text stringlengths 1 2.12k | source dict |
|---|---|
# Symbolic area calculation for a parametric self-intersecting closed curve
The parametric equation of the curve is:
$$\begin{cases} x &= -9 \sin (2 t)-5 \sin (3 t) \\[6pt] y & = 9 \cos (2 t)-5 \cos (3 t) \end{cases}\quad t\in[0,2\pi]$$
which can be easily visualized as:
The implicit form:
$$\begin{array}{rl} F(x,... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9793540656452214,
"lm_q1q2_score": 0.841915183867935,
"lm_q2_score": 0.8596637451167995,
"openwebmath_perplexity": 3799.1356710434197,
"openwebmath_score": 0.3483494520187378,
"tag... |
(This simple derivation is based on the solution by Michael E2 to question 33947, as highlighted by Shutao Tang in a comment above.) Then, following Belisarius, we apply Green's Theorem.
5/2 Integrate[(y[t] D[x[t], t] - x[t] D[y[t], t]), {t, Pi + t0, Pi - t0}]
// TrigExpand // FullSimplify
(* -(252/625) Sqrt[3 (-68561... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9793540656452214,
"lm_q1q2_score": 0.841915183867935,
"lm_q2_score": 0.8596637451167995,
"openwebmath_perplexity": 3799.1356710434197,
"openwebmath_score": 0.3483494520187378,
"tag... |
So:
(*Green's theorem, "vectorial" form*)
k[{t2_, t1_}] = Integrate[Last@Cross[r@t, r'@t], {t, t1, t2}]/2;
arean = k /@ pps;
area = Total@arean;
arean // N
area // N
(* {42.9706, 42.9706, 42.9706, 42.9706, 42.9706} *)
(* 214.853 *)
So area is composed by five numerically equivalent integrals. Let's get a symbolic fo... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9793540656452214,
"lm_q1q2_score": 0.841915183867935,
"lm_q2_score": 0.8596637451167995,
"openwebmath_perplexity": 3799.1356710434197,
"openwebmath_score": 0.3483494520187378,
"tag... |
# Computing the order of $[9]_{31}$ in $(\mathbb{Z}/31\mathbb{Z})^*$
A part of Aluffi's "Algebra: Chapter 0" exercise II.4.12 suggests computing the order of $$[9]_{31}$$ in $$(\mathbb{Z}/31\mathbb{Z})^*$$. Sure, I could just multiply $$9$$ a few times until I get $$1$$ as a remainder (and thus derive that the order i... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936370890583,
"lm_q1q2_score": 0.8419036462005337,
"lm_q2_score": 0.8539127585282745,
"openwebmath_perplexity": 169.5215730965175,
"openwebmath_score": 0.9478173851966858,
"tags... |
It is not very long to check that, $$\bmod 31$$, $$\begin{gather}9^2\equiv -12, \quad 9^3\equiv -12\cdot 9=-108\equiv 16,\quad 9^5\equiv -12\cdot 16=-192\equiv -6,\\ 9^6\equiv-6\cdot 9=-54\equiv8, \quad 9^{10}\equiv 36\equiv 5,\quad 9^{15}\equiv 5\cdot -6=-30\equiv 1, \end{gather}$$ so $$9$$ has order $$15$$.
• Since ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936370890583,
"lm_q1q2_score": 0.8419036462005337,
"lm_q2_score": 0.8539127585282745,
"openwebmath_perplexity": 169.5215730965175,
"openwebmath_score": 0.9478173851966858,
"tags... |
Note that $$9\cdot7=63\equiv1$$ mod $$31$$. Now $$9^2=81\equiv-12$$ mod $$31$$, so $$9^4\equiv144\equiv20$$. Since $$20$$ is neither $$9$$ nor $$7$$ mod $$31$$, we can conclude that neither $$9^3$$ nor $$9^5$$ is $$1$$ mod $$31$$.
• @OP "...it suffices to rule out $3,5$" is a special case of the Order Test. – Bill Dub... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936370890583,
"lm_q1q2_score": 0.8419036462005337,
"lm_q2_score": 0.8539127585282745,
"openwebmath_perplexity": 169.5215730965175,
"openwebmath_score": 0.9478173851966858,
"tags... |
# Probability of choosing two equal bits from three random bits
Given three random bits, pick two (without replacement). What is the probability that the two you pick are equal?
I would like to know if the following analysis is correct and/or if there is a better way to think about it.
$$\Pr[\text{choose two equal b... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363746096915,
"lm_q1q2_score": 0.8419036438781845,
"lm_q2_score": 0.8539127529517043,
"openwebmath_perplexity": 243.29030232128787,
"openwebmath_score": 0.9246839880943298,
"ta... |
In your listing of possibilities you missed 10. That would get $\Pr[\text{2nd bit} = 0] =\frac 12$
Then when you say $\Pr[\text{choose two equal bits}] = \Pr[\text{2nd bit} = 0 \mid \text{1st bit} = 0] + \Pr[\text{2nd bit} = 1 \mid \text{1st bit} = 1]$ you need to multiply the first term by $\Pr[\text{1st bit} = 0]=\f... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363746096915,
"lm_q1q2_score": 0.8419036438781845,
"lm_q2_score": 0.8539127529517043,
"openwebmath_perplexity": 243.29030232128787,
"openwebmath_score": 0.9246839880943298,
"ta... |
Hence your law of total probability should be something like
$$P(\text{sample two same bits})=P(\text{sample 2 same bits}|C1)P(C1) +2P(\text{sample 2 same bits}|C_2)P(C_2)\\ =1 \cdot \frac{2}{2^3} + 2 \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \binom{3}{1}\cdot \frac{1}{2^3}=\frac{1}{2}$$
Expanation: $\binom{3}{1} \cd... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363746096915,
"lm_q1q2_score": 0.8419036438781845,
"lm_q2_score": 0.8539127529517043,
"openwebmath_perplexity": 243.29030232128787,
"openwebmath_score": 0.9246839880943298,
"ta... |
# Does nth-roots with non-natural indexes exists?
My high school math teacher stated that roots with non-natural indexes are meaningless, just like $\frac{\infty}{\infty}$ or $0^0$, because "mathematicians decided so and so it is unless we change axioms".
It doesn't make sense to me.
Isn't $\sqrt[\frac 1a]{x}$ just ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363750229257,
"lm_q1q2_score": 0.8419036405656215,
"lm_q2_score": 0.8539127492339909,
"openwebmath_perplexity": 499.04684890444605,
"openwebmath_score": 0.9817405939102173,
... |
The final step needs the supreme axiom;
Let $\alpha\in\mathbb{R}$ and $x\in\mathbb{R}_+$, $$x^{\alpha}:=\text{sup}\lbrace x^r \mid r\in\mathbb{Q} \,\,\text{ and }\,\, r\leq\alpha\rbrace$$
One can prove that the definition given at the beginning of this post and this last one are equivalent.
-
Thank you! What is the ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363750229257,
"lm_q1q2_score": 0.8419036405656215,
"lm_q2_score": 0.8539127492339909,
"openwebmath_perplexity": 499.04684890444605,
"openwebmath_score": 0.9817405939102173,
... |
As an easy counter-example to "no non-natural roots", $\sqrt[\frac{1}{2}]{2} = 2^\frac{1}{\frac{1}{2}} = 2^2 = 4$. However, even irrational roots make sense. If you have a sequence of rational numbers ($(q_n)\in\mathbb{Q}^\mathbb{N}$) that converges to $\pi$, then $\sqrt[\pi]{a} = \lim_{n\rightarrow\infty} \sqrt[q_n]{a... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363750229257,
"lm_q1q2_score": 0.8419036405656215,
"lm_q2_score": 0.8539127492339909,
"openwebmath_perplexity": 499.04684890444605,
"openwebmath_score": 0.9817405939102173,
... |
# Conditional Probability?
I have a homework question that states that Bowl A has four red and two white chips and that Bowl B has three red and two white chips. A chip is drawn from random from bowl A and put into bowl B. After the chip is put into bowl B what is the probability that I draw a red chip from bowl B.
I... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363741964574,
"lm_q1q2_score": 0.8419036398598897,
"lm_q2_score": 0.8539127492339909,
"openwebmath_perplexity": 620.792493008877,
"openwebmath_score": 0.6168906092643738,
"tags... |
$P(R)$ can be found by conditioning on what type of chip was initially chosen from bowl $A$:
$\ \ \$Let $A_r$ be the event that the chip chosen from bowl $A$ and put into bowl $B$ was red.
$\ \ \$Let $A_w$ be the event that the chip chosen from bowl $A$ and put into bowl $B$ was white.
Then \eqalign{ P(R)&=P(R\cap A... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363741964574,
"lm_q1q2_score": 0.8419036398598897,
"lm_q2_score": 0.8539127492339909,
"openwebmath_perplexity": 620.792493008877,
"openwebmath_score": 0.6168906092643738,
"tags... |
# Statement about divisors of polynomials and their roots
I am a 10th grade student and there is a statement in my math book
If $a$ is a root of the polynomial $f(x)$ then $(x-a)$ is a divisor of $f(x)$
Why is $(x-a)$ a divisor of $f(x)$? Can you please tell me?
• Since a is a root it is $f(a)=0$. Therefore your po... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363700641144,
"lm_q1q2_score": 0.8419036381639438,
"lm_q2_score": 0.8539127510928476,
"openwebmath_perplexity": 328.9156964158226,
"openwebmath_score": 0.9282742738723755,
"tag... |
Now, try to prove your theorem. At first, assume that $a$ is a root of $f(x)$, set $g(x) = x-a$ and apply long division (I'm sure you can do it). The procedure is below, but try to do it by yourself at first.
If we apply long division, you get $q$ and $r$ such that $f = q\cdot (x-a) + r$ and $r$ has degree $0$ (why?),... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363700641144,
"lm_q1q2_score": 0.8419036381639438,
"lm_q2_score": 0.8539127510928476,
"openwebmath_perplexity": 328.9156964158226,
"openwebmath_score": 0.9282742738723755,
"tag... |
• I like this answer because it works from first principles and doesn't require the reader to know the factor theorem or remainder theorem or how to do polynomial. (Though Adi, if you're reading, those things are well worth knowing.) – Gareth McCaughan Jul 6 '16 at 16:47
This is essentially Factor Theorem, which is a ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363700641144,
"lm_q1q2_score": 0.8419036381639438,
"lm_q2_score": 0.8539127510928476,
"openwebmath_perplexity": 328.9156964158226,
"openwebmath_score": 0.9282742738723755,
"tag... |
# Prove that the sequence $(b_n)$ converges
Prove that if $(a_n)$ converges and $|a_n - nb_n| < 2$ for all $n \in \mathbb N^+$ then $(b_n)$ converges.
Is the following proof valid?
Proof
Since $(a_n)$ converges, $(a_n)$ must be bounded, i.e. $\exists M \in \mathbb R^+$ such that for each $n \in \mathbb N^+$, we have... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363708905831,
"lm_q1q2_score": 0.8419036370369615,
"lm_q2_score": 0.8539127492339909,
"openwebmath_perplexity": 76.54659135805679,
"openwebmath_score": 0.9976540207862854,
"tag... |
# Calculate $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$
I'm need of some assistance regarding a homework question:
"calculate the following: $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$"
Alright so since th... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363750229259,
"lm_q1q2_score": 0.8419036369001927,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 342.43735606087245,
"openwebmath_score": 0.9604335427284241,
"ta... |
# Uncountable sets in countable models of ZFC
If we assume ZFC to be consistent we have, by the Löwenheim-Skolem theorem, the existence of a countable model $$\mathcal{U}_0$$ of ZFC.
In $$\mathcal{U}_0$$ there is a infinite ordinal, that is a non-empty limit ordinal. Call the smallest one $$\omega$$. We can also cons... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363717170516,
"lm_q1q2_score": 0.8419036359099791,
"lm_q2_score": 0.8539127473751341,
"openwebmath_perplexity": 362.0433229134251,
"openwebmath_score": 0.8488641977310181,
"tag... |
• You mean countably infinite
– user81883
Jun 17 '17 at 22:05
• So the set $2^\omega \subset \mathcal{U}_0$ would still be countable if we'd compare it outside the context of $\mathcal{U}_0$? Jun 17 '17 at 22:05
• @Steven Exactly. Countability isn't about the set in-and-of itself: ultimately, it boils down to "there ex... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363717170516,
"lm_q1q2_score": 0.8419036359099791,
"lm_q2_score": 0.8539127473751341,
"openwebmath_perplexity": 362.0433229134251,
"openwebmath_score": 0.8488641977310181,
"tag... |
• Sort of related, wouldn't it be possible for the countable model $M$ to have an uncountable set (in $V$) as an element (clearly $M$ is not transitive)? I can't think of a simple argument to rule this out.
– user185596
Jun 18 '17 at 1:12
• @dav11 Yes, this will of course happen. For example, if $M\preccurlyeq V$ then ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9859363717170516,
"lm_q1q2_score": 0.8419036359099791,
"lm_q2_score": 0.8539127473751341,
"openwebmath_perplexity": 362.0433229134251,
"openwebmath_score": 0.8488641977310181,
"tag... |
Calculating permutations
# 1. Introduction
This article presents different algorithms to generate all permutations of a set. Oh no, just another tutorial about this topic, there are already million of them, you think? Maybe... but still I post it for two reasons:
• At first, it's more or less for myself. After never... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
# 3. Algorithms
At first, let's define some prerequisites which share all upcoming implementations:
• The set for which all permutations are generated is an array called items.
• The presented example implementations have char values as elements of the set. This is just for simplicity and no constraint - the algorith... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
Example of the backtracking algorithm for a set with three elements 'A', 'B' and 'C'
Call Generate(3) with items = ('A' 'B' 'C'): Enter loop with c = 0: Swap(items[0], items[2]) = Swap('A', 'C') -> items = ('C' 'B' 'A') (A is exchanged to the end, then permute the first two elements) Generate(2): Swap(items[0], items[... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
## 3.2 Lexicographic permutation algorithm (Algorithm L) (taken from [1])
Additional to above prerequisites, to find all permutations with this algorithm the items in the initial set must be increasing, so:
• items[1] <= items[2] <= ... <= items[n]
The main idea of this algorithm is to 'increase' the current permuta... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
Example of retrieving the next permutation of ('1' '0' '3' '2')
The largest non-increasing suffix is ('3' '2'), thus j = 1 (items[j] is '0'). The rightmost element greater than '0' is '2', so exchange '0' and '2'. This gives ('1' '2' '3' '0'). Note that the suffix ('3' '0') is decreasing. Reversing it gives finally ('... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
Example of the lexicographic permutation algorithm for a set with three elements 'A', 'B' and 'C'
Loop 1: Visit permutation: ('A' 'B' 'C') Find index j: j = 1 ('B', longest suffix is 'C') Find index l: l = 2 ('C', that's the rightmost element larger than 'B') Swap elements at indices j = 1 ('B') and l = 2 ('C'): -> ('... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
• Perform following steps n times, ranging from i = 0 to n - 1:
• Fix the last element.
• Generate the permutation for the first n-1 elements. These permutation have all the same last element.
• If n is even, swap the elements at index i with the last element.
• If n is odd, swap the elements at index 0 with the last e... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
# 4. Miscellaneous
## 4.1 Additional implemented algorithms
Beside to the three algorithms described above, the zip archive contains also implementations of some other permutation generation algorithms.
In total, following eight algorithms are implemented:
• Backtracking: Described in chapter 3.1.
• Backtracking rev... | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
Sunshine, January 2016
### References
[1] Donald E. Knuth - The Art Of Computer Programming Pre-Fascicle 2B. A Draft Of Section 7.2.1.2: Generating all permutations
[2] Robert Sedgewick - Permutation Generation Methods
[3] Heap's algorithm @ Wikipedia
[4] Recursive algorithm ("K-Level") by Alexander Bogomolny | {
"domain": "sunshine2k.de",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985936372130286,
"lm_q1q2_score": 0.8419036344301305,
"lm_q2_score": 0.8539127455162773,
"openwebmath_perplexity": 3299.044537869931,
"openwebmath_score": 0.2811872065067291,
"tags": n... |
$\kappa$-homogeneous topological spaces
Let $\kappa>0$ be a cardinal and let $(X,\tau)$ be a topological space. We say that $X$ is $\kappa$-homogeneous if
1. $|X| \geq \kappa$, and
2. whenever $A,B\subseteq X$ are subsets with $|A|=|B|=\kappa$ and $\psi:A\to B$ is a bijective map, then there is a homeomorphism $\varp... | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852383,
"lm_q1q2_score": 0.8418594603202268,
"lm_q2_score": 0.8577681086260461,
"openwebmath_perplexity": 282.2800184927178,
"openwebmath_score": 0.8759823441505432,
"tags... |
• Shouldn't you be asking for a space that is $\alpha$-homogeneous, but not $\beta$-homogeneous? It seems that $\alpha$ and $\beta$ are mixed up in your question. Jul 28 '16 at 14:08
• Set theorists would be inclined to make the definition for sets of size less than $\kappa$, rather than equal to $\kappa$, since this w... | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852383,
"lm_q1q2_score": 0.8418594603202268,
"lm_q2_score": 0.8577681086260461,
"openwebmath_perplexity": 282.2800184927178,
"openwebmath_score": 0.8759823441505432,
"tags... |
The real line $\mathbb{R}$ is $2$-homogeneous, but not $3$-homogeneous. For $2$-homogeneity, given any two pairs of reals $a,b$ and $x,y$, then no matter how you map these bijectively, you can extend to a homeomorphism of the line by affine translation (all pairs look alike). But the line is not $3$-homogeneous, since ... | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852383,
"lm_q1q2_score": 0.8418594603202268,
"lm_q2_score": 0.8577681086260461,
"openwebmath_perplexity": 282.2800184927178,
"openwebmath_score": 0.8759823441505432,
"tags... |
The real plane $\mathbb{R}^2$ appears to be $n$-homogeneous for every finite $n$, but not $\omega$-homogeneous (thanks again to Andreas). It is $n$-homogeneous, because given any two sets of $n$ points, we can imagine the plane made of stretchable latex and simply pull the points each to their desired targets, with the... | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852383,
"lm_q1q2_score": 0.8418594603202268,
"lm_q2_score": 0.8577681086260461,
"openwebmath_perplexity": 282.2800184927178,
"openwebmath_score": 0.8759823441505432,
"tags... |
• The plane $\mathbb{R}^2$ is obviously $3$-homogeneous, but it is also $4$-homogeneous, and I think also $5$-homogeneous. For higher finite levels, I can't quite see what happens yet. Jul 28 '16 at 16:10
• I think I can "see" that the plane is $n$-homogeneous for all finite $n$, but I don't see (yet) a clear proof. Th... | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852383,
"lm_q1q2_score": 0.8418594603202268,
"lm_q2_score": 0.8577681086260461,
"openwebmath_perplexity": 282.2800184927178,
"openwebmath_score": 0.8759823441505432,
"tags... |
The sort of space you describe is usually called strongly $\kappa$-homogeneous. If you google that phrase you will find some interesting results about these kinds of spaces (mostly concerning how this property relates to other homogeneity properties).
The earliest reference I could find to strongly $n$-homogeneous spa... | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852383,
"lm_q1q2_score": 0.8418594603202268,
"lm_q2_score": 0.8577681086260461,
"openwebmath_perplexity": 282.2800184927178,
"openwebmath_score": 0.8759823441505432,
"tags... |
# Math Help - How many triangles?
1. ## How many triangles?
How many triangles in the picture?
I counted over 140 then got lost..
Is there a way to calculate the number of triangles in the picture?
2. Originally Posted by metlx
How many triangles in the picture?
I counted over 140 then got lost..
Is there a way to... | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534354878827,
"lm_q1q2_score": 0.8418594552787222,
"lm_q2_score": 0.8577681068080749,
"openwebmath_perplexity": 463.92794030507144,
"openwebmath_score": 0.8221731781959534,
"ta... |
7. Originally Posted by metlx
How many triangles in the picture?
I counted over 140 then got lost..
Is there a way to calculate the number of triangles in the picture?
Counting by hand...
Starting at the bottom left-hand corner, using that corner as a vertex in each case,
there are 6 triangles with 2 vertices right n... | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534354878827,
"lm_q1q2_score": 0.8418594552787222,
"lm_q2_score": 0.8577681068080749,
"openwebmath_perplexity": 463.92794030507144,
"openwebmath_score": 0.8221731781959534,
"ta... |
$=6\left(12+10+8+6+4\right)+1-25$
$=6(40)-6(4)=6(36)=6^3$
8. Originally Posted by metlx
How many triangles in the picture?
I counted over 140 then got lost..
Is there a way to calculate the number of triangles in the picture?
The mathematics is much clearer when we avoid having to subtract double-counted triangles.
... | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534354878827,
"lm_q1q2_score": 0.8418594552787222,
"lm_q2_score": 0.8577681068080749,
"openwebmath_perplexity": 463.92794030507144,
"openwebmath_score": 0.8221731781959534,
"ta... |
# Prove that $2x^2 - 3xy + 2y^2 \geq 0$
Prove that $2x^2 - 3xy + 2y^2 \geq 0$.
This is a question on my homework assignment, but I don't even know where to begin as it is not factorable and that is my first instinct when I see this type of problem. Can I get a tip on where to begin at least? Thanks!
-
I find the que... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852384,
"lm_q1q2_score": 0.8418594549674646,
"lm_q2_score": 0.8577681031721325,
"openwebmath_perplexity": 979.2421385817358,
"openwebmath_score": 0.8952564001083374,
"tag... |
So it is enough to show that $x^2-\frac{3}{2}xy+y^2 \ge 0.$ Complete the square. We get $$x^2-\frac{3}{2}xy+y^2=\left(x-\frac{3}{4}y\right)^2 -\frac{9}{16}y^2+y^2=\left(x-\frac{3}{4}y\right)^2 +\frac{7}{16}y^2.$$
Now we are finished. The expression on the right is obviously non-negative, since both $(x-(3/4)y)^2$ and ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852384,
"lm_q1q2_score": 0.8418594549674646,
"lm_q2_score": 0.8577681031721325,
"openwebmath_perplexity": 979.2421385817358,
"openwebmath_score": 0.8952564001083374,
"tag... |
-
Nice solution, preserving the original symmetry. – André Nicolas Sep 19 '11 at 12:37
Factor $2y^2$ in order to get $E(x,y)=2x^2-3xy+2y^2=2y^2((\frac{x}{y})^2-\frac{3}{2}\frac{x}{y}+1)$. The case $y=0$ is solved because $E(x,0)=2x^2\geq 0$. The problem is reduced to determining the sign of $F(t)=t^2-\frac{3}{2}t+1$ ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852384,
"lm_q1q2_score": 0.8418594549674646,
"lm_q2_score": 0.8577681031721325,
"openwebmath_perplexity": 979.2421385817358,
"openwebmath_score": 0.8952564001083374,
"tag... |
# Does the collection of all finite subsets of the real numbers have the Finite Intersection Property?
Does F, the collection of all finite subsets of the real numbers, have the finite intersection property?
I'm not sure how to approach this questions but my intuition tells me that F does not have the finite intersec... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.981453437115321,
"lm_q1q2_score": 0.8418594548904328,
"lm_q2_score": 0.8577681049901037,
"openwebmath_perplexity": 146.00045325969282,
"openwebmath_score": 0.8495446443557739,
"tag... |
If $G$ does not have the $FIP$ then there is a finite $G' \subset G$ for which
$$\bigcap_{g \in G'} g = \emptyset$$
But given that $G$ is composed of the complements of sets in $F$, we get: $\forall g\in G'\ \ \exists f_g\in F:\ g = \mathbb{R} - f_g$
Rewriting the intersection, you get
$$\bigcap_{g \in G'} (\mathbb... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.981453437115321,
"lm_q1q2_score": 0.8418594548904328,
"lm_q2_score": 0.8577681049901037,
"openwebmath_perplexity": 146.00045325969282,
"openwebmath_score": 0.8495446443557739,
"tag... |
# Does the Gauss-Bonnet theorem apply to non-orientable surfaces?
I found statements of the Gauss-Bonnet theorem here, here, here, here, here, here, here, and here. None of them require that the surface be orientable. However, Ted Shifrin claims in a comment to this question that the Gauss-Bonnet theorem actually only... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
Given a (compact) nonorientable surface, say $M$, with metric $g$, consider its orientable double cover $\widetilde{M}$. The metric $g$ is naturally pulled back to a metric $\widetilde{g}$ on $\widetilde{M}$; i.e. locally, $\widetilde{M}$ is isometric to $M$. Then it is easy to see that the usual Gauss-Bonnet theorem o... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
In general, for any metric on the Möbius strip, the Gauss-Bonnet theorem should hold just as well, because the LHS remains constant under smooth deformation of metric. (Any deformation is a composition of local deformations, and for local deformations, it is a corollary of the orientable version of Gauss-Bonnet theorem... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
• I haven't worked it out, but I think the line integrals on the boundary cancel, rather than doubling. I did work out the Möbius strip with explicit computation (the exercise I posed to you) and the theorem most definitely did not check. [I think the boundaryless case is, in fact, convincing with the area density in t... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
Following the OP's request, I'm posting the details of my computations for the explicit embedding of the Möbius strip $$M$$ in $$\Bbb R^3$$. The orientation of (halves) of the boundary curve turns out to be the crucial matter, as I'd suspected. Although the theoretical arguments are compelling, I remain confused about ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
Consider the parametrization $$x(u,v) = \big((2+v\sin(u/2))\cos u,(2+v\sin(u/2))\sin u,v\cos(u/2)\big), \quad 0\le u\le 2\pi, -1\le v\le 1.$$ Note that $$B = x(0,1) = x(2\pi,-1)$$ and $$C=x(0,-1)=x(0,1)$$. As you can check, this is an orthogonal parametrization, and the first fundamental form has coefficients $$E = \|x... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
We can now use Mathematica to evaluate the geodesic curvature integrals. We note that $$ds = \frac{ds}{du}du$$, so $$\kappa_g\,ds = \big({-}\frac12 E_v/\sqrt E\big)u'(s)\,ds =\big({-}\frac12 E_v/\sqrt E\big)\,du$$. On half the boundary circle, going from $$B$$ to $$C$$, $$-\frac12\int_0^{2\pi} \frac{E_v}{\sqrt E}\Big|_... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
To double-check this computation, let's remove a tiny bit of our Möbius strip, say the region corresponding to $$0\le u\le \varepsilon$$. This leaves us an oriented surface, for sure. Its boundary has two extra pieces, $$u=0$$ (oriented downward) and $$u=\varepsilon$$ (oriented upward); these have no contribution, rega... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
EDIT: OK, I believe I've figured it out, much to my chagrin. We need to think about geodesic curvature intrinsically (as one does in a fancier proof of the Gauss-Bonnet Theorem). If $$e_1$$ is the unit tangent vector along $$\partial M$$ and $$e_2$$ is the inward-pointing normal to $$\partial M$$ in $$M$$, then, by def... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
• an answer to this question with a feasible proof of the Gauss-Bonnet for the non-orientable case;
• and many more interesting things, of course :)
On a side note, the Pfaffian has nothing to do with the orientability, bit rather with the dimension: it is defined in even dimensions (thus, in dimension 2 as well). Se... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594533404034,
"lm_q2_score": 0.857768108626046,
"openwebmath_perplexity": 286.0825430524377,
"openwebmath_score": 0.9532164931297302,
"tags... |
# Factoring a hard polynomial
This might seem like a basic question but I want a systematic way to factor the following polynomial:
$$n^4+6n^3+11n^2+6n+1.$$
I know the answer but I am having a difficult time factoring this polynomial properly. (It should be $(n^2 + 3n + 1)^2$). Thank you and have a great day!
-
You... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534354878828,
"lm_q1q2_score": 0.8418594517102141,
"lm_q2_score": 0.8577681031721325,
"openwebmath_perplexity": 334.2100116332026,
"openwebmath_score": 0.9477685689926147,
"tag... |
-
+1, good analysis. – Rustyn Feb 20 '14 at 21:42
Great solution! Thank you very much @MJD :) – InsigMath Feb 20 '14 at 21:47
You are welcome. I am glad you are pleased. – MJD Feb 20 '14 at 21:47
It is fair to take $a=p=1$, but then it doesn't seem fair to assume $c=r=1$. It works this time. If we are factoring over th... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534354878828,
"lm_q1q2_score": 0.8418594517102141,
"lm_q2_score": 0.8577681031721325,
"openwebmath_perplexity": 334.2100116332026,
"openwebmath_score": 0.9477685689926147,
"tag... |
having essentially substituted $k = n + \frac{1}{n}$ and used $k^2 - 2 = n^2 + \frac{1}{n^2}$.
-
There's another approach: check that for, say $n=\pm 1,\, 0,\,\pm 2$ your polynomial is a perfect square:$$\begin{cases}25,&n=1\\1,&n=-1\\1,&n=0\\1,&n=-2\\121,&n=2.\end{cases}$$ Therefore, in five points a polynomial of f... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534354878828,
"lm_q1q2_score": 0.8418594517102141,
"lm_q2_score": 0.8577681031721325,
"openwebmath_perplexity": 334.2100116332026,
"openwebmath_score": 0.9477685689926147,
"tag... |
# Is there a (matrix) operation that can count elements in vector pairs?
For two vectors $$x \in \{0, 1, 2\}^{n}$$ and $$y \in \{0, 1, 2\}^{n},$$ I need to generate a matrix $$C\in \mathcal{R}^{3\times3}$$ where $$C_{i,j}$$ equals the number of indexes $$t$$ where $$x[t]=i$$ and $$y[t]=j$$.
I know I can always write ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534365728416,
"lm_q1q2_score": 0.8418594508566031,
"lm_q2_score": 0.8577681013541613,
"openwebmath_perplexity": 373.13791801491686,
"openwebmath_score": 0.6017631888389587,
"ta... |
$$\left(\mathbf{x}\, {\otimes}_f\ \mathbf{y}\right)_{ij} = f(x_i, y_j).$$
Let $$s$$ be any scalar (a potential component of a vector). The indicator of $$s,$$ written $$I(,\, s),$$ is the function that returns the value $$1$$ when applied to $$s$$ and otherwise returns $$0:$$
$$I(t,s) = \cases{1 & t=s \\0 & \text{oth... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534365728416,
"lm_q1q2_score": 0.8418594508566031,
"lm_q2_score": 0.8577681013541613,
"openwebmath_perplexity": 373.13791801491686,
"openwebmath_score": 0.6017631888389587,
"ta... |
The interest in this approach therefore would focus either on being able to apply algebraic rules to analyze the operation or to exploit built-in efficiencies on an array-oriented computing platform. At the end of this post I assess the latter possibility for R.
Here is an implementation in R.
f <- function(x, y, n=2... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534365728416,
"lm_q1q2_score": 0.8418594508566031,
"lm_q2_score": 0.8577681013541613,
"openwebmath_perplexity": 373.13791801491686,
"openwebmath_score": 0.6017631888389587,
"ta... |
• This crosstabulation, as I see, amounts to multiplying Gx' Gy where Gx is the dummy set (i.e. design matrix) from variable X and Gy likewise from variable Y. But in your example the values (0,1,2) are known/specified in advance and are same for both variables, therefore, to get Gx and Gy, one don't need to explore th... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534365728416,
"lm_q1q2_score": 0.8418594508566031,
"lm_q2_score": 0.8577681013541613,
"openwebmath_perplexity": 373.13791801491686,
"openwebmath_score": 0.6017631888389587,
"ta... |
# Center of mass of Right angle Trapezoid
Given bases $a$, $b$, and the height $h$.
Get the $M(x,y)$ coordinates formula from point $O(0,0)$, where $M$ is center of mass. Wiki has a formula for $M(y) = \frac{h}{3}\frac{2a+b}{a+b}$. And I'm interested in how to find that formula (also for $M(x)$). The median $c$ is di... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534376578004,
"lm_q1q2_score": 0.8418594500029919,
"lm_q2_score": 0.8577680995361899,
"openwebmath_perplexity": 209.88791149190806,
"openwebmath_score": 0.9362356662750244,
"ta... |
$$\bar y = \frac{b+2a}{3(a+b)} h.$$
You can find this in many on-line sources such as Wolfram Mathworld.
You can prove this by integrating $\int_0^h y(b + (a-b)\frac yh) dy$ and dividing by the area of the trapezoid. But here's a derivation without calculus, using the fact that the distance from a side of a triangle ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534376578004,
"lm_q1q2_score": 0.8418594500029919,
"lm_q2_score": 0.8577680995361899,
"openwebmath_perplexity": 209.88791149190806,
"openwebmath_score": 0.9362356662750244,
"ta... |
Solve for $\bar y$. This looks messy, but it can be simplified if you realize that $$\frac{\mathop{Area}(T)}{\mathop{Area}(A)} = \frac{b^2 - a^2}{a^2}.$$ If you divide both the numerator and denominator on the right-hand side of the weighted average by $\mathop{Area}(A)$, you get
$$\frac{b}{3(b-a)}h = \frac{ \frac{3b ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534376578004,
"lm_q1q2_score": 0.8418594500029919,
"lm_q2_score": 0.8577680995361899,
"openwebmath_perplexity": 209.88791149190806,
"openwebmath_score": 0.9362356662750244,
"ta... |
# Product of infinite terms $\lim_{n\to\infty}\left(\frac{(n+1)(n+2)…(3n)}{n^{2n}}\right)$
Let there be a series
$$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...(3n)}{n^{2n}}\right)$$
is equal to?
For this type of problem I am unable to approach.The numerator starts from $(n+1)$ to $3n$ where as the denominato... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534349454032,
"lm_q1q2_score": 0.8418594476763841,
"lm_q2_score": 0.8577680995361899,
"openwebmath_perplexity": 384.1152448404729,
"openwebmath_score": 0.979878842830658,
"tags... |
Hint:) With Stirling's approximation $$n!\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n$$ we have $$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...3n}{n^{2n}}.\dfrac{n!}{n!} \right)=\lim_{n\rightarrow \infty}\dfrac{(3n)!}{n!n^{2n}}=\lim_{n\rightarrow \infty}\sqrt{3}\left(\dfrac{27}{e^2}\right)^n=\color{blue}{\infty}$... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534349454032,
"lm_q1q2_score": 0.8418594476763841,
"lm_q2_score": 0.8577680995361899,
"openwebmath_perplexity": 384.1152448404729,
"openwebmath_score": 0.979878842830658,
"tags... |
# How can I find the coefficient of a term in this generating function by using the “old” method?
I'm trying to verify my answer to this problem by going back and solving the same problem using methods I've used before learning about generating functions, but I'm not quite sure how to do it in this case:
Find the coe... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594426348788,
"lm_q2_score": 0.8577680977182187,
"openwebmath_perplexity": 251.29406540813707,
"openwebmath_score": 1.0000048875808716,
"ta... |
Without the upper bounds, you have correctly calculated $\binom{18}{2}$ possibilities. However, a small number of these have $d_1 = 16$ ($1$ way) or $15 \le d_2 \le 16$ ($3$ ways). You have to subtract these to maintain the sum of $16$.
By the way, the change from $e$ variables to $d$ variables amounts to factoring \b... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594426348788,
"lm_q2_score": 0.8577680977182187,
"openwebmath_perplexity": 251.29406540813707,
"openwebmath_score": 1.0000048875808716,
"ta... |
• In full generality, you'd go for inclusion/exclusion – vonbrand Apr 3 '14 at 2:51
You want (freely simplifying by getting rid of terms that can't affect the result): \begin{align} [x^{47}](x^{10}+x^{11}&+\cdots+x^{25})(x+x^{2}+\cdots+x^{15})(x^{20}+\cdots+x^{45}) \\ &= [x^{47}] \; x^{10} (1 + x + \cdots + x^{10}) \c... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534311480467,
"lm_q1q2_score": 0.8418594426348788,
"lm_q2_score": 0.8577680977182187,
"openwebmath_perplexity": 251.29406540813707,
"openwebmath_score": 1.0000048875808716,
"ta... |
# Find the value of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots+2015}$
The question:
Find the value of $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots +2015}$$
If this is a duplicate, then sorry - but I haven't been able to find this question yet. To start, ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406003707396,
"lm_q1q2_score": 0.8418459671622136,
"lm_q2_score": 0.849971181358171,
"openwebmath_perplexity": 149.7112715287353,
"openwebmath_score": 0.9686574935913086,
"tags... |
And I'm not sure if this is right. How does one check whether their summation is correct?
• That is correct. – Robert Z Nov 26 '17 at 9:01
• You don't have to know that $\frac2{n(n+1)} = \frac2n - \frac2{n+1}$, but you do have to know the concept of telescoping sums/series. Once you suspect that this is one of those, ... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406003707396,
"lm_q1q2_score": 0.8418459671622136,
"lm_q2_score": 0.849971181358171,
"openwebmath_perplexity": 149.7112715287353,
"openwebmath_score": 0.9686574935913086,
"tags... |
D. Circular Dance
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
There are $n$ kids, numbered from $1$ to $n$, dancing in a circle around the Christmas tree. Let's enumerate them in a clockwise direction as $p_1$, $p_2$, ..., $p_n$ (all these numbers are f... | {
"domain": "codeforces.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406023459218,
"lm_q1q2_score": 0.8418459631949599,
"lm_q2_score": 0.849971175657575,
"openwebmath_perplexity": 590.8722971660109,
"openwebmath_score": 0.48349103331565857,
"tags":... |
Examples
Input
5
3 5
1 4
2 4
1 5
2 3
Output
3 2 4 1 5
Input
3
2 3
3 1
1 2
Output
3 1 2 | {
"domain": "codeforces.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406023459218,
"lm_q1q2_score": 0.8418459631949599,
"lm_q2_score": 0.849971175657575,
"openwebmath_perplexity": 590.8722971660109,
"openwebmath_score": 0.48349103331565857,
"tags":... |
# What exactly is the difference between weak and strong induction?
I am having trouble seeing the difference between weak and strong induction.
There are a few examples in which we can see the difference, such as reaching the $k^{th}$ rung of a ladder and proving every integer $>1$ can be written as a product of pri... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
Initial remarks: Good question. I think it deserves a full response (warning: this is going to be a long, but hopefully very clear, answer). First, most students do not really understand why mathematical induction is a valid proof technique. That’s part of the problem. Second, weak induction and strong induction are ac... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
Kenneth Rosen remark in Discrete Mathematics and Its Applications Study Guide: Understanding and constructing proofs by mathematical induction are extremely difficult tasks for most students. Do not be discouraged, and do not give up, because, without doubt, this proof technique is the most important one there is in ma... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
then for every $n\geq k$, the statement $S(n)$ is true.
Proof of strong induction from weak: Assume that for some $k$, the statement $S(k)$ is true and for every $m\geq k, [S(k)\land S(k+1)\land\cdot\land S(m)]\to S(m+1)$. Let $B$ be the set of all $n>m$ for which $S(n)$ is false. If $B\neq\varnothing, B\subset\mathbb... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
Inductive step: Fix some $m\geq 2$, and assume that for every $t$ satisfying $2\leq t\leq m$, the statement $S(t)$ is true. To be shown is that
$$S(m+1) : m+1 \text{ is a product of primes},$$
is true. If $m+1$ is a prime, then $S(m+1)$ is true. If $m+1$ is not prime, then there exist $r$ and $s$ with $2\leq r\leq m$ a... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
Base step ($n=1$): $S(1)$ is true because $1=\frac{1(1+1)}{2}$.
Inductive step: Fix some $k\geq 1$, and assume that for every $t$ satisfying $1\leq t\leq k$, the statement $S(t)$ is true. To be shown is that
$$S(k+1) : 1+2+3+\cdots+k+(k+1)=\frac{(k+1)(k+2)}{2}$$
follows. Beginning with the left-hand side of $S(k+1)$,
... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
Proof. For each $m\in\mathbb{Z^+}$, let $S(m)$ be the statement that for $m+1$ statements $p_i$,
$$S(m) : [p_1\to p_2]\land[p_2\to p_3]\land\cdots\land[p_m\to p_{m+1}]\Rightarrow[(p_1\land p_2\land\cdots\land p_m)\to p_{m+1}].$$
Base step: The statement $S(1)$ says
$$[p_1\to p_2]\Rightarrow [(p_1\land p_2)\to p_2],$$
w... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
Inductive step: Fix $k\geq 1$, and assume that for any statements $q_1,\ldots,q_{k+1}$, both
$$S(1) : [q_1\to q_2]\Rightarrow [(q_1\land q_2)\to q_2]$$
and
$$S(k) : [q_1\to q_2]\land[q_2\to q_3]\land\cdots\land[q_k\to q_{k+1}]\Rightarrow[(q_1\land q_2\land\cdots\land q_k)\to q_{k+1}].$$
hold. It remains to show that fo... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
By mathematical induction, for each $n\geq 1, S(n)$ holds. $\blacksquare$
Usually, there is no need to distinguish between weak and strong induction. As you point out, the difference is minor. In both weak and strong induction, you must prove the base case (usually very easy if not trivial). Then, weak induction assum... | {
"domain": "bootmath.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.990440601781584,
"lm_q1q2_score": 0.8418459627152889,
"lm_q2_score": 0.8499711756575749,
"openwebmath_perplexity": 199.96245207569135,
"openwebmath_score": 0.8994497060775757,
"tags": n... |
Math Help - solve x^3 - 12x + 16 = 0
1. solve x^3 - 12x + 16 = 0
Hi
I can't see how to factorise this. And can't think of another method. Can someone please help?
x^3 - 12x + 16 = 0
Angus
2. This polynomial is $\displaystyle P(x) = x^3 - 12x + 16$.
Notice that $\displaystyle P(2) = 2^3 - 12\cdot 2 + 16 = 8 - 24 ... | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406029102594,
"lm_q1q2_score": 0.8418459599105624,
"lm_q2_score": 0.8499711718571774,
"openwebmath_perplexity": 401.15466733676203,
"openwebmath_score": 0.8091345429420471,
"ta... |
Of course, the rational root theorem can only point out possible rational roots. It may be that a polynomial has no rational number roots but in that case, no method is going to be easy! | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406029102594,
"lm_q1q2_score": 0.8418459599105624,
"lm_q2_score": 0.8499711718571774,
"openwebmath_perplexity": 401.15466733676203,
"openwebmath_score": 0.8091345429420471,
"ta... |
Does there exist a continuous onto function from $\mathbb{R}-\mathbb{Q}$ to $\mathbb{Q}$?
Does there exist a continuous onto function from $\mathbb{R}-\mathbb{Q}$ to $\mathbb{Q}$? (where domain is all irrational numbers)
I found many answers for contradicting the fact that there doesnt exist a continuous function whi... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406000885706,
"lm_q1q2_score": 0.8418459575122084,
"lm_q2_score": 0.8499711718571774,
"openwebmath_perplexity": 260.93894354557983,
"openwebmath_score": 0.9451914429664612,
"ta... |
• This is amazing!! How could you create such a function? I mean how did you come up with this? – Landon Carter Sep 13 '15 at 18:14
• Can you give more explanation , what is enumeration of rational ? – Shubham Ugare Sep 13 '15 at 18:21
• I guess this construction can also be used to show that, if $X$ is any topological... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9904406000885706,
"lm_q1q2_score": 0.8418459575122084,
"lm_q2_score": 0.8499711718571774,
"openwebmath_perplexity": 260.93894354557983,
"openwebmath_score": 0.9451914429664612,
"ta... |
# What is the expected number of steps in a random walk from leaf to leaf in a full binary tree?
Let $h \geq 2$ be a natural number. Consider a complete binary tree of height $h$. Say we take a random walk starting from the "leftmost" leaf. What is the expected number of steps before the "rightmost" leaf is visited? I... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308782546027,
"lm_q1q2_score": 0.8418408722975804,
"lm_q2_score": 0.8519528038477824,
"openwebmath_perplexity": 111.70108168262418,
"openwebmath_score": 0.9141199588775635,
"ta... |
At level $h$, you always step to level $h-1$. At level $i$ where $1 < i < h$, you have a $\frac{2}{3}$ chance of stepping to a child at level $i+1$ and a $\frac{1}{3}$ chance of stepping to the parent at level $i-1$.
Let $x_i$ denote the expected number of steps to randomly walk from level $h-i+1$ to level 1. We are t... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308782546027,
"lm_q1q2_score": 0.8418408722975804,
"lm_q2_score": 0.8519528038477824,
"openwebmath_perplexity": 111.70108168262418,
"openwebmath_score": 0.9141199588775635,
"ta... |
• What are the transition probabilities? May 11, 2016 at 5:34
• Equal probability distributed among neighboring nodes. At a leaf node, always step to the parent. At the root node, step to the left child with $1/2$ probability and to the right child with $1/2$ probability. At an internal node, step to the left child wit... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308782546027,
"lm_q1q2_score": 0.8418408722975804,
"lm_q2_score": 0.8519528038477824,
"openwebmath_perplexity": 111.70108168262418,
"openwebmath_score": 0.9141199588775635,
"ta... |
$$b_0=1+\frac12\left(2^h-3+b_0\right)\;,$$
with solution $b_0=2^h-1$. From then on, at depth $k$, we have probability $\frac13$ to make progress and $\frac23$ to stray back into a graph with $2^h-2-\frac12\left(2^{h-k}-2\right)=2^h-2^{h-k-1}-1$ edges, of which $2$ lead back, so the return time in this case is $2^h-2^{... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308782546027,
"lm_q1q2_score": 0.8418408722975804,
"lm_q2_score": 0.8519528038477824,
"openwebmath_perplexity": 111.70108168262418,
"openwebmath_score": 0.9141199588775635,
"ta... |
The case you were interested in was $c=0$ and $a=b=h-1$.
• Fantastic! This makes sense to me. Question: how did you find $t_k=c_1+c_22^{-k}-3k$? I can see that it works, but is there a general way to solve equations of that form? (Also, minor typo: the sign is missing in "$c_2 = -c_1.$") May 11, 2016 at 7:17
• @TomTse... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308782546027,
"lm_q1q2_score": 0.8418408722975804,
"lm_q2_score": 0.8519528038477824,
"openwebmath_perplexity": 111.70108168262418,
"openwebmath_score": 0.9141199588775635,
"ta... |
+0
# How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?
0
116
2
How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?
Guest Aug 8, 2017
#1
+1
The difference between the two numbers is always a factor of the difference between t... | {
"domain": "0calc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308801770117,
"lm_q1q2_score": 0.8418408683638895,
"lm_q2_score": 0.8519527982093666,
"openwebmath_perplexity": 794.9697087592342,
"openwebmath_score": 0.7030652761459351,
"tags": null... |
$$\begin{array}{|rcll|} \hline (a-b)(a+b) &=& a^2-b^2 \quad & | \quad a = 24 \qquad b = 5 \\ (24-5)(24+5) &=& 24^2-5^2 \\ 19\cdot 29 &=& 24^2-5^2 \\ \mathbf{19\cdot 29} & \mathbf{=} & \mathbf{551} \\ \hline \end{array}$$
heureka Aug 9, 2017
### 7 Online Users
We use cookies to personalise content and ads, to provid... | {
"domain": "0calc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308801770117,
"lm_q1q2_score": 0.8418408683638895,
"lm_q2_score": 0.8519527982093666,
"openwebmath_perplexity": 794.9697087592342,
"openwebmath_score": 0.7030652761459351,
"tags": null... |
# Math Help - [SOLVED] Prove Composite
1. ## [SOLVED] Prove Composite
Problem:
Show that $8^{n}+1$ is composite for all $n\geq1$.
What I have so far:
In general,
$a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)$
but this only holds for $n$ odd.
Is there a way to factor when $n$ is even? Or would I ... | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9927672355545386,
"lm_q1q2_score": 0.8418346130455531,
"lm_q2_score": 0.847967764140929,
"openwebmath_perplexity": 440.2856123009289,
"openwebmath_score": 0.9603204131126404,
"... |
We can conclude that $8^{n} + 1$ is composite.
Then to generalize, any number that can be expressed as $a^{3} + b^{3}$ is composite because it can be broken down into the factors $(a+b)(a^{2}-ab+b^{2})$! | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9927672355545386,
"lm_q1q2_score": 0.8418346130455531,
"lm_q2_score": 0.847967764140929,
"openwebmath_perplexity": 440.2856123009289,
"openwebmath_score": 0.9603204131126404,
"... |
# Week 6
## Lecture 15: Eigensolver Algorithms: Companion Matrices, Ill-Conditioning, and Hessenberg Factorization
### Summary
Pointed out that an “LU-like” algorithm for eigenproblems, which computes the exact eigenvalues/eigenvectors (in exact arithmetic, neglecting roundoff) in a finite number of steps involving ... | {
"domain": "mit.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9927672349125561,
"lm_q1q2_score": 0.8418345991511923,
"lm_q2_score": 0.8479677506936879,
"openwebmath_perplexity": 686.4371384785054,
"openwebmath_score": 0.8574604392051697,
"tags": null,
... |
However, finding roots of the characteristic polynomial is generally a terrible way to find eigenvalues. Actually computing the characteristic polynomial coefficients and then finding the roots somehow (Newton’s method?) is a disaster, incredibly ill-conditioned: gave the example of Wilkinson’s polynomial. If we can co... | {
"domain": "mit.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9927672349125561,
"lm_q1q2_score": 0.8418345991511923,
"lm_q2_score": 0.8479677506936879,
"openwebmath_perplexity": 686.4371384785054,
"openwebmath_score": 0.8574604392051697,
"tags": null,
... |
Unshifted QR method: proved that repeatedly forming A=QR, then replacing A with RQ (as in Problem set 3) is equivalent to QR factorizing An. But since we do this while only multiplying repeatedly by unitary matrices, it is well conditioned and we get the eigenvalues accurately.
To make the QR method faster, we first r... | {
"domain": "mit.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9927672349125561,
"lm_q1q2_score": 0.8418345991511923,
"lm_q2_score": 0.8479677506936879,
"openwebmath_perplexity": 686.4371384785054,
"openwebmath_score": 0.8574604392051697,
"tags": null,
... |
### Summary
Brief discussion of shifted QR method.
There are a number of additional tricks to further improve things, the most important of which is probably the Wilkinson shift: estimating μ from a little 2×2 problem from the last two columns to avoid problems in cases e.g. where there are two equal and opposite eig... | {
"domain": "mit.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9927672349125561,
"lm_q1q2_score": 0.8418345991511923,
"lm_q2_score": 0.8479677506936879,
"openwebmath_perplexity": 686.4371384785054,
"openwebmath_score": 0.8574604392051697,
"tags": null,
... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.