What is the importance of eigenvalues/eigenvectors? What is the importance of eigenvalues/eigenvectors?

Short Answer

Eigenvectors make understanding linear transformations easy. They are the "axes" (directions) along which a linear transformation acts simply by "stretching/compressing" and/or "flipping"; eigenvalues give you the factors by which this compression occurs.

The more directions you have along which you understand the behavior of a linear transformation, the easier it is to understand the linear transformation; so you want to have as many linearly independent eigenvectors as possible associated to a single linear transformation.

Slightly Longer Answer

There are a lot of problems that can be modeled with linear transformations, and the eigenvectors give very simply solutions. For example, consider the system of linear differential equations
\begin{align*}
\frac{dx}{dt} &= ax + by\\\
\frac{dy}{dt} &= cx + dy.
\end{align*}
This kind of system arises when you describe, for example, the growth of population of two species that affect one another. For example, you might have that species $x$ is a predator on species $y$; the more $x$ you have, the fewer $y$ will be around to reproduce; but the fewer $y$ that are around, the less food there is for $x$, so fewer $x$s will reproduce; but then fewer $x$s are around so that takes pressure off $y$, which increases; but then there is more food for $x$, so $x$ increases; and so on and so forth. It also arises when you have certain physical phenomena, such a particle on a moving fluid, where the velocity vector depends on the position along the fluid.

Solving this system directly is complicated. But suppose that you could do a change of variable so that instead of working with $x$ and $y$, you could work with $z$ and $w$ (which depend linearly on $x$ and also $y$; that is, $z=\alpha x+\beta y$ for some constants $\alpha$ and $\beta$, and $w=\gamma x + \delta y$, for some constants $\gamma$ and $\delta$) and the system transformed into something like
\begin{align*}
\frac{dz}{dt} &= \kappa z\\\
\frac{dw}{dt} &= \lambda w
\end{align*}
that is, you can "decouple" the system, so that now you are dealing with two independent functions. Then solving this problem becomes rather easy: $z=Ae^{\kappa t}$, and $w=Be^{\lambda t}$. Then you can use the formulas for $z$ and $w$ to find expressions for $x$ and $y$..

Can this be done? Well, it amounts precisely to finding two linearly independent eigenvectors for the matrix $\left(\begin{array}{cc}a & b\\c & d\end{array}\right)$! $z$ and $w$ correspond to the eigenvectors, and $\kappa$ and $\lambda$ to the eigenvalues.  By taking an expression that "mixes" $x$ and $y$, and "decoupling it" into one that acts independently on two different functions, the problem becomes a lot easier.

That is the essence of what one hopes to do with the eigenvectors and eigenvalues: "decouple" the ways in which the linear transformation acts into a number of independent actions along separate "directions", that can be dealt with independently. A lot of problems come down to figuring out these "lines of independent action", and understanding them can really help you figure out what the matrix/linear transformation is "really" doing.

An eigenvector $v$ of a matrix $A$ is a directions unchanged by the linear transformation: $Av=\lambda v$.
An eigenvalue of a matrix is unchanged by a change of coordinates: $\lambda v =Av \Rightarrow \lambda (Bu) = A(Bu)$. These are important invariants of linear transformations.


Elegantly Proving that $~\sqrt[5]{12}~-~\sqrt[12]5~>~\frac12$ $\qquad$ How could we prove, without the aid of a calculator, that $~\sqrt[5]{12}~-~\sqrt[12]5~>~\dfrac12$ ?

I have stumbled upon this mildly interesting numerical coincidence by accident, while pondering on another curios approximation, related to musical intervals. A quick computer search then also revealed that $~\sqrt[7]{12}~-~\sqrt[12]7~>~\tfrac14~$ and $~\sqrt[7]{15}~-~\sqrt[15]7~>~\tfrac13.~$ I am at a loss at finding a meaningful approach for any of the three cases. Moving the negative term to the right hand side, and then exponentiating, is —for painfully obvious reasons— unfeasible. Perhaps some clever manipulation of binomial series might show the way out of this impasse, but I fail to see how...

An approach using binomial series could look as follows:

For small positive $x$ and $y$ one has
$$(1+x)^{1/5}>1+{x\over5}-{2x^2\over25},\qquad (1+y)^{1/12}<1+{y\over12}\ .$$
Using the ${\tt Rationalize}$ command in Mathematica one obtains, e.g., $12^{1/5}\doteq{13\over8}$. In fact
$$12\cdot(8/13)^5-{18\over17}={1398\over 6\,311\,981}>0\ .$$
It follows that
$$12^{1/5}>{13\over8}\left(1+{1\over17}\right)^{1/5}>{13\over8}\left(1+{1\over85}-{2\over 85^2}\right)\doteq1.64367\ .$$
In the same way Mathematica produces $5^{1/12}\doteq{8\over7}$, and one then checks that
$$5\cdot (7/8)^{12}-{141\over140}=-{136\,294\,769\over2\,405\,181\,685\,760}<0\ .$$
It follows that
$$5^{1/12}<{8\over7}\left(1+{1\over140}\right)^{1/12}<{8\over7}\left(1+{1\over12\cdot 140}\right)\doteq1.14354\ .$$
This solution is not as elegant as the solution found by Giovanni Resta, but the involved figures are considerably smaller.


group theory for non-mathematicians A very smart non-mathematician friend is looking to learn about groups, and I was wondering if people might have suggestions (this is NOT a duplicate of this question, since a textbook is not what I am looking for, at least not at first).

Groups and Symmetry: A Guide to Discovering Mathematics, by David W. Farmer.

The highlighted title may convince that it assumes not too much mathematics for the learner. It is very little book, not of the type Definition-Theorem-Proof.

At least (in on-line preview) I don't find a single concrete mathematical statement, but always beautiful pictures and only that.

I would suggest to read, Simon Singh's Fermat's Last Theorem. http://simonsingh.net/books/fermats-last-theorem/

It starts with the introduction to simple problem, solutions. And entire history as it evolves to solve the problem surrounding group theory. And then into advanced theory.


Distance Rate Time problem One morning, Ryan remembered lending a friend a bicycle. After breakfast, Ryan
walked over to the friend’s house at 3 miles per hour, and rode the bike back home at 7
miles per hour, using the same route both ways. The round trip took 1.75 hours. What
distance did Ryan walk?

HINT:

Let the distance between the friend’s house from Ryan's is $d$ miles

As we know speed $\displaystyle =\frac{\text{distance}}{\text{time}}\implies $ time $\displaystyle =\frac{\text{distance}}{\text{speed}}$

So, to go to the friend’s house, he took $\frac d3$ hours

While returning he took $\frac d7$ hours

So,  $\frac d3+\frac d7=1.75$

As given total time taken is 1.75 hrs
So assume total distance is d than
1.75=d/3+d/7

1.75=10d/21
d=3.675miles ans walking distance


6 periods and 5 subjects Question : There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? Is the answer 1800 or 3600 ? I am confused.
Initially this appeared as a simple question. By goggling a bit, I am stuck with two answers. Different sites gives different answers and am unable to decide which is right.
Approach 1 (Source)
we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1
6P5*5C1=3600
Approach 2 (Source)
subjects can be arranged in 6 periods in 6P5 ways.
Remaining 1 period can be arranged in 5P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800
Is any of these approach is right or is the answer different?

Approach 1 is incorrect.  It arranges any five of the courses into one period each, then assigns the left over course to some period.  It double counts each arrangement by having each of the two courses taught at the same time included in the original five.  So the arrangement $12,3,4,5,6$ is counted as $1,3,4,5,6$ plus $2$ in first period and as $2,3,4,5,6$ plus $1$ in first period.  Approach 2 is the same, but acknowledges the double counting in the division by $2!$

I cannot follow the argument in the paragraph starting "Alternatively"

i am a student of class 10 and even i had the same question in my book.

But neither of your answers are correct. The exact answer in 600. I can explain.

_ _ _ _ _ _ imagine these are the 6 space where you should put the 5 subjects.
First 5 periods can be filled in 5P5 ways. i.e. 5!=120 and then the remaining one period can be filled with any one subject from the 5. i.e. 5P1=5.
Then multiply both which is 120*5=600.
And Combination cannot be used because this is a problem based on Permutation.
Combination is mere selection of object. Permutation is an orderly arrangement of object. Here you have to arrange the subjects and not selected.
There you go!


Derivative of definite integral using Taylor's theorem So I want to obtain the following form:
For $x, p \in R^n$

$$\nabla f(x+p) = \nabla f(x) + \int^1_0 \nabla^2 f(x+tp)p \,dt$$ for $f$ twice continuously differentiable and $t \in (0,1)$.

Taylor's theorem in integral form is:

$$f(x) = f(a) + \int^x_a \nabla f(t) (x-t)\,dt$$ from here: https://www.math.upenn.edu/~kazdan/361F15/Notes/Taylor-integral.pdf

My approach is:

$$f(x+p) = f(x) + \int^{x+p}_x \nabla f(t) (x+p-t)\,dt$$
Reformulated as:
$$f(x+p) = f(x) + \int^1_0 \nabla f(x+tp)(1-t)p\,dt$$
$$= f(x) + \int^1_0 \nabla f(x+tp) p \,dt  - \int^1_0 \nabla f(x+tp) t p \,dt$$

But I am not sure how to take the derivatives of the integrals with respect to x; am I at a dead end? Should I instead, just begin with $\nabla f(x)$ in place of $f(x)$ when I apply Taylor's theorem?

Okay, I figured out the problem.
Taylor's theorem in integral form is actually $$f(x) = f(a) + \int^x_a \nabla f(t) \,dt$$

From there, we have $$f(x+p) = f(x) + \int_0^1 \nabla f(x+tp)p\,dt$$

And we can substitute $\nabla f(x)$ for f(x) into the above statement.

EDIT:

The above is incorrect. Using https://www.math.washington.edu/~folland/Math425/taylor2.pdf (page 3), (with the same notation regarding $\alpha$)

$$f(x+p) = f(x) + \sum_{|\alpha| = 1} \frac{p^{\alpha}}{\alpha!} \int_0^1 \partial^{\alpha} f(x+pt) \,dt $$

$$f(x+p) = f(x) + \sum_{j=1}^n p_j \int_0^1 \partial_j f(x+pt) \,dt = f(x) + p \cdot \int_0^1 \nabla f(x+pt)\,dt $$

Taking the gradient of the above:
$$\frac{\partial}{\partial_i} f(x+p) = \frac{\partial}{\partial_i} f(x) + \frac{\partial}{\partial_i} p \cdot \int_0^1 \nabla f(x+pt) \, dt = \frac{\partial}{\partial_i} f(x) + p \cdot \int^1_0 \frac{\partial}{\partial_i} \nabla f(x+pt) \,dt $$

We have overall that

$$\nabla f(x+p) = \nabla f(x) + \int^1_0 \nabla^2 f(x+pt) p \, dt$$


connected sum of surface When I read Massey's Algebraic Topology:An Introduction,page 9,he points out that the topological type of $S_1$#$S_2$(here $S_i$ is surface,# is connected sum,i.e.,cutting an open disc $D_i$ in each surface,and then gluing the boundary circle through a homeomorphism h) does not depend on the choice of discs $D_i$ or the choice of the homeomorphism h.

The independence of h is quite evident,but I don't know why the choice of discs is irrelevant.Is there any refference?

Wikipedia has an extremely short "Disc Theorem" entry which references Palais'  "Extending diffeomorphisms" paper. It says

Disc Theorem. For a smooth, connected, $n$-dimensional manifold $M$, if $f, f'\colon D^n \to M$ are two equi-oriented embeddings then they are ambiently isotopic.

This is one of the fundamental results in Differential Topology, and in particular it implies that connected sums are well-defined wrt choice of embeddings. There might be a simpler proof in $2$D, but this is the standard result which is typically cited for all dimensions.

(Here "ambiently isotopic" means there is an isotopy $H: M\times I \to M$ which begins at the identity map and induces an isotopy between $f$ and $f'$; "equi-oriented" means that $f$ preserves orientation iff $f'$ does. In fact the proof of Palais' theorem shows a bit more: the ambient isotopy can be chosen to be fixed outside of a compact, contractible subspace containing the images of $f$ and $f'$.)

It's been a while since I looked at it, but a proof sketch sort of goes like this: First choose a small open tube around the unit interval $I\subset U\subset \mathbb{R}^2$ and an embedding $\gamma\colon \bar U \to M$ where $\gamma(0) = f(0)$, $\gamma(1) = f'(0)$. Now pick a small disc  $D\subset U$ centred at $0$, and construct an ambient isotopy in the tube $F\colon U\times I \to U$ which transports $D$ to a disc centred at $1$. Then you have to construct ambient isotopies $H_1, H_2$ on $M$ which shrink $f(D^n)$ and $f'(D^n)$ down to $\gamma(D)$ and $\gamma(F_1(D))$ respectively (I guess this step uses a linearization trick). Then these three isotopies are pieced together to give the result.

If you consider two different locations/ sizes of the disc on one surface, you can define a homotopy on that surface that transforms one disc to the other. This works because all disks are homotopically trivial. Hence the topological type of the connected sum does not depend on the particular choice of discs.


You are not setting up your induction very well. You should not start with the equality you want to establish, namely that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$. Instead, you should start with the Induction Hypothesis, which is that $(-1)^n - 10^n$ is a multiple of $11$.

So: the Inductive Step is to show that if $(-1)^n - 10^n$ is a multiple of $11$, then $(-1)^{n+1} - 10^{n+1}$ is also a multiple of $11$.

Let's write out our Induction Hypothesis: it says that
$$\text{There exists an integer }k\text{ such that }(-1)^n - 10^n = 11k.$$

What we want to prove is that:
$$\text{there exists an integer }\ell\text{ such that }(-1)^{n+1}-10^{n+1}=11\ell.$$
(Note that the multiple may be different, that's why I used a different letter).

So now we can try manipulating the expression we want. One possibility is to use the following identity:
$$a^{n+1}-b^{n+1} = (a-b)(a^n + a^{n-1}b + a^{n-2}b^2 + \cdots + ab^{n-1}+b^n),$$
if you already know this identity.

So we have, with $a=-1$ and $b=10$,
$$
(-1)^{n+1} - 10^{n+1} = \Bigl( (-1) - 10\Bigr)\Bigl( (-1)^n + (-1)^{n-1}(10) + \cdots + (-1)10^{n-1} + 10^n\Bigr).$$
Now notice that you don't even need to use the induction hypothesis to conclude that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$ (as could be seen in mac's answer).

If you don't know the identity, then you can perform some purely algebraic manipulations. E.g.,
$$\begin{align*}
(-1)^{n+1} - 10^{n+1} &= -1\left( (-1)^n + 10^{n+1}\right)\\
 &= -\left( (-1)^n -10^n + 10^n + 10^{n+1}\right)\\
&= -\left( \Bigl((-1)^n - 10^n\Bigr) + 10^n\Bigl(1 + 10\Bigr)\right)\\
&= -\left( 11k + 10^n(11)\right) &\quad&\text{(by the induction hypothesis)}\\
&= -\left( 11(k+10^n)\right)\\
&= 11\left( -(k+10^n)\right),
\end{align*}$$
which gives that $(-1)^{n+1} - 10^{n+1}$ is a multiple of $11$, as desired, from the assumption that $(-1)^n - 10^n$ is a multiple of $11$.

But easier still is to use the following property of congruences:

Proposition. Let $a,b,c,d,k$ be integers. If
$$a\equiv b\pmod{k}\qquad\text{and}\qquad c\equiv d\pmod{k}$$
then $ac\equiv bd\pmod{k}$.

Proof. Since $a\equiv b\pmod{k}$, then $k|a-b$, so $k$ divides any multiple of $a-b$; for example, $k|(a-b)c = ac-bc$. Since $k$ divides $ac-bc$, then $ac\equiv bc\pmod{k}$.

Since $c\equiv d\pmod{k}$, then $k|c-d$, so $k|(c-d)b = cb-db$, hence $bc\equiv bd\pmod{k}$.

Since $ac\equiv bc\pmod{k}$ and $bc\equiv bd\pmod{k}$, then $ac\equiv bd\pmod{k}$. QED

Corollary. If $a_1\equiv b_1\pmod{k}$, $a_2\equiv b_2\pmod{k},\ldots, a_n\equiv b_n\pmod{k}$, then
$$a_1\cdots a_n\equiv b_1\cdots b_n\pmod{k}.$$

Proof. Induction on $n$. QED

(This is where you would want to use induction, rather than the specific case you are looking at).

Corollary. If $a\equiv b\pmod{k}$, then for all positive integers $n$, $a^n\equiv b^n\pmod{k}$.

Proof. Apply previous corollary with $a_i=a$ and $b_i=b$ for all $i$. QED


If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?

I know how to solve a problem like

"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$"

by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:

Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$  to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$

Where I can not make progress with the question

"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".

Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?

What I have tried:

If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by  by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?

We have
$$s^2=(\cos\alpha+\sin\alpha)^2=1+2\cos\alpha\sin\alpha=(0.2)^2$$
so we find
$$p=\sin\alpha\cos\alpha=-0.48$$
hence $\sin\alpha$ and $\cos\alpha$ are roots of the quadratic equation:
$$x^2-sx+p=x^2-0.2x-0.48=0$$

\begin{align}
  \sin \alpha+\cos \alpha &= 0.2 \\
  (\sin \alpha+\cos \alpha)^2 &= 0.04 \\
  \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos\alpha &= 0.04 \\
  1+\sin 2\alpha &= 0.04 \\
  \sin 2\alpha &= -0.96
\end{align}

and that would be straight forward to proceed


Convergence of $\sum_{n=1}^\infty \frac{1}{n^{a_n}}$ Let $\lim _{n\to\infty}a_n=l$. Show that $\sum_{n=1}^\infty \frac{1}{n^{a_n}}$ converges if $l>1$ and diverges if $l<1$. What happens if $l=1$?

I tried to use the ratio test, but could not get a good estimate, I have difficulties, that there is the series $a_n$ involved. I know that $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if $p>1$ and diverges if $p\geq 1$ but I am not sure how to use this exactly.

For $l=1$, I guess both things could happen? Definitely, we can take $a_n=1$ for all $n$ and then we get the harmonic series which is divergent. Is there an example where the series converges?

If $l=1$ nothing can be said, since $\sum_{n\geq 1}\frac{1}{n\log^2(n+1)}$ is convergent but $\sum_{n\geq 1}\frac{1}{n\log(n+1)}$ is divergent by Cauchy's condensation test.

Hints:
if $l > 1$ then proof that there is $N$ such that $a_n > 1$ for any $n>N$
same method for $l < 1$

for $l = 1$
consider the series $a_n = 1 + 1/n$


Direct sum seen as functor. We know that if $(M_i)_{i\in I}, (N_i)_{i\in\ I}$ and $(P_i)_{i\in\ I}$ are three families of $R$-modules, where $R$ is a ring with unity, then $$M_i\xrightarrow{f_i}N_i\xrightarrow{g_i}P_i$$ is an exact sequence if, and only if, $$\bigoplus_{i\in\ I} M_i\xrightarrow{\oplus f_i}\bigoplus_{i\in I}N_i\xrightarrow{\oplus g_i}\bigoplus_{i\in\ I}P_i$$ is exact.

I would like to know if it is true for an arbitray abelian category when the set $I$ is finite.

Yes, because homology is preserved by finite direct sums. If $M_i\to N_i\to P_i$
is a complex but not exact, it has nonzero homology $H_i$ at the middle terms. The direct sum $\bigoplus M_i\to\bigoplus N_i\to \bigoplus P_i$
will have homology $\bigoplus H_i$ at the middle term which doesn't vanish.

Of course if any of the $M_i\to N_i\to P_i$ isn't a complex then neither
is the direct sum.


Generating elements of the free group of rank two up to conjugacy I've tried to google how to generate words in two generators up to conjugacy (this means I only want one representative for each conjugacy class). Sadly, I come up with articles that have, in the introduction, things like "It is well known that the elements of the free group of rank two can be enumerated by the rationals".

I think I understood that there is a bijection between the conjugacy classes of elements of the free group on two generators and the rational numbers, and this bijection has something to do with the continued fraction associated with a given rational number. I couldn't find any reference on the topic though, so if anyone could give me one I would be grateful.

I would also appreciate any other method to compute the conjugacy classes of the free group on two generators. (I need an algorithm for a program I am writing)

Thank you!

Every word is conjugate to a cyclically reduced word, and the cyclically reduced conjugates of a cyclically reduced word are all cyclic permutations.

This comes directly from wikipedia.

From here on, when I say a cyclically reduced word, I mean the smallest cyclic reduced word in the lexicographical order in the set of all its cyclic permutations.

The quote shows:

You can find a cyclically reduced representative for a conjugacy class.
A conjugacy class is uniquely determined by one cyclically reduced word.

The question becomes how to find a list of cyclically reduced word over two alphabets.
A cyclically reduced word is called a necklace. One can search for "necklace algorithms" to get more infomation.

Practically speaking, you might just look at the answer to this question: Good simple algorithm for generating necklaces in Scheme?.

If you just want something easy to implement and doesn't care about the running time:

Generate the next word. (The alphabet is just 1,-1,2,-2, so one can represent a word in base 4 and count up)
Check if the word is cyclically reduced. If not, go back to 1.
Rotate the word, see if this word is lexicographically smallest in all its rotations. If not, go back to 1. If yes, print this word.


What is the derivative of a vector with respect to its transpose? I've already looked at Vector derivative w.r.t its transpose $\frac{d(Ax)}{d(x^T)}$, but I wasn't able to find the direct answer to my question in that question. What is the value of $$\frac{d}{dx} x^T\text{ ?}$$
My initial intuition is that it is $1$, but I'm not exactly sure of why that would be so.

What sort of object can be the derivative of a vector-valued function whose values are row vectors and whose arguments are column vectors?  Generally, what kind of object can be the derivative of a function whose values are members of one vector space $W$ and whose arguments are members of another vector space $V$?

$$
f: V\to W
$$

The answer is that the value of such a derivative at any point in $V$ is a linear transformation from $V$ into $W$, and it may be a different linear transformation at each point in $V$.  But if $f$ is itself linear, then it's the same linear transformation at each point in $V$: it's $f$ itself.

Transposition is linear.  Therefore the value of its derivative at each point in its domain is itself.

Often one represents a linear transformation by a matrix.  What would be the matrix in this case? No matter what basis you pick for the domain $V$, it seems natural to pick as a basis of $W$ the set of transposes of the basis vectors you chose for $V$.  In that case, the matrix would be the identity matrix.

That depends on how you define vector derivative. There are generally two ways. One is applying abstract index notation, then
$$\frac{d}{dx}x^T=\left(\frac{dx_i}{dx^j}\right)=(\delta_{ij})=(e_1\otimes\cdots\otimes e_n)^T$$
where $e_i$s are unit vector whose $i$ th component is one and zero otherwise.

Another way to look at it is to regard as directional derivative, then
$$\frac{d}{dx}x^T=\lim_{h\to0}\frac{(x+hx)^T-x^T}{h}=x^T$$


The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $f(x) f(y) = f(x + y) + xy$ for all real numbers $x$ and $y.$ Find all possible functions $f.$ I was trying to solve the problem
:

The function $f : \mathbb{R} \to \mathbb{R}$ satisfies
$f(x) f(y) = f(x + y) + xy$ for all real numbers $x$ and $y.$ Find all possible functions $f.$

I started by substituting in $0$ for both, to find that $f(0) = 1$, as it becomes $f(0)f(x) = f(x)$ which can only mean $f(0) = 1$. However past this point, I started struggling, as setting both to $1$, or $1$ and $0$ doesn't reveal anything new about the problem. Trying to set both $f(x)$ terms equal to each-other didn't help either. Thanks!

Here is a solution. $f(x)f(1)-x=f(x+1)$. So $f(-1)f(1)=0$ So either $f(1)=0$ or $f(-1)=0$.  Suppose $f(1)=0$. Then $f(x+1)=-x$. Hence $f(x)=-x+1$ for every $x$ which does  satisfy the condition. Assume $f(-1)=0$. Then $f(x-1) = x$ or $ f(x)=x+1$ which is Ok. Answer: $f(x)=x+1$.or $-x+1$.

As you said we can say f(0)=1 and then substituting y=-x, we get
$f(x)f(-x)=f(0)-x^2$
$f(x)f(-x)=1-x^2$
Considering polynomials, we can say that the function is a linear polynomial.
Assuming $f(x)=ax+b$ we get x+1, -x+1 as the possible functions.


How can be proven that any number X is greater,lesser or equal to any other number Y? I have looked for it on the internet, really, but all I have found are particular cases like 1 > 0, or such. Is there an algebraic proof for proving that x > y or, x = y, or x < y? I thought of using Euclid's Fundamental theorem of Arithmetic as a tool, and comparing the primes that produce a compound number, but I still would have to prove that those primes (and primes in general) are ones greater (or lesser) than others.

Different text may do these differently but they are all equivalent.

x < y;x=y; or y < x are mutually exclusive and exhaustive by definition.

Addition and multiplication (and the inverses subtraction and division) are given by fiat so that a+0=a 1xa =a, a+(-a)=0.  If a ne 0 then ax1/a =1 and a (b+c)=ab+bc.  These are axioms.

From these we can prove basic this like (-(-a))=a and (-a)(-b)=ab and 0xa=0 and so on.

We are then given two essential axioms of order.

x+y < x+z if y < z

And xy > 0 if x >0 and y > 0.

From there we can prove everything.

We can prove if x > 0 then - x < 0

If x >0 and y < z then xy < xz.

That $x^2 > 0$ is $x \ne 0$ and in particular $1^2 = 1 > 0$.

That if 0 < x < y then 0 < 1/y < 1/x

So that's how we can compare any two rational numbers.

But what of irrational.  Well when we define the reals it is as the limits of sequences of rationals and the laws of order are extended.

It seems that you're asking how to prove that $\le$ is a total ordering. The answer depends on the definition of $\le$. If I define $x\prec y$ iff $x$ perfectly divides $y$ then $2 \prec 4$ but $3\not\prec 4$ and $4\not\prec 3$ and $3\not= 4$.

Suppose I define '$x\le y$' based upon the sign of $(y-x)$.

$(y-x)$ zero or positive : $x\le y$

$(y-x)$ zero or negative : $y\le x$

Since the difference of two numbers must be at least one of these cases, the ordering has totality. To be a total order I'd also have to show

(antisymmetry) if $a\leq b$ and $b\leq a$ then $a=b$ and

(transitivity) if $a\leq b$ and $b\leq c$ then $a\leq c$


How to solve $2 \sin^2(x)+\cos^2(x)+\cos(x)=0$ I need to solve the following equation:

$$2 \sin^2(x)+\cos^2(x)+\cos(x)=0$$

I think that $\sin^2(x)+\cos^2(x) = 1$ is helpful, but I don't know how to apply it here.

You can rewrite the equation to

$$2(1-\cos^2x) + \cos^2 x + \cos x = 0$$

Now, introduce a new variable: $y=\cos x$ and first solve for $y$.

Then, for each solution $y$, every $x$ that solves the equation $\cos x = y$ solves the original equation.

$2sen^2(x)+cos^2(x)+cos(x)=sen^2(x)+sen^2(x)+cos^2(x)+cos(x)=sen^2(x)+cos(x)+1=(1-cos^2(x))+cos(x)+1=2+cos(x)-cos^2(x)=0$.

This last equation is simpler.

Take $z:=cos(x)$ then $z^2-z-2=0$


Find all incongruent solutions to $21x \equiv 14 \pmod{91}$ Find all incongruent solutions to $21x \equiv 14 \pmod{91}$.

I am able to work out the solution using Euclidean algorithm techniques,  but the signs on the expression do not match up with the initial expression when I check my work.  So by the linear congruence theorem, my solution has to satisfy: $$21x - 91y = 14$$  but after going through the process with the $\gcd(21, 91)$ my expression ends up as $$91 - 21(4) = 7$$ which I multiply by $2$ to get: $$91(2) - 21(8) = 14$$
Which would mean my solution has to have a negative somewhere in it. I can "put" a negative on one of my values and the original expression would be satisfied but that is not what I obtained through the work I did. Is the confusion in signs occurring on purpose or am I treating something wrong?

By definition, the congruence
$$21x \equiv 14 \pmod{91} \tag{1}$$
is equivalent to the equation
$$21x = 14 + 91t, t \in \mathbb{Z} \tag{2}$$
If we divide each term of equation 2 by $7$, we obtain the equivalent equation
$$3x = 2 + 13t, t \in \mathbb{Z}$$
which is equivalent to the congruence
$$3x \equiv 2 \pmod{13} \tag{3}$$
Hence,
$$21x \equiv 14 \pmod{91} \Longleftrightarrow 3x \equiv 2 \pmod{13}$$
Since $\gcd(3, 13) = 1$, the congruence $3x \equiv 2 \pmod{13}$ has a solution.  We can find it by applying the extended Euclidean algorithm.
\begin{align*}
13 & = 4 \cdot 3 + 1\\
3 & = 3 \cdot 1
\end{align*}
Solving for $1$ in terms of $3$ and $13$ yields
$$1 = 13 - 4 \cdot 3$$
Thus,
$$1 \equiv -4 \cdot 3 \pmod{13} \implies -4 \equiv 3^{-1} \pmod{13}$$
Therefore, if we multiply both sides of congruence 3 by $-4$, we obtain
$$x \equiv -8 \pmod{13}$$
To find all the solutions of congruence 1, we must find all the solutions of the inequality
$$0 \leq -8 + 13t < 91$$
in the integers.
\begin{align*}
0 & \leq -8 + 13t < 91\\
8 & \leq 13t < 99\\
\end{align*}
Hence, $1 \leq t \leq 7$.  Therefore, the solutions of the congruence $21x \equiv 14 \pmod{91}$ are
\begin{align*}
x & \equiv 5 \pmod{91}\\
  & \equiv 18 \pmod{91}\\
  & \equiv 31 \pmod{91}\\
  & \equiv 44 \pmod{91}\\
  & \equiv 57 \pmod{91}\\
  & \equiv 70 \pmod{91}\\
  & \equiv 83 \pmod{91}
\end{align*}
which you can check by direct computation.

$$21x\equiv 14\pmod{91}\stackrel{:7}\iff 3x\equiv 2\equiv 15\pmod{13}$$

$$\stackrel{:3}\iff x\equiv 5\pmod{13}$$

All integers of the form $13k+5$ for some $k\in\mathbb Z$ are the solutions.


Intersection of two finite abelian subgroups I have a question concerning a proof that a group of order 144 is not simple.

Given two Sylow 3-subgroups, $P$ and $Q$, we know that $P$ and $Q$ are both abelian as they are of order $p^2$. Let $M=N_G(P \cap Q)$. Then $P \cap Q$ is a normal subgroup and therefore normal in both $P$ and $Q$. This is where I am running in to a problem, the proof goes on to conclude that since $P$ and $Q$ are subset of $M$, then $PQ \subseteq M$. I am not sure how to prove this to myself.

Also, $|P \cap Q|=3$.

Let $G$ be a group of order 144 and assume that $G$ is simple. We will argue through an analysis of the Sylow 3-subgroups of $G$ to arrive at a contradiction. On the way we will use a result maybe less known.
Lemma Let $G$ be a group and $p$ a prime dividing the order of $G$. Assume that for every pair $P, Q \in Syl_p(G)$, $P=Q$ or $P \cap Q = 1$. Then $n_p=|Syl_p(G)| \equiv 1$ mod $|P|$.Note that this generalizes the well-known result that $n_p \equiv 1$ mod $p$. I will not prove the lemma, but to sketch it: fix a $P \in Syl_p(G)$ and let this group act on the set of all Sylow $p$-subgroups by conjugation. The orbit of $P$ itself is $\{P\}$ and if $Q \neq P$, the orbit of $Q$ has length $|P|$.Let's get on with the analysis. Since $G$ is simple there are at least two different $P, Q \in Syl_3(G)$. Put $D = P \cap Q$. Of course $D$ is a proper subgroup of $G$. We are going to show that in fact $D=1$. In that case we can apply the lemma, $n_3 \equiv 1$ mod 9 and together with the fact that $n_3 \in \{1,2,4,8,16\}$, this yields $n_3 =1$, which means that the Sylow 3-subgroup is normal, against our assumption $G$ being simple.Now suppose $D \neq 1$. Since $P \neq Q$, $D$ is a proper subgroup of $P$, but $|P|=9$, so $|D|=3$. Put $H=N_G(D)$. Indeed $P,Q \subset H$, because $P$ and $Q$ are abelian. Note that since $P$ and $Q$ are different subgroups, $P\neq H$ and index$[H:P] \neq 2$, so index$[H:P] \geq 4$. On the other hand, index$[G:H]$ cannot be $2$, otherwise $H$ would be normal in the simple group $G$, so it must be a divisor of $16$ and be at least $4$. All this can only be the case if $|H|=36$, or equivalently, index $[G:H]=4$. Of course core$_G(H)=1$ and this means that $G/$core$_G(H)=G$ can be embedded in $S_4$, which is absurd since $144 \nmid 24$. This is the final contradiction and hence $G$ cannot be simple.

As $P$ and $Q$ are both abelian, each normalizes $P\cap Q$, so by definition is in $M$.  Let me add that it may not be true that $PQ$ is a group.  In general, if $P$ and $Q$ are subgroups of $M$, then $\langle P, Q\rangle$ is a subgroup of $M$.  However, in this case $\langle P, Q\rangle= PQ$ because $P$ and $Q$ are abelian.


least value of $\lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{c+b}{a}\rfloor+\lfloor \frac{a+c}{b}\rfloor$ If $a,b ,c>0$ . Then least value of $$\bigg\lfloor \frac{a+b}{c}\bigg\rfloor+\bigg\lfloor \frac{c+b}{a}\bigg\rfloor+\bigg\lfloor \frac{a+c}{b}\bigg\rfloor$$

Where $\lfloor x\rfloor$ is floor function of $x$

Plan

Using $$x-1< \lfloor x\rfloor\leq x$$

$$\frac{a+b}{c}-1< \bigg\lfloor \frac{a+b}{c}\bigg\rfloor \leq   \frac{a+b}{c}$$

$$\frac{b+c}{a}-1< \bigg\lfloor \frac{b+c}{a}\bigg\rfloor \leq\frac{b+c}{a}$$

$$\frac{c+a}{b}-1< \bigg\lfloor\frac{c+a}{b}\bigg\rfloor \leq \frac{c+a}{b}$$

How do i solve it Help me please

$$= 3 \bigg\lfloor \frac{a+b}{c} \bigg\rfloor$$

And given all variables are positive, the minimum occurs when $a+b$ is as close to $0$ as possible and $c$ is as large as possible.


Linear programming solution in vertex I want to prove that if linear programming problem
$$\max \{\langle c,x\rangle \ \colon Ax\leqslant b, \ x\geqslant 0\}$$
has a solution, then atleast one of the solutions is in the vertex of
$$\Omega=\{x\  \colon Ax\leqslant b, \ x\geqslant 0\}.$$
Any ideas on how to approach this problem?

This is actually the main theorem of LP theory:

Given a problem in standard form
\begin{align}
\max\ &c^{\top}x\\
&Ax=b\\
&x\geq 0
\end{align}
let $\Omega =\{x\in \mathbb{R}^n\ |\ Ax=b,\ x \geq 0\}$ be the feasible set. Assume that $rank(k)=m$. Then

If $\Omega \neq \emptyset$ then there exists at least a basic feasible solution to the problem (i.e. $\Omega$ has at least a vertex).
If the problem is not unbounded then there exists an optimal basic solution.

Proof of statement 1

Let $x$ be a feasible solution. WLOG assume that $x_1, x_2,\ldots, x_p > 0$ and $x_{p+1},\ldots, x_n =0$. If $A_1,\ldots,A_p$ are linearly independent columns of $A$ then $x$ is a basic feasible solution. Otherwise $A_1,\ldots,A_p$ are linearly dependent and $$\sum_{i=1}^p\lambda_iA_i=0$$
holds with at least one coefficient $\lambda_i \neq 0$. Observe that the equation system can be written as
 $$\sum_{i=1}^px_iA_i=b$$
Multiplying the first equation by $\epsilon \in \mathbb{R}$ and subtracting it from the last one we get
$$\sum_{i=1}^px_iA_i - \epsilon\sum_{i=1}^p\lambda_iA_i=\sum_{i=1}^p(x_i - \epsilon\lambda_i)A_i=b$$

Therefore the vector
$$x-\epsilon \lambda=[x_1-\epsilon \lambda_1,\ldots,x_p-\epsilon\lambda_p,0,\ldots,0]^{\top}$$
will be feasible if
$$x_i-\epsilon\lambda_i \geq 0 \ \ \  i=1,\ldots,p$$
The solution of this system of inequalities is $\epsilon=\min \{\epsilon_1, \epsilon_2\}$ where
$$\epsilon_1 = \max_{1\leq i\leq p}\Big\{\dfrac{x_i}{\lambda_i}\ |\ \lambda_i < 0\Big\}$$
and
$$\epsilon_2 = \min_{1\leq i\leq p}\Big\{\dfrac{x_i}{\lambda_i}\ |\ \lambda_i > 0\Big\}$$
Taking $\epsilon=\epsilon_1$ or $\epsilon=\epsilon_2$ the vector $\bar{x}=x-\epsilon \lambda$ has at least one more null component. Now check if the columns of $A$ related to the non null components of $\bar{x}$ are linearly independent. If they are linearly independent $\bar{x}$ is a basic feasible solution; otherwise repeat the whole procedure starting from $\bar{x}$.

Proof of statement 2

Let $x$ an optimal solution. If it is a basic feasible solution then we get the proof. If it is not basic, as in the statement 1, we can always construct a new vector $\bar{x}=x-\epsilon \lambda$ which is basic.
The objective function value at $\bar{x}$ is
$$c^{\top}\bar{x}=c^{\top}x-\epsilon c^{\top}\lambda$$

All we need to show for $\bar{x}$ to be optimal is that $c^{\top}\lambda=0$. Observe that

if $c^{\top}\lambda>0$, taking $\epsilon=\epsilon_1<0$ we get
$$c^{\top}\bar{x} > c^{\top}x $$
if $c^{\top}\lambda<0$
taking $\epsilon=\epsilon_2>0$ we get
$$c^{\top}\bar{x} > c^{\top}x $$
In both cases we get a contradiction. Therefore it is $c^{\top}\lambda=0$, $c^{\top}\bar{x} = c^{\top}x $  and $\bar{x}$ is an optimal basic solution.
QED

REMARKS

The theorem refers to problems with equality constraints (standard form problems). As you well know every LP problem can be transformed in standard form, so the theorem applies to all LP problems.
The feasible sets of a generic LP problem and the corresponding standard form problem have the same shape, although they lie in different spaces. Thus there is a one-to-one correspondence between the vertexes of the two feasible sets.
The theorem uses the concept of basic solution, but a well-known theorem states that $x$ is a vertex of $\Omega$ if and only if $x$ is a basic feasible solution of the system $Ax=b$.

This is a consequence of the Extreme Value Theorem.

I think the best way to understand it is by looking at what happens in the $1$-dimensional case (obviously this is not a proof, but it gives you the intuition as to why it holds).

An example, assume you are dealing with
$$
\max\limits_{x \in \mathbb{R}}\{2x\; |\; x\le 2,\; x \ge 0 \}
$$

So here, $\Omega = [0,2]$. Since $y=2x$ is linear, it is clear from a quick drawing that the maximum is in $x=2$, a vertex of $\Omega$.

Try doing the same thing in $2$-D to convince yourself a little more.


Is it impossible to perfectly fit a polynomial to a trigonometric function on a closed interval? On a closed interval (e.g. $[-\pi, \pi]$), $\cos{x}$ has finitely many zeros. Thus I wonder if we could fit a finite degree polynomial $p:\mathbb{R} \to \mathbb{R}$ perfectly to $\cos{x}$ on a closed interval such as $[-\pi, \pi]$.

The Taylor series is

$$\cos{x} = \sum_{i=0}^{\infty} (-1)^i\frac{x^{2i}}{(2i)!} = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!}-\dots$$

Using Desmos to graph $\cos{x}$ and $1-\frac{x^2}{2}$ yields:

which is clearly imperfect on $[-\pi,\pi]$. Using a degree 8 polynomial (the first 5 terms of the Taylor series above) looks more promising:

But upon zooming in very closely, the approximation is still imperfect:

There is no finite degree polynomial that equals $\cos{x}$ on all of $\mathbb{R}$ (although I do not know how to prove this either), but can we prove that no finite degree polynomial can perfectly equal $\cos{x}$ on any closed interval $[a,b]\subseteq \mathbb{R}$? Would it be as simple as proving that the remainder term in Taylor's Theorem cannot equal 0? But this would only prove that no Taylor polynomial can perfectly fit $\cos{x}$ on a closed interval...

Yes, it is impossible.

Pick any point in the interior of the interval, and any polynomial. If you differentiate the polynomial repeatedly at that point, you will eventually get only zeroes. This doesn't happen for the cosine function, which instead repeats in an infinite cycle of length $4$. Thus the cosine function cannot be a polynomial on a domain with non-empty interior.

Although this is an incomplete answer unlike the ones I read here, I'd like to offer what I eventually thought of since the idea still seems original:  There can be no polynomial with rational coefficients that exactly approximates $\cos$ on $[0,1]$, because it will have a wrong integral over this interval ($\sin 1$ being irrational).  I believe this argument can be adapted to a different interval $[\alpha,\beta]$ by finding a sub-interval with rational endpoints $[a,b] \subset [\alpha,\beta]$ and using something like the notion of algebraic independence over $\mathbb{Q}$ (search for $a$ and $b$ such that $\sin b - \sin a$ be irrational?  Which should happen most of the time) and/or Niven's theorem, and possibly enhanced to real coefficients since a polynomial with such coefficients can be well-approximated by sequences of polynomials with rational ones.  Thank you for your question, it reminds me much of the kind I would've asked when younger!


When $f(0) = 0,$ is it always true that $G(0) = 0,$ where $G$ is the antiderivative of $f$? I have a hunch that it is, but it would be nice if somebody could confirm / disprove it for me.

Thank you.

Edit
Is it when the constant of integration is equal to zero?

No, this is not true in general. Let $F(x)$ be a anti-derivative of $f(x)$ then $$\int f(x) \, \mathrm{d}x = F(x) + k$$ now, even if $F(0) = 0$ there's still the $+ k$ constant to contend with.

An example would be $f(x) = x\exp x$ then we have $$F(x) = \int x\exp x \, \mathrm{d}x = e^x (x-1) + k$$ Now, even if $k = 0$, then $F(0) = -1$ whilst $f(0) = 0$.

No. The simplest example I can think of?
$$f(x) = x$$
Then $G(x) = \frac{x^2}{2} + C$, which is not guaranteed to be always $0$.


Let a, b be two complex numbers in the left half-plane. $\int f(z) \leq 1$ and M-T suggest \exists c$

The geometric view is that $|a-b|$ is the length of a curve between $e^{ia}$ and $e^{ib}$ - namely, the curve $e^{i\theta}$ for $\theta\in[a,b]$ (or $[b,a]$, depending on the order of the values.)

But the shortest distance between two points is the linear distance, which is for these points $\left|e^{ib}-e^{ia}\right|$.

Lemma: For $h>0,1-\cos h < h^2/2.$ Proof: Since $\sin t < t$ for $t>0,$ we have

$$1-\cos h = \int_0^h\sin t \, dt < \int_0^h t \, dt = h^2/2.$$

Thm: $|e^{ih}-1| < h$ for $h>0.$ Proof: $|e^{ih}-1|^2 = 2(1-\cos h) < h^2,$ where we have applied the lemma. Now take square roots.

Corollary: If $a<b,$ then $|e^{ib} - e^{ia}| < b-a.$ Proof:

$$|e^{ib} - e^{ia}| = |e^{ia}(e^{i(b-a)} - 1)| = |e^{i(b-a)} - 1| < b-a,$$

the last inequality coming from the theorem.


Using Ito's Lemma with more than one brownian motion term Question :

Let
$$
dY_t=c_tdt+d_tdW^1_t+e_tdW^2_t
$$
Where $W^1_t,~~W^2_t$ are standard independent brownian motions.

I am trying to apply Ito's formula to this, say for example trying to find $d(\frac {1}{Y_t})$

I'm not too sure about how to treat the variance term in Ito's formula.

Any hints would be appreciated,

Cheers

I'll assume that $c,d,e$ depend only on $t$. In integral form,
$$
Y_t=Y_0+\int_0^tc_s\,\mathrm ds+\int_0^td_s\,\mathrm dW_s^1+\int_0^te_s\,\mathrm dW_s^2.
$$
Then, for any $C^2$ function $f$ Itô's formula (in differential form) states
$$
\mathrm d\left(f(Y_t)\right)=f'(Y_t)\,\mathrm dY_t+\frac12f''(Y_t)\,\mathrm d\langle Y\rangle_t.
$$
I think your problem is finding $\mathrm d\langle Y\rangle_t$. To do this, use bilinearity and symmetry of $\langle \cdot,\cdot\rangle$:
\begin{multline*}
\langle Y\rangle_t=\langle \int_0^\cdot c_s\,\mathrm ds\rangle_t+\langle \int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t+\langle \int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t
+2\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t\\
+2\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t+2\langle\int_0^\cdot e_s\,\mathrm dW_s^2,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t.
\end{multline*}
Now, $\langle \int_0^\cdot c_s\,\mathrm ds\rangle_t=0$ and $\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot \varphi_s\,\mathrm dW_s^i\rangle_t=0$ because $\int_0^\cdot c_s\,\mathrm ds$ has finite variation. Also, $\langle\int_0^\cdot e_s\,\mathrm dW_s^2,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t=0$ because the two Brownian motions are independent.

The only terms that remain are
$$
\langle Y\rangle_t=\langle \int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t+\langle \int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t
=\int_0^t d_s^2\,\mathrm ds+\int_0^t e_s^2\,\mathrm ds=\int_0^t \left(d_s^2+e_s^2\right)\,\mathrm ds.
$$
Thus,
$$
\mathrm d\left(f(Y_t)\right)=f'(Y_t)\left(c_t\mathrm dt+d_t\mathrm dW^1_t+e_t\mathrm dW^2_t\right)+\frac12f''(Y_t)\left(d_t^2+e_t^2\right)\,\mathrm dt.
$$
Lastly, if you look around, you'll see that a general form of Itô's formula can be applied directly to $f(Y_t)=f\circ F(\int c_s\,\mathrm ds,\int d_s\,\mathrm dW_s^1,\int e_s\,\mathrm dW_s^2)$, making directly appear the quadratic variations of $W_t^1$ and $W_t^2$. The two approaches coincide.

$$
d(1/Y_{t}) = -Y_{t}^{-2}dY_{t} + Y_{t}^{-3}dY_{t}dY_{t}
$$
Substituting $dY_{t}$ into this equation.


How to solve the initial value problem $y'(x)=\lambda \sin(x+y(x))$, $y(0)=1$. For $\lambda \in \mathbb{R}$, consider the initial value problem $y'(x)=\lambda \sin(x+y(x))$, $y(0)=1$. Then this initial value problem has

  no solution in any neighbourhood of $0$.
  a solution in $\mathbb{R}$ if $|\lambda|<1$
  a solution in a neighbourhood of $0$.
  a solution in $\mathbb{R}$ only if $|\lambda|>1$.

This is the first time I have encountered this kind of IVP, and have no idea to proceed. The entanglement of $x$ and $y(x)$ in $\sin(x+y(x))$ is what causing me the trouble to make any headway. So please help me to solve this. Thanks.

Hint: Observe we have
\begin{align}
y' = F(x, y)
\end{align}
where $F$ is Lipschitz in $y$ variable since
\begin{align}
|F(x, y_1) -F(x, y_2)| = |\lambda| |\sin(x+y_1)-\sin(x+y_2)| \leq |\lambda||y_1-y_2|.
\end{align}
Note we have used the fact $|\sin u-\sin v| \leq |u-v|$.

Now, by Picard-Lindelof theorem, one can guarantee local existence, i.e. (3) holds if we do not know anything about $\lambda$.

Moreover, one can use a Banach fixed point argument to show that the ode has a global solution, i.e. solution on all of $\mathbb{R}$ if $|\lambda|<1$, which means (2) holds.

You first make a string of numbers for the range of $x$. Let's say $x\in[0,1]$. You could select $x=0, 0.1 , 0.2, ..., 1.0$.

Next, substitute $x=0$ to the equation to get:
$y^\prime(0) = \lambda \text{sin}(0+y(0)) = 0$.

Then use:

$y(0.1) = y(0) + 0.1 y^\prime(0) = 0$.

Next, substitute $x=0.1$ in the equation:

$y^\prime(0.1) = \lambda \text{sin}(0.1+y(0.1)) = \lambda \text{sin}(0.1)$

You continue the iteration up to $x=1.0$. The finer the meshing, the more precise the answer.


Integers, rationals and reals as sets? Natural numbers can be represented as pure sets by defining them to contain every number that is smaller than them. Arithmetic can be performed on them using the Peano axioms. Are there any similar definitions for integers, rationals and reals?

For example, I could define a rational to be an ordered pair of dividend and divisor. But that would leave the two rationals $\frac{1}{2}$ and $\frac{2}{4}$ not equal to each other, and it would be based on ordered things rather than pure sets.

Yes the procedure you outline is correct, but you must define equivalence classes,
$(a,b)\sim (b,c)$ if $ac=bd$. Thus a rational number is an equivalence class of pairs.
One then defines the arithmetic operations addition and multiplications and show that they are invariant under equivalence. The same can be done to define the integers from the natural numbers here the relation is $(n,m)\sim (r,s)$ if $n+s=r+m$.
The reals are then defined as Dedekind cuts of the rationals.

Yes.

Define the ordered pair $(a, b)$ as the set $\{a, \{a, b\}\}$
Define the integers as $(\{1, 0\} \times \mathbb N) \setminus (0, 0)$: ordered pairs where the first element represents the sign and the second represents the magnitude, excluding negative $0$. In the first slot, $1$ represents positive, $0$ represents negative.
Define the rational number $\frac{p}{q}$ as $\{(a, b): aq=bp\}$: the set of all ordered pairs that reduce to the same thing $(p, q)$ does in lowest terms. (That's not quite what this definition says, but it's easier to say and amounts to the same thing.)
Define real numbers as sets of rationals for which if $\frac{p}{q}$ is in the set, then all rationals lower than $\frac{p}{q}$ are in the set. (Also, the empty set and $\mathbb Q$ are specifically excluded from being real numbers.)


Prove that Bernstein Transformation is linear The definition of the transformation is $$(B_nf)(x) = \sum_{k=0}^n f(\frac{k}{n}) {n \choose k}x^k(1-x)^{n-k}$$

How can I show this is a linear map? I know the sum of $B_n$ should be 1, but can I consider $f(\frac{k}{n})$ as a constant?

Hint: $B_n$ associates to each function f another function defined as you wrote. In order to show that $B_n$ is linear you have to prove that $B_n(f+g)= B_n(f)+ B_n(g)$, and $B_n(af)=aB_n(f)$, for any functions f and g, and any real number a. But two functions are equal if they have equal values for each argument x.


What's an intuitive way to think about the determinant? In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?

Your trouble with determinants is pretty common. They’re a hard thing to teach well, too, for two main reasons that I can see: the formulas you learn for computing them are messy and complicated, and there’s no “natural” way to interpret the value of the determinant, the way it’s easy to interpret the derivatives you do in calculus at first as the slope of the tangent line. It’s hard to believe things like the invertibility condition you’ve stated when it’s not even clear what the numbers mean and where they come from.

Rather than show that the many usual definitions are all the same by comparing them to each other, I’m going to state some general properties of the determinant that I claim are enough to specify uniquely what number you should get when you put in a given matrix. Then it’s not too bad to check that all of the definitions for determinant that you’ve seen satisfy those properties I’ll state.

The first thing to think about if you want an “abstract” definition of the determinant to unify all those others is that it’s not an array of numbers with bars on the side. What we’re really looking for is a function that takes N vectors (the N columns of the matrix) and returns a number. Let’s assume we’re working with real numbers for now.

Remember how those operations you mentioned change the value of the determinant?

Switching two rows or columns changes the sign.
Multiplying one row by a constant multiplies the whole determinant by that constant.
The general fact that number two draws from: the determinant is linear in each row. That is, if you think of it as a function $\det: \mathbb{R}^{n^2} \rightarrow \mathbb{R}$, then $$ \det(a \vec v_1 +b \vec w_1 , \vec v_2 ,\ldots,\vec v_n ) = a \det(\vec v_1,\vec v_2,\ldots,\vec v_n) + b \det(\vec w_1, \vec v_2, \ldots,\vec v_n),$$ and the corresponding condition in each other slot.
The determinant of the identity matrix $I$ is $1$.

I claim that these facts are enough to define a unique function that takes in N vectors (each of length N) and returns a real number, the determinant of the matrix given by those vectors. I won’t prove that, but I’ll show you how it helps with some other interpretations of the determinant.

In particular, there’s a nice geometric way to think of a determinant. Consider the unit cube in N dimensional space: the set of vectors of length N with coordinates 0 or 1 in each spot. The determinant of the linear transformation (matrix) T is the signed volume of the region gotten by applying T to the unit cube. (Don’t worry too much if you don’t know what the “signed” part means, for now).

How does that follow from our abstract definition?

Well, if you apply the identity to the unit cube, you get back the unit cube. And the volume of the unit cube is 1.

If you stretch the cube by a constant factor in one direction only, the new volume is that constant. And if you stack two blocks together aligned on the same direction, their combined volume is the sum of their volumes: this all shows that the signed volume we have is linear in each coordinate when considered as a function of the input vectors.

Finally, when you switch two of the vectors that define the unit cube, you flip the orientation. (Again, this is something to come back to later if you don’t know what that means).

So there are ways to think about the determinant that aren’t symbol-pushing. If you’ve studied multivariable calculus, you could think about, with this geometric definition of determinant, why determinants (the Jacobian) pop up when we change coordinates doing integration. Hint: a derivative is a linear approximations of the associated function, and consider a “differential volume element” in your starting coordinate system.

It’s not too much work to check that the area of the parallelogram formed by vectors $(a,b)$ and $(c,d)$ is $\Big|{}^{a\;b}_{c\;d}\Big|$
either: you might try that to get a sense for things.

The determinant of a matrix gives the signed volume of the parallelepiped
that is generated by the vectors given by the matrix columns.

You can find a very pedagogical discussion at page 16 of
A Visual Introduction to Differential Forms and Calculus on Manifolds
Fortney, J.P.
google book link, click on "1 Background Material"
Given a parallelepiped whose edges are given by $ v_1 , v_2 , \dots, v_n \in \mathbb{R}^n $. Then if you accept these 3 properties:

$D(I)=1$, where $I=[e_1,e_2,\dots,e_n]$ (identity matrix)
$D(v_1,v_2,\dots,v_n)=0$ if $v_i=v_j$ for any $i\neq j$
$D$ is linear, $$\forall j,\ D(v_1,\dots,v_{j-1},v+cw,v_{j+1},\dots,v_n)=D(v_1,\dots,v_{j-1},v,v_{j+1},\dots,v_n)+cD(v_1,\dots,v_{j-1},w,v_{j+1},\dots,v_n)$$

you can show that $D$ is the parallelepiped signed volume and that $D$ is the determinant.


Line integration over a 3-dimensional curve Consider a curve C that is defined by the intersection of the following surfaces:
$$x+y+z=0$$ $$and$$ $$x^2+y^2+z^2=K$$ for some non-zero, positive real number K.

Find $$I=\int_Cy^2ds$$

I first tried to find the equation for the curve in terms of a vector r(t) and I obtained: $$r(t)=\left\langle \frac{\cos t}{\sqrt{ 2+\sin 2t}},\frac{\sin t}{\sqrt{ 2+\sin 2t}},\frac{-\cos t - \sin t}{\sqrt{ 2+\sin 2t}}\right\rangle$$
But then, to solve for I with this method, one must take the magnitude of the derivative of r, which becomes incredibly complicated. My professor claims there is an easier way to solve this problem, but I cannot figure anything out.

Here is a „non brute-force“ approach (assume WLOG $K=1$.)

Note that, by symmetry of the curve, we have $$\int_C x^2 \,\mathrm ds= \int_C y^2 \,\mathrm ds =\int_C z^2 \,\mathrm ds.$$

Hence,
$$3 \int_C y^2 \,\mathrm ds=\int_C x^2+y^2+z^2\,\mathrm ds = \int_C 1\,\mathrm ds=\text{Length of the curve } c.$$

It is known that the intersection of a sphere with a plane is a circle (under certain conditions which are fulfilled here.)

Also, as can be seen also from your parametrization (compute the distance of two points with time $\pi$ apart), our circle has a diameter of $2$.

So the length of the curve $c$ is simply the circumference of a circle with radius $2$ which is $2\pi$.

So, by our above argument,
$$\bbox[5px,border:2px solid #C0A000]{ I=\int_C y^2\,\mathrm ds = \frac{\text{Length of }c}3=\frac{2\pi}3.}$$

Here is a „brute-force approach“ (assume WLOG $K=1$.)

Even with your parametrization the calculations are doable:

Note that $$r'(t)=\left(-\frac{2 \sin (t)+\cos (t)}{(\sin (2 t)+2)^{3/2}},\frac{\sin (t)+2 \cos (t)}{(\sin (2 t)+2)^{3/2}},\frac{\sin (t)-\cos (t)}{(\sin (2 t)+2)^{3/2}}\right)$$

so that $$\|r'(t)\|_2=\sqrt{\frac{(\sin (t)-\cos (t))^2}{(\sin (2 t)+2)^3}+\frac{(\sin (t)+2 \cos (t))^2}{(\sin (2 t)+2)^3}+\frac{(2 \sin (t)+\cos (t))^2}{(\sin (2 t)+2)^3}}$$

which reduces to $$\|r'(t)\|_2=\sqrt{\frac{6+6\sin(t)\cos(t)}{(2+\sin(2t))^3}}=\sqrt{3} \sqrt{\frac{1}{(\sin (2 t)+2)^2}}.$$

So $$I=\int_0^{2\pi} \frac{\sqrt{3} \sin ^2(t)}{(\sin (2 t)+2)^2}\,\mathrm dt=2\sqrt{3}\int_0^{\pi} \frac{\sin ^2(t)}{(\sin (2 t)+2)^2}\,\mathrm dt.$$

It remains to calculate the last integral, which I will call $J$. $J$ is a standard integral and we may proceed as follows: $$J=\int_0^\pi \frac{\sin^2(t)}{(2\sin(t)\cos(t)+1)^2}\,\mathrm dt\overset{(1)}=\int_0^\pi \frac{\csc^2(t)}{4\cot^2(t)+4\csc^4(t)+8\cot(t)\csc^2(t)}\,\mathrm dt\overset{(2)}=\int_{-\infty}^\infty\frac{1}{4(u^2+u+1)^2}\,\mathrm du.$$

(1): Expand the fraction by $\csc^4$
(2): Use that $\csc^2=\cot^2+1$ and substitute $u=\cot(t)$.

So, by substituting $s=u+\frac12$, $$J=\frac14\int_{-\infty}^\infty \frac{1}{(s^2+\frac34)^2}\,\mathrm ds=\frac{4}9\int_{-\infty}^\infty \frac{1}{(\frac43 s^2+1)^2}\,\mathrm ds.$$

The last integral can be evaluated as here.

The final result should be $J=\frac{\pi}{3\sqrt 3}$ so that $$\bbox[5px,border:2px solid #C0A000]{I=\frac{2\pi}3.}$$


How do I identify the set of points satisfying $|z-1|+|z+1|\leq 2$? How do I identify the set of points satisfying $|z-1|+|z+1|\leq 2$?
My idea is:     $|z+1|^{2}=|z|^2+1+2x$
$|z-1|^{2}=|z|^2+1-2x$
$|z-1|+|z+1|\leq 2$ $\implies$ $|z+1|\leq 2 -|z-1|$
$(|z+1|)^2\leq (2 -|z-1|)^2$ (* is this step correct?)(because $-3<2$ but $9 \nleq 4$)
I am confused at this step of my solution. Can anyone suggest me how do I fix this problem?

Think of the triangle inequality.
Since the distance from $z=-1$ to $z=1$ is $2$,
the only points satisfying the (in)equality are on the line segment from $z=-1$ to $z=1$.

$|z+1|+|z-1|=2$ denotes a line segment. But $|z+1|+|z-1|<2$ is a null set and it represents nothing in the Argand plane.
And when $|z+1|+|z-1|>2$, represents region (point) out side  an ellipse.


Sum of averages vs average of sums I have essentially a table of numbers -- a time series of measurements.  Each row in the table has 5 values for the 5 different categories, and a sum row for the total of all categories.

If I take the average of each column and sum the averages together, should it equal the average of the rows' sums (ignoring rounding error, of course)?

(I've got a case where the two values keep coming out different by about 30% and I'm wondering just how crazy I am.)

Update: See below -- I was (slightly) crazy and had an error in my code.

The average of the entries in a column is the sum of the entries in that column, divided by the number of entries. The number of entries is the number of rows. So the sum of the averages is the sum of all the entries in the table, divided by the number of rows.

The average of the row sums is the sum of all entries in the table divided by the number of rows, so you should get the same number either way.

If I take the average of each column and sum the averages together,
  should it equal the average of the rows' sums (ignoring rounding
  error, of course)?

Generally, no.


Expectation of a Product of Independent Random Variables I was wondering if there was a formula for the expectation of a product of $n$ independent random variables. I have only seen one for two random variables.

I guess what I am asking is:

  Let $X_1, \dots, X_n$ be $n$ independent random variables. Is there a formula for $E\left(\prod\limits_{i=1}^n X_i\right)$?

I need this for a problem I am working on. It would be nice if it was equal to $\prod\limits_{i=1}^n E(X_i)$. While this is true when $n=2$, I'm not sure if this is true for general $n$.

If you are able to provide a formula for this I would appreciate a source or some sort of derivation.

If the values are independent then, it basically follows from the definition of expectation and the fact of independence of the series (that the joint probability function is the product of the marginal probability functions):

$$\begin{align}
\mathsf E(\prod_i X_i) & = \iiint (\;\prod_i x_i f_i(x_i)\;)\; \mathrm d x_n\cdots\mathrm d x_1
\\ & = \int x_n f_n(x_n)\mathrm d x_n \cdot \iiint\prod_{i=1}^{n-1}x_if_i(x_i)\mathrm d x_{n-1}\cdots\mathrm d x_1
\\ & = \prod_i \int x_i f_i(x_i)\mathrm d x_i
\\ & = \prod_i \mathrm E(X_i)
\end{align}$$

Or you might prefer to use the Law of Iterative Expectation: $$\begin{align}
\mathsf E(\prod_{i=1}^n X_i) & = \mathsf E(\mathsf E(X_n\prod_{i=1}^{n-1} X_i\mid X_n)) \\ & = \mathsf E(X_n\mathsf E(\prod_{i=1}^{n-1} X_i)) \\ & = \mathsf E(X_n)\mathsf E(\prod_{i=1}^{n-1} X_i) \\ & = \prod_{i=1}^n \mathsf E(X_i)\end{align}$$

Yes, if $X_1, \ldots, X_n$ are independent, the expected value of the product is the product of the expected values.


Show, there exists exactly one operator with $\int_A P_T(f)\, d\lambda=\int_{T^{-1}(A)}f\, d\lambda$ Let $T\colon\mathbb{R}\to\mathbb{R}$ be a non-singular function, i.e. a measurable function with the property that
    $$
\forall A\in\mathcal{B}: \lambda(A)=0 \implies \lambda(T^{-1}(A))=0.
$$
    Show: There exists exactly one linear operator  $P_T\colon L_{\lambda}^1\to L_{\lambda}^1$ so that for all $f\in L_{\lambda}^1$ and all $A\in \mathcal{B}$ it is
    $$
\int_A P_T(f)\, d\lambda=\int_{T^{-1}(A)}f\, d\lambda.
$$

Hello, I would really prefer to present you my own recent ideas but I do not have own ideas. To be honest, I am rather helpless. Can you pls give me help?

Greetings.

math12

New Edit:

The only thing I already know is that

$A\mapsto\int_{T^{-1}(A)}f\, d\lambda$

is a signed measure.

Now one can apply Radon-Nikodým or something like that?

Consider the signed measure
$$
  \mu(A)
  =
  \int_{T^{-1}(A)}f\, d\lambda.
$$
AFTER verifying that $\mu$ satisfies the Radon-Nikodým hypothesis,
use the theorem to find a unique $g$ such that
$$
  \mu(A)
  =
  \int_A g\, d\lambda.
$$
This $g$ is exactly $P_T(f)$.

For the hypotheses, you have to show that $\mu$ is $\sigma$-finite and that $\mu \ll \lambda$. Let me know if you need help with that.

Edit:
Contrary to what I have written, you do not need to show that $\mu$ is $\sigma$-finite. This hypothesis is for $\lambda$. And by the way, $\lambda$ does not have to be Lebesgue. It just has to be $\sigma$-finite for the proof to hold.

It seems that, in order to use the theorem, both measures, $\lambda$ and $\mu$ have to be $\sigma$-finite. If you know of a proof of the Radon-Nikodým theorem that does not require $\mu$ to be $\sigma$-finite, let me know... :-)

But anyway, $\mu$ is not only $\sigma$-finite. It is finite, because $f$ has finite integral. In fact
$$
  |\mu(A)|
  =
  \left|\int_{T^{-1}(A)} f d\lambda\right|
  \leq
  \int_{\mathbb{}} |f| d\lambda
  <
  \infty
$$
Had we missed the hypothesis that $f \in L^1_{\lambda}$? ;-)

Also, $\lambda$ does not have to be Lebesgue. It just has to be $\sigma$-finite for the proof to hold.

$ \int_{T^{-1}(A)}fd\lambda = \int_{A}dT\#(f\lambda) $ where $ T\#\nu $ is the push-forward measure of $ \nu $ by $T$. Then by change of variable you can find that
$ T\#(f\lambda) = f\circ T^{-1}|det(DT\circ T^{-1})|^{-1}\lambda $.

Let
$P_T(f) = f(T^{-1})|det(DT\circ T^{-1})|^{-1} $, clearly linear in f. Not sure this proves uniqueness and you need some more assumptions on T, such as derivability, but perhaps this might help you.


Prove that if $A$ is an $n \times n$ matrix such that $A^{4} = 0$ then $(I_n - A)^{-1}=I_n+A+A^2+A^3$ Prove that if $A$ is an $n \times n$ matrix such that $A^{4}$ = 0 then:
$$(I_n - A)^{-1}=I_n+A+A^2+A^3$$
My proof is as follows:
$$(I_n - A)(I_n - A)^{-1}=I_n$$
$$(I_n - A)^{-1}=I_n/(I_n - A)$$
$$I_n/(I_n - A)=I_n+A+A^2+A^3$$
$$I_n=(I_n - A)(I_n+A+A^2+A^3)$$
$$I_n=I_n+A+A^2+A^3-A-A^2-A^3-A^4$$
$$I_n=I_n-A^4$$
because we know that: $$A^4=0$$
therefore:
$$I_n=I_n$$
Is this an acceptable justification or have I made an error in my logic?
*I apologize for any poor formatting

The essence of your proof is correct, but the first few lines are very confusing. Why not something more clear like:
In order to show $(I_n-A)^{-1}=I_n + A +A^2 +A^3$, it suffices to show
$$
(I_n-A)(I_n + A +A^2 +A^3)=(I_n + A +A^2 +A^3)(I_n-A)=I_n.
$$
It's easy to show that the left equals the middle. Then by multiplying through and using the fact that $A^4=0$, it's easy to show that both of them equal the right.

To prove $a$ is inverse of $b$. Just find $ab$ and $ba$. If $b$ is inverse of $a$ then $ab=I=ba$. Thus Take $B=I_n+A+A^2+A^3$ and solve $AB$ and $BA$.


What does $ \lvert z-a \rvert = \mathit Re(z)+a $ look like? What does a loci with the equation look like?

$ \lvert z-a \rvert = \mathit Re(z)+a $

This is for the applying complex numbers topic of an advanced HSC maths course. I was asked to describe the loci.

I know that $ \lvert z-a \rvert $ would get me either a perpendicular bisector or a circle. I also know that $ \mathit Re(z) $ refers to the horizontal values on the complex plane. But I just can't imagine what it looks like.

Given $|z-a|$= $\Re(z)+a$
$$|x+iy-a|=x+a\\|(x-a)+iy|=x+a\\ $$

$$\sqrt{(x-a)^2+y^2}=x+a\\$$

taking square on both side

$$(x-a)^2+y^2=(x+a)^2\\x^2+a^2-2ax+y^2=x^2+a^2+2ax$$

we get $$y^2=4ax$$

This is a right handed parabola with focus
 $(a,0)$

We can treat complex numbers $z = x + iy$ as equations over $(x, y) \in \mathbb R^2$, and use a geometry plotter to plot them. In this case, the equation system is:

$$
\begin{align*}
|x + iy - a| &= Re(x+iy) + a \\
|(x - a) + iy| &= x + a\\
\sqrt{(x-a)^2 + y^2} &= x + a \\
(x-a)^2 + y^2 &= (x + a)^2
\end{align*}
$$

One can use a tool like Desmos for plotting curves like these. In this case, here is a playable version of the graph with $a$ as a parameter.

The image for one choice of $a$ is:


Deriving Conditional Expectations I am really stuck on deriving the basic conditional expectations equations.
First how does one prove this equation below?
$$
E[X\mid A] = \frac{E[X\mathbf{1}_A]}{P(A)}.
$$
Second, using the equation above how does one derive the the conditional expectations formula:

$$E(x|y)= \int_a^bx\frac{f(x,y)}{f(y)}dx$$

I can't seem to figure how to go from this step
$$
E(x|y)=E[X\mid A] = \frac{E[X\mathbf{1}_A]}{P(A)}=\frac{\int_{supp(z)}x\ \mathbf{1(y=y)}dx}{f(y)}.
$$

How does the indicator function dissapear and appear in the bounds of the intergration and then create the joint pdf $f(x,y)$ as in equation 1. Thank you.

Most non-measure theory textbooks define conditional expectation in terms of a sum over a conditional mass function (for discrete cases) or an integral of a conditional density (for continuous cases).

1)  Assuming $P[A]>0$, you can prove $E[X|A] = \frac{E[X 1_A]}{P[A]}$ according to the law of total expectation, from my above comment.

2) If you assume $X$ takes values in the interval $[a,b]$, then we define:
$$ E[X|Y=y] = \int_a^b x f_{X|Y=y}(x)dx = \int_a^b x \frac{f_{XY}(x,y)}{f_Y(y)}dy$$
where you can motivate the definition for the conditional
PDF $f_{X|Y=y}(x) = \frac{f_{XY}(x,y)}{f_Y(y)}$ through various demonstrations, likely found in your textbook.  The difficulty, of course, is that the event $\{Y=y\}$ typically has probability 0, and so conditioning on such things is not obvious and needs to be defined separately.

You can also motivate the above definition of $E[X|Y=y]$ according to a  demonstration similar to that given in your question (I will fix some of the issues with that demonstration below): Fix $y \in \mathbb{R}$ and $\delta>0$ and assume $P[Y \in [y, y+\delta]]>0$.  So for small $\delta>0$ we can imagine:
\begin{align}
E[X|Y=y] &\approx E[X|Y \in [y, y+\delta]] \\
&= \frac{E[X 1_{Y \in [y, y+\delta]}]}{P[Y \in [y, y+\delta]]} \\
&= \frac{\int_{x=a}^b\int_{v=y}^{y+\delta} xf_{XY}(x,v)dxdv}{P[Y \in [y, y+\delta]]}  \\
&= \frac{\int_{x=a}^bx \left[\int_{v=y}^{y+\delta} f_{XY}(x,v)dv\right]dx}{P[Y \in [y, y+\delta]]}  \\
&\approx \frac{\int_{x=a}^b x[f_{XY}(x,y)\delta] dx}{f_Y(y)\delta}\\
&=\frac{\int_a^b xf_{XY}(x,y)dx}{f_Y(y)}
\end{align}

This isn't a proof (see the comments to your question, and Michael's answer for details on that), but let me give a heuristic argument for why you should expect that

$$ \operatorname{E} [X \vert A] = \frac{\operatorname{E} [X \mathbf{1}_A]}{\operatorname{P}(A)} $$

I know this isn't exactly what you're asking for, but what I give below might hopefully still be at least somewhat enlightening.

Recall from Bayes' rule that, when $A$ and $B$ are events,

$$ \operatorname{P} (B \vert A) = \frac{\operatorname{P}(B \cap A)}{\operatorname{P} (A)} $$

Notice also that when $X = \mathbf 1_B$,

$$\operatorname{E}[X] = \operatorname{E}[\mathbf 1_B] = \operatorname{P}(B)$$

Given that $\mathbf 1_{B \cap A} = \mathbf 1_B \mathbf 1_A $, we can write the conditional probability above as

$$ \operatorname{P} (B \vert A) = \frac{\operatorname{E}[\mathbf 1_B \mathbf 1_A]}{\operatorname{P} (A)} $$

If we again let $X=\mathbf 1_B$, and by analogy with the fact that probabilities of events are expectations of indicator functions for those events, we can define the conditional expectation of indicator functions:

$$ \operatorname{E}[X \vert A] =\frac{\operatorname{E}[X \mathbf 1_A]}{\operatorname{P}(A)} = \operatorname{P} (B \vert A)$$

Using the linearity of expectation, we can extend the above definition to the so-called simple functions which are constant on finitely many events:

$$ X = \sum_{i=1}^n a_i \mathbf 1_{B_i} $$

Hence,

$$ \operatorname{E}[X \vert A] = \frac{\operatorname{E}[X \mathbf 1_A]}{\operatorname{P}(A)} = \sum_{i=1}^n a_i \operatorname{P} (B_i \vert A) $$

What about more general random variables?

It turns out that almost all random variables of interest can be approximated by simple functions. Hence, to take the conditional expectation of those, we can first find a sequence of increasingly accurate approximations via simple functions. We know how to find the conditional expectation of each of those approximations. The limit of these approximations is then the conditional expectation of the random variable.

(I know I've glossed over a lot of detail in the previous paragraph. However, if you know a little measure theory, the previous construction along with the details should be familiar. If you don't, I don't think the details are going to be particularly helpful.)


Completeness of the space of Riemann integrable functions under $\left \| . \right \|_{\infty}$ on $\left [ 0,1 \right ]$ I read a proof about this topic at this site and was convinced with it , then I tried to construct a cauchy sequence of functions to see how things apply . Now I'm confused.
The sequence I constructed is : $f_n(x)=\left\{\begin{matrix}
-lnx & if & x> e^{-n}\\
 n& if & x\leq e^{-n}
\end{matrix}\right.$
I proved that this sequence is cauchy and that it belongs to the mentioned space . For the limit I got that $f_n(x)\rightarrow f(x)=-lnx$
almost everywhere. But $f$ is not bounded on $\left [ 0,1 \right ]$
Did I miss or mess with something ??

First observe that the space $R[0,1]$ of all Riemann integrable functions is a subspace of the complete space $B[0,1]$ formed by all bounded functions.  So $R[0,1]$ is complete iff it is closed in $B[0,1]$.
Theorem. $R[0,1]$ is closed in $B[0,1]$, and hence complete.
Proof. Given a sequence $\{f_n\}_n$ in $R[0,1]$ converging to some $f$ in $B[0,1]$, let $D_n$ be the set of points where $f_n$ is discontinuous.
It is well known that $D_n$ has measure zero, and hence so does
$$
D := \bigcup_n D_n.
$$
Notice that if $x$ is not in $D$ then each $f_n$ is continuous at $x$, and hence so is $f$.  In other words the discontinuities of $f$ lie in $D$ so $f$ is Riemann integrable.  QED


Get $5$ by doing any operations with four $7$s How can one combine four sevens with elementary operations to get $5$? For example $$\dfrac{(7+7)\times7}{7}$$ (though that does not equal $5$). I am not able to do this. Can you solve it or prove that it's impossible?

How about:

$$7 - \frac{7+7}{7} = 5$$

$$7 - \log_7 (7·7) = 5$$

$$7 - \frac{\ln (7·7)}{\ln 7} = 5$$

$$\left\lfloor \sqrt{\frac{7^7}{7!}} - 7\right\rfloor = 5$$

$$\lfloor 7\sin 777^\circ\rfloor = 5$$

$$\lfloor 7\cos 7^\circ\rfloor - \frac{7}{7} = 5$$

$$\lfloor 7\cos 7\rfloor = 5 \text{ using radians}$$

You can also use base $174$ and write:

$$\sqrt{\frac{77}{7·7}} = 5$$

That can also reduce the amount of sevens by one if you write:

$$\frac{\sqrt{77}}{7} = 5$$

7-(7+7)/7

I started looking for a way to use the modulus operator, but this is too simple.


Orthogonal Transformations How can I show that given any two unit vectors in Euclidean space, there is an orthogonal transformation taking one to the other? I considered something like a reflection, but I don't know how to formalize it, or even if it's correct.

For completeness, here's a proof of that orthogonal transformations preserve inner products:

Take $\Bbb H$ to be a finite-dimensional, linear vector space with element vectors $u$ and $v$ and inner product $<u,v>\equiv u^Tv$.

$\underline{\text{Show}}$: $<u,v>=<Mu,Mv>$, for $M$ some orthogonal transformation.
$$
<u,v>=u^Tv\\
<Mu,Mv>=(Mu)^T(Mv)=u^TM^TMv
$$

By definition of orthogonal matrices (i.e. $M^TM=MM^T=1$),
$$
<Mu,Mv>=u^TM^TMv=u^T(M^TM)v=u^Tv=<u,v>
$$
QED.

Here's a proof:

Suppose $a$ and $b$ are $n$-dimensional unit vectors, then
$$
a^Ta=\sum_{i=1}^{n}(a_n)^2=1\\b^Tb=\sum_{j=1}^{n}(b_n)^2=1
$$
where $a^T$ and $b^T$ are the transposes of their respective vectors.

Take $M$ to be a linear transformation between $a$ and $b$. Without loss of generality:
$$
Ma=b
$$

By the properties of transposes, $(Ma)^T=a^TM^T$, so $a^TM^T=b^T$. Take the norm of both sides of the above equation:
$$
(a^TM^T)(Ma)=b^Tb\\
a^T(M^TM)a=1
$$

Since $a^Ta=1$, if $M^TM=1$, then $a^T(M^TM)a=1$. $M$ is orthogonal if and only if $M^TM=1$. So, we may take $M$ to be orthogonal while still satisfying the conditions of the problem. Therefore, there exists a (not necessarily unique) transformation $M$ between $a$ and $b$ (with $a$ and $b$ unit vectors) such that $M$ is orthogonal.


Homework question compound interest If $ \$ 6000$ are invested at 7% compounded continuously, what amount after 2 years?
I know how to set it up but at one point I get lost

$$A=Pe^rt$$
$$A=6000^{0.07}(2)$$

Somebody please help.

If compounded continuously as you stated, at a nominal annual rate of 7%, then the amount accumulated is $$A = Pe^{rt},$$ where $P$ is the principal, $r$ is the nominal annual rate of interest, and $t$ is the time in years.  Your error is that $t$ was not in the exponent along with $r$ when it should be.  For $P = 6000$, $r = 0.07$, and $t = 2$, $A = 6901.64$.  The other answers assume that 7% represents an effective annual rate, not a nominal one.

If the figure of 7% were assumed to be an effective rate of interest $i$, then specifying the way the interest is compounded, and hence the purpose of the question, becomes pointless.

Link to calculation in WolframAlpha

In order to calculate the interest over several years, the idea is to do the calculation for the first year and then continue the calculation for the next year with your new result. So let's do the calculation.
At first, you'd have $6000\$$ and $6000\$ \cdot 0.07 = 420\$$ interest, so a total of $6420\$$. After the 2nd year, you'd have your $6420\$$ and an additional $6420\$ \cdot 0.07 = 449.40\$$ interest, so a total of $6869.40\$.

Note that we actually calculated the following:
$$(6400 \$ + 6400 \$ \cdot 0.07) + (6400 \$ + 6400 \$ \cdot 0.07) \cdot 0.07 = \\
(6400 \$ \cdot 1.07) + (6400 \$ \cdot 1.07)\cdot 0.07 = \\
(6400 \$ \cdot 1.07) \cdot 1.07 = \\
6400 \$ \cdot 1.07^2
$$

which hopefully sheds some light on the given formular, you've tried to apply.


How to prove $\mathbb Z_3\rtimes(\mathbb Z_2\mathbb \times\mathbb Z_2) \cong S_3\times\mathbb Z_2$? I know that there is a unique semidirect product $\mathbb Z_3\rtimes(\mathbb Z_2\times \mathbb Z_2)$, defined by mapping two of the order two generators of $\mathbb Z_2\times \mathbb Z_2$ to the inversion automorphism of $\mathbb Z_3$.

However, I am not exactly sure how to proceed to show that $\mathbb Z_3\rtimes(\mathbb Z_2\mathbb \times\mathbb Z_2) \cong S_3\times\mathbb  Z_2$.

What exactly is the map I could construct?

Let $G = S_3 \times \{\pm 1\}$ (writing the cyclic group multiplicatively).

Now map
$$
G \to \{\pm 1\} \times \{\pm 1\}
$$
by the rule
$$
(\sigma, \epsilon) \mapsto (\epsilon \cdot \text{sign}(\sigma), \epsilon).
$$
The kernel is $A_3 \times \{1\}$, so we have an exact sequence
$$
1 \to A_3 \to G \to \{\pm 1\} \times \{\pm 1\} \to 1,
$$
and there is a splitting $\{\pm 1\} \times \{\pm 1\} \to G$ given by $(-1, 1) \mapsto ((12), 1)$ and $(1, -1) \mapsto (\text{id}, -1)$.

It follows that $G$ is some semidirect product of $A_3$ by $\{\pm 1\} \times \{\pm 1\}$ as desired (using $A_3 = \mathbb{Z}/3\mathbb{Z}$). We still must check that it is the non-trivial semi-direct product, i.e. that the lift $((12), 1)$ acts non-trivially on $A_3$ by conjugation. This is indeed true since e.g. $(12)(123)(12) \neq (123)$.

If $G$ is the group defined by the semidirect product you defined and $H$ is the direct product of $S_3\times \mathbb Z_2$, then all elements of $G$ take the form
$$g=(x_1,x_2,x_3)$$
where $x_1\in\mathbb Z_3$ and $x_2,x_3\in\mathbb Z_2$. All elements of $H$ take the form
$$h=(y_1,y_2)$$
where $y_1\in S_3$ and $y_2\in\mathbb Z_2$.

If I understand your definition correctly, then for the group $G$, the following should hold:
$$(x_1,0,0)(a,b,c)=(x_1+a,b,c)$$
$$(x_1,1,1)(a,b,c)=(x_1+a,b+1,c+1)$$
$$(x_1,0,1)(a,b,c)=(-x_1+a,b,c+1)$$
$$(x_1,1,0)(a,b,c)=(-x_1+a,b+1,c)$$

Then you should be able to find an isomorphism between $G$ and $H$ by mapping the elements of $G$ that take the form $(x_1,0,0)$ and $(x_1,1,1)$ to elements of $H$ that take the form $(y_1,0)$ and elements of $G$ in the form $(x_1,0,1)$ and $(x_1,1,0)$ to elements of $H$ in the form $(y_1,1)$.


Help understanding the Feynman-Kac formula From wikipedia:

Suppose we wish to find the expected value of the function
$e^{-\int_0^t V(x(\tau)) d\tau}$ in the case where $x(\tau)$ is some realization of a diffusion process starting at $x(0) = 0$.  The Feynman–Kac formula says that this expectation is equivalent to the integral of a solution to a diffusion equation. Specifically, under the conditions that $uV(x) \ge 0$,

$$E[e^{-u\int_0^t V(x(\tau)) d\tau}] = \int_{-\infty}^\infty w(x, t) dx$$

where $w(x, 0) = \delta(x)$ and

$$\frac{\partial w}{\partial t} = \frac{1}{2} \frac{\partial^2w}{\partial x^2} - uV(x)w $$

My Questions:

(1) What is $\delta(x)$ in this context?  It's not mentioned by the rest of the wiki article.

(2) The "randomness" of the variable whose expectation is being measured is entirely contained within the term $x(\tau)$.  But it looks to me like the value of the expectation is independent of the function $x$.  It's simply an integral of the function $w$, and the PDE that defines $w$ uses $x$ as a parameter to $w$ but not a function (right?).  What am I missing?

Thanks for your help.

Check here:

http://www.math.sunysb.edu/~ajt/Teaching/560spring2011/PDF/Covariance%20Operator,%20Support%20of%20Wiener%20Measure.pdf

This should be helpful.


Is this distribution binomial distribution? Suppose that three persons (A, B and C) throw at a target. And A throws 10 times with the probability 0.3 to hit the target; and B throws 15 times with the probability 0.2; and C throws 20 times with the probability 0.1. Now Determine the probability that the target will be hit at least 12 times.

My solution is as follows:

For each throw, the probability of hitting the target is

Pr(H=1) = Pr(A)Pr(A Hit) + Pr(B)Pr(B Hit) + Pr(C)Pr(C Hit)

which is

Pr(H=1) = (10/45)*0.3 + (15/45)*0.2 + (20/45)*0.1 = 8/45

So, the throw distribution can be seemed as binomial distribution H ~ Bin(45, 8/45)
then can get the answer.

Am I right to consider the target hit variable as a binomial distribution?

And please give me a help to get the correct answer, any hints will be appreciated. Thanks.

No.  To see why, note that the MGF of the sum $W = X_1 + X_2$ of two independent binomial random variables $$X_i \sim \operatorname{Binomial}(n_i, p_i), \quad i = 1, 2,$$ is $$M_W(t) = M_{X_1}(t) M_{X_2}(t) = (1 + (e^t - 1) p_1)^{n_1} (1 + (e^t - 1) p_2)^{n_2}.$$  This is not in general equal to the MGF of a single binomial random variable $Y$ with parameters $n = n_1 + n_2$, $p = (n_1 p_1 + n_2 p_2)/n$, which would be $$M_Y(t) = \left(1 + (e^t - 1)\frac{n_1 p_1 + n_2 p_2}{n_1 + n_2} \right)^{n_1 + n_2},$$ except in the case where $p_1 = p_2$.

It is worth noting that the correct exact probability of the event described in the question is $$\frac{10229891531523289867038696518983728647}{119209289550781250000000000000000000000} \approx 0.0858145.$$  However, the probability described by your solution would be around $0.0905153$.

A random variable $H$ is distributed binomially if it represents the number of successes $H=h$ in $n$ independent and identical trials where $n\ge h$

Hence, $H$ is distributed binomially assuming the independence and identicalness of the throws in which case we have:

$$P(H = h) = \frac{45}{h} p^h (1-p)^{n-h}$$

As for the calculation of $p$, it looks like right but you'll have to clarify what exactly is meant by $P(A)$ and so on. How exactly did you come up with the formula? Did you use law of total probability?

I think

$$p = P(hit) = P(hit | A \ throws)P(A \ throws) + P(hit | B \ throws)P(B \ throws) + P(hit | C \ throws)P(C \ throws)$$

where $P(X \ throws)$ is the proportion of throws of $X$ to the total number of throws assuming independence, identicalness or something


Why does $e^{i\pi}=-1$? I will first say that I fully understand how to prove this equation from the use of power series, what I am interested in though is why $e$ and $\pi$ should be linked like they are.

As far as I know $\pi$ comes from geometry (although it does have an equivalent analytical definition), and $e$ comes from calculus.

I cannot see any reason why they should be linked and the proof doesn't really give any insights as to why the equation works.

Is there some nice way of explaining this?

Euler's formula describes two equivalent ways to move in a circle.

Starting at the number $1$, see multiplication as a transformation that changes the number $1 \cdot e^{i\pi}$.
Regular exponential growth continuously increases $1$ by some rate; imaginary exponential growth continuously rotates a number in the complex plane.
Growing for $\pi$ units of time means going $\pi\,\rm radians$ around a circle
Therefore, $e^{i\pi}$ means starting at 1 and rotating $\pi$ (halfway around a circle) to get to $-1$.

For more details explaining each step, read this article.

$$e^{ix} = \cos x + i\sin x$$

the $\pi$ is to some extent an arbitrary definition of the angle.


prove no two cycles in a graph will share an edge if all the cycles in the graph are of odd length I am trying to use induction but cant make any headway on this problem.
Was considering using the property that biparted graphs have no cycles of odd length but Im not sure if that helps

Suppose there are two (odd) cycles in your graph, say $C_1 = [u_1, u_2, \dots, u_n, u_1]$ and $C_2 = [v_1, v_2, \dots, v_m, v_1]$, that have k common edges. These edges lie between $k+1$ vertices in either cycle.

Now, we know that $n$ and $m$ are odd, so $n=2a-1$ and $m=2b-1$, for some $a,b \in \mathbb{N}$. Therefore the number of vertices in the big cycle (the one obtained by forgetting the common edges) is equal to $n-(k-1))+m-(k-1) = 2a-1+2b-1-2k+2=2(a+b-k)$, which is an even number, a contradiction.

Suppose that we have two cycles $C_1$ and $C_2$ that share an edge. Without loss of generality, these cycles intersect in a path. Why? If not, then we have two points $v, w$ in both $C_1$ and $C_2$, and two paths $v, v_1, \ldots, v_n, w$ and $v, w_1, \ldots, w_m, w$ in $C_1$ and $C_2$ respectively. Consider the least $i$ such that $v_i \neq w_i$, and the least $j \ge i$ such that $v_j$ belongs on the other path (i.e. is equal to $w_k$ for some $k \ge i$). Then
$$v_{i-1}, v_i, \ldots,v_j, w_{k-1},\ldots,w_{i-1}$$
is a cycle, where $v_{i-1} = w_{i - 1}$ is $v$ when $i = 1$, that intersects with $C_1$ (and $C_2$) in a path.

So, we have three paths $P_1, P_2, P_3$ such that $C_1 = P_1 \cup P_2$ and $C_2 = P_2 \cup P_3$. Consider the cycle $C_3$ made up of $P_1$ and $P_3$. Then,
\begin{align*}
|C_1| &= |P_1| + |P_2| \\
|C_2| &= |P_2| + |P_3| \\
|C_3| &= |P_1| + |P_3|.
\end{align*}
Each $|C_i|$ is odd, but when we sum these expressions,
$$|C_1| + |C_2| + |C_3| = 2(|P_1| + |P_2| + |P_3|),$$
which is even. This is a contradiction. Hence, the cycles cannot share an edge.


How to find Winding number Let $r:[0, 2\pi]\to\Bbb C$ be the circle given by $r(t)=1+2e^{2it}$, what is the winding number of $r(t)$ around $z=2$?

I struck at :

If $r(t)=1+2e^{it}$ then we can write as $|z-1|=2$, but
How can write $r(t)=1+2e^{2it}$,  in the circle form?

You can still write this as the circle described by $|z-1|=2$, the function just goes around the circle twice, since the exponent over $i$ ranges from $0$ to $4\pi$, which is twice $2\pi$

Just for fun, $\dfrac 1{2\pi i}\oint_r\dfrac {\operatorname dz}{z-2}=\dfrac 1{2\pi i}\int_0^{2\pi}\dfrac 1{1+2e^{2it}-2}4ie^{2it}\operatorname dt=\dfrac 1{2\pi i}\int_0^{2\pi}\dfrac 1{2e^{2it}-1}4ie^{2it}\operatorname dt=\dfrac  2{\pi}[-\dfrac 14\ln (2e^{2it}-1)]_0^{2\pi}=\dfrac{2\pi}{\pi}=2$, where at the end the principal branch of log was used.

This checks out, since $r(t)$ goes around $2$ twice as $t$ goes from $0$ to $2\pi$.  Note $r(t)$ is just the circle $2e^{2it}$ of radius $2$ centered at the origin, translated  $1$ unit to the right.


Showing $\lim_{n \to \infty} m(E_n) = 0$, assuming $f > 0$ a.e. and $\lim_{n \to \infty} \int_{E_n}f \,dm =0$ Let $f:[0,1] \to \mathbb{R}$ be Lebesgue measurable with $f > 0$ a.e. Suppose that $\{E_n\}$ is a sequence of measurable sets in $[0,1]$ with the property that $\displaystyle \lim_{n \to \infty}\int_{E_n} f \,dm = 0$. Prove that $\displaystyle \lim_{n \to \infty} m(E_n) = 0$.

This question is from an old analysis qual I am studying. So far I have tried a proof by contradiction: if $\displaystyle \lim_{n \to \infty} m(E_n) \neq 0$, then there is an $\epsilon > 0$ and a subsequence $\{n_k\}$ so that $m(E_{n_k}) \ge \epsilon$ for all $k$. I am trying to somehow use this subsequence and show $\displaystyle \lim_{k \to \infty}\int_{E_{n_k}} f \,dm \neq 0$, which would give me a contradiction.

Another fact I know from my measure theory course is that, for meausurable $E \subseteq [0,1]$, the map $\displaystyle \nu(E) = \int_E f \,dm$ defines a measure on the Lebesgue measurable subsets of $[0,1]$. Will this fact be useful to me?

Here is a direct proof: Let $k,n  \in \mathbb{N}$. Then

$$\int_{E_n} f \, dm \geq \int_{E_n \cap \left[f>\frac{1}{k}\right]} f \, dm \geq \frac{1}{k} \cdot m \left( E_n \cap \left[f> \frac{1}{k} \right] \right) \geq 0$$

Since $\int_{E_n} f \, dm \to 0$ as $n \to \infty$ we obtain

$$m \left( E_n \cap \left[f> \frac{1}{k} \right] \right) \to 0 \qquad (n \to \infty)$$

for all $k \in \mathbb{N}$. Thus

$$m(E_n) \leq m \left( E_n \cap \left[f> \frac{1}{k} \right] \right) + m \left( \left[f > \frac{1}{k} \right]^c \right) \to m \left( \left[f > \frac{1}{k} \right]^c \right) \qquad (n \to \infty)$$

We have

$$m \left( \left[f > \frac{1}{k} \right]^c \right) \to 0 \qquad (k \to \infty)$$

since $f>0$ a.s. and therefore conclude $m(E_n) \to 0$ as $n \to \infty$.

EDIT: Wrong!

If $\lim_{n\to\infty} m(E_n)>0$, then(*) you can find $E_{n_k}$ such that $m(E_{n_k})>\epsilon$ and $\bigcap E_{n_k}=E$ has positive measure, say $m(E)>\epsilon/2$.

Then $\int_{E_n}f(x)dm(x)\geq\int_E f(x)dm(x)>0$, so $\lim \int_{E_n}fdm\neq 0$.

(*)Wlog assume that $m(E_n)>\epsilon$ and consider $F=\bigcup E_n$. We can find a finite number $k$ such that $m(F\setminus (E_1\cup\ldots\cup E_k))<\epsilon/4$.

If for every $\delta>0$ there exists $h>0$ such that $m(E_j\cap E_l)<\delta$ for $j=1,\ldots, k$ and every $l>h$, then we can fix $\delta<\epsilon/4k$ and we get that
$$m((E_1\cup\ldots\cup E_k)\cap E_l)\leq \epsilon/4$$
so
$$m(E_l\cap(F\setminus (E_1\cup\ldots\cup E_k)))\geq 3\epsilon/4$$
but that's impossible, because $m(F\setminus (E_1\cup\ldots\cup E_k))\leq \epsilon/4$.


How to reverse digits of an integer mathematically? Is there a mathematical way to reverse digits of an integer mathematically? for example:
 $18539 \rightarrow 93581$

Let $n = \lfloor \log_{10}x\rfloor$, then $x$ has $n + 1$ digits.

Let $x = \displaystyle\sum_{i=0}^nx_i10^i$ where $x_i \in \{0, 1, \dots, 9\}$. To obtain $x_k$ (i.e. the $(k+1)^{\text{st}}$ digit of $x$), first divide by $10^k$ to obtain

$$x10^{-k} = \sum_{i=0}^nx_i10^{i-k} = \sum_{i=0}^{k-1}x_i10^{i-k} + \sum_{i=k}^nx_i10^{i-k}.$$

Note that the first sum on the right hand side is a non-negative number less than $1$ and the second sum is an integer. Therefore

$$\lfloor x10^{-k}\rfloor = \left\lfloor\sum_{i=0}^nx_i10^{i-k}\right\rfloor = \sum_{i=k}^nx_i10^{i-k}.$$

Now consider the corresponding expression for $k+1$:

\begin{align*}
\lfloor x10^{-(k+1)}\rfloor &= \left\lfloor\sum_{i=0}^nx_i10^{i-(k+1)}\right\rfloor\\
&= \sum_{i=k+1}^nx_i10^{i-(k+1)}\\
&= 10^{-1} \sum_{i=k+1}^nx_i10^{i-k}\\
&= 10^{-1}\left(\sum_{i=k}^nx_i10^{i-k}-x_k\right)\\
&= 10^{-1}\left(\lfloor x10^{-k}\rfloor - x_k\right).
\end{align*}

Rearranging for $x_k$ we find $x_k = \lfloor x10^{-k}\rfloor - 10\lfloor x10^{-(k+1)}\rfloor$. Therefore, the number you are after is

\begin{align*}
\sum_{k=0}^nx_{n-k}10^k &= \sum_{k=0}^n\left(\lfloor x10^{-(n-k)}\rfloor - 10\lfloor x10^{-(n-k+1)}\rfloor\right)10^k\\
&= \sum_{k=0}^n\lfloor x10^{-(n-k)}\rfloor10^k - \sum_{k=0}^n\lfloor x10^{-(n-k+1)}\rfloor10^{k+1}
\end{align*}

or if you prefer

\begin{align*}
\sum_{k=0}^nx_k10^{n-k} &= \left(\lfloor x10^{-k}\rfloor - 10\lfloor x10^{-(k+1)}\rfloor\right)10^{n-k}\\
&= \sum_{k=0}^n\lfloor x10^{-k}\rfloor10^{n-k} - \sum_{k=0}^n\lfloor x10^{-(k+1)}\rfloor10^{n-k+1}.
\end{align*}

We can simplify this even further. Note that $\lfloor x10^{-n+1}\rfloor = 0$ so the final term in the second sum is zero. Now shift the index in the second sum so that it begins at $k = 1$:

\begin{align*}
&\sum_{k=0}^n\lfloor x10^{-k}\rfloor10^{n-k} - \sum_{k=0}^n\lfloor x10^{-(k+1)}\rfloor10^{n-k+1}\\
&= \sum_{k=0}^n\lfloor x10^{-k}\rfloor10^{n-k} - \sum_{k=1}^n\lfloor x10^{-k}\rfloor10^{n-k+2}\\
&= \lfloor x10^0\rfloor10^n + \sum_{k=1}^n\lfloor x10^{-k}\rfloor10^{n-k} - 100\sum_{k=1}^n\lfloor x10^{-k}\rfloor10^{n-k}\\
&= x10^n - 99\sum_{k=1}^n\lfloor x10^{-k}\rfloor10^{n-k}.
\end{align*}

Therefore, given a positive integer $x$, the integer which has the same digits in the reverse order is $$x10^{\lfloor \log_{10}x\rfloor} - 99\sum_{k=1}^{\lfloor \log_{10}x\rfloor}\lfloor x10^{-k}\rfloor10^{\lfloor \log_{10}x\rfloor-k}.$$

Here's an example calculation. Let $x = 123$, then $\lfloor \log_{10}x\rfloor = 2$. As $\lfloor x10^{-1}\rfloor = \lfloor 12.3\rfloor = 12$ and $\lfloor x10^{-2}\rfloor = \lfloor 1.23\rfloor = 1$, we have

$$123\times 10^2 - 99\left(12\times 10 + 1\times 1\right) = 123\times 100 - 99\times 121 = 12300 - 11979 = 321$$

as expected.

let number be 123.
perform 123%10 = 3 = a
perform 123/10 =12
if not 0,then multiply a with 10, so a = 30
perform 12%10 = b
perform 12/10 = 1
if not 0, then multiply b with 10 & a with 10,so a = 300 & b = 20
perform 1%10 = c
perform 1/10 = 0
if not 0, continue....however, as this is 0, you stop the process.
just add a+b+c. = 300+20+1 = 321

& in terms of pure mathematics:

sum of [{floor of n/10^(N-1-k)}*{10^k] for k= 0 to N-1. where n is integer, & N is no of digits.


Test for convergence $\sum_{n=1}^{\infty} \frac{1}{2^\sqrt{n}}$ Possible Duplicate:
  convergence of a series involving $x^\sqrt{n}$

Test for convergence
$$\sum_{n=1}^{\infty} \frac{1}{2^\sqrt{n}}$$
My first thought was to use the ratio test but it's inconclusive since it yields $1$.
Are there some easy means to test the sum for convergence? Thanks!

Hint: For $n>2^{10}$, we have that $$\sqrt{n}\geq 2\log_2(n),$$  and so for $n>2^{10}$ $$2^{\sqrt{n}}\geq n^2.$$  Now try using the comparison test.

So, you could use the comparison test, that one would yield results.


Counting the number of squares on an $n\times n$ Board Yesterday I was asked by a friend how many squares are in a chess board of $8\times 8$.
I thought about 64 immediately, but of course this is not the solution, there are much more.

So the number of squares is: $$8\times 8 + 7\times 7 + 6\times 6 + 5\times 5 + 4\times 4 + 3\times 3 + 2\times 2 + 1\times 1=1^2 + 2^2 + 3^2 + 4^2...+ 8^2$$

I came across this formula: $$\frac{n(n + 1)(2n + 1)} 6$$

It produces the sum of squares in $n\times n$ board.

My question is, how did he reached this formula? Did he just guessed patterns until he reached a match, or there is a mathematical process behind?

If there is a mathematical process, can you please explain line by line?

Thanks very much.

Btw: Couldn't find matching tags for this question, says I can't create.

The first step is to recognize that there are $8^2$ squares of size $1$ by $1$, $7^2$ squares of size $2$ by $2$ and so on.  That justifies the total number being, as you say, $1^2+2^2+3^2+\ldots 8^2$.  Sums of powers are calculated by Faulhaber's formula.  There are several ways to derive them.  One way is to know or suspect that $\sum_{k=1}^n k^p$ should be of degree $p+1$.  So for squares, we want $\sum_{k=1}^n k^2=an^3+bn^2+cn+d$.  Then if we evaluate it at $n+1$, we get $\sum_{k=1}^{n+1} k^2=a(n+1)^3+b(n+1)^2+c(n+1)+d$.  Subtracting, we get $(n+1)^2=a((n+1)^3-n^3)+b((n+1)^2-n^2)+c((n+1)^1-n^1)$ and equating the coefficients gets the desired formula.  You can prove the formula rigorously by induction

formula to calculate the square in $m \cdot n$  order:
$[n(n+1)x(3m-n+1)]/6$


Infinite oscillation of random signs Suppose that $\left(a_n\right)$ is a sequence of real numbers and that $\left(\varepsilon_n\right)$ is a sequence of IID RVs with
$$P\left(\varepsilon_n = \pm 1\right) = \frac{1}{2}$$

According to Williams (Probability with Martingales (1991), Section 12.3 "Random signs", pp. 113-114), the results below show that

i. $\sum \varepsilon_n a_n$ converges (a.s.) if and only if $\sum a_n^2 < \infty$, and that

ii. $\sum \varepsilon_n a_n$ (a.s.) oscillates infinitely if $\sum a_n^2 = \infty$

I understand why (i) follows from the results below, but I don't see how to prove (ii).

Theorem 12.2 (Sums of zero-mean independent variables in $\mathcal{L}^2$, pp. 112-113). Suppose that $\left(X_k : k \in \mathbb{N}\right)$ is a sequence of independent random variables such that, for every $k$,
$$E\left(X_k\right) = 0,\ \sigma_k^2 := \mathrm{Var}\left(X_k\right) < \infty$$

a. Then
$$\left(\sum \sigma_k^2 < \infty\right)\ \textrm{implies that }\left(\sum X_k\ \textrm{converges a.s.}\right)$$

b. If the variables $\left(X_k\right)$ are bounded by some constant $K$ in $\left[0,\infty\right)$ in that $\left|X_k\left(\omega\right)\right| \leq K$, $\forall k, \forall \omega$, then

$$\left(\sum X_k\ \textrm{converges, a.s.}\right)\ \textrm{implies that }\left(\sum \sigma_k^2 < \infty\right)$$

Notes to the theorem

The Kolmogorov $0$-$1$ law implies that
$$P\left(\sum X_k\ \mathrm{converges}\right) = 0\ \mathrm{or}\ 1$$
The proof of part b given in the book shows in fact that if $\left(X_k\right)$ is a sequence of independent zero-mean RVs uniformly bounded by some constant $K$, then
$$\left(P\left\{\textrm{partial sums of }\sum X_k\ \textrm{are bounded}\right\} > 0\right) \implies \left(\sum X_k\ \textrm{converges a.s.}\right)$$

The events

$$A := \left\{\sum \varepsilon_k a_k\ \textrm{diverges to }+\infty\right\}$$

and

$$B := \left\{\sum \varepsilon_k a_k\ \textrm{diverges to }-\infty\right\}$$

are both tail events, hence by Kolmogorov's $0$-$1$ law $P\left(A\right),P\left(B\right)\in\left\{0,1\right\}$. However by symmetry $P\left(A\right)=P\left(B\right)$ and since $A \cap B = \emptyset$, we must have $P\left(A\right) = P\left(B\right) = 0$.

Since $ P(ε_n=±1)=\frac{1}{2}, var(ε_n ) = 1$
$\Sigma var(ε_k a_k)= \Sigma a^2_k \times var( ε_k )= \Sigma a^2_k $


Infinitely differentiable of Fourier solution to heat equation Given $f (x,t) = \sum_{n=1}^{\infty} a_{n}e^{-n^2t}\phi_{n}(x)$ with $\phi_{n}(x) =\sqrt{\frac{2}{\pi}}\sin(nx)$ and $a_n = \int_{0}^{\pi} f_0(x)\phi_{n}(x)dx$ is the Fourier solution to the heat equation $f_t - f_{xx} = 0$ on $(0,\pi)$ with Dirichlet condition $f(0,t) = f(\pi, t) = 0$ and initial condition $f(x,0) = f_{0}(x)$. In addition, assume $|a_{n}|\leq M$ for all $n$.

Question: Prove that $f(x,t)$ is $C^{\infty}$ in $x$ for each $t > 0$.

My attempt: I was able to show that at each $t = t_0$ ($t_0$ is fixed), $f(x,t_0)$ converges uniformly in $x$, thus $f(x,t)$ is $C^0$. But when I differentiate with respect to $x$, I got $f_x = \sum_{n=1}^{\infty} a_{n}e^{-n^2t_0}\sqrt{\frac{2}{\pi}}\ n\cos(nx)$. But I can't see how to show $\sum_{n=1}^{\infty} n\cos(nx) e^{-n^2t_0}$ converges, as the extra term $n$ causes me big trouble in bounding this series by a convergent series to apply the Weirstrass M-test. The more I differentiate, the higher degree of the term $n$ I would get.

Can anyone please help prove the later case (1st and 2nd derivative) to show that $f(x,t_0)$ is $C^{1}$ and $C^{2}$ in $x$?

For any fixed $t_0>0$, the exponential decay induced by $e^{-n^2 t_0}$ eats up any polynomial weight that arises from differentiation (of any order) in $x$. This can be expressed formally by an inequality such as the following:
$$\lvert n^k e^{-n^2 t_0}\rvert\le C_{t_0, k}\,e^{-n^2 \frac{t_0}2}. $$
In particular, the series for $\partial^k_x f$ converges uniformly in $x$ for any $k$.