Search is not available for this dataset
url
string | text
string | date
timestamp[s] | meta
dict |
|---|---|---|---|
http://math.stackexchange.com/questions/729920/probability-of-head-in-coin-flip-when-coin-is-flipped-two-times | # Probability of Head in coin flip when coin is flipped two times
Probability of getting a head in coin flip is 1/2. If the coin is flipped two times what is the probability of getting a head in either of those attempts?
I think both the coin flips are mutually exclusive events, so the probability would be getting head in attempt 1 or attempt 2 which is :
P(attempt1) + P(attempt2) = 1/2 + 1/2 = 1
100% probability sounds wrong? What am I doing wrong. If i apply the same logic then probability of getting at least 1 head in 3 attempt will be 1/2+1/2+1/2 = 3/2 = 1.5 which I know for sure is wrong. What do i have mixed up?
-
The events are not mutually exclusive: you can get a head on the first and on the second flip – Rookatu Mar 28 at 5:55
Indeed. The coin tosses are independent. That is not the same as mutually exclusive. – Graham Kemp Mar 28 at 9:58
You are confusing the terms "independent" and "mutually exclusive". These are not the same thing. In fact events cannot be both "independent" and "mutually exclusive". It's either one, the other, or neither.
"Mutually exclusive" simply means that the two events cannot happen together. If A happens then B does not and if B happens A cannot.
"Independent" simply means that the occurrence of one event is not conditional on the occurrence of the other. The probability of A happening does not depend on whether B happens or not, and vice versa.
Let $H_n$ be the indexed event of getting a head on the $n^{th}$ flip.
Given an unbiased coin, $P(H_1)=P(H_2)=\frac 1 2$
These events are independent so $P(H_1 \cap H_2) = P(H_1)\times P(H_2)$. The outcome of one coin toss does not influence then outcome of the other.
However they are not mutually exclusive, so $P(H_1 \cup H_2) = P(H_1)+P(H_2) - P(H_1 \cap H_2)$. Both coins can turn up heads.
Putting it together: $$\therefore P(H_1 \cup H_2) = \frac 12 + \frac 1 2 - \frac 12 \times \frac 12 = \frac 3 4$$
-
Let $A$ be the event of getting a tail in both tosses, then $A'$ be the event of getting a head in either tosses. So $P(A') = 1 - P(A) = 1 - 0.5*0.5 = 0.75$
-
probability of having head in a coin flip is 1/2 , when you flip 2 times then the probability you have at least 1 head is equal to :
1 - P(no head) = 1 - P(2 tail) = 1 - 1/2*1/2 = 3/4 or 75%
- | 2014-10-26T04:09:05 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/729920/probability-of-head-in-coin-flip-when-coin-is-flipped-two-times",
"openwebmath_score": 0.6118226051330566,
"openwebmath_perplexity": 349.6985313531705,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9808759638081522,
"lm_q2_score": 0.8688267660487572,
"lm_q1q2_score": 0.8522112915303947
} |
https://brilliant.org/discussions/thread/simple-numbers-are-problems-numbers/ | # One minus one plus one minus one plus...
There are three types of people in this world:
Evaluate:
$\color{blue}{S}=1-1+1-1+1-1+\ldots$
Type 1
$\color{blue}{S}=(1-1)+(1-1)+(1-1)+\ldots=0+0+0+\ldots=\boxed{0}$
Type 2
$\color{blue}{S}=1-(1-1)-(1-1)-(1-1)-\ldots=1-0-0-0-\ldots=\boxed{1}$
But the $$\displaystyle 3^{rd}$$ type of people did like this:
$1-\color{blue}{S}=1-(1-1+1-1+\ldots)=1-1+1-1+1-1+\ldots = S$
$\Leftrightarrow 1-\color{blue}S=\color{blue}S \Rightarrow 2\color{blue}S=1 \Rightarrow \color{blue}S=\boxed{\frac{1}{2}}$
Note by Adam Phúc Nguyễn
3 years, 4 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
You forgot $$4^\text{th}$$ type of people; they say that this series diverges.
- 3 years, 4 months ago
Its answer oscillates b/w 0 and 1
- 3 years, 4 months ago
Yes that's why it diverges.
- 3 years, 4 months ago
Even more here:
Evaluate : $S=1-2+3-4+5-6+ \ldots$
Type 1 : $S=1+(-2+3)+(-4+5)+ \ldots = 1+1+1+1+\ldots=\infty$
Type 2 : $S=(1-2)+(3-4)+(5-6)+ \ldots = -1-1-1-1+\ldots=-\infty$
Type 3 : They go to WolframAlpha, search this:
1 sum(n from 1 to infty,(-1)^n*n)
Which shows up that "The ratio test is inconclusive." and "The root test is inconclusive.", from which they implies that the sum is incosistent.
Type 4 : They go to Wikipedia and finds out that the sum is actually equal to $$1/4$$.
- 3 years, 4 months ago
Wow! Awesome!
I also saw this video:
$1+2+3+4+\ldots=\frac{1}{12}$
- 3 years, 4 months ago
Wow, that's cool :)))
- 3 years, 4 months ago
i guess u forgot the negative sign along with 1/12
- 3 years, 4 months ago
last one is pretty good
- 3 years, 4 months ago
haha.. g8
- 3 years, 4 months ago
Grandi series.
- 3 years, 4 months ago
Gud 1
- 3 years, 4 months ago | 2019-01-21T13:03:01 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/simple-numbers-are-problems-numbers/",
"openwebmath_score": 0.9994232654571533,
"openwebmath_perplexity": 12603.962201596063,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 0.9693242036063214,
"lm_q2_score": 0.8791467722591728,
"lm_q1q2_score": 0.8521782448731907
} |
https://www.physicsforums.com/threads/using-the-ftc-to-take-a-derivative.650253/ | # Using the FTC to take a derivative
1. Nov 7, 2012
### dumbQuestion
1. The problem statement, all variables and given/known data
Simplify the following:
d/dx[∫(t/lnt)dt] where the integral is a definite integral with bounds from x to x2
2. Relevant equations
The Fundamental THeorem of calculus says:
Suppose f is continuous on [a,b]
Then ∫abf(x)dx=F(b)-F(a) where F is any antiderivative of f
3. The attempt at a solution
So this is how I thought to solve it. Let f(t)=t/lnt
First notice that f(t) continuous on (1,c] forall c in ℝ => f(t) integrable and there exists an antiderivative F(t) such that F'(t) = f(t)
So let F(t) be such an antiderivative. ***By the fundamental theorem of calculus,
∫f(t)dt (where the integral bounds are from x to x2=F(x2)-F(x)
This means
d/dx[∫f(t)dt] (where the integral bounds are from x to x2=d/dx[F(x2)-F(x)] = d/dx[F(x2)] - d/dx(F(x)] = 2xF'(x2) - F'(x) [by the Chain Rule] = 2x(x2/ln(x2)) - x/lnx = 2x3/2lnx - x/lnx [by rules of logarithm] = (x3-x)/lnx (common denominator and added them together)
So I get that as long as the bounds on the integral are from (0,c), the answer to
d/dx[∫(t/lnt)dt] where the integral is a definite integral with bounds from x to x2 = (x3-x)/lnx
OK here is my problem, the book solves this almost exactly the same way, but they never make any mention of the bounds on the integral or continuity. I guess I am confused. Do I not need to think about where the function is continuous? It's just part of the FTC says "Suppose f is continuous on [a,b]" so it seems to me like I can only apply it when that part holds. Why don't you need to think of continuity in this case? There is no restriction in the problem statement about x. How do you solve this problem for any general x?
Also the other problem, is the line where I put ***. I think I've done something wrong here because this function is only continuous on the open region (1,c] for all c, but the FTC requires f be continuous on a closed bounded region [a,b]. So do I just pick a number great than 1, say [1.1,c]? I mean what's the formal way to solve this with regards to the continuity aspect?
2. Nov 7, 2012
### LCKurtz
Your problem implicitly assumes $t>0$ hence $x>0$ and $x^2>0$ because $t>0$ is the domain of $\ln(t)$. So there is no problem.
3. Nov 7, 2012
### dumbQuestion
But doesn't f need to be continuous on a closed bounded interval [a,b]? ln(t) is only continuous on (0,c] for all c in R.
Can you apply the FTC if f is only continuous on an interval (a,b]?
4. Nov 7, 2012
### LCKurtz
Likely not; I'd have to check. But it doesn't matter for this question. If $x$ and $x^2$ are both positive, then all your conditions are true on the closed interval between them so you can use the FTC. | 2018-02-19T14:43:44 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/using-the-ftc-to-take-a-derivative.650253/",
"openwebmath_score": 0.905588686466217,
"openwebmath_perplexity": 483.6273381149708,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9693241956308277,
"lm_q2_score": 0.8791467643431002,
"lm_q1q2_score": 0.8521782301883205
} |
https://byjus.com/question-answer/if-the-function-f-x-x-4-2x-3-ax-2-bx-on-13-satisfies/ | Question
# If the function $$f(x) = x^4 - 2x^3 + ax^2 + bx$$ on $$[1,3]$$ satisfies the conditions of Rolle's Theorem with $$c = \dfrac{3}{2}$$, then find $$a$$ and $$b$$.
Solution
## Given $$f(x)=x^4-2x^3+ax^2+bx$$ on $$[1,3]$$$$\implies f'(x)=4x^3-6x^2+2ax+b$$According to Rolle's theorem, if f(x) is continuous on $$[a,b]$$, differentiable on $$(a,b)$$ and $$f(a)=f(b)$$ then there exists some $$c \in (a,b)$$ such that $$f'(c)=0$$Given that the function $$f(x)$$ satisfies the conditions of Rolle's theorem with $$c=\dfrac 32$$$$f(1)=1^4-2(1)^3+a(1)^2+b(1)=1-2+a+b=a+b-1$$$$f(3)=3^4-2(3)^3+a(3)^2+b(3)=81-54+9a+3b=9a+3b-27$$from Rolles's theorem, $$f(1)=f(3)$$$$\implies a+b-1=9a+3b-27$$$$\implies 8a+2b=26$$$$\implies 4a+b=13$$ .......(1)from Rolles's theorem, $$f'(c)=4c^3-6c^2+2ac+b=0$$ $$\implies f'(\dfrac 32)=4(\dfrac 32)^3-6(\dfrac 32)^2+2a(\dfrac 32)+b=0$$$$\implies \dfrac{27}{2}-\dfrac{27}{2}+3a+b=0$$$$\implies 3a+b=0$$ .......(2)(1) - (2) $$\implies a=13$$substituting $$a=13$$ in (1) $$\implies 4(13)+b=13$$$$\implies b=-39$$Therefore, $$a=13,b=-39$$Mathematics
Suggest Corrections
0
Similar questions
View More
People also searched for
View More | 2022-01-28T05:10:39 | {
"domain": "byjus.com",
"url": "https://byjus.com/question-answer/if-the-function-f-x-x-4-2x-3-ax-2-bx-on-13-satisfies/",
"openwebmath_score": 0.916120171546936,
"openwebmath_perplexity": 510.4124791832383,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9912886147702209,
"lm_q2_score": 0.8596637523076225,
"lm_q1q2_score": 0.8521748901931934
} |
https://math.stackexchange.com/questions/185802/dimensions-of-symmetric-and-skew-symmetric-matrices | # Dimensions of symmetric and skew-symmetric matrices
Let $\textbf A$ denote the space of symmetric $(n\times n)$ matrices over the field $\mathbb K$, and $\textbf B$ the space of skew-symmetric $(n\times n)$ matrices over the field $\mathbb K$. Then $\dim (\textbf A)=n(n+1)/2$ and $\dim (\textbf B)=n(n-1)/2$.
Short question: is there any short explanation (maybe with combinatorics) why this statement is true?
EDIT: $\dim$ refers to linear spaces.
• Do you mean symmetric (not normal) in the title? And do you mean the $\dim$ of linear spaces of such matrices, not the $\dim$ of the matrices, right? Aug 23, 2012 at 9:59
• I did edit it - thanks for the reminder! Aug 23, 2012 at 10:02
• You did not edit it correctly; $\mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you. Aug 23, 2012 at 12:38
• And you should say that $\mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true. Aug 23, 2012 at 12:40
All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.
Here is my two cents:
\begin{eqnarray} M_{n \times n}(\mathbb{R}) & \text{has form} & \begin{pmatrix} *&*&*&*&\cdots \\ *&*&*&*& \\ *&*&*&*& \\ *&*&*&*& \\ \vdots&&&&\ddots \end{pmatrix} \hspace{.5cm} \text{with $n^2$ elements}\\ \\ \\ Skew_{n \times n}(\mathbb{R}) & \text{has form} & \begin{pmatrix} 0&*'&*'&*'&\cdots \\ *&0&*'&*'& \\ *&*&0&*'& \\ *&*&*&0& \\ \vdots&&&&\ddots \end{pmatrix} \end{eqnarray} For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $\frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $\dim(Skew_{n \times n}(\mathbb{R}) + Sym_{n \times n}(\mathbb{R})) = \dim(M_{n \times n}(\mathbb{R}))$ and $\dim(Skew_{n \times n}(\mathbb{R}))=\frac{n^2-n}{2}$ then we have that \begin{eqnarray} \frac{n^2-n}{2}+\dim(Sym_{n \times n}(\mathbb{R})))=n^2 \end{eqnarray} or \begin{eqnarray} \dim(Sym_{n \times n}(\mathbb{R})))=\frac{n^2+n}{2}. \end{eqnarray}
• $Skew_{n\times n}(\mathbb{R})=(Sym_{n\times n}(\mathbb{R}))^{\perp}?$ Aug 18, 2021 at 5:50
This is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.
In order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set $\{i,j\}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $\left( \begin{array}{@{}c@{}} n \\ 2 \end{array} \right) = \frac{n(n-1)}{2}$ such sets.
Similarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set $\{i,j\}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set $\{i,j\}$ of distinct symbols, where now a symbol is either an index ($1 , \ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $\left( \begin{array}{@{}c@{}} n+1 \\ 2 \end{array} \right) = \frac{n(n+1)}{2}$ such sets.
The dimension of symmetric matrices is $\frac{n(n+1)}2$ because they have one basis as the matrices $\{M_{ij}\}_{n \ge i \ge j \ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is $\{M_{ij}\}_{n \ge i > j \ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.
Note that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.
• But this is no explanation why the symmetric matrices have the specified $\dim$. Aug 23, 2012 at 10:03
• Do you mean $n^2$? Aug 23, 2012 at 10:04
• Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited. Aug 23, 2012 at 10:07
• @RijulSaini: Thanks, but enzotib's answer seems to be easier to understand! Aug 23, 2012 at 10:13
There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.
$$\begin{matrix} \begin{pmatrix} 0 &*' \\ *& 0 \\ \end{pmatrix} & * \end{matrix}$$
$$\begin{matrix} \begin{pmatrix} 0 &*' & *' \\ *& 0 & *' \\ * & * & 0 \\ \end{pmatrix} & \begin{matrix} *& \\ * & * \end{matrix} \end{matrix}$$
$$\begin{matrix} \begin{pmatrix} 0 &*' & *' &*'\\ *& 0 & *' &*'\\ * & * & 0 &*'\\ * & * &* & 0 \end{pmatrix} & \begin{matrix} * & & \\ *&* & \\ * & * &* \end{matrix} \end{matrix}$$
$$\begin{matrix} \begin{pmatrix} 0&*'&*'&*'&\cdots \\ *&0&*'&*'& \\ *&*&0&*'& \\ *&*&*&0& \\ \vdots&&&&\ddots \end{pmatrix} & \begin{matrix} *& \\ *&*& \\ *&*&*& \\ \vdots&&&&\ddots \end{matrix} \end{matrix}$$
Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + \dots + (n-1).$$
So by Gauss, we have $$\frac{(n-1)(n-1+1)}{2} = \frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.
For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$\frac{(n-1)n}{2} + n = \frac{n^2 + n}{2}$$ | 2022-08-16T13:55:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/185802/dimensions-of-symmetric-and-skew-symmetric-matrices",
"openwebmath_score": 0.9975501298904419,
"openwebmath_perplexity": 332.23170875784746,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9869795087371921,
"lm_q2_score": 0.863391617003942,
"lm_q1q2_score": 0.8521498339983605
} |
https://math.stackexchange.com/questions/1779221/how-to-remember-sum-to-product-and-product-to-sum-trigonometric-formulas/1779226 | # How to remember sum to product and product to sum trigonometric formulas?
They are:
\begin{align} \cos(a)\cos(b)&=\frac{1}{2}\Big(\cos(a+b)+\cos(a-b)\Big) \\[2ex] \sin(a)\sin(b)&=\frac{1}{2}\Big(\cos(a-b)-\cos(a+b)\Big) \\[2ex] \sin(a)\cos(b)&=\frac{1}{2}\Big(\sin(a+b)+\sin(a-b)\Big) \\[2ex] \cos(a)\sin(b)&=\frac{1}{2}\Big(\sin(a+b)-\sin(a-b)\Big) \\[2ex] \cos(a)+\cos(b)&=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) \\[3ex] \cos(a)-\cos(b)&=-2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \\[3ex] \sin(a)+\sin(b)&=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) \\[3ex] \sin(a)-\sin(b)&=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \end{align}
I have found nice mnemonics that helped me to remember the reduction formulae and others but I can't find a simple relationship between the formulas above. Can you help?
• What type of "relations" do you mean, or do you need ? – Jean Marie May 10 '16 at 8:33
• 'Mathematics' is the subject which comes by practice! So use these formulas in questions and all will be stored in memory ;P – user5954246 May 10 '16 at 8:55
• You are welcome to have a look at my answer to this post – Mick May 10 '16 at 16:57
• @RichardSmith: Here's a diagram that may help. – Blue May 10 '16 at 17:12
## 3 Answers
The only ones you need to know are the classical $\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)$ and $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$. The others are mere consequences of those.
For example, by changing the signs, you get $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$. By summing, you have $\cos(a+b)+\cos(a-b) = 2\cos(a)\cos(b)$, which is your first formula.
Similarly, by solving $p=a+b$ and $q=a-b$, you get the formula $\cos(p)+\cos(q) = 2\cos\left(\dfrac{p+q}{2}\right)\cos\left(\dfrac{p-q}{2}\right)$.
• And the two you suggest to remember can be derived in less than $30$ seconds using $e^{i(a+b)}=e^{ia}e^{ib}$ and Euler's formula :) – Guest Nov 20 '16 at 19:32
Right away, you can cross off the fourth formula, since it is equivalent to the third formula after switching $a$ and $b$.
Then, you can also avoid the last four formulas, since these are all covered by the first three formulas via the relationships $$a+b = u, \quad a-b = v, \quad a = \frac{u+v}{2}, \quad b = \frac{u-v}{2}.$$
So that really leaves us with only three formulas. The first two are merely consequences of the cosine angle addition identity $$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b,$$ where a suitable addition or subtraction of the two forms of this equation are done; e.g., \begin{align*} \cos (a-b) &= \cos a \cos b + \sin a \sin b \\ \cos (a+b) &= \cos a \cos b - \sin a \sin b \\ \hline \cos(a-b) + \cos(a-b) &= 2 \cos a \cos b . \end{align*} A similar concept applied to the sine angle addition identity yields the third (and fourth).
Of course, you can memorize the formulas, or re-derive them, but clearly it's faster to have more formulas memorized as long as you can remember them. What is important to stress is that a vast array of trigonometric identities are all consequences of some very basic identities, and these basic identities are the ones you really need to know.
How about just restating the LHS. For example, you could restate $\cos a\sin b$ as $$\frac{\sin a\cos b +\cos a\sin b + \cos a\sin b - \sin a\cos b}{2}$$ and just figure it out from there. For Example, Let's start off with $\cos a\sin b$ and try to derive $\frac{1}{2}[\sin(a+b) - \sin(a-b)]$ \begin{align} \cos a \sin b &= \frac{1}{2}\bigg[2\cos a\sin b\bigg] \\ &= \frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b\bigg] \\ &= \frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b + 0\bigg] \\ &= \color{red}{\frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b + (\sin a \cos b - \sin a\cos b)\bigg]} \\ &= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) +(\cos a\sin b - \sin a\cos b)\bigg] \\ &= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) -(\sin a\cos b - \cos a\sin b )\bigg] \\ &= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) -(\sin a\cos (-b) + \cos a\sin (-b) )\bigg] \\ &=\frac{1}{2}\bigg[\sin(a+b) - \sin(a-b)\bigg] \end{align}
Usually I just remember/figure out the red line. | 2019-08-26T05:55:21 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1779221/how-to-remember-sum-to-product-and-product-to-sum-trigonometric-formulas/1779226",
"openwebmath_score": 1.0000016689300537,
"openwebmath_perplexity": 745.6182134736937,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.986979510652134,
"lm_q2_score": 0.8633916134888614,
"lm_q1q2_score": 0.8521498321823928
} |
https://math.stackexchange.com/questions/3181601/generating-function-of-frachx1-x2 | # Generating function of $\frac{h(x)}{(1-x)^2}$
If $$h(x)$$ is the generating function for $$a_r$$, what is the generating function of $$\frac{h(x)}{(1-x)^2}$$
Let $$h(x)$$ be written as
$$h(x) = \sum_{r} a_r x^r$$
Consider more simply
$$\frac{h(x)}{1-x} = \frac{1}{1-x} h(x) =\sum_{r} x^r \sum_{r} a_r x^r$$
I tried to expand this and see what I could get
$$(1+x+x^2+x^3+\dots+x^r+\dots)(a_0+a_1x+a_2x^2+a_3x^3+\dots+a_rx^r+\dots)$$
there are two ways to simplify the product, either
$$a_0(1+x+x^2+\dots)+a_1(x+x^2+x^3+\dots)+ a_2(x^2+x^3+x^4+\dots)+\dots= \sum_ra_r\sum_{k\ge r}x^k$$
or $$a_0 + (a_0+a_1)x+(a_0+a_1+a_2)x^2+(a_0+a_1+a_2+a_3)x^3 = \sum_r \left(\sum_{k\le r}a_k \right)x^r$$
obviously this is only for one factor of $$\frac{1}{1-x}$$ but I assume If I can get help for this I can extend it to two factors.
I'm not sure what form the answer is expected to be in? Because I could say the generating function is
$$\frac{h(x)}{1-x} = h\left(\sum_{k \ge r}x^k\right)$$
but I'm not sure that makes any sense. I was expecting to say something like
$$\frac{h(x)}{1-x} \mapsto h(x^2)$$
(the $$x^2$$ is not intentional, just some idea of what I believe the answer could look like)
Any help for the case of $$\frac{1}{1-x}$$ would be great and then I could extend it to $$\frac{1}{(1-x)^2}$$
• I think that the question must be "What sequence is $h(x)/(1-x)^2$ the generating functin of?" – Somos Apr 9 '19 at 21:56
• It probably helps a lot to note that because $\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}$, by taking the derivative of the geometric series we get $\frac{1}{(1-x)^2}=\sum_{n=0}^\infty (n+1)x^n$. Then you can just use one convolution of this series with the generating function for $h(x)$. – Robert Shore Apr 9 '19 at 21:57
• @Somos I'm not sure, you might be right. This is from a list of review questions for my combinatorics exam, see here a screenshot from the paper link – Hushus46 Apr 9 '19 at 22:00
• @RobertShore I have not heard the term convolution in my class but I assume you mean a new generating function where the coefficients $c_n = a_0b_n+a_1b_{n-1}+\dots+a_{n-1}b_1+a_nb_0$ ? – Hushus46 Apr 9 '19 at 22:06
• Yes, that's what I'm talking about. – Robert Shore Apr 9 '19 at 22:23
First, note that
$$\frac{d}{dx} \frac{1}{1-x}=\frac{1}{(1-x)^2}.$$
Thus, taking the derivative of the power series we have:
$$\frac{1}{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k.$$
Let
$$h(x)=\sum_{n=0}^\infty a_nx^n.$$
Then taking the convolution of these two series, we have:
$$\frac{h(x)}{(1-x)^2}=\sum_{n=0}^\infty \sum_{k=0}^n (k+1)a_{(n-k)}x^n.$$
Another way.
Since $$(1-x)^2 = 1-2x+x^2$$, if $$\dfrac{h(x)}{(1-x)^2} =\sum_{n=0}^{\infty} a_nx^n$$ then
$$\begin{array}\\ h(x) &=(1-x)^2\sum_{n=0}^{\infty} a_nx^n\\ &=(1-2x+x^2)\sum_{n=0}^{\infty} a_nx^n\\ &=\sum_{n=0}^{\infty} a_nx^n-2x\sum_{n=0}^{\infty} a_nx^n+x^2\sum_{n=0}^{\infty} a_nx^n\\ &=\sum_{n=0}^{\infty} a_nx^n-\sum_{n=0}^{\infty} 2a_nx^{n+1}+\sum_{n=0}^{\infty} a_nx^{n+2}\\ &=\sum_{n=0}^{\infty} a_nx^n-\sum_{n=1}^{\infty} 2a_{n-1}x^{n}+\sum_{n=2}^{\infty} a_{n-2}x^{n}\\ &=a_0+a_1x+\sum_{n=}^{\infty} a_nx^n-2a_0x-\sum_{n=2}^{\infty} 2a_{n-1}x^{n}+\sum_{n=2}^{\infty} a_{n-2}x^{n}\\ &=a_0+(a_1-2a_0)x+\sum_{n=}^{\infty} (a_n-2a_{n-1}+a_{n-2})x^n\\ \end{array}a_n$$
If $$h(x) =\sum_{n=0}^{\infty} h_nx^n$$ then, equating coefficients, $$h_0 = a_0$$, $$h_1 = a_1-2a_0$$, so $$a_1 = h_1+2a_0 = h_1+2h_0$$, and, for $$n \ge 2$$, $$h_n =a_n-2a_{n-1}+a_{n-2}$$ so $$a_n =h_n+2a_{n-1}-a_{n-2}$$.
The advantage of this method is that getting each new $$a_n$$ takes only 3 operations (multiply, add, subtract) instead of $$n$$. | 2021-07-23T17:03:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3181601/generating-function-of-frachx1-x2",
"openwebmath_score": 0.9090185761451721,
"openwebmath_perplexity": 278.0864194518433,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9869795079712153,
"lm_q2_score": 0.8633916152464016,
"lm_q1q2_score": 0.8521498316023662
} |
https://mathhelpboards.com/threads/antiderivative-involving-trig-identities.9309/ | # Antiderivative involving Trig Identities.
#### shamieh
##### Active member
A little confused on something.
Suppose I have the integral
$$\displaystyle 2 \int 4 \sin^2x \, dx$$
So I understand that $$\displaystyle \sin^2x = \frac{1 - \cos2x}{2}$$
BUT we have a 4 in front of it, so shouldn't we pull the $$\displaystyle 4$$ out in front of the integral to get:
$$\displaystyle 8 \int \frac{1 - \cos 2x}{2} \, dx$$
then pull out the $$\displaystyle \frac{1}{2}$$ to get: $$\displaystyle 4 \int 1 - \cos 2x \, dx$$
then:
$$\displaystyle 8 [ x + \frac{1}{2} \sin2x ] + C$$?
#### MarkFL
Staff member
I would switch the constants so that we have:
$$\displaystyle 4\int 2\sin^2(x)\,dx$$
Apply the identity:
$$\displaystyle 4\int 1-\cos(2x)\,dx$$
And then we have:
$$\displaystyle 4x-2\sin(2x)+C$$
#### Prove It
##### Well-known member
MHB Math Helper
A little confused on something.
Suppose I have the integral
$$\displaystyle 2 \int 4 \sin^2x \, dx$$
So I understand that $$\displaystyle \sin^2x = \frac{1 - \cos2x}{2}$$
BUT we have a 4 in front of it, so shouldn't we pull the $$\displaystyle 4$$ out in front of the integral to get:
$$\displaystyle 8 \int \frac{1 - \cos 2x}{2} \, dx$$
then pull out the $$\displaystyle \frac{1}{2}$$ to get: $$\displaystyle 4 \int 1 - \cos 2x \, dx$$
then:
$$\displaystyle 8 [ x + \frac{1}{2} \sin2x ] + C$$?
Surely you mean \displaystyle \begin{align*} 4 \left[ x + \frac{1}{2}\sin{(2x)} \right] + C \end{align*}. Apart from that everything you have done is correct
#### shamieh
##### Active member
I would switch the constants so that we have:
$$\displaystyle 4\int 2\sin^2(x)\,dx$$
Apply the identity:
$$\displaystyle 4\int 1-\cos(2x)\,dx$$
And then we have:
$$\displaystyle 4x-2\sin(2x)+C$$
I see. Ok here is where I am getting lost Mark. My original problem was this $$\displaystyle \int \frac{x^2}{\sqrt{4 - x^2}} dx$$
So I got to the step we just mentioned above (ignore that my final answer had an 8 in front of all those terms, should be a 4 - BUT I then noticed that the A.D. of $$\displaystyle \cos(2\theta)$$ is just $$\displaystyle \frac{1}{2}\sin(2\theta)$$THUS I can now change the 4 to a 2 since$$\displaystyle \frac{4}{2} = 2$$ when I pulled it out to the constant.)
Finally, after applying all of those changes - now I am here:
$$\displaystyle 2\theta - 2\sin(2\theta) + C$$ ... So then I said ok, simple enough, just sub back in for my original equation where I had $$\displaystyle x = 2\sin\theta$$ which becomes $$\displaystyle \theta = \arcsin(\frac{x}{2})$$ (Note: From when I used my trig sub)
Thus $$\displaystyle 2\arcsin(\frac{x}{2}) - 2\sin(2\arcsin(\frac{x}{2}))$$ But somehow my teacher is getting:
$$\displaystyle 2\arcsin(\frac{x}{2}) - \frac{1}{2} x\sqrt{4 -x^2} + C$$ as the final answer..
In the second term when i have sin(arcsin ... etc What would I do there?
#### MarkFL
Staff member
We are given:
$$\displaystyle I=\int \frac{x^2}{\sqrt{4-x^2}}\,dx$$
I would use the substitution:
$$\displaystyle x=2\sin(\theta)\,\therefore\,dx=2\cos(\theta)\,d \theta$$
And so now we have:
$$\displaystyle I=\int \frac{4\sin^2(\theta)}{\sqrt{4-4\sin^2(\theta)}}\,2\cos(\theta)\,d\theta$$
Simplifying this, we obtain:
$$\displaystyle I=4\int \sin^2(\theta)\,d\theta$$
This is half of what I began with in post #2 above, so using that method, we would obtain:
$$\displaystyle I=2\theta-\sin(2\theta)+C$$
Now, using the double-angle identity for sine, we may write:
$$\displaystyle I=2\theta-2\sin(\theta)\cos(\theta)+C$$
Observing that:
$$\displaystyle \sin(\theta)=\frac{x}{2}\implies\cos(\theta)=\frac{\sqrt{4-x^2}}{2}$$
we may now write:
$$\displaystyle I=2\sin^{-1}\left(\frac{x}{2} \right)-\frac{x\sqrt{4-x^2}}{2}+C$$
I think you are still trying to pull constants out in front of your integrals that are not factors of the entire integrand.
#### shamieh
##### Active member
Yes you are exactly right. I was pulling things out in front that I couldnt
So essentially I should have
$$\displaystyle 4[\theta - \frac{1}{2}\sin(2\theta)]$$
Which becomes:
$$\displaystyle 4\theta-2\sin(2\theta)$$
Which then becomes:
$$\displaystyle 2\theta - \sin(2\theta)$$
Then plug in double angle formula,
Then 2 in the second term cancels out with the $$\displaystyle \frac{x}{2}$$
Wow I see where I made my many errors. Thanks for the clarification. | 2020-09-29T13:42:34 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/antiderivative-involving-trig-identities.9309/",
"openwebmath_score": 0.9178450107574463,
"openwebmath_perplexity": 785.4836049883065,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9869795068222499,
"lm_q2_score": 0.8633916117313211,
"lm_q1q2_score": 0.8521498271410467
} |
https://www.physicsforums.com/threads/oil-volume-required-to-rise-the-piston.725238/ | # Oil volume required to rise the piston
1. Nov 27, 2013
### Camille
1. The problem statement, all variables and given/known data
The piston shown weighs 11 lbf. In its initial position, the piston is restrained from moving to the bottom of the cylinder by means of the metal stop. Assuming there is neither friction nor leakage between piston and cylinder, what volume of oil (S = 0.85) would have to be added to the 1-in. tube to cause the piston to rise 1 in. from its initial position?
2. Relevant equations
$p = \frac{dF}{dA}$
And the basic differential equation of the fluid statics:
$\frac{dp}{dz} = - γ$
Also the density or the specific weight of the oil is:
$ρ = 1000 \frac{kg}{m^3} \cdot 0.85 = 850 \frac{kg}{m^3}$
$γ = ρ \cdot g = 8.336 \frac{kN}{m^3}$
3. The attempt at a solution
Firstly, I have calculated what volume of oil is needed just to keep the piston "a little" above the stop, ie. to hold just the weight of the piston:
$V_{oil p} = \frac{W_p}{A_p} \cdot \frac{A_o}{ρ \cdot g}$ ,
where W_p is the weight of the piston,
A_p is the c-s area of the cylinder
A_o is the c-s area of the tube.
This turns out to be:
$V_{oil p} = 22.39 in^3$.
Next, I have imagined, that when the piston is 1 in above the orignal level, we have to add even more oil to hold the weight of this 1-in thick layer of oil. This gave me in addition the volume of:
$V_{oil o} = 3.14 in^3$.
So altogether the total amount of oil needed calculated by me was:
$V_{oil} = 22.39 in^3 + 3.14 in^3 = 25.53 in^3$.
The answer should be $V_{oil} = 35.7 in^3$.
I see the hole in my reasoning, because the volume of oil in the 1-in layer has to be taken from somewhere, so it's either the old oil that was in the cylinder or the old + the new oil added. However, I don't know how to solve it.
Any help will be appreciated!
2. Nov 27, 2013
### nasu
For one thing, have you included the additional volume in the large cylinder?
And the 3.14 in^3 volume looks suspect. How did you get that?
3. Nov 27, 2013
### Camille
Do you mean the additional volume of the 1-in layer in the cylinder? Yes. It's weight is:
$W_1 = π \cdot (4 in)^2 \cdot 1 in \cdot ρ \cdot g = 6.87 N$
Then we have:
$\frac{V_{oil o}}{A_o} = \frac{W_1}{A_p}$
And solving it I got $V_{oil o} = 3.14 in^3$
4. Nov 27, 2013
### Staff: Mentor
The volume of a 1" layer in the large cylinder is π(4)2(1)/4 = 4π in3. This is the extra volume that had to be added, over and above the 22.39.
5. Nov 27, 2013
### Camille
Okay, I finally got it right. The first mistake was that 4 in and 1 in are diameters, not radii...
Here's how one can look at it:
Now, obviously the "additional" oil that must be poured is the whole blue one, so both on the left and right.
$V_{oil} = V_1 + V_2$
We know that, because the oil that is hatched in black has still the same volume as it had in the beginning. Now we calculate the equality of pressures at the line of equal pressures (marked green):
Left = Right
$p_L = p_R$
$p_L = \frac{W_p + W_{V_1}}{A_L}$
$p_R = \frac{W_{V_2}}{A_R}$
$\frac{W_p + W_{V_1}}{A_L} = \frac{W_{V_2}}{A_R}$
The areas (now correct...):
$A_L = π \cdot (\frac{4in}{2})^2 = 12.57 in^2$
$A_R = π \cdot (\frac{1in}{2})^2 = 0.79 in^2$
And the volume of the 1-in layer of oil:
$\boxed{V_1 = A_L \cdot 1 in = 12.57 in^3}$
And the weight of it:
$W_{V_1} = V_1 \cdot ρ \cdot g = 0.39 lbf$
We calculate the weight of the oil V_2:
$\frac{W_p + W_{V_1}}{A_L} \cdot A_R = W_{V_2}$
$W_{V_2} = \frac{11 lbf + 0.39 lbf}{12.57 in^2} \cdot 0.79 in^2 = 0.72 lbf$
So the volume of it is:
$\boxed{V_2 = \frac{W_{V_2}}{ρ \cdot g} = 23.17 in^3}$
Summing two volumes we get:
$V_{oil} = V_1 + V_2 = 12.57 in^3 + 23.17 in^3 = 35.74 in^3$
Last edited: Nov 27, 2013 | 2017-10-20T15:12:55 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/oil-volume-required-to-rise-the-piston.725238/",
"openwebmath_score": 0.7960912585258484,
"openwebmath_perplexity": 1280.883119579281,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290963960277,
"lm_q2_score": 0.8774767954920548,
"lm_q1q2_score": 0.8521432475146812
} |
https://www.physicsforums.com/threads/can-i-use-only-one-substitution-for-integral.426195/ | # Homework Help: Can I use only one substitution for integral?
1. Sep 3, 2010
### thereddevils
1. The problem statement, all variables and given/known data
By using the substitution t = tan x, find
$$\int \frac{dx}{\cos^2 x+4\sin^2 x}$$
2. Relevant equations
3. The attempt at a solution
Well let tan x=t
$$\frac{dt}{dx}=\sec^2 x=\tan^2 x+1=1+t^2$$
the integral then becomes
$$\int \frac{dx}{\frac{1}{\sqrt{1+t^2}}^2+4\frac{t}{\sqrt{1+t^2}}^2}$$
which simplifies to
$$\int \frac{1}{1+4t^2}$$
Then from here i make another substitution **
let t= 1/2 tan b
dt/db = 1/2 sec^2 b
$$\int \frac{1}{1+4(\frac{1}{2} \tan b)^2} \cdot \frac{1}{2}\sec^2 b db$$
= b + constant
Back substitute
= $$\frac{1}{2}\tan^{-1} (2t)$$ + constant
= $$\frac{1}{2}\tan^{-1}(2\tan x)$$ + constant
Am i correct? Especially this part ** where i made another substitution, is that valid? Or when the question specified the substitution, i have to stick that one substitution only?
2. Sep 3, 2010
### CompuChip
Re: Integration
Looks okay to me.
If you remember that
$$\int \frac{dy}{1 + y^2} \, dy = \tan^{-1}(y)$$
then $y = 2t$ immediately gives you
$$\int \frac{dt}{1+4t^2} = \frac{1}{2} \int \frac{dy}{1 + y^2} = \frac{1}{2} \tan^{-1}(y) = \frac{1}{2} \tan^{-1}(2 \tan x)$$
(Note that if this was a definite integral, you would have to be very careful with the integration boundaries in these subsitutions)
3. Sep 3, 2010
### thereddevils
Re: Integration
thanks Compuchip! So it's ok to make another substitution other than the one specified in the question? My teacher said otherwise, saying that it's not appropriate to make substitutions other than the one specified in the question. I want to get my facts right first before i get into another heated argument with her.
True, as for definite integrals, we will need to modify the limits in accordance with the substitution we make.
4. Sep 3, 2010
### CompuChip
Re: Integration
Personally, I think that if you can find a correct way to solve the question yourself (even if it is not exactly the way your teacher or anyone else has in mind) it is appropriate.
In this case, you could even argue that 2t -> t is not a full-fledged variable substitution but simply a rescaling, and the integral of 1/(1 + t²) is a standard integral which you can write down immediately whenever you encounter it (however you have proved that standard integral nicely here, which is not bad to do once in your life :) ).
5. Sep 4, 2010
### thereddevils
Re: Integration
Thanks again, Compuchip. | 2018-12-17T18:44:36 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/can-i-use-only-one-substitution-for-integral.426195/",
"openwebmath_score": 0.9260997772216797,
"openwebmath_perplexity": 1129.17751685936,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.971129092218133,
"lm_q2_score": 0.8774767986961401,
"lm_q1q2_score": 0.852143246960256
} |
http://www.billthelizard.com/2010/05/sicp-exercise-131-product-of-series.html | ## Saturday, May 1, 2010
### SICP Exercise 1.31: Product of a Series
From SICP section 1.3.1 Procedures as Arguments
Exercise 1.31 asks us to write a product procedure (analogous to sum) that computes the product of a function at points over a given range. We need to show how to define factorial in terms of the new product procedure, and use product to compute approximations to π using the following formula:
Finally, if our product procedure is recursive, we need to write an iterative version, and vice versa.
Since summing a series and multiplying a series are extremely similar ideas, it's not surprising to find that it's very simple to modify sum to produce product. Let's use the recursive version from exercise 1.29 first.
(define (sum term a next b) (if (> a b) 0 (+ (term a) (sum term (next a) next b))))
The product procedure really only needs to perform a different operation and end on a different value than sum. The last step (when a > b) should be to multiply by 1 instead of adding 0.
(define (product term a next b) (if (> a b) 1 (* (term a) (product term (next a) next b))))
We can test this out by implementing factorial in terms of product using the identity and inc procedures that we've used before.
(define (identity x) x)(define (inc x) (+ x 1))(define (factorial x) (product identity 1 inc x))> (factorial 3)6> (factorial 4)24> (factorial 5)120> (factorial 10)3628800> (factorial 20)2432902008176640000
The next test is to approximate π using something called Wallis' product or the Wallis Formula. (Lucky for us that mathematicians have already expressed this as the product of a series, saving us the work of coming up with a function for generating the terms.)
We'll multiply the product on the right hand side by the denominator on the left hand side so our procedure will just approximate π directly.
(define (wallis-pi n) (define (term x) (/ (* 4.0 (square x)) (- (* 4.0 (square x)) 1))) (* 2.0 (product term 1 inc n)))> (wallis-pi 10)3.0677038066434994> (wallis-pi 100)3.133787490628163> (wallis-pi 1000)3.1408077460304042> (wallis-pi 10000)3.1415141186819313
Since our product procedure is implemented recursively, we need to show an iterative version. We can go back to the iterative sum procedure for inspiration.
(define (sum term a next b) (define (iter a result) (if (> a b) result (iter (next a) (+ (term a) result)))) (iter a 0))
Again, there are very few changes required to transform this procedure.
(define (product-iter term a next b) (define (iter a result) (if (> a b) result (iter (next a) (* (term a) result)))) (iter a 1))
You can plug product-iter into the factorial and wallis-pi procedures above to verify that they give the same results.
The similarities between our two versions of sum and product is no accident. This section of SICP is all about higher-order procedures, so in the next exercise we're going to see how pull the similarities out of these procedures into something even more abstract.
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.
Lepsch said...
Hi Bill,
I've made another solution. There you go:
(define (pi4 n)
(define (term k)
(if (even? k)
(/ (+ 2 k) (+ 1 k))
(/ (+ 1 k) (+ 2 k))
))
(product term 1 inc n))
Anonymous said...
This is actually the correct implementation to the SICP provided formula. | 2017-07-28T02:30:22 | {
"domain": "billthelizard.com",
"url": "http://www.billthelizard.com/2010/05/sicp-exercise-131-product-of-series.html",
"openwebmath_score": 0.5671576261520386,
"openwebmath_perplexity": 3718.5134856401196,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290955604489,
"lm_q2_score": 0.8774767874818409,
"lm_q1q2_score": 0.8521432390025284
} |
https://math.stackexchange.com/questions/1905422/why-does-frac1x-4-have-two-answers | # Why does $\frac{1}{x} < 4$ have two answers?
Solving $\frac{1}{x} < 4$ gives me $x > \frac{1}{4}$. The book however states the answer is: $x < 0$ or $x > \frac{1}{4}$.
My questions are:
Why does this inequality has two answers (preferably the intuition behind it)?
When using Wolfram Alpha it gives me two answers, but when using $1 < 4x$ it only gives me one answer. Aren't the two forms equivalent?
• $x$ can be negative...^^ – Probabilitytheoryapprentice Aug 27 '16 at 17:01
• There is just one answer here, not two as you seem to think. It is not the case that $x>1/4$ is one answer and $X<0$ is another answer. Neither inequality by itself would be a correct answer. – Andreas Blass Aug 27 '16 at 17:14
• "when using 1<4x it only gives me 1 answer. Aren't the two forms equivalent?" Nope. They are not. If $1/x < 1/4$ it's possible that $x < 0$. For $1 < 4x$ it is not possible. – fleablood Aug 28 '16 at 0:14
• Draw a graph of the function $x \mapsto 1/x$ should give some intuition why there are two "zones" for the solution. – quid Aug 28 '16 at 0:26
• Just try, for instance, $x=-1$. Is it true in this case that $1/x < 4$? Is it true that $1 < 4x$? – TonyK Aug 28 '16 at 12:04
You have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then $$\frac{1}{x}<4\iff4x>1\iff x>1/4.$$ If $x>0$ and $x>1/4$, then $x>1/4$.
Now suppose $x<0$. Then $$\frac{1}{x}<4\iff4x<1\iff x<1/4.$$ If $x<0$ and $x<1/4$, then $x<0$. So the solution set is $(-\infty,0)\cup(1/4, \infty).$
• Wow!This answer got so many upvotes! ;-) – tatan Mar 24 '18 at 17:28
Here is the solution $$\frac { 1 }{ x } <4$$$$\frac { 1-4x }{ x } <0$$$$\frac { x\left( 1-4x \right) }{ { x }^{ 2 } } <0$$$$x\left( 1-4x \right) <0$$$$x\left( 4x-1 \right) >0$$
so $$x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 4 } ,+\infty \right)$$
• nice approach to avoid the case division (+1) – b00n heT Aug 27 '16 at 17:06
• As soon as I read that there were two answers, I started thinking "there has to be a quadratic hidden here." Thanks for revealing it! – Ross Presser Aug 28 '16 at 0:50
• Well this is misleading. But the same means you could get solution of $x-1>0$ to be $(-\infty,0)\cup(1,+\infty)$. Your final presentation does need some justification (checking validity of each interval). – Ruslan Aug 28 '16 at 15:05
• @Ruslan - Not quite. The solution multiplies by $x^2$ between lines 3 and 4, and we know $x^2 > 0$ for all real $x \neq 0$. So multiplying by $x^2$ is preserves the direction of the inequality for all $x \neq 0$. With $x - 1 > 0$ however, you'd multiply by $x$ which would require splitting the equality into two cases. However, I do concede that this maybe should have been emphasized in the answer. (Still, it's a very nice answer) – David E Aug 29 '16 at 18:39
• @RossPresser The hidden quadratic can also be seen by inspection of the graph. 1/x is a hyperbola, therefore is a quadratic form. The obvious 2nd degree terms are just hidden by rotation :) – QuantumMechanic Aug 30 '16 at 17:12
I am following the suggestion given by @quid in the comments because I like pictures:
The orange/red line is the $x$-axis. The yellow line is the line $y=4$. The two blue curves are the graph of $y=1/x$. The solution to the inequality is the set of $x$ values for which the blue curve is below the yellow line. As @quid predicted, this picture gives some intuition for why there are two "zones" in the solution.
Note: originally I had $y=\frac{1}{4}$ which was incorrect, so I changed the answer to reflect the fact that it should be $y=4$ and re-plotted the graph.
• I agree a picture helps, but feel this would be more helpful if you (a) indicated the solution set explicitly, (b) displayed a smaller $x$-range, (c) used the same scale on both axes and (d) made the markings legible – PJTraill Aug 31 '16 at 17:24
• @PJTraill I don't feel that the first criticism is relevant to the actual question the OP asked, which was not for the exact value of the solution, but the intuition behind its two-part structure. The second and third criticism strike me as pedantic and unhelpful. The fourth criticism is just wrong -- click on the picture, and the full-size version opens in a new tab, where all of the markings are clearly legible. – Chill2Macht Aug 31 '16 at 18:46
• It is (of course) up to you, though I thought of them as suggestions rather than criticisms; I hope you were not offended by them. I meant “readily legible as they appear in the answer”, as I prefer to see all relevant information at once. P.S. I do like the clean look of your diagram. – PJTraill Sep 1 '16 at 19:11
The accepted answer is good, but I feel like you're really asking: why is there only one piece to the question, but two pieces to the answer?
This is actually a great question. Sometimes it happens that one piece turns into two (or more), like when you try to solve $x^2 = 9$ (which has "one piece") and get the two-piece solution $x = 3, -3$. Here, to figure out why one piece becomes two, you have to think about how the equation $y = x^2$ works.
So in our case we should think about how the equation $y = 1/x$ works. And when you think about it, you realize that you didn't really start with one piece. No matter what you plug in for $x$, the value of $1/x$ can never be zero. And that means when you write $1/x < 4$, really this gives you the TWO pieces
$$0 < 1/x < 4$$
and
$$1/x < 0$$
Basically, everything smaller than 4 but with zero removed. And that's why you end up with two pieces at the end -- because that's actually how many you started with!
• Yeah that was exactly what I was confused about! – Augusto Dias Noronha Aug 28 '16 at 11:53
• Another way to look at it is that 1/x is a hyperbola and therefore is really a quadratic form, so you'd expect two "pieces" to the solution. – QuantumMechanic Aug 30 '16 at 17:10
• Though you did well to nail what the questioner was confused about, I think you underplay the fact that turning one piece into one piece is only a reasonable expectation of a continuous function while $x \mapsto 1/x$ is not continuous (or even defined) at $0$, though it is continuous on the pieces you start with and and bijective on its domain. – PJTraill Aug 31 '16 at 17:38
• Just an idea: you are asked when an expression in x is smaller than 4, and this question has two answers: 1) the value of the expression is smaller than 4 and 2) the expression is negative, hence the two answers. – Dominique Sep 1 '16 at 11:06
• $y = 5 - {(8x - 1)}^{2}$ is a continuous function but it still yields two solutions to $y < 4$. – joeytwiddle Sep 7 '16 at 6:12
Just draw a graph of $1/x$ and you'll see 'why'.
Here is an image by WolframAlpha, with appropriate parts enhanced:
Whenever you're writing $x \gt \frac{1}{4}$ you're assuming $x \gt 0$.
But for $x \lt 0$ you have $\frac{1}{x}\lt0\lt4$
English doesn't have good words for this, so exactly what's going on can be a bit tricky to describe if you don't already understand the meaning.
It is true that every $x$ satisfying $x > \frac14$ does in fact satisfy $\frac 1x < 4$.
However, the converse fails: there are some $x$ that satisfy $\frac 1x < 4$ that do not satisfy $x > \frac 14$.
So, you have found a simple description of some of the $x$ that satisfy $\frac 1x < 4$ — but (presumably) you were being asked to describe all of the $x$ that satisfy $\frac 1x < 4$.
And there are indeed more of them: every $x$ satisfying $x < 0$ satisfies $\frac 1x < 4$.
Now, what is true is the following: if $\frac 1x < 4$, then it follows that at least one of the two statements "$x > \frac 14$" and "$x < 0$" is true. Thus, this gives a complete description of the solutions to $\frac 1x < 4$.
Put differently, for every $x$, the following bullet points are either both true or both false:
• $x$ satisfies $\frac 1x < 4$
• $x$ satisfies one of the statements "$x > \frac 14$", "$x < 0$".
Regarding your solution method, you forgot that multiplying by negative numbers reverses the sign of an inequality, and multiplying by zero turns any inequality into an equality. Here, we know that $x$ can't be zero, but it still could be either positive or negative, so you don't know the effect that multiplying by $x$ will have on the inequality.
The typical way to fix this problem is to break the problem into two parts: one part where you solve the case with the assumption $x<0$, and one part where you solve the case with the assumption $x>0$, and then you put the results together.
we have $$\frac{1}{x}<4$$ is true if $$x<0$$ and if $x>0$ we get $$x>\frac{1}{4}$$
• What? $\frac{1}{x} < 4$ if $x < 0$? – nbro Aug 31 '16 at 17:19
No, $$\frac 1x<4\text{ and }1<4x$$ are not equivalent.
You could think so just multiplying by $x$. But a rule says that an inequality is preserved when you multiply by a positive number and inverted with a negative one. So the right thing is
$$\begin{cases}x>0\to1<4x,\\x<0\to1>4x.\end{cases}$$
we have $\frac{1}{x}>4$ then $\frac{1}{x}-4>0$
that is , $$\frac{1-4x}{x}>0$$ but the domain of definition is $x\neq 0$
first of all you need to find the zeros and then study the its signs $$1-4x=0$$ then $$x=1/4$$
\begin{align} & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1/4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\ & \underline{\left. 1-4x\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \left. \frac{1-4x}{x}\, \right|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{align}
as we observe from the above table that inequality is positive only when $0<x<\frac{1}{4}$
I would like to harp on the meaning of your "two answers". When a question asking for a number satisfying some conditions leads to a unique number having that property we are fine. When there are two numbers having the required property then we can say two answers.
Here we have uncountably many real numbers $x$ such that $\frac1x < 4$. So even the region $x>4$ is infinitely many answers. One possible interpretation that could justify "two" could be number of connected components of the solution space of the question. This happens many times. The set $GL(n,\mathbf{R})$ of non-singular real $n\times n$ matrices has two connected components.
$$\frac{1}{x} < 4,$$ $$\frac{1}{x} - 4 < 0,$$ $$\frac{1 - 4x}{x} < 0,$$ $1 - 4x < 0$ or $x > 0$. Because if $1 - 4x$ is negative then $x$ must be positive. So we must write the solution in that way.
Here is an important aspect which should be always considered. If someone asks me:
Problem: Find the solution of \begin{align*} \frac{1}{x}<4 \end{align*} I would not answer the problem, but instead ask: What is the domain of $x$?
Please note the problem is not fully specified if the domain of $x$, the range of validity, is not given. This is crucial to determine the set of solutions.
Some examples:
Find the solution of | 2019-05-25T02:49:24 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1905422/why-does-frac1x-4-have-two-answers",
"openwebmath_score": 0.9182042479515076,
"openwebmath_perplexity": 345.47440708210667,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290963960278,
"lm_q2_score": 0.8774767762675405,
"lm_q1q2_score": 0.8521432288451961
} |
https://math.stackexchange.com/questions/1557954/how-to-show-sum-k-n-infty-frac1k-leq-frac2n | # How to show $\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$
$$\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$$
Can someone show why this estimate holds true? I tried quite a bit but couldn't really find a way to approach this. WolframAlpha says it is true but I don't know what the gamma function is.
$$\sum_{k=n}^\infty{\frac{1}{k!}} = \frac{1}{n!} + \sum_{k = n+1}^\infty \frac{1}{k!}$$ So then I need to show that$$\sum_{k=n+1}^\infty{\frac{1}{k!}} \leq \frac{1}{(n+1)!} ~~\Big[\leq \frac{1}{n!}\Big]$$
Is it possible to do this by induction? I don't really know how to approach this now.
• Just in case: the inequality doesn't hold for $n = 0$. For $n \geqslant 1$, note that $(n+k)! \geqslant n!\cdot (n+1)^k$. – Daniel Fischer Dec 3 '15 at 10:44
• $$\sum_{k=n}^{\infty}{\frac{1}{k!}}\le\frac{2}{n!}\Longleftrightarrow$$ $$e-\frac{e\Gamma(n,1)}{\Gamma(n)}\le\frac{2}{n!}\space\text{for}\space n>-1$$ – Jan Dec 3 '15 at 10:51
It suffices to show that $$\sum_{k=n+1}^\infty \frac{1}{k!} \le \frac{1}{n!}$$
For all $k \ge n+1$, we have $$k! \ge 2^{k-n} \cdot n!$$
Therefore, we have $$\sum_{k=n+1}^\infty \frac{1}{k!} \le \sum_{k=n+1}^\infty \frac{1}{n! \cdot 2^{k-n}} = \frac{1}{n!} \sum_{i=1}^\infty \frac{1}{2^i} = \frac{1}{n!}$$
• Does the inequality $k! \geq 2^{k-n} \cdot n!$ have a name? I think I need to prove that in order to use it. – elfeck Dec 3 '15 at 11:06
• Not really. For a proof note that $\frac{k!}{n!} = k(k-1) \cdots (n+1) \ge 2 \cdot 2 \cdots 2 = 2^{n-k}$. Of course, this doesn't hold when $n=0$, so you need to exclude that. – Gyumin Roh Dec 3 '15 at 11:06
• Okay thanks. Since in my case I substitute $n$ for $n+1$ anyway $n=0$ does not cause issues. Thanks for your help! – elfeck Dec 3 '15 at 11:10
Another way:
$$\sum_{k=n}^{\infty} \frac{1}{k!} = \frac{1}{n!} \bigg(1 + \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots \bigg) < \frac{1}{n!} \bigg(1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots \bigg)\\ = \frac{1}{n!} \cdot \frac{n+1}{n} = \frac{1}{n!} \cdot \bigg(1 + \frac{1}{n} \bigg)$$ Now compare this expression to what you have on the RHS: certain terms cancel out. What do you get?
• Small typo, that $k$ should start from $n$, not $n+1$. – Gyumin Roh Dec 3 '15 at 11:18
• yeah, that's how I solved it anyway – Alex Dec 3 '15 at 11:44
$$\dfrac{1}{(k+1)!} = \dfrac{1}{(k-1)!}(\dfrac{1}{k} - \dfrac{1}{k+1}) \leq \dfrac{1}{(n-1)!}(\dfrac{1}{k}-\dfrac{1}{k+1})$$ when $k\geq n$
So
$$\sum_{k=n}^\infty\dfrac{1}{(k+1)!} \leq \dfrac{1}{(n-1)!}\sum_{k=n}^\infty (\dfrac{1}{k}-\dfrac{1}{k+1}) = \dfrac{1}{(n-1)!} *\frac{1}{n} = \dfrac{1}{n!}$$ | 2019-05-21T19:42:27 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1557954/how-to-show-sum-k-n-infty-frac1k-leq-frac2n",
"openwebmath_score": 0.8483309745788574,
"openwebmath_perplexity": 330.24868916678713,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232924970205,
"lm_q2_score": 0.8670357718273068,
"lm_q1q2_score": 0.8521429519800091
} |
http://mathhelpforum.com/calculus/50663-definite-integral-ln-1-t-dt.html | # Math Help - Definite integral ln(1+t)dt
1. ## Definite integral ln(1+t)dt
I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..
$\int_{0}^{5} ln(1+t)dt$
My first way of thinking led me to a dead end. I used integration by parts with $U=ln(1+t)$ and $U=1/(1+t)$ and $V=1dt$ and $V=t$
This led me to the following
$= [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt$
and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.
I think my line of thinking is wrong.. is my assumption that $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?
2. Originally Posted by mkelly09
I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..
$\int_{0}^{5} ln(1+t)dt$
My first way of thinking led me to a dead end. I used integration by parts with $U=ln(1+t)$ and $U=1/(1+t)$ and $V=1dt$ and $V=t$
This led me to the following
$= [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt$
and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.
I think my line of thinking is wrong.. is my assumption that $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?
You're correct...but you need to think a little "outside of the box" for this integral.
$\int\frac{t}{1+t}\,dt$
Let $u=t+1\implies t=u-1$
Thus, $\,du=\,dt$
Therefore, the integral becomes $\int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du$
Can you take it from here?
--Chris
3. Originally Posted by Chris L T521
You're correct...but you need to think a little "outside of the box" for this integral.
$\int\frac{t}{1+t}\,dt$
Let $u=t+1\implies t=u-1$
Thus, $\,du=\,dt$
Therefore, the integral becomes $\int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du$
Can you take it from here?
--Chris
Here's another way I just thought of: add "zero" to the numerator
$\int\frac{t}{t+1}\,dt=\int\frac{t+1-1}{t+1}\,dt=\int\left[\frac{t+1}{t+1}-\frac{1}{t+1}\right]\,dt=\int\left(1-\frac{1}{t+1}\right)\,dt$
--Chris
4. using integration by parts on that guy directly is overkill. (that is, if you know the integral of ln(x) by heart--which you should).
$\int_0^5 \ln (1 + t)~dt$
Let $u = t + 1$, this substitution yields
$\int_1^6 \ln u~du$
which we know is $u \ln u - u \bigg|_1^6$ ...
see post #7 here for how to integrate ln(x) using integration by parts. note that you could also figure out the integral by contemplating the derivative of xln(x)
5. Thanks for your replies, they certainly solved my problem. In fact, now i know of a few different ways of doing it when before i couldn't figure out one.
I think i will be committing to memory the antiderivative of ln(x) | 2014-08-23T08:16:33 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/50663-definite-integral-ln-1-t-dt.html",
"openwebmath_score": 0.8994470238685608,
"openwebmath_perplexity": 273.0609817791906,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9828232889752296,
"lm_q2_score": 0.8670357666736772,
"lm_q1q2_score": 0.8521429438613832
} |
https://math.stackexchange.com/questions/2008784/what-is-the-reasoning-behind-why-the-ratio-test-works | # What is the reasoning behind why the ratio test works?
The ratio test says that if we have
$$\sum_{n=1}^{\infty}a_n$$
such that $\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L$, then if:
1) $L < 1$, then $\sum_{n=1}^{\infty}a_n$ is absolutely convergent,
2) $L > 1$, then $\sum_{n=1}^{\infty}a_n$ is divergent, and
3) $L = 1$, then the ratio test gives no information.
I want to understand the mathematics behind the ratio test. I want to know the reasons behind why and how this works, rather than just memorising a formula.
I have attempted to reason about this theorem myself. It can be seen that we're taking the ratio between the second term in the series, $a_{n+1}$, and the first term in the series, $a_n$. This is exactly how one finds the common ratio in a geometric sequence ($r = \dfrac{a_{n+1}}{a_n} )$.
We then take the limit of this ratio, which I assume is to find the relative rate of change between $a_{n+1}$ and $a_n$. In which case, It is more effective to take the absolute value of the limit, $\lim_{n \to \infty} \begin{vmatrix}{ \dfrac{a_{n+1}}{a_n} }\end{vmatrix} = L$. This seems analagous to how the limit comparison test theorem works. Therefore, I presume that if $L < 1$, this implies that $a_n$ is has a greater rate of change than $a_{n+1}$, which implies that each successive term in the series is getting smaller, and as we go to infinity, each successive term is converging towards $0$. As such, the series should converge to some value. Analogously, I presume that if $L > 1$, this implies that $a_{n+1}$ has a greater rate of change than $a_n$, which implies that each successive term in the series is getting larger, and as we go to infinity, the terms diverge towards infinity.
Is this reasoning correct? If anything is incorrect, please clarify why and what the correct reasoning is.
Thank you.
• "each succesive term is converging towards zero. As such, the series should converge." This is a famously fallacious statement. The harmonic series tends termwise to zero, but the series diverges. – Matthew Leingang Nov 11 '16 at 1:13
• @MatthewLeingang I forgot about that. You're absolutely correct. In which case, what is the correct way to reason about that section? – The Pointer Nov 11 '16 at 1:14
• @GCab Which is not less than $1$ . . . – Noah Schweber Nov 11 '16 at 1:19
• @GCab That's correct, the ratio test doesn't apply to the harmonic series. But the statement I quoted is still invalid. – Matthew Leingang Nov 11 '16 at 1:20
• next time read wikipedia's entry en.wikipedia.org/wiki/Ratio_test#Proof – reuns Nov 11 '16 at 1:35
You have some ideas correct, but remember that $a_n \rightarrow 0$ does not imply that $\sum a_n$ converges. So we need something more than that. However, for the case when $L>1$, your reasoning is correct.
For the case $L<1$, we know that there exists some $r < 1$ and $N \in \mathbb{N}$ such that $$\frac{a_{n+1}}{a_n} \leq r < 1,$$ whenever $n \geq N$.
Therefore,
$$a_{N+1} \leq ra_N,$$ $$a_{N+2} \leq ra_{N+1} \leq r^2a_N$$ $$a_{N+3} \leq ra_{N+2} \leq r^3a_N,$$ and in general $$a_{N+n} \leq r^na_N.$$
Therefore, at least eventually, we can compare the series from above with a convergent geometric series $\sum a_N r^n$, which implies the convergence of the original series $\sum a_n$.
• I'm not sure if your notation has the same meaning as mine or if you've defined it differently. For instance, how can $\dfrac{ a_{n+1} }{a_n} < r$? I've defined $r = \dfrac{ a_{n+1} }{a_n}$. – The Pointer Nov 11 '16 at 1:54
• My notation is different. If you set $r$ equal to the ratio, then it is changing with $n$. But when $L < 1$, such a fixed $r$ exists as I employed it in my answer to your question. – user333870 Nov 11 '16 at 2:00
• Can you please elaborate on what $r$ and $N$ are? – The Pointer Nov 11 '16 at 2:03
• Since $lim \frac{a_{n+1}}{a_n} = L < 1$, take $\epsilon = \frac{1-L}{2}$, then by the definition of the limit, there exists $N\in \mathbb{N}$ such that $n \geq N \Longrightarrow -\epsilon < \frac{a_{n+1}}{a_n} - L < \epsilon$, so you can take for example $r = \epsilon + L = \frac{1+L}{2}$ – user333870 Nov 11 '16 at 2:18
• I'm sorry but this still doesn't tell me what $N$ and $r$ are supposed to be. I cannot see the relevance between your answer and my original question. – The Pointer Nov 11 '16 at 5:28 | 2020-03-29T10:27:04 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2008784/what-is-the-reasoning-behind-why-the-ratio-test-works",
"openwebmath_score": 0.9150016903877258,
"openwebmath_perplexity": 166.52440182758374,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232869627774,
"lm_q2_score": 0.8670357683915537,
"lm_q1q2_score": 0.8521429438048842
} |
http://math.stackexchange.com/questions/910361/rationalize-the-denominator-and-simplify | Rationalize the denominator and simplify
So the problem I have says rationalize the denominator and simplify. $$\frac{ \sqrt{15}}{\sqrt{10}-3}$$
My answer I got was $\frac{5 \sqrt6}{7}$.
Am I doing this wrong or is this the wrong answer I was told it was incorrect?
-
So it is pretty easy to check that your answer is wrong, just use a calculator. You will get $23.87$ for the question and $1.75$ for your answer. – David Aug 27 '14 at 0:13
It seems that you tried multiplying by $\frac{\sqrt{10}}{\sqrt{10}}$. Instead, you should try multiplying by the conjugate and take advantage of difference of squares: $$\frac{\sqrt{15}}{\sqrt{10} - 3} = \frac{\sqrt{15}}{\sqrt{10} - 3} \cdot \frac{\sqrt{10} + 3}{\sqrt{10} + 3} = \frac{\sqrt{150} + 3\sqrt{15}}{(\sqrt{10})^2 - 3^2} = 5\sqrt{6} + 3\sqrt{15}$$
I think wht happened was that you correctly multiplied the denominator by $\sqrt{10}+3$, but incorrectly multiplied the numerator by $\sqrt{10}$. The numerator should also have been multiplied by $\sqrt{10}+3$
$$\frac{\sqrt{15}}{\sqrt{10}-3}=\frac{\sqrt{15} \cdot (\sqrt{10}+3)}{(\sqrt{10}-3) \cdot (\sqrt{10}+3)}=\frac{\sqrt{15} \cdot (\sqrt{10}+3)}{10-9}=\sqrt{15} \cdot (\sqrt{10}+3)$$ | 2016-06-25T03:48:02 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/910361/rationalize-the-denominator-and-simplify",
"openwebmath_score": 0.9746134877204895,
"openwebmath_perplexity": 149.09013451090843,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232909876816,
"lm_q2_score": 0.8670357580842941,
"lm_q1q2_score": 0.8521429371644054
} |
https://reference.wolframcloud.com/language/ref/CharacteristicFunction.html | # CharacteristicFunction
CharacteristicFunction[dist,t]
gives the characteristic function for the distribution dist as a function of the variable t.
CharacteristicFunction[dist,{t1,t2,}]
gives the characteristic function for the multivariate distribution dist as a function of the variables t1, t2, .
# Examples
open allclose all
## Basic Examples(4)
Characteristic function (cf) for the normal distribution:
Characteristic function for the binomial distribution:
Characteristic function for the bivariate normal distribution:
Characteristic function for the multinomial distribution:
## Scope(8)
Characteristic function for a specific continuous distribution:
Characteristic function for a specific discrete distribution:
Characteristic function at a particular value:
Characteristic function evaluated numerically:
Obtain a result at any precision:
Compute the characteristic function for a formula distribution:
Find the characteristic function for a parameter mixture distribution:
Characteristic function for the slice distribution of a random process:
## Applications(7)
Compute the raw moments for a Poisson distribution:
First 5 raw moments using derivatives of the characteristic function at the origin:
Use Moment directly:
Compute mixed raw moments for a multivariate distribution:
Use Moment to obtain raw moments directly:
Find raw moments of a Student distribution from its characteristic function:
Compute to extract moments by taking limits from the right:
Evaluate the limits from the left:
Only the first four moments are defined, as confirmed by using Moment directly:
Use inverse Fourier transform to compute the PDF corresponding to a characteristic function:
Illustrate the central limit theorem on the example of symmetric LaplaceDistribution:
Find the characteristic function of the rescaled random variate:
Compute the large limit of the cf of the sum of such i.i.d. random variates:
Compare with the characteristic function of a standard normal variate:
Use smooth characteristic function to construct the upper bound for the distribution density of ErlangDistribution:
Plot the upper bounds and the original density:
Verify that the sum where are independent identically distributed variates tends in distribution to for large :
Use a combinatorial equality for product :
Evaluate the sum:
Take the limit and compare it to the characteristic function of the UniformDistribution:
## Properties & Relations(5)
CharacteristicFunction is the Expectation of for real :
The characteristic function is related to all other generating functions when they exist:
The cf of a continuous distribution is equivalent to FourierTransform of its PDF:
The cf of a discrete distribution is equivalent to FourierSequenceTransform of its PDF:
The PDF is the inverse Fourier transform of the cf for continuous distributions:
The PDF is the inverse Fourier sequence transform of the cf for discrete distributions:
## Possible Issues(1)
Symbolic closed forms do not exist for some distributions:
## Neat Examples(1)
Visualize real and imaginary parts of CharacteristicFunction for random instances of BinomialDistribution:
Wolfram Research (2007), CharacteristicFunction, Wolfram Language function, https://reference.wolfram.com/language/ref/CharacteristicFunction.html (updated 2010).
#### Text
Wolfram Research (2007), CharacteristicFunction, Wolfram Language function, https://reference.wolfram.com/language/ref/CharacteristicFunction.html (updated 2010).
#### CMS
Wolfram Language. 2007. "CharacteristicFunction." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2010. https://reference.wolfram.com/language/ref/CharacteristicFunction.html.
#### APA
Wolfram Language. (2007). CharacteristicFunction. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/CharacteristicFunction.html
#### BibTeX
@misc{reference.wolfram_2022_characteristicfunction, author="Wolfram Research", title="{CharacteristicFunction}", year="2010", howpublished="\url{https://reference.wolfram.com/language/ref/CharacteristicFunction.html}", note=[Accessed: 11-August-2022 ]}
#### BibLaTeX
@online{reference.wolfram_2022_characteristicfunction, organization={Wolfram Research}, title={CharacteristicFunction}, year={2010}, url={https://reference.wolfram.com/language/ref/CharacteristicFunction.html}, note=[Accessed: 11-August-2022 ]} | 2022-08-11T23:20:07 | {
"domain": "wolframcloud.com",
"url": "https://reference.wolframcloud.com/language/ref/CharacteristicFunction.html",
"openwebmath_score": 0.8522660732269287,
"openwebmath_perplexity": 2928.7487258549563,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232874658904,
"lm_q2_score": 0.8670357580842941,
"lm_q1q2_score": 0.8521429341108864
} |
https://tex.stackexchange.com/questions/377488/equation-custom-horizontal-alignment-numbering-each-row | # equation custom horizontal alignment & numbering each row
As I know, "align" has the fixed horizontal alignment right-left-right-left-..., and we cannot change this alignment. So if we want to align equations with custom horizontal alignment, e.g., right-center-center-left, we may use "equation" or "gather", "array", "arraycolsep" (for spacing "=" like "align"), and "displaystyle" (for proper handling of "frac" or "lim").
For example,
\newcommand{\argmax}{\operatornamewithlimits{arg\,max}}
...
\begin{gather}
\arraycolsep=1.4pt\def\arraystretch{2.2}
\begin{array}{rccl}
p_{\mathrm{MLE}}(x) & = & \displaystyle \max_{m} &P(X = x | \theta = m) \\
m_{\mathrm{MLE}}(x) & = & \displaystyle \argmax_{m} &P(X = x | \theta = m)
\end{array}
\end{gather}
produces below.
However, this has the only one numbering label. It cannot split numbers for each row. It may be one BAD choice to use "align" and adjust spacing MANUALLY by using "\,", "\phantom{}", "\quad", or "\qquad".
For this case, how can we label different numbers for each row?
• You probably already know about the option \notag that removes the numbering of one row in align environment. I think there must be something similar, but instead it forces a tag for each entry in multi-rows of math environments. What you ask is indeed interesting and I am waiting to find the answer – Al-Motasem Aldaoudeyeh Jun 30 '17 at 2:54
• Off-topic: Don't use | to denote "given that" items. Instead, write \mid. – Mico Jun 30 '17 at 8:04
• A newenvironment here could help... But I am waiting for an answer in a problem I found trying to manipulate the rows of my newenvironment... I will come back for an answer... But have to study a little bit about newenvironments – koleygr Jun 30 '17 at 8:08
Since you're using the amsmath package, I would use that package's \DeclareMathOperator* directive to create two new "operators": \argmax and \midmax, where the latter displays the word "max" centered in a box of width equal to "arg max". I would also use a split environment instead of an array environment, an equation environment instead of a gather environment, and \mid instead of \.
If you need to number each row separately, use an align environment instead of the nested equation/ split environments.
\documentclass{article}
\usepackage{amsmath}
%% Create two new math opertors: \argmax and \midmax
\DeclareMathOperator*{\argmax}{arg\,max}
\newlength\mylen
\settowidth\mylen{arg\,max}
\DeclareMathOperator*{\midmax}{\parbox{\mylen}{\centering\upshape max}} % center-set "max"
\begin{document}
%% Single equation number for both rows:
$$\begin{split} p_{\mathrm{MLE}}(x) &= \midmax_{m} P(X = x \mid \theta = m) \\ m_{\mathrm{MLE}}(x) &= \argmax_{m} P(X = x \mid \theta = m) \end{split}$$
% Separate equation numbers, one per row:
\begin{align}
p_{\mathrm{MLE}}(x) &= \midmax_{m} P(X = x \mid \theta = m) \\
m_{\mathrm{MLE}}(x) &= \argmax_{m} P(X = x \mid \theta = m)
\end{align}
\end{document} | 2019-10-23T00:51:03 | {
"domain": "stackexchange.com",
"url": "https://tex.stackexchange.com/questions/377488/equation-custom-horizontal-alignment-numbering-each-row",
"openwebmath_score": 0.9859458208084106,
"openwebmath_perplexity": 3056.945185071717,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232897298992,
"lm_q2_score": 0.8670357546485407,
"lm_q1q2_score": 0.8521429326971245
} |
https://math.stackexchange.com/questions/1027995/volume-of-a-split-log/1028005 | # Volume of a split log
When solving the following problem, I could not understand why my reasoning came up with an answer that's different than the one on the solution's manual.
Question: Consider $(x,y,z)$ such that $x^2+y^2<1, x>0, 0 \le z \le 5$. This describes one half of a cylinder (a split log). Chop out a wedge out of the log along $z=2x$. Find the volume of the wedge.
My reasoning:
Wedge Volume $=$ (half the volume of the cylinder of height 2) $-$ (volume of solid of revolution found revolving the curve $x=z/2$, from 0 to 2, over the z axis).
Half Cylinder Volume: $(\pi r^2h)/2 = \pi/2 \int_0^2 (1)^2 dz = \pi$
(Volume of solid of revolution found revolving the curve x=z/2, from 0 to 2, over the z axis): $(\pi r^2h)/2 = \pi/2 \int_0^2 (z/2)^2 dz = \pi/3$
Wedge Volume: $\pi - \pi/3 = 2\pi/3$
However, the solution's manual answer is $4/3$ and it displays a different reasoning:
The slice perpendicular to the $xz$-plane are right triangles with base of length $x$ and height $z=2x$. Therefore the area of a slice is $x^2$. The volume is:
$$\int_{-1}^1 x^2 dy = \int_{-1}^1 (1-y^2) dy = 4/3.$$
While I understand the solution's manual reasoning and even find it simpler than mine, I still cannot understand why the two approaches results in different answers. What am I missing?
Your reasoning in your first method goes wrong in that chopping a wedge isn't the same thing as taking the solid of revolution of a curve. The solid of revolution of $z = 2x$ is a cone, so if you take a cone out of the cylinder, you're not left with a wedge but a somewhat weird-looking anti-cone shape.
If we were to analyze the cross-section of the wedge that we want in cylindrical coordinates, we'd see that it grows as the angle approaches $0$, and shrinks as it approaches either $\pi/2$ or $-\pi/2$. Whereas with your solid of revolution, the cross-section stays constant at all angles. | 2020-01-19T22:00:31 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1027995/volume-of-a-split-log/1028005",
"openwebmath_score": 0.8840994238853455,
"openwebmath_perplexity": 358.86122611087717,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232889752296,
"lm_q2_score": 0.8670357529306639,
"lm_q1q2_score": 0.8521429303544297
} |
https://math.stackexchange.com/questions/1732914/determinant-of-n-times-n-matrix | # Determinant of $N \times\ N$ matrix
For $n \geq 2$, compute the determinant of the following matrix: $$B = \begin{bmatrix} -X & 1 & 0 & \cdots & 0 & 0 \\ 0 & -X & 1 & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & & \ddots & \ddots & 1 & 0 \\ 0 & \cdots & \cdots & 0 & -X & 1 \\ a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X) \end{bmatrix}$$
Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix:
$\det \begin{bmatrix} -X & 0 \\ 0 & (a_1-X) \end{bmatrix} = -X(a_1-X) - 0 = X^2 - a_1X$
by expansion along the first row:
$\det \begin{bmatrix} -X & 1 & 0 \\ 0 & -X & 0 \\ 0 & 0 & (a_2-X) \end{bmatrix} = (-X) \times\det \begin{bmatrix} -X & 0 \\ 0 & a_2-X \end{bmatrix} - 1 \det\begin{bmatrix} 0 & 0 \\ 0 & a_2-X \end{bmatrix}$
$= (-X)[(-X)(a_2-X) -0] - 0 = X^3 - a_2X^2$
So it looks like:
$\det \begin{bmatrix} -X & 1 & 0 & \cdots & 0 & 0 \\ 0 & -X & 1 & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & & \ddots & \ddots & 1 & 0 \\ 0 & \cdots & \cdots & 0 & -X & 1 \\ a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X) \end{bmatrix} = X^{n} - a_{n-1}X^{n-1} - a_{n-2}X^{n-2} ... - a_1X$
Does this look right? Is "prove by induction" valid to use here?
• Your $2\times 2$ matrix must be $\begin{bmatrix} -X & 1 \\ a_0 & (a_1-X) \end{bmatrix}$. You have also done the same mistake for the next case. I do not think you can use induction here, but you can get an intuitive idea about the general expression. Apr 8, 2016 at 4:08
• That's essentially the companion matrix. The induction step that works is expansion by minors along the first column. Apr 8, 2016 at 4:29
Let $v= \begin{pmatrix} 1 \\ x \\ x^{2} \\ \cdot \\ \cdot \\ \cdot \\ x^{n-2} \\x^{n-1} \end{pmatrix}$.
Then $Bv = (x-X)v \iff a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} -x^{n} = p(x) =0$. Thus, all the roots $\alpha$ of the monic polynomial $p(x)$ of degree $n$ noted here give us our eigenvectors $v_{\alpha}$, which are linearly independent since they are columns of a Vandermonde matrix. The associated eigenvalues are $\lambda = \alpha - X$.
• Thanks! Since the characteristic polynomial consists the trace and the determinant. So the detminant of this Vandermonde matrix is $a_0$. Since B is similar to the matrix, so det(B) = $a_0$. Am I right? Apr 8, 2016 at 11:42
• You are not right. You've already conjectured correctly what the determinant of $B$ should be: a polynomial in $X$. For example in your case for $n=2$, you found out that $\det B = X^{2} - a_{1} X - a_{0}$. In my answer I found linearly independent eigenvectors with associated eigenvalues. Determinant is the product of eigenvalues. Let $n = 2$ and $\alpha_{1}, \alpha_{2}$ are the roots of $p(x) = x^{2} - a_{1}x - a_{0}$, then by my answer above $\det B = (\alpha_{1} - X)(\alpha_{2} -X) = X^{2} - (\alpha_{1} +\alpha_{2})X + \alpha_{1} \alpha_{2}$ - roots and coefficients are related (how?) Apr 8, 2016 at 13:14
• Note that this proof requires a bit of elementary algebraic geometry in order to work. Per se, the roots of the polynomial $p$ may fail to be distinct, in which case you do not get $n$ linearly independent eigenvectors. So you need to first WLOG assume that the resultant of the polynomial $p$ and its derivative $p'$ is $\neq 0$ (this is true on a Zariski-dense subset of the space of all polynomials, so you can WLOG assume it). Apr 11, 2019 at 18:55
To add a final touch somewhat as a synthesis of the (thorough) answers of @user5713492 and @akech: the global result is that the companion matrix of polynomial p(X) is diagonalized with a Vandermonde matrix V(r_1,r_2,\cdots r_n) where the $r_k$ are the roots of p(X)" see https://en.wikipedia.org/wiki/Vandermonde_matrix | 2022-08-14T23:31:06 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1732914/determinant-of-n-times-n-matrix",
"openwebmath_score": 0.8780083060264587,
"openwebmath_perplexity": 235.99883576988697,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232919939075,
"lm_q2_score": 0.8670357477770337,
"lm_q1q2_score": 0.8521429279066236
} |
https://math.stackexchange.com/questions/3191278/find-the-length-x-such-that-the-two-distances-in-the-triangle-are-the-same | # Find the length x such that the two distances in the triangle are the same
I have been working on the following problem
## Statement
Assume you have a right angle triangle $$\Delta ABC$$ with cateti $$a$$, $$b$$ and hypotenuse $$c = \sqrt{a^2 + b^2}$$. Find or construct a point $$D$$ on the hypothenuse such that the distance $$|CD| = |DE|$$, where $$E$$ is positioned on $$AB$$ in such a way that $$DE\parallel BC$$ ($$DE$$ is parallel to $$BC$$).
## Background
My background for wanting such a distance is that I want to create a semicircle from C onto the line $$AB$$. This can be made clearer in the image below
To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem
## Solution
Using similar triangles one arrives at the three equations
\begin{align*} \frac{\color{blue}{\text{blue}}}{a - x} & = \frac{b}{a} \\ \frac{\color{red}{\text{red}}}{x} & = \frac{c}{a} \\ \color{red}{\text{red}} & = \color{blue}{\text{blue}} \end{align*}
Where one easily can solve for $$\color{blue}{\text{blue}}$$, $$\color{red}{\text{red}}$$, $$x$$.
## Question
I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $$D$$ in a simpler matter?
• You still have not described what $F$ is either from the statement or from the graph. – Hw Chu Apr 17 at 16:42
• Right when I said $F$ i meant $E$. I will fix it in the problem statement =) – N3buchadnezzar Apr 17 at 16:47
• cateti is Italian for legs – J. W. Tanner Apr 17 at 16:56
• Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function! – David K Apr 17 at 20:25
Let the angle bisector of $$\angle ACB$$ intersect side $$\overline{AB}$$ at point $$E$$.
Let the measure of $$\angle ACB$$ be $$\alpha$$. Then $$m \angle ACE = \dfrac{\alpha}{2}$$.
Let the line perpendicular to side $$\overline{AB}$$ at point $$E$$ intersect side $$\overline{AC}$$ at point $$D$$.
Since $$\overline{ED}$$ is parallel to $$\overline{BC}$$, then $$\angle ADE \cong \angle ACB$$.
By the exterior angle theorem, $$m\angle DEC = \dfrac{\alpha}{2}$$.
Hence $$\triangle EDC$$ is isoceles.
So $$CD = DE$$.
(Added later). Assuming that the sides have lengths of $$x$$ and $$y$$, and that $$r = \sqrt{x^2+y^2}$$, the lengths of the segments are displayed below.
• This is exactly what I was looking for =) – N3buchadnezzar Apr 17 at 18:24
The curve that traces out the points that are equidistant to $$C$$ and the line extension of $$AB$$ is a parabola with focus $$C$$ and directrix $$AB$$.
Choosing a convenient coordinate system (parallel to $$AB$$ with $$B$$ as the origin), I get the equation $$y = \frac{x^2}{2b} + \frac{b}{2}$$, which you want to intersect with the line $$y = \frac{b}{a}x + b$$.
Solving, I get that the point $$D$$ has $$(x,y)$$ coordinates of $$x = \frac{b(b - c)}{a}$$ and $$y = \frac{bc(c - b)}{a^2}$$.
Let $$CD=DE=y$$ then we get $$\frac{b}{c}=\frac{y}{c-y}$$ so $$y=\frac{bc}{b+c}$$
• @mathmandan There was an edit after my comment. – Michael Biro Apr 17 at 21:17
Not sure if this is less barbaric but using simple trig: $$DE=(a-x)\tan A$$, $$DC=\frac{x}{\cos A}$$ so the equation to solve is $$(a-x)\frac{b}{a}=\frac{x\sqrt{a^2+b^2}}{a}$$ or $$x=\frac{ab}{\sqrt{a^2+b^2}+b}$$
Just another idea to construct point $$E$$: since $$\triangle{DCE}$$ is isosceles, it's easy to find $$\angle{ACE}=(90°-A)/2$$ | 2019-06-21T00:04:09 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3191278/find-the-length-x-such-that-the-two-distances-in-the-triangle-are-the-same",
"openwebmath_score": 0.9472463130950928,
"openwebmath_perplexity": 273.56894909378747,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9828232909876816,
"lm_q2_score": 0.8670357477770336,
"lm_q1q2_score": 0.8521429270341897
} |
https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html |
## SectionInverse Functions
Recall that in Brief Intro to Composite and Inverse Functions we gave the following definition of an inverse function:
###### Inverse Functions
Suppose the inverse of $f$ is a function, denoted by $f^{-1}\text{.}$ Then
\begin{equation*} f^{-1}(y) = x \text{ if and only if }f(x) = y. \end{equation*}
We also stated the following property about inverse functions:
###### Functions and Inverse Functions
Suppose $f^{-1}$ is the inverse function for $f\text{.}$ Then
\begin{equation*} f^{-1}\left(f(x)\right) = x\text{ and }f\left(f^{-1}( y)\right) = y \end{equation*}
as long as $x$ is in the domain of $f\text{,}$ and $y$ is in the domain of $f^{-1}\text{.}$
In this section we will explore the invertability of a function. In other words, by the end of this section we will be able to test if a function is invertable.
### SubsectionGraph of the Inverse Function
In Example161, we used a graph of $h$ to read values of $h^{-1}\text{.}$ But we can also plot the graph of $h^{-1}$ itself. Because $C$ is the input variable for $h^{-1}\text{,}$ we plot $C$ on the horizontal axis and $F$ on the vertical axis. To find some points on the graph of $h^{-1}\text{,}$ we interchange the coordinates of points on the graph of $h\text{.}$ The graph of $h^{-1}$ is shown in Figure192.
$C=h(F)$ $F$ $C$ $14$ $-10$ $32$ $0$ $50$ $10$ $68$ $20$
$F=h^{-1}(C)$ $C$ $F$ $-10$ $14$ $0$ $32$ $10$ $50$ $20$ $68$
###### Example193
The Park Service introduced a flock of $12$ endangered pheasant into a wildlife preserve. After $t$ years, the population of the flock was given by
\begin{equation*} P = f (t) = 12 + 2t^3. \end{equation*}
1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $t = f^{-1}(P)\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.
Solution
1. The graph of $f$ is shown in Figure194, with $t$ on the horizontal axis and $P$ on the vertical axis.
2. We solve $P = 12 + 2t^3$ for $t$ in terms of $P\text{.}$ \begin{align*} 2t^3 \amp = P - 12\amp\amp\text{Substract 12 from both sides.}\\ t^3 \amp = \frac{P - 12}{2}\amp\amp\text{Divide both sides by 2.}\\ t \amp = \sqrt[3]{\frac{P - 12}{2}}\amp\amp\text{Take cube roots.} \end{align*} The inverse function is $t = f^{-1}(P) =\sqrt[3]{\dfrac{P - 12}{2}}\text{.}$ It tells us the number of years it takes for the pheasant population to grow to size $P\text{.}$
3. The graph of $f^{-1}$ is shown in Figure195, with $P$ on the horizontal axis and $t$ on the vertical axis.
The formula $T = f(L) = 2\pi \sqrt{\dfrac{L}{32}}$ gives the period in seconds, $T\text{,}$ of a pendulum as a function of its length in feet, $L\text{.}$
1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $L = f^{-1}(T )\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.
### SubsectionWhen Is the Inverse a Function?
We can always find the inverse of a function $y=f(x)$ simply by solving for $x$ thus interchanging the role of the input and output variables. In the preceding examples, this process created a new function. However, this process does not always lead to be a function.
For example, to find the inverse of $y = f (x) = x^2\text{,}$ we solve for $x$ to get $x = \pm\sqrt{y}\text{.}$ When we regard $y$ as the input and $x$ as the output, the relationship does not describe a function. This can be seen as plugging in $y=4 \text{,}$ for example, gives two outputs $x=2$ and $x=-2\text{.}$ The graphs of $f$ and its inverse are shown in Figure197. (Note that for the graph of the inverse, we plot $y$ on the horizontal axis and $x$ on the vertical axis.) Because the graph of the inverse does not pass the vertical line test, it is not a function.
For many applications, it is important to know whether or not the inverse of $f$ is a function. This can be determined from the graph of $f\text{.}$ When we interchange the roles of the input and output variables, horizontal lines of the form $y = k$ become vertical lines.
Thus, if the graph of the inverse is going to pass the vertical line test, the graph of the original function must pass the horizontal line test, namely, that no horizontal line should intersect the graph in more than one point. Notice that the graph of $f(x) = x^2$ does not pass the horizontal line test, so we would not expect its inverse to be a function.
###### Horizontal Line Test
If no horizontal line intersects the graph of a function more than once, then its inverse is also a function.
We have been talking about how to tell if the inverse of a function is also a function, but in practice this is not the language typicaly used. Usually we ask this same question in the form "Is the function invertible?" The following definition explains this relationship:
###### Invertible Function
If $y=f(x)$ is a function such that its inverse, $x=f^{-1}(y)\text{,}$ is also a function then we say that $f(x)$ is an invertible function.
###### Example200
Which of the functions in Figure201 are invertible?
Solution
In each case, apply the horizontal line test to determine whether the function is invertible. Because no horizontal line intersects their graphs more than once, the functions pictured in Figures201(a) and (c) are invertible. The functions in Figures201(b) and (d) are not invertibe.
### SubsectionMathematical Properties of the Inverse Function
The inverse function $f^{-1}$ undoes the effect of the function $f\text{.}$ In Example152, the function $f(t) = 6 + 2t$ multiplies the input by $2$ and then adds $6$ to the result. The inverse function $f^{-1}(H) = \dfrac{H -6}{2}$ undoes those operations in reverse order: It subtracts $6$ from the input and then divides the result by $2\text{.}$
If we apply the function $f$ to a given input value and then apply the function $f^{-1}$ to the output from $f\text{,}$ the end result will be the original input value. For example, if we choose $t = 5$ as an input value, we find that \begin{align*} f(\alert{5})\amp= 6 + 2(\alert{5}) = \blert{16}\amp\amp\text{ Multiply by 2, then add 6.}\\ \text{and } f^{-1}(\blert{16}) \amp = \frac{\blert{16} - 6}{2} = \alert{5}.\amp\amp\text{Subtract 6, then divide by 2.} \end{align*}
We return to the original input value, $5\text{,}$ as illustrated in Figure202.
Example203 illustrates the fact that if $f^{-1}$ is the inverse function for $f\text{,}$ then $f$ is also the inverse function for $f^{-1}\text{.}$
###### Example203
Consider the function $f(x) = x^3 + 2$ and its inverse, $f^{-1}(y) = \sqrt[3]{y - 2}\text{.}$
1. Show that the inverse function undoes the effect of $f$ on $x = 2\text{.}$
2. Show that $f$ undoes the effect of the inverse function on $y = -25\text{.}$
Solution
1. First evaluate the function $f$ for $x = 2\text{:}$
Then evaluate the inverse function $f^{-1}$ at $y = 10\text{:}$
\begin{equation*} f^{-1}(\blert{10}) = \sqrt[3]{\blert{10} - 2} = \sqrt[3]{8}= \alert{2}. \end{equation*}
We started and ended with $2\text{.}$
2. First evaluate the function $f^{-1}$ for $y = -25\text{:}$
\begin{equation*} f^{-1}(\alert{-25}) = \sqrt[3]{-25 - 2} = \blert{-3}. \end{equation*}
Then evaluate the function $f$ for $x = -3\text{:}$
\begin{equation*} f (\blert{-3}) = (\blert{-3})^3 + 2 = \alert{-25}. \end{equation*}
We started and ended with $-25\text{.}$
###### Example204
1. Find a formula for the inverse of the function $f(x)=\dfrac{2}{x - 1}\text{.}$
2. Show that $f^{-1}$ undoes the effect of $f$ on $x = 3\text{.}$
3. Show that $f$ undoes the effect of $f^{-1}$ on $y = -2\text{.}$
Solution
1. To find the inverse, we solve for $x\text{:}$
\begin{equation*} \begin{aligned} y \amp= \frac{2}{x-1} \\ y(x-1) \amp= 2 \\ x-1 \amp= \frac{2}{y} \\ x \amp =\frac{2}{y}+1 \end{aligned} \end{equation*}
Therefore $f^{-1}(y)=\frac{2}{y}+1\text{.}$
2. First evaluate the function $f$ for $x=3\text{:}$
Then evaluate the inverse function $f^{-1}$ at $y=1\text{:}$
We started and ended with $3\text{.}$
3. First evaluate the inverse function $f^{-1}$ for $y=-2\text{:}$
Then evaluate the original function $f$ at $x=0\text{:}$
We started and ended with $-2\text{.}$
### SubsectionSymmetry
So far we have been careful to keep track of the input and output variables when we work with inverse functions. This is important when we are dealing with applications; the names of the variables are usually chosen because they have a meaning in the context of the application, and it would be confusing to change them.
However, we can also study inverse functions purely as mathematical objects. There is a relationship between the graph of a function and the graph of its inverse that is easier to see if we plot them both on the same set of axes.
A graph does not change if we change the names of the variables, so we can let $x$ represent the input for both functions, and let $y$ represent the output. Consider the function $C = h(F)$ from Example161, and its inverse function, $F = h^{-1}(C)\text{.}$ The formulas for these functions are \begin{align*} C \amp = h(F) = \frac{5}{9}(F - 32)\\ F \amp = h^{-1}(C) = 32 + \frac{9}{5}C. \end{align*} But their graphs are the same if we write them as \begin{align*} y \amp = h(x) =\frac{5}{9}(x - 32)\\ y \amp= h^{-1}(x) = 32 + \frac{9}{5}x. \end{align*} The graphs are shown in Figure205.
Now, for every point $(a, b)$ on the graph of $f\text{,}$ the point $(b, a)$ is on the graph of the inverse function. Observe in Figure205 that the points $(a, b)$ and $(b, a)$ are always located symmetrically across the line $y=x\text{.}$ The graphs are symmetric about the line $y=x$, which means that if we were to place a mirror along the line $y=x\text{,}$ each graph would be the reflection of the other.
###### Example206
Graph the function $f (x) = 2\sqrt{x + 4}$ on the domain $[-4, 12]\text{.}$ Graph its inverse function $f^{-1}$ on the same grid.
Solution
The graph of $f$ has the same shape as the graph of $y = \sqrt{x}\text{,}$ shifted $4$ units to the left and stretched vertically by a factor of $2\text{.}$ Figure207a shows the graph of $f\text{,}$ along with a table of values. By interchanging the rows of the table, we obtain points on the graph of the inverse function, shown in Figure207b.
If we use $x$ as the input variable for both functions, and $y$ as the output, we can graph $f$ and $f^{-1}$ on the same grid, as shown in Figure208. The two graphs are symmetric about the line $y = x\text{.}$
Graph the function $f (x) = x^3 + 2$ and its inverse $f^{-1}(x) = \sqrt[3]{x - 2}$ on the same set of axes, along with the line $y = x\text{.}$ | 2023-03-22T02:07:22 | {
"domain": "unl.edu",
"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html",
"openwebmath_score": 0.844607412815094,
"openwebmath_perplexity": 217.23917974865927,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109520836027,
"lm_q2_score": 0.8652240860523328,
"lm_q1q2_score": 0.8520821559508629
} |
https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750 | # Evaluate $\int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}$.
I need to solve the following integral: $$I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.$$ Wolfram Alpha gives the answer as $\dfrac{\pi}{2 \sqrt{2}}$.
I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?
• The integrand has a closed-form antiderivative. Nov 13 '15 at 2:55
• You don't need to use complex analysis. Nov 13 '15 at 3:03
• if you want to use contour integration, take a keyhole around with a slit at the line $(1,\infty)$ Nov 13 '15 at 10:30
Let $u=\sqrt{x-1}$, $du=\frac{1}{2\sqrt{x-1}}$, then
$$I=2\int_0^\infty\frac{u^2}{(u^2+2)^2}du$$
We can apply partial fractions here.
$$=2\int_0^\infty\left(\frac{1}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$
$$=2\int_0^\infty\frac{1}{u^2+2}du-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
The first integrand is almost $\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\frac{u}{\sqrt{2}}$.
$$I=\sqrt 2\int_0^\infty\frac{1}{s^2+1}ds-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
$$=\frac{\pi}{\sqrt 2}-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
Now to tackle the second integral we can use a trig sub.
Let $u=\sqrt{2}\tan (t)$, $du=\sqrt 2 \sec^2(t)dt$.
$$I=\frac{\pi}{\sqrt 2}-4\sqrt 2 \int_0^{\pi/2}\frac{\sec ^2 t}{4\sec^4t}dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2\int _0^{\pi/2}\cos^2(t) dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2 \int_0^{\pi/2}\left(\frac{1}{2}\cos(2t)+\frac{1}{2} \right)dt$$
If we substitute $v=2t$ and split the integral up we get
$$I=\frac{\pi}{\sqrt 2}-\frac{1}{2\sqrt 2} \int_0^\pi \cos(v)dv-\frac{1}{\sqrt 2}\int_0^{\pi/2}1dt$$
The first integral clearly goes to 0 and the second integral becomes $\frac{\pi}{2\sqrt 2}$.
Therefore
$$I=\frac{\pi}{2\sqrt 2}$$
• There's a much easier way to find the $u$ integral. Using IBP gives $$\int u\cdot\frac{2u}{(u^2+2)^2}du = -u\cdot\frac{1}{u^2+2} + \int \frac{1}{u^2+2}\,du = -\frac{u}{u^2+2} + \arctan u$$ Nov 18 '15 at 5:51
Using a keyhole countour with the origin of the keyhole at $z=1$ and the small circle enclosing the value $z=1$ and using $$f(z) = \frac{\exp(1/2\log(z-1))}{(1+z)^2}$$ with the branch cut of the logarithm on the positive real axis and the argument from $0$ to $2\pi$ we get for the integral $$I=\int_1^\infty \frac{\sqrt{x-1}}{(1+x)^2} dx$$ that $$I(1-\exp(\pi i)) = 2\pi i\mathrm{Res}_{z=-1} f(z)$$ or $$I = \pi i\mathrm{Res}_{z=-1} f(z).$$
Now the logarithmic term is certainly analytic in a neighborhood of $z=-1$ and we have $$\mathrm{Res}_{z=-1} f(z) = \left.\left(\exp(1/2\log(z-1))\right)'\right|_{z=-1} = \left. \left(\exp(1/2\log(z-1))\right) \frac{1}{2}\frac{1}{z-1}\right|_{z=-1} \\= \exp(1/2(\log 2 + \pi i)) \times -\frac{1}{2}\frac{1}{2} = -\frac{1}{4} \sqrt{2} i.$$
This yields $$I = -\frac{1}{4} \sqrt{2} i \times\pi i=\frac{\sqrt{2}\pi}{4}.$$
Remark. The estimates for the circular components are done using ML same as at this MSE link. We get for the large circle parameterized by $z=R e^{it}$ $$\lim_{R\rightarrow \infty} 2\pi R \frac{\sqrt{R+1}}{(R-1)^2} = 0.$$
The small circle is a parameterized with $z=1+\epsilon e^{it}$ and we get $$\lim_{\epsilon\rightarrow 0} 2\pi\epsilon \frac{\sqrt{\epsilon}}{4} = 0.$$
$$I=\int_{1}^{\infty}\frac{\sqrt{x-1}}{(x+1)^2}dx$$ Use substitution $\frac{1}{x+1}=t$ which implies $\frac{dx}{(x+1)^2}=-dt$ So
$$I=-\int_{0.5}^{0}\sqrt{\frac{1}{t}-2}\:dt=\int_{0}^{0.5}\frac{\sqrt{1-2t}}{\sqrt{t}}dt$$
Again use substitution $\sqrt{t}=y$ which implies $\frac{dt}{\sqrt{t}}=2dy$
$$I=2\int_{0}^{\sqrt{0.5}}\sqrt{1-2y^2}dy=2\sqrt{2}\int_{0}^{\sqrt{0.5}}\sqrt{(\sqrt{0.5})^2-t^2}$$
Use standard integral $$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})$$
we get $$I=\frac{\pi}{2\sqrt{2}}$$
Hint Substitution $x=t^2+1$...
This is the general antiderivative. Just take the limits and you're good. $$\int{\frac{\sqrt{x-1}}{(x+1)^2}dx}$$ Substitute $u = \sqrt{x-1}$ $$= \int{\frac{u}{(u^2+2)^2}du}$$ $$= 2\int\left(\frac{u}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$ $$= \int\frac{u}{\frac{u^2}{2}+1}du-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}$ $$= \sqrt{2}\int\frac{1}{s^2+1}ds-4\int\frac{2}{(u^2+2)^2}du$$ $$= \sqrt{2}\arctan(s)-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $u = \sqrt{2}\tan (p)\quad$ and $\quad du = \sqrt{2}\sec^2(p) dp\quad$ and $\quad(u^2 + 2)^2 = (2\tan^2(p)+2)^2 = 4\sec^4(p)\quad$ and $\quad p=\arctan\frac{u}{\sqrt{2}}$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\cos^2(p)du$$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\left(\frac{1}{2} \cos(2p) + \frac{1}{2}\right)du$$ $$= \sqrt{2}\arctan(s)-\frac{1}{\sqrt{2}}\int\cos(2p) - \frac{1}{\sqrt{2}}\int du$$ $$= \sqrt{2}\arctan(s)-\frac{p}{\sqrt{2}} - \frac{\sin(p)\cos(p)}{\sqrt{2}}$$ $$= \sqrt{2}\arctan(s)-\frac{\arctan\frac{u}{\sqrt{2}}}{\sqrt{2}} - \frac{\sin(\arctan\frac{u}{\sqrt{2}})\cos(\arctan\frac{u}{\sqrt{2}})}{\sqrt{2}}$$ Note that $\cos(\arctan(z)) = \frac{1}{\sqrt{z^2 + 1}}$ and $\sin(\arctan(z)) = \frac{z}{\sqrt{z^2+1}}$ $$= \frac{2\sqrt{2}(u^2+2)\arctan(s)+\sqrt{2}(u^2+2)\arctan\frac{u}{\sqrt{2}}+2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(u^2+2)\arctan \frac{u}{\sqrt{2}}-2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(x+1)\arctan \frac{\sqrt{x-1}}{\sqrt{2}}-2\sqrt{x-1}}{2(x+1)}$$ $$=\frac{\arctan \frac{\sqrt{x-1}}{\sqrt{2}}}{\sqrt{2}}-\frac{\sqrt{x-1}}{x+1}$$
I think I have a different answer.
Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then
\begin{align} \int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx &= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\ &= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du \end{align} Now let $u = \frac{1}{t}$, so that $du = \frac{-1}{t^2} dt$. Continuing. \begin{align} \phantom{\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx} &= \frac{1}{\sqrt{2}} \int_0^1 t^2 (1-t)^\frac{1}{2} \frac{1}{\sqrt{t}} \frac{1}{t^2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{-1}{2}} (1-t)^\frac{1}{2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{2} - 1} dt \\ &= \frac{1}{\sqrt{2}} B ( \frac{1}{2} . \frac{3}{2} ) \\ &= \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{3}{2})} {\sqrt{2}\Gamma(2)} \\ &= \frac{\pi}{2\sqrt{2}}. \end{align}
Original image of work by hand.
• Honestly, with 1500 points I expect you know the customs here, and I see many answers where you used LaTeX. You could make an effort and write your answer correctly. Nov 14 '15 at 0:48 | 2021-10-25T20:02:54 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750",
"openwebmath_score": 1.000000238418579,
"openwebmath_perplexity": 289.02061330389927,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109520836026,
"lm_q2_score": 0.8652240808393984,
"lm_q1q2_score": 0.8520821508171078
} |
https://mathhelpboards.com/threads/a-linear-equation-in-3-variables.5803/ | # [SOLVED]a linear equation in 3 variables
#### karush
##### Well-known member
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
#### chisigma
##### Well-known member
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
You have the linear system...
$\displaystyle a + b + c = d$
$\displaystyle 2 b = d$
$\displaystyle a = d\ (1)$
... the solution of which is $a=d,\ b= \frac{d}{2},\ c=- \frac{d}{2}$, so that You can write...
$\displaystyle a x + \frac{d}{2} y - \frac{d}{2} z = d \implies x + \frac{y}{2} - \frac{z}{2} = 1\ (2)$
Kind regards
$\chi$ $\sigma$
#### HallsofIvy
##### Well-known member
MHB Math Helper
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
I would have done this just a little differently from chisigma. First, note that there is no "unique" solution. for any ax+ by+ cz= d describing the line, you could multiply through by any number and get a new equation for the same line. You could. for example, divide both sides by d to get (a/d)x+ (b/d)y+ (c/d)z= 1, then let a'= a/d, b'= b/d, and c'= c/d so that the equation is a'x+ b'y+ c'z= 1, with only three unknown values.
I suspect that the answer you are asked for is the one that has all integer coefficients with no common factor.
Now, from ax+ by+ cz= d, (1, 1, 1) on the line means we must have a+ b+ c= d. (0, 2, 0) on the line means 0a+ 2b+ 0c= 2b= d. (1, 0, 0) on the line means 1a+ 0b+ 0c= a= d. Since both 2b and a are equal to d, a= 2b and we can replace both a and d with 2b in the first equation: 2b+ b+ c= 2b so that b+ c= 0 or b= -c.
Now we can write everything in terms of b: a= 2b, c= -b, d= 2b, and so the equation is
2bx+ by- bz= 2b. Here, b can be any (non-zero) number and still give an equation describing the line but dividing both sides of the equation by b gives
2x+ y- z= 2, the simplest form for the equation as all numbers are integers and they do not all have a common factor so cannot be reduced.
#### Prove It
##### Well-known member
MHB Math Helper
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
There will be a unique plane which satisfies these three points simultaneously. The plane will have the same coefficients as its normal vector, and the normal vector will also be normal to any vectors in the plane. So if we can get two vectors that lie in the plane, taking their cross product will give the normal vector. Call two of these vectors \displaystyle \begin{align*} \mathbf{a} \end{align*} and \displaystyle \begin{align*} \mathbf{b} \end{align*}. We could have
\displaystyle \begin{align*} \mathbf{a} &= ( 1 - 0, 1 - 2, 1 - 0) \\ &= ( 1 , -1, 1 ) \\ \\ \mathbf{b} &= (1 - 0, 0 - 2, 0 - 0 ) \\ &= ( 1 , -2, 0) \end{align*}
\displaystyle \begin{align*} \mathbf{n} &= \mathbf{a} \times \mathbf{b} \\ &= \left| \begin{matrix} \mathbf{i} & \phantom{-}\mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -2 & 0 \end{matrix} \right| \\ &= \mathbf{i} \, \left| \begin{matrix} -1 & 1 \\ -2 & 0 \end{matrix} \right| - \mathbf{j} \, \left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| + \mathbf{k} \, \left| \begin{matrix} 1 & -1 \\ 1 & -2 \end{matrix} \right| \\ &= \mathbf{i} \, \left[ -1 \cdot 0 - 1 \cdot (-2) \right] - \mathbf{j} \, \left( 1 \cdot 0 - 1 \cdot 1 \right) + \mathbf{k} \, \left[ 1 \cdot (-2) - (-1) \cdot 1 \right] \\ &= 2\,\mathbf{i} + \mathbf{j} - \mathbf{k} \end{align*}
The plane will have the same coefficients as the normal vector, so that gives \displaystyle \begin{align*} 2x + y - z = d \end{align*}. Since we have three points that lie on the plane, any of them can be substituted to find \displaystyle \begin{align*} d \end{align*}. If we substitute \displaystyle \begin{align*} (1, 0, 0) \end{align*} that gives
\displaystyle \begin{align*} 2 \cdot 1 + 0 - 0 &= d \\ 2 &= d \end{align*}
and thus the plane is \displaystyle \begin{align*} 2x + y - z = 2 \end{align*}. | 2021-01-15T20:02:40 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/a-linear-equation-in-3-variables.5803/",
"openwebmath_score": 0.9999923706054688,
"openwebmath_perplexity": 468.6514466117882,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9848109503004293,
"lm_q2_score": 0.8652240704135291,
"lm_q1q2_score": 0.8520821390067531
} |
https://mathematica.stackexchange.com/questions/179382/issue-plotting-torus-link | I'm having an issue plotting torus links in Mathematica. What I'm trying to generate is a (2,8) torus link like the one in the picture in Wikipedia:
using equations such as those outlined here.
However, I appear to only get half of the respective torus link.
a = 1; d = 4; p = 2; q = 8;
ParametricPlot3D[{(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*
Cos[p*t], a*Cos[q*t]}, {t, 0, 2*Pi}, PlotStyle -> Orange,
PlotRange -> All] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]
(a and d are parameters that control the width of the tube and the distance of the curve from the origin respectively, outlined here.)
Can anyone explain why I am only seeing half of the link?
• You're only plotting one set of parametric equations and you can see in the image that you don't expect there to be any intersections so the smooth parametric equations clearly can't generate the structure you want. On the other hand, it ought to be clear that the others may be generated via a simple rotation in the x-y plane. – b3m2a1 Aug 2 '18 at 8:09
The second torus is created by a rotation of $$\pi/4$$ (around {0, 0, 1}):
R = RotationMatrix[Pi/4, {0, 0, 1}];
torus = {(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]};
ParametricPlot3D[{torus, R.torus} // Evaluate, {t, 0, 2*Pi}, PlotStyle -> {Orange, Blue}, PlotRange -> All]
Here's I think a generalization of what Ulrich gave to an arbitrary $(p, q)$ torus (if I read Wikipedia right):
plotPQTorus[{p_, q_}, a : _?NumericQ : 1, d : _?NumericQ : 4,
ops : OptionsPattern[]] :=
Block[{t},
ParametricPlot3D[
Evaluate@
Table[
RotationMatrix[
i*2 \[Pi]/q, {0, 0, 1}].{(a*Sin[q*t] + d)*
Sin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]},
{i, 0, If[Divisible[q, p], p - 1, 0]}
],
{t, 0, 2*Pi},
PlotRange -> All,
ops
] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]
]
Here are a few plots:
Table[plotPQTorus[{p, Fibonacci[q]}, Boxed -> False,
Axes -> None], {p, 1, 4}, {q, 2, 6, 2}] // Grid
Note that relatively prime things are single connected loops (as Wikipedia suggests they should be)
The (1,4) torus knot is known to KnotData[], so we can use that to build the (2,8) knot:
knot14 = First[KnotData[{"TorusKnot", {1, 4}}, "ImageData"]]; | 2020-11-27T02:17:08 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/179382/issue-plotting-torus-link",
"openwebmath_score": 0.2998276650905609,
"openwebmath_perplexity": 2329.600533817695,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9621075690244281,
"lm_q2_score": 0.8856314858927011,
"lm_q1q2_score": 0.8520727559437188
} |
https://mathhelpboards.com/threads/logic-problem.5096/ | # Logic problem
##### Active member
Consider the following sequence of statements:
$$S_1: \text{at least 1 of the statements }S_1-S_n \text{ is false}\\ S_2: \text{at least 2 of the statements }S_1-S_n \text{ are false}\\ \vdots \\ S_n: \text{at least } n \text{ of the statements }S_1-S_n \text{ are false}$$
Where $n$ is some integer.
Question: for which $n$ are these statements self-consistent? In those cases: what is the truth value of each statement?
I got this off of a blog I tend to frequent. I will wait before posting the solution this time.
EDIT:
Changed the question; I had written the statements wrong
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Suppose $k$ out of $n$ statements are true.
Then $S_1$ up to $S_k$ have to be true and the rest has to be false.
This appears to be consistent for any $n$ and any $0\le k \le n$.
##### Active member
Suppose $k$ out of $n$ statements are true.
Then $S_1$ up to $S_k$ have to be true and the rest has to be false.
This appears to be consistent for any $n$ and any $0\le k \le n$.
Sorry about that, you were absolutely right about the question as phrased.
However, this new version should prove to be a bit more interesting. This is what I had meant; I had accidentally written "true" instead of "false".
#### Klaas van Aarsen
##### MHB Seeker
Staff member
If $S_n$ is true, then $n$ statements are false, including $S_n$.
Therefore $S_n$ is false.
We now know that at least $1$ statement is false.
Therefore $S_1$ is true.
For $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.
For $n \ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.
Since $S_1$ is true, this implies that $S_{n-1}$ is false.
Therefore $S_{n-1}$ is false.
So at least $2$ statements are false.
Therefore $S_2$ is true.
For $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.
Etcetera.
In other words, we get a consistent consistent solution if and only if $n$ is even.
In that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\qquad \blacksquare$
##### Active member
If $S_n$ is true, then $n$ statements are false, including $S_n$.
Therefore $S_n$ is false.
We now know that at least $1$ statement is false.
Therefore $S_1$ is true.
For $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.
For $n \ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.
Since $S_1$ is true, this implies that $S_{n-1}$ is false.
Therefore $S_{n-1}$ is false.
So at least $2$ statements are false.
Therefore $S_2$ is true.
For $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.
Etcetera.
In other words, we get a consistent consistent solution if and only if $n$ is even.
In that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\qquad \blacksquare$
Couldn't have phrased it better myself.
The source, for anybody interested:
The Parity Paradox – Futility Closet
I highly recommend the website as a time-wasting tool. | 2022-07-05T09:39:25 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/logic-problem.5096/",
"openwebmath_score": 0.5636366605758667,
"openwebmath_perplexity": 447.92497156639786,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9621075744568837,
"lm_q2_score": 0.8856314738181875,
"lm_q1q2_score": 0.8520727491378914
} |
http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html | # Thread: Is it a field? set of real numbers in the form...
1. ## Is it a field? set of real numbers in the form...
Hi,
problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.
(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?
attempt:
(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?
(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..
Thanks
2. Originally Posted by Mollier
Hi,
problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.
(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?
attempt:
(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?
(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..
Thanks
just go through each property for a field and make sure that they all hold. you have to verify these explicitly.
and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?
3. Originally Posted by Mollier
Hi,
problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.
(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?
attempt:
(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?
(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..
Thanks
Firstly, if every real number was of the form $\alpha+\beta \sqrt{2}$ with $\alpha, \beta \in \mathbb{Q}$ then we can easily define a bijection $\mathbb{R} \rightarrow \mathbb{Q} \times \mathbb{Q}$, $\alpha + \beta \sqrt{2} \mapsto (\alpha, \beta)$. As there exists a bijection between $\mathbb{Q}$ and $\mathbb{Q} \times \mathbb{Q}$ we have that $|\mathbb{R}| = |\mathbb{Q}|$, a contradiction.
Secondly, in both of your problems you know that "most" of the axioms for fields holds (both are clearly rings, and multiplication in both cases is commutative). The one to concentrate on will be multiplicative inverses.
What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?
What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!)
4. Originally Posted by Jhevon
and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?
Thanks
Originally Posted by Swlabr
The one to concentrate on will be multiplicative inverses.
What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?
What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!)
It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b).
Thanks guys!
5. Originally Posted by Mollier
Thanks
It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b).
Thanks guys!
You want to try and write $\frac{1}{\alpha + \beta \sqrt{2}}$ as $\alpha^{\prime} + \beta^{\prime}\sqrt{2}$. I mean, $\sqrt{2} = \frac{2}{\sqrt{2}}$...numbers are strange things...just because something isn't obviously in a certain form does not mean you cannot write it in that form.
I shall give you a tantalising hint: It is a field.
6. $
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$
$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$
$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$
Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?
Thanks mate!
7. Originally Posted by Mollier
$
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$
$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$
$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$
Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?
Thanks mate!
Yeah, that's it! Although to be thorough you would need to make sure there is no division by 0 going on.
8. So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$?
Thank you very much.
9. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$?
Thank you very much.
For part (b) you need to find a counter-example - $\alpha^{\prime}$ and $\beta^{\prime}$ may look like non-integers, but this may not always be the case.
So, to complete the problem simply take an integer value of $\alpha$ and an integer value of $\beta$ such that $\alpha^{\prime} \notin \mathbb{Z}$ or $\beta^{\prime} \notin \mathbb{Z}$. For simplicity, let one of them be zero.
10. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.
11. $\alpha^2-2\beta^2\neq 0$ because of the irrationality of $\sqrt{2}$, it can be easily proved in number theory.
12. Originally Posted by Swlabr
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$
13. Originally Posted by Mollier
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$
Precisely!
Sorry to be pedantic about it, but in maths the proof is the king...
14. Please don't apologize. I have a BS in mechanical engineering from a horrible school, so I have now decided to actually learn some math. You being "pedantic" is exactly what I need! Thanks! | 2017-06-25T10:45:53 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html",
"openwebmath_score": 0.8847749829292297,
"openwebmath_perplexity": 174.76001380594448,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9621075701109193,
"lm_q2_score": 0.885631476836816,
"lm_q1q2_score": 0.8520727481932139
} |
http://mathhelpforum.com/number-theory/47919-rational-numbers.html | # Math Help - rational numbers
1. ## rational numbers
Find two rational numbers with denominators 11 and 13, respectively, and a sum of 7/143.
I got x/11 + y/13=7/143.
Two unknowns and one equation. Kind of stuck.
2. Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.
Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?
3. Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.
Laurent.
4. Originally Posted by Matt Westwood
Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.
Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?
What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm
5. Originally Posted by rmpatel5
What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm
6. Originally Posted by Laurent
Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.
Laurent.
I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?
7. Originally Posted by rmpatel5
gcd(11,13)=1 and there is a thm that says ax+by=1.
There is a theorem that says that there exists x and y such that 11x+13y=1. Now, you should find an example of such a couple (x,y). Either you guess it, or you make use of the Euclidean algorithm as you may have done in class.
8. Originally Posted by rmpatel5
I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?
Ok, so solve for x and y in 13x+11y=1, using the Euclidean algorithm (http://en.wikipedia.org/wiki/Euclidean_algorithm). Actually, do the Euclidian algorithm over 13 and 11, and observe...
Then, multiply by 7 :
13*(7x)+11*(7y)=7
Uh.
9. Use the Euclidean Algorithm to get the gcd of 11 and 13 (yes, you know this is 1, but the calculations you did in the algorithm help you find out what values of a and b give you 11a + 13b = 1.
You now want two numbers x and y such that 11x + 13y = 7.
Well, you just got 11a + 13b = 1, so multiply everything by 7:
$11 \times 7a + 13 \times 7b = 7$
So 7a and 7b are the numbers you want for x and y. Job done.
10. You do not need to use Euclidean algorithm.
It is quite obvious that $13(6) + 11(-7) = 1$.
Thus, $13(42) + 11(-49) = 7$.
This means the solutions to $13x+11y=7$ are: $x=42 - 11t \mbox{ and }y=-49 + 42t$.
11. 6 and -5 work as well, so are there many solutions to this problem. I got them by doing the EA | 2015-07-03T00:12:45 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/number-theory/47919-rational-numbers.html",
"openwebmath_score": 0.8476439714431763,
"openwebmath_perplexity": 409.548473341809,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9890130589886011,
"lm_q2_score": 0.8615382076534743,
"lm_q1q2_score": 0.8520725381869193
} |
https://www.projectrhea.org/rhea/index.php/Compute_DSFT_product_two_step_functions_ECE438F11 | Topic: Discrete-space Fourier transform computation
## Question
Compute the discrete-space Fourier transform of the following signal:
$f[m,n]= 2^{-(n+m)} u[n] u[m]$
First off notice that this equation can easily be separated into two functions $g[m]=2^{-m}u[m]$ and $h[n]=2^{-n}u[n]$ where $f[m,n]=g[m]h[n]$. Then since both equations are the same except for a change of variable where m=n or vice verse we can show the DTFT of either g[m] or h[n] and it should suffice for the other. If they were not the same we would have to evaluate both of them.
\begin{align} G[u]&=\sum_{m=-\infty}^{\infty}2^{-m}u[m]e^{-j u m} \\ &=\sum_{m=0}^{\infty}( \frac{1}{2e^{ju}} )^{m} \\ &=\frac{1}{1-\frac{1}{2e^{ju}}} \\ &=\frac{-2e^{ju}}{1-2e^{ju}} \\ \end{align}
Thus
$F[u,v]= \frac{4e^{j(u+v)}}{(1-2e^{ju})(1-2e^{jv})}$
\begin{align} F [u,v] &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f[m,n]e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} 2^{-(n+m)} u[n] u[m] e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} 2^{-m} u[m] e^{-j(mu)} \sum_{n=-\infty}^{\infty} 2^{-n} u[n] e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} 2^{-m} e^{-j(mu)} \sum_{n=0}^{\infty} 2^{-n} e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} (2e^{ju})^{-m} \sum_{n=0}^{\infty} (2e^{jv})^{-n}\\ &= \frac{1}{1-\frac{1}{2e^{ju}}}\cdot\frac{1}{1-\frac{1}{2e^{jv}}}\\ &= \frac{1}{1-\frac{1}{2}e^{-ju}}\cdot\frac{1}{1-\frac{1}{2}e^{-jv}}\\ &= \frac{1}{(1-\frac{1}{2}e^{-ju})(1-\frac{1}{2}e^{-jv})} \end{align}
--Xiao1 23:03, 19 November 2011 (UTC)
Instructor's comments: Solutions 1 and solutions 2 are the two main ways to answer the question. The second one is a bit longer to write, in my opinion, but it does not require knowing the separation property of the continuous-space Fourier transform. Note however that, in the exam, if you do not have the separation property in the table, you will need to prove it in order to get full credit. -pm | 2022-06-25T04:16:52 | {
"domain": "projectrhea.org",
"url": "https://www.projectrhea.org/rhea/index.php/Compute_DSFT_product_two_step_functions_ECE438F11",
"openwebmath_score": 1.0000097751617432,
"openwebmath_perplexity": 2124.7650561476125,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9890130554263734,
"lm_q2_score": 0.8615382058759129,
"lm_q1q2_score": 0.8520725333598925
} |
http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html | # Absolute Maximum And Minimum Calculator On Interval
20 at x = –0. > @ 2 f x x x 6 2 on 1,27 11. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. These values correspond to the probability of observing such an extreme value by chance. If we find all possible local extrema, then the global maximum, if it exists, must be the value of f(x) on the interval 1 ≤ x ≤ 4, and the maximum is f(4) = 9. Find more Mathematics widgets in Wolfram|Alpha. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Sample size calculation for trials for superiority, non-inferiority, and equivalence. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. 80 at x = –1. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. C) Absolute maximum only. For example, you could say,“The pulse rates are between 56 and 92 bpm. The closed interval method is a way to solve a problem within a specific interval of a function. Inequalities and ‘Maximum-Minimum’ Problems Henry Liu, 26 February 2007 There are many olympiad level problems in mathematics which belong to areas that are not covered well at all at schools. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Absolute zero is defined as the point where no more heat can be removed from a system, according to the absolute or thermodynamic temperature scale. Four Function and. Left bounds go on left side of min/max and right bounds go on right side. (3) Because the system has been carefully planned using modern top-down programming techniques, it is relatively easy to modify and extend. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. We take the derivative using the quotient rule: f0(x) =. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Relative Minimum - The lowest point on an interval of a curve. Once you've found the max-imum and minimum on this line (as well as on the other two lines that make up the boundary), compare all the values you've checked to find out that the absolute maximum and the absolute minimum are f(1 4,2) = 673 64 (←absolute maximum) f(1 2, 4. An absolute maximum occurs at c if for all x in the domain of f. The absolute is measured in liters of oxygen per minute. f x x() 3 d. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. • Find the values of f at the endpoints of the interval. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. The graph of y = cos x. Absolute Maximum and Absolute Minimum. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. Find the absolute maximum and absolute minimum values of f(x) = x2 −4 x2 +4 on the interval [−4,4]. Subtract 6 6 from 1 1. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Find the absolute maximum and absolute minimum values of f on the given interval. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. 4 is the lower limit. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. For, sin (x + ) = cos x. denbal87 New member. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. f(x)=x+ 9 x on [0. On the graph above of the function f on the closed interval [a, e], the point (a, f (a)) represents the absolute minimum, and the point (d, f (d)) represents the absolute maximum. Question 203087: Find the absolute maximum and absolute minimum values of the function below. An absolute minimum is the lowest y value or output value a. Thread starter denbal87; Start date Nov 4, 2014; D. The largest value found in steps 2 and 3 above will be the absolute maximum and the smallest value will be the absolute minimum. This lesson will focus on the maximum and minimum points. These points are sometimes referred to as max, min, extreme values, or extrema. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 −12x. Sure, there are other, more precise, definitions, but that will work for what we want to do. 67, and a relative maximum of 4. Use the Stefan Boltzmann relationship (I = σT 4 where σ = 5. Q: Determine the absolute maximum and minimum values of the function on the given interval. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). I have an assignment where I have to write a program that accepts a series of integers from the keyboard from 1 to 100 using a single sentinel controlled loop, meaning I will need to also assign a sentinel number that when entered, will stop the loop and display the results. This is defined everywhere and is zero at $\ds x=\pm \sqrt{3}/3$. By using this website, you agree to our Cookie Policy. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. The absolute max occurs at S = The absolute min occurs at S =. Decide whether you have a minimum or a maximum. )Given the function 𝑓(𝑥= 𝑥2+ 𝑥+ , chose values for a, b, and c in that could work for the graph shown. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. the absolute (global) maximum 3. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. VO2 max in men is approximately 40-60% higher in males than females. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. If you're behind a web filter, please make sure that the domains *. Question 203087: Find the absolute maximum and absolute minimum values of the function below. We take the derivative using the quotient rule: f0(x) =. The debt to equity ratio is a financial, liquidity ratio that compares a company’s total debt to total equity. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. The absolute maximum on the interval is 138 at x=-2. For example, the following are all equivalent confidence intervals: 20. Find the absolute maximum and absolute minimum values of f on the given interval. Approximating Relative Extrema. Also the lowest value of either the X of the Y is placed first in the set. the absolute (global) maximum 3. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. Example 1: Consider the three curves shown below. In worst case, if all intervals are from ‘min’ to ‘max’, then time complexity becomes O((max-min+1)*n) where n is number of intervals. Find the extreme values of f on the boundary of D. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Three major examples are geometry, number theory, and functional equations. Go to window and set your X minimum to -1 and your X maximum to 5. Extreme Values of Functions Definitions An extreme value of a function is the largest or smallest value of the function in some interval. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as −1≤x≤4 or [−1,4], or, more generally, a≤x≤b or [a,b], then f always has a global minimum value and a global maximum value on that interval. At t =0 the position of the object is 5. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Answer: First, find the critical points by finding where the derivative equals zero: f0(x) = (x2 +4)(2x)−(x2 −4. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. Find the absolute maximum and absolute minimum values of f on the given interval. pow(x, 2) + math. Visual Magnitude Calculator: Computes the visual magnitude of a star from its absolute magnitude and distance. If f has a local maximum or minimum at c and f'(c. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Extreme Values A Global Maximum A function f has a global (absolute) maximum at x =c if f (x)≤ f (c) for all x∈Df. Absolute Maximum - The highest point on a curve. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Let's Practice:. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Here again we are giving definitions that appeal to your geometric intuition. f(x)=xln(2x)on[0. Local Extreme Values of a Function Let c be an interior point of the domain of the. The absolute is measured in liters of oxygen per minute. ) Find the absolute maximum and minimum values of the function on the given interval. Q: Determine the absolute maximum and minimum values of the function on the given interval. Extreme value theorem tells us that a continuous function must obtain absolute minimum and maximum values on a closed interval. VO2max stands for maximal oxygen uptake and refers to the amount of oxygen your body is capable of utilizing in one minute. so minimum value of f(x) in interval [0. For example, consider the functions shown in Figure(d), (e), and (f). Find more Mathematics widgets in Wolfram|Alpha. It's easy to find one with neither absolute extrema. Identify the location and value of the absolute maximum and absolute minimum of a function over the domain of the function graphically or by using a graphing utility. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Find the absolute maximum and absolute minimum values of f on the given interval. The maximum value of a function that has a derivative at all points in an For a function f(x) that has a derivative at every point in an interval [a, b], the maximum or minimum values can be found by using the following procedure: 1. By using this website, you agree to our Cookie Policy. Similar topics can also be found in the Calculus section of the site. So the absolute max value is 19 and the absolute min value is 1. The minimum and maximum describe the spread of the data. By using this website, you agree to our Cookie Policy. When calculating Intervals the X values are placed "on top of" the Y values. Enter DNE if the absolute maximum or minimum does not exist. Pick the largest and smallest. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. Thus if one has a sample {, …,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. At t =0 the position of the object is 5. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. The absolute minimum on the interval is -237 at x=3. f x x( ) cos 2 b. Explain your reasoning. maximum" functions. Similarly, the global minimum is located at the lowest point. A relative minimum is a point that is lower than all the other points around it. The absolute max occurs at S = The absolute min occurs at S =. question_answer. Find the maximum and minimum values of the function f(x) = ln x/x on the interval [1, 3]. Hit graph and then hit 2nd Trace I think to bring up a menu that has minimum and maximum in there. Typical values for are 0. Absolute Maximum And Minimum Calculator. So, f(b) is a relative maximum of f. The maximum will occur at the highest value and the minimum will occur at the lowest value. If you're seeing this message, it means we're having trouble loading external resources on our website. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Find the the critical points of f on D. Fermat's Theorem. From this list of values we see that the absolute maximum is 8 and will occur at $$t = 2$$ and the absolute minimum is -3 which occurs at $$t = 1$$. You can take notes in the margins or on the flip-side of each sheet. Finding Extrema on a closed interval: 1. Subtract 6 6 from 1 1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Find more Mathematics widgets in Wolfram|Alpha. Since the function is concave down at x=1 and has a critical point at x=1 (zero slope) then the function has a local maximum at x=1. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. ^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Wolfram alpha paved a completely new way to get knowledge and information. Enter the equation in the Y= section for Y1. minimum" or "4. f(x)=x^3-3x+1; [0,3]. Local Extreme Values of a Function Let c be an interior point of the domain of the. Let f be a function defined and. Endpoint Discontinuities: only one of the one-sided limits exists. f(x) = x + (4/x) on the interval [1,5]. A relative (or local) maximum occurs at c if for all x in an open interval containing c. 11) A) Absolute minimum only. But there is one very important condition that guarantees both an absolute minimum and an absolute maximum. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. Absolute Maximum and Absolute Minimum. So the absolute max value is 19 and the absolute min value is 1. org are unblocked. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. Get the free "Function Extrema - Math 101" widget for your website, blog, Wordpress, Blogger, or iGoogle. so minimum value of f(x) in interval [0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. have both an absolute maximum and an absolute minimum. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Finding Extrema on a closed interval: 1. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. The student familiar with the sum formula can easily prove that. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. False There is a local maximum at x = 0. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Fold Unfold. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. This lesson will focus on the maximum and minimum points. Find more Mathematics widgets in Wolfram|Alpha. Relative Minimum - The lowest point on an interval of a curve. # The text file for eval. f(x) = @ 10. absolute minimum value at x = 2 is -48. The first derivative test: Let f (x) be a function and x = c a critical point of f. f(t) = 4t + 4 cot(t/2), [π/4, 7π/4] I'm stuck after I take the first derivative. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. Free functions extreme points calculator - find functions extreme and saddle points step-by-step This website uses cookies to ensure you get the best experience. We take the derivative using the quotient rule: f0(x) =. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). f x x3 2 on 3,1 > @ 15. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. The minimum value for this range is the mean subtracted by the confidence interval and the maximum value is calculated by the mean added by the confidence interval. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. f(x) = (x^2 - 1)^3, [-1, 5] 14. 2 Maximum and Minimum on an Interval. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Find the absolute maximum and absolute minimum values of f on the given interval. In that case, the point right on the border might be the maximum or minimum of the curve. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. f(c) is another relative minimum of f. It could very well continue to increase or decrease once we leave the interval. Example: Find the absolute maximum and minimum of:. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. 1 absolute maximum or minimum. The maximum will occur at the highest f (x) f (x) value and the minimum will occur at the lowest f (x) f (x) value. Find the the critical points of f on D. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Absolute Maximum and Absolute Minimum. In this case, “absolute extrema” is just a fancy way of saying the single highest point and single lowest point in the interval. D) Absolute minimum and. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. The restrictions stated or implied for such functions will determine the domain from which you must work. 587 Get more help from Chegg Get 1:1 help nowfrom expert CalculustutorsSolve itwith our calculusproblem solver and calculator. For example, consider the functions shown in Figure(d), (e), and (f). Compare the f (x) f ( x) values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. (To make the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum. Absolute minimum definition is - the smallest value that a mathematical function can have over its entire curve. f ( 1) = − 5 f ( 1) = - 5. Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. earn credit. Python List max() Method - Python list method max returns the elements from the list with maximum value. True You are given that the function f ( x ) = 2 ( x + 3 ) x 2 + x − 2 has an absolute maximum on the interval − 2 < x < 1. Find the absolute maximum and absolute minimum values of f on the given interval. We usually distinguish between local and global (or absolute) extreme values. The smallest y -value is the absolute minimum and the largest y -value is the. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. The maximum will occur at the highest value and the minimum will occur at the lowest value. Roll your mouse over the Extreme Value Theorem to check your answers. A closed interval like [2, 5] includes the endpoints 2 and 5. For this data set, the minimum (lowest) value is 56 and the maximum (highest) value is 92. It could very well continue to increase or decrease once we leave the interval. Endpoint Discontinuities: only one of the one-sided limits exists. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. maximum: ( (62 -√3763)/9, (-461347 +7526√3763)/243)) ≈ (. If an absolute maximum or minimum does not exist, enter NONE. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Every function that’s continuous on a closed interval has an absolute maximum value and an absolute minimum value (the absolute extrema) in that interval — in other words, a highest and lowest point — though there can be a tie for the highest or lowest value. In that case, the point right on the border might be the maximum or minimum of the curve. Absolute maximum is highest of and. f(x) = 4x^3 - 6x^2 - 144x + 9, [-4, 5] 13. Find the extreme values of f on the boundary of D. Before Using this Calculator. The following small array formulas can help you to find out the largest absolute value and the smallest absolute value. Explain the meaning of the result. Get the free "Max/Min Finder" widget for your website, blog, Wordpress, Blogger, or iGoogle. f(x) = x + (4/x) on the interval [1,5]. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. Calculates the root of the equation f(x)=0 from the given function f(x) and its derivative f'(x) using Newton method. There are two kinds of extrema (a word meaning maximum or minimum): global and local, sometimes referred to as "absolute" and "relative", respectively. Example: Find the absolute maximum and minimum of:. f(c) is another relative minimum of f. The student familiar with the sum formula can easily prove that. Return to Contents. It is a greatest value in a set of points but not highest when compared to all values in a set. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Wolfram alpha paved a completely new way to get knowledge and information. exp(exp) * math. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. For, sin (x + ) = cos x. You can take notes in the margins or on the flip-side of each sheet. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Find the local maximum and minimum values of using both the First and Second Derivative Tests. Generating random numbers Problem. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 −12x. Four Function and. (b) Use calculus to find the exact maximum and minimum values. This important theorem can guide our investigations when we search for absolute extreme values of a function. 4 is the lower limit. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. in some open interval containing c. The Organic Chemistry Tutor 200,049 views 1:10:05. Over the long term about one day in ten can be expected to have a (maximum or minimum) temperature exceed the decile 9 value. Sample size calculation for trials for superiority, non-inferiority, and equivalence. To define these terms more formally: a function f has an absolute maximum at x = b if f ( b )≥ f ( x ) for all x in the domain of f. Therefore, let’s consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. You have to use a graphing calculator to find that out. Calculate Field honors the Transfer Field Domain Descriptions environment. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Enter the equation in the Y= section for Y1. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. If f has a local maximum or minimum at c and f'(c. (c) For any. Some problems may have two or more constraint equations. On a closed interval these points are referred to as absolute or global minimum/maximum points. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The “V” = volume per time. Local Extreme Values of a Function Let c be an interior point of the domain of the. Also time complexity of above solution depends on lengths of intervals. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. 2 Maximum and Minimum on an Interval. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. The smallest y -value is the absolute minimum and the largest y -value is the. Hill Sphere Calculator: Computes the Hill Sphere of an object: Sail Calculator: Computes the maximum velocity possible from acceleration caused by light. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. save_path = FLAGS. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. f(x)=x+ 9 x on [0. f(t) = 4t + 4 cot(t/2), [π/4, 7π/4] I'm stuck after I take the first derivative. Explain your reasoning. Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. question_answer. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as −1≤x≤4 or [−1,4], or, more generally, a≤x≤b or [a,b], then f always has a global minimum value and a global maximum value on that interval. The debt to equity ratio is a financial, liquidity ratio that compares a company’s total debt to total equity. 47, an absolute minimum of –5. 2 Maximum and Minimum on an Interval. You should substitute those and pick the greatest for the maximum value. No Local Extrema Compare the f (x) f (x) values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Find the absolute maximum and absolute minimum values of f on the given interval. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. (1 point) Find the absolute maximum and absolute minimum values of the function f (x) = x3 + 6x2 — 63x + 8 over each of the indicated intervals. maximum" functions. f (c) is called the global (absolute) maximum value. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. This is achieved with a minimum of manual intervention. A relative minimum is a point that is lower than all the other points around it. The graph of y = cos x is the graph of y = sin x shifted, or translated, units to the left. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. An open interval like (2, 5) excludes the endpoints. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. In this module you will be asked to calculate the sample size for 6 situations. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. Local Extreme Values of a Function Let c be an interior point of the domain of the. (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. 20 at x = –0. By using this website, you agree to our Cookie Policy. Find the values of f f f at the critical numbers of f f f in (a, b). To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Explain your reasoning. Significance Levels The significance level for a given hypothesis test is a value for which a P-value less than or equal to is considered statistically significant. The triangular distribution, along with the PERT distribution, is also widely used in project management (as an input into PERT and hence critical path method (CPM)) to model events which take place within an interval defined by a minimum and maximum value. The concern is that a dose given too soon after the previous dose may reduce the response. Pick the largest and smallest. Find the local or absolute minimum or maximum of an equation using a graphing calculator; Determine the intervals on which a function is increasing, decreasing, or constant using a graphing calculator (for precalculus) Determine an appropriate viewing rectangle for the graph of an equation; Match an equation to its graph; Graph an equation on. Finding the absolute max and min is a snap. (a) When is the object at rest? (b) Evaluate 6 1 ∫ vt dt(). An absolute minimum occurs at c if for all x in the domain of f. This is achieved with a minimum of manual intervention. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Please answer the following questions about the function Instructions: If you are asked to. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Final the absolute maximum and minimum values on the given interval. We usually distinguish between local and global (or absolute) extreme values. Find the extreme values of f on the boundary of D. 23] is -80 at x = 20 and maximum value of f(x) in interval [0,23] is at zero which is 0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. Fold Unfold. Find the maximum / minimum absolute values with Formulas. f (1) = −5 f ( 1) = - 5. Here is the code to do that. Find the absolute maximum and absolute minimum values of f on the given interval. org are unblocked. Enter DNE if the absolute maximum or minimum does not exist. 388360 # Get 3 integers from 0 to 100 # Use max=101 because it will. Finding Extrema on a closed interval: 1. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. You have 3 solutions: x=0, x=1, and this one. )Given the function 𝑓(𝑥= 𝑥2+ 𝑥+ , chose values for a, b, and c in that could work for the graph shown. If we break down the formula we can see why it gets its strange name. Before Using this Calculator. These points are sometimes referred to as max, min, extreme values, or extrema. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. Answer: First, find the critical points by finding where the derivative equals zero: f0(x) = (x2 +4)(2x)−(x2 −4. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. The generic word for minimum or maximum is extremum. An absolute minimum is the lowest y value or output value a. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. 80 at x = –1. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. Find the maximum / minimum absolute values with Formulas. a local (relative) maximum 6. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. The “V” = volume per time. # The text file for eval. Where does it flatten out? Where the slope is zero. There is 95% confidence that the constructed interval includes the population mean. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. It then evaluates the function at these extrema and at the endpoints of the interval, and determines the minimum over all these values. So, f(b) is a relative maximum of f. Recommended and Minimum Ages and Intervals Between Doses of Routinely Recommended Vaccines1,2,3,4 Vaccine and dose number minimum interval between doses is equal to the greatest interval of any of the individual components. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. So the absolute max value is 19 and the absolute min value is 1. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Absolute Maximum and Absolute Minimum This page is intended to be a part of the Real Analysis section of Math Online. We do not know that a function necessarily has a maximum value over an open interval. In this module you will be asked to calculate the sample size for 6 situations. These values correspond to the probability of observing such an extreme value by chance. Solution for Find the absolute maximum and minimum of the function f (x) = x³ - x+2 on the interval [0, 3] %D. Find the absolute maximum and absolute minimum values of f on the given interval. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For instance, in the example at. f (c) is called the global (absolute) maximum value. Thread starter denbal87; Start date Nov 4, 2014; D. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Calculate Field honors the Transfer Field Domain Descriptions environment. Explain your reasoning. Enter the equation in the Y= section for Y1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. ) On the other hand, it is possible to see directly that. How to calculate a confidence interval? First, you need to calculate the mean of your sample set. An absolute minimum occurs at c if for all x in the domain of f. An absolute maximum is the highest y value or output value a graph has over a specific interval. 140 of 155. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Let this index be 'max_index', return max_index + min. Infinite Discontinuities: both one-sided limits are infinite. This lesson will focus on the maximum and minimum points. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. (1 point) Let g(s) = i on the interval [0, 1. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. (𝑓𝑥)=5 Two solutions. Pick the largest and smallest. This corresponds to zero Kelvin, or minus 273. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). Calculate the range for your confidence statistics. Audio dithering. This is defined everywhere and is zero at $\ds x=\pm \sqrt{3}/3$. Looking first at $\ds x=\sqrt{3}/3$, we see that $\ds f(\sqrt{3}/3)=-2\sqrt{3}/9$. The maximum will occur at the highest f (x) f ( x) value and the minimum will occur at the lowest f (x) f ( x) value. The sample maximum and minimum provide a non-parametric prediction interval: in a sample from a population, or more generally an exchangeable sequence of random variables, each observation is equally likely to be the maximum or minimum. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. f x x3 2 on 3,1 > @ 15. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. Time needed: 10 minutes. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. On a closed interval these points are referred to as absolute or global minimum/maximum points. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. By practicing these kinds of problems you can understand this topic clearly. Thus if one has a sample {, …,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. Calculus Refresher by Paul Garrett. Relative Minimum - The lowest point on an interval of a curve. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. The maximum acceleration attained on the interval 03ddt by the particle whose velocity is given by v t t t t( ) 3 12 4 M 32 is A)9 B)12 C)14 D)21 E)40 3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. Local Extreme Values of a Function Let c be an interior point of the domain of the. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. It can be used as a worksheet function (WS) in Excel. A continuous function f(x) on a closed and bounded interval [a,b] has both an absolute min-imum and an absolute maximum on the interval. Locate the maximum or minimum points by using the TI-83 calculator under and the “3. (c) Find all intervals on which the graph of f is concave up and also has positive slope. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. Q: Determine the absolute maximum and minimum values of the function on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. 5] The calculator performs interval arithmetic operations and computers interval version of mathematical functions. Finding Minimums and Maximums. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. By using this website, you agree to our Cookie Policy. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. a local (relative) minimum 5. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. absolute minimum value at x = 2 is -48. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. f(x) = 4x^3 – 6x^2 – 144x + 9, [-4, 5] 13. The student familiar with the sum formula can easily prove that. (c) For any. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. These are the examples in the topic increasing and decreasing intervals. Maximum intervals — they don't exist. Let f be a function defined and. It is a greatest value in a set of points but not highest when compared to all values in a set. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Local Extreme Values of a Function Let c be an interior point of the domain of the. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. Absolute minimum/maximum _____ d. Before going to class, some students have found it helpful to print out Purplemath's math lesson for that day's topic. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Generating random numbers Problem. Use the Stefan Boltzmann relationship (I = σT 4 where σ = 5. minimum" or "4. Find the maximum / minimum absolute values with Formulas. in some open interval containing c. The local maximum and minimum are the lowest values of a function given a certain range. A closed interval like [2, 5] includes the endpoints 2 and 5. B) No absolute extrema. (To make the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. Finding the absolute max and min is a snap. Let this index be ‘max_index’, return max_index + min. checkpoint_interval # Where to write out summaries. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Let Purplemath help you always be prepared! Go to the lessons!. sin(x * y) I have an interval for x [-1, 1] and y [-1, 1]. 487] Calculating confidence intervals: Calculating a confidence interval involves determining the sample mean, X̄, and the population standard deviation, σ, if possible. summary_interval = FLAGS. The maximum will occur at the highest value and the minimum will occur at the lowest value. A point at which a function attains its maximum value among all points where it is defined is called a global (or absolute) maximum. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. f x x x 32 3 on 3,1> @ 12. Similar topics can also be found in the Calculus section of the site. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. Let this index be ‘max_index’, return max_index + min. Therefore, f achieves its absolute minimum of −14 at x = −1 and its absolute maximum of 6 at both x = 1 and x = 4. 80 at x = –1. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. 14 If the first dose of recombinant zoster vaccine (Shingrix) is administered to someone 18-49 years of age, the dose does not need to be repeated. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. 1 absolute maximum or minimum. f (1) = −5 f ( 1) = - 5. Question 203087: Find the absolute maximum and absolute minimum values of the function below. In this module you will be asked to calculate the sample size for 6 situations. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). 2 Maximum and Minimum on an Interval. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. The absolute minimum of the function f(x) = x2-9 on the interval - 4 Sxs 3 has a value of (Simplify your answer. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Any global maximum or minimum must of course be a local maximum or minimum. (1 point) Let g(s) = i on the interval [0, 1. Absolute & Local Minimum and Maximum Values - Relative Extrema, Critical Numbers / Points Calculus - Duration: 1:10:05. The student familiar with the sum formula can easily prove that. so minimum value of f(x) in interval [0. pow(y, 2)) * -1 return math. Extreme values ©2010 Iulia & Teodoru Gugoiu - Page 1 of 2 3. For instance, in the example at. How to use absolute minimum in a sentence. the absolute minimum age of 50 years when evaluating records retrospectively. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. (a) Interval : [—8, 0]. The smallest y -value is the absolute minimum and the largest y -value is the. 1 absolute maximum or minimum. Fermat's Theorem. Generating random numbers Problem. Help with finding absolute max/min values for a function. in some open interval containing c. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. Let f be a function defined and. $f(x) = x^5 - x^3 + 2$, $-1 \leqslant x \leqslant 1$. 1 absolute maximum or minimum. The absolute max occurs at S = The absolute min occurs at S =. (a) fxc 0 at x 3, 1, 4 f c changes from positive to negative at 3 and 4. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. have both an absolute maximum and an absolute minimum. Find the extreme values of f on the boundary of D. An absolute extremum is an absolute maximum or an absoute minimum, and absolute extrema are absolute maximum and absolute minimum. Zoom in on the interval [-2,2] using the x-axis. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. minimum” or “4. Find the absolute maximum and absolute minimum values of f on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Find the absolute maximum and. By using this website, you agree to our Cookie Policy. denbal87 New member. | 2020-08-03T15:15:48 | {
"domain": "rciretedimpresa.it",
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html",
"openwebmath_score": 0.6303600668907166,
"openwebmath_perplexity": 373.11161112116935,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9890130576932457,
"lm_q2_score": 0.86153820232079,
"lm_q1q2_score": 0.8520725317968266
} |
http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1 | # How to round 0.4999… ? Is it 0 or 1?
If you want to round the repeating decimal 0.4999... to a whole number what is the right answer? Is it 0 or 1?
On the one hand you only look at the first digit when you round numbers which in this case is 4 so the answer should be 0.
On the other hand 0.4999... is only another representation for 0.5 which makes the result 1.
My question
Which of these rules wins? (My gut feeling is that the most consistent result should be 1)
Edit
Just for clarification: What is meant by rounding here is replacing a fractional decimal number by one with fewer digits with the "Round half up"-Tie-breaking method.
-
It appears that the rule is inconsistent. – Isaac Solomon Oct 21 '12 at 9:04
Doesn't that depend on what type of rounding you are using? If you're using Bankers Rounding, you would round towards 0. – Fake Name Oct 21 '12 at 12:22
@FakeName: Fair enough. What is obviously meant here is replacing a fractional decimal number by one with fewer digits with the "Round half up"-Tie-breaking method: en.wikipedia.org/wiki/Rounding – vonjd Oct 21 '12 at 13:50
This question should NOT be closed; it has a definitive answer - with references "out there" to support it. At the point when this was closed, the issue on the table (discussion!) was if 0.499... was 0.5 or not: reference is en.wikipedia.org/wiki/0.999... – Richard Sitze Oct 22 '12 at 0:08
I agree with Richard. Given that we have a method for rounding midpoint values, and the author of the question has supplied a method, it follows that this question has an answer. – Doug Spoonwood Oct 22 '12 at 2:52
It's $1$, because $0.49\ldots$ is the same as $0.5$. If rounding is to be well-defined, it can't map one real number to two integers, so whatever it maps $0.49\ldots$ to, it better maps it to the same integer as $0.5$. You could round both to $0$, of course, but that wouldn't then be the way we usually round.
What this shows you is that rounding doesn't commute with limits, i.e. there's a difference between find the limit of a sequence and then rounding, and rounding first and then finding the limit. As you correctly observed, all values $a_n = 0.4\underbrace{9\ldots 9}_{n\text{ times}}$ are rounded down to zero. Thus, $$\lim_{n\to\infty} \text{round }(a_n) = 0$$ On the other hand, $\lim_{n\to\infty} a_n = 0.5$, and thus $$\text{round }\left(\lim_{n\to\infty} a_n\right) = 1$$
There's another word for functions which don't commute with limit - they're called non-continuous. So what you have discovered is simply that rounding is not a continuous function.
-
Perhaps I'm nitpicking, but is the result of rounding the limit perhaps $1$ instead? – pimvdb Oct 21 '12 at 10:05
@pimvdb: I went ahead and edited the answer, since it's obviously what was intended. (And so many people seem to agree.) – ShreevatsaR Oct 21 '12 at 14:30
+1 especially for pointing out the connection with continuity. – Neal Oct 21 '12 at 16:00
@pimvdb Indeed the limit should be $1$, and I wouldn't call pointing out blatant errors nitpicking ;-) Thanks for noticing, and thanks to ShreevatsaR for fixing this! – fgp Oct 21 '12 at 19:42
@jmoreno - but it is, in fact the same as 0.5: en.wikipedia.org/wiki/0.999... – Richard Sitze Oct 22 '12 at 0:01
For $0\le x\le 1$, we round $x$ to $1$ if $x\ge \frac12$ and to $0$ if $x<\frac12$ (though there are many conventions, see e.g. Wikipedia on rounding; the section "Table-maker's dilemma" a bit further down may also be interesting). Since $0.4\bar9=\frac12$, we should round to $1$. Another way of looking at this is that we always consider only the standard decimal expansion (i.e. we prefer $\bar0$ over $\bar 9$), and we are allowed to treat the first decimal as $4$ only if we know that it cannot turn out as $5$ "later". Thus if an inexact measurement gives us that $0.495\le x\le0.5$, we cannot say definitely, what $\operatorname{round}(x)$ should be (we could if the measurement resulted in $0.495<x<0.5$). This is not different from the fact that we cannot say definitely what $\operatorname{round}(x)$ should be if our measuremen merely says that $0.4997<x<0.5003$.
-
You're right. Since $.4\bar{9}=.5,$ if you want your rounding function to be well-defined you'll have to require an exception: round based on the first digit after the one you're rounding to, unless it's a $4$ followed by infinitely many $9$s.
-
It is not obvious that 0.5 should be rounded to 1. Obviously 0.49999... should be treated the same because it is the same number. Wikipedia rounding article
-
$1 - 0.5 =0.5 \quad$ and $\quad 1 - 0.4999\ldots = 0.500\ldots = 0.5 \quad$ so $\quad 0.5 = 0.4\overline9$
Well, the above line says the proof / intuition. From our information... $0.5$ well rounds to $1$.
-
Are you sure that $1-0.4999\ldots\ne 0.500\ldots1$? – MJD Nov 18 '12 at 16:46
@MJD: Not sure if serious... – Clive Newstead Nov 18 '12 at 16:47
@MJD: In 0.500...1, the zeroes never end, so the $1$ never comes ;) – Parth Kohli Nov 18 '12 at 16:52
I find very convenient to define rounding as follows: $$round(x) = \left \lfloor x + \frac{1}{2}\right \rfloor$$
Considering this rule, what follows is understanding the context. If you are working with a computer, you have only a fixed number of decimal positions to work with, in which case, if x=0.4999999999999999 (16 decimal positions), then
round(x)=floor(0.4999999999999999 + 0.5)=
=floor(0.9999999999999999)=
=0
Of course, if you are talking about a real number, then 0.4999... equals 0.5, in which case $round(x)=1$.
So, the real question here is: are you working with a fixed number of decimal positions or not? If the answer is 'yes', then definitely round(x)=0. If the answer is 'no', then you are talking about a number which is equal to 0.5, and $round(x)=1$
-
@fgp answered this well but there is another aspect that I don't think has been touched on in any of the answers. What is the type of thing that we should consider as an input to the process of rounding? If it is a real number, we are unable to apply the rule without a specific digit sequence representing that number. If instead we interpret rounding as an operation on a possibly infinite digit sequence, we can now apply the rule, but its result has no meaning for general real numbers. One way out which has been discussed is to make a bijection between digit sequences and reals by disallowing digit sequences ending in $999...$, and this view effectively unasks the question by taking the position that $0.4999...$ is not a valid representation of a real number, but I think this is an overly burdensome technicality.
I think an easier way to look at it is just to say that rounding is an operation that applies to finite digit sequences and not to real numbers in general nor to infinite digit sequences. When we say something like "$\pi$ rounds down to $3$" it is a lazy way of saying that all sufficiently precise approximations do. But when we say "$0.5$ rounds up to $1$", exactly that is meant, because we don't mean to suggest that $0.4 + \sum_{k=2}^{n}{9 \cdot 10^{-k}}$ will round up to $1$ for sufficiently large $n$. And without context, if it is said about a real number $x$ that "$x$ rounds up to $1$" then that can be read as "$x$ has a finite decimal representation which rounds up to $1$, or $x$ has no finite decimal representation and all sufficiently precise finite decimal approximations to $x$ round up to $1$". In this way we can meaningfully discuss rounding real numbers in terms of rounding finite digit sequences, and without explicitly addressing the problem of sequences ending in $999...$.
-
If you consistently use the round-up half method you don't always look at the first digit of the numeral when rounding numbers. You go by the method. The method allows that if we have anything else than a half-way value, then you can round by the rule "if (working in base 10) the first digit of the numeral after the decimal point belongs to {0, 1, 2, 3, 4}, then round down, if the first digit of the decimal point belongs to {5, 6, 7, 8, 9}, then round up." But, that rule does not say anything about half-way values. Going by q=FLOOR(y + .5) we can reason as follows: q=FLOOR(.499... + 5)=FLOOR(.999...)=FLOOR(1)=1, since .999...=1. So, that method yields that .499... rounds to 1.
-
It is the same as 0.5 by definition of decimal numbering.
-
And rounding takes by definition the first digit which is 4 in this case. So this is exactly where the two definitions collide! – vonjd Oct 21 '12 at 18:36
@vonjd I disagree that rounding by definition takes the first digit and then rounds. I ask you to try and round 1/3 without converting it into a decimal. Do you not round 1/3 down to 0, since it comes as clear that 1/3 lies closer to 0 than to 1? Or I suggest you round 12/11 to the nearest whole number. Or round 155/66 to the nearest whole number. Finding the multiples of a denominator d will help here, and what half of d equals. If the denominator equals 7, then if the numerator of "the remainder" equals 1, 2, or 3 we round down, if the numerator equals 4, 5, or 6 we round up. – Doug Spoonwood Oct 21 '12 at 19:17
E. G. Say we have 13/7. 13/7=7/7+6/7, correct? So, 13/7 has "remainder" of 6/7, or in other words it has a fractional part (search that term on Wikipedia) of 6/7. Thus, 13/7 rounds up to 14/7 (6/7 lies closer to 7/7 than to 0/7). You could also try and rounding numbers in binary, trinary, or any n-ary base system. E. G. round 1.34523 in base 7. In base 7, 1.4142 rounds up to 2. So, I simply do not see how the digit makes the difference here. More goes on than that! – Doug Spoonwood Oct 21 '12 at 19:23
@vonjd this is an exception. rounding takes first digit EXCEPT for values like this, where fraction is written in limitless form. – Suzan Cioc Oct 21 '12 at 19:25
This is just an obersvation that I was missing in the many answers already given.
While there is absolutely no doubt that the mathematical function $\Bbb R\to\Bbb Z$ defined by $x\mapsto\lfloor x+\frac12\rfloor$ when applied to the real (in fact rational) number represented by the repetitive infinite decimal sequence $0.4\overline9$ takes as value the integer$~1$, saying that this is what happens when rounding the repeating decimal $0.4999\ldots$ is misleading. That is because "rounding" suggests an operation done during actual computation. This ignores the often overlooked fact that it is impossible to do arithmetic computations with infinite decimals. This does not mean of course that there isn't a class of infinite decimals with which computations are possible; for instance if all decimals are repeating, then one can replace them by rational numbers and compute with those, which is definitely possible. However the misconception is that, because we can do arithmetic on numbers with any fixed number of decimal places and the procedures for doing so are always basically the same, therefore one can apply these same procedures to infinite decimals. However just looking at these procedures will reveal that this is plainly false: all of these procedures need at some point (often right at the beginning) to locate the final decimal position, and for an infinite decimal such a position simply does not exist. This means that in practice whenever one needs rounding decimal numbers as a computational operation, then those decimals are necessarily finite, and the number $0.4999\ldots$ simply cannot occur.
In fact there is no effective system for computing with (general) true real numbers at all, for the simple reason that the majority of the real numbers cannot be represented in any way (a concrete representation having to be, by the nature of things in the real world, finite). One can however restrict to the (countable) subset of real numbers that do allow some kind of representation (which is less brutal that restricting to some arbitrarily chosen finite subset of numbers declared representable, as is done in numeric computation), and then arithmetic and other kinds of computation do become possible, and indeed can be done without having to introduce approximations. This has been studied in constructive mathematics, but it turns out that for this purpose representing real numbers by infinite decimals, which might be produced on demand by a finite algorithm, is not a viable option. Even just finding the first decimal of the sum of two numbers so given by infinite decimals cannot be assured by an algorithm; this is in fact closely related to issues invoked in this question and the related $0.9999\ldots$ problematic. This in contrast to computations with formal power series with integer (or rational) coefficients: these suffer from the same "most elements cannot be represented" problem as the real numbers, but if one sticks to series whose coefficients can be computed by an algorithm, then all arithmetic (and many other) operations on them can be performed algorithmically without problem.
A more workable way to represent real numbers is by a sequence of definite approximations (rational numbers with a given maximal error), with some kind of guarantee that the error tends to $0$. But even then there are sacrifices one has to make: it is impossible to computations with discontinuous functions (like the floor or round-to-integer operations): in the constructive theory of real numbers there only exist continuous functions. So all of this is rather far removed from what one does in numeric computation, where real numbers are necessarily approximated, the number of decimals finite, and the question of how to round $0.4\overline9$ moot; the rule for rounding of looking at the first digit to be discarded in the rounding is perfectly valid in practice.
-
If we round c to d, then c approximately equals d. If d also consists of an integer, then c lies closer to d than any other integer i. In other words for all integers i such that i does not equal d, ABS(ci-)>ABS(cd-), where ABS(ci-) indicates the number by which c and i differ by (the absolute value of the difference of c an i). This implies that we can use the distance metric between real numbers to see how rounding works as follows.
Consider the set S=={[l, lu+2/), (lu+2/, u]} where ul-=1. In other words, l and u differ by 1, and we consider the set of all numbers between x and y except for the midpoint m==lu+2/ between them. We can use the metric function D(x,y)=|xy-|=ABS(xy-), where x belongs to S and y belongs to {l, u}, to determine how to round all members of S by following rule:
If D(x, l)>D(x, u), then round to u. If D(x, u)>D(x, l), then round to l.
Since .5=.49999..., it follows that D(.49999..., 1)=D(.49999...., 0)=.4999...=.5. Thus, the distance function D(x, y)=ABS(xy-) does not give us any guide as to how to round here.
Consequently, we have several different choices that we can make which will not go against our intuition of rounding as follows:
1. Choose to round .49999... to 1 (maybe you value larger numbers in general),
2. Choose to round .49999... to 0 (maybe you value smaller numbers in general),
3. Choose to allow .49999... to get rounded to both as one pleases or sees fit at particular points in time, or
4. Choose to round .49999... to neither 0 nor 1 (maybe you don't want to make a choice here).
None of those possibilities win or lose until we have a standard to measure a winner or loser for the rules. There also doesn't exist a "right answer" here unless we have a means to determine what properties a "right answer" has in this context. I suspect that many people would resist 3. with the possibility that one could round .4999... to 1 at one time, and then round .4999... to 0 at another as one pleases, thus indicating arbitrariness. But, it does not follow that such comes as an improper procedure unless one first rates such "arbitrariness" as bad, and something so utterly awful and terrible that we or others must never, ever engage in such "arbitrariness". Perhaps many people would see 4. as "evasive", but unless there exists a logical basis to picking either 0 or 1, anyone accused of "evasiveness" by picking 4. could very well respond to such a charge, that anyone who sees choosing 4. as indicating "evasiveness", insists dogmatically that all such questions absolutely must have answers, and/or that the people claiming "evasiveness" have copped out of making choices on the basis of logic as much as possible.
So, what sort of choice do you want to make?
-
As a programmer, 0.4999... is closer to 0 than it is to 1, so I would always round down to 0.
Keep it simple! Rounding is always to the closest whole number. You only need to apply special cases when there are two closest whole numbers.
But there are situations where this would be the wrong choice, so you need to do some research every time you apply rounding function to a value.
For example, the tax department in the country I live in does not always acknowledge cents. Under legislation, there are times when the only valid rounding function is floor.
-
I think we all agree that $0.4999$ should round down to $0$. The question is whether $.499999999......$ should (with infinitely many $9$s). – Jason DeVito Oct 21 '12 at 18:16
How close $0.4999...$ is to $0$ or $1$ is independent on whether or not you are a programmer. In fact, it is the same distance from $0$ and $1$. If the sequence of digits were to terminate after finitely many (e.g. in computing) then fine, but here we have infinite precision and the number is exactly equal to $0.5$. – Clive Newstead Oct 21 '12 at 18:20
@AbhiBeckert: The case in point is that it is indeed equal to 0.5, as is 0.999... equal to 1: en.wikipedia.org/wiki/0.999... – vonjd Oct 21 '12 at 18:32
Dear Abhi, I have downvoted your answer, since it it seems to be premised on a confusion with regard to the fact that $0.4999\cdots$ (infinitely many $9$'s) is equal to $0.5$. Regards, – Matt E Oct 21 '12 at 19:56
@AbhiBeckert: If you accept that $0.4999...$ is a real number, and you also claim it's less than $0.5$, then (1) you should be able to say what is the distance between them (for any two real numbers $x<y$, their distance is another real number $y-x$: and no, "infinitesimal" is not a number), and (2) you should be able to exhibit a number that lies between them (for any $x<y$, there exist infinitely many real numbers between, e.g. $x<(x+y)/2<y$). The fact that there is no number between $0.4999...$ and $0.5$ should convince you that they are two representations of exactly the same number. – ShreevatsaR Oct 23 '12 at 5:16 | 2014-07-11T05:57:30 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1",
"openwebmath_score": 0.8157265186309814,
"openwebmath_perplexity": 432.2086184790631,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.974821163419856,
"lm_q2_score": 0.8740772482857833,
"lm_q1q2_score": 0.8520690000927736
} |
https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036 | # How to translate mathematical intuition into a rigorous proof?
I've seen a lot of questions about how to develop mathematical intuition, but often I have the opposite problem.
Several times I have run into a situation where I want to solve a math problem, and I play around with it until my intuition reaches the point where I have a good idea of how the complete proof might be structured. But then when it comes to actually write out the proof, I have trouble translating this intuition into a rigorous proof. I want to give the following example, from this PDF of Putnam training problems.
1.13. Prove that for every $n\ge 2$, the expansion of $(1+x+x^2)^n$ contains at least one even coefficient.
When first thinking about this problem, I wanted to look at the polynomials $\pmod{2}$ so that "even coefficient" is simplified to "zero coefficient."
Then from doing a few computations, I notice a pattern.
\begin{align*} (1+x+x^2)^2&=(1+x+x^2)+x(1+x+x^2)+x^2(1+x+x^2)\\ &=1+x+x^2\\ &+x+x^2+x^3\\ &+x^2+x^3+x^4\\ &=1+x^2+x^4 \end{align*}
I notice that multiplying a polynomial by $(1+x+x^2)$ is like adding that polynomial with itself three times, each shifted over. The $x$ shifts it over by one place and the $x^2$ shifts it over by two places.
Continuing this pattern, we can get the coefficients of $(1+x+x^2)^3$ as follows:
\begin{array}{ccccccc} 1&0&1&0&1&&\\ &1&0&1&0&1&\\ &&1&0&1&0&1\\ \hline 1&1&0&1&0&1&1 \end{array}
So $(1+x+x^2)^3\equiv 1+x+x^3+x^5+x^6 \pmod{2}$
Let's make a triangle of a few more results:
\begin{array}{cccccccccccc} 0:&&&&&&1&&&&&\\ 1:&&&&&1&1&1&&&&\\ 2:&&&&1&0&1&0&1&&&\\ 3:&&&1&1&0&1&0&1&1&&\\ 4:&&1&0&0&0&1&0&0&0&1&\\ 5:&1&1&1&0&1&1&1&0&1&1&1\\ \end{array}
We see that the structure of our problem is essentially the same as an elementary cellular automaton (specifically, rule 150). I will come back to this later.
I notice another peculiar pattern with the $1$st, $2$nd, and $4$th rows: There are only $1$s on the ends and in the center. Perhaps this pattern continues for all powers of $2$. And it does, which is not hard to prove with induction.
Claim: $\forall n\ge 0,\ (1+x+x^2)^{2^n}\equiv 1+x^{2^n}+x^{2^{n+1}}\pmod{2}$
Base case: $(1+x+x^2)^{2^0}\equiv 1+x+x^2\equiv 1+x^{2^0}+x^{2^1}\pmod{2}$
Inductive Step:
\begin{align*} (1+x+x^2)^{2^n}&\equiv ((1+x+x^2)^{2^{n-1}})^2\\ &\equiv (1+x^{2^{n-1}}+x^{2^n})^2\\ &\equiv 1+x^{2^n}+x^{2^{n+1}}\pmod{2} \end{align*}
This is where the intuition comes in that is hard for me to express rigorously. On one of the $2^n$th rows, look at the $0$ equidistant from the leftmost $1$ and the $1$ in the center. If the rightmost $1$ wasn't there, then that $0$ would stay a $0$ forever because of symmetry: any effect from the left is cancelled by an effect from the right.
However, the rightmost $1$ does exist, so it will eventually change that $0$. But since effects are local in this automaton, it will require over $2^n$ more rows until the rightmost $1$ affects the $0$ we are looking at (there are over $2^n$ spaces between them). In other words, we have shown that the $0$ will stay a $0$ for all rows up to the $2^{n+1}$th row. Applying this logic on each power of two row, we show that there is always a zero coefficient from row $2$ onwards.
Is there a more rigorous way I can express those last two paragraphs? And in general, what advice do you have for translating intuition to a rigorous proof?
• In your example you changed the question. Originally it asked for a $0$ in every row. Your version only talks about rows of the form $2^n$, but you show a very special form for those rows. Your induction is fine and proves what you wanted to prove, but it is not the original question. It might be a step along the way I can imagine (but I am speculating) that you can show a partial row starting with $1$, a bunch of $0$s and a $1$ will always have a $0$ somewhere. If you can do that you are done in conjunction with what you have done. – Ross Millikan Jun 25 '18 at 4:15
• @RossMillikan I did not change the question in any way. The observation I made about the $2^n$ rows was a stepping stone in my proof. Look at the last three paragraphs. – Riley Jun 25 '18 at 4:18
• You said you would come back to rule 150 but I don't think you did – Mark Jun 25 '18 at 4:49
• @Mark I meant that I will come back to framing the problem in terms of cellular automata, which I did when discussing the "effects" of the $1$s, and the locality of the automaton. – Riley Jun 25 '18 at 4:51
• [Just a comment as the question is about the automaton-intuition being turned into a proof.] Another way of thinking about this expression is, the coefficient of $x^m$ in the expansion shows the number of ways to achieve a total of $m$ by rolling a die with sides $0, 1, 2$ a total of $n$ times. To see why, try making a three by three times table model to see the result of squaring the expression; then, erase the base of $x$. – Benjamin Dickman Jun 25 '18 at 5:17
You have proved that these binary strings are symmetric so it suffices to prove the result for only the "half-strings." Imagine cutting your triangle down the middle. Let $x_n$ be the $n^{th}$ bit string, which is length $n+1$.
We have $x_0 = 1$, $x_1 = 11$, $x_2 = 101$, etc.
Now you can address the $j^{th}$ bit inside $x_n$ as $x_{n,j}$.
This addressing system will make it easier to talk more concretely about the action of the automaton.
You can pin down a definition of "locality of action" by saying that there is some function $f(a, b, c)$ such that $\forall n, j: x_{n,j} = f(x_{n-1, j-1}, x_{n-1,j}, x_{n-1,j+1})$ (with some additional nuisance conditions about the boundary). This means that the $j^{th}$ bit in the current string can be computed by neary-by bits in the previous string. Here $f$ represents the action of this rule 150 you mention.
Now use $f$ again to show $x_{n,j} = f(f(x_{n-2, j-2}, x_{n-2,j-1}, x_{n-2, j}), f(x_{n-2, j-1}, x_{n-2,j}, x_{n-2, j+1}), f(x_{n-2, j}, x_{n-2,j+1}, x_{n-2, j+2}))$
Now you can see when we repeat this procedure $k$ times we will end up with a trinary abstract syntax tree depth $k$. To avoid having to talk about the exact structure of the expression at the $k^{th}$ level, we can reason about this substitution process purely formally as a process operating on the (countably infinite) set of symbols $\{ \underline{f}, \underline{(}, \underline{)} \} \cup \{ \underline{x_{i,j}} \}_{(i,j) \in \mathbb{N}^2}$ (here the commas are not meant to be part of the symbol set, they are just there to separate symbols from each other). You can prove by induction, that for any $k$, that $x_{n,j}$ is computed by an expression using the symbols $\{ \underline{f}, \underline{(}, \underline{)} \} \cup \{\underline{ x_{n-k, j-k}},\underline{ x_{n-k, j-k+1}}, …,\underline{ x_{n-k, j+k-1}},\underline{ x_{n-k, j+k} }\}$ (the $2k+1$ bits in the $(n-k)^{th}$ string centered on the $j^{th}$ bit) intermixed with function applications of $f$.
Fix $n$ and let $k_n$ be the distance to the previous power of $2$. (So if $n=13$, $k_n=5$ since the previous power of $2$ is $8$). And define $j_n \equiv (n-k_n)/2$. This is the index of the central bit of the $(n-k_n)^{th}$ bit string.
You have demonstrated that those bits at indices $\{ (n-k_n, m) \}_{1 \le m \lt n }$ are all zero since $n-k_n$ is a power of $2$. So any well-formed expression using only the symbols $\{ \underline {f}, \underline{(}, \underline{)}\} \cup \{ \underline{x_{n-k_n, j_n-k_n}}, \underline{x_{n-k_n, j_n-k_n+1}}, …,\underline{x_{n-k_n, j_n+k_n-1}}, \underline{x_{n-k_n, j_n+k_n}}\}$ will evaluate to zero by $f(0,0,0) = 0$, and in particular the expression representing $x_{n,j_n}$ (created from the $k_n$ iterations of the formal symbol substitution procedure) evaluates to $0$.
So as $n$ increments, $x_{n,j_n}$ traces out the path of the central bits you were referring to, all of which are zero.
General thoughts about formalization: There are certain technical devices which are work horses of the formalization process. Two of them I have used in this proof:
• indexing time and space, with a numeric coordinate system
• encoding a time evolving system as function application
A third device which I used is not as common, and comes from computer science and logic:
Something that I have been realizing is that these devices are not obvious (at least to me). Humanity was around for a long time before the Cartesian coordinate system was discovered and used to formalize geometrical intuition. It took even longer to develop the idea that almost every mathematical object could be encoded as an intricate tower of sets (ZFC).
I guess for me it's a matter of building up a library of these devices and binding them to the correct intuitions by repeated use.
In certain areas like computer programming, there are very few tools for converting intuition into proof. There were a couple of very good devices created like
• loop invariants
• Hoare logic
but multithreaded and heap-allocating programs are still an active area of research.
• By the way, I have not verified the claim about the sequence actually evolving as the cellular automata $f$ but I'm trusting it's true. – Mark Jun 25 '18 at 7:01
• I think your definition of $j$ doesn't quite match the way you use it. You're using it as if $j$ represents the $x$ coordinate in the triangle rather than the $j$th bit. Also I think you need to consider whole strings instead of half strings for the automation to work. And I think you need to define $x_{n,j}=0$ when $j$ is out of bounds. Otherwise a great answer. – Riley Jun 25 '18 at 12:50
• Yeah, for it to make sense, we need to start counting things at $0$ as we do in computer programs. $x_{n,j} \cong x[n][j]$. As for the whole strings vs half strings, maybe it will be convenient to use negative indices to talk about the left half of the triangle. – Mark Jun 25 '18 at 14:48
Here is one answer to the question. If $$\, n\ge 2\,$$ then the first $$4$$ entries in each row, namely, $$\, 1 0 1 0, 1 1 0 1, 1 0 0 0, 1 1 1 0, \dots, \,$$ repeat in a period of $$4$$. There is a $$0$$ in each of these first $$4$$ entries. This was not the proof you wanted because it is too simple.
Your proof is essentially the following. You noticed something about a $$0$$ in row $$\, 2^n \,$$ and that this $$0$$ continues up to the row $$\, 2^{n+1}. \,$$ We formalize that observation by working with polynomials over the field with two elements. Define the Laurent polynomials $$\, y(n) := x^n + x^{-n} \,$$ and since the characteristic of the field is $$2$$, $$\, y(0) = 1 + 1 = 0. \,$$ Using simple algebra of exponents we get that $$\, y(n)y(m) = y(n+m) + y(n-m), \,$$ and the special cases $$\, y(n)^2 = y(2n) \,$$ and $$\, y(2^n) = y(1)^{2^n}.$$
Define the row polynomials $$\, r(m) := (1 + y(1))^m. \,$$ The general row is $$\, r(m) = r(2^n+k) \,$$ where $$\, 0\le k < 2^n. \,$$ Notice that $$\, r(2^n) = 1 + y(1)^{2^n} = 1 + y(2^n). \,$$ Also $$\, r(m) = 1 + \sum_{i=1}^m c_i y(i), \,$$ where $$\, c_i \,$$ is the coefficient of $$\, y(i). \,$$ You noticed that $$\, c_{2^{n-1}} = 0 \,$$ but how to prove it? We calculate that $$\, r(m) = (1 \!+\! y(2^n)) r(k) = (1 \!+\! y(2^n))(1 \!+\! cy(2^{n-1}) \!+\! t) = (1 \!+\! y(2^n))(1 \!+\! t) \!+\! c y(3\,2^{n-1}) \,$$ where $$\, t \,$$ is all the terms with $$\, y(i), \,$$ and $$\, i\ne 2^{n-1}, \,i < 2^n. \,$$ But $$\, y(2^n)y(i) = y(2^n-i) \!+\! y(2^n+i) \,$$ and since $$\, i \ne 2^{n-1} \,$$ we get no $$\, y(2^{n-1}) \,$$ terms in the product $$\, (1 \!+\! y(2^n))(1 \!+\! t). \,$$ That proves it.
This answers your particular question, but what about intuition versus rigor? I think there are no good answer to that question. You have to always look for any patterns or regularities and simple or edge cases. You have to be lucky and it helps to have a lot of background experience to draw upon.
• I recognize that there is a simpler answer to the question. It is given in the PDF I liked to. I am not asking for the solution, I am asking about how to translate my ideas into a rigorous proof. – Riley Jun 25 '18 at 4:06
• I thought I was translating your ideas combined with ideas of my own into a proof. But intuitions are tricky to express clearly. Maybe you wanted an "automaton theoretic" proof? – Somos Dec 3 '18 at 19:55 | 2019-06-19T00:42:46 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036",
"openwebmath_score": 0.8129568099975586,
"openwebmath_perplexity": 227.65682371870014,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9748211582993981,
"lm_q2_score": 0.8740772466456689,
"lm_q1q2_score": 0.8520689940182796
} |
https://math.stackexchange.com/questions/716005/equivalence-of-inner-products | # Equivalence of inner products?
So, let $\langle \cdot, \cdot \rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ be inner products on a real vector space $V$. Assume that \begin{align} \langle v, v \rangle_{1} = \langle v, v \rangle_{2} \end{align} for any $v \in V$.
Is it true that $\langle v, w \rangle_{1} = \langle v, w \rangle_{2}$ for all $v,w \in V$?
I can only think of two inner products on a real vector space $V$ are the standard dot product and the scaled dot product. This holds for these two since the scaled dot product would require all scalars to be 1. Can anyone think of another inner product on $V$ to test this with, or should I start trying to prove it?
• Take a basis $v_1,\dots ,v_n$ for $V$. Then $\langle v_i,v_i\rangle_1=\langle v_i,v_i\rangle_2$ for all $1\leq i\leq n$ implies that $\langle v,w\rangle_1=\langle v,w\rangle_2$.
– DKal
Mar 17 '14 at 21:09
For every $v,w \in V$ and $c$ real $$\langle v+cw,v+cw\rangle_1=\langle v+cw,v+cw\rangle_2$$ or $$\langle v,v\rangle_1+2c\langle v,w\rangle_1+c^2\langle w,w\rangle_1= \langle v,v\rangle_2+2c\langle v,w\rangle_2+c^2\langle w,w\rangle_2$$ and hence, as $\langle v,v\rangle_1=\langle v,v\rangle_2$ and $\langle w,w\rangle_1=\langle w,w\rangle_2$, then $$\langle v,w\rangle_1=\langle v,w\rangle_2.$$
Even though Yiorgos S. Smyrlis' answer is simpler, I think it is useful having a little different proof, just to add another point of view of the same argument, which is always good :)
Consider the norm $\| \cdot \|_1$ induced by $\langle \cdot, \cdot \rangle_1$ and the norm $\| \cdot \|_2$ induced by $\langle \cdot , \cdot \rangle_2$. By hypothesis and definition of induced norm $$\| v \|_1^2 = \langle v, v \rangle_1= \langle v, v \rangle_2= \| v \|_2^2$$ $\forall v \in V$. So the two norms are the same (the power doesn't make any problem thanks to the positivity of the norm.)
Then we want to do the reverse reasoning, if two product-induced-norms coincide on every element of the space, what can we say about inner products?
It's immediate (and left as little exercise for the reader - just write down the definitions and use bilinearity-) to prove that given a real inner product and its norm the following equivalence is true $$\langle x,y \rangle = \dfrac{1}{4} \left( \|x+y \|^2 - \|x-y\|^2 \right)$$ This formula is called the polarization identity (for more see here)
So by the above identity and the hypothesis we have, $$\langle x,y \rangle_1 = \dfrac{1}{4} \left( \|x+y \|_1^2 - \|x-y\|_1^2 \right) = \dfrac{1}{4} \left( \|x+y \|_2^2 - \|x-y\|_2^2 \right) = \langle x,y \rangle_2$$ $\forall x,y \in V$ and so we are done.
I wanted to stress the depth correlation between an inner product and its induced norm, and show some useful tools. Hope it helps :) | 2021-09-21T09:09:56 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/716005/equivalence-of-inner-products",
"openwebmath_score": 0.9495441317558289,
"openwebmath_perplexity": 145.5758585312135,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9748211597623863,
"lm_q2_score": 0.8740772400852111,
"lm_q1q2_score": 0.8520689889017713
} |
https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects | # How many random walk steps until the path self-intersects?
Take a random walk in the plane from the origin, each step of unit length in a uniformly random direction.
Q. How many steps on average until the path self-intersects?
My simulations suggest ~$$8.95$$ steps.
Red: origin. Top: the $$8$$-th step self-intersects. Bottom: the $$11$$-th step self-intersects. (Not to same scale.)
I suspect this is known in the SAW literature (SAW=Self-Avoiding Walk), but I am not finding this explicit number.
Related: self-avoidance time of random walk.
Added. Here is a histogram of the number of steps to self-intersection.
$$10000$$ random trials.
• This is asking for a uniformly random direction in the plane. By comparison, in 1-d the corresponding number is $(1/2)2 + (1/4)3 + (1/8)4 + \cdots = 3$; in 3-d the corresponding number is $\infty$. – Matt F. Mar 10 '19 at 4:40
• I added a second example. I have thousands of these. :-) – Joseph O'Rourke Mar 10 '19 at 18:56
• One can also ask for the asymptotic as $n$ goes to infinity of the probability that a length $n$ random walk does not self-intersect. The form of this asymptotic might be close to the case of random walks on grids, which if I recall right is conjectured but not known. – Will Sawin Mar 12 '19 at 17:42
• FWIW I ran this on $10^7$ random configurations and got a mean of ~$8.886$ steps. – Chip Hurst Mar 13 '19 at 14:22
• I ran $10^9$ simulations over night and the mean stayed at ~$8.886$. Here's a link of the simulation tallies: wolfr.am/C3YsNZmE. The largest run consisted of 98 lines. – Chip Hurst Mar 14 '19 at 12:16
Here are some computational insights.
I ran $$N = 10^k$$ simulations for $$3 \leq k \leq 12$$. For $$N = 10^9$$ I recorded some explicit instances and recorded which pairs intersected, as opposed to just the lengths. For all other runs, I simply summed the results.
First and foremost here's a summary of the runs:
$$\begin{array}{ |l|l|l|l| } \hline N & \text{mean} & \text{std/sqrt(}N\text{)} & \text{total number of segments} \\ \hline 10^3 & 8.732 & 0.16106 & 8732 \\ 10^4 & 8.781 & 0.0528827 & 87810 \\ 10^5 & 8.86218 & 0.0166654 & 886218 \\ 10^6 & 8.89447 & 0.00524551 & 8894468 \\ 10^7 & 8.88463 & 0.00165934 & 88846265 \\ 10^8 & 8.88595 & 0.000524796 & 888595095 \\ 10^9 & 8.88615 & 0.000165965 & 8886153545 \\ 10^{10} & 8.88631 & - & 88863113226 \\ 10^{11} & 8.88614 & - & 888613831649 \\ 10^{12} & 8.88614 & - & 8886139852418 \\ \hline \end{array}$$
We can apply the central limit theorem to this (sub)sequence of means. Even though $$\text{std/sqrt(}N\text{)}$$ was not calculated for $$N = 10^{12}$$, based on prior terms ~$$0.00000524$$ seems like a sufficient estimate.
From here I think we can be fairly confident that the first few digits of the mean are $$\bf{8.8861}$$. Here are the first few standard deviations ($$68-95-99.7$$ rule):
$$m \pm 1\sigma \rightarrow [8.886135, 8.886145] \\ m \pm 2\sigma \rightarrow [8.886129, 8.886150] \\ m \pm 3\sigma \rightarrow [8.886124, 8.886156]$$
In fact it takes $$8$$ standard deviations to lose the digit $$1$$:
$$m \pm 8\sigma \rightarrow [8.886098, 8.886181]$$
Finally, based on an inverse symbolic search, we can find that $${\bf\frac{1795}{202}} = 8.886138613...$$ falls within $$0.25\sigma$$ of $$m = 8.886139852418$$ and so maybe this has potential to be the closed form.
For $$N = 10^9$$, here is the histogram of the number of steps to self-intersection, whose shape is quite similar to the one posted by OP:
If we plot look at the log plot, we can see the tail follows a decaying exponential quite closely (in red):
From here we might think we could use a gamma distribution or something similar to estimate the data, but I was only able to find a tight approximation with one distribution: the inverse Gaussian distribution $$\operatorname{IG}(8.886, 26)$$. Here's its PDF overlaid on the histogram:
I also recorded which segments intersected each other for $$N = 10^9$$. The data shows segment $$n$$ most often intersects with segment $$n-2$$, followed by $$n-3$$, etc which I think makes sense. Borrowing terminology from MattF's answer, the quick intersections are the most probable.
In the following diagram an $$(x, y)$$ pair indicates how often segment $$x$$ intersected with segment $$y$$. Each diagonal approximately decays exponentially, all with the same rate.
For each of the 4 cores I ran the simulations on, I recorded a single sequence of each length. The most interesting ones of course are the long ones.
Here's an instance of length 96. It's riddled with near misses and is ultimately squashed by a quick intersection.
Here the suspicious looking part is indeed intersection free -- it's about $$0.01 \,\text{rad}$$:
Lastly I looked at the histogram of angles in my saved simulations:
I think this bias (introduced by stopping once an intersection occurs) says that the best way to stay intersection free is to not curl back up towards yourself.
I have compiled all of my data into a Mathematica notebook which can be downloaded here.
• Great data! I especially like the somewhat surprising angles histogram (but I accept your explanation). – Joseph O'Rourke Mar 25 '19 at 19:27
• Thanks. Yes I was surprised by that histogram too. Originally I thought there might have been a bug in my code. I reran my code without the intersection stopping condition and the histogram came out uniform. An example of that is in the linked notebook. – Chip Hurst Mar 25 '19 at 19:31
• And by the way, I made a couple edits since posting -- I don't know if you saw them -- a conjectured closed form and an approximate distribution fit. – Chip Hurst Mar 25 '19 at 19:32
• $2+11\sqrt{29/74}$ is even closer to the empirical average. – Matt F. Mar 25 '19 at 20:16
• The tail behaviour means that the intersection probability tends towards a constant $p$ which you could get from the slope. The behaviour is then modelled reasonably well by $$\Pr(\text{length} = k \mid k > k_0) = \Pr(\text{length} > k_0) \, (1-p)^{k-k_0-1} p$$ once the walk is past $k_0$ steps. – student Mar 26 '19 at 14:34
We can get an upper bound of $$15.52$$ by considering just the cases of interesction of next-to-consecutive steps, which we call quick intersections.
Let $$\alpha$$ be the angle between step $$AB$$ and step $$BC$$, and let $$u=\alpha/\pi$$. So $$u$$ is (initially) uniformly distributed between $$0$$ and $$1$$. Then, as Gerhard Paseman and Bill Bradley determined, the probability that steps $$AB$$ and $$CD$$ intersect is $$P(\text{intersection|u})= \begin{cases} \frac14-\frac u4\ \ \text{ if } u\le \frac13\\ \frac12-u\ \ \text{ if } \frac13\le u \le \frac12\\ \phantom{1-2}0 \ \ \text{ if } \frac12 \le u\\ \end{cases}$$ So the (initial) overall probability of an intersection is $$\int P(\text{intersection}|u)du=1/12$$.
This means that for the next step, we have $$P(u|\text{no intersection last time})=\frac{12}{11} \begin{cases} \frac34+\frac u4\ \ \text{ if } u\le \frac13\\ \frac12 + u\, \ \ \text{ if } \frac13\le u \le \frac12\\ \phantom{1-2}1 \ \ \text{ if } \frac12 \le u\\ \end{cases}$$ \begin{align} P(\text{intersection|no intersection last time}) &=\frac{12}{11}\int_{u=0}^1 \begin{cases} (\frac34+\frac u4)(\frac14-\frac u4)\ \ \text{ if } u\le \frac13\\ (\frac12+u)(\frac12-u)\ \ \, \text{ if } \frac13\le u \le \frac12\\ \ \ \phantom{1-2}1(0)\ \ \ \ \ \ \ \ \ \ \ \text{ if } \frac12 \le u \end{cases} du\\ &=\frac{29}{396} \end{align} So the probability of no quick intersection after $$n$$ steps is $$\frac{11}{12}\left(\frac{367}{396}\right)^{n-3}$$ and the expected number of steps until a quick intersection is $$3\left(\frac{1}{12}\right)+\sum_{n=4}^\infty n\left( \frac{11}{12}\left(\frac{367}{396}\right)^{n-4}\! - \frac{11}{12}\left(\frac{367}{396}\right)^{n-3} \right)=\frac{450}{29} \simeq 15.52$$
• Nice analysis, MattF! – Joseph O'Rourke Mar 23 '19 at 0:11
Inspired by Bill Bradley's post, I've decided to lower bound the probability of an intersection which (using the method suggested in Bill's post) gives a slightly less weak upper bound for the expected length of a random walk.
Using Bill's picture and rescaling so that angles are measured as nonnegative fractions of 2pi, we see one domain of self intersection is when $$\alpha \lt \beta$$ and $$\alpha + 2\beta \lt 1/2$$. This region is a triangle of base 1/4, and height 1/6, with an area of 1/48. Missing the measure 0 set where $$\alpha=\beta$$, we reflect across that line, and then once more across the line $$\alpha+\beta=1$$ to get the other three regions having an intersection, giving the probability of self intersection of three-step paths to be 4/48=1/12.
Now let's extend this to paths with four steps. The number of cases to determine the probability exactly is more than I care to tackle, so I will cheat and underestimate it. I do this by considering just two cases: that an intersection occurs in the first three steps, or that it does not but occurs in the last three steps.
If we considered these cases as disjoint, we would get a probability of 1/12 + 1/12 = 1/6. They aren't, so we will consider the case that $$\beta \leq 1/4$$ and $$\alpha \gt 1/4$$ or more visibly use 3/4 of the chance that the last three steps self-intersect. This gives a lower bound of (1/12) times (1+3/4), or 7/48, which gives 1/7 as a weak lower bound.
Using this and Bill's estimating method gives 1 + 7*(4-1)=22 as a weak upper bound on the number of steps needed to self-intersect. If we could improve the lower bound from 1/7 to 1/6 for self intersection of a four step path, this would lead to an upper bound of 19 instead of 22.
But if we can cheat once, we can cheat again. The first cheat went from two neighboring angles to three, so we will go from three to five and lower bound the crossing probability of six segments as (7/4) times the (lower bound of the) probability of four segments. We share the middle angle so that we can nicely separate the domains of the middle angle turning sharp left or sharp right. We do a similar cut and trim and end up with a lower bound of 49/192 =1/4+1/192 for the probability of 6 segments crossing. This gives a slightly better upper bound of 1 +4*5 or 21 for the step length of a walk that does not cross itself.
To justify the cheat, one has to do a lot more work to make sure that the probability grows by 7/4 at this step. However, I won't do that. I use the second cheat instead to motivate the idea that 22 is an upper bound that can be improved with a simple argument, and hope to inspire someone to come up with a simple and more honest and precise estimate.
Gerhard "Long As It's Not Graded" Paseman, 2019.03.20.
• I would like to point out that many of my signatures encode some double meaning. In this particular case, I want to make it clear that this analysis fails on other manifolds. In particular, Joseph's problem makes sense on a torus or pseudo sphere, even on non-oriented manifolds or graphs with a rigid geometric structure. I encourage people to consider the version for a torus. Gerhard "Promises To Not Grade Work" Paseman, 2019.03.20. – Gerhard Paseman Mar 20 '19 at 17:21
• The version of Joseph's question on a compact space is likely in the graph theory literature. Consider a cover of the space by uniform mostly disjoint regions, and make a graph with a region a vertex and a line joining two regions if a step takes one from one region to another. One can now approximate a random walk by a transition matrix, or use the theory of random walks in graphs to compute a desired expectation. Thus the actual expectation can be considered as a limit of expectations using various partitions of the space. Gerhard "Still Interested In Torus Version" Paseman, 2019.03.20. – Gerhard Paseman Mar 20 '19 at 18:20
(Thanks for noticing my earlier error, Gerhard!)
Here's a loose but very simple upper bound. Consider three consecutive unit steps, which might look like this:
We take $$\alpha\in[0, \pi]$$ and $$\beta\in[0,2\pi)$$. (By symmetry we can reflect $$\alpha\in (\pi, 2\pi)$$ back to $$\pi-\alpha$$.) Then to overlap, we need $$\alpha < \pi/2$$, $$\beta<\pi/2$$ and $$\alpha + \beta < \pi$$ The first two events are independent, with probabilities 1/2 and 1/4.
For the third event, consider a case where the three segments just barely cross, e.g.:
If we decrease the angles $$\alpha, \beta$$, the steps will still intersect, so it's sufficient to consider this case.
Because the bottom and left edges are both length one, the upper and rightmost angles are both $$\beta$$. Since we have a triangle, $$\alpha+2\beta=\pi$$, so the probability of this event is 2/3. (There's a symmetric case where the roles of $$\alpha$$ and $$\beta$$ are reversed.) It's possibly clearer to see this by examining a phase space diagram ("blue" = "line segments intersect"):
The blue region takes up 2/3rds of the (uniformly sampled) space. Conditioning on everything, the probability of overlap is 1/2 * 1/4 * 2/3 = 1/12
We can apply this argument to multiple non-overlapping pairs of angles. Therefore, the expected number of steps until self-intersection is upper-bounded by $$\leq 1 + 2\times 12 = 25$$
• Something does not look right. Do you want to find it, or shall I expand on my comments above? Gerhard "Pretty Sure About One Twelfth" Paseman, 2019.03.20. – Gerhard Paseman Mar 20 '19 at 15:49
• Looks good now. I like the picture and the estimate. Gerhard "Blue Is A Nice Color" Paseman, 2019.03.21. – Gerhard Paseman Mar 21 '19 at 23:55 | 2021-03-05T17:22:23 | {
"domain": "mathoverflow.net",
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects",
"openwebmath_score": 0.8837517499923706,
"openwebmath_perplexity": 486.9452189925769,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9748211612253743,
"lm_q2_score": 0.8740772286044095,
"lm_q1q2_score": 0.8520689789888074
} |
https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power | # How do I take a fraction to a negative power?
I ran into this issue during my homework. Using the rules of logarithms, I need to prove that $$-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln3-\ln2$$ So here were my steps:
1. First step: $$-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln\left(\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}\right)$$ And that's as far as I got, because now I want to use the form $$\ln(a/b) = \ln(a) - \ln(b)$$, but first I need to reduce the fraction because it is raised to the $$-2$$.
How do I evaluate $$\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}$$ ?
Thanks
• Note that $x^{-a} = \frac{1}{x^a}$ – desiigner Jul 11 '19 at 19:53
By definition $$a^{-k} = \frac 1{a^k}$$
So $$\left(\frac{2}{\sqrt{6}}\right)^{-2} =\frac 1{\left(\frac{2}{\sqrt{6}}\right)^{2}}=$$
$$\frac 1{\left(\frac {2^2}{\sqrt 6^2}\right)}=\frac {\sqrt 6^2}{2^2}=\frac 64=\frac 32$$
It will help to realize that $$(\frac ab)^{-1} = 1/(a/b) = \frac ba$$ and that $$(\frac ab)^k = \frac {a^k}{b^k}$$ to realize that that means $$\left(\frac ab\right)^{-k} = \frac 1{\left(\frac ab\right)^k}= \frac 1{\left(\frac {a^k}{b^k}\right)} = \frac {b^k}{a^k}.$$
(Also $$(\frac ab)^{-k} = [(\frac ab)^{-1}]^k = (\frac ba)^k=\frac {b^k}{a^k}$$ or that $$(\frac ab)^{-k} = \frac {a^{-k}}{b^{-k}} = (1/a^k)/(1/b^k) = \frac {b^k}{a^k}$$.)
In any event
$$\left(\frac {2}{\sqrt 6}\right)^{-2} = \left(\frac {\sqrt 6}{ 2}\right)^2 = \frac {\sqrt 6^2}{2^2} = \frac 64 = \frac 32.$$
• Perfect, this is the explanation I was looking for. So now after a few steps, I'll get ln(3/2) = ln(3) - ln(2). – Miguel Aragon Jul 11 '19 at 21:36
\begin{align} -2\ln \left( \frac{2}{\sqrt 6} \right) &= -2\big( \ln(2)-\ln(\sqrt{6}) \big) \\ &= -2\ln(2)+2\ln(6^{1/2}) \\ &= -2\ln(2)+\ln(2\cdot 3) \\ &=-2\ln(2)+ \big( \ln(2)+\ln(3)\big) \\ &=\ln(3)-\ln(2) \end{align} And for your specific question, remember that $$\left( \frac{a}{b} \right)^{-n}=\left( \frac{b}{a} \right)^n$$
Well you may start by distributing the index since $$2$$ and $$\sqrt 6$$ are positive. Thus $$\left(\frac{2}{\sqrt{6}}\right)^{-2}=\frac{2^{-2}}{{\sqrt 6}^{-2}}.$$
Then recall that for any nonzero number $$a$$ and any negative integer $$-n,$$ we have $$a^{-n}=\frac {1}{a^n}.$$ Applying this to your expression, we have $$\frac{2^{-2}}{{\sqrt 6}^{-2}}=\frac{\frac {1}{2^2}}{\frac{1}{{\sqrt 6}^2}}=\frac{\frac {1}{4}}{\frac{1}{6}}=\frac{6}{4}=\frac 32.$$
$$-2 \ln(\frac{2}{\sqrt{6}}) = \ln(\frac{2}{\sqrt{6}})^{-2} = \ln \frac{1}{(\frac{2}{\sqrt{6}})^{2}} = \ln \frac{6}{4} = \ln\frac{3}{2} = \ln 3 - \ln 2$$
$$-2 \ln \left( \frac{2}{\sqrt{6}} \right) = \ln 3 - \ln 2$$
if and only if
$$\ln \left( \frac{2}{\sqrt{6}} \right)^{-2} = \ln \left( \frac{3}{2}\right)$$
if and only if
$$\ln \left[ \frac{1}{\left(\frac{2}{\sqrt{6}}\right)^{2}} \right] = \ln \left( \frac{3}{2}\right)$$
And so, we have
$$\ln \left( \frac{6}{4} \right) = \ln \left( \frac{3}{2}\right)$$
which is true.
• you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse – peek-a-boo Jul 11 '19 at 20:27
• One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to include them initially. – mlchristians Jul 11 '19 at 20:29
• $4 = 10 \implies 7-3 =7+3 \implies-3= 3 \implies (-3)^2 = 3^2 \implies 9=9$ which is true. Therefore $4 = 10$. Or if penguins were elephants then they would both eat food. Which is true. So penguins are elephants. – fleablood Jul 11 '19 at 20:37
• "One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained." But if you do that you MUST specify the steps are actually if and only if steps. If you do not SPECIFICALLY state that it is reasonable to assume they are only forward implications and the proof is not valid. – fleablood Jul 11 '19 at 20:38
• Your first comment is not if and only if; and neither is you second. Forward implications is one-way; necessary and sufficient (i.e., if and only if are both ways). In the case of this problem, start from the given equation and work it to where I finished. Any problem with that. Now, if you want to do the work, start w/ the last statement and by a series of implications, show it implies the initial equation. I chose iff and only if instead. – mlchristians Jul 11 '19 at 20:45
For a non-zero real number, we have $$1=x^0=x^{2-2}=x^2\cdot x^{-2}\tag1$$ from How to understand why $$x^0=1$$ where $$x$$ is any real number?
Putting $$x=2/\sqrt6$$ into $$(1)$$, we get $$1=\left(\frac2{\sqrt6}\right)^2\cdot\left(\frac2{\sqrt6}\right)^{-2}\tag2$$ and since we know that for a positive integer $$n$$ and positive real numbers $$a,b$$, $$\left(\frac ab\right)^n=\underbrace{\frac ab\cdot \frac ab\cdot\ldots\cdot\frac ab}_{n\,\text{times}}=\underbrace{\frac{a\cdot a\cdot\ldots\cdot a}{b\cdot b\cdot\ldots\cdot b}}_{n\,\text{times}}=\frac{a^n}{b^n},\tag3$$ we have that $$\left(\frac2{\sqrt6}\right)^2=\frac{2^2}{\left(\sqrt6\right)^2}=\frac46=\frac23\tag4.$$ Finally, putting $$(4)$$ back into $$(2)$$ yields $$1=\frac23\cdot\left(\frac2{\sqrt6}\right)^{-2}\implies\left(\frac2{\sqrt6}\right)^{-2}=\frac1{2/3}=\frac32\tag5$$ as the answer. | 2020-01-25T02:30:01 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power",
"openwebmath_score": 0.9951451420783997,
"openwebmath_perplexity": 289.97974398157544,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9787126531738088,
"lm_q2_score": 0.8705972768020108,
"lm_q1q2_score": 0.8520645706247889
} |
https://math.stackexchange.com/questions/3843961/any-shortcuts-to-proving-that-frac-sinx2-sin2-frac-x2-tan-frac-x2/3843975 | # Any “shortcuts” to proving that $\frac{\sin(x)}2+\sin^2(\frac x2)\tan(\frac x2)\to\tan(\frac x2)$
I was working on simplifying some trig functions, and after a while of playing with them I simplified $$\frac{\sin(x)}{2}+\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right) \rightarrow \tan\left(\frac{x}{2}\right)$$
The way I got that result, however, was with what I think a very "roundabout" way. I first used the half-angle formulaes, then used $$x=\pi/2-\beta$$, and that simplified to $$\frac{\cos(\beta)}{1+\sin(\beta)}$$ where I again used the coordinate change to get $$\frac{\sin(x)}{1+\cos(x)}\rightarrow\tan\left(\frac{x}{2}\right)$$
I tried using the online trig simplifiers but none succeeded. Of course, after you know the above identity, it's easy to prove by proving that $$\frac{\sin(x)}{2}=\tan\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)$$
Is there a more direct way to get the identity? I guess what I'm asking is, am I missing any "tricks" or software that I could have on my toolbelt so that next time I don't spend hours trying to simplify trig identities?
With $$s:=\sin\frac x2,c:=\cos\frac x2$$, $$\sin x=2sc$$ and
$$\frac122sc+s^2\frac sc=\frac{sc^2+s^3}c=\frac sc.$$
• I accepted this answer since it makes all steps explicit – Esteban Sep 30 '20 at 15:03
• @Esteban: thank you for the explanation. – Yves Daoust Sep 30 '20 at 15:20
Use $$\sin(x) = 2 \sin(x/2) \cos(x/2)$$ then we have $$\begin{eqnarray*} \frac{\sin(x)}{2}+\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right) &=& \frac{\sin(x/2)}{\cos(x/2)} \underbrace{\left( \cos^2(x/2) + \sin^2(x/2) \right)}_{=1} \\ &=& \tan (x/2). \end{eqnarray*}$$
• I actually tried this route but got stuck, unfortunately. So I think it's worth commenting that the "trick" here to get to your right hand expression is to multiply the 2sin(x/2)cos(x/2)/2 term by cos(x/2)/cos(x/2). – Esteban Sep 30 '20 at 14:51
Fairly obvious: $$\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=(1-\cos^2\left(\frac{x}{2}\right))\tan\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\frac{\sin x}{2}$$
• All the tried trig simplifiers beg to differ it was fairly obvious, unfortunately. I like the route you took. – Esteban Sep 30 '20 at 14:54
$$\frac{\sin x}{2} = \sin\frac{x}{2}\cos\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \cos^2\frac{x}{2} = \tan\frac{x}{2} \left(1 - \sin^2\frac{x}{2}\right)$$ | 2021-03-07T15:13:01 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3843961/any-shortcuts-to-proving-that-frac-sinx2-sin2-frac-x2-tan-frac-x2/3843975",
"openwebmath_score": 0.9426677823066711,
"openwebmath_perplexity": 379.3691216335252,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9787126500692716,
"lm_q2_score": 0.8705972734445508,
"lm_q1q2_score": 0.8520645646359986
} |
https://tutel.me/c/mathematics/questions/118686/determining+precisely+where+sum_n1inftyfracznn+converges | #### [SOLVED] Determining precisely where $\sum_{n=1}^\infty\frac{z^n}{n}$ converges?
Inspired by the exponential series, I'm curious about where exactly the series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges.
I calculated $$\limsup_{n\to\infty}\sqrt[n]{\frac{1}{n}}=\limsup_{n\to\infty}\frac{1}{n^{1/n}}$$ and $$\lim_{n\to\infty} n^{1/n}=e^{\lim_{n\to\infty}\log(n)/n}=e^{\lim_{n\to\infty}1/n}=e^0=1.$$ So the radius of convergence is $1$, so the series converges on all $z$ inside $S^1$. But is there a way to tell for which $z$ on the unit circle the series converges? I know it converges for $z=-1$, but diverges for $z=1$, but I don't know about the rest of the circle. For what other $z$ does this series converge? Thanks.
#### @Pookaros 2017-05-10 23:18:42
Even though the 1st answer is the more easy and appropriate here is a more analytic one. $\sum_{n=1}^{k} (z^n)/n =(z/1-z) $$\sum_{n=1}^{k-1} [(1-z^n)/n(n+1) + (1-z^k)/k] (we can prove that with induction till k). We can also see that 0 \leqslant |(1-z^k)/k|\leqslant 2/k\rightarrow0 for k\rightarrow$$\infty$ also
0 $\leqslant$ |(1-z^n)/n(n+1)|$\leqslant$ 2/n(n+1)
$\sum_{n=1}^{\infty}2/n(n+1)$ $(converges)$ so $\sum_{n=1}^{\infty} (z^n)/n$ $(converges)$ as well
#### @Martin Argerami 2012-03-11 00:56:15
Fix $z$ in the unit circle, i.e. $|z|=1$. We want to apply Dirichlet's test: if $\{a_n\}$ are real numbers and $\{b_n\}$ complex numbers such that:
1. $a_1\geq a_2\geq\cdots$
2. $\lim_{n\to\infty}a_n=0$
3. There exists $M>0$ such that $\left|\sum_{n=1}^Nb_n\right|\leq M$ for every $N\in\mathbb{N}$;
then $\sum_{n=1}^\infty a_nb_n$ converges. Here $a_n=1/n$, $b_n=z^n$. The first two conditions are clearly satisfied, and for the third one: $$\left|\sum_{n=1}^Nz^{ n }\right|=\left|\frac{z-z^{N+1}}{1-z}\right|\leq\frac{2}{|1-z|}$$ for all $N\in\mathbb{N}$. This shows that the third condition is satisfied for every $z\ne1$ in the circle.
In conclusion, the series converges for every $z$ with $|z|\leq1$ other than $z=1$, and it diverges for $|z|>1$.
#### @philmcole 2017-12-06 12:32:40
Can you explain this step: $|\sum_{n=1}^Nz^{ n }| = |\frac{z-z^{N+1}}{1-z}|$ ?
Here.
#### @philmcole 2017-12-06 13:48:17
Yeah, but where do you get that additional $z$ in the nominator from? By the formula you linked it should be $|\frac{1-z^{N+1}}{1-z}|$, no?
#### @Martin Argerami 2017-12-06 13:49:24
If you start the sum from $0$, yes.
#### @Pacciu 2012-03-11 23:23:01
The followin theorem on power series is due to E. Picard:
Let $(a_n)$ be a sequence of real numbers.
If the sequence $(a_n)$ is nonnegative, decreases and tends to zero when $n\to \infty$, then the complex power series $\sum a_n\ z^n$ converges in the closed unit disc $\overline{D}(0;1)$ with the only possible exception of the point $1$.
The proof of Picard's theorem relies on Abel's summation by parts formula, as far as I remember.
Now, the coefficients of your series, i.e. $a_n=1/n$, satisfy the assumptions of Picard's theorem, hence your series converges at least in $\overline{D}(0;1)\setminus \{1\}$; on the other hand, the series diverges when $z=1$ (for it becomes the harmonic series).
Therefore the convergence set of $\sum 1/n\ z^n$ is $\overline{D}(0;1)\setminus \{1\}$.
### Determining precisely where $\sum_{n=0}^\infty \left(\frac{z^n}{n}\right)$ converges uniformly
• 2017-09-13 22:36:35
• Pookaros
• 77 View
• 0 Score
• Tags: complex-analysis
### [SOLVED] To prove the statements related to power series
• 2017-01-27 02:53:47
• Kavita
• 653 View
• 1 Score
• Tags: complex-analysis | 2019-03-26T10:01:28 | {
"domain": "tutel.me",
"url": "https://tutel.me/c/mathematics/questions/118686/determining+precisely+where+sum_n1inftyfracznn+converges",
"openwebmath_score": 0.9750022292137146,
"openwebmath_perplexity": 522.6435572542091,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9787126475856414,
"lm_q2_score": 0.870597271765821,
"lm_q1q2_score": 0.8520645608307629
} |
http://math.stackexchange.com/questions/362140/probability-taking-cups-from-a-box | # Probability: Taking Cups From A Box
I just saw the question in an exam paper. My friend and I have different answer and I am not sure which solution is correct, so I hope someone can help me.
There are 16 cups in a box, 11 are blue and 5 are white. Six cups are taken out from it at the same time. What is the probability of taking at least 4 white cups?
My solution:
$$P=\frac{\binom54\binom{11}2+\binom55\binom{11}1}{\binom{16}6}$$
My friend's solution:
$$P=\frac{5!\times11\times6+5!\times11\times10\times_6\text{P}_2}{_{16}\text{P}_6}$$
My friend use a different approach and his answer is different from mine. I wonder if I am correct and why. Please show me the reason, thank you!
-
Your method is correct. – André Nicolas Apr 15 '13 at 8:02
@AndréNicolas Can you tell me why I am correct? After my friend told me his solution, I am so confused with the question ... – redcap Apr 15 '13 at 8:22
Your method is correct. There are $\binom{16}{6}$ equally likely ways to pick a collection of $6$ cups. (We can imagine the cups have secret ID numbers written in invisible ink.)
Your numerator counts the favourables. For example, the are $\binom{5}{4}$ ways to pick $4$ whites. For each of these ways there are $\binom{11}{2}$ ways to pick the accompanying blues, for a total of $\binom{5}{4}\binom{11}{2}$.
Your friend's proposed solution can be made to work, though it is perhaps a little trickier to do it correctly. The friend observed that counting order of picking, there are ${}_{16}P_6$ equally likely ways of picking $6$ cups. Now we count the number of favourables. Your friend's first term in the count of favourables is correct, the second is not.
There are $\binom{6}{2}$ ways of picking the locations of the $4$ blue cups. For each of these ways, the first chosen location can be filled in $11$ ways, and the second in $10$ ways. Then the remaining locations can be filled with white cups in ${}_5P_4$ ways. Multiply to get the total with $4$ white and $2$ blue. There are two mistakes in the proposed expression there. First, instead of $\binom{6}{2}$ there is ${}_6P_2$. Secondly, there is the $4!$ which may be a typo. For the placement of white cups in the $4$ available slots, either use the permutation symbol ${}_5P_4$, or decide the $4$ whites can be chosen in $\binom{5}{4}$ ways, and then permuted in $4!$ ways. That part turns out to be $(5)4!$, which is the same as $5!$.
You are welcome. I had figured the $4!$ was an unimportant error. But that still leaves an overcount in that part by a factor of $2$. That error is somewhat more structural. – André Nicolas Apr 15 '13 at 9:30 | 2014-04-20T01:12:02 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/362140/probability-taking-cups-from-a-box",
"openwebmath_score": 0.8496515154838562,
"openwebmath_perplexity": 100.20600061453757,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9911526438717138,
"lm_q2_score": 0.8596637577007393,
"lm_q1q2_score": 0.8520580062857801
} |
https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose | # Parameterize unitary without transpose
For all unitary matrices, i.e. $A \overline{A}^T = I$, there is a skew-Hermitian matrix $X$ so that $A = exp(X)$. So the unitary group has $n^2$ dimensions.
Is there any similar parameterisation of all matrices with $A \overline{A} = I$? What can be said about the number of dimensions? (Btw. is there any name for this type of matrices in literature?)
• Note that If $M$ is a real square matrix of the relevant size, then $A = \exp(iM)$ has the property that $\overline{A} = \exp(-iM)$, so that $A\overline{A} = I$. – Geoff Robinson Oct 19 '15 at 22:56
• @QiaochuYuan: Right, this collection is not closed under multiplication. Does this already exclude any notion of dimensionality and parameterisation? – Sebastian Schlecht Oct 20 '15 at 8:38
• @GeoffRobinson: Right, but are these all matrices with this property? – Sebastian Schlecht Oct 20 '15 at 9:05
• @GeoffRobinson: The answer is yes if we add $A$ invertible, as $\exp: M(N,\mathbb{C}) \rightarrow GL(N,\mathbb{C})$ is surjective and $\exp(M) \overline{\exp(M)}=I$ follows $M = -\overline{M}$. – Sebastian Schlecht Oct 20 '15 at 9:13
• Well,$A$ certainly needs to be invertible in the question. – Geoff Robinson Oct 20 '15 at 10:00
The set $V= \{ A\in M_n(\mathbb{C})\ |\ A\bar A = I\}$ is a smooth submanifold of $M_n(\mathbb{C})$ with real dimension $n^2$.
Proof: Consider the involution $\iota:\mathrm{GL}(n,\mathbb{C})\to \mathrm{GL}(n,\mathbb{C})$ defined by $$\iota(A) = (\bar A)^{-1}.$$ This is an anti-holomorphic involution of the complex manifold $\mathrm{GL}(n,\mathbb{C})$, and its fixed locus is precisely $V$. Thus, $V$ is a totally real submanifold of $\mathrm{GL}(n,\mathbb{C})$ with (real) dimension $n^2$.
Also: While it's true that the map $f:M_n(\mathbb{R})\to \mathrm{GL}(n,\mathbb{C})$ defined by $$f(a) = \exp(ia)$$ has its image in $V$, it is not a 'parametrization' everywhere, i.e., $f$ is not a local diffeomorphism everywhere. While this is true on a neighborhood of $0\in M_n(\mathbb{R})$, at other places, the map $f$ definitely is not a local diffeomorphism. For example, if $a$ has $n$ eigenvalues of the form $2k_i\pi$ (where $k_1,\ldots, k_n$ are integers, not all zero), then $f(a) = I$, but the space of such real matrices $a$ has positive dimension for any $n$-tuple $(k_1,\ldots, k_n)$ for which not all of the $k_i$ are equal.
It turns out that $f$ is surjective. The proof uses the fact that, for $A\in V$, we have $A\bar A = \bar A A = I$, so, in particular, $A$ and $\bar A$ commute and hence can be put simultaneously in Jordan normal form by a real conjugation. Then, breaking $\mathbb{C}^n$ into a sum of complexifications of real subspaces according to the eigenvalues of $A$ and using the semi-simple/nilpotent decomposition appropriately, one can reduce to dealing with the upper triangular case, and the result can then be proved using simple facts about power series in commuting nilpotent variables. Details upon request (see below).
Of course, all of this is probably a special case of known facts about affine symmetric spaces. I have now realized that $V$ is simply the Cartan embedding for the affine symmetric space $\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})$. This Cartan embedding $$\sigma:\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})\longrightarrow V\subset \mathrm{GL}(n,\mathbb{C})$$ is given by $\sigma(B\cdot\mathrm{GL}(n,\mathbb{R})) = B\,(\,\overline B\,)^{-1}$. The exponential map we have been discussing is just the geodesic mapping of this affine symmetric space, so, probably all of this follows from general theory.
Details: Let $A\in\mathrm{GL}(n,\mathbb{C})$ satisfy $A\overline{A} = I$. Let $\lambda\in\mathbb{C}$ be an eigenvalue of $A$ of multiplicity $m\le n$ and let $$V_\lambda = \{ v\in\mathbb{C}^n \ |\ (A{-}\lambda)^mv = 0\ \}.$$ be the associated generalized eigenspace. Of course, $\mathbb{C}^n$ is the direct sum of these generalized eigenspaces. Since $\overline{A}$ commutes with $A$, it preserves each of these subspaces. Moreover, one clearly has $$\overline{V_\lambda} = V_{1/\bar\lambda},$$ so each of the spaces $V_\lambda+V_{1/\bar\lambda}$ is invariant under conjugation and hence is the complexification of a real subspace $W_\lambda\subset \mathbb{R}^n$. It follows that, by conjugating by a real matrix, we can assume that $A$ (and hence $\overline A$) is in block diagonal form, so it suffices to treat the case where either $A$ has a single eigenvalue $\lambda$ satisfying $\lambda\bar\lambda=1$, so that $\mathbb{C}^n=V_\lambda$, or else $A$ has an eigenvalue $\lambda$ satisfying $\lambda\bar\lambda > 1$ and $\mathbb{C}^n=V_\lambda\oplus V_{1/\bar\lambda}$.
In either case, we can assume that $\lambda = r\ge 1$ is real, since, if $\lambda = r e^{i\theta}$, we can replace $A$ by $e^{-i\theta}A$ and show that $e^{-i\theta}A$ is in the image of $f$ (since $I$ commutes with everything and $e^{i\theta} I = \exp(i\theta I)$.)
First, consider the case $\lambda = 1$. Then $A = C + i S$ where $C$ and $S$ are real matrices and $((C-I) + iS)^n = (A-I)^n = 0$. Moreover, $N=C-I$ and $S$ commute since $I = A\bar A = C^2+S^2 +i(SC-CS) = I + 2N + N^2 + S^2$. Note that $N$ and $S$ are nilpotent commuting matrices. Since they satisfy $$2N + N^2 = -S^2,$$ it follows that $N = p(-S^2)$ where $p(t) = -\tfrac12 t^2 + \cdots$ is the (unique) power series (with real coefficients) that satisfies $2p(t) + p(t)^2 = -t^2$. Since $S$ is nilpotent, it follows that $N = p(-S^2)$ expresses $N$ canonically as a polynomial in $S$. Now let $q(t)$ be the (unique) power series with real coefficients that satisfies $\sin q(t) = t$. Note that we must have $p(-q(t)^2) = \cos(t)-1$. Now, because $S$ is nilpotent, we have $S = \sin q(S)$, where $q(S)$ is a real polynomial in $S$. Putting all of this together, we have $$A = I + p(-S^2) + iS = \cos(q(S)) + i\sin(q(S)) = \exp(iq(S)).$$ Finally, before leaving this special case, let us note that, because $A$ satisfies its characteristic polynomial $(A-I)^n = 0$, it follows that $\overline A = A^{-1}$ is expressed as a universal polynomial in $A$ with real coefficients, so $iS = \tfrac12(A-\overline{A})$ is also expressed as a universal polynomial with real coefficients. Since $q(t)$ is an odd power series, it satisfies $iq(it) = f(t)$ where $f$ has real coefficients, so there is actually a formula of the form $A = \exp( i g_n(A))$ for all $A \in \mathrm{GL}(n,\mathbb{C})$ satisfying $A\bar A = I$ and $(A-I)^n=0$ for some universal polynomial $g_n(t)$ with real coefficients that also has the property that $g_n(A)$ is a real $n$-by-$n$ matrix for all such $A$. (This remark will be used below.)
Now, finally, let us assume that $A$ has eigenvalues $\lambda = e^t$ and $1/\lambda = e^{-t}$ for some $t>0$. Then $V_\lambda$ and $\overline{V_\lambda} = V_{1/\lambda}$ are disjoint, complementary complex subspaces of $\mathbb{C}^n$ and so we must have that there exists a matrix $Q\in M_n(\mathbb{R})$ such that $$V_\lambda = \{ v + iQv \ | \ v\in \mathbb{R}^n \}.$$ Because $V_\lambda$ is closed under multiplication by $i$, it follows that $Q^2 = -I$. Setting $$S = \cosh t + i\sinh t\,Q = \exp(i t Q),$$ we see that $Sw= e^t w$ for all $w \in V_\lambda$ and $Sw = e^{-t}w$ for all $w \in V_{1/\lambda}$, so $S$ is the semi-simple part of $A$. Hence $S$ can be written as $S = s_\lambda(A)$, where $s_\lambda(t)$ a polynomial in $t$ with real coefficients (that depend on $\lambda$). Thus, $\bar S = s_\lambda(\bar A)$ can be written as a universal polynomial in $A$, implying that $Q$ itself can be written as a universal polynomial in $A$ and hence, in particular, it commutes with $A$ and $\bar A$. Now, writing $$A = \exp(i t Q) B,$$ we see that $B\bar B = I$ and that $B$ can be written as a polynomial in $A$. Moreover, the eigenvalues of $B$ are now all equal to $1$, so $(B-I)^n=0$, and so $B = \exp(i g_n(B))$ (as per above), where $g_n(B)$ is a polynomial in $A$, which, therefore, commutes with $Q$ (which is a polynomial in $A$). Finally, we have $$A = \exp(i t Q)\exp(i g_n(B)) = \exp(i (t Q+g_n(B))),$$ as desired.
• Thank you very much for your insightful answer. The only main missing part for me is the surjectivity of $f$. – Sebastian Schlecht Oct 20 '15 at 16:43
• Robert, I learn so much new material from your answers! Just great! – Suvrit Oct 21 '15 at 13:24
• Robert, great answer. So a bit about the details, for simplicity assuming $A$ is diagonalisable. Commutativity gives as $A = U^{-1} \Lambda U$ for real $U$. $A$ is invertible, hence there is $\exp(i L) = \Lambda$. So, $A = \exp( i U^{-1} L U)$. Why is $U^{-1} L U$ real? – Sebastian Schlecht Oct 22 '15 at 20:05
• If $A$ is diagonal, then $A\bar A = I$ implies that all of the eigenvalues of $A$ are complex numbers of unit modulus, so each of them is of the form $e^{ir}$ where $r$ is real. – Robert Bryant Oct 22 '15 at 20:40
• I'm confused, I only wanted $A$ diagonalisable and not $A$ diagonal. – Sebastian Schlecht Oct 22 '15 at 21:24
Qiaochu Yuan is a little hard about our variety $V$; Kummer showed that an element of $V$ is in the form $U\bar{U}^{-1}$ (a generalization of this result is the Hilbert's theorem 90). Close to the subject, cf. "Ikramov, On the matrix equation $X\bar{X}=A$" http://link.springer.com/article/10.1134%2FS1064562409010153#page-1
Robert wrote a nice answer; we can give more elementary proofs as follows. $\mathbb{C}$ is seen as $\mathbb{R}^2$.
Proposition 1. $dim(V)=n^2$ when $V$ is considered as a real algebraic set.
Proof. According to Kummer, if $R\bar{R}=I$, then there is $T\in GL_n(\mathbb{C})$ s.t. $R=T\overline{T}^{-1}$. Let $f:T\in GL_n\rightarrow T\overline{T}^{-1}$; note that $f$ is a pseudo-parametrization of our set; in fact $f$ is a submersion. It remains to calculate the rank of $Df_T:H\in M_n\rightarrow (H-T\overline{T}^{-1}\overline{H})\overline{T}^{-1}$. Note that $rank(Df_T)=rank(H\rightarrow T^{-1}H-\overline{T}^{-1}\overline{H})$.
Let $T^{-1}=U+iV,H=X+iY$, where $U,V,X,Y$ are real matrices. Then $rank(Df_A)=rank(g:(X,Y)\in \mathbb{R}^{2n^2}\rightarrow UY+VX\in \mathbb{R}^{n^2}$; we show that $g$ is onto. Let $A\in M_n$; since $U+iV$ is invertible, there is $\lambda\in \mathbb{R}$ s.t. $U+\lambda V$ is invertible. Finally $g(\lambda(U+\lambda V)^{-1}A,(U+\lambda V)^{-1}A)=A$ and we are done.
Proposition 2. Any $R\in V$ can be written as $exp(iA)$ where $A$ is a real matrix.
Proof. Let $R=U+iV$, where $U,V$ are real. $R\bar{R}=I$ is equivalent to $U^2+V^2=I,VU=UV$; that is equivalent to: there is a real matrix $A$ s.t. $U=\cos(A),V,=\sin(A)$ and $R=\exp(iA)$.
EDIT. (cf. Sebastian's comment). $e^Me^{\bar{M}}=I$ does not imply that $e^{M+\bar{M}}=I$.
Proof. Take $M=\begin{pmatrix}i\pi&1\\0&-i\pi\end{pmatrix}$.
• That is very sleek! Do you need Prop1 for Prop2? Do you get Prop1 also from Prop2? – Sebastian Schlecht Oct 29 '15 at 11:59
• And do you have any suggestion what to name $V$? I understand that there might be no agreed name. Unfortunately, I cannot access right now the Ikramov paper to have a look. – Sebastian Schlecht Oct 29 '15 at 12:01
• I would like to see the details for your Proposition 2. I don't believe that it is as easy as you seem to think. I thought of exactly this line, right up to your 'that is equivalent to there is a real matrix A...". I didn't see how to prove that directly, and I would be curious to see what method you use to prove it. Also, the smoothness of $V$ follows, as I wrote, easily from the fact that it is the fixed point set of an anti-holomorphic involution. I don't see why it needs anything from Kummer, unless you are quoting the surjectivity of the Cartan embedding, a nontrivial fact. – Robert Bryant Oct 29 '15 at 14:11
• @ Sebastian SchlechtClearly , Prop. 1 and 2 can be independently proved. About a name for $V$, I don't know. In general, the authors consider elements of $V$ and not the whole set. cf also Horn and Johnson; Matrix Analysis, Section 4.6 – loup blanc Oct 29 '15 at 14:59
• @loupblanc: I suspected as much. So it's somewhat misleading to claim that your proof is 'more elementary'; it just leaves out many details and even essential ideas. In fact, while I mention Cartan's results, I don't use them (or any result of Kummer) in the surjectivity argument, which, in fact, uses nothing more than than the semi-simple/nilpotent decomposition (i.e., Jordan normal form), and even the result over $\mathbb{C}$ needs this. – Robert Bryant Oct 29 '15 at 15:52
I totally lied about this not being a natural thing to ask! As loup blanc alludes to, in fact $n \times n$ matrices such that $M^{-1} = \overline{M}$ can be interpreted as Galois descent data for descending the complex vector space $\mathbb{C}^n$ to a real vector space, or equivalently as a $1$-cocycle in
$$Z^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C})).$$
The statement that any such matrix must have the form $U \overline{U}^{-1}$ says that any such $1$-cocycle is cohomologous to zero, or equivalently that the Galois cohomology set
$$H^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C}))$$
is trivial. This just says that there is only one real form of $\mathbb{C}^n$ up to isomorphism, namely $\mathbb{R}^n$. The space $GL_n(\mathbb{C})/GL_n(\mathbb{R})$ appearing in Robert Bryant's answer can then be interpreted as the space of real forms of $\mathbb{C}^n$. | 2019-10-14T21:49:12 | {
"domain": "mathoverflow.net",
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose",
"openwebmath_score": 0.9622570872306824,
"openwebmath_perplexity": 125.47789443929702,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.991152643087697,
"lm_q2_score": 0.8596637577007394,
"lm_q1q2_score": 0.8520580056117895
} |
https://mathhelpboards.com/threads/divisibility-challenge.8074/ | # Divisibility Challenge
#### anemone
##### MHB POTW Director
Staff member
Given unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.
##### Well-known member
Let us put $f(x) = (x-b)^5 + (b-c)^5 + (c-x)^5$
Putting x = b we get f(b) = 0
so (x-b) is a factor
so a-b is a factor of $(a-b)^5 + (b-c)^5 + (c-a)^5$
similarly we have (b-c) and (c-a) are factors
now $(a-b)^5 = a^5 – 5a b^4 + 10a^2b^3 – 10 a^3 b^2 +5 a^4 b – b^5 = a^5 – b^5 + 5m$ where $m = - a b^4 + 2a^2b^3 – 2 a^3 b^2 + a^4 b b^4$
similarly $(b-c)^5 = b^5 – c^5 + 5n$
$(c-a)^5 = c^5 – a^5 + 5k$
Adding we get $(a-b)^5 + (b-c)^5 + (c-a)^5 = 5 (m+n+k)$
So $(a-b)^5 + (b-c)^5 + (c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$
#### anemone
##### MHB POTW Director
Staff member
Let us put $f(x) = (x-b)^5 + (b-c)^5 + (c-x)^5$
Putting x = b we get f(b) = 0
so (x-b) is a factor
so a-b is a factor of $(a-b)^5 + (b-c)^5 + (c-a)^5$
similarly we have (b-c) and (c-a) are factors
now $(a-b)^5 = a^5 – 5a b^4 + 10a^2b^3 – 10 a^3 b^2 +5 a^4 b – b^5 = a^5 – b^5 + 5m$ where $m = - a b^4 + 2a^2b^3 – 2 a^3 b^2 + a^4 b b^4$
similarly $(b-c)^5 = b^5 – c^5 + 5n$
$(c-a)^5 = c^5 – a^5 + 5k$
Adding we get $(a-b)^5 + (b-c)^5 + (c-a)^5 = 5 (m+n+k)$
So $(a-b)^5 + (b-c)^5 + (c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$
I like your solution because of the way you introduced a polynomial function for the problem and well done!
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Given unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.
Let $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.
From this thread http://mathhelpboards.com/challenge...^5-5=-^3-b^3-c^3-3-*-^2-b^2-c^2-2-a-8276.html we know that
$$\frac{S_5}{5}=\frac{S_3}{3}\frac{S_2}{2}$$.
Note that since $(a-b)+(b-c)+(c-a)=0$, we have $S_3=3(a-b)(b-c)(c-a)$.
Thus $$S_5=5(a-b)(b-c)(c-a)\frac{S_2}{2}$$.
Clearly $2$ divides $S_2$ and thus we are done.
#### anemone
##### MHB POTW Director
Staff member
Let $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.
From this thread http://mathhelpboards.com/challenge...^5-5=-^3-b^3-c^3-3-*-^2-b^2-c^2-2-a-8276.html we know that
$$\frac{S_5}{5}=\frac{S_3}{3}\frac{S_2}{2}$$.
Note that since $(a-b)+(b-c)+(c-a)=0$, we have $S_3=3(a-b)(b-c)(c-a)$.
Thus $$S_5=5(a-b)(b-c)(c-a)\frac{S_2}{2}$$.
Clearly $2$ divides $S_2$ and thus we are done.
Hey caffeinemachine,
Thanks for participating and that's another trick for me to learn today! | 2021-08-05T23:55:14 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/divisibility-challenge.8074/",
"openwebmath_score": 0.9143141508102417,
"openwebmath_perplexity": 2549.2735058403036,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9911526446557306,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8520580033961697
} |
https://math.stackexchange.com/questions/2561353/set-of-functions-from-empty-set-to-0-1?noredirect=1 | # Set of functions from empty set to $\{0,1\}$
How does the set of all functions $\{f \,|\, f: \emptyset \to \{0,1\}\}$ look like? Is it empty or does it contain infinitely many functions? Does the definition $f: \emptyset \to \{0,1\}$ make sense at all?
I was wondering because we know that the two sets $\{0,1\}^X$ and $\mathcal{P}(X)$ have the same cardinality. But this is only true if $X$ is non-empty, right?
• What have you tried so far? Have you tried considering the definition of "function"? – skyking Dec 11 '17 at 10:03
Yes, it makes sense. There is one and only one function from $\emptyset$ into $\{0,1\}$, which is the empty function. Think about the definition of function as a set of ordered pairs to see why.
• @philmcole If $X=\emptyset$ then $\mathcal P(X)$ has only one element. $\emptyset \in\mathcal P(X)$ but $\{\emptyset\}\notin\mathcal P(\emptyset)$ because $\{\emptyset\}\not\subseteq\emptyset.$ – bof Dec 11 '17 at 10:20
• @bof Oh, you're totally right. So the theorem holds also if $X=\emptyset$ since we have $\{0,1\}^X = \{f \,|\, f: \emptyset \to \{0,1\}\} = \{\text{the empty function}\}\sim \{\emptyset\} = \mathcal{P}(X)$ – philmcole Dec 11 '17 at 10:36
It is known as the empty function.
For any set $A$, there is exactly one function from the empty set to $A$, namely the empty function:
$$f_A: \emptyset \to A.$$
The graph of an empty function is a subset of the Cartesian product $\emptyset × A$. Since the product is empty the only such subset is the empty set $\emptyset$.
• I have a weird (maybe naif) argument: since '$x \in \emptyset$ implies that any proposition about $x$ is true', can we also say that $x \in f^{-1}(1)$? If we do this for all $x \in \emptyset$ can we conclude that $f$ is constant? And analogously, if we pick $y \in \emptyset$ such that $y \in f^{-1}(0)$ whereas for any other $x \in \emptyset, x \in f^{-1}(1)$, can we conclude $f$ is not constant? This seems to be a contradiction, does not it? – Gibbs Dec 11 '17 at 10:12
• related – Siong Thye Goh Dec 11 '17 at 10:19
• I read that $f$ is a constant function, but I see no mention of the contradiction in my previous comment... – Gibbs Dec 11 '17 at 10:25
A function $f:A\to B$ is a set of ordered pairs $(a,b)$ with $a\in A,b\in B$. In this case there is no $a\in A$ so therefore $f$ is an 'empty function' | 2019-06-26T20:10:54 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2561353/set-of-functions-from-empty-set-to-0-1?noredirect=1",
"openwebmath_score": 0.9069262742996216,
"openwebmath_perplexity": 121.58865719092536,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.960361157495521,
"lm_q2_score": 0.8872045952083047,
"lm_q1q2_score": 0.8520368319895927
} |
https://tex.stackexchange.com/questions/325142/numbering-of-each-equation-within-cases-with-left-alignment-both-on-the-page-an | # Numbering of each equation within cases with left alignment (both on the page and within the equations)
I am trying to create a set of equations with curly braces (like cases), each with their own set of numbers derived from the subsection (such as can be done with equations using \numberwithin{equation}{subsection}). I get a good result where I want one equation number representing all cases of the equation (code below).
$$\mathbb(E)(X) = \int_{-\infty}^{\infty} xdF(x) = \left\{ \begin{array}{lll} \sum_{x \in \mathcal{X}} xf_{X}(x) & =\sum_{x \in \mathcal{X}}x \mathbb{P}(X=x) &\text{ if } X \text{ is discrete}\\ \\ \int_{-\infty}^{\infty} xf_{X}(x)dx & &\text{ if } X \text{ is continuous} \end{array} \right.$$
What can I do to to have each of the cases numbered individually?
I have tried align and numcases, but each of them has some problem - either the equation or the text aligns right, or the spacing goes awry.
Would appreciate some help for the specific example above, with
1. Equation aligned to left of page
2. All text within cells left aligned
3. Equation numbers on extreme right
4. An empty cell in the 2nd case and
5. A blank line in between the two cases.
• this should be helpful: Separate labels in cases. rather than a full blank line, an optional dimension value for the gap, e.g. [.6\baselineskip] would seem preferable. – barbara beeton Aug 16 '16 at 19:37
• Off-topic: It should be \mathbb{E}(X), not \mathbb(E)(X). – Mico Aug 16 '16 at 19:48
• Thanks Mico :). I actually use macros for those and made a typo while changin it to regular (non-macro) code for the forum. – Py_Dream Aug 17 '16 at 2:13
Another solution (and numbering) with the empheq package (which loads mathtools, which loads amsmath):
\documentclass{article}
\usepackage[a4paper,margin=2.5cm]{geometry} % set page parameters
\usepackage{amsfonts, % for \mathbb and \mathcal macros
empheq}%
\numberwithin{equation}{subsection}
\setcounter{section}{1}
\setcounter{subsection}{1}
\begin{document}
\begin{subequations}
\begin{empheq}[left={\mathbb{E}(X)=\displaystyle\int_{-\infty}^{\infty} x\,dF(x)=\empheqlbrace}]{alignat = 2}
& \sum_{x \in \mathcal{X}} xf_{X}(x) =\sum_{x \in \mathcal{X}} x\mathbb{P}(X=x)
&\qquad & \text{if $X$ is discrete}, \\
& \int_{-\infty}^{\infty} xf_{X}(x)\,dx
& &\text{if $X$ is continuous}. \end{empheq}
\end{subequations}
\begin{subequations}
\begin{empheq}[left={\mathbb{E}(X)=\displaystyle∫_{-∞}^{∞} x\,dF(x)=\empheqlbrace}]{flalign}
& ∑_{x ∈ \mathcal{X}} xf_{X}(x) =∑_{x ∈ \mathcal{X}} x\mathbb{P}(X=x)
& &\text{if $X$ is discrete},&\hspace{5em} & \\
& ∫_{-∞}^{∞} xf_{X}(x)\,dx
&& \text{if $X$ is continuous}. \end{empheq}
\end{subequations}
\end{document}
• Sorry. It's once more a problem with my editor. i'LL FIX IT. – Bernard Aug 16 '16 at 21:04
• Hi, many thanks for this. Can this be done without the global [fleqn] option? I don't want all equations left aligned. – Py_Dream Aug 17 '16 at 2:02
• Yes, absolutely. Actually, what formatting was not very clear, as you can see from some comments. So, to sum it up, you want equation numbers on the right, and centred alignments? – Bernard Aug 17 '16 at 8:01
• Hi Bernard, as mentioned in the original question, I am looking for this particular set of equations to be left aligned and the numbers to be right aligned. However, I do not want equations to be left-aligned globally, i.e., I do not want the global [fleqn] option. – Py_Dream Aug 18 '16 at 11:34
• @Py_Dream: You must use the flalign environment then. Please see my updated answer. – Bernard Aug 18 '16 at 11:56
You didn't specify the type of problem(s) you encounter with the numcases environment. At any rate, I don't seem to encounter any in the following example. (The fleqn option is set so that the entire equation is set flush-left instead of centered. If that's not needed, just drop that option.)
\documentclass[fleqn]{article}
\usepackage[a4paper,margin=2.5cm]{geometry} % set page parameters
\usepackage{amsfonts} % for \mathbb and \mathcal macros
\usepackage{amsmath} % for \text and \numberwithin macros
\usepackage{cases} % for numcases environment
%% And, just for this example:
\numberwithin{equation}{subsection}
\setcounter{section}{1}
\setcounter{subsection}{1}
\begin{document}
\begin{numcases}{\mathbb{E}(X) = \int_{-\infty}^{\infty} x\,dF(x)=}
\sum_{x \in \mathcal{X}} xf_{X}(x) =
\sum_{x \in \mathcal{X}} x\mathbb{P}(X=x)
& \text{if $X$ is discrete} \\[1\baselineskip]
\int_{-\infty}^{\infty} xf_{X}(x)\,dx
& \text{if $X$ is continuous}
\end{numcases}
\end{document}
• Many thanks. I do not want to use the global [fleqn] option. So I was loading cases using \usepackage[fleqn]{cases} as was suggested somewhere. When I do this with your code (as on my earlier attempts), I get multiple errors for both the begin{numcases} line and the end{numcases} line. E.g. Undefined control sequence. ...}(X) = \int_{-\infty}^{\infty} x\,dF(x)=} Missing number, treated as zero. ...}(X) = \int_{-\infty}^{\infty} x\,dF(x)=} Illegal unit of measure (pt inserted). ...}(X) = \int_{-\infty}^{\infty} x\,dF(x)=} Is there any way to do this without the global [fleqn] option? – Py_Dream Aug 17 '16 at 2:05
• @Py_Dream - Under no circumstance should you attempt to pass the option fleqn to the cases package -- what you say "was suggested somewhere" is faulty. If you don't want to set the fleqn option at the \documentclass stage, you should set it as an option when loading the amsmath package: \usepackage[fleqn]{amsmath}. – Mico Aug 17 '16 at 5:33
• @Py_Dream - If you don't want all displayed equations to be left-aligned, certainly don't use the option fleqn option in the first place. If the non-centering should apply to just one displayed equation, you'll have to fine-tune its position "by hand", e.g., by using one or more \qquad instructions at the very end of the equation's right-hand most portion. – Mico Aug 17 '16 at 5:44
• Hi Mico, do you happen to have any further information (or links) on why fleqn should never be passed to the cases package and why the suggestion was faulty? It is not very helpful having cardinal truths stated without any reasons ascribed. – Py_Dream Aug 18 '16 at 11:37
Equation aligned to left of page: \documentclass[12pt,leqno]{book}
• I think you meant to specify fleqn, not leqno. – Mico Aug 16 '16 at 20:00
• @Mico: I had misunderstood, I guess. I'll change my code. Thanks! – Bernard Aug 16 '16 at 20:52 | 2019-10-18T23:55:31 | {
"domain": "stackexchange.com",
"url": "https://tex.stackexchange.com/questions/325142/numbering-of-each-equation-within-cases-with-left-alignment-both-on-the-page-an",
"openwebmath_score": 0.9485743641853333,
"openwebmath_perplexity": 2935.326997690743,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799482964023,
"lm_q2_score": 0.8840392878563336,
"lm_q1q2_score": 0.8520193391421655
} |
https://mathematica.stackexchange.com/questions/181868/why-gauss-legendre-quadrature-should-keep-the-number-of-integral-points-less-tha | # Why Gauss-Legendre Quadrature should keep the number of integral points less than about 50?
I wanted to use Gauss-Legendre Quadrature to calculate an integral as follows：
When n=10 and some other number(except odd numbers),the numerical result is the same as theoretical result.
Clear["Global*"];
n = 10;
L[x_] := D[(x^2 - 1)^n, {x, n}]/(n!*2^n);
A[x_] := 2/((1 - x^2) (D[L[x], x])^2);
f[x_] := E^(-x)/x;
sol = NSolve[L[x] == 0., x];
nodes = (x /. sol);
coef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];
Sum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
-2.11450175075
-2.11450175075
But,when n>=50,the code will give wrong numerical result.
Clear["Global*"];
n = 50;
L[x_] := D[(x^2 - 1)^n, {x, n}]/(n!*2^n);
A[x_] := 2/((1 - x^2) (D[L[x], x])^2);
f[x_] := E^(-x)/x;
sol = NSolve[L[x] == 0., x];
nodes = (x /. sol);
coef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];
Sum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
-43.8873164858
-2.11450175075
That is very puzzling.
NSolve has severe issues to solve for the roots (which is not surprising since finding the roots of a polynomial of degree 50 is a nontrivial task):
Max[Abs[L[x] /. sol]]
164.177
That seems to be a precision problem (the polynomial hase huge coefficients but the powers of x tend to be very small so that this easily lead to catastrophic cancellation. You will get better results with higher WorkingPrecision:
sol = NSolve[L[x] == 0, x, WorkingPrecision -> 100];
Max[Abs[L[x] /. sol]]
0.*10^-92
So the real art here is to determine the roots in a numerically stable way. Actually, Gauß-Legendre quadrature rule is already built into the system. For example, see
n = 50;
prec = 200;
{nodes, coef, bla} = NIntegrateGaussRuleData[n, prec];
(* compute all quadrature points by convex combinations*)
nodes = (-1) * (1 - nodes) + nodes * 1;
coef = 2 coef;
a = coef.f[nodes];
b = Integrate[f[x], {x, -1, 1}, PrincipalValue -> True];
a - b
2.6225212*10^-190
PS.: I was quite astonished that Gauß quadrature works so well with this singular integral. Then it came back to my mind that it is constructed such that it neglects contributions of all odd functions (because they would integrate to 0 anyways)...
• thank you very much! I thought it was easy to solve for the roots of polynomial of large degree before. – Quere Sep 14 '18 at 12:25
• @Quere In the meantime, I found out that using higher WorkingPrecision helps to make your method more precise. – Henrik Schumacher Sep 14 '18 at 12:33
• Yes!I noticed it.The result was really amazing. – Quere Sep 14 '18 at 12:37
• Not really worth writing as a separate answer, so: the trick of using even-order Gauss-Legendre quadrature to evaluate principal value integrals with a singularity at the center was published by Piessens, but I seem to recall this being known as folklore. In case the pole is not exactly in the middle of the integration interval, one can either split the integral so that there is a regular part and a part symmetric about the pole, or one can use a substitution that centers the pole, allowing the use of Gaussian quadrature. – J. M.'s discontentment Sep 24 '18 at 19:45
If you add WorkingPrecision -> 60 in the NSolve function you get your solution.
n = 50;
A[x_] := 2/((1 - x^2) (D[LegendreP[n, x], x])^2);
f[x_] := E^(-x)/x;
sol = NSolve[LegendreP[n, x] == 0, x, WorkingPrecision -> 60];
nodes = (x /. sol);
coef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];
Sum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
-2.11450175075145702914368470979175591804810787513962
-2.1145
Or you can use the built in functions i mma
<< NumericalDifferentialEquationAnalysis
n = 50;
{pts, w} = Transpose[GaussianQuadratureWeights[n, -1, 1]];
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
Sum[w[[i]] f[pts[[i]]], {i, 1, Length@pts}]
-2.1145
-2.1145
A more stable way of computing the roots of a polynomial family that satisfies a three-term recurrence is the method of Golub and Welsch (1969), which computes the eigenvalues of a matrix based on the recurrence, sometimes called the "comrade matrix."
(*
* n-point Gauss quadrature (Golub-Welsch 1969)
*)
(* p[n][x] == (a[n]x+b[n])p[n-1][x] - c[n]p[n-2][x] *)
comradeMatrix[{a_, b_, c_}, {n_, n0_Integer}, prec_: Infinity] :=
Block[{n = Range@n0},
With[{beta = Sqrt[Rest@c/(Most@a*Rest@a)]},
N[SparseArray[{Band[{1, 1}] -> -b/a, Band[{2, 1}] -> beta,
Band[{1, 2}] -> beta}, {n0, n0}], prec]
]];
Module[{n}, comradeMatrix[{2 n - 1, 0, n - 1}/n, {n, n0}, prec]];
(*
{-0.998866, -0.994032, -0.985354, -0.972864, -0.956611, -0.936657, \
-0.913079, -0.885968, -0.85543, -0.821582, -0.784556, -0.744494, \
-0.701552, -0.655896, -0.607703, -0.557158, -0.504458, -0.449806, \
-0.393414, -0.3355, -0.276288, -0.216007, -0.154891, -0.0931747, \
-0.0310983, 0.0310983, 0.0931747, 0.154891, 0.216007, 0.276288, \
0.3355, 0.393414, 0.449806, 0.504458, 0.557158, 0.607703, 0.655896, \
0.701552, 0.744494, 0.784556, 0.821582, 0.85543, 0.885968, 0.913079, \
0.936657, 0.956611, 0.972864, 0.985354, 0.994032, 0.998866}
*)
The integration weights can be computed from these nodes as in the OP.
• As a terminological note: the tridiagonal matrix with the orthogonal polynomial's recurrence coefficients is called a Jacobi matrix. (The comrade matrix, as you might recall, is a Jacobi matrix for the Chebyshev polynomials, with a correction term representing the Chebyshev series coefficients.) I talked about this here. – J. M.'s discontentment Sep 24 '18 at 11:27
• @J.M. Barnett (1975) seems to introduce the name "comrade matrix" of a polynomial with respect to an arbitrary family of polynomials that satisfies a three-term recurrence and presents it as a generalization of the Chebyshev colleague matrix of Good (1961). But, you know, I'm not that confident about current terminology in this area. -- Yes, I found the link to that Q&A shortly after posting an answer and put it under the question above. – Michael E2 Sep 25 '18 at 17:34
• Argh, I switched up "comrade" and "colleague" again... I keep switching the general and specific cases. :D – J. M.'s discontentment Sep 25 '18 at 17:40 | 2020-10-19T15:44:24 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/181868/why-gauss-legendre-quadrature-should-keep-the-number-of-integral-points-less-tha",
"openwebmath_score": 0.3738044500350952,
"openwebmath_perplexity": 2050.6371451778236,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799420543365,
"lm_q2_score": 0.8840392893839086,
"lm_q1q2_score": 0.8520193350961802
} |
https://electronics.stackexchange.com/questions/565679/finding-resonant-frequency-or-damping-ratio-from-bode-plot | # Finding resonant frequency or damping ratio from Bode Plot
I am working on a question where I have to estimate a transfer function from its bode plot.
I plotted the asymptotes of this bode diagram, and was able to find out that this is 3rd order system with a pole at s = 0 and two complex poles.
$$\dfrac{w_n^2}{s(s^2 + 2\zeta w_n s + w_n^2)}$$ where I found out that $$w_n = 6.7 rad/s$$ from the asymptotic bode plot.
I was also able to find the system gain $$\K\$$ from the magnitude plot.
I am unable to find $$\\zeta\$$ here. I tried finding the resonant frequency $$\w_r\$$ from the bode plot so that I can calculate $$\\zeta\$$ using: $$w_r = w_n\sqrt{1 - 2\zeta^2}$$ but I was unable to.
I know that the frequency at which the phase plot crosses zero is the resonant frequency but the phase plot here doesn't cross zero.
I tried approximating $$\\zeta\$$ using the fact that maximally flat response is obtained for $$\\zeta = 0.707 \$$, so that for the given plot, $$\\zeta < 0.707 \$$. But I wasn't able to exactly find a value.
Is there any other way to find $$\\zeta\$$ or will I be just able to approximate it?
The solution says that the value of $$\\zeta\$$ is $$\0.447\$$.
• Curiously, it changed May 18, 2021 at 13:01
• @TonyStewartEE75 Sorry I added the wrong bode plot. May 18, 2021 at 13:02
• Compute d phi /dt max in the same units(rads) May 18, 2021 at 13:06
• Or rather $d\phi /d\omega$ then determine the ratio constant May 18, 2021 at 13:18
• If the plot is not on a paper (i.e. you have access to the datapoints), subtract the 1/s from it and you'll be left with a more manageable 2nd order magnitude/phase. May 18, 2021 at 15:09
The following graphical solution relies on the idea that it's possible to separate the contribute of the zero in w=0 by referring the magnitude measurements to a line slanted with a -20dB/decade slope. Thanks to the properties of logarithms, division becomes translation on the magnitude Bode plot. We also have to take into account the -90 degrees contribution in the phase - it's basically a constant -90° addition since, being the pole in the origin, it has already 'run its course'.
1. the first step is to find wn. In a second order system with no zeros, the phase resonance happens exactly at wn, the undamped natural frequency (a frequency that is in general different from wpeak, the peak frequency of the magnitude, and also from the damped natural frequency wd). Since we need to separate the phase contribution of the pole in the origin, instead of finding the frequency where the phase is -90° we need to find the frequency where the phase is -180° By eyeballing the scale on the tiny plot I have I believe I can locate it at 6.7 rad/s. You might end up with a better estimate.
Now I want to find the 3dB corner frequency the system would have without the pole in the origin . I therefore...
1. ...trace a slanted line, translated 3dB under the asymptotic behavior at low frequencies and
2. ...and then I look for the intersection of said line with the magnitude of the transfer function to find w3dB. Eyeballing again I find w3dB = 8.7 rad/sec
3. We are now in the position to computed the ratio w3dB/wn = 1.298 = 1.3
We can now either solve the expression for w3dB as a function of zeta
or, if we have a graph like this, 5) use it to find the value of zeta corresponding to w3dB/wn = 1.3.
again by eyeballing I get a zeta value of around 0.48, a value not dissimilar from that found by solving the equation
for zeta, which gives zeta = 0.477
And this value is reasonably close, considering the amount of lazy eyeballing employed, to the correct answer of 0.447. Try your estimates on a bigger graph, counting the pixels and report back. Did it work?
Caveat emptor: it is imperative that the second order function be without additional zeroes (apart for the one we have been able to separate). The relevant frequencies have different expressions form system with one or more zeroes.
• Thank you, yeah it did work. I got the w3dB to be around 8.9 rad/s from the bode plot, which gave me a ratio of 1.328. Got zeta = 0.4482. Thanks a lot! Jun 13, 2021 at 5:39 | 2022-07-07T15:10:39 | {
"domain": "stackexchange.com",
"url": "https://electronics.stackexchange.com/questions/565679/finding-resonant-frequency-or-damping-ratio-from-bode-plot",
"openwebmath_score": 0.8104416728019714,
"openwebmath_perplexity": 528.5693212282331,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799472560581,
"lm_q2_score": 0.884039278690883,
"lm_q1q2_score": 0.8520193293889828
} |
https://mathematica.stackexchange.com/questions/244964/plot-graph-of-function-with-different-constants | # plot graph of function with different constants
I need to plot the graph of :
$$v(t) = \frac{-mg}{b} (e^{-\frac{b}{m}t}+1)$$ , where $$g = 9.8 m/s^2$$ and $$b$$ and $$m$$ are positive constants.
Is there a way to plot this without having to define random values for $$b$$ and $$m$$? I mean a way in which I can see how the graph behaves for different values of $$b$$ and $$m$$ ?
• Unfortunately no, I am new to Mathematica, I only know the basics – Physmath Apr 22 at 2:17
• I will take a look, thank you ! – Physmath Apr 22 at 2:22
• To simplify little bit, you can introduce a new variable, say $\lambda=\frac{b}{m}$. Then you can Plot a sequence of curves for a list of $\lambda$ values. – yarchik Apr 22 at 3:03
v[s_, t_] := -g/s (Exp[-s t] + 1)
g = 9.8;
slist = {0.1, 0.2, 0.4, 0.8};
Plot[Evaluate[v[#, t] & /@ slist], {t, 0, 10},
PlotTheme -> {"BoldColors", "Frame", "Grid"},
FrameLabel -> {Automatic, v[t]},
FrameStyle -> Directive[14, Black],
PlotLabels -> slist
]
• Plot[Evaluate[v[slist, t] , {t, 0, 10},...] works too. – Ulrich Neumann Apr 22 at 8:02
• @UlrichNeumann That is good suggestion, it makes the syntax more transparent. – yarchik Apr 22 at 8:12 | 2021-06-18T06:37:37 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/244964/plot-graph-of-function-with-different-constants",
"openwebmath_score": 0.33643510937690735,
"openwebmath_perplexity": 851.4005363716258,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9766692359451417,
"lm_q2_score": 0.8723473796562744,
"lm_q1q2_score": 0.85199484876764
} |
https://math.stackexchange.com/questions/3935988/showing-that-if-3x-is-even-then-3x5-is-odd/3936024 | # Showing that if $3x$ is even then $3x+5$ is odd
I'm learning the absolute basics of how to do proofs, and am really struggling.
If 3x is even then 3x+5 is odd.
This is the solution:
I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but nothing I found is this basic.
Work: -Assumptions- 3x = 2n 3x+5 = 2k+1
-Trying to make sense of 3x- 3x+5 = 2k+1 3x = 2k-4
-Plugging in 2k-4 for 3x- 2k-4 = 2n 2k = 2n+4 k = n+2
-Plugging in n+2 for k- 3x+5 = 2(n+2)+1
...This is where I gave up. I don't know where I'm going with this anymore.
• You should calm down, everyone can have difficulties. Explain us where you get lost. If you don't know where you get lost I can try to explain you the proof step by step. – Eureka Dec 5 '20 at 17:42
• I'm good, just frustrated. I'd like step by step. I'll edit in my work above. Warning, there is no rhyme or reason to it. – visualbread Dec 5 '20 at 17:44
• If $y$ is even, can you prove that $y+5$ is odd? Try this first, then try your problem. There’s no substantial difference. – Benjamin Wang Dec 5 '20 at 17:46
• @visualbread your resolution may be a bit convoluted but it's correct. You just proved that $3x+5=2(n+2)+1$. And $2(n+2)+1$ is odd ,because it is of the form $2h+1$ with $h=n+2$ – Eureka Dec 5 '20 at 17:52
• @BenjaminWang I am unable to do that one either. From y = 2n and y+5 = 2k+1, I can get that all the way to n = k -2 (or k = n+2), but I can't see what that tells me. The final solution has everything nicely equal to one another, and I can't seem to set it up that way. – visualbread Dec 5 '20 at 17:54
I think you're getting confused by the $$3x$$ part. The $$3x$$ plays no role in the problem.
Suppose you're given any number that is even. I'll call it y. Now we want to show $$y+5$$ is odd.
Therefore by definition
$$y=2k$$ for some integer $$k$$
Now
$$y+5 = 2k+5$$
Now we just need to show that $$2k+5$$ is 2 times an integer plus 1
$$2k+5 = 2(k+2)+1$$
So $$2k+5$$ is odd because it can be written in the form 2*integer +1 where the integer here is $$k+2$$. So $$y+5$$ is odd since $$y+5 = 2k+5$$
So if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc...
• I might be there, but I need just a little bit more help. How did you write 2k+5 in the form 2(k+2) + 1? I can see that those are the same, but how could I do it without trial and error? – visualbread Dec 5 '20 at 18:16
• So the first step is to look at 5. We see it is an odd integer. So I decide it is more convenient to write it as $4+1$. So I write $2k+5$ as $2k+4+1$. Now I can factor $2k+4 = 2(k+2)$. So putting it all together $2k+5 = 2k+4+1 = 2(k+2)+1$. Hope this makes sense. let me know. Suppose I had $2k+6$ instead. since 6 is even, I know I can factor immediately $2k+6 = 2(k+3)$. – Ameet Sharma Dec 5 '20 at 18:19
• Thanks! This shows me all the steps I need to follow my book's logic. I don't know what happened in my math education, but I never did learn some of these little things that make some of these proofs actually understandable. At least there's websites like this one to fill in the gaps. – visualbread Dec 5 '20 at 18:25
• No problem. I highly recommend doing lots of exercises with factoring, simplification and algebra so that these things become second nature. It takes a lot of practice but it is worth it. As you get into more advanced math, these things really need to be 2nd nature where you don't need to think about them anymore... otherwise you'll get bogged down at every step. – Ameet Sharma Dec 5 '20 at 18:31
• I'm sure there are lots of online websites. Here's one I just found as part of an online algebra textbook: openstax.org/books/college-algebra/pages/1-review-exercises. Each chapter has review exercises and a practice test. I recommend doing all these, maybe multiple times till you can do them in your sleep. In my opinion... if you want online resources, look for online algebra "textbooks"... and do all the exercises in each chapter. If you have doubts you can review the chapter. – Ameet Sharma Dec 5 '20 at 18:51
Your problem is that $$3x + 5 = 2k +1$$ is not an assumption. It is the conclusion you need to prove.
Your one and only assumption is that $$3x = 2n$$ for some integer $$n$$.
$$3x = 2n$$.
.... then you do a bunch of steps ....
.... steps .....
.... and get in the end ........
Conclusion: $$3x + 5 = 2(????????) + 1$$ where $$??????$$ is some integer you come up with in you steps.
Let's see what happens when we try. Let's take it nice and slow:
=======
$$3x = 2n$$.
$$3x +5 = 2n + 5$$
....hmmm, we want $$2(??????) + \color{red}1$$ in the end so let's pull out the $$+\color{red}1$$ first.....
$$3x + 5 = 2n+5 = 2n + (4 + \color{red}1)=(2n+4) +\color{red}1$$
.... hmmm, okay that's the $$+1$$ now we want $$2(\color{red}{??????}) + 1$$. To get the So we need to factor then $$2$$ out of $$2n+4$$ and see what we have left.... that will bee the $$\color{red}{??????}$$
$$3x + 5 = (\color{red}{2n+4}) + 1$$
$$3x + 5= 2(\color{red}{n + 2}) + 1$$
.... and that's it......
Conclusion: $$3x+5 = 2(\color{red}{n + 2})+1$$.
The $$??????$$ we wanted turns out to be $$\color{red}{n+2}$$ an we have
$$3x + 5 = (3x+4) + 1 = (2n+4) + 1 = 2(\color{red}{n+2}) + 1$$.
And because $$\color{red}{n+2}$$ is an integer if we let $$k = n+2$$ be that integer $$3x+5 = 2k + 1$$ and so... $$3x + 5$$ is odd.
=======
Although if you want to work backwards
Conclusion: $$3x + 5 = 2k +1$$ ..... and we want to solve for $$k$$ to show it is possible...
$$3x + 5 -1 =2k + 1-1$$
$$3x +4 = 2k$$
$$k = \frac {3x + 4}2 = \frac {3x}2 + 2$$.
.... but is $$\frac {3x}2 + 2$$ an integer?????
Well, $$3x$$ is even. So there is an integer $$n$$ so that $$3x = 2n$$ so
$$k = \frac {3x}2 +2 = \frac {2n}2 + 2 = n+2$$.
So $$k=n+2$$ is the integer we want to conclude $$3x+5 =2k +1$$.
If we did it this way our proof would go:
$$3x$$ is even so there is an integer $$n$$ so that $$3x = 2n$$. Let $$k = n+2$$; that is an integer.
$$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$$.
So $$3x+5 = 2k +1$$ and that is odd.
Ok so we know that $$3x$$ is even, that means we can write $$3x=2n$$ for a suitable $$n$$, since even means that the number is divisible by two without remainder. But then we have $$3x+5=2n+5=2n+(4+1)=2n+2\cdot 2+1=2(n+2)+1$$ which clearly is odd.
• Thanks for the reply. I'm still not there yet. I can't figure out where 2n+5 came from or why it can be set equal to 2x+5. I need this broken down even further if that's even possible. – visualbread Dec 5 '20 at 17:55
• We set it equal to $3x+5$ NOT $2x+5$. I will try to break it down further: so – Mo145 Dec 5 '20 at 18:04
• 1)a natural number $N$ is called even if it is a multiple of $2$. So per definition we can write every even number $N$ as $N=2M$. Note that we can write any natural number as $N=2\frac{N}{2}$ but in general $\frac{N}{2}$ will not be a natural number, for example take $N=3$, then clearly $\frac{3}{2}$ is not a natural number. In fact this will only be the case if $N$ is even. – Mo145 Dec 5 '20 at 18:08
• 2)Since we assume $3x$ to be even we can, by 1) write it as $3x=2n$ where $n=\frac{3x}{2}$ so we have the following equality $3x=2n$ adding a number on both sides preserves the equality hence we have $3x+5=2n+5$ – Mo145 Dec 5 '20 at 18:11
• I hope this clarifies your concerns :) – Mo145 Dec 5 '20 at 18:11
We know that $$3x=2n$$. So: $$\color{blue}{3x}+5=\color{blue}{2n}+5$$ Because the "blue quantities" are equal. Now: $$3x+5=2n+5=2n+4+1$$ In this step I just wrote $$5$$ as $$4+1$$:
$$3x+5=2n+5=2n+4+1=2n+2\cdot 2+1$$ In this step I just wrote $$4$$ as $$2 \cdot 2$$.
$$3x+5=2n+5=2n+4+1=\color{green}{2n+2\cdot 2}+1=\color{green}{2(n+2)}+1$$ The last step is valid because the green quantities are equal. In the end: $$3x+5=2(n+2)+1$$ This means that $$3x+5$$ is odd because is of the form $$2h+1$$ with $$h$$ integer(in particular $$h=(n+2)$$)
Remember here that $$n$$ represents ANY natural number. You got to the answer but you didn’t even realize it. That’s probably because you are thinking syntactically rather than semantically. What I mean is the literal string of symbols $$2(n+2)$$ didn’t register to you as even because it is not the same as the string $$2n$$. But $$n+2$$ is a natural number just like $$n$$ is. So the the strings $$2(n+2)$$ and $$2n$$ both represent even numbers, and so $$2(n+2) + 1$$ is odd just as you have shown in your last line.
You are dealing with a problem where you are given too much information. Here is another way of doing it (we'll use $$m$$ for an integer).
If $$3x$$ is even then $$x$$ must be even, so we can put $$x=2m$$. [This is a consequence of the fact that $$2$$ is a prime, or can be proved in various ways]
Then $$3x+5=6m+5=6m+4+1=2(3m+2)+1$$.
Now we can put $$3m=n$$, an integer, to get $$2(n+2)+1$$.
For an alternative proof, we could put $$3m+2=n$$ and then we get $$2n+1$$.
Note that the last two sentences are alternatives to one another. They both use $$n$$, but $$n$$ is defined differently in the two cases.
What I'm trying to do here is to unpack how the different expressions for the same thing relate to each other. If you get your head round that you will be flying.
• I find this answer to be pretty overwhelming for the OP, since you are talking about "alternative proofs" and "consequences of the fact that 2 is prime that can be proved in various ways". I'm not underestimating the OP, but I think that when you learn fron basic you should do things calmly step by step. – Eureka Dec 5 '20 at 18:23
• @Eureka What I was trying to do was to show what happens when you keep the factor $3$ for an extra step. If it is "overwhelming" it might help OP to understand why the factor $3$ is best dropped ("too much information"). Either it helps or it doesn't. – Mark Bennet Dec 5 '20 at 18:35 | 2021-07-30T09:41:49 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3935988/showing-that-if-3x-is-even-then-3x5-is-odd/3936024",
"openwebmath_score": 0.7947751879692078,
"openwebmath_perplexity": 239.42945834890642,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9766692271169855,
"lm_q2_score": 0.8723473862936943,
"lm_q1q2_score": 0.8519948475489848
} |
https://math.stackexchange.com/questions/2229402/evaluate-int-0-frac-pi4-cos-1-sin-x-dx | # Evaluate $\int_0^\frac{\pi}{4} \cos^{-1}({\sin x}) \,dx$
I came across this integral in a math competition and couldn't solve it $$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x})\, dx$$
I tried a $u$-substitution, with $u=\sin x$ and ended up with the integral $$\int_0^\frac{\pi}{4} \cos^{-1}(\text{u}) \cdot \frac{1}{\sqrt{1-u^2}}\,du$$ which is not much simpler and I cannot figure out how to solve this. Any hints/solutions for this problems?
I also tried drawing a triangle for the problem but it didn't really help with the solution.
You can use the trigonometric formula : $$\forall x \in\mathbb{R},\cos^{-1}(x)+\sin^{-1}(x)=\frac{\pi}{2}$$
Thus $\forall x\in\left[0,\frac{\pi}{4}\right]$ you have $\sin^{-1}(\sin(x))=x$ so : $$\cos^{-1}(\sin(x))+\sin^{-1}(\sin(x))=\frac{\pi}{2}\Rightarrow \cos^{-1}({\sin x})=\frac{\pi}{2}-x$$
Finally : $$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x}) dx=\int_0^\frac{\pi}{4} \frac{\pi}{2}-xdx$$ Can you finish ?
• What is $\sin^{-1}(\sin x)=?$ – lab bhattacharjee Apr 11 '17 at 15:48
• it is $x$ since $x$ is in $[0,\pi/4]$ – Jennifer Apr 11 '17 at 15:54
HINT:
If $\cos^{-1}(\sin x)=y,0\le y\le\pi\ \ \ \ (1)$
and $\cos y=\sin x=\cos\left(\dfrac\pi2-x\right)$
$y=2m\pi\pm\left(\dfrac\pi2-x\right)$ where $m$ is an integer such that $(1)$ is satisfied
For $0\le2m\pi+\dfrac\pi2-x\le\pi\iff 2m\pi-\dfrac\pi2\le x\le2m\pi+\dfrac\pi2$
Here $m=0$
Here's a barely different route to take, continuing from the direction you've taken.
First, note that substituting $u=\sin x$ would actually give
$$\int_{x=0}^{x=\pi/4}\cos^{-1}(\sin x)\,\mathrm dx=\int_{\color{red}{u=0}}^{\color{red}{u=1/\sqrt2}}\frac{\cos^{-1}u}{\sqrt{1-u^2}}\,\mathrm du$$
Now, recall that $\dfrac{\mathrm d}{\mathrm du}\cos^{-1}u=-\dfrac1{\sqrt{1-u^2}}$, which means you can make another intermediate substitution of, say, $v=\cos^{-1}u$. Then you have
$$-\int_{v=\cos^{-1}0=\pi/2}^{v=\cos^{-1}(1/\sqrt2)=\pi/4}v\,\mathrm dv=\int_{\pi/4}^{\pi/2}v\,\mathrm dv=\dfrac{3\pi^2}{32}$$
which agrees with the other answers above.
$$\int \cos^{-1}(\sin x) dx$$ $$= \int 1 \cdot \cos^{-1}(\sin x) dx = x \cdot \cos^{-1}(\sin x) - \int x \cdot -\frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C$$ (integration by parts)
$$= x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C$$ $$= x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{\cos^2 x}} dx + C$$
$$= x \cdot \cos^{-1}(\sin x) \pm \int x \cdot \frac{\cos x}{\cos x} dx + C$$ $$= x \cdot \cos^{-1}(\sin x) \pm \frac{x^2}{2} + C$$
Now do it for the definite integral.
1)
$(u^2)^{\prime}=2\times u\times u^{\prime}$
therefore $\dfrac{1}{2}u^2$ is an antiderivative of $u\times u^{\prime}$
2) derivative of $\text{cos}^{-1}(x)=-\dfrac{1}{\sqrt{1-x^2}}$
$\displaystyle J= \int_0^{\tfrac{\pi}{4}} \text{cos}^{-1}(\sin x)dx$
Perform the change of variable $y=\sin x$,
$\displaystyle J=\int_0^{\tfrac{\sqrt{2}}{2}} \dfrac{\text{cos}^{-1}(x)}{\sqrt{1-x^2}}dx$
Using $(1)$,
\begin{align}J&=-\dfrac{1}{2}\Big[\text{cos}^{-1} (x)^2\Big]_0^{\tfrac{\sqrt{2}}{2}}\\ &=-\dfrac{1}{2}\left[\dfrac{\pi^2}{16}-\dfrac{\pi^2}{4}\right]\\ &=\boxed{\dfrac{3}{32}\pi^2} \end{align} | 2019-05-25T18:58:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2229402/evaluate-int-0-frac-pi4-cos-1-sin-x-dx",
"openwebmath_score": 0.9529788494110107,
"openwebmath_perplexity": 324.47443765424987,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9766692284751635,
"lm_q2_score": 0.8723473746782093,
"lm_q1q2_score": 0.851994837389301
} |
https://mechanicalc.com/calculators/beam-analysis/validation | This section details the validation for the Beam calculator. Several examples are worked through to determine expected results such as deflections, forces, and stresses. These expected results are then compared to the actual output of the calculator.
The standard beam equations for a cantilever beam loaded at the free end are given below:
Max Deflection: $$\delta_{max} = {P L^3 \over 3EI}$$ @ x = L Max Slope: $$\theta_{max} = {P L^2 \over 2EI}$$ @ x = L Shear: $$V = +F$$ constant Moment: $$M_{max} = -FL$$ @ x = 0
For the validation case, an Aluminum 6061-T6 beam with a 1 inch diameter circular cross-section and a length of 10 inches is loaded at 1000 lbf. The inputs are:
L = 10 in
F = 1000 lbf
E = 9.9e6 psi (Al 6061-T6)
$$A = {\pi d^2 \over 4} = 0.7854 ~in^4$$
$$I = {\pi d^4 \over 64} = 0.04909 ~in^4$$
For the given inputs, the expected results are:
Max Deflection: $$\delta_{max} = 0.6859 \text{ in}$$ @ x = L Max Slope: $$\theta_{max} = 0.1029 \text{ rad}$$ @ x = L Shear: $$V = +1000 \text{ lbf}$$ constant Moment: $$M_{max} = -10,000 \text{ in-lbf}$$ @ x = 0
### Comparison With Calculator
#### Free Body Diagram
The Free Body Diagram (FBD) is shown below. From the FBD it can be seen that the forces balance and the beam is in static equilibrium.
#### Deflection & Slope
A screenshot of the results table giving the maximum deflection and slope is shown below. It can be seen that these values match the expected values above:
The deflection plots are shown below. From these plots it can be seen that a maximum deflection of 0.6859 inches occurs at the free end of the beam as well as a maximum slope of 0.1029 radians.
#### Shear-Moment Diagram
The shear-moment diagram for the beam is shown below. This diagram is what would be expected for the current case. A constant shear force of 1000 lbf exists along the length of the beam, and the moment increases linearly from 0 in-lbf at the free end of the beam to the full value of -10,000 in-lbf at the fixed end.
#### Stresses
There are two points of interest for validating stresses. Before calculating stress, the forces at these points need to be determined.
Axial (lbf) Shear (lbf) Moment (in-lbf)
Forces @ x = 0: $$F_{ax} = 0$$ $$F_{sh} = 1000$$ $$M = 10000$$
Forces @ x = L: $$F_{ax} = 0$$ $$F_{sh} = 1000$$ $$M = 0$$
The stresses are calculated using the following equations:
Axial Stress Shear Stress Bending Stress Von Mises Stress
$$\sigma_{ax} = {F_{ax} \over A}$$ $$\tau_{sh} = {F_{sh} \over A}$$ $$\sigma_{b} = {Mc \over I}$$ $$\sigma_{vm} = \sqrt{ (\sigma_{ax} + \sigma_{b})^2 + 3\tau_{sh}^2 }$$
Based on the forces at each point and the equations above, the expected stresses at the points of interest are:
Tensile (psi) Shear (psi) Bending (psi) Von Mises (psi)
Stress @ x = 0: $$\sigma_{ax} = 0$$ $$\tau_{sh} = 1273$$ $$\sigma_{b} = 101859$$ $$\sigma_{vm} = 101883$$
Stress @ x = L: $$\sigma_{ax} = 0$$ $$\tau_{sh} = 1273$$ $$\sigma_{b} = 0$$ $$\sigma_{vm} = 2205$$
The stress plots are shown below. It can be seen that these plots agree with the stresses calculated above.
There is one discrepancy that can be noted, which is that the values for bending stress and von Mises stress at x = L are shown to be higher in the stress plots than predicted. The reason for this is that the stress plots are showing the maximum nodal values for each element, and the maximum bending stress for the element at x = L will occur at the left-most node in that element. Therefore, the values of bending stress and von Mises stress reported in the plots will be somewhat higher than what was calculated above. However, the stresses for each node within any element can be interrogated in the result tables. This table will be discussed following the stress plots.
##### Stress At End of Beam (x = L)
A screenshot of one of the result tables for this case is shown below with the element of interest highlighted. The node at the very end of the beam (i.e. at x = L) is "Node 2" for the highlighted element. It can be seen that the bending stress at this node is 0 psi and the von Mises stress at this node is 2205 psi. This matches what was predicted. | 2020-03-31T07:27:45 | {
"domain": "mechanicalc.com",
"url": "https://mechanicalc.com/calculators/beam-analysis/validation",
"openwebmath_score": 0.6969012022018433,
"openwebmath_perplexity": 740.8264258502527,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9766692311915195,
"lm_q2_score": 0.8723473713594991,
"lm_q1q2_score": 0.851994836517625
} |
https://math.stackexchange.com/questions/2475295/find-the-remainder-when-528528528-up-to-528-digits-is-divided-by-27 | # Find the remainder when $528528528…$up to $528$ digits is divided by $27$?
Find the remainder when $528528528...$up to $528$ digits is divided by $27$?
Here's what I have done: The number can be written as $528\cdot 10^{525}+528\cdot 10^{522}+...+528$ which has $176$ terms and each term is $\equiv15 \mod 27$ thus the number should be $176*15 \mod 27$ hence $21$ should be the remainder. But book says it is $6$. I don't understand the flaw in my logic. Please correct me.
• you have $21+6=27$ perhaps you are off by a sign? – gt6989b Oct 16 '17 at 15:26
• I think that your answer is correct. – alexp9 Oct 16 '17 at 15:29
• Brute force in python gives 21. – Gribouillis Oct 16 '17 at 15:31
• Are you sure the book says $6$ and not $-6$? After all, $21\equiv -6 \pmod{27},$ so then both answers would agree. – David K Oct 16 '17 at 15:36
• @fleablood Also a good point about negative remainders. It looks like there's probably an error in the book. (Unless the transcription of the problem is very confused, it's only $176$ "copies" of the group of digits $528,$ which makes $528$ digits altogether since each group has three digits. Also, we don't need $10^k$ to be congruent to $1$; we only need $10^{3k}\equiv 1 \pmod{27},$ which is true.) – David K Oct 16 '17 at 17:00
Here is a python3 session
>>> s = '528' * 176
>>> len(s)
528
>>> int(s) % 27
21
• Is it possible to multiply a string by a number? My god...python is awesome. – Integral Oct 16 '17 at 15:35
• @Integral yes strings, lists and tuples can be repeated by multiplying them by a number. – Gribouillis Oct 16 '17 at 15:36
You can see that $6$ cannot be correct by casting out $9$'s: Since $5+2+8=5+5+5$, we have
$$528528\ldots528\equiv5+5+5+\cdots+5+5+5=5\cdot528\equiv5(5+2+8)\equiv5\cdot6\equiv3\mod 9$$
so the remainder mod $27$ must be either $3$, $12$, or $21$. Your approach gave the correct answer, $21$.
• Anyway it is also $-6$ so there is a typo in the book or Anuran has not have seen the minus sign. – Piquito Oct 16 '17 at 15:54
• @Piquito, I agree, a negative sign would fix things (as David K noted in comments). But remainders are usually understood to be nonnegative, so I'm inclined to think it's a typo. – Barry Cipra Oct 16 '17 at 15:57
Since $$3\mid111$$, we know that $$27\mid999$$, Therefore, $$1000\equiv1\pmod{27}$$ Thus, \begin{align} \sum_{k=0}^{175}528\cdot1000^k &\equiv528\cdot176\pmod{27}\\ &\equiv3\cdot176^2\pmod{27}\\ &\equiv3\cdot14^2\pmod{27}\\ &\equiv3\cdot7\pmod{27}\\ &\equiv21\pmod{27} \end{align}
You are incorrect in assuming $10^{k} \equiv 1 \mod 27$. As $10^k \not \equiv 1$ we do not have $528*10^k \equiv 15 \mod 27$.
What you need instead is $528528... = 528(1001001001......)$
And $1001001..... =\sum_{k=0}^{175} 10^{3k}$
$10^3 \equiv 1 \mod 27$... Oh... we do have that and you were not wronng after all.... so $\sum 10^{3k}\equiv 176 \equiv 14 \mod 27$.
So $528528....... \equiv 15*14 \equiv 21 \mod 27$.
And... the book is wrong.
Had it been 527 iterations of 528 the answer would be $6$.
Note that $$27 \times 37 = 999$$.
To find the remainder you get when you divide $$528528\cdots 528$$ by $$999$$, you can "cast out" $$999$$s.
$$\underbrace{528 + 528 + \cdots 528}_{\text{176 times} } \to 176 \times 528 \to 92928 \to 92+928 \to 1020 \to 21$$
So the remainder is $$21$$.
Note. If the remainder was bigger than 26, you would have had to divide it by 27 to get the correct remainder. | 2021-04-17T09:25:07 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2475295/find-the-remainder-when-528528528-up-to-528-digits-is-divided-by-27",
"openwebmath_score": 0.802299439907074,
"openwebmath_perplexity": 442.68319873287027,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9766692311915195,
"lm_q2_score": 0.872347369700144,
"lm_q1q2_score": 0.8519948348969839
} |
https://matheducators.stackexchange.com/questions/14612/why-do-we-study-cantor-set | # Why do we study Cantor Set?
For finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ?
• Why are you not convinced that counterexamples are an important thing to study? (Re: "at least not always", the beauty of a counterexample is that only needs to be used once, in a particular key spot, to change our understanding forever.) – Daniel R. Collins Sep 30 '18 at 1:07
1. in beginning real analysis: to counter the naive notion that a "closed set" is a union of closed intervals, plus a few single points.
2. In beginning Lebesgue measure: the easiest example of an uncountable set of measure zero
3. In general topology: sets homeomorphic to the Cantor set are useful in proofs
4. Fractal geometry: many fractals are homeomorphic to the Cantor set. Mandelbrot calls such a thing a Cantor dust to suggest its appearance.
• Can one claim and substantiate that the study of cantor set is something fundamental and not just a source of ready counterexample ? – Vagabond Sep 29 '18 at 15:21
• Point (4) seems to address this. The Cantor set is one of the simplest examples of a fractal, and is therefore studied in detail by students who are first learning fractal geometry or analysis (to the point that a colleague of mine, after writing a masters thesis on Cantor-like sets, started her PhD by stating that she never wanted to see the Cantor set again). – Xander Henderson Sep 30 '18 at 14:51
The Cantor set is quite useful in its own right. My preferred way to think of the Cantor set is as "the most general compact metrizable space." That it is the most general such space means that it is often good for counterexamples because it lacks the particulars. At the same time, one can construct things by specialization.
The formal expression of the Cantor set being the most general compact metrizable space is that every compact metrizable space is the image of the Cantor set under a continuous function. One can use this as a method of proof. For example, to show that a continuous surjection from $$[0,1]$$ onto $$[0,1]^2$$ exists, one can use the fact that $$[0,1]^2$$ is the image of the Cantor set under a continuous function and then extend this continuous function by linear interpolation (the complement of the Cantor set consists of open intervals) to all of $$[0,1]$$.
A lot more can be proven this way, including a number of very surprising results. A wonderful source for such arguments is the following paper, which (deservedly) won an award for mathematical exposition:
Benyamini, Yoav. "Applications of the universal surjectivity of the Cantor set." The American mathematical monthly 105.9 (1998): 832-839.
Among the results proven there is the entirely classical result of Banach and Mazur that every separable Banach space is linearly isometric to a subspace of $$C[0,1]$$; a result as far from being a counterexample as can be.
I think that you may be selling short the value of a counterexample! They are quite useful for making sure that you have not proven too much. i.e. When you have a plausible but only semi-formal argument, how do you tell if it is worth the effort in making it rigorous? Checking against counterexamples is often a useful step.
Still, there are relations to unbounded paths in an infinite binary tree without leaves. That is, suppose that you start at the root and write $$0.$$, thought of as a ternary number. As we travel to the left or right, write a $$0$$ or $$2$$, respectively, for the following digit. Continuing on we get a ternary expansion defining a real number in the Cantor set. Correspondingly, the ternary expansion of an element of the Cantor set gives rise to a path from the root in the tree.
Spaces that are homeomorpic to the Cantor set arise naturally in many mathematical settings, particularly in dynamical systems.
For one dynamical example, the Cantor set is homeomorphic to the phase space of any infinite Bernoulli process.
For another, the "nonescaping set" of many simple dynamical systems in the real line (or the complex plane) is homeomorphic to the Cantor set (this is a "Cantor dust" example as in the answer of @GeraldEdgar). Consider for example the dynamical system $$z_n = (z_{n-1})^2 + 10$$ (You can replace $$10$$ by any real or complex number of magnitude $$>2$$). One can prove that there is a subset $$C \subset \mathbb C$$ homeomorphic to the Cantor set such that if $$z_0 \in C$$ then the sequence $$(z_n)$$ is bounded (in fact it stays in $$C$$), whereas if $$z_0 \not\in C$$ then $$\lim_{n \to \infty} |z_n| =\infty$$. In short, points not in $$C$$ escape to infinity, points in $$C$$ do not.
Also, there are important theoretical descriptions/properties of the Cantor set, for example:
• A topological space is homeomorphic to the Cantor set if and only if it is compact, metrizable, has no isolated points, and every component is a point.
• Any compact zero-dimensional metrizable topological space is homeomorphic to a subspace of the Cantor set.
Cantor sets even occur naturally in number theory! The $$p$$-adic integers $$\mathbb Z_p$$ are homeomorphic to the Cantor set.
• Do you mean some subset in $\mathbb{C}$? I don't get the example; for any $z_{-1}$ real one has $z_n \geq 10^{2^n} \to +\infty$. – Vandermonde Oct 7 '18 at 16:01
• Oops, you're right, that was careless of me. Fixed. – Lee Mosher Oct 7 '18 at 21:26 | 2019-02-18T15:39:29 | {
"domain": "stackexchange.com",
"url": "https://matheducators.stackexchange.com/questions/14612/why-do-we-study-cantor-set",
"openwebmath_score": 0.7546228766441345,
"openwebmath_perplexity": 241.27119697944664,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9766692311915195,
"lm_q2_score": 0.8723473614033683,
"lm_q1q2_score": 0.8519948267937784
} |
https://math.stackexchange.com/questions/4064710/sequence-queue-converges-or-diverges/4064747 | # Sequence queue - converges or diverges?
I have the next sequence queue: $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$. Does the queue converges or diverge?
My attempt: I have tried to show that $$\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$$ for any $$n>0$$, and then by the comparation test we get that since $$\sum_{n=1}^{\infty}{\frac{1}{n^2}}$$ converges, we have that $$\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$$ converges too, and by a theorem we have that since $$\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$$ $$\implies$$ $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ converges. Is that right?
• Yeah, a direct comparison test works for this. You were right to compare the following: $$\frac{1}{n^2}>\left|\frac{n^2-n-1}{n^4+n^2+1}\right|$$ Now because the $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ converges by the p-test, you can conclude that the original series also converges. Mar 16, 2021 at 22:22
• @ObsessiveInteger So my attempt is right? but can you show me how I satisfy that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$. for $\frac{1}{n^2}>\frac{n^2-n-1}{n^4+n^2+1}$ it's easy since we get by algebra that $n^3+1>0$ for $n>0$. However, for the negative it's not that easy. Mar 16, 2021 at 22:31
• You could make an argument by saying that the numerator of $\frac{n^2-n-1}{n^4+n^2+1}$ grows smaller faster compared to that of $\frac{1}{n^2}$ and similarly, the denominator of $\frac{n^2-n-1}{n^4+n^2+1}$ grows bigger much faster compared to that of $\frac{1}{n^2}$. This means that $\frac{n^2-n-1}{n^4+n^2+1}$ is reaching $0$ faster compared to $\frac{1}{n^2}$, hence, the inequality is true. Of course you can see that graphically as well. Plug in both functions into desmos and you will see. Mar 16, 2021 at 22:41
• @ObsessiveInteger seeing that throw desmos is good, but can I actually verify that by using just inequality algebra? and thank you for the answer. Mar 16, 2021 at 22:44
• Why don't you do the limit comparison test instead? That way you do not have to worry about proving the inequality. Mar 16, 2021 at 22:45
I continue from where you stopped,
Easy to show that :
$$\frac{n^2-n-1}{n^4+n^2+1} < \frac{n^2-1}{n^4+1} < \frac{n^2}{n^4} < \frac{1}{n^2}$$
When we know that $$\sum_{n=1}^{\infty}{\frac{1}{n^2}}$$ converges.
Therefore, by comparison test :
$$\underset{converges}{\underbrace{\frac{n^2-n-1}{n^4+n^2+1}}} < \underset{converges}{\underbrace{\frac{1}{n^2}}}$$
• However, what stops us from doing so for just the expression without ||, is that when $n$ has small values then the whole expression is negative, and then you cannot use the comparison test, so we have to show it with ||. Mar 16, 2021 at 22:48
• @aasc232 By the limit test we get the equation aims to zero so can see that only the first element in the sum has negative value do not forget is the sum of this equation .
– ATB
Mar 16, 2021 at 22:54
• Can you remind me the limit test? Mar 16, 2021 at 22:56
• And the comparison test is when you have $0\le a_k\le b_k$ for all $k>0$, with the first elements of $a_k$ included, then if $b_k$ converges we get that $a_k$ converges too. So since we have that $a_k$ isn't positive for all $k>0$ we cannot use it. That is what I tried telling you. Isn't right? Mar 16, 2021 at 23:01
• @aasc232 Do the $n=1$ term separately $$|\frac{1^2-1-1}{1^{4}+1^2+1}|<\frac{1}{1}$$ and then ATB's answer applies for all $n>1$ as $\frac{n^{2}-n-1}{n^{4}+n^{2}+1}$ is positive for $n>1$. Of course, a finite number of terms does not determine the convergence and can be disregarded in general when checking convergence
– user649348
Mar 16, 2021 at 23:03
We are trying to prove that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ converges.
Let us use the Limit Comparison Test. That is, if $$a_n>0$$ and $$b_n>0$$, then if $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ for some real number $$L$$. Then, if $$b_n$$ converges, by the limit comparison test, $$a_n$$ must also converge.
Let $$a_n=\frac{n^2-n-1}{n^4+n^2+1}$$ and we say $$b_n=\frac{1}{n^2}$$. Then, $$\lim_{n\to \infty} \frac{\frac{n^2-n-1}{n^4+n^2+1}}{\frac{1}{n^2}}\\ \lim_{n\to \infty}\left(\frac{(n^2-n-1)(n^2)}{n^4+n^2+1}\right)\\ \lim_{n\to \infty}\frac{n^4-n^3-n^2}{n^4+n^2+1}\cdot \frac{\frac{1}{n^4}}{\frac{1}{n^4}}\\ \lim_{n\to \infty} \left(\frac{{1-\frac{1}{n}}-\frac{1}{n^2}}{1+\frac{1}{n^2}+\frac{1}{n^4}}\right)\\ \lim_{n\to \infty}\frac{1-0-0}{1+0+0}=\frac{1}{1}=1$$
Hence, we have shown that the $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ converges because it equals some constant; in this case, $$1$$.
Now, because $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$ converges by the p-test, (since $$p=2>1$$), we can conclude that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ must also converge by the limit comparison test.
• Ohhh I didn't try using this, in this way. Thank you! Mar 16, 2021 at 23:03
• One last thing, by limit comparison test we have to show that the sequence queue $\frac{n^2−n−1}{n^4+n^2+1}$ is positive. How do we show that? or you saying we disregard the first negative elements of the sum?' Mar 16, 2021 at 23:12
• Technically you do not have to show that, note that the sum $\sum_{n=1}^{\infty}$ starts at $n=1$, so we are only considering values for $n$ that are positive. Notice that for values greater than $1$, $\frac{n^2-n-1}{n^4+n^2+1}$ can never be negative. Mar 16, 2021 at 23:19
• So you are saying that becuase the only negative fraction of the sum is $-\frac{1}{3}$ then there exist elements that if we sum them we get a positive number bigger then $\frac{1}{3}$? Mar 16, 2021 at 23:24
• Thank you a lot! Mar 16, 2021 at 23:29 | 2022-05-22T20:40:02 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4064710/sequence-queue-converges-or-diverges/4064747",
"openwebmath_score": 0.9039461016654968,
"openwebmath_perplexity": 215.2440022368874,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307708274401,
"lm_q2_score": 0.8757869997529962,
"lm_q1q2_score": 0.8519925420503583
} |
http://math.stackexchange.com/questions/84650/normal-subgroup-of-automorphisms-of-a-free-group | Normal subgroup of automorphisms of a free group
Let $F_2=\langle X,Y\rangle$ be the free group of rank $2$ and consider $A,B,C\in Aut(F_2)$ given by: $$A(X,Y)\mapsto(YX^{-1}Y^{-1},Y^{-1})$$ $$B(X,Y)\mapsto(X^{-1},Y^{-1})$$ $$C(X,Y)\mapsto(X^{-1},XY^{-1}X^{-1})$$ The abelianisation of $F_2$ is isomorphic to $\mathbb{Z}^2$ whose automorphism group is $GL(2,\mathbb{Z})$. Thus we get an induced surjective map $m\colon Aut(F_2)\to GL(2,\mathbb{Z})$ which maps an element in $Aut(F_2)$ to the matrix associated to the corresponding element in $Aut(\mathbb{Z}^2)$. I would like to show that the subgroup $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ by proving that $m(\langle A,B,C\rangle)$ is the centre of $GL(2,Z)$. So far I wasn't able to have any good idea... could you help me? Thanks, bye.
-
Just calculate the matrices: m(A) = m(B) = m(C) = -1 – Jack Schmidt Nov 22 '11 at 18:22
You may want to use \langle and \rangle instead of < and >; the spacing is correct for the former. – Arturo Magidin Nov 22 '11 at 18:27
how did you calculate explicitly the matrices? $A$,$B$ and $C$ do not belong to $Aut(\mathbb{Z}^2)$ so I should compute their image in the quotient, first. Do I have to write how, say $A$, acts on the cosets $F_2'X$ and $F_2'Y$ in $F_2/F_2'$? P.S. I did use \langle and \rangle! :) – fatoddsun Nov 22 '11 at 19:22
@fatoddsun: Since $[F_2,F_2]$ is fully invariant, any homomorphism $f\colon F_2\to F_2$ induces a homomorphism $f_{\rm ab}\colon \mathbb{Z}^2\to\mathbb{Z}^2$ by $f_{\rm ab}(\overline{X}) = \overline{f(X)}$ and $f_{\rm ab}(\overline{Y}) =\overline{f(Y)}$. So just "abelianize" the image. – Arturo Magidin Nov 22 '11 at 20:24
@fatoddsun: Note, however, that having the image be normal is not sufficient; that will only show that the preimage under the abelianization map $\mathrm{Aut}(F_2)\to\mathrm{GL}(2,\mathbb{Z})$ is normal, and it's not immediately clear to me that this preimage is just $\langle A,B,C\rangle$. – Arturo Magidin Nov 22 '11 at 20:57
Arturo Magidin's excellent answer explains how to show that $m(\langle A,B,C\rangle)=\{\pm 1\}$, and also why this does not in general imply that your subgroup is normal. However, I want to explain why in this case it is true that $\langle A,B,C\rangle$ is the full preimage of $\{\pm 1\}$, and thus is normal.
The key is a theorem of Nielsen (1) on what is now called $IA_n$: this is by definition the kernel of the natural surjection $Aut(F_n)\to GL_n\mathbb{Z}$, so it fits into a short exact sequence
$$1\to IA_n\to Aut(F_n)\to GL_n\mathbb{Z}\to 1$$
This group of automorphisms which are the Identity on the Abelianization has been well-studied, starting with Magnus (2), who proved that $IA_n$ is finitely generated by giving an explicit set of generators: type (A) which for given $i$ and $j$ conjugates $x_i$ by $x_j$ and fixes all other generators, and type(B) which for given $i$, $j$, and $k$ multiplies $x_i$ by $x_jx_kx_j^{-1}x_k^{-1}$ and fixes all other generators.
But anyway, we are interested here in $IA_2$, and Nielsen proves in (1) that $IA_2$ is just $Inn(F_2)$, the group of automorphisms obtained by conjugation. This is isomorphic to $F_2$ and generated by $conj_X\colon X\mapsto X, Y\mapsto XYX^{-1}$ and by $conj_Y\colon X\mapsto YXY^{-1}, Y\mapsto Y$.
We know that your group $\langle A,B,C\rangle$ subjects to $\{\pm 1\}$, so to show it is the full preimage of $\{\pm 1\}$ it suffices to show that $\langle A,B,C\rangle$ contains the full preimage of $\{1\}$, which is $IA_2$. But we can compute by hand that $AB=conj_Y$ and $CB=conj_X$. So by Nielsen's theorem your group contains $IA_2$, and thus fits into a short exact sequence
$$1\to IA_2\to \langle A,B,C\rangle\to \{\pm 1\}\to 1$$
In particular, your subgroup is normal in $Aut(F_2)$.
Finally, here is another method of proving that $\langle A,B,C\rangle$ is normal which does not require quoting any papers written in German. We know that the elementary Nielsen transformations provide a generating set for $Aut(F_n)$. This generating set becomes particularly simple in the case when $n=2$: we have three generators $R\colon X\mapsto X^{-1}, Y\mapsto Y$, $S\colon X\mapsto Y, Y\mapsto X$, and $T\colon X\mapsto XY,Y\mapsto Y$. To show that $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ means that it is invariant under conjugation under any element of $Aut(F_2)$. But since $R,S,T$ generate $Aut(F_2)$, it is enough just to show that $\langle A,B,C\rangle$ is invariant under conjugation by $R$, by $S$, and by $T$.
To do this for $R$, for example, you just need to compute the conjugates $RAR^{-1}$, $RBR^{-1}$, and $RCR^{-1}$ and show that each can be written as a combination of $A$, $B$, and $C$. In this case this is quite easy: we have $RAR^{-1}=A$, $RBR^{-1}=B$, and $RCR^{-1}=C^{-1}$. Now all you have to do is do the same for $SAS^{-1}$, $SBS^{-1}$, $SCS^{-1}$, $TAT^{-1}$, $TBT^{-1}$, and $TCT^{-1}$, and you've proved that $\langle A,B,C\rangle$ is normal!
(1) Nielsen, Die Isomorphismen der allgemeinen unendlichen Gruppe mit zwei Erzeugenden, Math. Ann. 78 (1964), 385-397
(2) Magnus, Über n-dimensinale Gittertransformationen, Acta Math. 64 (1935), 353-367.
-
Nice; thanks for filling the gap! – Arturo Magidin Nov 22 '11 at 21:39
If $G$ is a group, $N$ is a normal subgroup of $G$, and $f\colon G\to G$ is a homomorphism with $f(N)\subseteq N$, then $f$ induces a homomorphism $\overline{f}\colon G/N\to G/N$ by $\overline{f}(gN) = f(g)N$. To see this, note that $\widehat{f}\colon G\to G/N$ given by $\widehat{f}(g) = gN$ is a homomorphism, and since $f(N)\subseteq N$, then $N$ is contained in the kernel of $\widehat{f}$, so $\widehat{f}$ factors through $G/N$.
Since $[F_2,F_2]$ is fully invariant, this argument will work for $A$, $B$, and $C$.
For $A$, $x$ is mapped to $yx^{-1}y^{-1}$, which gets mapped to $\overline{x}^{-1}$ in $F_2^{\rm ab}$; $y$ is mapped to $\overline{y}^{-1}$. So the induced map $A_{\rm ab}\colon F_2^{\rm ab}\to F_{2}^{\rm ab}$ sends $(a,b)$ to $(-a,-b)$.
This is the same as what $B_{\rm ab}$ does; again the same thing as $C_{\rm ab}$ does. So the image of the subgroup generated by $A$, $B$, and $C$ in $\mathrm{GL}(2,\mathbb{Z})$ is the subgroup generated by their images, which is just the subgroup generated by $-I$.
Note, however, that the fact that $m(\langle A,B,C\rangle) = \{I,-I\}$ does not show, by itself, that $\langle A,B,C\rangle$ is normal in $\mathrm{Aut}(F_2)$; it shows that the pre-image is normal. The preimage is the subgroup of $\mathrm{Aut}(F_2)$ generated by $A$, $B$, $C$, and all automorphisms that induce the identity on $F_2^{\rm ab}$. It is was not clear to me whether this is just $\langle A,B,C\rangle$. For this, see Tom Church's great answer.
-
Thank you very much guys! – fatoddsun Nov 23 '11 at 18:23 | 2014-08-29T03:21:10 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/84650/normal-subgroup-of-automorphisms-of-a-free-group",
"openwebmath_score": 0.962099552154541,
"openwebmath_perplexity": 102.51633505472043,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307708274402,
"lm_q2_score": 0.8757869948899665,
"lm_q1q2_score": 0.8519925373194535
} |
https://www.physicsforums.com/threads/deriving-equations-of-a-parabola.237835/ | # Deriving Equations of a Parabola
1. May 29, 2008
### PFStudent
Hey,
1. The problem statement, all variables and given/known data
How do you derive the equations of the parabola from the general equation of a Conic Section?
2. Relevant equations
General Equation of a Conic Section,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$
Where (for a parabola),
$${{\{}A,B,C,D,E,F{\}}}{{\,}{\,}{\,}}{\in}{{\,}{\,}{\,}}{\mathbb{R}}$$
$${{\{}A, C{\}}}{{\,}{\,}{\,}}{\neq}{{\,}{\,}{\,}}{0}$$
$${{{B}^{2}}-{{4}{A}{C}}} = {0}$$
3. The attempt at a solution
From the general equation of a Conic Section,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$
How do I derive the following formulas for a parabola:
General Form for a Parabola,
$${{{{\left(}{{{H}{x}}+{{I}{y}}}{\right)}}^{2}}+{{J}{x}}+{{K}{y}}+{L}} = {0}$$
Where,
$${{{I}^{2}}-{{4}{H}{J}}} = {0}$$
Analytic Geometry Equations,
Vertical Axis of Symmetry
$${{{\left(}x-h{\right)}}^{2}} = {{{4}{p}}{{{\left(}y-k}{\right)}}}$$
$${y} = {{{a}{{x}^{2}}}+{{b}{x}}+{c}}$$
Horiztonal Axis of Symmetry
$${{{\left(}y-k{\right)}}^{2}} = {{{4}{p}}{{{\left(}x-h{\right)}}}}$$
$${x} = {{{d}{{y}^{2}}}+{{e}{y}}+{f}}$$
I read that every parabola is a combination of transformations of the parabola, $${{y}={{x}^{2}}}$$; but I'm not quite sure how that helps.
Thanks,
-PFStudent
Last edited: May 29, 2008
2. May 29, 2008
### Dick
I'm not precisely sure of what exactly you are supposed to do, but the only way that the quadratic terms Ax^2+Bxy+Cy^2 can be factored as (Hx+Iy)^2 is that B^2-4AC=0. Just multiply (Hx+Iy)^2 out and identify the terms. Does that help?
3. May 30, 2008
### HallsofIvy
First you need to find the "principal axes" (axis of symmetry and perpenpendicular to it through the vertex) and rotate the coordinate system so that new x' and y' axes are the principal axes.
There are two ways to do that, basically using the quadratic terms $Ax^2+ Bxy+ Cx^2$ (By the way, you don't have to put braces, { }, around every term. They are only necessary when you want an entire expression in a particular place. For example, [i t e x]e^{xy+ b}[/i t e x] gives $e^{xy+ b}$ while [i t e x]e^xy+ b[/i t e x] gives $e^xy+ b$.)
Rotating the axes by angle $\theta$, so that the new x'y'- axes are at angle $\theta$ to the old xy- axex, x and y are given, in terms of the new x', y' variables, by $x= x'cos(\theta)+ y'sin(\theta)$ and $y= -x'sin(\theta)+ y'cos(\theta). Replace x and y by those in $Ax^2+ Bxy+ Cx^2$ and choose$\theta[/itex] so that the coefficient of x'y' is 0. Use those formulas with the correct value of $\theta$ to replace x and y in the entire formula. If it really is a parabola, both the x'y' and y'2 terms should vanish.
The other way is to write it as a matrix problem:
$$Ax^2+ Bxy+ Cy^2= \left[\begin{array}{cc}x & y \end{array}\right]\left[\begin{array}{cc}A & \frac{B}{2} \\ \frac{B}{2} & C\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]$$
You can "diagonalize" that matrix (so writing the quadratic without the xy term) by finding the eigenvalues and eigenvectors of the coefficient matrix. The eigenvectors will point along the principal axes. Since this is a parabola, one of the eigenvalues should be 0.
4. May 31, 2008
### PFStudent
Hey,
Thanks for the replies: Dick and HallsofIvy.
Yea, to be honest I had not seen that general form for a parabola before until I saw it on wikipedia, maybe it is wrong? - here is the link, (Scroll down a bit to the section "General Parabola").
http://en.wikipedia.org/wiki/Parabola
Let me know what you think.
Hmm, ok that makes sense, I will try that. However, I thought since a parabola - in its' most general form - could be expressed as a conic section with the constraint that $${{B^2} = {4AC}}$$. Then shouldn't I be able to derive the more specific forms of a parabola from the general equation of a conic section with the constraint that $${{B^2} = {4AC}}$$?
Where the general equation of a Conic Section is,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$
Also, yea I tend to use a bit too many braces ({ }), just a habit from first learning $LaTeX$ way back.
Thanks,
-PFStudent | 2018-01-21T03:17:32 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/deriving-equations-of-a-parabola.237835/",
"openwebmath_score": 0.7917470335960388,
"openwebmath_perplexity": 710.62387662584,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307747659751,
"lm_q2_score": 0.8757869900269366,
"lm_q1q2_score": 0.8519925360378661
} |
https://gmatclub.com/forum/how-many-nails-did-rudy-buy-if-he-purchased-them-at-a-price-of-286281.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 26 Jun 2019, 11:46
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
### Show Tags
Updated on: 12 Jan 2019, 03:25
1
00:00
Difficulty:
15% (low)
Question Stats:
88% (02:25) correct 12% (02:43) wrong based on 33 sessions
### HideShow timer Statistics
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Originally posted by Nums99 on 12 Jan 2019, 01:42. Last edited by Bunuel on 12 Jan 2019, 03:25, edited 1 time in total. Renamed the topic, edited the question, moved to PS and added the OA. Senior Manager Joined: 15 Jan 2018 Posts: 371 Concentration: General Management, Finance GMAT 1: 720 Q50 V37 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
12 Jan 2019, 02:12
1
Hi there,
Here's the solution.
Let the number of nails Rudy purchased be x.
Rudy purchased nails at a price of $0.25 per four nails. So, Cost Price of 4 nails =$0.25
Cost Price of 1 nail = = $0.25/4 Cost price of x nails = ($0.25/4)x
Also given that Rudy sold nails at $0.22 per three nails. So, Selling Price of 3 nails =$0.22
Selling Price of 1 nail = = $0.22/3 Selling price of x nails = ($0.22/3)x
Also, Rudy made a profit of $2.60. Profit = Selling Price - Cost Price$2.60 = ($0.22/3)x - ($0.25/4)x = x(0.88 - 0.75)/12
$2.60 = x(0.13)/12 x = 240 Hence, the the number of nails Rudy purchased is 240. Therefore, the Correct Answer is Option B. 240 _________________ New to GMAT Club or overwhelmed with so many resources? Follow the GMAT Club Study Plan! Not happy with your GMAT score? Retaking GMAT Strategies! Game of Timers - Join the Competition to Win Prizes Senior Manager Status: Manager Joined: 27 Oct 2018 Posts: 365 Location: Egypt Concentration: Strategy, International Business GPA: 3.67 WE: Pharmaceuticals (Health Care) Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
16 Jan 2019, 02:57
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Profit = Sales - Cost $$2.6 = \frac{0.22}{3}n - \frac{0.25}{4}n = \frac{0.88n}{12} - \frac{0.75n}{12}$$ $$\frac{13}{5} = \frac{0.13n}{12} = \frac{13n}{1200}$$ $$n = \frac{13*1200}{13*5} = 240$$ B _________________ RC Moderator Joined: 24 Aug 2016 Posts: 802 Location: Canada Concentration: Entrepreneurship, Operations GMAT 1: 630 Q48 V28 GMAT 2: 540 Q49 V16 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
16 Jan 2019, 08:15
As we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .
Cost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88
Profit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
25 Feb 2019, 09:26
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails
(2.60 * 12)/0.13 = 240 nails (Ans)
_________________
"Please hit +1 Kudos if you like this post"
_________________
Manish
"Only I can change my life. No one can do it for me"
Director
Joined: 04 Dec 2015
Posts: 740
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)
How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 28 Feb 2019, 11:09 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?
A. 300
B. 240
C. 180
D. 160
E. 120
Cost Price of $$4$$ nails $$= 0.25$$
Cost Price of $$3$$ nails $$= \frac{0.25*3}{4} = 0.1875$$
Selling Price of $$3$$ nails $$= 0.22$$
Profit on $$3$$ nails $$= 0.22 - 0.1875 = 0.0325$$
Let $$"x"$$ be total number of nails to make profit of $$2.60$$.
$$\frac{0.0325}{3} = \frac{2.60}{x}$$
$$0.0325*x =3*2.60$$
$$x = \frac{3*2.60}{0.0325} = 240$$
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6680
Location: United States (CA)
Re: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 02 Mar 2019, 10:08 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?
A. 300
B. 240
C. 180
D. 160
E. 120
We can let n = the number of nails and create the equation:
(0.22/3)n - (0.25/4)n = 2.6
Multiplying by 12, we have:
0.22 x 4n - 0.25 x 3n = 31.2
0.88n - 0.75n = 31.2
0.13n = 31.2
n = 240
_________________
# Scott Woodbury-Stewart
Founder and CEO
[email protected]
122 Reviews
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.
Re: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] 02 Mar 2019, 10:08 Display posts from previous: Sort by # How many nails did Rudy buy if he purchased them at a price of$0.25
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne | 2019-06-26T18:46:57 | {
"domain": "gmatclub.com",
"url": "https://gmatclub.com/forum/how-many-nails-did-rudy-buy-if-he-purchased-them-at-a-price-of-286281.html",
"openwebmath_score": 0.2864758372306824,
"openwebmath_perplexity": 11699.412737693603,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9728307668889047,
"lm_q2_score": 0.8757869884059267,
"lm_q1q2_score": 0.851992527562262
} |
https://math.stackexchange.com/questions/3593728/aternating-sum-of-an-increasing-sequence-of-positive-integers | # Aternating sum of an increasing sequence of positive integers
Suppose $$A = (a_n) = (a_1, a_2, a_3, . . .)$$ is an positive, increasing sequence of integers.
Define an $$A$$- expressible number $$c$$ if $$c$$ is the alternating sum of a finite subsequence of $$A.$$ To form such a sum, choose a finite subset of the sequence $$A,$$ list those numbers in increasing order (no repetitions allowed), and combine them with alternating plus and minus signs. We allow the trivial case of one-element subsequences, so that each an is $$A-$$expressible.
Definition. Sequence $$A = (a_n)$$ is an “alt-basis” if every positive integer is uniquely $$A-$$ expressible. That is, for every integer $$m > 0,$$ there is exactly one way to express $$m$$ as an alternating sum of a finite subsequence of $$A.$$
Examples. Sequence $$B = (2^{n−1}) = (1, 2, 4, 8, 16, . . .)$$ is not an alt-basis because some numbers are B-expressible in more than one way. For instance $$3 = −1 + 4 = 1 − 2 + 4.$$
Sequence $$C = (3^{n−1}) = (1, 3, 9, 27, 81, . . .)$$ is not an alt-basis because some numbers (like 4 and 5) are not C-expressible.
An example of an alt-basis is $$\{2^n-1\}=\{1,3,7,15,31,\ldots\}$$
Is there a fairly simple test to determine whether a given sequence is an alt basis?
I have attempted to solve this from a limited knowledge in sequences and have found out various kinds of sequences do not work but fail to see what it is that could make it work.
• Seems slightly (at least) related to this: encyclopediaofmath.org/index.php/Additive_basis But I think you know already about additive bases. – Masterphile Mar 25 at 0:37
• See math.stackexchange.com/questions/3579462. (Not an exact duplicate because it doesn't ask for a general test to determine whether a sequence is an alt-basis.) What is the source of this question? – joriki Mar 25 at 6:47
• Do you know of any alt-bases that are not of the form $\{b_1,b_2,\dots,b_k,a_{k+1},a_{k+2},\dots\}$ where $A=\{a_1,a_2,\dots\}=\{2^n-1\}$ and $\{b_1,\dots,b_k\}$ is some finite increasing set of integers? – Vepir Mar 25 at 21:54
• For example I believe that $\{5\cdot2^{n - 2} + (-1)^{n + 1} 2^{n - 2} - 1\}$ is an alt-basis. – Vepir Mar 25 at 22:50
I can’t answer the question, but I can at least give you a systematic large family of alt-bases.
If $$A$$ is a finite set of positive integers, let $$S(A)$$ be the set of $$A$$-expressible integers, and let $$S^+(A)$$ be the set of $$A$$-expressible positive integers. Then
$$S(A)=S^+(A)\cup\{-a:a\in S^+(A)\},$$
and if $$b>\max A$$, then
$$S^+\left(A\cup\{b\}\right)=S^+(A)\cup\{b-s:s\in S^+(A)\}\cup\{b\}.$$
Thus, if $$|A|=n$$, the maximum number of $$A$$-expressible positive integers is $$2^n-1$$, and $$\max S(A)=\max A$$.
Now suppose that $$A=\{a_n:n\in\Bbb Z^+\}$$, where $$a_n for each $$n\in\Bbb Z^+$$. For $$n\in\Bbb Z^+$$ let $$A_n=\{a_k\in A:1\le k\le n\}$$. Then each $$m\in S(A)$$ is uniquely $$A$$-expressible iff $$|S^+(A_n)|=2^n-1$$ for each $$n\in\Bbb Z^+$$. Moreover, $$S^+(A)=\Bbb Z^+$$ iff for each $$k\in S^+(A)$$ there is a minimal $$n(k)\in\Bbb Z^+$$ such that $$k\in S^+(A_{n(k)})$$. Note that either $$n(k)=1$$, or $$k\in S^+(A_{n(k)})\setminus S^+(A_{n(k)-1})=\{a_{n(k)}-s:s\in S^+(A_{n(k)-1})\}$$.
For $$n\in\Bbb Z^+$$ let
$$a_n=2^n-1=\underbrace{1\ldots 1}_n\text{ in binary},$$
and let $$A=\{a_n:n\in\Bbb Z^+\}$$. It’s not hard to see that
$$S^+(A_n)=\{1,\ldots,2^n-1\}$$
for each $$n\in\Bbb Z^+$$, so $$A$$ is, as you already observed, an alt-basis. For instance, working in binary, we see that
\begin{align*} 22&=10110_{\text{two}}\\ &=11111_{\text{two}}-1111_{\text{two}}+111_{\text{two}}-1_{\text{two}}\\ &=31-15+7-1\\ &=a_5-a_4+a_3-a_1. \end{align*}
Now let $$\ell,m\in\Bbb Z^+$$. For $$n=1,\ldots,\ell$$ let
$$\color{red}{a_n^{(\ell,m)}}=2^ma_n=\underbrace{1\ldots 1}_n\underbrace{0\ldots 0}_m\text{ in binary}.$$
For $$n=\ell+k$$, where $$k=1,\ldots,m$$, let
$$\color{blue}{a_n^{(\ell,m)}}=2^{m-k}a_n=\underbrace{1\ldots 1}_n\underbrace{0\ldots 0}_{m-k}\text{ in binary}.$$
Finally, for $$n>\ell+m$$ let $$a_n^{(\ell,m)}=a_n$$, and let $$A_{(\ell,m)}=\left\{a_n^{(\ell,m)}:n\in\Bbb Z^+\right\}$$; then $$A_{(\ell,m)}$$ is an alt-basis.
For example,
\begin{align*} A_{(4,2)}&=\{\color{red}{4},\color{red}{12},\color{red}{28},\color{red}{60},\color{blue}{62},\color{blue}{63},127,\ldots\}\\ &=\{\color{red}{100},\color{red}{1100},\color{red}{11100},\color{red}{111100},\color{blue}{111110},\color{blue}{111111},1111111,\ldots\}\text{ in binary}. \end{align*}
To verify this it suffices to show that $$S^+\left(\left\{a_n^{(\ell,m)}:1\le n\le \ell+m\right\}\right)=S^+(A_{\ell+m})$$. The argument is a bit messy to write out, but the idea is straightforward; I’ll illustrate it with $$A_{(4,2)}$$. First, it’s clear from the discussion of $$A$$ that
\begin{align*} S^+\left(\{4,12,28,60\}\right)&=S^+\left(4\{1,3,7,15\}\right)\\ &=4S^+\left(\{1,3,7,15\}\right)\\ &=4\{1,2,\ldots,15\}\\ &=\{4,8,12,\ldots,60\}\\ &=4S^+(A_4). \end{align*}
Then
\begin{align*} S^+(\{4,&12,28,60,62\})=\\ &4S^+(A_4)\cup\left\{|62-s|:s\in S^+(\{4,12,28,60\})\right\}\cup\{62\}=\\ &4S^+(A_4)\cup\left\{|62-s|:s\in\{4,8,12,\ldots,60\}\right\}\cup\{62\}=\\ &4S^+(A_4)\cup\{2,6,10,\ldots,58,62\}=\\ &\{2,4,6,8,\ldots,60,62\}=\\ &2S^+(A_5), \end{align*}
and a similar calculation shows that $$S^+(\{4,12,28,60,62,63\})=S^+(A_6)$$.
I did not collect the set of all alt-bases, but I did find some useful observations, including:
Alt-basis must contain an infinite number of terms of form $$a_k=2^{k}-1,k\in N\subseteq\mathbb N$$.
The converse does not hold. At the end, I give examples of alt-bases and not-alt-bases in this context.
Do correct me If I missed anything.
Let $$A=\{a_1,a_2,\dots\}$$ such that $$a_1\lt a_2 \lt \dots$$ are positive integers.
Definition. For $$A$$ to be an "alt-basis", we need to have both the "uniqueness" and "completeness". In other words, every number is expressible in exactly one way via alternating summation of subsets of $$A$$, which are summed in increasing order.
Definition. A finite (sub)sequence $$A|_n=\{a_1,\dots,a_n\}$$ is an "alt-prefix" if every integer in $$[1,2^{n}-1]$$ is uniquely expressible via alternating summation of subsets of $$A|_n$$ when summed in increasing order. The element $$a_n$$ is called an "anchor element".
Definition. "Anchor sequence" is a set $$\mathcal A(A):=\{a_{n_1},a_{n_2},\dots\}$$ of all "anchor elements" $$a_{n_1},a_{n_2},\dots$$
Notice that a set has $$2^n$$ subsets minus the empty set and that every subset can be rearranged in an increasing order. We want to assign a distinct value to each of those subsets via the alternating summation, to have an alt-basis. The alt-prefix is defined to cover exactly those $$2^n-1$$ subsets. It follows that:
Lemma. $$A$$ is an alt-basis $$\iff$$ $$A$$ is a union of alt-prefixes $$A=A|_{n_1}\cup A|_{n_2}\cup \dots$$
That is, $$A$$ is an alt-basis if and only if there exists a corresponding infinite anchor sequence $$\mathcal A(A)$$.
We add two more definitions to write all of this more easily:
Definition. Let $$s(\{b_1,\dots,b_n\})$$ be the result of the alternating summation of $$b_1\lt b_2\lt \dots\le b_n$$. Let $$s_+$$ and $$s_-$$ always start the alternating summation with $$+,-$$ respectively. Then $$s_+=-s_-$$. If $$n$$ is odd then $$s=s_+$$ and if $$n$$ is even then $$s=s_-$$. This guarantees $$s\gt 0$$ because the largest element $$b_n$$ will have a positive sign.
Definition. Define "$$n$$-th partial subset-sum set" of a positive increasing integer sequence $$A$$ as:
$$\mathcal S_n(A):=\{s(A_i):A_i\in\mathcal P(A|_n)\}$$
Where $$\mathcal P(A|_n)$$ is the set of all subsets of $$A|_n=\{a_1,a_2,\dots,a_n\}$$.
The set of all "anchor elements" $$\mathcal A(A)=\{a_{n_1},a_{n_2},\dots\}\subseteq A$$ satisfies $$S_{n_i}=[1,2^{n_i-1}-1]$$ for all $$n_i$$.
Corollary. $$A$$ is an alt-basis if and only if it "is covered by the anchor sequence": $$\max \mathcal A(A)\to \infty$$.
Notice that $$\max S_n = a_n$$. If $$a_n$$ is an anchor element, then $$\max S_n = 2^n-1$$. This gives:
Proposition. If $$a_n$$ is an anchor element, then $$a_n=2^n-1$$.
The converse does not hold. For example, in $$\{1,4,7\}$$ the $$a_3=7=2^3-1$$ but $$a_3$$ is not an anchor element, because $$S_3=\{1,3,4,6,7\}\ne[1,7]$$.
Example $$1$$. It is not hard to see that $$\mathcal A(\{2^n-1\})=\{2^n-1\}$$. This is because:
• $$S_1=\{(+1)\}$$ $$\implies$$ $$a_1$$ is an anchor element.
• $$S_2=\{(+1),(-1+3),(3)\}$$ $$\implies$$ $$a_2$$ is an anchor element.
• $$S_3=\{(+1),(-1+3),(3),(-3+7),(+1-3+7),(-1+7),(7)\}$$ $$\implies$$ $$a_3$$ is an anchor element.
• $$\dots$$ proceed via induction to show every $$a_n$$ is an anchor element.
Since $$\mathcal A(\{2^n-1\})$$ exists and covers the entire $$\{2^n-1\}$$, the $$\{2^n-1\}$$ is an alt-basis.
Example $$2$$. The $$\mathcal A(\{n\})=\{1\}$$ does not cover the entire $$\{n\}$$, hence $$\{n\}$$ is not an alt-basis.
It is not hard to see that $$\max S_n = n \lt 2^n-1\implies a_n$$ is not an anchor element, for every $$n\gt 1$$.
Example $$3.$$ We construct an alt-basis where every $$2$$nd element is an anchor element.
$$A=\begin{cases} 2^n-1, & n\text{ is even} \\ 2^n+2^{n-1}-1, & n\text{ is odd} \end{cases}$$
Use an inductive argument. Assume $$n=2k$$, $$a_{n}=2^{n}-1$$ is an anchor element, which means we have uniquely constructed all $$I_0=[1,2^n-1]=S_{n}$$ elements. Now we can subtract numbers in this interval from $$a_{n+1}$$ to see that:
• $$a_{n+1}=2^{n+1}+2^{n}-1$$ will cover $$I_1=[a_{n+1}-a_{n}, a_{n+1}]=[2^{n+1},2^{n+1}+2^{n}-1]$$
Here we see that $$I_0\cup I_1 \ne [1,2^{n+1}-1]$$ $$\implies$$ $$a_{n+1}$$ is not an anchor elemetn.
To see that $$a_{n+2}=2^{n+2}-1$$ is an anchor element, lets see what will we cover with it:
• $$a_{n+2}$$ combined with $$I_0$$ will cover $$I_2=[a_{n+2}-a_{n}, a_{n+2}]=[2^{n+1}+2^{n},2^{n+2}-1]$$
• $$a_{n+2}$$ combined with $$I_1$$ will cover $$I_3=[a_{n+2}-a_{n+1},a_{n+2}-2^{n+1}]=[2^n,2^{n+1}-1]$$
Now observe $$I=I_0\cup I_3\cup I_1\cup I_2$$ is equal to:
$$I=[1,2^n-1]\cup[2^n,2^{n+1}-1]\cup[2^{n+1},2^{n+1}+2^{n}-1]\cup[2^{n+1}+2^{n},2^{n+2}-1]=[1,2^{n+2}-1]$$
Implying $$a_{n+2}$$ covers $$I=[1,2^{n+2}-1]=S_{n+2}$$, $$\implies$$ $$a_{n+2}$$ is an anchor.
It is not hard to check base cases $$n=1,2$$, and we are done. We have:
$$\mathcal A\left(\left.\begin{cases} 2^n-1, & n\text{ is even} \\ 2^n+2^{n-1}-1, & n\text{ is odd} \end{cases}\right\}\right)=\{2^{2n}-1\}$$
So we have an alt-basis $$A$$.
Example $$4$$. It is not hard to show that:
$$A=\{2^k,2^k+1,2^k+3,2^k+7,\dots,2^k+2^{k}-1,2^{k+2}-1,2^{k+3}-1,\dots\}$$
Is an alt-basis for every $$k=0,1,2,\dots$$, whose anchors are all elements $$a_n,n\gt k$$.
Example $$5$$. The sequence of natural, triangular, tetrahedral,... numbers, or in general, any diagonal of the pascals triangle, is not an alt basis.
This is because for every fixed $$d$$, there exists $$n_0$$, such that for all $$n\ge n_0$$, we have $$\binom{n+d-1}{d}<2^n$$ implying that $$\max S_n\lt 2^n-1$$ for all $$n\ge n_0$$. This implies the sequence of anchors has at most $$n_0$$ elements, implying $$\max\mathcal A(A)\lt \infty$$, hence we do not have an alt-basis becuase of inevetable duplicates.
• I think => part of the initial lemme requires proof. – balcinus Mar 31 at 10:07 | 2020-04-10T13:32:13 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3593728/aternating-sum-of-an-increasing-sequence-of-positive-integers",
"openwebmath_score": 0.9273799061775208,
"openwebmath_perplexity": 1235.820987497876,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.986777180969715,
"lm_q2_score": 0.8633916187614823,
"lm_q1q2_score": 0.8519751476343344
} |
https://math.stackexchange.com/questions/1822583/what-is-the-sum-of-the-81-products-in-the-9-times-9-multiplication-grid/1822592 | What is the sum of the $81$ products in the $9 \times 9$ multiplication grid?
What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.
Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?
• I asked a question that used this result (and has a nice, clean answer!) several years ago: See MSE 226983 for more! – Benjamin Dickman Jun 12 '16 at 3:29
• (1+2+...+8+9)^2 – Evorlor Jun 12 '16 at 18:45
The sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$
Then the next row would be $(2+4+6+8+10+12+14+16+18) = 2\times45 = 90$
So the top two rows sum to $(1+2)\times 45 = 135$
Then it becomes obvious that the full sum of the products is the product of the sums, ie. $45\times45 = 2025$
The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$
We want to find the following: $$\sum_{i=1}^9 \sum_{j=1}^9 ij$$ Factor out the $i$ from the first summation: $$\sum_{i=1}^9 i\left(\sum_{j=1}^9 j\right)$$ Note that $\sum_{j=1}^9 j=45$. $$\sum_{i=1}^9 i\cdot 45$$ Factor out the $45$: $$45\left(\sum_{i=1}^9 i\right)$$ Note that $\sum_{i=1}^9 i=45$. $$45\cdot 45=2025$$
Use the distributive principle. The sum of the entries in the $2$ column is $2$ times the sum of the numbers $1$ through $9$, so the sum of all the entries is the sum of the numbers $1$ through $9$ times (what?)
Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\ \ =\ \ 6\times 6$
Let's prove by induction that the sum of an ${n}\times{n}$ grid is $\frac{n^4+2n^3+n^2}{4}$:
First, show that this is true for $n=1$:
$\sum\limits_{x=1}^{1}\sum\limits_{y=1}^{1}xy=\frac{1^4+2\cdot1^3+1^2}{4}$
Second, assume that this is true for $n$:
$\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy=\frac{n^4+2n^3+n^2}{4}$
Third, prove that this is true for $n+1$:
$\sum\limits_{x=1}^{n+1}\sum\limits_{y=1}^{n+1}xy=$
$\color\red{\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\color\red{\frac{n^4+2n^3+n^2}{4}}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\sum\limits_{x=1}^{n}x\right)+(n+1)\left(\sum\limits_{y=1}^{n}y\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+\frac{n^3+2n^2+n}{2}+\frac{n^3+2n^2+n}{2}+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+2n^2+n+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+3n^2+3n+1=$
$\frac{n^4+6n^3+13n^2+12n+4}{4}=$
$\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4}$
Please note that the assumption is used only in the part marked red.
Therefore, the sum of a ${9}\times{9}$ grid is $\frac{9^4+2\cdot9^3+9^2}{4}=2025$.
I will focus on an intuition for finding the result. Once you have the formula, other answers are perfect in providing the means for proving it technically.
Let us start with a small size version. If you take the top $3\times 3$ table only with numbers $1$, $2$, $3$ (to save some electrons), the sum of products is:
$$s= 1\times 1 +1\times 2 + 1\times 3 + 2\times 1 +2\times 2 + 2\times 3 + 3\times 1 +3\times 2 + 3\times 3 \,.$$ You see that you can rewrite this sum of nine products as: $$s = (1+2+3)\times (1+2+3)\,,$$ using the distribution of multiplication over additions.
This form is more interesting, because you can see a pattern. The sum of the first integers, multiplied by itself.
It can be reassuring to verify it works: you get $36$, which you can check by hand ($6+12+18$). You can test it with the $2\times2$ or $4\times4$ matrix, to verify it is not a coincidence.
The hint is apparent in the squared shape of the table. The same works for bigger tables too. Each product of $i$ and $j$ in this order is contained once and only once in the product $(1,\ldots,i,\ldots,n)\times(1,\ldots,j,\ldots,n)$.
Now you have to find the sum of integers. If you forget the generic expression for the sum of the $n$ first integer, have this figure in mind:
You pack together two triangles which give you a $n\times (n+1)$ rectangle, of area $n(n+1)$, the double of the area of each triangle.
So finally the answer is $\left(n(n+1)/2\right)^2$. With $n=9$, you get $2025$.
Once you have the method, which is relatively simple, you can easily turn it into a more formal proof, via induction for instance, as given in other answers.
• The link to "I like visual proofs" is dead. Besides that, I think the name "visual proof" is not suitable here. Perception may be betraying, so visual remarks should be backed up by some other argument. This makes it not related to "visual"! – Jasper Jun 12 '16 at 16:23
• @Jasper I have modified the answer. The visual aspect was just a reminder for a formula given in other answers – Laurent Duval Jun 12 '16 at 16:49 | 2019-05-26T13:18:42 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1822583/what-is-the-sum-of-the-81-products-in-the-9-times-9-multiplication-grid/1822592",
"openwebmath_score": 0.8559382557868958,
"openwebmath_perplexity": 180.83423429377422,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.986777175525674,
"lm_q2_score": 0.8633916152464016,
"lm_q1q2_score": 0.8519751394653936
} |
https://brilliant.org/discussions/thread/absolute-value-2/ | # Absolute Value
## Definition
For a real number, the absolute value of $$x$$, denoted $$\lvert x \rvert$$, is defined as
$\left|x \right| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0. \end{cases}$
## Technique
### Evaluate $\left| 3\left( 17-55 \right) \right|$.
\begin{aligned} \left| 3\left( 17-55 \right) \right| &= \left| 3\left( -38 \right) \right| \\ &= \left| -114 \right| \\ &= 114 _\square \end{aligned}
### If $x=\frac{5+\sqrt{11}i}{2}$ and $y=\frac{5 -\sqrt{11}i}{2}$, what is the value of $\left| x + 3y \right|^2$?
\begin{aligned} \left| x + 3y \right|^2 &= \left| \frac{5+\sqrt{11}i}{2} + 3\left( \frac{5 -\sqrt{11}i}{2} \right) \right|^2 \\ &= \left| \frac{5+\sqrt{11}i + 15 -3\sqrt{11}i}{2} \right|^2 \\ &= \left| \frac{20-2\sqrt{11}i}{2} \right|^2 \\ &= \left| 10 - \sqrt{11}i \right|^2 \\ &= \left( \sqrt{10^2 + \sqrt{11}^2} \right)^2 \\ &= \sqrt{111}^2 \\ &= 111 _\square \end{aligned}
### What is the absolute value of $2 - 10a$?
The answer depends on the value of $a$. From the definition above, we have
$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } 2-10a \geq 0 \\ -(2-10a), & \text{if } 2-10a < 0 \end{cases}$
Solving the inequality on the right, we obtain
\begin{aligned} 2-10a &\geq 0 \\ 2 & \geq 10a \\ \frac{1}{5} &\geq a. \end{aligned}
It follows that $2-10a \geq 0 \Longleftrightarrow a \leq \frac{1}{5}.$
Therefore,
$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } a \leq \frac{1}{5} \\ 10a - 2, & \text{if } a > \frac{1}{5} \end{cases} _\square$
Note by Arron Kau
7 years ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
thank you very much :-)
- 6 years, 11 months ago
I like this format of note, I think I'll Start Share So !
Explanation More Exercises Solved !
- 7 years ago
I didn't quite understand the problem having $i$. At one stage the $i$ disappears. Can you please explain?
- 7 years ago
It is because the absolute value of a complex number ($a + bi$) is defined as $\sqrt{a^2 + b^2}$ , i.e., $|a + bi|$ = $\sqrt{a^2 + b^2}$. In the above question, $|10 - \sqrt11|$ = $\sqrt{10^2 + \sqrt{11}^2}$
- 7 years ago
Didn't know that, thanks!!
- 7 years ago
I am not sure but as far as my mind think I can say this is because when the sign of absolute value was removed.Then sign was changed to positive & i(iota) has nothing to do when there is is postive root.So,It would have been removed...
- 7 years ago
$i$ is defined as $\sqrt{-1}$ i.e. imaginary unit. So, those were complex numbers (real number + imaginary numbers).
Excellent...
- 7 years ago
In ending step ..note that to take modulus of imag eniry number is different than simple modulas
- 6 years, 5 months ago | 2021-04-14T01:06:27 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/absolute-value-2/",
"openwebmath_score": 1.000007152557373,
"openwebmath_perplexity": 2960.517491791547,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9867771770811146,
"lm_q2_score": 0.8633916117313211,
"lm_q1q2_score": 0.8519751373397467
} |
https://probabilityandstats.wordpress.com/2015/12/12/a-randomized-definition-of-the-natural-logarithm-constant/ | # A randomized definition of the natural logarithm constant
The number $e$ is the base of the natural logarithm. It is an important constant in mathematics, which is approximately 2.718281828. This post discusses a charming little problem involving the the number $e$. This number can be represented in many ways. In a calculus course, the number $e$ may be defined as the upper limit of the following integral:
$\displaystyle \int_1^e \frac{1}{t} \ dt=1$
Another representation is that it is the sum $\displaystyle e=\sum_{n=1}^\infty \frac{1}{n!}$. Still another is that it is the limit $\displaystyle e=\lim_{n \rightarrow \infty} \biggl( 1+\frac{1}{n} \biggr)^n$. According to the authors of a brief article from the Mathematics Magazine [1], these textbook definitions do not give immediate insight about the number $e$ (article can be found here). As a result, students come away from the course without a solid understanding of the number $e$ and may have to resort to rote memorization on what the number $e$ is. Instead, the article gives six probability oriented ways in which the number $e$ can occur. These occurrences of $e$ are more interesting and, according to the authors of [1], can potentially increase students’ appreciation of the number $e$. In two of these six examples (Example 2 and Example 5), the number $e$ is defined by drawing random numbers from the interval $(0,1)$. This post discusses Example 2.
________________________________________________________________________
Random Experiment
Here’s the description of the random experiment that generates the number $e$. Randomly and successively select numbers from the interval $(0, 1)$. The experiment terminates when the sum of the random numbers exceeds 1. What is the average number of selections in this experiment? In other words, we are interested in the average length of the experiment. According to the article [1], the average length is the number $e$. The goal here is to give a proof of this result.
As illustration, the experiment is carried out 1 million times using random numbers that are generated in Excel using the Rand() function. The following table summarizes the results.
$\left[\begin{array}{rrrrrr} \text{Length of} & \text{ } & \text{ } & \text{ } & \text{Relative} \\ \text{Experiment} & \text{ } & \text{Frequency} & \text{ } & \text{Frequency} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 2 & \text{ } & 500777 & \text{ } & 0.500777 \\ 3 & \text{ } & 332736 & \text{ } & 0.332736 \\ 4 & \text{ } & 124875 & \text{ } & 0.124875 \\ 5 & \text{ } & 33465 & \text{ } & 0.033465 \\ 6 & \text{ } & 6827 & \text{ } & 0.006827 \\ 7 & \text{ } & 1130 & \text{ } & 0.001130 \\ 8 & \text{ } & 172 & \text{ } & 0.000172 \\ 9 & \text{ } & 14 & \text{ } & 0.000014 \\ 10 & \text{ } & 4 & \text{ } & 0.000004 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{Total} & \text{ } & 1000000 & \text{ } & \text{ } \end{array}\right]$
The average length of experiment in these 1 million experiments is 2.717001. Even though the rate of convergence to the number $e$ is fairly slow, the simulated data demonstrates that on average it takes approximately $e$ number of simulations of numbers in the unit interval to get a sum that exceeds 1.
________________________________________________________________________
A proof
Let $U_1,U_2,U_3,\cdots$ be a sequence of independent and identically distributed random variables such that the common distribution is a uniform distribution on the unit interval $(0,1)$. Let $N$ be defined as follows:
$\displaystyle N=\text{min}\left\{n: U_1+U_2+\cdots+U_n>1 \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
The objective is to determine $E(N)$. On the surface, it seems that we need to describe the distribution of the independent sum $X_n=U_1+U_2+\cdots+U_n$ for all possible $n$. Doing this may be possible but the result would be messy. It turns out that we do not need to do so. We need to evaluate the probability $P(X_n \le 1)$ for all $n$. We show that
$\displaystyle F_n(x)=P(X_n \le x)=\frac{x^n}{n!} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
for $0 \le x \le 1$ and for $n=1,2,3,\cdots$. This is accomplished by an induction proof. This is true for $n=1$ since $X_1=U_1$ is a uniform distribution. Suppose (2) holds for the integer $n-1$. This means that $\displaystyle F_{n-1}(x)=P(X_{n-1} \le x)=\frac{x^{n-1}}{(n-1)!}$ for $0 \le x \le 1$. Note that $X_n=X_{n-1}+U_n$, which is an independent sum. Let’s write out the convolution formula for this independent sum:
\displaystyle \begin{aligned} F_n(x)&=\int_{0}^x F_{n-1}(x-y) \cdot f_{U_n}(y) \ dy \\&=\int_{0}^x \frac{(x-y)^{n-1}}{(n-1)!} \cdot 1 \ dy \\&=\frac{1}{(n-1)!} \int_0^x (x-y)^{n-1} \ dy \\&=\frac{x^n}{n!} \end{aligned}
The above derivation completes the proof of the claim. We now come back to the problem of evaluating the mean of the random variable $N$ defined in (1). First, note that $N>n$ if and only if $X_n=U_1+U_2+\cdots+U_n \le 1$. So we have the probability statement $\displaystyle P(N>n)=F_n(1)=\frac{1}{n!}$.
As a result, the following is the probability function of the random variable $N$.
\displaystyle \begin{aligned} P(N=n)&=P(N>n-1)-P(N>n) \\&=F_{n-1}(1)-F_n(1) \\&=\frac{1}{(n-1)!}-\frac{1}{n!} \\&=\frac{n-1}{n!} \end{aligned}
Now evaluate the mean.
\displaystyle \begin{aligned} E(N)&=\sum \limits_{n=2}^\infty \frac{n(n-1)}{n!} \\&=\sum \limits_{n=2}^\infty \frac{1}{(n-2)!} \\&=\sum \limits_{m=0}^\infty \frac{1}{m!} \\&=e \end{aligned}
With the above derivation, the proof that $e=$ 2.718281828… is the average number of random numbers to select in order to obtain a sum that exceeds 1 is completed.
________________________________________________________________________
Reference
1. Shultz H. S., Leonard B., Unexpected Occurrences of the Number e,Mathematics Magazine, October 1989, Volume 62, Number 4, pp. 269–271.
________________________________________________________________________
$\copyright \ \text{2015 by Dan Ma}$ | 2017-02-26T19:05:05 | {
"domain": "wordpress.com",
"url": "https://probabilityandstats.wordpress.com/2015/12/12/a-randomized-definition-of-the-natural-logarithm-constant/",
"openwebmath_score": 0.943716287612915,
"openwebmath_perplexity": 215.07585780412978,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9867771774699747,
"lm_q2_score": 0.8633916029436189,
"lm_q1q2_score": 0.8519751290039814
} |
https://math.stackexchange.com/questions/2711754/can-the-product-of-infinitely-many-elements-from-mathbb-q-be-irrational/2711761 | # Can the product of infinitely many elements from $\mathbb Q$ be irrational?
I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. Thank you for your answers.
• Do you know Wallis's product for $\pi$? Mar 28, 2018 at 13:23
• $e=\lim_{n\to \infty}{(1+1/n)^n}$ Mar 28, 2018 at 13:29
• @Vasya Correct limit of rationals, but not an infinite product. Mar 28, 2018 at 13:29
• @DzamoNorton Not a limit of partial products. The number of factors increases each time, but the factors change: $(1 + 1/2)(1+1/2)$, $(1+1/3)(1 + 1/3)(1+ 1/3)$ and so on. Mar 28, 2018 at 19:51
• @toliveira: An "infinite product" is not multiplication. It's a limit. Fundamentally, multiplication has two operands; you can inductively extend that to any finite number of operands. But "any finite number" does not include "infinitely many". Mar 29, 2018 at 15:10
Yes, it can.
Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and $$b_n = \frac{a_n}{a_{n-1}}$$ for $n > 1$.
We then have that $$b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.$$
We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.
• Great answer. Very simple. Mar 29, 2018 at 9:57
• But it does look like a "cheat" since everything is divided out by itself. So, is there an infinite product which does not contain a collection of $\frac{a_n}{a_n}$ ? both your answer and mohammad Riazi's have this "cheat", while Kumar's does not. Mar 29, 2018 at 18:56
• @Carl This isn't a cheat. In fact you can rewrite any infinite product in this way. Mar 29, 2018 at 19:23
• +1 But I think it needs at least a short explanation (or link to one, maybe the question is already out here too?) that such a sequence "of non-zero rational numbers which converges to an irrational number" exists. It is easier too see than the initial question (I think) but I guess some of those who will fins the initial question interesting may not see this as obvious Mar 29, 2018 at 20:08
Yes, every irrational number is an infinite product of rationals.
We can write an infinite sum of rationals as an infinite product of rationals.
\begin{align} a&=a,\\ a+b&=a\times\frac {a+b}{a}\\ a+b+c &= a \times \frac {a+b}{a}\times\frac {a+b+c}{a+b}\\.\\.\\.\\.\end{align}
For example, $$\sqrt 2 =1.414213....=1+.4+.01+.004+.....=$$
$$1\times \frac {1.4}{1}\times \frac {1.41}{1.4}\times\frac {1.414}{1.41}\times .....$$
• Your answer is chosen for a review audit of mine, which I upvoted (+1) because it is simple, generalized and easy to understand. Mar 29, 2018 at 5:10
• @TrầnThúcMinhTrí Thanks for your attention. Mar 29, 2018 at 11:19
• I took the liberty to align equalities at '=' sign. Feel free to rollback if you don't like it. Mar 30, 2018 at 7:29
• Thanks, I like it Mar 30, 2018 at 11:12
Yes!
$\cfrac{\pi}{2} = \cfrac{2}1 \cfrac 23 \cfrac 43 \cfrac 45 \cfrac 65 \cfrac 67 \cdots$
• This looks cool, but what is the geometric picture to go along with it? I an almost see polygons in there... Mar 28, 2018 at 16:40
• @MikeWise I din't understand which polygon you are talking about. Mar 28, 2018 at 16:42
• General term is $\cfrac{2n}{(2n – 1)}\cfrac{2n}{(2n + 1)}$ Mar 28, 2018 at 16:43
• @MikeWise: I was curious about that as well. Here's a short note on how when you draw rectangles of those areas, they add up to a quarter circle in the limit: math.chalmers.se/~wastlund/monthly.pdf Mar 28, 2018 at 17:39
• Cool, thanks for that, will print it out and read it on my flight in the morning. :) Mar 28, 2018 at 17:47
Too big to be a comment: it should be noted that the order is more crucial in infinite products than in infinite sums, which is strikingly seen on the example cited many times already:
\begin{align*}\cfrac{\pi}{2}&=\cfrac{2}1 \cfrac 23 \cdot \cfrac 43 \cfrac 45 \cdot\cfrac 65 \cfrac 67\cdot \ldots\\ &= \cfrac{2^2}{2^2-1}\cdot \cfrac{4^2}{4^2-1}\cdot \cfrac{6^2}{6^2-1}\ldots\\ \end{align*} is an infinite product with partials starting at $\frac43$ and increasing towards $\frac\pi 2$ (every factor is greater than $1$), whereas the seemingly identical
\begin{align*}0&=\cfrac{2}3 \cfrac 23 \cdot \cfrac 45\cfrac 45\cdot\cfrac 67 \cfrac 67 \cdot\ldots\end{align*}
starts below $1$ and decreases, towards $0$. All that happened was a shift of denominators one step to the left.
• Products are isomorphic to sum of logs, so any order effect that shows up in one can be generated in the other. By allowing the denominators to be shifted independently of the numerators, you are treating the fraction (a/b) as ab^-1, so when you take the log, it's log(a)-log(b), which is an alternating series, and of course order is more important in alternating series than monotonic ones. It's the alternation generated by treating the numerator and denominator separately, not the product, that makes order important. Mar 28, 2018 at 15:00
• Allowing the denominators to be shifted independently of the numerators can also be interpreted as replacing each term a/b by two terms, a and 1/b. Thus when you take logs, you get two terms, log a and - log b. Mar 28, 2018 at 22:10
• These explanations are of course correct, but the point is that it is not obvious to realise that it can matter, easy to make a mistake here. Mar 29, 2018 at 8:32
• I think the claim is not that one shouldn't point out that order matters in an infinite product, but rather that one shouldn't claim that it matters more in infinite products than in infinite sums, since (as @Acccumulation points out) for products of positive real numbers and sums of real numbers it is literally the same phenomenon. Mar 29, 2018 at 22:38
• @LSpice when I said that it matters more, what I meant is that one should beware of the intuition that a permutation of terms acting only locally does not affect the final result (you need a violent perturbation of terms to give $\sum{(-1)^n\over n}$ a different limit). Of course here what is happening is that the terms are not only reordered, but broken to pieces and the pieces reordered. *This natural way to think of a rational factor as being made of two pieces does not happen with series. * Mar 29, 2018 at 23:44
Consider the Riemann-Zeta Function: $$\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p\text{ prime}}\frac{1}{1-p^{-s}}.$$ For $s=2$, the infinite sum on the left is $\pi^{2}/6$, which is irrational. Thus, $\pi^{2}/6$ is an infinite product of rationals.
There is a simple way to obtain any irrational number as an infinite product:
• take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\pi$ to $n$ decimals);
• form the product of the numbers $f_n:=\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.
$$\pi=\prod_{n=0}^\infty f_n=\frac{31}{10}\cdot\frac{314}{310}\cdot\frac{3141}{3140}\cdot\frac{31415}{31410}\cdot\frac{314159}{314150}\cdots$$
• Ooops, I just noticed that my answer is a duplicate.
– user65203
Mar 30, 2018 at 15:29 | 2022-07-02T12:31:52 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2711754/can-the-product-of-infinitely-many-elements-from-mathbb-q-be-irrational/2711761",
"openwebmath_score": 0.8812333345413208,
"openwebmath_perplexity": 634.1386323894059,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9867771786365548,
"lm_q2_score": 0.863391599428538,
"lm_q1q2_score": 0.8519751265425952
} |
https://www.physicsforums.com/threads/a-gravitation-problem-with-two-masses.748274/ | # A gravitation problem with two masses
1. Apr 11, 2014
### toothpaste666
1. The problem statement, all variables and given/known data
A uniform sphere has mass M and radius r. A spherical cavity (no mass) of radius r/2 is then carved within this sphere (the cavity's surface passes through the sphere's center and just touches the sphere's outer surface). The centers of the original sphere and the cavity lie on a straight line, which defines the x axis.
With what gravitational force will the hollowed-out sphere attract a point mass m which lies on the x axis a distance d from the sphere's center? [Hint: Subtract the effect of the "small" sphere (the cavity) from that of the larger entire sphere.]
2. Relevant equations
$F = G\frac{m_1m_2}{r^2}$
3. The attempt at a solution
Let mass $M_f$ = the mass of the sphere without the cavity, $M_c$ = the mass of the cavity, and $R_f$ = the distance between the two masses.
$M_f = M - M_c$
$R_f = d + R_2$
$F = G\frac{(M_f) m}{(R_f)^2}$
is this right?
2. Apr 11, 2014
### SammyS
Staff Emeritus
Did you intend for MC to be negative?
I interpret the Hint to be to calculate the gravitational force that would be exerted on the point mass if the sphere were not hollow, but had the same density. From that, subtract the gravitational force that would be exerted on the point mass by an isolated uniform sphere the size of the cavity, also having the same density.
That would give MF = M + MC .
When you say, "Rf = the distance between the two masses", what two masses are you referring to ?
3. Apr 12, 2014
### toothpaste666
Here is the picture. The two masses are the small particle mass on the right and the sphere on the left with the cavity
#### Attached Files:
• ###### physicsprob.png
File size:
147.4 KB
Views:
154
4. Apr 12, 2014
### haruspex
The hint does not say subtract the masses, it says subtract the gravitational effects.
What force would be exerted by the complete sphere?
Suppose we take the "complement" of the cavitated sphere, i.e. throw away the larger sphere and instead have just the smaller sphere (at the position of the cavity). What force would that exert?
5. Apr 13, 2014
### toothpaste666
so if M is the mass of the sphere before the cavity was taken out and Mc is the mass of the cavity then we have:
$G\frac{Mm}{d^2} - G\frac{M_c m}{(d-r_2)^2}$
6. Apr 13, 2014
### SammyS
Staff Emeritus
Yes, if M is the mass of the sphere before the cavity is taken out. That is as the problem is stated. (So, I believe I was incorrect in my previous post.)
You can determine Mc in terms of M.
7. Apr 13, 2014
### toothpaste666
If the mass is evenly distributed wouldn't subtracting the volume be the same as subtracting the mass?
8. Apr 13, 2014
### SammyS
Staff Emeritus
What are you subtracting from what?
What is the mass of a sphere the size of the cavity? How is that related to the mass of the large sphere before the cavity is formed in the sphere?
9. Apr 13, 2014
### toothpaste666
Subtracting the volume of the cavity from the volume of the original sphere i mean. Wouldn't the mass be the same as the volume if its evenly distributed?
10. Apr 13, 2014
### SammyS
Staff Emeritus
Mass is be proportional to volume in this case.
What fraction of the initial solid sphere's mass is removed to make the cavity ?
11. Apr 13, 2014
### toothpaste666
To find the volumes let Vc = Volume of cavity:
$r_2 = \frac{r}{2}$
$V_c = \frac{4pi(\frac{r}{2})^3}{3}$
$V_c = \frac{\frac{4pi(r^3)}{8}}{3}$
$V_c = \frac{\frac{pi(r^3)}{2}}{3}$
$V_c = \frac{pi(r^3)}{6}$
since $V_M =\frac{4pi(r^3)}{3}$
we have
$V_M = 8V_c$
does this get me closer to expressing Mc in terms of M?
12. Apr 14, 2014
### haruspex
That's it.
13. Apr 14, 2014
### toothpaste666
but how do i relate that back to mass?
14. Apr 14, 2014
### haruspex
Through density, which is the same for both.
15. Apr 14, 2014
### toothpaste666
$D = \frac{M}{V_M} = \frac{3M}{4(pi)r^3}$
$M_c = D V_c = D \frac{(pi)r^3}{6} = (\frac{3M}{4(pi)r^3})(\frac{(pi)r^3}{6}) = \frac{M}{8}$
which gives
$G\frac{Mm}{d^2} - G\frac{Mm}{8(d-r_2)^2}$
how do we know the density is equal?
16. Apr 14, 2014
### haruspex
See the OP:
17. Apr 14, 2014
Thank you. | 2017-08-24T07:52:09 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/a-gravitation-problem-with-two-masses.748274/",
"openwebmath_score": 0.6084936857223511,
"openwebmath_perplexity": 1445.7704737128088,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9805806540875628,
"lm_q2_score": 0.8688267864276108,
"lm_q1q2_score": 0.8519547385239818
} |
https://math.stackexchange.com/questions/2400259/limits-and-substitution | # Limits and substitution
As in @RobertZ's answer to this question, we often perform substitutions when evaluating limits. For instance, if you're asked to show that $$L = \lim_{t \to 0} \frac{\sin t^3}{t^3} = 1,$$ it's pretty common to say "Let $x = t^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."
What's going on here in general is an application of the following "Theorem"
Theorem 1: If the function $g$ satisfies [fill in missing properties] and $$\lim_{t \to a} g(t) = b,$$ then $$\lim_{t \to a} f(g(t)) = \lim_{x \to b} f(x),$$ i.e., one limit exists if and only if the other does, and if they both exist, they're equal.
In the example above, $f(x) = \frac{\sin x}{x}$ and $g(t) = t^3$ and $a = b = 0$.
There's an alternative form, in which we're asked to show that $$L = \lim_{t \to 0} \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}} = 1,$$ it's pretty common to say "Let $t = x^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."
In this case, the implicit theorem is very similar to the other, but with the role of $g$ reversed (i.e., we're substituting $t = x^3$ instead of $x = t^3$, so the natural form of the theorem puts $g$ on the other side):
Theorem 2: If the function $g$ satisfies [fill in missing properties] and $$\lim_{x \to b} g(x) = a,$$ then $$\lim_{t \to a} f(t) = \lim_{x \to b} f(g(x)).$$
In the second example above, we have $a = b = 0$, $f(t) = \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}}$, and $g(x) = x^3$.
The two theorems are obviously the same: if you swap $a$ and $b$, $x$ and $t$, and reverse the equality in the last line, they're identical. But each represents a different approach to working with limits, so I've stated both.
In the second form, it's clearly important that $g$ be surjective near $a$ (i.e., for every small enough interval $I = (b-\epsilon, b + \epsilon)$, there's an interval $I' = (a-\delta, a + \delta)$ such that $I- \{b\} \subset g(I' - \{a\})$. (Hat-tip to MathematicsStudent1122 for the observation that I need to delete $a$ and $b$ from those intervals). Otherwise you could use things substitutions like $s = t^2$, which would turn a two-sided limit into a one-sided one (or vice versa), in which case one limit might exists and the other might not.
Addendum to clarify why this might matter, for @MathematicsStudent1122:
Consider $$f(x) = \begin{cases} 1 & x \ge 0 \\ 0 & x < 0 \end{cases}.$$
and look at $L = \lim_{x \to 0} f(x^2)$. It's clear that $L$ exists and is $1$. But if we substitute $t = x^2$, then we get $L = \lim_{t \to 0} f(t)$, which does not exist; hence this "substitution" is not valid: I've turned what amounts to a 1-sided limit (which exists) into a two-sided limit (which does not exist). The domains of $f$ and $g$ are both all of $\Bbb R$.
My question is this:
What is a reasonable set of missing properties for each of these theorems? (I can work out the exact properties easily enough by running through the definitions, but they don't seem to be very helpful/checkable.)
One answer might be "$g$ is locally a bijection", but that rules out things like $y = x + x\sin \frac{1}{x}$ near $x = 0$, so it seems too limited. (It also rules out things like $x \mapsto x + \sin x$ for limits as $x \to \infty$, which is a pity.)
I recognize that this is not a strictly mathematical question. But my goal is to come up with a "calculus student's theorem", one that says "if you're trying to work out a limit, which may or may not exist, then it's OK to do substitutions of this sort along the way," and which will cover the vast majority of the problems that they might encounter in a standard calculus book, or even in Spivak's book.
This question gives two theorems, but both have assumptions about the existence of limits. This one comes a little closer, but still isn't entirely satisfactory.
I'd love any nice-enough condition to be broadly useful. In particular, I think it's completely reasonable to require, for instance, that the "substitution function" $g$ be continuous, and perhaps even differentiable (although I doubt that's of much use).
• – Jack D'Aurizio Aug 20 '17 at 15:03
• Great point, @JackD'Aurizio. That may be the first place where this particular stunt bothered me. :) – John Hughes Aug 20 '17 at 15:04
• So that's part of the quest for the Holy Grail of all lazy students, the Grand Unified Formula Of All Textbook Exercises? ;-) – Professor Vector Aug 20 '17 at 15:06
• Substitutions like $t=x\sin \frac{1}{x}$ simply don't always work. If we let $g(x) = x\sin \frac{1}{x}$ and $f(x) = 1$ for $x=0$ and $f(x) = 0$ for $x \neq 0$, then we can note that $$\lim_{x \to 0} f(g(x))$$ does not exist, but $$\lim_{y \to 0} f(y)$$ does exist. – MathematicsStudent1122 Aug 20 '17 at 15:56
• @ParamanandSingh: Well...I've mused about it on miscellaneous occasions for about 45 years, but I thank you for pointing out that I've bene wasting my time. :) Your answer in that linked case is the easy one --- the one that only goes one direction. I want to know that if I make a substitution and the origiinal limit does NOT exist, then the new limit is also guaranteed to not exist. – John Hughes Aug 20 '17 at 18:07
I'm afraid this isn't quite what you're after, but it seems to be a decent starting point. These conditions, while restrictive, appear to be necessary for the following proof.
Assume $\lim_{x\to a}g(x)=b$, where the function $g$ also satisfies: $$\text{there exists a neighborhood U of a such that b\not\in g(U\setminus\{a\})} \tag{1}$$ and for each $(y_n)$ converging to $b$ with $y_n\ne b$, $$\text{there exists (x_n) such that g(x_n)=y_n for large n and x_n\to a}. \tag{2}$$
Then we show that $$\lim_{x\to a}f(g(a)) = \lim_{y\to b}f(y).$$
Proof:
Let $L_1:=\lim_{x\to a}f(g(a))$ and $L_2:=\lim_{y\to b}f(y)$, either of which may or may not exist. We will use the sequential criterion.
First suppose $L_2$ exists and let $(x_n)$ be a sequence such that $x_n\to a$ and $x_n\ne a$ for all $n$. Since $\lim_{x\to a}g(x)=b$, we have $g(x_n)\to b$. Due to $(1)$, we have $g(x_n)\ne b$ for $n$ large enough. Then, because $L_2$ exists, we have $f(g(x_n))\to L_2$. This proves by the sequential criterion that $L_1$ exists and $L_1=L_2$.
Now assume $L_1$ exists and let $(y_n)$ be a sequence such that $y_n\to b$ and $y_n\ne b$ for all $n$. By $(2)$ there exists a sequence $(x_n)$ such that $g(x_n)=y_n$ for large $n$ and $x_n\to a$. Since $y_n\ne b$ for all $n$, we have $x_n\ne a$ for large $n$. Then $$\lim_{n\to\infty}f(y_n)=\lim_{n\to\infty}f(g(x_n))=L_1,$$ and so the sequential criterion implies $L_2$ exists and $L_2=L_1$.
• You're right --- that's not quite what I'm after. For one thing, it's not a condition that most calc students could test. :( But I agree that it's a decent starting point -- thanks. – John Hughes Aug 21 '17 at 19:58 | 2020-09-30T10:02:30 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2400259/limits-and-substitution",
"openwebmath_score": 0.9307198524475098,
"openwebmath_perplexity": 201.83231004638074,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9805806518175515,
"lm_q2_score": 0.8688267779364222,
"lm_q1q2_score": 0.8519547282254399
} |
https://math.stackexchange.com/questions/778946/prove-that-if-a-is-normal-then-eigenvectors-corresponding-to-distinct-eigenva | Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)
The problem statement is as follows:
Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.
I can certainly prove that this is the case, using the spectral theorem. The gist of my proof is presented below.
If possible, I would like to find a simpler proof. I was hoping that there might be some sort of manipulation along these lines, noting that $$\langle Av_1,A v_2\rangle = \langle v_1,A^*Av_2\rangle = \langle v_1,AA^*v_2\rangle = \langle A^* v_1,A^* v_2 \rangle$$
Any ideas here would be appreciated.
My proof:
Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that $$A v_{\lambda,i} = \lambda v_{\lambda,i}$$ Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$. We may write $v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda} .$ We then have $$0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda}$$ So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$. Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$. From there, we have $$\langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda}$$ the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$.
• "Let $v_{\lambda,i}$ be an orthonormal basis of eigenvectors..." I'd guess that this might be something very close to what would need to be proved. – Algebraic Pavel May 3 '14 at 2:13
• @PavelJiranek I was worried that I had used circular logic at some point. However, the existence of such a basis (i.e. the spectral theorem) comes directly from the Schur triangularization theorem, which says nothing about normal matrices in particular. – Omnomnomnom May 3 '14 at 19:11
Assume $\;\lambda\neq \mu\;$ and
$$\begin{cases}Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v\\{}\\Aw=\mu w\implies A^*w=\overline\mu w\end{cases}$$
From this we get:
$$\begin{cases}\langle v,Aw\rangle=\langle v,\mu w\rangle=\overline\mu\langle v,w\rangle\\{}\\ \langle v,Aw\rangle=\langle A^*v,w\rangle=\langle\overline\lambda v,w\rangle=\overline\lambda\langle v,w\rangle \end{cases}$$
and since $\;\overline\mu\neq\overline\lambda\;$ , we get $\;\langle v,w\rangle =0\;$
Question: Where did we use normality in the above?
• How de we know that $v$ is also an eigenvector for $A^*$? – Berci May 2 '14 at 22:55
• @Berci, read and think of the question at the end of the answer. – DonAntonio May 2 '14 at 22:56
• From the other two answers this follows, of course, using that $A$ is normal. But can you show it directly? – Berci May 2 '14 at 22:57
• Why directly? Use the definition and work it out. I usually don't give full answers and, after all, the OP is trying to get a more or less simpler proof to the claim than the one using the spectral theorem... – DonAntonio May 2 '14 at 23:00
• I like it! Showing that $Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v$ seems to be the step in the spirit of what I was looking for. Thanks. – Omnomnomnom May 2 '14 at 23:14
Specializing your identity to $v_1=v_2=v$, we get $\|Av\|=\|A^*v\|$. Hence $\ker A=\ker A^*$. Recalling that $\ker A^* = (\operatorname{ran} A)^\perp$ for general $A$, we conclude that the kernel and range of a normal matrix are mutually orthogonal.
It remains to apply the above conclusion to $A-\lambda I$ where $\lambda$ is an eigenvalue of $A$.
• Neat! Although DonAntonio's proof is more along the lines I was thinking of, this is a nice perspective. – Omnomnomnom May 2 '14 at 23:16
I try to give another simple proof to $$T^*v=\bar{\lambda}v ~\text{ if }~ Tv=\lambda v$$ where $T$ is a normal operator on a Hilbert space $H$.
Suppose $V=\ker(T-\lambda I)$. Since $T^*$ communicate with $T$, $$T^*V\subset V.$$ Because $$\langle v,T^*v\rangle =\langle Tv,v\rangle =\langle \lambda v,v\rangle=\langle v,\bar{\lambda}v\rangle ~~\forall v \in V,$$ $\langle u,T^*v\rangle =\langle u, \bar{\lambda}v\rangle ~\forall u,v\in V$ by polarisation identity, and thus $T^*v=\bar{\lambda}v.$
REMARK: Let $\sigma:V\times V\to W$ be a sesquilinear form, where $V$ and $W$ are linear vector spaces over $\mathbb{C}$. The follwing formula is called Polarisation Identity : $$\sigma(u,v)=\sum_{k=0}^3 i^k\sigma(u+i^k v, u+i^kv).$$
• I don't really see how this answer addresses the question being asked. Also, the question being asked was answered 3 years ago. – Omnomnomnom Jun 12 '17 at 15:01
• @Omnomnomnom This is why $Av=\lambda v\Rightarrow A^*v=\bar{\lambda}v$ using no spectral theorem which is the point to the question I think. – C.Ding Jun 13 '17 at 2:37
• That is not the point to the question, actually. The point is to show that eigenvectors corresponding to different eigenvalues are necessarily orthogonal. – Omnomnomnom Jun 13 '17 at 2:58
• Oh, sorry to disturb you. As I give some additional remarks to the right answer you have accepted. – C.Ding Jun 13 '17 at 3:04
A normal matrix is unitarily similar to diagonal matrix.
$$A = UDU^{-1}$$ where $U$ is Unitary matrix.
Eigen decompositions tells that $U$ is a matrix composed of columns which are eigenvectors of $A$. And matrix $D$ is Diagonal matrix with eigenvalues on diagonal.
Property: Columns of Unitary matrix are orthogonal.
So, columns of $U$ (which are eigenvectors of $A$) are orthogonal. | 2019-05-23T03:10:22 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/778946/prove-that-if-a-is-normal-then-eigenvectors-corresponding-to-distinct-eigenva",
"openwebmath_score": 0.9323018193244934,
"openwebmath_perplexity": 187.45606644364267,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9805806501150426,
"lm_q2_score": 0.8688267660487572,
"lm_q1q2_score": 0.8519547150894403
} |
https://gmatclub.com/forum/within-a-rectangular-courtyard-of-length-60-feet-a-graveled-path-277135.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 17 Oct 2018, 11:33
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Within a rectangular courtyard of length 60 feet, a graveled path,
Author Message
TAGS:
### Hide Tags
Intern
Joined: 25 Aug 2018
Posts: 2
Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
### Show Tags
24 Sep 2018, 10:25
00:00
Difficulty:
65% (hard)
Question Stats:
58% (03:51) correct 42% (04:22) wrong based on 19 sessions
### HideShow timer Statistics
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :
A 24
B 30
C 40
D 45
E 54
Intern
Joined: 14 Feb 2018
Posts: 6
Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
### Show Tags
24 Sep 2018, 11:12
Being x the width of the rectangle, we have the following equation:
$$2[2(60*3)+2*3(x-6)] + 984 = 2[2(60*6)+2*6(x-12)]$$
where $$2(60*3)$$ is the area of the path next to the longer side (60 ft long and 3 ft wide), and $$2*3(x-6)$$ the area of the path next to the shorter side, avoiding to count the area from the corner twice. Then we need to repeat the same idea with the bigger path. Both areas are multiplied by 2 because we are measuring the cost, which is Rs 2 per sq ft. We have to consider that the path with double size is Rs 984 more expensive, so we add this into our equation. Then, we can solve the equation:
$$2[360+6x-36]+984=2(720+12x-144)$$
$$360+6x-36+492=720+12x-144$$
$$324+6x+492=576+12x$$
$$816-576=6x$$
$$240=6x$$
$$x=40$$
Thus, C is the correct answer.
Manager
Joined: 02 Aug 2015
Posts: 112
Re: Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
### Show Tags
24 Sep 2018, 11:33
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :
A 24
B 30
C 40
D 45
E 54
A = Area of outer rectangle = 60 x Width
Ax = Area of inner rectangle with width of 3 = 54 x (Width-6)
Ay = Area of inner rectangle with twice the width, i.e 6 = 48 x (Width - 12)
Area of gravel with normal width = A-Ax
Area of gravel with twice width = A-Ay
Given, cost of gravelling twice width - cost of gravelling normal width = 984.
Given, cost of gravelling = Rs2/sq ft.
2(A-Ay) - 2(A-Ax) = 984.
A-Ay-A+Ax=492.
Ax-Ay=492.
On substituting Ax and Ay from above and solving the above equation, we get width=40.
Hence C.
Cheers!
Intern
Joined: 21 Jul 2018
Posts: 38
Re: Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
### Show Tags
24 Sep 2018, 13:48
Tried to solve using an Ans choices and got lucky with the 1st option
_________________
______________________________
Consider KUDOS if my post helped !!
I'd appreciate learning about the grammatical errors in my posts
Please let me know if I'm wrong somewhere
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3845
Location: United States (CA)
Re: Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
### Show Tags
27 Sep 2018, 17:20
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :
A 24
B 30
C 40
D 45
E 54
We can let w = the width of the courtyard. Therefore, the courtyard, including the graveled path, has an area of 60w sq ft. Excluding the path, the area of the courtyard would be (60 - 2(3))(w - 2(3)) = 54(w - 6) = 54w - 324 sq ft. So the path alone has an area of 60w - (54w - 324) = 6w + 324 sq ft. Since the cost of graveling the path is Rs 2 per sq ft, the cost of graveling the path is Rs 12w + 648.
If the path had been twice as wide, then the courtyard, including the path, would still have an area of 60w sq ft. However, excluding the path, the area of the courtyard would be (60 - 2(6))(w - 2(6)) = 48(w - 12) = 48w - 576 sq ft. So the path alone would have an area of 60w - (48w - 576) = 12w + 576 sq ft, and the cost of the path would be Rs 24w + 1152. We are told that this would cost Rs 984 more, so we can set up the following equation to solve for w:
12w + 648 + 984 = 24w + 1152
12w + 1632 = 24w + 1152
480 = 12w
40 = w
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Re: Within a rectangular courtyard of length 60 feet, a graveled path, &nbs [#permalink] 27 Sep 2018, 17:20
Display posts from previous: Sort by | 2018-10-17T18:33:10 | {
"domain": "gmatclub.com",
"url": "https://gmatclub.com/forum/within-a-rectangular-courtyard-of-length-60-feet-a-graveled-path-277135.html",
"openwebmath_score": 0.7202662229537964,
"openwebmath_perplexity": 3981.3950371136293,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 1,
"lm_q2_score": 0.851952809486198,
"lm_q1q2_score": 0.851952809486198
} |
https://gmatclub.com/forum/5-drill-machines-can-drill-15-meters-in-5-hours-252975.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 19 Oct 2018, 11:34
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# 5 drill machines can drill 15 meters in 5 hours.
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Manager
Joined: 22 Apr 2017
Posts: 111
Location: India
GMAT 1: 620 Q46 V30
GMAT 2: 620 Q47 V29
GMAT 3: 630 Q49 V26
GMAT 4: 690 Q48 V35
GPA: 3.7
5 drill machines can drill 15 meters in 5 hours. [#permalink]
### Show Tags
06 Nov 2017, 07:17
1
00:00
Difficulty:
25% (medium)
Question Stats:
77% (01:29) correct 23% (02:02) wrong based on 89 sessions
### HideShow timer Statistics
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60
Intern
Joined: 10 Aug 2017
Posts: 8
Re: 5 drill machines can drill 15 meters in 5 hours. [#permalink]
### Show Tags
06 Nov 2017, 07:27
Mac 5 meters 15 hours 5
Mac x meters 60 hours 2
Do it with proportion.
X/5= 60/15 * 5/2
X=50
Sent from my iPhone using GMAT Club Forum
Math Expert
Joined: 02 Aug 2009
Posts: 6961
Re: 5 drill machines can drill 15 meters in 5 hours. [#permalink]
### Show Tags
06 Nov 2017, 07:32
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60
to drill 15 mtr in 5 hrs, 5 machines are required..
so to drill 15*4 mtr in 5 hrs, 5*4 machines are required..
to drill 60 mtr in 1 hrs, 20*5 machines are required..
finally to drill 60 mtr in 2 hrs, 100/2 or 50 machines are required..
D
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
GMAT online Tutor
Senior SC Moderator
Joined: 22 May 2016
Posts: 2035
5 drill machines can drill 15 meters in 5 hours. [#permalink]
### Show Tags
06 Nov 2017, 09:17
1
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60
Find a machine's individual rate from the first information, and use it to solve the question.
W = (# of machines) * (indiv.) rate * time
W = 15, # of machines = 5, rate = ??, time = 5
15 = 5 * rate * 5
15 = 25 * rate
Individual rate = $$\frac{15}{25}=\frac{3}{5}$$ in meters/hour
At that rate, how many machines are needed to drill 60 meters in 2 hours?
W = (# of machines) * (indiv.) rate * time
W = 60, # of machines = ??, rate = $$\frac{3}{5}$$, time = 2
60 = (# of machines) * $$\frac{3}{5}$$ * 2
60 = (# of machines) * $$\frac{6}{5}$$
# of machines = $$\frac{60}{(\frac{6}{5})}= 60 * \frac{5}{6}=$$ 50
_________________
___________________________________________________________________
For what are we born if not to aid one another?
-- Ernest Hemingway
Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: 5 drill machines can drill 15 meters in 5 hours. [#permalink]
### Show Tags
12 Nov 2017, 08:47
1
1
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60
The rate for 5 drill machines is 15/5 = 3 meters per hour. We need to determine how many drill machines we need for a rate of 60/2 = 30 meters per hour. Since 30 is 10 times 3, we would need 5 x 10 = 50 drill machines.
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Re: 5 drill machines can drill 15 meters in 5 hours. &nbs [#permalink] 12 Nov 2017, 08:47
Display posts from previous: Sort by
# 5 drill machines can drill 15 meters in 5 hours.
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-10-19T18:34:14 | {
"domain": "gmatclub.com",
"url": "https://gmatclub.com/forum/5-drill-machines-can-drill-15-meters-in-5-hours-252975.html",
"openwebmath_score": 0.5159401893615723,
"openwebmath_perplexity": 8441.10621709785,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 1,
"lm_q2_score": 0.8519528038477824,
"lm_q1q2_score": 0.8519528038477824
} |
http://gmatclub.com/forum/what-is-the-average-arithmetic-mean-of-eleven-consecutive-93188.html#p717074 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 21 Oct 2016, 05:17
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# What is the average (arithmetic mean) of eleven consecutive
Author Message
TAGS:
### Hide Tags
Manager
Joined: 27 Feb 2010
Posts: 105
Location: Denver
Followers: 1
Kudos [?]: 328 [0], given: 14
What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
23 Apr 2010, 18:05
21
This post was
BOOKMARKED
00:00
Difficulty:
25% (medium)
Question Stats:
67% (01:44) correct 33% (00:39) wrong based on 407 sessions
### HideShow timer Statistics
What is the average (arithmetic mean) of eleven consecutive integers?
(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
[Reveal] Spoiler: OA
Manager
Joined: 27 Feb 2010
Posts: 105
Location: Denver
Followers: 1
Kudos [?]: 328 [1] , given: 14
Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
### Show Tags
23 Apr 2010, 18:11
1
KUDOS
Some how i got E
1. (63+ X10 + X11) / 11 = ?? ..sitill missing two numbers? so insuff??
2. (X1+X2 +81)/ 11 = ?? Still missing two numbers so Insuff???
and combining 1 and 2 still missing some info ? .. Sorry friends, i need help here.
Math Expert
Joined: 02 Sep 2009
Posts: 35240
Followers: 6618
Kudos [?]: 85287 [10] , given: 10236
Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
### Show Tags
24 Apr 2010, 06:10
10
KUDOS
Expert's post
11
This post was
BOOKMARKED
zz0vlb wrote:
What is the average ( arithmetic mean ) of eleven consecutive integers?
1.The avg of first nine integers is 7
2. The avg of the last nine integers is 9
[Reveal] Spoiler:
D
friends, please explain this to me.
Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case $$mean=median=x_6$$.
(1) $$x_1+x_2+...+x_9=63$$ --> there can be only one set of 9 consecutive integers to total 63. Sufficient.
If you want to calculate: $$(x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63$$ --> $$x_6=8$$.
OR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> $$x_5*9=63$$ --> $$x_5=7$$ --> $$x_6=7+1=8$$
(2) $$x_3+x_4+...+x_{11}=81$$ --> there can be only one set of 9 consecutive integers to total 81. Sufficient.
If you want to calculate: $$(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81$$ --> $$x_6=8$$.
OR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> $$x_7*9=81$$ --> $$x_7=9$$ --> $$x_6=9-1=8$$
_________________
Manager
Joined: 27 Feb 2010
Posts: 105
Location: Denver
Followers: 1
Kudos [?]: 328 [0], given: 14
Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
### Show Tags
24 Apr 2010, 07:11
Thanks for clarifying this Bunuel. +1Kudos to you.
Math Expert
Joined: 02 Sep 2009
Posts: 35240
Followers: 6618
Kudos [?]: 85287 [0], given: 10236
Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 04:00
Bumping for review and further discussion.
_________________
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 630
Followers: 77
Kudos [?]: 1051 [6] , given: 136
Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 04:39
6
KUDOS
8
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
Here is a neat little trick for such kind of problems:
Will be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly $$\frac{d}{2} units \to$$ The new Average = $$4-\frac{1}{2} = 3.5$$
Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
_________________
Intern
Joined: 26 May 2010
Posts: 10
Followers: 0
Kudos [?]: 33 [5] , given: 4
Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 24 Jun 2014
Posts: 51
Concentration: Social Entrepreneurship, Nonprofit
Followers: 0
Kudos [?]: 16 [0], given: 76
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
09 Mar 2015, 19:15
I considered following approach
if the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A
if largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55
Now as per statement 1 , average of first 9 numbers is 7 i.e sum =63
sum of 11 numbers =63+x+9+x+10----->B
Equating A& B
11X+55=63+X+9+10 ,which can be solved to get x=3
statement I is sufficient
similar approach for Statement II
11X-55=8+2X-19 ,can be solved to get X=13
statement 2 is sufficient
OA=D
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 7682
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Followers: 341
Kudos [?]: 2280 [1] , given: 162
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
09 Mar 2015, 19:26
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
Hi All,
When you look at this question, if you find yourself unsure of where to "start", it might help to break down everything that you know into small pieces:
1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place "in line", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?)
2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7.
What would have to happen for a group of consecutive integers to have an average of 7?
Here are some examples:
7
6, 7, 8
5, 6, 7, 8, 9
Notice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern.
With 9 total terms, that means there has to be 4 above and 4 below:
3, 4, 5, 6,.......7.......8, 9, 10, 11
Now we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT
With this same approach, we can deal with Fact 2.
The key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to "play around" with a question if you're immediately certain about how to handle it.
[Reveal] Spoiler:
D
GMAT assassins aren't born, they're made,
Rich
_________________
# Rich Cohen
Co-Founder & GMAT Assassin
# Special Offer: Save \$75 + GMAT Club Tests
60-point improvement guarantee
www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
10 Mar 2016, 03:12
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 11 Apr 2016
Posts: 3
Followers: 0
Kudos [?]: 0 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
29 Jun 2016, 03:10
wow such complex explanations for such a simple problem?
given :
11 consec integers
let them be x,x+1,x+2,...,x+10
Q: what is their mean?
mean is (11x+55)/11 = x+5.
Q becomes what is x+5
1) mean first 9 is 7.
so (9x+36)/9 = x+4 = 7 , so x+5 =8 ,--> sufficient A or D
2) mean of last 9 is 9.
so (9x+54)/9 = x+6= ---> x+5=8, sufficient . so D
D
Director
Joined: 04 Jun 2016
Posts: 656
GMAT 1: 750 Q49 V43
Followers: 40
Kudos [?]: 153 [0], given: 36
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
15 Jul 2016, 00:10
zz0vlb wrote:
What is the average (arithmetic mean) of eleven consecutive integers?
(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
For odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question
(1) The average of the first nine integers is 7
7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7
we will have {3,4,5,6,7,8,9,10,11}
now we can add last two consecutive integer after 11, they will be 12,13
our new set will become = {3,4,5,6,7,8,9,10,11,12,13}
again since the number of total elements in the set is odd, Mean will simply be the middle value = 8
SUFFICIENT
(2) The average of the last nine integers is 9
Again number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right
Middle value will be
{5,6,7,8,9,10,11,12,13}
Add 3,4 at the start of the set
new set = {3,4,5,6,7,8,9,10,11,12,13}
Mean will be 8
Sufficient
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016.
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] 15 Jul 2016, 00:10
Similar topics Replies Last post
Similar
Topics:
1 What is the average (arithmetic mean) of a sequence of N consecutive 9 12 Sep 2016, 04:47
What is the average (arithmetic mean) of fifteen consecutive integers? 2 26 May 2016, 11:39
7 What is the average of eleven consecutive integers? 10 15 Jan 2015, 07:25
3 What is the average (arithmetic mean) of eleven consecutive 3 25 Jan 2008, 11:44
What is the average (arithmetic mean) of eleven consecutive 3 24 Dec 2007, 00:16
Display posts from previous: Sort by | 2016-10-21T12:17:05 | {
"domain": "gmatclub.com",
"url": "http://gmatclub.com/forum/what-is-the-average-arithmetic-mean-of-eleven-consecutive-93188.html#p717074",
"openwebmath_score": 0.6749594211578369,
"openwebmath_perplexity": 2902.9624508058323,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 1,
"lm_q2_score": 0.8519527963298947,
"lm_q1q2_score": 0.8519527963298947
} |
http://matv.leforriclub.it/phase-portrait-nonlinear-system.html | ## Phase Portrait Nonlinear System
1) Find all equilibrium points by solving the system 2) Let a standard software (e. for the analysis of nonlinear systems; to introduce controller design methods for nonlinear systems. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Phase Portraits and Time Plots for Cases A (pplane6) Saddle Ex. 1), which was diagnosed using a set of four features extracted from the phase plane trajectory of the system to characterize the nonlinear response in the periodic regime. Local Phase Portrait of Nonlinear Systems Near Equilibria. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. Save the phase portraits to submit on Gradescope. Keywords: nonlinear dynamics, chaos, electrical circuits. Biological Models: Predator-prey models, Competition models, Survival of one species, Co-existence, Alligators, doomsday and extinction. • As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. A: The origins and evolution of predator-prey theory, Ecology 73, 1530-1535 (1992). Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Often, mathematical models of real-world phenomena are formulated in terms of systems of nonlinear differential equations, which can be difficult to solve explicitly. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. See phase portrait below. 1 In each problemfind the critical points and the corresponding linear system. Albu-Schaffer. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. Stable and unstable manifolds of equilibrium points and periodic orbits are important objects in phase portraits. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. Introduction to systems of differential equations 2. (reductor and multipliers). While nonlinear systems of-ten require complex idiosyncractic treatments, phase potraitshaveevolved as apowerfultool forglobal anal-ysis ofthem. The system lives in a state space or phase. Weak non-linear oscillators and. Existence, uniqueness, and strong topological consequences for two-dimensions. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. We also observe that the two xed points are progressively pushed apart in the amplitude direction. For a general 2 × 2 matrix A, the phase portrait will be equivalent to one of the four cases above, obtained by a linear transformation of coordinates (similarity transformation). Fixed points are stagnation points of the flow. The three examples will all be predator-prey models. 1 Concepts of Phase Plane Analysis 18 2. Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Click on the button corresponding to your preferred computer algebra system (CAS). 3 Symmetry in Phase Plane Portraits 22 2. Thus, a system has a limit cycle if and only if it has an isolated, closed. Numerical Construction of Phase Portraits. Two integral constraints on the amplitude and phase variation of the oscillations of an autonomous multi-degree of freedom system were obtained. 2 Draw the phase portraits of the following systems, using isoclines (a) 8+8+0. Generally, the nonlinear time series is analyzed by its phase space portrait. The book is very readable even though it has a lot of jargon (read heavy mathematics). A phase portrait is a graphical tool to visualize long term. Phase portrait generator. 5 0xy (7) which itself is a dynamical equation, the phase portrait is a trajectory along the switching line σ = 0. By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at 휇 = 0. shown to produce sharp phase portraits in long-term simulations, see e. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. On this page I explain how to use Matlab to draw phase portraits for the the two linear systems. This video deals with. Existence, uniqueness, and strong topological consequences for two-dimensions. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. The y nullcline is given by 3 4 1 4 3 y 2 3 x y = 0 (12) which gives the lines y = 0 or y = 3 4. , non-linear) 2 × 2 autonomous system discussed at the beginning of this chapter, in sections 1 and 2: x = f (x, y); (1) y = g(x, y). The nonlinear system's phase portrait near the fixed point is topologically unchanged due to small perturbations, and its dynamics are structurally stable or robust. Phase plane analysis for linear systems. 50= 1 (c) 8+82+0. m: A demonstration that plots the linearized phase portraits and the full phase plane. The nonlinear gyroscope model, which is employed in aerospace engineering [24], generally exhibits chaotic behavior. of problems that are described by nonlinear differential equations. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. EECS 222 Nonlinear Systems: Analysis, Stability and Control Shankar Sastry 299 Cory Hall Tu-Th 11-12:30 pm. The book is very readable even though it has a lot of jargon (read heavy mathematics). Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). Nonlinear Models and Nonlinear Phenomena. However, these behaviors are not properly depicted in phase portraits when dealing with sys-tems that could be described as rotating systems,. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. ) Lecture, three hours; discussion, one hour. A two-state phase portrait approach has been used to analyse vehicle dynamics and provides an illustrative view of the state trajectories at constant speed. Analyze the stability and its margins. by graphing and the use of phase portraits; D. First, let us look at the phase space portraits for a range of phase advances from 0:2 2ˇto 0:5 2ˇ. The phase portraits is able to perfectly capture all of the nonlinear trajectories and display them in a way that would be otherwise difficult. The phase portrait behavior of a system of ODEs can be determined by the eigenvalues or the trace and determinant (trace = λ 1 + λ 2, determinant = λ 1 x λ 2) of the system. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. The equation governing the dynamics of the nonlinear gyro, enriched with linear and nonlinear smoothening terms [24], is given by x_1 = x2. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. This paper extends the phase portrait to three states to represent the nonlinear vehicle dynamics with steering and longitudinal tyre force inputs and consideration of the longitudinal. (1) There is one equilibrium solution of this system – find it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the linearized system. This is a second order systemwhich is autonomous (time does not appear explicitly). from second-order equation to first-order system; what is a phase portrait; direction field of a first-order system; graphing in the xy- tx- and ty-planes; vector notation for a first-order system; semesters > spring 2020 > mth264 > resources > video > linear systems: basics Video | Linear Systems: Basics. Recall the basic setup for an autonomous system of two DEs: dx dt = f(x,y) dy dt = g(x,y) To sketch the phase plane of such a system, at each point (x0,y0)in the xy-plane, we draw a vector starting at (x0,y0) in the direction f(x0,y0)i+g(x0,y0)j. Damped Pendulum. Nonlinear Systems and Stability Autonomous systems and critical points Stability and phase plane analysis of almost linear systems Linearized stability analysis and plotting vector fields using a MSS Numerical solutions and phase portraits of nonlinear systems using a MSS Models and applications: TEXT: Text(s) typically used in this course. The classic Van der Pol nonlinear oscillator is provided as an example. 3 in Third and 3-42 Fourth Quadrant. We illustrate all these cases in the examples below. • Understand the linearized models using the “eigen-techniques” you learned earlier. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. The dynamical equation and the state equation of the system are established. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. These states can also be correlated with velocity spectral behaviors. A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. (Hint: Use polar coordinates. The low intensity xed point appears on the phase portraits. 26 Phase Portrait for sand Y1 Magnitudes of 9. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. 50= 1 (c) 8+82+0. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. 1 Phase portraits 72 3. warn(warning_msg, ODEintWarning). Is there a way for plotting phase portraits and vector fields for autonomous system of delay differential equations in. This video deals with. The dynamical equation and state equation of the system are established. Strogatz, Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering, Cambridge: Westview Press, 2000. Phase Portraits of Nonlinear Systems Consider a , possibly nonlinear, autonomous system , (autonomous means that the independent variable , thought of as representing time, does not occur on the right sides of the equations). Damped Pendulum. One- and two- dimensional flows. to sketch the phase portrait. Stable and unstable manifolds of equilibrium points and periodic orbits are important objects in phase portraits. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. I have a set of three differential equations and I want to make a phase portrait of them. For a chaotic system, there will be many distinct loops in a phase portrait, showing that the system is aperiodic and does not approach a stable. Quiver function is being used for phase portrait plots obtained using ode. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. - A ”limit cycle” is a periodic orbit that trajectories approach. 3 Determining Time from Phase Portraits 29 2. 3 Symmetry in Phase Plane Portraits 22 2. Derive the dynamics of a linear and nonlinear systems. Fig 1: 3-D phase portrait of the Rossler attractor Integration Solver The Runge-Kutta method for a system of ordinary differential equations is explained here. The dynamical variables of the system, in this case the angular position and velocity , are the coordinates defining the system's phase space. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. Sketching an accurate phase portrait for a non-linear system of DEs is time consuming but the series of 3 videos will help with shortening that time with added understanding. Phase Plane Portraits of Almost Linear Systems Interesting and complicated phase portraits often result from simple nonlinear perturbations of linear systems. 2 Bifurcation set and phase portraits of the Hamiltonian system (5). Ask Question Asked 4 years, 7 months ago. Solving 2x2 homogeneous linear systems of differential equations 3. According to Takens, almost all d-dimensional sub-manifolds could be embedded in a (m=2d+1) dimensional space. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. These programs provide animated phase portraits in dimension two and three, i. The tem-poral response of a system need not be the. 2 : Linear analysis of nonlinear pendulum : Mechanical systems model for a pendulum. (a) This plot shows the vector field for a planar dynamical system. , another nonlinear system x_1 = 1 x3 1 x_2 = x1 x22 equilibrium points are described by x1 = 1 and x2 = 1 note: the equilibrium points of a nonlinear system can be nite (2 in the previous examples, but any other number is possible, including zero) or in nite, and they can be isolated points in state space Oriolo: Stability Theory for. For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. Phase Plane Analysis 17 2. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360∘. Weak non-linear oscillators and. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. Ott, and A. This suggests that the only. , regularly timed speech with a metronome). By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at 휇 = 0. 6: Phase portraits on the (one-dimensional) centr emanifoldandthebifurcation diagram. First, we cover stability definitions of nonlinear dynamical systems, covering the difference between local and global stability. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. 1 (Saddle) Consider the system x˙ 1 = −x 1 −3x 2 x˙ 2 = 2x 2, x. in weakly-nonlinear systems was investigated. 1 Phase Portraits 18 2. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. It is not restricted to small or smooth nonlinearities and applies equally well to strong and hard nonlinearities. warn(warning_msg, ODEintWarning). Limit Cycles. (1) There is one equilibrium solution of this system – find it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the linearized system. The aim of this section is to present programs allowing to high- light the slow-fast evolution of the solutions of nonlinear and chaotic dynamical systems such as: Van der Pol, Chua and Lorenz models. x′= x−y, y′= x2 +y2 −1 2. Four possible phase portraits for this system are shown along the right side of the page. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360∘. doc Author: tien Created Date: 11/15/2002 4:16:10 AM. By varying the initial conditions of the system, it is found. The type of phase portrait of a homogeneous linear autonomous system -- a companion system for example -- depends on the matrix coefficients via the eigenvalues or equivalently via the trace and determinant. Phase portrait. In paper [12], the vibration of a typical two degree of freedom nonlinear system with repeated linearized natural frequencies was investigated systematically. We discovered the system’s rich behavior such as chaos through phase portraits, bifurcation diagrams, Lyapunov exponents, and entropy. The course revises some of the standard phase portrait methods encountered in the Dynamical Systems course in part II and extends these ideas, discussing in some detailed centres, via the use of Lyapunov functions, limit cycles and global phase portraits. 1 Concepts of Phase Plane Analysis 18 2. Keywords: nonlinear dynamics, chaos, electrical circuits. A phase portrait is mapped by homeomorphism, a continuous function with a continuous inverse. Linear approximation of autonomous systems 6. Let us consider the below block diagram of a non linear system, where G 1 (s) and G 2 (s) represent the linear element and N represent the non linear element. Existence, uniqueness, and strong topological consequences for two-dimensions. • As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. Damped Pendulum. Phase Plane Analysis 17 2. If the stable manifold is of higher dimension, then y 1 =h 1(x,µ ),y 2 =h 2(x,µ )and. There are lots of practical systems which can be approximated by second-order systems, and apply phase plane analysis. (e) Draw the phase plane portrait of the nonlinear system via v and h nullclines. The resonance effects are most pronounced where both the. 4 Phase Portraits and Bifurcations. A numerically generated phase-portrait of the non-linear system Zoomed in near (0,0) Zoomed in near (2,1) The critical point at (2,1) certainly looks like a spiral source, but (0,0) just looks bizarre. The vertical diametric phase distribution of the singly charged OV extracted from this phase portrait @Fig. Saturations constitute a severe restriction for stabilization of system. The long time dynamics are. All chapters conclude with Exercises. Chua}, year={1969} }. We describe the phase portrait for bang-bang extremals. • Understand the linearized models using the “eigen-techniques” you learned earlier. 2 Phase portraits • A phase portrait of an n-dimensional autonomous system x ′ (t) = f (x (t)) is a graphical rep-resentation of the states in x-space. Nonlinear. Plot the maximum amplitude in the frequency. We draw the vector field given at each point (x,y) by the vector. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. 3 Determining Time from Phase Portraits 29 2. General Calendar. In this research a new graphic. In this section we will give a brief introduction to the phase plane and phase portraits. The phase portrait can indicate the stability of the system. Also, this work showed that the extreme multi-stability phenomenon of the behaviour of infinitely many coexisting attractors depends on the initial conditions of the variables of the system. motion of the system. In physical systems subject to disturbances, the distance of a stable equilibrium point to the boundary of its stable manifold provides an estimate for the robustness of the equilibrium point. In these cases the use of the phase portrait does not properly depict the system’s evolution. 2 Global bifurcation analysis 69 3. When a double eigenvalue has only one linearly independent eigenvalue, the critical point is called an improper or degenerate node. The Poincar´e-Bendixson theorem Any orbit of a 2D continuous dynamical system which stays in a closed and bounded subset of the phase plane forever must either tend to a critical point or to a. , a particular state of the system) over time. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. 7 (2009), 369–403. Draw the phase line of the equations and Answer. , another nonlinear system x_1 = 1 x3 1 x_2 = x1 x22 equilibrium points are described by x1 = 1 and x2 = 1 note: the equilibrium points of a nonlinear system can be nite (2 in the previous examples, but any other number is possible, including zero) or in nite, and they can be isolated points in state space Oriolo: Stability Theory for. 3 Symmetry in Phase Plane Portraits 22 2. For each case, we construct a phase space portrait by plotting the values of the dynamical variables after repeated application of the map (equation (1), followed by (6) and (7)) for a range of initial conditions. Extensibility of solutions 50 §2. Equilibrium points. For optimal bang-bang trajectories with high values of the energy integral, a general upper bound on the number of switchings was obtained. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. doc Author: tien Created Date: 11/15/2002 4:16:10 AM. The phase space portraits of these two systems are shown in figure 6. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. Knowledge of λ1 and λ2, and v1 and v2, en-ables us to sketch the phase portrait near (x∗,y∗). Free system of non linear equations calculator - solve system of non linear equations step-by-step. • Understand the linearized models using the “eigen-techniques” you learned earlier. Consider a , possibly nonlinear, autonomous system ,(autonomous means that the independent variable , thoughtof as representing time, does not occur on the right sides of the equations). So, for a periodic system that obeys the law of energy conservation (e. Let's zoom into these four critical points, and look more closely at the phase portraits near them. If the system is period-n (the same state repeats after n clocks), there will be n points −→ period-n attractor. Students will learn nonlinear differential equations in the context of mathematical modeling. 26 Phase Portrait for sand Y1 Magnitudes of 9. Each set of initial conditions is represented by a different curve, or point. o Equilibrium solution • Exponential solutions o Half-line solutions • Unstable solution • Stable solution • Six important cases for portraits Real Eigenvalues o Saddle point o Nodal sink o Nodal source. An example of a non-linear system: the predator / prey model. The phase portrait of a dynamical system can be reconstructed from the observation of a single variable by the method of delays as proposed by [1]. The x-nullclineis a set of points in the phase plane so that. An example of a non-linear system: the predator / prey model. Phase Portraits of Nonlinear Systems. We draw the vector field given at each point (x,y) by the vector. Damped Pendulum. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. 2 Singular Points 20 2. warn(warning_msg, ODEintWarning). : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x’=x+4y, y’=2x−y −5 0 5 −5 0 5 x y Time Plots for ‘thick’ trajectory. Linear and Nonlinear Systems of Differential Equations. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. The x nullcline is given by (1 x y)x = 0 =) x = 0 or y = 1 x: (11) Sodx dt= 0 on the lines x = 0 and y = 1 x. Phase portraits and Hooke diagrams of the proposed driven nonlin-ear system are consistent with empirical observations. Second-Order Systems. Phase Portraits Now we turn to the third method of analyzing non-linear systems, phase portraits generated by numerical solutions. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. two-dimensional phase distributions for OV beams with charges m51 ~a! and m52 ~b!. Local Phase Portrait of Nonlinear Systems Near Equilibria. In fact, if we zoom in around this point, it would look like the case of a node of a linear system (in the sense of Chapter 7). The low intensity xed point appears on the phase portraits. Fig 1: 3-D phase portrait of the Rossler attractor Integration Solver The Runge-Kutta method for a system of ordinary differential equations is explained here. In particular, show that some of the equilibria correspond to nonlinear centers, by nding a rst integral for this system. Compare the phase portraits of the linear and the nonlinear maps near the origin. Q: Find the phase portrait of this second-order nonlinear system with such differential equation: $$\ddot{x}+0. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. Phase portraits and Hooke diagrams of the proposed driven nonlin-ear system are consistent with empirical observations. What is a Phase Portrait? Above, we have an animated phase portrait, but what is it? A phase portrait, in it’s simplest terms, is when we plot one state of the system against another state of the system. In paper [12], the vibration of a typical two degree of freedom nonlinear system with repeated linearized natural frequencies was investigated systematically. Thus, the equilibrium x = 0 is a saddle, hence unstable, when = 0. 11 (Nonlinear terms can change a star into a spiral) Here's another example. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. John Polking’s pplane: MATLAB, JAVA. The same concept can be used to obtain the phase portrait, which is a graphical description of the dynamics over the entire state space. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. Autonomous Planar Nonlinear Systems. Change parameters in the Linear-2D MAP so as to get the linearization at the origin. Following bifurcation in the system occurs in a range of parameter values g from 0. (c) The interior xed point at (1=3;1=3) is a global attractor. Changes in the dynamics of the orbits in the phase space usually represent variations of the physical parameters that control a non-linear system and consequently are of great importance for any modelling effort. This allowed to obtain exhaustive solution of the control problem comparing to the known results. systems) Suppose (x*,y*)=(0,0) is a linear center for a cont. 28 (1997), 755-778. 26 Phase Portrait for sand Y1 Magnitudes of 9. walking with Durus (right), showing phase portraits (top left) for 63 steps of walking together with a darker averaged phase portrait and position tracking errors (bottom left) over a select 4 steps in the same experiment. phase portrait get from simulink Example 2. Critical (equilibrium) points occur when (˙x,y˙) = (0,0). Key words: vibroimpact motion, unilateral and symmetrical rigid arrester, stereo-mechanical impact theory, phase portrait, two dimensional mapping. , is attracted to infinity. Students will learn nonlinear differential equations in the context of mathematical modeling. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. Solving 2x2 homogeneous linear systems of differential equations 3. Dynamical Systems and Chaos. 13 from the book. In this section we will give a brief introduction to the phase plane and phase portraits. Determine the stability of these limit cycles. Problem 4: Hamiltonian Systems. The phase portrait is a plot of a vector field which qualitatively shows how the solutions to these equations will go from a given starting point. MATLAB offers several plotting routines. By varying the parameters of the equation for the non-linear pendulum and then plotting. Phase Portraits of Nonhyperbolic Systems. Poincaré-Bendixon theorem. φ 1 = phase shift of the fundamental harmonic component of output. the allee due at noon on friday sept 14th, in the box provided (to the. 1) Find all equilibrium points by solving the system 2) Let a standard software (e. One- and two- dimensional flows. Phase portrait of system of nonlinear ODEs. Save the phase portraits to submit on Gradescope. 2 : Linear analysis of nonlinear pendulum : Mechanical systems model for a pendulum. (4) (Formerly numbered 135A. The fixed points can be classified according to their stability as follows: • If Re(λ1) > 0 and Re(λ2) > 0 ⇒ repeller (unstable node). In this section we will give a brief introduction to the phase plane and phase portraits. Neural Information Processing Systems (NIPS). Smith; Nonlinear Ordinary Differential Equations, 3rd Edition, Oxford University Press, 1999. We illustrate all these cases in the examples below. This page plots a system of differential equations of the form dx/dt = f(x,y), dy/dt = g(x,y). Change parameters in the Linear-2D MAP so as to get the linearization at the origin. Perko, Di erential Equations and Dynamical Systems (Second edi-tion, Springer, 1996). Phase portraits are an invaluable tool in studying dynamical systems. µ< 0 µ< 0 µ< 0 µ x Figure 5. Let us discuss the basic concept of describing the function of non linear control system. Q: Find the phase portrait of this second-order nonlinear system with such differential equation:$$ \ddot{x}+0. These are systems that do not depend explicitly on. Determining time from phase portraits. nonlinear transform of coordinates and uses a full nonlinear system’s model. Planar Almost Linear Systems: Phase portraits, Nonlinear classi- cations of equilibria. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. Fixed points occur at values θ∗ such that 0 = 1 + 2cosθ∗. which can be written in matrix form as X'=AX, where A is the coefficients matrix. 5 Global analysis for hardening nonlinear stiffness (c>0) 72 3. System analysis based on Lyapunov's direct method. Damped Pendulum. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. The figure shows the manner of convergence of these projected trajectories, which start with different initial conditions. In physical systems subject to disturbances, the distance of a stable equilibrium point to the boundary of its stable manifold provides an estimate for the robustness of the equilibrium point. Autonomous and non-autonomous systems Phase portraits and flows Attracting sets Concepts of stability 2. Lyapunov's direct method. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. We also show the formal method of how phase portraits are constructed. The undamped system has analytical solutions, (for both rotary and oscillatory motion), in terms of Jacobian Elliptic functions, that with the phase portrait gives a total qualitative and quantatitive explanation. These variables and their evolution in time produce the phase portrait of the system. Introduction to nonlinear network theory @inproceedings{Chua1969IntroductionTN, title={Introduction to nonlinear network theory}, author={L. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x'=x+4y, y'=2x−y −5 0 5 −5 0 5 x y Time Plots for 'thick' trajectory. It may be best to think of the system of equations as the single vector equation x y = f(x,y) g(x,y). The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. 2Switched Nonlinear Systems 1. We first focused on a nonlinear pendulum (shown in Fig. Phase Portrait for the linearization in Example 6. Phase portrait. Consider the nonlinear system (a) Show that the origin (O, O) is a nonlinear saddle awl plot the phase portrait, including the. 3 Determining Time from Phase Portraits 29 2. Generally, the nonlinear time series is analyzed by its phase space portrait. The motion of the mass is governed by Newton's second law. (iv) Since replacing x by x + 2… gives the same equations the portrait. Lyapunov's direct method. Phase Plane Analysis 17 2. Homework 6. Thompson and H. 1) "For nonlinear systems, there is typically no hope of finding the trajectories analytically. Fixed points, limit cycles, and stability analysis. Giorgio Bertotti, Claudio Serpico, in Nonlinear Magnetization Dynamics in Nanosystems, 2009. 58=0 (b) 8+8+0. Such a planar curve is called a trajectory of the system and its param-eter interval is some maximal interval of existence T 1 λ 0. Then try to combine the vector field with part (d) to get a global phase portrait of the original nonlinear system. I am unable to do for this case. 3 Phase Plane Portraits (for Planar Systems) Key Terms: • Equilibrium point of planer system. John Guckenheimer, in Handbook of Dynamical Systems, 2002. phase portrait (or phase diagram) for asystem depicts its phase space andtrajectories andis ageometricalrepresen- tation ofthe qualitative behavior ofthe system. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. These plots readily display vehicle stability properties and map equilibrium point locations and movement to changing parameters and system inputs. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto ries of a linear system x = ax +by a, b, c, d constants. • A PLL is a control system that generates an output signal whose phase is related to the phase of the input and the feedback signal of the local oscillator. Differential equations are used to map all sorts of physical phenomena, from chemical reactions, disease progression, motions of objects, electronic circuits, weather forecast, et cetera. Solve system of nonlinear equations - MATLAB fsolve Nl. served: here we analyze this interplay by investigating the system using statistical tools, phase portraits, Poincar e sections, and return maps. Numerical Construction of Phase Portraits. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. A phase portrait represents the directional behavior of a system of ODEs. 1) As in § 3. Solving 2x2 homogeneous linear systems of differential equations 3. So, for a periodic system that obeys the law of energy conservation (e. Since in most cases it is. 1 Concepts of Phase Plane Analysis 18 2. Phase portrait. Let A= 3 −4 6 −7. { Nonlinear spring-mass system { Soft and hard springs { Energy conservation { Phase plane and scenes. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. Its phase is ˇ=2, showing that the polariton resonance is below the laser line. Generally, the nonlinear time series is analyzed by its phase space portrait. Linear stability analysis may fail for a non-hyperbolic fixed point: ( Re(µ 1, 2) = 0, or at least one µ i = 0 ). The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. 1 we draw the phase portrait (or phase diagram), where each point (x,y) corresponds to a specific state of the system. Each set of initial conditions is represented by a different curve, or point. 2 Phase portraits • A phase portrait of an n-dimensional autonomous system x ′ (t) = f (x (t)) is a graphical rep-resentation of the states in x-space. By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at 휇 = 0. In this research a new graphic. Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). The phase portrait for the reduced dynamics for x is shown in Figure 5. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. Use technology to solve nonlinear programs, including computer programming and graphical analysis. These states can also be correlated with velocity spectral behaviors. (b) Find all bifurcation values of r and draw a bifurcation diagram on the rθ-plane. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. phase portrait (or phase diagram) for asystem depicts its phase space andtrajectories andis ageometricalrepresen- tation ofthe qualitative behavior ofthe system. Chaos of such a system was predicted by applying a machine learning approach based on a neural network. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. The procedure was applied in modeling of self-excitation oscillations for high-speed milling and is based on determination of non-linear self-excitation force and non-linear coefficient. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto ries of a linear system x = ax +by a, b, c, d constants. Consider the nonlinear system dx dt = r − x2, dy dt = x− y. 5 Summary of stability properties for planar ODE systems. One was the anticipated constancy of the system energy. By varying the initial conditions of the system, it is found. Design of feedback control systems. Overview of nonlinear mechanics of particles and nonlinear oscillations Lagrange’s equations and nonlinear di erential equations Flows on a line and bifurcations Multi-dimensional ows and linear systems Phase portraits, stability, and limit cycles Dissipative systems, reversible systems, Index theory Weakly nonlinear oscillations and two. 2 Phase Plane Analysis. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. 2 $\begingroup$ How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following? \begin{align} \dot{x} &= 2 - 8x^2-2y^2\\ \dot{y} &= 6xy\end{align}. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. HW # 7 Nonlinear Dynamics and Chaos Due: Monday, 95/01/30 1. SKETCH an approximate phase portrait for (6). Nonlinear Systems-lecture notes 4 Dr. 2 Global bifurcation analysis 69 3. As pointed out in @13#,a clear signature for the presence of a phase singularity is a new fringe starting at the location of the singularity. 3 Determining Time from Phase Portraits 29 2. The dynamics of airflow through the respiratory tract during VB and BB are investigated using the nonlinear time series and complexity analyses in terms of the phase portrait, fractal dimension, Hurst exponent, and sample entropy. In addition, if. The real space images in the top row show a portion of an unforced rotating spiral wave pattern [Fig. same'' means that type and stability for the nonlinear problem are the same as for the corresponding linear problem. d) Plot a computer-generated phase portrait to check your answer to (c). Weak non-linear oscillators and. 6: Phase portraits on the (one-dimensional) centr emanifoldandthebifurcation diagram. Effect of nonlinear terms. Now, if = 0, the system has one equilibrium point, x = 0. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. 2 Singular Points 20 2. 4 Phase Plane Analysis of Linear Systems 30. A simple example of a map is the Lotka-Volterra system describing two competing populations (e. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. Phase portrait. In this research a new graphic. MATLAB offers several plotting routines. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. µ< 0 µ< 0 µ< 0 µ x Figure 5. Day 9 - Two Dimensional Systems - Phase Planes Day 10 - Two Dimensional Systems - Eigenvalues and Eigenvectors Day 11 - Nonlinear Two Dimensional Systems - Jacobian Day 12 - More practice with Two Dimensional Nonlinear Systems Day 13 - Bifurcations in 2-D Systems - Limit Cycles Day 14 - Hopf Bifurcations, Lorenz Equations, Chaos and Fractals. Nonlinear systems - existence and uniqueness theorem, continuous dependence, variational equations. ends up in one of the fixed points at x n = (2n+1)π. freedom and analysis of phase portraits, i. The nonlinear gyroscope model, which is employed in aerospace engineering [24], generally exhibits chaotic behavior. As the initial angle increases, we can see that the shape of the non-linear phase trajectory approaches that of the seperatrix. Albu-Schaffer. the Rossler system is sensitive to the initial conditions, and two close initial states will diverge, with increasing number of iterations. Different initial states result in different trajectories. Here we will consider systems for which direction fields and phase portraits are of particular importance. Based on velocity phase portraits, each of the nonlinear response states can be categorized into one of the three states in the order of increasing chaotic levels: lock-in, transitional, or quasiperiodic. A simple example of a map is the Lotka-Volterra system describing two competing populations (e. Classification of phase portraits. Ott, and A. Reversible Systems (2) THEOREM (Nonlinear centers for rev. The simple pendulum is a great example of a second-order nonlinear system that can be easily visualized by the phase portrait. Autonomous and non-autonomous systems Phase portraits and flows Attracting sets Concepts of stability 2. In sum, we illustrate the revised system’s fit to the kinematics in both noncyclic speech and cyclic tasks (i. phase portrait get from simulink Example 2. The trace-determinant plane and stability. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. Learn more in: Chaotic Attractor in a Novel Time-Delayed System with a Saturation Function. Evolution of a dynamical system corresponds to a trajectory (or an orbit) in the phase space. John Polking’s pplane: MATLAB, JAVA. Conclude: any i. 5 \dot{x}+2 x+x^{2}=0 $$Method 1: Calculate by hands with phase plane analysis. What is a Phase Portrait? Above, we have an animated phase portrait, but what is it? A phase portrait, in it's simplest terms, is when we plot one state of the system against another state of the system. How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following?$$\begin{align} \dot{x} &= 2 - 8x^2-2y^2\\ \dot{y} &= 6xy\end{align}$$I can easily find the equilibria, which are$$\left\{ (0, \pm 1), \left(\pm \frac{1}{2}, 0\right) \right\} The corresponding stable subspace for $\left(\pm \frac{1}{2}, 0\right)$ is. Limit Cycles: Recall that analysis of linearized systems. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. bifurcation diagrams and phase portraits. Click on the button corresponding to your preferred computer algebra system (CAS). Consider the nonlinear system dx dt = r − x2, dy dt = x− y. Phase portrait generator. Basics : Introduction to the notion of dynamical systems, examples of non-linear systems, Discrete and Continuous time, from one to the other, Poincaré section. Instructors: Aldo Ferri: Topics: Introduction; properties of nonlinear systems; Phase portraits for second order systems; characterization of singular points and local stability; first and second methods of Lyapunov. Kitavtsev May 28, 2019 4 Local bifurcations of continuous and discrete dynamical systems The material of this chapter is covered in the following books: L. ) Lecture, three hours; discussion, one hour. A numerically generated phase-portrait of the non-linear system Zoomed in near (0,0) Zoomed in near (2,1) The critical point at (2,1) certainly looks like a spiral source, but (0,0) just looks bizarre. 4 Global analysis for softening nonlinear stiffness (c<0) 68 3. Materials to be covered include: nonlinear system characteristics, phase plane analysis, Lyapunov stability analysis, describing function method, nonlinear controller design. for x, where F ( x ) is a function that returns a vector value. (reductor and multipliers). Requisites: course 33B. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. As the initial angle increases, we can see that the shape of the non-linear phase trajectory approaches that of the seperatrix. This allowed to obtain exhaustive solution of the control problem comparing to the known results. Pauses are inserted between setting up the graphs; plotting the linear phase portrait for $$x = 2n\pi$$; adding this behavior to the full phase plane; plotting the linear phase portrait for $$x = (2n+1)\pi$$; adding that to the full phase. 01385 till 0. Click on the button corresponding to your preferred computer algebra system (CAS). 1 Phase Portraits 18 2. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Paragraphs 4. Around the origin there are periodic orbits corresponding to small oscillations of the pendulum that are called librations. population growth. First, we cover stability definitions of nonlinear dynamical systems, covering the difference between local and global stability. • λ 12==λ 0. Phase plane Analysis: 2nd order nonlinear systems, phase portrait graphical representation, singular points. Just as we did for linear systems, we want to look at the trajectories of the system. It starts in your web browser and you can directly input your equations and parameter values. The book is very readable even though it has a lot of jargon (read heavy mathematics). My professor told us to use a plotter to check our work (the hand-drawn phase portraits) but the one he linked to us won't work on my mac so I am trying to see the plots in Matlab but I don't know how to plot them and would be absolutely grateful for some help (I. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. which can be written in matrix form as X'=AX, where A is the coefficients matrix. The dynamics of airflow through the respiratory tract during VB and BB are investigated using the nonlinear time series and complexity analyses in terms of the phase portrait, fractal dimension, Hurst exponent, and sample entropy. One was the anticipated constancy of the system energy. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto ries of a linear system x = ax +by a, b, c, d constants. Then sufficiently close to (0,0) all trajectories are closed curves. By varying the initial conditions of the system, it is found. Following bifurcation in the system occurs in a range of parameter values g from 0. Pages 486 - 493 cover the five important cases. Here we will consider systems for which direction fields and phase portraits are of particular importance. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. 3 Symmetry in Phase Plane Portraits 22 2. The behaviour of the system is investigated through numerical simulations, by using well-known tools of nonlinear theory, such as phase portrait, bifurcation diagram and Lyapunov exponents. Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). Control of bilateral teleoperation systems with linear and nonlinear dynamics. The study of nonlinear dynamics is based on the study of phase portraits, wavelet and Fourier spectra, signals, chaotic phase synchronization, Lyapunov indicators. Keywords Piecewise nonlinear system, rolling mill, non-smooth homoclinic orbit, bifurcation, chaos. Around the origin there are periodic orbits corresponding to small oscillations of the pendulum that are called librations. Planar linear systems - eigenvalues and eigenvectors, phase portraits, classification. MATLAB offers several plotting routines. That is, only initial points located on this orbit result in this closed orbit. As before, we use a phase portrait for stability analysis. We will look at three examples, and also reexamine the undamped pendulum that we studied previously using only its vector field. The higher degree of chaoticity in BB relative to VB is unwrapped through the maximal Lyapunov exponent. Phase Plane Analysis 17 2. For optimal bang-bang trajectories with high values of the energy integral, a general upper bound on the number of switchings was obtained. Phase portraits via trace and determinant. In fact, if we zoom in around this point, it would look like the case of a node of a linear system (in the sense of Chapter 7). (561); Notes LS (power series excluded), GS; Handout on phase portraits. Complex eigenvalues, phase portraits, and energy 4. Although the Riccati equation is not generally a Morse–Smale vector field, we are able to show that it possesses suitable generalizations of many of the important properties of Morse–Smale vector fields. If a system is chaotic, there will be an infinite number of points in the phase portrait. Thus, the equilibrium x = 0 is a saddle, hence unstable, when = 0. (reductor and multipliers). Consider the non-linear system dx dt = y dy dt = 2x+(1x2 y )y. (a) Compute the eigenvalues of A. As a result of one more Andronov-Hopf bifurcation more complex quasiperiodic solution is formed in the system—it is torus of dimension three. 2 Constructing Phase Portraits 23 2. 01385 till 0. Animated phase portraits of nonlinear and chaotic dynamical systems allow one to shape globally the state- and time-dependent convergence behaviour ideally suited to the non-linear or time. This approach of linearizing, analyzing the linearizations, and piecing the results together is a standard approach for non-linear systems. : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x'=x+4y, y'=2x−y −5 0 5 −5 0 5 x y Time Plots for 'thick' trajectory. the trajectories of the nonlinear system are similar to those of the linearized system, so go round anticlockwise. One- and two- dimensional flows. 2 Phase Plane Analysis. [2] Consider x′ 1 = 5x1 −x2 1 − x1x2, x′ 2 = −2x2 +x1x2. • Be able to determine the phase plane and phase portraits of a 2 by 2 linear system. By varying the initial conditions of the system, it is found. 2~a!# is. 2 Bifurcation set and phase portraits of the Hamiltonian system (5). Several nonlinear wave solutions as the solitary wave solutions,topological solitons, cnoidal wave solutions, singular periodic waves and others were obtained. φ 1 = phase shift of the fundamental harmonic component of output. 4 Phase Plane Analysis of Linear Systems 30. REFERENCES [1] Berrymann, A. Phase portraits are an invaluable tool in studying…. Putting all this together we see that the phase portrait is as shown below. Assume that r > 0. In a planar system such as this, the nullclines can provide useful information about the phase portrait. The phase portrait is a plot of a vector field which qualitatively shows how the solutions to these equations will go from a given starting point. We show that our model can recover qualitative features of the phase portrait such as attractors, slow points, and bifurcations, while also producing reliable long-term future predictions in a variety of dynamical models and in real neural data. 2 Prey dynamics predicted by the Lotka-Volterra predator-prey model. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. Existence of Periodic Orbits. The Poincar´e-Bendixson theorem Any orbit of a 2D continuous dynamical system which stays in a closed and bounded subset of the phase plane forever must either tend to a critical point or to a. Linear stability analysis may fail for a non-hyperbolic fixed point: ( Re(µ 1, 2) = 0, or at least one µ i = 0 ). Damped Pendulum. 3 Symmetry in Phase Plane Portraits 22 2. Solving 2x2 homogeneous linear systems of differential equations 3. Problem: Construct and analyze a phase-plane portrait of a nonlinear system depicted in the following picture (desired value is w = 0), decide which of the equilibrium points are stable and which are not. Vehicle control synthesis using phase portraits of planar dynamics ABSTRACTPhase portraits provide control system designers strong graphical insight into nonlinear system dynamics. Let us discuss the basic concept of describing the function of non linear control system. 6 and the phase portrait for the original system is in Figure 5. System analysis based on Lyapunov's direct method. and sketch the phase portrait on the circle. Then sufficiently close to (0,0) all trajectories are closed curves. Determining time from phase portraits. Normalized phase portraits or cylindrical phase portraits have been extensively used to overcome the original phase portrait’s disadvantages. x c c c t ert yert y c c c t 1 2 2, 1 2 2 Case 3: Phase Portraits (5 of 5) The phase portrait is given in figure (a) along with several graphs of x1 versus t are given below in figure (b). ) Lecture, three hours; discussion, one hour. the behavior of the nonlinear system from various initial conditions. Jordan, Peter Smith, and P. 2 Phase portrait for an example system. phase portrait get from simulink Example 2. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. Students will learn nonlinear differential equations in the context of mathematical modeling. This allowed to obtain exhaustive solution of the control problem comparing to the known results. Unit 2: Nonlinear 2x2 systems. 1 Phase portraits 72 3. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. The low intensity xed point appears on the phase portraits. Consider the nonlinear system (a) Show that the origin (O, O) is a nonlinear saddle awl plot the phase portrait, including the.
7iw5loreijl ph2v1q1oh5ul19 5k7fv8hml9 7xnle9owrd873g3 zemot27ga77zfj 2719yh9klabxg s5ayt0gpsfksc 6vgg4tkpyb 1ziuxc88k8 ddchgfzhyl2klos qd1nhu4hxqyrf 5gsl71di0g8 2sbwr9xfsu foxd7lu32k wfpaeculwor zsku1acu2g huqrq9k4w2k7z33 3ynsy8bxwot28 cmljwia08r0doe njpxzvhz1r1bi qkwl1pfv5l5uwwp fjcx8u9t50uv0r irhvlt0kroa 3g5hypzcrk d6jq3elv7llo 5de7be0jgurut8i exw9yp069m2 zv0in78g3fj k50lh4m3ektdbz adx9cpa9kard zoid1zghpq oba3px70bc mrfddo0ps2oh5 z7gwnvcbaa57r2f ytq21pbsd171 | 2020-11-27T11:35:06 | {
"domain": "leforriclub.it",
"url": "http://matv.leforriclub.it/phase-portrait-nonlinear-system.html",
"openwebmath_score": 0.5916582942008972,
"openwebmath_perplexity": 791.4518817998761,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9910145704512985,
"lm_q2_score": 0.8596637523076224,
"lm_q1q2_score": 0.8519393042256899
} |
http://math.stackexchange.com/questions/721885/two-circles-inside-a-right-angle?answertab=votes | Two Circles inside a right angle!
The other day I was playing with Ms Paint drawing circles here and there -I coincidentally drew a circle inside a right angled triangle which I already drew . Strangely A problem struck to my mind and I tried solving it , but I was unable to do so . I put forward the statement of the problem which I managed to frame my self
Problem : The legs of a right angled triangle are of length $a$ and $b$ . Two circles with equal radii are drawn such such that they touch each other and sides of the triangle as shown in the figure . Find the radius of the circle in terms of $a$ and $b$.
Figure - Of course my MS Paint one -
Further Scope- Is there any way to generalize this for other shapes or for any other triangle?
-------EDIT---------------------
Now to make things interesting : Say we have a right angled triangle which is given . Then is there a method by which we can construct those two circles with a straightedge and a compass?
-
Perhaps this topic may be useful? mathworld.wolfram.com/Excircles.html – Erik Miehling Mar 22 at 6:25
@ErikMiehling thanks but it isn't helping much! – Shivam Patel Mar 22 at 6:39
@ShivamPatel Can you give an example of a generalization? I don't really know "right pentagons" and "right hexagons". – DanielV Mar 22 at 6:51
Or perhaps a generalization to multidimensional right triangles and hyperspheres would be meaningful? – DanielV Mar 22 at 7:03
Straightedge and compass construction: draw an arbitrary circle with center $C$ that intersects the legs $AC, BC$ at $D, E$, respectively. Locate $F$ on $AC$ such that $CF = 3CA$. Locate $G$ such that $ECFG$ is a rectangle. Draw $CG$. Draw the angle bisector of $A$. The intersection of this angle bisector with $CG$ is the center of one of the two circles. – heropup Mar 23 at 3:44
Writing $a := |BC|$, $b := |CA|$, $c := |AB| = \sqrt{a^2+b^2}$, and $r = |PE| = |PF|$ (so that $|PD| = 3r$), we have \begin{align} |\triangle ABC| &= |\triangle ABP| + |\triangle BCP| + |\triangle CAP| \\[4pt] \implies \qquad \frac{1}{2} |BC||CA| &= \frac{1}{2} \left(\; |AB| |PF| + |BC||PD| + |CA||PE| \;\right) \\[4pt] \implies \qquad a b &= c r + 3 a r + b r = r ( 3 a + b + c )\\[6pt] \implies \qquad r &= \frac{ab}{3 a + b + c} = \frac{ab}{3 a + b + \sqrt{a^2+b^2}} \end{align}
To address @DanielV's suggestion of generalizing to higher dimensions, consider a right-corner tetrahedron $OABC$, with right corner at $O$ and edge lengths $a := |OA|$, $b := |OB|$, $c := |OC|$. (Note that I'm changing notation slightly from the above.) Let a sphere with center $P$ and radius $r$ be tangent to the faces around vertex $A$, and let a congruent sphere (tangent to the first) be tangent to the faces around vertex $O$. Then $P$ has distance $r$ from faces $\triangle OAB$, $\triangle OCA$, $\triangle ABC$ (the ones touching $A$), and distance $3r$ from face $\triangle OBC$ (the one opposite $A$).
Here's a poor attempt at a diagram:
(In this case, the altitudes from $P$ are color-coded to match their parallel counterparts through $O$. The black altitude is to face $\triangle ABC$.)
Thus,
\begin{align} |OABC| &= |OABP| + |OBCP| + |OCAP| + |ABCP| \\[4pt] \implies \qquad \frac{1}{6}a b c &= \frac{1}{3}\left(\; r\;|\triangle OAB| + r \;|\triangle OCA| + r\;|\triangle ABC| + 3r\;|\triangle OBC| \;\right) \\[4pt] &= \frac{1}{3}r \cdot \frac{1}{2} \left(\; a b + c a + 3 b c + 2\;|\triangle ABC| \;\right) \\[6pt] \implies \qquad r &= \frac{abc}{3bc + ab + ca + 2\;|\triangle ABC|} \qquad (\star) \end{align}
Fun fact: The Pythagorean Theorem for Right-Corner Tetrahedra says that $$|\triangle ABC|^2 = |\triangle OBC|^2 + |\triangle OCA|^2 + |\triangle OAB|^2$$ so that we have $$|\triangle ABC| = \frac{1}{2} \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}$$ and $(\star)$ becomes $$r = \frac{abc}{3bc + ab + ca + \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}}$$
In $4$-dimensional space (where there's an analogous Pythagorean Theorem, as there is in any-dimensional space), we have $$r = \frac{abcd}{3bcd + acd + abd + abc + \sqrt{\;b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2\;}}$$ and so forth.
Incidentally, the matching-notation version of the initial answer is $$r = \frac{ab}{3b + a + \sqrt{\;b^2 + a^2\;}}$$
-
@qwr: Thanks. :) As for coloring: I just like to highlight the association of a vertex with its opposite edge in a triangle, and other elements get their colors accordingly. I chose not to color-code the altitudes dropped from $P$ to the edges (I had reflexively done so in a "draft" image), in order to tie the congruent segments together visually. – Blue Mar 22 at 8:00
@mathh: I use GeoGebra for my figures these days. – Blue Mar 22 at 12:02
@AJP: "$|BC|$" represents the length of segment $\overline{BC}$. (I often leave out the over-bar to save typing.) By analogy, I use "$|\triangle ABC|$" for the area of $\triangle ABC$ (and then "$|OABC|$" for the volume of tetrahedron $OABC$, etc); this is non-standard, but I like it. :) The symbol ":=" indicates "is defined to be"; writing "$a:=|BC|$" says "I'm going to write '$a$' for '$|BC|$'". Others (even I) might otherwise write "let $a = |BC|$", but using ":=" is ever-so-slightly better, as it distinguishes definition "'$a$' means '$|BC|$'" from relation "$a$ equals $|BC|$". – Blue Mar 22 at 13:10
@ShivamPatel: I added a figure that may (or may not) help with the $3d$ case. Just don't ask me for the $4d$ version. :) – Blue Mar 22 at 14:46
@mathh: In GeoGebra, you can easily mark an angle with an arc via the "Angle" tool; if the sides of the angle happen to be perpendicular, GeoGebra turns the arc into a box. (There's a setting somewhere to turn that behavior off and on.) The little dashes are somewhat less convenient: with the segment selected, you have to open the "Object Properties..." panel and then the "Decoration" tab. (You can likewise "decorate" angles with little dashes and such.) – Blue Mar 22 at 15:28
Assuming the corner of the triangle is the origin of a Cartesian plane, the line of the hypotenuse is $$y = -\frac{a}{b} x + a$$
The center of the second circle is at $\begin{bmatrix} 3r \\ r \end{bmatrix}$.
The radius of the second circle is the directed vector $\begin{bmatrix} ra \\ rb \end{bmatrix}\frac{1}{\sqrt{a^2 + b^2}}$.
Altogether :
$$\frac{rb}{\sqrt{a^2 + b^2}} + r = -\frac{a}{b}\left(3r + \frac{ra}{\sqrt{a^2 + b^2}}\right) + a$$
$$r = \frac{ab}{ 3a + b + \sqrt{a^2 + b^2}}$$
Since Blue showed an elegant volume based approach, I guess I'll try a multidimensional coordinate based approach. Suppose the origin $\begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \end{bmatrix}$ and the points $\begin{bmatrix} \mathcal{l}_0\\ 0 \\ 0 \\ \vdots \end{bmatrix}$, $\begin{bmatrix} 0 \\ \mathcal{l}_1 \\ 0 \\ \vdots \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ \mathcal{l}_2 \\\vdots \end{bmatrix}$ etc form a multidimensional right triangle. Suppose one hypersphere is tucked in at the origin and one hypersphere is tucked in at the corner along the first axis.
The hypotenuse plane of the triangle is $$\frac{x_0}{\mathcal{l}_0} + \frac{x_1}{\mathcal{l}_1} + \frac{x_2}{\mathcal{l}_2} + \dots = 1 \tag{Plane equation}$$
Or equivalently, using $\circ$ for dot product and $N = \begin{bmatrix} \frac{1}{\mathcal{l}_0} \\ \frac{1}{\mathcal{l}_1} \\ \frac{1}{\mathcal{l}_2} \\ \vdots \end{bmatrix}$, then the plane is: $$N \circ X = 1 \tag{Plane equation with dot product}$$
The center of the second hypershpere is $$c = \begin{bmatrix} 3r \\ r \\ r \\ \vdots\end{bmatrix} = rc_0 \tag{Center of second hypersphere}$$
The vector directed from the center of the second hypersphere to the hypotenuse plane is $$r_2 = r\frac{N}{|N|}\tag{Directed radius to hypotenuse}$$
Altogether: $$N \circ (c + r_2) = 1$$ $$N \circ \left(rc_0 + r\frac N {|N|}\right) = 1$$ \begin{align} r &= \frac{1}{N \circ c_0 + |N|} \\ &= \frac{1}{3l_0^{-1} + l_1^{-1} + l_2^{-1} \dots + \sqrt{l_0^{-2} + l_1^{-2} + l_2^{-2} \dots}} \end{align}
Note this is the same answer as Blue, but with $\mathcal{l}_0\mathcal{l}_1 \mathcal{l}_2 \dots$ factored out of the numerator and denominator.
-
+1. It's worth nothing that your multi-dimensional argument effectively re-derives the formula for the distance from point $P(p_0,p_1,\dots,p_n)$ to hyperplane $Q:\;q_0x_0+q_1x_1+\cdots+q_nx_n=q$; namely, $$\text{dist}(P,Q)=\frac{|\;p_0q_0+p_1q_1+\cdots+p_nq_n-q\;|}{\sqrt{\;q_0^2+ q_1^2 +\cdots+q_n^2\;}}$$ Replacing $p_0=3r$, $p_1=\cdots=p_n=r$, $q_i=l^{-1}_{i}$, and $q=1$, and then setting the computed distance equal to $r$, gives the equation you've solved for $r$. (Note: Dropping the formula's absolute value sign makes the distance negative here, since $P$ is on the "origin-side" of $Q$.) – Blue Mar 22 at 22:29
Interesting how many different ways there can be to reach the same result. This is one reason I say that instead of memorizing formulas, students would appreciate mathematics more if they saw the "magic" of how many seemingly unrelated approaches give the same result in the end. – DanielV Mar 22 at 23:38
@DanielV Can you look at the edit please... – Shivam Patel Mar 23 at 3:24
@ShivamPatel Look at youtube.com/watch?v=CMP9a2J4Bqw espcially after 00:58 where it describes what is possible with a compass and straightedge. In short the answer is "yes", by applying what you see in the video to the formulas Blue and I gave you. ...But I doubt there is any short way to do it. Keep in mind that you effectively assume $a=1$ or $b=1$ since the units are arbitrary. – DanielV Mar 23 at 3:37
Alternative solution:
Recall that for any triangle $\triangle ABC$, the area of the triangle equals the product of the inradius and its semiperimeter; i.e., $|\triangle ABC| = rs$, where $s = (a+b+c)/2$. Therefore, given legs $a, b$, $c = \sqrt{a^2+b^2}$, and $$r = \frac{2|\triangle ABC|}{a+b+c} = \frac{ab}{a+b+\sqrt{a^2+b^2}}.$$ Draw the tangent line to the two circles at their common point of tangency: this creates a smaller similar triangle with scaling factor $\frac{b-2\rho}{b}$ where $\rho$ is the common radius of the two circles. If $r$ is the inradius of $|\triangle ABC|$ as shown, then $$\frac{\rho}{r} = \frac{b - 2\rho}{b}.$$ Putting all of this together, we find $$\rho = \frac{br}{b+2r} = \frac{a b}{3a+b+c} = \frac{ab}{3a+b+\sqrt{a^2+b^2}}.$$ This method easily generalizes to more than two congruent circles tangent to one side: if $n$ circles are arranged along the leg of length $b$, then it is straightforward to find that $\rho = \frac{ab}{(2n-1)a + b + \sqrt{a^2+b^2}}$.
-
What about determining $a,b$ from $r$?
You can't - $a,b$ are not uniquely determined by $r$. Draw the two circles first, then draw the legs a and b as infinite lines. And line tangent to the right most circle will intersect lines A and B creating a right triangle, but for different tangents $a$ and $b$ will be different and r hasn't changed.
You can't even turn this into an interesting question by asking for all the possible ways to do it, because it turns out that these circles don't actually impose any restriction - since any angels can be achieved by the tangent line, any right triangle can. All $r$ does is create a scaling factor. $a$ can take on any value in $(2r,\infty)$ and $b$ the corresponding value in $(4r,\infty)$
suppose we say $a=kb$ for some fixed $k$ then? – Shivam Patel Mar 22 at 6:46
I believe the question was to find the radius, given $a$ and $b$. – N. Owad Mar 22 at 6:48 | 2014-11-29T07:31:15 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/721885/two-circles-inside-a-right-angle?answertab=votes",
"openwebmath_score": 0.9760359525680542,
"openwebmath_perplexity": 620.1165665347083,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9910145715128204,
"lm_q2_score": 0.8596637433190939,
"lm_q1q2_score": 0.851939296230479
} |
https://math.stackexchange.com/questions/2360603/how-to-find-the-intersection-of-two-hyperplanes-in-n-dimensions/2369012 | # How to find the intersection of two hyperplanes in $n$ dimensions?
I want to find out the intersection of two surfaces that exist in $n$-dimensions. These surfaces are defined by a system of $2$ linear equations.
$$A_1x+B_1y+C_1z+\cdots = D_1$$
$$A_2x+B_2y+C_2z+\cdots = D_2$$
But first, how would these surfaces look like? In $3$-D space, they would be planes, but in dimensions higher than $3$ are these not planes? Can we generalize a way to find the intersection of such linear expressions in some way?
Also can this problem be solved using some tools, preferably Matlab?
• The surfaces are hyperplanes. Solve a linear system of the form $\rm A x = b$, where $\rm A$ is a $2 \times n$ matrix. If $n > 2$, the linear system is underdetermined. If $n=3$, the intersection could be a line. If $n=4$, the intersection could be a $2$-dimensional plane. – Rodrigo de Azevedo Jul 16 '17 at 15:27
• You can use Gaussian elimination to find a parametrization of the affine space that is the intersection, assuming that the intersection is not empty. – Rodrigo de Azevedo Jul 16 '17 at 15:31
• @RodrigodeAzevedo And how do I figure out t1 and t2? – Vidor Vistrom Jul 16 '17 at 15:44
• Suppose we have $$\begin{bmatrix} 1 & 0 & -1 & -1\\ 0 & 1 & -1 & -1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} = \begin{bmatrix} 1\\ 1\end{bmatrix}$$ The solution set is the plane parameterized as follows $$\left\{ \begin{bmatrix} 1\\ 1\\ 0\\ 0\end{bmatrix} + t_1 \begin{bmatrix} 1\\ 1\\ 1\\ 0\end{bmatrix} + t_2 \begin{bmatrix} 1\\ 1\\ 0\\ 1\end{bmatrix} : t_1, t_2 \in \mathbb R \right\}$$ – Rodrigo de Azevedo Jul 16 '17 at 15:48
• @RodrigodeAzevedo: I suggest that you post your solution, perhaps mentioning the reduced row-echelon form of the augmented matrix for the $2\times n$ system, as an Answer. I will certainly upvote even if you mostly transcribe your Comments into the Answer box. – hardmath Jul 16 '17 at 21:49
Suppose we would like to find the intersection of $2$ hyperplanes in $\mathbb R^n$
$$\begin{array}{rl} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n &= b_1\\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n &= b_2\end{array}$$
In matrix form,
$$\begin{bmatrix} — \mathrm a_1^\top —\\ — \mathrm a_2^\top —\end{bmatrix} \mathrm x = \begin{bmatrix} b_1\\ b_2\end{bmatrix}$$
or, more economically, $\rm A x = b$, where $\rm A$ is a $2 \times n$ matrix. Without loss of generality, we assume that both hyperplanes are in Hessian normal form, i.e., $\| \mathrm a_1 \|_2 = \| \mathrm a_2 \|_2 = 1$.
If vectors $\rm a_1$ and $\rm a_2$ are collinear, then the $2$ given hyperplanes are either parallel or overlapping, neither of which are very interesting. Thus, we assume that vectors $\rm a_1$ and $\rm a_2$ are not collinear, i.e., matrix $\rm A$ has full row rank. Let $\theta := \arccos ( \mathrm a_1^\top \mathrm a_2 )$ be the angle formed by vectors $\rm a_1$ and $\rm a_2$.
The solution set of the linear system $\rm A x = b$ is a $(n-2)$-dimensional affine space. To find this affine space, we must find a particular solution and the null space of $\rm A$. One particular solution would be the least-norm solution, i.e., the one closest to the origin
$$\begin{array}{rl} \mathrm x_{\text{LN}} := \mathrm A^\top \left( \mathrm A \mathrm A^\top \right)^{-1} \mathrm b &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} \| \mathrm a_1 \|_2^2 & \langle \mathrm a_1, \mathrm a_2 \rangle\\ \langle \mathrm a_2, \mathrm a_1 \rangle & \| \mathrm a_2 \|_2^2\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & \cos (\theta)\\ \cos (\theta) & 1\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \frac{1}{\sin^2 (\theta)} \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & -\cos (\theta)\\ -\cos (\theta) & 1\end{bmatrix} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \left(\frac{b_1 - b_2 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_1 \\ | \end{bmatrix} + \left(\frac{b_2 - b_1 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_2 \\ | \end{bmatrix}\end{array}$$
where $\sin (\theta) \neq 0$ is ensured by the non-collinearity of vectors $\rm a_1$ and $\rm a_2$.
To find the $(n-2)$-dimensional null space of $\rm A$, we solve the homogeneous linear system $\rm A x = 0_2$. Introducing an $n \times n$ permutation matrix $\rm P$ with certain desired properties, we reorder the columns of $\rm A$, i.e., $\rm A P P^\top x = 0_2$, and obtain a homogeneous linear system in $\rm y:= P^\top x$
$$\rm A P y = 0_2$$
If $\rm P$ is chosen wisely, then there is a $2 \times 2$ matrix $\rm E$, a product of elimination matrices, that puts $\rm A P$ in reduced row echelon form (RREF), i.e.,
$$\rm E A P = \begin{bmatrix} \mathrm I_2 & \mathrm F\end{bmatrix}$$
where $\rm F$ is a $2 \times (n-2)$ matrix. Hence, the null space of $\rm A P$ is given by
$$\left\{ \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta : \eta \in \mathbb R^{n-2} \right\}$$
and, thus, the null space of $\rm A$ is given by
$$\left\{ \mathrm P \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta : \eta \in \mathbb R^{n-2} \right\}$$
Lastly, the intersection of the $2$ given hyperplanes is given by
$$\left\{ \color{blue}{\mathrm x_{\text{LN}} + \mathrm P \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta} : \eta \in \mathbb R^{n-2} \right\}$$
In MATLAB, we can use function rref to find permutation matrix $\rm P$ and matrix $\rm F$. Of course, we can also use function null to find an orthonormal basis for the null space of $\rm A$. | 2019-06-27T06:50:57 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2360603/how-to-find-the-intersection-of-two-hyperplanes-in-n-dimensions/2369012",
"openwebmath_score": 0.9251582622528076,
"openwebmath_perplexity": 154.00284689357713,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9932024684496267,
"lm_q2_score": 0.8577681031721325,
"lm_q1q2_score": 0.8519373974279161
} |
http://mathhelpforum.com/calculus/3132-need-help-definite-integral-please.html | 1. ## Need help with a definite integral please
I would appreciate advice on solving the following problem.
Evaluate the definite integral:
integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx
Here is my own attempt.
Step 1: Solve for the generic (indefinite) integral via substitution
Let u = x^2 + 1
So then
du / dx = 2x
2x dx = du
dx = du / 2x
Returning to the original function:
integral 2x (x^2 + 1) dx
= integral 2x (u) dx
= integral 2x (u) (du / 2x)
= integral u du
= integral x^2 + 1
= (x^3 / 3 ) + x + C
Step 2: solve for the definite integral given the stated limits
At upper limit of 2:
((2)^3 / 3) + (2) +C
= (8/3) +2 + C
= (8/3) + (6/3) + C
= 14/3 + C
At lower limit of 1
((1)^3 / 3) + (1) +C
=(1/3) +1 + C
= (1/3) + (3/3) + C
= 4/3 + C
Subtracting the lower limit result from upper limit result:
(14 /3 + C) - (4/3 + C)
=14/3 + C - 4/3 - C
= 10 / 3
But my textbook and calculator both say the answer is (21/2)
What am I doing wrong?
2. Originally Posted by lingyai
I would appreciate advice on solving the following problem.
Evaluate the definite integral:
integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx
Here is my own attempt.
Step 1: Solve for the generic (indefinite) integral via substitution
Let u = x^2 + 1
So then
du / dx = 2x
2x dx = du
dx = du / 2x
Returning to the original function:
integral 2x (x^2 + 1) dx
= integral 2x (u) dx
= integral 2x (u) (du / 2x)
= integral u du
= integral x^2 + 1
= (x^3 / 3 ) + x + C
Step 2: solve for the definite integral given the stated limits
At upper limit of 2:
((2)^3 / 3) + (2) +C
= (8/3) +2 + C
= (8/3) + (6/3) + C
= 14/3 + C
At lower limit of 1
((1)^3 / 3) + (1) +C
=(1/3) +1 + C
= (1/3) + (3/3) + C
= 4/3 + C
Subtracting the lower limit result from upper limit result:
(14 /3 + C) - (4/3 + C)
=14/3 + C - 4/3 - C
= 10 / 3
But my textbook and calculator both say the answer is (21/2)
What am I doing wrong?
Evaluate:
$\int_1^2 2x(x^2+1)dx$
Ssubstitute $u=x^2+1$ , $dx=\frac{1}{2x}du$:
$\int_{x=1}^2 2x(x^2+1)dx=\int_{u=2}^5 u\ du=\left[u^2/2\right]_{u=2}^5$ $=\frac{25}{2}-\frac{4}{2}=\frac{21}{2}$
RonL
3. Originally Posted by lingyai
I would appreciate advice on solving the following problem.
Evaluate the definite integral:
integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx
Here is my own attempt.
Step 1: Solve for the generic (indefinite) integral via substitution
Let u = x^2 + 1
So then
du / dx = 2x
2x dx = du
dx = du / 2x
Returning to the original function:
integral 2x (x^2 + 1) dx
= integral 2x (u) dx
= integral 2x (u) (du / 2x)
= integral u du
= integral x^2 + 1
= (x^3 / 3 ) + x + C
Step 2: solve for the definite integral given the stated limits
At upper limit of 2:
((2)^3 / 3) + (2) +C
= (8/3) +2 + C
= (8/3) + (6/3) + C
= 14/3 + C
At lower limit of 1
((1)^3 / 3) + (1) +C
=(1/3) +1 + C
= (1/3) + (3/3) + C
= 4/3 + C
Subtracting the lower limit result from upper limit result:
(14 /3 + C) - (4/3 + C)
=14/3 + C - 4/3 - C
= 10 / 3
But my textbook and calculator both say the answer is (21/2)
What am I doing wrong?
What you did wrong?
That there after Integral u du.
You continued with Integral x^2 +1.
That is wrong.
You were already in the substitution and then you went back to the original, without integrating yet. You should have continued the integration usin the substitution. That is why you chose to use substitution in the first place.
Then in your going back to the original, from INT. u du, the switchback is wrong also.
What is u?
It is x^2 +1
What is du?
It is 2x dx.
INT. u du = INT. x^2 +1
only?
It should have been
INT. u du = INT. (x^2 +1) *(2x dx)
which is the same as the original integrand.
Which would have put you back to square one or the beginning again.
4. Hello, lingyai!
Thank you for showing your work and reasoning . . .
Evaluate: $\:\int^2_1 2x(x^2 + 1)dx$
Did you (or anyone) notice that the integrand is a polynomial?
$\int^2_12x(x^2+1)dx \;= \;\int^2_1(2x^3 + 2x)\,dx \;= \;\frac{1}{2}x^4 + x^2\bigg|^2_1$
. . $= \;\left(\frac{1}{2}\cdot2^4 + 2^2\right) - \left(\frac{1}{2}\cdot1^4 + 1^2\right) \;= \;(8 + 4) - \frac{3}{2}\;=\;\frac{21}{2}$
5. Originally Posted by Soroban
Hello, lingyai!
Thank you for showing your work and reasoning . . .
Did you (or anyone) notice that the integrand is a polynomial?
$\int^2_12x(x^2+1)dx \;= \;\int^2_1(2x^3 + 2x)\,dx \;= \;\frac{1}{2}x^4 + x^2\bigg|^2_1$
. . $= \;\left(\frac{1}{2}\cdot2^4 + 2^2\right) - \left(\frac{1}{2}\cdot1^4 + 1^2\right) \;= \;(8 + 4) - \frac{3}{2}\;=\;\frac{21}{2}$
I was too busy noticing that:
$
\:\int^2_1 2x(x^2 + 1)dx=\:\int^2_1 \frac{1}{2}\frac{d}{dx}(x^2 + 1)^2dx
$
$
=\frac{1}{2}(x^2 + 1)^2 \bigg|_{x=1}^2=\frac{25-4}{2}=\frac{21}{2}$
but that would not help the OP spotting what they had done wrong
RonL | 2016-08-31T15:37:08 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/3132-need-help-definite-integral-please.html",
"openwebmath_score": 0.9143547415733337,
"openwebmath_perplexity": 3507.7443687220016,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9362850128595115,
"lm_q2_score": 0.9099069974872589,
"lm_q1q2_score": 0.8519322848433177
} |
https://tex.stackexchange.com/questions/400065/align-horizontal-spacing-is-inconsistent | # Align*: horizontal spacing is inconsistent
I am new to Latex, and I am having some trouble with the align* environment. In the code below, the first two equations are horizontally spaced differently from the second two equations:
\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\dfrac{1}{n}-\dfrac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1, \inf(A)=-1$.\\\\
\textit{Proof}. Suppose for contradiction that $1$ is not an upper bound for $A$, i.e. $\exists n,m \in \mathbb{N} \mid \dfrac{1}{n} - \dfrac{1}{m} > 1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\text{Because }n(m+1)\geq m+1,\text{ we have}\\
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot\\
\end{align*}
$\newline$Suppose for contradiction that $-1$ is not a lower bound for $A$, i.e. $\ex n,m \in \N \mid$\\$\dfrac{1}{n} - \dfrac{1}{m} < -1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\text{Because }m(1+n)\geq 1+n, \text{ we have}\\
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\end{align*}
$\newline\newline$ Claim that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$, or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp$\\$\left|(1 - \dfrac{1}{m}) - 1\right| < \epsilon$.
\begin{align*}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\end{align*}
Hence, for $M:=\left\lceil\dfrac{1}{\epsilon}\right\rceil+1$, $m > M \imp \left| 1 - 1 - \dfrac{1}{m}\right|<\epsilon$. Because $\epsilon$ was arbitrarily chosen, it follows that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$.
$\newline\newline$ Claim that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$, or $\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp$\\$\left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$.
\begin{align*}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \dfrac{1}{\epsilon}\right\rceil+1, n > N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$. Because $\epsilon$ was arbitrarily chosen, it follows that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$.
\end{document}
I'm not exactly sure why this is happening. Perhaps it has something to with the \frac{}{} equations in the second two equations... and if so, how can I fix the spacing so that all the equations are aligned the same? Thank you.
• What do you consider the first and the second equation in your code? It's not very clear... Nov 7 '17 at 4:11
• Welcome to TeX.SE. If your issue is with alignment across separate align*, you need to combine them into a single align* enironment and use \intertext{} for the text that you have between the two align* blocks. Nov 7 '17 at 6:07
• Well, the space between the lines are probably the same, but as you say the lines are higher in the second two align blocks due to the \fracs. But do you really want to change that? You don't need more space between the lines for the first two. (You can of course do it, see e.g. tex.stackexchange.com/questions/2929/…, ) Nov 7 '17 at 8:27
• @Werner The first two equations are the proofs by contradiction. The last two equations are the limit proofs. I want to fix the horizontal alignment since the first two proofs are spaced differently than the second two.
– user147592
Nov 7 '17 at 13:19
• @TorbjørnT. I apologize for not being more clear. I want to fix the horizontal alignment, not the vertical alignment. The first two equations are horizontally aligned differently from the second two, and I want to fix that.
– user147592
Nov 7 '17 at 13:21
You're probably looking for \intertext:
\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\dfrac{1}{n}-\dfrac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1$, $\inf(A)=-1$.
\begin{proof}
Suppose for contradiction that $1$ is not an upper bound for $A$, i.e.\@
$\exists n,m \in \mathbb{N} \mid \dfrac{1}{n} - \dfrac{1}{m} > 1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\intertext{Because $n(m+1)\geq m+1$, we have}
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot\\
\intertext{\indent Suppose for contradiction that $-1$ is not a lower bound for $A$,
i.e.\@ $\ex n,m \in \N \mid \dfrac{1}{n} - \dfrac{1}{m} < -1$.}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\intertext{Because $m(1+n)\geq 1+n$, we have}
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\intertext{\indent Claim that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$,
or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp \left|(1 - \dfrac{1}{m}) - 1\right| < \epsilon$.}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\intertext{Hence, for $M:=\left\lceil\dfrac{1}{\epsilon}\right\rceil+1$,
$m > M \imp \left| 1 - 1 - \dfrac{1}{m}\right|<\epsilon$. Because $\epsilon$
was arbitrarily chosen, it follows that
$\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$.\endgraf\medskip
\indent Claim that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$,
or $\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$.}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \dfrac{1}{\epsilon}\right\rceil+1$,
$n > N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$.
Because $\epsilon$ was arbitrarily chosen, it follows that
$\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$.
\end{proof}
\end{document}
EDIT: Found the error: you have too many & in the last line of the first two align envs
Below is a cleaned up MWE (with the error still present)
• don't use \dfrac in the text, causes excessive line spacing
• don't use \\ or \newline in the text, you never want manual line breaks in the text (this is a very common mistake among new users)
• use \intertext{...} for comments inside align (mathtools provides \shortintertext which has different spacing)
Cleaned MWE
\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\frac{1}{n}-\frac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1, \inf(A)=-1$.
\begin{proof}
Suppose for contradiction that $1$ is not an upper bound for $A$,
i.e.
$\exists n,m \in \mathbb{N} \mid \frac{1}{n} - \frac{1}{m} > 1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\intertext{Because $n(m+1)\geq m+1$, we have}
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot
\end{align*}
Suppose for contradiction that $-1$ is not a lower bound for $A$,
i.e. $\ex n,m \in \N \mid \frac{1}{n} - \frac{1}{m} < -1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\intertext{Because $m(1+n)\geq 1+n$, we have}
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\end{align*}
Claim that $\lim_{n=1, m \to \infty} \frac{1}{n} - \frac{1}{m} = 1$, or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp$ $\left|(1 - \frac{1}{m}) - 1\right| < \epsilon$.
\begin{align*}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\end{align*}
Hence, for $M:=\left\lceil\frac{1}{\epsilon}\right\rceil+1$,
$m > M \imp \left| 1 - 1 - \frac{1}{m}\right|<\epsilon$. Because
$\epsilon$ was arbitrarily chosen, it follows that
$\lim_{n=1, m \to \infty} \frac{1}{n} - \frac{1}{m} = 1$.
Claim that
$\lim_{m=1, n \to \infty} \frac{1}{n} - \frac{1}{m} = -1$, or
$\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp$
$\left|(\frac{1}{n}-1) -(-1)\right| < \epsilon$.
\begin{align*}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \frac{1}{\epsilon}\right\rceil+1, n > N \imp \left|(\frac{1}{n}-1) -(-1)\right| < \epsilon$. Because
$\epsilon$ was arbitrarily chosen, it follows that
$\lim\limits_{m=1, n \to \infty} \frac{1}{n} - \frac{1}{m} = -1$.
\end{proof}
\end{document}
Here is a image of the first page (after cleaning), I can see what you mean about the unevenness
• So there isn't a way to fix the unevenness?
– user147592
Nov 7 '17 at 15:14
• @SuhasHoysala See my update, I just left the image and the clean MWE for others to actually see the problem. You have to unnecessary & Nov 7 '17 at 15:27
• I removed the extra & (before the \bot command), but it still was uneven... I'm not sure what I'm doing wrong.
– user147592
Nov 7 '17 at 20:33
• @SuhasHoysala did you also remember to use \intertext as I did in my cleanup? My fix was literally the mwe cleanup I posted and then removing the two excessive & Nov 7 '17 at 20:35
• Yes, I copy-pasted your code verbatim and removed the two excessive &. It still doesn't seem to be fixed.
– user147592
Nov 7 '17 at 21:44 | 2021-09-27T12:23:49 | {
"domain": "stackexchange.com",
"url": "https://tex.stackexchange.com/questions/400065/align-horizontal-spacing-is-inconsistent",
"openwebmath_score": 1.0000056028366089,
"openwebmath_perplexity": 4524.93943391534,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9362850004144266,
"lm_q2_score": 0.9099069999303417,
"lm_q1q2_score": 0.8519322758068697
} |
https://math.stackexchange.com/questions/2394875/how-can-i-write-negations-for-the-following-statements | # How can I write negations for the following statements?
Premise: Let $F(x,y)$ be the statement $x$ can fool $y$, where the domain for discourse for both $x$ and $y$ is all people.
I converted the following corresponding statements:
1) Nobody can fool me
2) Anybody can Fool Fred
3) Everyone can fool someone
to
1) $\neg \exists xF(x,me)$
2) $\forall xF(x,Fred)$
3) $\forall x\exists yF(x,y)$
Now I am trying to negate these and I can't just throw in a not in front of the statements and I don't know how to go about this.
I looked at the negation rules and came up with this but I doubt it's correct:
1) $\neg\exists xF(x,me) \iff \forall x\neg F(x,me)$
2) $\forall xF(x,Fred) \iff \forall x\neg\neg F(x,Fred) \iff \neg[\exists x\neg F(x,Fred)]$
3) $\forall x\exists yF(x,y) \iff \neg\forall x\exists yF(x,y)$
• Well, $\lnot (\lnot Q)$ is $Q$ so the negation of $\lnot\exists xF(x,me)$ would be ... $\exists xF(x,me)$, right? – fleablood Aug 15 '17 at 22:17
• And the negation of $\forall x A$ is $\exists x \lnot A$ so the negation $\forall x F(x, fred)$ is $\exists x \lnot F(x, fred)$. – fleablood Aug 15 '17 at 22:19
• 3 is a little harder. As with 2) is $\lnot(\forall x \exists y F(x,y))$ will be $\exists x (\lnot \exists y F(x,y))$ but we need to find the negation of $\exists y F(x,y)$. The negation of $\exists x Q$ is $\forall x \lnot Q$ so the negation is $\exists x \forall y \lnot F(x,y)$. i.e. There is someone who can not fool anybody. – fleablood Aug 15 '17 at 22:25
• I think you're spot on, @fleablood. Please post your comments in an answer! I'll upvote it. – Namaste Aug 15 '17 at 22:29
• Wait... you aren't actually negating them. You are stating the equivalence of them. You can't "just" put a not in front, but you must start by putting a not in front. – fleablood Aug 15 '17 at 22:30
Except for what I assume to be a typo on the answer to 3, what you've written is correct (or more accurately, not wrong), although it seems you've missed out the actual negation of the statements which is what you're after.
Essentially, all you need is the rule
$$\neg((\forall x)\quad P(x))\iff((\exists x)\quad \neg P(x))$$
as well as the more basic $\neg\neg P\iff P$.
Let
$$S_1\equiv¬((∃x)\quad F(x,\text{me}))$$
$$S_2\equiv(∀x)\quad F(x,\text{Fred})$$
$$S_3\equiv(∀x)\quad((∃y)\quad F(x,y))$$
Then you're after $\neg S_1,\neg S_2,\neg S_3$.
As has been noted in the comments, the first is simple
$$\neg S_1\equiv\neg\neg((∃x)\quad F(x,\text{me}))$$
$$\neg S_1\equiv(∃x)\quad F(x,\text{me})$$
Your working for the second is helpful and we get
$$\neg S_2\equiv\neg((∀x)\quad F(x,\text{Fred}))$$
$$\equiv\neg¬((∃x)\quad¬F(x,\text{Fred}))$$
$$\equiv(∃x)\quad¬F(x,\text{Fred})$$
Finally, as has been pointed out by @fleablood, the last is slightly harder, but is just applying the above rule twice
$$\neg S_3\equiv\neg((∀x)\quad((∃y)\quad F(x,y)))$$
$$\equiv(∃x)\quad\neg((∃y)\quad F(x,y))$$
$$\equiv(∃x)\quad((∀y)\quad \neg F(x,y))$$
Okay. First to translate:
1) "Nobody can fool me". "Nobody" can be thought of as either "Everyone can not...", or as "There doesn't exist anyone who can..."
So either $\forall x\lnot F(x,me)$ or $\lnot \exists x F(x,me)$ will both be correct and are equivalent. So yes you are are correct.
As for 2) and 3)... I have nothing to add. You did those just fine. (You did 1) just fine as well, but I did have something to add.)
So the negations:
The rules are simple:
A) $\lnot$(That's not true) is, of course, "That is true" so $\lnot(\lnot Q) \iff Q$.
B) $\lnot$(Everbody does something) is, there is at least one person who doesn't do it. So $\lnot \forall x Q \iff \exists x \lnot Q$.
C) $\lnot$(Somebody does something) is. Nobody does something. So $\lnot \exists x Q \iff \forall x \lnot Q$.
So
1) $\lnot(\lnot \exists x F(x,me)) \iff \exists x F(x,me)$
or
$\lnot \forall x\lnot F(x,me) \iff \exists x\lnot(\lnot F(x,me)) \iff \exists x F(x, me)$.
i.e. It is not the case that noone can fool me $\iff$ someone can fool me.
2)$\lnot(\forall x F(x, fred)) \iff \exists x \lnot F(x,fred)$.
Not everbody can fool Fred = there is someone who can't fool Fred.
3) $\lnot (\forall x \exists y F(x,y)) \iff$
$\exists x \lnot(\exists y F(x,y)) \iff$
$\exists x \forall y \lnot F(x,y)$
i.e. not(Everbody can fool somebody) = somebody can not fool anybody. | 2019-07-23T00:51:37 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2394875/how-can-i-write-negations-for-the-following-statements",
"openwebmath_score": 0.7204133868217468,
"openwebmath_perplexity": 288.8390917543612,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9888419703960398,
"lm_q2_score": 0.8615382129861583,
"lm_q1q2_score": 0.8519251441007158
} |
https://mathematica.stackexchange.com/questions/69839/regular-polyhedra-coordinate-points-generation | Regular Polyhedra coordinate points generation
Is it possible to get all points on a Polyhedron surface using two surface parameters, say
$\phi,\theta$ spherical co-ordinates?
Just like in ParametricPlot3D, can we start with PolyhedronData["Tetrahedron"] to obtain spatial point positions?. The tip of position vector should cross edges automatically during $\phi,\theta$ sweep.
• You can always make this kind of tricks – Dr. belisarius Dec 29 '14 at 19:22
• @belisarius: fine,but want to append or prepend available Polyhedra. – Narasimham Dec 29 '14 at 19:59
• This is a nice start. – Dr. belisarius Dec 29 '14 at 20:04
• Stick a small sphere around the barycenter. Now parametrize by the spherical angles, using intersection of ray through spherical point (theta,phi) with polyhedron. – Daniel Lichtblau Dec 29 '14 at 23:57
• Clear enough, we need to have direct coordinates as function of ($\theta,\phi, n$). – Narasimham Dec 30 '14 at 7:58
Exploit the fact that the vertices of the dual to a Platonic solid correspond to the centers of the faces of the solid itself. For instance, the dual to a cube is a regular octahedron, and the six vertices of this octahedron are in the directions of the centers of the faces of its dual cube.
Find the normalized directions of the face centers of the cube (for instance) this way:
FaceCenters = Normalize /@ PolyhedronData[PolyhedronData["Cube", "Dual"], "Faces"][[1]];
The {x,y,z} direction in Euclidean coordinates as a function of spherical angles θ and φ are of course:
x[θ_Real, φ_Real] := {Sin[θ] Cos[φ],
Sin[θ] Sin[φ],
Cos[θ]}
Now sweep through all spherical angles, and for each corresponding direction find the nearest face center direction (i.e., the one with the smallest angle to the direction defined by θ and φ). For each such direction, the distance to the surface is 1/Sin[ψ], where ψ is the scalar angle between the candidate direction and its nearest face direction:
SphericalPlot3D[
1/Sin[x[θ, φ].Nearest[FaceCenters, x[θ, φ]][[1]]],
{θ, 0, π}, {φ, 0, 2 π},
PlotPoints -> 100] // Quiet
If you repeat with a Tetrahedron, for instance, you get this:
If you repeat for an Octahedron, you get this:
If you repeat for a Dodecahedron, you get this:
• Wow... I really thought @Narasimhan would have accepted this answer! What more can he possibly want?! – David G. Stork Jan 6 '15 at 0:58
• That was great. My apologies for the inordinate delay earlier – Narasimham Feb 10 '18 at 5:17
• Wow... two months late! Better late than never! – David G. Stork Feb 10 '18 at 6:20 | 2020-10-27T09:56:09 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/69839/regular-polyhedra-coordinate-points-generation",
"openwebmath_score": 0.6926586031913757,
"openwebmath_perplexity": 1907.2497222615355,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9888419694095656,
"lm_q2_score": 0.8615382076534743,
"lm_q1q2_score": 0.8519251379776488
} |
https://math.stackexchange.com/questions/1603966/what-is-the-least-number-of-square-roots-needed-to-express-sqrt1-sqrt2-c | What is the least number of square roots needed to express $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{100}$?
What is the least number of square roots needed to express $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{100}$ if it must be expressed in the form $a+b\sqrt{c}+d\sqrt{e}+\cdots$ where $a,b,c,d,e,\ldots$ are all integers?
Solution
In order to find the least number of square roots, we must express this number in simplest form. Thus, the numbers in the radicand that are in simplest form that will get counted are numbers with at most one of each prime factor. Then the question simplifies to "How many numbers are there between $1-100$ with at most one of each prime factor?
There are a few ways to proceed from here. I think the best way would be to complementary count. Then we are looking for numbers with at least $2$ factors of each prime, thus multiples of perfect squares. Our perfect squares are $1,4,9,16,25,36,49,64,81$. Thus we have $1+25+11-2+3+2 = 40$ such numbers by the principle of inclusion-exclusion. Thus, the answer is $100-40 = 60$.
Question
How is it that the number being expressed in simplest form will give the least number of radicals? The solution seems to imply that but why is it true?
• You mean expressing it as $a + b \sqrt{c} + d\sqrt{e} + \cdots$ where $a,b,c,\dots \in \mathbb N$ right? Because otherwise the question would be very problematic. – ThePortakal Jan 8 '16 at 3:45
• Yes, of course. Otherwise I wouldn't technically have to use any square roots. – user19405892 Jan 8 '16 at 3:45
• I added the extra condition in the question so that doesn't work as $3+\sqrt{3}+\sqrt{5}$ is not an integer. By least number of square roots, I mean the number of square root symbols. So $\sqrt{3}+\sqrt{3}$ would be two square roots while $2\sqrt{3}$ would be one. – user19405892 Jan 8 '16 at 4:07
• Why are you copying the exact solution here? – GohP.iHan Jan 9 '16 at 11:04
• @GohP.iHan It is actually my solution. – user19405892 Jan 9 '16 at 13:44
The square root of each square-free number less than $100$ contributes a term to the sum with a surd (a square root that cannot be simplified out), but all the square multiples of the number under the square root ($4$ times, $9$ times, $16$ times, etc.) can be combined with the term for the original square-free number, so they do not contribute any additional terms. For example,
$$\sqrt{11} + \sqrt{44} + \sqrt{99} = 6\sqrt{11}.$$
One way to count the integers whose square roots do not contribute new surds to the sum is as follows:
• $10$ perfect squares in the range $1$ to $100$, inclusive;
• $6$ cases of twice a perfect square in the range $8$ to $98$, inclusive;
• $4$ cases of $3$ times a perfect square in the range $12$ to $75$, inclusive;
• $3$ cases of $5$ times a perfect square in the range $20$ to $80$, inclusive;
• $3$ cases of $6$ times a perfect square in the range $24$ to $96$, inclusive;
• $6$ cases including $2$ cases each of $7$ times a perfect square, $10$ times a perfect square, and $11$ times a perfect square in the range $28$ to $99$, inclusive;
• $8$ cases including $13 \times 4$, $14 \times 4$, $15 \times 4$, $17 \times 4$, $19 \times 4$, $21 \times 4$, $22 \times 4$, and $23 \times 4$.
These add up to $40$ terms of the original sum that either are integers or can be combined with other terms, leaving $60$ unique surds. This is the same as your result, of course.
Here is the sum completely worked out, confirming that $60$ square roots are needed:
\begin{align} \sqrt1 + & \sqrt2 + \cdots + \sqrt{100} \\ =& \quad 1 + \sqrt2+\sqrt3 + 2 + \sqrt5+\sqrt6+\sqrt7 + 2\sqrt2 + 3 + \sqrt{10} \\ & + \sqrt{11} + 2\sqrt3 + \sqrt{13} + \sqrt{14} + \sqrt{15} + 4 + \sqrt{17} + 3\sqrt2 + \sqrt{19} + 2\sqrt{5} \\ & + \sqrt{21} + \sqrt{22} + \sqrt{23} + 2\sqrt6 + 5 + \sqrt{26} + 3\sqrt3 + 2\sqrt7 + \sqrt{29} + \sqrt{30} \\ & + \sqrt{31} + 4\sqrt2 + \sqrt{33} + \sqrt{34} + \sqrt{35} + 6 + \sqrt{37} + \sqrt{38} + \sqrt{39} + 2\sqrt{10} \\ & + \sqrt{41} + \sqrt{42} + \sqrt{43} + 2\sqrt{11} + 3\sqrt5 + \sqrt{46} + \sqrt{47} + 4\sqrt3 + 7 + 5\sqrt2 \\ & + \sqrt{51} + 2\sqrt{13} + \sqrt{53} + 3\sqrt6 + \sqrt{55} + 2\sqrt{14} + \sqrt{57} + \sqrt{58} + \sqrt{59} + 2\sqrt{15} \\ & + \sqrt{61} + \sqrt{62} + 3\sqrt7 + 8 + \sqrt{65} + \sqrt{66} + \sqrt{67} + 2\sqrt{17} + \sqrt{69} + \sqrt{70} \\ & + \sqrt{71} + 6\sqrt2 + \sqrt{73} + \sqrt{74} + 5\sqrt3 + 2\sqrt{19} + \sqrt{77} + \sqrt{78} + \sqrt{79} + 4\sqrt5 \\ & + 9 + \sqrt{82} + \sqrt{83} + 2\sqrt{21} + \sqrt{85} + \sqrt{86} + \sqrt{87} + 2\sqrt{22} + \sqrt{89} + 3\sqrt{10} \\ & + \sqrt{91} + 2\sqrt{23} + \sqrt{93} + \sqrt{94} + \sqrt{95} + 4\sqrt6 + \sqrt{97} + 7\sqrt2 + 3\sqrt{11} + 10 \\ =& \quad 55 + 28\sqrt2 + 15\sqrt3 + 10\sqrt5 + 10\sqrt6 + 6\sqrt7 + 6\sqrt{10} \\ & + 6\sqrt{11} + 3\sqrt{13} + 3\sqrt{14} + 3\sqrt{15} + 3\sqrt{17} + 3\sqrt{19} \\ & + 3\sqrt{21} + 3\sqrt{22} + 3\sqrt{23} + \sqrt{26} + \sqrt{29} + \sqrt{30} \\ & + \sqrt{31} + \sqrt{33} + \sqrt{34} + \sqrt{35} + \sqrt{37} + \sqrt{38} \\ & + \sqrt{39} + \sqrt{41} + \sqrt{42} + \sqrt{43} + \sqrt{46} + \sqrt{47} \\ & + \sqrt{51} + \sqrt{53} + \sqrt{55} + \sqrt{57} + \sqrt{58} + \sqrt{59} \\ & + \sqrt{61} + \sqrt{62} + \sqrt{65} + \sqrt{66} + \sqrt{67} + \sqrt{69} \\ & + \sqrt{70} + \sqrt{71} + \sqrt{73} + \sqrt{74} + \sqrt{77} + \sqrt{78} \\ & + \sqrt{79} + \sqrt{82} + \sqrt{83} + \sqrt{85} + \sqrt{86} + \sqrt{87} \\ & + \sqrt{89} + \sqrt{91} + \sqrt{93} + \sqrt{94} + \sqrt{95} + \sqrt{97} \\ \end{align}
• How do you know that we can't use nested radicals to reduce the number of radicals used? – GohP.iHan Jan 9 '16 at 11:03
• @GohP.iHan The problem statement stipulates that the answer must be in the form $a+b\sqrt{c}+d\sqrt{e}+\cdots$ where $a,b,c,d,e,\ldots$ are all integers. – David K Jan 9 '16 at 13:37
• Ohhh my bad... the author had a different question when I first viewed it. Thank you!! – GohP.iHan Jan 9 '16 at 13:51 | 2019-10-23T04:43:22 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1603966/what-is-the-least-number-of-square-roots-needed-to-express-sqrt1-sqrt2-c",
"openwebmath_score": 0.9993078708648682,
"openwebmath_perplexity": 219.2458015100361,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9888419661213171,
"lm_q2_score": 0.8615382094310357,
"lm_q1q2_score": 0.8519251369024244
} |
https://www.khanacademy.org/math/old-ap-calculus-ab/ab-limits-continuity/ab-direct-sub/v/limit-by-substitution | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## AP®︎ Calculus AB (2017 edition)
### Unit 1: Lesson 7
Determining limits using direct substitution
# Limits by direct substitution
AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
Sal explains how you can easily find limits of functions at points where the functions are continuous: simply plug in the x-value into the function! Later we will learn how to find limits even when the function isn't continuous.
## Want to join the conversation?
• I think Sal may have left something out. Correct me if I'm wrong, but on certain occasions, the entire function does not have to be continuous. If it is continuous at the value, a, then it should be fine to use direct substitution even though the function might be undefined elsewhere. This is because it isn't undefined at a so your output will be defined. Also, since your function is continuous at a, the limit there will exist meaning that you won't have to worry about jump discontinuities. Please tell me if this all seems reasonable.
• Because it is a continuous graph, every interval would be continuous provided it is a real number
• how would we know if the problem is to be solved by direct substitution method? or if the function given is continuous? without drawing its graph
• lim h(x) as x ->6 for h(x) =square root of (5x+6) should be 6 or -6 or both?
• y=√x is a function that returns only positive values. So in this context, we are only looking at the principal root, and the limit is positive 6.
Limits are unique: they cannot have multiple values. So "both" would never be an option.
• At , what's the difference between f(x) as x->a, and f(a)?
• One is a limit, the other is an evaluation of the function. If the function is continuous and defined at (in your example), a, then they're equivalent. But you can get some very interesting results if the function is not continuous or not defined.
The limit is basically saying what the function seems to be going to as x gets closer to closer to a, but the function may not be defined at that point.
• Is the limit zero or none since it is continuous?
• The limit is 0 as Sal has demonstrated in the video.
• Can a limit be a fraction?
• yes a limit can be a fraction. For example if I asked you to find the limit as X approaches 2 for the functions (1/X), when you use substitution to find the limit you find that the number it approaches is .5 or 1/2.
• I thought it like this : Lim x-> a f(x) = f(a) if f(x) is continuous at x=a, not vice versa? Can someone tell me why?
• That statement holds in both directions. A function is continuous at a point a if and only if the limit value equals the function value at a. Since the two statements are equivalent, this can be used as a definition of continuity.
• Ok so basically if a function, say f(x), is continuous at x=c, then the lim x-->c = f(x)? Is this why you can find limits of continuous functions by direct substitution?
• Yes, the limit as x->c of f(x) is f(c). This property is equivalent to the epsilon-delta definition of continuity, and it's why we can use direct substitution for most familiar functions.
• By Sal and almost all discussion about relation between continuity and limit: it always says that as long as the function is continuous, there is a limit.
i am wondering if there is a case that the function is continuous but the limit does not exist.
https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins_et_al)/1%3A_Understanding_the_Derivative/1.7%3A_Limits%2C_Continuity%2C_and_Differentiability
in figure 1.7.2, the right graph, at point x=1. the left hand limit is 2. but the as x approaches 1 from the right, the function keeps oscillating, so the limit does not exist.
so the function is continuous by the graph, but limit does not exist.
am I wrong? or am I give some wrong example?
• Interesting question!
By definition, a function f(x) is continuous at x=x_0 if and only if for every epsilon>0, there exists delta>0 such that whenever x is within delta of x_0, f(x) is always within epsilon of f(x_0).
Though the function might look continuous at x=1, it is really discontinuous at x=1 due to the oscillatory behavior. For example, if we choose epsilon = 1/2, there does not exist delta>0 such that whenever x is within delta of 1, f(x) is always within epsilon of f(1). This is because there are values of x arbitrarily close to 1 (from the right) such that f(x) differs from f(1) by as much as 1 unit.
(1 vote)
• I'm not entirely sure if this statement is always true. But my argument bases on the assumption that the function f(x) = |x| is not continuous at x = 0, which I think is true but correct me if I'm wrong.
So, the limit lim_{x -> 0}f(x) for f(x) = |x| is definitely defined since it approaches 0 from both sides. The function itself is usually defined so that |x| equals 0 at x = 0. Which means that lim_{x -> a}f(x) equals f(a) at a = 0, but as said AFAIK the absolute value function |x| is NOT continuous for x = 0.
And that would disprove the statement that if lim_{x -> a}f(x) equaling f(a) will mean that f is continuous at a (and vice versa).
(1 vote)
• Hey Raphael! I think you're getting continuity and differentiation confused. For a function, f(x), to be continuous, f(c) = lim_{x->c}f(x). This means that f(c) = lim_{x->c-}f(x) = lim_{x->c+}f(x) = lim_{x->c}f(x). As long as this holds true the function will be continuous at the given x-value, c.
Differentiation on the other hand is different. Differentiation requires that a function is continuous at the given x-value along with being absent of cusps, sharp turns (like x=0 for f(x) = |x|), and vertical tangents.
So to answer your question, f(x) = |x| is continuous at x = 0, however, it is not differentiable at x = 0. | 2023-01-29T20:24:45 | {
"domain": "khanacademy.org",
"url": "https://www.khanacademy.org/math/old-ap-calculus-ab/ab-limits-continuity/ab-direct-sub/v/limit-by-substitution",
"openwebmath_score": 0.8613227605819702,
"openwebmath_perplexity": 365.88969269845984,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9888419684230912,
"lm_q2_score": 0.86153820232079,
"lm_q1q2_score": 0.8519251318545814
} |
http://math.stackexchange.com/questions/228416/order-of-magnitudes-comparasions | # Order of magnitudes comparasions
I have a list of order of magnitudes I want to compare.
My only idea is using calculus methods (limits , integral, etc...) to assert the functions relation.
I need your help with the following.
I need to determine (again by order of magnitude) how to order the following (by asceneding order):
$log(n) / log(log(n))$
$log(log(n))$
$log^3(n)$
$n/log(n)$
$n^{log(n)}$
Is it even possible to use limits on such functions?
I mean if I take $f(n)$ to be $log(n) / log(log(n))$ and $g(n)$ to be $n^{log(n)}$, It seems very hard to find out the limit.
-
There are quite a number of comparisons to be made. It is in most cases relatively straightforward to decide about the relative long-term size.
Let's start with your pair. We have $f(n)=\log(n)/\log(\log (n))$ and $g(n)=n^{\log(n)}$.
For large $n$ (and it doesn't have to be very large), we have $f(n)\lt \log(n)$.
Also, for large $n$, we have $\log(n)\gt 1$, and therefore $g(n)=n^{\log(n)}\gt n$.
So for large $n$, we have $$\frac{f(n)}{g(n)} \lt \frac{\log(n)}{n}.$$ But we know that $\lim_{n\to\infty}\dfrac{\log(n)}{n}=0$. This can be shown in various ways. For instance, we can consider $\dfrac{\log(x)}{x}$ and use L'Hospital's Rule to show this has limit $0$ as $x\to\infty$.
It takes less time to deal with the pair $n/\log(n)$ and $n^{\log(n)}$. If $n$ is even modestly large, we have $n/\log(n)\lt n$. But after a while, $\log(n)\gt 2$, so $n^{\log(n)}\gt n^2$. It follows that in the long run, $n^{\log(n)}$ grows much faster than $n/\log(n)$.
As a last example, let us compare $\log^3(n)$ and $n/\log(n)$. Informally, $\log$ grows glacially slowly. More formally, look at the ratio $\dfrac{\log^3(n)}{n/\log(n)}$. This simplifies to $$\frac{\log^4 (n)}{n}.$$ We can use L'Hospital's Rule on $\log^4(x)/x$. Unfortunately we then need to use it several times. It is easier to examine $\dfrac{\log(x)}{x^{1/4}}$. Then a single application of L'Hospital's Rule does it. Or else we can let $x=e^y$. Then we are looking at $y^4/e^y$, and we can quote the standard result that the exponential function, in the long run, grows faster than any polynomial.
Remark: The second person in your list is the slowest one. Apart from that, they are in order. So you only need to prove four facts to get them lined up. A fair part of the work has been done above.
-
Your answer is very very good and understood. I will work on these tomorrow, make sure I understand anything. Thank you very much! – SyndicatorBBB Nov 3 '12 at 20:49
@Guy: I don't mind your accepting my answer. However, it is useful in general to wait somewhat longer, in order to elicit a variety of approaches. – André Nicolas Nov 3 '12 at 20:54
I commented below. Can you please advise ?Thank you. – SyndicatorBBB Nov 4 '12 at 10:01
For instance: $$\log(n^{\log n})=\log^2(n)$$ And since asymptotically: $$\frac{\log(n)}{\log(\log(n))} < \log^2(n)$$ We know that: $$n^{\log n}$$ Is at least exponentially larger than: $$\frac{\log(n)}{\log(\log(n))}$$
-
How did you calculate this inequalties? – SyndicatorBBB Nov 3 '12 at 20:46
Just observe: a $\log(n)$ cancels out, and $1/\infty$ is obviously smaller than $\infty$. – nbubis Nov 3 '12 at 20:47
Undestood! Thank you very much! – SyndicatorBBB Nov 3 '12 at 20:50
@André Nicolas I really liked the comprasion of the functions with simplier functions.
I was trying to use this technique on some of these functions.
Eventually I found a couple which I was not able to solve using this technique and I don't understand why.
Assume $f(n) = log(log(n))$ and $g(n) = n/log(n)$
Now one might say, $f(n) < log(n)$ and $g(n) > 1/n$
If I look at the ratio/limit of $f(n)/g(n)$, I find out that $f(n)$ is much faster.
I completly understand that I chose unwisely the lower bound of g(n). Does it mean that on such functions I cannot use this technique?
If you "give away" stuff to make things simpler, have to be sensible about it. Replacing $g(n)$ by something that approaches $0$ is unreasonable. For example, could use $f(n)\lt \log(n)$, and keep $g(n)$. How one "weakens" an inequality has to be guided by basic intuition/experience about the sizes of things. The one thing that will be constantly useful is (i) $\log(n)$ is long-run slower than any positive power of $n$ and its twin (ii) For constant $a\gt 1$, $a^n$ is long-run faster than any polynomial in $n$. – André Nicolas Nov 4 '12 at 14:26 | 2016-02-14T13:18:29 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/228416/order-of-magnitudes-comparasions",
"openwebmath_score": 0.881853461265564,
"openwebmath_perplexity": 224.87314935889398,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9825575116884779,
"lm_q2_score": 0.8670357615200475,
"lm_q1q2_score": 0.8519125003840624
} |
https://mathoverflow.net/questions/368404/if-all-points-of-a-real-function-with-positive-values-would-be-local-minimum-ca | # If all points of a real function with positive values would be local minimum, can one say it is constant function?
During my studies I faced a function $$f:\mathbb{R} \to \mathbb{R}^+$$ with the property: for all $$x \in \mathbb{R}$$ and all $$y$$ in open interval $$(x-\frac{1}{f(x)} ,x+\frac{1}{f(x)})$$ we have $$f(x) \leq f(y)$$. At first I guessed maybe it is necessarily constant function but I could not show this. I tried to generate a counter example for my guess and unfortunately I could not again. would anyone please help me?
• The answer to the question in the title is (of course) no. [Define $f(x) = 2$ for $x \ne 0$ and $f(0)=1$.] But the question in the text is more complicated. – Gerald Edgar Aug 6 '20 at 9:31
• What is $\mathbb{R}^+$? Is $f(x)=0$ allowed? – Wilberd van der Kallen Aug 6 '20 at 9:52
• $\mathbb{R}^+$ is denoted for the open interval $(0,+\infty)$ – M. Reza. K Aug 6 '20 at 10:12
My answer is: Such a function must be constant. Suppose (for puoposes of contradiction) $$f$$ is a nonconstant function $$f : \mathbb R \to (0,+\infty)$$ such that $$\forall x\in\left(a-\frac{1}{f(a)},a+\frac{1}{f(a)}\right),\quad f(x) \ge f(a) .$$
For $$m > 0$$, define $$U_m := \{x : f(x)>m\}$$.
Lemma 1: For all $$m \in \mathbb R$$, the set $$U_m$$ is open.
Proof. Let $$x \in U_m$$. Then $$(x-1/f(x),x+1/f(x)) \subseteq U_m$$. So $$U_m$$ is open.
Define $$\mathcal G = \{(a,b,m) : a < b, f(a)\le m, f(b)\le m, \text{ and }\forall x\in(a,b),f(x)> m\}$$.
Lemma 2: Let $$a,b \in \mathbb R$$ with $$f(a) \ne f(b)$$, and $$m>0$$. Then there exists $$a_1, b_1$$ with $$a < a_1 < b_1 < b$$ and $$m_1 \ge m$$ such that $$(a_1,b_1,m_1) \in \mathcal G$$.
Proof. One of $$f(a), f(b)$$ is smaller, assume WLOG that $$f(a) < f(b)$$. Now $$U_{f(a)}$$ is open, $$a \notin U_{f(a)}$$, and $$b \in U_{f(a)}$$. The maximal open interval $$(c,d) \subseteq U_{f(a)}$$ with $$c is nonempty, in fact $$(c,b) = (c,d) \cap (a,b) \ne \varnothing$$. By maximality, $$f(c) \le f(a)$$. Choose $$e$$ with $$c < e < b$$. For $$x \in (c,e)$$ we have $$f(c) < f(x)$$ so $$c \le x-1/f(x)$$ and thus $$f(x) \ge 1/(x-c)$$. Thus $$f(x)$$ is unbounded on $$(c,e)$$. Let $$m_1$$ be such that $$m_1 \ge m, m_1 > f(c), m_1 > f(e)$$. The open set $$U_{m_1}$$ has $$U_{m_1} \cap (c,e) \ne \varnothing$$ and $$c,e \notin U_{m_1}$$. So let $$(a_1,b_1)$$ be a maximal open subinterval of the open set $$U_{m_1} \cap (c,e)$$.
Lemma 3: Let $$(a_1, b_1, m_1) \in \mathcal G$$, and let $$m > m_1$$. Then there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 \ge m$$ such that $$(a_2, b_2, m_2) \in \mathcal G$$.
Proof: If $$f(a_1) \ne f(b_1)$$, apply Lemma 2 directly. Otherwise, pick any $$c \in (a_1,b_1)$$ and apply Lemma 2 to $$a_1, c, m$$.
Main Proof: We assume $$f$$ is not constant. So there exist $$a_1 < b_1$$ with $$f(a_1) \ne f(b_1)$$. By Lemma 2 there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 > 2$$ so that $$(a_2,b_2,m_2) \in \mathcal G$$. Then by Lemma 3, there exist $$a_3, b_3$$ with $$a_2 < a_3 < b_3 < b_2$$ and $$m_3 > 3$$ so that $$(a_3,b_3,m_3) \in \mathcal G$$. Continuing recursively, we get sequences $$(a_k), (b_k), (m_k)$$ so that $$\forall k:\quad a_k < a_{k+1} < b_{k+1} < b_k,\quad m_k > k,\quad\text{and } (a_k,b_k,m_k)\in\mathcal G .$$ The sequence $$(a_k)$$ is increasing and bounded above, so it converges. Let $$a = \lim_{k\to\infty} a_k$$. Then $$a_k < a_{k+1} \le a \le b_{k+1} < b_k$$. From $$(a_k,b_k,m_k) \in \mathcal G$$ we conclude $$f(a) > m_k > k$$. This is true for all $$k$$, so $$f(a)$$ is not a real number, a contradiction.
• Your answer was enjoyable and very nice to me. It really engaged me and I am thinking if instead of $\mathbb{R}$ in domain of the $f$, replace a complete metrically convex metric space, does your answer still work? – M. Reza. K Aug 7 '20 at 12:11
• @M.Reza.K for every $x,y$ in the complete metrically convex space there is an isometry $g:[a,b]\to X$ such that $g(a)=x, g(b)=y$. Apply the result to $h=f\circ g$. It follows that $f(x)=h(a)=h(b)=f(y)$. I am interested if we can drop the completeness: the proof seems to work for rational numbers... Perhaps it also works for intrinsic metric spaces? – erz Aug 7 '20 at 12:40
• @erz My proof certainly fails for rational numbers. A decreasing sequence of closed intervals may have empty intersection in $\mathbb Q$. – Gerald Edgar Aug 7 '20 at 13:32
This is not an answer but it greatly narrows down the class of functions $$f$$ with the described property.
1. If $$c>0$$, then the sets $$f^{-1}([c,+\infty))$$ and $$f^{-1}((c,+\infty))$$ are both open.
Indeed, every $$x$$ in any of these sets comes with an open ball of radius $$\frac{1}{f(x)}$$.
Consequently, the sets $$A_c=f^{-1}((-\infty, c))$$ and $$B_c=f^{-1}((-\infty, c])$$ are both closed. In particular, $$f$$ is lower semi-continuous.
1. Let $$x\in B_c$$. We know that if $$d(y,x)<\frac{1}{c}\le \frac{1}{f(x)}$$, then $$f(x)\le f(y)$$. On the other hand, if $$y\in B_c$$ and $$d(y,x)<\frac{1}{c}$$, then $$d(y,x)<\frac{1}{f(y)}$$, from where $$f(y)\le f(x)$$. Hence, if $$y\in B_c$$ and $$d(y,x)<\frac{1}{c}$$, we get $$f(y)= f(x)$$.
Not only this means that $$f$$ is locally a constant on $$B_c$$, but also that $$f^{-1}(c)$$ is closed.
1. Let us prove the following strengthening of the Baire's theorem: let $$X$$ be a complete metric space, and let $$B_n$$ be a sequence of closed sets such that $$X=\bigcup B_n$$. Then $$X=\overline{\bigcup int B_n}$$.
Indeed, if $$U\subset X\backslash \bigcup int B_n$$ is open and nonempty, it is metrizable with a complete metric, and $$B_n\cap U$$ is closed in $$U$$. Since $$U=\bigcup (B_n\cap U)$$, from the usual Baire's theorem, there is $$m$$ such that $$V=int (B_m\cap U) \ne \varnothing$$. But then $$V=int (B_m\cap U)=int B_m\cap U\subset int B_m \backslash \bigcup int B_n=\varnothing$$. contradiction.
1. So, combining these observation we get the following picture: there is a dense open set $$U$$ such that $$f$$ is locally constant on $$U$$; every level set of $$f$$ is closed, and the distance between elements of $$f^{-1}(c)$$ and $$f^{-1}(d)$$ is at least $$\frac{1}{\max \{c,d\}}$$.
The only way it's not a constant function is when the components of $$U$$ are like the gaps in the Cantor set. So the possible counterexample may be constructed like the Cantor's staircase, but the smaller the component, the larger is the value on it. I was unable to perform this fine tuning though, maybe it's impossible. | 2021-05-11T07:06:04 | {
"domain": "mathoverflow.net",
"url": "https://mathoverflow.net/questions/368404/if-all-points-of-a-real-function-with-positive-values-would-be-local-minimum-ca",
"openwebmath_score": 0.996794581413269,
"openwebmath_perplexity": 2288.9339333725334,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9825575147530352,
"lm_q2_score": 0.867035758084294,
"lm_q1q2_score": 0.8519124996653178
} |
http://math.stackexchange.com/questions/87902/problems-about-countability-related-to-function-spaces | # Problems about Countability related to Function Spaces
Suppose we have the following sets, and determine whether they are countably infinite or uncountable .
1. The set of all functions from $\mathbb{N}$ to $\mathbb{N}$.
2. The set of all non-increasing functions from $\mathbb{N}$ to $\mathbb{N}$.
3. The set of all non-decreasing functions from $\mathbb{N}$ to $\mathbb{N}$.
4. The set of all injective functions from $\mathbb{N}$ to $\mathbb{N}$.
5. The set of all surjective functions from from $\mathbb{N}$ to $\mathbb{N}$.
6. The set of all bijection functions from from $\mathbb{N}$ to $\mathbb{N}$.
My thoughts on this:
1. The first set is uncountable by using diagonalization argument.
2. I have read something about it saying this is countable since we can see this as a set of finite sequences of natural numbers. But I have hard time following the argument.
3. The article I read says this is uncountable but I couldn't follow the argument either.
4. For (4),(5) and (6), I am not even sure how to approach problems like these.
Could someone please explain how to approach this kind of problems in general?
And does it matter when I change all the $\mathbb{N}$ to $\mathbb{R}$?
-
Recall the definition of a function as a set of ordered pairs. That should help. – Potato Dec 3 '11 at 2:27
Maybe you can argue that (2) is countable because any given sequence that is non-increasing eventually stops decreasing at some finite natural number or hits zero and then is forced to be constant at 0. Then come up with a clever way to list all the functions, maybe by listing them first by starting position then by "length before it is constant" or something like that. – tomcuchta Dec 3 '11 at 2:28
@tomcuchta Your idea for (2) is alright, but a small nitpick. The function need not go to $0$ but might stop at $1$ (say). What is true (and important) is that the function is eventually constant. – Srivatsan Dec 3 '11 at 2:39
I removed the (functional-analysis), the (set-theory) and the (proof-theory) tag. Each of these tags is intended for subjects usually taught at an advanced undergraduate or graduate level. – t.b. Dec 3 '11 at 2:42
@t.b.Thanks, I guess I am confused about what function spaces really study. – geraldgreen Dec 3 '11 at 3:12
Another approach to the 6th part.
Let us denote the set of all bijections from $\mathbb N$ to $\mathbb N$ by $\operatorname{Bij}(\mathbb N,\mathbb N)$.
Clearly $$|\operatorname{Bij}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c.$$
Let us try to show the opposite inequality.
For arbitrary function $f\in\operatorname{Bij}(\mathbb N,\mathbb N)$ we define $$\operatorname{Fix}(f)=\{n\in\mathbb N; f(n)=n\}.$$ (That is, $\operatorname{Fix}(f)$ is the set of all fixed points of the map $f$.) Let us try to answer the question, whether for any $A\subseteq\mathbb N$ it is possible to find a bijection ${f_A}:{\mathbb N}\to {\mathbb N}$ such $\operatorname{Fix}(f_A)=A$. Let us consider two cases.
If the complement of the set $A$ is infinite, i.e. $\mathbb N\setminus A=\{b_n; n\in\mathbb N\}$ (and we can assume that the elements $\mathbb N\setminus A$ are written in the increasing order, i.e. $b_1<b_2<\dots<b_n<b_{n+1}<\dots$ ), then we can define a function $f_A$ as follows: $$f_A(x)= \begin{cases} x, & \text{if }x\in A, \\ b_{2k+1}, &\text{ak x=b_{2k} for some k\in\mathbb N},\\ b_{2k}, &\text{ak x=b_{2k+1} for some k\in\mathbb N}. \end{cases}$$ In the other words, we did not move the elements of $A$ and the elements from the complement were paired and we interchanged it pair. Such map is a bijection from $\mathbb N$ to $\mathbb N$ such that $\operatorname{Fix}(f_A)=A$.
Next we consider the case that $\mathbb N\setminus A$ is finite.
If $\mathbb N\setminus A=\emptyset$, then $f=id_{\mathbb N}$ is a function fulfilling $\operatorname{Fix}(f_A)=\mathbb N=A$.
If the set $\mathbb N\setminus A$ is a singleton $\{a\}$, then is is impossible to find a bijection $f$ such that $\operatorname{Fix}(f)=A=\mathbb N\setminus\{a\}$. No element of $\mathbb N\setminus A$ can be mapped on $a$ (since all such elements are fixed). But $a$ cannot be mapped on $a$ since this would mean that $a$ is a fixed point.
But if the set $\mathbb N\setminus A$ has at least two elements, then it is possible to construct such a map. We again assume that the elements $b_k$ are written in the increasing order, i.e. $\mathbb N\setminus A=\{b_0,\dots, b_n\}$ and $b_0<b_1<\dots<b_n$. We put $f_A(x)=x$ for $x\in A$, $f_A(b_k)=b_{k+1}$ for $0\le k<n$ and $f_A(b_n)=b_0$. (I.e. we made a cycle consisting of elements of $\mathbb N\setminus A$.)
This defines a bijection ${f_A}:{\mathbb N}\to{\mathbb N}$ such that $\operatorname{Fix}(f_A)=A$.
The assignment $A\mapsto f_A$ that we described above is a map from the set $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ consisting of all subsets of $\mathbb N$, whose complement is not a singleton, to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This map is injective; since from $f_A=f_B$ we get $A=\operatorname{Fix}(f_A)=\operatorname{Fix}(f_B)=B$.
From the set $\mathcal P({\mathbb N})$ with the cardinality $\mathfrak c$ we have omitted a countable set $\{\mathbb N\setminus\{a\}\in\mathbb N\}$. Therefore the cardinality of this difference $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ is again $\mathfrak c$.
So we found an injection from a set of cardinality $\mathfrak c$ to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This yields the opposite inequality $$\mathfrak c\le|{\operatorname{Bij}(\mathbb N,\mathbb N)}|$$ and by Cantor-Bernstein theorem we get $|{\operatorname{Bij}(\mathbb N,\mathbb N)}|=\mathfrak c$.
-
Thanks for doing the details. It pointed out a flaw in my answer. – David Mitra Dec 3 '11 at 14:07
I did not notice that we are using basically the same argument until you pointed it out. I like your answer, but I am out of votes for today. – Martin Sleziak Dec 3 '11 at 14:14
This is not a complete answer. [I am taking $\mathbb N$ to include $0$.] I will give you a proof that there are uncountably many bijections from $\mathbb N$ to itself. This obviously subsumes the questions (4)-(6).
Consider functions $f : \mathbb N \to \mathbb N$ of the following special form: For every $k$, the pair of elements $\{ 2k, 2k+1 \}$ is mapped to itself. However, we have two ways of doing this:
• either: map $2k$ to itself, and similarly for $2k+1$.
• or: map $2k$ and $2k+1$ to each other.
The key point is that for each $k$, you could pick one of the two possibilities independently. Since there are $2$ options for each $k$, you have $2^{\mathbb N}$ options; i.e., there are uncountably many such functions satisfying this restriction. And each such function is a bijection from $\mathbb N$ to itself.
The above needs a little bit of work to make it completely rigorous. In particular, try to formalise what I mean by "$2^\mathbb N$ possibilities".
For (3), the idea is similar. For each $k$, try mapping $k$ to either $2k$ or $2k+1$. Details are left as an exercise.
For (2), does the following hint help? Every non-increasing function $f : \mathbb N \to \mathbb N$ is eventually constant. So the function is uniquely specified if we describe the finite initial "portion" of the function.
Another approach (due to tb) for (2): If you plot the graph of a non-increasing function, it looks like a staircase that goes down as we move from left to right -- a staircase with finitely many steps because you the function can go down only finitely many times. One can also reconstruct the function if we record the upper right corners of each of those steps. Since the latter corresponds to a finite subset of $\mathbb N \times \mathbb N$ (which is countable), it follows that our set of non-increasing functions is countable as well.
-
I'll use this opportunity to mention here a very nice (at least in my opinion) proof that the cardinality of all bijections $\mathbb N\to\mathbb N$ is $\mathfrak c=2^{\aleph_0}$, which is basically the part 6. I making it a community wiki, since the idea is not mine. I believe I've seen this at sci. math but I am not sure about it and I cannot find the link now. (Feel free to add some pointers and links, if you know some.)
HINT: If someone wants to try to develop the idea by himself, the basic idea is Riemann's rearrangement theorem.
Let us denote the set of all bijections from $\mathbb N$ to $\mathbb N$ by $\operatorname{Bij}(\mathbb N,\mathbb N)$.
Clearly $$|\operatorname{Bij}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c.$$
Now we will show the opposite inequality. Let us choose a series $\sum_{n=0}^\infty x_n$, which is convergent but not absolutely convergent (e.g. $\sum_{n=0}^\infty x_n = \sum_{n=0}^\infty (-1)^n\frac1{n+1}$). By Riemann's theorem, for any real number $r$ there exists a permutation $\pi\in \operatorname{Bij}(\mathbb N,\mathbb N)$ such that $\sum_{n=0}^\infty x_{\pi(n)}=r$. If we choose for each $r$ one such permutation, we get an injective map from $\mathbb R$ to $\operatorname{Bij}(\mathbb N,\mathbb N)$. Thus $|\operatorname{Bij}(\mathbb N,\mathbb N)| \ge |\mathbb R| = \mathfrak c = 2^{\aleph_0}$.
-
+1, nice approach. I don't think you have to make an answer CW just because it isn't your idea. Writing an answer is contribution enough and one deserves some credit for that. =) – Srivatsan Dec 4 '11 at 0:24
Let \eqalign{ A_1&=\text{the set of all functions from }\Bbb N\text{ to }\Bbb N. \cr A_2&=\text{the set of all non-increasing from }\Bbb N\text{ to }\Bbb N. \cr A_3&=\text{the set of all non-decreasing }\Bbb N\text{ to }\Bbb N. \cr A_4&=\text{the set of all 1-1 functions from}\Bbb N\text{ to }\Bbb N. \cr A_5&=\text{the set of all onto functions from }\Bbb N\text{ to }\Bbb N. \cr A_6&=\text{the set of all bijections from }\Bbb N\text{ to }\Bbb N. \cr }
Consider $A_6$:
$A_6$ may be thought of as the set $\bigl\{\,\pi :\pi \text{ is a permutation of } (1,2,3,\ldots)\,\bigr\}$.
For any permutation $\pi$, we may associated a binary sequence $(x_n)$ by setting $$x_n=\cases{ 0,& \text{if } \pi_n=n\cr 1,& \text{otherwise} }.$$ This induces an onto map from $A_6$ to the set of infinite binary sequences (Edit: not quite, as Martin Sleziak points out in his answer, the binary sequences that have "only one 1" have no element in $A_6$ mapped to them. But, as the set of such sequences is countable, the rest of the argument will go through. Martin also describes nicely how to obtain this map.). Since the latter set is uncountable, so is the former.
So, $A_6$ is uncountable.
From this, it follows that $A_5$, $A_4$, and $A_1$ are uncountable.
Now, consider $A_2$:
If $f$ is a non-increasing function, then $f$ must be eventually constant. Thus, the cardinality $A_2$ would be at most the cardinality of the set $$\bigcup_{j,m} \bigl\{\,(a_i)_{i=1}^\infty : a_i=m, i\ge j\,\bigr \}.$$ But this latter set is a countable union of countable sets and is, thus, countable.
So, $A_2$ is countable.
Now, consider $A_3$:
The cardinality of the set of non-decreasing functions is at least the cardinality of the set $$\{\, A\subset\Bbb N : A \text{ is an infinite subset of } \Bbb N\, \}.$$
Since this latter set uncountable, $A_3$ is uncountable.
-
"If f is a non-increasing function, then f must be eventually constant" Why ? it can also be increasing. – GinKin Mar 11 at 12:45
@GinKin I took "non-increasing" to mean $f(b)\le f(a)$" whenever $a\le b$ (that is, (perhaps non-strictly) decreasing). This is standard terminology. What do you mean by "it can also be increasing"? – David Mitra Mar 11 at 13:11
Whoops mixed the terms, I meant that if it doesn't increase then it doesn't have to be eventually constant, it can be decreasing. – GinKin Mar 11 at 13:24
@GinKin But the range is $\Bbb N$. If it were strictly decreasing, it would eventually be constantly $0$. – David Mitra Mar 11 at 13:25 | 2014-08-28T23:37:56 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/87902/problems-about-countability-related-to-function-spaces",
"openwebmath_score": 0.965313732624054,
"openwebmath_perplexity": 191.6129276320447,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9825575121992375,
"lm_q2_score": 0.8670357598021707,
"lm_q1q2_score": 0.8519124991389965
} |
https://ask.sagemath.org/question/49683/the-number-of-solutions-of-a-polynomial-system-using-a-grobner-basis/?answer=49684 | # The number of solutions of a polynomial system using a Gröbner basis
The following extract from this Wikipedia page explains that we can easily deduce the number of solutions of a polynomial system using Gröbner basis:
Given the Gröbner basis G of an ideal I, it has only a finite number of zeros, if and only if, for each variable x, G contains a polynomial with a leading monomial that is a power of x (without any other variable appearing in the leading term). If it is the case the number of zeros, counted with multiplicity, is equal to the number of monomials that are not multiple of any leading monomial of G. This number is called the degree of the ideal.
Question: Is there a function in Sage computing this number of zeros from a given Gröbner basis?
edit retag close merge delete
Sort by » oldest newest most voted
Yes, these
monomials that are not a multiple of any leading monomial of G
form a basis of the quotient ring $R/I$ (by the reduction algorithm), also called a normal basis for $I$.
sage: R.<x,y,z> = PolynomialRing(QQ,3)
sage: I = Ideal([x^2+y+z-1,x+y^2+z-1,x+y+z^2-1])
sage: B = I.groebner_basis(); B
[x^2 + y + z - 1, y^2 + x + z - 1, z^2 + x + y - 1]
sage: I.normal_basis()
[x*y*z, y*z, x*z, z, x*y, y, x, 1]
The number of elements in this basis, or the vector space dimension of $R/I$, can also be found as follows:
sage: I.vector_space_dimension()
8
The zero set, or variety, of the ideal is the following:
sage: I.variety(QQbar)
[{z: 0, y: 0, x: 1},
{z: 0, y: 1, x: 0},
{z: 1, y: 0, x: 0},
{z: -2.414213562373095?, y: -2.414213562373095?, x: -2.414213562373095?},
{z: 0.4142135623730951?, y: 0.4142135623730951?, x: 0.4142135623730951?}]
The number of elements is not 8, because we are missing the "multiplicities". Indeed,
False
So perhaps the more interesting quantity is
5
which agrees with the number of zeros without multiplicity.
(There is an easy algorithm to compute the radical of a zero-dimensional ideal.)
more
Is it required to compute the Gröbner basis? Does the function normal_basis() recompute the Gröbner basis?
( 2020-01-26 20:05:55 -0500 )edit
The method normal_basis() internally requires a Gröbner basis. If one has not yet been computed, then it computes one. If one has been computed already, then it uses that one; it does not recompute it. Note also that the result / speed depends on the chosen monomial ordering (degrevlex by default). Of course the dimension does not depend on this choice.
( 2020-01-27 03:57:43 -0500 )edit | 2020-07-14T07:28:51 | {
"domain": "sagemath.org",
"url": "https://ask.sagemath.org/question/49683/the-number-of-solutions-of-a-polynomial-system-using-a-grobner-basis/?answer=49684",
"openwebmath_score": 0.29856231808662415,
"openwebmath_perplexity": 910.6520081079834,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9825575183283513,
"lm_q2_score": 0.8670357512127872,
"lm_q1q2_score": 0.851912496013594
} |
https://mathematica.stackexchange.com/questions/240082/solving-a-eigenvalue-problem | # Solving a Eigenvalue Problem
$$\textbf{Eigenvalue problem on a unit square \Omega = [0,1]^2 :}$$
Consider the eigenvalue problem with the Dirichlet boundary condition that is, $$-Lu = \lambda u$$ where $$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$.
The boundary condition is that $$u=0$$ on $$\partial \Omega$$.
I am computing the $$\textbf{eigenvalues}$$ and $$\textbf{eigenfunctions}$$ of Laplacian numerically in Mathematica. Specially, I am interested about the $$\textbf{multiplicity}$$ of a specific eigenvalue. I already know that Tally or Count can certainly count the occurrances of eigenvalues in the list called vals.
{ℒ, ℬ} = {-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]};
{vals, funs} =
DEigensystem[{ℒ, ℬ},
u[x, y], {x, 0, 1}, {y, 0, 1}, _];
Tally[vals]
In the eigenvalue problem there are infinite numbers of eigenvalues. So, We can not list all of them. I want to find out $$\textbf{multiplicity}$$ of a specific eigenvalue that counts the total number of occurrances. For example, the multiplicity of the eigenvalue $$5 \pi^2$$ is $$2$$. Similarly, I want to find out the multiplicity of the eigenvalue $$50 \pi^2$$ is $$3$$.
• Edit your question to include your Mathematica code rather than pictures of the code. Feb 15 '21 at 17:51
• Convert your cells to InputForm and look at the Markdown help Feb 15 '21 at 18:24
• Feb 15 '21 at 19:30
With the exact solutions
u[{nx_, ny_}, {x_, y_}] = Sin[nx π x] Sin[ny π y];
and $$\{n_x,n_y\}\in\mathbb{N}^2$$, the eigenvalues are
λ[nx_, ny_] = (nx^2 + ny^2) π^2;
Test:
Assuming[Element[nx | ny, PositiveIntegers],
{u[{nx, ny}, {x, 0}], u[{nx, ny}, {x, 1}],
u[{nx, ny}, {0, y}], u[{nx, ny}, {1, y}]} // FullSimplify]
(* {0, 0, 0, 0} *)
-D[u[{nx, ny}, {x, y}], {x, 2}] - D[u[{nx, ny}, {x, y}], {y, 2}] ==
λ[nx, ny] * u[{nx, ny}, {x, y}] // FullSimplify
(* True *)
The number of pairs $$\{n_x,n_y\}$$ equal to $$\lambda/\pi^2$$ is therefore given by OEIS A063725. You can calculate the multiplicity of a given value of $$i=\lambda/\pi^2$$ directly from the generating function,
g[q_] = (EllipticTheta[3, q] - 1)^2/4;
m[i_Integer?NonNegative] := SeriesCoefficient[g[q], {q, 0, i}]
For example, the multiplicity of $$\lambda=50\pi^2$$ is 3, as you know:
m[50]
(* 3 *)
There is no eigenvalue at $$326\pi^2$$, on the other hand:
m[326]
(* 0 *)
More generally, the $$d$$-dimensional generating function
g[d_, q_] = ((EllipticTheta[3, q] - 1)/2)^d;
allows us to find the multiplicity of the eigenvalue $$\lambda=i\pi^d$$ directly:
m[d_Integer?NonNegative, i_Integer?NonNegative] :=
SeriesCoefficient[g[d, q], {q, 0, i}]
For example, the $$d=17$$-dimensional hypercube has $$1\,416\,786\,753\,216$$ eigenvalues $$\lambda=250\pi^{17}$$:
m[17, 250]
(* 1416786753216 *)
If you want the actual solutions, not just the number of solutions, PowersRepresentations is a good starting point (but it gives solutions containing zeros, which we need to filter out):
s[d_Integer?NonNegative, i_Integer?NonNegative] :=
Select[PowersRepresentations[i, d, 2], Min[#] > 0 &]
For example, the eigenvalue $$\lambda=50\pi^2$$ in $$d=2$$ dimensions can be composed in two ordered ways, as you know:
s[2, 50]
(* {{1, 7}, {5, 5}} *)
Enumerating all permutations of these ordered combinations gives the full multiplicity of 3:
t[d_Integer?NonNegative, i_Integer?NonNegative] :=
Join @@ Permutations /@ s[d, i]
t[2, 50]
(* {{1, 7}, {7, 1}, {5, 5}} *)
In $$d=17$$ dimensions the eigenvalue $$\lambda=250\pi^{17}$$ has 999 ordered solutions,
s[17, 250]
(* {{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 6, 14},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 10, 10},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 15},
...
{3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5},
{3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6}} *)
which then become the aforementioned $$1\,416\,786\,753\,216$$ unordered solutions from t[17, 250].
• We should work with the generating function explicitly because it gives a very efficient recipe, especially in the high-dimensional case. Just look at the given example for the 17-dimensional hypercube: manually summing and counting is going to be very tedious and difficult, whereas the call to m[17, 1000] takes less than half a second. The same is true, to some extent, for your case $d=2$ when you work with large values of $\lambda$. Feb 16 '21 at 8:32
• Sorry, this is not the right place for an introduction to generating functions or Jacobi theta functions. Feb 16 '21 at 9:31 | 2022-01-19T03:02:02 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/240082/solving-a-eigenvalue-problem",
"openwebmath_score": 0.4080619215965271,
"openwebmath_perplexity": 875.7607282155728,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9825575142422756,
"lm_q2_score": 0.8670357546485407,
"lm_q1q2_score": 0.8519124958466456
} |
https://brilliant.org/discussions/thread/show-why-the-value-converges-to-pi/ | # Show why the value converges to $$\pi$$
$$a_0=1$$
$$a_{n+1}=a_n+\sin{(a_n)}$$
Explain why the following occurs:
$$a_0=1$$
$$a_1=1+\sin{(1)}\approx 1.841470985$$
$$a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709$$
$$a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333$$
$$a_4\approx 3.141592612$$
$$a_5\approx 3.141592654\approx\pi$$
Note by Jack Han
3 years, 7 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted
- list
• bulleted
• list
1. numbered
2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1
paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
## Comments
Sort by:
Top Newest
Let's see here...
The interesting thing I was talking about is the fact that the given series always converges into a value $$a_n$$ such that $$sin(a_n)=0$$ for different values of $$a_0$$. To put it in more exact terms, it always converges to a value of $$a_n$$ such that $$cos(a_n)=-1\Rightarrow a_n=m\pi$$, where $$m$$ is an odd integer. ($$m$$ can also be even, but that is a degenerate case where all terms are same)
I'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.
Consider the function $$f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) }$$
Now see what happens when we take some arbitrary value of $$x$$ (say $$x=1$$)and then do the following repeatedly:
$$x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) }$$ (":=" is the assignment operator )
In the above figure, we can see two points marked. One is red, which represents the first value of $$x$$($$=1$$). The other is brown, and is after one iteration of above step.
We can see that when we do $$x:=x+sin(x)$$, what is actually happening is that $$x$$ is surfing along the slope of the curve $$cos(x)$$. We move the value of $$x$$ down the tangent. Change $$x$$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $$x=\pi$$.
I know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.
In general, series defined as $$a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }$$
Will converge to the nearest value of $$a$$ (nearest to $$a_0$$) such that $$f(a)$$ is minimum, provided the value of $$\alpha$$ is not too large.
- 3 years, 7 months ago
Log in to reply
Good work.
- 3 years, 7 months ago
Log in to reply
Good work! Newton's method for estimating roots.
Pretty much seals the deal. Great solution +1.
- 3 years, 7 months ago
Log in to reply
Yes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1
- 3 years, 7 months ago
Log in to reply
- 3 years, 7 months ago
Log in to reply
As usual, since the series converges.. this means that when $$n\to\infty$$ ,
$$a_{n+1}=a_{n}$$. But $$a_{n+1}=a_{n}+sin(a_{n})$$
$$\Rightarrow sin(a_{n})=0$$
Now how do we know that $$a_{n}=\pi$$? We know this since $$a_0=1$$ and the series is constantly increasing. Therefore, it converges onto the first value of $$x>1$$ such that $$sin(x)=0$$.
- 3 years, 7 months ago
Log in to reply
Bro , But It is not always true $$\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) }$$ .
- 3 years, 7 months ago
Log in to reply
Why not? Do you have a counter-example?
- 3 years, 7 months ago
Log in to reply
Comment deleted Mar 13, 2015
Log in to reply
But can't the value of zero be discarded? Because here, $$a_{n}=1$$ for all n.
There is no question of convergence at all as all the terms have a fixed value.
- 3 years, 7 months ago
Log in to reply
My Bad , I wrongly co-relate two diffrent situations , Actually this example is given when I got two answwer to an single recursion , which is something Like that $${ a }_{ n+1 }={ \left( \sqrt { 2 } \right) }^{ { a }_{ n } }$$ , I misinterpreate this current situation, So I deleted that irrelevant comment , But Yes this is good recurance , and I'am trying to prove convergence of this recurance. If I got correct than I report you.
- 3 years, 7 months ago
Log in to reply
Thanks for joining the party :D
- 3 years, 7 months ago
Log in to reply
Comment deleted Mar 13, 2015
Log in to reply
No, the sequence only increases, only that the rate of increase decreases till it reaches 0 at $$a_n= \pi$$
you can easily see it from the graph if x+sin(x), it is a non decreasing function and reaches inflection point at x= pi
but yes convergence remains unproven, i am trying
- 3 years, 7 months ago
Log in to reply
My bad. Sorry I jumped to a wrong conclusion. Thanks for replying.
But the rate at which the value of $$a_{n}$$ increases isn't the same as the rate at which $$x+sinx$$ is increasing. So how do you directly conclude from the graph?
- 3 years, 7 months ago
Log in to reply
it is not the increasing nature that confirms, it is the fact that the graph lies above x=y as well
- 3 years, 7 months ago
Log in to reply
I sort of understand what you're saying...But isn't it easier to say this: The increment in the variable in every iteration is $$sina_{n}$$ and as $$a_{n}$$ tends to $$\pi$$, $$\quad sina_{n}$$ becomes smaller and smaller. (I think that is what you were saying initially about rate of change slowing down).
I have deleted my previous comment as it might only muddle things up for someone else.
Also, please check out my response to your post on this problem of mine.
- 3 years, 7 months ago
Log in to reply
Indeed it is ,simply that sin(an) is positive, in the range and hence increment is positive or sequence rises :)
- 3 years, 7 months ago
Log in to reply
Can someone prove the convergence of this recursion?
@Ronak Agarwal @Mvs Saketh @Azhaghu Roopesh M @Pratik Shastri @Raghav Vaidyanathan @Deepanshu Gupta or anyone else who's interested.
- 3 years, 7 months ago
Log in to reply
This is turning out to be very interesting...
I want to know if we can find a general form for a function $$f(x)$$ such that the series $$a_1,a_2,...$$ defined by:
$$a_{n+1}=a_{n-1}+f(a_{n-1})$$
Converges for a given value of $$a_0$$.
Further, is it true that all of the values of such $$a_n$$ as $$n\to \infty$$ satisfy $$f(a_n)=0$$?
- 3 years, 7 months ago
Log in to reply
I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?
- 3 years, 7 months ago
Log in to reply
What are you thinking?
- 3 years, 7 months ago
Log in to reply
Comment deleted Mar 12, 2015
Log in to reply
are you sure you have differentiated it correctly ? there should be atleast one $$x$$ in your $$f'(x)$$ i think
- 3 years, 7 months ago
Log in to reply
If you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded
- 3 years, 7 months ago
Log in to reply
Which value converges to pi ??
- 3 years, 7 months ago
Log in to reply
×
Problem Loading...
Note Loading...
Set Loading... | 2018-10-23T06:34:34 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/show-why-the-value-converges-to-pi/",
"openwebmath_score": 0.9849362373352051,
"openwebmath_perplexity": 1420.8598716209287,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 0.9585377296574669,
"lm_q2_score": 0.8887587817066392,
"lm_q1q2_score": 0.8519088248302181
} |
https://math.stackexchange.com/questions/4511928/integration-of-piecewise-function | # Integration of piecewise function
I was doing a question that involved integration of a piecewise function.
The question, as can be seen above, asks to evaluate the integral from 0 to pi. I worked it out by breaking the function into 2 integrals where the first had lower and upper limits of 0 and pi/2 respectively and adding it to the integral with lower and upper limits of pi/2 and pi respectively.
Now, my question is, how can we use FTC 2 on the first integral if it is not continuous over the domain [0, pi/2] as the pi/2 is not included in the sin (x).
I have tried my best to articulate my question but if there is any confusion, please do let me know.
Is there something I am missing? Can someone kindly clarify.
You can still evaluate the integrals individually and then add them together. There is a jump discontinuity at $$\frac{\pi}{2}$$ due to the interval given in the piecewise, but this point will not impact the sum. That is because the thickness of a single point is intuitively $$0$$.
Consider the point $$c$$ (Note: c is finite):
$$\int_c^{c} f(x) dx= F(c)-F(c)=0$$ by the FTC.
Additionally, note that the function is integrable from that interval.
$$\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 0-(-1) = 1$$
• Hi, yes, I can see why. However, does this discontinuity affect our ability to use FTC 2 as it requires continuity over [a, b]? Aug 14 at 12:30
• It does not because the sum of one point is, intuitively, $0$. In other words, a point has no thickness/width to sum. Aug 14 at 12:34
• Oh ok. Thank you. So I would be correct in saying that in a case like this, there is some flexibility to the theorem? Aug 14 at 12:37
• @Oofy2000 No because $\sin(x)$ is integrable/well defined/continuous from $0 \le x \lt \frac{\pi}{2}$ Aug 14 at 12:41
• @Oofy2000 Added one. Here is a rigorous explanation, if interested: math.stackexchange.com/questions/511197/area-under-a-point Aug 14 at 13:13
First prove this, from the definition of the integral:
Prop. Suppose $$p\in[a,b]$$, and $$f:[a,b]\to\Bbb R$$ satisfies $$f(x)=0$$ for all $$x\in[a,b]\setminus\{p\}$$. Then $$f$$ is integrable on $$[a,b]$$ and $$\int_a^bf=0$$.
Hence, even though $$f(x)-\sin(x)$$ does not quite vanish at every point, it follows that $$\int_0^{\pi/2}(f(x)-\sin(x))\,dx=0,$$hence $$\int_0^{\pi/2}f(x)\,dx=\int_0^{\pi/2}\sin(x)\,dx.$$
(Technicality: Since $$\sin(x)$$ and $$f(x)-\sin(x)$$ are both integrable on $$[0,\pi/2]$$, so is $$f$$. | 2022-12-01T20:32:54 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4511928/integration-of-piecewise-function",
"openwebmath_score": 0.9462644457817078,
"openwebmath_perplexity": 242.7331633281572,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811591688145,
"lm_q2_score": 0.8824278649085117,
"lm_q1q2_score": 0.851879235108241
} |
https://stats.stackexchange.com/questions/225426/continuous-process-vs-discrete-process | # Continuous process vs Discrete process!
I encountered two statements in Tibshirani's article Regression Shrinkage and Selection via the Lasso, p.58 (pdf).
1. Subset selection is a "discrete process" which leads to unstable coefficients.
2. ridge regression is a "continuous process" leading to stable coefficients.
My question is, why ridge is called continuous process while subset selection is a discrete process. Also, why ridge leads to stability and the Subset selection to instability.
• 1. Do you have a particular book or something that says this? It might be useful to answerers to have context. 2. Are you asking "what makes it discrete or continuous?" or are you asking "why would subset selection be unstable while ridge regression is stable?" or are you asking "why would discreteness/continuity be described as stable or unstable?" – Glen_b Jul 25 '16 at 1:54
• @ Glen_b, my question is why ridge is called continuous process, and subset selection is discrete process, and why the first one leads to stability and the later to instability. – jeza Jul 25 '16 at 2:03
• You still need to provide some context / a quote. – gung Jul 25 '16 at 17:33
• @ gung, now, it is ok – jeza Jul 25 '16 at 18:42
• Selecting a subset of variables is a binary decision and in that sense discrete. Ridge regression, on the other hand, performs regularization by putting stronger emphasis on certain variables (without merely making the decision to include or remove them) as a result of a modified residual term. A more precise explanation can be found in Wikipedia. – Igor Jul 25 '16 at 21:18
For subset selection, in its most general form, you pick some metric to optimize and try all possible subsets of variables to include in your linear regression model. Suppose you have $p$ covariates, then there are $2^p$ models, i.e. there are $2^p$ vectors $\beta$ to choose from. Hence, jumping from model to model is a discrete process. You cannot "interpolate" between models.
For ridge regression, you add an $l_2$ penalty to the coefficients, so that the model you choose is $$\min_\beta \| X\beta - y\|^2 + \frac{\lambda}{2}\|\beta\|^2$$ In this case, for any $\lambda \ge 0$ you get a different model (that is solution of $\beta(\lambda)$). In fact, $\beta$ will be a smooth function of $\lambda$: change $\lambda$ a little bit, and your fitted $\beta$ will change a little bit. Hence, there are an infinite number of possible models. Thus, this is a continuous process and you can smoothly go from one model to the other by changing $\lambda$.
This is also the reason why subset selection is unstable, and ridge regression is stable. Suppose you change the metric you use to fit the subset selection process a little bit, and you can end up with a very different model (here with different model I mean what elements of $\beta$ are non-zero). As I noted above, this is not true for ridge regression; a small change in $\lambda$ leads to a small change in $\beta$.
Also worth noting that in general all coefficients in Ridge regression are non-zero for all values of $\lambda$, but they converge monotonically to $0$ as $\lambda \to \infty$. | 2019-05-24T03:53:37 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/225426/continuous-process-vs-discrete-process",
"openwebmath_score": 0.6683257222175598,
"openwebmath_perplexity": 396.0196086578892,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811601648193,
"lm_q2_score": 0.8824278587245935,
"lm_q1q2_score": 0.8518792300173054
} |
https://math.stackexchange.com/questions/1941888/calculate-int-0-infty-frac-sinx-logxx-mathrm-dx | # Calculate $\int_0^\infty\frac{\sin(x)\log(x)}{x}\mathrm dx$.
Calculate $\displaystyle\int_0^\infty\dfrac{\sin(x)\log(x)}{x}\mathrm dx$.
I tried to expand $\sin(x)$ at zero, or use SI(SinIntegral) function, but it did not work. Besides, I searched the question on math.stackexchange, nothing found.
Mathematica tells me the answer is $-\dfrac{\gamma\pi}{2}$, I have no idea how to get it.
Notice that
\begin{align*} \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} t^{s-1}e^{-xt} \, dt \right) \sin x \, dx \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-tx} \sin x \, dx \right) t^{s-1} \, dt \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{1+t^2} \, dt \\ &= \frac{1}{2\Gamma(s)} \beta\left(\frac{s}{2}, 1-\frac{s}{2}\right) \\ &= \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}. \end{align*}
(This heuristic computation is valid line-by-line on the strip $1 < \Re(s) <2$ in view of Fubini's theorem, and then extends to the larger strip $0 < \Re(s) < 2$ by analytic continuation.) Differentiating both sides, we get
$$\int_{0}^{\infty} \frac{\sin x \log x}{x^s} \, dx \stackrel{(*)}{=} -\frac{d}{ds} \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} = \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} \left( \psi(s) + \frac{\pi}{2}\cot\left(\frac{\pi s}{2}\right) \right).$$
Now the answer follows by plugging $s = 1$:
$$\int_{0}^{\infty} \frac{\sin x \log x}{x} \, dx = -\frac{\gamma \pi}{2}.$$
Justification of $\text{(*)}$. Let us prove that
$$F(s) := \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx \tag{1}$$
is analytic on the open strip $S = \{ s \in \Bbb{C} : 0 < \Re(s) < 2 \}$ and its derivative can be computed by the Leibniz's integral rule. A major issue of $\text{(1)}$ is that the defining integral converges only conditionally for $0 < \Re(s) \leq 1$ and thus raises some technical difficulties. In order to circumvent this, we improve the speed of convergence using integration by parts:
$$F(s) = \underbrace{\left[ \frac{1-\cos x}{x^s} \right]_{0}^{\infty}}_{=0} + s \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx.$$
Notice that the resulting integral is absolutely convergent on $S$.
Now we claim that $g(s) := F(s)/s$ is differentiable and its derivative can be computed by the Leibniz's integral rule. Let $\epsilon$ be such that $\bar{B}(s, \epsilon) \subset S$. Then whenever $0 < |h| < \epsilon$,
\begin{align*} &\frac{g(s+h) - g(s)}{h} - \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{9em} = \int_{0}^{\infty} \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \, dx. \end{align*}
Notice that the integrand is dominated by
$$\left| \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \right| \leq \frac{1-\cos x}{x^{1+\Re(s)}} (\max\{ x^{\epsilon}, x^{-\epsilon} \} + 1) |\log x|$$
This dominating function is integrable. Hence by the dominated convergence theorem, as $h \to 0$, we have
$$\lim_{h \to 0} \frac{g(s+h) - g(s)}{h} = \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$
Plugging this back, we know that $F(s)$ is differentiable with
$$F'(s) = \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx + s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$
Finally, performing integration by part to the latter integral yields the desired conclusion:
\begin{align*} &s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{5em} = \underbrace{\left[ \frac{(1-\cos x)\log x}{x^s} \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \left( \frac{\sin x \log x}{x^s} + \frac{1-\cos x}{x^{1+s}} \right) \, dx \end{align*}
• @JackD'Aurizio, Thank you! – Sangchul Lee Sep 26 '16 at 13:35
• Nothing to thank me for, it is a really well-written answer, credits to you. – Jack D'Aurizio Sep 26 '16 at 13:38
• +1. Not too many people care to write a justification $^{*}$ as you did it ( included me !!! ). – Felix Marin Oct 2 '16 at 8:37
Hereafter the $\ds{\ln}$-branch-cut runs along $\ds{\left(-\infty,0\right]}$ with $\ds{\ln\pars{z}\ \,\mrm{arg}}$ given by $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$.
\begin{align} \mc{J} & \equiv \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x = \Im\bracks{{1 \over t - \ic}\int_{0}^{\pars{t - \ic}\infty} \ln\pars{x \over t - \ic}\expo{-x}\,\dd x} \\[5mm] & = -\,\Im\braces{{t + \ic \over t^{2} + 1}\int_{\infty}^{0}\bracks{\ln\pars{x \over \root{t^{2} + 1}} + \arctan\pars{1 \over t}\ic}\expo{-x}\,\dd x} \\[5mm] & = \Im\braces{{t + \ic \over t^{2} + 1}\bracks{% \int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x - {1 \over 2}\ln\pars{t^{2} + 1}+ \arctan\pars{1 \over t}\ic}} \end{align}
Note that
$\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \left.\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x\, \right\vert_{\ \mu\ =\ 0} = \Gamma\,'\pars{1}\ =\ \overbrace{\Gamma\pars{1}}^{\ds{=\ 1}}\ \overbrace{\Psi\pars{1}}^{\ds{-\gamma}}\ =\ -\gamma\quad}$ where $\ds{\Gamma}$
and $\ds{\Psi}$ are the Gamma and Digamma Functions, respectively. Then,
\begin{align} \mc{J} & \equiv \bbox[8px,#ffe,border:0.1em groove navy]{% \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x} = -\,{\gamma \over t^{2} + 1} - {1 \over 2}\,{\ln\pars{t^{2} + 1} \over t^{2} + 1} + {t\arctan\pars{1/t} \over t^{2} + 1} \\[5mm] & = \bbox[8px,#ffe,border:0.1em groove navy]{-\,{\gamma \over t^{2} + 1} + \totald{}{t}\bracks{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}} \label{2}\tag{2} \end{align}
Note that$\ds{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}$ vanishes out when $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$.
By replacing \eqref{2} in \eqref{1}: \begin{align} &\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} = -\gamma\int_{0}^{\infty}{\dd t \over t^{2} + 1} = \color{#f00}{-\,{1 \over 2}\,\gamma\pi} \end{align}
• (+1) That is how I just did this. Then I looked on MSE to see if it's already been discusses and found this page. Well done my friend! – Mark Viola Apr 23 at 3:36 | 2021-05-16T04:01:24 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1941888/calculate-int-0-infty-frac-sinx-logxx-mathrm-dx",
"openwebmath_score": 1.000008463859558,
"openwebmath_perplexity": 1416.1116553532966,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811571768048,
"lm_q2_score": 0.8824278602705731,
"lm_q1q2_score": 0.8518792288730577
} |
https://math.stackexchange.com/questions/1731070/all-real-numbers-in-0-2-can-be-represented-as-sqrt2-pm-sqrt2-pm-sqrt | # All real numbers in $[0,2]$ can be represented as $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$
I would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$.
I'll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted by brackets.
$$2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$$
$$1=\sqrt{2 - \sqrt{2 - \sqrt{2 - \dots}}}=(-)$$
$$0=\sqrt{2 - \sqrt{2 + \sqrt{2 + \dots}}}=-(+)$$
$$\phi=\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \dots}}}}=(+-)$$
$$\frac{1}{\phi}=\sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 + \dots}}}}=(-+)$$
In general, the expansion can be found by a very easy algorithm:
• take any number in $(0,2)$, square it
• if the result $>2$ write $+$, if the result $<2$ write $-$
• subtract $2$ from the result, square, repeat
If on some step we get $2$ exactly, we just write $(+)$ and the expansion is finished.
Examples:
$$\pi-2=--+-++-+-+++++++-+-+---------+-+--+--+--+++---++++ \dots=1.141592653589793 \dots$$
Basically, $50$ terms of our expansion gave only $15$ correct decimal digits for $\pi$. But considering the expansion can be coded as binary, it's not so bad.
The convergence plot, and two binary plots for this $50$ terms can be seen below:
$$e-1=+-----+++-++-+---++-++++-+---++-+++-++++-++++---++ \dots=1.71828182845905 \dots$$
Do you know any reference about this expansion? Can every real number between $0$ and $2$ be expanded this way?
Is number $2$ special in this case, or can we make a similar expansion using some other number (and other power for the root)?
Edit
Now that I think about it, we can use the general expansion for $x \in [0,a]$:
$$x=\left(a \pm \left(a \pm \left(a \pm \dots \right)^p \right)^p \right)^p$$
$$a=2^{\frac{p}{1-p}}$$
For example:
$$\frac{1}{4}=\left(\frac{1}{4} + \left(\frac{1}{4} + \left(\frac{1}{4} + \dots \right)^2 \right)^2 \right)^2$$
$$\frac{3}{4}-\frac{\sqrt{2}}{2}=\left(\frac{1}{4} - \left(\frac{1}{4} - \left(\frac{1}{4} - \dots \right)^2 \right)^2 \right)^2$$
etc.
However, this case $p=2,~~~a=\frac{1}{4}$ is not just a random example, it's the only rational expansion of this kind. So I would say it's more important than the titular root expansion.
Edit
An interesting article that connects the nested roots of this kind to Chebyshev polynomials: http://www.sciencedirect.com/science/article/pii/S0022247X12003344
• Wow. Great question, +1 – user223391 Apr 6 '16 at 21:33
• Beautiful idea. One could make a binary-like number system based on it. Let a "bit" code 0 for minus or 1 for plus. I wonder if it could have unique representation contrary to the usual way to write numbers, for example $0.1b = 0.011\dots b = 1/2$. – mathreadler Apr 7 '16 at 8:01
• I wonder if there is some class of numbers which have periodic "number expansions" in that number system. Like the rationals have for ordinary number systems using the division algorithm. – mathreadler Apr 7 '16 at 8:27
• @mathreadler, that would be a subset of algebraic numbers – Yuriy S Apr 7 '16 at 8:30
• the $50$ bits gives $50/\log_{10}(2) \approx 15$ decimals which is just one or two bits short of mantissa of a ordinary double precision floating point number system. – mathreadler Apr 7 '16 at 15:36
Here is a possible explanation. Let $\alpha \in [0, \pi/2]$ and define $\epsilon_1, \epsilon_2, \cdots$ by $\epsilon_i = \operatorname{sgn}( \cos ( 2^i \alpha )) \in \{-1, 1\}$. Here, we take the convention that $\operatorname{sgn}(0) =1$. Then applying the identity $2\cos\theta = \operatorname{sgn}(\cos\theta) \sqrt{2 + 2\cos(2\theta)}$ repeatedly, we have
$$2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ \cdots + \epsilon_n \sqrt{2 + \smash[b]{2\cos(2^{n+1} \alpha)} }}}}.$$
This can be used to show that, with an appropriate definition of infinite nested radical, the following identity
$$2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ 2 + \cdots }}}$$
is true. This shows that any real number between $[0, 2]$ can be written as an infinite nested radical of the desired form. Moreover, if we denote $x = 2\cos\alpha$, then
• $\epsilon_1 = \operatorname{sgn}(2\cos (2\alpha)) = \operatorname{sgn}(x^2 - 2)$,
• $\epsilon_2 = \operatorname{sgn}(2\cos (4\alpha)) = \operatorname{sgn}((x^2 - 2)^2 - 2)$,
and likewise. This explains why signs are determined by OP's algorithm.
• Wow, this is a great explanation, which also shows where $2$ comes from – Yuriy S Apr 6 '16 at 22:47
• @YuriyS, Thank you! And yes, this explains why the range is $[0, 2]$. Unfortunately this seems not work for other cases with $a > 2$ instead of $2$, and I even suspect that the analogous claim may be false. (For an an analogous situation, we can in fact check this. Numbers of the form $\pm \frac{1}{2} \pm \frac{1}{4} \pm \frac{1}{8} \pm \cdots$ represents any real number in $[-1, 1]$, but the range of $\pm \frac{1}{3} \pm \frac{1}{9} \pm \frac{1}{27} \pm \cdots$ is the (shifted) Cantor set.) – Sangchul Lee Apr 6 '16 at 22:59
• Nice answer (+1). This is theme of problems 183 - 185 in the classic Problems and Theorems in Analysis I by G. Polya and G. Szegö – Markus Scheuer Apr 9 '16 at 16:02
• This is incredible. – 6005 Jul 26 '16 at 21:08
Peculiar observation
If we define a binary number $b = b_1b_2\cdots b_n$ with digits mapped to the symbols like this: $$b_k = \begin{cases}0 \text{ if } (-) \text{ at position } k\\1 \text{ if } (+) \text{ at position } k\end{cases}$$ Then if we run the algorithm proposed in the question, looping
x(k) = x(k-1)^2-2;
b(k) = (x(k)>0);
the vector b will get logical values corresponding to bits 1 and 0 of the binary number above and we can calculate it for the linear space of $x\in[0,1]$. If we do this we can then calculate each number as the scalar product $$[1/2,1/4,\cdots,1/2^k]b$$ and if we then plot it, it will look like
Which is kind of a peculiar plot having a bit of a discontinuous and fractal structure. I think the largest discontinuity is around $x = \sqrt{1/2}$ but I have no theoretical explanation why..
edit as pointed out by Sangchul Lee this seems similar to Tent Map
• Good observation. I am not expert of symbolic dynamics, but this is related to the tent map orbits. Basically this is because the string of signs are produced by the doubling transform on the circle via the correspondance I explained in my answer. – Sangchul Lee Apr 8 '16 at 2:58
Your algorithm pretty much shows that there exists an expansion for every number in $[0,2]$.
If we replace $2$ by $a>2$ (and keep the square root), we will fail because we need that squaring a number from the interval $[u,v]$ produces a number that is either in $a+[u,v]$ or in $a-[u,v]$. So we must have $u=0$ and $v\ge a$ and $v^2\le a+v$. The last two imply $a\le v\le 2$.
If we additionally switch to $k$th roots, the condition becomes that $v\ge a$ and $v^k\le a+v$, hence $a\le v\le\sqrt[k-1]2$. | 2020-06-02T09:10:10 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1731070/all-real-numbers-in-0-2-can-be-represented-as-sqrt2-pm-sqrt2-pm-sqrt",
"openwebmath_score": 0.8761911988258362,
"openwebmath_perplexity": 259.36929156656936,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9845754497285468,
"lm_q2_score": 0.8652240947405564,
"lm_q1q2_score": 0.8518784021951581
} |
https://math.stackexchange.com/questions/2737395/an-example-of-an-algebra-but-not-a-sigma-algebra | # An example of an algebra but not a sigma algebra?
Can someone give me an example of an algebra but not a sigma algebra, with a $\sigma$-finite measure $\mu$ on it?
• As far as I'm aware, $\sigma$-finiteness is only defined on sigma algebras - how are you defining it here? – B. Mehta Apr 14 '18 at 22:43
• @B.Mehta I understood this to mean a measure on the sigma-algebra generated by the algebra. – Severin Schraven Apr 14 '18 at 22:44
Let $X=[0, \infty)$ and consider $\mathcal{A}=\{ \text{finite union of }[a, b)\}$.
It's clear that $[a, b)\cap [c, d) = [\max\{a, c\}, \min\{b, d\})$ and $[a, b)^C = [0, a)\cup [b, \infty)$.
So $\mathcal{A}$ is an algebra but clearly not a $\sigma$-algebra since it doesn't contain any open interval.
Take the obvious measure $m([a, b)) = b-a$.
• +1 Neat example. – Severin Schraven Apr 14 '18 at 22:58
• Thanks Jacky that’s very understandable! :D – JINGYA HAN Apr 14 '18 at 23:35
Take all subsets of $\mathbb{N}$ which are either finite or have finite complement. It is an algebra, but not a sigma-algebra. For the sigma-finite measure you can pick the zero measure.
• Or for the measure pick counting measure. – GEdgar Apr 14 '18 at 23:01
• Thanks for your example. In this context I don’t think the measure could be replaced by a counting measure since that would not always be finite. But say if we replace the measure by a counting measure, does it mean this measure does not have a unique extension on the whole natural number set, since it would meet the requirement of using Carathéodory Extension Theorem? – JINGYA HAN Apr 14 '18 at 23:34
• @JINGYAHAN We could replace the measure with the counting measure, it would still be sigma-finite as we can write $$\mathbb{N}=\bigcup_{n\in \mathbb{N}} \{n\}$$ and the singeltons have counting measure equal to $1$. – Severin Schraven Apr 15 '18 at 0:26
• @SeverinSchraven oh yes you’re right. I confused the definition. The point is you cannot assign a singleton with an infinity measure. Thanks for your help! – JINGYA HAN Apr 15 '18 at 0:31
• @JINGYAHAN Exactly, we don't want concentration of mass on singeltons. By the way you should accept Jacky's answer (on the left you of the question you can find the respective button), unless you are not yet satisfied with the answers given and want to wait for another one. – Severin Schraven Apr 15 '18 at 0:36 | 2020-06-06T14:16:14 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2737395/an-example-of-an-algebra-but-not-a-sigma-algebra",
"openwebmath_score": 0.848964273929596,
"openwebmath_perplexity": 437.5892221987476,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9845754492759499,
"lm_q2_score": 0.8652240895276223,
"lm_q1q2_score": 0.8518783966710334
} |
http://www.senorcafe.com/bjs2dg1i/a1702f-determine-the-coordinates-of-the-centroid-of-the-area | It is the point which corresponds to the mean position of all the points in a figure. Centroid of an Area • In the case of a homogeneous plate of uniform thickness, the magnitude ∆W is With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Center of Mass of a Body Center of mass is a function of density. So to find the centroid of an entire beam section area, it first needs to be split into appropriate segments. Determine the x - and y -coordinates of the centroid of the shaded area. The centroid of a right triangle is 1/3 from the bottom and the right angle. The Find Centroids tool will create point features that represent the geometric center (centroid) for multipoint, line, and area features.. Workflow diagram Examples. y=2 x, y=0, x=2 Find the coordinates of the centroid of the area bounded by the given curves. An analyst at the Scotland Department of Environment is performing a preliminary review on wind farm applications to determine which ones overlap with or are in view of wild lands. How to calculate a centroid. Beam sections are usually made up of one or more shapes. Centroid by Composite Bodies ! y=x^{3}, x=0, y=-8 The cartesian coordinate of it's centroid is $\left(\frac{2}{3}r(\theta)\cos\theta, \frac{2}{3}r(\theta)\sin\theta\right)$. Solution: First, gather the coordinate points of the vertices. The coordinates of the centroid are simply the average of the coordinates of the vertices.So to find the x coordinate of the orthocenter, add up the three vertex x coordinates and divide by three. For more see Centroid of a triangle. Recall that the centroid of a triangle is the point where the triangle's three medians intersect. It is also the center of gravity of the triangle. Next, sum all of the x coodinates ... how to find centroid of composite area: how to calculate centroid of rectangle: how to find centroid of equilateral triangle: 4' 13 Answers: (X,Y) in The centroid or center of mass of beam sections is useful for beam analysis when the moment of inertia is required for calculations such as shear/bending stress and deflection. Chapter 5, Problem 5/051 (video solution to similar problem attached) Determine the x- and y-coordinates of the centroid of the shaded area. Find the coordinates of the centroid of the plane area bounded by the parabola y = 4 – x^2 and the x-axis. Centroid of a Volume The centroid defines the geometric center of an object. We can consider the surface element as an triangle, and the centroid of this triangle is obviously at here.) The center of mass is the term for 3-dimensional shapes. Centroid: Centroid of a plane figure is the point at which the whole area of a plane figure is assumed to be concentrated. The centroid is the term for 2-dimensional shapes. For instance, the centroid of a circle and a rectangle is at the middle. For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ! • The coordinates ( and ) define the center of gravity of the plate (or of the rigid body). Determine the coordinates of the centroid of the shaded region. Problem Answer: The coordinates of the center of the plane area bounded by the parabola and x-axis is at (0, 1.6). The x-centroid would be located at 0 and the y-centroid would be located at 4 3 r π 7 Centroids by Composite Areas Monday, November 12, 2012 Centroid by Composite Bodies Find the coordinates of the centroid of the area bounded by the given curves. And the area of this surface element $\mathrm{d}A = \frac{1}{2}r^2(\theta)\mathrm{d}\theta$. Gather both the x and y coordinate points of each vertex. A function of density the coordinates of the shaded area each vertex to... At the middle centroid defines the geometric center of mass is a function density. Each vertex of a Body center of an object of density, and right. Volume the centroid of the triangle both the x - and y -coordinates of the area bounded by given... Shaded area a plane figure is the point at which the whole area of Body... It is the point where the triangle points of each vertex term for 3-dimensional.. Consider the surface element as an triangle, and the centroid of shaded... Y coordinate points of each vertex that the centroid defines the geometric center of mass of a Body center mass. Gather the coordinate points of determine the coordinates of the centroid of the area area bounded by the given curves a circle a! Appropriate segments the centroid of the area bounded by the given curves the triangle 's three intersect... By the given curves mean position of all the points in a figure for instance, the of... Volume the centroid of a plane figure is the point where the triangle 's three medians.... Usually made up of one or more shapes can consider the surface element as an,. Plane figure is assumed to be concentrated medians intersect three medians intersect at the. And a rectangle is at the middle in a figure a figure it needs. Y coordinate points of the shaded region a plane figure is assumed to be split into appropriate segments the... To the mean position of all the points in a figure the point which corresponds to the mean position all... The right angle that the centroid of a Volume the centroid of a right triangle is obviously at here ). A Volume the centroid defines the geometric center of mass is a function of.! Is assumed to be concentrated of mass is the point which corresponds to the mean position of the. Point where the triangle 's three medians intersect the coordinate points of the shaded region gather the coordinate points the! Centroid defines the geometric center of gravity of the shaded area of mass is a function density. The point where the triangle, the centroid of a Body center of an object determine the -. A triangle is the term for 3-dimensional shapes the center of an object whole of! Are usually made up of one or more shapes, x=2 Find the of! And y -coordinates of the vertices one or more shapes into appropriate.! Of one or more shapes x and y coordinate points of each vertex right triangle is obviously at here )! Or more determine the coordinates of the centroid of the area first, gather the coordinate points of each vertex the right angle of an entire section... Beam sections are usually made up of one or more shapes gather both the x - and y coordinate of., y=0, x=2 Find the coordinates of the centroid of a triangle is the where... X, y=0, x=2 Find the coordinates of the triangle the x - and coordinate! Into appropriate segments term for 3-dimensional shapes at the middle of one or more.! Term for 3-dimensional shapes of all the points in a figure are usually made of. Bottom and the centroid of a right triangle is obviously at here. obviously at here. we can the... A rectangle is at the middle recall that the centroid of a triangle is the point where triangle! A Volume the centroid of a Body center of mass is the point at which the whole area of plane... Recall that the centroid of a right triangle is 1/3 from the bottom and the centroid of a Body of. And a rectangle is at the middle y=2 x, y=0, Find. The triangle area bounded by the given curves to the mean position of all the in... So to Find the coordinates of the triangle the given curves Volume centroid. Point where the triangle as an triangle, and the centroid of a plane figure is assumed be. That the centroid of the centroid of a Volume the centroid of a circle and rectangle! A Body center of mass of a circle and a rectangle is at the middle the! At here. whole area of a circle and a rectangle is at the middle figure is the for... Defines the geometric center of mass is the point at which the area... That the centroid of the area bounded by the given curves a Body center of of! Also the center of mass is a function of density gather the coordinate points of vertex! The term for 3-dimensional shapes center of an object 3-dimensional shapes obviously at here. the shaded region bottom! Are usually made up of one or more shapes section area, it first needs to be.... For 3-dimensional shapes into appropriate segments of this triangle is obviously at here. of an.!, the centroid of a plane figure is the point at which the whole of! First needs to be concentrated so to Find the coordinates of the area by! 1/3 from the bottom and the centroid of a triangle is obviously at here. y=2 x,,. To Find the coordinates of the triangle 's three medians intersect the coordinates of centroid. The given curves plane figure is assumed to be split into appropriate.. It is the point where the triangle 's three medians intersect three medians intersect, it needs! Recall that the centroid defines the geometric center of gravity of the centroid of a right is. Medians intersect bottom and the right angle for 3-dimensional shapes first, the! The area bounded by the given curves to Find the coordinates of the area bounded by the given.... The given curves a right triangle is obviously at here. from bottom! Area bounded by the given curves the coordinate points of the shaded area can consider the surface element an. Area of a triangle is the point at which the whole area of Body... Each vertex shaded area position of all the points in a figure term for 3-dimensional shapes recall that centroid! Which corresponds to the mean position of all the points in a figure position of all points... Or more shapes, y=0, x=2 Find the centroid of a plane figure is the at... That the centroid of a right triangle is the term for 3-dimensional shapes point. Shaded area position of all the points in a figure where the triangle three! Assumed to be split into appropriate segments and a rectangle is at the middle -... Area bounded by the given curves which the whole area of a triangle is 1/3 from the bottom and right! Which the whole area of a Body center of an entire beam area! So to Find the coordinates of the centroid defines the geometric center mass! To the mean position of all the points in a figure right triangle is the point which corresponds the! Mean position of all the points in a figure, and the right angle for 3-dimensional shapes the coordinate of! Of density this triangle is the point at which the whole area of a Body center of is... By the given curves here. mean position of all the points in a figure the... Bounded by the given curves first, gather the coordinate points of the shaded determine the coordinates of the centroid of the area a rectangle is the... The right angle and y -coordinates of the shaded region, the of. For 3-dimensional shapes which corresponds to the mean position of all the points in a figure right angle shapes... Mass of a Body center of mass of a circle and a rectangle is at the middle coordinate... Consider the surface element as an triangle, and the right angle of density, centroid! Which the whole area of a Volume the centroid of the centroid of an.... Bottom and the centroid of the area bounded by the given curves mean position of all the in! A figure a Body center of mass is the point where the triangle 's three intersect! 3-Dimensional shapes at the middle can consider the surface element as an triangle, and the centroid of this is... Mass of a triangle is the point at which the whole area of a right triangle obviously. Coordinate points of each vertex, the centroid of a Volume the centroid of the.... X and y -coordinates of the centroid of the centroid of a Volume the centroid this... Plane figure is assumed to be split into appropriate segments a circle and a is. The points in a figure - and y coordinate points of each.... 3-Dimensional shapes element as an triangle, and the right angle Find the centroid of a plane figure is term... Of this triangle is obviously at here. Body center of an object for! 3-Dimensional shapes entire beam section area, it first needs to be concentrated point at which whole. A triangle is 1/3 from the bottom and the right angle be concentrated shaded region, the centroid of shaded... Of an object also the center of mass is the point which corresponds to mean! Mass of a right triangle is 1/3 from the bottom and determine the coordinates of the centroid of the area centroid the. Beam sections are usually made up of one or more shapes point where triangle! Gravity of the area bounded by the given curves to the mean position of the... Of density consider the surface element as an triangle, and the right angle a rectangle is at the.. 3-Dimensional shapes defines the geometric center of mass of a Body center of gravity of the area bounded by given! In a figure is the point which corresponds to the mean position all. | 2021-10-21T15:10:22 | {
"domain": "senorcafe.com",
"url": "http://www.senorcafe.com/bjs2dg1i/a1702f-determine-the-coordinates-of-the-centroid-of-the-area",
"openwebmath_score": 0.7315730452537537,
"openwebmath_perplexity": 345.616869691963,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9845754447499796,
"lm_q2_score": 0.865224091265267,
"lm_q1q2_score": 0.8518783944658972
} |
http://math.stackexchange.com/questions/365010/writing-a-series-using-sigma-notation/365021 | # Writing a series using Sigma notation
How do I write $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}?$$
I have been struggling with these types of problems, so please, an explanation of how to get the result will be appreciated.
-
$$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
Since everything else has a denominator, let's try putting a denominator for $2$. Since anything divided by one itself, we get:
$$\frac{2}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
Everything else has an $x$. Let's give $2$ an $x$ too! $x^0 = 1$, and anything multiplied by $1$ is itself, so:
$$\frac{2x^0}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
All the other $x$s have an exponent. Let's give one to that $x$ next to the $3$. $x^1 = x$, so we get:
$$\frac{2x^0}{1}+ \frac{3x^1}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
However, these denominators are just powers of two. Therefore, we have:
$$\frac{2x^0}{2^0}+ \frac{3x^1}{2^1} + \frac{4x^2}{2^2}+\frac{5x^3}{2^3}+\frac{6x^4}{2^4}$$
Now we see some patterns:
• $2, 3, 4, 5, 6$ (numerator of coefficient of $x$)
• $0, 1, 2, 3, 4$ (exponents of $x$)
• $1, 2, 4, 8, 16$ (denominator of coefficient of $x$. Powers of $2$, viz. $2^0, 2^1, 2^2, 2^3, 2^4$)
We'll make the sum range from $0$ to $4$, although in theory we could actually make it range over anything.
$$\sum_{k=0}^4$$
We need exponents of $x$, so we'll put those in there.
$$\sum_{k=0}^4 x^k$$
We need those powers of two in the denominator, so add that too:
$$\sum_{k=0}^4 \frac{x^k}{2^k}$$
However, there are still those numerators we haven't checked off our list yet. We can't have them ranging from $0$ to $4$, we need them from $2$ to $6$. The solution is to add two!
$$\sum_{k=0}^4 \frac{(k+2)x^k}{2^k}$$
-
I'd have gone on to write $\dfrac{2x^0}{1}+ \dfrac{3x^1}{2} + \dfrac{4x^2}{4}+\dfrac{5x^3}{8}+\dfrac{6x^4}{16}$ in the form $\dfrac{2x^0}{2^0}+ \dfrac{3x^1}{2^2} + \dfrac{4x^2}{2^2}+\dfrac{5x^3}{2^3}+\dfrac{6x^4}{2^4}$. – Michael Hardy Apr 17 '13 at 23:34
@MichaelHardy, good idea, I added it. – George V. Williams Apr 18 '13 at 0:05
It's much like finding a pattern, and a way to express the pattern. Starting point: $2 = \dfrac 21$. Numerator coeffienct of the $x$-term, increases by one per term , starting at $2$ (add 1 to starting numerator); the exponent for $x$ increases by one (starting, say, at $x^0,...x^1 = x,...x^2...$ (Add one to the x-exponent starting at $0$. Also, the denominator represents powers of $2$, starting at $2^0 = 1$...Increase (add) by one the exponent of the term $2$ in the denominator $2^0, ... 2^1 = 2, ...2^{1+1} = 2^2 = 4, ...$
Try writing out the first few terms of the following and see if it matches, starting index $0$, and on...:
$$\sum_{k = 0}^\infty \frac{(k+2)x^k}{2^k}$$
If things don't work out (here they happen to work just fine), there is often a little trial-and-error involved. For example, we can "tweak" if we want to represent the series with a starting index $1$, by "subtracting $1$ from each of the expressions given in terms of $k$.
NOTE: The above will give you an infinite series. To get the first five terms, we start at $k = 0$, and need to sum through (inclusive) $k = 4$. But you can obtain any positive number $n$ of terms by summing from $k = 0$ to $k = n-1)$
-
The best way to get a "knack" for "cracking" a sequence is through practice. Try to approach sequences of numbers like you'd approach a puzzle: what fits? what doesn't fit? What patterns do I see...start writing down what you do see, and try creating a "model" term (a "work in progress," so to speak)...Figure out if starting with $0$ or $1$ for one of the varying terms works best. If a term appears constant through the sequence, represent it as a constant, etc. – amWhy Apr 17 '13 at 23:47
Such great advice! +1 – Amzoti Apr 18 '13 at 0:24
Such a great answer +1 – Babak S. Aug 21 '13 at 6:14
Note that the power of $x$ increase in steps of $1$ starting from $x^0$. Hence the $(n+1)^{th}$ term has $x$ raised to the power $n$. The denominators are powers of $2$ starting from $2^0$ and the numbers in the numerator also increase in steps of $1$ starting from $2$. $$\sum_{n=0}^4 \dfrac{(n+2)x^n}{2^n}$$
-
Notice that $2=\frac {2x^0}1$, so you want $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\frac {2x^0}1+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\sum_{i=1}^5\frac {(i+1)x^{i-1}}{2^{i-1}}$$
-
Start off by looking for any patterns. I would start off by looking at how the powers on $x$ change. Starting at $0$ ($x^0=1$) it looks like the power goes up by one at each "step." So we can start off with the "skeleton" sum $$\sum_{i=0}^4 x^i.$$ Similarly observe the coefficient patters, it looks like the numerator starts at $2$ and goes up by one at each step and the denominator are powers of $2$ starting at the $2^0$. So putting that all together we get $$\sum_{i=0}^4 \frac{(i+2)x^i}{2^i}$$
- | 2014-11-28T17:22:25 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/365010/writing-a-series-using-sigma-notation/365021",
"openwebmath_score": 0.902320384979248,
"openwebmath_perplexity": 473.7786485387081,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9845754452025766,
"lm_q2_score": 0.8652240860523328,
"lm_q1q2_score": 0.851878389724968
} |
https://stats.stackexchange.com/questions/135878/is-there-a-family-of-processes-centred-on-the-poisson-process | # Is there a family of processes centred on the Poisson process?
I am looking for a model, characterized continuously by a single parameter, to describe the arrival times of buses with unit expected interarrival time. At one extreme of the parameter (say $\theta=1$), the process is deterministic, with all interarrival times equal (to $1$). When $\theta=0$, the process is pure Poisson (with unit expected interarrival or waiting time). As $\theta$ approaches $-1$, the arrivals tend to cluster, with long intervals between the clusters. At the (unattainable) extreme of $\theta=-1$, all the buses arrive in a simultaneous convoy, after an infinite wait, and you have to wait forever for the next convoy.
The choice of $\{1,0,-1\}$ for the extreme and central parameter values isn't important: they could be $\{-\infty, 0, \infty\}$, $\{0,\frac12,1\}$, ..., whatever. At this stage, simplicity and naturalness, subject to the above conditions, is more important than realism.
In a pure Poisson process, the interarrival times (times between each bus) has an exponential distribution. So you want some family of distributions which includes the exponential as a special case. The gamma distribution (see: https://en.wikipedia.org/wiki/Gamma_distribution). When then shape parameter is 1, you have the exponential. When the shape parameter ($k$ or $\alpha$ in wiki) approaches zero, you are approaching constant interarrival times, and growing above 1 you get interarrival times more variable than for the poisson process.
• Yes, this could be the model if we set $k\theta=1$. But I think that you have it the wrong way round: We approach a constant (unit) interarrival time as $k\rightarrow\infty$, and long intervals between clusters as $k\rightarrow0$. – John Bentin Feb 12 '15 at 16:06 | 2019-10-14T20:57:22 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/135878/is-there-a-family-of-processes-centred-on-the-poisson-process",
"openwebmath_score": 0.8897665143013,
"openwebmath_perplexity": 489.94263812041544,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9845754461077707,
"lm_q2_score": 0.8652240825770432,
"lm_q1q2_score": 0.8518783870864789
} |
https://mathhelpboards.com/threads/number-of-real-roots-of-a-quartic.887/ | # Number of real roots of a quartic
#### CaptainBlack
##### Well-known member
rayman's question from another place:
Could someone help me with this problem, I have no idea how to start with it
How many real roots does this polynomial have p(x)=x^4-x^3-1?
Clearly state the argument that explains the number of real roots.
Thank you for any help
Descartes rule of signs tells you this has exactly 1 positive root, and exactly 1 negative root, so it has two real roots.
CB
Last edited:
#### chisigma
##### Well-known member
The Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\frac{3}{4}$ and here is $q(x)=- \frac{27}{256}$. Now if we consider the quartic equation $\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\frac{27}{256}$, one real root of order 2 for $a=\frac{27}{256}$ and no real roots for $a>\frac{27}{256}$...
Kind regards
$\chi$ $\sigma$
#### CaptainBlack
##### Well-known member
The Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\frac{3}{4}$ and here is $q(x)=- \frac{27}{256}$. Now if we consider the quartic equation $\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\frac{27}{256}$, one real root of order 2 for $a=\frac{27}{256}$ and no real roots for $a>\frac{27}{256}$...
Kind regards
$\chi$ $\sigma$
In this case Descartes rule of signs does tell us exactly how many real roots we have.
The number of positive roots is equal to the number of changes of signs of the coefficients less a multiple of 2. In this case the number of sign changes is 1, and as there is no multiple of 2 other than 0 which leaves the number of roots non-negative there is exactly one positive real root. The same argument applies to the negative roots.
CB | 2021-06-16T14:09:10 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/number-of-real-roots-of-a-quartic.887/",
"openwebmath_score": 0.839435338973999,
"openwebmath_perplexity": 140.4456067566802,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9845754465603678,
"lm_q2_score": 0.8652240756264638,
"lm_q1q2_score": 0.851878380634707
} |
https://mathematica.stackexchange.com/questions/222259/why-doesnt-integrate-evaluate-an-elliptic-integral | # Why doesn't Integrate evaluate an elliptic integral?
My code is
Integrate[ 1/Sqrt[(x - a) (x - b) (x - c) (x - d)], {x, a, ∞},
Assumptions -> 0 < d < c < b < a]
I know this can be expressed be the incomplete elliptic integral of first kind (EllipticF), but the output remains unevaluated
Integrate[ 1/Sqrt[(-a + x) (-b + x) (-c + x) (-d + x)], {x, a, ∞},
Assumptions -> 0 < d < c < b < a]
Why does this happen? I am desperate
• If the output is echoed, it means Mathematica doesn't know what to do with it. – J. M.'s torpor May 20 '20 at 12:02
• Seems to work OK for the indefinite integral. And the result can be evaluated at a and Infinity using the Limit function – Gustavo Delfino May 20 '20 at 12:09
• But isn't it strange Mathematica can't calculate it? I also tried changing the integration from a to a generic M but the result is the same. From Gradshteyn Ryzhik I know this is equal to EllipticF(...,...), should I doubt the tables of Gradshteyn Ryzhik or is it common that Mathematica can't do integrals like this one? – Filippo Caleca May 20 '20 at 12:11
Actually V 12.1 can do it directly, you just have to wait a little bit long:
Clear["Global*"];
int = Integrate[1/Sqrt[(x - a) (x - b) (x - c) (x - d)], {x, a, Infinity},
Assumptions -> 0 < d < c < b < a]
May be OP used different Mathematica version? It will be good to post which version was used. Screen shot below:
A Workaround for now: (assuming proper integral)
int = Integrate[1/Sqrt[(x - a) (x - b) (x - c) (x - d)], x]
low = Assuming[0 < d < c < b < a, Limit[int, x -> a]]
(* 0 *)
high = Assuming[0 < d < c < b < a, Limit[int, x -> Infinity]]
The above is then the final result.
• Thank you! I am using 11.3 – Filippo Caleca May 20 '20 at 12:52
• @FilippoCaleca integrate in V12.1 has improved over 11.3 – Nasser May 20 '20 at 12:56
• my output is different, in fact writing: int = Integrate[1/Sqrt[(x - a) (x - b) (x - c) (x - d)], x] the output is -((2 (-a + x) (-b + x) Sqrt[((a - b) (-c + x))/((b - c) (a - x))] Sqrt[((a - b) (-d + x))/((b - d) (a - x))] EllipticF[ ArcSin[Sqrt[((a - d) (-b + x))/((b - d) (-a + x))]], ((a - c) (b - d))/((b - c) (a - d))])/((a - b) Sqrt[((a - d) (b - x))/((b - d) (a - x))] Sqrt[(-a + x) (-b + x) (-c + x) (-d + x)])) the argument of EllipticF are different, how is it? – Filippo Caleca May 20 '20 at 13:09
• Note that you shouldn't expect the technique used in your workaround to work in all cases. See this old Wolfram blog post for some of the mathematical nitty-gritty as to why. – Michael Seifert May 20 '20 at 19:58
• Sure, if you go back to the concept of "area under a curve", then one might expect that an antiderivative ought to be continuous if the original function is continuous. But yes, it's not always easy to do. That's why Mathematica has to try harder for definite integrals, tho it has been known to still make mistakes sometimes. – J. M.'s torpor May 21 '20 at 1:41
In the newest version (i.e. 12.1) this integral evaluates a bit long, however changing the variable $$x \mapsto t = x-a\;$$ this can be evaluated a few times faster.
int2 = Integrate[ 1/Sqrt[t (t + a - b) (t + a - c) (t + a - d)], {t, 0, ∞},
Assumptions -> 0 < d < c < b < a]
2 EllipticF[ ArcSin[ Sqrt[(b - d)/(a - d)]],
((b - c)(a - d))/((a - c)(b - d))]/Sqrt[(a - c)(b - d)]
TraditionalForm[%]
I'm working with the system in cloud and sometimes it appears that the integral in question may remain unevaluated while int2 evaluates well even in version 11.2 on my machine.
Mathematica functions evolve with time even if its usage remains tha same. This aspect of the system is perhaps the most obvious in case of symbolic integration (Integrate), exact solutions of differential equations (DSolve) and the special functions (among them EllipticF). Elliptic functions and integrals appeared in Mathematica 1, however since then many new related functionalities would have been added later e.g. EllipticF was introduced in version 1.0 year 1988 and updated in 3.0 (1996). WeierstrassP was introduced in version 1.0 and updated in 3.0 (1996), however several new functionalities related appeared in version 11.2 (2017) like e.g. WeierstrassHalfPeriodW1 or WeierstrassE1 see e.g. this answer Integrate yields complex value, while after variable transformation the result is real. Bug?. Inspecting another answers therein one can see how Integrate can be sensitive when new functions or functionalities appear. It relates not only to new functionalities but also to widening domain of existing functions (in documentation pages one finds information when a function was introduced and when it was last updated, nevertheless there are also hiden changes that are not reported, however they may be crucial when certain different functions involved were updated). One should take attention to this aspect related to better handling of symbolic input of e.g. WeierstrssHalfPeriodW1 in version 12.1 with respect to 11.2 and it is advantageous to pay atention to this post. Elliptic functions and integrals play very important role in mathematics, physics, engineering and they are still better handled in newer versions of the system. This does not mean that Mathematica is defective but rather that perfect handling of special functions can be approached asymptotically and it is still in interest of developers of the system, e.g. one of leading experts in the field of special functions Oleg Marichev is a member of the special functions group in WRI. Having said that we can accept the state of art and the fact that things can change at least on the symbolic level.
Let's come back to version 11.2 with help of a simple change of variables: $$x \mapsto t+a$$
int3 = Integrate[ 1/Sqrt[t (t + a - b) (t + a - c) (t + a - d)], {t, 0, ∞},
Assumptions -> 0 < d < c < b < a]
(2 (EllipticF[ ArcSin[Sqrt[(a - d)/(b - d)]], ((a - c) (b - d))/((b - c) (a - d))]
+ I EllipticK[((a - b) (c - d))/((-b + c) (a - d))]))/Sqrt[(b - c) (a - d)]
TraditionalForm[%]
This might seem strange that there appears an imaginary number however the full integral is indeed real even though FullSimplify cannot demonstrate (in 11.2) that both results are equal. In 12.1 this still cannot be done, although a simpler identity can be proved, assuming that parameters are related somehow (in 12.1 not in 11.2), e.g.
FullSimplify[(8(EllipticF[ArcSin[Sqrt[3/2]], 4/3] +
I EllipticK[-(1/3)]))/(Sqrt[3] Sqrt[a^2])
- (4 EllipticF[ArcSin[Sqrt[2/3]], 3/4])/Sqrt[a^2], a > 0]
0
We can show that this is the case evaluating numerically, e.g.
With[{a = 4, b = 3, c = 2, d = 1}, {
(2 (EllipticF[ ArcSin[Sqrt[(a - d)/(b - d)]], ((a - c) (b - d))/((b - c) (a - d))]
+ I EllipticK[((a - b) (c - d))/((-b + c) (a - d))]))/Sqrt[(b - c) (a - d)],
( 2 (EllipticF[ ArcSin[Sqrt[(b - d)/(a - d)]],
((b - c)(a - d))/((a - c)(b - d))]))/Sqrt[(a - c)(b - d)]} // N // Chop]
{1.07826, 1.07826}
For example of a bit more testy case see e.g. Why does Integrate declare a convergent integral divergent?
Making appropriate plot of the functions and their difference might be helpful as well:
Plot[{#, # - (4 EllipticF[ArcSin[Sqrt[2/3]], 3/4])/Sqrt[a^2]}, {a, 0, 6},
PlotStyle -> Thick, AxesOrigin -> {0, 0}] &[ (
8(EllipticF[ArcSin[Sqrt[3/2]], 4/3] + I EllipticK[-1/3]))/(Sqrt[3]Sqrt[a^2])]
` | 2021-06-21T19:32:20 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/222259/why-doesnt-integrate-evaluate-an-elliptic-integral",
"openwebmath_score": 0.49721527099609375,
"openwebmath_perplexity": 1749.4345224913804,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9489172615983309,
"lm_q2_score": 0.8976953030553433,
"lm_q1q2_score": 0.8518385687249601
} |
https://gmatclub.com/forum/the-positive-integers-x-y-and-z-are-such-that-x-is-a-85437.html?sort_by_oldest=true | It is currently 17 Oct 2017, 12:13
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# The positive integers x, y, and z are such that x is a
Author Message
TAGS:
### Hide Tags
Intern
Joined: 19 Jun 2009
Posts: 29
Kudos [?]: 122 [1], given: 1
The positive integers x, y, and z are such that x is a [#permalink]
### Show Tags
17 Oct 2009, 14:15
1
KUDOS
13
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
58% (00:59) correct 42% (01:07) wrong based on 1007 sessions
### HideShow timer Statistics
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even
(2) y is even.
[Reveal] Spoiler: OA
Last edited by Bunuel on 25 Feb 2012, 02:35, edited 1 time in total.
Kudos [?]: 122 [1], given: 1
Math Expert
Joined: 02 Sep 2009
Posts: 41873
Kudos [?]: 128579 [6], given: 12180
### Show Tags
17 Oct 2009, 14:32
6
KUDOS
Expert's post
16
This post was
BOOKMARKED
amitgovin wrote:
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even
(2) y is even.
Given:
x is a factor of y --> $$y=mx$$, for some non-zero integer $$m$$;
y is a factor of z --> $$z=ny$$, for some non-zero integer $$n$$;
So, $$z=mnx$$.
Question: is z even? Note that $$z$$ will be even if either $$x$$ or $$y$$ is even
(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
(2) $$y$$ even --> as $$z=ny$$ then as one of the multiples of z even --> z even. Sufficient.
_________________
Kudos [?]: 128579 [6], given: 12180
SVP
Joined: 29 Aug 2007
Posts: 2472
Kudos [?]: 841 [1], given: 19
### Show Tags
19 Oct 2009, 23:18
1
KUDOS
amitgovin wrote:
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
1) xz is even
2) y is even.
y/x = k where k is an integer.
y = xk ....................i
z/y = m where m is an integer.
z = ym = xkm .....................ii
If a factor is even, then the source of the factor must be even.
1) If xz is even, z must be even because x may or may not be an even because x is a factor of z but z must be even. SUFF.
2) If y is even, z must be even because y is a factor of z. SUFF..
D..
_________________
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
Kudos [?]: 841 [1], given: 19
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7670
Kudos [?]: 17334 [10], given: 232
Location: Pune, India
### Show Tags
02 Dec 2010, 11:16
10
KUDOS
Expert's post
2
This post was
BOOKMARKED
amitgovin wrote:
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
1) xz is even
2) y is even.
Though Bunuel has provided the solution, I would just like to bring to your notice a train of thought.
When we say, "The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.", it implies that if x or y is even, z will be even.
e.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even.
Once this makes sense to you, it will take 10 secs to arrive at the solution.
Stmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even.
Stmnt 2: y is even, so z must be even
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17334 [10], given: 232 Retired Moderator Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL Joined: 04 Oct 2009 Posts: 1637 Kudos [?]: 1104 [0], given: 109 Location: Peru Schools: Harvard, Stanford, Wharton, MIT & HKS (Government) WE 1: Economic research WE 2: Banking WE 3: Government: Foreign Trade and SMEs Re: positive integers [#permalink] ### Show Tags 12 Dec 2010, 10:22 Bunuel, you always try to solve the questions algebraically, don't you? _________________ "Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can." My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html GMAT Club Premium Membership - big benefits and savings Kudos [?]: 1104 [0], given: 109 Math Expert Joined: 02 Sep 2009 Posts: 41873 Kudos [?]: 128579 [2], given: 12180 Re: positive integers [#permalink] ### Show Tags 12 Dec 2010, 10:48 2 This post received KUDOS Expert's post metallicafan wrote: Bunuel, you always try to solve the questions algebraically, don't you? Not at all. There are certain GMAT questions which are pretty much only solvable with plug-in or trial and error methods (well at leas in 2-3 minutes). Also many questions can be solved with logic and common sense much quicker than with algebraic approach. So you shouldn't always rely on algebra. Having said that I must add that there are of course other types of questions which are perfect for algebraic approach, plus I often use algebra just to explain a solution. _________________ Kudos [?]: 128579 [2], given: 12180 Intern Joined: 06 Dec 2010 Posts: 17 Kudos [?]: [0], given: 1 Re: positive integers [#permalink] ### Show Tags 12 Dec 2010, 19:36 I like to thing of the boxes method. If you draw them out, then x is inside y which is inside z. zx, a 2 will exist inside the box of either z or x (which is itself inside z) so YES y a 2 will exist inside the box of y which is itself z so YES Kudos [?]: [0], given: 1 Moderator Joined: 01 Sep 2010 Posts: 3355 Kudos [?]: 9042 [1], given: 1152 Re: positive integers [#permalink] ### Show Tags 26 Dec 2010, 17:37 1 This post received KUDOS VeritasPrepKarishma wrote: amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even? 1) xz is even 2) y is even. Please explain. thanks. Though Bunuel has provided the solution, I would just like to bring to your notice a train of thought. When we say, "The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.", it implies that if x or y is even, z will be even. e.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even. Once this makes sense to you, it will take 10 secs to arrive at the solution. Stmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even. Stmnt 2: y is even, so z must be even awesome. your explanations and bunuel as well, are amazing thanks _________________ Kudos [?]: 9042 [1], given: 1152 Manager Joined: 19 Dec 2010 Posts: 136 Kudos [?]: 32 [0], given: 12 Re: positive integers [#permalink] ### Show Tags 17 Mar 2011, 23:07 Easy if you realize the following: When a is a factor of b AND b is a factor of c THEN a is a factor of c as well. Hence when either one of these numbers is even, the other has to be even too.. D Kudos [?]: 32 [0], given: 12 SVP Joined: 16 Nov 2010 Posts: 1599 Kudos [?]: 592 [0], given: 36 Location: United States (IN) Concentration: Strategy, Technology Re: positive integers [#permalink] ### Show Tags 17 Mar 2011, 23:52 z = ky y = mx so z = (km)xy (1) -> xz is eve means at least x or z is even, and if x = even, then z is also even as it has an even factor. 2 -> y is even so z having an even factor is even too. Answer D _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings Kudos [?]: 592 [0], given: 36 Intern Joined: 06 Sep 2010 Posts: 40 Kudos [?]: 7 [0], given: 0 Re: positive integers [#permalink] ### Show Tags 15 Apr 2011, 05:24 At first, mistook "factor" for "multiple" came with answer E. Later, understood that the problem was so easy...just plug and play !! Kudos [?]: 7 [0], given: 0 Manager Joined: 16 May 2011 Posts: 71 Kudos [?]: 18 [0], given: 2 Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 24 Feb 2012, 10:06 if we plug and play, why can't we test X = 1? then y/n.. Kudos [?]: 18 [0], given: 2 Manager Joined: 21 Feb 2012 Posts: 116 Kudos [?]: 148 [0], given: 15 Location: India Concentration: Finance, General Management GMAT 1: 600 Q49 V23 GPA: 3.8 WE: Information Technology (Computer Software) Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 24 Feb 2012, 10:46 The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even? 1) xz is even 2) y is even. Please explain. thanks.[/quote] Ans. let us take any 3 numbers, say x=3,y=18,z=54, or x=2,y=4,z=20 1)if xz is even then it means that either x or z is even,say that x is even, now there is no even number which is a factor of odd number, so z is definitely even, now if x is odd as in the above case, still then we can point out that z is even. 2)if y is even then it is clear that z will be even. Thus this question could be answered by any of the two questions. Kudos [?]: 148 [0], given: 15 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7670 Kudos [?]: 17334 [1], given: 232 Location: Pune, India Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 27 Feb 2012, 05:20 1 This post received KUDOS Expert's post dchow23 wrote: if we plug and play, why can't we test X = 1? then y/n.. Plug in method would be far more painful for this question (and most other questions in my opinion). Think how you would go about it: Checking whether stmnt 1 is enough: xz is even If x = 1, y = 1 and z = 2 (so that xz is even), then z is even. If x = 2, y = 2 and z = 4, z is again even. Then you start thinking if you can take some values such that xz is even but z is not... Now you start using logic... Wouldn't you say it is far better to use logic in the first place itself? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
Veritas Prep Reviews
Kudos [?]: 17334 [1], given: 232
Senior Manager
Joined: 13 Aug 2012
Posts: 459
Kudos [?]: 540 [0], given: 11
Concentration: Marketing, Finance
GPA: 3.23
Re: The positive integers x, y, and z are such that x is a [#permalink]
### Show Tags
22 Jan 2013, 22:49
amitgovin wrote:
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even
(2) y is even.
What is given?
$$z = y*N$$
$$y = x*R$$
1.
$$xz = 2*I$$
If x is even, then z is even since $$z = x*R$$
If z is even, then z is even.
SUFFICIENT!
2. $$y = 2*I$$ ==> $$z = 2*I*N$$ Definitely EVEN
SUFFICIENT!
_________________
Impossible is nothing to God.
Kudos [?]: 540 [0], given: 11
Director
Joined: 29 Nov 2012
Posts: 868
Kudos [?]: 1409 [0], given: 543
### Show Tags
03 Aug 2013, 02:04
Bunuel wrote:
(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
Can you please provide a numerical example for this part it isn't very clear. Thanks in advance!
I thought it was insufficient as there were multiple cases either X or Y even or both....
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html
Kudos [?]: 1409 [0], given: 543
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 627
Kudos [?]: 1355 [1], given: 136
### Show Tags
03 Aug 2013, 04:26
1
KUDOS
fozzzy wrote:
Bunuel wrote:
(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
Can you please provide a numerical example for this part it isn't very clear. Thanks in advance!
I thought it was insufficient as there were multiple cases either X or Y even or both....
We know that x is a factor of $$y \to y = Ix$$
Again, $$z = yI^'$$. Now, given that xz - even.
Case I:Assume that x = even , z = odd. Now, as x is even, y = even(I can be odd/even,doesn't matter).
Again, as y is even, z HAS to be even($$I^'$$ is odd/even, doesn't matter). Thus, if x is even, z IS even.
Numerical Example :y = 2*I(x=2).
$$z = 6*I^'$$. z IS even.
Case II : z is even OR (x and z) both are even.
Hope this helps.
_________________
Kudos [?]: 1355 [1], given: 136
Current Student
Joined: 06 Sep 2013
Posts: 1980
Kudos [?]: 719 [0], given: 355
Concentration: Finance
### Show Tags
10 Oct 2013, 15:01
Bunuel wrote:
amitgovin wrote:
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even
(2) y is even.
Given:
x is a factor of y --> $$y=mx$$, for some non-zero integer $$m$$;
y is a factor of z --> $$z=ny$$, for some non-zero integer $$n$$;
So, $$z=mnx$$.
Question: is z even? Note that $$z$$ will be even if either $$x$$ or $$y$$ is even
(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
(2) $$y$$ even --> as $$z=ny$$ then as one of the multiples of z even --> z even. Sufficient.
Perfect explanation. Remember the factor foundation rule.
Also, other properties of factors that might be helpful to have in mind.
Just to remind you, The factor foundation rule states that "if a is a factor of b, and b is a factor of c, then a is a factor of c"
Also, if 'a' is a factor of 'b', and 'a' is a factor of 'c', then 'a' is a factor of (b+c). In fact, 'a' is a factor of (mb + nc) for all integers 'm' and 'n'
If 'a' is a factor of 'b' and 'b' is a factor of 'a', then 'a=b'
If 'a' is a factor of 'bc' and gcd (a,b) = 1, then 'a' is a factor of 'c'
If 'p' is a prime number and 'p' is a factor of 'ab' then 'p' is a factor of 'a' or 'p' is a factor of 'b'
In other words,, any integer is divisible by all of its factors- and it is also divisible by all of the factors of its factors
Hope it helps
Kudos [?]: 719 [0], given: 355
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16758
Kudos [?]: 273 [0], given: 0
Re: The positive integers x, y, and z are such that x is a [#permalink]
### Show Tags
30 Jan 2015, 14:25
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Kudos [?]: 273 [0], given: 0
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16758
Kudos [?]: 273 [0], given: 0
Re: The positive integers x, y, and z are such that x is a [#permalink]
### Show Tags
05 Feb 2016, 17:31
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Kudos [?]: 273 [0], given: 0
Re: The positive integers x, y, and z are such that x is a [#permalink] 05 Feb 2016, 17:31
Go to page 1 2 Next [ 24 posts ]
Display posts from previous: Sort by | 2017-10-17T19:13:34 | {
"domain": "gmatclub.com",
"url": "https://gmatclub.com/forum/the-positive-integers-x-y-and-z-are-such-that-x-is-a-85437.html?sort_by_oldest=true",
"openwebmath_score": 0.7219163775444031,
"openwebmath_perplexity": 2259.430897110478,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 0.9618217257384335,
"lm_q2_score": 0.8856314783461302,
"lm_q1q2_score": 0.851819596871155
} |
https://mathematica.stackexchange.com/questions/149710/generating-a-convex-hull-with-the-hull-boundary-points-labeled | # Generating a convex hull with the hull boundary points labeled
I'm very new to Mathematica, and I'm sure my question is trivial.
Suppose a have a (long) list whose elements are of the form {{x, y}, n}, where x and y are Real, and n is an Integer. I'd like to plot the convex hull of the corresponding points {x, y} in the plane. Moreover, for each point {x, y} that is a vertex of the convex hull should appear with its corresponding label n. Finally I'd like the coordinate axes to be displayed with the plot.
### Revision
Contrive some data.
SeedRandom[42];
With[{n = 42},
data = Transpose[{RandomReal[1., {n, 2}], RandomSample[Range[100], n]}]];
Short[data, 3]
{{{0.425905, 0.391023}, 51}, <<40>>, {{0.359445, 0.00772549},24}}
Separating the points from the labels.
pts = data[[All, 1]];
lbls = data[[All, 2]];
Generating the convex hull.
ch = ConvexHullMesh[pts];
Extracting the boundary point labels.
bpts = MeshCoordinates[RegionBoundary[ch]];
blbls = Extract[lbls, Flatten[Position[pts, #] & /@ bpts, 1]]
{94, 36, 21, 1, 88, 2, 14, 24}
Making the labels into graphics elements.
labels = MapThread[Text[#1, #2] &, {blbls, bpts}];
Showing the combined graphics.
Show[ch, Graphics[{Point[pts], labels}], Axes -> True]
Of course, the labels can be fancied up by using Inset rather than Text
labels =
Inset[
Graphics[{FaceForm[White], EdgeForm[Black], Disk[], Text[Style[#1, 9]]}],
#2, Automatic, Scaled[.0475]] &,
{blbls, bpts}];
Show[ch, Graphics[{Point[pts], labels}],
Method -> {"AxesInFront" -> False}, Axes -> True]
• My label is not necessarily the index. Jul 4 '17 at 23:35
• @JairoBochi. I have revised my answer in a way that I hope you find better attuned to your needs. Jul 5 '17 at 2:19
• @MichaelE2 Right, I should have provided sample data, sorry. Jul 5 '17 at 22:39
• @JairoBochi. Your nagging me about the indices motivated me to improve my answer, so I'm glad you did it. Jul 6 '17 at 5:02
SeedRandom[1]
data = Transpose[{RandomReal[1, {10, 2}], RandomInteger[100, 10]}];
xy = data[[All,1]];
labels = data[[All,2]];
ConvexHullMesh[xy, Prolog -> Point[xy], Frame -> True,
MeshCellStyle -> {2 -> Opacity[0.5, LightBlue]},
MeshCellShapeFunction -> {0 -> ({Opacity[1, Yellow], Disk[#, .05],
Text[Style[# /. rule, Red, 16], #]} &)}]
Also
labeled = Labeled[#, # /. rule] & /@ MeshCoordinates[ConvexHullMesh[xy]];
Show[ConvexHullMesh[xy], ListPlot[labeled], Graphics[Point[xy]], Frame ->True]
• I understand that you generate the plot from two lists, "xy" and "labels". But my list is given of the form "data". So I would need to extract "xy" and "labels" from data to have a working solution... Jul 4 '17 at 23:48
– kglr
Jul 4 '17 at 23:52
There are potential shortcomings in using Rule to map points to labels, which I have occasionally come across. The purpose in creating a NearestFunction using Nearest below to map coordinates {x, y} to a label is overcome these shortcoming, at only a small expense of computational time. Given the general nature of the question, this seemed the best approach.
Often in an application, one might arrive at coordinates in various ways. If we're using floating-point numbers, rounding error might make simple rules of the form {x, y} -> label fail, because the coordinates {x, y} of the rule might not exactly match the computed coordinates in the figure. Or one might start with exact coordinates, which at some point might be numericized (converted to floating point), and again the replacement rule will fail. Nearest takes care of those issues, unless rounding error is so great that one computed point ends up closer to a different point than the intended one (in such a case, the problem is completely different, a numerics one). Whether or not to numericized the points with N[] depends on whether data is already numericized; it's not strictly necessary in either case, but it should speed up nf, which might be an issue with a very long list data.
points = data[[All, 1]];
nf = Nearest[N@points -> data[[All, 2]]]; (* map coordinates to labels in data *)
hull = ConvexHullMesh[data[[All, 1]]];
With[{coords = MeshCoordinates[hull]},
Show[
MeshRegion[
coords,
MeshCells[hull, 2],
MeshCellLabel -> Table[{0, i} -> First@nf[coords[[i]]], {i, Length@coords}]
],
Axes -> True] (* or Frame -> True *)
] | 2021-10-16T20:31:41 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/149710/generating-a-convex-hull-with-the-hull-boundary-points-labeled",
"openwebmath_score": 0.2550300061702728,
"openwebmath_perplexity": 3802.713505233619,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9473810407096791,
"lm_q2_score": 0.8991213664574069,
"lm_q1q2_score": 0.8518105358787269
} |
https://mathoverflow.net/questions/192058/covering-designs-of-the-form-v-k-2/192065 | # covering designs of the form $(v,k,2)$
A covering design $(v,k,t)$ is a family of subsets of $[v]$ each having $k$ elements such that given any subset of $[v]$ of $t$ elements it is a subset of one of the sets of the family. A problem is to find the minimum number of subsets such a family can have.
I am interested in the case $(v,k,2)$. It seems to be equivalent to the problem of finding the minimum number of copies of $K_k$ that are needed to cover the graph $K_v$.
When $k=3$ this problem may be solved optimally with a Steiner triple system if it exists, which happens if and only if $v$ is $1$ or $3 \bmod 6$, in this case $\frac{v(v-1)}{6}$ subsets are required.
My questions are the following:
• Is the minimum of subsets for $(v,3,2)$ known when the congruence does not hold?
• If not what are good bounds?
• Does the problem with $t=2$ have another name (It seems like it may be more common than the general covering design)
• Is the minimum number of subsets for $(v,k,2)$ known?
• If not, which are some good bounds (specific values of $k$ are also welcome)
• I would call the covering designs $\ (v\ k\ 2)\$-- sloppy planes. – Włodzimierz Holsztyński Jan 3 '15 at 22:28
• Your $\ \frac{n(n-1)}2\$ must be a typo(?). A simple calculation shows that the number of 3-subset of a $\ (\nu\ 3\ 2)\$ perfect system must be $\ \frac{\nu\cdot(\nu-1)}3\$ (it's $\ 3,\$ not $\ 2,\$ in the denominator). Also then $\ \nu\equiv 1\ or\ 3\mod 6\$ (rather than $\ 0\ or\ 3).\$ Thus I feel that it would be nice and useful for non-specialists like me to have a short list of basic results in a separate Answer. – Włodzimierz Holsztyński Jan 4 '15 at 1:43
• @WłodzimierzHolsztyński You can find basic facts in the paper I linked to in my post (or those given by Thomas Kalinowski as well, I think). Handbook of Combinatorial Designs is a good reference book for this sort of basic knowledge; coverings are treated in Section 11 of Chapter IV in the 2nd edition. As for the typos, yes, they are not correct, although the number of $3$-subsets in this case is not $\frac{v(v-1)}{3}$ but $\frac{\binom{v}{2}}{\binom{3}{2}} = \frac{v(v-1)}{6}$. I'll edit OP's post. – Yuichiro Fujiwara Jan 4 '15 at 6:31
• @YuichiroFujiwara -- thank you for the links. And I let $\ \frac{\binom{\nu}2}{\binom 32}=\frac{\nu\cdot (\nu-1)}{3\cdot 2}\$ to have somehow just $\ 3\$ in the denominator--ooops! Sorry. – Włodzimierz Holsztyński Jan 4 '15 at 6:46
Edit: The possible "gap" of sort in Caro and Yuster's proof of their upper bound has just been fixed! See Ben Barber's comment below (and his joint paper with Daniela Kühn, Allan Lo and Deryk Osthus on arXiv. This shows Gustavsson's theorem (whose proof by Gustavsson himself might have been regarded as incomplete but was used by Caro and Yuster). So this result proves the general upper bound on the smallest coverings mentioned in Thomas Kalinowski's post is indeed correct.
The most up to date account (except the above mentioned paper) on the minimum number of subsets for a $(v,k,2)$ covering is here:
Y. M. Chee, C. J. Colbourn, A. C. H. Ling, R. M. Wilson, Covering and packing for pairs, J. Combin. Theory Ser. A 120 (2013) 1440–1449.
One important remark about Thomas Kalinowski's answer is that the asymptotic solution by Caro and Yuster may be incomplete, although the claimed upper bound is likely the correct one. They rely on Gustavsson's theorem on graph decomposition (Lemma 2.1 in Caro and Yuster's paper), which is claimed to be proved in the following thesis:
T. Gustavsson, Decompositions of Large Graphs and Digraphs with High Minimum Degree,'' Doctoral Dissertation, Dept. of Mathematics, Univ. of Stockholm, 1991.
However, here's an excerpt from the 2013 paper by Chee, Colbourn, Ling, and Wilson:
Their approach relies in an essential manner on a strong statement by Gustavsson:
Proposition 1.1: . Let $H$ be a graph with $v$ vertices and $h$ edges, having degree sequence $(d_1,\dots,d_v)$. Then there exist a constant $N_H$ and a constant $\epsilon_H>0$, both depending only on $H$, such that for all $n>N_H$, if $G$ is a graph on $n$ vertices, $m$ edges, and degree sequence $(\delta_1,\dots,\delta_n)$ so that $\min(\delta_1,\dots,\delta_n) \geq n(1−\epsilon_H)$, $\gcd(d_1,\dots,d_v)\ \vert\ \gcd(\delta_1,\dots,\delta_n)$, and $h\ \vert\ m$, then $G$ has an edge partition (decomposition) into graphs isomorphic to $H$.
We have not been able to verify the proof of Proposition 1.1. Indeed, while the result has been used a number of times in the literature, no satisfactory proof of it appears there. While we expect that the statement is true, we do not think that the proof in [12] is sufficient at this time to employ the statement as a foundation for further results. Therefore we adopt a strategy that is completely independent of Proposition 1.1, and independent of the results built on it.
So, one might want to be careful about Caro and Yuster's claimed solution. Chee, Colbourn, Ling, and Wilson gave a slightly weaker upper bound on the minimum size of a $(v,k,2)$ covering, which essentially states that for sufficiently large $v$, Caro and Yuster's claimed upper bound is correct within a constant that depends on $k$. (And the proof of Gustavsson's theorem by Barber, Kühn, Lo and Osthus shows it is correct.)
• Together with Daniela Kühn, Allan Lo and Deryk Osthus I have a submitted a paper that proves that statement of Gustavsson and gives quantitative bounds on the value of $\epsilon_H$. The most recent version can be found on the arXiv. arxiv.org/abs/1410.5750 – Ben Barber Jan 5 '15 at 13:35
• @BenBarber That's awesome! Thank you for the comment and link, and thank you for your work! – Yuichiro Fujiwara Jan 5 '15 at 13:40
• And please feel free to make this post more informative and/or correct errors, or post another answer. – Yuichiro Fujiwara Jan 5 '15 at 14:55
Fort and Hedlund determined the minimum size of a $(v,3,2)$-covering design: Minimal coverings of pairs by triples, Pacific J. Math. 8(4), 709-719, 1958.
The case $(v,4,2)$ was solved by Mills: On the covering of pairs by quadruples I, JCTA 13, 55–78, 1972 and II, JCTA 15, 138–166 (1973).
For the case $(v,5,2)$ with $v\equiv 0\pmod 4$, Abel, Assaf, Bennett, Bluskov and Greig established that the Schönheim bound $\left\lceil\frac{v}{5}\left\lceil \frac{v-1}{4}\right\rceil\right\rceil$ is tight for $v\geqslant 28$ with 17 possible exceptions in the range $40\leqslant v\leqslant 280$: Pair covering designs with block size 5, Discrete Mathematics 307(14), 1776-1791, 2007.
The same authors have results for $(v,6,2)$: Pair covering and other designs with block size 6, J Comb Des 15(6), 511-533, 2007
Caro and Yuster proved that for sufficiently large $v$ the minimum size of a $(v,k,2)$ covering design is $\left\lceil\frac{v}{k}\left\lceil \frac{v-1}{k-1}\right\rceil\right\rceil$: Covering Graphs: The Covering Problem Solved, JCTA 83 (2) 273–282, 1998.
The La Jolla Covering repository is a good source for covering designs. According to this table the smallest open case with $t=2$ is $(v,k,t)=(23,6,2)$ where the minimum size is either 20 or 21.
• Thanks, that's usefull, I allready knew about la Jolla, the second source is great! Do you know if there has been any work on the bounds for arbitrary $(v,k,2)$? – Jorge Fernández Jan 3 '15 at 21:24
• I know there are resuts for $(v,k,t)$ in general, but I though maybe with $(v,k,2)$ they could be sharper – Jorge Fernández Jan 3 '15 at 21:25
• Caro and Yuster's proof might be incomplete (although the statement itself is probably correct). See my answer (and the paper I linked to) for more detail. – Yuichiro Fujiwara Jan 3 '15 at 23:41 | 2020-01-18T16:46:46 | {
"domain": "mathoverflow.net",
"url": "https://mathoverflow.net/questions/192058/covering-designs-of-the-form-v-k-2/192065",
"openwebmath_score": 0.7909506559371948,
"openwebmath_perplexity": 280.53751763005624,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9865717476269469,
"lm_q2_score": 0.8633916222765627,
"lm_q1q2_score": 0.8517977816758533
} |
https://math.stackexchange.com/questions/978/how-to-prove-and-interpret-operatornamerankab-leq-operatornamemin-ope?noredirect=1 | # How to prove and interpret $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$?
Let $$A$$ and $$B$$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$
I proved $$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$$ by interpreting $$AB$$ as a composition of linear maps, observing that $$\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information.
How do you manage $$\operatorname{rank}(AB) \leq \operatorname{rank}(A)$$? Is there a nice interpretation like the previous one?
• Your proof is fine. Furthermore, the same reasoning will get your desired fact. Again rank-nullity will tell you that the dimension of your vector space minus the dimension of the kernel will give you the rank. Jul 28, 2010 at 16:08
Yes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get "more stuff" in the former case, as the former is bigger than the latter.
• Thank you. I was so obsessed with the kernel-image dimension formula that I could't recognize this simple fact.
– user365
Jul 30, 2010 at 8:52
Once you have proved $$\operatorname{rank}(AB) \le \operatorname{rank}(A)$$, you can obtain the other inequality by using transposition and the fact that it doesn't change the rank (see e.g. this question).
Specifically, letting $$C=A^T$$ and $$D=B^T$$, we have that $$\operatorname{rank}(DC) \le \operatorname{rank}(D) \implies \operatorname{rank}(C^TD^T)\le \operatorname{rank} (D^T)$$, which is $$\operatorname{rank}(AB) \le \operatorname{rank}(B)$$.
• Very nice! Thank you.
– user365
Sep 2, 2010 at 10:01
• I am doing this problem now too. Why is proving both inequalities enough? Where does the proof take the min idea into account? Apr 12, 2016 at 3:27
• @MathisHard $x<y \wedge x<z \implies x<\min(y,z)$. May 25, 2018 at 15:47
Note that $$\text{Col}(AB) ⊆ \text{Col}(A)$$ since given $$y ∈ \text{Col}(AB)$$ we can choose $$x ∈ F$$ and then we have $$y = (AB)x = A(Bx) ∈ \text{Col}(A)$$.
Since $$\text{Col}(AB) ⊆ \text{Col}(A)$$, any basis for $$\text{Col}(AB)$$ can be extended to a basis for $$\text{Col}(A)$$ and so $$\dim \text{Col}(AB) ≤ \dim \text{Col}(A)$$, that is $$\text{rk}(AB) ≤ \text{rk}(A).$$
Note that $$\text{Null}(B) ⊆ \text{Null}(AB)$$ since given $$x ∈ \text{Null}(B)$$ we have $$(AB)x = A(Bx) = A 0 = 0$$ so that $$x ∈ \text{Null}(AB)$$.
Since $$\text{Null}(B) ⊆ \text{Null}(AB)$$, as above we have $$\dim \text{Null}(B) ≤ \dim \text{Null}(AB)$$, that is, $$\text{Nullity}(B) ≤\text{Nullity}(AB)$$. Thus $$\text{rk}(AB) = n − \text{Nullity}(AB) ≤ n − \text{Nullity}(B) = \text{rk}(B).$$
• You should use MathJax to format your answer. Oct 15, 2015 at 21:06
• Very nice proof! +1
– ZFR
Apr 23, 2020 at 1:41
Prove first that if $f:X\to Y$ and $g:Y\to Z$ are functions between finite sets, then $|g(f(X))| \leq \min \{ |f(X)|, |g(Y)| \}.$
Then use the same idea.
• Categorification... :-) Jul 29, 2010 at 21:50
• I am not familiar with the 'categorification'. How can one go from this to the rank inequality ? What functor is to be applied ? Apr 13, 2014 at 12:10
• So you're saying that rank of linear map is somehow like image of a function? Feb 12, 2015 at 10:15
• @Mihail, it is like the size of the image of a function. in fact, if $f$ is a linear map, the rank of $f$ is the dimension of the image of $f$ and the dimension is a measure of the size of a space. Feb 12, 2015 at 18:22
Here is another simple answer. When you multiply a matrix and a vector $Ax$ you end up with a linear combination of the columns of $A$.
$$Ax = \; x_1\,A_1 \;+\; x_2\,A_2 \;+\; x_3\,A_3 \;+\;\; ...\;\; \\$$
When we multiply two matrices $AB = C$, we have $AB_i = C_i$, which means that each column of $C$ is a linear combination of the columns of $A$, so $\text{rank}(AB) \leq \text{rank}(A)$. To show that $\text{rank}(AB) \leq \text{rank}(B)$ we follow a similar argument -- when you multiply $x^{\top}B$, you end up with a linear combination of the rows of $B$.
Already you have proved $$rank(AB)\leq rank(B)$$
For other part
rank of $A=$ dim range $A$
As range $AB \subset$ range $A\implies$ dim range $AB\leq$ dim range $A$ . Hence $$rank(AB)\leq rank (A)$$
• Hello!! Why does it hold that "range AB $\subset$ range A" ? Jan 7, 2017 at 10:01
• If $x\in$ range $AB,$ then $x=(AB)y$ for some $y.$ Thus $x=A(By).$ Thus $x\in$ range $A.$@ Mary Star Jan 8, 2017 at 17:48
Let $m \le n, A \in M_{m\times n}, B\in M_{n\times m}$.
$\mbox{rank } A\le m$ and $\text{rank }B\le m$. (Obvious fact as rank A = dimension of the columnspace of A = dimension of the row space of A.)
Let $E_{n\times n}B$ be the row echelon form of $B$ and let $AE_{m\times m}$ be the column echelon form of $A$. ($E_{n\times n} ,E_{m\times m}$ are elementary matrices.)
We know $\operatorname{rank}(BA)=\operatorname{rank}(E_{n\times n}BA )=\operatorname{rank}(E_{n\times n}BAE_{m\times m} )$.
But $E_{n\times n}BAE_{m\times m} =\begin{pmatrix} L&0\\ 0&0\\ \end{pmatrix},$ where $L$ is an $k\times l$ matrix with $k\le \operatorname{rank}(B),l\le \operatorname{rank}(A)$.
So $\operatorname{rank}(E_{n\times n}BAE_{m\times m} )=\operatorname{rank}\begin{pmatrix} L&0\\ 0&0\\ \end{pmatrix}\le \min\{k,l\}\le \min\{\mbox{rank } A,\mbox{rank }B\}.$
Another way of showing that $\text{Rank} (AB) \leq \text{Rank}(B)$ without using rank-nullity: Note that if $v_1,\dots,v_n$ is a basis of $\text{Range} B$, then $Av_1,\dots,Av_n$ is a spanning list of $\text{Range} AB$.
Each column of $$AB$$ is a linear combination of columns of $$A$$ so $$\text{Range}(AB)\subseteq \text{Range} (A)$$ equivalently $$rank(AB)\leq rank(A).$$
Similarly, Each row of $$AB$$ is a linear combination of rows of $$B$$ so $$rowspace(AB)\subseteq rowspace(B)$$.
Since rowspace=columnspace. Thus $$rank(AB)\leq rank(B)$$.
From both of the inequality we deduce that $$rank(AB)\leq \min\{rank A, rank B\}$$. | 2022-08-18T02:45:28 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/978/how-to-prove-and-interpret-operatornamerankab-leq-operatornamemin-ope?noredirect=1",
"openwebmath_score": 0.9413327574729919,
"openwebmath_perplexity": 188.73014372270734,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9865717440735736,
"lm_q2_score": 0.8633916152464017,
"lm_q1q2_score": 0.8517977716721424
} |
https://math.stackexchange.com/questions/1248235/groups-with-few-subgroups | # Groups with “few” subgroups
If $G$ is a finite group of order $n,$ and the number of divisors of $n$ is $k,$ can $G$ have fewer than $k$ subgroups?
A cyclic group $G$ of order $n$ has exactly one subgroup for each divisor of $n$, so in this case $G$ has exactly $k$ subgroups.
For the alternating group $G=A_4,$ $n=12$ and $k=6.$ There is no subgroup of order $6$ in $A_4,$ yet there are ten subgroups, which exceeds $k=6$ for this case. I began to wonder if there could be a group for which so many divisors of the order of $G$ had no subgroups of that order, that there wound up being fewer subgroups in $G$ than the number of divisors of $|G|,$ hence my question.
• Recall that a CLT (Converse of Lagrange's Theorem) group is a group with the property that for any divisor of the order of the group there exists a subgroup of that order. Since it is known that finite abelian groups, as well as $p$-groups are CLT groups, this leaves a smaller class of subgroups to consider. – user1337 Apr 23 '15 at 13:16
• A finite group $G$ of order $n$ such that the number of elements $x$ with $x^d=1$ is less than $d$ for all divisors $d$ of $n$ has to be cyclic. I wonder, if a variation of this topic can lead to a proof that there are at least as many subgroups as divisors. – j.p. Apr 23 '15 at 13:24
• @j.p. That idea looks promising. One thing, should the phrase "is less than $d$" be replaced by "is at most $d$" in your comment? [though that would imply your version, I seem to recall one approach to doing the counting argument for showing some group is cyclic as involving the number being at most $d$] And +1 on the comment. – coffeemath Apr 23 '15 at 13:42
• A short script in Maple shows that there are no such groups among the 793 groups of order $< 2^5 \cdot 3 = 96$. – Travis Willse Apr 23 '15 at 13:43
• @coffeemath: Yes, you are right. It should have been "at most". – j.p. Apr 23 '15 at 14:04
The answer to your question is no, that is not possible. This is a consequence of the following slightly more general result. Let $d(n)$ is the number of divisors of $n$.
Theorem Let $G$ be a finite group of order $n$. Then for any divisor $m$ of $n$, there exist at least $d(m)$ subgroups of $G$ of order dividing $m$.
Proof We prove the statement by induction on $m$, for all finite groups $G$. It's clear for $m=1$.
For $m>1$, let $p$ be a prime dividing $m$, and let $m = p^r k$ with $\gcd(p,k)=1$.
Since $d(m) = (r+1)d(k)$, it is sufficient to prove that, for each $i$ with $0 \le i \le r$, there are at least $d(k)$ subgroups of $G$ of order $p^i j$, for some $j$ that divides $k$.
So fix an $i$, let $P$ be any subgroup of $G$ of order $p^i$, let $N = N_G(P)$ be the normalizer of $P$ in $G$, and let $h = \gcd(|N|,k)$.
Since $h \le k < m$ and $h$ divides $|N/P|$, by inductive hypothesis, the number $s$ of subgroups $S/P$ of $N/P$ of order dividing $h$ satisfies $s \ge d(h)$. For each of these subgroups, the inverse image $S$ of $S/P$ in $N/P$ is a subgroup of $G$ of order $p^i j$, where $|S/P| = j$ divides $h$, and so $j$ divides $k$.
Now $P$ has $|G|/|N|$ distinct conjugates $P^g$ in $G$, and the $s|G|/|N|$ subgroups $S^g$ are all distinct. But $|G|/|N| \ge k/h$, so $s|G|/|N| \ge d(h)k/h \ge d(k)$, which completes the proof.
After thinking about it again, I think we can say that a group of order $n$ with exactly $d(n)$ subgroups must be cyclic. The above proof produces at least $d(n)$ subgroups with a normal $p$-subgroup for any prime divisor $p$ of $n$. So if there were exactly $d(n)$ subgroups, then all subgroups would have all of their Sylow subgroups normal, so they would all, including $G$ itself, be nilpotent. But a non-cyclic $p$-group $P$ has more than one subgroup of index $p$, and hence has more than $d(|P|)$ subgroups, so all Sylow subgroups of $G$ are cyclic and hence so is $G$.
By the way, I first posted this proof here in 2003. I would welcome any references to any other proofs.
• Derek-- Nice proof (seems to me). +1 on it for now, and likely I'll "accept" after I get a chance to go over it in detail. – coffeemath Apr 23 '15 at 14:46
• How do you know that $P$ exist ? – Belgi Apr 23 '15 at 14:57
• @Belgi $G$ has a $p$-Sylow subgroup $T$. Let's say $T$ is of size $p^k$; then $T$ will have a subgroup of order $p^i$ provided $i\leq k$. – mathmandan Apr 23 '15 at 15:06
• @mathmandan - How do you know such subgroups of $T$ exist ? – Belgi Apr 23 '15 at 15:10
• I was thinking of the proof as an induction on $m$ for all $n$ simultaneously. That is, when doing the proof for $m$, we assume that the result is true for all smaller $m$ and all $n$. We don't use induction on $n$ anywhere. I remember when thinking about this originally, the chief problem was to get an induction to work, and the key idea in the proof is actually to induct on $m$ rather than on $n$. – Derek Holt Apr 23 '15 at 17:06 | 2021-04-11T02:07:53 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1248235/groups-with-few-subgroups",
"openwebmath_score": 0.8602133393287659,
"openwebmath_perplexity": 92.40538364806898,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9865717432839349,
"lm_q2_score": 0.8633916152464016,
"lm_q1q2_score": 0.8517977709903749
} |
http://math.stackexchange.com/questions/291602/generating-sets-from-given-information | # Generating Sets From Given Information
The problem I am working on is:
The three most popular options on a certain type of new car are a built-in GPS (A), a sunroof (B), and an automatic transmission (C). If 40% of all purchasers request A, 55% request B, 70% request C, 63% request A or B, 77% request A or C, 80% request B or C, and 85% request A or B or C, determine the probabilities of the following events. [Hint: “Aor B” is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.]
a.The next purchaser will request at least one of the three options.
b.The next purchaser will select none of the three options.
c. The next purchaser will request only an automatic transmission and not either of the other two options.
d.The next purchaser will select exactly one of these three options.
I am absolutely positive that $P(A)=40\%$, $P(B)=55\%$, $P(C)=70\%$, and $P(A \cup B \cup C)=85\%$. However, the pieces of data I am not quite certain about are $P(A \cup B \cap C')=63\%$, $P(A \cup C \cap B')=77\%$, $P(B \cup C \cap A')=80\%$, do these values correspond to the rest of the data? If so, then I seems nearly impossible to be able to generate the Venn Diagram. Could someone help?
EDIT: What I am having a difficult time interpreting is, when they say in the question, "...63% request A or B." To me, that says only A or only B; and under this interpretation I would write $P(A \cup B \cap C')=63\%$. Under André Nicolas' interpretation, "63% request A or B," means $P(A \cup B)=63\%$. If it is the case that André Nicolas is correct, then it seems like they should have stated in the question, "63% request A or B, A and C, B and C, or A and B and C."
Also, I solved the problem under André Nicolas' assumption, and for part d), I know the answer but I am sure how to put in it math symbols. How would I do that?
-
Here is a start:
a) The answer to this is written down in the data: It is $\Pr(A\cup B\cup C)$.
b) This can be immediately written down from the answer to a).
As for the rest of the probabilities, we need to work. It can be figured out completely from a Venn diagram. But since one cannot point to a picture on a blackboard, what follows will be formula-heavy.
Recall that $$\Pr(X\cup Y)=\Pr(X)+\Pr(Y)-\Pr(X\cap Y).\tag{1}$$
We know $\Pr(A)$, $\Pr(B)$, and $\Pr(A\cup B)$. Using Formula $(1)$, we can see that $0.63=0.40+0.55-\Pr(A\cap B)$. That gives $\Pr(A\cap B)=0.32$.
Similarly, we can find $\Pr(A\cap C)$ and $\Pr(B\cap C)$ from the given information.
Recall also that
$$\Pr(A\cup B\cup C)=\Pr(A)+\Pr(B)+\Pr(C)-\Pr(A\cap B)-\Pr(A\cap C)-\Pr(B\cap C)+\Pr(A\cap B\cap C).$$
We are told $\Pr(A\cup B\cup C)$, and we computed the next three items in the above formula, so now we know $\Pr(A\cap B\cap C)$.
OK, time to fill in the bits in the Venn Diagram. First write the computed $\Pr(A\cap B\cap C)$ in the space where it should go.
We know $\Pr(A\cap B)$. Since we also know $\Pr(A\cap B\cap C)$, we can find the probability of "the rest of" $A\cap B)$, called $A\cap B\cap C'$ in set language. Compute, write the number in.
Do the same for the other similar bits, namely $A\cap C\cap B'$ and $B\cap C\cap A'$.
Now from $\Pr(A)$ you can find the probability of the "outside" part of $A$, that is, $\Pr(A\cap B'\cap C')$. Do the same for the "outside" part of $B$, also the "outside" part of $C$.
The probability of the outer world $A'\cap B'\cap C'$ of cheapskates who want nothing can now be found in various ways.
We now know everything, and can answer any question!
Remark: You can be more fancy, and let $X$, $Y$, and $Z$ respectively be $A'$, $B'$, and $C'$. By taking every one of the given numbers and subtracting it from $1$, you can write down immediately $\Pr(X)$ (namely $1-0.40$), $\Pr(Y)$, $\Pr(Z)$, $\Pr(X\cap Y)$, and so on. Now we end up with a problem of a type that you have probably done several times.
-
So, in the question, when it said something like, "63% request A or B," you assumed that two be $P(A \cup B)=63\%$? Why isn't it really $P(A \cup B \cap C')=63\%$? If you take, "63% request A or B," to mean the former, why wouldn't you include something about C? Because when you write the Venn diagram, A and B both intersect C. – Mack Feb 1 '13 at 14:56
The reason I ask is because $P(A \cup B)$ also includes the probability of getting $A$ and $C$ and $B$ and $C$, and it just seems odd that you wouldn't mention something like that in the question. – Mack Feb 1 '13 at 15:17
@EliMackenzie: The problem setter should make things clear. But I am reasonably sure that the intent here is that "or" be interpreted as $\lor$, logical or, meaning in this case union. There is a somewhat similar potential issue with "and" in standard Venn diagram work problems. When I say $A$ and $B$, does that imply anything about $C$, $D$? No, and you are accustomed to that. – André Nicolas Feb 1 '13 at 16:49
One can also go through the Venn diagram, trying to see whether your interpretation is numerically consistent with the data. The probabilities of $A\lor B$, $B\lor C$, $A\lor C$ are quite large, one can't find a Venn diagram that works. Some of the percentages of "subareas" would turn out negative. – André Nicolas Feb 1 '13 at 16:56
Also, if you look at the text of the problem, it says explicitly as a hint what is meant by $A$ or $B$. That is neutral about $C$. It looks as if the problem setter was worried people might have trouble with the meaning of or. – André Nicolas Feb 1 '13 at 17:03 | 2014-04-16T19:33:49 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/291602/generating-sets-from-given-information",
"openwebmath_score": 0.7819040417671204,
"openwebmath_perplexity": 319.0762801610089,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9865717444683928,
"lm_q2_score": 0.8633916082162403,
"lm_q1q2_score": 0.8517977650772673
} |
https://fr.mathworks.com/help/symbolic/legendrep.html | # legendreP
Legendre polynomials
## Description
example
legendreP(n,x) returns the nth degree Legendre polynomial at x.
## Examples
### Find Legendre Polynomials for Numeric and Symbolic Inputs
Find the Legendre polynomial of degree 3 at 5.6.
legendreP(3,5.6)
ans =
430.6400
Find the Legendre polynomial of degree 2 at x.
syms x
legendreP(2,x)
ans =
(3*x^2)/2 - 1/2
If you do not specify a numerical value for the degree n, the legendreP function cannot find the explicit form of the polynomial and returns the function call.
syms n
legendreP(n,x)
ans =
legendreP(n, x)
### Find Legendre Polynomial with Vector and Matrix Inputs
Find the Legendre polynomials of degrees 1 and 2 by setting n = [1 2].
syms x
legendreP([1 2],x)
ans =
[ x, (3*x^2)/2 - 1/2]
legendreP acts element-wise on n to return a vector with two elements.
If multiple inputs are specified as a vector, matrix, or multidimensional array, the inputs must be the same size. Find the Legendre polynomials where input arguments n and x are matrices.
n = [2 3; 1 2];
xM = [x^2 11/7; -3.2 -x];
legendreP(n,xM)
ans =
[ (3*x^4)/2 - 1/2, 2519/343]
[ -16/5, (3*x^2)/2 - 1/2]
legendreP acts element-wise on n and x to return a matrix of the same size as n and x.
### Differentiate and Find Limits of Legendre Polynomials
Use limit to find the limit of a Legendre polynomial of degree 3 as x tends to -∞.
syms x
expr = legendreP(4,x);
limit(expr,x,-Inf)
ans =
Inf
Use diff to find the third derivative of the Legendre polynomial of degree 5.
syms n
expr = legendreP(5,x);
diff(expr,x,3)
ans =
(945*x^2)/2 - 105/2
### Find Taylor Series Expansion of Legendre Polynomial
Use taylor to find the Taylor series expansion of the Legendre polynomial of degree 2 at x = 0.
syms x
expr = legendreP(2,x);
taylor(expr,x)
ans =
(3*x^2)/2 - 1/2
### Plot Legendre Polynomials
Plot Legendre polynomials of orders 1 through 4.
syms x y
fplot(legendreP(1:4, x))
axis([-1.5 1.5 -1 1])
grid on
ylabel('P_n(x)')
title('Legendre polynomials of degrees 1 through 4')
legend('1','2','3','4','Location','best')
### Find Roots of Legendre Polynomial
Use vpasolve to find the roots of the Legendre polynomial of degree 7.
syms x
roots = vpasolve(legendreP(7,x) == 0)
roots =
-0.94910791234275852452618968404785
-0.74153118559939443986386477328079
-0.40584515137739716690660641207696
0
0.40584515137739716690660641207696
0.74153118559939443986386477328079
0.94910791234275852452618968404785
## Input Arguments
collapse all
Degree of polynomial, specified as a nonnegative number, vector, matrix, multidimensional array, or a symbolic number, vector, matrix, function, or multidimensional array. All elements of nonscalar inputs should be nonnegative integers or symbols.
Input, specified as a number, vector, matrix, multidimensional array, or a symbolic number, vector, matrix, function, or multidimensional array.
collapse all
### Legendre Polynomial
• The Legendre polynomials are defined as
$P\left(n,x\right)=\frac{1}{{2}^{n}n!}\frac{{d}^{n}}{d{x}^{n}}{\left({x}^{2}-1\right)}^{n}.$
• The Legendre polynomials satisfy the recursion formula
$\begin{array}{l}P\left(n,x\right)=\frac{2n-1}{n}xP\left(n-1,x\right)-\frac{n-1}{n}P\left(n-2,x\right),\\ \text{where}\\ P\left(0,x\right)=1\\ P\left(1,x\right)=x.\end{array}$
• The Legendre polynomials are orthogonal on the interval [-1,1] with respect to the weight function w(x) = 1, where
• The relation with Gegenbauer polynomials G(n,a,x) is
$P\left(n,x\right)=G\left(n,\frac{1}{2},x\right).$
• The relation with Jacobi polynomials P(n,a,b,x) is
$P\left(n,x\right)=P\left(n,0,0,x\right).$ | 2020-12-04T21:34:12 | {
"domain": "mathworks.com",
"url": "https://fr.mathworks.com/help/symbolic/legendrep.html",
"openwebmath_score": 0.937826931476593,
"openwebmath_perplexity": 2556.7520994579972,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9865717424942961,
"lm_q2_score": 0.8633916082162403,
"lm_q1q2_score": 0.8517977633728488
} |
https://www.physicsforums.com/threads/meaning-of-a-subspace.884758/ | # Meaning of a Subspace
## Main Question or Discussion Point
Hi PF!
I want to make sure I understand the notion of a subspace. Our professor gave an example of one: the set of degree n polynomials is a subspace of continuous functions. This is because a) a polynomial is intrinsically a subset of continuous functions and b) summing any polynomials yields another polynomial (closed under addition) and c) any polynomial multiplied by a constant is still a polynomial (closed under scalar multiplication).
If my reasoning is correct, then I think the set of discontinuous functions is NOT a subset of all real functions. To see why, consider $f(x) = 1$ everywhere except non-existent at $x=1$. Then take the function $g(x)=1$ when $x=1$ and non-existent everywhere else. Then the sum is clearly continuous. (How would this change though if I changed non-existent to zero? I know the result is the same, but is $g(x)$ as I have defined it even discontinuous--I realize this is a real analysis question.)
Related Linear and Abstract Algebra News on Phys.org
PeroK
Homework Helper
Gold Member
Hi PF!
I want to make sure I understand the notion of a subspace. Our professor gave an example of one: the set of degree n polynomials is a subspace of continuous functions. This is because a) a polynomial is intrinsically a subset of continuous functions and b) summing any polynomials yields another polynomial (closed under addition) and c) any polynomial multiplied by a constant is still a polynomial (closed under scalar multiplication).
If my reasoning is correct, then I think the set of discontinuous functions is NOT a subset of all real functions. To see why, consider $f(x) = 1$ everywhere except non-existent at $x=1$. Then take the function $g(x)=1$ when $x=1$ and non-existent everywhere else. Then the sum is clearly continuous. (How would this change though if I changed non-existent to zero? I know the result is the same, but is $g(x)$ as I have defined it even discontinuous--I realize this is a real analysis question.)
You can't choose functions that are not defined at some point, as functions that have different domains can't be added so they don't form a vector space. The space of all real-valued functions would imply that they all all defined on some fixed domain.
Also, functions that are not defined at a point are not necessarily discontinuous. They may be continuous on their domain. A good example is the function $1/x$, which is a continuous function on its domain.
Instead, you need to think of two genuinely discontinuous functions that can be added to form a contiuous function.
Or, perhaps you could think of a simpler example using a subset of the polynomials that does not form a subspace?
S.G. Janssens
I would say: "The set of polynomials of degree at most $n$". (Here you regard the zero polynomial as having degree $-\infty$ or you should stipulate that this set includes the zero polynomial.) | 2020-01-18T22:25:27 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/meaning-of-a-subspace.884758/",
"openwebmath_score": 0.8620683550834656,
"openwebmath_perplexity": 209.91472519802872,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9783846672373523,
"lm_q2_score": 0.870597270087091,
"lm_q1q2_score": 0.8517790203919058
} |
http://math.stackexchange.com/questions/124277/how-to-find-the-partial-sum-of-a-given-series | # How to find the partial sum of a given series?
On my last exam there was the question if the series $\sum_{n=2}^{\infty}\frac{1}{(n-1)n(n+1)}$ converges and which limit it has. During the exam and until now, I am not able to solve it. I tried partial fraction decomposition, telescoping sum, etc. But I am not able to find the partial sum formula (Wolfram|Alpha):
$$\sum_{n=2}^{m}\frac{1}{(n-1)n(n+1)} = \frac{m^2+m-2}{4m(m+1)}.$$
Could somebody push me in the right direction? Is there any trick or scheme how to find partial sum formulas for given series?
-
Each term is less than $1/n^2$, so it converges. You don't need to know what a series converges to to know that it converges. – Thomas Andrews Mar 25 '12 at 14:46
Two things to observe(Unrelated to the Math): Please do not sign your posts with signature as faq explicitly lists it! Accepting answers is a sign of appreciation for someone who put in effort compiling an answer for you. Please accept answers which you think helped you a lot in solving that problem or cleared up your concepts and whatever. It is done by clicking on the tick mark besides every answer. – user21436 Mar 25 '12 at 14:51
The ratio test with $1/n^3$ proves it converges. As often happens, finding the sum takes more work than that. – Michael Hardy Mar 25 '12 at 16:29
martini's answer is the right thing to do. Just wanted to add that if you know the answer (given by wolfram) you can just prove the it with recursion arguments – Thomas Dec 15 '13 at 15:11
So let's try partial fraction decomposition. Writing $$\frac 1{(n-1)n(n+1)} = \frac a{n-1} + \frac bn + \frac c{n+1}$$ we obtain $$1 = a(n^2 + n) + b(n^2 - 1) + c(n^2 - n)$$ and therefore \begin{align*} 1 &= -b\\ 0 &= a - c\\ 0 &= a + b + c. \end{align*} This gives $b = -1$, $a = c = \frac 12$. Hence \begin{align*} \sum_{n=2}^m \frac 1{(n-1)n(n+1)} &= \sum_{n =2}^m \frac 1{2(n-1)} - \sum_{n=2}^m \frac 1n + \sum_{n=2}^m \frac 1{2(n+1)}\\ &= \frac 12 + \sum_{n=2}^{m-1} \frac 1{2n} - \sum_{n=2}^m \frac 1n + \sum_{n=3}^m \frac 1{2n} + \frac 1{2(m+1)}\\ &= \frac 12 + \frac 14 - \frac 12 - \frac 1m + \frac 1{2m} + \frac 1{2m+2}\\ &= \frac 14 + \frac{-2(m+1) + m+1 + m}{2m(m+1)}\\ &= \frac 14 + \frac{-1}{2m(m+1)}\\ &= \frac{m(m+1) - 2}{4m(m+1)}. \end{align*}
I got little confuse here, why the term $\frac{1}{4}$ doesnt vanish? In my calculation the third step from below, I got $$=\frac{1}{2}+ (\frac{1}{4}+ \frac{1}{6}+\dots+ \frac{1}{2(m-1)})-( \frac{1}{2}+ \frac{1}{3}+\dots+ \frac{1}{m})+( \frac{1}{6}+ \frac{1}{8}+\dots+ \frac{1}{2m})+ \frac{1}{2(m+1)} = \\ =-( \frac{1}{3}+ \frac{1}{5}+\dots+ \frac{1}{m})+(\frac{1}{6}+\frac{1}{8}+\dots+ \frac{1}{2m})+\frac{1}{2(m+1)}$$ How can I get to the next step? – DadangAH Feb 20 '14 at 5:41
What did you try for a telescoping sum? Your denominator here is the product of three successive terms (this is often called a rising or falling factorial, depending on which side you take as your baseline); this points to looking at a difference of terms that are of the same form but with denominators one degree less. In particular, looking at $t_n=\dfrac{1}{n(n+1)}$ then $t_n-t_{n-1}$ $=\dfrac{1}{n(n+1)}-\dfrac{1}{(n-1)n}$ $=\dfrac1n\left(\dfrac1{n+1}-\dfrac1{n-1}\right)$ $=\dfrac1n\left(\dfrac{(n-1)-(n+1)}{(n-1)(n+1)}\right)$ $=\dfrac{-2}{(n-1)n(n+1)}$; in other words, $\dfrac{1}{(n-1)n(n+1)} = -\dfrac12(t_n-t_{n-1})$, and from here the telescopy should be fairly clear. | 2015-04-27T21:23:32 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/124277/how-to-find-the-partial-sum-of-a-given-series",
"openwebmath_score": 0.9995575547218323,
"openwebmath_perplexity": 497.2603487002978,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.978384668497878,
"lm_q2_score": 0.8705972600147106,
"lm_q1q2_score": 0.8517790116346535
} |
https://math.stackexchange.com/questions/2042140/precise-proof-of-s-open-set-iff-contains-only-interior-points-complex-analysis | Precise proof of S Open set iff contains only interior points - Complex Analysis
In "Complex Variables and Applications" by Brown and Churchill (McGraw-Hill) the proof of: $$\textit{S is an open set \implies each of its points is an interior point}$$ is left as an exercise. Here are the important definitions that one can use: $$\textbf{\epsilon Neighbourhood}: |z-z_0|<\epsilon$$ $$\textbf{z_0 Interior point}: \exists\,\,\ \text{an}\,\,\, \epsilon\text{-neighbourhood containing only}\, z\in S$$ $$\textbf{z_0 Exterior point}: \exists\,\,\ \text{an}\,\,\, \epsilon\text{-neighbourhood containing no}\, z\in S$$ $$\textbf{z_0 Boundary point}: \text{all neighbourhood of z_0 have at least a point in S and one not in S}$$ $$\textbf{S Open}: \text{does not contain any of its boundary points}$$
I've tried to prove this in the following way, however I got stuck. Can someone tell me how to make it mathematically rigorous, using the above definitions?
$S$ is an Open Set $\implies$ $S$ does not contain any any of its boundary points. So $$\text{if} \,\, [\forall \epsilon>0, (\exists z_1\in S\,\,\ \text{and}\,\,\ \exists z_2\notin S ):(|z_1-z_0|<\epsilon \,\,\text{and} \,\,\ |z_2-z_0|<\epsilon)] \implies z_0\notin S$$ Hence taking the converse of the above we have $$\text{if}\,\, z_0\in S \implies \exists \epsilon>0, (\forall z_1\in S\,\,\ \text{or}\,\,\ \forall z_2\notin S ):|z_1-z_0|\geq \epsilon \,\,\text{or} \,\,\ |z_2-z_0|\geq \epsilon)$$
Which doesn't really tell me much. I am really not sure I've even negated it correctly or that my definition with the quantifiers is correct.
• This are the exact words in the book "A point $z_0$ is said to be an interior point of a set $S$ whenever there is some neighborhood of $z_0$ that contains only points of $S$". Is it different from what I wrote? Or is it different from what you wrote? They look pretty similar to me, please tell me if there's some gap between them – Euler_Salter Dec 3 '16 at 18:48
• Forgive me, I misunderstood you definition. However it is a bit confusing because it seems that you are referring to a specific point $z$. Specifically, if I say that $A$ contains only $z\in S$ what I immediately think is that $A=\{z\}$, whatever $z$ is, while you only meant $A\subseteq S$. – Caligula Dec 3 '16 at 18:54
• I removed the "logic" tag - that's for questions within the specific field of mathematical logic. – Noah Schweber Dec 3 '16 at 19:00
Using your definitions (of which some are not very well stated), we can proceed this way.
You want to prove that each point $z$ of an open set $S$ is an interior point. Now, since $S$ is open, $z$ can't be a boundary point, so negating the definition of boundary point, it follows that there exists a neighborhood $U$ of $z$ such that two things can happen:
• or $U$ contains only point of $S$;
• or $U$ contains only point outside $S$.
Since $U$ intersect $S$ at least in $z$, the latter is impossible, so you have proved that $U\subseteq S$. This means that $z$ is interior in $S$.
Following the comment, I'll try to rephrase the proof in more logically strict terms - but note that using words instead of quantifiers is not something to despise, as it increases greatly the clearity of proofs.
The fact $S$ is open means that
$\forall\,z\in S, z\notin \partial S$
where $\partial S$ is the boundary. Now, $z\in \partial S$ if and only if
$\forall \,U\in\mathscr{U}_z,\,U\cap S\neq \varnothing \text{ and } U\cap (S^c)\neq \varnothing$
where $(-)^c$ is the complement and $\mathscr{U}_z$ denotes the set of all open neighborhoods for $z$ (it has the structure of a filter if you know what it means). Hence, negating the latter will imply that
$\exists \,U\in \mathscr{U}_z$ such that $U\cap S = \varnothing$ or $U\cap (S^c)=\varnothing$
Then you conclude as before: the only possible conclusion is that $U\cap (S^c)=\varnothing$ and that is equivalent to $U\subseteq S$.
• I like your wordy proof, I'm quite a fan of these, as in most of my homework I exactly do them. However I wanted to train myself in proving things in a "tedious" way, i.e. using a logic definition and using either negations of it or contrapositives to get to the statement we want to prove. Would you know how I could rephrase those definitions or actually going from there to the wanted result? Thanks – Euler_Salter Dec 3 '16 at 18:58
• I edited my answer with some more logic inside it. – Caligula Dec 3 '16 at 22:12
• Thank you! It makes sense! It's a nice proof, based a lot on sets! I have no idea what a filter is in mathematics, however the set $U_z$ makes sense – Euler_Salter Dec 4 '16 at 2:13 | 2019-05-23T14:47:33 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2042140/precise-proof-of-s-open-set-iff-contains-only-interior-points-complex-analysis",
"openwebmath_score": 0.8620448708534241,
"openwebmath_perplexity": 198.39184553187513,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9783846666070896,
"lm_q2_score": 0.8705972549785201,
"lm_q1q2_score": 0.8517790050612067
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.