Dense sequence definition A subset $A$ of a space $X$ is called dense if every point $x$ in $X$ either belongs to $A$ or is a limit point of $A$ ; that is, the closure of $A$ constitutes the whole set $X$.
That the definition of a dense set, what about sequences? what is the meaning of dense sequence?

A sequence is dense if the set whose elements are the sequence elements, is dense according to your definition.

A sequence is not dense. It’s a common abuse of terminology that identifies the sequence, which is a function from the natural numbers to the real numbers (usually), with its image set, which is a subset of the codomain of the sequence.
A sequence $(x_n)_{n}$ is a function from the natural numbers to a set $X$ and, formally, can be written as $(x_n)_{n} \in X^{\mathbb{N}}$. The image of any function is a subset of the codomain, and so we write $\{x_n : n \in\mathbb{N}\}$. Hence, to be precise, we should say that the image set of the sequence is dense in $X$, that is $\overline{\{x_n : n \in\mathbb{N}\}}=X$
The same Rudin, Principles of Mathematical Analysis, definition 2.7: "By a sequence, we mean a function $f$ defined on the set $J$ of all positive integers".


A String Tied Around The Earth Say you're standing on the equator and you have a string below you tied around the equator (40,075 km) that is the length of the equator + 1 meter (40,075.001 km). What is the maximum height you can you lift the string off the ground? Can you create a function of both circumference of the circle (earth) and string to output the distance between the two if pulled tight?

Assumptions:

For illustration, the result would be pulled from a single point, making a triangle until it met with the earth, in which it would follow the curvature of the earth. Similar to a snow-cone or O>
The string does not stretch
The earth can be assumed to be a perfect sphere

We have to solve (with respect to $d$) the trascendental equation:
$$ 2\sqrt{(R+d)^2-R^2}+2R\arcsin\frac{R}{R+d}=\pi R+1 $$
or solve (with respect to $x=\frac{d}{R}$) the trascendental equation:
$$ 2\sqrt{x^2+2x}+2\arcsin\frac{1}{1+x}=\pi+\frac{1}{R} $$
that is equivalent to:
$$ \sqrt{x^2+2x}-\arctan\sqrt{x^2+2x}=\frac{1}{2R}.$$
In order to solve $z-\arctan z=\frac{1}{2R}$ we may apply Newton's method, then $x=-1+\sqrt{z^2+1}$.

Since $z$ is close to zero, $z-\arctan z\approx\frac{z^3}{3}$, hence:
$$ d\approx \sqrt[3]{\frac{9R}{32}}\approx \color{red}{121.5\,m}.$$

Half a meter.  You take the slack extra meter and lift it half a meter high.


How many cycles, $C_{4}$, does the graph $Q_{n}$ contain? Let $n \geq 2$ be an integer. How many cycles of length $4$ (that is, subgraphs isomorphic to $C_4$) does the hypercube graph $Q_n$ contain?

Examples: For $n = 2$, the hypercube graph $Q_2$ is a single $4$-cycle, so the answer in this case is $1$. For $n = 3$, the hypercube graph $Q_3$ is a cube, and the answer is $6$.

I was told that $n$-dimensional cube graphs, $Q_{n}$ can be represented by binary but how can that help me approach this question?

How many $C_4$s pass through a given vertex $v$? There are  two edges in this cycle through
$v$. There are $n$ edges through $v$ so $\binom {n}2$
pairs of edges. Once we have these two edges there is
just one way to complete the $C_4$. Don't forget there are $2^n$
vertices overall and each $C_4$ features four of them.

Here is another way to look at it: $Q_n$ is a two-dimensional projection of an $n$-dimensional hypercube, so each distinct $C_4$ in $Q_n$ is a face of the $n$-dimensional hypercube.

When $n\ge 3$ the $n$-dimensional cube satisfies the generalized Euler polytope formula
$$
V-E+F=1+(-1)^{n-1}.
$$
For $Q(n)$ we have $V=2^n$ and $E=n2^{n-1}$ so that
$$
F=n2^{n-1}-2^n+1+(-1)^{n-1}.
$$

As to how $Q_n$ is represented with binary strings: Each vertex in $Q_n$ corresponds with each of the possible $n+1$-digit binary string and an edge exists between two vertices if their corresponding strings differ in exactly one digit.


Essential singularity This is an exercise from Gamelin.

If $f(z)$ is a complex function with a not removable singularity in $ z_{0} \ $, then $e^{f(z)} \ $ has an essential singularity in $z_{0} $.

Any hint?

If your function $f$ has a pole at $z_0$ write $(z-z_0)^m f(z) = p(z) + (z-z_0)^m h(z)$ where $h$ is holomorphic and $p$ is a polynomial of degree $<m$.  Then $f(z)= \displaystyle \frac{p(z)}{(z-z_0)^m} + h(z)$.  Now take $e$ to both sides, note that $e^{h(z)}$ is holomorphic, and use the power series expansion of the exponential to show that $e^{\frac{p(z)}{(z-z_0)^m}}$ has infinitely many negative powers of $z-z_0$.

I haven't looked at this stuff in a while, but Casorati-Weierstrass might help.

In particular there is a sequence $(z_k)$ converging to $z_0$ such that $(f(z_k))$ converges to $w_0$ (this is page 175 of your book). Since $w_0$ is arbitrary can you think of a particular value that helps solve this problem?


What is $a+b$ if $ x^2-y^2+11x+7y+18$ can be factorized as $(x+y+a)(x-y+b)$? If the polynomial $x^2-y^2+11x+7y+18$ can be factorized as $(x+y+a)(x-y+b)$ what is the value of $a+b$?

Observe that the expression can be rearranged with respect to  (x) and factorized as
\begin{align} x^2-y^2+11x+7y+18 &= x^2+11x-(y^2-7y-18) \\ &= x^2+11x-(y+2)(y-9) \\ &= (x+(y+2))(x-(y-9)) \\ &= (x+y+2)(x-y+9). \end{align}
This implies  (a=2) and  (b=9), hence  (a+b=11).

I don't understand how they got from step 2,
$ \\ x^2+11x-(y+2)(y-9)\\$, to step 3,  $\\ (x+(y+2))(x-(y-9)) \\ $

https://brilliant.org/practice/algebraic-manipulation-rearranging/?problem=algebra-problem-56641&subtopic=advanced-polynomials&chapter=algebraic-manipulation

Hint:  consider $x^2+11x-y^2+7y+18$ as a quadratic in $\,x\,$, then its discriminant is:

$$
\Delta = 11^2 - 4(-y^2+7y+18) = 4y^2-28y+49=(2y-7)^2
$$

The roots in $\,x\,$ are $x_{1,2}=\big(-11 \pm (2y-7)\big)/2\,$, then the quadratic factors as $(x-x_1)(x-x_2).$

The solution is on this image Simplify like this


set of values taken by the density of the random variable $X$ is continuous random variable. $f_X(x)$ is density of $X$, $x\in\mathbb{R}$.
What are possible values of expression $f_X(x)$  ?

To be more precisely, the question is about image: $f_X(\mathbb{R})$.

$X$ is not normal, we don't anythhing about it. The only thing we know is continuity, so $P(x\in [a, b])=\int_{a}^{b}f_X(x)dx$

Well, if that is all you know then this is a very general question.

You can easily look up the properties of probability densities on Wikipedia https://en.wikipedia.org/wiki/Probability_density_function.

Densities can take any non negative real value. That and the fact that the integral over the whole real line should be one is all you can say without more information.

If I'm right in recalling that the density of an impulse is a zero-width spike of infinite height, then the value in general, i.e. across all possible density functions, can be any non-zero real number.


The frustum equations; how much can you tilt a frustum? "Mathematicians are machines for turning coffee into theorems".
  -Paul Erdòs

I'm pretty sure almost everyone reading this has drunk a cup of coffee before. If you get coffee from Starbucks or Dunkin' Donuts, it comes in a brown cylindrical frustum cup full of the delicious good stuff to keep you up as you work on typing the answer to that one Calculus problem that has stumped everyone else. A few days ago I was cleaning up after I spilled a cup of Joe and wondered, 'is there a formula to the limit of tipping a coffee (or otherwise frustum shaped) cup?' It seems likely, since there is a formula for doing the same thing with a cylinder.

(For those of you who are confused, a frustum is a cylinder that has one of its circles larger than the other, and all parallel cross sections line up on a line perpendicular to the base)
I may not be experienced with Calculus, but I do know all the Trigonometry stuff, and using it and simple math, I deduced that if a=radius of cylindrical cup, and $h_1$=height of the cup, and $h_2$=height of liquid, then the maximum angle you can tip the cup, assuming $0$ surface friction, is
$$\sin^{-1}
\frac{\sqrt{a+(h_1-h_2)}}
{2(h_1-h_2)}
$$
I think, anyway. But frustums are much harder. Considering all the variables are the same, and $a_1$ is the radius of the top of the frustum and $a_2$ is the smaller base, the one issue I have is, should I calculate the radius $a_3$ which would be the average, or the average radius of the area directly between $h_1$ or $h_2$. I've tried a lot over the last few days, and they are all flawed.

So, just to recap, if the top and bottom of the frustum cup are $a_1$ and $a_2$ respectively, $h_1$ is the height of the cup, and $h_2$ is the height of the liquid in the cup, then what is the formula for how much you can tilt it before the coffee spills out? Or am I providing insufficient information? Thanks in advance.

In diagram below I reproduced a section of the given frustum, with $HA=a_1$, $KC=a_2$, $HK=h_1$ and $MK=h_2=$ height of liquid when the cup is not tilted. We can extend the frustum to a full cone of vertex $V$ and set $VK=h$ and $\theta=\angle CVK$. Some easy geometry gives then $h=a_2h_1/(a_2-a_1)$ and $\tan\theta=a_2/h$.

If the cup is tilted to the limit point, the surface of the liquid is $AE$ and the tilt angle is $\alpha$. I will deal here only with the case when the frustum base is completely covered by the liquid, that is when
$$
h_2\ge{h_1\over1+\sqrt{a_1/a_2}}.
$$
In this case we can then apply the result given in this answer to a similar question:
$$
EV\cdot AV = LV^2.
$$
By applying then the sine rule to triangle $VAE$ we get after some algebra:
$$
\tan\alpha=\cot\theta\,{AV^2-LV^2\over AV^2+LV^2}.
$$
It is now only a matter of expressing $AV$ and $LV$ in terms of our data:
$$
AV={h_1+h\over\cos\theta},\quad LV={h_2+h\over\cos\theta}
$$
and then
$$
\tan\alpha=
{h\over a_2}{(h_1-h_2)(h_1+h_2+2h)\over h_1^2+h_2^2+2h^2+2(h_1+h_2)h}
={h_1(h_1-h_2)\big(a_2(h_1-h_2)+a_1(h_1+h_2)\big)
\over h_1^2(a_1^2+a_2^2)+2h_1h_2a_2(a_1-a_2)+h_2^2(a_1-a_2)^2}.
$$
In particular, letting $a_2=a_1=a$ we recover the analogous formula for a cylinder:
$$
\tan\alpha={h_1-h_2\over a}.
$$

The coffee cup tips when $\bar z \sin \theta > b$

We just need a formula for $\bar z$

Center of mass

$\bar z = \frac {\pi\int_0^h z y^2(z) dz}{\pi\int_0^h y(z) dx}$

$y(z) = \frac {b-a}{h} z + b$

$\int_0^h (\frac {b -a}{h}z + b)^2 dz\\
(\frac {(b-a)^2}{3} + \frac {2(b-a)b}{2} + b^2) h$

$\int_0^h z(\frac {b -a}{h} z + b)^2 dz\\
(\frac {(b-a)^2}{4} + \frac {2(b-a)b}{3} + \frac {b^2}{2}) h^2$

$\bar z = \frac {3(b-a)^2 + 8(b-a)b + 6b^2}{4(b-a)^2 + 12(b-a)b+ 12 b^2} h$


Revised GRE Math Probability section The table shows the distribution of a group of $40$ college students by gender and class
$$
\begin{array}{c|lcr}
 & \text{Sophomores} & \text{Juniors} & \text{Seniors} \\
\hline
\text{Males} & 6 & 10 & 2 \\
\text{Females} & 10 & 9 & 3 \\
\end{array}
$$

If one student is randomly selected from this group, find the probability that the student chosen is (i) a female or a sophomore?

How many students are either females or sophomores (or both)? First count the  number of females, then count the number of male sophomores (female sophomores need not be counted since you already counted them when you counted the number of females). What percent of the total number of students is this?

There are $10 + 9 + 3 = 22$ females and $6$ male sophomores, so there are $22 + 6 = 28$ students out of $40$ who are either female or sophomore.

$28/40 = 0.7$

The number of sophomores= 16

Number of females=10+9+3=22
Number of female sophomores=6

We get a venn diagram between sophomores and females, with common area as 6
Total people who are sophomore or female= 10+6+16=32

Probability=32/40=0.8


What does the symbol $\operatorname{Tr}$ in the Yang-Mills action mean? I find that many authors write the Yang-Mills action as follows:
$$\mathcal{J}= \int \operatorname{Tr}(F \wedge  \star F).$$
I have yet to find a formal description of the symbol $\operatorname{Tr}$ and am especially interested in its relation to differential geometry. How should I understand it?

We choose the physics convention
$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu - i g [A_\mu,A_\nu]$, where the generators of the Lie algebra are Hermitian.
In this case the integral has a minus sign,
$\mathcal{J} = -\int \operatorname{Tr}(F \wedge  * F)$.

The first thing to note is that $F$ has vector and Lie algebra indices.
The trace is over the Lie algebra, not over the vector indices.
The vector indices are just like those of the field strength in QED.
In Yang-Mills the field strength (also called the curvature form) is Lie algebra valued.
In this case $F_{\mu\nu} = F_{\mu\nu}^a T^a$, where the summation convention is used, and where the $T^a$ are the generators of $\mathfrak{su}(n)$.
To be explicit, $F$ has not only tensor components but matrix components
$$(F_{\mu\nu})_{ij} = F_{\mu\nu}^a T^a_{ij}.$$

The inner product of $F$ with itself is by definition $(F,F) = \int F\wedge * F$ where $*$ is the Hodge $*$ operator.
Thus, we are calculating $\mathcal{J} = -\operatorname{Tr}(F,F)$.
It is a standard exercise to find the exterior product of two $r$-forms.
We find
$$F\wedge * F  = \frac{1}{2!} F_{\mu\nu} F^{\mu\nu} dx^1\wedge\cdots\wedge dx^4.$$
Note that the differential forms don't "know" the Lie algebra.
The algebra hasn't come into the calculation yet.

From here we must calculate the trace of $F_{\mu\nu}F^{\mu\nu}$,
$$\begin{eqnarray}
\operatorname{Tr} \left(F_{\mu\nu}F^{\mu\nu}\right)
&=& \operatorname{Tr} \left( F_{\mu\nu}^a T^a F^{\mu\nu b}T^b\right) \\
&=& \operatorname{Tr}\left(T^a T^b\right)
 F_{\mu\nu}^a F^{\mu\nu b} \\
&=& \frac{1}{2} \delta^{a b}
 F_{\mu\nu}^a F^{\mu\nu b} \\
&=& \frac{1}{2} F_{\mu\nu}^a F^{\mu\nu a},
\end{eqnarray}$$
where we have used the standard normalization convention for the $T^a$,
$\operatorname{Tr} T^a T^b = \frac{1}{2}\delta^{ab}$.
(This comes from the fact that we want $\mathfrak{su}(2)$ to live naturally in $\mathfrak{su}(n)$, and the generators of $\mathfrak{su}(2)$ are taken to be $T^a = \sigma^a/2$, where $\sigma^a$ are the Pauli matrices.)
Thus, we find
$$\begin{eqnarray}
\mathcal{J} &=& -\int \operatorname{Tr}(F \wedge  * F) \\
&=& -\frac{1}{4} \int d^4x \, F_{\mu\nu}^a F^{\mu\nu a}.
\end{eqnarray}$$

Tr is the trace. Usually defined as
$$ \operatorname{Tr}\left(F_{\mu\nu}\right) = \sum_\eta F_{\eta\eta} $$
With corresponding generalizations for continuous function etc. In your case this definition should be sufficient though, as $F$ is something as the following?
$$ F_{\mu\nu} = \frac{1}{ig}\left[D_\mu,D_\nu\right] $$


Compactness of $C([0,1])$ I have to verify if the $C([0,1])$, space of all continuous functions defined on interval $[0,1]$ with supremum metric is compact.

As I know, we have to check if every sequence of functions $f_{n}(x)$ has subsequence that $f_{n_{k}}(x)$ is convergent.
In this metric of course conervgence implies uniform convergence, so there won't be a problem of showing continuity of the limit function.
But I really don't know if we can make subsequence convergent.

The sequence $f_n(x)=x^n$ does not have a convergent subsequence since $f_n$ converges pointwise towards $f(x)=0$ if $x\neq 1$ and $f(1)=1$ which is not continuous. Hence, the space is not compact.

A Banach space is finite-dimensional if (and only if) it is locally compact. It's easy to see that $C[0,1]$ is a Banach space with respect to the supreme norm. And, trivially, the property of compactness implies local compactness.
To sum it up, there is not infinite-dimensional Banach space (such as $C[0,1]$ ) that is also compact.


Classification of a PDE $$u_{xxxx}+4u_{xy}=\tan x$$
Is it linear? (I think it is linear)

And what is the order of the PDE? (it is $4$?)

And other question:
$$u_{xx}+4u_{xyy}=0$$
the order of the last PDE is $3$ (not $2$), right?

The PDE
$u_{xxxx}+4u_{xy}=\tan x$ has order $4$, but is not linear. If $u$ and $v$ are solutions and $w:=u+v$, then  $w_{xxxx}+4w_{xy}=2 \tan x \ne \tan x$.
Hence $w$ is not a solution.
The PDE
$u_{xx}+4u_{xyy}=0$ has order $3$.


Show that a continuous function on a compact set satisfies this bound. Problem: Suppose $K\subseteq \mathbb{R}^n$ is compact, $f: K\to \mathbb{R}$ is continuous, and $\epsilon>0$. Show that there is a number $A>0$ such that
$$|f(x)-f(y)| \leq A\|x-y\| + \epsilon,\;\forall x,y\in K. $$

Attempt at solution: By uniform (since $K$ is compact) continuity, there is a $\delta>0$ such that $\|x-y\| < \delta$ implies $|f(x)-f(y)|<\epsilon$. By compactness we can cover $K$ with finitely many, say $N$, such $\delta$-balls. Since the image of $K$ under $f$ is compact, can we do something like cover the image with finitely many $\epsilon$-balls? I'm just not sure how I can get a bound in terms of $\|x-y\|$.

Another method: Let $\epsilon>0$. By uniform continuity, take $\delta>0$ so that $|f(x)-f(y)|< \epsilon$ whenever $\|x-y\|< \delta$. Define $$A:= \max_{\|x-y\|\geq \delta} \frac{|f(x)-f(y)|}{\|x-y\|}$$ Note that this maximum is finite (and achieved) since $\{(x,y)\in K \times K : \|x-y\| \geq \delta\}$ is a compact set (being a closed subset of the compact set $K \times K$), and the expression being maximized is a continuous function on that set.

Then for $\|x-y\| \leq \delta$, we have that $|f(x)-f(y)| \leq \epsilon$, and for $\|x-y\| \geq \delta$, we have that $|f(x)-f(y)| \leq A\|x-y\|$, thus in general $$|f(x)-f(y)| \leq \max\{\epsilon, A\|x-y\|\} \leq \epsilon + A\|x-y\|$$

Suppose it is not true, for every integer $n$, there exists $x_n,y_n\in K$ such that $|f(x_n)-f(y_n)|\geq n\|x_n-y_n\|+\epsilon$. Since $K$ is compact, we can extract a sequence $(x_{n_k})$ from $x_n$ which converges towards $x$ and a sequence $({y_{n_k}}_l)$ from $(y_{n_k})$ which converges towards $y$, remark that $({x_{n_k}}_l)$ converges also towards $x$. To simplify the notation, we write ${x_{n_k}}_l=x_l$. We have $|f(x_l)-f(y_l)|\geq l\|x_l-y_l\|+\epsilon$. Since $f$ is continue, we deduce that $|f(x)-f(y)|\geq l\|x-y\|+\epsilon$ for an infinite number of distinct integers $l$. This implies that $\|x-y\|=0$ and $x=y$. We deduce that $|f(x)-f(x)|\geq l\|x-x\|+\epsilon$ where $\epsilon>0$. Contradiction.


Can we solve this recurrence relation using recursion tree method The recurrence relation is given as follows:

$T(n) = 2T(\sqrt{n})+1$

$T(1) = 1$

I tried to solve it with recursion tree as follows:

But to find the number of levels that may occur, I have to solve:

$\sqrt[2^x]{n} = 1$

$2x=\log{_1}{n}$

$\log{1} = 0$

So I am stuck here and cant move further.

What I am doing wrong?

PS:
I know we can solve this using master theorem. But I am specifically interested in solving this using recursion tree method.

You are looking at this recurrence relation:

$T(n) = 2T(\sqrt{n})+1$

$T(1) = 1$

Suppose T is a solution.  Then we have $T(1) = 2T(1) + 1$, so $1 = 3$.  That is, this recurrence relation has no solution.


L1 Minimization with Non-negativity constraints. Is there any fast approach to solving l1 minimization problem with non-negativity contraints? The problem is to minimize $|x_1|+|x_2|+...+|x_n|_{l_1}$ subject to $Ax = b$ and $Ax \geq 0$.

You can convert problems like this into linear programs by introducing variables $t_i, i = 1,\ldots,n$ that satisfy
\begin{equation}
-t_i \leq x_i \leq t_i
\end{equation}
for $i = 1,\ldots, n$.
(Note that this is equivalent to requiring $t_i \geq | x_i| \, \forall \, i$.)

Just minimize $t_1 + \cdots + t_n$ subject to these constraints, and to whatever other constraints you have on $x$.  Efficient methods are available to solve linear programs.

This trick (which is standard) is described in chapter 6 ("Approximation and fitting") of Vandenberghe's 236b notes.  See slide 6-3.

Indeed, this can be solved as a quadratic program by any QP solver (I do this routinely).

The problem is (if I understood correctly):

$min||Ax-b||^2_2+c||x||_1$ such that $x>=0$

In your case $c=1$ but in general this is a regularization parameter (for more information read about LASSO or L1-regularized least squares).

Start with the fact that $min||Ax-b||^2_2$ such that $x>=0$ is clearly a QP since:

$||Ax-b||^2_2=x^TA^TAx-2(b^TA)^Tx+b^Tb$

Now the additional constraint won't change much, and the problem is still a QP:

$||Ax-b||^2_2+c||x||_1=x^TA^TAx-2(b^TA)^Tx+b^Tb+c\underline{1}^Tx$

I hope I got all those nasty transpose right :)


Induced representation of $G=SL_2$ by $\chi _ w$ is irreducible if $w^2\neq 1$. This is the question from Serre's book #7.4. $G=SL_2(k)$, where $k$ is a finite field and $H\leq G$ such that $H$ consists of matrices $\begin{pmatrix} a & b \\ 0 & d  \end{pmatrix}$. Let $w:k^* \rightarrow \mathbb{C}^*$ be a homomorphism and $\chi_w$ be the character of degree 1 of $H$ defined by $\chi_w \begin{pmatrix} a & b \\ c & d  \end{pmatrix}=w(a)$. Show that the representation of $G$ induced by $\chi_w$ is irreducible if $w^2\neq 1$.

How can i start this problem?

Compute the inner product of the induction with itself, using Frobenius reciprocity and Mackey's formula. For the latter, begin by showing that the set of double cosets $H\backslash SL_2(k)/H$ has size 2, with representatives $\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$.

Sorry to resurrect such an old thread, but in the interests of completeness, we provide a complete solution.

Choose $X = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \in G \setminus H$. A routine calculation gets us$$A := \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}\begin{pmatrix}
x & y \\
0 & z
\end{pmatrix}\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}^{-1} = \begin{pmatrix}
adx - acy - bcz & a(bz + ay-bx) \\
c(dx - cy - dz) & adz + acy - bcx
\end{pmatrix}.$$This is in $H$ exactly when $y = {{d(z-x)}\over{c}}$. In this case, we can rewrite the above matrix as $$A = \begin{pmatrix}
z & a(bz + ay - bx) \\
0 & x
\end{pmatrix},$$ so this describes all matrices in $H_X$. Also, $$A^{-1} = \begin{pmatrix}
x & a(bz + ay - bx) \\
0 & z
\end{pmatrix}.$$ Because $x = z^{-1}$, each matrix in $H_X$ is completely described by its top left entry. Now we can compute$$|H_X|\langle\rho^X,\text{Res}_X(\rho)\rangle_{H_X} = \sum_{Y \in H_X} \chi_w(Y^{-1})\chi_w(X^{-1}YX) = \sum_{x \in k^\times} w(x)w(x).$$Because $k$ is a finite field, $k^\times$ is a cyclic group, hence $w$ maps elements of $k^\times$ to roots of unity. Because $w^2 \neq 1$, a generator of $k^\times$ is mapped to an $n$th root of unity with $n >2$. Therefore, the last sum above cycles through all $(n/2)$th roots of unity, and each one appears the same amount of time. In particular, the sum is $0$, hence $\rho^X$ and $\text{Res}_X(\rho)$ are disjoint. Because $\chi_w$ is a degree $1$ representation, it is irreducible. Thus, by Proposition 23 of Serre's Linear Representations of Finite Groups, the induced representation of $G$ is irreducible.


probability and expected value Hey I am not sure if I thinking correctly on this question? In a carnival, there is game which charges you $3$ dollars to play a game. You win $1$ dollar for every consecutive head you get and you you can play till you get tail. if you get head head tail you get back two dollars. What is expected profit?

my theory is your probability of t,ht,hht,hhht and so on will be $1/2,1/4,1/8,1/16$ and so on, and income will be $0,1,2,3$ and so on.

so average income will be
$$ 0 * \frac{1}{2} + 1 * \frac{1}{4} + 2 * \frac{1}{8} + 3 *\frac{1}{16} + ....$$

which will come to a dollar?

so average profit will be, $1-3 = -2$??

okay I tried to solve this one,

$$ [a+(n-1)d] r^{n-1} $$ is term used to show nth term of Arithmetico-geometric sequence

if I take $1/2$ common from the whole series I get

$$ 0 * 1  + 1 * \frac{1}{2} + 2 * \frac{1}{4} + 3 *\frac{1}{8} + ....$$

which gives me$$ a = 0, d=1, r=1/2$$

now the sum is

$$\lim_{n \to \infty}S_{n} = \frac{a}{1-r}+\frac{rd}{(1-r)^2}$$

which is $2$.

and after multiplying with previous $1/2$ we took out, we get $1$.

so average profit is $$1-3 = -2$$

reference Wikipedia https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence

The problem states that you CAN play until you get tails, so according to my understanding you can stop even if you don't get tails. So your calculations are correct but you should add in the factor of stopping of getting multiple consecutive heads. So the probability of getting 1 head is 1/2 and you get $1, two heads is 1/4 and $2, and so on. Adding this all up gets $2, and adding this to your calculation gets $3, which means the expected profit is $0-- according to my understanding.


Evaluate a finite sum with four factorials Given positive integers $k, m, n$ such that $1 \leq  k \leq  m \leq  n$. Evaluate

  $$ \sum^{n}_{i\mathop{=}0}\frac{1}{n+k+i}\cdot\frac{(m+n+i)!}{i!(n-i)!(m+i)!}$$

Any hints? I'm stuck on this one.

Just a partial answer to collect some ideas.

Since
$\binom{m+n+i}{m+i}$
is the coefficient of $x^n$ in the Taylor series of $\frac{1}{(1-x)^{m+i+1}}$ around $x=0$, we have:
$$ S = [x^n]\left(\frac{1}{(1-x)^{m+1}}\cdot\sum_{i=0}^{n}\frac{1}{n+k+i}\binom{n}{i}\frac{1}{(1-x)^{i}}\right),\tag{1}$$
where $\frac{1}{n+k+i}=\int_{0}^{1}y^{n+k+i-1}dy$ leads to:
$$ S = [x^n]\left(\frac{1}{(1-x)^{m+1}}\cdot\int_{0}^{1}y^{n+k-1}\sum_{i=0}^{n}\binom{n}{i}\frac{y^{i}}{(1-x)^{i}}dy\right),$$
$$ S = [x^n]\left(\frac{1}{(1-x)^{m+1}}\cdot\int_{0}^{1}y^{n+k-1}\left(1+\frac{y}{1-x}\right)^n dy\right),$$
$$ S = [x^n]\left(\frac{(1-x)^{n+k}}{(1-x)^{m+1}}\cdot\int_{0}^{\frac{1}{1-x}}y^{n+k-1}(1+y)^n dy\right).\tag{2}$$
By setting $y=\frac{z}{1-z}$
$$ S = [x^n]\left(\frac{(1-x)^{n+k}}{(1-x)^{m+1}}\cdot\int_{0}^{\frac{1}{2-x}}\frac{z^{2n+k-1}}{(1-z)^{n+k+1}}dz\right)\tag{3}$$
we can see that the integral is just a value of the incomplete beta function, but I bet that other manipulations are more useful.

The sum is $$\frac{(m+n)!}{(n+k)n!m!}{}_3F_2(-n,n+k,m+n+1; m+1,n+1+k; -1)$$. Unfortunately there are not many closed form expressions known for these hypergeometric series at -1. A few like         http://dx.doi.org/10.1016/S0377-0427(96)00154-9 may be able to map the format to $_3F_2(;;+1)$$ and be handled by Milgram's table in http://arxiv.org/abs/1105.3126 .


Given two vectors, how can I denote the angle between them? Given the vectors $\vec{a}$ and $\vec{b}$, how can I denote the angle between them?

I like the following method:

  ..., where $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$...

Seriously, words can really improve proof legibility, and should be used.

What's wrong with using the identity that $a\cdot b = |a||b|\cos{\theta}$ and solving for theta?


monty hall question with 4 doors Explain the Monty Hall problem in the case of 4 doors computing specific probabilities.

I got that you have 1/4 chance of picking the door with the goat.
1/4 chance to pick the door with the prize and so on.

if I pick an empty door you have a 1/2 chance of doing this in this case you have 1/2 chance of winning the prize. if you don't switch

if you don't switch your chance is 1/4

switching is better

I think Im  missing something

I assume there are 4 doors - 3 with goats and a single one with the prize:

FIRST CASE: Show host opens 2 doors for you:

If you had picked the right door to begin with, you would lose from changing. However, this will only be the case 1/4 of the time. So 1/4 of the time you will win by staying.

Then show opens two doors with goats -

3/4 of the time, you will be at a goat to begin with, but the showing of two goat makes it so that the remaining one (if you are on a goat) is the prize. Now since 3/4 of the time you will be on a goat - 3/4 of the time you will win by changing.

SECOND CASE: Show host opens just 1 door for you:

If you had picked the right door to begin with, you would lose from changing. However, this will only be the case 1/4 of the time. So 1/4 of the time you will win by staying.

Then show opens a door with a goat -

3/4 of the time, you will choose a door with a goat to begin with. The revealing of a goat makes it so that 3/4 of the time the remaining 2 doors must hold the prize! But wait, now it's down to pure luck, good old 50/50. So your chance of winning in this case will be (3/4) * (1/2) or 3/8.

So overall your three options: there is a 2/8 (1/4) chance of winning by not changing, a 3/8 chance of losing by changing (moving to lets say door x) and a 3/8 chance of winning from changing (lets say moving to door y). The sum is fortunately =1.

Cheers!

First part of experiment:

probability of wining gift is $.25$.

2nd part of experiment:

without switching the probability of prize is $.25$
with switching the probability of wining the prize is $.75 \times .5 = .375$, which is greater as compared to without switching case.


Let $p_n$ denote the $n$th prime and $c_n$ the $n$th composite number. Determine all positive integers $n$ such that $|p_n − c_n| = 1$. The only ones with the property in title I could find are $n = 5$ and $n = 6$: $|11 − 10| = 1$  and $|13 − 12| = 1$. Past the $20$th prime, the list of primes grows too fast for composite numbers. Is there more general way to show this?

If $|p_n-c_n| = 1$ then there must exist a $N$ such that the set $\{2,3,\ldots,N\}$ contains an equal number of primes and composite numbers.

The number of composite numbers in the $N-1$ element set $\{2,\ldots, N\}$ is larger than (numbers divisible by $2$ + numbers divisible by $3$ - numbers divisible by $2\cdot 3$ since these have been counted twice)

$$M = \left\lfloor\frac{N-1}{2}\right\rfloor + \left\lfloor\frac{N-1}{3}\right\rfloor - \left\lfloor\frac{N-1}{6}\right\rfloor - 2$$

where we have subtracted $2$ for the two primes $2$ and $3$.

When $N>25$ we have $M \geq \frac{4N-21}{6} > \frac{N-1}{2} + 1$ so there will always be at least two more composite numbers than prime numbers in the set $\{2,3,\ldots,N\}$ and since $p_{10} = 29 > 25$ it follows that $|p_n-c_n|=1$ has no solutions for $n\geq 10$.

In general $p_n \sim n \ln n$. This can prove useful for your question.


Evaluating: $\int 3x\sin\left(\frac x4\right) \, dx$. $\displaystyle\color{darkblue}{3\int x\sin\left(\dfrac x4\right)\,\mathrm dx}$

  $$\begin{align}
\dfrac{-4x}{x}\cos\dfrac x4 \,\,\boldsymbol\Rightarrow\,\, & -4\cos\left(\dfrac x4\right)-\int \dfrac{-4}{x}\cos\left(\dfrac x4 \right)\,\mathrm dx\\\,\\
& 3\left(-4\cos\left(\dfrac x4\right)+\int\dfrac4x\cos\left(\dfrac x4\right)\,\mathrm dx\right)\\\,\\
&\int\dfrac{4\cos\left(\frac x4\right)}{x}\mathrm dx
\end{align}$$ $\displaystyle \color{darkblue}{uv-\int v\dfrac{\mathrm du}{\mathrm dx}\,\mathrm dx }$

  $\displaystyle\boxed{\displaystyle\,\,-4\cos\left(\dfrac x4\right)+4\int\dfrac{\cos x/4}{x}\,\mathrm dx\,\,}$

  $\displaystyle 3\left[-4\cos\left( \dfrac x4\right) +4\left(\cos\left( \dfrac x4\right)\ln(x)\right)\right]$

  $\displaystyle -12\cos (x/4) + 12\cos (x/4) \ln(x) \rightarrow \text{ wrong.}$

  $\displaystyle\color{darkblue}{\int 3x\sin\left(\dfrac x4\right)\,\mathrm dx}$
  $\quad\quad\quad\quad\displaystyle\int\dfrac{\cos(x/4)}{x}\rightarrow\dfrac1x\int\cos(x/4)\mathrm dx$

  $\displaystyle3\left[-4\cos(x/4)+\dfrac4x\sin(x/4)\right]$

  $\displaystyle -12\cos(x/4)+\dfrac{48}{x}\sin(x/4)$

In the text above is my work done to solve the following question:

    Find the indefinite integral of: $\left[3x\sin\left(\dfrac x4\right)\right]$

The bordered area is the furthest I got (there should be a 3 at the front to multiply the whole equation but I usually remember to add that at the end) The part where I wrote "wrong" is what I thought the answer was, I assumed it would be such. What I'm having troubles with is integrating the $\frac{\cos(x/4)}{x}$. Would I need to make it $\frac{\cos(x/4)}{x}$ and then integrate by parts to get that integral? Thanks in advance, I hope I made some sense in what I'm trying to achieve.  I guess what I'm looking for, is a way to integrate $$ \frac{\cos(x/4)}{x}$$

I think you have made a mistake in the application of integration by parts. Taking $u$ as $x$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ as $\sin \frac{x}{4}$, we should get
\begin{align*} 3 \int {x} \sin \frac{x}{4} \mathrm{d}x &= 3 ( -4x \cos \frac{x}{4}- \int -4 \cos \frac{x}{4} \mathrm{d}x) \\ &= -12x \cos \frac{x}{4} + 48 \sin \frac{x}{4} + C.
\end{align*}

I know this wasn't exactly asked for - but this is a "nice to have" add on in my opinion.

Another method which I've found rather interesting: differentiation under the integral sign (popularised by Feynman).

Consider the integral: $\displaystyle I = \int x\sin{kx} dx$, where $k$ is a constant.

Note that $\displaystyle I = \frac{\partial}{\partial k}\int (-\cos kx)dx = \frac{\partial}{\partial k}(-\frac{1}{k}\sin kx + c) = \frac 1{k^2}\sin kx - \frac xk\cos kx$

Now to get the desired integral, substitute $\displaystyle k= \frac 14$ and remember to multiply by $3$ to get the desired integral (and don't forget the constant):

$\displaystyle \int 3x \sin{\frac x4}dx = 3(\frac 1{(\frac 14)^2}\sin (\frac 14)x - \frac x{\frac 14}\cos (\frac 14)x) + c = 48\sin{\frac x4}- 12\cos{\frac x4} + c$

To address your other sub-question, note that functions like $\frac{\cos x}{x}$ and $\frac{\sin x}{x}$ (give or take a constant multiplier in the argument) do not have elementary anti-derivatives. That means that you need to use special functions - either you have to define a special function to cover this case or compute it in terms of other special functions. Since defining these special functions often involves integrals anyway, the whole process is rather circular.

Long story short, there is no elementary way to compute the indefinite integral you're now asking about. But note that the definite integral is solvable for particular bounds, but that's another story.


How to rephrase these delta epsilon inequalities? When we say that $|f(x) - L| < \epsilon$ can this be restated as $L - \epsilon < f(x) < L+\epsilon$? (removing absolute value to make this easier to understand)

If so I don't know how to restate $0 < |x - c| < \delta$ without the absolute value. because there's a $0 <$ on the end and I don't know how to incorporate it.

I don't think I can ignore the $0$ and just say $c-\delta < x < c+\delta $

No. It is equivalent to$$-\varepsilon+L<f(x)<\varepsilon+L.$$

By the definition of a limit, you know that for every $\epsilon > 0$ there exists a $\delta > 0$ such that for every $x \in D$ ($x \neq c$) exists $0 < | x - c| < \delta$, then $|f(x) - L| < \epsilon$.

Because you know that $\delta > 0$, then you could rewrite $$0 < | x - c| < \delta$$ as $$-\delta < 0 < | x - c| < \delta$$

and from here to get:
$$-\delta < x - c < \delta$$


Is a convex cone a convex polyhedron? Say that I have a convex cone

$C=\{t|Ax = t, x\geq 0\}$.

where $x\in R^n$, and $t\in R^m$, $A\in R^m\times R^n$.

Can I say that this is a convex polyhedron? and why?

EDIT:
Just in order to avoid confusion, the definition of convex polyhedron I am thinking of is: the set of solutions to a finite number of linear equalities and inequalites.

Yes, or more precisely (as polyhedron may imply three dimensions) a convex polytope. You may readily verify that $C$ is convex, and also that its boundary is piecewise a part of a hyperplane.

No.  A cone is not a "polyhedron".  A polyhedron, by definition has planar (straight) sides, not curved sides.


Why isn't finite calculus more popular? I'm reading through Concrete Math, and learning about finite calculus. I've never heard of it anywhere else, and a Google search found very few relevant sources. It seems to me an incredibly powerful tool for evaluating sums, essentially a systematization of the use of telescoping sums. Why isn't it more widely known and used? (Related question)

I'm reading through Concrete Math, and learning about finite calculus. I've never heard of it anywhere else, and a Google search found very few relevant sources.

The name "finite calculus" is unusual.  The traditional term is calculus of finite differences or variants such as difference calculus. A search for those terms will be more productive.

  It seems to me an incredibly powerful tool for evaluating sums, essentially a systematization of the use of telescoping sums. Why isn't it more widely known and used?

It is widely known and used, most obviously in numerical analysis, but also in many other subjects.

Using finite differences involves a considerable complication of the procedures of the calculus. Such an approach may be constructively more satisfactory but may not be appropriate for a first acquaintance with the subject. The technical complications of the epsilon-delta approach are involved enough, and it is how certain things are defined (and it is way easier to do things rigorously with) and furthermore it is the received way of doing things.


Explain proof by contradiction. Would anyone mind explaining me how proof by contradiction works? I have a very vague understanding of it so I always avoid using it when it comes to discrete math. From what I've seen its something that is extremely useful when it comes to proofs.

From what I understand proof by contradiction is the following. Given an example I have to prove it to be true. I take the complete opposite from it and prove it to be false(?), from then on I can assume that the original example is true(?). If anyone would mind correcting my understanding, please do. Also, what would be a good step-by-step guide I can have in my mind when using it?

As Sherlock Holmes said, "When you have eliminated the impossible, whatever remains, however improbable, must be the truth".

If you can prove that not-$P$ is impossible, then you conclude that $P$ must be true.

Proof by contradiction is simple: If $\;\neg A, \Sigma \vdash \bot\;$ then $\;\Sigma \vdash A\;$.

we add the negation of what-is-to-be-shown to our premises
we demonstrate that doing this leads to nonsense (that is, a contradiction).
we conclude the premises entail what-is-to-be-shown.


Coordinate Geometry - Area of a Quadrilateral What is the area in square units, of a quadrilateral whose vertices are $$(5,3), (6,-4), (-3,-2), (-4,7)?$$

I have tried creating the triangles, but didn't know how to find the diagonal. I wanted to try the shoelace method but I thought it only worked for triangles. The answer that was provided is $69$.

Let $A(5,3), B(6,-4), C(-3,-2), D(-4,7)$.

The area of $\triangle ABC$ is given by
$$\frac{1}{2}\sqrt{\left|\vec{AB}\right|^2\left|\vec{AC}\right|^2-\left(\vec{AB}\cdot \vec{AC}\right)^2}.$$
One can get the area of $\triangle{ACD}$ in the same way as above.

Added : Since $$\vec{AB}=(6-5,-4-3)=(1,-7)$$$$\vec{AC}=(-3-5,-2-3)=(-8-5)$$$$\vec{AB}\cdot \vec{AC}=1\cdot(-8)+(-7)\cdot (-5)=27,$$one has
$$\frac{1}{2}\sqrt{\left|\vec{AB}\right|^2\left|\vec{AC}\right|^2-\left(\vec{AB}\cdot \vec{AC}\right)^2}=\frac{1}{2}\sqrt{(1+(-7)^2)\cdot((-8)^2+(-5)^2)-27^2}.$$

A(x1,y1), B(x2,y2), C(x3,y3), D(x4,y4) are verticies of a quadrilateral either convex or concave (one of the internal angle greater than 180 degrees) taken in order,  then we can use the following elegant formula for calculating it's area.

A= 1/2*abs((xi-x3) *(y2-y4) -(x2-x4) *(y1-y3))


Unique Factorization intuitive understanding What is an easy and intuitive explanation to the unique factorisation problem:

If two prime numbers a and b are multiplied, the product (a*b) cannot be divided by a different prime number c.

I think that in this case contradiction is the most intuitive way to understand this property. So let's consider two prime numbers $p$ and $q$. As you mentioned, if you take the product $pq$ the only two positive divisors, not equal to $1$ are $p$ and $q$.

How can you grok this intuitively?

  It was pointed out to me that the following does not provide a necessarily intuitive rationale for why $x$ in this case must divide into $p$ or $q$. I will breakout my intuition further here to show this.
  Well, what would happen if there was a third prime, call it $x$ that divided into $y$? If $x | y \implies x | pq$. But we know that if some prime number divides into a product of two numbers it must divide either one, or the other, (or perhaps both). This is a relatively straightforward factor, and should make sense based on how division is defined!

A prime number cannot be broken down into further factors, by the definition of primes, and as such $pq$ is the unique way to write this as a product of prime factors.

If it could be written in some other manner [$a, b \neq 0$ are prime], say $(ab)q$ then we would of course have $(ab) = p$ and as such, $p$ would not be prime. If we had that $ab = pq$, then we know that $a$ must divide both sides, as must $q$ so we have that: $\frac{b}{q} = \frac{p}{a}$. But the only two factors that divide $b$ are $b$ and $1$. So then we must have that either $q = b$ or $q = 1$. Similarly, we get that $a = p$ or $a = 1$. Because we know that by definition $a, b, p, q$ are prime we must have that $a = p$ and $b = q$. As such, the factorization is unique.

If the uniqueness of the factorization is not quite enough to fully ensure that the intuition is clear, consider if we had a third prime $x$, which did divide into $pq$. We know from the above logic that $x\times c = pq$ must mean one of the following things:

$x = p \implies x$ is not a third divisor.

$x = q \implies x$ is not a third divisor.

$x = pq \implies x$ is not prime.

$x = 1 \implies x$ is not prime.

The remaining options would be that $x | p$ or $x | q$, since it cannot be that $x$ divides some combination of the factors of $p$ and $q$ since the only factors of $p$ and $q$ are one and themselves! However, for $x | p$ then we know that $x = 1$ or $x = p$ which are options above and for $x | q$ we have the same issues.

So this says that we cannot possibly have a third prime $x$ which divides $pq$ unless $x$ is in fact either $p$ or $q$.


Proving that $\sqrt 3\not\in Q(\sqrt[4]2)$ I came across this problem while solving another one. I will show how far I could get on my own:

Suppose that $\sqrt 3 \in Q(\sqrt[4]2)$. Since $Q(\sqrt3)$ is the smallest field containing both $Q$ and $\sqrt3$, thus $Q(\sqrt3)\subset Q(\sqrt[4]2)$.

Now $[Q(\sqrt[4]2):Q] =4 $ and $[Q(\sqrt3):Q]=2$ (I've already proven both), so$[Q(\sqrt[4]2):Q(\sqrt 3)] =2$. Therefore by the definition of degree extension, there exist $p(x)\in Q(\sqrt 3)[X]$ of degree 2 realizing $\sqrt[4]2$, that is, exists: $a,b,c\in Q(\sqrt 3)$ such that:

$$a(\sqrt[4]2)^2+b \sqrt[4]2+c =0.$$

How to proceed now? I should use the fact that those extensions lie in $\mathbb R$ to get a contradiction? I only know that $a\neq 0$, but what about the other coefficients?

EDIT: I could find an argument below, check the answers. Check it out if you agree! The argument is just a continuation of the above one.

I think that I could find an answer. Check it out if it's correct:

As $p(x) = ax^2+bx+c\in Q(\sqrt3)[X]$ is a polynomial of smallest degree killing $\sqrt[4]2$ (by the very definition of being algebraic of degree 2 over $Q(\sqrt3)$), we must have $a\neq 0$. Also we can conclude that $c\neq 0$, otherwise $p(x) = xq(x)$, where degree of $q(x) $ $<$ degree of $p(x)$; and therefore $q(\sqrt[4]2) = 0,$ which again contradicts the minimality of the degree of $p(x)$.

Now we see that $b\neq 0$. If not, noticing that those field extensions all  lie in $\mathbb R$, we find that $a\sqrt 2+c=0$ and hence that $\sqrt2\in Q(\sqrt3), $ which does not happen.

Therefore, $p(x)$ is such that $a,b,c$ are all non-zero. Solving for $\sqrt 2$, we find that $$\sqrt2 = \frac{-b\sqrt[4]2-c}{a} \,.$$

Since $\sqrt 2 \not\in Q(\sqrt 3)$, we must have the denominator of the above equation equal to zero and hence $\sqrt[4]2\in Q(\sqrt 3)$. By the minimality argument of fields, we see that $Q(\sqrt 3) =  Q(\sqrt[4]2)$ (noticed that we have already assumed that $Q(\sqrt 3) \subset Q(\sqrt[4]2)$).  But again this cannot happen, since the degree of $Q(\sqrt[4]2)$ over $Q$ is $4$ and the degree of $Q(\sqrt 3)$ over $Q$ is $2$.

With all this such contradictions, we conclude that $\sqrt 3 \not \in Q(\sqrt[4]2)$.

$\sqrt3=p/q\sqrt[4]{2} \text{ with } gcd(p,q)=1\Rightarrow 9=(p^4/q^4) 2\Rightarrow 9q^4=2p^4\Rightarrow 3|p \text{ and }  2|q.$ Suppose $p=3a$ and $q=2b$. Then $9\times 16 b^4=2\times 81 a^4\Rightarrow 8b^4=9a^4\Rightarrow 3|b\Rightarrow 3|q\Rightarrow gcd(p,q)>1.$


Quotient of monotone functions is monotone? Suppose $f,g$ are monotone( say increasing) and differentiable and nonnegative. Both go from $\mathbb{R}^{\geq 1} \to \mathbb{R} $

Is $\frac{f}{g}$ also monotone ?

$f(x)=2x+\sin x,\ g(x)=x,\ 0\lt x\lt\infty$

Counterexample:
Let $f(x) = e^x+5, g(x) = x$.
$$\frac{f}{g}(x) = \frac{e^x+5}{x}$$

Can you tell me why $\displaystyle{\frac{f}{g}}$ is not necessarily monotone?


Conditions for $\frac{a_{n+1}}{a_n}$ to converge? Given a sequence $a_n \to 0$. I'm looking for a necessary and sufficient condition for $\frac{a_{n+1}}{a_n}$ to converge.

In general, the form $\lim _{n\rightarrow \infty}\frac{a_{n+1}}{a_n}$ will converge if the limit of this is $<1$. If the above holds, then write $a_{n+1}<a_n$ for every index $n$. This is the definition of a monotonically decreasing function.
A necessary condition would be that $a_n \neq 0.$ Monotonically decreasing would be a sufficient condition. Also, non-negativity of the sequence follows from $a_n\neq 0$ since $a_n=0\Rightarrow  a_{n+1}<0.$


How to solve a cryptarithm? Given multiplication is
$$\begin{array}{cccccc}
  &   &   & P & E & N \\
  &   &   & I & N & K \\\hline
  &   & L & K & P & R \\
  & I & R & T & N &   \\
E & A & K & N &   &   \\\hline
L & E & T & T & E & R
\end{array}$$

how to find the values of the letters?

Start from N*N ends in N, so N must be 5 or 6 (can't be 0 or 1 from the second line in the sum).  Then I*N ends in N (and I !=1) says N is 5 and I is odd.  Then K is even and R is 0.  Continue


Checking whether a point is on an elliptic curve I have the point $(2,1)$ and I would like to check whether it is on my elliptic curve.

The elliptic curve is defined by $y^2 = x^3 + 3x + 1$ in $GF(7)$.

My solution to check whether or not the point is on the curve was to substitute the point into the equation:

$$ (1)^2 = (2)^3 + 3(1) + 1. $$

From inspection one may observe the left side does not equal the right side, therefore, the point is not on the curve.

However, I suspect this is incorrect I was just looking for some guidance. Thank you so much.

You have a small error there. Given the curve equation in $\operatorname{GF}(7)$ check the point $P(2,1)$
$$y^2 = x^3 + 3x + 1$$
put $x= 2, y=1$
\begin{align}
(1)^2 &= (2)^3 + 3(\mathbf{\color{blue}{2}}) + 1 & & \pmod{7}\\
1 &= 8 + 6 + 1 & &\pmod{7}\\
1 &= 15 = 2\cdot7+1 & &\pmod{7}\\
1 &= 1 & &\pmod{7}\\
\end{align}
Therefore, the point is on the curve.
Your mistake was $(1)^2 = (2)^3 + 3(\mathbf{\color{red}{1}}) + 1 \pmod{7}$


Calculating the limit of a function? I've been trying to calculate the limit of the following, as $x$ tends to $0$:
$$f(x) = \left(\frac{e^x-1}x\right)^{1/x}$$

I've tried writing it as $e$ raised to the power of its log, but I am unable to solve it. Any tips on how to proceed will be appreciated!

$\lim_{x \to 0}\left( \dfrac{e^x-1}x \right)^\frac1x=\lim_{x \to 0}\left( \dfrac{(1+x+\frac{x^2}2+o(x^2))-1}x \right)^\frac1x=\lim_{x \to 0}\left( {1+\frac{x}2+o(x)} \right)^\frac1x=\lim_{x \to 0}e^{\frac1x \ln\left( {1+\frac{x}2+o(x)} \right)}=\lim_{x \to 0}e^{\frac1x \left( {\frac{x}2+o(x)} \right)}=\sqrt e$


If $\sigma=(a_1 a_2 ... a_n)$ and $|\sigma|$ is odd, then what is $\sigma^2$? I'm trying to understand the way to infer the power of a permutation.

If $\sigma=(a_1 a_2 ... a_n)$ is a $k$-cycle and $k=|\sigma|$ is odd, then how can I infer what $\sigma^2$ is?

HINT: Look at some examples:

$$\begin{align*}
&(1)^2=(1)\\
&(123)^2=(132)\\
&(12345)^2=(13524)\\
&(123457)^2=(1357246)
\end{align*}$$

There’s a pretty obvious pattern developing here. We can guess, for instance, that

$$(a_1a_2a_3a_4a_5a_6a_7a_8a_9)^2=(a_1a_3a_5a_7a_9a_2a_4a_6a_8)\;.$$

Try to express this pattern in a general form and then prove that it actually holds for all cycles of odd length.

It has nothing to see with the length of the cycle:
$$\sigma^2=(a_1\,a_3\,a_5\dots a_{2i-1\bmod n}\dots a_{2n-1\bmod n}).$$


Can I prove this using modular congruence? $3 \mid 5^n + 2\cdot11^n$ [EDIT] I'd prove this by induction, and I'd been thinking in how prove that using congruence. Please help.

[EDIT]
I can got to in (−1)^n+(−1)^n+1 mod 3, but I don't know if this it's a congruent mod 3 AAAAAAAAA

If you look at the remainders when each number in your expression is divided by $3$ then,

$5^n + 2\cdot11^n \equiv (-1)^n + (-1)\cdot (-1)^n \equiv (-1)^n + (-1)^{n+1} \equiv 0 \mod 3$

This is because if $n$ is odd, $(-1)^n = -1$ and $(-1)^{n+1} = 1$ hence their sum is zero which is divisible by three. Similarly if $n$ is even, $(-1)^n = 1$ and $(-1)^{n+1} = -1$ and their sum is again zero which is divisible by $3$.


Fixed-point iteration for $x=\tan(x)$ I've been trying to find an accurate $g(x)$ in order to find a solution for $x=\tan(x)$ in the interval $[4,5]$. However, no matter what, all of them end up converging to zero which is not the answer. So far, I've tried:

$$g(x)=\tan(x)$$
$$g(x)=\tan^{-1}(x)$$
$$g(x)=\sin^{-1}(x\cos(x))$$
$$g(x)=\sqrt{x\tan(x)}$$

I thought maybe I was having trouble with the interval since I was working with angles, but even if I just use $[\pi/45,\pi/36]$, the functions still converge to 0. Can anybody help me out? I need to solve this problem using fixed-point iteration.

The $\arctan$ function is more nicely behaved than $\tan$. Let
$$g(x)=\arctan x+\pi.$$
Be sure your calculator is in radian mode, and start with $x_0$ not too far from $4.5$. Use the iteration $x_{n+1}=g(x_n)$. Convergence should be acceptably quick.

Added: The reason we get nice behaviour with our choice of $g(x)$ is that near the root, the derivative of $\arctan x$ has rather small absolute value. By way of contrast, the derivative of $\tan x$ near the root is much larger than $1$. That means that with the choice $g(x)=\tan x$, the root is a repelling fixed point.


Set-theoretic equality of splitting fields within a fixed algebraic closure Let $F$ be a field and let $f(x)\in F[x]$ be a polynomial. Recall the following two facts:

(1) algebraic closures are unique up to isomorphism

(2) splitting fields are unique up to isomorphism

Fix an algebraic closure $\overline F$ of $F$. Is it true that any two (necessarily isomorphic) splitting fields $E,E'\subseteq \overline F$ for $f$ are equal $as$ $sets$ in addition to being isomorphic?

My intuition tells me that this should be the case: fixing an algebraic closure allows us to fix roots of $f$ within this particular ambient field, and since any splitting field is the smallest field containing these roots, any two such fields must actually be equal as sets, in addition to being isomorphic. Does this sound about right? Is this totally trivial?

Yes, this is correct.  The only splitting field for $f$ over $F$ in $\overline{F}$ is the subfield of $\overline{F}$ generated by $F$ and all the roots of $f$ in $F$.

(To be clear, for this to be true, by "splitting field for $f$ over $F$ in $\overline{F}$" we must mean a splitting field for $f$ over $F$ which is a subfield of $\overline{F}$ (so the field operations are the same) and whose copy of $F$ is the same as the copy of $F$ inside $\overline{F}$.)

I don't think they are equal as sets. Remember that in our way to get splitting field for $\;f(x)\in  F[x]\;$ , we first get (assuming $\;f\;$ is irreducible, otherwise we take one of its irreducible factors) the quotient ring (field) $\;K:=F[x]/\langle f(x)\rangle\;$ .

Here, we already have no more $\;F\;$ but an isomorphic copy of $\;F\;$ within $\;K\;$ , and there could be several ways to obtain that copy...


Why is the area under a curve the integral? I understand how derivatives work based on the definition, and the fact that my professor explained it step by step until the point where I can derive it myself.

However when it comes to the area under a curve for some reason when you break it up into an infinite amount of rectangles, magically it turns into the anti-derivative. Can someone explain why that is the definition of the integral and how Newton figured this out?

First: the integral is defined to be the (net signed) area under the curve. The definition in terms of Riemann sums is precisely designed to accomplish this. The integral is a limit, a number. There is, a priori, no connection whatsoever with derivatives. (That is one of the things that makes the Fundamental Theorems of Calculus such a potentially surprising things).

Why does the limit of the Riemann sums actually give the area under the graph? The idea of approximating a shape whose area we don't know both from "above" and from "below" with areas we do know goes all the way back to the Greeks. Archimedes gave bounds for the value of $\pi$ by figuring out areas of inscribed and circumscribed polygons in a circle, knowing that the area of the circle would be somewhere between the two; the more sides to the polygons, the closer the inner and outer polygons are to the circle, the closer the areas are to the area of the circle.

The way Riemann tried to formalize this was with the "upper" and "lower" Riemann sums: assuming the function is relatively "nice", so that on each subinterval it has a maximum and a minimum, the "lower Riemann sum" is done by taking the largest "rectangle" that will lie completely under the graph by looking at the minimum value of the function on the interval and using that as height; and the "upper Riemann sum" is done by taking the smallest rectangle for which the graph will lie completely under it (by taking the maximum value of the function as the height). Certainly, the exact area under the graph on that interval will be somewhere between the two. If we let $\underline{S}(f,P)$ be the lower sum corresponding to some fixed partition $P$ of the interval, and $\overline{S}(f,P)$ be the upper sum, we will have that
$$\underline{S}(f,P) \leq \int_a^b f(x)\,dx \leq \overline{S}(f,P).$$
(Remember that $\int_a^bf(x)\,dx$ is just the symbol we use to denote the exact (net signed) area under the graph of $f(x)$ between $a$ and $b$, whatever that quantity may be.)

Also, intuitively, the more intervals we take, the closer these two approximations (one from below and one from above) will be. This does not always work out if all we do is take "more" intervals. But one thing we can show is that if $P'$ is a refinement of $P$ (it includes all the dividing points that $P$ had, and possibly more points) then
$$\underline{S}(f,P)\leq \underline{S}(f,P')\text{ and } \overline{S}(f,P')\leq \overline{S}(f,P)$$
so at least the approximations are heading in the right direction. To see why this happens, suppose you split one of the subintervals $[t_i,t_{i+1}]$ in two, $[t_i,t']$ and $[t',t_{i+1}]$. The minimum of $f$ on $[t_i,t']$ and on $[t',t_{i+1}]$ are each greater than or equal to the minimum over the whole of $[t_i,t_{i+1}]$, but it may be that the minimum in one of the two bits is actually strictly larger than the minimum over $[t_i,t_{i+1}]$. The areas we get after the split can be no smaller, but they can be larger than the ones we had before the split. Similarly for the upper sums.

So, let's consider one particular sequence of partitions: divide the interval into 2 equal parts; then into 4; then into 8; then into 16; then into 32; and so on; then into $2^n$, etc. If $P_n$ is the partition that divides $[a,b]$ into $2^n$ equal parts, then $P_{n+1}$ is a refinement of $P_n$, and so we have:
$$\underline{S}(f,P_1) \leq\cdots \leq \underline{S}(f,P_n)\leq\cdots \leq\int_a^b f(x)\,dx \leq\cdots \leq\overline{S}(f,P_n)\leq\cdots \leq \overline{S}(f,P_2)\leq\overline{S}(f,P_1).$$

Now, the sequence of numbers $\underline{S}(f,P_1) \leq \underline{S}(f,P_2)\leq\cdots \leq \underline{S}(f,P_n)\leq\cdots$ is increasing and bounded above (by the area). So the numbers have a supremum; call it $\underline{S}$. This number is no more than $\int_a^b f(x)\,dx$. And the numbers $\overline{S}(f,P_1) \geq \overline{S}(f,P_2)\geq\cdots \geq \overline{S}(f,P_n)\geq\cdots$ are decreasing and bounded below, so they have a minimum; call this $\overline{S}$; again, it is no less than $\int_a^bf(x)\,dx$. So we have:
$$\lim_{n\to\infty}\underline{S}(f,P_n) = \underline{S} \leq \int_a^b f(x)\,dx \leq \overline{S} = \lim_{n\to\infty}\overline{S}(f,P_n).$$
What if we are lucky? What if actually we have $\underline{S}=\overline{S}$? Then it must be the case that this common value is the value of $\int_a^b f(x)\,dx$. It just doesn't have a choice! It's definitely trapped between the two, and if there is no space between them, then it's equal to them.

What Riemann proved was several things:

If $f$ is "nice enough", then you will necessarily get that $\underline{S}=\overline{S}$. In particular, continuous functions happen to be "nice enough", so it will definitely work for them (in fact, continuous functions turn out to be "very nice", not just "nice enough").
If $f$ is "nice enough", then you don't have to use the partitions we used above. You can use any sequence of partitions, so long as the "mesh size" (the size of the largest subinterval in the partition) gets smaller and smaller, and has limit of $0$ as $n\to\infty$; if it works for the partitions "divide-into-$2^n$-equal-intervals", then it works for any sequence of partitions whose mesh size goes to zero.

So, for example, we can take $P_n$ to be the partition that divides $[a,b]$ into $n$ equal parts, even though $P_{n+1}$ is not a refinement of $P_n$ in this case.
In fact, you don't have to do $\underline{S}(f,P)$ and $\overline{S}(f,P)$. For the partition $P$, just pick any rectangle that has as its height any value of the function in the subinterval (that is, pick an arbitrary $x_i^*$ in the subinterval $[t_i,t_{i+1}]$, and use $f(x_i^*)$ as the height). Call the resulting sum $S(f,P,x_1^*,\ldots,x_n^*)$. Then you have
$$\underline{S}(f,P) \leq S(f,P,x_1^*,\ldots,x_n^*)\leq \overline{S}(f,P)$$
because $\underline{S}(f,P)$ is computed using the smallest possible values of $f$ throughout, and $\overline{S}(f,P)$ is computed using the largest possible values of $f$ throughout. But since we already know, from 1 and 2 above, that $\underline{S}(f,P)$ and $\overline{S}(f,P)$ have the same limit, then the sums $S(f,P,x_1^*,\ldots,x_n^*)$  also get squeezed and must have that same limit, which equals the integral.

In particular, we can always take the left endpoint (and get a "Left Hand Sum") or we can always take the right endpoint (and get a "Right Hand Sum"), and you will nevertheless get the same limit.

So in summary, you can pick any sequence of partitions, whichever happens to be convenient, so long as the mesh size goes to $0$, and you can pick any points on the subintervals (say, ones which make the calculations simpler) at each stage, and so long as the function is "nice enough" (for example, if it is continuous), everything will work out and the limit will be the number which must be the value of the area (because it was trapped between the lower and upper sums, and they both got squeezed together trapping the limit and the integral both between them).

Now, (1) and (2) above are the hardest part of what Riemann did. Don't be surprised if it sounds a bit magical at this point. But I hope that you agree that if the lower and upper sums for the special partitions have the same limits then that limit must be the area that lies under the graph.

Thanks to that work of Riemann, then (at least for continuous functions) we can define $\int_a^b f(x)\,dx$ to be the limit of, say, the left hand sums of the partitions we get by dividing $[a,b]$ into $n$ equal parts, because these partitions have mesh size going to $0$, we can pick any points we like (say, the left end points), and we know the limit is going to be that common value of $\underline{S}$ and $\overline{S}$, which has to be the area. So that, under this definition, $\int_a^b f(x)\,dx$ really is the net signed area under the graph of $f(x)$. It just doesn't have a choice but to be that, when $f$ is "nice enough".

Second, the area does not turn into "the" antiderivative. What happens is that it turns out (perhaps somewhat magically) that the area can be computed using an antiderivative. I'll go into some more details below.

As to how Newton figured this out, his teacher, Isaac Barrow, was the one who discovered there was a connection between derivatives and tangents; some of the basic ideas were his. They came from studying some simple functions and some simple formulas for tangents he had discovered.

For example, the tangents to the parabola $y=x^2$ were interesting (there was generally geometric interest in tangents and in "squaring" regions, also known as finding the "quadrature" of a region, that is, finding a way to construct a square or rectangle that had the same area as the region you were considering), and led to associating the parabola $y=x^2$ to lines of the form $y=2x$. It does not take too much experimentation to realize that if you look at the area under $y=2x$ from 0 to a, you end up with $a^2$, establishing a connection. Barrow did this with arguments with infinitesimals (which were a bit fuzzy and not set on entirely correct  and solid logical foundation until well into the 20th century), which were generally laborious, and only for some curves. When Newton extended Barrow's methods to more general curves and tangents, he also extended the discovery of the connection with areas, and was able to prove what is essentially the Fundamental Theorem of Calculus.

Now, here is one way to approach the connection. We want to figure out the value of, say,
$$\int_0^a f(x)\,dx$$
for some $a$. This can be done using limits and Riemann sums (Newton and Leibniz had similar methods, though not set up quite as precisely as Riemann sums are). But here is an absolutely crazy suggestion: suppose you can find a "master function" $\mathcal{M}$, which, when given any point $b$ between $0$ and $a$, will give you the value of $\int_0^b f(x)\,dx$. If you have such a master function, then you can use it to find the value of the integral you want just by taking $\mathcal{M}(a)$!

In fact, this is the approach Barrow had taken: his great insight was that instead of trying to find the quadrature a particular area, he was trying to solve the problem of squaring several different (but related) areas at the same time. So he was looking for, for instance, a "master function" for the region was like a triangle except that the top was a parabola instead of a line (like the area from $0$ to $a$ under $y=x^2$), and so on.

On its face, this is a completely ludicrous suggestion. It's like telling someone who is trying to know how to get from building A to building B that if he only memorizes the map for the entire city first, then he can use that knowledge to figure out how to get form A to B. If we are having trouble finding the integral $\int_0^a f(x)\,dx$, then the "master function" seems to require us to find not just that area, but also all areas in between! It's like telling someone who is having trouble walking that he should just run very slowly when he wants to walk.

But, again, the interesting thing is that even though we may not be able to say what the "master function" is, we can say how it changes as b changes (remember, $\mathcal{M}(b) = \int_0^b f(x)\,dx$ is a number that depends on $b$, so $\mathcal{M}$ is a function of $b$). Because figuring out how functions change is easier than computing their values (just think about derivatives, and how we can easily figure out the rate of change of $\sin(x)$, but we have a hard time actually computing specific values of $\sin(x)$ that are not among some very simple ones). (This is also something Barrow already knew, as did Newton).

For "nice functions" (if $f$ is continuous on an interval that contains $0$ and $a$), we can do it using limits and some theorems about "nice" functions:
Using limits, we have:
\begin{align}
\lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}}{h} &= \lim_{h\to 0} \frac{1}{h}\left(\int_0^{b+h}f(x)\,dx - \int_0^bf(x)\,dx\right)\\\
&= \lim_{h\to 0}\frac{1}{h}\int_b^{b+h}f(x)\,dx.
\end{align}
Since we are assuming that $f$ is continuous on $[0,a]$, it is continuous on the interval with endpoints $b$ and $b+h$ (I say it this way because $h$ could be negative). So it has a maximum and a minimum (continuous function on a finite closed interval). Say the maximum is $M(h)$ and the minimum is $m(h)$. Then $m(h) \leq f(x) \leq M(h)$ for all $x$ in the interval, so we know, since the integral is the area, that
$$hm(h) \leq \int_b^{b+h}f(x)\,dx \leq hM(h).$$
That means that
$$m(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq M(h)\text{ if $h\gt 0$}$$
and
$$ M(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq m(h)\text{ if $h\lt 0$.}$$

As $h\to 0$, the interval gets smaller, the difference between the minimum and maximum value gets smaller. One can prove that both $M$ and $m$ are continuous functions, and that $m(h)\to f(b)$ as $h\to 0$, and likewise that $M(h)\to f(b)$ as $h\to 0$. So we can use the Squeeze Theorem to conclude that since the limit of $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ is squeezed between two functions that both have the same limit as $h\to 0$, then $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ also has a limit as $h\to 0$ and is in fact that same quantity, namely $f(b)$. That is
$$\frac{d}{db}\mathcal{M}(b) = \lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}(b)}{h} = \lim_{h\to 0}\frac{1}{h}\int_{b}^{b+h}f(x)\,dx = f(b).$$

That is: when $f$ is continuous, the "Master function" for areas turns out to have a rate of change equal to $f$. This is not that crazy, if you think about it: how is the area under $y=f(x)$ from $x=0$ to $x=b$ changing? Well, it's changing by whatever $f$ is.

This means that, whatever the "Master function" turns out to be, it will be an antiderivative of $f(x)$.

We also know, because we are very good with derivatives, that if $\mathcal{F}(x)$ and $\mathcal{G}(x)$ are two functions, and $\mathcal{F}'(x) = \mathcal{G}'(x)$ for all $x$, then $\mathcal{F}$ and $\mathcal{G}$ differ by a constant: there exists a constant $k$ such that $\mathcal{F}(x) = \mathcal{G}(x)+k$ for all $x$.

So, we know that the "Master function" is an antiderivative. If, by some sheer stroke of luck, we happen to find any antiderivative $F(x)$ for $f(x)$, then we know that the only possible difference between $\mathcal{M}(b)$ and $F(b)$ is a constant. What constant? Well, luckily we know one value of $\mathcal{M}(b)$: we know that $\mathcal{M}(0) = \int_0^0f(x)\,dx$ should be $0$.  So, $M(0) = 0 = F(0)-F(0)$, which means the constant has to be $-F(0)$. That is, we must have $M(b) = F(b)-F(0)$ for all $b$.

So, if we find any antiderivative $F$ of $f$, then $\mathcal{M}(b) = F(b)-F(0)$ is in fact the "Master function" we were looking for, the one that gives all the integrals between $0$ and $a$, including $0$ and including $a$. So that we have that two very different processes (computing areas using limits of Riemann sums, and derivatives) are connected: if $f(x)$ is continuous, and $F(x)$ is any antiderivative for $f(x)$ on $[0,a]$, then
$$\int_0^a f(x)\,dx = \mathcal{M}(a) = F(a)-F(0).$$

But the integral did not "magically turn" into an antiderivative. It's that the "Master function" which can be used to keep track of all integrals of $f(x)$ has rate of change equal to $f$, which gives us a "back door" to computing integrals.

Newton was able to prove this because he had the guide of Barrow's insight that this was happening for the functions he worked with. Barrow's insight was achieved because he had the brilliant idea of trying to come up with a "Master function" instead of trying to rectify lots of different areas one at a time, and he noticed the connection because he had already worked with tangents/derivatives for those functions. Leibniz likewise had access to Barrow's ideas, so the connection between the two was also known to him.

heres a very simple explanation of how an integral works. Forget about things needed for the proof. This is just for the concept

take the function f'(x) and its antidervative, f(x)

We can find the area under a graph by taking the average of all the y values and multiplying by  Delta X, creating a rectangle of equal area.

 (1)Area = Yaverage of f'(x)* DeltaX
  average y for f'(x) is average slope of f(x)

finding average slope of f(x) is easy. Simply go to the original function and do the average slope calculation

 (2)average slope =(y1-y2)/(deltaX)

Now substitute formula 2 into 1:

  area of f'(x) = (y1-y2)/delaX  * deltaX
  area of f'(x) = (y1-y2)

y1 = f(x1)

 area of f'(x)=f(x1)-f(x2)

Took me a while to figure this out. IMO the mean sum of tiny squares hinders understanding. mean value theorem is good to find average slope. Its convoluted and imprecise though when people say there is a value c that does this or that. Irrelevant to finding the average slope. Just supply the formula for finding average slope. Best of luck

Such a simple concept to be so over complicated. I hope they change the way it is taught in canadian schools


Cancelling differential terms Can we cancel two differential terms while they are in a ratio. For example if we have (dx/dt) / (dy/dt), can we just directly cancel dt by dt and write it as dx/dy. I mean is is this step allowed?

Yes
The reason for that is quite simple because in differential calculus the Letter "d" denotes a certain amount of change in a quantity
So "dx/dy" means "change in x with respect to change in y". By that logic, "dt" in both terms represent change in quantity "t". Being the same "term" they can be cancelled to create a ratio which gives the proportion of change in one quantity with respect to the another.


$\lim_{x\to 0^+}[1+[x]]^{2/x}$,where [x] denotes greatest integer function less than or equal to x Find the value of $$\lim\limits_{x\to 0^+}[1+[x]]^{\frac2x}$$where $[x]$ denotes greatest integer function less than or equal to $x$.

My attempt:
I calculated $[1+[x]]$ to be $1$ as $x\to 0^+$.  Now I am stuck.  Please help me.

Substitute $y=\frac1x$, so as $x\to0^{+}, y\to\infty$.
So we get $$\lim\limits_{y\to\infty}[1+[\frac{1}{y}]]^{2y}=\lim\limits_{y\to\infty}([1+[\frac1y]]^{y})^2=1^{2.\infty}=1$$


systems of first degree equation, mathematics. which solutions of the
system?
observation: this system is determined.
  https://i.stack.imgur.com/QeLWz.png

By substituting $ x = -4y$ in the rest of the two equations you will get the below as the answer.The solution is $$x=0,y=0,z=0$$


How am I to interpret this result? Prove of give counter example: Let A and B be sets.

$$ A \backslash ( A \backslash B) = B \backslash ( B \backslash A) $$

An attempt at a proof:

If $ x \in  A \backslash ( A \backslash B)$ then $x \in A$ and $ x \notin A \backslash B$.

But if $ x \notin A \backslash B$ then either $x \notin A,B$ or $x \in B$ which is a contradiction.

Am I wrong?  Is there an obvious counter example?

$$A\backslash (A\backslash B)=A\cap (A\cap B^c)^c=A\cap(A^c\cup B)=(A\cap A^c)\cup (A\cap B)=A\cap B,$$

$$B\backslash (B\backslash A)=B\cap (B\cap A^c)^c=B\cap(B^c\cup A)=(B\cap B^c)\cup (B\cap A)=B\cap A= A\cap B,$$

and thus both are equal.

To answer to your question:

There is no contradiction. If $x\in A\backslash (A\backslash B)$, then $x\in A$ but $x\notin A\backslash B$. Therefore $x\in B$ too. Indeed, if $x\notin A\backslash B$, then $x\in A^c$ or $x\in B$. But if $x\in A$, then $x$ can't be in $A^c$, therefore $x$ must be in $B$. We conclude that $x\in A$ and $x\in B$, what we can traduce by $x\in A\cap B$.

There is no contradiction.   In point of order, $x\notin A\setminus B$ means either $x\notin A$ or $x\in B$.   That is not an "and".

So if we take any $x$ in $A\setminus(A\setminus B)$, then we have "any $x$ in A except those not also in $B$".  Which is simply "any $x$ only in both $A$ and $B$".

$$A\setminus(A\setminus B) = A\cap B$$


Evaluating the decimal equivalent of binary numbers in; sign and magnitude, one's complement and two's complement For example, i have this binary number : 1011 1101

Now i wish to evaluate the decimal equilant using sign and magnitude, one's complement and two's complement.

Now for sign and magnitude, i know the answer would be like, take the left most bit as sign, in this case it is 1, so negative number. Take the rest of the bits as magnitude, which is 61. So the decimal equilant is -61.

How about the one's and two's complement. Do i need to flip all the bits, or do i need to keep the left most bit as it is, and flip the rest?.

Like for ones complement of the binary number above, would it be 11000010 or 01000010.

There are a plethora of (poorly-tagged) questions on math.SE about twos-complement binary numbers. In summary, the usual definition of twos complement is, to convert a negative number $-N$ to twos complement, you write $N$ (the magnitude of the number) in binary, "flip" (take the complement of) all the bits, then add $1$ (a number with $1$ in the least significant place and $0$ everywhere else).

The conversion in the reverse direction is: "flip" all the bits
and add one, treat the result as a positive number, convert it to decimal,
and then flip the sign of the number so that you get back the original
negative number. So, for example, $1011\,1101$
in eight-bit twos-complement is $-0100\,0010_2 +1 = -0100\,0011_2 = -67_{10}.$

For ones complement you just flip all the bits. So, for example, $1011\,1101$
in eight-bit ones-complement is $-0100\,0010_2 = -66_{10}.$

Personally, I prefer to think of a twos-complement negative number $-N$ in $w$ bits as simply $2^w - N$. The ones-complement representation of $-N$ is then
$(2^w - 1) - N$.


Show that $\sum\limits^{\infty}_{n=1} (-1)^n \frac {x^2+n} {n^2}$, $x \in \mathbb R$, converges uniformly on every bounded interval Show that  $\sum\limits^{\infty}_{n=1} (-1)^n \frac {x^2+n} {n^2}, x \in \mathbb R$ converge uniformly on every bounded interval $I \subseteq \mathbb R$.

I've already shown that the series does not converge absolutely for $x \in \mathbb R$. I guess, I proved this to conclude that the series doesn't have a majorant series which would imply that the series converged uniformly (viewed as a series of functions) and absolutely as a tradional series for $x \in \mathbb R$.

I know I must show that the $s_k = \sum^{k}_{n=1} (-1)^n \frac {x^2+n} {n^2}$ converge uniformly, but I haven't had any luck in showing that.

Well, the simpler the better...
$$
\sum_{n\geqslant N}(-1)^n \frac {x^2+n} {n^2}
=
x^2\cdot\sum_{n\geqslant N}\frac{(-1)^n}{n^2}
+
\sum_{n\geqslant N}\frac{(-1)^n}n
$$
Edit: Both series in the RHS are alternating series hence the absolute value of a given rest is bounded by the absolute value of the first term in the rest, that is,
$$
\left|\sum_{n\geqslant N}\frac{(-1)^n}{n^2}\right|\leqslant\frac1{N^2},\qquad
\left|\sum_{n\geqslant N}\frac{(-1)^n}n\right|\leqslant\frac1N.
$$


Given a range [a, b], how to find the x middle numbers? Given a range [$a$,$b$], how can I find the $x$ middle numbers?

For example:

[$1$,$10$]

Now I know that the middle $2$ numbers start with "$5$", but is there any way I can find the starting number, which in this case is "$5$", given a range doing some basic calculation$?$

If you meant the middle point of the interval then it is $\;M:=\cfrac{a+b}2\;$ . Why? Because (check this)

$$(1)\;\;a\le M\le B\;\;,\;\;\;(2)\;\;M-a=b-M$$

In fact, (2) above is the explanation for the formula for $\;M\;$ in the first line.

If $a$ and $b$ are integers, and you want the $x$ integers closest to the centre of $[a,b]$, then these are
$$\frac{a+b-x+1}{2},\frac{a+b-x}{2}+1,\ldots,\frac{a+b+x-1}{2}$$

assuming $a+b-x+1$ is even.

If $a+b-x+1$ is odd, then there is no unique solution. What are the two middle numbers in the range $[1,9]$, for instance?


% of % - Please Help Me Prove My Friend Wrong Here is the situation:  My friend and I are at an impasse.  I believe I'm correct, but he's so damn stubborn he won't believe me.  Also, I'm not the most articulate at explaining things.  Hopefully some of you guys can help me explain this to him in a way he'll understand.  Here is the problem:

A DVD is either at his parent's house or his own.  The probability that it's at his house is 30%.
If the DVD is at his own house, there is a 90% chance it's on the porch, and a 10% chance it's in the living room.
What is the % chance the DVD is on the porch?

My friend says you take 90% of 30% which is 27 and that is the % chance it's on the porch.  Is this correct?  I don't believe so.

I believe that regardless of where the DVD is, the chance of it being anywhere in his house is still 30% overall.  Location inside his house won't change those odd because the porch and the living room are both part of the house.  If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.

Now, if you rephrase the question and ask, "The DVD Is either at my parents, my porch, or my living room.  What is the % chance it's on my porch?", the answer is 33%.  If there are three places it could be, then there is 33.333% chance it's on the porch.  Even if it's a 90% chance it's at his house, if there are only three places it can be, it remains the same.

I think the correct way of answering the question is:  There is a 30% chance the DVD is at my house.  If it is at my house, there is a 90% chance it's on my porch.  They are two separate odds and you can't take a percentage of the overall odds since the locations are inside the house.

Is this correct or am I wrong?  And regardless, please give me your explanation.

Your friend is correct and I'll give you an experiment you can try:

Take two boxes labeled "my house" and "my friend's house" and in the first box put two bags labeled "porch" and "living room". You will also need a marble and a die. Take one die and roll it (we're going to approximate 70% ~ 2/3 here). If it's a 1,2,3 or 4, put your hand in the "my house" box but don't pick a bag yet. If it's a 5 or 6 put it anywhere in the "friend's house" box. If your hand is in the "my house" box then you need to roll again to figure out which bag to put it in. If it's a 1,2,3,4 or 5 (let's say 90% ~ 5/6) put it in the "porch" box, and if it's a 6 put it in the "living room" box. Repeat this exercise until you have a feel for how often it lands in "my porch." Record the trials and see what the odds are. You should be convinced now.

Your reasoning is incorrect here:

  If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.

No, you changed things. There is not a 90% chance it's on the porch. If - if! - it is in your house, then (and only then!) there is a 90% chance it's on the porch. The magic of the 27% comes from the fact that mathematically - and experimentally, as I hope the above box/bag/marble exercise shows - we know how the "in your house" 30% and the "on the porch" 90% interact. Namely, they interact multiplicatively.

How about this? There is a 30% chance you'll go to New York and a 100% chance you'll go to the Empire State Building if you go to New York (because why else would you go to New York? kidding...). Does this mean there's a 100% chance you'll go to the Empire State Building? Well, since you have to go to NY to go to the ESB, that would mean there's a 100% you'll go to New York - and now we're being contradictory! So this interpretation makes no sense and is never what we mean mathematically or in plain English.

  "The DVD Is either at my parents, my porch, or my living room. What is the % chance it's on my porch?", the answer is 33%.

This is definitely wrong. I'm pretty sure you either have ebola or you don't have ebola. Up to you whether you need to call 911 and get yourself quarantined right away because there's a 50% chance you have ebola. Or perhaps the doctors gave your sick relative 6 months to live, but your relative might live a year or two years or three years or four years or five years, which means there's at least an 86% chance the doctor is wrong.

Now put 10 red m&ms in a bag and 1 blue one. Grab one without looking. Since there's two possibilities, there's a 50% chance it's a blue one, right? So I'll bet you a dollar that it's red and you bet me a dollar that it's blue, and we'll see who's paying for lunch later.

If you have already dictated the odds, you can't then change them and treat the house and places as isolated cases. Your friend is right.


$\prod_{n=1}^{+\infty} \left(1-\frac{1}{2n^2}\right)>0$? How to prove that the infinite product $\prod_{n=1}^{+\infty} \left(1-\frac{1}{2n^2}\right)$ is positive ?

Thanks

Hint: Only the lower limit is a problem. Look e.g. at the product for $n\geq 2$, you could prove:
$$ \prod_{n\geq 2} (1-\frac{1}{2 n^2}) \geq  \prod_{n\geq 2} (1-\frac{1}{ n^2})=\prod_{n\geq 2} \frac{(n-1)(n+1)}{ n \cdot n}=\frac12$$
(the last being a telescopic product). So your product is $\geq \frac14$.

$1-\dfrac{1}{n^2}<1-\dfrac{1}{2n^2}<\left(1-\dfrac{1}{n^2}\right)^2$,So you can see Its limit exist.

Sorry For I cannot use LaTex and My poor English.


Find the solution for the spring-mass problem $y′′+9y=\cos(3t)$. Solve with initial conditions $y(0) = 0$, $y′ (0) = 0$. Using Laplace transform I first took the Laplace transform of each part then getting $s^{2}Y+9Y=\frac{s}{s^{2}+9}$ then solving for Y, I got $Y=\frac{s}{(s^{2}+9)^{2}}$ but don't know how to simplify that to something that can be transformed back.

If you let k=3, then you can use that fact that L.T. t*sin(kt) = 2ks/(s^2+k^2)^2  or in your case, 2*3*s/(s^2+3^2)^2  -> 6s/(s^2+3^2)^2    which if you invert should lead to (1/6)t*sint(3t). Try that.