What is the probability of flipping at least 3 heads or 3 tails in a row over the course of 10 flips? I want to find the chance of a streak of 3 or more of the same side occurring within 10 flips.

My idea:

Number of favorable outcomes / Number of possible outcomes
Number of favorable outcomes / 2^10

To find the number of favorable outcomes:

First, I find all possibilities of the 7 non-streaking flips:

2^7

Then I find all the possibilities of the streak of 3 flips:

It could be HEADS or TAILS
It could be in any of the 8 positions between the non-streaking flips
2 sides * 8 positions = 16

My wrong answer:

2^7*16 / 2^10 = 2

What did I do wrong?

It's easier to work out the complementary probability, that all streaks
have length at most $2$. Then there will be $j$ streaks of length $2$
and $10-2j$ streaks of length $1$, $10-j$ streaks altogether. Here $0\le j\le 5$. Out of the $10-j$ streaks you must choose $j$ to have length $2$
and you can choose to start with a head or tail. Therefore there are
$$2\binom{10-j}j$$
sequences with $10-2j$ streaks of length $1$ and $j$ of length $2$.
The probability you seek is then
$$1-\frac1{2^9}\sum_{j=0}^5\binom{10-j}j.$$

Let ${\mathscr{S}}_{n}$ be the set of sequences of $n$
flips that don't contain a streak of $3$. We are going
to compute $\left|{\mathscr{S}}_{n}\right|$ by induction on $n$. First
note that ${\mathscr{S}}_{1} = \left\{H , T\right\}$ and
${\mathscr{S}}_{2} = \left\{H H , H T , T H , T T\right\}$, so that
$\left|{\mathscr{S}}_{1}\right| = 2$ and $\left|{\mathscr{S}}_{2}\right| = 4$.

In order to give an induction formula, we partition

$${\mathscr{S}}_{n} = {\mathscr{V}}_{n} \cup  {\mathscr{W}}_{n}$$

where ${\mathscr{W}}_{n}$ is the subset of sequences that end with a double
streak, such as $H H$, and ${\mathscr{V}}_{n}$ is the set of sequences
that don't end with a double streak. Obviously, $\left|{\mathscr{V}}_{2}\right| = 2$
and $\left|{\mathscr{W}}_{2}\right| = 2$.

We create elements of ${\mathscr{S}}_{n+1}$ by appending a flip to an element
$s \in  {\mathscr{S}}_{n}$. If $s \in  {\mathscr{W}}_{n}$, there is only
one way to complete it without creating a streak of $3$ and
the result an element of ${\mathscr{V}}_{n+1}$. If $s \in  {\mathscr{V}}_{n}$, there
are two ways to complete it, one way creates an element of ${\mathscr{V}}_{n+1}$,
the other way creates an element of ${\mathscr{W}}_{n+1}$.

It follows that we have the following induction relations

$$\left\{\begin{array}{rcl}\left|{\mathscr{V}}_{n+1}\right|&=&\left|{\mathscr{V}}_{n}\right|+\left|{\mathscr{W}}_{n}\right|\\
\left|{\mathscr{W}}_{n+1}\right|&=&\left|{\mathscr{V}}_{n}\right|
\end{array}\right.$$

It is now obvious by induction that $\left|{\mathscr{V}}_{n}\right| = 2 {F}_{n}$ and
$\left|{\mathscr{W}}_{n}\right| = 2 {F}_{n-1}$, where ${F}_{n}$ is the
$n$th Fibonacci number. Indeed the result is true when
$n = 2$ and the relation ${F}_{n+1} = {F}_{n}+{F}_{n-1}$ ensures
the induction.
The conclusion is that

$$\boxed{\left|{\mathscr{S}}_{n}\right| = \left|{\mathscr{V}}_{n}\right|+\left|{\mathscr{W}}_{n}\right| = 2 {F}_{n}+2 {F}_{n-1} = 2 {F}_{n+1}}$$

In particular, $\left|{\mathscr{S}}_{10}\right| = 2\times{F}_{11} = 178$.

It follows that the required probability is

$$1-\frac{178}{{2}^{10}} = \frac{423}{512}$$

Incidentally, combining this with @LordSharkTheUnknown's reasoning yields the conjectural result (which proof by induction looks easy).
$$
\boxed{\sum_{j\ \le\ n-j}\binom{n-j}{ j} = F_{n+1}}
$$
This formula is also given in the Wikipedia page about binomial coefficients.


How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement. The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.

I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.

I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.

Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards  $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$  which is what I want $\to 720/6 = 120$.

So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.

I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting
$A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states,  $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.

Your solution is correct.

Here is an alternative approach:

Line up the six players in some order, say alphabetically.  The first person in line can choose a partner in $5$ ways.  Remove those two players from the line.  That leaves four players.  The first person left in the line can choose a partner in $3$ ways.  Remove those two players from the line.  The remaining two players must play each other.  Hence, there are
$$5!! = 5 \cdot 3 \cdot 1 = 15$$
ways to form three pairs of chess partners.  The expression $5!!$ is read $5$ double factorial.

Yes, you are absolutely right.

The other approach might be selecting $2$ players out of $6$ since arrangement doesn't matter ($_6C_2=15$).


I think there is an error in the solution. Is it correct? I read the solution and think there is an error in the solution.

I think $z_1$ in red boxes should be changed to m.

Am I correct? Or did I misinterpret the solution?

I think the key is that m(the lowest f value)) can't be an interior point!

I'm a bit confused about this solution to start with.  We are talking about f on an open set, which is certainly not compact, so there is no guarantee it will attain a max or min there; you can close the interval of course and get your max and min.  That said, I think we can do this solution better.

Let us suppose f is not monotonic on the interval (a,b).  Then it has either a minimum or a maximum inside (a,b). Call that point c. If c is the absolute max or min, whatever f maps (a,b) into, it will contain f(c) as an endpoint and cannot be open.

Suppose c is only a local max or min.  Then consider the problem on an interval (q,r) around c where c is the absolute max or min.  So as per the above paragraph whatever f maps (q,r) into it must contain f(c) as an endpoint and cannot be open.

The arguments below can be applied to (a,b) if c is an absolute max/min or (q,r) if c is a local max/min.

Turning it around, suppose the mapping of (a,b) is not open and its included endpoint is f(c).  Then f(c) must be ≤ f(a) or f(c) ≥ f(b) -- either way  f is not monotonic on (a,b).

Note that if f is indeed monotonic on (a,b), then its map of (a,b) will certainly be open, because it will be exactly (f(a), f(b)).

What you need to do to finish up is: if f is an open mapping and so maps every open interval into an open interval,  it has to be monotonic on  on the entire x-axis.  I suspect you have to fuss around a little to nail down that last statement. For example, you might start with (-1,1) and keep expanding to larger intervals -- no matter how large an interval you choose, f will be monotonic, so this must apply to the real line.


On Hatcher's proof that $H_0(X)$ is a direct sum of $\mathbb{Z}$s? I'm confused about part of the proof of Proposition 2.7 in Hatcher.

  If $X$ is nonempty and path-connected, then $H_0(X)\simeq\mathbb{Z}$. Hence for any space $X$, $H_0(X)$ is a direct sum of $\mathbb{Z}$s, one for each path-component of $X$.

Proof: By definition, $H_0(X)=C_0(X)/\operatorname{Im}\partial_1$. Define a homomorphism $\epsilon\colon C_0(X)\to\mathbb{Z}$ by $\epsilon(\sum_i n_i\sigma_i)=\sum_i n_i$. This is surjective, so the proof proceeds to show $\operatorname{Im}\partial_1=\ker\epsilon$.

To show $\ker\epsilon\subset\operatorname{Im}\partial_1$, suppose $\epsilon(\sum_i n_i\sigma_i)=0$, so $\sum_i n_i=0$. The $\sigma_i$ are singular $0$-simplicies, which are simply points of $X$. Choose a path $\tau_i\colon I\to X$ from a basepoint $x_0$ to $\sigma_i(v_0)$, and let $\sigma_0$ be the singular $0$-simplex with image $x_0$. We can view $\tau_i$ as a singular $1$-simplex $\tau_i\colon [v_0,v_1]\to X$ and then we have $\partial\tau_i=\sigma_i-\sigma_0$.

This final sentence is confusing to me. I know we can think of $1$-simplicies as line segments, so it sort of makes sense to think of $\tau_i$ as a singular $1$-simplex. But I thought then $\partial\tau_i=\tau_i|[v_1]-\tau_i|[v_0]$, why does Hatcher say $\partial\tau_i=\sigma_i-\sigma_0$?

You are almost all the way there. You have $\partial \tau_i=\tau_i|[v_1]-\tau_i|[v_0]$, but $\tau_i|[v_j]$ is just mapping $v_j$ to a point of $X$, i.e. it is a zero simplex. By construction, it coincides with $\sigma_i(v_j)$.

Because $\tau_i| [v_1]$ is $\tau_i$ evaluated at $v_1$, which is just $\sigma_1$.


How to solve the equation $k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$ I've been struggling a little on this problem for a while. I'm not sure what I'm missing. It goes like this:
$$k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$$
Here $k, a$ are constants and $v, t$ represent velocity and time respectively. I want to know $v$ as a function of $t$. Is it possible to solve this? Any help would be appreciated!

Set $V(t)=\int_0^tv(s)ds$. Then your equation reads as
$$
k=V''(a+2V)^2.
$$
this can be separated and integrated once
$$
V''=\frac{k}{(a+2V)^2}\implies V'^2=c-\frac{k}{a+2V}.
$$
In principle you can now apply separation again, the resulting integral with square root might be difficult or impossible to solve.

If $v$ is velocity and is given by $v=\frac{ds}{dt}$ then
$$\int_0^tv(\tau)d\tau=\int_0^t\frac{ds}{d\tau}d\tau$$
By the fundamental theorem of calculus and Leibniz's rule:
$$\int_0^t\frac{ds}{dt}d\tau=\frac{d}{dt}\int_0^tsd\tau=\frac{d}{dt}st=s$$
Note: We are only allowed to do this because these specific limits of integration(zero at the left part, variable/parameter on the right part)
Substituting back in the original equation:
Which is 1st order in $v$
$$k=\frac{dv}{dt}(a+2s)^2$$
But because $v$ depends on $s$ we need to integrate it wrt. to $s$, then it's a 2nd order linear ODE:
$$k=\frac{d^2s}{dt^2}(a+2s)^2$$
$$\iint kdtdt=\iint(a+2s)^2dsds$$
$$k\frac{t^2}{2}=\int \frac{(a+2s)^3}{3a}ds+C=\frac{(a+2s)^4}{4a^2}+C$$
Now solve for $s$ and differentiate wrt. $t$ get back your solution to $v$.
Why doesn't distance depend on time?
To OP: try going through each step, i'll leave you to differentiate it on your own. Tell me if you got stuck.


Find solutions of the inequality Find all real values of $x$ satisfying the inequality: $x^{\log_3
 (x+4)} \lt 3^5$.

Hint: take the logarithm w.r.t to $3$ on both sides of the inequality.

We need to solve $f(x)<5$, where $f(x)=\log_3x\log_3(x+4)$.

But $f$ is an increasing function on $[1,+\infty)$ and all $0<x<1$ are solutions.

Thus, the answer is $(0,x_0)$, where $x_0$ is a root of the equation $f(x)=5$:

$x_0=9.9024...$


How to solve $x^2 = e^x$ The question is to find $x$ in:

  \begin{equation*}
x^2=e^x
\end{equation*}

  I know Newton's method and hence could find the approx as $x\approx -0.7034674225$ from

\begin{equation*}
x_{n+1} = x_n - \dfrac{x_n^2-e^{x_n}}{2x_n-e^{x_n}}
\end{equation*}

According to WolframAlpha:

  They also say that $x=-2W(\dfrac{1}{2})$ which shows that it can be solved using some Lambert-W function...Can anyone tell me how to do this?

Thanks a lot!

P.S. - I studied a li'l bit of Lambert-W ... So i guess a detailed explanation would not be needed ... just the initial steps!

$$
x^2=e^x\implies x/2=\pm\tfrac12e^{x/2}\implies-x/2\,e^{-x/2}=\pm\tfrac12
$$
Therefore,
$$
x=-2\mathrm{W}\!\left(\pm\tfrac12\right)
$$
Since $\mathrm{W}(x)$ is real only for $x\ge-\frac1e$, we only have one real solution:
$$
x=-2\mathrm{W}\!\left(\tfrac12\right)=-0.70346742249839165205
$$

You can write $\mathrm{f(x)=e^{-x}.x^2}$
Now differentiate this function w.r.t x.
$$\mathrm{f`(x)=e^{-x}(2x-x^2)}$$.
Now set $$\mathrm{f'(x)=0}$$.
It yields the solutions $$\mathrm{x=0 and x=2}$$.
The values of f(x) at these x values are 0 and $\mathrm{4/e^2}$ respectively. Obviously $\mathrm{4/e^2}$ is less than 1.
Now plot the graph of this function. Find the number of intersection points of this function and $\mathrm{y=1}$.


Can one deduce whether a given quantity is possible as the area of a triangle when supplied with the length of two of its sides? Recently I have found a question like following:

  In triangle $ABC$, $AB=AC=2$. Which of the following could be the area of
  triangle $ABC$? Indicate all possible areas:

  [A] $0.5$ [B] $1.0$ [C] $1.5$ [D] $2.0$ [E] $2.5$ [F] $3.0$

From my points of view, I can only guess one answer if I assume the triangle is a right angle triangle. In that case, the area will be $2$.

  But the answer showed the result, [A][B][C][D].

So, my question this is there any axiom that the area of the right angle will be the highest area of any type of triangle with the expressing two lengths of it?

Thanks in advance.

Not an axiom, but yes:  The area of a triangle is $\frac{1}{2}ab \sin \theta$, where $a$ and $b$ are two known side lengths and $\theta$ is the angle between the two given side lengths.  Since $\sin \theta$ reaches its maximum value when $\theta=90°$ (in which case we have $\sin 90° = 1$), the largest possible area is indeed attained for the case of a right angle.

No it is not possible to deduce just by two sides, because we don't know the angle between those two sides, which will decide how much area triangle will cover.
Take two lines of  $5$cm each and keep on changing the angle $(b,w)$ from $0$ to $180°$, you'll find that just two sides isn't enough.


Calculate sin(x) over sin(y) I'm given the equation $$\frac{\sin(\delta)}{\sin(102.9^\circ-\delta)}= b\cdot\frac{\sin(\alpha)}{a\sin(\beta)}$$ where I know the values of $a$, $b$, $\alpha$ and $\beta$.
How can I solve the equation for $\delta$?

To expand @Piquito comment your problem can be written as:
$$\sin(\delta)=A \sin(a-\delta)$$
One way to solve such problem is to write $\sin(a-\delta)=\sin(\delta) \cos(a)-\cos(\delta) \sin(a)$ in order to have:
$$(1-A\cos(a))\sin(\delta)=-A\sin(a) \cos(\delta)$$
which is easier to solve.

I would just do Mathematica here. The solutions are non-trivial. The code

FullSimplify[Solve[Sin[delta]/Sin[angle-delta]==b Sin[alpha]/(a Sin[beta]),delta]]

yields

$$\left\{\left\{\delta \to -\cos ^{-1}\left(\frac{-a-b \sin (\alpha ) \cos
   (\text{angle}) \csc (\beta )}{\sqrt{a^2+b \sin (\alpha ) \csc (\beta ) (2 a \cos
   (\text{angle})+b \sin (\alpha ) \csc (\beta ))}}\right)\right\},\left\{\delta \to
   \cos ^{-1}\left(\frac{-a-b \sin (\alpha ) \cos (\text{angle}) \csc (\beta
   )}{\sqrt{a^2+b \sin (\alpha ) \csc (\beta ) (2 a \cos (\text{angle})+b \sin
   (\alpha ) \csc (\beta ))}}\right)\right\},\left\{\delta \to -\cos
   ^{-1}\left(\frac{a+b \sin (\alpha ) \cos (\text{angle}) \csc (\beta )}{\sqrt{a^2+b
   \sin (\alpha ) \csc (\beta ) (2 a \cos (\text{angle})+b \sin (\alpha ) \csc (\beta
   ))}}\right)\right\},\left\{\delta \to \cos ^{-1}\left(\frac{a+b \sin (\alpha )
   \cos (\text{angle}) \csc (\beta )}{\sqrt{a^2+b \sin (\alpha ) \csc (\beta ) (2 a
   \cos (\text{angle})+b \sin (\alpha ) \csc (\beta ))}}\right)\right\}\right\}$$
Here the "angle" is, of course, your $102.9^{\circ}$.


Infinite integral that seems to diverge but question requires it does not. The formula for half space solutions to the Laplace equation can be given:
\begin{equation}
u(x,y) = \int_{z\in \mathbb{R}} \frac{1}{\pi}\frac{y}{(x-z)^2+y^2}f(z)
\end{equation}

Use this formula to solve the initial value problem:
\begin{equation}
\nabla^2u=0. \quad u(x,0)=f(x)=\frac{1}{1+x^2}
\end{equation}
Now I have gotten the integral into the form (as a hint in the question asks):
\begin{equation}
\int_{z\in \mathbb{R}}\frac{a(z-x)+b}{(z-x)^2+y^2} + \frac{cz+d}{1+z^2}
\end{equation}
Where a,b,c,d do not depend on z.
But have been unable to compute this integral. Wolfram Alpha says the integral diverges. The integral can be Lebesgue integration. Any hints/similar eamples would be great,thanks.


Is complex analysis more "real" than real analysis? In physics, in the past, complex numbers were used only to remember or simplify formulas and computations. But after the birth of quantum physics, they found that a thing as real as "matter" itself had to be described by complex wave functions and there's no way to describe it using only real numbers.

In mathematics, in real analysis, there's examples like the function $f(x)=\frac{1}{1+x^2}$, and why this function does not have the "smoothtness" of the exponential function, polynomials, sine and cosine functions, why it has a radius of convergence equals 1 despite the fact that this function is infinitely differentiable. You can't see the reality of this function until you see it through the field of complex analysis, where you can observe that $f$ is not that smooth because it has 2 singularities in the complex plane.

I am just asking for examples like this such that when you see it in the narrow "window" of real analysis, you can't see the "reality" until you view it from the window of complex analysis. I am just starting to self learn complex analysis and I find it more natural than real analysis and it tells you the "truth" behind a lot of things.

Not all real degree $n$ polynomials have $n$ roots (counting multiplicity) because some of the roots are complex. In the real domain a matrix can have no eigenvalues, e.g. 2-dimensional rotation matrix, but any real matrix has a complex eigenvalue. These are manifestations of  $\mathbb{C}$ unlike $\mathbb{R}$ being algebraically closed, i.e. every polynomial equation having a solution. In the real domain $\sqrt{x}$ and $\ln{x}$ are only defined for positive $x$ because for negative $x$ the value is a complex number, and it is not unique.

In the real domain exponents and trigonometric functions are completely different functions, but in the complex domain they are related by simple Euler formulas. The same goes for logarithms and inverse trigonometric functions. This is the main reason why identities for hyperbolic functions are almost the same as familiar trigonometric identities. Many definite integrals of functions that do not have elementary antiderivatives can be computed in elementary terms by extending the path of integration to the complex plane and using residues, e.g. $\int_0^\infty\frac{\ln x}{(1+x^2)^2}\,dx=-\frac{\pi}{4}$. More generally, integral and series representations of many real functions can be converted into each other because these functions extend into the complex plane and contour integrals there reduce to sums over residues. The Riemann zeta function is a typical beneficiary. These manifest another advantage of complex analysis over real one. Many commonly used real functions extend to holomorphic functions in the complex plane, and for holomorphic functions calculus tools are much stronger than for smooth ones, which is what real analysis mostly treats them as.

In the real domain ellipses and hyperbolas are different types of curves, but in the complex plane they are related by a rotation of axes, i.e. they are the 'same' (more precisely, we are looking at two different projections of the same complex curve). In a similar way spherical and hyperbolic geometries are related by a complex rotation. The Schrödinger equation of quantum mechanics and the heat equation of classical physics are also related by a complex rotation called Wick rotation. Path integral interpretation of quantum mechanical solutions can be made precise using this relation and the Feynman–Kac formula.

Heaviside developed operational calculus for solving ordinary differential equations with constant coefficients by treating time derivative as a 'variable' $p$ and writing solutions in terms of symbolic 'functions' of it. It turned out that the magic worked because $p$ is in fact a complex variable, and Heaviside's symbolic solutions can be converted into real ones by taking the inverse Laplace transform, which is a contour integral in the complex plane.

Harmonic functions, solutions to the Laplace equation, have many nice analytic properties like being sums of convergent power series, attaining extrema at the boundary of their domains, being equal at any point to the average of values on any circle centered at it, etc. The underlying reason is that harmonic functions are exactly the real and imaginary parts of holomorphic functions. If the potential of a vector field is a harmonic function $\varphi$ then its flow lines are level curves of another harmonic function $\psi$, exactly the one that makes $\varphi+i\psi$ holomorphic. Solution formulas for the Dirichlet boundary problem for the Laplace equation in some special domains are reflections of the Cauchy integral fomula for holomorphic functions that works in 'any' domain.

Some things about complex analysis seem special when in fact those results are better viewed through the lens of vector calculus.

For instance, any holomorphic function can be viewed instead as a vector field with vanishing divergence and curl.  Consider the Cauchy integral theorem and Stokes' theorem:

$$\begin{align*}\text{Cauchy integral theorem: }& \oint_C f(z) \, dz = 0, \quad \left(\frac{\partial f}{\partial \bar z} = 0 \right)\\
\text{Stokes' theorem: } & \oint_C F \cdot d\ell = 0, \quad (\nabla \times F = 0) \\
\text{Stokes' theorem corollary: }& \oint_C F \times d\ell = 0, \quad (\nabla \cdot F = 0)
\end{align*}$$

Now, you may be tempted to say that the complex analysis version of this result is more elegant because it's one line instead of two.  You can package the vector calculus result into one line with the proper tools (clifford algebra) as well, though, and more importantly, when you properly identify the derivatives as exterior and interior derivatives, the result generalizes to arbitrarily high-dimensional real vector spaces.

Similarly, meromorphic functions correspond to vector fields with point sources, but that's usually where complex analysis stops, while vector calculus has the tools to treat vector fields with arbitrary sources.

That quantum mechanics is written in terms of complex numbers is, in my opinion, a pedagogical mistake.  It would be clearer to do quantum-style probability in terms of vectors in $\mathbb R^2$; of course, the math will ultimately come out the same, but I think the picture of things would be considerably cleaner.

Complex analysis also puts a lot of emphasis on complex differentiability, but as you can see from the above argument about the CIT and Stokes' theorem, it's actually differentiation with respect to $\bar z$ that relates very directly to vector calculus.  Complex differentiation is utterly insignificant when considering more general applications.


Calculate: $\lim _{x\to \infty }\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$. Calculate:
$$\lim _{x\to \infty }\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$$

Here's my attempt:
I first tried to the statement up. I first let $f(x)=\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$, $f_1(x)=\left(\frac{\sqrt{x^3+4}}{\sqrt{x^3+2x^2+7x+11}}\right)$ and $f_2(x)=\left(\frac{\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$. We know that $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f_1(x)+\lim_{x \to \infty}f_2(x)$. Now we can find the limit of $f_1(x)$ and $f_2(x)$, which will give us our solution.Therefore, we have:
\begin{align}
\lim_{x \to \infty}f_1(x)&=\lim_{x \to \infty}\left(\frac{\sqrt{x^3+4}}{\sqrt{x^3+2x^2+7x+11}}\right)\\
&= {\sqrt{\lim_{x \to \infty} \frac{x^{3} + 4}{x^{3} + 2 x^{2} + 7 x + 11}}} \\
&= \sqrt{\lim_{x \to \infty} \frac{1 + \frac{4}{x^{3}}}{1 + \frac{2}{x} + \frac{7}{x^{2}} + \frac{11}{x^{3}}}} \\
&= \sqrt{\frac{\lim_{x \to \infty}\left(1 + \frac{4}{x^{3}}\right)}{\lim_{x \to \infty}\left(1 + \frac{2}{x} + \frac{7}{x^{2}} + \frac{11}{x^{3}}\right)}} \\
&= \ ... \\
&= 1
\end{align}
\begin{align}
\lim_{x \to \infty}f_2(x)&=\lim_{x \to \infty}\left(\frac{\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right) \\
&= \ ... \\
&= 0
\end{align}
For the limit of $f_2(x)$, I left out the actuall working out but you can either use the squeeze theorem or the fact that since $\sin(x)$ is divergent and the denominator converges to $0$, which means that the overall function will also converge to $0$. According to the Squeeze Theorem, since $- \frac{1}{\sqrt{x^{3} + 2 x^{2} + 7 x + 11}} \leq \frac{\sin{\left(x \right)}}{\sqrt{x^{3} + 2 x^{2} + 7 x + 11}} \leq \frac{1}{\sqrt{x^{3} + 2 x^{2} + 7 x + 11}}$ and both the functions on the left and right converge to $0$, so does $f_2(x)$.
Therefore: $$\lim _{x\to \infty }\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)=1$$
However, I'm not sure if this is $100%$ since neither Symbolab nor EMathHelp could determine the solution.

When $x$ is very large the most dominant term in Num. is $x^{3/2}$ and so is in the Den. Hence  the limit as $x$ tends to $\infty$ is 1.

Result from WolframAlpha for Verification


Can one cancel $\mathbb R$ in a bi-Lipschitz embedding? Let $X$ be a metric space. Suppose that the product $X\times\mathbb R$ admits a bi-Lipschitz embedding into $\mathbb R^{n}$. Does it follow that $X$ admits a bi-Lipschitz embedding into $\mathbb R^{n-1}$?

Definitions and comments:

A map $F\colon (Y,\rho)\to\mathbb R^n$ is a bi-Lipschitz embedding if there exists a constant $L$ such that $L^{-1}\rho(a,b)\le |F(a)-F(b)|\le L\rho(a,b)$ for all $a,b\in Y$.
The choice of a product metric on $X\times \mathbb R$ makes no difference. For definiteness, let the distance between $(x,t)$ and $(x',t')$ be $d_X(x,x')+|t-t'|$.
The cancellation fails if bi-Lipschitz is replaced by topological or smooth embedding. For example, $S^1\times \mathbb R$ embeds into $\mathbb R^2$ smoothly, but $S^1$ does not embed into $\mathbb R$ in any sense.
The cancellation is possible when $n=1$, according to $\mathbb R^0=\{0\}$. :) I'm pretty sure it's also possible when $n=2$, but don't have a proof. For $n\ge 3$ I don't have a clue.

I think the answer to your question does not depend at all on the target space, but only on the domain of your function and how you define the metric on it.
If you use the metric you suggested, then the answer is positive for whichever target space $Y$. For, suppose $F\colon X\times\mathbb{R}\to Y$ is a bi-Lipschitz embedding into any metric space $Y$. Define a new map $\hat F\colon X\to Y$ by setting $\hat F(x)=F(x,0)$. Then you get: $d_{Y}(\hat F(x),\hat F(y))=d_Y(F(x,0),F(y,0))\le L d_{X\times\mathbb{R}}((x,0),(y,0))=d_X(x,y)$
and similarly for the other inequality. I hope this helps!


Can the scalars of a vector space be of a different field than the entries of the vector space? When denoting a particular vector space can the field that denotes the entries in the vectors of a vector space be different than the field that denotes the scalars that can be used on that vector space. For example if $V$ is a vector space denoted by $M_{m\times n}(R)$ do the scalars $c$ that act on that vector space have to be elements of $R$? I was reviewing the text "Linear Algebra" by Friedberg, Insel and Spence, and was a little confused by the following question:

Let $V$ denote the set of all $m\times n$ matrices with real entries; so $V$ is a vector space over  $R$ by example 2 (note this was the general definition of matrices over $R$). Let $F$ be the field of rational numbers. Is $V$ a vector space over $F$ with the usual definitions of matrix addition and scalar multiplication?

If $c\in R$ then no because you could do scalar multiplication with an irrational number and that would take you out of the vector space over $Q$ because an irrational number times a rational number is an irrational number. Now if $c$ is in $Q$ then yes it is a vector space. Without $c$ being explicitly denoted do I assume it belongs to the field that was defined with the vector space? Also, can $c$ and $V$ be different fields when denoting a vector space?

Yes, it is a vector space over F.

A vector space over F is a set V (in your case $m\times n$ matrices with real entries) with operations $+:V\times V \to V$ and $\cdot:F\times V \to V$ satisfying certain axioms, as found for example on the wikipedia vector space page.

That these axioms are satisfied is easily verified. The first four amount to V being an abelian group, and that last four must be satisfied by F since it is subfield of R.

It also follows easily from these observations that if V is a vector space over a field K, then it is a vector space over any subfield F of K (i.e., the example above generalizes).

It is important to note, though, that V is NOT a finite dimensional vector space over F. The dimensional isn't even countable, in fact.

I think you're misinterpreting the question. Let's take a look at it:

  Let $V$ denote the set of all $m\times n$ matrices with real entries; so $V$ is a vector space over  $R$ by example 2 (note this was the general definition of matrices over $R$). Let $F$ be the field of rational numbers. Is $V$ a vector space over $F$ with the usual definitions of matrix addition and scalar multiplication?

Recall that a vector space is a set of indexed arrays (vectors) whose entries come from a field (scalars) that satisfy the vector space axioms. The field is required in the definition of the vector space just as much as the operations. It should not be taken for granted that addition is generally just to mean addition in the elementary sense. Certain fields can be paired with very odd operations that still satisfy the vector space axioms, and so we call the operation "addition" in the context of discussion of the vector space. Really, one might consider a vector space itself as a tuple of three elements($\mathbb{F},[\;]^{m \times n},+)$ with three entries that are first a field of scalars, an array structure to hold the scalars, and third an operation to combine the arrays that behaves like addition in certain ways.

In the first sentence of the question, the author is establishing that the set of $m \times n$ arrays of real entries with Euclidean vector addition satisfies the axioms of a vector space. In the tuple notation, $(\mathbb{R},[\;]^{m \times n},+)$ is a vector space.

In asking "Is V a vector space over F?", the author cannot possibly mean "Is the set of $m \times n$ matrices with real entries a vector space over $F$?" because to say "a vector space over $F$" means that the entries come from $F$, so the answer is a trivial no. It is trivial that $\mathbb{R}^{m \times n}$ is not a vector space over $F$, because by it's very definition it has as its field of scalars the real numbers.

I don't believe that the author intended to ask a trivial question. Instead, the task is to determine if in the definition of $V$ one were to replace "real" with "rational", would the resulting $V$ given $(F,[\;]^{m \times n},+)$ also be a vector space. The question is non-trivial because $F$ is a subset of $\mathbb{R}$, so in this case if $(F,[\;]^{m \times n},+)$ is a vector space, then it is a subspace of $(\mathbb{R},[\;]^{m \times n},+)$ so the result can be demonstrated by showing that the set is closed under linear combinations.


Cohomology of augmented double complex with exact rows using spectral sequences I am learning about spectral sequences through Vakil's notes and wanted to try out a simple spectral sequence argument on a double complex. I recalled this little paragraph in Bott & Tu, p.97:

  If all the rows of an augmented double complex are exact, then the total cohomology of the complex is isomorphic to the cohomology of the initial column.

(In their case this applies to the Čech-de Rham complex.) I think this is amenable to a proof via spectral sequences.

So the situation is that we have some double complex $C^{p,q}$ with $p \geq -1$ and $q \geq 0$. (The column $p = -1$ is the augmented one.) Since all the rows are exact, the first page using the horizontal spectral sequences already vanishes. Now we should compare this to the vertical spectral sequence, but this is where I get stuck. Since we have a first quadrant spectral sequence, all the terms will eventually stabilize to zero, but it seems that this might take some time very far into the double complex. Even for the very first terms, for instance, I find that $E_2^{-1,0} = 0$, but this only tells me that the $0$th cohomology of the augmented column injects into $H^{0,0}(C)$, not even that this is an isomorphism.

Can you help me finish this argument?

They could have made it clearer, but I think that Bott and Tu are making a distinction between "the double complex" and the "augmented double complex" with an extra initial column.

So in terms of spectral sequences, all they're saying is that if the rows of the augmented double complex are exact, then the cohomology of the rows of the unaugmented double complex just gives the initial column of the augmented complex. So the "rows first" spectral sequence for the unaugmented complex just degenerates at the $E_2$ page to the cohomology of the initial (augmenting) column.

The "columns first" spectral sequence isn't used here. All you need is that for a first quadrant double complex, either of the two spectral sequences converges  to the cohomology of the total complex.

I'm currently studying that part of the book Differential Forms in Algebraic Topology and I think that the claim you are citing is not completely correct. To be more clear...Look at the the Lemma $4$ of these notes. You can easly prove the lemma by using spectral sequences. Indeed, take the s.s. ${}^IE^{pq}$ obtained by filtring the rows of the double complex, then you see that ${}^IE^{pq}$ collapses at the first sheet, i.e. the cohomology of the of the first non-exact row is isomorphic to the cohomology of the total complex. Bott and Tu want show that $H^\bullet_{dR}(M)\simeq H^\bullet(\mathcal{U},\mathbb{R})$, see the figure below.

By applying the Corollary $6$ of the notes what you get is exactly that Bott and Tu want check!

Notice that the topological assumption to apply the corollary is that $\mathcal{U}$ is a good cover, i.e. (algebraically) the columns eccept $\Omega^\bullet (M)$ are exact.


Non-rotatable dominoes Imagine matching dominoes end-to-end such that only like numbers are joined.  For a collection of dominoes, one could ask whether it is possible to join them all into a single strand.

I'm interested in a slightly different problem.  What if dominoes could be translated but not rotated?  That is, [2 | 3] is distinct from [3 | 2] so there are $n^2$ dominoes with numbers up to $n$ rather than ${n\choose2}+n$ dominoes.  In particular, for a given multiset of dominoes (a non-negative integer-valued $n\times n$ matrix, if you prefer), is it possible to choose an order such that the second and first numbers, respectively, of adjacent dominoes are equal?  (The first number of the first domino need not equal the second number of the last domino: they are in a line not a ring.)

I'd also be interested in related questions:

the number of solutions (if there are solutions)
an algorithm for finding a solution (if there are solutions)
the length of a maximal solvable submultiset (if there are no solutions)
an algorithm for finding a maximal solvable submultiset (if there are no solutions)

if anyone has a reference (or is looking for brownie points).

Clarification: I do not have all $n^2$ dominoes, but a multiset of dominoes.  Some appear more than once, others (most!) do not appear at all.

I don't quiet like @RossMillikan's answer since it reduces to a hard problem (Hamiltonian path!). Here is another reduction to an easier question: Eulerian path in the directed multi-graph (if you want your non-rotation condition and allow repeated dominos).

If you dominos have ends numbered from $1$ to $n$, then create a graph on $n$ vertices. If a domino is labeled $[x | y]$ then add the directed edge $(x,y)$ to your graph (add multiple edges if the domino appears multiple times). To make a chain you need to find a Eulerian path in this graph (if you want your dominos to form a circle, then you need a Eulerian cycle). Further, note that the reduction in the opposite direction works too, every dircted multigraph can be encoded as a dominos problems. Thus this question is equivalent to the Eulirean path problem, and hence a reference is the literature on Eulerian paths.

If you said domino could only appear once, then the graph would not be a multi-graph. If you said we could rotate dominos, then the graph would be undirected.

How do you check if a multi-graph has a Eulirean path? If you have a set (instead of a multi-set) of dominos or in terms of graphs you have a directed multi-graph then the condition is simple:

The graph is connected
And each vertex has the same in-degree and out-degree. With the possible except of two vertexes with one having one more in-degree than out-degree and the other having one more out-degree than in-degree.

There are always many solutions.  You can think of it as a directed graph.  Each domino is a vertex.  There is a path from vertex $A$ to vertex $B$ if domino $B$ can follow $A$ in the chain.  Each domino has $n$ incoming paths and $n$ outgoing paths, and you are looking for a Hamiltonian path.  One way to prove it possible is by induction.  It works for $n=1$ as there is only one domino.  Given a solution for $n$, create a solution for $n+1$ by joining all dominoes $[a,n+1]$ to $[n+1,a]$ to make a new $[a,a]$.  These can be inserted anywhere in the chain where there is an $a$.  You also need to hide the $[n+1,n+1]$ in one of the joints.

There are clearly many more solutions.


If $T$ is a stopping time, prove that $X_T$ is $\mathcal F_T$ measurable. Let $(X_t)$ a progressively measurable process, i.e. $[0,t]\times \Omega \ni (t,\omega )\mapsto X_t(\omega )\in (\mathbb R, \mathcal B(\mathbb R))$ is $\mathcal B([0,t])\otimes \mathcal F$ measurable. Prove that $X_T$ is $\mathcal F_T$ measurable on $\{T<\infty \}$.

Q1) What does it mean that $X_T$ is $\mathcal F_T$ measurable on $\{T<\infty \}$ ? Does it mean that for all Borel set $B$, we have $(X_T\boldsymbol 1_{\{T<\infty \}})^{-1}(B)\in \mathcal F_T$ ?

Suppose my definition is correct. Then that's enough to prove

$\{X_T\boldsymbol 1_{\{T<\infty \}}\leq x\}\in \mathcal F_T$ for all $x$. ($*$)

Q2) Any idea on how to prove ($*$) ?

I don't see how I can prove that

$\{X_T\boldsymbol 1_{\{T<\infty \}} \} \in \mathcal F_\infty  $ and $\{X_T\boldsymbol 1_{\{T<\infty \}}\leq x\}\cap \{T\leq t\}\in \mathcal F_t$ for all $t$.

Hint

Let $B$ a Borel set.

$$\{X_{T}\in B\}\cap \{T\leq t\}=\{X_{T\wedge t}\in B\}\cap \{T\leq t\}.$$

Therefore, if you can prove that the stopped process is progressively measurable, the claim follow.

Proof: provided that $T$ takes a finite value, for every positive value $t$ it will take, we will be able to decide whether $X_t$ belongs to a Borel set $B$, whatever set we choose.

Formaly, we need to prove for all Borel set $B$ and $t\ge0$ that
$$
\left\{\omega\in\Omega,\;X_{T(\omega)}(\omega)\in B\right\}\cap\left\{\omega\in\Omega,\;T(\omega)\le t\right\}
$$
is $\mathcal F_t$-measurable. But
\begin{alignat*}{2}
\left\{X_{T}\in B\right\}\cap\left\{T\le t\right\} & = \left\{\omega\in\Omega:\; T(\omega) = s, s \le t, X_s(\omega) \in B \right\}
\end{alignat*}
which is $\mathcal F_t$-mesurable.


Formal definition of Mathematical Induction & Strong Induction I have been reading some notes on Induction and Strong Induction and fully understand how they work. However I was interested in a formal/mathematical way of expressing their definition and was wondering if it is correct!

For regular induction I have in my notes:

  let $p(n)$ be a proposition such that: $p(1)$ holds and for all $n \in N$,  $p(n)$ $\implies p(n+1)$. Then $p(n+1)$ holdes for all $n \in N$.

Is this correct? I understand induction but I feel like this definition is wrong :/

Also if it is correct, how can I build on it to make a definition for Strong Induction?
Thank you!!

Ordinary induction need not start at $1$; it can start at any integer, positive, negative, or $0$. It’s the following principle:

  Let $n_0$ be any integer, and let $P(n)$ be a proposition (about integers) such that $P(n_0)$ is true, and for each $n\ge n_0$, if $P(n)$ holds, then so does $P(n+1)$; then $P(n)$ holds for all integers $n\ge n_0$.

Here your induction hypothesis is $P(n)$, and the induction step consists in proving that $$P(n)\to P(n+1)\;.$$

Strong induction is the following principle:

  Let $n_0$ be any integer, and let $P(n)$ be a proposition (about integers) such that $P(n_0)$ is true, and for each $n\ge n_0$, if $P(k)$ holds for all integers $k$ such that $n_0\le k\le n$, then so does $P(n+1)$; then $P(n)$ holds for all integers $n\ge n_0$.

Here the induction hypothesis is $P(n_0)\land P(n_0+1)\land\ldots\land P(n)$, and the induction step consists in proving that

$$P(n_0)\land P(n_0+1)\land\ldots\land P(n)\to P(n+1)\;.$$

In both cases the conclusion is that $P(n)$ holds for all integers $n\ge n_0$; one cannot conclude anything about the truth or falsity of $P(n)$ for integers $n<n_0$.

The two principles are logically equivalent: each can be proved from the other.

http://mathcircle.berkeley.edu/BMC4/Handouts/induct/node6.html

http://en.wikipedia.org/wiki/Mathematical_induction#Complete_induction

These links should answer your question.


Turning product of multiple variable factors with constants into Big-O I have a function which requires $(m+1)k(n+1)$ operations. How would I describe it in Big-O notation? I guess I need to keep the constants in $(m+1)$ and $(n+1)$ since they ensure that at best the complexity is $O(k)$ which is still not constant.

If you have no information about $k, m, n$, then the best you can do is $O(kmn + km + kn + k)$. I'll argue that you can have a case where each of the terms in this sum could dominate the others:

If $m, n = \frac{1}{\sqrt{k}}$, then you have $O(1 + \sqrt{k} + \sqrt{k} + k) = O(k)$.
If $n = m = k = w(1)$, then it should be $O(kmn)$.
If $m > 1, n = \frac{1}{m}$, then you have $O(k + km + \frac{k}{m} + k) = O(km)$.
You can construct similar to point 3 for the result to be $O(kn)$.


Darboux's sums of $f(x)=\frac{1}{x}$ Consider the function $f(x)=\frac{1}{x}$ on $[1,2]$.

How to find $\epsilon>0$, s.t for all partition $P$ of $[1,2]$ with $\lambda(P)<\epsilon$, we will have that

$$|U_{f,P}-L_{f,P}|<0.01$$

Where $U_{f,P}$ is upper Darboux's sum, and $L_{f,P}$ lower Darboux's sum.

You have that $$f(x)=\frac 1 x$$ is uniformly continuous over $[1,2]$, because it is continuous on $[1,2]$. Thus, given $\delta >0$, pick $\rho>0$ such that whenever $$|x-y|<\rho$$ we have $$|f(x)-f(y)|<\delta$$ for any choice of $x,y$.

This means in particular that if $$\lambda(P)<\rho$$ then $$\sup_{I_k} f-\inf_{I_k} f<\delta$$ over each interval $I_k=[x_{k-1},x_k]$ of the partition. It follows that $$U(f,P)-L(f,P)<\sum_{i=1}^n \delta (x_{i}-x_{i-1})=\delta (2-1)=\delta$$

So, it all reduces to finding this $\rho$. Your function is differentiable over $(1,2)$, and by the mean value theorem, we have that for each $x,y\in(1,2)$ there is a $c$ in between for which $$\left|\frac 1 x -\frac 1 y\right|=\frac{1}{c^2}|x-y|$$ But since $x,y>1$, so is $c>1$, so over $[1,2]$

$$\left|\frac 1 x -\frac 1 y\right|=\frac{1}{c^2}|x-y|<|x-y|$$

Can you finish this off?

The function is decreasing, so the $sup$ and $inf$ over the $[p_i,p_{i+1}]$ are really easy to express.


if $x^2=y^3$, then $x=a^3$ and $y=b^2$ We are given that $x^2 = y^3$ and that $x,y$ are positive integers.
We need to show that there are $a,b$ positive integers such that
$x=a^3$, and $y=b^2$.
I tried to do modular arithmetic in order to solve this problem but I got nowhere. I can't see how one would proceed for this question.
Any help is greatly appreciated

As $x^2=y^3$, $y^3$ is a perfect square. Hence all its prime factors have an even multiplicity. At the same time, these multiplicities are multiple of $3$, hence they are multiple of $6$. Both $x^2$ and $y^3$ are perfect sixth powers, from which the claim follows.

If your question is correct, then:
$$
\begin{align}
x^2 &= y^3 \\
\left(a^3\right)^2 &= \left(b^3\right)^2 \\
a^6 &= b^6 \\
a &= \pm b
\end{align}
$$
Because $a, b \in \mathbb{N}$, then $a = b$.
So in conclusion: $x^2 = y^3$ if and only if $a = b$ with $a, b \in \mathbb{N}$.


Derive an explicit formula for a power series $\sum_{n=1}^\infty n^2x^n$ Could anyone help me find an explicit formula for:

$$
\sum_{n=1}^\infty n^2x^n
$$

We're supposed to use: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}  \qquad  |x| <1 $$

We know that the power series for:

$$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}  \qquad  |x| <1 $$

In order to find the power series for:

$$
\sum_{n=1}^\infty n^2x^n
$$

We differentiate the first summation above:

$$\left(\sum_{n=1}^\infty nx^n\right)' = \sum_{n=1}^\infty n^2x^{n-1} =  -\frac{x+1}{{(x-1)}^3}  \qquad  |x| <1$$

Finally, we multiply by $x$:

$$x\sum_{n=1}^\infty n^2x^{n-1} = \sum_{n=1}^\infty n^2x^{n} = -\frac{x(x+1)}{{(x-1)}^3}$$

Note that if $$s(x)=\sum f(n)x^n$$ then $$xs'(x)=\sum nf(n)x^n$$

Apply with $f(n)=1$, and then with $f(n)=n$. Just take care with the lower limit and the early terms.


expected value of a $e^{-2|x|}$ Let $f(x)=e^{-2|x|}$

Find:

$E(X)$
$E(|X|)$
$E(x')$ where $x'$ denotes the largest integer not greater than $x$.

I'm stuck on this question and am confused about how to use the modulus sign. I feel I need to show it separately for cases when $x\ge0$ and $x\le0$ but I'm not sure.

Any help is appreciated

By definition,
$$E(X)=\int_{-\infty}^\infty xe^{-2|x|}\,dx.\tag{1}$$

To evaluate this integral is easy. First, a technical thin, the integral does converge. The function $xe^{-2|x|}$ is an odd function, so the integral is $0$.

We have
$$E(|X|)=\int_{-\infty}^\infty |x|e^{-2|x|}\,dx.\tag{2}$$
Now the integrand is symmetric about the $y$-axis, so we find the integral from $0$ to $\infty$, and double.

To find the integral from $0$ to $\infty$, we need
$$\int_{-\infty}^\infty xe^{-2x}\,dx.$$
We can use integration by parts, or recall that the exponential with parameter $2$ has mean $\frac{1}{2}$. That tells us that $\int_0^\infty (x)2e^{-2x}=\frac{1}{2}$.

The greatest integer one takes a bit more woek to analyze. The integral breaks up into parts. From $0$ to $1$ we are integrating $0\cdot e^{-2x}$. From $1$ to $2$ we are integrating $1\cdot e^{-2x}$, and so on.

So we get
$$(1)\frac{1}{2}(e^{-2}-e^{-4})+(2)\frac{1}{2}(e^{-4}-e^{-6})+(3)\frac{1}{2}(e^{-6}-e^{-8})+\cdots.$$
This is a geometric series, not hard to evaluate.

However, we can be tricky, and do the negative part, hoping for cancellation. We get
$$(-1)\frac{1}{2}(e^{0}-e^{2})+(-2)\frac{1}{2}(e^{-2}-e^{-4})+(-3)\frac{1}{2}(e^{-4}-e^{-6})+\cdots.$$
Add the two, simplify.

Hint:

  $$\operatorname{E}[X] = \int_{-\infty}^\infty x f(x)\, \mathrm{d}x,  $$

and more generally

  $$\operatorname{E}[g(X)] = \int_{-\infty}^\infty g(x) f(x)\, \mathrm{d}x. $$

To deal with integrals contain the floor function $\lfloor x\rfloor$ see here.


Separate zeros of a smooth function This question occurred to me while working on paper. I would like to separate a zero from a smooth function similar to what you can do with polynomials. So here is my full question:

If I am given a smooth function $f: \mathbb{R}^n \to \mathbb{R} $  which vanishes in $0$ and in some further derivatives, i.e.
$$ f(0) =0 \text{ and  }  \frac{\partial^\alpha f}{ \partial x^\alpha } (0) =0,$$
Can I conclude that there is a multiindex $\tilde\alpha \neq \alpha $ such that I can write $f$ in terms of
$$ f = x^\tilde\alpha  h $$ for a smooth function $h : \mathbb{R}^n \to \mathbb{R}$?

Can anyone explain why this is possible or why not?

No you cannot conclude that. A smooth function may vanish to infinite order at a single point, which means that all the derivatives can vanish at once, without the function itself being identically zero. The classic example is $\exp(-1/x^2)$ for positive $x$ and 0 for non-positive $x$.


What do I miss on this function? $f(t) = (t-1)^{s/2}-t^{s/2}+1$ I would like to know the sign  the function $f: [1,\infty)\to \Bbb R$ given by
$$f(t) = (t-1)^{s/2}-t^{s/2}+1$$ with $0<s<1.$ Differentiating I have
$$f(t) = \frac{s}{2}\left[(t-1)^{s/2-1}-t^{s/2-1}\right]\ge 0$$
indeed, $0<t-1<t$ and $-1<\frac{s}{2}-1<-\frac{1}{2}<0.$

Therefore the function $f$ is increasing and we have $ f(1) =0$

Hence I could conclude that $$ f(t)\ge 0~~~~t\ge 1. $$

  Problem:
  Patently in the following I found a contradictory argument. Since
  $$ \lim_{t\to \infty} f(t)=\lim_{t\to \infty}[(t-1)^{s/2}-t^{s/2}+1]= \lim_{t\to 0^+}\frac{(1-t)^{s/2}-1}{t^{s/2}}+1\\\sim \lim_{t\to 0^+}\frac{-{s/2}}{t^{s/2-1}}+1 = -\infty$$

The problem is that, according to the above result, this limit  of $f$ at infinity  limit should instead  $\lim_{t\to \infty} f(t)=\infty$. Rather this give a contradiction.

Which of the following  is false ? $ f(t)\ge 0~~~~t\ge 1. $ or $\lim_{t\to \infty} f(t)=-\infty$. and why ? what did I miss.

Patently I am facing the same paradox with the function $g(x) = (x+1)^{s/2}-x^{s/2}-1~~~x>0. $ where I also got $ g(x)\le 0~~~~x\le 0. $ and  $\lim_{x\to \infty} g(x)=\infty$.

You found the result of the limit of $f(x)$ wrong.

In infinity, we have:
$\lim_{t\to \infty} (t-1)^{c}\sim\lim_{t\to \infty} t^{c}$

So for the limit we can simplify like below:
$\lim_{t\to \infty} f(t)=\lim_{t\to \infty}[(t-1)^{s/2}-t^{s/2}+1]= \lim_{t\to \infty}[t^{s/2}-t^{s/2}+1]=1 >0$


Whether the function is surjective or not? How to check whether a function is surjective or not?
  function is $$f(x)=(x^{2}+1)^{35}$$ where $x\in \mathbb{R}$
   How to write $F(x)$ in terms of $y$?

If you consider it as a function $f:\Bbb R\to\Bbb R$, or even $f:\Bbb R\to[0,\infty)$, then it is not surjective. Indeed, $(x^2+1)^{35}\geq (0+1)^{35}=1^{35}=1$, so the range of $f$ is contained in $[1,\infty)$.

Actually, the range is precisely $[1,\infty)$, so the function
$$f:\Bbb R\to [1,\infty),\quad f(x)=(x^2+1)^{35}$$
is surjective.

For this particular case, you can exhibit numbers which are not included in the image of $f$, 0 for instance :

$\forall x\in \mathbb R, (x^2+1)^{35}\geqslant 1$

so any $x<1$ is not attained by f. Hence f is not surjective.


Find number of ways to seat $n$ boys and $n$ girls in a row so that every boy has at least one girl sitting beside him. My attempt: I am getting $2^n(n!)^2$ .

First I paired $n$ boys and $n$ girls in $n!$ ways then these pairs can be arranged in $n!$ ways and in each of these pairs boy and girl can arrange themselves in $2!$ ways.

Begin with seating $n$ indistinguishable girls leaving ample space between and around  them, like so:
$$\ \underline{\ \  }\ G\ \underline{\ \ }\ G\ \underline{\ \ }\ G\ \underline{\ \ }\ \ldots\ \underline{\ \ }\ G\ \underline{\ \ }\ G\ \underline{\ \ }\quad .$$
An admissible seating pattern can then be constructed as follows:

Choose an $r$ with $0\leq2r\leq n$.
Choose $r$ of the $n-1$ inner spaces between the $G$s, and write $B^2$ there. This can be done in ${n-1\choose r}$ ways.
Choose $n-2r$ of the remaining $n+1-r$ spaces, and write a single $B$ there. This can be done in ${n+1-r\choose n-2r}$ ways.

The total number $P(n)$ of admissible seating patterns therefore comes to
$$P(n)=\sum_{r=0}^{\lfloor n/2\rfloor}{n-1\choose r}\>{n+1-r\choose n-2r}\qquad(n\geq1)\ ,$$
producing the sequence $(2,4,10,26,70,192,\ldots)$, as in Markus Scheuer's answer. Making girls as well as boys distinguishable then gives the number
$$N=(n!)^2 P(n)$$
of personalized seatings.

Start with two lines, one of boys and one of girls, and imagine constructing the arrangement by picking the first person in one of the lines. You get a factor of 2 every time the choice can be from either line.

This construction is useful because no seating arrangement is double counted, which can be seen by noting that you can retrieve the order the boys (or girls) were in in their monogender lines by skipping the people of another gender.


For function satisfying given conditions prove that $f(x). f(\frac{1}{x})=1$ If $f(x \times f(y)) =\frac{f(x)}{y}$ for every $x,y \in R$ and $y \neq0$, then prove that $f(x)\times f(\frac{1}{x})=1$

Is it possible to find the function $f(x)$ here using the given data? I tried it by using $x=0$ and $y=k$ ($k$ is any non-zero number), to get $f(0)$ then tried using partial differentiation with respect to x but the $f(y)$ is creating problem. Could someone please help me with it?

Write $a:=f(1)$, assumed non-zero for non-triviality; then putting $y=1$ gives $f(ax)=f(x)$, from which follows that $f(a)=a$. Putting $y=a$ gives $f(xa)=f(x)/a$, so $f(a)=1$. Hence $a=1$, that is, $f(1)=1$.

Putting $x=1$ gives $f\circ f(y)=1/y$. Thus $f(1/x)=f\circ f\circ f(x)=1/f(x)$, which is the required relation.


Separation of variables and well posedeness Let $k$ be a real constant. Prove that the equation
$$
u_t = u_{xx} +ku_x
$$
for $x\in [0,1]$ and $t\geq 0$ given with zero dirichlet boundary condition is well posed. The hint is to use separation of variables and Fourier expansion to show the norm of the solution is bounded in finite time horizon.

I have only seen separation of variables for the wave equation but this seems more complicated to me because of the ODE found for $X$, namely
$$
X'' +kX'+\lambda X =0
$$
which I cannot solve easily given I do not know the discriminant's sign of the characteristic polynomial.
Any help or reference?

EDIT: I add the definition of well posedeness. Consider for all $t \geq 0$ the operator $E(t)$ that sends the initial condition $u_0$ to the function $x \rightarrow u(t,x)$. Then we say the PDE is well posed if for all $T>0$ there exists a constant $C_T$ such that $||E(t)|| \leq C_T $ for $0\leq t \leq T.$

We are working on the function space $L^2$ and the norm of an operator is defined as usual by $||E(t)||= \sup_{u\in L^2, u\neq 0} \frac{||E(t)(u)||}{||u||}$

In this particular setting we want to show that if we fix $T>0$ then there is a constant $C_T$ such that $||u(t,\cdot)|| \leq C_T||u_0||$ for $0 \leq t \leq T$. where $u_0$ is the intial condition.


How to rotate any graph by any angle about any point? How can I do exactly as the title asks? I'm tearing my hair out trying to figure this out. I watched this video, and I now understand how to rotate about the origin, but I tried using the same steps to find the equation for any point, using h and k to represent the coordinates of the "new origin", and I got lost. I got to this point, and have no idea how to simplify, or if its even possible at all.

Any help at all will be greatly appreciated. Thank you.

To use this method one should convert $y=F(x)$ form into two separate parametric equations of form $x=f(t), y = g(t) $. For example
$$ y^2 = 4 f x ; \rightarrow x= ft^2, y=2 ft $$

Next, a translation and a rotation  through angle $\theta$ can be done as you know$ (x,y)$ shifts $(h,k)$ between required points.

$$ x_1= x \cos \theta - y \sin \theta+ h\, ; y_1= x \sin \theta + y \cos \theta + k. $$

Next, you can see yourself if eliminating $\theta$,( which can be done) is desirable or worthwhile.

Do you have a function you're working with? i.e $f(x)$ = something?

If so, you can translate the point you're rotating about to the origin, rotate and then translate back.

To translate vertically by $\pm k$, simply look at $f(x) \pm k$. To translate horizontally by $\pm h$, simply look at $f(x \mp h)$

As an example of translating, say we have the function $f(x) = (x-2)^2 + 4$ and we want to translate it so the point $(2, 4)$ coincides with the origin. Using the notation in the previous paragraph, $k = -4$ and $h = -2$. To take care of the vertical translation we look at $g(x) = f(x) - 4 = (x-2)^2$. To take care of the horizontal translation we look at $h(x) = g(x+2) = x^2$. Now our function has been translated to the origin. You can rotate $h(x)$ and then translate its rotated version vertically by $+4$ and horizontally by $+2$


Second Covariant Derivative Leibniz Rule I’ve been trying to understand why

  $$\nabla_{u,v}^2 w := (\nabla(\nabla w) )(u, v) = \nabla_u{(\nabla_v w)}-\nabla_{\nabla_u v}w$$

Should be true. Is there some easy way to prove this identity? I'd like to see a proof that doesn't expand out the product in terms of it's coordinates if possible as well. Thanks in advanced!

Remember that we define the covariant derivative of arbitrary tensors so that it satisfies the Leibniz rule and commutes with contractions.

The term $\nabla W(V)$ is the contraction of $\nabla W\otimes V.$ The Leibniz rule tells us that
$$\nabla_U (\nabla W \otimes V) = (\nabla_U(\nabla W)) \otimes V +\nabla W\otimes\nabla_UV.$$

Contracting this equation yields

$$\nabla_U (\nabla W (V)) = (\nabla_U(\nabla W))(V) +(\nabla W)(\nabla_UV).$$

Recalling the definition $(\nabla W)(X) = \nabla_X W,$ we can rewrite this as the desired

$$ \nabla_U(\nabla_V W) = (\nabla (\nabla W))(U,V)+ \nabla_{\nabla_U V} W.$$

Use the semilatus theorem that states $\delta^i_j=\delta^j_i$ if and only if $i=j$, followed by $$\Gamma_i\oplus \Gamma_j=\sum_{i=1}^{k}\hat{\Gamma}_i$$ where $\hat{\Gamma}$ is a half-metric tensor under a morphic dilational phase. If we use this on gradient fields, we have $$\nabla \Gamma_i\oplus \Gamma_j=\sum_{i=1}^{k}\nabla_i \hat{\Gamma}_i$$
Hence, use the same for your argument. If you combine each psuedoterm correctly, you'll get $$\nabla_{u,v}^2 w := (\nabla(\nabla w) )(u, v) = \nabla_u{(\nabla_v w)}-\nabla_{\nabla_u v}w$$ as you wanted to prove!

That being said, there are alternative methods. This is the method I learnt when I did tensor field theory. Hope this helps!


Annihilators of elements of a finitely generated faithful module over a noetherian reduced ring Lately I've been thinking to annihilator of modules and I've conjectured  a proposition I can't prove, so I'll expose my claim.

  Let $A$ be a noetherian reduced (commutative) ring and let $M$ be a faithful finitely generated module on $A$. Is it true that there exists an element $m \in M$ such that $\operatorname{Ann}(m)=0$?

Let's recall that $\operatorname{Ann}_{S^{-1}R}(S^{-1}M)=S^{-1}\operatorname{Ann}_R(M)$ for any finitely generated $R$-module $M$ and every multiplicative set $S\subset R$.

Now let $S$ be the set of all non-zerodivisors in $R$, and consider the $S^{-1}R$-module $S^{-1}M$. This is finitely generated and faithful. It is well known that $S^{-1}R$, the total ring of fractions of $R$, is isomorphic to a finite direct product of fields. Set $S^{-1}R\simeq K_1\times\cdots\times K_n$ where $K_i$ are fields. Then $S^{-1}M$ decomposes as $N_1\times\dots\times N_n$ with $N_i$ a $K_i$-module. If there is $N_i=0$ then $\operatorname{Ann}_{S^{-1}R}(S^{-1}M)\ne0$, a contradiction. Thus for each $i=1,\dots,n$ we can pick $x_i\in N_i$, $x_i\ne 0$. Setting $x=(x_1,\dots,x_n)$ we get $\operatorname{Ann}(x)=0$.
In terms of rings and modules of fractions we have found a non-zero element $\frac ms\in S^{-1}M$ with $\operatorname{Ann}_{S^{-1}R}(\frac ms)=0$. If $a\in R$ is such that $am=0$, then $\frac a1\frac ms=\frac 01$, and therefore $\frac a1=\frac 01$ hence there is $t\in S$ such that $ta=0\Rightarrow a=0$. This gives us $\operatorname{Ann}(m)=0$, and we are done.

(since $A$ is reduced we can) Assume that $0= p_1\cap ...\cap p_n$. Since $M_{p_i}\neq 0$ there exists $m_i$ such that $Ann\ m_i\subset p_i$. Now $Ann\ (m_1+...+ m_n)\subset  p_1\cap ...\cap p_n=0$


Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$ Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$

I tried assuming it was equal to $\frac p q $.

I get $2^nq^2-q^2 = p^2 $

But I don't see where to go from there.

For $n\geq2$, $2^n-1\equiv 3\pmod 4$, unlike a perfect square.

It is clearly equivalent to prove that the diophantine equation $$2^n=x^2+1$$ has no integer solution for $n\gt 1$ Notice that $x$ must be odd so $x=2m+1$ hence
$$2^n=(4m^2+4m+1)+1=4m^2+4m+2\iff2^{n-1}=2m^2+2m+1$$ This is absurde.


Regex for strings with no three identical consecutive characters I wanna ask what the regular expression for the strings having the property in the title should be. For binary string with no three consecutive 0, it's quite a simple regex (1|01|001)*|(0|00|$\epsilon$) but I found its very difficult for a normal alphabet string. Do you have any suggestion/solution for that?

Note: Those strings don't contain any three identical consecutive characters. For example, they do not have $aaa$, $bbb$, $ccc$, etc. in their content.

Thanks,

It isn't easy to product a complement of regular expression. Very long and exhausting calculations :(

For alphabet $\{a,b\}$ (and if I didn't miscalculate) it is something like

  ((a|ba|bba)(aba|abba|ba|bba)*(abb|ab|bb|a|b|$\varepsilon$))|bb|b|$\varepsilon$

which is ... bulky.

You need to convert your regex to DFA, then turn all non-terminal state into terminal states and vice versa, then convert the resulting DFA to regex back.

And here there is a well-done example of it.

\b([a-zA-Z0-9])\1\1+\b
checks any character that appears more than 3 times


connectedness of uniform topology This question can also be found in this link, but because of dull brain I could not understood that One line proof.

Let $\mathbb{R}^{\omega}$ be the set of all (infinite) sequences of real numbers. Determine whether $\mathbb{R}^{\omega}$ is connected or not in the uniform topology.
Basis elements of uniform topology are $B_{\rho}(x,\epsilon)=\{ y\mid \rho(x,y)<\epsilon \}$, where $\rho(x,y)=\sup\{ \min(d(x_\alpha,y_\alpha),1)\mid \alpha \in \mathbb{Z} \}$

From the answer in the link, I understood that it is not connected.
What I found:
Note that $B_{\rho}(x,1)=\overline{B_{\rho}(x,1)}$. So I need to find $x_1,x_2\in \mathbb{R}^{\omega}$ such that $B_{\rho}(x_1,1)\cap B_{\rho}(x_2,1)=\phi$
How to find such $x_1$ and $x_2$?

So we consider on $\mathbb{R}^\omega$ what Munkres calls the uniform topology, i.e. the topology induced by the metric
 $$d(x,y) = d((x_n)_n,(y_n)_n) = \sup_n \{\min(|x_n - y_n|,1): n \in \mathbb{N}\}$$

Define the following relation $\sim$ on $X$: $x \sim y$ iff the sequence $(x_n - y_n)_n$ is bounded in $\mathbb{R}$.

This is an equivalence relation on $X$:

$x \sim x$ because the $0$-sequence is bounded.
$x \sim y$ implies $y \sim x$ because $(x_n -y_n)_n$ is bounded iff
its negative $(y_n - x_n)_n$ is.
$x \sim y$ and $y \sim z$ imply $x \sim z$ because if the sequences
$(x_n - y_n)_n$ and $(y_n - z_n)_n$ are bounded then their sum
$(x_n -z_n)_n$ is bounded too.

So, we have a partition into equivalence classes.
Denote the equivalence class of $x=(x_n)\in \mathbb{R}^\omega$ by $[x]$.
Also note that all classes are mutually homeomorphic:

If we have a fixed $x \in \mathbb{R}^\omega$, then the map $f_x$ that sends $(y_n) \in [x]$ to
$(y_n - x_n)$ in $[0]$ (the class of the $0$-sequence, i.e, all
bounded sequences in $X$) is a homeomorphism.
($X$ is a topological group under $+$, or check continuity
with the definition; it's even an isometry.)

Also, every set $[x]$ is open in $X$. For, let $y \in [x]$.
Then the open ball $B_d(y, {1 \over 2}) \subset [x]$: if $z  \in B_d(y,{1 \over 2})$ this implies that for
all $n$,  $|y_n - z_n| < {1 \over 2}$, so $y_n - z_n$ is bounded, so $z$ is in $[y]=[x]$.

So we have a partition of open sets, so they are all closed as well, as the complement of a class is
the union of the other classes, hence also open.

Note also that $[0]$ is a topological vector space, it is homeomorphic to
$\ell_{\infty}$; or see in another way that $[0]$ (and hence all classes) are path-connected:

Every $x$ in $[0]$ is connected by a path $F_x:[0,1] \to [0]$ from $0$ to $x$:
$F(t) = ((t*x_n)$, for $t \in [0,1]$. This is continuous on $[0]$
because there $d(F_x(t_1),F_x(t_2)) = \sup_n \{\min(1, |t_1-t_2|*|x_n|)\}$;
now let $M$ be a bound on all the $|x_n|$ (by boundedness of $x$), and let $|t_1-t_2| < {1 \over M}$. Then:
$d(F_x(t_1),F_x(t_2)) < M\cdot |t_1-t_2|$, which shows directly that $F_x$ is continuous.

So every point in $[0]$ is connected by a path to $0$, hence $[0]$ is path-connected,
and as said, so are all other classes as well. Because the classes are clopen, there can be
no larger path-connected or connected set that has points in more than one class.
So sets of the form $[x]$ are maximal (path-)connected subsets.
So the $[x]$ ,$x \in X$ form the (path-)components of $X$.

As to your last part: take one sequence $x_n=0$, and the other $x_n =n$. Their balls of radius $\frac{1}{2}$ or $1$ do not intersect. These sequences clearly lie in different (and hence disjoint clopen) classes.

Not connected,  because there is set of bounded sequences that is both open and closed, in the uniform topology.
If $A$ is this set of bounded sequences and $X - A$ be the set of unbounded sequence, then both seem to be open in uniform topology.
So, that makes a separation on  $\mathbb{R}^\omega$ in the uniform topology.


Can I say $\log{O(n)}$? Suppose $e^{f(x)} = O(x)$, or equivalently, there exists a $c$ such that for all $x$, $f(x) \lt \ \log{x} + c$.  Is $f(x) = \log{O(x)}$ generally understood to mean the same thing?

The standard is to write O(log n).

See also: https://stackoverflow.com/questions/2307283/what-does-olog-n-mean-exactly


Proving that the empty set is unique I haven't been able to figure out if the following reasoning is correct, so I'd like to have the opinion of other people on that.

The goal is to prove that the empty set is unique.

In order to do that, let $E$ be a set, $A$ be an empty set and  $B$ be an empty set. I want to prove that $A = B$.

First, I can try proving that $A \subset B$.

I know that $\forall x \in E, x \notin A$. Now I can consider this proposition : $x \in A \Rightarrow x \in B$

Since $x \notin A$, I can safely say that that the proposition above is true.

Hence, I can conclude that $A \subset B$

Am I right in my conclusion?

Your conclusion is correct. You can also resort directly to the Axiom of Extension

  $A=B\iff \forall z\colon(z\in A\leftrightarrow z\in B)$

For the given conditions give us $z\notin A$ and (equivalently) $ z\notin B$ for arbitrary $z$, hence the right hand side in the axiom is true.

Here's an approach that's probably cleaner:

Per Axiom of Existence, there exists a set which has no elements. Let $A$ and $B$ each be a set which has no elements. Since $ x \in A \iff x \in B $ is vacuously true, by Axiom of Extensionality $A=B$.

Equivalently, that $x \in A \iff x \in B $ is true can be established using the truth table of the biconditional, by recognizing that $x \in A$ and $x \in B$ are in each case false (ex falso quodlibet).


What's exactly the deal with differentials? (Confessions of a desperate calculus student) So I don't know if I'm the only one to feel this, but ever since I was introduced to Calculus, I've had a slight (if not to say major) aversion to differentials.

This sort of "phobia" started from the very first moment I delved into integrals. Riemann sums seemed to make sense, though for me they were not enough for justifying the use of "dx" after the integral sign and the function. After all, you could still do without it in practice (what's the need for writing down the base of these rectangles over and over?). I was satisfied by thinking it was something merely symbolic to remind students what they were doing when they calculated definite integrals, and/or to help them remember with respect to what variable they were integrating (kinda like the reason why we sometimes use dy/dx to write a derivative). Or so I thought.

Having now been approached to differential equations, I'm starting to realize I was completely wrong! I find "dy" and "dx" spread out around equations! How could that be possible if they are just a fancy way of transcribing derivatives and integrals? I imagined they had no meaning outside of those particular contexts (i.e.: dy/dx, and to indicate an integration with respect to x or whatever).

Could anybody help me out? I'm really confused at the moment. I'd really appreciate it :)  (P.S.: Sorry to bother you all on Thanksgiving - assuming some of you might be from the US.)

EDIT: I don't think my question is a duplicate of Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?, as that one doesn't address its use in integrals and in differential equations. Regardless of whether dy/dx is a ratio or not; what I'm really asking is why we use dx and dy separately for integration and diff. equations. Even if they're numbers, if they tend to 0, then dx (or dy) * whatever = 0. Am I wrong in thinking that way?

If you read a real analysis textbook such as Calculus by Spivak, they manage to develop calculus rigorously while avoiding differentials like "$dx$" and "$dy$" entirely.  This is the standard way to make calculus rigorous -- you just avoid using differentials.  And indeed, in undergrad differential equations classes, arguments that involve manipulating $dx$ and $dy$ as individual quantities can easily be rephrased to avoid doing this.

For example, if a differential equations textbook says:
\begin{align}
& y \, dy = dx \\
\implies & \int y \, dy = \int \, dx \\
\implies &  \frac{y^2}{2} = x + C \\
\end{align}
we can rephrase this argument as
\begin{align}
& y \frac{dy}{dx} = 1 \\
\implies & \frac{y^2}{2} = x + C,
\end{align}
where in the second step we simply took antiderivatives of both sides, using the chain rule in reverse to find an antiderivative of $y \frac{dy}{dx}$.

But note: even though a rigourous approach might avoid using differentials entirely, there is no need to throw "differential intuition" out the window, because it makes perfect sense if we just think of $dx$ and $dy$ as being extremely tiny but finite numbers, and if we replace $=$ with $\approx$ in the equations we derive.  Perhaps the word "infinitesimal" could be thought of as meaning "so tiny that the errors in our approximations are utterly negligible".  We can plausibly obtain exact equations "in the limit" (if we are careful).
There is something aesthetically appealing about treating $dx$ and $dy$ symmetrically, which can perhaps in some situations give us a feeling that the approach using differentials is the "right" way or more beautiful way to do these computations.  Compare these two ways of writing an "exact" differential equation:

\begin{equation}
I(x,y) \,dx + J(x,y)\, dy = 0
\end{equation}
vs.
\begin{equation}
I(x,y) + J(x,y) \frac{dy}{dx} = 0.
\end{equation}
The first version is aesthetically compelling, because it's more symmetrical; this might help explain why the second version is not seen more often (despite its being easier to understand, in my opinion).

Of course, for any results derived using "differential intuition", we must later find a rigorous proof to confirm there is no mistake.

Note also: There are other approaches to making calculus rigorous (based on nonstandard analysis I think) that actually make infinitesimals rigorous.  So they manage to embrace $dx$ and $dy$ as legitimate quantities, rather than avoiding them.

Additionally, in differential geometry, quantities like $dx$ are defined precisely as "differential forms", and some treatments of calculus (like Hubbard & Hubbard) embrace differential forms at an early stage.  But you can understand calculus rigorously without using differential forms.

If you want to know what a mathematical thing is, you need two things:

How it works - what the rules of using it are.
A way of expressing it in terms of mathematical things you know and
trust.

For differentials, we know what the rules of using them are. And I'm told some clever folks have worked out how to model them (notably Abraham Robinson with non-standard analysis).

There is a whole pyramid of mathematical things that we accept without demur these days that were once as suspect as differentials probably still are. What tends to happen is that we know how the thing works, then we have to find what it is or might be.

If you're used to counting, what's a negative number? ... a rational number?

Our confidence in real numbers is perhaps as naive as that in differentials.


Finding the range of a function without graph? I'm asked to find the range of the function :

$
(-x^2  , +1)
$

Taking R as a real number which represents a quantity along a line and graphing :

then the range for this function from viewing the graph appears to be :

$
R = (-\infty  , +\infty)
$

Is there an alternative method of finding the range of this function without using a graph ?

Watching the khan academy tutorial suggests using graphs : https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/domain-and-range-from-graphs but is there a pure algebraic method instead of using graphs ?

It is not clear what the range is when the range appears infinite as how do we know that at some point on the axis the range functions stops tending towards infinity ?

Notice that $x^2 \ge 0$ for all $x \in \Bbb R$, which means that $1 - x^2 \le 1$ for all $x \in \Bbb R$, with equality for $x=0$. Notice, too, that $x^2$ increases towards $\infty$, therefore $1-x^2$ will decrease towards $-\infty$. Finally, notice that $x^2$ is surjective (for every $y \ge 0$, there exist $\sqrt y$). Therefore, the values of $f$ lay between $- \infty$ and $1$, meaning that the range is $(-\infty, 1]$.

You see that if we put x as infinity we get f(x)=-∞.
Now to find the maximum value of functions just differentiate it and put the derivative equal to zero to find the point where the function stops increasing and starts decreasing.
$\frac{d}{dx}(-x^2 + 1) = -2x$
$-2x=0 =>$ f(x) is maximum when x=0
Therefore maximum value of f(x) is +1
$f(x)\epsilon (-∞,+1) $ here is the max value on graph


$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$ for $a,b,c >0$ and $a+b+c=3$ Let $a,b,c >0$ with $a+b+c=3$. Prove that
$$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52.$$

What I did:
It is cyclic inequality so I assume $c= \min\{ a,b,c \}$.
I consider the first case where $a\ge b\ge c$ then
$$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$$
$$\Leftrightarrow \frac{a^a}{a^b} +\frac{b^b}{b^c}+\frac{c^a}{c^c}> \frac52$$
I check function $f(x) =x^x$ to see if it is a strictly monotonic function or not. It turns out that it is a concave up function so I get stuck here.

Not a bounty candidate. Just a pictorial comment.Make an isoline/contour plot in the $(a,b)$-plane of the function:
$$
f(a,b) = a^{|b-a|}+b^{|c-b|}+c^{|a-c|} - \frac52 \quad \mbox{with} \quad c=3-a-b
$$
Then this is what we get. The blue spots are where $\,|f(a,b)| < 0.02$ . There seem to be several of these minimum values. I wish the rigorous proof producers among us good luck.

Perhaps it would help to take $a=1+x$, $b=1+y$, $c=1+z$, so that $x+y+z=0$

Now the inequality looks like:

$$(1+x)^{|y-x|} + (1+y)^{|z-y|} + (1+z)^{|x-z|} \geq \frac{5}{2}$$

Using bernoulli's inequality, the expression is more or less

$$1 + x|y-x| + 1 + y|z-y| + 1 + z|x-z| \geq \frac{5}{2}$$
$$x|y-x| + y|z-y| + z|x-z| \geq \frac{-1}{2}$$

And we can split it in two cases depending on how x, y and z are sorted.

For example, if $x<y<z$, we must prove that:
$$x(y-x) + y(z-y) + z(z - x) = -x^2 - y^2 + z^2 + xy + yz - xz \geq \frac{-1}{2}$$

We recall that $x+y+z=0$, and thus $z = -x-y$:

$$-x^2 - y^2 + z^2 + xy + yz - xz = -x^2 - y^2 + x^2 + y^2 - 2xy + xy - (y+x)(y-x) = x^2 - y^2 - xy$$

It should not be hard to check if this is more that $\frac{-1}{2}$ given our restraints.


Proving the formulas for $\sin 3\theta$ and $\cos 3\theta$. (Check my work.) Substitute in the equation $3\theta=(2\theta+\theta)$, prove that:
a. $\sin 3\theta = 3\sin\theta-4\sin^3\theta$
b. $\cos 3\theta = 4\cos^3\theta-3\cos\theta$

Is this the answer to a)?
$$\sin 2x = 2\sin x\cos x,\quad\cos 2x = 1-2\sin^2 x$$
\begin{align*}
\sin 3x & = \sin(2x+x)\\
& = \sin 2x \cos x + \cos 2x \sin x\\
& = 2\sin x \cos^2x + (1-2\sin^2 x)\sin x\\
& = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x\\
& = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x\\
& = 3\sin x - 4\sin^3 x
\end{align*}
b)
\begin{align*}
\cos 3x & = \cos(2x)\cos(x) - \sin(2x)\sin x\\
        & = (2\cos^2x - 1) \cos x - (2\sin x \cos x) \sin x\\
        & = 2\cos^3x - \cos x - 2\sin x \cos x\\
        & = 2 \cos^3 x - \cos x - 2(1 - \cos^2x) \cos x\\
        & = 2\cos^3x - \cos x- (2\cos x - 2\cos^3x)\\
        & = 4\cos^3x - 3\cos x
\end{align*}
Is b) also correct?

Additional material:

Using complex numbers,

$$\cos3\theta+i\sin3\theta=e^{i3\theta}=(e^{i\theta})^3=(\cos\theta+i\sin\theta)^3.$$

Then by the binomial formula, and with abbreviations,

$$c^2+3ic^2s-3cs^2-is^3.$$

From this

$$\cos3\theta=c^2-3cs^2=4c^3-3c$$

and

$$\sin3\theta=3c^2s-s^3=3s-4s^3.$$


I'm stuck with this summation problem How can I show that

  $$\displaystyle\sum_{n=1}^\infty \left(-\frac{1}{2^n}\left(\bigg| {\rm sgn} \left( {\rm round}\ (n \cdot \frac{1}{3})-n \cdot \frac{1}{3}\right)\right)-1\bigg|\right) = \frac{1}{7}?$$

Hint:  Unpack it.  The result of the sgn function depends only on $n \pmod 3$.  Take three terms of the sum and evaluate it.  You will get a geometric sum.  You might notice  that $\frac 17=\frac 1{8(1-\frac 18)}$

Looking at $\bigg| {\rm sgn}\ ( {\rm round}\ (n \cdot \frac{1}{3})-n \cdot \frac{1}{3})\bigg|$, if n is divisible by 3 then the result is 0, otherwise it's 1. Then after subtracting 1 if n is divisible by 3 it's -1, otherwise it's 0.
Then the whole equation becomes an infinite sum of $\frac{1}{8} + \frac{1}{64} + ... = \frac{1}{7}$.


"Simple" beautiful math proof Is there a "simple" mathematical proof that is fully understandable by a 1st year university student that impressed you because it is beautiful?

Here's a cute and lovely theorem.

  There exist two irrational numbers $x,y$ such that $x^y$ is rational.

Proof. If $x=y=\sqrt2$ is an example, then we are done; otherwise $\sqrt2^{\sqrt2}$ is irrational, in which case taking $x=\sqrt2^{\sqrt2}$ and $y=\sqrt2$ gives us: $$\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt2^{\sqrt2\sqrt2}=\sqrt2^2=2.\qquad\square$$

(Nowadays, using the Gelfond–Schneider theorem we know that $\sqrt2^{\sqrt2}$ is irrational, and in fact transcendental. But the above proof, of course, doesn't care for that.)

I like the proof that not every real number can be written in the form $a e + b \pi$ for some integers $a$ and $b$.  I know it's almost trivial in one way; but in another way it is kind of deep.


Proving equivalence of two definitions of Lebesgue measure. I got two definitions of Lebesgue measure of open sets. One is $$m_1(A)=\inf\Big\{\sum^{\infty}_{k=1}l(I_k):A\subset \bigcup_k I_k\Big\}$$ and the other is $$m_2(A)=\sup\Big\{\sum^{N}_{j=1} l(I_k):  \bigcup_k I_k \subset A \text{ & } I_i^\circ \cap I_j^\circ= \emptyset \Big\}$$ where $I_k$ is rectangle in $\mathbb{R}^n$.

How can we show these two difinitions are equivalent? I want to prove this by writing $A$ as the union of countable rectangles with their interiors pairwise disjoint. But I'm not sure whether we can do this.

Any hints? Thanks in advance!

I believe it is possible, indeed here is my argument.

Let $A \in \mathbb{R}^n$ be open.

Since A is open, for every $x \in A$ there is a ball $B_x$ with center $x$ that is inside $A$. Since the rationals $\mathbb{Q}^n$ are dense in $\mathbb{R}^n$, I can choose a (n-dimensional) rational $q_x$ and a radius $r_x$ such that the (n-dimensional) square $R_x$ with side $r_x$ and center $q_x$ is inside $B_x$ (and thus inside A) and $x \in R_x$.

So by choosing for each $x$ one such rectangle $R_x$ I can define a function $F: x \mapsto R_x$ from $A$ to the set $S$ of all squares of rational center and rational radius. This set has a trivial bijection with $\mathbb{Q}^{n+1}$, so it is countable. $R := F(A)$ inside S is also countable, so let $R_n$ be an enumeration of this set.

It is obvious that

$$
\bigcup_{n=0}^{\infty}{R_n} = A
$$

since every $R_n$ is inside A and all points of A are in some $R_n$. So we just wrote A as a countable union of rectangles.

We can now make this union with emtpy intersection of the interiors if we just define

$$
B_n' := R_n \setminus \bigcup_{k=0}^{n-1}{R_k}
$$

and $B_n = \bar{B_n'}$. This works since the set that results from taking a finite number of rectangles from a rectangle can be written as a union of rectangles. We take care of boundaries by taking the closure.


Does a group $G$ with an element of finite order $> 1$ and two conjugacy classes necessarily have $|G|=2$? Very close to this question but instead of specifying finite $G$, we are given a group element with finite order $>1$.

First, $e$ is its own conjugacy class. Let the element $a \neq e$
 have order $n$. All elements of $G$ other than $e$ belong to $C(a)$.

Since $o(a)=n$, we can prove that all the other elements in $C(a)$ also have order $n$. Moreover, if $n$ was composite $n=mk$ then $o(a^k)$ would divide $m$ which is a contradiction. Therefore $n$ is prime.

I am stuck at this point. If $a^2\neq e$, $ a^2= xax^{-1}$. This should lead to a contradiction but I am unable to arrive at it. Any hints would be helpful.

Here is a way to complete the proof:

Define $I_g(x) = gxg^{-1}$ (conjugation map).

Since $a^2$ and $a$ are in the same conjugation class, $\exists g\in G$ such that
$$ I_g(a) = gag^{-1}=a^2$$
$$ I_{g^2}(a)= g(gag^{-1})g^{-1} = ga^2g^{-1} =gaag^{-1}=gag^{-1}gag^{-1} = a^4$$ and so on:

$$ I_{g^n}(a) = g^nag^{-n} = a^{2^n}$$

Since $g^n = g^{-n} = e,$
$$ a^{2^n}=a$$
$$ \implies 2^n = 1 \mod{n}$$
But using Fermat's little theorem, for prime n:
$$ 2^n = 2 \mod{n} $$
Which leads to a contradiction. Thus $a^2=e$. This means $n=2$ and all elements are self-inverses. Thus for any two $a,b \in G$:
$$(ab)^{-1}=b^{-1}a^{-1}=ab $$
$$ab=ba$$
G is abelian. This means $C(a) = \{a\} $ and hence $|G|=2$.

$G$ is the union of $C(a) $ and the neutral element. The fact the order if $a$  is odd implies that the order of the the group is even. By Lagrange you have an element of order 2. Contradiction.


Does the formula $\sqrt{ 1 + 24n }$ always yield prime? I did some experiments, using C++, investigating the values of $\sqrt{1+24n}$.

 n: 1 -> 5
 n: 2 -> 7
 n: 5 -> 11
 n: 7 -> 13
 n: 12 -> 17
 n: 15 -> 19
 n: 22 -> 23
 n: 35 -> 29
 n: 40 -> 31
 n: 57 -> 37
 n: 70 -> 41
 n: 77 -> 43
 n: 92 -> 47

I wonder,  if
$$\sqrt{1+24n}$$
is an integer, will it also be a prime?

Is there any interesting theory about this formula?

Thanks,
Chan

How about $n=26$?

In general, take a composite number of the form $12k+1$ and take $n = k + 6k^2$ to arrive at a contradiction for your statement.

For instance,

$k=2 \Rightarrow n=26 \Rightarrow \sqrt{1+24n} = 25$

$k=4 \Rightarrow n=100 \Rightarrow \sqrt{1+24n} = 49$

$k=7 \Rightarrow n=301 \Rightarrow \sqrt{1+24n} = 85$

and so on.

There are infinite composite numbers of the form $(12k+1)$ which gives infinite counterexamples to your claim.

Your observation though is a nice one, since $24 | (p^2-1)$, $\forall \text{ primes } p > 3$. So you will find that all the primes $>3$ can be written as $\sqrt{1+24n}$.

Right ! I've proven recently this formula which at least englobes all prime numbers for sure. until now I didn't find a prime number that is not written as √24n+1.


Is $\mathbb{N} = \mathbb{Z}^+$? Because $\{1,2,3,4,\ldots\}$ contains all natural numbers, which are also all positive integers.

Depends. In a lot of modern mathematics, $\mathbb N=\{0,1,2,\dots\}$. In particular, since you've tagged the question "set theory," set theorists almost always include $0$ in $\mathbb N$.

If $\mathbb N$ does not include $0$, you are correct.

Often, mathematicians choose to write $\mathbb Z^+$ simply to avoid ambiguity.

There is an infuriating practice in some linear programming to include $0$ in $\mathbb Z^+$, and using $\mathbb Z^{++}$ to mean the "really positive integers."

In other words, it depends on your definitions. $\mathbb N$ and $\mathbb Z^+$ are both just notation. What they mean varies from source to source.

There's one last pedantic wrinkle, actually. Usually, the integers are defined in terms of ordered pairs of natural numbers. In that sense, it is not even true that $\mathbb N\subset \mathbb Z$. Rather, there is an embedding of $\mathbb N$ into $\mathbb Z$. We usually skip this sort of pedantry, and when we are talking about integers, we treat the symbol $\mathbb N$ to be the image of this embedding of the natural numbers into the integers.

In some reference it is true,or $\Bbb{Z}^+=\Bbb{N}\cup\{0\}$


Attitude towards exercises in mathematics I'm doing self study on a couple of topics in mathematics, such as real analysis, abstract algebra, and linear algebra. From time to time, there are always a couple of exercises which I found too difficult to solve. I spend quite some time to think about them. When I fail, I google to find the solutions. Most of the time, I get the solutions.

However, there are some downsides to this attitude. First, I would spend too much time on a single problem. So I feel that my progress is a little bit too slow. Other than that, when I read a solution, I don't get the real understanding of the problem. When I read a proof, my brain is working mechanically to check every statement, so I don't know what exactly is going on.

I start to wonder whether I'm doing this correctly. What I want to ask here is, what should I do when I encounter difficult exercises? Should I think about them myself until I get the answers? Or should I skip the difficult ones and move on to the next chapters, then go back after I gain more understanding?

It depends on the textbook. However, most authors like to include a few exercises that they feel will challenge even very good students (and there is at least one respected author who deliberately includes open research problems among his exercises), so it's not necessarily grounds for panic if you fail to solve some of the exercises.

It's always a good idea to at least try all of the exercises. Some authors use the exercises to present useful facts (typically "prove that such-and-such", and sometimes entire well-known concepts that are peripheral to the main development are defined only in an exercise); trying to solve it will help you remember the result later even if you didn't succeed in proving it.

If you get stuck, however, just leave it be and carry on -- unless it's a majority of the exercises for a chapter or section that fail you. Return to them later, and if they still seem interesting and hard even then, you can consider asking for help here.

Mathematical concepts or theorems is like a tool, it will be awkward using it first but with time & practice, the better you become!

I was a 4th year math & stats when NCSA Mosiac, Gopher and Kermit was still around (read: 2400baud modem). No googling for answers and we relied on Teaching Assistant or the smartest guy in class for help (sorry, very few gals). My own recipe for approaching a question is:-

Take a theorem and apply it to the problem, does it fit?
If not,take another theorem and try fitting it
What is the result of applying this theorem?
Repeat from 1) until you are satisfied with the answer

Exams test how well verse you are in applying these concepts, and you only have 3 hours in the finals to show-off. It's like Lego, only that the blocks comes in different shapes and sizes and each term, the throw in new blocks just to confused you.


Integration by parts $ \int \sqrt{x-x^2} dx $ I have to integrate $$ \int \sqrt{x-x^2} dx $$
The answer on my textbook is $ \frac 14 \left( \arcsin(\sqrt x)-(1-2x)\sqrt{x-x^2} \right) $ but I want to solve this by myself so can you please give me a clue ? :) Thank you.

Hint: Complete the square:
$$\int\sqrt{x-x^2}dx=\int\sqrt{-x^2+2\frac x2-\frac14+\frac14}dx=\int\sqrt{-\left(x-\frac 12\right)^2+\frac14}dx$$
Set $u=x-\frac 12$ to obtain
$$\int\sqrt{-u^2+\frac14}du$$
I think you now how to deal with this one.

for calculation of Integral $\displaystyle \int \sqrt{a^2-x^2}dx$

We will Use Integration by parts method

Let $\displaystyle \mathbb{I} = \int \sqrt{a^2-x^2}.xdx = \sqrt{a^2-x^2}.x-\int\frac{1}{2\sqrt{a^2-x^2}}.-2x.xdx$

$\displaystyle \mathbb{I}= x.\sqrt{a^2-x^2}-\int\frac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx$

$\displaystyle \mathbb{I} = x.\sqrt{a^2-x^2}-\mathbb{I}+a^2.\sin^{-1}\left(\frac{x}{a}\right)$

So $\displaystyle \mathbb{I}=\sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}.\sin^{-1}\left(\frac{x}{a}\right)+\mathbb{C}$


What is the derivative of the expected value of a continuous random variable? Say we have a random variable $X$, with density function $f(x)$, and moment generating function $M(t) = E[e^{tX}]$, and we take the derivative of $M(t)$

$$\frac{d}{dt} M(t) = \frac{d}{dt}E\left[e^{tX} \right] = E\left[\frac{d}{dt}e^{tX} \right]$$

I can see how this works in the discrete case, as we are bringing the derivative inside a summation and the derivative of sums is the sum of the derivatives. But why are we allowed to bring the derivative inside the integral in the continuous case?

$$\frac{d}{dt}\int_X e^{tX}f(x)dx = \int_X \frac{d}{dt}[e^{tX}f(x)]dx$$?

I believe what you need is a direct application of Leibniz's Rule.  It requires continuity on the integrand and its partial derivative, so I believe that distributions can be constructed in which this does not hold (though perhaps such cases can be handled by the measure-theoretic formulation).

For $X$ a cont. random variable, in any case when $E |X| < \infty$, your equation holds.

The following proof for characteristic functions can be transferred to the moment gen. function.


Find $\lim_{n \to \infty}f_n(x)$ Find $$\lim_{n \to \infty}f_n(x)$$ where $$f_n(x)=n^2x(1-x)^n$$ $0 \lt x \lt 1$

My try:

By symmetry $$\lim_{n \to \infty}f_n(x)=\lim_{n \to \infty}f_n(1-x)=\lim_{n \to \infty}n^2(1-x)x^n=(1-x)\lim_{n \to \infty}n^2 x^n$$

Now $$\lim_{n \to \infty}n^2 x^n=\lim_{n \to \infty}\frac{x^n}{\frac{1}{n^2}}$$

Now can we use L'hopital's rule here?

It might be easier to use the Ratio Test:
$$
\begin{align}
\lim_{n\to\infty}\frac{f_{n+1}(x)}{f_n(x)}
&=\lim_{n\to\infty}\frac{(n+1)^2}{n^2}(1-x)\\
&=1-x
\end{align}
$$
For $0\lt x\lt1$, we have $0\lt1-x\lt1$. Therefore, the series
$$
\sum_{n=0}^\infty f_n(x)
$$
converges. Thus, the terms go to zero. That is,
$$
\lim_{n\to\infty}f_n(x)=0
$$


question about the limit $\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}$ Because $\sin'(x)=\cos(x)$ we can prove that $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$. but, by definition we have
$$\arcsin'(x)=\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}\tag{1}$$
therefore,
$$\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}\tag{2}$$
My question: Can we prove $(2)$ without using $(1)$? How?

Let $\arcsin(x+h)=u$ and $\arcsin x=v$

$\displaystyle \implies (i) x+h=\sin u, x=\sin v$

and $\displaystyle (ii) -\frac\pi2\le u,v\le \frac\pi2$ (using the definition of principal value)

$\displaystyle\implies \cos u,\cos v\ge0$

$$\displaystyle\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin x}h=\lim_{u\to v}\frac{u-v}{\sin u-\sin v}$$

$$\displaystyle=\lim_{u\to v}\frac{u-v}{2\sin\frac{u-v}2\cos\frac{u+v}2}$$

$$\displaystyle=\lim_{u\to v}\frac{\frac{u-v}2}{\sin\frac{u-v}2} \cdot\frac1{\lim_{u\to v}\cos\frac{u+v}2}$$

$$\displaystyle=1\cdot\frac1{\cos v}=+\frac1{\sqrt{1-x^2}}$$

Can be prove by the following:

$y=arc\sin x\to y'=y'_x=\frac{1}{x'_y}=\frac{1}{cosy}=\frac{1}{\sqrt{1-sin^2y}}=\frac{1}{\sqrt{1-x^2}}$


Does almost sure convergence implies convergence of the mean? I asked a slightly similar question here: Does Convergence in probability implies convergence of the mean?, but now I wish to examine a stricter scenario:
Let $\{X_n\}_{n=1}^\infty$ be a sequence of random variables converging a.s to a const $c$. Is it required for the sequence to be uniformly integrable in order to imply  $\lim_{n\to \infty} EX_n = c$?

And what about $\lim_{n\to \infty} EYh(X_n) = E[Y]h(c)$ for some random variable $Y$ and a continuous function $h$?  Under which regularity conditions does the last equality holds?

For the first question, try $P[X_n=0]=1-1/n^2$, $P[X_n=2^n]=1/n^2$, then $X_n\to0$ almost surely but $E[X_n]$ does not converge to $0$. (The second question is unclear.)

Think again here $~X_n~$ not converges to $~0~$ almost surely. When you are taking $~X_n=2^n~$ then probability approaches towards $~0~$ but not zero.

Definition is $~P[\lim X_n=0]=1~$, but here when $~X_n=0~$ probability is not exactly $~1~$, lies near $~1~$; hence contradicts the definition.


Prove that if $Hg_1^{-1}=Hg_2^{-1}$ then $g_1 H \subseteq g_2 H$ Prove that if $H$ is a subgroup of a group $G$ and $g_1,g_2 \in G$ then if $Hg_1^{-1}=Hg_2^{-1}$ then $g_1 H \subseteq g_2 H$.
My attempt:

Obviously $g_2^{-1} \in Hg_2^{-1}=Hg_1^{-1}$ so $g_2^{-1} = hg_1^{-1}$ for some $h\in H$. Then $g_2^{-1}g_1=h\Rightarrow g_1 = g_2 h \Rightarrow g_1 h' = g_2 h h' $ for some $h' \in H$. As $hh' \in H$ this means that $g_1 H \subseteq g_2 H$. Is there any flaw in this?

you need multiply g1 from right:
$$ Hg_{1}^{-1}*g_{1} = Hg_{2}^{-1}* g_{1} $$
$$H=Hg_{2}^-{1}*g_{1}$$
so the group is normal and the property hold


How to convert between Sum Of Products and Product of sums? I have a Boolean expression. we'll call it F.

for instance, F = ab' + ad + c'd + d'.

Assuming I did all the necessary steps too get F complement , i.e. F'.

I got: F' = b'd + ac'd'.

How do I get the Product of sums form of F?

$$F=(F')'=(b'd+ac'd\,')'=(b'd)'(ac'd\,')'=(b+d\,')(a'+c+d)\;.$$

(Note: I did not check your $F'$.)

Because of the way the De Morgan laws work, the complement of a product of sums is always a sum of products, and the complement of a sum of products is always a product of sums.

If the boolean expression is AB+BC.Then the product of sum expression is (A+B)(B+C).To find this, the complement of sum product expression is the product of sum expression.                                                                        (AB+BC)'=A'B'+B'C'
           By applying DEMORGANS LAW
            =(A'+B')(B'+C')
       Now,
      (A'+B')'(B'+C')'=(A+B)(B+C).


Prove that $\mathbb{R^*}$, the set of all real numbers except $0$, is not a cyclic group Prove that $\mathbb{R^*}$ is not a cyclic group. (Here $\mathbb{R^*}$ means all the elements of $\mathbb{R}$ except $0$.)

I know from the definition of a cyclic group that a group is cyclic if it is generated by a single element. I was thinking of doing a proof by contradiction but then that ended up nowhere.

Another way to see it: Assume $\Bbb{R}$ is cyclic and $a$ is its generator. Then $1=na$ for some $n\in\Bbb{Z}$ and $\sqrt{2}=ma$ for some $m\in\Bbb{Z}$. These imply that $a=1/n=\sqrt{2}/m$ and hence $\sqrt{2}=n/m$, contradiction.


Proving that a limit does not exist Given the function
$$f(x)= \left\{\begin{matrix}
  1 & x \gt 0 \\
  0 & x =0 \\
  -1 & x \lt 0
 \end{matrix}\right\}$$

What is $\lim_{a}f$ for all $a \in \mathbb{R},a \gt 0$?

It seems easy enough to guess that the limit is $1$, but how do I take into account the fact that $f(x)=-1$ when $x \lt 0$?
Thanks

For $a>0$, let $\delta=a$. Then for any $\epsilon>0$, we have that
$$|x-a|&lt\delta\implies x>0\implies f(x)=1\implies |f(x)-1|&lt\epsilon$$
hence $\lim\limits_{x\to a}f(x)=1$, as we have satisfied the definition of limit.

Here $\lim_{x\to 0^+}f(x)=1$ & $\lim_{x\to 0^-}f(x)=-1$ both limit are not equal.

therefore limit is not exists


Closed form for finite summation of sequence $\sum^n_{i=1}{e^i/i}$ Does there exist a closed form for finite summation of the sequence  $\sum^n_{i=1}{e^i/i}$ ?

Take the derivative of the function $f(x) = \sum_{i = 1}^n \frac{x^i}{i}$ which is easier to compute.

Maple writes it as $$\sum_{i=1}^n \frac{{\rm e}^i}{i} = -\Phi \left( {{\rm e}},1,n \right)\; {{\rm e}^{n}}-\ln
 \left( -{{\rm e}}+1 \right) +{\frac {{{\rm e}^{n}}}{n}}$$

where $\Phi$ is the Lerch Phi function.


Logical equivalence + Quantifiers (Universal and Existential) I'm taking a Discrete Mathematics course and we're using Rosen's book (which I hate because it seems like it makes difficult material to understand even more incomprehensible). Trouble is, I am lost when it comes to combining what I learned about logical equivalency and quantifiers.

Sample problem from homework:

Show that ∃x( P(x) v Q(x) ) and ∃xP(x) v ∃xQ(x) are logically equivalent.

What is the step by step process to answering this question?

Edit: Does anyone have a recommendation for a better textbook on Discrete Math?

Hint: You are over complicating this problem.

Think about what both expressions mean. The "proof" that
$$
\exists x(P(x)\lor Q(x)) \Longleftrightarrow \exists xP(x)\lor \exists xQ(x)
$$
is remarkably simple.

Proof. Both expressions are true precisely when at least one of $P(x)$ and $Q(x)$ is true for at least one value of $x$ in the universe of discourse. $\Box$

Of course, if you want to prove it in a different manner, then you are welcome to try, but that explanation is really all you need.

Here's a simple example with trivial constructions, but it may help you see why the two expressions are equivalent. Assume we are working with integers as the domain of discourse.

Start by constructing statements $P(x)$ and $Q(x)$.

$P(x) : x\neq x$
$Q(x) : x=1$

Now think about what both expressions mean.

Expression 1: $\exists x(P(x)\lor Q(x))$ means that there exists an $x\in\mathbb{Z}$ where at least one of $P(x)$ and $Q(x)$ is true. Well, $x\neq x$ is never true for any integer $x\in\mathbb{Z}$. Thus, we know $P(x)$ is not true. However, $Q(x)$ is true. Certainly there exists an $x\in\mathbb{Z}$ such that $x=1$, namely $x=1$. Okay cool. So expression 1 is true for our sample constructions of $P$ and $Q$.

Expression 2: $\exists xP(x)\lor \exists xQ(x)$ means that (1) there exists an $x\in\mathbb{Z}$ such that $P(x)$ is true, OR (2) there exists an $x\in\mathbb{Z}$ such that $Q(x)$ is true. Well, we just saw that $x\neq x$ is never true in $\mathbb{Z}$; thus, we can throw out $\exists xP(x)$, but what about $\exists xQ(x)$? Again, certainly there exists an $x\in\mathbb{Z}$ such that $x=1$, namely $x=1$. Thus, $\exists xQ(x)$ is true, thereby making $\exists xP(x)\lor \exists xQ(x)$ a true statement.

Notice that Expression 1 and Expression 2 are really saying the same thing; that is, they are actually logically equivalent.

$\exists{x}(P(x)\lor{Q(x)})\equiv\neg \forall{x}\neg(P(x)\lor{Q(x)})\equiv\neg \forall{x}(\neg P(x)\land\neg{Q(x)})\equiv \neg\forall{x}\neg P(x) \land \neg\forall{x}\neg{Q(x)} \equiv \exists{x}P(x)\land \exists{x}Q(x) $