How can I calculate the total number of possible anagrams for a set of letters? How can I calculate the total number of possible anagrams for a set of letters?

For example:

"Math" : 24 possible combinations.

math
maht
mtah
mtha
mhat
mhta
amth
amht
atmh
athm
ahmt
ahtm
tmah
tmha
tamh
tahm
thma
tham
hmat
hmta
hamt
hatm
htma
htam
Total: 24

I generated this by actually generating each combination, one by one, but I want to know the formula for calculating the total possible number of combinations of any group of letters.

Suppose you have word with $n_1$ letters $a_1$, $n_2$ letters $a_2$, $\cdots$, $n_k$ letters $a_k$.

First give each letter unique pair: you have first letter  $a_1$, it will be $(a_1,1)$, second $a_1$ will be $(a_1,2)$ ,$\cdots$ ,$n_1$th letter $a_1$ will be $(a_1,n_1)$, first letter $a_2$ will be $(a_2,1)$ etc...

Now you have $(n_1+n_2+\cdots+n_k)!$ possible combinations of pairs.

Next it's easy to get letters combinations from pair combination: note that for each $s$ there is $s!$ combinations of pair $(a_s,i)$, but only one combinations of letters, so finally you have:

$$\frac{(n_1+n_2+\cdots+n_k)!}{n_1!n_2!\cdots n_k!}$$

Possible combinations of letters.

$n!$, that is factorial of $n$, where $n$ is the number of digits.

If $n=4$, then $4! = 1\times 2\times 3\times 4$

$n! = 1\times 2\times 3\dots\times n$


Set of all inner automorphisms is a normal subgroup In order to prove this, I first proved that the set of all automorphisms from a group $G$ to $G$ form a group under composition:
The identity homorphism is an automorphism because sends $x$ from $G$ to $x$ from $G$:

$\phi_e(x) = x \in G$

The inverse of an homomorphism exists:

well...

The associativity works because composition of functions is always associative.

Closure property:

$$\phi\circ\gamma(a+b) = \phi(\gamma(a+b)) = \phi(\gamma(a)+\gamma(b)) = \phi(\gamma(a)) + \phi(\gamma(a)) = \phi\circ\gamma(a)+\phi\circ\gamma(b)$$

So, an Inner automorphism is defined as a function $f$ such that

$f(x) = a^{-1}xa$

for a fixed element $a$ from $G$.

I'm supposed to prove that the set of all these automorphisms form a normal subgroup of $G$, that is:

$$gfg^{-1}\in N$$ for all $g$

where $N$ is the set of all inner automorphisms of $G$, and $g$ is an automorphism of $G$.

  UPDATE:

  Ok, so what I learned from this is that

  we have a group $Aut(G)$ made of all the automorphisms of $G$, and we
  want to show that the subgroup of $Aut(G)$ made of all the inner
  automorphisms, is normal. That is, given an inner automorphism
  $\phi_a(x) = a^{-1}xa$ for a fixed $a$ in $G$ and $x\in G$.

  So, we need to show that, given $g$ as an automorphism from $Aut(G)$
  and $\phi_a$ an inner automorphism from the subgroup $N$ of inner
  automorphisms, we must have:

  $$g^{-1}\phi_ag \in N$$

  for all $g$

  Am I rigth?

For any given $a \in G$, conjugacy by $a$, namely the map $\varphi_a \colon G \to G$ defined by $\varphi_a(g):=aga^{-1}$, is an automorphism of $G$. In fact, $\varphi_a(g)=\varphi_a(g') \Rightarrow aga^{-1}=ag'a^{-1} \Rightarrow g=g'$ (injectivity); then, for evey $g\in G, \varphi_a(a^{-1}ga)=g$ (surjectivity); finally, $\varphi_a(gg')=agg'a^{-1}=aga^{-1}ag'a^{-1}=(aga^{-1})(ag'a^{-1})=\varphi_a(g)\varphi_a(g')$ (homomorphism). Therefore, $\Phi:=\{\varphi_a, a \in G\} \subseteq \operatorname{Aut}(G)$.

Now, $(\varphi_a\varphi_b)(g)=\varphi_a(\varphi_b(g))=a(bgb^{-1})a^{-1}=(ab)g(ab)^{-1}=\varphi_{ab}(g)$, whence $\varphi_a\varphi_b=\varphi_{ab} \in \Phi$ (closure); then, $(\varphi_a\varphi_{a^{-1}})(g)=\varphi_a(\varphi_{a^{-1}}(g))=a(a^{-1}ga)a^{-1}=g$, whence $\varphi_a^{-1}=\varphi_{a^{-1}}\in \Phi$ (closure by inverses). Therefore, $\Phi \le \operatorname{Aut}(G)$.

Let's now prove that $\Phi \unlhd \operatorname{Aut}(G)$. Well, $\forall a,b \in G, \forall \sigma \in \operatorname{Aut}(G)$, we get: $(\sigma^{-1}\varphi_a\sigma)(b)=\sigma^{-1}(\varphi_a(\sigma(b)))=\sigma^{-1}(a\sigma(b)a^{-1})=$ $\sigma^{-1}(a)b\sigma^{-1}(a^{-1})$; call $\tau:=\sigma^{-1} \in \operatorname{Aut}(G)$, then $(\sigma^{-1}\varphi_a\sigma)(b)=\tau(a)b\tau(a^{-1})=\tau(a)b\tau(a)^{-1}=\varphi_{\tau(a)}(b)=\varphi_{\sigma^{-1}(a)}(b)$, so that $\sigma^{-1}\varphi_a\sigma=\varphi_{\sigma^{-1}(a)} \in \Phi$. $\Box$

$\Phi$ is precisely $\operatorname{Inn}(G)$.

First, you used additive notation for the group operation, but we needn't suppose that the group $G$ is abelian, the result holds in the general case.

Second, group automorphisms $Aut(G)$ is a different group from $G$ itself (so, inner automorphisms do not form a subgroup in $G$...).

The set/class of all inner automorphisms $Inn(G)$ form a normal subgroup in $Aut(G)$, not in $G$ itself!

Conjugation operation, you called it $f(x)$, more precisely $f_a(x)$ or $x^a$, is a group homomorphism $G\rightarrow Aut(G)$ (not necessarily isomorphism or even mono- or epi-).
And the whole image of a group homomorphism is always a normal subgroup in the codomain (the proof is a bit verbose).

In this case the whole image of the conjugation operation is exactly the set of all inner automorphisms (forming a subgroup in $Aut(G)$).

$$ f: G\rightarrow Aut(G) $$

$$ f_a(x) := a^{-1}xa $$

$$ f_G = Inn(G) $$

$$ Inn(G) \triangleleft Aut(G) $$

So we can indeed always speak about outer automorphisms $Out(G)$ of a group $G$, which is possibly trivial, because $Inn(G)$ exhaust the whole $Aut(G)$ (all automorphisms are inner), or, conversely, equal (isomorphic) to $Aut(G)$, because $Inn(G)$ is trivial (in case of abelian groups).


Probability that 3 points in a plane form a triangle This question was asked in a test and I got it right. The answer key gives $\frac12$.

  Problem: If 3 distinct points are chosen on a plane, find the probability that they form a triangle.

Attempt 1: The 3rd point will either be collinear or non-collinear with the other 2 points. Hence the probability is $\frac12$, assuming that collinearity and non-collinearity of the 3 points are equally likely events.

Attempt 2: Now suppose we take the midpoint (say $M$) of 2 of the points (say $A$ and $B$). We can draw an infinite number of lines passing through $M$, out of which only 1 line will pass through $A$ and $B$. Keeping this in mind, we can choose the 3rd point $C$ on any of those infinite lines, excluding the one passing through $A$ and $B$. Now it seems as if the probability will be tending to 1.

What is wrong with attempt 2? Or is the answer actually 1 and not $\frac12$?

There is no such thing as a uniform distribution on the plane.  Without specifying how the points are chosen, the question is not properly stated.  However, if the points are chosen independently from some continuous distribution (absolutely continuous with respect to Lebesgue measure), the probability of the third point lying exactly on the line through the first two is $0$.

A possibly silly way to finesse the ambiguity in the initial distribution: put the points on the projective plane. Then uniform distribution is well-defined and clearly, in the presence of uniform distribution, it is legitimate to use it to define a set of three points chosen randomly. This, of course, again leads to the probability of three points forming a triangle being 1.


How to prove or disprove that $\gcd(ab, c) = \gcd(a, b) \times \gcd(b, c)$? I'm new to proofs, and am trying to solve this problem from William J. Gilbert's "An Introduction To Mathematical Thinking:  Algebra and Number Systems". Specifically, this is from Problem Set 2 Question 74. It asks:

  How to prove or disprove that $\gcd(ab, c) = \gcd(a, b) \times \gcd(b, c)$?

What I've tried is to use the proposition that $\gcd(a, b) = ax + by$ to rewrite the whole equality, but I can't manage to equate the two statements.

Any help would be appreciated.

Notice if $a = b = c = 3$, then

$$ \gcd(ab,c) = \gcd(9,3) = 3 $$

while

$$ gcd(a,b) \times gcd(b,c) = gcd(3,3) \times gcd(3,3) = 3 \times 3 = 9 $$

$$  \therefore gcd(ab,c) \neq gcd(a,b)\times gcd(b,c) $$

just a few remarks:

note: i use the standard abbreviation $(a,b)$ for the GCD of $a$ and $b$

a. it is possible to approach a problem in the wrong way. occasionally this may lead to startling insight, but usually it leads to a waste of time, and can even be demotivating. here, OP's introduction of:

$$
(a,b) = ax+by
$$
suggests that someone should explain why this property is unlikely to be helpful in the present case, or, alternatively, find a proof which is based on this approach.

b. sometimes an assertion smells fishy - an intuition which can be wrong, but is often correct. it directs us to look for a counterexample. finding counterexamples is a highly useful ability, which some have more of than others. psychologically, the heuristics of finding counter-examples are different from those of finding a proof of a result, and different again from those of making a discovery.

c. the assertion in this problem smells fishy, and it might be more interesting if, say, the question had been a request to evaluate:
$$
\frac{(a,c)(b,c)}{(ab,c)}
$$
d. sometimes the sense of pain or aversion one feels on examining a problem (if you introspect the workings of the mind, which is inherently lazy, at least in my case) can actually serve as a guide to how the problem should be approached. i think the sense of "Oh, no!" i feel when i see the terms GCD or LCM is due to the fact that i like working with fields, rings, ideals and so on, but feel uncomfortable with boolean algebra. questions to do with division arise in an algebraic context of the ideals of $\mathbb{Z}$, but are often really about lattice properties.

e. so, given my mental unease, and OP's evident lack of experience, perhaps the following approach may generate some insight. i will assume the prime decomposition properties of $\mathbb{Z}$. if anyone objects i suggest they prove the necessary assertions as a lemma, and then continue as here.

let $\mathbb{P}$ be the set of prime numbers, and let $\mathbb{N}^+$ denote the set of positive integers. we define the map:
$$
\psi :\mathbb{P} \rightarrow \mathbb{N}^{\mathbb{N}^+}
$$
we use the abbreviation $\mathfrak{p}_n$ for the more unwieldy $\psi(\mathfrak{p})(n)$. intuitively $\mathfrak{p}_n$ is the exponent of the highest power of $\mathfrak{p}$ which divides $n$.

LEMMA  we use the symbol $m \land n$ for min$(m,n)$

for any $\mathfrak{p} \in \mathbb{P}$
$$
\mathfrak{p}_{(a,b)} = \mathfrak{p}_a \land  \mathfrak{p}_b
$$
and
$$
\mathfrak{p}_{ab} = \mathfrak{p}_a +  \mathfrak{p}_b
$$
the claim in the question can now be expressed as a spurious assertion of distributivity, but which can be improved to:
$$\\ $$
THEOREM

$\forall \mathfrak{p} \in \mathbb{P}$, and $a,b,c \in \mathbb{N}^+$ we have:
$$
(\mathfrak{p}_a+\mathfrak{p}_b) \land \mathfrak{p}_c \le
\mathfrak{p}_a \land \mathfrak{p}_c + \mathfrak{p}_a \land \mathfrak{p}_c
$$
perhaps someone can produce an aesthetically satisfying demonstration?


Is there a standard way of quantifying "closeness to being symmetrical"? Consider these three polygons:

Object #1 is clearly symmetrical. Object #2 is not, but in some sense it seems "close" to being so (in that it is not very different from object #1). Object #3 appears to be very far from being symmetrical.

Is there a standard way of quantifying these observations, i.e., a measure for "closeness to symmetry"?

The work of Prof. David Avnir, which works in the meaurement of shape, symmetry and chirality in molecules and related objects (orbitals, electron density,...) might interest you:

http://chem.ch.huji.ac.il/avnir/p_symmetry.html

http://www.csm.huji.ac.il/new/

http://www.ee.ub.edu/index.php?option=com_content&view=article&id=72&Itemid=469

Let's pretend we have defined rigorously what we mean by a symmetric shape. Let's also pretend we have some sensible metric metric on the set of shapes in $\mathbb R^2$. Then we can measure how symmetric the shape $A$ is using the non-negative number. . .

$|A|= \inf \{d(A,B): B$ is symmetric$\}$

Symmetric shapes will always satisfy $|A|=0$ because $d(A,A)=0$. Whether there are non symmetric shapes will depend on the measure and the exact notion of symmetry.

So what kind of metrics could we use? Well there's always the Hausdorff metric on $\mathbb R^2$: For any two shapes (subsets) $A$ and $B$ we define the distance. . .

$d(A,B) = \inf \{\varepsilon : $every point of $A$ is within distance $\varepsilon$ of some point of $B$ and vice-versa$\}$

This is a good metric because according to it a 'shape' can be any closed subset of the plane. However the Hausdorff metric might give $(1)$ and $(3)$ as being very far apart because the tip of $(3)$ is very far from the edge of the square would be.

Another metric for shapes is by taking the symmetric difference. .  .

$A \Delta B = \{x \in \mathbb R^2 : x$ is an element of exactly one of $A$ or $B$ $\}$

and calculating its area:

$d(A,B) =$ The area of $A \Delta B$

This metric is more likely to give $d((1),(3))< d((1),(2))$. However it only allows for a narrower definition of what a 'shape' is. For example a disc will be distance $0$ from a 'lollipop' shape because the symmetric difference is just the stick, which has measure zero. So we might restrict to, for example, finite unions of polygons. Maybe that's enough for what you want.

Edit: For either of the two metrics you probably want to be allowed move the shapes around before comparing them. In that case you actually want to consider the distance. . .

$D(A,B) = \inf \{d(A,B+x): x \in \mathbb R^2\}$

where $B+x$ just means $\{b + x : b \in B\}$


Proving $n! = n \Rightarrow (n = 1 \quad or \quad n = 2)$ I want to know whether my proof is correct. Any elegant proofs are welcome.

  $n\in\mathbb{N}.\quad$Prove $ (n!=n) \Rightarrow (n=1\quad or\quad n=2)$

$ (n!=n) \Rightarrow (n=1\quad or\quad n=2) \equiv$

$(n\neq1\quad and\quad n\neq2) \Rightarrow (n!\neq n) \equiv$

$n>2 \Rightarrow (n!\neq n)$

Suppose $n>2$. Suppose $n!=n$.

Then $n!=n \equiv $

$(n-1)!=1 \equiv$

$(n-1)(n-2)!=1 \quad, (n-2)!$ is defined since $n>2$

Thus, $(n-2)! = 1/(n-1)$

but $(n-2)!\in\mathbb{N}$ and $1/(n-1)\in\mathbb{Q}\setminus\mathbb{N}$, so contradiction.

Therefore $ (n!=n) \Rightarrow (n=1\quad or\quad n=2)$

If $n!>n$, then
$$
(n+1)! = (n+1)n! > (n+1)n > n+1
$$
as long as $n>1$. Now, $3!>3$, and by induction we have proved that $n!>n$ for all $n \geq 3$. Since $1!=1$ and $2!=2$, the assertion is proved.

This seems shorter:

If $n = 1$ or $n = 2$, then  $n! / n = 1$ and thus $n! = n$.

If $n > 2$, then $n! / n = (n-1)! \geq 2$ and thus $n! \neq n$.

Therefore, $n \in \mathbb{N} \backslash \{ 0 \}$, $n! = n \Rightarrow n = 1 \vee 2$.


Properties of trace $0$ matrices: similarity, invertibility, relation to commutators $1.$ Are trace $0$ matrices always of the form $AB-BA$?

$2.$ Is a trace $0$ matrix over the complex field always similar to a matrix with $0$ as a diagonal element?

$3.$ Is a trace $0$ matrices over any field always similar to a matrix with $0$ as a diagonal element?

$4.$ Is a trace $0$ matrix not invertible if it is upper triangular.?

I solved one problem in hoffman kunze saying : $W$ be the span of $n\times n$ matrices over the field $F$ and $W_0$ be the subspace spanned by the matrices $C$ where $C=AB-BA$. Then we proved there that $W_0$ is the exactly subspace of matrices which have trace $0$, so from this result can we say $1$ is true?

For (1), see the citation in my answer to a previous question. In particular, yes, the set of all traceless matrices are precisely the set of all commutators, regardless of the underlying field.

The exercise in Hoffman and Kunze asks whether the subspace of all traceless matrices is equal to the subspace spanned by all commutators. This is different from asking whether the subspace of all traceless matrices is equal to the set of all commutators. Put it another way, the exercise in Hoffman and Kunze evades the question of whether all commutators form a matrix subspace.

For (2), see my aforementioned answer again.

For (3) and (4), consider $I_2$ over $\mathbb{F}_2$.

If it is the matrix algebra $\mathfrak{sl}_n$ over a field then yes, because $\mathfrak{sl}_n$ is a simple Lie algebra and $[\mathfrak{sl}_n,\mathfrak{sl}_n]$ is a nontrivial ideal of $\mathfrak{sl}_n$ thus equal to it.
No, look at the matrix with diagonal $(1, -1)$.
See 2.
See 2.


Area of rectangle and triangle derivation I was wondering about the derivation for the area of a triangle and the area of a rectangle. Of course, we all know them to be $\dfrac{1}{2}bh$ and $bh$ respectively, but where is the derivation of these? If we derive the rectangle one then the triangle one obviously follows, but I don't recall ever seeing a derivation for the area of a rectangle. Is it like an axiom that we don't prove?

Area is defined as the number of smallest blocks of 1 unit that are required to fill in any 2-d geometry.
Considering that, it is very easy to prove any area formula.

One way to define area is to start with a $b \times h$ rectangle and require its area to be $bh$.  See for example https://en.wikipedia.org/wiki/Area#Formal_definition which includes:

  Every rectangle $R$ is in $M$. If the rectangle has length $h$ and breadth $k$ then $a(R) = hk$.

Then half the rectangle (a right-angled triangle) has area $\frac12 bh$.

Any acute triangle can be split into two right-angled triangles with a perpendicular from one vertex to the opposite side (giving the perpendicular height), while an obtuse triangle may be the difference between two right-angled triangles.

I have more on areas of triangles at http://www.se16.info/hgb/triangle.htm


BOX Problem A combinatorics question A big box contains 10 small boxes. Each of these small
boxes is either empty or contains 10 boxes which are even
smaller. Again, each of these smaller boxes is either empty
or contains 10 boxes smaller than itself and so on. After
opening all the boxes it turned out that exactly 2008 boxes
were nonempty. So, how many boxes were empty ?

This a math competition question.  I solved it and think that it is 2007.  Could anyone confirm that

There’s no reason to introduce probability, and I see no reason for the assumption that a box is non-empty with probability $\frac12$. Suppose that there are $e$ empty boxes and $n$ non-empty boxes. Imagine that each box except the big one is connected by a string to the box that contains it. There are $e+n$ boxes altogether, but one of them is the big one, so there are $e+n-1$ strings. On the other hand each non-empty box has $10$ strings attaching it to the smaller boxes inside it, and this accounts for all of the strings, so there are $10n$ strings. Thus, $e+n-1=10n$, so $e=9n+1$. In our case $n=2008$, so $e=18073$.

This is really a problem in graph theory. The boxes form a rooted tree, the big box being the root. The non-empty boxes correspond to interior nodes of the tree and the empty boxes to leaves. Each interior node has $10$ daughter nodes. The strings are the edges of the tree.

Probability of a box being empty = 1/2

Probability of a box being non-empty = 1/2

Level 1 Let n1 be the number of  boxes empty, then 10-n1 will be the numbe non-empty

P[x=n1] = 10Cn1 (1/2)^10
P[x=10-n1] = 10C(10-n1) (1/2)^10

Level 2 For each box that is non-empty, let there be n2 boxes empty and 10-n2 boxes non empty.

P[x=n2] = 10Cn2 (1/2)^10
P[x=10-n2] = 10C(10-n2) (1/2)^10

Recursively,

A => No of boxes empty = P[x=n1]P[x=n2]...(10*100*1000...

B => No of boxes non-empty = P[x=(10-n1)]P[x=(10-n2)] ... *(10*100*1000...

Using the property => nCr = nCn-r

A = B

In which case, 2008 were non-empty including the root box.  If we opened that will be the levels 1 , 2 , 3....  Thus the number of empty will be 2008-1 = 2007 ( deducting the root).

Do you think the approach is right?

Thanks

Satish


How to find the value with which I need to divide a unitary matrix such that its first component lies within the range of cosine? In quantum mechanics, we generally deal with unitary matrices. In the IBM cloud computers and Qiskit, the general $2\times 2$ unitary is defined as

$$U(\theta, \phi, \lambda) =
\begin{pmatrix}
\cos\left(\frac{\theta}{2}\right) & -e^{i\lambda} \sin\left(\frac{\theta}{2}\right) \\
e^{i\phi} \sin\left(\frac{\theta}{2}\right) & e^{i(\lambda + \phi)} \cos\left(\frac{\theta}{2}\right)
\end{pmatrix}.
$$

Now say I want to find the parameters $\theta, \phi$ and $\lambda$ corresponding to the unitary $$\begin{pmatrix}-\frac{1}{2}+\frac{i}{2}& -\frac{1}{2}-\frac{i}{2}\\ -\frac{1}{2}+\frac{i}{2} & -\frac{1}{2}+\frac{i}{2}\end{pmatrix}.$$ Then I can't apply this code directly as the first component of the unitary does not lie in the range of $\cos$. So I'll first have to divide the whole matrix by $\frac{1+i}{\sqrt{2}}$ first, to get to

$$\begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}.$$

Note that $|\frac{1+i}{\sqrt{2}}|=1$ and so the two unitaries are equivalent in the quantum mechanics sense. Furthermore, now the first component $\frac{1}{\sqrt 2}$ lies within the range of cosine i.e. $[-1,1]$.

Now for any general unitary matrix how do I find the value with which I have to divide it, such that the first component lies within the range of cosine? In this case, the value was $\frac{1+i}{\sqrt 2}$, but how do I find that for a general $2\times 2$ unitary matrix?

Àccording to Wikipedia, the general form of a unitary $Z$ can also be written as $$Z=e^{i\varphi/2}\begin{bmatrix}e^{i\varphi_1}\cos\theta & e^{i\varphi_2}\sin\theta \\ -e^{-i\varphi_2}\cos\theta & e^{-i\varphi_1}\cos\theta \end{bmatrix}$$ where the determinant of the unitary is $\det Z=e^{i\varphi}$. So the required value is the principle value of $\sqrt{\det Z}$ multiplied by the phase factor $e^{i\varphi_1}$ of the first component of $Z$, which can be easily obtained by representing it in polar form.

I am sorry, but I cannot comment on the question. I have doubts that the matrix mentioned in the question is actually unitary:
$$\begin{pmatrix}-\frac{1}{2}+\frac{i}{2}& -\frac{1}{2}-\frac{i}{2}\\ -\frac{1}{2}+\frac{i}{2} & -\frac{1}{2}+\frac{i}{2}\end{pmatrix}.$$
Maybe you meant one of the square roots of NOT:
$$\begin{pmatrix}-\frac{1}{2}+\frac{i}{2}& -\frac{1}{2}-\frac{i}{2}\\ -\frac{1}{2}-\frac{i}{2} & -\frac{1}{2}+\frac{i}{2}\end{pmatrix}$$
and just made a mistake in one sign?
P.S. This certainly does not question the answer.


Is a $\Omega$-Spectrum a connective one? I can't find this result anywhere, but it seems pretty straightforward. I want to avoid silly mistakes, but I can't see any fault. I'd love to receive some feedback

Let $X$ be a $\Omega$-spectrum (of compactly generated topological spaces). By definition for $q<0$, $$\pi_q(X)=\lim_k \pi_{k+q <k}X_k$$

But since $\pi_i(X_k)\cong \pi_{i}(\Omega X_{k+1})\cong \pi_{i+1}(X_{k+1})$ the spaces are more and connected, in particular, $X_k$ is $k-1$-connected and therefore $\pi_q(X)=0$.

Is that correct?

A $\Omega$-spectrum is not necessarily a connective one.  In your argument, there is no reason "the spaces are more and [more] connected".  For a concrete example, consider the $K$-theory spectrum $KU$.  In its usual incarnation (with $BU \times \mathbb{Z}$ even and $\Omega(BU \times \mathbb{Z})$ odd), it is already a $\Omega$-spectrum by Bott periodicity.  But it's 2-periodic, so its homotopy groups extend infinitely in negative degrees.

In general, remember then every spectrum can be converted into a $\Omega$-spectrum by a telescope construction, without changing its homotopy groups.  So if every $\Omega$-spectrum was connective, this would mean then every spectrum is connective, which should strike you as patently false.  This gives another bunch of counterexamples: take any spectrum, desuspend it a bunch of times so that it has negative homotopy groups, and then "$\Omega$-spectrify".

Direct answers would be helpful. I believe you are saying for connective $\Omega$-spectrum $X$, $\pi_i(X)=0$ for all $i\leq -1$. For non connective $\Omega$-spectrum, there are nontrivial negative homotopy groups.


Let $x \in \mathbb R : \sin(\sin x )=1/2$ and $2<x<3$. What is the value of $\cos(-\sin x)$? Let $x \in \mathbb R : \sin(\sin x )=1/2$ and $2<x<3$.
What is the value of $\cos(-\sin x)$ ?

So far I have tried to solve:
$$\sin(\sin x)=1/2 \Leftrightarrow \sin x=\sin^{-1}(1/2) \Leftrightarrow \sin x=\pi/6 \Leftrightarrow x= \sin^{-1}(\pi/6) $$
Then I plugged it in :
$$\cos(-\sin(\sin^{-1}(\pi/6)))$$

Since $|\sin x|\le1$ for all $x$, and $1\lt\pi/2$, we have $\cos(\sin x)\gt0$, so that

$$\cos(-\sin x)=\cos(\sin x)=\sqrt{1-\sin^2(\sin x)}=\sqrt{1-\left(1\over2\right)^2}={\sqrt3\over2}$$

simply apply pythagorean identity and property of cosine function which is an even function.


GCD with Gaussian integers Let $\alpha,\beta\in\mathbb{Z[i]}$, show that if
  $gcd(N(\alpha),N(\beta))=p$ where $p$ is a prime $\in\mathbb{N}$ then
  $p\not\equiv 3\pmod 4$.

I suppose $gcd(N(\alpha),N(\beta))=p$ where $p\in\mathbb{N}$ is prime.

$N(gcd(\alpha,\beta))|gcd(N(\alpha),N(\beta))=p$ and $N(gcd(\alpha,\beta))=a^2+b^2$ for some $a,b\in\mathbb{Z}$

So $p=d*(a^2+b^2)$ for some $d\in\mathbb{Z}$

But $p$ is prime, so either $d=1$ and $a^2+b^2=p$ or $d=p$ and $a^2+b^2=1$

Suppose $d=1\implies a^2+b^2=p$, then $p$ is the sum of two squares and p must be $\not\equiv 3\pmod 4$.

How can I get $p\not\equiv 3\pmod 4$ in the case where $a^2+b^2=1$ and $d=p$? I know that $\alpha$ and $\beta$ are coprime, should that help me?

Assume that $p|a^2+b^2$. If $p\not |a$ then for some $c$ $$ac\equiv 1(\mod p)$$

This implies that $$(bc)^2\equiv -1(\mod p)$$ this is impossible since $p\equiv 3(\mod 4)$. Thus $p|a$ and so $p|b$ thus $p^2|a^2+b^2$.


A relation between gcds and determinants While working on a proof, I stumbled on the following fact.

Let $M:=(x_{i,j})_{i,j}\in \mathrm {GL}_{n,d}(\mathbb Z)$. Let's denote by $\alpha_i:=\gcd(x_{1,i},\ldots,x_{n,i})$ for all $i\in \{1,\ldots,d\}$.
Let's also define $N:=\binom nd$ and $\eta_1,\ldots,\eta_N$ be all the minor of $M$ of size $d\times d$.
Then
$$\alpha_1\cdots\alpha_d=\gcd(\eta_1,\ldots,\eta_N).$$

What I tried.
Let $k$ be an integer.
Assume $k$ divides $\alpha_1\cdots\alpha_d$. Let $j\in \{1,\ldots,N\}$. Then $\eta_j$ is a minor of the matrix $M$ of size $d\times d$, extracted by taking only the lines $i_1,\ldots,i_d$. So we have
$$\eta_j=\sum_{\sigma\in \mathfrak S_d}\epsilon(\sigma)\prod_{\ell=1}^d x_{i_\ell,\sigma(\ell)}.$$
But $k\mid \alpha_1\cdots\alpha_d$, so for all $d$-uple $(i_1<\cdots<i_d)$, we have
$$k\mid x_{i_1,1}\cdots x_{i_d,d}= x_{i_1,\sigma(1)}\cdots x_{i_d,\sigma(d)}$$
for all permutation $\sigma\in\mathfrak S_d$. So $k$ divides $\eta_j$ for all $j$, so
$$ k\mid \gcd(\eta_1,\ldots,\eta_N).$$
Reciprocally, I am unable to do it.
Any help would be much appreciated.

This statement seems to be very connected to the theorem in this recent publication:
http://nntdm.net/volume-26-2020/number-3/5-7/
"Equalities between greatest common divisors involving three coprime pairs"
that goes as:
For $a_i$ and $b_i$ positive integers such that $\gcd(a_i, b_i) = 1$ for $i ∈ \{1, 2, 3\}$ and $d_{ij} = |a_ib_j − a_jb_i|$, then $\gcd(d_{32}, d_{31}) = \gcd(d_{32}, d_{21}) = \gcd(d_{31}, d_{21})$.
Basically $a_i$ and $b_i$ can form a 3x2 matrix (your $x$)  and $d_{ij}$ are its minors (your $\eta$). I think this can be helpful.


Can I assume this data will be normally distributed? If I were to get a sample of random people to solve a Sudoku puzzle, recording the time is takes them to complete it, can I model the range of times as a normal distribution with a mean $\mu$ and variance $\sigma^2$?

If it can't be modelled this way, how else could I do it?

@Ben Harris: drhab is right. The normal distribution is unlimited along the $x$-Axis, whereas your time data are limited by $0$ on the left. How about a Poisson Distribution(url=https://en.wikipedia.org/wiki/Poisson_distribution)? This is reasonable, since if I start my puzzle at time $0$, and I am done at time $t\geq0$, then I am also done at every $t+\epsilon$ for any $\epsilon\geq 0$.


Induction on Real Numbers One of my Fellows asked me whether total induction is applicable to real numbers, too ( or at least all real numbers ≥ 0) . We only used that for natural numbers so far.
Of course you have to change some things in the inductive step, when you want to use it on real numbers.

I guess that using induction on real numbers isn't really possible, since $[r,r+\epsilon]$ with $\epsilon > 0$, $r \in \mathbb R$ is never empty.

Can you either give a good reason, why it isn't possible or provide an example where it is used?

Okay, I can't resist: here is a quick answer.

I am construing the question in the following way: "Is there some criterion for a subset of $[0,\infty)$ to be all of $[0,\infty)$ which is (a) analogous to the principle of mathematical induction on $\mathbb{N}$ and (b) useful for something?"
The answer is yes, at least to (a).

Let me work a little more generally: let $(X,\leq)$ be a totally ordered set which has
$\bullet $a least element, called $0$, and no greatest element.
$\bullet$ The greatest lower bound property: any nonempty subset $Y$ of $X$ has a greatest lower bound.

Principle of Induction on $(X,\leq)$: Let $S \subset X$ satisfy the following properties:
(i) $0 \in S$.
(ii) For all $x$ such that $x \in S$, there exists $y > x$ such that $[x,y] \subset S$.
(iii) If for any $y \in X$, the interval $[0,y) \subset S$, then also $y \in S$.
Then $S = X$.

Indeed, if $S \neq X$, then the complement $S' = X \setminus S$ is nonempty, so has a least upper bound, say $y$.  By (i), we cannot have $y = 0$, since $y \in S$.  By (ii), we cannot have $y \in S$, and by (iii) we cannot have $y \in S'$.  Done!

Note that in case $(X,\leq)$ is a well-ordered set, this is equivalent to the usual statement of transfinite induction.

It also applies to an interval in $\mathbb{R}$ of the form $[a,\infty)$.  It is not hard to adapt it to versions applicable to any interval in $\mathbb{R}$.

Note that I believe that some sort of converse should be true: i.e., an ordered set with a principle of induction should have the GLB / LUB property.  [Added: yes, this is true.  A totally ordered set with minimum element $0$ satisfies the principle of ordered induction as stated above iff every nonempty subset has an infimum.]

Added: as for usefulness, one can use "real induction" to prove the three basic Interval Theorems of honors calculus / basic real analysis.  These three theorems assert three fundamental properties of any continuous function $f: [a,b] \rightarrow \mathbb{R}$.

First Interval Theorem: $f$ is bounded.

Inductive Proof: Let $S = \{x \in [a,b] \ | \ f|_{[a,x]} \text{ is bounded} \}$.  It suffices to show that $S = [a,b]$, and we prove this by induction.
(i) Of course $f$ is bounded on $[a,a]$.
(ii) Suppose that $f$ is bounded on $[a,x]$.  Then, since $f$ is continuous at $x$, $f$ is bounded near $x$, i.e., there exists some $\epsilon > 0$ such that $f$ is bounded on
$(x-\epsilon,x+\epsilon)$, so overall $f$ is bounded on $[0,x+\epsilon)$.
(iii) If $f$ is bounded on $[0,y)$, of course it is bounded on $[0,y]$.  Done!

Corollary: $f$ assumes its maximum and minimum values.
Proof: Let $M$ be the least upper bound of $f$ on $[a,b]$.  If $M$ is not a value of $f$,
then $f(x)-M$ is never zero but takes values arbitrarily close to $0$, so $g(x) = \frac{1}{f(x)-M}$ is continuous and unbounded on $[a,b]$, contradiction.

(Unfortunately in the proof I said "least upper bound", and I suppose the point of proofs by induction is to remove explicit appeals to LUBs.  Perhaps someone can help me out here.)

Second Interval Theorem (Intemediate Value Theorem): Suppose that $f(a) < 0$ and $f(b) > 0$.  Then there exists $c \in (a,b)$ such that $f(c) = 0$.

Proof: Define $S = \{x \in [a,b] \ | \ f(x) \leq 0\}$.  In this case we are given that $S \neq [a,b]$, so at least one of the hypotheses of real induction must fail.  But which?
(i) Certainly $a \in S$.
(iii) If $f(x) \leq 0$ on $[a,y)$ and $f$ is continuous at $y$, then we must have $f(y) \leq 0$ as well: otherwise, there is a small interval about $y$ on which $f$ is positive.
So it must be that (ii) fails: there exists some $x \in (a,b)$ such that $f \leq 0$ on
$[a,x]$ but there is no $y > x$ such that $f \leq 0$ on $[a,y]$.  As above, since $f$ is continuous at $x$, we must have $f(x) = 0$!

Third Interval Theorem: $f$ is uniformly continuous.

(Proof left to the interested reader.)

Moreover, one can give even easier inductive proofs of the following basic theorems of analysis: that the interval $[a,b]$ is connected, that the interval $[a,b]$ is compact (Heine-Borel), that every infinite subset of $[a,b]$ has an accumulation point (Bolzano-Weierstrass).

Acknowledgement: My route to thinking about real induction was the paper by I. Kalantari "Induction over the Continuum".

His setup is slightly different from mine -- instead of (ii) and (iii), he has the single axiom that $[0,x) \subset S$ implies there exists $y > x$ such that $[0,y) \subset S$ -- and I am sorry to say that I was initially confused by it and didn't think it was correct.  But I corresponded with Prof. Kalantari this morning and he kindly set me straight on the matter.

For that matter, several other papers exist in the literature doing real induction (mostly in the same setup as Kalantari's, but also with some other variants such as assuming (ii) and that the subset $S$ be closed).  The result goes back at least to a 1922 paper of Khinchin, who in fact later used real induction as a basis for an axiomatic treatment of real analysis.  It is remarkable to me that this appealing concept is not more widely known -- there seems to be a serious PR problem here.

Added Later: I wrote an expository article on real induction soon after giving this answer: it's very similar to what I wrote above, but longer and more polished.  It is available here if you want to see it.

In the case of $[0,\infty[$, perhaps this looks like induction:

Let $A\subseteq [0,\infty[$ such that

$i)$ $[0,1]\subset A$

$ii)$ if $x\in A$ then $x+1\in A$.

Let $x\in[0,\infty[$. Then $x - \lfloor x\rfloor\in A$ by $i)$, and by repeating $ii)$ some  times, we get $$(x - \lfloor x\rfloor)+\underbrace{1+\cdots+1}_{\text{$\lfloor x\rfloor$ times}}=x\in A.$$ Therefore $A=[0,\infty[$.


What is $\mathbb{E}\!\left(X^2 \mid X + 2Y \right)$ if $X,Y \sim \mathcal{U}\left(\left[-1,1\right]\right)$ independently? $X$ and $Y$ are independent random variables of the uniform distribution on $\left[-1,1\right]$, $X,Y \sim \mathcal{U}\!\left(\left[-1,1\right]\right)$.
What is $\mathbb{E}\!\left(X^2 \mid  X + 2Y \right)$?
My attempt:
We need the joint pdf of $X$ and $X+2Y$ (as suggested). So we do a change of variables, let $W = X,~Z = X + 2Y$. This is a linear transformation, so the jacobian will be a constant. We need to find the image of the square $-1 \leq x, y \leq 1$ under this transformation. Since it is a linear transformation, the four edges of boundaries will be mapped to line segments. So we can find them by computing where the four corners of the square are mapped:
$$\begin{align}
\left(x,y\right)&|\left(w,z\right)\\
(-1,-1) &| (-1, -3)\\
(-1,+1) &| (-1,+1)\\
(+1,+1) &| (+1,+3)\\
(+1,-1) &| (+1,-1)
\end{align}$$
The image of the square is the parallelogram with vertices as above and in the picture below:

The joint density of $W$ and $Z$ will be uniform on this region, let A denote the interior of the parallelogram. Since it has area of 8, we can conclude
$f_{W,Z}\left(w,z\right) = \frac{1}{8}\unicode{x1D7D9}\left(\left(w,z\right) \in A\right) = \begin{cases}
\frac{1}{8},  & \text{if $-1\leq w \leq 1,~-2+w\leq z\leq2+w$} \\
0, & \text{otherwise}
\end{cases}.$
The computations for $\mathbb{E}\left(X|X+2Y\right)$ and $\mathbb{E}\left(X^2|X+2Y\right)$ are below.

(for $\mathbb{E}\left(X|X+2Y\right)$, $Z = X + 2Y$)
Now we can figure out what $f_{X|Z}\left(x|z\right)$ is. We must consider three cases:

$-3 \leq z \leq -1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $z+2$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{-1+2+z}{2} = \frac{z}{2} + \frac{1}{2}$

$-1 \leq z \leq 1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $1$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{-1+1}{2} = 0$

$1 \leq z \leq 3$, given $Z = z$, $X$ is uniformly distributed between $1$ and $z-2$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{1-2+z}{2} = \frac{z}{2} - \frac{1}{2}$

Cases $1,2,3$ (with $0$ otherwise) gives us $\mathbb{E}\!\left(X|Z=z\right)$.

(for $\mathbb{E}\left(X^2|X+2Y\right)$, $Z = X + 2Y$)
We must consider three cases:

$-3 < z \leq -1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $z+2$ (so probability density function $f_{X|Z}(x|z) = \frac{1}{(z+2) - (-1)} = \frac{1}{z+3}$). So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_{-1}^{z+2}x^2 \cdot \frac{1}{z+3}dx = \frac{1}{3} \cdot \frac{(z+2)^3+1}{z+3}$

$-1 \leq z \leq 1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $1$ (probability density function $f_{X|Z}(x|z) = \frac{1}{2}$).
So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_{-1}^{1}x^2 \cdot \frac{1}{2}dx = \frac{1}{3}$

$1 \leq z < 3$, given $Z = z$, $X$ is uniformly distributed between $1$ and $z-2$. So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_1^{z-2}x^2\frac{1}{z-3}dx = \frac{1}{3}\cdot\frac{1}{z-3}\cdot\left((z-2)^3-1\right)$

Cases $1,2,3$ (with $0$ otherwise) gives us $\mathbb{E}\!\left(X^2|Z=z\right)$.

Define $h:\mathbb{R} \to \mathbb{R}$ such that $h(x)=\frac 1 3 \chi_{[-3,3]}$.
The claim is that $\mathbb{E}\left(X^2 | X + 2Y \right)=h(X+2Y)$.

Suppose $B \in \mathscr{B}([-3,3])$.
Let us prove $\int \chi_{B}(x+2y)x^2dx dy=\int \chi_{B}(x+2y)h(x+2y)dxdy$.
If $g(x,y)=(x+2y,x)$ then $|\det g (x,y)|=2$ and thus we have to prove
$$\int_{[-3,3]\times [0,1]} \chi_{B}(z)x^2dz dx=\int_{[-3,3]\times [0,1]} \chi_{B}(z)h(z)dzdx$$
thanks to the change of variable.
Now $\int_{[-3,3]\times [0,1]} \chi_{B}(z)x^2dz dx=\mu(B)\frac 1 3$ and $\int_{[-3,3]\times [0,1]} \chi_{B}(z)h(z)dzdx=\int_{[-3,3]\times [0,1]} \chi_{B}(z)\frac 1 3 dzdx=\mu(B)\frac 1 3$ which conclude the proof.


Number of matrices , such that sum in each row and each column is 0 To find number of 4 cross 4 matrices, such that each element is 1 or -1. Also sum of elements in each row and each column should be zero.

I am able to think that each row and each column shall consist of two 1 and two -1. But then how to proceed further

answer is 90

Hint: across any row and down any column, your choice in the first three elements uniquely determines the fourth. As such, you can consider the number of $3 \times 3$ matrices with entries in $\{1,-1\}$ such that no row/column $\mathbf v$ has $v_1 = v_2 = v_3$ i.e. all entries are the same.

Arrange two columns. 6*6 which is 36. Now for each such arrangement other two columns gets fixed in two ways. So 72


How can I solve $\tiny\begin{cases} \dfrac{dx}{dt} = t^2 x^2 y\\ \dfrac{dy}{dt} = -t^2 y^2 x \end{cases}$? I have:

$$\begin{cases}
\dfrac{dx}{dt} = t^2 x^2 y\\
 \dfrac{dy}{dt} = -t^2 y^2 x
\end{cases}$$

How do I go about solving this system of equations?

I tried separation of variables however because of the extra $y$ term in the $\frac{dx}{dt}$ and $x$ term in $\frac{dy}{dt}$ I am unsure how to start.

As suggested by Jack, you could consider eliminating $t$ from both equations leaving you with the following:

$$ t^2 = \frac{1}{x^2y}\frac{dx}{dt} = -\frac{1}{y^2x}\frac{dy}{dt}$$

$$ \frac{dy}{dx} = -\frac{y}{x}$$ which is now variable separable.

First observe that
$$
\frac{d}{dt}(xy)=x\frac{dy}{dt}+y\frac{dx}{dt}=0.
$$
Thus $x(t)y(t)=c_1$, for some constant $c_1\in\mathbb R$. The system now becomes
$$\begin{cases}
\dfrac{dx}{dt} =  c_1t^2x\\
 \dfrac{dy}{dt} = -c_1t^2y.
\end{cases}$$
General solution of the system is hence
$$
x(t)=c_2\mathrm{e}^{c_1t^3/3},\quad y(t)=c_3\mathrm{e}^{-c_1t^3/3},
$$
where $c_2c_3=c_1$.


If $\gamma$ is a closed path, then $\int_{\gamma}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\in2\pi\mathbb{Z}$ As the title says:
Let $\gamma$ be a smooth closed path that avoids $(0,0)$. I need to show that the line integral: $$\int_{\gamma}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ is a multiple of $2\pi$.
Note that $\gamma$ is allowed to self-intersect itself.

Intuitively this value should represent the "number of rotations around $(0,0)$", as they are counted with orientation, but this intuition is all I have.

Any ideas?

Take $\gamma : [0,1] \to \mathbb C$ a complex valued function, so that the quotient $\gamma'/\gamma$ is exactly what you obtain when replacing $\gamma$ into
$$\frac{dz}{z} = \frac{x-iy}{x^2+y^2}(dx+i dy)$$
and multiplying out. Thus, consider the function $h(t) = \int_0^t \frac{\gamma'(s)}{\gamma(s)}ds$. Set $H(t) =\gamma(t) \exp(-h(t))$, and observe that $H'(t) = 0$. This means $H(t)$ is constant, and then that $H(1) = H(0)$. All in all, we have $\gamma(1) \exp (-h(1)) = \gamma(0) \exp (-h(0))$
which gives that $\exp h(1) = 1$ or, what is the same, that $h(1) = 2\pi i n$ with $n\in \mathbb Z$.

Remark: this elementary argument is to be found in Apostol's Mathematical Analysis.

Write $x=rsin(u), y=rsin(u)$, the form  $\alpha={{-y}\over{x^2+y^2}}dx+{{x}\over{x^2+y^2}}dy=du$ then take a circle $C$ contained in $\gamma$ and apply Stokes formula, since $d\alpha=0$, you obtain $\int_C\alpha=\int_{\gamma}\alpha=2\pi$.


$\max (X_t, a)$ is not a martingale, counterexample It is quite easy to prove that if $[X_t \}_{t \in [0,T]}$ is a martingale, then for any number $a \in \mathbb{R}$ $ \{\max (X_t,a)\} _t$ is a submartingale and  $ \{\min (X_t, a)\} _t$ but I cannot come up with a counterexample which proves that those need not be martingales.

That is, I want to find a martingale $\{X_t\}_{t \in [0,T]}$ and a constant $a \in \mathbb{R}$ such that $$\mathbb{E}\left(\max (X_t,a) | F_s \right) < \max(X_s, a)$$ for all $s \le t$ and adequately for minimum: $$\mathbb{E}\left(\min (X_t,a) | F_s \right) > \min(X_s, a)$$ for all $s \le t$

Could you tell me how/where to look for such examples?

Hint: An easier way to show that $(Z_t)$ is not a martingale  is to find some $t$ with $\mathbb{E}(Z_0)\neq \mathbb{E}(Z_t)$.

Let $(\Omega, \mathscr F, \{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)$ be a filtered probability space where $T > 0$.

  If we have two $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingales $X$ and $Y$, it does not follow that $\max(X,Y)$ is a $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingale. $\tag{*}$

Now, $\forall a \in \mathbb R$ the stochastic process $A = (A_t)_{t \in [0,T]}$ given by $a = A_t \ \forall t \in [0,T]$ is a $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingale.

Following the proof of $(*)$, we can show that it does not follow that $\max(X,A) = (\max(X_t,a))_{t \in [0,T]}$ is a $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingale.

Pf based on answers here:

Let $(\Omega, \mathscr F, \mathbb P)$ be our probability space.

Let $X_0, X_1, X_2, ...$ be iid random variables s.t.

$$P(X_i = 1) = 1/2 = P(X_i = -1)$$

Define $W^X_n := \sum_{i=0}^{n} X_i$.

Consider the filtration $\mathscr F_{n} := \mathscr F_{n}^X$ for $n = 0,1,...$

Note that $W^X_n$ is a martingale w/rt $\mathscr F_{n}$.

Let $a \in \mathbb R$. Consider $M = (M_n)_{n \in \mathbb N}$ where

$$M_n := \max(W_n^X, a) = W_n^X 1_{A_n} + a 1_{A_n^C}$$

where $A_n = \{W_n^X \ge a\}$ and $0 < P(A_n) < 1 (**)$

Let us try to show (and hopefully fail in trying) that $E[M_n|\mathscr F_{n-1}] = M_{n-1}$:

$$E[M_n|\mathscr F_{n-1}] = E[W_n^X 1_{A_n} + a 1_{A_n^C}|\mathscr F_{n-1}] $$

$$= E[W_n^X 1_{A_n}|\mathscr F_{n-1}] + E[a 1_{A_n^C}|\mathscr F_{n-1}] $$

$$= E[W_n^X 1_{A_n}|\mathscr F_{n-1}] + a E[1_{A_n^C}|\mathscr F_{n-1}] $$

Now consider this. If at time 1, we have that $W_1^X = a + 0.5$ it's possible that at time 2, we have $W_2^X = a - 0.5$. Thus, $M_2 = a$ but $M_1 = W_1^X$

Formally:

If at time n, we have $\omega \in A_n$ (ie $W_n^X(\omega) \ge a$), then

$$E[M_n|\mathscr F_{n-1}] = E[W_n^X 1_{A_n}|\mathscr F_{n-1}] + 0$$

$$= E[W_n^X|\mathscr F_{n-1}] = W_{n-1}^X$$

But is it that $W_{n-1}^X = M_{n-1}$?

Not necessarily. It could be that $\omega \in A_{n-1}^C$ (ie $W_{n-1}^X(\omega) \le a$). Hence, $M_{n-1} = a$

$\therefore,$ it is not necessarily true that $E[M_n|\mathscr F_{n-1}] = M_{n-1}$. QED

I hope you don't mind that I used $\mathbb N$ instead of $[0,T]$ :P

$(**)$

Oh also $P(A_n) \notin \{0,1\}$ is important:

If $P(A_j) = 0$ or $1 \ \forall j \in \mathbb N$, then $M_n$ is $W_n^X$ or a a.s. Hence, $M$ is a $(\{\mathscr F_n\}_{(n \in \mathbb N)}, \mathbb P)-$martingale (at least almost surely) because such implies $1_{A_1} = 1_{A_2} = ...$ a.s., in particular, $1_{A_n} = 1_{A_{n-1}}$ a.s..

This might be relevant: Is a probability of 0 or 1 given information up to time t unchanged by information thereafter?


Prove $\sqrt{1+x}$ can be represented by a power series I need to show that $\sqrt{1+x}$ can be represented as a power series. I need to prove the equality between the function and its Taylor series, not to prove that the Taylor series of the function, $\sum\limits_{n=0}^{\infty}{\frac{1}{2}\choose n}x^n$ converges in $(-1,1)$.

My lead is to use the Lagrange form of the remainder, by I do not know how to show the remainder tends to zero.

Edit:

The $n$-th derivative is
$$
f^{(n)}(0) = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdot\ldots\cdot\left(\frac{1}{2} - n + 1\right)}{n!}
$$
and hence the Lagrange form of the remainder is
$$
R_n(x) = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdot\ldots\cdot\left(\frac{1}{2} - n + 2\right)}{(n+1)!}(1+c)^{\frac{-1-2n}{2}}x^n
$$

But I do not know how to show this tends to $0$ as $n\to\infty$

Start by simplifying the expression for $R_n(x)$. The large fraction becomes, after taking absolute values:

$$
\frac{1}{(n+1)!} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdots \frac{2n - 3}{2} = \frac{(2n - 3)(2n - 5) \cdots (3)(1)}{2^{n+1}(n+1)!}
$$

Observe that the numerator is bounded above by $(n + 1)!$, and so the large fraction is smaller than $2^{-n-1}$. So we have:

$$
|R_n(x)| \le \frac{(1+c)^{ \frac{1-2n}{2} } } {2^{n+1}} x^n = \frac{(1+c)^{1/2}}{2} \left( \frac{x}{2(1+c)} \right) ^n
$$

For any $c$ between $0$ and $x$, the right hand side tends to $0$ as $n \to \infty$.


General expression for geometric product of blades in terms of scalar and exterior products I am trying to find a general expression for the geometric product of two blades in terms of the scalar and exterior products of vectors. Some preamble to be clear on conventions:

In a geometric algebra on a finite-dimensional real vector space, we have the geometric product between two vectors defined by:
$$vw=\langle v,w\rangle + v\wedge w$$
where $\langle v,w\rangle$ is the scalar product, and $v\wedge w$ is the exterior product. A $k$-blade is an exterior product of $k$ vectors, $v_1\wedge \cdots\wedge v_k$. We can extend the scalar product on vectors to a scalar product on $k$-blades by defining it as the following determinant:
$$\langle v_1\wedge \cdots\wedge v_k,w_1\wedge \cdots\wedge w_k\rangle = \begin{vmatrix}
     \langle v_1,w_1\rangle & \cdots & \langle v_1,w_k\rangle \\
     \vdots  & \ddots & \vdots \\
     \langle v_k,w_1\rangle & \cdots & \langle v_k,w_k\rangle
     \end{vmatrix}$$
(This, of course, is independent of how the blade is factored into vectors.)

Note that because of the associativity of both the geometric product and the exterior product and their commutativity properties, the exterior product of $k$ vectors (a $k$-blade) can be expressed as the totally antisymmetrized geometric product of $k$ vectors:
$$v_1\wedge \cdots\wedge v_k=\sum_{\sigma}{1\over k!}(\operatorname{sgn}\sigma)v_{\sigma(1)}\cdots v_{\sigma(k)}$$
where $\sigma$ ranges over all permutations of $1\ldots k$ and $\operatorname{sgn}$ is the sign of the permutation ($+1$ for even and $-1$ for odd).

Now, after playing around with it in my head and on paper for a bit, I am guessing that the geometric product between a $j$-blade $U=u_1\wedge \cdots\wedge u_j$ and a $k$-blade $V=v_1\wedge \cdots\wedge v_k$ can be found by the following rather involved expression:
$$UV=\sum_{i=0}^{\min j,k}\sum_{\sigma_i,\tau_i}(\operatorname{sgn}\sigma_i)(\operatorname{sgn}\tau_i)\langle u_{\tau_i(j)}\wedge \cdots\wedge u_{\tau_i(j-i+1)},v_{\sigma_i(1)}\wedge \cdots\wedge v_{\sigma_i(i)}\rangle u_{\tau_i(1)}\wedge \cdots\wedge u_{\tau_i(j-i)}\wedge v_{\sigma_i(i+1)}\wedge \cdots\wedge v_{\sigma_i(k)}$$
where $\sigma_i$ and $\tau_i$ range over all the permutations of $(1\ldots k)$ and $(1\ldots j)$, respectively, which satisfy the following constraints:
$$\sigma_i(1)\lt\cdots\lt\sigma_i(i)$$
$$\sigma_i(i+1)\lt\cdots\lt\sigma_i(k)$$
$$\tau_i(1)\lt\cdots\lt\tau_i(j-i)$$
$$\tau_i(j-i+1)\lt\cdots\lt\tau_i(j)$$
Essentially, this expression is the sum over all the ways of factorizing $U=U_{outside}\wedge U_{inside}$ and $V=V_{inside}\wedge V_{outside}$, taking the inner product of $U_{inside}$ (reversed) with $V_{inside}$, and multiplying it to the exterior product of $U_{outside}$ with $V_{outside}$.

However, I don't know how to prove this.

My question, then, is: does anyone know the best way to go about proving this equation, or can someone find a counterexample to it? Thank you!

For a blade $U = a_i \wedge   \ldots \  a_n  \rightarrow  \prod_{k_i}  a_{k_i}  $, i.e. simply the geometric product in the order the vectors are given.
Let $[k] = \{ k_i \}$ is the index list.
So then
$$
U V =  sgn( [k]) \prod\limits_{k_i < k_j} a_{k_j}
$$
where $a$ traverses both $U$ and $V$.
Then you simplify all common vectors, which happen to be neighbors in the  sorted order.
So then the scalar product $U \star V = <UV>_{0}$.

On the other hand
$A_k \wedge B_l =  A B \quad k+ l \leq dim \quad  | \quad 0 \quad  k+ l> dim $
therefore your formula can not be completely correct.


Why does adding a scalar multiple of row $i$ to row $j$ not change the determinant? I'm trying to arrive at the computational definition of the determinant of a square matrix $M$ by first defining it as the signed volume of applying the corresponding linear transformation to the unit cube.

In doing so it seems essential to understand the effect each elementary row operation has on the determinant.

What geometrical argument could then be made to show that adding a scalar multiple of row $i$ to row $j$ does not change the determinant?

I would appreciate any help.

Since the determinant is a multilinear map on its rows, we have:
$$M = [R_1,...,R_n]^T$$
$$\det [R_1,...,R_n+sR_1]^T = \det [R_1,...,R_n]^T + s\det [R_1,...,R_1]^T = \det(M) + 0 = \det(M)$$
The second $\det$ annihilate because it has a repeated row.

For $i,j$ it's the same.


Are the limit points of a sequence and of a set defined differently? Are limit points of a sequence and an interval differently defined? I ask because I've been given the following definition of a limit point:

  A limit point is one such that every of its neighborhoods contains infinitely many points of the set other than itself.

If that is so then this criteria applies only when we talk about an interval and it fails to give limit points of any sequence - for instance $(-1)^n$ has limit points at $1$ and $-1$, but these are not limit points according to the above definition.

What's the discrepancy between the limit points of a sequence and the definition given above?

It's true that, taken naively, this definition does conflict with that of the limit points of a sequence. The problem is that one is counting the values $1$ and $-1$ only once, whereas they occur infinitely many times in the sequence. That is to say that we really should distinguish between the set $\{1,-1\}$ (which has no limit points) and the sequence $(-1)^n$ (which has two limit points).

A more analogous statement for sequences would be:

  A limit point is a point such that the sequence has infinitely many terms in any neighborhood of it.

Which just basically repairs the statement for sequences by changing "distinct terms" to just "terms" - so that factors which are repeated infinitely often get included as limit points. So, all we need to do is be a bit careful about how we could when we have a set vs. how we count when we have a sequence.

A limit point of a sequence $(a_n)_{n\to \infty}$ is defined as the point the sequence itself gets close. An interesting example of this is the sequence $(1)_{n\to \infty}$ approaches $1$.

While the limit point of a set is seen as a point in which every neighborhood of that point (the limit-point) contains points other than itself. More interesting the set $\{1\}$ has no limit point and the set, the set $\{1+\frac{1}{n}: n\in \mathbb{N}\}$ has $1$ as it's only limit point, and the set $\mathbb{R}$ has itself has it's set of limit points.

Conceptually they hold the same idea of closeness to a point. But mechnically they work a bit different. Sequences can only have one limit point (if we are talking about metric spaces), while Sets can have an infinite set limit points.


evaluating some limits - calculus Problems with the following limits:

$$
1. \quad \quad  \lim_{x\to0^+} e^{1/x} + \ln x \, .
$$

Substitutions such as $e^{1/x}=t$ and $1/x = t$ don't yield any useful results.

Pretty much the same with
$$
2. \quad \quad \lim_{x\to 0^+} e^{1/x} - 1/x \, ,
$$
Common denominator doesn't help much.

Thanks.

If you accept that $\ln u\lt u$ for all $u$, it follows that $\ln(1/x)\lt1/x$, hence

$$e^{1/x}-1/x\lt e^{1/x}-\ln(1/x)=e^{1/x}+\ln x$$

Consequently, it suffices to show

$$\lim_{x\to0^+}(e^{1/x}-1/x)=\infty$$

Now we certainly have $\lim_{x\to0^+}e^{1/x}=\infty$.  Let's write

$$e^{1/x}-1/x=e^{1/x}\left(1-{1/x\over e^{1/x}} \right)$$

and take a look at

$$\lim_{x\to0^+}\left(1-{1/x\over e^{1/x}} \right)=1-\lim_{x\to0^+}{1/x\over e^{1/x}}=1-\lim_{x\to0^+}{-1/x^2\over(-1/x^2)e^{1/x}}=1-\lim_{x\to0^+}{1\over e^{1/x}}=1-0=1$$

This feeds into the general theorem that if $\lim f(x)=\infty$ and $\lim g(x)\gt0$, then $\lim f(x)g(x)=\infty$.

first let us show that $$  \lim_{u \to \infty} \frac{\ln u}{e^u} = 0$$ this follows because $$ \ln u < u < \frac{u^2}{2} < e^u \text{ so that } \frac{\ln u}{e^u} < \frac{2u}{u^2} \rightarrow 0 \text{ as } u \to \infty. $$
let us make a change of variable $u = \frac 1 x, x = \frac 1 u.$ with that we nee to find the $$\lim_{u\to \infty} e^u - \ln u = \lim_{u \to \infty}e^u\left(1- \frac{\ln u}{e^u}\right) =\infty \times 1 = \infty. $$


Find the probability that A or B occurs but not both. Question: Let A and B be events on a probability space. Find the probability that A or  B occurs but not both. Express your answer in terms of P(A), P(B), and $ P(A\cap B)$.

Soln:

$P(A\cap {B^c})+P({A^c}\cap B)$

$=P(A\cup B) - P(B) + P(A\cup B) - P(A)$

$=[P(A)+P(B)-P(A\cap B)-P(B)]+[P(A)+P(B)-P(A\cap B)-P(A)]$

$=P(A)+P(B)-2P(A\cap B)$

So did I get it right?

Yes, it's right. You can draw a Venn's diagram to see it


Construct a compact set of real numbers whose limit points form a countable set? (Rudin Exercise) I think $S = \{r \in \Bbb Q: 0 \leq r \leq 1\}$ in $\Bbb R$ can be a set that satisfies the conditions. First, it is compact by the Heine-Borel theorem since it is closed and bounded. It is closed because it contains all its limit points. Every rational number is a limit point; if we take an open neighbourhood of radius s, we can find some rational number contained in the open ball. Finally, the rationals are countable.

Is this correct?

Thank you.

What about $\{\frac 1n:n\in Z^+\} \cup \{0\}$?


multiplying and dividing in congruence $$x\equiv2\ (\text{mod }6)$$

This one has solution x=2, 8, 14, ...

By multiplying 2 to both sides,

$$2x\equiv4\ (\text{mod }6)$$

By dividing by $2$,
$x\equiv 2\ (\text{mod } 3)$  (because $\text{gcd}(2, 6)=2$)

And the solution for this congruence is x=2, 5, 8, ...

What's wrong with this calculation?

Note that
$$x \equiv y\ (\text{mod }m) \implies ax \equiv ay\ (\text{mod }m)$$ This means that any solution of $x \equiv y\ (\text{mod }m)$ is also a solution of $ax \equiv ay\ (\text{mod }m)$. But, the implication doesn't go both ways, i.e. a solution of $ax \equiv ay\ (\text{mod }m)$ isn't necessarily a solution of $x \equiv y\ (\text{mod }m)$ as well. That is why, in your approach, you got more solutions than the correct answer.


How do I rationalize higher index roots? Often I've found exercises like

$$\frac{2-\sqrt[6]{3x+64}}{5x}$$

Where I need the rationalize. But I am not sure how to do it with indexes greater than $2$. How do I rationalize higher index roots? (the above one is just an example).

That is, perform

$$\frac{2-\sqrt[6]{3x+64}}{5x} \cdot \frac{2+\sqrt[6]{3x+64}}{2+\sqrt[6]{3x+64}}$$

(See my comments on original for links that explain this in detail... effectively, if we have a sum or difference involving an $n^{th}$ root we take either the sum of difference of $n^{th}$ roots and evaluate, using the given terms as the first factor and then plug in values for the remaining factor... it's messy, but it does work)
$$a^6 - b^6 = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5 )$$
We now set $a = 2$ and $b =\sqrt[6]{3x+64}$
$$\frac{2-(3x+64)^{1/6}}{5x} * \frac{(3x+64)^{5/6}+2 (3 x+64)^{2/3}+4 \sqrt{3 x+64}+8 (3 x+64)^{1/3}+16 (3 x+64)^{1/6}+32}{(3x+64)^{5/6}+2 (3 x+64)^{2/3}+4 \sqrt{3 x+64}+8 (3 x+64)^{1/3}+16 (3 x+64)^{1/6}+32}$$

This yields
$$\frac{-3x}{5x[(3x+64)^{5/6}+2 (3 x+64)^{2/3}+4 \sqrt{3 x+64}+8 (3 x+64)^{1/3}+16 (3 x+64)^{1/6}+32]}$$

I hope you're happy... all that TeXing :P


Show that logistic regression with squared loss function is non-convex How would you show that if you do logistic regression with a squared loss function, it is not a convex optimization problem (in parameters)?

In other words, your loss function for an individual observation is $(y - p)^2$, where $y$ is the dependent variable and $p$ is your prediction, and
$$p = \frac{1}{1+\exp(-w^Tx)},$$
where $x$ is the vector of predictors and $w$ is the vector of weights.

As long as $x$ is not zero, the squared error loss with respect to $w$ will be non-convex.

Here is a graph that shows both the squared error loss and the log loss of the sigmoid function: https://www.desmos.com/calculator/kxz6lzszf9

You can see that the squared error loss (red and orange curves) is non-convex, whereas the log loss (green and blue curves) is convex.

To follow up with Qiang Chen's answer, the red and orange curves are the squared loss functions in question, and they are both non-convex (not fully convex). They are also non-concave (not fully concave). They are half convex on one side and half concave on the other, with a middle inflection point where $f(x)=0$. Convex and concave functions are ones that always curve in the same direction, either up or down, they can not have that kind of "S" shape that both the red and orange curves have.

All the $x_1$ and $x_2$ values on the concave side of the function do not satisfy that definition of a convex function.

I think it's convex, from Elliot Waite's demo: https://www.desmos.com/calculator/kxz6lzszf9, you can not find any data avoid the definition of convex function


Find the number of ways to get from one point on a cartesian place to another Find the number of ways to get from point (0, 0) to point (15, 17). You can move (0, 1), (1, 0), (1, 1) but you have to move by (1, 1) exactly 6 times.

You have to make 6 NE moves, 9 E moves and 11 N moves. A total of 26 moves and you can make them in any order, but each order gives a different valid path, so the answer is just $$\frac{26!}{11!9!6!}$$


Would you ever stop rolling the die? You have a six-sided die. You keep a cumulative total of your dice rolls. (E.g. if you roll a 3, then a 5, then a 2, your cumulative total is 10.) If your cumulative total is ever equal to a perfect square, then you lose, and you go home with nothing. Otherwise, you can choose to go home with a payout of your cumulative total, or to roll the die again.

My question is about the optimal strategy for this game. In particular, this means that I am looking for an answer to this question: if my cumulative total is $n$, do I choose to roll or not to roll in order to maximize my cumulative total? Is there some integer $N$ after which the answer to this question is always to roll?

I think that there is such an integer, and I conjecture that this integer is $4$. My reasoning is that the square numbers become sufficiently sparse for the expected value to always be in increased by rolling the die again.

As an example, suppose your cumulative total is $35$. Rolling a $1$ and hitting 36 means we go home with nothing, so the expected value of rolling once is:

$$E(Roll|35) = \frac 0 6 + \frac {37} 6 + \frac {38} 6 + \frac{39} 6 + \frac {40} {6} + \frac{41}{6} = 32.5$$

i.e.

$$E(Roll|35) = \frac 1 6 \cdot (37 + 38 + 39 + 40 + 41) = 32.5$$

But the next square after $35$ is $49$. So in the event that we don't roll a $36$, we get to keep rolling the die at no risk as long as the cumulative total is less than $42$. For the sake of simplification, let's say that if we roll and don't hit $36$, then we will roll once more. That die-roll has an expected value of $3.5$. This means the expected value of rolling on $35$ is:

$$E(Roll|35) = \frac 1 6 \cdot (40.5 + 41.5 + 42.5 + 43.5 + 44.5) = 35.42$$

And since $35.42 > 35$, the profit-maximizing choice is to roll again. And this strategy can be applied for every total. I don't see when this would cease to be the reasonable move, though I haven't attempted to verify it computationally. I intuitively think about this in terms of diverging sequences.

I recently had this question in a job interview, and thought it was quite interesting. (And counter-intuitive, since this profit-maximizing strategy invariably results in going home with nothing.)

How to use your observation in general:

Just checking things for $35$ isn't indicative of the general case, which for large values is different. Take your argument and use a general square $n^2$.

Suppose you're at $n^2-1$. You can leave with $n^2-1$, or you can roll. On a roll, you lose everything with probability $\frac{1}{6}$. With probability $\frac{5}{6}$ you'll get at least $(n+1)^2-6 = n^2 + 2n - 5$ by rolling until it isn't safe to any more. So, for a simple lower bound, we want to know if $\frac{5}{6}(n^2 + 2n - 5)$ is greater than $n^2-1$. For large $n$, this is not the case, and we can actually get an upper bound by similar reasoning:

An upper bound:

[A tidied up version of the original follows. Keeping the original upper bound would have required a few extra lines of logic, so I've just upped it slightly to keep it brief.]

An upper bound: Suppose we're between $n^2-5$ and $n^2-1$. If we roll, the best things could go for us would be to lose $\frac{1}{6}$ the time, and, when we don't lose, we get in the range $(n+1)^2-6$ to $(n+1)^2-1$ (the highest we could get without taking another risk). Just comparing current valuation, you're trading at least $n^2-5$ for at most $\frac{5}{6}(n^2 + 2n)$ by the time you get to the next decision. The difference is $-\frac{1}{6}n^2 + \frac{5}{3}n + 5$. For $n \geq 13$ this is negative. So if you're at $13^2-k$ for $1 \leq k \leq 6$, don't roll. (Higher $n$ making this difference even more negative gives this conclusion. See next paragraph for details.)

Extra details for the logic giving this upper bound: Let $W_n$ be the current expected winnings of your strategy for the first time you are within $6$ of $n^2$, also including times you busted or stopped on your own at a lower value. The above shows that, if your strategy ever includes rolling when within $6$ of $n^2$ for $n \geq 13$, then $W_{n+1}$ is less than $W_n$. Therefore there's no worry about anything increasing without bound, and you should indeed never go above $168$.

An easy lower bound (small numbers):

Low values: For $n = 3$ we have $(5+6+7+8)/6 > 3$ so roll at $n = 3$. Except for the case $n = 3$, if we roll at $n$, we'll certainly get at least $\frac{5}{6}(n+3)$. So if $\frac{5}{6}(n+3) - n > 0$, roll again. This tells us to roll again for $n < 18$ at the least. Since we shouldn't stop if we can't bust, this tells us to roll again for $n \leq 18$.

An algorithm and reported results for the general case:

Obtaining a general solution: Start with expected values of $n$ and a decision of "stay" assigned to $n$ for $163 \leq n \leq 168$. Then work backwards to obtain EV and decisions at each smaller $n$. From this, you'll see where to hit/stay and what the set of reachable values with the optimal strategy is.

A quick script I wrote outputs the following: Stay at 30, 31, 43, 44, 45, 58, 59, 60, 61, 62, and 75+. You'll never exceed 80. The overall EV of the game is roughly 7.2. (Standard disclaimer that you should program it yourself and check.)

Let's deal with the trivial straight up: If you are not within 6 of the next square then you will keep rolling.

If you are within 6 of $n^2$ then your decision is to take the money or roll the dice. If you win, then you proceed to the next decision point (which may be immediately if you are still less than $n^2$). The cases where $n\le3$ are special because you can hit 1 of 2 perfect squares with your next roll.

Each roll is a new game with your current winnings as the stake - you do not stand to lose zero as you did with the first roll because now you have something of value.

The following gives the expected value of the next roll for numbers up to 49. For most numbers greater than 9 this is negative.

$$0: 2.67$$
$$1: N/A$$
$$2: 2.5$$
$$3: 0.33$$
$$4: N/A$$
$$5: 1.17$$
$$6: 1$$
$$7: 0.83$$
$$8: 0.67$$
$$9: N/A$$
$$10: -0.83$$
$$11: -1$$
$$12: -1.17$$
$$13: -1.33$$
$$14: -1.5$$
$$15: -1.67$$
$$16: N/A$$
$$17: 3.5$$
$$18: 3.5$$
$$19: -3.83$$
$$20: -4$$
$$21: -4.17$$
$$22: -4.33$$
$$23: -4.5$$
$$24: -4.67$$
$$25: N/A$$
$$26: 3.5$$
$$27: 3.5$$
$$28: 3.5$$
$$29: 3.5$$
$$30: -7.5$$
$$31: -7.67$$
$$32: -7.83$$
$$33: -8$$
$$34: -8.17$$
$$35: -8.33$$
$$36: N/A$$
$$37: 3.5$$
$$38: 3.5$$
$$39: 3.5$$
$$40: 3.5$$
$$41: 3.5$$
$$42: 3.5$$
$$43: -11.83$$
$$44: -12$$
$$45: -12.17$$
$$46: -12.33$$
$$47: -12.5$$
$$48: -12.67$$


Does $\lim_{x \to 5} \sqrt{x-5}$ exist, even though $5$ cannot be approached from the left? A basic limit question I don't quite understand... The question is as $$\lim_{x \to 5} \sqrt{x-5}$$ and solve for the limit.
When I plug in $5$ I get the answer zero, but when I tackle this question graphically the limit does not exist as nothing comes from the left of $5$.

Could anyone clear this up for me? Thanks.

It is a common mistake to forget an essential aspect of the definition of the limit: the constraint $|f(x)-L|<\epsilon$ must hold $\forall x\color{red}{\in\text{dom}(f)\setminus\{c\}}:|x-c|<\delta$.

I would say "Yes" but I prefer the answer "It depends upon the definition/notation of limit we are referring to".
With reference to the simplified definition, often used in high school context, that is
$$\lim_{x \to x_0} f(x)=L \iff \lim_{x \to x_0^-} f(x)=\lim_{x \to x_0^+} f(x)=L$$
we should conclude that limit doesn't exists but, according to the more general definition of limit, excluding from the domain the points where the expression is not defined, we have that

$$\lim_{x \to 5} \sqrt{x-5}=:\lim_{x \to 5^+} \sqrt{x-5}=0$$

and we can conclude that the limit exists.
Refer also to the related

What is $\lim_{x \to 0}\frac{\sin(\frac 1x)}{\sin (\frac 1 x)}$ ? Does it exist?
A limit exists iff and only the left limit and the right limit exist and are equal to each other

Here is an example of "simplified" definition

(credits)


Solving an exponential inequality $4^x + 25^x + 0.01^x \lt 2$ without using logarithms I need to solve this inequality without logarithms , 10th class

  $$4^x + 25^x + 0.01^x \lt 2$$

I got $$2^{2x} + 5^{2x} + 0.1^{2x} <2$$ then $$5^{2x} + 0.1^{2x} < 2(1-2^{2x-1}),$$ so after this I realized that the left side is greater than $0$, and so $1-2^{2x-1}$ should be greater than $0$ , and by logic I got that $x$ is less than $0.5$.  This is all I could do.
P.S thank u all

I think I realized how he wants you to do the problem but it is by no means obvious:

$4^x +25^x+10^{-2x} = 2^{2x}+5^{2x}+10^{-2x}= 2^{2x}+5^{2x}+5^{-2x}2^{-2x}$

Now let $u=2^{2x}$ and $v=5^{2x}$

Essentially you are solving the equation:
$$ u+v +u^{-1}v^{-1} < 2 \Rightarrow u+v<2uv \Rightarrow u-2uv+v <0 \Rightarrow (u-v)^2<0$$

But this cannot be true for any real numbers u and v. Since I am assuming you are only searching for real solutions for x the same holds for $(2^{2x}-5^{2x})^2<0$

And as such there are no real solutions.


Generate evenly spaced points on 2D graph I want to draw dots on an image that is W by W pixels. The image is stored as a 1-D array of pixels. The pixel (x,y) is at array index x + y * W.  I am thinking that I can use a fixed step size, N, and draw a dot at every Nth pixel. What value of N will produce equilateral triangles?

Here is the result using W = 100 and N = 673

As you can see the triangles formed are nearly equilateral. Given W, how can I find "good" values for N, which will form near equilateral triangles?

I doubt if there is a simple answer to this question. Below is a plot of the "goodness" of the 10 triangles that can be formed using the first five points. Goodness is min side length over max side length. W = 100, and the first point is at (50,0). The horizontal axis is N, from 600 to 1200.

You can see there are 10 peaks above 0.9 goodness. These correspond to N = 628, 674, 726, 759, 826, 860, 926, 1026, 1137, 1165. The very last "good" spot is at N = 2850 where the points 0,1,2 form a nearly equilateral triangle. I have a theory that for any W, there will be an nearly equilateral formed when N = 0.2885 * W^2 (approximately).


Convergence proof on Newton's method from Boyd & Vandenberghe's book on convex optimization From Boyd & Vandenberghe's book, page 491:

Applying the Lipschitz condition,we have
$$\begin{aligned}
\left\|\nabla f\left(x^{+}\right)\right\|_{2} &=\left\|\nabla f\left(x+\Delta x_{\mathrm{nt}}\right)-\nabla f(x)-\nabla^{2} f(x) \Delta x_{\mathrm{nt}}\right\|_{2} \\
&=\left\|\int_{0}^{1}\left(\nabla^{2} f\left(x+t \Delta x_{\mathrm{nt}}\right)-\nabla^{2} f(x)\right) \Delta x_{\mathrm{nt}} d t\right\|_{2} \\
& \leq \frac{L}{2}\left\|\Delta x_{\mathrm{nt}}\right\|_{2}^{2} \\
&=\frac{L}{2}\left\|\nabla^{2} f(x)^{-1} \nabla f(x)\right\|_{2}^{2} \\
& \leq \frac{L}{2 m^{2}}\|\nabla f(x)\|_{2}^{2}
\end{aligned}$$
to get the inequality assumption (9.33), which is:

$$\frac{L}{2 m^{2}}\left\|\nabla f\left(x^{(k+1)}\right)\right\|_{2} \leq\left(\frac{L}{2 m^{2}}\left\|\nabla f\left(x^{(k)}\right)\right\|_{2}\right)^{2}$$

I dont understand how he get the first two lines of the derivation?

My understanding is that $\nabla f\left(x^{+}\right)$ equals $\nabla f\left(x+\Delta x_{\mathrm{nt}}\right)$ already when t=1, so where do the rest of two terms come from?
please anyone help me, really appreciate.

$-\nabla f(x)-\nabla^{2} f(x) \Delta x_{\mathrm{nt}}=0$ since $\Delta x _{nt}$ is the Newton search direction.


Existance of a Twice Differentiable Function Does there exist a twice differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$f'(x) = f(x+1)-f(x)$$ for all $x$ and $f''(0) \ne 0$.

Some things that I have tried is that obviously any linear function satisfies the first condition but not the second. Any hints/motivation would be helpful. Thanks.

Since we are just looking for existence, this will do.

Assume $f(z)$ is a function on complex plane instead, and assume it have the form $f(z)=e^{cz}$ for some constant $c$. The condition $f''(z)\not=0$ simply means $c\not=0$, while the other condition give: $ce^{cz}=(e^{c}-1)e^{cz}$ which means we need to solve for $e^{c}-c-1=0$. Applying standard technique to reduce to: $-(c+1)e^{-(c+1)}=-\frac{1}{e}$ which allow us to use Lambert W-function to get $c=-1-W(-\frac{1}{e})$. Hence found the solution in complex number.

To reduce this back to real number just take the real part, so we get: $f(x)=e^{Re(c)x}\cos(Im(c)x)$ which satisfy all condition.

Since the question is about existence: you can think of the right side as a difference quotient.
$$f^{'}(x) =\frac{f(x+1)-f(x)}{(x+1)-x}.$$
The right hand side is a slope of the secant line between the points with $x-$coordinates $x$ and $x+1$. So any affine function (with non-zero slope and non-zero constant term) will satisfy this relation. For sure they are twice differentiable.

By affine I mean: $f(x)=ax+b$


Explanation on derivation of this equation? I'm extremely stuck on how my book was able to derive this equation. Basically, it says:

Let $V = W = P_2(\mathbb{R})$. A basis for V is $1, 1+x, 1+x+x^2$. Define the linear transformation $T$ such that $T(1) = x$, $T(1+x) = x^2$ and $T(1+x+x^2) = 1$. If we express some $p(x)$ in terms of the basis we have:

$p(x) = (a_0 - a_1)1 + (a_1 - a_2)(1+x) + a_2(1+x+x^2)$

And it doesn't provide much of an explanation on how it was able to express this in terms of the basis and I've been pondering how for the past hour. Can someone please enlighten me?

Given $p(x) = a_0 + a_1 x + a_2 x^2$, need to find $\alpha_1, \alpha_2, \alpha_3$, such that
$$p(x) = \alpha_1 * (1) + \alpha_2 (1 + x) + \alpha_3 (1 + x +x^2) = a_0 + a_1 x + a_2 x^2$$
Expand LHS,
$$(\alpha_1 + \alpha_2 + \alpha_3) + (\alpha+2 + \alpha_3) x + \alpha_3 x^2 = a_0 + a_1 x + a_2 x^2 $$

We have
$$\alpha_3  = a_2$$,
$$\alpha_2 + \alpha_3 = a_1$$,
$$\alpha_1 + \alpha_2 + \alpha_3 = a_0$$
Solve $\alpha$'s, get the answer.

The RHS can be written as a general quadratic by simplifying,$$p(x) =a_0+a_1x+a_2x^2$$

Since we already know that $\{1,1+x,1+x+x^2\}$ for a basis we know that we can write any polynomial of degree $2$ in terms of the elements of the above set.


Which sets defines a function Let $B$ and $C$ be two arbitrary sets. A set $A\subset B\times C$  defines a function $f: G(f) \to Y$, if:  $(x,y_1),(x,y_2) \in A \Rightarrow y_1 = y_2$.

$G(f) = \{x \in X \mid \exists y \in Y$ such that $ (x,y) \in A\}$ and $f(x) = y \to (x,y) \in A$.

$a) A_1 := \{(x,y) \in \mathbb{R}\times\mathbb{R}: x = y^2\}\text{ or } A_2 := \{(x,y)\in \mathbb{R}\times\mathbb{R}: x = y^3\}$

Which set in $a)$ defines a function and is it surjective or injective?

Every time I think I understood something in math, it hits me so hard for being so naive... Intuitively I would say set $A_1$ is a function. I do understand the symbols and terms but I do not see the relations..

No. $A1$ is not a function. Note that both $(1,1)$ and $(1,-1)$ are elements of $A1$.


Area of octagon constructed in a square The following picture is constructed by connecting each corner of a square with the midpoint of a side from the square that is not adjacent to the corner. These lines create the following red octagon:

The question is, what is the ratio between the area of the octagon and the area of the square. One is supposed to find the solution without a ruler.

By removing some lines, I find it easy to see that the ratio between the yellow area and the square is 1/4. But I am not sure if this helps.

I think the following image will say more than any text. You can divide the image into smaller squares that will allow you immediately to calculate the ratio.

The ratio of the red area within the whole square is the same as the red area in the big green square to the area of the whole green square. And this is (counting in units of the 9 small squares): $(1+1/4+1/4) : 9 = 1.5 : 9 = 1:6$.

I was frustrated by the solution being 1/6 because that is not the result if one calculates the area of a regular octagon with a radius that is L/4, where L is the length of large square.

The gridded image above makes clear that red shape is close to, but not actually, a regular octagon.

Instead, the horizontal distance from the center of the large square to a vertex is 1.5 times the length of the smallest green square. The angled distance from the center of the large square to a vertex is 'sqrt(2)' times (not 1.5 times) the length of the smallest green square. The vertices do not lie on a circle, so it is not a regular octagon (with equal all the included angles being equal), even though the lengths of all the sides are equal.


Show that there is a solution of the Laplace equation $(\mu-A)p=f$ Let $C_0(\mathbb R)$ denote the space of continuous functions vanishing at infinity equipped with the supremum norm, $B$ be a contractive linear operator on $C_0(\mathbb R)$ and $$Af:=\lambda(Bf-f)\;\;\;\text{for }f\in C_0(\mathbb R).$$

  Are we able to show that for some $\mu>0$, there exists a dense subspace $D\subseteq C_0(\mathbb R)$ (maybe even $D=C_0(\mathbb R)$) such that for all $f\in D$, $$\underbrace{(\mu-A)}_{=:\:A_\mu}p=f\tag1$$ has a solution $p\in C_0(\mathbb R)$?

EDIT: Maybe it is as simple as follows: By the Lumer-Phillips theorem, $A$ is the generator of a strongly continuous contraction semigroup $(T(t))_{t\ge0}$ and maybe we can show that $$p:=\int_0^\infty e^{-\mu t}T(t)f\:{\rm d}t$$ is the desired solution of $(1)$ for a given $f$.

The equality $(\mu-A)p=f$ is equivalent to (assume $\lambda + \mu \neq 0$)

$$ p= \frac{\lambda}{\lambda + \mu} Bp + \frac{f}{\mu+ \lambda}.$$

Let $\Phi : X\to X$, $\Phi(p) = \frac{\lambda}{\lambda + \mu} Bp + \frac{f}{\mu+ \lambda}$. Then

$$\| \Phi (p_1) - \Phi (p_2)\| = \frac{\lambda}{|\lambda + \mu|} \| B p_1 - B p_2\|< \frac{\lambda c}{|\lambda + \mu|}\| p_1 - p_2\| $$
for some $c<1$ since $B$ is contractive.

Note that $\Phi$ is also contractive if

$$ \frac{|\lambda|c}{|\lambda + \mu|} < 1 \Leftrightarrow \mu > c|\lambda | - \lambda \text{ or } \mu < -c|\lambda| - \lambda$$

so for these choices of $\mu$, there is $p\in X$ so that $\Phi(p)= p$, which is the same as the equality you want.


How to evaluate the limits of $\int (\sin y) dy$? How to evaluate this integral of $$\int^{\arctan2}_{\arctan \frac{1}{2}} \sin y ~dy$$?

Now I know this is $$-[\cos(\arctan 2)-\cos(\arctan 0.5)]=[\cos(\arctan 0.5)-\cos(\arctan 2)]$$ but I have no idea how to figure this out without a calculator.

Look at a $1,2,\sqrt{5}$ triangle and all will reveal itself.


True or false: The contrapositive statement of "$\text{not } A\implies \text{not } B$" is "$A\implies B$"? Question : State whether 'True' or 'False' and justify your answer giving reason.
The contrapositive statement of "$\text{not } A\implies \text{not } B$" is "$A\implies B$" where $A$ and $B$ are two statements.

I know that when $A\implies B$ is equivalent to it's Contrapositive namely "$\text{not } B\implies \text{not } A$" that is if $B$ doesn't hold then $A$ doesn't hold.
I asked this question because I have never been taught this by my teacher and text given in my book is just given in 3 - 4 lines which is totally insufficient. Kindly help at least a bit.
Even if know that the answer is true but I don't know how to justify it.
I tried to take this example :
$ x\neq -2 \implies x^2\neq4$ is equivalent to it's Contrapositive $x = -2\implies x^2 = 4$

Well, the contrapositive of
$$\neg A\longrightarrow \neg B$$
is $$\neg(\neg B)\longrightarrow \neg(\neg A)$$
but I think we're allowed to know that $\neg \neg A \equiv A, \neg\neg B \equiv B$, hence
the contrapostive is
$$B\longrightarrow A$$
which is not the same as $A\longrightarrow B$. So the answer is false.

In general, the law of contraposition says that the $P\longrightarrow Q$ and its contrapositive
$$\neg Q\longrightarrow \neg P$$
are logically equivalent.

Using the example given, the contrapostive of
$ x\neq -2 \implies x^2\neq4$
is
$$x^2 = 4\implies x = -2.$$

You can think like this :

The contrapositive of ' not A => not B ' is ' not( not A => not B ) ' which simplify to ' A not(=>) B ' which is ' A <- B '


Conclusion that can be drawn from 4 points in space whose angles are $90^{\circ}$ Four points A, B, C and D are in space such that angles $A\hat BC, B\hat CD, C\hat DA$ and $D\hat AB$ are all right angles, then

A, B, C, D cannot be coplanar
A, B, C, D are necessarily coplanar.
A, B, C, D may or may not be coplanar.
No such points exist.

I got an answer by pure imagination. Please provide a proof/reasoning so as to why you feel your answer is correct.

From $\angle{ABC}=90^\circ$, we may suppose that
$$A(0,a,0),\quad B(0,0,0),\quad C(c,0,0)$$
It follows from $\angle{BCD}=90^\circ$ that the $x$-coordinate of $D$ is $c$.

It follows from $\angle{DAB}=90^\circ$ that the $y$-coordinate of $D$ is $a$.

Let $D(c,a,d)$. By the law of cosines,
$$AC^2=DA^2+DC^2-2\cdot DA\cdot DC\cos\angle{CDA},$$
$$\Rightarrow \quad c^2+a^2=c^2+d^2+a^2+d^2\quad\Rightarrow\quad d=0.$$

The four points are on the plane $z=0$.

As ABC is right angle, line segment
1) line segment AB & BC must be on same plane.
2) line segment BC & CD must be on same plane
3) line segment CD & DA must be on same plane
4) line segment DA & AB must be on same plane.
(any two lines are always on same plane)
From points 1 and 2, AB,BC and DA has to share a common plane. From either point 2 or 3, CD must also be in same plane.

So all the line segments must be on same plane and so all points are coplanar.


Finite quotient ring of $\mathbb Z[X]$ Since userxxxxx (I don't remember the numbers) deleted his own question which I find interesting, let me repost it:

  Let $f,g\in\mathbb Z[X]$ with $\mathrm{gcd}(f,g)=1$. Prove that the ring $\mathbb Z[X]/(f,g)$ is finite.

The already-present answers show that $(f,g)$ necessarily contains a non-zero integer $n$, and so the quoitient is equal to $(\mathbb Z/n\mathbb Z)[x]/(f,g).$  So it remains to show that this quotient is finite.

There are various ways to proceed at this point.  Here is one.

First, suppose $n$ is a prime $p$.  Since $f$ and $g$ don't have a non-trivial common factor in $\mathbb Z[x]$, at least one of them is non-zero mod $p$, and the quotient of $\mathbb F_p[x]$ by any non-zero ideal is finite.
Now if $n$ is a prime power, say $p^e$, then by what we just proved, the ideal $(f,g)$ has non-zero image in $\mathbb F_p[X]$, hence it contains a polynomial which is monic when reduced mod $p$.  An induction proves that some power of this polynomial is actually monic mod $p^e$.  Thus $(f,g)$ contains a monic polynomial when reduced mod $p^e$, and so $(\mathbb Z/p^e\mathbb Z)[x]/(f,g)$ is finite.
Factor $n = p_1^{e_1}\cdots p_r^{e_r}$.  Then by CRT, there is an isomorphism

$$(\mathbb Z/n\mathbb Z)[x]/(f,g) \cong \prod_{i =1}^r (\mathbb Z/p^{e_i}_i
\mathbb Z)[x]/(f,g),$$
and so by what we have already proved, $(\mathbb Z/n\mathbb Z)[x]/(f,g)$ is a product of finite rings, hence finite.

Added: Now that some solutions from first principles have been put up, it may be worth describing a way to deduce this statement from general principles of commutative algebra.

First, if $J$ is any ideal in a Noetherian ring $A$, and $I = $rad$(J)$, then since $J \subset I,$ we have that $A/I$ is a quotient of $A/J$.  Thus if $A/J$ is finite, so is $A/I$.  On the other hand, by Noetherianness (more precisely, by the fact that $I$ is f.g.), we have
$I^n \subset J$ for some $n$, and so $A/J$ is a quotient of $A/I^n$, which is filtered by $A\supset I \supset I^2 \supset \cdots \supset I^n.$ Each quotient $I^i/I^{i+1}$ is a f.g. $A/I$-module, and so if $A/I$ is finite, so is each $I^i/I^{i+1}$, hence so is $A/I^n$, and hence so is $A/J$.  Thus $A/J$ is finite if and only if $A/I$ is.

Now take $A = \mathbb Z[x]$.  This is a UFD of Krull dimension two, and so the prime ideals of $A$ are either $0$, height one and hence principal, or maximal.  Let $J = (f,g)$.  Since $f$ and $g$ have no non-trivial common divisor, the ideal $(f,g)$ cannot be contained in a principal prime ideal, and so each minimal prime of $J$ must be a maximal ideal of $A$.  Thus if $I$ is the radical of $J$, then $I = \mathfrak m_1 \cap \cdots \mathfrak m_r$ for some maximal
ideals $\mathfrak m_i$, and so $A/I$ embeds into the product $\prod_{i=1}^r A/\mathfrak m_i$.  Thus to show that $A/I$ is finite, it suffices to show that $A/\mathfrak m$ is finite for any maximal ideal of $A$.  This last fact follows from the general version of the Nullstellensatz for Jacobson rings applied to $\mathbb Z[x]$ (using that $\mathbb Z$ is Jacobson).

Consider $\mathbb{Z}[x]$ as a subring of $\mathbb{Q}[x]$ which is a PID. Since $\rm{gcd}(f,g) = 1$ in $\mathbb{Z}[x]$ the same is true in $\mathbb{Q}[x]$ so $\mathbb{Q}(f,g) = \mathbb{Q}[x]$ which means that $(f,g)$ contains an element that is invertible inside $\mathbb{Q}[x]$, so it contains some integer. Now we know that the quotient is actually a quotient of $(\mathbb{Z}/n\mathbb{Z})[x]$ which means that unless $(f,g)$ consists of just constants, we get a finite ring, as we get a bounded degree. But if this is the case, then both $f$ and $g$ are constants, and $(f,g) = \mathbb{Z}[x]$ since they have gcd equal to $1$.

Edit: The last part is missing some ingredient as noted in the comment.


If $h(x,t)=\int_{\mathbb R}\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy$ why can I permute $\partial _t$, $\partial _x$ and integral? Let $$h(x,t)=\int_{\mathbb R}\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy.$$
Why $$\partial _th(x,t)=\int_{\mathbb R}\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy$$
and $$\partial _xh(x,t)=\int_{\mathbb R}\partial _x\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy\ \ ?$$

I know that if for a.e. $y$, $x\mapsto p(x,y)$ is differentiable and if there is $k\in L^1$ s.t. $|\partial _xp(x,y)|\leq k(y),$

then $$\partial _x\int_{\mathbb R}p(x,y)dy=\int_{\mathbb R}\partial _x p(x,y)dy.$$

Unfortunately, if $p(x,y,t)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}$, I can't bound uniformly in $t$ the function $|\partial _t p(x,y,t)|$ neither bound $|\partial _x p(x,y,t)|$ uniformly in $x$. Any idea ?

In the book PDE of Evans they say : since $\Phi:(x,t)\mapsto \frac{1}{\sqrt{t2\pi}}e^{-\frac{x^2}{2t}}$ is infinitely differentiable on $\mathbb R\times [\delta ,\infty )$ for all $\delta >0$, we have that $$h_t=\int_{\mathbb R}\Phi_t\quad \text{and}\quad h_{xx}=\int_{\mathbb R}\Phi_{xx},$$
but I don't really understand the argument.

For $0 < \delta_1 \leqslant t  \leqslant \delta_2$ we have

$$\left|\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}\right| = \left|\frac{-1}{2\sqrt{2\pi}t^{3/2}}e^{-\frac{(x-y)^2}{2t}} + \frac{(x-y)^2}{2\sqrt{2\pi}t^{5/2}}e^{-\frac{(x-y)^2}{2t}}\right| \\\leqslant \frac{1}{2\sqrt{2\pi}\delta_1^{3/2}}e^{-\frac{(x-y)^2}{2\delta_2}} + \frac{(x-y)^2}{2\sqrt{2\pi}\delta_1^{5/2}}e^{-\frac{(x-y)^2}{2\delta_2}}$$

The RHS is integrable with respect to $y$ over $\mathbb{R}$ and by the Weierstrass M-test we have uniform convergence for $t \in [\delta_1, \delta_2]$ of

$$\int_{\mathbb R}\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}\,dy$$

Clearly, the uniform convergence holds for $t$ in any compact interval contained in $(0,\infty)$ and that is enough to prove that for all $t >0 $,

$$\tag{*}\partial _th(x,t)=\int_{\mathbb R}\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}} \, dy$$

The switching of the partial $x$-derivative and the integral can be justified in a similar way.

We also have

$$\left|\partial _x\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}\right| = \left|\frac{-2(x-y)}{\sqrt{2\pi}t^{3/2}}e^{-\frac{(x-y)^2}{2t}} \right| = \frac{2|x-y|}{\sqrt{2\pi}t^{3/2}}e^{-\frac{|x-y|^2}{2t}}$$

Note that for $\delta_1 \leqslant |x| \leqslant \delta_2$ we have $|x-y| \leqslant |x| + |y| \leqslant \delta_2 +|y|$ and $-|x- y|^2 \leqslant -||x| - |y||^2 = -|y|^2 + 2|x||y| - |x|^2$, and, thus,

$$\exp\left(-\frac{|x-y|^2}{2t}\right) \leqslant \exp\left(-\frac{|y|^2}{2t}\right)\exp\left(\frac{|x||y|}{t}\right)\exp\left(-\frac{|x|^2}{2t}\right) \\ \leqslant \exp\left(-\frac{|y|^2}{2t}\right)\exp\left(\frac{\delta_2|y|^2}{t}\right)\exp\left(-\frac{\delta_1^2}{2t}\right)$$

Again, we can find an integrable upper bound independent of $x$ is a compact interval and can apply the Weierestrass M-test to obtain uniform convergence of the integral of the partial $x$-derivative.  This justifies the integral-derivative switch for all $x \in \mathbb{R}$.

THEOREM 1.(Leibnitz)
Let $f(t,x)$ be continuous in $D=\left[t_1,t_2\right]\times\left[a,b\right]$. If exists the partial derivative $\partial f/\partial t$ and is continuous in $D$, then the function $g:\left[t_1,t_2\right]\rightarrow \textbf{R}$, which is defined as
$$
g(t)=\int^{b}_{a}f(t,x)dx,
$$
is differentiable in $(t_1,t_2)$ and we have
$$
g'(t)=\int^{b}_{a}\frac{\partial f(t,x)}{\partial t}dx\textrm{, }\forall t\in\left(t_1,t_2\right)
$$

In case the integral is generalized, for example
$$
\int^{\infty}_{a}f(t,x)dx,
$$
we have the next theorem

THEOREM 2.
We set $D_N=\left[t_1,t_2\right]\times\left[a,N\right]$ and
$$
g_N(t)=\int^{N}_{a}f(t,x)dx,
$$
for $N$ arbitrary large. If
$$
f(t,x)\textrm{ is continuous in }D_{\infty}\tag 1
$$
and
$$
\partial f/\partial t\textrm{ exists and is continuous in }D_{\infty}\tag 2
$$
and
$$
g(t)=\lim_{N\rightarrow\infty}g_{N}(t)<\infty\textrm{, uniformly in }D_{\infty}\tag 3
$$
and
$$
\exists t_0\in\left[t_1,t_2\right]:
\int^{\infty}_{a}f(t_0,x)dx<\infty.\tag 4
$$
Then we have
$$
\frac{d}{dt}\int^{\infty}_{a}f(t,x)dx=\int^{\infty}_{a}\frac{\partial f(t,x)}{\partial t}dx.
$$
PROOF. From Leibnitz theorem we have
$$
g_N'(t)=\int^{N}_{a}\frac{\partial f(t,x)}{\partial t}dx
$$
Taking the limits (since $g_N(t)$ is term by term differentiatable from (3),(4))
$$
g'(t)=\lim_{N\rightarrow\infty}g_N'(t)=\int^{\infty}_{a}\frac{\partial f(t,x)}{\partial t}dx.
$$
QED

NOTE. The integral you investigate is always constant (for all reals $x,t$, with $t>0$) since if we set
$$
I(x,t):=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{-\infty}\exp\left(-\frac{(x-y)^2}{2t}\right)dy.
$$
For the integral to converge we must have $t>0$.
Then with change of variable $y=x+w$, we get $dy=dw$. Hence
$$
I(x,t)=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{-\infty}\exp\left(-\frac{w^2}{2t}\right)dw=\frac{2}{\sqrt{2\pi t}}\int^{\infty}_{0}\exp\left(-\frac{w^2}{2t}\right)dw
$$
Set now $w^2=u$. Then
$$
I(x,t)=\frac{2}{\sqrt{2\pi t}}\int^{\infty}_{0}e^{-u/(2t)}\frac{1}{2\sqrt{u}}du=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{0}e^{-u/(2t)}u^{-1/2}du.
$$
Now set $u=2vt$. Hence $du=2tdv$ and the integral becomes
$$
I(x,t)=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{0}e^{-v}(2t)^{-1/2}v^{-1/2}2tdv=\frac{\sqrt{2t}}{\sqrt{2\pi t}}\int^{\infty}_{0}e^{-v}v^{1/2-1}dv.
$$
But we have
$$
\Gamma(s)=\int^{\infty}_{0}e^{-v}v^{s-1}dv\textrm{, }s\neq 0,-1,-2,\ldots.
$$
Hence
$$
I(x,t)=\sqrt{\frac{1}{\pi}}\Gamma\left(\frac{1}{2}\right)=1=const
$$

However for pedagogically reasons I will try to apply Theorem 2.

In your case for $x,t$ you have to show that
$$
I^{(1)}_{N}(x,t):=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{N}\exp\left(-\frac{(x-y)^2}{2t}\right)dy>0
$$
is such that
$$
\lim_{N\rightarrow\infty}I^{(1)}_N(x,t)=0\textrm{, uniformly
 }
$$
and the same with
$$
I^{(2)}_N(x,t)=\frac{1}{\sqrt{2\pi t}}\int^{-N}_{-\infty}\exp\left(-\frac{(x-y)^2}{2t}\right)dy=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{N}\exp\left(-\frac{(x+y)^2}{2t}\right)dy
$$
The first of which ($I^{(1)}_{N}(x,t)$) for $y=x+w$ and $N$ large $N>x$, gives
$$
I^{(1)}_N(x,t)=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{N-x}\exp\left(-\frac{w^2}{2t}\right)dw.
$$
As above notes I make the change of variables $z=w^2$. Hence $dw=\frac{dz}{2\sqrt{z}}$, $z>0$. Then
$$
I^{(1)}_N(x,t)=\frac{1}{\sqrt{2\pi t}}\int^{\infty}_{(N-x)^2}\exp\left(-\frac{z}{2t}\right)\frac{dz}{2\sqrt{z}}.
$$
Make now the change of variable $u=z/(2t)$. This will give us:
$$
I^{(1)}_N(x,t)=\frac{1}{\sqrt{2\pi t}}\frac{2t}{2\sqrt{2t}}\int^{\infty}_{(N-x)^2/(2t)}e^{-u}\frac{du}{\sqrt{u}}.
$$
Hence
$$
I^{(1)}_N(x,t)=\frac{1}{\sqrt{2\pi t}}\sqrt{\frac{t}{2}}\int^{\infty}_{(N-x)^2/(2t)}e^{-u}u^{1/2-1}du=\frac{1}{2\sqrt{\pi}}\Gamma\left(\frac{1}{2},\frac{(N-x)^2}{2t}\right).
$$
Hence for $N$ large such that $\frac{(N-x)^2}{2t}=M>1$, then
$$
0<I^{(1)}_{N}(x,t)\leq\int^{\infty}_{M}e^{-u}du=e^{-M}=\exp\left(-\frac{(N-x)^2}{2t}\right)\rightarrow 0 \textrm{, when }N\rightarrow\infty.
$$

The same holds and for $I^{(2)}_N(x,t)$.

Hence we can write according Theorem 2
$$
\frac{d}{dx}\int^{\infty}_{0}\exp\left(-(x-y)^2/(2t)\right)dy=-\int^{\infty}_{0}\frac{x-y}{t}\exp\left(-(x-y)^2/(2t)\right)dy\tag 5
$$
Also in the same way
$$
\int^{0}_{-\infty}\exp\left(-(x-y)^2/(2t)\right)dy=-\int^{0}_{\infty}\exp\left(-(x+y)^2/(2t)\right)dy=
$$
$$
=\int^{\infty}_{0}\exp\left(-(x+y)^2/(2t)\right)dy.
$$
Hence
$$
\frac{d}{dx}\int_{-\infty}^{0}\exp\left(-(x-y)^2/(2t)\right)dy=-\int^{\infty}_{0}\frac{x+y}{t}\exp\left(-(x+y)^2/(2t)\right)dy=
$$
$$
\int^{-\infty}_{0}\frac{x-y}{t}\exp\left(-(x-y)^2/(2t)\right)dy=-\int^{0}_{-\infty}\frac{x-y}{t}\exp\left(-(x-y)^2/(2t)\right)dy\tag 6
$$
From (5) and (6) (adding them) we get
$$
\frac{d}{dx}\int^{\infty}_{-\infty}\exp\left(-\frac{(x-y)^2}{2t}\right)dy=-\int^{\infty}_{-\infty}\frac{x-y}{t}\exp\left(-\frac{(x-y)^2}{2t}\right)dy=
$$
$$
=-\int^{\infty}_{-\infty}\frac{w}{2}e^{-w^2/(2t)}dw=0
$$
Because $we^{-w^2/(2t)}$ is odd function.

About $t$ you have
$$
\frac{d}{dt}\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}\frac{\exp\left(-(x-y)^2/(2t)\right)}{\sqrt{t}}dy=
$$
$$
\int^{\infty}_{-\infty}\exp\left(-(x-y)^2/(2t)\right)\frac{(x-y)^2}{2t^{5/2}}dy-\int^{\infty}_{-\infty}\exp\left(-(x-y)^2/(2t)\right)\frac{dy}{2t^{3/2}}=
$$
$$
=\frac{1}{t^{5/2}}\int^{\infty}_{0}e^{-w^2/(2t)}w^2dw-\frac{1}{t^{3/2}}\int^{\infty}_{0}e^{-w^2/(2t)}dw=
$$
$$
=\frac{1}{t^{5/2}}\int^{\infty}_{0}e^{-z/(2t)}z\frac{dz}{2\sqrt{z}}-\frac{1}{t^{3/2}}\int^{\infty}_{0}e^{-z/(2t)}\frac{dz}{2\sqrt{z}}=
$$
$$
=\frac{2t 2 t}{2t^{5/2}\sqrt{2t}}\int^{\infty}_{0}e^{-u}u^{1/2}du-\frac{2t}{2t^{3/2}\sqrt{2t}}\int^{\infty}_{0}e^{-u}u^{-1/2}du=0
$$
QED


If we extract $3$ marbles randomly from an urn that contains... If we extract $3$ marbles randomly from an urn that contains $6$ white marbles and $5$ black marbles, what is the probability that one is white and the other two black?

Solution. If we take into account the order of extraction and suppose that every possible outcome of the sample space is equiprobable, the probability required is

$\frac{(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)}{11 · 10 · 9}=\frac{3 · 120}{990}=4/
11= 0.3636$

I don't understand the numerator $(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)$. The denominator $11 · 10 · 9$ are the possible cases, you have $11$ possibilities for the first marbles, $10$ for the second, $9$ for the last one.

There are $3$ mutual exclusive possibilities: WBB, BWB, BBW.

This with:

$P(WBB)=\frac6{11}\frac5{10}\frac4{9}$
$P(BWB)=\frac5{11}\frac6{10}\frac4{9}$
$P(BBW)=\frac5{11}\frac4{10}\frac6{9}$

More precise we can write: $$P(BWB)=P(B_1\cap W_2\cap B_3)=P(B_1)P(W_2\mid B_1)P(B_3\mid B_1\cap W_2)=\frac5{11}\frac6{10}\frac4{9}$$where e.g. $W_2$ stands for the event that the second chosen marble is white.

Addition of these probabilities gives you the probability of the mentioned event.

They are considering the three possible orders of selecting one white and two blacks: $WBB, BWB$ and $BBW$. E.g. for $WBB$ (first marble you pick is white, then next two are black), the number of ways to do this is $6\times 5\times 4$, since you can pick the white marble in $6$ ways, then pick a black marble in $5$ ways, then pick the second black marble in $4$ ways.


What is wrong with this proof? -3 = 3 What is wrong with this proof?

$-3 = \sqrt[3]{-27} = {(-27)}^{\frac 13} = {(-27)}^{\frac 26} = \sqrt[6]{{(-27)}^2} = \sqrt[6]{{27}^2} = {(27)}^{\frac 26} = {(27)}^{\frac 13} = \sqrt[3]{27} = 3$

This is obviously false since $-3 \neq 3$.

But still I can't figure out which equation is the wrong one and why that is.

Thanks in advance for anyone who will help.

As suggested in comments, the $(-27)^{2/6} \neq ((-27)^2)^{1/6}$, because $a^{bc}=(a^b)^c$ does not generally hold for $a<0$.

You can find more related info in @mrf's answer here: Is $(-1)^{ab} = (-1)^{ba}$ true? => $(-1)^{ab} = ((-1)^a )^b$ is true?.

you could aswell prove this:

$-1 = \sqrt {(-1)^2}=\sqrt{1}=1$

Clearly the first equality is wrong (and your's third equality is wrong for similar reason).


Is such a condition possible for the provided group? So I came across this question in Gallian's book where we have to prove that a group G is abelian when for every x belonging to G | $x^2=e$ where e is an identity element.

So while thinking about $x^2=e$ , I stumbled upon the idea that -
=> $x^2=e$
=>  $x\circ x=e$ | $\circ$ = binary operation
But $x\circ x=e$ will only be possible if $x=x^{-1}$ ($x$ inverse) because only then
$x\circ x = x\circ x^{-1} = e$

So possible that $x=x^{-1}$ in this case ?

You are correct! By definition, the element $x^{-1}$ is the unique element such that $xx^{-1} = x^{-1}x= e$. Since $x(x) = (x)x = e$, we have that $x=x^{-1}$.

To show that this insight is instrumental in proving that $G$ is abelian, I’ll give you a hint!

  Let $x$ and $y$ be any two elements. We want to show that $xy=yx$. Since $xy\in G$, what can we say about $(xy)^{-1}$?

How about $\mathbb Z_2^ n$?  By which I mean a direct product of $n$ copies of $\mathbb Z_2$...


Erlang distribution and Central Limit Theorem Let $X\sim\textrm{Erlang}(k,\lambda)$. Use the Central Limit Theorem to evaluate

$\underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|>\epsilon \cdot \mathbb{E}\left[X\right]\right)$

where $\epsilon>0$ but small.

According to Larsen and Marx, the central limit theorem states that:

$$\underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|>\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= 1 - \underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|\le\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

Let us try to use the central limit theorem on

$$\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|\le\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= \mathbb{P}\left( -\epsilon \cdot \mathbb{E}\left[X\right] \le X-\mathbb{E}\left[X\right] \le \epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= \mathbb{P}\left( -\epsilon \cdot \frac{k}{\lambda} \le (W_1 + ... + W_k)-\frac{k}{\lambda} \le \epsilon \cdot \frac{k}{\lambda} \right)$$

$$= \mathbb{P}\left( \frac{-\epsilon \cdot \frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \frac{\epsilon \cdot \frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \right)$$

$$= \mathbb{P}\left( -\epsilon \sqrt{k} \le \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \epsilon \sqrt{k} \right)$$

Unfortunately, our 'a' and 'b' here depend on k. Nevertheless, assuming I am interpreting my stochastic calculus class notes correctly, CLT is still applicable (if the random variable didn't depend on k, I think we could've used continuity of probability):

$$= \lim_{k \to \infty} \mathbb{P}\left( -\epsilon \sqrt{k} \le \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \epsilon \sqrt{k} \right)$$

$$= \mathbb{P}\left( \lim_{k \to \infty} -\epsilon \sqrt{k} \le \lim_{k \to \infty} \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \lim_{k \to \infty} \epsilon \sqrt{k} \right)$$

$$= \mathbb{P}\left( -\infty < Z < \infty \right) = 1$$

Hence

$$\underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|>\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= 1 - \underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|\le\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= 1 - 1 = 0$$