What are complex numbers, actually? What are complex numbers, actually? You can prove $1=-1$ and a complex cosine function can have value greater than $1$ and so on, there are many unexpected results when we use complex numbers. So, what are they actually? Do, they have any physical meaning or are they just a method in mathematics to manipulate numbers?

A complex number is a number that has both Real and Imaginary Components, It is the the form $$a+bi$$ where $i$ is defined as $$i=\sqrt{-1}$$


Long modular arithmetic going wrong I have attempted to do a question that I crafted by hand: 6239^5 mod 15367

I know that 15367 is made up of prime numbers, notably 121 and 127.

Hence, I break it down by doing 6239^5 mod 121 and 6239^5 mod 127 individually. This yields the following results

6239^5 mod 121 = 109
6239^5 mod 127 = 64

Now I know that x is congruent in the following way:

x ≡ 109 (mod 121)
x ≡ 64 (mod 127)

As such this can then be calculated as:

109 + 121k ≡ 64 (mod 127)
121k (mod 127) ≡ (64-109) (mod 127)
121k (mod 127) ≡ -45 (mod 127)
-6k ≡ -45 (mod 127)
k = 45/6

So we substitute the k into the first formula:
109 + 121k
= 109 + 121(45/6 + 127u)
= 109 + 907.5 + 127u.

I'm stuck now because 907.5 is a decimal.

How do I now work out the final answer?

One never writes fractions in modular arithmetic. So where you went wrong is on the line $k = 45/6$.

Instead, one computes the multiplicative inverse of $6$ modulo 127. Let's call that number $x$ (you can compute it for yourself or look it up somehow).

Then you would instead have
$$-6k  \equiv -45 \pmod {127}
$$
$$6k \equiv 45  \pmod {127}
$$
$$k \equiv 45x  \pmod {127}
$$
and carry on from there.


An Inequality Problem with not nice conditions How to show that $\dfrac{a^3}{a^2+b^2} + \dfrac{b^3}{b^2+c^2} + \dfrac{c^3}{c^2+a^2} \ge \dfrac32$, where $a^2+b^2+c^2=3$, and $a,b,c > 0$ ?

As has been mentioned in some comments too, this question does not seem to allow a pre-calculus solution. Anyway, it is again a question of the symmetric type, such as listed in:

Why does Group Theory not come in here?

Completed with:

Do symmetric problems have symmetric solutions?

From the latter article comes the following
Theorem (The Purkiss Principle). Let $f$ and $g$ be symmetric functions with continuous second derivatives in the neighborhood of a point $P = (r, \cdots, r)$.
On the set where $g$ equals $g(P)$, the function $f$ will have a local maximum or minimum at $P$ except in degenerate cases.

Our function $f$ in this case is:
$$ f(a,b,c) = \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}$$
Applying the Purkiss Principle gives (with $r > 0$):
$$ g(r,r,r)=r^2+r^2+r^2=3 \quad \Longrightarrow \quad r=1 \quad \Longrightarrow \quad f(r,r,r) = \frac{3}{2}$$
The main problem with the Purkiss Principle, most of the time, is to prove that the minimum found is global,
making this part of the answer only a partial answer.

Another possibility would have been to introduce spherical coordinates, thereby eliminating the equation $a^2+b^2+c^2=3$ :
$$
a = \sqrt{3}\cos(\theta)\cos(\phi) \qquad b = \sqrt{3}\cos(\theta)\sin(\phi) \qquad c = \sqrt{3}\sin(\theta)
$$
But this destroys the beautiful symmetry in the first place. And it doesn't help much in the second place, because
the partial derivatives to $\theta$ and $\phi$ are too messy to determine the values where they are zero, which
would be required for establishing a minimum (at least MAPLE - as steered by me - refuses to do the job).
EDIT. Which is a bit too pessimistic view; one can do something with those spherical coordinates:
$$
a = \sqrt{3}\cos(\theta)\cos(\phi)=1 \quad , \quad b = \sqrt{3}\cos(\theta)\sin(\phi)=1 \quad , \quad c = \sqrt{3}\sin(\theta)=1 \\
\Longrightarrow \quad \theta = \arcsin(1/\sqrt{3})=\arctan(1/\sqrt{2}) \quad \Longrightarrow \quad \phi = \arcsin(1/\sqrt{2})=\arctan(1)=\pi/4
$$
Substitution of the spherical coordinates in our function $f$ defines $\;g(\phi,\theta)=f(a,b,c)-3/2\;$.After some tedious calculations (or rather trusting MAPLE :-) for $(\phi_0,\theta_0) = \arctan(1,1/\sqrt{2})$ :
$$
g(\phi_0,\theta_0) = 0 \quad ; \quad
\left[ \begin{array}{c} \frac{\partial g}{\partial \phi} \\ \frac{\partial g}{\partial \theta} \end{array} \right](\phi_0,\theta_0) = 0 \quad ; \quad
\left[ \begin{array}{cc} \frac{\partial^2 g}{\partial \phi^2} & \frac{\partial^2 g}{\partial \phi \, \partial \theta} \\
\frac{\partial^2 g}{\partial \theta \, \partial \phi} & \frac{\partial^2 g}{\partial \theta^2} \end{array} \right](\phi_0,\theta_0) =
\left[ \begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array} \right]
$$
Proving that there is indeed a local extreme at $\;g(\phi_0,\theta_0)\;$ and that this extreme is a minimum.The problem is, again, to prove that the minimum found is global, making this part of the answer a partial answer as well.

How $a^2+b^2+c^2=3$, then is equivalent to proof that
$$\frac{a^3}{3-a^2}+\frac{b^3}{3-b^2}+\frac{c^3}{3-c^2}\geq\frac{3}{2}$$
Now the function $f(x)=\frac{x^3}{3-x^2}$ have $f''(x)>0$ in $(0,\sqrt{3})$ then $f(x)$ is convex in this interval, then:

$$f(a)+f(b)+f(c)\geq3f\left(\frac{a+b+c}{3}\right)$$

how $(a+b+c)^2=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$, for the Rearrangement Inequality,  $2ab+2ac+2bc\leq 2a^2+2b^2+2c^2$, then $(a+b+c)^2=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\leq 3(a^2+b^2+c^2)=9\Longrightarrow a+b+c\leq3$

A hint for conclude the proof is use the  "smoothing principle", ($a=b=c$), so in this case $f(a)+f(b)+f(c)$ is minimized,  (but I do not understand this topic), and then:

$$f(a)+f(b)+f(c)\geq 3f(1)=\frac{3}{2}\geq3f\left(\frac{a+b+c}{3}\right)$$
God  bless


Doubt in epsilon-delta limit proof (more specifically, the inequalities) For example:
We shall prove that
$$ \lim_{x \to 2} x^2 = 4.$$
Then we shall prove the statement
$$ \forall \varepsilon >0, \,\, \exists \delta >0 \,\, \mid \,\, 0 < \left| x - 2\right| < \delta \implies \left|x^2 - 4 \right| < \varepsilon$$

Draft:
$$ \left|x+2 \right| \left|x-2 \right| < \varepsilon \qquad (1) $$
Let $\delta \leq 1$:
$$ \left|x - 2 \right| + |2| < 1 + |2|$$
$$ |(x - 2) + (2)| \leq \left|x - 2 \right| + |2| $$
$$ |x| < 1 + |2|$$
$$ x < 3 $$
So
$$ |x-2||x+2| < 5|x-2| \qquad (2)$$

And here comes the doubt!

$$ 5 \left|x - 2 \right| < \varepsilon \qquad (3) $$

Proof:

Let
$$\delta = \min \left(1, \, \dfrac{\varepsilon}{5} \right)$$
[The proof continues, but I don't have trouble with the remainder part.]

Why can we say that $ 5 \left|x - 2 \right| < \varepsilon $ ? If we add the inequalities (1) and (2), we won't get the inequality (3).
Can anyone help me?
Thank you.

Sorry for my bad English.

We know that $\lim\limits_{x\rightarrow2}{x}=2 \tag{1}$(Hint: Try $\delta=\varepsilon$)

Note that $x^2-4=\left(x-2\right)\left(x-2\right)+4(x-2)\tag{2}$

Using $(1)$ we can say that there exist a $\delta_1$ such that $|x-2|<\sqrt{\frac{\varepsilon}{2}}$. Again using $1$ we can say that there exist a $\delta_2$ such that $|x-2|<\frac{\varepsilon}8$.

Now,there exist a $\delta=\min\{\delta_1,\delta_2\}$ such that $(2)$(with the help of triangular inequality) can be reduced to $$\left|x^2-4\right|<\left|\left(x-2\right)\right|\left|\left(x-2\right)\right|+\left|4(x-2)\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$

There are several issues going on here.  To answer your first question, we can't conclude $5|x-2| < \epsilon$.

Pretend I have chosen my favorite small number $\epsilon$.  To find $\delta$, I need to figure out how close $x$ must be to $2$ in order to make $|x^2 -4| < \epsilon$.  So from this, hopefully you can see that $\delta$ depends on $\epsilon$.  The smaller $\epsilon$ is, the smaller $\delta$ will need to be.  Therefore, you should not have the line "Let $\delta \leq 1$:" in your proof.  Rather, you should have

"Let $\delta = $ (some expression involving $\epsilon$)"

So maybe try something like "Let $\delta = \sqrt{\epsilon}$" and see what you can come up with.  (Note: this is not a correct value of $\delta$, it's just an example to get you going.)


How do I solve this matrix equation $X^3=\begin{pmatrix}19&30\\ -45&-71\end{pmatrix}$? I have to solve for the $2\times 2$ matrix $X$ the following equation

$$X^3=\begin{pmatrix}19&30\\ -45&-71\end{pmatrix}$$

To be honest I have no idea how to start.

I know how to solve equations like
$$A \cdot X=C => X=A^{-1} \cdot C$$

$$X \cdot B=C => X=C \cdot B^{-1}$$

$$A \cdot X \cdot B=C => X=A^{-1} \cdot C \cdot B^{-1}$$

but I don't know how I could apply it in here

The given matrix $m$ has determinant $1$ and trace = $-52$. Assuming $X$ is a real matrix, $det(X)=1$. Let $t= tr(X)$. Now $X$ satisfies the equation $x^2-tx+1=0$.
From this it is easy to see that $m=X^3=(t^2-1)X-t*I_2$ , where $I_2$ is the identity matrix. So $(t^2-1)X=m+t*I_2$. Taking trace of both sides, we see that $t$ is a root of the cubic equation $t^3-3t=-52$ whose only real solution is $t=-4$. Therefore $15X=m-4I_2$. This gives

$$X=\begin{pmatrix}1&2\\ -3&-5\end{pmatrix}$$

I'm not providing here a method to provide the result directly, but I would decompose the resulting matrix into simple geometrical transformation : dilatation, rotation ..., then I would divide the dilatation coefficient by 3, the angle by 3 etc ...

That would be my approach.


Proof the quotient and remainder exists in $\mathbb{Z}^+$. If $a$ and $b$ are positive integers, prove that there exists an integer $q$ called
the quotient and an integer $r$ called the remainder such that $a = q b + r$ and
$0 \leq r < b$.

I've seen proofs for this problem and I have been given a solution using induction but I've half thought of a proof that seems much simpler but I'm not sure if it is correct.

There are 3 cases, case 1 and 2 are simple so I'll only half prove case 3.

Case 1: $a=b$

Case 2: $a < b$

Case 3: $b<a$. Then there exists an integer $q=gcd(a,b)$. This step worries me since we use the above theorem to calculate the $gcd$ but I don't think the above theorem says anything about its existence. So even without the theorem above we can be sure of its existence. Therefore we can let $r=a-gcd(a,b)b$. So if I could prove $0\leq r < b$ would the proof be correct?

Hint: prove the statement by induction on $a$.

The case $a=0$ is easy. Suppose the statement holds for $a$ and try it for $a+1$. By inductive hypothesis
$$
a=bq+r
$$
with $0\le r<b$. Then $a+1=bq+(r+1)$, but $r+1$ might fail to be less than $b$: when?

Alternately, given $a,b > 0$, you can consider the set $S = \{a - bq : q \in \mathbb{Z}, a - bq \geq 0\}$. This set is nonempty, and thus has a least element by the Well-Ordering Principle. Call this least element $r$.

It is now very easy to prove that $r$ satisfies the desired inequality (otherwise, it wouldn't be the least element in $S$) and this gives you your unique $q,r$. $\diamondsuit$


Finding an intersection between 2 lines I got 2 (2d) lines, one of them is defined by 2 points $(x_1, y_1)$ $(x_2, y_2)$, and the other one is represented by the normal $y = ax + b$ equation.

Now, I want to calculate the $(x, y)$ of the point of intersection (given that $x_1$ is not equal to $x_2$)

What is the formula to this point's $x$ and $y$?

Thanks.

Hints:

Any point on the first line has coordinates $(tx_1+(1-t)x_2, ty_1+(1-t)y_2)$ for some $t$.

A particular $t$ will make this lie on the second line

Here’s a way to do this using homogeneous coordinates: The line through two points is given by the cross product of the points, and the intersection of two lines is also given by their cross product. This method works quite well when you have concrete values for the points/coefficients, but can get messy if you try to derive general formulas.

Given two points $(x_1,y_1)$ and $(x_2,y_2)$, the line through them is $$(x_1,y_1,1)\times(x_2,y_2,1)=(y_1-y_2,x_2-x_1,x_1y_2-x_2y_1).$$ For the equation $y=mx+b$, we get $$(0,b,1)\times\left(-b,0,m\right)=\left(bm,-b,b^2\right).$$ After computing the cross product of these two results, you get something that doesn’t have a $1$ as its last coordinate, so you have to divide by that value, finally resulting in $$\begin{align}
x &= {(x_1y_2-x_2y_1)+b(x_2-x_1) \over (y_2-y_1)-m(x_2-x_1)}\\
y &= {m(x_1y_2-x_2y_1)+b(y_2-y_1) \over (y_2-y_1)-m(x_2-x_1)}
\end{align}$$ for the intersection point. This is exactly the result you get by solving the system of equations for the two lines.

As a concrete example, take the line given by $y=-3x+5$ and the line through the points $(-3,1)$ and $(1,3)$: $$
(-3,1,1)\times(1,3,1) = (-2,4,-10) \\
(0,5,1)\times(-5,0,-3)=(-15,-5,25) \\
(-2,4,-10)\times(-15,-5,25) = (50,200,70),
$$ so the lines intersect at $\left(\frac57,\frac{20}7\right)$, which you can verify by solving the system of equations for these lines.


Why does this innovative method of subtraction from a third grader always work? My daughter is in year $3$ and she is now working on subtraction up to $1000.$ She came up with a way of solving her simple sums that we (her parents) and her teachers can't understand.

Here is an example: $61-17$

Instead of borrowing, making it $50+11-17,$ and then doing what she was told in school $11-7=4,$ $50-10=40 \Longrightarrow 40+4=44,$ she does the following:

Units of the subtrahend minus units of the minuend $=7-1=6$
Then tens of the minuend minus tens of the subtrahend $=60-10=50$
Finally she subtracts the first result from the second $=50-6=44$

As it is against the first rule children learn in school regarding subtraction (subtrahend minus minuend, as they cannot invert the numbers in subtraction as they can in addition), how is it possible that this method always works? I have a medical background and am baffled with this…

Could someone explain it to me please? Her teachers are not keen on accepting this way when it comes to marking her exams.

So she is doing
\begin{align*}
61-17=(60+1)-(10+7)&=(60-10)-(7-1)\\
 & = 50-6\\
& =44
\end{align*}
She manage to have positive results on each power of ten group up to a multiplication by $\pm 1$ and sums at the end the pieces ; this is kind of smart :)

Conclusion : If she is comfortable with this system, let her do...

Your daughter is clever. She's breaking the numbers into smaller and easily subtractable numbers:
$$61-17= (50+11)-(10+7)$$

Rearranging the order:$$(50-10)+(11-7)=44$$


Is all of mathematics based on logic? I do not understand the discipline in mathematics that is called "mathematical logic". For me, all of mathematics is based on logic and that is what makes it the exact science. Every theorem or lemma or result should be based on correct reasoning, and there should be some true logic behind it. Or am I mistaken?

Are there some results that cannot be treated with logic? What is the difference between the discipline of "mathematical logic" and the logic used in mathematics? Are there known results in mathematics that are not based on logic?

For example:

  Find $x$ in $(1)$: $$(1)\quad x+1=0.$$
  The logic is to find $x=-1$ by subtracting $-1$ and no one else can say otherwise.

For me, this is mathematics. Even though a lot of theorems are hard to understand (at least for me), there must be some kind of logic behind them. Am I right?

Generally speaking, mathematical logic investigates the nature and limitations of logical systems. For example, consider the question: "can every true statement be proved?" This is a problem about logic. When we study geometry, algebra  and other branches of mathematics we are interested in using logical thinking, but the purpose is to understand something about shapes or symmetries.

I think the same way as you think as if we see, mathematics all trick until now has been concerned with "=" , ">" , "<" , etc have a meaning and that is for = it means that what is written on left side is exact same as what is on the right side and so on. So we'll Yes actually, mathematics is completely logical as you can not reach a statement that is true by just moving numbers here and there but you can get multiple forms of a true statement by displacing numbers and that's what math is. Understanding multiple forms of the same truth. For example :
If we write :

X-2 = 0

Then it means that any number in which we subtract 2 happens to be 0 then exactly we can also say that

X = 2

Because 2 is such a number .

And both of these statements refer to the same number , 2 , hence we can see clearly that by twisting numbers, we just see the same truth other way.


Isomorphisms and Automorphisms An automorphism is a mapping of a graphs nodes onto it's own nodes. Whereas an isomorphism is the mapping of a graphs nodes onto another graphs nodes.

Doesn't this mean the are fundamentally the same thing, it's just a matter of what the nodes are labelled (for example in one graph the nodes may numbered sequentially, but in the other graph they are labelled alphabetically)?

So will the number of automorphisms be the same as the number of isomorphisms?

If two graphs $X$ and $Y$ are isomorphic (that is, there exists an isomorphism between them) then it is true that the number of automorphisms of $X$ is equal to the number of isomorphisms from $X$ to $Y$.

Let's recall the definition of an isomorphism, as I don't know what notation you're used to. Let $X$ and $Y$ be graphs with respective vertex sets $V(X)$ and $V(Y)$. An isomorphism from $X$ to $Y$ is a bijection $f:V(X)\to V(Y)$ such that each pair of vertices $u,v\in V(X)$ are adjacent in $X$ if and only if $f(u),f(v)$ are adjacent in $Y$.

Let $\operatorname{Iso}(X,Y)$ denote the set of all isomorphisms from $X$ to $Y$. (This is the empty set if $X$ and $Y$ are not isomorphic, by definition.)

Let $\operatorname{Aut}(X)$ denote the set of automorphisms of $X$. That is, $\operatorname{Aut}(X) = \operatorname{Iso}(X,X)$.

Suppose $X$ and $Y$ are isomorphic. Then by definition there exists some $f\in \operatorname{Iso}(X,Y)$. We can then use this element $f$ to construct a bijection
$$\Phi:\operatorname{Aut}(X)\to\operatorname{Iso}(X,Y)$$ by definining $\Phi(\alpha) = f\circ\alpha$ for each automorphism $\alpha$ of $X$. (As an exercise, verify that $\Phi(\alpha)$ is indeed an isomorphism.)

So if $X$ and $Y$ are isomorphic, then indeed the number of automorphisms of $X$ (or of $Y$, for that matter) is the same as the number of isomorphisms from $X$ to $Y$. However, note that the bijection we constructed above depended on the choice of isomorphism $f$. There is no canonical bijection $\operatorname{Aut}(X)\to\operatorname{Iso}(X,Y)$, unless of course each of these sets has just one element.

Your understanding matches mine.  The distinction is somewhat academic.  I am not aware of any theorems that hinge on the distinction.

Does the identity count as an automorpism?  If not, there is probably one more isomorphism than automorphism.  Also, intuition breaks down when you are talking about infinite sets.


Least squares estimator of mu The question is:

  Assuming that $y_i = \mu + \epsilon_i $,$i = 1,\ldots,n$ with independent and identically distributed errors $\epsilon_i$ such that $E[\epsilon_i] = 0$ and $Var[\epsilon_i] = \sigma^2$, find the least squares estimator of $\mu$. Find its variance.

I'm not sure how to go about doing this.

I know that the least squares bit means that I minimize the sum of the errors, and so I would have to use the formula:

$$\sum_i (y_i - \mu)^2$$

and then differentiate (wrt to $\mu$?) and then let it equal 0.

Is that correct?

Once I've done this, I would I calculate its $E[\mu]$, because I don't have any definition for $\mu$. Or is $\mu = \beta_0 + \beta_1 \cdot x_i$? If it is, then isn't the estimator the same?

Some hints, but not quite the full answer:

There is a difference between a parameter $\mu$ and an estimator of that parameter.  So if we call the estimator $\hat{\mu}$ then you want to minimise $$\sum_i (y_i - \hat{\mu})^2$$ which is $$\sum_i y_i^2  - \sum_i 2 y_i \hat{\mu} +\sum_i \hat{\mu} ^2$$ and (as you suggest) this will be when its derivative with respect to $\hat{\mu}$ is zero.  Strictly speaking you should check this is a minimum, but since the derivative is monotone increasing that is obvious.

Since $y_i = \mu + \epsilon_i$, you know $E[y_i] = E[\mu] + E[\epsilon_i]$, so it will be easy to find $E[\hat{\mu}]$.

As for $Var(\hat{\mu})$, you again have to multiply out a square, looking at $$E\left[\left(\hat{\mu}-E[\hat{\mu}]\right)^2\right].$$  You might want to use the fact that  $y_i^2 = \mu^2 + 2 \mu \epsilon_i +\epsilon_i^2$ implies $E[y_i^2] = \mu^2  + \sigma^2$.

$$
\begin{align}
\sum_{i=1}^n (y_i-\mu)^2 & = \sum_{i=1}^n ((y_i - \bar y) + (\bar y -\mu))^2 \\[8pt]
& = \sum_{i=1}^n \left((y_i-\bar y)^2 + 2(y_i-\bar y)(\bar y-\mu) + (\bar y -\mu)^2 \right) \\[8pt]
& = \left(\sum_{i=1}^n (y_i-\bar y)^2 \right) + \left(2(\bar y -\mu)\sum_{i=1}^n (y_i-\bar y)\right) + n(\bar y -\mu)^2 \\[8pt]
& = (\text{something not depending on }\mu) + (0) + n(\bar y -\mu)^2.
\end{align}
$$
(The second sum is $0$: if you subtract the average of the numbers in a list from each of the numbers, then the sum of the resulting numbers is $0$.)  So you're looking for the value of $\mu$ that minimizes $(\bar y - \mu)^2$, and that's obviously $\mu=\bar y$.

None of this relies on the fact that the errors are normally distributed.

Next question: variance of $\bar y$:
$$
\operatorname{var}(\bar y) = \operatorname{var}\left(\frac{y_1+\cdots+y_n}{n}\right) = \frac{1}{n^2}\operatorname{var}(y_1+\cdots+y_n)
$$
$$
= \frac{1}{n}(\sigma^2 + \cdots+\sigma^2) = \frac{1}{n^2}\cdot n\sigma^2 = \frac{\sigma^2}{n}.
$$
Here we relied on independence to conclude that the sum of the variances equals the variance of the sum, but we still didn't need the fact that the errors are normally distributed.


Why is this rotation "incorrect"? I've been trying to use the following formula for the rotation of a point around the origin:

$$
\begin{bmatrix}
x' \\ y'
\end{bmatrix} =
\begin{bmatrix}
\cos{\theta} & -\sin{\theta} \\
\sin{\theta} & \cos{\theta}
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
$$

Now, I'm trying to apply this formula to the coordinate $(5,3)$ and rotating it $90$ degrees clockwise, and I ended up with the following result:

$$
\begin{bmatrix}
x' \\ y'
\end{bmatrix} =
\begin{bmatrix}
\cos{90} & -\sin{90} \\
\sin{90} & \cos{90}
\end{bmatrix}
\begin{bmatrix}
5 \\ 3
\end{bmatrix}
\\
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
5 \\ 3
\end{bmatrix}
\\
=
\begin{bmatrix}
0(5) -1(3) \\
1(5) + 0(3)
\end{bmatrix} \\
=
\begin{bmatrix}
-3 \\ 5
\end{bmatrix}
$$

I ended up with the rotated coordinates $(-3,5)$. Unfortunately, this was wrong. Can anyone tell me what I'm doing wrong, and how I can do it correctly? I tried this method on other coordinate points, and all of them were wrong as well.

It seems right to me. Ah, maybe you're thinking of a clockwise rotation. That matrix gives a counterclockwise rotation through an angle $\theta$.

It is correct. What makes one think that it is wrong? You mean the convention that CCW is positive?

EDIT1:

What I meant is with Clockwise rotation I obtain $(-3,5)$ by matrix multiplication and Counterclockwise rotation gives  $(3,-5)$ for a diametrically opposite point. So only association with sign convention of rotation could be the source of error.


What does $x^n=y^n$ imply for $x=\pm y$ in a group? I am wondering if the following statement is true:

  If $G$ is a group, then for all $x,y \in G, x^n=y^n$ implies $x=y$ or $x=-y$.

I have tried using an induction argument and it might work by using casework on whether $n$ is odd or even, but I'm not sure.

EDIT: If $G$ is a group, then if $x,y \in G$ and $ x^n=y^n$, then what information does this give us about $x$ and $y$?

No. If the group has order $n$ then $x^n = e$ (the group identity) for every $x$.

If $x^n = y^n$, then $x^n y^{-n} = 1$, and hence $(xy^{-1})^n = 1$. In other words, $x$ times the inverse of $y$ is idempotent.


proof that the sequence satisfies weak law of large numbers strong textHow to check if this sequence satisfies the weak law of large numbers?
$$P(X_n=1)= {{1}\over{3^n}} ,P(X_n=-1)= {{1}\over{3^n}}, P(X_n=0)= {1 -{{2}\over{3^n}}}  $$
I calculated that:
$$E(X_n)=0 $$
so according to the definition of this law: $${\sum_{n=1}^{N}{(X_n-0)}}\over{N} $$and from this one we can deduce that this expression converges to 0 so the sequence satisfies the weak law of large numbers. Is that correct?

$\sum P(X_n \neq 0)=\sum \frac 2 {3^{n}}<\infty$. By Borel Cantelli Lemma $P(\lim \sup (X_n \neq 0))=0$. Hence, with probaility $1$, $X_n=0$ for all $n$ suffciently large. But this implies that the averages $\frac 1n \sum\limits_{k=1}^{n} X_k$ tends to $0$ with probability $1$. Thus, we actually have strong law for this sequence.

Simply using the definition of WLLN
$$\lim\limits_{n\to +\infty}\mathbb{P}[|X_n|>\epsilon]=\frac{2}{3^n}=0$$
this means that
$$X_n\xrightarrow{\mathcal{P}}0$$
This if $\epsilon$ is chosen $|\epsilon|\leq1$. If it is chosen >1 the probability is exactly zero. Anyway the sequence converges.


Flatness of torus and surfaces of higher genus For the very first sight it may be surprise that the ordinary torus $S^1 \times S^1$ is flat: one argument to see this is the following. One can imagine a torus as a square with opposite sides identified and the square is obviously flat. However there are surfaces with higher genus (higher numbers of "holes") which also can be represented as polygons (4$g$ polygon for a surface of genus $g$) with sides indetified. However higher genus surfaces are negatively curved. So my question is

  Why higher genus surfaces are negatively curved while they can be represented as flat polygons with sides identified?

If you try to turn those polygon identifications into flat metrics you run into a problem at the vertices: the angles don't add up. This issue can only be addressed for the square.

Theorem Benzecri. A closed surface is flat if and only it is euler number is zero.

Goldman, W. Two papers which change my life: Milnor seminal work on flat manifolds and bundles.

http://arxiv.org/pdf/1108.0216.pdf

p.4 which has a good outline of the proof.

To be more precise, a surface of genus $g>1$ than cannot be endowed with a differentiable metric whose curvature vanishes identically, since it cannot be endowed with a Koszul derivative which is flat, here flat means that the curvature and the torsion form vanish identically, since we know that the torsion form of the Levi-Civita connection vanishes, a differentiable manifold endowed with a flat differentiable metric has a flat structure; i.e it is an affine manifold. It is a well-known theorem that a  Riemannian closed flat manifolds are finitely covered by the torus $T^n$, their fundamental groups are crystallographic groups, so again this theorem also implies that the only oriented surface endowed with a flat metric is the torus.

Remark that in the case of dimension 2, Milnor in his paper On the existence of a connection with curvature zero, Comm. Math.
Helv. 32 (1958), 215-223
Milnor has generalized the work of Benzecri, by computing an equality which represents an obstruction for a bundle defined on a surface to be flat.


How to treat Dirac delta function of two variable? We can treat one variable delta function as

$$\delta(f(x)) = \sum_i\frac{1}{|\frac{df}{dx}|_{x=x_i}} \delta(x-x_i).$$

Then how do we treat two variable delta function, such as $\delta(f(x,y))$?

for example, how calculate $\int\int \delta(x-y)$ ?

I first thought using  $\int f(x)\delta(x-y)dx = f(y)$

$$\int\int \delta(x-y)dxdy = \int\int1 * \delta(x-y)dx dy = \int dy.$$

but this is nonsense, since we can also think it as $\int\int \delta(x-y)dxdy = \int dx$

The simplest way to start is to think about how the delta function is defined in 1D,
$$\int_{-\infty}^{\infty} f(x)\delta(x-a)dx = f(a)$$
which you correctly identified above.  This is saying that some function $f(x)$ multiplied by the a shifted delta function $\delta(x-a)$ and integrated everywhere returns the same function evaluated at the value given by the shift, $f(a)$.  We want the same sort of definition to apply when we integrate everywhere in 2D, so we have to have a 2-variable function $f(x,y)$ and it has to have a 2-variable delta function $\delta(x,y)$.  (Notice that $\delta(x-y)\neq \delta(x,y)$ in general.) Applying it to the 2D case in this way we should give
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\delta(x-a,y-b) \ dx dy = f(a,b)$$

which should be true whether you integrate either $x$ or $y$ first as shown below.

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\delta(x-a,y-b) \ dx dy = \int_{-\infty}^{\infty} f(a,y)\delta(y-b) \ dy  = f(a,b)$$

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\delta(x-a,y-b) \ dx dy = \int_{-\infty}^{\infty} f(x,b)\delta(x-a) \ dx  = f(a,b)$$

These are easiest to solve when $f(x,y)$ is a separable function like $f(x,y)=g(x) h(y)$, but some harder functions are still quite doable.

I think something is still wrong. Because according to your answer in the integration limits $(-\infty,+\infty)$ it doesn't matter which is the function inside the $\delta$, the answer is always the same.


How to decide limits of integral in Riemann's sum I need to calculate
$$\lim_{n \to \infty}\frac{1}{\sqrt{n^2 -0^2}}+\frac{1}{ \sqrt{n^2 -1^2}} + \frac{1}{ \sqrt{n^2 -+2^2}}+.....+ \frac{1}{ \sqrt{n^2 -(n-1)^2} }$$
It can be written as
 $$ \frac{1}{n}\sum_{r=0}^{n-1}{ \frac{1}{\sqrt{1-(r^2/n^2)}} }$$ and in integral form as  $$ \int_?^? \frac{1}{ \sqrt{1-(x^2) }}dx
$$ How to decide limits?

The evaluation of the limit of the sum uses lower Riemann sum where each $x_i = \dfrac{i}{n}, i = \overline{0,n-1}$. Thus: $\displaystyle \lim_{n\to \infty} S_n = \displaystyle \int_{0}^1 \dfrac{1}{\sqrt{1-x^2}}dx = \arcsin x|_{x=0}^1 = \dfrac{\pi}{2}$


$\frac{1}{n}/\frac{1}{n^2}$ understanding The limit of $\frac{1}{n}/\frac{1}{n^2}$ when $n \rightarrow +\infty$ is $+\infty$.

By calculation this is an obvious result, yet I don't understand why this is true.
$1/n$ will never reached the value $0$ because the équation $1/n=0$ hasn't any solutions. It's exactly the same for $1/n^2$.
Moreover the difference $1/n-1/n^2=0$ when $n \rightarrow +\infty$.
So why $\frac{1}{n}/\frac{1}{n^2} \ne 1$ when $n\rightarrow +\infty$.

It's all about the speed of convergence. Clearly $\lim_{n\rightarrow \infty}\frac{1}{n}=0$ and $\lim_{n\rightarrow \infty}\frac{1}{n^2}=0$, but $\frac{1}{n^2}$ tends to zero faster than $\frac{1}{n}$.

Thus when we consider the ratio $\frac{1}{n}/\frac{1}{n^2}$ the denominator will then to zero faster than the nominator. Indeed, just pick a large value for $n$ and compare: $n=10$ yields $\frac{1}{n}=\frac{1}{10}$ and $\frac{1}{n^2}=\frac{1}{100}$.

In many such cases you can use l'Hopitals rule, and if you think about it, all this rule does is compare the speed of convergences of the nominator to that of the denominator.

Adding to EdMo38, take examples of n getting smaller and smaller, eg. n=1 : lim=1, n=0.1 : lim=10, n=0.01 : lim=100, n=0.001 : lim=1000, etc.


Trig substitution; why can we ignore the absolute value? If we have to integrate $$f(x)=\frac{x}{\sqrt{1-x^2}}$$ and we substitute $x=\sin \theta$ then we eventually have to take the square root of $\cos^2x$ which is equal to $|\cos x|$.  But in my textbook and class lectures, we simply remove the absolute value sign and replace it with $\cos x$.

Why do we do this? I don't understand how we can do this.

This is a good question. It is a very easy mistake to assume that $\sqrt{u^2} = u$ (whether the $u$ in question is $\cos^2 \theta$ or something else).
Of course the equation is true if $u$ is positive, but then we must have a
good reason to say $u$ is positive.

So we started with the integral
$$\int \frac{x}{\sqrt{1-x^2}}\, dx,$$
and we decide to try the trig substitution $x = \sin\theta$.
This gives us
$$ \int \frac{\sin \theta}{\sqrt{\cos^2\theta}}\, \cos\theta \, d\theta. $$
But before we go any further, let's consider what kind of values $x$
might have and how our substitution might relate values of $x$ with
values of $\theta$ in this new integral.

I'm assuming you are not doing complex analysis here, that is,
only real numbers are allowed, not complex numbers.
So $\sqrt{1-x^2}$ is defined only for $-1 \leq x \leq 1,$
because if $|x| > 1$ then $1 - x^2$ is negative.

Now if we allow $x$ to vary within the range $-1 \leq x \leq 1,$
what is the correct range of $\theta$?
We want a one-to-one relationship between the
values of $x$ in $-1 \leq x \leq 1$
and our chosen range of values of $\theta$.
For example, $x = -1$ should correspond to exactly one value of $\theta$
within the range of values we allow for $\theta$.

Clearly if $x = -1$ and $\sin\theta = x$, we must have
$\theta = -\frac\pi2 + 2n\pi$ where $n$ is an integer.
Suppose we choose $n = 0$, so $\theta = -\frac\pi2$.
If we increase $\theta$ toward $\frac\pi2$, then $x$ increases toward $1$.
So let's let $\theta$ vary within the range
$-\frac\pi2 \leq \theta \leq \frac\pi2.$

But if $-\frac\pi2 \leq \theta \leq \frac\pi2$,
then $\cos\theta \geq 0$, and the statement
that $\sqrt{\cos^2\theta} = \cos\theta$ is justified.
Everything proceeds smoothly from there in the way you have already
(presumably) seen in your textbook and lectures.

How can we justify the assumption that $-\frac\pi2 \leq \theta \leq \frac\pi2$?
It isn't so much an assumption as a definition on our part.
When we substitute one variable for another variable,
we get to say what range of values of the new variable we want to relate to
the range of values the old variable might have, provided that these
are consistent with the function we chose to use to relate the two variables.

Now just out of curiosity, suppose we had done the substitution just
a little bit differently?
We could have tried letting $\theta$ vary within the interval
$\frac\pi2 \leq \theta \leq \frac{3\pi}2$.
These values of $\theta$ also give us
all the values of $x$ we might need.
For these values of $\theta$, $\cos \theta \leq 0$,
so $\sqrt{\cos^2\theta} = -\cos\theta$.
Our integral then becomes
$$\int \frac{\sin \theta}{\sqrt{\cos^2\theta}}\, \cos\theta \, d\theta
= \int -\sin \theta \, d\theta = \cos\theta + C.$$
Now we want to reverse the substitution in order to get the
answer to our original integral in terms of $x$.
Since $\cos\theta \leq 0$, the only substitution consistent with
our original substitution is
$$\cos\theta = -\sqrt{1-x^2}.$$
We conclude that
$$\int \frac{x}{\sqrt{1-x^2}}\, dx = -\sqrt{1-x^2} + C,$$
which is the same result we got when we let
$-\frac\pi2 \leq \theta \leq \frac\pi2.$

We would also get the same result if we tried
$-\frac{3\pi}2 \leq \theta \leq -\frac\pi2$
or $\frac{7\pi}2 \leq \theta \leq \frac{9\pi}2$.
That's good; the solution to the integral is not a mere accident of
exactly which substitution we chose.

Although in most cases positive sign is adequate to see how the integrand magnitude varies, we in fact neither ignore nor remove the double sign $ \pm $ in front of  radical sign. Either sign is implicitly understood. The above integral is strictly

$$ \pm \sqrt{ 1-x^2} + C $$

which you may like to call minus or plus to correspond with + or - of original integrand .

If you integrate $ \frac {1}{\sqrt {x^2}} $ in such a situation you get (up to constant of integration)
 $ \; \log x $ and $ \dfrac {1}{\log x}. $

EDIT 1:

Without reference to $\theta$ substitution two curves can be drawn one above x-axis and another below x-axis. For the former, area under curve is > 0, and for latter, < 0.


Prove that $f'(0)=L$. Let $f$ be continuous at $0$. Suppose $\displaystyle \lim _{x\rightarrow 0} \frac{f(2x)-f(x)}{x} =L$. Prove that $f'(0)=L$.

My Work:

$\displaystyle \Bigg|\frac{f(x)-f(0)}{x}-L\Bigg|=\Bigg|\frac{f(x)-f(2x)+f(2x)-f(0)}{x}-L\Bigg|$

$\displaystyle \leq \Bigg|\frac{f(x)-f(2x)}{x}-L\Bigg|+\Bigg|\frac{f(2x)-f(0)}{x}\Bigg|$. Since $f$ is continuous at $0, |f(2x)-f(0)|$ goes to $0$ as $x$ goes to $0$. Now I want to get rid of $|x|$ of the second term. How can I do it? Can somebody please help me?

Let $\epsilon > 0$ be fixed. Since $\displaystyle\lim_{x \to 0}\dfrac{f(2x)-f(x)}{x} = L$, there exists a $\delta > 0$ such that $(L-\epsilon)y \le f(2y)-f(y) \le (L+\epsilon)y$ for all $0 < |y| < \delta$.

Since $f(x) - f\left(\dfrac{x}{2^n}\right) = \displaystyle\sum_{k = 1}^{n}f\left(\dfrac{x}{2^{k-1}}\right)-f\left(\dfrac{x}{2^k}\right)$, by applying the above bounds for $y = \dfrac{x}{2^k}$, we have:

$\displaystyle\sum_{k = 1}^{n}(L-\epsilon)\dfrac{x}{2^k} \le \displaystyle\sum_{k = 1}^{n}f\left(\dfrac{x}{2^{k-1}}\right)-f\left(\dfrac{x}{2^k}\right) \le \displaystyle\sum_{k = 1}^{n}(L+\epsilon)\dfrac{x}{2^k}$

$(L-\epsilon)\left(1-\dfrac{1}{2^n}\right)x \le f(x) - f\left(\dfrac{x}{2^n}\right) \le (L+\epsilon)\left(1-\dfrac{1}{2^n}\right)x$

for all $0 < |x| < \delta$.

Now, what happens if you take the limit as $n \to \infty$?

   In case you didn't see it, I'll work it out. By taking the limit as $n \to \infty$, we get $(L-\epsilon)x \le f(x)-f(0) \le (L+\epsilon)x$ for all $0 < |x| \le \delta$. Hence $f'(0) := \displaystyle\lim_{x \to 0}\dfrac{f(x)-f(0)}{x} = L$, as desired.


angle between minute and hours hands at $7:35$ P.M I want to know what is the angle between minute and hours hands at $7:35$ P.M

I have no Idea. could any one tell me?

In $60$ minutes the hour hand moves 30 degree so in 35 minutes it would have moved

$$\frac{30}{60} \times 35 $$
$$=17.5^\circ$$

Hour hand at $17.5^\circ$ from 7 . minute hand exactly points at 7 , so angle between minute hand and hour hand = $17.5^\circ$

I would point you to an online tool

  clock angle calculator


Sum of four squares not a prime Let $ a, b, c, d $ be natural numbers such that $ ab=cd $. Prove that $ a^2+b^2+c^2+d^2 $ is not a prime.

I am clueless on this one. I tried contradiction, but didn't get anywhere.

Can you help?

Edit: I understand natural numbers to be strictly positive, excluding $0 $.

Since $d=\frac{ab}{c}$, we obtain

$$
a^2+b^2+c^2+d^2=\frac{(a^2+c^2)(b^2+c^2)}{c^2}=\left(\left(\frac{a}{a'}\right)^2+b'^2\right)\left(\left(\frac{b}{b'}\right)^2+a'^2\right).
$$
where $c=a'b'$ such that $a' \mid a$ and $b' \mid b$.

A rather odd way to perhaps get there..

ab = cd

no prime > 2 is even.

sum of any even number of odd numbers is even.

square of odd numbers is odd and of even numbers is even.

so either 1 or 3 of a, b, c, d must be odd if we are possibly to get a prime.

It cannot be that only one of them is odd or ab != cd.  So we need 3 of them odd.

If a,b,c are odd then ab/c = d which is a contradiction.  QED
:)


constructing nondegenerate polygon from n sides by making cuts We are given the length of n sticks and we need to make a polygon using all the n sticks. If it is impossible to construct a nondegenerate polygon from the given n sticks, we can cut one or more sticks into two parts (length of cut part need not be integer).
How many cuts do we need to make for given n values ?

 3(n=3)

 1 2 3

We cannot construct a triangle so we try to make a square and we cut stick 3 into two parts of length 1 and 2.
thus answer is 1.
How can we solve this for any n numbers given to us ?

You always need at most one cut. The sticks can be used to form a polygon as long as there's no stick that's as long as all the others put together. If originally there is such a stick, just cut it in half.

I just got the correct solution in some discussion forum of another platform

For a polygon  to be non degenerate polygon,each side of the polygon must satisfy this condition   sum=a$1$ +a$2$+........+a$n$
and a$i$<1/2(sum) wher  1<=i<=n
.Here a$1$,a$2$..... ,a$n$ are the lengths of sticks given to us and every time the above condition a$i$<1/2(sum) is violated for any i,increase the count by 1.


reduction of order on nonhomogenious ODE Use reduction of order to find the homogeneous and associated particular solution. $y''-3y'+2y=5e^{3x}$ given $y_1=e^x$

For homogeneous:
$y_2=y_1 \int \frac{e^{-\int P(x) dx}}{y_1^2}dx=e^x \int \frac{e^3x}{e^2x} dx=e^{2x}$
Now what do I do?

for a particular solution to the non homogeneous problem $$y''-3y'+2y=5e^{3x}, $$ try $y = Ae^{3x}, y'= 3Ae^{3x}, y''= 9Ae^{3x}.$  you get $$A\left(9-9+6\right)e^{3x} = 5e^{3x}  \to A = \frac 56.$$  so a particular solution is $$ y= \frac 56 e^{3x}$$


Using the epsilon-delta to prove $\lim_{x \rightarrow 1} \frac{x^2}{x+2} = \frac{1}{3} $ I'm trying to use the epsilon-delta definition to prove
$\lim_{x \rightarrow 1} \frac{x^2}{x+2} = \frac{1}{3} $.
Could someone please check if i got this right?

$ \forall  \epsilon >0, \exists  \delta >0$ s.t. if $0< | x-1 | < \delta$ , then  $|  \frac{x^2}{x+2} - \frac{1}{3}  | < \epsilon $.

$|  \frac{3x^2 - x - 2}{3(x+2)} | =  |  \frac{(3x+2)(x-1)}{3x+6}|< \epsilon $

Let $\delta=\frac{1}{3}$. Then, $|x-1|<\frac{1}{3} \Rightarrow |x|<\frac{4}{3}, |3x+2|<6$..

Then, $|  \frac{(3x+2)(x-1)}{3x+6}|<|  \frac{6(x-1)}{6}|<\epsilon $

$|  x-1|<\epsilon $, so  $\delta$ =min{1/3,  $\epsilon$ }

i think this feels correct, but to be sure... i thank in advance!

Your calculations are correct, but that alone does not make a correct solution. It should be clear what the connection between the different lines is. The first line is what you want to prove, so say that.

  We have to show: $ \forall  \epsilon >0, \exists  \delta >0$ s.t. if $0< | x-1 | < \delta$ , then  $|  \frac{x^2}{x+2} - \frac{1}{3}  | < \epsilon $.

I personally would have used more English words (“We have to show that for each $\epsilon>0$ there is a $\delta$ such that...”), but that is a matter of taste.

If I was grading this as homework, then regarding your next line I would ask: Where does the $\epsilon$ come from? So far you have not introduced an epsilon. Of course, I understand that you want that inequality to hold, but then you have to say that. I would prefer to just state what you are actually doing, namely a quick calculation:

  We have $|  \frac{x^2}{x+2} - \frac{1}{3}  | = |  \frac{3x^2 - x - 2}{3(x+2)} | =  |  \frac{(3x+2)(x-1)}{3x+6}|$.

You begin the next line with “let $\delta=\frac13$”. This is not good, because actually you want to make a statement that holds whenever $\delta\le\frac13$. But so far there is no need to introduce $\delta$, just state what you are claiming.

  We note that for $|x-1|<\frac{1}{3}$ we have $0<|x|<\frac{4}{3}$ and hence $|3x+2|<6$ and $|  \frac{(3x+2)(x-1)}{3x+6}|<|  \frac{6(x-1)}{6}|=\lvert x-1\rvert$.

I have added $0<\lvert x\rvert$, because you are using this for $\lvert 3x+6\rvert>6$.

Now we can wrap things up. What I do not like about your formulation is the “so  $\delta =\mathrm{min}\{1/3,  \epsilon \}$”. Since $\delta$ has not yet been defined, you cannot claim that it equals something else. What you mean is that setting $\delta$ to this value will work. So let's just do this.

  Now let $\epsilon>0$ be given. We set $\delta =\mathrm{min}\{1/3,  \epsilon \}$. Then for all $x$ with $\lvert x-1\rvert<\delta$ we have
  \begin{align*}
\left\lvert\frac{x^2}{x+2} - \frac{1}{3}\right\rvert
&<\lvert x-1\rvert&&\text{because $|x-1|<\frac{1}{3}$}
\\&<\delta\le\epsilon.
\end{align*}
  This is what we had to show.

Now I am sure that you can still improve some of these formulations. The important thing is that you really ask yourself what your claims are and that you express these clearly.

There is a slight mistake you made 3x+2 < 6 and 3x+6<6

All in all this is a correct proof except that slight mistake.

But if I were you I would use some theorems about limits to handle this much more easier.Here is how I would have done that.

It is true that if $$\lim_{x \to a} f(x) = l$$ $$and$$ $$\lim_{x \to a}g(x) = m$$ then following are also true $$\lim_{x \to a } f(x)g(x) = lm$$ and $$\lim_{x \to a}(f+g)(x) = l + m $$

Using these facts we can then transform your limit into something like this $$\frac{\lim_{x \to 1} x^2}{\lim_{x \to 1} x + \lim_{x \to 1}2}$$

Since 2 would be constant function its limit will be 2.
Since x approaches 1 near one so does sqaure of x

Thus we can conclude that:$$\frac{\lim_{x \to 1} x^2}{\lim_{x \to 1} x + \lim_{x \to 1}2} = \frac{1}{1+2}=\frac13$$


How do we prove this kind of inequality? Let $1\le p<2$, we want to show that for any $a,b,c,d\in\Bbb R$ the following holds:
$$
|a-b|^p + |b-c|^p + |c-d|^p + |d-a|^p \ge |a-c|^p + |b-d|^p
$$
The equation is symmetric in $a,c$ and $b,d$.

Since I am quite bad at solving this kind of inequality in general, I would really love you could explain the thought process behind solving inequalities like this one.

PS. The tag functional analysis is because I encountered this in a context related to $L^p$ spaces.

Fact: For $x,y \in \mathbb{R}$ and $0 < p \leq 1$ we have $(|x+y|)^p\le|x|^p+|y|^p.$

For proof: $L^p$ norm and triangle inequality

Using the fact above you get the following 4 inequalities,

$$(|a-c|)^p\le|a-b|^p+|b-c|^p,$$ $$(|a-c|)^p\le|c-d|^p+|d-a|^p,$$ $$(|b-d|)^p\le|b-c|^p+|c-d|^p,$$ $$(|b-d|)^p\le|a-b|^p+|d-a|^p.$$

Now just combine all the above inequalities to get the result that you are after.

Have to work out the proof of the inequality when $1 \leq p < 2$. But nonetheless the said inequality is true even for $0 < p \leq 1$ as proved above.


Probability distributions - Exam paper question - $\mathrm{Cov}(X,Y)$, PDF Two people have decided to meet at a certain point in a forest sometime between noon and 2pm.
Their respective independent arrival times are $X$ and $Y$ such that $X \sim \mathrm{Unif}(0,2)$ and $Y \sim \mathrm{Unif}(0,2)$.

Hence the joint density of $X$ and $Y$ is
$$f_{X,Y} {(x,y)} = \begin{cases} 1/4, &  0< x <2 , 0< y <2 \\ 0, & \text{otherwise.} \end{cases} $$

They have agreed that whoever arrives first will wait for at most $20$ minutes for the arrival of the other.

a) Sketch the region in the $xy$ plane of times values for which they will meet and specify precisely the appropriate bounds (in terms of $x$ and $y$) for this region; then find the probabilty that they will meet by integrating the joint PDF $f_{X,Y} {(x,y)}$ over this region.

b) Since $X$ and $Y$ are independent, what value must $\mathrm{Cov}(X,Y)$ have?

c) Calculate explicitly $\mathrm{Cov}(X,Y)$ starting from its definition. Recall that
$$\mathrm{Cov}(X,Y) = E[(X - E(X))(Y - E(Y))].$$

I know this is quite a long question but I didn't know how to break it down into smaller parts without just having to type it into three different questions to ask.
If you could give me an idea about the sketch great! I'm not sure how to integrate the PDF as it's only $1/4$? Would it not just be $x/4 + C$?

Also for $\mathrm{Cov}(X,Y)$, I don't seem to have any notes on this, so detail would be good too.

Test is in the morning and you guys have been a big help so far!

I forgot part a)

a- When you shade the square in $\{[0,2],[0,2]\}$ then $20 min=1/3$ and they can not meet only if you start waiting from $2-1/3=5/3$ and it follows that you have the probability that thay cannot meet

$$\int_0^{5/3}\int_0^{5/3}\frac{1}{4} \, dx \, dy=\frac{25}{36}$$

and they meet with probability

$$1-\frac{25}{36}=\frac{11}{36}$$

b- If $X$ and $Y$ are independent then their covariance should be zero as follows

$$E[XY]=E[X]E[Y]$$

$$Cov(X,Y)=E[(X-E(X))(Y-E(Y))]=E[XY]-E[X]E[Y]=0$$

c- $$E[XY]=\int_0^2\int_0^2 \frac{1}{4}xy \, dx \, dy=1$$
and you have

$$f_X(x)=\int_{-\infty}^\infty\frac{1}{4}f_{XY}(x,y)\,dy=\int_0^2 \frac{1}{4}1 \, dy=\frac{1}{2} \in [{0,2}]$$

you get

$$E[X]=E[Y]=\int_0^2f_X(x)xdx=\int_0^2\frac{1}{2}x \, dx=1$$

If X and Y are independent then they must be uncorrelated which means Cov(X,Y)=0.  Another way to show that is by using the fact that if X and Y are independent then E(XY) =E(X) E(Y).
Now Cov(X,Y) = E[(X-EX) (Y-EY)]= E(X-EX) E(Y-EY) by independence
and then=0 since both E(X-EX) and E(Y-EY)=0.

The integral will depend on the region of integration and you would be integrating out both x and y. Also keep in mind to determine the region of integration that X and Y must be within 20 minutes of each other.  So -20/60<=X-Y<=20/60.

The answer is obtained by taking ∫$_0$$^2$∫$_a$ $^b$  1/4 dx dy where a=max(0, y-1/3) and b=min(2, y+1/3).


Was Fermat's last theorem proved based on Peano's postulates? Is the proof of Fermat's last theorem solely based on the Peano's postulates $+$ first order logic? Or it contains other axiomatic systems as well? What does it mean from foundations of math perspective to use several axiomatic systems to prove a conjecture? Do we know these axiomatic systems are consistent with one another? I'm not sure if I am asking it the right way, but I think logicians only prove the consistency of the axioms of one system not two different systems.

Logicians also prove the consistency of a formal system relative to each other. For instance, if ZFC is consistent, then PA (Peano arithmetic) is consistent. So in that sense ZFC is stronger than PA.
Fermat's last theorem was proved in a system stronger that PA. In that sense, it might be the case that the proof is inconsistent (we'll likely never know) while PA is still consistent. Plus is still unclear if Fermat's last theorem could be proved only using PA (it might be undecidable on PA but I am not aware of that).


Natural invertible functions Can you give examples of two functions f: N -> N and g: N -> N that are invertible in which g is not injective? And how did you proceed to find them? Also, is f(x) = x^2 [0, ∞[ a real or natural function?

Is the answer will be the same if it only said (g o f) = iN and g is not injective? forget about the invertible

Consider the function $f(x)=2x$ mapping $\mathbb{N}\mapsto \mathbb{N}$.  Clearly, $f(x)$ is one-to-one and hence invertible.  Note that the range of $f$ is just the even numbers.  $f$ is injective (not surjective).

Take $g(x)=x/2$ if $x$ is even, and $g(x)=1$ if $x$ is odd.  Since $g$ maps many values to 1, $g$ is not injective, and hence, not invertible.  $g$ is surjective.  $g$ is not injective.

But, $g(f(x))=x$ for any $x\in \mathbb{N}$.  So, $g(f(x))$ is invertible.  In fact, $g(f(x))$ is both injective and surjective.


Help proving that $\lim(s_n+t_n)=+\infty$ If $\lim s_n=+\infty$ and if $(t_n)$ is a bounded sequence, then $\lim(s_n+t_n)=+\infty$.

My proof:
Suppose $\lim s_n=+\infty$ and $(t_n)$ is bounded.
Since $\lim s_n=+\infty$, by definition we have that for all $M\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies that $s_n>M$.
Since $(t_n)$ is bounded, we have that for all $n\in\mathbb{N}$, $k\leq t_n\leq k'$ for some $k,k'\in\mathbb{R}$.
This is where I get stuck. I'm not sure how to put these two definitions together coherently. It's obvious to me that $t_n$ being bounded means that $s_n + t_n >M$. However, I'm having trouble arriving at this result from the definitions that I've stated.
Maybe since $\lim s_n=+\infty$, I could say that $s_n>M-t_n$ which implies that $s_n+t_n>M$. But I'm not sure how the definition of $t_n$ being bounded plays into this.

Am I on the right track? Any hints/advice would be helpful. Thank you!

You want to prove that $\lim (s_n + t_n) = \infty$.
Given some $A \in \mathbb{R}$.
We know that there exists $K$ such that $|t_n| < K$ for all $n$.
Since $\lim s_n = \infty$, we can pick $N$ such that $n\ge N\implies s_n \ge A+K$.
Thus, $n\ge N \implies s_n+t_n\ge A \ \ \blacksquare$.
Comment: I used to struggle with these proofs as well. Make sure to think about what exactly you want to prove. I always start these proofs with "given some $A$ / given some $\epsilon$ (as appropriate)". Consider the definition of the statement as a "challenge": if an adversary gives you some $A$, how do you come up with $N$ that beats the challenge?
Further comment on your attempt:
The way you did it almost worked (you got $s_n>M-t_n$). You just have to see that $|t_n|<K$.

You got a definition of a limit to infinity,
check for $s_n+t_n$, that is
$\forall _{M \in \mathbb{R}} \exists _{N \in \mathbb {N}} \forall _{n \geq N} \; s_n + t_n > M \;$ Because...
$\forall _{M \in \mathbb{R}} \exists _{N \in \mathbb {N}} \forall _{n \geq N} \; s_n > M \hspace{30px} \exists _{k,k' \in \mathbb{R}} \forall _{n \in \mathbb {N}} \; k \geq t_n \geq k' \Rightarrow \exists _{K \in \mathbb{R}} \forall _{n \in \mathbb{N}} \; |t_n| < K$
$\Rightarrow \forall _{M \in \mathbb{R}} \exists _{N \in \mathbb{N}}\forall _{n \geq \mathbb{N}} \; s_n > M + K \Rightarrow s_n + t_n > s_n - K > M \quad$ Since $\quad -K < t_n < K$


Questions on factor analysis and linear algebra Is it possible that $A*A^T=B*B^T$ while A and B have different ranks?

Thanks in advance!

(My answer assumes that your matrices are $\in M(m \text{ x }n, \mathbb{R})$.)

No, it is not. The key here is to prove that $\text{rank}(A) = \text{rank}(AA^T)$. For this, prove that $$A^TAx = 0 \iff Ax = 0\ \forall x \in \mathbb{R}^n$$

(then they have the same kernel, thus dimension formula provides that $\text{rank}(A) = \text{rank}(A^TA)$ and thus $\text{rank}(A) = \text{rank}(A^T) = \text{rank}(AA^T)$)

"$\Leftarrow$" is obvious. For "$\Rightarrow$" assume $A^TAx = 0$ for $x \in \mathbb{R}^n$. Then $$x^TA^TAx = 0 \implies (Ax)^TAx = 0 \implies \langle Ax,Ax \rangle = 0 \implies Ax = 0$$, where "$\langle\ , \rangle$" is the Euclidian scalar product (this doesn't hold for $\mathbb{C}$, for example, thus my assumption that the matrices are elements of $M(m \text{ x }n, \mathbb{R})$).

So now we have $\text{rank}(A) = \text{rank}(AA^T)$, thus $AA^T \neq BB^T$ if $A$ and $B$ have different ranks.

The rank of $A*A^T$ = $m*m$ Where $m$ is the no. of rows in $A$. So, as long as $A$ and $B$ have same no. of rows, it is possible


Linear dependence of a set of linear combinations If the elements of $A=\{a_1,a_2,...,a_n\}$ are linear combinations of the elements of another set, say $B=\{b_1,b_2,...,b_n\}$, then how can I show that $A$ is not linearly independent?

You can't because that statement is not always true. Counterexample: If $B = \{ (1,0,0), (0,1,0), (0,0,1) \}$, then $A = \{ (1,1,0), (1,0,1), (0,1,1) \}$ is a set of linear combinations of the elements of $B$, but $A$ is linearly independent.

As it was said before, A must not be lineary dependent. The easy way to define it is to make a matrix, where coefficients are the coefficients from linear combinations of vectors from A through B, and then calculate its determinant. If det = 0 $ \Rightarrow$ A is not linearly independent. For example, If B={(1,0,0),(0,1,0),(0,0,1)}, A={(1,1,1),(1,0,1),(0,1,0)} then  matrix is \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, it's det = 0, so A is not lineary independent


The set of real numbers and power set of the natural numbers I have learnt that the cardinality of the power set of the natural numbers is equal to the cardinality of the real numbers. What is the function that gives the one-to-one correspondence between these two sets?

I have also learnt that there exists no set whose cardinality is strictly between the natural numbers and the real numbers. Is there a proof of this or at least some intuitiveness behind it?

It's not quite correct to ask for "the function", because if there is one then there are many. Moreover, explicit bijections are highly overrated. We can write one, but it's much much oh so much easier to use the Cantor-Bernstein theorem, and simply exhibit two injections.
If you do insist on writing an actual bijection, let me identify $\mathcal P(\Bbb N)$ with infinite binary sequences (which is quite standard). Now let me describe the steps. We would like to take a binary sequence to the real number in $[0,1]$ which has this binary string as an expansion. However some numbers, e.g. $\frac12=0.1\bar0_2=0.0\bar1_2$, one sequence with finitely many $1$'s and the other has finitely many $0$'s.

First enumerate all the strings which contain finitely many $0$'s the strings containing finitely many $1$'s. One can show that both sets are countably infinite, one can even enumerate them in a very nice way. Write them as $p_n$ for the $n$-th sequence with finitely many zeros and $q_n$ for the $n$-th sequence with finitely many $1$'s.
The next step is to take $f\colon2^\Bbb N\to2^\Bbb N$ defined as: $$f(x)=\begin{cases} q_{2k} & x=p_k\\ q_{2k+1} & x=q_k\\ x &\text{otherwise}\end{cases}$$ Easily this is an injection whose range is $2^\Bbb N\setminus\{p_n\mid n\in\Bbb N\}$.

Now map $x\in2^\Bbb N$ to $r\in[0,1)$ such that, $$r=\sum_{n\in\Bbb N}\frac{f(x)}{2^{n}}$$ that is the real number whose binary expansion is $f(x)$. One can show that this is a surjective function, since if a number has a binary expansion then it has one which has infinitely many $0$'s. It is also injective since if a real number one has two different binary expansions then we can show that exactly one of them has finitely many $0$'s and the other finitely many $1$'s. But since we use $f(x)$, this is impossible.

Find a bijection between $[0,1)$ and $\Bbb R$. Usually one does that by first "folding $0$ in" and having a bijection between $[0,1)$ and $(0,1)$ and then using something like $\frac{2x-1}{x(x-1)}$ or a similar function for a bijection with $\Bbb R$.

Using the Cantor-Bernstein theorem is much easier.

First note that that $\Bbb R$ can inject into $\mathcal P(\Bbb Q)$ by mapping $r$ to $\{q\in\Bbb Q\mid q<r\}$. Since $\Bbb Q$ is countable there is a bijection between $\cal P(\Bbb Q)$ and $\cal P(\Bbb N)$. So $\Bbb R$ injects into $\cal P(\Bbb N)$.

Then note that we can map $x\in2^\Bbb N$ to the continued fraction defined by the sequence $x$. Or to a point in $[0,1]$ defined by $\sum\frac{x(n)}{3^{n+1}}$, which we can show is injective in a somewhat easier proof.

Finally, as mentioned the last part is false. From the usual axioms of modern set theory (read: $\sf ZFC$) we cannot prove nor disprove that there are no intermediate cardinalities between $\Bbb N$ and $\Bbb R$. The proof of that is difficult and require a deep understanding of modern [read: axiomatic] set theory, as well logic.
If the last part somehow confused you, perhaps my answer to this question can help, Why is the Continuum Hypothesis (not) true?.

The easiest method I know of is to relate elements in $\mathcal{P}(\mathbf{N})$ to binary sequences in $(0,1)$. For any $A \subset \mathbf{N}$ there is a natural sequence $(a_n)$ such that $a_n = 1$ if $n \in A$, $a_n=0$ if $n \notin A$. Then, we can view any $x\in (0,1)$ as $\sum_{n=1}^{\infty}\frac{a_n}{2^n}$, where $a_i \in \{0,1\}$.

Once you have this, use the map $x \mapsto f(x)=\tan(\pi (x-1/2))$ to produce all real numbers. You must be slightly careful, though, since rational numbers yield two real representations. However, the idea should be clear and give you some insight for why this is true.


Proving $A_1, A_2, \dotsc, A_k$ are real numbers in an arbitrary partial fraction decomposition. I have two extra credit assignments for my Calculus 2 class and after attempting them for several days and unsuccessfully trying to find any resources on how to do these questions I've decided to ask here.

The first question is as follows:

Let $\alpha$ be an arbitrary real number. Let k be one of the following numbers:
$1, 2, 3, \dotsc$. Let $a_{k-1}, a_{k−2}, \dotsc, a_{0}$ be real numbers. Let $p(x)= a_{k-1}x^{k-1} + a_{k-2}x^{k-2} + \dotsb + a_{0}x^{0}$. Prove that there exists real numbers $A_{1}, A_{2}, \dotsc, A_{k}$ such that

$$\frac{p(x)}{(x-\alpha)^{k}} = \frac{A_{1}}{(x-\alpha)} + \frac{A_{2}}{(x-\alpha)^{2}} + \dotsb +\frac{A_{k}}{(x-\alpha)^{k}}$$

The second question is as follows:

Let $\beta$ and $\gamma$ be arbitrary real numbers. Let $m$ be one of the following numbers: $1, 2, 3, \dotsc$. Let $b_{2m-1}, b_{2m-2}, \dotsc, b_{0}$ be real numbers. Let $q(x) = b_{2m-1}x^{2m-1} + b_{2m-2}x^{2m-2} +\dotsb +b_{0}x^{0}$. Prove there exist real numbers $B_{1}, C_{1}, B_{2}, C_{2}, \dotsc, B_{m},C{m}$ such that

$$\frac{q(x)}{(x^{2}+\beta x+\gamma)^{m}} = \frac{B_{1}x+C_{1}}{x^{2}+\beta x+\gamma} + \frac{B_{2}x+C_{2}}{(x^{2}+\beta x+\gamma)^{2}} + \dotsb + \frac{B_{m}x+C_{m}}{(x^{2}+\beta x+\gamma)^{m}}$$

EDIT: I must also state that all $\alpha$, $k$, $a_{k-1}$, $\beta$, $\gamma$, $m$, $b_{2m-1}$, etc must remain arbitrary in the solutions and I cannot plug-in any numbers for them and solve for that to prove the statement based on just that number that was plugged-in.

Any help would be extremely helpful in understanding and solving these problems. I just can't seem to find anything in my book or on the web dealing with a problem like these, only the basic equations for a partial fraction.

For the first one, multiply the equation by $(x-\alpha)^k$. Then
$$ p(x) = \sum_{n=0}^{k-1} A_{k-n} (x-\alpha)^{n}. $$
Now you can either expand the brackets and look at the coefficients of $x^r$, or differentiate both sides $r$ times and set $x=\alpha$. The latter gives you a formula for the $A_r$ directly.

The second one is worse because the differentiation trick doesn't apply. Multiply up in the same way, expand the brackets, and show that the system of equations has a unique solution.

Hint: For the first problem, if $\alpha=0$ simply expand $\frac{p(x)}{x^k}$ into a sum of fractions. Otherwise, note that $a_{k-1}x^{k-1} = a_{k-1}(x-\alpha)^{k-1} + \text{lower-order terms}$; continuing (technically, using induction), you can write $p(x)$ as a polynomial in $(x-\alpha)$. Now divide both sides by $(x-\alpha)^k$.

The second problem is similar.


Wieferich prime-Lang-Trotter conjecture connection? Crandall-Dilcher-Pomerance prediction states that the number of Wieferich primes $<x$ is $log\ logx $
N.Katz in "WIEFERICH PAST AND FUTURE" states;

The Crandall-Dilcher-Pomerance prediction is analogous to the Lang-Trotter
conjecture [L-T] for an elliptic curve $E$ over $Q$.

So my question is; if it is proven that there are only 2 Wieferich primes base 2, does that mean the Lang-Trotter conjecture is false?

I digged in the reference:The reference is to Serge Lang, Old and new conjectured diophantine inequalities, Bulletin of the American Mathematical Society, 23(1):37–75, 1990.

On page 42 it is stated:

Actually, in line with Lang-Trotter conjectures, the probability
that $2^{p-1} = 1 + pk\pmod{p^2}$ with a fixed residue class $k \pmod p$
should be $O(1/p)$, so the number of primes $p\leq x$ such that
$2^{p-1} = 1 \pmod {p^2}$ should be $O(\log\log x)$.

So, I translate this as: "If there are only 2 Wieferich primes base 2, the Lang-Trotter conjecture is false". Any comments?


Complex function’s injection(solve it please......) Let $D=\{z| |z-z_0|<1\}$
If $f$ is analytic function on $D$ and $|f’(z)-f’(z_0)|<|f’(z_0)|$,
Then $f$ is injective

First, I’m sorry
This is my first time to use this page
So, I want to solve this question.
I heard this question is very hard
But I believe you can solve it.
I have been studying Complex analsis
I spent time to solve this question
I couldn’t prove this question..
Please answer me, How to solve it?

Assume wlog $z_0=0$ (change variables $z-z_0=w$) and rotate and scale s.t $f'(0)=1$ (this doesn't change injectivity or lack of it); then the condition becomes $|f'(z)-1|<1$ for all $|z|<1$  and that implies $\Re f'(z) >0$  (since $|a-1|<1$ implies by squaring $|a|^2-2\Re a <0$).

But now the unit disc is convex so the easy Noshiro Warschavski theorem (discussed here in other posts and obtained by integration on segments) applies so $f$ is injective

From the statement it follows that, on the disk, $f’$ doesn’t vanish.

But we can actually state more: define on the disk $g(z)=f’(z_0)(z-z_1)$; then $|(f-f(z_1))’-g’| < |g’|$. Since $(f-f(z_1))(z_1)=g(z_1)=0$, and since $g’$ is constant, we can integrate and it follows $|(f-f(z_1))-g| < |g|$.

So by Rouché, in any disk of $D$, $f-f(z_1)$ and $g$ have the same number of roots. As a consequence, $\{f=f(z_1)\}=\{z_1\}$.

Edit: in answer to your comment, the result is the following. Let $U$ be a convex open subset of $\mathbb{C}$ and let $f,g$ be holomorphic functions on $U$ such that $g$ is affine, $|f’| < |g’|$ and for some $p$ in $U$, $f(p)=g(p)=0$. Then, everywhere in $U$ except at $p$, $|f| < |g|$.

Proof: let $z \in U$, $z \neq p$, $$|f(z)|=\left|(z-p)\int_0^1{f’((1-t)p+tz)\,dt}\right|=|z-p|\int_0^1{|f’((1-t)p+tz)|\,dt} < |z-p|\int_0^1{|g’((1-t)p+tz)/,dt}=|z-p|\left|\int_0^1{g’((1-t)p+tz)\,dt}\right|=|g(z)|.$$


Compute $\lim\limits_{x \to \infty} \frac{\ln (x!)}{x \ln (x)}$ I was just playing with logarithms and factorials, and then realized that $f(x) = \frac{\ln (x!)}{\ln(x)}$ is slower than $f(x) = x$. From there I got $\frac{\ln x!}{(x \ln (x))}$. In most calculators, I can't find the limit, and when I put a ridiculously large number it doesn't go below 0.6. So is the limit 0?

If the limit is considered along $\mathbb{N}$ it is enough to invoke Cesàro-Stolz:
$$\lim_{n\to +\infty}\frac{\log(n!)}{n\log n}=\lim_{n\to +\infty}\frac{\log(n+1)}{(n+1)\log(n+1)-n\log n}=\lim_{n\to +\infty}\frac{\log(n+1)}{\log(n+1)+\underbrace{n\log\left(1+\frac{1}{n}\right)}_{\to 1}}=1. $$
If the limit is considered along $\mathbb{R}$ it is enough to invoke De l'Hospital:
$$ \lim_{x\to +\infty}\frac{\log\Gamma(x+1)}{x\log x} = \lim_{x \to +\infty}\frac{\psi(x+1)}{1+\log x}=\lim_{x\to +\infty} x\cdot\psi'(x+1)=\lim_{x\to +\infty}x\sum_{n\geq 1}\frac{1}{(n+x)^2} $$
where
$$\sum_{n\geq 1}\frac{1}{(n+x)^2}>\sum_{n\geq 1}\frac{1}{(n+x)(n+x+1)}=\frac{1}{x+1}, $$
$$\sum_{n\geq 1}\frac{1}{(n+x)^2}<\frac{1}{(x+1)^2}+\sum_{n\geq 1}\frac{1}{(n+x)(n+x+1)}=\frac{x+2}{(x+1)^2}.$$

Stirling's approximation:
$$\frac{\ln (x!)}{x \ln (x)}\sim \frac{\ln (x!)}{\ln (x^x)}\sim \frac{x!}{x^x}\sim \frac{(x/e)^x\sqrt{2\pi x}}{x^x}\sim\frac{\sqrt{2\pi x}}{e^x}\to 0.$$


Why isn't the area of a square always greater than the length of one of its sides? Intuitively, it seems like the area of a square should always be greater than the length of one of its sides because you can "fit" one of its sides in the space of its area, and still have room left over.

However when the length of a side, $s$, is less than $1$, then the area $s^2 < s$, which doesn't make sense to me for the reason above.

I think your intuition is failing you because you are trying to compare a 1-dimensional object (the length of a side) with a 2-dimensional object (the area of the interior of the square).  You can fit loads of segments into a square of any size -- infinitely many, in fact!  That comparison doesn't really mean anything.

On the other hand, here's a comparison that does make sense:  Set a square of side length $s$ side-by-side with a rectangle whose sides are $s \times 1$.  Now you are comparing area to area.  The rectangle's area will fit inside the square if and only if $s>1$.

First physics and then maths.

You cannot compare an area and length while attempting to make sense.

If you are talking about areas of different side lengths in the numbering system based on $10$ as the most common scale of notation adopted here then the square whose side has magnitude  $a<1$ i.e., $ a <10^0$ units produces an area of magnitude $a^2<a $  and when magnitude of its side is $a>1$ unit, then it produces an area whose magnitude is $a^2>a.$


Why is the Plancherel measure interesting? One can average a class function $f:G\to\Bbb C$ for a finite group $G$ by interpreting $f$ as a complex-valued function on the space ${\rm cl}(G)$ of conjugacy classes and computing the expectation $\Bbb E(f(C))$, where $C$ is a class-valued random variable whose distribution is pulled from the uniform one on $G$.

In light of $\sum(\dim\rho)^2=|G|$ (where the sum is over complex irreps $\rho$), we could equally attach the weights $(\dim\rho)^2/|G|$ to irreps and turn $\widehat{G}$ into a probability space. This is how we get the Plancherel measure on a finite group. But what question is the Plancherel measure an answer to? Historically, we know the Lebesgue and more generally the Haar measure answer the question of translation-invariant measures, and do so nicely (existence and uniqueness). Is there some a priori system of nice properties we might have liked a measure on irreps to have, which we could have later discovered the Plancherel measure uniquely satisfies?

For $G$ locally compact abelian, the complex irreps comprising $\widehat{G}$ are $1$-dimensional and form a group under tensoring - this is just the dual group, which has its own topology (the compact-open topology), and the Plancherel measure is just the Haar measure. However if $G$ is nonabelian, the irreps can be higher-dimensional and so aren't closed under tensor product. (Pontryagin duality generalizes to Tannaka-Krein duality, where we look at symmetry of the tensor operation on the full category of reps to recover $G$, at least as I understand it.) I may be muddying the waters with infinite nonabelian $G$ though, as I don't know how to define the Plancherel measure in that context.

We know conjugacy classes are dual to irreducible representations (see Qiaochu's answer here; the tl;dr version is that they are identified with the maxspecs of two algebras in a dual pairing). They are not generally in bijection, but one situation where they are in bijection is with finite symmetric groups. There might be some process that is "dual" to averaging class functions (which is where the boring measure on the set of conjugacy classes appears) in which the Plancherel measure is useful. This could also potentially explain why there are applications of the Plancherel  measure on symmetric groups to combinatorial/probabilistic questions but I haven't seen any applications with other groups.

What you ultimately want to study is (upon fixing a Haar measure in the noncompact case) the left regular representation $\lambda\colon G\to U(L^2(G,\mathrm{Haar}))$. Now, general theory tells us that while it's not always possible to decompose $L^2(G)$ as a direct sum of irreducible reprenentations (this already fails for $G=\mathbb Z$), it is always possible to decompose it as a direct integral of irreducible representations (which are parametrised by the unitary dual $\widehat G$ of $G$). Now, if $G$ is unimodular and type I,  the direct integral decomposition (with respect to both left and right actions of $G$) is as follows:
$$
L^2(G) \cong \int_{\widehat G} H_\pi\,d\mu(\pi),
$$
where $H_\pi = \pi\otimes \pi^*$, and its understanding requires, in particular, to determine the measure $\mu$ on $\widehat G$ such that the above becomes an isometric isomorphism. The unique measure with this property is called the Plancherel measure of $G$ (associated to a given Haar measure). Equivalently, it's the unique measure such that
$$
\|f\|_2^2 = \int_{\widehat{G}} \|\pi(f)\|_{\mathrm{HS}}^2 \mathrm{d}\mu(\pi),\quad f\in L^1(G,\mathrm{Haar})\cap L^2(G,\mathrm{Haar}).
$$

In the case, $G = S^1$ we have the class functions $f \in L^2(G) = L^2(S^1)$ can be written as the sum over irreudicible representations $\theta \mapsto e^{2\pi i n \theta}$ giving Fourier series:

$$ f(x) = \sum_{n \in \mathbb{Z}} \hat{f}(n) e^{2\pi i n x} = \mathbb{E}[\hat{f}(n)e^{2\pi i n x}]$$

In a sense we reconstructing our function by the "average" over all representations.

We could do without any probabilistic interpretation at all, but has proven beneficial in certain contexts.


Defining closed and open sets with quantifiers. I'm taking an intro to analysis course, and, given the following definitions of closed and open sets, I was wondering whether my attempt at quantifying them is correct.

Definition A set $A\subset R$ is closed iff every accumulation point of $A$ belongs to $A$  $$\overset{{?}}{\iff}\forall{x}\left(\forall\varepsilon>0\exists{a\in{N(x,\varepsilon)\cap (A\setminus{\{x\})}}}\implies x\in{A}\right)$$
A set $E\subset R$ is open iff for every $x\in A$ there is a neighborhood  $N(x,\varepsilon)$ of $x$ such that $N\subset A$ $$\overset{{?}}{\iff}\forall{x}\in{A}\exists\varepsilon>0:N(x,\varepsilon)\subset{A}$$

Let $A\subset \mathbb R $.

$A $ is colsed in $\color {red}{\mathbb R} \iff $

$$\overline {A}\subset A \iff $$

$$(\forall x\in \color {red}{\mathbb R}) $$

$$\Bigl ( \forall \epsilon>0 \; (x-\epsilon,x+\epsilon)\cap A \neq \emptyset \implies x\in A\Bigr) $$


Confusion Between For Each and For All in Nested Quantifiers I am asked to translate the following into English:

$$\exists x \forall y \forall z\left((F(x,y) \wedge F(x,z) \wedge (y \neq z) \implies \neg F(y,z)\right)$$

The solution says:

  This expression says that if students $x$ and $y$ are friends, and students $x$ and $z$ are friends, and furthermore, if $y$ and $z$ are not the same student, then $y$ and $z$ are not friends.

But I don't understand why $x$ and $y$ are for all and not existential quantifiers. Because it's not saying there is a student who is friends with all students -- just for one.

I would translate it as follows as the solution you quoted doesn't treat the quantifiers very explicitly:

  There exists a student $x$ such that for any choice of students $y$ and $z$, where $y$ and $z$ are distinct, if $x$ is friends with both $y$ and $z$, then $y$ and $z$ are not friends.

Note that this is not saying $x$ is friends with every $y$ and $z$, but that if $x$ is friends with both, then $y$ and $z$ are not friends.

Given that $F(x,y)$ stands for "$x$ and $y$ are friends", and $St(x)$ stands for "$x$ is a student", the solution given is translation of the following formula:

$\forall x \forall y \forall z [St(x)\land St(y)\land St(z)\land F(x,y)\land F(x,z)\land y\neq z \rightarrow \neg F(y,z)]$


What is the secret of number '2520'? At first sight, this number will be considered a normal number. But the strange thing is 2520 is able to be divided at even or odd number. Like  1=2520, 2=1260 , 3=840, 4=630, 5=504, 6=420, 7=360, 8=315, 9=280, and by 10=252, which is hard to find integers with the same characteristics.

Also, you can get this number by (7 * 30 * 12 = 2520) which I think is 7 days in the week, 30 days in a month and 12 months in a year! Nothing special, but it is a weird thing, right.

Are there any numbers that can be divided from 1-10 without any fraction? If yes, what are they? And is there any mathematical explanation of such a phenomenon?

If you want a number divisible by $a, b,$ and $c$, you can just multiply them together.  So one number that is divisible by $1$ through $10$, is the product $1\cdot 2\cdot \cdots \cdot 10 = 3628800.$  The smallest such number will be the least common multiple of $1, 2, \ldots, 10$ which is your $2520.$

Is it "weird" that $2520 = 7\cdot 30 \cdot 12$?  Not really.  Before metric ruined everything, we chose our units to be divisible by small numbers.  There are lots of $12$'s and $30$'s (and $60$'s, etc) in our old measuring systems.  And since this problem depends on having lots of small factors, it would be expected that the least common multiple of the first few numbers would contain the same factors as many of the ancient units.

To add to this one observation: refactor 7 * 30 * 12 = 2520 to 7 * 360 = 2520 (as of course 12 * 30 = 360). Your number here, 2520 is equal to 7 circles, in terms of degrees (that would bring up the question: is 360° just a convention or a true mathematical reality, I believe it is the latter, but won't defend that here). So that makes it a very special number. That should answer part of your question at least:

Also, you can get this number by (7 * 30 * 12 = 2520) which I think is 7 days in the week, 30 days in a month and 12 months in a year! You were already turning a year into 360 days (in effect at least), so if we were to allow that, it would just be a circular statement, 360 (days in this case) * 7 ('years' in this case) of course simply equates to 2520.


How to see that existence non-trivial parallel $p$-form implies $b_p\geq 1$ using De Rham cohomology? Suppose $(M,g)$ be a (for simplicity consider closed) Riemannian manifold. Because every parallel $p$-form $\omega$ is harmonic so the $p$-th Betti number should be positive i.e. $b_p\geq 1$. How to see this using De Rham cohomology (of course without using properties of Hodge theory like $\delta$, 1-1 correspondence between De Rham and Harmonic forms, etc.)? It is well-known that parallel forms are closed. What is the next step?
Such an easy argument based on harmonic forms I think there must be similar easy argument based on De Rham cohomology or it needs somewhat difficult argument?

You should read a reference on Hodge theory.
The conclusion is not hard, that the map $\mathcal H^k \to H^k_{dR}$ is injective (that is, that harmonic forms are closed and exact harmonic forms are zero). The hard part is arguing that this map is surjective.

Lemma: A $p$-form is harmonic, $\Delta \alpha = 0$, if and only if $d \alpha = 0$ and $d^* \alpha = 0$.

Proof: Recall that $\Delta = \pm (d^* d + d d^*)$, where the sign depends on dimension of manifold and degree $p$ of the form. $$\langle \pm \Delta \alpha, \alpha \rangle = \langle d^* d \alpha, \alpha \rangle + \langle d d^* \alpha, \alpha \rangle = \langle d^* \alpha, d^*\alpha \rangle + \langle d\alpha, d\alpha \rangle = |d^*\alpha|^2 + |d\alpha|^2,$$ here using that $d^*$ is the $L^2$-adjoint of $d$; the inner product is $\langle \alpha, \beta \rangle = \int \alpha \wedge *\beta$.
The RHS can only be zero if $d^* \alpha$ and $d\alpha$ are both zero. QED.

Corollary: if $\alpha$ is an exact harmonic form, then $\alpha = 0$.

Proof: we know that $d^*\alpha = 0$. If $\alpha = d\beta$, it follows that $d^* d\beta = 0$. But then $$0 = \langle d^* d\beta, \beta \rangle = \langle d\beta, d\beta \rangle = |d\beta|^2,$$ so that $0 = d\beta = \alpha,$ as desired. The same argument shows that coexact harmonic forms ($\alpha = d^* \beta$) are also zero.


Find the maximal domain and range of $f(x) = \ln(x^2 - 7)$ Find the maximal domain and range of:

  $$f(x) = \ln(x^2 - 7)$$

I reasoned that the domain can be found by finding the values of $x$ such that $x^2 - 7 > 0$ so that the logarithm is defined. So:

$$x^2 > 7\\
x > \pm \sqrt{7}$$

Shouldn't I have arrived at $x > \sqrt{7}$ and $x < -\sqrt{7}$ somehow? From inspection I can see this, but I couldn't seem to understand why I didn't arrive at that algebraically.

Also, by inspection, I would guess that the range is $\{y \in \mathbb{R}\mid y > 0\}$. Is it standard to just derive this from a graph? or should I discuss the limiting behaviour as $x \rightarrow \pm~ \infty$ and as $x \rightarrow \pm\sqrt{7}$ from $+$ and $-$ respectively?

For the domain, your reasoning is correct, but
$$\begin{align*}x^2 > 7 \implies |x| > \sqrt{7} & \implies -x > \sqrt{7} \; \text{or }\; x > \sqrt{7}\\
& \implies x < -\sqrt{7} \; \text{or } \; x > \sqrt{7},\end{align*}$$ as you were expecting.  You just skipped a step when you wrote $x < \pm \sqrt{7}$, forgetting to reverse the inequality when you moved the negative across.

For the range, this is all the values that $\ln(x^2 -7)$ can take, so you could say that it goes to $-\infty$ as $x \to \sqrt{7}$ from the right and as $x \to -\sqrt{7}$ from the left.  It also goes to $\infty$ as $x \to \pm \infty$.  So the range is all of $\mathbb{R}$.

Your domain appears to be in the set of Real Numbers - the domain is x is an element of the Real Numbers less than the negative square root of seven or greater than the postive square root of seven.

If you wish your domain to be in set of Complex Numbers, try it out and see what Complex values of x must be excluded. What is the range?

Try it for x as an element of the Integers. What is the range?

What does "maximal" mean in these cases? Maximal domain is the domain which is the super set of all posible domains - in other words, the maximal domain is the domain which has all possible domains as its subsets. For example, you may choose the domain with only the positive values. This is a subset of the maximal domain.


pdf of sum of two dependent random variables What is the pdf of sum of two dependent random variables given we know their joint
pdf and individual pdfs. I have seen already some posts but none of them answered when they are dependent. Every one solved for only the independent case but i need for dependent case in terms of the joint pdf and individual pdfs in an explicit form.

If $f(x,y)$ denotes the PDF of $(X,Y)$ and $Z=X+Y$ then:
$$f_Z(z)=\int^{\infty}_{-\infty}f(x,z-x)dx$$

Substituting $y=u-x$ we find:$$F_Z(z)=\int\int_{x+y\leq z}f(x,y)dxdy=\int^{\infty}_{-\infty}\left[\int^{z-x}_{-\infty}f(x,y)dy\right]dx=$$$$\int^{\infty}_{-\infty}\left[\int^{z}_{-\infty}f(x,u-x)du\right]dx=\int^{z}_{-\infty}\left[\int^{\infty}_{-\infty}f(x,u-x)dx\right]du$$

Taking the derivative of $F_Z$ we come to the mentioned result.

Let $X,Y$ be dependant variables with PDF $f(x,y)$.

Then sum PDF is: $$f_{X+Y}(k)= \int f(x,k-x) dx$$


If $\,f^{7} $ is holomorphic, then $f$ is also holomorphic. I need some help with this problem:

  Let $ \Omega $ be a complex domain, i.e., a connected and open non-empty subset of $ \mathbb{C} $. If $ f: \Omega \to \mathbb{C} $ is a continuous function and $ f^{7} $ is holomorphic on $ \Omega $, then $ f $ is also holomorphic on $ \Omega $.

Thanks in advance.

We may assume $\Omega$ connected.

Let $g(z)\colon = [f(z)]^7$, a holomorphic function. Let $N$ the set of zeroes of $g$. If $N$ has an  accumulation point  in $\Omega$ then $g$ is identical $0$ and so is $f$.
Otherwise, the set $N$ is discrete ( it may have an accumulation point in $\partial \Omega$).

The set $\Omega \backslash N$ is open and connected. The function $g$ maps $\Omega \backslash N$  to $\mathbb{C}\backslash\{0\}$. Let $p \colon \mathbb{C}\backslash\{0\} \to \mathbb{C}\backslash\{0\}$, $p(z) = z^7$.

$p$ is a holomorphic covering . We have
$$p\circ f = g$$
that is, $f$ is a lift of $g$. Moreover, $f$ is continuous. It is a general fact that any continuous lift of a  holomorphic map for a holomorphic covering is also holomorphic. We conclude that $f$ is holomorphic on $\Omega \backslash N$. The singularities $N$ are removable because $f$ is continuous on $\Omega$. So $f$ holomorphic on $\Omega$.

Obs: In fact it works with any nonconstant holomorphic function $\phi(z)$ instead of $z^7$. The problem can be treated locally in a similar way.

There exist exactly $7$ holomorphic functions $g_0,...g_6$ such that $g_i^7=f^7$ (there are seven holomoprhic $7^{th}$ root functions which we can compose with $f^7$, or $7$ branches if you prefer). We need to show $f$ is equal to one of them. We know that everywhere it takes a value that is equal to one of the $7^{th}$ roots of $f^7$; Hence we can argue by continuity that its values lie on one of the branches only, and we are done.


What is wrong with this proof of chain rule? Let $U$ and $V$ be open subsets of $\mathbb{R}$. Let $f:U\rightarrow
V$ be differentiable at $x_0$ and $g:V\rightarrow\mathbb{R}$ be
  differentiable at $f(x_0)$. Then $g\circ f$ is differentiable at
  $x_0$.

What is wrong with the following "proof" of the above fact:

$$(g\circ f)'(x)=\lim_{x\rightarrow x_0} \Big(\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}\cdot \frac{f(x)-f(x_0)}{x-x_0}\Big)=$$ $$\lim_{x\rightarrow x_0} \frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}\cdot \lim_{x\rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}=g'(f(x_0))\cdot f'(x_0)$$

Just note that if $f$ is constant you cannot write
  $$\frac{g(f(x))-g(f(x_{0}))}{f(x)-f(x_{0})}$$
  because it is not well defined.

First, you should read the "proof" from the back to the front because you do not know whether the first limit in your proof exists or not. By assumption (and the fact that $f$ is continuous at $x=x_0$) you know that the two limits in the second line exist. Therefore the limit in the first row exists.

Now, recall the definition of a limit: The limit is defined as: $\lim_{x\to a} f(x) = L$ if and only if for all sequences $(x_n)_{n\in \mathbb N}$, $x_n\neq a$ and $x_n\to a$, the sequence $f(x_n)$ converges to $L$ (https://en.wikipedia.org/wiki/Limit_of_a_function).

Therefore $f(x)-f(x_0) \neq 0$ for all sequences and for all $n$. Therefore you can simplify the expression in the first line and you are done.

To wrap up, I would only change the order of your equations and start from the end. Then the proof should be fine.


Series that converges while the sequence doesn't necessarily decrease This might be a kind of stupid question but I can't seem to get my mind around it so I hope someone can help me.

There's this statement that says:

  "If $a_n > 0$ for every $n\in\Bbb N$, the series
  $\sum\limits_{n=1}^\infty(-1)^n\cdot a_n$ converges if and only if
  $a_n$ is a decreasing sequence".

I know this is false but I can't seem to find any example that contradicts it.

Thank you in advance to everyone.

Consider the nondecreasing sequence:

$$1, 2, 1, \frac{1}{2}, 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{2}, \frac{1}{4}, \ldots$$

It's easiest to see the pattern and the convergence if you look at the terms 3 at a time.

First, your title, "series that converges while the sequence doesn't" is impossible.  One of the first things one should learn about series is that "a necessary (but not sufficient) condition that the series $\sum a_n$ converges is that the sequence $\{a_n\}$ converges to 0.  That is not a "sufficient condition" because there exist examples of sequences that converge to 0 while the series does not converge $\sum\frac{1}{n}$ is such an example.  However, for alternating sequences, $\sum (-1)^n a_n$ it is both necessary and sufficient that $a_n$ converge to 0.


Proving $\sup(|f|) - \inf(|f|) \leq \sup(f) - \inf(f)$ I appreciate if you could give me some hints on how to prove that :

$$\sup(|f|) - \inf(|f|) \le \sup(f) - \inf(f)$$

Here is a strategy. Since you asked for hints, I let you check the details.

This is not really a fact about functions. It is rather a property of subsets $S$ of $\mathbb{R}$. So take such a set $S$ and denote $|S|:=\{|s|\;;\;s\in S\}$. We want
$$
\sup |S|-\inf|S|\leq \sup S-\inf S.
$$

We have an alternative.

Case 1: $\sup |S|=\sup S$. Clearly $\inf S\leq \inf |S|$. So the inequality holds.

Case 2: $\sup|S|=-\inf S$. Check that $-\inf|S|\leq \sup S$. The inequality follows.

Let $c = \pm 1$,then
$sup(f) - inf(f) \leq \left| sup(f)-inf(f) \right| \leq c*[sup(f)-inf(f)] = c*sup(f)-c*inf(f) = sup(c*f)-inf(c*f) \leq  sup(\left| f \right|) - inf(\left| f \right|) $


Projection of a point on standard unit complex Let $$\Delta_{n-1}:=\{x\in R^{n}:x_1+x_2+....x_n=1, x_1, x_2,....x_n\geq0\}$$ and $$a\in R^n$$

Let $$z:=P_{\Delta_{n-1}}(a)$$ be the projection of point a onto $\Delta_{n-1}$. Show that $z$ satisfies the system of inequalities- $$z-y=a-\mu\textbf{e}, z\ge0, y\ge0, z^Ty=0$$ where $\textbf{e}$ is the vector of all ones. $y,z\in R^n, \mu \in R $. One can use obtuse angle condition of the projection theorem over the convex set along with Farkas Lemma.

I don't know how to approach this problem. Please help.

$z = \mbox{argmin} \{(x-a)^T(x-a) : x^Te = 1, x \ge 0\}$ use KKT


Calculated rotated point coordinate: is my solution correct Is my calculation correct for this rotation around a point?

A point a(-19.94,392.11) is rotated -49.45 degrees, what is the new coordinates of point a?

My solution:

x' = x*cos(0) - y*sin(0)
y' = x*sin(0) + y*cos(0)

x' = (-12.961) - (-298.0036)
y' = (15.15) + (254.92)

x' = 285.04
y' = 270.07

I don't know why this old point emerged... Anyway let's try this using complex numbers :
$$(-19.94+392.11 i)\cdot e^{2\pi i\dfrac{-49.45}{360}}\approx  284.98 + 270.07i$$

So that the OP's answer looked not so bad!

Result of my calculation is slightly different than yours .

Let's denote :

$\theta=-49.45^\circ $

$x'=r\cos \alpha ~\text{ and }~y'=r\sin \alpha$

where $~r=\sqrt{x^2+y^2} =392.5$

Note that :

$\alpha=\arctan\left|\frac{y}{x}\right|+\frac{\pi}{2}-|\theta|$

$\alpha=42.07^\circ$

hence :

$(x',y')=(291.69,262.64)$


Book recommendation for proof? I now realize that a big part of higher mathematics is proof construction, and it is an area I would like to get familiar with.

I have found the book ‘how to prove it’ by Velleman, but am wondering if anyone has any other suggestions for proof.

In terms of difficulty I am looking for a book of relative ease as I am new to proof. Any suggestions would be appreciated, thanks.

Here is a list of freely available books I know, but since I've personally learnt proofs before most of them were written, I am no good judge of their quality as proof tutorials:

Eric Lehman, F Thomson Leighton, Albert R Meyer, Mathematics for Computer Science. Part I is specifically about proofs, and parts II and III should provide good practice. It's written for MIT students, so it might be a bit fast-going, but it's written in an enjoyable lively style (most math books either avoid or fail at jokes; this one is an exception).
Martin V. Day, An Introduction to Proofs and the Mathematical Vernacular (archive.org mirror). Big upside (to me) is that it does determinants in detail. A downside is that it mis-defines polynomials as functions.
Daniel J. Velleman, How to prove it, 2nd edition 2006. This seems to be slow and systematic, but (as a consequence) doesn't get far enough to prove anything really exciting. This is, of course, a common problem with introductions to proofs, particularly when they are written for 1-semester courses.
Richard Hammack, Book of proof. Roughly same level as Velleman, from its look.
Clive Newstead, An Infinite Descent into Pure Mathematics. Still a work in progress. This goes deeper than most other comparable texts into combinatorics (e.g., the "path-dependent product rule" for counting is formalized and proven!).
Paolo Aluffi, Introduction to Advanced Mathematics. These are lecture notes (though they seem reasonably polished), and also suffer from the "not getting anywhere too interesting" syndrome (see Velleman), but Aluffi is known for enjoyable writing.
James Aspnes, Notes on Discrete Mathematics. This is written from a TCS point of view (e.g., page 41 is a table of the rules of natural deduction), but it covers the usual ground along with some graph theory and enumerative combinatorics. Aspnes writes well (here are notes on introductory programming, C++ and Linux).
Matthias Beck and Ross Geoghegan, The Art of Proof: Basic Training For Deeper Mathematics.
Alistair Savage, MAT1362 Mathematical reasoning and proofs. These notes are written for a course that follows Beck and Geoghegan, so can be regarded as an expanded version of parts of the latter, including proofs that are omitted from the latter. See Savage's teaching page for more notes.
Ted Sundstrom, Mathematical Reasoning: Writing and Proof.
Franco Vivaldi, Mathematical writing for undergraduate students. This is likely a draft of his 2014 book.

These are just the ones I have found in my personal library and freely available on the internet; if I had listed the paywalled ones, this post would be again as long. As I said, I have not actually learned proofs myself from any of these, so all I can offer are quick judgments based on leafing through.

Proofs from the Book by Aigner and Ziegler comes to mind.

It contains many elegant proofs spanning different topics.


Number Theory: m>1000, m ends in 007, m is relatively prime to n Let $n$ be a positive integer. Prove that there exists a positive integer $m > 1000$ with the following two properties: $m$'s last 3 digits are 007, and $m$ is relatively prime to $n$.
I know that I should first solve the case that $n$ is relatively prime to 1000 and then try to reduce the second case to the first but I don't really know how to do that.
ALso I know that I can express $m$ as $1000k+007$.

You can use the Chinese Remainder Theorem.  Having the last three digits  of $m$ be $007$ means $m \equiv 7 \pmod {1000}$  As long as $n$ has no factors of $2$ or $5$ you can just form a set of congruences $m \equiv 1 \pmod p$ where $p$ ranges over the primes that divide $n$.  That set of congruences along with $m \equiv 7 \pmod {1000}$ has a solution and you are done.  I leave patching up the cases where $2$ or $5$ divides $n$ to you.

As Lord Shark has pointed, Dirichlet's theorem implies that there are infinitely many primes that end with $007$. Choose one greater than $n$.

Actually, the series
$$\sum_{p\text{ prime that ends with $007$}}\frac 1p$$
diverges.


If a sub-C*-algebra does not contain the unit, is it contained in a proper ideal? Suppose that $R$ and $S$ are unital rings and that $S$ is a subring of $R$ in the weak sense where the multiplicative identities $1_R$ and $1_S$ are not assumed to be the same. In fact, assume $1_R \neq 1_S$. Then $S$ cannot contain any invertible element of $R$ (if $s \in S$ and there is an $r \in R$ with $sr=1_R$ then $0 = (1_Rs - 1_Ss)r = 1_R - 1_S$ so $1_R = 1_S$).  This is sort of reminiscent of the statement a proper ideal in $R$ cannot contain any invertible element of $R$. My question is whether there must actually be some proper ideal $I$ of $R$ with $S \subset I \subset R$. Actually, I would like to ask the question with the rings replaced by C*-algebras - but a positive answer to the corresponding question about rings would clearly suffice.

  Let $A$ and $B$ be unital C*-algebras with $B$ a sub-C*-algebra of $A$, but suppose $1_A \neq 1_B$. Does it necessarily follow that there is a proper closed ideal $I$ of $A$ satisfying $B \subset I \subset A$?

I'm also curious what happens if the assumption that $B$ is unital is dropped. In this case I'm not even certain whether $B$ can contain an invertible element of $A$.

  Let $B$ be a non-unital sub-C*-algebra of a unital C*-algebra $A$. Can $B$ contain an invertible element of $A$? If not, does there have to exist a proper closed ideal $I$ of $A$ with $B \subset I \subset A$?

For the first question, consider $2$-by-$2$ matrices, and the subalgebra of matrices of the form $\begin{bmatrix}a&0\\0&0\end{bmatrix}$.

For the first part of the second question, the answer is no.  Suppose that $a\in B$ is invertible in $A$.  Then by spectral permanence, $a^{-1}$ is in $B+\mathbb C\cdot 1_A$, so $a^{-1}=b+\lambda\cdot 1_A$ for some $b\in B$ and $\lambda\in \mathbb C$.  This means that $1_A=a^{-1}a=ba+\lambda a\in B$.

(Note that this is special of C*-algebras, and is not true for general algebras or even general Banach algebras.  Consider $X\in X\cdot\mathbb C[X]\subset \mathbb C[X,X^{-1}]$.  For a similar Banach algebra example, consider $f(z)=z$ in the algebra $C(\mathbb T)$ of continuous functions on the unit circle in $\mathbb C$, which is invertible in $C(\mathbb T)$ but not in the closed subalgebra generated by $f$.)

For the second part of the second question, the answer is no.  Let $C_0(\mathbb R)$ act on $L^2(\mathbb R)$ as multiplication operators.  Then it forms a nonunital C*-subalgebra of $B(L^2(\mathbb R))$ contained in no proper ideal (its nonzero elements are noncompact).

The second question has a more general answer. Let $ \mathcal{H} $ be any infinite-dimensional Hilbert space. Then the set $ \mathcal{K}(\mathcal{H}) $ of compact operators on $ \mathcal{H} $ has the following three properties:

It is a non-unital $ C^{*} $-subalgebra of $ \mathcal{B}(\mathcal{H}) $.
It does not contain any invertible element of $ \mathcal{B}(\mathcal{H}) $.
It is a maximal two-sided ideal of $ \mathcal{B}(\mathcal{H}) $.


The cross product of two sets Just reading about the cross product. It says here:
$$A\times B= \{(x,y)|x\in A, y\in B\}$$
Does this mean that the result of $A\times B$ is a new set of ordered pairs? If so, what happens if both A and B share a common member? What would the result be, since an ordered pair cannot include duplicates?

Edit: Quoting the free textbook I'm looking at online,

  If we are given that A is a set and no other information about A, then there is no
  ordering to the elements of A. Thus, we cannot speak of “the second element of the set A”
  unless we have specified an ordering of the elements of A. If we wish to regard A as ordered in some way, then we specify this fact explicitly: “The elements of A are ordered a, b, c,” or “A = (a, b, c).” The latter notation replaces the braces with parentheses and designates that A is ordered, left to right, as indicated. We call this an ordered set. An ordered set is also called a linear order. Various other names are also used: list, vector, string, word — all with no repeated elements. Of course, you’ve seen repeated elements in vectors, for example the point in the plane at the coordinates (1,1). That’s fine, it’s just not an ordered set.

  Why is it okay to specify a set S = {a, b, c, a} where the element a has been repeated,
  but it is not okay to have repeated elements in an ordering of S?   When we say S  =
  {a, b, c, a}, we know that S contains just the three elements a, b and c.  If we were to talk
  about the ordered set (a, b, c, a) it would not make sense because it would say that the
  element a is in two places at once: the first position and the last position.

Yes. $A \times B$ is a set of ordered pairs of elements from $A$ and $B$. If $x \in A$ and $x \in B$, then the ordered pair generated will be $(x,x)$ which is a valid ordered pair and is an element of $A \times B$.

For instance, if $A = \{0,1\}$ and $B = \{1,2\}$, then $A \times B = \{(0,1),(0,2),(1,1),(1,2)\}$. Hope this clarifies.

No,This is no the real multiplication of sets.The real multiplication of sets is:
f(x)   multiplicity of x
A*B={x+y:f(x)f(y)|x:f(x)∈A,y:f(y)∈B}
when f(x)=f(y)=1
AB={x+y|x∈A,y∈B}
why?because:
P({a,b})={0,{a},{b},{a,b}}
P({c,d})={0,{c},{d},{c,d}}
P({a,b,c,d})={0,{c},{d},{c,d},{a},{a,c},{a,d},{a,c,d},{b},{b,c},{b,d},{b,c,d},{a,b},{a,b,c},{a,b,d},{a,b,c,d}}
P(A+B)=P(A)*P(B)
People do not know the real multiplication of sets.


Prove that $X,Y,Z$ lie on a single line. Let $ABCD$ be a convex quadrilateral such that no two opposite sides are parellel to each other. Denote by $Q$ the intersection of lines $AD$ and $BC$ and by $R$ the intersection of lines $AB$ and $CD$. Let $X,Y,Z$ be midpoints of $AC, BD$ and $QR$ respectively. Prove that $X,Y,Z$ lie on the same line.

I am not getting any approach to solve this question. Please help.

This is also known the Newton-Gauss line of $ABCD$. The usual proof is either by area considerations or by Menelaus' theorem. However, I present a purely synthetic approach.

Let $L,M,N$ be the midpoints of $\overline{AB},\overline{AR},\overline{BR}$. By considering midpoints in $\triangle ABC$, we have $XL\parallel BC$, and with $\triangle BQR$, we have $BQ\parallel NZ\implies XL\parallel NZ$. In a similar way, we can obtain $LY\parallel MZ$ and $MX\parallel NY$. But now Pappus' theorem on the hexagon $XMZNYL$ implies that $X,Y,Z$ are collinear.

Here is a simple solution using coordinates:

Consider $A = (0, 0)$, $D = (D_x, 0)$ , $B = (B_x, B_y)$ and $C = (C_x, C_y)$.

We will get $X = \left( \dfrac {C_x} 2, \dfrac {C_y} 2 \right)$ and $Y = \left( \dfrac {B_x + D_x} 2, \dfrac {B_y} 2 \right)$. After some computation we will get the coordinates $Z$ in terms of $B_x, C_x, D_x$ and $B_y, C_y, D_y$. After some simplification, the solution becomes clear (I will not provide details unless OP requests so). Then, we can just check if they are on the same line and we are done!