Weakening demands on Law of Large Numbers I'm reading about the Law of Large Numbers in the book Probability Theory: Independence, Interchangeability, Martingales, by Yuan Shih Chow and Henry Teicher:

Theorem 4 (Feller): If $\{X_n\}$ are i.i.d then $ (S_n-C_n)/n $ converges in probability to $0$ for some choice of $(C_n)$ iff $nP(|X_1|>n) \rightarrow 0$. And then $\lim C_n/n=E(X_1)$.

  But what if $X_n$ has no expected value?

Assume that $ nP(|X_1|>n)\rightarrow 0$. Is $C_n/n$ still converging to some limit, and what is the limit?

My first though was "this must be wrong", so I constructed a simple, positive valued distribution
$$
X=\frac{1}{U\cdot\ln(1+1/U)}\text{ where }U\sim\text{Uniform}[0,1]
$$
which satisfies $n\cdot\Pr[X>n]\rightarrow 0$ as $n\rightarrow\infty$, but has $\text{E}[X]=\infty$.

The following are observations, not proofs, but should be sufficient to understand what is going on.

If you sample values of $S_n/n=(X_1+\cdots+X_n)/n$, they tend to be clustered together around a value $C_n/n$, but with some extreme values in the upper tail.

As $n$ increases, the centre of the cluster $C_n/n$ will increase slowly without converging, while the distribution will become narrower (in the sense that for any $a>0$ you have $\Pr[|S_n/n-C_n/n|>a]\rightarrow 0$ as $n$ increases). In other words, $S_n/n-C_n/n$ will converge in probability to zero, but $C_n/n\rightarrow\infty$ as $n\rightarrow\infty$.

The criterion $n\cdot\Pr[X>n]\rightarrow 0$ is important because it ensures (almost surely) that the jumps in the sequence $S_n/n$ will be smaller and smaller as $n$ increases. However, when the expected value is infinite (or undefined), these "jumps" caused by the extreme values of the distribution are sufficiently frequent to ensure $S_n/n$ will keep increasing rather than converge.

If $X_1$ does not have an expected value, then the law of large numbers does not apply and you will find  $nP(|X_1|>n) \not \rightarrow 0$ in probability.

For example, with a standard Cauchy random variable it converges to $2/\pi$.


Orthogonality proof by row-column partioning I'm stuck at the following problem:

Let $A$ be a $n \times n$ matrix. Show that if $x \in Null(A)$ and $b \in col(A^{T})$, then $b$ and $x$ are orthogonal.

Here is my approach:

$x \in Null(A)$ means: $Ax=0$ and also $x^{T}A^{T} = 0$.

By $Ax=0$ we have $\displaystyle\sum_{k=1}^{n} col_{k}(A) row_{k}(x) = 0 ~~~~~~~~~~~~~~~~~(*)$.

By $x^{T}A^{T} = 0$ we have $\displaystyle\sum_{k=1}^{n} col_{k}(x^{T}) row_{k}(A^{T}) = 0 ~~~~~~(**)$.

Since $b \in col(A^{T})$, we can conclude $b^{T} \in row(A)$.

This is the point that I cannot proceed anymore, since $b$ and $b^{T}$ correspond to $col(A^{T})$ and $row(A)$, but $(*)$ and $(**)$ don't include them, so I have no idea how to replace $b$ or $b^{T}$ into $(*)$ or $(**)$.

Since I am looking for orthogonality, I must end up with something like this in reverse:

$b.x=0 \Leftarrow b^{T}x=0 \Leftarrow \displaystyle\sum_{k=1}^{n} col_{k}(b^{T}) row_{k}(x) = 0 ~~~~~~(***)$, but I can't build $(***)$.

Any help?

Let me make it clear. Define:

$Col_k(A)$ is k-th column vector of matrix A, n-by-1 vector
$Row_k(A)$ is k-th row vector of matrix A, 1-by-n vector
$A_{ij}$ is i-th row, j-th column element of matrix A
$Col(A)$ is column space of matrix A
$Row(A)$ is row space of matrix A.

the basis of $Col(A^T)$ are n $Row_k(A)$ vectors. Since $b \in Col(A^T)$, so b is linear combination of these n $Row_k(A)$ vectors. That is:

$b=\sum_{k=1}^nc_kRow_k(A)$  (b is 1-by-n vector)

Since $x \in Null(A)$, so $Ax=0$, where $x$ is n-by-1 vector. Therefore

$\sum_{k=1}^nA_{kj}x_j=0$ for every k. That is $Row_k(A).x=0$ for every k.

So

$b.x=\sum_{k=1}^nc_kRow_k(A).x$=$\sum_{k=1}^nc_k*0=0$

b and x are orthogonal.


Find the Maclaurin series representation of the function: $\ln(\frac{1+x}{1-x})$ Find the Maclaurin series representation of the function: $\ln(\frac{1+x}{1-x})$ I know the answer, but i want to know why my approch did not work.

I started with the series
$$\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n $$
then i put in $x=\frac{2x}{1-x}$ and get:
$$\ln(1+x)=\ln(1+\frac{2x}{1-x})=\ln(\frac{1+x}{1-x}))=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(\frac{2x}{x-1})^n$$
It seems like this is not correct based on wolfam alpha. Where did i make a mistake?

This is not formally incorrect (although you would have to define in what sense the series converges), it is simply not a Maclaurin series. A Maclaurin series is expressed in powers of $x$ with constant coefficients, here you express it in inverse powers of $1-x$ which is not the form of the series. You could try to expand each function $\left(\frac{2x}{x-1}\right)^n$ but that is unnecessarily complicated and dangerous.
Much simpler, write $\ln \left( \frac{1+x}{1-x} \right)=\ln (1+x)-\ln (1-x)$, develop each function separately and check which terms cancel out to find the answer. I will let you conclude!


How many different sudoku puzzles are there? The answer: 6, 670, 903, 752, 021, 072, 936, 960, according to this site:
https://www.technologyreview.com/2012/01/06/188520/mathematicians-solve-minimum-sudoku-problem/
I have tried to get this number using direct methods but basically I have found the question too hard. There are too many possibilities, but I am likely missing some strategic ways to solve the problem. Any help would be greatly appreciated, thanks.

Based on the equation (9!^2(9)), the number of possible sudoku puzzles is exactly 1,185,137,049,600.


Is it appropriate to call the circular edges of a cylinder "corners"? I recently corrected someone when they were describing a machining operation on a lathe. They referred to the circular edge of a cylinder as a "corner" and I got confused.

Then, thinking more about this, I realized that maybe it is indeed a corner mathematically speaking and maybe they have a better intuition about geometry than I do.

The ambiguity of the term 'corner' is why we tend not to use it in mathematics. Instead we talk about 'vertices' and 'edges'. A cylinder does not have any vertices, but it does have two edges.

Before deciding whether it is mathematically appropriate to call the edge a 'corner' you would have to define in terms of mathematics what you mean by a corner. Once you have done that, you could then start deciding whether or not the edge is a corner. Until then the question has no mathematical answer.

I don't think that's the correct terminology. For any figure of that sort we generally use the term vertex, edge etc. The question is more language based than mathematical. More essentially a cylinder has no vertices. So that may not be the correct term.


If the square root of two is irrational, why can it be created by dividing two numbers? I'm doing pre-calculus and got a bit caught on this question… I looked online and it said that a number is rational if it can be the quotient of two integers. So I did this:

I know that it is irrational because if this was actually rational, mathematicians would have figured out. So my question is, why is the square root of two irrational?

Squaring the number you posted:

$$1.41421356237 \cdot 1.41421356237 = 1.9999999999912458800169$$

so the result is not $2$, so the number is not $\sqrt{2}$ (but rather a rational approximation thereof).

"Does that mean all irrational numbers are infinite?"

The irrational numbers are exactly those numbers that require infinitely many digits right of the radix point, no matter which base you are using. (Thanks Beanluc: Normally we say "decimal point" but that would obviously wrong in bases other than base 10, so the generic term is "radix point". )

A number like 1/3 needs infinitely many digits in base 2, 10, 16 but not in base 3, 6, 9, 12. $2^{1/2}$ needs infinitely many digits in every base.


$Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$ $Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$

We know:

$E(Z) = 0$

$\operatorname{Var}(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$

Thus, we need:

$E(Z^2) = \int_{-\infty}^{\infty} z^2 \frac{1}{\sqrt{2\pi}}e^{-1/2(z^2)}dz$

Using int by parts, $u = z, du = dz$, then $dv = z \frac{1}{\sqrt{2\pi}}e^{-1/2z^2}dz$, $v = -\frac{1}{\sqrt{2\pi}}e^{-1/2z^2}$

Then we have

$$\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-1/2z^2}dz$$

I'm aware that $\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} = (0-0) = 0$

No clue how to proceed.

Using Fubini's theorem you can compute $(\int e^{-x^2})^2$=$\pi$ and in the same way that $\int e^{-\frac{x^2}{\alpha}}=\sqrt{\alpha\pi}$, which gives you the result you were looking for.

The answer is by definition of the law 1.

Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.


Estimating P(X $\ge$ k) with Chebyshev's inequality I have managed to derive non-rigorously that

P(|X - E[X]| $\ge$ a) $\le$ $\frac{E[X - E[X]|^2}{a^2}$. for a random variable X.

Now let X be a random variable with Poisson distribution, with mean and variance equal to $\lambda$.
I am supposed to use this inequality to place an estimate on P(X $\ge$ k) where k $\ge$ 0.

Initially I tried to do this by noting that P(X $\ge$ k) = 1 - P(X < k) = 1 - P(0 $\le$ X < k) and then writing this as an inequality, but this hasn't led anywhere.

Any help would be greatly appreciated.

Chebyshev's inequality will be useful if $k \gg \lambda = E(X)$. In this case, you can write
$$
P(x \ge k) = P(X -\lambda \ge k-\lambda)\le \frac{E(X - \lambda)^2}{(k-\lambda^2)} = \frac{\lambda}{(k-\lambda)^2},
$$
which gives something nontrivial if $k \gg \lambda.$ If $k $ is much smaller than $\lambda$ then its not possible to get an upper bound which is useful. For example if $k=0$ then $P(X \ge 0) =1$. It would be helpful to know what kind of estimate you want exactly to give a more precise answer.

You want a more general inequality: $P(|X|\geq a)\leq E[|X|]/a$ or $P(|X|^k\geq a^k)\leq E[|X|^k]/a^k$. Chebyshev's inequality is a consequence of this.


When is a sum of consecutive squares equal to a square? We have the sum of squares of $n$ consecutive positive integers: $$S=(a+1)^2+(a+2)^2+ ... +(a+n)^2$$ Problem was to find the smallest $n$ such, that $S=b^2$ will be square of some positive integer. I found an example for $n=11$. Now, I'm trying to prove, that if $2<n<11$ there is't any solution. So I need some help with this: If $n,a,b \in \mathbb{N}$ and $n<11$ prove that equation

$$
n\cdot a^2 + n(n+1)\cdot a + \frac{1}{6}n(n+1)(2n+1)=b^2
$$

can't be solved. Or if I'm wrong, find counterexample.

The only idea I have is: to consider the remains $\operatorname{Mod}[b^2,n]$ and $\operatorname{Mod}[\frac{1}{6}n(n+1)(2n+1),n]$. For each $2<n<11$, but it is very long.

Below is a reasonable (but not very illuminating) proof. Put $S_n(x)=\sum_{k=1}^{n} (x+k)^2$. Note that it is also true that for $2<n<11$, $S_n(x)$ is never a square modulo $n^2$, and $S_n(x)$ is also never a square modulo $900$. Perhaps this will inspire others to produce more intelligent proofs.

Here it goes :

$$
\begin{array}{ccll}
S_3(x) & \equiv & 3x^2+12x+14 \equiv 2 & {\sf mod} \  3 \\
S_4(x) & \equiv & 4x^2+20x+30 \equiv 2 & {\sf mod} \  4 \\
S_5(x) & \equiv & 5x^2+30x+55 \equiv 2 \ \text{or} \ 3 & {\sf mod} \  4 \\
S_6(x) & \equiv & 6x^2+42x+91 \equiv 3 & {\sf mod} \  4 \\
S_7(x) & \equiv & 7x^2+56x+140 \equiv 3,8,11 \ \text{or} \ 12 & {\sf mod} \  16 \\
S_8(x) & \equiv & 8x^2+72x+204 \equiv 2,5,6 \ \text{or} \ 8 & {\sf mod} \  9 \\
S_9(x) & \equiv & 9x^2+90x+285 \equiv 6 & {\sf mod} \  9 \\
S_{10}(x) & \equiv & 10x^2+110x+385 \equiv 5,10 \ \text{or} \ 20& {\sf mod} \ 25  \\
\end{array}
$$

Each time, we see that $S_n(x)$ is never a square for the given modulus.

There is a solution for n=2, namely:
n=2; a=2,b=5.
since 3 squared + 4 squared = 5 squared,
proof: 9+16=25.
However I noticed n=2 was excluded in the equality stating n less than 11. It is not clear to me why n=2 is excluded.


Plane through the intersection of two given planes. If I have two planes $$\mathbf r\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r\cdot\mathbf {\hat n_2}=p_2$$
If they intersect somewhere then the intersection will form a line, if $\mathbf r'$ is the position vector of a point on the intersection line then I have $$\mathbf r'\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r'\cdot\mathbf {\hat n_2}=p_2$$
If I combine these two equations I would get $$\mathbf r'\cdot(\mathbf {\hat n_1}+\mathbf{\hat n_2})=p_1+p_2$$
This looks like a vector equation of a plane.
But if we see in internet or in books we would find there is scalar $\lambda$  multiplied.
$$\mathbf r'\cdot(\mathbf {\hat n_1}+\lambda\mathbf{\hat n_2})=p_1+\lambda p_2$$

So my question is why there is $\lambda$ there? Please also give me the geometrical image of the plane which is passing through the intersection of those two planes. I'm not able to imagine how would that plane look like.

We have to make some considerations:

The intersection of two planes could be empty if the planes are parallel, but you already state that you assume an intersection, so this is not a problem.
The intersection of two planes which are the same is just the plane itself. We will deal with this case later.
Suppose now you have two distinct, non-parallel planes. You write the equations of each plane as $\vec{r}\cdot\hat{n}_1=p_1$ and $\vec{r}\cdot\hat{n}_2=p_2$. Now, if I multiply each of these equations by a constant, the equations remain true. For instance, $\vec{r}\cdot\hat{n}_1=p_1$ implies $\vec{r}\cdot(A\hat{n}_1)=Ap_1$, and similarly I can get $\vec{r}\cdot(B\hat{n}_2)=Bp_2$.
These two planes are distinct and non-parallel, so they intersect in a line. As you say, points on this line have to satisfy both plane equations simultaneously, so I can describe the line by the system of equations
$$
\begin{cases}
\vec{r}\cdot(A\hat{n}_1)=Ap_1
\\\vec{r}\cdot(B\hat{n}_2)=Bp_2
\end{cases}.
$$
As you also pointed out, the combination of these equations
$$\vec{r}\cdot(A\hat{n}_1+B\hat{n}_2)=Ap_1+Bp_2$$
looks like the equation of a plane for given $A$ and $B$, because it is so. This is not the equation of a line, many more points satisfy it. Let us divide through by $A$ (assuming it is not zero), and denote $B/A$ by $\lambda$. We obtain:
$$\vec{r}\cdot(\hat{n}_1+\lambda\hat{n}_2)=p_1+\lambda p_2.$$
Once again, for a fixed $\lambda$, this is the equation of a plane.
However, we obtain a line if we require that this equation is true for any value of $\lambda$. Remember that the values of $A$ and $B$ were arbitrary, and therefore so is $\lambda$. The intersecting line should satisfy this new equation for all $\lambda$. For instance, if I choose $\lambda=0$ and $\lambda=1$, I recover the system:
$$
\begin{cases}
\vec{r}\cdot\hat{n}_1=p_1
\\\vec{r}\cdot(\hat{n}_1+\hat{n}_2)=p_1+p_2
\end{cases},
$$
which is equivalent to the system we had in point 4.
This is all still true if the two planes are the same, but in this case the system of equations is redundant because $\hat{n}_1=\hat{n}_2$ and $p_1=p_2$. This gives
$$
\begin{cases}
\vec{r}\cdot(A\hat{n}_1)=Ap_1
\\\vec{r}\cdot(B\hat{n}_1)=Bp_1
\end{cases}
$$
and
$$\vec{r}\cdot[(\lambda+1)\hat{n}_1]=(\lambda+1) p_1,$$
which only define the same plane again, not a line.
As for the picture, I found this image online:

The red and green planes intersect in a line, and the blue one is another plane which passes through the line.

I hope this helps!

Lambda is used for direction here
We can see the plane passes through there intersection can be perpendicular to them or at 120 degrees, so its purpose is to give different planes for different values of lambda.


how to show that a linear transformation exists between two vectors? Is there a linear transformation $T$ from $\Bbb{R^3}$ into $\Bbb{R^2}$ such that $T(1,-1,1)=(1,0)$ and $T(1,1,1)=(0,1)$

My try:
According to me i think that I have to show that $(1,-1,1),(1,1,1)$ form a basis but according to my computations they don't. Now since they don't form a basis i cannot write the vector (1,0,0) in terms of these two vectors. So because of this at least there will  one vector in $\Bbb{R^3}$ who wont have a linear transformation in $\Bbb{R^2}$ since by the given mapping i cannot represent the standard basis vectors. Therefore there isnt a linear transformation with such kind of a mapping. Am i right?

sorry but i found a duplicate over here
 :-is there any transformation T from $\Bbb R^3 \to \Bbb R^2$ for this

Extend the two given vectors in $\mathbb R^3$ to a basis $B$ of $\mathbb R^3$ by arbitrarily adding a third linearly independent vector (say, $\vec e_1$) that is mapped to any other vector (say, $\vec 0$) by $T$. Express each vector in the standard basis for $\mathbb R^3$ as a linear combination of the vectors in $B$ and use the linearity of $T$ to find each column of the standard matrix for $T$. For example, notice that since:
$$
\frac{1}{2}\begin{bmatrix}
1 \\ -1 \\ 1
\end{bmatrix}
+ \frac{1}{2}\begin{bmatrix}
1 \\ 1 \\ 1
\end{bmatrix}
- \begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 1
\end{bmatrix}
$$
It follows that the third column of the standard matrix is:
$$
\frac{1}{2}\begin{bmatrix}
1 \\0
\end{bmatrix}
+ \frac{1}{2}\begin{bmatrix}
0 \\ 1
\end{bmatrix}
- \begin{bmatrix}
0 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
1/2 \\ 1/2
\end{bmatrix}
$$
Indeed, it's not too hard to verify that the linear tranformation $T\colon \mathbb R^3 \to \mathbb R^2$ defined by $T(\vec x) = A\vec x$ where:
$$
A = \begin{bmatrix}
0 & -1/2 & 1/2 \\
0 & 1/2 & 1/2
\end{bmatrix}
$$
satisfies the desired property.

define $T$ by $$T(1,-1,1)=(1,0), T(1,1,1) =(0,1),T(1,0,0)= (0,0).$$ observe that $a=(1,-1,1),b=(1,1,1),c=(1,0,0)$ forms a basis for $R^3.$ now extend $T$ by defining $$T(\alpha a + \beta b + \gamma c)= \alpha T(a) + \beta T(b)$$ and you have your linear transformation $T.$


Law of total expectation and Information. I am reading a paper by Glostom and Milgrom on "Bid/Ask transaction prices". Here is the gist of the issue I am having.  $F_t$ is the current information / filtration. $Z_t$ is a random variable. $A$ is the ask price. $V$ is the random unknown fair value of a stock.  The claim is below.

$E[V 1_{\{Z_t > A\}} | S_t] = E[V | S_t, \{Z_t > A\}] P(Z_t > A | S_t)$

I kinda see that this is the law of total expectation. I have seen and know how to prove when the conditioning is over single sets, I am not sure how I can extend a measure theoretic argument over to this situation. In general how does one prove this to conditioning over any set of information sets.

What is $S_t$ in this situation?

Edit: If $S_t$ is also a filtration and $Z_t$ is $S_t$-measurable I don't really see what the conditioning $S_t,\{Z_t>A\}$ is supposed to mean. I guess they probably mean the random variable $Y=E[V|S_t,\{Z_t>A\}]$ which is $0$ outside of $\{Z_t>A\}$ and equal to $c\cdot E[V|S_t]$ inside of $\{Z_t>A\}$. Then the question of how to choose this $c$ value is directly related to your question. $c$ needs to make $Y$ a random variable, i.e. $E[Y]=1$, and for that you need to inflate it to account for the "lost" expectation by setting $Y=0$ outside of $\{Z_t>A\}$. Therefore $c=\frac{1}{P(Z_t>A|S_t)}$. If you enter this into your equation you will see that the equation holds.


Integral of $xe^y \,dy$ I tried integration by parts but that leaves me with $xe^y$ - integral of $e^y \,dx$.

As far as I know that should leave me with $xe^y - xe^y$ which is $0$. My book says its $xe^y$.

It doesn't seem like substitution is going to work because I have two variables.

If you can explain this to me I'd appreciate it.

Of course it is $xe^y$:$$\int xe^y\,\mathrm dy=x\int e^y\,\mathrm dy=xe^y.$$In this problem, $x$ is a constant.

If $x$ does not depend on $y$,

$$\int xe^ydy=xe^y+C$$

and if it does

$$\int x(y)\,e^ydy$$

cannot be solved by lack of information on the relationship.


For each positive integer $n$ , show that there are more than $n$ positive primes. Suppose a picked the positive integer 7. Of course there are more than 7 prime numbers in Z. I just don't know how to show that. How do you show that?

The Wikipedia article on "Euclid's Theorem" (https://en.wikipedia.org/wiki/Euclid%27s_theorem) gives several proofs of the theorem which are very accessible. Some of them were entirely new to me. (Does everyone but me know of the beautiful proof by Furstenberg?)

P.S. I can't comment on this site, but i just saw your question about the contradiction at the heart of Euclid's proof. It's very simple: at the beginning you assume that there are k primes -- what is meant is that there are EXACTLY k primes (k is some particular number). But then Euclid produces a number which has a non-zero remainder when divided by every one of the k primes. If there were only exactly k primes, then that new number would have to be divisible by one of them. But it's not -- so either there is some OTHER number, not among the list of k primes, that is a prime divider, or that number itself is prime. Either way, it's just not true that there are exactly k primes.

P.P.S. Not that I particularly care, but I notice that within seconds of posting this answer, there were two down votes. I found that strange: I actually answered his actual question, which was "how do you show that?" There are many many ways to show the result, only one of which is ascribed to Euclid. I gave a link to a wide variety of ways to "show" that. And then I answered his particular question about how Euclid showed that. I can only assume that the down votes were because the original hyperlink led to some error page. Well, that error page was erroneous (you can't believe everything you read in Wikipedia :)) -- there IS an article in Wikipedia with the exact name I gave. In any case, the link is now correct. Perhaps the hater-voters can unhate their previous votes, or explain the hate. Thx!


Optimization problems: Finding the optimal path I'm still trying to get the hang of optimization problems in calculus and I'm looking for a little help. I'm having trouble finding equations to model the following problem: I'm fairly sure I need to combine two variables into a single variable and set up a related rates problem, but I'm not sure how to get started.

The problem is as follows: Alaina wants to get to the bus stop as quickly as possible. The bus stop is across a grassy park, 2000 feet west and 600 feet north of her starting position. She can walk west along the edge of the park on the sidewalk at a speed of 6 ft/s. She can also travel through the grass in the park, but only at a rate of 4 ft/s. What path will get her to the bus stop the fastest?

Could someone help set it up for me? Thanks!

EDIT

To solve your question:

The path you should take is a combination of both traveling on the sidewalk and through the grass.

Let's start:

$$T(x) = \frac{\text{Distance through grass}}{\text{Rate through grass}} + \frac{\text{Distance on sidewalk}}{\text{Rate on sidewalk}}$$

$$T(x) = \frac{2000-x}{6} + \frac{\text{Distance through grass}}{4}$$

And

$$600^2 + x^2 = (\text{Distance through grass})^2$$

$$\text{Distance through grass} = \sqrt{360000 + x^2}$$

$$T(x) = \frac{2000-x}{6} + \frac{\sqrt{360000 + x^2}}{4}$$

$$T'(x) = \frac{-1}{6} + \frac{x}{4\sqrt{360000 + x^2}} = 0$$

Solving, we get $x = 240\sqrt{5}$

And our distances are:

$$\text{Through Grass} = 360\sqrt{5}$$
$$\text{On sidewalk} = 2000 - 240\sqrt{5}$$

Comment if you need further explanation.

Also take a look at this, it might help.

Compare the travel time along the two sides of the right triangle (2000 ft+600 ft @ speed of 6 ft/s) versus the travel time along the hypotenuse of the right triangle (2088.06 ft @speed of 4 ft/s).

It is faster to travel along the edge of the park on the sidewalk, besides being grass-friendly!


Nonmeasureable subset of ${\mathbb{R}}^2$ such that no three points are collinear? I'm exploring the properties of sets in the plane that do not contain a set of three collinear points.  In particular, I'm interested in the "largest" they can be.

Things I know so far:

Assuming the axiom of choice, one can use Zorn's lemma to show that there are maximal subsets of ${\mathbb{R}}^2$ with this property.

A circle is an easy example of such a set that is uncountable.

If a Lebesgue measurable set $S \subseteq {\mathbb{R}}^2$ has this property, then it contains at most two points on any vertical line.  So the integral of $\chi_S$ in the $y$-direction is 0, and by Tonelli's theorem $m(S) = 0$.  But this argument doesn't work if $S$ is nonmeasurable.

So the question arises:  Is there such a set that is Lebesgue nonmeasurable?  (Since there are models of ZF in which every subset of ${\mathbb{R}}$ is Lebesgue measurable, this might actually depend on the choice of set theory axioms.)

edit:  One line of thought I am pursuing is thinking of Lebesgue nonmeasurable sets as sets with positive outer measure.  Also, one can generalize the question to consider subsets of ${\mathbb{R}}^n$ that do not contain a set of 3 collinear points, or perhaps $n+1$ co-hyperplanar points.  For these questions the case ${\mathbb{R}}^1$ is trivial, and maybe there's something different about dimensions higher than 2 à la Banach-Tarski.

I didnt understand many of the things written by you, like why the integral of $\chi_S = 0$. Anyways, as far as non-measurable set of $\mathbb{R}^2$ with no three points collinear is concerned, consider this:

Let $N$ be a non-measurable set of $[0,1)$. Let $\phi:[0,1)\longrightarrow \mathbb{R}^2$ such that $\phi(t) = (\cos(2\pi t), \sin(2\pi t))$, then $\phi(N)$ is non-measurable too. Also no three points of $\phi(N)$ are collinear. See if this example helps you.


Maximization of Utility Function Kabir’s utility is $U(c,d,h) = 2c+5d-d2-2h$, where $d$ is the number
  of hours per day that he spends driving around, $h$ is the number of
  hours per day spent driving around by the other people in his home
  town and $c$ is the amount of money he has left to spend on other
  stuff besides petrol and auto repairs. Petrol and auto repairs cost
  Rs.$50$ per hour of driving . All the people in Kabir’s home town have
  the same tastes. If each citizen believes that his own driving will
  not affect the amount of driving done by the others, they will all
  drive $D1$ hours per day. If they all drive the same amount,they would
  all be best off if each drove $D2$ hours per day, where

  (a) $D1=2$ and $D2=1$

  (b) $D1=D2=2$

  (c) $D1=4$ and $D2=2$

  (d) $D1=5$ and $D2=0$

I used the formula $$U=2(\bar{M}-50d)+5d-d^{2}-2h$$

and maximized w.r.t $d$ but it's giving an answer: $D1=\frac{5}{102}$

Any help will be appreited.

petrol and auto repairs cost 0.5 (not 50). This means $U=2(M−0.5d)+5d−d^2−2h$ and maximization w.r.t "$d$" gives $D1=2$. The formula for $D2$ is $U=2(M−0.5d)+5d−d^2−2d$, where "$h$" is replaced by "$d$". Maximization gives $D2=1$. Both formulas assume the budget constraint $0.5d+c=M$.


We have 1000 apples and 900 of them are "Good".If we distribute these apples into some equal groups ,what's the minimum percentage of groups ... We have 1000 apples and 900 of them are "Good".If we distribute these apples into some equal groups ,what's the minimum percentage of groups in which the majority of apples are "Good"?

I think the answer must be 90%,but the formal answer says it is: 80% !! I don't get it! because overall 90% of apples are "Good"! I tried many different equal groupings...
(Origin: Iranian National Mathematical Olympiad,1st round,2016)

We divide the 1000 apples in $n$ groups. In order for a group to not have a majority of good apples, there must be $\lceil \frac{1000}{n} \times \frac{1}{2} \rceil = \lceil \frac{1000}{2n} \rceil$ bad apples in the group. Since we have 100 apples, the maximum number of groups for which there are at least $\lceil \frac{1000}{2n} \rceil$ bad apples equals $\bigg\lfloor \frac{100}{\lceil \frac{1000}{2n} \rceil} \bigg\rfloor = \frac{n}{5}$ for $n | 500$ and $5 | n$. Since there are $n$ groups, the maximum percentage of groups for which there is no majority of good apples thus equals $\frac{\frac{n}{5}}{n} = \frac{1}{5} = 20\%$. As such, at least $80\%$ of groups contain a majority of good apples.

The following table shows the ratio of groups not containing a majority of bad apples, for all divisors of 1000 (multiply row and column value):

$$
\begin{array}{r | c c c c}
 & 1 & 2 & 4 & 8 \\
\hline
  1 & 0.000 & 0.000 & 0.000 & 0.125\\
  5 & 0.200 & 0.200 & 0.200 & 0.175\\
 25 & 0.200 & 0.200 & 0.200 & 0.167\\
125 & 0.200 & 0.200 & 0.200 & 0.100
\end{array}
$$

Let $g$ and $b$ represent the numbers of good and bad apples in one box (group). There could be 3 cases: $g>b$, $g=b$ or $g<b$. The minimum percent of good boxes means maximum bad boxes. Note that a bad box is when $b\ge g$. Having $b>g$ in a box is not efficient because extra bad apples can be switched to a good box to make it a bad box. Thus, $b=g=100$, that is 200 in $B$ bad boxes leaving 800 in $G$ good boxes. There are several solutions: $B$=divisors of 100.


If $u,v\in V$ then there is a linear transformation $T:V\to V$ such that $T(u)=v$. Suppose $V$ is a vector space, not necessarily finite dimensional. If $u,v\in V$, is there a linear transformation $T:V\to V$ such that $T(u)=v$.

I proved the result for the finite dimensional case where I used a basis for $V$. What should one do for the infinite dimensional case?

Not always. If $u=0$, then $v$ must be $0$.

If $u\ne0$ and $V$ has an inner product, then $\displaystyle T(x)=\dfrac{\langle x,u\rangle}{\langle u,u\rangle}v$ works.

Geometrically, $T$ projects $V$ onto the subspace generated by $u$, which is canonically isomorphic to the subspace generated by $v$.


How do I convince someone that $1+1=2$ may not necessarily be true? Me and my friend were arguing over this "fact" that we all know and hold dear. However, I do know that $1+1=2$ is an axiom. That is why I beg to differ. Neither of us have the required mathematical knowledge to convince each other.
And that is why, we decided to turn to Math Stackexchange for help.
What would be stack's opinion?

It seems that you and your friend lack the mathematical knowledge to handle this delicate point. What is a proof? What is an axiom? What are $1,+,2,=$?

Well, let me try and be concise about things.

A proof is a short sequence of deductions from axioms and assumptions, where at every step we deduce information from our axioms, our assumptions and previously deduced sentences.
An axiom is simply an assumption.
$1,+,2,=$ are just letters and symbols. We usually associate $=$ with equality; that is two things are equal if and only if they are the same thing. As for $1,2,+$ we have a natural understanding of what they are but it is important to remember those are just letters which can be used elsewhere (and they are used elsewhere, often).

You want to prove to your friend that $1+1=2$, where those symbols are interpreted as they are naturally perceived. $1$ is the amount of hands attached to a healthy arm of a human being; $2$ is the number of arms attached to a healthy human being; and $+$ is the natural sense of addition.

From the above, what you want to show, mathematically, is that if you are a healthy human being then you have exactly two hands.

But in mathematics we don't talk about hands and arms. We talk about mathematical objects. We need a suitable framework, and we need axioms to define the properties of these objects. For the sake of the natural numbers which include $1,2,+$ and so on, we can use the Peano Axioms (PA). These axioms are commonly accepted as the definition of the natural numbers in mathematics, so it makes sense to choose them.

I don't want to give a full exposition of PA, so I will only use the part I need from the axioms, the one discussing addition. We have three primary symbols in the language: $0, S, +$. And our axioms are:

For every $x$ and for every $y$, $S(x)=S(y)$ if and only if $x=y$.
For every $x$ either $x=0$ or there is some $y$ such that $x=S(y)$.
There is no $x$ such that $S(x)=0$.
For every $x$ and for every $y$, $x+y=y+x$.
For every $x$, $x+0=x$.
For every $x$ and for every $y$, $x+S(y)=S(x+y)$.

This axioms tell us that $S(x)$ is to be thought as $x+1$ (the successor of $x$), and it tells us that addition is commutative and what relations it bears with the successor function.

Now we need to define what are $1$ and $2$. Well, $1$ is a shorthand for $S(0)$ and $2$ is a shorthand for $S(1)$, or $S(S(0))$.

Finally! We can write a proof that $1+1=2$:

  $S(0)+S(0)=S(S(0)+0)$ (by axiom 6).
  $S(0)+0 = S(0)$ (by axiom 5).
  $S(S(0)+0) = S(S(0))$ (by the second deduction and axiom 1).
  $S(0)+S(0) = S(S(0))$ (from the first and third deductions).

And that is what we wanted to prove.

Note that the context is quite important. We are free to define the symbols to mean whatever it is we want them to mean. We can easily define a new context, and a new framework in which $1+1\neq 2$. Much like we can invent a whole new language in which Bye is a word for greeting people when you meet them, and Hi is a word for greeting people as they leave.

To see that $1+1\neq2$ in some context, simply define the following axioms:

$1\neq 2$
For every $x$ and for every $y$, $x+y=x$.

Now we can write a proof that $1+1\neq 2$:

$1+1=1$ (axiom 2 applied for $x=1$).
$1\neq 2$ (axiom 1).
$1+1\neq 2$ (from the first and second deductions).

If you read this far, you might also be interested to read these:

How would one be able to prove mathematically that $1+1 = 2$?
What is the basis for a proof?
How is a system of axioms different from a system of beliefs?

Here's an argument to consider. Take the following three propositions:

(A) 1 + 1 = 2
(B) If there is exactly one $F$ and exactly one $G$ (and no $F$ is $G$), then there are exactly two things which are $F$-or-$G$.
(C) $(\exists x)(Fx \land \forall y(Fy \to y = x)$,  $(\exists x)(Gx \land \forall y(Gy \to y = x) \vdash$
$\quad\quad\quad(\exists x)(\exists y)(\neg x = y \land (Fx \lor Gx) \land (Fy \lor Gy) \land (\forall z)(Fz \lor Gz \to (z = x \lor z = y))$

Now, it is a vexed question what exactly the relation is between (A) and (B). But it is very plausible to say that if (A) is indeed interpreted as a statement of ordinary schoolroom arithmetic (and not as a mere de-intepreted string of formal symbols), then so understood, (A) is true iff, for any $F$, $G$, (B) is true. But (B), with numerical quantifiers, can be regimented as (C), with ordinary quantifiers. And (C) is a straight logical theorem of first-order logic.
So (C) is true as a matter of logical necessity, hence (B) is. And it looks a priori that (A)'s truth necessarily goes with the truth of (B) for any $F$, $G$. So it follows that (A), understood the ordinary way, is indeed necessarily true. too.


You are making cookies and add N chips to dough randomly, and split it into 100 equal cookies, again at random. How many chips should go into dough? Question: You are making chocolate chip cookies. You add N chips randomly to the dough and you randomly split the dough into 100 equal cookies. How many chips should go into the dough to give a probability of at least 90% that every cookie has at least one chip?

I tried to attempt to solve this using IID random variables. I am not sure how to set the problem up. I know that there should at least be 100 chocolate chips or else the cookies will not meet the "at least 1 chip per cookie" requirement and that there is 10% chance that the cookies do not have a chip.

There seem to be several interpretations of what you are looking for.
I have revised my answer to make it clear what I am answering, and appended a comment for a different interpretation.

I'd suggest an approximate Poisson model.  If there are $N$ chips in the dough,
then the number of chips in a random cookie is $X \sim Pois(\lambda = N/100).$
You want a 90% probability that a random cookie has at least
one chip, then solve for $N$ in $$1 - P\{X = 0\} = 1 - e^{-N/100} = .90.$$

Addendum: If you want a 90% chance that every one of the 100 cookies has
at least one chip, then that's another problem, to which the
Coupon Collection approach suggested in another Answer is a specific idea that may be
appropriate. There are Comments with
strong opinions about approaches, but no specific suggestions yet.

If there is a requirement that the chocolate chips are evenly (or "equally") distributed, then you need at least 100 chips in order for there to be a non-zero probability of having at least 1 chip in every cookie. But then there will be a 100% probability that every cookie will have a chip in it because the chips are evenly distributed, so $N=100$.

On the other hand, if you believe that the chips were "randomly" distributed, you would need to know the maximum amount of chips which could fit into 1 cookie, and assume that every cookie is the same size. You would also need a function to describe the "randomness" - i.e. if some cookies will get more chips than others and by how much.


Why does linearly independent spanning set imply minimal spanning set for a vector space? Suppose β is a linearly independent spanning set of some vector space V. Why must it be the minimal spanning set?

In other words, why can there not be two linearly independent spanning sets of a vector space V with different sizes?

Suppose $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$ are bases for $V$ where $n<m$.

By writing the $w_i$ in terms of the $v_j$, you create a system $w_i=\sum_j \alpha_{ij}v_j $.

In terms of coordinates, this means that the vector $w_i$ has coordinates $[\alpha_{i1},\alpha_{i2},\ldots,\alpha_{in}]$ in the basis of $v_i$'s. In other words, you have $m$ length $n$ vectors that you believe are linearly independent (since the $w_i$ are linearly independent.) You can put these in a matrix, and keep in mind you believe the rows to be linearly independent.

By using elementary row operations, though, you can put this matrix into row-echelon form, where it must have all zeros in the last row. This says that the last row is a linear combination of the rows above it, and that means that the $w_i$ weren't linearly independent after all. This contradiction shows why the assumption in the first line can never occur.

Just a quick explanation from dimension theorem of vector space.

http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

Suppose you have $v_1$ and $v_2$; they are linearly independent and span $V$. Then they form the basis for $V$. As such the dimension of $V$ is $2$ and according to the dimension theorem, all set of linearly independent vectors that span $V$must be of size $2$ as well.


Help with set proof: $A \cap B = A $ if and only if $A \subseteq B $. $A \cap B = A $ if and only if $A \subseteq B $

It's been a while since I've done this sort of proof. I can't think of how I would prove this statement. I'm too used to numerical proofs.

What kind of proof would I use and how?

Suppose $A\not\subset B$, then there exists an $a\in A$ such that $a\notin B$. Hence $a\notin A\cap B$. Thus $A\cap B\neq A$.

Conversely, if $A\cap B=A$ then $A\subset A\cap B$, so for each $a\in A$ we have that $a\in A\cap B$, which implies that $a\in B$. Thus $A\subset B$.

Let $A\cap B=A$ Show that $A\subseteq B$.

Let $A\subseteq B$, show that $A\cap B=A$.

Where is your problem?


Group Action by Inversion I was reading through the chapter on Cohomology in Dummit and Foote, and I constantly came across the term "action of the cyclic group $C_{2}$ on $G$ by inversion". What does this refer to? Is it an action such that:

$g \cdot e= g \hspace{0.2cm} \forall g \in G \quad g \cdot x=g^{-1} \hspace{0.2cm} \forall g \in G$

For $x$ being the non-trivial element in $C_{2}$? This is a very simple question, but I'm not so sure on what is the correct terminology here.

In general, an action of $C_n$ on an object $G$ of a category (for example a group) is a homomorphism $C_n \to \mathrm{Aut}(G)$, which, by the universal property of $C_n = \langle x : x^n = 1 \rangle$, amounts to an element $\sigma \in \mathrm{Aut}(G)$ with $\sigma^n=1$. Therefore, one usually only specifies $\sigma$. The action is then $x^k \mapsto \sigma^k$.


If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$? If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.

So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer and longer. Am I on the right track? Thanks!

We can write:

$$1+ x + x^2 = \frac{1-x^3}{1-x}$$

Therefore:

$$f(x) = \frac{1-x}{1-x^3} $$

We can then expand this in powers of $x$:

$$f(x) = (1-x)\sum_{k=0}^{\infty}x^{3 k}$$

which is valid for $\left|x\right|<1$. The coefficient of $x^{36}$ is thus equal to $1$, so the 36th derivative at $x = 0$ is $36!$ .

A non power/taylor series method using Leibniz' generalize product rule:

$$ (fg)^{n}(x) = \sum_{r = 0}^{n} \binom{n}{r} f^{n-r}(x) g^{r}(x)$$

Let $f(x) = \dfrac{1}{x^2 + x + 1}$ and $g(x) = x^2 + x +1 $

Then $f(x) g(x) = 1$. We apply the rule to $fg$.

If $a_k = f^{k}(0)$ the we get the recurrence relation

$$a_k + ka_{k-1} + k(k-1)a_{k-2} = 0$$

With $a_{0} = 1$ and $a_1 = -1$

Now let $b_k = \dfrac{a_k}{k!}$, then we get

$$b_k + b_{k-1} + b_{k-2} = 0$$

with $b_{0} = 1$ and $b_{1} = -1$.
Thus we get that the $b_i$ are

$$1, -1, 0, 1, -1, 0, 1, \dots$$

We see that $b_{3k} = 1$, $b_{3k+1} = -1$ and $b_{3k+2} = 0$.

Thus $$a_{36} = b_{36}\cdot 36! = 36!$$


Unique factorization of an element in an UFD By definition:

  An integral domain $R$ is a unique factorization domain if the following conditions are satisfied:

  Every element $a \in R$, $a \neq 0$ that is not a unit can be factored into a product $a = c_1 \cdots c_n$ where $c_1,\dots,c_n \in R$ are irreducible elements.
  If $c_1,\dots,c_n$ and $d_1,\dots,d_m$ are two factorizations of the same element of $R$ into irreducibles, then $n = m$ and $d_j$ can be renumbered so that $c_i$ and $d_i$ are associates.

I need to prove that every element $a \in R$, $a \neq 0$ which is not a unit can be written uniquely as:
\begin{equation}
 a = up_1^{e_1} \cdots p_s^{e_s}
\end{equation}
where $u \in R$ is a unit, $p_1,\dots,p_s \in R$ are irreducible elements mutually not associate and $e_1,\dots,e_s \in \mathbb{N} \setminus \{0\}$. I think I need to start with an arbitrary factorization $a = c_1 \cdots c_n$, then use the following result, but honestly I don't know how to put it formally.

  Let $R$ be an integral domain and let $a,b \in R$. If $a$ and $b$ are associate elements, then $a,b \neq 0$ and $a = b \cdot u$ for some unit $u \in R$.

The action you take is identical to the following situation where you consider words that are monomials in several variables: for example $$(\frac{3}{4}x)(5y)(x)(\frac{5}{3}x)(\frac{2}{5}z)(\frac{1}{4}y) $$
First group all associates: $$(\frac{3}{4}x)(x)(\frac{5}{3}x).(5y)(\frac{1}{4}y). (\frac{2}{5}z)$$
Then for each group of accociates extract a unique unit: $$(\frac{3}{4}\frac{5}{3})(x)(x)(x).(5\frac{1}{4})(y)(y).(\frac{2}{5})(z) $$
Then bring all units together and simplify them and exponentiate the rest:
$$ \frac{5}{8}x^3y^2z $$


What is the prerequisit to studying spectral graph theory and hyper graphs? If one knows only the basics of graph theory Can anyone suggest to me some books to learn spectral graph theory, hyper graph theory, and tensors used to study hyper graphs please? Preferribly books for beginners and then some books for advanced learning.

As a prerequisite to Spectral Graph Theory, you should take a look at the more general Algebraic Graph Theory first and have a solid foundation of Linear Algebra. A book I used is Biggs, N. (1993) ‘Algebraic Graph Theory’ (2nd Ed.). UK: Cambridge University Press.

Below are the books which I think are very useful in learning Algebraic Graph Theory especially the first one (As I learnt a lot form it ).
1. An Introduction to the Theory of Graph Spectra Dragoš M. Cvetković, Peter Rowlinson, Slobodan Simić-  -Cambridge University Press (2010).

Introduction to Graph and Hypergraph Theory  Vitaly I. Voloshin


single valued function and multi valued function square root function is a multi valued function. for example 16 have two square root 4 and -4. but it is violation of the definition of function that each element of domain has one and only one image. then how it is a function. please clear my dout.

for example 16 have two square root 4 and -4.

Yes, that is correct.

  square root function is a multi valued function.

Not quite.  Square roots are different from the square root function.  The definition of $\sqrt{x}$ is not all numbers $y$ satisfying $y^2=x$, but the unique nonnegative number $y$ such that $y^2=x$.

Thus $\sqrt{16} = 4$, not $4$ and $-4$.

With this restriction, $f(x) = \sqrt{x}$ is a function with domain $[0,\infty)$.

$f:[0,\infty)\to[0,\infty)$ defined by $f(x)=\sqrt{x}$ assigns to each $x$ a single value, the positive square root. This is a function.

Similarly, $-f$ is a function.

Functions cannot output more than one value for a given input.


LLN question: convergence of sequence average This is a question relation to the law of large numbers. Let $(X_n:N\in\mathbb{N})$ is as sequence of independent random variables such that $\mathbb{E}(X_n) = \mu$ and $\mathbb{E}(X_n^4) \leq M$ for all n, for some constants $\mu \in \mathbb{R}$ and $M<\infty$. Now set $P_n = X_1X_2+X_2X_3+...+X_{n-1}X_n$. Then I wish to show that $P_n$/$n$ converges almost surely as $n\to\infty$, and I wish to find this limit.

Thanks you so very much!!

This is a homework question, and in class we have looked at the condition $\mathbb{E}(X_n^4) \leq M$ in the context of the law of large numbers, so I'm quite sure this should come in at some point.

My thoughts are the series $P_n$ / $n$ converges to $\mu^2$, but I am uncertain what I need to state to show this.

Recall the following theorem:

  Let $(X_n)_{n \in \mathbb{N}} \subseteq L^2(\mathbb{P})$ be a sequence of independent random variables such that $$\sum_{n=1}^{\infty} \frac{\mathbb{V}X_n}{n^2} < \infty$$ where $\mathbb{V}X_n$ denotes the variance of $X_n$. Then, $$\frac{1}{n} \sum_{i=1}^n (X_i-\mathbb{E}X_i) \to 0 \quad \text{a.s.}$$

Since the random variables are independent and $\mathbb{E}X_j = \mu$, we have

$$\mathbb{E}((X_j+X_{j-1})^2) = \mathbb{E}(X_j^2)+\mathbb{E}(X_{j+1}^2)+2\mu^2$$

hence,

$$\begin{align*}2 X_j X_{j+1} &= (X_j+X_{j+1})^2 - X_j^2-X_{j+1}^2 \\ &= \big[ (X_j+X_{j+1})^2 - \mathbb{E}((X_j+X_{j+1})^2) \big] - \big[ X_j^2-\mathbb{E}(X_j^2) \big] - \big[ X_{j+1}^2 - \mathbb{E}(X_{j+1}^2) \big] +2 \mu^2\end{align*}$$

Therefore, we can write

$$P_n = Q_n^1+Q_n^2+Q_n^3+(n-1)\mu^2 \tag{1} $$

where

$$\begin{align*}
Q_n^1 &:= \frac{1}{2} \sum_{j=1}^{\lfloor \frac{n}{2} \rfloor} \bigg[(X_{2j}+X_{2j+1})^2-\mathbb{E}((X_{2j}+X_{2j+1})^2) \bigg] \\
Q_n^2 &:=\frac{1}{2} \sum_{j=1}^{\lfloor \frac{n}{2} \rfloor} \bigg[(X_{2j+1}+X_{2j+2})^2-\mathbb{E}((X_{2j+1}+X_{2j+2})^2) \bigg] \\
Q_n^3 &:= \sum_{j=1}^n \bigg[ X_j^2- \mathbb{E}(X_j^2) \bigg]+ \frac{1}{2} (X_n^2-\mathbb{E}(X_n^2)) \end{align*}$$

By the previous theorem and the assumption $\mathbb{E}(X_n^4) \leq M$, we know that

$$\lim_{n \to \infty} \frac{1}{\lfloor \frac{n}{2} \rfloor} Q_n^1 = \lim_{n \to \infty} \frac{1}{\lfloor \frac{n}{2} \rfloor} Q_n^2 = \lim_{n \to \infty} \frac{1}{n} Q_n^3 = 0 \quad \text{a.s.} \tag{2}$$

Combining $(1)$ and $(2)$, we find

$$\frac{P_n}{n} \to \mu^2 \quad \text{a.s.}$$

$E(X_nX_{n+1})=E(X_n)E(X_{n+1})=\mu^2$ by the independence.
Then

$E(P_n)=n\mu^2$


basis of neighborhoods of zero in Schwartz Space I have the following question:

  In $S(\mathbb{R}^n)$, and for $\alpha, \beta\in \mathbb{N}^n $, we defined
  \begin{align}
ρ_{\alpha,\beta}(\phi) = \sup_{x\in\mathbb{R}^n}|x^{\alpha}D^{\beta}\phi(x)|.
\end{align}
  We defined a distance as follows: we take a bijection $\sigma: \mathbb{N}\mapsto \mathbb{N}^n\times \mathbb{N}^n$, and set
  \begin{align}
d(\phi,\varphi) = \sum_n 2^{-n}\frac{ρ_{\sigma(n)}(\phi-\varphi)}{1 + ρ_{\sigma(n)}(\phi − \varphi)}
\end{align}
  This distance induces a topology in $S(\mathbb{R}^n)$. Starting from these definitions, prove that the family of sets
  \begin{align}
N_{\epsilon, l, m} = \Big\lbrace \phi\in S\mid \sum_{|\alpha|\leq l, |\beta|\leq m}p_{\alpha, \beta}(\phi)<\epsilon\Big\rbrace
\end{align}
  defines a basis of neighborhoods of $0 \in S$.

I know that what we have to do is to assume that we have a open set $U$ in $S(\mathbb{R}^n)$ containing zero, that is, for every $x\in U$ there exists a disc
\begin{align}
B_{r,x}=\lbrace \phi\in S\mid d(x,\phi)<r\rbrace
\end{align}
such that $x\in B_{r,x}\subseteq U$. Then we have to look if there exist some $N_{\epsilon,l,m}$ that defines a basis for $U$. Thats the point I get lost...

I appreciate any help :)

Writing the basis distance in base 2, we can see that only the first finite number of digits matter for various basis diameters ($\frac{p}{1+p} \in[0,1)$ and increasing in $p$). Those digits are controlled by the magnitudes of the initial subset of $\{p\}$.


List four elements of each of these sets I am currently having trouble understanding these sets.

I am on letter a and I worked out that k is an element of P, where p is composed of positive integers, such that "k-1 is a multiple of 7." No clue what to do with this part.

$\mathbb{P}$ is weird notation for positive integers. (I would read it as the primes). Anyway, you just need to find four positive integers $k$ such that $k-1$ is a multiple of 7. This means $k$ is one more than a multiple of 7. 7 is a multiple of 7, so $k=8$ is in the set. Just think of a few other multiples of 7 and proceed accordingly.

This is my answer:

a)
$\mathbb{P}$ is a notation used for positive integers (and a lot of times a notation for prime numbers), I would take it as prime numbers. So you need to find $4$ prine numbers $p_i,~i\in\mathbb{N},1\le i\le 4$ such that $7\mid(p_i+1)$ so you need to $7q+1,~q\in\mathbb{N}$ be a prime number.

Obviously if $q$ is odd then $7q+1$ is even then it is prime if $7q+1=2\Rightarrow7q=1$ and this is not possible because $q\in\mathbb{N}$.

So $q$ is even lets find $p_i$:
$q=2\Rightarrow7k+1=7\cdot2+1=15$ and it is not prime $(3\mid15)$.
$q=4\Rightarrow7k+1=7\cdot4+1=29$ and it is prime, so $p_1=29$.
$q=6\Rightarrow7k+1=7\cdot6+1=43$ and it is prime, so $p_2$=43.
$q=8\Rightarrow7k+1=7\cdot8+1=57$ and it is not prime $(3\mid57)$.
$q=10\Rightarrow7k+1=7\cdot10+1=71$ and it is not prime $(3\mid57)$.
$q=12\Rightarrow7k+1=7\cdot12+1=85$ and it is not prime $(5\mid85)$.
$q=14\Rightarrow7k+1=7\cdot14+1=99$ and it is not prime $(3\mid99)$.  $q=16\Rightarrow7k+1=7\cdot16+1=113$ and it is prime, so $p_3=113$.  $q=18\Rightarrow7k+1=7\cdot18+1=127$ and it is prime, so $p_4=127$.

Then this are your four elements: $k\in\{29,43,113,127\}$.

b)
$x\in\{$apple, strawberry, pear, peach$\}$

c)
$\displaystyle x\in\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\right\}$

d)
$n\in\{-1,-2,-3,-4\}$

e)
Again $\mathbb{P}$, lets take $n\in\{2,3,5,7\}$.


Given that $x^y=y^x$, what could $x$ and $y$ be? It's not too difficult to figure out that $x$ and $y$ can both be 1, and also $x$ can be 2 and $y$ can be 4 (and vice versa). But I can't rule out if there are other solutions. Does it have anything to do with inverse functions? Is there a way to see the solutions by graphing, or algebraically?

Here's a good way to look at this:
$$
x^y = y^x \implies \\
y \ln x = x \ln y \implies\\
\frac{\ln x}x = \frac{\ln y}y
$$
So one way of solving this is looking at the graph of $y = \frac{\ln x}{x}$ and seeing where it hits a particular value twice.

However, I'm going to go a step further.  I will say that I want my solution $(x,y)$ of $x^y = y^x$ to be of the form $(x,ax)$ so that $y = ax$ for some value $a$.  Furthermore, I'll assume that $x \neq y$ so that $a \neq 1$ (of course, $x = y$ is always a solution if $x^x$ is defined). With that, we have
$$
\frac{\ln x}x = \frac{\ln (ax)}{ax} \implies (\text{assume } x\neq 0)\\
\ln x = \frac{\ln (ax)}{a} \implies\\
\ln x = \frac{\ln (a)}{a} + \frac{\ln x}{a} \implies\\
\frac{a-1}{a}\ln x = \frac {\ln(a)}{a} \implies (\text{we assumed }a \neq 1)\\
\ln x = \frac{\ln(a)}{a-1} \implies\\
x = a^{\frac{1}{a-1}}
$$
So, for any value $a \neq 1$, the pair $x = a^{1/(a-1)},$ $y = a \cdot a^{1/(a-1)} = a^{a/(a-1)}$ will give you a solution to the original equation.

For example, plugging in $a = 2$ gives you $x=2$ and $y = 4$.  Try some other values.

all numbers of the form $x^{1/(x-1)}$, $x>1$ satisfy this.


Finite group with at most one subgroup for each order. Let $G$ be a finite group.
I want to prove the following propositon.

  If for every positive integer $n$, the number of subgroups of $G$ of order $n$ is $1$ or $0$, then $G$ is
  cyclic.

Is my proof right?

$$\begin{align*}
f(d)&＝\text{The number of element of }G\text{ of order }n\\
&\leq(\text{number of subgroups of order }d)\times(\text{number of generators of }G)
\end{align*}$$

Above inequality holds for any finite group $G$.

From assumption, we can deduce
$$f(d)\leq \text{number of generators of }G = \varphi(d)$$

On the other hand, $n$ is the sum of $f(d)$ where $d$ divides $n$.

And the sum is less than the sum of $\varphi(d)$ where $d$ divides $n$. And the last sum is $n$, so, $f(d)=\varphi(d)$.

In particular, $f(n)=\varphi(n)\geq 1$, so $G$ has an element $a$ which has order $n$. Therefore, $G=\langle a\rangle$.

The argument has a bunch of typos and omissions,  and is not very clear, but is essentially correct.

You define $f(d)$ to be the number of elements of order exactly $d$ in $G$. Then we note that if $x$ is an element of order $d$, then so are $x^a$ for any $a$, $1\leq a\leq d$, $\gcd(a,d)=1$; so each element of order $d$ gives you $\varphi(d)$ other elements of order $d$. Now, if $H$ is a subgroup of order $d$, then if it is cyclic it contains exactly $\varphi(d)$ elements of order $d$, and if it is not cyclic, then it contains $0$, and hence fewer than $\varphi(d)$ elements. Each element of order $d$ is counted this way.

That means that $f(d)$ is, at most, the number of subgroups of order $d$ times $\varphi(d)$. The typos in your first equation are:

The $n$ on the first line; and
It's not "generators of $G$", in the second line, it's generators of the subgroup of order $d$.

Now, since by assumption there is at most one subgroup of order $d$, that means that $f(d)\leq\varphi(d)$ for each $d$. That's correct.

Since $|G|=\sum_{d|n}f(d)$, then
$$n = |G|=\sum_{d|n}f(d) \leq\sum_{d|n}\varphi(d).$$
But it is known that $\sum_{d|n}\varphi(d) = n$, so we have equality throughout. In particular, for every $d|n$, $f(d)=\varphi(d)$. That's correct, though poorly worded.

And then you conclude that since $\varphi(n)\geq 1$, then $f(n)\geq 1$, which means $G$ contains at least one element of order $n$, and thus is cyclic. Again, correct.

Let $p_1^{r_1}\cdot p_2^{r_2}\cdots p_n^{r_n}$ be the prime factorization of $m:=\text{exp}(G)$. Each cyclic $G$ subgroup $\langle g\rangle$ must also be a normal $G$ subgroup as it is the unique cyclic subgroup of that order. Choose $g_1,g_2,\dots ,g_n\in G$ such that $\text{ord}(g_i)=p_i^{r_i}$ and compose the internal direct sum $$\langle g_0\rangle:=\langle g_1\rangle\oplus\langle g_2\rangle\oplus\cdots\oplus\langle g_n\rangle$$ Finally, given any $g\in G$ then $m=\text{ord}(g_0)$ is a multiple of $|\langle g\rangle|$ which must equal the size of a $\langle g_0\rangle$ subgroup so that we may conclude $\langle g\rangle\subseteq\langle g_0\rangle$. Therefore $G=\langle g_0\rangle$.


Differentiate $y=x$ with respect to $x^2$? Let $y=x$. Differentiate $y$ with respect to $x^2$.

My intuition tells me to go down a substitution route here. So, let $u=x^2~~ \Leftrightarrow~~x=\sqrt{u}$, then what we are looking to find is simply

$$\frac{d}{du}\sqrt{u}$$
which is
$$\frac{1}{2\sqrt{u}}$$
as $\sqrt{u}=x$,
$$\frac{d}{dx^2}~x=\frac{1}{2x}$$

is this approach/solution correct, or is there something fundamentally wrong here? Any help is appreciated.

Yes it is correct.

Alternatively, from the Chain Rule, $$y=x\implies y^2=x^2\implies \frac{dy^2}{d(x^2)}=\frac{d}{d(x^2)}(x^2)\implies 2y\frac{dy}{d(x^2)}=1\implies \boxed{\frac{dy}{dx^2}=\frac1{2y}=\frac1{2x}}$$

Note that this method is used very often in statistics (e.g. maximum likelihood estimates for the variance since we estimate $\sigma^2$ rather than $\sigma$; hence we prefer to differentiate the former).

Set $z=x^2$. Then $x= \sqrt z$ and:
$$\frac{d}{d(x^2)}y=\frac{d}{d(x^2)}x=    \frac{d}{dz}\sqrt z=   \frac{1}{2 \sqrt z}=  \frac{1}{2x}  $$


Group homomorphism that maps to one element Is there some way to construct a group homomorphism $G \to H$ that maps everything in $G$ to just one non-identity element of $H$ (besides mapping $0$ to $0$)? For arbitrary (finite) $H$ but you have your choice of $G$?

The only example of this is the identity homomorphism $C_2\rightarrow C_2$ (where $C_2$ is the cyclic group of order $2$). Otherwise, if the image of a homomorphism consists of just one non-identity element and the identity, there are multiple elements mapping to the identity.

In particular, let $f:G\rightarrow H$ be a homomorphism. Then, it must hold that, for any $x\in H$ (and letting $1$ be the identity of $H$) such that $f^{-1}(x)$ is not empty:
$$|f^{-1}(x)|=|f^{-1}(1)|$$
- that is, the set of elements of $G$ mapping to $x$ is as large as the set of elements mapping to $1$ under any homomorphism. This is simple to prove, since, if we let $z$ be some element in the preimage $f^{-1}(x^{-1})$ - that is, some element satisfying $f(z)=x^{-1}$ - then it holds that
$$z\cdot f^{-1}(x)=f^{-1}(1)$$
where $z\cdot f^{-1}(x)$ is a coset of $f^{-1}(x)$. This can be proven since, if we apply $f$ to the left side, we would conclude that $f(z\cdot f^{-1}(x))=f(z)x=x^{-1}x=1$ - meaning every element of $z\cdot f^{-1}(x)$ is in $f^{-1}(1)$. Proving the converse - that every element in $f^{-1}(1)$ is in $z\cdot f^{-1}(x)$ is a simple exercise which I will leave to you.

However, what we can prove from this is that if the image of a homomorphism $f:G\rightarrow H$ consists of just two elements, then exactly half of the elements of $G$ map to the identity, and half the elements of $G$ map to the other element. If only one element is permitted to map to the identity, this gives that $1$ is half the order of $G$ - hence $G$ has order two.

Let such a homomorphism $f$ exist and all the non identity elements of $G$ are mapped to the non -identity element $b$ of $H$ .then Im$f$ will consist of $b$ and $e_H$ and ker$f$ will consist only of $e_G$.then by 1st isomorphism theorem

$|G|/|kerf|=|Imf|$

i.e$|G|=|Imf|$ snce ker$f$ consists only of $e_G$

i.e. $|G|=2$ which is the only possibility


How sin(90°+ θ) is equal to M'P'/OP' or Cos θ? I'm learning Trigonometry right now with myself and at current I'm understanding how to find the trigonometric ratio of the angle (90°+ θ) in those of θ. I'm little bit confused right now in the Figure, I didn't get how angle sin(90°+ θ) is equal to M'P'/OP'.And how sin(90°+θ)=sin θ of triangle P'OM']1
[]2Please help. Sorry If I'm asking foolish question. Thankyou in advance.

Just another way we know $sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ so here $a=\theta,b=90$ thus $$\sin(\theta+90)=\sin(\theta)\cos(90)+\cos(\theta)\sin(90)=\cos(\theta)$$


Find all possible values of $(-i)^{i}$. Find all possible values of $(-i)^{i}$.

Here is my attempt:

$(-i)^{i}=\mathit{e}^{ilog(-i)}=\mathit{e}^{i(log|-i|+iarg(-i))}=\mathit{e}^{-\frac{3\pi}{2}}=\mathit{e}^{\frac{\pi}{2}+2\pi n}$, where $n$ is an integer.

However, the answer is $\mathit{e}^{\frac{\pi}{2}-2\pi n}$. Why is $2\pi n$ being subtracted?

Note that the solutions

$$e^{\frac\pi2 - 2\pi n}\>\>\>\>\> \text{and} \>\>\>\>\>e^{\frac\pi2 + 2\pi n}$$

are equivalent for $n \in \mathbb{Z} $.

Sorry, I'm having trouble with Math.Jax but -i is exp(((4n+3)(pi)/2)i) so the answer is exp((-(4n+3)(pi)/2), where n is any integer.


Is $f:\Bbb Q\rightarrow \Bbb R$ continuous? Define $f:Q\rightarrow R$ such that $f= 1$ if  $x^2<2$  or  $f=0$ is  $x^2>2 $. Is this function continuous? All proofs are welcomed but i am more interested using the definitions  maybe the definition that the preimage of a continuous function sis an  οpen set or the ε,δ definition.My problem is that these definitions are for a single point.How can i use the definitions for every point?Trying to practice on using definitions and get a better understanding for continuity rather than that the graph of f is interrupted or has "holes".

It is continuous.

One way to see this is by using the fact that $x \mapsto x^2$ is continuous in $\mathbb{Q}$. Hence, the sets $A=\{x \in \mathbb{Q} \mid x^2 <2\}, B=\{x \in \mathbb{Q} \mid x^2 >2\} $ are open. Since the pre-images of open sets of your function $f$ are $A,B,\emptyset$ and $\mathbb{Q}$, it follows that $f$ is continuous.

Realize that your function f is continuous in R. Therefore, it is definitely continuous on a subset of R, which Q is. This is what you call f restricted to Q. Also, from a metric space standpoint, we know that Q is not a closed set since its complement R\Q is not open (you can prove this easily). Therefore, just use the definition of continuity as you pointed out. Take any open set, its pre image will be in Q, which is not closed.


Find continuous functions on $[0,1]$ satisfying this inequality. For a real valued function $f \in C([0,1])$ define $s(f) =\max _{x \in [0,1]}\mid x \cdot f(x) \mid$. One can easily check that $s$ defines a norm on $C([0,1])$. Let $n \in \mathbb{N}$. I am trying to find  functions $f, g \in C([0,1])$ such that  $s(fg) > n \cdot s(f)s(g)$.

I am having trouble finding these $f,g$ explicitly. I messed around with $f,g$ beeing polynomials of degree $n$, but I am not making a lot of progress. Some help would be appreciated.

Take $f(x) = \frac{|\sin (\pi k x / 2)|}{x}$ on $[0,1]$, with the understanding that $f(0) = \frac{\pi k }{2}$, so that $f$ is continous on $[0,1]$, where we take $k > n$. Then $s(f) = 1$ and $s(f^2) = \sup_{x \in [0,1]} |x f(x)^2| \geq |1/k \cdot f(1/k)^2| = k > n$, so taking $g = f$ will give you examples for any $k > n$.

Let's take $f(x) = g(x) = 1-x$. Then $s(f)=s(g)=max \left|x \cdot (1-x) \right| = \frac{1}{4}$ for $x=\frac{1}{2}$.

If we take $s(fg)=max \left|x \cdot (1-x)^2 \right| = \frac{4}{27}$ for $x=\frac{1}{3}$.

At last $\frac{4}{27} > n \cdot \frac{1}{16}$ is correct for $n=1$ and $n=2$.

p.s. Feeling a certain guilt and not having an answer, I hurry to help, as far as I can. Once again, I apologize if something was/is wrong.


Verification of whether $S=\{(x,y,z)\in \mathbb{R^3}:x=z^2\}$ is a vector space Let's say that we have a set $S=\{(x,y,z)\in \mathbb{R^3};x=z^2\}$
To prove that S is a Vector Space I must prove the 8 properties of a Vector Space, such as:

$X,Y,Z \in S$

A1) $X+Y=Y+X$

A2)$(X+Y)+Z = X+(Y+Z)$

A3)$X+0 = X$

A4)$X + (-X) = 0$

M1)$r(sX) = (rs)X$ where $r,s \in \mathbb{R}$

M2)$(r+s)X = rX + sX$

M3)$r(X+Y) = rX + rY$

M4)$1X = X$

My doubt is when proving the multiplicative properties M1 and M2.
My development is the following, but I don't know if it is right.

\begin{align}
X:(x,y,z)=(z^2,y,z)\\
sX = (s \cdot z^2, s \cdot y, s \cdot z)=\\
r(sX) = r \cdot (s \cdot z^2, s \cdot y, s \cdot z) = (rs \cdot z^2, rs \cdot y, rs \cdot z)=\\
=rs(z^2,y,z)
\end{align}

Is that correct?

You have to also show that S is closed under vector addition. This means that if you take two arbitrary vectors from S and add them together, the resulting vector must still be in S. If we take $(z^{2}, y, z)$ and $(w^{2}, v, w)$ and add them we get $(z^{2} + w^{2}, y + v, z + w)$ and since $z^{2} + w^{2}$ is not equal to $(z + w)^{2}$ the vector is no longer in S thus S is not closed under vector addition and not a vector space.

You are missing three important rules for it to be a vector space in its own right:

It has to be non-empty.
It has to be closed under vector addition.
It has to be closed under scalar multiplication.

All the other rules will actually be ok since it's already a subset of a larger vector space. These three are all we need to worry about.

Note that all subspaces of $\mathbb{R}^n$ have linear equations so it is highly doubtful this set is a subspace. The fact that the equation involves different powers hints that probably scalar multiplication will be a problem.

Consider $(1,0,1) \in S$. The vector $2(1,0,1) = (2,0,2) \not\in S$ since $x=2$ but $z^2 = 4 \neq x$. Thus $S$ is not closed under scalar multiplication and so it not a subspace.


Using implicit differentiation on trigonometric fraction and logarithm I am supposed to solve the following using implicit differentiation.

$$\sin\left(\frac{x}{y}\right)+\ln(y)=xy$$

This is what I have so far:

$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}$$

My problem here is differentiating the fraction as well as the $\ln y$, especially since $\ln y$ does not contain $x$ to differentiate towards.

Thanks for your help in understanding this.

Keep doing what you did on the right-hand side to get

$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}\;.\tag{1}$$

You have

$$\frac{d}{dx}\left(\frac{x}y\right)=\frac{y\cdot1-x\frac{dy}{dx}}{y^2}=\frac1y-\frac{x}{y^2}\frac{dy}{dx}$$

and

$$\frac{d}{dx}\ln y=\frac1y\frac{dy}{dx}\;;$$

both of these are just applications of the chain rule (together with the quotient rule and the rule for differentiating the natural log). Now for convenience I’ll write $y'$ for $dy/dx$; then $(1)$ becomes

$$\cos\left(\frac{x}y\right)\left(\frac1y-\frac{x}{y^2}y'\right)+\frac{y'}y=y+xy'\;,$$

and all that remains is to solve for $y'$.


How do I find the derivative of $\frac{1}{2\sqrt{x}}$ using the definition of the derivative? Using the definition of the derivative and not the power rule, how would I find $f(x) = \frac{1}{2\sqrt{x}}$

Let $h$ be an arbitrary real number, then
$$\frac{1}{2}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) = \frac{1}{2}\left(\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}\right) = \frac{1}{2}\left(\frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x+h}+\sqrt{x})}{\sqrt{x+h}\sqrt{x}(\sqrt{x+h}+\sqrt{x})}\right)$$
so
$$\lim_{h\to 0} \frac{1}{2h}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) = \lim_{h\to 0}\frac{1}{2}\left(\frac{-1}{\sqrt{x+h}\sqrt{x}(\sqrt{x+h}+\sqrt{x})}\right).$$


A tricky probability question. I have been asked the following question, and unfortunately I have no idea how to proceed. Here is the question:

Suppose we have 99 empty papers and we wrote numbers from 1 to 99(using each number) on one side of the papers randomly. Then we mixed all the papers randomly and started to write numbers from 1 to 99 on the other sides (empty sides) of each paper randomly. What is the probability of having the same number on both sides of at least one paper?

Thanks!

the answer is very simple . say in one piece of paper it is '5' chance that we end up  with wrong number on backside  out of 99 possible number is 98/99. if we write 5 ,we will be right. so chance of writing wrong number on backside in all 99 paper  is (98/99)^99 . so probability of at least one correct  = 1-(98/99)^99= 63.34%


Prove that this is a metric space? I'm supposed to show that If X is the set of all functions on the interval $[a,b]$ and $\displaystyle d(f,g)= \int^{b}_{a}|f(x)-g(x)|dx\,$, then $(X, d)$ is a metric space.

But I don't think it is. The problem I'm having is that in order for $(X,d)$ to be a metric space, it must be true that $d(f,g) = 0$ iff $f=g$. But d could also be $0$ if, for instance,

$f(x) - g(x)$ is odd and is symmetrical across $x=(a+b)/2$

Am I right about this or am I missing something?

Sorry, about the format by the way. I don't know how to type the actual symbols on here!

No: you are taking the integral of the absolute value of the difference  $$\displaystyle d(f,g)= \int_a^b \left|f(x)−g(x)\right|dx$$ while your counterexample would apply if you were taking the absolute value of the integral of the difference $$\displaystyle \left|\int_a^b (f(x)−g(x))\,dx\right|.$$

You may need to watch out for the possibility that $f(x)$ and $g(x)$ differ on a set of measure zero, but apart from that you will have $\left|f(x)−g(x)\right|$ as non-negative by definition, and positive if they differ, making the integral positive if they differ on a large enough subset of $[a,b]$.

I think you are confusing that the limits of integration is arbitrary. The $a$ and $b$ are to be any number, you don't determine them, they are given to you.


Convergent subsequences I'm studying for a real analysis exam tomorrow and I am curious about something:

Two true false questions:
a) Every sequence in the interval (0,1) has a convergent subsequence.
b) Every sequence in the interval (0,1) has a subsequence that converges to a point in (0,1)

This dude on Yahoo Answers says both are false and uses the counterexample {1/n}. I think he is wrong because the sequence {1/n} isn't even in my interval! (It fails for 1). In fact I think both are true, and I don't know the proper reasoning but all I know is that I am not convinced by this supposed "counterexample"

Secondly, my book describes the following proposition: "Let the sequence {An} converge to the limit a. Then every subsequence of {An} also converges to the same limit a." In feel like this confirms the validity of (b).

Any input on the truth of these two would be very much appreciated.

For $(a)$, you get sequences that are bounded. Use the Bolzano-Weierstrass theorem, and the Monotone Convergence theorem.

For $(b)$, consider the sequence $a_n=\frac 1 {n+1}$. Then $a_n\in(0,1)$ but $\lim a_n=0$.

ADD You say

  Secondly, my book describes the following proposition: "Let the sequence $\{a_n\}$ converge to the limit $a$. Then every subsequence of $\{a_n\}$ also converges to the same limit $a$." In feel like this confirms the validity of $(b)$.

Note that that proposition talks about a convergent sequence, and the behaviour of its subsequences, while $(b)$ considers any sequence and it's subsequence, and the value it converges to. If you have an exam tomorrow, consider writing down the argument you "feel" confirms the validity of something, so that you know wether you're right or not, and reading propositions to understand what they are really stating. In this case, that proposition tells you  nothing about $(b)$, unless you consider the convergent sequences in $(0,1)$ that converge to a point in $(0,1)$. But then you're missing a lot of other sequences that don't converge, but are bounded, so have a convergent subsequence, which may, or may not converge to a point in $(0,1)$. In particular, this is because $0$ and $1$ are limit points of $(0,1)$. If we were talking about $[0,1]$, the answer would be affirmative, because $[0,1]$ us a closed subset of $\mathbb R$.

The dude is almost right. Both are false. Take $a_n = \frac{1}{2n}$.

By the way, are you sure the question is stated with open intervals?

Edit: I have misunderstood your question, my answer applies only if your topological space is just $(0,1)$.


Prove that $\det(\varphi(\sigma)) = \operatorname{sgn}(\sigma)$ for all $\sigma \in S_n$. Fix a positive integer $n$.

  For each $\sigma \in S_n$ define an $n\times n$ matrix, $\varphi(σ)$ by

  $$\varphi (\sigma)_{i,j}= \begin{cases} 1 & \text{if } i = \sigma (j), \\ 0 & \text{if } i \neq \sigma (j). \end{cases}
$$

I have proved that $\varphi$ is a homomorphism and found $\ker \varphi.$

  Prove that $\det(\varphi(σ)) = \operatorname{sgn}(σ)$, for all $σ \in S_n$

There should be some nice way to do this considering that $\det,$ $\operatorname{sgn},$ and $\varphi$ are all homomorphisms.
I tried writing it out for a single transposition but wasn't sure how to generalize it?

I want to link it to this question Prove that for each $\phi \in S_n$, $\det A=\sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)\prod_{j=1}^n a_{\phi(j)\sigma(j)}$ cause it feels very similar but i cant seem to figure out how (and i dont understand the answer given here)

If $\sigma$ is a product of $k$ transpositions, then $\operatorname{sgn}(\sigma) = (-1)^k,$ so if $\sigma$ is a transposition, then $\operatorname{sgn}(\sigma) = -1.$

Suppose $\sigma = \tau_1 \circ \cdots \circ \tau_k$ where $\tau_1,\ldots,\tau_k$ are transpositions.

Then
\begin{align}
& \overbrace{ \varphi(\tau_1)\cdots \cdots\varphi(\tau_k) = \varphi(\sigma) }^{\Large\text{since $\varphi$ is a homomorphism}}
\end{align}
and therefore
\begin{align}
\det \varphi\left(\sigma\right)
= (\det \varphi(\tau_1)) \cdots (\det \varphi(\tau_k))
= (-1)^k
\end{align}
provided that one can show $\det(\varphi(\tau_i)) = -1$ whenever $\tau_i$ is a transposition. But all we need for that is that the determinant of the identity matrix is $1$ and $\varphi(\tau_i)$ comes from interchanging two rows of the identity matrix, and interchanging two rows multiplies the determinant by $-1.$

Then we have
$$
\varphi(\rho\sigma) = (-1)^{\text{a certain power}}
$$
and you can think about what that exponent is.

$\operatorname{sgn}$ is by definition the only nontrivial morphism $\mathfrak{S}_n \to \mathbb{C}^{\times}$.

${\det}\circ \phi$ is a morphism $\mathfrak{S}_n \to \mathbb{C}^{\times}$. Clearly it is nontrivial, so it is $\operatorname{sgn}$.


How to evaluate $\lim\limits_{s\to\infty}\log s$? I am stuck with applying limit at the following step,
limit
$$
\lim\limits_{s\to\infty}\log s.
$$
Now I am unable to do anymore steps(I cant figure out how do I  apply the limit and get a valid answer). Please help me out.

The statement $$\lim_{x\to\infty}\log x=+\infty$$

is true if and only if for each $M>0$ there exists $N>0$ such that whenever $x>N$ we have $\log x>M$. Given $M$, take $N=e^M$. Since the logarithm is an increasing function, whenever $x>N=e^M$ we will have $\log x>\log N=M$, thus $$\lim_{x\to\infty}\log x=+\infty$$

If you observe graph below which represents $\log(s)$ you may conclude that:

$\lim\limits_{s\to+\infty}\log s=+\infty$

$\lim\limits_{s\to-\infty}\log s$  isn't defined since $s>0$


Any way to compute this serie? Let $(u_n)_{n \ge1}$ such as $u_1=\frac12$ and $(\forall n \ge2):u_n=(1-\frac3{2n})u_{n-1}$. Find the value of $\displaystyle \sum_{n=1}^{\infty}u_n$. This is my try :
We can rewrite it as $\displaystyle u_n=(-1)\cdot\frac{\frac32-n}{n}\cdot u_{n-1}\Rightarrow\prod_{k=2}^nu_k=(-1)^{n-1}\cdot\prod_{n=2}^n\left(\frac{\frac32-n}n\right)\cdot\prod_{k=2}^nu_{k-1}$.
It is clear that $(\forall n \ge 0): u_n \neq 0$, thus we have $ \displaystyle u_n=(-1)^{n-1}\prod_{k=1}^n\left(\frac{\frac32-k}k\right)$.
Using the fact that $\displaystyle \binom{p}{q}=\prod_{k=1}^q \frac{p+1-k}k$ with $p=\frac12$, we now have $\displaystyle u_n=(-1)^{n-1}\binom{\frac12}{n}$.
Thus $\displaystyle \sum^\infty_{n=1}u_n=-(-1)^{-1}\binom{\frac12}{0}+\sum^\infty_{n=0}(-1)^{n-1}\binom{\frac12}{n}=1+\sum^\infty_{n=0}(-1)^{n-1}\binom{\frac12}{n}$, but here I'm stuck because i wanted to use the Newton's generalized binomial theorem yet it is not possible here....
Any help please ?

Hint (stolen from your idea of using Newton's generalized binomial theorem)
$$(1+(-1))^{1/2}=\sum_{n=0}^\infty \binom{1/2}{n}1^{1/2-n} (-1)^n=\sum_{n=0}^\infty (-1)^n\binom{1/2}{n}$$


Area of the Limiting Polygon Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one.

Continue this process ad infinitum.

What is the area of the limiting "polygon"?

My first thought was that it would be the circle tangent to the midpoint of each side. However, by making some pictures, it becomes clear quite quickly that the limiting polygon is not a circle and, in fact, might perhaps be an ellipse. The problem with that is that while there is a unique circle passing through any three points in general position, I need five generic points to specify a conic.

Can anyone suggest any hints as to how to find the area? By "hint" I don't mean "what you think might work", I mean "what you have tried and you know will work". Thanks in advance.

We calculate explicitly that the area of the limiting polygon is $\dfrac{4}{7}A$ where $A$ is the area of the original triangle.

Note that on the $(n+1)$-th iteration we cut off twice as many triangles as we did in the $n$-th iteration. Consider the triangles we cut off in the $(n+1)$-th iteration. Each of these has $\dfrac{1}{3}$ the height and $\dfrac{1}{3}$ the base of a triangle cut off in the $n$-th iteration. Hence the ratio of the areas of these triangles is $\dfrac{1}{9}$. Since the total area of the first triangles we cut off is $\dfrac{1}{3}A$, we have that the area of the limiting polygon is

$$A-\sum\limits_{n=0}^\infty2^n\left(\frac{1}{3}A\right)\left(\frac{1}{9}\right)^n=A-\frac{1}{3}A\frac{1}{1-\frac{2}{9}}=\frac{4}{7}A$$

Whether this actually helps us determine the shape of the limiting figure I am not so sure.

This was motivated by Hagen von Eitzen's calculation and Michael's subsequent observation.

$\\$

EDIT: Explanation of $\dfrac{1}{3}$ base and $\dfrac{1}{3}$ height statement.

Consider the polygon at a particular vertex $V$ just before making the $n$-th iteration cuts. Choose a particular edge bordering $V$ and let it have length $9x$.

Now take the $n$-th and $(n+1)$-th iteration cuts. In the figure above, the green triangle is one that is cut off in the $n$-th iteration cuts, and the red triangle is one cut off in the $(n+1)$-th iteration cuts.

Because we trisect each time, one side of the green triangle has length $3x$ and one side of the red triangle has length $x$. Call these sides the base of each triangle. Consider now the length of the altitudes corresponding to these bases, letting the altitude of the green triangle be of length $3y$. By similar triangles we find that the altitude of the red triangle has length $y$ (the similar triangles are outlined in black).

Hence the ratio of area of the red triangle to the area of the green triangle is
$$\frac{\frac{1}{2}xy}{\frac{1}{2}(3x)(3y)}=\frac{1}{9}$$
as claimed.

Fun note: we did not use the fact that the triangles in question are isosceles.

Snipping off corners as you've described you get an infinite-sided regular polygon: a circle.

It will be tangent to the original triangle at the midpoints of the sides.

If the length of the side of your triangle is $1$, the radius of the inscribed circle is $\tfrac{1}{2}$ and the area is $\tfrac{\pi}{4}$.


Suppose that every sequence in A has a convergent sub-sequence. Prove that A is bounded. My attempt:

We will prove by contradiction, assume to the contrary that A is unbounded. Then we will show that if every sequence in A has a convergent subsequence then A is unbounded. But if A is unbounded, then we have a sequence $f_n$ that diverges to $\infty$ (1), diverges to $-\infty$ (2), or diverges to $\infty$ and $-\infty$ (3). But $f_n$ diverges to infinity then every subsequence of $f_n$ also diverges to infinity. Therefore, there is a sequence that has no convergent subsequence which is a contradiction. This holds for (2) and (3). Thus A must be bounded.

Is this correct?

Your proof needs to be cleaned up a little. You say you are using proof by contradiction, but then say

  ... assume to the contrary that A is unbounded. Then we will show that if every sequence in A has a convergent subsequence then A is unbounded.

That's not what contradiction means. What you mean to say is

  ... assume to the contrary that $A$ is unbounded. Then we will find a sequence in $A$ with a non-convergent subsequence.

This clearly states the aim of the argument, and presents a clear contradiction to the hypotheses.

The remainder of the proof is, I believe, a bit confused and longer than necessary. For instance, you asset the existence of a sequence $f_n$ in $A$ which diverges in one of three ways. How do you know such a sequence exists? Once you know that it exists, why deal with three separate cases? What does divergence to $\infty$ and divergence to both $\infty$ and $-\infty$ mean? If a sequence diverges to $\infty$, how do you know every subsequence diverges to $\infty$? Some of these questions have good answers, but regardless, they should be addressed in a rigorous proof. I think a simpler argument is

  By the definition of unbounded, for every $n\in\mathbb N$, there exists $a_n\in A$ so that $|a_n|>n$.

Now consider the sequence $(a_n)$. It is a subsequence of itself. If you can show that the (sub)sequence $(a_n)$ does not converge, then you are done.

Your second sentence doesn't make sense, and should be removed. "Diverges to infinity" is imprecise; see this question: Diverging to Positive and Negative Infinity . You should probably be more explicit in your reasoning that "But [if] fn diverges to infinity then every subsequence of fn also diverges to infinity."


What does it mean to have a determinant equal to zero? After looking in my book for a couple of hours, I'm still confused about what it means for a $(n\times n)$-matrix $A$ to have a determinant equal to zero, $\det(A)=0$.

I hope someone can explain this to me in plain English.

For an $n\times n$ matrix, each of the following is equivalent to the condition of the matrix having determinant $0$:

The columns of the matrix are dependent vectors in $\mathbb R^n$
The rows of the matrix are dependent vectors in $\mathbb R^n$
The matrix is not invertible.
The volume of the parallelepiped determined by the column vectors of the matrix is $0$.
The volume of the parallelepiped determined by the row vectors of the matrix is $0$.
The system of homogenous linear equations represented by the matrix has a non-trivial solution.
The determinant of the linear transformation determined by the matrix is $0$.
The free coefficient in the characteristic polynomial of the matrix is $0$.

Depending on the definition of the determinant you saw, proving each equivalence can be more or less hard.

If a matrix A has an inverse, it means that the matrix multiplication AB = I, the identity matrix.  In plain English then, if a matrix is invertible then it may have a solution.

If a matrix's determinant is nonzero, the matrix may have a solution. If the determinant is zero, then the matrix is not invertible and thus does not have a solution because one of the rows can be eliminated by matrix substitution of another row in the matrix.

Common reasons for matrix invertibility are that one or more rows in the matrix is a scalar of the other.

EXAMPLE for a 3x3 matrix: Row 1: x + y + z = P  versus Row 2: Qx + Qy + Qz = QP.  Row two can be rewritten as Q*(x + y + z = P) or Q*(Row 1)]

OR one of the rows can be created by addition of two or more of the other rows.

EXAMPLE for a 3x3 matrix:  Row 1: x + 2y + 3z = P, Row 2: 4x + 5y + 6z = Q, Row 3: 7x + 8y + 9z = P+Q. You see that Row 3 can be duplicated by adding Row 1 and Row 2.

In short, if the determinant of a matrix is zero, the matrix does not have a solution because the matrix cannot be inverted.


How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$ How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$

I know $x^3-c^3=(x-c)(x^2+xc+c^2)$ but I can't figure out how to pull this off with $1/3$ instead of $3$


Number pattern problem I need some help to figure out the following number pattern:

Find the next number in the following sequence:
$3, 2, 3, 12, 35, 78$.

I tired to figure out the difference by add/subtract/multiple/division also tried to find the existing known pattern still could not work it out.

According to the Online Encyclopedia of Integer Sequences, this is the first 6 values of $n$ of the sequence $n^3 - 5n^2 + 7n$

183, because I say so.

(This makes it more than 30 characters, so there!)


I need help understanding Frege's definition of number I have really been trying to understand Frege's definition of a number or at least gain a strong intuition of it. However, my attempts have not been fruitful. If someone could help me it would be much appreciated. I will write his definition down explicitly, then give write down a little bit of what he writes around this definition.

Number - The content of a statement of number is an assertion about a concept.

[source : The Foundations of Arithmetic, §55, Engl.transl. page 67]

Here is what he writes before and immediately after stating his definition:

Immediately Before: While looking at one and the same external phenomenon I can say with equal truth both "It is a copse" and "It is five trees," or both "Here are four companies" and "Here are 500 men." Now what changes here from one judgement to the other is neither any individual object, nor the whole, the agglomeration of them, but rather my terminology. But that is itself only a sign that one concept has been substituted for another.

[source : The Foundations of Arithmetic, §46, Engl.transl. page 59]

Immediately After: This is perhaps clearest with the number $0$. If I say "Venus has $0$ moons", there simply does not exist any moon oor agglomeration of moons for anything to be asseted of; but what happens is that a property is assigned to the concept "moon of Venus", namely that of including nothing under it. If I say "the King's carriage is drawn by four hourses", then I assign the number four to the concept "horse that draws the King's carriage"

[source : The Foundations of Arithmetic, §46, Engl.transl. page 59]

Edit: For those who are more inclined to reading logic. On part 2.5 of the following link it explains what Frege was doing. Unfortunately, I'm having difficulties understanding this as well. I have been exposed to some first order logic, but no second order logic whatsoever. I am not sure if that is necessary though.

http://plato.stanford.edu/entries/frege/#NatNum

Thank you in advance for anyone who can help!

Some hints

The context of Frege's assertion is his analysis of the "number assertions"; see :

Die Grundlagen der Arithmetik: eine logisch-mathematische Untersuchung über den Begriff der Zahl (1884), translated as The Foundations of Arithmetic: A logico-mathematical enquiry into the concept of number by J.L. Austin (1950).

Roughly speaking, Frege denies that when we use number-words in natural language assertions we express a property of objects; i.e. when I say "my hands are white" I'm asserting that each of my hands is white, while when I'm saying  "I have two hands", I'm not asserting that each of my hands "is two".

Thus, Frege's conclusion is that the "number-words" are like adjectives for concepts, i.e. they express a property of a concept.

Example : there are two satellites of Mars : Phobos and Deimos. When we say "Mars has two satellites", we express the fact that the concept "satellite of Mars" is instantiated by two objects.

In modern expression, we have that :

  $\exists x \exists y [Sat_M(x) \land Sat_M(y) \land x \ne y \land \forall z(Sat_M(z) \to (z=x \lor z=y))]$.

This is also clear from the analysis of the number $0$ :

  "if I say "Venus has $0$ moons", there simply does not exist any moon or agglomeration of moons for anything to be asseted of; but what happens is that a property is assigned to the concept "moon of Venus", namely that of including nothing under it."

i.e.

  $\lnot \exists x Moon_V(x)$.

This point has fundamental consequences in Frege's view : numbers are akin to quantifiers.

In Frege's terminology, both number and quantifiers are second-level concepts, i.e. they are concepts that express a property of first-level concepts (like the concept : "moon of Venus", i.e. concepts "applying" to objects).

This fact can be highlighted by the "formal" analysis; we have that Phobos satisfy the concept "Satellite of Mars", i.e. $Sat_M(Phobos)$ holds.

In Frege's terminology, the object Phobos "falls under" the concept "Satellite of Mars".

The concept "Satellite of Mars" is instantiated, i.e. there are objects that "fall under" it. Thus the second-level concept "existence" applies to the first-level concept "Satellite of Mars", i.e. $\exists x Sat_M(x)$ holds.

We can "formalize" Frege's view (in a "philological" inaccurate way) as follows :

1) assume the universal quantifier $\forall$ as primitive;

2) define the existential quantifier $\exists$ as usual, as : $\lnot \forall \lnot$;

3) introduce the "numerical" quantifiers :

i) $\exists^0$ defined through : $\lnot \exists Fx$ : "there are no $F$'s"

ii) $\exists^1$ defined as the "usual" $\exists !$, i.e. as $\exists x Fx \land \forall y (Fy \to x=y)$ : "there is (exactly) one $F$"

iii) $\exists^2$ as $\exists x \exists y (Fx \land Fy \land x \ne y \land \forall z(Fz \to z=x \lor z=y))$ : "there are two $F$'s"

and so on.

These "numerical" quantifiers are the formal counterpart of "number-words".

You can see :

Kevin Klement, Frege’s Changing Conception of Number, Theoria (2012).

Frege's description of number (which is a philosophical and not mathematical question, yet essential to mathematics) is problematic because it appeals to the relationship "falls under". Yet there is no relationship, at least not a logical one, that is given by "falls under". There is merely a grammatical tradition. We say "objects fall under concepts", but this only expresses the way we casually talk about objects and concepts. It leaves us in the dark regarding both concepts and objects and their relationship. Yet it is their supposed relationship that Frege uses in his definition of number.

But the question of what Frege meant by number can be dismissed as secondary to a higher problem. The question we should be asking is not what sort of things numbers are, or in what domain they exist, but what are the conditions for the creation of numbers. Frege takes numbers to be independent of any constructional conditions. He takes what is called a trancendentally real line, a posture common to all scientists and mathematicians, and most philosophers. Instead, he should be taking (like Wittgenstein) a transcendentally ideal line, which shows that there are no independent numbers, nor any domain that supports their existence. Logic and mathematics are unable to distinguish between transcendentally real and transcendentally ideal objects. The syntax of their expression looks the same in both models, a fact that astonishingly has gone unnoticed throughout the history of science and logic.


Show that $f$ attains its minimum in $\mathbb R$. Let $f: \mathbb R \to \mathbb R$ be a continuous function, such that $f(x)\ge |x|$ for all $x \in \mathbb R$. Show that the function $f$ attains its minimum in $\mathbb R$.

My thoughts:
When $|x|$ is big, then $f(x)$ is also big, since it's continuous it reaches its minimum in $\mathbb R$ by extreme value theorem. How do I show this, and do I need to go in the specifics a bit more?

Let $f(0)=0$.

Then $f$ attains its minimum at $0$ because $f(x)\ge 0\forall x$

So ,we assume that Let $f(0)>0 \forall x$.

Then consider the function $g(x)=\dfrac{1}{f(x)}$ which is also continuous.

Also $g(x)\le \dfrac{1}{|x|}$ forall $x\in \Bbb R$ and $g(0)>0$.

So consider the interval $A=[-\dfrac{1}{g(0)},\dfrac{1}{g(0)}]$ which is compact ,So $g$ being bounded attains its maximum in $A$ say at the point $x_0$.
Hence $g(x)\le g(x_0)\forall x\in A$.

For $x\in A^c;|x|>\dfrac{1}{g(0)}\implies g(x)\le \dfrac{1}{|x|}<g(0)$.

Also $g(0)\le g(x_0) $ since $x_0\in A$. Hence $g(x)\le g(x_0)\forall x$

So $g$ attains its maximum and hence $f$ attains its minimum at $x_0$.


Identification of a quadrilateral as a trapezoid, rectangle, or square Yesterday I was tutoring a student, and the following question arose (number 76):

My student believed the answer to be J: square. I reasoned with her that the information given only allows us to conclude that the top and bottom sides are parallel, and that the bottom and right sides are congruent. That's not enough to be "more" than a trapezoid, so it's a trapezoid.

Now fast-forward to today. She is publicly humiliated in front of the class, and my reputation is called into question once the student claims to have been guided by a tutor. The teacher insists that the answer is J: square ("obviously"... no further proof was given).

Who is right? Is there a chance that we're both right?
How should I handle this? I told my student that I would email the teacher, but I'm not sure that's a good idea.

Clearly the figure is a trapezoid because you can construct an infinite number of quadralaterals consistent with the given constraints so long as the vertical height $h$ obeys $0 < h \leq 9$ inches.  Only one of those infinite number of figures is a square.

I would email the above statement to the teacher... but that's up to you.

As for the "politics" or "pedagogy" of drawing a square, but giving conditions that admit non-square quadralaterals... well, I'd take this as a learning opportunity.  The solution teaches the students that of course any single drawing must be an example member of a solution set, but need not be every example of that set.  In this case:  a square is just a special case of a trapezoid.

The solution goes further and reveals that the vertexes (apparently) lie on a semi-circle... ("obvious" to a student).  A good followup or "part b" question would be to prove this is the case.

One might also point out that a square is also a special case of trapezoid, rectangle, rhombus, quadrilateral and parallelogram, so any of those answers would technically be correct, assuming the object in question was, indeed, a square.

However, I think what everyone is missing is that while in a classroom the "right" answer is the one the teacher is looking for, even if that answer isn't the "correct" one...  I've found in my experience that teachers can be obstinate, and aren't likely to be pleased by being corrected by a student in front of the class.  Better (for ones academic success) to do so after class and let the teacher correct him/herself in a future class, or not...


Double Summation: Need help to handle $ i \neq j $ : $ \sum_{i=0 \to 7,\ j=1 \to 8,\ i\neq j} (8i + j) $ [Q1]. Can I ? ( write the same summation as ) :
  $$ \sum_{i=0, i \neq j}^7 \sum_{j=1}^8 (8i + j)   \tag{1}$$

I tried to solve the following Summation as follows:

Let i = m-1 then, $ \sum_{i=0,\ i \neq j}^7 $ becomes
$\sum_{m=1,\ j\neq m-1}^8 $

Now applying change of base equation (1) i.e.
$ \sum_{i=0, i \neq j}^7 \sum_{j=1}^8 (8i + j)    $ becomes
$$ \sum_{m=1,\ j\neq m-1}^8 \sum_{j=1}^8 \Big(8(m-1) + j \Big) $$

[Q2]. Can we do the next step? By what rule?

$$\text{Above}= \sum_{m=1,\ j=1, \ j\neq m-1}^8 (8m - 8 +j)
\\
= (\sum_{m=1,\ j=1, \ j\neq m-1}^8 8m )- (\sum_{m=1,\ j=1, \ j\neq m-1}^8 8) + (\sum_{m=1,\ j=1, \ j\neq m-1}^8 j)
\\
= 8\ \Big(\sum_{m=1}^8 m \Big)-  8 \ \Big(\sum_1^8 1 \Big) + \Big(\sum_{j=1}^8 j\Big)
\\
= 8 \Big( 8. \frac{8+1}{2} \Big)  -  8(8)+ \Big( 8. \frac{8+1}{2} \Big)
\\
=260 $$

[Q3]. Don't we need to handle $ i \neq j $ ? If so, then in which situation does $ i \neq j $ matter and how ? Please explain what iff. $ i ==j $ then ?

Please clear my concept by answering all my 3 questions above(Q1,Q2 and Q3).

For your first question, the answer is yes, you can change indeed rewrite your outer sum by substituting $m-1$ for $j$ and changing the limits of that sum.

You can't, however, make the step indicated in your second question. Consider, for example, your middle summation. That sum is over all pairs $(m, j)$ satisfying by the predicate
$$
P(m, j) = (1\le m\le 8) \land (1\le j\le 8) \land (j\ne m-1)
$$
There will $64-8 = 56$ terms in this sum. The first step in your argument was correct:
$$
\sum_{P(m, j)} 8 = 8\left(\sum_{P(m, j)} 1\right)
$$
but your next step wasn't, since
$$
\sum_{P(m,j)} 1 \ne \sum_1^8 1
$$
To see this, just observe that the sum on the left has 56 terms and the sum on the right has only 8.

Finally, a useful technique for problems like this is to ignore for the moment the excluded diagonal terms, sum all the terms, and then subtract the excluded ones. Using your chessboard idea, you noticed that the squares in the chessboard contain the numbers $1, \dots , 64$ so the sum of all the entries is
$$
\frac{64\cdot65}{2} = 2080
$$
Now all we have to do is subtract the sum of the 7 diagonal elements in positions
$$
(k-1, k) = (2, 1), (3, 2), (4, 3), (5, 4), (6, 5), (7, 6), (8, 7)
$$
where the values are $9, 18, 27, 36, 45, 54, 63$. The sum of these is 252, so the answer to the problem is $2080-252=1828$.

Q1 yes you can .

Q2 yes you can .

Q3 you need to subtract all the options of j to be i .
You need to cover all the i in the sigma were j===I like this :

Sigma of m from 1 to 8 that divide (8m-8+j) because m-1= j >>> (8m-8+m-1)
(9m-9)


Geometric position of a complex set. I'm quite struggling with this question and help would be appreciated.

Define the geometric position of a complex set:

$|z-4i| +|z+4i|=10$

Translate to geometry: the sum of the distances from $z$ to the two points $4i$ and $-4i$ is $10$. We recognize this an ellipse with foci $\pm 4i$.


How do you evaluate $\bigcup\bigcup S$ if $S = \{\{a\},\{a,b\}\}$? How do you evaluate $\bigcup\bigcup S$ if $S = \{\{a\},\{a,b\}\}$? I understand that $\bigcup S = \{a,b\}$ but how can I find $\bigcup\{a,b\}$ if I don't even know what the elements of $a$ and $b$ are? Here is a picture of the exercise from the book I saw this in

$S$ has only two elements $\{a\}$ and $\{a,b\}.$ The double union of $S$ is still $S$ itself. It is NOT $\{a,b\}$