Upper triangular is when all entries below the main diagonal are zero: An upper triangular matrix. When we multiply a matrix by a scalar value, then the process is known as scalar multiplication. If you add the m × n zero matrix to another m × n matrix A, you get A: In symbols, if 0 is a zero matrix and A is a matrix of the same size, then A + 0 = A and 0 + A = A A zero matrix is said to be an identity element for matrix addition. C. Unit matrix. Zero Matrix (Null Matrix) Zeros just everywhere: Zero matrix. Null Matrix A “null matrix” is one which has the value zero for all of its elements. It is a matrix with 0 in all its entries. A special kind of diagonal matrix in which all diagonal elements are the same is known as a scalar matrix. If we want to make A = a null matrix we need to multiply it by 0. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. The null space may also be treated as a subspace of the vector space of all n x 1 column matrices with matrix addition and scalar multiplication of a matrix as the two operations. Yes. Null matrix. C. is orthogonal. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. An identity matrix is a scalar matrix with diagonal elements equal to one. Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix. A square null matrix is also a diagonal . C. diagonal. For example, are null matrices of order 2x3 and 2x2. Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix. Lower triangular is when all entries above the main diagonal are zero: A lower triangular matrix. Example 4.1 (Special Matrices) Some examples follow: 1. Note that the denominator of the fraction (just before the pivot's column vector) is the pivot itself (in this case “3”). Null Matrix Watch more videos at https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Er. By matrix-vector dot-product definition (a and u are vectors) $\begin\left\{bmatrix\right\} \begin\left\{array\right\}\left\{c\right\} a_1 \\ \hline \vdots \\ \hline a_n \\ \end\left\{array\right\} \end\left\{bmatrix\right\} * u = \left[a_1 * u, \dots, a_m * u\right]$ . Example 1. A scalar matrix has all main diagonal entries the same, with zero everywhere else: A scalar matrix . It is denoted by O. A unit matrix plays the role of the number 1 in numbers. 10. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array There are several types of matrices, but the most commonly used are: Rows Matrix Columns Matrix Rectangular Matrix Square Matrix Diagonal Matrix Scalar Matrix Identity Matrix Triangular Matrix Null or Unit matrix and scalar matrix are special case of a diagonal matrix. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. The null matrix is the one in which all elements are zero. Z is a scalar matrix with lamda = 0. B. Scalar matrix. Null matrix or Zero-matrix : A matrix is said to be a null matrix or zero-matrix if each of its elements is zero. Is the scalar matrix is always a identity matrix? A square matrix m[][] will be diagonal matrix if and only if the elements of the except the main diagonal are zero. EASY. Two examples of a scalar matrix appear below. Let [([a.sub.ij]).sub. In other words, the square matrix A = [a ij] n×n is an identity matrix, if a ij = 1, when i = j and a ij = 0, when i ≠j. are all zero matrices. A diagonal matrix is a square matrix in which all the elements other than the principal diagonal elements are zero. Diagonal matrix. Then n x n matrix V = VX . D. has rank n. Solution: QUESTION: 9. A diagonal matrix, in which all diagonal elements are equal to same scalar, is called a scalar matrix. 11. Let us put into practice the knowledge gained about the properties of matrix scalar multiplication and solve the next example exercises. Answer. Related Video. Any basis for the row space together with any basis for the null space gives a basis for . (i,j) [member of] AxB] be a scalar matrix with positive entries, and denote its columns by [[alpha].sub.j] = [([a.sub.ij]).sub.i[member of]A] and its rows by [[beta].sub.i] = [([a.sub.ij]).sub.j[member of]B]. It is a diagonal matrix with equal-valued elements along the diagonal. Properties of matrix addition & scalar multiplication. 6.5k VIEWS. Example : Identity Matrix. Question 3: Explain a scalar matrix? However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. null. Null space of zero matrix. The types of matrices you have checked here are scalar matrix, unit and identity matrix, null or zero matrix, triangular matrix, with both options lower and upper triangular matrices. Program to check diagonal matrix and scalar matrix. Then diagonal matrix, rectangular matrix, row and column matrices. (i) A zero-matrix need not be a square matrix. Holder's inequality: some recent and unexpected applications. Program for scalar multiplication of a matrix. Triangular Matrix. A. Since the zero matrix is a small and concrete concept in itself which can be used through many of our lessons in linear algebra, we are now forced once more to enter into the topic of a later lesson: the null space of a matrix. Solution: QUESTION: 8. Rotation, coordinate scaling, and reflection. To show that the null space is indeed a vector space it is sufficient to show that , ∈ ⇒ + ∈ and ∈ ⇒ ∈ These are true due to the distributive law of matrices. However, in our case here, A 2 is not zero, and so we continue with Step 3. The dimension of the null space of a matrix is the nullity of the matrix. Step 3. In the special case when M is an m × m real square matrix, the matrices U and V * can be chosen to be real m × m matrices too. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I. Here is the call graph for this function: Here is the caller graph for this function: m_matrix::m_matrix (const m_matrix & : that) D. Skew symmetric matrix. Matrix multiplication also known as matrix product . Google Classroom Facebook Twitter. When we add or subtract the 0 matrix of order m*n from any other matrix, it returns the same Matrix. We know that a matrix can be defined as an array of numbers. If a matrix A is symmetric as well as skew-symmetric, then A is a (A) Diagonal matrix (B) Null matrix asked Dec 6, 2019 in Trigonometry by Rozy ( 41.8k points) matrices We use the notation I p to denote a p×p identity matrix. 0, a matrix composed entirely of zeros, is called a null matrix. Just a note on separate Qs & As here. A submatrix of the given matrix can be obtained by deleting . Even a single number is stored as a matrix. 9. Let M be an arbitrary square matrix and Z be a zero matrix of the same dimension. A. has rank zero. It is a binary operation that produces a single matrix by taking two or more different matrices. A = 3: 0: 0: 3: B = 5: 0: 0: 0: 5: 0: 0: 0: 5: The identity matrix is also an example of a scalar matrix. Learn what a zero matrix is and how it relates to matrix addition, subtraction, and scalar multiplication. Symmetric. A matrix is a two-dimensional, rectangular array of data elements arranged in rows and columns. The most basic MATLAB® data structure is the matrix. We have to find whether the given square matrix is diagonal and scalar matrix or not, if it is diagonal and scalar matrix then print yes in the result. Examples: etc. Java Scalar Matrix Multiplication Program example 2. Answer: The scalar matrix is similar to a square matrix. 6.5k SHARES. If M is a square matrix, is a scalar, and x is a vector satisfying then x is an eigenvector of M with corresponding eigenvalue . EDIT The question is this: Scalar multiplication is defined as B = A * s, where B and A are equally sized matrices (2D array of numbers, in this example let's use integers) and s is a scalar value. 263.1k SHARES. The elements can be numbers, logical values (true or false), dates and times, strings, or some other MATLAB data type. A square matrix A is symmetric if a ij = a ji for all i, j. The null matrix is also called the zero matrix. Then Z*M = Z = 0*M = 0 => Z = 0. If A 2 happens to be a null matrix, then the process terminates and the rank of A 1 is 1, which is then the largest subscript of a nonzero matrix. Diagonal matrix: A square matrix is said to be diagonal matrix if the elements of matrix except main diagonal are zero. The matrices » » » ¼ º « « « ¬ ª 0 0 0 0 0 Z and » » » ¼ º « « « ¬ ª 0 0 0 0 N are both null matrices. 263.1k VIEWS. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. Select any nonzero element of A 2. Scalar Matrix. is said to be a scalar matrix if b ij = 0, when i ≠j b ij = k, when i =j, for some constant k. (vi) A square matrix in which elements in the diagonal are all 1 and rest are all zeroes is called an identity matrix. If M has n columns then rank(M)+nullity(M)=n. The product of any matrix by the scalar 0 is the null matrix. If equals (A) (B) zero matrix (C) a scalar quantity (D) identity matrix 1:36 35.6k LIKES. Scalar matrix and identity matrix 305.4k LIKES. Examples: Input : mat[][] = {{2, 3} {5, 4}} k = 5 Output : 10 . D. unit. Its effect on a vector is scalar multiplication by λ. It is a matrix with 0 in all its entries. Intro to zero matrices. B. has rank 1. Each element of A is multiplied to s, which is then stored in the corresponding element in matrix B. Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title "Prince of Wales" originally claimed for the English crown prince via a trick? X = [X 1, X 2, ... , X n] is an n — tuple non — zero vector. Ridhi Arora, Tutorials Point India Private Limited. View All. However a scalar matrix need not be a unit matrix. Email. A rectangular matrix 1234 5678 2. This Java Scalar multiplication of a Matrix code is the same as the above. In view of (7) like of the case of scalar matrix games the following theorem is proving. Example : Properties of Zero Matrix. Identity matrix is a scalar matrix in which all diagonal elements are 1. A 1 ×1 matrix is a scalar. A matrix is known as a zero or null matrix if all of its elements are zero. A scalar matrix is a special kind of diagonal matrix. 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# Probability a sample mean will fall in a range

1. Nov 21, 2015

### toothpaste666

1. The problem statement, all variables and given/known data
A random sample of size n = 81 is taken from an infinite population with the mean μ = 128 and the standard deviation σ = 6.3. With what probability can we assert that the value we obtain for the sample mean X will fall between 126.6 and 129.4?

3. The attempt at a solution
z = (x-μ)/(σ/sqrt(n))
so we have
z = (126.6-128)/(6.3/9) = -2 and z = (129.4-128)/(6.3/9) = 2
so the probability it will fall in the range is
F(2) - F(-2) = .9772 - .0228 = .9544

is this correct?

2. Nov 21, 2015

### Orodruin

Staff Emeritus
This depends on the actual distribution in the population. You can only do what you did if this distribution is assumed to be Gaussian.

3. Nov 21, 2015

### toothpaste666

Gaussian means "normal" right? I am confused a bit about that. In my book they seem to use "z" for the test statistic and use "t" when the population is known to be normal. From what I can tell they are the same thing except that with z you use the standard normal table and with t you use a different table with a certain amount of degrees of freedom. I don't think I fully get it.

4. Nov 21, 2015

### Ray Vickson

I do not actually believe you; I think you are mis-reading your book (although, to be honest, I am making this judgement sight-unseen). Typically, for an independent random sample from an underlying normal (=Gaussian) distribution with mean $\mu$ and variance $\sigma^2$: (1) we use $z$ and normal tables when we KNOW the value of $\sigma$; but (2) use $t$ and t-tables when we do not know $\sigma$, but have estimated it from the sample data itself.

In case (2), we estimate
$$\text{estimator of }\: \sigma^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2$$
where the sample values are $x_1, x_2, \ldots, x_n$ and $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the sample mean. In that case the jargon is that there are $n-1$ "degrees of freedom".

In the limit as $n \to \infty$ the t-distribution with (n-1) degrees of freedom becomes the standard normal, so using $z$ is like having infinitely many degrees of freedom.

5. Nov 21, 2015

### toothpaste666

so for either of the two statistics to work, the distribution must be normal?

6. Nov 21, 2015

### Ray Vickson

Theoretically, yes, but for a large sample-size, using the "normal" results give a "reasonably accurate" approximation. This is based on the so-called Central Limit Theorem; see, eg.,
https://en.wikipedia.org/wiki/Central_limit_theorem
or http://davidmlane.com/hyperstat/A14043.html
or http://www.statisticalengineering.com/central_limit_theorem.htm .

For a "reasonable" non-normal underlying distribution, a sample size of n = 81 is likely large enough that normal-based estimates will be informative, if not absolutely accurate.

7. Nov 21, 2015

### toothpaste666

ahh ok what my book actually says is use z for samples of n>30 with σ known and if σ is not known replace σ with s and if the sample is n<30 And the population is normal use t. so since my sample is large enough, my solution to this problem should be close enough?

8. Nov 21, 2015


# Double Sum Involving Condition

I would like to compute the dimensions of some small free nilpotent Lie algebras. However, I am totally new to this and I could not figure out how to write the double sum which gives the dimension of the free nilpotent Lie algebra $L(c,n)$ of step $c$ with $n$ generators.

Long story short, I would like to learn how I can write the sum

$$\dim L(c,n) = \sum_{m=1}^{c}\frac{1}{m}\sum_{d \mid m}\mu(d)n^{m/d}$$

where $\mu$ is the Möbius function, in Mathematica.

• Maybe diml[c_, n_] := Block[{j}, Sum[DirichletConvolve[MoebiusMu[j], n^j, j, m]/m, {m, c}]]? If this does what you want, I'll write an answer. – J. M. will be back soon Feb 18 '16 at 6:32
• @J.M. Hello. I tried this with some known values and it produced the correct values. I am sorry, since I do not know much about Mathematica this is the only way of me knowing that it works. Thank you! So please go ahead and write an answer, – Can Hatipoglu Feb 18 '16 at 7:11
• No need to apologize! You are fortunate here that Mathematica has the necessary stuff for the computations you want to do. Give me a few, and I'll write something up. – J. M. will be back soon Feb 18 '16 at 7:14
• FWIW, your inner sum equals JordanTotient[k, n] but unfortunately that function is not in Mathematica. – Chip Hurst Jun 2 '16 at 18:10

## A story of incremental improvement

Let's look at the OP's original expression again, for reference:

$$\sum_{m=1}^{c}\frac{1}{m}\sum_{d \mid m}\mu(d)n^{m/d}$$

Most people here are familiar with Sum[], and would not have much trouble translating the outer summation into Mathematica syntax. The inner part,

$$\sum_{d \mid m}\mu(d)n^{m/d}$$

is not terribly familiar to those who do not do much number theory. What this basically is saying is that the terms of the sum are indexed over the divisors of $m$. Appropriately enough, Mathematica does have a Divisors[] function. The inner sum can thus be written as

Sum[MoebiusMu[d] n^(m/d), {d, Divisors[m]}]

Summing over the divisors of a number, however, is so common a number-theoretic operation that Mathematica has seen it fit to provide a DivisorSum[] function. Thus, the inner sum can also be written as

DivisorSum[m, MoebiusMu[#] n^(m/#) &]

The story does not end here. The operation

$$\sum_{d \mid m}f(d)g(m/d)$$

frequently turns up in number-theoretic and other contexts, that it has been given a special name: Dirichlet convolution. Mathematica, fortunately, also provides a function for evaluating convolutions, called DirichletConvolve[]. Thus, the inner sum is most compactly expressed as

DirichletConvolve[MoebiusMu[j], n^j, j, m]

The function in the OP can now be implemented like so:

diml[c_Integer?Positive, n_Integer?Positive] := Block[{j},
Sum[DirichletConvolve[MoebiusMu[j], n^j, j, m]/m, {m, c}]]

or, using formal symbols as suggested by The Doctor,

diml[c_Integer?Positive, n_Integer?Positive] :=
Sum[DirichletConvolve[MoebiusMu[\[FormalJ]], n^\[FormalJ], \[FormalJ], m]/m, {m, c}]

(They look messy here on SE, but should look nice when pasted into Mathematica.)

Here is the function in action:

Table[diml[c, n], {c, 5}, {n, 5}]
{{1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 5, 14, 30, 55},
{1, 8, 32, 90, 205}, {1, 14, 80, 294, 829}}

• Thank you for introducing me to DirichletConvolve and for this instructive answer . This is my vote :) – ubpdqn Feb 18 '16 at 8:55

Please see J.M.'s answer (in comments) using DirichletConvolve. I would vote for this as an answer.

l[c_, n_] :=Module[{r = Range[c]},
Total[(1/r) Total /@MapThread[MoebiusMu@#1 n^(#2/#1) & ,{Divisors /@ r, r}]]]

I will happily delete and vote for JM answer.

I like J.M.'s answer and voted for it. However, it's hard to format comments, so I've posted this as an answer instead. Rather than using Block[{j},...], you can use formal symbols instead, used to represent a formal parameter that will never be assigned a value. Moreover, if you enter

DivisorSum[m, Function[d, MoebiusMu[d] n^(m/d)]] // TraditionalForm

you will see that Mathematica uses formal symbols automatically. Also, in many cases, I prefer to use the "map notation" for Function instead of # and &. For example,

is executable code for this function.

Finally, in TraditionalForm notation, the sum can be expressed as

which corresponds nicely to the original sum posted in the question. However, using DirichletConvolve should be more efficient.


# Give an example of a linear transformation whose kernel is the line spanned by vector: $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$

I know that a linear transformation could be a projection onto the plane with normal vector $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$, but finding the projection would be too difficult.

I could easily think up a matrix where multiplied by $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T = 0$, but I'm not sure on how to choose a matrix where $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$ is the only element of the kernel.

Also, can you please explain this Hint: "to describe a subset as a kernel means to describe it as an intersection of planes?"

• it's not too hard to cook up a matrix with rank 2 that kills this vector – qbert Jan 31 '18 at 4:53
• @qbert Yeah but how can I ensure nothing else send the matrix to 0? – Goldname Jan 31 '18 at 4:59
• "cook up" a $3\times 3$ matrix, making sure that (-1,1,2) is in the kernel and then show that the rank of this matrix is 2. And you know that rank + nulity = 3, so $(-1,1,2)$ must span the kernel. – Doug M Jan 31 '18 at 5:00
• if it has rank 2, it's kernel is rank 1. If it kills your given vector, then that spans the kernel – qbert Jan 31 '18 at 5:00
• ah I see, I forgot about that theorem, thanks – Goldname Jan 31 '18 at 5:18

Guide:

• $$2(-1) + 0(1) +1(2)=0$$
• $$0(01) + (-2)(1)+1(2)=0$$

• Verify that $\begin{bmatrix} 2 & 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 0 & -2 & 1 \end{bmatrix}$ are linearly independent.

• Now think of someway to form your matrix and prove that span of $\{\begin{bmatrix} -1 & 1 & 2 \end{bmatrix}^T\}$ is a basis to the kernel.

As for the explanation of the hint: To describe $x$ which satisfy $Ax=0$ where $a_i^T$ are the $i$-th row means $x$ is in the intersection of $\{ a_i^Tx = 0 : i \in \{ 1, \ldots, m\}\}$.

• I don't understand what your equations are supposed to mean – Goldname Jan 31 '18 at 4:59
• I can easily think of a matrix, but I don't understand what is required so that [-1 1 2] is the only element in the kernel – Goldname Jan 31 '18 at 4:59
• $[-1, 1, 2]$ is not the only element, any multiple of $[-1, 1, 2]$ is in the kernel. – Siong Thye Goh Jan 31 '18 at 5:01
• yeah that's what I meant – Goldname Jan 31 '18 at 5:03
• rank-nullity theorem is useful. geometrically, two non-parallel 3D plane intersect at a line. – Siong Thye Goh Jan 31 '18 at 5:03

Expanding upon the comment: Choosing the following matrix works $$\begin{bmatrix} 1&-1&1\\ 2&0&1\\ 1&1&0 \end{bmatrix}$$ It is clear that $(-1,1,2)$ is in the kernel. The dimension of the image is $2$, so the dimension of the kernel is just $1$, by rank nullity. So the only vectors that map to $0$ under the matrix are your vector and multiples of it.

Pick any two vectors that together with $[-1,1,2]^T$ form an ordered basis of $\mathbb R^3$. Relative to this basis, the matrix $\operatorname{diag}(1,1,0)$ represents a transformation with the required properties. Apply a change of basis to it to get a matrix relative to the standard basis. For that matter, since a linear transformation is completely determined by its action on a basis, the description $T(v_1)=v_1$, $T(v_2)=v_2$, $T([-1,1,2]^T)=0$, where $v_1$ and $v_2$ are the two vectors you chose, is a complete description of the transformation, but that’s likely not the answer that whoever gave you this problem is looking for.

As far as the hint goes, the kernel is a line through the origin, which can be described as the intersection of a pair of planes. The normals to those planes span the orthogonal complement of the line, so you can use them to construct the above basis.

• How did you conclude that "the matrix $diag(1,1,0)$ represents a transformation with the required properties"? Thanks – gbox Jan 31 '18 at 9:53
• @gbox The columns of a transformation matrix are the images of the basis vectors. In fact, for any linear transformation one can find bases for the domain and codomain such that the matrix has the form $\tiny{\left[\begin{array}{c|c}I&0\\\hline0&0\end{array}\right]}$. – amd Jan 31 '18 at 19:39
• @gbox BTW, $\operatorname{diag}(1,1,0)$ gives you a projection. If you instead choose some other pair of linearly independent vectors for the first two columns, you’ll get a different type of rank-2 transformation. – amd Jan 31 '18 at 19:47


# Is there a name for the curve traced by the midpoint of two moving points on a circle with different speed?

I'm curious about what I did in Geogebra while I was experimenting with animation. Here it is:

1. Construct a circle with radius $$r$$
2. Define two points $$A(r\cos a\theta , r \sin a\theta)$$ and $$B(r\cos b\theta, r \sin b\theta)$$. Let $$(r,0)$$ be the initial position of both $$A$$ and $$B$$.
3. Get the midpoint of $$AB$$. Name this point $$M$$.
4. Change $$\theta$$ continuously starting from $$0$$ until $$A$$ and $$B$$ are both at $$(r,0)$$ again.

The midpoint, by doing the fourth step, seem to make some form of curve. I noticed that the curved traced by the midpoint does not change as long as $$a/b$$ does not change, even if $$a$$ and $$b$$ does so. Also, it doesn't matter if the values of $$a$$ and $$b$$ are swapped, but to avoid such ambiguities, we will avoid swapping of values and $$a$$ is always greater than $$b$$.

For the file, see this graph in Desmos.

For the angle, let $$p$$ be the numerator of $$a/b$$ in lowest terms. Then, we have $$0 \leq \theta \leq 2p\pi$$. (I don't know how to this in Desmos, by the way)

## Examples

For $$a = 2.1$$ and $$b = 1.4$$,

For $$a = 3$$ and $$b = 1$$,

And for $$a = 1.1$$ and $$b = 0.007$$ $$(0 \leq \theta \leq 98\pi)$$,

As for the midpoint, the coordinates is $$\left(\frac{r}{2}\left(\cos(a\theta) + \cos(b\theta)\right), \frac{r}{2}\left(\sin(a\theta) + \sin(b\theta)\right)\right)$$

This means that the parametric equation is \begin{align*} x(\theta) &= \frac{r}{2}\left(\cos(a\theta) + \cos(b\theta)\right) \\[10pt] y(\theta) &= \frac{r}{2}\left(\sin(a\theta) + \sin(b\theta)\right) \end{align*} Is there a name for these curves?

Update 1: It seems like this is somehow related to the mathematical basis of a spirograph.

... and therefore the trajectory equations take the form \begin{align*} x(t) &= R\left[(1-k)\cos t+lk\cos {\frac{1-k}{k}}t\right],\\y(t)&=R\left[(1-k)\sin t-lk\sin {\frac {1-k}{k}}t\right]\end{align*}

• It looks like what they call the evolute of a cardioid in this page (second figure) mathcurve.com/courbes2d.gb/largeur%20constante/… Jul 7 at 13:31
• Depending on a and b the curves can take many forms with multiple loops (b=1, a = 3) or spirals (a=2.3, b=2) so not cardioid, though one may be a special case.
– Paul
Jul 7 at 14:31
• This has the properties of several of the so-called transcendental curves, such as the sinusoidal spirals, Archimedean spiral, involute of a circle, and cochleoid. Curves are defined by equations, Cartesian and polar. You need to express this curve as an equation in order to define it, rather than a set of instructions. My go-to reference for plane curves is A Catalog of Special Plane Curves, by J. Dennis Lawrence, Dover Publications, 1972. Jul 9 at 14:59
• @CyeWaldman The parametric equation can be found before Update 1. Jul 9 at 16:24
• @soupless Thank you. From those equations, I would conclude that this is an unnamed curve. On a quick search the closest curve I can find to this is the hypotrochoid, described by $x=n\cos t+h\cos(n/b\cdot t)$ and $y=n\sin t-h\sin(n/b\cdot t)$. Jul 9 at 20:25

I am posting this since my updates that can answer my question should be posted as an answer, not as a part of the question.

We restrict the values of $$a$$ and $$b$$ to be integers where $$a$$ and $$b$$ are both nonzero. Then, the curve seems to represent an epitrochoid given by the parametric equation \begin{align*} x(t) &= m \cos t - h \cos\left(\frac{mt}{c}\right) \\ y(t) &= m \sin t - h \sin\left(\frac{mt}{c}\right). \end{align*}

where $$m = 1$$, $$h = 1$$, and $$c = \frac{a}{b}$$ for $$0 \leq t \leq 2a\pi$$.. For the parametric equation in the question, we let $$r = 2$$.

We can say that the two curves match if they are in the same magnitude and angle, and that they don't match if they don't have the same angle and magnitude. However, we'll say that they are a pseudomatch if they have the same magnitude but not having the same angle.

Now, for odd $$a$$, the two curve are a pseudomatch regardless of the value of $$b$$. For even $$a$$, there are two cases:

1. If $$a$$ is a power of $$2$$, then the curves are a pseudomatch if $$b$$ is a multiple of $$a$$. Otherwise, the curves are a match.
2. If $$a$$ is not a power of $$2$$, then the curves are a pseudomatch if $$b$$ is a multiple of the greatest power of $$2$$ that divides $$a$$. An example would be $$a = 28$$, where the curves are a pseudomatch if $$b$$ is a multiple of $$4$$.

As of the moment, I don't know how to prove that this works. I am just relying on graphs from the file.


# Using functions to find indeterminate forms

The question is to use two functions f(x) and g(x) to show that ∞ - ∞ (infinity - infinity) is indeterminate.
I don't really know how to get started on this. I think I need to find two limits for f(x) and g(x) that are both infinity. But how to I show that subtracting them is indeterminate?

Edit: In the example we were shown, lim x --> 0 was used. But that was for the indeterminate form 0∞. I don't know if that applies here or not.

"Indeterminate" means "not having a unique value" as $x$ tends to whatever the problem says (I am guessing $x \rightarrow 0$ or $x \rightarrow \infty$, but you really should specify and not make the community you are asking for help guess). I will assume $x \rightarrow \infty$.

So, to show indeterminacy, I will exhibit three examples of the function pair $f(x), g(x)$ with different limiting behaviors of the difference $[f(x)-g(x)]$ as $x \rightarrow \infty$.

Example 1: $f(x) = g(x) = x$. Here $\lim_{x \rightarrow \infty}[f(x) - g(x)] = 0$.

Example 2: $f(x) = x^2, \; g(x) = x$.

Example 3: See Mark Fischler's answer.:)

• Thank you for the reply. The part that I don't understand is that when you use different values for f and g. Wouldn't you expect a different answer? I don't really understand how that makes it indeterminate. – user372991 Sep 28 '16 at 21:58
• Well consider a form that is not indeterminate: $1 / \infty$. In detail: if you have $$\lim_{x \rightarrow \infty} f(x) = \mbox{(some real cosntant)}, \quad \lim_{x \rightarrow \infty} g(x) = \infty,$$ then, no matter which $f, g$ you use, as long as they satisfy the above conditions, you will get $$\lim_{x \rightarrow} {f(x) \over g(x)} = 0.$$ In a high-level summary: The goal is, knowing the asymptotic behavior of $f, g$, to be able to ascertain the asymptotic behavior of an expression involving these two. When we cannot do that, the latter expression is indeterminate. – user8960 Sep 28 '16 at 23:18

Consider $$f(x) = \csc^2x \\ g(x) = \cot^2 x \\ h(x) = 1/x^2$$ and look at the limit as $x\to 0$ of the expressions $f(x)-g(x)$ and $f(x)-h(x)$.

Both of these expressions are of the form $\infty - \infty$

But $$\lim_{x\to 0} (f(x) - g(x)) = 1 \\ \lim_{x\to 0} (f(x) - h(x)) = \frac13$$ This shows that $\infty - \infty$ could be $1$ or $\frac13$; for that matter it could be $0$ (consider $f(x)-f(x)$).

• If you wouldn't mind, could you explain how f(x) -g(x) = 1 and f(x) - h(x) = 1/3? I thought that they all would equal infinity? – user372991 Sep 28 '16 at 22:18
• @Jordan He is just giving you the answers for those limits. You can verify them with L'Hospital's Rule. It's a good exercise – imranfat Sep 28 '16 at 22:52
• On the other hand, the first one is particularly easy: Since $csc^2x - \cot^2x = 1$ for all values of angles where they are both finite, "obviously" the limit as $x\to$ any real value (or for that matter even any complex value) must be $1$. – Mark Fischler Oct 1 '16 at 21:20


# how many subsets of $\{1,2,3,4,5,6,7\}$ has $6$ as largest?

Let $A =\{1,2,3,4,5,6,7\}$, The how many subsets have 6 as largest element?

My approach -

Total subsets are $2^7 = 128$. Then First I have to exclude the $1$ empty subset i.e. $\{\}$, $6$ sets with $1$ element except $\{6\}$, then two elements not containing $6$ as its element(I am not getting a way to count these element) and the $2$ subsets $\{6,7\},\{7,6\}$. Then I am stuck here as I am not getting the way to count these numbers. Any help?

• Well, none of the sets have 7 so this is the same as asking what are the subsets of B = {1.... 6} they all have 6 so all sets are equal to $C \cup \{6\}$ for some set C. C is a subset of $\{1,...., 5\}$ so we just need to find the subsets of {1,2...5} and add {6} to all of them. There are $2^5$ subset of {1,2....5} so there are the same number of subsets of {12,...5,6} that must have 6 which is the same number of subsets of {1,2,....7} that must have 6 and must not have 7. – fleablood Aug 25 '16 at 19:58
• 128 subsets. Half have 6 as an element. 1/4, contain both 6 and 7. So 1/4 have 6 and do not have 7. – Doug M Aug 25 '16 at 20:23

Remember how we determined that the number of subsets was 2 to the power of the number of elements.

We did that because for each element, $a$, either $a$ could be in a subset. So the total number of sets was the product of all the choices each of which was 2.

This is the same. Either 1 is in a subset or not. That 2 choices. Either 2 is in the subset or not. That's $2*2$ choice. Keep it up EXCEPT notice $6$ must be in the subset so that is only $1$ choice and $7$ must not be in the set so that's only one set.

So the number of sets is $2*2....*2*1*1$ which is what.

....

Or

....

Don't eliminate the sets one at a time. Remove ALL the subsets that do not have $6$ in them. How many do not have $6$. Remove ALL the subsets that do have $7$. How many is that? Then to avoid double counting add back the ones that had $6$ and didn't have $7$. How many is that?

....

Or

....

Figure 1/2 of the 128 have 6 and half do not. That only leaves half of them acceptable. Then half of those have 7 and half to not. That leaves only half of those acceptable.

....

Or

....

No set has $7$. So all the elements are taken from {$1,2,3,4,5,6$}.

All sets have $6$. So all elements that aren't $6$ are taken from {$1,2,3,4,5$} and $6$ is always added to it.

There are $2^5$ subsets of {$1,2,3,4,5$} and we are only taking those and sticking a $6$ into them. There are $2^5$ such sets.

• I like your way of dividing it up. That makes good sense! – John Aug 25 '16 at 20:16
• You have to be careful in assuming that your condition is exactly 1/2 though. In this case you can but if you were asked say How many sets have the numbers add up to an even number and contain the number 5 you can't assume half the sets add up to an even number and of those half of them contain 5. – fleablood Aug 25 '16 at 20:23

Hint:

Include $6$. How many ways are there to include zero through five members of $\{1,2,3,4,5\}$?

(Order doesn't matter with sets.)

• okay now i got. There are total subsets = 2^6 = 64. – Brij Raj Kishore Aug 25 '16 at 19:58
• The element $6$ has to be there. The element $7$ can't be in there. But any, all, or none of $1$ through $5$ can be. In other words, if $S_5 = \{1,2,3,4,5\}$, then the union of $\{6\}$ and any subset of $S_5$ will have $6$ as the largest element. You figured out the number of subsets for a set of $7$ elements. How about for $5$ elements? – John Aug 25 '16 at 20:02
• Regarding your answer, not quite. It's not optional whether $6$ is in there or not. – John Aug 25 '16 at 20:03
• So the answer is 2^6 -1 - 5. Am i correct? But in option there are 4 options as - 32,31,64,128. – Brij Raj Kishore Aug 25 '16 at 20:05
• No set has 7. All sets have 6. So all sets are a subset of {1,2,3,4,5} unioned up with {6}. How many sets are there. There is {6} = {} + {6}; there is {1,4,5,6} = {1,4,5} + {6}. They are all C + {6} where $C \subset$ {1,2,3,4,5}. How many such sets are there. – fleablood Aug 25 '16 at 20:14

Think about it we need $6$ as an element but we can't have $7$ because $7$ is larger than $6$, we can have anything else.

What we're looking for: How many subsets have $6$ as an element but not $7$?

We must choose to include $6$, that gives $1$ choice. We may choose to to include $1$ or not, that gives $2$ choices. We may choose to include $2$ or not, that gives $2$ choices. We may choose to include $3$ or not, that gives $2$ choices. We may choose to include $4$ or not, that gives $2$ choices. We may choose to include $5$ or not, that gives $2$ choices. We must choose not to include $7$, that gives $1$ choice. By the multiplication principle the number subsets with $6$ but not $7$ is:

$$1•2•2•2•2•2•1=2^5$$


# How to determine number with same amount of odd and even divisors

With given number N, how to determine first number after N with same amount of odd and even divisors? For example if we have N=1, then next number we are searching for is :

2
because divisors:
odd : 1
even : 2

I figured out that this special number can't be odd and obviously it can't be prime. I can't find any formula for this or do i have just compute it one by one and check if it's this special number ? Obviously 1 and number itself is divisors of this number. Cheers

-
Is this homework? How many powers of 2 can divide your special numbers? – user641 Jul 1 '11 at 11:27
We have holiday, i don't go to school :-) Is your question is a hint? – Spinach Jul 1 '11 at 11:52
Here's a hint. Try and pair up the even and odd factors in a natural way. You gave 2 = 1*2 as an example. How about factors of 2p, where p is an odd prime? – hardmath Jul 1 '11 at 12:05
Whenever you have two finite sets of the same size, you can think about whether there is a one-to-one correspondence (a bijection) between them. You might think whether you could apply this observation to the divisors of a number ... – Mark Bennet Jul 1 '11 at 12:07

To get some idea of what's going on, we do like other scientists do, we experiment.

Special numbers will be even, so we write down the number of odd divisors, even divisors, for the even numbers, starting with $2$. If a number turns out to be special, we put a $\ast$ in its row.

So we make a table, giving the number, how many odd divisors it has, how many even. Calculations are easy, but we must be very careful, since errors could lead us down the wrong path.

$2 \qquad 1 \qquad 1\quad\ast$

$4 \qquad 1 \qquad 2$

$6 \qquad 2 \qquad 2\quad\ast$

$8 \qquad 1 \qquad 3$

$10 \qquad 2 \qquad 2\quad\ast$

$12 \qquad 2 \qquad 4$

$14 \qquad 2 \qquad 2\quad\ast$

$16 \qquad 1 \qquad 4$

$18 \qquad 3 \qquad 3\quad\ast$

We could easily go on for a while. It is definitely not a waste of time, since it is useful to be well-acquainted with the structure of the smallish numbers that we bump into often.

A pattern seems to jump out: every second even number seems to be special. It looks as if "special" numbers are not all that special! It can be dangerous to jump to conclusions from data about small integers. But in this case, the conclusion turns out to be correct.

The special numbers, so far, all have the shape $2a$, where $a$ is an odd number. They are divisible by $2$ but not by $4$. The even numbers in our list that are not special are all divisible by $4$.

Now we try to prove that every number that is divisible by $2$ but not by $4$ is special, and that the others are not.

Take an odd number $b$, and look at the number $2b$. Think about the divisors of $2b$. If $k$ is an odd divisor of $2b$, then $2k$ is an even divisor of $2b$, and vice-versa.

If $k$ is an odd divisor of $2b$, call $2k$ the friend of $k$. Split the divisors of $2b$ into pairs of friends. For example, if $b=45$, we have the following pairs of friends.

$$(1,2)\qquad (3,6) \qquad(5,10)\qquad(9,18)\qquad(15,30) \qquad (45,90)$$

We have split the divisors of $2b$ into pairs of friends. Each pair has one odd number and one even number, so $2b$ has exactly as many odd divisors as even divisors.

Now let's show that no number divisible by $4$ can be special. The idea is that if a number is divisible by $4$, then it has "too many" even divisors. I will not write out the details, but you should. The idea goes as follows. Take a number $n$ that is divisible by $4$, like $36$ or $80$. Split the divisors of $n$ into teams. If $k$ is an odd divisor of $n$, put into the same team as $k$ the numbers $2k$, $4k$, and so on however far you can go.

Here are the teams for $n=36$. $$(1,2,4) \qquad (3,6,12)\qquad (9,18,36)$$

Each team has more even numbers than odd numbers, so $n$ has more even divisors than odd divisors. That means $n$ can't be special.

Now let's get to your question: what is the first special number after $N$?

If $N$ is divisible by $4$, the first special number after $N$ is $N+2$. If $N$ is divisible by $2$ but not by $4$, the first special number after $N$ is $N+4$. If $N$ has remainder $1$ on division by $4$, the first special after $N$ is $N+1$, and if the remainder is $3$, the first special is $N+3$. These facts follow easily from what we have discovered about special numbers.

Formulas: We have been operating without formulas, just straight thinking. But I should mention a relevant formula. Let $n$ be an integer greater than $1$, and express $n$ as a product of powers of distinct primes. In symbols, let $$n=p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}$$ Then the number of divisors of $n$ is given by $$(a_1+1)(a_2+1) \cdots(a_k+1)$$

For example, $720=2^43^25^1$. The number of (positive) divisors of $n$ is $(4+1)(2+1)(1+1)$.

The formula that gives the number of divisors of $n$ is not hard to prove. Try to produce a proof! The formula could be adapted to give a count of the odd divisors of $n$, and of the even divisors. Then we could use these formulas to identify the special numbers. But formulas cannot do the thinking for you. So as a first approach, the way we tackled things is much better than trying to use a formula.

-
Thanks a lot ! Seems like experimenting is very useful, thanks for teaching me something :) And thanks for an so full answer. Chris – Spinach Jul 1 '11 at 14:20
+1; this answer embodies a very important lesson. A lot of questions here on math.SE could have been answered by the posters if they had decided to experiment first... – Qiaochu Yuan Jul 2 '11 at 4:11
@Qiaochu Yuan: A very common problem with students is that they do not really understand that mathematical questions, even those that seem quite abstract, are about concrete things. Unlike working mathematicians, students, even pretty good ones, are often naive formalists. – André Nicolas Jul 2 '11 at 4:28

I wrote a C++ application. My results are all the following numbers:

0, 2, 6, 10, 14, 18, 22, 26, 30, 34, ....

So, first zero, then two and then each time plus 4.

This means in general that if $n \equiv 2 (mod 4)$ is true, your number is one of the kind you're searching. Except from the first zero.

Here is the code of the application:

for (int i = 0; i < 1000; i += 2)
// Always +2 because of odd numbers don't have any even divisors
{
int even = 0;
int odd = 0;
for (int d = 1; d <= i; ++d)
{
if (i % d == 0)
{
if (d % 2 == 0)
{
even++;
} else
{
odd++;
}
}
}
if (even == odd)
{
printf("%d\n", i);
}
}

-
We know that there's always 1 divisor odd ( 1 ) and 1 divisor even for sure ( N ), so we need only to check to n/2 or even to sqrt(n) ( not sure if we won't miss some of divisors expect n/2 ). But nice application. – Spinach Jul 2 '11 at 14:01

For a given integer $n$, every divisor larger than $\sqrt{n}$ is paired with a divisor smaller than $\sqrt{n}$. Use this to figure out a general principle.

-
I don't think that's very much help in this case. There is a much simpler way, hinted at in the comments to the OP. – TonyK Jul 1 '11 at 13:46
@TonyK - I think this works as a pairing for this problem, though it involves making sure that you know that $n$ is not a square, so that there is a pairing - but since the number of distinct factors is known to be even, n can't be a square. – Mark Bennet Jul 1 '11 at 16:25
The pairing works of ncmathsadist works just fine, both for the proof that we have equality when $n \equiv 2 \pmod{4}$, and that there are more even divisors than odd when $n \equiv 0 \pmod{4}$. The pairing is in effect the same as mine when $n \equiv 2 \pmod 4$. But the mathematical description of the pairing is substantially better than mine, because the ncmathsadist pairing has a number of other uses. More detail would undoubtedly have been useful to the student. – André Nicolas Jul 2 '11 at 4:43
I didn't want to give away the whole thing. – ncmathsadist Jul 2 '11 at 13:25


# Is it possible for an eigenspace to have more than one vector in its basis? What would that imply? Would every vector in the basis be an eigenvector?

So far I've been seeing that the vector that makes up a basis for the null space of the matrix $$A-Ix$$ (where x is an eigenvalue) is the eigenvector corresponding to the eigenvalue $$x$$. But I've never run into a situation where the basis had more than one vector in it.

Let's say the basis of the null space of $$A-2*I$$ was made up of two vectors. Would that mean those two (and only those two) vectors are the eigenvectors corresponding to the eigenvalue $$2$$ for the matrix $$A$$?

One thing of note is that, from what I understand, the identity transformation (in any space) has only one eigenvalue, $$1$$, but it has infinitely many eigenvectors (not sure if there is a vector space where this wouldn't be true, but it's true for $$R^n$$ at least). Not sure how this fact fits in here.

Just wanting to be sure I'm not misunderstanding something. I'm grateful for any help.

• Every nonzero vector in an eigenspace is an eigenvector. – amd Mar 9 at 20:10

Yes of course, you can have several vectors in the basis of an eigenspace.

First, when you have only one vector $$v$$ in a basis for a matrix $$A$$, with eigenvalue $$\mu$$, then any multiple of this vector is in the basis : $$\forall c \in \mathbb{R}, \ A(cv)=\mu(cv)$$

Then if you have two vectors in a basis, $$v$$ and $$w$$, any linear combination of these two vector will be an eigenvector for $$\mu$$: $$\left\{ \begin{array}{l} Av=\mu v\\ Aw=\mu w\\ \end{array} \right. \Rightarrow A(cv+dw) = \mu (cv+dw)$$

For exemple, let $$A=J-I$$ a matrix $$n\times n$$ of all 1, except 0 in the diagonal (this exemple comes from graph theory and the complete graph $$K_n$$). Then $$A$$ has eigenvalues $$n-1$$ and $$-1$$, and while the eigenspace associated with $$n-1$$ has dimension $$1$$ (there is only 1 vector in its basis), the eigenspace associated with the eigenvalue $$-1$$ has dimension $$n-1$$. Hence you can find an orthogonal basis of $$n-1$$ vectors. Any vector $$v$$ verifying $$Av=-v$$ can be written as a linear combination of the vectors in this basis.

You are understanding things just right.

For example, the $$3 \times 3$$ matrix $$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ has two eigenvalues, $$2$$ and $$3$$.

Every nonzero vector on the $$x$$-axis is an eigenvector for eigenvalue $$2$$, and a basis for that eigenspace.

Every nonzero vector in the $$y$$-$$z$$ plane is an eigenvector for eigenvalue $$2$$. Any two linearly independent vectors in that plane form a basis of eigenvectors for that eigenvalue.


Question

# Let $$a$$ denote the element of the $${i^{th}}$$ row and $${j^{th}}$$ column in a $$3 \times 3$$ matrix and let $${a_{ij}} = \, - {a_{ji}}$$ for every i and j then this matrix is an -

A
Orthogonal matrix
B
singular matrix
C
matrix whose principal diagonal elements are all zero
D
skew-symmetric matrix

Solution

## The correct option is B skew-symmetric matrixIf $$a_{ij}$$ is the element of $$i^{th}$$ row and $$j^{th}$$ column and $$a_{ij}=-a_{ji}$$then lets assume$$A=\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{bmatrix}$$Where, $$a_{ij}=-a_{ji}$$$$a_{11}=-a_{11}\implies a_{11}=0$$, similarly $$a_{22}=a_{33}=0$$and $$a_{21}=-a_{12}, a_{31}=-a_{13}, a_{32}=-a_{23}$$Putting the values$$A=\begin{bmatrix} { 0 } & { a }_{ 12 } & { a }_{ 13 } \\ { -a }_{ 12 } & { 0 } & { a }_{ 23 } \\ { -a }_{ 13 } & { -a }_{ 23 } & { 0 } \end{bmatrix}$$$${ A }^{ T }=\begin{bmatrix} { 0 } & { -a }_{ 12 } & { -a }_{ 13 } \\ { a }_{ 12 } & { 0 } & { -a }_{ 23 } \\ { a }_{ 13 } & { a }_{ 23 } & {0 } \end{bmatrix}\\$$and$$-A=\begin{bmatrix} { 0 } & -{ a }_{ 12 } & -{ a }_{ 13 } \\ { a }_{ 12 } & { 0 } & -{ a }_{ 23 } \\ { a }_{ 13 } & { a }_{ 23 } & { 0 } \end{bmatrix}$$thus$$A^T=-A$$$$\therefore$$ The matrix is a skew-symmetric matrix.Mathematics

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Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges

I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:

Monotonic:

The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$
$$a_1=1\leq 1+\frac{1}{2^2}=a_2$$

Need to show that $a_{n+1}\leq a_{n+2}$
$$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$
Thus the sequence is monotone and increasing.

Boundedness:

Since the sequence is increasing it is bounded below by $a_1=1$.
Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don’t know what my thinking process should be to find an upper bound.

Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?

Solutions Collecting From Web of "Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges"

Your work looks good so far. Here is a hint:
$$\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} – \frac{1}{n}$$

To elaborate, apply the hint to get:
$$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} – \frac{1}{2}\right) + \left(\frac{1}{2} – \frac{1}{3}\right) + \left(\frac{1}{3} – \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-1} – \frac{1}{n}\right)$$

Notice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.

Also notice that all terms on the right-hand side cancel out except for the first and last one. Thus:
$$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le 1 – \frac{1}{n}$$

Add $1$ to both sides to get:
$$a_n \le 2 – \frac{1}{n} \le 2$$

It follows that $a_n$ is bounded from above and hence convergent.

It is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.

Besides to Ayman’s neat answer, you may take $f(x)=\frac{1}{x^2}$ over $[1,+\infty)$ and see that $f'(x)=-2x^{-3}$ and then is decreasing over $[1,+\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $$\sum_{k=1}^{+\infty}\frac{1}{n^2}$$ to see the series is convergent. Now your $a_n$ is the $n-$th summation of this series.

You can show this geometrically too.

If you take a square, and divide the height into $\frac 12$, $\frac 14$, $\frac 18$, and so forth, doubling the denominator on each deal.

Now take the squares $\frac 12$ and $\frac 13$: these go onto the top shelf.

On the second shelf go $\frac 14$ to $\frac 17$. These fractions are all less than $\frac 14$, so fit onto the second shelf. Likewise, squares from $8$ to $15$ go onto the third shelf, each $\frac 1n$ is smaller than $\frac 18$, and so forth.

Therefore $\sum_{n=2}^{\infty}\frac 1n < 1$, and therefore the whole lot is less than two squares.

by the integral test :

$\int^\infty_1 \frac{1}{n^2} dn \le \sum_{i=1}^\infty \frac{1}{n^2}\le 1+\int^\infty_1 \frac{1}{n^2}dn$ .

you can compute the integral so the answer is :

$1 \le \sum_{i=1}^\infty \frac{1}{n^2}\le 2$ .

because : $\int^\infty_1 \frac{1}{n^2} dn =1$

Hint: Prove the the following holds for all $n$ by induction.

$$\sum_1^n \frac{1}{k^2} \le 2 – \frac{1}{n}.$$

Is it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work.

You can use the same technique to bound other values of the zeta function. For example, try showing $\zeta(3)$ is bounded above by $\frac{3}{2}$.

(I am copying my answer from a duplicate question that was closed as a copy of this one since the induction approach is not available here.)

This here should work with $n \geq 1$ :

$$s_n = \sum\limits_{n=1}^\infty \frac{1}{n^2}= \frac{1}{1} + \frac{1}{4} +\frac{1}{9} + \frac{1}{16}+ …+ \frac{1}{n²}$$$$b_n = \sum\limits_{n=1}^\infty \frac{1}{2^{n-1}} = \frac{1}{1} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8} + …+ \frac{1}{2^{n-1}}$$

$b_n$ is directly compared greater than $s_n$ :
$$s_n < b_n$$
and $b_n$ converges, because of its ratio test :
$$\frac{1}{2^{n-1+1}} / \frac{1}{2^{n-1}} = \frac{2^{n-1}}{2^{n}} = \frac{1}{2} < 1$$

Notice that $2k^2 \geq k(k+1) \implies \frac{1}{k^2} \leq \frac{2}{k(k+1)}$.

$$\sum_{k=1}^{\infty} \frac{2}{k(k+1)} = \frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \ldots$$

$$\sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2\Big(\, \Big(1 – \frac{1}{2}\Big) + \Big(\frac{1}{2} – \frac{1}{3} \Big) + \Big(\frac{1}{3} – \frac{1}{4} \Big) + \ldots \Big)$$

$$\sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2 (1) = 2$$.

Therefore $\sum_{k=1}^{\infty} \frac{1}{k^2} \leq 2$.

$a_n=\frac{1}{n^2}$ and because $a_n>0$ we have $|a_n|=a_n$
First: check the necessary condition
$$\lim_{n\to +\infty} na_n=\lim_{n\to +\infty}\frac{1}{n}=0$$
Second: check D’Alembert’s ratio test
$$\lim_{n\to +\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to +\infty}(\frac{n}{n+1})^2=1$$
Third: Because the answer of D’Alembert’s test is $1$, you should use Raabe’s test:
$$\lim_{n\to +\infty}n(1-|\frac{a_{n+1}}{a_n}|)=\lim_{n\to +\infty}n(1-\frac{a_{n+1}}{a_n})=\lim_{n\to +\infty}n(\frac{2n+1}{n^2+2n+1})=2\gt1$$
so the series is convergent


# Thread: Show that 64 divides 9^n + 56n + 63

1. ## Show that 64 divides 9^n + 56n + 63

Show that 64 divides 9^n + 56n + 63 for all positive integers n.

Based off of basic computation I can see that for numbers like n=1, n=2, etc this holds but how would I show it for all positive integers?

2. ## Re: Show that 64 divides 9^n + 56n + 63

are you familiar with induction?

$9 + 56 + 63 = 128 = 2 \cdot 64$

so this is true for $n=1$ and we would say $P_1 = True$

next, we assume that $P_n=True$ and we use this to show that $P_{n+1} = True$

here we'd do

\begin{align*} &9^{n+1} + 56(n+1) + 63 \\ = &~9^n\cdot 9 + 56n + 56 + 63 \\ = &~9^n+ 56n + 63 + 8\cdot 9^n + 56 \\ = &~64k + 8\cdot 9^n + 56 \text{; because }P_n=True \\ =&~64k + 8(7+9^n) \end{align*}

clearly $64k$ is divisible by $64$ so we need to show that $8(7+9^n)$ is divisible by $64$ or alternatively that

$(7+9^n)$ is divisible by $8$

Using the same method we are using to show the original statement

$n=1 \Rightarrow 7+9^n = 16 = 2\cdot 8$ so $\tilde{P}_1=True$

Now you show that $\tilde{P}_n \Rightarrow \tilde{P}_{n+1}$ for this and thus that $7+ 9^n$ is divisible by 8

and you will have shown that

$9^{n+1} + 56(n+1) + 63$ is divisible by $64$

and thus that for the original statement

$P_n \Rightarrow P_{n+1}$

and thus it is True for all $n\geq 1$

so see if you can prove that $\tilde{P}_n \Rightarrow \tilde{P}_{n+1}$ it's not hard

3. ## Re: Show that 64 divides 9^n + 56n + 63

You can also note:

$\displaystyle 9^n + 56n + 63 =$

$\displaystyle 9^n + 56n + (8n - 8n) + 63 + (1 - 1) =$

$\displaystyle 9^n + (56n + 8n) + (63 + 1) - 1 - 8n =$

$\displaystyle 9^n + (64n + 64) - 1 - 8n$

64 divides (64n + 64), so it remains to show that 64 also divides $\displaystyle \ \ 9^n - 1 - 8n$.

It has been shown or mentioned it's true for n = 1 in at least one prior post. Let's look at $\displaystyle \ \ n \ge 2$:

$\displaystyle 9^n - 1 - 8n =$

$\displaystyle (8 + 1)^n - 1 - 8n =$

$\displaystyle \bigg(8^n \ + \ n\cdot8^{n - 1} \ + \ \dfrac{n(n - 1)}{2}8^{n - 2} \ + \ . . . \ + \ \dfrac{n(n - 1)}{2}8^2 \ + \ n\cdot8^1 \ + \ 1\bigg) \ - \ 1 \ - \ 8n =$

$\displaystyle 8^n \ + \ n\cdot8^{n - 1} \ + \ \dfrac{n(n - 1)}{2}8^{n - 2} \ + \ . . . \ + \ \dfrac{n(n - 1)}{2}8^2 \ + \ 8n \ - \ 8n \ + \ 1 \ - \ 1 =$

$\displaystyle 8^n \ + \ n\cdot8^{n - 1} \ + \ \dfrac{n(n - 1)}{2}8^{n - 2} \ + \ . . . \ + \ \dfrac{n(n - 1)}{2}8^2$ . . . . . . . . . **

The fractions in front of the powers of eights are combinations and are thus positive integers. The smallest power of eight is equal to 64,
so this expression is also divisible by 64.

** . . . . If n = 2, there will only be one term here. If n = 3, there will only be two terms here. And so on.

4. ## Re: Show that 64 divides 9^n + 56n + 63

The induction step can be done as follows:

\displaystyle \begin{align*}9^{n+1} + 56(n+1) + 63 &= 9\cdot 9^n + 56n + 56 + 63 \\ &= 8( 9^n + 7) + (9^n + 56n + 63) \end{align*}

where the final parenthesis is a multiple of 64 by the inductive hypothesis. So all we need is for $\displaystyle 8$ to divide $\displaystyle (9^n + 7)$.

$\displaystyle 9^n + 7 \equiv 1^n - 1 \pmod{8} = 1-1 \pmod{8} = 0 \pmod{8}$

QED


# Proving an expression is composite

I am trying to prove that $n^4 + 4^n$ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.

This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.

I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.

-
Write it as a difference of squares. –  Adam Hughes Jul 1 '14 at 19:54
How? It's $n^4 + 4^n$ not $n^4 - 4^n$ –  Mathmo123 Jul 1 '14 at 19:55
@Mathmo123: see my answer below. –  Adam Hughes Jul 1 '14 at 20:03

$(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...

-
Ah. Very neat!! –  Mathmo123 Jul 1 '14 at 20:05
Originally, I was not sure how I should go about writing it as a difference of squares. Thanks for your help. –  pidude Jul 1 '14 at 20:12

Hint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?

Let me know if you need further hints.

-
I originally tried doing that but to no avail. Evaluated at 3 I obtain 145 which has factors of 5 and 29. At 5, the expression equals 1649, which has factors of 17 and 97. I will keep looking for a pattern and let you know if I need more hints. Thanks for your help. –  pidude Jul 1 '14 at 19:59
Ah... using this method, there will be a difference between multiples of 5 and other odd numbers. You will find that for odd numbers that are not a multiple of 5, 5 will be a divisor –  Mathmo123 Jul 1 '14 at 20:01
Ok, so I was able to prove that. Now I'm working numbers which are multiples of 5. –  pidude Jul 1 '14 at 20:09

I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.

• We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by
1. $n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law
2. Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication
3. $(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression
4. $(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression
5. $(n^2+2^n)^2-2(n^2)(2^n)$ from step 2
6. $(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential
• We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again
1. Treat $n^2+2^n$ as $a$
2. Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer
3. So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$
4. $2^{2m}n^2=(n2^m)^2$ associative law
5. Treat $n2^m$ as $b$
• It implies that both $a$ and $b$ are both positive integer
• From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$
• Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$
-

Some interesting factorizations of a polynomial of type $x^4+\text{const}$: $$x^4+4=(x^2+2x+2)(x^2-2x-2) \tag{1}$$

$$x^4+1=(x^2+\sqrt[]{2}x+1)(x^2-\sqrt[]{2}x+1) \tag{2}$$

So one can ask, how to select the coefficients $a,b,c,d$ in

$$(x^2+ax+b)(x^2+cx+d) \tag{3}$$

such that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.

If we expand $(3)$ we get

$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$

And the coefficients disappear, if

$$\begin{eqnarray} c+a &=& 0 \\ d+ ac +b &=& 0 \\ ad+bc &=& 0 \end{eqnarray}$$

When solving for $b,c,d$ we get

$$\begin{eqnarray} c &=& -a \\ b &=& \frac{a^2}{2} \\ d &=& \frac{a^2}{2} \end{eqnarray}$$

and therefore

$$x^4+\frac{a^4}{4} = (x^2+a x+\frac{a^2}{2})(x^2-ax+\frac{a^2}{2})$$

For $a=2$ this gives $(1)$, $a=\sqrt[]{2}$ this gives $82)$ . Substituting $a=2^{t+1}$ we get

$$x^4+4^{2t+1} = (x^2+2\cdot 2^t x+2^{2t+1})(x^2-2\cdot 2^tx+2^{2t+1})$$

Substituting $x=n=2t+1$ gives the required result for odd $n$.

-

@Mathmo123 As I observe the behavior of the multiply operation, I see..

• $v \times v$ produce $v$
• $v \times o$ produce $v$
• $o \times o$ produce $o$

and for the plus operation, I see..

• $v+v$ produce $v$
• $v+o$ produce $o$
• $o+o$ produce $v$

where $v$ is even number and $o$ is odd number.

I do not know that anyone notice about this properties, so I personally name them even-odd multiply properties and even-odd sum properties. So, the expression $n^4 + 4^n$, since n is odd($o$). can be evaluated follow by that properties it will be isolated into:

1. $n^4 = n \times n \times n \times n \equiv o \times o \times o \times o$ produces $o$
2. $4^n \equiv v \times v \times \cdots \times v$ for $n$ times produces $v$
3. $n^4+4^n \equiv o+v$ produces $o$

hence $2|(n^4 + 4^n)$, since $n$ is odd.
May I prove by this way? Any suggestions are kindly pleased.

So, I give up with this way.

-
1. is wrong.... –  lhf Jul 2 '14 at 11:25


# Coefficients of homology

I am wondering why people use different coefficients when defining homology of simplicial complex, like homology over $R$, $Z$, $Z/2$, etc? Is one better then the other and why?

Moreover, which one(s) is(are) most useful in practical computation other than abstract theory?

Thanks a lot.

• – Watson Jun 5 '16 at 16:00
• In some sense, if you don't go to deeply in the theory, knowing with $\mathbb{Z}$ coefficients is everything. This is the content of the universal coefficient theorem. It often occurs that one can compute homology with other coefficients more easily (e.g. with Z/2 coeffients poincare duality works directly for non orientable manifolds). Sometimes the way (co) homology is defined leads one directly to different coefficients. A great example is the de Rham cohomology. – Thomas Rot Jun 5 '16 at 20:54

There is no best coefficient in general, but there is indeed a best coefficient for specific case.

Let me show you some examples.

Suppose you want to compute the Euler Characteristic of a space $X$. Suppose that for some reason, counting cell is not a viable option, let's say you don't have an explicit cell structure of your space. You are left with trying to compute the rank of the homology groups. Now you can verify that $\text{rank}H_*(X;\mathbb{Z})=\dim H_*(X,\mathbb{Q})$. The latter group is indeed a vector space, so it has no torsion and it is easier to compute (every short exact sequence splits for example). Even more is true, computing Euler Characteristic is invariant from the chosen field coefficients. In conclusion, it is natural to trying to compute $H_*(X,\mathbb{Q})$ in this case.

Suppose you have a non-orientable closed manifold $M$: When you want to study closed manifolds (from an homology viewpoint), Poincaré Duality is a powerful tool. A powerful tool with some hypothesis to verify, namely orientability. If you don't have orientability you are left with two option, trying to compute $H_*(M;\mathbb{Z}_{\rho})$ (homology with local coefficient - something pretty technical which was let's say created for dealing with this kind of situation) or more easily $H_*(M;\mathbb{Z}_2)$, why? well because every manifold is $\mathbb{Z}_2$ orientable, and therefore you have P-D with these coefficients.

Clearly, homology with $\mathbb{Z}$ coefficients contains all the homological information, since we have UCT which means that if you know homology with $\mathbb{Z}$ for a space, then you know the homology for that space with coefficient any abelian group.

What's the problem: You don't always know how to compute homology with $\mathbb{Z}$-coefficient. By its very property that it bring a lot of informations, computations in general are cumbersome: you have torsion to deal with for instance which is a pain in the ass. So we came up with some weaker version of it, easier to compute but still interesting. This is how research is done usually, you find some great invariant and then realise that it's impossible to compute (very common) and then he must find something weaker but computable and still interesting (very hard).

A last example to back up what I've said. When you will deal with characteristic classes, in particular Stiefel-Whitney ones, you will realise what I mean by sometimes computing cohomology with integer coefficients is hard. But if you turn to $\mathbb{Z}_2$ coefficients, the cohomology ring turns out to be a free graded ring and the generators are the universal classes.

• I can't really see any way of making the assertion "$\operatorname{rank} H_*(M;\mathbb{Z}) = \dim H_*(M;k)$" true if the characteristic of $k$ is positive... – Najib Idrissi Jun 6 '16 at 8:41
• maybe I rushed a little, but if I recall correctly, $\text{rank}H_*(M;\mathbb{Z})=\dim H_*(M;\mathbb{Z})\otimes \mathbb{Q}$ and then one verifies at once that euler char doesn't depend on the field you are computing it. But you are right, I need to modify it a little, since what I wrote it's formally wrong – Riccardo Jun 6 '16 at 8:44
• Yes, by the UCT $\operatorname{rank} H_k(M;\mathbb{Z}) = \dim H_k(M;\mathbb{Q})$, and yes, the Euler characteristic does not depends on the field. But for example $H_2(\mathbb{RP}^2;\mathbb{Z}) = 0$ has rank zero whereas $H_2(\mathbb{RP}^2;\mathbb{F}_2) = \mathbb{F}_2$ has dimension one. ("Formally wrong" is just another way of saying "wrong"...) – Najib Idrissi Jun 6 '16 at 8:46
• yes, and thank you for pointing out this:) – Riccardo Jun 6 '16 at 8:52


# Math Help - Simplify this expression...

1. ## Simplify this expression...

$
({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}
$

2. Originally Posted by Alex2103
$
({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}
$
Get ready for a ride....

Here are the highlights:
Let $\displaystyle x = ( 2 + \sqrt{5} )^{1/3} + (2 - \sqrt{5} )^{1/3}$

Now
$\displaystyle x^3 = (2 + \sqrt{5}) + 3( 2 + \sqrt{5} )^{2/3}(2 - \sqrt{5} )^{1/3} + 3(2 - \sqrt{5} )^{1/3}(2 - \sqrt{5})^{2/3} + (2 - \sqrt{5})$

This reduces to
$\displaystyle x^3 = 4 -3(2 + \sqrt{5})^{1/3} - 3(2 - \sqrt{5})^{1/3}$

or, reminding ourselves of what x is equal to:
$\displaystyle x^3 = 4 -3x$

Now solve:
$\displaystyle x^3 + 3x - 4 = 0$

(Hint: There is only one real solution.)

-Dan

3. Hello, Alex2103!

$\text{Simplfy: }\;\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}$

A binomial of the form $a + b\sqrt{5}$ makes me suspect
. . that the problem involves the Golden Mean: . $\phi \:=\:\dfrac{1+\sqrt{5}}{2}$

$\text{And sure enough: }\;\phi^3 \;=\;\left(\frac{1+\sqrt{5}}{2}\right)^3 \;=\;2+\sqrt{5}$

. . . $\text{and we find that: }\;\left(\frac{1-\sqrt{5}}{2}\right)^3 \;=\;2-\sqrt{5}$

Therefore:

. . $\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} \;\;=\;\;\sqrt[3]{\left(\frac{1+\sqrt{5}}{2}\right)^3} + \sqrt[3]{\left(\frac{1-\sqrt{5}}{2}\right)^3}$

. . $=\;\;\dfrac{1 + \sqrt{5}}{2} + \dfrac{1-\sqrt{5}}{2} \;\;=\;\;1$

I love your solution, Dan!

4. Originally Posted by Soroban
I love your solution, Dan!
But yours is so more elegant and less time consuming!

-Dan

5. Thanks for help!

6. Heres' yet another way to do that. Cardano's formula for the reduced cubic equation says that a solution to $x^3+ mx= n$ is of the form $\sqrt[3]{\frac{n}{2}+ \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}+ \sqrt[3]{\frac{n}{2}- \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}$.

Looks familiar, doesn't it? That is precisely the same as $\sqrt[3]{2+ \sqrt{5}}+ \sqrt[3]{2- \sqrt{5}}$ with $\frac{n}{2}= 2$ and $\frac{n^2}{4}+ \frac{m^3}{27}= 5$.

From $\frac{n}{2}= 2$, n= 4 and then $\frac{n^2}{4}+ \frac{m^3}{27}= 4+ \frac{m^3}{27}= 5$ so that $\frac{m^3}{27}= 1$, $m^3= 27$, and $m= 3$.

That means that this number is a real root of $x^3+ 3x= 4$ or $x^3- 3x- 4= 0$. It is easy to see that $x^3+ 3x- 4= (x- 1)(x^2+ x+ 4)$. Since the discriminant of $x^2+ x+ 4$ is $1- 16= -15$ the only real root of that equation is 1 so we must have $\sqrt[3]{2+ \sqrt{5}}+ \sqrt[3]{2- \sqrt{5}}= 1$.


# Can an empty set be in relation?

We were given this problem for homework:

Let it be $A=\{a,b\}$.

a) Find the powerset $\mathcal P(A)$.

b) Construct an equivalence relation on $\mathcal P(A)$ so that it's quotient set has exactly two elements.

a) $\mathcal P(A)= \{\emptyset\ , \{a\},\{b\},\{a,b\}\}$

b) I constructed the following relation in which an empty set can be an element of ordered pair. (But I'm not sure if an empty set can be in relation. Can an empty set be an element of ordered pair?)

Let's say the relation is R. $$R = \{(\emptyset,\emptyset),(\{a\},\{a\}),(\{b\},\{b\}),(\{a,b\},\{a,b\}),(\{a\},\{b\}),(\{b\},\{a\}),(\emptyset,\{a,b\}),(\{a,b\},\emptyset)\}$$

I think this is correct because:

$[\{a\}] = \{\{a\},\{b\}\} = [\{b\}]$

$[\{\emptyset\}]=\{\emptyset,\{a,b\}\}=[\{a,b\}]$

which means that $\mathcal P(A)/_R=\{\{\{a\},\{b\}\}, \{\emptyset,\{a,b\}\}\}$.

Is this correct or should i remove the empty set from the relation?

• I do not see a problem. Its fine, I suppose. – hemu Nov 23 '17 at 18:55
• It’s perfectly correct. Don’t get confused by having a set of sets. To become more at ease with that, abstract a bit: $X = \mathcal P (A)$ is just a set with four elements, the empty set being one of it, technically. If that gives you a headache, just rename the elements $x_1 = ∅$, $x_2 = \{a\}$, … and work with the renamed elements. – k.stm Nov 23 '17 at 18:57
• By the way, the equivalence relation you have found can be described in words: Two sets $x, y ∈ \mathcal P (A)$ are equivalent (with respect to $R$) if and only if they are of the same parity. – k.stm Nov 23 '17 at 19:03
• @k.stm Thanks! Also, I have another question. I have to prove that $A\neq \emptyset$ ( for $A=\{(a,b) \in \Bbb R^2 : x+y=c \}$, for every $c$). How do I prove this exactly? I understand why this is true but I don't know how to prove it. – i dont know much about algebra Nov 23 '17 at 19:05
• You mean $A = \{(a,b) ∈ ℝ^2;~a + b = c\}$. Well fix a $c$ and find an element in $A$. You can always explicitly write one down. If you can’t manage that, try for $c = 0$ first and then see how you could generalise. – k.stm Nov 23 '17 at 19:07


# Do these matrices have a name?

I am looking for matrices $A\in \mathbb{R}^{m\times n}$ with $m>n$, $A^TA=\frac{m}{n}I$ and $diag(AA^T)=(1\ \dots\ 1)$ where $diag$ denotes the diagonal. Do such matrices have a name? An example for such a matrix would be an $8\times 3$ matrix with the coordinates of a cube centered at the origin as rows. Is the matrix $A$ for every pair $(m,n)$ uniquely determined up to right multiplication with an orthogonal matrix? How are the $m$ points given by the rows of $A$ distributed over the $(n-1)$-dimensional sphere?

My motivation is that I have an unknown $x\in \mathbb{R}^n$. I assume that I can measure the scalar product $a^Tx$ with any unit vector $a\in S^{n-1}$. The measurements are assumed to be i.i.d. Now we assume we have $m$ measurements and write the corresponding vectors as the rows of a matrix $A\in \mathbb{R}^{m\times n}$ and the vectors as a matrix $b\in \mathbb{R}^m$. The best possible reconstruction (see Gauss-Markov) of $x$ is given by the least square solution $(A^TA)^{-1}A^Tb$ where $b\in \mathbb{R}^n$ denotes the measurements. The covariance matrix for the reconstruction is $(A^TA)^{-1}$. This means that if $A$ satisfies the properties above then the components of the reconstructed $x$ are uncorrelated. Also I think that $A$ gives us the best possible way to measure $x$, i.e. minimizing the standard deviation.

• Is $m$ a power of $2$? It would be very nice if it were. – Rodrigo de Azevedo Dec 8 '16 at 16:28
• In general no. But if you have some insights for that case it would definitely be interesting. – user35593 Dec 8 '16 at 23:08
• In general if A is such a matrix then e can build new matrices by concatenating copies of A. Also we can permute the rows of A to get a new solution. – user35593 Dec 9 '16 at 7:29
• If we have solutions with the same n we can build new solutions by concatenating. One could introduce the notion of a 'prime' matrix with this property. I.e. a matrix which does not contain a submatrix with n columns with the same property – user35593 Dec 9 '16 at 7:44
• The rows of this matrix form a tight frame of unit vectors. See e.g. ams.org/notices/201306/rnoti-p748.pdf – Terry Tao Dec 9 '16 at 17:11

Assuming that the Hadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \times n$ matrix that satisfies the given constraints is given by

$$\boxed{\mathrm A := \frac{1}{\sqrt n} \mathrm H_m^{\top} \mathrm S_n}$$

where

• $\mathrm H_m \in \{\pm 1\}^{m \times m}$ is a Hadamard matrix. Thus, the $m$ rows of $\mathrm H_m$ are orthogonal, i.e., $$\mathrm H_m \mathrm H_m^{\top} = m \mathrm I_m$$

• $\mathrm S_n$ is a thin $m \times n$ matrix whose $n$ columns are chosen from the $m$ columns of the $m \times m$ identity matrix. Thus, the $n$ columns of $\mathrm S_n$ are orthonormal, i.e.,

$$\mathrm S_n^{\top} \mathrm S_n = \mathrm I_n$$

Hence,

$$\mathrm A^{\top} \mathrm A = \frac{1}{n} \mathrm S_n^{\top} \mathrm H_m \mathrm H_m^{\top} \mathrm S_n = \frac{m}{n} \mathrm S_n^{\top} \mathrm S_n = \frac{m}{n} \mathrm I_n$$

as desired. Let $\mathrm e_k$ and $\mathrm h_k$ denote the $k$-th columns of $\mathrm I_m$ and $\mathrm H_m$, respectively. Hence,

$$\mathrm e_k^{\top} \mathrm A \mathrm A^{\top} \mathrm e_k = \| \mathrm A^{\top} \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm H_m \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm h_k \|_2^2 = \frac 1n \sum_{k=1}^n (\pm 1)^2 = \frac nn = 1$$

for all $k \in \{1,2,\dots,m\}$, as desired. Note that we used the fact that the entries of $\mathrm h_k$ are $\pm 1$.

If $m$ is a power of $2$, then $\mathrm H_m$ can be built recursively using the Sylvester construction

$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_1 = 1$$

which builds (symmetric) Walsh matrices. If $m$ is not a power of $2$, we can use the Paley construction instead.

Example

Let $m = 8$ and $n = 3$. Since $8$ is a power of $2$, we can use the Sylvester construction to build $\mathrm H_8$.

Using MATLAB,

>> H1 = 1;
>> H2 = [H1,H1;H1,-H1];
>> H4 = [H2,H2;H2,-H2];
>> H8 = [H4,H4;H4,-H4]

H8 =

1     1     1     1     1     1     1     1
1    -1     1    -1     1    -1     1    -1
1     1    -1    -1     1     1    -1    -1
1    -1    -1     1     1    -1    -1     1
1     1     1     1    -1    -1    -1    -1
1    -1     1    -1    -1     1    -1     1
1     1    -1    -1    -1    -1     1     1
1    -1    -1     1    -1     1     1    -1

Let the $3$ columns of $\mathrm S_3$ be the first $3$ columns of $\mathrm I_8$

>> I8 = eye(8);
>> H8 * I8(:,[1,2,3])

ans =

1     1     1
1    -1     1
1     1    -1
1    -1    -1
1     1     1
1    -1     1
1     1    -1
1    -1    -1

Note that the last four rows are copies of the first four rows. Hence, let the $3$ columns of $\mathrm S_3$ be the 2nd, 3rd and 5th columns of $\mathrm I_8$

>> H8 * I8(:,[2,3,5])

ans =

1     1     1
-1     1     1
1    -1     1
-1    -1     1
1     1    -1
-1     1    -1
1    -1    -1
-1    -1    -1

Note that the $8$ rows are now the $8$ vertices of the cube $[-1,1]^3$.

We build matrix $\mathrm A$ by normalizing the rows

>> A = inv(sqrt(3)) * H8 * I8(:,[2,3,5])

A =

0.5774    0.5774    0.5774
-0.5774    0.5774    0.5774
0.5774   -0.5774    0.5774
-0.5774   -0.5774    0.5774
0.5774    0.5774   -0.5774
-0.5774    0.5774   -0.5774
0.5774   -0.5774   -0.5774
-0.5774   -0.5774   -0.5774

Is the constraint $\mathrm A^{\top} \mathrm A = \frac 83 \mathrm I_3$ satisfied?

>> A' * A

ans =

2.6667         0         0
0    2.6667         0
0         0    2.6667

It is. Are the diagonal entries of $\mathrm A \mathrm A^{\top}$ equal to $1$?

>> A * A'

ans =

1.0000    0.3333    0.3333   -0.3333    0.3333   -0.3333   -0.3333   -1.0000
0.3333    1.0000   -0.3333    0.3333   -0.3333    0.3333   -1.0000   -0.3333
0.3333   -0.3333    1.0000    0.3333   -0.3333   -1.0000    0.3333   -0.3333
-0.3333    0.3333    0.3333    1.0000   -1.0000   -0.3333   -0.3333    0.3333
0.3333   -0.3333   -0.3333   -1.0000    1.0000    0.3333    0.3333   -0.3333
-0.3333    0.3333   -1.0000   -0.3333    0.3333    1.0000   -0.3333    0.3333
-0.3333   -1.0000    0.3333   -0.3333    0.3333   -0.3333    1.0000    0.3333
-1.0000   -0.3333   -0.3333    0.3333   -0.3333    0.3333    0.3333    1.0000

They are.


Why does L'Hôpital's rule work for sequences?

Say, for the classic example, $\frac{\log(n)}{n}$, this sequence converges to zero, from applying L'Hôpital's rule. Why does it work in the discrete setting, when the rule is about differentiable functions?

Is it because at infinity, it doesn't matter that we relabel the discrete variable, $n$, with a continuous variable, $x$, and instead look at the limit of $\frac{\log(x)}{x}$?

But then what about the quotients of sequences that go to the indeterminate form $\frac{0}{0}$? Why is it OK to use L'Hôpital's rule, as $n$ goes to zero?

I haven't found anything on Wikipedia or Wolfram about the discrete setting.

Thanks.

• The counterpart of L'Hospital's rule for sequences is the so-called Stolz Theorem. – Zhanxiong Aug 16 '15 at 0:28
• Thanks for the link, @Zhanxiong. – User001 Aug 16 '15 at 0:52
• possible duplicate of Is there a discrete version of de l'Hôpital's rule? – J. M. isn't a mathematician Aug 16 '15 at 21:26
• @Guesswhoitis.I don't think that's the same question. He asks "why do people differentiate sequences?". – user223391 Aug 16 '15 at 22:07
• I disagree. Some dots would need to be connected in order to be able to say this question is a duplicate of the other one. – Robert Soupe Aug 17 '15 at 2:03

There IS a L'Hospital's rule for sequences called Stolz-Cesàro theorem. If you have an indeterminate form, then:

$$\lim\limits_{n\to\infty} \frac{s_n}{t_n}=\lim\limits_{n\to\infty} \frac{s_n-s_{n-1}}{t_n-t_{n-1}}$$

$$\lim\limits_{n\to\infty}\frac{\ln(n)}{n}=\lim\limits_{n\to\infty}\frac{\ln\left(\frac{n}{n-1}\right)}{n-n+1}=\lim\limits_{n\to\infty}\ln\left(\frac{n}{n-1}\right)=0$$

But that isn't your question. Your question is, why do people "differentiate"? Basically because the real case covers the discrete case.

Recall the definition of limits for real and discrete cases.

Definition. A sequence, $s_n\colon \Bbb{N}\to \Bbb{R},$ converges to $L$ as $n\to\infty$, written $\lim\limits_{n\to\infty} s_n=L$ iff for all $\epsilon>0$ there is some $N$ such that for all $n\in \Bbb{N}$ with $n>N$, $|s_n-L|<\epsilon$.

Definition. A function, $f(x) \colon \Bbb{R}\to \Bbb{R}$ converges to $L$ as $x\to\infty$, written $\lim\limits_{x\to\infty} f(x)=L$ iff for all $\epsilon>0$ there is some $X$ such that for all $x\in \Bbb{R}$ with $x>X$, $|f(x)-L|<\epsilon$.

So if $f(x)$ is a real valued function that agrees with a sequence, $s_n$ on integer values, then $\lim\limits_{x\to\infty} f(x)=L$ implies $\lim\limits_{n\to\infty} s_n=L$.

• Thanks so much @avid19 - this was an awesome explanation. – User001 Aug 16 '15 at 0:51
• My +1. I want to add here that there is a catch. There are cases when $f(n) \to L$ as $n \to \infty$ but $f(x)$ has no limit as $x \to \infty$. – Paramanand Singh Aug 16 '15 at 5:10
• @ParamanandSingh: The most obvious example would be $f(x) = \sin(2\pi x)$ – R.. GitHub STOP HELPING ICE Aug 16 '15 at 14:19

Your explanation is not really precise, it does matter whether you use the discrete or continuous variable. However, there is a theorem in mathematical analysis that states that the following is equivalent:

• $\lim_{x \rightarrow c} f(x) = A$
• For every sequence $\{x_n\}$ such that $\forall n \in \mathbb{N} : x_n \in D(f), x_n \neq c$ and that $\lim_{n \rightarrow \infty} x_n = c$ it is true that $\lim_{n \rightarrow \infty} f(x_n) = A$.

In simpler words, once you know the limit of a function in continuous variable, like $\lim_{x \rightarrow \infty} \frac{\log x}{x} = 0$, you also know the limit of any sequence you get by "picking out" points of this function's domain, in your case specifically you take the sequence $\{x_n\} = \{n\}$. Notice that the conditions of the second statement are met, since $\frac{\log x}{x}$ is defined for every $n$, $\lim_{n \rightarrow \infty} n = \infty$ and $n \neq \infty$ for every $n$ as well.

This also solves the problem for the limits of indeterminate form "$\frac{0}{0}$", or any other.

Hope that helps :),

Epsiloney

Fact: If a function $f:\Bbb R\to\Bbb R$ is continuous, $x_n\to x$ then $f(x_n)\to f(x)$.

You can see here Stolz–Cesàro theorem for the "l'Hôpital's rule" for sequences.

This uses the fact that if $\displaystyle\lim_{x\to\infty}f(x)=L, \text{ then } \lim_{n\to\infty}f(n)=L$ since

$\displaystyle\lim_{x\to\infty}f(x)=L\iff$ for every $\epsilon>0,$ there is an M such that if $x\ge M,$ then $\big|f(x)-L\big|<\epsilon$ and

$\displaystyle\lim_{n\to\infty}f(n)=L\iff$ for every $\epsilon>0,$ there is an N such that if $n\ge N,$ then $\big|f(n)-L\big|<\epsilon$ $\;\;$(with $n\in\mathbb{N}$)


# Finding the probability density function of $Z = X + Y$

### The problem

Let there be two independent random variables given. Let's name them $X$ and $Y$. Suppose the variables have both normal distribution $N(\mu, \lambda^2).$

Let's define third random variable $Z = X + Y$. The task is to find the density function of $Z$.

### My attempt

I know that the density function can be easily calculated using the distribution function (I would need to differentiate it). That's why I started with finding the distribution. $$F_Z(t) = Pr(Z \le t) = Pr(Y + X \le t) = Pr(Y \le t - X) = \int \limits_{-\infty}^{\infty}Pr(Y \le t-x)f_X(x) dx$$ The equation above leads us to the following one: $$\int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{t-x} f_Y(y)f_X(x) \mbox{d} y\mbox{d}x \tag{1}.$$ Using the knowledge about $f_X$ and $f_Y$ we can write $(1)$ in this form: $$\theta\int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{t-x}e^{-\frac{(y- \mu)^2}{\lambda^2}} e^{-\frac{(x- \mu)^2}{\lambda^2}} \mbox{d} y\mbox{d}x,$$ where $\theta = \frac{1}{\sqrt{2 \lambda^2}}.$
I wonder if my attempt was correct because the integral is quite horrible. Is it possible to calculate it in a "nicer" way?

Is $(1)$ equivalent to this expression $(f_X*f_Y)(x)$?
If yes I would get the following integral: $$\int \limits_{\mathbb{R}} e^{-\frac{(y- \mu)^2}{\lambda^2}} e^{-\frac{(x- y - \mu)^2}{\lambda^2}} \mbox{d}y.$$
Is there a method of calculating such things?

• en.wikipedia.org/wiki/… – saulspatz Jun 4 '18 at 18:50
• @Hendrra In general, sum of random variables $X$ and $Y$ with densities $f$ and $g$ has density being convolution of $f$ and $g$, you are correct – Jakobian Jun 4 '18 at 19:04
• @Adam Thank you. What about my first approach? – Hendrra Jun 4 '18 at 19:33

From your expression, $F_Z(t) = \int_\mathbb{R}P(Y \leq t-x)f_X(x)dx$, differentiate to obtain the density (you may need to justify why you can differentiate under the integral),

$$f_Z(t) = \int_\mathbb{R}f_Y(t-x)f_X(x)dx = \int_\mathbb{R}\frac{1}{\sqrt{2\pi}}e^{-(t-x-\mu)^2/2\lambda^2}\frac{1}{\sqrt{2\pi}}e^{-(x-\mu)^2/2\lambda^2}dx$$

which we recognize as $(f_Y*f_X) (x)$. To actually evaluate the integral, expand the exponent terms inside the integrand,

$$-\frac{1}{2\lambda^2}\left[(t-x-\mu)^2 + (x-\mu)^2\right] = -\frac{1}{2\lambda^2}(t^2-2\mu t + 2\mu^2 + \underbrace{2x^2-2tx}_{(*)})$$

Complete the square on the $(*)$ term, $2x^2 - 2tx = 2\left(x-\frac{t}{2}\right)^2-\frac{t^2}{2}$. Putting things together,

$$-\frac{1}{2\lambda^2}(t^2-2\mu t + 2\mu^2 + 2x^2-2tx) = -\frac{1}{4\lambda^2}(t^2-4\mu t + 4\mu^2)-\frac{1}{\lambda^2}(x-t/2)^2$$

So, we can pull out the $t$ terms from the integrand to be left with,

$$f_Z(t) = \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}\underbrace{\int_\mathbb{R}\frac{1}{\sqrt{\pi\lambda^2}}e^{-(x-t/2)^2/\lambda^2}dx}_{=1} = \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}$$

The integral above equals $1$ because the integrand is the density function of a $N(t/2,\lambda^2/2)$ distribution.

• At the end, did you intend $N(2 \mu,2 \lambda^2)$ for the mean and variance rather than $t/2, \lambda^2 /{2}$? – Henry Jun 5 '18 at 16:31
• Sorry, I was referring to the density function inside the final integral (the one that equals 1). The distribution of $Z$ is definitely $N(2\mu,2\lambda^2)$. – Flowsnake Jun 5 '18 at 16:33
• So was I. I think a density $f_Z(t)= \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}$ suggests a normal distribution for $Z$ with a mean of $2 \mu$ and a variance of $2 \lambda^2$ – Henry Jun 5 '18 at 16:37
• The density function I was referring to was, $\frac{1}{\sqrt{\pi\lambda^2}}e^{-(x-t/2)^2/\lambda^2}$, which is the density of a $N(t/2,\lambda^2/2)$. – Flowsnake Jun 5 '18 at 16:49

The problem can also be solved by using the following property: If $X\sim N(\mu_1,\sigma_1^2)$ and $Y \sim N(\mu_2,\sigma_2^2)$ and are independent, then $aX + bY \sim N(a\mu_1 + b\mu_2,a^2\sigma_1^2 +b^2\sigma_2^2)$ and thus in your case $Z \sim N(2\mu,2\lambda^2)$.

Edit: I just noticed the comment by @saulspatz above who already pointed this out.


# The coupon collector's most collected coupon

Suppose a coupon collector is collecting a set of $$n$$ coupons that he receives one-by-one uniformly at random.

If the collector stops exactly when the collection is complete, we know the expected number of coupons in his collection is $$n*H[n]$$. What is the expected number of copies, $$M$$, of his most collected coupon?

When $$n = 2$$, then $$M = 2$$ because after the first coupon he will collect the other coupon with probability p ~ Geom(1/2). So he will collect the same coupon one additional time on average, and that coupon is certain to be his most collected coupon.

I don't know the exact value for any $$n$$ larger than 2, but found some approximate values by simulation:

n M
3 2.8415
4 3.4992
5 4.0259
6 4.4633
7 4.8377
8 5.1649
9 5.4560

EDIT 1:

For small n enumerating the small possibilities converges faster than simulation, and from this approach I hypothesize that $$M[3] = (15 + 6 \sqrt{5})/10$$ but don't have any real argument to support that claim.

EDIT 2:

With some more thought, I find that M has an explicit sum formula. The probability the collection terminates with a given distribution of coupons $$v = \{c_1, ..., c_{n-1}\}$$ other than the final collected coupon is just the multinomial coefficient $$(c_1; ...; c_{n-1})$$ over $$n^\text{total # of coupons in the collection}$$. This is an infinite sum over n-1 variables. Here is Mathematica code the computes the sum for terms where the collection has no more than k copies of any coupon:

M[n_, k_] := Sum[Max[v]*(Multinomial @@ v)/n^Total[v], {v, Tuples[Range[1, k], n-1]}]

Mathematica can't actually evaluate this though for $$k \rightarrow \infty$$ though.

• You might be interested in Exercise 2.20 of Boucheron, Lugosi, and Massart's ""Concentration Inequalities: A Nonasymptotic Theory of Independence", which asks to prove an upper bound of $$e\log n \cdot ce^{W((1-c)/(ce))}$$for this expectation, when collecting $m = cn\log n$ coupons. ($W$ being the Lambert function.) May 21 at 6:52

Value of $$M_3$$. I will denote $$M$$ as $$M_n$$ to emphasize the dependence on $$n$$. Then by using A230137, we get

\begin{align*} M_3 &= \sum_{n=2}^{\infty} \frac{\sum_{k=1}^{n-1} \binom{n}{k} \max\{k,n-k\}}{3^n} \\ &= \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)n - 2(2n)}{3^{2n}} + \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)(2n+1) - 2(2n+1)}{3^{2n+1}} \\ &= 3\left(\frac{1}{2} + \frac{1}{\sqrt{5}}\right) \\ &\approx 2.84164, \end{align*}

which is the same as what OP conjectured.

Asymptotic Formula of $$M_n$$. A heuristic seems suggesting that

$$M_n \sim e \log n \qquad \text{as} \quad n \to \infty,$$

The figure below compares simulated values of $$M_n$$ and the values of the asymptotic formula.

Heuristic argument. First, we consider the Poissonized version of the problem:

1. Let $$N^{(1)}, \ldots, N^{(n)}$$ be independent Poisson processes with rate $$\frac{1}{n}$$. Then the arrivals in each $$N^{(i)}$$ model the times when the collector receives coupons of type $$i$$.

2. Let $$T$$ be the first time a complete set of coupons is collected. Then we know that $$T$$ has the same distribution as the sum of $$n$$ independent exponential RVs with respective rates $$1$$, $$\frac{n-1}{n}$$, $$\ldots$$, $$\frac{1}{n}$$. (This is an example of the exponential race problem.) For large $$n$$, this is asymptotically normal with mean $$\sum_{i=1}^{n}\frac{n}{i} \sim n \log n$$ and variance $$\sum_{i=1}^{n}\frac{n^2}{i^2} \sim \zeta(2)n^2$$. This tells that $$T \sim n \log n$$ in probability, and so, it is not unreasonable to expect that

$$M_n = \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_T \Bigr] \sim \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \Bigr]. \tag{1}$$

holds as well.

3. Note that each $$N^{(i)}_{n\log n}$$ has a Poisson distribution with rate $$\log n$$. Now, we fix $$\theta > 1$$ and choose an integer sequence $$(a_n)$$ such that $$a_n > \log n$$ and $$a_n/\log n \to \theta$$ as $$n \to \infty$$. Then

\begin{align*} \mathbf{P}\Bigl( N^{(i)}_{n\log n} > a_n \Bigr) &\asymp \frac{(\log n)^{a_n}}{a_n!}e^{-\log n} \\ &\asymp \frac{(\log n)^{a_n}}{(a_n)^{a_n} e^{-a_n+\log n} \sqrt{a_n}} \\ &\asymp \frac{1}{n^{\psi(\theta) + 1 + o(1)}}, \qquad \psi(\theta) = \theta \log \theta - \theta, \end{align*}

where $$f(n) \asymp g(n)$$ means that $$f(n)/g(n)$$ is bounded away from both $$0$$ and $$\infty$$ as $$n\to\infty$$. This shows that

$$\mathbf{P}\Bigl( \max_{1\leq i\leq n} N^{(i)}_{n\log n} \leq a_n \Bigr) = \biggl[ 1 - \frac{1}{n^{\psi(\theta) + 1 + o(1)}} \biggr]^n \xrightarrow[n\to\infty]{} \begin{cases} 0, & \psi(\theta) < 0, \\[0.5em] 1, & \psi(\theta) > 0. \end{cases}$$

So, by noting that $$\psi(e) = 0$$, we get

$$\max_{1\leq i\leq n} N^{(i)}_{n\log n} \sim e \log n \qquad \text{in probability}. \tag{2}$$

This suggests that $$M_n \sim e \log n$$ as well.

4. I believe that this heuristic argument can be turned into an actual proof. To this, we need the following:

1. We have to justify the relation $$\text{(1)}$$ in an appropriate sense. We may perhaps study the inequality

$$\max_{1\leq i\leq n} N^{(i)}_{(1-\varepsilon)n\log n} \leq \max_{1\leq i\leq n} N^{(i)}_T \leq \max_{1\leq i\leq n} N^{(i)}_{(1+\varepsilon)n\log n} \qquad \text{w.h.p.}$$

and show that the probability of the exceptional event is small enough to bound the difference of both sides of $$\text{(1)}$$ by a vanishing qantity.

2. Note that $$\text{(2)}$$ alone is not enough to show this. So, we need to elaborate the argument in step 3 so that it can actually prove the relation $$\mathbf{E}\bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \bigr] \sim e \log n$$.