# Matrix group isomorphic to $\mathbb Z$.

The set $G=\left\{\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\mid n\in \Bbb Z\right\}$ with the operation of matrix multiplication is a group. Show that $$\phi:\Bbb Z \to G,$$ $$\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$$ is a group isomorphism (where the operation on $\Bbb Z$ is ordinary addtion).

TO show it's isomorphism: I know I must show one-to-one, onto and homomorphism. I've done these examples before but never with matrices.

How can I show if $\phi(a)=\phi(b)$ then $a=b$? Same question for onto and operation preserving with matrices.

Thank you!

• Try it. What happens when you multiply $\phi(a)$ and $\phi(b)$? – hardmath Oct 23 '15 at 14:15
• Homomorphism, check. Generator? – user190080 Oct 23 '15 at 14:27

Hint:

$\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\begin{pmatrix}1 & m \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix}1 & n+m \\ 0 & 1 \\ \end{pmatrix}$

That should help with proving $\phi$ is a homomorphism.

Once you have proven that a homomorphism exists, you must prove it is bijective to prove the mapping is a isomorphism. You already have that the mapping in injective, you must prove it is surjective.

• This is what I used. Thank you! – maidel b Oct 23 '15 at 16:44
• Glad it helped. – kleineg Oct 23 '15 at 17:31
• If $\phi$ is a homomorphism from $\Bbb Z \to G$, shouldn't $\phi^{-1}$ take matrices as input? – pjs36 Mar 16 '17 at 19:43
• OK, yes, that's better now – pjs36 Mar 16 '17 at 20:09
• This is not the inverse. As in function inverse. This is not even a different function from $\phi$. What the hell? In what world the inverse of a mapping $X\to Y$ is a mapping $X\to Y$? How changing $n$ to $-n$ proves that something is bijective? Come on. The proper inverse is defined by $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\mapsto n$. I don't believe there are 2 answers with the same mistake. – freakish Mar 16 '17 at 20:11

Hint: For your first question, write down $\phi(a)$ and $\phi(b)$ (go ahead, write down the matrices on a sheet of paper). Now, if those two are equal, what does it tell you?

The other parts are obviously different, but the idea is the same: just look at the matrices involved and use what you know about matrix multiplication (for the last part).

• Thank you. I understand one-to-one after writing out the steps. I'm struggling with onto. Let me make sure I'm thinking correctly. I'm trying to find $\phi(?)=n$. Is there a different way? I'm not able to see what I'm supposed to do. – maidel b Oct 23 '15 at 17:50
• @ShayAbbott No. You want to take an arbitrary $g\in G$ and find an $n\in Z$ so that $\phi(n)=g$. But if $g\in G$, then you know what $g$ looks like... – Teepeemm Oct 23 '15 at 18:09

So let me straight things up, since there are multiple answer which get this incorrectly. First of all you need to show that $\phi$ is a homomorphism (for that see other answers, they are fine). Then you can show that it is an isomorphism by constructing the inverse. And by the inverse we mean the function inverse, i.e. a function

$$g:G\to\mathbb{Z}$$

such that $\phi\circ g=\mbox{id}_{G}$ and $g\circ \phi=\mbox{id}_{\mathbb{Z}}$. By general property if such inverse exists then it is a homomorphism as well and so $\phi$ is an isomorphism.

So you can easily check that

$$g\bigg(\begin{bmatrix} 1 & n\\ 0 & 1 \end{bmatrix}\bigg)=n$$

is the inverse of $\phi$.

Just note that $$\begin{pmatrix}1 & 1 \\ 0 & 1 \\ \end{pmatrix}^n = \begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$$ This implies that $\phi$ is a surjective homomorphism because exponents add when you multiply powers. It is clear that $\phi$ is injective because $\phi(n)_{12}=n$.

Hint:

There are other ways to show that this map is one-to-one.

Use that the $ker(\phi)=0$.

Does there exist an $n$ such that the rank of $\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$ is not $2$?

Alternatively, construct $\phi^{-1}(n)=\begin{pmatrix}1 & -n \\ 0 & 1 \\ \end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.

• That's not the inverse. The proper inverse is a function on matrices. It maps $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\mapsto n$. Also if $\phi$ is a homomorphism then so is $\phi^{-1}$. So the existance of $\phi^{-1}$ is enough, no need to check that it is a homomorphism. – freakish Mar 16 '17 at 20:07


How to know if there is a extraneous solution in a radical expression

I was trying to solve this problem:

$$\sqrt{3x+13} = x+ 3$$

$$(\sqrt{3x+13})^2 = (x+ 3)^2$$ $$(3x+13) = (x+ 3)^2$$ $$3x+13 = x^2 + 6x + 9$$ $$0 = x^2 + 3x - 4$$ $$0 = (x+4)(x-1)$$

So my final answer was $x = -4$ and $x = 1$. However, it was incorrect because when I plug back in -4 into the original equation I get a extraneous solution. My question is do I always need to plug back in my answers into a radical expression and check if they are valid? Or is there any other way to deduce that there will be a extraneous solution?

• In this case you can tell there is only one solution simply by thinking about the graphs of the functions on the left and right sides of your original equation. This tells you one of your roots is extraneous. It is also clear graphically that that solution is positive. So the extraneous root must be $-4$. Alternatively, recall that the range of the square root function is the set of nonnegative reals. So your original equation implies $x\geq-3$. – symplectomorphic Feb 18 '17 at 22:58
• @Khosrotash I am trying to avoid checking the final answers because it is a timed test and I need to move really quickly – Pablo Feb 18 '17 at 22:59

You don't need to plug in values, if you always ensure not to add extraneous solutions.

The equation forces two conditions, namely $3x+13\ge0$ and $x+3\ge0$, which together become $x\ge-3$.

Why $x+3\ge0$? Because $\sqrt{3x+13}\ge0$ by definition (when it exists, of course).

With this condition, you can safely square, because you have an equality between nonnegative numbers. You get (your computations are good) $$\begin{cases} (x+4)(x-1)=0 \\[4px] x\ge-3 \end{cases}$$ and therefore you know what roots are a solution of the original equation, in this case only $x=1$.

On its domain ($A\ge 0$), note that $$\sqrt A=B\iff A=B^2\enspace\textbf{and}\enspace B\ge 0,$$ since the symbol$\sqrt{\phantom{h}}$ denotes the non-negative square root of a non-negative real number.

Note that when you squared both sides in the first step, you ended up with the equation

$3x+13 = (x+3)^2$,

which could just as well come from the same original equation but with a negative square root instead:

$-\sqrt{3x+13} = x+3$.

So, at the end of the day, you've solved both equations. To check which solution(s) correspond to which equation(s), you need to plug back in and check.

• Amm so does that mean that every time I square a radical then I would get something like this $$±(3x+13)^2$$ – Pablo Feb 18 '17 at 23:01
• No, because the square of a number is always positive. It just means that when you square both sides to try and solve the original equation, you're actually solving two equations at once. – J Richey Feb 18 '17 at 23:02

I think the best or reliable solving is to plot the equations . in your case like below


# Minkowski metric in spherical polar coordinates

1. Mar 4, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

Consider Minkowski space in the usual Cartesian coordinates $x^{\mu}=(t,x,y,z)$. The line element is

$ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}$

in these coordinates. Consider a new coordinate system $x^{\mu'}$ which differs from these Cartesian coordinates. The Cartesian coordinates $x^{\mu}$ can be written as a function of these new coordinates $x^{\mu}=x^{\mu}(x^{\mu'})$.

(a) Take a point $x^{\mu'}$ in this new coordinate system, and imagine displacing it by an infinitesimal amount to $x^{\mu'}+dx^{\mu'}$. We want to understand how the $x^{\mu}$ coordinates change to first order in this displacement $dx^{\mu'}$. Argue that

$dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$.

(Hint: Taylor expand $x^{\mu}(x^{\mu'}+dx^{\mu'})$.)

(b) The sixteen quantities $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$ are referred to as the Jacobian matrix; we will require this matrix to be invertible. Show that the inverse of this matrix is $\frac{\partial x^{\mu'}}{\partial x^{\mu}}$. (Hint: Use the chain rule.)

(c) Consider spherical coordinates, $x^{\mu'}=(t,r,\theta,\phi)$ which are related to the Cartesian
coordinates by

$(t,x,y,z)=(t,r\ \text{sin}\ \theta\ \text{cos}\ \phi,r\ \text{sin}\ \theta\ \text{sin}\ \phi,r\ \text{cos}\ \theta)$.

Compute the matrix $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$. Is this matrix invertible everywhere? Compute the displacements $dx^{\mu}$ in this coordinate system (i.e. write them as functions of $x^{\mu'}$ and the infinitesimal displacements $dx^{\mu'}$).

(d) Compute the line element $ds^{2}$ in this coordinate system.

2. Relevant equations

3. The attempt at a solution

(a) By Taylor expansion,

$x^{\mu}(x^{\mu'}+dx^{\mu'}) = x^{\mu}(x^{\mu'}) + \frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$

$x^{\mu}(x^{\mu'}+dx^{\mu'}) - x^{\mu}(x^{\mu'}) = \frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$

$dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$

Am I correct so far?

Last edited: Mar 4, 2016
2. Mar 4, 2016

### stevendaryl

Staff Emeritus
Yes, but you should go ahead and calculate the nine components:

$\frac{\partial x}{\partial r}$, $\frac{\partial y}{\partial r}$, $\frac{\partial z}{\partial r}$
$\frac{\partial x}{\partial \theta}$, $\frac{\partial y}{\partial \theta}$, $\frac{\partial z}{\partial \theta}$
$\frac{\partial x}{\partial \phi}$, $\frac{\partial y}{\partial \phi}$, $\frac{\partial z}{\partial \phi}$

3. Mar 4, 2016

### spaghetti3451

Isn't that in part (c)?

Shouldn't I do (b) first?

4. Mar 4, 2016

### stevendaryl

Staff Emeritus
Yeah, I guess you should, even though part c doesn't actually depend on part b.

5. Mar 4, 2016

### spaghetti3451

(b) Via the chain rule,

$\frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\mu'}}{\partial x^{\nu}}=\delta_{\nu}^{\mu}$,

where we are using the summation convention only over $\mu'$.

Therefore, the inverse of the matrix $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$ is the matrix $\frac{\partial x^{\mu'}}{\partial x^{\nu}}$.

Is this correct?

6. Mar 4, 2016

### stevendaryl

Staff Emeritus
Yes.

7. Mar 4, 2016

### Staff: Mentor

What's wrong with taking the differentials of x, y, and z (expressed in terms of the spherical coordinates in post #1), evaluating their differentials (in terms of the spherical coordinates and their differentials), and then taking the sum of their squares? This should give the Minkowski metric in spherical coordinates, correct?

Chet

8. Mar 5, 2016

### spaghetti3451

I know that this is a correct and shorter approach, but I'm trying to follow the instructions of the question.


# Is there a better way of finding the order of a number modulo $n$?

Say I wish to find the order of 2 modulo 41.

The way I've been shown to compute this is to literally write out $2^k$ and keep going upwards with $0 \leq k \leq 41$, or until I observe periodicity in the sequence, and then state the order from there.

Is there a better way of deducing the order of a number with respect to some modulo?

This seems highly inefficient, especially if we are working with respect to some large modulo $N$ where it will take at most $N$ computations to determine when periodicity occurs.

• Since $41$ is prime, the order must be a factor of $40$ Nov 17, 2014 at 6:49
• I did not know this, but thanks for that. Very useful. Perhaps a bad example, what if $n$ is NOT prime? Are there any results that can further 'filter' it out? Also, say we know the order must be a factor of 40, then is there a systematic way of computing the order without having to bash out any numbers? Nov 17, 2014 at 6:51

For any $$a$$ and $$N$$ with $$\gcd(a,N)=1$$, the order of $$a$$ modulo $$N$$ must be a divisor of $$\varphi(N)$$. So if you know the prime factorization of $$N$$ (or $$N$$ is already prime) so that you can compute $$\varphi(N)$$ and also know the prime factorization of $$\varphi(N)$$, you can proceed as follows:

If we know an integer $$m>1$$ with $$a^m\equiv 1\pmod N$$ and know the prime divisors of $$m$$, for all primes $$p$$ dividing $$m$$ do the following: Compute $$a^{m/p}\bmod N$$ and if the result is $$\equiv 1\pmod N$$, replace $$m$$ with $$m/p$$ and repeat (or switch to the next prime divisor if $$m$$ is no longer divisible by $$p$$). When you have casted out all possible factors, the remaining $$m$$ is the order of $$a$$. Note that the computations $$a^m\bmod N$$ do not require $$m$$ multiplications, but rather only $$O(\log m)$$ multiplications mod $$N$$ if we use repeated squaring.

If $$N$$ is large and the fatorization of $$\varphi(N)$$ is known (and especially if you suspect the order of $$a$$ to be big), this is in fact a fast method.

Note that a couple of computations can be saved even beyon what is descibed above: In the case $$p=2$$, we may end up computing $$a^m, a^{m/2}, a^{m/4},\ldots$$ to cast out factors of $$2$$. But the later numbers were in fact intermediate results of computing $$a^m$$ by repeated squaring! Also, once we notice for some $$p$$ with $$p^k\mid m$$ that $$a^{m/p}\not\equiv 1\pmod N$$, we can save a few squarings and so speed up the task for the remaining primes if we replace $$a$$ with $$a^{p^k}\pmod N$$ and $$m$$ with $$m/p^k$$ - just remember to multiply the factor $$p^k$$ back into the final answer!

In your specific example, we know that $$N=41$$ is prime and that $$\varphi(N)=40=2^3\cdot 5$$. We check $$p=5$$ and note that $$2^{40/5}=256\equiv 10\pmod{41}$$, hence the factor $$5$$ cannot be eliminated. After that we check how many $$2$$'s we have to use: $$2^5\equiv 32\equiv -9$$, hence $$2^{10}\equiv 81\equiv -1$$, $$2^{20}\equiv (-1)^2\equiv 1$$. We conclude that $$2$$ has order $$20$$ modulo $$41$$.

The order must be one of $1,2,4,8,5,10,20,40$.

Calculate $2^2,2^4=(2^2)^2,2^8=(2^4)^2,...\\2^5=2^4*2,2^{10}=(2^5)^2,2^{20}=(2^{10})^2,2^{40}=(2^{20})^2$ which is seven calculations.

If 41 is not prime, say $41=a^2b^3c$,then

i) calculate the order of $2\pmod{a^2}$, and the order $\pmod{b^3}$ and the order $\pmod c$
ii) calculate the lowest common multiple of the separate orders.

Here we calculate the order of $$2$$ in $$(\mathbb{Z}/{41}\mathbb{Z})^\times$$.

Note: When calculating you can use the previous work to your advantage. To really cut down on the calculations, if you know the order is greater than or equal $$10$$, once you calculate $$2^{10} \pmod{41}$$ you are done,

$$\quad 2^{10} \equiv 1 \pmod{41} \quad \text{order is } 10$$
$$\quad 2^{10} \equiv 40 \pmod{41} \quad \text{order is } 20$$
$$\quad \text{NOT }[2^{10} \equiv 1, 40 \pmod{41}] \quad \text{order is } 40$$

Work Summary : $$2^1 \equiv 2 \pmod{41}$$.

The order of $$2$$ is one of $$2,4,8,5,10,20,40$$.

Work Summary : $$2^2 \equiv 4 \pmod{41}$$.

The order of $$2$$ is one of $$4,8,5,10,20,40$$.

Work Summary : $$2^4 \equiv 16 \pmod{41}$$.

The order of $$2$$ is one of $$8,5,10,20,40$$.

Work Summary : $$2^5 = \equiv 32 \pmod{41}$$.

The order of $$2$$ is one of $$8,10,20,40$$.

Work Summary : $$2^8 \equiv 10 \pmod{41}$$

The order of $$2$$ is one of $$10,20,40$$.

Work Summary : $$2^{10} \equiv 40 \pmod{41}$$

The order of $$2$$ is one of $$20,40$$.

Work Summary : $$2^{20} \equiv 1 \pmod{41}$$

The order of $$2$$ is equal to $$20$$.


Quantitative Comparison: “The Relationship Cannot Be Determined from the Information Given” Answer Choice

1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given

Many people dread choosing answer choice (D) on Quantitative Comparison (QC) Some feel it may be conceding defeat. Others think that the GRE is trying to trick them by making them pick (D). After all, they think, there must be some pattern that I’m not getting.

The truth is answer (D) comes up often. And to determine whether an answer cannot be determined is actually not too difficult.

#1 Determine a relationship

Say you find an instance, in which the answer is (A) the information in column A is greater. If that is the case, then the next step is to disprove that.

#2 Disprove that relationship

Meaning, see if you can come up with an instance, either through plugging in different variables, manipulating algebra, or manipulating a geometric figure, in which the answer is not (A). As soon you do that, you can stop. The answer is (D).

If you can’t disprove your answer, then it must be correct: it must be (A), (B) or (C).

1. $$-100 < x < 0$$

Column A Column B
$$x^{-4}$$ $$x^{-3}$$

1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given

Explanations:

Question #1

After choosing a few numbers you should note something: that Column A will always be greater than Column B.

Why? Will, whenever, you have an even exponent, positive or negative, that exponent will always yield a positive number.

Odd exponents, on the other hand, give you a negative output if the base (the number below the exponent) is negative. Remember x has to be negative. So no matter what number you plug in Column A will always be positive, Column B negative. This is a definite not (D). The answer is (A).

32 Responses to Quantitative Comparison: “The Relationship Cannot Be Determined from the Information Given” Answer Choice

1. lokesh nandni sood May 22, 2019 at 9:59 pm #

hi chrish i put all the different values if i put 1 , they both are equal , if i put -1 B is greater . same way if i put 2 value is 0.0625 A is smaller and 0.125 B is greater . didn’t get that . pls help

• Magoosh Test Prep Expert May 24, 2019 at 8:36 am #

Hi Lokesh,

Remember that we have an important limitation to which numbers we can choose. The question tells us that -100 < x < 0. This means that x must be a negative number. You are plugging in positive numbers, which will give you a different answer 🙂

2. Karan June 28, 2017 at 4:54 am #

Hi Chris,
I have a doubt related to quantitative comparison type questions. What option do we choose if both columns turn out to be infinity. For eg:
Column A Column B
The number of numbers from 1 to 2 The number of numbers from 1 to 5

D or C?

• Magoosh Test Prep Expert June 28, 2017 at 11:43 am #

Hi Karan,

Great and interesting question! Honestly, you will not see a question like this on the GRE, in which you are comparing two quantities, each with the solution of “infinite.” That being said, the comparison of infinite to infinite is equal, so I’d go with (C). But again, you will not see something like this. Still, it’s a very interesting and awesome question! Have a great day! 🙂

3. Prasanth January 5, 2017 at 11:43 pm #

Hi Chris,

In the above example
-100<x<0

Qty A : x^-4
Qty B : x^-3

Now multiply both sides by a positive number relation will not change. x^4 is a positive number.So

Qty A becomes : 1
Qty B becomes : x.

Now it is easy to plug -ve numbers for x.

In all definite ways, column A will be greater.

Can i approach this way or plugging numbers will be easy in test ?

I usually make mistakes while plugging -ve numbers .. Pls help

• Magoosh Test Prep Expert January 6, 2017 at 1:32 pm #

Hi Prasanth,

No, I do not recommend taking this approach. For example, imagine if you manipulated the problem by multiplying by x^5. So,

Qty A becomes: x
Qty B becomes: x^2

Now, in this case, Qty B is greater than Qty A, which is not true. It is best to approach this problem rephrasing this question.

Qty A: x^-4 = 1/(x^4)
Qty B: x^-3 = 1/(x^3)

Here you know that any negative number raised by an even exponent will be positive, and that same number raised by an odd exponent will be negative. This is how we know that column A is greater. I hope that helps! 😀

• fatima January 12, 2018 at 7:44 am #

Hi Chris, regarding this problem :

-100<x<0
Qty A : x^-4
Qty B : x^-3

Prasanth multiplied both sides by a positive number (x^4 is a positive number),which will be at all times positive since (-100<x<0), and it is legal to multiply both sides of an inequality by a positive number..
But it wil not be true to multiply both sides by x^5 since (-100<x<0),so x^5 will give a negative number. And it will change the inequality..
i think Prasanth's approach is right. Does it ?
thanks,,

• Magoosh Test Prep Expert January 18, 2018 at 12:40 pm #

Hi Fatima,

Prasanth’ approach is correct, in the sense that it really does work for this particular problem. Here, multiplying Quantity A and Quantity B by x to an even power will help simplify things and may help make them clearer.

The reason I’d still recommend against Prasanth’s approach is that it adds an extra layer of complexity that isn’t necessary. Under real testing conditions, every second counts, and simply recognizing the number properties at play here, without plugging in new numbers, is important. In addition under the pressure of test conditions, it becomes easier to make a mistake like choosing x^5 instead of x^4 without noticing that the inequality is reversed.

4. Yogesh kurade December 13, 2016 at 9:56 pm #

Thank you! Sir

5. Priyam June 13, 2016 at 9:11 am #

In lesson intro to Quantitative comparison, there’s this example question
N is not an integer
6<N<10
Quantity A Quantity B
N 8

Here, the answer is mentioned as D) cannot be determined reason stated as it can be a decimal or a fraction, but integer itself means "Whole nos which can be -ve as well as +ve".

• Magoosh Test Prep Expert June 18, 2016 at 12:23 pm #

Hi Priyam 🙂

Thanks for your message 🙂 In that example, we’re told that 6 < N < 10. So, we know that N is a positive number between 6 and 10. However, we cannot determine whether N is greater than 8 using only the information given, that N is not an odd integer. N could be any positive number between 6 and 10 except 7, which is the only odd integer within this range. For example, N could equal 6.5 or 8.3. Neither of these numbers is an odd integer and therefore fits the description. Because N may be less than, equal to, or greater than 8, the answer to that example is (D).

I hope this clears up your doubts 🙂

• Bhavna Sharma June 26, 2016 at 3:14 am #

I have a doubt regarding this. In the above explanation, you have mentioned that “N could equal 6.5 or 8.3”. But, in the question it has been specified that N is an integer and 6.5 or 8.3 are not integers.

• Magoosh Test Prep Expert June 26, 2016 at 4:28 am #

Hi Bhavna 🙂

In that example, the prompt states that “N is not an odd integer.” This is not the same as saying that N is an even integer. N could be an even integer, but N could also be a decimal or fraction, since decimals and fractions are not odd integers. I hope this clears things up 🙂

• Bhavna Sharma June 27, 2016 at 8:25 am #

Thank you so much.. 🙂 I missed this point.

6. Lars July 30, 2012 at 4:55 am #

Hi Chris,

In your explanation to the 1st question, in the 2nd last sentence you wrote

“So no matter what number you plug in Column A will always be negative, Column B positive.” Isn’t this the exact opposite because Column A wiil be positive and column B will be negative?

I got confused the first time I read this.

7. Sammy May 30, 2012 at 3:54 pm #

Chris,
For these types of problems I have a trick that allows me to correctly deduce what the answer is. I wanted to get your thoughts on my method.
An example problem that I fabricated –

n is an integer
Which is greater?
COL A – n^3 + 6
COL B – n^2 + 5n -30

I would attack this problem in this way – I would choose 5 arbitrary numbers (0,-0.6, 0.6,
3,-3) and plug each in till i get an explicit answer. The only problem I see with my this is that it can get to be time consuming.
Let me know what you think!

PS – I love it when you answer in a witty way with your sophisticated words!

• Chris June 1, 2012 at 12:58 pm #

Hi Sammy,

You put me in a little bit of a tough spot, because it’s kind of difficult to be witty with Quantitative Comparison :).

Your method definitely works well. Picking numbers can help you determine which column is bigger as long as you make sure to include 0, 1, -1, 2, -2 (or 3, -3) and 1/2 and -1/2 (I think 0.6 is a little unwieldy :)).

But in general this is a very effective method and one that I too employ. Often, esp. with algebraic equations, you may want to manipulate the equations and see if you can simply them.

Sorry, I didn’t drop any sophisticated words, but I hope that helped :).

• Sammy June 1, 2012 at 1:06 pm #

Thanks Chris, my only qualm would be that the method can be somewhat time consuming.

• Chris June 1, 2012 at 1:20 pm #

Oh, that’s right – you mentioned that…if you have a strong facility with numbers, that is you can quickly plug-in, then it shouldn’t be time consuming. Also you can come up with a tiny column grid:

A B
0
1
-1
2
-2

By the time you get the 2 you can usually know for certain what column it is.

Hope that helps :).

• Amal June 1, 2012 at 1:43 pm #

Hi Chris,

Would it be correct to say that if n is an integer, one need not test decimals that can be written as a fraction. For example, 0.6 = 6/10, which means that 0.6 is actually a fraction, aka an integer divided by another integer, or two integers. And the same thing for 0.5, since it equals 1/2. In fact, when it comes to “n is an integer” type questions, we should be able to exclude all decimals, since according to the “official GRE guide,” only real numbers are involved (e.g. we would not test 3.14 which is a decimal but can’t be written as a fraction. This would mean that every decimal number is a real number (which implies it can be written as a fraction) for GRE purposes, as far as quantitative comparison type questions are concerned. Great question by the way.

• Chris June 1, 2012 at 2:04 pm #

Yes, you are exactly right. I was speaking more broadly when giving that spread of numbers. One should always obey the constraints of the question, i.e. if x is a positive integer, you have to plug in accordingly.

Thanks for catching that :).

• Sammy June 1, 2012 at 2:06 pm #

Yes you are right, it was an egregious error made on my part. I meant to say ‘a real number’

• Chris June 4, 2012 at 2:44 pm #

No problem :). Glad I could help.

• harsha July 7, 2016 at 1:46 am #

hey chris i am your big fan of u can u give me links for practcicng gre full lenth tests of free costs

• Magoosh Test Prep Expert July 7, 2016 at 10:37 am #

Hi Harsha 🙂

I’d recommend that you first take the free PowerPrep tests from ETS, if you haven’t already. Next, here’s a free test from Manhattan (their materials are great!): Manhattan Free GRE Practice Test. Lastly, if you want access to more practice tests, I’d recommend purchasing one of the Manhattan books–each book comes with a code to access 6 online tests.

Hope this helps 🙂

8. Denis May 23, 2012 at 9:57 am #

Hi Chris,

I have a question regarding a concept in Question #2: you state “Three equal sides equal three equal angles.” Is this a characteristic of all polynomials, i.e., can we say “x equal sides equal x equal angles” for any figure?

Best regards,
Denis

• Chris May 23, 2012 at 11:40 am #

Hi Denis,

The measure of an angle corresponds to the length of the side opposite that angle, in a triangle. With quadrilaterals, or figures with more than four sides an angles, that relationship between which side corresponds to which angle is not as clear cut. However, we can say with certain that if there are x equal sides then there are x equal angles.

Hope that helps :).

• Denis May 23, 2012 at 1:34 pm #

Hi Chris,

I greatly appreciate the expeditious reply. Thanks for the help.

9. Amal May 20, 2012 at 11:58 am #

I’d say the answer for question 1 is (D). If you pick -1, then you get 1 for both column A and column B, since you have 1 over 1 raised to some positive power. If you pick, say, -2, then column A and column B are obviously different.

• Chris May 21, 2012 at 3:22 pm #

Hi Amal,

If you take any negative to an even exponent it will always yield a positive number. Therefore, if you plug in -1, you will get 1 for Col. A and -1 for Col B.

Hope that helps :).

• Amal May 21, 2012 at 4:11 pm #

Yeah you’re right. I forgot the negative sign on the coefficient. When you raise a negative coefficient to a negative odd power, the coefficient (which winds up in the denominator) is still negative. The coefficients are always negative, but the even power in column A yields a positive quantity.

• Chris May 22, 2012 at 1:41 pm #

No problem 🙂

Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.

We highly encourage students to help each other out and respond to other students' comments if you can!

If you are a Premium Magoosh student and would like more personalized service from our instructors, you can use the Help tab on the Magoosh dashboard. Thanks!


More- over, if the partition is in fact an all-square partition and A, B, and D are all invertible, then (3.2) We can assume that the matrix A is upper triangular and invertible, since $$\displaystyle A^{-1}=\frac{1}{det(A)}\cdot adj(A)$$ We can prove that $$\displaystyle A^{-1}$$ is upper triangular by showing that the adjoint is upper triangular or that the matrix of cofactors is lower triangular. Suppose the n × n matrix R is upper triangular and invertible, i.e., its diagonal entries are all nonzero. Clearly, the inverse of a block upper triangular matrix is block upper triangular only in the square diagonal partition. (a) if U is upper triangular and invertible then U^-1 is upper triangular. It fails the test in Note 5, because ad bc equals 2 2 D 0. An LU factorization refers to the factorization of A, with proper row and/or column orderings or permutations, into two factors – a lower triangular matrix L and an upper triangular matrix U: =. Taking transposes leads immediately to: Corollary If the inverse L 1 of an lower triangular matrix L exists, Show that R1 is also upper triangular. It's obvious that upper triangular matrix is also a row echelon matrix. Proposition If a lower (upper) triangular matrix is invertible, then its inverse is lower (upper) triangular. Hint. Let A and B be upper triangular matrices of size nxn. (b) The inverse of a unit lower triangular matrix is unit lower triangular (c) The product of two upper or (two lower triangular) matrices is upper or (lower) triangular It goes like this: the triangular matrix is a square matrix where all elements below the main diagonal are zero. An example is the 4 4 matrix 4 5 10 1 0 7 1 1 0 0 2 0 0 0 0 9 . The inverse element of the matrix [begin{bmatrix} 1 & x & y \ 0 &1 &z \ 0 & 0 & 1 end{bmatrix}] is given by [begin{bmatrix} 1 & -x & xz-y \ 0 & 1 & -z \ 0 & 0 & 1 end{bmatrix}.] Inverse of matrix : A square matrix of order {eq}n \times n{/eq} is known as an upper triangular matrix if all the elements below principle diagonal elements are zero. Let A be a square matrix. 11.7 Inverse of an upper triangular matri. Let $b_{ij}$ be the element in row i, column j of B. Let A be a n n upper triangular matrix with nonzero diagonal entries. For a proof, see the post The inverse matrix of an upper triangular matrix with variables. The proof for upper triangular matrices is similar (replace columns with rows). Use back substitution to solve Rsk-en for k 1, , n, and argue that (sk)i -0 for i > k. The inverse of a triangular matrix is triangular. It's actually called upper triangular matrix, but we will use it. In the next slide, we shall prove: Theorem If the inverse U 1 of an upper triangular matrix U exists, then it is upper triangular. In this problem, you will Example of upper triangular matrix: 1 0 2 5 0 3 1 3 0 0 4 2 0 0 0 3 A triangular matrix (upper or lower) is invertible if and only if no element on its principal diagonal is 0. Solving Linear Equations Note 6 A diagonal matrix has an inverse provided no diagonal entries are zero: If A D 2 6 4 d1 dn 3 7 5 then A 1 D 2 6 4 1=d1 1=dn 3 7 5: Example 1 The 2 by 2 matrix A D 12 12 is not invertible. In the lower triangular matrix all elements above the diagonal are zero, in the upper triangular matrix, all the elements below the diagonal are zero. In general this is not true for the square off-diagonal partition. Let $a_{ij}$ be the element in row i, column j of A. An upper triangular matrix is a square matrix in which the entries below the diagonal are all zero, that is, a ij = 0 whenever i > j. 82 Chapter 2.


# Vertically oriented spring and energy transfers

1. Feb 20, 2007

### CaptainZappo

I am posing this question due to a statement made by my TA:

Suppose there is a spring pointed vertically upwards. Further, suppose that the spring is compressed a distance x so that it stores some potential energy. A mass is then placed on top of the spring. Assume all energy is conserved.

Here's the question: upon releasing the spring, what is the kinetic energy of the ball as it passes the equilibrium position of the spring? Is it equal in magnitude to (1/2)kx^2 where x is the amount the spring is compressed, or is it (1/2)kx^2 - mgx?

My TA said the former is correct, while I cannot figure out why it is not the latter. My reasoning: as the ball is rising from its initial height (the point at which the spring is fully compressed) some of that energy is being converted to gravitational potential energy. The rest goes to kinetic energy.

Any insight will be greatly appreciated,
-Zachary Lindsey

2. Feb 20, 2007

### Staff: Mentor

Your reasoning is correct; you cannot ignore gravitational PE in calculating the KE of the mass.

3. Feb 20, 2007

### CaptainZappo

Thank you.

4. Feb 20, 2007

### Parlyne

Who is correct depends on whether x is the amount the spring is compressed from its free equilibrium length or that from its equilibrium length with the mass sitting on it. If it is the former, you are correct. If it's the later, your TA is correct.

To see this, let's let y be the distance the mass is sitting above the spring's free equilibrium position. Then, the potential energy is:

$$V(y) = \frac{1}{2} k y^2 + mgy$$

We can rearrange this:

\begin{align*} V(y) &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y\right ) \\ &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y + \frac{m^2g^2}{k^2} \right ) - \frac{m^2g^2}{2k} \\ &= \frac{1}{2} k \left (y + \frac{mg}{k} \right )^2 - \frac{m^2g^2}{2k} \end{align*}

From this, we can see that the mass' height above the compressed equilibrium length is $$x = y + \frac{mg}{k}$$.

So, $$V(x) = \frac{1}{2} k x^2 - \frac{m^2g^2}{2k}$$.

The difference between the potential energy when the spring is compressed by a distance x from the compressed equilibrium position ($$y = \frac{-mg}{k}$$) and at the compressed equilibrium is:

\begin{align*} \Delta V &= V(x) - V(0) \\ &= \frac{1}{2} k x^2 - \frac{m^2g^2}{2k} - \left (\frac{-m^2g^2}{2k} \right ) \\ &= \frac{1}{2} k x^2 \end{align*}

5. Feb 21, 2007

### Staff: Mentor

Parlyne is certainly correct. From the way the conditions were specified in the OP:
I presumed that x measures the compression from the uncompressed position of the spring. Only later is a mass involved:
The answer to that question depends on the meaning of "equilibrium postion" and where x is measured from. If, as I assumed from your description, x is measured from the uncompressed position of the spring (and equilibrium means x = 0), then 1/2kx^2 represents spring PE only and does not include gravitational PE.

But, as Parlyne explained, if x is measured from the equilibrium position of the mass-spring system (which is not the uncompressed position of the spring), then 1/2kx^2 represents changes in both spring PE and gravitational PE together.

CaptainZappo: Better tell us exactly what the TA stated, not your interpretation of it (if you can recall).

Regardless of what the TA says, learn what Parlyne explained and you'll be ahead of the game.


# Thread: Running Around a Track

1. ## Running Around a Track

Hello, helpful people. I'm in some need of help.

I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:

"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"

Thanks.

2. ## Re: Running Around a Track

Hello, MyHappyHarmony!

This one is tricky to set up.
I have a rather clunky approach . . .

Brooks and Avery are running laps around the outdoor track, in the same direction.
Brooks completes a lap every 78 seconds, while Avery completes a lap every 91 seconds.
Brooks (the faster runner) has just passed Avery.
How much time will it take for Brooks to overtake Avery again?"

We will use: . $\text{Distance} \:=\:\text{Rate} \times \text{Time} \quad\Rightarrow\quad r \,=\,\frac{d}{t}$
Let $d$ = distance in one lap (say, in feet).

Brooks' rate is $\frac{d}{78}$ ft/sec.

Avery's rate is $\frac{d}{91}$ ft/sec.

In $t$ seconds, Avery runs $\frac{d}{91}t$ feet.

In the same $t$ seconds, Brooks runs $\frac{d}{78}$ feet,
. . which is $d$ feet (one lap) more than Avery's distance.

There is our equation! . . . $\frac{d}{78}t \;=\;\frac{d}{91}t + d$

Divide by $d\!:\;\frac{t}{78} \:=\:\frac{t}{91} + 1$

Got it?

3. ## Re: Running Around a Track

Yes, I do! Thanks. Quick question: Would it be right to solve with a common denominator from there? The big numbers do look quite intimidating.

4. ## Re: Running Around a Track

Originally Posted by MyHappyHarmony
Hello, helpful people. I'm in some need of help.

I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:

"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"

Thanks.
Try to construct your solution from this figure

cross posting

5. ## Re: Running Around a Track

I'm sorry. I'm not drawing anything from that figure.

6. ## Re: Running Around a Track

another method
assume track is 1200ft B rate 1200/78=15.38 fps A rate 1200/91= 13.19fps
after B runs 1 lap A is 1029ft ahead of B 91 * 13.19 = 1029 ft
Rate of closure = 15.38-13.19 = 2.2 fps
closure time 1029/2.2 = 467.7 sec
total elapsed time = 467.7 +78= 555.7 sec

7. ## Re: Running Around a Track

Originally Posted by bjhopper
total elapsed time = 467.7 +78= 555.7 sec
typo; 545.7 ; 546 really: LCM(91,78)

These are quite simple: have the faster runner start behind the slower,
the distance behind being the track length; how long does it take to catch up?

8. ## Re: Running Around a Track

Hello Wilmer,
Long 'time no see.How are you?
Soroban' s equation gives elapsed time until runners meet. 546 sec exactly
My answer 555.7 sec.There is no typo error.
Can you show what I did wrong?

Here is another solution
B rate 1/78 laps per sec A rate 1/91 laps per sec
(1/78 -1/91) = rate of closure laps per sec when B is behind A
after 1 lap B is 78/91 =0.857 laps behind A
(1/78 -1/91)t = 0.857
(0.00183)t =0.857 t=468.3sec add 78 for first lap Total elapsed time = 556.3sec

What say you?

9. ## Re: Running Around a Track

Hey BJ. Nice to "type" with you again!!

Originally Posted by bjhopper
My answer 555.7 sec.There is no typo error.
But you show this in your previous post:
"total elapsed time = 467.7 +78= 555.7 sec"

But 467.7 + 78 = 545.7, not 555.7

That's what I was trying to tell you.


# Countable union of countable sets vs countable product of countable sets

I know that a countable union of countable sets is countable, and that a finite product of countable sets is countable, but even a countably infinite product of countable sets may not be countable.

Let $$X$$ be a countable set. Then $$X^{n}$$ is countable for each $$n \in N$$.

Now it should also be true that $$\bigcup^{\infty}_{n=1} X^{n}$$ is countable. How is this different from $$X^{\omega}$$, which is uncountable?

Ah, very good question. Let's take a look at this question with $$X=\mathbb{N}$$. You first mentioned

$$\bigcup{X^n}$$

The elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...).

So it is easily seen that every element of $$\bigcup{X^n}$$ can be represented by an element of $$X^\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...

But, and here comes the clue: there are much more elements in $$X^\omega$$. For example: (1,1,1,...) does not come from an element of $$\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\omega$$!! And so $$X^\omega$$ really is different than $$\bigcup{X^n}$$.

Of course, this is not a proof. But it merely gives an indication why the two sets are different...

Ah, very good question. Let's take a look at this question with $$X=\mathbb{N}$$. You first mentioned

$$\bigcup{X^n}$$

The elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...).

So it is easily seen that every element of $$\bigcup{X^n}$$ can be represented by an element of $$X^\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...

But, and here comes the clue: there are much more elements in $$X^\omega$$. For example: (1,1,1,...) does not come from an element of $$\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\omega$$!! And so $$X^\omega$$ really is different than $$\bigcup{X^n}$$.

Of course, this is not a proof. But it merely gives an indication why the two sets are different...

Yes, after thinking about it I came to the same conclusion. You can't put them in bijective correspondence because if you wanted to map the union onto $$X^\omega$$ you could do so injectively by adding 0s but there's no way to make this mapping surjective.

mathwonk
Homework Helper
i believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.

i believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.

Exactly, hence why this was a seeming contradiction.

Exactly, hence why this was a seeming contradiction.
What's the contradiction that you're referring to?

It's already been stated that the needed bijection doesn't exist. I'm not following you're argument...


# Polynomial riddle

1. Jun 8, 2014

### dislect

Hi guys,

My boss gave me a riddle.
He says that you have a "black box" with a polynomial inside it like f(x)=a0+a1x+a2x^2+a3x^3 ....
you don't know the rank of it or the coefficients a0, a1, a2 ....
You do know:
all of the coefficients are positive
you get to input two x numbers and their f(x) value

How can you find all of the coefficients a0,a1,a2 .... and the rank?

Thanks !

2. Jun 8, 2014

### hilbert2

Knowing f(x) for two values of x is not enough to determine the coefficients if the polynomial is of higher than first order.

3. Jun 8, 2014

### economicsnerd

It can be done if you know the polynomial has nonnegative integer coefficients, and if the second input is allowed to depend on the first output.

Put in $1$. The output gives you the sum $s$ of the coefficients.
- If $s=0$, you know $f=0$.
- If $s>0$, put in any integer $k>s$. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-$k$.

4. Jun 8, 2014

### dislect

Hi economicsnerd,

Couldn't follow "put in any integer k>s. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-k."
Could you give me an example? lets say f(x)=5+2x+4x^2 is in the 'black box'.
i put in x=1 and receive f(x)=s=a0+a1+a2=11 > 0 so i put in x=k=15 and get f(x)=a0+15a1+900a2=935
organize:
1. a0+a1+a2=11 2. a0+15a1+900a2=935
2 equations with 3 unknowns.

Whats the rational behind putting a k > s, and how can i find all coefficients with just 2 equations ?

5. Jun 8, 2014

### economicsnerd

Okay. So you know $f(1)=11$ and $f(15)=935$.
- First note that $15^3 > 935$, so that the polynomial can't be of degree $>2$. That is, $f(x)=a_0 + a_1 x + a_2 x^2$ for some $a_0,a_1, a_2\in\mathbb Z_+$.
- Next, note that $a_0,a_1\in\mathbb Z_+$ with $a_0+ a_1 \leq 11 < 15$, so that $a_0 + a_1(15)<15^2$. This means that $935 - 15^2< a_2 (15)^2 \leq 935$. It's easy to check that there's a unique integer $a_2$ that accomplishes this, namely $a_2 = 4$.
- Substituting the solution for $a_2$ in, we now know that: $a_0+a_1 < 15$ and $a_0 + a_1 (15) = 15 35$. [Of course, you have two equations and two unknowns now, but we don't actually need to know that $a_0 + a_1 =7$, because knowing they're integers gives us enough information.] Now we can do the same trick again. $35 - 15< a_1 (15) \leq 35$, which pins down $a_1=2$.
- Then substitution yields $a_0 = 5$.

We can always work backward in this way.

To get some better intuition for why, consider the following question:
Say I tell you I have a polynomial $g$ such that every coefficient of $g$ is a nonnegative integer $<10$. If I tell you $g(10)$, will you then be able to tell me what $g$ is?

For example:
- If $g(10)=5021$, then do we know $g(x) = 5x^3 + 2x + 1$?
- If $g(10)=60007$, then do we know $g(x) = 6x^4 + 7$?

Nothing is magical about $10$ that makes this trick work. It's just that we usually express numbers in a way (base-10) that makes it easy to answer this question. All the first question is needed for is to find a $k$ such that we can be sure that every coefficient of the polynomial is a nonnegative integer $<k$.

6. Jun 8, 2014

### D H

Staff Emeritus
The rationale behind using asking for f(k) where k>f(1) is because otherwise the magic might not work. Suppose the function in the black box is f(x)=5+2x+4x2. The black box cranks out 11 as a response to f(1). Suppose I use 4 next. The black box churns out f(4)=77, which is 1031 when expressed base 4. That suggests the polynomial is 1+3x+5x3, which is wrong.

To see what's going on, here are f(k) from k=2 to 16, expressed in base 10 and in base k:
Code (Text):

k f(k)  base k  Polynomial
2   25   11001  1+x^3+x^4
3   47    1202  2+x^2+x^3
4   77    1031  1+3x+x^3
5  115     430  3x+4x^2
6  161     425  5+2x+4x^2
7  215     425  5+2x+4x^2
8  277     425  5+2x+4x^2
9  347     425  5+2x+4x^2
10  425     425  5+2x+4x^2
11  511     425  5+2x+4x^2
12  605     425  5+2x+4x^2
13  707     425  5+2x+4x^2
14  817     425  5+2x+4x^2
15  935     425  5+2x+4x^2
16 1061     425  5+2x+4x^2
Can you see the pattern? It appears that f(k) when expressed in base k is 425 for all k>5. This is indeed the case. You might want to try proving it.

Note that in this case reading off the polynomial from the base-k representation of f(k) works for all k>5. That's because the largest coefficient is five. Using k=f(1)+1 (or higher) ensures that k is larger than the largest possible coefficient.

Suppose the black box function was instead f(x)=11x3. Once again you'll get f(1)=11, but now the base k representation of f(k) keeps changing until you reach k=12, at which point it becomes b00 ('b' means 11) and stays that way for all k>11. The black box knows what the largest coefficient is. You don't, but you do know that it cannot be larger than 11.

Last edited: Jun 8, 2014
7. Jun 9, 2014

### dislect

Thank you all :)


# Possible mathematical induction problem

• Dec 2nd 2013, 02:03 PM
Possible mathematical induction problem
Course: Foundations of Higher MAth

Prove that $24|(5^{2n} -1)$ for every positive integer n.

This is a question from my final exam today.

P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.

Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$

P(k+1): $24|(5^{2(k+1)}-1)$

$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$

Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
• Dec 2nd 2013, 03:09 PM
Plato
Re: Possible mathematical induction problem
Quote:

Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?

A strict grader may like to have seen more grouping symbols.
However, the argument is correct.


Tricky Triangle Area Problem

This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.

Edited to add that it was a multiple choice question, with possible answers:

a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$

-
Are you familiar with the Law of Cosines? – 2012ssohn Apr 28 '14 at 23:37
Additionally, it was Multiple choice, and the four answers were $2\sqrt{6}$, $5$, $3\sqrt{6}$, and $5\sqrt{6}$ – Asimov Apr 28 '14 at 23:37
@2012ssohn Of course i am, but how can i apply that formula to it? I even tried it at the time, but couldnt get anything useful about area from it – Asimov Apr 28 '14 at 23:38
Did you actually try to use Heron's formula? The formula simplifies pretty nicely. – Nate Apr 28 '14 at 23:41
When i tried it i got an ugly glob of roots and addition that just wouldn't simplify for me. Maybe i just need to practice simplification – Asimov Apr 28 '14 at 23:44

From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$.

Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$

Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle \sin^2 \theta = \frac{24}{25}$. This means that $\displaystyle \sin \theta = \frac{2\sqrt{6}}{5}$ (note that, because $0 \le \theta \le \pi$, $\sin \theta \ge 0$).

The area of the triangle is $\displaystyle \frac{1}{2} A B \sin \theta = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{2\sqrt{6}}{5} = 2\sqrt{6}$.

-
Thank you, this is clear, concise, and could be done quickly and easily in the situation. This is probably what the judges/writers of it wanted readers to do. – Asimov Apr 28 '14 at 23:48

Use the $\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.

\begin{align} & \frac{1}{4}\sqrt{4\cdot4\cdot25-(4+25-33)^2} \\ = & \frac{1}{4}\sqrt{4^2\cdot25-4^2} \\ = & \sqrt{25-1} \\ = & 2\sqrt{6} \end{align}

-
This is the way I would have done it, nice. – Sawarnik Apr 29 '14 at 6:40

You could've used Heron's formula straight away, actually.

\begin{align} T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\ & = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7+\sqrt{33})(7-\sqrt{33})(\sqrt{33}+3)(\sqrt{33}-3)} \\ & = \tfrac{1}{4} \sqrt{(49-33)(33-9)} \\ & = \tfrac{1}{4} \sqrt{16 \cdot 24} \\ & = \tfrac{1}{4} \sqrt{64 \cdot 6} \\ & = \tfrac{8}{4} \sqrt{6} \\ & = 2 \sqrt{6} \end{align}

-

Since this was multiple-choice, I think it's worth noting that you could have guessed the right answer without doing (much) arithmetic: the diagonal of a right triangle with sides $2$ and $5$ has length $\sqrt{2^2+5^2} = \sqrt{29}$; since $\sqrt{33}$ is fairly close to this, the answer should be close to the area of a $2-5-\sqrt{29}$ triangle, which is of course $\frac12(2)(5)=5$. If you imagine how to 'stretch out' the $\sqrt{29}$ diagonal to $\sqrt{33}$, it's clear that the right angle will have to become obtuse, and this in turn means that the area of the $2-5-\sqrt{33}$ triangle will have to be less than $5$; of the provided answers, only $2\sqrt{6}\approx4.472$ is less than $5$ (and of course very close to it).

-
Where do you see the answer options (multiple choices)? Is the original problem linked to? – Jeppe Stig Nielsen Apr 29 '14 at 9:05
@JeppeStigNielsen It's now there in the question. – Ramchandra Apte Apr 29 '14 at 14:06
I like this response, and used a similar estimation method, using the areas of slightly bigger and smaller triangles that were easier to calculate – Asimov Apr 30 '14 at 0:05
I hate it when there are two good answers, and both are right and good, and you wish you could accept both as correct – Asimov Apr 30 '14 at 1:41
@JohnJPershing For what it's worth, I think 2012ssohn's answer is much better than mine; it explains how to actually derive the correct value, rather than merely how to answer the multiple-choice question. Both are valuable, but that one's likely to be more broadly applicable; I just wanted to offer an alternate approach for the test. – Steven Stadnicki Apr 30 '14 at 3:24

Let be $ABC$ the triangle. Consider the altitude $AH$ over the greatest side, the one whose length is $\sqrt{33}$. Now call $x=BH$, so $\sqrt{33}-x=CH$. Apply Pythagoras' theorem to get $$\left\{ \begin{array}{rcl} x^2+h^2&=&4\\ \left(\sqrt{33}-x\right)^2+h^2&=&25 \end{array} \right.$$

Solve for $h$ and you are (almost) done.

-


# Permutation with Condition

The numbers 1-9 are drawn from a hat at random. Without repetition of numbers, how many numbers (with all digits drawn-- eg. 987654321) would be divisible by 11?

First I recalled that a number is divisible by 11 if the number's alternating digits minus the other digits are equal to a number divisible by 11.

Honestly, I did a bit of guess and check to follow and found that 948576321 was an answer divisible by 11. Therefore if I rearranged 9,8,7,3, and 1 (the leading alternating numbers) it would bring the same result. So I solved the permutation to receive 120 possible combinations. I then took the permutation of 4,5,6, and 2 and received a total of 24 combinations.

I multiplied these two answers together to get a grand total of 2880 combinations that would be divisible by 11.

-
I ran a quick python program, and the answer is 31680. I'll try to prove it in a few minutes. –  Alfonso Fernandez Feb 20 '13 at 22:33
I would appreciate it :) –  Hannah Feb 20 '13 at 23:06

The correct answer is 31680. Your combination of alternating numbers (9, 8, 7, 3, 1) is only one out of 11 possibilities:

1. 1, 2, 3, 4, 7
2. 1, 2, 3, 5, 6
3. 1, 3, 7, 8, 9
4. 1, 4, 6, 8, 9
5. 1, 5, 6, 7, 9
6. 2, 3, 6, 8, 9
7. 2, 4, 5, 8, 9
8. 2, 4, 6, 7, 9
9. 2, 5, 6, 7, 8
10. 3, 4, 5, 7, 9
11. 3, 4, 6, 7, 8

So we end up with $11 \cdot 5! \cdot 4! = 11 \cdot 120 \cdot 24 = 31680$.

-
But how do you come up with these 11 without checking all $\binom{9}{4}=126$ possibilities? –  Alfonso Fernandez Feb 21 '13 at 0:21
@AlfonsoFernandez I did it by checking all of them with a script. Brian's approach is more rigorous in that regard. –  ralph Feb 23 '13 at 14:40
(This kind of problem, finding a set of distinct numbers to add up to a certain number, is a key part of Kakuro puzzles. The list of valid combinations for this can be found in Kakuro tables such as brainbashers.com/combinations.asp under 5-digit combinations with a sum of either 17 or 28. Brian's answer proves why the sum must be 17 or 28.) –  ralph Feb 23 '13 at 14:47

Here’s a systematic approach; it’s still a bit tedious in spots, but not unreasonably so.

The sum of all $9$ digits is $45$. You need to split the digits into a set $A$ of $5$ digits, those in the odd-numbered positions, and a set $B$ of $4$ digits, those in the even-numbered positions, whose sums differ by a multiple of $11$. If $a$ is the sum of the digits in $A$ and $b$ the sum of the digits in $B$, then $a-b$ is a multiple of $11$, and $a+b=45$.

Clearly $b=45-a$, so $a-b=a-(45-a)=2a-45$. The smallest possible value of $a$ is $1+2+3+4+5=15$, and the largest is $9+8+7+6+5=35$, so $-15\le a-b\le25$. The multiples of $11$ in that range are $-11,0,11$, and $22$. However, $2a-45$ is clearly odd, so in fact the only possible values of $a-b$ are $-11$ and $11$.

• If $a-b=2a-45=-11$, then $a=17$.
• If $a-b=2a-45=11$, then $a=28$.

What sets of $5$ digits sum to $17$? Since $1+2+3+4+5=15$, we either increase $5$ by $2$ to get the set $\{1,2,3,4,7\}$ or increase $4$ and $5$ by $1$ each to get the set $\{1,2,3,5,6\}$; anything else produces either too big a sum or duplicated digits.

It’s a bit harder to enumerate the sets that sum to $28$, because there are more of them. A good starting point is the fact that $9+8+7+6+5=35$, which is $7$ too large. First note that we need to have at least one of $8$ and $9$: $7+6+5+4+3=25$ is too small. What can we do without the $9$? $8+7+6+5+4=30$, so, much as in the previous case, our only choices are to subtract $2$ from $4$ to get the set $\{8,7,6,5,2\}$ or subtract $1$ each from $4$ and $5$ to get $\{8,7,6,4,3\}$. Every other set will include the $9$.

Let’s try next for the ones that include $9$ but not $8$. $9+7+6+5+4=31$; we can subtract $3$ from the $4$, the $5$, or the $6$ to get the sets $\{9,7,6,5,1\}$, $\{9,7,6,4,2\}$, and $\{9,7,5,4,3\}$. A little experimentation shows that trying to reduce the sum by $3$ by subtracting $2$ from one digit and $1$ from another produces nothing new.

The last batch will be those containing both $9$ and $8$. If we keep the $7$ as well, we already have a total of $9+8+7=24$, and the only two digits that supply the missing $4$ are $1$ and $3$, giving us the set $\{9,8,7,3,1\}$. If we start with $9,8$, and $6$, we can get the missing $5$ from either $1$ and $4$ or $2$ and $3$, giving us the sets $\{9,8,6,4,1\}$ and $\{9,8,6,3,2\}$. If start with $9,8$, and $5$, the only way to get the missing $6$ is with $2$ and $4$, giving us the set $\{9,8,5,4,2\}$. And $9+8+4+3+2=26$ is too small, so we’ve found all of the possibilities for the set $A$.

As you already observed, we can permute $A$ and $B$ arbitrarily, so each choice of $A$ yields $5!4!$ numbers, and there are $11$ possibilities for $A$, for a grand total of $11\cdot5!4!=31,680$ numbers.

-
Every number of the form $x*11$ is divisible by 11. Now you must think about $x$.
$x$ must be such that $(x)*11$ is a number with 9 digits! If it is useful for you I can explain it more.


# The Asymptotic Bode Diagram for Non-Minimum Phase Poles and Zeros

## A Real Pole with Negative ω0

Elsewhere we have discussed how to make Bode plots for a real pole. You should be familiar with that analysis. The discussion there assumed that the value of ω0 was positive; here we discuss the case if ω0 is negative. We start with

$$H(s)=\frac{1}{1+\frac{s}{\omega_0}}$$

#### Magnitude

If you carefully examine the analysis (here) of the "Magnitude" plot you'll see that the only time ω0 is used, it is squared (e.g., $\left( \frac{\omega}{\omega_0} \right)^2$). Therefore, the magnitude plot does not depend on the sign of ω0, only its absolute value, so we don't need to change anything to accomodate a negative value of ω0.

#### Phase

The phase however does change. The phase of a single real pole is given by is given by

$$\angle H\left( {j\omega } \right) = \angle \left( {{1 \over {1 + j{\omega \over {{\omega _0}}}}}} \right) = - \angle \left( {1 + j{\omega \over {{\omega _0}}}} \right) = - \arctan \left( {{\omega \over {{\omega _0}}}} \right)$$

Let us again consider three cases for the value of the frequency, but we assume ω0 is negative:

Case 1) ω<<ω0.  This is the low frequency case with ω/ω0→0, and doesn't depend on the sign of ω0.  At these frequencies We can write an approximation for the phase of the transfer function

$$\angle H\left( {j\omega } \right) \approx -\arctan \left( 0 \right) = 0^\circ = 0\;rad$$

Case 2) ω>>ω0.  Here we will consider the cases of positive and negative ω0 side by side.

ω0 > 0    (the minimum phase case, discussed previously)

This is the high frequency case with ω/ω0 → +∞.  We can write an approximation for the phase of the transfer function

$$\angle H\left( {j\omega } \right) \approx - \arctan \left( \infty \right) = - 90^\circ$$

The high frequency approximation is at shown in green on the diagram below.  It is a horizontal line at -90°.

ω0 < 0    (the non-minimum phase case)

This is the high frequency case with ω/ω0 → -∞.  We can write an approximation for the phase of the transfer function

$$\angle H\left( {j\omega } \right) \approx - \arctan \left(- \infty \right) = + 90^\circ$$

The high frequency approximation is at shown in green on the diagram below.  It is a horizontal line at +90°.

Case 3) ω=ω0. Again consider the cases of positive and negative ω0 side by side.

ω0 > 0    (the minimum phase case, discussed previously)

$$\angle H\left( {j\omega } \right) = - \arctan \left( 1 \right) = - 45^\circ$$

ω0 < 0    (the non-minimum phase case)

$$\angle H\left( {j\omega } \right) = - \arctan \left( -1 \right) =+ 45^\circ$$

From the above discussion you can see that the only effect of the pole having a negative value of ω0 is that the phase is inverted (it increases from 0 → +90° as ω increases from 0 → ∞. The images below show the Bode plots for

$$H_1(s)=\frac{1}{1+\frac{s}{10}}, \quad\quad H_2(s)=\frac{1}{1-\frac{s}{10}}$$

H1(s) has a positive ω00=+10, so H1(s)=1/(1+s/10)) and H2(s) has a negative ω00=-10, so H2(s)=1/(1-s/10)). The pole of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero)
H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.

The same conclusion holds for first order poles and second order poles and zeros (see below).

## A Real Zero with Negative ω0

The images below show the Bode plots for two functions, one with a positive ω00=+10) and one with a negative ω00=-10). The zero of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero).

$$H_1(s)=1+\frac{s}{10}, \quad\quad H_2(s)=1-\frac{s}{10}$$

H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites. Recall that 360° is equivalent to 0° so you can think of the plot for the angle of H2(s) as starting at 0° and dropping by 90° (though the plot shows it as starting at 360°).

## A Second Order Pole with Negative ζ

The images below show the Bode plots for (note the sign of the middle term in the numerator is different)

$$H_1(s)=\frac{1}{1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}, \quad\quad H_2(s)=\frac{1}{1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}$$

The poles of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+0.5±j9.987 (a positive real part, the right half of the s-plane; a non-minimum phase pole). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that, again, the magnitudes are identical, but the phases are opposites.

## A Second Order Zero with Negative ζ

The images below show the Bode plots for (note the sign of the middle term in the numerator is different)

$$H_1(s)=1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2, \quad\quad H_2(s)=1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2$$

The zeros of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+0.5±j 9.987(a positive real part, the right half of the s-plane; a non-minimum phase zero). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.

##### Key Concept: For poles and zeros with positive real parts, phase is inverted

If a 1st order pole has a positive real part (i.e., a nonminimum phase system, pole is in right half of s-plane) at say s=+5, so

$$H(s)=\frac{1}{1 - \frac{s}{5}}$$

the magnitude of the Bode plot is unchanged from the case of a corresponding pole with negative value, at s=-5 (pole is in left half of s-plane)

$$H(s)=\frac{1}{1 + \frac{s}{5}}$$

but the phase of the plot is inverted.

The same rule holds Bode plots for 2nd order (complex conjugate) poles, and for 1st and 2nd order zeros.

References

Replace


# Number of ways to two-color a sectored circle so that opposite sectors are different colors.

Given a circle divided into 2n equal sectors, how many ways can it be painted with two colors so that opposite sectors are not the same color? I'm interested in the number of solutions that are unique under rotation.

I've tried to enumerate the early situations and see patterns but haven't seen a path to a general solution. I think I only have three rules of thumb at the moment:

1. There will be n sectors of each color
2. An alternating paint job is only legal when n is odd.
3. I think we're bounded on top by 2^(n-2) for n>1.

I may have missed some examples but this link has my work so far: http://i.imgur.com/ZQylapu.png

If I've gotten everything, then the series goes:

• n=1, 1
• n=2, 1
• n=3, 2
• n=4, 2
• n=5, 4
• n=6, 6

n=6 is notable as two of the combinations are left- and right-handed versions of a similar arrangement. I imagine this becomes much more common beyond n=6. I counted these separately, but a general solution that wouldn't would be nice as well.

• How do you define opposite sector? If each sector has just one opposite then why wouldn't they answer by $2^n$? – fleablood Oct 12 '15 at 0:31
• Oh, I guess the unique under rotation puts a damper on it. – fleablood Oct 12 '15 at 0:33
• Huh, I have difficulty putting words to what I mean by opposite. What comes to mind is two sectors are opposites if the number of sectors between them in the clockwise direction is equal to the number in the counter-clockwise direction. – Polyhog Oct 12 '15 at 0:37
• No I get your definition. As a pair of sectors has no relation to other sectors the answer would be $2^n$ as each pair of sectors has two choices. But if they are unique to rotation, we have to calculate how many are equivalent under rotation. Good question. – fleablood Oct 12 '15 at 0:54
• Here's a thought. Fix sector 1. then for sector i = 2 to n give a binary value for color and add $2^i$ for green and nothing for yellow. Thus you'll have a 1-1 correspondence between colorings and the numbers from 1 to $2^{n - 1}. Now, we have to come up with a formula for determining if two numbers represent the same coloring under rotation. – fleablood Oct 12 '15 at 1:00 ## 1 Answer This can be done using Burnside's lemma. We need to count the assignments of colors being fixed by each type of permutation from among the$2n$permutations total in the cycle index$Z(C_{2n})$of the cyclic group acting on the sectors. There are $$\varphi(d)$$ permutations having cycle structure $$a_d^{2n/d}$$ in the cycle index$Z(C_{2n})$where$d|2n.$For a permutation to fix an assignment it needs to be constant on the cycles. Suppose that the cycle length$d$is even and pick one of its elements. The slot opposing this element is also located on this cycle. But they must have different colors, so there are no valid assignments fixed by a permutation of shape$a_d^{2n/d}$when$d$is even. On the other hand when$d$is odd the$2n/d$cycles of length$d$are grouped into$n/d$pairs which are reflections of each other and hence must have opposite colors. Therefore there are two possible assignments of colors to these cycles forming a pair (as opposed to four if there were no constraint). This yields a contribution of $$\varphi(d) 2^{n/d}.$$ Averaging this over the total$2n$permutations we get $$\frac{1}{2n} \sum_{d|2n,\;d\;\mathrm{odd}} \varphi(d) 2^{n/d}$$ which is $$\frac{1}{2n} \sum_{d|n,\;d\;\mathrm{odd}} \varphi(d) 2^{n/d}.$$ This yields the sequence $$1, 1, 2, 2, 4, 6, 10, 16, 30, 52, 94, 172, 316, 586, 1096, \\ 2048, 3856, 7286,\ldots$$ which is OEIS A000016 where additional material awaits. I found the OEIS entry by using a simple total enumeration algorithm (definitely not optimized) to compute the first few values (practical to about$n=10\$ which is enough to conclusively identify the sequence). This was the Maple code:

with(numtheory);

sectors :=
proc(n)
option remember;
local d, ind, orbits, orbit, rot, pos;

orbits := {};

for ind from 2^(2*n) to 2*2^(2*n) - 1 do
d := convert(ind, base, 2);

for pos to n do
if d[pos] = d[pos+n] then
break;
fi;
od;

if pos = n+1 then
orbit := {};
for rot from 0 to 2*n-1 do
orbit := {op(orbit),
[seq(d[q], q=1+rot..2*n),
seq(d[q], q=1..rot)]};
od;

orbits := {op(orbits), orbit};
fi;
od;

nops(orbits);
end;

Q :=
proc(n)
1/2/n*
select(d->type(d,odd), divisors(n)));
end;

• That is brilliant. The equation you reached is also is the same for number of binary sequences coming from a shift register complementing the last output (equivalent to going through the circle and outputting the sector's color as binary number). Thank you! – Weaam Oct 12 '15 at 3:04
• This was not original work. The calculation is just about included in the OEIS entry, it only needed filling in the details. – Marko Riedel Oct 12 '15 at 3:06
• Still! It is a great answer. – Weaam Oct 12 '15 at 3:22


# Computing a Limit using the Limit Definition

I've just started Real Analysis. In the textbook (Real Analysis and Applications, by Davidson and Donsig) they have defined the limit of a sequence. I am working on one of the provided exercises. No suggested solutions are provided for any questions, so I am looking for help checking my work to make sure I understand what I am doing. The exercise is as follows.

Compute the limit. Then, using $\epsilon =10^{-6}$, find an integer $N$ that satisfies the limit definition. $$\lim_{n \to \infty} \frac{1}{ln(ln(n))}$$

Firstly, I just want to address the limit definition. In the text, they define a real number $L$ to be the limit of a sequence of real numbers $(a_n)^{\infty}_{n=1}$ if for every $\epsilon > 0$ there is an integer $N=N(\epsilon)>0$ such that

$$|a_n - L|< \epsilon$$

for all $n\geq N$. Now for my informal interpretation.

I believe the idea is that after some point the distance between $a_n$ and the limit $L$ can be made arbitrarily small by choosing a sufficiently large $N$. But why does $N$ have to be a function of $\epsilon$? Any clarification of the definition would be appreciated.

Now, for the exercise. I know the divisor of the fraction of the given sequence approaches infinity as $n$ approaches infinity and as such the sequence approaches zero as $n$ approaches infinity. So I claim that $L=0$. I observe that

$$\left | \frac{1}{ln(ln(n))} - 0 \right | = \frac{1}{ln(ln(n))}$$

So now I need an $N$ such that, for all $n \geq N$, I get $|a_n - 0|<10^{-6}$. Now, I am not sure if my next steps are correct. I need an $N$ such that

$$\frac{1}{ln(ln(N))}<10^{-6}$$

So I just solve for N in the equality, which yields

$$N > e^{e^{1000000}}$$

Hence, for any $n \geq N$,

$$\frac{1}{ln(ln(n))}<10^{-6}$$

Do I find the closest integer to $e^{e^{1000000}}$? Any clarification would be appreciated.

-
It looks to me like you are done. You could compute $N$, but keep in mind it doesn't hurt for $N$ to be higher. It doesn't have to be the minimal one over $e^{e^{1000000}}$ –  rschwieb Jan 10 '13 at 21:46
You can just write your integer as $\lceil e^{e^{1000000}} \rceil$ to make it clear you're talking about an integer. Other than that you have found a value that works and so you are done. –  AvatarOfChronos Jan 10 '13 at 21:49
I suspect that computing $e^{e^{1000000}}$ would be a major project, requiring quite a bit of programming skill. It doesn't make sense to try to compute the exact value in this context at all. –  Henry B. Jan 10 '13 at 22:33
it's a very big number, even computers are afraid to compute it. –  Santosh Linkha Jan 10 '13 at 22:43

I think your confusion arises from the phrase "there exists an $N = N(\epsilon)$. You already showed why $N$ has to be a function of $\epsilon$: intuitively, the smaller the epsilon, the larger the $N$ has to be in order for the inequality to be satisfied.

You could just say then: choose $N > e^{e^{\epsilon^{-1}}}$ for $N(\epsilon)$; if you show such an $N$ exists then this implies that such a function exists. You don't have to exhibit a specific one, unless you want to (or asked on a question).

But, if you want to be explicit, you could use:

• $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil$ where $\lceil - \rceil$ denotes the least integer greater than its argument, as you thought
• $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil + 8434$
• $N(\epsilon) = \lceil 9434\pi e^{e^{\epsilon^{-1}}}\rceil$

See, all of them work, as long as the function gives you an integer sufficiently larger so that the $|a_n - L| < \epsilon$. However, I should stress once more that as long as you show that there exists such an integer for every $\epsilon$, this already shows that the function exists. (Modulo your philosophy on mathematics; you might actually need to construct the function for your peace of mind.)

-


# Why does this not seem to be random?

I was running a procedure to be like one of those games were people try to guess a number between 1 and 100 where there are 100 people guessing.I then averaged how many different guesses there are.

from random import randint

def averager(times):
tests = list()
for _ in range(times):
l = [randint(0,100) for _ in range(100)]
tests.append(len(set(l)))
return sum(tests)/len(tests)

print(averager(100))

For some reason, the number of different guesses averages out to 63.6

Why is this?Is it due to a flaw in the python random library?

In a scenario where people were guessing a number between 1 and 10

The first person has a 100% chance to guess a previously unguessed number

The second person has a 90% chance to guess a previously unguessed number

The third person has a 80% chance to guess a previously unguessed number

and so on...

The average chance of guessing a new number(by my reasoning) is 55%. But the data doesn't reflect this.

-
Shades of (1-1/e)? Note that $1 - 1/e = 0.63212$ Can you see why it has to be this? I thought I had the answer but not sure anymore. Got to look at it some more – user44197 Jan 14 '14 at 3:33
If you want more info about this problem, there's a classical one involving the same maths. It's the odds of having two people(or more) with the same birthday in a classroom. – Feu Jan 14 '14 at 3:42
@Feu an important one in cryptography too :) – Cruncher Jan 14 '14 at 14:16
A related question I asked before. Turns out $e^{-1}$ is a notable probability. Here's an older related question. – badroit Jan 14 '14 at 14:57
From the programming point of view, don't forget that python's basic random functions are not true random, or even cryptographic random. It is a seeded PRNG. If you want better random numbers, try a cryptographic number generator source such as /dev/random (full hardware generated randomness, runs out easily). – Linuxios Jan 14 '14 at 16:40

Suppose that $n$ guesses are made. For $i=1$ to $100$, let $X_i=1$ if $i$ is not guessed, and let $X_i=0$ otherwise. If $$Y=X_1+X_2+\cdots +X_{100},$$ then $Y$ is the number of numbers not guessed.

By the linearity of expectation, we have $$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_{100}).$$ The probability that $i$ is not chosen in a particular trial is $\frac{99}{100}$, and therefore the probability it is not chosen $n$ times in a row is $\left(\frac{99}{100}\right)^n$. Thus $$E(Y)=100 \left(\frac{99}{100}\right)^n.$$

In particular, let $n=100$. Note that $\left(1-\frac{1}{100}\right)^{100}\approx \frac{1}{e}$, so the expected number not guessed is approximately $\frac{100}{e}$. Thus the expected number guessed is approximately $63.2$, a result very much in line with your simulation.

In general, if $N$ people choose independently and uniformly from a set of $N$ numbers, then the expected number of distinct numbers not chosen is $$N\left(1-\frac{1}{N}\right)^N.$$ Unless $N$ is very tiny, this is approximately $\frac{N}{e}$, and therefore the expected number of distinct numbers chosen is approximately $N-\frac{N}{e}$. Note that the expected proportion of the numbers chosen is almost independent of $N$.

-
I think you are missing a "not": "then the expected number of distinct numbers [not] chosen is" – aaaaaaaaaaaa Jan 14 '14 at 19:19
The expected number not chosen is approximately $\frac{N}{e}$. Since there are $N$ numbers overall, the expected number chosen is approximately $N-\frac{N}{e}$. – André Nicolas Jan 14 '14 at 19:32
Yes, and you wrote the opposite in the second to last paragraph. – aaaaaaaaaaaa Jan 14 '14 at 19:34
Thank you, I finally found where the missing not was (not), had been looking for it towards the very end. Will fix. – André Nicolas Jan 14 '14 at 19:42

In your 1 to 10 example, it's not true in general that the third chooser has an 80% chance on choosing a new number. It's only the case if the second one has guessed a different number than the first one.

-

This is an example of the Birthday Paradox / Birthday Problem.

Birthday problem - Wikipedia, the free encyclopedia

I was just looking at my Online Cryptography class video lecture today on this very problem.

Coursera.org: crypto-009

There is an apparent paradox that there is more duplication of numbers than expected when the random numbers are supposedly independent.

But the Birthday Paradox is just one example of when our intuitive statistical sense is dead wrong.

-
Here is a cool video from Numberphile about the Birthday problem. youtube.com/watch?v=a2ey9a70yY0 – Rik Jan 14 '14 at 12:04
The birthday problem is about the probability of having at least one collision. It says very little about the number of different values found in a large number of independent samples (as in "how many different birthdays on average in a random group of $365$ people"). – Marc van Leeuwen Jan 14 '14 at 13:27


# Algorithm To Compute The Gaps Between A Set of Intervals

Problem

Given a set of intervals with possibly non-distinct start and end points, find all maximal gaps. A gap is defined as an interval that does not overlap with any given interval. All endpoints are integers and inclusive.

For example, given the following set of intervals:

$$\{[2,6], [1,9], [12,19]\}$$

The set of all maximal gaps is:

$$\{[10,11]\}$$

For the following set of intervals:

$$\{[2,6], [1,9], [3,12], [18,20]\}$$

The set of all maximal gaps is:

$$\{[13,17]\}$$

because that produces the maximal gap.

Proposed Algorithm

My proposed algorithm (modified approach taken by John L.) to compute these gaps is:

1. Order the intervals by ascending start date.
2. Initialise an empty list gaps that will store gaps
3. Initialise a variable, last_covered_point, to the end point of the first interval.
4. Iterate through all intervals in the sorted order. For each interval [start, end], do the following.
1. If start > last_covered_point + 1, add the gap, [last_covered_point + 1, start - 1] to gaps.
2. Assign max(last_covered_point, end) to last_covered_point.
5. Return gaps

I have tested my algorithm on a few cases and it produces the correct results. But I cannot say with 100% guarantee that it works for every interval permutation and combination. Is there a way to prove this handles every permutation and combination?

• @JohnL. yes the question is correct now - thanks
– Mojo
Dec 18 '20 at 21:25

The key to prove your algorithm is correct is to find enough invariants of the loop, step 4 so that we apply use mathematical induction.

Let $$I_1, I_2, \cdots, I_n$$ denote the sorted intervals. When the algorithm has just finished processing $$I_i$$, we record the values of gaps and last_covered_point as $$\text{gaps}_i$$ and $$\text{last_covered_point}_i$$ respectively.

Let us prove the following proposition, $$P(i)$$, for $$i=1, 2, \cdots, n$$.

$$\text{gaps}_i$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_i$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$.

When $$i=1$$, $$\text{gaps}_1$$ is the empty set and $$\text{last_covered_point}_1$$ is the right endpoint of $$I_1$$. So $$P(1)$$ is correct.

For the sake of mathematical induction, assume $$P(i)$$ is correct, where $$1\le i\lt n$$. Let $$I_{i+1}=[s, e]$$. There are two cases.

1. If $$s\gt\text{last_covered_point}_i+1$$, then $$\text{gaps}_i\cup[\text{last_covered_point}_i +1, s-1]=\text{gaps}_{i+1}.$$
Let $$m$$ be any point between the start point of $$I_1$$ and the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$. Suppose $$m$$ not covered by any of $$I_1, I_2, \cdots, I_{i+1}$$.

• If $$m\le\text{last_covered_point}_i$$, the induction hypothesis says that $$m$$ is covered by some interval in $$\text{gap}_i$$.
• Otherwise, $$m\gt\text{last_covered_point}_i$$. Since $$m$$ is not covered by $$I_{i+1}$$, we know $$m. So $$m$$ is covered by $$[\text{last_covered_point}_i +1, s-1]$$.

In both cases, $$m$$ is covered by some interval in $$\text{gaps}_{i+1}$$. Since $$\text{last_covered_point}_i$$ is the largest point covered by one of $$I_1, I_2, \cdots, I_i$$ and $$s$$ is the smallest point covered by $$I_{i+1}$$, $$[\text{last_covered_point}_i +1, s-1]$$ is a maximal gap.

2. Otherwise, we have $$s\le\text{last_covered_point}_i$$+1. We can also verify that $$\text{gaps}_{i+1}=\text{gaps}_{i}$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_{i+1}$$.

Finally, since step 4.2 says $$\text{last_covered_point}_{i+1}=\max(\text{last_covered_point}_i, e)$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$, $$\text{last_covered_point}_{i+1}$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$.

So, $$P(i+1)$$ is correct. $$\quad\checkmark$$.

• I probably didn't make my post precise enough and so I think you may have misunderstood what I meant by a gap. I have cleaned up my original post to define the problem better but I also took your example and customised it to what I think the algorithm should be. The main issue I have is that I am not sure if I have covered all possible interval permutations.
– Mojo
Dec 17 '20 at 15:19
• @Mojo, thanks for noticing that last_covered_point should be updated in step 4.2 as well. Dec 18 '20 at 16:30
• @Mojo, I just wrote a proof. The proof is not completely formal, since "maximal gap" is given by a definition (although it is easy to define) and the case 2 is not proved in detail. However, the idea should be clear enough. Dec 20 '20 at 5:24
• In my last comment, 'since "maximal gap" is given by a definition' should have been 'since "maximal gap" is not given by a definition'. Jan 4 at 1:35

I'd like to propose a quite simple algorithm as well. The idea is this: we're going to place open and close parentheses on the number line at the boundaries of each interval. For example, for the intervals $$(1,5), (2,7), (9, 10)$$, the number line would look like this:

1 2 3 4 5 6 7 8 9 10
( (     )   )   ( )

Then we'll just scan left to right, counting parentheses. When all the parentheses get closed, we start a gap. Note in particular that in the above diagram, we have lost the information about which close parenthesis is paired with which open parenthesis -- because it doesn't actually matter. So:

1. Convert each interval $$(a,b)$$ to pairs $$(a,\mathtt o),(b,\mathtt c)$$. ($$\mathtt o$$ and $$\mathtt c$$ are for open and close, respectively.)
2. Sort all the pairs you get from this process. (When sorting, use lexicographic ordering and $$\mathtt o < \mathtt c$$.)
3. Iterate through them, keeping a counter that starts from $$0$$.
• When you see a pair with $$\mathtt o$$ in the second part, increment the counter.
• When you see a pair with $$\mathtt c$$ in the second part, decrement the counter.
• When you decrement the counter, if that causes it to drop to $$0$$, then look at the next element of the list to decide what to do; empty list means you're done, otherwise if the next pair's first part is just one bigger than the current one's you do nothing, and in the last case you record a maximal gap between the current end point and the next open point.


# Why is $x^n\approx \left(n(x^{1/4096}-1)+1\right)^{4096}$?

There's an old-school pocket calculator trick to calculate $x^n$ on a pocket calculator, where both, $x$ and $n$ are real numbers. So, things like $\,0.751^{3.2131}$ can be calculated, which is awesome.

This provides endless possibilities, including calculating nth roots on a simple pocket calculator.

The trick goes like this:

1. Type $x$ in the calculator
2. Take the square root twelve times
3. Subtract one
4. Multiply by $n$
6. Raise the number to the 2nd power twelve times (press * and = key eleven times)

Example:

I want to calculate $\sqrt[3]{20}$ which is the same as $20^{1/3}$. So $x=20$ and $n=0.3333333$. After each of the six steps, the display on the calculator will look like this:

1. $\;\;\;20$
2. $\;\;\;1.0007315$
3. $\;\;\;0.0007315$
4. $\;\;\;0.0002438$
5. $\;\;\;1.0002438$
6. $\;\;\;2.7136203$

The actual answer is $20^{1/3}\approx2.7144176$. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.

Question:

So the question is now, why does this trick work? More specifically, how do we prove that: $$x^n\approx \Big(n(x^{1/4096}-1)+1\Big)^{4096}$$

Note: $4096=2^{12}$.

I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.

I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:

$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$

This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. $a=12$ is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with $a\to\infty$ can be useful in proving why this trick works, however I still can't of think a formal proof for this.

Thank you for your time :)

• Which one would you like to prove, the approximation for $a = 12$, or just the limit expression for $a \to \infty$? – Xiangxiang Xu Sep 7 '18 at 2:01
• In my opinion the one with the limit with $a\to\infty$ is better to prove because it is more general. From there, the specific case of $a=12$ can be simpler to prove (I think) – KKZiomek Sep 7 '18 at 2:04

A standard trick is to calculate the natural logarithm first to get the exponent under control:

$$\log(\lim_{a\to\infty}(n(x^{1/a}-1)+1)^a)=\lim_{a\to\infty}a\log(nx^{1/a}-n+1)$$ Set $u=1/a$. We get $$\lim_{u\to 0}\frac{\log (nx^u-n+1)}{u}$$ Use L'Hopital: $$\lim_{u\to 0}\frac{nx^u\log x}{nx^u-n+1}=n\log x=\log x^n$$ Here we just plugged in $u=0$ to calculate the limit!

So the original limit goes to $x^n$ as desired.

• I was confused first at L'Hopital step, but then I realized that I have to differentiate with respect to $u$ not $x$ :). I'm giving this answer the best even though all other answers were equally good, because if you haven't mentioned L'Hopital, I would never think of using it. Thank you for the answer. – KKZiomek Sep 7 '18 at 2:19
• Glad to help. It's a neat problem. – Cheerful Parsnip Sep 7 '18 at 2:24

For fixed $x > 0$ and $n$, let $t = 1/2^a \to 0$. Then we need to prove that $$\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t} = x^n.$$ In fact, we have $$\ln \left[\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t}\right] = \lim_{t \to 0} \frac{\ln (1 + n(x^t - 1))}{t} = \lim_{t \to 0} \frac{n(x^t - 1)}{t} = n\ln x,$$ where the first equality follows from the continuity of $\ln(x)$, and the second equality has used the fact that $\ln(1 + x) \sim x$ when $x \to 0$.

If $x$ (actually $\ln x$) is relatively small, then $x^{1/4096}=e^{(\ln x)/4096}\approx1+(\ln x)/4096$, in which case

$$n(x^{1/4096}-1)+1)\approx1+{n\ln x\over4096}$$

If $n$ is also relatively small, then

$$(n(x^{1/4096}-1)+1)^{4096}\approx\left(1+{n\ln x\over4096}\right)^{4096}\approx e^{n\ln x}=x^n$$

Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,

$$\left({1\over3}(20^{1/4096}-1)+1\right)^{4096}=2.7150785662\ldots$$

is more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\ldots$, both of which are too large. Very curiously, if you average the two calculator results, you get

$${2.7136203+2.7152613\over2}=2.7144408$$

which is quite close to the true value!


## Wednesday, 2 December 2020

### On the Colour of the Third Card

Three cards are dealt from a normal deck.  You don't see them.  You are told that the first two are the same colour, but not what colour they are.  What is the probability that the next card is the same colour?

$\frac{1}{2},\;\frac{1}{4},\;\frac{{12}}{{25}},\;or\;\frac{4}{{17}}\;?$

This beautiful, onthefaceofit innocuous little problem comes from the great Martin Gardner's 'Modelling Mathematics with Playing Cards'.  It's a problem that never fails to invoke heated discussion and vehement argument when I share it with students to play with.  The four solutions and reasoning sketched out below are those typically proffered (and invariably staunchly defended) by students.  Which solution would you go with, and why?  Do you have a different solution?  And what argument would you give to those who firmly hold the solution to be one of the others to show them that they are wrong and you are right?  (Note that Gardner's solution was solution 2 [1])

## Solution 1

The last card could either be 1) the same colour by being red, or 2) a different colour by being red, or 3) the same colour by being black, or 4) a different colour by being black.  So there are four possibilities and two of these result in the final card being the same colour as the previous two.  So:

$\frac{2}{4} = \frac{1}{2}$

## Solution 2

There are eight ways that the arrangement of the colours of the three cards can occur, namely RRR, RRB, RBR, BRR, RBB, BRB, BBR, or BBB (where R = Red and B = Black), and two of these arrangements have the final card as the same colour as the previous two.  So:

$\frac{2}{8} = \frac{1}{4}$

## Solution 3

After the first two cards, fifty cards remain in the deck.  If the first two cards are both red, there remain twenty-four cards that are red.  Similarly, if the first two cards are both black, there remain twenty-four cards that are black.  So:

$\frac{24}{50} = \frac{12}{25}$

## Solution 4

For the last card to be the same colour as the first two, the colours of all three cards are either RRR or BBB.  You have fifty-two cards to choose from for your first card, and twenty-six of these are red and twenty-six are black.  After taking the first card from the deck, you have fifty-one cards left to choose from for your second card, and assuming that the first card was red, twenty-five of the fifty-one cards left are also red.  After taking the second card from the deck, you have fifty cards left to choose from for your third and final card, and assuming that the first two cards were red, twenty-four of the fifty cards left are also red.  The same would be true if the cards pulled from the deck were black.  So:

$\left( {\frac{{26}}{{52}} \times \frac{{25}}{{51}} \times \frac{{24}}{{50}}} \right) \times 2 = \frac{{31200}}{{132600}} = \frac{4}{{17}}$