# Is $K_1$ bipartite?

I'm thinking that it could be trivially bipartite since it only has one vertex and no edges but I am still a little bit unsure about it being trivially bipartite.

• Ummm... what is "k sub 1"? Do you instead mean $K_1$? – David G. Stork Jun 10 at 19:01
• not only is it bipartite, it is also unipartite – Yorch Jun 10 at 19:02
• @DavidG.Stork Yes, did not know how to make it like that in my title. Seemed to fix itself. – ThreeRingBinder Jun 10 at 20:23

Depends on your definitions. (As pointed out in Matthew Daly's answer, $$K_1$$ should be bipartite, because it has no odd cycles, so it would otherwise be an awkward exception. But not all definitions will play nicely with this desired property.)

West's Introduction to Graph Theory says

1.1.10. Definition. A graph $$G$$ is bipartite if $$V(G)$$ is the union of two disjoint (possibly empty) independent sets called partite sets of $$G$$.

So under this definition, if $$V(K_1) = \{v\}$$, then we let $$\{v\}$$ be one partite set, and $$\varnothing$$ be the other; $$K_1$$ is bipartite.

Bondy and Murty write

A graph is bipartite if its vertex set can be partitioned into two subsets $$X$$ and $$Y$$ so that every edge has one end in $$X$$ and one end in $$Y$$

which still works just fine, setting $$X = \{v\}$$ and $$Y = \varnothing$$. They do not specifically point out that $$X$$ or $$Y$$ could be empty, but they do not rule it out either. (Elsewhere, Bondy and Murty talk about "nontrivial partitions" or "partitions into nonempty parts", which make it clear that a partition without this qualifier is allowed to have an empty part.)

They are in better shape than Diestel, who has

A graph $$G = (V, E)$$ is called $$r$$-partite if $$V$$ admits a partition into $$r$$ classes such that every edge has its ends in different classes

with an earlier qualification that the classes of a partition may not be empty. Since $$K_1$$ should be bipartite by any reasonable definition, Diestel is in the wrong here. (Diestel later claims that all graphs with no odd cycles are bipartite, with no mention of $$K_1$$ as a special exception.)

If we treat "bipartite" as synonymous with "$$2$$-colorable", then $$K_1$$ is happily bipartite, since any function on its vertex set is a $$2$$-coloring (and also a $$1$$-coloring).

• Quick question. Lets say we have 𝑋={𝑣} and 𝑌=∅, in reference to Bondy and Murty's definition. Would the edge connecting both subsets have vertices that are the empty set? Ie one empty set from X and the other from Y – ThreeRingBinder Jun 11 at 4:36
• The definition says every edge has a certain property; it doesn't say there exist any such edges. Since $K_1$ has no edges, the definition holds. – Misha Lavrov Jun 11 at 5:24
• Ah, makes sense now. Thanks! – ThreeRingBinder Jun 11 at 6:51

My graph theory professor spent a while teaching on this, because it's a mess either way.

It probably should be trivially considered bipartite. Otherwise you have to put disclaimers in all of your theorems (like the fact that $$K_1$$ doesn't contain any odd cycles). But $$\{\{v\},\emptyset\}$$ is not a valid partition of the vertex set, since it contains the empty set as a member.

• A reasonable definition of "bipartite" does not run into trouble if one partite set is empty. – Misha Lavrov Jun 10 at 19:24
• @MishaLavrov On the other hand, requiring the parts of a partition to be non-empty isn't entirely unheard of either. So defining bipartite as a partition of the set of vertices into two can run into issues with $K_1$. – Arthur Jun 10 at 19:33
• Yeah, I think that Dr. Schäffer's solution was to have us write in the margin of Bondy and Murty "... or $K_1$". – Matthew Daly Jun 10 at 19:44
• But Bondy and Murty are totally fine with empty parts in a partition, as far as I can tell... – Misha Lavrov Jun 10 at 20:22
• @MishaLavrov As I say, my professor assumed that "partitioned into two subsets $X$ and $Y$" means that $\{X,Y\}$ was a partition of the vertex set, which could not be legally done if one of the sets was empty. Bondy and Murty clearly either thought that it was okay or they hadn't considered the edge case, but pedantry like that got under my professor's skin. ^_^ – Matthew Daly Jun 10 at 20:37


# Do we have to expand the brackets before applying De Morgan's Laws?

If we have the example:

$$\overline{(A + B + C)D}$$

Then can we apply De Morgan's Law as is, or do we first need to expand out the brackets?

If I expand the terms first, I get:

$$original: \overline{(A + B + C)D}$$ $$expanded: \overline{AD + BD + CD}$$ $$applying De Morgan's: \overline{AD}.\overline{BD}.\overline{CD}$$ $$simplifying: \overline{A}\overline{B}\overline{C}\overline{D}$$

In contrast, if I don't expand the brackets, I get something like:

$$original: \overline{(A + B + C)D}$$ $$applying De Morgan's: \overline{(A + B + C)}+\overline{D}$$ $$applying De Morgan's: (\overline{A} . \overline{B + C})+\overline{D}$$ $$applying De Morgan's: (\overline{A} . \overline{B} . \overline{C})+\overline{D}$$ $$simplifying: \overline{A} . \overline{B} . \overline{C}+\overline{D}$$

So you can see, I'm getting different answers. I'm curious which method is correct - I think the first seems correct, but perhaps the 2nd is correct or both are wrong.

If anyone is able to explain why one particular method is wrong, I'd appreciate it too.

• MathJax tip: when writing text in mathematics code, use the \text{} command to get proper spacing and formatting. For example, $$\text{applying De Morgan's} : \overline{(A + B + C)} + \overline{D}$$ produces$$\text{applying De Morgan's} : \overline{(A + B + C)} + \overline{D}$$ Aug 22, 2021 at 0:54
• Thanks. Now I will do that in future :) Aug 22, 2021 at 5:58
• @RandomUser123 You can also do that now. Questions and answers can be edited, and indeed should be edited if they can be improved. Aug 22, 2021 at 8:45

Both lead to the same result; your error is in the first one when you simplify from $$\overline{AD}.\overline{BD}.\overline{CD}$$ to $$\overline{A}\overline{B}\overline{C}\overline{D}$$.
If you apply De Morgan's law to each pair, you get: $$(\overline{A}+\overline{D})(\overline{B}+\overline{D})(\overline{C}+\overline{D}).$$ From here, you can apply the distributive law to get what you got by the second method in the next to last line. The simplifcation there is also incorrect because you cannot simply drop the parenthesis; you must apply the distributive property.
• Makes sense. I was viewing it as $$\overline{ADBDCD}$$ then just cancelling removing the unneccessary duplicates. Because the NOTs were over multiple variables, I shouldn't be able to do that. I also tried myself, expanding out those 3 sets of brackets and surpisingly managed to simplify down that mess to the correct answer...seeing that all of the brackets contain D' and using your suggestion of using the distributive law is much easier though...I guess it's a good way to check...since I worry if I use these shortcuts, I will apply it incorrectly/in a situation that's invalid or something Aug 22, 2021 at 6:17


# What's the probability that a given permutation has exactly $k$ fixed points. [duplicate]

Given a random permutation $$\sigma \in S_n$$ from $$[n] \to [n]$$ in a uniform probability space, what is the probability that $$\sigma$$ has exactly $$k$$ fixed points for a given $$k$$ between $$1$$ and $$n$$?

In other words: what is the probability that $$\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i$$ for $$\ i\in \{1,...,k\}$$ and for every $$y \notin \{x_1 , ... , x_k\}$$ we get $$\sigma(y) \neq y$$.

I saw that $$\lim_{n \to \infty } prob(A_0) = e^{-1}$$ using Inclusion–exclusion principle and i belive that for a given k : $$\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$$ but I am not sure how to show it.

*$$A_k$$ stands for the event "k".

• The number $D_{n,k}$ of permutations with exactly $k$ fixpoints of an $n$-set is called a rencontres number. – Jeppe Stig Nielsen Oct 10 '17 at 11:36

There are $\binom{n}{k}$ possibilities for the $k$ fixed points.

If they are established then there are $!(n-k)$ derangements for the other points.

That gives $\binom{n}{k}\left[!(n-k)\right]$ permutations with exactly $k$ fixed points on a total of $n!$ permutations.

Also we have the formula: $$!(n-k)=(n-k)!\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$ and we end up with probability: $$\frac1{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$

• Very cool! As $n$ grows (for fixed $k$), this approaches $\frac1{k! e}$ – MichaelChirico Nov 6 '19 at 11:31
• And also by corollary, the probability of at least one fixed point approaches $1-\frac1{e}$ – MichaelChirico Nov 6 '19 at 11:38
1. Number of dearrangements of $k$-elements set is $$!k = k!\sum\limits_{m=0}^{k}\frac{(-1)^m}{m!}$$
2. In n-elements set we can select $(n-k)$ elements to dearrange them (all remaining $k$ points are fixed) in $\binom{n}{k}$ ways

Thus $$A_k^n = \binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$ And probability is $$P(A_k^n) = \frac{\binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{\frac{n!}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{1}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$

By way of enrichment here is an alternate formulation using combinatorial classes. The class of permutations with fixed points marked is

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) + \textsc{CYC}_{=2}(\mathcal{Z}) + \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$

This gives the generating function $$G(z, u) = \exp\left(uz + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \frac{z^5}{5} + \cdots\right)$$

which is

$$G(z, u) = \exp\left(uz-z+\log\frac{1}{1-z}\right) = \frac{\exp(uz-z)}{1-z}.$$

Now for $$k$$ fixed points we get

$$[u^k] \frac{\exp(uz-z)}{1-z} = [u^k] \frac{\exp(uz)\exp(-z)}{1-z} = \frac{z^k}{k!} \frac{\exp(-z)}{1-z}.$$

This is the EGF of permutations having $$k$$ fixed points. We extract the count by computing (the factor $$n!$$ is canceled because we require the average)

$$[z^n] \frac{z^k}{k!} \frac{\exp(-z)}{1-z} = \frac{1}{k!} [z^{n-k}] \frac{\exp(-z)}{1-z}.$$

We find

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{k!} \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}.}$$

We can identify this as choosing the $$k$$ fixed points and combining them with a derangement of the rest:

$$\frac{1}{n!} {n\choose k} (n-k)! \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}$$

which is the combinatorial class

$$\textsc{SET}_{=k}(\mathcal{Z}) \times \textsc{SET}(\textsc{CYC}_{\ge 2}(\mathcal{Z})).$$


# Show that $\sqrt{1+x}<1+\frac{x}{2}$ for all $x>0$

I am a little stuck on this question and would appreciate some help. The question asks me to prove that $$\sqrt{1+x}<1+\frac{x}{2}$$ for all $$x>0$$.

I squared both sides of the question to get $$1+x<\frac{x^2}{4}+x+1$$ for all $$x>0$$. Then, I multiplied both sides by $$4$$ to get $$4+4x for all $$x>0$$.

I am a little stuck and was wondering what to do after this step and how to actually provide sufficient proof to say that this statement is true.

You did well. Let's finish it: we have $$x^2>0$$ thus $$\dfrac{x^2}{4}>0$$. Adding $$x+1$$ to both sides, we have $$\dfrac{x^2}{4}+x+1> x+1$$ or $$(\frac{x}{2}+1)^2>x+1$$ or $$\frac{x}{2}+1>\sqrt{x+1}$$

Squaring the equation was sufficient. After you cancel the $$1+x$$ on each side, you have

$$\frac{x^2}{4} > 0$$

which is true for all real $$x \ne 0$$ since $$x^2 \ge 0$$ (with equality only when $$x=0$$). Thus, the inequality is proved.

You can also prove the inequality using the Mean Value Theorem. Define $$f(t)=\sqrt{t+1}$$ and apply the theorem on the interval $$[0,x]$$. Then, there exists $$c\in(0,x)$$ s.t. $$\frac{\sqrt{x+1}-\sqrt{1}}{x-0}=\frac{1}{2\sqrt{c+1}}>\frac{1}{2}.$$

I am answering less to provide a list of steps to follow for proving the inequality, and more to discuss how I believe that the argument should be presented. This seems appropriate, as the question has been tagged with and . If I asked a student to prove some inequality and their first step was to assume the inequality and square both sides, I would almost certainly deduct points unless the student were careful about justifying such a step. The goal here is to start with known true statements and to then deduce the desired result.

Claim: For any $$x > 0$$, $$\sqrt{1+x} < 1 + \frac{x}{2}.$$

Proof: Assume that $$x > 0$$. The square of any real number is positive, hence $$0 < \left( \frac{x}{2} \right)^2.$$ By additive cancelation (i.e. by "adding the same number to both sides"), this implies that $$x + 1 < \left( \frac{x}{2} \right)^2 + x + 1 = \left( \frac{x}{2} + 1 \right)^2.$$ The square root function is increasing on $$[0,\infty)$$ (that is, if $$0 \le a < b$$, then $$0 \le \sqrt{a} \le \sqrt{b}$$), and the assumption that $$x > 0$$ ensures that $$1+x > 0$$, thus $$1 + x$$ is in the domain of the square root function (i.e. it has a well-defined real square root). Hence $$\sqrt{x+1} < \sqrt{ \left( 1 + \frac{x}{2} \right)^2 } = \left|1 + \frac{x}{2}\right| = 1 + \frac{x}{2},$$ where the final equality follows from the assumption that $$x > 0$$.

### Discussion

Remember that if you start with false assumptions, you can prove anything—false implies true. Thus it is poor practice to begin a proof by asserting the statement which you are trying to prove and then applying algebraic manipulation. You can do this if you are careful—be very observant of two-sided implications and check the hypotheses at each step. Indeed, this is a very reasonable way of deducing a correct proof in the first place.

However, when presenting an argument, such a proof is typically poor style. Personally, I think that it is better (stylistically, which is a matter of taste) for every statement in a proof to follow from previous statement, and not depend on future statements via "if and only ifs".

I will also note that the proof presented above is very elementary—it doesn't rely on any deep theorems. It is, however, somewhat tedious. More powerful theorems give quicker, more elegant proofs. For example, El31's proof, via the mean value theorem, is quite slick. However, when first learning a topic, I think that one should focus on finding an elementary proof first: find such a proof first. Then, if need be, look for something more elegant.

The important principle here is that if $$a,b$$ are nonnegative real numbers, then $$a iff $$a^2 This property is used all the time. If we grant this, then all we need to show is that

$$(\sqrt{1+x})^2 < (1+x/2)^2,$$

which is the same as saying $$1+x < 1+x + x^2/4.$$ That is obviously true, and we're done.

Observe that : $$\left(\forall x>0\right),\ \sqrt{1+x}-1-\frac{x}{2}=-\frac{x^{2}}{4}\int_{0}^{1}{\frac{1-t}{\left(1+tx\right)\sqrt{1+tx}}\,\mathrm{d}t} < 0$$

• No offense meant, but this is perhaps one of the most obtuse correct answers I've seen on this site. – jawheele Mar 30 at 5:51


# Math Help - Solve Equations 0< or = x < or = 2pi

1. ## Solve Equations 0< or = x < or = 2pi

How would you solve these

Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.

Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?

2. Originally Posted by j0nath0n3
How would you solve these

Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.

Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
Note that $\sin^2 x=1-\cos^2 x$

So we can rewrite the equation as $\cos^2 x -\left[1-\cos^2 x\right]=-\cos x$

This simplifies to $2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$

Now, if we make the substitution $z=\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\implies (2z-1)(z+1)=0$

This tells us that $z=-1$ or $z=\tfrac{1}{2}$

But $z=\cos x$, so now that means either $\cos x=\tfrac{1}{2}$ or $\cos x=-1$

Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?

3. Originally Posted by Chris L T521
Note that $\sin^2 x=1-\cos^2 x$

So we can rewrite the equation as $\cos^2 x -\left[1-\cos^2 x\right]=-\cos x$

This simplifies to $2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$

Now, if we make the substitution $z=\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\implies (2z-1)(z+1)=0$

This tells us that $z=-1$ or $z=\tfrac{1}{2}$

But $z=\cos x$, so now that means either $\cos x=\tfrac{1}{2}$ or $\cos x=-1$

Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
Thats really good and I understood however my answear choices are in radian... and I dont understand why..

4. Originally Posted by j0nath0n3
Thats really good and I understood however my answear choices are in radian... and I dont understand why..
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\text{ degrees}=2\pi\text{ radians}$.

5. ## Re :

Originally Posted by Chris L T521
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\text{ degrees}=\pi\text{ radians}$.
Chris , is it a typo , i thought $360\text{ degrees}=2\pi\text{ radians}$

Chris , is it a typo , i thought $360\text{ degrees}=2\pi\text{ radians}$


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In Section VO:Vector Operations we defined vector addition and scalar multiplication. These two operations combine nicely to give us a construction known as a linear combination, a construct that we will work with throughout this course.

## Linear Combinations

Definition LCCV (Linear Combination of Column Vectors) Given $n$ vectors $\vectorlist{u}{n}$ from $\complex{m}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n$, their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n} \end{equation*}

So this definition takes an equal number of scalars and vectors, combines them using our two new operations (scalar multiplication and vector addition) and creates a single brand-new vector, of the same size as the original vectors. When a definition or theorem employs a linear combination, think about the nature of the objects that go into its creation (lists of scalars and vectors), and the type of object that results (a single vector). Computationally, a linear combination is pretty easy.

Example TLC: Two linear combinations in $\complex{6}$.

Example ABLC: Archetype B as a linear combination.

With any discussion of Archetype A or Archetype B we should be sure to contrast with the other.

Example AALC: Archetype A as a linear combination.

There's a lot going on in the last two examples. Come back to them in a while and make some connections with the intervening material. For now, we will summarize and explain some of this behavior with a theorem.

Theorem SLSLC (Solutions to Linear Systems are Linear Combinations) Denote the columns of the $m\times n$ matrix $A$ as the vectors $\vectorlist{A}{n}$. Then $\vect{x}$ is a solution to the linear system of equations $\linearsystem{A}{\vect{b}}$ if and only if $\vect{b}$ equals the linear combination of the columns of $A$ formed with the entries of $\vect{x}$, \begin{equation*} \vectorentry{\vect{x}}{1}\vect{A}_1+ \vectorentry{\vect{x}}{2}\vect{A}_2+ \vectorentry{\vect{x}}{3}\vect{A}_3+ \cdots+ \vectorentry{\vect{x}}{n}\vect{A}_n = \vect{b} \end{equation*}

In other words, this theorem tells us that solutions to systems of equations are linear combinations of the $n$ column vectors of the coefficient matrix ($\vect{A}_j$) which yield the constant vector $\vect{b}$. Or said another way, a solution to a system of equations $\linearsystem{A}{\vect{b}}$ is an answer to the question "How can I form the vector $\vect{b}$ as a linear combination of the columns of $A$?" Look through the archetypes that are systems of equations and examine a few of the advertised solutions. In each case use the solution to form a linear combination of the columns of the coefficient matrix and verify that the result equals the constant vector (see exercise LC.C21).

## Vector Form of Solution Sets

We have written solutions to systems of equations as column vectors. For example Archetype B has the solution $x_1 = -3,\,x_2 = 5,\,x_3 = 2$ which we now write as \begin{equation*} \vect{x}=\colvector{x_1\\x_2\\x_3}=\colvector{-3\\5\\2} \end{equation*} Now, we will use column vectors and linear combinations to express all of the solutions to a linear system of equations in a compact and understandable way. First, here's two examples that will motivate our next theorem. This is a valuable technique, almost the equal of row-reducing a matrix, so be sure you get comfortable with it over the course of this section.

Example VFSAD: Vector form of solutions for Archetype D.

This is such an important and fundamental technique, we'll do another example.

Example VFS: Vector form of solutions.

Did you think a few weeks ago that you could so quickly and easily list all the solutions to a linear system of 5 equations in 7 variables?

We'll now formalize the last two (important) examples as a theorem.

Theorem VFSLS (Vector Form of Solutions to Linear Systems) Suppose that $\augmented{A}{\vect{b}}$ is the augmented matrix for a consistent linear system $\linearsystem{A}{\vect{b}}$ of $m$ equations in $n$ variables. Let $B$ be a row-equivalent $m\times (n+1)$ matrix in reduced row-echelon form. Suppose that $B$ has $r$ nonzero rows, columns without leading 1's with indices $F=\set{f_1,\,f_2,\,f_3,\,\ldots,\,f_{n-r},\,n+1}$, and columns with leading 1's (pivot columns) having indices $D=\set{d_1,\,d_2,\,d_3,\,\ldots,\,d_r}$. Define vectors $\vect{c}$, $\vect{u}_j$, $1\leq j\leq n-r$ of size $n$ by

\begin{align*} \vectorentry{\vect{c}}{i}&= \begin{cases} 0&\text{if $i\in F$}\\ \matrixentry{B}{k,n+1}&\text{if $i\in D$, $i=d_k$} \end{cases}\\ \vectorentry{\vect{u}_j}{i}&= \begin{cases} 1&\text{if $i\in F$, $i=f_j$}\\ 0&\text{if $i\in F$, $i\neq f_j$}\\ -\matrixentry{B}{k,f_j}&\text{if $i\in D$, $i=d_k$} \end{cases}. \end{align*}

Then the set of solutions to the system of equations $\linearsystem{A}{\vect{b}}$ is \begin{equation*} S=\setparts{ \vect{c}+\alpha_1\vect{u}_1+\alpha_2\vect{u}_2+\alpha_3\vect{u}_3+\cdots+\alpha_{n-r}\vect{u}_{n-r} }{ \alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_{n-r}\in\complex{\null} } \end{equation*}

Note that both halves of the proof of Theorem VFSLS indicate that $\alpha_i=\vectorentry{\vect{x}}{f_i}$. In other words, the arbitrary scalars, $\alpha_i$, in the description of the set $S$ actually have more meaning --- they are the values of the free variables $\vectorentry{\vect{x}}{f_i}$, $1\leq i\leq n-r$. So we will often exploit this observation in our descriptions of solution sets.

Theorem VFSLS formalizes what happened in the three steps of Example VFSAD. The theorem will be useful in proving other theorems, and it it is useful since it tells us an exact procedure for simply describing an infinite solution set. We could program a computer to implement it, once we have the augmented matrix row-reduced and have checked that the system is consistent. By Knuth's definition, this completes our conversion of linear equation solving from art into science. Notice that it even applies (but is overkill) in the case of a unique solution. However, as a practical matter, I prefer the three-step process of Example VFSAD when I need to describe an infinite solution set. So let's practice some more, but with a bigger example.

Example VFSAI: Vector form of solutions for Archetype I.

This technique is so important, that we'll do one more example. However, an important distinction will be that this system is homogeneous.

Example VFSAL: Vector form of solutions for Archetype L.

## Particular Solutions, Homogeneous Solutions

The next theorem tells us that in order to find all of the solutions to a linear system of equations, it is sufficient to find just one solution, and then find all of the solutions to the corresponding homogeneous system. This explains part of our interest in the null space, the set of all solutions to a homogeneous system.

Theorem PSPHS (Particular Solution Plus Homogeneous Solutions) Suppose that $\vect{w}$ is one solution to the linear system of equations $\linearsystem{A}{b}$. Then $\vect{y}$ is a solution to $\linearsystem{A}{b}$ if and only if $\vect{y}=\vect{w}+\vect{z}$ for some vector $\vect{z}\in\nsp{A}$.

After proving Theorem NMUS we commented (insufficiently) on the negation of one half of the theorem. Nonsingular coefficient matrices lead to unique solutions for every choice of the vector of constants. What does this say about singular matrices? A singular matrix $A$ has a nontrivial null space (Theorem NMTNS). For a given vector of constants, $\vect{b}$, the system $\linearsystem{A}{b}$ could be inconsistent, meaning there are no solutions. But if there is at least one solution ($\vect{w}$), then Theorem PSPHS tells us there will be infinitely many solutions because of the role of the infinite null space for a singular matrix. So a system of equations with a singular coefficient matrix never has a unique solution. Either there are no solutions, or infinitely many solutions, depending on the choice of the vector of constants ($\vect{b}$).

Example PSHS: Particular solutions, homogeneous solutions, Archetype D.

The ideas of this subsection will be appear again in Chapter LT:Linear Transformations when we discuss pre-images of linear transformations (Definition PI).


Question

# Obtain all other zeroes of $$3x^4+6x^3-2x^2-10x-5$$, if two of its zeros are $$\sqrt{\dfrac{5}{3}}$$ and $$-\sqrt{\dfrac{5}{3}}$$.

Solution

## Since the two zeroes of the given polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 } } ,-\sqrt { \dfrac { 5 }{ 3 } }$$, therefore, $$\left( x-\sqrt { \dfrac { 5 }{ 3 } } \right) \left( x+\sqrt { \dfrac { 5 }{ 3 } } \right) =x^{ 2 }-\dfrac { 5 }{ 3 }$$ is a factor of the given polynomial.now, we divide the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ by $$x^{ 2 }-\dfrac { 5 }{ 3 }$$ as shown below:Therefore, $$3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5=\left( x^{ 2 }-\dfrac { 5 }{ 3 } \right) (3x^{ 2 }+6x+3)$$Now, we factorize $$3x^{ 2 }+6x+3$$ as follows:$$3x^{ 2 }+6x+3=0\\ \Rightarrow 3(x^{ 2 }+2x+1)=0\\ \Rightarrow x^{ 2 }+2x+1=0\\ \Rightarrow (x+1)^{ 2 }=0\quad \quad \quad \quad (\because \quad (a+b)^{ 2 }=a^{ 2 }+b^{ 2 }+2ab)\\ \Rightarrow (x+1)(x+1)=0\\ \Rightarrow x+1=0,\quad x+1=0\\ \Rightarrow x=-1,\quad x=-1$$Hence, the zeroes of the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 } } ,-\sqrt { \dfrac { 5 }{ 3 } } ,-1,-1$$.Maths

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### TheScrasse's blog

By TheScrasse, history, 7 weeks ago,

Hello everyone,
here is a very simple idea that can be useful for (cp) number theory problems, especially those concerning multiples, divisors, $\text{GCD}$ and $\text{LCM}$.

Prerequisites: basic knowledge of number theory (divisibility, $\text{GCD}$ and $\text{LCM}$ properties, prime sieve).

## Idea

Let's start from a simple problem.

You are given $n$ pairs of positive integers $(a_i, b_i)$. Let $m$ be the maximum $a_i$. For each $k$, let $f(k)$ be the sum of the $b_i$ such that $k | a_i$. Output all pairs $(k, f(k))$ such that $f(k) > 0$.

An obvious preprocessing is to calculate, for each $k$, the sum of the $b_i$ such that $a_i = k$ (let's denote it as $g(k)$). Then, there are at least $3$ solutions to the problem.

#### Solution 1: $O(m\log m)$

For each $k$, $f(k) = \sum_{i=1}^{\lfloor m/k \rfloor} g(ik)$. The complexity is $O\left(m\left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{m}\right)\right) = O(m\log m)$.

#### Solution 2: $O(n\sqrt m)$

There are at most $n$ nonzero values of $g(k)$. For each one of them, find the divisors of $k$ in $O(\sqrt k)$ and, for each divisor $i$, let $f(i) := f(i) + g(k)$.
If $m$ is large, you may need to use a map to store the values of $f(k)$ but, as there are $O(n\sqrt[3] m)$ nonzero values of $f(k)$, the updates have a complexity of $O(n\sqrt[3] m \log(nm)) < O(n\sqrt m)$.

#### Solution 3: $O(m + n\sqrt[3] m)$

Build a linear prime sieve in $[1, m]$. For each nonzero value of $g(k)$, find the prime factors of $k$ using the sieve, then generate the divisors using a recursive function that finds the Cartesian product of the prime factors. Then, calculate the values of $f(k)$ like in solution 2.

Depending on the values of $n$ and $m$, one of these solutions can be more efficient than the others.

Even if the provided problem seems very specific, the ideas required to solve that task can be generalized to solve a lot of other problems.

Hint 1
Hint 2
Hint 3
Solution

## agc038_c - LCMs

Hint 1
Hint 2
Hint 3
Solution

Implementation (C++)

## abc191_f - GCD or MIN

Hint 1
Hint 2
Hint 3
Hint 4
Solution

Implementation (C++)

## Conclusions

We've seen that this technique is very flexible. You can choose the complexity on the basis of the constraints, and $f(k)$ can be anything that can be updated fast.

Of course, suggestions/corrections are welcome. In particular, please share in the comments other problems that can be solved with this technique.

I hope you enjoyed the blog!

• +105

 » 7 weeks ago, # |   +15 Nice blog! I like solving problems which use such ideas, so here are a few more problems with a similar flavour:1436F - Sum Over SubsetsChefsums1493D - GCD of an Array
 » 10 days ago, # |   +8 The Solution spoiler for 1154G - Minimum Possible LCM seems to be broken, I see this:
•  » » 10 days ago, # ^ |   +6 Fixed. I don't know the reason of the failed parsing, though. The original text was Spoiler$\text{GCD}(a_i, a_j) = h$.and I had to change it to Spoiler$\text{GCD}(a_i, a_j) = h$ .
 » 10 days ago, # |   +13 Solution 1 can be easily optimized to $O(m \log{\log{m}})$ using the same idea as calculating sums-over-submasks. The code looks something like this: vector f = g; for (int p : primes) for (int i = m / p; i > 0; --i) f[i] += f[i*p]; I suspect there are very few (if any) problems in practice for which solution 3 is directly applicable, but this optimized solution 1 is not.


# Testing for Linearity of Vectors in a Subspace - Examples with Solutions

Given a set of vectors in a subspace, how do we test whether the vectors are linearly independent ?

## What are Linearly Dependent or Independent Vectors?

We first use vectors in two and three dimensional spaces to visualize the concept of dependence and independence of vectors.
In figure 1 below we have have two vectors that are parallel such that $v_2 = 2 v_1$. We say that these two vectors are dependent because we can express one vector in terms of the other as follows:
$v_2 = 2 v_1$ or $v_1 = \dfrac{1}{2} v_1$.

In figure 2 below we have have two vectors that are parallel such that $v_2 = - v_1$. These two vectors are dependent because we can express one vector in terms of the other.

In figure 3, all vectors are parallel to each other. All These vectors are dependent because we can express one vector in terms of the other(s).

In figure 4, we can never express one vector in terms of the other because they are not parallel. These vectors are independent because we cannot express one vector in terms of the other.

In figure 5, using the geometrical sums of vectors, we can write
$3 v_4 = 2 v_1 + 3 v_2 + v_3$
and therefore these vectors are linearly dependent because we can express one vector in terms of the others.

In figure 6, the 3 vectors $v_1$ , $v_2 \}$ and $v_3$ are in the same plane $P$ and are therefore dependent because we can express any of these vectors in term of the other two vectors using linear combinations.

In figure 7, the pairs of vectors $\{v_1 , v_2\}$ , $\{v_2 , v_3\}$ and $\{v_1 , v_3\}$ are in different planes and are therefore independent because we cannot express any of these vectors in term of the other two vectors using linear combinations.

## Formal Definiton of Linearity of Vectors

Vectors $v_1, v_2 .... v_n$ are linearly dependent when at least one of the vectors may be expressed as a linear combinations of the remaining vectors as follows
$v_1 = r_2 v_2 + r_3 v_3 + .... + r_n v_n$
Writing the above with the zero on the right side, we obtain
$v_1 - r_2 v_2 - r_3 v_3 - .... - r_n v_n = 0$
Hence the following definition
Given a set of vectors $S = \{\textbf{v}_1 , \textbf{v}_2, ... , \textbf{v}_n \}$ ,
If the equation
$r_1 \textbf{v}_1 + r_2 \textbf{v}_2 + ... + r_n\textbf{v}_n = \textbf{0}$         (I)
has only one trivial solution $r_1 = 0 , r_2 = 0 , ... , r_n = 0$, we say that $S$ is a set of linearly independent vectors.
If the above equation has other solutions where not all $r_i$ equal to zero, then $S$ is a set of linearly dependent vectors.
In the examples below, matrices are row reduced in order to test for linearity. This may done using the row reduce augmented matrices calculator included.

## Examples with Solutions

Example 1
Are the vectors in the set $\left \{ \begin{bmatrix} -2 \\ 1 \end {bmatrix} , \begin{bmatrix} 6 \\ -3 \end {bmatrix} \right \}$ linearly independent?

Solution to Example 1
Let $v_1 = \begin{bmatrix} -2 \\ 1 \end {bmatrix}$ and $v_2 = \begin{bmatrix} 6 \\ -3 \end {bmatrix}$
The question could be answered by noticing that $v_2 = - 3 v_1$ and therefore the given vectors $v_1$ and $v_2$ are dependent (not independent)
The above was possible because we are dealing with vectors of small dimension. (2 components only)

We will now use a method that may be applied in any situation.
According to the definition given above, we need to find $r_1$ and $r_2$ such that
$r_1 v_1 + r_2 v_2 = 0$
The above may be written in matrix form as follows
$[ v_1 \;\; v_2] \begin{bmatrix} r_1 \\ r_2 \end {bmatrix} = 0$
where $[ v_1 \;\; v_2]$ is a matrix whose columns are $v_1$ and $v_2$

Write the system of equations in augmented matrix form
$\begin{bmatrix} -2 &6&|&0\\ 1 & -3&|&0 \end{bmatrix}$
Use the Gauss Jordan method to row reduce the above augmented matrix and obtain
$\begin{bmatrix} 1 &-3&|&0\\ 0 & 0&|&0 \end{bmatrix}$       (I)
The solution to the above reduced system (which is also a solution to the system before reduction) is found as follows
The second equation gives: $0 r_2 = 0$ therefore $r_2$ may take any real value
The first equation gives: $r_1 = 3 r_2$
The solution set may be written as: $\left \{ \begin{bmatrix} 3 r_2\\ r_2 \end{bmatrix} \right \} , r_2 \in R$
which therefore means that we have an infinite number of solutions and therefore the two vectors are linearly dependent.

Note that you do not have to solve the system in order to decide whether the given vectors are dependent or independent.
Construct the augmented matrix using the vectors as columns of the matrix and the constant column on the right is all zeros which in fact may be omitted. We then row reduce the augmented matrix. Then the following conclusions may easily be drawn:
1) The columns with a pivot are independent of each other
2) The columns with no pivots are dependent on the ones with the pivot.
3) The coefficients in the columns without pivot gives the coefficients of dependence on the independent columns.

Let us apply the above to the reduced matrix (I) above
1) Column 1 has a pivot and may be considered as independent
2) Column 2 has no pivot and may therefore be considered as dependent on column (1)
3) The coefficient $- 3$ in column 2 is telling us that $v_2 = - 3 v_1$

Example 2
Are the vectors in the set $\left \{ \begin{bmatrix} -2 \\ 1 \\ 4 \end {bmatrix} , \begin{bmatrix} 1\\ 0 \\ 5 \end {bmatrix} , \begin{bmatrix} 1\\ 2 \\ -1 \end {bmatrix} \right \}$ linearly independent?

Solution to Example 2
step 1: Construct the augmented matrix
Construct the augmented matrix whose columns are the given vectors and zeros on the right column
$\begin{bmatrix} -2 &1&1&|&0\\ 1 & 0&2&|&0 \\ 4 & 5 & -1& |& 0 \end{bmatrix}$

step 2: Row reduce the above matrix (you may use the row reduce augmented matrices calculator included).
$\begin{bmatrix} 1 &0&0&|&0\\ 0 & 1&0&|&0 \\ 0 & 0 & 1& |& 0 \end{bmatrix}$

step 3: Draw conclusions from the row reduced matrix
All 3 columns have a pivot each and therefore all 3 given vectors are independent

Example 3
Are the vectors in the set $\left \{ \begin{bmatrix} 2 \\ -1 \\ 3\\ 1 \end {bmatrix} , \begin{bmatrix} 0 \\ 2 \\ -1\\ 1 \end {bmatrix} , \begin{bmatrix} 4 \\ -8 \\ 9\\ -1 \end {bmatrix} \right \}$ linearly independent?

Solution to Example 3
step 1: Construct the augmented matrix using the given vectors as columns and zeros on the right column
$\begin{bmatrix} 2&0&4&|&0\\ -1&2&-8&|&0\\ 3&-1&9&|&0\\ 1&1&-1&|&0 \end{bmatrix}$

step 2: Row reduce the above matrix
$\begin{bmatrix} 1&0&\color{red}{2}&|&0\\ 0&1&\color{blue}{-3}&|&0\\ 0&0&0&|&0\\ 0&0&0&|&0 \end{bmatrix}$

step 3: Draw conclusions
Only the first and second columns (from the left) have a pivot and therefore the given vectors are not independent.
The coefficients $\color{red}{2}$ and $\color{blue}{-3}$ in the third column give the dependence of the original third columns as a linear combination of the other first and the second columns as follows
$\begin{bmatrix} 4 \\ -8 \\ 9\\ -1 \end {bmatrix} = \color{red}{2} \begin{bmatrix} 2 \\ -1 \\ 3\\ 1 \end {bmatrix} - \color{blue}{3} \begin{bmatrix} 0 \\ 2 \\ -1\\ 1 \end {bmatrix}$


# Big List of Erdős' elementary proofs

Paul Erdős was one of the greatest mathematicians of all time and he was famous for his elegant proofs from The Book. I posted a question about one of his theorem and got a reference, and I have other questions I want to know the answer to too. But, instead of requesting a reference for each theorem he gave with an elementary proof, I've decided to make a thread for a big list of all his elementary proofs.

I'm excited. Let's make an index of the pages of the Book shown to us!

To get you guys started, I will make a wish list of his theorems who's references I want to see. I encourage you to add to my wish list if you so desire.

Wish list :

• The product of two or more consecutive positive integers is never a square or any other higher power.
• A connected graph with a minimum degree $d$ and at least $2d+1$ vertices has a path of length at least $2d+1$.
• Let $d(n)$ be the number of divisors of $n$. Then the series $\sum_{n=1}^\infty d(n)/2^n$ converges to an irrational number
• Let $g(n)$ be the minimal number of points in the general position in the plane needed to ensure a subset exists that forms a convex $n$-gon. Then $$2^{n-2} + 1 \leq g(n) \leq \frac{(2n-4)!}{(n-2)!^2} + 1$$
• Erdős-Mordell inequality
• You need to define what you mean by "elementary". I think the list of elementary proofs will be very big, e.g. it will contain all his results in set theory, all his results on finite and infinite graphs. – bof Jul 6 '16 at 9:31
• @bof Hi bof. It will be big but I don't think all of them will be posted here. For starters, I'm looking for the questions in the wish list. Please post if you have any results – user230452 Jul 6 '16 at 9:35
• Is it fair that 4 people get to close a thread when 50 people have upvoted it ? Please open the thread for further posts – user230452 Jul 7 '16 at 10:59
• I voted to reopen. – Spenser Jul 7 '16 at 13:00
• @Spenser Thanks, Spencer. I appreciate it. I have collected another theorem of Erdos' to post. – user230452 Jul 7 '16 at 13:24

I think this is worth posting here, mostly because I really enjoy the simplicity of this proof but also because I have no idea how well it is known. The result is not deep or important, so the main interest is in the simplicity of the argument. Erdős proved a lower bound on the number of primes before an integer $n$.

Wacław Sierpiński, in his Elementary Theory of Numbers, attributes to Erdős the following elementary proof of the inequality $$\pi(n)\geq\frac{\log{n}}{2\log{2}}\quad\text{for }n=1,2,\ldots.$$

Please note that I have adapted the argument from the text of the book to make things, in my opinion, a bit clearer. Note also that the only tools used in the below proof are some basic combinatorial facts and some results about square-free numbers, which can, for example, be proved with the Fundamental Theorem of Arithmetic.

Let $n\in\mathbb{N}=\{1,2,3,\ldots\}.$ Consider the set $$S(n) = \{(k,l)\in\mathbb{N}^{2}:l\text{ is square-free and }k^{2}l\leq n\}.$$ It is a standard fact that every natural number has a unique representation in the form $k^{2}l,$ where $k$ and $l$ are natural numbers and $l$ is square-free. This gives $\lvert S(n)\rvert = n.$

Now if we have a pair $(k,l)$ with $k^{2}l\leq n,$ then we must have $k^{2}\leq n$ and $l\leq n$, since $k$ and $l$ are positive. Note that this gives $k\leq\sqrt{n}.$ Since $l$ is square-free, $l$ can be expressed as a product of distinct primes, each of which must be not-greater-than $n$ since $l\leq n$. That is, $l$ can be expressed as a product of the primes $p_{1},p_{2},\ldots,p_{\pi(n)}.$ There are $2^{\pi(n)}$ such products.

Therefore, if we know $(k,l)\in S(n)$ then there are at most $\sqrt{n}$ possibilities for what $k$ might be and at most $2^{\pi(n)}$ possibilities for what $l$ might be (independent of $k$, of course). It follows that $\lvert S(n)\rvert \leq 2^{\pi(n)}\sqrt{n},$ so $n\leq2^{\pi(n)}\sqrt{n}.$

Taking $\log$s and rearranging gives the result.

• How can you say a lower bound on the Prime Number Theorem is not important ! – user230452 Jul 6 '16 at 6:05
• @user230452: Better bounds are known through elementary arguments (I think the most famous is by Chebyshev, and has the advantage of providing an upper bound of the same order of growth). So this result is not so important since a little more work can give much better results, with still only elementary ideas. – Will R Jul 6 '16 at 17:23
• @Therkel: I think I understand, but, in this situation, I felt it best to make clear a) how I knew this argument was from Erdos and b) that chasing up the source would lead readers to a superficially different-sounding argument. – Will R Jul 6 '16 at 17:27
• This is brilliant! – Vinícius Novelli Jul 20 '16 at 1:41

Here is an exposition of the proof that made Erdos famous by David Galvin. An elementary proof of Bertrand's postulate, which states that there is a prime number in between every $n$ and $2n$. The essence of this proof is in noticing that the lower bound of $$\binom{2n}{n} \geq \frac{4^n}{2n + 1}$$ The binomial expression is the middle term (and the largest) of $(1+x)^{2n}$. The lower bound is the average of the sum of all binomial coefficients. This is obtained by putting $x=1$, and then dividing by the number of terms. This gives us the average. Obviously, the largest term should be bigger than the average. If the postulate does not hold, there is an upper bound that is smaller than this lower bound for large $n$. The postulate can easily be verified for the smaller values of $n$.

https://www3.nd.edu/~dgalvin1/pdf/bertrand.pdf

One very simple, and yet one of my favorites is the Erdős-Anning theorem:

Let $A \subseteq \mathbb C$ be an infinite set of points, such that

$$\forall x, y\in A \quad |x-y| \in \mathbb N$$

then there exists some $c,k \in \mathbb C$, such that all $a \in A$ is of the form $a = cx + k$ for some $x \in \mathbb R$.

It was proved in 1945 in the American Mathematical Bulletin. http://www.ams.org/journals/bull/1945-51-08/S0002-9904-1945-08407-9/S0002-9904-1945-08407-9.pdf

A more colorful formulation is that an infinite sets of points on the Cartesian plain with mutual integer distances must lie on a straight line.

To prove it, we need an upper bound on the number of non collinear points possible in a set with integral distances. More specifically, if there is a set of non collinear points have integer distances, all at most $d$, then at most $4(d + 1)^2$ points with integer distances can be added to the set.

• I have a doubt. Can you help me ? The theorem says that for any $n$, it is possible to have $n$ points on a plane, not all collinear such that they have integral distances but not infinitely many points. As I understand it, infinity is not a number, but a concept. To say that infinitely many points are possible it means that for any $n \geq l$, it is possible, where $l$ is an arbitrarily large natural number. It seems to me the first part of the theorem is stating it, and the second is contradicting it. – user230452 Jul 6 '16 at 3:45
• @user230452 No, to say that there is an infinite set with integral distances is not the same as saying for any $n\ge l$ there is a set with $n$ points and integral distances. Compare: For any $n\ge l$ there is a bounded subset of $\Bbb R$ with integral distances, but there is no bounded infinite subset of $\Bbb R$ with integral distances. $\Bbb N$ is an unbounded infinite subset of $\Bbb R$ with integral distances. – Mario Carneiro Jul 6 '16 at 8:05
• @bof There was a typo; the constraint $0 \in A$ was lacking. It is equivalent to adding some constant, s.t. $a = cx + k$. – user347499 Jul 6 '16 at 16:23

Erdos' proof of Infinite primes

The following proof is taken from the book - "Proofs from THE BOOK" by Martin Aigner and Gunter Ziegler. This proof is attributed to Erdos.

This proves that there are infinitely many primes and that the series of the sum of prime reciprocal steps diverges. Let us assume that the infinite series $\sum\frac{1}{p}$, where $p$ denotes the prime numbers is a convergent one. Then, there must exist some natural number $k$ such that, $$\sum_{i > k} \frac{1}{p_i} < \frac{1}{2}$$

For an arbitrary natural number $N$, we get the inequality $$\sum_{i > k} \frac{N}{p_i} < \frac{N}{2}$$

Now, we call all the primes $p_1, p_2, \dots , p_k$ the small primes, and all the other primes the big primes.

Let $N_b$ denote the number of positive integers $n \leq N$ that have at least one divisor that is a big prime. And $N_s$ denote the number of positive integers less than $N$ that have only small prime divisors.

We will show that for a suitable $N$, $N_b + N_s < N$, which is a contradiction because their sum should equal $N$.

First, we estimate $N_b$ $$N_b \leq \sum_{i > k}\big\lfloor \frac{N}{p_i} \big\rfloor < \frac{N}{2}$$

Now, we look at $N_s$. We write every $n \leq N$ as $a_nb_n^2$ where $a_n$ is the square free part. $a_n$ is a product of different small primes. There are only $k$ small primes, and each prime may either be chosen or not chosen. So, there are only $2^k$ different values of $a_n$. Furthermore, \begin{align} b_n \leq \sqrt n \leq \sqrt N \\ \text{There are at most $\sqrt N$ different values of $b_n$} \\ \implies N_s \leq 2^k\sqrt N \end{align}

Since, $N_b$ is always less than $N/2$, we just have to find a value of $N$ such that $N_s \leq N/2$. This happens when $N = 2^{2k + 2}$. When $N = 2^{2k + 4}$, $N_s < N/2$. This proves our contradiction that $$N_b + N_s < N$$ which proves that the series diverges and that there are infinitely many primes.

Erdos' favourite questions. The following are not research papers but popular questions Erdos used to ask children.

• If $n+1$ integers are chosen from the first $2n$ integers, there will always be two that are co prime.

There will be two numbers that are consecutive. These two numbers will be relatively prime. To see that this is not true when $n$ integers are chosen, just select all the even numbers.

• If $n+1$ integers are chosen from the first $2n$ integers, there will always be two such that one divides the other.

Every number is expressible as the product of a power of two and an odd number. There are only $n$ odd numbers in the first $2n$ integers. Two of the numbers are multiplied by the same odd number. One of them has a smaller power of two. This number divides the other. To show that it is not necessarily true for $n$ integers, choose $n+1,n+2, \dots 2n$.

• Suppose we have $n$ natural numbers none of which is greater than $2n$ such that the least common multiple of any two is greater than $2n$. Then, all $n$ numbers are greater than $2n/3$.
• Regarding the first one, you can easily see that "all the even numbers" is actually the only solution -- we can't select 1, because it's coprime to everything; and we can't have any consecutive numbers. If we take $n-1$ numbers, a bit of thinking/excluding a few small cases shows that for $n>3$, the only solutions are "all even numbers but one". For $n=3$, we have $\{3,6\}$. In general, any set of $n-k$ must be all even unless $n-k \ge 2n/3$, because we have an odd number at least $3$, which immediately narrows us to $1/3$ of the numbers. I wonder how # of solutions grows with $n$ and $k$? – Alex Meiburg Jul 7 '16 at 23:21
• @Alex, your final conclusion is false. In fact, for all $\epsilon > 0$ there exists an $n$ and a set $S \subset \{1,2,\ldots, 2n\}$ with $|S| > (1 - \epsilon)n$ such that $S$ contains an odd integer and for all $x,y \in S$ we have $\gcd(x,y) > 1$. For example, choose $n$ to be the product of the first $m$ odd primes and define $S = \{2k | k < n \text{ and } \gcd(k, n) > 1 \} \cup \{n\}$. You will see that $|S| = n - \varphi(n) + 1 > n(1 - \frac{c}{\log \log(n)})$ for some absolute constant $c$. For $m \ge 8$ we have $|S| > \frac{2n}{3}$. – Woett Oct 11 '18 at 12:00
• @Woett, right, how silly of me. I figured that if $k$ is odd then the density of numbers sharing a factor with $k$ is at most $1/3$ -- totally neglecting the possibility of multiple factors. :) Taking $n=105=3\cdot 5\cdot 7$ gives a density of $0.54$, of course. – Alex Meiburg Oct 12 '18 at 2:46

Erdos answered the following question in the affirmative - Are there infinitely many odd numbers that are not expressible as the sum of a prime number and a power of $2$. The proof is explained in this paper : http://www.maa.org/sites/default/files/3004416309960.pdf.bannered.pdf

The essence of the proof is in showing that for every integral value of $k$, there is a $2^k$ which has a certain residue with certain moduli. By the Chinese remainder theorem, there are infinitely many odd numbers that satisfy those same congruences. Whenever they do, $a - 2^k$ is composite for all integer values of $k$.

• Surely this is clear because the set of all numbers $p+2^k$ is density zero? Although perhaps that is not elementary. – Mario Carneiro Jul 7 '16 at 1:20
• The density should be positive, not zero. There are n/log(2) choices of (prime, power of 2) both less than n. If most, or a positive fraction, of those are unique representations then a positive density subset of integers less than 2n have such a representation. – zyx Jul 7 '16 at 2:08
• @zyx Oh, indeed if $|A\cap[n]|\sim f(n)$, and $B=\{a+2^k:a\in A,k\in\Bbb N\}$, then $|B\cap[n]|\le\sum_k f(n-2^k)\le f(n)\log_2n$, which is not strong enough to conclude density zero with $A=\Bbb P$, $f(n)=n/\log n$. A simulation of the quantity $\sum_{p\le n}\log_2(n-p)$ appears to converge to about $1.468n$, which is obviously useless as an upper bound on $|B\cap[n]|$ since it's larger than the trivial bound. – Mario Carneiro Jul 7 '16 at 4:59
• That is close to n/log(2) = 1.443 n . – zyx Jul 7 '16 at 5:08
• Does the simulation appear to converge to 1/log(2) when using integer logarithms, $\lfloor \log_2 ( \dots ) \rfloor$? @MarioCarneiro – zyx Jul 7 '16 at 19:10

The proof of the Littlewood-Offord lemma for sums of real numbers.

Erdos noticed that under the correspondence between sequences of $\pm$ signs and finite sets, Sperner's theorem applies and gives the optimal bound for the Littlewood-Offord lemma in dimension $1$ (on how many signed sums of $n$ given numbers of absolute value at least $1$, can have absolute value at most $1$). This observation is a few lines of very basic algebra and combinatorics.

The rest of the paper is more difficult, and is an extension of Sperner's theorem to handle the case of larger bounds on the sums, once again reading optimal bounds from the combinatorics. This gave evidence that the higher dimensional Littlewood-Offord problem on sums of vectors might also be essentially combinatorial, which was later shown to be true.

Erdos - Mordell Inequality : For a point $O$ inside a given triangle $ABC$, the perpendiculars $OP$, $OQ$ and $OR$ are drawn to the side $$OA + OB + OC \geq 2(OP + OQ + OR)$$

The product of consecutive integers is never a power

This was proved by Erdos and Selfridge.

http://www.renyi.hu/~p_erdos/1975-46.pdf

Suppose $A$ is a set of $r$ subsets on the set $\{1,2,\dots,n\}$ such that any two sets in $A$ have a non empty subset and $n/2\geq r$, the maximal size of $A$ is $\binom{n-1}{r-1}$ i.e., $$|A| \leq \binom{n-1}{r-1}$$with equality holding if and only if all the sets share a common element.
A family of sets may also be called a hypergraph. When every set in $A$ has the same size $r$, it is called a $r$-uniform hypergraph.
Note : The condition of $n/2 \geq r$ is imposed because if $r$ is more than half of $n$, then any two sets have a non empty intersection. The maximal size of the family is then $\binom nr$.


Sum of sin waves with same frequency and different amplitudes and phase

For the equation:

$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$

I've been able to show that the amplitude of the sum is (I believe this is a standard problem):

$$A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta_1 - \theta_2)}$$

and the new phase is:

$$\theta_3 = \arctan \left(\frac{A_1 \sin \theta_1 + A_2 \sin \theta_2}{A_1 \cos \theta_1 + A_2 \cos \theta_2}\right)$$

My question is what happens when the phases $\theta_1$ and $\theta_2$ are zero (or just equal to each other). Most books and internet sites (eg http://mathworld.wolfram.com/HarmonicAdditionTheorem.html) state that under these conditions:

$$A_3 = \sqrt{A_1^2 + A_2^2}$$

However, if I set $\theta_1$ and $\theta_2$ to zero one gets instead (Since $cos(0) = 1$):

$$A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2}$$

I've been staring at this for a while and I don't understand why the published answer is different from mine. I am sure it is something simple but I've not been able to spot it (Addendum: Wikipedia site: https://en.wikipedia.org/wiki/List_of_trigonometric_identities has the same result under section Linear Combinations).

• I realized the problem: $A_3 = \sqrt{A^2_1 + A^2_2 + 2 A_1 A_2}$ applies to the relation $cos (wt + \theta_1) + sin(wt + \theta_2)$ not $sin (wt + \theta_1) + sin(wt + \theta_2)$. For the latter it works fine. This is an error on my part. ok to delete this question, as its not a valid one? – rhody Jul 13 '18 at 18:23
• Although your Question contained and was motivated by an error, it brings up some useful content, and I'd like to retain it. Even the error, articulated in your Comment, bears pointing out as it is the sort of thing anyone might overlook at first glance. – hardmath Jul 14 '18 at 13:02

The formula $\sqrt{A_1^2+A_2^2}$ applies to sinusoids in quadrature ($\frac\pi2$ phase difference), such as when summing a sine and a cosine.

Signals in phase just add up their amplitudes, $A_1+A_2$, while signals in opposition subtract, $|A_1-A_2|$.

• I've been staring at this for hours yesterday, and then this morning after I posted the question and suddenly realized my error. – rhody Jul 13 '18 at 18:26

You wrote:

$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$

The Wolfram article you cite says:

It is always possible to write a sum of sinusoidal functions $$f(\theta)=acos(\theta)+bsin(\theta )$$
as a single sinusoid the form $$f(\theta)=ccos(\theta+\delta)$$ $$A_3 = \sqrt{A_1^2 + A_2^2}$$

So you have sine plus sine, while Wolfram has sine plus cosine. To convert Wolfram's cosine to a sine, you need to shift the phase by $\frac{\pi}2$, which then makes the cosine of the phase difference equal to zero.

Waves with no phase difference (or even pi's) directly add up their amplitudes to form a new wave.

$$A_1 \sin (\omega t) + A_2 \sin (\omega t) = (A_1+A_2) \sin (\omega t )$$

The $A_3$ you prescribed is for waves with phase difference $(\theta_1 - \theta_2)=\frac{\pi}{2}$.

The equation you got putting $\theta_1=\theta_2=0$ is correct and simplifies to $A_3=(A_1+A_2)$.

The result given in Equation (21) of Wolfram site

for adding in-phase waves of same frequency with zero relative phase angle gives

$$A_3= \sqrt{A_1^2+A_2^2+ 2 A_1 A _2}= \pm (A_1+A_2)$$

The other result came about when the phase difference is $\pi/2$ obtained directly by Pythagoras thm as diagonal length of a rectangle of vectors or phasors.

The parallelogram of amplitudes can be drawn and the situation becomes clear. There are three situations. In-phase,add; Out of phase, subtract, or take the difference; when relative phase difference is $\pi/2$ take root square sum.


# How to find the elongation of a spring when it is tied to a brass bar over a tilted wall?

The problem is as follows:

An homogeneous brass bar which goes from point $$A$$ to point $$B$$ has a weight of $$300\,N$$. The bar is suported over two frictionless surfaces as it is shown in the picture from below. The system is in static equilibrium by the action of the force excerted by the spring tied to the end at point $$B$$ on the bar which in turn is fixed to the ceiling. Using this information find the elongation of the spring. Assume that the constant of the spring is $$k=500\,\frac{N}{m}$$.

The alternatives given in my book are as follows:

$$\begin{array}{ll} 1.&28\,cm\\ 2.&23\,cm\\ 3.&25\,cm\\ 4.&26\,cm\\ 5.&30\,cm\\ \end{array}$$

What I did to solve this problem was to use a "trick" which I seen in a similar problem about a ladder standing next to a wall.

The rationale was that when an horizontal object is leaning against two surfaces there will be a reaction from both. But if it is asked a force in between. You may use any point to establish the torque about that point and use Varignon's theorem? to find the unknown force. This is done to "cancel out" the reaction force from both surfaces.

Therefore:

Torque about the point specified in the figure from below results into:

$$\sum_{i=1}^{n}\tau=0$$

$$(l\cos 30^{\circ})\times(300)+500\times x \times 2l =0$$

simplifying terms:

$$-\frac{\sqrt {3}}{2}\times 3 + 10 x = 0$$

$$x= \frac{3 \sqrt {3}}{20}\approx .2598076212 \,m$$

which transformed into cm would be about $$26\,cm$$ and this appears in the alternatives as option four. But is my answer the right approach to this problem?.

My source of confusion is that in the problem is not indicated if the spring is paralell to the incline or if it does make another angle.

It is not indicated in the problem but how can I find let's say the Reaction in both the floor and in the wall?. How can I find these?. Am I using Varignon's theorem in this problem?. Can someone help me with this matter please?.

• @Cesareo I'm sorry but I still don't get to the answer you reached. This is not homework. Can you post a solution or more hints?. How exactly you got to $30\,cm$.? – Chris Steinbeck Bell Jul 11 '20 at 3:27

First, if you are referring to Varignons' theorem about quadrilaterals and the parallelogram formed by the midpoints of the sides, I don't think you are usinng it.

Second, to determine both the reactions in the floor and the wall we can apply Newton's law and impose that the sum af all forces is zero, and then impose that the torque about any point we want is zero. You already did the second part, so you could just apply Newton's law: $$\vec{N_A}+\vec{N_B}+\vec{P}+k\vec{\Delta l}=0$$ If the spring is parallel to the wall we know the directions of all this vectors and by decomposing along two axes $$x$$ and $$y$$ (I choose the $$x$$-axis parallel to the ground, but it's arbitrary) we obtain two equations that allow us to find the moduluses of the reaction forces: $$N_A+\frac{N_B}{2}-P+\frac{\sqrt{3}}{2}k\Delta l=0$$ $$-\frac{\sqrt{3}}{2}N_B+\frac{1}{2}k\Delta l=0$$

Personally I solved the problem calculating the torque about the point B because there are 4 forces and two are applied in B, so the equations are particularly simple, but any other point is just as good and gives the same solutions. I too found that the answer is 26cm

Fig 2 has made things easy for taking force/moment equilibrium to evaluate forces.

Take moments about top left corner call it $$I$$

$$k x \cdot 2 l = l \cos 30^{\circ} 300$$

$$x = 0.2598$$ meters i.e., the fourth option:

$$x= 25.98 cm$$

Btw, I is the center about which the brass bar is instantaneously rotating in dynamic problems.

Calling $$\alpha = 30^{\circ},\ \beta = 60^{\circ}$$ and

$$\cases{ A=(0,0)\\ B=A+l(\cos\alpha,\sin\alpha)\\ G = \frac 12\left(B+A\right) }$$

with the forces

$$\cases{ f_A = (0,N_A)\\ f_B = k \Delta x (\cos\beta,\sin\beta)+N_B(-\cos\alpha,\sin\alpha)\\ f_G = (0,-W) }$$

we have

$$\cases{ f_A+f_B+f_G = 0\\ (A-B)\times f_A+(G-B)\times f_G = 0 }$$

from which we obtain

$$N_A = \frac W2,\ N_B = \frac W4,\ \Delta x = \frac{\sqrt{3}W}{4k}$$

so $$\Delta x\approx 0.26$$ cm

Another way using virtual displacements.

Calling

$$\cases{ f_{B_1} = k \Delta x (\cos\beta,\sin\beta)\\ f_{B_2} = N_B(-\cos\alpha,\sin\alpha) }$$

In equilibrium, we have

$$\delta A\cdot f_A + \delta G\cdot f_G + \delta B \cdot\left(f_{B_1}+f_{B_2}\right) = 0$$

and due to orthogonality

$$\delta A\cdot f_A = \delta A\cdot f_G = \delta B\cdot f_{B_2} = 0$$

and following

$$\frac 12\left(\delta A + \Delta x(\cos\beta,\sin\beta)\right)\cdot f_G + k (\Delta x)^2\left(\cos\beta,\sin\beta\right)\cdot \left(\cos\beta,\sin\beta\right)=0$$

or

$$-\frac 12 W\sin\beta + k\Delta x = 0\Rightarrow \Delta x = \frac{\sqrt{3}W}{4k}$$

• The bar should be $2l$ long, not $l$. – enzotib Jul 11 '20 at 8:06


# Why both data have same standard deviation?

But I am curious why the standard deviation for both data is same. I think because data 2 is "100 - data 1".

• $var(X+a)=var(X)$ and $var(aX)=a^2var(X)$ for a real number $a$ and a random variable $X$. With those, can you prove why $var(100-X)=var(X)$ where X is your data set? (Equal variances is the same as equal standard deviations.) – Dave Feb 15 '20 at 0:24

I still would like you to try to prove it, but I’ll give some intuition about those two identities I gave.

Remember that variance (and standard deviation) have something to do with how spread out your data are.

$$var(X+a)=var(X)$$ means that if you take data with some amount of spread and slide them up or down the number line, you do not change the spread.

$$var(X)=var(-X)$$ is not quite what I gave, but it’ll be part of your proof. The intuition here is that you’re just spinning it around, taking a mirror image. The data are as spread out as their reflection.

Together, $$var(100-X)=var(X)$$ means that you flip the data to the mirror image and then slide along the number line, but you do not change the spread.

Capisce?

• what is the meaning of Capisce? – aan Feb 16 '20 at 11:41
• that's Italian for "do you comprehend ?" – IrishStat Feb 16 '20 at 13:20

I believe the answer to this question can be expressed in more elemental terms. Beginning with the formula for the sample standard deviation

$$s_{x} = \sqrt{\frac{ \sum_{i=1}^{n}{(x_i - \overline{x})^2} }{n - 1}},$$

I want to draw your attention to the expression $$(x_i - \overline{x})$$ in the numerator. Ignore the summation notation for the purposes of this answer. Note, we decrement the sample mean (i.e., $$\overline{x}$$) from each realization of $$x_{i}$$. If we were to increase or decrease each $$x_{i}$$ by the same amount, the mean will change. However, the distance of each realization of $$x_{i}$$ from that central tendency remains the same. In other words, each deviation from the mean is the same.

Subtracting each $$x_{i}$$ from a constant (e.g., $$100 - x_{i}$$) flips the vector of values to its mirror image, then slides it along the number line by a constant amount. Suppose the first realization of $$x_{1}$$ = 20 and $$\overline{x}$$ = 4. In keeping with the foregoing expression,

$$(20 - 4) = 16,$$

which is the deviation from the sample mean. Now, assume we don't move along the number line just yet and we simply put a negative sign in front of each $$x_{i}$$, such that we have $$-(x_{i})$$; this flips the sign of the mean as well. The first realization, $$x_{1}$$, is now $$-20$$. Again, substituting $$-20$$ into the expression,

$$(-20-(-4)) =-16,$$

which is the same number of units away from the mean. Note, squared deviations result in a positive number, so the numerator remains the same. You can run this code in R which builds upon @knrumsey's insightful response, but breaks it down further:

x <- rnorm(10000, 30, 5) # Simulates 10,000 random deviates from the normal distribution

par(mfrow = c(2, 2))
hist(x, xlim = c(0, 100), col = "blue")
hist(-x, xlim = c(-100, 0), col = "red")  # Note the -x, it simply flips the sign (mirror image)
hist(x, xlim = c(0, 100), col = "blue")
hist(100-x, xlim = c(0, 100), col = "red")  # Subtracting x from 100 shifts values along the x-axis

The first row of plots shows how the realizations translate when we negate each $$x_{i}$$. The second row shows what happens when each $$x_{i}$$ is subtracted from 100; the translation first flips then slides across the number line. The spread of the distribution is unaffected by this.

To supplement @Dave's answer, take a look at the following histograms which have the same x-axis. Data2 is just a shifted version of Data1 and therefore the standard deviation, which is a measure of spread, shouldn't change.

R code to generate histograms:

x <- 100*rgamma(1000, 6, 2) #Simulate some data
hist(x, xlim=c(0,100))
hist(100-x, xlim=c(0,100))

• Thanks. well-plotted histogram. May I know how do you plot the histogram? – aan Feb 15 '20 at 5:29
• @aan see the update. – knrumsey Feb 15 '20 at 17:49
• @knrumsey Good illustration. However, simulating from the gamma distribution would produce realizations that cluster around a mean that is much lower than what I see here given the parameters you specified. Right? – Thomas Bilach Feb 16 '20 at 20:24
• @Tom, that is correct. I didn't put too much thought into which distribution I simulated from. Just wanted draws between 0 and 100. – knrumsey Feb 16 '20 at 20:45
• @Thomas Bilach, realizing that I misunderstood your question. Thanks for pointing this out, I've updated the post to reflect the fact that I multiplied the data by 100. – knrumsey May 3 '20 at 0:48

changes period to period in DATA 1 identical to the negative changes in DATA 2

OBS DATA 1 FIRST DIF DATA 2 FIRST DIFF

1 30 NA 70 NA

2 40 10 60 -10

3 80 40 20 -40

4 30 -50 70 50

5 20 -10 80 10

6 33 13 67 -13

7 33 0 67 0

8 33 0 67 0

9 33 0 67 0

CONTINUING STEPS TO UNCOVER THE BASIC RELATIONSHIP OF DATA1 AND DATA2:

I took the series DATA1 and computed the ACF ... here it is ..

AND for DATA2 here

...note that not only are the standard deviations the same but the ACF's are the same.

• This is an interesting perspective, but I believe it would need more explanation to be understood by many readers. – whuber Feb 15 '20 at 22:53
• @IrishStat, thanks. do you mind to explain more? – aan Feb 16 '20 at 11:42
• i added some material to my response detailing how differenced data period to period ( val2-val1,val3-val2,.. val9-val8) for both original series DATA1 and DATA2 were complements of each other – IrishStat Feb 16 '20 at 13:19
• @IrishStat. Noted. what is the meaning of Q.E.D and OBS? – aan Feb 16 '20 at 15:58
• An explicit connection between those assertions and the result is needed. What is interesting is your use of successive differences ("changes period to period" in an otherwise unordered dataset) to characterize the standard deviation of the dataset. In other words, you appear to be claiming that the set of equalities $$x_{i+1}-x_i = -(y_{i+1}-y_i)$$ for $i=1, 2, \ldots, n-1$ implies $$\sum_{i=1}^n(x_i-\bar x)^2=\sum_{i=1}^n(y_i-\bar y)^2.$$ That's true, but it does need to be shown: you can't just exhibit a table of numbers and declare "QED"! – whuber Feb 16 '20 at 20:15


# Wanna practice Induction ?

The principle of mathematical induction is a very very useful tool in many proofs, and also in proving some useful formulas. $\color{#D61F06}{\text{The principle states that any given set of positive integers}}$ has ALL natural numbers if it follows the following conditions -

$(i)$ The first natural number, i.e. $1$ is in the set.

$(ii)$ Whenever the integer $k$ is in the set, then $k+1$ is also in the set.

This looks so obvious, see that if you have been given the two statements simultaneously, then by using $(ii)$ on $(i)$, you can say that $2$ is in the set. Then again applying the statement $(ii)$ for $2$, we say that $(3)$ will be in the set, so on.

$\color{#3D99F6}{\textbf{Thus if you prove a certain result for 1}}$ , $\color{#3D99F6}{\textbf{and then assume that the result is true}}$ for $k$, just prove that it is true for $k+1$ and you're done !

Same thing you can do for 0, then prove for whole numbers !!!

Problems for practice:-

Prove that the following results hold $\forall n \in \mathbb{N}$

$(a) \displaystyle \sum_{k=0} ^n 2^k = 2^{n+1}-1$

$(b) \displaystyle \sum_{k=1} ^n k = \dfrac{n(n+1)}{2}$

$(c) \displaystyle \sum_{k=1} ^n k^2 = \dfrac{n(n+1)(2n+1)}{6}$

$(d) \displaystyle \sum_{k=1}^n k^3 = \biggl( \sum_{k=0} ^n k \biggr) ^2$

$(e) \displaystyle 24 \mid 2\cdot 7^n + 3\cdot 5^n -5$

$(f) \displaystyle 1\cdot 2+2\cdot 3+3\cdot 4+ .... + n(n+1) = \dfrac{n(n+1)(n+2)}{3}$

I felt like sharing because though it's very well known to all, I am willing to find some good level problems for practicing this foundation builder concept once more... $\color{#20A900}{\text{Isn't this a good revision ?}}$

5 years, 4 months ago

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\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

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@Aditya Raut this is really a nice note for practice...... could you please add more divisibility kinda problems to give a more diverse experience to readers? ........ thanks :)

- 5 years, 4 months ago

- 5 years, 3 months ago

Thanks - they look like a good selection of induction problems

- 4 years, 10 months ago

I actually haven't tried $(d)$ and $(f)$ before since doing math XD

Here's the solution for $(e)$ if someone got stuck.

$2\times 7^{m+1} + 3 \times 5^{m+1} - 5$

$= 2\times 7^{m}\times 7 + 3\times 5^{m}\times 5 - 5$

$= 14\times 7^{m} + 15 \times 5^{m} - 5$

$= 12\times 7^{m} + 12\times 5^{m} + 2\times 7^{m} + 3\times 5^{m} - 5$

$= 12(7^{m} + 5^{m}) + 24k$ for some $k$

$= 24n + 24k$ for some $n$

- 5 years, 3 months ago

$(a)$ Base Case $n = 1$

$\displaystyle \sum_{k = 0}^{1}2^{k} = 3$

$2^{n+1} - 1 = 3$

True.

Propose it is true $\forall n \in N$

Then it must be true for $n + 1$

$\displaystyle\sum_{k = 0}^{n+1}2^{k} = \sum_{k = 0}^{n}2^{k} + 2^{n+1}$

Substituting the value from our proposal,

$\sum_{k = 0}^{n}2^{k} + 2^{n+1} = 2^{n+1} - 1 + 2^{n+1}= 2^{n+2} - 1$

Thus, $LHS = RHS$

Thus, $LHS = RHS \forall n \in N$

Sorry for ugly arrangement.

- 5 years, 3 months ago

@Aditya Raut Can you add these to the Induction Wiki page? Thanks!

Staff - 5 years, 1 month ago

Nice! I'm too into creating notes and problems set for inequality :) Why not take a look at it @Aditya Raut ? :D

- 5 years, 3 months ago

What is e) ?... in d) second summation is it k=0 or k=1 ? Nice collection.

- 5 years ago

Thanks, but Just think ! Will adding $0$ make any change? In that summation, starting from $k=0$ will yield same thing as starting with $k=1$, won't it ? @Niranjan Khanderia

- 5 years ago

Thanks.

- 5 years ago

And e) means prove that for all positive integers $n$, $(2\cdot 7^n + 3\cdot 5^n -5)$ is divisible by $24$.

$a\mid b$ is the symbol of "a divides b" or "b is divisible by a".

Latex code for the symbol $\mid$ is "\mid"

- 5 years ago

Thanks. I did not get it since I did not assume the parenthesis.

- 5 years ago

/(good/)

- 4 years, 10 months ago

A good level problem is here: Fro every positive integer $n$, prove that $\sqrt{(4n+1)}<\sqrt n +\sqrt{(n+1)}<\sqrt{(4n+2)}$. Hence or otherwise, prove that [$\sqrt{n}+\sqrt{n+1}$]=[$\sqrt{4n+1}$] where [.] denotes floor function. It is an IITJEE problem.

- 4 years, 7 months ago


all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:

$$\begin{eqnarray} \lvert \frac{x+1}{x-2} \rvert &<& 3\\ \Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac{x+1}{x-2} < 3\\ x+1&>&-3(x-2)\hspace{2cm}\text{(Solving for the left inequality)}\\ x+1&>&-3x+6\\ 4x&>&5\\ x&>&\frac{5}{4} \end{eqnarray}$$ which cannot be true, as $$x$$ can equal $$2$$ with these restrictions which is obviously not allowed. Doing the left side the correct way:

$$\begin{eqnarray} \frac{x+1}{x-2} &>& -3\\ \frac{x+1}{x-2} + 3 &>& 0\\ \frac{x+1}{x-2} + \frac{3x-6}{x-2} &>& 0\\ \frac{4x-5}{x-2} &>& 0\\ \Rightarrow x < \frac{5}{4}\lor x>2\\ \end{eqnarray}$$

leads to the solution set $$(-\infty,\frac{5}{4})\cup(2,\infty)$$. My question is, what happened with the first method where I failed to come up with a solution for $$(2,\infty)$$, (and why the signage for $$x>\frac{5}{4}$$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.

• The use of the Union symbol $\cup$ is certainly not the correct way, as it is used to denote union of sets. You can use a $\textrm{or}$, or the or symbol $\vee$ (\vee). Jul 7, 2021 at 23:31
• @ultralegend5385: When writing in the context of mathematical logic, I think \lor, \land and \lnot are easier to remember for $\lor,\land,\lnot$ respectively. Then again, sometimes I prefer writing \neg over \lnot for the logical not $\neg$ Jul 7, 2021 at 23:44
• @ultralegend5385 fixed. Jul 7, 2021 at 23:54

In the first method, when you go from $$\dfrac{x+1}{x-2}\gt -3$$ to $$x+1\gt -3(x-2)$$, what you're doing is multiplying both sides of the inequality by $$x-2$$ and you assume that $$x-2\gt 0$$ when doing that because otherwise the inequality sign flips to $$\le$$

So, assuming $$x\gt 2$$, you get from that inequality that $$x\gt 5/4$$ and the second inequality $$\dfrac{x+1}{x-2}\lt 3$$ gives $$x\gt 7/2$$

So, assuming $$x\gt 2$$, you get $$x\gt 5/4$$ and $$x\gt 7/2$$; together they imply $$x\gt\max\{2,5/4,7/2\}=7/2$$

Similarly, assume $$x-2\lt 0$$, ie, $$x\lt 2$$ and solve $$x+1\lt -3(x-2)$$ and $$x+1\gt 3(x-2)$$ simultaneously to obtain the other solution set. You get $$x\lt\min\{2,5/4,7/2\}=5/4$$

The complete solution set is thus $$(-\infty,5/4)\cup (7/2,\infty)$$

The answer you arrived at is wrong. Note that $$x=3\in (2,\infty)$$ doesn't satisfy the original equation.

In the second method, you're only solving one of the two inequalities that are supposed to hold. If you solve the second inequality $$\dfrac{x+1}{x-2}\lt 3$$ in a similar way, you get $$\frac{x+1}{x-2}-3\lt 0\iff\frac{7-2x}{x-2}\lt 0\iff x\in (-\infty, 2)\cup (7/2,\infty)$$

and the solution set to your original problem is the intersection of this with $$(-\infty,5/4)\cup (2,\infty)$$ so that

$$[(-\infty,2)\cup (7/2,\infty)]\cap[(-\infty,5/4)\cup (2,\infty)]=(-\infty,5/4)\cup (7/2,\infty)$$

which is the solution set to the original problem $$\left|\dfrac{x+1}{x-2}\right|\lt 3$$

• I understand now. Thank you for explaining it so well. Jul 7, 2021 at 23:55
• @js2822: You're welcome. Also, nice work for asking a good question and being responsive to others! +1 ;) Jul 7, 2021 at 23:58

HINT

Since both sides are nonnegative, you can square them in order to obtain an equivalent inequation: \begin{align*} \left|\frac{x+1}{x-2}\right| < 3 & \Longleftrightarrow \left(\frac{x+1}{x-2}\right)^{2} < 9\\\\ & \Longleftrightarrow \frac{(x^{2} + 2x + 1) - 9(x^{2} - 4x + 4)}{(x-2)^{2}} < 0\\\\ & \Longleftrightarrow \frac{-8x^{2} + 38x - 35}{(x-2)^{2}} < 0\\\\ & \Longleftrightarrow \begin{cases} 8x^{2} - 38x + 35 > 0\\\\ x\neq 2 \end{cases} \end{align*}

Can you take it from here?

• One question: Does squaring an absolute value add any extraneous solutions in certain circumstances? Or do they never appear since the absolute value is always nonnegative? Jul 8, 2021 at 0:00
• As long as both sides are nonnegative, you can proceed as suggested so that there won't be any extraneous solutions. Jul 8, 2021 at 0:46

We can only "flip the sign" if what is inside the absolute value is negative. And similarly, we "keep the sign", if what is inside the absolute value bracket is positive.

We could write it this way...

$$-3<\frac {x+1}{x-2}<0$$ or $$0\le \frac {x+1}{x-2}<3$$

Or say, the left-hand equation only applies when $$-1

Then we can solve the two sets of inequalities:

$$-1 or $$x > \frac {7}{2}$$ or $$x \le -1$$

Now we can combine the intervals

$$(-\infty, \frac 54)\cup (\frac 72,\infty)$$


# Why doesn't the derivative of integration by trig substitution match the original function?

For one of my assignments in Calc II, I had to solve ${\displaystyle\int} \frac{\sqrt{x^2-4}}{x} \text{ d}x$. By using Trig substitution where $x = 2 \text{ sec } \theta$, my solution was $\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right) + C$. I tested my solution by graphing $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$, and it only matched the original function where $x \geq 2$. To match for $x \leq -2$, I had to use $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} + 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$.

Assuming I solved this correctly, what causes this discrepancy? How can we know if, when, and where this kind of thing will happen? Also, how would we notate the difference? Is it necessary to write the solution as a piecewise function, or does none of this matter anyway? I would think that, if it was because of the square root, the change in sign would be in front of it, but this seems to be because of the arc secant.

Edit Since a comment mentioned using WolframAlpha, I decided to double-check through there, and they do have my same answer under the Step-by-Step Solution, although they then simplified it to arctan somehow. Graphing the derivative of their solution with arctan matches the original function even less, with both a different C and shape. For explicitness, here's my work: $x = 2\sec \theta \Rightarrow \text{d}x = 2 \sec\theta \tan\theta$

$\int \frac{\sqrt{4 \sec^2\theta - 4}}{2 \sec\theta}(2 \sec\theta \tan\theta \text{ d}\theta) = \int 2 \sqrt{\sec^2 \theta-1} \tan\theta \text{ d}\theta = 2 \int \tan^2 \theta \text{ d}\theta$

With trig identity replacement: $2 \int \sec^2 \theta \text{ d}\theta - 2 \int 1 \text{ d}\theta = 2 \tan\theta - 2\theta$

With replacement of $\theta$ back to its original value: $\sqrt{x^2 - 4} - 2\text{ arcsec} \left(\frac{x}{2}\right) + C$

Edit 2 I understand that the square root causes sign issues, but even switching to absolute value doesn't seem to have any effect in the graphing. Here's a link to everything graphed on desmos.

• I beat you to it, deleting my comment! – Namaste Oct 21 '16 at 20:17
• According to Wolfram, your antiderivative is incorrect. – Ken Duna Oct 21 '16 at 20:17
• I hadn't checked with Wolfram before, but in their Step-By-Step, their answer is mine, but with different work, so I updated my answer in response and showed my work. – BrainFRZ Oct 21 '16 at 20:53

The trouble is when getting rid of the square root. In general we have $\sqrt{x^2} = |x|$, so $$\sqrt{\sec^2\theta-1} = \sqrt{\tan^2\theta} = \lvert\tan\theta\rvert.$$ So the integrand is really $$\lvert\tan\theta\rvert \tan \theta = \begin{cases} \sec^2\theta - 1 & 0\le\theta<\pi/2,\\ -(\sec^2\theta - 1) & \pi/2<\theta<\pi\\ \end{cases}$$ (where you take $\theta = \operatorname{arcsec} (x/2)$ with $0 \le \theta < \pi$). When reversing the substitution there is also the issue of $$\tan\theta = \begin{cases} \sqrt{\sec^2\theta - 1} & 0\le\theta<\pi/2,\\ -\sqrt{\sec^2\theta - 1} & \pi/2<\theta<\pi\\ \end{cases}$$ so the correct integral has two parts: $$\int \frac{\sqrt{x^2-4}}{x}\,dx = \begin{cases} \sqrt{x^2-4} - 2 \operatorname{arcsec}(x/2) + C_1 & x \ge 2, \\ \sqrt{x^2-4} + 2 \operatorname{arcsec}(x/2) + C_2 & x \le -2. \\ \end{cases}$$
• Sorry, I had an error in my post which I just fixed. And generally you will have to deal with things piecewise when sign changes occur (e.g. across $x=0$ in $\sqrt{x^2} = |x|$). – arkeet Oct 21 '16 at 21:25


# Given $A^2+cA+cI=0$, how to find inverse of $A+(c-1)I$?

Suppose a square matrix $A$ such that $A^2+cA+cI=0$ for all $c \in \mathbb{Z}$. How can I show that $A+(c-1)I$ is invertible and find its inverse?

I started off this way: $A+(c-1)I = A+cI-I$

Then $(A+cI-I)(d_1A+d_2I)=I$, where $d_1, d_2 \in \mathbb{Z}$.

Expand it and it becomes: $d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$

$\Rightarrow (c-1)d_2=1 \;\; and \; \; d_1A^2+d_1cA-d_1A+d_2A=0$

$\Rightarrow d_2=\frac{1}{(c-1)}$

Continue from $d_1A^2+d_1cA-d_1A+d_2A=0$, after some manipulation, I got $d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$.

Since given that $A^2+cA+cI=0$,

$d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$

$\Rightarrow -d_1cI-d_1A+d_2A=0$

$\Rightarrow -d_1cI-d_1A+\frac{1}{c-1}A=0$

$\Rightarrow -d_1cI-d_1A=-\frac{1}{c-1}A$

$\Rightarrow d_1(cI+A)=\frac{1}{c-1}A$

At this point, since I don't know if $A$ is invertible yet, I cannot do it as $d_1=\frac{1}{c-1}\frac{A}{(cI+A)}$.

Even if I did this, I still cannot find a value for $d_1$ and $d_2$ to find the inverse of $A+(c-1)I$. How should I continue from here?

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Did you really mean for all $c \in \mathbb{Z}$? Wouldn't then $A^2 = A+I = 0$, hence $I^2 = 0$? – Niels Diepeveen Sep 25 '11 at 18:23

Instead of all this, rewrite your starting equation as $A^2+(c-1)A+A+(c-1)I=-I$. Then factor the left-hand-side...

But if you favor a more systematic approach you can also proceed as you do until $$d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$$ At this point you decide to set $(c-1)d_2=1$, but that is too early! First get rid of the $A^2$ using $A^2=-c(A+I)$, to get $$-d_1c(A+I)+d_1cA - d_1A + d_2A + (c-1)d_2I = I$$ which simplifies to $$(d_1c-d_1+d_2-d_1c)A + ((c-1)d_2-d_1c-1)I = 0$$ $$(d_2-d_1)A + ((c-1)d_2-cd_1-1)I$$ Now set both coefficients to 0. This gives immediately $d_1=d_2$, and we then need to solve $$(c-1)d-cd-1=0$$ in which the $cd$'s cancel and give us $d=-1$. So the sought inverse is $-A-I$.

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hmm...What happens if I couldn't see this and didn't start from this equation? I remember I had another problem to find inverses and I used a similar method and worked. I had the problem posted here too at math.stackexchange.com/questions/58841/… – xenon Sep 25 '11 at 15:23
I've added a more systematic approach. You have to start from the given relation, though -- otherwise you have no way to know there is even an inverse anywhere. – Henning Makholm Sep 25 '11 at 15:36
Thanks! But how's $A^2=-(A+I)$ and not $A^2=-(cA+cI)$? – xenon Sep 25 '11 at 15:57
At one part, you set both the coefficients to be $0$. Is there a reason for doing this? Because I was thinking why wouldn't it be $((c-1)d_2-cd_1-1)=1$ and $(d_2-d_1)A=-I$? Of course, what you have done by having the coefficients zero makes the whole thing simpler but I was just thinking what if I set it something else other than zero. – xenon Sep 25 '11 at 16:21
It follows from some experience with quotients of polynomial rings. Your original relation can be used to prove that any linear combination of powers of $A$ can be rewritten into the form $pA+qI$, and if that is all we know about $A$, then there is only one such form. So if we have $pA+qI=0$ then either it must be because $p=q=0$ or else $A$ happened to satisfy a nicer property than the one we can depend on. (In your example the nicer property is "is a multiple of $I$"). It's always worth it to try a general method first and try more complex tactices if that fails. – Henning Makholm Sep 25 '11 at 16:31

Perhaps a more transparent way: "change of variables". Let $B = A + (c-1) I$, or $A = B - (c-1) I$. Then $0 = A^2 + c A + c I = (B - (c-1) I)^2 + c (B - (c-1) I) + c I= B^2 + (2-c) B + I$. Multiply by $B^{-1}$ to get $B + (2-c) I + B^{-1} = 0$, i.e. $B^{-1} = -B + (c-2) I$ or $(A + (c-1) I)^{-1} = -A - (c-1) I + (c-2) I = -A - I$.

You may find it a bit dodgy to multiply by $B^{-1}$ before you know that $B^{-1}$ exists, but once you have the result $-A-I$ it's easy to verify that this works by multiplying it by $A + (c-1) I$.

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$$\begin{eqnarray*} x^2+cx+c = \left[x+(c-1)\right](x+1)+1\\ \therefore 0 = A^2+cA+cI = \left[A+(c-1)I\right](A+I)+I\\ \therefore \left[A+(c-1)I\right](A+I)=-I\\ \therefore \left[A+(c-1)I\right]^{-1} = -(A+I). \end{eqnarray*}$$

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NEW ANSWER. In the "long" division $$X^2+cX+c=(X+c-1)(X+1)+1,$$ replace $X$ by $A$: $$0=A^2+cA+cI=\Big(A+(c-1)I\Big)\Big(A+I\Big)+I,$$ $$\Big(A+(c-1)I\Big)^{-1}=-A-I.$$

[EDIT. I'm realizing that this answer is the same as user1551's, who posted it before. Sorry...]

OLD ANSWER. Let $A$ be an $n$ by $n$ matrix with coefficients in a field $K$, let $f\in K[X]$ be a polynomial annihilating $A$, and let $g\in K[X]$ be any polynomial.

If $g$ is prime to $f$, then $g(A)$ is invertible, and the inverse of $g(A)$ is given by $h(A)$ where $h\in K[X]$ is an inverse of $g$ mod $f$.

Moreover, there is a closed formula for such an $h$ (see this answer).

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