Why the definition of topology is what it is? I recently attended an interview for PhD. One of panel members asked me definition of topology which I answered and next question was why the definion supposed to be that way.
   I explained through an example. Is it right to explain definition using example? What could be a better answer?
Any help will be deeply acknowledged.

I don't think one example is enough for justifying such a definition. If you only have one specimen it's premature generalization to introduce a classification of that specimen.

As for the set of properties that you use for classification it will depend on which properties that are actually used in proving corresponding propositions for different spaces.

Also note that for a classification to be meaningful you need to have examples that addresses specimina that lies outside subclasses. For example for topological spaces you need at least one example for a space that is not metric (ie if all your examples are metric then the definition of metric space would suffice).

In addition the examples need to be of general importance (for the definition is to be of general interrest).

Consider real line with usual metric, then open sets are union of open intervals.
we have following results:

Any union of open sets is open
Finite intersection of open sets is open
Empty set and real line are open

Now,  we can ask a question,  For any set X,  and a collection of subsets of X satisfying the above results.  If it satisfy,  is called a topology on X


Solving second order IVP I want to solve this IVP but the general solution should be the linear combination of $~y_1,~y_2~$ and $~y_3~$, but I only have $~2~$ initial data,how can I solve for $~3~$ unknown ?

And $~y_1, ~y_2,~ y_3~$ are linearly independent on $~(-\infty,1)~$,is it impossible to find $~2~$ vectors to span $~y_1, ~y_2,~ y_3~$ ?

thanks!!!

Consider the homogeneous equation
$$y''(t) + p_1(t)y' + p_2(t)y = 0$$
Then by subtraction of the (linear) inhomogeneous equations, we see that $y_2(t) - y_1(t) = 2e^t$ and $y_3(t) - y_2(t) = 2e^t - t$ solve this homogeneous equation.
Subtracting again we see that $t$ solves the equation, so that $e^t$ and $t$ are independent solutions to the homogeneous equation. Hence the general solution to the homogeneous equation is given by $Ae^t + Bt$ where $A$ and $B$ are arbitrary constants.

The solution to your original inhomogeneous equations is therefore $y_p(t) + Ae^t + Bt$ where $y_p(t)$ is a particular solution. Looking at any of the three given solutions we see $y_p(t) = e^{t^2}$, so the general solution to this equation is given by
$$e^{t^2} + Ae^t + Bt$$
Here $A$ and $B$ are arbitrary constants. If $y(0) = 1$ and $y'(0) = 2$, we plug in and see
$$1 + A = 1$$
$$A + B = 2$$
Hence $A = 0$ and $B = 2$ and we have the solution
$$e^{t^2} + 2t$$

Each of the given solutions has an $e^t$ and an $e^{t^2}$. Thus, the general solution $y(t)$ has "$c_1 e^t + c_2 e^{t^2}$" in it where $c_1, c_2 \in \mathbb{R}$.

The third solution has $-t$ in it. Thus, the general solution $y(t)$ has a polynomial in it call it $At + B$ where $A, B \in \mathbb{R}$ (Q: Why did I choose degree 1?).

There is a first derivative and second derivative. I expect terms of the form $te^{t^2}$ and $t^2 e^{t^2}$. Thus, the general solution $y(t)$ has "$c_3 te^{t^2} + c_4 t^2 e^{t^2}$" in it where $c_3, c_4 \in \mathbb{R}$.

Thus, the general solution is $y(t) = c_1 e^t + c_2 e^{t^2} + c_3 te^{t^2} + c_4 t^2 e^{t^2} + At + B$.

Back these up with propositions in your textbook like linear combinations of solutions are solutions, etc.

Other stuff: If one of the solutions had a $\sin(t)$ in it, then the general solution would have $c_1 \sin(t) + c_2 \cos(t)$ in it.

Note that these coefficients may be zero. So, the general solution could lack a $e^t$ in the sense that its coefficient is zero.


Slope of tangent when $|f(x_1)-f(x_2)|\leq(x_1-x_2)^2$ for all $x_1,x_2 \in R$ If $|f(x_1)-f(x_2)|\leq(x_1-x_2)^2$ for all $x_1,x_2 \in R$, then find the slope of tangent to the curve at $y=f(x)$ at the point $(1,2)$.

Could someone hint me as how to approach this question. Is data sufficient to find the required slope?

I hope this helps. Next time, learn your definitions


Inequality of function on interval [a,b]. I want to show, that for every $f\in C([a,b])$ inequality

$\sup|f(x)| \le (b-a)(|f(a)| + \sup|f'(x)|$ holds.

When $b-a \ge 1$ this inequality is a simple consequence of Mean value theorem and triangle inequality.

If $0 < b-a < 1$ I can construct function $g\colon [0,1]\to\mathbb{R}$ like this:

$g(t) = f\left((1-t)a + tb\right)$, and the inequality holds for $g$.

Now, because $g(t) = f(a + (1-t)(b-a))$ it also holds for $f$.

Is that correct?

Let $x\in [a,b]$ we have $$\frac{f(x) -f(a)}{x-a} =f'(t) $$ for some $t\in [a,b].$
Thus $$|f(x)- f(a) |\leqslant |f'(t) (x-a)| \leqslant (b-a) |f'(t)| \leqslant (b-a)\sup_{t\in [a,b]} |f'(t)|$$ hence $$|f(x) |\leqslant |f(x)-f(a)| +|f(a)|\leqslant |f(a)| + (b-a)\sup_{t\in [a,b]} |f'(t)|$$ so

$$\sup_{x\in [a,b]}|f(x) |\leqslant |f(a)| + (b-a)\sup_{t\in [a,b]} |f'(t)|.$$


Does Liebniz-Criteria become a necessary condition for convergence if $a_n$ is monotonically decreasing? The Leibniz Criterion says that if the sequence $a_n$ is monotonically decreasing then the following statements are equivalent:

\begin{align}
1) & & & \sum_{n=0}^\infty (-1)^na_n \text{ converges} \\[6pt]
2) & & & \lim_{n\rightarrow\infty}(a_n) = 0
\end{align}

so this has to mean that if $a_n$ is monotonically decreasing, the following is true

$$  \sum_{n=0}^\infty (-1)^na_n \text{ converges} \Leftrightarrow \lim_{n\rightarrow\infty}(a_n) = 0 $$

I know that $ \lim_{n\rightarrow\infty}(a_n) = 0 $ is not a necessary condition for convergence if $a_n$ is not monotonically decreasing, but the way I am reading the rule stated in my script, it becomes a necessary condition if $a_n$ is monotonically decreasing.

For example to prove the non-convergence of

$$ \sum_{n=0}^\infty (-1)^n(1)_n $$

I can just point out to the fact that

$$ \lim_{n\rightarrow\infty}(1)_n = 1$$

because $(1)_n$ is a monotonically decreasing sequence.
Is this interpretation correct?

As Giuseppe Negro has noticed, for the convergence of $\displaystyle \sum a_n$,$$ \lim_{n\rightarrow\infty}(a_n) = 0 $$ is a necessary condition anyway.

The convergence test of Leibniz is only applicable to alternating series, i.e. series where every positive term is followed by a negative term, and every negative term is followed by a positve term ( + - + - + - + - ...).

For a general series, the condition $\displaystyle\lim_{n \to \infty} a_n= 0$ is a necessary condition for convergence, but not sufficient. This is the main point of Leibniz test. He could prove that for alternating series this condition is both necessary and sufficient. Thus if you've got an alternating series $\sum (-1)^n a_n$ with $\displaystyle\lim_{n \to \infty} a_n= 0$, than you are certain it converges.

I don't know why you are referring to the condition of "monotonically decreasing". This doesn't matter for Leibniz test. It's only an issue in the integral test.


Why does the inequality $1-x^5 \leq 5(1-x)$ hold? If $x \in [0,1]$. It is obviously true, by analyzing the plot of these two functions. But how can one prove it analytically?

For $x\in[0,1]$ we have $0\leq x^n\leq 1$, $n\in\mathbb N$. Then we simply get that

$$
\frac{1-x^5}{1-x}=\frac{(1-x)(1+x+x^2+x^3+x^4)}{1-x}=1+x+x^2+x^3+x^4\leq 5.
$$

Multiplying by $1-x$ we show that the given relation holds true.

Replacing $x$ by $1+y$ gives
$$ (1+y)^5 - 5y - 1 = y^5 + 5y^4 + 10y^3 + 10y^2 > 0 $$
when $y>0$, i.e. when $x>-1$, so $x^5 + 5(1-x) - 1 > 0$, equivalent to your inequality. The same works if $5$ is replaced any positive integer.


If $ A$ is open and $ B$ is closed, is $B\setminus A$ open or closed or neither? If $ A$ is open and $ B$ is closed, is $B\setminus A$ open or closed or neither?

I think it is closed, is that right? How can I prove it?

Yes, if $A$ is open and $B$ is closed, then $B\setminus A$ is closed. To prove it, just note that $X\setminus A$ is closed (where $X$ is the whole space), and $B\setminus A=B\cap(X\setminus A)$, so $B\setminus A$ is the intersection of two closed sets and is therefore closed.

Alternatively, you can observe that $X\setminus(B\setminus A)=(X\setminus B)\cup A$ is the union of two open sets and therefore open, so its complement, $B\setminus A$, is closed.

Given $A$ is open and $B$ is closed.
\begin{eqnarray*}
B-A&=&\{x|x\in B \ \& \ x\notin A\}\\
&=&\{x|x\in B \ \& \ x\in A^c\}\\
&=&\{x|x\in B\cap A^c\}\\
B-A&=&B\cap A^c
\end{eqnarray*}
which is closed since it is the intersection of closed sets.


If $F(x)$ is onto function such that $F: \mathbb R \to [-1,2]$ , $F(x)= \frac {x^2+9bx+17}{ax^3+x^2+bx+33}$ and $F'(d)=F'(e)=0$ where $d<e$ then The following question appeared in entrance and i din't solve it because i thought it will take lot of time.Is there any "smart" way to solve this?

  If $F(x)$ is onto function such that $F: \mathbb R \to [-1,2]$ , $F(x)= \frac {x^2+9bx+17}{ax^3+x^2+bx+33}$ and $F'(d)=F'(e)=0$ where $d<e$ then

$(1)a+b+d+e=4$
$(2)a+b+d+e=-4$
$(3) d+e=2$
$(4) d+e=-2$

More than one options can also be correct.Any hints/ideas?

Hint: $F'(d) = F'(e) = 0$ gives you two equations.

Surjectivity allows you to assert that $F(x) = 2$ for some $x$, and $F(x) = -1$ for some $x$, and any other real number between $-1$ and $2$. This gives you some more information. Just calculate the derivative and insert the points. The result should fall out easily.

$a$ must be  zero given that the function is defined for all real.

Moreover subjectivity can be used to determine value of $b$ along with definition of the function.
Then proceed to find $d$ and $e$ with the so formed quadratic expression.


Does every irrational number have continued fraction representation? Are CFs fastest method so far? (Rational approximations of $ e,$ and $\gamma$) What are rational approximations to $e$ and Euler-Mascheroni constant $\gamma$ ?

(Like $ 22/7 , 355/113, ...$ are for  $\pi.$)

EDIT 1:

(Number theory is not my cup of tea..). I wish to be able to learn if for every irrational number there exists an infinite set of progressively reducing
error rational numbers as approximation and a procedure by which it is obtained.

EDIT 2:

Are continued fraction the fastest or the only method to obtain rational successive  approximations to irrationals?

You can calculate continued fraction representations, where you can choose to terminate after you've reached some level of accuracy:

$$e=[2;1,2,1,1,4,1,1,6,1,\dots]$$

Here are some of the first results:

$$e\approx\frac{19}{7}, \frac{87}{32}, \frac{106}{39 },\dots$$

Another (beautiful, but non-simple) continued fraction representation of $e$ is $$e=2+\frac{1}{1+\frac{1}{2+\frac{2}{3+\frac{3}{3+\cdots}}}}.$$

Same for $\gamma$:

$$\gamma=[0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40,\dots],$$

where the first few (useful) convergents are
$$\gamma \approx \frac{11}{19}, \frac{15}{26},\frac{71}{123},\dots$$

Edit as a response to OPs edit:

From the Wikipedia-page: every irrational number $\alpha$ is the value of a unique infinite continued fraction, whose coefficients can be found using the non-terminating version of the Euclidean algorithm applied to the incommensurable values $\alpha$  and 1.

Euler's approximation (with function in link):
19/7, 87/32, 106/39, etc...
http://www.johndcook.com/blog/2013/01/30/rational-approximations-to-e/

Euler-Mascheroni constants can be found here with references to the original findings:
http://mathworld.wolfram.com/Euler-MascheroniConstantApproximations.html


Express $4\ln(x)+2\ln(x^4y^3)+5\ln(z)$ as a single logarithm The problem is to express $4\ln(x)+2\ln(x^4y^3)+5\ln(z)$ as a single logarithm.

Our teacher has shown us examples for the same base and when it's both add and subtract. But I'm not sure how to do this.

$4\ln(x)+2\ln(x^4y^3)+5\ln(z)$

By using identity, $a\times ln x=ln x^a$

$\implies \ln(x)^4+\ln{(x^4y^3)}^2+\ln(z)^5$

By using identity, $\ln x+ \ln y=\ln {xy}$

$\implies \ln(x)^{4+8}y^{3\times2}(z)^5$

Final Answer,

$\implies \ln(x)^{12}y^{6}(z)^5$


Prove that vectors in Euclidean space are linearly independent Could you help me with an idea of solving the following problem? I think that proof involves the positive definiteness of Gram matrix, but I don't know how.

Consider a system of vectors $e_1, e_2, ..., e_n, e_{n+1}$ in some Euclidean space such that dot product $(e_i, e_j) <0$ for all $i \neq j.$ Prove that any $n$ vectors of the system are linearly independent.

Let us try by recurrence and suppose that $n$ is the dimension of the space.

for $n=2$ we have $(e_1, e_2, e_3)$ in an Euclidian space of dimension 2, such that the scalar products are non positive.
Let us suppose that $e_1$ and $e_2$ (without loss of generality) are linearly dependent, i.e that there exist $(\lambda_1, \lambda_2) \neq (0,0) $ such that $$\lambda_1 e_1 + \lambda_2 e_2 = 0$$ If $\lambda_1 = 0$, then $\lambda_2 = 0$ therefore we can suppose that $\lambda_1 > 0$.
Then $\lambda_1 (e_1|e_1)+\lambda_2(e_1|e_2) =0$ and therefore $\lambda_2>0$.
Then you can also write that $0=\lambda_1(e_1|e_3)+\lambda_2(e_2|e_3)<0$ which is absurd.
You conclude that $e_1$ and $e_2$ are linearly independent.

Let $n>2$, and suppose that the property we want to prove is right at the rank $n-1$. Let $e_1, \dots e_{n+1}$ be vectors verifying the hypothesis from the question. We can suppose that each of these vectors have unit norm (by dividing them by their norm).
Let us take $n$ of these vectors, let us say $e_1, \dots, e_n$ and suppose they are linearly dependent, i.e. that there exist $(\lambda_1, \dots, \lambda_n) \neq (0, \dots, 0)$ such that $$\sum_{i=1}^n \lambda_i e_i =0$$ Then $$\sum_{i=1}^n \lambda_i (e_i|e_n)e_n =0$$ If we substract the two previous equalities we obtain $$\sum_{i=1}^n \lambda_i f_i =0$$ where we note $f_i=e_i-(e_i|e_n)e_n$.
We notice that $f_n=0$ (since $\|e_n\|^2=1$). This shows that $$\sum_{i=1}^{n-1} \lambda_i f_i =0$$
Let us know compute $(f_i|f_j)$ for $i \neq j$ (and $i,j \neq n$). $$\begin{align}(f_i|f_j)&=(e_i-(e_i|e_n)e_n|e_j-(e_j|e_n)e_n)\\&=(e_i|e_j)-(e_i|(e_j|e_n)e_n)-(e_i|e_n)(e_n|e_j)+(e_i|e_n)(e_j|e_n)\|e_n\|^2\\&=(e_i|e_j)-(e_i|e_n)(e_j|e_n)<0\end{align}$$
Moreover all the $f_i$ ($i \neq n$) are non-zero vectors located in $(e_n)^{\perp}$ space, which is of dimension $n-1$.
We have now $n$ vectors verifying the hypothesis at rank $n-1$, therefore any $n-1$ combination of the vectors $(f_i)_{i \in \{1, \dots, n-1, n+1 \}}$ is linearly independant. In particular $f_1, \dots, f_{n-1}$ are independant and consequently $\lambda_1, \dots, \lambda_{n-1} = 0$.
This proves that $(e_1, \dots, e_n)$ are independant. And the property is true at rank n.

Here is an overkill, but it has an interesting connection to another area of matrix theory.

Let $E=[e_1|\cdots|e_n|e_{n+1}]$. By the given conditions, $P=E^\top E\in M_{n+1}(\mathbb R)$ is a positive semidefinite irreducible $Z$-matrix. Therefore $P+tI$ is an $M$-matrix for every $t>0$, so that $(P+tI)^{-1}\ge0$ entrywise. As $P+tI$ is irreducible, so is $(P+tI)^{-1}$. By Perron-Frobenius theorem, the largest eigenvalue of $(P+tI)^{-1}$ is simple. Thus the smallest eigenvalue of $P$ is simple too. Hence the rank deficiency of $P$ is at most $1$ because $P$ is positive semidefinite.

It follows that $\operatorname{rank}(E)=\operatorname{rank}(P)\ge n$. That is, some $n$ vectors of $\{e_1,e_2,\ldots,e_n,e_{n+1}\}$ are linearly independent. Without loss of generality, suppose the subset $\{e_2,\ldots,e_{n+1}\}$ is linearly independent. We want to prove that $\{e_1,e_2,\ldots,e_n\}$ (and similarly for other subsets of size $n$ containing $e_1$) is also linearly independent. Suppose
$$
\sum_{i=1}^k p_ie_i = \underbrace{\sum_{i=k+1}^n p_ie_i}_{:=\, x},
$$
where $k\ge1$ and each $p_i$ is nonnegative. By taking inner products on both sides with $x$, we see that $x$ must be zero. As $e_{k+1},\ldots,e_n$ are linearly independent, $p_i$ must be zero for every $i\ge k+1$. In turn, $\sum_{i=1}^k p_ie_i = 0$. Take inner products again on both sides with $e_{n+1}$, we see that $p_i$ must also be zero for each $i\le k$. Hence all coefficients are zero and $e_1,\ldots,e_n$ are linearly independent. QED

Our conclusion also translates to the following

  Proposition. The rank deficiency of a positive semidefinite irreducible $Z$-matrix is at most $1$. Furthermore, if the matrix has strictly negative off-diagonal entries, then its proper principal submatrices are all nonsingular.


A mother's age is $19$ years more than the sum of the ages of her sons A mother's age is $19$ years more than the sum of the ages of her two sons. $5$ years ago, the mother's age was $4$ times than the sum of the ages of her two sons. What is the age of the older child?

Let's say the sum of the ages of her sons is $x$, the mother's age will be $x+19$.
$5$ years ago, $x-5$ = $x+19-5$. However, I believe that I've gone too wrong.
What kind of methods can I use to solve this question?

I'll be waiting for your professional helps.

My Kindest Regards!

There isn't enough information to solve for the age of the older child, even if there are two children. I can only solve for the following:

$\text{Ages of sons combined} = x$

$\text{Mother's age}=y$

$$x+19=y$$
$$4(x-10)=y-5$$
$$4x-40=x+19-5$$

$$3x=54$$

$$x=\dfrac {54}{3}=18$$

You know the sums of the ages of the two sons is $18$. But, you need to find the age of one of the children to find the age of the other child.

Note that the younger child must be older than $5$ years but younger than $9$ years. For example, if the younger child is $7$ years old, then the older child is $11$ years old.

hint

Today
$$y=x+19$$
five years ago

$$y-5=4 (x-5n) $$
$$14=3x-20n$$
$n $ is the number of sons.

The sum $x $ must satisfy $x>5n $

thus $n=2$ and
$x=18$ years .
the mother has
$37$ years .


How to show an inner product is unique? Show there exists an inner product on V such that V's basis is an orthonormal basis.

If i let the inner product be $\langle v_i,v_j \rangle$ such that

$\langle v_i,v_j \rangle=1$ if $i=j$ and $\langle v_i,v_j \rangle=0$ if $i \neq j$ then this makes the basis orthonormal, correct?

Now I am asked:

"show that the inner product is unique"

Do I start with "suppose $[v_i,v_j]=1$ if $i=j$ and $\langle v_i,v_j \rangle=0$ if $i \neq j$?

Did i do the first part incorrectly? How can I show an inner product is unique?

What you're done is correct so far. Because of linearity, it suffices to define an inner product in $V$ to pick up a basis $\{v_1,..., v_n\}$ and define $\langle v_i, v_j\rangle$, for $i, j \in \{1,...,n\}$.


How's it possible for a bounded subset of the reals not to contain it's $\inf$ or $\sup$? Definition
By the real numbers we mean $\Bbb R:=\{\gamma:\gamma \text{ is a cut of $\Bbb Q$}\}$.

In Principles of Mathematical Analysis Rudin proves that if $A\subseteq \Bbb R$ is bounded, and we let $$\alpha :=\bigcup_{\gamma \in A}\gamma$$
Then $\alpha = \sup A$.
From this construction, I don't understand how could there be a set $A$ in which $\bigcup_{\gamma \in A}\gamma\not \in A$, that is : a set that doesn't contain it's supremum.
Could someone clarify my confusion?
E:
It seems I wasn't clear enough: I'm aware such sets exist, what I don't understand is how, with this construction, could $\alpha$ not be an element of $A$ (as I understand it, $\alpha$ is "the biggest cut": one of the $\gamma$s in $A$, am I wrong?).
E2:

Definition
By a cut $\beta$ we mean a proper subset of $\Bbb Q$ which satisfies:

$p\in \beta$ and $q<p$ implies $q \in \beta$.
If $p \in \beta$ then there exists a $r \in\beta$ such that $p<r$.

I don't have a proof or anything for my above "claim", I just thought Rudin's argument would imply that $\alpha\in A$: I imagine cuts (informally, of course) as intervals of rational numbers $(-\infty,r)$ where the $r$ is determined by some condition (for example $\{x:x^2<2\}$ determine $r=\sqrt 2$, etc).
So looking at it that way, it looks like that when I say "$\alpha=\bigcup_{\gamma\in A} \gamma$", the resulting set (the union) must be just a single one of those $\gamma$s, more clearly, it looks like (to me) that $\alpha=\gamma_0$ where $\gamma_0\in A$ and $\gamma \subseteq \gamma_0\forall \gamma \in A$. Maybe I'm just interpreting wrongly what a cut is.

In general, when you take a union of the sets in a collection of sets,
you can easily end up with a new set that is not in the original collection.

I'll assume that when we take $\bigcup_{\gamma\in A} \gamma$, where each
$\gamma$ is a "cut", we are actually referring to a union of the lower sets
in each cut; that is, each $\gamma$ is a set of rationals that does not
contain its own supremum.

Let $A$ be the countably infinite collection of Dedekind cuts named
$1/2, 2/3, 3/4, 4/5, \ldots$.
That is, $A = \left\{ 1 - \frac1n \mid n \in \mathbb N \right\}$.
Now let $q$ be any rational number less than $1$;
for a large enough $n$, $q < 1 - \frac1n$, so $q$ is a member of the
cut named $1 - \frac1n$, and therefore $q \in A$.
That is, $A$ is the set of all rational numbers less than $1$,
and therefore $A$ is just exactly the Dedekind cut for the real number $1$.

In short, $\sup A = 1$ and $\bigcup_{\gamma\in A} \gamma = 1$,
but $1 \not\in A$.


Is there a difference between $\mathbb R^\infty$, $\mathbb{R^N}$, and $\mathbb R^\omega$? I've seen these three notations and I'd like to know the difference:

$\mathbb{R}^\infty$
$\mathbb{R}^\mathbb{N}$
$\mathbb{R}^\omega$

$\mathbb{R}^{\mathbb{N}}$ and $\mathbb{R}^{\omega}$ are almost always synonyms, and both refer to the set of functions $\mathbb{N}\to\mathbb{R}$ (aka sequences of real numbers indexed by $\mathbb{N}$).  The symbol $\omega$ is an alternate name for the set of natural numbers (or more specifically, for one particular set-theoretic construction of the natural numbers) which is commonly used in set theory.  Conceivably someone might distinguish $\mathbb{R}^\mathbb{N}$ and $\mathbb{R}^\omega$ as, for instance, $\omega$ is always taken to include $0$ but some authors exclude $0$ from $\mathbb{N}$.  I've never seen anyone actually intentionally make such a distinction, though.
$\mathbb{R}^\infty$ is sometimes a synonym for $\mathbb{R}^\mathbb{N}$, but sometimes it instead refers to the subset of $\mathbb{R}^\mathbb{N}$ consisting of those functions $\mathbb{N}\to\mathbb{R}$ that are $0$ on all but finitely many inputs.

Hints : A basic open set of $\mathbb{R}^{\omega}$ is of the form
$$
U = U_1\times U_2\times \cdots \times U_n \times \mathbb{R} \times \mathbb{R} \times \cdots
$$
where $U_i$ are open sets in $\mathbb{R}$
similarly  for $\mathbb{R}^\infty$ and  $\mathbb{R}^\mathbb{N}$


Which is the correct Hessian matrix (the standard matrix of a bilinear form)? Please note the typo in the first entry: $\frac {\partial^2f} {\partial x_1\partial x_2}$ should instead be $\frac{\partial^2f} {\partial x_1\partial x_1}$.
Also, this Hessian matrix need not be symmetric as the partials need not all be continuous.
The convention in use is: $\frac {\partial^2f} {\partial x_1\partial x_2}$ means $\partial_{x_1} (\partial_{x_2}f)$.

Question: Shouldn't the correct matrix be the transpose of what is
given?
My reasoning is this:
Let $\{e_1, ..., e_n\}$ be the standard basis for $\mathbb R^n$.
Because the matrix is for the bilinear form
$D^2f(x)$, by definition, the $ij$-entry of the matrix is given by
$$[D^2f(x)](e_i,e_j)=[[D(Df)(x)] (e_i)](e_j) =\frac {\partial^2f} {\partial x_j\partial x_i} \neq \frac {\partial^2f} {\partial x_i\partial x_j}.$$ In other words, in the $ij$-entry, the $i$th variable should be partialled before the $j$th variable, not the other way round.

Referring to the proof is not helpful as it is cryptic and makes no sense:

Information on how $D^2f(x)$ is defined:

$\newcommand{\hom}{\operatorname{Hom}}$Let us write $A \subseteq U=V=\mathbb{R}^n$. We have
\begin{align*}
f:& A \to \mathbb{R}\\
Df:& A \to \hom(V,\mathbb{R})\\
D^2f:=D(Df):& A \to \hom(U, \hom(V,\mathbb{R}))
\end{align*}
Note that the codomain of $Df$ is $\hom(V,\mathbb{R})$ (the set of all linear maps from $V$ to $\mathbb{R}$), which is not a Euclidean space. So, what do we mean by $D(Df)$? Since $\hom(V,\mathbb{R})$ is isomorphic to $\mathbb{R}^n$, the answer is that we first identify each linear operator $L_i: e_j\mapsto\delta_{ij}$ with the vector $e_i$. Hence the linear map $Df(x)$ is identified with the gradient vector $\nabla f(x)=\left(\frac{\partial f}{\partial x_1},\,\frac{\partial f}{\partial x_2},\,\ldots,\,\frac{\partial f}{\partial x_n}\right)^\top$ before taking a second derivative.

Having this identification, the matrix of $D^2f$, as a linear operator, is the Jacobian matrix of $\nabla f$, i.e. $J(\nabla f)(x)=\left(\frac{\partial^2 f}{\partial x_j \partial x_i}\right)$. Therefore, $D^2f(x)(u)$ is represented by the vector $\left(\frac{\partial^2 f}{\partial x_j \partial x_i}\right)u$ and
$$
\left(D^2f(x)(u)\right)(v) = \left[\left(\frac{\partial^2 f}{\partial x_j \partial x_i}\right)u\right]^\top v = u^\top \left(\frac{\partial^2 f}{\partial x_i \partial x_j}\right)v.\tag{1}
$$
Technically, $D^2f(x)$ is not a bilinear form, but a linear operator that maps vectors to linear operators. Yet, the mapping $B_x:U\times V\to\mathbb{R}$ by $B_x(u,v)=\left(D^2f(x)(u)\right)(v)$ is bilinear. Therefore we can identify $D^2f(x)$ with the bilinear form $B_x$. And when we speak of the matrix of $D^2f(x):U\times V\to\mathbb{R}$, we actually mean the matrix of $B_x$ (and $D^2f(x)$ is not really a function from $U\times V$ to $\mathbb{R}$ in the first place).

By convention, if $\mathbf{u},\mathbf{v}$ and $\mathbf{B}$ represent respectively two vectors $u,v$ and a bilinear form $b(u,v)$ with respect to some basis, then $b(u,v)=\mathbf{u}^\top \mathbf{B}\mathbf{v}$. Therefore, from $(1)$, we see that the matrix of $B_x$ with respect to the standard basis is $\left(\frac{\partial^2 f}{\partial x_i \partial x_j}\right)$.

If $f$ is twice differentiable (in sense of totally / Frechet differentiable), then the Hessian is symmetric.

Notice that there is little merit to using the Hessian if $f$ is not twice differentiable but only twice partially twice differentiable.

Proof of Schwarz's theorem (for twice differentiable function)

Without loss of generality assume the domain $A = [-1, 1]^2$ is a square, $f$ differentiable on $A$, $\nabla f$ is differentiable at $0$, and $f(0) = 0$ and $\nabla f(0) = 0$. We show that $\partial_y \partial_x f(0) = \partial_x \partial_y f(0)$.
The proof is little bit ugly. The key idea is that you can expand $f(x,y)$ in two way:
\begin{align}
f(x,y) &= \frac12 \partial_x \partial_x f(0) x^2 + \frac12 \partial_y \partial_y f(0) y^2 +  \partial_y \partial_x f(0) x y + \mathcal o(\|(x,y)\|^2), \\
f(x,y) &= \frac12 \partial_x \partial_x f(0) x^2 + \frac12 \partial_y \partial_y f(0) y^2 +  \partial_x \partial_y f(0) x y + \mathcal o(\|(x,y)\|^2).
\end{align}
Thus, it follows
$$ \partial_y \partial_x f(0) - \partial_x \partial_y f(0) = \mathcal o(1). $$
That is, $\partial_y \partial_x f(0) = \partial_x \partial_y f(0)$.

Let us see, how to get the first expansion.
Since $\nabla f$ is continuous at $0$, we may assume $\nabla f$ is bounded on $A$ without loss of generality.
Thus, by the Taylor theorem, we have
\begin{align}
f(x,y) &= f(0,y) + \int_0^1 \partial_x f(t x, y) \;\mathrm dt \cdot x \\
&= \frac12 \partial_y\partial_y f(0) y^2 + \mathcal o(y^2) + \int_0^1 \partial_x f(t x, y) \;\mathrm dt \cdot x \\
&= \frac12 \partial_y\partial_y f(0) y^2 + \mathcal o(y^2) + \int_0^1 \partial_x \partial_x f(0)tx + \partial_y \partial_x f(0)y \;\mathrm dt \cdot x + \mathcal o(x^2 + y^2) \\
&= \frac12 \partial_y\partial_y f(0) y^2 + \frac12\partial_x \partial_x f(0)x^2 + \partial_y \partial_x f(0)xy + \mathcal o(x^2 + y^2).
\end{align}
Analogously, we have
\begin{align}
f(x,y) &= f(x,0) + \int_0^1 \partial_y f(x, ty) \;\mathrm dt \cdot y \\
&= \frac12 \partial_x\partial_x f(0) x^2 + \frac12\partial_y \partial_y f(0)y^2 + \partial_x \partial_y f(0)xy + \mathcal o(x^2 + y^2).
\end{align}
Which had to be proven.


How would I solve this trigonometric equation $\ln[(\sin(x)+\cos(x))^{1+\sin(2x)}]=2$? How would I simplify this? I can't understand which log property to use to simplify this to get the solution of this trigonometric equation

$$\ln[(\sin(x)+\cos(x))^{1+\sin(2x)}]=2$$

I know that $1+\sin(2x) = (\sin(x)+\cos(x))^2$, but I cant understand which property of logarithms to apply to get the solution here.

The equation can be written as:

$$t^{t^2}=e^2$$

where $t = \sin x+\cos x$. The function $f:(0,\sqrt{2}]\to \mathbb{R}, f(t)=t^{t^2}$, has one local minimum at $t=\frac{1}{\sqrt{e}}$, so:

$$f(t)\leq \max \{\lim_{x\to 0^+} f(x),f(\sqrt{2})\}=f(\sqrt{2})=2<e^2$$

because $\lim\limits_{x\to 0^+} f(x)=1$. The equation does not have any solution.

If we use ln a^b = b ln a then we get (sin x + cos x)^2 ln (sin x + cosx) = 2.............but max value of sin x + cos x is root 2 and if we put that we get ln 2 so I dont think any real solution exists. If you want to solve for unreal solution put cos x = (e^ix + e^-ix)/2 and sin x = (e^ix - e^-ix)/2i.


number of possibilities to interview students One semester includes 13 weeks, Given 8 students we want to interview them with the following conditions:

1) Each student should be interviewed at least once.

2) None can be interviewed more than 3 times (3 is the maximum).

Calculate the number of possibilities to interview the students.

Note: As usual, the order of interviews matters so interviewing A at the first week and B at the second is not the same as interviewing B at the first and A at the second week.

Hint: Split by cases.

   Each student has to be interviewed once, so how can the remaining 5 interviews be done? It has to be either $1+1+1+1+1$, $1+1+1+2$, or $1+2+2$.

Can you take it from here?

After getting Hint here's my suggested answer, hope someone can review it:


Looking for a pattern in a math riddle Looking to find a pattern but no idea how:

$12\mathop{\square}21 = 86$,

$13\mathop{\square}31 = 192$,

$14\mathop{\square}58 = 389$,

$14\mathop{\square}94 = \ ?$

This problem is not clearly defined. One of infinitely many ideas is the ansatz

$$x\square y := x\cdot a +b+ c\cdot y$$

leading to 3 equations, which can be solved giving

$$a=\frac{892}{17}, b=\frac{-11153}{17}, c=\frac{91}{17}.$$

Now, what is $$14\square 94=?$$


Why don't we define "imaginary" numbers for every "impossibility"? Before, the concept of imaginary numbers, the number $i = \sqrt{-1}$ was shown to have no solution among the numbers that we had. So we declared $i$ to be a new type of number. How come we don't do the same for other "impossible" equations, such as
$x = x + 1$, or $x = 1/0$?

Edit:
OK, a lot of people have said that a number $x$ such that $x = x + 1$ would break the rule that $0 \neq 1$. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining $-1$ to be the number such that $-1 + 1 = 0$. Note that this breaks the "rule" that "if $x \leq y$, then $ax \leq ay$", which was true for all $a, x, y$ before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing.

Here's one key difference between the cases.

Suppose we add to the reals an element $i$ such that $i^2 = -1$, and then include everything else you can get from $i$ by applying addition and multiplication, while still preserving the usual rules of addition and multiplication. Expanding the reals to the complex numbers in this way does not enable us to prove new equations among the original reals that are inconsistent with previously established equations.

Suppose by contrast we add to the reals a new element $k$ postulated to be such that $k + 1 = k$ and then also add every further element you can get by applying addition and multiplication to the reals and this new element $k$. Then we have, for example,  $k + 1 + 1 = k + 1$. Hence -- assuming that old and new elements together still obey the usual rules of arithmetic -- we can cheerfully subtract $k$ from each side to "prove" $2 = 1$. Ooops! Adding the postulated element $k$ enables us to prove new equations flatly inconsistent what we already know. Very bad news!

Now, we can in fact add an element like $k$ consistently if we are prepared to alter the usual rules of addition. That is to say, if we not only add new elements but also change the rules of arithmetic at the same time, then we can stay safe. This is, for example, exactly what happens when we augment the finite ordinals with infinite ordinals. We get a consistent theory at the cost e.g. of having cases such as  $\omega + 1 \neq 1 + \omega$ and $1 + 1 + \omega = 1 + \omega$.

i should not be defined as a solution to x^2=-1 (i.e. as an artificial way to find a "solution" to a previously considered "impossible" equation. Complex numbers are defined as orderer pairs (a,b) of real numbers. You define arithmetic rules (+,x,-,/) for such ordered pairs which can be done in a way that makes (a,0) equivalent with the ordinary real number a. Complex numbers (0,b) will be found to satisfy (0,b)^2=(-b^2,0). Voila, we have found a certain kind of numbers (0,b) also called "imaginary" numbers which when squared will give a complex number (-b^2,0) equivalent with the negative real number -b^2. A similar construction of a new type of number x satisfying x=x+1 cannot be made with preservation of established arithmetic rules.


Composition of translation and rotation is a rotation, but what is its center? Consider the rotation $r_{\Omega,\alpha}$ of center $\Omega$ and angle $\alpha$.Furthermore let $t_{\vec{v}}$ be the translation by vector $\vec{v}$.
Then $$t_{\vec{v}}\circ r_{\Omega,\alpha}=r_{\Omega',\alpha}.$$
I am wondering how $\Omega'$ is defined.

For $r_{\Omega,\alpha}\circ t_{\vec{v}}=r_{\Omega',\alpha}$ we have the relation $\Omega+\vec{v}=\Omega'$.
Is there a similar relation for $$t_{\vec{v}}\circ r_{\Omega,\alpha}=r_{\Omega',\alpha}.$$
Thanks in advance!

The formula you give is not correct, for instance, if you evaluate both side at $\Omega' = \Omega + \vec{v}$:

$$(r_{\Omega,\alpha} \ \circ \ t_{\vec{v}} \ )(\Omega + \vec{v}) = r_{\Omega,\alpha}(\Omega + 2\vec{v})\neq \Omega + \vec{v} = r_{\Omega',\alpha}(\Omega + \vec{v})$$

On the other hand there is a formula for the conjugation of a rotation with a traslation:

$$(r_{\Omega,\alpha})^{t_{\vec{v}}} := t_{\vec{v}} \ \circ \ r_{\Omega,\alpha} \ \circ \ t_{-\vec{v}} = r_{\Omega + \vec{v}, \ \alpha}$$


Convergence of $\sum2^{-\sqrt{k}} $ So I am going to determine whether this series converges or not:
$$\sum_{k=0}^\infty 2^{-\sqrt{k}} $$

Since this chapter is about the ratio test, I applied that test to this series.
I end up with this limit $$\lim_{k \to \infty} 2^{\sqrt{k}-\sqrt{k+1}}$$

I'm stuck here, don't know how to calculate this limit. I could simplify to: $$\sqrt{k}-\sqrt{k+1} = \frac{1}{\sqrt{k}+\sqrt{k+1}} $$
But I doubt this will help me.

Could anyone help me?

The integral test works. We calculate \begin{align*} \int^\infty_0 2^{-\sqrt x} &= \int^\infty_0 e^{-\sqrt{x}\log 2} dx \\
&= \frac{ 2 }{(\log 2)^2}\int^\infty_0 te^{-t} dt \,\,\,\,\,\,\, [t = \sqrt x \log 2].
\end{align*} The latter converges (easily shown via integration by parts) so the sum converges as well.

EDIT: As suggested in the above comments, a comparison should work as well. For sufficiently large $k$, we see $\sqrt{k} \ge 2\log_2 k$. This follows since asymptotically $\sqrt{k} \gg \log_2 k$. For such $k$, we have $$2^{-\sqrt k} \le 2^{-2\log_2 k} = \frac 1 {k^2}.$$

Hint:

$$\lim\limits_{k \to \infty} 2^{\sqrt{k} -\sqrt{k+1}} = 2^{\lim\limits_{k \to \infty} \sqrt{k} -\sqrt{k+1}}$$

Can you finish from here?


W is normal conditional on a normal variable Z. Does it follow that W is unconditionally normal? $Z$  is normal with mean $\mu_z$ and standard deviation $\mu_z$. Contitional on $Z=z, W$ is normal with mean $z$ and standard deviation $\sigma_w$.

Does it follow from these hypotheses that $W$ is a normal variable with mean $\mu_z$ and standard deviation $\sigma_w$?

My work: Using law of iterated expectations I can take $$E[W]=E[E[W|Z]]=E[Z]=\mu_z.$$ Similarly, I can compute $E[W(2)]$ with iterated expectations to get $\sigma w^2 + \mu_z^2$, implying that the variance of $W$ is just $\sigma_w^2$.

Now can I conclude that $W$ is a normal variable unconditionally? My reasoning is that if I take the product of the conditional density and multiply by the pdf of $Z$, and integrate out the variable $Z$, the resulting functional form of the pdf will be $A\exp{\left(\frac{(w−c)^2}b\right)}$ where A,b,c are constants, implying that $W$ is a normal variable (whose mean and expectations were already derived).

[I presume that the standard deviation of $Z$ should be $\sigma_Z$.] You have the right idea -- iterated expectations. Use this method to the compute the characteristic function of $W$. Notice that you will end up with the variance of $W$ equal to $\sigma^2_Z+\sigma^2_W$ ("variance equals the mean of the conditional expectation plus the expectation of the conditional mean").

No. Let $p(z) = .5\delta(z-1)+.5\delta(z+1)$, where $\delta$ is the Dirac delta function. Then $$p(w)=\int\!\mathrm{d}z\,\mathcal{N}(w; z, \sigma_w^2) p(z)=.5\mathcal{N}(w; -1, \sigma_w^2)+.5\mathcal{N}(w; 1, \sigma_w^2),$$ which is not Gaussian (instead, it is a mixture of Gaussians).

The flaw in your reasoning is that when you integrate out $z$, the result will in general not be Gaussian; in fact, it will be Gaussian only if $p(z)$ was itself Gaussian.


What is the limit of $n\sin \left(\frac{1}{n}\right)$ as $n \to \infty$ The textbook I'm reading gives no proof of the limit being 1 apart from a flimsy 0/0, and has covered only the "epsilon" and sandwich test for convergence to a particular limit.

Do not rearrange to turn $n$ into $\frac{1}{n}$, that is the source of the flimsy $0/0$ argument.

Hint:
$$\lim_{x\to 0}\frac{\sin x}x=1.$$

Hint:  $\sin \left(\dfrac{1}{n}\right) \approx_{0} \dfrac{1}{n} - \dfrac{1}{2!}\cdot \dfrac{1}{n^2} + \dfrac{1}{3!}\cdot \dfrac{1}{n^3}- ......\Rightarrow n\sin \left(\frac{1}{n}\right) \approx_0 1 - \dfrac{1}{2n} + \dfrac{1}{6n^2} - ....--$


Expanding partial derivatives Let $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$  Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$. I want to show that $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac\partial{\partial r}+\frac1{r^2}\frac{\partial^2}{\partial \theta^2}.$$

I use the chain rule to compute $$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$ and $$\frac{\partial}{\partial y} = \frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta}$$

So $$\Delta=\left (\frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta} \right )^2 + \left (\frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta} \right )^2$$

Expanding, I get $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1{r^2}\frac{\partial^2}{\partial\theta^2}$$

Where does the term $\frac1r\frac\partial{\partial r}$ disappear?

You have to rewrite everything in terms of $r$ and $\theta$, so where you are leaving $x$ and $y$ behind you have to clear up. Also, these are operators, so you have to have them act on a test function $f$ -- this also makes you be a bit more careful.

What you'll find is that, since $\Delta = \nabla \cdot \nabla$, you need to compute the Jacobian for the transformation, that is
\begin{equation}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial x}
\end{equation}
etc. This gets you how to rewrite the $\nabla$ operator in polar coordinates. Then you need to figure out what $\nabla \cdot (\nabla f)$ is to figure out what $\Delta f$ is.

There is also a metric tensor way to rewrite the Laplacian, see under http://mathworld.wolfram.com/Laplacian.html for example.


Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$

I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...

Let $a = 2^x, b = 2^y$. Use some algebraic manipulation to arrive at $(a - 1)^2 + (b - 1)^2 = 2$

This is a circle of radius $\sqrt{2}$ centered at $(1, 1)$, and we want to find the minimum and maximum values of $a + b$.

So $a + b = M \rightarrow b = -a + M$, which is the equation of a line with slope $-1$. The maximum and minimum will be when we find the maximum and minimum y-intercept of this line such that it intersects at least one point on the circle with $a, b > 0$

It's tangent to the circle at the top right when $a = b = 2$ which gives a value of $M = 4$. The smallest y-intercept occurs at $m = 2$, however, we do not include $2$ because it makes one of the coordinates zero. This establishes $(2, 4]$ as the range.

Let $\delta x = y-x$ and $x\ge y$. Then

$$4^x(1+4^{\delta x}) = 2^{x+1}(1+2^{\delta x})$$

then from factorization we can claim that:

$$2x = x+1 ~\text{and}~~ 4^{\delta x} = 2^{\delta x}$$

So, we get $x = y = 1$.


Prove that this is no real number such that $x \leq a$ for all real $x$. Prove that this is no real number such that $x \leq a$ for all real $x$.

I want to know if the way I proved it is valid or not.

Proof. We first prove that there is no real number $a$ such that $x=a$ for all real $x$. If this were true, then all real numbers would be equal to $a$. This is not possible, because the axiom that garantees the existence of identity elements tells us that the set of real numbers has the numbers $0$ and $1$. Therefore, the set of real numbers has more than only one element.

If $x<a$ for all real $x$, then, of course $a>0$. It can be proven that $1>0$. There is an axiom that tells that if two real numbers are in $R^{+}$, the sum of the two is in $R^{+}$. So, $a+1>0$ is in in $R^{+}$ and in $R$. But we assumed that $a$ is greater than all the real numbers. Thus, $a>a+1$, which is an absurd.

Is this proof correct? Can I improve it? Thank you.

Let me propose an alternative proof.

Proof:  By contradiction, suppose such an $a \in \mathbb{R}$ exists.  Then $a + 567 \le a$, which implies $567 \le 0$, a contradiction.

This is a consequence of the Archimedean property of the reals. (I assume that you have some familiarity with real analysis; in particular, I assume that you know that the reals have the supremum property.)

Theorem: For every real $\alpha > 0$ and real $\beta$, there exists an integer $n$ (in the naive sense of integers) such that

$$n \alpha > \beta$$

Proof: Suppose that there were to exist counterexamples $\alpha_0$ and $\beta_0$ (i.e., $\alpha_0$ such that for every integer $n$, $n\alpha_0 < \beta_0$). If we form $S = \{s \in \mathbb{R}: s = n\alpha \text{ for } n \in \mathbb{N}\}$, then this would say that every member of S is smaller than $\beta_0$ by hypothesis.

But clearly $S$ is not empty if $n\alpha_0 < \beta_0$ for every $n \in \mathbb{N}$; since it is bounded above (and since these are real numbers!) there exists $\lambda = \mathrm{sup}(S)$.

Now, for every $\epsilon > 0$, we must have some $s' \in S$ such that $s' > \lambda - \epsilon$. (Otherwise, every $s \in S$ is smaller than $\lambda - \epsilon$, so $\lambda - \epsilon$ would be an upper bound of $S$ strictly smaller than $\lambda$, which was by definition the smallest upper bound. This would be absurd.)

But this means that for some $s' \in S$, $s' > \lambda - \alpha$ (since we put $\alpha > 0$). Since $s' \in S$ if, and only if, $s'$ is an integer multiple of $\alpha$, this says

$$ s' = n\alpha > \lambda - \alpha $$

But then we get

$$ (n+1)\alpha > \lambda $$

so that some member of $S$ is greater than $\lambda$. This is absurd; so we conclude that there are no counterexamples to the theorem in $\mathbb{R}$.

Q.E.D.

Now, the result you want is an immediate corollary.

Theorem: There is no $\gamma \in \mathbb{R}$ such that $\delta < \gamma$ for all $\delta \in \mathbb{R}$.

Proof: Suppose there were a counterexample $\gamma_0$. Pick any real $\delta_0$. By the Archimedean property,

$$\exists n: n\delta_0 > \gamma_0.$$

This directly contradicts the hypothesis, since $n\delta_0$ is a real greater than $\gamma_0$.

Q.E.D.


Categorical meaning of the action of a group on itself by conjugation For a group $G$ there are always two actions of $G$ on itself: $g*h=gh$ and $g\star h=ghg^{-1}$. As objects in the category $G$-$Set$, the former is the free $G$-set on a singleton set. When $G$ is considered as a one-object category, this same $G$-action is also the unique representable functor on $G$. Is there any similar way to understand the action by conjugation in categorical terms?

One way to see this rather abstractly is as the diagonal of the Yoneda bimodule $G\times G^{\mathrm{op}}\to \mathbf{Set}$. That is, compose the $G$-biset $(g,h)*k=gkh$ with the diagonal homomorphism/functor $G\to G\times G^{\mathrm{op}}$, $g\mapsto (g,g^{-1})$. Under this formulation we find a similar conjugation action of any groupoid, or even any dagger-category, on itself, or generalizing in a different direction, on any bi-$G$-set.

The question was raised in the comments of constructing this action as the internal hom of $G$ in $G-\mathbf{Set}$. This is trickier than it first appears, however. Let's calculate $G^G$ from first principles. We have isomorphisms of underlying sets: $G^G\cong G-\mathbf{Set}(G,G^G)\cong G-\mathbf{Set}(G\times G,G)\cong \mathbf{Set}(G,G)$. So the internal hom is actually the set of all functions $G\to G$, with the conjugation action, $f:G\to G\mapsto g*f, g*f(g_1)=gf(g^{-1}g_1)$. The odd thing is that we can't construct the conjugation action this way. Indeed, the equivariant maps between any two $G$-sets are precisely the fixed points of the action!  Or perhaps this isn't odd-we certainly don't want to start claiming we have conjugation actions on arbitrary $G$-sets. One way to make something closer to this work is to transfer the conjugation action along the natural isomorphism between $G-\mathbf{Set}(G,G)$ and $G^{\mathrm{op}}-\mathbf{Set}(G,G)$, which more or less brings us back to the suggestion of the previous paragraph.

The way Galois came up with the operation of conjugation is that he asked himself how does a group act on its subgroups. The only subgroup closed under the actions by multiplication (be it on the left or on the right) is the whole group. The nontrivial equivariant subobjects of the group acting on itself are precisely the normal subgroups, acted on by conjugation.


Mean & Standard deviation given probability of range in normal distrubition On SAT,scores range from 2000 to 2400, with two thirds of the scores falling in the range of 2200 to 2300. If we further assume that test scores are normally distributed in this range from 2000 to 2400, determine the mean and standard deviation.

I believe there is a mistake in the statement of your problem.
By symmetry, and for things to work out as intended '2200' has to be '2100'.

The Empirical Rule says that 68% of observations fall within one standard deviation of the mean. What you request can be done
only approximately, so we will use 68% instead of $2/3.$
Then the population mean needs to be halfway between 2100 and 2300; $\mu = 2200.$ Also, that implies that $\sigma = 100.$

Then using the CDF function pnorm in R, we can find the probability of a score less than 2000 and greater than 4000
for a test with scores distributed $\mathsf{Norm}(\mu=2200, \sigma=100).$ About 2.27% of students will have scores in the
'tails' of the distribution.

pnorm(2000, 2200, 100)
[1] 0.02275013
1 - pnorm(2400, 2200, 100)
[1] 0.02275013

The figure below shows the normal distribution of SAT scores. Vertical red lines mark values in the discussion above.

No matter how you fit a normal distribution to this data, it will always put positive probability outside of 2000 to 2400 range. So it is not possible to fix the range to a bounded set and assume that the underlying distribution is normal.
However, we can approximate a normal distribution assuming that the probability of scores being less than 2000 is equal to the probability of scores being above 2400.

Now, let $\mu$ denote the mean and $\sigma^2$ denote the variance. Then,
$$\Phi(\frac{2300-\mu}{\sigma})-\Phi(\frac{2200-\mu}{\sigma})=\frac{2}{3}$$
 and
$$1-\Phi(\frac{2400-\mu}{\sigma})=\Phi(\frac{2000-\mu}{\sigma})$$

where $\Phi(.)$ is the CDF of standard normal. Solving these two equations give $\mu
$ and $\sigma$.


If $3^{33}+3^{33}+3^{33}=3^{x}$. Solve for $x$. If $3^{33}+3^{33}+3^{33}=3^{x}$. Solve for $x$.

So we have:

$$3^{33}+3^{33}+3^{33}=3^{x}$$

I added the left side and obtained:

$3(3^{33})=3^{x}$

The problem I have is that extra $3$. If not, I could have said $x=33$. Any hints in how to proceed with this problem?


x is divided by 3 and 4 implies its also divided by 6 I want to show that if $[3$ divides $x$ and $4$ divides $x]$ then $[6$ divides $x]$.

I guess my starting point is something like
$x=3n$
$x=4m$

But how do i show that there is also a $c$ with
$x=6c$?

Is this the correct approach?

You can do it like this, which is generalisable:

$x=3n$ implies $4x=12n$

$x=4m$ implies $3x=12m$

Subtract to obtain $x=12(n-m)$

If $p$ and $q$ are both factors of $x$ we can write:

$x=pm=qn$

Now suppose that $p$ and $q$ are coprime. There is a theorem that we can find integers $a, b$ such that $ap+bq=1$.

We multiply this by $x$ to obtain $apx+bqx=x$ and we substitute carefully for $x$ on the left-hand side so that $$apqn+bpqm=pq(an+bm)=x$$And we see that $x$ is divisible by $pq$. You don't need to be able to prove the theorem to use it in a particular case - it tells you that if you look carefully to will find the $a$ and $b$ you need.

Since $3$ and $4$ divide $x$, $x$ must therefore equal $4 * 3 * n$ = $2*2*3*n$ for some $n$ (since 3 and 4 are relatively prime). Taking a new integer $m = 2n$, it is easy to see that $x = 2*3*m = 6*m$ for the integer $m$. Therefore, 6 divides $x$.


Chess board probability problem. Three random squares are chosen from a regular chess board. Find the probability that they form the letter 'L'.
I cannot think about a general way to go about these type of questions. Need hints or solutions.

Note: I'm assuming that an "L" is three squares that fall in a $2\times 2$ box. If you want the sides of the L to be able to have any size, please clarify.

Why don't you try counting the number of $L$s that are possible? We can count the number of "successes" (the number of times three squares form an $L$) and divide that by the number of ways to choose three squares.

If we fix a direction on the chessboard as "up", then there are four orientations our $L$ can take. By symmetry, there are the same number of ways to make each $L$, so let's count the ones that are oriented the way the letter $L$ usually is. If we think of an $L$ as being a $2\times 2$ box with one square removed, we are looking at the case where the upper right square is the one removed.

The location of the $L$ is uniquely determined by the location of the lower left square. This square can be placed anywhere except on the right-hand column or the top row, as that would cause the $L$ to not fit on the chessboard. Thus there are $7\cdot7=49$ ways to place the $L$ in the orientation we have chosen. There are four orientations, so the total number of $L$s is $4\cdot49=14^2=196$

You don't specify how the squares are chosen, but if we assume you want to choose three squares uniformly at random, there are ${64\choose 3}$ ways to do that, since there are $64$ squares on a chessboard. Since I'm assuming every choice is equally likely, the ratio of these two numbers is the answer: $\frac{7}{1488}\approx 0.0047$ or a $0.47\%$ chance.

There is a 1.92 % chance of making an "L" without space in between. If you mean that they can form an "L" when they are far from each  other is a different story.

POSSIBLE Solution: 64/100=0.64    0.64x3=1.92,
And this was in percentage form. If you need the answer not in percentage and in something else i can`t help you right NOW.


Why do these "equal" logarithms give different answers This came across a discussion amongst Algebra 2 teachers at my school.
We know $a\log x= \log x^a$
Say
$2\log x=5$
$\log x^2 =5$
When $\log x=\log_{10} x$
Solving the first equation yields $x=10^{5/2}$ while the second equation yields $x=\pm\sqrt{10^5}$.
Which solution is correct or does it depend on the equation?

The equations "$2\log x = 5$" and "$\log x^2=5$" are not equivalent.

The reason is that the first equation implies that $x>0$ while the second does not.

The correct way to move from the first to the second is to conjoin the condition $x>0$. So instead, one can say that "$2\log x = 5$" and "$\log x^2 = 5 \textrm{ where }x>0$" are equivalent.

Both equations are different and have different solutions. It matters whether or not the $x$ is squared initially. That's a subtle point. Any change to an equation makes it a different equation, technically.


Does $x,y,z>0$ and $x+y+z=1$ imply $\left(1+\frac 1x\right)\left(1+\frac 1y \right)\left(1+\frac 1z \right)\ge 64$? If $x,y,z$ are positive real numbers such that $x+y+z=1$ then is it true that

$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$ ?

By using Jensen's inequality with the convex function $f(x) = \log(1+\frac1x)$, we get
$$f(x)+f(y)+f(y) \ge 3f\left(\frac{x+y+z}3\right) = 3\log4$$
which on exponentiation gives the inequality.

Use the first equation to rewrite $z$in terms of $x$ and $y$, and then consider
Consider the function $\mathbb R^2\to\mathbb R$:
$$
f(x,y)= \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{\text{your expression for $z$}}\right)
$$
you are then looking for it's minimum which should be findable by elementary analytic methods.


What's wrong with me? Over the past few days I have been pondering about this: I enjoy technical things (like programming and stuff) and try to find the patterns and algorithms in everything. My life is number oriented. I'll spend all day working on a programmatic problem. I'll spend however much time is needed to think of an elegant/efficient solution. Yet I have am not as good with high school Algebra (I am sixteen)... I don't get it. How can I love numbers but not be very good at math? I think I slacked off in my earlier years in math classes because I found them to be boring. I feel like I am behind a wall. On the other side is math. I need to lower down this wall, but at the moment I am unable to identify it or its cause. I feel like I love math, but am no good at it or have been taught poorly. I understand that I can get no where in Computer Science if I am bad at math. I find some aspects of math to be boring... perhaps teachers don't do well at explaining it in the US (I've researched). Since I assume there are some great mathematicians here, I would like to request some advice.

Thanks.

In recent years teachers in the USA have been under heavy pressure to teach only what can be tested on standardized tests.  That has been corrupting the education system to the point where teachers orchestrate widespread cheating on standardized tests to get federal funding for the school.

There's also the fact that high schools are required to teach math to everyone, including those who are not interested.  That results in courses of a kind that almost everyone can pass, thereby holding back those who are interested and could do well in the topic.

I would recommend two books by C. Stanley Ogilvy to high-school students interested in mathematics: Excursions in Number Theory and Excursions in Geometry.  Those are addressed to students who don't yet know much but are intelligent and want to understand the subject, as opposed to wanting to pass standardized tests in it.

PS: People's interests and abilities differ greatly.  Failing to match some standard ideal prescribed by a system that suffers from the sort of corruptions I've mentioned above doesn't mean something is wrong with you.

If you truly enjoy numbers and programming, yet you're having trouble in a high school math course, it is a safe bet that there is nothing wrong with you. The problem is that your math teachers for the past 4 years (at least) have almost certainly been blithering idiots. This is unfortunately quite common (especially in math), and the only solution is to learn it on your own, as our supposed educational system has failed you.

The book that really got me interested, when I was slightly younger than you are now, is Douglas Downing's Algebra the Easy Way. It's pretty silly at first glance, but it's got a lot of great material and it's extremely accessible. And it taught me some things I never learned elsewhere, even in college. Downing's other two books (Trigonometry and Calculus) are supposed to be excellent as well, but I have no personal experience.

Whatever book or text or whatever you get to help you, remember that it (probably) will not help just to read the thing cover to cover. Any decent book will come with practice exercises, maybe even a CD full of tests, trainers, and such. You actually have to do these, or you're just fibbing to yourself about trying to learn. They should be marked by difficulty; start with the easy ones to make sure you have the basic concepts under control, then work up to the tough ones. At least one author I know of likes to reserve his highest difficulty ranking for questions that are actually open research problems; don't try to do those! But you shouldn't find that in an introductory text.

The most important thing is not to get discouraged and assume you're just plain incompetent, because it's not true. I have never met anyone who was simply unable to understand a high school level of mathematics, just people who either weren't trying or who had never had a competent teacher. It seems like you're trying, so I suspect the latter cause.


School Admissions Lottery An elementary school admits students on a basis of a random lottery.  There are 50 slots available, and 100 families (students) have applied.  Names will be drawn from a hat.  Seems pretty straightforward that an individual student has a 1 in 2 chance of getting a slot or 50%.  However, the school has a policy that if twins apply, BOTH names are put in the hat, and if and EITHER twin is picked then BOTH twins are accepted.  Although each twin has the a same probability of having his/her name picked from the hat as anyone else, intuitively families with twins seem to have an enormous advantage, but I can't figure how to quantify it. This might be a simple problem, and I would  think it should be easy to find on the  internet, but I'm not getting anywhere with it.

We make a simplified model. Among the $100$ applicants, there is one pair of twins, named Alpha and Beta (the parents are mathematicians, and Alpha was born first).

The $100$ names are thrown in a hat, and $50$ names are drawn.  If one of Alpha or Beta is drawn, but not the other, then one of the $49$ non-twin "winners" is bounced.

What is the probability that neither Alpha's  name nor Beta's  is chosen? There are $\binom{100}{50}$ equally likely ways to choose $50$ names. There are $\binom{98}{50}$ ways to choose $50$ names none of which is Alpha or Beta. So the probability the twins strike out is
$$\frac{\binom{98}{50}}{\binom{100}{50}}.$$
This simplifies to $\frac{49}{198}$, close to $1/4$.

More informally, each name has probability $\frac{1}{2}$ of not being drawn. But the probability neither name is drawn is a little bit smaller than $\left(\frac{1}{2}\right)^2$, because of (mild) dependencies.

So Alpha has probability about $3/4$ of being admitted to the school.

Remark: The addition of a small  number of additional twin pairs to the applicant mix does not change the analysis at all: Alpha, and any member of a pair of twins, still has probability close to $3/4$ of getting in, as long as the kids bounced because of application of the twin rule are non-twins.

I think that symmetry can be used to get an answer here.

Suppose you do the draw as described, settle the edge cases, so that you have divided the 100 students into two lists of fifty names each, with the two twins in the same list.

Then you flip a (fair) coin to see which list is the rejects.  Admit the others.

EDIT:

This solution(?) assumes a certain methodology:

Fifty names are chosen at random.

If this list satisfies the "twins together" requirement, the list is used.

If the twins are not together, the list is rejected, and a new set of fifty names is selected.


Why do we study Polish spaces? In descriptive set theory, a lot of space is devoted to properties of Polish spaces. (A Polish space is a topological space, which is separable and completely metrizable.)

I would like to know why there is so much interest in this class of spaces. Why are they useful? What are interesting applications? Which important tools do they give us? Do they bring new insights into other areas?

TL;DR: What is the motivation for studying Polish spaces?

You ask a very general question which is hard to answer completely... Here's a loose and unfinished collection of thoughts that might be used for fleshing out a more polished answer.

I suppose one main reason for the interest in Polish spaces is a combination of the simplicity of the concept and the fact that one can achieve reasonable generality and usefulness outside of the realm of descriptive set theory itself.

The definition itself is easy to understand: one barely needs more than a first course in real analysis to have the necessary background to be able to grasp the definition.
The setting is convenient: topological and measure-theoretic pathologies are kept to a minimum. Many objects of interest are covered.
Polish spaces allow for powerful results, quite sufficient for 'everyday mathematics'.

One reason certainly is that Polish spaces have reasonable closure properties, for example, they are closed under taking countable products and passing to open, closed or, more generally, $G_{\delta}$-subspaces. It was also mentioned in another answer that in many fields one never really cares about objects that aren't Polish spaces, so results about Polish spaces are usually 'good enough'.

A further reason is that Polish spaces are "well-behaved": they cannot be 'too large' or 'too intangible', so various 'annoying' pathologies are excluded. The facts that they admit an open surjection from the Baire space and that they can be embeded into the Hilbert cube can be used to reduce questions about general Polish spaces to spaces that one understands relatively well.

Especially the 'universality' of the Baire space allows for powerful, rather explicit, coding techniques, like Lusin and Suslin schemes. These codings can be exploited to give explicit constructions (thus guaranteeing measurability) of sections for surjections, or selections of points in families of sets, which is very useful in various applications where the functions obtained directly from the Axiom of Choice aren't good enough due to their potential lack of measurability.

Surely, the Borel sets of Polish spaces are as important in applications as Polish spaces themselves. In probability theory there is often an explicit or implicit assumption that the underlying measure spaces are standard Borel spaces.

Since you ask about applications, let me mention a few of those.

Outside of descriptive set theory, Polish spaces are ubiquitous: the best-understood objects are Polish: most spaces arising in geometric contexts and well-behaved things like locally compact second countable spaces and groups, manifolds, separable Banach spaces and algebras, and so on. On the other hand, they are an almost indispensable tool for handling 'tame' measurable spaces (standard Borel spaces) in probability theory and ergodic theory.

Probability theory certainly is one major field of application, Polish or standard Borel spaces allow the formulation of Kolmogorov's consistency theorem, the construction of conditional expectations and Prokhorov's theorem in useful generality. Moreover, measures on Polish spaces have decent regularity properties and one has a painless formulation if the Riesz representation theorem. Applications include the construction of sequences of random variables with prescribed distributions and the treatment of discrete Markov processes. See for example Parthasarathy's book on Probability measures on metric spaces, for an exposition of these results that doesn't require much background in probability.

People interested in probability theory, ergodic theory or functional analysis (operator theory and representation theory) probably aren't too interested in Polish spaces as such, but it crystallized over time that many of the deeper results in their domain are quite satisfactorily covered by general theorems on Polish spaces (or their underlying Borel spaces).

A number of famous mathematicians used Polish spaces in part of their work and contributed to the recognition of the concept as central to a non-negligible part of mathematics:

Blackwell argued successfully that many of the pathologies inherent in Kolmogorov's approach to probability theory can be avoided by restricting attention to standard Borel spaces.
Rokhlin accomplished similar things in his thesis on ergodic theory.
Choquet in his work on potential theory (as exposed in the standard treatises by Dellacherie-Meyer or Doob).
Mackey (and later Effros) in representation theory of locally compact groups and Banach algebras and their applications to quantum mechanics.
Zimmer used Polish spaces crucially in his approach to Margulis's theory of lattices in semisimple Lie groups ("super-rigidity").

All these mathematicians not only applied existing results from descriptive set theory, but they also contributed back to the general theory with their own insights (partition theorems, the theory of capacities, selection theorems, etc.). This brief list already shows that Polish spaces are of interest to mathematicians with otherwise rather disjoint focus.

One prominent central result is the theorem on disintegration of measures. Apart from the probabilistic applications linked on the Wikipedia page (e.g. conditional probabilities), it is also a central part of the 'Mackey machine' in representation theory (loosely speaking, the decomposition of representations into a direct integral of irreducible representations) and is used throughout ergodic theory in guise of 'ergodic decomposition'.

To sum this attempt at an answer up: there is a lot of interest from outside descriptive set theory which motivates further study.

The books by Kechris and Srivastava both give a good overview of numerous basic applications to other branches of mathematics while they develop the fundamental machinery.

In more recent days there has been considerable interest in Borel equivalence relations, see e.g. the work of Hjorth and various surveys by Kechris.

I haven’t seen anyone say optimal control theory. Surely, this is one of the biggest reasons we study Polish spaces.


Proof: $\tan(x)$ is surjective from $(-\pi/2,\pi/2)$ onto $\mathbb R$ To prove $\tan(x)$ defined on $]-π/2;π/2[$ is injective I take the derivative of $\tan(x)$ to get $\sec(x)^2$.

This shows that $\tan(x)$ is monotonic (strictly) increasing which implies it is injective.

However how do I show it is surjective ? That every single real number corresponds to some number in the domain of $\tan(x)$ ?

I can see that a rigorous analytical proof for the surjectivity of $f\colon(-\pi/2,\pi/2)\to\mathbb{R}$, where $f(x)=\tan(x)$, is a way far off. However, I wanted to give a picture of what @achillehui mentioned in a comment, as it is rather beautiful in my opinion:

$\color{white}{Put it in the center!!!!!!!}$

Given the picture above, we note that
$$
\tan(\theta)=\frac{x}{1}=x\qquad\text{and}\qquad\tan(-\theta)=-\frac{x}{1}=-x.
$$
Hence, we can "clearly" see that
$$
\lim_{\theta\to\pi/2^-}\tan(\theta)=\infty\qquad\text{and}\qquad\lim_{-\theta\to-\pi/2^+}\tan(-\theta)=-\infty.\qquad\approx\blacksquare
$$

Given that $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\rightarrow \mathbb{R}, f(x) = \tan(x)$.

Proof of Surjectivity

Suppose $\alpha \in \mathbb{R}$. We have to prove that $\exists \beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$  for which $f(\beta) = \alpha$.

We will use the Intermediate Value Theorem to prove that.

Intermediate Value Theorem

Let $f:[a, b] \rightarrow \mathbb{R}$ be continuous, and assume that $s$ lies between $f(a)$ and $f(b)$. Then $\exists$ at least one $c \in [a, b]$ such that $f(c) = s$.

I omit the proof of the fact that $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\rightarrow \mathbb{R}, f(x) = \tan(x)$ is continuous.

Let $-\infty < \alpha < \infty$, where $\infty = \lim_{x \to {\frac{\pi}{2}}^-} f(x)$ and, $-\infty = \lim_{x \to {-\frac{\pi}{2}}^+} f(x)$. Note that $\lim_{x \to {\pm \frac{\pi}{2}}} f(x)$ do not exists.

Then by the Intermediate Value Theorem, $\exists$ at least one $\beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $f(\beta) = \alpha$.

It follows that $f$ is surjective.

This answer is inspired by the comments above.


Solution of this system of linear equations What are the solutions of the following system of linear equations for $a\in\mathbb{R} $?

$a^2x+2ay=3a$

$2ax+a^2y=-3a$

Hint

If $a=0$, the system becomes $0=0$ and admits an infinite amount of solutions. Assuming $a \ne 0$ we can divide both sides by $a$ to get
$$
\begin{pmatrix}a & 2 \\ 2 & a \end{pmatrix}
\begin{pmatrix}x \\ y \end{pmatrix}
= \begin{pmatrix}3 \\ -3 \end{pmatrix}
$$
so now apply Gaussian elimination. Can you finish the problem? You should see immediately that when $a=2$, the equations on the LHS become the same...

If $a=0$ infinitely many solutions. If $a=-2$, the same thing happening. If $a \ne 0$ and $a \ne -2$, we can get $x=-y$ and then we can solve easily.


Radius of convergence of $ \sum_{k=1}^{\infty}(\frac{z^4}{4} - \frac{\pi}{7})^k$ I have the complex power series $ \sum_{k=1}^{\infty}(\frac{z^4}{4} - \frac{\pi}{7})^k$.

Through algebraic manipulation I obtain $ \sum_{k=1}^{\infty}(\frac{1}{4})^k(z^4 - \frac{4}{7}\pi)^k$. I now argue that this is a power series around $\frac{4}{7}\pi$ with radius of convergence R = 4, using the euler root test, asserting all the while that the power 4 in the argument z doesn't affect those two quantities. However, I'm not sure how to prove or even heuristically show that last bit. In fact, I'm not even sure I'm right in asserting that the power doesn't matter, it's really more of a hunch. Does anyone know how to handle this scenario?

Hint: This is a geometric series
  \begin{align*}
\sum_{k=1}^\infty\left(\frac{z^4}{4}-\frac{\pi}{7}\right)^k
=\frac{1}{1-\left(\frac{z^4}{\pi}-\frac{\pi}{7}\right)}-1
\end{align*}

  converging for $\left|\frac{z^4}{4}-\frac{\pi}{7}\right|<1$

Put $t=z^4$ so the series is $\sum_{k=0}^\infty \dfrac{1}{4^k}(t-t_0)^k$ where $t_0=\dfrac{4\pi}{7}$

Then the radius of convergence $R=\lim \sup \dfrac{a_k}{a_{k+1}}=4$ where $a_k=\dfrac{1}{4^k}$

Hence the series converges $\forall t$ such that $|t|<4\implies |z|^4<4\implies |z|<\sqrt 2$

Note: the radius of convergence of a power series remains unaltered if you change the point about which the series is given.So here $t_0$ has no role to play.It is needed only when you need the circle of convergence which is not needed here

So the radius of convergence is $\sqrt 2$


Propositional logic: how to show if tautology using proof by contradiction I have been shown how to prove a propositional formula is a tautology by using contradiction. However, I'm not fully understanding the logic.

Could someone help me understand the logic behind the below steps.

$$((P \to Q) \land (R \to S) \land (\lnot Q \lor \lnot S)) \to (\lnot P \lor \lnot R) $$

We assume the right hand side of the implication is false. Which mean P = 1, and R = 1
$$((P \to Q) \land (R \to S) \land (\lnot Q \lor \lnot S)) \to (P \land R) $$

We now see if the left hand side is true or false.
$$((1 \to Q) \land (1 \to S) \land (\lnot Q \lor \lnot S))$$

$$ = Q \land S \land (\lnot Q \lor \lnot S)$$

This can't be true since if $Q = 1, \lnot Q = 0,$ if $S = 1, \lnot S = 0$.

All of the above means that the main formula is a tautology. But why?

Wait, is it because, for the formula as whole to be false, the premise (Left side) has to be true and the consequence(right side) has to be false. However, making the consequence false leads to the premise also being false, in which case the implication formula is true.

So there is no state in which the formula is false. A.K.A tautology.

Wait, is it because, for the formula as whole to be false, the premise (Left side) has to be true and the consequence(right side) has to be false. However, making the consequence false leads to the premise also being false, in which case the implication formula is true.

Yes this is the correct reason. However, the way you expressed the earlier part of your question is not correct.

You want to prove:

  $((P \to Q) \land (R \to S) \land (\lnot Q \lor \lnot S)) \to (\lnot P \lor \lnot R)$

To do so you consider the situation in which its negation is true:

  $((P \to Q) \land (R \to S) \land (\lnot Q \lor \lnot S)) \land (P \land R)$.

Since in this situation you deduce that $Q$ is true and $S$ is true, and hence $(\lnot Q \lor \lnot S)$ is false, you have reached a contradiction, and therefore this situation does not occur.

The only remaining possibility is that the original sentence is (always) true.

I cannot comment yet. So I write an answer.

The left side must always be true. When you assume that $\neg p \vee\neg q$ is false you end up in a situation where left side is false.

If $p\wedge q$ is true, then $p$ and $q$ must be true. Now, the implications $p\rightarrow q$ and $r\rightarrow s$ can be eather true or false, but also $\neg q \vee \neg s$ can be true or false. Both lead into contradiction.


Evaluating the limit of the given integral Evaluate the following:
  $\displaystyle\lim_{n\to\infty} \displaystyle\int_0^\infty \dfrac{n\sin x}{1+n^2x^2}\ dx$

Attempt: My idea to use the Dominated Convergence Theorem. It can easily be shown that $f_n(x)=\dfrac{n\sin x}{1+n^2x^2}$ converges pointwise to $0$ on $\left(0,\infty\right)$ but I'm having difficulties on trying to get an integrable function $g$ that would satisfy $|f_n|\leq g$. I tried using the inequalities $|\sin t|\leq t$ for small $t$ and of course the obvious one which is $|\sin t|\leq 1$ but the function I get is not integrable. Any help would be great.

Hint: Let $u=nx$, then your integral becomes
\begin{align}
\int^\infty_0 \frac{\sin\frac{u}{n}}{1+u^2}\ du
\end{align}

This is not a complete answer:

The integral from Jacky Chong can be evaluated with Mathematica:
$$\int_0^{\infty } \frac{\text{Sin}\left[\frac{u}{n}\right]}{1+u^2} \
\, du=\frac{\sqrt{\pi } \text{MeijerG}\left[\{\{0\},\{\}\},\left\{\{0,0\},\left\{-\frac{1}{2}\right\}\right\},\frac{1}{4n^2}\right]}{4 n}$$
Therefore the solution can be expressed in form of a MeijerG-Function and can  be further analyzed by an asymptotic series expansion. This leads to the same result as from Claude Leibovici.


How to determine the periods of a periodic function? I am aware of the other similar questions but was not able to figure out what I want to know from those question thus posting it here.
Given a periodic function $f(x)=sin(x)$,

Why is the period of above function becomes $2\pi$?
for function $sin(nx)$ , I am being told to use this formula $2\pi/n$ to get the period. But I am not able to fully understand the concept behind it.
I am using sin function for the sake of simplicity. My question is how do I quickly work out what are the periods of a certain periodic function(preferably without drawing out the graph)?
for example, $cos2t$ has periods $\pi,2\pi,3\pi,...$ (I have no idea why)
Also, there are many different types of periodic waves apart from sine and cosine. So, what would be the best way to get their periods if the only thing I am being given is the formula of the function.
For your info, I am supposed to find the smallest period so that I can work out the Fourier series of that function.

First, let's discuss what the definition of a period is for a periodic function.  A function $f$ is periodic with period $T$ means $f(t+T) = f(t)$ for all $t$.

The period of $\sin$ is $2\pi$ by definition.  (You might ask why $\sin$ is defined this way, but that question may be outside the scope of this thread.)  This means that $\sin(t+2\pi)=\sin(t)$ for all $t$.

Now that we know that the period of $\sin$ is, what is the period of $\sin(kt)$?  Let us define a function $g$ as $g(t)=\sin(kt)$.  We are asking, what is the period of $g$.  That is, what value $T_g$ satisfies $g(t+T_g)=g(t)$ for all $t$.

We know $\sin(kt+2\pi)=\sin(kt)$ for all $t$.  So what value of $T_g$ satisfies $k(t+T_g)=kt+2\pi$?  Solving for $T_g$, we see that $T_g=\frac{2\pi}{k}$.

Because the sine function is defined to be the ratio of the $y$ value to the radius in a circle of radius $r$ the values will repeat every $2\pi$.  Thus the period of $\sin{x}$ is $2\pi$.  From there it reduces to recognizing, for any function $g(x)$, $g(nx)$ will compress the graph in the $x$ direction by a factor of $1/n$.


Probability Question relating prison break I am stuck in a question regarding a prisoner trapped in a cell with 3 doors that actually has a probability associated with each door chosen(say $.5$ for door $A$, $.3$ for door $B$ and $.2$ for door $C$). The first door leads to his own cell after traveling $2$ days, whereas the second door leads to his own cell after $3$ days and the third to freedom after $1$ day.

"A prisoner is trapped in a cell containig three doors. The first door leads to a tunnel that returns him to his cell after two days of travel. The second leads to a tunnel that returns him to his cell after three days of travel. The third door leads immediately to freedom.
a)  Assuming that the prisoner will always select doors 1,2,and 3 with probability 0.5, 0.3, 0.2 what is teh expected number of days until he reaches freedom?
b)  Assuming that the prisoner is always equally likely to choose among those doors that he not used, what is the expected number of days until he reaches freedom? (In this version, for instance, if the prisoner initially tries door1, then when he returns to the cell, he will now select only from doors 2 and 3)
c)    For parts (a) and (b) find the variance of the number of days until the prisoner reaches freedom.
 "

In the problem I was able to find the $E[X]$ (Expected number of days until prisoner is free where X is # of days to be free). Where I get stuck is how to find the variances for this problem. I do not know how to find $E[X^2$ given door $1$ (or $2$ or $3$) chosen$]$. My understanding is that he does not learn from choosing the wrong door. Could anyone help me out? Thank you very much

Let's go through it for part (a) first. Let $X$ denote the number of days until this prisoner gains freedom. I think you already have $E[X|D=1], E[X|D=2], E[X|D=3]$:

$E[X|D=1] = E[X + 2]$

$E[X|D=2] = E[X + 3]$

$E[X|D=3] = E[0]$

So we have

$E[X^2|D=1] = E[(X + 2)^2] = E[X^2] + 4E[X] + 4$

$E[X^2|D=2] = E[(X + 3)^2]= E[X^2] + 6E[X] + 9$

$E[X^2|D=3] = E[0^2] = 0 $

and now you can solve it b/c you have $E[X]$ already?

Let $J$ be the number of days spent in jail. Let $D$ be the chosen door. We know that if the prisoner chooses door $1$. His
      expected time will be $2 + E[J]$ because the process restarts. If he chooses door $2$, his expected time is now $3 + E[J]$

$$
  \begin{eqnarray}
    E[J] & = & \sum _{i=1}^3 E[J | D = i] P(D = i) \\
    & = & (2 + E[J]) 0.5 + (3 + E[J]) 0.3 + (0) 0.2 \\
    & = & 1.9 + 0.8 E[J]
  \end{eqnarray}
  $$

which results in $E[J] = 9.5$

Now in order to find the variance let's use the second moment,

$$
    \begin{eqnarray}
      E[J^2] & = & \sum _{i=1}^3 E[J^2 | D = i] P(D = i) \\
    & = & (2 + E[J])^2 0.5 + (3 + E[J])^2 0.3 + (0)^2 0.2 \\
       & = & (4 + 4E[J] + E[J^2]) 0.5 + (9 + 6E[J] + E[J^2]) 0.3\\
      & = &  2 + 2E[J] + 0.5E[J^2] + 2.7 + 1.8E[J] + 0.3E[J^2]\\
      & = & 4.7 + 3.8 E[J] + 0.8 E[J^2] \\
      & = & 4.7 + 3.8 (9.5) + 0.8 E[J^2]
    \end{eqnarray}
    $$

Which means that:
$0.2 E[J^2] = 40.8 \Leftrightarrow E[J^2] = 204$.

And the Variance equals: 204 - $E[J]^2$ = 204 - 90.25 = 113.75.


Suppose that $a$ and $b$ are relatively prime natural numbers such that $ab$ is a perfect square. Show that $a$ and $b$ are each perfect squares. Is there any other way of proving this besides showing that all of their respective factors or representation have an even number of exponents?

Let $$ab =d^2$$

Let $m = \max \{a,b\}$. If $m=1$ then $a=b=1$ and we are done. Now suppose there exist $m>1$. Then among all $m$ take minimal one. There exit a prime $p$ such that $p\mid m$. But then $p\mid d$ and so $p^2\mid m$. Say $a>b$. Put $a' = a/p^2$ and $d' = d/p$, so we have $$a'b = d'^2$$
and $m' = \max \{a',b\} < m$. A contradiction.

Hint:

Consider their prime factorization knowing that the powers of the prime factors of any square number are all even.


meaning of the symbol $Z_n^*$ in discrete mathematics I was reading discrete Mathematics, and i found a symbol $$Z_n^*.$$ I don't know what it means.
The text says that the "image" with the multiplication operator is an abelian group.
can any one explain.

maybe the author is talking about the group of unit elements of $\mathbb{Z}_n$ ?

it means modulo n i.e after doing binary operation on a,b dividing the ans by n, then remainder is d final ans


Calculate the distance of an object based on its size My problem is this:

There is an object whose height is $30$cm. At some distance it looks like it is $0.5$cm high. Can I calculate the distance based on this information?

Thanks in advance.

My interpretation:

If the height is h, the distance is d, and it looks like k,
then h/d = k.

What do you mean by "looks like"?


Is a correlation matrix with positive determinant PSD? Please note: I'm not interested in the difference between positive definiteness and semi-definiteness for this question.

A correlation matrix is a symmetric positive semi-definite matrix with 1s down the diagonal and off-diagonal terms $ -1 \leq M_{ij} \leq 1$.

Since a correlation matrix must be positive semi-definite, it must have a positive (or zero) determinant, but does a positive determinant imply positive definiteness? In other words, if I have a matrix with 1s down the diagonal, off-diagonals satisfying $ -1 \leq M_{ij} \leq 1$ and positive determinant, is that enough to prove that the matrix is positive definite (and thus an acceptable correlation matrix)?

Thank you.

The answer to your question, as I now understand it, is no.  In particular, we can construct a matrix of you particular pattern with a positive determinant that fails to be positive definite.

In particular, consider the matrix

$$
M =
\pmatrix{
1&-1&-1&0&0&0\\
-1&1&-1&0&0&0\\
-1&-1&1&0&0&0\\
0&0&0&1&-1&-1\\
0&0&0&-1&1&-1\\
0&0&0&-1&-1&1\\
}
$$
which has eigenvalues $-1-1,2,2,2,2$

Slight variation to original poster's question, would the conjecture hold if the determinants of leading principal minors are non-negative? I.e.

Take a matrix with the following properties:
a) real symmetric
b) all entries between -1 and 1
c) diagonal elements are all equal to 1
d) determinants of leading principal minors are non-negative

Is there an example of such a matrix which is not positive semi-definite? I wasn't able to construct any such example yet.

This is very much related to Sylvester's criterion and already holds for positive definite matrices, whereas in the case of positive semi-definite matrices, all principal minors must have non-negative determinants. The conjecture (which seems what the original poster was after) really means that for such "nicely" structured matrices, the stronger property holds and we only have to check the leading principal minors?


Using induction to prove sum of certain number-theoretic function. For any positive integer $ n > 1$, let $P(n)$ denote the largest prime not exceeding $n$.
Let $N(n)$ denote the next prime larger than $P(n)$. (For example $P(10) = 7$ and
$N(10) = 11$, while $P(11) = 11$ and $N(11) = 13$.) If $n + 1$ is a prime number, prove
that the value of the sum

$$ \frac{1}{P(2)N(2)} + \frac{1}{P(3)N(3)} + \cdot\cdot\cdot + \frac{1}{P(n)N(n)} = \frac{n-1}{2n+2}$$

Can I use induction on two consecutive primes $k+1$ and $k+q+1$?

If so, then here is my proof:

Since the result holds true for $n=2$ and $n=4$, suppose that it holds true for $n=k$ with $k+1$ being prime. Then,

$$\sum_{2}^{k}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)}$$

Then suppose that $k+q+1$ is the next prime greater than $k+1$, then $\forall m$ such that $ k+1 \leq m < k+q+1$,

$P(m) = k+1$ and $N(m) = k+q+1$.

Therefore,

$$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}} = \sum_{2}^{k}{\frac{1}{P(n)N(n)}} + \sum_{k+1}^{k+q}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)} +\frac{q}{(k+1)(k+q+1)}$$

Which, on manupulation yields

$$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}}=\frac{k+q-1}{2(k+q+1)}$$ Hence completes the proof.

Is this correct or am I missing something?

I checked it and didn't find any mistakes. Nice identity!

I don't think you used any facts about primes anywhere, so if it's true it would be true for any sub-sequence of natural numbers.


Is every subalgebra of $F[x]$ one-dimensional? Let $F$ be a field. Let $A$ be a subalgebra of the polynomial ring $F[x]$. Does $A$ necessarily have Krull dimension $\leq 1$?

Alternative proof: WLOG let $A\neq F$, since otherwise the claim is trivial. Then, $A$ contains a nonconstant polynomial $p$. As a consequence, $p\left(X\right) = p$ is a nonzero polynomial equation with $A$-coefficients satisfied by $X=x$. Thus, $F\left[x\right]$ is an integral ring extension of $A$. Since Krull dimension is invariant under integral extensions (e. g. Theorem 3 in http://www.math.rwth-aachen.de/~Eva.Zerz/ast10/dim1.pdf ), this yields $\dim A = \dim F = 1$.

Such an algebra is a domain. If it is finitely generated, then its Krull dimension is the same as its trascendence degree, and then it is clear that it is at most that of $k[x]$.


Find least positive integer $n$ which satisfies the condition Find least positive integer $n$ which satisfies the condition

$| {0.5}^{\frac{1}{n}}-1|<10^{-3} $
$0.9^{n}<10^{-3}$

What I have tried is I have split the first inequality  took log on both side .. but I came to know it is getting complicated

for the second one we get by taking the logarithm $$n\ln\left(\frac{9}{10}\right)<\ln(10^{-3}$$ therefore $$n>\frac{10^{-3}}{\ln(0.9)}$$
we get $$n\geq 66$$
for the second one we get
$$n>\frac{\ln(0.5)}{\ln(1-10^{-3})}$$ therefore $$n\geq 693$$


permutation of 5 digit numbers divisible by 3 "The total number of possible combination of 5 digits numbers formed from the digits(0,1,2,3,4,5,6,7,8,9) which are divisible by 3?"

This was the question given to me by my mathematics teacher during out permutation and combination lessons;I was able to solve this with ease but later I was thought of modifing the problem a little bit to

"The total number of possible combination of 5 digits numbers formed from the digits(0,1,2,3,4,5,6,7,8,9) which are divisible by 3 without repetation of any digits?  eg 12345 not 33120"

I asked my teacher about the solution of the problem because i was unable to solve it with certain accuracy, he was unable to give me a satisfactory answer.

Can anyone help me in solving this problem
thank you.

Update

Among the $10$ digits there are $4$ having remainder $0$ mod $3$, and $3$ each having remainder $1$ or $2$ mod $3$. We have to select $x_0$, $x_1$, $x_2$ of these digits, such that $$x_0+x_1+x_2=5,\qquad x_1-x_2=0\quad {\rm mod}\ 3\ .\tag{1}$$

We can partition $5$ into three nonnegative parts in the following $5$ ways:
$$p_1:\  5+0+0,\quad p_2:\ 4+1+0,\quad p_3:\ 3+2+0,\quad p_4:\ 3+1+1,\quad p_5:\ 2+2+1\ .$$
The first and the second of these cannot be used. Then $p_3$ together with $(1)$ leads to the solutions $(x_0,x_1,x_2)=(2,3,0)$ and $(x_0,x_1,x_2)=(2,0,3)$. For $p_4$ we obtain $(x_0,x_1,x_2)=(3,1,1)$, and for $p_5$ we obtain $(x_0,x_1,x_2)=(1,2,2)$.

It follows that we can select the five digits in
$$2{4\choose2}{3\choose3}+{4\choose 3}{3\choose1}{3\choose1}+{4\choose 1}{3\choose2}{3\choose2}=84\tag{2}$$
ways. Multiply this by $5!=120$ to account for their arrangement, and obtain $10\,080$.

In a comment later on the OP has excluded the strings beginning with $0$. In order to account for this we first compute the admissible selections of five digits containing no $0$. To this end it suffices to replace $4$ by $3$ in $(2)$:
$$2{3\choose2}{3\choose3}+{3\choose 3}{3\choose1}{3\choose1}+{3\choose 1}{3\choose2}{3\choose2}=42\ .$$
Therefore $84-42=42$ admissible selections of five digits contain a zero, and these give rise to $42\cdot{5!\over5}=1008$  strings beginning with a $0$. Subtract these from $10\,080$, and obtain $9072$ as final result.

There are in total $9*9*8*7*6$ 5 digit numbers with distinct digits.  The probability that a number is divisible by 3 in all these cases is $\frac {1}{3}$

Hence the number of numbers with 5 distinct digits satisfying the given conditions is
$$9*9*8*7*6* \frac {1}{3}=9072$$


Difference between multi-task lasso regression and ridge regression The question is: what is the difference between multi-task lasso regression and ridge regression?
The optimization function of multi-task lasso regression is
$$
min_w \sum_{l=1}^L1/N_t\sum_{i=1}^{N_t} J^l(w,x,y) + \gamma\sum_{l=1}^L{||w^l||_2}
$$
while ridge regression is
$$
min_w \sum_{l=1}^L1/N_t J^l(w,x,y) + \gamma{||w^l||_2}
$$

which looks the same as the ridge regression. As for me, the problem of multi-task lasso regression is equivalent to solve global ridge regression. So what is the difference between these two regression methods? Both of them use L2 function.
Or does it mean that in multi-task lasso regression, the shape of W is (1,n)?

LASSO (Least Absolute Shrinkage and Selection Operator) uses $L^1$ regularization
$$\min_x f(x) + \lambda \|x\|_1.$$

See, for instance this presentation.


In how many points is tangent of $f(x)=\frac{e^x-2}{(e^x-1)^2}$ parallel with $y=x$ In how many points is tangent of $f(x)=\frac{e^x-2}{(e^x-1)^2}$ parallel with $y=x$

I found first derivative and put $$f'(x)=\frac{-e^x(e^x-3)}{(e^x-1)^3}=1$$

I can't solve it. I think that trick is only to find how many solutions are there, not finding exact ones.

Hint:

Let $y=e^x$.

Then the equation symplifies to a polynomial equation of $-y(y-3)=(y-1)^3$.

Hint2:

You don't need to solve this equation, you only need to show how many solutions it has.

One possibility to show how many solutions an equation has, is to use the intermediate value theorem.

https://en.wikipedia.org/wiki/Intermediate_value_theorem

you know your function is continous and look for points where your new function $g(x):=f(x)-x$ is $>0$ and $<0$; intermediate value theorem says then there must be a solution.


Why do $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$ not have the same domain of definition, but are the same? Let $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$.

It is clear that $f(x)=g(x)$ for all real numbers $x$.

My question: Why do $f$ and $g$ not have the same domain of definition, however $f(x)=g(x)$?

Note: I know only that $f$ extended by continuity at $x=1$ as a reason, I hope to know more reasons if it is possible.

The two functions are not the same because $f(x)$ is not defined for $x=1$, so its graph is a stright line that has a ''hole''.

The Domain of definition is the subset of $\mathbb{R}$ where the function is defined and this means that has values in the codomain. Division by $0$ is not defined so the function $f(x)$ cannot be defined for $x=1$. You can extend it by continuity but this means that you have a new function $f_1(x)$ that coincide with $f(x)$ for $x \ne 1$ and is defined also in this point in such a way that
$$
f_1(1)=\lim_{x\rightarrow 1} f(x)
$$
and, in this case, you really have $f_1 =g$.

I personally think it's silly to say that $ f(x) = \frac{x^{2}-1}{x+1} $ at $x = 1$ is undefined. I've been giving this some thought lately. It's like saying that $ f(x) = x$ is undefined at $x = 0$ just because it's equal to $\frac{x^{2}}{x}$ (multiply $\frac{x}{1}$ by $\frac{x}{x}$; if you don't agree that they're equal then you also don't agree that $x \times 1 = x$).
I completely agree that $\frac{1}{x}$ at $x = 0$ is undefined, but this is a different scanario. Just look at the graphs:

With $\frac{1}{x}$, the line completely swerves off into positive and negative infinity. $\frac{x^2}{x}$ looks completely identical to $f(x) = x$. With $\frac{x^2}{x}$ you can approach $0$ from both directions and both limits will be $0$. The notion of discontinuity doesn't seem particularly useful in this case.
Just my opinion, though. Clearly $\frac{x^2}{0}$ is undefined, which definitely indicates something strange about notation.


If $A$ is a real matrix, then $det(AA^T)\geq0$, where $A$ is $m\times n$ matrix without using eigenvalues? I searched the http://math.stackexchange.com but most of the answer for this question use eigenvalues etc, but I need more elementary ways to show that why $det(AA^T)\geq0$ for any real matrix, and $A^T$ is the transpose of $A$. I could easily prove that when $m=n$. I need to show that when $m<n$ or $m>n$.

I guess also when $m>n$, then $det(AA^T)=0$. Is it right? If yes, how?

Thanks!

Without eigenvalues:

If $A^T$ has rank deficit, so does $AA^T$. Thus, in that case we have $\det(AA^T) = 0$. Now, assume $m \le n$ and rank of $A$ equals $m$. Let
$$ A^T = Q \begin{bmatrix} R \\ 0 \end{bmatrix} $$
be a QR decomposition of $A^T$ (for example given by gram schmidt), that is, $Q$ is square and orthongal and $R$ is invertible and upper triangular. Then, weh have
$$ \det(AA^T) = \det(R^T R) = \det(R)^2 > 0. $$

Why without using eigenvalues? For $A\in \mathbb{C}^{n\times m}$, $AA^H$ is a Hermitian $n\times n$ matrix. Thus, its eigenvectors form an orthonormal basis for $\mathbb{C}^n$ and their corresponding eigenvalues are real. Let $(\lambda,x)$ be an eigen-pair of $AA^H$, then
$$AA^Hx=\lambda x \Longrightarrow 0\leq\|Ax\|_2^2=\lambda$$
where last equation by taking inner product with $x$. Since $\lambda$ was arbitrary eigenvalue, det$(A^HA)\geq 0$.


Is there a function such that f(x)= the decimal part of x ( sister function of integer part function). I'm looking for a little formula that would allow me to convert the degree mesure of an angle into its radian mesure.

In case the degree mesure is greater than 360° the complete 360° turns contained in the angle do not count.

So the idea is as follows: (1) first dividing the degree mesure by 360, and keeping only the decimal part of the quotient.

Then, (2) multiplying the decimal part by 360 and (3) multiplying the result by pi/180 in order to get the redian mesure.

Is there a known function that allows me to keep only the decimal part of a number?

I mean, sure this function exists, but does it have a name? Is it possible to find it on some calculators?

Yes. The name for such a function is the fractional part function, and the fractional part of $x$ is usually written as $\{ x \}$. So in this case, the formula would be $\{ \frac{\pi x}{180} \}$.


A closed ball in a metric space is a closed set Prove that a closed ball in a metric space is a closed set

My attempt: Suppose $D(x_0, r)$ is a closed ball. We show that $X \setminus D $ is open. In other words, we need to find an open ball contained in $X \setminus D$.

Pick $$t \in X-D \implies d(t,x_0) > r \implies d(t,x_0) - r > 0 $$ Let $B(y, r_1)$ be an open ball, and pick $z \in B(y,r_1)$. Then, we must have $d(y,z) < r_1 $. We need to choose $r_1$ so that $d(z,x_0) > r$. Notice by the triangle inequality

$$ d(x_0,t) \leq d(x_0,z) + d(z,t) \implies d(z,x_0) \geq d(x_0,t) - d(z,t) > d(x_0,t) - r_1.$$

Notice, if we pick $r_1 = d(t,x_0)-r$ then we are done.

Is this correct?

Your proof consists of some correct steps done in the wrong order, which makes it something other than a valid proof. It looks more like scratchwork done in preparation for a proof.  I  rewrite it below, with some of the more important additions in bold. I will also change $t$ to $y$ throughout; when you wrote "$y$" you probably meant the same thing as "$t$".

Suppose $D(x_0, r)$ is a closed ball. We show that $X\setminus D(x_0,r) $ is open. In other words, for every point $y\in X\setminus D(x_0,r)$ we need to find an open ball contained in $X \setminus D$ with center $y$.
Since $y \in X\setminus D(x_0,r)$, it follows that $d(y,x_0) > r$, so $d(y,x_0) - r > 0 $. Let $r_1 = d(y,x_0)-r$.
I claim that the open ball $B(y, r_1)$ is contained in $X\setminus D(x_0,r)$. To prove this, consider any $z \in B(y,r_1)$.  Notice by the triangle inequality
$$ d(x_0,y) \leq d(x_0,z) + d(z,y) \implies d(z,x_0) \geq d(x_0,y) - d(z,y) > d(x_0,y) - r_1 =r.$$
This shows $z\in X\setminus D(x_0,r)$, which completes the proof.

No I do not think this is correct. The idea seems correct, but the execution was poor. You should specify that $y\in X\backslash D$. I am also not sure how you justify your last inequality.  If $t$ is arbitrary in $X\backslash D$ we cannot conclude $d(z,t)<r_1$ and $-d(z,t)>-r_1.$

Here is how I would solve the problem:

Let $X$ be a metric space, $p\in X$, $r>0$.

Let $A=\{q\in X :d(p,q)\leq r\}$.  Let $\{q_k\}\in A$ with $d(q_k,q)\rightarrow 0$.

We want to show $q\in A$.

By the triangle inequality we have  $$d(q,q_k)+d(q_k,p)\ge d(q,p)$$. $\Rightarrow$ $$d(q,q_k)+r\ge d(q,p)$$ (Since $\{q_k\}\subseteq A)$

Now take the limit as $k\rightarrow \infty$ of both sides and we have $$0+r\ge d(p,q)$$
$$\Rightarrow q\in A$$ $\Rightarrow A$ is closed $$QED$$


If $A \in C^{nxn}$ , $A \ge 0 $ and A is sing., there exists a sequence of matrices $C_k$, that $C_k \ge 0$,$|C_k| = 1$ and trace $AC_k \le 1/k$ Question: Show that if $A \in C^{n \times n}$ , $ A \ge 0 $ and A is singular, then there exists a sequence of matrices $C_k$, $k = 1,2,...$ such that $C_k \ge 0$, det $C_k = 1$ and trace $AC_k \le 1/k$

My approach: I have no idea how to even approach this exercise, any help is highly appreciated.

Hint. If $A=B\oplus0_{(n-r)\times(n-r)}$ where $B>0$ and $r=\operatorname{rank}(A)$, consider $C=\left(\varepsilon I_r\right)\oplus\left(\varepsilon^{-r/(n-r)}I_{n-r}\right)$ for some positive scalar $\varepsilon$. What are $\det(C)$ and $\operatorname{trace}(AC)$ in this case?