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Basic math problem with Integrating Factors: Differential Equations I was watching a youtube tutorial on Integrating Factors and I'm lost in a part where the derivative of: xy' + 1y becomes xy.
Please I need some clarification on that part.
Secondly.
I was watching a another youtube tutorial on Integrating Factors and I also got lost in this part.
https://i.stack.imgur.com/pmcP6.png
I don't understand how the first line converts to the other, Please I also need some clarification here.
Thanks.
| Suppose you have
$$xy'+1y=0$$
Notice that we have
this is that we have $$\frac{d}{dx}(xy)=x\frac{dy}{dx}+\frac{dx}{dx}y=xy'+y$$
by product rule.
Similarly, for $$\frac{dy}{dx}+\frac{2x}{1+x^2}y=0$$
By multiplying integrating factor of $(1+x^2)$, we have
$$(1+x^2) \frac{dy}{dx}+2xy=0$$
which can be written as
$$(1+x^2) \frac{dy}{dx}+\frac{d(1+x^2)}{dx}\cdot y$$
or
$$\frac{d}{dx}(y(1+x^2))=0$$
| {
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intersection of all neighborhoods of a point in zariski topology. Consider the affine space $A^n$ with the Zariski topology,
V a variety (with the induced topology) and let $P\in V$ a point. Let B be the set of all neighborhoods of the point P in V. Is it true that $P=\bigcap\limits_{U_i\in B} U_i$?
Obviously this is not true on a general topological space (e.g consider the trivial topology where the only open sets are $\emptyset$ and the whole space), so the fact that we have Zariski topology is important.
Apart from that, I' m not sure on how to proceed. Intuition tells me that this is indeed true, but given the fact that open sets here are very big in size (they 're dense, so no two open sets can have an empty intersection), I 'm not confident of the result.
Any hint would be welcome.
| Follows from the fact that $\Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.
| {
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Show that each connected component of a topological space $X$ is connected.
Show that each connected component of a topological space $X$ is connected.
Here is my proof
My Proof: Choose a connected component $$C = [x] = \{y \in X \mid \exists \text{ a connected set containing both $x$ and $y$}\}$$
Now for each $y \in C$ let $C_y$ be the connected set containing $x$ and $y$. We claim that $C = \bigcup_{y \in [x]}C_y$. To prove this claim pick $\alpha \in \bigcup_{y \in [x]}C_y$, then $\alpha \in C_y$ for some $y \in C$. Since $C_y$ is a connected set in $X$ containing both $\alpha$ and $x$, it follows that $\alpha \in C$ hence $\bigcup_{y \in [x]}C_y \subseteq C$.
Conversely pick $\beta \in C$, then by construction $\beta \in C_{\beta}$ which is the connected set containing both $x$ and $\beta$. Thus $\beta \in \bigcup_{y \in [x]}C_y$ and we have $C = \bigcup_{y \in [x]}C_y$.
Now since $\bigcup_{y \in [x]}C_y$ all contain $x$, their intersection is nonempty and since each $C_y$ is connected it follows that $\bigcup_{y \in [x]}C_y$ is connected. Thus $C$ is connected. $\square$
First of is my proof correct? Can I improve it in any way? I'm not too happy about my construction of $\bigcup_{y \in [x]}C_y$, because I said
"for each $y \in C$ let $C_y$ be the connected set containing $x$ and $y$"
but there's no need for $C_y$ to be unique and that made matters a bit hairy when I tried to show that $C \subseteq \bigcup_{y \in [x]}C_y$
Is there a better way to prove this?
| I don't think it matters too much that the sets $C_y$ may not be unique. The proof rests on the notion that a union of connected sets with common intersection is connected, which seems plausible (I haven't tried to prove it though). If that isn't an established proposition in your text though, I think it should be proved.
But the construction seems a little complicated. You might just appeal to the definition of connected. If $C = [x]$ were not connected, you could find disjoint sets $X$ and $Y$ such that $C \cap X \neq \varnothing$, $C \cap Y \neq \varnothing$, and $C \subseteq X \cup Y$. Choose an $x\in X \cap C$ and $y \in Y \cap C$, and a connected set $D$ containing $x$ and $y$. Can you argue to a contradiction from here?
| {
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Determining the null space of the matrix
Determine the null space of the matrix:$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$$
My try:
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}_{R_2\rightarrow R_2-2R_1\\R_3\rightarrow R_3-R_1}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}_{R_3\rightarrow 5R_3-2R_2}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 0 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{5}}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$
From this I got $$x-y=0\implies x=y\\y=0$$
$$(x,y,z)^T=(y,0,z)^T=y(1,0,0)^T+z(0,0,1)^T$$
So, $(1,0,0)^T$ and $(0,0,1)^T$ is the null space. Is this correct?
| Recall that by definition the nullspace is the subspace of all vectors $\vec x$ such that $A\vec x=\vec 0$ and in that case we have ony the trivial solution $(x_1,x_2)=(0,0)$ then $Null(A)=\{\vec 0\}$.
Notably to solve $Ax=0$ we can proceed by RREF to obtain
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}\to
\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}\to
\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$
that is
$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
that is $x_1=x_2=0$.
| {
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How to write a polynomial in mod p Consider the polynomial $ \ x^3 - 3\ x^2 +2\ x -1$
How can this polynomial be written in mod 3 ?
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
|
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
Your first thought is correct: we don't change any power. Even though $3$ is in the exponent, we don't take the exponents mod 3, but only the coefficients.
So, taking a polynomial mod 3 just means taking each coefficient mod 3 (and leaving the exponents). For example, $6x^3 + 4x^2 - 2 \mod 3$ would be $0x^3 + x^2 + 1$, which is the same as $x^2 + 1$ since we can drop $0$s. This is because $6 \mod 3$ is $0$, $4 \mod 3$ is $1$, and $-2 \mod 3$ is $1$.
| {
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Is there a fundamental mathematical function that requires 3 inputs or more? So a mathematical operation can be represented as a function that maps inputs to outputs. For example "sin(x)" is a function that maps 1 input to 1 output, and "a + b" maps 2 inputs to 1 output. My question is is there a function that requires a minimum of 3 inputs on a fundamental level? I'm not talking about something like "a + b + c" even though that has 3 inputs because the steps of solving that function are evaluated 2 variables at a time ie "(a + b) + c". Is there something that needs 3 or more inputs to make sense in the same way addition needs 2? I don't have very strong math background so I didn't know how to google my question properly or how to tag it here properly. Thank you for your patience.
| Thanks for your answers and comments! A conditional statement is to me a satisfactory answer. In order for a conditional statement to make sense it requires (a) the condition, (b) the yield if the condition evaluates to true, and (c) the yield if the condition evaluates to false. To the best of my knowledge all three components are required. Sorry if this answer is not satisfactory to your interpretation of the question.
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show that $\operatorname{rank}(g\circ f) \leq \operatorname{rank}(f)+\operatorname{rank}(g)-n$ Let $E$ a vector space and $\dim(E)=n$
and let $f,g \in L(E)$
show that $\operatorname{rank}(f\circ g) \leq \operatorname{rank}(f)+\operatorname{rank}(g)-n$
I can see that $\operatorname{Ker}(g) \subset \operatorname{Ker}(f\circ g)$
so $\dim \operatorname{Ker}(g) \leq \dim \operatorname{Ker}(f\circ g)$
by the rank-nullity theorem $\operatorname{rank}(g) \leq \operatorname{rank}(f\circ g)$
I am stuck here.
| The reverse inequality is true. To see it, apply the rank-nullity theorem twice. $\DeclareMathOperator{\Im}{Im}$
Observe in the first place that $$\DeclareMathOperator{\rk}{rank}\rk(g\circ f)=\rk\Bigl(g_{\,\bigm\vert_{\,\scriptstyle\Im f}}\Bigr)\quad\text{and}\quad
\ker\Bigl(g_{\,\bigm\vert_{\,\scriptstyle\Im f}}\Bigr)=\ker g\cap\Im f, $$
so, by the rank-nullity theorem
$$\rk(g\circ f)= \dim(\Im f)-\dim(\ker g\cap\Im f)=\rk f-\dim(\ker g\cap\Im f). $$
Now $\;\ker g\cap\Im f\subset \ker g$, whence
$$\dim(\ker g\cap\Im f)\le \dim(\ker g)=n-\rk g, $$
and therefore
$$\rk(g\circ f)\ge\rk f-(n-\rk g).$$
| {
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Uniform Convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n} x^n$ Prove that $\sum_{n=1}^\infty \dfrac{(-1)^n}{n} x^n$ converges uniformely on $[0,1]$.
Using the Leibiz criteria, I could show, that the series converges for $x\in[0,1]$. But how can I show the unifom convergence? I know about the Weierstrass M-test but that doesn't apply here...
| The Dini Theorem: If each $f_n:[0,1]\to \Bbb R$ is continuous and if $(f_n(x))_n$ converges monotonically to $f(x)$ for each $x\in [0,1],$ and if $f$ is continuous, then $f_n\to f$ uniformly on $[0,1].$
Apply this, first to $(g_n)_n,$ where $g_n(x)=\sum_{j=1}^{2n}(-1)^jx^j/j,$ and second to $(h_n)_n$ where $h_n(x)=\sum_{j=1}^{2n-1}(-1)^j x^j/j.$
The series in the Q converges to $f(x)=-\ln (1+x),$ which is continuous on $[0,1].$
Proof of the Dini Theorem. Given $\epsilon >0:$
For each $x\in [0,1]$ take $n_x$ such that $0\leq |f(x)-f_{n_x}|\leq\epsilon /3.$ Then take $a_x>0$ such that $(y\in [0,1] \land |y-x|<a_x)\implies |f_{n_x}(y)-f_{n_x}(x)|\leq\epsilon /3.$ And take $b_x>0$ such that $(y\in [0,1]\land |y-x|<b_x)\implies |f(y)-f(x)|\leq\epsilon /3.$
Let $c_x=\min (a_x,b_x).$
Now $\{(-c_x+x,c_x+x):x\in [0,1]\}$ is an open cover of $[0,1] ,$ and $[0,1]$ is compact . So there exists a finite non-empty $S\subset [0,1]$ such that $\cup_{x\in S}(-c_x+x,c_x+x)\supset [0,1].$
Let $M=\max \{n_x:x\in S\}.$
Now for $y\in [0,1]$ take $x\in S$ such that $y\in (-c_x+x,c_x+x).$ For any $m\geq M$ we have $$|f_m(y)-f(y)|\leq |f_{n_x}(y)-f(y)|\leq$$ $$\leq |f_{n_x}(y)-f_{n_x}(x)| +|f_{n_x}(x)-f(x)|+|f(x)-f(y)|\leq$$ $$\leq \epsilon /3+\epsilon /3 +\epsilon /3=\epsilon.$$ Therefore $\sup_{m\geq M}\sup_{y\in [0,1]}|f(y)-f_m(y)|\leq \epsilon.$
Remarks: The Dini Theorem holds if the domain is any compact space (not just for $[0,1].$) Compactness is used to ensure that $M$ exists, because $S$ is finite. Monotonicity of each sequence $(f_n(y))_n$ is used to ensure that $m\geq M\geq n_x\implies |f_m(y)-f(y)|\leq |f_{n_x}(y)-f(y)|.$
Further remarks. In the Dini Theorem we cannot omit the continuity of $f.$ For example if $f_n(x)=x^n$ for $x\in [0,1].$ And we cannot omit the monotonicity of each sequence $(f_n(x))_n$. For example if $f_n:[0,1]\to [0,n]$ where $f_n(x)=0$ when $x\in [0,1/3n]\cup [1/n,1]$ and $f_n(1/2n)=n.$ Interestingly, in this example each sequence $(f_n(x))_n$ is eventually monotonic but the convergence is not uniform.
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Sum of n terms of this series $\frac{1}{1.3} + \frac{2}{1.3.5} +\frac {3}{1.3.5.7} + \frac{4}{1.3.5.7.9}........ n $ Terms.
I Know the answer to this problem but I couldn't find any proper way to actually solve this question. I thought the denominators were the product of n odd natural numbers. So I wrote the nth term as:
$\frac{2^{n+1}(n+1)!.n }{[2(n+1)]!}$
But I don't know if that's correct or even what to do next.?
| The $n$th term has $n$ as its denominator, and the odd factors in $(2n+1)!$ as its numerator. Or, if we multiply both by the even factors, which are $\prod_{i=1}^n 2i=n!2^n$, the denominator becomes $(2n+1)!$. The $n$th term is therefore $\dfrac{n! n2^n}{(2n+1)!}$. (The formula you've obtained is also correct, by the way; you've just added a factor of $2(n+1)=2n+2$ into the two parts of the fraction.)
Alternatively, we could leave the numerator and denominator as they were originally, and write the $n$th term as $\dfrac{n}{(2n+1)!!}$, where the subfactorial $k!!$ is defined to only include the factors from $1$ to $k$ of the same parity as $k$. (In particular, the rescaling I suggested above used a factor of $(2n)!!$.)
As for the sum of the first $n$ terms, I think you can prove by induction it's $\frac{a_n}{2a_n+1}$ with $2a_n+1=(2n+1)!!$.
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$ABC$ is a triangle there are points $D, E$ and $F$ on sides $AC, AB$ and $BC$ respectively if $AF = 4$ and $BD = 12$ find the minimum value of $EC$ Does this use any specific inequality?
here's my approach:
since a line interior of a triange drawn from a vertice cannot exceed the neighbouring sides we can set an inequality
and also all the measurements are integers
| It’s simple if you allow points A&D, C&F, and B&E to coincide. Then, a variation of (assuming angle BAC is not right) the Pythagorean theorem will yield [a variation of] $(EC)^2=(AF)^2+(BD)^2$. Whatever the approach, the minimum length of EC will require the segments AF, BD, and EC each to be the entire length of their respective sides.
| {
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Prove by induction that $n^2 + n + 1 \forall n\geq 1$ given the following recurrence relation My question is as follows:
Consider the following recurrence relation:
$a_{n} = a_{n-1}+2n$, with $a_{1}=3$
Prove by induction that
$a_{n}=n^{2}+n+1 \forall n\geq 1$
I have no idea how to even approach this question. I am quite familiar with the process of induction but am just unsure how to draw that in for this question? I am not very familiar with recurrence relations.
| Induction is usually achieved in three steps.
1) Initialization. You just need to check that your property is valid for the lowest integer required (here $n = 1$). So what you need to check is that indeed, $a_1 = 1^2 + 1 + 1$.
2) Heredity. This step is usually the most difficult, and you have to be careful when writing it. At this point, you assume that your property is valid for some $n \geq 1$ (and certainly not for all $n$, in which case there is nothing left to show).
Therefore, you assume that, for some $n \geq 1$, $a_n = n^2 + n + 1$. Now use the recurrence relation to deduce that $a_{n+1} = (n+1)^2 + (n+1) + 1$.
3) Conclusion. In step 2, you have shown that if your property is valid at rank $n$, it is also valid at rank $n+1$. In step 1, you have shown that the property is valid at rank 1. Thus, the property is valid at rank 1+1 = 2, then at rank 2+1 = 3,... and finally for any $n \geq 1$. This step is usually skipped once one is familiar with the whole induction process.
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Quantifying a free variable in an example from "How To Prove It" by Velleman This is an example from How To Prove It by Daniel J. Velleman (2nd Ed., p. 71):
Example 2.2.3. Analyze the logical forms of the following statements.
*
*Statements about the natural numbers. The universe of discourse is $\mathbb N$.
a. $x$ is a perfect square.
And here is the solution:
*
*a. This means that $x$ is the square of some natural number, or in other words $\exists y(x = y^2)$.
My question is regarding the free variable $x$. Is it correct to leave it free, as the author has done? I was under the impression that, by convention, a free variable may be assumed to be universally quantified, which I think wouldn't make sense in this case.
If I wanted to fully formalize the statement, how would I do that? My best guess is as follows:
$$\exists x\exists y(x = y^2)$$
Is this correct?
| You shouldn't confuse variables (the only ones that can be quantified) with individual constants.
The logical form of the sentence "$x$ is a perfect square" is indeed $\exists y (x = y^2)$. In this formula, $x$ plays the role of an individual constant, not of a variable, because it refers to one specific individual in the universe of discourse $\mathbb{N}$.
The meaning of $\exists x \exists y (x = y^2)$ is "there is a perfect square", which is completely different because it does not refer to one specific individual in the universe of discourse $\mathbb{N}$.
Remark. The starting point of my answer above is the fact that the exercise claims that "$x$ is a perfect square" is a statement, that is a a meaningful declarative sentence that is true or false because every sign in it is interpreted, hence $x$ should refer to a specific individual in the universe of discourse.
If you see "$x$ is a perfect square" just as a phrase which is not a sentence (and so it is not true or false because not all its signs are interpreted), then you can see $x$ as a free variable (waiting for an interpretation to assign a truth value to the phrase). Even in this case the logical (but meaningless) form of the phrase is $\exists y (x = y^2)$.
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How to find the coefficient of $a$ in a quadratic equation?
If a root of the equation $$3x^2 + 4x + 12a + 9ax = 0$$ is greater than 6, then the correct statement of the coefficient $a$ is:
a) $a = 2$
b) $a> -2$
c) $a = -2$
d) $a <-2$
e) $-2\le a\le 2$
What I did was solve it as if a root was exactly $6$, and I find that $a$ is equal to $2$ ... But the premise says that a root has to be greater than $6$, not equal to $6$ ... and I do not know how find that, for now I would know that $a$ does not have to be $2$ ...
| Hint
\begin{align} 3x^2+4x+12a+9ax = 0&\iff 3x^2+(4+9a)x+12a= 0\\ &\iff x=\dfrac{-4-9a\pm \sqrt{(4+9a)^2-144a}}{6}.\end{align} So,
$$\dfrac{-4-9a\pm \sqrt{(4+9a)^2-144a}}{6}\ge 6 \iff -4-9a\pm \sqrt{(4+9a)^2-144a}\ge 36. $$
That is
$$-4-9a\pm \sqrt{16+72a -63a^2}\ge 36$$
So, you can check what is the correct answer.
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Need help finding a sum I found this problem that I'm not sure how to solve. I would appreciate if anyone could point me to the right direction.
I need to find the following sum:
$\sum_{i+j+k=7} (-1)^i(-1)^j\frac{7!}{i!j!k!}$ where $i,j,k$ are elements of $\mathbb{N_0}$
There must be a much better way to solve this other than writing down all possible combinations and sum them (please correct me if I'm wrong).
The thing I believed could be useful was the multinomial theorem (I apologize if this is not the correct term in English), which states that:
$(\sum_{i=1}^k x_i)^n=\sum_{\sum_{i=1}^k n_i =n}\frac{n!}{\prod_{i=1}^k n_i!}\prod_{i=1}^k x_i^{n_i}$
But I'm missing $(-1)^k$.
I'm a bit new to this subject, but couldn't find any example like this in my books. Thank you in advance!
EDIT: My apologies to anyone who read the original question, I've misplaced $i$ and $j$, didn't see it while revising what I wrote. It's fixed now.
| There is no $(-1)^k$ because $x_3$ is going to be $1$: $$-1=((-1)+(-1)+1)^7=\sum_{i+j+k=7} (-1)^i(-1)^j(1)^k\frac{7!}{i!j!k!}$$
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How to construct a matrix given the null basis of A? Construct a $4\times4$ matrix $A$ such that $\{(1,2,3,4),(1,1,2,2)\}$ is a basis of $N(A)$.
So I know that $A$ will have two pivot columns and two free columns, but beyond this I'm not sure how to approach/solve.
| The row space of a matrix is the orthogonal complement of its null space. So, you can construct the required matrix by finding a basis for this orthogonal complement. In this case, this will give you two of the rows, and the other two rows can be any linear combinations of those two rows, including rows of all zeros.
Calling the two given vectors $\mathbf n_1$ and $\mathbf n_2$, the orthogonal complement of their span is the set of all vectors $\mathbf x$ that satisfy $\mathbf n_1\cdot\mathbf x=\mathbf n_2\cdot\mathbf x=0$. This is a pair of homogeneous linear equations in the components of $\mathbf x$, so $\mathscr N(A)^\perp$ is the null space of the matrix $\small{\begin{bmatrix}\mathbf n_1 & \mathbf n_2\end{bmatrix}}^T$. I’m sure you know how to compute that.
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Discriminant (in the context of PDE classification): $b^2 - 4ac$ or $b^2 - ac$? I'm reading two textbooks on partial differential equations. In their respective sections on classification of PDEs (hyperbolic, parabolic, elliptical), they differ in what they describe as being the discriminant. One textbook says that the discriminant is $b^2 - 4ac$, while the other describes it being $b^2 - ac$. Are these both correct, or is one correct and the other incorrect?
EDIT: What are the $a$, $b$, and $c$? The second-order linear PDE in two independent variables is $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants.
| A few points -
*) If we are discussing the second order linear PDE of two independent variables, which say is of the form - $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants, then we must be careful about the coefficients of the PDE, they are $A,B,C,D,E,F,G$ and not $a,b,c,d,e,f,g$, (very simple yet much important).
*) If we are considering the PDE $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants, then their classification is as follows -
We check the Discriminant which is $d = B^2(x_{0},y_{0}) - 4 A(x_{0},y_{0})C(x_{0},y_{0})$.
At $(x_{0},y_{0})$, the equation is said to be -
1) Elliptic if $d <0$
2) Parabolic if $d = 0$
3) Hyperbolic if $d>0$
If this is true for all points $(x_{0},y_{0}) \in $ domain $\Omega$, then the equation is said to be Elliptic, Parabolic or Hyperbolic in that domain.
Extra point -
Now if there is the case of $n$ independent variables like $x_{1},x_{2},...,x_{n}$and second order linear PDE (of the form $\sum \sum a_{i,j} u_{x_{i},x_{j}}+ $ lower order terms $= 0$ then the classification depends on the signature of the eigenvalues of the coefficient matrix.
*) Elliptic if the eigenvalues are all positive or all negative.
*) Parabolic - The eigenvalues are all positive or all negative, save one which is zero
*) Hyperbolic - There is only one negative eigenvalue and all the rest are positive, or
there is only one positive eigenvalue and all the rest are negative.
Few references similar to this question -
*) How to determine where a non-linear PDE is elliptic, hyperbolic, or parabolic?
*) Characterizing 2nd order partial differential equations
*)Classification of a system of two second order PDEs with two dependent and two independent variables
Perhaps by checking the second reference above and the answers to your questions above you will know the difference of usage of $B^2 -4AC$ and $B^2 - AC$!
| {
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What's the equation of the graph $y=x^3-x^2+x-2$ after it is reflected in the x axis, and then the y axis as well? I don't fully understand the steps to reach the answer.
I've plotted the graph and I found out that the graph is an odd graph and that after reflecting in both axes, the only thing that changes on the graph are the points where it intersects the x & y axis.
I would like to see the algebraic steps one would take on how to solve this equation to better grasp what is going on, thanks.
| 1)Reflection about $x-$axis:
$(x,y) \rightarrow (x,-y)$;
2)Reflection about $y$-axis:
$(u,v) \rightarrow (-u,v)$;
Combining:
3) $(x,y) \rightarrow (x,-y) \rightarrow (-x,-y)$.
Given : $y=f(x) =x^3-x^2+x-2$;
We have $(x,f(x)) \rightarrow (-x,-f(x))$, the reflected graph .
$(-x)$ is mapped to $-f(x)$, or
$x = -(-x)$ is mapped to $-f(-x)$.
The reflected graph is $(x,-f(-x))$, or
$y = -( (-x)^3 -(-x)^2+(-x) -2)$.
| {
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Prove that if $f$ differentiable in $x^*$ and $f(x^*)=0$, then $\liminf_{x\to x^*}\frac{|f(x)|}{||x-x^*||}=0$ if $n>1$ Let $f\colon\mathbb{R}^n\to\mathbb{R}$ be differentiable at $x^*$ and $f(x^*)=0$. Prove that if $n>1$, then $$\liminf_{x\to x^*}\frac{|f(x)|}{||x-x^*||}=0.$$
Is this true for $n=1$?
I know that because $f$ is differentiable in $x^*$, there is a linear transformation $L\colon \mathbb{R}^n\to\mathbb{R}$ such that $$0=\lim_{x\to x^*}\frac{|f(x)-f(x^*)-L(x-x^*)|}{||x-x^*||}=\lim_{x\to x^*}\frac{|f(x)-L(x-x^*)|}{||x-x^*||}.$$ How to proceed from there? And what is special about $n=1$?
| Why the difference between the cases $n=1$ and $n>1$?
For $n=1$, the kernel of a linear map can be reduced to the zero vector. This isn't the case for $n>1$.
The result is true for $n>1$
In that case $L$ is a linear form and its kernel is not reduced to the zero vector. Take $a \in \ker L \setminus \{0\}$. For $m \in \mathbb N$ and $x_m = \frac{a}{m} + x^*$, you have $L(x_m-x^*) = 0$ and $\lim\limits_{m \to \infty} \dfrac{|f(x_m)-L(x_m-x^*)|}{||x_m-x^*||}= \dfrac{|f(x_m)|}{||x_m-x^*||}=0$. As $\lim\limits_{m \to \infty} x_m = x^*$ you have $\lim \inf_{x\to x^*}\dfrac{|f(x)|}{||x-x*||}=0$.
The result is wrong for $n=1$
Just take $f(x)=x$ and $x^*=0$. You have $\dfrac{|f(x)|}{||x-x*||}=1$ for all $x \in \mathbb R \setminus \{0\}$.
| {
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Ask about global maximum and global minimum? The temperature distribution in a metal rod given by the following function of the position $x \in \mathbb{R}$: $$T(x) = \frac{1 + 2x}{2 + x^2}$$
What is the maximal and minimal temperature in the metal rod?
$T'(x) = 0$ when $x = 1$ or $x = -2$. But I can't calculate the global maximum and global minimum because $T(x)$ does not belong to any intervals? Am I correct?
Thank you.
| Assuming your work on the derivative is correct (I haven't checked), since $T$ is continuous:
\begin{align}
T(1)&=1\\
T(-2)&=-\frac{1}{2}\\
T(-\infty)&=T(+\infty)=0
\end{align}
So there you have: global max at $x=1$ and global min at $x=-2$.
| {
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How to argue that a series that isn't a power series is analytic? Let $f(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{1-z^{n}}$. Then $f$ converges in the unit disc. I want to show that it is analytic in this region as well, but since it is not a power series I don't have any theorems to apply. My only thought was to try and use similar reasoning as with power series: show that converges is uniform so that we can switch integral and sum to show that the integral of the series is 0, then apply Morera's Theorem. But the problem is, according to Wolfram Alpha, $\int_{\gamma}\frac{z^{n}}{1-z^{n}}=\int_{0}^{2\pi}\frac{e^{int}ie^{it}}{1-e^{int}}dt$ doesn't converge.
EDIT: To apply Morera's Theorem $\gamma$ should be closed path contained in the unit circle. (It probably should have been obvious that the integral doesn't converge since the unit circle contains 1...). So the above strategy does works (yay!) since the function is holomorphic in any closed disk with radius less than 1.
| Let $C$ be a compact subset of the open unit disk and let $M=\sup_{z\in C}|z|$. Then $M<1$ and, for each $n\in\mathbb N$,$$\left|\frac{z^n}{1-z^n}\right|=\frac{|z|^n}{|1-z^n|}\leqslant\frac{M^n}{1-|z|^n}\leqslant\frac{M^n}{1-M^n}.$$Since the series $\sum_{n=1}^\infty\frac{M^n}{1-M^n}$ converges, your series converges uniformly on $C$. Since this holds for every compact subset of $D_1(0)$, the sum of the series is analytic.
| {
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What does it mean for points to be picked uniformly and independently? I saw a which said that points were picked uniformly and independently. I have a feeling this is important for the solution but I am not sure what they mean by uniformly and independently. Any help would be appreciated. I will post the original question here.
Question:
Four points are chosen uniformly and independently at random in the interior of a given circle. Find the probability that they are the vertices of a convex quadrilateral.
Note: I am not looking for a solution to the problem, just for clarification on what uniformly and independently mean.
| Sometimes a picture helps more than words and formulas, at least for the term "uniform":
"Independent" means that a single one of the random points does not care where the other points are, but its position is picked independently of the others, as if they were not there. E.g. to be not independently chosen, a point could try to avoid being placed at a position where many other points have been placed already. This can give a uniform pattern, but they are not chosen independently (now the order in which the points are chosen matters).
As Hurky said, when one (mostly a non-mathematician) says "random", what he means most of the time includes the properties uniform and independent. E.g. throwing a dice gives uniformly and independently chosen random numbers from $\{1,2,3,4,5,6\}$. Let's see what the same experiment might look like when one of the properties is missing:
*
*not uniform: the dice is biased, e.g. it was manipulated to show the $6$ more often (on average) than any other number.
*not independent: the dice "has a memory". E.g. it tries to not show the same number as on the last throw. Sometimes people think that because there was no $6$ for a long time, the probability that there will come a $6$ is now higher. This can only be true if the dice does not generate independent numbers, which is usually not the case.
| {
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Why is the following statement impossible for a sequence $\{p_n\}$?
A sequence is said to be convergent if there is a point to which it converges. A convergent sequence cannot converge to two distinct limits. For if $\{p_n\}$ were to converge to both $p$ and $q$ with $p \ne q$, then we could choose spherical neighborhoods about $p$ and $q$ that are disjoint; but, all but a finite number of the $p_n$ would have to be inside each of these neighborhoods, which is impossible.
I'm a little confused by the following statement.
but, all but a finite number of the $p_n$ would have to be inside each of these neighborhoods, which is impossible.
Is it impossible by definition of a convergent sequence?
| It is impossible to converge to two different points by definition of convergence described by @copper.hat on the comment above. You cannot have only finite number of values outside two disjoint neighborhoods
| {
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Given $k$ points in $n$-dimensional space, is there always a continuous $n-1$ surface that can divide the points into two arbitrary groups? Say we have $k$ points in set $P\mid x_i\in\mathbb{Z}^n$, such that $k=\mathcal{O}(n!)$. We now arbitrarily divide the points into two sets, $A$, $B$. Note that $A\cup B = P$ and $A \cap B = \varnothing$.
Is there always $n-1$ dimensional surface in $\mathbb{R}^n$ that perfectly separates these two sets of points? Is there some transformation $T:\mathbb{R^n}\rightarrow\mathbb{R^n}$, such that this $n-1$ dimensional surface is a function $f(\vec x)$? If so is it always possible to find this function?
I know this is similar to the idea of SVN machine learning, but I'm more interested in the theoretical ideas about existence of these divides.
| Solve the poisson equation,
$\Delta u = f$
with $f$ consisting of delta function sources at points in $A$ and sinks at points in $B$. The zero set of the solution $u$ separates the points. One should be able to prove that either this zero set is smooth, or the zero set associated with arbitrarily small perturbations of the points is smooth, by using elliptic regularity and Sard's theorem.
You could also compute the sensitivity of the surface to movement of the points with the implicit function theorem.
The surface may consist of several disconnected components though.
| {
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Proving $f'$=$g'$ has some c such that $g=f+c$ Suppose that $f$ and $g$ are differentiable functions on $(a,b)$ and suppose that $g'(x)=f'(x)$ for all $x \in (a,b)$. Prove that there is some $c \in \mathbb{R}$ such that $g(x) = f(x)+c$.
So far, I started with this:
Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then $h'(c)=0 \implies h(c)=c$
After this i'm not sure where to go, or if this is correct at all, any hints?
This is also my first post in Latex so sorry if there's any mistakes!
| You say that
Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in
> \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then
$h'(c)=0 \implies h(c)=c.$
This is not correct. If $h'(c)=0$ then you have $h(b)=h(a).$
But you are in the correct way. Instead of $a,b$ consider $x,y\in [a,b], x\ne y.$ Then you have
$$h'(c) = \frac{h(y)-h(x)}{y-x} =0.$$ So $h(x)=h(y)$ from where you get that $h$ must be constant.
| {
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When "If $A$ is true then $B$ is true", is it valid to assert that "If $B$ is false, $A$ must also be false"? If it is given that:
"People that ride buses, also ride planes"
then is the statement
"people that don't ride planes, also don't ride buses"
necessarily true?
I don't think so, but the explanation to a problem in textbook I'm using uses that as logical proof to the answer provided.
It is based on the rule (according to this textbook) that, if this is true : ( if $A$ is true, then $B$ is also true), then it follows that if $B$ is false, then $A$ must also be false.
| What you are asking is does:
A ⟹ B (A implies B)
mean
Not B ⟹ Not A (The converse of B implies the converse of A)?
The answer is yes.
In fact this is the basis of what is known as a Proof By Contradition in Maths.
The way I explain it to my A-Level students (18 year old Mathematicians in the UK) that meet this for the first time is with the following.
Let Statement A = "It is 12th of August 2018"
and
Let Statement B = "It is a Sunday"
Here
A ⟹ B
However, if it is not a Sunday (ie Not B, or the converse of B) is true, then we do not know what date it is with an absolute certainty, but we can say with absolute certainty it is NOT the 12th of August 2018, (ie we can say it is Not A, or we can say it is definitely the converse of statement A). It might be the 11th, it might be the 10th, but it definitely is not the 12th.
The famous one in Maths is the proof of the √2 cannot be expressed exactly as a fraction, ie √2 is irrational. This is done by a Proof by Contradition.
For contradition assume:
A = "√2 is rational"
If that is true, then
B = "√2 = a/b, where a and b are whole numbers, and 'a' and 'b' have no factors"
So we assume B is true and with a bit of basic maths we find if √2 = a/b, then both a and b have a factor of 2. So we have establish NOT B.
But from what we have said above:
If A ⟹ B, then Not B ⟹ Not A
So we have established that Not A is true, ie √2 is irrational, ie cannot be expressed as a fraction.
| {
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Zeroth-homology of a complex of $n$ connected components I am new at Algebraic topology and am reading Basic Concepts of Algebraic Topology of Croom.
I have a question. In Theorem 2.4/page 25, it states that if $K$ is a complex with $n$ connected components, then $H_0(K)$ is isomorphic to $\mathbb{Z}^n$.
Anyway, in the proof, Croom showed that "Applying this result to each connected component $K_1,..., K_n$
of $K$,
there is a vertex $a_i$
of $K_i$
such that any $0$-cycle on $K$
is homologous to a $0$-chain
of the form $\sum h_i \langle a_i \rangle$ where $h_i$ is a integer and $\langle a_i \rangle$ denotes the $0$-cycle that maps $(a_i)$ to 1 and other $0$-simplicies to $0$.
Hence it suffices to show that the representation here is unique, which means if we have $\sum (g_i- h_i) \langle a_i \rangle=\partial (c)$, then $g_i=h_i$. This part is clearly trivial to Croom, but I do not understand.
Can you clarify this part for me? Thank u
| The boundary of a $1$-chain is a linear combination of
boundaries of $1$-simplex. A $1$-simplex in $K$ has endpoints in the
same component of $K$, so in the same $K_i$. If $\partial(c)=\sum_v r_v
\langle v\rangle$ then the sum of the $r_v$
over the vertices $v$ in the same $K_i$ is zero. In your example this sum
is $g_i-h_i$.
| {
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Try to learn more about linear operator,subspace and dimension Prove that for every subspace $F$ and every linear transformation $L$ of linear space $V$ and $\dim V=n$ is $\dim L(F)+\dim\ker L=\dim(F+\ker L)$
I know that $F$ and $\ker L$ are subspaces of the same vector space, so we have $\dim(F+\ker L)=\dim F +\dim\ker L -\dim(F\cap\ker L)$. And if we have some basis for subspace $F, B=\{f_1,f_2,\ldots,f_k\}\ k\le n,$ and they are linearly independent, then $L(f_1), L(f_2),\ldots,L(f_n)$ are linearly independent if $N(L)=0$, because this is an isomorphic linear transformation (iff they have same dimension, but here we have) then $\dim L(F)=\dim F$,then we have $\dim(F\cap\ker L)=0$, but I do not know is this good how I explain?
| This is a consequence of the isomorphism theorems for modules (in particular, vector spaces).
We have $$ L(F) \simeq \frac{F}{F\cap kerL} \simeq \frac{F+kerL}{kerL}.$$
Computing the dimensions we have $$dimL(F) = dim(F+kerL) - dimKerL. $$
| {
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Expected least distance between closest two points out of $n$ drawn from a distribution Suppose I draw $n$ points in $\mathbb{R}$ from a distribution $p$. What is the expected least distance between two of the points drawn? I am particularly interested in the uniform distribution $\texttt{Unif}(a,b)$.
I'd be interested in other statistics about this random variable besides just its expectation.
| With the uniform distribution, this can be discretized nicely, and then it leads to a standard combinatorial problem. For simplicity, I am working on the interval $[0,1]$.
So divide the interval into $N$ equal subintervals ($N$ is large). Then we pick $n$ of these (the points picked are identified by the small intervals they are contained in), and imagine that the distance of points is about the distance of the midpoints of the intervals.
So by rescaling with a multiple $N$, we really have numbers $1, \ldots, N$ representing chairs, pick $n$ of them randomly (we let $n$ people sit down), and the random variable is the distance of the closest two people.
Number of cases: about $\binom{N}{n}$ (people are twins; also we may assume that no two people sit in the same chair, as it doesn't make a difference for large $N$).
Number of cases when minimal distance is at least $k$ for $1\leq k\leq (N-1)/(n+1)$?
To obtain this, make the people stick out their left arm, so that they not only cover their own chair, but also the next $k-1$ to the left. In order for the last chairs to be available, we have to insert $k-1$ extra chairs on the left.
That means that the resulting picture after the people are seated will contain $n$ people with extended arms and $N+k-1-nk$ unoccupied chairs. So the number of possibilities is the number of choosing $n$ objects out of $N+k+n-1-nk$ ones, that is $\binom{N-(n-1)(k-1)}{n}$.
So the expected value is $$\frac{1}{\binom{N}{n}}\sum\limits_{k=1}^{(N-1)/(n+1)} \binom{N-(n-1)(k-1)}{n}= \frac{1}{\binom{N}{n}}\sum\limits_{k=0}^{(N-n-2)/(n+1)} \binom{N-k(n-1)}{n}$$
The estimation $\binom{N-k(n-1)}{n}\approx \frac{1}{n!} (N-nk)^n$ introduces little error in the relevant region. So the sum is about $\frac{n!}{N^n}\frac{1}{n!}\sum\limits_{k=0}^{(N-n-2)/(n+1)} (N-k(n-1))^n = \sum\limits_{k=0}^{(N-n-2)/(n+1)} (1-k\frac{n-1}{N})^n = \frac{N}{n-1}\sum\limits_{k=0}^{(N-n-2)/(n+1)} \frac{n-1}{N}(1-k\frac{n-1}{N})^n$.
This last summation is (close to) an upper Riemann approximation of the integral $\int\limits_{0}^{1} x^n = \frac{1}{n+1}$, so we obtain $\frac{N}{n^2-1}$. After compensation by the rescaling factor $N$ we have that the answer to the original question is $\frac{1}{n^2-1}$.
I will re-check the estimations tomorrow, it is quite late here. Maybe some errors need a bit more care, but I am sure the general argument is fine.
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How does one create a polynomial equation based on graphic features? So as many of you know, graphs of polynomial equations with degree greater than 2 have what are known (to me) as local minima and local maxima (the point(s) on its graph where the derivative of the function is zero, the point to which output values converge inside a certain range of input values). Based on the coordinates of these points, and other information taken from the graph, is it possible to create an equation that describes it? For example, if I asked, "Create the equation of a cubic function with positive-positive end behavior that has a local maximum at (3,5) and a local minimum at (6,2), with the graph passing through the origin. Is there a systematic way to do this? Also, if there is not enough information, I would like to know, and why.
Thanks!
| Taking the points with the maximum and minimum, which are sufficient to define the cubic, we can just write the cubic as $y=ax^3+bx^2+cx+d$ and plug in the data we have.
$$5=a3^3+b3^2+c3+d\\2=a6^3+b6^2+c6+d\\0=3a3^2+2b3+c\\0=3a6^2+3b6+c$$
This is four equations in four unknowns which can be solved by the usual techniques to yield $a,b,c,d$.
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Are there any 2 primitive pythagorean triples who share a common leg? So is it possible for:
$\gcd(a,b,c)=1$
$a^2+b^2=c^2$
and
$\gcd(a,d,e)=1$
$a^2+d^2=e^2$
?
| This system of equations:
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&z^2+b^2=x^2\end{aligned}\right.$$
Solutions have the form:
$$b=4tkp^2s^2$$
$$a=2(t^2-k^2)p^2s^2$$
$$z=4k^2s^4-t^2p^4$$
$$c=2(t^2+k^2)p^2s^2$$
$$x=4k^2s^4+t^2p^4$$
$t,k,p,s$ - integers asked us.
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Prove that $a_n
So I got this question and I was able to do up to (III) (a) however part b has been pretty difficult for me to complete and I can't seem to get it right . Any help ?
I tried considering $a_n - a_{n+1}$ and I got the required proof but I don't feel as though this method is logical enough
$a_n - a_{n+1} < 0$
$a_n < a_{n+1}$
| Guide:
*
*Prove that $a_n>0$ if $a_1>0$.
*Compute $a_{n+1}-a_n$ in terms of $a_n$, your goal is to check that the expression is positive.
*check that the denominator is positive.
*Show that the numerator is positive using the property that $0<a_n<2$.
| {
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Solve $\log_a(\log_a x^n)$ if $n=a^2$ ; $x= e^2$ My brother challenged me to solve this problem. Trying since 2 days. I came up with $a^{a^y}= x^n$ assuming $y$ is $\log_a(\log_a x^n)$. There's no solution available on net as well. If someone can solve it, it would be of great help! Thanks
Trial 1:
$\Rightarrow\log_a(\log_a (e^2)^{a^2})$
$\Rightarrow\log_a(\log_a \exp(2\cdot a^2))$ ...By $\exp$ property
$\Rightarrow\log_a(2*(a^2)\cdot\log_a (e))$ ... By log property $\log x^a= a \log x$
$\Rightarrow\log_a(2\cdot(a^2)\cdot(\frac{1}{\log_e(a)}))$ ... By $\log$ property
$\Rightarrow\log_a(2\cdot(a^2)) + \log_a(\frac{1}{\log_e(a)})$ ...By $\log$ property
$\Rightarrow 2\cdot\log_a(a^2)+\log(\frac1{\log_e(a)})$
$\Rightarrow 4 + \log(\frac{1}{\log_e(a)})$
Couldn't solve beyond that
Trial 2:
Considering $y=\log_a(\log_a (x^n))$
$\Rightarrow a^y= \log_a (x^n)$
so, $a^{a^y}= x^n$
| First notice that your expression is
$$
\log_a(n\log_a x)=\log_a n+\log_a\log_ax
$$
For $n=a^2$ you have $\log_an=\log_a(a^2)=2$; for $x=e^2$,
$$
\log_a\log_a x=\log_a(2\log_a e)=\log_a2+\log_a\log_ae
$$
You could observe that
$$
\log_a2=\frac{\log 2}{\log a}
$$
and
$$
\log_ae=\frac{1}{\log a}
$$
so
$$
\log_a\log_ae=\log_a\frac{1}{\log a}=-\log_a\log a=-\frac{\log\log a}{\log a}
$$
so you finally get
$$
2+\frac{\log2}{\log a}-\frac{\log\log a}{\log a}
$$
Notes.
*
*The unadorned $\log$ symbol denotes the natural logarithm.
*I skipped over the checks for existence; the computations make sense for $a>1$ as the final formula shows.
| {
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Associative laws with negation Is it possible to simplify a statement like the one below with the associative law despite the negation? I can't seem to find a law that outlines this. The associative law is the closest I could find.
\begin{align}
\ (\lnot p \land \lnot q) \ \lor \ q
\end{align}
If the above is possible, is it also possible if only one variable in the brackets were to be negated like the below?
\begin{align}
\ (p \land \lnot s) \ \lor \ s
\end{align}
Thanks.
| Associative low holds only when three formulas are put together by two occurrences of the same connective $\land$ or $\lor$. For instance:
\begin{align}
(\lnot p \land \lnot q) \land q &\equiv \lnot p \land (\lnot q \land q) &&\text{or} & (\lnot p \lor \lnot q) \lor q &\equiv \lnot p \lor (\lnot q \lor q)
\end{align}
So, you cannot use associative law in your formula $(\lnot p \land \lnot q) \lor q$.
Indeed, you can easily check that the the truth table of your formula $(\lnot p \land \lnot q) \lor q$ is different from the one of $\lnot p \land (\lnot q \lor q)$.
However, you can simplify your formula $(\lnot p \land \lnot q) \lor q$, but using other logical equivalences instead of associative law (I refer to the ones listed here):
\begin{align}
(\lnot p \land \lnot q) \lor q &\equiv (\lnot p \lor q) \land (\lnot q \lor q) &&\text{distributivity} \\
&\equiv \lnot p \lor q &&\text{identity}
\end{align}
where in the last equivalence the identity law can be applied since $\lnot q \lor q$ is a tautology (i.e. it is true under any valuation).
Note that in the simplification above the fact that there is a negation $\lnot$ in front of $p$ does not play any role, so you can simplify in the same way also your second formula:
\begin{align}
(p \land \lnot s) \lor s &\equiv (p \lor s) \land (\lnot s \lor s) &&\text{distributivity} \\
&\equiv p \lor s &&\text{identity.}
\end{align}
| {
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Solve: $(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$ Solve: $(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$
Is my solution correct:
Answer given in the book : $y(x+1)=c_1+c_2\log(x+1)+\log3(x+1)$
For my later reference: link wolfram alpha
| Robert pointed out your mistake..
Here is another approach
$$(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$$
divide by $x+1$
$$(x+1)^2y''+3(x+1)y'+y=6\frac {\log(x+1)}{x+1}$$
on the left there is a derivative
$$((x+1)^2y')'+((x+1)y)'=6\frac {\log(x+1)}{x+1}$$
Integrate
$$(x+1)^2y'+(x+1)y=6\int \frac {\log(x+1)}{x+1} dx$$
$$(x+1)^2y'+(x+1)y=3\log^2(x+1)+K_1$$
Divide again by $x+1$
$$(x+1)y'+y=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$
$$((x+1)y)'=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$
Integrate again
$$(x+1)y=3\int\frac {\log^2(x+1)}{x+1}+K_1 \int \frac {dx} {x+1}$$
$$(x+1)y=\log^3(x+1)+K_1 \ln |{x+1}|+K_2$$
$$ \boxed {y=\frac {\log^3(x+1)}{x+1}+K_1 \frac {\ln |{x+1}|}{x+1}+\frac {K_2}{x+1}}$$
| {
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Can any metric on $\Bbb R^n$ be bounded above and below for any other metric? Let $d_1(x,y)$ and $d_2(x,y)$ be any two metrics on $\mathbb{R}^n$. Can it be shown that,
$$c\cdot d_2(x,y) \le d_1(x,y) \le C\cdot d_2(x,y)$$
for all $x,y \in \mathbb{R}^n$ for some fixed positive constants $c,C$? If not, under what conditions could such a relation hold (for a compact set it seems straightforward)? If yes, does this result hold for two arbitrary topological metric spaces as well?
Thanks!
| If $d_1$ is the usual metric and $d_2$ is the discrete metric, then there are no such constants.
However, your statement is true if the metrics are induced by norms in $\mathbb{R}^n$.
| {
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If $\Omega\subset\mathbb{R}^n$ is convex and $f$ is differentiable, then $f(x)-f(y)\geq f'(y)(x-y),\;\forall \;x,y\in \Omega$ Let $\Omega$ be a convex set in $\Bbb{R}^n$. We say that that $f:\Omega\to \Bbb{R}$ is convex if $$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y),\;\forall\;0\leq t\leq 1,\;\&\;\forall\;x,y\in \Omega.$$ I want to show that if $f$ is convex and differentiable on $\Omega,$ then $$f(x)-f(y)\geq f'(y)(x-y),\;\forall \;x,y\in \Omega.$$
I'm thinking of using Partial derivatives but don't know how to go about it. Please, can anyone help out?
| Clearly You have $$\frac{f(tx+(1-t)y)-f(y)}{t}\leq f(x)-f(y)$$
and for $t\rightarrow 0$ the LHS numerator and denominator tend to zero so that De L´Hospital applies and the partial derivative $\partial/\partial t$ of the numerator is $$f^´(tx+(1-t)y)(x-y)$$ and that of the denominator $1$ and now take $t\rightarrow 0$
| {
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Writing a Cartesian Equation for a plane. A plane $\pi_2$ intersects $\pi_1$: $4x-2y+7z-3=0$ at right angles. Two points lie on $\pi_2$: $A(3,2,0)$ and $B(2,-2,1)$. Write a cartesian equation for $\pi_2$.
I know that the normals of these two planes must be perpendicular since the planes are perpendicular. So if $n_2=(a,b,c)$ is the normal for plane $2$ and $n_1=(4,-2,7)$ is the normal for plane $1$ then the dot product between $n_1$ and $n_2$ is $0$.
Not sure where else to go with this question.
| The plane $\pi_2$ has a cartesian equation of the type $ax+by+cz=d$. And you know that $(3,2,0),(2,-2,1)\in\pi_2(\iff3a+2b=2a-2b+c=d)$. Furthermore, $\pi_1$ and $\pi_2$ intersect at right angles, which is equivalent to the assertion$$(a,b,c).(4,-2,7)=0(\iff4a-2b+7c=0).$$Can you take it from here?
| {
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On the proof of Weils hyperelliptic theorem Let $p$ be a large prime. Consider $F_p$
Theorem:
Let $P$ be an element in $F_p[x]$ of degree $k$, assume that $P$ is not a constant multiple of a square. Then the number of solutions $(x,y)$ in $(F_p)^2$ to $y^2 = P(x)$ is $p+O_k(\sqrt p)$
A proof of this result can be found in:
https://terrytao.wordpress.com/2009/08/18/the-least-quadratic-nonresidue-and-the-square-root-barrier/#more-2664
The proof uses the polynomial method, defining the polynomial $Q(x) = P^l(R(x,x^p)+P^{(p-1)/2}S(x,x^p))$ for $l\cong \sqrt p$ and polys $R(x,z),S(x,z)$ satisfying some degree condition. Then one shows you can choose such $S,R$ so that for each $x$ with $P(x)$ a square, $Q$ vanishes at least $l$ times.
I don't understand why we need to multiply by $P^l$, why doesn't Terry's argument go through even without it?
Thanks
| Yeah it's actually stupid:
We want the derivative to also be of the form with $P^((p-1)/2)$, and a way to promise that is multiply by $P^l$ and give copies of it when taking the derivative.
| {
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Which of the following intervals contains integers satisfying the following three congruences: Question: Which of the following intervals contains integers satisfying the following three congruences:
$x\equiv 2\pmod 5, x\equiv 3\pmod 7$ and $x\equiv 4\pmod {11}$,
(i) $[401,600]$, (ii) $[601,800]$, (iii) $[801,1000]$, and (iv) $[1001,1200]$
After solving I have found $x=2292\equiv 367\pmod {385}$. But this does not belong to any of the above option. How can I find this solution?
| The point is that there is no reason to focus on the smallest non negative solution.
Adding/subtracting an arbitrary amount of times the number $385 $ will yield other solutions. In fact, all of them, by the Chinese Remainder Theorem.
| {
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Optimal number of elements in a subset The following problem appears in many Combinatorics books:
Prove that among 70 positive integers less than or equal to 200 there are two whose difference is 4,5, or 9. The solution is straightforward using Pigeonhole principle. I was wondering is 70 the optimal number? I have tried to prove that we can get by with 64.
Let $A$ be any subset of $\{ 1, 2, 3, \ldots, 200\}$ such that no two integers in $A$ differ by 4, 5 or 9.
Let for any integer $k$, $A+k$ represent the set $\{a+k: a \in A\}$. Note that
\begin{equation*}
A \cap A+4 = A \cap A+5 = A \cap A+9 = \phi \\
A+4 \cap A+ 9 = A+5 \cap A+9 = \phi
\end{equation*}
Thus
\begin{equation*}
|A \cup A+4 \cup A+5 \cup A+9| = 4|A| - |A+4 \cap A+5)|
\end{equation*}
Note that $A$ can not contain more than 4 consecutive numbers. Also, if it contains $i, i+1, i+2, i+3$, then $i+4, i+5, i+6, i+7, i+8, i+9, i+10, i+11, i+12$ can not belong to $A$. Thus in any set of 13 consecutive numbers, it can not contain more than 4 consecutive numbers. Note that $A+4 \cap A+5$ is maximum when $A$ contains as many consecutive numbers as possible. Thus when the maximum of $|A+4 \cap A+5|$ happens, we have
$|A+4 \cap A+5|$ is at most $3 \times \lfloor \frac{200}{13}\rfloor + 4 = 49$. Hence
\begin{equation*}
4|A| - 49 \leq |A| - |A+4 \cap A+5| \\
= |A \cup A+4 \cup A+5 \cup A+9| \\
\leq 209
\end{equation*}
and $|A| \leq 64$.
A set with 64 elements satisfying the property is
$$\{13k + 1, 13k+2, 13k+3, 13k+4: k = 0, 1, 2, \ldots, 15 \}$$
My question: Is the above proof correct?
| Your proof looks O.K. to me. Anyway, your result is correct. Some comments:
*
*Your introduction is slightly misleading. If you have quoted the problem correctly from the (unnamed) combinatorics books, then you have improved the number $70$ to $65,$ not $64.$ You have proved that, among $65$ positive integers less than or equal to $200,$ there are two whose difference is $4,$ $5,$ or $9.$
*Perhaps a simpler approach is to prove that if $A$ is a set of integers, no two of which differ by $4,$ $5,$ or $9,$ then any set of $13$ consecutive integers contains at most $4$ elements of $A$ (regardless of whether those elements of $A$ are consecutive or not). To see this, suppose that the set $[13]=\{1,2,3,\dots,13\}$ contains $5$ elements of $A.$ Since each of the five sets $\{1,6,10\},\{2,7,11\},\{3,8,12\},\{4,9,13\},\{5\}$ contains at most one element of $A,$ we must have $5\in A.$ A similar argument shows that we must also have $9\in A,$ but this is absurd.
*You have improved the $70$ to $65,$ but (apparently) you have not considered improving the $200?$ It seems that, among $65$ positive integers less than or equal to $208$ (or more generally $4n+1$ positive integers less than or equal to $13n$), there are two whose difference is $4,$ $5,$ or $9.$
| {
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Inscribed trapezoids problem Let $ABCD$ be inscribed trapezoid with $(AB) \parallel (CD)$ and let $P$ be the point where its diagonals meet.
The circumcircle of $\triangle APB$ meets line $(BC)$ (again) at $X$.
$Y$ is a point on $(AX)$ such that $(DY)\parallel (BC)$.
Prove that $\angle YDA = 2 \angle YCA $.
Any hints would be appreciated.
| Some hints:
*
*Prove that $ADPY$ is cyclic.
*Prove that $PC=PY$.
| {
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How to calculate (x,y) position of number in square I am looking for a way to determine the X & Y position of a number to draw following square:
1 2 5 10 17
3 4 6 11
7 8 9 12
13 14 15 16
What kind of algorithm / formula can I use ?
I have tried rounding the square root of the number to determine the X position.
To be clear, here is what I want to achieve, numbers with their corresponding X and Y position:
# X Y
1: 1 1
2: 2 1
3: 1 2
4: 2 2
5: 3 1
6: 3 2
| Here is an algorithm (written for Matlab) that works:
n = 5;
x = zeros(n*n,1);
y = zeros(n*n,1);
x(1) = 1; % set the first 3 points
y(1) = 1;
x(2) = 2;
y(2) = 1;
x(3) = 1;
y(3) = 2;
for i = 4:n*n
if (x(i-1) == y(i-1)) % jump all the way down (and 1 to the right)
x(i) = x(i-1)+1;
y(i) = 1;
elseif (x(i-1) == y(i-1)+1) % jump all the way to the left (and 1 up)
x(i) = 1;
y(i) = y(i-1)+1;
elseif (x(i-1) > y(i-1)) % move 1 up
x(i) = x(i-1);
y(i) = y(i-1)+1;
else
x(i) = x(i-1)+1; % move 1 to the right
y(i) = y(i-1);
end
end
If at the end you call plot(x,y) you get the following figure:
| {
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Finding the Jordan Form of a transformation defined by $T(X)=AX$ when $A,X \in M_{4\times 4}(\mathbb C)$
Given $A=\left(\matrix{0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&-1}\right)$ and define $T(X)=AX$. when $A,X \in M_{4\times 4}(\mathbb C)$
Find the Jordan Form of T
I found that the minimal polynomial is $(t-1)(t+1)$ therefore $T$ is diagonalizable. However, I'm not sure how to find the amount of each eigenvalue to put in the diagonal without explicity finding the $16\times16$ representative matrix of T.
| I identify a $4\times 4$ matrix by a vector of length $16$ by reading it from top to bottom, from left to right. So the first four coordinates are from the first column, etc.
Then $T$ is diagonal in the last $8$ coordinates: identical on coordinates $9-12$, and negative of the identity on $13-16$, so that part is covered.
Furthermore, $T$ switches the first four coordinates by the second four as a product of four transpositions. You can view these transpositions separately. The matrix of a transposition of two coordinates is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, whose Jordan normal form is $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
So the Jordan normal form of $T$ is diagonal with eigenvalues:
$$1,-1,1,-1,1,-1,1,-1,1,1,1,1,-1,-1,-1,-1$$
| {
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Convergence in probability, mean and almost surely $X$, $Z_n$, $Y_n$ are independent random variables, where X is integrable, $Y_n$ has Bernoulli distribution $b(1,n^{-2})$ , $Z_n$ has Poisson disribution with parameter $n^2$. I need to check convergence of $V_n = Y_nZ_n + (1-Y_n)X$ to $X$ in mean, probability and a.s.
I can't prove that it is convergent in probability or not. It seems to me it is not.
| Observe that for $\epsilon>0$: $$\{|V_n-X|>\epsilon\}\subseteq\{V_n\neq X\}\subseteq\{Y_n\neq 0\}$$so that:$$\mathsf P(|V_n-X|>\epsilon)\leq\mathsf P(V_n\neq X)\leq\mathsf P(Y_n\neq 0)=n^{-2}$$
| {
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Is combination of elementary row operation considered elementary row operation? I came across this question in my homework if $R_1-R_2-R_3$ considered elementary row operation. My opinion is that it should not be an elementary row operation since it contains three rows which violates the rule "Add a multiple of one row to another row" which only contains two rows. I understand that $R_1-R_2-R_3$ is simply the combination of $R_1-R_2 \to R_1$ and $R_1-R_3 \to R_1$.
| I wouldn't consider it an elementary row operation. In much the same way, when we write 'transposition' (for instace), we mean a very specific kind of permutation, and the result of composing two transpositions is not a transposition itself.
| {
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Prove that, $\left\lfloor{\frac{x}{n}}\right\rfloor=\left\lfloor{\frac{\lfloor{x}\rfloor}{n}}\right\rfloor$ where $n \in{\mathbb{N}}$
Prove that
$$\left\lfloor{\frac{x}{n}}\right\rfloor=\left\lfloor{\frac{\lfloor{x}\rfloor}{n}}\right\rfloor,$$ where $n \in{\mathbb{N}}.$
My Attempt:
Let $x=nt$.
Then, I need to prove,
$$\lfloor{t}\rfloor=\left\lfloor{\frac{\lfloor{nt}\rfloor}
{n}}\right\rfloor.$$
Let $t=n\lambda+r+f$,
where $0\leq{r}\leq{n-1}$ and $0\leq{f}\lt{1}.$
This gives R.H.S as,
$$\left\lfloor{\frac{\lfloor{n^2\lambda+nr+nf}\rfloor}
{n}}\right\rfloor.$$
How do I continue?
I also know the property,
$$\lfloor{x}\rfloor=\left\lfloor{\frac{x}{n}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{n}}\right\rfloor+\left\lfloor{\frac{x+2}{n}}\right\rfloor \cdots \left\lfloor{\frac{x+(n-1)}{n}}\right\rfloor.$$
Can I use that here?
| It is actually much simpler.
See, $\left\lfloor \frac xn\right\rfloor$ is the unique integer $l$ such that $l \leq \frac xn \leq l+1$, or in other words, $nl \leq x \leq nl+n$.
Note that $nl$ is an integer less than or equal to $x$, and $\lfloor x \rfloor$ is the greatest such integer, so $\lfloor x \rfloor \geq nl$. Of course, $\lfloor x \rfloor \leq x \leq nl + n$.
Consequently, $nl \leq \lfloor x \rfloor \leq nl + n$. Dividing by $n$, one sees the conclusion.
| {
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If $ab = e$, then $ba = e$? $x\cdot (a\cdot b) = x\cdot e$
$(x\cdot a)\cdot b = x$
$e\cdot b = x$
$b = x$
Are my steps correct?
What I wanted to prove is that if $ab = e$, then $ba = e$
$x$ is inverse of $a$ and $e$ is identity element .
| I don't see why you need to introduce $x$ for $a^{-1}$. Since $ab=e$ we have
\begin{align*}
e & = a^{-1}ea\\
& = a^{-1}(ab)a\\
& = ba
\end{align*}
by multiplying
by $a^{-1}$ from the left and then by $a$ from the right. So conjugating $ab=e$ gives $ba=e$.
| {
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Shortest distance between two non-intersecting differentiable curves is along their common normal I want to formally prove the following property:
The shortest distance between two differentiable non-intersecting curves is along their common normal.
I looked at the discussion on this on Quora, (Link: https://www.quora.com/How-can-I-show-that-the-shortest-distance-between-2-non-intersecting-curves-always-lies-along-their-common-normal), but the proofs were not formal there.
My approach:
Let the curves be denoted by f and g respectively. And let AB be a line segment with A lying on f and B lying on g.
WLOG, take A as the origin of the coordinate axis, and the tangent to f at A to be the x-axis.
Let $B \equiv (b,g(b))$, $A = (0,0)$. Let the angle AB makes with the x-axis be $\theta \in (0,\pi)$.
Let 2 rays from A making an angle of $\theta + \epsilon$ and $\theta - \epsilon$ ($\epsilon > 0$) with the x-axis intersect $y = g(x)$ at $B_1$ and $B_2$ respectively.
I wish to show that $\theta \neq \frac{\pi}{2} \Leftrightarrow |AB| > \min\{|AB_1|, |AB_2| \}$.
Suppose $\theta \neq \frac{\pi}{2} $. Then let $B_1 \equiv (b_1,g(b_1))$ and $B \equiv (b_2,g(b_2))$.Then
\begin{align}
g(b) &= b\tan(\theta)\\
g(b_1) &= b_1\tan(\theta + \epsilon)\\
g(b_2) &= b_2\tan(\theta - \epsilon) \\
\Rightarrow |AB| &= |b \sec(\theta) | \\
|AB_1| &= |b_1 \sec(\theta + \epsilon) | \\
|AB_2| &= |b_2 \sec(\theta - \epsilon) |
\end{align}
I wanted to relate $b_1$ and $b_2$ to $b$ using the continuity of $g$. However I am not able to do so. How do I proceed from here? - or is there an easier formal approach?
| Suppose $a(s),b(t)$ are two curves in $\mathbb R^2$ with parameter interval $(0,1).$ Assume $a'(s),b'(t)$ never vanish, otherwise normal vectors make no sense. Define
$$f(s,t)= |a(s)-b(t)|^2.$$
Suppose $f(s_0,t_0)>0$ is the minimum value of $f$ (hence $\sqrt {f(s_0,t_0)}$ is the distance between the two curves). Then $(s_0,t_0)$ is a critical point of $f.$ Thus
$$\frac{\partial f}{\partial s}(s_0,t_0) = 2a'(s_0)\cdot(a(s_0)-b(t_0)) = 0$$
and
$$\frac{\partial f}{\partial t}(s_0,t_0) = 2b'(t_0)\cdot(b(t_0)-a(s_0)) = 0.$$
(Here $\cdot$ denotes the dot product.) Thus both $a'(s_0),b'(t_0)$ are perpendicular to the vector $b(t_0)-a(s_0).$ Thus the vector $b(t_0)-a(s_0)$ is perpendicular to $a$ at $a(s_0),$ and is perpendicular to $b$ at $b(t_0).$ This is the desired conclusion.
| {
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How can I plot a straight line which is in normal form? Take a look at this link.
Suppose, I have the following equation of a straight line:
$$x \cos\theta+ y \sin \theta=p$$
If we rearrange the equation as follows:
$$y = \frac{p - x\cos\theta}{\sin\theta}$$
we are again stuck with infinite values of $x$. So, what is the benefit of introducing $\theta$ and $p$ here?
How can/should I plot this?
| The normal form can be intepreted as follow, let
*
*$p$ the distance between the line and the origin
*$n=(\cos \theta, \sin \theta)$ the normal unitary vector to the line
therefore for any point $OP=(x,y)$ on the line by dot product we have
$$OP\cdot n=(x,y)\cdot (\cos \theta, \sin \theta)=x\cos \theta+y\sin \theta=p$$
Note that for a given line $\theta$ and $p$ are fixed therefore by the following
$$y = \frac{p - x\cos\theta}{\sin\theta}$$
for $\sin \theta \neq 0$ we can find all the points on the line.
| {
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Differential forms and partial derivatives If I have a vector valued function $f(\vec R)$ and I imagine that $R$ is also a function of time, so that $R(t)$ Then I'm struggling to prove that:
$$ df \left({\partial R \over \partial t}\right) = {df \over dt}$$
Intuitively, it makes some sense, that if df is a differential form, and a covector; and $\partial R \over \partial t$ is the velocity vector, then a covector and vector product will give a scalar. Since $df \over dt $ is also a scalar output function differentiated in terms of a scalar, it too would be a scalar.
What I'm struggling to see is why this should be true.
| The function $f$ takes in a point $(x,y,z)$ in space and outputs a real number $f(x,y,z)$; provided that it is regular enough, you can write by the total differential theorem that
$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz \tag1$$
On the other hand, $\mathbf R$ is a function that takes in a real number (the time $t$) and outputs a point $\mathbf R(t)$ in space; if it is regular enough, its derivative is well-defined and we may say that the vector $\mathbf R$ defines a trajectory in space. The derivative of $\mathbf R$ (total or partial is the same) is the vector
$$\frac{d\mathbf R}{dt} = \left(\frac{dR_x}{dt}, \frac{dR_y}{dt}, \frac{dR_z}{dt} \right)\tag2$$
If you imagine formally dividing $(1)$ by $dt$, you get
$$\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}, $$
and you can think of “evaluating” this expression on the trajectory of $\mathbf R$ by plugging in the components of $(2)$ into the total derivatives of $x$,$y$, and $z$. I think that is what is meant by the LHS of your formula.
| {
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Problem regarding convergency: $\sum_{k=0}^{\infty}\frac{x^k}{k!} \to e^x$
Question. Let $X$ be the space of all polynomials in one variable,with real coefficients. If $p=a_0+a_1x+\dots+a_nx^n\in X$, define $$|p|=|a_0|+|a_1|+\dots+|a_n|,$$ which gives the metric $d(p,q)=|p-q|$ on $X$.
Does the metric space $X$ is complete?
Now, this question has been answered 'false' here.
According to that answer let $p_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}$
...
I know this is due to the fact of Taylor series (theorem) $$\sum_{k=0}^{\infty} \frac{x^k}{k!}= e^x ,\ \text{*as*} \ n \longrightarrow \infty \ ,\forall \ x \in \mathbb{R}: ~p_n \to e^x$$
But still I have some questions:
*
*When we say $p_n \to e^x$ as $n \to \infty$.then In which space this convergency is happening...?? I mean w.r.to which norm? Does the space here is $X=C(\mathbb{R}):= \text{Space of continuous functions on}~ \mathbb{R}$ ...But then which norm?? I mean "sup-norm" shouldn't work here as $e^x$ is not bounded...!!
*Now how does this convergency (i.e. $p_n\to e^x$) remains independent of norm put on the space?
Please help me to clarify these confusions. Correct me if I'm wrong. Thank you.
| The statement that $p_n$ converges to $e^{x}$ and hence it does not converge in the given space is just suppose to be a hint and not a complete proof. No norm is used in saying that $p_n$ converges to $e^{x}$. You have to give a rigorous proof along the following lines: if a sequence of polynomials converges in the given norm then the coefficients converge. So if $p_n$ converges to a polynomial $p$ then its coefficients have to be $\frac 1 {k!}$, $k=0,1,2,...$. In particular no coefficient can be $0$ which contradicts the fact that $p$ is a polynomial. To be more explicit suppose $p$ has degree $m$. Since each each $p_n$ with $n >m$ has $\frac 1 {(m+1)!}$ as the $m+1$-st coefficient the same must be true of $p$ but $p$ does not have this term.
| {
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How do you explain to a 5th grader why division by zero is meaningless? I wish to explain my younger brother: he is interested and curious, but he cannot grasp the concepts of limits and integration just yet. What is the best mathematical way to justify not allowing division by zero?
| You shouldn't try to do that. Instead make counter question.
"What should it be, then?" and let them think about it.
(Lengthy) justification: There are many important concepts in math you can come up with if you start experimenting with multiplication. Take for example area of a rectangle. You multiply the sides. Area of a curve? You take the integral. What is an integral? Well Riemann imagined thin thin slices, almost infinitely thin, actually. The idea that we can calculate area of these slices where one side is so tiny it almost is 0. If we disqualify limits, or the idea of multiplying something "almost 0" to be 0 then we would have a tougher time coming up with an excuse to investigate integrals, which have been veeery important to the development of modern technology.
Any kid who could come up with some new interpretation of this could be very valuable.
| {
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Average of Vectors squared minus the square of the averaged vectors I have already asked this question in Physics, but as someone pointed out it is more appropriate to place it here.
Say I've obtained a series of 3D vectors, (x, y, z), with a different vector at different points in time. I want to obtain the second moment of this set of vector's fluctuations in time. This is given by:
$<$ M $^2> - <$ M $>^2$,
where M is the vector in 3D, and the average is over time. To me, you can solve this simply by taking:
$<$ M $^2> - <$ M $>^2 = <$ (x, y, z ) $ \cdot $ (x, y, z ) $> - <$ (x, y, z ) $> \cdot <$ (x, y, z ) $>$.
However, I have seen this solved by decomposing the vector along different Cartesian axis:
$<$ M $^2> - <$ M $>^2 = < $ M $_x^2 > - < $ M $_x >^2 + < $ M $_y^2 > - < $ M $_y >^2 + < $ M $_z^2 > - < $ M $_z >^2 $.
This is shown on page 2548 in Pitera et al., Biophysical Journal, 2001, 20, 2546–2555. Paper Link.
I just wanted some clarification if this second method is correct, and I guess whether the first is also. To me, the second method is not possible as you cannot decompose $<$ M $>$ as they have. Less so if you then square the decomposed parts.
Thank you for any help, and apologies for any grievances about the formatting.
| The average of the square is the average of the square of the components by linearity of expectations:
$$
\langle\mathbf M^2\rangle=\langle M_x^2+M_y^2+M_z^2\rangle=\langle M_x^2\rangle+\langle M_y^2\rangle+\langle M_z^2\rangle
$$
The square of the average can be decomposed into components, and no cross terms will survive, since we're evaluating an inner product in an orthonormal basis. (In a non-orthonormal bases, this decomposition would not be valid.)
$$
\langle\mathbf M\rangle^2=\langle M_x\hat x+M_y\hat y+M_z\hat z\rangle^2=\left(\langle M_x\rangle \hat x+\langle M_y\rangle \hat y+\langle M_z\rangle \hat z\right)^2=\langle M_x\rangle^2+\langle M_y\rangle^2+\langle M_z\rangle^2
$$
Does this clear up the logic behind the decomposition?
| {
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Find the $n$-th derivative of $f(x)=\frac{x}{\sqrt{1-x}}$ Find the $n$-th derivative of
$$f(x)=\frac{x}{\sqrt{1-x}}$$
First I just calculated the first, second and 3-th, 4-th derivatives and now I want to summarize the general formula. But it seems too complicated. Then I want to use binomial theorem or Taylor expansion... Also got no more clues.
| We have $f(x)=g(x)h(x)$ where $g(x)=x$ and $h(x)=(1-x)^{-1/2}$. Hence using Leibniz's rule we have
$$f^{(n)}(x)=\sum_{k=0}^n{n\choose k}h^{(k)}(x)g^{(n-k)}(x)={n\choose n}h^{(n)}(x)g(x)+{n\choose n-1}h^{(n-1)}(x)g'(x) \quad(*)$$
It remains to compute
$$h^{(n)}(x)=\frac{1}{2}\frac{3}{2}\cdots \frac{2n-1}{2}(1-x)^{(2n-3)/2} =\frac{(2n)!}{2^{2n}n!}(1-x)^{(2n-3)/2}$$
and then substitute it in $(*)$ and simplify.
| {
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Stuck with boundary value PDE problem The last time I posted this I got many down votes and I don't know why. Maybe because I forgot to include my work?
I got the PDE $\dfrac{\partial^2 u}{\partial x^2} -2 \dfrac{\partial^2 u}{\partial x \, \partial y} - 3\dfrac{\partial^2 u}{\partial y^2} = 0$
General solution found to be $u(x, y) = F_1(x - y) + F_2(3x + y)$ these are arbitrary functions.
I want to get a solution to the equation that satisfies the boundary conditions $u(x,0) = x$, $u_y(x,0) = 0$.
$$u(x, 0) = x: u(x, 0) = F_1(x) + F_2(3x) = x$$
$$u_y(x, y) = - F_1'(x - y) + F_2'(3x + y)$$
$$u_y(x,0) = 0: u_y(x, 0) = - F_1'(x) + F_2'(3x) = 0$$
And also get a solution to the equation with the general boundary conditions $u(x,0) = g_0(x)$ and $u_y(x,0) = g_1(x)$.
$$u(x,0) = g_0(x): u(x, 0) = F_1(x) + F_2(3x) = g_0(x)$$
$$u_y(x,0) = g_1(x): u_y(x, 0) = -F_1'(x) + F_2'(3x) = g_1(x)$$
I am stuck. Please show me how this is done. Thank you.
| You have $F_1(x) + F_2(3x) = x$, differentiating by $x$ we'll get $F_1'(x)+3F_2'(3x)=1$. From the last boundary condition we have $-F_1'(x)+F_2'(3x)=0$. Adding two equations together, we get $4F_2'(3x)=1$. Now we can integrate and find $F_2$: $4\int{F_2'(3x)}dx=\int{dx}=x+c$; $F_2(x)=\frac{x}{12}+c$. Now substitute $F_2$ back in the initial condition and find $F_1$: $F_1(x)+\frac{3x}{12}+c=x$; $F_1=...$
| {
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Convergence of $\sum \frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-\frac{1}{e}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}$ Study the convergence of the following series as $\alpha \in \mathbb{R}$
$$\sum_{n=1}^{\infty}\frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-\frac{1}{e}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}$$
Maybe this series is quite simple but gives me hard times how to interpret it asymptotically. Could comparison test work?
| Hint for the denominator, you can write:
\begin{align*}
\left(1+\frac{1}{n}\right)^n
&= \exp\left( n \log\left(1+\frac{1}{n}\right)\right) \\
&= \exp\left( n\left(\frac{1}{n} - \frac{1}{2 n^2} + o\left(\frac{1}{n^2} \right)\right)\right) \quad (n\rightarrow \infty) \\
&= e^1\left( \exp\left( - \frac{1}{2 n} + o\left(\frac{1}{n} \right)\right) \right)\quad (n\rightarrow \infty) \\
&= e^1\left(1 - \frac{1}{2n} + o\left(\frac{1}{n} \right) \right) \quad (n\rightarrow \infty) \\
\end{align*}
Therefore:
$$ e - \left(1+\frac{1}{n}\right)^n \underset{n\rightarrow\infty}{ \sim \frac{e}{2n} } $$
Using the same method for the numerator, you might be able to discuss the convergence of the series.
| {
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How can variance and mean of brownian motion be stated I have seen in the construction of brownian motion that the Its mean is 0 and variance t. i.e
$E(B(t))=0$ , $Var(B(t))=t$ Why in a problem sheet does it state $B_{j}(t)$ for $j=1,...,n$ are brownian motions with variances $\sigma^{2}_{j}$. How can these be brownian motion with these variances... is it saying $Var(B_{j}(t)) \neq t$
| If $B(t)$ is a standard Brownian motion, then indeed $\mathbb E[B(t)]=0$ and $\mathsf{Var}(B(t))=t$ for $t\geqslant0$. Let $X(t) = \sigma B(t) + \mu t$ for some $\mu\in\mathbb R$ and $\sigma>0$. Then $X(t)\sim\mathcal N(\mu t, \sigma^2 t)$, and is called a Brownian motion with drift $\mu$ and volatility (or variance) $\sigma^2$.
In your example, it seems that each $B_j(t)$ is a Brownian motion with zero drift and volatility $\sigma^2_j$.
| {
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Is $\limsup |C_n|^{\frac{1}{n}} = \limsup |C_{n+j}|^{\frac{1}{n}}? $ where $j\geq0 $ $C_n$ is a sequence of complex numbers.
I would like to prove that the radius of convergence of a differentiated power series is the same as the original series.
i.e. $\limsup |(n+j)(\cdots)(n+1)C_{n+j}|^{\frac{1}{n}} = \limsup |C_n|^{\frac{1}{n}}$
I know that $\limsup |(n+j)(\cdots)(n+1)C_{n+j}|^{\frac{1}{n}} = \limsup|C_{n+j}|^{\frac{1}{n}}$ since $\lim((n+j)(\cdots)(n+1))^{\frac{1}{n}}=1$
But I am stuck at the above question.
Any help?
| I'll assume $j=1$ (the case of arbitrary $j$ is analogous, and actually directly addressed by the proposition in the block below).
The question is then equivalent to asking whether, for a positive sequence $a_n$,
$$\limsup a_n^{1/(n-1)}=\limsup a_n^{1/n}.$$
By putting the left side as $(a_n^{1/n})^{(n/(n-1))}$, we have that the question will be answered if we can prove the following statement:
If $b_n$ is a sequence which converges to $1$ and $a_n>0$, then
$$\limsup a_n^{b_n} =\limsup a_n. $$
For that, we can proceed as follows: fix $\epsilon>0$. For sufficiently large $n$, it holds that
$$a_n^{1-\epsilon}\leq a_n^{b_n} \leq a_n^{1+\epsilon}. $$
Therefore,
$$\limsup a_n^{1-\epsilon} \leq \limsup a_n^{b_n} \leq \limsup a_n^{1+\epsilon}, $$
which implies
$$(\limsup a_n)^{1-\epsilon} \leq \limsup a_n^{b_n} \leq (\limsup a_n)^{1+\epsilon}.$$
Since this holds for every $\epsilon>0$, we have
$$\limsup a_n\leq \limsup a_n^{b_n} \leq \limsup a_n,$$
which implies $\limsup a_n^{b_n} = \limsup a_n.$
| {
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Evaluation of the limit $\lim_{q\rightarrow 1} \frac{\phi^5(q)_{\infty}}{\phi(q^5)_{\infty}}$ Given the Euler function $\phi(q)=\prod_{n = 1}^{\infty}(1-q^{n})$ which is a modular form where $q=\exp(2\pi i \tau)$, $|q|\lt1$
Then what is the limit $\lim_{q\rightarrow 1}\frac{\phi^5(q)_{\infty}}{\phi(q^5)_{\infty}}$ ?
| When $q\to 1^-$, $\tau \to 0$, hence $-1/\tau \to i\infty$,
$$\begin{aligned}\frac{{{\phi ^5}{{(q)}_\infty }}}{{\phi {{({q^5})}_\infty }}} = \frac{{{\eta ^5}(\tau )}}{{\eta (5\tau )}} &= \frac{{{{\left( {\frac{{ - 1}}{{\tau i}}} \right)}^{5/2}}{\eta ^5}(\frac{{ - 1}}{\tau })}}{{{{\left( {\frac{{ - 1}}{{5\tau i}}} \right)}^{1/2}}\eta (\frac{{ - 1}}{{5\tau }})}} \\ &= -\frac{{\sqrt 5 }}{{{\tau ^2}}}\frac{{{e^{ - \frac{{5\pi i}}{{12\tau }}}}\prod\limits_{n = 1}^\infty {{{\left( {1 - {e^{ - \frac{{2\pi ni}}{\tau }}}} \right)}^5}} }}{{{e^{ - \frac{{\pi i}}{{60\tau }}}}\prod\limits_{n = 1}^\infty {\left( {1 - {e^{ - \frac{{2\pi ni}}{{5\tau }}}}} \right)} }}\\ &\sim -\frac{{\sqrt 5 }}{{{\tau ^2}}}{e^{ - \frac{{2\pi i}}{{5\tau }}}} \to 0\end{aligned}$$
Note that the two infinite products both $\to 1$.
| {
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Let $f : [1,2] → \mathbb {R}$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and compute $\int_1^2f(x)dx$ Let $f : [1,2] → \mathbb {R}$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and
compute $$\int_1^2f(x)dx$$ as the limit of upper (and lower) sums.
I tried solving this but I don't know if my answer is right.
Let $P_n$ be the uniform partition of $[1,2]$ given by
$$1 \lt 1+\frac1{n} \lt 1+\frac2{n} \lt \cdots \lt 1+\frac{n-1}{n}\lt2$$
The function $f(x)=x$ is increasing, hence
$$m_i=inf_{x \in [x_{i-1},x_i]}f(x)=f(x_{i-1})=1+\frac{i-1}{n}$$
and
$$M_i=sup_{x \in [x_{i-1},x_i]}f(x)=f(x_i)=1+\frac{i}{n}$$
$$L(f,P)=\sum_1^n(1+\frac{i-1}{n})\frac{i}{n}=\frac{3n-1}{2n}$$ and $$U(f,P)=\sum_1^n(1+\frac{i}{n})\frac{i}{n}=\frac{3n+1}{2n}$$
Therefore $$\lim_{n\to\infty}L(f,P_n)=\frac32 \quad \lim_{n\to\infty}U(f,P_n)=\frac32$$
By the Criterion of Integrability $$\int_1^2f(x)dx=\lim_{n\to\infty}L(f,P_n)=\lim_{n\to\infty}U(f,P_n)=\frac32$$
| Yes, your proof is fine, everything is o.k.
| {
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Find Laplace transform of $(sin^2 2t)$ Find Laplace transform of $(\sin^2 2t)$
How do I go about this ?
do I spilt them up like
$ L ( \sin 2t) \cdot L (\sin 2t) $ ?
| Given $\sin^22t$
$$\sin^22t=\dfrac{1-\cos4t}{2}$$
$$L(\sin^22t)=\dfrac{L(1)-L(\cos4t)}{2}=\dfrac{1}{2s}-\dfrac12\left(\dfrac{s}{s^2+16}\right)$$
| {
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Convergence of sequence of functions in metric spaces Let $f_n(x)=\frac{nx}{nx+1}$
$a)$ Show that in $C_{[0,2]}$ $$\lim_{n \to \infty}{f_n(x)}=1$$
$b)$ Does $f_n$ converge in $C_{[0,1]}?$
Here is my attempt:
$a)$ We need to show that $d(f_n,f)<\epsilon$ where $f$ is the constant function $f(x)=1$. Following previous example in my book I assume that the metric here is the supremum metric (Is this good guess if the metric is not specified?). Using the definition of the metric:
$$\sup_{x \in C_{[0,2]}} {\left \lvert {f_n(x)-f(x)} \right \rvert} = \sup_{x \in C_{[0,2]}} {\left \lvert {\frac{nx}{nx+1}-1} \right \rvert}= \sup_{x \in C_{[0,2]}} {\frac{1}{nx+1}}=1$$
And we get a constant as a result. Not sure how to go from here. I was hoping to get something dependent on $n$ and that way we can find the appropriate term after which $d(f_n,1)<\epsilon$ for all $\epsilon>0$.
Exactly the same can be done for $b)$ although not sure whether that proves that it converges (since we have to guess the limit first).
| Note that for $x\neq 0$ we have that
$$
\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}\frac{x}{x+n^{-1}}=1
$$
and clearly $f_n(0)\to 0$ as $n\to \infty$. To note that the convergence is not uniform observe that
$$
f_n(1/n)=\frac{1}{2}.
$$
| {
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Proving lim inf means "all but a finite number" for measure theory Full disclosure: a philosophical logic person trying to learn math. Here is the problem from Gallant(1997), a measure-theoretic intro to econometrics:
Let $F_i$ where $i = 1, 2, ...$ be an infinite sequence of events from
the sample space $\Omega$. Let $F$ be the set of points that are in
all but a finite number of the events $F_i$. Prove that
$F=\bigcup_{k=1}^\infty \bigcap_{i=k}^\infty F_i$.
Per the suggestion of the problem, I'm starting by trying to prove $\omega \in F \rightarrow \omega \in \bigcup_{k=1}^\infty \bigcap_{i=k}^\infty F_i$
Importantly, the notion of "all but a finite number" isn't formalized in the text (beyond what we're trying to prove). I'm starting with something like $\exists m \in \mathbb N$ s.t. $\omega \in \bigcap_{i=n}^\infty F_i$ and $m \le n$ but also $\omega \notin \bigcap_{i=1}^\infty F_i.$ Not sure if this gets me to where I want to go, however, and not sure of the next steps. Any proof I can think of that doesn't rely on an explicit formalism of the notion basically just looks like I'm describing how the $\bigcup_{k=1}^\infty \bigcap_{i=k}^\infty F_i$ structure works, which doesn't feel like a proof to me.
This is a problem early in the text, so it should be provable by fairly primitive notions. I think mentioning the philosophical logic part is relevant because maybe the somewhat more informal proof writing rules of math are throwing me for a loop. (For instance, earlier in the text they "proved" DeMorgan's laws for pairwise union and intersection by assuming DeMorgan's laws for 'or' and 'and,' which seemed like a cop-out, not that proving DM's laws for those notions are challenging).
Any help on this proof?
| If $\omega\in F$, then $\omega$ is in all but a finite number of the events $F_i$. Therefore, there is a natural number $N$ such that $\omega$ is in every $F_i$ when $i\geqslant N$. So, $\omega$ belongs to each$$\bigcap_{i=k}^\infty F_i$$as long as $k\geqslant N$ and therefore$$\omega\in\bigcup_{k=1}^\infty\bigcap_{i=k}^\infty F_i.$$
| {
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Properties of $\{x\in X\mid f(x)=||f||\}$ Let $X$ be a normed space, $f\in X^*\setminus\{0\}$ (the continuous dual), $E:=\{x\in X\mid f(x)=\|f\|\}$. Prove that $E$ is a nonempty closed set and that $\inf \{\|x\|\mid x\in E\}=1$.
I have no idea how to prove that $E$ is non empty and that $\inf \{\|x\|\mid x\in E\}=1$.
$E$ is closed because it is the inverse image of a point ($\|f\|$) and $f$ is continuos, so inverse images of closed sets are closed.
| Since $f \neq 0$ we have $f(x^*) \neq 0$ for some $x^*$, so then $f({\|f\| \over f(x^*)} x^*) = \|f\|$ so $E$ is not empty.
Since $f$ is continuous, $f^{-1} ( \{ \|f\| \})$ is closed.
Note that $f(x) = \|f\|$ for all $x \in E$. Since $\|f\| = f(x) \le \|f\| \|x\|$ we see that $\|x\| \ge 1$.
| {
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$xy'-y=0$ when $x<0$
How to solve $xy'-y=0$ when $x<0$?
It seem to be a simple equation, but it is confusing when $x<0$
$$\frac{y'}{y}=\frac{1}{x}$$
Now to integrate the both sides must I assume that $y$ is positive then
$$\ln y =- \ln(-x)+c$$so
$$y(x)=e^{-\ln(-x)}$$
Am I right? what if the assumption on $y$ is false?
This equation has a solution over all $\mathbb{R}$?
Thanks in advence
| Without log function
$$y'x-y=0$$
for $x \ne 0$
$$\frac {(y'x-y)}{x^2}=0$$
$$\left(\frac yx \right)'=0$$
Integrate
$$\frac yx= K$$
$$y(x)=Kx$$
| {
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Proof by induction that you can order natural numbers where the average isn't between any pair of numbers Consider a list of natural numbers like
$$1, 2, 3, \dots, n$$
Prove using strong induction that you can order the list for any Natural n, in a way where if you pick any pair of numbers, the average of the numbers you choosed isn't between the numbers you choosed, consider the example
$$1, 3, 2$$
Where $1, 2, 3$ is not an acceptable list since the average of $1$ and $3$ is $2$, who is between $1$ and $3$.
Until now I have found that first you may separate the odd numbers, let them in the beginning or the end of the list, then separate from the sublist of odd numbers, the numbers who has odd index, and then make that again recursively with the sublist you create, this because the average of an odd with odd index with an odd with pair index is pair, and pairs are in the list of pairs you separated at the beginng
| The method you mentioned is for all intents and purposes a complete solution. Here is way to write it as a proof by strong induction.
The base case $n=1$ is obvious.
For any $n>1$, $d=\lceil n/2\rceil$ and let $e=\lfloor n/2\rfloor$. Note $d,e<n$. Also, $d$ is the number of odd numbers in $\{1,2,\dots,n\}$, while $e$ is the number of even numbers.
By induction, the sets $\{1,2,\dots,d\}$ and $\{1,2,\dots,e\}$ can be ordered so no two numbers surround their average. Let these orderings be $(a_i)_{i=1}^{d}$ and $(b_i)_{i=1}^{e}$. Now, consider the list
$$
(2a_1-1,2a_2-1,\dots,2a_{d}-1,2b_1,2b_2,\dots,2b_{e})
$$
You can show that this contains every number in $\{1,2,\dots,n\}$ exactly once, and the average property still holds.
| {
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Does there exist a right triangle with area 7 and perimeter 12? This question is really trivial. I can prove that there is no right triangle with area 7 and perimeter 12, but what I do is solve the following system: if $a$, $b$ and $c$ are, respectively, the two legs and hypotenuse of such a triangle, then
$$a^2 + b^2 = c^2,$$
$$a + b +c = 12,$$
$$ab = 14$$
It is easy (although a bit boring and long) to see that there are no real solutions to this system.
But I feel that there is a simple answer to this question - perhaps using the triangle inequality - but I just cannot see it.
| I'm not sure if this counts as a "simple answer", but let $x$ be the length of one leg of a right triangle of area $7$; then the other leg is $\frac{14}x$, and the hypotenuse is $\sqrt{x^2 + \left(\frac{14}x\right)^2}$, so the perimeter is given by the function
$$P(x) = x + \frac{14}x + \sqrt{x^2 + \left(\frac{14}x\right)^2}$$
Asymptotically, we have $\lim_{x\to 0} P(x) = +\infty$ and $\lim_{x\to\infty} P(x) = +\infty$, and intuitively it seems clear that the absolute minimum occurs when the two legs of the triangle are equal in length, i.e. when $x=\sqrt{14}$. At this value, we have
$$P(\sqrt{14}) = 2\sqrt{14} +2\sqrt{7}$$
This is the smallest possible perimeter for a right triangle of area $7$. It remains only to convince yourself that this number is greater than $12$.
| {
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How to find point in Descartes Folium with slope of -1/3? I stumbled upon a problem in my calculus book that asked to find the point in $x^3 + y^3 = 3xy$ that had a slope perpendicualr to $y = 3x + 1$ and also was in the first quadrant.
I began by getting the derivative of the equation and got $\dfrac{x^2-y}{x-y^2}$.
Given that the slope is $-\frac13$, I wrote $\dfrac{x^2-y}{x-y^2}=-\frac13
$, which got me to $3x^2+x=y^2+3y$, and that is where I got stuck, because I cannot solve for $x$ or $y$, to substitute in the original equation.
I looked for similar examples on the internet, and only found some where the slope was $0$, which can be solved because $y = x^2$, which can be substituted back.
Hope this is redacted well enough, any help is really appreciated.
| Let's find a parametric equation of the curve defined by $x^3+y^3=3xy$.
Let $y=xt$, then
$$x^3(1+t^3)=3x^2t$$
Hence
$$x=\frac{3t}{1+t^3}$$
$$y=\frac{3t^2}{1+t^3}$$
Now, the tangent vector is given by
$$x'(t)=\frac{3(1+t^3)-3t(3t^2)}{(1+t^3)^2}=\frac{3-6t^3}{(1+t^3)^2}$$
$$y'(t)=\frac{6t(1+t^3)-3t^2(3t^2)}{(1+t^3)^2}=\frac{6t-3t^4}{(1+t^3)^2}$$
You want $t$ such that $y'(t)=-\frac13x'(t)$, that is
$$6t-3t^4=2t^3-1$$
You have thus a quartic equation to solve: $3t^4+2t^3-6t-1=0$. There does not seem to be a trivial solution, so you'll have to go through Descartes' or Ferrari's method to solve this, or a numerical method.
There are two real roots (hence two points), with numerically
$$t_1=-0.1678460300438111, x_1=-0.5059304363344097, y_1=0.0849184152170638$$
$$t_2=1.1301467382482941, x_2=1.3875575405757694, y_2=1.5681436286135308$$
| {
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Counting problem - fault in my reasoning. The problem is as follows:
The dean of science wants to select a committee consisting of mathematicians and physicists to discuss a new curriculum. There are $15$ mathematicians and $20$ physicists at the faculty; how many possible committees of $8$ members are there, if there must be more mathematicians than physicists (but at least one physicist) on the committee?
The given solution splits up the choices into the cases where there is $1$ physicist, $2$ physicists, and $3$ physicists. Then it's simply a matter of summing up $\binom{15}{5}\cdot\binom{20}{3}$ and $\binom{15}{6}\cdot\binom{20}{2}$ and $\binom{15}{7}\cdot\binom{20}{1}$, giving $4503070$.
This is what I tried to do:
first I pick a physicist, making sure I have at least one on the team. I have $20$ choices for this. Then I pick $5$ mathematicians, making sure there are more mathematicians than physicists on the team. I have $15\cdot14\cdot13\cdot12\cdot11$ choices for this. Then I pick two people from the $29$ people that are left because it doesn't matter whether they are mathematicians or physicists. I have $29\cdot28$ choices for this. After I multiply all those numbers, I need to divide by the number of ways I can permutate them, because order doesn't matter. I get $$\dfrac{20\cdot15\cdot14\cdot13\cdot12\cdot11\cdot29\cdot28}{8!} = 145145,$$ which is obviously not the right solution. Where did I go wrong?
Thanks in advance.
| You don't have $8!$ possible permutations. For instance, the first person you chose can never be a mathematician and the second, third, fourth, fifth and sixth persons you chose can never be physicists.
| {
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Does this mathematical technique have a formal name? And why does it work? When I split a number in the powers of 2. I am able to make any combination of any number that is less than it by taking each number of the series only once.
For example:
$7=1+2+4$
I can construct any number less than or equal to seven using the $3$ numbers $1,2,4$.
Or take $10$ for instance:
$10 = 1 + 2 + 4 + 3$
I can construct any number that is less than or equal to $10$ using the 4 numbers.
How this decomposition is done is via breaking the number in powers of 2 until you can't break it anymore. More examples:
5 -> (1, 2, 2)
10 -> (1, 2, 4, 3)
15 -> (1, 2, 4, 8)
1000 -> (1, 2, 4, 8, 16, 32, 64, 128, 256, 489)
Does this method of decomposition have a name? And how does it allow us to make any number that is less than or equal to it?
| Countrary to most of your feedback, this is not at all binary representation, because you split from the small end and not from the big end.
For binary representation to represent 7 you start with:
*
*8 no it's too big
*4, ok it fits, 7-4=3 left,
*2, ok it fits 3-2=1 left
*1, ok it fits 1-1=0 left, done!
What you are doing is splitting from the small end, so you will end up with a number $2^k-1 + b$, where $b$ is some remainder. I don't know if this has any name, but it sure is not a binary representation of the number.
| {
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If H(b) $\leq$ b then $\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b) - [1-b]db $? $H$ is a CDF over $[0,1]$
I want to prove that if $H(b) \leq b$ then :
$$\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b) - [1-b]db $$ for all $k \in [0,1]$
I think this is true, but a rigorous proof does seem difficult to me. Can anyone help me to prove this? or does any theorem helps?
| I assume $h(b)=H'(b)$ is a density, $v$ is a constant and $H(b) \leq b \quad\forall b$
\begin{align}
&H(b) \leq b \implies H(b)-1 \leq b-1 \implies 1-H(b)\geq1-b \implies -(1-H(b))\leq-(1-b)
\end{align}
Substituting in the equation you get the following inequality.
$\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b)h(b) - [1-b]db$
Now we need to show that substituting $h(b)$ for $1$ gives us the desired inequality.
We know that $\int_0^k(v-b)db = vk -\frac{k^2}{2}$ then:
\begin{align}
\int_0^k (v-b)h(b)db &= v\int_0^kh(b)db-\int_0^kbh(b)db\\
&= v[H(k)-H(0)]-\int_0^kbh(b)db\\
&= v(H(k)-H(0))-bH(b)|_0^k +\int_0^kH(b)db\\
&\leq(v-k)H(k)-H(0)v +\int_0^kbdb\\
&H(b)\leq b\; \forall b\; \wedge \text{H(.) is a cdf}\implies H(0)=0 \\
&\leq (v-k)k+\frac{k^2}{2}\\
&=vk-\frac{k^2}{2}
\end{align}
then the following holds whenever $v \geq k$
$$\int_0^k (v-b)h(b)db \leq vk-\frac{k^2}{2} = \int_0^k(v-b)db$$
| {
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Why should the error of a local linearization for $x$ near $a$ be small relative to $(x-a)$? I am reading a calculus book. It says
The fact that the graph of $f$ looks like a line as we zoom in means that not only is $E(x)$ small for $x$ near $a$, but also that $E(x)$ is small relative to $(x−a)$
where $E(x) = f(x) - f(a) - f(a)^{'}(x-a)$ is an error of the local linearization, and then introduces a theorem that $$\lim_{x\to a}{\frac{E(x)}{x-a}}=0$$
I just can't understand why $E(x)$ should be small relative to $(x-a)$. What is the need to be so if we want a function to be linear when we zoom in?
P.S. My question is not about why the theorem holds. Rather I can't understand the quoted statement above. Why relative to $(x-a)$ not other value. What is special about this relation?
| I'm going to focus on the intuition. I make no claims that my answer is rigorous.
Write $$E(x) = f(x)-f(a)-f^{\prime}(a)(x-a)\text{.}$$
$f^{\prime}(a)$ is the "slope" of $f$ at $x = a$. Make sure you understand that derivatives are just slopes. For sake of convenience, let's call this $m$.
Way back in your Pre-Calculus days, remember that the computational formula for the slope of a linear equation (hence why there's so much focus on linearity) is
$$m = \dfrac{y_2 -y_1}{x_2 - x_1}\text{.}$$
Now, you can think of $x$ and $a$ as corresponding to $x_2$ and $x_1$ respectively. Hence we have
$$m(x_2 - x_1) = y_2-y_1 \implies m(x-a)=y_2-y_1\text{.}$$
For a linear function $f$, $y_2$ is the $y$-value of $f$ at $x$ and $y_1$ is the $y$-value of $f$ at $a$. Hence, we have
$$m(x-a) = f(x)-f(a)$$
when $f$ is linear, but when $f$ is non-linear, this is only an approximation. So,
$$m(x-a) \approx f(x)-f(a)$$
and hence, for $x$ near $a$, we would hope that
$$E(x) \approx f(x)-f(a)-[f(x)-f(a)] = 0\text{.}$$
This explains the first claim.
I think @callculus explained the second claim sufficiently well: note that in order to even begin to compute $f^{\prime}(a)$, $f$ must be differentiable at $a$. Notice that
$$\begin{align}
\lim\limits_{x \to a}\dfrac{E(x)}{x-a} &= \lim_{x \to a}\dfrac{f(x)-f(a)-f^{\prime}(a)(x-a)}{x-a} \\
&= \lim\limits_{x \to a}\dfrac{f(x)-f(a)}{x-a}-\lim\limits_{x \to a}\dfrac{(x-a)f^{\prime}(a)}{x-a} \\
&=\lim\limits_{x \to a}\dfrac{f(x)-f(a)}{x-a}-f^{\prime}(a)
\end{align}$$
The limit $$\lim\limits_{x \to a}\dfrac{f(x)-f(a)}{x-a}$$
must exist for $f$ to be differentiable at $a$ (this is by the definition of a derivative); otherwise, you wouldn't be able to compute $f^{\prime}(a)$.
As long as $f$ is differentiable at $a$, the resulting limit above is
$$\lim\limits_{x \to a}\dfrac{E(x)}{x-a} = f^{\prime}(a)-f^{\prime}(a) = 0\text{.}$$
| {
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Reduction Formula for $I_n=\int \frac{dx}{(a+b \cos x)^n}$ Reduction Formula for $$I_n=\int \frac{dx}{(a+b \cos x)^n}$$
I considered $$I_{n-1}=\int \frac{(a+b \cos x)dx}{(a+b \cos x)^n}=aI_n+b\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$
Let $$J_n=\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$ using parts for $J_n$ we get
$$J_n=\frac{\sin x}{(a+b \cos x)^n}-nb \int \frac{\sin^2 x\:dx}{{(a+b \cos x)^{n+1}}}$$
Can we proceed here?
| This is not a full answer the the original question but instead evidence that any such answer will not lead to the underlying problem at hand: calculating the given integral.
Mathematica yields:
$$\frac{\csc (x) \sqrt{\frac{b (\cos (x)+1)}{b-a}} \sqrt{\frac{b-b \cos (x)}{a+b}} (a+b
\cos (x))^{1-n} F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{a+b \cos
(x)}{a-b},\frac{a+b \cos (x)}{a+b}\right)}{b (n-1)}$$
which strongly suggests you won't find a simple such reduction. So in principle there may be a way to proceed as you ask, but it seems virtually certain that such an approach will lead to a dead end.
| {
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Rectangle on sphere Is it true that every set of positive measure on a 2-sphere contains four points which form a rectangle? Note that I am not asking about if there exists a "filled-in" rectangle, or even a "rectangle" in the product of measurable sets sense, but simply four points which form a rectangle in $\mathbb{R}^3.$
| Suppose $X$ is a subset of the sphere $S$ and measure $\mu(X)>0$. It is straightforward to show that one can find a point $x\in S$ and a disk $D(x,r)\subseteq S$ centered at $x$ of some radius $r>0$ such that
$$\dfrac{\mu(X \cap D(x,r))}{\mu(D(x,r)} \approx 1.$$
Project the set $X\cap D(x,r)$ on the plane through the origin that is parallel to the circle boundary of $D(x,r)$. This will be a subset of $B(0,r)$, the disk of radius $r$ on the 2-dimensional plane. So the problem follows from the following claim: (throughout the proof by $a \approx b$ we mean $a$ can be made arbitrarily close to $b$).
Claim: Let $Y \subseteq B(0,1)$ such that $\mu(Y) \approx \pi$. Then there exists a point $y\in Y$ such that $y,\theta(y),\theta^2(y), \theta^3(y) \in Y$, where $\theta$ is the rotation by 90 degrees around the origin.
Proof: Let $Y_i$ denote the intersection of $Y$ with the open $i$th quadrant, $i=1,2,3,4$. Consider $\theta(Y_1)$ is a subset of the second quadrant. Then $\mu(\theta(Y_1)) \approx \pi/4$ and $\mu(Y_2) \approx \pi/4$. So $\mu(\theta(Y_1) \cap Y_2) \approx \pi/4$. Next consider $Y_2'=\theta(Y_1) \cap Y_2)$ and note that the measure of this set is close to $\pi/4$ (as close as we want, this argument can be made precise). Therefore $\mu(\theta(Y_2') \cap Y_3)$ can be made arbitrarily close to $\pi/4$. Finally, for the set $Y_3'=\theta(Y_2') \cap Y_3)$, we have $\mu(\theta(Y_3') \cap Y_4) \approx \pi/4$, so certainly nonempty. For any $z$ in this nonempty set, $z, \theta(z), \theta^2(z), \theta^3(z) \in Y$ are vertices of a rectangle.
| {
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If $S(n)$ is an odd integer, what is the sum of all possible $\frac1n?$ If $n$ is a positive integer, let $S(n)$ be the sum of all the positive divisors of $n$.
If $S(n)$ is an odd integer, what is the sum of all possible $\frac1n?$
The function $S$ is multiplicative and so, if we have the prime factorisation $n = p_1^{a(1)}p_2^{a(2)} \cdots p_m^{a(m)}$, where $p_1,p_2,...,p_m$ are distinct primes, then how will I continue to solve this?
| $S(n)$ is multiplicative, so $S(n) = \prod_{i=1}^m S(p_i^{a_i})$. If any of the $S(p_i^{a_i})$ are even, the whole product will be even. So what is the condition on $S(p^a)$ being even for some prime $p$ and natural number $a$? What does that tell you about the set of numbers $n$ for which $S(n)$ is odd?
Once you've figured this out, you're going to need to sum the series. You'll want to factor out a geometric series involving powers of $2$, and you'll need to use the fact that $\sum_{k=0}^\infty k^{-2} = \pi^2/6$.
| {
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Solutions of a differential equation I'm trying to solve the following differential equation and I'm stuck at what it appears to be simple calculations. I'm terribly sorry if this turns out to be really simple.
$(1)$ $X(f)=2f$
where $X=x_1^2 \frac \partial {\partial x_1}-x_2^2 \frac \partial {\partial x_2}$ in $\Bbb R^2$ with the identity chart $Id_{\Bbb R^2}=(x_1,x_2)$
and $f:\Bbb R^2 \to \Bbb R$,
$(2)$ $f(cosθ,sinθ)=cosθ+sinθ$.
Let $φ^Χ_t(p)=(\frac {x}{1-tx},\frac {y}{1+ty})$, where $p=(x,y)$, be the flow of $Χ$ and by denoting $h(t)=f(φ^Χ_t(p))$ we can make $(1)$ look like $h'(t)=2h(t)$ which can be easily solved to:
$e^{2t}f(x,y)=f(\frac {x}{1-tx},\frac {y}{1+ty})$
Then by use of the initial condition $(2)$ we have
$e^{2t}(cosθ+sinθ)=f(\frac {cosθ}{1-tcosθ},\frac {sinθ}{1+tsinθ})$ (this is as far as I can go)
I tried setting $u = \frac {cosθ}{1-tcosθ}, v=\frac {sinθ}{1+tsinθ} $
but I haven't been able to isolate $u,v$ from $θ, t$
Can you give me any hints? Is there any trick I'm not thinking of?
| $$x^2 \frac {\partial f}{\partial x}(x,y)- y^2 \frac {\partial f}{\partial y}(x,y)=2f(x,y).$$
Search for the general solution (without taking account of the boundary condition) with the method of characteristics :
The characteristic ODEs are :
$$\frac{dx}{x^2}=\frac{dy}{-y^2}=\frac{df}{2f}$$
A first characteristic equation comes from $\frac{dx}{x^2}=\frac{dy}{-y^2}$ :
$$\frac{1}{x}+\frac{1}{y}=c_1$$
A second characteristic equation comes from $\frac{dx}{x^2}=\frac{df}{2f}$ :
$$e^{2/x}f=c_2$$
The general solution expressed on the form of implicite equation is :
$$\Phi\left(\frac{1}{x}+\frac{1}{y}\:,\:e^{2/x}f \right)=0$$
where $\Phi$ is an arbitrary function of two variables. This function has to be determined later according to boundary conditions.
Or, equivalently on explicit form : $e^{2/x}f=F\left(\frac{1}{x}+\frac{1}{y} \right)$
$$f(x,y)=e^{-2/x}F\left(\frac{1}{x}+\frac{1}{y} \right)$$
where $F$ is an arbitrary function. This function has to be determined later according to boundary conditions.
BOUNDARY CONDIION :
In the original wording of the question, the boundary condition is not clearly defined. A discussion took place in the comments.
To the question : Is the boundary condition $f(x,y)=x+y$ on the unit circle $x^2+y^2=1$ ? the OP answered "that should be it", which is not really affirmative. So, this supposed boundary condition can be suspected to be mistaken.
Supposing that the boundary condition is $f(x,y)=x+y$ on the unit circle $x^2+y^2=1$, thus $y=\pm\sqrt{1-x^2}$ , my comment is :
The function $F$ has to be determined from :
$$x\pm\sqrt{1-x^2}=e^{-2/x}F\left(\frac{1}{x}+\frac{1}{\pm\sqrt{1-x^2}} \right)$$
In fact, it is theoretically possible to find the function $F$ but the calculus is rather arduous and the function $F$ is very complicated. This draw to think that something might be wrong in the wording of the question. The OP should re-examine what is really the boundary condition. To help him, it should be necessary that the OP re-edit his question with a detailed explanation how he got the above boundary condition.
| {
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Find a simple formula of k numbers which gives output a number not among these k numbers. This question just came to my mind while thinking. I just wanted to know if it is even possible. Anyway, the question goes as follows.
You are given k numbers $a_{i}$ (k < n) from 1 to n.
Your job is to give a simple formula represented as a function f($a_{1}, a_{2}, .., a_{k}$) which gives a number N $\in$ {1, 2, .., n} such that N $\neq$ $a_{i}$ for any i $\in$ {1, 2, .., k}
My approaches(which obviously have failed :P) :
*
*($\sum_{i=1}^{k}a_{i})$ % n + 1
*($\prod_{i=1}^{k}a_{i}$) % n + 1
*(xor all of them) % n + 1
*(and all of them) % n + 1 && (or all of them) % n + 1
I tried above approaches because they are similar to Euclid's proof that primes are infinite. This question seems to be related to abstract algebra, but I am still learning it so pardon me if I may have made a dumb mistake.
By "simple" I mean operations that can be performed in O(1) like +, -, x, /, xor, |, &, ...
| You can do $f(K) = \min\{\{1,\dots,n\} \setminus K\} $ where $K=\{a_1,\dots\}$.
| {
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calculating a higher order derivative My task is to find the values $f^{(2017)}(0)$ and $f^{(2018)}(0)$ for $f(x)=\frac{\arccos(x)}{\sqrt{1-x^2}}$.
Basically, it's about finding the $n^{th}$ derivative of $f$.
So I noticed if I let $g(x)=arccos(x)$, then $f(x)=-g(x)\cdot g'(x)$. I was able to prove by induction that for all $n\geq 2$ and $k\in \Bbb{N}$ the $n^{th}$ derivative of $g'$ is $$[(g')^{k}]^{(n)}(0)=k\cdot(n-1)\cdot[(g')^{k+2}]^{(n-2)}(0)$$
But even with applying the Leibniz rule to $f=g\cdot g'$ I don't understand how to get the final result. Did I make the wrong approach to the problem or is the general formula above useful? If so, how should I apply it?
| Let $ n\in\mathbb{N} $, if $ f $ is a $ \mathcal{C}^{n+1} $ function on $ \left(-1,1\right) $, then we can use Leibniz formula for the $ n $-th derivative of the product :
$$ \left(\forall x\in\left(-1,1\right)\right),\ \left(ff'\right)^{\left(n\right)}\left(x\right)=\sum_{k=0}^{n}{\binom{n}{k}f^{\left(k\right)}\left(x\right)f^{\left(n+1-k\right)}\left(x\right)} $$
Setting $ f:x\mapsto\arccos{x} $, and $ g:x\mapsto\frac{1}{\sqrt{1-x^{2}}} $, we have : \begin{aligned} \left(\forall p\geq 1\right)\left(\forall x\in\left(-1,1\right)\right),\ f^{\left(p\right)}\left(x\right)&=-\frac{\mathrm{d}^{p-1}}{\mathrm{d}x^{p-1}}\left(\frac{1}{\sqrt{1-x^{2}}}\right)\\ &=-\sum_{k=0}^{p-1}{\binom{p-1}{k}\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}\left(\left(1-x\right)^{-\frac{1}{2}}\right)\frac{\mathrm{d}^{p-1-k}}{\mathrm{d}x^{p-1-k}}\left(\left(1+x\right)^{-\frac{1}{2}}\right)}\\ &=-\sum_{k=0}^{p-1}{\binom{p-1}{k}\left(-1\right)^{k}\prod_{j=0}^{k-1}{\left(-\frac{1}{2}-j\right)}\left(1-x\right)^{-\frac{1}{2}-k}\prod_{j=0}^{p-2-k}{\left(-\frac{1}{2}-j\right)}\left(1+x\right)^{\frac{1}{2}+k-p}}\\ f^{\left(p\right)}\left(x\right)&=-\left(p-1\right)!\sum_{k=0}^{p-1}{\left(-1\right)^{k}\binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}}{p-1-k}\left(1-x\right)^{-\frac{1}{2}-k}\left(1+x\right)^{\frac{1}{2}+k-p}}\end{aligned}
Thus : $$ f^{\left(p\right)}\left(0\right)=-\left(p-1\right)!\sum_{k=0}^{p-1}{\left(-1\right)^{k}\binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}}{p-1-k}}=-\frac{1-\left(-1\right)^{p}}{2}\times\frac{\Gamma\left(\frac{p}{2}\right)\Gamma\left(p\right)}{\sqrt{\pi}\Gamma\left(\frac{p+1}{2}\right)} $$
Hence : \begin{aligned} \left(ff'\right)^{\left(n\right)}\left(0\right)&=f\left(0\right)f^{\left(n+1\right)}\left(0\right)-\sum_{k=1}^{n}{\binom{n}{k}\frac{1-\left(-1\right)^{k}}{2}\times\frac{\Gamma\left(\frac{k}{2}\right)\Gamma\left(k\right)}{\sqrt{\pi}\Gamma\left(\frac{k+1}{2}\right)}\frac{1-\left(-1\right)^{n+1-k}}{2}\times\frac{\Gamma\left(\frac{n+1-k}{2}\right)\Gamma\left(n+1-k\right)}{\sqrt{\pi}\Gamma\left(\frac{n+2-k}{2}\right)}}\\ &=\fbox{$\begin{array}{rcl}-\displaystyle\frac{\pi\left(1+\left(-1\right)^{n}\right)}{4}-\frac{n!\left(1-\left(-1\right)^{n}\right)}{2\pi}\sum_{k=1}^{n}{\left(\frac{1-\left(-1\right)^{k}}{2k}\right)\frac{\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{n+1-k}{2}\right)}{\Gamma\left(\frac{k+1}{2}\right)\Gamma\left(\frac{n+2-k}{2}\right)}}\end{array}$} \end{aligned}
| {
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Show that $u\times v+v\times w+w\times u=\textbf{0}$ implies that $u,v,w$ are linearly dependent I have what I think might be a solution, but I'm not sure it's formal enough. I begin by noting that $u,v,w$ are linearly dependent iff. they lie on the same plane. Then I construct the following chain of equivalences:
\begin{align*}
u\times v+v\times w+w\times u=\textbf{0}&\Longleftrightarrow u\times v+v\times w=-w\times u\\&\Longleftrightarrow v\times (w-u)=u\times w
\end{align*}
Then my reasoning is that since $w-u$ is obviously on the plane that $u,w$ spans, then for $v\times (w-u)$ to form the same normal vector as $u\times w$, v must lie on the same plane. I think the reasoning is correct, but how do I formalize this last part using mathematical notation? Is it necessary to do so?
| Note that $$v \times w + w \times u + u \times v =v \times w - u \times w + u \times v $$
$$=(v-u) \times (w-v)$$
Thus $$ v \times w + w \times u + u \times v = 0$$
$$\implies (v-u) \times (w-v)=0$$
Which implies $u$,$v$,and $w$ are linearly dependent.
| {
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Definition of orthogonal subspace In the book I'm using to study there is this definition: given a vectorial space $E$ with an internal product:
*
*given $u \in E$, the set of all vectors in $E$ that is orthogonal to $u$ is a subspace of $E$;
*given a subset $M$ of $E$, the set of all v in E such that v is orthogonal to $u$, for any $u$ in $M$ is a subspace of $E$.
Question: can $M$ be any set? Doesn't it need to be, itself, a subspace of $E$? And them, the newly created subspace would called $M$-orthogonal?
| As my analysis prof used to say "you can define whatever you want as long as you are consistent" ;) note that the orthogonal set does not change, when you replace $M$ by its span. This means, you can take the vector space generated by $M$ and everything reduces to the case you are familiar with.
Added: We want to prove the following equality of sets
$$ \{ x\in E \ \vert \ \forall m\in M : \langle x, m\rangle=0 \} = \{ x\in E \ \vert \ \forall m\in span(M) : \langle x, m\rangle=0 \}$$
The inclusion $\supseteq$ is clear. Let us prove the other inclusion. Let $x$ be orthogonal to $M$, we want to prove that it is orthogonal $span(M)$. Let $m\in span(M)$, then we can write $m=\alpha_1 m_1 + \dots + \alpha_n m_n$ where $\alpha_i \in \mathbb{C}$ and $m_i \in M$ and we get
$$ \langle x , m \rangle = \alpha_1 \langle x, m_1 \rangle + \dots + \alpha_n \langle x, m_n \rangle = \alpha_1 \cdot 0 + \dots + \alpha_n \cdot 0 = 0. $$
The first equality is the bilinearity of the inner product and the second equality is the assumption that $x$ is orthogonal to $M$. Thus, $x$ is orthogonal to $span(M)$.
| {
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Conjecture that $ \frac{\gcd(a+b,ab)}{\gcd(a,b)} \mid \gcd(a,b)$ I have discovered some exercise type conjectures which I can't prove and this is one of them:
Given positive integers $a,b$, then
$$ \frac{\gcd(a+b,ab)}{\gcd(a,b)}\ \bigg|\ \gcd(a,b)$$
Can this be proved or disproved?
From time to time, when testing my growing math packages BigZ and Forthmath, I recognize some patterns which I can't prove or disprove (or even have the ambition to). I post them here with the hope that it will not annoy too much. I hope you can bear with it.
| Write $\gcd(a,b)=d$, then $a=da',b=db'$ and thus $\frac{\gcd(a+b,ab)}{d}=\gcd(a'+b',a'b'd)$ where $\gcd(a',b')=1$. Notice now that $\gcd(a'+b',a'b')=1$ since $a'(a'+b')-a'b'={a'}^2$ and thus $\gcd(a'+b',a'b')|{a'}^2 \implies \gcd(a'+b',a'b')|a'$ and $\gcd(a'+b',a'b')|a'+b' \implies \gcd(a'+b',a'b')|a',b' \implies \gcd(a'+b',a'b')=1$. This means that $\gcd(a'+b',a'b'd)=\gcd(a'+b',d)$ and thus it divides $d$ by definition. So your conjecture is indeed true.
| {
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Derivative of inverse function proof verification Can someone verify whether my attempt to prove this theorem is correct?
Notice that I use a generalized definition of derivative:
Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Let $p$ both a point and a limit point of $E$. Then we define the derivative of $f$ at $p$ as the limit $f'(p) = \lim_{x \to p, x \in E \setminus \{p\}} \frac{f(x)-f(p)}{x-p}$, provided the limit exists.
Theorem: Let $f: E \subseteq \mathbb{R} \to F \subseteq \mathbb{R}$ be an invertible function that is differentiable at $p \in
E$. Suppose that $f^{-1}: F \to E$ is continuous at $f(p)$ and that
$f'(p) \neq 0$. Then $f^{-1}$ is differentiable at $f(p)$, and we have
$$(f^{-1})'(f(p)) = \frac{1}{f'(p)}$$
Proof: Before proving the theorem, we have to check that differentiation at $f(p)$ makes sense: we have to show that $f(p)$ is a limit point of $Y$.
Let $\epsilon > 0$. Because $f$ is differentiable at $p$, $f$ is continuous at $p$ and it follows that $|f(p)-f(x)| < \epsilon$ whenever $x \in (p- \delta, p + \delta) \cap E \setminus \{p\}$ for some $\delta > 0$. Notice: $f'(p) \neq 0$ implies that $f$ is not constant on $(p- \delta, p + \delta) \cap E \setminus \{p\}$ (if it were constant, we would have $f'(p) = \lim_{x \to p} \frac{f(x)-f(p)}{x-p} = \lim_{x \to p, x \in E \cap (p- \delta, p + \delta)\setminus \{p\}}\frac{f(x)-f(p)}{x-p} = 0)$. Combining these facts, we deduce that $0 < |f(x)-f(p)| < \epsilon$ for some $x \in E$, and $f(p)$ is a limit point of $F = f(E)$.
Define $$F: E \to \mathbb{R}: x \mapsto \begin{cases} \frac{f(x)-f(p)}{x-p} \quad x \neq p \\ f'(p) \quad x = p\end{cases}$$
Clearly, $F$ is continuous at $p$.
The theorem now follows from the following easy calculation:
$$(f^{-1})'(f(p)) = \lim_{y \to f(p)} \frac{f^{-1}(y)- f^{-1}(f(p))}{y-f(p)}$$
$$= \lim_{y \to f(p)}\frac{1}{\frac{f(f^{-1}(y))-f(p)}{f^{-1}(y)- p}}$$
$$ = \lim_{y \to f(p)} \frac{1}{F(f^{-1}(y))}$$
$$= \frac{1}{F( \lim_{y \to f(p)} f^{-1}(y))} = \frac{1}{F(p)} = \frac{1}{f'(p)}$$
However, some equalities need some explanation. The second equality is justified by noticing that $f^{-1}(y) - p = f^{-1}(y) - f^{-1}(f(p))$ is zero only when $y = f(p)$, so there are no trouble with dividing by zero. The fourth equality uses the continuity of $F$ at $p$ and the fifth equality follows from the continuity of $f^{-1}$ at $f(p) \quad \square$
| Your work will be done faster if not make use of the First Principle of Derivative.
We know that $f^{-1}(f(x))=x$, differentiate it combining with Chain Derivative, we have $$\begin{align}(f^{-1}(f(x)))'&=1\\f^{-1}{'}(f(x))f'(x)&=1\\\therefore f^{-1}{'}(f(x))&=\dfrac1{f'(x)}\end{align}$$
I am not too used in rigorous proof in derivative (such like the first principle) but I think your proof is good enough.
| {
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Existence of a transitive model is strictly stronger than consistency? It seems like we should be able to prove that the existence of a transitive model for ZFC is strictly stronger than Con(ZFC), but I can't find anything saying so / giving an argument for it. Is there a standard way of demonstrating this?
An example of what I'm looking for, if it existed: given Con(ZFC), is there a way of generating a model which models Con(ZFC) but which believes that no models of ZFC are transitive?
| Any model of ZFC+Con(ZFC)+$\neg$Con(ZFC+Con(ZFC)) should do. (This theory is, by the second incompleteness theorem, consistent if ZFC+Con(ZFC) is).
Because the model believes Con(ZFC) and also believes that no model of ZFC can believe Con(ZFC), the only models of ZFC it can know will be ones that it considers to have non-standard integers. Non-standard models of arithmetic are never well-founded, so such an internal model cannot be transitive.
| {
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Properties of locally compact metric spaces.
Let $(X,d)$ be a locally compact metric space. Then for each $x \in X$ $\exists$ $\epsilon_x > 0$ such that $B[x;\epsilon_x] = \{y \in X : d(x,y) \leq \epsilon_x \}$ is compact.
How do I proceed to prove it? Please help me in this regard.
Thank you very much.
| Pretty simple!
Let us choose $x \in X$ arbitrarily. Since $X$ is locally compact so $\exists$ a compact neighbourhood $C_x$ of $x$ in $X$ i.e. $\exists$ $\epsilon >0$ such that $x \in B(x;\epsilon) \subset C_x$. Consider $0 < \epsilon_x < \epsilon$ then clearly $B[x;\epsilon_x] \subset B(x;\epsilon) \subset C_x$. Now $B[x;\epsilon]$ (being a closed subset of a compact set $C_x$) is compact. Which proves your claim.
| {
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Checking Continuity of functions Consider a function $\sqrt {x-1}$+$\sqrt{1-x}$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions
| What is your definition for $l=\lim\limits_{x\to 1}f(x)$ if it exists?
If this is it:
$$\forall\varepsilon>0,\exists\delta>0,\forall x\in D_f,|x-1|<\delta\implies|f(x)-f(1)|<\varepsilon$$
then take ANY $\varepsilon>0$ and ANY $\delta>0$. $D_f=\{1\}$ so $|x-1|=0$ for any $x\in D_f$ and thus $f(x)=f(1)$.
| {
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Does the series for $\cos(x)/x$ converges? The sequence of
$$
a_x ={\cos (x)\over x}
$$
does converge to zero.
As a result, intuitively
$$
\sum_{x=1}^\infty {\cos (x)\over x}
$$
should also converge right? But I've been told that the series diverges. This shouldn't be true... right?
| It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. For example $\sum_{n=1}^{\infty} \frac1n$ is well known to diverge even though $\frac1n\to 0$. Or, as an even easier example, consider
$$ 1 + \underbrace{\frac12+ \frac12}_{2\text{ halves}} +
\underbrace{\frac13 + \frac13+ \frac13}_{3\text{ thirds}} +
\underbrace{\frac14 + \frac14 + \frac14+ \frac14}_{4\text{ fourths}}+
\underbrace{\frac15 + \frac15+ \frac15+ \frac15+\frac15}_{5\text{ fifths}}+
\cdots $$
It does look like your particular series converges (conditionally), by Dirichlet's test, though.
| {
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Prove the inequality Using Mean Value theorem Show that : $$x < \log\Bigl(\frac{1}{1-x}\Bigr) < \frac{x}{1 - x}\,;$$
If $$0 < x < 1$$
Solution:
If $$f(x) = \log\Bigl(\frac{1}{1 - x}\Bigr)$$
Then the function is continuous in [0, x]
And also differentiable in (0, x)
So We can apply Lagrange's MVT on f(x).
Fine!
So $$f'(x) =\frac{1}{1 - x}$$
As per LMVT : $$f(x) - f(0) = xf'(θx) ;$$
Where : $$0 < θ < 1$$
=> $$xf'(θx) = - \log(1 - x) $$
=> $$\frac{x} {1 - θx} = - \log(1 - x)$$
=> $$\frac{x} {1 - θx} = - \log(1 - x)......... (1)$$
So, $$0 < θx < x$$
Or $$0 > - θx > - x $$
Or $$1 > 1 - θx > 1 - x $$
Or $$1 < \frac{1} {1 - θx} < \frac{1} {1 - x} $$
Or $$x < \frac{x} {1 - θx} <\frac{x}{1 - x} $$
Or $$x < - \log(1 - x) < \frac{x} {1 - x}... \text{ from } (1)$$
Or
$$x < \log\Bigl(\frac{1}{1 - x}\Bigr) <\frac{x}{1 - x}$$
| Hint:
The inequalities should be reversed.
Start from the well-known inequality
$$\log u\le u-1\quad\text{for all }u>0\qquad(< \text{ if } u\ne1),$$
and make the relevant substitutions.
Some details:
First set $\;u=\dfrac1{1-x}$ (which is positive since $0<x<1$). You obtain instantly
$$\log\frac1{1-x} <\frac1{1-x}-1=\frac x{1-x}.$$
Now set $u=1-x$: you get
$$\log(1-x)<(1-x)-1=-x\iff \log\frac1{1-x}>-(-x)=x.$$
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What does $f(x,y)$ mean? I know from the chapter "functions" that $f(x)$ is a function of $x$ and to roughly put it, it maps $x$ values to another set called co-domain where all the $y$ values are.
But I also sometimes see $f(x,y)$ on internet. I can guess that it means some expression in $x$ and $y$.
I'm not familiar with them yet and they aren't in my high school syllabus but I'd love to know more about and I have a few questions,
*
*What type of function is this? What's it called?
*What does is represent? Can you also represent $f(x,y)$ using arrow diagram between $2$ sets?
*Is $(x,y)$ in $f(x,y)$ an ordered pair? Or $f(x,y)$ is same as writing $f(y,x)$
| Here is a short answer to your questions:
1) The function can be called a bivariate function; it is a function that depends on two variables $x$ and $y$ that may assume different domains. The function is defined on the union of those domains. An example is
$$ f(x,y) : = x^2 + y^2$$
If you fix $x$ to any value say $\bar{x}$, then $f(\bar{x}, y)$ is a function in $y$. The same holds if you fix $y$ instead, then the function becomes a function in $x$.
2) it represents a rule of mapping values of $x$ and $y$; you can still use arrow diagrams yes.
3) When you define the function, the pair $(x,y)$ should be ordered. But it is a matter if notation which argument u want to appear first.
| {
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Analogues of the elementary symmetric polynomials for the alternating group In the case of three variables, the elementary symmetric polynomials are
$$ \begin{align}
e_1(X_1,X_2,X_3)&:=X_1+X_2+X_3, \\
e_2(X_1,X_2,X_3)&:=X_1 X_2+X_1 X_3+X_2 X_3, \\
e_3(X_1,X_2,X_3)&:=X_1 X_2 X_3. \\
\end{align}$$
Knowledge of the values of $e_1,e_2,e_3$ determines the variables $X_1,X_2,X_3$ up to any permutation of $S_3$. That is, if
$$
\begin{align} e_1(X_1,X_2,X_3)&=e_1(Y_1,Y_2,Y_3), \\ e_2(X_1,X_2,X_3)&=e_2(Y_1,Y_2,Y_3) ,\\
e_3(X_1,X_2,X_3)&=e_2(Y_1,Y_2,Y_3),\\ \end{align}$$
then there exists a permutation $\sigma \in S_3$ such that $$X_i=Y_\sigma(i) ,$$
for all $1 \leq i \leq 3$.
I'm curious as to whether there exist other "less symmetric" polynomials, say $\{P_n(X_1,X_2,X_3)\}_n$ such that having
$$P_n(X_1,X_2,X_3)=P_n(Y_1,Y_2,Y_3) $$for all $n$ implies that there exists an even permutation $\sigma \in A_3 \subsetneq S_3$ for which $X_i=Y_\sigma(i)$ for all $1 \leq i \leq 3$.
I have tried keeping two of the elementary symmetric polynomials, replacing the third, but that didn't work out.
I would appreciate help with finding such polynomials $P_n$ (if they exist). Thank you!
| Here’s one way to do it: to get a list of polynomials for any $n$, start with the list of elementary symmetric polynomials on $n$ variables and add in the Vandermonde polynomial on $n$ variables.
| {
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Are two real, two variable polynomials, satisfying the Cauchy-Riemann equations, a complex polynomial? Let $u, v \in \mathbb{R}[x,y]$ satisfying $u_{x} = v_{y}$ and $u_{y} = -v_{x}$ everywhere in $\mathbb{C}$. Is the function $f(x + iy) = u(x,y) + iv(x,y)$ a polynomial in the variable $z = x + iy$?
I really don't know where to start with this. I tried to build a counterexample, but I had no success.
| Assuming that (as in the title) $u$ and $v$ are polynomial functions, then yes, it is true. The function $f$ is holomorphic and therefore analytic. And, since $u$ and $v$ are polynomial functions $f^{(n)}=0$ is $n$ is large enough. Therefore, $f$ is polynomial too.
| {
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G be a finite group and H be a non trivial proper subgroup of index 3, then G is not simple. If the index of a subgroup is the smallest prime dividing the order of group then that subgroup is normal, with this, I am done with the case order of group is odd. How to proceed when order is even.
| In general: if $G$ is a finite non-abelian simple group and $H$ is a subgroup, then $|G:H| \geq 5$.
Proof (Sketch) Assume $|G:H| \leq 4$. Let $G$ act by left multiplication on the left cosets of $H$. The kernel of this action is $core_G(H)=1$, since $G$ is simple. Hence $G$ can homomorphically be embedded in $S_{|G:H|}$. Since $S_4$ is solvable and $G$ is simple, it follows that $G=1$, a contradiction.
Observe that $A_5$ is the first example of a non-abelian simple group and that $|A_5:A_4|=5$.
| {
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When is the dirac delta function taken as 1 and when it is taken as infinity I understand that the dirac delta function basically describes a pulse of area one, if the pulse is very narrow, the height will be infinity. However, Im confused because sometimes, people consider the dirac delta function as being one at t=0 (AKA the unit impulse function) rather than infinity, so when is it considered to be one rather than infinity?
P.S. In MATLAB the dirac delta function is infinity at t=0
| As you say, the "value" of the delta "function" is infinite at $0$ and the "area under it" is $1$. I am not aware of any time we take the value to be $1$. The unit impulse refers to the total impulse delivered, which is the area under the force-time curve.
| {
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An example that $f,g$ is Riemann integrable on $[a,b]$, the range of $f$ is $[a,b]$, but $g\circ f$ is not Riemann integrable. I can easily find an example that $f,g$ is Riemann integrable but $g\circ f$ is not integrable, put $f(x)=R(x)=\begin{cases} \dfrac{1}{q}, & x=\dfrac{p}{q},\\ 0,& x\in\mathbf{Q}^{C}\end{cases}, g(x)=\begin{cases} 1, & x\neq 0,\\ 0, & x=0\end{cases}$, for example, then $g(f(x))$ is the Dirichlet function.
However, if we add the condition
the range of $f$ is $[a,b]$,
this is not so trival. I presume, we could construct a function $f$ with Intermediate Value Property but non-continuous, then search for an appropriate $g$.
| Why not define:
$$f : [-1, 1] \to [-1, 1] : x \mapsto \begin{cases} 2x + 1 & \text{if } -1 \le x < 0 \\ \frac{1}{q} & \text{if } x \in [0, 1] \text{ and } x = \frac{p}{q} \text{ with } p, q \in \mathbb{Z} \text{ and } \operatorname{gcd}(p, q) = 1 \\ 0 & \text{if } x \in [0, 1] \setminus \mathbb{Q}\end{cases},$$
and $g$ as before, defined over $[-1, 1]$.
Then, if $g \circ f$ is Riemann integrable, then $g \circ f|_{[0, 1]}$ ought to be Riemann integrable, but it too is just the Dirichlet function.
| {
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Need help understanding the derivation of Hydrostatic Equilibrium in a star First and foremost I am sorry for this; in order to make my questions clear I must first upload the lecture notes from my institution $^\zeta$ for the derivation of Hydrostatic Equilibrium:
$^\zeta$ Lecture notes courtesy of Imperial College London, Astrophysics dept, 2017-2018 edition.
Firstly, in equation $(3.1)$ why is there no negative sign for $F_g$? Last time I checked, Newton's law of gravitation told us that
$$F=-\frac{GMm}{r^2}$$
Nextly, I have a conceptual problem with equation $(3.2)$; when they use the term 'pressure' I am going to assume that they are referring to the radiation pressure that is released due to nuclear fusion reactions taking place in the sun. In that formula it must be the case that $P(r) \gt P(r+\delta r)$ if this is the case then $(3.2)$ should read $$F_p=-\left(\frac{dP}{dr}\right)\delta r\,\delta A $$
Correct me if I'm wrong but Hydrogen nuclei are fused together to generate the more stable Helium and hence lots of radiation pressure acting outwards from the centre of the star. If this is correct then $P(r+\delta r)=0$, since the outside face of that cylindrical volume element in Figure $3.1$ will not 'feel' any of this radiation pressure because this pressure only acts on the inner face. What am I not understanding here?
Moving on to equation $(3.3)$, let's suppose the sign of the gravitational force in $(3.1)$ really is neglected. Does this mean that the negative sign in $(3.3)$ is due to $P(r+\delta r) \lt P(r)$ and hence $$\frac{GM(r)\rho(r)}{r^2}=-\left(\frac{dP}{dr}\right)\,?$$ I know this is the same equation as in $(3.3)$ but I have written it with the negative sign next to the pressure differential as I would like to know where this minus sign originates from. Does anyone know?
My final concern is the sign in $(3.4)$ as last time I checked $$g=-\frac{GM}{r^2}$$ and not $$g=\frac{GM}{r^2}$$
Hence, equation $(3.4)$ should read
$$\left(\frac{dP}{dr}\right)=\rho(r)\,g\,?$$
I forgot to mention that I have done some research and this is the closest derivation I could find, which, needless to say is no less helpful than the notes I have.
| In the star, light does not travel far before being scattered. Its pressure acts in all directions, not just outward.
| {
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$f(x) \equiv 1$ mod $(x-1)$ and $f(x) \equiv 0$ mod $(x-3)$ then is there any $f(x)$? Let $S$ be the set of polynomials $f(x)$ with integer coefficients satisfying
$f(x) \equiv 1$ mod $(x-1)$
$f(x) \equiv 0$ mod $(x-3)$
Which of the following statements are true?
a) $S$ is empty .
b) $S$ is a singleton.
c)$S$ is a finite non-empty set.
d) $S$ is countably infinite.
My Try: I took $x =5$ then $f(5) \equiv 1$ mod $4$ and $f(5) \equiv 0$ mod $2$ . Which is impossible so $S$ is empty.
Am I correct? Is there any formal way to solve this?
| We have that
$$f(x)\equiv 1\bmod{(x-1)}\iff \exists a(x)|f(x)=(x-1)a(x)+1.$$
$$f(x)\equiv 0\bmod{(x-3)}\iff \exists b(x)|f(x)=(x-3)b(x).$$
So, if $f$ exists then it must be
$$(x-1)a(x)+1=(x-3)b(x).$$ We have for $x=1:$
$$1=-2b(1)\implies b(1)=-\dfrac 12,$$ which is not possible, since $b(x)$ is a polynomial with integer coefficientes and $1\in\mathbb{Z}$. (Note taht in a similar way we have $a(3)=-\frac12$.) Thus we conclude that there is no a polynomial with integer coefficients satisfying the given conditions.
As it was said in comments your proof is essentially correct. If $f$ exists then it must be
$$f(5)=4a(5)+1\equiv 1\bmod{4}$$ and $$f(5)=2b(5)\equiv 0\bmod{2}.$$ Since both congruences can't hold together you can conclude that the polynomial doesn't exist.
| {
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Strange isomorphism: $R/(A + B) \cong (R/B)/\bar{A}$. $R$ is a ring and $A, B$ are ideals of $R.$
I was playing around with some stuff and I was wondering if $R/(A + B) \cong (R/B)/\bar{A}.$ for $\bar{A}$ being the image of $A$ under $R \rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?
| This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B \leq A+B \leq R$ implies that
$$
R/(A+B) \cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $\pi : R \to R / B$
| {
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Derive the following identity $1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$.
Count the elements of the following set
$$A=\{(x,y,z): 1\leq x,y,z \leq n+1, z>\max\{x,y\}\}.
$$
From this derive the following identity:
$$1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$
In the same manner find the formula for $1^k + 2^k + \ldots + n^k$ for $k=3,4$.
It is easy to see that $|A| = 1^2 + 2^2 + \ldots + n^2$, since from the sum rule we have $$|A| = \sum_{i=0}^{n+1} |\{(x,y,i): 1\leq x,y,z \leq n+1, i> \max\{x,y\}\}| = \sum_{i=0}^{n+1} i^2$$ (as we can choose $x$ and $y$ in $i \times i$ ways for each $i$).
However I can't see why is $|A|$ equals $\dfrac{n(n+1)(2n+1)}{6}$.
| We can prove it by induction:
Step 1: check it for base case (n=1)
If we replace $n$ by one, we get: $$ 1^2=\frac{1\times 2 \times 3}{6}$$
which holds $\checkmark$
Step 2: assume it is true for $n=k$ and check if holds for $n=k+1$
If it holds for $n=k$, then we have:
$$1^2+2^2+…+k^2=\frac{k(k+1)(2k+1)}{6}$$
Now we show it also holds for $n=k+1$. By replacing $n$ by $k+1$ we should show:
$$1^2+2^2+…+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
or
$$1^2+2^2+…+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$$
(I want to skip some parts here, but its straight forward). Now the right hand side can be reshaped as:
$$\frac{(k+1)(k+2)(2k+3)}{6}=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
which proves it holds for $n=k+1$ if it holds for $n=k$ $\checkmark$
This completes the proof.
| {
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Can there be other multiplications on $\mathbb{R}^2$ making it a ring? Together with addition $+ : \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$
$$(x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1 + y_2)$$
the multiplication $\cdot : \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$
$$ r_1 e ^{i\phi_1} \cdot r_2 e ^{i\phi_2} = (r_1 r_2)e^{i(\phi_1 + \phi_2)}$$
make the plane $\mathbb{R}^2$ a ring. (Please forgive me making use of $\mathbb{C}$-notation, I guess it's clear what I mean.)
I wonder if there may be other multiplications – i.e. functions $\times : \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$ – which together with $+$ make the plane $\mathbb{R}^2$ a ring.
| Yes, there are. One is the component-wise multiplication:
$$
(x_1, y_1)\cdot (x_2, y_2) = (x_1x_2, y_1y_2)
$$
Another (if you don't require rings to have a multiplicative unit) is the trivial multiplication
$$
(x_1, y_1)\cdot (x_2, y_2) = (0,0)
$$
and I am sure there are many others.
As a side-note, you could have something like
$$
(x_1, y_1)\cdot (x_2, y_2) = (x_1x_2, x_1y_2 + x_2y_1 + 2y_1y_2)
$$
which is actually the complex numnbers again, only camouflaged: $(1, 0) = 1$ and $(0,1) = 1+i$.
| {
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Convergence of $\sum_{n=1}^{\infty}\left(\frac{n}{n^2+1}\right)^{k(n)}\,\,\,\,\,;\,\,k(n)=\frac{1}{\cos\left(\frac{1}{\ln^{a}(n)}\right)}$ Study the convergence of the series as $a > 0$
$$\sum_{n=1}^{\infty}\left(\frac{n}{n^2+1}\right)^{k(n)}\,\,\,\,\,;\,\,k(n)=\frac{1}{\cos\left(\frac{1}{\ln^{a}(n)}\right)}$$
In the text there's the integer part of $\left(\frac{n}{n^2+1}\right)$, but I think it is a typo beacuse it would be identically zero. How to handle $k(n)$ in this case?
| Hint First you can write :
$$\left(\frac{n}{n^2+1}\right)^{k(n)} = \exp \left(k(n)\log\left(\frac{n}{n^2+1}\right)\right) $$
And you can try to approximate $k(n)$ and $\log\left(\frac{n}{n^2+1}\right)$ as $n \rightarrow \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Chromatic number in a union of planar graphs I am trying to solve the following problem:
Let $G_1, G_2, \dots,G_{100}$ be $100$ planar graphs on the same vertex set $V$ , with edge sets $E_1, E_2,...,E_{100}$, respectively, and consider the graph $G = (V, \bigcup_{i=1}^{100}E_i)$ which is the union of the graphs $G_1, G_2,\dots,G_{100}$. Prove that $\chi(G) ≤ 600$.
I tried to play around with the inequality $E(G) \leq 3 |G| - 6$ that holds for any planar graph, and also with the fact that we can colour any planar graph with 4 colours but I didn't manage to advance after that. I also though of induction on the number of vertices so that I could do that for an arbitrary set of vertices but this didn't help me either.
Can anyone provide a hint/direction so I can make some progress?
| I think I have the answer.
the $G_1, ..., G_{100}$ are planar graphs on the same set of vertices, thus |V| is the same for all $G_i$.
We know $|E(G_i)|≤3|V|−$6 $\forall i $, since they are all planar thus $ |\bigcup_{i=1}^{100}E_i| \leq |\sum_{i=1}^{100} E_i| ≤100 (3|V|-6)= 300|V|-600$
For convenience let $ E = \bigcup_{i=1}^{100}E_i$
By the degree sum formula or handshake lemma,$\sum_{v} deg(v) = 2|E| \leq 2*(300|V|-600) = 600 |V|-1200 $, thus there exists (proof by contradiction or average degree) a vertex $v$ with degree strictly less than 600, i.e. degree less than or equal to 599. Thus since every subgraph of the union of 100 planar graphs has a vertex of degree less than or equal to 599, the graph is by definition 599- degenerate and therefore 600-colourable.
Is this correct?
| {
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