Prove $f_n(x)=\sqrt{x+\frac{1}{n}}-\sqrt{x}$ converges uniformly on $\mathbb{R}$. From Spivak's Calculus 24-2:

  Prove $f_n(x)=\sqrt{x+\frac{1}{n}}-\sqrt{x}$  converges uniformly on
  $\mathbb{R}$.

My issue is whether this makes sense since square root of negative numbers are not yet defined.  In the book complex numbers have not been covered.  In any case can anyone offer up a solution?

Correct me if wrong:

Let

$f_n (x):= \sqrt{x+1/n} -\sqrt{x}$, $x\ge 0.$

$|\sqrt{x+1/n}-\sqrt{x}| =$

$(1/n)|\dfrac{1}{\sqrt{x+1/n} +\sqrt{x}}| \le$

$(1/n)\sqrt{n} = \dfrac{1}{\sqrt{n}}.$

Let $\epsilon \gt 0$ be given.

Choose $n_0 \gt \dfrac{1}{\epsilon^2}.$

For $n \ge n_0 :$

$|f_n(x)| \lt \dfrac{1}{\sqrt{n}} \le \dfrac{1}{\sqrt{n_0}} \lt \epsilon,$ I.e.

$f_n(x)$ converges uniformly to $0$  on $\mathbb{R_+}.$

If you use
$(1+x/2)^2
=1+x+x^2/4
\gt 1+x
$,
you get
$\sqrt{1+x}
\lt 1+x/2
$.

Therefore

$\begin{array}\\
\sqrt{x+1/n}-\sqrt{x}
&=\sqrt{x}\sqrt{1+1/(nx)}-\sqrt{x}\\
&\lt\sqrt{x}(1+1/(2nx))-\sqrt{x}\\
&\lt\sqrt{x}(1+1/(2nx))-\sqrt{x}\\
&=\sqrt{x}+1/(2n\sqrt{x})-\sqrt{x}\\
&\dfrac1{2n\sqrt{x}}\\
\end{array}
$

and this should gove you what you need.


Calculate an unconditional probability given only the conditional probability Let X and Y be jointly absolutely continuous random variables. Suppose X~Exponential(2) and P(Y > 5 | X = x) = $$e^{-3x}$$. Compute P(Y > 5).

I've computed P(Y > 5) as follows:

P(Y > 5) = P(Y > 5 | X = $x_1$) + ... + P(Y > 5 | X = $x_n$) = $\int_0^\infty e^{-3x} dx = \frac13$

However, I'm not sure the answer is correct, since I did not have to use the fact that X~Exponential(2). Can someone help?

If $P\{Y > 5\mid X = x\} = \exp(-3x)$, then
$$P\{Y > 5\} = \int_{-\infty}^\infty P\{Y > 5\mid X = x\} f_X(x)\,\mathrm dx
= \int_0^\infty \exp(-3x) \cdot 2\exp(-2x)\,\mathrm dx$$
and so the answer does depend on the distribution of $X$. The above
is  an application of
the law of total probability for continuous random variable as @Patrick says in
his answer, but your answer of $\frac{1}{3}$ is incorrect. You are
adding up conditional probabilities conditioned on different events,
and that is not what the law of total probability does.

Your integrated needs to be: $e^{-3x}f_X(x)$ and so the result follows from the law of total probability. Also, the sum you wrote as being equal to the integral is not correct.  I am not the best person to explain the theory behind the integral being equal to a limit of Riemann sums but just know that your sum does not make sense because the sample space for $X$ is not discrete.


Does $\sum_{n\ge1}\frac1n\sin\left(\frac xn\right)$ converges uniformly? On each bounded interval $[a,b]$ :
$\left|\frac1n \sin\left(\frac xn\right)\right|\le \frac{\max\{|a|,|b|\}}{n^2}$,
the series $\sum_{n\ge 1}\frac {\max\{|a|,|b|\}}{n^2}$ converges,

therefore
$\sum_{n\ge1}\frac1n \sin\left(\frac xn\right)$ converges uniformly by the Weierstrass M-Test.

May I conclude from this statement that this series converges uniformly on $\mathbb{R}$?

Thanks.

The series does not converge uniformly on $\mathbb{R}$.

Consider $x_m = m\cdot \frac{\pi}{2}$ for $m \in \mathbb{Z}^+$. Then

$$\sin \frac{x_m}{n} \geqslant \frac{1}{2}$$

for $m \leqslant n \leqslant 3m$, and so

$$\sum_{n=m}^{3m} \frac{1}{n}\sin \frac{x_m}{n} \geqslant \frac{1}{2} \sum_{n=m}^{3m}\frac{1}{n} > \frac{1}{2} \log 3.$$

Choose a $\:x\:$ that will create a contradiction such as $\:x=n.$

$$\implies \sum_{n\in\mathbb N}\frac{1}{n}\sin(1)\overset{\text{Harmonic}}{=}\infty$$
So that it does not converge uniformly on any bounded set since it doesn not converge on $\:\mathbb R_+.$


What, Exactly, Is a Tensor? I've repeatedly read things that reference tensors, and despite reading the wiki page and other answers here on stackexchange I still don't know what a tensor is. I'm fine with hearing things in the framework on linear algebra, abstract algebra, and physics-but wiki just says it's a geometric object and in physics my professors all ways say it builds spaces out of other spaces. Can somebody clarify things?

At the lowest level of understanding a tensor $T$ of rank $r$ is an $r$-dimensional array (think of a spreadsheet) whose "side-lengths" are all equal to a given $n\geq1$. Therefore $T$ has $n^r$ entries, which we assume to be real numbers in the following.

When we are setting up such a tensor we have some application in mind, say in geometry or physics. That's where the difficulties come in. The tensor is meant to be "applied" to one or several (variable) vectors, and the result will be a number or a vector of interest in the context at hand. E.g., the value $T(x,y)$ could be the scalar product of $x$ and $y$, or the area of the parallelogram spanned by $x$ and $y$, or the image of $x$ under $T$ when $T$ is considered as a linear map, or the retaliatory force felt when moving in direction $x$, and on and on. For the computation of actual values we need the coordinates of $x$ and $y$.  Now these depend on the choice of basis in the ground space ${\mathbb R}^n$, and when we change the basis the coordinate values of the points $x$ change. But  the scalar product or some induced force, being "well defined" geometric or physical quantities, should not change. This in turn implies that the entries in our tensor (spreadsheet) $T$ will have to change, albeit in a characteristic way, called "contravariant" or "covariant", depending on the case at hand.

(As an aside: When a certain Excel-spreadsheet is meant to be a price list for various fabrics, then its entries will change as well in a characteristic way when the currency or the units of measurement are changed.)

But we have a definite feeling that there is some hidden "robust identity" incorporated in $T$ that is independent of the more or less accidental values appearing in the spreadsheet. It is only in the second half of the last century that mathematics has found a universal (and abstract!) way to express and to deal with this "hidden identity" of $T$. The field of mathematics concerned with this is called multilinear algebra. Only in this realm it then makes sense to talk about the tensor product. But I won't go into this here.

Think of a tensor as a matrix transforming one vector into another (or one-form into another one form). For example in linear algebra we learned that given a vector $\nu$ and the matrix $A$, we can get a new vector by multiplying the matrix by the vector $\nu$, i.e.
$$
\nu\cdot A = \omega
$$
In linear algebra $A$ is just a matrix, in tensor analysis $A$ would be a dyad tensor (tensor of order two), in linear algebra we have $A_{i,j}$ in tensor analysis we have $A_{ij}$.
In other words a tensor is a matrix.


Linearly dependent vector fields that are not spanned by fewer (continuous) vector fields. Consider the following continuous vector fields in $\mathbb{C}^3$:
\begin{equation*}
w_1(x,y,z):=(x,y,0), \; \; w_2(x,y,z):=(-z,0,y), \;\; w_3(x,y,z):= (0,z,x).
\end{equation*}
At every point $(x,y,z) \in \mathbb{C}^3$, the vectors are linearly dependent, as the matrix
\begin{pmatrix}
x & y & 0 \\ -z & 0 & y \\ 0 & z & x
\end{pmatrix}
has determinant zero. In particular, this means that at each point there are two vectors $v_1, v_2$ that linearly generate $w_1,w_2,w_3$ at that point. Choosing such a pair of vectors at each point we end up with two vector fields $v_1(x,y,z),v_2(x,y,z)$ which span the other three as $w_i(x,y,z) = \alpha_i(x,y,z)v_1(x,y,z) + \beta_i(x,y,z) v_2(x,y,z)$ for some functions $\alpha_i, \beta_i : \mathbb{C}^3 \rightarrow\mathbb{C}$.
The question is, can there be two continuous vector fields $v_1, v_2$ with that property? (that they span $w_1,w_2,w_3$ at all points). I strongly suspect that this is not the case, but I am not sure how to approach this. Am I missing some obvious condition the vector fields should satisfy for this to be true? I am actually interested in a higher dimensional example, so I am looking for a general proof strategy or hints on how to attack this problem in general rather than a trick that works for this particular example.

So the following strategy seems to work (in some situations) to prove that such vector fields do not exist. I'll explain it for the vectors that I gave as an example.
If they did, we can first restrict to the space $\mathbb{R}^3$. Then, note that at every point in $\mathbb{R}^3 \setminus\{(0,0,0)\}$ the fields $w_1, w_2, w_3$ span a two dimensional subspace. Therefore, this gives a rank $2$ vector bundle over the sphere $S^2$. In the example that I give, it turns out to be isomorphic to the tangent bundle, and so it doesn't have two linearly independent sections $v_1, v_2$.
Unfortunately, this does not seem to work the other way around: after removing the "singularities" (the points at which the vector fields do not span a vector space of generic dimension) you could in principle be left with a trivial bundle in the remaining space but still no vector fields in the whole space that span the original ones.

One way is to find an easy map directly and check. As you write, the matrix is linearly dependent, so we can take first and third vectors and generate the second one:
$$
(-z,0,y) = \frac{y}{x} (0,z,x) - \frac{z}{x} (x,y,0)
$$
which seems to be continuous away from $(0,0,0)$


Every positive integer can be expressed uniquely as $N = a_0 + a_1 2 + a_n 2^n $ and all $ a_i $ belongs to $\{0, 1\}$ I am trying assignment problems of an Institute in which I don't study. I am unable to think about this problem.

  Problem ->Prove that every positive integer can be expressed uniquely as $N = a_0 + a_1 2 + a_n 2^n $ and all $ a_i $ belongs to $\{0, 1\}$.

Can someone please help.

The statement is false. For example, the number $N=12$ cannot be written as $12=a_0 + 2a_1 + 2^n a_n$ where $a_i\in\{0,1\}$, and it is easy to show this since:

$a_n$ must be $1$, since otherwise, $a_0 + 2a_1 + 2^n a_n=a_0+2a_1 < 3 < 12$.
$n$ must be equal to $3$, since $n\leq 2$ means that $a_0 + 2a_1 + 2^n a_n$ is at most $7$, and $n\geq 4$, then $a_0 + 2a_1 + 2^n a_n$ is at least $16$.
$12=a_0 + 2a_1 + 2^n a_n$ means that $a_0=0$, since the left side is even, and if $a_0=1$, then the right side is odd.
Therefore, $12=2a_1 + 2^3 a_3 = 2a_1 + 8a_3$, meaning $6=a_1+4a_3$, which is impossible (either $a_1=0$, in which case $6=4a_3$, which is impossible since $a_3$ must be $0$ or $1$, or $a_1=1$, in which case $a_1+4a_3$ is odd, while $6$ is even).


Why is $g(x,y)=\frac{xy}{(x^2+y^2)^{1/2}}$ not differentiable at $(0,0)$? I need to know why the following is not true
Let $g(x,y)=\frac{xy}{(x^2+y^2)^{1/2}}$ when $(x,y)$ not equal to $(0,0)$
and $g(x,y)=0$ when $(x,y)=0$
then  $g(x,y)$ on the $x$ axes$(y=0)$ and the $y$ axes $(x=0)$ is $0$ so the partial derivatives at $(0,0)$ exist and they are zero and since zero is constant function it is continuous hence $g(x,y)$ has continuous partial derivatives and therefore is differentiable at $(0,0)$

Use definition of the differentiability in 2D – Function $g(x,y)$ to be differentiable at $(x_0, y_0)$
$$
\lim_{(x,y) \to (x_0, y_0)} \frac {g(x, y) - g(x_0, y_0)}{\sqrt{\left( x-x_0\right )^2+\left( y-y_0\right )^2}}
$$
should exist. Latter means that the limit should be at least the same when approaching $(x_0, y_0)$ from all directions. Simplest case if you approach $(0,0)$ along rays $y = kx$, so
$$
\lim_{(x,y) \to (x_0, y_0)} \frac {g(x, y) - g(x_0, y_0)}{\sqrt{\left( x-x_0\right )^2+\left( y-y_0\right )^2}} = \lim_{x \to 0} \frac {x^2 k}{x^2 \left( k^2 + 1\right )} = \lim_{x \to 0} \frac k{k^2+1} = \frac k{k^2+1}
$$
which means that limit in fact depends on path you approach $(0,0)$(i.e. on the value of $k$), hence doesn't exists.

Use polar coordinates $x=r\cos(\phi)$ $y=r\sin(\phi)$, your $g(x,y)$ becomes

$$\frac{1}{2}r\sin(2\phi)$$

This function is differantiable.


Maximum number of edges in no 3-matching Let $G$ be a graph on $n$ vertices. Find the maximum possible number of edges if $G$ has no matching of size $3$.

Also, what happens with other sizes?

Hint:
It is easy to see that the "best" case (i.e. to have maximum no. of edges) with a minimum amount of matchings possible is when the graph is completely connected. Now convince yourself that the completely connected graph $K_5$ with $n = 5$, i.e. with $10$ edges has no matching with $3$ edges

$\quad$

Edit:
My answer is just a specific case ($n = 5$). For a general $n\geq 6$, you will have to use the result suggested by user W.R.P.S in the comment above for $s = 3$. It gives $\max\{10, 2n - 3\}$ as the answer to your question.


Compute $\int_{|z|=1}\frac{\log z}{z}dz$. Here is a question about contour integration in complex analysis:

  Compute
  $$\int_{|z|=1}\frac{\log z}{z}dz$$

I am not sure if I understand the question since the logarithm must be defined in a simply connected domain $\Omega$ with $0\notin\Omega$ and $1\in\Omega$. Hence, we cannot define $\log$ for every value in the circle $|z|=1$; there must be at least one point where it is not defined. However, if I forget this and consider the principal branch, I can compute
$$\int_{|z|=1}\frac{\log z}{z}dz = \int_{-\pi}^\pi \frac{\log e^{i\theta}}{e^{i\theta}}ie^{i\theta}d\theta = \int_{-\pi}^\pi i\theta id\theta=-\frac{\theta^2}{2}\bigg\vert_{-\pi}^\pi=0$$
Is this right?

The analysis is not quite complete because, generally, the contour must not enclose the branch point at $z=0$.  Because the example above does, you are getting any number of "answers."  Consider the case when $\arg{z} \in [0,2 \pi)$.  A naive evaluation of the integral produces

$$i \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{i \theta}{e^{i \theta}} = -2 \pi^2 \ne 0$$

The problem of course lies in the multivaluedness of the integrand.  A correct approach excludes the log from the interior of the contour.  One approach is to consider a keyhole contour $C$ of outer radius $1$ and inner radius $\epsilon$ about the positive real axis:

$$\oint_C dz \, \frac{\log{z}}{z}$$

This contour integral is equal to

$$i \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{i \theta}{e^{i \theta}} + \int_1^{\epsilon} dx \frac{\log{x}+i 2 \pi}{x}  + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log{\epsilon}+i \phi}{\epsilon e^{i \phi}} + \int_{\epsilon}^1 dx \frac{\log{x}}{x}$$

which in turn is equal to

$$-2 \pi^2 + i 2 \pi \log{\epsilon} -i 2 \pi \log{\epsilon} + 2 \pi^2 = 0$$

as expected, as there are no poles in the interior of $C$.  This type of analysis works for any branch cut definition for the log.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\oint_{\verts{z}\ =\ 1}
{\ln\pars{z} \over z}\,\dd z}
\\[5mm] = &\
\lim_{\epsilon\ \to\ 0^{+}}\,\,\left[-\int_{-1}^{-\epsilon}{\ln\pars{-x} + \ic\pi \over x}\,\dd x -
\int_{\pi}^{-\pi}{\ln\pars{\epsilon} + \ic\theta \over \epsilon\expo{\ic\theta}}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta\right.
\\[2mm] &\
\phantom{\lim_{\epsilon\ \to\ 0^{+}}\,\,\left[\,\,\right.}\left.
-\int_{-\epsilon}^{-1}{\ln\pars{-x} - \ic\pi \over x}\,\dd x
\right]
\\[5mm] = &\
\lim_{\epsilon\ \to\ 0^{+}}\,\,\braces{%
-2\pi\ic\ln\pars{\epsilon} + \ic\int_{-\pi}^{\pi}
\bracks{\ln\pars{\epsilon} + \ic\theta}\dd\theta} =
\bbx{0} \\ &
\end{align}


What's the intuition behind Pythagoras' theorem? Today we learned about Pythagoras' theorem. Sadly, I can't understand the logic behind it.

$A^{2} + B^{2} = C^{2}$

$C^{2} = (5 \text{ cm})^2 + (7 \text{ cm})^2$
$C^{2} = 25 \text{ cm}^2 + 49 \text{ cm}^2$
$C^{2} = 74 \text{ cm}^2$
${x} = +\sqrt{74} \text{ cm}$

Why does the area of a square with a side of $5$ cm + the area of a square with a side of $7$ cm always equal to the missing side's length squared?

I asked my teacher but she's clueless and said Pythagoras' theorem had nothing to do with squares.

However, I know it does because this formula has to somehow make sense. Otherwise, it wouldn't exist.

Ponder this image and you will see why your intuition about what is going on is correct:

The accepted answer provides no explanation for the diagrams:

Each diagram shows an outer square, $\color{red}{(a+b)}$ units on a side, divided into other figures. The diagrams suggest different expressions for the area of the outer square. Equating these expressions leads to the equation $a^2+b^2=c^2$.

Total area of $\color{red}{outer}$ square = Area not covered by triangles + Area covered by triangles

$\color{red}{(a+b)^2} = c^2 + 4(\dfrac1{2}ab)$  ________$\color{red}{(a+b)^2} = a^2 +b^2 + 4(\dfrac1{2}ab)$

$c^2 + 4(\dfrac1{2}ab)= a^2 +b^2 + 4(\dfrac1{2}ab)$

$\therefore c^2 = a^2+b^2$

source: Algebra Structure and Method Book $1$ by Brown, Dolciani et al.

Here is a visual aid to help the new learner move from one diagram to the other:

or


Cancellation of integrals Suppose we have two continuous functions $f,g: \mathbb{R}\rightarrow\mathbb{R}$

Then, suppose $$\forall a,b \in \mathbb{R:}\int\limits_{a}^{b} f(x)dx = \int\limits_{a}^{b} g(x)dx$$

Can we conclude that $f =g$?

If not, under what conditions can we conclude that $f=g$?

If yes, can you provide a proof and is the same result possible with weaker assumptions?

Can we generalise this to other kinds of integrals (line integrals, surface integrals, double integrals, ...)

Thanks in advance!

hint

let $a\in \mathbb R $.

$f ,g$ continuous at  $\mathbb R\;\;\implies $

$$F:x\mapsto \int_a^x f (t)dt $$
$$=G (x)=\int_a^x g (t)dt$$
differentiable at $\mathbb R\;$ and

$$F'(x)=G'(x)=f (x) =g (x)$$

$$\implies f=g $$

Take f and g to be any two odd functions.  Then $\int_{-a}^a f(x)dx= \int_{-a}^b g(x)dx= 0$.

In general just the fact that the integral between two fixed points of two functions are the same number tells us nothing about the relationship between the two functions.  If $\int_a^b f(x)dx= \int_a^b g(x)dx$ for all a and b, then f(x)= g(x) for all x.


Probability involving the average of a simple random sample So here's the question: Maria is shopping for wireless routers and is overwhelmed by the number of options. In order to get a feel for the average price, she takes a random sample of $75$ routers. The average price for this sample is $\$76$ and the standard deviation is $\$30$.

A consumer information website claims that the average price of all routers is $\$80$. If we assume that the population standard deviation of prices is $\$30$ (i.e., if we assume Maria's sample standard deviation is a good estimate of $\sigma$), what is the probability that a random sample of $75$ routers would have an average price of less than or equal to $\$76$? Give your answer to three decimal places.

How do I solve this problem? I'm completely stuck and would really appreciate any help.

Here are some hints to help you structure your thinking:

If $\sigma_{X_i} = 30$ what is the standard deviation of $\frac{1}{75}\sum_{1}^{75} X_i$? (Let's call this $\sigma_{\bar X}$)
What is the $Z$ score of the difference $D=76-80$?
What is the probability that a standard normal random variable will take on values less than $Z$?

To put OP out of his misery...

We are told that the router prices are normally distributed with a standard deviation of 30 and a mean of 80. What this problem is asking you to do is determine how likely it is that Maria's sample of 75 routers came from a normally distributed population with mean 80, standard deviation 30.

The first thing to do is calculate the standard deviation of the sample average of 75 iid observations from this population. As you correctly calculated, that would be $\frac{30}{\sqrt{75}} \approx 3.46$. So, 95% of the time, a sample average of 75 iid observations from a normally distributed population of mean 80, standard deviation 30 will fall i the interval $80 \pm 2\times 3.46$.

Maria's sample average is 76. The question is asking you for the probability of getting less than or equal to 76 for the sample average if it came from a population with mean 80 and standard deviation 30.

How many standard deviations below the mean is the sample average of $76$?

We already calculated that the standard deviation of a 75 observation mean to be 3.46, so 76 is $\frac{76-80}{3.46}\approx -1.15$ standard deviations below the expected sample average.

How likely is this? We simply need to look up the above value (-1.15) in a standard normal table (you should have been taught how to do this already) or you can use excels NORMSDIST(-1.15) to get a lower tailed probability. This will give you 0.12 or $12\%$

8.) A population has to pass a quality test based on data from past runs there is a probability that 93.9% percent will pass and 4.1% will fail, there are 176 units in the population.  What is the standard deviation for a sample of 50 and 75 units, what is the probability of having 2 failed units in a sample of 50 and 75.


Modular quadratic equation question- Where did I go wrong? \begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N >0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$,  $\frac{N^2+N}{2}$ has a remainder of 2
I'd like a correct solution as well. Thanks.

Your first error is that you you scaled by $\,2,\,$ which is not invertible $\!\bmod 4,\,$ so this will not yield an equivalent congruence. Rather, it yields necessary but not sufficient conditions on roots (so possibly  extraneous roots). See here for more on the insufficiency of unidirectional inferences.
To get an equivalent congruence we need to scale the modulus too, since $\,4\mid a/2\iff 8\mid a,\,$ so
$$(n^2+n)/2\equiv 0\!\!\!\pmod{4}\iff n^2+n\equiv 0\!\!\!\pmod{8}\qquad$$
Now we can complete the square as you did, but since this too involves scaling the modulus, this will end up being fruitless, leading back to where we started, namely
$$\begin{align} n^2+n&\equiv 0\pmod{8}\\
\iff\ \ \ \ \ \ \ 4n^2+4n&\equiv 0\pmod{32}\\
\iff 4n^2+4n+1&\equiv 1\pmod{32}\\
\iff \ \ \ \ \ \, \color{#c00}{(2n+1)^2}&\:\color{#c00}{\equiv 1}\pmod{32}\\
\iff (2n)(2n+2)&\equiv 0\pmod{32}\\
\iff\ \ \ \ \ \ \  n(n+1)&\equiv 0\pmod{8}
\end{align}\qquad$$
where we solved $\,\color{#c00}{a^2\equiv 1}\,$ by factoring the difference of squares $\,0\equiv a^2-1\equiv (a-1)(a+1).\,$  You can't simply take square roots as you did, e.g. $\,x^2\equiv 1\pmod{8}\,$ has $4$ roots $\,x\equiv 1,3,5,7$.
Instead, by $\,n,\,n\!+\!1\,$ coprime, $\,8\mid n(n\!+\!1)\iff 8\mid n\,$ or $\,8\mid n\!+\!1,\,$ thus we conclude that  $\,n(n+1)/2\equiv 0\pmod{4}\iff n\equiv 0,7\pmod{8}$
Finally, beware that modular fractions are well-defined (uniquely exist) only when they are writable with denominator coprime to the modulus, when $\,a/b := ab^{-1}.\,$ For more on modular fractions see here and here.

$(2N+1)^{2} \equiv 1  $
$2N+1 \equiv 1$
This step is wrong, for example: $(2*3+1)^2 \equiv 1$.


proving a set is not path-connected Let
$$A=\{(x,\sin(1/x)) \mid x > 0 \} \cup \{(0,y) \mid y\in[-1,1]\} \subset\mathbb{R}^2$$

how to prove that this set is not path-connected(or just not connected)?

Thank you

Let $C$ be that sine curve part, and $S$ be the vertical segment. Suppose there exists a path $\omega: [0, 1] \to A$ that connects a point $(0, 1) \in S$ with $(\frac{1}{\pi}, 0) \in C$. Consider the set $U = \{t \in [0, 1]: t \in C \} = \omega^{-1}(C)$. As sine curve is open in $A$, $U$ is an open subset of $[0, 1]$. Let $t_0 = \inf U$. As $U$ is open and $0 \not \in U$ (because $\omega(0) = (0, 1) \in S$, $\omega(t_0) \in S$, but for some $t_1 > t_0$, $\omega((t_0, t_1)) \subset C$. Consider a small open neighbourhood $V \subset A$ of $t_0$ that has infinitely many disconnected components. By continuity of $\omega$, there's some interval $(t_0 - \epsilon, t_0 + \epsilon)$ such that $\omega((t_0 - \epsilon, t_0 + \epsilon)) \subset V$. For some point in $t \in (t_0, t_0 + \epsilon)$, $\omega(t)$ is in a different component of $V$ than $t_0$, which is a contradiction, because $\omega((t_0 - \epsilon, t_0 + \epsilon))$ is connected, as a continuous image of a connected set.

Suppose this set is path connected. Consider the point $(0,0)$ and the point $(1,\sin(1))$. The only way to link these two points is along the path $(x,\sin(\frac{1}{x}))$. However $\lim_{x \rightarrow 0}\sin(\frac{1}{x})=$DNE. Hence there is no path connecting $(0,0)$ to $(1,\sin(1))$, a contradiction. Therefore the set is not path connected.


Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:

$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$

so now I have
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$

Now,

$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$

and

$$\int\frac{2}{(1+t^2)^2}dt$$

using partial integration we have:

$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$

and $$dt=dv \Rightarrow t=v$$

so now we have:

$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$

so, the final solution should be:

$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$

since the original variable was $x$ we have

$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$

But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$

I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?

You have mistake here:

$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{\color{red}(1+t^2\color{red}{)^3}}$

Notice, you can continue from here without using partial fractions $$\int \frac{2t^2}{(1+t^2)^2}\ dt$$
 $$=\int \frac{2t^2}{t^4+2t^2+1}\ dt$$
$$=\int \frac{2}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}+\int \frac{\left(1-\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}$$
$$=\int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+4}+\int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2}$$
$$=\frac 12\tan^{-1}\left(\frac{t-\frac 1t}{2}\right)-\frac{1}{\left(t+\frac{1}{t}\right)}+c$$
$$=\frac 12\tan^{-1}\left(\frac{x-1}{2\sqrt x}\right)-\frac{\sqrt x}{x+1}+c$$

let $\sqrt x=\tan\theta\implies \theta=\tan^{-1}(\sqrt x)$,
$$=\frac 12\tan^{-1}\left(\frac{\tan^2\theta-1}{2\tan\theta}\right)-\frac{\sqrt x}{x+1}+c$$
$$=\frac 12\tan^{-1}\left(\tan\left(\frac{\pi}{2}+2\theta\right)\right)-\frac{\sqrt x}{x+1}+c$$
$$=\frac{\pi}{4}+\theta-\frac{\sqrt x}{x+1}+c$$

$$=\theta-\frac{\sqrt x}{x+1}+C$$
$$=\color{red}{\tan^{-1}(\sqrt x)-\frac{\sqrt x}{x+1}+C}$$


GCD and LCM number theory problem I am having some trouble finishing a solution to this problem, so I thought I might ask you all for help.
The problem goes as follows:
Find all natural numbers $a,b>0$ for which holds:
$$lcm(x,y)-gcd(x,y)=\frac{xy}{2021}$$
The problem is annoted with the hint, that 2021=43*47.
I have tried using the fact, that the $gcd=d$ from which follows $x=a*d$ and $y=b*d$. The $lcm(x,y)$ therefore must be $lcm(x,y)=a*b*d$.
Given this and the fact that the product of two numbers $x,y$ and their $gcd$ and $lcm$ is the same, meaning
$$xy=lcm(x,y)*gcd(x,y)=abd*d$$ I arrive at the equation:
$$(abd)-d=(abd^2)/2021$$
which equals
$$2021(ab-1)=abd$$
Where do I go from here? Any help or correction of mistakes that I may have made is greatly appreciated!

You rewrite it as $(2021 - d)ab = 2021$. Now use the factorization of $2021$ given in the hint.

Hint: write it as $\,f(a)=0\,$ so $\,a\mid f(0)\,$ since $\,f\,$ is a polynomial in $\,a\,$ with integer coef's. Thus the integer values that $a$ can take are constrained to the finite number of factors of $\,f(0)\,$ (if $\,f(0)\neq 0)$
Remark $ $ This can be viewed as special (trivial) case of the ubiquitous Rational Root Test, i,e., an integer root of $f$ divides its constant term.


Probability of getting all faces of a die an equal number of times I have a question:

A die is rolled 36 times.

What is the probability of getting each number 6 times?

I think the answer is: $6\cdot\left(\frac16\right)^6$

Am I wrong?

The answer you give is incorrect. There are 6 events you want to have happen, rolling each number 6 times, so you multiply their probabilities. Additionally, there are many ways this could happen, so we multiply by the number of ways of arranging the die rolls. This gives:

$$ \left(\frac{1}{6}\right)^{36} \frac{36!}{(6!)^6}$$

This sort of question is something that could easily be set by someone with no knowledge of nor understanding of probability or asked carelessly while dealing with something else.  If you are 8(ish) then the person asking the question probably wont be expecting an answer that uses "factorials" = the "!", of course;

2! = 2 x 1

3! = 3 x 2 x 1

4! = 4 x 3 x 2 x 1

If, however, you are taking a university degree in mathematics then factorials (or even more interesting) are likely to be expected.  It is difficult to know what level of accuracy is expected.  Sometimes questions like this get asked to create social friction with more accurate answers getting derided for "over-thinking it" or "over-complicating it" or "taking too long" whereas giving a quicker answer might be considered too flippant, or hasty or ill-thought out.  If your boss has already given an answer then giving something different might be considered "insubordinate".

Most of the answers in this thread give a 'reasonable' approximation.  Some veer towards what most non-mathematicians might find easier to understand whereas others veer towards being far more accurate.  The answer that is given in the question is perfectly accurate enough for almost all cases where this question might arise.  There are complexities it doesn't address but the use-case might not need those to be taken into account.  So it's not a 'wrong' answer as such, just as many other answers in this thread are also not really wrong = it's just that they are all approximations with a varying level of accuracy.

To understand why this is a non-trivial question first think of a simpler case where you get 6 rolls and need to get each number but only once each.

On the first throw you need any number 1-6, so it's fairly certain = 1.  Some people might say you don't even need to throw the die because a good outcome is so certain - unless the die lands on it's edge, or corner, or disappears into an alternate reality or some other thing we can disregard as being "not a proper throw".
On the 2nd throw one of the numbers has become unsuitable because it was already thrown in the first throw.  So you've got 5/6 chance of getting something suitable.
On the 3rd throw there are now 2 numbers that would disqualify you.  So that leaves 4 'good' results = 4/6

So it's only the 6th throw where you have a 1/6 chance of getting the right number.  I have a nagging feeling about 2 more complexities that make this not quite true but it's a good approximation.

Ok, so now another slightly less simple experiment, but still simpler than the original question.  Say 36 throws but you need to get 6 results that are 6.

On the 1st throw there is an approximately 6/36 chance of getting a 6.  Lets say this does happen, that you do get a 6 on the 1st throw ...
On the 2nd throw (given that you got a 6 already on the first throw) the chance of getting a 6 would be approximately 5/35.  Lets say this throw is also a 6.
Ok, so given the first 2 throws were 6 then we only need 4 more 6s but we only have 34 more throws so the chance of getting a 6 is 4/34.

But doesn't this contradict the fact that each time a die is rolled it has a 1/6 chance of getting a particular number?  (assuming a perfect world for dice-rolling)  Not really because we are looking at the probability of all 36 dice rolls.  Not just the first throw.  So we have to consider all the 36 rolls as being 'one' "event".

In the last experiment we could see a factorial pattern forming;

6/36 x 5/35 x 4/34

So on the top its 6 x 5 x 4 x 3 x 2 x 1, which is written "6!" as you probably know.

I think the permutations (or is it combinations) work out pretty well.  I think the extra complexities, to do with which order the 6s appear in, obligingly cancel each other out.

So all in all i think the initial answer was close enough for most cases.  It's not really a case of being "right or wrong" because all answers have to make some assumptions such as generally ignoring "edge cases".  I think the more accurate answers tend towards it being more unlikely.

I really like the answers that use factorials but my fav is the one that starts "fun fact".  I'm not keen on results shown as percentages but it gets rid of a couple of zeroes and i guess it does make it more human-readable.

Good luck and regards from
Tom :)


Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$? $\{\emptyset\}$ is a set containing the empty set. Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$?

My hypothesis is yes by looking at the form of "the superset $\{\{\emptyset\}\}$" which contains "the subset $\{\emptyset\}$".

It's easier to just replace everything by variables, then it's clearer.

Set $a=\varnothing, b=\{\varnothing\}=\{a\}$ and $c=\{\{\varnothing\}\}=\{\{a\}\}=\{b\}$.

Now the question is whether or not $\{a\}\subseteq\{b\}$. But since $a\neq b$, it's easy to see that the answer is negative.

Think of {∅} as a box containing an empty box, then
{{∅}} is a box containing a box which contains an empty box, in which case
{{∅}} is box containing {∅}, which means

{∅} is subset of {{∅}}


Is a homeomorphism which maps lines to lines (and fixes zero) necessarily linear? We know the homeomorphism $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ maps straight lines to straight lines and the zero vector to the zero vector. Is it Linear?? If so, how can we prove it?

Are you asuming $n>1$? If not a good counter-example could be the function $\phi(x)=x^3$ ( $x\in\mathbb{R}$) . It's a homeomorphism, maps $\mathbb{R}$ to $\mathbb{R}$ (the only line in $\mathbb{R}$) and obviously it maps $0$ to $0$

This is not enough to suggest linearity w/o it being a homeomorphism... It has satisfy $\phi(a*v_1 + b*v_2) = a\phi(v_1)+b\phi(v_2)$.  So lines to lines does not promise linearity, though it is a requirement as lines are a linear combination of vectors in $\mathbb{R}^n$.  Assuming you are looking at $\mathbb{R}^n$ as a combination of vectors in the typical sense.

The homeomorphism takes away the discontinuous cases and the situations with a kernel.  My bad I did not read the problem correctly initially.

Lines can be expressed as a linear combination of elements in a vector space with an indeterminate to the first power.  Just express a matrix with such a basis as the mapping and you have you linearity in the form of a matrix.


Union of any finite collection of bounded sets is bounded If you could evaluate my proof on scale of 1-5 in clarity, conciseness, and validity, I would appreciate your 2 cent!!!

Sketch of Proof in words:

  Consider for some $n\in\mathbb{N}$, $\bigcup^{n}_{i=1} P_i$ where $P_i$ is a bounded set in arbitrary metric space $(X,d)$.
   By boundedness, for each $i$, $\exists x^*\in X,\exists M_i\in\mathbb{R}$, $d(x^*,p)\leq M_i$ $\forall p\in P_i$.
  Let $M^*= \max\{M_i:i=1,...,n\}$.
  If $q$ is in the union, then for some $i$, $q\in P_i$.
  $d(q,x^*)\leq M_i\leq M^*$.
   The union of any finite collection of bounded sets is bounded. QED

Assume it is unbounded. It has an unbounded sequence. It has a monotone unbounded subsequence (that is, it goes to infinity). This sequence has infinitely many terms from one of your bounded sets. This set is unbounded.


Ways of making 10 letter combinations out of 3 given letters Find the number of 10 letter combinations from ${a, b, c}$ that contain:

At least one of each letter
At least two of each letter
At least one A, two Bs, three Cs
Any number of each letter

(Also repetition is allowed)
For number 1 I think that the answer is $\binom{9}{2}$, since there's n-1 places that I can place a bar from stars&bars.
For number 2 though I'm not sure how I can apply this in order to place the bars so that I get at least 2 of each element $a,b,c$.
Edit 2: I did some thinking about question 2 and I think I might have an answer which is correct. Since I need at least 2 of each letter I decided to group 1 of $AA, BB, CC$ so that $AA$ was one object meaning that now in total I had 7 objects. Which then means that I would have $\binom{6}{2}$ places to divide this group up. Please let me know if this approach is correct. Or if I've made a mistake. Thank you.

Let frequency of $a,b,c$ be $x,y,z$ respectively. You're looking for integral solutions of following cases :

$x \ge 1, y \ge 1, z \ge 1$ and $x+y+z=10$
$x \ge 2, y \ge 2, z \ge 2$ and $x+y+z=10$

This is same as $u \ge 1, v \ge 1, w \ge 1$ and $u+v+w=7$

$x \ge 1, y \ge 2, z \ge 3$ and $x+y+z=10$

This is same as $x \ge 1, p \ge 1, q \ge 1$ and $x+p+q=7$

$x \ge 0, y \ge 0, z \ge 0$ and $x+y+z=10$

Hence answers would be $\binom{9}{2}, \binom{6}{2},\binom{6}{2}, \binom{12}{2}$ respectively.

For (1), start with "abc" so that you have fulfilled  the "at least one of each letter" requirement.  You have 7 letters that can be any of the three "a", "b", or "c" so there are 3^7 possible choices.  For each such 10 letter combination there are 10! possible arrangements of the letters.  There are 10!(3^7) such combinations.
For (2), start with "aabbcc".  There are then 4 letters left so 3^4 possible choices.  Since there are again 10! different arrangements, there are 10!(3^4) such combinations.
For (3), start with "abbccc" and finish in the same way.
For (4), with just 10 letters chosen from "a", "b", and "c", there are 3^10 such combinations.


Positive and negative complex numbers? Can there be such a thing as positive and negative complex numbers? Why or why not?

What about positive or negative imaginary numbers?

It seems very tempting to say $+5i$ is a positive number while $-2i$ is a negative number. On an Argand diagram (complex plane) $+5i$ would be represented by a point above the horizontal axis while $-2i$ is a point below the horizontal axis.

[begin crackpot theory]

Mathematicians have committed a great blunder by thinking that a certain imaginary number is $i$ and another is $-i$, when it's really the other way around!!!!

[end crackpot theory]

If one were to take the position above, and try to rewrite all of mathematics consistently with the theory propounded above, the result would be that nothing at all would change.  Which one is called $i$ and which is called $-i$ doesn't matter.

It does in some contexts make sense to pay attention to whether the real part of a complex number is positive.  A Dirichlet series $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n^s}$ converges if the real part of $s$ is greater than the real part of the abscissa of convergence of the series, and diverges if the real part of $s$ is less than the abscissa of convergence.

We may partially order the complex numbers in accord with their real parts so that
   a+bi < c+di iff a < c.
Thus the line of imaginary numbers divides positive complex numbers from negative complex numbers.  However, if b and d are not equal, a+bi and c+di are incomparable.  Thus it is no longer true that every positive number is greater than every negative number.  Most positive-negative pairs are incomparable.  This partial ordering has some neat features.  For example, in the Riemann sphere representation of the complex plane, a point at infinity is added to the imaginary line, transforming it into a great circle.  This circle thus divides the Riemann sphere into positive and negative hemispheres.  The imaginary numbers, 0, and infinity, which lie between the hemispheres, are all “neutral”—that is, neither positive nor negative—but all other complex numbers are either positive or negative.


Integrating Factorials I feel like I'm doing something wrong here: $$\frac{d^n}{dx^n}(x^n)=n!$$

$$ 5!=\frac{d^5}{dx^5}(x^5)$$

$$ \int{5! dx}=\int{\frac{d^5}{dx^5}(x^5)}dx=x\frac{d^4}{dx^4}(x^4)=x*4!$$

Please explain what I'm doing wrong.


What is $\gcd(0,0)$? What is the greatest common divisor of $0$ and $0$?
On the one hand, Wolfram Alpha says that it is $0$;
on the other hand, it also claims that $100$ divides $0$,
so $100$ should be a greater common divisor of $0$ and $0$ than $0$.

The word "greatest" in "Greatest Common Divisor" does not refer to being largest in the usual ordering of the natural numbers, but to being largest in the partial order of divisibility on the natural numbers, where we consider $a$ to be larger than $b$ only when $b$ divides evenly into $a$. Most of the time, these two orderings agree whenever the second is defined. However, while, under the usual order, $0$ is the smallest natural number, under the divisibility order, $0$ is the greatest natural number, because every number divides $0$.

Therefore, since every natural number is a common divisor of $0$ and $0$, and $0$ is the greatest (in divisibility) of the natural numbers, $\gcd(0,0)=0$.

From Wikipedia:

  The greatest common divisor of $a$ and $b$ is well-defined, except for the
  case $a=b=0$, when every natural number divides them.


Show that $M_n = X_n^2 - n$ is a martingale Suppose $X_n$ is a symmetric random walk on $\mathbb{Z}$. To show that it is a martingale I need to show
$$
\mathbb{E}[M_{n+1}|X_{0:n}] = M_n
$$

$$
\begin{align}
\mathbb{E}[M_{n+1}|X_{0:n}] &= \mathbb{E}[X_{n+1}^2 - (n+1)|X_{0:n}]\\
&= \mathbb{E}[X_{n+1}^2|X_{0:n}] - (n+1)
\end{align}
$$

I'm stuck here.

Assume that the question without the -n is a martingale. The latter is easy to prove M (n) =X^2 (n). The expectation of X^2 (n+1) given the past information up to n is E(X2n) = n because it is like a wiener process that is Normal with mean zero and variance the time  (t) then the process you have has an expectation of zero and variance of 2(....)by the use of the definition of expectation and variance


How to solve $x^2 + \ln(x) = 0$ I was just investigating $y = f(x) = e^{-x^2}$ and then went ahead to plot $x=f(y), y=-f(x), and x=-f(y)$, and what I got was interest rounded square shape, and I think we can calculate this area using integration.
However to get the bounds I must solve the equation $x^2 + \ln(x) = 0$. Is there a way to solve this equation without using the graph?
Shape

Not really. Just as you can't solve $2^x=3$ using algebra (you need to create a new kind of function; logarithms), you can't solve tihs one without creating a new kind of special function. The one we have for this is called the Lambert W function, the inverse function of $xe^x$. Then we can solve
$$ x^2+\ln x=0 $$
$$ x^2+\frac{1}{2}\ln x^2=0 $$
$$ 2x^2+\ln x^2=0 $$
$$ 2x^2+\ln(2x^2)=\ln(2) $$
$$ 2x^2e^{2x^2}=2 $$
$$ 2x^2=W(2) $$
$$ x=\sqrt{\frac{W(2)}{2}}. $$


Illustrative examples of a phenomenon in the logic of mathematical induction It is said (and I myself have said) that in some cases the easiest way to prove a statement by mathematical induction is to prove a stronger statement by mathematical induction, because then one has a stronger induction hypothesis to use.

But then I ask myself: if a bright student were to ask me for some typical examples of that phenomenon, what would I say?

The only example that came to mind immediately when I thought of this question is Łos's theorem: A first-order sentence $\varphi$ is true in an ultraproduct $\left(\prod_{i\in I} A_i\right)/F$, where $F$ is an ultrafilter on the index set $I$, if and only if the set $\{i\in I : \varphi\text{ is true in}A_i\}$ is a member of $F$.  The stronger statement speaks of first-order formulas (which may contain free variables) rather than of first-order sentences (which have no free variables).  The proof is by induction on the formation of first order formulas, and it works since the class of first-order formulas is closed under certain operations and the class of first-order sentences is not.

That's not a great example for the situation I imagined.

Looking around m.s.e. a bit, I find Steven Stadnicki's answer to this question and my answer to this question, and maybe Martin Brandenburg's answer to this question.

This is not a great list of examples for illustrative purposes at an elementary level (although Steven Stadnick's answer would fit into such a list).

If the purpose is to illustrate this phenomenon, which examples should be used, both at the most elementary levels and at more advanced levels?
Is there a logician's viewpoint on this phenomenon?  Might there be, for example, some idempotent mapping $T$ from the class of statements-that-are-weaker-versions-of-things-provable-by-induction to the class of things-provable-by-induction, where $T\varphi$ is in each case a generalization of $\varphi$?

Consider the simple continued fraction $\langle a_0;a_1,a_2,\dots\rangle$, where all the $a_i$ are integers, and all positive except possibly $a_0$.

Define the sequences $p_i$, $q_i$ by

$p_{-1}=0$, $p_0=a_0$, and $p_k=a_kp_{k-1}+p_{k-2}$, and

$q_{-1}=0$, $q_0=1$, and $q_k=a_kq_{k-1}+q_{k-2}$.

We want to show that $\langle a_0;a_1,\dots,a_k\rangle=\frac{p_k}{q_k}$.

The standard way to prove the result by induction is to prove the stronger result that for any positive $x$,
$$\langle a_0;a_1,\dots,a_{k-1},x\rangle=\frac{xp_{k-1}+p_{k-2}}{xq_{k-1}+q_{k-2}}.$$

As a small additional example, a recent question asked for a proof that the denominator of
$\frac{1}{\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_n}}$ can be rationalized. An induction proof used the stronger induction hypothesis that $\frac{1}{\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_n}+t}$ can be rationalized, where $t$ is a free parameter.

For early induction arguments, however, I think inequalities are quite persuasive, since it is clear that for example knowing that $1+\frac{1}{2^2}+\frac{1}{n^2}\lt 2$ cannot by itself imply that $1+\frac{1}{2^2}+\frac{1}{n^2}+\frac{1}{(n+1)^2}\lt 2$

Formation rules:

1) Each lower case letter of the Latin alphabet qualifies as a formula.

2) If $\alpha$ as well as $\beta$ qualify as a formula, so does C$\alpha$$\beta$.

3) Nothing else will get consider as a formula for the following.

Since I'll only talk about something in two-valued logic here, C will stand for material implication.  That said, what gets stated applies to more than just two-valued logic.

Suppose we wanted to prove that in classical propositional logic with substitution and detachment, (1) if $\vdash CsCrCpCqp$, then $\vdash C\alpha_n C\alpha_{(n-1)} \dots C\alpha_2C\alpha_1\alpha_2$, where $n>1$, lower case letters are atomic formulas, and subscripted $\alpha$'s indicate any formulas.  We could do this, but it comes as easier to prove the stronger meta-statement that (2) if $\vdash CpCqp$, then $\vdash C\alpha_n C\alpha_{(n-1)} \dots C\alpha_2 C\alpha_1\alpha_2$ first, and it then quickly follows that if $\vdash CsCrCpCqp$, then $\vdash C\alpha_n C\alpha_{(n-1)} \dots C\alpha_2 C\alpha_1 \alpha_2$.  Or at least proving if $\vdash CsCrCpCqp$, then $\vdash C\alpha_n C\alpha_{(n-1)} \dots C\alpha_2 C\alpha_1\alpha_2$ takes longer to do than the other case.

Also, I do think that logicians would regard $\vdash$CpCqp as a stronger statement than $\vdash CsCrCpCqp$ (which implies 2) stronger than 1)), since CpCqp qualifies as an organic theorem in that it has no proper subformulas which qualify as tautologies.  On the other hand, $CsCrCpCqp$ qualifies as non-organic since it has a proper subformula which qualifies as a tautology.


Why $\hat{\mu}\in L^2(\mathbb{R}^n)$ implies $\mu$ absolutely continuous? Let $\mu$ be a compactly supported probability measure on $\mathbb{R}^n$.  Suppose that $\hat{\mu}\in L^2(\mathbb{R}^n)$, then $\mu$ must be absolutely continuous with respect the Lebesgue measure. How to prove ?

The Fourier transform $\hat{\mu}$ is given by
 $$
 \hat{\mu}(\xi)=\int e^{-2\pi i\langle\xi,x\rangle}d\mu(x).
 $$

Thank you and sorry if my English wasn't correct.

Hint

Prove that $\mu$ has a $L^2$ Lebesgue density.
Then conclusion is trivial since

$$\mu(E) = \int_E f_\mu \ \mathrm d\lambda$$

If you provide the definition of absolutely continuous measures, you will be able to prove it.


What free tools can I use to plot complex functions on the complex plane? I would like to plot $y = e^{ix}$ to view the imaginary unit circle and then change different parameters to see what happens. I am having trouble getting wolfram alpha to do this though. Is there a good way?

First you must define your complex function  as a curve in $\Bbb R^3$ using a parameter, by example $t$, and separating each coordinate.

In our case we have that

$$f(x):=e^{ix}=\cos (x)+i\sin (x)$$

then we can transform the above function in a parametric curve in $\Bbb R^3$ writing

$$f(t)=t\cdot{\bf i}+\cos(t)\cdot{\bf j}+\sin(t)\cdot{\bf k}$$

Using Geogebra we can write in the algebra view

Curve[t,cos(t),sin(t),t,-5,5]

and the graph can be viewed in the 3DView tab. Indeed Geogebra is an extraordinary tool by it simplicity and portability, we can define applets easily as this (with little work we can add text boxes or buttons for any kind of interactivity!)

Using the Wolfram language (in wolfram alpha or the Wolfram programming lab) we can write

ParametricPlot3D[{t,Cos[t],Sin[t]},{t,-5,5}]

And with a very similar code we can write in SageMathCell

t = var('t'); parametric_plot3d([t,cos(t),sin(t)],(t,-5,5), aspect_ratio=[1,1,1], zoom=1.5)

(the result can be seen here).

There are a lot of different tools to graph online, by example I discovered today the library plotly that can be used in many programming languages and online too! The homepage is full of tutorials for any kind of plot, in our case we have a tutorial for a curve in $\Bbb R^3$ here.

Try https://treena.org/lesson/complex-grapher
It should do most of what you want

disclaimer: I helped build it.


Prove that $\forall d > 0$ there exists $x_1, x_2 \in \mathbb{R}$ such that $x_1 - x_2 = d$ and $f(x_1) = f(x_2)$ Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with the property that $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x \rightarrow -\infty} f(x)$ exist and are equal. Prove that $\forall d > 0$ there exists $x_1, x_2 \in \mathbb{R}$ such that $x_1 - x_2 = d$ and $f(x_1) = f(x_2)$.

I applied Rolle's theorem on $\mathbb{R}$ since $\lim_{x \rightarrow \infty} f(x)= \lim_{x \rightarrow -\infty} f(x)$. So, there exists $c \in \mathbb{R}$ such that $f'(c) = 0$. This means that our function has a minimum or a maximum. Now, since the limits at $\infty$ and $-\infty$ are equal we can conclude that $\forall d > 0$ there exists $x_1, x_2 \in \mathbb{R}$ such that $x_1 - x_2 = d$ and $f(x_1) = f(x_2)$.

Is this a correct solution? If not, please help me find a good one.

Thank you!

Since $f$ is not assumed differentiable you cannot use Rolle's theorem; but the IVT is available, and does the job.

After substracting a suitable constant from $f$ (which is irrelevant for the problem at hand) we may assume that $\lim_{x\to-\infty}f(x)= \lim_{x\to\infty}f(x)=0$, and that there is an $a\in{\mathbb R}$ with $f(a)>0$. Using a standard argument we then can infer that there is a $\xi\in{\mathbb R}$ with $$f(x)\leq f(\xi)\qquad(-\infty<x<\infty)\ .$$
Given a $d>0$ put $g(x):=f(x)-f(x+d)$. Then
$$g(\xi-d)=f(\xi-d)-f(\xi)\leq0,\qquad g(\xi)=f(\xi)-f(\xi+d)\geq0\ .$$
As $g$ is continuous it follows that there is a point $x_1\in[\xi-d,\xi]$ with $g(x_1)=0$. Put $x_1+d=:x_2$; then $|x_2-x_1|=d$ and
$$f(x_1)-f(x_2)=f(x_1)-f(x_1+d)=g(x_1)=0\ .$$

Since you can't use Rolle's theorem because your function is not continuous, maybe we can try something else. The idea is this: you take a point $x_0$ and calculate $x_0+d$. Now you are certain that the two points are at the right distance from each other.

Then calculate $f(x_0)$ and $f(x_0+d)$. Then define $g(x)=f(x)-f(x+d)$. If this function is zero, we are done. We have found our points $x_1$ and $x_2$.

So, start out with one point $x_0$ and calculate $g(x_0)$. It will either be positive or negative, say without loss of generality that it's positive. We have
\begin{equation}
0<g(x_0)=f(x_0)-f(x_0+d).
\end{equation}
Now, we know that at some other point we must have
\begin{equation}
0>g(x_{3})=f(x_{3})-f(x_{3}+d),
\end{equation}
otherwise we would have a monotonically decreasing sequence and then the limits at infinity can not be the same.

You know that g(x) is a continuous function, so you can apply the Intermediate Value Theorem, which is true for all continuous functions on a bounded interval.

The IVT gives you a point $x_1$ at which $g(x_1)=f(x_1)-f(x_1+d)=0$. This gives you your $x_1$ and $x_2$.


Orthogonal complement in a finite field ${\mathbb Z}^{n}_{q}$ When $V=\mathbb{Z}^n_q$ is a vector space, where $\mathbb Z_q$ is the set of integers modulo prime $q>2$, are the following statements true?

  If $U ⊂ V$ is a $k$-dimensional subspace, then
  $$
U^⊥= \{ x∈V∣x^Tu=0 \ \mathrm {for\ all}\ u∈U \}
$$
  is of dimension $n-k$.
  (If statement 1 is true) $U^⊥$ can be written by $n-k$ orthogonal vectors.

For an example of statement 1, if $V=\mathbb{Z}^3_3$, and $U=\{ a_1(1,2,1)^T+a_2(0,1,1)^T\ |\ \forall a_1,a_2\in\mathbb Z_3\} $  then $U^⊥$ can be expressed with basis $(1,2,1)^T$.

I'm not familiar with using linear algebra in finite fields, so I don't know how I should handle "orthogonal" vectors. I wrote down some examples and am pretty sure that statement 1 holds in finite fields, as it does in real case, but I'm not so certain of statement 2. Furthermore, I don't have a clue on how to prove these statements.

I'm not used to using English, so if there's anything wrong with my explanation, don't hesitate to ask.
Thanks for your help!

You probably want to apply rank-nullity, so find a linear transformation which expresses $U^\perp$ as an image or kernel. For instance, consider the linear transformations $T_u \colon V \to \mathbb{Z}_q$ given by $x \mapsto x^T u$. Then $U^\perp = \cap_{u \in U} \ker T_u$. Fix a basis $u_1, \ldots, u_k$ of $U$; it's clear $U^\perp = \cap_{u_i} \ker T_{u_i}$. Glue the transformations together by considering $T \colon V \to \mathbb{Z}_q^k$ given by $x \mapsto (x^T u_i)_{i=1}^k$. Hence $U^\perp = \ker T$, so $\dim U^\perp = n - \dim {\rm im}(T)$. So, we just need to show $\dim {\rm im}(T) = k$. For that, notice that $T$ is given by right-multiplying $x^T$ by the matrix $(u_1, \ldots, u_k)$ ($u_i$ being column vectors here). By definition, this has column-span of dimension $k$, so it has row-span of dimension $k$, so $\dim {\rm im}(T) = k$. (For reference on this part, see here.)
Since it's possible for $u^T u = 0$ for $u \neq 0$, one can't interpret $u^T u$ as (square) "length" and divide by it. I can't think of anything else "orthonormal" would refer to. Edit: Ah, you actually mean "orthogonal", in the sense that you want a basis $\{u_i\}$ for $U^\perp$ where $u_i^T u_j = 0$ for all $i \neq j$. It's a little hard to find counterexamples, since the Gram-Schmidt process actually works with the usual inner product replaced by $u^T v$, at least whenever you don't divide by $0$, i.e. whenever none of your intermediate vectors satisfies $u^T u = 0$. A problem with Mike's subspace is that it's a subset of $Z = \{u \in \mathbb{Z}_q^n : u^T u = 0\}$. Interestingly, it seems $Z$ is itself a vector subspace if and only if $q=2$ (more generally, in characteristic $2$). Maybe this will help guide the construction of a counterexample/a proof that no counterexamples exist for $q>2$?

You haven't said what your definition of $U^\perp$ is, so it's rather hard to check whether it's equal to something.  You also don't say what goal or application you have in mind, so it's also hard to suggest how to get there.

But orthogonality over finite fields is a very strange concept, and certainly does not perfectly translate from the real case.  For example, suppose that $n=q$, and $U=\langle u \rangle$, where $u=(1,1,\ldots,1)$.  Then $u^T u = 1+1+\ldots+1 = q = 0$, so $u\in U^\perp$ if we use the above definition!

Also, you should keep in mind that, just like over $\mathbb{Q}$, you might have to extend the field to scale a basis to be orthonormal.  For instance, take $(1,1)$ over $\mathbb{F}_3$.  Then the normalized vector is $\frac{1}{\sqrt{2}} (1,1)$.  But $2$ isn't a square in $\mathbb{F}_3$!  We need to look in $\mathbb{F}_9^2$, the vector space over the field with 9 elements, to find the normalized vector.

With these caveats, much of the same theory works.  For example, it seems to be true that $\dim{U} + \dim{U^\perp} = \dim V$, if we use your Statement #1 as a definition.  But it's no longer true that $V=U+U^\perp$, since we may have $U\cap U^\perp \neq 0$.

As a general rule, things that you can prove using the basic algebra of matrices and determinants will still be true, while anything requiring the inner product might fail if it uses properties of $\mathbb{R}$.


How to prove that a function is Riemann integrable if and only if it is Darboux integrable? (Pg. 16 Exercise 1.1.22. in Terry Tao’s Measure Theory) Show that if $f:[a,b]\to\mathbb{R}$ is a bounded function, then it is Riemann integrable if and only if it is Darboux integrable such that its Darboux and Riemann integrals are equivalent. In other words
$$\int_a^b f(x) \ dx=\lim_{\|P\|\to 0}\mathcal{R}(f,\mathcal{P})\Longleftrightarrow\sup_{g\le f;\kern 0.1em\mathrm{piecewise \kern 0.1em constant}}\textrm{p.c.}\int_a^b g(x) \ dx =\inf_{h\ge f,\kern 0.1em\mathrm{piecewise \kern 0.1em constant}}\textrm{p.c.}\int_a^b h(x) \ dx.$$
Could anyone give me any hints?

I can provide a brief outline and you can develop that into a proper proof. First you need to understand that the definition of Darboux integral as given in your post is equivalent to the formulation given in most textbooks. Thus given a bounded function $f:[a, b] \to\mathbb {R} $ we take an arbitrary partition $$P=\{a=x_0,x_1,x_2,\dots,x_n=b\}, x_{i-1}<x_{i},i=1,2,\dots,n$$ of $[a, b] $ and define Darboux sums $$L(f, P) =\sum_{i=1}^{n}m_{i}(x_i-x_{i-1}),\, U(f,P)=\sum_{i=1}^{n}M_{i}(x_i-x_{i-1})$$ where $$M_i=\sup_{x\in[x_{i-1},x_i]}f(x),\, m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$$ Let $\mathcal{P}[a, b] $ denote the set of all possible partitions of interval $[a, b] $. The Darboux sums are themselves bounded ($L(f, P) \leq M(b-a), U(f, P) \geq m(b-a) $ where $M, m$ are supremum and infimum of $f$ on $[a, b] $) and thus $$\overline{J} =\inf_{P\in\mathcal{P} [a, b]} U(f, P), \, \underline{J} =\sup_{P\in\mathcal{P} [a, b]} L(f, P) $$ exist. You need to show that $$\underline{J} =\sup_{g\leq f, \text{ piecewise constant}}\text{p.c.}\int_{a}^{b}g(x)\,dx,\,\overline{J}=\inf_{h\geq f, \text{ piecewise constant}} \text{p.c.}\int_{a}^{b}h(x)\,dx$$ which establishes the equivalence of definitions of Darboux integral as given by Tao and as given in other textbooks.

Once this is done you need to show that $$\underline{J} =\lim_{|P|\to 0}L(f,P),\,\overline{J}=\lim_{|P|\to 0}U(f,P)$$ This is already done in this answer.

Since a Riemann sum is always sandwiched between upper and lower Darboux sums, it follows that if $f$ is Darboux integrable then both upper and lower Darboux sums tend to a common limit as $|P|\to 0$ and therefore so do the Riemann sums and the value of the integral defined by these approaches is also same.

To go the reverse way (from Riemann to Darboux) one just needs to prove that one can find a Riemann sum as near to a Darboux sum as we please by choosing the tag points $t_i\in[x_{i-1},x_i]$ such that $f(t_i) $ is near $M_i$ (or $m_i$ as needed). And thus if Riemann sums tend to a given value then the Darboux sums also do the same.

The approximation for a Riemann integral is bounded above and below respectively by the Darboux sup and inf.  Therefore Darboux implies Riemann.

Riemann sums can be constructed as Darboux upper and lower sums.  Therefore Riemann implies Darboux.


Is an object located within the field of view of a robot? Background:

Assume we have three robots as given in the figure.

The robot in the middle has camera in the front and in the back.

How to calculate whether:

the grey robot is in the field-of-view (FOV) -- marked with orange line -- of the front camera of the green robot?
the blue robot is in the FOV of the back camera of the green robot?

Current knowledge:

Ignore the rectangular form of the robots, (if that helps).
Coordinates of all of the robots are known, they are given as Coord(x,y)
The length of the maximum range of the FOV is 100m --> that's the length of the orange lines.
The half-angle of the FOV is 45 degrees (the angle between the blue arrow and the orange line)
The coordinate system used to obtain the heading of the robots ranges between -pi and pi (as given in the figure below). So the heading for green robot is -pi

What has been done so far:

So far, I tried to adapt some implementation to my case but I am not sure whether it is correct, so I need validation on that front, or comments from your side.

double distanceFromGreen2Grey = greenPos.distance(greyPos);

// Angles
double greenHeading = r0->getHeading();
double greyHeading = r1->getHeading();

// Calculate normalized vectors
Coord vectorFromGreen2Grey = (greyPos - greenPos) / distanceFromGreen2Grey;
Coord greenHeadingVector = Coord(cos(greenHeading), sin(greenHeading)) * 1; // <-- 1 used for front
Coord greyHeadingVector = Coord(cos(greyHeading), sin(greyHeading)) * -1; //  <-- -1 used for back

// why use acos here?
double angle = acos(vectorFromGreen2Grey * greenHeadingVector) / M_PI * 180;
double bearing = acos(vectorFromGreen2Grey * greyHeadingVector) / M_PI * 180;

bool inRange = distanceFromGreen2Grey <= 100;
bool inAngle = (vectorFromGreen2Grey * greyHeadingVector) > cos(M_PI/4);
bool inAngle2 = ((greyPos - greenPos)/cos(M_PI/4))*greyHeadingVector >= distanceFromGreen2Grey;
bool withinBearing = (vectorFromGreen2Grey * greenHeadingVector) < 0;

Extra question: How to find the end points for the orange lines so I can draw lines from the coords of the green robot to the endpoints of the orange lines. The FOV lines should move as the robots move too.

References:

I found other questions on SE which address this topic, but they don't use acos as in my code.

https://stackoverflow.com/questions/22542821/how-to-calculate-if-something-lies-in-someones-field-of-vision mentions something like velocity vector but I could not map that to my case.
Something similar has also been asked here: How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?

This is almost exactly the same problem as one of the ones you cite. The only interesting difference is that instead of having to determine the camera’s facing from the endpoints of a line segment (robot arm section), you already know it. In addition, the field of view has a limited range, but that’s not a significant complication, nor is the addition of a rear-facing camera.

To review, in the cited question the target point is in the field of view if $$(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta,\tag{*}$$ where $Q$ is the target point, $P_4$ is the location of the camera, $P_3$ the other end of the robot arm segment, and $\theta$ is the half-angle of the field of view. Since we don’t have a robot arm holding the camera, let’s just call the camera’s location $P$ and let’s call the green robot’s heading $\phi$. $P_4-P_3$ gives us a vector in the direction of the camera’s facing, so assuming that the camera is facing directly forward, we can use the unit vector in that direction: $(\cos\phi,\sin\phi)$. Using a unit vector also eliminates the factor of $\|P_4-P_3\|$, since that came from normalizing the camera-direction vector. For this camera, $\theta=\pi/4$, so $\cos\theta=1/\sqrt2$. Putting this all together and moving the $\sqrt2$ to the left side, we get $$\sqrt2\,(Q-P)\cdot(\cos\phi,\sin\phi)\ge\|Q-P\|.\tag{1}$$ The range check is, of course, $\|Q-P\|\le100$. To check the rear view, just reverse the unit facing vector to $(-\cos\phi,-\sin\phi)$.

Since you have two conditions that both involve testing against $\|Q-P\|$, it’s more efficient to test the one that requires less work first. I suggest testing the square of the distance first, i.e., $\|Q-P\|=(Q-P)\cdot(Q-P)\le100^2$, to avoid computing a square root if the target is too far away. I can’t tell from your code snippet whether or not that’s practical, though. If the target isn’t too far away, then you can go ahead and take the square root and compute the rest of test (1).

Looking at your code snippet, it looks like you’re trying to do something along these lines, but there’s an error. The gray robot’s heading is wrong thing to use for inAngle. The target’s heading is irrelevant to deciding whether or not it’s visible. In the original formula, $(P_4-P_3)/\|P_4-P_3\|$ corresponds to your greenHeadingVector instead. Observe, too, that at least for this computation, there’s no need to normalize vectorFromGreen2Grey. That saves you a division operation. Of course, if you need the normalized value for other things, it’s almost certainly more efficient to normalize once, as you do.


Real numbers having disjoint neighborhood How does one prove that each point in real line has disjoint neighborhoods?

HINT: Do you mean that any two points on the real line have disjoint neighborhoods? If so, consider that for any real numbers $x$ and $y$ with $x\ne y$, the average $\frac12(x+y)$ is between $x$ and $y$. (Exactly halfway between them, in fact, though you don’t need that.)

If you really meant that there is a pairwise disjoint collection $\{V_x:x\in\Bbb R\}$ of open sets such that $x\in V_x$ for each $x\in\Bbb R$, that’s simply false: every pairwise disjoint family of non-empty open sets in $\Bbb R$ is countable, because each of them must contain a different rational number, and there are only countably many rational numbers.

Since $\mathbb{R}$ is Hausdorff, distinct points in $\mathbb{R}$ have disjoint neighbourhoods.


Let S be the set of all real numbers in the interval (0; 1) whose decimal expansions contain only 0's, 4's and 8's. Prove that S is uncountable. I just prove
$$S_1=\{0.4,0.8\}$$
$$S_2=\{0.04,0.08,0.44,0.48,0.84,0.88\}$$
...
So I can calculate the number of elements in $S_n = 2 \times 3^{n-1}$

I just prove it is countably infinite.

Let $f(x)$ be the number whose ternary representation has a zero wherever $x$ has a zero, a $1$ wherever $x$ has a $4$, and a $2$ wherever $x$ has an $8$.  Then $f$ defines a bijection between $S$ and $(0,1)$; hence $S$ has the same cardinality as $(0,1)$.

First of all, $\sum_{k=0}n #{S_n}$ is not $2\cdot 3^{n-1}, it is $3^n -1$.

But that sort of summing is moot.  You can show the union of all $S_n$ to be uncountable as follows:

List all the entries in some order; if the set is countable you can do that.  Let's call the $k$-th number in that list $L_k$.  Now consider the number $X$
formed as a decimal containing only zeros, fours and eights according to the following rules for each digit:

If the $n$-th digit of $L_n$ is 0, then set the $n$-th digit of $X$ to 4.
If the $n$-th digit of $L_n$ is 4, then set the $n$-th digit of $X$ to 8.
If the $n$-th digit of $L_n$ is 8, then set the $n$-th digit of $X$ to 0.

Then $X$ cannot be in the list $\{L_k | \k \in Bbb{N} \}$ since if $X$ is in that set, then $X = L_s$ for some integer $s$, and consider the $s$-th digit of $X$, which is by construction different than the $s$-th digit of $L_s$.  So we have a contradiction:  If the list is complete, there is a number that is not in the list.  Therefore, such a counting order cannot exist, and the set is uncountable.


How to work out how many combinations in a $16\times 16$ grid with $256$ values in each? I have a $16\times16$ grid. $256$ squares.

In each square there can be a value from $1$-$256$.

How do I work out how many total combinations the complete grid could have?

I would like the know the formula so I can work it out for different grid sizes.

EDIT (thanks for @RSerrao's comment): If, the values are distinc, you can think like this: the first cell can have 256 posibilities, the second cell can have 255 posibilities and so on. Then, you can get the formula.

If not, i.e. the values can be repeated, so each cell has 256 posibilities.

if you have a n*n grid then the total numbers you can use is n^2 , now at a first square u have chances (n^2)chances ,in the next you have (n^2-1) chances , in such a process the total set of combination you have (n^2)!.


Find the maximum value of $\sqrt{x - 144} + \sqrt{722 - x}$ Find the maximum value of $\sqrt{x - 144} + \sqrt{722 - x} .$

What I Tried :- Let me tell you first . What I Tried is absolutely silly , but you may check for it .
I thought AM-GM would do the trick and got :-
$\sqrt{x - 144} + \sqrt{722 - x} \geq 2 \sqrt{\sqrt{(x - 144)(722 - x)}}$
$\rightarrow \sqrt{x - 144} + \sqrt{722 - x} \geq 2 \sqrt{\sqrt{(-x^2 + 866x + 103968)}}$
From here I realised that there's no useful information I can find for $x$ .
Back to Another Attempt :- let $P = \sqrt{x - 144} + \sqrt{722 - x}$ . Then :-
$P^2 = 578 + 2\sqrt{(x - 144)(722 - x)}$
As $P \geq 0$ , we have that $P^2 \geq 578 \rightarrow P \geq 17\sqrt2 .$
This Attempt seemed reasonable and I thought I already found the solution, but then I realised that this is the minimum value of $P$ , and now I am hopeless .
Wolfram Alpha gives the answer to be $34$ , which can also be checked by Trial and Error of integer values , but that does not seem to be a proof of this problem .
Can anyone help me with this?

By Cauchy-Schwartz:
$a=x-144,b=722-x$
$\sqrt{a}+\sqrt{b}\le \sqrt{(a+b)(1+1)}=\sqrt{1156}$
(You can read https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality)
Alternbatively
${(\sqrt{a}+\sqrt{b})}^2=a+b+2\sqrt{ab}$
recall by AM-Gm
$2\sqrt{ab}\le a+b$
thus
${(\sqrt{a}+\sqrt{b})}^2\le 2(a+b)$
So finally,
$\sqrt{a} + \sqrt{b} \le \sqrt{2(a+b)}$
Equality holds for $a=b$.

By C-S $$\sqrt{x-144}+\sqrt{722-x}\leq\sqrt{(1+1)(x-144+722-x)}=34.$$
The equality occurs for $(1,1)||(x-144,722-x),$ which says that we got a maximal value.


What does the Cartesian equation of a Parametric function tell us? I've been taught that a Parametric function can be converted into a Cartesian one by eliminating the parameter $t$ but I've never been taught of how it specifically relates to the Parametric. Does it form an exact replica or what?

The parametric curve converted into a cartesian curve, when it is possible to do so, passes through all of the same points as the original. In that sense, it is an "exact replica." The information lost is how fast the curve gets traced out. Depending on the situation, this piece of information may or may not be important.

parametric equn of parabola

$$ x= 2 f t ;\, y = f t^2 $$

Find $t$ from first part

$$  t = x/(2 f) $$

Plug it into second part

$$ y = f (x/(2f) ) ^2 $$
simplify

$$ 4f y = x ^2 $$
Both ways it represents the same curve.

If $t$ represents time we know how the speed of x,y components are individually
represented. Here x speed is constant but y speed increase with time.


Solve $\lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}$ Prove that $$\lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}=\frac{1}{2}$$

$$
\lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}=\lim_{x→0} \frac{\tan^{-1}\frac{x}{\sqrt{1-x^2}}-\tan^{-1}x}{x^3}
\\
=\lim_{x→0} \frac{\tan^{-1}\bigg[\frac{\frac{x}{\sqrt{1-x^2}}-x}{1+\frac{x^2}{\sqrt{1-x^2}}}\bigg]}{x^3}
=\lim_{x→0} \frac{\tan^{-1}\bigg[\frac{x(1-\sqrt{1-x^2})}{\sqrt{1-x^2}+x^2}\bigg]}{x^3}\\
\text{since, }\frac{x^2}{\sqrt{1-x^2}}>-1
$$

I think I am getting stuck here, do not really see how to proceed further with the steps.

Note: I don't want to use L'Hospital's rule as I 'd like to solve it without using the derivatives of $\sin^{-1}x$ and $\tan^{-1}x$.

hint

Sorry, i understood that $\sin^{-1}(x)$ means $\frac{1}{\sin(x)}$ instead of $\arcsin(x)$.


Generalizing Bernoulli's inequality We are already familiar with Bernoulli's inequality: $(1+px)\le(1+x)^p$ for $x\ge-1, p\ge1$.

Can this be generalized to say something useful about $(1+p_1x_1+\cdots+p_nx_n)$? For instance, one would hope that something like this could hold:
$$(1+p_1x_1+\cdots+p_nx_n) \le (1+x_1+\cdots+x_n)^{p_1+\cdots+p_n}$$
However, this doesn't seem obvious. If you have any other ideas than using Bernoulli, that's great too, of course.

$$(1+p_1x_1+\cdots+p_nx_n) \leq (1+\frac{p_1x_1+\cdots+p_nx_n}{p_1+...+p_n})^{p_1+...+p_n}  \leq (1+x_1+\cdots+x_n)^{p_1+...+p_n}\,.$$

whenever

$$p_1x_1+\cdots+p_nx_n < (p_1+..+p_n)(x_1+..x_n) \,.$$

This is true for example if $x_1,..x_n \geq 0$, but also in many other cases.

Negative $x_i$ I n general the inequality is not true if we allow for negative $x_i$, the problem is that $p_1,..,p_n$ are independent.

For example, pick $x_1>0, x_2 <0$ so that $x_1+x_2 <0$. Then for all $p_1, p_2$ we have

$$(1+x_1+x_2)^{p_1+p_2} <1 \,.$$

but, if $p_1 > \frac{-p_2x_2}{x_1}$ then

$$(1+p_1x_1+p_2x_2)>1 \,.$$

Note that $x_1+x_2<0$ can be made as close to 0 as one wants.

We can use the binomial theorem to prove that $ (1+x)^p \geq ( 1 + p x )\,,$ $p\geq 1$ (finite natural number), $x \geq 0\,, $ that is

$$ (1+x)^p = 1 + p x + \frac{p(p-1)}{2} x^2 + \dots+x^p \geq 1 + p x \,,$$

for $ x\geq 0\,, p \geq 1\,. $

Now apply the binomial theorem to $ ( 1 + x_1 + x_2 )^{p_1+p_2} $ for $x \geq 0\,, p_i \geq 1\,, i=1,2 \,, $ ($p_1\,,p_2$ are finite natural numbers)

$$(1+x_1+x_2)^{p_1+p_2} = 1+(p_1+p_2)(x_1+x_2)+\dots+(x_1+x_2)^{p_1+p_2} ) = 1 + p_1 x_1 + p_2 x_2 + p_1
x_2 + p_2 x_1 + \dots \geq 1 + p_1 x_1 + p_2 x_2 \,, $$
for $ x_i\geq 0\,, p_i \geq 1 \,,i=1\,,2 \,. $

One can generalize this process to the general case in your problem. I'd like you to investigate the case when $ x_i \geq -1\,, i = 1,2,\dots, n\,.$


Proving uniqueness of solution in Boolean algebra System:
$a+x=1 \land a\cdot x=0$ has unique solution for x, for all values of $a \in B$. It is obvious that $x=a'$ is one solution , but how to prove the it is only one? I have tried assuming that there is another solution $b\neq a'$, but I keep getting nowhere with using that and cycling axioms.

Try making the truth table for this problem. Assume $0$ and $1$ as False and True respectively. Also:

$$P:a+x=1$$

$$Q:a.x=0$$

$$R:P\wedge Q$$
For instance when $a=0,x=0$ then then the statement $P:a+x=1$ is false so the value of $P=0$. So  now let us consider every case.
$$$$\begin{array}{|c|c|c|} \hline
a & x & P:a+x=1&Q:a.x=0&R:(a+x=1)\wedge (a.x=0) \\ \hline
0 & 0 & 0&1&0 \\ \hline
0&1&1&1&1\\ \hline
1&0&1&1&1\\ \hline
1&1&1&0&0\\ \hline
\end{array}$$$$
Now you can see there are only two cases for the statement $R$ to be True. The cases are $a=0,x=1$ and $a=1,x=0$. And you want to find the solution of $x$ in terms of $a$ so it is obvious that:
$$x=\bar a$$

The solution is unique because we have considered every value of $a$ and $x$ in the truth table.Satisfying both the statements $P$ and $Q$ you find there is only one possibility.

Hope this helps...

There are a couple of ambiguities in the question.  First, "Boolean Algebra" can mean a general complemented distributive lattice, or the special case of this with just the two elements $\ 0\ $ and $\ 1\ $.  Evidently, SNEHIL SANYAL's answer has (perfectly reasonably, I think) has adopted the latter interpretation.

Second, while the symbol "+" is occasionally used to represent the join operation of a Boolean algebra,  it is more commonly used—in my experience, at least—to represent the symmetric difference operation—that is, $\ a+b = a.b' \vee a'.b\ $, where "$\vee$" is here used to denote the join.  In the following, I assume it was intended to denote the join, and I will replace it with "$\vee$" .

The equations to be solved are
$$ a\vee x = 1,\ a.x = 0 $$
Taking the join of both sides of the second equation with $\ a'.x\ $, we get:
$$ a'.x = a'.x \vee a.x = (a' \vee a).x = 1.x = x\ ,$$
by using the distributivity of meet over join.  Now taking the meet of both sides of the first equation with $\ a'\ $, we get:
$$ a' = a'.(a\vee x) = a'.a \vee a'.x = 0 \vee a'.x = a'.x\ ,$$
again using the distributivity of meet over join.  Combining these two results, we get $\ a' = a'.x = x\ $.


No of n digits no in incresing order i have a set{1,2,3}, now i want to find no of n digits no such that they always start with 1 and ends with 3 in  non decreasing order and all nos from set must be included atleast by 1.
for ex. if n=4 then, 1123,1223,1233 are only combination, so ans is 3.

Now, in general terms, consider a set(1,2,3....,k} ,then what should be the answer such that they starts with 1 and ends with k and they must be in non decreasing order.

The numbers in the $n$ digits must be ordered as $1,2,\cdots,k$. Let $a_t$ be the number of $t$s in the $n$ digits. Then $a_1+a_2+\cdots+a_k=n$ and $a_t\geq1$ for all $1\leq t\leq k$. And for each such combination of $a_t$s, there corresponds exactly one $n$ digit number satisfying the condition.(112222333 corresponds to 2+4+3=9)

So the number of possible $n$ digit numbers are same as the number of possible $a_t$s. Now $(a_1-1)+(a_2-1)+\cdots+(a_k-1)=n-k$ and each term in the summation should be a nonnegative integer. Each solution of this equation corresponds to one arrangement of $(k-1)$ '$+$'s and $(n-k)$ '$1$'s and vice versa.(For example, 1+3+2=6 corresponds to 1+111+11) Thus the number of solutions to the equation is $\binom{n-1}{k-1}$.

As karvens pointed out, my original answer on this was wrong. It is, in fact, a "stars and bars"  situation, as per his solution. We have "bars" representing the divider "|" between each type of digit and "stars" representing each digit. E.g. the string "$|*|**|**|$" represents the number "12233".

We must have a bar at the end of the string and each of the $k$ digits must appear once so we are arranging $k-1$ lots of "$|*$" and $n-k$ lots of "$*$". So we have $k-1 + n-k = n-1$ objects in total and we need to choose places for the $k-1$ lots of "$|*$".

The number of ways to do this is $$\binom{n-1}{k-1}$$


Odd numbers expressible as a sum of 3 composite odd integers. How many positive odd integers between $0$ and $999$ inclusive can be
written as sum of $3$ odd composite positive integers?

I am trying to approach this by looking at equivalence classes $\mod 6$ but the calculations are a bit tedious.

Let's start with integers equivalent to $3\pmod6$.  The smallest is obviously $27=9+9+9$.  Every following odd multiple of $3$ also works as it can be written as $9+9+3(2n+3)$.  $27=6(4)+3$ and $999=6(166)+3$, so we have $163$ odd multiples of $3$.

The first odd number not a multiple of $3$ is $5(5)=25\equiv1\pmod6$.  So the first value equivalent to $1\pmod6$ is $9+9+25=43=6(7)+1$.  The same trick in the first step works here, so we have $160$ values here.

The first odd number equivalent to $5\pmod6$ is $5(7)=35$.  The smallest value that works equivalent to $5\pmod6$ is then $9+9+35=53<9+25+25$.  $53=6(8)+5$.  We have $158$ numbers here that work.

So our final tally is $163+160+158=481.$

Let the three odd numbers be $2a + 1$, $2b + 1$, $2c + 1$.

The sum of these numbers is $6(a + b + c) + 3$. Our goal is to find out the number of integers of that form.

Equating it with 999, we find $a + b + c$ equals 166. This implies there should exist 166 odd numbers that are the sum of three odd numbers.


Is the base case $m=0$ in this proof by induction correct? Let $\Bbb J(n)$ the set of bounded intervals in $\Bbb R^n$, that is, if $I\in\Bbb J(n)$ there are bounded intervals $I_k\in\Bbb R$ such that $I=\prod_{k=1}^n I_k$, where the product is the cartesian product (we consider the empty set an interval in $\Bbb R$ and consequently also in $\Bbb J(n)$).
Then let any finite collection $I,J_1,J_2,\ldots, J_m\in\Bbb J(n)$, and I want to prove using strong induction that
$$I\setminus\left(\bigcup_{k=1}^m J_k\right)\text{ can be written as a union of finite pairwise disjoint intervals}\tag1$$
If I choose the base case $m=1$ then the proof becomes messy, but if I choose the base case $m=0$ then I find that $(1)$ is just $I$, so the base case holds trivially, and the rest of the proof is very simple.

Proof: assume that a base case $m=0$ or $m=1$ holds, then we choose $(1)$ as hypothesis for all $1\le k\le m$ for some $m$. Then the induction step gives
$$I\setminus\left(\bigcup_{k=1}^{m+1}J_k\right)=\left(I\setminus\left(\bigcup_{k=1}^m J_k\right)\right)\setminus J_{m+1}\\=\left(\bigsqcup_{j=1}^r H_j\right)\setminus J_{m+1}=\left(\bigsqcup_{j=1}^r H_j\right)\cap J_{m+1}^\complement\\=\bigsqcup_{j=1}^r(H_j\cap J_{m+1}^\complement)=\bigsqcup_{j=1}^r(H_j\setminus J_{m+1})$$
for $H_j\in\Bbb J(n)$, and using again the hypothesis we can write $H_j\setminus J_{m+1}$ as a union of disjoint intervals, so we are done.

There is a huge difference in difficulty in this proof choosing $m=0$ or $m=1$ as the base case, so I'm not sure if the proof for $m=0$ is completely right. Can someone confirm that we can choose $m=0$ in this proof without harm?
Also, I would like to know if someone knows how to handle easily the base case $m=1$. I tried but I lost myself in a mountain of intersections, unions and abstract thinking.
Of course, if someone knows how to prove $(1)$ in a different way I would like to know it.

Equivalently $(1)$ can be stated as: for any finite collection $J_1,J_2,\ldots,J_m\in\Bbb J(n)$ then $\bigcup_{k=1}^m J_k$ can be written as a pairwise disjoint union of intervals.
However, with this assertion, the induction step is as hard as the base case of the above when $m=1$.

EDIT:
Amazingly it seems that the base case $m=0$ is valid. Observe what happens with this reformulation of $(1)$:

Let $I\in\Bbb J(n)$ and $\mathcal S\subset\Bbb J(n)$ finite. Then there is some finite $\mathcal B\subset\Bbb J(n)$ such that $$I\setminus\left(\bigcup\mathcal S\right)=\bigsqcup\mathcal B\tag2$$

Then from $(2)$ we can do induction over $|\mathcal S|$ and start with zero, so it seems that the base case of $m=0$ on $(1)$ is perfectly valid.

In the forum Rincón matemático professor Carlos Ivorra answered the question clearly so I will copy here a resume:

If $P(0)$ is true and $P(n)\implies P(n+1)$ then the proof by induction is not automatically correct (there is a classical example), we also need to check that $P(n)\implies P(n+1)$ can be shown true for all $n\ge 0$.

In this case we can't prove $P(0)\implies P(1)$ so the base case $m=0$ is not useful for a proof by induction.


Example of the semisimple ring $R$ but $R^{{\rm op}}$ is not. Is there any example of this kind of rings? i don't have any imagine of this rings, if they are exist!

By the Wedderburn-Artin theorem, a semisimple ring is a product of full matrix rings over division rings. Conversely, a product of simple artinian rings (like full matrix rings over division rings) is semisimple. This is left-right symmetric.

No. Every semisimple ring is isomorphic to its opposite ring, and this fact is a corollary of Artin-Wedderburn theorem. So if $R$ is semisimple, then $R^{op}$ is semisimple.


Implementing backward Euler with a nonlinear system I'm trying to implement a backward Euler method in a situation when I have a nonlinear system of equations, and I'm having some trouble. My set up is as follows:

I have $\vec{f}(\vec{u}, t) = \begin{pmatrix} f_1(p(t), v(t), t) \\
f_2(p(t), v(t), t)\end{pmatrix}$.

$f_1, f_2$ are both nonlinear functions of $t, p(t)$ and $v(t)$, where $p, v$ are unknown functions of $t$.

The update scheme for backward Euler is $U^{n+1} = U^n + \Delta t f(t^{n+1}, U(t^{n+1}))$

It was suggested that I Taylor expand $f$ around $t^n, U(t^n)$.

Then my linearized update scheme, I believe, should be:

$U^{n+1} = U^{n} + \Delta t\big{(} f(t^n, U^n) + \Delta t \frac{Df}{Dt}\big{)}$.

I think my issue is the last term. I am computing it as $\frac{df}{dt} + J(t^n, U^n)\cdot f(t^n, U^n)$ where $J(t^n, U^n)$ is the Jacobian evaluated at $t^n, U^n$, and $\frac{df}{dt}$ is the function with the derivative taken only with respect to explicit $t$.

I have two issues:
1) It was suggested I would need to invert something to find the update. As it stands, no inversion is necessary. I simply add up the vectors on the right to get the left hand side
2) My answer is wrong; obviously a big problem.

I obviously don't want the answer (and haven't provided the details of the functions in any case, so you can't give it to me), but I do want to make sure I'm going about deriving this system the correct way. Am I missing something?

You have the wrong idea of the Taylor expansion. It is not the total time derivative, as that would also have to be applied to the left side. It simply is
$$
u_{n+1}\approx u_n+Δt\left(f(u_n,t_n)+\frac∂{∂u}f(u_n,t_n)(u_{n+1}-u_n)+\frac∂{∂t}f(u_n,t_n)Δt\right)
$$
This should give a result that is close to the most simple Rosenbrock method.

For better precision, solve the implicit equation for
$u_{n+1}$ using a Newton-like method. You look for the solution of the equation $$0=F(U)=U-hf(U,t_{n+1})-u_n.$$ The derivative of that function is approximately $F'(U)\approx I-hJ_n=I-hf_u(u_n,t_n)$ which is good enough for the simplified Newton method
$$
U^{[j+1]} = U^{[j]}-(I-hJ_n)^{-1}F(U^{[j]}).
$$
One might be able to keep the same derivative matrix and its factorization for multiple steps, as the simplified Newton method only demands a half-ways good approximation of the derivative to yield linear convergence to the solution. Observing the speed of convergence can be used to tell of when it is necessary  to refresh the derivative matrix.


$P_n=\{X\subseteq \mathbb{N}:|X|=n\}$ is countable. Let $n\in \mathbb{N}$ and $P_n=\{X\subseteq \mathbb{N}:|X|=n\}$.
How can I prove that $P_n$ is countable?

Let $(p_i)_{i \ge 0}$ be the sequence of prime numbers with $p_0 = 2$. To each $X = \{x_1, \dots, x_n \} \subseteq \Bbb N$ you may associate a natural number by the formula $X \mapsto p_{x_1} \dots p_{x_n}$. Note that if $p_{x_1} \dots p_{x_n} = p_{y_1} \dots p_{y_s}$ then necessarily $s = n$ and $x_i = y_i$ (modulo a possible reordering). Therefore, the mapping $X \mapsto \Bbb N$ described above is injective, therefore $P_n$ injects into $\Bbb N$, so it is at most countable.

Let us now see that $P_n$ is not finite. Well, this is easy, because it must contain all the subsets of the form $\{k+1, \dots, k+n\}$ for all $k \in \Bbb N$, and these are infinitely many. Therefore, $P_n$ is infinite.

Since $P_n$ is infinite and at most countable, it follows that it is countable.

To show that a set is countable a one to one correspondance of the elements of the set with the elements of N must be  established.
Is countable the union of two countable sets?
Let's see

C ={ c1, c2, c3,...
D ={ d1, d2, d3,...

C U D ={c1, d1, c2, d2, c3, d3,.....cj ,   dj ,
  N   ={ 1,  2,  3,  4,  5,  6,.....2*j-1,  2*j, ...

here we have the one to one rule, between C U D and  N

Now we only have to show that P2 can be expressed as a union of sets, it is isomorphic to N X N = Unions of 1 X N , 2 X N, 3 X N,...
And P3 is built from P2 X N,


$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$  ?

Let's use the arithmetic-geometric inequality.

That gives us $(xyz)^{\frac{1}{4}}≤\frac{1}{4}(x+y+z)\leq\frac{3}{4}$

thus $(xyz)^4\leq(\frac{3}{4})^{16}$

Now all you have to prove is that $x^3+y^3+z^3\leq3*(\frac{4}{3})^{16}$

But $x^3+y^3+z^3\leq 3^3+3^3+3^3=81$ because $x\leq3$ and $y\leq3$ and $z\leq3$.

Thus you have to prove that $3^{15+4}\leq4^{16}$.


Demystify integration of $\int \frac{1}{x} \mathrm dx$ I've learned in my analysis class, that

  $$ \int \frac{1}{x} \mathrm dx = \ln(x). $$

I can live with that, and it's what I use when solving equations like that.
But how can I solve this, without knowing that beforehand.

Assuming the standard rule for integration is

  $$ \int x^a \, \mathrm dx = \frac{1}{a+1} \cdot x^{a+1} + C .$$

If I use that and apply this to $\int \frac{1}{x} \,\mathrm dx$:

$$ \begin{align*}
  \int \frac{1}{x}\mathrm  dx &= \int x^{-1} \,\mathrm dx \\
                      &= \frac{1}{-1+1} \cdot x^{-1+1} \\
                      &= \frac{x^0}{0}
\end{align*} $$

Obviously, this doesn't work, as I get a division by $0$. I don't really see, how I can end up with $\ln(x)$. There seems to be something very fundamental that I'm missing.

I study computer sciences, so, we usually omit things like in-depth math theory like that. We just learned that $\int \frac{1}{x} dx = \ln(x)$ and that's what we use.

If you want to try to prove $\int\frac{\mathrm dx}x=\ln x + C $ (for $x \gt 0$), try the substitution

$$ \begin{align}
  x &= e^u \\
  \mathrm dx &= e^u \mathrm du
\end{align} $$

This substitution is justified because the exponential function is bijective from $\mathbb{R}$ to $(0,\infty)$ (hence for every $x$ there exists a $u$) and continuously differentiable (which allows an integration by substitution).

$$\int\frac{\mathrm dx}x=\int\frac{e^u\mathrm du}{e^u}=u+C$$

Now just use the fact that natural log is the inverse of the exponential function.  If $x=e^u,u=\ln x$.

I did sign up just to say some answers here are misleading.

$$\int \frac{1}{x} dx = \ln(x). $$

is not  a calculation but a definition. This is how ln(x) is defined. ln(x) is defined before exp(x)

http://en.wikipedia.org/wiki/Natural_logarithm#Definitions


Show that $C_0([a, b], \mathbb{R})$ is not $\sigma$-compact $C_0([a, b], \mathbb{R})$ is the space of real-valued continuous function on $[a, b]$. The hint says think Baire. So I assume that $C_0([a, b], \mathbb R)$ is $\sigma$-compact. Then it is the countable union of compact, and hence closed, subsets $M_i$. Since $C_0([a, b], \mathbb R)$ is a complete metric space, by Baire's theorem at least one of the $M_i$'s has a non-empty interior. But I got stuck here. I think I can derive a contradiction from this, but I'm not sure how. Any hints on how to solve this (using contradiction or not) would be much appreciated.

Hint: Show that the closed unit ball is not compact. Now conclude that if $U$ is a compact set then it has an empty interior.

Hint : show that a subspace of a vector space that has not empty interior is equal to the whole space.


Closed form of recurrence relation $F(n) = 2 + F(n-1) + F(n-2)$ I was figuring out an answer to the question,

  How many Boolean arrays of length $n$ could be formed if there are to
  be no two falses in a row?

I could see that it boils down to a Fibonacci like equation,

$$F(n) = 2 + F(n-1) + F(n-2)$$

But I am unable to find a closed form for this recursion. May be I am following the wrong direction here.

Please help.

Given that
$$
F(n + 2) = 2 + F(n + 1) + F(n)
$$
we have
$$
G(n + 2) = G(n + 1) + G(n)
$$
where $G(n) = F(n) + 2$. Because $G(1) = 3$ and $G(2) = 5$, we have
$$
G(n) = H(n + 3)
$$
where $H(n)$ is Fibonacci sequence with $H(1) = 1$ and $H(2) = 1$. The closed form then can be derived from the closed form of Fibonacci numbers.

I like to do these recurrences in terms of generating functions:

Let $A(x) = \sum_{n\geq1} F(n)x^n$ for some $x<1$. Now take your recurrence,
$$
F(n+2) = 2 + F(n+1) + F(n),
$$
multiply it by $x^n$ and sum over all $n$. We get
$$
\frac{A(x) - x^2 F(2) - xF(1)}{x^2} = \frac{2x}{1-x} + \frac{A(x)-xF(1)}{x} + A(x),
$$
which, after solving for $A(x)$, gives
$$
A(x) = \frac{x(x+1)}{(1-x)(1-x-x^2)} = \frac{x+2}{1-x-x^2} - \frac{2}{1-x}.
$$
It is well known that the generating functions for the Fibonacci and Lucas numbers are
$$
f(x) = \frac{x}{1-x-x^2}, \quad L(x) = \frac{2-x}{1-x-x^2}
$$
Hence,
$$
A(x) = 2f(x) + L(x) - \frac{2}{1-x}.
$$
Recall now that $F(n)$ are the coefficients of the series expansion of $A(x)$, and thus
$$
F(n) = 2f_n + L_n - 2,
$$
where $f_n$ and $L_n$ are the $n$th Fibonacci and Lucas numbers respectively.


Change of variables to make this differential equation separable: $\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$ I'm trying to define a change of variables to make the next differential equation separable:  $$\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$$ assuming that $ae-bd=0$. I tried setting $u=ax+by+c$ and $v=dx+ey+f$ to make the right side of the equation homogeneous-like, but then i got confused trying to compute $\frac{dv}{du}$ (using that $ae-bd=0$). I also tried the substitution $u=\frac{ax+by+c}{dx+ey+f}$, and in this case I used the hypothesis that $ae-bd=0$ to deduce the next expresions for the derivatives of $u$:
$$\frac{du}{dx}=\frac{af-cd}{(dx+ey+f)^2}$$ and
$$\frac{du}{dy}=\frac{bf-ce}{(dx+ey+f)^2}$$
but I don't know how to use this info. Any hints will be appreciated.

$$\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$$
For $e \ne 0$
$$\frac{dy}{dx}=F\left(\frac{aex+bey+ce}{e(dx+ey+f)}\right)$$
Substitute $ae=bd$
$$\frac{dy}{dx}=F\left(\frac{bdx+bey+ce}{e(dx+ey+f)}\right)$$
$$\frac{dy}{dx}=F\left(\frac{b(dx+ey)+ce}{e(dx+ey+f)}\right)$$
Substitute
 $z=dx+ey+f\implies z'=d+ey'$
$$\frac {(z'-d)}e=F\left(\frac{bz+ce-bf}{ez}\right)$$
This last equation is separable
$$z'=e\underbrace{F\left(\frac{bz+ce-bf}{ez}\right)}_{=h(z)}+d$$
$$\int \frac {dz}{eh(z)+d}=\int dx=x+K$$
You can treat also the case $e =0 \implies bd=0$

$$
u = a x + b y + c \to u' = a + b y'\\
v = d x + e y + f \to v' = d + e y'\\
\frac{du}{dv} = \frac{a+b y'}{d+ e y'}\to y' = \frac{d\frac{du}{dv}-a}{b-e\frac{du}{dv}}
$$

hence

$$
\frac{d\frac{du}{dv}-a}{b-e\frac{du}{dv}} = F\left(\frac uv\right)\to \frac{du}{dv}=\frac{a+bF\left(\frac uv\right)}{d+eF\left(\frac uv\right)}
$$

finally making $u = \lambda v\to u' = \lambda' v + \lambda$

$$
\lambda' v + \lambda = \frac{a+bF\left(\lambda\right)}{d+eF\left(\lambda\right)}
$$

so we arrive to the separable form

$$
\frac{dv}{v} = \frac{d\lambda}{\frac{a+bF\left(\lambda\right)}{d+eF\left(\lambda\right)}-\lambda}
$$


sets theory-Realation proof question Let $A,B,C$ and $R,S,T$ be sets.
And assume that

$$ R \subseteq A \times B, ~~ S \subseteq B \times C, ~~
T \subseteq B \times C.$$

Then, I want to show that
$$
\begin{equation}
(S \circ R) \cap (T \circ R) \subseteq (S\cap T) \circ R, \\
(S \circ R) \cup (T \circ R) = (S \cup T) \circ R.
\end{equation}
$$

I tried solving this for hours.
I have no idea to even handle this.
I would love any kind of help or assistence.
Thank you.

If a(R o (S $\cap$ T))c, then exists
b in B with aRb and b(S $\cap$ T)c.
Thus aRb, bSc and bTc;  a(RoS)c, a(RoT)c.
In conclusion, R o (S $\cap$ T) subset RoS $\cap$ RoT.

a(R o (S $\cup$ T))c iff exists b in B with
aRb, b(S $\cup$ T)c iff aRb and (bSc or bTc)
iff (aRb and bSc) or (aRb and bTc)
iff a(RoS)c or a(RoT)c.
In conclusion R o (S $\cup$ T) = RoS $\cup$ RoT.

Note that I'm using the conventions aRb for (a,b) in R and
for composition RoS ={ (a,c) : exists b with (aRb and bSc) }


Are adjugates of similar matrices similar? Is it true in general, that if $A \sim B$, i.e. $\exists C: A =C^{-1}BC$, then $\mathrm{adj}(A) \sim \mathrm{adj}(B)$? It's quite obvious if the matrices in question are invertible (that is, $\det A \neq 0,\ \det B \neq 0$); if they are not, however, it's not as easy. I couldn't prove nor disprove this fact – could someone here do this?

We need the identities $adj(MN)=adj(N)adj(M)$ and $adj(M)=\det(M)M^{-1}$. See the wiki for a proof. See the answers in this site for another source. So we have \begin{align}adj(A)&=adj(C^{-1}BC)\\&=adj(BC)adj(C^{-1})\\&=adj(C)adj(B)adj(C^{-1})\\&=C^{-1}adj(B)C\end{align}So $adj(A)$ similar to $ adj(B)$.