What's the best way to optimize this energy function, and is it convex? I have an energy function

$E({\bf y})=||\,g({\bf Ay+c})-{\bf d}\,||^2_2 + ||\,{\bf y-e}\,||^2_2 + \alpha\,|{\bf y}|_1$

I need to minimize this with respect to $\bf y$, all other variables being constant.

$g(\bf u)$ is a non-linear scalar function that acts on the elements of $\bf u$ independently. Typically this function is also monotonic and convex. In the simplest case, $g(\bf u) = u$, but other choices include $g({\bf u})=\log(1+\exp(\bf u))$.

My first question is whether this is a convex problem. I think that it is convex if $g(\bf u) = u$, but guessing it also might be convex if this function is nonlinear but has convex form, but I'm really not sure.

Then I would please like a suggestion as the quickest way to optimize for $\bf y$. I don't really know how to deal with the L1 term. The dimensionality of $\bf y$ is of the order 10,000 and $\bf A$ and $\bf B$ are typically sparse.

Turns out this is not convex - see comments.
Any suggestions on the best non-convex approach to use?

If $g(u)$ is convex, then you problem is convex as well: [EDIT by mcg: the linear/affine case is probably the only useful one for which this is convex.]

Other cases can be convex as well. There are several simple rules to be used to detect convexity. You may want to take a look to a classic book:

http://stanford.edu/~boyd/cvxbook/

The $|\cdot|_1$ is a simple regularization term that you an easily convert in a set of linear constraints: $\min |x|_1$ is equivalent to

$$
\min \sum t_i\\
t_i\geq |x_i| \quad i=1,\ldots,n
$$

and hence

$$
\min \sum t_i\\
t_i\geq x_i \quad i=1,\ldots,n
t_i\geq -x_i \quad i=1,\ldots,n
$$

More details are in the book I mentioned.

Than you can use any QP solver. If work on MATLAB and you do not want to work on the formulation, I would suggest you to use YALMIP or CVX.

You can find


Mathematical function to convert two numbers into one? Is there a mathematical function that converts two numbers into one so that the two numbers can always be extracted again?

Consider the map $$(x, y)\to 2^x3^y$$
For example, $$(\color{blue}{7}, \color{green}{3})\mapsto 2^\color{blue}{7}3^\color{green}{3} = 128\cdot27 = 3456.$$

To extract the original two numbers, divide 3456 by 2 and count how many times you can do this before you are left with an odd number: $$3456 \to 1728\to 864\to 432\to 216\to108\to 54\to 27$$  That's $\color{blue}{7}$ times.  Then divide 27 by 3 until you are left with 1: $$27\to9\to3\to 1$$ That's $\color{green}{3}$ times, so the original numbers were $\color{blue}{7}$ and $\color{green}{3}$.

Unlike the other answer in this thread, this works for any whole numbers, of unlimited size. But the process for extracting the original two numbers by this method is slow.

It is possible if you omit the condition "always". For example, for natural numbers $n,x,y \leq 10^n$  consider  the function $f(x,y)=10^{2n}x+y.$ It this case we are able  to extract $x$ and $y$  from $f(x,y).$


Let $G$ a group, with ... Show that $G$ is a cyclic group. Let $G$ be a group with $|G|=455$. Show that $G$ is a cyclic group.

It is well-known that if $n$ is a natural number, there is only one group of order n if and only if $\gcd(n,\varphi(n))=1$. Here $\varphi$ is the Euler totient function. For $n=455$ this applies. If there is only one group of a particular order it must necessarily be cyclic.

Notice $455 = 13*7*5$ and we know $13$, $7$ and $5$ are prime. Now use Lagrange theorem to show that G is cyclic. This is a hint.


Prove $a ≤ x ≤ b ⇒ x ≤ a +b ∀a,b ∊ R^+$ It seems too obvious to be proved algebraically.
What i thought was:
We know that the sum of two positive numbers is a positive number so:
$a + b > b$ and $a + b > a$
Using the hypotesis of $a > x$ and $b > x$ then
$a + b > b > x$ and $a + b > a > x$
Is it correct? If you know, please tell me another way to prove it.

It seems you are on the right track, but you made one error in the final line with the $a + b > a > x$. I would solve this as:
$\forall a,b \in \mathbb{R}^+$, $b<a+b$ and $a<a+b$ (definition of addition between two positive numbers)
Given $x, a, b$ such that $a \leq x \leq b$ and $b<a+b$ then we can conclude through the transitive property $a \leq x < b$ or
$$x<a+b \ \ \  \text{and} \ \ \ x \geq a$$
The only item I can't seem to prove is that $x \leq a+b$. I would say by the above proof, $x$ is strictly less than $a+b$.


Solving $e^x + x = 5$ for $x$ without using a numerical method? Canadian economist Mike Moffat asks on Twitter:

  Math nerd Q: Is there a way to solve $e^x + x = 5$ for $x$, without using a numerical method?

Write $y=5-x$. Then $ye^y=e^5$. Using the Lambert W function, this gives $x=5-W(e^5)$.

Step 1. Transform the equation $e^x+x=5$ into $e^x=5-x$

Step 2. Write each side of the new equation as function : $y_1=e^x$ and $y_2=5-x$

Step 3. Graph these two functions in Desmos Graphing Calculator (you find its application in internet)

Step 4. Find the intersection of the two graphs. It has an $x$ and $y$ value.

Step.5 The $x$-value is the solution of our equation : $x=1.307$ (approximately)


Total Curvature of 4 pi What does it mean for a surface to have a total curvature of $4\pi $?

I have seen that both the catenoid and Enneper surface are the only minimal surfaces that have this total curvature, but I don't really understand what significance this has?

Could anyone please explain this?

For calculate the finite total curvature of a complete minimal surface $\phi:M\longrightarrow\mathbb{R}^3$, by start from the topology of $\bar{M}$ and the geometry of ends.

An $\mathbf{end}$ of a completle minimal surface of finite total curvature is the image $c_i=\phi(D_i-p_i)$ of a sufficiently small punctured disk $D_i-p_i$ on $\bar{M}$ with centered at the point $p_i$.

$\mathbf{Theorem}$. If $g$ is the genus of $\bar{M}$, $M\cong{\bar{M}-\{p_1,...,p_r\}}$
and $d_1,...,d_r$ are the multiplicities of the ends $c_1,...,c_r$, and $\tau$ the total curvature

$$\tau(M)=2\pi(2-2g-r-\sum_{j=1}^rd_j)=2\pi(\chi(\bar{M})-r-\sum_{j=1}^rd_j).$$

--M=Catenoid has two ends. $r=2, d_1=d_2=1$, and $\bar{M}=\mathbb{S}^2$. Consequently

$$\tau(M)=2\pi(2-2-2)=-4\pi.$$

--The M=Enneper has one end of multiplicitiy $d_1$, and $\bar{M}=\mathbb{S}^2$, $\tau(M)=-4\pi$ Consequently $d_1=3$.

Well first of all these have total curvature $-4\pi$, not $4\pi$.

The Gauss-Bonnet theorem tells us that the total curvature of a surface is equal to $2\pi$ times the Euler characteristic of the surface. Moreover we know that the Euler characteristic of a connected surface is always an integer that is at most 2.

So in general we might want to ask: What are all connected minimal surfaces without boundary in $\mathbb{R^3}$ of with total curvature $2\pi n$? We see this is the same as asking that the Euler characteristic be n.

Well, if $n=1,2$ the answer is none, since a minimal surface has nonpositive Gaussian curvature everywhere. Now we can go down from there trying to understand the most topologically simple minimal surfaces in $\mathbb{R^3}$.

For $n=0$ it's not hard to figure out that a plane is the only connected minimal surface without boundary in $\mathbb{R^3}$ that has total curvature 0.

What you are saying is that these are the only two minimal surfaces in $\mathbb{R^3}$ with Euler characteristic $-2$. So we see this case still has a relatively nice answer.

But as we decrease $n$ even further the story becomes increasingly complicated, and a general answer is unknown.


What Exactly Are Quotient/Factor Groups and Rings? I'm having a lot of trouble wrapping my head around this one. The mathematical definition is that $G'$ is a factor group of $G$ if $G'$ = $G$/$N$, where $N$ is some normal subgroup of $G$. $R'$ is a factor ring of $R$ if $R'$  = $R$/$I$, where $I$ is an ideal.

My question is, just "what the hell" are they exactly? Are we taking all elements of $N$ (and similarly $I$) and taking them out of $G$ and $R$? If so, why? Why is that important? If not, what exactly are we doing when we create a factor group/ring? I could really use some help visualizing these. Thank you.

Another important way to think about quotient groups: they are precisely the images of group homomorphisms.

This is a consequence of the first isomorphism theorem: if $G$ and $H$ are groups and $\phi : G \to H$ is a group homomorphism, then $G/K \cong \operatorname{im}\phi$, where $K = \operatorname{ker}\phi$.

Conversely, if $N$ is any normal subgroup of $G$, then there is a natural homorphism $\phi : G \to G/N$ given by $\phi(g) = gN$. The image of this homomorphism is $G/N$, and the kernel is $N$.

Similar remarks apply to quotient rings.

The basic idea is that whenever you have a sub-group(ring), this sub-object allows you to classify the elements of the big group(ring) into different set-theoretical classes or "boxes".

How? In the easiest way you can imagine: consider a sub-group $H\subset G$ and $a,b\in G$. We say these elements are in the same box (class) iff the sets $aH=bH=\{bh:h\in H\}$. The same is done with rings, but using the commutative structure (if $I\subset R$ and $a,b\in R$, then $a\sim b$ iff $a-b\in I$).

Now, these are just set-theoretical decompositions of your original algebraic objects, but if additionally the sub-group(ring) is normal(ideal) these boxes inherit the algebraic structure from the original group(ring)!


Does there exist a vector space $V$ which is isomorphic to $\mathbb{R}^m$, but is not a topological/smooth manifold of dimension $m$? Does there exist a vector space $V$ which is isomorphic to $\mathbb{R}^m$, but is not a topological/smooth manifold of dimension $m$?

Suppose $V$ is isomorphic to $\mathbb{R}^m$.

In the category of vector spaces, $V$ is equivalent to $\mathbb{R}^m$, but $V$ may not have a topology on it, so it may not be a topological manifold, and hence it can definitely not be a smooth manifold (since all smooth manifolds are just topological manifolds endowed with a smooth structure).

If $V$ is homeomorphic to $\mathbb{R}^m$, then we can conclude in that $V$ is equivalent to $\mathbb{R}^m$ in the category of topological spaces, and $V$ must also be a topological manifold, however since every topological manifold is not a smooth manifold (see the Exotic Sphere as an example), we can't necessarily conclude that $V$ is diffeomorphic to $\mathbb{R}^m$.

So my question is the following, does an isomorphism between $V$ and $\mathbb{R}^m$ induce a homeomorphism between $V$ and $\mathbb{R}^m$, and furthermore does a homeomorphism between $V$ and $\mathbb{R}^m$ (along with the isomorphism between the two vector spaces) induce a diffeomorphism between $V$ and $\mathbb{R}^m$?

If not could someone provide me with an example of a vector space $V$ which is  isomorphic to $\mathbb{R}^m$, but not homeomorphic, and also of an example of a vector space $V$ which is homeomorphic and isomorphic to $\mathbb{R}^m$ but not diffeomorphic to $\mathbb{R}^m$?

Note: Isomorphism here means isomorphism of vector spaces.

You can give whatever topology/smooth structure to $\mathbb R^n$ as you like without affecting it's vector space structure.

Consider $X=\mathbb R^n$ with the trivial topology, i.e. only two open subsets: the empty set and $X$. Then, this is not homeomorphic to the $Y=\mathbb R^n$ with the usual Euclidean topology. Then these are not homeomorphic spaces for many reasons (eg: $X$ is compact but $Y$ is not).

Again, let $X'=\mathbb R^n$ and find a bijection with $\mathbb R$ and use this bijection to make $X'$ a $1$ dimensional manifold. Of course this cannot be homeomorphic to $Y=\mathbb R^n$ which is an $n$-dimensional manifold for $n>1$.

No. Change of base in V gives you a smooth function in $\mathbb{R}^n$. So you have a smooth structure which is diffeomorphic to $\mathbb{R}^n$


Yearly contribution to reach given stock (compound interest) I'm working on a calculator to compute the final stock given a yearly contribution (either a fixed contribution or with a linear growth). The user provides the contribution for the first year, if he wants to increase this contribution yearly (the growth value is fixed, the user only chooses if he wants it or not), the interest rate and the number of years (compounding frequency is always yearly).

Calculate the final stock given those imputs isn't difficult, I simply use the Compound Interest formula for each year from $zero$ to $t$ and accumulate the results:
$$F = Py(1+i)^t$$
Where:

$F =$ Final value

$Py =$ Contribution on year $y$ (same each year or increased by a percentage)

$i =$ Interest rate

$t =$ Remaining years

But I don't have a clue about how to calculate (or at least approximate) the other way around. I.e. calculate the needed yearly contribution to reach a desired final stock, given the interest rate and the number of years, either with fixed contributions or increasing ones.

Let g be the (constant) growth rate and  $G=1+g$ the growth factor. And let r be the interest rate and $q=1+r$. Then the future value of the growing contributions  is

$F_n=C\cdot  q^{n-1}+ C\cdot q^{n-2}\cdot G+\ldots + C\cdot q\cdot G^{n-2}+C\cdot  G^{n-1}$

$C\cdot  q^{n-1}:$ The first contribution has to be compounded n-1 times and no growth.

$C\cdot  G^{n-1}:$ The n-th contribution has grown n-1 times and no compounding.

$F_n=C\cdot \left[  q^{n-1}+  q^{n-2}\cdot G+\ldots + q\cdot G^{n-2}+  G^{n-1} \right]$

$q\cdot F_n=C\cdot \left[  q^{n}+  q^{n-1}\cdot G+\ldots + q^2\cdot G^{n-2}+ q \cdot G^{n-1} \right] \quad \quad \quad (1)$

$G\cdot F_n=C\cdot \left[ \ \ \quad  q^{n-1}\cdot G+  q^{n-2}\cdot G^2+\ldots + q\cdot G^{n-1}+  G^{n} \right] \quad (2)$

Substracting (2) from (1)

$(q-G)\cdot F_n=C\cdot (q^n-G^n)$

$\boxed{F_n=C\cdot\frac{q^n-G^n}{q-G}}$

Your formula is $\sum_{t=0}^n P_y(1+ i)^t$.  That is the same as the "geometric sum" $\sum_{t= 0}^n ar^t$ for which a formula is known:  $\frac{a(1- r^{n+1}}{1- r}$.    Here $a= P_y$ and r= 1+ I so that becomes
$F=\frac{P_y(1- (1+ i)^n)}{i}$.  So $P_y= \frac{Pi}{1- (1+i)^{n+1}}$


Proving a function is continuous for all real numbers This a homework question:

Prove  $f(x) = 2x^3 - 4x^2 + 5$ is continuous for all real numbers.

Which proof technique do I use for this? I don't really know where to start.

Hint: If $f$ and $g$ are continuous, then so are $f + g, f - g, f \cdot g, f / g$ (the later provided $g$ isn't equal to $0$).

As  $f(x)=2x3−4x2+5$ is a polynomial function, it is continuous at every real number (it is a theorem).


Does $\lim_{k\to\infty} \sum_{n=1}^{k} f(n)$ converge? I am wondering if there is a rigourous way to show the following series converges:
$$ f(x) = \begin{cases}
\dfrac{1}{x^2} & x \in \mathbb{N} \\[1ex]
g(x) & x \not\in \mathbb{N}\in \mathbb{R}
\end{cases}$$
Here, $g(x)$ could be anything, but for interest, wouldn't converge on its own.
When we do the following:
$$\lim_{k\to\infty}\sum_{n=1}^{k}f(n).$$
Do we get a limit that exists?
Intuitively we would, because summation is only defined on the natural numbers (unless we redefined it here, doing some interpolation between points). However, I am not certain this function is integrable, which would suggest the function does not converge (unless my understanding of the Integral Test is wrong, after all, it says nothing of whether the integral exists, just if it converges the sum converges and vice-versa).

To address your confusion, we note that the integral test requires that $f(x)$ must be monotone decreasing over the interval $\left[1, \infty\right)$ to be applicable.
Check this link: https://en.wikipedia.org/wiki/Integral_test_for_convergence
Hence, if we have
$$f(n) = \frac{1}{n^2}$$
for all $n \in \mathbb{N}$, and you want to extend it to $\mathbb{R}$ by setting $f(x) = g(x)$ for all $x \in \mathbb{R} \setminus \mathbb{N}$, then you have to make sure that the selection of $g(x)$ keeps $f(x)$ monotone decreasing over the interval $\left[1, \infty\right)$ so that the integral test can kick in.
If the ultimate goal of extending $f(n)$ to the domain $\mathbb{R}$ as $f(x)$ is to hopefully construct $f(x)$ so that we can use the integral test to prove that the series
$$\sum_{n = 1}^{\infty} f(n) = \lim_{k \to \infty} \sum_{n = 1}^{k} f(n)$$
converges, then in this case, the way you extend $f(n)$ to $f(x)$ matters. If you can prove that there exists at least one way to extend $f(n)$ to $f(x)$ such that $f(x)$ is monotone decreasing over the interval $\left[1, \infty\right)$, then this is sufficient to invoke the integral test to prove that
$$\sum_{n = 1}^{\infty} f(n) < \infty$$
For instance, you can trivially take $g(x) = 1/x^2$ for $x \in \mathbb{R} \setminus \mathbb{N}$ so we have $f(x) = x^2$ for all $x \in \mathbb{R}$ which is monotone decreasing.
But if you just just want to know if the limit exists, then you can do so without extending $f(n)$ to $\mathbb{R}$. I mean, you can do so without invoking the integral test. Evaluating this sum is known as the Basel problem and the result is $\pi^2/6$.
Check this link: https://en.wikipedia.org/wiki/Basel_problem

About the series,
$$\sum_{n\ge 1} f(n)=\sum_{n\ge 1}\frac{1}{n^2}$$
converges as a Riemann series $(2>1)$.
As an integral, we can say nothing. It depends on the expression of $ f(x).$
Take for example $ f(x)=1$ if $ x\notin \Bbb N$, then
$$\int_1^{+\infty}f(x)dx \;\; diverges$$
and If $$(\forall x\ge 1)\;\; f(x)=\frac{1}{x^2}\; $$
the integral $$\int_1^{+\infty}f(x)dx\;\; converges$$


How to prove a subspace? Let X be a Banach space and let M be a subset of X. Then M is itself
a Banach space (using the norm from X) if and only if M is closed subspace.

How to prove (=>) to show that M is closed subspace?

Question:If I have M is subset of X then I have M is a subspace???? If this statement is true, I can prove it.

Thanks. :-)

Suppose $M$ is closed. You have to show $M$ is complete. Let $x_n$ be a Cauchy Sequence in $M$. Since $X$ is a Banach space, $x_n$ converges to some $x\in X$. Since $M$ is closed, $x\in M$. So $M$ is complete.

Suppose $M$ is a Banach space. You have to show $M$ is closed in $X$. Since $X$ is a normed space it suffices to show that, if $x_n$ is a sequence in $M$ which converges in $X$ to some $x\in X$, then if follows that $x\in M$. So suppose $x_n$ is convergent to $x$. Then $x_n$ is a Cauchy Sequence in $X$, but since $M$ carries the same norm, in $M$ as well. Since $M$ is a Banach space, $x_n$ converges in $M$ to $x^\prime \in M$. Since $M\subset X$ with the induced topology, $x_n\rightarrow x^\prime$ in $X$. By uniqueness of the limit, $x = x^\prime$, so $M$ is closed.

To $M$ be a subspace you need that $\forall x,y,\in M $ and $\forall \alpha$ scalar

$x+y\in M$
and
$\alpha x \in M$.

But, it doesn't mean that $M$ is a Banach space. To be a Banach space one needs that $M$ be a complete space, that's why one requires that $M$ be closed.

To show that $M$ is closed you need to show that every Cauchy sequence converges to a point in $M$.


Does the Cauchy Schwarz inequality hold on the L1 and L infinity norm? So i am wondering if the Cauchy Schwarz inequality holds for all p-norms, not just when p=2, which is the euclidean space.

Thank you.

Cauchy Schwarz inequality can be generalized as follows:
\begin{equation}\label{d}
|x^\top y|\leq \|x\|\|y\|_{\star}, \forall x,y \in \mathbb{R^{n}}
\end{equation}
where $\|\|_\star$ is called the dual norm.
Since dual of $l_2$ norm is itself, using $l_2$ in above we reach to Cauchy Schwarz inequality. Since $l_1$ and $l_\infty$ are dual, we obtain:
$$|x^\top y|\leq \|x\|_1\|y\|_\infty$$

Among other things, this question is easily construed as a reasonable one, about the continuity of pairings $\langle, \rangle \colon V\times V^*\to \mathbb C$. Among other true things that can be said, it is a general fact that separately continuous pairings on Frechet spaces turn out to be jointly continuous for general reasons. ... thus... what is the real question?


Reference for Functional Analysis I am in search for Functional Analysis Books.

I need a book which have lots of elementary problems.

I mean it should force us to think more which will give a deeper insight for functional analysis.

To be precise, I need lots of problems like Is $A$ open in the space $\ell^1$ with respect to norm $\|.\|_2$. Find a counterexample to this statement and etcetera.

I mean it shouldn't ask us trivial questions. It should ask us questions which will force us to think more and more from functional analysis

The most elementary book for functional analysis I've seen is Introductory functional analysis of E.Kreyszig, really really understandable and its problems are 'basics'.

If you're new, try Simmons's book. Other than that both books of Walter Rudin and John B. Conway consists of many non-trivial questions. P.R. Halmos's "A Hilbert Space Problem Book" by springer is difficult to me atleast.


Solving Recurrences using Telescoping/Backwards Substitution Specifically,

$$T(n)=3T(n-1)+1; \quad T(1)=1.$$

I have
\begin{align*}
T(n) & = 3T(n-1)+1 \\
& = 3(3T(n-2)+1)+1 \\
& = 9T(n-2)+4 \\
& = 9(3T(n-3)+1)+4 \\
& = 27T(n-3)+13 \\
& = \cdots \\
& = (3^k)T(n-k)+(3^k - 1).
\end{align*}

Am I on the right track?  I feel like something is off, b/c it doesn't work for the base case, but I don't know where I went wrong? and is there any hard and fast rule about how far you should "telescope" backwards?

Thanks in advance

Here is a general method. You are given $T_n=A(T_{n-1})$ for some affine function $A$, defined by $A(x)=ax+b$, and you want to iterate $A$ because you know that $T_n=A^{n-1}(T_1)$ for every $n\geqslant1$. This is doable because:

Affine functions are conjugate to linear functions.
Linear functions are easy to iterate, to wit, if $L:x\mapsto cx$, then $L^n(x)=c^nx$.

Thus, the task is to find $L$ linear such that $A$ and $L$ are conjugate. Here is the only bit to remember:

  The function $A$ is conjugate to a linear function $L$ through a translation, for every $a\ne1$, and the translation should send the fixed point of $L$, which is $0$, to the fixed point of $A$.

Hence the first task is to find the fixed point $z_A$ of $A$: this solves $z_A=A(z_A)=az_A+b$, that is, $z_A=b/(1-a)$ (since $a\ne1$).

And now, behold, for every $t$, $A(z_A+t)=A(z_A)+at=z_A+at$. Iterating, one sees that, for every $n$, $A^n(z_A+t)=z_A+a^nt$ for every $t$, that is, $A^n(x)=z_A+a^n(x-z_A)$. (Once again, this algebraic miracle occurs due to the conjugation of $A$ to $L:t\mapsto at$.)

Here, $(a,b)=(3,1)$ hence $z_A=-\frac12$ and $T_n=A^{n-1}(1)=z_A+3^{n-1}(1-z_A)=\frac12(3^n-1)$.

There is a simple way to make this kind of recurrence homogenous. Let $A(n)=T(n) +a$
such that the recurrence satisfied by it is homogenous. We get, $a=3a+1$,$a=-\frac{1}{2}$

The question is simplified to:

$A(n)=3A(n-1)$

which is easy to solve.


Average of Random variables converges in probability. Let $(\Omega, F, P)$ be probability theory.

Suppose that $X_1, X_2, X_3,...$ be sequence of random variable and $E(X_i)=0$ for all $i\in \mathbb{N}$.

Let $Y_n=\frac{X_1+X_2+... +X_n}{n}$.

claim . $Y_n$ converges to 0 in probability.

Let $A_n=\{w : |Y_n(w)| > \epsilon >0\}$

$$\int _{A_n} |Y_n| dP \geq \int_{A_n} \epsilon dP \geq P(A_n) \epsilon$$

But $|Y_n| \leq |X_1|/n + |X_2|/n +... |X_n|/n$ so $P(A_n)=0$.

is it right?

If you assume the $X_i$ are iid, this is the Weak Law of Large Numbers.  Without that assumption (or some slight generalizations of it), the statement is false.

The result does not hold. Robert has a fine counterexample.

You are stating that

$|Y_n| \leq |X_1|/n + ... + |X_n|/n$

and concluding that therefore, $P(A_n)=0$.

How do you arrive at this conclusion? I suppose you evaluate

$\int_{A_n} |Y_n| \, \mathrm{d}P
\leq
\frac{1}{n} \sum_{i=1}^n  \int_{A_n} |X_i| \, \mathrm{d}P$,

but how do you argue hereafter?

Maybe useful: $EX_i=0$ does not imply $E|X_i|=0$. Actually $E|X_i|=0$ iff $X_i=0$ almost surely, in which case your statement is trivial.


Intuition for pullback operation In manifolds and complex geometry there is this thing called the pullback. Usually when I see it, its going backwards on maps that are going forwards. I've been told that it is just a composition of functions. I just need help understanding this concept.

Sometimes an example is better than a thousand words. Let's say you have these forms on $\mathbb{R}^{4}$:
    $$\omega^{1}= y\mbox{d}x-x\mbox{d}y-t\mbox{d}z+z\mbox{d}t,$$
$$\omega^{2}= t\mbox{d}x-z\mbox{d}y+y\mbox{d}z-x\mbox{d}t,$$
$$\omega^{3}= -z\mbox{d}x-t\mbox{d}y+x\mbox{d}z+y\mbox{d}t,$$
$$\omega^{4}= x\mbox{d}x+y\mbox{d}y+z\mbox{d}z+t\mbox{d}t.$$
And you want to pull them back into $S^3$. Then you need an application from $S^3$ into $\mathbb{R}^{4}$. Let's take this one
$$F\left(\psi,\,\theta,\,\phi\right)=\left(\sin\psi\sin\theta\cos\phi,\sin\psi\sin\theta\sin\phi,\sin\psi\cos\theta,\cos\psi\right).$$ This application sends a point of coordinates $\left(\psi,\,\theta,\,\phi\right)$ on the sphere to a point of coordinates $(x,y,z,w)$ on $\mathbb{R}^{4}$ where of course
$$x=\sin\psi\sin\theta\cos\phi,$$
$$y=\sin\psi\sin\theta\sin\phi,$$
$$z=\sin\psi\cos\theta,$$
$$w=\cos\psi.$$
Now to do the pullback you have by definition
$$F^{*}(\omega^{i})=\omega^{i}(F_{*})$$
Then what you have to do is just differentiate
$$dx = \cos\psi\sin\theta\cos\phi d\psi+\sin\psi\cos\theta\cos\phi d\theta-\sin\psi\sin\theta\sin\phi d\phi$$
$$dy = \cos\psi\sin\theta\sin\phi d\psi+\sin\psi\cos\theta\sin\phi d\theta+\sin\psi\sin\theta\cos\phi d\phi$$
$$dz = \cos\psi\cos\theta d\psi-\sin\psi\sin\theta d\theta$$
$$dw = -\sin\psi d\psi.$$

Then you substitute inside the original formula (substituiting $dx,dy,dz,dw$ along with $x,y,z,w$ of course and you obtain the desired forms pulled back
$$\omega^{1}= \cos\theta\mbox{d}\psi-\sin\psi\cos\psi\sin\theta\mbox{d}\theta-\sin^{2}\psi\sin^{2}\theta\mbox{d}\phi,$$
$$ \omega^{2}= \sin\theta\cos\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\sin\phi-\cos\psi\cos\theta\cos\phi\right)\mbox{d}\theta-\sin\psi\sin\theta\left(\cos\psi\sin\phi+\sin\psi\cos\theta\cos\phi\right)\mbox{d}\phi,$$
$$ \omega^{3}= -\sin\theta\sin\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\cos\phi+\cos\psi\cos\theta\sin\phi\right)\mbox{d}\theta -\sin\psi\sin\theta\left(\cos\psi\cos\phi-\sin\psi\cos\theta\sin\phi\right)\mbox{d}\phi,$$
$$ \omega^{4}= 0.$$
You notice that the last one is 0 because the form was variating only along the radius of the sphere. Anyhow you cannot obtain more than 3 indipendent form on $S^3$ and the first three are indipendent.
I think this example is complicated enough to let you practice, but simple enough to le you grasps what's going on here (which is effectively just the formalization of a change of variables).

The general definition of a pullback:

Given two morphisms $f:X\to Z$ and $g:Y\to Z$, then a pullback of $f$ and $g$ (if it exist) is two morphisms $p_X:P\to X$ and $p_Y:P\to Y$
$\require{AMScd}$
\begin{CD}
    P @>p_Y>> Y\\
    @V p_X V V= @VV g V\\
    X @>>f> Z
\end{CD}
such that given morphisms $p'_X:P'\to X$ and $p'_Y:P'\to Y$ with
\begin{CD}
    P' @>p'_Y>> Y\\
    @V p'_X V V= @VV g V\\
    X @>>f> Z
\end{CD}
there exist a unique morphism $\varphi:P'\to P$ such that
$p'_X=p_X\varphi$ and $p'_Y=p_Y\varphi$.

I suspect that the morphism $\varphi$ sometimes is called pullback, but in category theory it's the two morphisms $p_X:P\to X$ and $p_Y:P\to Y$.

See also: Universal Definition for Pullback


About amicable numbers My question might seem wrong but I will explain it now:

We will define $\phi(a)$ as the sum of divisors of $a$ except $a$.

A pair $(a,b)$ is an amicable pair $\iff ((\phi(a) = b)  \land  (\phi(b) = a)) \land (a \neq b)$

Obviously if $(a,b)$ is an amicable pair $(b,a)$ is also an amicable pair. So, we will say $a$ is an amicable number $\iff$ There exist an $b$, such that $(a,b)$ is an amicable pair and $b < a$. Also we will say $a$ is a perfect number $\iff$ $\phi(a) = a$.

So my question is if we pick an arbitrary number, is it more likely an amicable number or perfect number? This question might be wrong because they might be finite but my question is simple: "Are there more amicable numbers than perfect numbers?"

I check with computer and believed that it is more likely an amicable number.

Let's formalize this as follows:  Let $P(n)$ be the number of perfect numbers $\leq n$, and let $A(n)$ be the number of amicable numbers less than $n$.  Does $\lim_{n\rightarrow\infty} \frac{P(n)}{A(n)}$ exist, and if so what is its value?

I got the limits that I'm basing this on mainly from Wikipedia and WolframAlpha, so feel free to correct if something's wrong.

According to Wikipedia, $P(n)<c\sqrt n$, where c is a constant. I assume that this is a rough asymptotic limit, correct me if I'm wrong. It also says that $A(n)<ne^{-{\ln(n)}^{1/3}}$, so $\frac {P(n)}{A(n)}\sim\frac {c\sqrt n}{ne^{-{\ln(n)}^{1/3}}}$. The limit is easier to evaluate after taking the logarithm, so we have $\ln ({\frac {P(n)}{A(n)}})\sim\ln(n)^{1/3}-\frac{\ln(n)}2+\ln(c)$, so the limit is $-\infty$, since the $-\ln(n)\over 2$ dominates the function as n approaches $\infty$. To convert this back into its original form, we take $e^{-\infty}=0$. Therefore $\lim_{n\to \infty} \frac{P(n)}{A(n)}=0.$


Ambiguous conditions for $x$ in logarithm? I have just realised that there is a little ambiguity in defining conditions for $x$ in logarithm. Let me illustrate it on a simple example:

$\log{x}$ is valid for $x>0$ ,

$2\log{x}$ is also valid only for $x>0$ ,

but $\log{x^2}$ is valid for both $x>0$ and $x<0$ .

How is that possible when $2\log{x} = \log{x^2}$ ?

Note that logarithm identities are true only when both sides of the equation are defined, so $2\log x = \log x^2$ for $x > 0$, and not when $x < 0$.

This thinking could lead to things like $-1 = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1$, which is obviously false. The problem here is that $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ only when $a,b>0$.

As well known language do not follow the rules of logics. "If you behave well
you will have an ice cream" according to logic does not imply that if you behave badly you will not have the ice cream, but any child knows that in language this is precisely what it means. One could argue that language reflect the way we reason, logics the way we should reason. No students, even the ones in math would not understand what you mean when you say $\log(x)$ is defined for positive $x$ and then $\log x^2 =2 \log x$.


Does a 15-puzzle always have a solution For those that are not familiar with (this type of) sliding puzzles, basically you have a number of tiles on a board equal to (n * m) - 1 (possibly more holes if you want). The goal is to re-arrange the tiles in such a way that solves the puzzle.

The puzzle could be anything, from number games to images.

While writing a small app for this, I found that if I were to initialize the puzzle by randomly shuffle all of the pieces, I could end up in a situation where there is no solution if my puzzle was 2x2.

So the problem I have is: given a sliding puzzle with n-by-m dimensions, is there always a solution if there are a sufficient number of tiles (eg: a 3x3 board)? How would I even begin to prove this, or simply convince myself that it is the case?

If it is possible that a random shuffle could result in no-solution, then I'd have to figure out how to verify that there exists a solution, which is a completely different beast.

Martin Gardner had a very good writeup on the 14-15 puzzle in one of his Mathematical Games books.

Sam Loyd invented the puzzle.  He periodically posted rewards for solutions to certain starting configurations.  None of those rewards were claimed.

Much analysis was expended, and it was finally determined, through a parity argument (as mentioned in the comment above), that half of the possible starting configurations were unsolvable.  Interestingly, ALL of Loyd's reward configurations were unsolvable.

SO: No, every possible configuration is not solvable.  If you START with a solved puzzle, and apply only legal transformations (moves) to it, you always wind up at a solvable configuration.

For the GENERAL nxm question, you'd probably have to expand the parity argument.

Here, we consider the following problem:
For a given position on the board to say whether there is a sequence of moves leading to a decision or not.

Let there be given a position on the board:

+------------------+.
| 11 | 15 | 2 | 13 |.
|----|----|---|----|.
| 14 | 1  | 8 | 3  |.
|----|----|---|----|.
| 7  | 6  | 0 | 10 |.
|----|----|---|----|.
| 4  | 12 | 9 | 5  |.
+------------------+.

wherein one of the elements is zero and indicates an empty cell.
Consider the numbers in the matrix serially (rowwise):

11 12 2 13 14 1 8 3 7 6 0 10 4 12 9 5

Denote N - the number of inversions in the permutation (i.e. the number of such elements a[i] and a[j] that i < j but a[i] > a[j]).

Next, let K- line number in which there is an empty element (i.e. in our notation K = 3).

Then, a solution exists if and only if N + K is even.


Is it true that the only matrix that is similar to a scalar matrix is itself A square matrix $A$ is said to be similar to a square matrix $B$, of the same order, if there is an invertible matrix $P$ such that
$A = PBP^{-1}$.
Is this true?
The only matrix that is similar to a scalar matrix is itself.
I.e., if $C$ is a scalar matrix and $T = P CP^{−1}$, then $T = C$.
If yes, how do I prove it?

If $C$ is a scalar matrix, then $C = \mu I_{d}$, then $T = PCP^{-1} = P(\mu I_{d})P^{-1} = \mu I_{d} (P P^{-1}) = \mu I_{d}$
Note : This proof relies on a simple detail and funny exercise, but that should not be underestimated; In the third equality I used that the center of matrices are only the scalar ones, i.e $Z(M_{n}(\mathbb{K})) = \mu I_{d}$


is the set of integers as big as P(prime numbers)? I have a function $f:\mathbb{N}\to P(\mathbb{P})$ where $f(x) = \{prime\ factors\ of\ x\}$

$f$ is not $1:1$ (easy)

I proved that $f$ is on $P(\mathbb{P})$ which is probably wrong since it means $|\mathbb{N}|\geq |P(\mathbb{P})|$ but $|\mathbb{N}|=|\mathbb{P}|$.

Every subset of primes has a source in $\mathbb{N}$ (the multiplication of all the primes)

What's wrong with this?

Every finite set of primes is in the range of $f$.

However, what about any infinite set of primes? Say, the set of all primes, itself?

No number has infinitely many prime factors, so no infinite set of primes is in the range of $f$. Indeed, the set of finite sets of primes is countable, while the set of infinite sets of primes is uncountable - "most" sets of primes are infinite!

As someone pointed out in the comments, $\mathbb{N}$ has the same cardinality as $\mathbb{P}$ because they are both infinite subsets of the countably infinite set $\mathbb{Z}$, hence they are both countably infinite sets. If you want to see this explicitly, just define $f : \mathbb{N} \to \mathbb{P}$ letting $f(n)$ be the $n$th prime, i.e. $f(1) = 2, f(2) = 3, f(3) = 5,\ldots$.


If N is a quadratic residue modulo p for all primes p<N, is N a perfect square? It is known that $N$ is a perfect square if and only if $N$ is a quadratic residue for every prime $p$.  This gives a good probabilistic algorithm for testing if a randomly chosen positive integer is a perfect square - simply compute it's Legendre Symbol for a sufficiently large set of randomly chosen primes.  A single result of $-1$ definitively tells you that $N$ is not a perfect square, while repeated results of only $+1$ or $0$ increase the probability that $N$ is a perfect square.

This probabilistic test could be much quicker with lower probability of a false positive if we could check only primes less than the integer $N$.  The proofs I've seen that $N$ is a perfect square IFF $N$ is a QR mod p for all primes $p$ cannot be used to prove the result using only the finitely many primes less than $N$.

Has anyone seen or know of a proof of this more specific result, or alternatively, know of a counter-example?  I've burned more than a few CPU cycles on Mathematica looking for a counterexample and have not found one yet.

No. For example, $\;6\;$ is a quadratic residue mod $\;2,3\;\text{ and }\,5\;$ .

A related problem: It is true that if $n$ is a quadratic residue modulo $p$ for all primes $p$ then $n$ is a perfect square.


Defining addition in second order logic (before saying it's duplicate, read whole question)

I was told by someone that we can define addition and multiplication purely in terms of successor function, provided that we work in second order arithmetic. When trying to find a way in which it's defined, I found this M.SE question, and in the first answer a definition is given. I found it, however, quite unsatisfactory, because this definition uses quantification over functional symbols (which can be rewritten to use binary predicate quantification), and I expected a definition which would only require quantification over subsets of $\Bbb N$ (equivalently unary predicates). I expect something similar to answer to this question.

My question is: how to define addition in second order arithmetic using quantification over sets only? Is that even possible?

Thanks in advance.

As it says in the answer to Can equinumerosity by defined in monadic second-order logic?, Büchi showed that the monadic second order of theory of the successor function is decidable. This implies that you cannot define addition in this theory, because if you could, then  you could also define multiplication (as shown in the answer to How to define multiplication in addition terms in monadic second order logic?) and then the theory would be at least as strong as PA, which is not decidable.

I can't find the original work by Büchi online, but this paper by Rabinovich and Thomas has the relevant references.

Not sure what you mean by "using quantification over sets only," but this is how you define addition and multiplication in terms of the successor function.

Addition:

$\forall x \in N: x+0=x$

$\forall x,y\in N: x+S(y) = S(x+y)$

Multiplication:

$\forall x \in N: x\cdot0=0$

$\forall x,y\in N: x\cdot S(y) = x\cdot y + x$

You can construct these functions using Peano's axioms and the axioms of set theory.

For the addition function, we would define the set of ordered triples $A$ such that

$\forall x,y,z:[(x,y,z)\in A \iff (x,y,z)\in N^3 $

$\land \forall B\subset N^3:[\forall a\in N: (a,0,a)\in B \land \forall a,b,c: [ (a,b,c)\in B \implies (a,S(b),S(c))\in B]$

$ \implies (x,y,z)\in B]] $

Similarly, for the multplication function (having defined +), we would define the set of ordered triples $M$ such that

$\forall x,y,z:[(x,y,z)\in M \iff (x,y,z)\in N^3 $

$\land \forall B\subset N^3:[\forall a\in N: (a,0,0)\in B \land \forall a,b,c: [ (a,b,c)\in B \implies (a,b+1,c+a\in B]$

$ \implies (x,y,z)\in B]] $


How to solve the differential equation $yy' =-x$? I just started learning about Differential Equations and there's this equation that I had problems with:
$$yy' =-x$$
I know that it's a separable ODE and that I can integrate both sides. The right side is then going to be $-\frac{x^2}{2}+c$ but I'm not sure how to go about the left side. I know that the solution is $\frac{y^2}{2} = -\frac{x^2}{2}+c$ but why is the left side equal to that? Usually $\frac{y^2}{2}$ is the anti-derivative of $y$ and not $yy'$. What am I missing here ?
Thanks.

$$y y' = -x$$
$$y \frac{dy}{dx} = -x$$
$$y dy = -x dx$$
This is what it means to separate it. Now only terms involving $y$ are on the left, and terms involving $x$ are on the right. So integrate...
$$y^2/2 + C_1 = - x^2/2 + C_2$$
$$y^2 = -x^2 + C$$
$$y = \pm \sqrt{C-x^2}$$
As another user pointed out, you can also just recognize that $$(y^2)' = 2y y'$$

hint
You can easily check that The antiderivative of
$\sin(x)$ is not $\frac{\sin^2(x)}{2}$
But, you can be sure that the antiderivative of
$\sin(x)\cos(x) $ is $\frac{\sin^2(x)}{2} $.
do not forget to add a constant $ C$.
$$\int x.1.dx =\frac{x^2}{2}$$
and by the same
$$\int y(x).y'(x)dx=\frac{y^2(x)}{2}$$


Why is a circle not a convex set? Why is a circle not a convex set?

My attempt: I was googling it, but I didn't find the correct answer, as I didn't understand the explanation why a circle is not a convex set as defined here:

As shown in diagram 1, points A and B and points C and D lie inside the circle, thus I think the circle must be a convex set, as clearly shown in figure 1.

Why isn't the circle a convex set?

They mean just the edge of the circle, without the interior. That is not a convex set.

Because the chord is not placed on the circle.


Products of adjugate matrices Let $S$ and $A$ be a symmetric and a skew-symmetric $n \times n$ matrix over $\mathbb{R}$, respectively. When calculating (numerically) the product $S^{-1} A S^{-1}$ I keep getting the factor $\det S$ in the denominator, while I would expect to get the square $$S^{-1} A S^{-1} = \frac{(\text{adj }S) A (\text{adj }S)}{(\det S)^2},$$ where $\text{adj }S$ is the adjugate of $S$.

Is there a way to prove that the combination $(\text{adj }S) A (\text{adj }S)$ already contains a factor of $\det S$?

Assume that $\det(S)=0$. Then $adj(S)$ has rank $1$ and is symmetric; then $adj(S)=avv^T$ where $a\in \mathbb{R}$ and $v$ is a vector. Thus $adj(S)Aadj(S)=a^2v(v^TAv)v^T$. Since $A$ is skew-symmetric, $v^TAv=0$ and $adj(S)Aadj(S)=0$. We use the Darij's method; here, the condition is that $\det(S)$ is an irreducible polynomial when $S$ is a generic symmetric matrix; if it is true, then $\det(S)$ is a factor of every entry of $adj(S)Aadj(S)$.

EDIT 1. For the proof  that $\det(S)$ is an irreducible polynomial when $S$ is a generic symmetric matrix, cf. https://mathoverflow.net/questions/50362/irreducibility-of-determinant-of-symmetric-matrix
and we are done !

EDIT 2. @ darij grinberg , hi Darij, I read quickly your Theorem 1 (for $K$, a commutative ring with unity) and I think that your proof works; yet it is complicated! I think (as you wrote in your comment above) that it suffices to prove the result when $K$ is a field; yet I do not kwow how to write it rigorously...

STEP 1. $K$ is a field. If $\det(S)=0$, then $adj(S)=vw^T$ and $adj(S).A.adj(S)=v(w^TAw)v^T=0$ (even if $char(K)=2$). Since $\det(.)$ is irreducible over $M_n(K)$, we conclude as above.

STEP 2. Let $S=[s_{ij}],A=[a_{i,j}]$. We work in the ring of polynomials $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i<j}]$ in the indeterminates $(s_{i,j}),(a_{i,j})$. This ring has no zero-divisors, is factorial and its characteristic is $0$ and even is integrally closed. Clearly the entries of $adj(S).A.adj(S)$ are in $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i<j}]$; moreover they formally have $\det(S)$ as a factor.

Now, if $K$ is a commutative ring with unity, we must use an argument using a variant of Gauss lemma showing that the factor $\det(S)$ is preserved over $K$. What form of the lemma can be used and how to write it correctly ?

I just see that the OP takes for himself the green chevron; we are our own best advocates

Because $S\mbox{adj}(S)=\mbox{adj}(S)S=\mbox{det}(S)I$, then $S$ is invertible iff $\mbox{det}(S)\ne 0$, and, in that case
$$
                         S^{-1} = \frac{1}{\mbox{det}(S)}\mbox{adj}(S).
$$
Therefore,
$$
                         S^{-1}AS^{-1} = \frac{1}{\mbox{det}(S)^{2}}\mbox{adj}(S)\,A\,\mbox{adj}(S).
$$


Prove that for all sets $A$ and $B$, $( A \cup B ) - B = A$ implies $A \cap B =\varnothing$. In the following proof we harness properties such as:
For all set $A$,
(i) $A = A - \varnothing$
(ii) $A \cap\varnothing=\varnothing$
Proof:
Let $A$ and $B$ be arbitrary sets and suppose $( A \cup B ) - B = A$.
Given (i) then $$( A \cup B ) - B = A - \varnothing.$$
So then, $( A \cup B ) = A$ and $B = \varnothing$.
That being so,
$$A \cap B = ( A \cup B ) \cap \varnothing = \varnothing$$
If we apply (ii) to the foregoing equivalence relation. Therefore
$$A \cap B = \varnothing$$
Is this proof right? if it is I would like to see alternative ones.

Suppose that $x \in A \cap B$ existed. Then $x \in A$ but also $x \in A \cup B$ and $x \in B$ so that $x \notin (A \cup B) - B$. This contradicts the starting assumption that $A=(A\cup B)-B$, so $A \cap B = \emptyset$.

Suppose $(A\cup B) \setminus B = A$.
Let $x \in A\cap B$.
Then $x \in A$ and $x\in B$.  Then $x \in B$ so $x\not \in M\setminus B$ for any set $M$ so $x\not \in (A\cup B) \setminus B$.  So $x \not \in A$. But that's a contradiction.
So if $(A\cup B)\setminus B =A$ then $x \in A\cap B$ is impossible.  So $A\cap B=\emptyset$.
Not this in no way means $B$ is empty; just $A$ and $B$ are dijoint.
.....
Alternatively.  If $(A\cup B)\setminus B = A$ then if $x \in A$ then $x\in (A\cup B) \setminus B$ so $x \not \in B$.  So $x$ in both $A$ and $B$ is impossible so $A\cap B=\emptyset.
Or....
If $(A\cup B)\setminus B = A$ and if $x \in B$ then $x\not \in (A\cup B)\setminus B = A$ so $x$ in both $B$ and $A$ is impossible.
....
An alternative way would be to prove that $(M\cup N)\setminus N = M\setminus N$ always.  I'll leave that proof to you.
That means $(A\cup B) \setminus B= A\setminus B= A$.  So if $x\in B$ the $x$ is removed for $A$ to get $A\setminus B$ so $x\not \in A$.
Etc.


Found $\sin x$ partial limits How to find the sequences of $\sin n$ (n=natural number) sub-sequence limits? I know that it is a $[-1;1]$, but how to proof?

Edit:
Is it true, that sin(n), with all natural numbers have different value? How to proof? If that is true, it is posible found bijection between N->[-1;1]

Outline: Consider the points $P(n)=(\cos n,\sin n)$ as $n$ ranges over the positive integers. It is enough to show that this set of points is dense in the unit circle.

Since $\pi$ is irrational, we have $P(m)\ne P(n)$ if $m\ne n$. It follows by the Pigeonhole Principle that given any $\epsilon \gt 0$, there exist $m_0$ and  $n_0$, with $m_0\lt n_0$,  such that $0\lt d(P(m_0),P(n_0))\lt \epsilon$. In particular, $0 \lt |\sin(m_0) -\sin(n_0)|\lt \epsilon$.

Then for any point $P$ on the unit circle, there are infinitely many positive  integers $k$ such that the distance of $P(m_0+k(n_0-m_0))$ from $P$ is less than $\epsilon$. In particular, for any $x$ there exist infinitely many $k$ such that
$|\sin x -\sin(m_0+k(n_0-m_0))|\lt \epsilon$.

Remark: The modified question asks whether $\sin m$ and $\sin n$ can be equal if $m\ne n$. Note that $\sin x=\sin y$ if and only if $y=x+2k\pi$ or $y=(2k+1)\pi -x$ for some integer $k$. Since $\pi$ is irrational, this cannot happen if $x=m$ and $y=n$, where $m$ and $n$ are distinct integers. In essence this fact was used in the proof outlined in the answer.

It's easy to prove that $\forall n,m\in\mathbb{N}, \sin n \neq \sin m$ if $n \neq m  $.

If $\sin{n}=\sin{m}$, then $n=m+2\pi k (k\in\mathbb{Z})$ by periodicity. So, $\pi=\dfrac{n-m}{2k}\in\mathbb{Q}$ But since the left side is clearly irrational number, this contradiction implies the above fact.


Tangent Lines- h(x)=f(g(x)) Assume $f$ and $g$ are differentiable on their domains with
$h(x)=f(g(x))$.
Suppose the equation of the line tangent to the graph
of $g$ at the point $(4,70)$ is $y=3x-5$ and the equation of the line
tangent to the graph of $f$ at $(7,9)$ is $y= -2x+23$
a. calculate $h(4)$ and $h'(4)$.
b. Determine an equation of the line tangent to the graph of $h$ at
the point of the graph where $x=4$.

I'm lost on a, can anyone give me a hint on how to approach it?

We have insufficient information about $f(x)$ to solve part a. All we know about $f(x)$ is that it passes through $(7,9)$ and that it has derivative $f'(7)=-2$. This could be satisfied by $f(x)=-2x+23$, in which case $h(4)$ would be $-117$ but it could also be satisfied by $f(x)=x^2-16x+72$, in which case, $h(4)$ would be $3852$. Presumably the value of $70$ was a typo and should have been $7$, in which case $h(4)$ would have been: $4\underbrace{\rightarrow}_{g}7\underbrace{\rightarrow}_{f}\underline{9}$.

For part b, it suffices to use the chain rule.


Evaluating a sum $\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}$ How should I evaluate the sum: $\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}$ ?

I have no relevant observations or partial results so far.

Any kind of help or advice would be truly appreciated!

Use binom of Newton:

$$(1+x)^n=\sum_{k=0}^n \binom nk x^k\stackrel{\text{integ.}}\implies\frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\binom nk\frac{x^{k+1}}{k+1}+C\;,\;\;C=\text{ constant}$$

Now susbtitute $\;x=0\;$ to find the constant, and then $\;x=1\;$ and do some little algebra.


In the equation $y = ax^2 + bx + c$ of a parabola, what do "$a$", "$b$", "$c$" represent? I have trouble grasping some basic things about parabolas. (This should be easily found on Google, but for some reason I couldn't find an answer that helped me).
I know one simple standard equation for a parabola:

$$y = ax^2 + bx + c$$
My problem is: I'm not sure what the following letters represent: $a$, $b$, and $c$.

Please try to explain to me what each of these letters represent in the equation, in a simple manner so I will understand, since I have very basic knowledge in math.
Thank you

It would be worth your while to learn another standard form of the equation of a parabola, and you can complete the square, given $y = ax^2 + bx + c$, to obtain this form:

$$4p(y - k) = (x-h)^2$$

The vertex of the parabola is given by $(h, k)$.
$$h = \frac{-b}{2a};\quad  k = \frac{4ac - b^2}{4a}$$

$$4p = \frac 1a$$

So, we have f(x) = ax^2 + bx + c, where a is distinct from  0.

'A' is related with growth. If 'a' is negative, function decrease, if 'a' is positive, function increase. 'C' is of course y-argument of point (0, f(x)) (let's imagine this as a place where function cross OY axis).


Euclidean norm second derivative I really need Your help.

I need to prove that Euclidean norm is strictly convex. I know that a function is strictly convex if $f ''(x)>0$. Can I use it for Euclidean norm and how? $||x||''=\frac{||x||^2-x^2}{||x||^3}$

Thank You!

The function
$$
\lVert \mbox{ }\rVert:\mathbb{R}^n\to \mathbb{R},\, f(x)=\lVert x\rVert
$$
is certainly convex. In fact, given $t\in [0,1]$ and $x,y\in \mathbb{R}^n$ we have:
$$
\lVert tx+(1-t)y\rVert=\lVert tx+(1-t)y\rVert\le \lVert tx\rVert+\lVert (1-t)y\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert.
$$
But is it STRICTLY CONVEX? i.e. if $t\in (0,1)$ and $x,y \in \mathbb{R}^n$, with $x\ne y$, then
$$
\lVert tx+(1-t)y\rVert<t\Vert x\rVert+(1-t)\lVert y\rVert.
$$
The answer is NO. In fact, if $t\in (0,1)$, and $y=sx$, with $0<s \ne 1$,  and $x\in \mathbb{R}^n$, we certainly have $x\ne y$. But
\begin{eqnarray}
\lVert tx+(1-t)y\rVert&=&\lVert tx+(1-t)sx\rVert=\lVert (t+(1-t)s)x\rVert=(t+(1-t)s)\lVert x\rVert\\
&=&t\lVert x\rVert+(1-t)\lVert sx\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert
\end{eqnarray}
Hence the function $\lVert\mbox{ } \rVert$ is CONVEX but NOT STRICTLY CONVEX.

By the definition of convexity and the triangle inequality we have

  $$ ||ax+(1-a)y|| \leq |a| ||x||+ |1-a|||y||,\quad 0\leq a\leq1. $$


Product of two regular spaces Def: A space $X$ is regular if $\forall x \in X$, and for every closed set $C \subset F$ with $x \notin C$, there exists disjoint open sets $U$ and $V$ s.t $x \in U$ and $C \subset V$.

Problem: A subspace of a regular space is regular; $X \times Y$ is regular iff each of $X$ and $Y$ is regular.

So, the first part I think I figure out.

A Subspace of a regular space is regular.

Proof: Let $Y$ be a subspace of a regular space $X$ and let $x \in Y$ and $F$ be a closed set in $Y$ not containing $x$.  Since $Cl_{X}(F) \cap Y = F$ then $x \notin Cl_{X}(F)$.  Then because $X$ is regular $\exists U,V$ disjoint open set in $X$   such that $  x \in U$ and $Cl_{X}(F) \subset V$. Then $U \cap Y$ and $V \cap Y$ are open sets in Y such that $x \in U \cap Y$ and $F \subset V \cap Y$.  Therefore $Y$ is a regular subspace of $X$.

The second half of the question is where I'm having trouble.  I feel like it should follow straight forward from the definition of a regular space and work out component wise on $X \times Y$, but at the same time I know how these questions go and that the question wouldn't be phrased how it is if I didn't need to use the fact that a subspace of a regular space is regular.  Also,  talking to a friend made me feel this to be even more true, but I can't seem to get it or see why it is really needed. Here's my attempt though.

$X \times Y$ is regular iff each of $X$ and $Y$ is regular.

Proof:

$\Rightarrow$) Assume $X \times Y$ is regular.  Then $\forall (x,y) \in X \times Y$ and for every closed $(F_{1}, F_{2}) \subset X \times Y$ with $(x,y) \notin (F_{1}, F_{2})$ there exists disjoint open sets $(U_{1}, U_{2}), (V_{1}, V_{2}) $in $ X \times Y$ such that $(x,y) \in (U_{1}, U_{2})$ and $(V_{1}, V_{2}) \subset (F_{1}, F_{2})$.  Then $x \in U_{1} \subset X$, $y \in U_{2} \subset Y$, and $V_{1} \subset F_{1}$ in $X$ and $V_{2} \subset F_{2}$ in $Y$.  Therefore, $X$ is regular and $Y$ is regular.

$\Leftarrow$) I'm not really sure here, and the proof I have now for this part is fairly analogous to the first implication.

As far as using the subspace result for proving the second half of the question the only thing I can really think to do is to consider, the subspaces $X$ and $Y$ in $X \times Y$ but I wasn't able to quite get the proof when messing around with that and my notation is probably a little off since I have not done much with products in topology, so I'll spare everyone and not post that. This question seemed like it would be fairly easy on first glance I'm not sure why it's getting me so confused. Anyways, thank you very much for any help/insight you can provide.

Here are some suggestions; I’ve left a lot of details for you to complete.

The product question is much easier if you first prove this little

  Proposition: A space $X$ is regular if and only if for each point $x\in X$ and each open set $U$ containing $x$ there is an open set $V$ such that $x\in V\subseteq\operatorname{cl}_X V\subseteq U$.

For the proof of $(\Leftarrow)$, suppose that $\langle x,y\rangle\in X\times Y$ and $U$ is an open set in $X\times Y$ such that $\langle x,y\rangle\in U$. Then by the definition of the product topology there are open sets $V\subseteq X$ and $W\subseteq Y$ such that $\langle x,y\rangle\in V\times W\subseteq U$. Clearly $x\in V$ and $y\in W$, so you can use the regularity of $X$ and $Y$ and the proposition to get open sets $G\subseteq X$ and $H\subseteq Y$ such that $x\in G\subseteq\operatorname{cl}_X G\subseteq V$ and $y\in H\subseteq\operatorname{cl}_Y H\subseteq W$. Now consider the open set $G\times H$, which clearly contains $\langle x,y\rangle$; what is its closure in $X\times Y$?

For the proof of $(\Rightarrow)$ you need to assume that $X\times Y$ is regular and then prove that $X$ and $Y$ are regular. To show directly that $X$ is regular, you must start start with a point $x\in X$, not a point in $X\times Y$. If you’re going for a direct proof, and you use the proposition, you’ll also start with an open set $U$ containing $x$. Let $y$ be any point of $Y$, and consider the open neighborhood $U\times Y$ of $\langle x,y\rangle$. Get an open set $V\subset X\times Y$ such that $$\langle x,y\rangle\in V\subseteq\operatorname{cl}_{X\times Y}V\subseteq U\times Y\;,$$ and find a basic open set in the product topology (i.e., an open ‘box’, like $V\times W$ in the proof of $(\Leftarrow)$) containing $\langle x,y\rangle$ and contained in $V$. You should have little trouble using that ‘box’ to get an open set in $X$ containing $x$ whose closure is contained in $U$.

However, for this direction you can, as you suspected, use the first part of the problem. Pick any $y\in Y$; then $X\times \{y\}$ is a subspace of the regular space $X\times Y$, so it’s regular, and it’s also homeomorphic to $X$, so $X$ is regular as well. You can handle $Y$ similarly.

($\Leftarrow$) Suppose $X\times Y$ is regular. Pick points $x_0\in X$ and $y_0\in Y$ then the subspaces $X\times\{y_0\}$ and $\{x_0\}\times Y$ are regular (by your first result). Now note that these subspaces are homeomorphic to $X$ and $Y$ respectively hence $X$ and $Y$ are regular.


How do we prove that something is unprovable? I have read somewhere there are some theorems that are shown to be "unprovable". It was a while ago and I don't remember the details, and I suspect that this question might be the result of a total misunderstanding. By the way, I assume that unprovable theorem does exist. Please correct me if I am wrong and skip reading the rest.

As far as I know, the mathematical statements are categorized into: undefined concepts, definitions, axioms, conjectures, lemmas and theorems. There might be some other types that I am not aware of as an amateur math learner. In this categorization, an axiom is something that cannot be built upon other things and it is too obvious to be proved (is it?). So axioms are unprovable. A theorem or lemma is actually a conjecture that has been proved. So "a theorem that cannot be proved" sounds like a paradox.

I know that there are some statements that cannot be proved simply because they are wrong. I am not addressing them because they are not theorems. So what does it mean that a theorem is unprovable? Does it mean that it cannot be proved by current mathematical tools and it may be proved in the future by more advanced tools that are not discovered yet? So why don't we call it a conjecture? If it cannot be proved at all, then it is better to call it an axiom.

Another question is, how can we be sure that a theorem cannot be proved? I am assuming the description might be some high level logic that is way above my understanding. So I would appreciate if you put it into simple words.

Edit- Thanks to a comment by @user21820 I just read two other interesting posts, this and this that are relevant to this question. I recommend everyone to take a look at them as well.

When we say that a statement is 'unprovable', we mean that it is unprovable from the axioms of a particular theory.

Here's a nice concrete example.  Euclid's Elements, the prototypical example of axiomatic mathematics, begins by stating the following five axioms:

  Any two points can be joined by a straight line

  Any finite straight line segment can be extended to form an infinite
  straight line.

  For any point $P$ and choice of radius $r$ we can form a circle
  centred at $P$ of radius $r$

  All right angles are equal to one another.

  [The parallel postulate:] If $L$ is a straight line and $P$ is a point not on the line $L$ then there is at most one line $L'$ that passes through $P$ and is parallel to $L$.

Euclid proceeds to derive much of classical plane geometry from these five axioms.  This is an important point.  After these axioms have been stated, Euclid makes no further appeal to our natural intuition for the concepts of 'line', 'point' and 'angle', but only gives proofs that can be deduced from the five axioms alone.

It is conceivable that you could come up with your own theory with 'points' and 'lines' that do not resemble points and lines at all.  But if you could show that your 'points' and 'lines' obey the five axioms of Euclid, then you could interpret all of his theorems in your new theory.

In the two thousand years following the publication of the Elements, one major question that arose was: do we need the fifth axiom?  The fifth axiom - known as the parallel postulate - seems less intuitively obvious than the other four: if we could find a way of deducing the fifth axiom from the first four then it would become superfluous and we could leave it out.

Mathematicians tried for millennia to find a way of deducing the parallel postulate from the first four axioms (and I'm sure there are cranks who are still trying to do so now), but were unable to.  Gradually, they started to get the feeling that it might be impossible to prove the parallel postulate from the first four axioms.  But how do you prove that something is unprovable?

The right approach was found independently by Lobachevsky and Bolyai (and possibly Gauss) in the nineteenth century.  They took the first four axioms and replaced the fifth with the following:

  [Hyperbolic parallel postulate:] If $L$ is a straight line and $P$ is a point not on the line $L$ then there are at least two lines that pass through $P$ and are parallel to $L$.

This axiom is clearly incompatible with the original parallel postulate.  The remarkable thing is that there is a geometrical theory in which the first four axioms and the modified parallel postulate are true.

The theory is called hyperbolic geometry and it deals with points and lines inscribed on the surface of a hyperboloid:

In the bottom right of the image above, you can see a pair of hyperbolic parallel lines.  Notice that they diverge from one another.

The first four axioms hold (and you can check this), but now if $L$ is a line and $P$ is a point not on $L$ then there are infinitely many lines parallel to $L$ passing through $P$.  So the original parallel postulate does not hold.

This now allows us to prove very quickly that it is impossible to prove the parallel postulate from the other four axioms: indeed, suppose there were such a proof.  Since the first four axioms are true in hyperbolic geometry, our proof would induce a proof of the parallel postulate in the setting of hyperbolic geometry.  But the parallel postulate is not true in hyperbolic geometry, so this is absurd.

This is a major method for showing that statements are unprovable in various theories.  Indeed, a theorem of Gödel (Gödel's completeness theorem) tells us that if a statement $s$ in the language of some axiomatic theory $\mathbb T$ is unprovable then there is always some structure that satisfies the axioms of $\mathbb T$ in which $s$ is false.  So showing that $s$ is unprovable often amounts to finding such a structure.

It is also possible to show that things are unprovable using a direct combinatorial argument on the axioms and deduction rules you are allowed in your logic.  I won't go into that here.

You're probably interested in things like Gödel's incompleteness theorem, that say that there are statements that are unprovable in a particular theory called ZFC set theory, which is often used as the foundation of all mathematics (note: there is in fact plenty of mathematics that cannot be expressed in ZFC, so all isn't really correct here).  This situation is not at all different from the geometrical example I gave above:

If a particular statement is neither provable nor disprovable from the axioms of all mathematics it means that there are two structures out there, both of which interpret the axioms of all mathematics, in one of which the statement is true and in the other of which the statement is false.

Sometimes we have explicit examples: one important problem at the turn of the century was the Continuum Hypothesis.  The problem was solved in two steps:

Gödel gave a structure satisfying the axioms of ZFC set theory in which the Continuum Hypothesis was true.
Later, Cohen gave a structure satisfying the axioms of ZFC set theory in which the Continuum Hypothesis was false.

Between them, these results show that the Continuum Hypothesis is in fact neither provable nor disprovable in ZFC set theory.

The most common way (AFAIK) to establish $A \not \vdash B$ is to find some statement $X$ such that:

$A,~X \vdash \lnot B$
$A \land X$ is consistent

Then you can conclude $A \not \vdash B$ because constructively if $A \vdash B$ then $A ,~ X \vdash B$ then $A ,~ X \vdash B \land \lnot B$ which contradicts the assumption that $A \land X$ are consistent.

For example, if you wanted to show that from "$n$ is divisible by $3$" you cannot prove "$n$ is divisible by $6$" you do the obvious thing of pointing out the $X$ of $n = 9$.  Since

$3|n , ~ n = 9 \vdash \lnot 6|n$
$(3|n) \land (n = 9)$ is consistent

So $3|n \not \vdash 6|n$.  This simple approach can get very complex, mainly because establishing $A \land X$ as being consistent can be extremely demanding, usually an informal model theoretic approach is used for this.

It is also a well known way to establish undecidability, if you can find $X_1$ and $X_2$ such that:

$A, ~ X_1 \vdash B$
$A, ~ X_2 \vdash \lnot B$
$A \land X_1$ is consistent
$A \land X_2$ is consistent

Then for the same reason $A \not \vdash B$ and $A \not \vdash \lnot B$, so $B$ is independent of $A$.


Problem about complex integration The question is to find $\int \frac{z^2-z+1}{z-1}dz$ over $|z|=1$.

My solution is:

Using Cauchy's integral formula we have:

$$f(1) = \frac{1}{2\pi i}\int \frac{f(z)}{z-1}dz,$$

but $f(1) = 1$. Therefore:

$$\int \frac{f(z)}{z-1}dz = 2\pi i,$$

where $f(z) = z^2-z+1$. Is my solution correct? If not then please help me to rectify it. Thanks a lot.

Your solution is incorrect because you are using Cauchy's integral formula incorrectly. Using Cauchy's formula we can evaluate the function at any point inside the domain bounded by the contour of integration and in this case, the point lies on the contour.
The correct approach is to find the Principal Value of the integral as pointed out by @MhenniBenghorbal.

Let $C$ denote the unit circle. Note that
$\int_C \dfrac{z^2-z+1}{z-1}\, \mathrm{d} z = \int_c z \, \mathrm{d} z + \int_C \dfrac{\mathrm{d} z}{z-1} = \int_C \dfrac{\mathrm{d} z}{z-1}$ where the first integral is zero by Cauchy's theorem. Now we find the Principal Value of the second integral.

$ \mathrm{PV} \int_C \dfrac{\mathrm{d} z}{z-1} = \lim_{\epsilon \rightarrow 0} \left[\int_{C_1} \dfrac{dz}{z-1} + \int_{C_2} \dfrac{dz}{z-1}\right]$.

Here C is approximated by the boundary of the region given by $\{|z|<1 \}\backslash \{|z-1|<\epsilon \}$ and $C_1$ is the bigger arc while $C_2$ is the arc of radius $\epsilon$. Hence $C_1: z=e^{i\theta}; \epsilon < \theta < 2\pi-\epsilon$ in the counter-clockwise direction and $C_2: z = 1+\epsilon e^{i\phi}; \pi/2 < \phi < 3\pi/2$ in the clockwise direction.

Therefore $ \mathrm{PV} \int_C \dfrac{\mathrm{d} z}{z-1} =2\pi i \times \left( \text{Residue of }1/(z-1) \text{ inside the region bounded by } C_1 \cup C_2 \right) = 0$.

$0 =  \lim_{\epsilon \rightarrow 0} \left[\int_{\epsilon}^{2\pi-\epsilon} \dfrac{ie^{i\theta}d\theta}{e^{i\theta}-1} - \int_{\pi/2}^{3\pi/2}\dfrac{i \epsilon e^{i\phi}d\phi}{\epsilon e^{i\phi}} \right]$,

$\lim_{\epsilon \rightarrow 0} \int_{\epsilon}^{2\pi-\epsilon} \dfrac{ie^{i\theta}d\theta}{e^{i\theta}-1} = i\pi $;

that is $\int_C \dfrac{dz}{z-1} = i\pi$

If we just take $z=\cos x+i\sin x$ and solve the question and instead of $dz$ we make i

Is that "i" an "integral" which was cut off or an $i$, imaginary unit?


Prove or disprove that in an 8-element subsets of $\{1,2…,30\}$ there must exist two $4$-element subsets that sum to the same number. How can I show that for any set of $8$ distinct positive integers not exceeding $30$, there must exist two distinct $4$-elements subsets that same up to the same number?

I tried using pigeon hole principle, but i still don't get it.

There are $$\binom {8}4=70$$ four-elements subsets of an $8$-element set.

The least possible sum is $1+2+3+4=10$ and the greatest possible sum is $27+28+29+30=114$. Hence, there are $105$ sums.

I have no idea how to continue because the number of possible integer sums is greater than the number of four-element subsets.
The $4$-element subsets are not necessarily non-overlapping.

Edit:
For example, from $X=\{1,3,9,11,15,20,24,29\}$ , we can choose two different subsets $\{1,3,15,24\}$ and $\{3,9,11,20\}$ because they both sum up to $43$.

Let the elements of $X$ be $a_1<a_2<...<a_8$ and denote the seven successive differences by $d_i=a_{i+1}-a_i.$
Consider the subsets of size $4$ which contain either $2$ or $3$ elements of $\{a_5,a_6,a_7,a_8\}$. There are $$\begin{pmatrix}4\\1\\\end{pmatrix}\begin{pmatrix}4\\3\\\end{pmatrix}+\begin{pmatrix}4\\2\\\end{pmatrix}\begin{pmatrix}4\\2\\\end{pmatrix}=52$$ of these subsets and the possible sums of their elements range from $a_1+a_2+a_5+a_6$ to $a_4+a_6+a_7+a_8$. So, by the pigeon-hole principle, we are finished unless $$a_4+a_6+a_7+a_8-(a_1+a_2+a_5+a_6)+1\ge 52$$ $$\text {i.e.} 2(a_8-a_1)\ge51+d_1+d_4+d_7.$$
Since $a_8-a_1\le 29$ we must have $d_1+d_4+d_7\le7$. Using the observations given below, $d_1,d_4,d_7$ are all different and no two can add to the third and so  $\{d_1,d_4,d_7\}=\{1,2,4\}$ and $\{a_1,a_{8}\}=\{1,30\}.$

Some observations about the $d_i$.
(1) Any two non-adjacent differences are unequal.
(2) Given three non-adjacent differences, none is the sum of the other two.
(3) Given two adjacent differences, the sum of these differences can replace one of the differences in observations (1) and (2). (We still require the 'combined difference' to be non-adjacent to the other differences involved.)

The proofs of these are all elementary and of the same form. As an example, suppose we have $d_2+d_3=d_5+d_7$, which is a combination of (2) and (3). Then $$a_4-a_2=a_6-a_5+a_8-a_7.$$
The sets $\{a_4,a_5,a_7\}$ and $\{a_2,a_6,a_8\}$ then have the same sum and $a_1$, say, can be added to each.

To return to the main proof where we know that the differences $\{d_1,d_4,d_7\}=\{1,2,4\}$.

Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $1$. Then, by the observations, $\{d,d+1\}\cap\{2,4,6\}$ is empty. So $d\ge7$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $2$. Then, by the observations, $\{d,d+2\}\cap\{1,3,4,5\}$ is empty. So $d\ge6$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $4$. Then, again by the observations, $\{d\}\cap\{1,2,3\}$ is empty. So $d\ge4$.
The sum of the differences (which is $29$) is now at least $(1+2+4)+(7+6+4)+d$, where $d$ is the 'other' difference adjacent to $d_4$. Therefore $d_4=4$ and the two differences adjacent to it (which cannot be equal) are $4$ and $5$. The differences adjacent to the differences of $1$ and $2$ are thus forced to be $7$ and $6$, respectively. Then $a_1+a_8=a_3+a_5$ and we are finished.

The statement is false.
Take for example a subset with 7 odd numbers and 1 even number.
Then we divide this subset into two 4-element subsets. One of them will have 4 odd numbers whose sum will be an even number while the other will have 1 even number and 3 odd numbers which will add up to an odd number.

Example: 1,3,5,7,9,11,13,14
Total sum is 63 and if we were to form 2 subsets, sum in each would not be a whole number.


What is the smallest $n$ such that the condition holds? What is the smallest $n$ such that for all $a,b$ with $12 \mid a+b$, it is the case that if $n\mid ab$, then $12 \mid a$ and $12 \mid b$?

I think I see how it works if $n = 144$, but I don't know how to prove that it is the smallest (or that there is another value of $n$ which is).

Let's see. Apparently, $n$ is divisible by 2 and 3, otherwise it couldn't enforce divisibility by 12. Now, the product $ab$ is divisible by 3, which means that at least one of the numbers $a$ and $b$ is divisible by 3, and so is their sum, hence so is the other number. By similar reasoning, both are divisible by 2. So $a=6x,\;b=6y$. Now the question changes to: given that $x+y$ is even, what is the condition on $n$, such that ${n\over36}\mid xy$, that would make both $x$ and $y$ even? The answer is obvious: the product must be even as well, so $n=72$ will suffice.

If I get you right, we have to find $n,a,b$ so that the conditions

$12 \mid a+b $
$n \mid ab $
$12 \mid a $
$12 \mid b $

hold true.

Well, if you are searching for the smallest possible values, than you might be able to simply check the first numbers with a simple, however inefficient, computer program. This is a Python example:

import itertools
for i in itertools.product(range(1,100),range(1,100),range(1,100)):
    n,a,b = i
    if ((a + b) % 12 == 0) and ((a * b) % n == 0):
        if (a % 12 == 0) and (b % 12 == 0):
            print n,a,b

The results are:

1 12 12
1 12 24
1 12 36
1 12 48
1 12 60
1 12 72
...


How to prove that a connected graph with $|V| -1= |E|$ is a tree? I could neither show myself nor find a proof of the following result:  if $G=(V,E)$ is a connected graph with $|E|=|V|-1$ then $G$ is a tree.

Could somebody please provide an argument to establish this.

An empty graph on $n$ vertices has $n$ connected components, Suppose you have a graph and add an edge, then the number of connected components is reduced by at most one ( since this edge touches at most two connected components). Therefore a connected graph on $n$ vertices has at least $n-1$ edges).

Suppose a connected graph on  $n$ vertices has $n-1$ edges, we must prove no cycle exists, suppose it does, if we remove an edge from the cycle we get a connected graph (because if we have a path that uses this edge, instead of using the edge we can go around the remaining part of the cycle). This is a contradiction, since the graph would have $n-2$ edges, and would hence not be connected.

Suppose the graph contains cycles, then we can remove an edge without removing a vertex and the graph is still connected. After removing, it will has $n$ vertices and $n - 2$ edges. Continue to do so until we reach a tree (because there is no cycle) with $n$ vertices and $n - k$ edges $(k > 1)$. However it contradicts with the fact that a tree with $n$ vertices must has $n - 1$ edges.
Therefore the graph must not contain cycles, and then it is a tree by definition.


How many positive 3-digit integers can be formed using the digits 0, 1, 2, 3, 4, and 5? My Question says, Using the digits (0, 1, 2, 3, 4, 5), how many positive 3-digit integers can be formed?

So my work so far is for positive 3 digit integers you would have the choice of 6 digits for the first digit, the choice of 6 digits for the second digit, and the choice of 6 digits for the third digit: (6)(6)(6)

But I'm not sure where to go from here, do I simply multiply them together? Is there another step afterwards? Am I missing something? Please help!

The first digit cannot be $0$. Therefore the answer is $5\times6^2$.


Show that $(\mathcal{M},d)$ is complete metric space Let $(\Bbb{R},\mathcal{M},\mu)$ be the Lebesgue measure space modulo the equivalence relation $A\sim B$ if $\mu(A\bigtriangleup B)=0$. Let $d(A,B)=\mu(A\bigtriangleup B)$.
Show that $(\mathcal{M},d)$ is complete metric space.

I could show that $d$ is a metric on the equivalence classes but how can I show that this metric is complete?

Note that $d(A,B) = \| 1_A-1_B\|_1$, so we can use properties of $L^1(\mathbb{R})$ to show completeness.

As Michael noted in the comments above, one needs to be careful with infinities.
Suppose we have a $d$-Cauchy sequence of sets $A_n$. Without loss of generality
we can presume that
$d(A_n,A_1) < 1$ for all $n$.

Let $f_n = 1_{A_n} - 1_{A_1}$. We have $f_n \in L^1(\mathbb{R})$ for all $n$
and $f_n$ is Cauchy. Since $L^1(\mathbb{R})$ is complete, we know that there
is some $f \in L^1(\mathbb{R})$ such that $f_n \to f$ (in the $\|\cdot\|_1$ norm, of course).

The only issue remaining is to 'extract' a set $A$ such that $f(x) = 1_A(x)- 1_{A_1}$ ae. $[\mu]$.

A standard result (see https://math.stackexchange.com/a/716328/27978, for example) is that if $f_n \to f$ in $L^1(\mathbb{R})$, then there is a
subsequence such that $f_{n_k}(x) \to f(x)$ ae. $[\mu]$. We note that
$f_n(x) \in \{-1,0,+1\}$ for all $n$, hence we have
$f(x) \in \{-1,0,+1\}$ ae. $[\mu]$. Let $N= f^{-1}(\{-1\})$, $Z= f^{-1}(\{0\})$ and $P= f^{-1}(\{+1\})$ and let $A = (Z \cap A_1) \cup P$, then
it is easy to check that $f(x) = 1_A(x) - 1_{A_1}(x)$ ae. $[\mu]$ and
$d(A,A_n) \to 0$. (Measurability follows from the fact that $f \in L^1(\mathbb{R})$.)

You need to appeal to the definition of completeness. Suppose you are given a Cauchy sequence of sets $A_n$. That is, for any $\epsilon > 0$ one can find a sufficiently large $N$ for which $m,n \geq N$ implies $\mu(A_{n} \bigtriangleup A_{m}) < \epsilon$. We would like to show that a limit set $A_{\infty}$ exists, and is measurable.

Here is a hint: Given a sequence of measurable sets $A_n$, there are a number of different sets one can associate to it, all of which are measurable. One example is the union $\bigcup A_n$, but there are others. Consider these.


Expected value of independent random variables So I have this problem where I need to find $E(4X+3Y-2Z^2-W^2+8)$ where $W,X,Y,Z$ are all standard normal and I'm kind of confused on how to find the expected value here. I thought to do it we just had to add the means together like this: $E(X_1+X_2+X_3)=\mu_1+\mu_2+\mu_3$ as long as they are iid of course. How come the answer is $5$?

I tried doing $$\begin{align}E(4X+3Y-2Z^2-W^2+8)\\ =E(4X)+E(3Y)-E(2Z^2)-E(W^2)+E(8)\\ =4E(X)+3E(Y)-2E(Z^2)-E(W^2)+8\\=(4*0)+(3*0)-(2*0)-0+8\\=8\end{align}$$ And quite obviously $8$ doesn't equal $5$. So I'm just not sure where I went wrong.

The variance of $Z$ is $E(Z^2)-(E(Z))^2$. Since the variance is $1$, and $E(Z)=0$, we have $E(Z^2)=1$. Same for $W^2$. That gets us $5$. The error was in thinking that $E(Z^2)=0$ and $E(W^2)=0$.

You cant open $E(X+Y^2 ) = E(X) + E(Y^2)$.
It is not linear at all .


Questions about cosets etc. I don't understand my notes that my teacher gave me, so... please answer those questions in bold.

From my teacher's notes: "Suppose $H$ is a subgroup of $G$. For any $a\in G$, we define its associated left coset and right coset to be $$aH = \{ab | b \in H\}$$ $$Ha = \{ba | b\in H\}$$ respectively."

What is a coset?
What are the differences between left and right?

.

Also from the notes: "Suppose $H$ is a subgroup of $G$. The left cosets $\{aH | a\in G\}$ define a partition of $G$, so are the right cosets."

What does "define a partition of $G$" mean?

.

Also: "A function $f:G\rightarrow H$ where $G$, $H$ are groups is called a group homomorphism if $$f(a\cdot b)=f(a)\cdot f(b)$$ for all $a,b \in G$. It's called an isomorphism if it's bijective."

Can anyone explain the notation of "$f:G\rightarrow H$"?
What does bijective mean?

The definitions of left / right coset are right there, and the different definitions make it clear that left and right coset are in general different.

A partition of a set $X$ is a collection of (non-empty) subsets of $X$ such that these subsets are pairwise disjoint while the union of all these subsets is $X$.

If $X,Y$ are sets, the notation $f\colon X\to Y$ expresses that $f$ is a function from $X$ to $Y$; that is, for each $x\in X$, $f$ defines some $f(x)\in Y$.
A function can be injective (or one-to-one) meaning that $f(x_1)=f(x_2)$ holds only when $x_1=x_2$; it can be surjective (or onto) if for every $y\in Y$ there is at least one $x\in X$ with $f(x)=y$; and if a function is both injective and surjective, it is called bijective.

In my opinion, Wikipedia is a good enough source to read & understand these basic concepts in Algebra.

(Cosets) https://en.wikipedia.org/wiki/Coset

(Partitions of a group) https://groupprops.subwiki.org/wiki/Left_cosets_partition_a_group

(Group Isomorphism)https://en.wikipedia.org/wiki/Group_isomorphism


How to find the last one digit of $8^{97}$? What will be the units digit of:
  $8^{97}$

I know I will have to start like:

$8^1$, then $8^2, 8^3$ etc...

is there a shorter method to finding the units/digit?

Hint for proof:

Notice that when you take a product of two numbers the final digit of the result only depends on the final digit of each original factor. Why should that be the case?

Why does this imply that taking powers of 8 will give a pattern in the final digit? In fact this is true for any natural number, not just 8, and this pattern will always be of a specific sort.

You can use the geometric progression formula which is in the form a*r^(n-1)

"a" represents the first digit in the sequence.

"r" represents the ratio between every 2 successive numbers in the sequence

"n" represents the position of the number you are looking for.

So......

a=8^1=1

r=8^2/8^1=8^3/8^2=(well, you get the idea)

r=8

n=97

8*8^(97-1)

Unfortunately I use a 6 bit calculator which will give me the wrong value so just calculate for yourself 8^96 then multiply by8.


Longest cylinder of specified radius in a given cuboid Find the maximum height (in exact value) of a cylinder of radius $x$ so that it can completely place into a $100 cm \times 60 cm \times50 cm$ cuboid.

This question comes from http://hk.knowledge.yahoo.com/question/question?qid=7012072800395.

I know that this question is equivalent to two times of the maximum height (in exact value) of a right cone of radius $x$ so that it can completely place into a $50 cm \times$ $30 cm \times 25 cm$ cuboid whose the apex of the right cone is placed at the corner of the cuboid, but I still have no idea until now.

Here's a possibly-wrong approach ...

Center a "short" cylinder at the cuboid's center, oriented along a diagonal. Elongate the cylinder ---in each direction--- until it collides with a face, say the "top" (and, at the other end, the "bottom"). Letting the cylinder scrape against the top (and bottom) face(s), continue elongating the cylinder ---such that the projection of its axis into the top face coincides with the face's diagonal--- until it collides with a second (pair of) face(s). Finally, letting the cylinder scrape those faces, elongate in the only dimension allowable until it collides with the final (pair of) face(s).

Symbolically ...

Let the cuboid have dimensions $2a$, $2b$, $2c$; center it at the origin, and let its edges be axis-aligned. Let the cylinder have radius $s$. (I use "$r$" below for a different parameter.)

Taking $P = p \; ( a, b, c )$ to be the center of one end of the "short" cylinder, we elongate the cylinder (that is, we increase $p$) until it collides with the walls $x=\pm a$; this happens when

$$P_x + s \frac{P_x}{|P|} = a \qquad (1)$$

Let $p_\star$ be the value of $p$ solving $(1)$.

Now, let $Q := p_\star \; ( a, b, c ) + q \; ( 0, b, c )$ take over as the center of the end of the cylinder; we elongate until the cylinder hits the walls $y=\pm b$:

$$Q_y + s \frac{Q_y}{|Q|} = b \qquad (2)$$

Writing $q_\star$ for the appropriate value of $q$ solving $(2)$, we finish with cylindrical endpoint $R := p_\star \; ( a, b, c ) + q_\star\; ( 0, b, c ) + r ( 0, 0, c )$ until

$$R_z + s \frac{R_z}{|R|} = c \qquad (3)$$

when $r = r_\star$. Then

$$\ell = 2|R| = 2\sqrt{\; p_\star^2 \; a^2 + \left( p_\star + q_\star \right)^2 \; b^2 + \left( p_\star + q_\star + r_\star \right)^2 \; c^2 \; }$$

may (or may not) be the length of the longest cylinder in the cuboid. At least, it should be a "local maximum".

As for solving $(1)$, $(2)$, $(3)$ ...

For $(1)$, defining $d := \sqrt{a^2+b^2+c^2}$, we have

$$p a + s \frac{p a}{p d} = a \qquad \to \qquad p_\star = 1 - \frac{s}{d}$$

For $(2)$, we have

$$(p_\star+q) b + \frac{s(p_\star + q )b}{\sqrt{p_\star^2 a^2 + (p_\star + q )^2(b^2+c^2)}} = b$$
$$\to \qquad s(p_\star + q ) = \left(1-p_\star-q\right)\;\sqrt{p_\star^2 a^2 + (p_\star + q )^2(b^2+c^2)}$$
$$\to \qquad s^2 (p_\star + q )^2 = \left(1-p_\star-q\right)^2\;\left( p_\star^2 a^2 + (p_\star + q )^2(b^2+c^2) \right)$$

which makes $q_\star$ the root of a quartic polynomial. The same is true for $r_\star$. I'll leave the sorting-out of those roots to the reader.

Does the $x$ axis-aligned? Would it possible to have $x=0$? The only thing want to mention is the maximum length of a line within the bounded space would be the diagonal, based on the metric property.


Evaluate $\lim_{x\rightarrow\mathrm\pi}\frac{\sin(mx)}{\sin(nx)}$ I have to evaluate $\lim_{x\rightarrow\mathrm\pi}\frac{\sin(mx)}{\sin(nx)}$ where $m,n \in\mathbb{N*}$. At first I thought I could just use the remarkable limit $\lim_{x\rightarrow0}\frac{\sin(x)}x = 1$ and the answer could just be $\frac {m}{n}$ but this is not the answer.... I mean it's a part of it but I don't understand why.

Just since $mx \to 0$ does not hold when $x\to \pi$.


Sum from infinity to infinity How does one evaluate the following limits?

1) $\lim_{n \rightarrow \infty} \sum_{k=n}^\infty 1$

2) $\lim_{n \rightarrow \infty} \sum_{k=n}^\infty k^{-1}$

3) $\lim_{n \rightarrow \infty} \sum_{k=n}^\infty 2^{-k}$

Do all three limits evaluate to $0$? If so, why? Perhaps only the third limit evaluates to $0$, while the first two are undefined. Again, why? @PeterTamaroff pointed out that the first two limits are undefined for fixed $n$, but does that necessarily imply they are undefined in the limit?

Consider the sequence $$\left(\sum_{k=n}^\infty 1\right)_{n \geq 1}.$$ This sequence is $$(\infty,\infty,\infty,\ldots).$$  Hence, if we wish to define a limit of this sequence, it should be $$\lim_{n \rightarrow \infty} \sum_{k=n}^\infty 1=\lim_{n \rightarrow \infty} \infty=\infty.$$

The second example is resolved in the same way (since the series is also divergent).

In the third case, the sequence is $$\left(\sum_{k=n}^\infty \frac{1}{2^k}\right)_{n \geq 1}.$$  Since $$\sum_{k \geq 1} \frac{1}{2^k}=1,$$ the sequence is $$(1,\tfrac{1}{2},\tfrac{1}{4},\ldots)$$ so $$\lim_{n \rightarrow \infty} \sum_{k=n}^\infty \frac{1}{2^k}=0.$$

Although your second sum

  2) $\lim_{n \rightarrow \infty} \sum_{k=n}^\infty k^{-1}$

converges to 0, it is intresting that

$\lim_{n \rightarrow \infty} \sum_{k=n}^{2n} k^{-1} = \ln 2$

looking for other cases that base index starts from some infinity?


The positive integer $k$ has the property that for all $m \in \mathbb{N}$, $k \mid m \implies k \mid m_r$. Show that $k \mid 99$. Question from Engel's book Problem solving strategies.

The positive integer $k$ has the property that for all $m \in \mathbb{N}$, $k \mid m \implies k \mid m_r$, where $m_r$ is the reflection of $m$, i.e. if $m=1234$ then $m_r = 4321$. Show that $k \mid 99$.

I start with a small case, say $k$ divides some 2 digit number $ab$. Then $k$ divides $ba$ also. Since $ab = 10a+b$ and $ba=10b+a$, I eliminate $a$ to get $k \mid -99b$. $b$ is a 1 digit number, so I think I need to use this fact somehow, but I am stuck here.

For the 3 digit case, let the number be $abc$. By similar reasoning, I get that $k$ divides $99a-99b = 99(a-b)$. Again, a-b is small, so maybe I can brute force this.

I have considered trying to show that $k$ must be a palindrome? Since $k \mid k$, we have $k \mid k_r$, and maybe try to get something? Thanks for the help!

Let's just look at the divisibility rules for 3, 9, and 11. If 3 or 9 divides a number's sum of its digits then it's divides the number. If 11 divides the number you get from the alternating sum of its digits, then 11 divides the number.

Let's enumerate the digits of $m$ with subscripts so that $m_1$ is the first digit and $m_n$ is the last where $m$ is an $n$ digit long number.

The sum of the digits is $S=\sum_{i=1} ^n m_i$, and the alternating sum $S_a=\sum_{i=1} ^n (-1)^{n+1}m_i$.

If $\ 9 \ | \ S$, then $ \ 9\  | \ m_r \ $ just by the law of communicative addition.

Similarly if $\ 11\ |\ S_a$, then $ \ 11 \ | \ m_r $ .  The reason being that if $n$ is odd the same digits will be negative in the sum $S_a$ regardless of the order being reversed, and if $n$ is even then the sum will be opposite in parity but the same magnitude.

Ex:  Odd $n$: the number $94572$

$m_r: 94572 \to 27549$

$9-4+5-7+2=2-7+5-4+9$

Ex:  Even $n$:the number $9457$

$m_r: 9457 \to 7549$

$9-4+5-7 = - (7-5+4-9)$

HINT

Write $$k = \sum_{i=0}^{r_k}10^ix_i$$
where $x_i \in \{0,...,9\}$.

Since $k | k \implies k | k_r$, we have

$$\sum_{i=0}^{r_k}10^ix_i \quad \biggr{|} \quad\sum_{i=0}^{r_k}10^ix_{(r_k-i)},$$
$$\implies \lambda\sum_{i=0}^{r_k}10^ix_i = \sum_{i=0}^{r_k}10^ix_{(r_k-i)},$$
$$\implies 0 = \sum_{i=0}^{r_k}10^i(x_{(r_k-i)}-\lambda x_i),$$
$$\implies 0 = x_{(r_k-i)}-\lambda x_i \quad \forall i.$$

We therefore have that $k$ is a palindrome.


How to add some accents in Microsoft Word Equation for geometry I hope this will be the right place for my question. I'm working on a Geometry book and I have to type some accents and I cannot find the way to do it. Why Microsoft Word, because I tried with MathType and MathMagic but they make a huge mess on my files, even corrupted them. Also, I don't have time to learn Latex or money to pay another person to do it. I need these to be text, not images.

I'm desperate, and will be grateful of any hint.

The accents are the following:

Triangle notation over a letter
Arc notation over letters
Semiplane symbol

You could use \above.

For $\overset{\Delta}{A}$, use A \above("Delta Symbol"), it'll become A┴Delta and then press Enter for $\overset{\Delta}{A}$.

For $\overset{\cap}{AB}$, use AB\above("cap symbol").

Information was really helpful. To get complete list of all the shortcut for accents please see Shortcut for putting accents in MS Word equation editor


Has $n^{n+1}+(n+1)^{n+2}$ other obvious factors than that I found? Has the number $$f(n):=n^{n+1}+(n+1)^{n+2}$$ "obvious" factors (algebraic, aurifeuillan or similar kinds) apart from those , I mention below ?

I only managed to find out forced factors for odd numbers $\ n\ $ :

If $\ n\ $ is of the form $\ 6k+1\ $ , then $\ f(n)\ $ is divisible by $\ 3\ $.
If $\ n\ $ is of the form $\ 6k+3\ $ , then $\ f(n)\ $ is divisible by $\ n^2+n+1\ $
and finally, if $\ n+2\ $ is prime, then $\ f(n)\ $ is divisble by $\ n+2\ $.

For even $n$, I did not find forced factors.

The smallest not completely factored number of this form is $f(62)$. It has the composite cofactor $$29645851324749161395794060252012567916992450650017954$$
$$8416412620499302880901240095492001218810908429181608669479$$ with $111$ digits.

The full factorization of $f(62)$ is

$$\begin{eqnarray}
f(62) & = & 97\times 503\times 434254837008211200040837849611255155960657\times \\ & & 682683272545530598298287751380982048464896752734487326295497775449847\end{eqnarray}$$

Somewhat disappointingly, it was reached using raw computing power rather than any smart ideas; just by running YAFU, the factoring tool.


Find all integer solutions of the equation $x^2+15^a=2^b$ I found an integer solution for $$x^2+15^a=2^b$$ Which is $x=7$, when $a=1$ and $b=6$

In general: $x$ must be odd, $a$ must be odd and $b$ must be even

The problem is that I don't know wheater there are more solutions. Otherwise, how can I prove there is no other solution?

I'll do the case where $x,a$ and $b$ are positive, the other one is easier but a bit intricate.

Working mod $3$ we have that $b$ must be even.

Let $b=2v$

We now have: $15^a=2^b-x^2=(2^v-x)(2^v+x)$

Clearly $x$ must be coprime to $15$, and so $(2^v-x)=3^a$ and $(2^v+x)=5^a$. or  $2^v-x=1$ and $2^v+x=15^a$.

We deal with the first case first ($(2^v-x)=3^a$ and $(2^v+x)=5^a$):

So $3^a+5^a=2(2^v)$.

So in fact finding solutions to your equation is equivalent to $3^a+5^a$ being a power of $2$. When $a$ is even it is not possible (work mod $4$). When $a$ is odd it only works with $a=1$, this is easy to observe with the lifting the exponent lemma.

Since we have $2^v-x=3$ and $2^v+x=5$ we have $x=1,v=2,a=1$ or $x=1,b=4,a=1$.

We now deal with the second case: ($2^v-x=1$ and $2^v+x=15^a$):

We get $15^a+1=2(2^v)$.

So by Catalan's theorem $a=1$, and so $v=3$. Hence we get $x=7,b=6,a=1$. And this gives all solutions with $x,a,b$ positive.

Clearly the sign of $x$ is irrelevant, and simple arguments show there are no solutions with $a$ or $b$ negative.

However, there remains one other possibility, $x=0$ (neither positive not negative).  For this case $15^a=2^b $, and as $15$ is not any rational power of $2$ there can be no nonzero solutions for $a $ and $b $.  Only $a=b=0$ works for the case $x=0$.

The complete solution set is then

$(x,a,b) \in \{(0,0,0),(1,1,4),(7,1,6)\} $

Another solution is $x=1,a=1,b=4$

$b$ must be even for the right to be $1 \bmod 3$ as $x^2 \equiv 1 \pmod 3$.  Write $c=b/2$ and we are looking for $$(2^c-x)(2^c+x)=15^a$$  The two factors on the left are coprime, so we have the possibilities $$2^c-x=3^a\\2^c+x=5^a\\\text {and} \\2^c+x=15^a\\2^c-x=1$$
The solution $x=1,a=1,b=4,c=2$ is the first of these and $x=7,a=1,b=6,c=3$ is the second.  We cannot get any more from the second because we would have to have $2^{c+1}=15^a+1$ and the only perfect powers that differ by $1$ are $8,9$.  We cannot get any more from the first because we would have to have $2^{c+1}=5^a+3^a$ which can never be true $\bmod 16$ for $a \gt 1$


On the arithmetic differential equation $n''=n'$ If $n'$ denotes the arithmetic derivative of non-negative integer $n$, and $n''=(n')'$, then solve the following equation $$n''=n'.$$

What I have found, you can read in one minute! I have tried to explain it very detailed so anyone, even with a little knowledge of elementary number theory (like me), can follow the steps.

$n=0$ and $n=1$ are solutions. It is known that for a natural number $n>1$ with prime factorization of $\prod_{i=1}^{k} p_i^{a_i}$ arithmetic derivative is $$n'=n \sum_{i=1}^{k} \frac{a_i}{p_i}. \tag{1}$$

Let $m=n'$, then equation becomes $m'=m$. Let prime factorization of $m$ be $\prod_{j=1}^{l} q_j^{b_j}$. Then from equation $(1)$ we get $$\frac{b_1}{q_1}+ \frac{b_2}{q_2}+... + \frac{b_l}{q_l}=1. \tag{2}$$

This equation implies that $q_j \ge b_j$. Multiply both sides of the equation $(2)$ by $q_1 q_2 ... q_{l-1}$. It follows that $q_1 q_2 ... q_{l-1}\frac{b_l}{q_l}$ is an integer. Thus $q_l | b_l$. Hence $b_l \ge q_l$ and $b_l=q_l$. Subsequently $b_1=b_2=...=b_{l-1}=0$ and $m=q^q$ for some prime number $q$.

Thus we have $n'=m=q^q$ and $n\sum_{i=1}^{k} \frac{a_i}{p_i}=q^q$ or $$\prod_{i=1}^{k} p_i^{a_i-1}\sum_{i=1}^{k} \left( p_1 p_2 ... p_k \frac{a_i}{p_i} \right)=q^q. \tag{3}$$ Notice that if $p_i \neq q$ is a prime divisor of $n$, then $a_i=1$. We claim that if $q$ is a prime divisor of $n$, then its the only one. If $q \mid n$ then $n$ is in the form $$n=p_1p_2...p_kq^a,$$

Where $\gcd(q, p_i)=1$. Now its easy to see from equation $(3)$ that $a \le q$ and dividing both sides of it by $q^{a-1}$ gives $$q\sum_{i=1}^{k} \left( \frac{p_1 p_2 ... p_k}{p_i} \right)+p_1 p_2 ... p_k a=q^{q-a+1}.$$Therefore, $q|a$, which leads to $a \ge q$ and $q=a$. Thus $$\sum_{i=1}^{k} \left( \frac{p_1 p_2 ... p_k}{p_i} \right)+p_1 p_2 ... p_k=1,$$Which is a contradiction and $n=q^q$. Thus $n=q^q$ is a solution to the original equation, where $q$ is a prime number.

If $q \nmid n$, then equation $(3)$ gives

$$\sum_{i=1}^{k} \left( \frac{p_1 p_2 ... p_k}{p_i} \right)=q^q,$$Where I am stuck with.

Edit: According to @user49640 comment, there are some solutions of the form $n=2p$, where $p=q^q-2$ is a prime. For example for $q=7$ and $q=19$. See also @Thomas Andrews answer for an another solution not in the form $n=p^p$.

Look at this solution I found:

$$(2\times17431\times147288828839626635378984008187404125879)'=29^{29}$$

We have that

$$(3\cdot 29\cdot 25733)'=3\cdot 29 + 3\cdot 25733 + 29\cdot 25733=7^7$$

So you are going to get non-trivial solutions. It's probably a difficult problem to come up with all solutions.

I was looking for "3 prime" solutions. So, if $n=abc$ with $n'=ab+ac+bc=(a+c)(b+c)-c^2$. Trying to solve with $q=5$ gives:

$$(a+c)(b+c)=5^5+c^2$$

But $5^5\equiv 1\pmod{4}$ and thus $5^5+c^2$ cannot be divisible by $4$. so $a+c$ and $b+c$ cannot both be even, so one of $a,b,c$ must be $2$. We can assume $c=2$. Then we want $(a+2)(b+2)=3129=3\cdot 7\cdot 149$. No way to factor this as $mn$ with $m-2$ and $n-2$ prime. So there is no $3$-prime counter-example with $q=5$.

So I tried $q=7$ and found the above solution. It helped that $7^7+9$ has divisible by $256$, which gave me a lot of possibilities for factorizations.

There are two-prime solutions if $q^q-2$ is an odd prime.

For all $k\geq 2$, with $\{p_k\}$  any set of $n+1$ distinct primes,
and $a_k$ any set of $k$ positive integers such that no $a_k$ is a multiple of the corresponding $p_k$,
$$
\sum{\frac{a_k}{p_k} \neq 1}
$$
The proof is by induction, starting with a basis at $n=1$:  $1-\frac{a_1}{p_1}$ is a fraction with denominator $p_1$, say $\frac{r_1}{p_1}$ with $0<r_1<p_1$.  So for $\sum_1^2{\frac{a_k}{p_k} = 1}$ to hold, you must have
$$
\frac{a_2}{p_2}=\frac{r_1}{p_1}\\ a_2 p_1 = r_1 p_2
$$
Since $p_2$ is coprime with $p_1$, for this to hold  $r_1 = mp_1$. But $r_1 < p_1$, so this is a contradiction, and there cannot be such a combination.

the proof of the induction step is very similar, relying on the facts that the product of several primes none of which match $p_{n+1}$ cannot be divisible by $p_{n+1}$ and the "deficiency" of $\sum_1^n\frac{a_k}{p_k}$ must have a denominator of the form of the product of the $p_k$.

Therefore, the only solutions to
$$
n'=n
$$
are of the form $n=p^p$, where $p$ is prime.

So for your equation, we must lok for number whose aritmetic derivatives are of the form $p^p$.
We see immediately that for such a number w, if
$w=p^t\prod p_k^{a_k}$ with all the $p_k$ distinct from $p$,then all the $a_k = 1$ (otherwise, a prime other than $p$ will creep into $w'$).

THus the problem of finding a solution of a form other than $p^p$ comes down to finding a collection of primes $\{p_k\}, 1\leq k \leq s$, a distinct prime $p$, and an integer exponent $t>0$ such that
$$
\sum_{k=1}^s\frac1{p_k}+\frac1t = p^{p-t}
$$
Clearly $p^{p-t}< p^p$, otherwise we get back our solution $p^p$.

Let's see how this works, by trying to do this for $p=3$. If $t=2$, then we want
$$
\sum_{k=1}^s\frac1{p_k}+\frac12 = p^{3-2}=3
$$
It is easy to find a set of primes wose reciprocals add to more than $\frac52$ but then the denominator of that sum of reciprocals is a large number having all those primes as factors.  In order to get the full sum to be the simple fraction $\frac52$, $t$ would have to be at least half as big as that large number, thus it would need to be more than $3$. Since $t$ must be less than $p$, this won't work.

So for this to work, $t$ must be a large number, forcing $p$ to be a large prime, thus requiring very many terms in the sum of reciprocal primes, in turn requiring $t$ to be a much larger number.  This argument can be formalized, to show that
$$
n=p^p$$
is the only way to have
$$
n''=n'
$$


Calculus - inequality problem. I have this inequality :

$$|g(x)-B|<\frac{|B|}{2}$$
$$-\frac{|B|}{2}<g(x)-B<\frac{|B|}{2}$$
$$B-\frac{|B|}{2}<g(x)<B+\frac{|B|}{2}$$

I don't understand how it possible to conclude from the inequality that

$$\frac{|B|}{2}<|g(x)|$$

Thanks.

If $B>0$ then the first inequality implies $B/2 < g(x)< 3B/2$.
When $B<0$, the first inequality implies $B/2>g(x)>3B/2$