Is the Cartesian product of sets associative? Is it associative? In general sense, if we compare $A \times (B \times C)$ and $(A \times B) \times C$, then $(a,(b,c))$ does not equal $((a,b),c)$, but both of those pairs may be interpreted as one triplet $(a,b,c)$. Is it correct to do that and can we use associativity of Cartesian product in set theory problems?

Equality in $(A\times B) \times C=A\times (B\times C)$ holds if, and only if, at least one of the sets is empty. As mentioned already, there is a canonical isomorphism $(A\times B) \times C\to A\times (B\times C)$, mapping $((a,b),c)$ to $(a,(b,c))$. Many authors thus treat $\times $ as if it is an associative operation on sets, though it is not. However, the true reason why one can safely pretend $\times $ is associative and not get into any problems is due to coherence. As a warm-up, assume you have $127$ sets and you multiply them, using $\times $ in some way. You get an incredibly complicated set whose elements contain lots and lots of parenthesis in all sorts of places. You would like to say that this set is essentially just the set of tuples of length $127$ where the $k$-th coordinate is taken from the $k$-th set, but are you now really sure that's the case? The expressions can be so complicated that maybe it is now not so obvious anymore. And besides, do we really only care whether two sets admit a bijection between them in order to identify them? Of course not. So, what is going on here is that it is not so much the sets we should care about as much as it is the functions between them.

Consider again the canonicity of the functions used for the identification $(A\times B) \times C \cong A\times (B\times C)$ but now take four sets. When you have four sets you can multiply them together (in a given fixed order) in five different ways. These five different sets are, we would like to say, essentially the same, but, minding the above, that is not what we wish to say. We don't want to identify them, since they are different. What we want to know is that the canonical functions between them compose coherently. The precise meaning of that is a bit technical, and you may wish to read Mac Lanes's "Categories for the working mathematician" to properly understand coherence, but basically you will discover the definition yourself if you take four sets, multiply them to get five sets, place them at the vertices of a pentagon, connect the dots using the canonical maps, and claim that the different compositions along the perimeter are the same. The point then is Mac Lane's coherence theorem: starting with any finite number of sets, any two compositions of the canonical functions between one way to multiply the sets and another way to multiply the sets are the same.

So, we can safely pretend $\times $ is associative on sets because there are canonical maps between different choices which (and this is really important) always compose to give the same function when used on any number of sets, multiplies in any way you like.

An illustrative example where coherence fails for the 'wrong' choice of canonical maps is when you consider signed elements. Suppose that all your sets have elements, each of which is designated as either positive or negative, and that each element $x$ can change sign to $-x$. Now, you can map $A\times (B\times C)\to (A\times B)\times C$ by $(a,(b,c))\mapsto -((a,b),c)$. Mathematically, it's just as canonical as the other possibility. However, this one is not coherent (you can check that with the pentagon).

There is an abstract notion of products in category theory. A category where every finite collection of objects has a product is called a cartesian category. In those categories (and Set is one of them) the product is associative. That means that you can identify the two different sets $A\times(B\times C)$ and $(A\times B)\times C$ in any discussion where isomorphism is accepted as synonymous with identity.


Probability Confusion/Clarification Two pizzas are each divided into eight slices and placed inside separate boxes with the
lids closed, and left in an empty room. People randomly arrive at the room to take a slice.
Each person who arrives randomly chooses a box, opens it, takes a slice of pizza and closes
the box. If all eight slices from a box are taken then the lid is not closed.
The first person to discover an empty box opens the other box. Calculate the probability
that it contains all eight slices.
The answer says $\left(\dfrac{1}{2}\right)^8$. This is what I intuitively thought, until my friend posed the solution $\left(\dfrac{1}{2}\right)^7$ as he is looking at the problem from the standpoint that all the others have to do is pick the same box as the first person, and it doesn't matter which box the first person chooses. I have a feeling that perhaps his situation is more conditional probability, but I'm struggling to get my head around what is correct! Can anyone clarify?

The difference between $\left(\frac{1}{2}\right)^8$ versus $\left(\frac{1}{2}\right)^7$ depends on how we interpret the problem.
Problem 1:

We have two pizzas, each with eight pizza slices.  Each time someone enters the room, they open one of the two boxes uniformly at random, independent of each other persons' choices, and will take a slice if able and then close the lid except in the case where they have taken the last slice in which case they throw out the box.  We ask what the probability is that the first person to come in the room after a box has been thrown out will find a full eight slices of pizza in the remaining box.

For this problem, let us label the two pizza boxes $H$ and $T$ and let the people make the decision which box to eat from based on a coin flip.  Here, we have the two sequences HHHHHHHH and TTTTTTTT both result in the ninth person coming in and seeing one of the boxes missing and the remaining box having all of the slices of pizza in it, with a probability of $\left(\frac{1}{2}\right)^8 + \left(\frac{1}{2}\right)^8 = \left(\frac{1}{2}\right)^7$
The calculation could have been made simpler as your friend had noted by not caring which of the two boxes was chosen by the first person and then for the next seven all of them choosing the same box, yielding the same $\left(\frac{1}{2}\right)^7$

Problem 2:

We have two pizzas, each with eight pizza slices.  Each time someone enters the room, they open of of the two boxes uniformly at random, independent of each other persons' choices, and will take a slice if able and then close the lid... even in the case that they closed the lid on a now empty box.  We ask what the probability is that the first time that someone comes into the room and opens an empty box (that they did not cause to be empty themselves), that the remaining box has all eight slices still in it.

For this problem, similarly to the last, we describe via a sequence of coin flips.  Here however, after the first box becomes empty, we require that the empty box be discovered before slices from the still full box be taken.  In this case we required nine heads in a row or nine tails in a row.  This occurs with probability $\left(\frac{1}{2}\right)^9 + \left(\frac{1}{2}\right)^9 = \left(\frac{1}{2}\right)^8$

Which of these was the intended problem?  That is up to the person who asked the problem, but I interpret it as the first as per the included phrase "If all eight slices from a box are taken then the lid is not closed." which implies to me that no random event needs to occur for the ninth person in the event that the first eight people all ate from the same box since the contents of the empty box are plainly visible due to the lid remaining open.
As for why the linked Matchbox Problem gives the other answer, that is because it follows the style of problem 2 where we do not remove the newly emptied box once it becomes empty and we leave it until we attempt to retrieve a match from it and discover there are none to be had.

The only possible way for a person to discover an empty box and a full one, is if  eight slices have been taken (not more). The second condition is, that every slice was taken from the same box. This adds up to your result $(\frac{1}{2})^8$ since every person must take a slice from the same box.


Probability of selecting consecutive floors in an elevator Three people get into an empty elevator at the first floor of a building that has 10 floors. Each presses the button for their desired floor (unless one of the others has already pressed the button). Assume that they are equally likely to want to go to floors 2 through 10 (independently of each other). What is the probability that the buttons for 3 consecutive floors are pressed? (question from 'introduction to probability' by Bliztstein)

I thought the answer might be $$\frac{7 \cdot3!}{9^3}$$... but then I thought maybe the total number of ways is the bose-einstein value $n+k-1 \choose k$ (how many ways there are to choose k times from a set of n objects with replacement, if order doesn't matter).. so then the answer could also be $$\frac{7}{165}$$ Which is the correct answer, and why is the logic for the other answer wrong?

I think there is only one reasonable interpretation of the problem.  It leads to a probability model that produces your first answer.

Let our three passengers be A, B, and C. If A, B, C wish respectively to go to floors $p$, $q$, and $r$, record that fact as an ordered triple $(p,q,r)$. By the equally likely and independent part of the problem statement, all ordered triples $(p,q,r)$, where $p$, $q$, and $r$ range from $2$ to $10$, are equally likely. There are $9^3$ such triples. Whether a passenger, say C, happens not to press the elevator button to Floor $6$ because it has already been pressed is irrelevant.

We now count the favourables. There are $7$ ways to choose a collection of $3$ consecutive numbers in the interval from $2$ to $10$. For each of these collections, there are $3!$ ways in which A, B, and C might wish to get off at the floors in this collection, for a total of $7\cdot 3!$. For the probability, divide.

In the sequence $2,3,4,\cdots,10$ there are $7$ sequences of three consecutives numbers:
$$(2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), (8,9,10)$$
and by outher side there are ${9 \choose 3}=84$ distinc modes od choose $3$ numbers in set ${2,3,\cdots,10}$, Then, in my opinion, the probability is $P=\frac{7}{84}$.


Do we know if there exist true mathematical statements that can not be proven? Given the set of standard axioms (I'm not asking for proof of those), do we know for sure that a proof exists for all unproven theorems? For example, I believe the Goldbach Conjecture is not proven even though we "consider" it true.

Phrased another way, have we proven that if a mathematical statement is true, a proof of it exists? That, therefore, anything that is true can be proven, and anything that cannot be proven is not true? Or, is there a counterexample to that statement?

If it hasn't been proven either way, do we have a strong idea one way or the other? Is it generally thought that some theorems can have no proof, or not?

Relatively recent discoveries yield a number of so-called 'natural
independence' results that provide much more natural examples
of independence than does Gödel's example based upon
the liar paradox (or other syntactic diagonalizations).
As an example of such results, I'll sketch a simple example due
to Goodstein of a concrete number theoretic theorem whose proof
is independent of formal number theory PA (Peano Arithmetic) (following [Sim]).

Let $\,b\ge 2\,$ be a positive integer. Any nonnegative integer $n$
can be written uniquely in base $b$
$$\smash{n\, =\, c_1 b^{\large n_1} +\, \cdots + c_k b^{\large n_k}} $$

where $\,k \ge 0,\,$ and  $\, 0 < c_i  < b,\,$ and $\, n_1  > \ldots > n_k
\ge 0,\,$ for $\,i = 1, \ldots, k.$

For example the base $\,2\,$ representation of $\,266\,$ is
$$266 = 2^8 + 2^3 + 2$$

We may extend this by writing each of the exponents $\,n_1,\ldots,n_k\,$
in base $\,b\,$ notation, then doing the same for each of the exponents
in the resulting representations, $\ldots,\,$ until the process stops.
This yields the so-called 'hereditary base $\,b\,$ representation of $\,n$'.
For example the hereditary base $2$ representation of $\,266\,$ is
$$\smash{266 = 2^{\large 2^{2+1}}\! + 2^{2+1} + 2} $$

Let $\,B_{\,b}(n)$ be the nonnegative integer which results if we take the
hereditary base $\,b\,$ representation of $\,n\,$ and then syntactically
replace
each $\,b\,$ by $\,b+1,\,$ i.e. $\,B_{\,b}\,$ is a base change operator
that 'Bumps the Base'
from $\,b\,$ up to $\,b+1.\,$ For example bumping the base from $\,2\,$ to
$\,3\,$ in the prior equation yields
$$\smash{B_{2}(266) = 3^{\large 3^{3+1}}\! + 3^{3+1} + 3\quad\ \ \ }$$

Consider a sequence of integers obtained by repeatedly applying
the operation:  bump the base then subtract one from the result.
For example, iteratively applying this operation to $\,266\,$ yields
$$\begin{eqnarray}
266_0 &=&\ 2^{\large 2^{2+1}}\! + 2^{2+1} + 2\\
266_1 &=&\ 3^{\large 3^{3+1}}\! + 3^{3+1} + 3 - 1\ =\ B_2(266_0) - 1 \\
    ~ \ &=&\ 3^{\large 3^{3+1}}\! + 3^{3+1} + 2 \\
266_2 &=&\ 4^{\large 4^{4+1}}\! + 4^{4+1} + 1\qquad\! =\ B_3(266_1) - 1 \\
266_3 &=&\ 5^{\large5^{5+1}}\! + 5^{5+1}\phantom{ + 2}\qquad\   =\ B_4(266_2) - 1 \\
266_4 &=&\ 6^{\large 6^{6+1}}\! + \color{#0a0}{6^{6+1}\! - 1}  \\
    ~ \ &&\ \textrm{using}\quad  \color{#0a0}{6^7\ -\,\ 1}\ =\ \color{#c00}{5555555}\, \textrm{ in base } 6 \\
    ~ \ &=&\ 6^{\large 6^{6+1}}\! + \color{#c00}5\cdot 6^6 + \color{#c00}5\cdot 6^5 + \,\cdots + \color{#c00}5\cdot 6 + \color{#c00}5 \\
266_5 &=&\ 7^{\large 7^{7+1}}\! + 5\cdot 7^7 + 5\cdot 7^5 +\, \cdots + 5\cdot 7 + 4 \\
&\vdots & \\
266_{k+1} &=& \ \qquad\quad\ \cdots\qquad\quad\ = \ B_{k+2}(266_k) - 1 \\
\end{eqnarray}$$

In general, if we start this procedure at the integer $\,n\,$ then
we obtain what is known as the Goodstein sequence starting at $\,n.$

More precisely, for each nonnegative integer $\,n\,$ we recursively define
a sequence of nonnegative integers $\,n_0,\, n_1,\, \ldots ,\,
n_k,\ldots\,$ by
$$\begin{eqnarray}
n_0\ &:=&\ n \\
n_{k+1}\ &:=&\
\begin{cases}
B_{k+2}(n_k) - 1 &\mbox{if }\ n_k > 0 \\
\,0 &\mbox{if }\ n_k = 0
\end{cases} \\
\end{eqnarray}$$

If we examine the above Goodstein sequence for $\,266\,$ numerically
we find that the sequence initially increases extremely rapidly:

$$\begin{eqnarray}
2^{\large 2^{2+1}}\!+2^{2+1}+2\ &\sim&\ 2^{\large 2^3} &\sim&\,  3\cdot 10^2 \\
3^{\large 3^{3+1}}\!+3^{3+1}+2\ &\sim&\ 3^{\large 3^4} &\sim&\,  4\cdot 10^{38} \\
4^{\large 4^{4+1}}\!+4^{4+1}+1\ &\sim&\ 4^{\large 4^5} &\sim&\,  3\cdot 10^{616} \\
5^{\large 5^{5+1}}\!+5^{5+1}\ \ \phantom{+ 2}  \ &\sim&\ 5^{\large 5^6} &\sim&\,  3\cdot
10^{10921} \\
6^{\large 6^{6+1}}\!+5\cdot 6^{6}\quad\!+5\cdot 6^5\ \:+\cdots +5\cdot 6\ \ +5\
&\sim&\ 6^{\large 6^7} &\sim&\,  4\cdot 10^{217832} \\
7^{\large 7^{7+1}}\!+5\cdot 7^{7}\quad\!+5\cdot 7^5\ \:+\cdots +5\cdot 7\ \ +4\
&\sim&\ 7^{\large 7^8} &\sim&\,  1\cdot 10^{4871822} \\
8^{\large 8^{8+1}}\!+5\cdot 8^{8}\quad\!+5\cdot 8^5\ \: +\cdots +5\cdot 8\ \ +3\
 &\sim&\ 8^{\large 8^9} &\sim&\,  2\cdot 10^{121210686} \\
9^{\large 9^{9+1}}\!+5\cdot 9^{9}\quad\!+5\cdot 9^5\ \: +\cdots +5\cdot 9\ \ +2\
&\sim&\ 9^{\large 9^{10}} &\sim&\,  5\cdot 10^{3327237896} \\
10^{\large 10^{10+1}}\!\!\!+5\cdot 10^{10}\!+5\cdot 10^5\!+\cdots +5\cdot 10+1\ &\sim&\
10^{\large 10^{11}}\!\!\!\! &\sim&\,  1\cdot 10^{100000000000} \\
\end{eqnarray}$$

Nevertheless, despite numerical first impressions, one can prove that this
sequence converges to $\,0.\,$ In other words, $\,266_k = 0\,$ for all
sufficiently large $\,k.\,$ This surprising result is due to Goodstein
$(1944)$ who actually proved the same result for all Goodstein
sequences:

Goodstein's Theorem $\ $  For all $\,n\,$ there exists $\,k\,$ such
that $\,n_k = 0.\,$
In other words, every Goodstein sequence converges to $\,0.$

The secret underlying Goodstein's theorem is that hereditary
expression of $\,n\,$ in base $\,b\,$ mimics an ordinal notation for
all ordinals less than epsilon nought $\,\varepsilon_0 = \omega^{\large \omega^{\omega^{\Large\cdot^{\cdot^\cdot}}}}\!\!\! =\, \sup \{ \omega,\, \omega^{\omega}\!,\, \omega^{\large \omega^{\omega}}\!,\, \omega^{\large \omega^{\omega^\omega}}\!,\, \dots\, \}$. For such
ordinals, the base bumping
operation leaves the ordinal fixed, but subtraction of
one decreases the ordinal. But these ordinals are well-ordered, which allows us to conclude that a Goodstein sequence
eventually converges to zero. Goodstein actually proved his
theorem for a general increasing base-bumping function $\,f:\Bbb N\to \Bbb
N\,$
(vs. $\,f(b)=b+1\,$ above). He proved that convergence of all such
$f$-Goodstein sequences is equivalent to transfinite induction
below  $\,\epsilon_0.$

One of the primary measures of strength for a system of logic is
the size of the largest ordinal for which transfinite induction
holds. It is a classical result of Gentzen that the consistency
of PA (Peano Arithmetic, or formal number theory) can be proved
by transfinite induction on ordinals below $\,\epsilon_0.\,$ But we know
from Godel's second incompleteness theorem that the consistency
of PA cannot be proved in PA.  It follows that neither can
Goodstein's theorem be proved in PA. Thus we have an example of
a very simple concrete number theoretical statement in PA whose
proof is nonetheless independent of PA.

Another way to see that Goodstein's theorem cannot be proved in PA
is to note that the sequence takes too long to terminate, e.g.

$$ 4_k\,\text{ first reaches}\,\ 0\ \,\text{for }\, k\, =\,
3\cdot(2^{402653211}\!-1)\,\sim\, 10^{121210695}$$

In general, if 'for all $\,n\,$ there exists $\,k\,$ such that $\,P(n,k)$'
is
provable, then it must be witnessed by a provably computable
choice function $\,F\!:\, $ 'for all $\,n\!:\ P(n,F(n)).\,$'  But the
problem
is that $\,F(n)\,$ grows too rapidly to be provably computable in PA,
see [Smo] $1980$ for details.

Goodstein's theorem was one of the first examples of so-called
'natural independence phenomena', which are considered by most
logicians to be more natural than the metamathematical
incompleteness results first discovered by Gödel. Other finite
combinatorial examples were discovered around the same time,
e.g. a finite form of Ramsey's theorem, and a finite form of
Kruskal's tree theorem, see [KiP], [Smo] and [Gal]. [Kip]
presents the Hercules vs. Hydra game, which provides an
elementary example of a finite combinatorial tree theorem
(a more graphical tree-theoretic form of Goodstein's sequence).

Kruskal's tree theorem plays a fundamental role in computer
science because it is one of the main tools for showing that
certain orderings on trees are well-founded. These orderings
play a crucial role in proving the termination of rewrite rules
and the correctness of the Knuth-Bendix equational completion
procedures. See [Gal] for a survey of results in this area.

See the references below for further details, especially
Smorynski's papers. Start with Rucker's book if you know no
logic, then move on to Smorynski's papers, and then the others,
which are original research papers. For more recent work, see
the references cited in Gallier, especially to Friedman's school
of 'Reverse Mathematics', and see [JSL].

References

[Gal] Gallier, Jean. What's so special about Kruskal's theorem and the ordinal $\Gamma_0$?
A survey of some results in proof theory,
Ann. Pure and Applied Logic, 53 (1991) 199-260.

[HFR] Harrington, L.A. et.al. (editors)
Harvey Friedman's Research on the Foundations of Mathematics,  Elsevier 1985.

[KiP] Kirby, Laurie, and Paris, Jeff.
Accessible independence results for Peano arithmetic,
Bull. London Math. Soc., 14 (1982), 285-293.

[JSL] The Journal of Symbolic Logic,* v. 53, no. 2, 1988
 This issue contains papers from the Symposium "Hilbert's
 Program Sixty Years Later".

[Kol] Kolata, Gina. Does Goedel's Theorem Matter to Mathematics?
Science 218 11/19/1982, 779-780; reprinted in [HFR]

[Ruc] Rucker, Rudy. Infinity and The Mind, 1995, Princeton Univ. Press.

[Sim] Simpson, Stephen G. Unprovable theorems and fast-growing functions,
Contemporary Math. 65 1987, 359-394.

[Smo] Smorynski, Craig. (all three articles are reprinted in [HFR])
Some rapidly growing functions, Math. Intell., 2 1980, 149-154.
The Varieties of Arboreal Experience, Math. Intell., 4 1982, 182-188.
"Big" News from Archimedes to Friedman, Notices AMS, 30 1983, 251-256.

[Spe] Spencer, Joel. Large numbers and unprovable theorems,
Amer. Math. Monthly, Dec 1983, 669-675.

Euclid's fifth postulate:

Two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. This postulate is equivalent to what is known as the parallel postulate.

Euclid's fifth postulate cannot be proven as a theorem, although this was attempted by many people. Euclid himself used only the first four postulates ("absolute geometry") for the first 28 propositions of the Elements, but was forced to invoke the parallel postulate on the 29th. In 1823, Janos Bolyai and Nicolai Lobachevsky independently realized that entirely self-consistent "non-Euclidean geometries" could be created in which the parallel postulate did not hold. (Gauss had also discovered but suppressed the existence of non-Euclidean geometries.)


Conditions for f(x,y) to coincide with some g(x) + h(y) Given a function of two variables, say f(x,y), how to know/check whether it can be simplified into some g(x) + h(y)?

Some property of some condition that condition that f(x,y) satisfies to know this.

A necessary and sufficient condition is that, for every $(x,y,z,t)$,
$$
f(x,y)+f(z,t)=f(x,t)+f(z,y).
$$

For purposes of analysis proofs, and algebraicly, I think the most useful way of formulating that property is how you already wrote it, i.e. that it can be written as the sum of two functions, one of which doesn't depend on $x$ and the other of which doesn't depend on $y$.

Assuming you're looking for a different formulation, I don't know one in general, but for differentiable $f$
$$
\frac{\partial f}{\partial x}(x,y) \text{ is independent of } y
$$
$$
\frac{\partial f}{\partial y}(x,y) \text{ is independent of } x
$$

Also, if $f$ is twice differentiable, and the second derivative is continuous, a nice formulation is the condition that
$$
\frac{\partial f}{\partial x \partial y} (x,y) = 0.
$$

(Can you prove that the above is both necessary and sufficient in the continuous, twice differentiable case?)


Finding range of values of trigonometric function $f(n)= (\sin x)^n + (\cos x)^n$
I wanted to find the range of the values of this function in terms of n .

What I tried
I tried to use various inequalities like AM, GM ,HM but failed to derive s definite value due to the restriction on domain of sin x and cos x.
I was hoping someone would help me derive an expression for range in terms of n or more specifically help me calculate it for n=4,6.

For all $n \ge 2$,
$$
 \sin^n(x) + \cos^n(x) \le \vert \sin(x) \vert^n + \vert \cos(x) \vert^n
 \le \sin^2(x) + \cos^2(x) = 1
$$
with equality for $x = 0$ (as already stated in another answer).

For odd $n \ge 3$,
$$
 \sin^n(x) + \cos^n(x) \ge -\vert \sin(x) \vert^n - \vert \cos(x) \vert^n
 \ge - \sin^2(x) - \cos^2(x) = -1
$$
with equality for $x = \pi$.

For even $n = 2k \ge 2$, substitute $u = \sin^2(x)$, then
$$
\sin^n(x) + \cos^n(x) = u^k + (1-u)^k \ge 2 \left( \frac {u + (1-u)}{2} \right)^k = \frac{1}{2^{k-1}}
$$
because $t \to t^k$ is a convex function for $t \ge 0$.
Equality holds for $x = \frac \pi 4$, corresponding to $u = \frac 12$.

Putting it together, the range is
$$
 [-1, 1] \quad \text{for odd $n \ge 3$,}
$$
and
$$
 [\frac{1}{2^{n/2-1}}, 1] \quad \text{for even $n \ge 2$.}
$$

Finally, for $n=1$ the range can be deduced from the identity
$$
 \sin(x) + \cos(x) = \sqrt 2 \sin(x + \frac \pi 4) \, .
$$

Graph for $n=4$ and $n = 5$ (created with wxMaxima):

For $n\geq2$ we obtain:
$$\sin^nx+\cos^nx\leq\sin^2x+\cos^2x=1.$$
The equality occurs for $x=0$, which says that $1$ is a maximal value.
$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x=1-\frac{1}{2}\sin^22x\geq\frac{1}{2}.$$
The equality occurs for $x=\frac{\pi}{4}$, which says that $\frac{1}{2}$ is a minimal value.
$$\sin^6x+\cos^6x=1-3\sin^2x\cos^2x=1-\frac{3}{4}\sin^22x=\frac{1}{4}.$$
The equality occurs for $x=\frac{\pi}{4}$ again.

Since $f$ is a continuous function we obtain the answer for $n=4$:
$$\left[\frac{1}{2},1\right]$$ and for $n=6$:
$$\left[\frac{1}{4},1\right].$$
Done!


rate of change/water poured into cone Water is poured into an inverted cone at a rate of π-units per second. If the radius of the base of the cone is r and its height is 2r, what is the rate at which the depth of the water is changing when the height of the water is 1/2r?

Let h, and R  be the height of the water and the radius of the water surface. Then:

h/2r = R/r ==> R = h/2.

The volume V of the water in the cone is: V = 1/3*pi*R^2*h = 1/12*pi*h^3.

So: pi = dV/dt = 1/4*pi*h^2*dh/dt = 1/4*pi*(1/(4*r^2))*dh/dt ==> dh/dt = 16r^2


At which height an element belongs Suppose there is a triangle as follows where first row contains $1$ element,
the second row contains $2$ elements and so on like below diagram.

$$
\begin{array}{rccccccccc}
& & &1& & & \\
& &2& &3& & \\
&4& &5& &6& \\
& & & \vdots& & & \\
\end{array}
$$

Then what will be the row number of an element?

For example for $n=3$ the row number is $2$.

Let, $$x=\frac{\sqrt{8n+1}-1}{2}$$
Then the row the $n$-th number will be in is given by;
$$R_n=n-\frac{\lceil x \rceil(\lceil x \rceil-1)}{2}$$

Since in the question you stated that $n^{th} $ row contains n elements , the answer is , well....

Edit : Since you have edited the question , the new answer is :

Hint:

The first element in the $n^{th} $ row is given by $1+\frac{(n-1)n}{2} $ and the last element is given by $1+\frac{(n-1)(n+2)}{2}.$

Can you figure out the rest ?


Limit from UPB exam book 2014 $$\lim_{x \to 0^{+}} (\sin x)^{\cos x}\left(\frac{\cos^{2}x}{\sin x} - \sin x\log(\sin x)\right)$$

The answer is one, but i dont know how to proceed in solving the problem.

First note that instead of $x \to 0$ you must have $x \to 0^{+}$ otherwise both $\log\sin x$ and $(\sin x)^{\cos x}$ are undefined. The given expression can be expressed as $$(\sin x)^{\cos x - 1}\cos^{2}x - (\sin x)^{\cos x}\sin x\log\sin x$$ and limit of each term will be evaluated separately.

Now note that as $x \to 0^{+}$, $\cos x \to 1$ and $\log\sin x \to -\infty$ hence $\cos x \log \sin x \to -\infty$ and therefore $$(\sin x)^{\cos x} = \exp(\cos x \log \sin x) \to 0\tag{1}$$ Further note the standard result $$\lim_{t \to 0^{+}}t\log t = 0$$ and hence on putting $t = \sin x$ we get $$\lim_{x \to 0^{+}}\sin x \log \sin x = 0\tag{2}$$ It follows from $(1), (2)$ that $$\lim_{x \to 0^{+}}(\sin x)^{\cos x}\sin x\log \sin x = 0\tag{3}$$ We are now left with the term $$\lim_{x \to 0^{+}}(\sin x)^{\cos x - 1}\cos^{2}x$$ and since $\cos^{2}x \to 1$ we just need to calculate the limit of $(\sin x)^{\cos x - 1}$. This limit will be the final answer. Let it be $L$. Then
\begin{align}
\log L &= \log\lim_{x \to 0^{+}}(\sin x)^{\cos x - 1}\notag\\
&= \lim_{x \to 0^{+}}\log(\sin x)^{\cos x - 1}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0^{+}}(\cos x - 1)\log\sin x\notag\\
&= \lim_{x \to 0^{+}}(\cos x - 1)\log\left(\frac{\sin x}{x}\cdot x\right)\notag\\
&= \lim_{x \to 0^{+}}(\cos x - 1)\log\left(\frac{\sin x}{x}\right) + (\cos x - 1)\log x\notag\\
&= 0\cdot \log 1 + \lim_{x \to 0^{+}}\frac{\cos^{2}x - 1}{\cos x + 1}\cdot\log x\notag\\
&= -\frac{1}{2}\lim_{x \to 0^{+}}\sin^{2}x\log x\notag\\
&= -\frac{1}{2}\lim_{x \to 0^{+}}\frac{\sin^{2}x}{x^{2}}\cdot x\cdot x\log x\notag\\
&= -\frac{1}{2}\cdot 1\cdot 0\cdot 0 = 0\notag
\end{align}
It follows that $L = e^{0} = 1$.

$$\lim_{x\to 0 }x^{cos x-1}=1\\\lim_{x\to 0 }\frac{x^{cos x}}{x}=1\\ \xrightarrow[x \to 0]{sin x \approx  x} \lim_{x\to 0 }\frac{sin x^{cos x}}{x}=1 $$ name
 $u(x)=sinx ^{cos x}\\u'(x)=sinx ^{cos x} (\frac {cos^2x}{sinx}-sin x \ln sinx )$
now
$$ \lim_{x\to 0 }\frac{sin x^{cos x}}{x}=1\\\lim_{x\to 0 }\frac{u(x)}{x}=\frac{0}{0}=1 $$ l'hosptial because ($\lim_{x\to 0 }sin x^{cos x}=0$ )

$$\lim_{x\to 0 }\frac{u(x)}{x}=\frac{U'(x)}{1}=1 \ \to \\ \lim_{x\to 0 }u'(x)=1 \\\lim_{x\to 0 }u'(x) =\lim_{x\to 0 }sinx ^{cos x}\\u'(x)=sinx ^{cos x} (\frac {cos^2x}{sinx}-sin x \ln sinx )=1 $$


Is the Matrix Diagonalizable if $A^2=4I$ I have two question:

Let $A$ be a non-scalar matrix, $A_{k \times k} \in \Bbb R$, and $A^2=4I$. Is the matrix A, always diagonalizable in $\Bbb R$?
Answer

I know that the answer is yes:
$$A^2=4I \rightarrow A^2-4I=0$$
Then by the Cayley Hamilton theorem I know that the matrix satisfies the equation above.
Now I don't know how to explain the fact that the characteristic polynomial is $P_A=(\lambda-2)(\lambda+2)$, then the characteristic polynomial has two different roots, and no more. what's the reason for it?
Second Question - Irrelevant to the first question
When we say that a matrix is diagonalizable if it has different linear roots, it means that for if I have the following characteristic polynomial (for example) $(t-1)^2(t-2)$ then the matrix is not diagonalizable since the root 1 appears twice in the characteristic polynomial?

There are only two roots since a polynomial of degree $n$ has exactly $n$ roots (in $\mathbb{C}$). Here, $n=2$, so there are only $2$ roots.
The statement in your second question is not correct. For example, consider the identity matrix of dimension $n$. Clearly the identity matrix is diagonalizable (as it is diagonal), but it has characteristic polynomial $(t-1)^n$.

I'll try to answer without using the minimal and characteristic polynomials.

Consider the Jordan normal form of $A$,
$$
A = Q J Q^t,
$$
with $Q$ orthogonal, and split $J$ into its diagonal part $D$ plus its strictly upper triangular part $U$, namely
$$
J = D + U.
$$
Then,
$$
4I = A^2 = Q J Q^t QJ Q^t = Q J^2 Q^t \Longrightarrow J^2 = 4 I.
$$
Exploiting the splitting, we obtain
$$
4I = (D + U)(D+U) = D^2 + DU + UD + U^2.
$$
Notice that $D^2$ is diagonal, whereas $DU + UD + U^2$ is upper triangular, with zeros on the main diagonal. Hence,
$$
D^2 = 4I \Longrightarrow D = \pm2I.
$$
Moreover, the upper triangular part of $4I$ is zero, hence
$$
0 = DU + UD + U^2 = \pm 4 U + U^2 = U ( U \pm 4 I).
$$
This is possible only for $U = 0$.


$n$-th derivative change of variable I have a change of variables from $\alpha$ to $\beta$ such such that $$\frac{d}{d\beta} = \frac{1}{f(\alpha)}\frac{d}{d\alpha}$$
therfore, the second derivative will read
$$\frac{d^2}{d\beta^2} =\frac{1}{f(\alpha)}\frac{d}{d\alpha}\times \frac{1}{f(\alpha)}\frac{d}{d\alpha}= \frac{1}{f^2(\alpha)}\frac{d^2}{d\alpha^2}-\frac{f'(\alpha)}{f^3(\alpha)}\frac{d}{d\alpha}$$
and the third order reads
$$\frac{d^3}{d\beta^3} =-\frac{f''(\alpha)}{f^4(\alpha)}\frac{d}{d\alpha}+\frac{3 f'^2(\alpha)}{f^5(\alpha)}\frac{d}{d\alpha}-\frac{3 f'(\alpha)}{f^4(\alpha)}\frac{d^2}{d\alpha^2}+\frac{1}{f^3(\alpha)} \frac{d^3}{d\alpha^3}$$
I need to find the $n$-th derivative in $\beta$:
$$\frac{d^n}{d\beta^n} =\, ?$$
is there a general formula for this case?

Yes, known as Leibnitz formula. It gives the $n^{\text{th}} $ derivative of a product
$$(F.G)^{(n)}=\sum_{k=0}^n\binom{n}{k}F^{(k)}G^{(n-k)}$$


Calculate coil diameter using length and thickness of the material I'm developing a software for a stainless steel plant, and I need to calculate the diameter of steel coils so the software assigns a position inside a deposit according to the size of the coil.

I have access to a lot of data from the coil, and tried thinking of a way to calculate using the inner hole diameter and the length and thickness of the steel strip.

In a research I've found this formula:

L = 3,141[(D2/2)² - (D1/2)²]/T

Where

  L = Length of the coil

  D1 = Diameter of the inner hole

  D2 = Diameter of the coil

  T = Thickness of the material

But in the software I have to put in this format:

D2 = (formula)

I'm no good with math, how can I "convert" the formula to suit my needs?

What you need to do is called changing the subject of the formula.
$$\begin{align}
L = \frac{3,141\left[(D2/2)^2 + (D1/2)^2\right]}{T}\\
\\
\hline\\
\text{Multiplying by}\; \frac{T}{3,141}\; \text{on both sides}:\frac{TL}{3,141} &= \left[(D2/2)^2 + (D1/2)^2\right]\\
\text{Adding} \;(D1/2)^2 \;\text{on both sides}:\frac{TL}{3,141}+ (D1/2)^2 &= \left[(D2/2)^2 \right]\\
\text{Taking square roots on both sides}:\sqrt{\frac{TL}{3,141}+ (D1/2)^2 }&= \sqrt{\left[(D2/2)^2 \right]}=D2/2\\
\text{Multiplying both sides by}\; 2:2\times\sqrt{\frac{TL}{3,141}+ (D1/2)^2 }&=D2\\
\end{align}$$

L=([(OD+ID)/2]3.14)[(OD-ID)/2]/T
Figure OD being 30 and ID being 20
Figure 24 gauge thick or .024” *note I always
Figure the thickness as being slightly thicker
Due to wraps not ever being perfectly tight so for
.024 I use .026.
Example: 30+20=50/2=25 (25 is the average
Diameter of coil the multiply by pi)
25x3.14=78.5 then round down to nearest inch
Average wrap of coil is 78” long
Then 30-20=10/2=5
5” is the thickness of coil
5/ thickness of steel .026 = 192.3 wraps of coil
Multiply average wrap length 78” x 192.3 wraps
= 14,999.4” of coil / 12 = 1249.9 lineal feet of coil
I’ve used this for years in my business and real world application I’m usually within 10 lineal feet. The trick is the rounding down of the average length and the adding about .02” to the thickness of material.


Definition of increasing at point of a function I came across two definitions which are as follows:

First definition:

A function f is increasing at a point $x_0$ if there is a δ > 0 such that $f(x)⩽f(x_0)⩽f(y)\,\forall x∈(x_0−δ,x_0)\cap D_f$ and $\forall y∈(x_0,x_0+δ)\cap D_f$

Second definition:

A function f is increasing at a point $x_0$ if there is a δ > 0 such that $f(x)<f(x_0)<f(y)\,\forall x∈(x_0−δ,x_0)\cap D_f$ and $\forall y∈(x_0,x_0+δ)\cap D_f$

As per first definition constant function is increasing at every point but as per second it is not which one is correct definition?

Or should we define two terms increasing at point and strictly increasing at point as we use to do for intervals?

One more doubt at end point of domain can we discuss f is increasing/decreasing. I think we can discuss as in this case we will only take one side of neighborhood.

The first definition is increasing function. The second one is strictly increasing function and both are the correct definitions. However, a constant function falls in increasing functions not in strictly increasing functions. That is my opinion.


Riemann problem of non-homogeneous Burgers equation $u_t+uu_x=u$ How to solve $u_t+uu_x=u$ with initial condition $u(x,0)=u_l$ if $x<0$ and $u(x,0)=u_r$ if $x>0$ with $u_l$ and $u_r$ being constant?

This PDE is a particular scalar balance law. Following this post, the entropy condition would imply that the solution is a shock if $u_l>u_r$, and a rarefaction otherwise. Firstly, let us compute the classical solution by applying the method of characteristics. We are left with $\frac{\text d x}{\text dt} = u =\frac{\text d u}{\text dt}$ along the characteristic curves, leading to the sets of curves
$$
x =
\left\lbrace
\begin{aligned}
u_l (e^t - 1) + x_0, \quad x_0<0 \\
u_r (e^t - 1) + x_0, \quad x_0>
0\end{aligned}
\right.
$$
along which $u=u_l e^t$ if $x_0<0$, and $u=u_re^t$ if $x_0>0$.

Case $u_l>u_r$. The solution is classical on both sides of the discontinuity located at the abscissa $x_s$. The discontinuity satisfies the Rankine-Hugoniot condition $\dot x_s(t) = \frac{1}{2}(u_l+u_r)e^t$ with the initial location $x_s(0)=0$, i.e., $x_s(t) = \frac{1}{2}(u_l+u_r)(e^t - 1)$. Therefore, the entropy solution reads
$$
u(x,t) = \left\lbrace
\begin{aligned}
&u_le^t &&\text{if}\quad x< x_s(t) \\
&u_re^t &&\text{if}\quad x> x_s(t)
\end{aligned}\right. ,\qquad
x_s(t) = \frac{u_l+u_r}{2}(e^t - 1) .
$$

Case $u_l<u_r$. The solution is classical on both sides of the rarefaction located between the curves $x=u_l (e^t - 1)$ and $x=u_r (e^t - 1)$ along which $u = u_l e^t$ or $u_r e^t$ is growing exponentially. We look for smooth solutions of the PDE of the form $u(x,t) = v(\xi,t)$ where $\xi$ is a function of $x$ and $t$ to be determined. This assumption yields $$
v_t + v_\xi \left(\xi_t + v \xi_x\right) = v \, ,
$$ and the problem can be reduced by imposing $\xi_t + v \xi_x = 0$ with $v_t = v$. One concludes that $\xi$ is constant along each characteristic curve starting at the origin, while $v$ is exponentially increasing in time. In other words, solving the PDE $v_t = v$ tells us that $v = A(\xi)e^t$ where $A$ is an arbitrary function. Now, we use the method of characteristics to solve the transport equation satisfied by $\xi$ with the boundary condition $A(\xi) = u_l$ at the left side $x = u_l (e^t - 1)$ of the rarefaction wave. Along the characteristic curves, we have
\begin{aligned}
 \bullet\; &\dot t(s) = 1 \text{ and } t(0) = t_0 \text{ yields } t=s+t_0 \\
 \bullet\; &\dot \xi(s) = 0 \text{ and } \xi(0) = \xi_0 \text{ yields } \xi = \xi_0 \text{ s.t. } A(\xi_0) = u_l \\
 \bullet\; &\dot x(s) = A(\xi) e^t \text{ and } x(0) = u_l (e^{t_0} -1) \text{ yields } x = u_l (e^{s+t_0} -1)
\end{aligned}
and therefore, we know $A(\xi) = x/(e^t - 1)$. Thus, the entropy solution reads
$$
u(x,t) = \left\lbrace
\begin{aligned}
&u_le^t &&\text{if}\quad x< u_l (e^t - 1) \\
&\frac{x e^t}{e^t - 1} && \text{if}\quad u_l (e^t - 1)\leq x\leq u_r (e^t - 1) \\
&u_re^t &&\text{if}\quad x> u_r (e^t - 1)
\end{aligned}\right.
$$

For generalizations, see e.g. D. Fusco, N. Manganaro: "A reduction approach for determining generalized simple waves", Z. angew. Math. Phys. 59 (2008) 63-75. doi:10.1007/s00033-006-5128-1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=u$ , letting $u(0)=u_0$ , we have $u=u_0e^s=u_0e^t$

$\dfrac{dx}{ds}=u=u_0e^s$ , letting $x(0)=f(u_0)$ , we have $x=f(u_0)+u_0(e^s-1)=f(ue^{-t})+u(1-e^{-t})$ , i.e. $u=e^tF(x+u(e^{-t}-1))$

$u(x,0)=\begin{cases}ul&\text{if}~x<0\\ur&\text{if}~x>0\end{cases}$ :

$F(x)=\begin{cases}ul&\text{if}~x<0\\ur&\text{if}~x>0\end{cases}$

$\therefore u(x,t)=\begin{cases}ule^t&\text{if}~x<0\\ure^t&\text{if}~x>0\end{cases}$


Find to how many digits the value $\frac{355}{113}$ is an accurate approximation of $3.1415929204$. Find to how many digits the value $\frac{355}{113}$ is an accurate approximation of $3.1415929204$.

What i did was i computed it using a calculator and got the of $\frac{355}{113}$ to be

$3.14159265$

Now i see that up to the $6$th decimal place they are same. So i write as up to $6$ digits the value of $\frac{355}{113}$ is the same as a given approximation of $\pi$.

Am i right? Is this the way to do these type of questions?

Thanks.

It looks like the matching sequence is "3.141592".

You are correct that the approximation is accurate up to 6 decimal places, but I believe the number of digits should be 7, if that is what the question is asking.

I agree with Sam.

I want to add that that number is quite famous (Buffon's needle). You can see it here: https://en.wikipedia.org/wiki/Buffon%27s_needle#Estimating_.CF.80


condition for $f(x)$ to be irreducible in $\mathbb{Q}$ when it is reducible in $\mathbb{Z}$ I have to find a condition on a polynomial $f(x) \in \mathbb{Z}[x]$ such that $f(x)$ is reducible in $\mathbb{Z}[x]$ but irreducible over $\mathbb{Q}$.
My attempt:        Let $f(x)$ be reducible over $\mathbb{Z}$ then for some  $a(x), b(x) \in \mathbb{Z} [x]$, we have $$f(x)=a(x). b(x)$$
           where $0\leq$deg($a(x)$), deg($b(x)$) $<n$ ,and neither $a(x)$ nor $b(x)$ unit in $\mathbb{Z}[x]$
and I know that units in $\mathbb{Z}[x]$ are $1, -1$.
To ensure that this polynomial is reducible over $\mathbb{Z}$ but not over $\mathbb{Q}$ at least one of $a(x)$ or $b(x)$ has to be a constant different from $1,-1$
which clearly means that $f(x)$ cannot be primitive, because if it were so then it wasn't possible to write it as a product of a non unit constant integer and a polynomial with the same degree.
So, $f(x)$ has to be a non primitive polynomial
But, at the back of the textbook I'm using the answer is given"$f(x)$ is primitive"
Can someone tell me what I did wrong to arrive  at the opposite conclusion?

Thanks a lot!

P.S.: Alternate solutions/hints are welcome.

The polynomial $f(x)=2x-2$ is reducible in $\mathbb{Z}[x]$, because both $2$ and $x-1$ satisfy the definition.

On the other hand, $2x-2$ is irreducible in $\mathbb{Q}[x]$.

So it should be easy to find the condition: can there be other cases?

(Probably the book missed a “not”.)

Let's assume that $f(x)\in\mathbb{Z}[x]$ is irreducible when considered in $\mathbb{Q}[x]$, but not in $\mathbb{Z}[x]$. Then we can write a proper factorization
$$
f(x)=a(x)b(x)
$$
in $\mathbb{Z}[x]$, with neither $a(x)$ nor $b(x)$ being invertible. Since this is also a factorization in $\mathbb{Q}[x]$, one of the two factors (we can assume it is $a$) is invertible in $\mathbb{Q}[x]$, so it is a non zero constant in $\mathbb{Z}$.

Therefore the constant $a$ divides all coefficients of $f$ in $\mathbb{Z}$ and so $f$ is not primitive.

Conversely, a non primitive polynomial of positive degree is always reducible in $\mathbb{Z}[x]$.

It is true that the only way an irreducible polynomial over the rationals can be reducible over the integers is that it is not primitive.

Maybe check if you did reproduce the question exactly. The answer in the textbook would answer: give a criterion for an irreducible polynomial over the rationals to be irreducible over the integers, which is almost but not quite the same.


Why is $x^0 = 1$ except when $x = 0$? Why is any number (other than zero) to the power of zero equal to one? Please include in your answer an explanation of why $0^0$ should be undefined.

For non-zero bases and exponents, the relation $ x^a x^b = x^{a+b} $ holds. For this to make sense with an exponent of $ 0 $, $ x^0 $ needs to equal one. This gives you:

$\displaystyle x^a \cdot 1 = x^a\cdot x^0 = x^{a+0} = x^a $

When the base is also zero, it's not possible to define a value for $0^0$ because there is no value that is consistent with all the necessary constraints. For example, $0^x = 0$ and $x^0 = 1$ for all positive $x$, and $0^0$ can't be consistent with both of these.

Another way to see that $0^0$ can't have a reasonable definition is to look at the graph of $f(x,y) = x^y$ which is discontinuous around $(0,0)$. No chosen value for $0^0$ will avoid this discontinuity.

Because when you say 'to the power of zero' it just means that you are dividing that number by itself. Therefore zero to the power of zero is zero divided by zero which is undefined.


The Burger's vortex in 2 Dimension - solving Differential equation After simplifying the vortex equation, I get to this equation:

$$
-\alpha y \partial_y \omega = \alpha \omega + \nu \partial_{yy} \omega
$$

where the $\alpha$ and $\nu$ are constant values and $\omega(y)$. I want to solve the equation for $\omega(y)$. How can I do it? I appreciate in advance for kind helps.

Regards,

Ehsan

$$
-\alpha y \dfrac{d\omega}{dy} = \alpha\omega + \nu \dfrac{d^2\omega}{dy^2}
$$

this is the same as

$$
\nu \dfrac{d^2\omega}{dy^2} + \alpha \dfrac{d}{dy}(y\omega) = 0
$$
thus

$$
\nu \omega ' + \alpha y \omega + C = 0
$$

now if $C=0$ then the solution is separable but if not then we have to solve
$$
\frac{d\omega}{dy} + \frac{\alpha}{\nu} y \omega= -C
$$
using integrating factor
$$
\omega \mathrm{e}^{\frac{\alpha}{\nu}\frac{y^2}{2}} = -\int_0^y C \mathrm{e}^{\frac{\alpha}{\nu}\frac{y'^2}{2}}dy'
$$

or

$$
\omega(y) = -\left(\int_0^y C \mathrm{e}^{\frac{\alpha}{\nu}\frac{y'^2}{2}}dy'\right)\mathrm{e}^{-\frac{\alpha}{\nu}\frac{y^2}{2}}
$$

Maple finds solutions
$$ \omega(y) = \left(c_1 + c_2\; \text{erfi}\left(\sqrt{-\alpha/(2\nu)} y\right)\right) e^{-\alpha y^2/(2 \nu)}$$


Radon-Hurwitz Number I found this interesting segment about the Radon-Hurwitz numbers $\rho (n)$ on Wikipedia:

  "For $N$ written as the product of an odd number A and a power of two $2^B$, write $$B=c+4d, 0 \le c <4.$$
  Then, $$\rho(N)=2^c+8d.$$ In matrix theory, the Radon–Hurwitz number counts the maximum size of a linear subspace of the real n×n matrices, for which each non-zero matrix is a similarity transformation, i.e. a product of an orthogonal matrix and a scalar matrix. "
  Wikipedia link

I was not able to prove this, or to find a proof online. Can someone help prove this?

That passage is clearly wrong, as the union of $\{0\}$ and the set of all scalar multiples of orthogonal matrices does not constitute a real linear subspace. E.g. $\operatorname{diag}(2,0)=\operatorname{diag}(1,-1)+I_2$ is neither zero nor a scalar multiple of a real orthogonal matrix.

$\rho(n)$ is actually the maximal possible dimension of a subspace of real $n\times n$ invertible matrices (i.e. a subspace in which every nonzero matrix is invertible). See Vector subspace of $M_n(\mathbb{R})$ with invertible matrices for some relevant literature.


Length of an arc of a circle when the angle is infinitesimally small The task is to express the length of an arc of a circle trapped between two radii named $r$ if the angle between them is infinitesimally small, named $d\theta$.

The solution to this problem is supposed to be:

$$l=r \cdot {d\theta}$$

but I do not understand why this would be the case.

I've provided a simple Paint sketch for more details:

Since $d\theta$ is infinitesimally small, line $\overline{BE}$ is infinitesimally small as well, so we note that with $\overline {BE} = dx$. Since $dx$ is so small, its projection $\overline {BD}$ and circular chord $\overline{BC}$ can be approximated as:
$$\overline{BD} \approx \overline{BC} \approx dx$$
The similar can be concluded for the radii:
$$\overline{AB} = \overline{AC} = r$$
$$r \approx \overline{AD} \approx \overline {AE}$$

The task was to find the length of an arc limited with the chord $\overline {BC}$, that is, trapped between $\overline{AB}$ and $\overline{AC}$. The solution should be $l=r \cdot {d\theta}$, but how?

Formula for deducing arc length is:
$$l=2r\pi \frac{d\theta}{360°} = r\pi \frac{d\theta}{180°}$$

The only way for this to become $l=r\cdot d\theta$ is if $\pi$ and $180°$ would somehow cancel each other out. $\pi rad$ indeed has the value of $180°$, but this $\pi$ has the meaning of length, that is $\pi \approx 3.14$, and not radians.

If there I did not specify something enough, please let me know, so I can explain myself better. Thank you in advance.

Consider a circle has radius $r$ units.
length of circumference is $2\pi r$ which produces total angle  $2 \pi$ radians.
unit radian angle produce arc length$= \frac{2\pi r}{2\pi} = r$ units
"$\theta$ radian (no matter how small or big it is )" angle produces arc length $= \theta r$.
ANOTHER WAY
Two end points of the arc is very close to each other forming an angle $dA$ in radians. For small angle $\tan dA = dA$. you can also draw a tangent at that point.
For right-angle triangle:
$$\tan dA  = dA =  \text{length of arc / radius} $$
so length of arc $= r dA$ ...


Proof for primitive pythagorean triples I was wondering if there was a proof that any integer pythagorean triple can be represented as a positive integer multiple of a primitive pythagorean triple. This seems quite related to the fundamental theorem of arithmetic, so I was wondering if the proof would be similar.

Hint: If you have a pythgorean triple $(a,b,c)$ divide each number by $gcd(a,b,c)$ and remember if $p$ divides any two of the new numbers it must divide the third.

A multiple of a Pythagorean triple is also a Pythagorean triple,
$$a^2+b^2=c^2\iff(na)^2+(nb)^2=(nc)^2.$$

There is nothing more to say.


Converting nonlinear second order ODE to first order The second order ODE is:
$$
y'' + ky' + (mg/c)\sin(y) = 0
$$
$k$, $m$, $g$ and $c$ and constants.

What is the first order ODE and how do I do it?

Let $x_1 = y$ and $x_2 = \dot{y}$. Then $\dot{x}_1 = x_2$ and $\dot{x}_2 = -kx_2-\frac{mg}{c}\sin(x_1)$. Now stack $x_1$ and $x_2$ into a column vector, i.e. $x = [x_1 \ x_2]^\top$. Then, $\dot{x} = f(x)$ where $f(x)$ is a vector valued function which follows from what I wrote above. This is the first order form.

If y is "small" then sin y ~y
This called linearzation

This is special case .


Can any finite digit of the non-computable function always be a computable? Can any finite digit of the non-computable function always be computable? If so, does this create a contradiction?

Let $\left\{a_n\right\}$ be a non-computable binary sequence and $f:\mathbb{N^+}\longrightarrow \left\{0,1\right\}$ be a  $n'$th digit of the sequence $a_n$ or $f(n):=\left\{a_n\right\}_{n\in\mathbb{N^+}}$. Of course, we have no $n'$th digit formula for $f(n)$. Does this imply, any finite digit of the non-computable binary sequence is also not computable ? Otherwise, for any $n\in\mathbb N^+$, if $f(n)$ is computable, this implies $f(n)$ is computable which gives a contradiction. What are the points I miss?

If I correctly understand your question, each individual digit of your sequence $f$ is 0 or 1. 0 is computable, namely by the program "print 0" (which ignores it input). Similarly, 1 is computable. So each individual digit in your sequence is computable.  None of these trivial facts has any bearing on whether the whole function $f$ is computable.

Hint: A function $\ g:\mathbb{N}^+\rightarrow\{0,1\}\ $ is computable if there exists a Turing machine or equivalent that will eventually return the value of $\ g(n)\ $ when given the value of $\ n\ $ as input.  Suppose $\ h:\mathbb{N}^+\rightarrow\{0,1\}\ $ is non-computable and $\ g:\mathbb{N}^+\rightarrow\{0,1\}\ $ is computable.  Define
$$
a_n=f(n)=\cases{g(n)& if $\ n\in\mathbb{N}^+\ $ is odd,\\
             h\left(\frac{n}{2}\right)& if $\ n\in\mathbb{N}^+\ $ is even.}
$$
Is $\ \left\{a_n\right\}\ $ computable or non-computable?


Given the function, minimize the function subject to the constraint. (Lagrange's Multipliers) Given the function $f(x, y, z) = (x-1)^2 + y^2 + (z+1)^2$:

Minimize $f(x, y, z)$ subject to the constraint $x + y +z = 1$

I know that you need to use Lagrange's Multipliers and thus would need to set $g(x, y, z) = x + y + z$ and find the gradient of both $f(x, y, z)$ and $g(x, y, z)$, but I'm not sure what to do next.

Let $g(x,y,z,\lambda)=(x-1)^2+y^2+(z+1)^2-\lambda(x+y+z-1)$.

Set $\dfrac{\partial  g}{\partial x}=2(x-1)-\lambda=0$,

$\dfrac{\partial  g}{\partial y}=2y-\lambda=0$,

$\dfrac{\partial  g}{\partial z}=2(z+1)-\lambda=0$,

and  $\dfrac{\partial  g}{\partial \lambda}=x+y+z-1=0$.

You should find that $\lambda=2(x-1)=2y=2(z+1)$,

so $y=x-1=z+1$, so $y+1=x$ and $y-1=z$.

Given $x+y+z-1=0$, this means $3y-1=0$, so $y=\dfrac13$, so $x=\dfrac43$, and $z=-\dfrac23$.

(Check that $x+y+z=1$.)

Therefore, the extremum is $(x-1)^2+y^2+(z+1)^2=\left(\dfrac13\right)^2+\left(\dfrac13\right)^2+\left(\dfrac13\right)^2=\dfrac13$.

Note that it occurs when $x-1=y=z+1$.


Modeling path of a rolling ellipse I'm trying to solve Project Euler problem 525. My approach is to find a parametric equation that can model the path of the center point as it rolls, then take the arc length of that function for one rotation.

A rotated ellipse can be expressed with the equation: $$\frac{(x\cos\theta-y\sin\theta)^2}{a^2}+\frac{(x\sin\theta+y\cos\theta)^2}{b^2}=1$$

And in order to make this ellipse lie tangent or "rest" on the x axis when rotated $\theta$ degrees, there is a y shift of: $\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$. Here is a visual representation of this. Therefore the y component of my final equation will be: $y(\theta) = \sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$.

The x component of the equation is harder. After $\theta$ degrees of rotation, the point of the ellipse touching the x axis will have traveled a distance equal to the arc length of the ellipse from 0 to $\theta$. This can be expressed with the arc length formula using the parametric form of an ellipse:

$$L(\theta) = \int_{0}^{\theta} \sqrt{y'(\theta)^2 + x'(\theta)^2} d\theta$$

This almost solves it, but there is an additional horizontal distance between the tangent point and the center point as seen in the picture below.

How do I express this extra x-distance? I think it might not even require any calculus.

(Note, I don't want a solution to the problem, that takes the fun out of the problem. Just some help to get through this step or a point in the right direction.)

One year after you, i'm triyng to solve this problem. I hope you have not yet found the solution...
Please see here a picture: https://imgur.com/a/pcvfBrz

Coordinates of M (acos(t);b(1+sin(t))
Tangent vector T(-a.sin(t);b.cos(t))
Normal vector N(b.cos(t), a.sin(t))

You can then search the coordinates of I = intersection of (MI) and (CI)

With w = (a²sin²(t)+b²cos²(t)) then
I = ( (a-ab/w).cos(t) , (b-ab/w).sin(t) + b )

after you can easily calculate MI and CI which are the values you are looking for.

But after, when you try to calculate the length , it's more complicated.

(dMI/dt)²=a²b²(b²-a²)²sin²(t)cos²(t)\w^3

(dOM/dt+dIA/dt)²=(sqr(w)+(w²-a²b²)/w^(3/2))²

and then... i'm sucked...


What is the smallest integer with 100 non trivial factors? Excluding 1 and itself as factors, what is the smallest integer with 100 different factor values?

Example: 20 has 4 non trivial factors (2*10 and 4*5)

Assuming you mean "exactly $100$ non-trivial divisors".

If the natural number $n>1$ has prime decomposition $n=\prod p_i^{a_i}$, the total number of factors is $$d(n)=\prod (a_i+1)$$

As you wish to exclude $1$ and $n$, you are interested in $F(n)=d(n)-2$.  We see that you are requiring $F(n)=100$ or, equivalently, $d(n)=102$

Now, $102=2\times 3\times 17$.  We therefore seek $\{a_i\}$ with $\prod (a_i+1)=2\times 3\times 17$.  We remark that our $n$ can not have more than three prime divisors.

Case I:  One prime.  Then the best we can do is $n=2^{101}$ which seems unlikely to be minimal.

Case II.  Two primes  Then the best we can do is $2^{16}3^5$, certainly a lot better than $2^{101}$

Case III. Three primes.  Now the best we can do is $2^{16}3^25$, which is only $\boxed {2949120}$.

If it has to be 100 different prime factors, the 10 year olds in my class looked at the pattern of answers starting at 1 prime factor, 2,3,4, 5,6 etc and the size of the numbers (powers of $10$). We decided that as a minimum the answer must have at least $98$ zeros.


Wiener Process $dB^2=dt$ Why is $dB^2=dt$? Every online source I've come across lists this as an exercise or just states it, but why isn't this ever explicitly proved? I know that $dB=\sqrt{dt}Z$, but I don't know what squaring a Gaussian random variable means.

$$dB_t^2 = dt, \qquad (dt)^2 = 0, \qquad dB_t \, dt = 0 \tag{1}$$ are basically rules to simplify the calculation of the quadratic (co)variation of Itô processes - and nothing more:

Let $(B_t)_{t \geq 0}$ a one-dimensional Brownian motion and $(X_t)_{t \geq 0}$ an Itô process, i.e.

$$dX_s = \sigma(s) \, dB_s + b(s) \, ds$$

Then, by Itô's formula,

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, \sigma^2(s) \, ds. \tag{2}$$

The point is: If we simply apply the rules in $(1)$, we obtain

$$dX_s^2 = (\sigma(s) \, dB_s + b(s) \, ds)^2 = \sigma^2(s) \underbrace{dB_s^2}_{ds} + 2b(s) \sigma(s) \underbrace{ds B_s}_{0} + b^2(s) \, \underbrace{ds^2}_{0} \\ = \sigma^2(s) \, ds.$$

Therefore, we can rewrite $(2)$ in the following way:

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, dX_s^2$$

i.e. Itô's formula justifies the calculation rules in $(1)$.

The (mathematical) reason why this works fine can be seen while proving Itô's formula. Actually, one can show that

$$\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2 \tag{3}$$

converges to

$$\int_0^t g(B_s) \, ds$$

as the mesh size $|\Pi|$ of the partition $\Pi=\{0=t_0<\ldots<t_n=t\}$ tends to $0$. This convergence is based on the fact that $B_t^2-t$ is a martingale. On the other hand, comparing $(3)$ with the definition of Riemann-Stieltjes integrals, it's natural to define

$$\int_0^t g(B_s) \, dB_s^2 := \lim_{|\Pi| \to 0}\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2.$$

Consequently,

$$\int_0^t g(B_s) \, dB_s^2 = \int_0^t g(B_s) \, ds.$$

A similar reasoning applies to Itô processes. Note that these integrals, i.e. integrals of the form

$$\int_0^t g(X_s) \, dX_s^2,$$

are exactly the integrals appearing in Itô's formula $(2)$ ($g \hat{=} f''$).

Obviously $dB_t^2 \neq dt$, since $dB_t \sim \mathcal{N} (0, dt)$ is a random variable, while $dt$ is deterministic.

As Michael Hardy said, they really meant to say $\mathbb{E} \left[ dB_t^2 \right] = dt$.  To convince yourself, compute
$$ \mathbb{E} \left[ dB_t^n \right] = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi dt}} \exp\left(-\frac{x^2}{2  dt}\right) x^n dx \, .$$


Why is the nullity of an invertible matrix 0? I am aware that nullity is the dimension of the null space. So I assume that the null space is empty? But why?

The unique vector space of dimension zero is the trivial space $\{0\}$ which includes the zero vector. This space is nonempty, and in fact the empty set is not a vector space because any vector space must have the zero vector.

By the invertible matrix theorem, one of the equivalent conditions to a matrix being invertible is that its kernel is trivial, i.e. its nullity is zero. I will prove one direction of this equivalence and leave the other direction for you to prove.

$(\Rightarrow)$ Suppose $A$ is an invertible $n\times n$ matrix. Let $v\in\ker A$ so that $Av=0$. Then $A^{-1}Av=A^{-1}0\iff Iv=0\iff v=0$. Thus, $\ker A=\{0\}$ so $A$ has nullity zero.

$(\Longleftarrow)$ Suppose that the nullity of a square matrix $(n\times n)$ A is equal to 0.

So this means that for any vector $(n\times 1)$, $Av=0$ has only $1$ solution, which is the trivial one: $v=0$.

And we know that this is equivalent to say that the determinant of A is nonzero.

And finally, because the determinant of A is nonzero, it means it is invertible

This completes the proof of equivalence as Dave already posted the $(\Longrightarrow)$.


Here is an arithmetic sequece find the 20 th term Find the 20th term of this sequence.
$1, 5, 4, 8, 7, 11, 10, 14.........$
it is like add 4 to the st term then subtract 1
i need to find a recursive formula that can help me get the 20th term. Help please.

There are two arithmetic series in the given series. The numbers at odd position constitute the following series

$1,4,7,10,\cdots$

The numbers at even position constitute the following series:

$5,8,11,14,\cdots$

Since $20$ is an even number, the $20^{th}$ term will be the $10^{th}$ term of the second series. And it is $32$.

In Mathematica:

FindSequenceFunction[{1, 5, 4, 8, 7, 11, 10, 14}, 20]

(* 32 *)

and the function derived is:

$\frac{1}{4} (-1)^n \left(3 (-1)^n (2 n+1)+5\right)$


If a linear transformation has domain = range = codomain , is its kernel trivial? Suppose that a linear transformation $L$ has the same domain, range, and codomain. Then $\dim(\text{domain}) = \dim(\text{range}) + \dim(\text{kernel})$ implies that $\dim(\text{domain}) = \dim(\text{range})$.

Can we conclude from here that  $\ker L$ = $\{0\}$?

It is quite important to say whether you work in a finite dimensional space. (In this case what you wrote is true. See, for example, Show that if $W$ is a finite dimensional $K$-vector space any linear surjective map $f:W\to W$ is bijective.)

If you allow infinite dimensional spaces, this is not necessarily true.

Take the space $V=\mathbb R^{\mathbb N}$ of all real sequences. Define $L\colon V\to V$ as
$$L \colon (x_1,x_2,x_3,\dots) \mapsto (x_2,x_3,x_4,\dots).$$
Then the range is the whole space $V$. But the map is not injective. For example, the sequence $(1,0,0,\dots)$ is in the kernel.

Yes. It means the transformation is bijective.


Seven people are interviewed for a possible promotion. In how many orders can the seven candidates be interviewed? Seven people are interviewed for a possible promotion. In how many orders can the
seven candidates be interviewed?

I know the answer is 5040 but don't know how to get to it.

Hint: How many possibilities are there for the first interview? The second? Third? Etc.

The requires input into your scientific calculator.

7!= 5040


Combinatorial proof that binomial coefficients are given by alternating sums of squares? A student recently asked whether there was a combinatorial proof of the following identity:

$\begin{equation*}
\sum^n_{k=1}(-1)^{n-k}k^2 = {n+1 \choose 2}.
\end{equation*}$

I was in a rush and couldn't come up with anything nice off the top of my head. Any ideas or pictures to make this clearer?

More a visual proof than combinatorial, perhaps:
$$ 5^2 - 4^2 + 3^2 -2^2 +1^2 = {6 \choose 2}$$

On the left, you have the alternating sum as an inclusion-exclusion of squares: the total sum is the number of coloured cells.
On the right, you have those L shaped shapes rearranged in the top left of a 6x6 grid.
If you think of each cell as a coordinate $(x_1,x_2)$ that gives two elements chosen from the set $\{1, 2 \cdots 6\}$, it's seen that the elements are choosen  with $ x_2 > x_1$, what corresponds to a combination (no repetition, and no order).
Added: The others are well known, but, just for the sake of completeness...
$$\displaystyle \sum_{k=1}^n (-1)^{n-k} k^2 = {n+1 \choose 2} = \sum_{k=1}^n \; k = \frac{(n+1) \; n}{2}$$

By induction. The formula holds for $n=0$ because both sides vanish.

You can also increase $n$ by one:
$$LHS_n = n^2 - LHS_{n-1} = \dots $$
I had to add the last term $n^2$. However, the signs of all the previous terms were reverted because of the $(-1)^n$ factor, so I had to add a minus sign in front of $LHS_{n-1}$.

Assuming that the equation holds for $n-1$, i.e. that $LHS_{n-1}=RHS_{n-1}$, the displayed formula is
$$ \dots = n^2 - RHS_{n-1} = n^2 - (n-1)n/2 = n(n+1) / 2 = RHS_n$$
which you wanted to be proved.


Definition of class of continuously differentiable functions on a closed interval This is a technical simple question.  The space $C^1([a,b])$ is defined as

$$C^1([a,b])=\{f:[a,b]\to R: \exists\; f' \textrm{ and is continuous on } [a,b]  \}.$$

I think there is some explaining to do here, and is the fact that $f'$ cannot be defined at $a$ or $b$ because those points do not belong to the interior of the domain. The question is:

In the above definition, $f'(a)$ and $f'(b)$ are understood in the sense of right and left derivatives respectively? If not, in which sense?

A derivative can in fact be defined on the edge of a domain. For example, let $f(x): [0, 1] \to [0,1]$ be defined by $x \mapsto x$. Then $f'(0) = \lim\limits_{x \to 0}\frac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0}\frac{f(x)}{x}$. This limit exists and is equal to 1. Proof: let $\epsilon > 0$ be arbritrary. Choose $\delta = \epsilon$. Let $x$ satisfy $0 < |x - 0| < \delta$ and $x \in [0, 1]$, which is $x \in ]0, \delta[$. Then $|f(x)/x - 1| = |x/x - 1| = 0 < \epsilon$.


Is $\mathbb{Z}[√13] $ a Unique factorization domain? I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.

Hint for the title question: We have $${\displaystyle (1+{\sqrt {13}})^{2}=2(7+{\sqrt {13}})}.$$
Since $2$ is irreducible in ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ but does not divide ${\displaystyle 1+{\sqrt {13}}}$, we see that ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ is not a UFD.

$\mathbb{Z}[\sqrt{13}]$ is not even integrally closed (consider for instance $\displaystyle \frac{3+\sqrt{13}}{2}$), Hence it can't be a principal ideal domain. It also can't be a unique factorization domain because a Dedekind domain is a PID if and only if it is a UFD, but $\mathbb{Z}[\sqrt{13}]$ is not even integrally closed thus not Dedekind. At any rate: if you're not integrally closed then you're not a UFD.


Associativity and De Morgan's for more than 2 literals Do logical operators have meaning when used with more than 2 literals "associatively", e.g.:
$(A \land B \land C)$?

I.e., are statements such as $(A \land B \land C)$ meaningful, as opposed to $((A \land B) \land C)$, which is meaningful?

If yes, is it possible to prove De Morgan's for any finite n by induction like so:

For $n = 2$, base De Morgan's applies: $\neg(A_{1} \lor A_{2}) \Leftrightarrow   \neg A_{1} \land \neg A_{2}$.
Assume that for $ n = k $,  $\neg (A_{1}\lor A_{2}\lor \ldots \lor A_{k}) \Leftrightarrow    (\neg A_{1}\land \neg A_{2} \land \ldots \land \neg A_{k})$.
Then for $ n = k + 1 $: $\neg (A_{1}\lor A_{2}\lor \ldots \lor A_{k} \lor A_{k+1})    \Leftrightarrow \neg ((A_{1}\lor A_{2}\lor \ldots \lor A_{k})\lor A_{k+1})\Leftrightarrow \neg    (A_{1}\lor A_{2}\lor \ldots \lor A_{k}) \land \neg A_{k+1} \Leftrightarrow \neg A_{1} \land \neg   A_{2} \land \ldots \land \neg A_{k} \land \neg A_{k+1}$
Therefore, $\neg (A_{1}\lor A_{2}\lor \ldots \lor A_{n}) \Leftrightarrow (\neg A_{1}\land \neg A_{2} \land \ldots \land \neg A_{n})$ is true for any finite n.

?

Thank you.

Certainly $\land$ and $\lor$ can be used meaningfully in this way. This is because we can prove (e.g. by truth tables, a straightforward exercise) that they are associative, i.e.:

$$A \land (B \land C) \iff (A \land B) \land C \qquad A \lor (B \lor C) \iff (A \lor B) \lor C$$

so that we can argue that it doesn't really matter whether we use parentheses or not. From there on, you have given a proper proof that De Morgan's laws generalise to these expressions.

Do note, however, that this does not apply to every logical operator out there. This question deals with associativity of $\to$; it is shown not to be associative:

$$A \to (B \to C) \not\iff (A \to B) \to C$$

No.  For example, (A$\land$B$\lor$C) is not meaningful, even though both $\land$ and $\lor$ associate.

If we have that for any binary operations B1, B2, for all x, y, z, ((xB1y)B2z)=(xB1(yB2z)), then we can drop parentheses as you've suggested.  Thus, if we just have a semigroup, then we can drop all parentheses.  But, if we have more than one binary operation, we can't drop parentheses.

Also, if (A∧B∧C) is an expression, than the rule of uniform substitution fails. The rule of uniform substitution is important for having an axiomatic system of propositional logic. If say A∧B is meaningful (which you implied), and we have the rule of uniform substitution, then substituting B with CvD we obtain A∧CvD. Suppose A=0, C=0, D=1. Then [A∧(CvD)]=0, while [(A∧C)vD]=1. So, I started with an expression assumed as meaningful and deduced a contradiction, which renders either the expression or the rule of uniform substitution as not meaningful.


Proving $(a+b)\log_2(a+b) \ge a\log_2 a +b\log_2 b +2a $ for positive integers $b \ge a$ On the 6th page of the paper "On Induced Subgraphs of the Cube" (PDF link via usdc.edu) (precisely: the last line of Lemma 4.1) the author uses an inequality:
$$(a+b)\log_2(a+b) \ge a\log_2 a +b\log_2 b +2a $$ where $b \ge a$ is known.
I'm not sure how that holds. Can someone explain this calculation to me?
Additional information: $a$ and $b$ are number of vertices of graph (respectively $|V_1|$ and $|V_2|$ in the paper) so they are both positive integers.

With the substitution $b=ka$, $k \ge 1$, the inequality becomes
$$
 (k+1)a \left(\log_2 (k+1) + \log_2 a \right) \ge a \log_2 a + k a \left( \log_2 k + \log_2 a \right) + 2a
$$
which reduces to
$$
 (k+1) \log_2(k+1) \ge k \log_2 k + 2
$$
or
$$
 \frac{(k+1)^{k+1}}{k^k} \ge 4 \, .
$$
The last inequality is true for $k\ge 1$ because
$$
\frac{(k+1)^{k+1}}{k^k} = (k+1) \left( 1 + \frac 1k \right)^k \ge 2 \cdot 2 = 4
$$
(using the binomial formula or Bernoulli's inequality for the second factor).
The fact that $a$ and $b$ are integers is not needed for the estimate, only that $b \ge a > 0$.

This is  true for naturals (The cited paper is on graph theory.), as the result of the command of Mathematica
NMinimize[{(a + b)*Log[2, a + b] - a*Log[2, a] - b*Log[2, b] - 2 a,
b >= a &&{a, b} \[Element] PositiveIntegers}}, {a, b}]

$\{3.552713678800501\cdot 10^{-15},\{a\to 4,b\to 4\}\}$
shows.


How many ways are there to arrange the letters in STATISTICAL? My Work:
This is clearly a permutation so
I used $P(11; 2, 3, 2, 2, 1, 1) = 831600$ ways
Is my reasoning correct?

Well, STATISTICAL is a word of type 1232412456, i.e., $1^22^33^24^25^16^1$, and so the multinomial coefficient $$11\choose {2,3,2,2,1,1}$$ gives the result. Your reasoning is a bit short.

It is a permutation of multiset . There are three T's, two S's, two A's two I's,one C and one L.
$\dfrac{11!}{3!2!2!2!1!1!}$


triple vector product: vector vs gradient I think there's a simple explanation for this, but could not find one from a few online searches. The triple vector product and the curl of $\mathbf{A}\times \mathbf{B}$ have very similar forms, however there are additional terms in the differentiation case:

$
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \bullet \mathbf{C}) - \mathbf{C} (\mathbf{A} \bullet \mathbf{B}) \\
  \nabla \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\nabla \bullet \mathbf{C}) - \mathbf{C} (\nabla \bullet \mathbf{B}) + (\mathbf{C} \bullet \nabla)\mathbf{B} - (\mathbf{B} \bullet \nabla)\mathbf{C}
$

Can someone explain why? I'm familiar with Einstein notation if that helps. Thank you for your help!

I thought it might be useful to present way forward that I prefer.  Here, we begin by introducing notation.

Let $(x_1,x_2,x_3)$ be Cartesian coordinates.  We designate by $\hat x_i$, a unit vector along the $x_i$ axis and by $\partial_i$ the partial derivative with respect to $x_i$.

Then, using the convention of summing over repeated indices, the $i$'th component of the vector $\nabla \times (\vec B\times \vec C)$ can be written as

$$\begin{align}
\hat x_i\cdot \left(\nabla \times (\vec B\times \vec C)\right)&=\hat x_i\cdot \left(\hat x_j\times (\hat x_k\times \hat x_\ell)\right)\partial_j(B_kC_\ell)\tag 1\\\\
&=\left(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}\right)\left(B_k \partial_j(C_\ell)+C_\ell \partial_j(B_k)\right)\tag 2\\\\
&=B_i\partial_j(C_j)-B_j\partial_j(C_i)+C_j\partial_j(B_i)-C_i\partial_j(Bj)\tag 3\\\\
&=B_i(\nabla \cdot \vec C)-(\vec B\cdot \nabla)C_i+(\vec C\cdot \nabla)B_i-C_i(\nabla \cdot \vec B) \tag 4
\end{align}$$

In going from $(1)$ to $(2)$, we used the vector triple product rule (presumed established) while in going from $(2)$ to $(3)$ we used the sifting property of the Kronecker Delta.

Since $(4)$ is true for all $i$, then upon multiplying by $\hat x_i$ and summing over $i$, we find the coveted identity

$$\nabla \times (\vec B\times \vec C)=\vec B(\nabla \cdot \vec C)-(\vec B\cdot \nabla)\vec C+(\vec C\cdot \nabla)\vec B-\vec C(\nabla \cdot \vec B)$$

As stated above, the first expression given is simply product of vectors, which can be expressed in terms of the dot product. The second involves differentiation, acting on a product. The product rule for vector differentiation will inevitably lead to the extra terms.


Is there a solution to this recurrence relation Let $a(n+2) =  -a(n+1) - a(n) - 1$ for all $n\geq 1$. In this case, is there a solution for a(n) and if so, what is it and how do you find it?

It depends of course on $a(0),a(1)$. Suppose $a(0)=a$ and $a(1)=b$ then

$a(2) = -a-b-1$

$a(3) = -(-a-b-1)-b-1 = a+b+1-b-1 = a$

$a(4) = -a-(-a-b-1)-1 = b$

and then the sequence repeats $a(5)$ is again $-a-b-1$.

So we have

$$a(n) = \begin{cases} \displaystyle
  a & n\text{ mod } 3 = 0\\
  b & n\text{ mod } 3 =1\\
-a-b-1 & n\text{ mod } 3 =2\\
\end{cases}$$

A formal proof is by induction on $n$.

Note that this solution makes sense because you can rewrite the equation as $a(n+2)+a(n+1)+a(n)=-1$


$C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$. Let $M_n(\mathbb{C})$ denote the vector space over $\mathbb{C}$ of all $n\times n$ complex matrices. Prove that if $M$ is a complex $n\times n$ matrix then $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.

My Try:

I proved that $C(M)$ is a subspace. But how can I show that it is of dimension at least $n$. No idea how to do it. I found similar questions posted in MSE but could not find a clear answer. So, please do not mark this as duplicate.

Can somebody please help me how to find this?

EDIT: Non of the given answers were clear to me. I would appreciate if somebody check my try below:

If $J$ is a Jordan Canonical form of $A$, then they are similar. Similar matrices have same rank. $J$ has dimension at least $n$. So does $A$. Am I correct?

Your approach is correct. Let $PMP^{-1}=J$ with invertible $P$ and Jordan form $J$. Then
$$
\begin{align}
AM=MA &\Longleftrightarrow PAMP^{-1}=PMAP^{-1} \\
&\Longleftrightarrow PAP^{-1} PMP^{-1} = PMP^{-1} PAP^{-1}\\
&\Longleftrightarrow PAP^{-1} J=   J PAP^{-1}.
\end{align}
$$
Thus, we have
$$
\phi_P : C(M) \rightarrow C(J)
$$
given by $\phi_P(A) = PAP^{-1}$, is an invertible linear transformation with $\phi_P^{-1}(B)= P^{-1}BP$. So, $C(M)$ and $C(J)$ are isomorphic via the isomorphism $\phi_P$. Therefore, $C(M)$ and $C(J)$ have the same dimensions over $\mathbb{C}$.

Then if you prove that $C(J)$ has dimension at least $n$, then the same is true for $C(M)$ too.

Alternatively, as I commented last year, you can also use the general formula given in Centralizer of a Matrix. This gives the following info:

Let $\mathbb{F}$ be a field and $M\in M_n(\mathbb{F})$. Denote by $C(M)=\{A\in M_n(\mathbb{F}) | AM=MA\}$ the centralizer of $M$. The dimension of
$C(M)$ over   $\mathbb{F}$ is given by
$$
\mathrm{dim}_{\mathbb{F}} C(M) = \sum_p (\mathrm{deg}(p))\sum_{i,j} \min (\lambda_{p,i}, \lambda_{p,j}),
$$
where $p$ is any irreducible polynomial that divides the characteristic polynomial of $M$, and $\lambda_p= \sum \lambda_{p,i}$ is the exact power of $p$ in the characteristic polynomial of $M$. Here, $\lambda_{p,i}$ are the powers of $p$ in the primary decomposition of the $\mathbb{F}[x]$-module $\mathbb{F}^n$ in which $x$ acts by $M$-multiplication on the left.

By taking $i=j$ only in the double sum, we obtain that
$$
\mathrm{dim}_{\mathbb{F}}C(M) \geq \sum_p (\mathrm{deg}(p)) \sum_i \lambda_{p,i} = n.
$$

The equality occurs if and only if the minimal polynomial and the characteristic polynomial of $M$ coincide. Moreover, we have in this case,
$$
C(M)=\{f(M) | f\in \mathbb{F}[x]\} = \mathrm{span}_{\mathbb{F}} \{ I, A, \ldots , A^{n-1}\}.
$$

In the case that $M$ has $n$ unique eigenvalues and eigenvectors, there is a fairly straightforward argument. For the eigenvalue $\lambda$, there are vectors $v_L$ and $v_R$ which are the left and right eigenvectors. Construct the $n\times n$ matrix $C_\lambda = v_L v_R^T$: $C_v$ is such that all its rows are scalar multiples of $v_R$ and all its columns are scalar multiples of $v_L$.

Therefore $M C_\lambda = C_\lambda M = \lambda C_\lambda$.

I'd expect to be able to make good headway working with generalised eigenvectors.


Convolution of maximum and minimum of uniform random variables Let $X_1,\ldots, X_n$ be $n$ independent random variables uniformly distributed on $[0,1]$.
Let be $Y=\min(X_i)$ and  $Z=\max(X_i) $. Calculate the cdf of $(Y,Z)$ and verify $(Y,Z)$ has independent components.

Please help me out

Even without computing anything, one can guess at the onset that $(Y,Z)$ is probably not independent since $Y\leqslant Z$ almost surely. Now, the most direct way to compute the distribution of $(Y,Z)$ might be to note that, for every $(y,z)$,
$$
[y\lt Y,Z\leqslant z]=\bigcap_{i=1}^n[y\lt X_i\leqslant z],
$$
hence, for every $0\lt y\lt z\lt1$,
$$
P(y\lt Y,Z\leqslant z)=P(y\lt X_1\leqslant z)^n=(z-y)^n.
$$
In particular, using this for $y=0$ yields, for every $z$ in $(0,1)$,
$$
F_Z(z)=P(Z\leqslant z)=z^n.
$$
Thus, for every $(y,z)$ in $(0,1)$,
$$
F_{Y,Z}(y,z)=P(Y\leqslant y,Z\leqslant z)
$$
is also
$$
F_{Y,Z}(y,z)=P(Z\leqslant z)-P(y\lt Y,Z\leqslant z)=z^n-(z-y)^n\cdot\mathbf 1_{y\lt z}.
$$
One may find more convenient to describe the distribution of $(Y,Z)$ by its PDF $f_{Y,Z}$, obtained as
$$
f_{Y,Z}=\frac{\partial^2F_{Y,Z}}{\partial y\partial z}.
$$
In the present case, for every $n\geqslant2$,
$$
f_{Y,Z}(y,z)=n(n-1)(z-y)^{n-2}\mathbf 1_{0\lt y\lt z\lt1}.
$$

Hint: What is the definition of the cumulative distribution function? The CDF of a random variable $X$ is defined as $$P(X \leq x) \,\,\,\text{for all $x \in \mathbb{R}$}\,\,.$$
Additionally, take advantage of the independence of the $X_i$ (as a set), however, be sure to note that $Y$ and $Z$ are not independent - this is clear, since the minimum of a set is dependent on the other elements of the set and similarly for the maximum of a set.


Definition of the category Sets In the category Sets consider the object (i.e., set) $A$ defined as the positive real numbers, another object (set) $B$ defined as the negative reals, and a function $f$ defined as the negative square root.  So, $f(25)=-5$ and $f(9)=-3$ , etc.

Now define a functor $F$ from Sets to Sets such that $F(A)=-A$, $F(B)=-B$, and $F(f)=f$.

Strangely, we now we have a "function" that takes the negative square root of a negative number and gives a positive real number.  This is clearly impossible in any normal sense of sets and functions.

As stated in Wikipedia, "the category Set is the category whose objects are Sets.  The arrows or morphisms between sets $A$ and $B$ are all triples $(f,A,B) $ where $f$ is a function from $A$ to $B$."

Here is my problem: The functor above from Sets to Sets shows that f need NOT be a function from one set to another set (because the negative square root function cannot map a negative number to the positive reals).  All this violates the definition of the category Sets, and/or the definition of a function, and/or the definition of a set (of numbers).

What am I doing wrong?  Is $F$ not a functor?  (And if not, how would I be able to tell it's not a valid functor?)

EDIT:  Let me restate the second part of my question (in parentheses above), since it has not been answered so far:  Suppose we have a given category $D$ and we are given a pair of functions (f,g) where the domain of f is the objects of $D$ and the domain of g is the arrows of $D$.  This pair may, or may not, constitute a functor from $D$ to some other (unspecified) category.

My question concerns what methods/tests/procedures one can use to determine whether such a pair of functions constitutes a functor, or not.  (Perhaps no general methods exist, or perhaps the question is too broad, but if that is the case, then that should be the answer to my question.)

$F$ is not a functor. By definition of a functor, $F(f)$ should be a map from $F(A)$ to $F(B)$ i.e.
$$F(f): F(A) \rightarrow F(B)$$
$f:A\rightarrow B$ cleary doesn't satisfy this as $F(A) = -A \neq A$ and same for $B$, so you cannot have $F(f)=f$ if $F$ is a functor

Here is a category D and a functor F that allows the kind of "nonsense" morphisms/functions I was trying to demonstrate in my example.  (Thanks to Jens' comments I can see that Sets was not the right choice for the codomain since that implies morphisms must be functions.)

Let D be a 3 object category with A={25}, B={-5}, and C={-10}. We define f:A$\rightarrow$B where f is the negative square root function.  Also g:B$\rightarrow$C where g(x)=2x.

Now consider the following functor that maps D to its dual: Let the functor F map A to C, B to B, C to A, f to g, and g to f.  This means that in the dual, F(g)=f: -5 $\rightarrow$ 25.  But by definition f is a function that takes the negative square of its argument, so how can f=F(g) take -5 to 25? Obviously this does not make sense in our "normal" way of thinking about functions.

This shows that F(g)=F(A)$\rightarrow$F(B) does not have to be "reasonable" because here we have f=F(g):-5 $\rightarrow$ 25.


Prob. 11, Chap. 3 in Baby Rudin: If $a_n > 0$ and $\sum a_n$ diverges, then how do we show that $\sum \frac{a_n}{1+a_n}$ too diverges? Here's Prob. 11, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $a_n > 0$, $s_n = a_1 + \cdots + a_n$, and $\sum a_n$ diverges.
(a) Prove that $\sum \frac{a_n}{1+ a_n}$ diverges. [ I have no clue of how to prove this!]
(b) Prove that $$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq 1- \frac{s_N}{s_{N+k}}$$ [I can show this.] and deduce that $\sum \frac{a_n}{s_n}$ diverges. [How to?]
(c) Prove that $$ \frac{a_n}{s_n^2} \leq \frac{1}{s_{n-1}} - \frac{1}{s_n}$$ and deduce that $\sum \frac{a_n}{s_n^2}$ converges. [This I can show, I think.]
(d) What can be said about (the convergence or divergence of) $$\sum \frac{a_n}{1+ n a_n} \ \ \ \mbox{ and } \ \ \ \sum \frac{a_n}{1+ n^2 a_n}?$$ [ How to answer this?]

I would prefer those answers that use only the machinary developed by Rudin himself upto this point in the book.
Here's what I can show:
Since $a_n > 0$, we have $0 < s_{N+1} < \cdots < s_{N+k}$ and so
$$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq \frac{a_{N+1} + \cdots + a_{N+k}}{s_{N+k}} =  1- \frac{s_N}{s_{N+k}}.$$
As $a_n > 0$, so, for all $n = 2, 3, 4, \ldots$, we have $0 < s_{n-1} < s_n$ and therefore $$ \frac{a_n}{s_n^2} = \frac{s_n - s_{n-1}}{s_n^2} \leq \frac{s_n - s_{n-1}}{s_n s_{n-1}} = \frac{1}{s_{n-1}} - \frac{1}{s_n},$$ and hence $$ 0 \leq \sum_{k=1}^n \frac{a_k}{s_k^2} = \frac{1}{a_1} + \sum_{k=2}^n \frac{a_k}{s_k^2} \leq \frac{1}{a_1} + \sum_{k=2}^n \left( \frac{1}{s_{k-1}} - \frac{1}{s_k} \right) = \frac{1}{a_1} + \frac{1}{s_1} - \frac{1}{s_n} \to \frac{2}{a_1} + 0 $$ as $n \to \infty$ because $a_n > 0$ and $\sum a_n$ diverges and hence $s_n = a_1 + \cdots + a_n \to \infty$ as $n \to \infty$. So if $\lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} $ exists [ but how to show this?}, then we must have $$ 0 \leq \lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} \leq \frac{2}{a_1}.$$
Am I right?

Consider two cases: either there are infinitely many $k$ such that $a_k \geq 2$, or there are only finitely many such $k$.

In the first case, the series $\sum_k \frac{a_k}{1+a_k}$ has infinitely many terms which are at least $\frac23$, so it diverges.

In the second case, except for finitely many $k$, we have that
$$
\frac{a_k}{1+a_k} \geq \frac{a_k}3,
$$
so the series also diverges.

we take $x_n$=$\frac{a_n}{1+a_n}$ . then if $\sum x_n$ is convergent then lim$x_n$=0.

so for $\epsilon$=$\frac{3}{2}$ >0 there exists k$\in$ N s.t $\forall$ n$\geqslant$k  we have|$x_n$| < $\epsilon$

so |$\frac{a_n}{1+a_n}$| <$\epsilon$ so $\frac{a_n}{1+a_n}$<$\epsilon$ as $a_n$ >0 $\forall$ n$\in$ N....so

$a_n$ < $\frac{\epsilon}{1-\epsilon}$=-3 <0 $\forall$ n$\geqslant$k .but we have $a_n$ >0 $\forall$ n$\in$ N.

we have the conclution that if $a_n$ >0 then $\sum\frac{a_n}{1+a_n}$ is divergent. you need not use $\sum a_n$ is divergent.


By definition two sets are equal if and only if they have the same members: $x=y$ if and only if $\forall z(z\in x\leftrightarrow z\in y)$. In your example the sets $\{x\}$ and $\{z\}$ are assumed to be equal, so they must have exactly the same elements. The only element of $\{x\}$ is $x$, and the only element of $\{z\}$ is $z$, so it must be the case that $x=z$.

As was noted in the comments, your example is incorrect: if $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$, then $x\ne z$, so $\{x\}\ne\{z\}$.

$x$ consists of $4$ elements, $z$ consists of $3$ elements. Obviously, they cannot be equal.


Finding the value of Inverse Trigonometric functions beyond their Real Domain I wanted to ask how can we calculate the values of the inverse of trigonometric functions beyond their domain of definition, for example $\arcsin{2}$ beyond its domain of $-\frac{\pi}{2}<x<\frac{\pi}{2}$.
I tried to use the Euler form but did not get much. Thanks.

@Neelesh Vij, he makes an interesting point.  I will do my best to explain.

We have as follows:$$\sin(z)=x, z=\arcsin(x)$$$$e^{iz}=\cos(z)+i\sin(z)$$$$e^{-iz} =\cos(-z)+i\sin(-z)=\cos(z)-i\sin(z)$$And we are trying to solve for an unknown z, given x.

Made clear, the last line has the equation $e^{-iz}=\cos(z)-i\sin(z)$.  This was found with trigonometric identities.

Now subtract the last two equations to get the following:$$e^{iz}-e^{-iz}=2i\sin(z)$$$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$Recall from the very top that $\sin(z)=x$.$$x=\frac{e^{iz}-e^{-iz}}{2i}=\frac1{2i}e^{iz}-\frac1{2i}e^{-iz}$$Multiply both sides by $e^{iz}$ to get a solvable quadratic.$$xe^{iz}=\frac1{2i}e^{2iz}-\frac1{2i}\to0=\frac1{2i}e^{2iz}-xe^{iz}-\frac1{2i}$$We can now use quadratic formula to solve for $e^{iz}$.  If you are confused, try substituting $e^{iz}=y$.$$e^{iz}=\frac{x\pm\sqrt{x^2-1}}{\frac1i}=i(x\pm\sqrt{x^2-1})$$Now, solve for z.$$iz=\ln[i(x\pm\sqrt{x^2-1})]=\ln(i)+\ln[(x\pm\sqrt{x^2-1})]=\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}],n=\pm(0,1,2,3,4,\ldots)$$

We get a grand and wondrous solution:$$z=\frac{\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}]}i,n=\pm(0,1,2,3,4,\ldots)$$So for $x=2$, we can find $z=\arcsin(2)$.$$\arcsin(2)=\frac{\pi(\frac12+2in)+\ln[2\pm\sqrt{2^2-1}]} i,n=\pm(0,1,2,3,4,\ldots)$$

Lastly, take note that $\arcsin(x)=\arccos(x)-\frac{\pi}2\pm2\pi m, m=0,1,2,3,4,\ldots$ so that we have:$$\arccos(x)=\frac{\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}]}i+\frac{\pi}2\pm2\pi m, m=0,1,2,3,4,\ldots,n=\pm(0,1,2,3,4,\ldots)$$

Let us start with the well known Euler form:$$e^{i\theta}=\cos\theta+i\sin\theta$$Now we solve for $cos\theta$.$$\cos\theta=e^{i\theta}-isin\theta$$Now take the $arccos$ of both sides.$$\theta=\arccos(e^{i\theta}-i\sin\theta)=\arccos[f(\theta)]$$Attempt to find the inverse of $f(\theta)$.$$f(\theta)=e^{i\theta}-i\sin\theta=x$$I do not have the skills to do this, if it is possible.  I will assume I have the answer to be $\theta=f^{-1}(x)$.  You would then substitute it up above.$$\theta=\arccos[f(\theta)]$$$$f^{-1}(x)=\arccos[f(f^{-1}(x))]=\arccos(x)$$

And that is your solution.


Different definitions of a limit of a function in $\mathbb R$. Are they equivalent or not? If a limit $L$ of a function $f:A\to\mathbb R$ exists at a point $a\in \mathbb R$, where $A\subset\mathbb R$ is a proper subset of the set of real numbers, is there any difference between the statements "the limit exists when $a$ is an accumulation point" and "the limit exists when $f$ is defined on a deleted neighbourhood of $a$"?

My motivation for asking this question is because of an answer to a previous question of mine.

You cannot approaches to any $a \in Dom(f)$. $a$ must be a limit point but this  not implies that the limit exists. e.g., $f(x)=\frac{1}{x}$, $x \in [-1,1]=A$,  $x=0$  is a limit point for $A$ but as $x\to 0$, $f\to \pm\infty$.


How to prove this function is increasing $\frac{e^x - x -1}{x^2}$ How to prove that the function $\displaystyle h(x) = \frac{e^x - x - 1}{x^2} = \sum_{k=2}^{\infty} \frac{x^{k-2}}{k!}$ is increasing?
Edit for context: I was asking because this function appears in Wainwright's high dimensional statistics textbook, as part of proving a Bernstein type probability inequality for random variables that are bounded on one side.

The function should be extended by continuity at $0$ with $h(0)=1/2$ and it becomes differentiable at $0$ with $h'(0)=1/6$.
For $x\ne0$, it holds
$$
h'(x)=\dfrac{(x-2)e^x+x+2}{x^3}
$$
and we can consider $g(x)=(x-2)e^x+x+2$. Then $g'(x)=(x-1)e^x+1$ and $g''(x)=xe^x$. Thus we infer that $g'$ has a minimum at $x=0$, with $g'(0)=0$. Therefore we have $g'(x)>0$ for $x\ne0$, which proves that $g$ is strictly increasing.
Since $g(0)=0$, we conclude that $g(x)<0$ for $x<0$ and $g(x)>0$ for $x>0$.
Therefore $h'(x)>0$ for every $x$.

For $x\not =0$ we have :
$$\int_0^1 v\exp(uvx)du=\frac{\exp(vx)-1}{x}$$ and
$$\int_0^1 \frac{\exp(vx)-1}{x}dv =\frac{1}{x}(\frac{\exp(vx)}{x}-v)_0^1=\frac{\exp(x)-1-x}{x^2}$$
Hence with $D=[0,1]\times [0,1]$
$$\int \int_Dv\exp(uvx)dudv= \frac{\exp(x)-1-x}{x^2}$$ and this is also true for $x=0$. Now use that the function $x\to v\exp(uvx)$ for fixed $u,v\in [0,1]$is increasing.


Total distance traveled when visiting all rational numbers My students found an old problem given in my school in 2007 (probably from a Honor Calculus class) and had been trying to solve for some time. Here is the problem:

  Prove or disprove: there exists a bijection $a$ from $\mathbb N$ onto $\mathbb Q$ such that $\sum_{n=1}^\infty (a_n-a_{n+1})^2$ is convergent. (With $a_n=a(n)$)

I have to confess that I am clueless on the method to deal with this problem. My intuition is that the sum represents the square of the distance traveled when visiting all rational numbers, but there is so many rationals that the sum should be infinite. On the other hand, the density of Q means that I can travel the distance between consecutive rationals of my travel can be arbitrarily small, so I am confused...

The only thing we could prove is that $\sum(a_n-a_{n+1})$ is divergent (otherwise, $\{a_n\}$ would be bounded, a contradiction).

Any clue is welcome.

My intuition says yes (there exists). Choose $a$ so that $a_n-a_{n+1}\leq \frac{1}{n}$; since $\sum_n\frac{1}{n^2}$ converges, if you can indeed choose one such $a$ you should be fine. Now, because $\sum_n\frac{1}{n}$ diverges, you could in principle take steps that are roughly that length and still get to 'cover all of $\mathbb{Q}$'. I might try and make this more precise.

EDIT: Making it precise.

Choose any bijection $f:\mathbb{N}\longrightarrow \mathbb{Q}$. We will obtain $a$ from $f$ inductively as follows.

First, let $a(1)=f(1)$. Now, if $|f(2)-a(1)|\leq \frac12$, we set $a(2)=f(2)$. Otherwise, we set $a(2)=a(1)\pm\frac12$, whichever sign brings $a(2)$ closer to $f(2)$. Then, if |$f(2)-a(2)|\leq \frac13$, we set $a(3)=f(2)$; otherwise we set $a(3)=a(2)\pm \frac13$, whichever signs brings $a(3)$ closer to $f(2)$. We proceed in this manner until we each $f(2)$, then we move onto $f(3)$, and so on.

Observations:

Because $\sum_n\frac{1}{n}$ diverges, we will always reach the next $f(n)$ in finite time.
Of course, since $f$ is a bijection, our intermediate $a(k)$'s may step onto some of the $f(n)$'s prematurely. If, after reaching some $f(n)$, $f(n+1)$ has already been previously assigned to some $a(k)$, we simply skip $f(n+1)$ and proceed to $f(n+2)$. Notice that, at any point in the induction, $a$ will be defined at a finite number of points so there may be at most a finite number of skips, and the process can always continue.
Finally, if when defining $a(k+1) = a(k) \pm \tfrac1k$ the RHS is already taken, simply choose another rational $r$ near $a(k) \pm \tfrac1k$, say $\left|r-\left(a(k) \pm \tfrac1k\right)\right|\leq\tfrac{1}{2k}$, and put $a(k+1)=r$.

These observations guarantee that this inductive process can always continue and will eventually reach every rational. It uses the axiom of choice near the end, and I believe there's a way not to use it: look for

\begin{align}
&a(k) \pm \tfrac1k\\
&a(k) \pm \tfrac1k \mp\tfrac{1}{2k}\\
&a(k) \pm \tfrac1k \mp\tfrac{1}{2k}\pm\tfrac{1}{4k}\\
&a(k) \pm \tfrac1k \mp\tfrac{1}{2k}\pm\tfrac{1}{4k}\mp\tfrac{1}{8k}\\
&\dots
\end{align}

until you find one that's not yet been used. But honestly I'm not too worried about it.

My intuition says no.

Let $a_{n_1} := 0, a_{n_q} := q \in \mathbb{N}$

Then $ q^2 = (a_{n_1} - a_{n_q})^2 \leq \sum_{n = 1}^{\max(n_1,n_q)} (a_n - a_{n+1})^2 $ By the triangle inequality.

We know that $n_q \to \infty$ as $q \to \infty$ (because $a$ is a bijection, and the rationals are unbounded)

Hence $\sum_{n = 1}^{\infty} (a_n - a_{n+1})^2$ diverges.


Question involving some roots of the equation $31z^{15} - z^{10} + 32 = 0 $ Consider the equation $$31z^{15} - z^{10} + 32 = 0.$$ What would be the sum of all those roots of the equation whose real part is positive? Only trivially trying to solve the equation I find not helpful. Even factorizing we get , by putting $z^5=x $ ,

$31x^3 - x^2 + 32 = (x+1)(31x^2 - 32x + 32) =0 $ which is of no help .

$31z^{15}-z^{10}+32=0\implies 31(z^{15}+1)-(z^{10}-1)=0$

$31(z^5+1)(z^{10}-z^5+1)-(z^5+1)(z^5-1)=0$

$(z^5+1)\{31(z^{10}-z^5+1)-(z^5-1) \}=0 $

$(z^5+1)\{31z^{10}-32z^5+32 \}=0 $

If $z^5+1=0, z^5=-1=e^{(2m+1)\pi i} $ where $m$ is any integer.

So, $z=e^{\frac{(2m+1)\pi i}5}$ where any $5$ in-congruent values of $m\pmod 5$ will give us essentially the same set of $5$ distinct solutions, find the explanation here, the simplest example can be $0,1,2,3,4$.

Now, $e^\frac{(2m+1)\pi i}5=\cos \frac{(2m+1)\pi}5+i\sin \frac{(2m+1)\pi}5$

For the real part to be positive, $-\frac \pi 2<\frac{(2m+1)\pi}5< \frac \pi 2$

$\implies -\frac74<m<\frac 3 4\implies m=-1,0$

So, those roots are $\cos \frac{\pi}5\pm  \sin \frac{\pi}5,$ the sum being $2\cos \frac{\pi}5$

If $31z^{10}-32z^5+32=0,z^5=\frac{16\pm12\sqrt 5 i}{31}$

Let $r\cos A= \frac{16}{31}--->(1), r\sin A=  \frac{12\sqrt 5 }{31}--->(2)$ where $r>0$

so that $z^5=r(\cos A\pm i\sin A)=re^{\pm iA}=re^{i(2n\pi\pm A)}$ where $n$ is any integer.

$z=r^\frac 15 e^{\frac{i(2n\pi\pm A)}5}$ where $0\le n<5$

Dividing $(2)$ by $(1),\tan A=\frac{12\sqrt 5}{16}=\frac{3\sqrt5}4$

So, $A=\arctan \frac{3\sqrt5}4$ where $A$ lies $\in(0,\frac \pi 2)$ as $\cos A,\sin A>0$

Squaring and adding  $(2)$ and $(1), r^2=\left(\frac{16}{31}\right)^2+\left(\frac{12\sqrt 5 }{31}\right)^2$

The equation $31x^3-x^2+32=0$, has the roots,
$$
-1\qquad \frac{4}{31(4 + i\sqrt{46})}\qquad \frac{4}{31(4 - i\sqrt{46})}
$$
so take all them to the $1/5$-th power and sum there real parts.


Are there two non-congruent quadrilaterals with same sets of sides and angles? Are there two non-congruent quadrilaterals with same sets of sides and angles?

It is relatively easy to construct such pentagons. Trying to construct quadrilaterals allows for seemingly multiple degrees of freedom, yet my every attempt ends up not working out.

Yes.  There are two degrees of freedom in the solution space, so we can add two extra constraints to make the computation easier.  Here’s a solution with a 90° angle and a 45° angle:

The side lengths, from longest to shortest, are
$$\begin{align*}
1&, \\
α &\approx 0.838045, \\
(2 + \sqrt2) + (6 + 4\sqrt2) α + \sqrt2 α^2 - (12 + 8\sqrt2) α^3 &\approx 0.454531, \\
(3 + 3\sqrt2) + (9 + 7\sqrt2) α - (2 + 2\sqrt2) α^2 - (16 + 12\sqrt2) α^3 &\approx 0.284498,
\end{align*}$$
where $α$ is the positive root of
$$1 - \sqrt2 - 2α - 2α^2 + (4 - \sqrt2)α^3 + 4α^4 = 0.$$

No, two non-congruent quadrilaterals with same sets of sides and angles don't exist.
Let's prove this:

Suppose that there exist such two quadrilaterals $ABCD$ and $A^{'}B^{'}C^{'}D^{'} $.
We have $AB=A^{'}B^{'}, AD=A^{'}D^{'}$ and angles $A$ and $A^{'}$ are congruent, therefore $\triangle ABD \equiv \triangle A^{'}B^{'}D^{'}$, therefore $BD = B^{'}D^{'}.$
Analogously, we have $AC=A^{'}C^{'}$

Therefore, both quadrilaterals have same length of sides, diagonals, and same angles between any two un-opposed sides and, because of the earlier-proven congruences, same angles between sides and diagonals. Therefore, the two quadrilaterals are congruent, which is contradiction. Q.E.D


No.  The problem is very simple: there are more huge numbers than abbreviated forms.  Suppose that I allow you to use the numerals 0-9, the English alphabet, and spaces to describe whatever numbers you want; that's still only $37$ characters. There are only $37^{140}$ expressions you can write down using these characters which is $140$ characters or shorter (the length of a tweet!), so there are only $37^{140}$ numbers that you could conceivably tweet to someone else.  In particular, since $\log_{10} 37^{140} = 219.5$, this implies that there is at least one number with $220$ digits which is impossible to tweet to someone else without using something other than numerals and English.

All right, I'll be even more generous: I'll let you use all $127$ ASCII characters. That means there are $127^{140}$ possible tweets, some of which mathematically describe some kind of number, and since $\log_{10} 127^{140} = 294.5$, there is at least one number with $295$ digits which is impossible to tweet to someone else (again, assuming that Twitter uses ASCII).  I could give you the entire Unicode alphabet and you still won't be able to tweet, for example, the kind of prime numbers that public key cryptosystems use.

These kinds of questions are studied in computer science and information theory; you will find the word Kolmogorov complexity tossed around, as well as the word entropy.

The way I understand this question, the answer is "yes." Simply write the number in, say, hexadecimal notation, and it will always be shorter than in decimal (if it is >=100000). If you are looking for the shortest representation, however, that is well defined only if you specify exactly what counts as a valid representation of a number. After all, every number you come up with can technically be represented in just one digit if you choose a sufficiently large base (at least one higher than the number itself).

About the standard way to reduce it: I'd say the decimal system is the standard way of reducing numbers from just writing the appropriate number of dashes. :-) In the sense pointed out by Bill Dubuque, there is no representation that is more space efficient in principle, that is if you transfer the problem to a more abstract domain. If, instead, you are looking for the shortest representation using a given set of mathematical symbols, then no, there is no standard way to find it. I also suspect no algorithm can do better than brute force search (that is, first check all numbers representable using 1 character, then those representable in 2 characters, and so on). In any case, nobody would write a number as "loge 56 (log 56 base e) when we discard first 6 numbers and then consider first 200 digits," so no conceivable set of standard symbols will account for that anyway.