If two angles do not form a linear pair, then they are not supplementary. Linear Pair Two adjacent angles form a linear pair if their non-common sides form a straight angle. Axiom 1: If a ray stands on a line then the adjacent angles form a linear pair of angles. They just need to add up to 90 degrees. Contrapositive: If the angles are not adjacent, then the two angles do not form a linear pair. B A V CLinear Pair Postulate: If two angles form a linear pair, then they are supplementary. Therefore, all the exterior angles are equal, and can be found by dividing 360° by the number of angles . Linear A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. The sides of the angles do not form two pairs of opposite rays. [Image will be Uploaded Soon] In the figure given above, all line segments are passing through the point O, as shown in the figure. Name a linear pair. 9 years ago. where ∠DBC is an exterior angle of ∠ABC and, ∠A and ∠C are the remote interior angles. Related Questions to study. Write. A) vertical angles B) complementary C) acute D) a linear pair * please answer only if you know As the ray OF lies on the line segment MN, angles ∠FON and ∠FOM  form a linear pair. Question 1. Don't two angles ALWAYS have to be adjacent in order to form a line (linear pair)? Assignment Directions: Solve the following word problems involving linear pairs. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. Are two angles in the same plane with a common vertex and a common side, but no common interior points. The measure of one angle is 24 more than the measure of the other angle. As the ray OA lies on the line segment CD, angles ∠ AOD and ∠ AOC form a linear pair. B. Can you explain this answer? Linear PairA linear pair consists of two adjacent angles whose noncommon sides are opposite rays. For #1-6, use the figure at the right. Flashcards. a. Tell whether the angles are only adjacent, adjacent and form a linear pair, or not adjacent. 14. In which diagram do angles 1 and 2 form a linear pair. (iii) When two lines intersect opposite angles are equal. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary means add … The sum of the linear pair of angles is always equal to 180 degrees. A Linear Pair is two adjacent angles whose non-common sides form opposite rays. Match. Tell whether the angles are only adjacent, adjacent and form a linear pair, or not adjacent. The other two pairs of angles are adjacent, but they are not forming a linear pair. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. Name two acute vertical angles. Add your answer and earn points. If two angles are not adjacent, then they do not form a linear pair. ii) When non-common sides of a pair of adjacent angles form opposite rays, then the pair forms a linear pair. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. In the figure, ∠ 1 and ∠ 2 form a linear pair. However, just because two angles are supplementary does not mean they form a linear pair. A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. Favorite Answer. Name an angle complementary to . View Answer . Basically, a linear pair of angles always lie on a straight line. Similarly, ∠GON and ∠HON form a linear pair and so on. The measure of one angle is 15 less than half the measurement of its supplement. For △ABC above, ∠A+∠C+∠ABC=180°. "Complementary Angles" These are angles that add up to 90 degrees. If the two complementary angles are adjacent then they will form a right angle. They share a common vertex, but not a … Consider a Ray OP Stand on the Line Segment  as Shown: The angles which are formed at E are ∠QEM and ∠QEN. Two angles may be supplementary, but not adjacent and therefore not form a linear pair. A linear pair forms a straight angle of 180 degrees, so you have two angles whose sum is given as 180, which means they are supplementary. Therefore, the exterior angles measure = = 72° each. Name a pair of adjacent angles. Learn. Example 2: If two adjacent angles form a linear pair, show that their angle bisectors are perpendicular. When two lines intersect each other at a common point then, a linear pair of angles are formed. Such angle pairs are … Linear-pair-angles are always supplementary. <3 and <4 form a linear pair. Hope this helps! In the diagram above, ∠ABC and ∠DBC form a linear pair. Hence vertical angles are not adjacent. Linear Pair Of Angles. Related Questions to study. Two angles are called supplementary angles if Answer Save. How to Find Adjacent Angles. | EduRev Class 7 Question is disucussed on EduRev Study Group by 2108 Class 7 … B Supplementary angles and linear pairs both add to 180°. Two angles that add to 180 degrees and when adjacent form a straight line. If two angles are not adjacent, then they do not form a linear pair. Two adjacent angles are said to form a linear pair angles , if their non-common arms are two opposite rays. Adjacent angles do not overlap. S_Pinkston. If the angles are adjacent to each other after the intersection of the lines, then the angles are said to be adjacent. "Congruent Angles" They have the same angle in degrees or radian. As the adjacent angles form a linear pair and they are supplementary. <1 and <2 are adjacent angles. Linear Pair Angles. C. If two angles are not supplementary, then they do not form a linear pair. Pro Lite, Vedantu Example 2: Determine if the following pair of angles are vertical angles, linear pairs, or neither. ∴    Yes, ∠1 is vertically opposite to ∠4, And angle vertically opposite to ∠5 is ∠BOC. 1 See answer chanybonilla is waiting for your help. What can you say about its other angle? Example 2: Determine if the following pair of angles are vertical angles, linear pairs, or neither. Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. 12. The sum of their angles is 180° or π radians. ! In the figure above, all the line segments pass through the point O as shown. what are the two factors that make you decide whether a pair of adjacent a angles from a linear pair 2 See answers saryunahar141005 saryunahar141005 Answer: All adjacent angles do not form a linear pair. In the diagram below, ∠ABC and ∠DBE are supplementary since 30°+150°=180°, but they do not form a linear pair since they are not adjacent. Pro Subscription, JEE In which diagram do angles 1 and 2 form a linear pair. In the diagram above, ∠ABC and ∠DBC form a linear pair. By: Carol H. answered • 02/22/18. Also, the ray is that part of the line which has only one endpoint. Test. A linear pair of angles is formed when two lines intersect. Thus, the sum of the exterior angles is: For regular polygon, all of the angles of a are equal. so, a pair of straight angles can also be adjacent angle. Join Yahoo Answers and get 100 points today. The pair of adjacent angles are constructed on a line segment, but not all adjacent angles are linear. Learn how to identify angles from a figure. <2 and <4 <2 and <4. This can be shown by using linear pairs. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. It can't think of a contradiction. Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. Example 2 : In the diagram shown below, Solve for x and y. Get the answers you need now. 14. (ii) Adjacent complementary angles. Substituting the second equation into the first equation we get. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. D. If two angles form a linear pair, then they are supplementary. Linear pair. Supplementary angles are two angles whose same is 180o Linear pairs are adjacent angles who share a common ray and whose opposite rays form a straight line. A circle of the infinite radius can be visualized on the straight line. The figure shows the design on an outdoor fence. Gravity. Use ½ sheet of paper. Ans : (d) Bisectors of the adjacent angles forming a linear pair form a right angle. A linear pair is a pair of adjacent angles whose non-adjacent sides form a line. See tutors like this. 3. The sum of their angles is 180 ° or π radians. Learn vocabulary, terms, and more with flashcards, games, and other study tools. In the regular pentagon above, n = 5. 0 0. A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. PLAY. Supplementary angles do not have to be adjacent, whereas a linear pair must be adjacent and create a straight line. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. An angle is 18c less than its complementary angle. 1 See answer chanybonilla is waiting for your help. (iv) Unequal supplementary angles. a. b. c. Exploration #1: Use this link to explore the relationship with vertical angles. Linear pair is a pair of adjacent angles where non-common side forms a straight line. Answer:Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pairStep-by-step explanation:see A linear pair forms a straig… 2 0. If two angles are supplementary, then they form a linear pair. They have common vertex O. i tried searching it but i wasn't sure cause it was so confusing at first i thought i was getting it but i wasn't. Correct answer to the question: 5. Also, there is a common arm that represents both the angles of the linear pair. A line segment with A and B as two endpoints is represented as AB. Spell. They do not form a straight line. Easily fill out PDF blank, edit, and sign them. They have common side OB. If a linear pair is formed by two angles, the uncommon arms of the angles forms a straight line. In a diagram angle 1 and angle 2 form an linear pair and angle 2 and angle 3 form vertical angles. The above discussion can be stated as an axiom. ii) When non-common sides of a pair of adjacent angles form opposite rays, then the pair forms a linear pair. If two angles form a linear pair and the measure of one angle is 47 degrees, what is the measure of the other angle. If two angles are adjacent, they form a linear pair? The measure of an angle is three times the measurement of its complement. Linear pair of angles - definition Two angles are said to be in a linear pair if they are adjacent to each other, lie on the same side of the line and the sum of their measures is 1 8 0 o. ∠BOC and ∠AOC are linear-pair-angles. Therefore, AB represents a line. Created by. 13. 1 Answer. A Linear Pair is two adjacent angles whose non-common sides form opposite rays. Vertically Opposite Angles When a pair of lines intersect, as shown in the fig. Also, ∠ABC and ∠DBC form a linear pair so. (iv) Unequal supplementary anglesFor line BD ∠ BOC and ∠ COD are in linear pair Therefore, ∠BOC + ∠COD = 180° So, their sum is 180° And, ∠BOC is not equal to ∠COD ∴ ∠BOC & ∠COD are unequal supplementary angles Ex 5.1, 14 In the adjoining figure, name the following pairs of angles. Such angle pairs are called a linear pair.. Angles A and Z are supplementary because they add up to 180°.. Vertical angles: When intersecting lines form an X, the angles on the opposite sides of the X are called vertical angles. Find the values of the angles l, p, and q in each of the following questions, ∠DOC = ∠AOE           (vertically opposite angles), Pair of Linear Equations in Two Variables, Vedantu This statement is correct. (iii)   Does the angles ∠COE and ∠EOD form a linear pair? ∠BOC and ∠AOC are linear-pair-angles. It is known that the angle between the two line segments ME and EN is 180°. Angles ∠ Z W I and ∠ H W I are adjacent angles. (vi)  Find the vertically opposite angle of ∠5? In the figure, clearly, the pair $$\angle BOA$$ and $$\angle AOE$$ form adjacent complementary angles. Complete Identify Each Pair Of Angles As Adjacent, Vertical, Complementary, Supplementary, Or A Linear Pair online with US Legal Forms. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary … Add comment More. Anonymous. Supplementary angles: Two angles that add up to 180° (or a straight angle) are supplementary. Relevance. Angle Pairs (Vertical, linear pair, adjacent, complementary & supplementary) STUDY. The two axioms mentioned above form  Linear Pair Axioms and are helpful in solving various mathematical problems. In a linear pair, the arms of the angles that are not always common or collinear i.e., they lie on a  straight line. A: If two angles form a linear pair, then angles are supplementary. 1 and 2 are adjacent angles. If two parallel lines are intersected by a transversal, then each pair of corresponding angles so formed is (a) Equal (b) Complementary (c) Supplementary (d) None of these Ans : (a) Equal 4. If a point O is taken anywhere on the line segment AB as shown, then the angle between the two line segments AO and OB is a straight angle i.e., 180°. Two adjacent supplementary angles form a linear pair. Here, these angles are in linear pair as. Report 1 Expert Answer Best Newest Oldest. If two angles are a linear pair, then they are supplementary. T or F? There are n angles in the polygon, so there are n linear pairs. If Two Angles Form A Linear Pair, The Angles Are Supplementary. Two adjacent supplementary angles form a linear pair. A linear pair is a pair of adjacent, supplementary angles, whihc means they add up to 180 degrees. The measure of an angle is three times the measurement of its complement. 12. Linear Pairs. If two angles form a linear pair, the angles are supplementary. Hence, we can also say that a linear pair of angles is the adjacent angle whose non-common arms are basically opposite rays. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. the term adjacent means next to each other. 1 decade ago. Two adjacent angles are said to form a linear pair angles , if their non-common arms are two opposite rays. In the figure, clearly, the pair $$\angle AOE$$ and $$\angle EOD$$ form adjacent angles that do not form a linear pair. The angles are adjacent, sharing ray BC, and the non-adjacent rays, BA and BD, lie on line AD. (add up to 180 0) ∠BOC + ∠AOC = 180 0 Examples : 1) One of the angles forming a linear-pair is a right angle. Then, find the angle measures. A straight angle is formed when the angle between two lines is 180 degrees. What are not adjacent angles? Two vertical angles are always the same size as each other. adjacent angles are the angles which have same pair of vertex and have one common side. Two angles form a linear pair. Sorry!, This page is not available for now to bookmark. Covid-19 has led the world to go through a phenomenal transition . Similarly, ∠GON and ∠HON form a linear pair and so on. Converse statement inverses statement Contrapositive statement Key Concepts: Terms in this set (12) Adjacent complementary angles (Image) angles that are "next to" each other. a. b. c. Exploration #1: Use this link to explore the relationship with vertical angles. Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. Adjacent Angles• Share a common side• Share the same vertex• Do not share any interior points B A C V Bistro 72 Reviews, 5 Gallon Air Compressor, Clementine Churchill Images, How Old Is Chopper, Allegiant Air Flight Schedules, Counting Video For Kindergarten, Langerhans Cells Function Quizlet, Eine Kleine Nachtmusik Mr Bean, Improvisation Of Melody Brainly, Dynamite Definition Synonyme, Kazimir Malevich Art,


Prove that the set of $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$ is countable.

I have seen proofs of $\mathbb{N}\times\mathbb{N}$. The proof by list where you list each element in a pair of elements and then counting them diagonally is the most convincing. I have two thoughts in mind:

1. Write it as a composition of functions. Show that both functions are bijections, so the composition is also a bijection.
2. Show that $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$ has the same cardinality as $\mathbb{N}\times\mathbb{N}$, and $\mathbb{N}\times\mathbb{N}$ has the same cardinality as $\mathbb{N}$.

Generally, I'm having difficulty going from $\mathbb{N}\times\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}\times\mathbb{N}$, since it easy to prove the bijection from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$.

• Youre idea in 1. seems like a good approach. – JJC94 Oct 31 '17 at 0:48
• It's not a proof-explanation if you don't point at a specific proof and ask us to explain it. – Asaf Karagila Oct 31 '17 at 0:51
• Consider the Cantor pairing. – Wuestenfux Oct 31 '17 at 9:42

If you already know that $\Bbb{N\times N}$ is countable, fix a bijection $f\colon\Bbb{N\times N\to N}$. Now consider $g\colon\Bbb{N\times N\times N\to N\times N}$ defined as: $$g(n,m,k)=(n,f(m,k)),$$ or better yet $h\colon\Bbb{N\times N\times N\to N}$ defined as $$h(n,m,k)=f(n,f(m,k))$$ and cut out the middle-man.

As my freshman discrete mathematics professor used to tell us, go home and convince yourself this is correct.

• I can convince myself this is correct, but how can I prove it to a stubborn skeptic? – tigerustin Nov 1 '17 at 2:11
• Slowly and carefully by verifying the definitions of a bijection. – Asaf Karagila Nov 1 '17 at 6:53
• I proved injectivity. How would I prove surjectivity? – tigerustin Nov 1 '17 at 20:21
• Exactly the same. Slowly, carefully and by using the fact that $f$ is already surjective itself. – Asaf Karagila Nov 1 '17 at 20:23
• I really don't understand. Can you explain further please? I don't want to have to start another thread on this question. – tigerustin Nov 2 '17 at 2:10

Why not think in stages? Let $\varphi : \mathbb{N}\times\mathbb{N} \to \mathbb{N}$ denote your favorite bijection. But then

$$\mathbb{N} \times \mathbb{N} \times \mathbb{N} = (\mathbb{N} \times \mathbb{N}) \times \mathbb{N} \approx \varphi(\mathbb{N}\times \mathbb{N}) \times \mathbb{N} = \mathbb{N} \times \mathbb{N} \approx \varphi(\mathbb{N}\times \mathbb{N}) = \mathbb{N}$$ (where, in this context, $=$ means equality, and $\approx$ denotes "of the same cardinality"). You can do this for any number of copies of $\mathbb{N}$ via an induction argument.

Consider $f:\mathbb N \times \mathbb N \times \mathbb N \to \mathbb N$ given by $f(a,b,c)=2^a3^b5^c$

and use the Fundamental theorem of arithmetic for a cute solution.

• Gives an injective mapping. – Wuestenfux Oct 31 '17 at 9:41
• And it is clearly infinite, so... – Andres Mejia Oct 31 '17 at 14:06
• It would be difficult to prove bijectivity of this function? – tigerustin Nov 1 '17 at 2:12
• It’s not bijective. It is an infinite subset of the natural numbers – Andres Mejia Nov 1 '17 at 2:15

I like to think of 'countable' as 'listable' ... That is, as long as you can create a list of objects and ensure that all elements of some set $S$ are somewhere on that list, then $S$ is countable. To me, the advantage of that kind of thinking is that elements from $S$ may be repeated on the list, and that the list may even contain objects that are not in $S$; as long as every element of $S$ is at least once in the list, it's clear that $S$ is countable. As such, this method avoids having to construct explicit and possibly complicated bijections.

So, how can we create a list of all elements of $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$?

Here is one way:

First, get all triples with elements that sum to $0$: which is just $<0,0,0>$

Now get all those that sum to $1$: $<1,0,0>,<0,1,0>,<0,0,1>$

Then those that sum to $2$, then $3$, etc.

Thus, the list is:

$<0,0,0>,<1,0,0>,<0,1,0>,<0,0,1>,<2,0,0>,<1,1,0>,...$

Since for any $n$ there can only be finitely many triples whose elements sum to $n$, you will eventually get to every triple.

Note that this is of course just the same thing as diagonally going through an array of elements, just now in $3$ dimensions.

• Reminiscent of the method I know for showing the algebraics are countable. Nice. – Kaj Hansen Oct 31 '17 at 2:34

Cantor-Bernstein's Theorems states that if $A$ and $B$ are two sets and there exists two injections $f:A \to B$ and $g:B \to A$ then there is a biyection from $A$ to $B$ (i.e. $A$ and $B$ have the same cardinality).

Let $f:\mathbb{N} \times \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ be defined as $f(m,n,k)=2^n \cdot 3^m \cdot 5^k$, then if $f(m_1,n_1,k_1) = f(m_2,n_2,k_2) \Leftrightarrow 2^{m_1}3^{n_1}5^{k_1}=2^{m_2}3^{n_2}5^{k_2}$, the Fundamental Theroem of Arithmetic shows that $m_1=m_2$, $n_1=n_2$ and $k_1=k_2$. Then $f$ is a injective function.

Next, consider the function $g:\mathbb{N} \to \mathbb{N} \times \mathbb{N} \times \mathbb{N}$ defined by $g(n)=(0,0,n)$, then $$g(n)=g(m) \Leftrightarrow (0,0,n) =(0,0,m) \Leftrightarrow n=m$$ and hence $g$ is injective.

Applying Canotr-Bernstein's Theorem gives that $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$ is countable.

Using induction, you can easily prove this result for the product of any finite number of copies of $\mathbb{N}$. The hardest part of the proof is the base case, which is that $\mathbb{N} \times \mathbb{N}$ is countable. It seems you already have this result.

For the induction hypothesis, assume that $\displaystyle \prod^n \mathbb{N}$ is countable. Now we just need to show that $\displaystyle \prod^{n+1} \mathbb{N}$ is countable. First, note that can write the $(n\!+\!1)$-fold product in the form $\mathbb{N} \times \left( \displaystyle \prod^n \mathbb{N} \right)$. A bijection $\displaystyle f:\prod^n \mathbb{N} \rightarrow \mathbb{N}$ is guaranteed by the induction hypothesis, which we can use to construct a bijection between the $(n\!+\!1)$-fold product and $\mathbb{N} \times \mathbb{N}$:

$$(x_1, x_2, \cdots, x_{n+1}) \mapsto (x_1, f(x_2, x_3, \cdots, x_{n+1}))$$

And $\mathbb{N} \times \mathbb{N}$ is countable per the base case.


# Ways to fill a $n\times n$ square with $1\times 1$ squares and $1\times 2$ rectangles

I came up with this question when I'm actually starring at the wall of my dorm hall. I'm not sure if I'm asking it correctly, but that's what I roughly have:

So, how many ways (pattern) that there are to fill a $n\times n:n\in\mathbb{Z}_{n>0}$ square with only $1\times 1$ squares and $1\times 2$ rectangles?

For example, for a $2\times 2$ square:

• Four $1\times 1$ squares; 1 way.

• Two $1\times 1$ squares and one $1\times 2$ rectangle; $4$ ways total since we can rotate it to get different pattern.

• Two $1\times 2$ rectangles; 2 ways total: placed horizontally or vertically.

$\therefore$ There's a total of $1+4+2=\boxed{7}$ ways to fill a $2\times 2$ square.

So, I'm just wondering if there's a general formula for calculating the ways to fill a $n\times n$ square.

Thanks!

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+1 for coming up with your own math question. – Sammy Black Apr 11 '13 at 20:58
For $1\times n$ boards, you get the Fibonacci numbers. For $2\times n$ boards it's already a bit harder. You get the recurrence relation $a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$, see oeis.org/A030186 – azimut Apr 11 '13 at 21:49

A hand count for $n=3$ yields $131$ ways, and a search for $1,7,131$ yields OEIS sequence A028420, the "number of monomer-dimer tilings of an $n\times n$ chessboard". The entry doesn't provide a formula (so it's likely that none is known), but some references.

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so it's like a program? – user67258 Apr 11 '13 at 21:18
See also oeis.org/A210662 which gives all the rectangles, not just the squares. – Charles Apr 11 '13 at 21:32
@herderp: Sorry, I don't understand that question. What's like a program? – joriki Apr 11 '13 at 22:12
I thought the sequence is computed by a computer program – user67258 Apr 11 '13 at 22:24

If you just used $1\times2$ rectangles, then this is same as finding the number of matchings in the $m \times n$ rectangle graph, and a formula for that has been given by Kastelyn:

$$\sqrt{\left|\prod_{j=1}^{m}\prod_{k=1}^{n} \left(2 \cos \frac{\pi j}{m+1} + 2i\cos \frac{\pi k}{n+1}\right)\right|}$$

This was done, by mapping the number to the square root of the determinant of a weighted adjacency matrix of the graph.

If you include $1 \times 1$ squares, you essentially need to find the sum over all subgraphs (think of placing a $1\times 1$ square as deleting that vertex) of the $m \times n$ graph, and for each subgraph, the determinant approach still works, I believe, but might not have a nice closed form.

You can find a nice exposition for the $1\times 2$ case in the first chapter here: http://users.ictp.it/~pub_off/lectures/lns017/Kenyon/Kenyon.pdf

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We can probably give some upper and lower bounds though. Let $t_n$ be the possible ways to tile an $n\times n$ in the manner you described. At each square, we may have $5$ possibilities: either a $1\times 1$ square, or $4$ kinds of $1\times 2$ rectangles going up, right, down, or left. This gives you the upper bound $t_n \leq 5^{n^2}$.

For the lower bound, consider a $2n\times 2n$ rectangle, and divide it to $n^2$ $2\times 2$ blocks, starting from the top left and putting a $2\times 2$ square, putting another $2\times 2$ square to its right and so on... For each of these $2\times 2$ squares, we have $5$ possible distinct ways of tiling. This gives the lower bound $t_{2n} \geq 7^{n^2}$. Obviously, $t_{2n+1} \geq t_{2n},\,n \geq 1$, and therefore $t_n \geq 7^{\lfloor \frac{n}{2}\rfloor ^2}$.

Hence, \begin{align} 7^{\lfloor \frac{n}{2}\rfloor ^2} \leq t_n \leq 5^{n^2}, \end{align} or roughly (if $n$ is even), \begin{align} (7^{1/4})^{n^2} \leq t_n \leq 5^{n^2}. \end{align} BTW, $7^{1/4} \geq 1.6$. So, at least we know $\log t_n \in \Theta(n^2)$.

Note: Doing the $3\times 3$ case for the lower bound, we get $(131)^{1/9} \geq 1.7$ which is slightly better.

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# Why is the eigenvector of a covariance matrix equal to a principal component?

If I have a covariance matrix for a data set and I multiply it times one of it's eigenvectors. Let's say the eigenvector with the highest eigenvalue. The result is the eigenvector or a scaled version of the eigenvector.

What does this really tell me? Why is this the principal component? What property makes it a principal component? Geometrically, I understand that the principal component (eigenvector) will be sloped at the general slope of the data (loosely speaking). Again, can someone help understand why this happens?

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Short answer: The eigenvector with the largest eigenvalue is the direction along which the data set has the maximum variance. Meditate upon this.

Long answer: Let's say you want to reduce the dimensionality of your data set, say down to just one dimension. In general, this means picking a unit vector $u$, and replacing each data point, $x_i$, with its projection along this vector, $u^T x_i$. Of course, you should choose $u$ so that you retain as much of the variation of the data points as possible: if your data points lay along a line and you picked $u$ orthogonal to that line, all the data points would project onto the same value, and you would lose almost all the information in the data set! So you would like to maximize the variance of the new data values $u^T x_i$. It's not hard to show that if the covariance matrix of the original data points $x_i$ was $\Sigma$, the variance of the new data points is just $u^T \Sigma u$. As $\Sigma$ is symmetric, the unit vector $u$ which maximizes $u^T \Sigma u$ is nothing but the eigenvector with the largest eigenvalue.

If you want to retain more than one dimension of your data set, in principle what you can do is first find the largest principal component, call it $u_1$, then subtract that out from all the data points to get a "flattened" data set that has no variance along $u_1$. Find the principal component of this flattened data set, call it $u_2$. If you stopped here, $u_1$ and $u_2$ would be a basis of the two-dimensional subspace which retains the most variance of the original data; or, you can repeat the process and get as many dimensions as you want. As it turns out, all the vectors $u_1, u_2, \ldots$ you get from this process are just the eigenvectors of $\Sigma$ in decreasing order of eigenvalue. That's why these are the principal components of the data set.

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Phenomenal answer. Thank you. – Ryan Mar 28 '11 at 11:55

Some informal explanation:

Covariance matrix $C_y$ (it is symmetric) encodes the correlations between variables of a vector. In general a covariance matrix is non-diagonal (i.e. have non zero correlations with respect to different variables).

But it's interesting to ask, is it possible to diagonalize the covariance matrix by changing basis of the vector?. In this case there will be no (i.e. zero) correlations between different variables of the vector.

Diagonalization of this symmetric matrix is possible with eigen value decomposition. You may read 'A Tutorial on Principal Component Analysis'by Jonathon Shlens to get a good understanding. (pages 6-7)

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# Using mathematical induction prove that $11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8

Prove that $\phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8 for all $n \in N$.

Base: $n = 0$

$8 | 11 + 3 - 6$ is obvious.

Now let $\phi(n)$ be true we now prove that is also true for $\phi(n+1)$.

So we get $11 \cdot 3^{n+1} + 3 \cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8.

For instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here.

Suppose $11*3^n + 3*7^n - 6 = 8k$

The $11*3^{n+1} + 3*7^{n+1} - 6 = 11*3^n*3 + 3*7^n*7 - 6$

$=3(11*3^n + 3*7^n-2) + 4*3*7^n$

$= 3(11*3^n + 3*7^n - 6) + 4*3*7^n + 12$

$= 3(8k) + 4(3*7^n + 3)$; $3*7^n$ is odd and $3$ is odd so $(3*7^n + 3)$ is even.

$= 3(8k) + 8(\frac{3*7^n + 3}2) = 8(3k + \frac{3*7^n + 3}2)$.

======

Actually I like and am inspired by Bill Dubuques answer.

We want to prove $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 0 \mod 8$

And we know $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 3*3^n + 3*(-1)^n -6 = 3^{n+1} + 3*(-1)^n - 6 \mod 8$.

So it's a matter of showing $f(n) = 3^{n+1} + 3(-1)^n \equiv 6 \mod 8$.

And if we notice $f(n+2) = 3^{n+3} + 3(-1)^{n+2} = 3^{n+1}*9 + 3(-1)^{n} \equiv 3^n + 3(-1)^{n}= f(n) \mod 8$.

So it's now just a matter of showing for $f(0) \equiv f(1) \equiv 6 \mod 8$.

Which is easily verified $3^1 + 3*(-1)^0 =3+3= 6$ and $3^2 + 3*(-1)^1 = 9 -3 = 6$

• Yap, that's it. Thank you, just what I was looking for. – stigma6 Jan 27 '17 at 7:15
• You want a secret confession? When I began answering I didn't know how it would turn out. But I know I had to factor $f(n+1) = manipulate(f(n)) = manipulate(8k)$ and I had faith that $manipulate(8k) = 8j$ and I just chewed on it to see what would happen. That's how a lot of figuring induction proofs. You know $f(n) = Property(t)$ and $f(n+1) = manipulate(f(n))=manipulate(Property(t))$ and know you want $manipulate(Property(t))=Property(s)$. Then you just gum and chew away until you get it. – fleablood Jan 27 '17 at 7:36

${\rm mod}\ 8\!:\ f(n\!+\!2)\equiv f(n)\,$ by $\,a\equiv 3,7\,\Rightarrow\,a^{\large 2}\equiv 1\,\Rightarrow\,a^{\large n+2}\equiv a^{\large n}.\,$ Thus $\,8\mid f(n)\iff 8\mid f(n\!+\!2),\,$ hence by (strong/parity) induction, it is true for all $n$ $\iff$ it is true for the base cases $\,n=0,1.$

• @Downvoter The downvote is puzzling. If something is not clear then please feel welcome to ask questions and I will be happy to elaborate. This is one of the easiest ways to prove it, – Bill Dubuque Jan 26 '17 at 22:40
• If congruences are unknown then we can eliminate them, viz. show that $8$ divides $\,f(n\!+\!2)-f(n)\,$ because $8$ divides $\,3^{\large n+2}-3^{\large n} = (3^{\large 2}-1)3^{\large n} = 8\cdot 3^n,\,$ and similarly $8$ divides $\,7^{\large n+2}-7^{\large n}.\ \$ – Bill Dubuque Jan 26 '17 at 23:04

Setup the same as your current work:

$\dots$

$\dots = 11\cdot 3^{n+1}+3\cdot 7^{n+1}-6 = 11\cdot 3\cdot 3^{n}+ 3\cdot 7\cdot 7^n - 6$

$=33\cdot 3^n + 21\cdot 7^n - 6 = (11+22)\cdot 3^n + (3 + 18)\cdot 7^n - 6$

$=\underbrace{11\cdot 3^n + 3\cdot 7^n - 6}_{\text{should be familiar}} + \underbrace{22\cdot 3^n + 18\cdot 7^n}_{\text{unknown}}$

Now, what can we say about $22\cdot 3^n+18\cdot 7^n$? Anything? You say in a previous example, you had to run a second induction proof to finish, might that be useful here?

• fyi: your answer was downvoted 1 second before mine, so it doesn't appear that the downvoter took the time to carefully consider the answers. – Bill Dubuque Jan 26 '17 at 22:50


Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?

Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?

So far I have seen the following results:

1. If $f\colon U\to\mathbb C$ has an antiderivative $F$ on $U$ then $\displaystyle\int_\gamma f(z)~\mathrm dz=F(\gamma(b))-F(\gamma(a))$ along a smooth curve $\gamma\colon[a,b]\to\mathbb C$.
2. If $f$ has an antiderivative then $\displaystyle\int_\gamma f(z)~\mathrm dz$ depends only on the start and end point of the curve $\gamma$.
3. If $f$ has an antiderivative, then $\displaystyle\oint_\gamma f(z)~\mathrm dz=0$ for all closed curves $\gamma$.

My problem is that I could use those results to show a function has no antiderivative (like $1/z$ yielding $2\pi\mathrm i\neq 0$ on $\partial B_r(0)$) but I am unsure what might work the other way round. What might be useful here in terms of complex analysis?

EDIT

There is a follow-up question based on the problem above.

Use the result from above to calculate

$$\int_{-\infty}^\infty\exp(-x^2-\mathrm ikx)\,\mathrm dx, k\in\mathbb R.$$

I guess something constructive like GEdgar's suggestion might be useful here, isn't it?

EDIT 2

Here is my solution for the follow-up question

\begin{align*} \int_{-\infty}^\infty \exp(-x^2-\mathrm ikx)\,\mathrm dx &= \int_{-\infty}^\infty \exp(-(x^2+\mathrm ikx))\,\mathrm dx \\ &= \int_{-\infty}^\infty \exp\left(-\left(x^2+\mathrm ikx-\frac{k^2}{4}+\frac{k^2}{4}\right)\right)\,\mathrm dx \\ &= \int_{-\infty}^\infty \exp\left(-\left(\left(x+\frac{1}{2} \mathrm ik\right)^2+\frac{k^2}{4}\right)\right)\,\mathrm dx \\ &= \exp\left(-\frac{k^2}{4}\right)\int_{-\infty}^\infty \exp\left(-\left(x+\frac{1}{2} \mathrm ik\right)^2\right)\,\mathrm dx \\ &= \exp\left(-\frac{k^2}{4}\right)\int_{-\infty}^\infty \exp(-t^2)\,\mathrm dt \\ &= \exp\left(-\frac{k^2}{4}\right)\sqrt{\pi} \end{align*}

Another way.

$\exp(-z^2)$ is an entire function. Its power series at the origin converges for all $z$.
Take the anti-derivative of that series term-by-term. It still converges for all $z$. (Use the formula for radius of convergence.) So the series of antiderivatives is an antiderivative for $\exp(-z^2)$.

That argument works for all entire functions. In this case: $$f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!} z^{2n} \\ F(z) :=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\;\frac{z^{2n+1}}{2n+1} \\ F'(z) = f(z)$$

The composition of functions with an antiderivative has an antiderivative.

$\exp$ has an antiderivative because it is its own derivative (and hence its own antiderivative) everywhere.

$z \mapsto -z$ has an antiderivative $z \mapsto \frac{-z^2}{2}$.

$z \mapsto z^2$ has antiderivative $z \mapsto \frac{z^3}{3}$.

• Really? $\exp(\exp(x))$ doesn't have an antiderivative though, does it? – Lundborg May 1 '16 at 12:17
• @Neutronic Not in closed form, certainly, but there is a function which differentiates to $\exp \circ \exp$ everywhere. – Patrick Stevens May 1 '16 at 12:18
• @PatrickStevens ... Still, we need a proof of that fact. – GEdgar May 1 '16 at 13:38

Define, for some $z_0\in \mathbb{C}$ $$g(z)=\int_{z_0}^z e^{-t^2}dt$$ Now for any $2$ homotopic (within the domain where $e^{-z^2}$ is analytic) curves the value of this integral will be the same. Since $e^{-z^2}$ is analytic and $\mathbb{C}$ is simply connected, we find that the value of $g(z)$ is independent of the chosen contours. Hence this is a well defined function of $z$, and clearly $g'=f$.

more details: We know that holomorphic differentials are closed, i.e. $d(fdz)=0$. We also know that on a simply connected domain closed forms are exact. This means we can write, using stokes theorem, $$\int_{\gamma}f dz = \int_{\gamma}dg=\int_{\partial\gamma}g=g(\gamma(1))-g(\gamma(0))=g(z)-g(z_0)$$ which shows that the integral does not depend on the path $\gamma$ we chose.

• So far we haven't been talking about homotopy so I find it difficult to fully understand your proof. I have a feeling you are trying to pick arbitrary curves which start and end point are fixed and showing that the integrals along those curves is always the same. Can you confirm this? I am struggling to understand how you deduce this exactly. – Christian Ivicevic May 1 '16 at 13:04
• Yes that is what I am doing. I'll include a bit more detail in my answer – user2520938 May 1 '16 at 13:07


# solve linear congruence

We need now aplly the above recursive relation: Finally, solutions to the given congruence are, $$x \equiv 61, 61 + 211, 61 \pmod{422} \equiv 61, 272 \pmod{422}.$$. Necessary cookies are absolutely essential for the website to function properly. This says we can take x = (105*7 + 65)/50 = 16. Solving linear congruences is analogous to solving linear equations in calculus. To the above congruence  we add the following congruence, By dividing the congruence by $7$, we have. The solution to the congruence $ax \equiv b \pmod m$ is now given with: $$x \equiv v + t \cdot m’ \pmod m, \quad t= 0, 1, \ldots, d-1.$$. So if g does divide b and there are solutions, how do we find them? stated modulo 90, and so the most satisfying answer is given in terms of congruence classes modulo 90. The solution of a linear congruence can be found in the Wolfram Language using Reduce[a*x == b, x, Modulus -> m]. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. We must now see how many distinct solutions are there. solve the linear congruence step by step. Let x 0 be any concrete solution to the above equation. However, if we divide both sides of the congru- Solve the linear system sa+ tm= 1: Then sba+ tbm= b: So sba b (mod m) gives the solution x= sb. We find y = 4. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. This problem has been solved! Linear Congruence Video. The algorithm says we should solve 100y â¡ -13(mod 7). A Linear Congruence is a congruence mod p of the form where,,, and are constants and is the variable to be solved for. most likely will be coming back here in the future, Thank you! Here, "=" means the congruence symbol, i.e., the equality sign with three lines. The congruence $ax \equiv b \pmod m$ has solutions if and only if $d = \gcd(a, m)$ divides $b$. If (a;m) = 1, then the congruence ax b mod mphas exactly one solution modulo m. Constructive. Then the solutions to ax â¡ b (mod m) are x = y + tm/g where t = 0, 1, 2, â¦,Â g-1. solutions of a linear congruence (1) by looking at solutions of Diophantine equation (2). We can repeat this process recursively until we get to a congruence that is trivial to solve. Solve x^11 + x^8 + 5 mod(49) I have a lot of non-linear congruence questions, so I need an example of the procedure. If u 1 and u 2 are solutions, then au 1 b (mod m) and au 2 b (mod m) =)au 1 au 2 (mod m) =)u 1 u Solve The Linear Congruence Step By Step ; Question: Solve The Linear Congruence Step By Step . Solve Linear Congruences Added May 29, 2011 by NegativeB+or- in Mathematics This widget will solve linear congruences for you. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. $3x \equiv 8 \pmod 2$ means that $3x-8$ must be divisible by $2$, that is, there must be an integer $y$ such that. Find all solutions to the linear congruence $5x \equiv 12 \pmod {23}$. Solve the following system of linear congruences: From the first linear congruence there exists a such that: Substituting this into the second linear congruence gives us: Notice that , and so there exists a solution. Then x 0 ≡ … Let d = gcd(c,m), and choose q, r 2Z such that c = dq and m = d r. If b is a solution to (1), then it is also a If $d \nmid b$, then the linear congruence $ax \equiv b \pmod m$ has no solutions. 1 point Solve the linear congruence 2x = 5 (mod 9). Example 1. Existence of solutions to a linear congruence. // Example: To solve € … Since 7 and 100 are relatively prime, there is a unique solution. Thanks a bunch, Your email address will not be published. Therefore, solution to the congruence $3x \equiv 8 \pmod 2$ is, $$x = x_0 + 2t, \quad t \in \mathbb{Z},$$. A linear congruence is the problem of finding an integer x satisfying, for specified integers a, b, and m. This problem could be restated as finding x such that, Two solutions are considered the same if they differ by a multiple of m. (It’s easy to see that x is a solution if and only if x + km is a solution for all integers k.). Linear Congruence Calculator. First, suppose a and m are relatively prime. Thus: Hence our solution in least residue is 7 (mod 23). The algorithm can be formalized into a procedure suitable for programming. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Solving the congruence a x ≡ b (mod m) is equivalent to solving the linear Diophantine equation a x – m y = b. See the answer. Your email address will not be published. The result is closely related to the Euclidean algorithm. One or two coding examples would’ve been great, though =P, this really helpful for my project. Since $\gcd(6,8) = 2$ and $2 \nmid 7$, there are no solutions. We can calculate this using the division algorithm. The calculations are somewhat involved. This is a linear Diophantine equation and it has a solution if and only if $d = \gcd(a, m)$ divides $b$. Theorem 1. If y solves this new congruence, then x = (my + b)/ a solves the original congruence. In particular, (1) can be rewritten as This field is denoted by $\mathbb{Z}_p$. This is progress because this new problem is solving a congruence with a smaller modulus since a < m. If y solves this new congruence, then x = (my + b)/a solves the original congruence. Example. By the Euler’s theorem, $$a^{\varphi (m)} \cdot b \equiv b \pmod m.$$, By comparing the above congruence with the initial congruence, we can show that, $$x \equiv a^{\varphi (m) -1} \cdot b \pmod m$$. The linear congruence The result is closely related to the Euclidean algorithm. Solving Linear Congruence by Finding an Inverse. Although Bill Cook's answer is completely, 100% correct (and based on the proof of the Chinese Remainder Theorem), one can also work with the congruences successively; we know from the CRT that a solution exists. This category only includes cookies that ensures basic functionalities and security features of the website. Recall that since $(31,24)=1$ and $1|12$ there is exactly one incongruent solution modulo $24.$ To find this solution let’s use the definition of congruence, … Multiply the rst congruence by 2 1 mod 7 = 4 to get 4 2x 4 5 (mod 7). The CRT is used solve systems of congruences of the form $\rm x\equiv a_i\bmod m_{\,i}$ for distinct moduli $\rm m_{\,i}$; in our situation, there is only one variable and only one moduli, but different linear congruences, so this is not the sort of problem where CRT applies. Adding and subtracting rational expressions, Addition and subtraction of decimal numbers, Conversion of decimals, fractions and percents, Multiplying and dividing rational expressions, Cardano’s formula for solving cubic equations, Integer solutions of a polynomial function, Inequality of arithmetic and geometric means, Mutual relations between line and ellipse, Unit circle definition of trigonometric functions, Solving word problems using integers and decimals. The linear congruence equation ax = b (mod n) may be rewritten as ax1 = b - nx2 where x1, x2 -E- Z. That works in theory, but it is impractical for large m. 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If we need to solve the congruence $ax \equiv b \pmod p$, we must first find the greatest common divisor $d= \gcd(a,m)$ by using the Euclidean algorithm. So, we restrict ourselves to the context of Diophantine equations. Rather, this is linear algebra. The brute force solution would be to try each of the numbers 0, 1, 2, â¦,Â m-1 and keep track of the ones that work. Solve the following congruence: We must first find $\gcd(422, 186)$ by using the Euclidean algorithm: Therefore, $\gcd(422, 186) = 2$. Proposition 5.1.1. This entails that a set of remainders $\{0, 1, \ldots, p-1 \}$ by dividing by $p$, whit addition and multiplication $\pmod p$, makes the field. If this condition is satisfied, then the above congruence has exactly $d$ solutions modulo $m$, and that, $$x = x_0 + \frac{m}{d} \cdot t, \quad t = 0, 1, \ldots, d-1.$$. Previous question Next question Get more help from Chegg. Thanks :) If the number $m =p$ is a prime number, and if $a$ is not divisible by $p$, then the congruence $ax \equiv b \pmod p$ always has a solution, and that solution is unique. Now substitute for x in the second congruence: 3(6+7t) 4 (mod 8). First, let’s solve 7x â¡ 13 (mod 100). Example 2. In the table below, I have written x k first, because its coefficient is greater than that of y. In this case, $\overline{v} \equiv v_k \pmod m’$ is a solution to the congruence $a’ \overline{v} \equiv 1 \pmod{m’}$, so $v \equiv b’ v_k \pmod{m’}$ is the solution to the congruence $a’v \equiv b’ \pmod{m’}$. You can verify that 7*59 = 413 so 7*59 â¡ 13 (mod 100). By finding an inverse, solve the linear congruence $31 x\equiv 12 \pmod{24}.$ Solution. It is possible to solve the equation by judiciously adding variables and equations, considering the original equation plus the new equations as a system of linear … But opting out of some of these cookies may affect your browsing experience. 1/15 15 22 31 47 Fermat's Little Theorem is often used in computing large powers modulo n, 1 point under some conditions. By subtracting obtained equations we have: It follows: $x – x_0 = 2t, t \in \mathbb{Z}$. I intend to write posts in the future about how to solve simultaneous systems of linear congruences and how to solve quadratic congruences. For example, 8x â¡ 3 (mod 10) has no solution; 8x is always an even integer and so it can never end in 3 in base 10. Let $a$ and $m$ be natural numbers, and $b$ an integer. We also use third-party cookies that help us analyze and understand how you use this website. Let , and consider the equation (a) If , there are no solutions. Email: donsevcik@gmail.com Tel: 800-234-2933; The most important fact for solving them is as follows. Then x = (100*4 + 13)/7 = 59. My colleagues and I have decades of consulting experience helping companies solve complex problems involving data privacy, math, statistics, and computing. It turns out x = 9 will do, and in fact that is the only solution. Which of the following is a solution for x? We assume a > 0. Since gcd(50, 105) = 5 and 65 is divisible by 5, there are 5 solutions. So we first solve 10x â¡ 13 (mod 21). Then $x_0 \equiv b \pmod m$ is valid. For example, we may want to solve 7x â¡ 3 (mod 10). Solution to a linear congruence equation is equivalent to finding the value of a fractional congruence, for which a greedy-type algorithm exists. the congruences whose moduli are the larger of the two powers. These cookies do not store any personal information. Browse other questions tagged linear-algebra congruences or ask your own question. Solve the following congruence: $$x \equiv 5^{\varphi(13) -1} \cdot 8 \pmod{13}.$$, Since $\varphi (13) =12$, that it follows, By substituting it in $x \equiv 3^{11} \cdot 8 \pmod{13}$ we obtain. Since $\frac{m}{d}$ divides $m$, that by the theorem 6. Proof. Example 1. For example 25x = 15 (mod 29) may be rewritten as 25x1 = 15 - 29x2. Our rst goal is to solve the linear congruence ax b pmod mqfor x. Unfortu-nately we cannot always divide both sides by a to solve for x. A linear congruence is an equation of the form. This simpli es to 5t 2 (mod 8), which we solve by multiplying both sides by The answer to the first question depends on the greatest common divisor of a and m. Let g = gcd(a, m). We have $a’ = \frac{186}{2} = 93$, $b’ = \frac{374}{2} = 187$ and $m’ = \frac{422}{2} = 211$. In this way we obtain the congruence which also specifies the class that is the solution. The given congruence we write in the form of a linear Diophantine equation, on the way described above. This website uses cookies to improve your experience while you navigate through the website. Now what if the numbers a and m are not relatively prime? The method of  transformation of coefficients consist in the fact that to the given equation we add or subtract a well selected true congruence. Solutions we can write in the equivalent form: $$x_1 = 61 + 422t, \quad x_2 = 272 + 422t, \quad t \in \mathbb{Z}.$$, The Euler’s method consist in the fact that we use the Euler’s theorem. In the second example, the order is reversed because the coefficient of the x k is smaller than the coefficient of the y. and that is the solution to the given congruence. However, linear congruences don’t always have a unique solution. If b is not divisible by g, there are no solutions. Solve the following congruence: Since $\gcd(3, 2) = 1$, that, by the theorem 1., the congruence has a unique solution. With modulo, rather than talking about equality, it is customary to speak of congruence. 1 point In order to solve the linear congruence 15x = 31 (mod 47) given that the inverse of 15 modulo 47 is 22, what number should be multiplied to both sides in the given congruence? Get 1:1 help now from expert Advanced Math tutors (b) If , there are exactly d distinct solutions mod m.. Observe that Hence, (a) follows immediately from the corresponding result on linear … Solution: We have gcd(42,90) = 6, so there is a solution since 6 is a factor of 12. Here we use the algorithm to solve: 5x−3y=1 (5x≡1 (mod 3), which is easily solved by testing. first place that I’ve understood it, after looking through my book and all over the internet This means that there are exactly $d$ distinct solutions. The one particular solution to the equation above is $x_0 = 0, y_0 = -4$, so $3x_0 – 2y_0 = 8$ is valid. The complete set of solutions to our original congruence can be found by adding multiples of 105/5 = 21. This was really helpful. You also have the option to opt-out of these cookies. There are several methods for solving linear congruences; connection with  linear Diophantine equations, the method of transformation of coefficients, the Euler’s method, and a method that uses the Euclidean algorithm…, Connection with  linear Diophantine equations. It is mandatory to procure user consent prior to running these cookies on your website. Theorem. Linear CongruencesSimultaneous Linear CongruencesSimultaneous Non-linear CongruencesChinese Remainder Theorem - An Extension Theorem (5.6) If d = gcd(a;n), then the linear congruence ax b mod (n) has a solution if and only if d jb. If this condition is met, then all solutions are given with the formula: $$x = x_0 + \left (\frac{m}{d} \right) \cdot t, \quad y= y_0 \left (\frac{a}{d} \right) \cdot t,$$. Solving the congruence 42x ≡ 12 (mod 90) is equivalent to solving the equation 42x= 12+90qfor integers xand q. Suppose a solution exists. If we need to solve a system of three linear congruences with one unknown, then we need first solve a system of two linear congruences, and then see which of the obtained solutions also satisfy the third congruence. linear congruences (in one variable x). Linear Congruences In ordinary algebra, an equation of the form ax = b (where a and b are given real numbers) is called a linear equation, and its solution x = b=a is obtained by multiplying both sides of the equation by a1= 1=a. I enjoyed your article but impore you to give more examples in simpler forms, thank you for explaining this thoroughly and easy to understand Expert Answer . For this purpose, we take any two solutions from that set: $$x_1 = x_0 + \left( \frac{m}{d}\right) \cdot k_1,$$, $$x_2 = x_0 + \left (\frac{m}{d}\right) \cdot k_2.$$, $$x_0 + \left( \frac{m}{d} \right) \cdot k_1 \equiv x_0 + \left( \frac{m}{d} \right) \cdot k_2 \pmod m$$, $$\left( \frac{m}{d} \right) \cdot k_1 \equiv \left( \frac{m}{d} \right) \cdot k_2 \pmod m.$$. 24 8 pmod 16q. This website uses cookies to ensure you get the best experience on our website. Example 3. Substituting this into our equation for yields: Thus it follows that , so is the solution t… Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Now let’s find all solutions to 50x â¡ 65 (mod 105). For daily tweets on algebra and other math, follow @AlgebraFact on Twitter. If it is now $x_1$ any number from the equivalence class determined with $x_0$, then from $x_1 \equiv x_0 \pmod m$ follows that $ax_1 \equiv ax_0 \pmod m$, so $ax_1 \equiv b \pmod m$, which means that $x_1$ is also the solution to $ax \equiv \pmod m$. How do I solve a linear congruence equation manually? This reduces to 7x= 2+15q, or 7x≡ … Let $x_0$ be any concrete solution to the above equation. Therefore, if $ax \equiv b \pmod m$ has a solution, then there is infinitely many solutions. The algorithm can be formalized into a procedure suitable for programming. A linear congruence  $ax \equiv b \pmod m$ is equivalent to. In an equation a x ≡ b (mod m) the first step is to reduce a and b mod m. For example, if we start off with a = 28, b = 14 and m = 6 the reduced equation would have a = 4 and b = 2. The notion of congruences was first introduced and used by Gauss in his Disquisitiones Arithmeticae of 1801. Lemma. In case the modulus is prime, everything you know from linear algebra goes over to systems of linear congruences. Systems of linear congruences can be solved using methods from linear algebra: Matrix inversion, Cramer's rule, or row reduction. x ≡ (mod )--- Enter a mod b statement . Construction of number systems – rational numbers. is the solution to the initial congruence. This simpli es to x 6 (mod 7), so x = [6] 7 or x = 6 + 7t, where t 2Z. Linear Congruences. Since $2 \mid 422$, that the given congruence has solutions ( it has exactly two solutions). If not, replace ax â¡ b (mod m) with –ax â¡ –b (mod m). For instance, solve the congruence $6x \equiv 7 \pmod 8$. Featured on Meta “Question closed” … The proof for r > 2 congruences consists of iterating the proof for two congruences r – 1 times (since, e.g., € ([m 1,m 2],m 3)=1). Section 5.1 Solving Linear Congruences ¶ Our first goal to completely solve all linear congruences $$ax\equiv b$$ (mod $$n$$). A modular equation is an equation (or a system of equation, with at least one unknown variable) valid according to a linear congruence (modulo/modulus). In general, we may have to apply the algorithm multiple times until we get down to a problem small enough to solve easily. These cookies will be stored in your browser only with your consent. We can repeat this process recursively until we get to a congruence that is trivial to solve. Find more at https://www.andyborne.com/math See how to solve Linear Congruences using modular arithmetic. Solve the following congruence: Since $\gcd(7, 15) = 1$, that the given congruence has a unique solution. Linear Congruence Calculator. where $k$ is the least non-zero remainder and $q_i$ are quotients in the Euclidean algorithm. Since 100 â¡ 2 (mod 7) and -13 â¡ 1 (mod 7), this problem reduces to solving 2y â¡ 1 (mod 7), which is small enough to solve by simply sticking in numbers. 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# Monotonically decreasing/increasing

1. Oct 11, 2014

### BOAS

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

Plot each function and decide, based on the plot whether or not it is monotonically increasing/decreasing or strictly monotonically increasing/decreasing.

f(x) = -5

2. Relevant equations

3. The attempt at a solution

I can plot this function no problem, and show algebraically why it fits the definition of monotonically increasing and monotonically decreasing.

My question is, how do I justify this 'based on the plot'?

The next part of the question is to show it algebraically, but graphically i'm not really sure why my answer is correct...

2. Oct 11, 2014

### Ray Vickson

You really need to show more: what exact definitions are you using for "monotonically increasing/decreasing"? You seem to be making a distinction between monotonically increasing and strictly monotonically increasing, etc. Your answer could be right or wrong, depending on these details.

3. Oct 11, 2014

### Staff: Mentor

The question distinguishes between monotonically increasing/decreasing and strictly monotonically increasing/decreasing. In the latter, the graph of the function would have to be trending up (mon. increasing) or down (mon. decreasing). From your plot, which should show a horizontal line, which you could characterize as both monotically decreasing and monotonically increasing.

4. Oct 11, 2014

### BOAS

I have attached the definitions I am using as a picture. I will explain how i'm using them to come to my result.

From this, I say that the function $f(x) = -5$ is monotonically increasing and monotonically decreasing because for all $x_{1}, x_{2} \in \mathbb{R}$ where $x_{1} < x_{2}, f(x_{1}) \leq f(x_{2})$.

Likewise, for all $x_{1}, x_{2} \in \mathbb{R}$ where $x_{1} < x_{2}, f(x_{1}) \geq f(x_{2})$.

5. Oct 11, 2014

### Staff: Mentor

I figured that these were the definitions you were using. I agree with your answer.

6. Oct 11, 2014

### vela

Staff Emeritus
Is your question essentially "How can a flat graph be considered increasing or decreasing?"

7. Oct 11, 2014

### BOAS

that is at the heart of it.

8. Oct 11, 2014

### vela

Staff Emeritus
It's just how it is. As long as the function satisfies the definition, we say it's monotonically increasing or decreasing, even if it may run counter to our everyday sense of the words. Your intuition aligns with strictly monotonically increasing or decreasing.

9. Oct 11, 2014

### Staff: Mentor

Note that the definitions for monotonically increasing and decreasing include the possibility that f(x1) = f(x2). IOW, they include the possiblity of graphs that are flat. The "strictly" definitions don't include this.


# Area of region bounded by two parabolas using double integrals

"Let R be the region in the first quadrant bounded by the graphs of the parabolas $y=2x^2$, $y=9-x^2$ and the line x=0. Express the area of region R:
(i) Integrating first with respect to y, and then with respect to x
(ii) Integrating first with respect to x, and then with respect to y "

I have tried sketching the two curves, and expressing the double integral. However, my sketch and the double integral itself is quite different from the solution. Please have a look at my attempt and the solution below:

The Solution:

Could someone please help me understand this solution? I feel like I am missing a very important concept about double integrals. Any help would be highly appreciated.

• The area that you drew is not correct. It should extend upward until the $9-x^2$ parabola. And that is only one of two solutions. Mar 31 '16 at 9:01
• Thanks for the reply. But after extending the graph, I would still get a similar shape to the one I currently have. The solution's graph is different. Could it be incorrect? Mar 31 '16 at 19:30

If I well understand your question maybe that this intuitive interpretation can help.

When we calculate an area by a double integral, we subdivide this area in a sum of little ''area elements''. If we chose these elements as $dy dx$ or $dx dy$, this means that the elements are little rectangles of sides $dy$ and $dx$.

If we chose the order $dydx$ this means that we want to count these elements starting from ''vertical'' strips in which (in your case) the side $dy$ start from $2x^2$ and goes to $9-x^2$, so these values are the limits of the sum, and becomes the limits of the integration in $dy$, than we sum all these strips summing for the oter side $dx$ has limits $o$ and $\sqrt{3}$, and this gives the double integral: $$\int_0^{\sqrt{3}}\int_{2x^2}^{9-x^2}dydx$$

If we change the order to $dx dy$ we count the area elements starting from horizontal stripes, and in this case the side $dx$ start from $x=0$ and goes to $\sqrt{y/2 }$ if $y\le 6$, and to $\sqrt{9-y }$ if $6<y\le 9$. So you have the other double integral that is divided in two parts.

• Thank you very much for this explanation; it makes perfect sense to me. But how do I go about sketching the projection? I have a habit of always sketching the region before setting up the double integrals, Apr 1 '16 at 13:34
• I've added a figure. I hope it's useful. Apr 1 '16 at 13:50
• Perfect! Thanks you :) Apr 3 '16 at 8:26

The first graph you draw is good, but you paint incorrectly. You should have painted from left: $x=0$, from up: $y = 9-x^2$, and from down: $y = 2x^2$.

The graph in the solution is not correct, i think, but the double integrals are true.

• I thought so, the solution graph is incorrect. Thanks for responding Mar 31 '16 at 19:31


# Rock drawing possibility [closed]

In the latest episode of Survivor there were 6 people and they had to draw rocks from a bag. In the bag were 6 rocks, 5 white rocks and 1 black rock. First person drew a rock and kept it in his hand. Then the second person drew a rock. Last person took the only rock that left.

Then they all showed what they got at the same time.

Was the possibility to draw a black rock for each person 1/6 or what was it?

First point of view is they can all draw the rocks at the same time without changing the outcome.

Another way of seeing this is to compute the probability that the $k$th person picks the black stone. The $k-1$ persons before must have picked a white stone, and the $k$th picks the black. The probability is $$\frac56\times\frac45\times\dots\times \frac{6-(k-1)}{6-(k-1)+1}\times \frac{1}{6-(k-1)} = \frac16$$ (telescopic product).

Yes.

For the first person it's obviously $1/6$.

For the second it is either $1/5$ if the first person drew a white ($5/6$), or $0$ if the first person drew black already ($1/6$). So $p($Person 2 draws black$)=5/6 \times 1/5+1/6 \times 0=1/6$.

Logic continues and gives $1/6$ for each person.

Yes, the probability of drawing the black rock is 1/6 for each person. This is obvious for the first person. Intuitively "by symmetry," it should be clear that everyone has the same probability 1/6 of drawing the black rock.

One more formal way to look at it is that there are $6! = 720$ orders in which the six rocks can be drawn. Let's number the rocks 1 through 6, with #1 being the black rock. Some examples among the 720 ways are 123456, 321456, 135246, 654132, and so on.

Suppose you are third in line to draw. Then you would get the black rock for outcomes such as 321456 and 541326.

How many ways are there to arrange the rocks so that the black rock is third? Put the black rock in the 3rd position, and then there are 5! ways to arrange the other five rocks. So the number of ways for you to get the rock are 5! = 120.

Then the probability you get the black rock is $5!/6! = 120/720 = 1/6.$

Addendum: While I have been typing this and taking holiday phone calls, I see that two other answers have appeared. I have up-voted them both, as being reasonable alternative methods to show the same thing.


# Rate of flow Question

How would you solve a general problem of a steady stream of leakage:

Water leaks at a rate

$$r(t)= 20 \sqrt{3} \sec^2 (2t) \, \frac{\text{gallons}}{\text{hour}}.$$

At time 0, there are 50 gallons of water.

So how should I find a function that represents the amount remaining at a certain time, say $\pi/6$ hours?

I understand to take the integral of $r(t)$, but how would you proceed then?

As a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity...

• Hi! Welcome to MSE. I've editted your post for clarity without (hopefully) altering meaning. Best wishes. – Mark Fantini Feb 3 '14 at 0:32
• Thanks. This makes it better. – user235059 Feb 3 '14 at 3:40
• You are welcome. :) – Mark Fantini Feb 3 '14 at 9:29
• As a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity... – user235059 Feb 5 '14 at 17:55

Hint: You are given the if $W(t)$ is the amount of water in the tank, your function $r(t)$ is $W'(t)$. So if you integrate $$\int W'(t) dt=\int r(t) dt$$ that will give you $W(t)$, the amount of water at time $t$. Use the fact that $W(0)=50$. Then the question is just asking what is $W(\frac{\pi}{6})$.

• Not quite: since $\ r(t) \$ is a volume rate of change, the integral represents the amount of water added to or lost from the volume in the tank. This is why we must include an "initial condition" for the actual amount of water in the tank in order to obtain a specific function for the tank's contents. – colormegone Feb 3 '14 at 0:42
• That's why I said to include $W(0)=50$! – Kyle L Feb 4 '14 at 7:23
• The rate function is a way of writing a differential equation, $\ \frac{dV}{dt} \ = \ r(t) \ .$ This does not give us exclusively a function for the rate of water volume change in the tank, but only the rate at which water is flowing. What I was taking issue with is that the integral can also describe the amount of water accumulating elsewhere, such as another tank, the outside environment, etc. Since the rate function given by OP is described as leakage, you need to write your integral as $\int - r(t) \ dt \$ in order to produce the function you call $\ W(t) \ .$ – colormegone Feb 4 '14 at 8:29

The "Net Change Theorem" tells us that the volume of water lost from the container is

$$\Delta V (T) \ = \ \int_0^{T} \ - r(t) \ \ dt \ ,$$

the negative sign being inserted since we are told this is "leakage", and thus a reduction of water volume in the container. What remains in the container at time $\ T \$ is then

$$V(T) \ = \ 50 \ + \ \Delta V(T) \ .$$

[We add the change in volume, which is a negative change for this problem.]


Example Sequence $\{ b_n \}$ with $\lim_{n \rightarrow \infty} nb_n = 0$ but $\sum_{n=1}^{\infty} b_n$ diverges.

As the title states, I'm looking for an example of a strictly decreasing sequence of positive numbers with the properties that $$\lim_{n \rightarrow \infty} nb_n = 0$$ but $$\sum_{n=1}^{\infty} b_n$$ diverges.

My efforts have been unsuccessful so far. I know that nothing of the form

$$b_n = \frac{1}{n^p}$$

works, as $\lim_{n \rightarrow \infty} nb_n = 0$ if $p>1$, also implying that the series will converge via p-test. I've also tried more creative sequences like $$b_n = \frac{\sin(\frac{1}{n})}{n}$$ but still no luck.

More than a specific example, is there a certain strategy I should employ to find such an example? I was thinking the sequence must go to zero must faster than $n$ goes to infinity, but not fast enough for the series to converge.

• Check out $1/(n \log(n))$. Unfortunately there is not a lot of strategy with some of these questions. It requires knowing a good handful of examples that you can assemble for future use. This is one of the most classic examples in calculus / analysis. – Cameron Williams Jun 5 '18 at 21:47

Hint :

Look at Bertrand series.

Bertrand series are the series of the form : $$\frac{1}{n^{\alpha}(\log n)^{\beta}}$$ and you know that this serie converges only if : $\alpha > 1$ or ($\alpha = 1$ and $\beta > 1$).

With $$s_n:=\sum_{k=1}^n b_k,$$ try to achieve that $s_n\to \infty$, but sloooowly.

If $s_n= n$, then $b_n=1$ which is too large.

If $s_n=\sqrt n$, then $b_n\sim \frac1{\sqrt n}$, which makes $nb_n\sim\sqrt n$, still too large.

If $s_n=\ln n$, then $b_n\sim \frac 1n$, still too large: $nb_n\sim 1$.

In general, if $s_n=f(n)$ then $b_n\sim f'(n)$.

Now what if $s_n=\ln\ln n$? Then $b_n\sim \frac1{n\ln n}$ and $nb_n\sim \frac1{\ln n}\to 0$!

Let $p_n$ denote the $n$-th prime number. Then $$\sum_{n\in\Bbb N}\frac{1}{p_n}=\infty\tag{1}$$

But $$\lim_{n\to\infty}\frac{n}{p_n}=0\tag{2}$$

For $(1)$, see Divergence of the sum of the reciprocals of the primes.

For $(2)$, this is the Prime Number Theorem: $p_n\sim n\log n$. See for instance this discussion. This is also another way to prove $(1)$.

Using the idea in this answer: Is there a slowest rate of divergence of a series?

We set $D_n = 1/n$, so $\sum_{n = 1}^{\infty}D_n$ converges. Write $H_n = \sum_{k = 1}^{n}D_n$, the $n$th harmonic number. Finally, let $d_n = D_n/H_{n-1} = \frac{1}{nH_{n-1}}$. We can easily show that $d_n$ is positive and decreasing and $nd_n \rightarrow 0$, and that answer shows that $\sum_{n = 2}^{\infty}d_n$ diverges.

Asymptotically this answer is no different from $b_n = \frac{1}{n\ln(n)}$ given in other responses, but I thought this would be a nice fact to share.


# Fourier Transform Integration

I have a Fourier transform to complete with the definition of the Fourier Transform.

Let $$\phi$$ be defined as follows. $$\tag{1} \phi(x) = Ne^{-\frac{(x-x_0)^2}{a^2}}e^{ik_0x}$$

I must complete the Fourier transform of the function.

The definition of a Fourier transform is as follows.

$$\tag{2} \hat f(k) = \frac{1}{\sqrt{2\pi}}\int f(x) e^{-ikx}dx$$

To compute the Fourier transform we must evaluate the following integral with $$\phi$$ substituted into (2).

$$\tag{3} \hat\phi(k) = \frac{1}{\sqrt{2\pi}}\int \phi(x) e^{-ikx}dx$$

$$\tag{4} \hat\phi(k) = \frac{1}{\sqrt{2\pi}}\int Ne^{-\frac{(x-x_0)^2}{a^2}}e^{ik_0x} e^{-ikx}dx$$

I have tried completing this integral with completion of squares. I cannot find a way to finish this integral. How can I solve this integral?

• You were correct to pursue completing the square. You will need to evaluate a Gaussian integral thereafter. And to be rigorous, you need to deform the contour back to the real line by appealing to Cauchy's Integral Theorem. And that is about it. – Mark Viola Nov 20 '17 at 2:00
• After completing the square, $\hat\phi$ will become the following integral. $$\frac{1}{\sqrt{2\pi}}\int Ne^{-(x-\frac{2x_0 + ik_0a^2}{2}}e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ – Jeremy Nov 20 '17 at 2:33
• There is an error in the previous comment. $\hat\phi$ will be defined as follows. $$\frac{1}{\sqrt{2\pi}}\int Ne^{-(x-\frac{2x_0 + ik_0a^2}{2})^2}e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ – Jeremy Nov 20 '17 at 2:39
• After evaluating the Gaussian integral $\hat\phi$ is defined as follows. $$\hat\phi(k) = \frac{1}{\sqrt{2}}\int e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ – Jeremy Nov 20 '17 at 2:41
• I substituted $u = x-\frac{(2x_0 + ik_0a^2)}{2}$ in. After squaring both sides and changing the integral to polar coordinates, I found that the integral evaluated to $\sqrt{\pi}$. This equation I have arrived at does not match the provided solution. I may have made an error in evaluating the Gaussian integral. Where did I go wrong? – Jeremy Nov 20 '17 at 2:46

First, enforce the substitution $x-x_0\to x$ so that

\begin{align} \int_{-\infty }^\infty e^{-\frac1{\alpha^2}(x-x_0)^2-i(k-k_0)x}\,dx&=e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-\frac1{\alpha^2}x^2-i(k-k_0)x}\,dx \end{align}

Then, enforce the substitution $x/\alpha\to x$ so that

$$e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-\frac1{\alpha^2}x^2-ikx}\,dx=\alpha e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-x^2-i(k-k_0)\alpha x}\,dx$$

Completing the square reveals

$$\alpha e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-x^2-i(k-k_0)\alpha x}\,dx=\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty }^\infty e^{-(x-i(k-k_0)\alpha /2)^2}\,dx$$

Enforcing the substitution $x-i(k-k_0)\alpha/2\to x$ yields

\begin{align}\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty }^\infty e^{-(x-i(k-k_0)\alpha /2)^2}\,dx&=\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\\\\ &\times \int_{-\infty-i(k-k_0)\alpha/2 }^{\infty-i(k-k_0)\alpha/2} e^{-x^2}\,dx\end{align}

Applying Cauchy's Integral Theorem, we can deform the contour back onto the real line to obtain

$$\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty-i(k-k_0)\alpha/2 }^{\infty-i(k-k_0)\alpha/2} e^{-x^2}\,dx=\alpha e^{-((k-k_0)\alpha/2)^2}\underbrace{\int_{-\infty}^{\infty} e^{-x^2}\,dx}_{=\sqrt\pi}$$

Putting it all together, we find that

$$\int_{-\infty }^\infty e^{-\frac1{\alpha^2}(x-x_0)^2-i(k-k_0)x}\,dx=\alpha e^{-i(k-k_0)x_0}\sqrt\pi e^{-((k-k_0)\alpha/2)^2}$$

• I think the same techniques you've applied in your solution can be applied to the original problem. I think you have misread the problem statement. You have two exponentials with $(x-x_0)$ in both. Please note in the original problem statement that one exponential contains $(k-k_0)$ – Jeremy Nov 20 '17 at 3:45
• Jeremy, I edited on 20 November in response to your posted comment herein. Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to both up vote and accept an answer as you see fit. Happy holidays! -Mark – Mark Viola Dec 18 '17 at 18:40
• Please feel free to up vote and accept an answer as you see fit. And Happy Holidays! – Mark Viola Dec 24 '17 at 17:29
• Your solution is incorrect by a factor of $\frac1{\sqrt{2}}$. – Jeremy Dec 29 '17 at 22:48
• @Jeremy Why do you say that? – Mark Viola Dec 29 '17 at 22:50


# Number of closed walks on an $n$-cube

Is there a known formula for the number of closed walks of length (exactly) $r$ on the $n$-cube? If not, what are the best known upper and lower bounds?

 Note: the walk can repeat vertices.

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Yes (assuming a closed walk can repeat vertices). For any finite graph $G$ with adjacency matrix $A$, the total number of closed walks of length $r$ is given by

$$\text{tr } A^r = \sum_i \lambda_i^r$$

where $\lambda_i$ runs over all the eigenvalues of $A$. So it suffices to compute the eigenvalues of the adjacency matrix of the $n$-cube. But the $n$-cube is just the Cayley graph of $(\mathbb{Z}/2\mathbb{Z})^n$ with the standard generators, and the eigenvalues of a Cayley graph of any finite abelian group can be computed using the discrete Fourier transform (since the characters of the group automatically given eigenvectors of the adjacency matrix). We find that the eigenvalue $n - 2j$ occurs with multiplicity ${n \choose j}$, hence

$$\text{tr } A^r = \sum_{j=0}^n {n \choose j} (n - 2j)^r.$$

For fixed $n$ as $r \to \infty$ the dominant term is given by $n^r + (-n)^r$.

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Thanks! that's what I needed. –  Lev Reyzin Jul 31 '11 at 19:24
I'm guessing not, but is there any chance this expression has a closed form? –  Lev Reyzin Aug 2 '11 at 17:12
@Lev: you mean without a summation over $n$? I doubt it. Is fixed $n$ as $r \to \infty$ not the regime you're interested in? –  Qiaochu Yuan Aug 2 '11 at 17:36
In some sense, I'm more interested in fixed r as n gets large. The summation is certainly quite helpful, but of course if a closed form existed, it would even be nicer :) –  Lev Reyzin Aug 2 '11 at 18:06

The number of such walks is $2^n$ (the number of vertices of the $n$-cube) times the number of walks that start (and end) at the origin. We may encode such a walk as a word in the letters $1, -1, \dots, n, -n$ where $i$ represents a positive step in the $i$th coordinate direction and $-i$ represents a negative step in the $i$th coordinate direction. The words that encode walks that start and end at the origin are encoded as shuffles of words of the form $i\ -i \ \ i \ -i \ \cdots\ i \ -i$, for $i$ from 1 to $n$. Since for each $i$ there is exactly one word of this form for each even length, the number of shuffles of these words of total length $m$ is the coefficient of $x^m/m!$ in $$\biggl(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\biggr)^{n} = \left(\frac{e^x + e^{-x}}{2}\right)^n.$$ Expanding by the binomial theorem, extracting the coefficient of $x^r/r!$, and multiplying by $2^n$ gives Qiaochu's formula.

Let $W(n,r)$ be the coefficient of $x^r/r!$ in $\cosh^n x$, so that $$W(n,r) = \frac{1}{2^n}\sum_{j=0}^n\binom{n}{j} (n-2j)^r.$$ Then we have the continued fraction, due originally to L. J. Rogers, $$\sum_{r=0}^\infty W(n,r) x^r = \cfrac{1}{1- \cfrac{1\cdot nx^2}{ 1- \cfrac{2(n-1)x^2}{1- \cfrac{3(n-2)x^2}{\frac{\ddots\strut} {\displaystyle 1-n\cdot 1 x^2} }}}}$$ A combinatorial proof of this formula, using paths that are essentially the same as walks on the $n$-cube, was given by I. P. Goulden and D. M. Jackson, Distributions, continued fractions, and the Ehrenfest urn model, J. Combin. Theory Ser. A 41 (1986), 21–-31.

Incidentally, the formula given above for $W(n,r)$ (equivalent to Qiaochu's formula) is given in Exercise 33b of Chapter 1 of the second edition of Richard Stanley's Enumerative Combinatorics, Volume 1 (not published yet, but available from his web page). Curiously, I had this page sitting on my desk for the past month (because I wanted to look at Exercise 35) but didn't notice until just now that this formula was on it.

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thanks - this is nice. –  Lev Reyzin Aug 2 '11 at 23:08

Assuming a "closed walk" can repeat vertices, we can count closed walks starting at $0$ by counting the $r$-sequences of $[n]$ so that each number appears an even number of times. The bijection is given by labeling edges by the coordinate that is toggled between the vertices. You can probably count these sequences by inclusion/exclusion and then multiply by $2^n/r$ to account for the choice of start position.

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If we assume the path moves in each dimension 0 or 2 times, you can select ${n \choose r/2}$ dimensions and then permute them $r^1/2^r$ ways. This is a lower bound on the number of walks and is likely the right asymptotics. –  Derrick Stolee Jul 31 '11 at 17:03
That should be $r!/2^r$. –  Derrick Stolee Jul 31 '11 at 17:03


# Question about summation notation index

I'm working through a proof involving the sum of covariances but the notation is tripping me up. What does it mean when you are taking a summation over the index $i < j$? For instance $\sum_{i < j}\mathrm{Cov}(X_{i},X_{j})$ where Cov is covariance.

I'm sure it's nothing complicated I just want to make sure I understand.

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You run over all $j$ and for each $j$ you run over all $i$ such that $i < j$. For instance is $\sum_{i < j} i = 1+2+3+\cdots+n$ if $j$ can be maximal $n$. In particular this means here that you have $\sum_{i<j} Cov(x_i,x_j) = \sum_{i<1} Cov(x_i,x_1) + \sum_{i<2} Cov(x_i,x_2) + \sum_{i<3} Cov(x_i,x_3)+ \cdots +\sum_{i<n} Cov(x_i,x_n)$ if $j$ is maximal $n$. –  André Oct 20 '12 at 16:04
In general $\sum_{i<j} = \sum_{j=1}^n \sum_{i=1}^j$ –  André Oct 20 '12 at 16:13
This is an old post. If your question is solved, maybe you'd like to pick an answer you'd like best. Regards. –  FrenzY DT. Dec 4 '12 at 17:03

### An Illustration

Suppose you have $i\in I=\{1,2,\cdots,5\}$ and $j\in J=\{1,2,\cdots,6\}$. The following picture shows you which of the terms $\sum\limits_{i<j} a_{i,j}$ include.

According to the picture, if you choose a certain value of $i$, then $j$ should at least be $i+1$, otherwise the condition $i<j$ isn't met (falling on or below the line $i=j$).

In conclusion, $$\sum\limits_{i<j} a_{i,j} = \sum_{i} \sum_{\text{All j's that's >i}} a_{i,j}.$$

If you sum $j$ last, you get $$\sum\limits_{i<j} a_{i,j} = \sum_{j} \sum_{\text{All i's that's <j}} a_{i,j}.$$

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The notation means that you sum over all allowed $i$ and $j$ such that $i < j$. The notation is compact and protects you from errors that may come from doing the end-points incorrectly.

Here is an example. Suppose that $1\leq i \leq m$ and $1\leq j \leq n$ where $m$ and $n$ are not the same. If $m < n$, the sum becomes

$$\sum_{i < j} a_{ij} = \sum_{j = i+1}^n \sum_{i=1}^{m}a_{ij}.$$

If $n < m$ we have

$$\sum_{i < j} a_{ij} = \sum_{j = i+1}^n \sum_{i=1}^{n-1}a_{ij}.$$

Formally, these are two different expressions and to determine which one you are using can take a few seconds here but it could take a lot more time for more complicated sums.

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# Mentally finding factors of an integer of up to three-digits.

I’m looking for the easiest methods to find factors, not necessarily prime, of positive integers $<1000$, preferably mentally, but using pen and blank paper when required.

Although these techniques might be useful for a certain televised Numbers Game, I’m just looking for methods faster than trial division.

I take for granted full knowledge of the twelve by twelve multiplication tables, and recognition of primes up to $101$

This, with the addition of $7\cdot13=91$ should cover much of the factorisation up to two digits.

Notation Using the idea of hundreds, tens and units, so $n=100H+10T+U$ is the number, I’ll use $H,T,U$ for the individual digits, $HT$ for the first two digits and $TU$ for the last two.

For example, $n=456$ gives $H=4$, $T=5$, $U=6$, $HT=45$ and $TU=56$

Some easy published divisibility tests

If $U=0$ then $10|n$

If $(TU)\equiv 0{\pmod{25}}$ then ${25}|n$

If $U=5$ then $5|n$

If $U\equiv 0{\pmod 2}$ then $2|n$

If $(H+T+U)\equiv 0{\pmod 9}$ then $9|n$

If $(H+T+U)\equiv 0{\pmod 3}$ then $3|n$

Merged published divisibility tests for primes $p=7$ to $47$

With $z$ taken from the following, if $(HT+zU)\equiv 0{\pmod p}$ then $p|n$

$$(p,z)$$ $$(7,-2)$$ $$(11,-1)$$ $$(13,4)$$ $$(17,-5)$$ $$(19,2)$$ $$(23,7)$$ $$(29,3)$$ $$(31,-3)$$ $$(37,-11)$$ $$(41,-4)$$ $$(43,13)$$ $$(43,-30)$$ $$(47,-14)$$

For some examples in the range, the test may need to be repeated on the result.

As $31^2=961$, the last few test are of limited use.

Alternative divisibility test for $11$

If $(H+U-T)=0$ or $11$ then $11|n$

Difference of two squares

Using $a^2-b^2\equiv(a+b)(a-b)$ helps in a few cases. For example,

$$899=900-1=(30+1)(30-1)=31\cdot29$$

$$91=100-9=(10+3)(10-3)=13\cdot7$$

$$391=400-9=(20+3)(20-3)=23\cdot17$$

My question

I was tempted to ask what other divisibility or factorisation techniques exist that are potential easier than trial division, but “easier” is subjective; so I questions is:.

What other mental divisibility or factorisation techniques exist for an integer up to three digits?

Clarification update 11 May 2018

Sorry, I don’t feel I’ve sufficiently emphasised the importance of time, just assuming that easier was faster.

Full factorisation is ideal; partial factorisation is much better than nothing, so $663=3\cdot13\cdot17$ is best, but $663=3\cdot221$ is better than nothing.

Obtaining more than one small prime factor with one test is desirable, where the prime factors are obvious, as per the tests for $10,25$ and $9$, but a full list of all possible factors of $n$ is not needed.

• For three digit numbers you only have to do 11 trial divisions to find factors (for the primes from 2 up to 31). If none of them divides your number, no number does (except 1 and the number itself, of course). Trial divisions for 2,3,5,7 and 11 are easy. – YukiJ May 11 '18 at 10:58
• @YukiJ The OP wants to find factors, not necessarily prime. For instance, 6 is a factor. – scaaahu May 11 '18 at 11:35
• @scaaahu Yes, but if we observe that 2 and 3 are factors in a number then their product $2\cdot 3 =6$ is necessarily also a factor. Hence, if you find the prime factors, you also found the remaining other factors. – YukiJ May 11 '18 at 11:37
• @YukiJ Having a factor 9 implies having a factor 3, not the other way around. – scaaahu May 11 '18 at 11:52
• @scaaahu True but if you know that 3 divides $n$ then you can also check if 3 also divides $n/3$. If so, 9 divides $n$. In the particular case with 9 you could even check if the sum of digits of $n$ is divisible by 9. – YukiJ May 11 '18 at 11:55

As you have stated:

Checking a number is even will automatically give you the factor of $2$.

Summing the digits, if the digit sum is a multiple of $3$ or $9$, the number is also a multiple of $3$ or $9$.

Multiples of $5$ end in $5$, multiples of $10$ end in $0$.

For my extra method, let's consider $473$. It has no obvious factors based on those rules, so here I would use prime multiple subtraction. Start with $7$, the multiple of $7$ that ends in $3$ is $7\cdot9=63$, so subtract $63$ to get $410$. $41$ is not a multiple of $7$ or $9$, so neither are factors. Then consider $11$. $3\cdot 11=33$ so subtract $33$ to get $440$. $440=40\cdot 11$, hence we can see that $473=43\cdot 11$.

Note for $11$ that if you have $ab\cdot11$ , where $a,b$ represent the digits, the result is $a(a+b)b$, again to mean digits: e.g. $35\cdot11=385$. When $a+b>10$, it is harder to spot but still possible. What you should look for then is: $a(a+b-11)b$, and this is $(a-1)b \cdot 11$. E.g. for $528$, $5+8-11=2, \therefore 528=11*48$

If you can identify numbers as concatenations of multiples of the same number, they will be a multiple of that number. E.g. $654$ contains $6$ and $54$, both of which are multiples of $6$, and so we can do $6\div6=1$, and $54\div6=09$, therefore $654\div6=109$

• Thanks for your answer, which I’ve up-voted. I’ve used a similar method with numbers such as $721$. – Old Peter May 11 '18 at 14:42
• Thanks for the upvote. Note that $721$ is $7\cdot103$, you can see this by the subtraction method, or notice that $7$ and $21$ are both multiples of $7$, and so we can simply divide both by $7$ for $1$ and $03$ and then put these back together. A different example might be $726$, we see $72=12\cdot 6$ and $6=1\cdot 6$, putting back together we get $121$. – Rhys Hughes May 11 '18 at 14:56