Prove that $\log_9 15$ is irrational Im having trouble with the following proof... Ill post what I have completed so far..

  Prove that $\log_915$ is irrational.

Ill attempt by contradiction assuming $\log_915$ is rational.

So,

$\log_915 = \frac ab$

$15 = 9^{\frac ab}$

$15^b = 9^a$
   (This is where I'm getting stuck)

Any hints/tips/advice would be great. Thanks

$a$ and $b$ are positive integers. $15^b$ and $9^a$ are positive integers. $5$ definitely does not divide $9^a$, so what must $b$ be?

$$15^b = 9^a$$

After this, you can write it as
\begin{align*}
\frac{15^b}{9^a} & = 1\\
\frac{3^b5^b}{3^{2a}} & = 1\\
\frac{5^b}{3^{2a-b}} & = 1
\end{align*}

To fulfill this equation, numerator should be equal to denominator but no power of $3$ and $5$ will be same because power of $3$ will end with $1$,$3$,$7$,$9$ and that of $5$ will end with $5$ only, so their power can only be $0$.

$2b=0$ and $2a-b=0$

so $b$ equals zero but it is denominator in our first assumption of rationality, which contradicts our assumption.
Therefore, $\log_9 15$ is irrational.


Is it possible to find $\limsup\limits_{n\to\infty} \frac{2×3^n-3×2^n}{2^{\alpha(n)}-3^n}$? I need help to solve this problem:

$$\lim_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n}{2^{\alpha(n)}-3^n}$$ or
$$\limsup_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n } {2^{\alpha(n)}-3^n}$$

where,
$\alpha(n)-$ is the smallest positive integer, such that $2^{\alpha(n)}-3^n>0$.
I tried $\alpha (n)= \lfloor n\log_23+1 \rfloor$. Then, we have
$$\lim\limits_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n}{2^{\alpha(n)}-3^n}= \lim\limits_{n\to\infty} \frac{2}{\frac{2^{\lfloor n\log_23+1 \rfloor}}{3^n}-1}$$
I'm stuck here. I can not continue..Can MSE help me? Are my steps wrong?
Can I take $\alpha (n)= \lfloor n\log_23+1 \rfloor$ ?
Thank you.

We have
$$\alpha(n) = \lceil n \log_2 3 \rceil = n \log_2 3 + 1 - \{n \log_2 3\}, \\
\frac {2 \cdot 3^n - 3 \cdot 2^n} {2^{\alpha(n)} - 3^n} =
\underbrace {3^{-n} (3 \cdot 2^n - 2 \cdot 3^n)}_{\to -2} \cdot
 \frac 1 {1 - 2^{1 - \{n \log_2 3 \}}} \,.$$
Since $\{n \beta\}$ is dense in $[0, 1]$ for irrational $\beta$, we can choose a subsequence that doesn't converge and can choose a subsequence that converges to $1$.

I don't think the limit exists. Write $3^n=2^{n\log_2(3)}$, then the latter simplifies to
$$ \frac{2}{2^{\lfloor n\log_2(3) + 1 \rfloor - n\log_2(3)}-1} =  \frac{2}{2^{\lfloor n\log_2(3) + 1 \rfloor - n\log_2(3)}-1} $$
I believe there exists an increasing sequence $(a_n)_{n=1}^{\infty}\in\mathbb{N}^{\mathbb{N}}$ such that
$\lfloor a_n\log_2(3) + 1 \rfloor - a_n\log_2(3)\to 0$. The limit under this subsequence will thus converge to infinity. Moreover, we can also pick the $(a_n)$ such that $\lfloor a_n\log_2(3) + 1 \rfloor - a_n\log_2(3)\to1$, which means the subsequence converges to $1$. Thus, the limit can't exist.

I am $99\%$ sure that this is a valid argument. However, to be precise, you have to define the $a_n$ in both cases and work out their respective limits.


Balls and bins problem with 5 labelled balls and 3 labelled bins Suppose we throw 5 labelled balls randomly into 3 labelled bins.
a) What is the probability that the first bin is empty?
b) What is the probability that at least two bins are empty?

I know that the sample space is k^n, in this case being 3^5, but I can't figure out how I can solve these 2.

Label the bins A,B,C

example a/ grants P(No balls land in A)=P(All balls land in B or C)=$(2/3)^5$

example b/ is a little more tricky, P(at least two bins are empty) = P(A is full) + P(B is full) + P(C is full) $= 3 \cdot (1/3)^5$

a) the first bin is empty:

you can fill the other two bins in 6 ways (we will consider only bin 2 because after choosing balls for bin 2, we have to put the rest of the balls in bin 3):

i) 5 balls in bin 2 and 0 balls in bin 3, you can do it in $^5C_5 = 1$ way

ii) 4 balls in bin 2 and 1 balls in bin 3,you can do it in $^5C_4 = 5$ way

iii) 3 balls in bin 2 and 2 balls in bin 3,you can do it in $^5C_3 = 10$ way

iv) 2 balls in bin 2 and 3 balls in bin 3,you can do it in $^5C_2 = 10$ way

v) 1 balls in bin 2 and 4 balls in bin 3,you can do it in $^5C_1 = 5$ way

vi) 0 balls in bin 2 and 5 balls in bin 3,you can do it in $^5C_0 = 1$ way

In total = (1+5+10+10+5+1) = 32 ways to keep the 1st bin empty.

Now the probability will be $$\frac{32}{3^5}$$.

(b) at least two bins are empty
it has two cases:

case-1 : only two bins are empty

case-2 : more than two bins (three here) are empty.

case-1: two bins are empty that means one bin is full. We can choose one bin from the 3 bins in $^3C_2 = 3$ way.

case -2: All the three bins are empty this can be done by only $1$ way.

so the probability of at least two bins empty $$=\frac{3+1}{3^5}=\frac{4}{3^5}$$


What is the intuition behind the unit normal vector being the derivative of the unit tangent vector? I've seen the math, but... It just doesn't make sense to me. How is the slope going to point perpendicular to the vector that is clearly a straight line not going in that direction?

Take some point $P_1 = C(s_1)$ on a curve $C$. Let $T_1$ be the unit tangent at this point. Now move a tiny distance along the curve, to get to the point $P_2 = C(s_2)$, and let $T_2$ be the unit tangent at this point. Now draw the vector $V = T_2 - T_1$. If $P_1$ and $P_2$ are very close together, $T_1$ and $T_2$ will be almost equal, and $V$ will be roughly perpendicular to both of them (because the triangle with sides $T_1$ and $T_2$ will be almost isosceles and very skinny). Here's a picture.

But, by the definition of derivative,
$$
\frac{dT}{ds} = \lim_{s_2 \to s1} \frac{T_2 - T_1}{s_2 - s_1}
$$
and this limit will also be roughly perpendicular to $T_1$.

This can be very confusing. A simple explanation is as follows: The (principle) unit Normal vector (dT/dt) is always orthogonal to the curve and, therefore, also to the unit Tangent vector T, and is contained in the vector subspace defined by the two unit vectors T and N. The acceleration of a particle on a curve is partioned between the unit Tangent and the unit Normal vector. Thus, acceleration = kT + mN. For example, a stright line has k=0, and there is  no acceleration in the normal direction. However, the unit Normal vector to the line is still well defined. The point is that the unit Normal vector is part of a coordinate system that may be used to define a component of acceleration.


Probability of presence of not defective products in a pack. In a pack of 50 products there mustn't be more than 2 defective products.
The probability of presence of 0, 1 or 2 defective products in the pack is the same.
Five randomly chosen products are not defective.
What's the probability of the rest of the products in the pack being not defective too?

My attempt to solution:
$$
   P_0 = P_1 = P_2 = \frac{1}{3} \\
   P_0 + P_1 + P_2 = 1
$$
Where $ P_i $ is the probability of presence of i defective products in the pack with 50 products.
I assumed that there could be present conservation law (hence the second equation).

The 5 selected products that are taken out of the pack form $ \frac{5}{50} = \frac{1}{10}\text{th} $ of the initial number of products in the pack.

If the 5 selected products are not defective, the $ P_{0} $ must have decreased in favour of $ P_1 \text{ and } P_2$ (the conservation law).

The question is: what's the new probability $ P'_{0} $ of no defective products being present in the pack?

Based on previous assumptions, I thought that $ P'_{0} = \frac{1}{3} - (\frac{1}{10}\cdot\frac{1}{3} ) = \frac{3}{10} $.

Let's denote the events of $0$, $1$, or $2$ defectives by $D_0$, $D_1$, $D_2$; and the event of getting $5$ goods by $G$.

We are interested in the conditional probability $P(D_0 \mid G)$.

We have
$$P(D_0 \mid G)=\frac{P(D_0\cap G)}{P(G)}$$
Noting that $D_0\subseteq G$, we obtain
$$P(D_0 \mid G)=\frac{P(D_0)}{P(G)}$$
We know $P(D_0)=\frac13$.

We can also calculate
$$ \begin{array}{lcl}P(G)&=&P(G\mid D_0)\cdot P(D_0)+P(G\mid D_1)\cdot   P(D_1)+P(G\mid D_2)\cdot P(D_2)\\
&=&1\cdot \frac13+\frac{9}{10}\cdot\frac13+\frac{48\choose 5}{50\choose 5}\cdot \frac13
\end{array}$$
You should be able to finish it from there.


Tricky Differential Equation Problem I am unsure of how to tackle the following differential equation:

$$ dx+ x\,dy = e^{-y}\sec^2y\,dy$$

I have done the following so far:

$$dx + x\,dy = e^{-y} \sec^2y \, dy$$

$$=>dx = e^{-y} \sec^2y \, dy - x \, dy$$

$$=>dx = (e^{-y} \sec^2y - x) \, dy$$

$$=>dx/dy = (e^{-y}) \sec^2y - x $$

Is this the correct approach? How can the problem be solved after this?

(Not homework: preparing for a test)

Hint: Rewrite/Simplify it as:

$$e^y~dx + (x e^y - \sec^2 y)~ dy = 0$$

Now, check this as being an Exact Equation to proceed.

Yeah! But if you just want to, you can also use the other method which is Linear Differential equation, where in the standard solution goes like this: dx/dy+xP(y)=Q(y)--> if eq. is linear in x; dy/dx+yP(x)=Q(x)--> if eq. is linear in y.

since the given problem is linear in x, use dx/dy+xP(y)=Q(y)
where P(y)=1 and Q(y)=e^(−y)sec^(2)y
and IF (Integrating factor): e^(integral of P(y) or e enter image description here

then use the formula:
x*IF=enter image description here(IF * Q(y))dy + C
thus,
x*e^(y)=tan^(2)y+C [this is the final answer in general form ]


Some help with sin and cos I'm having trouble to understand the following equalities in these two equations, i.e. how to apply the addition formulas.

Firstly:

$$ \frac {1- \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} {1+ \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} = \frac {cos^2(\frac x2) - sin^2(\frac x2)} {cos^2(\frac x2) + sin^2(\frac x2)}= cos^2(\frac x2) - sin^2(\frac x2)$$

and secondly:

$$ \frac {2tan(\frac x2)} {1+tan^2(\frac x2)} = 2* \frac {sin(\frac x2)} {cos(\frac x2)(1+\frac {sin^2(\frac x2)}{cos^2(\frac x2)})} = 2* \frac {sin(\frac x2)cos(\frac x2)} {cos^2(\frac x2)+sin^2(\frac x2)}= 2sin(\frac x2)cos(\frac x2) $$

I don't understand all the steps in those equations unfortunately. Can someone help me out here?

To get the first equality

$$\frac {1- \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} {1+ \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} \underbrace{=}_{(1)} \frac {cos^2(\frac x2) - sin^2(\frac x2)} {cos^2(\frac x2) + sin^2(\frac x2)}\underbrace{=}_{(2)} cos^2(\frac x2) - sin^2(\frac x2)$$

in step $(1)$ you multiply numerator and denominator by $\cos^2\frac x2$ (note that $\frac ab=\frac{ac}{bc}, \forall c\ne 0.$)

In step $(2)$ it is used the equality $\sin^2 t+\cos^2 t=1.$

To get the second equality

$$\frac {2tan(\frac x2)} {1+tan^2(\frac x2)} \underbrace{=}_{(1)} 2\frac {sin(\frac x2)} {cos(\frac x2)(1+\frac {sin^2(\frac x2)}{cos^2(\frac x2)})} \underbrace{=}_{(2)}2\frac {sin(\frac x2)cos(\frac x2)} {cos^2(\frac x2)+sin^2(\frac x2)}\underbrace{=}_{(3)} 2sin(\frac x2)cos(\frac x2)$$ note that:

$(1)$ is just the definition of tangent $\left(\tan t=\frac{\sin t}{\cos t}\right).$ In $(2)$ multiply numerator and denominator by $\cos \frac{x}{2}$ (note that, $\frac{a}{b}=\frac{ac}{bc}, \forall c\ne 0.$) In step $(3)$ it is used again the equality $\sin^2 t+\cos^2 t=1.$

(2) that's multiply numerator and denominator by $\left(\cos\left(\tfrac x2\right)\right)^2$.


Does $\lim_{x\to 0}\sqrt{x}$ exist? Since I'm an engineering student I have learned mathematics more in the intuitive way rather than formally. But I really like to understand mathematics formally and I'd like to take some courses with mathematicians. I've spent all my last semester trying to understand logic and set theory, and now I'm restudying again my course of calculus by myself. In this question I'm trying to understand the concept of limit. Then here is this thing that arose about the existence of $\lim_{x\to 0}\sqrt{x}$. In my course I learned that this doesn't exist because the funcion $f:(0,\infty)\longrightarrow \mathbb{R}$ such that $f(x)=\sqrt{x}$ is not defined  on the the values before $0$, but now I'm not sure if that is the case or if this is so for engineering but for mathematicians this is different. Apparentely the limit does exist and is $0$ for mathematicians. I'd like to hear about it and all your comments about the definition of limit related to this.

Edit: This is the formal definition for limit that I have so far from the answer to my quesiton I said above (from which I can conclude that the limit of the expression I'm asking about is $0$):

  Let $A, B\subseteq \mathbb{R}$ and $f:A\longrightarrow B$ a function such that $a\in \mathbb{R}$ is an acummulation point of $A$. Then we say that $l\in \mathbb{R}$ is the limit of the function $f$ when $x$ approches $a$ and is denoted by $\lim_{x\to a}f=l$ if and only if $\forall \epsilon\in \mathbb{R}(\epsilon>0)\exists\delta\in \mathbb{R}(\delta >0)\forall x\in A(0<|x-a|<\delta\longrightarrow |f(x)-l|<\epsilon)$.

Later LittleO told me that this is the definition found in Rudin's book of analyis: Definition 4.1

So, does $\lim_{x\to 0}\sqrt{x}$ exist or not? I'm really confused.

It is an unfortunate necessity that any practical notation has to leave some things implicit to be filled in by the reader.

This situation is one of those. There are two very useful things we can say here:

Every interval around $0$ has points where $\sqrt{x}$ is not defined.
On the domain where $\sqrt{x}$ is defined, it approaches 0 as $x$ decreases to zero.

If the author does not explain which he means, you have to infer his intention from context.

Yes this limit must be zero in all senses, even if you define multivalued functions, i.e., a function assuming two or more values.


Find $A$ given $A^2$ and $B$ and $(A^2)^{-1} = A^{-1}B$ This excersice took place in class I had today. The exercise was the following:

  Let the regular matrix $A$,$B$:

  $$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right]  \hspace{2cm}  B = \left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1&3&-1 \end{matrix} \right]$$

  Knowing that $(A^2)^{-1} = A^{-1}B$, find A.

My Attempt

During the exam I tried the following:

$(A^2)^{-1} = A^{-1}B \Leftrightarrow A^2(A^2)^{-1} = A^2A^{-1}B \Leftrightarrow I_n = AB \Leftrightarrow A = B^{-1}$

And procedeed to find the inverse of B, which is:

$A = B^{-1} = \left[ \begin{matrix} \frac{1}{2} &\frac{-5}{4} &  1\\ 0 & \frac{1}{2} & 0 \\ \frac{1}{2}& \frac{1}{4}&0 \end{matrix} \right]$

But, as I tend to make silly mistakes, I figured out that it would be a better way to do find $A$ just by using matrix products. We had $I_n = AB \Leftrightarrow A = A^2B$. Using this method I got that

$A = A^2B = \left[ \begin{matrix} 2 & 0 &  2\\ 2 & 12 & -6 \\ -4& -8&12 \end{matrix} \right]$

But as you can see the $2$ matrix are different. Where is the mistake in my logic?

Another thing I noticed is that $I_n = AB \Leftrightarrow A*I_n*B = A*(AB)*B \Leftrightarrow I_n = AB = A^2B^2$, but $A^2B^2$ isn't the identity matrix, so maybe the exercise is wrong, but I rather not rush into that thinking.

Use the properties of determinant to show that the given relationship does not hold.
$$\det ((A^2)^{-1})=-\frac1{64}=\det (A^{-1})\cdot(-4)$$
from the given relationship and properties of determinant. So $\det (A)=64\cdot 4$ and this cannot be true.

Long story short, matrix multiplication is not commutative.

Te equation $A=A^2B$ is not correct. From the equation $$(A^2)^{-1} = A^{-1}B$$

you have to multiply it by $A$ from the left to get
$$A(A^2)^{-1} = B$$

and the multiply it by $A^2$ from the right and you get

$$A=B\cdot A^2$$

which is not the same as $A=A^2B$.

However, even using $A=BA^2$ results in two different calculations of what $A$ is, and this seems to me like a mistake in the exercise. Indeed, from $(A^2)^{-1} = A^{-1}B$, it should follow that $I=A^2A^{-1}B=AB$, and from that, it should follow that $A=B^{-1}$. However, it is not true that $(B^{-1})^2=A^2$ which means that $A=B^{-1}$ cannot hold, which is a contradiction.


Expand in Taylor series $\frac{1}{1-\sin{x}}$ Expand in Taylor series $\frac{1}{1-\sin{x}}$

I have an idea that $\frac{1}{1-\sin{x}} = 1 + \sin {x} + \sin^2 {x} + \sin^3 {x} + \dots$

But I don't know what to do next. Every sine expands in infinity series...

Can anybody help?

$$\sin^n(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n=\frac1{(2i)^n}\sum_{k=0}^n\binom nke^{i(2k-n)x}=\\
\frac1{(2i)^n}\sum_{k=0}^n\binom nk\sum_{j=0}^\infty\frac{i^j(2k-n)^j}{j!}x^j,$$

and

$$\frac1{1-\sin(x)}=\sum_{n=0}^\infty\frac1{(2i)^n}\sum_{k=0}^n\binom nk\sum_{j=0}^\infty\frac{i^j(2k-n)^j}{j!}x^j.$$

Not the simplest on Earth, but usable for the first few terms.

You can use Maclaurin's expansion to expand in powers of x as follows:

$f(x)$=$f(0)$+$xf'(0)$+$(x^2$/$2!$)$f''(0)$+...

where $f(x)=1/(1-sinx )$


Proof of that in an integral domain, every prime element is irreducible. I would like to prove that in an integral domain $R$, every prime element $p$ is irreducible. I understand the case where $p = ab$ but the textbooks I have read do not address the case where $p \neq ab$, i.e.,$px = ab$.

I was wondering why they do not discuss the case where $px = ab$.

You don't need to examine the case $p\neq ab$.

In order to prove every prime element $p$ is irreducible you have to show IF $p=ab$, then $a$ or $b$ is an unit (see the definition of irreducible element).

However, if we have $p\neq ab$? It doesn't matter, we don't care, what matters to us is just the case whenever $p=ab$.

Proof:
suppose P is a prime element of an integral domain R, and p=ab
we know that, P|P => P|ab => P|a or P|b
              P|a and p=ab or a|P => a is unit
similarly P|b and p=ab ir b|p => b is unit,
so either a is a unit or b is a unit, hence p must be irreducible element,


How to prove $A \subset B \iff A \cup B = B$ Let A, B be two sets. Prove that $A \subset B \iff A \cup B = B$

I'm thinking of using disjunctive syllogism by showing that $\neg \forall Y(Y \in A).$ However, I'm not sure how the proving steps should proceed such that it leads me to that premise.
Edit: Thanks for the input. FYI, I need to prove this using predicate logic.

Taking the "longer" road. Let us review the definitions:

$A\subseteq B$ if and only if for every $x\in A$, $x\in B$.
$x\in A\cup B$ if and only if $x\in A$ or $x\in B$.
$A=B$ if and only if $A\subseteq B$ and $B\subseteq A$.
$P\iff Q$ means that if we assume that $P$ holds, then $Q$ must hold; and vice versa.

Now assume $A\subseteq B$. This means that for every $x\in A$ we have $x\in B$. We want to show that $A\cup B=B$.

So we take $x\in B$, then $x\in A$ or $x\in B$, and therefore $x\in A\cup B$.
Now take $x\in A\cup B$, we want to show that $x\in B$. By definition either $x\in A$ or $x\in B$.

If $x\in B$ we are done.
If $x\in A$ then by the assumption that $A\subseteq B$ we have that $x\in B$.

Either way we have that $x\in B$.

We have shown that if $A\subseteq B$ then $A\cup B\subseteq B$ and $B\subseteq A\cup B$, which is by fact number $3$ to say $A\cup B=B$.

Now we need to assume that $A\cup B=B$, and deduce that $A\subseteq B$. So we need to show that if $x\in A$ then $x\in B$.

Take $x\in A$ to be an arbitrary element. Because $x\in A$ we have that $x\in A$ or $x\in B$, and therefore $x\in A\cup B$. The assumption was, however, that $A\cup B=B$ and therefore we have that $x\in B$ as wanted.

Let  $A\subseteq B$.
Claim: We need to prove $A\cup B\subseteq B$ and $B\subseteq A\cup B$
Let  $x\in A\cup B$. Then  $x\in A$ or $x\in B$. But $A\subseteq B$. Hence $x\in B$. Therefore $A\cup B\subseteq B$.
Now conversely $x\in B$. Then $x\in A\cup B$. Therefore $B\subseteq A\cup B$. Thus $A\cup B=B$.
Consider $A\cup B=B$.
Claim: $A\subseteq B$
Let $x\in A$. Then  $x\in A\cup B=B$. Therefore $A\subseteq B$.


$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ I want to solve this limit:

$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$$

I have proved that $\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n} = 0$ and $\lim\limits_{n \rightarrow +\infty} \frac{1}{(1-\cos(1/n^2))}= \infty$ but I have indeterminate form. How can I solve that?

I would be inclined to write this as $\lim_{x\to\infty}\frac{\ln(1+ x+ x^3)- 3 \ln(x)}{x(1+ \cos(1/x^2)}$ where $x$ does not have to be an integer. That way the functions are differentiable and I can use "L'Hopital's rule".

We can write the numerator as $\ln(1+ x + x^3)- 3 \ln(x)= \ln(1+ x+ x^3)- \ln(x^3)= \ln\left(\frac{1 + x+ x^3}{x^3}\right)= \ln(x^{-3} + x^{-2} + 1)$.

Its derivative is $\frac{-3x^{-4}- 2x^{-3}}{x^{-3}+ x^{-2}+ 1}= \frac{-3- 2x}{x+ x^2+ x^3}$. Since the denominator has higher degree than the numerator, the limit as $n$ (and so $x$) goes to infinity is $0$.

We can write the denominator as $x(1+ \cos(1/x))= x(1+ \cos(x^{-1}))$.

Its derivative is $1+ \cos(x^{-1})+ x(-\sin(x^{-1}))(-x^{-2})= 1+ \cos(x^{-1})- x^{-1}\sin(x^{-1})$.

As $x$ goes to infinity all those "$x^{-1}$" terms go to $0$ so the denominator goes to $1+ 1- 0= 2$.

Since that denominator is nonzero, the limit of the original expression is $0$.


What is the name of the branch of Probability that deals with the experiments with Uncountably Infinite outcomes? I am having lectures that is repeatedly discussing the experiments like the following:

We draw two points $p$ and $q$ at random from the interval $[-1, 1]$. Let $x_1$ and $x_2$ denote the roots of the equation $x^2 + px + q = 0$.
Find the probability that:
(a) $x_1, x_2 \in \Bbb R$
(b) ${x_1}^2 + {x_2}^2 < 1$

My instructor says that, this is called Continuous Probability. But, I am not finding any chapter or topic of the name Continuous Probability in any books of introductory probability.
There are some topics named Continuous Random Variable, or, Continuous Probability Distribution. But, they seems to be not dealing with these types of problems. Am I missing something?
Where can I find materials to study and solve these Continuous Probability problems?
I am in trouble. Please, help me.

It's called a geometric probability, most of the introductory books cover this topic as an example but they don't dedicate much effort to it. If you want a dedicated book, you may look at (for example): Klain, Rota, Introduction to Geometric Probability, CUP (available online).


Draw plane via 3 points I have 3 points (1,1,1), (2,2,2), (2,2,0). It is enough to draw plane via these 3 points.
But how to do this in Matlab? I'm planning to use surf() function. I do:
x=[1 2 2 ]
y=[1 2 2 ]
[X,Y] = meshgrid(x,y)

it creates mesh :
X
1 2 2
1 2 2
1 2 2

y
1 1 1
2 2 2
2 2 2

Problems appeared while construct Z matrix:
Z
1 ? ?
? 2 ?
? ? 0

I know only 3 points , how to fill other ones defined as ? ?

I think you need the fill3 function. See below:
p1 = [1 1 1];
p2 = [2 2 2];
p3 = [2 2 0];

points=[p1' p2' p3'];
fill3(points(1,:),points(2,:),points(3,:),'r')
grid on


Predicate to prove a set is countably infinite Hi I have this question and have been struggling to find an answer.

  Prove that the set of numbers which are powers of 2 (i.e. $\{1, 2, 4, 8, 16, 32, \ldots\}$) is a countably infinite set.

Not sure if I've been over thinking it but I've been trying it for the last week and haven't got anywhere with it.

Hint: $f:\mathbb N \rightarrow A, f(n)=2^n$ is a bijection, where $A=\{1,2,4,8,..\}$

Assuming $\mathbb N$ contains zero.

This question has two parts:

Is the set infinite?
Is the set countable?

Number 1. you can do by contradiction, much like how Euclid proved that there are infinitely many primes. For 2., the definition of countable is that there exists a one-to-one function from your set to the integers. Inclusion is such a function in this case.


How to solve this functional equation given
$$ f(x)f(1/x) =f(x) + f(1/x) $$
and
$$  f(3) = 28 $$
then find the value of $f(4)$
the way i attempted the problem was that i found the value of $f(x)$ in terms of $f(1/x)$ which came put to be
$$ f(x) =\frac { f(\frac 1x)}{(f(\frac 1x) - 1)}$$
assuming that on putting$ x=3 \space \space \space , \text the \space \space RHS = 28$

So i assumed $f(\frac13)$ as $t$ and then solved for $t$ which gave me the value of $\frac {29}{27}$ so by judging this value ,  I  assumed
$$t =\frac{(9x+2)}{9x}$$
 now putting this value of $t$ which i assumed to be $f(1/x)$ in RHS the function simplified to become $$ f(x) = 9x + 1 $$ so putting $x =3 \space $,  I absolutely get $f(3) = 28$ but putting $x =4$ it doesn't gives $65$ which is the answer.

So can someone please help me which is the correct approach and method and tell where I went wrong ?

Give any $g:[1,\infty)\to\mathbb R\setminus\{1\}$ such that $g(1)=2,$ you can define:

$$f(x)=\begin{cases}g(x)&x\geq 1\\ \frac{g(1/x)}{g(1/x)-1}&x\in(0,1)\end{cases}$$

Then $f(x)$ satisfies your property.

In particular, you can have any value of $f(4)=g(4)$ given the known value of $f(3).$

If $f$ is known to be a polynomial, then if you define $p(x)=f(x)-1$ then you have:

$$p(x)p(x^{-1})=1$$

Then you can show that $p(x)=\pm x^m$ for some natural number $m.$

Note to OP: Check that $f(x)=9x+1$ does not satisfy the original functional equation, which it has to.

Let $f(x)=A, f(1/x)=B; A-1=g(x), B-1=g(1/x)$, then $$AB=A+B \implies (A-1)(B-1)=1 \implies g(x)=\frac{1}{g(1/x)} \implies g(x)=K x^m, K=\pm 1$$,
So $$g(x)=\pm x^m \implies f(x)=1\pm x^m$$ the given condition $f(3)=28$, fixes $m=3$ and  $+$ sign here. Finally, we get $f(x)=1+x^3.$


What is the nth house number if house numbers cannot end in 3 or 4? Newer Taiwan house numbers cannot end in 4. In order to not have one side of the
street grow faster than the other, we must also throw away those ending in  e.g., 3.

Is there a algebraic function f(n) that would give

  n   f(n)
-------------
  1   1
  2   2
  3   5
  4   6
  5   7
  6   8
  7   9
  8  10
  9  11
 10  12
 11  15

or is the only way to determine what number we should give a given nth
house is to run through the whole list up to n? (perl:)

my $t;
for ( 1 .. $ARGV[0] ) {
    $t += 2 if ++$t % 10 == 3;
}
printf "%3d %3d\n", $_, $t for $ARGV[0];

(Perhaps actually a plot to reserve numbers for invading Martian's pillboxes?!)

One might think this sounds like octal numbers, but alas we are only talking about the final digit here.

Perhaps you can use the relationship $h(n+8)=h(n)+10$ where $h(n)$ denotes the house number of the $n$-th house.  This then generalizes to $h(n+8k)=h(n)+10k$.

So for example, $h(79)=h(7+8\cdot 9)=h(7)+10\cdot 9=9+90=99$.

You should use this function:

$$f(n) = \begin{cases} n + 2 + 2 \lfloor \frac { n - 3} 8 \rfloor, & n \ge 3 \\
n, & 0<n<3 \end{cases} .$$


You use Rolle's theorem to prove n-polynomial has n distinct roots, but how would you do this question? Show that if a polynomial of degree n has n real roots (which don't have to be distinct) then its derivative has n-1 real roots (which also don't have to be distinct).

Let us consider the distinct roots $s_1,s_2,\ldots s_{\ell}$ with respective multiplicities $b_1,b_2,\ldots b_{\ell}$. Now see that $b_1+b_2+\ldots+b_{\ell}=n$ now the $\ell$ distinct roots give $\ell-1$ roots of the derivative and the multiplicities contribute $b_i-1$ each. Adding them up we get $\left(\displaystyle \sum_{i=1}^{\ell}b_i\right)+\ell-1=n-1$ roots.
$\blacksquare$

Suppose that $p(x)$ is a polynomial with $n$ ordered real roots, $a_1 \leq ...,\leq a_n$.  Then since $p$ is continuous on $[a_1, a_2]$, differentiable on $(a_1, a_2)$, and $p(a_1) = p(a_2)$ we can apply Rolle's theorem to conclude that there exists a real number, $c_1 \in (a_1, a_2)$ such that $f^\prime(c_1) = 0$.  Repeat this argument for the $n-2$ remaining intervals $(a_j, a_{j+1})$.


What does algebra $A \rtimes G$ stand for? Let $k$ be an algebraically closed field, $A$ a finite dimensional (unital associative) $k$-algebra of finite dimension, and $G$ a torus over $k$ acting on $A$.

What does $k$-algebra $A \rtimes G$ stand for? More specifically how to define addition, multiplication, and scaler multiplication?

The only usage of $\rtimes$ that I know is that of semidirect product and trivial extension algebra. But none of them make sense here. I came across this in the proposition 5.1 in this paper (alternative link); the author says the statement is classic and doesn't explain what he meant by the symbol. Could you help me?

In a semidirect product $A\rtimes B$, the group $B$ acts on $A$. That is, there is a homomorphism from $B$ into $\operatorname{Aut}(A)$.

Here, you have $G$ acting on $A$, but in a different context. So maybe that is motivating using the notation $A\rtimes G$.

With groups, $A\rtimes B$ is the collection of pairs $(a,b)$, and there is a new group operation $(a_1,b_1)(a_2,b_2)=(a_1a_2^{b_1},b_1b_2)$ where $a_2^{b_1}$ is the image of $a_2$ under the action from $b_1$.

In OP's setting $A\rtimes G$ is the collection of pairs $(a,g)$, with $k(a,g)=(ka,g)=(a,kg)$. You again can have $(a_1,g_1)(a_2,g_2)=(a_1a_2^{g_1},g_1g_2)$, making $A\rtimes G$ an algebraic group. I'm not sure you can get an algebra out of this though, since the $G$ has no addition.


Determine if matrices A and B are similar Determine if matrices $A$ and $B$ are similar.

$A=\begin{pmatrix}1 & 1 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 3 & 0 \\0 & 0 & 0 & 3\end{pmatrix}\: $

and

$B=\begin{pmatrix}1 & 1 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 3 & 1 \\0 & 0 & 0 & 3\end{pmatrix}$

I found the characteristic polynomials to be the same for both matrices

$(1-λ)(3-λ)(3-λ)(1-λ)$

so I went to plug in the values to find the eigenvectors for matrix $A$ and I think I may have screwed up because

$\begin{pmatrix}0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 2 & 0 \\0 & 0 & 0 & 2\end{pmatrix}$

I got an eigenvector for $λ=1$ to be $\begin{pmatrix}0 \\0 \\0 \\0\end{pmatrix}$ which I'm not even sure if I did it correctly or not because I don't think this is possible.

for $λ=3$ I got $\:\begin{pmatrix}1 \\\frac{1}{2} \\0 \\0\end{pmatrix}$

If I did do this correctly then is the geometric multiplicity for both = 1?

Both matrices are in Jordan normal form. This normal form is unique for all similar matrices. Henece $A$  and $B$ are not similar.

Besides, these Jordan normal forms tell us the eigenspace $E_1$ (for the eigenvalue $1$ has dimension $1$ for  both matrices, while the eigenspace $E_3$ has dimension $2$ for $A$ and $1$ for $B$.


Transition matrix (Markov chain) mean values If I have a n $\times$ n transition matrix and want to find the expected times passing the point j if the starting point is i and the amount of steps is m.

For example
$$P =\begin{bmatrix}0.5&0.5&0&0&0\\0.5&0&0.5&0&0\\0.5&0&0&0.5&0\\0&0&0.5&0&0.5\\0&0.5&0.5&0&0\end{bmatrix}$$

i = 1,
 j = 5,
 m = 100

Thank you.

The transition probabilities after $k$ steps become $\mathbf{P}^k$. Hence, the $(i, j)$ entry of matrix $\mathbf{N}$ below is the expected number of times the chain is in state $j$, given that the chain started in state $i$.

$$\mathbf{N}=\sum_{k=0}^m{\mathbf{P}^k}$$

If the matrix $\mathbf{I-P}$ is invertible, then the above sum can be expressed as $(\mathbf{I}-\mathbf{P}^{m+1})(\mathbf{I}-\mathbf{P})^{-1}$ using the geometric sum formula.
Otherwise, if $\mathbf{P}$ is diagonalizable
$$\mathbf{P}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}$$
where $\mathbf{\Lambda}$ is a diagnoal matrix with elements $\lambda_i$ that are eigen values of $\mathbf{P}$. The matrix $\mathbf{N}$ is then can be given by
$$\mathbf{N}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{-1}$$
where $\mathbf{D}$ is a diagonal matrix whose diagonal elements are given by $$d_i=\begin{cases}m+1,&\lambda_i=1\\\frac{1-\lambda_i^{m+1}}{1-\lambda_i},&\lambda_i\neq1\end{cases}$$


Sum of random decreasing numbers between 0 and 1: does it converge?? Let's define a sequence of numbers between 0 and 1. The first term, $r_1$ will be chosen uniformly randomly from $(0, 1)$, but now we iterate this process choosing $r_2$ from $(0, r_1)$, and so on, so $r_3\in(0, r_2)$, $r_4\in(0, r_3)$... The set of all possible sequences generated this way contains the sequence of the reciprocals of all natural numbers, which sum diverges; but it also contains all geometric sequences in which all terms are less than 1, and they all have convergent sums. The question is: does $\sum_{n=1}^{\infty} r_n$ converge in general? (I think this is called almost sure convergence?) If so, what is the distribution of the limits of all convergent series from this family?

Let $(u_i)$ be a sequence of i.i.d. uniform(0,1) random variables. Then the sum you are interested in can be expressed as
$$S_n=u_1+u_1u_2+u_1u_2u_3+\cdots +u_1u_2u_3\cdots u_n.$$
The sequence $(S_n)$ is non-decreasing and certainly converges, possibly to $+\infty$.

On the other hand, taking expectations gives
$$E(S_n)={1\over 2}+{1\over 2^2}+{1\over 2^3}+\cdots +{1\over 2^n},$$
so $\lim_n E(S_n)=1.$ Now by Fatou's lemma,
$$E(S_\infty)\leq \liminf_n E(S_n)=1,$$
so that $S_\infty$ has finite expectation and so is finite almost surely.

I just ran a quick simulation and I get a mean sum of one (standard deviation of 0.7)

Caveat: Not sure I coded it all right! Especially since I didn't test convergence.


Is there a sufficiently reachable plausibility argument that $\pi$ is irrational? I was teaching someone earlier today (precisely, a twelve-year-old) and we came upon a problem on circles. Little did I know in what direction it would lead. I was able to give a quick plausibility argument that all circles are similar and that they are related by some constant (most conventionally $π$). We came to the actual computation of $π$ and there was no way I could escape mentioning its irrationality. Indeed, being a somewhat bright student, this person asked, after I had written out the partial expansion $3.141592...$ whether she could not continue computing the digits, and how she could do this. Although I mentioned approximation by polygons as one possible approximation algorithm, she assured me that she would continue to find more digits, to which I asked, To what end? Indeed, I went on, it is impossible to finish computing the digits. But it must start to repeat after some digit, no matter how large, she retorted. No, I said. It is an irrational number (we had earlier talked about irrationality and it was easy enough showing a variant of the classic proof of the irrationality of $\sqrt 2,$ being a simple algebraic number). She then asked for reasons why this is the case. Suddenly, I found myself short of explanation as I had myself never bothered to study any of the known proofs in detail. I tried telling her that the known proofs were impossible for her to understand now, but she persisted nevertheless. Then I promised that I would come with it when next we met.

However, I am sure that this would not benefit her in any way. Therefore, I sought for a simple explanation (not necessarily a proof in the usual sense) that was sufficiently convincing, but so far I have found none. All I have found are variants on Lambert's or Hermite's proofs, and those are far from what I'm looking for.

In consequence, I thought to ask here. Perhaps someone has come across a similar situation and had found an argument sufficiently enlightening at that level (that is, a sketch of ideas or plausibility argument that can lead to a proof -- ideas can always be grasped by anyone, after all). In short, do you know any argument that I could present to this person that could at least slake their curiosity for now until they are ready for the classic proofs (if they continue to be interested)? If so please present them.

Thank you.

But it must start to repeat after some digit, no matter how large, she retorted.

This sounds like a misunderstanding that you could do something about by showing her a concretely defined different irrational where it is clear to see that the digits cannot repeat, such as
$$ \sum_{n=1}^\infty 10^{-n^2} $$
If you manage to convince her that is is possible for a number to have a decimal expansion that will never start (and continue!) repeating, it might become easier to accept that $\pi$ could be one of those numbers too.

Consider regular polygons with $2^n$ sides inscribed in a circle. To determine the ratio of the area of a $2^n$-gon to that of a square, one truncates Viète’s formula at an appropriate number of factors:

$$ \frac{A_4}{A_{2^n}} =\underbrace{{\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdots}_\text{$n-2$ factors} $$

The limit of the ratio is of course $\dfrac2\pi$. To the student, the factors should look obviously irrational. And while it takes a rigorous proof (Mathologer’s take on Lambert’s proof is probably the most accessible) to exclude the possibility that the product might somehow end up rational, it should be apparent that it would be exceptional for $\pi$ to be rational if its natural approximations are all irrational. At least the student should see that other quantities, clearly related to $\pi$, are indisputably irrational and it can’t be claimed that digits will eventually repeat, and the same can very likely be said for $\pi$ as well.


Can divergent series be useful? As explained by Terence Tao on his blog for example, it is possible to give a value to some divergent series using analytic continuation. For instance, that allows identities like

$$\sum\limits_{n \geq 1} 1 =  - \frac{1}{2}, \ \ \sum\limits_{n \geq 1} n = - \frac{1}{12} \ \ \text{and} \ \ \sum\limits_{n \geq 1} n^2 = 0.$$

My question is:

  Is it just an artificial definition or can it be really used in some computation or argument?

Stirling's approximation, so useful for approximating $n!$, is actually a divergent series.  The terms decrease initially, but for a given $n$ they start to increase after a while.  The error of the approximation is bounded by the first ignored term, so if you want the best for a given $n$ you just keep adding until they start to grow.  For most purposes the first term or two is all you need.

In addation to the post above:
You might reconise this as stirlings appoximation but it is actualy the first term of a divergent product.
$$n!= (n/e)^n  \sqrt{2n\pi}  \prod_{c=1}^{\infty}(1-c/n)$$

$$\prod_{c=1}^{\infty}(1-c/n)=e^{\sum_{c=1}^{\infty} ln(1-c/n)}$$
$$ e^{\sum_{c=1}^{\infty}\frac{\zeta(-c)(-1)^c}{(cn^c)}}$$
$$n!= (n/e)^n  \sqrt{2n\pi}   e^{\sum_{c=1}^{\infty}\frac{\zeta(-c)(-1)^c}{(cn^c)}}$$

This is the most ugly way to compute it, but also the most widely "accepted" i've the feeling. For me to just write out the product as sums is the most natural way.

$$\prod_{c=1}^{\infty}(1-c/n)=$$
$$1+\sum_{c=1}^{\infty}-c/n +$$
$$1/2!((\sum_{c=1}^{\infty}-c/n)^2-\sum_{c=1}^{\infty}(-c/n)^2)+$$
$$1/3!((\sum_{c=1}^{\infty}-c/n)^3-3*(\sum_{c=1}^{\infty}-c/n)*\sum_{c=1}^{\infty}(-c/n)^2+2*\sum_{c=1}^{\infty}(-c/n)^3)+... $$

And you will get$$ \prod_{c=1}^{\infty}(1-c/n)=1-1/(-12n)+1/(2*(-12n)^2)-1/6*(1/(-12n)^3+2*1/(120*n^3))+...$$


How to find the coefficients of an iteration of a polynomial function? Let $f(x)$ be a polynomial of degree $N$ with real coefficients $c_n$:

$$ f(x) = \sum_{n=0}^{N} x^n c_n$$

Further, let $f^{(1)}(x) =f(x)$, $f^{(2)}(x) =f(f(x))$ and so on ...

It is obvious that the degree of $f^{(i)}(x)$ is $N^i$.

Here is my question:
How do you calculate the coefficients of $f^{(i)}(x)$?
I dont know how to tackle this.

If for example $i=2$ then
$$f(f(x)) =  \sum_{n=0}^{N} c_n ( \sum_{k=0}^{N} x^k c_k)^n$$

But i dont get anywhere with this. Any help is appreciated.

Edit: See below for a much better way to derive the main result.
Represent the coefficients of a polynomial $p$ as a map $c_p$ from integers to real numbers, where $c_p(j)$ is the $j^{th}$ coefficient of $p$. Equivalently, $c_p$ is just the vector of coefficients of $p$.  Multiplying two polynomials $p$ and $q$ corresponds to taking a discrete convolution of the corresponding coefficient vectors:
$$
c_{pq}(i) = \sum_j c_p(i-j) c_q(j) = (c_p*c_q)(i)
$$
By the convolution theorem, the Fourier transform of a convolution is a product of the individual Fourier transforms.  Hence, the Fourier transform of the coefficient vector of $p^{k}$ (Caution! This is a regular power of $p$, not the same as $p^{(k)}$) is simply $\mathcal{F}[c_p^{*k}]=\mathcal{F}[c_p]^k$, where $\mathcal{F}$ is a discrete Fourier transform and $c_p^{*k}$ means $c_p$ convolved with itself $k$ times.
Now from the relation $p^{(k)}(x)=\sum_j c_p(j) \left(p^{(k-1)}(x)\right)^j$ we find that the coefficient vector $c_{p^{(k)}}$ satisfies $$c_{p^{(k)}}=\sum_j c_p(j) c_{p^{(k-1)}}^{*j}$$ hence the Fourier transform satisfies $$\mathcal{F}[c_{p^{(k)}}]=\sum_j c_p(j) \mathcal{F}[c_{p^{(k-1)}}^{*j}]=\sum_j c_p(j) \mathcal{F}[c_{p^{(k-1)}}]^j=p\left(\mathcal{F}[c_{p^{(k-1)}}]\right)$$
and inductively
$$\mathcal{F}[c_{p^{(k)}}]=p^{(k-1)}\left(\mathcal{F}[c_{p}]\right)$$
Hence $$c_{p^{(k)}}=\mathcal{F}^{-1}\left[ p^{(k-1)}\left(\mathcal{F}[c_{p}]\right) \right]$$
Using Fast Fourier Transforms, this is a very numerically efficient way to compute the coefficients you seek, as opposed to the brute force way of multiplying all the polynomials or convolving the coefficients directly.  (There may also be cases where the DFT can be computed analytically, so that this formula leads to closed form expressions for the coefficients, but this will not be the case in general.)

EDIT: A better derivation.
I recently found a much better way to derive the main result above.  As before, let $c_{p}(j)$ denote the $j^{th}$ coefficient of polynomial $p$.
By taking a (conventionally normalized) DFT we find $$\mathcal{F}[c_p](k) = \sum_j c_p(j) \left(e^{-2\pi i k/N}\right)^j = p\left(e^{-2\pi i k/N}\right)$$ where $N$ is the size of the DFT, and can be any integer larger than the degree of the polynomial $p$.  Hence, we have this expression for the coefficients of $p$:
$$c_p = \mathcal{F}^{-1}\left[ p\left(e^{-2\pi i k /N}\right) \right]$$
In particular, this works for iterated polynomials $p^{(n)}$:
$$\boxed{c_{p^{(n)}} = \mathcal{F}^{-1}\left[ p^{(n)}\left(e^{-2\pi i k /N}\right) \right]}$$
A few remarks:

A small variation on this works for analytic functions too.  You simply replace the DFT with a DTFT (i.e. let the size $N$ of the DFT $\rightarrow \infty$).  Then the expression for $c_p$ becomes $\mathcal{F}^{-1}\left[ p\left(e^{-2\pi i k}\right) \right]$, with $k\in [0,1]$ real valued.
The preceding expression can be viewed as Cauchy's integral formula.  Explicitly,
$$\begin{align*}
c_p(j) & = \mathcal{F}^{-1}\left[ p\left(e^{-2\pi i k}\right) \right] \\
& = \int_0^1 e^{2\pi i k j} p\left(e^{-2\pi i k}\right) dk \\
& = \frac{1}{2\pi i}\oint \frac{p(z)}{z^{j+1}} dz \\
& = \left. j! \frac{d^j p}{dz^j}\right|_{z=0}
\end{align*}
$$
where the third line follows from a change of variables and the last line is from Cauchy's derivative formula.
The boxed equation above can similarly be viewed as a discrete version of Cauchy's formula.

I implemented this in Mathematica to make sure it works, and sure enough it did. Something to watch out for if you try to implement this numerically is the normalization on your Fourier transforms.  We've assumed particular normalizations of the Fourier transform above, and if you use a different one (as is the default in Mathematica, it turns out), you need to multiply by a constant factor on every Fourier and inverse Fourier transform.  In Mathematica, that factor is $\sqrt{N}$, where $N$ is the length of vector you use to store the coefficients.  Define your own transform which is the default times $\sqrt{N}$, and your own inverse which is the default divided by $\sqrt{N}$, and everything will work out.  Also note that Mathematica's Fourier function is actually an inverse Fourier transform per standard conventions, and similarly InverseFourier is a Fourier transform.

The arbitrary polynomial can be presented via the vector of the coefficients
$\overrightarrow{C_n}$ in the form of
$$P_n\left(\overrightarrow{C_n},x\right) =
\begin{pmatrix}c_{n}\\c_{n-1}\\ \dots \\ c_2 \\c_1\\c_0\end{pmatrix}
\cdot
\begin{pmatrix}x^{n}\\x^{n-1}\\ \dots \\ x^2 \\ x\\ 1\end{pmatrix}
= \overrightarrow{C_n}\cdot \overrightarrow{X_n}.\tag1$$

On the other hand, the arbitrary polynomial can be presented via the roots in the form of
$$Q_m(\overrightarrow {R_m},x) = q\prod\limits_{j=0}^m (x-r_j).\tag2$$

If to consider the presentation $(1)$ as an operator, the presentation $(2)$ as an operand, and the superposition as the resulted operator, then the operator transformation can be presented as the production of the modified polynomials in the form of

$$P_{mn}\left(\overrightarrow{C_n},Q_m(\overrightarrow {R_m},x)\right)
 = q\prod\limits_{j=0}^m \left(
\begin{pmatrix}c_{n}\\c_{n-1}\\ \dots \\ c_2 \\c_1\\c_0-r_j\end{pmatrix}
\cdot
\begin{pmatrix}x^{n}\\x^{n-1}\\ \dots \\ x^2 \\ x\\ 1\end{pmatrix}
\right).\tag3$$

For example, let
$$f(x) = 2(x^2-3x+2),$$
then
$$P_2(x) = x^2-6x+4,\quad Q_2(x) = 2(x-1)(x-2),$$

$$f_2(x)=2(2x^2-6x+3)(2x^2-6x+2) = 8x^4 - 48x^3 + 92x^2 - 60x + 12,$$
$$f_3(x) = 2(8x^4 - 48x^3 + 92x^2 - 60x + 11)(8x^4 - 48x^3 + 92x^2 - 60x + 10)\\
= 4(32x^8 - 384x^7 + 1888x^6 - 4896x^5 + 7196x^4 - 6024x^3 + 2766x^2 - 630x + 55),$$
etc. (see also Wolfram Alpha calculations)

Therefore, the roots calculations looks the nesessary if we try to get useful closed form.

Now, let us use only the roots in the form $(2).$

  $$f_{(i)} = q^i\prod\limits_{j_1=0}^N (\prod\limits_{j_2=0}^N (\dots\prod\limits_{j_i=0}^N (x-r_{j_i})-\dots -r_{j_2})-r_{j_1}).\tag4$$

Ready

Calculations above gives result in the suitable form and does not require special algorithm.

At the same time, applying of the FFT algoithm has some features:

as was pointed in the previous answer, using of the convolution for the polynomials mulplying requires to reorder one of them;
since the convolution algorithm calculates cyclic convolution, then the initial arrays should be filled zeros from the end to reserve the place for all coefficients of the resulting polynomial;
since DFT transform is a linear one, then, applying FFT templates for the constant, is possible to avoid the waste calls.


How to solve $\lim_{x \to \ 0^+} (\mid \ln(x) \mid)^{\frac{1}{x}}$ I want to solve $$\lim_{x \to \ 0^+} (\mid \ln(x) \mid)^{\frac{1}{x}}$$.

Thanks for help.

Hint: $$\left(|\ln(x)| \right)^{\frac{1}{x}}=\exp\left(\dfrac{1}{x}\ln(|\ln(x)|) \right)$$
One can compute this limit as a composition of well known limits.


Contraction semigroup on $X$ with generator $A$ Let $\{S(t)\}_{t \ge 0}$ be a contraction semigroup on $X$, with generator $A$. Inductively define $D(A^k):=\{u\in D(A^{k-1}) \mid A^{k-1}u \in D(A)\}$ ($k=2,\ldots$). Show that if $u \in D(A^k)$ for some $k$, then $S(t)u \in D(A^k)$ for each $t \ge 0$.

This is from PDE Evans, 2nd edition: Chapter 7, Exerise 15.
For reference, here is Theorem 1. This is on page 435 in the textbook.

THEOREM 1 (Differential properties of semigroups). Assume $u \in D(A)$. Then
(i) $S(t)u\in D(A)$ for each $t \ge 0$
(ii) $AS(t)u=S(t)Au$ for each $t \ge 0$.
(iii) The mapping $t \mapsto S(t)u$ is differentiable for each $t>0$.
(iv) $\frac d{dt}S(t)u = AS(t)u \quad (t > 0)$.

May I receive feedback on my following proof to this problem?
My attempted proof:
If $u \in D(A^k)$, then $u \in D(A^{k-1})$ and $A^{k-1}u \in D(A)$. If $u \in D(A^{k-1})$, then $u \in D(A^{k-2})$ and $A^{k-2}u \in D(A)$. Inductively proceeding, we obtain: If $u \in D(A^2)$, then $u \in D(A)$ and $Au \in D(A)$. By parts (i) and (ii) of the theorem, $S(t)u \in D(A)$ and $AS(t)u=S(t)Au \in D(A)$, respectively; hence, $S(t)u \in D(A^2)$. Inductively proceeding back to $k$, we obtain: $S(t)u \in D(A^{k-1})$ and $A^{k-1}S(t)u \in D(A)$; hence, $S(t)u \in D(A^k)$.

Yes, basically that's it.

It might be only a typo, but using part (i) for $A^{k-1}u\in D(A)$ yields $S(t)A^{k-1}u\in D(A)$, and thus you need the exchange rule ($A^{k-1}S(t)u=S(t)A^{k-1}u$) to conclude $u\in D(A^k)$.

Note also that you implicitly used $S(t)A^{d}u\in D\left(A^{k-d}\right)$ for each $d=0\dots k$. This might need a proof.


Find the exact value of $\tan(\cos^{-1} (-\sqrt{3}/2))$. The question: Find the exact value of $\tan(\cos^{-1} (-\sqrt{3}/2))$.

The link is the image of the method I used. However it isn't the right answer, how come this method doesn't work?

Use the famous identity $$\tan(\arccos(x)) = \frac{\sqrt{1-x^2}}{x}$$ I think you can take it from your here. You can check the answer after you have done it. The answer is : $$\frac{-\sqrt{3}}{3}$$

Follow :
$$ \tan (\arccos(- \frac{\sqrt{3}}{2}))$$
Suppose $x = \arccos(- \frac{\sqrt{3}}{2})$ :
$$ \cos x = - \frac{\sqrt{3}}{2} = \frac {\text{Adjacent}}{\text{hypotenuse}}$$
Pythagorean theorem:
$$\text{opposite side} = \pm~\sqrt{4-3} = \pm 1$$
$$ \tan x = \frac {\text{Opposite}}{\text{Adjacent}} =  \pm \frac{1}{\sqrt{3}} $$
Considering $ 0 \le x \le \pi$ . Take only negative as $\tan$ is negative in second quadrant :
$$- \frac{1}{\sqrt{3}}$$


Proving $x\leq \tan(x)$ How can i prove that $x\leq \tan(x)$ for any $x$ in $[0,\frac{\pi}{2})?$

Define

$$f(x):=x-\tan x\implies f'(x)=1-\frac1{\cos^2x}=\frac{\cos^2x-1}{\cos^2x}\le0\implies f(x)$$

monotonic decreasing, and since $\;f(0)=0\;$, we get for $\;x\in[0,\pi/2)\;$ that

$$x-\tan x=f(x)\le f(0)=0\implies x\le \tan x$$

You can't prove it because it is false:$$
\frac{3\pi}{4} > \tan\left( \frac{3\pi}{4} \right)=-1
$$
And in case you want the $\leq$ sign as in the title,
$$
\frac{\pi}{4} < \tan\left( \frac{3\pi}{4} \right)=1
$$


What is the intuition behind uniform continuity? There’s another post asking for the motivation behind uniform continuity. I’m not a huge fan of it since the top-rated comment spoke about local and global interactions of information, and frankly I just did not get it.

Playing with the definition, I want to say uniform continuity implies there’s a maximum “average rate of change”. Not literally a derivative, but the rate of change between two points is bounded in the domain. I’m aware that this is essentially Lipschitz continuity, and that Lipschitz implies uniform. This implies there’s more to uniform continuity than just having a bounded average rate of change.

And also, how is it that $ f(x)=x$ is uniform yet $f(x)f(x)=g(x)=x^2$  is not? I understand why it isn’t, I can prove it. But I just don’t understand the motivation and importance of uniform continuity.

The real "gist" of continuity, in its various forms, is that it's the "property that makes calculators and measurements useful". Calculators and measurements are fundamentally approximate devices which contain limited amounts of precision. Special functions, like those which are put on the buttons of a calculator, then, if they are to be useful, should have with them some kind of "promise" that, if we only know the input to a limited amount of precision, then we will at least know the output to some useful level of precision as well.

Simple continuity is the weakest form of this. It tells us that if we want to know the value of a target function $f$ to within some tolerance $\epsilon$ at a target value $x$, but using an approximating value $x'$ with limited precision instead of the true value $x$ to which we may not have access or otherwise know to unlimited precision, i.e. we want

$$|f(x) - f(x')| < \epsilon$$

then we will be able to have that if we can make our measurement of $x$ suitably accurate, i.e. we can make that

$$|x - x'| < \delta$$

for some $\delta > 0$ which may or may not be the same for every $\epsilon$ and $x$.

Uniform continuity is stronger. It tells us that not only do we have the above property, but in fact the same $\delta$ threshold on $x'$'s accuracy will be sufficient to get $\epsilon$ worth of accuracy in the approximation of $f$ no matter what $x$ is. Basically, if the special function I care about is uniform continuous, and I want 0.001 accuracy, and the max $\delta$ required for that is, say, 0.0001, by measuring to that same tolerance I am assured to always get 0.001 accuracy in the output no matter what $x$ I am measuring. If, on the other hand, it were the case that the function is merely continuous but not uniformly so, I could perhaps measure at one value of $x$ with 0.0001 accuracy and that accuracy would be sufficient to get 0.001 accuracy in the function output, but if I am measuring at another, such a tolerance might give me only 0.5 accuracy - terrible!

Lipschitz continuity is even better: it tells us that the max error in approximating $f$ is proportional to that in approximating $x$, i.e. $\epsilon \propto \delta$, so that if we make our measurement 10 times more accurate, say (i.e. one more significant figure), we are assured 10 times more accuracy in the function (i.e. gaining a significant figure in the measurement lets us gain one in the function result as well).

And in fact, all the functions (that are real-analytic, not combinatorial functions like nCr and what not) on your real-life calculator are at least locally Lipschitz continuous, so that while this proportionality factor (effectively, absolutely how many sig figs you get for a given number of such in the input) may not be the same everywhere, you can still be assured that in relative terms, adding 10x the precision to your measurements, i.e. one more significant figure, will always make the approximation (however good or not it actually is) returned by your calculator 10x more accurate, i.e. also to one more significant figure.

And to top it all off, all these forms of continuity - at least in their local variants, that is, over any bounded interval - are implied by differentiability.

One interpretation that I love is the one using non-standard analysis:

Let $f : E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Then the followings are equivalent:

$f$ is uniformly continuous.
Its hyperreal version ${{}^*\!f} : {{}^*\!E} \subseteq {}^*\mathbb{R} \to {}^*\mathbb{R}$ is continuous, i.e., whenever $x, y \in {{}^*\!E}$ are infinitely close, ${{}^*\!f}(x)$ and ${{}^*\!f}(y)$ are infinitely close as well.

Here, the 'hyperreal version' refers to the $*$-transform of $f$. The catch is that the domain ${{}^*\!E}$ of ${{}^*\!f}$ contains hyperreal numbers which are either infinitely close to $E$ or infinitely large. So the continuity needs to be tested around those numbers as well to establish uniform continuity.
This statement can be recast in terms of real numbers only:

Let $f : E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Then the followings are equivalent:

$f$ is uniformly continuous.

For any sequences $(a_n)$ and $(b_n)$ in $E$ such that $|a_n - b_n| \to 0$, we have $|f(a_n) - f(b_n)| \to 0$.

In this version, the role of infinitely close hyperreal numbers is replaced by a pair of sequences which become arbitrarily close. Again, we see that uniformly continuity is really about enforcing continuity infinitesimally beyond $E$. Here are some examples that demonstrate this idea:
Example 1. Let $f(x) = 1/x$ on $(0, \infty)$. If we pick two 'infinitesimals' $a_n = \frac{1}{n}$ and $b_n = \frac{1}{n+1}$, then they become arbitrarily close but $|f(a_n) - f(b_n)| = 1 \not\to 0$. So $f$ is not uniformly continuous.
Example 2. Let $f(x) = x^2$ on $\mathbb{R}$. If we pick two 'infinitely large' $a_n = n$ and $b_n = n+\frac{1}{n}$, then $a_n$ and $b_n$ become arbitrarily close, but $|f(a_n) - f(b_n)| = 2 + \frac{1}{n} \not\to 0$. So $f$ is not uniformly continuous.


Simplify $(1+i^\frac{1}{2})^\frac{1}{2}$ How can I simplify $\sqrt{1+\sqrt{i}}$?

I thought about making $z^2=1+\sqrt{i}$ and then $w=z^2$

But I'm not really sure

$$(1+i^\frac{1}{2})^\frac{1}{2}=(1+(e^{i\frac\pi2+i2\pi n})^\frac12 )^\frac12
=(1+e^{i\frac\pi4+i\pi n} )^\frac12$$

$$=\left(1+\cos(\frac\pi4 +\pi n)+ i \sin(\frac\pi4 +\pi n) \right)^\frac12
=(re^{i\theta})^\frac12\tag 1$$

where,
$$r=\sqrt{\left(1+\cos(\frac\pi4 +\pi n)\right)^2+\sin^2(\frac\pi4 +\pi n) }
=\sqrt{2+2\cos(\frac\pi4 +\pi n) }=2| \cos(\frac\pi8 +\frac{\pi n}2) |$$

$$ \tan\theta = \frac{\sin(\frac\pi4 +\pi n)}{1+\cos(\frac\pi4 +\pi n)}
= \tan (\frac\pi8 +\frac{\pi n}2)\implies \theta =\frac\pi8 +\frac{\pi n}2+k\pi $$

Substitute $r$ and $\theta$ into (1),

$$(1+i^\frac{1}{2})^\frac{1}{2}
=r^\frac12 e^{i\frac{\theta}2} = \sqrt{2| \cos(\frac\pi8 +\frac{\pi n}2) |}
\>e^{i\pi(\frac1{16} +\frac{ n}4+\frac{k}2)}$$

which assumes multiple values. For the special case $n=k=0$,

$$(1+i^\frac{1}{2})^\frac{1}{2}
= \sqrt{2\cos\frac\pi8}\>e^{\frac{i\pi}{16}}
=\sqrt{2\cos\frac\pi8}\>(\cos\frac{\pi}{16} + i\sin\frac{\pi}{16})$$

Using polar coordinates $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt{1+\sqrt{e^{i \pi/2}}} = \sqrt{1 + e^{i \pi/4}} = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} \, $ from this we have

$\displaystyle r^2 = \left( 1 + \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 = 1 + \sqrt{2} + \frac{1}{2} + \frac{1}{2} = 2 + \sqrt{2} \implies r = \sqrt{2 + \sqrt{2}} $

and

$\displaystyle \tan \theta  = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1 \implies \theta = \arctan(\sqrt{2} - 1)$

So, $\displaystyle \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} = \sqrt{re^{i\theta}} = \sqrt{r} \, e^{i\theta/2}$

Finally, $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt[4]{2 + \sqrt{2}}  \, \left( \cos\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) + i\sin\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) \right)$

We can also obtain cartesian form:

$\displaystyle a + ib = \sqrt{1+\sqrt{i}}  = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}}$ yields

$\displaystyle a^2 + b^2 +i(-2ab) = 1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \implies a^2 + b^2 = 1 + \frac{\sqrt{2}}{2}$ and $\displaystyle -2ab = \frac{\sqrt{2}}{2}$. So we have

$\displaystyle (a-b)^2 = 1 + \sqrt{2} \,$ and $\displaystyle (a+b)^2 = 1 $ which gives $\displaystyle a = \frac{1 + \sqrt{1 + \sqrt{2}}}{2} $ and $\displaystyle  b = \frac{-1 + \sqrt{1 + \sqrt{2}}}{2}$


Integer solutions to $x^3=y^3+2y+1$? Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$

My approach:

I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.

Hint: if $y>0$, then $y^3< y^3+2y+1< (y+1)^3$, so the RHS expression cannot be a perfect cube. A similar idea works if $y$ is a small enough negative number, but some negative numbers close to $0$ (or indeed $0$ itself) can provide a solution.

Try to find a lower bound, and then check the remaining possible values.

$$x^3=y^3+2y+1\tag{1}$$
We are after integer solutions of $(1)$. On setting $y=x-a$, where $a\in\mathbb{Z}$, we have from $(1)$
$$x^3=(x-a)^3+2(x-a)+1=x^3-3a x^2+(3a^2+2)x-a^3-2a+1\tag{2}$$
which rearranges to
$$3a x^2-(3a^2+2)x+a^3+2a-1=0\tag{3}$$
Solve for $x$
\begin{align*}
x&=\frac{(3a^2+2)\pm\sqrt{(3a^2+2)^2-12a(a^3+2a-1)}}{6a}\\
&=\frac{(3a^2+2)\pm\sqrt{-3a^4-12a^2+12a+4}}{6a}\tag{4}
\end{align*}
The discriminant is
\begin{align*}
-3a^4-12a^2+12a+4&=-3a^4-4(3a^2-3a-1)\\
&=-3a^4-4[3\left((a-\tfrac12)^2-\tfrac7{12}\right)]\\
&=-3a^4-12(a-\tfrac12)^2+7\\
&=7-(3a^4+3(2a-1)^2)
\end{align*}
Now check various integral $a$ for solutions: $a=0$ gives division by zero in $(4)$; $a=1$ gives $x_+=1$ or $x_-=\frac23$; $a=-1$ gives  discriminant of $-23$; and for $|a|\ge2$ we get the discriminant $7-(3a^4+3(2a-1)^2)<0$ (since $3a^4+3(2a-1)^2$ is nonnegative).

Hence $x=1$, $y=0$ is the only solution.


L'Hospital's rule exception $$ \lim_{x\to-2}\frac{x^3-3x^2-10x}{(x+2)^2}$$ in this problem I tried using L'Hospital's rule but the outcome is not true why? Are there any exceptions to this rule?

The rule says that $\lim _{x \to a} \frac {f(x)} {g(x)}=\lim _{x \to a} \frac {f'(x)} {g'(x)}$ if the original limit is in the form $\frac 0 0$ and $\lim _{x \to a} \frac {f'(x)} {g'(x)}$ exists. This last condition is important. In this case this condition  is not satisfied and you cannot apply the rule.


Proof of an equivalence about powers of $2$ and divisors I found this equivalence:

  For all natural numbers $n > 1$,
  $\exists k\in\mathbf N, n=2^k \iff \neg (\exists p\in\mathbf Z, (2p+1)\ne\pm 1 \wedge (2p+1)\mid n)$

in my textbook. The proof has been omitted. I will like to see the proof and run through it myself. I have tried to search on the web but could not find any.

P.-S. I have translated the equivalence into English from the original language.

This is just not true. Take e.g. $n=6$.

$ n = 6 = 2 . 3$  but $n$ has an odd divisor other than $+1$ and $-1$.
This divisor is 3.

Probably in the statement should have been written $∃k \in N,n=2^k$ instead of $n=2k$.

Almost certainly it's a typo, that typo.

Proof:

1) If $n$ has no odd divisor other than $+1$ and $-1$, then n has no prime odd divisor. Then since n>1, it means that only the prime $2$ divides $n$. So the left side follows.

2) If $n=2^k$ then obviously it has no odd prime factor. So it cannot have an odd divisor other than $+1$ and $-1$.

All in all, this is too obvious if you have already studied the fundamental theorem of arithmetic.

Now... I don't know if you know that theorem or not. I used it in this proof (in fact it makes the proof trivial).

Maybe you're looking for a proof which does not use that theorem, not sure.

Some one has edited the equivalence.
A moderator maybe? I find this very odd and I'm not sure what to do.
The original equivalence was
$\exists k\in\Bbb N: n=2^k\iff\neg(n\text{ has other odd divisors than}\pm 1)$


What is the real thing when using Taylor to prove limits, such as $\lim_{x \to 0}\frac{\sin x-x}{x^3}=\frac16$? I see some answers in this site proving many limits by Taylor, for example

$$\lim_{x \to 0}\dfrac{\sin x-x}{x^3}=-\dfrac16$$

comes to my mind. However, my teacher didn't teach me this technique, also the popular calculus/analysis books I have do not even mention this technique. When people use Taylor to compute this limit, they often use big-$O$ operations, or omitting the terms having high order. However, what are the big-$O$ techniques? How does they work? Why is this rigorous? And how to choose where the omitting occurs? Should I use Taylor "expansions" (with Lagrange remainder) or Taylor "series"? I totally have no idea. Need your help.

Formally, a Taylor expansion up to the $k$-th term means that

$$f(x+h)=f(x)+hf'(x)+\cdots+\frac 1{k!}h^k f^{(k)}(x)+\color{red}{o(h^k)}$$

where the little $o$-term is a placeholder for a not further determined function $g(h)$ with

$$\lim_{h\to 0}\frac{g(h)}{h^k}=0$$

Because this fraction vanishes for small $h$, it vanishes in limits as the one in your example. This is useful in situations where we have an approximation for a certain function which is not perfect, but we want to express that we know this and that we can say how good/bad it approximates.

Example. Choose $f(x)=\sin(x)$ and observe

$$\sin(x)=\sin(0)+x\cos(0)-\frac{x^2}2\sin(0)-\frac{x^3}6\cos(0)+o(x^3)=x-\frac{x^3}6+o(x^3).$$

Now consider your example:

\begin{align}
\frac{\sin(x)-x}{x^3}
&=\frac{x-\frac{x^3}6+o(x^3)-x}{x^3}\\
&=\frac{-\frac{x^3}6+o(x^3)}{x^3}\\
&=-\frac16+\frac{o(x^3)}{x^3}\\
\end{align}

and we know that $o(x^3)/x^3\to0$ for $x\to 0$. So the limit is $-1/6$. See that we do not actually need to know what the function $o(x)$ stands for, as we only need that it vanishes in certain limits. This is the power of the Landau $\mathcal O$-notation (I could have used $\mathcal O(x^{4})$ instead, but I personally find $o(x^3)$ more appealing here). Of course the really powerful result here is Taylor's theorem which provides you with this $o(x^3)$ term in the first place.

Hint: $\sin(x)=x-x^3/6+ O(x^4)$ This implies that ${{\sin(x)-x}\over x^3}=-1/6+O(x)$.

https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions


Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:

Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:

$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sqrt{n^2+n} + (n+\frac{1}{2})}$

And I'm not sure how to go on from here. Help would be appreciated.

Hint:
\begin{align}
\sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}
\end{align}
and
\begin{align}
\left|\frac{n}{\sqrt{n^2+n}+n} -\frac{1}{2}\right| = \left|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+2n}\right| = \frac{n}{2(\sqrt{n^2+n}+n)^2} \leq \frac{n}{8n^2}
\end{align}

$$\sqrt{n^2+n}-n = n(\sqrt{1+1/n}-1)$$

Which by the binomial theorem is equal to

$$n\left(-1 + \sum_{k=0}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right) = n\left(\sum_{k=1}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right)$$

(With the product equal to 1 for $k = 0$.)  So now move the n inside the sum to get

$$\sum_{k=1}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}} = \frac{1}{2} + \sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}$$

So $\epsilon$ would be the absolute value of $\sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}$, and all that remains is to show this approaches zero for sufficiently large n.  Well, we'll start by saying...

$$\left|\sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}\right| = \left|\sum_{k=2}^\infty\frac{(-1)^{k-1}\cdot\frac{1}{2}\cdot\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}}\right| < \sum_{k=2}^\infty\frac{\frac{1}{2}\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}} < \sum_{k=2}^\infty\frac{(k-1)!}{2\cdot k!n^{k-1}} = \sum_{k=2}^\infty\frac{1}{2kn^{k-1}}$$

And I should hope you can take it from there.


Why is the eigenvector of a covariance matrix equal to a principal component? If I have a covariance matrix for a data set and I multiply it times one of it's eigenvectors.  Let's say the eigenvector with the highest eigenvalue.  The result is the eigenvector or a scaled version of the eigenvector.

What does this really tell me?  Why is this the principal component?  What property makes it a principal component?  Geometrically, I understand that the principal component (eigenvector) will be sloped at the general slope of the data (loosely speaking).  Again, can someone help understand why this happens?

Short answer: The eigenvector with the largest eigenvalue is the direction along which the data set has the maximum variance. Meditate upon this.
Long answer: Let's say you want to reduce the dimensionality of your data set, say down to just one dimension. In general, this means picking a unit vector $u$, and replacing each data point, $x_i$, with its projection along this vector, $u^T x_i$. Of course, you should choose $u$ so that you retain as much of the variation of the data points as possible: if your data points lay along a line and you picked $u$ orthogonal to that line, all the data points would project onto the same value, and you would lose almost all the information in the data set! So you would like to maximize the variance of the new data values $u^T x_i$. It's not hard to show that if the covariance matrix of the original data points $x_i$ was $\Sigma$, the variance of the new data points is just $u^T \Sigma u$. As $\Sigma$ is symmetric, the unit vector $u$ which maximizes $u^T \Sigma u$ is nothing but the eigenvector with the largest eigenvalue.
If you want to retain more than one dimension of your data set, in principle what you can do is first find the largest principal component, call it $u_1$, then subtract that out from all the data points to get a "flattened" data set that has no variance along $u_1$. Find the principal component of this flattened data set, call it $u_2$. If you stopped here, $u_1$ and $u_2$ would be a basis of the two-dimensional subspace which retains the most variance of the original data; or, you can repeat the process and get as many dimensions as you want. As it turns out, all the vectors $u_1, u_2, \ldots$ you get from this process are just the eigenvectors of $\Sigma$ in decreasing order of eigenvalue. That's why these are the principal components of the data set.

If we would project our data $D$ onto any vector $\vec{v}$, this data would be obtained as $\vec{v}^{\intercal} D$, and its covariance matrix then becomes $\vec{v}^{\intercal} \Sigma \vec{v}$.

Since the largest eigenvector is the vector that points into the direction of the largest spread of the original data, the vector $\vec{v}$ that points into this direction can be found by choosing the components of the resulting covariance matrix such that the covariance matrix $\vec{v}^{\intercal} \Sigma \vec{v}$ of the projected data is as large as possible.

Maximizing any function of the form $\vec{v}^{\intercal} \Sigma \vec{v}$ with respect to $\vec{v}$, where $\vec{v}$ is a normalized unit vector, can be formulated as a so called Rayleigh Quotient. The maximum of such a Rayleigh Quotient is obtained by setting $\vec{v}$ equal to the largest eigenvector of matrix $\Sigma$.

In other words; the largest eigenvector of $\Sigma$ corresponds to the principal component of the data.

If the covariances are zero, then the eigenvalues are equal to the variances:

If the covariance matrix not diagonal, the eigenvalues represent the variance along the principal components, whereas the covariance matrix still operates along the axes:

An in-depth discussion (and the source of the above images) of how the covariance matrix can be interpreted from a geometrical point of view can be found here: http://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/


Change a 3D plane to XY plane I have a 3D cloud of points from witch I determine an "average" plane.

The plane equation is Ax + By + Cz + D =0;

I would like to re-compute all the points to obtain their new position considering that "average" plane is the new XY plane.

I suppose I have to do first a Z translation of -D/C and then a rotation.

Can someone give me the [3*3] rotation matrix?

Thank you Aretino for your answer.

I test your solution but I have a problem in some cases.  In my test, I first do a Z translation of –d/c to obtain a plan eq of ax + by + cz = 0 form then I do the 3D rotation provided in your answer.

It seems that, in some cases, I need to invert the sign of all my a/b/c/d plane coef. I can't understand the need of that inversion.

For example if I use the 3 following points:

P1[-18.17;-15.97;26.33]

P2[-42.52;10.93;19.09]

P3[7.84;10.59;46.09]

The plane eq is: 0.013586 x + 0.005496 y -0.025271 z + 1.000000 = 0

The compute matrix WITHOUT plane coefficients’ inversion is:

-0.602725  -0.648415   -0.465060

-0.648415  0.737670    -0.188150

0.465060   0.188150    -0.865054

The WRONG resulted points are (WITHOUT plane coefficients’ inversion):

-9.341623  -12.398300  -68.463539

-8.740604  24.596144   -68.463539

-51.429983 -13.388881  68.463539

The compute matrix WITH plane coefficients’ inversion (a=-a, b=-b...) is:

-0.602725  -0.648415   -0.465060

-0.648415  0.737670    -0.188150

0.465060   0.188150    -0.865054

The GOOD resulted points are (WITH plane coefficients’ inversion):

15.334650   -2.415010   68.463539

-10.820679 23.754607   68.463539

46.271901  26.138410   68.463539

You can find a 3D view of the result in the image provided.

Another question / remark: why the Z coordinates are not null after the transformation?

$\vec{n}=(A,B,C)$ is a vector perpendicular to your average plane, forming an angle $\theta$ with the $z$-axis.
Consider the vector
$\vec{n}'=(B,-A,0)$, which is perpendicular to the plane formed by $\vec{n}$ and $z$-axis: a rotation of angle $\theta$ around $\vec{n}'$ will carry $\vec{n}$ along the $z$-axis and is thus what you are looking for. The rotation matrix is then:
$$
\left(
\begin{array}{ccc}
 \frac{B^2}{A^2+B^2}+\frac{\left(1-\frac{B^2}{A^2+B^2}\right) C}{\sqrt{A^2+B^2+C^2}} &
   -\frac{A B \left(1-\frac{C}{\sqrt{A^2+B^2+C^2}}\right)}{A^2+B^2} &
   -\frac{A}{\sqrt{A^2+B^2+C^2}} \\
 -\frac{A B \left(1-\frac{C}{\sqrt{A^2+B^2+C^2}}\right)}{A^2+B^2} &
   \frac{A^2}{A^2+B^2}+\frac{\left(1-\frac{A^2}{A^2+B^2}\right) C}{\sqrt{A^2+B^2+C^2}}
   & -\frac{B}{\sqrt{A^2+B^2+C^2}} \\
 \frac{A}{\sqrt{A^2+B^2+C^2}} & \frac{B}{\sqrt{A^2+B^2+C^2}} &
   \frac{C}{\sqrt{A^2+B^2+C^2}}
\end{array}
\right)
$$


Stopping Probabilities for Random Walk with Drift [EDIT] After the comments and answer I understood that my formation of the problem is incorrect. However I'm still interested in a stochastic solution to the original riddle. Please skip to the "Background" part of this question. And thanks for setting a bounty on this question!!
Let $X_1,X_2,...$ be independent and identically distributed random variable. $X_i = 2$ or $X_i=-1$ each with 50% probability. And let $S_n = X_1+\cdots+ X_n$ be the associated random walk. So we can think of this as a random walk with drift $\mu = 0.5$ and step-length = $1.5$
For a given constant $m$, suppose we define a stopping rule to stop when $S_n \leq -m$ or $S_n \geq K$. How do we find $K$, such that the probability of stopping at $S_n \geq K$ is equal to the probability of stopping at $S_n \leq -m$?
When there is no drift the solution is trivial $K = m$. I suspect this little modification should be a classical problem in stochastic process too?
The solution is apparently $K \sim O(m^2)$, but I'm looking for a stochastic explanation. see the background below.
Background: I found this riddle and its solution. I reproduce the riddle here:

At a pivotal moment in an epic battle between the living and the dead,
the Night King, head of the army of the dead, raises all the fallen
(formerly) living soldiers to join his ranks. This ability obviously
presents a huge military advantage, but how big an advantage exactly?
Forget the Battle of Winterfell and model our battle as follows. Each
army lines up single file, facing the other army. One soldier steps
forward from each line and the pair duels — half the time the living
soldier wins, half the time the dead soldier wins. If the living
soldier wins, he goes to the back of his army’s line, and the dead
soldier is out (the living army uses dragonglass weapons, so the dead
soldier is dead forever this time). If the dead soldier wins, he goes
to the back of their army’s line, but this time the (formerly) living
soldier joins him there. (Reanimation is instantaneous for this Night
King.) The battle continues until one army is entirely eliminated.
What starting sizes of the armies, living and dead, give each army a
50-50 chance of winning?

So we can think of this riddle as above problem. Let $m$ be the size of dead army. Let $S_i$ be (current  difference in army sizes - initial difference of army sizes). For each step $S_i$ increments by either −1 or 2. If $S_n=−m$, that means comparing to the initial state, the dead army is down by $−m$, the battle ended. If $S_n=K$, that means comparing to the initial state, the dead army is up by $K$, the battle ended. The solution is apparently $K \sim O(m^2)$. The combinatorial argument is nice, but the paper offers no stochastic explanation, which I am truly curious about.
I am fine with an approximation solution. So if we replace $X_i$ with a normal random variable with non-zero mean is fine to me too, if that helps with the estimation. But I think for a large enough $n$ this probably doesn't matter anyways.

This is too long for a comment, but the problem you formulated is not equivalent to the problem in the background material. The interpretation is supposed to be that $X_i=+2$ means that a dead soldier won a duel, and $X_i=-1$ means a living soldier won. You then say that $S_n=K$ means that the dead army has won. However, if the dead army initially wins $K/2$ battles in a row, then you will have $S_{n}=2(K/2)=K$, but the dead army would not have won yet.
Here is a correct formulation. Suppose there are initially $K$ living soldiers and $m$ dead ones. Let $Y_1,Y_2,\dots$ be i.i.d. and equal to $-1$ or ${\bf +1}$ with equal probability; the event $Y_i=+1$ represents a dead soldier winning a duel, and $Y_i=-1$ means a living soldier won (note the random walk is now symmetric).
Let $T_n=Y_1+\dots+Y_n$. Notice that $(n+T_n)/2$ is the number of dead soldier wins, and $(n-T_n)/2)$ is the number of living soldier wins. Now, we see that:

When $(n+T_n)/2=K$, then the dead soldiers have won $K$ duels, so they won the battle. That is, the winning condition for the dead soldiers is $T_n= 2K-n$. Whereas you had the "target value" as a constant $K$, the target value should change over time.

When $T_n=-m$, then the living soldiers have reduced the dead army by $m$, so the living soldiers have won (This part is the same as your setup).

Now, I do not think there is a good "stochastic" reason why $K\sim m^2$ produces a fair fight. I do know a good reason, but it has to do with carefully analyzing the walks in the 2D plane defined by the above random process, and using the reflection principle cleverly. Unfortunately, the same reflection principle argument does not work if you approximate the discrete walk by a Brownian motion, so the argument cannot be "stochastic" alone, it needs tricky combinatorics.

If you take $K=+\infty$ and $m=some\ large\ number$, then it is more likely to go right ad infinum then to finally hit $-m$. From the central limit theorem one can deduce that on step $n$ with high probability we will be somewhere within the range $\mu n \pm c\sqrt n$, where $c$ is a constant depending on our desired confidence with which we want to make such a statement (and it also includes that $1.5$ coefficient). And $\mu n$ asymptotically grows faster than $c\sqrt n$ independently of the value of $c$, so from some point $n$ this range is never hitting negative zone.
Therefore, for large enough $m$ there is simply no such $K$. And for small $m$ I suspect you can't really find this in any better way than with Monte-Carlo simulation.


Invariant Probability Vector I'm reading through my textbook, Introduction to Stochastic Processes (Lawler), before the semester begins in hopes of getting ahead, and I've run into something I just plain cannot figure out: How to compute the invariant probability vector for a transition matrix.  I was hoping that one (or many) of you would be able to walk me through how you would do this for just a simple matrix:

$$\begin{bmatrix} .4&.2&.4 \\\\ .6&0&.4 \\\\ .2&.5&.3 \end{bmatrix}$$

I know that you can compute it by raising the matrix to a large power, but this practice problem says to "compute the invariant probability vector as a left eigenvector."  How would one go about doing this?

Thanks for your help!

If the transition matrix is $A$ and the probability vector is $\mu$, "invariant" means that $\mu A = \mu$.  Another way of saying this is that $\mu$ is a left eigenvector of $A$ with eigenvalue 1.

$\mu A = \mu$ is really just a system of linear equations.  If we write $\mu = [\mu_1, \mu_2, \mu_3]$ then we have
$$[\mu_1, \mu_2, \mu_3] \begin{bmatrix} .4&.2&.4 \\\\ .6&0&.4 \\\\ .2&.5&.3 \end{bmatrix}= [\mu_1, \mu_2, \mu_3]$$
or in other words
$$\begin{align*}
.4 \mu_1 + .6 \mu_2 + .2 \mu_3 &= \mu_1 \\
.2 \mu_1 + 0 \mu_2 + .5 \mu_3 &= \mu_2 \\
.4 \mu_1 + .4 \mu_2 + .3 \mu_3 &= \mu_3.
\end{align*}
$$
Since $\mu$ is to be a probability vector we also have to have
$$\mu_1 + \mu_2 + \mu_3 = 1.$$
So you have a system of 4 linear equations in 3 unknowns.  Now you just have to solve this system.

If T is the transistion probability matrix the stationary (or invariant) distribution satisfies the matrix equation

P(X)= T P(X)


Application of Composition of Functions: Real world examples? Do you know of a real world example where you'd combine two functions into a composite function?  I see this topic in Algebra 2 textbooks, but rarely see actual applications of it.  It's usually plug and chug where you take f(g(4) and run it through both functions.  This leads to the idea of creating a composite function f(g(x).  But it's somewhat academic, and it's not like you're saving time b/c you need to run 50 different numbers through both functions.

While on this topic, where is this topic used in later math?  In Precalculus, you can determine the domain of the composite function.  In Calc, composition is used to describe the ideas behind the Chain Rule.  In Calc, you break down a function into the 2 components to show it's continuous. (If the components are continuous, so is the composite function)  Any other main areas?

Thanks!

First example of Algorythms: You have a list, compose by a head (an element) and a tail (a list). A composition of functions could return the second element of the list, let's say, L:

$ Head(Tail(L)) $

This is a simple examen in my field of study, I don't know if that's what you're looking for.

Many people believe that $+$ is a binary function. They are wrong. Since Haskell Curry it is known that $+$ is in fact an unary function, operating on numbers and returning another function, also operating on numbers. $a+b$ is a shorthand for a composition $(+(a))(b)$

Same goes for any other $n$-ary function, of course.


Express the following statements using quantifiers and atomic statements How do i approach this problem?

Let G represent a graph with vertices V. The only atomic statements about the graph are of the form G(v,w) where v and w are vertices in V and G(v,w) means there is an edge between v and w. The quantifiers ∀x and ∃x refer to the vertex set V. With this context, express the following statements about the graph G formally using quantifiers and atomic statements:

There is some vertex in the graph that has an edge connecting it to every other vertex.
The graph is the complete graph.
Any two vertices are connected by a path of length 3.
The graph contains no triangle.
Among any four vertices at least two are connected by an edge.

I'll reword each of your three constraints in a way that's closer to the atomic-statement formulation:

There exists a vertex in the graph, such that for every other vertex, there is an edge between the two.
Any two vertices have an edge between them.
For any two vertices that we want to connect, we can find two new vertices to form a path between them (i.e. the first original vertex is connected to the first new vertex, the two new vertices are connected, etc.)

Comment if you need more guidance.

∃x:∀y:G(x,y)
∀x:∀y:G(x,y)
∀x:∀y:∃w:∃z:G(x,w)∧G(w,z)∧G(z,y)

The way I approach these types of problems is by translating them into constrained English ("for all x, ..." "there exists a y, ...") and then rendering them with equivalent formal notation.  For example, 1. can be translated as "there exists a vertex x such that for all vertices y, x and y are connected by an edge", and 2. can be translated as "for all vertices x and y, x and y are connected by an edge".  When I can't figure out how to translate them as such, that is a strong hint that either the problem is ill-specified or else my own understanding of the definitions is lacking.


Direct proof of Gelfand-Zetlin identity Denote by $D(a_1,\dots,a_n)$ the product $\prod_{j>i}(a_j-a_i)$. Assuming that $a_i$ are integers s.t. $a_1\le a_2\le\dots\le a_n$, proove that $D(a_1,...,a_n)/D(1,...,n)$ is the number of Gelfand-Zetlin triangles (that is, triangles consisting of $\frac{n(n+1)}2$ integers, s.t. each number is greater it's lower-left neighbor but not greater than lower-right neighbor) with the base $a_i$.

For example, for n=3 one needs to prove that number of b1, b2, b' s.t. $a_1\le b_1<a_2\le b_2<a_3$, $b_1\le b'<b_2$ is exactly $\frac{(a_3-a_2)(a_3-a_1)(a_2-a_1)}{3}$.

As one can guess from the name “Gelfand-Zetlin”, this fact is well-known in representation theory (namely, in LHS we count dimension of a $gl_n$-representation by Weyl formula, and in RHS we count elements in Gelfand-Zetlin basis of the same representation). But maybe someone can come with more or less direct proof? (Some kind of bijective proof, maybe.)

Informal probabilistic argument

For simplicity, consider the case $n=3$: D(a_1,a_2,a_3) counts the number of triangles s.t. $a_1\le b_1<a_2\le b_2<a_3$, $a_1\le b'<a_3$ — and we're interested only in G-Z triangles, i.e. in triangles s.t. $b_1\le b'< b_2$. Now, mathematical expectation of the length of the interval $(b_1,b_2)$ is exactly one half of the length of the interval from which $b'$ is chosen. So one may expect that the probability that random triangle is G-Z is $1/2$ — and the answer is indeed $D(a_1,a_2,a_3)/D(1,2,3)$.

(The main problem with this computation is that we're multiplying probabilities for events that are clearly not independent. And although for n=3 it's not hard to transform this heuristic argument into a formal proof, even for n=4 I failed to do such thing.)

As Qiaochu points out the formula given by Gjergji Zaimi is certainly relevant.
It realizes $D(a_1,\ldots,a_n)/D(1,\ldots,n)$ as the
determinant of the $n$-by-$n$ matrix $M$ with $(i,j)$-entry
$${a_i\choose j-1}.$$
By row operations this equals the determinant of the $(n-1)$-by-$(n-1)$  matrix $N$
with $(i,j)$-entry
$${a_{i+1}\choose j}-{a_i\choose j}.$$
The $i$-th row of $n$ is the sum of vectors $v_{a_i},v_{a_i+1},\ldots,v_{a_{i+1}-1}$
where the $j$-th entry of $v_c$ is
$${c\choose j-1}.$$
By the multilinearity of the determinant as a function of its rows, $N$
is the sum the determinants with rows $v_{b_1},\ldots,v_{b_{n-1}}$
where $a_1\le b_i < a_2\le b_2 < a_3\le\cdots$. These tuples
$(b_1,\ldots,b_{n-1})$ are precisely the admissible penultimate rows
in GZ triangles with bottom row $(a_1,\ldots,a_n)$. Hence both sides of the
sought equality satisfy the same recurrence.

Just to make sure: you know that what you are asking for is actually Weyl's dimension formula for the highest weight representation in the case of SL(n,C) resp. SU(n)? I suppose there is no direct combinatorial proof that does not make use of this fact.


NBG conservative extension of ZFC? Wikipedia (https://en.wikipedia.org/wiki/Conservative_extension) says:

  Von Neumann–Bernays–Gödel set theory is a conservative extension of Zermelo–Fraenkel set theory with the axiom of choice (ZFC).

In which sense is NBG a conservative extension of ZFC? I think that in NBG one can prove the existence of a proper class (i.e. the class of all sets) whilst in ZFC one can't prove the existence of such an entity. So there are really statements in NBG that are not provable in ZFC. Why should it be conservative extension then?

What is meant by "conservative extension" in this case is:

  Take any sentence in the language of ZFC and translate it into NBG by rewriting every quantifier to restrict its variable to range over sets only. Then the rewritten sentence is a theorem of NBG if and only if the original sentence is a theorem of ZFC.

There are sentences in NBG that don't correspond to a ZFC formula under this translation, and the "conservative extension" property doesn't say anything about those sentences at all. That's why it is an extension.

(To prove the property, assuming the metatheory is a set theory, note that any model of ZFC can become a model of NBG by adding to it every definable collection that is not already a set to the model as proper classes. This gives a model of NBG where the translation of a ZFC sentence is true in the extended model exactly if the original sentence was true in the original model. Conversely, given a model of NBG, removing the proper classes from it produces a model of ZFC).

To add to the previous answers, here is a proof $NBG$ is conservative over $ZFC$.

First consider a transitive $(M,\in)\vDash\\$$ZFC$. Then it is easy to see $(2^M,\in)\vDash\\$$NBG-Global\,Choice$. To see choice, observe the formula "$z=(x,y)$" is $\Delta_0$ and so absoloute in $M$. Therefore $M^2\subseteq\\$$M$ and so if $R$ is a well order of $M$, $R\subseteq\\$$M$. Such a well order exists by normal choice.

Now, let $(N,E)$ be non-transitive. Then let $\pi(x)$ be an isomorphisim between $(N,E)$ and some transtive $(M,\in)$. Then if $R$ is a well order of $M$, $W=${$(\pi(x),\pi(y))|(x,y)\in\\$$R$} well orders $(N,E)$ and $W\subseteq_EN^2$.


Is every non-increasing function not increasing? I want to prove or disprove my hypothesis:
Let $f:(a,b)\rightarrow R$ be non-increasing function, then it is not increasing.

Definitions:
$$\text{Function is increasing iff $x<y\Rightarrow f(x)<f(y)$}\tag{1}$$

$$\text{Function is non-increasing iff $x<y\Rightarrow f(x)\geq f(y)$}\tag{2}$$
So, my idea was to just write the theorem as it is:
$$\{\forall x,y\in(a,b):x<y\Rightarrow f(x)\geq f(y)\}\Rightarrow \{\forall x,y\in(a,b):x<y\Rightarrow f(x)< f(y)\}$$
And make use of truth values of $x<y,f(x)<f(y)$ but I don't get a tautology, I couldn't make use of the quantifiers.

Secondly I would be interested in disproving the converse statement that any not inreasing function is non-increasing. As a counter example i would give $x\mapsto \sin{x}$ on $(0,2\pi)$

Non-increasing:

$$\forall\,x, y: x<y:f(x)\ge f(y).$$

Not increasing:

$$\overline{\forall\,x,y:x<y:f(x)<f(y)}\iff \exists\,x,y:x<y:f(x)\ge f(y).$$

Obviously,

$$\forall\,x, y: x<y:f(x)\ge f(y)\implies \exists\,x,y:x<y:f(x)\ge f(y).$$

Notice the change of the quantifier.

Hint:  By trichotomy $a < b \iff a \not \ge b$.


How do I transform a vertical line with a matrix If I have a vertical line defined as:$$x=c$$where $c$ is some constant, and I want to transform it with a matrix:$$\begin{bmatrix}m_{00} & m_{01}\\m_{10} & m_{11}\end{bmatrix}$$and end with a line with the format:$$y=ax+b.$$How would the math work out? I can easily see how to calculate the $a$ value, as it is just given by the right column vector of matrix's rise over run:$$a=\frac{m_{11}}{m_{01}}$$So now I'm left with the question of how to calculate b?

You would perform a rotation to get the gradient correct and then a scaling to change the y-intercept from the one you have after rotation to the desired value of b. You need to know the form of the matrix for a rotation of a given angle, a way to calculate the angle you want, then calculate where the y-intercept will end up, call it p, and know the form of a matrix to scale by a factor of b/p. Finally need to compose the matrices to get the answer in the form of a 2 x 2 matrix and be careful to do so in the correct order as matrices are not generally commutative!

Edit: As per the comment above, it will not work for c=0 as p will be zero in the above.

Just after posting i had a realization.

We know a point the old line went through $(c, 0)$. If we transform this like it was a vector, by multiplying with the matrix, it should end up a place the new line will go through. And by knowing a point it need to go through and a slope, b is given by:$$b=ax-y$$Edit: i will just crunch the algebra for any future readers:
Let $p$ be the point that the $\begin{bmatrix}c\\0\end{bmatrix}$ ends up on. That means:$$p=\begin{bmatrix}m_{00}&m_{01}\\m_{10}&m_{11}\end{bmatrix}\begin{bmatrix}c\\0\end{bmatrix}=\begin{bmatrix}m_{00}c\\m_{10}c\end{bmatrix}$$And then inserting the coordinates of $p$ into our equation for b we get:$$b=am_{00}c-m_{10}c$$


Problem with basic definition of a tangent line. I have just started studying calculus for the first time, and here I see something called a tangent. They say, a tangent is a line that cuts a curve at exactly one point. But there are a lot of lines that can cut the same point just like shown in the picture-

WHY aren’t we saying that lines M and C are also tangents?

What is the real definition of a tangent line?

“ A tangent line is a line which passes through two infinitesimally close points” Is this definition correct?
Thanks!

Animations showing that the limit of a secant as the variable point tends towards a fixed point becomes the tangent line.

First, when the variable point (in red) approaches the fixed point (in black) from below:

Second, when the variable point approaches the fixed point from above:

In both cases, the secant line becomes the same tangent line.  This suggests that the derivative is well-defined at the black point (which is highlighted in green when tangency occurs).

Formally, then, this gives us a definition of the derivative as follows.  For a real-valued function $f(x)$, the derivative at $x = a$ is the limit of the slope of the secant line through two points $$P_a = (a, f(a)), \quad P_b = (b, f(b))$$ as $b$ approaches $a$, whenever such a limit exists.  Since the slope is simply $$\frac{f(b)-f(a)}{b-a},$$ we then have $$f'(a) = \lim_{b \to a} \frac{f(b) - f(a)}{b-a}.$$  The equation of the tangent line at the point $P_a$ is therefore $$y - f(a) = f'(a)(x-a),$$ if the line does not have infinite slope; otherwise $1/f'(a) = 0$ and we can write the equation of the tangent line as $x = a$.

The tangent line is the unique line that "touches" the point without actually actually crossing it. For the other lines in your example, there's a crossover at the point; the line goes from being above the curve to below it. The tangent line remains beneath it.

This is obviously not the rigorous definition, but for someone just starting out with calculus this is probably the easiest way to conceptualise it.


Can we find a real number $a>0$ such that $f(t)>a>0$ and $a$ does not depend of the value of $t$. Let $f$ be a strictly positive real function defined on the interval $(0,1)$. Let $t∈(0,1)$ be fixed.

Can we find a real number $a>0$ such that $f(t)>a>0$ and $a$ does not depend of the value of $t$.

If $t$ is fixed as you say, the existence of a function $f$ defined on $(0,1)$ is irrelevant, and the requirement "does not depend on the value of $t$" is vacuous. All we have is that the real number

$$f(t)$$

is positive and obviously there is an $a$ such that

$$f(t)>a>0.$$

Anyway, $a$ must depend on the value of $f(t)$, for example as $a=\dfrac{f(t)}2$. You can't pick a suitable value for $a$ without knowing $f(t)$, because $\mathbb R_+^*$ has no minimum.

There is no such $a$. Fix $t \in (0,1)$ and imagine that there is some constant $a$ that doesn't depend on $t$ such that
$$f(t)> a > 0.$$

Since $a$ doesn't depend on $t$ we must have $f(t') > a > 0$ for any other $t' \in (0,1).$

Now consider any function which is strictly positive such that $f(t) \rightarrow 0$ as $t \rightarrow 0$ (e.g. $f(x) = x$ on $(0,1)$) then taking the limit as $t \rightarrow 0$ in
$$ f(t) > a > 0$$
gives us
$ 0 \geq a \geq 0.$ Therefore $a = 0$ which contradicts the fact that $a > 0.$


$G$-invariant complement to an infinite dimensional vector space Let $G$ be a finite group and let
$$\rho : G \to GL(V)$$
be a complex representation of $G$.

Suppose we have an internal direct sum decomposition
$$V=W \oplus U$$
where $W$ is infinite dimensional and $U$ is finite dimensional and nonzero. Assume further that $W$ is $G$-invariant.

Does there exist a subspace $U' \subseteq V$ such that $U'$ is $G$-invariant and $V=W \oplus U'$?

Yes, there is such a $U'$, by Maschke's Theorem, which states that if $V$ is a representation of a finite group over a field of characteristic zero, and $W$ is a subrepresentation, then it has a complement $U'$ that is also a subrepresentation. Some introductions to representation theory might only state Maschke's Theorem for finite dimensional representations, as they will often only be concerned with the finite dimensional case, but if you look at the usual proof then you'll see that it nowhere uses the fact that $W$ is finite dimensional.

Also, there is no need to assume that $W$ has a finite dimensional vector space complement $U$.

The answer is no, not in general.  Of course this is not true in the finite dimensional case, and making $W$ infinite dimensional does nothing to help the situation. Here's kind of a contrived example:

Let's say $G$ acts on trivially on the infinite dimensional vector space $W$.
Now let $K$ be a finite dimensional representation of $G$ with a $G$-invariant subspace $U$, but such that $U$ does not have a $G$-invariant complement.

Now $K\oplus W$ is an infinite dimensional representation of $G$ with an infinite dimensional $G$ invariant subspace $U\oplus W$, but this subspace has no $G$-invariant complement.


Could a repeated eigenvalues of zero still have independent eigenvectors? If a matrix $A$ has repeated eigenvalues, its eigenspace matrix would be dependent because there isn't enough eigenvector to tag along with the eigenvalue.

But if a matrix $A$ has repeated eigenvalues of $0$, that means the matrix $A$ is singular, then the dimension of the nullspace of matrix $A$ is the number of $0$ eigenvalues that matrix $A$ has. Although I read that the eigenvectors of the matrix A are dependent if there are repeated eigenvalues, I'm thinking why couldn't the eigenvectors be independent if the repeated eigenvalues were $0$?

Say in a matrix that has 2 eigenvalues of $0$, this implies that $dim(N(A))=2$, then wouldn't the eigenvector to each of this eigenvalue of $0$ be the 2 independent nullspace vectors? Then when putting all the eigenvectors together to form an eigenspace matrix, they will all be independent, wouldn't it?

What do I don't understand that lead me to this thought?
Thanks for any help.

Update:

With the example suggested by Didier Piau,
$$A=\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}$$
The $rank(A)=1$, $N(A)=\begin{bmatrix}
1\\
0
\end{bmatrix}$, $eigenvals(A)=0, 0$ and true enough, the eigenvector can only be $\begin{bmatrix}
1\\
0
\end{bmatrix}$ and therefore the eigenspace is dependent. There isn't enough eigenvectors to tag to the other eigenvalue of zero.

But say with another example of matrix $B$,
$$B=\begin{bmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
2 & 4 & 6
\end{bmatrix}$$
The $rank(B)=1$, the $N(A)=c_{1}\begin{bmatrix}
-2\\
1\\
0
\end{bmatrix}
+
c_{2}\begin{bmatrix}
-3\\
0\\
1
\end{bmatrix}, c_{1},  c_{2} \in \mathbb{R}$, $eigenvals(B)=9, 0,0$ then the eigenvectors are...
$$B\begin{bmatrix}
0.5\\
0.5\\
1
\end{bmatrix}=9\begin{bmatrix}
0.5\\
0.5\\
1
\end{bmatrix}$$
Then for the next two eigenvalues of $0$, which eigenvector do I choose to use as their eigenvector? Do I use one for each or only just one of the two or both of the two?
$$B\begin{bmatrix}
0.5\\
0.5\\
1
\end{bmatrix}=0\begin{bmatrix}
-2\\
1\\
0
\end{bmatrix}$$
and it could also be...
$$B\begin{bmatrix}
0.5\\
0.5\\
1
\end{bmatrix}=0\begin{bmatrix}
-3\\
0\\
1
\end{bmatrix}$$
Then for the eigenspace, should it be: $\begin{bmatrix}
0.5 & -2 & 3\\
0.5 & 1 & 0\\
1 & 0 & 1
\end{bmatrix}$ or $\begin{bmatrix}
0.5 & -2 & -2\\
0.5 & 1 & 1\\
1 & 0 & 0
\end{bmatrix}$ or $\begin{bmatrix}
0.5 & 3 & 3\\
0.5 & 0 & 0\\
1 & 1 & 1
\end{bmatrix}$?

One can test one's understanding of this question on the matrix A=[0,1|0,0], since this example already exhibits many of the relevant features of the problem. For eample, one cannot find two linearly independent vectors v and w such that Av=Aw=0 (try to solve for v the equation Av=0) although 0 is a double root of the characteritic polynomial det(XI-A)=X^2.

This is not specific to the eigenvalue 0. Recall that for any eigenvalue one distinguishes its algebraic multiplicity n (in this case, n=2) from its geometric multiplicity m (in this case, m=1). See here for more details and a worked out example.

you have 2 eigenvectors that represent the eigenspace for eigenvalue = 1 are linear independent and they should both be included in your eigenspace..they span the original space... note that if you have 2 repeated eigenvalues they may or may not span the original space, so your eigenspace could be rank 1 or 2 in this case.