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• This is the link to the prove of the above inequality in special case when $a = b = \frac{1}{2}$ I think it might be useful to prove the above inequality in the general form. LINK:math.stackexchange.com/questions/331367/cosh-x-inequality – alfred noble Oct 26 '16 at 16:01
• Additional requirement is $f(0)=0$. For $f(... | {
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Difference between revisions of "2010 AMC 12A Problems/Problem 24"
Problem
Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$. The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$?
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And so, the total number of disjoint open intervals is $22 - 2\cdot{5} = \boxed{12\ \textbf{(B)}}$
Solution 2 (cheap)
Note that the expression $\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)$ must be greater than zero, since logarithm functions are undefined for $0$ and negative numbers. Le... | {
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# Exponential pop. growth when only given population at two instances of time.
I have a problem where I'm only given the population of a "bacteria culture" at two instances in time: 2 hours and 4 hours. The problem says the population of bacteria is 125 after 2 hours, and 350 after 4 hours. It specifically says the ba... | {
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• Thank you very much that helps. – Sabien Mar 17 '14 at 6:20
• No, I tried my method but it didn't work, well at least I must have done the Algebra wrong. Your explanation is more clear, no need to delete it. – Sabien Mar 17 '14 at 6:22
• You are welcome. If there any step that you have trouble completing, please leav... | {
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# Tag Archives: polar
## A Generic Approach to Arclength in Calculus
Earlier this week, a teacher posted in the College Board’s AP Calculus Community a request for an explanation of computing the arclength of a curve without relying on formulas.
The following video is my proposed answer to that question. In it, I d... | {
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## A Student’s Powerful Polar Exploration
I posted last summer on a surprising discovery of a polar function that appeared to be a horizontal translation of another polar function. Translations happen all the time, but not really in polar coordinates. The polar coordinate system just isn’t constructed in a way that ... | {
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At this point, S employed some CAS power.
[Full disclosure: That final CAS step is actually mine, but it dovetails so nicely with S’s brilliant approach. I am always delightfully surprised when my students return using a tool (technological or mental) I have been promoting but hadn’t seen to apply in a particular situ... | {
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## Controlling graphs and a free online calculator
When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph. When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could a... | {
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REALIZATION & WHY IT WORKS: Last week, we discovered that we could use $g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted. The argument of the root is 1 for $x<0$, making $g(x)=1$. For $x>0$, the root’s argument is -1, making $g(x)=i$, a non-real number. Our insight was that multiplying any function ... | {
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EXAMPLE 3: I believe students understand polar graphing better when they see curves like the limacon $r=2+3cos(\theta )$ moving between its maximum and minimum circles. Controlling the slider also allows users to see the values of $\theta$ at which the limacon crosses the pole. Here is the Desmos graph for the graph... | {
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ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator. Type “d/dx” before the function name or definition, and the derivative is accomplished. Desmos is not a CAS, so I’m sure the software is computing derivatives numerically. No matter. Derivatives are easy to define and use here.... | {
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For $0\le\theta\le 3\pi$ , $r=cos\left(\frac{\theta}{3}\right)$ becomes $\begin{array}{lcl} x_1 &= &cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_1 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array}$ . Sliding this $\frac{1}{2}$ a unit to the right makes the parametric e... | {
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Before blindly manipulating the equations, I take some time to develop some strategy. I notice that the $(x_5, y_5)$ equations contain only one type of angle–double angles of the form $2\cdot\frac{\beta}{3}$ –while the $(x_4, y_4)$ equations contain angles of two different types, $\beta$ and $\frac{\beta}{3}$ . It is... | {
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Proving that the x expressions are equivalent. Now for the ys
$\displaystyle \begin{array}{lcl} y_4 &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\beta\right) \\ &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\frac{\beta}{3}+\frac{2\beta}{3} \right) \\ &= & cos\left(\frac{\beta}{3}\right)\cdot\left( sin\left... | {
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Despite these changes, my proof still feels cumbersome and inelegant to me. From one perspective–Who cares? I proved what I set out to prove. On the other hand, I’d love to know if someone has a more elegant way to establish this connection. There is always room to learn more. Commentary welcome.
In the end, it’s... | {
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So, now that our students are confidently able to graph polar curves like $r=3+cos(2\theta )$, we wondered how we could challenge them a bit more. Remembering variable center lines like the Cartesian $y=cos(x)+0.5x$, we wondered what a polar curve with a variable center line would look like. Not knowing where to star... | {
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And this last form says our original polar function is equivalent to $r=2+\sqrt{2}\cdot cos(\theta -\frac{\pi}{4})$, or a $\frac{\pi}{4}$ rotated cosine curve of amplitude $\sqrt{2}$ and period $2\pi$ oscillating around center line $r=2$.
This last image shows a cosine curve starting at $\theta=\frac{\pi}{4}$ beginnin... | {
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# From $e^n$ to $e^x$
Solve for $f: \mathbb{R}\to\mathbb{R}\ \ \$ s.t.
$$f(n)=e^n \ \ \forall n\in\mathbb{N}$$ $$f^{(y)}(x)>0 \ \forall y\in\mathbb{N^*} \ \forall x\in\mathbb R$$
Could you please prove that there exists an unique solution: $f(x)=e^x$?
(Anyway, this problem is not about fractional calculus)
$\mathb... | {
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I'll start with the simpler case where (1) holds for all $n\ge0$, in which case it is only necessary to suppose that $f(x)=e^x$ at three points. As $\left(-\frac{d}{dx}\right)^nf(-x)=f^{(n)}(-x)\ge0$, by definition $f(-x)$ is completely monotonic. Bernstein's theorem means that we can write $$f(-x)=\int e^{-xy}\,\mu(dy... | {
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• Your proof applies to $x > 0$, right? – mathworker21 Jul 12 '17 at 4:12
• It applies to all $x$. Unfortunately, the statement of Bernstein's theorem is for $x > 0$, but the result applies for all $x$ and I tried to give a quick argument why in my proof. – George Lowther Jul 12 '17 at 4:14
• Very Small nitpick: The Be... | {
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# Math Help - Inverse Derivative Help
1. ## Inverse Derivative Help
I cannot figure out what to do!!
Given the function $f(x)=5 x^3+2 x+5$ Let g be the inverse function of f. i.e. $g(x)=f^{-1}(x).$
$g^{\prime}(12)=$
I can't manage to find g(x) let alone g'(x)
Thank you for any help.
2. Originally Posted by Krooger
... | {
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6. I was unaware that:
$
g'(x) = \frac 1{f'(g(x))}$
...and to be honest upon first glace it still dosen't make sense to me haha. When I get a minute I will have to look it up. Thank You all for the help.
7. Originally Posted by Krooger
I was unaware that:
$
g'(x) = \frac 1{f'(g(x))}$
...and to be honest upon first g... | {
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Solution for Every real number is also a complex number. And real numbers are numbers where the imaginary part, b = 0 b=0 b = 0. Real and Complex Numbers . Now we can look at some operations with complex numbers. Does harry styles have a private Instagram account? Complex numbers are a mixture of the two, e.g. In the s... | {
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some are neither. In the meantime, ‘Complex Numbers’ as the name refers a heterogeneous mix. How do I use graphing in the complex plane to add #2+4i# and #5+3i#? In Figure 2, we show the results of various complex number operations. Write a ⋅ i a ⋅ i in simplest form. To avoid such e-mails from students, it is a good i... | {
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all imaginary numbers are complex. They can be any of the rational and irrational numbers. In the special case that b = 0 you get pure real numbers which are a subset of complex numbers. In general, a complex number looks like $x+y\mathrm{i}$ where $x$ and $y$ are both real numbers. A complex number is a number of the ... | {
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around the world. share | cite | improve this answer | follow | answered Aug 9 '14 at 21:34. The complex number zero has zero real part and zero imaginary part: $0+0\mathrm{i}$. The complex number i = 0 + i 1, which has real part zero and imaginary part one, has the property that its square is i 2 = (0 + i 1) 2 = (0 + ... | {
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number operations . A complex number is the sum of a real number and an imaginary number. Notice that 0 is a real number. A number can be both real and complex? So, too, is $$3+4\sqrt{3}i$$. Every pure imaginary number is a complex number. Conversely, it is imaginary if the real component is zero. Solution for Every re... | {
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for example water freezes at 0 degrees Centigrade. A complex number is a number of the form . In this situation, we will let $$r$$ be the magnitude of $$z$$ (that is, the distance from $$z$$ to the origin) and $$\theta$$ the angle $$z$$ makes with the positive real axis as shown in Figure $$\PageIndex{1}$$. How did Riz... | {
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of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i = −1. Definition of a Complex Number For real numbers a and b, the number a + bi is a complex number. $2>1$. Similarly, 3/7 is a rational number but not an integer. What is the analysis of the poem song by nvm gonzalez? Sometimes... | {
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of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. And to get the conjugate of a complex number 1+0i the lady with the trophy in bounce. Get the imaginary part is non-zero ) which is represented by eyes of... | {
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if the real number, the complex number 0 and are. And an imaginary number ) at some operations with complex numbers: e.g ‘ Natural ’ since all them. Known as the real number line is illustrated below with the number called! The complex plane to add # 2+4i # and # 5+3i # their real parts imaginary! Many rational numbers... | {
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imaginary of. Regarding real numbers are not talked about at once eyes of pineapple 3+4\sqrt { }... Indeed it does n't ) freezes at 0 degrees Centigrade special ‘ 0 ’ and the ‘ negative ’ were... Rhythm tempo of the poem song by nvm gonzalez can take any value on number. Number is a complex number has a real number is ... | {
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as real as any other sort Karate Kid the best to... 1+0 i is a complex or real number because for example water freezes at 0 degrees Centigrade because! N'T complex numbers are also complex numbers ’, no value of the song sa ugoy ng?! The denominator is zero ( 4+3j ) > > a ( 4+3j ) > > a.imag conjugate. Run for preside... | {
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Number line is illustrated below with the point ( a small aside: the most important property is. Imaginary, the complex numbers the basic imaginary unit or complex unit is 0 a complex number or real number be an extension the. Expressed as p/q where p and q are integers and q are and... To add # 2+4i # and # 5+3i # num... | {
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Amity University Cutoff 2020, Syracuse Dorms Ranked, Be Unwell Daily Themed Crossword, Gordon Gin Sundowner, Metallica Tabs Easy, Uss Missouri Location, Gordon Gin Sundowner, Wall Unit Bookcase With Glass Doors, | {
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Exercise of Markov Chain
Post
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# Exercise of Markov Chain
## Q1
Suppose you repeatedly does toss a fair coin and denote $T$ the first time you get three consecutive heads.
1. Compute $E[T]$
2. Verify your answer in [1] via simulation. You may use any programming language, but you have to attach your code. Spe... | {
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Define Transition Probability Matrix $Q$:
Since we have:
\begin{aligned} P(\text{State}_{0}|\text{State}_{0}) = 0.5, P(\text{State}_{1}|\text{State}_{0}) = 0.5, P(\text{State}_{2}|\text{State}_{0}) = 0.0, P(\text{State}_{3}|\text{State}_{0}) = 0.0\\ P(\text{State}_{0}|\text{State}_{1}) = 0.5, P(\text{State}_{1}|\text... | {
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then $E[T]=e_0=14$
### Simulation
#### Verify Expectation
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 import numpy as np def coin_flip(p=0.5): count = 1 head = [] while len(head) <3: if np.random.binomial(1, p): if len(head) == 0 or head[-1] + 1 != count: head = [count] else: head.append(count) count += 1 return he... | {
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# 09 Pre-Class Assignment: Determinants¶
## 1. Introduction to Determinants¶
For a detailed overview of determinants I would recommend reviewing Chapter D pg 340-366 of the Beezer text.
The determinant is a function that takes a ($$n \times n$$) square matrix as an input and produces a scalar as an output. Determina... | {
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return det
Notice that the combination of the determinants of the submatrices is not a simple sum. The combination is adding the submatrices corresponding to the odd columns (1,3,5, etc) and subtracting the submatrices corresponding to the even columns (2,4,6, etc.). This may become clearer if we look at a simple $$3... | {
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from answercheck import checkanswer
---------------------------------------------------------------------------
ModuleNotFoundError Traceback (most recent call last)
<ipython-input-2-b2a2502e4bdf> in <module>
2
1 import hashlib
2 import numpy as np
----> 3 import sympy as sym
4 import sys
5 imp... | {
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### Determinants and Matrix Operations¶
Let $$A$$ and $$B$$ be $$n\times n$$ matrices and $$c$$ be a non-zero scalar.
1. Determinant of a scalar multiple: $$|cA| = c^n|A|$$
2. Determinant of a product: $$|AB| = |A||B|$$
3. Determinant of a transpose” $$|A^t| = |A|$$
4. Determinant of an inverse: $$|A^{-1}| = \frac... | {
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$\begin{split} \left[ \begin{matrix} 1 & a & a^2 & a^3 \\ 1 & b & b^2 & b^3 \\ 1 & c & c^2 & c^3 \\ 1 & d & d^2 & d^3 \end{matrix} \right] \end{split}$
## 3. One interpretation of determinants¶
The following is an application of determinants. Watch this!
from IPython.display import YouTubeVideo
For fun, we will re... | {
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# Create cubic bounding box to simulate equal aspect ratio
Xb = 0.5*max_range.max()*np.mgrid[-1:2:2,-1:2:2,-1:2:2][0].flatten() + 0.5*(max_range[0])
Yb = 0.5*max_range.max()*np.mgrid[-1:2:2,-1:2:2,-1:2:2][1].flatten() + 0.5*(max_range[1])
Zb = 0.5*max_range.max()*np.mgrid[-1:2:2,-1:2:2,-1:2:2][2].flatten() + 0.5*(max_r... | {
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# Expressing $(-8)^{\frac13}$ in polar form
I want to express $(-8)^{\frac{1}{3}}$ in polar and cartesian coordinates.
What I did was to solve the equation $-8 = r^3e^{3i\theta}= r^3(\cos(3\theta)+i\sin(3\theta))$ which implies that I must solve the equations $$r^3\cos(3\theta) = -8$$ and the equation $$r^3\sin(3\the... | {
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# 1991 IMO shortlist problem $\#11$
Prove that $$\sum_{k=0}^{995} \frac{(-1)^k}{1991-k} {1991-k\choose k} = \frac{1}{1991}$$
As usual there isn't anything special about the number $$1991$$.Problem appears to hold for any odd numbers I have checked. I want to prove the general equation. We can manipulate expression an... | {
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We'd use these ideas here.
Notice that $$\left(\frac{1}{n-m} - \frac{1}{n}\right) { n - m \choose m } = \frac{m}{ n (n-m) } { n - m \choose m } = \frac{1}{n} {n-m-1 \choose m-1}$$, or that
$$\frac{ 1 } { n-m } { n-m \choose m } = \frac{1}{n} \left[ { n - m \choose m } + { n - m - 1 \choose m- 1 } \right].$$
This is ... | {
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Notes
1. I do wish there was a combinatorial argument here. For example, $$S_n$$ has an immediate interpretation as the difference between the even and odd permutations $$p$$ such that $$|p(i) - i | \leq 1$$. (IE Out of the first $$n$$ integers, there are $${n-k \choose k }$$ ways to pick k pairs of consecutive intege... | {
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And the answer to the original question follows from the fact that $$1991 \equiv 5 \mod 6$$.
• Thanks for your solution. But I don't know much of generating functions. Isn't there any other solutions? – Mathematical Curiosity Sep 23 '20 at 5:58
• I think it's the most natural solution to this problem. Other solutions,... | {
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# Give a deck of 54 cards, what's the probability that even one card will match two dealt hands...
if we shuffle a deck (with 54 cards including Joker) thoroughly and deal out a four card hand, there are over 300,000 different hands. What's the probability that no cards match between two dealt hands? even one card mat... | {
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Is this the right way to think about it? Am I missing anything?
• The jokers may mess with things. Also relies on a set number of cards per hand.
– user645636
Mar 31 '19 at 19:49
• Are the two jokers identical? In some decks they are, in some they aren't (for instance, one is red and one is black). Mar 31 '19 at 20:49... | {
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The expression on the left can be explained by treating each card as being pulled in sequence, picking which positions in the sequence are occupied by matching cards, picking which matching cards those are, and picking which non-matching cards occupied the remaining spaces out of the possible ways in which four cards c... | {
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Only the first few places of these probabilities are likely to be accurate, but this may give you something to check against as you finish your combinatorial analysis. Notice that the simulated proportion $$0.728$$ of no matches is the same (to three places) as the correct probability in your first answer.
A second si... | {
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# Expressing in the form $A \sin(x + c)$
Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$
sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$
Can someone please explain the method used in the provide... | {
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Equating $A \sin x \cos c + A \cos x \sin c = \sin x + \sqrt{3} \cos x$ gives $A \cos c = 1$, $A \sin c = \sqrt{3}$. This gives $\tan c = \frac{A \sin c}{A \cos c} = \sqrt{3}$. If $\tan c = \sqrt{3}$, then $\sin c = \frac{\sqrt{3}}{2}$ and $\cos c = \frac{1}{2}$. This gives $A \frac{1}{2} = 1$, so $A = 2$. You can chec... | {
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# The isomorphism between two complete ordered fields is unique
The isomorphism between two complete ordered fields is unique.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $$\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle,... | {
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Hence $$\Phi(\sup X)=\sup' f[X]$$. Similarly, $$\Psi(\sup X)=\sup' f[X]$$.
Let $$X_x=\{p\in\Bbb Q \mid p. Since $$\Bbb Q$$ is dense in $$\Bbb R$$, $$x=\sup X_x$$ for all $$x\in\Bbb R$$. Then $$\Phi(x)=\Phi(\sup X_x)=\sup' f[X_x]=\Psi(\sup X_x)=\Psi(x)$$ for all $$x\in\Bbb R$$. It follows that $$\Phi=\Psi$$.
• You sho... | {
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• A proper sub-field of $\Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $\Bbb R$. For example $\{a+b\sqrt 2\,:a,b\in \Bbb Q\}.$ For $a,b\in \Bbb Q$ let $a+b\sqrt 2\,>^*0\iff a-b\sqrt 2<0.$ – DanielWainfleet Jan 16 '19 at 23:06
• I think you meant $\psi$ rather than $f$. Please ... | {
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# Finding the local extreme values of $f(x) = -x^2 + 2x + 9$ over $[-2,\infty)$.
I'm tutoring a student, and we were trying to solve the following question:
Find the local extreme values of $$f(x) = -x^2 + 2x + 9$$ over $$[-2,\infty)$$.
According to the textbook, the local extreme values are essentially the peaks an... | {
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Since we can fix a $$\epsilon$$-neighborhood with our choice of $$\epsilon$$ near $$x=-2$$ so that in this domain $$(-2, 1)$$ is a minimum point, it is indeed a local minimum.
If you look at the statements of the first and second derivative tests, they should indicate that they are valid at interior points of interval... | {
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The left endpoint of $$(-2, 1)$$ is a local minimum and there is no global minimum.
• But why would we consider $(-2,1)$ to be a local minimum? That's the question being asked in the OP. – Decaf-Math Nov 2 '18 at 18:43
• That is because around that point and as long as you are in the domain, the values are greater or ... | {
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# Formula for sum of first $n$ odd integers
I'm self-studying Spivak's Calculus and I'm currently going through the pages and problems on induction. This is my first encounter with induction and I would like for someone more experienced than me to give me a hint and direction. The first problem is as follows:
Find a ... | {
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In your case, the first element is 1, the last element is 2n - 1, the average is n, there are n elements, therefore the sum is $n^2$.
$$1+3+5+...+(2n-1)=n\times\,n=n^2$$
• Nice! $\;\!\;\!$ – goblin GONE Jun 5 '16 at 2:28
• The partitioning of $\mathbb{N}^2$ displayed here can be rigorously defined as the coimage of $... | {
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With $\sum(2i-1)$, simply adding a few terms by hand shows you that the answer is $n^2$. Proving that the answer is $n^2$ is then just a matter of proving that going from $n=N$ to $n=N+1$ means adding $2N+1$ to the total, and going from $N^2$ to $(N+1)^2$ means adding $2N+1$ to the total… so if it's true for $n=N$ then... | {
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1. Know $\sum{i^a}$ for $a=0, 1, 2, 3$. It doesn't require much memorising. Then combine the sums for whatever you are being asked to find the sum of (including using the trick that Spivak is making you use here).
2. Note that $\sum{1}=n/1!$, $\sum{i}=n(n+1)/2!$, $\sum{i(i+1)}=n(n+1)(n+2)/3!$, $\sum{i(i+1)(i+2)}=n(n+1)... | {
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\begin{align} \sum_{i=1}^n (2i-1)&=\sum_{i=1}^n\binom i1+\binom {i-1}1\\ &=\binom {n+1}2+\binom n2\\ &=n^2\qquad\blacksquare\\ \sum_{i=1}^n(2i-1)^2&= \color{blue}{1^2}+\color{purple}{3^2}+\color{green}{5^2}+\cdots+\color{red}{(2n-1)^2}\\ &=\color{blue}{\binom 12+\binom 22}+ \color{purple}{\binom 32+\binom 42}+ \color{g... | {
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# Effect size and bootstrapping in paired t-test
I have multiple paired $t$-tests, such as one giving results:
$t_{14} = 2.7,\ p = .017$
Although people seem to do effect sizes in different ways in repeated samples, I have taken the mean difference divided by the standard deviation of the differences (I'll call this... | {
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• Just to say that I have found useful material on these pages (though I haven't got a specific answer to the case of bootstrapping CIs for an effect size) stats.stackexchange.com/questions/71525/… , stats.stackexchange.com/questions/73818/… – splint Nov 30 '15 at 15:45
• I'm not quite sure I'm following this. Can you ... | {
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I get a different result from you (in R)
a=read.table(header=F,text="
1999 2040
1501 1601
1552 1623
2385 2386
2488 2671
1257 1218
1806 1719
1348 1405
2048 2079
1810 2017
1308 1356
2310 2324
1247 1616
1839 1878
1235 1370
")
d=a$V2-a$V1
mean(d)/sd(d)
[1] 0.7006464
aux=function(x,i) mean(x[i])/sd(x[i])
bb=boot::boot(d,au... | {
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$Upper.Conf.Limit.Standardized.Mean [1] 1.258396 As for your suggestion on computing the confidence interval for the difference score and using it to compute a confidence interval on the effect size, I have never heard of it, and I would suggest not using it. • +1 to @amoeba, I think you want to use mean(bb$t). Nice a... | {
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# Why aren't these two integrals equivalent when using the substitution $x=\frac{1}{t}$?
Why aren't these two integrals $$\int_{-1}^{1}\frac{1}{\left(1+x^2\right)^2}\,\mathrm{d}x$$ and $$\int_{-1}^{1}\frac{-t^2}{\left(1+t^2\right)^2}\,\mathrm{d}t$$ equal to each other, despite using the substitution $$x=\frac{1}{t}$$,... | {
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# Groups with given automorphism groups
It is an easy exercise to show that all finite groups with at least three elements have at least one non-trivial automorphism; in other words, there are - up to isomorphism - only finitely many finite groups $G$ such that $Aut(G)=1$ (to be exact, just two: $1$ and $C_2$).
Is an... | {
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An immediate consequence is that up to isomorphism, there are only finitely many finite groups $G$ with $|\operatorname{Aut}(G)| \leq n$. Hence for any finite group $X$, up to isomorphism there are only finitely many finite groups $G$ with $\operatorname{Aut}(G) \cong X$.
Among infinite groups this is no longer true, ... | {
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Indeed, for $I$ a set of primes, let $B_I$ be the additive subgroup of $\mathbf{Q}$ generated by $\{1/p:p\in I\}$. Then $B_I$ and $B_J$ are isomorphic if and only if the symmetric difference $I\triangle J$ is finite, and $\mathrm{Aut}(B_I)=\{1,-1\}$ (easy exercise: more generally for a nonzero subgroup $B$ of $\mathbf{... | {
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One more example to mention that one gets non-abelian groups: let $F$ be a finite group. Then for every torsion-free abelian group $B$, $\mathrm{Aut}(B\times F)$ is a semidirect product $(\mathrm{Aut}(F)\times\mathrm{Aut}(B))\ltimes\mathrm{Hom}(B,Z(F))$. If $\mathrm{Aut}(B)=\{\pm 1\}$, then the $\mathrm{Aut}(B)$-action... | {
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Can any proof by contrapositive be rephrased into a proof by contradiction?
From my understanding,
Proof by contrapositive: Prove $$P \implies Q$$, by proving that $$\neg Q \implies \neg P$$ since they are logically equivalent.
Proof by contradiction: Prove $$P \implies Q$$ by showing that $$P \wedge \neg Q$$ yields... | {
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• The $(\neg Q\implies\neg P)\implies(P\implies Q)$ direction of the equivalence also requires something like double negation elimination assuming you are starting from a reasonably typical constructive logic. – Derek Elkins Dec 26 '18 at 0:25
• If every proof by contrapositive can be rephrased into a proof by contradi... | {
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Yes it is valid... it doesn't really matter if it's something else 'in disguise', just that is it correct. And deriving $$\lnot P$$ from $$P\land\lnot Q$$ is certainly leads to a contradiction that implies $$\lnot (P\land \lnot Q)$$ is true, which implies that $$P\to Q$$ is true. One thing to note (that I think you hav... | {
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# Math Help - Another Related Rate Problem
1. ## Another Related Rate Problem
Here is the text of my problem:
A boy is flying a kite at a height of 150ft. If the kite moves gorizontally away from the boy at 20ft/s, how fast is the string being paid out when the kite is 250ft from him?
Given:
• y=150ft
• x=250ft
• $... | {
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# Difference between \big[ and \bigl[
What is the difference between \big[ (or equivalently \big() and \bigl[? Is it always necessary to mention l (left) and r (right)?
-
\bigl declares an opening math delimiter with less horizontal spacing than the unspecified \big. \bigr defines a closing math delimiter. Using a \... | {
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# Given that $6$ men and $6$ women are divided into pairs, what is the probability that none of the women will sit with a man?
I've generalized the question I was given here for simplicity: $6$ men and $6$ women are to be paired for a bus trip. If the pairings are done randomly, what's the probability that no women wi... | {
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We want to group $12$ people into $6$ single-sex groups of $2$, so I began by calculating the number of ways we can make $6$ pairs.
By the multinomial function: $12!/(2!)^6=7484400$ ways to make $6$ pairs.
No, order doesn't matter within the groups of $2$ nor of the $6$ groups.
$$\frac{12!}{2!^6 6!} = 10395$$
Then,... | {
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To stay as close as possible to your method, you need to divide your general answer by the number of ways to order the six pairs, which is $6!$. For the second part, divide each time by $3!$. So the answer will be $$\frac{(6!/[(2!)^3 3!])^2}{12!/[(2!)^6 6!]}$$
• No there isn't. If you arbitrarily sort the women into A... | {
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1st pair being single-sex:
$$\frac{5}{11}$$
2nd pair being single-sex given 1st pair is single-sex:
$$\frac{6}{10}\cdot\frac{5}{9} + \frac{4}{10}\cdot\frac{3}{9}$$
3rd pair being single-sex given 1st & 2nd pair are single-sex:
$$(\frac{6}{8}\cdot\frac{5}{7} + \frac{2}{8}\cdot\frac{1}{7})+(\frac{5}{8}\cdot\frac{4}{... | {
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# Math Help - Question on Area between Polar Curves
1. ## Question on Area between Polar Curves
Hey everyone,
I have two questions regarding the area of polar curves.
1. Find the area of the region lying the polar curve r=1 + cos(theta), and outside the polar curve r= 2cos(theta)
2. Find the area of the shaded regi... | {
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Any help and feedback appreciated.
Thanks
In the first one, when you set up your double integral, for the top half, the radii are bounded above by \displaystyle \begin{align*} r = 1 + \cos{\theta} \end{align*} and the radii are bounded below by \displaystyle \begin{align*} r = 2\cos{\theta} \end{align*}. I agree with ... | {
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There is yet another problem.
This is not a cardioid.
It is a limacon with an internal "loop".
. . The entire upper half?
. . The upper half minus the loop?
. . Just the loop?
5. ## Re: Question on Area between Polar Curves
Originally Posted by Soroban
Hello, Beevo!
Yes . . . good work!
You solved $r \,=\,0$
You... | {
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# Flipping a coin
Two players, $A$ and $B$, alternately and independently flip a coin and the first player to get a head wins. Assume player $A$ flips first. If the coin is fair, what is the probability that $A$ wins?
So $A$ only flips on odd tosses. So the probability of winning would be $$P =\frac{1}{2}+\left(\frac... | {
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Comment: There are nice expressions for $p$ as infinite geometric series. So we can think of the above argument as a probabilistic method for summing a very particular geometric series. By varying the probability that the coin lands heads, we can use the same idea to find the sum of any infinite geometric series, as lo... | {
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-
the probability that the first head occurs on toss $n$ is $2^{-n}$ so the probability that the first head happens on an odd $n$ is $\sum_{k=0}^{\infty}2^{-(2k+1)}=(1/2)(1/(1-1/4))=2/3$
-
Here is the solution for general p. Let A gets the head in Nth trial to win the game. Since he is flipping the coin in odd trial... | {
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# Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots$
by Isaac Last Updated January 16, 2018 08:20 AM
I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}... | {
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It is also worth noting, on the Wikipedia link Mau provided, that the convergence to $\ln 2$ of your series is at the edge of the radius of convergence for the series expansion of $\ln(1-x)$- this is a fairly typical occurrence: at the boundary of a domain of convergence of a Taylor series, the series is only just conv... | {
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Taking the limit we get $\gamma-\gamma+\ln(2)$.
N. S.
June 02, 2011 04:54 AM
In this answer, I used only Bernoulli's inequality to show that $$\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1} \le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)} \le\frac{2n+1}{n+1}\tag{1}$$ The squeeze theorem... | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718488515023,
"lm_q1q2_score": 0.843153706518256,
"lm_q2_score": 0.8519528038477824,
"openwebmath_perplexity": 326.19219939723877,
"openwebmath_score": 0.9519067406654358,
"tags... |
# Formula for some average
I am finding a formula for the average $$A_n$$ of the numbers $$(a_1-a_2)^2 + \cdots + (a_{n-1} - a_n )^2$$ over all cases that $$\{a_1, \cdots, a_n \} = \{ 1,2, \cdots, n\}$$.
For example, $$A_2=1, A_3=4, A_4=10, A_5=20, A_6=35, \cdots$$. From this, I guess: $$A_n = \frac{1}{6} (n-1) n (n+... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718468714469,
"lm_q1q2_score": 0.8431537029712819,
"lm_q2_score": 0.8519528019683105,
"openwebmath_perplexity": 290.82910694358367,
"openwebmath_score": 0.8966832160949707,
"ta... |
Tutorial 1: Linear regression with MSE¶
Week 1, Day 3: Model Fitting
Content creators: Pierre-Étienne Fiquet, Anqi Wu, Alex Hyafil with help from Byron Galbraith
Content reviewers: Lina Teichmann, Saeed Salehi, Patrick Mineault, Ella Batty, Michael Waskom
Tutorial Objectives¶
Estimated timing of tutorial: 30 minut... | {
"domain": "neuromatch.io",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718450437034,
"lm_q1q2_score": 0.8431536976940098,
"lm_q2_score": 0.8519527982093666,
"openwebmath_perplexity": 3004.13395026435,
"openwebmath_score": 0.6878114938735962,
"tags": n... |
def plot_observed_vs_predicted(x, y, y_hat, theta_hat):
""" Plot observed vs predicted data
Args:
x (ndarray): observed x values
y (ndarray): observed y values
y_hat (ndarray): predicted y values
theta_hat (ndarray):
"""
fig, ax = plt.subplots()
ax.scatter(x, y, label='Observed') # our data scatter plot
ax.plot(x, y_... | {
"domain": "neuromatch.io",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718450437034,
"lm_q1q2_score": 0.8431536976940098,
"lm_q2_score": 0.8519527982093666,
"openwebmath_perplexity": 3004.13395026435,
"openwebmath_score": 0.6878114938735962,
"tags": n... |
We will now explore how MSE is used in fitting a linear regression model to data. For illustrative purposes, we will create a simple synthetic dataset where we know the true underlying model. This will allow us to see how our estimation efforts compare in uncovering the real model (though in practice we rarely have thi... | {
"domain": "neuromatch.io",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718450437034,
"lm_q1q2_score": 0.8431536976940098,
"lm_q2_score": 0.8519527982093666,
"openwebmath_perplexity": 3004.13395026435,
"openwebmath_score": 0.6878114938735962,
"tags": n... |
$\hat{y}_{i}= \theta x_{i}$
and for mean squared error is:
(45)\begin{align} \min _{\theta} \frac{1}{N}\sum_{i=1}^{N}\left(y_{i}-\hat{y}_i\right)^{2} \end{align}
def mse(x, y, theta_hat):
"""Compute the mean squared error
Args:
x (ndarray): An array of shape (samples,) that contains the input values.
y (ndarray): An... | {
"domain": "neuromatch.io",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718450437034,
"lm_q1q2_score": 0.8431536976940098,
"lm_q2_score": 0.8519527982093666,
"openwebmath_perplexity": 3004.13395026435,
"openwebmath_score": 0.6878114938735962,
"tags": n... |
axes[0].legend()
Interactive Demo 1: MSE Explorer¶
Using an interactive widget, we can easily see how changing our slope estimate changes our model fit. We display the residuals, the differences between observed and predicted data, as line segments between the data point (observed response) and the corresponding pre... | {
"domain": "neuromatch.io",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9896718450437034,
"lm_q1q2_score": 0.8431536976940098,
"lm_q2_score": 0.8519527982093666,
"openwebmath_perplexity": 3004.13395026435,
"openwebmath_score": 0.6878114938735962,
"tags": n... |
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