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a negative number is squared it also provides a positive number. A complex number is a number of the form . Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. If some of these functions seem difficult to understand, it's best to research the basic logic behind them. So, a Complex Number has a real part and an imaginary part. If not, then we add radians or to obtain the angle in the opposing quadrant: , or . and are allowed to be any real numbers. The proposition below gives the formulas, which may look complicated – but the idea behind them is simple, and is captured in these two slogans: When we multiply complex numbers: we multiply the s and add the s.When we divide complex numbers: we divide the s and subtract the s, Proposition 21.9. However, we have to be a little careful: since the arctangent only gives angles in Quadrants I and II, we need to doublecheck the quadrant of . A real number can store the information about the value of the number and if this number is positive or negative. A complex number is said to be a combination of a real number and an imaginary number. The real numbers are a subset of the complex numbers, so zero is by definition a complex number (and a real number, of course; just as a fraction is a rational number and a real number). Here ‘x’ is called the real part of z and ‘y’ is known as the imaginary part of z. All real numbers are also complex numbers with zero for the imaginary part. A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely imaginary if its real part is zero i.e., Re(z) = 0. If z1,z2,——zn are the complex numbers then z1.z2. The primary reason is that it gives us a simple way to picture how multiplication and division work in the plane. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. Complex numbers actually combine real and imaginary number (a+ib), where a and b | {
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number-- 0 plus i. Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. A complex number is the sum of a real number and an imaginary number. Complex numbers can be used to solve quadratics for zeroes. Therefore, all real numbers are also complex numbers. Let be a complex number. The set of real numbers is a proper subset of the set of complex numbers. Using the functions and attributes that we've reviewed thus far will aid in building programs that can be used for a variety of science and engineering applications. They can be any of the rational and irrational numbers. Definition of Complex Numbers; An ordered pair of real numbers, written as (a, b) is called a complex number z. Any real number is a complex number. (2 plus 2 times i) The complex numbers are referred to as (just as the real numbers are . Complex numbers can be multiplied and divided. VIDEO: Multiplication and division of complex numbers in polar form – Example 21.10. You can add them, subtract them, multiply them, and divide them (except division by 0 is not defined), and the result is another complex number. 2020 Spring – MAT 1375 Precalculus – Reitz. A complex numberis defined as an expression of the form: The type of expression z = x + iy is called the binomial form where the real part is the real number x, that is denoted Re(z), and the imaginary partis the real number y, which is denoted by Im(z). Because no real number satisfies this equation, i is called an imaginary number. Complex numbers which are mostly used where we are using two real numbers. Number line can be expressed as an actual geometric line where a point is chosen to be the origin. All imaginary numbers are also complex numbers with zero for the real part. So, too, is $$3+4\sqrt{3}i$$. In other words, if the imaginary unit i is in it, we can just call it imaginary number. Multiplying a Complex Number by a Real Number. Convert the complex | {
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just call it imaginary number. Multiplying a Complex Number by a Real Number. Convert the complex number to polar form.a) b) c) d), VIDEO: Converting complex numbers to polar form – Example 21.7, Example 21.8. If b is not equal to zero and a is any real number, the complex number a + bi is called imaginary number. This class uses WeBWorK, an online homework system. A complex number is a number that can be written in the form x+yi where x and y are real numbers and i is an imaginary number. Logged-in faculty members can clone this course. Let’s begin by multiplying a complex number by a real number. Don’t forget to complete the Daily Quiz (below this post) before midnight to be marked present for the day. Hi everyone! They have been designed in order to solve the problems, that cannot be solved using real numbers. Why is polar form useful? Then, the product and quotient of these are given by, Example 21.10. With this article at OpenG… The absolute value of , denoted by , is the distance between the point in the complex plane and the origin . You could view this right over here as a complex number. That’s it for today! Multiplying complex numbers is much like multiplying binomials. For the complex number a + bi, a is called the real part, and b is called the imaginary part. So, too, is $3+4\sqrt{3}i$. Consider √- 4 which can be simplified as √-1 × √ 4 = j√4 = j2.The manipulation of complex numbers is more complicated than real numbers, that’s why these are named as complex numbers. So, too, is 3 + 4i√3. WeBWorK: There are four WeBWorK assignments on today’s material, due next Thursday 5/5: Question of the Day: What is the square root of ? a, b ∈ R. a,b\in \mathbb {R} a,b ∈ R. Imaginary Numbers are the numbers which when squared give a negative number. If x and y are two real numbers, then a number of the form is called a complex number. In complex number, a is the real part and b is the imaginary part of the complex number. Topic: This lesson covers Chapter 21: | {
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real part and b is the imaginary part of the complex number. Topic: This lesson covers Chapter 21: Complex numbers. But in complex number, we can represent this number (z = … Complex Numbers are considered to be an extension of the real number system. Here r = √x2 + y2 = |z| is the modus of z and θ is called argument(or amplitude) of z is denoted by arg z. Subtracting Complex Numbers 1. (adsbygoogle = window.adsbygoogle || []).push({}); Copyright © 2021, Difference Between | Descriptive Analysis and Comparisons. Note that is given by the absolute value. New York City College of Technology | City University of New York. Image Courtesy: mathpowerblog.wordpress.comom, wikipedia.org. 3. A complex number is the sum of a real number and an imaginary number. Example 21.7. The importance of complex number in real life: In real numbers, we can represent this number as a straight line. A complex number is a number having both real and imaginary parts that can be expressed in the form of a + bi, where a and b are real numbers and i is the imaginary part, which should satisfy the equation i 2 = −1. A single complex number puts together two real quantities, making the numbers easier to work with. and are allowed to be any real numbers. Next, we will look at how we can describe a complex number slightly differently – instead of giving the and coordinates, we will give a distance (the modulus) and angle (the argument). Its algebraic form is z=x+i*y, where i is an imaginary number. This includes (but is not limited to) positives and negatives, integers and rational numbers, square roots, cube roots , π (pi), etc. You’ll see this in action in the following example. The real part of z is denoted by Re(z) and the imaginary part by Im(z). Definition 21.1. Required fields are marked *. Once they're understood, they're very simple and easy-to-use for just about anyone. It is provided for your reference. Complex Numbers: In mathematics, complex numbers are numbers that can be written in | {
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your reference. Complex Numbers: In mathematics, complex numbers are numbers that can be written in the form a + bi, where a and b are real numbers, and i is the imaginary number with value √−1 − 1. We distribute the real number just as we would with a binomial. Let and be two complex numbers in polar form. They're composed of real and imaginary numbers and are not necessarily the simplest to work with. This includes numbers like 3 – 2i or 5+√6i, as they can be written as the sum or difference of a real number and an imaginary number. The complex numbers are referred to as (just as the real numbers are . For example, 5 + 2i is a complex number. Definition 21.4. Yes, all real numbers are also complex numbers. Therefore we have: z = Re(z) + iIm(z). Comparison between Real Number and Complex Number: A real number is a number that can take any value on the number line. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. Infinity does not fall in the category of real numbers. The OpenLab is an open-source, digital platform designed to support teaching and learning at City Tech (New York City College of Technology), and to promote student and faculty engagement in the intellectual and social life of the college community. A complex number is any number that includes i. Therefore a complex number … The real part of the complex number is 6 and the imaginary part is 0 .So, the number will lie on the real axis. Difference Between | Descriptive Analysis and Comparisons, Counterintelligence Investigation vs Criminal Investigation. Terminologies often used in number Theory + 5.4 i, 2 + 5.4 i, 2 + 5.4 i and... 3 } i [ /latex ], we use two number lines, crossed form. Actual geometric line where a point is chosen to be an extension of complex... X x is called the real and imaginary numbers are referred to as ( just as the name a! B ) c ), VIDEO: multiplication and division of complex numbers are also numbers! The information about the value of, and | {
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Logical proof for a value derivable by a linear combination
Is there a logical basis to finding values provided possibly by a linear combination of integers. As the integers are closed under addition and hence their linear combination should be also. Hence, the integer points generated by two integers can be plotted on a lattice grid, extending infinitely. There can be multiple linear combinations, in other words, for two integer quantities. These two quantities can be, in the case of gcd computation, the divisor ($a$) and dividend ($d$).
If I want to dis-prove that a given linear combination can exist, then the approach is:
Say, the linear combination of two 'not' co-prime integers can never be $1$, as there will be no lattice point(s) generated as for : $2x + 4y = 1$. There will always be non-integral solutions, and any integer (lattice point) value of $x$ and $y$ will not work.
But, will the dis-proof above will work generally, i.e. if I want to prove that a given value can be generated or not by a linear combination. I request some better sort of way that will work generally for proving and disproving the feasibility of a given value for a given linear combination.
• I can't really understand what you're aiming at, but the integer combinations of $a$ and $b$ are exactly the multiples of $\operatorname{gcd}(a,b)$. This goes by the name of Bézout's lemma. – user228113 Nov 24 '17 at 4:57
• I hope my edit to OP makes it clear that similar to the dis-proof for the possibility of not co-prime integers having a value (for their linear combination) of 1, I want a similar proof for the opposite case. – jiten Nov 24 '17 at 5:17
• @DavidReed I have edited the OP to make it most clear. I would request some details on the feasible way to show existence of a linear combination, and for an efficient algorithm to find a given integer combination. – jiten Nov 24 '17 at 5:37
1 Answer
Let $d = gcd(a,b)$. Then there exists $x,y$ such that $ax+by=n$ if and only if $d|n$ | {
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1 Answer
Let $d = gcd(a,b)$. Then there exists $x,y$ such that $ax+by=n$ if and only if $d|n$
The extended Euclidean algorithm is the algorithm one would use to find $x$ and $y$.
To prove no such integer combination exists, one would again use the euclidean algorithm to find the gcd, and then simply show that n is not a multiple of the gcd.
EDIT-
Heres how you would prove an integer combination exists. Say you want to know if there are integers $x$ and $y$ such that $ax+by = n$. Let $d = gcd(a,b)$ You can always find integers $u$ and $v$ such that $au + bv = d.$ You can find $u$ and $v$ via the extended Euclidean algorithm, which also, as an added benefit, will compute $d$. If n is not a multiple of $d$, then no integer combination is possible.
Otherwise, let $k = n/d$ so that $n = kd$. Set $x= ku$ and $y = kv$, then
$$ax+by= aku+bkv = k(au+bv) = kd = n$$
$\\$
Here is an example. Lets say I want to find solutions to $$135x+48y = 3375$$
I use the Euclidean Algorithm as follows:
\begin{align} &(135,48) \to 135= 2*48 + 39 \to \quad 39 = 135 - (2)(48) \\ &(48,39) \to \quad 48 = 1*39 + 9 \to \quad 9 = 48 - (1)(39)\\ &(39,9) \to \quad 39 = 4*9+ 3 \to \quad 3 = 39 - (4)(9) \\ &(9,3) \to \quad 9 = 3*3 + 0 \\ &(3,0) \to \quad \text{done} \end{align}
So $3$ is the gcd. $3375/3 = 1125$ which is a whole number. Therefore a solution exists. Now to find it:
We now work backwards:
\begin{align} 3 =& \ 39-(4)(9) \\ =& \ 39 - (4)(48-(1)(39)) \\ =& \ 135 - (2)(48) - (4)(48-(1)(135-(2)(48))) \\ =& \ 135 - (2)(48) - (4)(48-135+(2)(48)) \\ =& \ 135 -(2)(48) - (4)(48) +(4)(135)-(8)(48) \\ =& \ (5)(135) + (-14)(48) \end{align}
Thus
$$(135)(5)+ (48)(-14) = 3$$
Now since $3375/3 = 1125$, we multiply the whole equation by $1125$ and get
$$(135)(5)(1125) + (48)(-14)(1125) = (3)(1125)$$
$$\\$$ $$\implies (135)(5625)+(48)(-15750) = 3375$$ | {
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$$\\$$ $$\implies (135)(5625)+(48)(-15750) = 3375$$
• @jiten I meant to change that. It should be 135x+48y and not 135x+6y. I first used six but the Euclidean alg was too short (like 1 step) so I changed it to one that provided a better example. Its fixed now. – David Reed Nov 24 '17 at 9:11
• Thanks for removing confusion. – jiten Nov 24 '17 at 9:14
• @jiten The Euclidean algorithm is a good algorithm to know. Its used in several other places in algebra and number theory. Equations like this, $ax+by = n$ are called "linear Diophantine equations.", in case you have any interest in researching further. – David Reed Nov 24 '17 at 9:17 | {
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# A stronger version of discrete “Liouville's theorem”
If a function $f : \mathbb Z\times \mathbb Z \rightarrow \mathbb{R}^{+}$ satisfies the following condition
$$\forall x, y \in \mathbb{Z}, f(x,y) = \dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$
then is $f$ constant function?
-
You probably wanto to add a boundedness condition. Otherwise $f(x,y)=x$ is a counterexample. – Julián Aguirre Jul 17 '11 at 10:24
@Julian Aguirre: since $x\in\mathbb Z$, we don't have $f(x,y)\geq 0$. – Davide Giraudo Jul 17 '11 at 11:08
@girdav You are right. The lower bound is probably enough. – Julián Aguirre Jul 17 '11 at 13:08
You can prove this with probability.
Let $(X_n)$ be the simple symmetric random walk on $\mathbb{Z}^2$. Since $f$ is harmonic, the process $M_n:=f(X_n)$ is a martingale. Because $f\geq 0$, the process $M_n$ is a non-negative martingale and so must converge almost surely by the Martingale Convergence Theorem. That is, we have $M_n\to M_\infty$ almost surely.
But $(X_n)$ is irreducible and recurrent and so visits every state infinitely often. Thus (with probability one) $f(X_n)$ takes on every $f$ value infinitely often.
Thus $f$ is a constant function, since the sequence $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge. | {
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-
I love probabilistic arguments in analysis! Very nice. – Jonas Teuwen Jul 17 '11 at 11:46
This is indeed very nice: Question: Is it possible to change this argument in such a way that it applies for $\mathbb{Z}^n$ instead of $\mathbb{Z}^2$ only? Is it even true that a non-negative harmonic function on $\mathbb{Z}^n$ is constant for $n \geq 3$? For bounded ones this seems clear by considering the Poisson boundary. – t.b. Jul 17 '11 at 11:53
@Byron: This paper contains the claim that it is true that "nonnegative nearest-neighbors harmonic function on $\mathbb{Z}^d$ are constant for any $d$" on page 2. – t.b. Jul 17 '11 at 12:08
The usual Liouville theorem also holds with just a one-sided bound. – GEdgar Jul 18 '11 at 13:56
If you know that the real part of an entire function $f(z)$ is non-negative on the complex plane, what can you say about the function $g(z)=f(z)/(1+f(z))$? – Jyrki Lahtonen Jul 18 '11 at 14:07
I can give a proof for the d-dimensional case, if $f\colon\mathbb{Z}^d\to\mathbb{R}^+$ is harmonic then it is constant. The following based on a quick proof that I mentioned in the comments to the same (closed) question on MathOverflow, Liouville property in Zd. [Edit: I updated the proof, using a random walk, to simplify it] | {
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First, as $f(x)$ is equal to the average of the values of $f$ over the $2d$ nearest neighbours of $x$, we have the inequality $f(x)\ge(2d)^{-1}f(y)$ whenever $x,y$ are nearest neighbours. If $\Vert x\Vert_1$ is the length of the shortest path from $x$ to 0 (the taxicab metric, or $L^1$ norm), this gives $f(x)\le(2d)^{\Vert x\Vert_1}f(0)$. Now let $X_n$ be a simple symmetric random walk in $\mathbb{Z}^d$ starting from the origin and, independently, let $T$ be a random variable with support the nonnegative integers such that $\mathbb{E}[(2d)^{2T}] < \infty$. Then, $X_T$ has support $\mathbb{Z}^d$ and $\mathbb{E}[f(X_T)]=f(0)$, $\mathbb{E}[f(X_T)^2]\le\mathbb{E}[(2d)^{2T}]f(0)^2$ for nonnegative harmonic $f$. By compactness, we can choose $f$ with $f(0)=1$ to maximize $\Vert f\Vert_2\equiv\mathbb{E}[f(X_T)^2]^{1/2}$.
Writing $e_i$ for the unit vector in direction $i$, set $f_i^\pm(x)=f(x\pm e_i)/f(\pm e_i)$. Then, $f$ is equal to a convex combination of $f^+_i$ and $f^-_i$ over $i=1,\ldots,d$. Also, by construction, $\Vert f\Vert_2\ge\Vert f^\pm_i\Vert_2$. Comparing with the triangle inequality, we must have equality here, and $f$ is proportional to $f^\pm_i$. This means that there are are constants $K_i > 0$ such that $f(x+e_i)=K_if(x)$. The average of $f$ on the $2d$ nearest neighbours of the origin is $$\frac{1}{2d}\sum_{i=1}^d(K_i+1/K_i).$$ However, for positive $K$, $K+K^{-1}\ge2$ with equality iff $K=1$. So, $K_i=1$ and $f$ is constant.
Now, if $g$ is a positive harmonic function, then $\tilde g(x)\equiv g(x)/g(0)$ satisfies $\mathbb{E}[\tilde g(X_T)]=1$. So, $${\rm Var}(\tilde g(X_T))=\mathbb{E}[\tilde g(X_T)^2]-1\le\mathbb{E}[f(X_T)^2]-1=0,$$ and $\tilde g$ is constant. | {
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-
Taxicab metric. Never heard that name (I learn maths in french at school). Funny! – Patrick Da Silva Jul 17 '11 at 20:24
@Patrick: Also called the Manhattan metric. – George Lowther Jul 17 '11 at 20:30
LAAAAAAAAAWL. Funnier. – Patrick Da Silva Jul 17 '11 at 20:39
Note: A similar proof will also show that harmonic $f\colon\mathbb{R}^d\to\mathbb{R}^+$ is constant. Interestingly, in the two dimensional case, Byron's proof can be modified to show that harmonic $f\colon\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^+$ is constant (as 2d Brownian motion has zero probability of hitting 0 at positive times). Neither of the proofs generalize to harmonic $f\colon\mathbb{R}^d\setminus\{0\}\to\mathbb{R}^+$ for $d\not=2$. In fact, considering $f(x)=\Vert x\Vert^{2-d}$, we see that $f$ need not be constant for $d\not=2$. – George Lowther Jul 17 '11 at 22:54 | {
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You have observations of n + m 6-sided dice rolls with each face numbered from 1 to 6n of the observations went missing, and you only have the observations of m rolls. Fortunately, you have also calculated the average value of the n + m rolls.
You are given an integer array rolls of length m where rolls[i] is the value of the ith observation. You are also given the two integers mean and n.
Return an array of length n containing the missing observations such that the average value of the n + m rolls is exactly mean. If there are multiple valid answers, return any of them. If no such array exists, return an empty array.
The average value of a set of k numbers is the sum of the numbers divided by k.
Note that mean is an integer, so the sum of the n + m rolls should be divisible by n + m.
Example 1:
Input: rolls = [3,2,4,3], mean = 4, n = 2
Output: [6,6]
Explanation: The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.
Example 2:
Input: rolls = [1,5,6], mean = 3, n = 4
Output: [2,3,2,2]
Explanation: The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.
Example 3:
Input: rolls = [1,2,3,4], mean = 6, n = 4
Output: []
Explanation: It is impossible for the mean to be 6 no matter what the 4 missing rolls are.
Example 4:
Input: rolls = [1], mean = 3, n = 1
Output: [5]
Explanation: The mean of all n + m rolls is (1 + 5) / 2 = 3.
Constraints:
• m == rolls.length
• 1 <= n, m <= 105
• 1 <= rolls[i], mean <= 6
## Solution: Math & Greedy
Total sum = (m + n) * mean
Left = Total sum – sum(rolls) = (m + n) * mean – sum(rolls)
If left > 6 * n or left < 1 * n, then there is no solution.
Otherwise, we need to distribute Left into n rolls.
There are very ways to do that, one of them is even distribution, e.g. using the average number as much as possible, and use avg + 1 to fill the gap.
Compute the average and reminder: x = left / n, r = left % n.
there will be n – r of x and r of x + 1 in the output array. | {
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e.g. [1, 5, 6], mean = 3, n = 4
Total sum = (3 + 4) * 3 = 21
Left = 21 – (1 + 5 + 6) = 9
x = 9 / 4 = 2, r = 9 % 4 = 1
Ans = [2, 2, 2, 2+1] = [2,2,2,3]
Time complexity: O(m + n)
Space complexity: O(1)
## C++
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Question
# Areas of square and rhombus are equal. A diagonal of a rhombus is twice of its other diagonal. If the area of rhombus is $$64$$ sq. cm find the ratio of perimeter of a square and rhombus.
A
3:1
B
2:5
C
2:3
D
5:2
Solution
## The correct option is B $$2 : \sqrt5$$Let $$a$$ be the side of a square and$$b$$ be the side of a rombus,and $$d_1, d_2$$ be the two diagonals of the rombus.Given,Area of square $$=$$ Area of rombus$$\Rightarrow a^2=\dfrac{d_1 d_2}{2}$$Also,$$d_1=2d_2$$$$\therefore a^2=\dfrac{2d_2 d_2}{2}$$$$\Rightarrow a^2=d_2$$$$\Rightarrow a=d_2$$ and$$\Rightarrow a=\cfrac{d_1}{2}$$Area of rhombus is half of the product of the diagonals$$\cfrac{d_1 d_2}{2}=64$$$$\Rightarrow d_1 d_2=128$$$$\Rightarrow (2a)a=128$$$$\Rightarrow a^2=64$$$$\Rightarrow a=8\ cm$$$$d_1=2a=8\times 2$$ $$=16\ cm$$$$d_2=a$$ $$=8\ cm$$Diagonals of a rhombus bisect each other at right angles. So, half of both the diagonals and the side of the rhombus make a right angled $$\triangle$$$$\therefore$$ Semi-Diagonal $$\dfrac{d_1}{2}=\dfrac{16}{2}$$ $$=8\ cm$$and semi-diagonal $$\dfrac{d_2}{2}=\dfrac82$$ $$=4\ cm$$$$\therefore 8^2+4^2=b^2$$$$\Rightarrow b^2=64+16$$$$\Rightarrow b^2=80$$$$\Rightarrow b=\sqrt{80}$$$$\Rightarrow b=4\sqrt5$$$$\therefore$$ Ratio of perimeter of a square and rhombus $$=4a:4b$$ $$=a:b$$ $$=8: 4\sqrt5$$ $$=2:\sqrt 5$$Maths
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# Matrix norm of Kronecker product
Is it true that $\| A \otimes B \| = \|A\|\|B\|$ for any matrix norm $\|\cdot \|$? If not, does this identity hold for matrix norms induced by $\ell_p$ vector norms?
• According to wikipedia, you can relate the eigenvalues of the Kronecker product to that of the operands. This should give you something for the spectral norm. – xavierm02 Jun 30 '17 at 17:40
• Yes, it's true for the spectral norm. That's the only case I know for certain. – Rob Jun 30 '17 at 18:08
• Compute everything for two arbitrary $2\times 2$ matrices (i.e., get both sides as an expression of $a_{11}, a_{12},\dots ,b_{22}$). I'd expect counter examples to be easy to find once you have done that. – xavierm02 Jun 30 '17 at 18:12 | {
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Theorem 8 here provides the answer: http://www.ams.org/journals/mcom/1972-26-118/S0025-5718-1972-0305099-X/S0025-5718-1972-0305099-X.pdf. As discussed on page 413, the identity holds for all matrix norms induced by $\ell_p$ vector norms. In fact, it seems to hold for any induced vector norm, or any submultiplicative norm.
Here is a proof for the lazy: Let $$A = \sum_i \sigma_i u_i v_i^T$$ and $$B = \sum_j \lambda_j x_j y_j^T$$ be the singular value decomposition (SVD) of the two matrices. Then, \begin{align} A \otimes B &= \sum_{i,j}\sigma_i \lambda_j (u_i v_i^T) \otimes (x_jy_j^T) \\ &= \sum_{i,j} \sigma_i \lambda_j (u_i \otimes x_j) (v_i^T \otimes y_j^T) \\ &= \sum_{i,j} \sigma_i \lambda_j (u_i \otimes x_j) (v_i \otimes y_j)^T \end{align} where the first equality is by the bilinearity of $$\otimes$$, the second by the "mixed product" property of Kronecker product and last one by $$(A\otimes B)^T = A^T \otimes B^T$$. It is not hard to see that $$\{u_i \otimes x_j, \forall i,j \}$$ is an orthonormal collection of vectors and similarly for $$\{v_i \otimes y_j, \forall i,j \}$$. It follows that the last line above is the SVD of $$A \otimes B$$, with singular values $$\{\sigma_i \lambda_j, \forall i,j\}$$. Hence, \begin{align} \| A \otimes B\| = \max_{i,j} \sigma_i \lambda_j = (\max_i \sigma_i)(\max_j \lambda_j) = \|A\| \|B\|. \end{align} for the $$\ell_2$$ operator norm. | {
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# Two bottles are partially filled with water. The larger bottle current
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Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
A. $$\frac{5}{6}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{7}{12}$$
E. $$\frac{1}{2}$$
_________________
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04 Aug 2016, 15:13
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
A. $$\frac{5}{6}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{7}{12}$$
E. $$\frac{1}{2}$$
Here the capacity of larger bottle be x and currently the larger bottle has x/3 capacity (given).
Now smaller bottle has 2/3(x) ...( GIVEN: The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle )
Currently the SB has 3/4( 2/3(x) ) = 1/2(x) capacity and this is added to larger bottle i.e. 1/2(x) + x/3 = 5/6(x).
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### Show Tags
04 Aug 2016, 21:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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12 Aug 2018, 07:18
Let the capacity of larger water bottle be "x"
Water filled in the larger bottle=x/3
Water filled in the smaller bottle=2x/3*3/4=x/2
If we pour the smaller bottle to bigger bottle i.e. x/3+x/2=5x/6=>5/6
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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04 Sep 2018, 22:21
1
Take larger bottle (L) to have capacity 9 L and since smaller bottle(S) has 2/3rd the larger bottles capacity, S = 6 L capacity.
Now we are told that L is filled up to 1/3rd its capacity i.e 3 L.
S is filled up to 3/4th its capacity i.e 4.5 L.
when we add 4.5 L to 3 L we get 7.5 L.
7.5/9 = 75/90 = 15/18 = 5/6
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when we add 4.5 L to 3 L we get 7.5 L.
7.5/9 = 75/90 = 15/18 = 5/6
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Re: Two bottles are partially filled with water. The larger bottle current &nbs [#permalink] 04 Sep 2018, 22:21
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# How do I get from log F = log G + log m - log(1/M) - 2 log r to a solution withoug logs?
I've been self-studying from Stroud & Booth's excellent "Engineering Mathematics", and am currently on the "Algebra" section. I understand everything pretty well, except when it comes to the problems then I am asked to express an equation that uses logs, but without logs, as in:
$$\log{F} = \log{G} + \log{m} - \log\frac{1}{M} - 2\log{r}$$
They don't cover the mechanics of doing things like these very well, and only have an example or two, which I "kinda-sorta" barely understood.
Can anyone point me in the right direction with this and explain how these are solved?
• Since each $\log$ is in the same base (ten) you can cancel them. That is, if $\log_b(x)=\log_b(y)$ then $x=y$. P.S. The converse is also true since $\log$ is an inyective function. Mar 19, 2019 at 17:17
Using some rules of logarithms you get $$\quad-\log\dfrac{1}{M}=+\log M$$ and $$-2\log r=-\log r^2=+\log \dfrac{1}{r^2}$$
So you have
$$\begin{eqnarray} \log{F} &=& \log{G} + \log{m} + \log M + \log{\dfrac{1}{r^2}}\\ \log{F} &=& \log{\left(GmM\cdot\dfrac{1}{r^2}\right)}\\ \log{F} &=& \log{\frac{mMG}{r^2}}\\ F &=&\frac{mMG}{r^2} \end{eqnarray}$$
The last step hinges upon the fact that logarithm functions are one-to-one functions. If a function $$f$$ is one-to-one, then $$f(a)=f(b)$$ if and only if $$a=b$$. Since $$\log$$ is a one-to-one function, it follows that $$\log A=\log B$$ if and only if $$A=B$$.
ADDENDUM: Here are a few rules of logarithms which you may need to review
1. $$\log(AB)=\log A+\log B$$
2. $$\log\left(\dfrac{A}{B}\right)=\log A-\log B$$
3. $$\log\left(A^n\right)=n\log A$$
4. $$\log(1)=0$$
Notice that from (2) and (4) you get that $$\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$$ | {
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Notice that from (2) and (4) you get that $$\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$$
• Nice clear answer. You may like to point out that $a = b \iff \log a = \log b$ so that $\log F = \log \frac {GMm}{r^2}$ would also mean $F = \frac {GMm}{r^2}$. Mar 19, 2019 at 18:27
• @fleablood Good point. Mar 19, 2019 at 20:04
Hint:
Product rule for Logarithms says that:
$$\log\prod_{k=1}^{n}a_k=\sum_{k=1}^{n}\log a_k$$
In the equation stated in your question: \begin{aligned}\log F&=\log G+\log m+\log M+\log\dfrac{1}{r^2}\\ \log F &= \log \dfrac{GMm}{r^2}\\ \exp\log F&=\exp\log\dfrac{GMm}{r^2}\\ F&=G\cdot\dfrac{Mm}{r^2}\end{aligned}
The SINGLE most important rule of logarithms is:
$$\log N + \log M = \log N\times M$$.
This is because if $$a = \log N; b=\log M$$ then $$10^a = N; 10^b = M$$ so $$10^{a+b} = 10^a\times 10^b = N\times M$$ and so, by definition, $$a+b = \log N\times M$$.
From this simple rule we get $$n \log M = \log (M^n)$$ and $$\log M -\log N = \log \frac MN$$ and so on.
So......
Well, the basic rules of combining logarithms: $$\log a + \log b = \log ab$$ will give us:
$$\log{F} = \log{G} + \log{m} - \log\frac{1}{M} - 2\log{r} = \log \frac {Gm}{\frac 1Mr^2}=\log \frac {GMm}{r^2}$$.
The log function is one to one so we know that for positive real numbers that $$a = b \iff \log a = \log b$$.
(If we need to convince ourselves of this: $$m= \log a = \log b = n\implies 10^m =10^{\log a} = a; 10^n = 10^{\log b}= b; 10^m = 10^n\implies a=b \implies \log a=\log b$$.)
So from here we get:
$$F = \frac {GMm}{r^2}$$. | {
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# Must an uncountable subset of R have uncountably many accumulation points? [duplicate]
This question is taken from problem 4.1.8 of "Real Analysis and Foundations" by Krantz
"Let S be an uncountable subset of $\mathbb{R}$. Prove that S must have infinitely many accumulation points. Must it have uncountably many?"
The first part of the question took some work but ended up coming out pretty smoothly; however, I'm at a complete loss as to how to go about addressing the second problem. Intuitively, I think the answer is yes.
My first attempt was to try to show that a countable number of accumulation points allowed one to order the elements of S in such a way that they are countable (ie prove the contrapositive), but I did not manage to get much further than that.
My second attempt was to show that $S-{s_{1},s_{2}...}$ where $s_{1},s_{2},...$ are countably many accumulation points of S is uncountable and thus must have an accumulation point, so S has an accumulation point that is not one of the countably infinite set. Hence, the number of accumulation points is uncountable.
My question is, is this logic valid? I would have to prove that an uncountable set minus a countable set is uncountable, which shouldn't be too difficult.
Any hints/points in the right direction/outright answers are greatly appreciated.
## marked as duplicate by Asaf Karagila♦, AlexR, user61527, egreg, M TurgeonFeb 5 '14 at 23:32
• You might like this blog post. – David Mitra Feb 5 '14 at 22:05
• Your second proof assumes that the accumulation points of $S$ belong to $S$. This isn't true unless your set is closed. – EuYu Feb 5 '14 at 22:05
• There are many duplicates of this question. – Asaf Karagila Feb 5 '14 at 22:15
• @AsafKaragila This particular one doesn't seem like it, note that this question asks about the cardinality of acc-pts, while the linked one only asks about existence. – AlexR Feb 5 '14 at 22:45
• @Alex: See Brian's answer there. – Asaf Karagila Feb 5 '14 at 22:46 | {
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Let $$T$$ be the set of elements of $$S$$ that are not accumulation points of $$S$$. Then for each $$x \in T$$ there exists $$\epsilon_x > 0$$ such that the interval $$I(x,\epsilon_x) = (x-\epsilon_x,x+\epsilon_x)$$ contains no other points of $$S$$. The intervals $$\{I(x,\epsilon_x/2) | x\in T\}$$ are disjoint, and each contains a rational number; so there can only be a countable number of them.
Hence $$T$$ is countable, and the set of accumulation points of $$S$$, which contains $$S-T$$, is uncountable.
• Why $\{I(x,\epsilon/2) | x\in T\}$ is disjoint? Thanks – StammeringMathematician Feb 19 at 15:27
• First before assuming that $x\in T$ you need to show that $T$ is non-empty! – SunShine Aug 8 at 17:58
• @SunShine: $T$ doesn't have to be non-empty. For instance, $S$ might be the whole of $\Bbb R$. But then there is nothing more to prove, is there? An empty set is certainly countable. – TonyK Aug 8 at 18:24
Assuming a finite number of accumulation points for $S$ would mean that the elements of $S$ that are not accumulation points are isolated, meaning $S$ is at most countable, a contradiction.
Assuming a countable number of accumulation points yields the same problem. The elements of $S$ that are not accumulation points would be isolated, and you still have that $S$ is at most a union of two countable sets which is countable, a contradiction.
I think both TonyK's and Darrin's answers are good ones. Here's another, possibly inferior, approach.
First trust the following lemma: for any uncountable set $S$ of real numbers, there exists an interval $(a,b)$ such that both $S\cap(-\infty,a]$ and $S\cap[b,\infty)$ are uncountable. ("You can divide an uncountable set into two with a buffer interval.") | {
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Given the lemma, you can find two uncountable subsets $S_1,S_2$ of $S$ that are separated by an interval. Iterating, you can find four uncountable subsets of $S$ all separated from one another by intervals, then 8, 16, etc. There are uncountably many branches down this repeated-bifurcation tree; each one results in an accumulation point; and the accumulation points can't coincide because of the separating intervals. Thus there are uncountably many accumulation points.
Why is the lemma true? Assume $S$ is contained in $[0,3]$ (an easy reduction), and consider the intersections of $S$ with $[0,1]$, $[1,2]$, and $[2,3]$. If the first and third intersections are uncountable, then we're done. Otherwise, we have an uncountable set in a smaller interval (perhaps length 1, perhaps length 2) and we divide that interval into three equal pieces. We keep doing this until either we find the first and last intersections uncountable (in which case we're done), or else we iterate countably often and end up with all the countable subsets contained in a sequence of nested closed intervals whose lengths tend to $0$. But this latter case can't happen, because then we will have written $S$ as a countable union of countable sets, together with the single point in the intersection of the nested intervals, contradicting the uncountability of $S$. | {
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# Pseudo-inverse of a matrix that is neither fat nor tall?
Given a matrix $A\in\mathbb R^{m\times n}$, let us define:
• $A$ is a fat matrix if $m\le n$ and $\text{null}(A^T)=\{0\}$
• $A$ is a tall matrix is $m\ge n$ and $\text{range}(A)=\mathbb R^n$
Using the finite rank lemma, we can find that:
• When $A$ is a fat matrix, its (right) pseudo-inverse is $A^\dagger = A^T(AA^T)^{-1}$
• When $A$ is a tall matrix, its (left) pseudo-inverse is $A^\ddagger = (A^TA)^{-1}A^T$
My question is what is the pseudo-inverse when $A$ is neither fat nor tall (in the sense of the above definitions), i.e. it is a matrix such that $\text{null}(A^T)\ne \{0\}$ (i.e. the null space is non-trivial) and $\text{range}(A)\ne\mathbb R^n$? An example of such a matrix is:
$$A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 3 & 0 \end{bmatrix}$$
which clearly does not map to full $\mathbb R^4$ and whose null space is $\text{span}\left\{\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right\}$.
• You should read about the Moore-Penrose inverse of a matrix. Or, in general, the generalized inverse of a matrix. – steven gregory Oct 16 '16 at 16:16
• A conceptual definition of the pseudoinverse of an $m \times n$ matrix $A$ is that the pseudoinverse takes a vector $b$ as input, projects $b$ onto the range of $A$ (call the projection of $b$, say, $\hat b$), then returns as output the vector $x$ with least norm such that $Ax =\hat b$. – littleO Mar 1 '17 at 23:13
There is no need to study both the fat matrix and the tall matrix. Let $B = A^t$. In that case, the tall matrix problem for $B$ is the fat matrix problem for $A$ and vice versa. | {
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Thus, we consider only the tall matrix case. Assume that the rank of $A$ is $r$ where $r \leq n$. Let the SVD (singular value decomposition) of $A$ is defined as $A = U \Sigma V^t$ where $U$ is an $m \times r$ matrix with orthogonal columns, $V$ is an $r \times n$ matrix with orthogonal columns and the diagonal $r \times r$ matrix $\Sigma$ contains the $r$ singular values in the diagonal. These diagonal values $\sigma_i$, $i=1 \colon r$ are ordered in the non-increasing order.
The Moore-Penrose pseudo inverse is then given by $$A^{+} = V \Sigma^{-1} U^t .$$ This pseudo inverse exists even when $A^tA$ is singular.
Numerical analysts will tell you to define the "zero singular values" as the singular values which are "tiny". Tiny means any number which is in the same order as $\|A\|*\epsilon$ {(any) norm of $A$ times the machine precision}. Unfortunately, the Moore-Penrose inverse often depends on the way "tiny" is defined. Mathematicians do not have this problem since zero means zero and nothing else.
For your matrix $A$, there are two non-zero singular values (3.7519 and 0.9610) and hence $r=2$. Since there are no tiny singular values, the Moore-Penrose inverse is well defined.
Fundamental Theorem of Linear Algebra
The matrix $\mathbf{A} \in \mathbb{C}^{m\times n}$ induces the four fundamental subspaces. In finite dimension, the domain $\mathbf{C}^{n}$, and codomain, $\mathbf{C}^{m}$ are \begin{align} % \mathbf{C}^{n} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} \\ % \mathbf{C}^{m} = \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A} \right)} % \end{align}
Least squares | {
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Least squares
Consider a data vector $b \in \mathbb{C}^{m}$ that does not lie in the null space, that is, $b \notin \color{red} {\mathcal{N} \left( \mathbf{A} \right)}$. We are solving for the least squares minimizers defined as $$x_{LS} = \left\{ x\in\mathbb{C}^{n} \colon \lVert \mathbf{A} x_{LS} - b \rVert_{2}^{2} \text{ is minimized} \right\}.$$ These solutions are $$x_{LS} = \color{blue}{\mathbf{A}^{\dagger} b} + \color{red}{\left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right) y}, \quad y\in\mathbb{C}^{n}$$ The Moore-Penrose pseudoinverse matrix is \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{cc} \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} & \color{red}{\mathbf{U}_{\mathcal{N}}}^{*} \end{array} \right] % \end{align}
SVD General case
The singular value decomposition provides an orthonormal basis for the four subspaces. The range spaces are aligned and the difference in length scales is expressed in the singular values. | {
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The decomposition for a matrix with rank $\rho$ is \begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cccccc} \sigma_{1} & 0 & \dots & & & \dots & 0 \\ 0 & \sigma_{2} \\ \vdots && \ddots \\ & & & \sigma_{\rho} \\ & & & & 0 & \\ \vdots &&&&&\ddots \\ 0 & & & & & & 0 \\ \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red}{\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % & = % U \left[ \begin{array}{cccccccc} \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{v_{1}^{*}} \\ \vdots \\ \color{blue}{v_{\rho}^{*}} \\ \color{red}{v_{\rho+1}^{*}} \\ \vdots \\ \color{red}{v_{n}^{*}} \end{array} \right] % \end{align} Let's look at your special cases.
Tall: $m>n$
The overdetermined case of full column rank $\rho = n$. | {
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Tall: $m>n$
The overdetermined case of full column rank $\rho = n$.
\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{c} \mathbf{S}_{\rho\times \rho} \\ \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \end{array} \right] \\ % \end{align} \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho}^{-1} & \mathbf{0} \end{array} \right] % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ \color{red} {\mathbf{U}_{\mathcal{N}}}^{*} \end{array} \right] % \end{align}
Wide: $m<n$
The underdetermined case of full row rank $\rho = m$.
\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho} & \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red} {\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % \end{align} \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{cc} \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red} {\mathbf{V}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho}^{-1} & \mathbf{0} \end{array} \right] % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \end{array} \right] % \end{align}
Square: $m=n$
The nonsingular case $\rho = m = n$. | {
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Square: $m=n$
The nonsingular case $\rho = m = n$.
\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{c} \mathbf{S} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \end{array} \right] % \end{align} \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{c} \mathbf{S}^{-1} \end{array} \right] % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \end{array} \right] % \end{align}
Original example
@Vini presents the solution clearly. To emphasize his conclusions, here are the background computations. The product matrix is $$\mathbf{A}^{\mathrm{T}}\mathbf{A} = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 14 & 0 \\ 0 & 0 & 0 \\ \end{array} \right],$$ which quickly leads to the characteristic polynomial $$p(\lambda) = \left( -\lambda^{2} + 15 \lambda-13\right) \lambda.$$ The eigenvalue spectrum is then $$\lambda \left( \mathbf{A}^{\mathrm{T}}\mathbf{A} \right) = \left\{ \frac{1}{2} \left( 15 \pm \sqrt{173} \right), 0 \right\}.$$ The singular values are the square root of the nonzero eigenvalues $$\left\{ \sigma_{1}, \sigma_{2} \right\} = \left\{ \sqrt{\frac{1}{2} \left( 15 + \sqrt{173} \right)}, \sqrt{\frac{1}{2} \left( 15 -\sqrt{173} \right)} \right\}$$ There are two nonzero singular values, therefore the matrix rank is $\rho = 2$. | {
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Let $$\mathbf{S} = \left( \begin{array}{cc} \sigma_{1} & 0 \\ 0 & \sigma_{2} \\ \end{array} \right).$$ This diagonal matrix is embedded in the sabot matrix, padded with zeros to insure conformability between the matrices $\mathbf{U}$ and $\mathbf{V}$: $$\Sigma = \left[ \begin{array}{ccc} \sqrt{\frac{1}{2} \left(\sqrt{173}+15\right)} & 0 & 0 \\ 0 & \sqrt{\frac{1}{2} \left(15-\sqrt{173}\right)} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] = \left( \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \\ \end{array} \right).$$ The pseudoinverse of the $\Sigma$ matrix is $$\Sigma^{\dagger} = \left[ \begin{array}{ccc} \sqrt{\frac{1}{2} \left(\sqrt{173}+15\right)}^{-1} & 0 & 0 \\ 0 & \sqrt{\frac{1}{2} \left(15-\sqrt{173}\right)}^{-1} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] = \left( \begin{array}{cc} \mathbf{S}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \\ \end{array} \right).$$ In this case where $m=n$, the matrix products are identical: $$\Sigma^{\dagger}\Sigma = \Sigma\, \Sigma^{\dagger} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] = \left[ \begin{array}{cc} \mathbf{I}_{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \\ \end{array} \right].$$
Another example of $\Sigma$ gymnastics is in here: Singular Value Decomposition: Prove that singular values of A are square roots of eigenvalues of both $AA^{T}$ and $A^{T}A$ | {
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The eigenvectors of the product matrix are $$% v_{1} = \left[ \begin{array}{c} \frac{1}{2} \left(-13+\sqrt{173}-13\right) \\ 1 \\ 0 \end{array} \right], \quad % v_{2} = \left[ \begin{array}{c} \frac{1}{2} \left(-13-\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right], \quad %$$ The normalized form constitute the column vectors of the domain matrix $$\mathbf{V} = % \left[ \begin{array}{cc} % \left( 1 + \left(\frac{-13 + \sqrt{173}}{4}\right)^2 \right)^{-\frac{1}{2}} % \left[ \begin{array}{c} \frac{1}{2} \left(-13+\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right] & % \left( 1 + \left(\frac{13 + \sqrt{173}}{4}\right)^2 \right)^{-\frac{1}{2}} % \left[ \begin{array}{c} \frac{1}{2} \left(-13-\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right] % \end{array} \right]$$ | {
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# Find an integer $n$ such that $\mathbb{Z}[\frac{1}{20},\frac{1}{32}]=\mathbb{Z}[\frac{1}{n}]$.
How can we find an integer $n$ such that $\mathbb{Z}[\frac{1}{20},\frac{1}{32}]=\mathbb{Z}[\frac{1}{n}]$?
-
Rapid answer: since $\frac1{32}=\frac1{2^5}=\frac{2\times5^3}{20^3}\in\mathbb Z[\frac1{20}]$, one can take $n=20$.
More generally the following considerations apply.
Let $R$ be a subring of $\mathbf Q$, and consider the set $D$ of positive integers $n$ such that $\frac1n\in R$. Then for all reduced fractions $\frac nd\in R$ one has $\frac1d\in R$ and therefore $d\in D$: since $n$ is relatively prime to $d$ there exists $s\in \mathbb Z$ with $sn\equiv1\pmod d$, and subtracting an integer from $s\times\frac nd$ will give $\frac1d$. Also the set $D$ is closed under taking arbitrary divisors (since we can multiply $\frac1d$ by any integer), and under multiplication (since we can multiply $\frac1d$ and $\frac1{d'}$). It easily follows that $D$ (and therefore $R$) is determined by the subset $P$ of prime numbers in $D$, as $D$ will be the set of all positive integers all of whose prime factors are in $P$. If $P$ is finite, one can write $R=\mathbb Z[\frac1n]$ for any $n$ such that the set of prime divisors of $n$ is $P$.
In your example $P=\{2,5\}$ and any $n$ with exactly those prime divisors will work; $n=10$ is the smallest positive example of such $n$.
-
HINT: In order to have $\Bbb Z\left[\frac1{20},\frac1{32}\right]\subseteq\Bbb Z\left[\frac1n\right]$, having integers $a$ and $b$ such that $\frac1{20}=\frac{a}n$ and $\frac1{32}=\frac{b}n$ will work nicely. This implies that $n=20a$ and $n=32b$, so $n$ must be a common multiple of $20$ and $32$. On the other hand, there must also be integers $a$ and $b$ such that $\frac1n=\frac{a}{20}+\frac{b}{32}$; rearrange that to get an equation without fractions involving $a,b$, and $n$, and see what it tells you about $n$. Between the two requirements, you should be able easily to find an $n$ that works. | {
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-
With the ususal sense of square brackets ("ring generated by") the "must" in your first sentence is not warranted. See my answer. In fact your answer works for "additive group generated by". – Marc van Leeuwen Sep 12 '12 at 7:11
@Marc: Fair enough. I was taking the easy route and then carelessly over-generalizing. – Brian M. Scott Sep 12 '12 at 7:16
Hint $\$ Let $\rm\,Z = \Bbb Z$ or any Bezout domain, i.e. a domain with gcds which are linear combinations $\rm\,(a,b) = ja+kb,\ j,k\in Z.\:$ Every ring $\rm\,R\,$ between $\rm\,Z\,$ and its fraction field $\rm\,Q\,$ is equal to the subring generated by $\rm\,Z\,$ and inverses of primes of $\rm\,Z\,$ occuring in denominators of reduced fractions in $\rm\,R.$
First $\rm\,a/b \in R\iff 1/b\in R,\,$ since, wlog $\rm\,(a,b)=1,\,$ thus $\rm\,ja+kb=1\,$ for some $\rm\,j,k,\in Z,\,$ hence $\rm\,j(a/b)+k = (aj+kb)/b = 1/b\in R.\:$ Therefore $\rm\,R\,$ is generated by $\rm\,Z\,$ and the inverses of the denominators of the reduced fractions in $\rm\,R.$
Also $\rm\:1/ab\in R\iff 1/a,\,1/b\,\in R,\:$ by $\rm\:a(1/ab) = 1/b.\,$ Thus if, further, $\rm\,Z\,$ is a UFD, then we can uniquely factor the denominators into primes, whittling the generating set down to $\rm\,Z\,$ and the inverses of said primes. Therefore, for your example
$$\rm \Bbb Z\left[\frac{1}{20},\frac{1}{32}\right] =\, \Bbb Z\left[\frac{1}{5\cdot 2^2},\frac{1}{2^5}\right] =\, \Bbb Z\left[\frac{1}2,\frac{1}5\right] =\, \Bbb Z\left[\frac{1}{10}\right]$$
Remark $\$ Much is known about rings enjoying this and similar properties, e.g. see
Gilmer, Robert; Ohm, Jack. Integral domains with quotient overrings.
Math. Ann. 153 1964 97--103. MR 28#3051 13.15 (16.00) | {
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An integral domain D with field of quotients K is said to have the QR-property if every ring D' between D and K is a ring of quotients of D. The article studies integral domains with the QR-property. Starting with some simple properties of integral domains with the QR-property, the authors prove the following theorem. Let D be an integral domain with the QR-property, and let P be its arbitrary prime ideal. Then D_P is a valuation ring. If A is a finitely generated ideal of D, then A is invertible and some power of A is contained in a principal ideal.
As for the converse, the authors prove that if every valuation ring of K containing D is a ring of quotients of D and if every prime ideal of D is the radical of a principal ideal, then D has the QR-property. As a particular case, it follows that a Noetherian integral domain D has the QR-property if and only if it is a Dedekind domain in which the ideal class group is a torsion group. Some properties of intermediate rings for such rings are observed.
{Reviewer's note: Though the authors added that the result on Noetherian integral domains with the QR-property was obtained independently also by E. D. Davis, the reviewer is afraid that this special case may be known by some other people. Substantially the same result is also contained in the paper of Goldman reviewed below [#3052].}
Reviewed by M. Nagata
Richman, Fred. Generalized quotient rings.
Proc. Amer. Math. Soc. 16 1965 794--799. MR 31#588013.80 (16.00) | {
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Let A be an integral domain with the quotient field K , and consider an intermediate ring A < B < K . The author calls an intermediate ring B a "generalized quotient ring of A" if B is A-flat. (For example, if B = S^{-1} A for some multiplicative subset S of A, then B is A-flat.) One of the main theorems reads as follows. Let A be an integral domain with the quotient field K. For an intermediate ring A < B < K, the following statements are equivalent: (i) B is a generalized quotient ring of A; (ii) (A:Ab)B = B for all b in B; (iii) B_M = A_{M/\A} for all maximal ideals M of B. The author then studies the correspondence of ideals in A and in its generalized quotient ring, and it is also shown that if B is a generalized quotient ring of A, then the integral closure of B is a generalized quotient ring of the integral closure of A. Finally, the author gives a new characterization of Prüfer rings; namely, an integral domain A is a Prüfer ring if and only if every extension of A in its quotient field is flat.
Reviewed by D. S. Rim
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Let $n=\operatorname{lcm}(32,20)=160$ then $R=Z[1/32,1/20]\subset Z[1/n]$ since 1/32=5/160, 1/20=8/160. On the other hand $1/160=4\cdot 1/32\cdot 1/20\in R$. So $n=160$ is a valid answer
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# How do “Dummy Variables” work?
I do not understand how dummy variables work in math.
Suppose we have:
$$I_1 = \int_{0}^{\infty} e^{-x^2} dx$$
How is this equivalent to:
$$I_2 = \int_{0}^{\infty} e^{-y^2} dy$$
How does this dummy variable system work?
Since $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$
??
Thanks!
• You are mistaken, $y$ is not a dependent variable for $I_1$. Why do you think that? – Mark Fantini Jan 5 '15 at 15:36
• The $y$ in $I_2$ is not necessary this one in $y = e^{-x^2}$. – Alex Silva Jan 5 '15 at 15:36
• See @AlexSilva's comment. $I_1$ is just a number, it has no functional relation. If you had $$I_1 = \int_0^{\infty} e^{-y x^2} \, dx,$$ then now you'd have $I_1$ as a function of $y$. But this would be the same as $$I_1 = \int_0^{\infty} e^{-y t^2} \, dt.$$ The dummy variable changed but all is the same. – Mark Fantini Jan 5 '15 at 15:40
• If you are not in a setting where you've stated explicitly that $y = e^{-x^2}$, then there is no connection between $x$ and $y$ at all. And even if you were, the integral $\int e^{-x^2}dx$ is equal to the integral $$\int e^{-y^2}dy = \int e^{-e^{-2x^2}}d(e^{-x^2})$$ This is what variable substitution is all about. (I cannot use limits here, since $e^{-x^2}$ never becomes $0$ or $\infty$.) – Arthur Jan 5 '15 at 15:41
• As I always tell my students: you need to break free from the chains of labels, and see them as a vehicle to speak of something, not the ontological manifestation of something. Whether $y$ is a dependent variable, independent variable, coordinate function or a scream of agony during finals depends entirely on context, not on the particular shape of the symbol or its common use in one course or another. – guest Jan 5 '15 at 21:39 | {
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$$\sum_{j=1}^3 j^4 = 1^4 + 2^4 + 3^4 = \sum_{k=1}^3 k^4.$$ In the first term, $1^4$, we can say that $j=1$; in the second term $2^4$ we have $j=2$, and in the third term, $3^4$ we have $j=3$. But when we call the index $k$ rather than $j$, then in the first term $k=1$, in the second $k=2$, and in the third $k=3$.
$j$ or $k$ is a bound variable (also sometimes called a dummy variable). The value of the whole expression $1^4+2^4+3^4$, which is $98$, does not depend on the value of a bound variable.
If we write $\displaystyle\sum_{j=1}^3 (j \cos (j+m))^4$, then $j$ is bound and $m$ is free. The sum is $$(1\cos (1+m))^4 + (2\cos (2+m))^4 + (3\cos(3+m))^4.\tag 1$$ Its value depends on the value of the free variable $m$, but not on the value of anything called $j$. Accordingly we do not see $j$ in $(1)$. We could rename $j$ and call it $k$, and the whole thing would still be equal to the expression $(1)$ in which we also do not see $k$.
Another example is expressions like $\displaystyle\lim_{h\to0}\frac{(x+h)^3-x^3}h=3x^3$. The value of this expression depends on the value of the free variable $x$, but not of the bound (or "dummy") variable $h$. We could have said $\displaystyle\lim_{k\to0}\frac{(x+k)^3-x^3}k=3x^3$ and it would still be $3x^2$. | {
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• Replace the last $h$ by $k$. – Alex Silva Jan 5 '15 at 15:54
• Thank you (+1), but what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \implies \int dF(x) = \int x dx$$ The LHS then does make sense after $\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? – anonymous Jan 5 '15 at 17:44
• In the notation $\int_a^b g(x)\,df(x)$, one is dealing with a Riemann–Stieltjes integral, defined a a limit of $\sum\limits_{x\in\text{partition}} g(x)\,\Delta f(x)$ $=\sum_x g(x_i^{*})(f(x_{i+1}-f(x_i))$ as the mesh of the partition approaches $0$. This is the same as $\int_a^b g(x)f'(x)\,dx$ in cases where $f$ is everywhere differentiable, but it is also defined in cases where $f$ may do a lot of its changing of values in places where it is not differentiable, including jump discontinuties and places where $f$ is continuous but not absolutely continuous. ${}\qquad{}$ – Michael Hardy Jan 5 '15 at 18:47
• But $\int_a^b g(x)\,df(x)$ is the same as $\int_a^b g(w)\,df(w)$, i.e. one can re-name a bound variable. But the functions $g$ and $f$ are particular functions; the value of the integral depends on which function $f$ is, so $f$ is not a bound variable. ${}\qquad{}$ – Michael Hardy Jan 5 '15 at 18:49
• @MichaelHardy thanks for replying. But I mean, in the above, we have $f(x) = e^{-x^2}$ right? The vertical axis is $f(x)$ then, we have $f(y) = e^{-y^2}$ right? Then we iterate the integral meaning now we have $dxdy$ then if both are independent variables, how will we look at this model in 3D? Since we cannot have the axes related to each other? I mean $dxdy$ how do we look at this in 3D? sicne $y$ and $x$ are both independent variables, we cannot have the traditional width, length $dxdy$ – anonymous Jan 6 '15 at 12:45 | {
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In both cases they are the independent variable. Note in your integrals that $x$ appears nowhere in $I_2$ and $y$ occurs nowhere in $I_1$.
It's common to write $y$ as the dependent variable and $x$ the independent variable. But there's no rule that says you have to. You can write $x = f(y) = y^2 + 2\sin y$ if you wanted to, or put the $y$-axis horizontal and $x$ vertical.
edit: note in any event that $I$ is constant $\left({I = \dfrac {\sqrt{\pi}} 2}\right)$ so it isn't "dependent" in the way you may be used to. Mathematicians will say things like "$I$ is a trivial function of $x$" which means that even though it's technically true that $I = f(x)$, the relationship isn't interesting.
Here's an interesting non-trivial function:
$\displaystyle I(z) = \int_0^z e^{-t^2} \, \mathrm dt$
• (wrote in Michael Hardy's answer too) what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \implies \int dF(x) = \int x dx$$ The LHS then does make sense after $\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? – anonymous Jan 5 '15 at 17:45
IMO this can be understood much better by writing it out in a proper context-free way, rather than standard mathematical notation. The integration symbol $\int_0^\infty$ is basically a higher-order-function, taking a real function1 and returning a single number. $$\int\limits_0^\infty : (\mathbb{R}\to \mathbb{R}) \to \mathbb{R}.$$ Now, people keep saying stuff like "$\sin(x)$ is a function..." but really this is incorrect. $\sin(x)$ is, for any value of $x$, just a single real number. What's a function is $\sin$ itself, i.e. not applied to anything. For instance, $$\int\limits_0^\pi \sin = 2$$ would be quite a reasonable statement. OTOH, it's nonsense to write $$\int\limits_{-\infty}^0 e^x = 1$$ because $e^x$ is not a function. | {
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Trouble is, most functions you need to integrate won't have a predefined name. You can't thus write "the function itself" directly, but have to specify those results of the function for general inputs, in form of some algebraic expression. You know, the equation-definitions $$f(x) = e^{-x^2}.$$ Here it should be quite clear that the $x$ symbol is just an arbitrary choice – it's something you use to signify "anything you can put into the function", and $f(y) = e^{-y^2}$ says obviously exactly the same thing.
Always having to give functions a name before integrating would be tedious, so we have "shortcut syntax" to define an "anonymous function" and use it right away. You can thus think of the $\mathrm{d}x$ symbol much as a lambda function.
1In fact, more accurately and generally, a differential form.
2Again, that's not all there is to it since $\mathrm{d}x$ really is a differential form.
When you are told to compute some mathematical quantity depending on given data $a$, $\ldots\>$, $q$ the slave (person or computer system) doing the computation for you will introduce certain auxiliary variables whose names, temporary values, etc., will be invisible to you, and will not affect the end result. The latter depends only on the "outer" quantities $a$, $\ldots\>$, $q$. When these auxiliary variables are really varying during the computational process you can call them dummy variables – they will be "burnt" at the end; and when they are uniquely determined by the given data you can call them accessory parameters to the question. | {
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# Probability of tossing a fair coin with at least $k$ consecutive heads
Tossing a fair coin for $N$ times and we get a result series as $HTHTHHTT\dots~$, Here '$H$' denotes 'head' and '$T$' denotes 'tail' for a specific tossing each time.
What is the probability that the length of the longest streak of consecutive heads is greater than or equal to $k$? (that is we have a $HHHH\dots~$, which is the substring of our tossing result, and whose length is greater than or equal to $k$)
I came up with a recursive solution (though not quite sure), but cannot find a closed form solution.
Here is my solution.
Denote $P(N,k)$ as the probability for tossing the coin $N$ times, and the longest continuous heads is greater or equal than $k$. Then (For $N>k$)
$$P(N,k)=P(N-1,k)+\Big(1-P(N-k-1,k)\Big)\left(\frac{1}{2}\right)^{k+1}$$
• Your example sequence reminded me of this brilliant Simpsons scene :-) – joriki Nov 10 '12 at 8:58
• I'm getting $(N-k+1)/2^k$ for the closed form. Would you like for me to post my solution, or do you want to think about it some more before I spoil the beans? – Braindead Nov 10 '12 at 8:59
• @Braindead: That can't be right; it's $\gt1$ for small $k$. – joriki Nov 10 '12 at 9:00
• Duh, you are right. I overcounted. – Braindead Nov 10 '12 at 9:08
• I tried to clarify some of the formulations; please check whether I preserved the intended meaning. In particular, I assumed that you had merely accidentally written "greater" once instead of "greater or equal". – joriki Nov 10 '12 at 9:12
Your recurrence relation is correct. I don't think you can do much better than that for general $k$, but you can find a closed form for specific values of $k$. For the first non-trivial value of $k$, the recurrence relation is
$$p_n=p_{n-1}+(1-p_{n-3})/8\;.$$ | {
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$$p_n=p_{n-1}+(1-p_{n-3})/8\;.$$
With $p_n=1+\lambda^n$, the characteristic equation becomes $\lambda^3-\lambda^2+1/8=0$. One solution, $\lambda=1/2$, can be guessed, and then factoring yields $(\lambda-1/2)(\lambda^2-\lambda/2-1/4)$ with the further solutions $\lambda=(1\pm\sqrt5)/4$. Thus the general solution is
$$p_n=1+c_1\left(\frac12\right)^n+c_2\left(\frac{1+\sqrt5}4\right)^n+c_3\left(\frac{1-\sqrt5}4\right)^n\;.$$
The initial conditions $p_0=0$, $p_1=0$, $p_2=1/4$ determine $c_1=0$, $c_2=-(1+3/\sqrt5)/2$ and $c_3=-(1-3/\sqrt5)/2$, so the probability is
\begin{align} p_n &=1-\frac{1+3/\sqrt5}2\left(\frac{1+\sqrt5}4\right)^n-\frac{1-3/\sqrt5}2\left(\frac{1-\sqrt5}4\right)^n\\ &=1-\frac4{\sqrt5}\left(\left(\frac{1+\sqrt5}4\right)^{n+2}-\left(\frac{1-\sqrt5}4\right)^{n+2}\right)\;. \end{align}
Thus, for large $n$ the probability approaches $1$ geometrically with ratio $(1+\sqrt5)/4\approx0.809$.
• I've added an explicit formula for the $P(N,k)$ which might be of interest to you. Note, that your $p_n=1-\frac{1}{2^n}F_{n+2}$ with $F_n$ the Fibonacci numbers. Regards, – Markus Scheuer Jan 18 '16 at 21:40
We can derive an explicit formula of the probability $P(N,k)$ based upon the Goulden-Jackson Cluster Method.
We consider the set of words $\mathcal{V}^{\star}$ of length $N\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the bad word $\underbrace{HH\ldots H}_{k \text{ elements }}=:H^k$ which is not allowed to be part of the words we are looking for. We derive a function $f_k(s)$ with the coefficient of $s^N$ being the number of wanted words of length $N$.
The wanted probability $P(N,k)$ can then be written as \begin{align*} P(N,k)=1-\frac{1}{2^N}[s^N]f_k(s) \end{align*} | {
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According to the paper (p.7) from Goulden and Jackson the generating function $f_k(s)$ is \begin{align*} f_k(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\tag{1} \end{align*} with $d=|\mathcal{V}|=2$, the size of the alphabet and with $\mathcal{C}$ the weight-numerator with \begin{align*} \text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[H^k]) \end{align*} We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[H^k])&=-\frac{s^k}{1+s+\cdots+s^{k-1}}=-\frac{s^k(1-s)}{1-s^k} \end{align*}
We obtain the generating function $f(s)$ for the words built from $\{H,T\}$ which don't contain the substring $H^k$ \begin{align*} f_k(s)&=\frac{1}{1-2s+\frac{s^k(1-s)}{1-s^k}}\\ &=\frac{1-s^k}{1-2s+s^{k+1}}\tag{2}\\ \end{align*}
Note: For $k=2$ we obtain \begin{align*} f_2(s)&=\frac{1-s^2}{1-2s+s^{3}}\\ &=1+2s+3s^2+5s^3+8s^4+13s^5+21s^6+34s^7+\mathcal{O}(s^8) \end{align*}
The coefficients of $f_2(s)$ are a shifted variant of the Fibonacci numbers stored as A000045 in OEIS.
Note: For $k=3$ we obtain \begin{align*} f_3(s)&=\frac{1-s^3}{1-2s+s^{4}}\\ &=1+2s+4s^2+7s^3+13s^4+24s^5+44s^6+81s^7+\mathcal{O}(s^8) \end{align*}
The coefficients of $f_3(s)$ are a shifted variant of the so-called Tribonacci numbers stored as A000073 in OEIS.
We use the series representation of $f_k(s)$ in (2) to derive an explicit formula of the coefficients. | {
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\begin{align*} [s^N]f(s)&=[s^N](1-s^k)\sum_{m=0}^{\infty}(2s-s^{k+1})^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m(2-s^k)^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\\ &=([s^N]-[s^{N-k}])\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{3}\\ &=\sum_{m=0}^{N}([s^{N-m}]-[s^{N-k-m}])\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{4}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ &\qquad-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N-k}\binom{m}{\frac{N-m}{k}-1}(-1)^{\frac{N-m}{k}-1}2^{m-\frac{N-m}{k}+1}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\tag{5}\\ &\qquad+\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ \end{align*}
Comment:
• In (3) we use the linearity of the coefficient of operator and $[s^N]s^mf(s)=[s^{N-m}]f(s)$
• In (4) we change the limit of the left hand sum from $\infty$ to $N$ according to the maximum coefficient $[s^N]$. According to the factors $s^{kj}$ we consider in the following only summands with $m\equiv N(\bmod k)$
• In (5) we reorganise the sums from the line above by extracting from the left hand sum the first summand and shifting in the right hand side the index by one and putting both sums together.
We conclude: An explicit representation of the probability $P(N,k)$ $(n\geq 0)$ is according to (5) \begin{align*} P(N,k)&=1-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}}\\ &\qquad-\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}} \end{align*}
A bit more practical focus than previous answers:
Main idea:
1. We build a "staircase": | {
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A bit more practical focus than previous answers:
Main idea:
1. We build a "staircase":
2. $p$ chance to get to next step, $1-p$ chance to fall down and have to restart (or "reflip"). Except for the highest step where we always stay (whenever we manage to get there).
3. We need $k+1$ steps to "remember" where on the stairs we are for $k$ flips in a row.
We can use this to build a matrix for a Markov chain:
We can build a block matrix:$$\frac{1}{2}\left[\begin{array}{cc}2&1&\bf 0^T\\\bf 0&\bf0&\bf I_k\\0&1&\bf 1^T \end{array}\right]$$
(For the special case $k = 5$). We build a stochastic matrix:
$${\bf P} = \frac{1}{2}\left[\begin{array}{cccccc}2&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\0&1&1&1&1&1\end{array}\right]$$
Now our answer will simply be $$[1,0,{\bf 0}]\, {\bf P}^k\, [{\bf 0},0,1]^T$$
Which is a scalar product using ${\bf P}^k$ as a gramian matrix.
Now to calculate this in practice one could probably in general (for general $p$) benefit from a canonical transformation of $\bf P$, but it's doubtable in this case as the matrix is already sparse with literally only sums and permuations and bit shifts required on the vector elements.
Feller considers this problem in section XIII.7 of An Introduction to Probability Theory and Its Applications, Volume 1, Third Edition. He shows that the probability of having no run of length $k$ in $N$ throws is asymptotic to $$\frac{1-(1/2)\;x}{(k + 1 - kx)\; (1/2)} \cdot \frac{1}{x^{N+1}}$$ where $x$ is the least positive root of $$1 - x +(1/2)^{k+1} x^{k+1} = 0$$
(I have changed Feller's notation to agree with the problem statement and have only considered the case of a fair coin; Feller considers the more general case of a biased coin. For more information see equation 7.11, p. 325 in the referenced document.) | {
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# Can an open set be covered by proper open subsets?
Let $(X, \tau)$ be a topological space, let $U\in\tau$. An open cover of $U$ is a set $\{U_i \ |\ i\in I\}$ (of open sets $U_i$) whose union $\bigcup U_i$ contains $U$.
If $U\subsetneq X$, then $U$ admits an open covering by open sets $\{U_i \ |\ i\in I\}$, where an arbitrary open set $U_i$ may not be a subset of $U$. (I think.)
Does $U$ admit an open covering by open sets $\{U_i \ |\ i\in I\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets "inside" of $U$?
If the answer is no, then what are the open sets $U$ that admit such "internal" coverings?
Is the answer any different if we replace $U$ by an arbitrary closed set? | {
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Is the answer any different if we replace $U$ by an arbitrary closed set?
• Have you thought about the integers? – Steve D May 15 '17 at 21:38
• For many topological spaces you can do this. For example, any open subset of $\mathbb R^n$ is a union of open balls, which can be made proper. However, @SteveD's hint shows that you can't always do this. – Cheerful Parsnip May 15 '17 at 22:20
• If $X$ is a $T_1$ space and if $U$ is open and has more than $1$ member then $\{U$ \ $\{p\}: p\in U\}$ is an open cover of $U$ by open proper subsets of $U.$ ....If $\tau=\{X,\phi\}$ then with U=X, no proper non-empty subset of $U$ is open.... Another example is Sierpinski space $S=\{0,1\}$ with $\tau=\{S,\phi, \{0\}\}$. – DanielWainfleet May 15 '17 at 23:57
• If U is open, of course. If U is open every point is an interior point so for every point x in U there is an open neighborhood of x completely in U. Call this N_x, then $\cup_{x\in U} Nx$ is an open cover. If U is not open then of course not, If U is not open there is a point that is not an interior point and any open set containing it must contain points not in U (else it'd be an interior point). So all open covers contain points not in U. – fleablood May 16 '17 at 1:25
• @fleablood This is the argument I was going to give, but (as others pointed out to me) it only goes through if the space is at least $T_1$. – Austin Mohr May 16 '17 at 1:31
A space is called $T_1$ if, for any two points in the space, each has an open neighborhood that misses the other. This is one of many separation axioms that measure how strongly we can separate points in the space. For some mathematicians, topological spaces aren't even worth considering until they are at least $T_2$ (Hausdorff), which is even stronger. (This is to say that all but the most pathological topologies are at least $T_1$.) | {
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We can construct a cover of the type you describe if the space is at least $T_1$ and $U$ contains more than one point. For each $x \in U$, construct an open neighborhood $U_x$ as follows:
1. Choose any $y \in U$ that is distinct from $x$.
2. Appeal to the $T_1$ property to get an open set $G_x$ that contains $x$ but does not contain $y$.
3. Set $U_x = G_x \cap U$, so that $U_x$ is a proper open subset of $U$ containing $x$.
Finally, we can take $\{U_x \mid x \in U\}$ as an open cover of $U$ by proper open subsets.
• But why is it guaranteed that the "first" $y$ we chose is an element of some element of $\{ U_x \ | \ x\in U \}$ ? Moreover, why is every such $y$ eventually in some of the cover elements? Because, by construction, $y$ is not in $U_x$ – étale-cohomology May 17 '17 at 5:52
• Let $y_0$ be the "first $y$" chosen in this process. Eventually we will need to construct the set $U_{y_0}$, which must contain $y_0$. Similarly, any $y \in U$ is an element of $U_y$ and therefore covered. – Austin Mohr May 17 '17 at 8:36
• Got it. Thank you. I wonder if it's possible to prove that $T_1$ is the weakest possible hypothesis that allows existence of these covers... – étale-cohomology May 18 '17 at 22:38
• One can push my argument through with something weaker (but I don't know if it has a name): For any open set $U$ and any point $x$ of $U$, there is an open neighborhood $V$ of $x$ that is a strict subset of $U$. This property is not as strong as $T_1$, since it still allows for the existence $x$ and $y$ that cannot be separated. However, you could take the $V$ in this property as the $U_x$ in my answer and it still get your desired refinement. This is property would be weakest possible, since otherwise the only way to cover $x$ is to use all of $U$. – Austin Mohr May 19 '17 at 1:01 | {
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It depends. Consider $\mathbb{R}$ with the usual topology and $U = (0,1)$, then we can write $(0,1) = \bigcup_{n=2}^\infty\left(\frac1n,1\right)$, for instance. Since every open set $U\subset\mathbb{R}$ is the union of intervals, it means that every open subset of $\mathbb{R}$ can be written as the union of proper open subsets of itself. It is easy to see that the same argument can be applied to general normed vector spaces.
However, if a singleton $\{x\}$ is an open set of your space then there are no proper open subsets of $\{x\}$ to unite.
Show that $K = [0,1]$ cannot be covered by open sets that are contained by $K$.
If $U$ is open, then $U$ can be covered by $\{U\}$. If $U$ is an open singleton, then there is no cover by proper subsets.
When $U$ is open, then for all $x$ in $U$, there is an open base set $U_x$ with $x$ in $U_x$ subset $U$. Often the base sets can be chosen to be proper subsets. | {
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# Why is $x+e^{-x}>0$ for all $x \in \mathbb{R}$?
Denote $$f(x) = x + e^{-x}$$. Note that $$f(0) = 1$$ and $$f'(x) = 1-e^{-x}$$. That means $$\lim_{x\rightarrow -\infty} f'(x) = -\infty.$$ So if the rate of change of $$f(x)$$ keeps decreasing exponentially fast to negative infinity, why is $$\lim_{x\rightarrow -\infty} f(x) = +\infty$$?
• Addressing the title, for all $x, e^{-x}\ge1-x$ so $x+e^{-x}\ge1>0$ – J. W. Tanner Mar 21 at 4:45
• To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}\ge t+1$ for all real $t$. So $e^{-x}\ge -x+1> -x$ for all real $x$, as desired. – Minus One-Twelfth Mar 21 at 4:46
• Roughly speaking - if you are visualising $x$ going from $0$ to $-\infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards". – David Mar 21 at 4:49
• Thank you. That's what I was missing. Going backward. :( – Paichu Mar 21 at 4:50
You have got your directions mixed up.
When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.
Take
$$x < y < 0; \tag 1$$
then
$$f(y) - f(x) = \displaystyle \int_x^y f'(s)\; ds = \int_x^y (1 - e^{-s})\; ds; \tag 2$$
it is clear that, for any $$M < 0$$, there exists $$y_0 < 0$$ such that
$$s < y_0 \Longrightarrow 1 - e^{-s} < M; \tag 3$$
if we now choose
$$y < y_0, \tag 4$$
then
$$f(y) - f(x) = \displaystyle \int_x^y (1 - e^{-s})\; ds < \int_x^y M \; ds = M(y - x); \tag 5$$
we re-arrange this inequality:
$$f(x) - f(y) > -M(y - x) = M(x - y), \tag 6$$
$$f(x) > f(y) + M(x - y); \tag 7$$
we now fix $$y$$ and let $$x \to -\infty$$; then since $$M < 0$$ and $$x - y < 0$$ for $$x < y$$,
$$\displaystyle \lim_{x \to -\infty} f(y) + M(x - y) = \infty, \tag 8$$
and hence
$$\displaystyle \lim_{x \to -\infty}f(x) = \infty \tag 9$$
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and hence
$$\displaystyle \lim_{x \to -\infty}f(x) = \infty \tag 9$$
as well.
We note that (9) binds despite the fact that $$f'(x) < 0$$ for $$x < 0$$; though this derivative is negative, when $$x$$ decreases we are "walking back up the hill," as it were; as $$x$$ decreases, $$f(x)$$ increases.
Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show
$$x + e^{-x} > 0, \forall x \in \Bbb R; \tag{10}$$
for
$$x \ge 0, \tag{11}$$
$$e^{-x} > 0 \tag{12}$$
as well, hence we also have
$$x + e^{-x} > 0; \tag{13}$$
for
$$x < 0, \tag{14}$$
we may use the power series for $$e^{-x}$$:
$$e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3} + \ldots; \tag{15}$$
then
$$x + e^{-x} = 1 + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \ldots > 0, \tag{16}$$
since every term on the right is positive when $$x < 0$$. End of Note.
Note that $$f''(x) = e^{-x} >0$$ so $$f$$ is strictly convex.
Since $$f'(0) = 0$$, we see that $$f(x) \ge f(0) = 1$$ for all $$x$$.
Note that
• $$\forall x \in \mathbb{R} : x+e^{-x}>0 \Leftrightarrow \forall x \in \mathbb{R} :e^{-x}>-x \Leftrightarrow \forall x \in \mathbb{R} :\color{blue}{\boxed{e^{x}>x}}$$
The last inequality follows directly by Taylor: $$\color{blue}{e^x} = 1+x + \underbrace{\frac{e^{\xi}}{2}x^2}_{\geq 0} \color{blue}{> x}$$ | {
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# How can you express the next two consecutive odd numbers in terms of x?
I'm trying to express the next two consecutive odd numbers in terms of x. How do go about this?
Precisely, given a number $x$ I want two smallest consecutive odd numbers $m,n$ such that $x<m<n$.
• Be aware that you wrote two very different things in your question and your body. Are you trying to express the next two consecutive odd numbers in terms of x, or are you trying to express x in terms of the next two consecutive odd numbers? If $x$ is, say $5$, do you consider "the next two consecutive odd numbers" to be $5$ and $7$ or do you consider the next two consecutive odd numbers to be $7$ and $9$? – JMoravitz Oct 19 '15 at 2:32
• My error. I'm trying t express the next two consecutive odd numbers in terms of x. Edited to reflect. – user214824 Oct 19 '15 at 2:45
• That still doesn't answer the question of what exactly you mean by "next two consecutive odd numbers" is. If $5$ is your number, do you want $7$ and $9$? or do you want $5$ and $7$ (despite $5$ not being bigger than $5$). – JMoravitz Oct 19 '15 at 2:47
• @JMoravitz Yes, that is it. 7 and 9. – user214824 Oct 19 '15 at 2:48 | {
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Presumably $x$ is an integer. We have two cases: either $x$ is even or $x$ is odd.
In the case that $x$ is odd, then $x+2$ and $x+4$ will both be odd and will be the next two consecutive odd numbers.
In the case that $x$ is even, then $x+1$ and $x+3$ will both be odd and will be the next two consecutive odd numbers.
Let $x$ an integer, then the next two odd integers are $$2\times\Bigl\lfloor \frac{x+1}{2}\Bigr\rfloor + 1 \qquad\text{and}\qquad 2\times\Bigl\lfloor \frac{x+1}{2} \Bigr\rfloor + 3.$$
Indeed, if $x = 2n$ where $n\in\mathbb{Z}$, then $$2\lfloor (x+1)/2\rfloor + 1 = 2 \lfloor n + 1/2 \rfloor + 1 = 2n + 1$$ and if $x = 2n + 1$, then $$2\lfloor (x+1)/2 \rfloor + 1 = 2\lfloor n + 1 \rfloor + 1 = 2n + 3.$$
• Maybe doesn't you know the symbol $\lfloor \cdot \rfloor$. If $x$ is a real number, $\lfloor x \rfloor$ is the biggest integer less or equal to $x$. For example $\lfloor 2.7 \rfloor = 2$, $\lfloor -3.8 \rfloor = -4$ and $\lfloor x \rfloor = x$ for $x$ integer. – Éric Guirbal Oct 19 '15 at 3:01 | {
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# Is this theorem equivalent to Transfinite Recursion Theorem?
Let $$V$$ be the class of all sets, $$\operatorname{Ord}$$ be the class of all ordinals, and $$G:V\to V$$ be a class function.
Transfinite Recursion Theorem:
There exists a class function $$F:\operatorname{Ord}\to V$$ such that $$F(\alpha)=G(F\restriction \alpha)$$ for all $$\alpha\in\operatorname{Ord}$$.
While I'm able to prove that Transfinite Recursion Theorem implies the below theorem, I have tried but to no avail in proving that the theorem implies Transfinite Recursion Theorem.
I would like to ask if it is possible to prove that this theorem implies Transfinite Recursion Theorem.
Theorem:
Let $$G_1,G_2,G_3$$ be class functions from $$V$$ to $$V$$. There exists a class function $$F:\operatorname{Ord}\to V$$ such that
(1) $$F(0)=G_1(\emptyset)$$
(2) $$F(\alpha+1)=G_2(F(\alpha))$$ for all $$\alpha\in\operatorname{Ord}$$
(3) $$F(\alpha)=G_3(F\restriction\alpha)$$ for all limit $$\alpha\neq 0$$
• Do you know how to prove strong induction (on $\mathbb{N}$) from weak induction? It's the same idea. – Eric Wofsey Oct 25 '18 at 2:46
• Hi @EricWofsey! Yes, I do. I will try your suggestion. – LE Anh Dung Oct 25 '18 at 2:54
The trick is to not construct $$F$$ itself, but to construct the function $$H$$ which sends $$\alpha$$ to $$F\restriction \alpha$$. That way, at successor steps you have access to the entire history of the recursion so far and not just the last step. | {
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In detail, given $$G:V\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: $$G_1(x)=\emptyset$$ $$G_2(x)=x\cup\{(\operatorname{dom}(x),G(x))\}$$ $$G_3(x)=\bigcup\operatorname{ran}(x)$$ By the theorem, we then get a function $$H:Ord\to V$$ such that $$H(0)=G_1(\emptyset)$$, $$H(\alpha+1)=G_2(H(\alpha))$$, and $$H(\alpha)=G_3(H\restriction\alpha)$$ for $$\alpha\neq 0$$ limit. It is then easy to prove by induction that $$H(\alpha)$$ is a function with domain $$\alpha$$ for each $$\alpha$$, with $$H(\alpha)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha$$. So, defining $$F=\bigcup\operatorname{ran}(H)$$, $$F$$ is a function on $$Ord$$ with $$F\restriction\alpha=H(\alpha)$$ for each $$\alpha$$. For each $$\alpha$$, we then have $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=G_2(F\restriction\alpha)(\alpha).$$ Since $$\operatorname{dom}(F\restriction\alpha)=\alpha$$, our definition of $$G_2$$ tells us that $$F(\alpha)=G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha),$$ as desired. | {
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• I'm feeling confused. You have shown how to derive the Q's 2nd theorem from its first. But the proposer seems to want to see the 1st derived from the second, although it is obvious that if the second one holds, then, for a given $G$, we can simply let $G_1=G_2=G_3=G$. So I'm thinking that the proposer stated the Q backwards. – DanielWainfleet Oct 25 '18 at 3:49
• @DanielWainfleet: I am deriving the first theorem from the second. I start with a function $G$ as in the first theorem, then use the second theorem to get a function $H$ and then define $F$ from $H$ which verifies the conclusion of the first theorem. It does not work to simply define $G_1=G_2=G_3=G$ since $G_2$ takes as its input just $F(\alpha)$, not $F\restriction\alpha$. – Eric Wofsey Oct 25 '18 at 4:04
• Thank you so much! You make my day :) – LE Anh Dung Oct 25 '18 at 8:14
• Thank you for clearing up my confusion. – DanielWainfleet Oct 25 '18 at 21:26
• Hi! I'm now struggling with a seemingly similar problem like this one. I have tried to replicate your proof but failed at the end. If you don't mind, please have a look at math.stackexchange.com/questions/2970783/…. Thank you for your help! – LE Anh Dung Oct 27 '18 at 14:16
I fill in @Eric Wofsey's proof with detail and post it here. All credits are given to @Eric Wofsey.
Given $$G:V\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: \begin{align}&G_1(x)=\emptyset\text{ for all }x\\&G_2(x)=\begin{cases} x\cup\{(\operatorname{dom}(x),G(x))\}&\text{if }x\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}\\&G_3(x)=\begin{cases} \bigcup\operatorname{ran}(x)&\text{if }x\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}\end{align}
By The theorem, there is a class function $$H:\operatorname{Ord}\to V$$ such that $$H(0)=G_1(\emptyset)$$, $$H(\alpha+1)=G_2(H(\alpha))$$, and $$H(\alpha)=G_3(H\restriction\alpha)$$ for $$\alpha\neq 0$$ limit. | {
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First, we prove by induction that $$H(\alpha)$$ is a function with domain $$\alpha$$ for all $$\alpha\in\operatorname{Ord}$$ and with $$H(\alpha)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha$$.
• $$H(0)=G_1(0)=\emptyset$$. Then the statement is trivially true for $$\alpha=0$$.
• Assume that the statement is true for $$\alpha$$. Then $$H(\alpha+1)=G_2(H(\alpha))=$$ $$H(\alpha)\cup\{(\operatorname{dom}(H(\alpha)),G(H(\alpha)))\}=H(\alpha)\cup\{(\alpha,G(H(\alpha)))\}$$. It follows that $$\operatorname{dom}(H(\alpha+1))=\operatorname{dom}(H(\alpha))\cup \{\alpha\}=\alpha\cup \{\alpha\}=\alpha+1$$. For $$\beta=\alpha$$, $$H(\alpha+1)\restriction \beta=H(\alpha+1)\restriction \alpha=H(\alpha)=H(\beta)$$. For $$\beta<\alpha$$, $$H(\alpha+1)\restriction \beta=$$ $$H(\alpha)\restriction \beta=H(\beta)$$. Thus $$H(\alpha+1)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha+1$$.
• Assume that the statement is true for all $$\beta<\alpha$$ where $$\alpha\neq\emptyset$$ is limit ordinal. Then $$H(\alpha)=G_3(H(\alpha))=\bigcup\operatorname{ran}(H(\alpha))=\bigcup\{H(\beta)\mid \beta<\alpha\}$$. For any $$\beta_1\le\beta_2<\alpha$$: $$H(\beta_2)\restriction\beta_1=H(\beta_1)$$and thus $$H(\beta_1)\subseteq H(\beta_2)$$. Then $$H(\alpha)=\bigcup\{H(\beta)\mid \beta<\alpha\}$$ is actually a function. It follows that $$\operatorname{dom}(H(\alpha))=\bigcup_{\beta<\alpha}\operatorname{dom}(H(\beta))=\bigcup_{\beta<\alpha}\beta=\alpha$$ since $$\alpha$$ is limit ordinal. Moreover, $$H(\alpha)\restriction \beta=\{(\gamma,H(\alpha)(\gamma))\mid \gamma<\beta\}=\{(\gamma,H(\gamma+1)(\gamma))\mid \gamma<\beta\}=$$ $$\{(\gamma,H(\beta)(\gamma))\mid \gamma<\beta\}=H(\beta)$$.
As a result, $$\forall\beta<\alpha:H(\alpha)\restriction \beta=H(\beta)$$ and thus $$\forall\beta<\alpha:H(\beta)\subsetneq H(\alpha)$$. | {
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Next we define $$F=\bigcup\operatorname{ran}(H)$$. Then $$F=\bigcup\{H(\alpha)\mid \alpha\in\operatorname{Ord}\}$$ is a function with domain $$\operatorname{Ord}$$ and with $$F\restriction\alpha=\{F(\beta)\mid\beta<\alpha\}=\{H(\beta+1)(\beta)\mid\beta<\alpha\}=\{H(\alpha)(\beta)\mid\beta<\alpha\}=H(\alpha)$$ for all $$\alpha\in\operatorname{Ord}$$.
Since $$F\restriction\alpha=H(\alpha)$$ and $$\operatorname{dom}(H(\alpha))=\alpha$$, $$\operatorname{dom}(F\restriction\alpha)=\alpha$$. Then $$G_2(F\restriction\alpha)=(F\restriction\alpha)\cup \{(\alpha,G(F\restriction\alpha))\}$$ and thus $$G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha)$$.
For each $$\alpha\in\operatorname{Ord}$$, we have $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha)$$.
Update: I have found another way to define $$F$$
We define $$F$$ as follows $$F(\alpha):=H(\alpha+1)(\alpha)$$
Then $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=(H(\alpha)\cup\{(\operatorname{dom}(H(\alpha)),G(H(\alpha)))\})(\alpha)=(H(\alpha)\cup\{(\alpha,G(H(\alpha)))\})(\alpha)=G(H(\alpha))$$.
Moreover, $$H(\alpha)=\{(\beta,H(\alpha)(\beta))\mid\beta<\alpha\}=\{(\beta,H(\beta+1)(\beta))\mid\beta<\alpha\}=\{(\beta,F(\beta))\mid\beta<\alpha\}=F\restriction\alpha$$.
Thus $$F(\alpha)=G(H(\alpha))=G(F\restriction\alpha)$$. | {
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# Envelope of Projectile Trajectories
For a given launch velocity $$v$$ and launch angle $$\theta$$, the trajectory of a projectile may be described by the standard formula $$y=x\tan\theta-\frac {gx^2}{2v^2}\sec^2\theta$$
For different values of $$\theta$$ what is the envelope of the different trajectories? Is it a parabola itself?
The standard solution to this "envelope of safety" problem is to state the formula as a quadratic in $$\tan\theta$$ and set the discriminant to zero. The resulting relationship between $$x,y$$ is the envelope.
This question is posted to see if there are other approaches to the solution.
Edit 1
Thanks for the nice solutions from Jack and Blue, received so far. From the solution of the envelope it can be worked out that the envelope itself corresponds to the right half of the trajectory of a projectile launched at $$(-\frac{v^2}{g^2},0)$$ at a launch angle $$\alpha=\frac{\pi}4$$ and a launch velocity $$V=v\sqrt2$$. This means that both vertical and horizontal components of the launch velociy are equal to $$v$$. It would be interesting to see if these conclusions can be inferred from the problem itself by inspection and without first solving it. If so, then this would form another solution. | {
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• I think it is more like an ellipse. – Archis Welankar Oct 24 '15 at 10:05
• @ArchisWelankar - What makes you think that might be the case? – hypergeometric Oct 24 '15 at 10:13
• When the angle is very small like 10 then the curve becomes elongated and more like an ellipse than a parabola. – Archis Welankar Oct 24 '15 at 10:25
• @ArchisWelankar - 1. The curve or trajectory of the projectile is always a parabola (even when elongated, at small launch angles) as given by the formula. 2. An ellipse doesn't have to be elongated. 3. The question refers to the envelope of different trajectories and not any one particular trajectory. – hypergeometric Oct 24 '15 at 10:34
• "The standard solution to this "envelope of safety" problem is to state the formula as a quadratic in tanθ and set the discriminant to zero. " I don't understand this approach, can someone explain ? – Dat Sep 11 at 9:38
Yes it is. To find the envelope, we just have to find the intersections between two trajectories associated to two slightly different angles. If we solve $$x\tan\theta -\frac{gx^2}{2v^2}\sec^2(\theta) = x\tan(\theta+\varepsilon) -\frac{gx^2}{2v^2}\sec^2(\theta+\varepsilon)$$ we get $x=0$ or $$x=\frac{2v^2}{g}\cdot\frac{\tan(\theta)-\tan(\theta+\varepsilon)}{\sec^2(\theta)-\sec^2(\theta+\varepsilon)}$$ and by letting $\varepsilon\to 0$ we get $x=\frac{v^2}{g}\cot(\theta)$, from which $y=\frac{v^2}{2g}\left(2-\frac{1}{\sin^2(\theta)}\right)$.
It follows that the equation of the envelope is given by: $$y = \frac{v^2}{2g}\left(1-\left(\frac{gx}{v^2}\right)^2\right)=\frac{v^2}{2g}-\frac{g}{2 v^2}\,x^2$$ that clearly is a parabola with vertex in $\left(0,\frac{v^2}{2g}\right)$ through the points $\left(\pm\frac{v^2}{g} ,0\right)$. | {
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We may notice that the envelope and the trajectory with $\theta=\frac{\pi}{4}$ are homothetic, and the dilation ratio is just $2$. The vertices of the trajectories lie on an ellipse that is tangent to the envelope parabola, with centre at $\left(0,\frac{v^2}{4g}\right)$, a vertex in the origin and a vertex at $\left(\frac{v^2}{2g},\frac{v^2}{4g}\right)$.
• Thank you for a beautiful and creative solution! And nice graphics! A big +1! – hypergeometric Oct 24 '15 at 13:29
• Please check out comments under Edit 1 in the question and see if you have any further thoughts. – hypergeometric Oct 24 '15 at 14:21
@Jack provides a very nice and intuitive derivation of the envelope as the points of intersection of infinitely-close members of the curve family. The Wikipedia "Envelope" entry provides this less-illuminating abstraction:
The envelope of the family [of curves parameterized by $t$ is] the set of points for which $$F(t, x, y) = \frac{\partial F}{\partial t}(t,x,y) = 0 \tag{\star}$$ for some value of $t$ [...].
In $(\star)$, $F$ is the function that, when set equal to $0$, defines each curve in the family. For the question at hand, we have (with parameter $\theta$ instead of $t$) $$F(\theta,x,y) = -y + x\tan\theta -\frac{g x^2}{2v^2}\sec^2\theta \tag{1}$$ Therefore, differentiating with respect to $\theta$ gives $$\frac{\partial F}{\partial \theta}(\theta,x,y) =x\sec^2\theta -\frac{g x^2}{2v^2}\cdot 2 \sec^2\theta\tan\theta = \frac{x\sec^2\theta}{v^2} ( v^2 - g x \tan\theta) \tag{2}$$ Solving $\partial F/\partial \theta = 0$ for $\theta$ (noting that $\sec\theta$ never vanishes) gives $$\tan\theta = \frac{v^2}{g x} \qquad\text{so that}\qquad \sec^2\theta = 1 + \tan^2\theta = \frac{g^2 x^2 + v^4}{g^2 x^2}$$
Substituting into $(1)$, and setting $F=0$, we have
$$y = x\;\frac{v^2}{gx} - \frac{g x^2}{2 v^2}\;\frac{g^2 x^2 + v^4}{g^2 x^2} = \frac{v^2}{2g} - \frac{g x^2}{2v^2} = \frac{v^4-g^2x^2}{2gv^2}$$ | {
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• Thanks for working out the solution from the article. Nice find! +1. – hypergeometric Oct 24 '15 at 13:55
• Please check out comments under Edit 1 in the question and see if you have any further thoughts. – hypergeometric Oct 24 '15 at 14:21
Here's another approach. Not as elegant as Jack's though. Perhaps similar to Blue's. (Edit: Revised slightly to use substitution of $H=v^2/2g$ instead of $\lambda=g/v^2$ for more convenient reference, per solution by Narasimham)
This approach maximises the value of $y$ for any given value of $x$.
For a given value of $x$, say $x=k$, \begin{align} y&=kT-\frac {k^2}{4H}(1+T^2)\qquad\text{where T=\tan\theta, H=\frac {v^2}{2g}}\\ \frac{dy}{dT}&=k-\frac {k^2}{4H}\cdot 2T=0\qquad\text{when T=\frac {2H}k}\\ \text{At T=\frac {2H}k}:\qquad\qquad y&=k\cdot\frac{2H}k-\frac{k^2} {4H}\left(1+\frac {4H^2}{k^2}\right) \color{lightgrey}{=2H-\frac {k^2}{4H}-H}\\ &=H-\frac {k^2}{4H} \end{align}
Hence the equation of the envelope is $$y=\frac {v^2}{2g}-\frac {gx^2}{2v^2}\quad\blacksquare$$
• This nicely exploits the fact that, in this particular case, we know that the envelope sits on top of the family of curves, and therefore consists of the points that maximize $y$ for a given $x$, over all $t$. (Of course don't always have such knowledge, but no matter.) Re-writing in terms of parameter $T$ is also a helpful simplification. And, to be explicit: the connection with the process in my answer is that the criterion for maximizing $y$ ---that is, setting $dy/dT = 0$--- corresponds directly to $\partial F/\partial \theta = 0$, because $y$ drops out of the latter equation. – Blue Oct 24 '15 at 21:38
• Thanks for your kind comment! – hypergeometric Oct 25 '15 at 5:57
I shall outline the method how we get the parabola. Usual notation
$$\ddot y = -g, \dot y = - g t + v \sin \theta , y =- g t^2/2 + v t \sin \theta\, +0 \tag {1}$$ $$\ddot x = 0, \dot x = v \cos \theta = const , x = v t \cos \theta +0 \tag{2}$$ | {
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Eliminating time $t$ between (1),(2) you got this parabola equation already.
Let $\tan \alpha = T$; Parabola equation in other words
$$y = x T - g/2 * ( x/ v \cos\theta)^2 = x T - ( g x^2/2 v^2) ( 1+T^2) \tag{3}$$
Differentiate partially with respect to $T$ and simplify, $T = v^2 / gx \tag{4}$
Eliminate $T$ between (3) and (4)
$$y = v^2/2g - g x^2 /(2 v^2) = H - x^2/(4H) \tag{5},$$
same as what was obtained before by Jack and Blue.
If you denote height reached by projectile on vertical firing $H = v^2/(2g) \tag{6}$ you would notice that the envelope is profiled exactly as a parabolic mirror with focus at gun delivery point, focal length is exactly H. Vertical force of gravity is acting like light :)..
The above procedure method is indicated by Blue in Wiki, is referred to as C-discriminant method to obtain envelopes and singular solutions.
Like what you said in your edit and I about mirror, they are ploys for remembering curves using similarities..
For the image I took values of $g=9.8 m/s^2 , v = 2 m/s$
The partial differentiation way and elimination is the right way to look at, perhaps not as what you said (standard solution.. way).
EDIT2:
The answer to your second question, i.e., to determine if it is going to be a parabola envelope without going through all of analysis... I can only reply with extended C-discriminant, strengthening the same result by another path.
$p$ discriminant method is also relevant, but I defer it, but best is to refer to differential calculus books of authors e.g., A.R. Forsythe.
I shall expand on the C-discriminant, wherein a two parameter system of equations variation of any of the two parameters leads to the same parabola envelope. This is in reply to your question , Why do you also plot partial concentric circles?
Well, they are circles alright, but not concentric circles. They expand and come down slowly with time. What you can see as plotted are the peripheries traced out descending with time . | {
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But first quickly watch some fireworks to see what I am discussing about:
It is common experience to see fire ball bright splinters expanding to bigger circles as the entire cluster comes down slowly with time. The periphery is a portion of a circle whose radius increases. Center of circle is always descending by gravity.
The two parameters are $\theta, t$ angle of elevation at first burst or fire, and time $t$.
By C-discriminant method the parabola envelope is the eliminant of either $\theta$ variable or
$$F(x,y,\theta) =0 ,\, F_{\theta } (x,y,\theta) =0 \tag{6}$$
or $t$ time variable.
$$F(x,y, t) =0 ,\, F_{t } (x,y, t) =0 \tag{7}$$
The first one is already discussed, the second one is expanding/descending fireworks circles as already stated.
In the latter case working is:
$$x = v t \cos \theta , y = v t \sin \theta - g t^2/2 \tag{8}$$
$$(\frac{x}{vt})^2 + (\frac{y+ gt^2/2}{vt}) ^2 = 1 \tag{9}$$
$$x^2 + ( y + g t^2/2)^2 = v^2t^2 \tag{10}$$
which is a Circle.
To find its envelope, as before partially differentiate with respect to time $t$ and cancel $2t$ on either side of equation , bring $v^2/g$ to right side :
$$y + gt^2/2 = v^2/g \tag {11}$$
$$x^2 + (v^2/g)^2 = 2 v^2/g * ( v^2/g-y) \tag {12}$$
$$x^2 + ( 2 H)^2 += 4 H ( v^2/g -y )\tag{13}$$
$$x^2 = 4 H ( H-y) \tag{14}$$
which is the same parabola envelope obtained earlier with $\theta$ as parameter. End points $( x=0, y=H ; x= 2 H, y= 0 )$
Although circle end traces are visible in a fireworks display it needs imagination as before with variable gun barrel angle to see that each circle is tangent to a fixed envelope. Hope you enjoyed it. | {
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• Thanks for your solution and insight on the parabolic mirror! (+1). Nice graphics too! In addition to different parabolas, why do you also plot partial concentric circles? – hypergeometric Oct 25 '15 at 5:56
• Also, is there an argument which would directly conclude that the envelope is a parabola with focus at launch point and tip at the peak of the vertical launch, without first going through the analysis? – hypergeometric Oct 25 '15 at 6:03
• The question in my comment above has been answered here. – hypergeometric Sep 30 '18 at 17:14 | {
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# Each of 60 cars is parked in one of three empty parking lots
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Each of 60 cars is parked in one of three empty parking lots [#permalink]
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Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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30 Sep 2019, 20:43
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EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
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Let the number of cars parked in largest lot = x
--> Middle lot = x - 8 & Smallest lot = x - 16
So, x + (x - 8) + (x - 16) = 60
--> 3x - 24 = 60
--> 3x = 84
--> x = 28 | {
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IMO Option D
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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30 Sep 2019, 21:25
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EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
let a,b,c be lot size in ascending order
a+b+c=60
c=b+8
c=a+16
Substituting
c-16+c-8+c=60
c = 28
D is correct
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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02 Oct 2019, 16:42
1
Let the number of cars parked in largest lot = L
Middle lot = M & Smallest lot = S
But M=L-8 & S=L-16
So, L + (L - 8) + (L - 16) = 60
3L - 24 = 60
3L = 84
L = 28
IMO Option D
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS... | {
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Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
Given: the largest lot holds 8 more cars than the middle lot.
So, x - 8 = number of cars in the MIDDLE lot
Given: the largest lot holds 16 more cars than the smallest lot.
So, x - 16 = number of cars in the SMALLEST lot
Given: Each of 60 cars is parked in one of three empty parking lots.
We can write: (# of cars in LARGEST lot) + (# of cars in MIDDLE lot) + (# of cars in SMALLEST lot) = 60
Substitute values: (x) + (x - 8) + (x - 16) = 60
Simplify: 3x - 24 = 60
Add 24 to both sides: 3x = 84
Divide both sides by 3: x = 28 | {
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Cheers,
Brent
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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10 Oct 2019, 22:39
1
The 3 slots A + B + C total to 60 cars
The highest slot C has 8 more than B and 16 more than A
Let us look at the second-largest answer choice which says C=28
As per question A=12 (16 less than C), B=20(8 less than C) ...This gives a total of 60. Hence Correct answer is D
EMPOWERgmatRichC: This is the correct way to do TTA ...right
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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11 Oct 2019, 11:30
Hi saukrit,
YES - that's a perfect use of TEST THE ANSWERS. In these sorts of situations, that Tactic (and some basic Arithmetic) is almost always faster and easier that a typical Algebraic approach - so it's something to keep in mind (especially if you have a pacing problem in the Quant section).
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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12 Oct 2019, 00:09
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5 | {
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### Show Tags
12 Oct 2019, 00:09
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
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To be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct
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$$L = M + 8 = S + 16$$
Again , L + M + S = 60
Now, Plug in the options and check
(A) Can be straightaway rejected as value of L must be > 16 , Since in this case if L = 12 = S + 16 , Or, S = - 4 ( Not Possible )
(B) If L = 20 , M = 12 & S = 4 , L + M + S = 36
(C) If L = 22 , M = 14 & S = 6 , L + M + S = 42
(D) If L = 28 , M = 20 & S = 12 , L + M + S = 60
(E) If L = 30 , M = 22 & S = 14 , L + M + S = 66
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Each of 60 cars is parked in one of three empty parking lots [#permalink]
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12 Oct 2019, 01:42
As the middle lot has 8 cars less than the larger it would be X-8
Smaller Lot is having 16 cars less than the larger that means X-16
As all the lots together having 60 it would be X+(X-8)+(X-16)=60
On solving this it becomes X= 28 Cars
Hence Option D would be apt.
Each of 60 cars is parked in one of three empty parking lots [#permalink] 12 Oct 2019, 01:42
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# Fibonacci sequence: Given $n$ and $\mathrm{Fib}(n)$, is it possible to calculate $\mathrm{Fib}(n-1)$?
Given $n$ and $\newcommand{\Fib}{\mathrm{Fib}} \Fib(n)$, is it possible to calculate the previous number in the Fibonacci sequence - $\Fib(n-1)$ using integer math in constant time? In other words, I believe I'm looking for a closed form formula.
For example: Given $n=10$ and $\Fib(10)=55$, I'd like to determine that $\Fib(9)=34$ without using $\Fib(7) + \Fib(8)$.
• Even output of $F_{n-1}$ cannot be done in constant time. (The length of binary representation of $F_{n-1}$ is linear in $n$.) But based on the wording in original post I would guess that you mean using constantly many operations with integers. Aug 20 '14 at 8:14
• @Martin Sleziak - you are correct that I meant "constantly many operations with integers" Aug 20 '14 at 13:23
By Binet's Formula, we know that $$F_n = \dfrac{1}{\sqrt{5}}\left(\phi^n-(-\phi)^{-n}\right) \approx \dfrac{1}{\sqrt{5}}\phi^n$$ where $$\phi = \dfrac{1+\sqrt{5}}{2}$$.
Using this formula, we can get that $$F_{n-1}$$ is the nearest integer to $$\dfrac{1}{\phi} F_n$$ for large $$n$$
More rigorously, we have $$F_{n-1} - \dfrac{1}{\phi} F_n = \dfrac{1}{\sqrt{5}}\left(\phi^{n-1}-(-\phi)^{-(n-1)}\right) - \dfrac{1}{\phi\sqrt{5}}\left(\phi^n-(-\phi)^{-n}\right)$$ $$= -\dfrac{(-\phi)^{-(n-1)}}{\sqrt{5}} - \dfrac{(-\phi)^{-(n+1)}}{\sqrt{5}}$$ $$= \dfrac{(-\phi)^{-n}}{\sqrt{5}}\left(\phi + \phi^{-1}\right) = (-\phi)^{-n}$$.
Thus, for $$n \ge 2$$, we have $$\left|F_{n-1} - \dfrac{1}{\phi} F_n \right| = \phi^{-n} < \frac{1}{2}$$, and so, $$F_{n-1}$$ is the nearest integer to $$\dfrac{1}{\phi}F_n$$.
Dividing $$F_n$$ by $$\phi$$ and rounding the result to the nearest integer shouldn't be too computational. | {
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• Here is a table that shows the pattern:$$\begin{array} {c|c|c} n & a_n & a_n / a_{n-1} \\ \hline 0 & 0 & \\ \hline 1 & 1 & \\ \hline 2 & 1 & 1.0000 \\ \hline 3 & 2 & 2.0000 \\ \hline 4 & 3 & 1.5000 \\ \hline 5 & 5 & 1.6667 \\ \hline 6 & 8 & 1.6000 \\ \hline 7 & 13 & 1.6250 \\ \hline 8 & 21 & 1.6154 \\ \hline 9 & 34 & 1.6190 \\ \hline 10 & 55 & 1.6176 \\ \hline 11 & 89 & 1.6182 \\ \hline 12 & 144 & 1.6180 \\ \hline 13 & 233 & 1.6181 \\ \hline 14 & 377 & 1.6180 \\ \hline \end{array}$$ Aug 20 '14 at 2:25
• @DanielV - agreed, this will work but as marty cohen points out in his anti-answer below, for large values of n simply rounding to the nearest integer will not necessarily yield the correct answer. Aug 20 '14 at 13:34
This is a dissenting note on some of these answers. I am purposly making this an answer, though it really is an anti-answer.
The problem with any answer that uses $\phi$ is that, as $n$ gets larger, $\phi$ has to be computed with increasingly great precision. This will not take constant time, either for the computation of $\phi$ or the multiplication by $\phi$.
Hint $\rm\ n\:$ is a Fibonacci number iff the interval $\rm\ [\phi\ n - 1/n,\ \phi\ n + 1/n]\$ contains a positive integer (the next Fibonacci number for $\rm\ n > 1)$. This follows from basic properties of continued fractions.
For example, $\rm\ 2 \to [2.7,\ 3.7],\ \ 3\to [4.5,\ 5.2],\ \ 5 \to [7.9,\ 8.3]\ \ldots$
For a proof see e.g. T. Komatsu: The interval associated with a fibonacci number. $\$ | {
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# Thread: Geometric series problem .
1. ## Geometric series problem .
A house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?
My attempt :
It forms a sequence as follows :
$\displaystyle 50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$
The 15th term , $\displaystyle T_15=50000(1.09)^14=167086.35$
which is the amount he has to pay by the end of 15 years .
Thus , every month he has to pay $\displaystyle \frac{167086.35}{15\times12}=928.26$
My answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..
2. ## Repayments
Originally Posted by mathaddict
A house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?
My attempt :
It forms a sequence as follows :
$\displaystyle 50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$
The 15th term , $\displaystyle T_15=50000(1.09)^14=167086.35$
which is the amount he has to pay by the end of 15 years .
Thus , every month he has to pay $\displaystyle \frac{167086.35}{15\times12}=928.26$ | {
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Thus , every month he has to pay $\displaystyle \frac{167086.35}{15\times12}=928.26$
My answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..
Thanks for showing us your working here. What you're forgetting is that
each year he pays not only the interest owing that year, but he also pays back some of the amount borrowed.
Are you allowed to use a spreadsheet to work out this answer? It's the easiest way.
If not, you'll have to let the monthly payment be x, and then say that, at the start of each new year, the amount owing
= amount owed at the start of the last year + interest payable during that year - 12x
Then work out how much interest he'll have to pay in the coming year to cover this new amount owing. Then repeat for the following year, and so on, up to the start of year 16. Then say that at the start of year 16 he owes nothing.
This is an Amortization problem and requires a special formula.
I've seen this type of problem posted many times before.
It seem that the students are never given the formula,
. . nor are they prepared for the (very) long derivation.
Could it be that all those teachers are ignorant of the difficulties
. . involved with time-payment problems?
I'm beginning to suspect that this is so . . . | {
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A house buyer borrows $50,000 from a bank to buy a house. The rate of interest 9% per annum. The buyer is required to repay his loan in monthly installments for 15 years. Assuming that the rate of interest is fixed for the entire duration of the loan, find the amount per month he has to repay the bank? The formula is: .$\displaystyle A \;=\;P\,\frac{i(1+i)^n}{(1+i)^n-1}$. . where: .$\displaystyle \begin{Bmatrix}P &=& \text{principal} \\ i &=& \text{periodic interst rate} \\ n &=& \text{number of periods} \\ A &=& \text{periodic payment} \end{Bmatrix}$We have: .$\displaystyle \begin{array}{ccc}P \:=\:50,\!000 \\
i \:=\:\frac{9\%}{12} \:=\:0.0075 \\ n \:=\:15\!\cdot\!12 \:=\:180 \end{array}$Hence: .$\displaystyle A \;=\;50,\!000\,\frac{0.0075(1.0075)^{180}}{1.0075^ {180}-1} \;=\;507.1332921$Therefore, the monthly payment is: .$\displaystyle \boxed{\$507.13}$
4. ## Re :
Very big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?
5. Originally Posted by mathaddict
Very big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?
Check this and especially this out. It is one way of deriving it. There are or course other ways.
Go through page 70 up to page 99.
Or you can just order this book if get tired of reading it online.
6. ## Repayments
In fact, the interest is calculated at the start of each year for the whole year, as if the whole amount is owing for the whole year. So you have to use the values
P = 50000
i = 0.09
n = 15
to get the total payment each year, and then divide by 12, to get a monthly payment of 516.91.
Incidentally, I don't think it's really that hard to work out the formula for yourself. After
$\displaystyle n$ years you'll find that the amount owing is
$\displaystyle P(1+i)^n - [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x$, where $\displaystyle x$ is the annual payment | {
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Inside the square brackets, it's a GP whose sum is
$\displaystyle \frac{(1+i)^n - 1}{i}$
Set this sum equal to zero and solve for
$\displaystyle x$ to obtain the formula.
7. Thanks all but i am having some problems with the series .
[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math
$\displaystyle T_1=x$
$\displaystyle T_2=xi+x=x(1+i)$
$\displaystyle T_3=[x(1+i)]\cdot{i}+x$ but this is not equal to $\displaystyle x(1+i)^2$
Why is it so ??
8. ## GP
Originally Posted by mathaddict
Thanks all but i am having some problems with the series .
[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math
$\displaystyle T_1=x$
$\displaystyle T_2=xi+x=x(1+i)$
$\displaystyle T_3=[x(1+i)]\cdot{i}+x$ but this is not equal to $\displaystyle x(1+i)^2$
Why is it so ??
Set your working out like this:
At the end of year 1, the amount owed is the original sum borrowed + the interest for the year - the total payments during the year
$\displaystyle = P + Pi - x$
$\displaystyle = P(1 + i) - x$
This, then is the amount owing at the start of year 2: the 'new $\displaystyle P$' if you like. So at the end of year 2, we'll replace $\displaystyle P$ by $\displaystyle P(1+i) - x$, to get that the amount owing is:
$\displaystyle [P(1 +i) - x](1+i) - x$
$\displaystyle = P(1+i)^2 - x[1 + (1 + i)]$
Continue like this, multiplying by $\displaystyle (1+i)$ and subtracting $\displaystyle x$ each time. So at the end of the year 3, the amount owing is:
$\displaystyle P(1+i)^3 - x[1 + (1 + i) + (1 +i)^2]$
And, in a similar way, at the end of year $\displaystyle n$, it is:
$\displaystyle P(1+i)^n - x[1 + (1 + i) + ...+ (1 +i)^{n-1}]$
So, inside the square brackets, [...], we have a GP whose first term is $\displaystyle 1$, common ratio $\displaystyle (1+i)$ and number of terms = $\displaystyle n$, and whose sum is $\displaystyle \frac{(1+i)^n - 1}{(1+i) - 1}= \frac{(1+i)^n - 1}{i}$.
Now if the whole sum is paid of after $\displaystyle n$ years, this total is zero. I.e. | {
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Now if the whole sum is paid of after $\displaystyle n$ years, this total is zero. I.e.
$\displaystyle P(1+i)^n - x\frac{(1+i)^n - 1}{i} = 0$
Hence $\displaystyle x = \frac{iP(1+i)^n}{(1+i)^n - 1}$
OK now? | {
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# Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X
Let $$X$$ be a random variable with a continuous and strictly increasing c.d.f. $$F$$ (so that the quantile function $$F^{−1}$$ is well-defined). Define a new random variable $$Y$$ by $$Y = F(X)$$. Show that $$Y$$ follows a uniform distribution on the interval $$[0, 1]$$.
My initial thought is that $$Y$$ is distributed on the interval $$[0,1]$$ because this is the range of $$F$$. But how do you show that it is uniform?
• This is not true in cases where there's a discrete component. For example, suppose $X=\left.\begin{cases} 1/2 & \text{with probability }1/2, \\ W & \text{with probability }1/2,\end{cases}\right\}$ and $W$ is uniformly distributed on $[0,1]$, and that the choice between whether $X=1/2$ or not is independent of $W$. Then the cdf of $X$ has no values between $1/4$ and $3/4$, so it cannot be uniformly distributed on $[0,1]$. It is, however, true of continuous distributions. Jul 15, 2014 at 21:41
• see the text of the question. X is continuous! Jul 15, 2014 at 21:44
• By the way, it is not necessary that $F$ is a strictly increasing CDF, continuity is sufficient. Just define the quantile function the usual way as a generalized inverse via $F^-(y)=inf\{x\in\mathbb{R}: F(x)\geq y\}$. See the proof of Proposition 3.1 in Embrechts, P., Hofert, M.: A note on generalized inverses. Mathematical Methods of Operations Research 77(3), 423-432 link for a very careful and detailed explanation. Jul 16, 2014 at 14:38
• Thanks @binkyhorse - that reference is really good. Apr 5, 2015 at 22:33
• @s0ulr3aper07 By Proposition 3.1 in the paper I linked above, yes. Prop. 3.1: Let F be a distribution function and X ~ F. (a) If F is continuous, F(X)∼U[0,1]. The paper includes a detailed proof. Mar 27, 2019 at 20:10
Let $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \in [0,1]$ we have:
$F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$. | {
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$F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$.
What distribution has this CDF?
• Are all CDF's of continuous densities invertible? Aug 22, 2017 at 7:43
• @tintinthong: not always completely, but enough. If you define $G(y)= \inf\{x:F_X(x) \ge y\}$ then $F_X(G(y))=y$ when $y \in (0,1)$ Aug 22, 2017 at 22:12
• Strictly increasing and continuous CDF is needed. Feb 20, 2019 at 15:36
$$Prob(Y\leq x)=P(F(X)\leq x)=P(X\leq F^{-1}(x))=x \\$$ The last equality is from the definition of the quantile function.
Let $y=g(x)$ be a mapping of the random variable $x$ distributed according to $f(x)$. In the mapping $y=g(x)$ you preserve the condition of probability density (namely you counts the same number of events in the respective bins)
$$h(y)dy=f(x)dx$$
where h(y) is the probability distribution of $y$
if $h(y)=1$ (uniform distribution) we have
$$dy=g'(x)dx=f(x)dx$$
This means that $$g(x)=\int f(x)dx$$
namely the function $g(x)$ that maps the random variable $x$ distributed according $f(x)$, into a random variable $y$ distributed uniformly is his own cumulative distribution function $\int f(x) dx$.
Here is an approach that does not use the quantile function whatsoever - the only property used is that independent copies of $$X$$ have zero probability of being equal. (The main ingredient in my argument is conditional expectation.) | {
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Consider the cumulative distribution function of $$X$$, namely $$F(t)=\mathbb P(X\leq t).$$ Your random variable - which I will suggestively call $$U$$ instead of $$Y$$ - can be described by starting with two independent and identically distributed random variables $$X,Z$$ and considering the conditional probability $$U=\mathbb P(X\leq Z\mid Z).$$ Then, for all integers $$n\geq 1$$, we can represent $$U^n$$ as follows. Let $$X_1,X_2,\ldots,X_n,Z$$ be independent and identically distributed. By independence, $$\mathbb P\bigl(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z\bigm\vert Z\bigr)=U^n,$$ and thus by the tower property $$\mathbb EU^n=\mathbb P(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z)=\mathbb P\bigl(Z=\max(X_1,X_2,\ldots,X_n,Z)\bigr).$$ Since $$X_1,\ldots,X_n,Z$$ are iid, each of them is equally likely to be the maximum and therefore $$\mathbb EU^n=\frac{1}{n+1}.$$ Thus $$U$$ has the same moments as a uniformly distributed random variable on $$[0,1]$$. Since $$U$$ is supported in $$[0,1]$$ as well, it follows (by the uniqueness of the Hausdorff moment problem) that $$U$$ is uniformly distributed, as desired.
Let $$y\in(0,1)$$. Since $$F$$ is continuous, there exists $$x\in\mathbb{R}$$ s.t. $$F(x)=y$$. Thus, $$\mathsf{P}(Y\le y)=\mathsf{P}(F(X)\le F(x))=F(x)=y,$$ i.e., $$Y\sim\text{U}[0,1]$$. In order to see the first equality we don't need continuity. Specifically, since any cdf is right-continuous, \begin{align} \{F(X)\le F(x)\}&=\{\{F(X)\le F(x)\}\cap\{X\le x\}\}\cup \{\{F(X)\le F(x)\}\cap\{X>x\}\} \\ &=\{X\le x\}\cup \{\{F(X)=F(x)\}\cap\{X>x\}\}, \end{align} and $$\mathsf{P}(\{F(X)=F(x)\}\cap\{X>x\})=0$$. | {
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For a proof of this problem when $$F_X(x)$$ is strictly increasing, refer to JimmyK4542's answer. Let's assume $$F_X(x)$$ is just non-decreasing (there are intervals such as $$[a,b]$$ where $$F_X(x') = c$$ for $$x'\in[a,b]$$). We define $$G(y)$$ similar to what Henry's comment suggests: $$G(y)=\inf\{x:F_X(x)\gt y\}$$ Now substituting this expression in what Jimmy has written will give us: $$F_Y(y) = \Pr[Y \le y] = \Pr[F_X(X) \le y] = \Pr[X \le G(y)] = F_X(G(y))= y \label{eq:I}\tag{I}$$
We need to show that:
1. $$F_X(x)\le y \rightarrow x \le G(y)$$
2. $$F_X(G(y))=y$$
The second argument is easier to prove. We have the following expression almost according to definitions: $$F_X(G(y))= F_X(\inf\{x:F_X(x)\gt y\})= y$$ Now for the first argument, we can still use what $$G(y)=\inf\{\cdots\}$$ implies; if $$F_X(x)\le y$$, then $$x\le \inf\{x:F_X(x)\gt y\};$$ hence $$x\le G(y)$$.
With the two arguments proved and a substitution in \ref{eq:I}, we have proved the main argument.
• I personally believe this problem is a severe case for abuse of notation, and a bad professor's problem. Feb 19 at 1:46 | {
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+0
# what is the probability of winning a lottery which requires selection of 5 numbers from1 to 40
0
895
7
what is the probability of winning a lottery which requires selection of 5 numbers from1 to 40
May 7, 2014
#2
+17
+8
Calculate the numerator: There is only one winning combination. So the numerator of the desired probability is 1. Calculate the denominator: There are "40 numbers, choose 5", or 40C5 possible combinations. So the denominator of the desired probability is 40C5. So the probability is 1/40C5=1/658008
May 7, 2014
#1
+1006
0
For this kind of probability problem, assuming the number is removed from play once chosen, the probability increases by one each time.
$$\left({\frac{{\mathtt{1}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right)$$
The bottom five numbers multiply to 78960960, which means that the odds are 1/78960960
May 7, 2014
#2
+17
+8
Calculate the numerator: There is only one winning combination. So the numerator of the desired probability is 1. Calculate the denominator: There are "40 numbers, choose 5", or 40C5 possible combinations. So the denominator of the desired probability is 40C5. So the probability is 1/40C5=1/658008
FoxTears May 7, 2014
#3
+17
0
Mine was assuming the numbers can be reused
May 7, 2014
#4
+115418
+5
The number of 5 digit combinations (order doesn't matter) from 40 is
40C5
$${\left({\frac{{\mathtt{40}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{40}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{658\,008}}$$
only one of those wins so the chance is $$\frac{1}{658008}$$
Just like FoxTears said!
Thanks Foxtears.
Foxtears, I don't think that yours was assuming that the numbers could be reused. | {
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Thanks Foxtears.
Foxtears, I don't think that yours was assuming that the numbers could be reused.
It was just assuming that the order they were chosen in was of no consequence.
GoldenLeaf I believe yours would have been correct if the numbers had to be chosen in a specific order.
May 7, 2014
#5
+32772
+5
GoldenLeaf's approach was almost right, but he should have had 5, 4, 3, 2 and 1 in the numerators.
There are 5 possible choices for the first number, so probability of this is 5/40. For each of these
there are 4 possible choices for the 2nd number, so probability of both 1st and 2nd is (5/40)*(4/39). For each of these ... etc. to get the overall probability as
$${\mathtt{p}} = \left({\frac{{\mathtt{5}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right) \Rightarrow {\mathtt{p}} = {\mathtt{0.000\: \!001\: \!519\: \!738\: \!361\: \!8}}$$
or p≈1.52*10-6
1/p = 658008
May 7, 2014
#6
+115418
0
Yes of course - that did not occur to me. Thanks Alan.
May 7, 2014
#7
0
Calling Bertie, CPhill and Rom (Troll alert! General quarters!)
Along with Melody and Alan, we need input from Bertie, CPhill, and Rom, too.
With the top brains represented here that Troll might comeback and bite one or more.
I know he’ll bite FoxTears for saying, “Mine was assuming the numbers can be reused” (your formula (40C5) contradicts this).
The Troll will follow-up, by asking what is the probability of (something) of (five, six or seven) mathematicians getting a probability question correct.
Then he’ll say about the same as walking, blindfolded, through a large buffalo heard and stepping in four piles with both feet.
What’s annoying: he’s usually right. | {
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What’s annoying: he’s usually right.
CPhill might not be able to make it. He might be trying to get the “Roman Zero” from inside Sisyphus’s boulder.
Maybe he could take a break and visit. He’s not been around for a day or two. I hope he wasn’t squashed by the boulder.
No matter: cartoon physics seem to apply in that realm. If he does visit, maybe one of those angels will help him get the “Roman Zero.”
By: Troll Detector
May 7, 2014 | {
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Sigma, Σ, is the standard notation for writing long sums. 3n 3 + 6n 2 + 6000 = Θ(n 3). ” - Sincerely, Dr. Commented: Roger Stafford on 22 Oct 2014 Accepted Answer: Star Strider. It's based on the upper case Greek letter S, which indicates a sum. Some basic ideas can be applied to creating process maps that make them easier to understand and use. Clears 1, 5, and 12, and if you remember the old notation the cardinality of Z is 3 because there are three elements in z. See Figure 3. So for this series, there is 7 terms The seven terms are: when n=0 -2x. 2040 - 3 sig fig 2. Okay, welcome back everyone. Most of the following problems are average. "My TI 83+ graphing calculator arrived faster than I expected. Although it can appear scary if you've never seen it before, it's actually not very difficult. Sigma Notation. This is the currently selected item. Define Sigma notation. In chemistry, a molecular orbital (MO) is a mathematical function describing the wave-like behavior of an electron in a molecule. Free online converters. APPENDIX E SIGMA NOTATION A35 Theorem If is any constant (that is, it does not depend on ), then (a) (b) (c) Proof To see why these rules are true, all we have to do is write both sides in expanded form. The variable n is called the index of summation. When WeBWorK gives a typeset version of your answer it only uses parentheses so for example it expresses your input of 2 [3 (4+5)+6] as 2 (3 (4+5)+6) but you can use whatever you want. Multiplication by a whole number can be interpreted as successive addition. Example ∑ = = + + + +. 00, but has features that will be used throughout algebra, trigonometry, and statistics. To see why Rule 1 is true, let's start with the left hand side of this equation, n i=1 cx i. 5 Sigma Notation And The Nth Term Ppt Video Online Download Index Of Wp Content Uploads 2017 02 Summation Notation Problems & Solutions Sigma Notation Sigma Worksheet 7 Probability And Sigma Notation Solutions(1) Maths Sigma Notation Guided | {
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Notation Sigma Worksheet 7 Probability And Sigma Notation Solutions(1) Maths Sigma Notation Guided Notesmath With Mrs Holst Tpt Document 20. Rules of Scientific Notation. These rules can be converted and applied to many log management or SIEM systems and can even be used with grep on the command line. If the series is multiplied by a. This involves the Greek letter sigma, Σ. The "a i " in the above sigma notation is saying that you sum all of the values of "a". In this unit we look at ways of using sigma notation, and establish some useful rules. The Organic Chemistry Tutor 270,467 views. With sequences, sometimes we like to start at n = 0 and sometimes we like to start at n = 1. We often use Sigma Notation for infinite series. Properties. For example, the Schrodinger equation, which has to do with dynamics in quantum systems and predates quantum computation by decades, is written using bra-ket notation. The Example shows (at least for the special case where one random variable takes only. The Sigma symbol can be used all by itself to represent a generic sum… the general idea of a. When you're calculating the big O. Use algebra rules of sums, along with the appropriate sum formulae, to evaluate the following: (a) X12 i=1 (i2 33i) (b) X10 i=1 (i2 2 1) (c) X8 i=1 (2 i ) 6. 2 Summation notation Summation notation. Control Chart with all Control Limits Visible QI Macros offers the following option to turn on and off the 1-2 sigma lines: After you have run a chart, click on the QI Macros Chart menu and select Show/Hide Sigma Lines. Go To Problems & Solutions Return To Top Of Page. Use sigma (summation) notation to calculate sums and powers of integers. Sigma questions are not rehashing of questions from a text book, each question is unique and has a slight twist. 3 – A service desk process map example. Sigma notation provides a way to compactly and precisely express any sum, that is, a sequence of things that are all to be added together. Today we're going to make it a | {
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sum, that is, a sequence of things that are all to be added together. Today we're going to make it a little bit more complicated, and we're going to go over some rules, For manipulating, Slash simplifying, Or making for complicated, if you like, sigma notation. Sigma Notation - Introduction to Summation 9:14. Write the expression 1 + 1 4 + 1 7 + 1 10 + + 1 3n+1 in P notation. These rules that make sense in simpler notation, such as or. Free algebra and math word problems. A business process model and notation diagram, or BPMN diagram for short, is used to build easy-to-read business process model flowcharts, which can be shared across organizations and industries. The concept of sigma notation means to sum up all terms and uses three parts to form math statements, like ∑ i a i. org are unblocked. In this unit we look at ways of using sigma notation, and establish some useful rules. A typical element of the sequence which is being summed appears to the right of the summation sign. Sigma Notation and Riemann Sums Sigma Notation: Notation and Interpretation of 12 3 14 1 n k nn k aaaaa a a (capital Greek sigma, corresponds to the letter S) indicates that we are to sum numbers of the form indicated by the general term. For example: 1. The Riemann sums and sigma notation exercise appears under the Integral calculus Math Mission. Binomial theorem refresher. Note how the subscripts in the FORTRAN example below exactly match the tensor notation for $$C_{ij} = A_{ik} B_{kj}$$. Posted 2/14/10 8:18 AM, 19 messages. Example of a Matrix. The sum of a series can be written in sigma notation. ) (Factor out the number 6 in the second summation. The example below shows three separate sets of control limits: Don't forget to rerun stability. ” D = { x: x is a river in a state} We should describe a certain property which all the elements. In this unit we look at ways of using sigma notation, and establish some useful rules. The resulting display string will be a string which has already | {
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and establish some useful rules. The resulting display string will be a string which has already been mapped according to the IDNA2003 rules. If you're seeing this message, it means we're having trouble loading external resources on our website. Contents 1 Algebra 8 1. As well as providing shorthand for mathematical ideas, this notation can aid students’ understanding of mathematics. Sigma notation. If , the series does not converge (it is a divergent series). If you want to use a different calculator, the TI-89 has an actual sigma (uses the greek sigma) that you enter the expression, variable, start, and stop values. Theorem: Sigma Notation Properties Suppose and are functions of , is any integer, and is any real number. It's based on the upper case Greek letter S, which indicates a sum. Question 486636: Expand and simplify. This calculus video tutorial provides a basic introduction into summation formulas and sigma notation. when n=2 6-2x. Summation rules Jochumzen. If x ≤ μ, then the pdf is undefined. Multiplication by a whole number can be interpreted as successive addition. In mathematical formula it is indeed the addition of many number or variable which represent to give concise expression for sum of the variable as Sigma or Summation. In Notes x4. Looking for Pi notation? Find out information about Pi notation. A series that converges has a finite limit, that is a number that is approached. The variable k is called the index of the sum. Okay, now that we've seen that fancy sigma notation let's work a really simple small example of numbers and then generalize. Geometric Series 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. 106L Labs: Sums and Sigma ( ) Notation Sequences, Sums, and Sigma ( ) Notation Sequences De nition A sequence is an ordered set of numbers de ned by some rule. Similarly, we can use sigma notation starting at different values of n to write the same series. Question 1 Question 2 Question 3 Question | {
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at different values of n to write the same series. Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10. This is true for all tensor notation operations, not just this matrix dot product. 1, we de ne the integral R b a f(x)dx as a limit of approximations. Use Fault Tree diagrams to document business processes, including Six Sigma and ISO 9000. These tell us the starting. Digits 1 to 9 always count 2. But with sigma notation (sigma is the 18th letter of the Greek alphabet), the sum is much more condensed and efficient, and you’ve got to admit it looks pretty cool: This notation just tells you to plug 1 in for the i in 5i, then plug 2 into the i in 5i, then 3, then 4, and so on all the way up to 100. In this unit rules for using sigma notation are established. Re: sigma notation (without lists or tables) mravi1 welcome to the forum As pointed out, unfortunately your post does not comply with Rule 2 of our Forum RULES. Sigma notation is essentially a shortcut way to show addition of series or sequences of numbers. For example, we often wish to sum a number of terms such as 1+2+3+4+5 or 1+4+9+16+25+36 where there is an obvious pattern to the numbers involved. Example ∑ = = + + + +. After multiplying everything out I got, = E (i^3 + 15i^2 +54i) Then I broke everything down to = E i^3 + (15)*E i^2 + (54)*E i. In quantum mechanics, the uncertainty principle (also known as Heisenberg's uncertainty principle) is any of a variety of mathematical inequalities asserting a fundamental limit to the precision with which the values for certain pairs of physical quantities of a particle, such as position, x, and momentum, p, can be predicted from initial conditions. And S stands for S um. $$S^2\sigma$$ is frequently over analyzed to the point where it is incorrectly concluded that a high $$S^2\sigma$$ is preferable even at the expense of lower $$zT$$. is the symbol for the sample mean. Alternatively, we may use ellipses to | {
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