Derivative of an even function is odd and vice versa This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even.
(Write the equation that says f is even, and differentiate both sides, using the
chain rule.)"

I already read numerous solutions online. This was the official solution but I didn't quite understand it (particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$).

Thanks in advance =]

Official, shmofficial: I think the following might prove to be easier to grasp for some: suppose $\,f\,$ is odd, then

$$f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{-f(x)+f(x_0)}{-x+x_0}=$$

$$=\lim_{x\to x_0}\frac{f(-x)-f(-x_0)}{(-x)-(-x_0)}\stackrel{-x\to y}=\lim_{y\to -x_0}\frac{f(y)-f(-x_0)}{y-(-x_0)}=:f'(-x_0)$$

The above remains, mutatis mutandis, in case $\,f\,$ is even.

Well, geometrically, even function means reflection along y axis, so any direction will reflect, that mean, the derivative on the right is the same as the derivative on the left, but the direction change. It means the value is the same, but with different sign.

Odd function means rotational symmetric, if you rotate an arrow, I.e. direction, you will change by 180 degree, so it is the same slope, hence the derivative of odd function is even.


Show that if $G$ is abelian, and $|G| \equiv2 \mod 4$, then the number of elements of order $2$ in $G$ is $1$. I've tried proving it by contradiction, assuming the number of elements is different than one, which, by Sylows $3$, implies that $|G|=2^xm$ with $m$ odd. With that I managed to show that $m\equiv1\mod4$, but kinda got stuck there...

Here's an alternative solution without using Sylow's theorems.

Note that $|G|\equiv 2 \pmod 4 \Rightarrow |G|=2(2k+1)$ for some $k\in \mathbb{Z}$.

Since $2$ divides $|G|$, by Cauchy's theorem there exists an element $g\in G$ of order $2$. Now $\langle g\rangle$ is a subgroup of order $2$. Now since $G$ is abelian we have that $\langle g\rangle $ is a normal subgroup, and hence $G/\langle g\rangle $ is a group. Now by Lagrange's theorem $|G/\langle g\rangle | = 2k+1$, which is odd.

Suppose there was another element $h\in G$ with order $2$ and $h\neq g$. Then we have that $ h + \langle g \rangle $ is an element of order $2$ in $G/\langle g\rangle$. Thus $\langle h + \langle g\rangle \rangle $ is a subgroup of $G/\langle g\rangle$ or order $2$. However, this is a contradiction since $|G/\langle g\rangle|$ is odd.

Hint: $|G|$ is $2(1+2p)$, a finite abelian group is a product of groups isomorphic to $\mathbb{Z}/p^l$ where $p$ is a prime this implies that the component associated to $2$ is $\mathbb{Z}/2$.

https://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Classification


relatively open sets Definition of a relatively open set:

$ D \subset K^N $ is a set.
$U \subseteq D $ is relatively open in D if
$$ U = \emptyset \quad$$ or
$$     \forall x \in U \quad \exists \quad r > 0 \quad | \quad B(x,r) \cap D \subseteq U$$

What I want to know is: is there a set U with $x \in U \subseteq D  $ such that  $ B(x,r) \cap D \nsubseteq U$.

Example:

If $D = (0,2] $ and $ U =[1,2]$
and $x = 2$, then $B(2,r) \cap D = (2-r,2]  \subseteq U$.

In the above example I dont see for any $x \in U$ where $ B(x,r) \cap D \subseteq U $ is not  satisfied.
Can someone please give examples of $D $ and $ U $ where $ B(x,r) \cap D \subseteq U $ is not  satisfied.

Definition of a relatively open set, U must be in D, but D=(0,2] and U=[1,2] , U is not in D, this is D in U, your example is not a counterexample


Can I use "$\iff$" symbol when I "transform" an expression to another form? I am writing a solution to prove that $\sqrt5$ is not rational. Here is my first half proof:

  Assume $\sqrt{5}$ is a rational number. By the definition of rational number, $\sqrt{5} = \frac{p}{q}$, where $p,q\in\mathbb{Z^+}, q\neq0$, and $gcd(p,q)=1$. We have $5=\frac{p^2}{q^2} \iff 5q^2=p^2$.

Can I use "$\iff$" symbol like this?

How about if a question is asking me to work backward from the desired conclusion and then prove:

If x and y are nonnegative integer, then $\frac{x+y}{2}\geq \sqrt{xy}$.

Can I do like something like:

  (first half proof) $\frac{x+y}{2}\geq \sqrt{xy} \iff x+y\geq 2\sqrt{xy} \iff (x+y)^2 \geq 4xy
\iff x^2+2xy+y^2 \geq 4xy \iff x^2-2xy+y^2 \geq 0 \iff (x-y)^2 \geq 0$.

Should I use $\iff$, $\Leftarrow$ or $\Rightarrow$?

Yes. "$\iff$" is used to denote logical equivalence, or necessary and sufficient conditions. It is read "if and only if," of which "iff" is a common abbreviation. $p \iff q$ is true if $p$ is true whenever $q$ is true and $q$ is true whenever $p$ is.

  Assume $\sqrt{5}$ is a rational number. By the definition of rational number, $\sqrt{5} = \frac{p}{q}$, where $p,q\in\mathbb{Z^+}, q\neq0$, and $gcd(p,q)=1$. We have $5=\frac{p^2}{q^2} \iff 5q^2=p^2$.

This is true, since $5=\frac{p^2}{q^2} \Rightarrow 5q^2=p^2$ and $5q^2=p^2 \Rightarrow 5=\frac{p^2}{q^2}$.

  If x and y are nonnegative integer, then $\frac{x+y}{2}\geq \sqrt{xy}$.

  Proof: $\frac{x+y}{2}\geq \sqrt{xy} \iff x+y\geq 2\sqrt{xy} \iff (x+y)^2 \geq 4xy
\iff x^2+2xy+y^2 \geq 4xy \iff x^2-2xy+y^2 \geq 0 \iff (x-y)^2 \geq 0.$

This is a fine proof. I would like to see it start where you end, but with $\iff$ used as you do here, it's perfectly valid as is.

You are moving forward in the proof, so you can use $\implies$ (this implies) sign which is more appropriate.


A problem on conditional geometric probability The point (x, y) is chosen randomly in the unit square. What is the conditional probability of $x^2+y^2 \leq 1/4$ given that $xy \leq 1/16$

I started solving this and while calculating I got some very unpleasant numbers but the problem is not marked as difficult in this book where I found it. So I start suspecting that I am having some conceptual mistake.

The two curves intersect at $x=\sqrt{2+\sqrt {3}} /4$ and at $x=\sqrt{2-\sqrt {3}} /4$

Now I have to find a few areas via definite integrals. Is that what this problem is about?!

The integral of $\sqrt{1/4-x^2} $ is very unpleasant, it seems. So?!

Is there some smarter / nicer approach?

$xy \leq \frac{1}{16}$ iff $2xy \leq \frac{1}{8}$.

Conditional on this,

$x^2 + y^2 \leq \frac{1}{4}$ iff $x^2 + 2xy + y^2 = (x+y)^2 \leq \frac{3}{8}$ iff $x+y \leq \sqrt{\frac{3}{8}}$.

Now do you know how to calculate the probability that the sum of two random numbers is less than a given number?

Suppose that $xy\le \frac{1}{16}.$ Then $x^2+y^2\le \frac14\iff (x+y)^2\le \frac38$  This should lead to an easier way to approach it.


Find the particular solution of the following equation my task is to Find the particular solution of the following equation at y(0) =1. the equation is
$$dy/dx + 4y = 7$$

I have made this much progress as far as separating the Ys and the Xs and taking the anti derivatives (I hope it's correct) but don't know how to reach an answer
dy/dx + 4y = 7    || .dx
$dy + 4y = 7 dx$
$4 y^2/2 = 7x +c$
.....
correct answer is:
$y = 7/4 - 3/4e^{-4x}$
update:
I managed to get 7/4 and Ce^-4x) but still don't get where -3/4 comes from
thank you!

We have
$$\frac{dy}{dx}=7-4y,$$ and therefore
$$\frac{dy}{7-4y}=dx.$$
Continue.

Added: Integrate. We get $-\frac{1}{4}\ln(|7-4y|)=x+C$. Put $x=0$. We get $C=-\frac{1}{4}\ln 3$. Thus
$$\ln(7-4y)=-4x+\ln 3.$$
Note this is valid only when $7-4y\gt 0$. Exponentiate. We get $7-4y=3e^{-4x}$.

Solve for $y$. We get $y=\frac{1}{4}\left(7-3e^{-4x}\right)$.

$\frac{dy}{dx}+4Y=7$ |
can be solved letting $y=Ae^{bx}$.
$y'=Abe^{-x}$ substituting  into the original equation and
solving the homogeneous part  gives
$(b+4)=0 .$
hence $b=-4 $ we have a solution $y=Ae^{-4x}$
called the complementary solution ,we need o find the particular solution

we check the right side of the equation ,the guess for a constant is $C$

$$Y(particular)=c$$
$$y'(particular)=0$$

substituting back into $\frac{dy}{dx}+4y=7$
we have $$4c=7$$
$$c=7/4$$

applying superposition whereby we have the general sol as $$y=Ae^{-4x}+7/4$$ the sum of complementary solution and particular solution .$y(0)=1$

$$1=A+7/4  \phantom{filler}\text{  (ANY NUMBER TO EXPONENT ZERO IS ONE)}$$
$$A =-3/4$$
$$Y=-3/4(e^{-4x})+7/4.$$


$\sup \left\| A x + B y + C z \right\|$ subject to $\left\|x\right\| = \left\|y\right\| = \left\|z\right\| = 1$ I'm interested in finding

$\sup \left\| A x  + B y + C z \right\|$ subject to
$\left\|x\right\| = \left\|y\right\| = \left\|z\right\| = 1$

where $A$, $B$, $C$ and $x$, $y$, $z$ are real matrices and vectors, respectively, of compatible sizes, and the norms are Euclidean.

What is this problem called in the literature? (If there is no established name, does this problem reduce to a well-known one?)

(If there was only one constraint, this would be the induced matrix norm)

Refutation of @zimbra314's answer (Can not fit in the comments)

  $[x, y, z]^{T}= \alpha_{1} v^{T}_{1} + \alpha_{2} v^{T}_{2}+\alpha_{3} v^{T}_{3}$, where $v_{i}$ is the $i$-the right singular vector of $[A,B,C]$ (i.e. singular vector corresponding to $i$-th largest singular value).

This assumes that the optimal $[x, y, z]^{T}$ is a linear combination of the first 3 right singular vectors of $[A, B, C]$. However, this assumption is incorrect. As a counterexample, consider

$
A = \begin{bmatrix}2&0&0\\0&2&0\\0&0&2\\0&0&0\\0&0&0\end{bmatrix}
$

$
B = \begin{bmatrix}0\\0\\0\\1\\0\end{bmatrix}
$

$
C = \begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}
$

Hint:

$\left \| Ax+By+Cz \right \| = \left \| \left [ A, B, C \right ] \left [ \begin{matrix}
x\\
y\\
z
\end{matrix} \right ]\right \|$

Detailed answer:

$[x, y, z]^{T}= \alpha_{1} v^{T}_{1} + \alpha_{2} v^{T}_{2}+\alpha_{3} v^{T}_{3}$, where $v_{i}$ is the $i$-the right singular vector of $[A,B,C]$ (i.e. singular vector corresponding to $i$-th largest singular value).  Denoting $[v_{ix},v_{iy},v_{iz}]^{T}=v_{i}$ to mark the partition of $v_{i}$ corresponding to $x$, $y$ and $z$.

Now solve: $||x||^{2}=1 \rightarrow ||(\alpha_{1} v_{1x} + \alpha_{2} v_{2x}+\alpha_{3} v_{3x})||=1$

Similarly ,
$||(\alpha_{1} v_{1y} + \alpha_{2} v_{2y}+\alpha_{3} v_{3y})||=1$

$||(\alpha_{1} v_{1z} + \alpha_{2} v_{2z}+\alpha_{3} v_{3z})||=1$

Three equations to solve for three unknowns $\alpha_{1}, \alpha_{2}$, and $\alpha_{3}$. Then, your answer is $\sqrt{\alpha^{2}_{1} \sigma^{2}_{1} + \alpha^{2}_{2} \sigma^{2}_{2} + \alpha^{2}_{3} \sigma^{2}_{3}}$, where $\sigma_{i}$ is the singular value corresponding to $i$-the largest singular vectors.


What is the average of no numbers? I have two programs that both behave nearly identically: they both take in any numbers you give them and can tell you the average and how many numbers were given. However, when you don't give them any numbers, one says the average is 0.0, and the other says it's NaN ("Not a Number"). Which of these answers, if any, is more correct, and why?

  Note: Although I use "programs" as a metaphor here, this isn't a programming question; I could've just as easily said "computers", "machines", "wise men", etc. and my question would be the same

From a statistical point-of-view, the average of no sample points should not exist. The reason is simple. The average is an indication of the centre of mass of the distribution. Clearly, for no observations there can be no way to prefer one location vs. another as their centre of mass since the the empty set is translation invariant.

More mathematically, taking the average is a linear operation, which means if you add a constant $c$ to each observation, then the average $a$ becomes $a+c$. Now if you add $c$ to each observation in the empty set, you get the empty set again, and thus the average will have to satisfy $a+c=a$ for all $c$, clearly nonsense.

You can have an average of whatever is input (assuming we are not looking at PI which I do not know as having an ending).  If you press Enter or whatever method is used to cause an input into the program, then the programming language or input design of the program will interpret that information, likely perform some calculation, then give the result of the calculation.

What I mean is, you can just press enter (with no other value entered) and (depending on the programming language or design of the program) the program can interpret your value (if reading integers) as Zero.  Some languages may interpret the input as a non value or nul and may output something similar to your 'NaN'.

However, in answer to your first question, the average of one thing is always that thing.  Unless you can only have an average until you have two or more of something to create an average.  So whatever is input, as long as it was only one input, would result in that input as being the average.

To answer your ending question, I would say the NaN is more correct because just pressing enter (assuming that is how the data is input) and is done so with the intent of at least not giving a specific value.  That is the answer for the person inputting the data.  The answer may also depend on the programmers intent on what type of input they are looking for --- numerical (0 may be correct) or non-numerical (Nan is correct).


Find $\iint \frac{1}{\sqrt{x^2+y^2}}\;\mathrm{d}y\;\mathrm{d}x$ Given that $x^2+y^2\leq a^2$, evaluate
$$
  \iint\limits_D \frac{1}{\sqrt{x^2+y^2}} \;\mathrm{d}A
$$

Now my initial idea is to use polar coordinates, but I do not get the right answer.

This is what I have done:

$$\int\limits_0^{2 \pi }\int\limits_0^a\frac{r}{a}\;\mathrm{d}r\;\mathrm{d}\theta=\pi  a$$

Now the answer I got I'm pretty sure is right, which makes me conclude that my setup of the initial integral is wrong since the answer should be $2\pi  a$ according to my textbook.

Could someone point out what I am missing?

$$\int\limits_0^{2 \pi }\int\limits_0^a\frac{r}{r}\;\mathrm{d}r\;\mathrm{d}\theta=2\pi  a$$

The integral is The integral of (1/r)r dr d theta and the limits are for from 0 to a and for theta from 0 to 2 pi giving the expected answer


What is a link between the topological and order-theoretic completeness? Take $\mathbb{R}$ as an example.

Order-theoretically, the set $\mathbb{R}$ of all real numbers can be developed as a complete totally ordered field, where "complete" indicates that the supremum axiom is imposed.

On the other hand, the set $\mathbb{R}$ happens to be topologically complete, in the sense that every Cauchy sequence in $\mathbb{R}$ converges in $\mathbb{R}$.

Is the order-theoretic completeness a notion developed independent of the concept of topological completeness, or is the converse true, or none of them is true? In general, what is a link between them?

The two concepts are strictly linked through the modern definitions of real number.

We have to consider Cantor's definition, based on Cauchy sequence, and Dedekind's one, based on Dedekind cuts.

Several axioms has been proposed to ensure the so-called "completeness" like :

metric completeness : every Cauchy sequence of points in a metric space $M$ has a limit in $M$
order-theoretic completeness.

See also Construction of the real numbers, Dedekind's Contributions to the Foundations of Mathematics and The Early Development of Set Theory; both Cantor and Dedekind was at the origins of :

foundations of analysis and definition of the structure of the set $\mathbb R$ of real numbers
point set topology
set theory.

In high school you learn to work with the system ${\mathbb Q}$ of rational numbers, or with the smaller system ${\mathbb D}$ of finite (binary or) decimal fractions. While these systems are unproblematic for daily business they are unsatisfactory from a mathematical standpoint: Many important numbers that we know should be there, are missing; e.g., ${1\over3}\notin{\mathbb D}$, $\sqrt{2}\notin{\mathbb Q}$, etcetera.

There are various "bigtime" abstract constructions that fill in these holes, so that a homogeneous continuum of real numbers ${\mathbb R}$ results.

One such construction uses "Dedekind cuts" and produces a system ${\mathbb R}_{\rm Ded}$ that is so called order complete. Another construction uses "Cauchy sequences" and produces a system ${\mathbb R}_{\rm Cau}$ that is  metrically complete.

Whichever approach you take, you can prove at the end that your ${\mathbb R}_{\rm Ded}$ is in fact also metrically complete, resp. that your ${\mathbb R}_{\rm Cau}$ is in fact also order complete – the reason being that  ${\mathbb R}_{\rm Ded}$ and ${\mathbb R}_{\rm Cau}$ are "isomorphic".


Bijection between partitions Give a bijective mapping from the set of partitions of $[n]$ with no cyclically consecutive integers in a block, and the set of partitions  of $[n]$ with no singleton blocks.

All the mappings that I come up with are injective. Can somebody please help?

[1,2,3]       [1] [2] [3]
[1] [2,3]     [2] [1,3]
[2] [1,3]     [3] [1,2]
[3] [1,2]     [1] [2,3]
[1] [2] [3]                      [1,2,3]

[1,2,3,4]     [1] [2] [3] [4]
[1] [2,3,4]   [2] [3] [1,4]
[2] [1,3,4]   [3] [4] [1,2]
[3] [1,2,4]   [1] [4] [2,3]
[4] [1,2,3]   [1] [2] [3,4]
[1,2] [3,4]   [1] [3] [2,4]
[1,3] [2,4]                     [1,3] [2,4]
[1,4] [2,3]   [2] [4] [1,3]
[1] [2] [3,4]  [3] [1,2,4]
[1] [3] [2,4]                   [1,2] [3,4]
[1] [4] [2,3]  [2] [1,3,4]
[2] [3] [1,4]   [4] [1,2,3]
[2] [4] [1,3]                   [2,3] [4,1]
[3] [4] [1,2]   [1] [2,3,4]
[1] [2] [3] [4]                 [1 2 3 4]

Above is a grouping for the $n=3$ and $n=4$ case. The way I was approaching it was to think of a way of flagging cyclic consecutive integers in a group with singletons, and the rule I came up with for this purpose was to say that if $[i,i+1]$ appear in any given block, then in the bijective map $[i]$ would be by itself, and $[i+1]$ would be set aside and combined with the others (if $[i+1,i+2]$ is not in the same block). I believe this is a bijection between bad sets to bad sets, but I suppose there is something to be done for the other way :).

Looks like it's not clear how to design the map on good sets... The current pairing for $n=4$ above works as follows: If there are no singletons, the map is an identity, otherwise for each singleton $[i]$, put $i+1$ in the same group. If $[i+1]$ is also a singleton, grab $[i+2]$ and place in the same group, etc. As this is homework, I don't want to spoil all the beans, only to give some ideas to push (it could be this doesn't work out, but it's an idea).

I'll try to give a bijection in the $n=2, 3, 4$ cases to see if we can see something.

{1 2}        {1,2}

{1 2 3}      {1,2,3}

{1 2 3 4}    {1,2,3,4}
{1 3,2 4}    {1 3,2 4}
{2 1,3 4}    {2,1 3,4}
{1 4,2 3}    {1,4 2,3}

These bijections look "natural", but I can't really think of a way to expand the thinking...


The number of homomorphisms from $S_5$ to $A_6$ I used the fact that $A_5$ is the only real normal subgroup of $S_5$. Then we have a 2 types of homomorphisms and the trivial homomorphism.

homomorphisms with a kernel $\{e\}$
homomorphisms with a kernel $A_5$

In the first case the first isomorphism theorem implies that the image of the homomorphism is isomorphic to $S_5$, therefore the number of homomorphism of that type is the number of injections of $S_5$ in $A_6$.

In the second case because the index of $A_5$ in $S_5$ is 2, the quotient group is isomorphic to $C_2$. Therefore  the image of the homomorphism is isomorphic to $C_2$.

That's as far as I got, I classified the homomorphisms but I don't know how to count them.

There is no injective homomorphism from $S_5$ to $A_6$.
Its image would have index $3$ in $A_6$. There is no such subgroup
of $A_6$. If there were, there would be a homomorphism from $A_6$ onto
a group of order $3$ or $6$, and there isn't.

This answer is wrong, I left it undeleted to remind me of my fault, please notice this and don't be misguided.

Another reason for there is no injective homomorphism from S5 to A6 is that, if so S5 can be seen as a subgroup of A6 but every element in A6 is even permutation while S5 has odd permutation.


Convergence in probability with subsequences after hours of trying I wonder whether someone of you could may help me by proving this claim :-)

Assume $X_n$ is a sequence of random variables which converges in probability to $X$. Now, let $X_{nk}$ be a subsequence of $X_n$. Then, there is a subsequence $X_{\overline{nk}}$ of $X_{nk}$ such that
$$ P( |X_{\overline{nk}} - X | \geq \frac{1}{k}) \leq \frac{1}{k^2} \quad \forall k \geq 1$$
Unfortunately, I do not how to prove this statement... The proof given was "It holds if one waits long enough". But obviously, this is not very mathematically...

Thanks in advance!

The definition of convergence in probability implies that for any $\epsilon > 0$ and $\delta > 0$, we can always find a $N_{\epsilon}$ such for all $n \geq N_{\epsilon}$, we have $P(|X_n - X| > \epsilon) < \delta$. Essentially, we are looking at convergence of the sequence of probabilities $P(|X_n - X| > \epsilon)$ to zero.

Now, every subsequence of the convergent sequence also converges to the same limit. The sequence of $P(|X_{nk} - X| > \epsilon)$ also converges to zero - which implies that we can pick an index from the sequence $X_{nk}$ which satisfies $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k}) \leq \frac{1}{k^2} $ for each $k \geq 1$ and construct a separate sequence $X_{\overline{nk}}$ from these picked indices. Note that for $k_2 > k_1$, $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k_2}) \leq \frac{1}{k_2^2} $ implies that $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k_1}) \leq \frac{1}{k_1^2} $. Therefore, we can pick an increasing set of indices.

Naturally, a sequence  $X_{\overline{nk}}$ constructed as above satisfies the property discussed in the question.

Let $Y_k=X_{n_k}$. Then $Y_k \to X$ in probability. Hence $P(|Y_k-Y| \geq\frac 1 j) \to 0$ for each $j$. So there exists $k_j$ such that $P(|Y_{k_j}-Y| \geq \frac 1 j) < \frac  1{j^{2}}$. We can choose the $k_j$ to be increasing. Now $P(|X_{n_{k_j}} -X| \geq \frac  1j) \leq \frac 1 {j^{2}}$ for all $j$.


Matrix positive definite Let $A$ and $B$ be both symmetric $ n \times n$ matrices, and $B \succ 0$;
$U$ be one $n \times q$ column orthogonal matrix ($n > q$).
Assume $$ 0 \preceq U^{T} A U \preceq U^{T} B U,$$
do we have the following inequality $$UU^{T}AUU^{T} \preceq B.$$

Nope. Let
$$
A=\begin{bmatrix}5 & 3 \\ 3 & 2\end{bmatrix},
\quad
B=\begin{bmatrix}5 & 1 \\ 1 & 1\end{bmatrix}
$$
(both are SPD). Let $U=[1,0]^T$. Then
$$
0\leq 5=U^TAU\leq U^TBU=5.
$$
But
$$
B-UU^TAUU^T=\begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix},
$$
which is indefinite.

Yes.  Just premultiply by $U^T$ and postmultiply by $U$ after bringing $B$ over to the other side.  If $C \leq 0$ then $U^T C U\leq 0$ also.


Limit involving exponentials Being bored, I recently started trying to prove the exponential derivative formula by difference quotient:

$\dfrac{d}{dx}n^x=\lim\limits_{\Delta x \to 0}\dfrac{n^{x+\Delta x}-n^x}{\Delta x} = n^x\log n$

Simple algebraic manipulation (exponent rule and factoring) brought me from the difference quotient to:

$\lim\limits_{\Delta x \to 0}n^x\dfrac{n^{\Delta x}-1}{\Delta x}$

Limit of a product:

$\Bigg(\lim\limits_{\Delta x \to 0}n^x\Bigg)\Bigg(\lim\limits_{\Delta x \to 0}\dfrac{n^{\Delta x}-1}{\Delta x}\Bigg)$

And finally limit of a constant.

$n^x\Bigg(\lim\limits_{\Delta x \to 0}\dfrac{n^{\Delta x}-1}{\Delta x}\Bigg)$

This limit is where I got stuck, however. Clearly it equals $\log n$ by the well-known formula, but how can the limit be evaluated? Apologies if this is somewhat basic.

Write $n^{\Delta x}=e^{\Delta x\cdot \log n}$. Then by the definition of $\exp$ by its Taylor development:

$$\frac{e^{\Delta x\log n}-1}{\Delta x}=\frac{(\Delta x\log n)+\frac{(\Delta x\log n)^2}{2}+\cdots}{\Delta x}=\log n+\Delta x\cdot \left(\frac{(\log n)^2}{2}+\cdots\right)\overset{\Delta x\rightarrow 0}{\longrightarrow} \log n$$

Alternatively one can use L'Hôpital's rule. But that makes use of knowing the derivative of $e^x$.

We need to show that there exists some value $e$ such that
$$\displaystyle\lim_{h\to0}\frac{e^h-1}{h}=1$$
suppose
$$\displaystyle f(n)=\lim_{h\to0}\frac{n^h-1}{h}$$
then consider $f(1)$ vs. $f(100)$. Clearly $f(1)=0.$
I'm assuming that the exponential function is an increasing function, and if I can't make that assumption, then I'll define it as such (with $e\gt1$). I would like to find $n$ such that $dn^x/dx=1$ to that end I would like to evaluate (and yes, I'm also assuming that $n^x$ is a convex function)
$$\frac{n^{x}-n^{x-\Delta x}}{\Delta x}=n^x\frac{1-n^{-\Delta x}}{\Delta x}=n^x\frac{1}{n^{\Delta x}}\cdot\frac{n^{\Delta x}-1}{\Delta x}\lt n^xf(n)$$
Evaluating for $n=100$ and $\Delta x=1/2$ we have
$$\frac{1}{100^{1/2}}\cdot\frac{100^{1/2}-1}{1/2}=\frac{18}{10}\lt f(100)$$
So if $f$ is continuous then by the intermediate value theorem there exists some $e$ such that
$$0=f(1)\lt 1=f(e)\lt 1.8\lt f(100)$$
$$\therefore \lim_{\Delta x\to0}\frac{n^{\Delta x}-1}{\Delta x}=\lim_{\Delta x\to0}\frac{{(e^{\ln n})}^{\Delta x}-1}{\Delta x}=\lim_{\Delta x\to0}\frac{{e^{\ln n\cdot\Delta x}}-1}{\Delta x}$$
$$=\lim_{\Delta x\to0}\frac{{e^{\ln n\cdot\Delta x}}-1}{\ln n\cdot\Delta x}\cdot\ln n=\lim_{\ln n\cdot\Delta x\to0}\frac{{e^{\ln n\cdot\Delta x}}-1}{\ln n\cdot\Delta x}\cdot\ln n$$
$$=f(e)\cdot\ln n$$
$$=\ln n$$


A conjecture: every power of $16$ greater then $16^4$ has at least one digit $1, 2, 4$, or $8$ when written as base $10$? Is there a proof for why the following is true? Does every power of $16$ greater then $16^4$ have a digit $1, 2, 4$ or $8$ when you write it as base $10$?


Is there a set for which no defining formula can be found? I saw this question, but since (1) it's a little unclear (in fact, it has even been closed as unclear), (2) I'm not familiar enough with the terminology of the answer to judge wheter or not it answers my question and (3) it's old enough for me to assume the commenters won't respond to it, I thought it would be better to ask.

In the book I've been using to study set theory, the author enunciates the Intuitive Principle of Abstraction in the very first chapter:

  A formula $P(x)$ defines a set $A$ by the convention that the members of $A$ are exactly those objects $a$ such that $P(a)$ is a true statement.

Thinking a bit more deeply into it, I noticed the author said nothing of the converse. That is, nothing was said if the following statement is true:

  For every set $A$ there is a formula $P(x)$ which defines it.

So, here I ask you all: is that statement true? Are there sets for which we can find no formula?

Fact 1: Since a formula must be of finite length, and our alphabets are finite, the are only countably many formulas.

Fact 2: We usually assume (mostly implicitly) that there are uncountably many sets.

Assuming these two facts, we can conclude that the answer is yes, there must be sets without a defining formula. But we can't describe a single one of them.


Generalize this notation for an ODE? I have

$$
\frac{dN(t)}{dt}=a(t)N(t) \tag 1
$$
where $N(0)=N_0 $ (constant).

Question 1:
From the information in $(1)$ I assume I have the functions $a, N:\mathbb R\rightarrow \mathbb R$? And $N_0\in \mathbb R$? Is that correct?
Suppose I now want to generalize the notation in $(1)$ by replacing the right hand side with a function $f$.
Question 2:
Does this mean I should write
$$
\frac{d N(t)}{dt}=f(t) \tag 2
$$
where $N, f:\mathbb R\rightarrow \mathbb R$ ?
Or maybe
$$
\frac{dN(t)}{dt}=f(a(t),N(t)) \tag 3
$$
where $f:\mathbb R \times \mathbb R \rightarrow \mathbb R$?
Or maybe
$$
\frac{dN(t)}{dt}=f(t,a(t),N(t)) \tag 4
$$
where $f:\mathbb R \times \mathbb R \times \mathbb R \rightarrow \mathbb R$ and $t \in \mathbb R$?
$\times$ is the Cartesian product.
Which one is correct? Thanks!

An explicit first order ODE has the following form (see wiki)
$$
\frac{dN(t)}{dt} = f(t, N(t)).
$$
The form (2) is too narrow, since does not allow dependence on $N(t)$ in the right hand side. Form (3) does not allow dependence on raw $t$, and the form (4) has extra $a(t)$ which may be easily built in the definition of $f(t, N)$ itself.
Few examples:

$\frac{dN(t)}{dt} = g(t)$ is a special case of $f(t, N) = g(t)$
$\frac{dN(t)}{dt} = a(t) N(t)$ is a special case of $f(t, N) = a(t) N$
$\frac{dN(t)}{dt} = \frac{a(t)}{N(t) + t}$ is a special case of $f(t, N) = \frac{a(t)}{N + t}$


Doubt in finding number of non-prime factors of an integer The question is:

  Find the number of non-prime factors
  of $4^{10} \times 7^3 \times 5^9$.

I represented the number as $2^{20} \times 7^3 \times 5^9$ then the number of factors of this integer is $21 \times 4 \times 10 = 840$ now I can only see that there could be only three prime factors here $2,7 \text{ and } 5$. So the number of non-prime factors should be $837$ but my module says the answer would be $437$, what exactly I am missing here?

You are missing making the error that your module made.  You are right, the module is wrong.

One could speculate about why the person hired to solve the problems made the mistake.
But note that
$$11 \times 4 \times 10=440$$
so the person doing the solutions may not have noticed that $4$ is not prime. You did notice.

If the question is "Find the number of non-prime factors of $~2^10 * 7^3 * 5^9~$"?

The answer will be $~440 - 3 =437~$


Constant function formal definition A generic function $F : X \rightarrow Y$ can be defined as $\forall x \in X \exists !y \in Y (x, y) \in F$. How about the formal definition of a constant function? The formula $\exists !y \in Y \forall x \in X (x, y) \in F$ does not seem right to me, since it does not prevent that the same element in the set $X$ is mapped to multiple different elements in the set $Y$. Instead, it only prevents that two different constant lines can be mapped from the same set $X$.

How about just combining the two statements together?
$$(\forall x\in X\ \exists!y\in Y\ (x,y)\in F)\land(\exists y\in Y\ \forall x\in X\ (x,y)\in F)$$
Note that uniqueness of the $\exists$ in ths second statement is not required: suppose two such $y$'s existed, $y_1$ and $y_2$. Then for any element $x\in X$, $(x,y_1)$ and $(x,y_2)$ are in $F$, which is not allowed by the first statement. Thus only one such $y$ may exist.

Maybe $\exists! y \in Y : ((x,y) \in F , \forall x \in X) \land ((x,y') \notin F , \forall y' \ne y, \forall x \in X)$?


Algebraic condition for a twist in 2D ball or a hypersphere? We define twisted ball here as a ball where only one point (the twist) separates the ball sides.
Simple implicit presentation for the 2D Ball is $x^2+y^2=r^2$. I am trying to find a general condition when the 2D ball has a crossing like the last parametric plot. Third one is just a line so it can also be considered as a naive twisted ball: is the twisted ball just a line with certain features? The extra 2 in the fourth plot results into a more apparent twisted ball: is there some other feature we should require for a twisted ball?

Questions

How to deduce the algebraic condition for the 2D twisted ball?

Easier to consider higher dimensional balls also, hyperspheres?

Please explain the twist in the ball.

Let us consider firstly the set of parametric plots in which the coordinates $x,y$ are given by $ [\sin(t), \cos(kt)]$. I will restrict my answer to the cases where $k$ is an integer.

In this set of parametric plots, the function crosses the $x$-axis for $2k$ different values of $t$, which are  those satisfying  $t=\pi(j-1/2)/k$, where $j=1,2,3...2k$. However, because of the symmetry of these $t$ values with respect to the $y$-axis, these zeros can be grouped in pairs with the same $x$-value, i.e. corresponding to the same point on the $x$-axis. As a result, the function crosses the $x$-axis in $k$ points when $k$ is even, and $k+1$ points when $k$ is odd.

For example, when $k=4$, the function crosses the $x$-axis for eight values of $t$, namely $\frac{\pi}{8}, \frac{3}{8}\pi, \frac{5}{8}\pi... \frac{15}{8}\pi$. However, these eight values correspond to only four points on the $x$-axis, since the $x$ value is equal for the pair $t=\frac{\pi}{8}$ and $t=\frac{7}{8}\pi$, for the pair $t=\frac{3}{8}\pi$ and $t=\frac{5}{8}\pi$, and so on. Accordingly, the parametric plot $ [\sin(t), \cos(4t)]$ is as follows:

On the other hand, for example, when $k=3$, the function crosses the $x$-axis for six values of $t$, namely $\frac{\pi}{6}, \frac{3}{6}\pi, \frac{5}{6}\pi... \frac{11}{6}\pi$. These six values correspond to only four points on the $x$-axis, since the $x$ value is equal for the pair $t=\frac{\pi}{6}$ and $t=\frac{5}{6}\pi$, and for the pair $t=\frac{7}{6}\pi$ and $t=\frac{11}{6}\pi$; the other two points are given by $t=\frac{3}{6}\pi$ and $t=\frac{9}{6}\pi$.  Accordingly, the parametric plot $ [\sin(t), \cos(3t)]$ is as follows:

Note then that, when $k$ is even, all $2k$ zeros of the function can be grouped in pairs, leading to $k$ points on the $x$-axis; conversely, when $k$ is odd, all $2k$ zeros of the function can be grouped in pairs, except two that are "unpaired" (those corresponding to $t=\frac{\pi}{2}$ and $t=\frac{3}{2}\pi$). This leads to $k+1$ points on the $x$-axis.  Also note other two things: 1) for $k$ odd, the two "unpaired" zeros given by $t=\frac{\pi}{2}$ and $t=\frac{3}{2}\pi$ correspond to the most lateral points (left and right) of the function; 2) for $k=1$, the function crosses the $x$-axis only in these two unpaired zeros, thus without crossing itself (in fact for $k=1$ it reduces to a circle).

Now it is also not difficult to show that, considering the behaviour of the function in these points laying on the $x$-axis, its derivative has a unique value when $k$ is even - which means that the function does not cross itself. On the other hand, when $k$ is odd, the derivative assumes two different values (positive and negative) in each of these points laying on the $x$-axis except for the two "unpaired", most lateral ones. This means that the function crosses itself in $k-1$ points on the $x$-axis. For example, we expect that for $k=14$ the parametric plot $ [\sin(t), \cos(kt)]$ crosses the $x$-axis in $14$ points and does not cross itself. This is demonstrated by the following figure showing $ [\sin(t), \cos(14t)]$:

On the other hand, for example we expect that for $k=9$ the parametric plot $ [\sin(t), \cos(kt)]$ crosses the $x$-axis in $9+1=10$ points, and that it crosses itself in $9-1=8$ of these points. This is demonstrated by the following plot of $ [\sin(t), \cos(9t)]$, showing the $8$ points where the function crosses itself and the remaining $2$ most lateral points ("unpaired" zeros) where the function crosses the $x$-axis:

As a result, within the set of parametric plots $x,y$ given by $ [\sin(t), \cos(kt)]$, there is no possibility to find a case that yields a unique crossing point dividing the function in two sides (left and right). For $k$ even, the function does not cross itself; among the odd $k$ values, for $k=1$ the function does not cross itself; for $k=3$, the function crosses the x-axis in four points (two of which are the lateral, "unpaired" ones), and therefore crosses itself in two points on the $x$-axis; lastly, as already shown, for higher odd values of $k$, the function crosses itself on the $x$-axis even more times.

All these considerations can be repeated for the set of parametric plots in which the coordinates $x,y$ are given by $ [\sin(t), \sin(kt)]$, with the only difference that the behaviour of the function for $k$ odd or even is inverted. More precisely, in this case the function crosses the $x$-axis $k$ times and never crosses itself when $k$ is odd; conversely, the function crosses the $x$-axis in $k+1$ points and crosses itself in $k-1$ of these points when $k$ is even.

As a result, within the set of parametric plots $x,y$ given by $ [\sin(t), \sin(kt)]$, the only case in which there is a unique crossing point dividing the function in two sides (left and right) occurs for $k=2$. In fact, in this case, the function crosses the $x$-axis in three points (two of which are the"unpaired" lateral ones) and crosses itself in a single central point, yielding a "twisted" ball according to the definition given in the OP. Accordingly, the plot of $ [\sin(t), \sin(2t)]$ is as follows:

I have some sort of intuitive touch on these twists based on Quantum physics - they like trigger sudden peaks which then again change the architecture of the object through Gibbs' phenomenon.
So the algebraic condition would be any sudden peak in a 2D twisted ball.


Number of ways to form three distinctive items Given a 6 by 5 array, Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column.

What I did was $C(30,1) \cdot C(20,1) \cdot C(12,1)$ however this is not the answer. They get 1200. How?

$1^{st}$ item: you will have $6\times5=30$ choices.

$2^{nd}$ item: you take out the row and column containing the $1^{st}$ chosen item, so you are left with $5\times4=20$ choices.

$3^{nd}$ item: you take out the row and column containing the $2^{nd}$ chosen item, so you are left with $4\times3=12$ choices.

However, note that the order of items doesn't matter (i.e choosing $ABC$ is the same as choosing $CBA$). Hence the desired answer is $(30\times20\times12)\div3!=1200$

Debes darle importancia a una de las dos, o la fila o la columna. Cuando quieras darle la importancia a una o a la otra, una va con permutación 3 y la otra con combinación 3, para hallar una especie de coordenada. Quedaría 5C3 * 6P3 o lo mismo 6C3 * 5P3. Esto teniendo en cuenta que las 3 respuestas quedarían en linea distinta y 2 de ellas no se encontrarían, porque una esta en la mitad. Con la permutación ya se eliminarían una de las lineas y también se eliminaría el problema de que se encuentren.


h Numbering $8$ vertices of cube from $1$ to $8$ such that there is no 2 consecutive number adjacent How many ways are there to number $8$ vertices of a cube using numbers
from $1-8$ such that there are no two consecutive numbers adjacent on
the cube ($1$ and $8$ are considered to be consecutive).

I know that the answer is $480$ (with symmetries/etc), however, I am looking for a solution. I see that there are $\frac{8!}{8\times 3}=1680$ ways (without symmetry/etc) to number the cube. I can't use graph theory for the problem. It was originally a probability problem.

Without symmetries, there are only ten solutions.  If the cube is $\{0,1\}^3$, put 1 at $(0,0,0)$ and the numbers at $(1,0,0),(0,1,0)$ and $(0,0,1)$ in increasing order.  Don't forget reflection symmetry.
There is another symmetry by reversing the order of the digits, from $1,2,...8$ to $1,8,7,...,2$.  You might assume the number at $(1,1,1)$ is less than 6.  The number at $(1,1,1)$ can't be 3 as there would be nowhere to put 2.
Now plough through the options.  With the extra symmetry of the second paragraph, there are between five and ten solutions to find.
If the number at $(1,1,1)$ is 2, then the neighbours of $1$ are $3$ and two out of $4,5,6,7$.  If the number at $(1,1,1)$ is 4, then the neighbours of $1$ are $3,5$ and another one.  If the number at $(1,1,1)$ is 5, then the neighbours of $1$ are $4,6$ and either 3 or 7.  There aren't that many possibilities to check.


Why do we start losing algebraic properties when dealing with hypercomplex numbers? Every form of hypercomplex number I have seen (including the complex numbers) lose some important algebraic property. Why is that? Is there a pattern to what we lose?

The complex numbers $\mathbb{C}$ do not lose any properties that $\mathbb{R}$ has. They are both fields and hence share all the properties of fields.

Now the simple answer to your question is because they do! Unlike real numbers (and maybe arguably complex numbers) is because we create them (not that you don't create the others but you can argue the real numbers rather boldly present themselves in nature the way other 'number' structures don't). The way that we create the quaternions, $\mathbb{H}$, is such that they lose commutativity of multiplication. It comes directly from how we define multiplication of quaternions.

Similarly, how we define the octonions means that they lose both associativity and commutativity of multiplication. For both groups of hypercomplex numbers, we form an operation called multiplication in terms of real numbers and check to see that they satisfy the properties we normally have when we work with real numbers. The way we have to define the operations to construct these hypercomplex numbers for $\mathbb{H}$ and $\mathbb{O}$ causes these properties to fail.

The more complex (and more realistic reason) that they lose the properties 'numbers' normally have is that we are trading the structures they lose to get something more. That is, we want to generalize $\mathbb{R}$ to bigger and bigger sets of 'numbers' that have certain properties. We call these bigger structures algebras (because they resemble the algebra we are used to). For example, we also want to be able to tell the size of these new 'numbers' (we call this the norm of a number). In order to generalize to bigger and bigger 'numbers' with these properties, essentially something has to give and what often is lost is commutativity and associativity of multiplication.

Of course, this is only the general idea. The mechanics of where and why things break down is quite complicated in the general case (involving Lie Algebras, Clifford Algebras, and general matrix groups). However, this gives a general insight into the question. In fact, there is still much work needed to be done to understand in full the structures for these objects and why they are the way that they are.


If $a \mid c$ and $b \mid c$, but $\gcd(a,b) = 1$, then $ab \mid c$. If $a | c$ and $b | c$ and $a$ and $b$ are relatively prime prove that
  $ab|c$.

What I did was since $(a,b)=1$ then we can find integers $m,n$ such that $ma + nb=1$. Now since $a|c$ then $a = mc$. Similary, since $b|c$ then $b=nc$. Then I got that:

$$ma+nb=1$$
$$c(m+n)=1$$
$$m +n = 1/c$$
$$abm +abn=\frac{ab}{c}.$$

Is this correct?

The key ideas are present in your proof.
We know that $a \mid c$ and $b \mid c$. As $a \mid c$, we have that $c = ak$ for some integer $k$. In this form, we see that $b \mid c = ak$, but $\gcd(a,b) = 1$. By the theorem below, we have that $b \mid k$, so that $k = b \ell$. Now $c = ak = ab\ell$, and now it's clear that $ab \mid c$.

Theorem
If $a \mid bc$ and $\gcd(a,b) = 1$, then $a \mid c$.

This can be proved from your application of Bezout's Theorem in your attempted solution. And in fact this is the content of this question and its answers.

If you want to avoid prime factorization, try saying $c = kab + r$ with $0 \leq r < ab$. Then, because $a | c$ and $b|c$, we have that $a | r$ and $b | r$. Since $(a,b) = 1$, what can you say about $r$?


Is $f(-x) = 1/f(x)$ under these conditions? Let $f\colon \mathbb{R}\to \mathbb{R}$, so that:

for $x\in \mathbb{R}$, $f(x)>0$
for $x,y \in \mathbb{R}$, $f(x+y) = f(x)f(y)$

Prove that $f(-x) = 1/f(x)$.

Using your second condition, you get

$$f(0)=f(x+(-x))=f(x)f(-x)$$
Dividing both sides by $f(x)$
$$\frac{f(0)}{f(x)}=f(-x)$$

Now, what is $f(0)$? Let's use that condition again:

$$f(x)=f(x+0)=f(x)f(0)$$
hence, dividing both sides by $f(x)$, it's evident that $f(0)=1$.

Substituting above, we get

$$\frac{1}{f(x)}=f(-x)$$

If we want to solve this problem in the straightforward way, start by putting $y = -x$, then

$f(0) = f(x)f(-x) => f(-x) = f(0) / f(x)$

To find out what f(0) would be, we can set $x = y =  0$ in the original equation and see, $f(0) = f(0)^{2} => f(0) = 0, 1$

So you can see, the only condition we need is, $f(0) \neq 0$

Now, we set y = 0 on the original equation, hence,
$f(x + 0) = f(x)f(0) => f(x) = f(x) f(0) => f(x)(1 - f(0)) = 0$

Now, according to condition 1, f(x) can't be zero. Hence, $f(0) = 1$

Now we can conclude, $f(-x) = f(0) / f(x) = 1 / f(x)$

(Q.E.D)

N.B.:
As an addition, if the function is instead defined as $f:R -> R^{+}$, which can be reduced to  Cauchy's Functional Equation.

From this pdf

  All continuous functions $f : R → (0,+∞)$ satisfying $f(x +y) = f(x)f(y)$ are of the form
  $f(x) = a^{x}$ . Namely the function $g(x) = log(f(x))$ is continuous and satisfies the Cauchy equation.

Now, if $g(x) = log(f(x))$, then,

$g(x+y) = log(f(x+y))$

Starting with your equation, taking log at both sides,

$log(f(x+y)) = log(f(x)) + log(f(y)) => g(x + y) = g(x) + g(y)$

Which is just Cauchy's Functional Equation.


Evaluate the following: $(1-i)^{1+i}$ My progression:

$(1-i)^{i+1} = e^{(i+1) * \ln(1-i)}$. I get stuck after this point.

The expression $z^w$ where $z, w \in \mathbb{C}$ is not uniquely determined. In fact, we define
$$z^w = e^{w\log z}$$
where $\log z$ is any logarithm of $z$. There are infinitely many choices of $\log z$, and for most values of $z$ and $w$ there will be infinitely many possible values for $z^w$.

To get something unique, you will have to specify a particular branch of the complex logarithm, but when you do so. $z^w$ won't be defined for all $z$ (or at the very least $z^w$ won't be continuous in $z$, depending on what your conventions with branches are).

In your particular case $\log(1-i) = \ln \sqrt 2 - \dfrac{i\pi}4 + 2\pi i k$ for some arbitrary integer $k$, and

\begin{align}
(1-i)^{1+i} &= e^{(1+i)\log(1-i)} \\
&= e^{(1+i)(\ln \sqrt 2 - \frac{i\pi}4 + 2\pi i k)} \\
&= e^{ \ln \sqrt 2 + \frac\pi4-2\pi k + i(\ln\sqrt 2 - i\frac{\pi}4 + 2\pi k)} \\
&= \sqrt 2 e^{\frac\pi4-2\pi k\pi}\cdot e^{i(\ln\sqrt2-\frac\pi4)} \\
&= \sqrt 2 e^{\frac\pi4-2\pi k\pi}\cdot \big( \cos (\ln\sqrt2-\frac\pi4) + i \sin(\ln\sqrt2-\frac\pi4) \big) \\
\end{align}


determine whether the vector is in the image of A I am having a really hard time trying to understand what "the image of A" means and how to start such a question.
I was thinking of creating a matrix with b1 as the augumented part of the matrix.

The image of a function $f: X \rightarrow Y$ is the set of elements of $Y$ that are the result of applying $f$ to some $x$ in $X$.

Any $3\times 3$ matrix $A$ gives rise to a linear transformation $T_A$ given by $T_A(\vec{x})=A \vec{x}$.

In other words, a vector $\vec{b}$ is in the image of $A$ if there exists a vector $\vec{x}$ such that $A\vec{x} = \vec{b}$.

Hint: What has this got to do with solving a system of linear equations?

Turn A into the upper echelon matrix U...the columns of A that correspond to the columns of U with the pivots of U are the vectors of the basis for the image of A


The ambigous definition of vacuous truth It is no doubt that the vacuous truth is related to material implication  "$P\Rightarrow Q$". We say the material implication statement is true when $P$ is false. However, seems that this is not the definition for vacuously truth. Do we call it vacuously true only when $P$ can never be true? (Seems that all examples are in this way).

More clearly, suppose we have a statement "$P(x)\Rightarrow Q(x)$", and at some domains of $x$ (we may denote the domain by $T_x$), $P(x)$ is true;  at some other domains of $x$ (denote the domain by $F_x$), $P(x)$ is false. Can we say the statement is vacuously true in the domain $F_x$? Is there any real example?

To make the problem more clear, we may check the statement "For all $x$, $P(x)\Rightarrow Q(x)$". Can we say this statement is vacuously true in the domain $F_x$ ($F_x$ is defined as above)?

I would frame the concept as pertaining to first order sentences of the form $$\forall x(P(x)\to Q(x)).$$ Vacous truth is the particular situation when there are no $x$ in the domain satisfying $P(x),$ in which case standard semantics says the statement is true. So yes, whether a first order sentence is vacuously true or not depends on the interpretation, and hence the domain, much as regular truth does. (However this doesn’t usually matter: when we’re talking about math, we generally are making statements relative to some fixed interpretation.)

Do we call it vacuously true only when P can never be true?

No. In classical logic, for any logical propositions P and Q (it doesn't matter if they are true or false), we have:

$P\implies [\neg P \implies Q]$

This is a tautology. See truth table here.


Find a polynomial of the specified degree that satisfies the given conditions. Find a polynomial of the specified degree that satisfies the given
  conditions.
  Degree $4$; zeros $-1$, $0$, $3$, $1/3$ ; coefficient of $x^3$ is $7$

My answer is...
$$
P(x)=3x^4 - 7x^3 - 7x^2 + 3x.
$$
When I entered this answer into the software (MindTap) for my class it was marked incorrect. mymathportal.com and wolframalpha.com both agree with my answer.

Is there another format that could be also correct or something I am missing?

Thanks for your time.

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  Since
  $\ds{\braces{\color{#f00}{-1},\color{#f00}{0},\color{#f00}{3},\color{#f00}{{1 \over 3}}}}$ are the polynomial roots, the polynomial we are looking for is proportional to

\begin{align}
\bracks{x - \color{#f00}{\pars{-1}}}\pars{x - \color{#f00}{0}}
\pars{x - \color{#f00}{3}}\pars{x - \color{#f00}{{1 \over 3}}} =
x^{4} - {7 \over 3}\,x^{3} - {7 \over 3}\,x^{2} + x
\end{align}
In order to 'adjust' the $\ds{x^{3}\!}$-coefficient, multiply it by
$\ds{-3 \implies \bbox[8px,#efe,border:0.1em groove navy]{-3x^{4} + \color{#f00}{7}x^{3} + 7x^{2} - 3x}}$

The answer was found in the comments to be
$$P(x)=-3x^4+7x^3+7x^2-3x$$


Definition of similarity mapping between ordered sets: why is a " strictly precedes" relation required on each side of the biconditionnal? As a definition of " similarity mapping" I read in Lipschutz, Set Theory :

the mapping f from A to B ( A and B being ordered sets) is a similarity mapping iff, (a) f is a bijection and (b) for any elements a and a' belonging to A :

                   a < a'  iff f(a) < f(a')

The definition is expressed in terms of < : "strictly precedes".

Would f also be a similarity mapping in case < were replaced by " precedes or is equal to" ?

What I do not understand is that (1) the author has defined ordered sets in terms of " precedes or is equal to " and (2) defines an " order preserving function" in terms of " strictly precedes".

Yes, the two notations are equivalent because $f$ is also required to be a bijection. This condition ensures that $f(a) = f(a') \iff a = a'$.

It follows that the strict and non-strict variants are equivalent.

As to the use of "strictly": one could argue it would have been better for the author to phrase the definition of "ordered set" in terms of "strictly precedes or is equal to" (which we would then abbreviate as "precedes").

The given condition implies $a\le a'\implies f(a) \le f(a')$, but not the other way around, as any constant function is a counterexample.


Is the determinant of a covariance matrix always zero? My understanding is that given matrix X, I can find its corresponding covariance matrix by:

finding the means of each column
subtracting each mean from each value in its respective column and
multiplying the resulting matrix by its own transpose. Let's call this matrix C.

Here is what it would look like in Python:

Y = X - numpy.mean(X, axis = 0)
C = numpy.dot(Y, Y.T)

If I do this, I can prove mathematically (and experimentally using some simple Python code) that det(C) = 0 always.

However, a colleague tells me that using the inverse of a covariance matrix is common in his field and he showed me some R code to demonstrate.

> det(cov(swiss))
[1] 244394171542

I notice that R has several ways of calculating the covariance matrix that leads to different results. I also notice from Googling that some people say the covariance matrix is always singular (eg here) whereas others say it is not.

So, my question is: why the differences of opinion and what's the true answer?

EDIT: I discovered that the determinant is only zero if the matrix is square. If anybody knows the proof for this or can throw some further light on the matter, I'd be grateful.

No, the covariance matrix is not always singular. Counterexample:
\begin{align}
X&=(\mathbf x_1,\mathbf x_2,\mathbf x_3)=\pmatrix{1&2&3\\ 1&1&4},\\
\bar{\mathbf x}&=\frac13(\mathbf x_1+\mathbf x_2+\mathbf x_3)=\pmatrix{2\\ 2},\\
Y&=(\mathbf x_1-\bar{\mathbf x},\ \mathbf x_2-\bar{\mathbf x},\ \mathbf x_3-\bar{\mathbf x})=\pmatrix{-1&0&1\\ -1&-1&2},\\
YY^T&=\pmatrix{2&3\\ 3&6},\text{ which is nonsingular}.
\end{align}
It is true, however, that when the data matrix $X$ is square or "tall", i.e. when $X$ is $n\times m$ with $n\ge m$ (or equivalently, when the number of data points $m$ does not exceed the number of variables $n$), $YY^T$ is always singular. This is evident because by definition, the sum of all columns of $Y$ is the zero vector. Hence $Y$ has deficient column rank and in turn, $\operatorname{rank}(YY^T)=\operatorname{rank}(Y)<m\le n$.

The result of matrix multiplication is only an approximation of the true covariance matrix and disregards distributions of random variables that are crucial for accurate calcualtion. Hope it helps.


Solve the following congruence: 2x ≡ 7 (mod 17) Question
Solve the congruence 2x ≡ 7 (mod 17).
I have tried working out this problem but I am stuck midway. Could someone help me out by showing me or explaining how to proceed further?
Here is the work that I have so far:
Inverse of a modulo m is an integer b for which ab ≡ 1 (mod m),
a = 2
m = 17
17 = 8 ⋅ 2 + 1
2 = 2 ⋅ 1 + 0
The greatest common divisor is the last non-zero remainder values, that is, gcd (a, m) = 1.
Expressing the greatest commmon divisor as a multiple of a and m,
gcd (a, m) = 1
= 17 - 8 ⋅ 2
= 1 ⋅ 17 - 8 ⋅ 2
The inverse would then be the coefficient of a = 2, which in this case, would be -8.
And, since, -8 mod 17 = 9 , 9 is also the inverse of a modulo m.
Solving the congruence 2x ≡ 7 (mod 17) by multiplying each side by the inverse 9,
9⋅2x  ≡ 9⋅7 (mod 17)
18x ≡ 63 (mod 17)
And, this is the part where I am stuck. Could anyone help out? Thank you.

you are basically done
$18x \equiv 63 \pmod{17}$ and $18x \equiv x \pmod {17}$ and $63\equiv 12 \pmod {17}$
so $x \equiv 12 \pmod{17}$.


Check of step in proof that open subsets of the real numbers can be expressed as unions of disjoint open intervals. The following is from "An Introduction to Lebesgue Integration and Fourier Series" by Howard J. Wilcox and David L. Myers:

7.2 Theorem: Every non-empty open set $G \subset \mathbb{R}$ can be expressed uniquely as a finite or countably infinite union of pairwise disjoint open intervals.

Proof: Suppose first that $G$ is bounded. Since $G$ is open, for each $x \in G$ there is an open subinterval of $G$ containing $x$. Let $b_{x} = \mathrm{lub} \{y \mid (x,y) \subset G\}$, and $a_{x} = \mathrm{glb} \{z \mid (z,x) \subset G\}$. Let $I_{x} = (a_{x}, b_{x})$, called the component of $x$ in $G$. Clearly $x \in I_{x}$.

Now $I_{x} \subset G$, for if $w \in I_{x}$, say $x < w < b_{x}$, then by definition of $b_{x}$, there is a number $y$ such that $w < y$ and $(x,y) \subset G$. Hence $w \in G$. The case where $a_{x} < w < x$ is handled similarly. (What about $w = x$?)

Also $a_{x} \notin G$ and $b_{x} \notin G$ (see exercise 9.10).

I am attempting Exercise 9.10: Prove that $a_{x} \notin G$, in the proof of Theorem 7.2.

This is my attempt:

Suppose for contradiction that $a_{x} \in G$. Since $G$ is open, then there exists an open subinterval $(\alpha, \beta) \subset G$ containing $a_{x}$. Since $(a_{x}, x) \subset I_{x}$, and $I_{x} \subset G$, then $(a_{x}, x) \subset G$. Then since $(\alpha, a_{x}) \subset G$, and $a_{x} \in G$ by assumption, then $(\alpha, x) \subset G$. Then $\alpha \in \{z \mid (z,x) \subset G\}$. Then since $\alpha < a_{x}$, and $a_{x}$ is a lower bound of $\{z \mid (z,x) \subset G\}$, this is a contradiction.

Is this correct? Is there a simpler or more elegant proof?

FWIW your proof seems fine to me. It's straightforward and it efficiently uses what is necessary for the proof without anything extraneous. Personally I can't think of a significantly better approach.

A simple proof.

Notation. For $x,y\in \Bbb R$ let $In[x,y]=[x,y]\cup [y,x]=[\min(x,y),\max (x,y)]$.

Let $G$ be an open subset of $\Bbb R.$ For $x,y\in G$ let $x\sim y$ iff $In[x,y]\subset G.$

Obviously $\sim$ is symmetric and reflexive on $G.$ Exercise: Show that $\sim$ is transitive on G. Hint: $In[x,y]\cup In[y,z]=In[\min(x,y,z),\max(x,y,z)]$ for all $x,y,z\in \Bbb R.$

So $\sim$ is an equivalence relation on $G.$ For $x\in G$ let $[x]_{\sim}=\{y\in G:y\sim x\}.$ Exercise: $[x]_{\sim}$ is convex for each $x\in G.$

Now the set $G_{/\sim}=\{[x]_{\sim}:x\in G\}$ of $\sim$-equivalence classes is a partition of $G.$  And each $[x]_{\sim}$  is open, because if $x\in G$ then $(-r+x,r+x)\subset G$ for some $r>0,$ so $y\sim x$ for all $y \in (-r+x,r+x).$

Therefore $G_{/\sim}$ is  a partition of $G$ into a family of pair-wise disjoint non-empty convex open sets.

$Any$ family $F$ of pair-wise disjoint non-empty open subsets of $\Bbb R$ is countable because $\Bbb R$ has a countable dense subset. For example, for each $f\in F$ let $\psi(f)\in f\cap \Bbb Q.$ Then $\psi:F\to \Bbb Q $ is injective, so $F$ is countable.

Therefore $G_{/\sim}$ is countable.

Remark. The notation $In[x,y]$ is ad hoc. I just found it convenient, that you don't need to distinguish the cases $x<y,x=y,$ or $x>y $.


Goldbach Conjecture Consequences I have been looking into the Goldbach Conjecture pretty recently and I have often heard that it would have far-reaching consequences. However, I haven't found many of the actual consequences.  I was wondering if you all could supply me with some of these consequences (theorems, etc.).

A proof of the Riemann Hypothesis will have security implications if that proof can be used to determine the distribution of primes. It is often claimed that the proof for the RH will verify the Goldbach Conjecture but this has not been proven.

I assume that this is where the idea that GC will have security implications comes from, since it is often presented with a confused relationship with the RH, but this doesn't necessarily mean that a proof of the GC will have any real consequences for mathematics.

Much like the GC, proving the RH will not have any major impact on mathematics aside from the phrase "assuming the Riemann Hypothesis" being removed from academic literature.


Two minus signs make a plus Why do two minus signs make a plus sign and is there a corresponding rule for division and multiplication signs?

Multiplication is a scaling action, on the number line. When you multiply by a negative number, you also reflect the scaling. Reflecting again - multiplying by another negative number - bring ths scaling back to the positive direction.

Note that you can always start the process with $1$ - so for example $-3 \times -4 = 1 \times -3 \times -4$.

There is a kind of corresponding process for division. If you divide $1$ by $4$, you get $\frac{1}{4}$, then if you divide $1$ by $\frac{1}{4}$, you get $4$ again. Not exactly the same but a similar concept of double-action reverting to the initial state.

One solution is to use complex numbers, using $z=e^{i\theta}$. $\theta$ in this representation represents the angle through which we rotate a unit vector anti-clockwise about the origin.

As the total internal rotation angle of a circle is $2\pi$, we arrive at $e^{i\pi}=-1$.

Multiplying two complex numbers $u\times v$, where $u=e^{i\alpha}$ and $v=e^{i\beta}$, gives $e^{i\alpha}e^{i\beta}=e^{i(\alpha+\beta)}$.

Therefore $-1\times-1=e^{i\pi}\times e^{i\pi}=e^{2i\pi}=e^0=1$.


Number having sexagesimal expansion end with infinitely many zeros? I am looking for all the real numbers whose sexagesimal expansion (base $60$) ends in infinite tail of zeros. Does they really exist?

It seems absurd to me or mm thinking it in a wrong manner?

A number $N$ has a finite sexagesimal expansion if it can be written as
$$
N=a_0+\frac{a_1}{60}+\frac{a_2}{60^2}+\dots+\frac{a_n}{60^n}
$$
with $a_0$ any integer and $0\le a_i<60$ for $i=1,2,\dots,n$.

Then we can write
$$
N=\frac{m}{60^n}
$$
The converse is obviously also true: take $a_0=\lfloor N\rfloor$ and write
$$
N-a_0=\frac{m'}{60^n}
$$
which is true for a unique $m'$ with $0\le b<60^n$. Then write the base $60$ expansion of $b$ and you're done.

You can notice that base $60$ has nothing special. A number $N$ has a finite base $b$ expansion if and only if it has the form
$$
N=\frac{m}{b^n}
$$
for some integer $m$ and nonnegative integer $n$.


Continuity and Differentiability and Analicity Relations between $f(z)$ and $\overline{f(\bar{z})}$ I know that if $f(z)$ is analytic in $D$ then $\overline{f(\bar{z})}$ is analytic in $\overline{D}$.

If $f(z)$ is continuous at $z_{0}$ then $\overline{f(z)}$ is continuos.

Is it true that  : $f(z)$ is continuous at $z_{0} $$ \Leftrightarrow$ $\overline{f(\bar{z})}$ is continuos ?

and $f(z)$ is differentiable at $z_{0}$ $\Leftrightarrow$ $\overline{f(\bar{z})}$ is differentiable

This question was in the Final Exam :

Prove that if $f(z)$ is continuous at a point $z_{0}$ ,then $g(z)=\overline{f(\bar{z})}$ is also continuous .Is it rue that differentiability of $f$ at $z_{0}$ implies differentiability of g ?

Thank You ...

If $f(z)$ is continuous at $z_0$, $\overline{f(\overline z)}$ is not necessarily continuous at $z_0$ :

Let $z_0 = ib \in \Bbb C, b \in \Bbb R \setminus \{0\}$. Consider $f(z) = \frac{1}{z+ib}$ if $z \neq -ib$ and $f(-ib) = \omega \in \Bbb C$. $f(z)$ is continuous at $z = ib = z_0$ but is not continuous at $z = -ib = \overline {z_0}$ because $\lim \limits_{z \rightarrow -ib} \lvert f(z) \rvert = \lim \limits_{z \rightarrow -ib} \lvert\frac{1}{z + ib}\rvert = + \infty$.  Similarly, $\overline{f(z)}$ is not continuous at $\overline {z_0}$ because $\lim \limits_{z \rightarrow -ib} \lvert f(z) \rvert = \lim \limits_{z \rightarrow -ib} \lvert \overline{f(z)} \rvert = + \infty$, i.e. $\overline{f(\overline z)}$ is not continuous at $z_0$ even though $f(z)$ is.

If $f(z)$ is complex-differentiable at $z_0$, then $\overline{f(z)}$ is usually not complex-differentiable at $z_0$ ($\iff$ $\overline{f(\overline z)}$ is usually not complex-differentiable at $\overline{z_0}$). A simple example is $f(z) = z$. It is differentiable but $\overline{f(z)} = \overline z$ is not (look at the Cauchy-Riemann equations to see it).

Obviously, $f(z)$ is continuous at $z_0$ $\iff$ $\overline{f(z)}$ is continuous at $z_0$ ($\iff$ $\overline{f(\overline z)}$ is continuous at $\overline{z_0}$). Indeed, if you write $f(z) = f(x + iy) = u(x, y) + iv(x, y)$, if $f$ is continuous, both $u$ and $v$ are continuous so the function $\overline{f(z)} = u(x, y) - iv(x, y)$ is also continuous and conversely.

Finally, $f(z)$ is complex-differentiable at $z_0$ and on an open neighborhood of $z_0$ $\iff$ $f(z)$ is analytic on a neighborhood of $z_0$ $\iff$ $\overline{f(\overline z)}$ is analytic on a neighborhood of $z_0$ $\iff$ $\overline{f(\overline z)}$ is complex-differentiable at $z_0$ and on an open neighborhood of $z_0$, but you seem to already know that.

Take $f(z)=\frac{1}{z+i}$ which is continuous at $z=i$ but $\overline{f(\bar z)}=\frac{1}{\overline{\bar z+i}}=\frac{1}{z-i}$ is not continuous at $z=i$.

Similarly you can check for differentiability of $f(z)$.

However, you may like to know the well-known result $f(z)=\overline{f(\bar z)}$ if $f(x)$ is real. (Reflection principle)


Why does the derivative graph of a curve look linear and not curvy? Why is the derivative graph of $y=x^2$ linear and not some sort of curve?  I know that $\frac{d}{dx}x^2=2x$, but I am not talking computing algebraically, but more conceptually, thinking in terms of how the slope of tangent changes.   When I look at the graph, why is the slope of $x^2$ changing at a linear rate?

Alternatively, why is the derivative of $y=ln(x)$ some kind of curve and not linear?   I assume the rate of "curviness" of $x^2$ is different than $ln(x)$ which gets flat in a hurry.

(Does 2nd derivative and concavity explain this?  Rate of change of the rate of change? )

My last example in the scan below summarizes my question.  Why is it one of those, and not the other 3?   All 4 start with a slope of 0, then 1, then back to 0.  What is the connection, graphically ?

Thank you!

It's easy to visually see that the derivative of $x^2$ is $2x$, by not thinking in terms of how the slope changes, but plotting the slope of the tangents at points and seeing.

As to 'why' it is $2x$... It is not clear what kind of answer you are looking for. I intepret your wish as wanting a more ontological answer.

Think about two consecutive square numbers. Well $(x+1)^2 =x^2+2x+1$.

So when we add one to $x$, we add $2x+1$ to $f(x)$. If we add $2$ to $x$ we add $4x+4$ to $f(x)$. That is when we add some $c$ to $x$, we add $2cx + c^2$ to $f(x)$. This gives us a slope term of $2x+c$.

But for $x$ much larger than $c$ this is $~2x$.

Now, when we take the tangent on the $x^2$ curve, we are making $c$ as small as possible to get a more accurate rate of change at $x$ and this can (atleast in my head) be thought of as making the tangent line intersect the $x^2$ curve as much as possible. And assuming all lines have same width (euclidean assumption), we get the largest area of intersection when $c$ is smallest because this results in the part of the tangent line leaving the $x^2$ curve having the smallest area possible. That is because a line is defined as having constant width, and a shortest $c$ implies shortest possible length of the tangent line that has left the $x^2$ curve between $x$ and $x+c$. So the area of tangent that has left the $x^2$ curve in this zone is smallest. Whilst the rest of the tangent line that is not intersecting the $x^2$ has infinite area no matter what $c$ .... hence my interpretation of taking a tangent at $x$ on $x^2$ is defined as increasing the intersection between tangent line and curve as much as possible about $x$ (of course this wouldnt work for all curves, but it would for $x^2$.

Anyways, in pure mathematics we assume that the smallest $c$ is in fact small enough to make $2x+c = 2x$ ignoring other physical variables.

In some sense, the slope being $2x$ for $f(x)=x^2$ is a consequence of our chosen deconstructions which give us nice aesthetical answers.

What you're attempting to reason is why a derivative to a graph $f(x)$ is linear or non-linear. If you do a simple test visually say, the first segment of your reference graph is concave up and positive slope. The positive slope notifies that the graph of the derivative will be in the positive terminal. The concave up quality of the initial part of the graph assumes that the derivative is increasing. However, the question is at what rate is the rate of change increasing? This gives that the graph of the derivative is concave up because the original function is increasing.


Disproving the proposition that $a_n = (-1)^nn$ has a limit. Is the following Proof Correct?

- Show that the sequence $a_n = (-1)^nn,\forall n\in\{1,2,3,...\}$ does not have a limit.

Proof. Assume on the contrary that $\lim a_n = L$ consequently there exists an $N\in\mathbf{R^+}$ such that given any $n\in\mathbf{Z^+}$ $|(-1)^nn-L|<|L|$ whenever $n>N$ the archimedian property guarantees that such an $n$ exists consequently for some $n>N$ we have $|(-1)^nn-L|<|L|$  but $|(-1)^nn-L| = |(-1)^{n+1}(n+L)| = |(-1)^{n+1}|\cdot|n+L|$ implying that $|n+L|<|L|$ resulting in a contradiction.

$\blacksquare$

The proof is missing $\varepsilon$ if you intend to use the $\varepsilon-N$ definition of the limit for sequences. If $(a_n)$ had a limit $L<+\infty$ then so would $(b_n)$ where $b_n:=|a_n|$. This follows from triangle inequality
$$|b_n-|L||=||a_n|-|L||\leqslant |a_n-L|<\varepsilon$$
whenever $n\geqslant N$ for some large enough $N$. But $\lim_nb_n=\lim_n |a_n|=\lim_nn=+\infty$.

if $n$ is even we get $$a_n=n$$ and this tends to infinity for $n$ tends to infinityif $n$ is odd, then we get $$a_n=-n$$ and this tends to $-\infty$ if $n$ tends to infinity, so no Limit exists


Why can we use mathematical induction to prove that $2^n \ge n^2$ for $ n \ge 5$ Why can we use mathematical induction to prove that $2^n \ge n^2$ for $ n \ge 5$

Normally, the proof via mathematical induction starts with $n =1$. Why does it still hold, when we start with other numbers?

if you insist, you can define $$ n = m+4 $$ and begin with $m=1$

multiplying $$2^k\geq k^2$$ by $2$ we get
$$2^{k+1}\geq 2k^2$$ now you have to prove that $$2k^2\geq (k+1)^2$$


can this integral be expressed in elementary functions? I have been trying to find the length of an arc of an ellipse and I have been stuck with this integral for a complete day : $$\int_{0}^{x} \sqrt{a^2\cos^2t+b^2\sin^2t} dt$$
And my question is : can this integral be expressed in terms of elementary functions ? If not then does this integral have a special function or something ?

The answer to the question in the title is

No, the integral cannot be expressed in terms of elementary functions. Elementary functions are a fairly restricted group (like polynomials, the exponential function and the natural logarithm).

But you go on to ask whether it can be expressed in terms of special functions or something.

The short answer is that it is easy to get arc lengths along an ellipse in terms of an incomplete elliptic function of the second kind. But you need software like Mathematica (where it is EllipticE[]) to be able to use it freely.

The incomplete elliptic function (the complete/incomplete distinction is about whether the function deals with the complete curve or only an arc) is certainly a "special function or something".

Whether it is a special function is more debatable.

Personally, I don't think I would refer to this function as a special function. Wikipedia disagrees, so does the NIST bible. But the term "special function" is often used for a relatively small group of functions which were mainly important in physics and applied maths and intensively studied in the century or more before WW2 (usually as functions of a complex variable).

In any case the term "special function" is starting to get archaic, and I don't think it is a helpful one today. Long ago they were special in the ordinary English language sense, because they had been extensively studied and people knew how to deal with them, although they required more expertise than the "elementary" functions. [Whereas many other functions were fairly intractable except by inequalities, bounds, crude approximations etc.]

Today I would rather call something like the functions used for ellipse arc length a "named function". The reason is that it is possible to use it (and indeed many of the old special functions) with much less special expertise.

What often matters in practice today is not so much whether a function is an "elementary" or "special" function, but whether it is easy to calculate with the function using software like Mathematica.

If you want more information on how to use the complete/incomplete elliptic functions, I suggest googling. You can find plenty of tutorial material about ellipse arc lengths and the functions used to deal with them. Eg here (about halfway through he turns to elliptic integrals).

There is also a fair amount of material in Wikipedia scattered over several articles.

Added a little later

Incidentally, do not think of math software as just a way of getting numerical answers or plotting functions with known parameters (ie $\sin 2x$ rather than $\sin kx$). It is much more powerful than that. For many functions you can easily get indefinite integrals, power series etc.

I remember spending a significant chunk of my undergraduate years at Cambridge University many decades ago learning endless tricks for integrating functions. That knowledge is of little use to me now. Mathematica easily outperforms most academics on this site at integration (and certainly me, but notably not Jack d'Aurizio, whose expertise with integrals often delights me).

In general, you can cheerfully manipulate its named functions just as easily as if they were elementary functions.

I mention Mathematica just because it is widely available and good for integration. But there is a good deal of both commercial and open source software available.


How do we know that Cantor's diagonalization isn't creating a different decimal of the same number? Edit: As the comments mention, I misunderstood how to use the diagonalization method. However, the issue I'm trying to understand is a potential problem with diagonalization and it is addressed in the answers so I will not delete the question.

Cantor's diagonalization is a way of creating a unique number given a countable list of all reals.

I can see how Cantor's method creates a unique decimal string but I'm unsure if this decimal string corresponds to a unique number. Essentially this is because $1 = 0.\overline{999}$.

Consider the list which contains all real numbers between $0$ and $1$:

$0.5000 \mathord\ldots \\
0.4586 \mathord\ldots \\
0.3912 \mathord\ldots \\
0.3195 \mathord\ldots \\
0.7719 \mathord\ldots\\
\vdots$

The start of this list produces a new number which to four decimal places is:

$0.4999 \mathord\ldots$

But $0.5$ was the first number and $0.4\overline{999} = 0.5$ so this hasn't produced a unique number.

Of course my list is very contrived, I admit that it's hard to imagine a list of the reals where numbers would align nicely to give a problem like this (since some numbers have no nines). However, I can't see a good reason why such an enumeration of numbers would be impossible.

This is in fact a potential problem if the proof is carelessly stated, but it’s easily avoided: if the $n$-th decimal digit of the $n$-th number on the list is $7$, we replace it by $6$, and if it’s not $7$, we replace it by $7$. The only numbers in $(0,1)$ with two decimal representations are those with one representation ending in an infinite string of nines and the other in an infinite string of zeroes, and this version of the argument clearly doesn’t produce a number of either of those forms.

Cantor's Diagonal proof was not about numbers - in fact, it was specifically designed to prove the proposition "some infinite sets can't be counted" without using numbers as the example set. (It was his second proof of the proposition, and the first was criticized because it did.)

What Cantor actually used, were infinite length strings that combine only two different characters. He used "m" and "w." But modern High School and Middle School students have experience with infinite decimal expansions, and it is easier to teach new concepts to them in terms of what is already familiar. The point is, that there is no ambiguity in saying that two strings are different if they have different characters in at least one position.

Another point that is taught poorly, is that the proof doesn't have to be a proof-by-contradiciton. Many people have a justifiable objection to assuming statement A, and then proving that not(A) follows from it. All it really proves is that there is something wrong in the circle that leads from A to not(A), not that it has to be the assumption of A. Which is why so many people try to invalidate one of the steps - what you claim here one of the ways they try (and yes, it can be gotten around easily, as stated in other answers).

Cantor only assumes that you have a countably infinite set of some of these strings. Diagonalization proves that there is a string not in your set. Since "If A then B" is logically equivalent to "If not(B) then not(A)," this proves that if you have a set of all such strings, then it can't be counted.

Elegant, no?


Continuous without restriction or Continuous with discrete interpretation I have a question , with contradictions with my teacher. Basically the question number is 7d .. And the instructions for number 7 questions are as follows:

  Determine whether each of the following function representation is
  discrete, continuous without restriction, or continuous with discrete
  interpretation.

And the graph for 7d question is:

The letter on top says Total Public expenditures since 1960. Ok so my teacher says its continuous with discrete interpretation while I say it's continuous. I say its continuous because dollars can be any point for example 12.5 billon dollars, 16.21313131 billon dollars, doesn’t matter. I don't understand why the teacher is saying continuous with discrete interpretation. Help Would be appreciated!

If you guys don't understand continuous with discrete interpretation , then my previous question explains it :

https://math.stackexchange.com/questions/1929805/what-is-discrete-with-continuous-interpretation

If you guys don't know what continuous with no restriction means , it basically just means continuous

I want to extend my comment:

After reading the definitions in the referenced question I conclude that your teacher wants to say, that "continous with discrete interpretation" means, it makes only sense to speak about the public expenditures in 1961, 1962, 1963,.... So $f$ is only meaningful at 1,2,3,.. (years after 1960), and not e.g. at 1.2345 or at $\sqrt{2}$. But I am not sure if I am of the same opinion. Even if the data were supplied in a tabel (year, expenditures) I think one can speak about the expenditures a an arbitrary point in time. A more clear example is something like
$$f:A\to \mathbb{R}$$
$$f(x)=y$$
where $x$ is the number of inhabitants of a city and $f(y)$ are  the expenditures for public transport. Clearly there is no city with $1234567.8$ inhabitants, nevertheles it make sense to set $A=\mathbb{R^+}$ in a model with out any problems.

What makes your counterargument wrong in this case is simply the fact that money amounts always have a minimum possible value - in this case, pennies.  Any fractional amount of billions-of-dollars which specifies a fraction of a penny is simply not a real amount of money.  The curve in the diagram has to be a continuous curve interpolated through a discrete point-set of actual financial data.


How to find the order of the ring I have some examples solved about order of an element in a ring.
In $\mathbb Z_{5}$  find order of element $2$ in the ring.   Identity is $1$
$2^1 ≠ 1$
$2^2 ≠ 1$
$2^3 ≠ 1$
$2^4 = 16 = 1$ <----------This is the part that I dont understeand

First of all, mention in the title that you would like to calculate order of an 'element' of a ring.
In $Z_5$, the multiplication is defined by multiplying two elements as integers and then reducing the result to modulo $5$ which means that you need to divide the result by $5$ and take remainder.
An example is $3\cdot 4=12$ which gives remainder $=2$ on dividing by $5$ and hence in $Z_5$, $3\cdot 4=2$.
Now in the last line of your question, the correct equation is $2^4=16\equiv 1 \pmod 5$ because $16$ yields remainder $=1$ when divided by $5$.

I think this is a matter of the rigor of definitions. Strictly speaking, $16$ is NOT an element of $\mathbb{Z}_5$, which as a set is $\{0,1,2,3,4\}$. Strictly speaking, we have a surjective homomorphism of rings $\pi:\mathbb{Z}\rightarrow\mathbb{Z}_5$ sending a number to its remainder modulo $5$, i.e., when dividing by $5$: $\pi(n):=r$ where $n=5q+r$.
So, strictly speaking we have $\pi(2^4)=\pi(16)=\pi(5\cdot 3+1)=1$ (in $\mathbb{Z}_5$), and also $2^4=\pi(2)^4=1$ in $\mathbb{Z}_5$ (since $\pi$ is a homomorphism such that $\pi(2)=2$), which we can write also as $2^4\equiv 16\equiv 1\pmod{5}$, where now $\equiv\pmod{5}$ is an equivalence relation for $\mathbb{Z}$. Nevertheless it is usual, but an abuse of notation, to write $16=1$ in $\mathbb{Z}_5$.


Finding the last root of $p(x) = x^5 + a_3 x^3 + a_2 x^2 + a_1x + a_0$ given that... The polynomial $p(x) = x^5 + a_3 x^3 + a_2 x^2 + a_1x + a_0$ has real coefficients and has 2
  roots of $x = -3$, and two roots of $x=4$. What is the last root, and
  how many times does it occur?

At first I expanded $(x+3)^2(x-4)^2$ to get a divisor to divide the polynomial with, but I could not get around the fact that the coefficients are not given, so I can't get any concrete number. Here's the expansion:
$$x^4 - 2x^3 - 23x^2 + 24x + 144$$
I don't quite get how to use it to help me find the last root.

Using Vieta formula  $$x_1+x_2+x_3+x_4+x_5=-{a_4\over a_5} =-{0\over 1}=0$$ you can get $x_5$. And then you can also calculate all other coefficients $a_0,a_1...$.

If $(x-r)$ is a repeated root of $p(x)$ then $p(r) =p'(r) =0$.

So $p(-3)=p(4)=p'(-3)=p'(4)=0$, allowing you to form a system of four linear equations in terms of four unknowns. Solve for the coefficients and find the last root (which has to be a single root as the polynomial is quintic).


Show that $\sum_1^\infty \frac{(-1)^{n-1}}{n} = \log 2 $ $$ \sum_1^\infty \frac{(-1)^{n-1}}{n} = \log 2 $$

I'm not sure how I can prove that.

I have tried changing the term to $-\cos(n\pi)/n$, but that doesn't help me much.  Any hint?  Thank you.

Recall the Taylor series for log
\begin{align*}
\log x=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}(x-1)^n.
\end{align*}
Plugging in $x=2$ solves your problem.

Here is an article that answers your question http://www.math.uconn.edu/~kconrad/blurbs/analysis/abelthm.pdf

It uses the power series for $\log(1+x)$ and Abel's theorem. If you are unfamiliar with this series, it comes from integrating the series $$\sum_{n=0}^{\infty} (-1)^nx^n=\frac{1}{1+x}.$$


$3$ and $5 $cent coins Prove that any amount of more that $7$ cents can be represented by $3$
and $5$ cent coins. (Assume $3$ cent coins exist.)

Let P(n) be true if we can find $n$ cents with $3$ and $5$ cent coins.

My basis step is:

$P(8)$ is true since $3 + 5 = 8$

Would this be correct?

As for the inductive step, I'm not sure how to show it.

I've looked around online and it seems to be:

$P(k) = 8 ≤ k ≤ n$

How would I prove this to be true?

Observe that $8 = 3+5$, $10 = 2\cdot 5$, and $9 = 3\cdot 3$.This means $P(8)$, $P(9)$ and $P(10)$ are true. Assume $P(k)$ is true for $10 \leq k < n$, we prove $P(n)$ is true: Observe that $n = (n-3) + 3$, and $P(n-3)$ is true so $n-3 = 3m+5p \to n = 3+3m+5p = 3(m+1) + 5p$

proof by induction
p(8): 3.(1)+5.(1)=8 holds
Assume p(k) holds => 3.a + 5.b = k
we must prove that p(k+1) holds.
p(k+1): 3x+5y = k+1=> 3x+5y = 3a+5b+1=> 3(x-a) + 5(y-a)
thus suffices to prove that 3z+5ω =1 has an integer solution. That is true because gcd(3,5)=1 | 1 (1 divides 1)
Calculating gcd(3,5) gives: 5 = 1*3 + 2
3 = 1*2 + 1
2 = 2*1 + 0
Then applying the Extended Euclidean Algorithm:
1 = (1 * 3) + (-1 * 2) = (-1 * 5) + (2 * 3)=> z0=2, ω0=-1 => 3.2+5.(-1) = 1=> p(k+1) holds as well


Evolution of angle between two triangles as they tilted towards each other My problem is as follows. I have two flat isosceles triangles (marked in yellow below) -- with sides of $L$ and base of width $w$ -- touching at a common point (marked in red), such that the angle between them is $\theta_i$. Note that $\theta_i$ is also the apex angle of another isosceles triangle, which I have depicted in lavender.
Now, I tilt these triangles (shown in deep blue) towards each other, by an angle $\alpha$. Then what is the equation relating the final angle $\theta_f$ to $\theta_i$ and $\alpha$? I know from measurements that $\theta_f \leq \theta_i$.
I realise that I only need to figure out how the lateral sides of the yellow triangles behave under the tilt $\alpha$, but I am unable to do so. Can someone please help?

Note: In the picture, the entire deep blue configuration is tilted off the horizontal plane, but this is just for viewing convenience.

Pick coordinates centred at the red point with the $z$-axis being the internal angle bisector of $\theta_i$, $x$-axis being the orthogonal on the plane and $y$-axis the vertical out-of-plane direction.  So the two initial apex are $(\pm L\sin\frac12\theta_i,0,L\cos\frac12\theta_i)$.  If the bases of the two triangles are on the same line, then we have $\sin\frac12\theta_i=w/(2L)$.  It isn't clear whether this is the case from your description.
Then rotate the triangles by angle $\pm\alpha$ about $z$-axis so $y$ becomes positive, we have the two points $(\pm L\sin\frac12\theta_i\cos\alpha,L\sin\frac12\theta_i\sin\alpha,L\cos\frac12\theta_i)$.  The angle they made is $\cos^{-1}(1-2\sin^2\frac12\theta_i\cos^2\alpha)$.
If you rotate about other axes parallel to this, you will get parallel line, so the angle is unchanged (except you get skewed lines instead of intersecting lines in general).


Correct usage of Stolz-Cesaro theorem in finding a limit Is this solution correct?

I've seen this solution before, but I don't know why it is so. And which rule?

The task is to find: $\omega=\displaystyle\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)$

I know that this $\lim$ equals $\frac{1}{e},$ but do you see if this solution is correct or not?

Solution is :

$\displaystyle\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\displaystyle\lim_{n\to +\infty}\left((n+1)\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}-n\sqrt[n]{\frac{n!}{n^{n}}}\right)$

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

$=\frac{1}{e}$

Notes & suggestions of mine:

I think he uses Stolz-Cesaro theorem:

See that

$\displaystyle\lim_{n\to +\infty}\frac{a_{n}}{b_{n}}=\displaystyle\lim_{n\to +\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}$

Chosen $a_{n}=n\sqrt[n]{\frac{n!}{n^n}}$

and $b_{n}=n$  clearly $b_{n}$ goes to $+\infty$ and $a_{n}$ goes to $+\infty$ because $n.\frac{1}{e}=+\infty$

we obtain:

$\omega=\displaystyle\lim_{n\to +\infty}\frac{n\sqrt[n]{n!}}{n}$

Then use Cauchy-d'Alembert

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

Is my explanation O.K.?
May I ask for a correction?

Any remark ?

The use of Cesaro-Stolz as you mention is invalid. In application of Cesaro-Stolz the existence of limit of $\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$ is a hypotheses and not the conclusion. Thus you have tried to use Cesaro-Stolz in reverse which does not work.

I don't think Stolz-Bolzano theorem is necessary here (:  , but I agree with everything you explained. Nevertheless, I'll write down everything in detail for everyone who might have not understood similar proofs to other (similar) posts.

Since $n\to\infty$, $n+1$ behaves (more or less) similarly:

Inductively: $$\lim_{n\to\infty}{\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}}=\lim_{n\to\infty}{\sqrt[n]{\frac{n!}{n^n}}}$$
$$\implies\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\lim_{n\to +\infty}\left((n+1)\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}-n\sqrt[n]{\frac{n!}{n^{n}}}\right)=\lim_{n\to\infty}\Bigg((n+1)\sqrt[n]{\frac{n!}{n^n}}-n\sqrt[n]{\frac{n!}{n^n}}\Bigg)=\lim_{n\to\infty}{\sqrt[n]{\frac{n!}{n^n}}}=\lim_{n\to\infty}\frac{n!}{n^n}$$
As proven here:
Why is $\lim\limits_{n \to +\infty }{\sqrt[n]{a_1 a_2 \ldots a_n}} =\lim\limits_{n \to +\infty}{a_n}$

You can apply the Cauchy-D'Alembert criterion already here:$$\lim_{n\to\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\lim_{n\to\infty}\frac{n^n(n+1)!}{(n+1)^{n+1}n!}=\lim_{n\to\infty}\frac{n^n(n+1)}{(n+1)^{n+1}}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{x\to\infty}\frac{1}{\Big(\frac{n+1}{n}\Big)^n}=\lim_{n\to\infty}\frac{1}{\Big(1+\frac{1}{n}\Big)^n}=\frac{1}{e}$$


Integrating a Poisson kernel in $n$ dimensional unit sphere Let
\begin{equation*}
P(x,y)=\frac{1}{\omega_n R} \frac{R^2-|x|^2}{|x-y|^n}
\end{equation*}
be a Poisson kernel where $x$, $y$ are in $R^n$, $|x|<R$, $|y|=R$, $\omega_n$ is area of n dimensional unit sphere.
Then what is $$\int_{\partial B_1(0)}P(x,y)dA_y=?$$

Hint: The function

$$g(x) = \int_{\partial B_1(0)} P(x, y) dA_y = \int_{\partial B_1(0)}  1 \cdot P(x, y) dA_y$$

satisfies $\Delta g = 0$ in $B_1(0)$ and $g(y) = 1$ on $\partial B_1(0)$. What can $g$ be?

"P(x,y)= (omega[n]R ())^(-1)(R^(2)-|x|( )^(2))|x-y|( )^(^(-n) ) ,

  |x-y(|)^(2)=|x|^(2)-2|x|R costheta +R^(2), for x*y =|x|R costheta "

"dsigma[n]=sin^(n-2)theta dtheta dsigma[n-1] for functions depending only on

theta, where dsigma[n] denotes Lebesgue measure on S( )^(n) , thus get

integral of type ∫(a+b*costheta)^(-n)sin^(n-2)theta dtheta from 0 to

Pi,, which can be evaluated by well known substitution tan(theta/(2))=u,

get ∫(1+u^(2))^(2) u^(n-2)(A+u^(2))^(-n)du from 0 to 1 which can be

evaluated by calculus of residues,  also note Claus Mueller, Spherical

harmonics, Lecture Notes v.42 ("?) ""

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsthm}
\usepackage[T1]{fontenc}

\usepackage{geometry}
\geometry{a4paper}

\begin{document}

$P(x,y)=(\omega_n R)^{-1}(R^2-|x|^2)|x-y|^{-n}$,$x*y=|x|R\cos\theta$,$ |x-y|^2=|x|^2-2|x|R\cos\theta+R^2$, $d\sigma_n=\sin^{n-2}\theta d\theta d\sigma_{n-1}$ for integrands depending on $\theta$ only, where $\sigma_n$ denotes Lebesgue measure on $S^{n-1}$, thus get integral of type $\int^\pi_0 (a+b\cos\theta)^{-n}\sin^{n-2}\theta d\theta$, (wlog $R=1$, and x in same direction as North pole of coordinate system (turn coordinate system around) so that $\theta$ also is angle between $x$ and $y$) which can be evaluated by $\tan \theta/ 2=u$, get $\int_0^\infty(1+u^2)^2u^{n-2}(A+u^2)^{-n}du$ which can be evaluated by residue calculus, also see Claus M\''uller, Spherical Harmonics (search Internet).
\end{document}


Why application of Sharkovskii's Theorem to Collatz is wrong? I thought about this years ago but never had it resolved. Sharkovskii's Theorem has a corollary according to Wikipedia that if there exists a 3-cycle on a map on $\mathbb{R}$, then the map has $n$-cycles for every $n\in\mathbb{N}$. Now, the Collatz map $f(x)$ has a 3-cycle: $f(f(f(4))) = 4$...

Now, I assume the conclusion that the Collatz map has cycles of every size is wrong (otherwise, it would have been solved) but why does it not apply here?

I don't understand how it could not be wrong?

As compu.sci. I work with base-2 and integers all the time.
Could you not map two to the power of n: $2^n$ in $\mathbb{R}$ ? and use those mappings
to map any $n$ to $f^3(n) = n$. where $n$ (the exponent) is mapped to all positive integers in $\mathbb{N^+}$ ?

With a slight modification to that latter mapping we get:

$f^3(2^{n\mod3}) = 2^{n\mod3}.$
so..

$$
\begin{array}{|c|c|}
\text{$n$} & \text{$f^3(2^{n\mod3})$} \\ \hline
 0 & 1 \\ \hline
 1 & 2 \\ \hline
 2 & 4 \\ \hline
 3 & 1 \\ \hline
 4 & 2 \\ \hline
 5 & 4 \\ \hline
 . & . \\ \hline
\end{array}$$
Consider this as mapping-function.

Any power of two number would reach the 3-cycle at any time this way.
Assuming all Collatz-numbers will eventually reach 4 this would indeed be correct. Not sure if the interpretation is easy to follow, so bear with me, im learning math every day. And this is not considered to be a solution, only an idea. -R


If $A$ and $B$ are square matrix such that $AB=I$ then $BA=I$ too. If $A$ is a square matrix which coefficients are the scalar of a field $\Bbb K$ then there exist a linear function $f:\Bbb K^n\rightarrow\Bbb K^n$ such that
$$
f(e_j):=A_j
$$
where $A_j$ is the $j$-th column of $A$ for any $j=1,...,n$.
Then if $I$ is the identity square matrix if $A$ and $B$ are two square matix such that $AB=I$ then there exist two function $f,g:\Bbb K^n\rightarrow\Bbb K^n$ such that the identity square matrix is the matrix of the function $(f\circ g)$ and this means that $(f\circ g)$ is the identity on $\Bbb K^n$. So to prove the statement of the question I have to prove that $(g\circ f)$ is the identity too. So how prove the statement? Could someone help me, please?

$f\circ g = \text{id}$ implies $g$ is injective (if a composition is injective then the "inner" function is always injective). Since $g:\Bbb{K}^n\to \Bbb{K}^n$ is a map between vector spaces of the same dimension, the rank-nullity theorem implies $g$ is also bijective. Thus, $f = g^{-1}$.
Another way to phrase the argument is to say that since $f\circ g = \text{id}$, then $f$ is surjective (again if a composition is surjective, then the "outer" function is surjective). Again, rank-nullity implies $f$ is bijective, so that again $g=f^{-1}$.

Theorem:
If $A$ and $B$ are two square matrices such that $A B=I$, then $B A=I$.
Proof:
By applying the properties of determinants of square matrices, we get that
$\det A\cdot\det B=\det(A B)=\det I=1\ne0$.
So it results that $\;\det A\ne0\;$ and $\;\det B\ne0\;$.
Now we consider the matrix
$C=\frac{1}{\det A}\cdot\text{adj}(A)$
where $\;\text{adj}(A)\;$ is the adjugate matrix of $A$ which is the transpose of the cofactor matrix of $A$.
$\text{adj}(A)=
\begin{pmatrix}
A_{1,1} & A_{2,1} & \cdots & A_{n,1} \\
A_{1,2} & A_{2,2} & \cdots & A_{n,2} \\
\vdots  & \vdots  & \ddots & \vdots  \\
A_{1,n} & A_{2,n} & \cdots & A_{n,n}
\end{pmatrix}$
where $\;A_{i,j}\;$ is the cofactor of the element $a_{i,j}$ of the matrix $A$.
So $\;A_{i,j}=(-1)^{i+j}\det M_{i,j}$
where $\;M_{i,j}\;$ is the submatrix of $A$ formed by deleting the $i^{th}$ row and the $j^{th}$ column.
We are going to use the Laplace expansions which are the following equalities:
$a_{i,1}A_{j,1}+a_{i,2}A_{j,2}+\ldots+a_{i,n}A_{j,n}=
\begin{cases}
\det A\;,\quad\text{ if } i=j\\
0\;,\quad\quad\;\,\text{ if } i\ne j
\end{cases}$
$A_{1,i}a_{1,j}+A_{2,i}a_{2,j} +\ldots+A_{n,i}a_{n,j}=
\begin{cases}
\det A\;,\quad\text{ if } i=j\\
0\;,\quad\quad\;\,\text{ if } i\ne j
\end{cases}$
By applying Laplace expansions, we get that
$A C = C A = I$.
Since for hypothesis $\;A B =I\;,\;$ then $\;C(A B)= C I\;,\;$ hence $\;(C A)B=C,\;$ therefore $\;I B = C\;$ and so we get that $\;B=C$.
Consequently,
$B A = C A = I\;,$
so we have proved that
$B A = I$.


Prove by induction: inequality Use induction to prove the inequality:

$$\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}\leq \frac{1}{\sqrt{n+1}}$$

I tried by multiplying $\frac{(2n+1)}{(2n+2)}$ to both sides but I don't know how to get $\frac{1}{\sqrt{n + 2}}$ on the RHS.

Hint:  For positive integers $n$ one has

$\frac{1}{\sqrt{n+1}}\cdot\frac{(2n+1)}{(2n+2)}\leq \frac{1}{\sqrt{n+2}}$

is true if and only if

$\frac{1}{n+1}\cdot \frac{4n^2+4n+1}{4n^2+8n+4}\leq \frac{1}{n+2}~~~~~$   (seen by squaring both sides)

is true if and only if

$(n+2)(4n^2+4n+1)\leq (n+1)(4n^2+8n+4)~~~~~$   (seen by cross multiplying)

We know that the inequality holds for n=1. Now let us assume the inequality holds for n, and prove it for value n+1. The LHS and RHS of the inequality at n+1 is divided by the LHS and RHS of the inequality at n, respectively. We see that the LHS is multiplied by (2n+1)/(2n+2) and the RHS by [(n+1)^(1/2)]/[(n+2)^(1/2)]. Now, to prove the inequality, we have to prove:

(2n+1)/(2n+2)  ≤  √(n+1)/√(n+2)

((2n+1)^2)/((2n+2)^2)  ≤(n+1)/(n+2)

4n^3+12n^2+9n+2 ≤ 4n^3+12n^2+12n+4

Hence, proved.


Which of two vertices has the wider angle? Given that:

two vertices from different paths share the same point;
the winding orientation of these paths is unspecified;
the angle at both vertices will be less than 180 degrees
these paths don't intersect

what's the most efficient way to determine which path is outside (has a wider angle than) the other?

I could use cross product formula to get the orientations of each path, to determine which of the respective adjacent vertices is on the left (relative to the other), and then perform a third cross product calculation using the left-most vertex from each path and the common (lower) vertex to determine which of the left vertices is outside the other. But is there a more efficient approach?

In other words, in the image above, how can I most efficiently verify that the blue path is outside the red path?

Edit:

It's also possible that either the left edges or the right edges are colinear (but both won't be colinear).

I would compare the angles, if you have x,y information of end points. First calculate the angles of each path, if you don't have the x,y of common vertice just give it a random x,y the inner line will remain inner no matter what.

Now that you have angular info of both paths,  you can compare them to each other. Find the angle between any two close end points from different paths, and find the angle between them from the shared vertice. Now you are questioning if A(b1-r1) + R(r1-r3) ?> B(b1-b3).

There is only one exception I guess you didn't mention, what if what if the order was like this: r1-b1-r3-b3? Now neither of them is inner or outer. Maybe you should also clarify that one. Sorry if I'm missing any point.


Taking the limit of an integral expression Hi guys I was doing the following question:-

let a,b ∈ $\mathbb{R}$ with a < b show the following

$$\lim_{n\to \infty}\int^a_b\frac{((b-a){e^x}^2)}{(b-a)n^2{e^x}^2+b^2} dx = 0 $$

so i started taking the lim as $\lim{n\to \infty}$ which gives the following:-

$$\lim_{n\to \infty}\frac{((b-a){e^x}^2)}{(b-a)n^2{e^x}^2+b^2}  $$

Since i am trying to evaluate the limit I decided to remove the top expression since its a constant and ended up getting

$$(b-a){e^x}^2\lim_{n\to \infty}\frac{1}{(b-a)n^2{e^x}^2+b^2}  $$

Now looking at the denominator expression can i directly substitute n as infinity

therefore this gives the following:-

$$(b-a){e^x}^2\lim_{n\to \infty}\frac{1}{(b-a)(\infty)^2{e^x}^2+b^2}  $$

Can i take this expression to be $\frac{1}{\infty} = 0 $

hence i get the following $$(b-a){e^x}^2(0) = 0 $$

Therefore when i go to the next part of the question to evaluate the integral well = 0 since the above shows that when the limit is evaluated f(x) = 0. I am not sure if i evaluated this question correctly when looking at the limit aspect and so hoping so one can confirm this for me.

You need, for example, Lebesgue Dominated Convergence Theorem to guarantee the swiping of the integral with the limit to move on.

Or you can do it in the following way, because the integrand has no singularity:
\begin{align*}
\dfrac{(b-a)e^{x^{2}}}{(b-a)n^{2}e^{x^{2}}+b^{2}}\leq\dfrac{(b-a)e^{b^{2}}}{(b-a)n^{2}e^{a^{2}}+b^{2}}\leq\dfrac{(b-a)e^{b^{2}}}{(b-a)n^{2}e^{a^{2}}}=\dfrac{e^{b^{2}}}{e^{a^{2}}}\dfrac{1}{n^{2}}.
\end{align*}
Taking integral both sides, the right sided is just $(b-a)\dfrac{e^{b^{2}}}{e^{a^{2}}}\dfrac{1}{n^{2}}$ and this goes to zero, now you use Squeeze Theorem to conclude that the integral also goes to zero.

$0$ is  the correct limit you can get it from the fact that the integrand lies between $0$ and $\frac 1 {n^{2}}$ because $b^{2} \geq 0$. Hence the integral itself lies between  $0$ and  $\frac {b-a} {n^{2}}$. You don't have to evaluate the integral or apply a theorem like DCT.