Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ  = 5$$

Find

$$4\cosθ -3\sinθ$$

Help me guys (PS:I'm not that good in maths)

HINT:

using Brahmagupta–Fibonacci identity

$$(4\sin\theta+3\cos\theta)^2+(4\cos\theta-3\sin\theta)^2=(4^2+3^2)(\sin^2\theta+\cos^2\theta)$$

A very fast solution
Maximum value of $\displaystyle4\sin\theta+3\cos\theta$ is $\displaystyle{5}$. This value is achieved when $\displaystyle\tan\theta=\dfrac{4}{3}$ by differentiating.So $\cos\theta=\dfrac{3}{5}$ and $\sin\theta=\dfrac{4}{5}$.
Hence $4\cos\theta-3\sin\theta=0$.


Is $\sqrt[4]{-1}=\sqrt(i)$? I get $\sqrt[4]{-1}=((-1)^{\frac{1}{2}})^{\frac{1}{2}}=\sqrt{i}$.

But on the one hand $\sqrt[4]{-1}$ has $4$ roots and on the other $\sqrt{i}$ has just $2$.

The function $\sqrt[n]{x}$ is usually defined for non-negative, real $x$ and returns the non-negative, real solution to $y^n=x$. It can be extended to other domains, but must be accompanied by clear restrictions on which roots are to be returned.

For example, we can define $\sqrt[4]{z}$ for $z\in\mathbb{C}$, where the root in the first quadrant (excluding the imaginary axis) is returned.

Similarly, we can define $\sqrt{z}$ for $z\in\mathbb{C}$, where the root in the first or second quadrant (excluding the negative real axis) is returned.

With these definitions, we get $\sqrt[4]{-1}=\sqrt{i\,}=\frac{1+i}{\sqrt2}$.

However, if we consider $\sqrt[4]{z}$ to be the set valued function $\left\{w\in\mathbb{C}:w^4=z\right\}$, and similarly define $\sqrt{z}$ to be the set $\left\{w\in\mathbb{C}:w^2=z\right\}$, then $\sqrt[4]{-1}$ has $4$ elements and $\sqrt{i\,}$ has only two elements, so they are not the same, but with this definition,
$$
\sqrt{i\,}\subset\sqrt[4]{-1}
$$

Indeed as you observed $\sqrt[4]{-1}$ has four roots while $\sqrt i$ has only 2 roots thus they are not the same.

What is true is that they have roots in common.


Check the invertibility of a matrix given a parameter Find all values of $c$, if any, for which the given matrix is invertible (using row operations -> gotta try simplifying to I).

Here is the matrix :
\begin{pmatrix}c&1&0\\1&c&1\\0&1&c\end{pmatrix}

$R_1,R_2,R_3$ are the first, second and third row

Let's assume that $c\neq0$
, as the case $c=0$ is very simple.

$\begin{pmatrix}c&1&0\\1&c&1\\0&1&c\end{pmatrix}$ $R_1\leftarrow \frac{R_1}c$

$\begin{pmatrix}1&1/c&0\\1&c&1\\0&1&c\end{pmatrix}$ $R_2\leftarrow R_2-R_1$

$\begin{pmatrix}1&1/c&0\\0&c-1/c&1\\0&1&c\end{pmatrix}$ $R_2\leftarrow R_2\cdot\frac{c}{c^2-1}$

$\begin{pmatrix}1&1/c&0\\0&1&\frac{c}{c^2-1}\\0&1&c\end{pmatrix}$ $R_3\leftarrow R_3-R_2$

$\begin{pmatrix}1&1/c&0\\0&1&\frac{c}{c^2-1}\\0&0&c-\frac{c}{c^2-1}\end{pmatrix}$ $R_1\leftarrow R_1-c\cdot R_2$

$\begin{pmatrix}1&0&\frac{c^2}{c^2-1}\\0&1&\frac{c}{c^2-1}\\0&0&\frac{c^3-2c}{c^2-1}\end{pmatrix}$

Can you do the last two steps by yourself ? :) :) :)

Find the determinant of this matrix and notice where it becomes zero. The matrix is invertible for every value except those.


Where are power series uniformly continuous? As far as I know, $f(x)=\sum\limits_{n=0}^\infty a_n(x-x_0)^n$ is continuous on the whole convergence interval $K:=\{x\in\mathbb R:|x-x_0|<r\}$. Is there anything we could say about uniform continuity?

Added: Would it be correct to claim the following?

  Power series is uniformly continuous on $\bar K:=\{x\in\mathbb R:|x-x_0|<\bar r\}$ where $0<\bar r<r$.

I'm asking for the clarification since we haven't defined compactness yet.

Check relevant theorems- Weierstrass M-test can be used to prove the uniform continuity.


The standard role of intuitive numbers in the foundations of mathematics In my career I've been formed mostly in the formal side of mathematics, that is, standard set theory and every classical branch of mathematics that uses set theory.
However, I am not quite sure about the common "rules" that are accepted in meta-math and, more specifically, in the foundations of mathematics.

I have several questions, but all of them are related to the nature of intuitive natural numbers.
I will start from the formal side, to explain my point, but I am really interested in the intuitive and metamathematical issues.

From the formal point of view, we have a "set" of natural numbers, $\mathbb N$, which can be obtained for example by the Von Neumann construction:

$$0 = \emptyset$$
$$1 = \{0\}$$
$$2 = \{0,1\}$$

and so on.

The existence of a set containing all these "formalized" natural numbers comes with the development of set theory. To be concrete, I will talk about ZFC.

However, ZFC has to be sustained in the first order logic, and both are formally expressed in the "language" of first order logic, which consists of a collection of symbols $\mathcal L = \{\wedge,\vee, \forall,\exists, ...\}$, and rules to form formulas with that symbols.

FIRST ISSUE. It is common to see in the books of logic that the collection $\mathcal L$ of symbols of the language is "finite". Moreover, the word "set" is used (I am using "collection" in order to not merge things of different contexts). The meaning of "finite" in the metamath context cannot be referring to any notion of "finite set" of ZFC, because this would be circular (the problem is that, if I am "constructing" set theory from "nothing", and the notion of "finite set" comes later, after some theory is done, I cannot talk about "finiteness" in the "ZFC sense"). So:

FIRST QUESTION. In what sense is it understood that the collection of symbols $\mathcal L$ (intended to construct first order logic) is considered "finite"? In what sense is it understood that the length of a formula of first order logic is finite? Is it implied by a previously accepted notion of intuitive natural number? Please, I need to know the standard point of view, and not opinions of a very personal kind.

SECOUND ISSUE. When methamath theorems are demonstrated, sometimes some "properties" of natural numbers are used, as for example the induction on the number of symbols in a given logical formula. Again, these properties are referring to the intuitive natural numbers.

SECOUND QUESTION. How can I be sure about what properties of natural numbers can be accepted or not, in the metamath context? Is there some kind of consensus about what intuitive properties of natural numbers can be used?

THIRD ISSUE. If a formal first order theory contains the Peano axioms, then on the semantic side there are a lot of non-standard models... However I am even more intrigued by the "standard model", which is, again, the intuitive natural numbers.

THIRD QUESTION. Are they actually a "model"? In order to prove that, what properties of intuitive natural numbers are commonly accepted as "true"?

FINAL QUESTION. Does there exist an agreement about what the intuitive natural numbers are and which of their properties are accepted and used in metamaths?

The usual answer (or dodge, depending on yuor philosophical position) is to talk about the intended model (more precisely, intended interpretation) of the natural numbers. If such a thing exists (and many mathematicians do believe it exists, including the constructivist Errett Bishop), then one can interpret references to "finite" at the meta-level as referring to things equinumerable with individuals in the intended model. If you don't want to believe in an intended model then you also have to give up hope of an absolutely rigorous development of mathematics "from scratch". Whether or not you actually lose anything in the process again depends on your philosophical position. Kronecker, by the way, never expressed any opinion on the matter in writing. Whatever is reported in his name is hearsay based on Weber (who certainly made a mistake when he mentioned "whole numbers" rather than "natural numbers").

Juat out of curiosity, I looked up "intended interpretation" on wiki, and was led to the following comment:

Intended interpretations. Many formal languages are associated with a particular interpretation that is used to motivate them. For example, the first-order signature for set theory includes only one binary relation, ∈, which is intended to represent set membership, and the domain of discourse in a first-order theory of the natural numbers is intended to be the set of natural numbers. The intended interpretation is called the standard model (a term introduced by Abraham Robinson in 1960).[8]

Note the definite article, which I guess begs the question (namely, yours). Here the reference is to the paper

Roland Müller (2009). "The Notion of a Model". In Anthonie Meijers. Philosophy of technology and engineering sciences. Handbook of the Philosophy of Science 9. Elsevier. ISBN 978-0-444-51667-1

which you may find useful (though I hasten to admit that I never read it). If you gain some insights do let me know.

The book that the article is in can be found here.

For a related question see What are natural numbers? where you will also find an accepted answer containing a passionate defense of the intended interpretation without mentioning the term ("categorical" and all).

For a detailed discussion of the intended interpretation see this post.

George Reeb's position was that the naive counting numbers do not exhaust $\mathbb N$; see this post for a bit of a discussion and this article for more details.

For the first question, it depends on the context. When talking about strict foundations, finite would usually mean finite in the intuitive sense. This is rather important in this context – firstly it allows us to avoid (or postpone) ontological and intuitive pitfalls related to infinity, and secondly it allows us to encode everything using some Gödel numbering.

For second question, I believe Peano arithmetic is generally accepted, and sufficient for the proof of most basic foundational facts.

For the third question, I don't think there is much controversy about what a natural number is. A natural number is an object which can be obtained by adding $1$ to itself some finite number of times (though “finite” may be controversial). If the set of all natural numbers exists, it is certainly a model of any interesting theory of arithmetic: they are all approximations of what the natural numbers are like.

The last question is covered in the previous one, I think.


The Dirichlet Eta Function and the limits interchange 1-) Let us handle the Dirichlet Eta Function and its partial sumshere which is convergent in the region
$0<\sigma< 1$  with  $s= \sigma +it $
$$ \eta (s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}},  \text{  and...... }\eta_N (s)=\sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{s}} $$
Meanwhile, we have already known that:
$$\eta (s_0)=0 ⇔ \eta (1-s_0)=0$$
.
Can we interchange the following limits?  If yes, how we can prove it ?
$$ \tag{1}\lim_{s\to s_0}\lim_{N\to \infty}\sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{s}}=\lim_{N\to \infty}\lim_{s\to s_0}\sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{s}} $$

2-) Let us suppose we have the following equality in the region
$0<\sigma< 1$
$$\tag{2}\lim_{s\to s_0}\frac {\lim_{N\to \infty}\eta_N (s)}{\lim_{N\to \infty} \eta_N (1-s)}=h(s_0)$$
$h(s)$ is continuous and finite and non-zero in the region
$0<\sigma< 1$
Thus, can we write the following equality depending on equation (2)? If yes, how we can prove it ?
$$\tag{3}\lim_{N\to \infty}\frac {\eta_N (s_0)}{ \eta_N  (1-s_0)}=h(s_0)$$

@reuns, you wrote as:

If $\eta(s)\ne 0$ and $\Re(s)\in (0,1)$ then $$\lim_{N \to \infty} \frac{\eta_N(s)}{\eta_N(1-s)} = \frac{\eta(s)}{\eta(1-s)}=h(s)$$

Thus now, let us consider some $s$ which very near $s_0$ as $ε$ sufficiently small as $0<ε$  as below:

$$s_0^-=(σ_0-ε)+it_0 \text{  and} ......s_0^+=(σ_0+ε)+it_0$$

$$\tag{4}\lim_{s\to s_0^-}\frac {\lim_{N\to \infty}\eta_N (s)}{\lim_{N\to \infty} \eta_N (1-s)}=h(s_0^-)$$

$$\tag{5}\lim_{N\to \infty}\frac {\eta_N (s_0^-)}{ \eta_N  (1-s_0^-)}=h(s_0^-)$$

We could write it also with $s_0^+=(σ_0+ε)+it_0$

Anyway now, we can see easly the follwing one by  $\lim {ε\to 0}$.

$$h(s_0^-)=h(s_0^+)=h(s_0)$$

Because   $h(s)$ is continuous and finite and non-zero in the region
$0<\sigma< 1$

Thus, can we write the following one depending equation (4) ? According to me YES !!.

$$\tag{6}\lim_{N\to \infty}\frac {\eta_N (s_0)}{ \eta_N  (1-s_0)}=h(s_0)$$


Let $n$ be a positive integer. Show that the smallest integer greater than $(\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}.$ Let $n$ be a positive integer. Show that the smallest integer greater than $(\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}.$

We know that $(\sqrt 3-1)^{2n}<1$

This has nothing to do with functional analysis. Let:
$$ A_n = (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n} = (4+2\sqrt{3})^n+(4-2\sqrt{3})^{n}\tag{1} $$
since $0<4-2\sqrt{3}<\frac{2}{3}$, we have that $A_n$ is the integer giving the ceiling of $(\sqrt{3}+1)^{2n}$.

Now we have:
$$ A_0 = 2, \quad A_1 = 8,\qquad A_{n+2} = 8 A_{n+1} - 4 A_{n}\tag{2} $$
and if we take:
$$ \nu_2(n) = \max\{m\in\mathbb{N}: 2^m\mid n\} \tag{3}$$
it happens that:
$$ \nu_2(A_n)\geq n+1\tag{4} $$
can be proved by induction from $(2)$. We can also factor a $2^n$ from the RHS of $(1)$ then just study the parity of the sequence given by:
$$ B_n = (2+\sqrt{3})^n+(2-\sqrt{3})^n,\tag{5}$$
for which:
$$ B_0=2,\quad B_1=4,\quad B_2=14,\quad B_{n+2}=4B_{n+1}-B_n\equiv -B_n\pmod{4}.\tag{6}$$
That gives:

  $$ \nu_2(A_n) = \left\{\begin{array}{rl}n+1 & \text{if }n\text{ is odd,}\\n+2 &\text{if }n\text{ is even.}\end{array}\right.\tag{7}$$


What is the tangent plane equation on the 3 spheres? 3 spheres are on $z=0$ plane and touch each other as shown in the picture. Coordinates of their centers are  $O_1=(0,0,5),O_2=(0,y_2,3),O_3=(x_3,y_3,2)$. What is the tangent plane equation on 3 spheres? ($ax+by+cz=d$)

Thanks for answers.

First solve some quadratic equations to find $y_2$, $x_3$, and $y_3$.

Then, since each of the spheres is tangent to $z=0$ as well as to the mystery plane, its center must lie on the plane that bisects the angle made between these two planes. We can compute an equation for bisecting plane, because that is defined by the centers.

Now reflect the plane $z=0$ about the bisecting plane.

Isn't $z=0$ a tangent plane, by construction?


Proving that $ω_1$ is locally compact I'm trying to show that $ω_1$ is locally compact, but when doing so, I need to show something else, which got me a bit stuck on.

I'm taking a $\alpha\in ω_1$, so $\{\alpha\}$ is an open set. Since $ω_1$ is Hausdorff, the closure of $\{ \alpha \}$ is $\{ \alpha \}$ itself. Now I want to show that $\{ \alpha \}$ is compact, and this is where I got stuck - I want to show that for every cover of $\{ \alpha \}$ I can find a finite sub-cover, but I don't see how to prove this, what made me wonder if I'm not trying to prove something which is false?

In order to prove that the space $\omega_1$ is locally compact, you have to prove that every point in $\omega_1$ has a compact neighborhood. If $\alpha\in\omega_1$ then the one-point set $\{\alpha\}$ is obviously compact, but it is not necessarily a neighborhood of $\alpha$; namely, it's not a neighborhood if $\alpha$ is a limit ordinal, such as $\omega$.

Hint: Prove that for each ordinal $\alpha$ the closed interval $[0,\alpha]=\{\xi:\xi\le\alpha\}$ is a compact neighborhood of $\alpha$.

More hints: Either $[0,\alpha]$ is always compact, or else there is a least counterexample. Suppose $\alpha$ is the least ordinal for which $[0,\alpha]$ is not compact, and let $\mathcal U$ be an open cover of $[0,\alpha]$ with no finite subcover. Choose $U\in\mathcal U$ so that $\alpha\in U$. Since $U$ is open, there is an ordinal $\beta\lt\alpha$ such that $(\beta,\alpha]\subseteq U$. Now $[0,\beta]$ is compact and

Every cover of $\{ a \}$ is actually $\{ a \}$ itself, so you can obviously take a finite sub-cover, which is simply $\{ a \}$! Thus, making $\omega_1$ locally compact.


Is the Collatz conjecture in $\Sigma_1 / \Pi_1$? Prompted by some of the comments on this question, I'm wondering if anything is known about the place of the Collatz Conjecture in the arithmetic hierarchy.  More specifically, is Collatz known to be equivalent to a $\Sigma_1$ or $\Pi_1$ statement?  It's obvious that it can be expressed in the second level of the hierarchy, as $\forall n\exists k . C^{(k)}(n) = 1$ (and I believe functional composition is $\Delta_0$, so the fact that we're using a slightly extended language here should be moot; we can encode the computational path involved in computing $C^{(k)}(n)$ as a finite sequence of integers and thus as a single integer by appropriate coding), but there doesn't seem to be any obvious way of expressing it as a first-level statement...

As suggested in my comment above, I will interpret your question as asking whether the Collatz conjecture is provably equivalent to a $\Pi_1$ or $\Sigma_1$ sentence over some base theory $T$. There are a variety of issues at stake depending on the outcome of the conjecture (in the standard model).

Case 0: The Collatz conjecture is false because some Collatz sequence is periodic with a period other than $4,2,1$. In that case, the Collatz conjecture is provably equivalent to $0=1$ over any theory with enough power to encode and decode the witnessing sequence up to its first repetition (e.g. PA or I$\Sigma_1$).

Case 1: The Collatz conjecture is false because some Collatz sequence never repeats itself. In that case, the conjecture is perhaps equivalent to a $\Pi_1$ sentence. To say that the Collatz sequence starting with some standard number $n_0$ is never repeating is a $\Pi_1$ statement. In the strange scenario where the Collatz sequences that don't repeat are precisely the ones that go through $n_0$, then the Collatz conjecture is equivalent to that $\Pi_1$ statement. More generally, if there is a (standard) finite list $n_0,n_1,\dots,n_k$ such that the Collatz sequences that don't repeat are precisely those that go through one of these, then we also have a $\Pi_1$ sentence equivalent to the Collatz conjecture. Still more generally, if there is a computable sequence $n_0,n_1,\dots$ with that property, we also have a $\Pi_1$ sentence equivalent to the Collatz conjecture. However, if no such computable sequence exists then it is unlikely that we can do better than $\Sigma_2$ unless we use a non-axiomatizable base theory like the $\Pi_1$ theory of the standard model.

Case 2: The Collatz conjecture is true. In that case, the Collatz conjecture is equivalent to the statement that $g(n) = (\mu k)(C^k(n)=1)$ is a total computable function. We can then compare $g$ to various levels of the fast-growing hierarchy for a more precise analysis. For example, if $g(n)$ is bounded above by some $f_k$ where $k$ is finite, then $g$ is actually primitive recursive and the Collatz conjecture is provably true in I$\Sigma_1$ (same as PA except that induction is restricted to $\Sigma_1$ formulas). If $g(n)$ is abounded above by some $f_\alpha$ with $\alpha \lt \varepsilon_0$, then the Collatz conjecture is provably true in PA. If $g$ grows faster than that then a still stronger theory is required to prove the Collatz conjecture.

See http://matwbn.icm.edu.pl/ksiazki/aa/aa91/aa9146.pdf
"A binomial representation of the 3x + 1 problem" by Maurice Margenstern and Yuri Matiyasevich.


Stieltjes Integral meaning. Can anybody give a geometrical interpretation of the Stieltjes integral:

$$\int_a^bf(\xi)\,d\alpha(\xi)$$

How would we calculate?
$$\int_a^b \xi^3\,d\alpha(\xi)$$
for example.

You can imagine $\alpha(\xi)$ as a non-uniform scaling applied to the $x$-axis.

Imagine, for example, that you're driving through the andes in peru, and your car's fancy GPS records the current elevation every kilometer (or mile if you prefer to keep it imperial). A few days later you compete in a marathon, and are asked to disclose the average altitude you were on, since your blood test shows an elevated count of red blood cells which is an indication of doping unless you spent a considerable amount of time at high altitudes.

Easy, you think, and download the data from your GPS, compute the sum, and divide by the number of data points. You report that value, and a prompty disqualified from this and any further events. What happened? It turns out your average speed was much higher at low altitudes than it was at high ones - perfectly reasonable, given the condition of some of the streets in those mountainous regions. Thus, you actually spent much more times at higher altitudes than the graph of altitude over distance that your GPS record shows. What you need is a graph of altitude over time, but you don't have that.

When you discover the problem, you check your car's logs as well, and are delighted to discover that your car maintens a log of the elapsed time, also recorded every kilometer (or mile). Can I use this to compute a corrected average, you ask a friend of yours, and thus rehabilitate my good name in the runner community? You can, he answers, and for once you're glad that you have a mathematician amongst your friends...

What does your mathematically inclined friend do? He aligns the list of altitudes from your GPS with the list of times from your car, and gets a table like the following $$
\begin{array}{lll}
\textrm{Distance} & \textrm{Altitude} & \textrm{Time} \\
\hline \\
0\textrm{km}      & 500\textrm{m}     & 0\textrm{s} \\
1\textrm{km}      & 550\textrm{m}     & 30\textrm{s} \\
2\textrm{km}      & 600\textrm{m}     & 65\textrm{s} \\
\ldots
\end{array}
$$

He then computes a new column "Time Delta" which contains the difference between each time and the one that immediatly precedes it. The new table looks like this
$$
\begin{array}{lll}
\textrm{Distance} & \textrm{Altitude} & \textrm{Time} & \textrm{Time Delta} \\
\hline \\
0\textrm{km}      & 500\textrm{m}     & 0\textrm{s}   & - \\
1\textrm{km}      & 550\textrm{m}     & 30\textrm{s}  & 30\textrm{s} \\
2\textrm{km}      & 600\textrm{m}     & 65\textrm{s}  & 35\textrm{s} \\
\ldots
\end{array}
$$

Each time delta then reflects the time it took to drive 1 kilometer (or mile), i.e. it reflects the velocity you were driving with (altough in seconds/kilometer, not kilometers/hour as usual. You could call that "slowness" instead of "speed", because while it refelects the velocity, it reflects higher velocities with lower values, and lower velocities with higher values). Each Time Delta also reflects the time you spend at the given altitude, since it's simply the time between the last altitude measurement and the current one.

All your friends needs to do then is to again average the altitude values, but weighted with their corresponding time deltas. In other words, he simply multiplies each altitude with the corresponding time delta, adds up all those producs, and divides by the sum of the time deltas (which is simply the delta between the first and the last logged time). And voila, the resulting average is much higher than the one you got, and upon notifying the authorities your ban is lifted and your name is cleared.

Stiltjes Integrals do exactly this kind of weighted summation. In the Stiltjes integral
$$
  \int f(\xi) \,d\alpha(\xi)
$$
$f(\xi)$ corresponds to the mapping from distance to altitude that your GPS gave you, while $\alpha(\xi)$ corresponds to the mapping of distance to time that your car logged. The integral weights each $f(\xi)$ with the rate of change of $\alpha$ at $\xi)$, i.e. it computes $$
  \int f(\xi) \alpha'(\xi) d\xi \text{.}
$$
Note how, compared to the tables above, the computation of differences between successive values has been replaced by taking the derivative. This is necessary since when $x$ ranges over a continuum, there's no longer such a thing as an immediate predecessor of a certain $x$.

Beware that this formula is only correct if $\alpha(\xi)$ is continously differentiable!

Since, Steiltjes integral is the generalization of Reimann integral so, basic ideas on R integral is must before getting into the R-S integral or simply Steiltjes integral.

Ok, i would like to give the geometrical interpretation of Steiltjes integral so that you will get the intuitive meaning of R-S integral.

GEOMETRICAL INTERPRETATION OF R-S INTEGRAL:

Here in R-S integral, let’s consider we are integrating the function $f(\xi )$ with respect to the monotonic function $\alpha (\xi )$within the interval$[a,b]$.

mathematically we represent as below,

$$\int_{a}^{b}f(\xi )d\alpha (\xi )$$

let’s get into the geometrical interpretation of above integral.

we generally take the area of the $f(\xi )$with respect to the x-axis within certain interval in Reimann integral. Similarly we are calculating the area but a bit complex than the Reimann integral.

let’s take three axes where we keep $\xi $ and the functions $f(\xi )$ and $\alpha (\xi )$ at x-axis,y-axis and z-axis respectively. Now, we erect the wall from the curve traced by those functions then we can have the junction curve(since the intersection of the planes is curve). let’s take a light source emitting the parallel rays of light along the x-axis which results in the formation of shadow behind the wall on $f− \alpha$ plane.
Hence, the R-S integral gives the area of the shadow.

For your another question(calculation), you can use the relation between the Reimann integral and Reimann-Steiltjes integral,

$$\int_{a}^{b}\xi ^{3}d\alpha (\xi ) = \int_{a}^{b}\xi ^{3}{\alpha }'(\xi )d\xi $$

thanks!


Nonempty, associative, and closed under inverses but not a group Given an example of a set $G$ and an operation $*$ on $G$ such that $*$ is not a binary operation on $G$ but associative, identity and inverses properties hold?

Basically, try to find an example to show that the closure property must be hold to be a group

How about the set $G = \{-1,0,1\}$ under usual addition?

All integers under addition.
Let Z= all integers:
where G= (Z,+,0)
Closure property:

consider the subset {1,2} under integers,
where 1+2=3 and 3 is an integer. Thus, proves the closure property holds for all integers under addition.


Independence of two non-negative integer valued random variables Let $X,Y$ be two non-negative integer valued random variables defined on a probability space $(\Omega,\cal F, \Bbb P)$. The question is,

  If $\Bbb P\{X=i,Y=j\}=\Bbb P\{X=i\}P\{Y=j\}$ for every $i,j\ge0$, then are $X,Y$ independent random variables? i.e. the events $\{X\le a\}$ and $\{Y\le b\}$ are independent for any $a,b\in\Bbb R$?

Anyone can help with a proof of this? Thank you!

We want to show that

$$P(X \le a, Y \le b) = P(X \le a) P(Y \le b)$$

If a or b is a negative number, then

$$LHS = 0 = RHS$$

If a and b are zero, then

$$LHS = P(X = 0, Y = 0)$$

$$RHS = P(X = 0) P(Y = 0)$$

If a and b are nonnegative integers not both zero, then

$$LHS = P\left([\bigcup_{x=0}^{a} X = x] \bigcap [\bigcup_{y=0}^{b} Y = y]\right)$$

$$ = P(X=0, Y=0 \ or ... \ or X=0, Y=b \ or X=1, Y=0 \ or ... \ or X=1, Y=b \ or ... \ or X=a, Y=0 \ or ... \ or X=a, Y=b)$$

$$ = P(X=0, Y=0)+ ... +P(X=0, Y=b) + P(X=1, Y=0) + ... + P(X=1, Y=b)$$

$$+ ... + P(X=a, Y=0) + ... + P(X=a, Y=b) \tag{*}$$

$$ = P(X=0) P(Y=0)+ ... +P(X=0) P(Y=b) + P(X=1) P(Y=0) + ... + P(X=1) P(Y=b) + ... + P(X=a) P(Y=0) + ... + P(X=a) P(Y=b)$$

$$RHS = P(\bigcup_{x=0}^{a} X = x) P(\bigcup_{y=0}^{b} Y = y) = [P(X=0) + ... + P(X=a)][P(Y=0) + ... + P(Y=b)] = (*)$$

If a and b are nonnegative numbers not both zero, then same as above but replace
$a$ with $\lfloor a \rfloor$
$b$ with $\lfloor b \rfloor$

except

Yes, that is the definition of independence of two discrete random variables.  And since $X$ and $Y$ are independent then if $A$ is any event for $X$ and $B$ is any event for $Y$ then $A$ and $B$ are independent.  In particular the two events you specify.


Trouble integrating $\int{\sqrt{1+x^{2}+x^{4}}} dx$ I'm confronted with the below integral:

$$\int{\sqrt{1+x^{2}+x^{4}}} dx$$

which has appeared within integrating a helio-central magnetic field.

The numerical evaluation just did work for our application. But I am curious to find a way for analytical calculation of it.

Does anybody have any idea?...

Best


Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality. Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.

I'm kinda stuck on how to do this problem. Any help would be appreciated. Thank you in advance!

By AM-GM, we have
$$
\sqrt{x}+\sqrt{x}+\frac{1}{x}\geq3\sqrt[3]{\sqrt{x}\sqrt{x}\frac{1}{x}}=3.
$$
Equality is iff $\sqrt{x}=\frac{1}{x}$ iff $x=1$.

We want to prove:

$(1)$ $$\forall x>0, \frac{1}{x} \geq 3-2\sqrt{x}  $$

Let's consider two cases:

Case $1$: $3-2\sqrt{x} \geq 0 \Rightarrow x \leq \frac{9}{4}$

Then both sides of the inequality are positive.

$\frac{1}{x} \geq 3 - 2\sqrt{x} \Leftrightarrow  3x - 2x^{1.5}\leq 1 \Leftrightarrow x> 0$

Therefore the inequality $(1)$ holds for $0< x\leq \frac{9}{4}$

Case $2$: $x>\frac{9}{4}$

Then we clearly see that $3-2\sqrt{x} < 0 $ and $\frac{1}{x} > 0 \Rightarrow \frac{1}{x} > 3-2\sqrt{x}$

So generalising to all $x>0$, we get that

$\frac{1}{x} \geq 3-2\sqrt{x}$ as required.


"Is not Lebesgue integrable" When one talks about Lebesgue integrability or lack thereof, what does one mean, exactly? Does it mean we cannot possibly take a Lebesgue integral of that function, because it simply isn't well-defined ... or does it mean the integral is finite? So if the latter, even a function like $f(x) = c$ for $c > 0$ is not Lebesgue integrable?

A function which is not Lebesgue integrable has (at least) one of the following properties:

It is not (Lebesgue) measurable. In this case, essentially all bets are off. Nonmeasurable functions can only be constructed using some form of the axiom of choice,so that all functions appearing in practice (piecewise continuous, monotonic,...) are Lebesgue measurable.
Note that it can happen for $|f|$ to be measurable (even integrable), although $f$ is not measurable. An example is $f=1_A - 1_{A^c}$, where $A$ is not measurable.

$f$ is measurable, but $\int |f|\,dx =\infty$. This means that $f$ is "too large" to admit a finite integral. In this case,it can still happen that $f$ is quasi integrable which means $\int f_+ \, dx <\infty$ or $\int f_-\, dx <\infty$, where $f_+,f_-$ denote the positive and negative parts of $f$. In these cases, we have $\int f \,dx =-\infty$ or $\int f \,dx =\infty$, respectively.

Finally, if $f$ is integrable, this means that $f$ is measurable and $\int |f|\,dx <\infty$, which then implies that $\int f \,dx$ is a well defined real number.

It means it doesn't satisfy the requirements to be Lebesgue integrable.

Expanding on that
If I say something isn't X it means it didn't satisfy the definition of X (or something logically equivalent to it)
if $P\implies $is lebesugue integrable then showing not $P$ isn't sufficient. If Lebesugue integrable $\implies P$ then $¬P\implies$ not lebesgue integrable.
As usual.


eigen vectors and linear independence I know that it is the case that if we have eigenvectors $v_1....v_r$ that correspond to distinct eigen values $λ_1....λ_r$ of an $n$ x $n$ matrix, then {$v_1....v_r$} is linearly independent.
But is is it possible to have $a_1$ linearly independent eigen vectors corresponding to $λ_1$, $a_2$ linearly independent eigen vectors corresponding to $λ_2$... $a_p$ linearly independent eigen vectors corresponding to $λ_n$, and the set of all $r$ eigenvectors where $r=a_1+a_2+...+a_p$ be linearly dependent?
For example: consider a $3x3$ matrix. Imagine we have one vector(call it $v_1$) forming a basis for the eigenspace of a one eigen value, but two vectors(call them $v_2$ and $v_3$) forming a basis for different eigen value. are we guaranteed that {$v_1,v_2,v_3$} are linearly independent?

Suppose  is any field. For example, it could be ℝ or ℂ.
Suppose $M$ is an n×n -matrix.
Suppose $\lambda_1, \dots, \lambda_r$ is a list of distinct eigenvalues of $M$.
Suppose:

$v_{1, 1}, \dots, v_{1, k_1}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_1$.
$v_{2, 1}, \dots, v_{2, k_2}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_2$.
$\vdots$
$v_{r, 1}, \dots, v_{r, k_r}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_r$.

Suppose $\alpha_{1, 1}, \dots, \alpha_{1, k_1}, \alpha_{2, 1}, \dots, \alpha_{2, k_2}, \dots, \alpha_{r,1}, \dots, \alpha_{r, k_r} \in \mathbb{F}$ are scalars such that
$$(\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}) +
  (\alpha_{2, 1} v_{2,1} + \dots + \alpha_{2, k_2} v_{2, k_2}) + \dots +
(\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r}) = 0.$$
Define $u_1 = (\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}), \dots,
u_r=(\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r})$.
Clearly, $u_1$ is either the zero vector or a $\lambda_1$-eigenvector of $M$. Analagously, $u_2$ is either the zero vector or a $\lambda_2$-eigenvector of $M$. And so on for $u_3, \dots, u_r$.
By the theorem you say you know, it must be the case that for each $i \in \{1, \dots, r\}$, $u_i$ is actually the zero vector.
But then for each $i \in \{1, \dots, r\}$, all the coefficients $\alpha_{i,1}, \dots, \alpha_{i, k_i}$ must be zero, because the vectors $v_{i,1}, \dots, v_{i, k_i}$ are linearly independent.
We have shown that any linear combination of $v_{1, 1}, \dots, v_{1, k_1}, \dots, v_{r, 1}, \dots, v_{r, k_r}$ equal to zero must actually be the trivial combination (i.e., with all coefficients zero). ∎

We claim that the set of all $r$ eigenvectors is linearly independent. The main proof ideas are already provided in the comments. For each $i=1,\dots, p$ let $v_{i1},\dots, v_{ia_i}$ be linearly independent system of eigenvectors corresponding to the eigenvalue $\lambda_i$ and  $\mu_{i1},\dots, \mu_{ia_i}$ be scalars such that $$\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$
For each $i$ put $v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij}$. Then $v_i$ is an eigenvectors corresponding to $\lambda_i$ and
$$\sum_{i=1}^p v_i=\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$
The linear independence property formulated in the first paragraph of the question implies that $0=v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij} $ for each $i$. Then the linear independence property formulated in the second  paragraph of the question implies that $\mu_{ij}=0$ for each $j$.


Binomial formula in $GF(2^m)$ there is a binomial formula:

$$(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$$

When operations are done in $GF(2^m)$ then all positive integers are reduced $\bmod2$, so binomial formula for $n=2^i$ in $GF(2^m)$ is:

$$(x+y)^{2^i}=x^{2^i} + y^{2^i} $$

So now the question. If there are reduced all binomial coefficients $\binom{n}{k}$, then why exponents like $2^i$ are not reduced?

Regard the simplest non-trivial example to get a better understanding.

For example $\{0,1,x,x+1\}$ with $2=0$ and $x^2=x+1$.

It's because multiplication by a coefficient is periodic in the coefficient with the right period $P$:
$$ (a+P)\cdot b = a\cdot b + P\cdot b \equiv a\cdot b \quad {\rm mod} \quad P$$
because the equivalence modulo $P$ is defined so that it allows me to subtract multiples of $P$ such as $P\cdot b$ above - while powers are not periodic functions of the exponent with the period $P$:
$$ a^{b+P} \neq a^b \quad {\rm mod}\quad P$$
For a simple example, look at powers of $2$ computed modulo $2$:
$$2^0 = 1, \quad 2^1 = 2 = 0, \quad 2^2 = 4 = 0, \dots $$
All the following powers are equal to $0$ modulo $2$ but the initial one, the zeroth power, is not. (Pick more complicated numbers to get the point - to see how complicated the powers may be especially if $P$ is large.) So the results of the exponentiation are not periodic as a function of the exponent, and you are never allowed to simplify the exponents by considering them modulo $P$.


Finding the area of a quadrilateral I have a quadrilateral whose four sides are given

$2341276$, $34374833$, $18278172$, $17267343$ units.

How can I find out its area? What would be the area?

Have you seen this (MathWorld article)? In particular, the area of a planar convex quadrilateral is given by $\frac{1}{4}(b^2+d^2-a^2-c^2) \tan \theta$ where $a,b,c$ and $d$ are the side lengths.

If  unlike sidelengths actually restrict formation any other quad then our given quad becomes unique and even geometrically we cannot change theta values i.e angles for the given quad.


Find the radius of convergence of the power series $\displaystyle\sum_{n=0}^{\infty}a_nz^n$, where $a_{2k+1} = 2^k$ and $a_{2k} = (1 + (1/k))^2$ for $k = 0, 1, 2, \dotsc$

I started off by doing the ratio test, but I know that the ration test is for $|a_{k+1} |/|a_k|$ and here they throw in the $2k+1$ so I am utterly lost and don't know how to start. can anyone help?

Cauchy-Hadamard makes it really simple here since

$$\lim_{n\to\infty}\sup\sqrt[n]{a_n}=\lim_{n\to\infty}\sup\begin{cases}\sqrt[n]{2^n}\\{}\\\sqrt[n]{\left(1+\frac1n\right)^2}\end{cases}\;\;=2$$

and the radius of convergence is thus $\;1/2\;$ .

Using the ratio test for odd/even-indexed subsequences:

$$\frac{a_{2n+2}}{a_{2n}}=\frac{\left(1+\frac1{n+1}\right)^2}{\left(1+\frac1n\right)2}\xrightarrow[n\to\infty]{}1$$

$$\frac{a_{2n+1}}{a_{2n-1}}=\frac{2^n}{2^{n-1}}=2$$

and we're done (why? Why can we split the sequence as above and do the ratio test separatedly) and then we choose the minimal one?)


(Soft Question) Largest known semiprimes with no known factors Is there a list, similar to prime numbers and probable primes, of the largest semiprimes with unknown factors? Is there a list of numbers that are either semiprime or prime, with no known factors? Is there literature on how to find large semi primes?

It doesn't exist. Any area of rectangle which sides are prime is a semiprime.


How to Show that $S=\{(x,y)\in R^2:x>y^2\}$ is open Show that $S=\{(x,y)\in R^2:x>y^2\}$ is open

quite simple one. We need to choose $\epsilon$ for open balls: $D((x_0,y_o),\epsilon)\subset S$ ,$\forall x_o,y_o\in S$.

we can take $\epsilon$ as the closest distance to the parabola. but couldnt do

You don't need to take $\epsilon$ as the closest distance to the parabola.

Take $(x,y) \in S$.  We have $x > y^2 \implies x - y^2 > 0$.

If $0 < r < 1$, note that
$$
(x-r) - (y+r)^2 = (x - y^2) - (r - 2yr + r^2) > (x - y^2) - (2 + 2|y|)r
$$
So, select and $r>0$ satisfying $r < \min\{(x - y^2)/(2 + 2|y|),1\}$.  How can we show that the ball of radius $r$ about $(x,y)$ is contained in $S$?

Property: A set is open if its complement is closed. Every open ball of radius d around any point of x<=y² intersects infinite points of x<=y² so the complement of S is closed and S is open.


Taylor Series First Order Method for $x' = x-2t$ with $x(0) = 0$ given: With the step sizes h = 1, 1/2, 1/4

My solution for all step sizes came out to be zero and not sure if I am doing this right.

Since this is first order taylor series method, I'll be using: $x_{n+1}$ = $x(n) + hx'(n)$ when $x(0) = 0$.

So because x'(0) = 0, no matter what step size I use, I end up getting 0.

Your solution is correct. You can check it  by direct method too.

$x'(t)-x=-2t$;  $x(0)=0$ is Linear first order DE, whose solution is easily obtained as:

$x(t)=2t+2-2e^t$. Expanding $e^t$ by Taylor's series;

$x(t)=2t+2-2\left(1+t+\frac{t^2}{2!}+...\right)$ which on approximating up to first order term(as you are using Taylor's series method of first order or Euler's method) gives $x(t)=2t+2-2(1+t)=0$ which is the same as obtained by you.


Determining whether there exists an integer $a$ such that $\text{ord}_{20}(a) = 8$. I am trying to determine whether there exists an integer $a$ such that $\text{ord}_{20}(a) = 8$. I know that if $(a,n) = 1$ and $n>0$, then $\text{ord}_{n}(a)\mid \phi(n)$. I cannot use any abstract algebraic techniques.  I am only expected to know what primitive roots are, Euler's Theorem, and some other elementary number-theoretic principles related to congruences, linear diophantine equations, and Fermat's Little Theorem.

I understand that I can answer this question by going through each of $1,2,3,\ldots,20$ and checking to see if each raised to the power of $8$ is congruent to $1$ mod 20. However, there is probably a faster way to do this and I am looking for help from the StackExchange community on how to do that. Thank you in advance.

Let $a$ be an integer coprime with $20$. Then $a$ is coprime with both $4$ and $5$.

By the Fermat—Euler theorem, we have $a^2 \equiv 1 \bmod 4$ and $a^4 \equiv 1 \bmod 5$.

By the Chinese Remainder theorem, these congruences imply $a^4 \equiv 1 \bmod 20$.

Hence, there is no integer having order $8$ mod $20$.

No there does not exists any such element. Because if $\exists~ a\in U_{20} $ then since order $U_{20} $ is $\phi(20)=\phi(2^2.5)=8$ which shows that $U_{20} $ is cyclic group. This is a contradiction since we know that only $\displaystyle U_{p^n}~\&~U_{2p^n}$ where $p$ is odd prime , are cyclic groups.


How to tell if a differential equation is homogeneous, or inhomogeneous? Sometimes it arrives to me that I try to solve a linear differential equation for a long time and in the end it turn out that it is not homogeneous in the first place.

Is there a way to see directly that a differential equation is not homogeneous?

Please, do tell me.

For a linear differential equation
$$a_n(x)\frac{d^ny}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y=g(x),$$
we say that it is homogenous if and only if $g(x)\equiv 0$. You can write down many examples of linear differential equations to check if they are homogenous or not. For example, $y''\sin x+y\cos x=y'$ is homogenous, but $y''\sin x+y\tan x+x=0$ is not and so on. As long as you can write the linear differential equation in the above form, you can tell what $g(x)$ is, and you will be able to tell whether it is homogenous or not.

Any function like y and its derivatives are found in the DE then this equation is homgenous

ex.  y"+5y´+6y=0 is a homgenous DE equation

But   y"+xy+x´=0 is a non homogenous equation becouse of the X funtion is not a function in Y or in its derivatives


Find all function $f(x)$ such that $f(x^2 + x + 3)f(3x + 1) = f(6x^3 + 7x^2 + 16x + 3), \forall x \in \mathbb R$. Find all function $f(x)$ such that $$\large f(x^2 + x + 3)f(3x + 1) = f(6x^3 + 7x^2 + 16x + 3), \forall x \in \mathbb R$$

I have provided my solution but there isn't a strong claim for why $f(x)$ is in the form of $(ax + b)^n$ where $a, b \in \mathbb R, n \ge 1$ (as Luca Bressan suggested).

I think the following can help. Define:$$a=x^2+x+{5\over 2}\\ b=6x+1$$and $g(x)=f(x+0.5)$, therefore$$g(a)g(b)=g(2ab)$$


Kalman decomposition using Hautus test In linear control theory, the Kalman decomposition is used as a similarity transformation to decompose a given linear time-invariant system

$$\dot{x}(t)=Ax(t)+Bu(t)$$
$$y(t) = Cx(t) + Du(t)$$

into its controllable & observable, controllable & not observable, not controllable & observable and not controllable & not observable subsystems.

A system is controllable if $\mathcal{C}=[B\quad AB \quad A^2B \ldots A^{n-1}B]^T$ has rank $n$ (number of rows/columns of the system matrix $A$).

A system is observable if $\mathcal{O}=[C \quad CA \quad CA^2 \ldots CA^{n-1}]^T$ has rank $n$ (number of rows/columns of the system matrix $A$).

It seems possible to construct the similarity transformation by using the eigenvectors of the Hautus test for controllability and observability.

It states a system is controllable if for all eigenvalues $\lambda$ of $A$ the matrix $[A-\lambda I\quad B]$ has full rank $n$ for a $n \times n$ system matrix $A$. A similar statement reads: A system is observable if for all eigenvalues $\lambda$ of $A$ the matrix $[A^T - \lambda I \quad C^T]$ has full rank $n$ for a $n \times n$ system matrix $A$.

The user fibonatic from Engineering Stack Exchange suggested that one could construct the similarity transformation for the Hautus test. I am wondering, how this is possible.

This is the same example as in the post on Engineering Stack Exchange.

$$\dot{x} = \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}x + \begin{bmatrix}0 & 1 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} u$$
$$y = \begin{bmatrix}1 & 1 & 1 \end{bmatrix} x$$

The similarity transfromation is given by
$$
M =
\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 & 0 \\
0 & 1 & -1
\end{bmatrix}.
$$
This is what I have tried. The three eigenvalues of $A$ are all $\lambda =1$ the eigenvectors are $v_1 = [1, 0,0]^T, v_2=[0,0,1]^T$ and $v_3=[1,0,1]^T$. The rank of the $[A-\lambda B]$ is 2. That means there is a deficit in rank for every eigenvector. If I include the eigenvectors to a "similarity" transformation

$$\tilde{M} =
\begin{bmatrix}
1 & 0 & 1\\
0 & 0 & 0\\
0 & 1 & 1\\
\end{bmatrix}
$$

it turns out that it is singular and cannot be the similarity transformation that I need.

  So how can I use the Hautus test to construct the similarity
  transformation for the Kalman decomposition?

A.Γ. is right that you need to use the generalised eigenvectors, so do a Jordan decomposition instead of an eigendecomposition. So in case of the example in your question the $A$ matrix has the following generalised eigenvectors

$$
V =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix}
$$

where the Jordan canonical form of $A$ consists out of two blocks of size two and one, both with the eigenvalue one.

However $\text{rank}\left[A-\lambda\,I \quad B\right]$ and $\text{rank}\left[A^\top-\lambda\,I \quad C^\top\right]$ tells you nothing about which block and corresponding generalized eigenvectors are causing those matrices to lose rank. So this would be the limitation of using the Hautus test in order to obtain the Kalman decomposition. The structure of a Jordan block actually always ensures that the largest rank loss caused by that block is at most one. So all this method would tell you is how many blocks have rank loss, but not which and by how much. For example when

$$
A =
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}, \quad
B_1 = \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}, \quad
B_2 = \begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
$$

where $A$ is one Jordan block of size three with eigenvalue zero, but for both $B_1$ and $B_2$ does $\left[A-\lambda\,I \quad B_i\right]$ lose rank. But for $B_1$ only one state is uncontrollable, while for $B_2$ two states are uncontrollable.

So this method fails when the matrix has an eigenvalue with algebraic multiplicity larger than one, but the rank loss (either for controllability, observability or both) is less then that. Because in that case it is not clear which subset of (generalised) eigenvectors needs to be taken.

Eigenvector $v_3=v_1+v_2$ which implies that your eigenvectors aren't independent.


Formula for standard deviation So I am studying for a final exam and I know, or thought I knew, the formula for standard deviation. I quadruple checked in my notes and the slides for the lecture and see that the formula for the standard deviation when $\sigma$ is unknown is:

$\sqrt((\Sigma d^2 - nd^-) /  n - 1)$

But for some reason on the final review solution sheet our professor somehow and some reason replaced $n$ in the numerator with $1/n$ and I have no clue why. So the formula is now:

$\sqrt((\Sigma d^2 - (1/n)d^-) / n - 1)$

My question is, how come we are able to use $1/n$ instead of $n$ and on the test, how can I tell to use $1/n$ as compared to $n$?

Thanks

Yes your professor has a typo.  The standard deviation for paired samples is usually written as follows:

$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}$$

What you professor is doing is using the trick (or arithmetic if you want to think of it that way) that we can rewrite the numerator as

$$\sum_{i=1}^n(d_i-\bar d)^2 = \sum_{i=1}^nd_i^2-n\bar d^2$$

And thus, the formula for the standard deviation should be

$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}= \sqrt{\frac{\sum_{i=1}^n(d_i)^2-n\bar d^2}{n-1}}$$

And finally, it should be $n$ and not $\frac{1}{n}$.

Your formulas are wrong as $\bar d$ needs to be squared.

Now, you can indeed meet valid formulas with $n$ and $\dfrac1n$, because

$$\color{blue}n{\bar d}^2=n\left(\frac1n\sum d_i\right)^2=\color{blue}{\frac1n}\left(\sum d_i\right)^2.$$

Confusion can arise from this.


Distance Formula Problem If two vertices of an equilateral triangle are $(1, -1)$ and $(-\sqrt{3}, - \sqrt{3})$, find the coordinates of the third vertex.

Step by Step procedure to get the answer.

Take $A=(1, -1)$, $B=(-\sqrt{3}, - \sqrt{3})$.
Let the third vertex be $C=(x, y)$.

The distance d between the two points $A=(x_1, y_1)$ and $B=(x_2, y_2)$ is given by the formula
$$
d= \sqrt{ (x_2-x_1)^2 + (y_2 - y_1)^2 }
$$
So we get the distance $AB = 2 \sqrt{2}$
. Similarly we have to find the distance between BC & AC using distance formula where we get the equation in the form of x & y even though the distance is $2\sqrt{2}$ because it is equilateral triangle. But the correct solution is not able to get. Please help.

Distance between two given points will be the side of the triangle $a=\sqrt{(1+\sqrt{3})^2+(\sqrt{3}-1)^2}=2\sqrt{2}$. Now you just need to draw two circles of radius $a$ and with centers at the known vertices. The equations of the circles are:
$(x-1)^2+(y+1)^2=8$
$(x+\sqrt{3})^2+(y+\sqrt{3})^2=8$
The circles will intersect at two points and to find the coordinates you need to solve this system:$$ \begin{cases}
(x-1)^2+(y+1)^2=(x+\sqrt{3})^2+(y+\sqrt{3})^2 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$

$$ \begin{cases}
y(2-2\sqrt{3})=x(2+2\sqrt{3})+4 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$

$$ \begin{cases}
y=\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}} \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$Finally, we get a quadratic equation for $x$: $$x^2-2x+1+\Big(\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}}+1\Big)^2=8 $$ The solutions are: $x=-1, x=2-\sqrt{3}$. We already have $y$ expressed via $x$ so just plug the $x$ values in that equation and find $y$.

After writing the equations of the two circles to find the points of intersection, I found the resulting quadratic equations to be too tedious for my patience.  Instead, I determined the (parametric) vector from $(-\sqrt{3}, - \sqrt{3} )$ to $(1, -1)$ to be

$$A = \left(-\sqrt{3} + \left(1+\sqrt{3}\right) T, -\sqrt{3} + \left(-1+\sqrt{3}\right) T\right)$$ where $0 \le T \le 1$.

I then added $(\sqrt{3}, \sqrt{3})$ to this to translate it to the origin, then multiplied it
by the rotation matrix $\begin{bmatrix} \cos &  -\sin \\ \sin & \cos \end{bmatrix}$, using $\theta = +60^{\circ}$ and $T=1$.  Lastly,
I subtracted $(\sqrt{3}, \sqrt{3})$ to translate back.  After MANY sign corrections, I got
one of the two possible points to be $(-1,1)$, which checks out using the distance formula.
No quadratics to solve, just vectors to add and rotate.

(Sorry about my lack of proper math notation formatting; if I find this site to be interesting, I will learn MathJax.)


Probability of grabbing blue ball We have a bag that has

  $3$ blue balls and $2$ white balls.

When we take a blue ball from the bag, we just remove it from the bag.
When we take a white ball from bag, we return that ball and add another white ball.

What is the probability that on our first take we take a blue ball when we know that after $2$ takes the amount of balls in the bag has changed?

My approach:

We know that amount of balls in bag changed after $2$ takes , so:

$( B , B ) = -2$ balls

$( B , W ) = 0$ balls

$( W, W ) = 2$ balls

$( W, B ) = 0$ balls

the only acceptable order is $BB$ or $WW$.

And now I am confused about how to decide what is the probability that on our first take we took a blue ball?

The probability for blue ball is $3/4$ and when we take that ball we have $2/4$ chance of another blue ball.

So is it $(3/4) \cdot ( 1/2)$

OR should I take in considering that we have $1/2$ chance, that $( B, B )$ occurs, and then multiply it by $3/4$ ?

I know that in this case, both results are the same, but if we had a different amount of balls, it would differ, and I am not sure which approach is correct.

Thanks for your help.

Let $E$ denote the event that after two takes the number of balls in the bag
has changed.

Let $B_{i}$ denote the event that at the $i$-th take a blue ball
is taken, and let $W_{i}$ denote the event that at the $i$-th take
a white ball is taken.

Then to be found is $P\left(B_{1}\mid E\right)$

We have the relation:

$$P\left(B_{1}\mid E\right)P\left(E\right)=P\left(B_{1}\cap E\right)\tag1$$

So after finding $P\left(E\right)$ and $P\left(B_{1}\cap E\right)$
we can also find $P\left(B_{1}\mid E\right)$.

First we will go for finding $P(E)$:

$$P\left(E\right)=P\left(\left(B_{1}\cap B_{2}\right)\cup\left(W_{1}\cap W_{2}\right)\right)=P\left(B_{1}\cap B_{2}\right)+P\left(W_{1}\cap W_{2}\right)=$$$$P\left(B_{2}\mid B_{1}\right)P\left(B_{1}\right)+P\left(W_{2}\mid W_{1}\right)P\left(W_{1}\right)=\frac{2}{4}\frac{3}{5}+\frac{3}{6}\frac{2}{5}=\frac{1}{2}$$

Now we will go for finding $P\left(B_{1}\cap E\right)$:

$$P\left(B_{1}\cap E\right)=P\left(B_{1}\cap B_{2}\right)=P\left(B_{2}\mid B_{1}\right)P\left(B_{1}\right)=\frac{2}{4}\frac{3}{5}=\frac{3}{10}$$

Substituting these results in $(1)$ we find: $$P\left(B_{1}\mid E\right)\frac{1}{2}=\frac{3}{10}$$
and we conclude that: $$P\left(B_{1}\mid E\right)=\frac{3}{5}$$


Let $ABCD$ be a rhombus Let  $ABCD$ be  a  rhombus and  $E$  and $F$ be  the  middle of the  sides  $AB$  and $BC$.  If  $P$  is  a point  in  the  interior   of  the  rhombus  such   that  $PE = CE$ and $ AP = AF$  then  $B, P, D $ are collinear points. I  have  no  idea. I  tried using coordinates. I  think is  necessary a construction.

$P$ has the same distance from $A$ and $F$ thus it lies on the perpendicular bisector of $AF$ and for similar reason lies on the perp bisector of $CE$.

It follows that $P$ is on the perpendicular bisector of $AC$ which contains both $B$ and $D$ so $B,P,D$ are collinear.

QED

Edit

It cannot be that $PF=AF$ and $B,P,D$ are collinear. Look at the second picture. The red circle is the locus of the point $P$ having distance $AF$ from $A$, the circle with centre $A$ and radius $AF$. As $P$ must be on the perpendicular bisector of the segment $CE$ it follows that line $BD$ does not contain $P$.

The hint.

Use  reflection $S$ of the rhombus in the line $BD$.

Thus, $S(P)=P$.


Mistake in the Algorithm Design Manual? Look at this excerpt from the Algorithm Design Manual by Skiena, 2nd Edition:

"The sum of the first $n$ even integers"? Surely, the two given versions of the sum given do not include only even numbers. Is this a mistake on the author's part, or am I missing something?

Expanding those 3 equation makes the author's statement true, i.e:

$ 1+2+3+...+n=\frac{n(i+(n-i+1))}{2}=\frac{(ni+n^2-ni+n)}{2}=\frac{n^2+n}{2}=\frac{n(n+1)}{2}=2+4+6+...+2n$

The confusing part could be the 2nd one though:

$ \sum_{i=1}^{n/2}(i+(n-1+1)) = 1/2(i+(n-i+1) \underbrace{...}_{n} i+(n-i+1) ) =  \frac{n(i+(n-i+1))}{2}$


Integral Related to Exponential Function and Modified Bessel Function 1 I'm looking for a closed form solution for the following integral:

$$S(x) = \displaystyle \int\limits_0^x r \,I_0(r) \,e^{-b^2r^2} \,dr $$

where $I_0()$ is the $0$-th order modified Bessel function of first kind,  $b>0$, and $x>0$.

Thanks.

Using the integral representation for $I_0$ and Fubini's theorem, the given integral equals

$$ \frac{1}{\pi}\int_{0}^{\pi}\int_{0}^{x} r \exp\left(r\cos\theta-b^2 r^2\right)\,dr\,d\theta $$
or
$$ \frac{1-I_0(x)e^{-b^2 x^2}}{2b^2}+\frac{1}{4b^3\sqrt{\pi}}\int_{0}^{\pi}\cos(\theta)e^{\frac{\cos^2\theta}{4b^2}}\left[\text{Erf}\left(\frac{\cos\theta}{2b}\right)-\text{Erf}\left(bx-\frac{\cos\theta}{2b}\right)\right]\,d\theta. $$
The second term is not really elementary, but such representation provides the asymptotics for $r\to 0^+$ or $r\to +\infty$ in a simple way. Would you mind sharing the original probabilistic problem?


Uniqueness of solution to 1st order pdes I am given a 1st order partial differential equation $y{\partial \psi\over\partial x}+x{\partial \psi\over\partial y}=0$ subjected to boundary condition $\psi(x,0)=\exp(-x^2)$. I have found that a solution is $\psi(x,y)=\exp(y^2-x^2)$. But I am asked when the solution is unique. Could someone please explain how to answer this? Thanks.

Consider the parametric curves $x = A e^t + B e^{-t}$, $y = A e^t - B e^{-t}$, which satisfy $x' = y$, $y' = x$.  Along such a curve any solution $\psi$ must be constant, according to the chain rule:
$$ \frac{d}{dt} \psi(x(t),y(t)) = \psi_x \frac{dx}{dt} + \psi_y \frac{dy}{dt} = 0$$
Now the curve intersects $y=0$ if and only if $A$ and $B$ are either both positive (i.e. $x > |y|$), both negative ($x < -|y|$), or both $0$ ($x=y=0$).  So a boundary condition on $y=0$ produces uniqueness only in the regions $|x| \ge |y|$. In the region $|y| > |x|$ the solution  is not unique.   For example, you could add $f(y^2 - x^2)$  to $\psi(x,y)$ where $f$ is differentiable with  $f(s) = 0$ for $s \le 0$.

Uniqueness can be addressed in the following way. Let us suppose that exist another solution $\phi(x,y)$ such that $\phi(x,0)=e^{-x^2}$ then, being your equation linear then also $\phi'(x,y)=a\psi(x,y)+b\phi(x,y)$, with two arbitrary coefficients $a$ and $b$ is a solution. The boundary condition will give $a+b=1$. So, unless you do not give another condition on $y$ your solution cannot be unique.

Of course, you have also the other condition in the given problem. From the fact that $\psi(x,0)=e^{-x^2}$ and from the other fundamental result that your equation has the general solution (characteristic method cited in the comments) $\psi(x,y)=\psi(x^2-y^2)$ for you is enough to set $\psi(0,0)=1$ and your solution is unique.

Finally, I would like to point out the simple way the solution OP proposed can be found. One have to search for a solution in the form

$$\psi(x,y)=\phi(y)e^{-x^2}$$

with $\phi(0)=1$ and the solution is immediately obtained, consistent with the characteristic method.


Pigeonhole principle (I think): colored points in the plane Suppose that each points in $\Bbb R^2$ is colored red, green or blue.

Prove that either there are two points of the same color a distance $1$ unit apart, or there is an equilateral triangle of side length $\sqrt3$ all of whose vertices are the same color.

I thought I would separate each point into cases. Like, the combinations you can have are $(R,R),(R,B),(R,G)$, etc. and something about there being an infinite number of points, so the ones that are $(R,R),(G,G)$ or $(B,B)$ have to not be empty? Not sure if this helps.

I also tried putting all the points on the unit circle and looked at the $\sqrt3$ triangles created by tangent circles all with centers of the same color.

You can actually prove more. You can prove that there are two points of distance $1$ part with the same color in all cases. Suppose not, and that every two points of distance one apart have different colors. Suppose the origin is colored blue. I claim that every point on a circle of radius $\sqrt{3}$ centered at the origin is also colored blue. Draw the radius out to any point and put two equilateral triangles on the radius as in BaronVT's picture where you have a dotted line (representing the radius) bisecting two back-to-back triangles. The two middle vertices are colored red and green (in some order) because the origin is colored blue. But this forces the vertex lying on the circle to be blue. Now that we have a circle of radius $\sqrt{3}$, just pick a chord of length $1$. This will give two points colored blue of distance one apart, which is a contradiction.

The right proof,

First, if there are two points A, B distance $2$ are different color, then the midpoint O, must be either same as A or B; O,A same color, then draw $2$ triangles using side OA, we have two points C, D with same color with B which is $\sqrt{3}$ side equilateral triangle
If there are no two points are different color, then easy deduct all the points must be same color. QED


Verifying proof for the following question I wanted to check my proof if it is correct or not. Using these field axioms:

(i)Trichotomy Property:  Exactly one of $x<y$, $y<x$, or $x=y$ hold.

(ii)Transitivity: if $x<y$ and $y<z$ (which we could write in shorthand as $x<y<z$), then $x<z$.

(iii)If $x<y$ then $x+z < y+z$.

(iv) If $x<y$ and $z>0$ then $xz<yz$.

  (A1) Addition is commutative
  (A2) Addition is associative
  (A3) Addition has a neutral element $0$
  (A4) Any element has an additive inverse
  (A5) Multiplication is commutative
  (A6) Multiplication is associative
  (A7) Multiplication has a neutral element $1$
  (A8) Any non-zero element has a multiplicative inverse
  (A9) Multiplication distributes over addition

Prove that for all $a,\, b,\, c\in\mathbb{R}$, if $a<b$ and $c<0$ then $bc < ac$.

Since $a<b$, by (A4), we have that $$
0 = a + (-a) < b + (-a) = b - a.$$
Since $b - a > 0$ and $c < 0$, then $c(b-a) < 0$. By (A9), this is equivalent to $cb-ca<0$. By (A3), $$
cb=cb-ca+ca<0+ca=ca.$$
Thus $cb<ca$.

"Since b−a>0 and c<0, then c(b−a)<0"

Why do you state this. Either this is what you wish to prove, or (and I suspect this) you are using  iv) but not saying so.

So you should say so.

Since $(b-a)> 0$ and $c < 0$ then by iv) $c(b-a) < 0\cdot (b-a)$

But notice. You don't have an axiom that $0\cdot x = 0$(!!!!)

You have to prove that.

Hint 1: $0d = (0+0)d$

Hint 2: $-0d$ must exist.

....

Once you prove that you have

$c(b-a) <0$

So by A9 $c(b-a) = c(b+(-a)) = cb +c(-a)$

So $cb + c(-a) < 0$ and

....

Now we need to prove $c(-a) = -ca$.

Hint: $c(-a) + ca = c(-a + a)$.

...

Now we have $cb - ca <0$ and ... you can finish your proof.

Your proof has a gap.  How do you know that $b-a \gt 0 \land c \lt 0 \Rightarrow c(b-a) \lt 0$?  I'm not saying you're wrong -- I'm just saying you haven't yet proved it.

Edit to add:

Lemma:  If $c \lt 0 \land x \lt y, \text{ then } xc \gt yc$.

Proof:  $c+(-c) = 0 \land c \lt 0 \Rightarrow -c \gt 0$.  (Use property (iii), adding $-c$ to both sides of the inequality.)

Thus, $-xc \lt -yc$.  Add $xc + yc$ to both sides:  $yc  \lt xc$ and the Lemma is proved.


Forming a conjecture about the relationship between $g(A \cup B)$ and $ g(A) \cup g(B)$ for an arbitrary function g. Question: Form and prove a conjecture about the relationship between $g(A \cup B)$  and $ g(A) \cup g(B)$ for an arbitrary function g.

I've already shown this:
For an arbitrary function $g$: $\mathbb{R} \to \mathbb{R}$, it is always true that\ $g(A\cap B)\subseteq g(A) \cap g(B)$ for all sets  $A, B \subseteq \mathbb{R}$.

by this method:

Suppose $y \in g(A \cap B)$. Then there exists $x \in A \cap B$ such that $g(x)=y$. So, $x \in A$ and $x \in B$. Thus, $g(x) \in g(A)$ and $g(x) \in g(B)$. So, $y\in g(A)$ and $y \in g(B)$. Therefore, $y \in g(A) \cap g(B)$. Thus,  $g(A\cap B)\subseteq g(A) \cap g(B)$.

-Really just need help forming the conjecture. I don't really know where to start.

Conjecture: Let $g(x)$ be a function. Then $g(A\cup B)=g(A)\cup g(B)$

For any $h \in g (A \cup B) $ then there is an $a \in A \cup B $ with $h = g (a) $ Since $a \in A \cup B$ then either $a \in A $ or $a \in B $ (or both) and $g(a) \in g (A) $ or $g (a) \in g (B) $ (or both) so $h=g(a) \in g (A) \cup g (B) $.

So $g (A \cup B) \subset g (A) \cup g (B)$.

If $h \in g (A) \cup g (B) $ then either $h \in g (A) $ and $h = g (a)$ for some $a \in A $ or $h \in g (B);h = g (a);a \in B $ (or both).  Either way $h = g (a) $ for some $a \in A \cup B$ so $h \in g (A \cup B) $.

So $g (A)\cup g (B) \subset g (A \cup B) $

So $g (A)\cup g (B) = g (A \cup B) $


How to solve this issue using basics of counting? We have 20 tutors in a mathematics college, 8 are specialized in algebra, 5 are specialized in discrete mathematics, 4 are specialized in applied mathematics and 3 are specialized in statistic.

How many ways as we can form a scientific committee consisted of 8 tutors in each of these situations:

It has 2 tutors from each section.
Three of them at least must be from discrete mathematics section.
One of them at least must be from algebra section.
Three of them must be from discrete math, but we have to take into consideration that 1 tutor from discrete math and 1 tutor from statistic mustn't be together in the committee.

Please explain your solutions.

You need to choose two (combinations - order doesn't matter) from each group, for a total of choosing 8.

Choose 2 from the 8 alebra, 2 from the 5 discrete math, 2 from the 4 applied math, and 2 from the 3 stats:

$8 \choose  2$$\times$ $5 \choose  2$ $\times$ $4 \choose  2$$\times$   $3 \choose  2$ = 5 040
You have to choose at least three from discrete math. That means that you have to choose 3, 4, or 5 (there are a maximum of 5 that can be chosen).

So, combinations for choosing three from discrete math is the 5 discrete math, choose 3, multiplied by what's left - the other 15 tutors, choose 5 to make 8 total chosen:

$5 \choose 3$ $\times$ $15 \choose 5$

Similarly for choosing four from discrete math, choose the 4 of 5 from discrete math, then choose the remaining 4 tutors to be chosen from the other 15 tutors:

$5 \choose 4$ $\times$ $15 \choose 4$

Then for choosing all 5 discrete math tutors:

$5 \choose 5$ $\times$ $15 \choose 3$

Now you have all the combinations for choosing 3, 4 and 5 discrete math tutors. Just add them all up to get the final answer of choosing at least three:

$5 \choose 3$ $\times$ $15 \choose 5$ +  $5 \choose 4$ $\times$ $15 \choose 4$ + $5 \choose 5$ $\times$ $15 \choose 3$ = 37 310
At least one of them must be from algebra. We can use the compliment here. At least one of them being from algebra is the same as NOT zero of them being from algebra. We can calculate what we don't want and then subtract that from the sample space (total number of all possible combinations) to get what we do want.

Sample space (20 tutors, choose 8 of them): n(s) = $20 \choose 8$

Then, the combinations to get ZERO tutors from algebra is 8 from algebra, choose 0 of them, multiplied the rest of the tutors, choose the 8 remaining chooses:

$8 \choose 0$ $\times$ $12 \choose 8$

And then finally, just subtract that from the sample space:

n(s) - $8 \choose 0$ $\times$ $12 \choose 8$ = $20 \choose 8$ - $8 \choose 0$ $\times$ $12 \choose 8$
                               = $20 \choose 8$ - $12 \choose 8$
                               = 125 475
Exactly 3 have to be chosen from discrete math, but one person from discrete math and one person from stats don't get along, so they can't be together.

So, let's just exclude the bad egg from discrete math:

5 tutors from discrete math, but take out 1, so it's 4, choose 3. Then multiply that by the 15 other tutors that are not from discrete math and choose the remaining 5 chooses:

$4 \choose 3$ $\times$ $15 \choose 5$

That, OR, exclude the bad egg from stats:

Choose 3 from all 5 in discrete math, but choose the remaining 5 chooses from only 14 of the remaining tutors, because you have to exclude the bad egg from the stats:

$5 \choose 3$ $\times$ $14 \choose 5$

And then you would just add those up:

$4 \choose 3$ $\times$ $15 \choose 5$ + $5 \choose 3$ $\times$ $14 \choose 5$ = 32 032

On number 4, I'm not so sure my answer is correct. The way I did it, the combinations of neither of them being on the committee may have overlapped, so you would have to subtract that from the answer, like this:

32 032 - $4 \choose 3$ $\times$ $14 \choose 5$

And the 14 choose 5 may be wrong as well. You may have to replace $14 \choose 5$ with:

$3 \choose 2$ $\times$ $12 \choose 3$ + $3 \choose 1$ $\times$ $12 \choose 4$ + $3 \choose 0$ $\times$ $12 \choose 5$

because 3 stats choose only 2 (excluding bad egg) and then choosing the other 3 from the remaining 12 tutors (not stats or discrete) and so on down to choosing zero from the stats.

I can't think right now, so someone else has to confirm this for me.

You need 8 tutors and must have 2 tutors from each group.

From algebra you need to select 2 tutor out of 8  so it can be done in $8\choose 2$ ways.

Similarly for discrete mathematics it can be done in $5\choose2$ ways.

For applied mathematics it can be done in $4\choose3$ ways.

And for statistics it can be done in $3\choose2$ ways.

Since each of the selections made for the committee are independent so all the above ways will be added as:

$8\choose2$+$5\choose2$+$4\choose2$+$3\choose2$=47

Hence committee can be formed in 47 ways.
Out of 8 members you need at least 3 from discrete mathematics.

It means that for selecting 3 from discrete mathematics you have $8\choose3$ ways.

Now for selecting other 5 members you have no conditions(means they cam be from any group).

Now since you have already selected 3 members so now you are left with only 17 members.

So for rest of the 5 members you can make selected in $17\choose5$ ways.

Now total number of ways would be $8\choose3$+$17\choose5$=6244 ways
Out of 8 members you have to select at least 1 from algebra.

It means you have $8\choose1$ ways of selecting tutor from algebra.

Now for selecting other 7 members you have no conditions(means they cam be from any group).

Now since you have already selected 1 members so now you are left with only 19 members.

So for rest of the 7 members you can make selected in $19\choose7$ ways.

Now total number of ways would be $8\choose1$+$19\choose7$=50396 ways.


Nonconstant harmonic function that is constant on open subset The following question arose while I was doing some homework - it's not a homework problem, just a question that might allow me to generalize the homework problem:

Let $h$ be a (real-valued) nonconstant harmonic function on an open connected set $U$. Is it possible that for some nonempty open subset of $U$ that $h$ is constant?

Reference: Theorems 1.27 and 1.28 (page 21) of Harmonic Function Theory by Sheldon Axler, Paul Bourdon, and Wade Ramsay.


ways to arrange positive integers from 1 to 100 on a circle In how many ways can the positive integers from 1 to 100 be arranged in a circle such that the
sum of every two integers placed opposite each other is the same? (Arrangements that are rotations
of each other count as the same.) Express your answer in the form a! · b^c
.

Since the question states "arrangements that are rotations of each other count as the same" , you need to be a bit careful.

After fixing the 1-100 diameter,

there are only 49 pairs left to be fixed and flipped, yielding $49!\cdot2^{49}$

If you flip the 1-100 diameter too, you will be duplicating previous arrangements looked at (say) clockwise.

For the sum of every opposing pair to be equal there will be only one potential partner for any given integer. How many ways can we arrange these fifty diameter pairs? Using circular permutation we know $49!$ is the number of ways we can arrange fifty items in a circle, but the pairs can be individually flipped. That means each potential flip doubles the possible configurations $2^{50}$

$49! \cdot 2^{50}$


What is the probability that P and Q have no common elements? A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of the subset of P. A subset Q of A is again chosen at random. Find the probability that P and Q have no common elements.

I tried to calculate in this way :

In set P we can have no element i.e.Φ, 1 element, 2 elements, ...... upto n elements. If we have no element in P, we will leave by all the elements and number of set Q formed by those elements will have no common element in common with P. Similarly, it there are r elements in P we are left with rest of (n - r) element to form Q, satisfying the condition that P and Q should be disjoint.

Now, my confusion is how can I find the Total number of ways in which we can form P and Q?

What you have done in the case $n=5$ is correct. If the first set has $k$ elements then the second set can be chosen in $2^{n-k}$ ways. Hence the total number us $\sum_{k=0}^{n} \binom {n} {k} 2^{n-k}=2^{n} (1+\frac 1 2)^{n}=3^{n}$.

$P,Q$ can both be one of the $2^n$ subsets of $A$. The total number of ways to form $P$ and $Q$ is $2^n\cdot2^n=2^{2n}$.


$S^{-1}R=0$ iff $0$ belongs to $S$. Let $R$ be a commutative ring with $1$ as identity and let $S$ be a multiplicative closed subset of $R$ containing $1$. Show that $S^{-1}R=0 \text{ iff } 0$ belongs to $S$.

My approach: Now $S^{-1}R=0$ implies if the $1$ in this ring is $zero$ i.e. $(1,1)~(0,1)$. This occurs if $x.1=0$ for some $x$ in $S$. That is if $0$ belongs to $S$. Now conversely let $0$ belongs to $S$. Then help me to procced in this direction.
Thanks for all your help in advance.

Recall that $(x,s)\simeq (x',s')$ iff $s'x=sx'$. The other way $(1,1)\simeq (0,0)$ since $1.0=0.1$


Infinitely many equations with infinitely many unknowns Let $$d_n=\sum_{m=1}^{\infty}a_ma_{m+n}$$ then can we find $\{a_n\}$ in terms of $\{d_n\}$? I appreciate any hints :-)

All values are real.
In general I am interested in $d_n$ such that $d_n \to 0$ and in general $d_n \neq 0$.
$\sum d_n < \infty$.

Not necessarily uniquely.

For example $d_0=1$ and $d_n=0$ for $n \not = 0$ can be generated by

\begin{equation}
    a_m=
    \begin{cases}
      \pm1, & \text{if}\ m=k \\
      0, & \text{otherwise}
    \end{cases}
  \end{equation}

given any positive integer $k$


What is the Geometric interpretation of $i^i$? We know that $i^i$ is real. But how to  explain it geometrically maybe in terms of rotation. like we can explain geometrically multiplication of two complex numbers and so on. Can someone show me a little bit  about geometric interpretation of $i^i$ and tell me if what I think below is true?

Note :(Some informations):
It is equal to $e^{i \cdot \log(i)}$, but $\log$ is only well-defined up to
adding integer multiples of $2 \pi i$. Thus in the correct setting $i^i$ is
each of the numbers $e^{-\frac{(4n+1)\pi}{2}}$, $n$ an integer.
Each of those are equally valid, thus finding a geometric interpretation
would by a bit silly.

Thank you for any help

Your note is precisely the point. There is a geometric meaning for addition and multiplication and real powers, but imaginary exponents don't have any because they must be defined in terms of $\exp$ and $\ln$ or something equivalent. And there is no geometric interpretation of $\exp$.

I don't have one, but maybe this helps.

I'd like to call $z^i$ the 'semiprocal' of $z$, because the $i^{\text{th}}$ power brings us 'halfway' to the reciprocal of $z$, being $\left(z^i\right)^i = z^{-1} = \frac1z$

Likewise, $i^i$ brings us halfway to the reciprocal of $i$, being $\left(i^i\right)^i = i^{-1} = \frac1i = -i$

Knowing that $i^i$ is 'halfway' between $i$ and $-i$, it is not that surprising that the answer is purely real.


How can I calculate "most popular" more accurately? I'm developing a website at the moment.

The website allows users to "rate" a post from 0 to 5.

Posts can then be displayed in order of popularity.

At the moment, my method of calculation is pretty primitive:

average_rating = total_rating/ratings

the problem is that a story with 1 rating of 5 is more popular than a story with 99 ratings of 5 and 1 of 4.

(5/1) > (499/100)

Could someone suggest a more accurate way to calculate popularity both on the number of votes and the quality of each vote?

A standard procedure (frequently -and loosely- called 'bayesian average') is to make a weighted average between the individual rating and the 'a priori' rating:

$R_a = W \; R + (1 - W ) \;  R_0$

where

$R_a = $ averaged ('bayesian') rating

$R = $ individual rating: average rating for this item.

$R_0 = $ a priori rating: global average rating, for all items in your database.

$W = $ weight factor: it should tend to $0$ if this items has few votes, and it should tend to $1$  if it has many.

Some choices: $W = \frac{n}{N_{max}}$, or $W = max( \alpha \frac{n}{N_{av}},1)$ , etc ($n=$ number of votes for this item, $N_{max}=$ maximum number of votes for all items, $N_{av}=$average, $\alpha=$ some number between 0.5 and 1... )
Also, frequently one discards items that have very low/big values when computing the statistics.

See some examples

Added: for another approach, specially for yes/no like/diskike votes, see here.

Yes, give every post an initial 3 to start with, so adjust your formula to

$${\rm average\,\ rating} = ({\rm total\ \,rating} + 3) / ({\rm ratings} + 1)$$

Alternatively, you may also amplify the posts with many ratings, e.g. by

$${\rm score} = \log ({\rm ratings}+1)  ({\rm total\ \,rating} + 3) / ({\rm ratings} + 1) $$

This score may grow by a "degree" every time the number of ratings gets multiplied by $e$, but this total degree will be reduced by the average rating.


Are there an infinite number of prime numbers where removing any number of digits leaves a prime? Suppose for the purpose of this question that number $1$ is a prime number.

Consider the prime number $311$. If we remove one $1$ from the number we arrive at the number $31$ which is also prime. If we removed $3$ instead of $1$ we would arrived at the number $11$ which is prime. If we remove $3$ from the number $31$ we arrive at the prime number $1$ and if we remove $1$ from $31$ we arrive at the number $3$ which is prime. And if we removed number $1$ from the number $11$ we would arrive at the prime number $1$.

So what property does this number $311$ have?

It has the property that if we remove one digit in any way we want we arrive at the prime number and if we remove two digits in any way we want we also arrive at the prime number.  (Of course, if the original prime number had $n$ digits then it should be prime whatever $k$ digits we remove, for $1\le k\le n-1$.)

The question would be:

  Are there an infinite number of prime numbers so that removing any number of digits leaves us with a prime number?

It seems to me that big enough prime numbers will rarely have the property that for every possible removal of some of the digits we will arrive at the prime number, but I see no reason that forbids an infinite number of these prime numbers.

It is clear that we cannot have digits $0,4,6,8,9$ in those prime numbers. There can be at most one $2$ because $22$ is composite,, at most one $3$ because $33$ is composite,at most one $5$ because $55$ is composite, at most one $7$ because $77$ is composite, at most two $1$`s because $111$ is composite, so such prime number can have at most $6$ digits so there is a finite number of such prime numbers.

I thought about what happens in bases b other than b = 10.

We cannot have the digit zero, or any digit that is a composite number (for example in hexadecimal: 4, 6, 8, 9, 10, 12, 14, 15). The possible digits are the number 1 and the primes p < b.

No digit other than 1 can appear more than once, since xx = x * 11 in every base. The digit 1 can appear at most three times, since 1111 = 11 * 101 in every base.

If the base b is odd, then a number de is even if d and e are both odd, so a solution x cannot contain two odd digits. The best we can ask for is 2d or d2 where d is an odd prime or 1.

Otherwise the base b is 6k, 6k+2, or 6k+4. If b = 6k+4 then x modulo 3 equals the sum of the digits, modulo 3. A solution x cannot contain both digits d = 1 (modulo 3) and digits d = 2 (modulo 3). It also cannot contain exactly three digits d = 1 or 2 (modulo 3). The largest possible solution has 3 digits, with one digit = 3, and the other two being equal (modulo 3). This case includes b = 10 (decimal numbers), so four or more digit solutions are impossible.

If b = 6k + 2, then for x = de we have x modulo 3 = (e - d) modulo 3. Therefore a solution x cannot have two digits that are the same (modulo 3). The largest possible solution has 3 different digits, with the first or the last digit = 3, and the other two being different (modulo 3).

The remaining case is b = 6k, where we can have at most three 1s, plus one of each prime < b. A digit 2 or 3 would have to be the first digit because d2 is divisible by 2 and d3 is divisible by 3 for any digit d. The number of digits in a solution x is therefore strongly limited, but I can't find any simple evidence that limits the number of digits.

In practice, if you take a five digit number x, there are ten 2-digit numbers, ten 3 digit numbers, five 4 digit numbers and one 5 digit number that would have to be primes, which is quite unlikely. And as the base b gets bigger, it gets more unlikely to find that many primes.

PS. I wrote a program systematically examining bases b up to about 4,000. There are thousands of four digit solutions. I found two five digit solutions: In base 12, 511b7 (b = 11) is a prime, and removing any digit or digits leaves a prime or the digit one. And in base 2730, the same for the five digit number (1)(1)(1423)(829)(631).

It looks like in base 2730 = 2 * 3 * 5 * 7 * 13, numbers where the last digit is a prime are primes much more often than in other bases.

Since a six digit solution requires several five digit solutions (at least four if the six digit solutions has three 1s and more if there are fewer 1s) for the same base b, and I found only two five digit solutions in total, a solution would be hard to find.

It's easy to show that if b is not a multiple of five, then for every six digit number x, that number or one of the numbers created by removing one to four digits is divisible by five. So to find a six digit solution, the base b must be a multiple of 30.

The chance that a random number x is prime is about 1 / ln x. But if we know that x is not divisible by 2, 3, or 5, then the chance is 3.75 / ln x. For x to be a six digit solution, the number x, plus six 5-digit, fifteen 4 digit, twenty 3-digit and fifteen 2-digit numbers must be prime. If we pick x where no digit is 2, 3, or 5, the chances for all these numbers to be prime is

$3.75^{57} / (ln p^6) / (ln p^5)^6 / ... / (ln p^2)^{15})$

or

$(3.75/ ln p)^{57} / (6 * 5^6 * 4^{15} * 3^{20} * 2^{15})$

On the other hand, the number of six digit numbers with all digits prime is about $(p / ln p)^6$. We can expect one solution x if the product is about 1. That's the case if the base is about b = 3.5 x $10^{15}$. At that point there are about $10^{85}$ six digit base-p numbers with all digits primes or 1. I don't think such a solution will ever be found.


Is chess Turing-complete? Is there a set of rules that translates any program into a configuration of finite pieces on an infinite board, such that if black and white plays only legal moves, the game ends in finite time iff the program halts?

The rules are the same as ordinary chess minus the 50 move rule, exchanges and castling.

And what is the minimum number of different types of pieces (i.e. the simplest game) that is needed for a chess-like game to be Turing-complete?
(Each type of piece having a set of allowed moves that is invariant under translations)

The classic reference on this is the paper "Computing a Perfect Strategy for nxn Chess Requires Time Exponential in n" by Aviezri Fraenkel and David Lichtenstein; unfortunately, I can't find the (30-year-old!) paper online anywhere and while I think I have the proceedings it came from, it's buried in a pile of boxes.  Of course, your board isn't finite, but your initial piece configuration is; this means that if there's any computable bound on the needed board size to resolve an initial position, then chess isn't Turing-complete.  (If the status of all positions of diameter $d$ could be determined using boards of size no more than $f(d)$, then they can be determined in $\mathrm{EXP}(f(d))$ time by the EXPTIME completeness result, and thus wouldn't be Turing-complete).  I vaguely recall seeing something about such a result, but unfortunately some quick searches aren't turning anything up.  The expert on the mathematics of chess is Noam Elkies, so hopefully he can spot this and chime in...

As chess is played on a finite board with finite many pieces, the number of which can not grow but decline during the game, chess is a game with finite many states and thus never Touring-complete. Though the number of states grows exponentally with the size of the board chess only represents a finite state machine and will be won by computers at some time in the future.


Function $f: \mathbb{R} \to \mathbb{R}$ that takes each value in $\mathbb{R}$ three times Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ that takes each value in $\mathbb{R}$ three times? If not, how could I prove that such a function does not exist?

Here's a continuous solution:

$$ f(x) = \begin{cases} 4x-3\lfloor x\rfloor & \text{when $\lfloor x\rfloor$ is even} \\
3\lfloor x\rfloor + 3 - 2x & \text{when $\lfloor x \rfloor$ is odd} \end{cases} $$

This takes on the values $0,4,2,6,4,8,6,10,8,\ldots$ at integer $x$ and interpolates linearly between them.

Any even number $2m$ is hit at $f(2m)$ (a local minimum), $f(2m-\frac32)$ (the middle of an increasing line segment), and $f(2m-3)$ (a local maximum). All other numbers are hit by a point in an increasing line segment, then a decreasing one, and finally another increasing one.

A smoother function with the same overall shape would be $g(x)=ax+\sin x$, except the coefficient $a$ has to be just right in order to get the local maxima and minima to coincide, and I'm too lazy to figure out what it should be.

$\text{cotan}(\frac{3\pi}{1+e^{-x}})$ will do. Here is the plot of the function:


gaussian integral from zero to infinty I want to calculate the following integral:
\begin{align} I_1'&= \int^\infty_0 e^{-\alpha^2r^2/2 + iwr} \ dr \\
    & = e^{-w^2/2\alpha^2} \ \int^\infty_0 e^{-\alpha^2\big{(} \frac{r}{\sqrt{2}} - \frac{iw}{\sqrt{2} \alpha^2} \big{)}^2} \ dr.\end{align}
Here I used completion of the square technique to get a Gaussian integral. However, if I substitute $u = \frac{r}{\sqrt{2}} - \frac{iw}{\sqrt{2} \alpha^2}$ the bounds go from $\frac{-iw}{\sqrt{2} \alpha^2}$ to $+\infty$ which I cannot calculate further using (half of) the formula
\begin{align} \int^{+ \infty}_{- \infty} e^{a(x+b)^2} dx = \sqrt{\frac{\pi}{a}}. \end{align}
Am I doing something wrong or is there a way ?

Matching your integral to your formula, you want $u = r / \sqrt{2}$.  Then your bounds are $0$ and $\infty$, $a = -\alpha^2$ (This minus sign makes me think something is missing from your formuala.), and $b = \mathrm{i}w/(\sqrt{2} \alpha^2)$.


Find $\beta$ as an element of $S_7$ Let $\beta^8 = (1$ $5$ $4$ $3$ $6$), a 5-cycle. Find $\beta$ as an element of $S_7$.

Any help would be appreciated on this problem. I do not have one similar to this. Thanks!

It is enough to find $\beta$ that is a $5$-cycle.

Then $\beta^{5}=1$ and so $\beta = \beta^{16} = (1 5 4 3 6)^2$.


Finding the integer $\le n$ with largest number of divisors As mentioned in an answer to this question an integer less than $n$ with largest number of divisors can be found exploring the numbers $x$ of the form
  $$ x = 2^{a_1} 3^{a_2} \dots p_k^{a_k} \dots $$
(where $p_k$ is the $k$-th prime number) with the conditions
  $$ x \le n < 2x \quad \text{and}\quad a_1 \ge a_2 \ge \dots \ge a_k \ge \dots$$
to determine the complexity of this algorithm I would like to know the asymptotic number of tuples $(a_1, a_2, \dots)$ verifying these conditions as $n\to \infty$. I suspect that this number is $\gg_l \log^l n$ for every $l$ and $\ll_\epsilon n^\epsilon$ for every $\epsilon > 0$, but I don't know how to prove it.

Thanks for your help.

For a given $n$, let $N(n)$ be the number of admissible exponent sequences, that is, the number of sequences $a_1 \geqslant a_2 \geqslant \ldots$ on nonnegative integers such that
$$\frac{n}{2} < \prod_{\nu = 1}^{\infty} p_{\nu}^{a_{\nu}} \leqslant n\,. \tag{1}$$
Then indeed we have $N(n) \ll_{\epsilon} n^{\epsilon}$ for every $\epsilon > 0$ and $N(n) \gg_l \log^l n$ for every $l$.

This is easy to prove if we write the product in $(1)$ as a product of primorials. The non-increasingness of the exponent sequence makes this possible. Define $k = k(n)$ as the index such that
$$p_k\# \leqslant n < p_{k+1}\#\,. \tag{2}$$
Then by the prime number theorem we have $k \sim \frac{\log n}{\log \log n}$. If
$$\prod_{\nu = 1}^{k} (p_{\nu}\#)^{b_{\nu}} \leqslant n\,,$$
then a fortiori we have
$$(p_{\nu}\#)^{b_{\nu}} \leqslant n \iff b_{\nu} \leqslant \frac{\log n}{\log (p_{\nu}\#)} = \frac{\log n}{\vartheta(p_{\nu})}$$
for each $\nu$. Hence
$$N(n) \leqslant \prod_{\nu = 1}^{k}\biggl(1 + \biggl\lfloor \frac{\log n}{\vartheta(p_{\nu})}\biggr\rfloor\biggr) < 2^k\prod_{\nu = 1}^{k} \frac{\log n}{\vartheta (p_{\nu})}\,.$$
Now
$$2^k = \exp (k\log 2) = \exp \frac{(1 + o(1))(\log 2)\log n}{\log \log n} = n^{o(1)}$$
and
$$(\log n)^k = \exp (k\log \log n) = n^{1 + o(1)}$$
are straightforward. To estimate the denominator, we use
\begin{align}
\sum_{\nu = 1}^{k} \log \vartheta(p_{\nu}) &= \sum_{\nu = 1}^{k} \bigl(\log p_{\nu} + O(1)\bigr) \\
&= \vartheta(p_k) + O(k) \\
&= \log n + O(\log \log n) + O(k) \\
&= \log n + O\biggl(\frac{\log n}{\log \log n}\biggr)
\end{align}
per the Chebyshev bounds and $(2)$, thus
$$\prod_{\nu = 1}^{k} \vartheta(p_{\nu}) = n^{1 + O\bigl(\frac{1}{\log \log n}\bigr)}\,.$$
Putting everything together gives $N(n) = n^{o(1)}$, whence $N(n) \ll_{\epsilon} n^{\epsilon}$ for all $\epsilon > 0$.

For the lower bound, fix an arbitrary $l > 0$ and assume $n$ is so large that $k > l+1$. Then we count the products
$$\prod_{\nu = 2}^{l+1} (p_{\nu}\#)^{b_{\nu}} \leqslant n\,.\tag{3}$$
For every such product there is a unique $b_1$ such that
$$\frac{n}{2} < \prod_{\nu = 1}^{l+1} (p_{\nu}\#)^{b_{\nu}} \leqslant n\,,$$
hence the number of products $(3)$ is a lower bound for $N(n)$. These products correspond to the lattice points in the simplex $S \subset \mathbb{R}^l$ with vertices $0$ and
$$\frac{\log n}{\vartheta(p_{\nu})}\cdot e_{\nu-1},\qquad 2 \leqslant \nu \leqslant l+1,$$
whose ($l$-dimensional) volume is
$$\frac{1}{l!}\prod_{\nu = 2}^{l+1} \frac{\log n}{\vartheta(p_{\nu})} = C_l \cdot \log^l n\,.$$
As $n \to \infty$ the number of lattice points in $S$ is asymptotically equal to the volume of $S$, hence $N(n) \gg_l \log^l n$ for every $l$.

As the question is posed, it is the largest m that 2^m is smaller than n. m is the answer.
In the discussion of the question, there is no mention the devisors are different from one another. If there is such requirement than the solution is the largest 2*3*5*7... all primes that maintain a value less than n.


Prove that for all positive reals $k,j$ with $2k>3j$, there exists positive integers $s,t$ such that $k > |\sqrt{s}-\sqrt{t}| > j$ Prove that for all positive reals $k,j$ with $2k>3j$, there exists positive integers $s,t$ such that

$k > |\sqrt{s}-\sqrt{t}| > j$

Let's show that there are positive integers $s, t$ such that:

$3j/2 > |\sqrt{s} - \sqrt{t}| > j$.

Note that $2k > 3j$ iff $k > 3j/2$, so that if we can find $s, t$ as above we'll be done.

Observe $\sqrt{n+1} - \sqrt{n} = 1/(\sqrt{n+1} + \sqrt{n}) \longrightarrow 0$ as $n$ tends to infinity. Pick $n$ large enough for this difference to be less than $j/2$. Then choose $a \in \mathbb{Z}^+$ such that $3j/2 > a(\sqrt{n+1} - \sqrt{n}) > j$.

(This will be possible, since increasing our prospective $a$ by $1$ amounts to taking "steps" of size less than $j/2$. Thus, you don't run the risk of totally "overstepping" the $j/2$ gap between $3j/2$ and $j$.)

Pick $s = a^2 (n+1)$ and $t = a^2 n$.

Another solution:

Select a rational $b/a$ with $a,b$ positive integers such that $2k>b/a>3j$. By multiplying the numerator and denominator by a constant integer, if necessary, we could ensure $3a^2>b$. Now we claim $s=a^2+b,t=a^2$ satisfy the condition. Proof:
$k>\frac{b}{2a}>\sqrt{a^2+b}-\sqrt{a^2}=\frac{b}{\sqrt{a^2+b}+a}>\frac{b}{3a}>j$


Need help with substitution for integration I'm trying to go through the solution for a SIS epidemic model, but I'm stuck figuring out the following steps.

So we have

$i′=r(k−i)i−αi$

And to solve this, they did a substitution $y=i^{−1}$. They integrated and came up with

$ye^{(rk−α)t}=∫re^{(rk−α)t}dt+C$

I have no idea how they ended up with this equation. So taking substitution $y = i^{-1}$, we get $\frac{dy}{dt} = -\frac{1}{i^2}$, but i don't know how to use that for the first equation.

Did they use separation of variables to solve this after the substitution? Anyone who could show me how they got to the second equation would be appreciated.

Thank you!

From $
-\mathrm{y}^{\prime}=(\mathrm{rk}-\alpha) \mathrm{y}-\mathrm{r}
$, we get $\mathrm{y}^{\prime}=-(\mathrm{rk}-\alpha) \mathrm{y}+\mathrm{r}$.
Thus we obtain the following linear first - order differential equation
$$ \mathrm{y}^{\prime}+(\mathrm{rk}-\alpha) \mathrm{y}=\mathrm{r}.$$
We get that the integrating factor is
$$
\mu(\mathrm{t})=\mathrm{e}^{\int(\mathrm{rk}-\alpha) \mathrm{dt}}=\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}}.
$$
Multiplying the differential equation by the integrating factor $\mu(\mathrm{t})$, we obtain
$$\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}} \mathrm{y}^{\prime}+\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}}(\mathrm{r} \mathrm{k}-\alpha) \mathrm{y}=\mathrm{re}^{(\mathrm{rk}-\alpha) \mathrm{t}},$$
which we can write as $$\left(e^{(\mathrm{rk}-\alpha) \mathrm{t}} \mathrm{y}(\mathrm{t})\right)^{\prime}=\mathrm{re}^{(\mathrm{rk}-\alpha) \mathrm{t}}.$$
Integrating both sides with respect to $t$,
$$
y e^{(r k-\alpha) t} = \int r e^{(r k-\alpha) t} d t+C
 $$

Regroup $$i′=r(k−i)i−αi = (rk-\alpha)i-ri^2$$

Divide both sides by $i^2$ to get $$i^{-2}i' =(rk-\alpha)i^{-1} -r$$

Let  $$y=i^{-1}$$

We get $$ -y'=(rk-\alpha)y -r$$

This is a linear equation with the given solution.


turing machine accept and reject state I am pretty new to Turing Machines and I am trying to understand the basic things first...so my question is , would this machine accept all words ending in 'a' ? if that's the case would the REJECT state be all string without 'a' and ending with 'b' ?

This Turing-machine's head will move to right on encounter of every 'a' and whenever it find the 'b', this will change it to 'a' and move to left and halt.

(a,a,R) : says that if tape symbol is 'a' keep it 'a' only and move R(Right).

(b,a,L) : says that if tape symbol is 'b' make it 'a' and move L(left) with halt.

So, this TM will accept all string ending with 'b', as it moves to halt state if and only if 'b' comes.

I hope this will help. good luck :)


Proving that Span{u, w} = Span{2u − w, w} Proving that Span{u, w} = Span{2u − w, w}, for all $u, w ∈ R^3$.

Let
$u=\begin{pmatrix} a \\ b \\ c \end{pmatrix}$
$w = \begin{pmatrix} d \\ e \\ f \end{pmatrix}$

$x_1, x_2, x_3, x_4 ∈ R$

$$x_1 \begin{pmatrix} a \\ b \\ c \end{pmatrix} + x_2 \begin{pmatrix} d \\ e \\ f \end{pmatrix} = x_3 \begin{pmatrix}2a-d\\ 2b-e\\ 2c-f\end{pmatrix} + x_4\begin{pmatrix}d\\ e\\ f\end{pmatrix}$$

When $x_1=2, x_2=0, x_3=1, x_4=1$, we get:

$\begin{pmatrix} 2a\\ 2b\\ 2c\end{pmatrix}$ = $\begin{pmatrix}2a-d+d\\ 2b-e+e\\ 2c-f+f\end{pmatrix}$

And therefore $\begin{pmatrix} 2a\\ 2b\\ 2c\end{pmatrix}$ = $\begin{pmatrix} 2a\\ 2b\\ 2c\end{pmatrix}$

Is this proof sufficient enough?

For any $x_1,x_2$ you need to find $x_3,x_4$ such that your equation is satisfied. Then you know $span(u,w)\subset span(2u-w,w)$ and still have to show $span(2u-w,w)\subset(u,w)$.

No, this is not a completed proof; you should show that, there exists a bijection between span(u,w) and span(2u - w, w)


Smallest possible area for triangle I'm solving the following question:

The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?

This is my attempt at solving the problem:
Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$
Let's find the tangent of the ellipse:
$2x^2 + y^2 = 1$
$4x + 2y \dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = \dfrac{-2x}{y}$
Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:
$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$
So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$
Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.
So now area is $1/2 a.b = 1/2 a^2$
Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.
I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ?  Unfortunately, the book I'm using doesn't provide
the answer to this question.

For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,  the equation of the tangent line at $(a\cos t, b \sin t)$ is given by
$$\frac{\cos t}ax+\frac{\sin t}by=1$$
which intersects the axes at $\frac a{\cos t}$ and $\frac b{\sin t}$, respectively. Then, the area of the triangle is
$$Area = \frac12 \frac a{\cos t}\frac b{\sin t}=\frac{ab }{\sin2t}\ge ab
$$
Here, $a=\frac1{\sqrt2}$ and $b=1$, allowing the smallest area $\frac1{\sqrt2}$.

Let $(h,k)$ be the point on ellipse from where you draw the tangent, then,
$$2h^2+k^2=1$$
You already know the slope of tangent is $$\frac{dy}{dx}=\frac{-2h}{k}$$
Constructing the equation of hypotenuse from this,
$$\frac{y-k}{x-h}=\frac{-2h}{k}$$
Putting $y=0$ , you get $x=a$,
$$a=\frac{1}{2h}$$
Similarly $$b=\frac{1}{k}$$
You need to minimise the area, i.e. , $$=\frac{ab}{2}$$
$$=\frac{1}{4hk}$$
$$=\frac{1}{4\cdot{k}\cdot{}\sqrt{\frac{1-k^2}{2}}}$$
$$=\frac{1}{2\sqrt{2}\cdot{k}\cdot{}\sqrt{1-k^2}}$$
Can you continue?