $U(a,b,c)=(a,b)\times(c,\to)$, where $a,b,c\in\Bbb R$ and $a<b$; $$U(a,b,c)=\{\langle x,y\rangle\in\Bbb R^2:a<x<b\text{ and }c<y\}\;.$$ You need to investigate how these basic open sets intersect $A$ and $B$. $A$ is just the graph of the line $y=-x$. Show that each $U(a,b,c)$ intersects $A$ in an open interval of the line $A$, and that each open interval on $A$ can be obtained in this way. This means that $A$ in its subspace topology is homeomorphic to what familiar space? $B$ is the graph of a rectangular hyperbola together with its centre point. Here again you should look at the intersections $B\cap U(a,b,c)$ of $B$ with basic open sets in the product. You’ll find that a lot of them are open intervals of $B$. However, you’ll find that $\langle 0,0\rangle$ is not an isolated point of $B$ in this topology, though it is in the usual topology; every $U(a,b,c)$ that contains $\langle 0,0\rangle$ also contains other points of $B$.
H: Finding points on graph with tangent lines perpendicular to a line Find all points $(x,y)$ on the graph of $y=\frac{x}{x-3}$ with tangent lines perpendicular to the line $y=3x-1.$ My thoughts on this problem: First I should find the slope of the given line and the tangent to the given curve. I'm unsure of how to proceed with this though. I know that the slope of the tangent line is equal to $\frac{dy}{dx}$ at any point on the curve. So the slope of the tangent line would be: $$y'=\frac{(x-3)(1)-(x)(1)}{(x-3)^2}=\frac{-3}{(x-3)^2}$$ I also know that the product of the slopes of two perpendicular lines is $-1.$ I'm not sure how to apply this to the problem, and also I'm not sure about how to find $x$- and $y$-coordinates. Overall I'm not sure about how to set this problem up step by step. Thank you. AI: The slope of a line perpendicular to $y = mx + b$ is equal to $-\frac 1m$, where slope $m$ of the equation $y = 3x - 1$ is $m = 3$. So solve for the points at which the first derivative is equal to $-\dfrac 1m = -\dfrac 13$. You'll get a quadratic equation, from which you can find the two solutions. $$y'=\frac{x-4}{(x-3)^2} = -\dfrac 13$$ $$ (x - 3)^2 = -3( x- 4),\;\;x \neq 3$$ $$x^2 - 6x + 9 = -3x + 12$$ $$ \iff x^2 - 3x - 3 = 0$$ Those solutions, (solving for $x$), will give you the $x$-coordinate of the points at which the tangent lines to the curve are perpendicular to the given line. We can simply use the quadratic equation to find the solutions, $x_1$ and $x_2$. To find the corresponding $y$ coordinate, just plug each solution $x_i$ into the original curve (not into $y'$) and compute $y$.
H: Strange results on removing lim in wolframalph Can anyone explain this wolframalpha result? $ f(x)=\lim\limits_{h \to 0} \frac{-1}{(3x-2)^2} = \frac{-1}{(2-3x)^2}$ [lim ((-1)/((3x-2)^2)) as h->0] = [((-1)/((2-3x)^2)) ] While this is not equal: $ f(x)=\lim\limits_{h \to 0} \frac{-1}{(3x-2)^2} = \frac{-1}{(3x-2)^2}$ [lim ((-1)/((3x-2)^2)) as h->0] = [((-1)/((3x-2)^2))] edit: addint $f(x)=$ to stop confusion about the lack of x in the limit. AI: You are taking the limit as $h \to 0$, but $h$ does not exist in the limit. Perhaps you meant $ \lim_{x \to 0} \frac{-1}{(3x-2)^2}$, where we are tending $x$ to $0$?
H: Prove that $f(x+h)-f(x) - \langle\nabla f(x), h\rangle\geq 0 \Rightarrow f $ convex At this link there is a demonstration that for $f$ continuously differentiable on $C \subseteq \mathbb{R}^n$ convex, $f(x+h)-f(x) - \langle\nabla f(x), h\rangle\geq 0 \Rightarrow f $ convex. This argument uses an intermediate step regarding "$x\mapsto \nabla f(x)$ is monotone", which I find tedious, and in general I find that this argument gets lost in alphas and betas and misses the geometry of the situation. Does anyone know a more elegant argument? AI: Since everything takes place along the line containing $x$ and $x+h$, it is sufficient to show this in one dimension (i.e. w.l.o.g. $n=1$). Now draw a picture, and it will be visually obvious. However, to make it a proof, consider $$ g(x) = \sup\{f(y) + (x-y)f'(y) : y \in C\} $$ By hypothesis $f(x) \ge g(x)$. By looking at the special case $y = x$ we see that $g(x) \ge f(x)$. Therefore $f(x) = g(x)$. And $g(x)$ is the suprema of linear functions, and the suprema of linear functions must be convex. And now we have a proof in one dimension, we can get a proof in $n$ dimensions: $$ g(x) = \sup\{f(y) + \langle x-y,\nabla f(y)\rangle : y \in C\} $$
H: Real Analysis: Show that the Taylor expression converges at every point. Suppose that the function $F:\mathbb{R}\to\mathbb{R}$ has derivatives of all orders and that \begin{cases} F'(x)-F(x)=0, & \text{for all $x$} \\ F(0)=2. \end{cases} Find a formula for the coefficients of the $n$th Taylor polynomial for F at $x=0$. Show that the Taylor expansion converges at every point. I have done the first half: Re-write the equation to be $F'(x)=F(x)$. Then we have $F^k(x)=F^{(k-1)}(x)$ and $F^k(0)=F^{(k-1)}(0)=2$. The coefficient of the $n$th Taylor polynomial at $x=0$ is $\frac{2}{k!}$. Which I hope is correct, but I need help showing that the Taylor expression converges at every point. Any help is appreciated! AI: So you have $$F(x) =\sum_{n=0}^\infty {2x^n\over n!}.$$ Apply the ratio test to see this converges for all real values (complex as well) values of $x$.
H: Prove that $f g$ is differentiable at $x_0$. $\newcommand{\R}{\mathbb{R}}$Let $f,g : \R \to \R$. Let $f(x_0) = 0$, $f(x)$ differentiable at $x_0$ and $g$(x) continuous at $x_0$. I need to prove that $fg$ is differentiable at $x_0$. Any ideas or hints about how to begin? Continuous doesn't mean differentiable... so if we don't know that $g$ is differentiable at $x_0$, then how are we supposed to solve it? AI: $$\lim_{x\to x_0}\frac{f(x)g(x)-0}{x-x_0}=\lim_{x\to x_0}g(x)\frac{f(x)}{x-x_0}=g(x_0)f'(x_0)$$
H: about possion gamma and exponential distribution about possion gamma and exponential distribution can someone explain how to sub in the numbers? i tried subbing in numbers and the outcome its not the same as the answer. maybe I'm not doing it correct . I'm getting 1/2 e^-1.5 for the first question . AI: Exponential distributions deal with the amount of time between events. The first problem is exponential with $\lambda = .5$. Since we are dealing with more than 3 hours between calls, we have $$P(X\ge{3})=\int_{3}^{\infty}.5e^{-.5x}dx=-e^{-.5x}\|_{3}^{\infty}=0-(-e^{-.5(3)})=e^{-1.5}=.2231$$ THe second problem, you need to know that the mean and standard deviation of an exponential distribution is $\frac1{\lambda}$. Since this number is $2$, $(.5^{-1})$ then the we can find $$P(X\ge\mu+\sigma)=P(X\ge{4})=\int_{4}^{\infty}.5e^{-.5x}dx=-e^{-.5x}\|_{4}^{\infty}=0-(-e^{-.5(4)})=e^{-2}=.1353$$ EDIT: Continued answer. For questions 3 and 4, we are talking about Poisson, since we are wanting the number of events given a mean and a particular interval. Since we know the mean in one hour is .5, the mean in a 3 hour interval is 1.5. Thus we will use $\lambda=1.5$. Now we are looking for exactly 2 calls in three hours, thus, using Poisson, $$P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}=\frac{1.5^2e^{-1.5}}{2!}=.2510$$ Now for 4, we have our $\lambda=1$ since it is in a 2 hour interval. The mean and variance for Poisson are both $\lambda$, so $\mu+2\sigma=1+2\sqrt1=3$. Since we need fewer than 3 we have $$P(X\le2)=\sum_{k=0}^{2}\frac{1^ke^{-1}}{k!}=e^{-1}\left(1+1+\frac1{2}\right)=.9197$$
H: Prove whether or not $H$ is a subgroup of $S_n$ $H$ is the set of permutations where $H$ = {$ID_{S_n}$,(12),(34),(12)(34),(13)(24),(14)(23),(1432),(1234)}. Is $H$ a subgroup of $S_4$? Is there a simpler way to do this than checking for combinations that may not be closed under the operation? (composition is the operation in permutation groups, right?) I find that (1432)(12) = (1)(243) = (243) $\notin H$. Is that enough to prove it's not a subgroup or am I testing the elements incorrectly? AI: Showing it isn't closed under the group operation with an example, such as you did, correctly shows it is not a subgroup. Alternatively, you can note that $H$ contains $(1,2)$ and $(1,2,3,4)$, and therefore generates $S_4$. In particular, if it were a subgroup, then $H=S_4$. But it clearly has less than $\|S_4\|=24$ elements.
H: Determining the Coordinates of the point on the x-axis that are equidistant So I've been doing my homework without any trouble so far and I came across this question that I did not understand how to do. Given the points A(-2,1,3) and B(4,-1,3), determine the coordinates of the point on the x-axis that are equidistant from these two points. My process: I figured out that equidistant means at equal distances, so point would be half way between A and B. SO I figured out the vector AB = (6,2,0) and then I though to divide it in 2. However this is wrong. The answer at the back of my text book is (1,0,0). How would I go about finding the answer? Is my thought process wrong? AI: Here's one way you can go about it. We begin by determining the plane of all points $(x,y,z)$ equidistant from those two points. The segment from $A$ to $B$ will be orthogonal to the plane, so the plane will have the equation $6x-2y+d=0$ for some constant $d$. Moreover, the plane will bisect this segment, so the midpoint of $A$ and $B$ will lie in the plane. That is, $(1,0,3)$ will lie in the plane, so $6+d=0,$ and so $d=-6.$ Hence, the plane has equation $$6x-2y-6=0.$$ To find the $x$-coordinate of the point in this plane lying on the $x$-axis, we set $y=z=0,$ and solve for $x,$ to find that $x=1,$ so that $(1,0,0)$ is the point in question.
H: Simplifying this logarithm series $$\sum_{i\; =\; 2}^{99}{\frac{1}{\log _{i}\left( 99! \right)}}$$ How would you evaluate (or at least simplify) this logarithm series? AI: Note that $\log_a b = \frac{\log b}{\log a}$, for any choice of base of the logarithm. (Mathematicians tend to mean the natural logarithm when they write $\log$.) In particular, $\frac1{\log_a b} = \frac{\log a}{\log b} = \log_b a$. Therefore your sum is: $$ \sum_{i=2}^{99} \log_{99!} i = \log_{99!} \left( \prod_{i=2}^{99} i \right) = \log_{99!} 99! = 1 $$ where in first equality I used the well-known fact that $\sum \log = \log \prod$.
H: Numerical integration of $\int_0^2 \frac{1}{x+4}dx.$ I have homework problem. Determine the number of intervals required to approximate $$\int_0^2 \frac{1}{x+4}dx$$ to within $10^{-5}$ and computer the approximation using (a) Trapezoidal rule, (b) Simpson's rule, (c) Gaussian quadrature rule. I think the phrase "within $10^{-5}$,"means that the error term. I know that the m-point Newton-Cotes rule is defined by $$Q_{NC(m)}=\int_a^b p_{m-1}(x)dx,$$ where $p_{m-1}$ interpolates the function on $[a,b].$ So when $m=2,$ we call $Q_{NC(2)}$ trapezoidal rule, ans $Q_{NC(3)}$ is simpson's rule. Can anyone explain what are these three rules and how I can proceed?? And what does $m$ represent?? Is it I am kind of lost in this class.. ans the text book is really really bad that I have no idea what it talks about... AI: yes, $10^{-5}$ is the error term. That is $$ \|I-I^{\prime}\| $$ where $I$ is the exact integrand and $I^{\prime}$ is an approximation. I will summarize the methods as follows Trapezoid is given by $$\int_{a}^{b}f(x)\,dx=(b-a)\frac{f(a)+f(b)}{2}$$ It approximates the integral by approximating the area under the curve like a trapezoid. Simpson is given by $$\int_{a}^{b}f(x)\,dx=\frac{(b-a)}{6}\left(f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right)$$ and it is really a quadratic interpolation to approximate the integral. Gaussian Quadrature is different in a sense that the integral is evlauted by picking certain points in interval with some weights. You an easily look up weights and points for GQ and it is given by $$\int_{-1}^{1} f(x)\,dx=\sum_{i=1}^{n} w_{i}f(x_{i})$$ For your exercise, start with a equally spaced points on the line and implement each of the above method. If the error is above the tolerance, increase the number of intervals and repeat the same process until the error is $\le 10^{-5}$. This will be few lines of code in your preferred language.
H: Cauchy Sequence if $\|s_{n+1} - s_n\| < 2^{-n}$ Let $s_n$ be a sequence such that $\|s_{n+1} - s_n\| < 2^{-n}$ for all $n \in N$. Prove $s_n$ is a Cauchy sequence and hence a convergent sequence. Here's what I've started with: Proof: Take $\epsilon > 0$. Let $N = -\log_2{\epsilon/2}$. Thus, $ \|s_{n+1} - s_n\| < 2^{\log_2{\epsilon/2}} = \epsilon/2$ I can't seem to show from this that: $ \|s_m - s_n\| <\epsilon$ Some ideas relating to geometric series of $\epsilon/2 + \epsilon/4 ... \epsilon/{2^n} < \epsilon$ are springing up in my mind, but I can't really pin it to the ground (and, I'm not sure if its fair play to use the infinite series formula in this case). AI: Here's a sketch; check the details carefully. You need to show that for all $\epsilon > 0$ that there is an $N$ such that for $n,m > N$, $\|s_n - s_m\| < \epsilon$. Let $\epsilon > 0$ be given. We have to find $N$. Assume that $m > n$ without loss of generality. Then by the triangle inequality and the given property of the sequence, $$\|s_m - s_n\| < \frac{1}{2^m} + \cdots + \frac{1}{2^n} = \frac{1}{2^m}\left(1+\frac{1}{2} + \cdots + \frac{1}{2^{n-m}}\right)$$ and so by the formula for a geometric series, $$\|s_m - s_n\| < \frac{1}{2^m} \frac{1-\left(\frac{1}{2}\right)^{n-m}}{1 - \frac{1}{2}} < \frac{1}{2^{m-1}}$$ Now just take $N = \log_2{\epsilon}$. Then for $m > n > N$, we have that $$\|s_m - s_n\| < \frac{1}{2^{m-1}} \leq \frac{1}{2^N} = \epsilon$$
H: Euler Totient Issues I was skimming again through Dummit & Foote's Abstract Algebra and I came across this exercise: Prove that for any given positive integer $N$ there exist only finitely many integers $n$ with $\varphi(n)=N$, where $\varphi$ denotes Euler $\varphi$-function. Conclude in particular that $\varphi(n)$ tends to infinity as $n$ tends to infinity. I don't doubt that $\varphi(n) \to \infty$ as $n \to \infty$. However, if I recall Erdös proved that if there is an $x_0$ such that $\varphi(x_0)=N$, then there must be infinitely many integers such that $\varphi(x)=N$. See this article for a mention of this (it's in the very first $2$ lines). Am I reading something wrong here? Has has Dummit and Foote actually asked the impossible? Or does the MathWorld site contain an error? AI: The result you refer to does not say what you think it does. What it says is that if there is an $m_0$ such that $\varphi(x)=m_0$ has exactly $k$ solutions, then there exist infinitely many $m$ such that $\varphi(x)=m$ has exactly $k$ solutions. For example, let $k=3$. There is an $m_0$, namely $2$, such that $\varphi(x)=m_0$ has exactly $k$ solutions. These are $3$, $4$, and $6$. We can conclude that there are infinitely many $m$ such that $\varphi(x)=m$ has exactly $3$ solutions.

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