e as limit with a rational sequence instead of n Hello and happy New Year to the Stackexchange community. I had trouble solving this seemingly trivial problem and it would be great if you could help me out.

$$\lim_{n \to \infty}(1+\frac{1}{n})^{n}=e$$

This definition of e is given and I want to show that the following identitiy is true as well.

$$\lim_{n \to \infty} (1+\frac{1}{q_{n}})^{q_{n}} = e$$

with $q_{n}$ as a positive rational sequence which diverges to $+\infty$ (therefore, $\lim_{n \to \infty} q_{n}=+\infty$).

I figured that since

$$\lim_{n \to \infty} n = +\infty = \lim_{n \to \infty} q_{n}$$

I could move $\lim_{n \to \infty}$ inside the given expression, so that we have

$$\lim_{n \to \infty}(1+\frac{1}{q_{n}})^{q_{n}}=(1+\frac{1}{\lim_{n \to \infty} q_{n}})^{\lim_{n \to \infty}q_{n}} = (1+\frac{1}{\lim_{n \to \infty} n})^{\lim_{n \to \infty} n} = \lim_{n \to \infty}(1+\frac{1}{n})^{n}=e.$$

However, apperantly, I need to show that the sequence is continuous in order to do this and my course didn't introduce continuity yet. Of course, I could first prove the therom needed to move the limit inside the expression and then show the sequence is continous, but I feel like there is a much more easier and elegant solution to this.

I'm sorry if this is a duplicate of a previously answered question (at least I couldn't find one) and any insight would be appreciated. Thank you very much.

Update — I think I managed to solve it by brute force, but it took me hours and my proof is 2+ pages long. I'm sure that my proof was not the intended one and there must be a simpler one. Anyway, thank you for the help and when I have time, I will update this question with my solution as reference.

It is not much difficult to establish what you seek. The function $$f(x) = \left(1 + \frac{1}{x}\right)^{x}$$ for $x \in \mathbb{Q}^{+}$ is an increasing function. This is easy to prove via Bernoulli's inequality. Let $x, y$ be positive rationals with $x > y$ then we have $$\left(1 + \dfrac{1}{x}\right)^{x/y} \geq 1 + \frac{x}{y}\cdot\frac{1}{x} = 1 + \frac{1}{y}$$ so that $f(x) \geq f(y)$.

Let $q_{n}$ be a sequence of rationals tending to $\infty$ and let $s_{n} = \lfloor q_{n}\rfloor$ and $t_{n} = s_{n} + 1$. Clearly we can see that from a certain value of $n$ the sequences $s_{n}, t_{n}$ are made of positive integers only and both tend to $\infty$. Now we know that $f(n) \to e$ as $n \to \infty$ and hence $f(s_{n}), f(t_{n})$ both tend to $e$ as $n \to \infty$. Since $s_{n}\leq q_{n}\leq t_{n}$ it follows that $f(s_{n}) \leq f(q_{n})\leq f(t_{n})$ and by Squeeze theorem $f(q_{n}) \to e$ as $n \to \infty$.

Note: The above proof requires the version of Bernoulli's inequality with rational exponents.

(long comment)

I suggest a way(not the only way) here. First you can treat $\lim_{n \to \infty}(1+\frac{1}{n})^{n}=e$ as a limit of a function $f:\Bbb R\to\Bbb R$, rather than a sequence, to infinity. So you may write $\lim_{x \to \infty}(1+\frac{1}{x})^{x}=e$ instead.

Then, if you have admit that $\lim_{x \to \infty}(1+\frac{1}{x})^{x}=e$ to be true, then you can directly use the famous theorem, Sequential Characterization of Limits(in particular, the version that related to limits as $x$ goes to $\infty$, if you don't know, just google it!). Then you can instantly get that $\lim_{n \to \infty} (1+\frac{1}{q_{n}})^{q_{n}} = e$ for all rational sequence $q_n$ that diverges to $\infty$.

PS: I'm afraid that if you are only allowable to admit $\lim_{n \to \infty}(1+\frac{1}{n})^{n}=e$ but not $\lim_{x \to \infty}(1+\frac{1}{x})^{x}=e$, then your claim can not be proved, though I'm not sure about this.


Maximising and minimising $f(x,y)$ on $x^2+y^2\leqslant 9$ Find the absolute minimum and maximum of $f(x,y):=x^2+y^2-8y+3$ on the
  disc $x^2+y^2\leqslant 9$.

I know what the answer is and how to obtain it.
What I do not understand is why we may assume that the extrema occur on the boundary of the disc.

I first looked at $Jf(x,y)=\mathbf{0}$ which occurs iff $(x,y)=(0,4)$, which is outside the disc.

One can then use the Lagrange Multiplier Theorem with the constraint $x^2+y^2=9$ to obtain the correct answers.
But I do not understand how we know this must be the constraint. How can we know for certain that the minimum and maximum do not occur in the interior of the disc?

You can  use a very elementary argument to argue that maximum and minimum must occur on the boundary. Suppose, if possible, the maximum occurs at a point $(a,b)$ with $a^{2}+b^{2} <9$. By increasing/decreasing $a$ slightly you can  stay within the disk but make the first term in $f$ larger. This contradiction shows that maximum occurs on the boundary. Similarly for the minimum.

Since $y\geq-3$, we obtain:
$$x^2+y^2-8y+3\leq9-8\cdot(-3)+3=36.$$
The equality occurs for $x=0$ and $y=-3,$ which says that we got a maximal value.

Also, since $y\leq3$, we obtain:
$$x^2+y^2-8y+3=y^2-8y+3=-12+x^2+(3-y)(5-y)\geq-12.$$
The equality occurs for $x=0$ and $y=3$, which says that we got a minimal value.


permutation and combination infinte supply Amika wants to buy 4 cupcakes from an infinite supply of three types of cupcakes: chocolate, vanilla & gems-laden. In how many
different ways can she buy cupcakes, if :

(1) order of buying the cupcakes is important, and

(2) arbitrary order is permitted

Suppose order is important.  first cupcake $3$ ways, second $3$ ways, third in $3$ ways, fourth in $3$ ways, So, $3^4$ in total

if order is not important, it is $\dbinom{3+4-1}{4}$
(this is similar to the stars and bars pattern I guess)

The number of different ways she can buy 4 cupcakes is $3^4=81$ ways because she can choose arbitrarily any cupcake from $3$ varieties. But, if order of choosing is unimportant, then it is $\binom{3+4-1}{3}$ which equals $15$. The formula in the general case is taken from combination with repititions


Cardinality of the set of all pairs of integers The set $S$ of all pairs of integers can be represented as $\{i \ | \ i \in \mathbb{Z} \} \times \{j\ | \ j \in \mathbb{Z}\}$. In other words, all coordinates on the cartesian plane where $x, y$ are integers.

I also know that a set is countable when $|S|\leq |\mathbb{N}^+|$. I attempted to map out a bijective function, $f : \mathbb{N}^+ \rightarrow S$.

$1 \rightarrow (1,1) \\
2 \rightarrow (1,2)\\
3 \rightarrow (1,3) \\
\quad \vdots
$

I determined from this that the natural numbers can only keep up with $(1,*)$. But there is the ordered pairs where $x=2,3,4,\cdots$ not to mention the negative integers. In other words, $|S|\geq |\mathbb{N}^+|$ and therefore $S$ is not countably infinite.

Is this correct? (I don't think it is... Something to do with my understanding of infinite sets)

Natural Numbers: There are many pairing functions that map $\mathbb{N}\times \mathbb{N}$ bijectively to $\mathbb{N}$. A simple example is the mapping $f$ such that $f(a,b)=2^{a-1}(2b-1)$.  For every positive integer $y$ can be uniquely expressed as a power of $2$ times an odd integer.

Integers: If you want a mapping $g(x,y)$ that maps $\mathbb{Z}\times \mathbb{Z}$ bijectively to $\mathbb{N}$, it is simplest to split the work into two parts.

Let $\phi$ be any mapping that maps $\mathbb{Z}$ bijectively to $\mathbb{N}$.  For a concrete example of such a mapping, let $\phi(t)=2t+2$ if $t \ge 0$, and let $\phi(t)=-(2t+1)$ if $t \lt 0$. The non-negative integers are sent to the even integers $\ge 2$, and the negative integers are sent to the positive odd integers.

Then the mapping $g(x,y)=f(\phi(x),\phi(y))$ works, where $f$ is any bijective map from $\mathbb{N}\times \mathbb{N}$ to $\mathbb{N}$. For example, we can use the mapping $f$ of the first paragraph, or the Cantor pairing function.

Remark: For most purposes, there is no particular virtue in having an explicit bijection, as long as we can prove that a bijection exists.

A more general result is a mapping between the set of integers to the set of finite tuples of integers of any size

If we order all positive ordered tuples of integers by their sum, we get the following:

$1\\
2\\
1, 1\\
3\\
2, 1\\
1, 2\\
1, 1, 1\\
4\\
3, 1\\
2, 2\\
2, 1, 1\\
1, 3\\
1, 2, 1\\
1, 1, 2\\
1, 1, 1, 1\\
etc.$

Clearly this set contains all possible finite ordered tuples of integers.

Next, we can define a mapping from the positive integers to these values as follows:

$0001 \rightarrow1 \\
0010 \rightarrow  2\\
0011 \rightarrow  1, 1\\
0100 \rightarrow 3\\
0101 \rightarrow 2, 1\\
0110 \rightarrow 1, 2\\
0111 \rightarrow 1, 1, 1\\
1000 \rightarrow 4\\
1001 \rightarrow 3, 1\\
1010 \rightarrow 2, 2\\
1011 \rightarrow 2, 1, 1\\
1100 \rightarrow 1, 3\\
1101 \rightarrow 1, 2, 1\\
1110 \rightarrow 1, 1, 2\\
1111 \rightarrow 1, 1, 1, 1\\
etc.$

And note that this mapping can be performed easily by adding one to every "1" for every zero after it, and then removing those zeros. This can be more formally described in terms of logarithms or dividing by/modulo 2, but that seems sufficient to get the point across. The interesting part is this has the result of proving that the cardinality of the set of integers is the same as the cardinality of the set of all finite ordered tuples of integers, the set of all finite ordered tuples of finite ordered tuples, etc.

Back to your question, we can just remove all tuples from that list above that aren’t of size two and then renumber it, giving us:

$
0001 \rightarrow 1, 1\\
0010 \rightarrow 2, 1\\
0011 \rightarrow 1, 2\\
0100 \rightarrow 3, 1\\
0101 \rightarrow 2, 2\\
0110 \rightarrow 1, 3\\
etc.$

which provides a one to one map of integers to pairs of integers, which is sufficient to prove the claim. You can do the same thing, but only keeping ordered tuples of size 3 to prove that the cardinality of the integers = the cardinality of all trios of integers, the same for 4, etc.

This answer does not include the negative numbers, but you can do that easily by padding the list with all the negative signs as well:

$1\\
-1\\
2\\
-2\\
1, 1\\
1,-1\\
-1,1\\
-1,-1\\
3\\
-3\\
2, 1\\
2,-1\\
-2,1\\
-2,-1\\
1, 2\\
1,-2\\
-1,2\\
-1,-2\\
1, 1, 1\\
1,1,-1\\
1,-1,1\\
1,-1,-1\\
-1,1,1\\
-1,1,-1\\
-1,-1,1\\
-1,-1,-1\\
4\\
-4\\
etc.$


Unknown parameter $a$ in inequality such that it has at least one positive or one negative solution Consider the Inequality $9^x-a\cdot 3^x-a+3\leq 0\;,$ Where $a\in \mathbb{R}$
Then The real values of $a$ so that
$(1)$ The given inequality has at least one real solution.
$(2)$ The given Inequality has at least one positive solution
$(3)$ The given Inequality has at least one negative solution.

$\bf{My\; Try::}$ For $(1)$
We can write Inequality as $$\displaystyle 9^x+3\leq a(3^x+1)\Rightarrow a\geq \frac{9^x+3}{3^x+1} = \frac{9^x-1+4}{3^x+1} = 3^x-1+\frac{4}{3^x+1}$$
So $$a\geq \underbrace{3^x+1+\frac{4}{3^x+1}}_{\bf{Using\; A.M\geq G.M}}-2\geq 4-2=2$$
So In first case we get $a\geq 2$
Now How can I solve $(2)$ and $(3)$ cases, Help required, Thanks

First denote t=3^x. Notice 9^x=t^2. Then solve inequality for t using vieta's formulas
With different requirements:
1. x is real iff t is positive.
2. x is positive iff t is greater than 1
3. x is negative iff t is less than one.
For each step add conditions of inequality for parameter a


In which senses can an integral exist? I asked about the value of an integral here:
Hard integral that standard CAS get totally wrong

The question got downvoted and voted to close because I didn't understand (and wasn't able to answer) the following question:

  In which sense is the integral supposed to exist?

So in what senses can integrals exist? What are the options here?

In addition to other useful answers... Before giving the standard example of an integral whose convergence is problemmatical, but which does have some sense... we should ask ourselves what it is we are expecting "integrals" to do, what properties the process should have. For example, "integration" should not just produce random numerical outcomes. And which functions, on which intervals, should be expected to be acceptable inputs? At the very least, if $f,g$ are acceptable, then linear combinations $af+bg$ should be, and, letting $I$ denote the integration procedure, $I(af+bg)=aI(f)+bI(g)$. ("Linearity".)

Riemann integration works best on finite intervals with nearly-continuous functions, while Lebesgue integration accommodates very-discontinuous functions, etc. In both cases, the integral of $f$ on a set or interval is a limit of finite sums, and the set-up for the game consists of proving these limits will exist (and be finite numbers!) under various assumptions on $f$.

A simpler standard example similar to one in your other post is $\int_0^\infty \sin(x^2)\;dx$. This has the disturbing feature that the function doesn't go to $0$ at infinity, so if we think of Cauchy's criterion for convergence of a series (rather than integral), we might conclude that this would diverge, meaning that $\lim_N \int_0^N \sin(x^2)\;dx$ might be $\pm\infty$? Or not exist? However, the oscillation produces enough self-cancellation so that this doesn't happen. In fact, changing variables, replacing $x$ by $\sqrt{x}$, gives integral
$\int_0^\infty {\sin(x)\over \sqrt{x}}\;dx$. Now, at least it looks like it decays at infinity, and there is still cancellation due to the oscillation. In fact, the limit can be evaluated by various tricks: I think it is $\pi/2$ or something similar.

In fancier circumstances, it often happens that "an integral" is not meant to be taken literally, but only to indicate the structure of some operation on functions. The basic case is with Fourier or Laplace transforms on the real line. Fourier transforms expressed as integrals $\hat{f}(\xi)=\int_{\mathbb R} e^{-i\xi x}\,f(x)\;dx$ make best sense for $\int_{\mathbb R} |f(x)|\;dx<\infty$, but, in fact, via the Plancherel theorem for Fourier transforms, we know that $\int_{\mathbb R} \hat{f}(\xi)\;\hat{g}(\xi)\;dx=\int_{\mathbb R}f(x)\,g(x)\;dx$ (maybe up to a constant multiple), so there is a unique extension of Fourier transform to square-integrable functions $f$, that is, such that $\int_{\mathbb R}|f(x)|^2\;dx<\infty$. This extension has the same properties as the Fourier transform that is literally an integral, but is not quite given by that integral.

Similarly, Fourier transforms can be extended to ("tempered") "generalized functions" (="distributions", not in the probabilistic sense), in a way that is completely sensible structurally, but in which the integrals are wildly not-convergent. For example, $\int_{\mathbb R} x^n\cdot e^{-i\xi x}\;dx$ doesn't converge at all, but by other means we can conclude that it is (a constant multiple of) the $n$th derivative of the Dirac delta distribution.

And, in case there were any doubt, computer algebra systems have their limitations, especially in dealing with "divergent" (not numerically docile!) integrals.

Well, there is a simple definition: an Integral $\int_{a}^{\infty} f(x)\;dx$ exists $\Leftrightarrow$ the series $\sum_{n=a}^{\infty} f(n)$ converges. The hard part might be proving this. Use a convergence test to proof, if it converges or not.

For Integrals with both bounds of integration infinite you should look up the Residue Theorem.


Show that $E = [0,1]$ is not open. Here is my proof. Not sure if I'm doing it correctly. We recall that $E$ is open provided every point of $E$ is an interior point. We then have $\forall p \in E$, there exists a neighborhood of $p$, $N$, such that $N \subset E$. Suppose that $E$ is open and consider the point $p=1$ in $E$. Since $E$ is open, then there exists a neighborhood, $N_r(1) = \{q : \ |q-1| < r\}$, such that $N_r(1) \subset E$. Here the topological space under consideration is $R$ with the usual real line topology.
Now, since $N_r(1) \subset E$ and $E$ is open, then each $p' \in N_r(1)$ is an interior point, that is, there exists a neighborhood, $N_{r'}(p') \subset E$. We will show that there exists a neighborhood, $N \subset N_r(1)$ for which $N \not\subset E.$ We consider three cases.
If $r \in (0,1)$, then we can take the neighborhood, $N_{r^2}(1+r^2)$. For example, is $r=\frac{1}{2}$, then
$$N_{\frac{1}{2}}(1)  =  \{q: \ |q-1| < \frac{1}{2}\} =\{q : \ q\in (\frac{1}{2}, \frac{3}{2})\},$$
$$N_{\frac{1}{4}}(5/4)  =  \{q: \ |q-\frac{5}{4}| < \frac{1}{4}\} =\{q : \ q\in (1, \frac{3}{2})\}.$$
Notice that $N_{\frac{1}{4}}(5/4) \subset N_{\frac{1}{2}}(1)$, but $N_{\frac{1}{4}}(5/4) \not\subset E$. On the other hand, if $r=1$, then we can take $N_{\frac{1}{4}}(\frac{3}{2})$, where
$$N_{\frac{1}{4}}(\frac{3}{2}) = \{q: \ |q-\frac{3}{2}| < \frac{1}{4}\} =\{q : \ q\in (\frac{5}{4}, \frac{7}{4})\}.$$
Clearly, $$N_{\frac{1}{4}}(\frac{3}{2}) \subset N_1(1) = \{q:  |q-1| < 1\} =\{q : \ q\in (0, 2)\},$$ but again $N_{\frac{1}{4}}(\frac{3}{2}) \not\subset E$. Finally, for $r > 1$, We can take $N_{\frac{1}{r^2}}(1+\frac{1}{r^2})$. For example, if $r=2$, then $N_2(1) =  \{q : |q-1|<2\} = \{q: q \in (-1,3)\}$ and $N_{\frac{1}{4}}(\frac{5}{4}) \subset N_2(1)$. Again, we have that $N_{\frac{1}{4}}(\frac{5}{4}) \not\subset E$. Does this argument shows that $E$ is not open?

This is way too complicated. Let $N_r(1)$ be any basic nbhd of $1$. Then $1+\frac{r}2\in N_r(1)\setminus E$, so $N_r(1)\nsubseteq E$. Thus, $1$ is a point of $E$ that is not an interior point of $E$, so $E$ is not open.

By the definition of open set. You can not find neighborhood around $0$ and $1$ which contained in $E$. I.e., it is not possible to find neighborhood $N$ around $1$ such that $1\in N\subset E$.


Floor function to the base 2 I'm not a math guy, so I'm kinda confused about this.
I have a program that needs to calculate the floor base $2$ number of a float.

Let say a number $4$, that base $2$ floor would be $4$. Other examples :

$5 \to 4$
$6 \to 4$
$8 \to 8$
$12 \to 8$
$16 \to 16$
$0.5 \to 0.5$
$0.6 \to 0.5$
$0.13 \to 0.125$

How would I do this and what is the name of this problem?

Let the number whose base 2 floor you want to find be $N$. We want to find the greatest $k\in \mathbb{Z}$ such that $2^k \leq N$. Take the base 2 log of both sides to get $k \leq \log_2{N}$. Since we want the maximum value of $k$ that still fulfills this inequality and that is an integer, we pick $k = \lfloor \log_2{N} \rfloor$. Then you just need to compute $2^k$, which is the actual value of your base 2 floor of $N$.

EDIT: So simply put, your wanted base 2 floor function $f_2$ looks like this:

$$ f_2(N) = 2^{\lfloor \log_2{N} \rfloor} $$

And for a floor in base $b$ (using the above logic), $f_b$ would be

$$ f_b(N) = b^{\lfloor \log_b{N} \rfloor} $$

Practical insight Floor of base two is nothing but the floor of a binary number. Floor of base 10 gives floor of decimals .
let's see examples
$$ floor_{10} 15.3=15$$
In binary-
$ floor_2 110$(decimal equivalent 6)$=100$ (decimal equivalent 4) so to do this function first convert it into binary then take only the left most 1 and put the rest as 0's

For values like 0.625 it can be broken down to binary form which is 0.101 then $floor_2 0.101 = 0.1$ in binary converting to decimal form we get 0.5


Prove $\exp{i\frac{\pi}{2}(-1+\sigma_{i})}=\sigma_{i}$ How do we prove

$e^{{i\frac{\pi}{2}(-1+\sigma_{i})}}=\sigma_{i}$ ?

where $\sigma_{i}:$Pauli matrix and $1=$ Identity matrix

Note: I understand that $i\frac{\pi}{2}(-1+\sigma_{i})$ is anti-hermitian since $(-1+\sigma_{i})$ is hermitian, hence the exponential of it is Unitary.

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  $\ds{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}} = \sigma_{k}:\ ?}$

Note that
$\ds{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}} =
-\ic\exp\pars{\ic\,{\pi \over 2}\,\sigma_{k}}}$. Lets consider
$\ds{\mrm{f}\pars{\mu} \equiv \exp\pars{\ic\mu\sigma_{k}}}$. Then,
\begin{align}
\mrm{f}'\pars{\mu} & = \ic\sigma_{k}\,\mrm{f}\pars{\mu}\,,\qquad
\left\{\begin{array}{rcl}
\ds{\mrm{f}\pars{0}} & \ds{=} & \ds{\mathbf{1}\ \pars{~identity~}}
\\[2mm]
\ds{\mrm{f}'\pars{0}} & \ds{=} & \ds{\ic\sigma_{k}}
\end{array}\right.
\\[5mm]
\mrm{f}''\pars{\mu} & = \ic\sigma_{k}\,\mrm{f}'\pars{\mu} = -\,\mrm{f}\pars{\mu}
\quad\implies\quad\mrm{f}''\pars{\mu} + \mrm{f}\pars{\mu} = 0
\end{align}

Then,
\begin{align}
\mrm{f}\pars{\mu} & = A\sin\pars{\mu} + B\cos\pars{\mu}\quad\implies\quad
\left\{\begin{array}{rcrcl}
\ds{0\,A}& \ds{+} &\ds{B} & \ds{=} & \ds{\mathbf{1}}
\\[2mm]
\ds{A}& \ds{+} &\ds{0\,B} & \ds{=} & \ds{\ic\sigma_{k}}
\end{array}\right.
\\[5mm]
\implies\,\mrm{f}\pars{\mu} & = \ic\sigma_{k}\sin\pars{\mu} + \mathbf{1}\cos\pars{\mu}
\implies
\color{#f00}{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}}} =
-\ic\,\mrm{f}\pars{\pi \over 2} = -\ic\pars{\ic\sigma_{k}} = \color{#f00}{\sigma_{k}}
\end{align}


Showing given function is real analytic on R I was reading about real analytic functions and I understood that to show a function is real analytic on $\mathbb{R}$, we need to show that there is a power series centered at $x_0$ which converges to an open interval around $x_0$  $\forall x_0 \in \mathbb{R}$

The example I am working on is the function $$f(x) = \frac{1}{1+x^2}$$
I can show that it converges for expansion around 0 by expanding it as a geometric series and this converges in (-1,1) when centered at x=0. How do I show it for other x $\in \mathbb{R}$ ?

I think I remember a fact that if power series centered around x=0 converges in (-1,1) then power series centered at any point in this interval would also converge(can someone give verify if this is indeed true?). But that still leaves us with the rest of the number line.

Try just centering a geometric series around a different point. For example, take $$\frac{1}{1-(-(x-x_0)^2))}$$

If this does have power series that converges, what would it look like? This would mean that you can get a radius 1 of converge around any point that you desire.


If there exists a bijection between two sets $A$ and $B$, there exists a bijection between $\wp(A)$ and $\wp(B)$ Where $A$ and $B$ are countably infinite and $\wp(A)$ and $\wp(B)$  are their powersets.

$\#$ represents cardinality.

My proof is this:
Let $c = \#\wp(A) = 2^{\aleph_0}$.
If there exists a bijection between $A$ and $B$, then this implies (this is given in the definition I am using) that $\#A = \#B = \aleph_0$
Thus, $2^{\aleph_0} = 2^{\#A} = 2^{\#B} = \#(2^A)=\#(2^B)=c$

Thus, because both $2^A$ and $2^B$ have the same cardinality, by definition, this means that there exists a bijection between $2^A$ and $2^B$ or $\wp(A)$ and $\wp(B)$.

Is this enough? I proved that $2^{\#A} = \#(2^A)$ for finite sets, is that enough for countably infinite sets?

Let $f:A \to B$ be a bijection and define $F:\wp(A) \to \wp(B)$ by

$F(M):=f(M)$.

Then $F$ is a bijection.


Prove that : $10^{5n+2}+(-1)^{n}\cdot 4 \equiv 0 \pmod {13}$ Prove that :
$$10^{5n+2}+(-1)^{n}\cdot 4 \equiv 0 \pmod {13}$$
I don't have enough skills in modular to do it
Please help


Find $\delta$ s.t. $|x-2|<\delta$ means $|x^2+2x-18|<\frac{1}{4}$ I need to find $\delta$ s.t. $|x-2|<\delta$ means $|x^2+2x-18|<\frac{1}{4}$ (as if I was asked to find $\delta$ when finding the limit at 2, with $\varepsilon = \frac{1}{4}$). My issue is that this does not factor nicely, so I am having trouble. When I try to factor out $(x-2)$ I get $x^2+2x-18=(x-2)(x+4)+10$, and this is not helping me to obtain $\delta$.
Thanks!


How many numbers of the form $m^2 + \sqrt{2} n^2$ are between $1 \times 10^6$ and $2 \times 10^6$? I have a purely computational question today.  How many numbers of the form $m^2 + \sqrt{2} n^2$ with $m,n \in \mathbb{Z}$ are between $1 \times 10^6$ and $2 \times 10^6$ ?

$$ \# \big\{ (m,n):  1 \times 10^6 < m^2 + \sqrt{2} \, n^2 < 2 \times 10^6 \big\} =\;  ? $$

The asymptotic answer for this sequence of numbers can be found using Weyl's law, for exmaple that:

$$ \# \big\{ (m,n) : 0 < m^2 + \sqrt{2} \, n^2 < X \big\} \sim \frac{\pi}{4 \sqrt{2}} \, X$$

This result is saing that numbers of the form $m^2 + \sqrt{2} \, n^2$ are roughtly linearly distributed on the real line $\mathbb{R}$.

The Weyl's law estimate gives $\frac{\pi}{4\sqrt{2}} \times 10^6 \approx 555,630 + 0.36726\dots$  Perhaps with a computer it's possible to obtain an exact answer?  E.g. using Python or sage.

COMMENT.- Without calculator the problem is
 tedious but not difficult.

The required points are enclosed between the ellipses of equations $m^2 + \sqrt{2} n^2=10^6$ and $m^2 + \sqrt{2} n^2=2\cdot10^6$. Define the following sets in the first quadrant:
$$A=\{(0,y)\text { with } y\gt0 \}\\B=\{(x,0)\text { with } x\gt0 \}\\C=\{(x,y)\text { with } x\gt0,y\gt0 \}$$ By symmetry we have for the total number $N$ of points
$$N=2A+2B+4C$$ $A$ has $573$ points and $B$ has $413$.

One has for the calculation of the set $C$ $$\frac{\sqrt{10^6-m^2}}{\sqrt[4]2}\lt y\lt\sqrt[4]2\sqrt{10^6-m^2}\text { where } 1\le m\le1000$$   Assuming that @rogerl's comment above is true, there are $4\cdot660074+1972=\color{red}{2\space642\space268}$ points in total.

First of all, I graphed your function in the terms of $x$ and $y$ instead of $m$ and $n$ and got this graph:

graphed using: Geogebra

The graph shows an elliptic Paraboloid which has a minimum value of zero, going up to $\infty$; However, the graph shows a function on $\mathbb R$ not $\mathbb Z$ (of course). so, we've to find the solution for $m$ and $n$ which are integers that have a result (after applying the formula) included between $1 \times 10^6$ and $2 \times 10^6$ only.

And to do this we have to find out the number of integers included in $[1 \times 10^6,2 \times 10^6]$ before doing anything, and they are simply $1 \times 10^6 +1$ (bec. million is included, we can't simply subtract). Then we must get the minimum and the maximum value that both $m$ and $n$ can have.

The minimum value for each of them must be zero; as if one of them reached its maximum value the other must be zero. but, getting the maximum value will be a bit more trickier; considering that each one of them is zero independently might help:
$$m^2 + \sqrt{2} \, n^2=2 \times 10^6  \\ m^2=2 \times 10^6 \\ m = \sqrt{2 \times 10^6} = 1414.213562  \ \ \ \ \ (m \notin \mathbb Z) \\ \lfloor m \rfloor = 1414$$
Having $m^2$ as the maximum value ($2 \times 10^6$) didn't help in getting an integer so flooring the result can help; as flooring the result will give an integer which is not more than $2 \times 10^6$. Now the same to get the maximum value for $n$ ($m=0$):
$$m^2 + \sqrt{2} \, n^2=2 \times 10^6 \\ \sqrt{2} \, n^2 = 2 \times 10^6 \\ n = \sqrt{\frac{2 \times 10^6}{\sqrt{2}}} = 1189.207115 \ \ \ \ \ (n \notin \mathbb Z) \\ \lfloor n \rfloor = 1189$$

Now we the minimum and the maximum values for each $m$ and $n$, and this can be summarized as follows:
$$0 \le m \le 1414 \\ 0 \le n \le 1189$$
Now we need the minimum value for each of them when the other is zero, and that's possible by considering $n=0$:
$$m^2 + \sqrt{2} \, n^2=1 \times 10^6  \\ m^2=1 \times 10^6 \\ m = \sqrt{1 \times 10^6} = 1000$$
and again by considering $m=0$:
$$m^2 + \sqrt{2} \, n^2=1 \times 10^6 \\ \sqrt{2} \, n^2 = 1 \times 10^6 \\ n = \sqrt{\frac{1 \times 10^6}{\sqrt{2}}} = 840.8964153 \ \ \ \ \ (n \notin \mathbb Z) \\ \lceil n \rceil = 841$$
However, $n$ can't be $1000$ (for example) when $m=800$. so, how can we solve this problem ?
It's easy we just have to find a relation between $m$ and $n$, but first we must find the number of possibilities we have (now) which is all possibile ordered pairs of $m$ and $n$ which is (by using Cartesian product):
$$\# \{ m: 0 \le m \le 1414 \} = 1414 + 1 \ \text{(bec. zero is included)} = 1415 \\ \# \{ n: 0 \le n \le 1189 \} = 1189 + 1 = 1190 \\ 1415 \times 1190 = 1683850$$
Now we know that the number of numbers of the form $m^2+\sqrt{2} \, n^2$ between $1 \times 10^6$ and $2 \times 10^6$ in the first quadrant is less than $1683850$, but what is the difference between the required number and this number ?

We can find out by using the graph, think of an area (on the plane $z=0$ in the 3d graph) that contains all the ordered pairs we need, it will be like an ellipse that we can graph as follows:$10^6<x^2+ \sqrt{2} \, y^2 < 2 \times 10^6$ (inequality graph).
I used $x$ instead of $m$ and $y$ instead of $n$ and got this:-

Two ellipses one inside other with bounds that we have calculated above, the colored area represents all ordered pairs that satisfy the formula, so we just have to calculate number of integer pairs in it.


Findinf the remaining eigenvalues of a $3\times 3$ matrix Let $n$ be a fixed natural number. We wish to compute the eigenvalues of the matrix below:
$$\begin{bmatrix}
a&b&c\\
d&e&f\\
g&h&i
\end{bmatrix}$$
We have a condition on the entries: $a+b+c=d+e+f=g+h+i=n$. It follows that one of the eigenvalues is $n$. I want the other eigenvalues in terms of the entries.

For example, if $\lambda_1 , \lambda_2,n$ are the eigenvalues, then $\lambda_1 + \lambda_2=n - a-e-i$.
Like this, I want both $\lambda_1 , \lambda_2$ in terms of the entries, not the sum of them.

Since you also have $\det A = n \lambda_1 \lambda_2$ (where $\det A$ can be expanded to your entries) see that
$$\lambda_2 = n-a-e-i-\lambda_1$$
and thus
$$\det A = \frac{(n-a-e-i-\lambda_1)\lambda_1}n = \frac1n \lambda_1^2 + (1-\frac1n(a+e+i))\lambda_1$$
So you can solve the quadratic to get explicit formulas (the two roots will be $\lambda_1$ and $\lambda_2$. The quadratic is
$$\lambda^2 + (n-\mathrm{tr}(A))\lambda - n\det A = 0$$
so
$$\lambda_{1,2} = \frac{\mathrm{tr}(A)-n}2 \pm \sqrt{\frac{(n - \mathrm{tr}(A))^2}4 + n\det A}$$

Replacing $\det A$ and $\mathrm tr A$ with the formula in terms of $a,b,\ldots, i$ will technically give you an explicit formula only dependent on your constants, but it's going to be a pain to write down or use.

I think the $\lambda_1 \lambda_2 n$ should be equal to $(-1)^n (-1)^n \det(M)$. Now putting them in a quadratic equation may help a little bit.


Inequality deduced from relatively prime numbers. If $a_n \text{ and }b_n$ are relatively prime for all $n$ and

$$\frac{a_n}{b_n}=\frac{1}{n}+\frac{1}{n(n+1)}+\frac{1}{n(n+1)(n+2)}+\cdots$$

Deduce that

$$b_n\geq b_{n+1}$$

CURRENT THOUGHTS

I can show that

$$\frac{a_{n+1}}{b_{n+1}}=\frac{na_n-b_n}{b_n}$$

and making $b_n$ the subject

$$b_n=(\frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$

so it would suffice to show that

$$na_n\geq a_{n+1}+b_{n+1}$$

which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?

I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.

i.e.

If

$$\frac{a_{n+1}}{b_{n+1}}=\frac{na_n-b_n}{b_n}$$

Then we must have that $b_n \geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.

Hope this helps!

If I am not mistaken, the next steps in this proof will be to conclude that

$$\frac{a_n}{b_n}>\frac{a_{n+1}}{b_{n+1}}\Rightarrow a_{n+1}<a_n$$

And we obtain a contradiction by

$$a_1>a_2>a_3>\cdots >0$$

Since it is not possible to have a infinite decreasing sequence of positive integers?

Very nice!

There is a simple straightforward proof that for each $n$ $$S(n)=\sum_{i=n}^\infty\prod_{j=n}^i \frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=\frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,\dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)\cdot\cdots\cdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<\sum_{i=m}^\infty m^{m-i-1}=\frac 1m\cdot\frac{1}{1-\frac 1{m}}=\frac1{m-1}<1,$$
a contradiction.


Higher interior regularity From PDE by Evans, 2nd edition, pages 332-333. My question and work shown are at the bottom of this post.

THEOREM 2 (Higher interior regularity). Let $m$ be a nonnegative integer, and assume $$a^{ij},b^i, c \in C^{m+1}(U) \quad (i,j = 1,\ldots,n) \tag{25}$$ and $$f \in H^m(U). \tag{26}$$ Suppose $u \in H^1(U)$ is a weak solution of the elliptic PDE $$Lu=f \quad \text{in }U.$$
$\qquad$Then $$u \in H_{\text{loc}}^{m+2} (U); \tag{27}$$ and for each $V \subset \subset U$ we have the estimate $$\|u\|_{H^{m+2}(V)} \le C(\|f\|_{H^m(U)}+\|u\|_{L^2(U)}),$$ the constant $C$ depending only on $m,U,V$ and the coefficients of $L$.
Proof. 1. We will establish $\text{(27), (28)}$ by induction on $m$, the case $m=0$ being Theorem 1 above.
$\qquad$2. Assume now assertions $\text{(27)}$ and $\text{(28)}$ are valid for some nonnegative integer $m$ and all open sets $U$, coefficients $a^{ij}, b^i, c$, etc., as above. Suppose then $$a^{ij},b^i,c \in C^{m+2}(U), \tag{29}$$ and $$f \in H^{m+1}(U) \tag{30},$$ and $u \in H^1(U)$ is a weak solution of $Lu=f$ in $U$. By the induction hypothesis, we have $$u \in H_\text{loc}^{m+2}(U) \tag{31},$$ with the estimate $$\| u \|_{H^{m+2}(W)} \le C(\|f\|_{H^m(U)}+\|u\|_{L^2(U)}), \tag{32}$$ for each $W \subset \subset U$ and an appropriate constant $C$, dpending only on $W$, the coefficients of $L$, etc. Fix $V \subset \subset W \subset \subset U$.
$\qquad$3. Now let $\alpha$ be any multiindex with $$|\alpha|=m+1, \tag{33}$$ and choose any test function $\tilde{v} \in C_c^\infty (W)$. Insert $$v :=(-1)^{|\alpha|} D^\alpha \tilde{v}$$ into the identity $B[u,v] =(f,v)_{L^2(U)}$, and perform some integrations by parts, eventually to discover $$B[\tilde{u},\tilde{v}]=(\tilde{f},\tilde{v}) \tag{34}$$ for $$\tilde{u} := D^{\alpha}u \in H^1(W) \tag{35}$$ and $$\tilde{f} := D^\alpha f-\sum_{\substack{\beta \le \alpha \\ \beta \not= \alpha}} {\alpha \choose \beta} \left[-\sum_{i,j=1}^n (D^{\alpha - \beta} a^{ij} D^\beta u_{x_i})_{x_j} + \sum_{i=1}^n D^{\alpha - \beta} b^i D^\beta u_{x_i} + D^{\alpha - \beta} c D^\beta u \right]. \tag{36}$$

My try:
The definition of $B[u,v]$, according to the definition on page 314,
$$B[u,v] := \int_U \sum_{i,j=1}^n a^{ij} u_{x_i} v_{x_j} + \sum_{i=1}^n b^i u_{x_i} v + cuv \, dx$$ and the definition of the inner product $(f,v)$ is given by the textbook as $$f(u,v) = \int_U fv \, dx$$ for $u,v \in H_0^1(U)$ and $f \in L^2(U)$.
So the identity $B[u,v]=(f,v)_{L^2(U)}$ can be expressed as $$\int_U \sum_{i,j=1}^n a^{ij} u_{x_i} v_{x_j} + \sum_{i=1}^n b^i u_{x_i} v + cuv \, dx = \int_U fv \, dx.$$
As per the textbook's instructions, I inserted the identity of $v :=(-1)^{|\alpha|} D^\alpha \tilde{v}$ into the identity and obtained this:
$$\int_U \sum_{i,j=1}^n a^{ij} u_{x_i} (-1)^{|\alpha|} D^\alpha \tilde{v_{x_j}} + \sum_{i=1}^n b^i u_{x_i} [(-1)^{|\alpha|} D^\alpha \tilde{v}] + cu [(-1)^{|\alpha|} D^\alpha \tilde{v}] \, dx = \int_U f [(-1)^{|\alpha|} D^\alpha \tilde{v}] \, dx.$$
I am trying to obtain the result of $$B[\tilde{u},\tilde{v}]=(\tilde{f},\tilde{v}) \tag{34}$$ Am I doing this right so far? Is this where I am ready to perform some integrations by parts?

To show that, you must start with $B[u,v]=(f,v)$, where $v=(-1)^{|\alpha |} D^\alpha \overline{v}.$ So, you need change every partial derivative of $v$ to $u$, by integration by parts.

You will need some theorems in multiindex notation, like Leibniz formula, during of integration by parts and later you will conclud that $B[\overline{u},\overline{v}]=(\overline{f},\overline{v}).$

(I'm studying PDE and finished calculation now!)


Vertex Reconstruction Conjecture For Asymmetric Graphs Is it known whether all graphs having trivial automorphism group are vertex reconstructible? And if so, what is the proof?

Consider the following passage from [1]:

  As pointed out in [22], graphs which are asymmetric should be easier to reconstruct, yet symmetric graphs (even those which are at least regular), which should present a stiffer challenge, are simple to reconstruct.

From that sentence, I would draw the conclusion that the reconstructibilty of asymmetric graphs was still not known to the authors of that survey in 2006.

While it is possible that further advancements have been made in the intervening years (or that the authors of that article were simply unaware of the result), the fact that none of the usual searches (including synonyms of the term "asymmetric graph" like "rigid" or "trivial automorphism group") turns up anything conclusive on the topic is a little suspicious. Of course, the absence of evidence is not necessarily any proof of absence but in my opinion, such a result would be of such significance (a blockbuster result, one might say) that the article that proved it should be very, very easy to find.

A few more recent sources that appear to omit mention of any result regarding the reconstructibility of asymmetric graphs are [2] and [3].

Asciak, K., et al. "A survey of some open questions in reconstruction numbers." Preprint (2006).
Hartke, Stephen G., et al. "Automorphism groups of a graph and a vertex-deleted subgraph." the electronic journal of combinatorics 17.1 (2010): R134.
Lauri, Josef, and Raffaele Scapellato. Topics in graph automorphisms and reconstruction. Cambridge University Press, 2003.

By a well known theorem of Erdös, the smallest non-trivial asymmetric graphs have V(G)=6. The Reconstruction Conjecture states that all finite graphs with 3 vertices or more are vertex reconstructible. So the vertex reconstructability of nontrivial, finite asymmetric graphs is entirely contingent on this fundamental, and yet unproved, result. The question of wether asymmetric graphs are all vertex reconstructible is essentially an iteration of the Reconstruction Conjecture itself.


Implications of mutual independence between random variables Consider the random vectors $Z_1, Z_2, Z_3, Z_4$ and assume they are i.i.d.

Moreover, $Z_i\equiv (Y_i, X_i)$, where $Y_i$ $X_i$ are scalar random variables, and $\epsilon_i\equiv g(Y_i, X_i)$ for $i=1,2,3,4$ for some function $g$

I want to show that $\epsilon_i \perp X_i $ $\forall i$ implies $\epsilon_i \perp(X_1, X_2, X_3, X_4)$ $\forall i$

Could you help me? ($\perp$ denotes independence)

My attempt (correct?)

Let $i=1$
$Z_1, Z_2, Z_3, Z_4$ i.i.d. and $\epsilon_1\equiv f(Y_1, X_1)$ $\Rightarrow$ $(\epsilon_1, X_1) \perp (X_2, X_3, X_4) $
By assumption $\epsilon_1 \perp X_1$
$Z_1, Z_2, Z_3, Z_4$ i.i.d.$\Rightarrow$ $X_1 \perp X_2, X_3, X_4$
Hence,
$$
f_{\epsilon_1, X_1, X_2, X_3, X_4}=f_{\epsilon_1, X_1} f_{X_2, X_3, X_4}=f_{\epsilon_1} f_{X_1} f_{X_2, X_3, X_4}= f_{\epsilon_1}f_{X_1,X_2, X_3, X_4}
$$
repeat for $i=2,3,4$

(2017.09.16) Is it me or is the plague of silent revenge downvotes spreading? Yet one more...

You might recognize that your question is a special case of the following:

  Let $(U,V,W)$ denote random variables such that $U$ is independent of $V$, and $(U,V)$ is independent of $W$, then $U$ is independent of $(V,W)$.

This seems direct, for example computing the characteristic function $$\varphi_{U,V,W}(u,v,w)=E[\exp(iuU+ivV+iwW)]$$


Projectile Motion, finding the optimal angle I've been studying multivariable calculus the last 2 weeks, and I understand (I think) how to optimize 2 variable equations through normal optimization and constrained optimization via Lagrange.

I couldn't draw the connection though when I tried to optimize the projectile motion equations, attempting to find the optimal angle to cover a constant distance in the minimum amount of time.

$$ = \cos()$$
$$ = \sin()t − \frac{1}{2}\mathrm{}^2$$

I also don't understand how to implement the constrain of covering a certain distance nor how to rewrite this equation minimizing time in the optimization. I am genuinely confused by the amount of variables and can't seem to 'make the leap' so to speak from what I know to what I don't.

Edit:

I apologize for the lack of clarity on this post.

The problem is covering a certain distance in the least amount of time by finding the optimal launch angle at a constant velocity.

Through your responses, I think this means that $d$, total displacement, and $v$, constant velocity have been defined in our equations. This leaves θ and $t$ as our two variables.

If we then rewrite the two main equations in terms of $t$ as a function of θ, I believe this would point us towards finding the solution.

$$t = \frac{x}{vcos()}$$ and $$ t = \frac{y}{vsin()-\frac{1}{2}gt}$$
This would mean $$ \frac{x}{vcos()} = \frac{y}{vsin()-\frac{1}{2}gt}$$

I am stumped however on what to do after this. Mainly, given a total displacement of $d$, how would I go about writing that in terms of $x$ and $y$. I think $x = dcos()$ and $y = dsin()$ through vector resolution, but how would I incorporate that into the equations to calculate the optimal angle for the least amount of time.

If anything is incorrect or still unclear, do let me know. I appreciate all of your responses; they have indeed helped expand my thinking about this.

If the goal is to cover the optimal distance in a given minimum time, then yes... shoot the projectile horizontally.  But that makes no physical sense... why would you care about that?  If on level ground, that would mean shooting the projectile horizontally and it stops immediately (hitting the ground).

Presumably you want to get the maximum distance or range.

Here are graphs of a projectile leaving a cliff with the same initial speed but at different angles.  Note especially that the (red) horizontal trajectory does NOT yield the greatest range.  That is because even though it has the highest horizontal speed, it does not stay aloft as long as some other projectiles.

Does this require a Lagrangian (or any calculus)?

The greater the horizontal component of motion the better.

If we are firing off of a cliff, then $\theta = 0.$

If we are firing over level ground, then we need to find the minimal angle to cover the distance.

There is some time $t^*$ such that $y(t^*) = y(0) = 0$

$0 = t^*(v\sin\theta - \frac 12 gt^*)\\
t^* = \frac {2v\sin\theta}{g}$

$x(t^*) = \frac {2v^2\sin\theta\cos\theta}{g} = d\\
\sin2\theta = \frac {dg}{v^2}\\
\theta = \frac 12\arcsin(\frac {dg}{v^2})$

Update: since there seems to be some confusion with the "firing off a cliff."

The shot fired horizontally will cover more horizontal distance in the same amount of time, then a shot with the same initial velocity but different angle of trajectory.


Logic problem involving sum of digits Good one guys!
I'm studying to the national maths olympiad (brazil)  by myself, and I ran up to the following question:

Let $S(n)$ be the sum of the digits of n. For example $S(77) = 14$   and   $S(2003) = 5 $ .
Tell if there is a n that is integer, and positive with :

$S(n) = 9 $

&

$S(n^2) = 81 $

I found a number, it is $111 111 111$  ( 9 times de digit 1),
but it was just a lucky guess.

I want to go deeper into the exercise and know if there are ways to calculate it (without lucky guesses) using logic or any concept from maths.

Thanks in advance

Only a sketch:
one can prove that

$$s(x+y) \le s(x)+s(y)$$
Equality holds if and only if no carry occurs when adding the digits.
So if $x=\sum 10^i x_i $ and  $y= \sum  10^i y_i$ then equality holds if and only if $x_i+y_i<=9$ for all $i$.

$$s(xy)<=s(x)s(y)$$
Equality holds if and only if: no digit product is larger than $9$ and no carry occurrs while adding the digit products, so $x_j y_{i-j}<=9$ for all possible $i,j$ and  $\sum_j x_j y_{i-j}<=9$.
Therefore $s(n^2)<=s(n)^2$
and  if $s(n)=9$ then $s(n^2)<=81$

So the number cannot contain a digit larger than $3$ because then the square of the digit is larger than $9$. Also if $x_i$ and $x_j$ are two digits and $x_i\ge 3$ and $x_j \gt 1$ then $x_i x_j+x_j x_i \ge 12$. This is larger than 9 and therefore cannot happen.

So we have only the following possibilities:
$n$ contains only the digits $0$ and $1$ and $2$
$n$ contains the digits $0$ and $1$ and exactly one digit $3$.

So besides $0$ we can have the following digits

$$1,1,1,1,1,1,1,1,1$$
$$1,1,1,1,1,1,1,2$$
$$1,1,1,1,1,1,2,2$$
$$1,1,1,2,2,2$$
$$1,2,2,2,2$$
$$1,1,1,1,1,1,3$$

If we position nonzero digits only at the digit positions
$$1,3,7,15,31,\ldots,2^k-1,\ldots$$
then we can guarantee that the product contains only digitsums of the form $(x_i)^2$ (at position $2i$) and $2x_i x_j$ (at position i+j). So no carry will occur.
So for each digit tuple above we can construct a number with $s(n)=9$ and $s(n^2)=81$, e.g for $1,2,2,2,2$ we can construct $$20000000000000002000000020002010$$
(we can also use a permutation of these digits, e.g $2,2,1,2,2$)
This will be a number with the desired property.If one adds $0$, it will be a valid number, too. But one can try to construct smaller numbers by removing some of the $0$. As long as no carry occurs, this will be a valid number.

Consider numbers with just three digits:

$n = a 100 + b 10 + c$, where $S(n) = a + b + c = 9$.

$n^2 = a^2 10^4 + a b 10^3 + (b^2 + 2 a c) 10^2 + 2 b c 10 + c^2$ where now $S(n^2) = a^2 + a b + (b^2 + 2 a c) + 2 b c + c^2 = 81$.

Set up some equations!

You can generalize this to larger numbers, if needed.

Good luck on the Olympiad!


Show a subring contains certain elements. Show that the set of all real numbers of the form
  $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$
  with $n≥0$ and $a_i ∈ \mathbb{Z}$ is a subring of $R$ that contains $\mathbb{Z}$ and $\pi$.

Proof. We show that the set of all real numbers of the form
$a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ (call it the set $A$) contains $\mathbb{Z}$, $\pi$, and the set is closed under subtraction and multiplication of its elements.

Take $a_1 = 1$ and $(a_n)_{n \in \{0\} \cup \mathbb{N} \setminus \{1\}}$. Then $\pi$ is in $A$.

Consider the set $D = \{\pi^n : n \in \mathbb{N}\}$. It becomes evident this can be re-written as $aD = Da : a \in A$. $A$ clearly has a bijection with $\mathbb{Z}$.

Another way to show this is to construct a function $g$ such that it sends $\pi^n$ to $1$. Thus we will make $g: D \twoheadrightarrow 1$. In this fashion I want to show I can use the function $g$ on $C$ to get it to correspond to $c_0 + c_1 + c_2 + \dots + c_n$. In this sense we can correspond to each $c_i$ a natural number. Since the naturals share the same cardinality with the integers, we are done.

Comment: I am annoyed at how I had to use set theory to arrive at my answer. Perhaps there is a better way, algebraically?

To show that $A$ is closed under subtraction, take any real number $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ in $A$ and subtract $b_0 + b_1\pi + b_2\pi^2 +\cdots+ b_n\pi^n$ (also in $A$) from it to get $$(a_0 - b_0) + (a_1 - b_1)\pi + (a_2 - b_2)\pi^2 +\cdots+ (a_n - b_n)\pi^n$$

Clearly subtraction of integers yields an integer, and since $\mathbb{Z}$ is contained in $A$, $A$ is closed under subtraction.

For multiplication, I have no idea. It gets quite messy in my head.

Questions. First off, what is it called when a set contains a ring? Is it called an ideal? Or a coset? For example, when I said "We show that the set of all real numbers of the form
$a_0 + a_1\pi + a_2\pi2 +\cdots+ a_n\pi^n$ (call it the set $A$) contains $Z$". I know $a_i \in Z$ and I have seen examples like $bZ = Zb : b \in B$ but I don't know what that's called.

Secondly, what is $\mathbb{Z}$ called? A ring? What is $\mathbb{Z} \mod 2$ called? An ideal in the ring? Just making sure.

Lastly, how was my proof written - subjectively, what did you like and dislike about the proof? Is it correct in its entirety?

You seem to be confused about various definitions :

The set
$$
A := \{a_0 + a_1\pi + a_2\pi^2 + \ldots + a_n\pi^n : a_i \in \mathbb{Z}\}
$$
forms a ring because, as you say, it forms a group under addition, and for any
$$
a = \sum_{i=0}^n a_i\pi^i, \text{ and } b:= \sum_{j=0}^m b_j\pi^j
$$
you have
$$
ab = \sum_{k=0}^{n+m} c_k\pi^k, \text{ where } c_k = \sum_{i=0}^k a_ib_{k-i}
$$
This is called a "Cauchy product" of two polynomials.
$\mathbb{Z}$ is a ring, $\mathbb{Z}/2\mathbb{Z}$ is a quotient ring (not an ideal).

For subtraction, you did it perfectly. For multiplication you need to do some algebra:

we know that $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ is in our set and, if we take $b_0 + b_1\pi + b_2\pi^2 +\cdots+ b_n\pi^n$ to be another member of our set,
we get that :
\begin{align}(a_0 + a_1\pi + a_2\pi^2 +& \cdots+ a_n\pi^n)(b_0 + b_1\pi + b_2\pi^2 +\cdots+ b_n\pi^n)\\
&=
(a_0b_0) + \pi(a_0 b1 + b_0 a_1) + \pi^2(a_2 b_0 + b_2a_0 + a_1b_1) + \cdots + (a_nb_n)\pi^{2n}
\end{align}

Verify that the coefficients of this product are integers, and that the product is in the set, and you are finished


The classical Gauss sum This is a problem from dummit and foote 14.7.11

I solved everything except b). I tried hard but couldn't solve it. I'm trying to using sum of pth root of unity is 0. But there are $p$ terms in lagrange resolvant. So this won't work.

Please help me with b) (why lagrange resolvant $(\zeta_p,1)=-1$).

I think the definition in Dummit and Foote is confusing. The following general definition makes more sense:

Definition Let $L/K$ be a Galois extension with cyclic Galois group $G$. Define the Lagrange Resolvent of an element $\theta \in L$ and a character $\chi : G \to K^\times$ to be $$R(\theta,\chi) = \sum_{g \in G} \chi(g) g(\theta).$$

Now we can do a worked example of the problem for $p = 7$. The squares mod 7 are 1,2 and 4.

I believe the characters in question are the principal character $1$ and the character "$-1$" that is $1$ for squares and $-1$ for nonsquares:

$\eta_0 = \zeta_7^1 + \zeta_7^2 + \zeta_7^4$ = -0.5 + 1.32287i
$\eta_1 = \zeta_7^3 + \zeta_7^5 + \zeta_7^6$ = -0.5 - 1.32287i
$R(\zeta_7,1) = \zeta_7^1 + \zeta_7^2 + \zeta_7^3 + \zeta_7^4 + \zeta_7^5 + \zeta_7^6$ = -1
$R(\zeta_7,-1) = -\zeta_7^1 - \zeta_7^2 + \zeta_7^3 - \zeta_7^4 + \zeta_7^5 + \zeta_7^6$ = 2.64575i

and $R(\zeta_7,1) + R(\zeta_7,-1) = 2 \eta_0$.

Update: We can relate back to the definition given in Dummit and Foote:

$$(\alpha, \omega) = \alpha + \omega \sigma(\alpha) + \omega^2 \sigma^2(\alpha) + \cdots + \omega^{n-1} \sigma^{n-1}(\alpha)$$

Now $\sigma$ is a generator of the cyclic Galois group, so in the case of 7 for example we will pick a primitive root of 7, 2 will work and map $\zeta_p^i \mapsto \zeta_p^{2i}$. Not only does this generate the Galois group but you may notice that every second iteration of $\sigma$ gives you a square. So $(\zeta_7, -1)$ agrees with $R(\zeta_7, -1)$ above.


bookstore, ordering of books A bookstore has 3 copies of the calculus book, 4 copies of the algebra book, and 5 copies of the statistics book.  How many different orders can they be arranged on a shelf?

My thinking:
12 books, so 12!

Let us start with a simpler example. Say we have the letters A,A,B. We need to find the total number of ways in which they can be arranged.

If we say "there are 3 letters so the total number should be $3!$", we are having repetitions such as AAB, AAB because we assumed that they were all distinct when they weren't. As you can see, a total of $2!$ arrangements are repeated among the $A$s alone, so we need to get rid of those by diving by $2!$, so we end up with $$\frac{3!}{2!}$$

Similarly, here we have 3 copies of one book, 4 copies of one, and 5 copies of one. We get rid of them by dividing for each of those, so we end up with $$\frac{12!}{3!\times 4!\times 5!}$$

Since the question mentions copies of the book, I will assume that copies are identical to each other. In this case, the number of ways is given by
$$N = \frac{12!}{3!\times4!\times5!}$$


Factoring of $x^3-3x^2+30x-1$ I need help factoring
\begin{align}
x^3-3x^2+30x-1=0.\tag{1}
\end{align}
Any thoughts? I've tried the old guess and check method with long division and $\left(x-\frac{1}{2}\right),\left(x-1\right),\left(x-12\right),\left(x-6\right)$, all to no avail.

Thank you for your time,

Do not look for a neat number(check wolframalpha)


Solve $\sqrt{7-3x}+3=x$ I have the equation $\sqrt{7-3x}+3=x$ and I need to find $x$.

I eventually found the answer to be $1$ and $2$.

In order to check my answer, I plugged $1$ first back into the equation.

I eventually got to the form: $\sqrt 4 +3 =1$. I know that the square root of $4$ has two answers: $-2$ and $2$. If used $-2$, the would count as a solution. If I use $2$, the solution would be considered extraneous. I was confused add search it up on wolframalpha, which stated that there are no answers. I was if the solution was extraneous or not and the reasoning behind it.

There is no solution to this equation. When you square it and solve the quadratic you get $x=1$ or $x=2$ as you have seen. But this does not guarantee that these are necessarily solutions. When you go back to the equation you finds that these do not satisfy the equation. So there is no solution.


How do I formalize this proof? "Prove that if $x\ne 1$ then there is a real number $y$ such that $\frac{y+1}{y-2} = x$" Consider trivial problem:

  Prove that if $x ≠ 1$ then there is a real number $y$ such that $\dfrac{y+1}{y-2} = x$

Let's solve for y

$$\frac{y+1}{y-2} = x$$
$$y+1 = (y-2)x$$
$$y+1 = yx - 2x$$
$$y - yx = - 2x - 1 $$
$$y(1-x) = - 2x - 1 $$
$$y = \frac{- 2x - 1}{1-x}$$

Now, although it is obvious that if $x ≠ 1$ then $y$ is defined, but how do I express it formally? I believe that something like "It can be seen that $y$ is defined for any $x$, provided that $x ≠1$ $\Box$" Won't be a rigorous conclusion.

I want to write formal proof of the conjecture above (including all the necessary steps), how should I write it?

When solving an equation you are actually assuming that such a y exists (which is what you wanted to prove!). In order to prove that a solution y to $\frac{y+1}{y-2}=x$ exists for all $x\neq 1$ you can simply start by setting $y=\frac{-2x-1}{1-x}$, and then show that this y actually solves the equation, which is essentially doing your calculations in reverse.

In my opinion, it is one step before the last one who needs the reminder, meaning: after you get $\;(1-x)y=-2x-1\;$ , then you bring up that principle (or axiom): "we can now divide both sides by $\;1-x\;$ if $\;x\neq1\;$" (as then we divide by a non-zero number!), and thus we get

$$\require{cancel}\frac{\cancel{(1-x)}y}{\cancel{1-x}}=\frac{-2x-1}{1-x}\implies y=\frac{-2x-1}{1-x}=\frac{2x+1}{x-1}$$


What happens if the greater limit becomes smaller when you change them for u-substitution? If I am doing u-substitution on a definite integral, and my upper limit is $b$ and my lower limit is $a$, and I change the limits to match the u, what happens if my $b$ turns out to be lower than my $a$ (so the now b is smaller than a)?
Does my new b still go at the top, or must I maintain the order?
For example, if $u = 1 + cosx$, and $b = \frac{\pi}{2}$, and $a = \frac{\pi}{3}$, when I change them, I get $b = 1$, and $a = 1.5$.
Now $a$ is greater than $b$.
So which goes at the top of the integral sign (which is the upper limit)?

$ \int_b^a u(x) dx= - \int_a^b u(x) dx.$


Finding a limit without using L'Hospital's rule $$\mathop {\lim }\limits_{x \to \infty } \frac{{2\ln (x) + 3}}{{3{x^{\frac{1}{3}}}}}$$

How do you find the limit without using LHR? It's the classic $\infty \over \infty$.
I think the denominator is getting large "faster" than the numerator. Other than that, I'm not sure how to calculate a limit in this form. Maybe manipulating the $ln(x)$ in some way?

Thanks

Writing $t = \log x$, we obtain

$$\lim_{t\to\infty} \frac{2t+3}{3e^{t/3}}.$$

The Taylor expansion of the exponential function immediately yields $e^s > \frac12 s^2$ for $s > 0$, and inserting that, we obtain the majorisation

$$\frac{6(2t+3)}{t^2},$$

that is easily seen to converge to $0$. On the other hand, $0$ is a lower bound for the expression, so

$$\lim_{t\to\infty} \frac{2t+3}{3e^{t/3}} = 0$$

follows.

You can use that $x^a$ grows faster than $\ln(x)$ for any $a>0$, and thus $$\lim_{x\to \infty}\frac{\ln(x)}{x^a}=0$$


Why does (1/3) mod 3016 = 2011? So I am taking a class where we are working on a cryptography section. Basically, the course says that:
$$\frac 1 3 \mod(3016) = 2011$$
or when run through Python - modified with SciPi:
$$\frac 1 3 \,\%\, 3016 = 2011$$

I don't understand how or why this happens. Does anyone know why the modulo division would produce this result?

What you're calling $\dfrac 13$ is just the multiplicative inverse of $3$ mod $3016$.  You can easily verify that $2011$ is the multiplicative inverse of $3$.  $$3\cdot 2011 = 6033 = (2\cdot 3016)+1$$

No one has mentioned that for any fixed N, {a | gcd(a,N)=1} forms a group. Closure under multiplication is easy to see with elementary manipulations.
The inverses usually are explained using the Euclidean Algorithm. At the very least if you want to understand inverses in modular systems, this desire could sustain you while you read/hear an explanation of the Euclidean Algorithm.


Finding basic probabilities of events Mike is sitting a multiple-choice exam, each question has four answer options, only one is correct. Mike takes $5$ guesses.

a) What is the expected number of correct answers Mike obtains from the 5 guesses?

I put $\frac54 = 1.25$, is this right?

b) What is the probability that all 5 of his guesses are incorrect?

I put $(\frac34)^5 = \frac{243}{1024}$, is this right?

c) What is the probability exactly $4$ or the $5$ are correct?

I put $(\frac14)^4 \times \frac34 = \frac{3}{1024}$, is this correct?

Everything is fine except for your last calculation. Think about how many different ways you can have 4 out of 5 correct, and multiply that by the probability that 4 out of 5 are correct after you have decided which 4. (Hint: It should have both $1/4$ and $3/4$ in the formula)

Yes. Your solution in the comment, $(\frac{1}{4})^{4}\cdot\frac{3}{4}$, is the correct answer on the last question.


Finding $P(X < Y)$ where $X$ and $Y$ are independent uniform random variables Suppose $X$ and $Y$ are two independent uniform variables in the intervals $(0,2)$ and $(1,3)$ respectively. I need to find $P(X < Y)$.

I've tried in this way:
$$
\begin{eqnarray}
P(X < Y) &=& \int_1^3 \left\{\int_0^y f_X(x) dx\right\}g_Y(y) dy\\
&=& \frac{1}{4} \int_1^3 \int_0^y dx dy\\
&=& \frac{1}{4} \int_1^3 y dy\\
&=& \frac{1}{8} [y^2]_1^3\\
&=& 1
\end{eqnarray}
$$
But I'm suspicious about this result. It implies that $X<Y$ is a sure event, which is not at all true.

This sketch might help.  You want the red area as a proportion of the red and blue areas.


The Second Hearts Problem According to the last part of these lecture notes, if we have a standard deck of playing cards and turn cards until the first heart appears,
the probability that the next card is a heart is $\color{red}{1/4}$.

Without using probability theory, this looks to me like we are computing:

$$E[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ ]$$

where:

$N$ is a discrete random variable whose range is $\{1,2,...,40\}$, indicating the first time a heart is turned over (I guess $N$ is a stopping time, but I don't believe I'm using probability theory)
$\{Y_i\}_{i \in \{1,2,...,52\}}$ are binary random variables indicating whether or not the $i$th card is a heart,

as follows:

$$E\left[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ \right]$$

$$= E\left[ \ \sum_{n=1}^{40} \mathbb P(Y_{n+1} = 1 | Y_n = 1, \{Y_i= 0\}_{\{i < n\}} ) 1_{\{N=n\}} \ \right]$$

$$= E\left[ \ \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}} \ \right]$$

$$= \sum_{n=1}^{40} \frac{12}{52-n} E\left[ \ 1_{\{N=n\}} \ \right]$$

$$= \sum_{n=1}^{40} \frac{12}{52-n} \mathbb P(N=n)$$

$$= \sum_{n=1}^{40} \frac{12}{52-n} \frac{13 \cdot P(39,n-1)}{P(52,n)} = \color{red}{1/4}$$

Questions:

So what then is the following quantity (in plain English):

$$\mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) = \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}}$$

?

That to me looks like

$$P((n+1) \text{th card is hearts} \ | \ n\text{th card is first heart})$$

So how is that different from what we are computing which again is

$$E[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ ]$$

?

The latter to me seems something like expected probability while the former is probability.

$\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})$ is a random variable (I hope that term exists in plain English?) that's a function of the random variable $N$.

You're comparing it to $P((n+1) \text{-th card is hearts} \mid n\text{th card is hearts})$, by which I suspect you mean more precisely $P((n+1) \text{th card is hearts} \mid n\text{-th card is the first hearts})$. This is similar, but with a small $n$: $\mathbb P(Y_{n+1} = 1 \mid Y_n = 1, \{Y_i= 0\}_{\{i < n\}})$. This is a (conditional) probability, not a random variable; it has a particular value for every particular value of $n$.

You ask how this differs from the quantity $E\left[\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})\right]$ being computed. This is the expected value of the random variable $\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})$, where $N$ takes all values of $n$, not a particular value.

In concrete terms, we have

\begin{align}
\mathbb P(Y_{n+1} = 1 \mid Y_n = 1, \{Y_i= 0\}_{\{i < n\}})&=\frac{12}{52-n}\;,\\
\mathbb P(Y_{N+1} = 1\mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) &= \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}}\;,\\
E\left[\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})\right]&=\sum_{n=1}^{40} \frac{12}{52-n} E\left[1_{\{N=n\}}\right]
\\&=\sum_{n=1}^{40} \frac{12}{52-n} \mathbb P(N=n)\;.
\end{align}

I don't feel sure that I've addressed your question, but if not, hopefully I've at least provided a basis on which we can address it further.

We should compute for

$$E[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ ]$$

rather than

$$\mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) = \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}}$$

because the latter computes

$$P((n+1) \text{th card is hearts} \ | \ n\text{th card is first heart})$$

meaning a certain card is already given to be the first heart while the question does not assume a certain card is given to be the first heart.


Does $n^{\log n}$ or $(\log n)^n$ grow faster? Which grows faster? $n^{\log n}$ or $(\log n)^n$ and how can we prove this?

This was presented as a "challenge question" for students to try ahead of the next class meeting. Any help would be appreciated!

Hint: Try to take n = 16.
explore the behavior of the both functions with n=16, and with n > 16.


What does $d_{\textbf{a}} f$ mean? I have a question regarding the differential $d_{\textbf a} f$.

Suppose we have the function $f(x,y)= xy$, and the vectors $\textbf a = (1,1)$ and $\textbf u = (2,1)$. Then, if I understand this correctly,
$$d_{\textbf a} f(\textbf u) = \nabla f(\textbf a) \cdot \textbf u = (1,1)\cdot (2,1) = 2+1 = 3,$$
where $\nabla f(\textbf a) = (\partial f/\partial x, \partial f/\partial y)$. But what if my assignment is to calculate $d_{\textbf a} f$? I don't know what it means. Do they want me to calculate $d_{\textbf a} f(x,y) = (1,1)\cdot (x,y) = x+y$, or something else?

Edit: Note that it is not the directional derivative that I'm asking about.

Essentially, you have worked out everything already, but there seems to be a bit of confusion about the definitions, so let me try to set this straight.

The differential of $f$ at the point $\mathbf{a} \in \mathbb{R}^2$ is the row matrix
$$ d_{\mathbf{a}}f = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}.$$

Now if you write $d_{\mathbf{a}}f (\mathbf{u})$ for $\mathbf{u} = \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} \in \mathbb{R}^2$ you're meaning the matrix product
$$d_{\mathbf{a}}f (\mathbf{u}) =
\begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 .$$

On the other hand, $\nabla f (\mathbf{a})$ is the column vector
$$
\nabla f (\mathbf{a}) =
\begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}$$
and when you're writing $\nabla f (\mathbf{a}) \cdot \mathbf{u}$ you're meaning the scalar product
$$\nabla f( \mathbf{a}) \cdot u = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot  \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 .
$$

So we see that for $f(x,y) = xy$
$$d_{\mathbf{a}}f = \begin{pmatrix} y & x \end{pmatrix} \qquad \text{while} \qquad
\nabla f (\mathbf{a}) = \begin{pmatrix} y \\\ x \end{pmatrix}.$$

Now the confused reaction was due to the fact that the notation used here for the derivative of $f$ at the point $\mathbf{a}$ is often used as the directional derivative, and as you rightly pointed out in a comment, we have the relations
$$ D_{\mathbf{u}} f (\mathbf{a}) : = d_{\mathbf{a}} f (\mathbf{u}) = \nabla f(\mathbf{a}) \cdot \mathbf{u},$$
and everything should be fine now, no?

Since you made the computations yourself already, I'll not repeat them here.

In the case of functions $f\colon \mathbb{R}^n \to \mathbb{R}$, like $f(x,y) = xy$ as you have, the differential $d_af$ is the same thing as the gradient $\nabla f(a)$.


What is $\int_b^\infty\frac{\ln x}{x^2+1}dx$? They ask me to evaluate
$$\int_b^\infty\frac{\ln x}{x^2+1}dx, \quad b\geq 0$$

Any suggestions please.

This is in the context of indefinite integrals that do not involve functions such as Li, constant Catalan G, only the elementary functions, even before they teach us Series.

It's important because it's a partial college exam question and I do not find suggestions for any $ b $, just for $ b = 1 $ it's easy to see that the integral mentioned is zero.

I tried to solve it with the method of integration by parts, but it does not work out, that is why I come to your help.

Let $f(x) =\log x$ and $g(x) =\frac{x}{1+x^2}$. Then $f(x) /g(x)$ tends to be $0$ as $x$ tends to infinity. Thus two functions behave alike. But  integration $g(x)$ to infinite divergent so $f(x)$ is divergent


"Dividing" matrix by a vector If I have $\mathbf{Ax}=\mathbf{c}$ where $\mathbf{x} =
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]$ and $\mathbf{c} = \left[\begin{array}{r}
2x_1-4x_2-x_3-3x_4+2x_5\\
-x_1+2x_2+x_3+x_5\\
x_1-2x_2-x_3-3x_4-x_5\\
-x_1+4x_2-x_3+5x_5
\end{array}\right]$.
To solve the equation for $\mathbf{A}$, I would therefore like to isolate $\mathbf{A}$ - is there a way to do this? Like "dividing" by $\mathbf{a}$. I know that $\mathbf{A}$ is obviously the coefficient matrix but I would like some justification to actually show this like by mathematically solving the equation for $\mathbf{A}$.


Frobenius method. Problem finding a recurrence relation The equation is $2z^2w''+3zw'-w=0$
$z_0=0$ is a regular singular point, so $w(z)=\sum_{n=0}^{\infty} a_nz^{n+r}$
then $w'(z)=\sum_{n=0}^{\infty} (n+r)a_nz^{n+r-1}$ and $w''(z)=\sum_{n=0}^{\infty} (n+r)(n+r-1)a_nz^{n+r-2}$
Replacing in the equation:
$2z^2\sum_{n=0}^{\infty} (n+r)(n+r-1)a_nz^{n+r-2}+3z\sum_{n=0}^{\infty} (n+r)a_nz^{n+r-1}-\sum_{n=0}^{\infty} a_nz^{n+r}=0$
$\sum_{n=0}^{\infty} 2(n+r)(n+r-1)a_nz^{n+r}+\sum_{n=0}^{\infty} 3(n+r)a_nz^{n+r}-\sum_{n=0}^{\infty} a_nz^{n+r}=0$
$\sum_{n=0}^{\infty} [2(n+r)(n+r-1)a_nz^{n+r}+ 3(n+r)a_nz^{n+r}- a_nz^{n+r}]=0$
$2(n+r)(n+r-1)a_n+ 3(n+r)a_n- a_n=0$
$(2(n+r)(n+r-1)+3(n+r)-1) a_n=0$
At this point, how can I construct a recurrence relation since I only got $a_n$?, I'd need $a_{n+1}$, right?
I already found the indicial equation and its roots, which are $r_1=1/2$ and $r_2=-1$

Your last equation is all you get. There's no recurrence, and you don't need one.
You necessarily have $\color{blue}{a_n=0}$ for $n>0$. Otherwise, if $a_n\neq 0$ for some $n>0$, you would have both $r$ and $n+r$ be roots of $2r^2+r-1=0$ (the indical equation). In our case, this is clearly impossible.

Notice that, for this problem, Frobelius method is not required. Let $w(z)=z u(z)$ to end with
$$6u''+13u'=0$$ Just reduction of order and a quite simple problem.


An example of a problem which is difficult but is made easier when a diagram is drawn I am writing a blog post related to problem solving and one of the main techniques used in problem solving is drawing a diagram. Essentially, I want to illustrate that some hard problems (for example, word problems) can be done fairly easily when diagrams are drawn.

There is a paper related to this question but the problems in that paper are much to simple. I was considering using the problem statement of the Langley's Problem but that question is not too easy even after the diagram.

Does anyone have any ideas?

Thanks a lot!

EDIT:

As pointed out in the comments, it's difficult to define an "easy" problem or a "difficult" problem. So, I would like to add that the best examples would be within the undergraduate mathematics range or word problems that non-math majors/mathematicians would understand.

EDIT 2:

I would like to clarify further (thanks to @ruakh) that "when a person trying to solve the problem draws a diagram" is what I am looking for as opposed to "when a person who already knows the solution draws a diagram that illustrates it"

Question:  Choose two points uniformly at random on the unit interval. What is the probability that these points are within distance $\frac{1}{2}$ of each other?

Solution:

The following picture is the set of points $(x,y)$ such that $|x−y|\leq\dfrac{1}{2}$:

So the probability is $\dfrac{3}{4}$.

Almost without words

coordinates change and tangent basis change on a surface


Allowed values of $x$ in $x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$ when $y=0$? Can someone please check the given explanation? I have this two-variable equation
$$x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$$
for $0\leqslant x,y<\infty$, and I want to check which values of $x$ satisfy this equation when $y=0$.
From the above equation, I see that when $y=0$, it is fulfilled for all  $0\leqslant x<\infty$.
Then, I rewrite the equation as
$$\frac{x}{3 x^2-5}=\frac{1}{2} \tan (3 y),$$
from here, I see that when $y=0$, there are only two solutions for $x$, i.e. $x=0$ and $x\to \infty$.

Then, what can I say in general about the solutions of $x$ when $y$ is zero?
And what can we say about the point $x=\sqrt{\frac{3}{5}}$? Is it an acceptable solution for $3y=\frac{\pi }{2}$? Since from the first equation, it can be a solution, but I have doubts about it from the second form of the equation.

When you "rewrite" an equation by multiplying or dividing by some factor, the new equation is equivalent to the old when that factor is not $0$.  When it is $0$, the two are generally not equivalent.
So, your first equation
$$ x \sin 6 y - (3 x^2 - 5) \sin^2 3 y $$
is indeed always true when $y = 0$, or more generally when $\sin 3 y = 0$.  When you divide by $(3 x^2 - 5) \sin 6 y$ to get your second equation
$$ \frac{x}{3x^2-5} = \frac{1}{2} \tan 3y $$
this is equivalent to the first as long as $3 x^2 - 5 \ne 0$ and $\sin 6 y \ne 0$.  When
$\sin 3y = 0$, the right side is $0$ but the left is not (unless $x=0$).  When $3x^2 - 5 = 0$ the left side is undefined but the right side is defined (unless $\cos 3 y = 0$).
When $\cos 3y = 0$ the right side is undefined but the left side is defined (unless $3 x^2-5 = 0$).


Sum to infinity of reciprocal of arithmetic sequence terms Is there a arithmetic sequence starting at any a, such that the sum of the reciprocals of its terms to infinity converges?
i.e.
$\sum_{n=0}^{\infty} \frac 1{a+nd}=\frac 1a+\frac 1{a+d}+\frac 1{a+2d} +\frac 1{a+3d}\cdots $ converges for any a or d

Simply observe that

$$\frac 1{a+nd}\sim \frac 1{nd}$$

then the given series diverges by limit comparison test with the harmonic series $\sum \frac 1{n}$.

Let the $n^{th}$ term be $a_n=\frac{1}{a+nd}$
Then, $a_n-a_{n-1}=\frac{-d}{a^2+n\left(n-1\right)d^2+\left(2n-1\right)}$
Which decreases at a very small rate which means the series diverges


Coordinates of a point at a given distance from (x,y) I seem to have forgotten my coordinate geometry and this is absolutely blank.

Say I have $(x_1,y_1)$ and a distance $d$.
Is it possible for me to solve for what the point $(x_2,y_2)$ is?

To explain more:
If I have

$(x_1,y_1) = (1,2)$, and

$(x_2,y_2) = (3,4)$

Using the distance formula, I can calculate the distance to be $d=2\sqrt2$.

But if I have $(x_1,y_1) = (1,2)$ and $d=2\sqrt2$, how do I recover $(x_2,y_2)$ ?

Edit
Yes, people are absolutely right when they say, multiple solutions are possible. Still, say, I also have an angle $\theta$ that tells me the orientation. Then, is it still not possible?

No you don't have enough information to recover the $(x_2,y_2)$. The reason is that there are many points around $(x_1,y_1)$ which are a distance $d$ apart from it. In fact all points on the circle of radius $d$ centered at $(x_1,y_1)$ satisfy this. (This in fact the definition of a circle if you think about it.)

However, if you know that this point makes an angle $\theta$ from the $x$-axis, say, then yes you can recover $(x_2,y_2)$. The formula is
$$x_2=x_1+d\times\cos\theta,\quad y_2=y_1+d\times\sin\theta,$$
where $d$ is the distance from $(x_1,x_2)$ to $(y_1,y_2)$.

If the billiard table is infinitely large and the ball can not reach to wall, then calculated, the coordinates $(X, Y)$ at which the ball stops can be calculated by the following formula.

$$
X = L * \cos (θ) + x \\
Y = L * \sin (θ) + y
$$

But if $a$ (horizontal length), $b$ (vertical length), $x$ (coordinate $x$), $y$ (coordinate $y$), $r$ (radius of the sphere), $θ$ (angle), $L$ (distance until the ball stops) are given then what?


How can you find the $x$-coordinate of the inflection point of the graphs of $f'(x)$ and $ f''(x)$? So I understand how to find the inflection points for the graph of $f(x)$.

But basically, I have been shown a graph of an example function $f(x)$ and asked the state the inflection points of the graph. (I am just shown a curve... not given an actual function in terms of $x$)

Easy enough... I can see by looking where the concavity of the curve changes. But here is the tricky part. I am then asked to find the inflection points of the curves of $f'(x)$ and $f''(x)$.

From the graph of $f(x)$, I can draw in the values it cuts the $x$-axis, and whether it is positive/negative, but I don't understand how you can also comment on the inflection points from it?

i.e. $f(x)$ has both absoulute and local max values at $x = 2$ and $x = 6$, and local min at $x = 4$. Hence, $f'(x)$ cuts $x$ axis at $2, 4$ and $6$. I would imagine the local min/max would therefore be $3$ and $5$ (middle of $2$ & $4$, and $4$ & $6$ respectively)

The book gives answers a) $(3,5)$ Got it!   b) $2, 4, 6$   c)$1, 7$

How do you find the point of inflection for $f'(x)$ and $f"(x)$ in this case?

An inflection point is where the second derivative is zero and changes sign. Thus, if the graph is of the second derivative then notice that the zeroes with sign change are at 1 and 7 on that graph.

If the graph is of the first derivative then the local min/max will be the inflection points since this is where if one were to look at the derivative of this graph it would be the second derivative that goes to zero at the extreme point as well as the sign change that is the criteria.

I've always thought that the points where the f(x) graph cuts the x-axis is where the points of inflection for f'(x) are.


Independence in counting problems I have been re-working some counting problems that I did in college (a long time ago in a galaxy far far away) and I just had an insight that I would like verification on.

Consider the following two problems:

  How many meals can be formed from 3 appetizers 2 mains and 3 desserts?

Everyone learns that this is simply,
$$3*2*3$$
and that this resulting set is unordered, meaning that $$\{Salad,Chicken,Pie\} = \{Pie,Chicken,Salad\} $$
People call this the fundamental counting rule sometimes.

Okay, now on the opposite end we have complete dependence,

  How many ways can a president and vice president be chosen from a group of 5 people?

Everyone's intuition here is again
$$5*4$$
which looks an awful lot like the above, except our result is ordered.
How can that be since in both cases we are just taking the product over all possibilities?

This confused me for a long time until I realized the crucial difference is independence. The reason the top is unordered is because appetizer is independent of main course which is independent of dessert, whereas the reason the bottom is ordered is because picking a vice president is not independent of picking a president because the sample space decreases by $1$

Is this true?

Yes. That is correct. You can refer to Rule of Product to understand why the first one order doesn't matter but the second one does.

  Rule of product. If there are $a$ ways of doing action $A$ and $b$ ways of doing action $B$ thing, then there are $ab$ ways of performing both actions.

For the first one, three actions are: choosing appetisers, choosing mains and choosing deserts. The combination of three actions will be choosing an appetiser, a main and a desert in which order doesn't matter.

For the second one, two actions are: choosing the first one out of $5$ people to be present and choosing one from the rest $4$ people to be vice president. With these actions, we can see that the order matters, where first person we choose must be president. Hence, we have $(A,B) \ne (B,A)$ since with $(A,B)$ then $A$ is the president, but with $(B,A)$ then $B$ is the president.

The  product rule for counting says that if a task consists of three steps, there are $n_1$ ways to do the first step of the task, there are $n_2$ ways to do the second step regardless of which choice was made in the first step, and there are $n_3$ ways to do the third step regardless of which choices were made in the first two steps, then the number of ways to complete the task is the product $n_1 n_2 n_3$.

This formula is applicable to both your examples.  There is some kind of independence here (and therefore some kind of independence in both your examples) because the number of ways $n_2$ to do the second step is independent of which choice was made in the first step, and similarly for $n_3$.


$a\equiv{b}\pmod{m},a\equiv{b}\pmod{n}\ \Rightarrow\ a\equiv{b}\pmod{lcm(m,n)}$ Show that if $a\equiv{b}\pmod{m},a\equiv{b}\pmod{n}\ \Rightarrow\ a\equiv{b}\pmod{lcm(m,n)}$
I think this must be a basic theorem in number theory but couldn't find it in my books,also it's very useful in solving many problems concerning LCM and GCD.
Please give a proof.I have no idea for it.

$a=b$ mod $m$ is equivalent $a-b= cm$; $a=b$ mod $n$ is equivalent $a-b=dn$. $a-b$ is a multiple of $m$ and $n$  implies that $a-b$  is a multiple of $lcm(m,n)$. This implies that $a-b=0$ mod $lcm(m,n)$ and $a=b$ mod $lcm(m,n)$.

$a\equiv b \pmod n$ and $a \equiv b \pmod m$ is equivalent to $n|(a-b)$ and $m|(a-b)$.

Let $n=P_1^{\alpha_1}P_2^{\alpha_2}...P_k^{\alpha_k}$ and $m=P_1^{\beta_1}P_2^{\beta_2}...P_k^{\beta_k}$ (with $\alpha_i$ and $\beta_i$ positive integers, possibly zero)

It is easy to show that $lcm(m,n)=P_1^{max(\alpha_1,\beta_1)}P_2^{max(\alpha_2,\beta_2)}...P_k^{max(\alpha_k,\beta_k)}$

We have, by hypothesis, $n|(a-b)$ and $m|(a-b)$.

Therefore, $lcm(m,n)|(a-b)$, i.e. $a\equiv b \pmod {lcm(m,n)}$


What is i to the power of 1/2 I am new to complex numbers and wondered, what is $i^{(1/2)}$ and how do I actually create a complex number, so that $i$ has a different power than $1$ in tuple form, like $(a, b) = a + b*i^2$? Is it just $(a, bi)$?

In the complex plane, $i$ is on the unit circle at an angle of $\frac{\pi}{2}$. Thus it has two square roots, which are obtained by taking half the angle and accounting for wraparound by $2\pi$.

Thus, the most obvious square root is the point on the unit circle with angle $\frac{\pi}{4}$. The less obvious square root is obtained by replacing $\frac{\pi}{2}$ with the equivalent angle $2\pi+\frac{\pi}{2}$, and taking half of it results in $\frac{5\pi}{4}$. Note that the second square root is diametrically opposite from the first square root on the unit circle.

To obtain the coordinates (real and imaginary parts) of these two square roots, you can use the fact that a point at angle $\theta$ on the unit circle has coordinates $(\cos \theta,\sin\theta)$ and substituting the values $\theta=\frac{\pi}{4}$ and $\theta=\frac{5\pi}{4}$ into this.

In reverse order:

$$a+bi^2 = a + b(-1) =  (a-b) + 0i = (a-b, 0)$$

Since the tuple form comes from the standard form of the complex number, there never is a higher power on the $i$ than $1.$

As for powers of $i$... We can find two possibilities for $i^{1/2}$ just by solving
$$(a+bi)^2 = i \\ (a^2-b^2) + (2ab)i = i \\ \begin{cases}a^2-b^2 = 0 \\ 2ab = 1\end{cases} \\ a = b = \pm \frac{\sqrt{2}}{2} \\ a+bi = \pm\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right)$$

A word of warning: While the powers of positive real numbers can be defined as a single number, you will almost certainly run into problems if you want a consistent definition of powers of complex numbers.  In the most general case, the power $(a+bi)^r$ is evaluated as $e^{r\ln(a+bi)},$ which means in order to find a power of a complex number, we first need to define the natural logarithm, which requires a lot of care.


How do you write, $(A∩B)\setminus (A∩C)$. How do you write, $(A\cap B)\setminus (A\cap C)$. I know $(A \cap B) = \{x: x \in A$ and $x \in B\}$, and $(A \setminus B) = \{x: x \in A$ and $x \not \in B)\}$, but I don't know how to combine these facts into one.

You are taking $A\cap B$ and excluding something from it. Anything you exclude from $A\cap B$ is automatically in $A$. So at that point, writing that you are excluding $A\cap C$ is a little redundant. You can just say you are excluding $C$. So you have the same thing as $$(A\cap B)\setminus C$$ It's hard to imagine anything simpler exists, since this references the three sets involved each precisely once. This is the same as $$A\cap(B\setminus C)$$ so you could just write $$A\cap B\setminus C$$

If you want set-builder notation, this is $$\{x\mid x\in A \text{ and } x\in B \text{ and } x\not\in C\}$$

You can learn a few simple laws of set theory in one week and use them always.

$(A\cap B)\setminus (A\cap C)=[\color{green}{\text{by Boby's comment}}]$

$(A\cap B)\cap (A\cap C)^c=$ [by De Morgan's laws]
$\color{red}{(A\cap B)}\cap (A^c\cup C^c)=$ [by this question]

$(\color{red}{A\cap B}\cap A^c)\cup (\color{red}{A\cap B}\cap C^c)=$

$(A\cap B\cap C^c)=[\color{green}{\text{by Boby's comment}}]$

$(A\cap B)\setminus C$


Show that the $\lim_{n\to \infty} \frac n{n+1} = 1$ Show that the $\lim_{n\to \infty} \frac n{n+1} = 1$ using the limit definition.
So far I have:
Let $\varepsilon > 0$, there exists $N \ge 0$ such that,
$|a_n-L|< \varepsilon$, for all $n \ge N$.
$\left|\frac n{n+1} -1\right| < \varepsilon$. I have ended up with $\left|\frac 1{n+1} \right|  \le \frac n n = 1$
(which is $< \varepsilon$, for all $n$)
I know there needs to be an n at the end, so for all $n >$ something $\varepsilon$. I am not sure what to do after $n/n$, or is that enough to show the limit of the sequence $= 1$.
Thanks

Using that for $A,B>0$
$$A<B \iff \frac 1 A > \frac 1 B$$
we have that $\forall \varepsilon>0$
$$\left| \frac1{n+1}\right|<\varepsilon \iff n+1 >\frac1 \varepsilon \iff n>\frac1 \varepsilon  -1$$
then we can take any $N\ge \frac1 \varepsilon  -1$ to comply the definition of limit.

" I have ended up with |1/(n+1)| ≤ n/n = 1": no, you end up with
$$\left|\frac1{n+1}\right|<\epsilon$$ and it is no problem to find an $N$ above which this holds.


Is there a name for this observation involving single-variable limit to infinity? Obviously,

$$\lim_{n\to \infty} \sqrt{n^2 +1} = \infty$$

However, the following (I am sure) is true:

$$\lim_{n\to \infty} \sqrt{n^2 +1} = n$$

Is there a name for this kind of behavior in limits? Such as a general rule or variable specific rule?

The statement

$$\lim_{n \to \infty} \sqrt{n^2 + 1} = n$$

is incorrect and meaningless, since $n$ is varying on the left, but is left constant on the right. Perhaps you meant that

$$\lim_{n \to \infty} \sqrt{n^2 + 1} - n = 0$$

which is true. It seems that in general, what you're looking for is asymptotic analysis, or the ideas involved in Big O notation.

It should be noted that I am a physicist, not a mathematician.  Also, i do spend a good deal of time revamping notations for the higher dimensions at my site http://www.os2fan2.com .

Much of the problems seem to lie in that if a term $A$ includes meanings $a, b, c$, it is then simply because the same term is being used, that arriving at $a$, one can suppose that $c$ is also true, or alternately, because $c$ is absent, so is $a$.

A simple example of this is that zero includes the concepts of empty bag (no elephants) and empty column (no dimes in the dollars and cents).  The assumption that the Egyptians had no zero is based on the absence of a symbol for the empty column, but no mention is made of the empty bag being present.

The serious effect is the use of $\infty$ to represent any unbounded number, and then to turn around and use this as a kind of equality.  So, while $x$ and $\log(x)$ are unbounded as $x$ goes large, they are not equal.  None the less, we see the basis of the Cantor diagonal theorm is based on the notion that $x=\log(x)$ somehow comes to contention as $x$ gets large: specifically, that one can have $x$ numbers, each $x$ places long.

One is better served by replacing $\infty$ by the main approximate, so the points in a 3D lattice $p(r)=r^3$ as $r \to \infty$.  The ones that have an $r^n$ distribution are what i call `class n', are often governed by an order-n isomorphism.  Class-2 numbers are also characterisied by being able to write all members in a correctly-spelt (ie sort by spelling) series (like a decimal: decimals are a class-2 system).  Putting $\infty$ here simply prevents this from being seen.

The particular comments by rschreib and T Bongers to my comments, are based on the notion that $\lim$ must be read as a function, and that it is meaningless and wrong to use the limit control outside the side of the equal-term the limit is on.  Meaningless and wrong are two entirely different things.  They both seemed to had been able to extract the correct meaning from the equation, so what was written is meaningful.  Wrong is an issue of convention, here the artificial one that $\lim$ is a function of lower order than Addition/subtraction.  Putting it at lower than equality makes it allowable to read the RHS too.  $\pmod{b}$ is read that way, so it's not unprecedented that a symbol can be lower than an equity.

Writing something like $\sqrt{n^2+1}=n$ as $n$ goes large is perfectly meaningful.  This is exactly the meaning that OP intended, and that both of the commentators read.  Of course, one can write 'as $n$ goes large' as $\lim_{n \to \infty}$ does exactly this.

So their justification that the statement is meaningless is more to do with supposing that the governing clause is broken by the equal sign, when set as a symbol, but is not broken when the thing is spelt out.

I have spent a good deal of time, devising symbols for merging polytopes against the same symmetry.  What this means is that when you write down a symmetry group down, and then the decorations for each polytope to it, you can only ensure you're talking of the same orientation, when the symbols are attached to the same symmetry.

So, here for example, to write something like this, one can not assure that the counters in the two limits run to the same tune: that eg when $x=3$ on the LHS, it is equal to 3 on the RHS.  $$\lim_{x \to \infty} \sqrt{x^2+1} = \lim_{x \to \infty} x $$

Writing $$\lim_{x \to \infty} \sqrt{x^2+1} = x$$ can be read as if the limit governs the whole statement, and that the $x$ on the LHS and RHS are governed by the same ticks (ie if $x=3$ on the LHS, then $x=3$ on the RHS.

It's considered bad form to use the same symbol for loop-counters and for variables outside the loop, and therefore it is apparent that the whole equation is governed by the limit.  This is why the meaning is self evident, regardless of whether it is 'right' or 'wrong'.  That is because the commentators chose to see the limit as a function governed by the equal sign, rather than governing the equal sign.  In fact, the limit governs the occurrance of its variable, wherever it falls, ie writing

$$lim_{x \to \infty} A(x)  \text{ is the same as } A(x) \; \text{ as } x \to \infty$$


Evans PDE chapter 2 problem 4 Problem is

  Give a direct proof that if $u \in C^{2}(U) \cap C(\overline{U})$ is harmonic within a bounded open set $U$, then
  $\max_{\overline{U}} u =\max_{\partial U} u$.

What I think is that;

  Let $u^{\epsilon} = u+ \epsilon |x|^{2}$, where $\epsilon >0 $. Then $\bigtriangleup u^{\epsilon} = \bigtriangleup u + 2\epsilon |x| = 2\epsilon |x| \geq 0$ since $u$ is harmonic. Equality holds if $x=0$.  Suppose $x_{0} \in U$ is maximum point of $u^{\epsilon}(\overline{U})$. Then $\bigtriangleup u^{\epsilon}(x_0) = 2 \epsilon |x_0| \leq 0$.

I want to show that $x_0 \neq 0$, so this is contradiction, but I don't know how to do that. Could anyone have an idea?

Let $u_{\varepsilon} = u + \varepsilon |x|^2, \forall \varepsilon > 0$. Then $\Delta u_{\varepsilon} = 2 \epsilon > 0$ since $u$ is harmonic. If $u_{\varepsilon}$ attained a local maximum at an interior point then $\Delta u \le 0$. Thus, $u_{\varepsilon}$ attains its maximum on the boundary and has no maximum at an interior point. If $|x| \le r, \forall x \in U$, it follow that
$$
\sup_U u \le \sup_U u_{\varepsilon} \le \sup_{\partial U} u_{\varepsilon} \le \sup_{\partial U} (u + \varepsilon r^2) .
$$
Let $\varepsilon \to 0$ we get $\sup_U u \le \sup_{\partial U} u$ and $\max_{\partial U} = \max_{\bar{U}} u$.

If the maximum is attained inside the domain, at point $x_0$, then the Hessian $H$ of $u$ at $x_0$ is symmetric definite negative. You first have that
$$Tr(H) = \Delta u$$
and because H is definite negative,
$$Tr(H) < 0$$
This implies a contradiction with $u$ is harmonic.


geometric question (About Pythagoras theorem) I might need two different method each Question 1:ABCD is a square and GDEF is a rectangle $BE=1,EC=3 $ find the length of GD

Question 2: ABCD is a right trapezoid, $CD=BC=10$ ,ABE is an equilateral triangle, please find the length of AB.

1) Triangle $ADG$ is similar to $EDC$. Hence:
$$
GD:DC=AD:DE,
\quad\text{that is:}\quad
GD:4=4:5.
$$

2) If $AD=x$ and $DE=y$ then by Pythagoras' theorem:
$$
x^2+y^2=10^2+(10-y)^2=10^2+(10-x)^2.
$$

Question 1:

Since BC=4=AB, the distance AC is $4\sqrt 2$. Also, CDE is a 3-4-5 triangle, of area 6.  So if DE is its base, then $A=\frac 12 bh$ gives its height to be 2.4.  That means that the distance GD, which is equal to the distance from A to the intersection of DE with AC, is $4\sqrt 2 - 2.4$.

Hint for question 2:

The bottom white triangle gives you:
$10^2 + CE^2 = ANS^2$

The top white triangle gives: $(10-CE)^2+AD^2=ANS^2$

You can figure out AD by setting up trigonometric identities.

Now, solve this system of equations.


If $a\ge 0$ and $b\ge 0$, then $\sigma(ab)\subset\mathbb{R}^+$. This is an exercise in Murphy's book:

  Let $A$ be a unital $C^*$-algebra and $a,b$ are positive elements in $A$. Then $\sigma(ab)\subset\mathbb{R}^+$.

The problem would be trivial if the algebra is abelian. On the other hand I do not have a clue for the non-abelian case. I guess one needs to use the fact $\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}$ and then maybe use some algebraic manipulation.

Anyway, I wonder whether someone has a hint on this. I guess I am missing a trick.

Better though, maybe someone has some general insight on the many techniques concerning positive elements and approximate identities. For me they all seem very tricky and mysterious. For instance, how can they think of those strange functions when proving the existence of approximate identities and quasicentral approximate identities.

Thanks!

A common trick is to write a positive element of a C*-algebra as a square of a positive element:  $b=\sqrt b\sqrt b$.  Thus $\sigma(ab)\cup\{0\}=\sigma(a\sqrt b\sqrt b)\cup\{0\}=\sigma(\sqrt b a\sqrt b)\cup \{0\}$.  If you haven't already seen it, you can think about why if $a$ is positive, then $xax^*$ is positive for all $x$.

$$\sigma(ab)\subset \sigma(a)\sigma(b)\subset \mathbb R^+$$


What is the pattern in these ordering properties? I'm not exactly sure what my question is, but I've observed something about orders and would like to see if there is more of a story to this.

Let's assume $\;\lt\;$ is an order on some unspecified domain, and let's define $$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
\tag{0}
a \le b \;\equiv\; a \lt b \lor a = b
$$  Now, as I recently discovered (when writing answer https://math.stackexchange.com/a/2251407/11994 where I used $\Ref{3}$), if the domain were the reals or the rationals then the following would hold:
\begin{align}
\tag{1} \langle \forall z :: z \le a \then z \le b \rangle \;\equiv\; a \le b \\
\tag{2} \langle \forall z :: z \lt a \then z \lt b \rangle \;\equiv\; a \le b \\
\tag{3} \langle \forall z :: z \lt a \then z \le b \rangle \;\equiv\; a \le b \\
\tag{4} \langle \forall z :: z \le a \then z \lt b \rangle \;\equiv\; a \lt b \\
\end{align}
Now visually it looks like $\Ref{4}$ is the odd one out, but let's see if that intuition is correct: I was wondering what kind of order we need to prove each of the above.

First let's look at the $\;\when\;$ direction: there, for all 4 equivalences, we only need the fact that $\;\lt\;$ is transitive.  For example, for $\Ref{3}$ we have that $\;z \lt a\;$ and $\;a \le b\;$ imply $\;z \lt b\;$ by transitivity.

For the $\;\then\;$ direction of $\Ref{1}$ and $\Ref{4}$ we see that we can simply choose $\;z := a\;$, and use the fact that $\;\le\;$ is reflexive (by definition $\Ref{0}$), in other words $\;a \le a\;$.

For $\Ref{2}$ we can choose $\;z := b\;$, and assuming $\;\lt\;$ is irreflexive (so $\;\lnot(b \lt b)\;$) we get $\;\lnot(b \lt a)\;$, and further assuming $\;\lt\;$ is trichotomous (so $\;a \lt b \lor a = b \lor b \lt a\;$) we get $\;a \le b\;$.

Finally the $\;\then\;$ direction of $\Ref{3}$ is the most complex one.  The best approach I found was to go existential:
$$\calc
    \langle \forall z :: z \lt a \then z \le b \rangle
\op\then\hint{see below, where $\;\lt\;$ is transitive and irreflexive}
    \langle \forall z :: z \lt a \then \lnot (b \lt z) \rangle
\op\equiv\hint{logic: DeMorgan}
    \lnot \langle \exists z :: z \lt a \land b \lt z \rangle
\op\then\hint{$\;\lt\;$ is dense, so $\;b \lt a\;$ implies there is a $\;z\;$ between them}
    \lnot (b \lt a)
\op\then\hint{$\;\lt\;$ is trichotomous}
    a \le b
\endcalc$$
where the first step above is proved as follows:
$$\calc
    z \le b \;\then\; \lnot(b \lt z)
\op\equiv\hint{logic}
    \lnot(z \le b \land b \lt z)
\op\when\hint{$\;\lt\;$ is transitive}
    \lnot(z \lt z)
\op\equiv\hint{$\;\lt\;$ is irreflexive}
    \true
\endcalc$$
So here we require that $\;\lt\;$ is transitive, irreflexive, trichotomous, and dense.

To summarize the requirements on $\;\lt\;$ that seem necessary to prove the statements above:
$$\begin{array}{cl}
\Ref{1}&\text{transitive}\\
\Ref{2}&\text{transitive, irreflexive, trichotomous}\\
\Ref{3}&\text{transitive, irreflexive, trichotomous, dense}\\
\Ref{4}&\text{transitive}\\
\end{array}$$

So by this table $\Ref{3}$ is the odd one out, since it requires the most properties of the order $\;\lt\;$: specifically it requires density, which makes it true for the reals and the rationals, but false for e.g. the integers.

So I think my questions are: Are all of the above indeed necessary, or can some be weakened?  And is there something more to this pattern?  Why does density only play a role in one of these equivalences?  Is there perhaps a more symmetrical way of organizing this information?

Assume transitivity.

As you've already made the case, lifting density will give you a straightforward counter-example for $(3) (\implies)$.

To lift trichotomy, one must lift one or more of anti-reflexivity, anti-symmetry, or completeness.

Lifting completeness gives a straighforward lattice counter-example for $(2) (\implies)$, so completeness is necessary for $(2) (\implies)$.

Lifting anti-reflexivity, take a domain with some $a, b$ such that $b < b < a \land \lnot (a < a) \land \lnot (a = b) \land (\forall z: z \ge a \lor z \le b)$.  Thus, $z \le b \implies z < b$, so $\forall z: z < a \implies z \le b \implies z < b$, yet $a > b \land \lnot (a = b)$, i.e. $\lnot (a \le b)$ -- so anti-reflexivity is necessary for $(2) (\implies)$ as well.

Anti-reflexivity together with transitivity implies anti-symmetry, so anti-symmetry is also necessary for $(2) (\implies)$.

With all of anti-reflexivity, anti-symmetry ans completeness necessary for $(2) (\implies)$, trichotomy is also necessary.  Because $LHS (2) \implies LHS (3)$, any premise necessary for $(2) (\implies)$ is also necessary for $(3) (\implies)$, so trichotomy is also necessary for $(3) (\implies)$.

In conclusion: assuming transitivity, all the other premises are necessary as listed.

(Not sure about the rest of your questions, but you may want to check if 42 fits...)

Domain D is linear is sufficient. Assume left premise.  If not a <= b,
then a < b, thus b < b.  Also if D is almost complete (partial) order, then
D holds.
D can be a dense linear order or an almost complete dense order for 3 to hold.


What is the cartesian product of a non empty set and a set containing an empty set (or $X_1 \times \{ \emptyset \}$ )? Let $X_1$ be a set. What is $X_1 \times \{ \emptyset \}$? I know that a the product of a set and an empty set is an empty set, but what is the product of a set and an empty set WITHIN a set?

The empty set, considered as an element of another set, is nothing special!  It is just an element.  You could do the problem by finding
$$X_1\times\{\,a\,\}$$
and then replacing $a$ by $\varnothing$.

Still empty. Cartesian product does not care about where your set lives. It's the set of all ordered pairs taken from the two sets, and when one of them is empty, there are no such pairs.