# Spring, weight and potential energy

1. Jan 30, 2014

### Bromio

1. The problem statement, all variables and given/known data
A particle is connected to a spring at rest. Because of weight, the mass moves a distance $\Delta x$. Calculate the value of the elastic constant $k$

2. Relevant equations
$U_{PE} = \frac{1}{2}k\Delta x^2$
$U = mgh$
$F = ma$ (in general)
$F = kx$ (Hooke's Law)

3. The attempt at a solution
I've tried to solve this problem from two points of view, but the result has been different in each one.

First, I've thought that, when the mass is at rest after being connected to the spring and got down to the new equilibrium position:
$mg = k\Delta x \Longrightarrow \boxed{k = \frac{mg}{\Delta x}}$

However, from the point of view of energies:
$mgh_0 = mgh + \frac{1}{2}k\Delta x^2$ (I use $\Delta x$ because the initial position is 0)

As $h_0 - h = \Delta x$, I've got:
$mg\Delta x = \frac{1}{2}k\Delta x^2 \Longrightarrow \boxed{k = \frac{2mg}{\Delta x}}$

Obviously there is a wrong factor of 2.

What's the problem here? Where is the mistake?

Thanks!

2. Jan 30, 2014

### BvU

Energy case is for the situation you let go at h=0. When it is at delta x, it's still moving. Until it is a 2 delta x, when it hangs still again. But now the spring is pulling harder and up she goes!
In other words, before there is rest, some energy has to be dissipated! You now know how much.

3. Jan 30, 2014

### Bromio

Intuitively I understand what you say, but I don't see how to write it analytically.

I mean, as we are in a conservative field, we only have to consider initial and final states. How should I have written the main equation ($mgh_0 = mgh + \frac{1}{2}k\Delta x^2$) to include what you say?

Thanks!

4. Jan 30, 2014

### BvU

That would be a little difficult, precisely because it describes an isolated case. It says there is energy left over, that has to go somewhere to get the "in rest" situation. E.g. in overcoming the air resistance which heats up the air a little.

5. Jan 30, 2014

### Bromio

Okay. So the dissipated energy has a value of $\frac{1}{2}k\Delta x^2$, hasn't it?

What I don't understand is why, if we are in a conservative field, we can't just consider initial and final states. Could you explain me the reason?

Thank you.

6. Jan 30, 2014

### BvU

You can, actually. It's just that it isn't a steady state. mg (h-h0) is converted into spring energy plus kinetic energy. If there is no friction, the oscillation goes on forever.

Is there a risk that we are misinterpreting the OP ? My idea was Δx is measured after the thing has come to rest.

7. Jan 30, 2014

### Bromio

No misinterpretation. Now I see all you're trying to explain. For some reason I thought there wasn't friction. It's obvious that, if there isn't it, I can't apply energy equations "at rest" situation, simply because there isn't that situation. Am I right?

So, when I study energies at the beginning ($h = 0$) and at the end ($h = \Delta x$, when entire system is at rest) I should include that dissipated energy.

If $TE_0$ and $PE_0$ are the total mechanical energy and the potential energy at the beginning respectively, $TE$ and $PE$ are the same energies but at the end, $SP$ is the spring energy at the end, and $Q$ is the dissipated energy (say heat), then:
$TE_0 = PE_0$
$TE = SP + PE$
$TE_0 = TE + Q$

Do I agree?

Thank you!

8. Jan 30, 2014

### BvU

Well, there is of course an "at rest" situation, it's just that it doesn't have the same energy as the situation where you attach the weight to the spring.

Yes. in fact, you can probably do that yourself: Suppose you exercise a force to keep the weight in position h0, then gently lower the weight to position h. If you do it slow enough, the force is mg - kx where x runs from 0 to h and h = mg/k. The weight does work, namely ∫ F(x) dx where the integral goes from 0 to h. F(x) = mg - kx. The integral (you can do it) is $\frac{1}{2}k h^2$. Looks familiar ?

9. Jan 30, 2014

### Bromio

When I say that there isn't an "at rest" situation I'm talking about the frictionless case. If there isn't friction, the spring will experiment an non-stop simple harmonic motion, won't it?

Thank you! I've solved that integral and I've got that result. Amazing!


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# A rectangular sheet of paper, when halved by folding it at the mid

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A rectangular sheet of paper, when halved by folding it at the mid  [#permalink]

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25 Oct 2018, 04:50
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Difficulty:

55% (hard)

Question Stats:

64% (02:10) correct 36% (02:13) wrong based on 47 sessions

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A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?

a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
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A rectangular sheet of paper, when halved by folding it at the mid  [#permalink]

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25 Oct 2018, 06:02
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?

a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3

$$\frac{l}{b} = \frac{b}{(l/2)}$$ where $$b = 2$$ given

i.e. $$b^2 = l^2 / 2$$

i.e. $$l = 2√2$$

Area of smaller rectangle $$= (l/2)*b = (2√2/2)*2 = 2√2$$

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Re: A rectangular sheet of paper, when halved by folding it at the mid  [#permalink]

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25 Oct 2018, 07:05
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?

a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3

OA: B

Initial Rectangle Dimensions
longer side $$= x$$
Shorter side$$= 2$$

Dimensions after folding
Longer side $$= 2$$
Shorter side $$= \frac{x}{2}$$

As per question
$$\frac{x}{2}=\frac{2}{\frac{x}{2}}$$
$$x^2=8 \quad;\quad x=2\sqrt{2}$$

Dimensions of Shorter Rectangle
Longer side $$= 2$$
Shorter side $$=\frac{2\sqrt{2}}{2}=\sqrt{2}$$
Area of Shorter Rectangle $$= 2\sqrt{2}$$
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Re: A rectangular sheet of paper, when halved by folding it at the mid  [#permalink]

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25 Oct 2018, 14:13
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?

a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3

$$?\,\, = 2x$$

The rectangles shown above are similar, hence:

$${{2x} \over 2} = {2 \over x}\,\,\,\,\, \Rightarrow \,\,\,{x^2} = 2\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\, > \,\,0} \,\,\,\,x = \sqrt 2 \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2\sqrt 2$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: A rectangular sheet of paper, when halved by folding it at the mid &nbs [#permalink] 25 Oct 2018, 14:13
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# For positive integers between 999 and 100 inclusive, how many contain the digit 5?

The question comes in two parts:

1. For positive integers between 999 and 100 inclusive, how many contain the digit 5 at least once?

2. For positive integers between 999 and 100 inclusive, how many contain the digit 5 exactly once?

Question 1

Numbers between 999 and 100 inclusive:

999 - 99 = 900

Numbers between 999 and 100 inclusive that do not contain digit 5:

8 * 9 * 9 = 648 (The 8 is because the first digit can't be 5 or 0)

Numbers that contain digit 5 at least once:

900 - 648 = 252

Question 2

Total numbers that contain digit 5 exactly once:

(1*9*9) + (8*1*9) + (8*9*1) = 225

I think I got it wrong but not sure which part.

• Well, a good way to check is to count the number with exactly two $5's$. $55X$ gives us $9$, $5X5$ gives us $9$, $X55$ gives us $8$. Thus there are $9+9+8=26$ such numbers. Of course $555$ is the unique number in the range with three $5's$. We check that $252=225+26+1$ which seems like pretty good evidence that you are right.
– lulu
Nov 1 '18 at 16:34
• @DavidG.Stork The OP is counting numbers with no digits equal to $5$ so that they can be subtracted from all three-digit positive integers. Nov 1 '18 at 16:38
• @DavidG.Stork OP is using a standard technique of counting how many numbers there are where we ignore the condition, giving $9\times 10\times 10 =900$ numbers in the range, and subtracting the number of numbers which violate the condition, violating the condition in this case meaning "does not have $5$ as a digit", there being $8\times 9\times 9$ such numbers, which will give the number of numbers that satisfy the condition as a result. Nov 1 '18 at 16:38
• @DavidG.Stork The purpose of that step is to get the number which does not contain digit 5 at all, which in this case is 648. My reasoning is that if I subtract 900 with 648, then all the remaining numbers would contain at least one digit 5. Nov 1 '18 at 16:40
• "I think I got it wrong but not sure which part" Looks good to me. Do you have specific reason to doubt the answer? Be more confident, you seem to be doing well from what I can see. Nov 1 '18 at 16:40

 There is one number with all three digits equal to 5.

There are 1x10x10 = 100 numbers with the first digit equal to 5.
There are 9x1x10 = 90 numbers with the second digit equal to 5.
There are 9x10x1 = 90 numbers with the third digit equal to 5.

There are 10 numbers with the first and second digit equal to 5.
There are 10 numbers with the first and third digit equal to 5.
There are 9 numbers with the second and third digit equal to 5.

There are 8x9x9 = 648 numbers with no digit equal to 5.

This is enough information to fill in the diagram above.

I got the same answers as you with the (javascript) code

var countFives = function(x){
var count = 0;
for (let c of x.toString(10)){
if (c==="5") count++;
}
return count;
};

var s = 0;
for (let i=100; i<=999; i++) if (countFives(i)>=1) s++;
console.log(s);

252

var s = 0;
for (let i=100; i<=999; i++) if (countFives(i)===1) s++;
console.log(s);

225


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In complex number, a is the 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 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! 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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 $d|n$

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

$$(135)(5)(1125) + (48)(-14)(1125) = (3)(1125)$$

$$\\$$ $$\implies (135)(5625)+(48)(-15750) = 3375$$

• @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


# 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.

-
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]

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.

-
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


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.

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

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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04 Aug 2016, 11:06
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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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Two bottles are partially filled with water. The larger bottle current  [#permalink]

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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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Re: Two bottles are partially filled with water. The larger bottle current  [#permalink]

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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
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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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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 $$\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$$

• 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:

$$F = \frac {GMm}{r^2}$$.


# 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

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.")

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 $S$.


# 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.

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

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.

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 $\rho = n$.

\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 $\rho = m = n$.

\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 $\rho = 2$.

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 $A^{T}A$

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 $$\mathbf{V} = % \left[ \begin{array}{cc} % \left( 1 + \left(\frac{-13 + \sqrt{173}}{4}\right)^2 \right)^{-\frac{1}{2}} % \left[ \begin{array}{c} \frac{1}{2} \left(-13+\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right] & % \left( 1 + \left(\frac{13 + \sqrt{173}}{4}\right)^2 \right)^{-\frac{1}{2}} % \left[ \begin{array}{c} \frac{1}{2} \left(-13-\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right] % \end{array} \right]$$


# 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 $\mathbb{Z}[\frac{1}{20},\frac{1}{32}]=\mathbb{Z}[\frac{1}{n}]$?

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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 $n$.

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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)

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)

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

$$\sum_{j=1}^3 j^4 = 1^4 + 2^4 + 3^4 = \sum_{k=1}^3 k^4.$$ 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 $3x^2$.

• 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

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.

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.