# distance between two points on a number line calculator

Line distance equation calculator solving for distance between two points given x1, x2, y1 and y2 To use the distance formula, we need two points. This web site owner is mathematician Miloš Petrović. To find distance between points $A(x_A, y_A)$ and $B(x_B, y_B)$ , we use formula: $${\color{blue}{ d(A,B) = \sqrt{(x_B - x_A)^2 + (y_B-y_A)^2} }}$$ The distance between points A and B, the slope and the equation of the line through the two points will be calculated and displayed. So it is a distance between two points calculator. Do the same with the second point, this time as x₂ and y₂. Find the horizontal and vertical distance between the points. Replace the values in the distance formula. Consider a number line with only one dimension. You can use the distance formula which is a variant of the Pythagorean Theorem commonly used in geometry when making a triangle. Take the first point's coordinates and put them in the calculator as x₁ and y₁. In such cases, you can either perform manual computations or use this distance formula calculator. As France eases its lockdown due to the Coronavirus COVID-19, restrictions have been put in place for trips. In this example, let’s use the distance in a line or 1D. In this example, let’s use the distance in a line or 1D. First, subtract y2 - y1 to find the … Find the distance from the line $3x + 4y - 5 = 0$ to the point point $( -2, 5)$. Calculator Use. It works for (easier to reason through) 1, 2, or 3 dimensions, plus 4, 5, and 6 dimensions as well. Depending on the dimension the distance between two points can be found using the following formulas: The formula for calculating the distance between two points A(x a, y a) and B(x b, y b) on a plane: AB = √ (x b - x a) 2 + (y b - y a) 2 Canada Distance Chart (Distance Table): For your quick reference, below is a Distance Chart or Distance Table of distances between some of the major cities in Canada. How to find the distance between two points. Draw lines which form a triangle with a right-angle while using these coordinates as the points of the triangle’s corners. These points can exist in any kind of dimension. ,$\color{blue}{ \text{ 2r(3/5) }= 2 \sqrt{\frac{3}{5}}}$. I designed this web site and wrote all the lessons, formulas and calculators . How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. Consider a number line with only one dimension. Click Calculate Distance, and the tool will place a marker at each of the two addresses on the map along with a line between them. We can redo example #1 using the distance formula. Distance between cities or 2 locations are measured in both kilometers, miles and nautical miles at the same time. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. This online calculator can find the distance between a given line and a given point. This calculator will find the equation of a line (in the slope-intercept, point-slope and general forms) given two points or the slope and one point, with steps shown. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. After you’ve made the computations by hand, you can use the distance between two points calculator to check if you performed the calculation correctly.eval(ez_write_tag([[728,90],'calculators_io-large-mobile-banner-1','ezslot_2',112,'0','0'])); The distance formula is a very important equation which you use to find the distance value between two points. How to use the distance formula calculator? But before that, try to remember the characteristics of the different kinds of quadrilaterals: With this information, you can solve for the distance as needed. 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# In a certain sculpture, coils of wire are arranged in rows.

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In a certain sculpture, coils of wire are arranged in rows.  [#permalink]

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02 Jan 2010, 14:44
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In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18
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02 Jan 2010, 15:01
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x13069 wrote:
Can anyone help with this one?

In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

1)22
2)21
3)20
4)19
5)18

We have arithmetic progression with common difference $$d=2$$, number of terms $$n=10$$ and the sum $$S=120$$.

The sum of AP $$S=120=n\frac{2a+d(n-1)}{2}=10\frac{2a+2(10-1)}{2}$$, where $$a$$ is the first term. --> $$120=10\frac{2a+2(10-1)}{2}=10a+90$$ --> $$a=3$$.

$$a_n=a_{10}=a_1+d(n-1)=3+2(10-1)=21$$

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Re: In a certain sculpture, coils of wire are arranged in rows.  [#permalink]

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19 Jan 2018, 07:40
x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18

Method 1:

Let $$n$$ be the number of rings in the final row. Each preceding row is 2 less than the current row going down to 10 rows in total. So,

$$n + n - 2 + n - 6 + n - 8 + n - 10 + n - 12 + n - 14 + n - 16 + n - 18 = 120$$
$$10n - 90 = 120$$
$$10n = 210$$
$$n = 21$$

Method 2:

We know that

Average * number of terms = Sum

putting in values

$$Average = \frac{120}{10} = 12$$

Since there are 10 terms and 2 apart, 5th and 6th term would be 11 and 13 respectively. adding $$4*2 = 8$$ to $$13$$ would give us the 10th term.
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Re: In a certain sculpture, coils of wire are arranged in rows.  [#permalink]

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20 Mar 2018, 03:26
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x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18

the coils are

x + x+ 2 + x+ 4 + x+ 6 + x+ 8 + x+ 10 + x+ 12 + x+ 14 + x+ 16 + x+ 18

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

= 10x + 2 * (9 * 10)/2

=) 10x + 90 = 120

=) x = 3

thus, the coils in the final row are
=x + 18 = 21 = B the answer

thanks
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Re: In a certain sculpture, coils of wire are arranged in rows.  [#permalink]

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21 Mar 2018, 15:11
x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18

We can let x = the number of coils in the first row, then x + 2 = the second row, x + 4 = the third row, and so on. Thus, x + 18 = the number of coils in the last (or tenth) row.

Since the number of coils in each row forms an arithmetic progression, the sum of all the terms equals the number of terms times the average of the terms.

We can calculate the average of the terms in this equally spaced set by averaging the first and last terms. Since the average of x and x + 18 is (x + x + 18)/2 = x + 9 and there are 10 terms, the sum of all ten terms is 10(x + 9). Setting this expression equal to 120, we have:

10(x + 9) = 120

x + 9 = 12

x = 3

So there are 3 + 18 = 21 coils in the last row.

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# The letters D, G, I, I , and T can be used to form 5-letter strings as

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14 Feb 2019, 08:42
# of ways in which I's are together: 4! (glue method). Usually with the glue method need to multiply by 2, but given I's are the same, don't need to. How many total ways can "digit" be arranged with no restriction? 5! = 120. Need to divide by 2! given the repetition. 60-24 = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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15 Feb 2019, 00:47
chetan2u
In the 2nd case ie when the 2 Is are together why are we not dividing the 4! by 2!, as the interchange of the 2 Is when they are together will lead to double the number of arrangements...
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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10 Mar 2019, 20:45
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The correct answer is Choice D.

First, determine the total number of ways of rearranging the letters in DIGIT.

5! / 2! (accounts for the repetition of the 2 letter "I"s) = (5 x 4 x 3 x 2) / 2 = 120 / 2 = 60

Next, consider all the ways two letter "I"s can be adjacent within the word. They can either be in the 1/2 slot, the 2/3 slot, the 3/4 slot, or the 4/5 slot (4 locations), and for each of those 4 locations there are 3 x 2 x 1 = 6 other ways of rearranging the final 3 letters, so multiply 6 by 4 to get 24.

Subtract those 24 instances from the total to get 60 - 24 = 36

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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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04 Jun 2019, 19:23
total number of ways that the letters can be arranged is (5!/2!) =60

it is easiest to find how many ways the I's can be together.

First we can treat both I's as a single entity ( I &I ).

This essentially this means we are arranging four entities which becomes 4!.

Thus the answer to the question is (5!/2!)- 4! = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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22 Jun 2019, 06:39
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Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

Let us calculate the total ways. Those would be (5!/2!) = 60.

Now since the question says "at least" let us find the number of arrangements when both I's are together. (Tie them up). so we have 4! ways to arrange such that I's always come together. 4! = 24

60 - 24 = 36.

I can never get this right! I always end up putting 4!*2! when i have to calculate ways of arranging DIGIT with both th I's combined, that because in my head i think both the I's can also be arranged in two ways. How do i get to correct this ! Bunuel Please help
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The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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Updated on: 07 Aug 2019, 22:03
We have letters: D, G, I, I, T
We need to have at least one letter between both the "I"s.
We can find the answer by finding: (Total Number of ways the 5 letters can be arranged) - (Number of ways the letters are arranged such that both the "I"s stay together) -> (a)

Number of ways the 5 letters can be arranged is $$\frac{5!}{2!}$$ = 60 ways

To calculate the number of ways the letters can be arranged such that both the "I"s stay together, we need to consider both "I"s as one element (can't be separated). Thus, we only have 4 elements to arrange now- "D", "G", "II", "T".
These 4 elements can be arranged in 4! ways = 24 ways [We do not have to worry about interchanging the two "I"s positions are they are not distinct elements].

Thus, From (a), the required number of arrangements = 60 - 24 = 36 ways.

Originally posted by Sayon on 01 Aug 2019, 21:03.
Last edited by Sayon on 07 Aug 2019, 22:03, edited 1 time in total.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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04 Aug 2019, 21:55
one of the easiest. We need to remember that 'Not together' in Permutation and Combinations can be achieved by
Total arrangements/combinations(Minus)together.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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17 Aug 2019, 00:19
varundixitmro2512 wrote:
IMO 36

Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24

no of ways at least one alpha is between two I's =60-24=36

Why isn’t 4 factorial divided by 2 factorial?

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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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17 Aug 2019, 01:17
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

Given: The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT.

Asked: Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

Total 5-letters strings that can be formed by using {D,G,I,I,T} = 5!/2! = 60
5-letter strings that can be formed by using II together = 4! = 24
5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter = 60 -24 =36

IMO D
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The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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22 Sep 2019, 22:58
ScottTargetTestPrep wrote:
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

This is a permutation problem because the order of the letters matters. Let’s first determine in how many ways we can arrange the letters. Since there are 2 repeating Is, we can arrange the letters in 5!/2! = 120/2 = 60 ways.

We also have the following equation:

60 = (number of ways to arrange the letters with the Is together) + (number of ways without the Is together).

Let’s determine the number of ways to arrange the letters with the Is together.

We have: [I-I] [D] [G] [T]

We see that with the Is together, we have 4! = 24 ways to arrange the letters.

Thus, the number of ways to arrange the letters without the Is together (i.e., with the Is separated) is 60 - 24 = 36.

Hi, Scott,

I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).

Regards,
Raxit.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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20 Nov 2019, 16:58
varundixitmro2512 wrote:
IMO 36

Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24

no of ways at least one alpha is between two I's =60-24=36

can you please explain why the No of ways two I's can be together is 4!=24?
thank you!!
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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20 Nov 2019, 21:08
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Hi A77777,

For the sake of calculating a permutation, the 'restriction' that the two "I"s must be 'next to one another' is essentially the same as saying that there is only one "I."

With 4 different letters, there are 4! = (4)(3)(2)(1) = 24 different arrangements

You can actually 'map' them all out if you choose; rather than list every option all at once, here are the 6 options that would start-off with the Is:

II D G T
II D T G
II G D T
II G T D
II T D G
II T G D

... and the 6 options that would start-off with the D:

D G II T
D G T II
D II G T
D II T G
D T G II
D T II G

There are two other 'sets of 6'; one that starts with a G and another that starts with a T.

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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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20 Nov 2019, 22:27
1
Hi, EmpowerGMATRich,

I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).

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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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20 Nov 2019, 22:36
Hi Raxit85,

You bring up a good question. Although the prompt does not explicitly state it, we are meant to assume that the two "I"s are identical. In simple terms, if you had just two letters (re: the two "I"s), then you could form just ONE distinct 'word': II (not two words - again, because the two "I"s are identical).

IF... the prompt did not want us to think in those terms, then we wouldn't have duplicate letters at all - it would just be 5 different letters and the question would ask something to the effect of "how many different ways are there to arrange the letters A, B, C, D and E so that there is at least one letter between the A and the B?"

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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as   [#permalink] 20 Nov 2019, 22:36

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# Find a sequence of measures and a measurable function

I am new to measure theory and I'm have a problem finding an example for the problem below:

If $$C$$ is a middle-thirds Cantor set; find a sequence $$\{ \mu_{n}\}_{n \geq 1 }$$ of measures on Borel $$\sigma -$$ algebra $$C$$ and a measurable function $$f:C \to R$$ such that $$\lim_{n \to +\infty} \mu_n(x) = 0$$ and $$\int f d\mu_{m} = +\infty$$ for every $$m \geq 1$$.

I think that this is the Cantor measure problem but I'm not sure.

I'm going to use the space of infinite binary strings (with the product topology) rather than the middle thirds cantor space. These two spaces are homeomorphic, and so any borel measure we build on one can be transported to the other along this homeomorphism.

Let $$\mu_0$$ be the standard coin-tossing measure on cantor space. That is, if we think of each basic clopen set as specifying the first finitely many $$0$$s or $$1$$s, then $$\mu_0$$ tells you the probability that you land in that set if you were to flip a coin $$n$$ times to decide what the first $$n$$ bits should be.

Let $$\mu_n$$ be the coin tossing measure conditioned on the event "the first $$n$$ bits are $$0$$".

Let's take a moment to see some examples. We'll write $$\mathcal{N}_{s}$$ for the set of strings starting with $$s$$.

• $$\mu_0 \mathcal{N}_{0010} = \frac{1}{16}$$, since if we flip $$4$$ coins to determine our first $$4$$ bits, we get $$0010$$ with probability $$\frac{1}{16}$$.

• $$\mu_1 \mathcal{N}_{0010} = \frac{1}{8}$$, since we're assuming the first bit is a $$0$$, but we still need to flip $$3$$ coins correctly to end up in this set.

• $$\mu_2 \mathcal{N}_{0010} = \frac{1}{4}$$, since now we're assuming the first $$2$$ bits are $$0$$s, and we need to flip $$2$$ coins correctly to end up in this set.

• $$\mu_3 \mathcal{N}_{0010} = 0$$, since we're assuming the first $$3$$ bits are $$0$$, which isn't the case for our set. Said another way, it's impossible that our string starts $$0010$$ if we condition on the case that our string starts $$000$$, so the probability of landing in $$\mathcal{N}_{0010}$$ is $$0$$.

It should be clear that $$\lim \mu_n = 0$$, since every point besides the all $$0$$s string is eventually excluded from the support of our measures.

Now let's define $$f$$. Let's put $$f(s) = 2^k$$, where $$k$$ is the index of the first $$1$$ in $$s$$.

Again, by examples,

• $$f(10010...) = 2$$, since the first $$1$$ is in the first position.
• $$f(00110...) = 8$$, since the first $$1$$ is in the third position.

Do you see why this function is measurable?

Notice $$f$$ is technically undefined at the all $$0$$s string, but that's a set of measure $$0$$, so ¯\_(ツ)_/¯. If you like, we can define $$f(00000...) = 0$$ to avoid this problem. The function will still be measurable and will do what we need it to do.

Ok, we're in the home stretch. Why does $$\int f \ d\mu_n = \infty$$ for each $$n$$?

Well, notice $$f(s) = 2^{k+1}$$ for every $$s \in \mathcal{N}_{0^k 1}$$. That is, $$f$$ is a countable sum of rescaled characteristic functions,

$$f = 2 \chi_{\mathcal{N}_{1}} + 4 \chi_{\mathcal{N}_{01}} + 8 \chi_{\mathcal{N}_{001}} + \ldots = \sum 2^{k+1} \chi_{\mathcal{N}_{0^k 1}}$$

But we know that

$$\mu_n \mathcal{N}_{0^k 1} = \begin{cases} \frac{1}{2^{k+1-n}} & k \geq n \\ 0 & k < n \end{cases}$$

and so (by the monotone convergence theorem) we find

\begin{aligned} \int f \ d\mu_n &= \sum_k \int 2^{k+1} \chi_{\mathcal{N}_{0^k 1}} \ d\mu_n \\ &= \sum_k 2^{k+1} \mu_n \mathcal{N}_{0^k 1} \\ &= \sum_{k \geq n} 2^{k+1} \frac{1}{2^{k+1-n}} \\ &= \sum_{k \geq n} 2^n \\ &= \infty \end{aligned}

I hope this helps ^_^

• I can't thank you enough. Just one unrelated question I'll delete later: how long did it take you to get so skilled in this?
– Pegi
Jul 14, 2021 at 22:41
• I'm not sure. I think I first did measure theory 2 or 3 years ago? But it's pretty related to a lot of things, and so I keep coming back to it. Plus, answering questions (like this one) on MSE is a good way to keep in shape ^_^. Jul 14, 2021 at 23:02
• Unrelated, but if this answered your question, you should mark it as such so that other users know how to spend their time. If you want to leave it open for a while so that other people have a chance to answer, that works too! Jul 14, 2021 at 23:03
• oh no I was still trying to go through it again to make sure I didn't miss any points. I have checked it now. Thank you!
– Pegi
Jul 14, 2021 at 23:04
• No worries at all! Happy to help ^_^ Jul 14, 2021 at 23:19


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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## What is the value of x? tagged by: BTGmoderatorLU ##### This topic has 2 expert replies and 0 member replies ### Top Member ## What is the value of x? in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x? A. 5 B. 6 C. 7 D. 11 E. 14 The OA is A. I'm confused with this PS question, can I say that the total area is (2x+1)^2=3^2+6^2+76 Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2666 messages Followed by: 125 members Upvotes: 1153 GMAT Score: 770 Top Reply LUANDATO wrote: in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x? A. 5 B. 6 C. 7 D. 11 E. 14 The OA is A. I'm confused with this PS question, can I say that the total area is (2x+1)^2=3^2+6^2+76 Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance. You're exactly right. The total area of the square is $$3^2 + 6^2 + 76 = 9 + 36 + 76 = 121$$ We know each side of the square must be 11, as 11*11 = 121 If a side is 11, then 2x + 1 = 11 --> 2x = 10 ---> x = 5. The answer is A _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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Quote:
in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x?

A. 5
B. 6
C. 7
D. 11
E. 14

The OA is A.

I'm confused with this PS question, can I say that the total area is

(2x+1)^2=3^2+6^2+76

Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance.
Hi LUANDATO,
Let's take a look at your question.

The length of the square is (2x+1), therefore the area of the bigger square is:
$$\text{Area of Big Square}=\left(2x+1\right)^2$$

Area of the shaded region and areas of two smaller squares add up to the area of the bigger square. Hence,
Area of Big Square = Area of Square with side 3 + Area of Square with side 6 + Area of shaded region
$$\left(2x+1\right)^2=3^2+6^2+76$$
$$\left(2x+1\right)^2=9+36+76$$
$$\left(2x+1\right)^2=121$$

Taking square root on both sides:
$$\sqrt{\left(2x+1\right)^2}=\sqrt{121}$$
$$2x+1=11$$
$$2x=11-1$$
$$2x=10$$
$$x=\frac{10}{2}$$
$$x=5$$

Therefore, Option A is correct.

Hope it helps.
I am available if you'd like any follow up.

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# possible number of sheets for a Moebius band covering

Let M be the Moebius band, identified by the quotient of $$[0,1]\times [0,1]$$ by the equivalence $$(x,0) \sim (1-x,1)$$.

Let $$p: M\to M$$ be a covering and $$n$$ its number of sheets.

Find the possible values of $$n$$.

what I did:

$$M$$ is path-connected so it is connected, all the fibers have the same cardinality.

$$M$$ is also compact so $$n$$ is finite.

My intuition is that any $$n\in \Bbb Z$$ would be valid, as the possible coverings are $$\Bbb R/n\Bbb Z\cong \Bbb R/\Bbb Z\times \Bbb Z/ n\Bbb Z$$.

Thank you for help and comments.

• Something of a side note, but that isomorphism you write is not correct. The left hand side is isomorphic to the circle, while the right is $n$ disjoint circles. – WSL Feb 23 at 2:50
• This is closely related to what we talked about in your other question. Remember that the universal cover of $M$ is $\tilde{M}=\mathbb{R}\times [0,1]$ and the deck transformations act by translating and flipping. All the connected coverings of $M$ are quotients of $\tilde{M}$ by subgroups of $\mathbb{Z}$, so you should try to determine for which values of $n$ the quotient $\tilde{M}/n\mathbb{Z}$ is homeomorphic to $M$. – William Feb 23 at 4:23

I think only an odd number of sheets are possible.

This is a great example where the general theory leads to interesting computational results in particular cases: we can determine possible coverings of the form $$M\to M$$ by first determining all connected coverings of $$M$$, and then detecting which ones have total space homeomorphic to $$M$$.

Classification Theorem: For any path-connected, locally path-connected, semi-locally simply-connected space $$X$$ there is a bijection between isomorphism classes of connected covering spaces of $$X$$ and conjugacy classes of subgroups of $$\pi_1(X)$$. (See for example Theorem 1.38, page 67.)

This works by constructing a universal covering $$\tilde{X}\to X$$ so that the quotients $$\tilde{X}/H$$ represent all the connected coverings of $$X$$ as $$H$$ varies over conjugacy classes. The covering $$\tilde{X}$$ is characterized up to covering space isomorphism by being simply-connected.

Universal covering space of $$M$$: Recall $$M\sim S^1$$ so $$\pi_1(M)\cong \mathbb{Z}$$. The universal covering space of $$M$$ is $$\mathbb{R}\times [0,1]$$ and the action of $$\mathbb{Z}$$ is given by $$n\cdot (x, t)= \big(x+n, f^{n}(t)\big)$$ for $$n\in\mathbb{Z}$$ and where $$f\colon [0,1] \to [0,1]$$ is the "flip" homeomorphism given by $$f(t)= 1-t$$. (Visually, think about $$\mathbb{R}\times[0,1]$$ as an infinite strip of tape that you're applying to the Möbius strip, which is alternating "front" and "back" sides.)

Quotients of $$\tilde{M}$$: Every connected covering of $$M$$ is a quotient of the form $$(\mathbb{R}\times [0,1])/n\mathbb{Z}$$. For each $$n$$ a fundamental domain of the quotient is $$[0,n]\times[0,1]$$, and the quotient only depends on how we identify the subspaces $$\{0\} \times [0,1]$$ and $$\{n\}\times [0,1]$$. When $$n$$ is odd then $$f^{n} = f$$ so we identify the ends using a flip, and hence the quotient is homeomorphic to $$M$$; on the other hand if $$n$$ is even then $$f^{n}=id$$ and so the quotient is actually the cylinder $$(\mathbb{R}/n\mathbb{Z})\times [0,1]$$.

Since these quotients make up all of the possible connected coverings of $$M$$, it follows that coverings of the form $$M\to M$$ can have any odd number of sheets.

• Thanks @William for your thorough answer. I see now why you say connected instead of path-connected. It boild down to the same as local path-connectedness is inherited by the coverings – PerelMan Feb 23 at 17:45
• Yes that's true. Since we already know $M$ is locally path-connected, so will be any covering. Therefore every connected covering of $M$ is already path-connected. – William Feb 23 at 19:40

Even though the question is already answered, I found an alternate argument that there are no even-sheeted covers using the first Stiefel-Whitney class of $$TM$$.

Recall that the tangent bundle of $$M$$ is non-orientable so $$w_1(TM)\neq 0 \in H^1(M;\mathbb{Z}/2\mathbb{Z})$$. An $$n$$-sheeted covering $$p_n\colon M\to M$$ is a local diffeomorphism and hence induces a bundle map $$TM\to TM$$, so by naturality of characteristic classes we have $$w_1(TM) = p_n^*(w_1(TM))$$. But the covering also restricts to an $$n$$-sheeted covering $$S^1\to S^1$$, which on cohomology $$H^1(S^1;\mathbb{Z}) \to H^1(S^1;\mathbb{Z})$$ induces multiplication by $$n$$. Since $$S^1 \to M$$ is a homotopy equivalence we also have

$$p_n^* = n\cdot(-)\colon H^1(M;R) \to H^1(M;R)$$

for any $$R$$. In particular if such a cover exists when $$n = 2k$$ is even then $$w_1(TM) = p_n^*(w_1(TM)) = 2k \cdot w_1(TM) = 0$$ which contradicts $$w_1(TM) \neq 0$$.

Edit: I'm currently trying to modify this argument into a proof of the following:

Conjecture: If $$M$$ is a non-orientable smooth manifold then there are no even-sheeted coverings of the form $$M\to M$$.

Update: This general conjecture is NOT true. Consider the double-cover $$p\colon S^1\times \mathbb{RP}^2\to S^1 \times \mathbb{RP}^2$$ given by $$p(z, x) = (z^2, x)$$.

• Thank you @William interesting alternate argument! I only studied some elementary De Rham cohomology, so not able to understand all of it but will read about the ring cohomology. I am interested in any references. – PerelMan Feb 23 at 21:09


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# How do we decide whether to visualize a matrix with its rows or columns?

Should one visualize a matrix by its rows, columns, or both depending on the situation? I see both used and it seems arbitrary. It would be nice if only one was used consistently. Shouldn't a graph of a matrix be denoted as being a row or column representation somehow to avoid confusion?

Example where author switches: https://intuitive-math.club/linear-algebra/matrices

[Example I] Given the transformation: $$\begin{bmatrix} 1 & 1\\ 2 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2\\ 3 & 1 \end{bmatrix}$$ The author represents the matrix after the transformation visually by its rows, using the following row vectors:

$$v_1 = \begin{bmatrix} 3\\ 2 \end{bmatrix} v_2 = \begin{bmatrix} 3\\ 1 \end{bmatrix}$$

[Example II] Given the transformation: $$\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \begin{bmatrix} 3 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1\\ -3 & 1 \end{bmatrix}$$

The author represents the matrix after the transformation visually by its columns, using the following column vectors:

$$v_1 = \begin{bmatrix} 1\\ -3 \end{bmatrix} v_2 = \begin{bmatrix} 1\\ -1 \end{bmatrix}$$

Question:

Why is did they author seemingly arbitrarily switch from a row → column visual representation? What is the intuition behind this – if any?

• I love this question! To give you a short, definitive “answer” that isn’t up to par with the expected quality on this site: Most people will default to looking at columns as vectors. If you break it into rows, most people will look at those as equations (as in a system) in multiple variables. Only occasionally do you interpret the rows as vectors—but it’s doable. I hope that helps. – gen-ℤ ready to perish Aug 13 at 2:02
• The real answer here, I think, is that the website you link to isn't choosing its visualizations in any sort of principled way. I wouldn't read anything into the the particular choices made on that site. – Will Orrick Aug 13 at 9:00

There's a lot of ways to interpret matrices, some of which involve reading it by rows and some by columns. But in this particular case, it is columns both times: you were misled by the fact that the matrix $$\begin{bmatrix}3 & 1 \\ 1 & 1\end{bmatrix}$$ is symmetric, so its columns are the same as its rows.

Here, the idea is that for any $$2 \times 2$$ (or more generally $$k \times 2$$) matrix $$A$$, we have $$A \begin{bmatrix}3 & 1 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}A \begin{bmatrix}3 \\ 1\end{bmatrix} & A\begin{bmatrix}1 \\ 1\end{bmatrix} \end{bmatrix}.$$ In other words, each column of the product is equal to $$A$$ times a column of the second matrix we multiplied.

In the picture you have, the vector $$\begin{bmatrix}3 \\ 1\end{bmatrix}$$ (in pink) gets sent to $$\begin{bmatrix}1 \\ -3\end{bmatrix}$$, and the vector $$\begin{bmatrix}1 \\1\end{bmatrix}$$ (in yellow) gets sent to $$\begin{bmatrix}1 \\ -1\end{bmatrix}$$, and all of these are columns of the respective $$2 \times 2$$ matrix.

• You are completely correct. That is my mistake for not noticing the symmetry. I may have to edit my questions example to get the correct point across. – Matthaeus Gaius Caesar Aug 12 at 23:06
• I completely edited the examples in my question. The example was mistakingly symmetric, which took away from the actual question at hand. I do not know if such a large edit is allowed, but the premise of the question remains the same. Should I make a new question with the new example? – Matthaeus Gaius Caesar Aug 12 at 23:07
• @Caesar I think the edit was fine as it’s not altering the question itself. You did the right thing by leaving a comment. The problem with major edits arises when they make an answer obsolete. – gen-ℤ ready to perish Aug 13 at 3:09
• The website's example of matrix addition has an illustration showing a vector $[2 0],$ which can only be a row of one of the matrices, not a column. I think it would have been better if it had said this explicitly. I don't see anything wrong with getting used to matrices being either columns of row vectors or rows of column vectors, but expecting you to guess while you're just learning is a bit much to ask. – David K Aug 13 at 12:05
• @MatthaeusGaiusCaesar, your edit almost completely invalidates this answer. While your question first had the two cases before and after transformation, now it has two different transformations. The sentence at the beginning of this answer "it is columns both times" no longer applies, and makes the answer look out of place. – ilkkachu Aug 13 at 12:17

As long as your main objects of study are column vectors, and you multiply matrix and (column) vector together by writing the matrix on the left and the vector on the right, a matrix is more naturally seen as a collection of columns rather than rows.

A matrix represents a linear transformation. The columns of the matrix are given by where this linear transformation sends your basis vectors. The result of a matrix-vector product similarly becomes a linear combination of the columns of the matrix (where the entries in the vector are the coefficients of this linear combination).

When multiplying two matrices, of course you can choose. Either you say "Apply the left-hand matrix to each column in the right-hand matrix, and collect the results in a new matrix" (in which case you see both matrices as collections of columns), or you say "Apply the right-hand matrix to each row in the left-hand matrix, and collect the results in a new matrix" (in which case both matrices are collections of rows). They both give the same result. Which one is most convenient comes down to whether one happens to be significantly easier to calculate than the other for some reason, and what you're going to do with the result afterwards.

Of course, the final answer is "it depends on the situation". Because what else could it be? But columns is much more common than rows.


# Math Help - Exact values

1. ## Exact values

Could someone help me with the following three problems?

1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).

2) cos[to the -1 power](cos(4[pi]/3)).

3) tan(sin[to the -1 power](-4/5)).

2. Originally Posted by mike1
Could someone help me with the following three problems?

1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
You are expected to know some special triangles, one of which is the
30, 60, 90 degree triangle, where if the side opposite the 30 degree angle
is 1 unit the side opposite the 60 degree angle is sqrt(3), and the other side
is 2. So using this right triangle we see that sin(30)=sqrt(3)/2, hence if:

x=asin(-sqrt(3)/2),

then

sin(x)=-sqrt(3)/2,

so:

sin(-x)=sqrt(3)/2.

Now there are multiple solutions to this, but we want x to be in the range
[-90,90], so we want -x=60, or x=-60 degrees, or -pi/3 radian.

Check: asin(-sqrt(3)/2)~=-1.57, -pi/2~=-1.57

2) cos[to the -1 power](cos(4[pi]/3)).
Now 4 pi/3= pi+pi/3, so cos(4 pi/3)=cos(pi)cos(pi/3)-sin(pi)sin(pi/3)=-cos(pi/3)
Now pi/3 radian is 60 degrees so we are back to our special triangle, and cos(pi/3)=1/2.

Now acos(-1/2)=pi-pi/3=2pi/3.

Check acos(cos(4pi/3))~=2.094, 2*pi/3~=2.094

3) tan(sin[to the -1 power](-4/5)).
Here we need to think 3-4-5 triangle, first:

tan(asin(-4/5))=-tan(asin(4/5))

(draw the triangle identify the angle whose sin is 4/5, then find its
tangent), so:

tan(asin(-4/5))=-4/3

Check: tan(asin(-4/5))~=-1.33.., -4/3~=1.33..

RonL

Note: when evaluating the inverse trig functions it is usual to give the answer
in a fundamental range such that the function has a unique value (the principle value).

That is evaluating x=asin(y) is not the same as finding x such that sin(x)=y,
since asin(y) gives a single value, while solving sin(x)=y has multiple solutions.

For asin the range is usually -pi/2 to pi/2 (or in degrees -90 to 90)
For acos the range is usually 0 to pi (0 to 180 degrees)
For atan the range is usually -pi/2 to pi/2 (or sometimes 0 to pi) (-90 to 90 degrees).

3. Originally Posted by mike1
Could someone help me with the following three problems?

1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).

2) cos[to the -1 power](cos(4[pi]/3)).

3) tan(sin[to the -1 power](-4/5)).
I assume sin[to the -1 power] is arcsine.

1) arcsin[-sqrt(3) /2]
That is an angle whose sine is -sqrt(3) /2. Let's call it theta.

In what quadrants is the sine negative? Since sine is y/r, and r is always positive, then sine is negative wherever y is negative. Below the origin, or below the x-axis. In the 3rd and 4th quadrants.
So theta is between pi and 2pi radians.

Then the angle.
If sin(theta) = sqrt(3) /2, then theta is 60degrees or pi/3 radians. What are the angles in the 3rd and 4th quadrants that are pi/3 from the x-axis?
Why, (pi +pi/3) in the 3rd quadrant and (2pi -pi/3) in the 4th quadrant.

----------------------
2) cos[to the -1 power](cos(4[pi]/3)).
That is "the cosine of an angle whose cosine is 4pi/3".

======
Ooppss, my mistake.
That is actually arccosine of cos(4pi/3).
An angle whose cosine is cos(4pi/3).

I do not know how to do that.

You sure you typed it right?
-------------------------------------
3) tan(sin[to the -1 power](-4/5)).
That is "the tangent of an angle whose sine is -4/5 "

First, about the negative sine value.
As in part (i) above, theta is in the 3rd or 4th quadrant, or theta between pi and 2pi radians.

Then the angle.
If sin(alpha) = 4/5, then y=4 and r=5 since sine = y/r.
tan is y/x. So we need to find x.
By Pythagorean theorem, x = sqrt(r^2 -y^2) = sqrt(5^2 -4^2) = 3.
Hence tan(alpha) = y/x = 4/3.

In the 3rd quadrant, x is negative, and y is negative also, hence,
tan(theta) = y/x = -4/ -3 = 4/3 ------***

In the 4th quadrant, x is positive, and y is negative, hence,
tan(theta) = y/x = -4/ 3 = -4/3 -----***

-----------------------
Note, in the reference triangle,
opposite side = y
hypotenuse = r --------always positive.

4. Hello, Mike!

ticbol did an excellent job of explaining.
Here are my versions . . .

I'll assume the angle are between $0$ and $2\pi$.

Find the exact value of:

$(1)\;\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

We want the angle whose sine is $-\frac{\sqrt{3}}{2}$

You are expected to know that: $\sin\left(\frac{\pi}{3}\right) \,=\,\frac{\sqrt{3}}{2}$

Sine is negative in quadrants 3 and 4.
So we have a reference angle of $\frac{\pi}{3}$ in quadrants 3 and 4.

Hence, the angles are: . $\frac{4\pi}{3},\:\frac{5\pi}{3}$

$(2)\;\cos^{-1}\left[\cos\left(\frac{4\pi}{3}\right)\right]$

Inside, we have: . $\cos\frac{4\pi}{3} \:=\:\text{-}\frac{1}{2}$

Then we have: . $\cos^{-1}\left(\text{-}\frac{1}{2}\right)$ . . . the angle whose cosine is $\text{-}\frac{1}{2}$

We know that: . $\cos\left(\frac{\pi}{3}\right) \,=\,\frac{1}{2}$
Cosine is negative in quadrants 2 and 3.

Therefore the angles are: . $\frac{2\pi}{3},\:\frac{4\pi}{3}$

$(3)\;\tan\left[\sin^{-1}\left(\text{-}\frac{4}{5}\right)\right]$

Inside, we have: . $\sin^{-1}\left(\text{-}\frac{4}{5}\right)$ . . . the angle whose sine is $\text{-}\frac{4}{5}$

Consider: $\sin\theta = \frac{4}{5}$
We don't know the exact value of this angle,
. . but we know it comes from this triangle:
Code:
                  *
* |
*   |
5 *     | 4
*       |
* θ       |
* - - - - - *
3

Since sine is negative in quadrants 3 and 4,
. . the two possible angles look like this:
Code:
              |                         |
|                         |
-3   |                         |   3
- + - - - + - - - - -       - - - - + - - - + -
:   θ / |                         | \ θ   :
-4:   /5  |                         |  5\   :-4
: /     |                         |     \ :
*       |                         |       *

Since $\tan\theta \:=\:\frac{\text{-}4}{\text{-}3}\text{ or }\frac{\text{-}4}{3}$

. . the answers are: . $\pm\frac{4}{3}$


# Help with complicated functional equation

Problem: Let $T=\{(p,q,r)\mid p,q,r \in \mathbb{Z}_{\geq0}\}$. Find all functions $f:T\to \mathbb{R}$ such that: $$f(p,q,r)=\\ =\begin{cases} 0, & \text{ if } pqr = 0 \\ 1 + \frac{1}{6}\left(f(p+1,q-1,r)+f(p+1,q,r-1)+f(p,q+1,r-1)+\\ \;\;\;\;\;\; f(p,q-1,r+1)+f(p-1,q+1,r)+f(p-1,q,r+1)\right) & \text{ otherwise.} \end{cases}$$

Progress so far: It's not hard to see that $f$ is symmetric in $p,q,r$, which is useful to know. From the recursive definition one can also infer that $f:T\to \Bbb{Q}^+$, so no trig functions or logs. That's all I could observe from the get-go. I've tried calculating some values of $f$ to have an idea on how the functions look like (if there are any) but having trouble calculating even small values of $f$, for example $f(1,2,3)$ or $f(2,2,2)$. All I know is that $f(0,a,b)=0$ and $f(1,1,1)=1$. I could guess a solution based on my initial observations but I can't see any obvious candidates.

Any help would be appreciated, thanks.

• $\displaystyle\;f(p,q,r) = \frac{3pqr}{p+q+r}\;$ is a solution. I believe this is the only solution but I can't prove it. – achille hui Jul 16 '14 at 19:56
• How did you get this solution? – Deathkamp Drone Jul 16 '14 at 20:00
• If you look at your equation, $(p,q,r)$ only connects to points with same $p+q+r$. This means you can look at each $p+q+r = \text{constant}$ layer separately. The intersection of $T$ with any such layer forms a triangular lattice. Your equation reduces to a discrete version of Poisson equation there. The boundary condition suggest the solution (at least in continuum limit) contains a factor proportional to $pqr$. Direct substitution shows that you can scale this to get a solution. Since we are dealing with some sort of "Poisson equation", that's why I suspect the solution is unique. – achille hui Jul 16 '14 at 20:09
• This is an IMO 2001 shortlist problem. – Calvin Lin Jul 18 '14 at 22:57
• Thank you, Calvin. I found this problem some lecture notes so I didn't know where it came from. Ironically, I looked up the solution on AoPS just to find out that they pulled out of nowhere the function Achille discovered and then proved that this is the only solution. – Deathkamp Drone Jul 19 '14 at 2:02

When I worked on this problem back in 2002, showing uniqueness was really easy through the "average of neighbors" observation (albeit on a slanted hexagonal board, instead of the regular chessboard).

Proof of uniqueness: Suppose we have 2 solutions $f(p,q,r)$ and $g(p,q,r)$. Let $h(p,q,r) = f(p,q,r) - g(p,q,r)$. Then, we get that

$$6 h(p,q,r) = h(p+1, q-1, r) + h(p-1, q+1, r) + h( p, q+1, r-1) + h( p, q-1, r+1) + h( p+1, q, r-1) + h(p-1, q, r+1).$$

Consider the plane $p+q+r = N$. Oberve that the neighbors of the cell $(p,q,r)$ are these 6 other cells with coordinates as given above. Hence, every cell is the average of it's neighbors. Through the standard argument (extremal principle), this implies that all cells on this finite board are equal.

We also have the boundary conditions that $h(p,q,r ) = 0$ for $pqr=0$, hence $h(p,q,r) = 0$. Thus, the function is unique $_\square$

Finding the solution was harder, but still motivated from the conditions.
Note: It is important to bear in mind that as an ('easy') Olympiad problem, it often has a nice solution that can be motivated.

Finding function: From the boundary condition that $pqr=0 \Rightarrow f(p,q,r) = 0$, we guess the initial function $F( p,q,r) = pqr$.

Observe that since $(p-1)(q+1) r + (p+1)(q-1)r = 2pqr - 2r$, so this guess gives us:

$F(p,q,r) = \frac{ p+q+r} { 3} + \frac{1}{6} [ F(p-1, q+1, r) + F(p+1, q-1, r) + F(p, q-1, r+1), F(p, q+1, r-1) + F( p-1, q, r+1), F(p+1, q, r-1) ]$.

Observe that since $p+q+r$ is a constant for all of these 7 terms, we should look at $$f(p,q,r) = \frac{ F(p,q,r) } { \frac{p+q+r} {3} } = \frac{3 pqr} { p+q+r}.$$

Indeed, this works. $_\square$

Note: Had $F(p,q,r) = pqr$ not worked, the next guess would have been $F(p,q,r) = p^2q^2r^2$

Achille Hui did the hardest part of the work by discovering the closed formula $\frac{3pqr}{p+q+r}$. The rest is a routine "maximum principle" argument that I explain below.

For a positive integer $k$, let $T_k$ be the finite set $\lbrace (p,q,r)\in T | p+q+r=k\rbrace$. For $x=(p,q,r)\in T$, define the neighborhood $N(x)$ of $x$ to be

$$\begin{array}{lcl} N(p,q,r)&=&\lbrace (p+1,q-1,r);(p+1,q,r-1);(p,q+1,r-1); \\ & & (p,q-1,r+1);(p-1,q+1,r);(p-1,q,r+1)\rbrace \end{array} \tag{1}$$

and the strict neighborhood $N'(x)$ of $x$ to be $\lbrace (u,v,w)\in N(x) | uvw>0\rbrace$. We say that $x\in T_k$ is interior if $N’(x)=N(x)$, and extremal otherwise.

Let $g(p,q,r)=f(p,q,r)-\frac{3pqr}{p+q+r}$ for $(p,q,r)\in T$. Then $g$ satisfies $g(p,q,r)=0$ if $pqr=0$, and

$$6g(x)=\sum_{y\in N'(x)}g(y) \tag{2}$$

for any $x\in T_k$ (note that $N(x)$ and $N’(x)$ stay in $T_k$ when $x\in T_k$).

Now, let $M$ ($m$) be the maximum (minimum) value of $g$ on $T_k$. There is some $x_M\in T_k$ such that $g(x_M)=M$. We now apply (2) to $x=x_M$, and obtain a formula (2').

If $M > 0$, then (2') is only possible when $x_M$ is interior and $g(y)=M$ for all $y\in N(x_M)$. If we put $L=\lbrace x\in T_k | g(x)=M\rbrace$, we would deduce that $L$ consists only of interior points but also satisfies $N(x)\subseteq L$ for any $x\in L$, which is impossible because when we move away from the interior of $T_k$ we are always forced to eventually reach extremal points.

So $M\leq 0$. A similar argument (or if you please, you may reuse the result just shown on $-g$ instead of $g$) shows that $m\geq 0$.

So $M\leq 0 \leq m$, but on the other hand $m\leq M$. This forces $m=M=0$, so $g$ is identically zero.

To conclude, $\frac{3pqr}{p+q+r}$ is the unique solution.


# what is the difference between bounded and convergent?

I know that bounded means to have an upper or lower bound.

Let $E \subset \mathbb{R}$ be nonempty.

1. The set $E$ is said to be bounded above if and only if there is an $M \in \mathbb{R}$ such that $a \leq M$ for all $a \in E$, in which case $M$ is called an upper bound of $E$.

2. A number $s$ is called a supremum of the set $E$ if and only if $s$ is an upper bbound of $E$ and $s \leq M$ for all upper bounds $M$ of $E$ (In this case we shall say that $E$ has a finite supremum $s$ and write $s=\sup E$.

Let $E \subset \mathbb{R}$ be nonempty.

1. the set $E$ is said to be bounded below if and only if there is an $m \in \mathbb{R}$ such that $a \geq m$ for all $a\in E$, in which case $m$ is called a lower bound of the set $E$.

2. A number $t$ is called an infimum of the set $E$ if and only if $t$ is a lower bound of $E$ and $t \geq m$ for all lower bounds $m$ of $E$. In this case we shall say that $E$ has an infimum $t$ and write $t = \inf E$

3. $E$ is said to be bounded if and only if it is bounded both above and below.

The meaning of convergence

A sequence of real number $\{ x_n \}$ is said to converge to a real number $a \in \mathbb{R}$ if and only if for every $\epsilon > 0$ there is an $N \in \mathbb{N}$ (which in general depends on $\epsilon$) such that $$n \geq N \mbox{ implies } \vert x_n - a \vert < \epsilon$$

In my previous classes, it was taught to take the the limit of the function or series to find the value which the function or series converges at the value. Is the value a function or series converges at the $\sup$ or $\inf$?

• The sequence defined by $a_n := \sin \left(\frac{\pi}{2} n\right)$, which has expansion $0, 1, 0, -1, 0, 1, 0, -1, \ldots$ is bounded but not convergent. – Travis Sep 12 '14 at 7:42
• Thanks, that example made the conflict more clear. – El Santi Sep 12 '14 at 7:59

I think there's two points to address here. The first is that a sequence can be bounded from both above and below, yet fail to be convergent. Consider the sequence $$\{x_n\} = \{(-1)^n\} = -1,1,-1,1, \cdots$$ Here we have $\displaystyle\sup_{n} \{x_n\} = 1$, $\displaystyle\inf_{n} \{x_n\} = -1$, and $\{x_n\}$ does not converge (if you pick say $\epsilon = 0.1$, then there are infinitely many $n$ for which $|x_n - 1| > 0.1$ and for which $|x_n - (-1)| > 0.1$).
Furthermore, when a sequence $\{x_n\}$ does converge, the limit of a sequence need not equal either $\displaystyle\sup_n\{x_n\}$ or $\displaystyle\inf_n\{x_n\}$. For example, we see with the sequence $\{x_n\} = \{\frac{(-1)^n}{n}\}$ that $$\displaystyle\sup_n\{x_n\} = \frac{1}{2}, \: \:\displaystyle\inf_n\{x_n\} = -1, \: \: \displaystyle\lim_{n \to \infty} \{x_n\} = 0$$
In that last example, I kind of cheated, by picking a sequence where the inf and sup of the elements of the sequence were the first two, and the convergence of a sequence talks about the behavior at the tail of the sequence. When we want to talk about upper and lower bounds of sequences, we usually like to talk about the limit superior, or $$\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\lim_{n \to \infty} \left(\sup_{N} \{x_N | \: N \geq n \} \right)$$ and the limit inferior, or $$\displaystyle\liminf_{n \to \infty} \{x_n\} = \displaystyle\lim_{n \to \infty} \left(\inf_{N} \{x_N | \: N \geq n \} \right)$$ This let's you bound the values of $x_n$ for arbitrarily large $n$, while allowing for a few badly behaved values of $x_n$ for small $n$. Then, we say that a sequence $\{x_n\}$ approaches a finite limit $M$ if and only iff $$\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\liminf_{n \to \infty} \{x_n\} = M$$ and you can read about the proof of this result here (Proof that a sequence converges to a finite limit iff lim inf equals lim sup).
• is there a typo at the end? $\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\limsup_{n \to \infty} \{x_n\}$ is it to be $\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\liminf_{n \to \infty} \{x_n\}$ ?? – El Santi Sep 12 '14 at 8:22


# Integration using Monte Carlo Method

I'm trying to solve this integral using the Monte Carlo Method.

$$I=\int_0^\pi \frac{1}{\sqrt{2\pi}}e^\frac{-sin(x)^2}{2}dx$$

Now it seems to me that there is a normal probability density function in there, but I'm not sure because of the sine function.

If there is a normal, then it's easy to simulate $N$ random numbers from the standard normal distribution and compute $I$. It's is also easy to find the size of the sample to get a 95% confidence interval.

So is there a normal probability density function? Or from what distribution should I get the random numbers to compute $I$?.

• Ok. How did you find the confidence intervals? – Fawcett512 Aug 31 '16 at 4:04
• I know how to get the confidence intervals for a sample of uniform random variables, but I'm not sure how to find them in this case given that I don't have a parameter but rather the approximate value of $I$ – Fawcett512 Aug 31 '16 at 4:06
• I think I'm making the problem more complicate that it should. So I could generate $N$ uniform random variables $x_i$ and then compute $\mathbb{E}=\frac{\pi}{N}\sum f(x_i)$. As $N$ tends to infinity, we should get closer to the integral. Yet I don't see how to get the confidence intervals. – Fawcett512 Aug 31 '16 at 4:40
• I think that I was trying to do some importance sampling (hence the question if we could use a normal probability density function), but seems that in this case that is not possible, Furthermore, if I just compute the above formula for $\mathbb{E}$,I should get a good estimate for $N$ provided that $N$ is big enough. The only thing that's not very clear is how to compute a 95% confidence interval. – Fawcett512 Aug 31 '16 at 4:47

Recall that if $Y$ is a random variable with density $g_Y$ and $h$ is a bounded measurable function, then $$\mathbb E[h(Y)] = \int_{\mathbb R} h(y)g_Y(y)\,\mathsf dy.$$ Moreover, if $Y\sim\mathcal U(0,1)$, then $a+(b-a)U\sim\mathcal U(a,b)$. So applying the change of variables $x=a+(b-a)u$ (with $a=0$, $b=\pi$) to the given integral, we have $$I = \int_0^1 \frac{\pi}{\sqrt{2\pi}} e^{-\frac12\sin^2 (\pi u) }\,\mathsf du=\int_0^1 h(u)\,\mathsf du,$$ with $h(u)=\sqrt{\frac\pi 2} e^{-\frac12\sin^2 (\pi u) }$. It follows then that $I=\mathbb E[h(U)]$ with $U\sim\mathcal U(0,1)$. Let $U_i$ be i.i.d. $\mathcal U(0,1)$ random variables and set $X_i=h(U_i)$, then for each positive integer $n$ we have the point estimate $$\newcommand{\overbar}[1]{\mkern 1.75mu\overline{\mkern-1.75mu#1\mkern-1.75mu}\mkern 1.75mu} \widehat{I_n} =: \overbar X_n= \frac1n \sum_{i=1}^n X_i$$ and the approximate $1-\alpha$ confidence interval $$\overbar X_n\pm t_{n-1,\alpha/2}\frac{S_n}{\sqrt n},$$ where $$S_n = \sqrt{\frac1{n-1}\sum_{i=1}^n \left(X_i-\overbar X_n\right)^2}$$ is the sample standard deviation.

Here is some $\texttt R$ code to estimate an integral using the Monte Carlo method:

# Define "h" function
hh <-function(u) {
return(sqrt(0.5*pi) * exp(-0.5 * sin(pi*u)^2))
}

n <- 1000 # Number of trials
alpha <- 0.05 # Confidence level
U <- runif(n) # Generate U(0,1) variates
X <- hh(U) # Compute X_i's
Xbar <- mean(X) # Compute sample mean
Sn <- sqrt(1/(n-1) * sum((X-Xbar)^2)) # Compute sample stdev
CI <- (Xbar + (c(-1,1) * (qt(1-(0.5*alpha), n-1) * Sn/sqrt(n)))) # CI bounds

# Print results
cat(sprintf("Point estimate: %f\n", Xbar))
cat(sprintf("Confidence interval: (%f, %f)\n", CI[1], CI[2]))

For reference, the value of the integral (as computed by Mathematica) is $$e^{-\frac14}\sqrt{d\frac{\pi }{2}} I_0\left(\frac{1}{4}\right) \approx 0.991393,$$ where $I_\cdot(\cdot)$ denotes the modified Bessel function of the first kind, i.e. $$I_0\left(\frac14\right) = \frac1\pi\int_0^\pi e^{\frac14\cos\theta}\,\mathsf d\theta.$$

• Thank you. This was very clear. – Fawcett512 Aug 31 '16 at 4:54
• You're welcome! The change of variables is not strictly necessary, but it simplifies the computations considerably. – Math1000 Aug 31 '16 at 4:58
• Just one more question because I have trouble understanding the concept of confidence intervals. How do I know the size of the sample if now I want to get a 98% confidence interval? – Fawcett512 Aug 31 '16 at 6:54
• I don't understand your question - the sample size ($n$) and confidence level ($\alpha$) are parameters which influence the width of the confidence interval, but are not directly related to each other. – Math1000 Aug 31 '16 at 7:49
• Never mind. That was a bad question. I was wondering how big the sample must be (because I was making a simulation) to get a 98% confidence interval, but as you said is a parameter which influences the width of the confidence interval. All is clear know (or at least I think so). – Fawcett512 Sep 1 '16 at 5:59


# 1 A simple differential equation

The examples in this section show how a single ordinary differential equation can be solved with wasora. Indeed this is one of its main features, namely the ability to solve systems of differential-algebraic equations written as natural algebraic expressions. In particular, the equation the examples solve is

\frac{dx}{dt} = -x

with the initial condition x_0 = 1, which has the trivial analytical solution x(t) = e^{-t}.

## 1.1 exp.was

As clearly defined in wasora’s design basis, simple problems ought to be solved by means of simple inputs. Here is a solid example of this behavior.

end_time = 1        # transient problem
PHASE_SPACE x       # DAE problem with one variable
x_0 = 1             # initial condition
x_dot .= -x         # differential equation
PRINT t x HEADER    # output
# exercise: plot dt vs t and see what happens
$wasora exp.was | qdp -o exp --pi 1$

By default, wasora adjusts the time step so an estimation of the relative numerical error is bounded within a range given by the variable rel_error, which has an educated guess by default. It can be seen in the figure that dt starts with small values and grows as the conditions allow it.

## 1.2 exp-dt.was

The time steps wasora take can be limited by means of the special variables min_dt and max_dt. If they both are zero (as they are by default), wasora is free to choose dt as it considers appropriate. If max_dt is non-zero, dt will be bounded even if the conditions are such that bigger time steps would not introduce large errors. On the other hand, if min_dt is non-zero, the time step is guaranteed not to be smaller that the specified value. However, it should be noted that wasora may need to take several internal times step to keep the error bounded. In the limiting case where min_dt = max_dt, the time step can be set exactly although, again, wasora may take internal steps.

This situation is illustrated with the following input, which is run for five combinations of min_dt and max_dt.

end_time = 2
min_dt = $1 max_dt =$2
rel_error = 1e-3
PHASE_SPACE x
x_0 = 1
x_dot .= -x
PRINT t x (x-exp(-t))/x
$wasora exp-dt.was 0 0 > exp-dt1.dat$ wasora exp-dt.was 0.1 0   > exp-dt2.dat
$wasora exp-dt.was 0 0.1 > exp-dt3.dat$ wasora exp-dt.was 0.1 0.1 > exp-dt4.dat
$wasora exp-dt.was 1 1 > exp-dt5.dat$ pyxplot exp-dt.ppl; pdf2svg exp-dt.pdf exp-dt.svg; rm -f exp-dt.pdf
$pyxplot exp-error.ppl; pdf2svg exp-error.pdf exp-error.svg; rm -f exp-error.pdf$

It can be seen that all the solutions coincide with the analytical expression. Even if the time step is set to a big fixed value, the error commited by the numerical solver with respect to the exact solution is the same as wasora iterates internally as needed. In general, the fastest condition is where dt is not bounded as wasora minimizes iterations by automatically adjusting its value. However, it is clear that controlling the time step can be useful some times. A further control can be obtained by means of the TIME_PATH keyword.


# Is every open and connected set in $\mathbb C$ the continuous image of the open unit disk?

Let $$\mathbb D=\{z\in\mathbb C\ |\ |z|<1\}$$ be the open unit disk in $$\mathbb C$$. It is well known that an open (nonempty) set $$U\subseteq\mathbb C$$ is simply connected if and only if it is homeomorphic to the unit disk. One can show that this is equivalent to the fact that there is a continuous injective function $$f:\mathbb D\to\mathbb C$$ such that $$f(\mathbb D)=U$$.

Therefore, my question is: If $$U$$ is a (nonempty) open and connected subset of $$\mathbb C$$, is there always a continuous (but not necessarily injective) function $$f:\mathbb D\to\mathbb C$$ such that $$f(\mathbb D)=U$$? It sufficices to assume $$U\subseteq\mathbb D$$.

If not, how do images of such functions look like?

I know that any compact connected and locally connected subset of $$\mathbb C$$ is the image of a continuous function defined on $$[0,1]$$ and hence also the image of a continuous function on $$\overline{\mathbb D}$$. However, the closure of an open and connected set does not have to be locally connected.

I believe that the answer to my first question is negative (although I hope that it is not ;) ).

Any help is highly appreciated. Thank you very much in advance!

• Not my field, but I'm pretty sure that the answer is positive to your first question. Here's my idea: first project $\Bbb{D}$ to the interval $(-1, 1)$, then expand to $\Bbb{R}$. From there, I'm pretty sure you can use a space-filling curve to fill any disk. Since $U$ is open and connected, it's path connected, and is made up of a countable union of open disks, which we can cover by a single space-filling curve (just join the endpoints for each of the countably many open disks by a path). The result will be heavily non-injective, but I think it works? Feb 14 at 11:46
• Feb 14 at 12:13

The Hahn-Mazurkiewicz theorem, as pointed out by Mathlover in the comments, is enough to show my educated guess actually works. In particular, it implies that closed balls in $$\Bbb{C}$$ are the continuous image of a compact interval. We will need a lemma:

Lemma $$\quad$$ Suppose $$\{C_n\}_{n=0}^\infty$$ is a countable collection of subspaces of a topological space $$X$$ (e.g. $$\Bbb{C}$$) which are continuous images of a compact interval and $$C := \bigcup_{n=0}^\infty C_n$$ is path-connected. Then $$C$$ is the continuous image of $$[0, \infty)$$.

Proof. Consider the intervals $$I_n := [2n, 2n + 1]$$ and $$J_n := [2n + 1, 2n + 2]$$ for $$n \in \Bbb{N} \cup \{0\}$$. Let $$\mathfrak{i}_n : I_n \to C_n$$ be surjective and continuous, and let $$\mathfrak{j}_n : J_n \to C$$ be a continuous function such that $$\mathfrak{j}_n(2n + 1) = \mathfrak{i}_n(2n + 1)$$ and $$\mathfrak{j}_n(2n + 2) = \mathfrak{i}_{n+1}(2n + 2)$$. Then, the map $$\phi : [0, \infty) = \bigcup_{n=0}^\infty (I_n \cup J_n) \to C$$ defined by $$\phi(x) = \begin{cases} \mathfrak{i}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n \le x < 2n + 1 \\ \mathfrak{j}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n + 1 \le x < 2n + 2. \end{cases}$$ This function is made up of countably infinitely many continuous pieces, joined at common points, making the function continuous. It's clearly surjective (even just restricting to $$\bigcup I_n$$), hence $$C$$ is the continuous image of $$[0, \infty)$$ under $$\phi$$. $$\square$$

Note that every ball in $$\Bbb{C}$$ is a countable union of closed balls, and recall that open subsets of $$\Bbb{C}$$ are a countable union of open balls. Also, in a locally path-connected space, open and connected implies path-connected. Thus, $$U$$ is the path-connected countable union of closed balls, so by the lemma, it is a continuous image of $$[0, \infty)$$.

So, to wrap up the proof, as in my comment, simply project the unit disk of $$\Bbb{C}$$ onto $$(-1, 1)$$, which is homeomorphic to $$\Bbb{R}$$. You can simply keep the map constant for points in $$(-\infty, 0]$$, and then use $$[0, \infty)$$ to map continuously onto $$U$$.

• This is amazing, thank you! Feb 14 at 13:45

One can prove a stronger claim: given a nonempty open connected $$U\subseteq \mathbb{C}$$, there's a holomorphic $$\varphi:\mathbb{D}\to \mathbb{C}$$ such that $$\varphi(\mathbb{D})=U$$.

First, let $$U\neq \mathbb{C},\neq \mathbb{C}-\{z_0\}$$. Thanks to the Riemann uniformization theorem, every such open connected subset of $$\mathbb{C}$$ is hyperbolic, i.e. has $$\mathbb{D}$$ as its universal cover. Thus there exists a surjective holomorphic map $$\mathbb{D}\to U$$. Note that there's a surjective holomorphic map $$(\exp(z)+z_0)$$ from $$\mathbb{C}$$ to $$\mathbb{C}-\{z_0\}$$, and thus it suffices to prove the result for $$\mathbb{C}$$. Now, let $$f(z)=z^3-1$$. It is easy to see that $$f:\mathbb{C}-\{1,\exp(2\pi i/3)\}\twoheadrightarrow\mathbb{C}$$. Precomposing with the covering map $$\pi:\mathbb{D}\to \mathbb{C}-\{1,\exp(2\pi i/3)\}$$, we get the claim.

One can also explore "how much" this function fails to be injective. Indeed, the uniformization theorem implies that, for $$U\neq\mathbb{C},U\neq \mathbb{C}-\{z_0\}$$, we have $$U\simeq \mathbb{D}/\Gamma$$, where $$\Gamma$$ is a discrete subgroup of $$Aut(\mathbb{D})$$ (complex automorphisms) isomorphic to $$\pi_1(U)$$. Thus two elements $$z_1,z_2$$ have the same image iff $$\exists \gamma\in \Gamma:\gamma(z_1)=z_2$$ (note that with $$U$$ simply connected we get injectivity).

• Beautiful answer, thank you very much! Feb 14 at 13:46


# $n$ Identical Balls Distributed Into $r$ Urns

• Question

In how many ways can at most $n$ identical balls be distributed into $r$ urns so that the $i$th urn contains at least $m_i$ balls, for each $i=1,...,r$?

Assume that $n\geq \sum_{i= 1}^r m_i$

• My Approach

$$x_1 + x_2 + \cdot\cdot\cdot + x_r = S$$

$$\sum_{i= 1}^r m_i \leq S \leq n$$

$$x_i \geq m_i$$

Let

$$y_i = x_i - m_i$$

then $$y_1+ y_2 + \cdot\cdot\cdot + y_r = S -\sum_{i= 1}^r m_i$$

Now I will proceed to find the number of non-negative integer-valued vectors $(y_1, y_2,...,y_r)$ such that $y_1+ y_2 + \cdot\cdot\cdot + y_r = S -\sum_{i= 1}^r m_i$ for every integer $S$ in the range of $\{\sum_{i= 1}^r m_i\quad,\quad n\}$ which is equal to $$\sum\limits_{S = \sum_{i= 1}^r m_i}^n {S - \sum_{i= 1}^r m_i + r -1\choose r-1}$$

Please have a look at my solution and give me any hints/suggestions you might have.

• What is your question actually? – user 170039 Oct 1 '14 at 13:22
• There is Question, My Approach, and then Please have a look at my solution and give me any hints/suggestions you might have.. What do you mean? – Kermit the Hermit Oct 1 '14 at 13:24
• I mean what question do you want the MSE to answer? Or is it that you want to verify your solution? – user 170039 Oct 1 '14 at 13:27
• Substitute $x_i = m_i + x_i'$ where $x_i' \ge 0$. Now subtract $m_1, m_2, \dots , m_i$ from both sides of the linear diophantine equation you have written. Finally, use stars and bars as usual. – Yiyuan Lee Oct 1 '14 at 13:28

You really seem to have the right idea, defining $y_i=x_i-m_i$, which basically says "If I put $x_i$ balls into urn $i$, this includes $m_i$ obligatory balls and then $y_i$ more that I can choose." This reduces you to solving:

$$x_1+x_2+\dots+x_r=(y_1+m_1)+(y_2+m_2)+\dots+(y_r+m_r)=n$$

which you note is the same as:

$$y_1+y_2+\dots+y_r=n-\sum_{i=1}^r m_i.$$

My question to you is: What is $S$? How is $S$ different from $n$? Unless I'm misreading this problem, $S$ seems to be extraneous. If $S$ is the number of balls you are distributing, the question seems to indicate that $S=n$ (not just $S\le n$).

So if you'll allow us to say $M=\sum m_i$, then you are correct in your reformulation of the problem, you are basically just distributing $n-M$ balls into $r$ urns, which is:

$$\binom{n-M+r-1}{r-1}.$$

That's the last term of your summation. You don't need to sum over values of this "$S$" because you need to distribute all $n$ of the balls. (If you had such a thing, it would be like counting ways of partially distributing the balls, which the question does not seem to indicate.)

You seem to know where the binomial formula for counting these tuples comes from, but for others reading, this formula is determined as explained here.

We basically reduced the problem "distributing $n$ balls into $r$ urns with restriction $m_i$ on each urn" to "distributing $n-M$ balls into $r$ urns" by translating solutions from one to the other. You can think of it physically as starting out by putting all of the $m_i$ balls into each urn as required -- you have $n-M$ left over to distribute freely, and you can use what you already know to count how to do that.

A simpler version of this problem might be "how many ways can you give your two children \$5 allowance (in whole dollar amounts) if each child needs at least \$1?" Rather than count some complicated configuration, you can really just think of this \$1 each as already handed out to them and then solve the easier problem "how many ways can you give your two children \$3 allowance?" (In this case, the answer is 4 ways.)

I am assuming from context (and from your use of the formula) that the urns are distinguishable (they are enumerated and each has a particular value of $m_i$, so philosophically -- and practically speaking, were you to attempt this exercise -- that makes the most sense).

• You just sparked something in me. For some reason I was thinking that you don't have to distribute all $n$ balls. Thus leading to $\sum_{i=1}^r m_i\leq x_1 + x_2 + ... + x_r \leq n$. – Kermit the Hermit Oct 1 '14 at 13:45
• If you are are allowed to use fewer than $n$ balls total (i.e. to just use $S$ balls and throw away the $n-S$ left over), the answer is just the sum (similar to what you have): $\sum_{S=M}^n\binom{S-M+r-1}{r-1}$. – Kellen Myers Oct 1 '14 at 14:17
• What is that $M$? Is that $\sum_{i=1}^r m_i$ ? – Kermit the Hermit Oct 1 '14 at 14:29
• Yes, it's shorthand. I think it's defined somewhere in the middle of my answer. – Kellen Myers Oct 1 '14 at 14:33
• Yes indeed, I missed that. I should pay more attention when reading it would seem. So taken in to consideration the way I interpreted the question, my solution would be correct? – Kermit the Hermit Oct 1 '14 at 14:37