Recent Questions - Mathematics Stack Exchange |
- Why are subclasses of first-order logic with the finite model property decidable?
- Solve $\frac{dy}{dx} \:= \:\frac{Ay}{By \:+ \:C} \:+ \:DP \:- \:EPy\:$ for $\:P$
- Limit of $\lim\limits_{x \to 1} \frac {3^\frac{x-1}{4}-1}{sin(5(x-1))}$, without L'Hopital's rule.
- Functoriality of sheaf Cohomology
- Integrate $\oint \frac{e^z}{z^2-1}dz$ over $\gamma$ being the circle centered at $0$ of radius $2$.
- Jordan canonical form $\mod p$ of a $p\times p$ cyclic permutation matrix
- If $X_n \nearrow X$, i.e., $X_n \subset X_{n+1}$ and $\bigcup_n X_n = X$, then $\operatorname{int} X_n \nearrow \operatorname{int} X$
- need help rewriting this arcsin function
- Is a topology determined by the convergence of nets?
- Why does polynomial regression yield worse results if you increase the order by too much?
- Determine the principal argument of a complex number
- How many $4$ letter words can be formed from the word "CORONAVIRUS".
- Finding Weight of three values based on Given Weight of Minimum Value
- algebraic properties of delta-function
- Smallest eigenvalue of weighted graph laplacian
- how $tx+(1-t)y \in S$ implies that the line joining $x$ and $y$ lies in $S$.
- Surface integral ( limits of surface)
- Halmos: Principle of induction shows only one function satisfies a recurrence relation.
- Orthogonality and completeness of spherical bessel functions
- Is it possible to generalize this equation more?
- Fourier transform of a continuous function from $L^2$.
- For a set like this, must there exist a function for which it is a level set?
- Does finitely generated groups have finitely many finite retracts?
- Understanding subgroups and how to (always) generate them.
- Probability of rolling exactly 1 number exactly 3 times in 6 rolls of a fair die
- A Pseudo-Derivative for $f(x)=|x|$
- Linear independence of the numbers $^m\ln(a),^n\ln(a)$?
- How to find locus of Q
- Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
- Perspective projection of a circle: what is the size of the semi-major axis?
| Why are subclasses of first-order logic with the finite model property decidable? Posted: 04 Jul 2022 12:38 PM PDT The title really says all: Why are subclasses of first-order logic with the finite model property decidable? |
| Solve $\frac{dy}{dx} \:= \:\frac{Ay}{By \:+ \:C} \:+ \:DP \:- \:EPy\:$ for $\:P$ Posted: 04 Jul 2022 12:35 PM PDT In the equation, $y\:$ is the dependent variable and $\:x\:$ is the independent variable. $\:A, \:B, \:C, \:D, \:E,\:$ and $\:P\:$ are constants. How can we find an optimum value for $\:P\:$ given the boundary conditions for $\:x\:$ and $\:y:$ $\:(x_0,y_0), \:(x_1,y_1)$? My solution: $\frac{dy}{dx} \:= \:\frac{Ay}{By \:+ \:C} \:+ \:DP \:- \:EPy$ $\implies dy\:\frac{B \:+ \:Cy}{Ay \:+ \:PBD \:- \:PEBy \:+ \:PCDy \:- \:PECy^2} \:= \:dx$ $\implies dy\:\frac{B \:+ \:Cy}{PBD \:+ \:(A \:- \:PEB \:+ \:PCD)y \:- \:PECy^2} \:= \:dx$ Equate $\:BD \:+ \:(A \:- \:PEB \:+ \:PCD)y \:- \:PECy^2\:$ with $\:ay^2 \:+ \:by \:+ \:c$, $\therefore \:a \:= \:- \:PEC$, $\:b \:= \:A \:- \:PEB \:+ \:PCD\:$ and $\:c \:= \:PBD$ Let $\:k_1\:$ and $\:k_2\:$ be the roots of $\:ay^2 \:+ \:by \:+ c \:= \:0$, $\therefore \:k_1,k_2 \:= \:\frac{- \:b \:\pm \:\sqrt{\Delta}}{2a}\:$ where $\:\Delta \:= \:b^2 \:- \:4ac$ $\frac{B \:+ \:Cy}{PBD \:+ \:(A \:- \:PEB \:+ \:PCD)y \:- \:PECy^2} \:= \:\frac{u}{y \:- \:k_1} \:+ \:\frac{v}{y \:- \:k_2}$ $\implies\frac{B \:+ \:Cy}{PBD \:+ \:(A \:- \:PEB \:+ \:PCD)y \:- \:PECy^2} \:= \:\frac{u (y \:- \:k_2) \:+ \:v (y \:- \:k_1)}{(y \:- \:k_1)\:(y \:- \:k_2)}$ $\implies\frac{B \:+ \:Cy}{PBD \:+ \:(A \:- \:PEB \:+ \:PCD)y \:- \:PECy^2} \:= \:\frac{(- \:k_2 u \:- \:k_1 v) \:+ \:(u \:+ \:v)y}{(y \:- \:k_1)\:(y \:- \:k_2)}$ $\therefore u \:+ \:v \:= \:C\:$ and $\:- \:k_2 u \:- \:k_1 v \:= \:B$ which results to $\:u \:= \:\frac{B \:+ \:C k_1}{k_2 \:- \:k_1}\:$ and $\:v \:= \:\frac{B \:+ \:C k_2}{k_1 \:- \:k_2}$ $\therefore dy\:(\frac{B \:+ \:Cy}{PBD \:+ \:(A \:- \:PEB \:+ \:PCD)y \:- \:PECy^2}) \:= \:dx\:$ becomes $\:dy\:(\frac{B \:+ \:C k_1}{(k_1 \:- \:k_2)\:(y \:- \:k_1)} \:- \:\frac{B \:+ \:C k_2}{(k_1 \:- \:k_2)\:(y \:- \:k_2)}) \:= \:dx$ On integration, $\frac{B \:+ \:C k_1}{k_1 \:- \:k_2}\:\ln(y \:- \:k_1) \:- \:\frac{B \:+ \:C k_2}{k_1 \:- \:k_2}\:\ln(y \:- \:k_2) \:= \:x \:+ \:constant$ We remove the constant term by putting in the boundary conditions and then solve the resulting equation numerically with methods like Newton-Raphson. Problem: I keep getting garbage values for P when I use this method with a bad initial guess. On using a near-correct guess, I get either of $\:k_1\:$ or $\:k_2\:$ equal to $\:y_1\:$ which gives a division by $\:0\:$ error. |
| Limit of $\lim\limits_{x \to 1} \frac {3^\frac{x-1}{4}-1}{sin(5(x-1))}$, without L'Hopital's rule. Posted: 04 Jul 2022 12:38 PM PDT
I first tried substituting $\frac{x-1}{4}=a$, to which I got $\lim\limits_{a \to 1}\frac{3^a-1}{\sin(20a)}$. Tried substituting again with $n=3^a-1$, but couldn't get any decent results. Can someone give me a tip? |
| Functoriality of sheaf Cohomology Posted: 04 Jul 2022 12:31 PM PDT Let $f:Y\rightarrow X$ be a morphism of schemes. I would like to construct a natural map of homologies $$H^{q}(X,F)\rightarrow H^{q}(Y,f^{*}(F))$$ where $F$ is a sheaf on X. My idea is to take an injective resolution $I^{*}$ of $F$ and $J^{*}$ an injective resolution of $f^{*}(F)$. Of course the resolution $I^{*}$ is flasque, so I think if I prove that $f^{*}(J^{*})$ is flasque then I am done because there is a map $I^{*}\rightarrow f_{*}f^{*} I^{*}$. But I am stuck on how to do this. Any ideas? |
| Integrate $\oint \frac{e^z}{z^2-1}dz$ over $\gamma$ being the circle centered at $0$ of radius $2$. Posted: 04 Jul 2022 12:30 PM PDT So I need to integrate $$\oint_\gamma \frac{e^z}{z^2-1}dz$$ where $\gamma$ is the circle of radius $2$ centered at $0$. So by the residue theorem I have that $$\oint_\gamma \frac{e^z}{z^2-1}dz = 2 \pi i \bigg( \text{Res}(\frac{e^z}{z^2-1};1)+ \text{Res} (\frac{e^z}{z^2-1};-1) \bigg)= \pi i (e-e^{-1})$$ So I calculated the residues at the two points of singularity. Namely, $z=\pm 1$. Did I do it correctly? |
| Jordan canonical form $\mod p$ of a $p\times p$ cyclic permutation matrix Posted: 04 Jul 2022 12:22 PM PDT Let $p$ be a prime number, Consider $V$ a vector space over $\mathbb{Z}/p$ with basis $e_0$, $e_1$, $\ldots$, $e_{p-1}$ and the permutation operator $T(e_i) = e_{i+1}$, $i$ modulo $p$. The characteristic polynomial of $T$ is $\lambda^p - 1 = (\lambda-1)^p$, and so $T$ has the eigenvalue $1$ with multiplicity $p$. The problem is to determine the Jordan canonical form of $T$. It probably is a single block $J_{p, 1}$. For $p=2$ it's easy, $p=3$ is just a bit of calculations. Note that we need to use $p$ is prime. If it worked $\mod n$ for arbitrary $n$, we would have $x^n-1 = (x-1)^n \mod n$, not true in general. Any feedback would be appreciated. Thank you for your attention! |
| Posted: 04 Jul 2022 12:20 PM PDT Let $X,X_n$ be subsets of a topological space $E$ such that $X_n \nearrow X$, i.e., $X_n \subset X_{n+1}$ and $\bigcup_n X_n = X$. Clearly, $$ \bigcup_{i=1}^n \operatorname{int} X_i = \operatorname{int} X_n = \operatorname{int} \left ( \bigcup_{i=1}^n X_i \right ). $$ I would like to ask if this property extends to countably infinite union, i.e., $$ \bigcup_n \operatorname{int} X_n = \operatorname{int} X = \operatorname{int} \left ( \bigcup_n X_n \right ). $$ |
| need help rewriting this arcsin function Posted: 04 Jul 2022 12:36 PM PDT How can i express the following function as $y$ as a function of $a$ and $x$, like this: $y(a, x) =\ ...$ $$ 2*\arcsin\left(\frac{\frac{x}{2}}{\frac{x^2}{8y}+\frac{y}{2}}\right)*\frac{x^2}{8y}+\frac{y}{2}=a*x $$ Reason for asking is: friends and I were laying a laminate floor and discussed how bad it would be if we left no room around the edge and the floor expanded a few %, my thesis was and is: pretty bad. But I'd love to be able to express how bad. I approximated the problem by disregarding 2 of the four walls and looking at a cross section of bulging floor. I then decided that it looks somewhat similar to a section of a circle (although in hind-sight a parabola might be more accurate). I know the distance between the walls $x$ and the expansion factor $a$ which will give me the length of the floor after it has bulged: $a*x$. I'm interested in the height of the bulge in the middle of the room $y$. Now I need to rewrite the above function that I obtained by performing more steps than I care to write out (unless people here want to know) in order to get an expression for $y$ that depends only on $a$ and $x$. But I'm stuck on the arcsin, I can't figure out how to get all terms with $x$ and $a$ to one side and $y$ to the other. Any help is greatly appreciated! |
| Is a topology determined by the convergence of nets? Posted: 04 Jul 2022 12:10 PM PDT Consider a space $X$ with two topologies $\tau_1$ and $\tau_2$. It is easy to see that nets remain convergent if the topology is made coarser, i.e. $$\tau_1 \subseteq \tau_2 \quad \Rightarrow \quad \Big\{ x_\alpha \overset{\tau_2}{\longrightarrow} x \, \Rightarrow \, x_\alpha \overset{\tau_1}{\longrightarrow} x \Big\}.$$ My question is: does the converse hold? i.e. is it true that $$\tau_1 \subseteq \tau_2 \quad \iff \quad \Big\{ x_\alpha \overset{\tau_2}{\longrightarrow} x \, \Rightarrow \, x_\alpha \overset{\tau_1}{\longrightarrow} x \Big\}$$ ? I think this is related to whether a topology is fully determined by the convergence of nets, but I couldn't find a clear answer. |
| Why does polynomial regression yield worse results if you increase the order by too much? Posted: 04 Jul 2022 12:38 PM PDT I'm aware this phenomenon is usually called overfitting, but since any polynomial can be expressed with the coefficients of a polynomial of higher order by setting the order higher-order coefficients to zero, increasing the order of polynomial regression should simply add "more" applicable regression functions to the model. How can the regression error get worse sometimes if you could simply re-take any best lower-order fits with a smaller regression error? |
| Determine the principal argument of a complex number Posted: 04 Jul 2022 12:32 PM PDT Given a non-zero complex number $z$, we define $Arg(z)=\theta$ if and only if $z=re^{i\theta}$ and $\theta\in(-\pi,\pi]$, $r>0$. For a real number $0<x<1$ and a positive integer $m$, my goal is to determine $$ Arg\big[\sin((m+1)x\pi) e^{-mx\pi i}\big]\qquad(\ast) $$ My attempt: If $\frac{2k}{m+1}<x<\frac{2k+1}{m+1}$, then $\sin((m+1)x\pi)>0$, so $(\ast)=Arg\big[e^{-mx\pi i}\big]$. If $\frac{2k+1}{m+1}<x<\frac{2k+2}{m+1}$, then $\sin((m+1)x\pi)<0$, so $(\ast)=Arg\big[-e^{-mx\pi i}\big]$. How to determine $Arg\big[\pm e^{-mx\pi i}\big]$ acording to the above intervals ? |
| How many $4$ letter words can be formed from the word "CORONAVIRUS". Posted: 04 Jul 2022 12:26 PM PDT Letters : C , O($2$ times), R($2$ times), N, A, V, I, U and S. $4$ letter words from C,N, A, V, I, U and S = $7 \times 6 \times 5 \times 4 $ = $840$ ways $4$ letter words from two O or two R while other $2$ letters are different = $^2C_1 \times ^8C_2 \times \frac{4!}{2!} $ = $672$ ways $4$ letter words from two O and two R together = $\frac{4!}{2! \times 2!}=6$ Total number of ways should be = $840 + 672 + 6 = 1518$ But this is not the right answer given. What am I doing wrong? What cases I am missing here? Please help !!! Thanks in advance !!! |
| Finding Weight of three values based on Given Weight of Minimum Value Posted: 04 Jul 2022 12:01 PM PDT I have three prices:
And W= (Weight) to find best value is (40%) is used to find best offer. The minimum value which is C is the best then it gets all W (40). Now how do I calculate A, C? |
| algebraic properties of delta-function Posted: 04 Jul 2022 12:27 PM PDT I'm studying partial derivatives equations, developing the case of solving Klaine-Gordon equation - after expansion of solution into Fourier integral there appears the algebraic equation $g(x)f(x) = 0, x$ in terms of generalized function. It is said that the solution is $f(x) = \delta(g(x))h(x), x\in\Bbb R^4$, where $h(x)$ - arbitrary function. Now I'm looking forward any proof of this result and will appreciate any help |
| Smallest eigenvalue of weighted graph laplacian Posted: 04 Jul 2022 11:59 AM PDT Is there some works about smallest eigenvalue of weighted (mix of positive, zero, and negative weights) graph laplacian? Thanks a lot for suggestions! |
| how $tx+(1-t)y \in S$ implies that the line joining $x$ and $y$ lies in $S$. Posted: 04 Jul 2022 12:14 PM PDT In the book "Linear Algebra, A Geometric Approach" by S Kumaresan the definition of Affine Space is given like this- "Let $V$ be an arbitrary vector space. An affine subspace is a nonempty set such that for all $x, y \in S$, the line joining $x$ and $y$ also lies in $S$. This means that if $x, y\in S$, then $tx +(1-t)y\in S$ for all $t \in \Bbb {R} $." |
| Surface integral ( limits of surface) Posted: 04 Jul 2022 12:03 PM PDT Calculate $$\iint_{S} \frac{xy}{\sqrt{1+2x^{2}}} dS$$ where $$S =\{(x, y, x^{2}+y ) , 0 \leq x \leq y, x+y \leq 1\} $$ My attempt is this i cannot figure out the limits of $x$ and $y$ In the ans it is given $ 0<y<1$ and $-y< x< 1-y$ While i think it should be $ 0<x<1/2$ and $x<y<1-x$ |
| Halmos: Principle of induction shows only one function satisfies a recurrence relation. Posted: 04 Jul 2022 12:18 PM PDT
(Book: Halmos, Paul. Naive Set Theory, Section 12, The Peano Axioms, p.48.) What might Halmos mean by the line in bold? What is the proof in question? I feel like I am missing something obvious. Context: The set of natural numbers is denoted $\omega$ here. The successor of $n$ is $n^{+}$ here. Principle of induction is expressed set-theoretically as $$ S\subseteq \omega \wedge 0\in S \wedge (n\in S \Rightarrow n^{+} \in S) \Rightarrow S = \omega$$ |
| Orthogonality and completeness of spherical bessel functions Posted: 04 Jul 2022 12:10 PM PDT I am interested in computing the following integral, which feels like something that must have been computed before: $$ \int \frac{k^2{\rm d} k}{2\pi^2}j_{\ell}(r k)j_{\ell'}(r' k) $$ From what I understand, there are two adjacent properties of Bessel functions that could be relevant. The first is $$ \int \frac{k^2{\rm d} k}{2\pi^2}j_{\ell}(r k)j_{\ell}(r' k) = \frac{1}{4\pi r^2}\delta(r - r')\,, $$ and the second is $$ \int \frac{{\rm d}z}{z} J_\alpha(z) J_\beta(z) = \frac{2}{\pi}\frac{\sin\left(\frac{\pi}{2}(\alpha - \beta)\right)}{\alpha^2 - \beta^2}\,. $$ Recalling that $$ j_\ell(k r) = \sqrt{\frac{\pi}{2 k r}}J_{\ell + 1/2}(k r)\,, $$ we see $$ \int {\rm d} k j_{\ell}(k r)j_{\ell'}(k r) = \frac{1}{r}\frac{\sin\left(\frac{\pi}{2}(\ell - \ell')\right)}{(\ell(\ell + 1)-\ell'(\ell ' + 1))}\,. $$ The first completeness relation clearly resolves the case of $\ell = \ell'$, while the orthogonality relation doesn't appear to directly help me. Is there any hope to compute the integral I posed? Are there special cases that have been worked out? Here are some explicit examples that I have computed $$ \int\frac{k^2{\rm d} k}{2\pi^2}j_0(k r)j_2(k r') = \Theta(r' - r)\frac{r'^2 - r^2}{8\pi r'^3}\,, $$ $$ \int\frac{k^2{\rm d} k}{2\pi^2}j_0(k r)j_4(k r') = \Theta(r' - r)\frac{r'^2 - r^2}{8\pi r'^5}\left(\frac34(r'^2 - r^2) - r^2\right)\,. $$ $$ \int\frac{k^2{\rm d} k}{2\pi^2}j_0(k r)j_6(k r') = \Theta(r' - r)\frac{r'^2 - r^2}{8\pi r'^7}\left(r^4 - \frac52(r'^2 - r^2)r^2 + \frac58(r'^2 - r^2)^2\right)\,. $$ $$ \int\frac{k^2{\rm d} k}{2\pi^2}j_0(k r)j_8(k r') = \Theta(r' - r)\frac{r'^2 - r^2}{8\pi r'^9}\left(\frac{35}{64}(r'^2 - r^2)^3 - \frac{35}{8}r^2(r'^2 - r^2)^2 +\frac{21}{4}r^4(r'^2 - r^2) -r^6\right)\,. $$ $$ \int\frac{k^2{\rm d} k}{2\pi^2}j_2(k r)j_4(k r') = \Theta(r' - r)\frac{r'^2 - r^2}{8\pi r'^5}\left(r^2-\frac{2}{3}(r'^2 - r^2)\right)\,. $$ |
| Is it possible to generalize this equation more? Posted: 04 Jul 2022 12:25 PM PDT This is one of my discoveries, and my question is: can it be more general? Let $x, y, z$ be three arbitrary complex numbers and set $x_1=3 x^2+y^2$, $x_2=1-z$, $x_3=1+z$, $x_4=3 x^2+y^2+4 x y z$, $x_5=3 x^2+y^2-4 x y z$, $x_6=3 x^2+y^2+z (3 x^2+2 x y-y^2)$, $x_7=3 x^2+y^2-z (3 x^2+2 x y-y^2)$, $x_8=3 x^2+y^2+z (3 x^2-2 x y-y^2)$, $x_9=3 x^2+y^2-z (3 x^2-2 x y-y^2)$. Then for $k=0,1,2,3,4,5$, $2(x_1^k+x_1^kx_2^k+x_1^kx_3^k)=x_4^k+x_5^k+x_6^k+x_7^k+x_8^k+x_9^k$. In case $x, y, z$ are integers, we get a Diophantine equation. For example, $2(7^1+21^1+(-7)^1)=(-9)^1+23^1+1^1+13^1+17^1+(-3)^1$, $2(7^2+21^2+(-7)^2)=(-9)^2+23^2+1^2+13^2+17^2+(-3)^2$, $2(7^3+21^3+(-7)^3)=(-9)^3+23^3+1^3+13^3+17^3+(-3)^3$, $2(7^4+21^4+(-7)^4)=(-9)^4+23^4+1^4+13^4+17^4+(-3)^4$, $2(7^5+21^5+(-7)^5)=(-9)^5+23^5+1^5+13^5+17^5+(-3)^5$. My discovery published at: https://demonstrations.wolfram.com/author.html?author=Minh%20Trinh%20Xuan |
| Fourier transform of a continuous function from $L^2$. Posted: 04 Jul 2022 11:56 AM PDT Good day everyone, I have the following question. If $f$ from $L^2(\mathbb{R})$ and continuous, is it true, that its Fourier transform is also continuous? It is definitely true and easy to show if $f$ in $L^1$. But what is for $L^2$? And what if $f$ is even differentiable? Is $\hat f$ also differentiable in this case? Thanks in advance. |
| For a set like this, must there exist a function for which it is a level set? Posted: 04 Jul 2022 12:36 PM PDT Our metric space is $X = [0,1]^n$. Consider a set of points $S \subset X$ such that for any $x, x' \in S, x'_i > x_i \implies x'_j \leq x_j$ for some $j \leq n$. $S$ can be uncountable. My question is, for any such $S$, does there exist a function $\phi: [0,1]^n \rightarrow \mathbb{R}$ with $\phi$ is increasing (weakly) in each of its components, such that $S \subseteq L$ for some level set of $\phi$? My question is related to this and this question. Though since in my case $S$ can be uncountable, one of them comes very close to answering it but doesn't. Edit: A previous version of this question said "analytic function". I don't need the function to be analytic. It was a mistake. I misunderstood analytic functions. My apologies. |
| Does finitely generated groups have finitely many finite retracts? Posted: 04 Jul 2022 12:19 PM PDT A group $H$ is called a retract of a group $G$ if there exists homomorphisms $f:H\to G$ and $g:G\to H$ such that $gf=id_H$. We know that a group $G$ is finite if and only if $G$ has finitely many subgroups. Now my question is that a finitely generated group $G$ has finitely many finite retracts? What I've tried: If $G$ is a finitely generated abelian group, then every retract of $G$ is a direct summand of $G$. So the number of finite retracts of $G$ is finite. |
| Understanding subgroups and how to (always) generate them. Posted: 04 Jul 2022 12:38 PM PDT In a bid to understand group theory, I would like to ask the following questions: What ways can subgroups of a group $G$ be generated (that is guaranteed to always work)? Is it (only) by the elements of $G$ (rule of $a^n$) or by the rule of subgroups generated by subsets? By the way, it does not seem like $a^n$ always work; please see the next section. This Wikipedia section here: https://en.wikipedia.org/wiki/Subgroup#Example:_Subgroups_of_Z8 mentions that the non-trivial subgroups of $G = \{0,4,2,6,1,5,3,7\}$ are $\{0,4\}$ and $\{0,4,2,6\}$. (Note that the operation here is Personally, I tried using the elements of $G$ to generate subgroups. I obtained the following (alleged$_0$) subgroups by raising each element to the power $k$; where $k$ $\in \mathbb{N}_1$ :
In my understanding, here, every $H_a$ do not qualify as nontrivial subgroups of $G$ except $H_4$; where $a$ is each element in $G$. If using elements of $G$ (can allegedly) generate a subgroup, why am I not able to generate the subgroup $\{0,4,2,6\}$ like in the Wikipedia example? Footnote: $_0$. Please bear with me. I say $_1$. Natural numbers starting from 1 |
| Probability of rolling exactly 1 number exactly 3 times in 6 rolls of a fair die Posted: 04 Jul 2022 12:22 PM PDT I'm trying to calculate the probability based on the size of the event space divided by the size of the sample space $P=\frac{|E|}{|S|}$ I know that $|S|=6^6$, but am not sure what exactly the event space consists. Currently my thoughts are that we have 6 choices for our favorable event(the triples) and for the remaining 3 numbers we have $5\times5\times4=100$, $4$ because we do not want to include the possibility of having 3 same numbers two times, and to consider all possible arrangements, there are then $\frac{6!}{3!1!1!1!}$ possibilities. This leads to our final equation of : $P=\frac{|E|}{|S|}=\frac{6!}{3!1!1!1!} \times \frac{6\times100}{6^6}$ But the problem is that this exceeds 1, which is clearly wrong but I couldn't really figure out what is the fix for my equation. Thanks:) |
| A Pseudo-Derivative for $f(x)=|x|$ Posted: 04 Jul 2022 12:00 PM PDT I was wondering what the significance of a function that gives the slope of $y=|x|$ at any $x$ is. If$$f(x)=|x|$$then we could do, as the derivative:$$\frac{\partial f}{\partial x}=\frac{x}{|x|}$$or$$\frac{\partial f}{\partial x}=\frac{|x|}{x}$$This would give us the slope at any $x$. What is wrong with my approach? How would this connect to limits? |
| Linear independence of the numbers $^m\ln(a),^n\ln(a)$? Posted: 04 Jul 2022 11:56 AM PDT Let's assume Schanuel's conjecture is true. Let $a$ be an algebraic number, $m,n\in\mathbb{N}_+$, $^n\ln(a)=f_1(f_2(...(f_n(a))...))$ $\ \ (f_1,...,f_n=\ln)$. Can we show that the numbers $^m\ln(a),^n\ln(a)$ $\ (a\neq 0,1)$ are linearly independent or algebraically independent for $1<n<m$? If yes, how can we show that? I can already show linear independence of the numbers $^m\ln(a),^n\ln(a)$ for $n=1$ and arbitrary $m>1$. Later I want to see, applying Schanuel's conjecture, if the numbers $^1\ln(a),^2\ln(a),...,^n\ln(a)$ are algebraically independent. We can show by applying Schanuel's conjecture that the numbers $\ln(\ln(a)),\ln(a),^1e^a,^2e^a,...,^ne^a$ $\ (a\neq 0)$ are algebraically independent $(^ne^a=g_1(g_2(...(g_n(a))...))\ \ \ (g_1,...,g_n=\exp))$. |
| Posted: 04 Jul 2022 12:03 PM PDT P is a variable point on a given straight line $p$ and A is a fixed point such that $A\notin p$. Q is a point on the segment $AP$ such that ratio AQ:AP is constant. Find locus of point Q. |
| Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$ Posted: 04 Jul 2022 12:36 PM PDT
My attempt, As I know from the original equation, $\alpha+\beta+\gamma=-a$ $\alpha\beta+\beta\gamma+\alpha\gamma=b$ $\alpha\beta\gamma=-c$ I've tried to expand $(\alpha+\beta+\gamma)^3$ which is equal to $\alpha^3+\beta^3+\gamma^3+3\alpha^2\beta+3\alpha\beta^2+3\alpha^2\gamma+6\alpha\beta\gamma+3\beta^2\gamma+3\gamma^2\alpha+3\gamma\beta$ Basically, I know I've to find what's the value of $\alpha^3+\beta^3+\gamma^3$, $\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3$ and $\alpha^3\beta^3\gamma^3$. But I m stuck at it. I would appreciate can someone explain and guide me to it. Thanks a lot. By the way, I would appreciate if someone provides another tactics to solve this kind of routine question. Thanks a lot. |
| Perspective projection of a circle: what is the size of the semi-major axis? Posted: 04 Jul 2022 12:00 PM PDT It can be proven that the perspective projection (or camera projection) of a circle is an ellipse. But I also need to prove that the semi-major axis has the same size as the radius of the original circle. Any idea on how to prove it? Cheers! |
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