Importance if Rolle's and Lagrange's theorem in daily life: for school children I am school teacher and teaching math. I know the Rolle's theorem and Lagrange's theorem. I can solve the problems numerically then and there. However, I was completely failed to explain the significance or applications of these theorems by practical.

If this site can help me to explain both of the theorems with good practical examples are enough and I can teach my students well. Awaiting your replies.

You can explain Rolle's theorem by saying that if your average speed during a journey from A to B was say 50kms/hour then there had to be a time when your instantaneous speed was 50kms/hour as well. Significance of LMVT.

Rolles Theorem Example: When I throw a ball vertically up, its initial displacement zero ($f(a)=0$) and when I catch it again its displacement is zero ($f(b)=0$). As displacement function satisfy criteria of rolles theorem of continuity and differentiability over $(a, b)$ interval. There exists at least one $f'(c)= 0$, and in fact velocity is zero at the highest point. Here $f(x)$ is the height of ball and $f'(x)$ is the vertical velocity of the ball.


Find triples $(a,b,c)$ of positive integers such that... Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$

I found this on a local question paper, and I am unable to solve it.

Any help will be appreciated.

We have $(1+\frac{1}{3})(1+\frac{1}{2})^2=3$, so that is one solution. It also shows that at least one of $a,b,c$ must be $<3$. wlog we may take $a\le b\le c$. So $a=1$ or $2$.

Suppose $a=1$. Then $(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}$. Since $(1+\frac{1}{5})^2<\frac{3}{2}$ we must have $b<5$. Obviously we need $b>2$. We find $b=3$ gives the solution $(a,b,c)=(1,3,8)$ and $b=4$ gives the solution $(1,4,5)$.

Suppose $a=2$. Then we have $(1+\frac{1}{b})(1+\frac{1}{c})=2$. Since $(1+\frac{1}{3})^2<2$ we must have $b<3$. That gives the solution already noted of $(2,2,3)$.


Continuity of $\sqrt{1-x^2-y^2}$ I saw this question on an interim exam; Let $f: \mathbb{R^2} \rightarrow \mathbb{R}$ be given by

$f(x,y)= \left\{
\begin{array}{}
             \sqrt{1-x^2-y^2}  &x^2+y^2 \leq 1, \ y \geq 0 \\
              0 &\text{Otherwise}
\end{array}
\right.$

Now, the question was: Is $f$ continuous in $(0,0)$, $(1,0)$, $(1,1)$. Using the fact that $f$ is continuous in $a=(a_1,a_2)$ iff $\lim_{(x,y)\rightarrow (a_1,a_2)}f(x,y) = f(a_1,a_2)$, I thought that $f$ is continuous in $(0,0)$ and $(1,0)$, although I am not sure about $(1,1)$.

Could someone verify that my assumption about the first two points is correct, and help me verifying the last point? Thanks in advance!


what's the real value of $\sqrt2^{\sqrt2^{\large \sqrt2^{\sqrt2^{\unicode{x22F0}}}}}$? I heard this problem from my friend today, but unfortunately we could not figure out what's happening here.

$$x^{x^{\large x^{x^{\unicode{x22F0}}}}}=2$$
$$x^2=2$$
$$x=\sqrt2$$

but

$$x^{x^{\large  x^{x^{\unicode{x22F0}}}}}=4$$
$$x^4=4$$
$$x=\sqrt2$$

I have a feeling that this might be a duplicate, but I could not find this problem (basically I couldn't figure out what to search). So what's happening here, and what's the real value of
$$\sqrt2^{\sqrt2^{\large  \sqrt2^{\sqrt2^{\unicode{x22F0}}}}}$$

?

Finding the value of an infinitely iterated function can be tricky. Usually, if $z$ is a fixed point of a function $f$ (that is, if $f(z)=z$), then it often happens that
$$\lim_{n\to\infty} f^n(x)=z, \forall x$$
...however, things tend to get a little bit complicated when $f$ has more than one fixed point, which is the case with your function
$$f(x)=\sqrt 2^x$$
that has fixed points
$$f(2)=2$$
$$f(4)=4$$
In this case, the value of $f^\infty(x)$ depends on $x$, which doesn't really make sense given the way you wrote it. Basically this means that its value depends on the number "on top" of your power tower. For example, if you start with $4$, and observe the sequence
$$a_0=4$$
$$a_{n+1}=\sqrt 2^{a^n}$$
then each $a_n=4$, and so
$$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=4$$
However, if you take $a_0=1$,
$$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=2$$
However, things are a little bit less complicated in this case, since $x=2$ turns out to be something called an "attracting fixed point". Basically, we end up with
$$f^\infty\left(x\right) = \left\{
     \begin{array}{lr}
       2 & : x \lt 4\\
       4 & : x = 4\\
       \infty & : x \gt 4\\
     \end{array}
   \right.\\$$

The problem here is that $4$ is not in the range (image) of the power tower function (which is only defined for $x\in [e^{−e},e^{1/e}]$). So, the assumption that the power tower equals 4 is false and thus gives a false "solution".


Trigonometric limit $\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$ The limit is
$$\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ I was able to solve it using L'hopital and the answer that I got was $2$.

Can you please confirm if the answer is right and suggest some other way to evaluate the limit without using L'hopital?

Multiply both deminator and numerator to the $\left( 1+\sqrt { 2 } \sin { x }  \right) $ and $\left( 1+\tan { x }  \right)  $ respectively
\begin{align*}
\lim_{x\to\frac{\pi}{4}}{\frac{1-\tan x}{1-\sqrt{2}\sin x}}={}&\lim_{x\to\frac{\pi}{4}}{\frac{\left(1-\tan^2\!x\right)\left(1+\sqrt2\sin x\right)}{\left(1-2\sin^2\!x\right)\left(1+\tan x\right)}}={} \\
{}={}&\lim_{x\to\frac{\pi}{4}}{\frac{\frac{\cos(2x)}{\cos^2\!x}\left(1+\sqrt2\sin x\right)}{\cos(2x)\left(1+\tan x\right)}}={} \\
{}={}&\lim_{x\to\frac{\pi}{4}}\frac{\left(1+\sqrt2\sin x\right)}{\cos^2\!x\left(1+\tan x\right)}=2.
\end{align*}

Let $f(x) = 1-\tan x, g(x) = 1-\sqrt 2\sin x.$ The expression equals

$$\frac{f(x) - f(\pi/4)}{g(x) - g(\pi/4)} = \frac{(f(x) - f(\pi/4))/(x-\pi/4)}{g(x) - g(\pi/4)/(x-\pi/4)} .$$

By definition of the derivative, this $\to f'(\pi/4)/g'(\pi/4)$ as $x\to \pi/4.$ This quotient of derivatives is easy to compute.


How many 5-letter words are possible out of a 6-letter word? How many 5-letter strings are possible out of the 6-letter word APPLES?

Here my attempt: if the 5-letter string has just 1 P. Then we have 5! ways of getting such strings. If the 5-letter string has 2 P's, then there are 4 ways this can be done namely:

P, P, A, L, E

P, P, A, L , S

P, P, A, E, S

P, P, L, E, S

Each of of the above 4 cases has: 5!/2! = 60 distinct 5-letter strings.
So the answer is : 120 + 4*60 = 360 such strings.

A generating function approach might work. It should be the exponent of $x^5$ in $$5! \left(1 + \frac{x}{1!}\right)^4 \left(1 + \frac{x}{1!} + \frac{x^2}{2!}\right)$$

This gives me 360.

Related: 6-letter permutations in MISSISSIPPI


Find the equations to the normal lines of a curve where $x = 4$. Find the equations to the normal lines for the curve defined by

  $$y^2+xy-x^2=5$$

  In the points where $x=4$.

I'm not entirely sure how to even tackle this kind of problem, but let's see:

When $x=4$, we would have

$$y^2+4y-16=5$$

And we would have two solutions: $y=3$ and $y=-7$, so we need the equations for the normal lines that pass through $(4,3)$ and $(4,-7)$.

So I'm already confused. How come this curve has two images for $x = 4$?
I don't think I even understand what I am being asked.

You are given an implicitly defined curve $F(x,y)=0$ and found points $(x_0,y_0)$ on this curve. The tangent at these points is defined by implicit differentiation
$$
∂_xF(x_0,y_0)·(x-x_0)+∂_yF(x_0,y_0)·(y-y_0)=0
$$
which also tells you the direction of the normal to the tangent and thus to the curve.

See, there are two tangent lines for the circle at $x=0$.


Let $ A={z\in \mathbb{C}: |z|>1 }$ and $ B={z \in \mathbb{C}: z \neq 0}$ then which of the following is true. Let $ A=\lbrace z\in \mathbb{C}: |z|>1 \rbrace$ and $ B=\lbrace z \in \mathbb{C}: z \neq 0\rbrace $ then which of the following is true.
The options are
a. There exists a continuous onto function from A to B

b. There exists a continuous one one function from B to A

c. There exists a non constant analytic function from A to B

d. There exists a non constant analytic function from B to A

I Can prove that C is true by taking just an inclusion map. Also i can prove that d is not true. As if it is true then say there is some non constant analytic function f from B to A. then clearly the function has either pole or essential singularity at 0 but then by picards theorem(in first case) and by casorati Weirstrass theorem(in second case) we get a contradiction to function being non constant. So c option is false.

Now topologicaly both the sets A and B are connected, non compact, open etc. so have same properties.So trying a few example it seems that 1 is true by considering $f(z)=e^z$. And i think option B is true as we can have f to be some specific branch of logarithm. or function can also be defined in such a way that it shifts all the points continuously 1 unit distance radially outward i.e. $f(z)=(r+1)e^{i \theta}$ where $z=re^{i\theta}$.Is this argument right?

For options a) and b), we may use the fact that the punctured plane is homeomorphic to the compliment of closed unit disc

1.$f(z)=$ \begin{cases}|z| &\text{if} |z|>1\\\dfrac{1}{|z|} &\text{if}|z|\le 1\end{cases}

2..$f(z)=$ \begin{cases}e^z &\text{if} |z|>1\\\dfrac{1}{e^z} &\text{if}|z|\le 1\end{cases}

3.You are correct

$(d)$ ;

If $f:B\to A$ then $|f(z)|>1\forall z$.

Hence define $g(z)=\dfrac{1}{f(z)}$ which is entire .Also $|g(z)|<1$ Hence $g$ is bounded entire function and so constant and hence so is $f$


Prove an upper bound for the binomials This is (supposed to be) an upper bound on the binomial coefficient:

$$ \binom{n}{k} \le \frac{n^n}{k^k(n-k)^{n-k}}$$

If we prove it by induction for all integers $0 \le k \le n/2$, we can easily show that it generalizes for $k \le n$, because $\binom{n}{k} = \binom{n}{n-k}$.

How can we prove it by induction? I get to:

$$ \binom{n}{k} \le \frac{n-k+1}{k}\frac{n^n}{(k-1)^{k-1}(n-k+1)^{n-k+1}} $$

And this is where I get stuck.

I personally thinks that there is an pitfall in Daniel Fischer's solution, but I can not comment directly because lack of reputation.

Daniel used Bernoulli inequality to prove $1-\frac{n-1}{{(n+1)}^2}\le{(1-\frac1{(n+1)^2})}^{n-1}$, but the actually form of Bernoulli inequality is $1 + rn \le {(1+r)}^n$, where r is a constant, but here $r = -\frac1{(n+1)^2}$ is a function of n.

So, when can't apply Bernoulli inequality directly, but follow the induction technique.
${(1-\frac1{(n+1)^2})}^{n} \ge  {(1-\frac1{(n+1)^2})} \times {(1-\frac{n-1}{(n+1)^2})}$
   $ = 1- \frac{n}{(n+1)^2} + \frac{n-1}{(n+1)^4}$, but what to prove is $ {(1-\frac1{(n+2)^2})}^{n} \ge 1-\frac{n}{{(n+2)}^2}$, it is trivial that $ {(1-\frac1{(n+2)^2})}^{n} \ge {(1-\frac1{(n+1)^2})}^{n} $, but it is easy verify that $ 1- \frac{n}{(n+1)^2} + \frac{n-1}{(n+1)^4} \not \ge 1-\frac{n}{{(n+2)}^2}$, so we  can't conclude that $ {(1-\frac1{(n+2)^2})}^{n} \ge 1-\frac{n}{{(n+2)}^2}$.


IMO 2016, Problem 3: Number Theory with the Area of a Polygon IMO 2016 (Problem 3). Let $P=A_1A_2\cdots A_k$ be a convex polygon in the plane. The vertices $A_1, A_2, \cdots , A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.

Also on http://www.artofproblemsolving.com/community/c6h1270467_imo_2016_problem_3
Here's my idea so far.
$k=3$ is a very easy case. My idea is to strong induct on $k$.
Assume that the statement for $3 \le k \le t$. We show this for $k=t+1$.
Start with a $t+1$-gon on a circle. If one of the main diagonals of this polygon has square of its length as a multiple of $n$, we can slice the polygon by that diagonal. Then we are left with two polygon with squares of all of its side lengths as a multiple of $n$, and the polygons have no more than $t$ vertices each. So if we denote the areas as $S_1, S_2$, we have $n|2S_1$ and $n|2S_2$, so $n|2S$ and we are good.
So for this induction to work, we need to prove that for every polygon that satisfies the condition, there is at least one main diagonal which has the square of its length as a multiple of $n$.
For example, we can prove the case $k=4$ with just Ptolemy.
Can someone help me finish the solution using this idea, or maybe introduce a different solution?

The convex polygon $P$ is inscriptible in a circle of radius $r$ and its $k$ vertices are integer solutions of the equation $ x ^ 2 + y ^ 2 = r ^ 2 $. This equation has in general more integer solutions than $k$ because if $ (a, b) $ is solution so are the four points $ (± a, ± b) $ and  P has been chosen such that their sides fulfill a certain condition of divisibility by $n$, odd integer.
The total area $S$ of $P$ is equal to the sum of  areas $S_{ij}$ of $k$  triangles (green in the figure below) and because of $\sin(\pi-x)=\sin(x)$ one has
$$2S_{ij}=\text{ area of the triangle }\triangle{A_iA'_iA_j}$$
Let us remember that if a triangle has its vertices with integer coordinates, then the double of its area is an integer. In the case of $S_{ij}$ one has

$$S_{ij}=\frac12\cdot\left|\begin{array}{cc}0 & 0& 1\\a_i & b_i&1\\a_j&b_j&1\end{array}\right|$$ so $2S_{ij}$ is an integer.

Since $\triangle{A_iA'_iA_j}$ is a  right triangle, we have
$$2S{ij}=\frac{\overline{A_iA_j}\cdot\overline{A_jA'_i}}{2}$$

$$\overline{A_iA_j}^2=(a_j-a_i)^2+(b_j-b_i)^2=2r^2-2(a_ia_j+b_ib_j)=2r^2(1-\cos(\alpha_{ij}))\\\overline{A_jA'_i}^2=(a_j+a_i)^2+(b_j+b_i)^2=2r^2+2(a_ia_j+b_ib_j)=2r^2(1+\cos(\alpha_{ij}))$$ (we have used the formula $\vec{OA_i}*\vec{OA_j}=a_ia_j+b_ib_j=|\vec{OA_i}|\cdot|\vec{OA_j}|\cos(\alpha_{ij})$).

The odd integer $n$ can not divide $1-\cos(\alpha_{ij})$ whose unique integer values can be $0$ and $2$. Therefore $n$ must divide $2r^2$ so must divide $r^2$ since $n$ is odd.

Now the radius $r$ can be integer or pure quadratic  (i.e.  $r=a+b\sqrt{\theta}$ with $a=0$).
The conclusion is that in both cases of $ r $ integer or quadratic, one has
$$2S{ij}=\frac{\overline{A_iA_j}\cdot\overline{A_jA'_i}}{2}=r^2\sqrt{1-\cos^2(\alpha_{ij})}=r^2\sin(\alpha_{ij})$$
Consequently $2S_{ij}$ is an integer divisible by $n$ and so is for $$2S=2S_{12}+2S_{23}+\cdots+2S_{k1}$$


Group of order $63$ has an element of order $3$, without using Cauchy's or Sylow's theorems I'm showing that any group of order $63$ has an element of order $3$, and can only use Lagrange's theorem not Cauchy's or Sylow's. I got it reduced to a case of having $62$ elements of order $7$ but I'm stuck now.

So let $|G|=63$, and consider an element $a \in G$, $a$ not equal to the identity. I'm allowed to use the theorem that if $|a|=n$, and $k$ divides $n$, then $|a^{n/k}|=k$.  Also, I know $G$ can only have elements of orders $1,3,7,9,21,63$.  If $|a|=63$, then $|a^{21}|=3$.  If $|a|=21$, then $|a^7|=3$.  If $|a|=9$, then $|a^3|=3$.  If $|a|=3$, then we are done. So $G$ must contain an element of order $3$ unless $G$ contains only the identity, and $62$ elements of order $7$. Now how do rule out this?

Hint: Let $G$ be your group of order $63$, and suppose every non-identity element has order $7$. Show that you can put an equivalence relation on $G\setminus\{e\}$, defined by $a\sim b$ when $a$ and $b$ generate the same subgroup of $G$. What are the sizes of the equivalence classes? Can they partition $G\setminus\{e\}$?

|G|=63 By Lagrange theorem if a belongs to G and a is non identity element
Possible orders of  a= 1,3,6,7,9,21,63
If | a | =63 , order of a^21= 3
If  | a |= 21, order of a^7 =3
If | a | =9, order of a^3= 3
If | a | =6 ,  order of a^2=  3
If a has order 63, 21, 9, 6 and of course 3 then there is an element of order 3 in G
So if a does not have order 3 all non identity(62 ) elments of G must have order 7
But we know for |G |= 63 no of elements of order 7= 6k ( a multiple of 6)
But 62 is not a multiple of 6 hence there must be a non identity element of G whose order is 3.


Prove or disprove validity: $(\forall x \exists y (P(x) \supset Q(y))) \supset(\exists y \forall x (P (x) \supset Q(y)))$ I have working on this formula $(\forall x \exists y (P(x) \supset Q(y))) \supset (\exists y \forall x (P (x) \supset Q(y)))$ to either prove or disprove it.

First, I tempted to disprove it, but I changed my mind.

I wrote down "for all x that there exists some y satisfies corresponding condition", and "there exists some y that for all x satisfies corresponding condition." I think these statements refer to the same idea.

Any suggestions?

In the usual context of nonempty structures, the statement is valid, meaning that it is true in all models. One way to see this is that if there is any $y$ in the structure for which $Q(y)$ holds, then the statement is true, since we may always choose that $y$, and this will make $Q(y)$ true and hence the final implication true, regardless of any $x$, and so the whole implication is true.  Otherwise, we are in the case where $Q(y)$ is always false, in which case $P(x)\to Q(y)$ is logically equivalent to $\neg P(x)$, and so the quantification over $y$ becomes irrelevent (provided the structure is nonempty). So again the statement is true. So the statement is true in any nonempty structure.

Edit. Meanwhile, in the context of first-order logic allowing the empty structure (which is a bit unusual, and which is only possible if your language has no constant symbols), then the statement is not valid, since it is false in the empty structure. This is because all universal statements hold vacuously in the empty structure and all existential statements fail in the empty structure, so the antecedent is true and the conclusion is false, and so the implication fails in the empty structure.

Conclusion: your statement is valid if the language has constant symbols; valid if your logic disallows the empty structure; but otherwise it is not valid.

$(\forall x \exists y (P(x) \supset Q(y))) \supset (\exists y \forall x (P(x) \supset Q(y)))$
The formula is not valid. Consider the following idea:
$P^M$ is the relation that {x : x moves first}
$Q^M$ is the relation that {x : x wins the game}
For the left part of implication $(\forall x \exists y (P(x) \supset Q(y)))$ it indicates that for all $x$ making the first move, there exists some $y$ that wins the game. However, considering right part of implication $(\exists y \forall x (P(x) \supset Q(y)))$ there exists some $y$ that beats all first moves of $x$. This is obviously different and harder than the left part of implication, which can be false while first part is true so that makes the theory false. The idea is from this page.


Rational Roots of a Quadratic Equation If $a,b,c$ are non zero, unequal rational numbers then prove that the roots of the equation
  $$(abc^2)x^2+ 3a^2cx+b^2cx-6a^2-ab+2b^2=0 $$
  are rational.

Use theory of equations and basic mathematics please. I am in high school only!

Here is a method based on how I was first taught to solve quadratics.

Note that $2b^2-ab-6a^2=(2b+3a)(b-2a)$

Then look for two expressions which have

product $abc^2(2b+3a)(b-2a)$ [coefficient of $x^2$ times the constant term]

and whose sum is $3a^2c+b^2c$ [coefficient of $x$]

The sum is homogeneous degree three and divisible by $c$, and there is no $ab$ term, so this needs to cancel. The only possibility is easily seen to be $ac(2b+3a) + bc (b-2a)=3a^2c+b^2c$

So the quadratic can be rewritten $$abc^2x^2+ac(2b+3a)x+bc(b-2a)x+(2b+3a)(b-2a)$$and this becomes $$acx \left(bcx+2b+3a\right)+(b-2a)(bcx+2b+3a)=(acx+b-2a)(bcx+2b+3a)$$

It is sometimes almost automatic to compute the discriminant and show that it is a rational square, or to use the quadratic formula. Here it is quite easy to check whether there is an easy spot for the sum/product procedure and fall back on the more general methods if you don't spot one. This method has easier manipulations than the others suggested.

it is easy to show . to prove this kind of thing the point to be remember is that the discriminant is perfect square more generally {b^2-4ac} is perfect suare
 now the discriminant is {(3a^{2}c−a^{2}c+4abc)^{2}}
 this sufficient to prove


How do I find the base when Log is given I'm trying to figure out how to calculate the base if:

$$ \log_b 30 = 0.30290 $$

How do I find $b$ ?

I've slaved over the Wikipedia page for logarithms, but I just don't get the mathematical notations.

If someone could let me know the steps to find $b$ in plain english, I'd be eternally grateful!

You need to think about the definitions.

Since $a^b=c$ can be rewritten as $\log_a c = b$.

That should tell you that,

$$b^{0.30290} = 30$$

and then,

$$b = \exp {\frac{\ln 30}{0.30290}} $$

In Excel: to quickly calculate the elusive "e":

To calculate "e" (the base of LN):
e = x^(1/LN(x))
Wherein: x = any number >or< 1 but > 0


Completion of rational numbers via Cauchy sequences Can anyone recommend a good self-contained reference for completion of rationals to get reals using Cauchy sequences?

Update 1

This should be the final edit. Lots of typos have been corrected and the presentation in the section on existence has been greatly improved. It should be very close to a form that can be used as a basis for a project for students.

Update 2

Even more typos corrected. Added example of non-archimedean ordered field. Fleshed out a few bits more.

Update 3

Fixed up the proof sketch in the last exercise. Also I keep finding these small annoying errors... I'm sorry that I keep up bumping this thread, I just hope that you don't mind too much.

Introduction

I'll do one better: I'll write you one (mostly because I actually have already written one - consider this sort of a Cliff's notes version with most of the details left as exercises). Recall that a first course in analysis typically asserts that there exists a field with an order and the least upper bound property, and you then develop the theory of sequences and series from there. Thus our goal should be to prove existence and uniqueness (up to some isomorphism notion) of such an object.

Uniqueness

Definition 1. An ordered field is a field, $K$ along with a total order $\leq$ on $K$ such that

$\forall x,y,z \in K : x \leq y \implies x+z \leq y+z$ (Transitivity).
$\forall x,y \in K : 0 \leq x \land 0 \leq y \implies 0 \leq xy$ (Positivity of product).

Moreover, define the absolute value function $\vert \cdot \vert: K \to K$ by
$$\vert x \vert = \begin{cases}
    x  & \text{if } 0 \leq x \\\\
    -x & \text{otherwise.}
\end{cases} $$

Exercise 1. There is an alternate definition of an ordered field: $K$ is said to be ordered if there is a subset $P$ of $K$ - called the positive elements of $K$ - such that

$K$ is the disjoint union of $P$, $-P$ and $\lbrace 0 \rbrace$.
$x,y \in P \implies x+y \in P$ and $xy \in P$.

Make formal the colloquial statement "The two definitions of ordered field are equivalent" and prove said equivalence.

Note that we could just as well require that the inequality in Definition 1 be strict since $x \lt y \lor x = y \iff x \leq y$ allows us to pass between strict and non-strict orderings as we please.

Definition 2. Let $K$ and $L$ be ordered fields. An order-preserving ($f$ is order-preserving means that $x\leq  y \implies f(x)\leq f(y)$) ringhomomorphism is called an embedding.

Note that any embedding must be injective since the kernel of a ringhomomorphism is and ideal and $0 \lt 1$.

Proposition 3. If $K$ is an ordered field, then $K$ has a subfield isomorphic to $\mathbb{Q}$.

Exercise 2. Prove Proposition 3 by showing that $\varphi: \mathbb{N} \to K$ defined by
$$ \varphi(n) = \underbrace{1+\dotsb+1}_{n \text{ } 1\text{s}} $$
extends to an embedding of $\mathbb{Q}$ in $K$.

Definition 4. An ordered field $K$ is called an archimedean field if for all nonzero $x$ in $K$ there exists a natural number $n$ such that $n|x| \gt 1$. Or equivalently: there exists a natural number $n$ such that $n \gt |x|$.

Proposition 5. Let $K$ be an ordered field. If $K$ has the least upper bound property, then $K$ is archimedean.

Exercise 3. Show the contrapositive of Proposition 5 by considering the set
$$ I = \lbrace x \in K \;\vert\; 0 \lt x \land \forall n \in \mathbb{N} : nx < 1 \rbrace. $$
Show that $I$ is nonempty, $1$ is an upper bound for $I$ and $I$ has no least upper bound [Hint: assume that $y$ is the least upper bound and split in to cases (i) for $y \in I$ look at $2y$, and (ii) for $y \notin I$ look at $y/2$].

In a bit of an abuse of language, the elements of $I$ can be said to be "infinitely small". For an example of an ordered field which is not archimedean, consider $\mathbb{Q}(X)$, the set of rational functions in one variable over $\mathbb{Q}$ (see [Wikipedia] for how the order is defined) - $0 \lt \frac{1}{X} \lt q$ for all positive rationals $q$.

Now it's time to consider the order topology on ordered and archimedean fields, i.e. the topology generated by the open intervals $(a,b)$ defined in the obvious way.

Exercise 4. Show that the open intervals in an ordered field $K$ are a basis for a topology on $K$.

Proposition 6. $\mathbb{Q}$ is dense in any archimedean field $K$.

Sketch of proof. Since the open intervals are a basis, we just need to verify that any open interval contains a rational. Define the ceiling function $\lceil \cdot \rceil : K \to \mathbb{Z}$ by
$$ \lceil x \rceil = \min \lbrace n \in \mathbb{Z} \;\vert\; n \geq x\rbrace, $$
which is well-defined since $K$ is archimedean.

Now let $x \in K$ and a $K \owns \varepsilon \gt 0$ be given. Since $K$ is archimedean we can find $n \in \mathbb{N}$ such that $n\varepsilon > 1$. Now we have:
$$ 0 \leq \left\lvert \frac{\lceil nx \rceil}{n} - x \right\rvert
     \leq \left\lvert \frac{\lceil nx \rceil - nx}{n} \right\rvert
     \leq \frac{1}{n} \lt \varepsilon
$$
Hence $\frac{\lceil nx \rceil}{n} \in (x-\varepsilon,x+\varepsilon)$ from which that result follows.

Corollary 7. $\mathbb{Q}$ is densely ordered in any archimedean field, i.e. if $K$ is archimedean, $x,y \in K$, and $x \lt y$, then there exists a $q \in \mathbb{Q}$ such that $x \lt q \lt y$.

Proposition 8. Any archimedean field is 2nd-countable.

Exercise 5. Prove Proposition 8 [Hint: Consider the set of intervals with rational endpoints. Show that it is a basis for a topology. Argue that it is coarser than the standard topology, and then show that any basis set for the standard topology (i.e. an interval) contains an interval with rational endpoints]

Proposition 9. Any archimedean field $K$ is a topological field, i.e. $K \times K \owns (x,y) \mapsto x+y \in K$, $K^\times \times K^\times \owns (x,y) \mapsto xy \in K^\times$ and $K^\times \owns x \mapsto x^{-1} \in K^\times$ are continuous.

Exercise 6. Prove Proposition 9 [Hint: Consider sequences $x_n$ and $y_n$ that converge to $x$ and $y$ respectively, then use that convergent sequences are bounded].

Note that proposition 9 in fact holds for ordered fields too. You just have to consider nets instead.

Definition 10. Let $G$ be a topological group. A sequence $(x_n)$ in $G$ is said to be a Cauchy sequence if for every neighbourhood $U$ of $e$ exists a $N$ such that $m,n \geq N \implies x_nx_m^{-1} \in U$. We shall denote the set of Cauchy sequences in $G$ by $G^\mathbb{N}_c$.

Note that this definition too can be used to define Cauchy nets in topological groups, but we shall not need that. Observe that the intervals with rational endpoints gives us a countable basis at $0$ in any archimedean field, so we see that for archimedean fields, the definition boils down to this:
$$ \forall \varepsilon \in \mathbb{Q}^+ \exists N : n,m \geq N \implies |x_n - x_m| < \varepsilon. $$

Proposition 11. Let $K$ be an archimedean field and $(x_n)$ a sequence in $K$. The following hold:

If $(x_n)$ is convergent, then $(x_n)$ is a Cauchy sequence, and any subsequence of $(x_n)$ converges to the same limit.
If $(x_n)$ is a Cauchy sequence, then it is bounded.
If $(x_n)$ is bounded, then $(x_n)$ has a Cauchy subsequence ("pre-Bolzano-Weierstraß")
If $(x_n)$ is a Cauchy sequence then any subsequence is a Cauchy sequence. Moreover if $(x_n)$ has a convergent subsequence, then $(x_n)$ converges to the same limit.
If $(x_n)$ is a Cauchy sequence and $(x_{k_n})$ is a subsequence, then $\lim_{n \to \infty} |x_n - x_{k_n}| = 0$.

Exercise 7. Prove Proposition 11.

Proposition 12. Let $K$ be an archimedean field. TFAE:

$K$ has the least upper bound property.
Any Cauchy sequence in $K$ is convergent ("$K$ is complete").

Exercise 8. Flesh out the following sketch of a proof of Proposition 12:
$1 \implies 2$: Let $(x_n)$ be a Cauchy sequence in $K$. If $(x_n)$ is eventually constant, then it is obviously convergent, so assume that it is not so, and consider the set
$$ X = \lbrace x \in K \;\vert\; \exists N \in \mathbb{N} : n \geq N \implies x \lt x_n \rbrace. $$

$X$ is nonempty and has an upper bound (why?) and thus $\sup X$ exists, and we claim that $x_n \to \sup X$. Given an $\varepsilon > 0$ find $N$ such that $|x_n - x_m| < \frac{\varepsilon}{2}$, and consider the interval $(x_N-\frac{\varepsilon}{2}, x_N + \frac{\varepsilon}{2})$. There are only finitely many elements of the sequence outside of this interval, so $x_N - \frac{\varepsilon}{2} \in X$. Moreover, $x + \frac{\varepsilon}{2}$ is an upper bound for $X$. The triangle inequality (note: prove the triangle inequality for archimedean fields) now gives
$$ |x_n - \sup X| \leq |x_n - x_N| + |x_N - \sup X| \lt \varepsilon. $$
$2 \implies 1$: Let $A \subset K$ be nonempty and bounded above by $y_0 \in K$. Let $x_0$ be some non-upper bound for $A$ e.g. $a-1$ for some $a \in A$, and recursively define a pair of sequences $(x_n)$ and $(y_n)$ over $\mathbb{N}_0$ by
$$ \begin{align}
y_{n+1} & = \begin{cases}
    \tfrac{y_n + x_n}{2}
        & \text{if } \tfrac{y_n+x_n}{2} \text{ is an upper bound for } A, \\\\
    y_n & \text{otherwise}
\end{cases} \\\\
x_{n+1} & = \begin{cases}
    \tfrac{y_n+x_n}{2}
        & \text{if } \tfrac{y_n+x_n}{2} \text{ is a non-upper bound for } A, \\\\
    x_n & \text{otherwise.}
\end{cases}
\end{align} $$
By induction we have that $(x_n)$ is increasing, $(y_n)$ is decreasing, and  for all $i,j$: $x_i \leq y_j$. Likewise by induction we get that for all $i$: $0 \leq y_i - x_i \leq 2^{-i}(y_0 - x_0)$. It is now easy to show that $(y_n)$ is a Cauchy sequence. Assume that $n \geq m$ and we have:
$$ \begin{align}
y_m-y_n & = y_m - y_{m+1} + y_{m+1} - \dotsb - y_{n-1} + y_{n-1} - y_n \\\\
        & \leq (2^{-m-1} + 2^{-m-2} + 2^{-n})(y_0 - x_0) \\\\
        & \leq 2^{-m}(y_0 - x_0)
\end{align} $$
from which it is easy to conclude that $(y_n)$ is Cauchy. Similarly $(x_n)$ is shown to be Cauchy. Thus the two sequences are convergent, and their limits must be the same since $y_n - x_n \leq 2^{-n}(y_0 - x_0)$ -- denote the limit by $s$. Finally observe that for all $a \in A$, $a \leq y_n$, so $a \leq s$ (why?) i.e. $s$ is an upper bound for $A$. Moreover if $u$ is an upper bound for $A$ then $x_n \leq u$ for all $n$, so $s \leq u$ (why?). Hence $s$ is the least upper bound for $A$. QED.

Proposition 12. Let $K$ and $L$ be archimedean fields. If $f: K \to L$ and $g: L \to K$ are embeddings, then $K$ and $L$ are isomorphic.

Proof. Let $g \circ f = \phi: K \to K$. By definition $\phi$ is an embedding. Assume that $\phi$ is not the identity. Then there exists an $x$ for which $\phi(x) \neq x$. Find a $q \in \mathbb{Q}$ between $x$ and $\phi(x)$ and observe that $\phi$ has to be the identity on $\mathbb{Q}$, but it can't preserve the ordering of $x$ and $q$. This gives a contradiction, and thus $\phi$ must be the identity. The proof that $f \circ g$ is the identity is similar. QED.

Lemma 13. Let $G$ and $H$ be topological groups and let $G'$ be a dense subgroup of $G$. If $\phi: G \to H$ is continuous and $\phi|_{G'}$ is a homomorphism, then $\phi$ is a homomorphism.

Proof. Let $x,y \in G$ and assume that $\phi(xy) \neq \phi(x)\phi(y)$. Since $G'$ is dense there exists nets $(x_\alpha)$ and $(y_\alpha)$ in $G'$ converging to $x$ and $y$ respectively. Since $\phi$ is a homomorphism on $G'$ we have for all $\alpha$ that $\phi(x_\alpha y_\alpha) = \phi(x_\alpha)\phi(y_\alpha)$, and since $\phi$ is continuous and $\cdot$ is continuous we have that $\phi(xy) = \phi(x)\phi(y)$.

Remark. This is not a vacuous use of reductio ad absurdum, since a direct proof would have to work with arbitrary nets converging to $x$ and $y$. Of course this lemma immediately generalizes to also apply to rings in an obvious manner.

Theorem 14. (Embedding theorem for archimedean fields). Let $K$ and $L$ be archimedean fields where $L$ is complete. Then there exists an embedding $\phi: K \to L$.

Proof. Define $\phi: K \to L$ by $\phi(x) = \sup \lbrace q \in \mathbb{Q} \;\vert\; q \lt x \rbrace$. It is easy to show that $\phi$ is order-preserving, so the only hard part is showing that it's a homomorphism. We wish to use the preceding lemma, so first we note that it's obvious that $\phi$'s restriction to $\mathbb{Q}$ is the identity, so the only hard part is verifying that it's continuous. Consider a $k \in K$ and let $V$ be a neighbourhood of $\phi(k)$. Find rationals $p$ and $q$ such that $\phi(k) \in (p,q) \subset V$. We claim that $\phi((p,q)) \subset (p,q)$. Let $k' \in (p,q) \subset K$ and find $q' \in \mathbb{Q}$ such that $p \lt k' \lt q' \lt q$. It's easy to see that $p \lt \phi(k') \leq q' \lt q$, so $\phi(k') \in (p,q)$, and we conclude that $\phi$ is continuous. QED.

Theorem 15. (Uniqueness of complete ordered field) There exists up to order-preserving isomorphism one and only one complete ordered field.

Proof. Let $K$ and $L$ be complete ordered fields. Theorem 14 gives embeddings $\phi: K \to L$ and $\varphi: L \to K$. Proposition 12 gives that $K$ and $L$ are isomorphic. QED.

Existence

Much more has been written on this subject than uniqueness, so I'll write a lot less here. Recall that $\mathbb{Q}$ under addition is a topological group, so we use the previously established notation for its set of Cauchy sequences: $\mathbb{Q}^\mathbb{N}_c$.

Proposition 16. $\mathbb{Q}^\mathbb{N}_c$ is a commutative unital ring under coordinatewise addition and multiplication.

Exercise 8. Prove Proposition 16.

Proposition 17. The set of rational Cauchy sequences that converge to zero, i.e.
$$ \mathfrak{z} = \left\lbrace (z_n) \in \mathbb{Q}^\mathbb{N}_c \;\bigg\vert\; z_n \to 0 \right\rbrace,$$
is a maximal ideal in $\mathbb{Q}^\mathbb{N}_c$.

Exercise 9. Prove Proposition 17. [Hint: Consider an ideal $\mathfrak{a} \supset \mathfrak{z}$ with a $(q_n) \in \mathfrak{a}\setminus\mathfrak{z}$. Show that there exists an $N$ such that $n \geq N \implies q_n \neq 0$ and consider the sequence $(z_n) = (1-q_1,\dotsc,1-q_N,0,0,\dotsc)$. Argue that $(q_n + z_n) \in \mathfrak{a}$ and conclude that $\mathfrak{a} = \mathbb{Q}^\mathbb{N}_c$.]

Thus $\mathfrak{z}$ is a maximal ideal and hence the quotient, which we shall denote by $R$, must be a field. From here on out the rest is sketchy as there are a lot of details to check and it's really rather boring.

We adopt the convention that we write sequences with bold, e.g. $(x_n) = {\bf x}$ and use the notation $[{\bf x}]$ for the equivalence class containing ${\bf x}$.

Define $\lt$ on $R$ by $[{\bf p}] \lt [{\bf q}] \iff \exists (p_n) \in [{\bf p}], (q_n) \in [{\bf q}], \alpha \in \mathbb{Q}^+ : p_n \lt q_n + \alpha$ eventually. In particular note that the existence of the positive $\alpha$ cannot be dropped; e.g. look at $(1/n)$, $(0)$ and $(-1/n)$. Without the requirement that such a positive $\alpha$ exists, you'd thus have $0 \lt 0$, which is clearly rubbish.

Exercise 10. Verify that if $(p_n) \in [{\bf p}]$ and $(q_n) \in [{\bf q}]$ are witnesses that $[{\bf p}] \lt [{\bf q}]$, then any pair of representatives will do just as well, i.e. if $(p'_n) \in [{\bf p}]$ and $(q'_n) \in [{\bf q}]$, then $(p'_n)$ and $(q'_n)$ are witnesses as well.

Exercise 11. Show that $\lt$ is a strict total order, i.e. $\lt$ is irreflexive, transitive and total [Hint (for totality): You need to show that exactly one of $[{\bf x}] \lt [{\bf y}]$, $[{\bf x}] = [{\bf y}]$ and $[{\bf y}] \lt [{\bf x}]$ holds, so first show that at least one holds by assuming that none of them holds and derive a contradiction, and then show that at most one holds likewise by contradiction and using that $[{\bf x}] \nless [{\bf y}] \iff [{\bf y}] \leq [{\bf x}]$. You'll probably need to pass to "nicer" subsequences many times, so Proposition 11(5) will come in handy].

And sit back and reap the reward with the following easy

Exercise 12. Show that $R$ is an ordered field.

Exercise 13. Show that $R$ is archimedean.

Exercise 14. Fill out the details of the following sketch of a proof that any Cauchy sequence in $R$ is convergent:
Proof. Starting with an arbitrary Cauchy sequence $([{\bf q}]^i)$ of equivalence classes, Proposition 11(4) allows us to assume WLOG that for all $i,j$ with $j \geq i$ that $\left\lvert[{\bf q}]^i - [{\bf q}]^j]\right\rvert \lt 4^{-i}$. Note that convergence means showing some inequalities hold, so we only need to show those for some representatives, so we are free to pick nice ones.
Pick representatives, $(q^i_n)$, such that $|q^i_n - q^i_m| \lt 4^{-n}$.
Take ${\bf x} = (q^1_1,q^2_2,\dotsc)$. This is obviously a Cauchy sequence (why?), and we claim that $[{\bf q}]^i \to [{\bf x}]$. Convince yourself that this is the same as this:
$$\forall \varepsilon \in \mathbb{Q}^+ \exists K,N : k \geq K, n \geq N \implies \left\lvert q^k_n - q^n_n \right\rvert \lt \varepsilon. $$
Note that the requirement of the existence of the positive $\alpha$ dropped out since the above has to hold for every $\varepsilon > 0$. So let $\varepsilon \gt 0$ be given. We use the standard "trick" of inserting a clever element into the inequality: $|q^k_n - q^n_n| \leq |q^k_n - q^k_k| + |q^k_k - q^n_n|$ and use that the both the sequence and the representatives have that "$4^{-m}$"-property to conclude that if we choose $N,K$ such that $N=K$ and $2^{-N} \lt \varepsilon$ then RHS is eventually smaller than $\varepsilon$.

CONGRATULATIONS! You're done and you now have a fully certified driver's license for the real numbers!

I just wanted to let people know that I finally got a chance to incorporate the work of Kahen's answer into my field theory notes.  It now appears as Section 12.7 of these notes, pp. 67-72.  (Yes, it takes five pages!)

I followed Kahen's suggestions mostly, but not completely.  In particular my treatment is a little "nettier".  I wanted one more little result to be completely convinced that convergence of Cauchy sequences implies convergence of Cauchy nets in an Archimedean ordered field (this is certainly no big deal).

Then in the proof of $2 \implies 1$ in Kahen's Proposition 12 (um, his first Proposition 12), I exploited the fact that I could use Cauchy nets more general than sequences to give a different proof that convergence of Cauchy nets implies Dedekind completeness: it is enough to show that any set which is bounded below has a greatest lower bound, and I actually show that the set of all lower bounds has a maximal element by applying Zorn's Lemma.  So that was amusing.  (I have no illusions that this proof could be both original and correct...)

Again, comments are welcome.  I wrote this all up pretty rapidly, so it is even more possible than usual that I made a mistake: please let me know.  If you have any other suggestions please let me know.

Once again I would prefer to acknowledge Kahen's contribution, if at all possible.

Added: Let me mention that my notes have much less to say than Kahen's answer about the construction of $\mathbb{R}$ using Cauchy sequences, this being something that was not the primary goal of these field theory notes (and in fact is described in other notes of mine in the context of completion of a normed field).  So this is not worth much as an answer to the current question (hence CW).  I decided to leave it as an answer and not as a comment so that it will be easier for others to comment on it, if they care to.


Why do I get a converging result when pressing cosine multiple times on a calculator? I'm trying to comprehend the following: If I choose any starting value (e.g. 1) and keep clicking on cosine on the calculator (in radian mode), it gives me a result of about 0.739085...(I believe it's the result of cos(x) = x), but when I repeat the same procedure using sin and tan, I get something completely different (looks like for sin it's converging to 0 while for tan I get very wild results).

Thanks for your help.

Let $\cos^n$ denote the $n$-fold composition of the cosine function with itself, e.g.
$$ \cos^3(\theta) = \cos(\cos(\cos(\theta))). $$
Note that this is not usually what this notation means in, for example, introductory calculus texts.  However, it is convenient in the current context.

What you are computing (assuming that it exists) is
$$ \lim_{n\to\infty} \cos^n(\theta). $$
Again, suppose that this limit exists and is equal to $L$.  That is, suppose that there is some $L\in\mathbb{R}$ such that
$$ \lim_{n\to\infty} \cos^n(\theta) = L. $$
Taking the cosine of each side of this, and using the continuity of the cosine, we get
$$ \cos(L) = \cos( \lim_{n\to\infty} \cos^{n}(\theta) )
= \lim_{n\to\infty} \cos^{n+1}(\theta)
= L. $$
Thus $L$, assuming that it exists, is a solution the equation
$$ \cos(x) - x = 0. $$
This confirms your suspicion that the number you are getting is a solution to this equation (presuming that such a number exists at all).

At this point, numerical methods are needed to get anything nicer out of this equation (hitting a button on the calculator repeatedly is such a numerical method), but you can prove that there is a solution via the intermediate value theorem:  $\cos(0) - 0 = 1 \ge 0$ and $\cos(\pi) - \pi = -\pi < 0$, so there is a solution between $0$ and $\pi$.

What remains is to show that this limit actually exists.  Note that the cosine function is a contraction on the interval $[-1,1]$.  That is, if $x,y\in [-1,1]$, then there is a constant $|C| < 1$ such that $|\cos(x)-\cos(y)| < C|x-y|$.  There are nice theorems that tell us that there must be a unique fixed point (specifically, theory that eventually leads to a theorem alternatively known as the "Contraction Mapping Principle" or the "Banach Fixed Point Theorem").

Note that cosine is a contraction on $[-1,1]$, and that $\cos(\mathbb{R})$ (i.e. the image of $\mathbb{R}$ under the cosine function, i.e. the range of the cosine function) is that interval.  Thus, from any starting point, after the first iteration we are in the regime where $\cos$ is contractive, and so we will eventually converge to our fixed point, which was "determined" above (in so much as we can describe it).  This pretty much says everything there is to know about the cosine function in this context.

The sine function is similar---it is contractive on the same interval, but it is much easier to find the fixed point:
$$ \sin(x) - x = 0 \implies x = 0. $$

The tangent function is not so nice.  It is actually a dilation on the symmetric unit interval, hence we can't apply the Banach Fixed Point Theorem as above.  There may be a fixed point, but to get at it, we'll have to work backwards, and consider iterative application of the $\arctan$ function.

These kinds of objects are studied as "dynamical systems."  In the language of dynamical systems, the fixed points of cosine and sine functions are attractive, in that everything eventually "falls into" those points.  The fixed point of the tangent function is repulsive (or repelent), in that things are pushed away from it.

According to Wikipedia (the greatest source ever), the fixed point at $x$ is attractive if the function is continuously differentiable in an open neighbourhood of $x$ and $|f'(x)|<1$ in that neighbourhood.

It is left as an exercise to the reader to confirm that it is indeed the case.


Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.

We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers.

I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.

This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$

Let $P(n)$ be summation shown above.

Base case for $n=1$, the first positive integer,

$$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$

so base case is true.

Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus

$$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$

This can be written as

$$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$

I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes.

I think that your notation is rather badly confused: I strongly suspect that you’re supposed to be showing that $$\sum_{k=1}^{2^n}\frac1k\ge 1+\frac{n}2\;,\tag{1}$$ from which one can conclude that the harmonic series diverges. The basis step for your induction should then be to check that $(1)$ is true for $n=0$, which it is: $$\sum_{k=1}^{2^n}\frac1k=\frac11\ge 1+\frac02\;.$$

Now your induction hypothesis, $P(n)$, should be equation $(1)$, and you want to show that this implies $P(n+1)$, which is the inequality $$\sum_{k=1}^{2^{n+1}}\frac1k\ge 1+\frac{n+1}2\tag{2}\;.$$ You had the right idea when you broke up the bigger sum into the old part and the new part, but the details are way off:

$$\begin{align*}\sum_{k=1}^{2^{n+1}}\frac1k&=\sum_{k=1}^{2^n}\frac1k+\sum_{k=2^n+1}^{2^{n+1}}\frac1k\\
&\ge 1+\frac{n}2+\sum_{k=2^n+1}^{2^{n+1}}\frac1k\tag{3}
\end{align*}$$

by the induction hypothesis $P(n)$. Now look at that last summation in $(3)$: it has $2^{n+1}-2^n=2^n$ terms, and the smallest of those terms is $\dfrac1{2^{n+1}}$, so $$\sum_{k=2^n+1}^{2^{n+1}}\frac1k\ge 2^n\cdot\frac1{2^{n+1}}=\frac12\;.$$ If you plug this into $(3)$, you find that $$\sum_{k=1}^{2^{n+1}}\frac1k\ge 1+\frac{n}2+\frac12=1+\frac{n+1}2\;,$$ which is exactly $P(n+1)$, the statement that you were trying to prove.

You’ve now checked the basis step and carried out the induction step, so you can conclude that $(1)$ is true for all $n\ge 0$.


Geometric Solution to Calculus Problem Given a rectangle $A$ of sides $L$ and $W$, such that $L$ is less than $W$, what is the average area of all circles entirely contained in $A$?  To be more explicit, what is the average area of all circles of all radii with centers located inside of $A$ such that each circle intersects any given side of A at most once?

I used calculus to find the solution to this problem and got that this average area is equal to

$$(4W-3L)(\pi L^2)(72W)^{-1} .$$

However, I was wondering if there exists a simpler, geometric interpretation to this problem.

EDIT:  Thanks to joriki, the 36 was corrected to be 72.

I have not thought about an analytic solution but a simulation approach seems straightforward.

First, sample a point by drawing a uniform random variable $c_x \sim U(0.L)$ and $c_y \sim U(0,W)$ where $(c_x, c_y)$ represents the center of the circle.

Second, given the center, sample a radius using a uniform distribution as: $r \sim U(0,min(c_x,L-c_x,c_y,W-c_y))$.

Third, compute the area of the circle and store it.

Fourth, repeat steps 1 and 2 lots, lots of times and compute the empirical average.

The empirical average should coincide with your calculus based solution provided there are no mistakes in your math and the simulation algorithm/code.


Is the domain of an one-to-one function a set if the target is a set? This is probably very naive but suppose I have an injective map from a class into a set, may I conclude that the domain of the map is a set as well?

If a function $f:A\to B$ is injective one, we can assume without loss of generality that $f$ is surjective too (by passing to a subclass of $B$), therefore $f^{-1}:B\to A$ is also a bijection.

If $B$ is a set then every subclass of $B$ is a set, so $f^{-1}:B\to A$ is a bijection from a set, and by the axiom of replacement $A$ is a set.

Define $g$ on the range of $f$ such that $g(f(x)) = x$.  This is well-defined because $f$ is injective. The domain of $g$ is equal to the range of $f$, which is a set.

Therefore by the axiom of replacement, or maybe the axiom of union applied to $\bigcup_{y\in{\rm Range}f} \{g(y)\} $, the range of $g$ is a set. But the range of $g$ is precisely the domain of $f$.


Difference in definition of differentiation Okay so quite often I see two different definitions of differentiation and I want to know when it is appropriate to use each one.

$$\lim_{h \to 0} \frac{f(x_{0}+h)-f(x_{0})}{h}$$
and
$$\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$

Why are their to definitions, how are they different and when do you use each one ?

It is almost routine in mathematics than one thing can be expressed into two different visible forms without any problem and in most cases it is a matter of convenience to prefer one form over another. For example we can write $1/2$ as $0.5$ also and it does not matter which form we choose.

In a similar manner both the definitions of derivative are equivalent. It should be easy to prove that $\lim_{x \to a}g(x) = \lim_{h \to 0}g(a + h)$ and choosing $a = x_{0}$ and $g(x) = \dfrac{f(x) - f(x_{0})}{x - x_{0}}$ we see that $$\lim_{x \to x_{0}}g(x) = \lim_{h \to 0}g(x_{0} + h)$$ or $$\lim_{x \to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}} = \lim_{h \to 0}\frac{f(x_{0} + h) - f(x_{0})}{h}$$

This is essentially the mean value theorem.
$$\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$

Note how similar this formula is to calculating the slope of a line.

$$m = \frac{y_2-y_1}{x_2-x_1}$$

Also note that $x-x_0$ is essentially the h in the definition of the derivative, as $\lim_{x\to x_0} x-x_0$ goes to zero, just as h does. The mean value theorem can be derived by approximating f'(x) with secant lines. Let $x$ go to $x_0$ and it will yield tangent lines. However, both should yield the same answer. They are two ways of expressing the same idea. I would recognize the first one as the true definition of a derivative, but the mean value theorem is also useful for differentiation.


how can i get a contradiction (number theory) i have some problems to get a contradiction:

problem: Does not exist $n > 0$ such that $a = p^{n}$. prove if $\log_{p}(a)$ is irrational (where $a \in \mathbb Z$ and $p$ prime)

i try this:

suppose $\log_{p} (a)$ is rational

then $\exists s,t$ such

$\log_{p} (a) =\frac{s}{t}$

$\implies p^{\frac{s}{t}}$ = a

$\implies \sqrt[t]{p^{s}}$ = a

$\implies p^{s} = a^{t}$ ( i think this is a contradiction)

if not how can i get a contradiction, can someone give me a hint

thanks for reading me

When you arrive at $p^s = a^t$, use the fundamental theorem of arithmetic to argue that the only prime factor of $a$ is $p$.

Hence $a$ must be an integer power of $p$, contradicting the non-existence of $n$ such that $a = p^n$.

Um...... $a = p^n \iff \log_p a = n$. That's a definition.

So if $\log_a p$ is irrational then $\log_a p\ne n$ for any integer and $a\ne p^n$ for any integer.

(Of course, we have to assume (prove) that the definition of "$\log_p a$" makes any sense.  That is that for $a>0, a\ne 1, p > 0$ there is a unique $x\in \mathbb R$ so that $a^x =p$.  And for that we have to define/figure out what $a^x$ means.  .... But none of that is in the realm of number theorem, I think.)


A simple permutation question - discrete math There are five distinct computer science books, three distinct mathematics books, and two distinct art books. In how many ways can these books be arranged on a shelf if one of the art books is on the left of all computer science book, and the other art book is on the right of all the computer science books?

For my answer I was thinking since there are $10$ books total and $5$ are computer science books, then I could place an art book on the first, second, third, fourth slot ... but I do not know how to finish it, I am sure it has something to do with rule of sum. Any advice appreciated. Thank you.

You are going in the right direction, but don't think in terms of slots. Just organize the books on the floor, then put them on the shelf.

First, order the computer science books. There are $5!$ ways of putting them in. Then put the two art books on both ends; there are $2$ ways of doing it.

Now you have $7$ books, with eight spaces between and to the left and right of them. We just need to put in the three math books. Let's say they are a Calculus I, a Calculus II, and a Calculus III textbooks.

We put in the Calc I book first. There are 8 places where it can go, relative to the other 7 books. Next, put the Calc II book in; there are 9 places where it can go, relative to the other 8 books that have already been placed. Finally, put in the Calc III book.

answer is $2520$

Number of permutations of $n$ things, taken all at a time, in which $p$ are of one type, $q$ of them are of second-type, $r$ of them are of third-type, and rest are all different is given by
$$
\frac{n!}{p!  q! r!}
$$
so
$$
\frac{10!}{5! 3! 2!} = 2520
$$


In need of tips/suggestions when to add or multiply probabilities I am having trouble deciding when to add or when to multiply probabilities as in the following example. I know that by constructing Probability tree diagrams we could multiply along branches and add vertically. However I could definitely use more suggestions/tips that
might help me decide when to multiply and when to add probabilities.

    A jar contains $4$ black and $3$ White balls. If you reach into the jar and pick two balls simultaneously , what is the probability that one is black and the other is white ?

This is how I am solving the above : Pr(Black from the total 7 balls)=$\frac{4}{7}$

Pr(White from the remaining 6 balls after choosing a Black ball) $= \frac{3}{6}$

So Ans = $\frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$

You add probabilities when the events you are thinking about are alternatives [Reading score 0 goals or 1 goal or 2 goals in their match] - you are looking for "mutually exclusive" events - things which could not happen at the same time (in the same match).

You multiply probabilities when you want two or more different things to happen "at the same time" or "consecutively" [Reading score 1 and Leeds score 1 and Arsenal score 2]. The key thing here is that the events are independent - they do not affect each other, or the second does not affect the first (etc).

In your example, to get a black ball and a white ball you have two "mutually exclusive" possibilities: $1.$ white first, black second; $2.$ black first, white second.

Possibility $1$. You can choose a white first with probability $\frac 3 7$ - and then you have 6 balls left, four of which are black, so $\frac 4 6$ of choosing a black one. These are independent events so multiply to get $\frac 3 7 \times \frac 4 6 = \frac 2 7$.

Possibility $2$.  Choosing black first then white gives $\frac 4 7 \times \frac 3 6 = \frac 2 7$.

Adding the two together gives $\frac 4 7$.

It takes some skill and practice to get these right all the time. Time spent working through and understanding key examples is time well spent.

Did anyone even consider the probability of picking 2 black (BB) or 2 white (WW) the same time - probability is a function of favorable outcomes upon Total possible outcomes, we all know this; the trick is to get this right. Get the basics and you will know when to multiply and when to add. Hence, in this case it is {BB, WW, BW, WB} = Favorable outcomes!, which is a set of 4; total possible outcomes {B,B,B,B,W,W,W} = 7;


Probability of getting a mirror image In a suitable font, the letters A, H, I, M, O, T, U, V, W, X, and Y are all mirror images of themselves. A string made from these letters will be a mirror image of itself if it reads the same backward and forward: for example, MOM, YUMMUY, MOTHTOM. If a four-letter string in these letters is chosen at random, what is the probability that this string is a mirror image of itself?

The first two letters can be anything.

The third letter must be the same as the 2nd.  That's 1/11.

The fourth letter must be the same as the 1st.  That's 1/11.

So the probability is $1/11^2 = 1/121$.


If $m, n$ be positive integers, prove that $\phi(mn)=\phi((m,n))\phi([m,n])$, where $(m,n)=$ gcd of $m, n$ and $[m, n]=$ lcm of $m, n$. If $m, n$ be positive integers, prove that $\phi(mn)=\phi((m,n))\phi([m,n])$, where $(m,n)=$ gcd of $m, n$ and $[m, n]=$ lcm of $m, n$.

I have no idea to solve this question. Please help me to solve the problem.

It is false. Take $m=n$.

You can prove that the statement holds if and only if we are in the trivial case $\text{gcd}(m,n)=1$.

To be precise, the claim should be modified to
$$
\varphi(mn)=\mathrm{gcd}(m,n)\cdot \varphi(\mathrm{lcm}(m,n)).
$$

Proof:

Since $mn=\mathrm{lcm}(m,n)\mathrm{gcd}(m,n)$ and a prime $p$ divides $mn$ if and only if it divides $\mathrm{lcm}(m,n)$ then
$$
\frac{\varphi(mn)}{mn}=\prod_{p\mid mn}\left(1-\frac{1}{p}\right)=\prod_{p\mid \mathrm{lcm}(m,n)}\left(1-\frac{1}{p}\right)=\frac{\varphi(\mathrm{lcm}(m,n))}{\mathrm{lcm}(m,n))}.
$$
It follows that
$$
\frac{\varphi(mn)}{mn}=\frac{\varphi(\mathrm{lcm}(m,n))}{mn} \cdot \mathrm{gcd}(m,n)
$$
hence the above claim..

First of all, we have $\operatorname{gcd}(a,b)\cdot\operatorname{lcm}(a,b)=ab$, then now you need to ascertain that Euler's Totient Function has the following property: $\phi(mn)=\phi(m)\phi(n)$. Hint: write a matrix with mn entries.


A simple question on limits Does the following equation hold:

$$ \lim_{x\to\infty} x^x = \left(\lim_{x\to\infty} x\right)^{\!x}?$$

I know that

$$ \lim_{x\to\infty} x^n = \left(\lim_{x\to\infty} x\right)^{\!n}$$

holds if $n$ is different from $x$, but I am not sure if it holds where $n=x$.

A notation like

$$\left(\lim_{x\to\infty}x\right)^x$$ is twice meanigless because

$x$ is used both as an independent variable (the exponent) and a dummy variable (argument of the limit), so which is which is ambiguous;
the limit diverges so that raising it to a power is not defined.

limit is similar to function, when you take limit, you let $x$ go to a specific point, or infinity, and you will get a exactly ONE answer if limit exist.( limit is unique if the space is Hausdorff, $R^n$ is Hausdorff).

I will slightly modify your question to :

$\lim_{x\to\infty} (\frac{1}{x}) ^x = (\lim_{x\to\infty} (\frac{1}{x}))^{x}$?

if we calculate the limit, we have   $0=0^x$.

for the right hand side of the equality, without any information on $x$, we cannot compute anything.


$ \int_0^af = 0, \forall a \in \mathbb{R} \Rightarrow f=0$ almost everywhere Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a Lebesgue integrable function and $\int_0^af=0, \forall a \in \mathbb{R}$.Prove that $f=0$ almost everywhere.

Here is my solution:

Let $0<a<b$, then $(0,b)=(0,a) \cup [a,b)$ thus $$\int_a^bf= \int_0^bf - \int_0^af \Rightarrow \int_a^bf=0$$

Now we have that $ \mathbb{R}= \bigcup_{n \in \mathbb{Z}}[n,n+1)$ which is a disjoint union, thus $$ \int_{\mathbb{R}}f=\sum_{n \in \mathbb{Z}} \int_n^{n+1}f=0$$.

Also we know that$$\{x:f(x)\neq 0\}= \{x:f(x)>0\} \cup \{x:f(x)<0\}=\{x:f(x)>0\} \cup \{x:-f(x)>0\}$$.

Now $\{x:f(x)>0\}=\{x:f(x)> \frac{1}{n}\})$ thus $m(\{x:f(x)> \frac{1}{n}\})\leqslant n \int_{\mathbb{R}}f=0$ therefore $m(\{x:f(x)> \frac{1}{n}\})=0 \Rightarrow m(\{x:f(x)>0\})=0$, from the subadditivity of the Lebesgue measure and from the Markov inequality.

With the same argument we can prove that $m(\{x:-f(x)>0\})=0$

Combining these we have that $$m(\{x:f(x) \neq 0\}) \leqslant m(\{x:f(x)>0\})+m(\{x:-f(x)>0\})=0$$

$\therefore$ $m(\{x:f(x) \neq 0\})=0$

$$Second-solution$$

Now $ \forall x \in \mathbb{R}$ we have in the same way as above that $\frac{1}{2 \delta}\int_{x-\delta}^{x+\delta}f(y)dy=0 ,\forall \delta>0$.

From Lebesgue's differentation theorem we have that $\lim_{\delta \rightarrow 0}\frac{1}{2\delta}\int_{x-\delta}^{x+\delta}f(y)dy =f(x)$ almost everywhere.

Thus $f(x)=0$ almost everywhere.

Are my solutions correct?

This is an exercise I found in a final exam, in undergraduate measure theory.
Although it is a final exam exercise, it seemed to me enough simple,that's why I want to verify my thoughts.

In case I am wrong I would appreciate any help to guide me to a correct solution.

Thank you in advance.

Following @Mario Gretsas 's thoughts. Please help me check if it's correct.

Suppose otherwise then $m\{f>0\}>0\ or\ m\{f<0\}>0$. Without loss of generality, we assume $m\{f>0\}>0$. Since $\{f>0\}=\lim_{n\to\infty}\{f\geq1/n\}$, and due to continuity of measurement and integrality of the function we have
$$\exists n_0\in\mathbb{N}\quad s.t.\quad 0<m\{f\geq1/n_0\}<\infty$$
For the measurablity of $\{f\geq1/n_0\},\exists\ an\ open\ set\ sequence\ \{G'_n\}_n\ $ such that

$$\bigcap_{n=1}^\infty G_n' \triangleq G\supset\{f\geq1/n_0\}\ \ \&\ \ m(G-\{f\geq 1/n_0\})=0\ \ \&\ \ \forall n,\ m(G_n')<\infty$$
Let $G_n=\bigcap_{k=1}^nG'_k$. Then we get a open set sequence $\{G_n\}$ monotonically decresing to $\{G_\delta\}$. Then $\lim_{n\to\infty}m(G_n-G)=0.$

Because of the uniform continuity of integrals, there is a real number $\delta>0$ such that for all measurable set $A$ satisfies $m(A)<\delta$ we have $|\int_Afdx|<\frac{m(G)}{2n_0}(>0)$.

Since $\lim_{n\to\infty}m(G_n-G)=0$, there is a $N$ such that $m(G_N-G)<\delta$.

$G_N$ is open, so it can be expressed as a countable union of open intervals, then $$\int_{G_N} fdx=0$$

But meanwhile, we have $$\int_G fdx=\int_{A_{n_0}}fdx\geq\frac{m\{f\geq1/n_0\}}{n_0}=\frac{m(G)}{n_0}>0$$

Thus $$\frac{m(G)}{2n_0}\geq|\int_{G_N-G}fdx|=|\int_G fdx-\int_{G_N} fdx|\geq\frac{m(G)}{n_0}$$

In contradiction to our hypothesis.

Take the contrapositive, if $f(x)\neq 0$ a.e. then $\int_{0}^{a} f\neq 0$.

Define the set $E=[0,a]$ (this is the domain of integration), note that we have $\mu(E)$ is non-empty and $\mu(E)=a$, as intervals have measure $b-a$. Now define $F=\{x\in E: f(x)\neq 0\}$. Clearly $F$ has positive measure. Now just use the definition of the Lebesgue integral to conclude: $\int_{E}f d\mu>0$. We have proven the contrapositive of the original statement, so indeed, we can conclude if $\int_{0}^{a} f=0$ $\forall a\in \mathbb{R}$ then $f=0$ a.e.

You can also try proving it by taking $\int_{0}^{a}f=\int_{0}^{b} f + \int_{b}^{a} f$ by choosing any arbitrary $b$ in $[0,a]$ and arguing by contradiction: if $f\neq 0$ a.e. then the sum of the integrals would not be zero. I haven't worked out the details but I think the argument carries through.


$\pi_n(X^n)$ free Abelian? I have encountered a problem which states that denote $X$ as an Eilenberg-MacLane space $K(G,1)$ and is a CW complex, show that $\pi_n(X^n)$ is free Abelian for $n \geqslant 2$.

However, I think I have a conceptual misunderstanding in this problem. According to cellular approximation, the map

\begin{equation}
f:  (S^n, s_0) \rightarrow (X,x_0)
\end{equation}

can be homotoped to a cellular map so that it maps $(S^n,s_0)$ into $X^n$. In this sense the $\pi_n(X^n)$ should be equal to $\pi_n(X)$, which is actually zero.

I would appreciate for clarification in this case. Moreover, how do we show that the homotopy group is free Abelian?

Joe Johnson 126's answer is incorrect; see Is it true that $\pi_n(X^{n-1})=0$ for $n>1$ if $X$ is $K(G,1)$ space?.  Indeed, the claimed statement is very much not true for all CW-complexes.  For instance, if $X=S^3$, then $\pi_4(X^4)=\pi_4(S^3)=\mathbb{Z}/2$.

Here is a correct answer.  First of all, to address the first question, all your argument shows is that the map $\pi_n(X^n)\to\pi_n(X)$ is surjective.  You know that every map $S^n\to X$ can be homotoped to a map $S^n\to X^n$, but it might be that two such maps are homotopic in $X$ but not in $X^n$.  A homotopy of such maps is a map $S^n\times I\to X$, which by cellular approximation can only be homotoped to $X^{n+1}$, since $S^n\times I$ is $(n+1)$-dimensional.  So you do have an isomorphism $\pi_n(X^{n+1})\to \pi_n(X)$.

Let's now solve the original problem.  Note that by the argument above, $\pi_k(X^n)\cong \pi_k(X)$ for all $k<n$, so since $X$ is a $K(G,1)$, $\pi_k(X^n)$ is trivial for $k<n$ except for $k=1$.  Let $Y$ be the universal cover of $X^n$; then $Y$ is also an $n$-dimensional CW-complex, and now $\pi_k(Y)=0$ for all $k<n$.  By Hurewicz, we have $\pi_n(Y)\cong H_n(Y)$.  But $H_n$ of any $n$-dimensional CW-complex is free abelian (because computing cellular homology, there are no nontrivial $(n+1)$-chains and hence no nontrivial $n$-boundaries, so $H_n$ is just the group of $n$-cycles, which is a subgroup of the free abelian group of $n$-chains).  Thus $\pi_n(Y)$ is free abelian, and hence so is $\pi_n(X^n)\cong \pi_n(Y)$.

This is true of any CW-complex.  We have an inclusion $X^{n-1}\hookrightarrow X^{n}$.  The long exact sequence of the pair gives
$$
0\rightarrow \pi_{n}(X^{n})\rightarrow \pi_{n}(X^{n},X^{n-1})\rightarrow \pi_{n-1}(X^{n-1})\rightarrow \cdots
$$
Since $X^n/X^{n-1}$ is a wedge of spheres, $\pi_n(X^n,X^{n-1})$ is free abelian.  Then $\pi_n(X^n)$ is a subgroup of a free abelian group, hence free abelian.

EDIT: To address the question in the comments below: The inclusion $X^n\hookrightarrow X$ induces an isomorphism on homotopy groups for $k<n$ and a surjection at $k=n$.  This is due to the same long exact sequence as above
$$
\cdots\to \pi_{k+1}(X,X^n)\to \pi_k(X^n)\to \pi_k(X)\to \pi_k(X,X^n)\to \pi_{k-1}(X^n)\to \cdots.
$$
If $k\leq n$ then cellular approximation tells us that all maps $S^k\to X$ can be homotoped into $X^n$.  Thus $\pi_k(X,X^n)=0$ for $k\leq n$.  Looking at the sequence above one sees that $\pi_k(X^n)\cong \pi_k(X)$ for $k<n$ and $\pi_n(X^n)$ surjects onto $\pi_n(X)$.  Similarly, $\pi_n(X^{n+1})\cong \pi_n(X)$.  Thus $\pi_n(X)$ only depends on the the $n+1$-skeleton.


Why doesn't adjoining $\sqrt{3}$ to $\mathbb{F}_{11}$ return $\mathbb{F}_{11}$? I am confused about a particular instance where adjoining an element of a field to itself makes it not equal to itself and I am asking for clarification. I can see the result is true, but I can not see why. We are not introducing any new element and we are not setting any new elements equal to zero.
In the finite field $\mathbb{F}_{11}$ we adjoin $\alpha$ where $\alpha^2 - 3 =0$. Because $(\pm 5)^2 -3 =0$ the two square roots of $3$ are already in $\mathbb{F}_{11}$, so we are either adjoining $5$ or $-5$. We do not know which, although both elements are invertible. However we cannot invert $\alpha +5$ because we don't know if $\alpha +5$ or $\alpha-5 = 0$ so $\mathbb{F}_{11}[\alpha]$ is not a field.
By adjoining an ambiguous element of the field to itself I thought maybe we were setting elements equal to zero, but it's not $5 = -5$ because then $10 = 0$ which makes every element $0$.
Source:  This was an example in Algebra by Artin where Artin says $\mathbb{F}_{11}[\alpha] \simeq \mathbb{F}_{11}[x]/(x^2-3)$ is not a field on page 366.
Sorry for edits
Edit 2: If I am understanding what is being said, $\alpha$ must assume a specific value in $\mathbb{F}_{11}[\alpha]$. So it is $not$ true $\mathbb{F}_{11}[\alpha] \simeq \mathbb{F}_{11}[x]/(x^2-3)$ because the kernel of the evaluation homomorphism $\phi: \mathbb{F}_{11}[x] \rightarrow \mathbb{F}_{11}[\alpha]$ is not the ideal $(x^2-3)$, but one of the ideals $(x-5)$ or $(x+5)$.
I will rewrite a comment here hopefully to clarify.
In this image Artin describes $R'$ as "obtained by adjoining an element $\alpha$ to $\mathbb{F}_{11}$". A page earlier Artin defined "$R[\alpha] = \text{ring obtained by adjoining} \ \alpha \ \text{to} \ R$". There are also examples of using the evaluation homomorphism to show results such as $\mathbb{R}[x]/(x^2+1)\simeq \mathbb{Q}$ and $R[x,y] \simeq R[x][y]$.
Artin writes
\begin{equation}
R' = \mathbb{F}_{11}[x]/(x^2-3)
\end{equation}
In the same paragraph he says "...procedure applied to $\mathbb{F}_{11}$ does not yield a field", and "But we haven't told $\alpha$ whether to be equal to $5$ or $-5$. We've only told that its square is $3$."
With this wording it sounds like the kernel of $\phi$ is not $(x-5)$ or $(x+5)$, but only $(x^2-3)$. Which again confuses me because then $\mathbb{F}_{11}[\alpha]$ is not a field.

There are two different rings being discussed here:

$\mathbb{F}_{11}[\alpha]$, the ring obtained by adjoining some specific $\alpha\in \overline{\mathbb{F}}_{11}$ with $\alpha^2 = 3$;
$\mathbb{F}_{11}[X] / (X^2 - 3)$.

In the first case, $\mathbb{F}_{11}$ already contains $\alpha$; as you point out, $\alpha = \pm 5\in \mathbb{F}_{11}$. In the latter case, we no longer have a field:
\begin{align*}
\mathbb{F}_{11}[X]/(X^2 - 3) = \mathbb{F}_{11}[X]/(X - 5)\oplus \mathbb{F}_{11}[X]/(X + 5).
\end{align*}
If $f\in \mathbb{F}_{11}[X]$ is an irreducible nonconstant polynomial, then the map $\mathbb{F}_{11}[X]/(f) \to \mathbb{F}_{11}[\alpha]$ given by $X \to \alpha$, where $\alpha$ is a zero of $f$ in $\overline{\mathbb{F}}_{11}$, is an isomorphism; for then $\alpha$ is not a zero of any polynomial of degree less than $\deg f$, and comparing dimensions gives the result. That result doesn't hold without the irreducibility assumption, though.

Your sequence
However we cannot invert $\alpha +5$ because we don't know if $\alpha +5$ or $\alpha-5 = 0$ so $\mathbb{F}_{11}[\alpha]$ is not a field.
is confusing.
Either $\alpha = 5$ and in this case $\alpha+5$ is invertible... And $\mathbb F_{11}[\alpha] = \mathbb F_{11}$ is a field. Or $\alpha = -5$ and in that case $\alpha+ 5=0$ is not invertible. But in that case still $\mathbb F_{11}[\alpha] = \mathbb F_{11}$ is a field.


Does the sequence $x_{n+1} = \frac{1}{2} ( (x_n)^2 + c ) $ diverge? Does this recursive sequence defined by the relation

$x_{n+1} = \frac{1}{2} ( (x_n)^2 + c ) $

diverge. Given that we know that , $x_0 > 1$ and c lies in the open interval (0,1)

If yes then please how ? If not then also please proof.

My attempt : My initial attempt was to assume that the sequence diverges , given that it is strictly  increasing. However, I realised that I could not actually proof that the series is going towards infinity. ( After one of my friends said that just saying that the series is monotonically increasing won't be enough )

So far my attempt to prove that the sequence goes to infinity has been to find the general term , and then show that as n tends to infinity the function also tends to infinity. However i could not find the general term .

So I tried another approach too , which was basically trying to find a  sequence $b_n$such that each term of the new sequence is smaller than the corresponding  term of the original sequence  {$a_n$} and then to show that $b_n$ diverges .

However, I haven't found such a sequence yet !.

Any help on whether I am proceeding in the right direction , and proof of whether the sequence converges or diverges will be appreciated.

If the sequence converges to some limit $L$, then we must have
$$L = \frac{1}{2}(L^2 + c)$$
so
$$L^2 - 2L  = -c$$
Completing the square on the left hand side, this becomes
$$(L - 1)^2 = 1-c$$
and therefore
$$L = 1 \pm\sqrt{1-c} <2$$
where the inequality holds because $0 < c < 1$.

Now consider $x_0 = 3$ and $c=1/2$. If $x_n \geq 3$ then
$$x_{n+1} = \frac{1}{2}\left(x_n^2 + \frac{1}{2}\right) \geq \frac{1}{2}\left(3^2 + \frac{1}{2}\right)= \frac{19}{4} \geq 3$$
so by induction we see that $x_n \geq 3$ for all $n$. Thus the sequence cannot converge to a limit smaller than $3$. But as shown above, if it does converge to a limit $L$, then $L < 2$. This means that the sequence diverges.

On the other hand, as pointed out by Logan Maingi in the comments, if $x_0 = 3/2$ and $c = 1/4$, then $x_n$ is strictly decreasing and bounded below by $0$, so it converges.

So, if all we know is that $x_0 > 1$ and $c \in (0,1)$, that is insufficient information to decide whether $x_n$ converges or diverges. It depends on the specific values of $x_0$ and $c$, and this dependence is shown precisely in the answer by mathguy.

Let $L = \lim_{n\to\infty}x_n$. Then $\lim_{n\to\infty}x_{n+1} = L$ as well. Therefore we have
    $$1 =\lim_{n\to\infty}\frac{x_n}{x_{n+1}} = \lim_{n\to\infty}\frac{x_n}{\frac{1}{2}((x_n)^2 +c)} = \frac{L}{\frac{1}{2}(L^2 + c)}$$

You should be able to solve for $L$, the solution will depend on $c$.


Alice and Bob play the determinant game Alice and Bob play the following game with an $n \times n$ matrix, where $n$ is odd. Alice fills in one of the entries of the matrix with a real number, then Bob, then Alice and so forth until the entire matrix is filled. At the end, the determinant of the matrix is taken. If it is nonzero, Alice wins; if it is zero, Bob wins. Determine who wins playing perfect strategy each time.

When $n$ is even it's easy to see why Bob wins every time. and for $n$ equal to $3$ I have brute-forced it. Bob wins. But for $n = 5$ and above I can't see who will win on perfect strategy each time. Any clever approaches to solving this problem?

I tried to approach it from Leibniz formula for determinants

$$\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n A_{i,\sigma_i}.$$

There are $n!$ factorial terms in this sum. Alice will have $\frac{n^2+1}{2}$ moves whereas Bob has $\frac{n^2-1}{2}$ moves. There are $n^2$ variables (matrix entries). Each of them taken alone appear in $(n-1)!$ terms in this summation. Whenever Bob picks a zero in his first move for any entry in the matrix, $(n-1)!$ factorial terms of this go to zero. For instance, consider a $5 \times 5$ matrix. So there are 120 terms. In first move, whenever Bob makes any matrix entry zero, he zeros out 24 of this terms. In his second move, he has to pick that matrix entry which has least number of presence in the first zeroed-out 24 terms. There can be multiple such matrix entries. In face, it can be seen that there is surely another matrix entry appearing in 24 non-zero terms in the above sum. Since $n$ is odd in this case, the last chance will always be that of Alice. Because of that, one doesn't have to bother about this terms summing to zero. What Bob has to do if he has to win is that

He has to make sure he touches at least once (in effect zeroes) each of this 120 terms. In the $n=5$ case, he has 12 chances. In this 12 chances he has to make sure that he zeros out all this 120 terms. In one sense, It means that he has to average at least 10 terms per chance of his. I looked at the $n=3$ case, bob has 4 chances there and 6 terms, he can zero out all of them in 3 moves.
He has to make sure that Alice doesn't get hold of all the matrix entries in one single term in 120 terms, because then it will be non-zero, and since the last chance is hers, Bob won't be able to zero it out, so she will win.

As per above explanation, in the $5 \times 5$, he has to just average killing 10 terms in each chance which seems quite easy to do. I feel this method is a bit easy to generalize and many really clever people in here can do it.

EDIT----------------

In response to @Ross Milikan, I tried to look at solving $5 \times 5$ case, this is the approach. Consider $5 \times 5$ matrix with its entries filled in by the english alphabets row-wise, so that the matrix of interest is

\begin{align}
\begin{bmatrix}
a & b & c & d& e \\ f& g & h &i& j \\k& l& m& n& o \\ p& q& r& s& t\\ u& v& w& x& y
\end{bmatrix}
\end{align}

Without loss of Generality (WLOG), let Alice pick up $a$ (making any entry zero is advantageous for her). Lets say Bob picks up $b$ (again WLOG, picking up any entry is same). This helps Bob to zero out 24 terms in the total 120. Alice has to pick up one entry in this first row itself otherwise she will be in a disadvantage (since then, Bob gets to pick the 3 terms in total from the first row and gets 72 terms zeroed out). So concerning the first row, Alice picks 3 of them, Bob picks 2 of them (say $b$ and $d$), and hence he zeros out 48 terms of the total 120. Now note that next move is Bob's. Let us swap the second column and first column. This doesn't change the determinant other than its sign. Look at the modified matrix

\begin{align}
\begin{bmatrix}
0 & \otimes & \otimes & 0 & \otimes \\ g & f & h &i& j \\l& k& m& n& o \\ q& p& r& s& t\\ v& u& w& x& y
\end{bmatrix}
\end{align}

where $0$ is put in entries Bob has modified and $\otimes$ has been put in entries modified by Alice. Now in the first column, lets say Bob gets hold of $g$ and $q$, and alice gets hold of $l$ and $v$. Again Alice has to do this and any other move will put her in a disadvantage. Bob had made 4 moves already, the next move is his and now the matrix will look like,

\begin{align}
\begin{bmatrix}
0 & \otimes & \otimes & 0 & \otimes \\ 0 & f & h &i& j \\ \otimes & k & m& n& o \\ 0 & p& r& s& t\\ \otimes & u& w& x& y
\end{bmatrix}
\end{align}

Now we are left with the lower $4 \times 4$ matrix, Bob is left with 8 chances, and the first move is his. Compare this with $4 \times 4$ case, it looks intuitively that Bob should win.

We know that the determinant of a matrix A is non-zero if the rank of the matrix is the same size of the matrix. So, in this case, rank(A) is odd. So, on the last column (or first column) all Alice has to do to guarantee a win is make sure the real number cannot be written as a linear combination of the others in that column, thereby making every column independent. This in turn makes the rank(A) = n, which is to say the determinant is non-zero.

May not be the correct answer as to what you are looking for, but it is a generalized result.


how to evaluate $\lim_{x \to \infty}(1+4/x)^\sqrt{x^2+1}$ $$
\lim_{x \to \infty}(1+4/x)^\sqrt{x^2+1}
$$
is like
$$
\lim_{x \to \infty}(1+1/x)^x = e
$$

I have replaced
$\sqrt{x^2+1}$ by $x$ but I haven't got the expected result ($e^4$).

$$\underset{\begin{smallmatrix}
 u\to 0 \\
 v\to \infty
\end{smallmatrix}}{\mathop{\lim }}\,{{\left( 1+u \right)}^{v}}={{\mathrm{e}}^{\underset{\begin{smallmatrix}
 u\to 0 \\
 v\to \infty
\end{smallmatrix}}{\mathop{\lim }}\,uv}}$$


finding the volume of the solid via disk or washer method the question is: $y = 1/4x^2$, $x = 2$, $y = 0$; about the $y$-axis

I tried to draw it out, but I can't figure this stuff out. The graphing is the hardest part for me because I don't know what to do to graph whatever information is given and then make some sort of reasonable deduction about what to plug in where for either disk or washer method.

The region that is being rotated is the part of the first quadrant that is below the easily drawn parabola $y=\frac{x^2}{4}$ and to the left of the vertical line $x=2$.

We are rotating about the $y$-axis. Imagine taking a slice of the solid perpendicular to the $y$-axis, "at" $y$. The cross-section is a disk with a circular hole in it. The outer radius of the cross-section is $2$, and the inner radius is $2\sqrt{y}$. So the area of cross-section is $\pi\left(2^2-(2\sqrt{y})^2\right)$. Note that when $x=2$, we have $y=1$. Thus the volume is
$$\int_0^1 \pi\left(4-4y\right)\,dy.$$
Integrate. We get $2\pi$.

Another way: Alternately, we can use the Method of Cylindrical Shells. Take a thin horizontal slice of width "$dx$" at $x$, and rotate it about the $y$-axis. We get a thin cylindrical shell of radius $x$, height $\frac{x^2}{4}$, and thickness $dx$. Thus the volume of the shell is approximately $(2\pi x)(x^2/4)\,dx$. Now "add up" (integrate) from $x=0$ to $x=2$. Our volume is
$$\int_0^2 (2\pi)\left(\frac{x^3}{4}\right)\,dx.$$
Calculate. Again we get $2\pi$.


Normal subgroup of prime index Generalizing the case $p=2$ we would like to know if the statement below is true.

Let $p$ the smallest prime dividing the order of $G$. If $H$ is a subgroup of $G$ with index $p$ then $H$ is normal.

This is a standard exercise, and the answer is that the statement is true, but the proof is rather different from the elementary way in which the $p=2$ case can be proven.

Let $H$ be a subgroup of index $p$ where $p$ is the smallest prime that divides $|G|$. Then $G$ acts on the set of left cosets of $H$, $\{gH\mid g\in G\}$ by left multiplication, $x\cdot(gH) = xgH$.

This action induces a homomorphism from $G\to S_p$, whose kernel is contained in $H$. Let $K$ be the kernel. Then $G/K$ is isomorphic to a subgroup of $S_p$, and so has order dividing $p!$. But it must also have order dividing $|G|$, and since $p$ is the smallest prime that divides $|G|$, it follows that $|G/K|=p$. Since $|G/K| = [G:K]=[G:H][H:K] = p[H:K]$, it follows that $[H:K]=1$, so $K=H$. Since $K$ is normal, $H$ was in fact normal.

In addition to the things answered here, I just want to add this one.
Proposition Let $G$ be a finite group and $H$ a subgroup of prime index $p$, with gcd$(|G|,p-1)=1$. Then $G' \subseteq H$.
 Note that this implies that $H \unlhd G$, and that it is in fact sufficient to prove that $H$ is normal, since then $G/H \cong C_p$ is abelian.
Proof  Firstly, we may assume by induction on $|G|$, that $H$ is core-free, that is core$_G(H)=\bigcap_{g \in G}H^g=1$. This means that $G$ can be homomorphically embedded in $S_p$. Let $P \in Syl_p(G)$ and note that because $|S_p|=p \cdot (p-1) \cdots \cdot 1$, $|P|=p$. By the $N/C$-Theorem, $N_G(P)/C_G(P)$ embeds in Aut$(P) \cong C_{p-1}$. By the assumption gcd$(|G|,p-1)=1$, we get that $N_G(P)=C_G(P)$. Since $P$ is abelian we have $P \subseteq C_G(P)$, whence $P \subseteq Z(N_G(P))$. We now can apply Burnside's Normal $p$-complement Theorem, which implies that $P$ has a normal complement $N$, that is $G=PN$ and $P \cap N=1$. Note that $|G/N|=p$. Look at the image of $H$ in $G/N$. Then $G=HN$, or $HN=N$. In the latter case $H \subseteq N$, and $|G:H|=|G:N|=p$, whence $H=N$ and we are done if we can refute the first case. If $G=HN$, then $|G:H \cap N|=|G:N|\cdot|N:H \cap N|=|G:N|\cdot |G:H|=p \cdot p=p^2$, contradicting the fact that $|G| \mid |S_p|$. The proof is now complete.
Corollary 1 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$, the smallest prime dividing the order of $G$. Then $G' \subseteq H$. In particular, $H$ is normal.
Corollary 2 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$ and gcd$(|H|,p-1)=1$. Then $H$ is normal.
Observe that this last result renders a well-known result for $p=2$! Finally for fun:
Corollary 3 Let $G$ be a finite group of odd order and $H$ a subgroup with $|G:H|=65537$. Then $H$ is normal.


Why is $1$ not a prime number? Why is $1$ not considered a prime number?

Or, why is the definition of prime numbers given for integers greater than $1$?

One of the whole "points" of defining primes is to be able to uniquely and finitely prime factorize every natural number.

If 1 was prime, then this would be more or less impossible.

The Question should be _"Why is the word $prime$ used only for a positive non-invertible indecomposable integer?"_The Answer is that Positive-NonInvertible-Indecomposable-Integer is too long, except in German. If you need a name for numbers that are either prime or $1$, you may invent one!


Legendre polynomials and primality testing Can you provide a proof or a counterexample for the following claim ?

  Let $n$ be an odd natural number greater than one . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \not\equiv 1 \pmod r$ . Let $P_n(x)$ be Legendre polynomial , then $n$ is a prime number if and only if $P_n(x) \equiv x^n \pmod {x^r-1,n}$ .

You can run this test here .

I have tested this claim up to $2 \cdot 10^4$ and there were no counterexamples .

This is a partial answer.

This answer proves that if $n$ is an odd prime number, then $P_n(x) \equiv x^n \pmod {x^r-1,n}$.

Proof :

Using the following expression
$$P_n(x)=\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2(x-1)^{n-k}(x+1)^k$$

we have, by the binomial theorem,
$$\begin{align}&2^n\left(P_n(x)-x^n\right)
\\\\&=-2^nx^n+\sum_{k=0}^{n}\binom{n}{k}^2(x-1)^{n-k}(x+1)^k
\\\\&=-2^nx^n+(x-1)^{n}+(x+1)^n+\sum_{k=1}^{n-1}\binom{n}{k}^2(x-1)^{n-k}(x+1)^k
\\\\&=-2^nx^n+\sum_{k=0}^{n}\binom nkx^{n-k}((-1)^{k}+1)+\sum_{k=1}^{n-1}\binom{n}{k}^2(x-1)^{n-k}(x+1)^k
\\\\&=-2^nx^n+2\sum_{j=0}^{(n-1)/2}\binom n{2j}x^{n-2j}+\sum_{k=1}^{n-1}\binom{n}{k}^2(x-1)^{n-k}(x+1)^k
\\\\&=(-2^n+2)x^n+2\sum_{j=1}^{(n-1)/2}\binom n{2j}x^{n-2j}+\sum_{k=1}^{n-1}\binom{n}{k}^2(x-1)^{n-k}(x+1)^k
\end{align}$$

Now, we have $-2^n+2\equiv 0\pmod n$ by Fermat's little theorem. Also, since $\binom nm\equiv 0\pmod n$ for $1\le m\le n-1$, we see that there is a polynomial $f$ with integer coefficients such that
$$2^n\left(P_n(x)-x^n\right)=nf$$
from which
$$P_n(x)=x^n+(x^r-1)\times 0+n\times \frac{1}{2^n}f$$
follows.

It follows from this and $\gcd(n,2^n)=1$ that
$$P_n(x) \equiv x^n \pmod {x^r-1,n}\qquad\square$$

See here for more information. This covers the test you are describing, however it is stated as a conjecture, not a primality test.


Funny thing. Multiplying both the sides by 0? Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way we may also say that $1=2=3$. I know it is wrong. But why? I mean if we can multiply both the sides by $2$ then why not by $0$.  For example, consider the following trigonometric identity :

Prove the identity : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta$

Usual way

To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $

$\displaystyle \implies {\sin^2 \theta \over \cos^2 \theta } = {\tan^2 \theta \cos ^2 \theta \over \cos^2 \theta}$ (multiplying both the sides by $\displaystyle 1 \over \cos^2 \theta$)

$\implies \tan ^2 \theta = \tan^2\theta$

$\implies LHS=RHS$

$\therefore proved$

Funny way

To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $

$\displaystyle \implies {\sin^2 \times 0} = {\tan^2 \theta \cos ^2 \theta \times 0}$ (multiplying both the sides by $0$)

$\implies 0 = 0$

$\therefore proved$

Please explain why is this wrong.

$a = b$ implies $ac = bc$, but $ac = bc$ doesn't imply $a = b$. (Not immediately. Read below.)

The way you usually get $a = b$ from $ac=bc$ is by multiplying both sides with $1/c$, which is only available when $c \ne 0$.

Zero differs from other numbers and mathematical entities which represent some sort of value because it is the absence of value, not a value itself. Thus, it would be incorrect to say $a \cdot 0 = b \cdot 0$ means that $a = b$. Similarly, in the case of infinity: let's say that $a$ does equal $b$. However, $a \cdot \infty \neq b \cdot \infty$ because infinity does not have a definite or specific value. Rather, it is a concept of something without limit.


Solving for inverse transformation in change of variables This is something probably silly, but I don't seem to see how this works. I have 2 independent random variables $X \sim Gamma(\alpha, \beta_x)$ and $Y\sim Gamma(\alpha, \beta_y)$, and I need to show that $\frac{X}{Y}$ and $X+Y$ are independent.

To show this, I let $U = X+Y$ and $V = \frac{X}{Y}$ and tried to express $X$ and $Y$ in terms of $U$ and $V$. I know that the solution is $X = \frac{U+V}{1+V}$ and $Y = \frac{U}{1+V}$, but I have no idea how to algebraically manipulate $U$ and $V$ to get to this point. Is there a systematic way to solve for inverse transformation?

For the ratio distribution of $X, Y,$ the distribution $  U=X/Y $ is

$$f_U(u) = \int_{-\infty}^\infty f_{X,Y}(uy,y)|y|\ dy$$

For the summation distribution of  $X, Y,$ the distribution $  V=X+Y $ is

$$f_V(v) =\int_{-\infty}^\infty f_{X,Y}(x,v-x)\ dx = Conv(f_X(x),f_Y(y))$$

The x solution is incorrect.

Solutions  should  be   x = uv/(1+v)   and   y = u/(1+v)

*Although post is a year old, others may benefit from knowing the correct answer. I couldn't post image of solution and I am too lazy to type it in this post. Click here for solution.


Solve $\dot v + \frac{\alpha}{m}v + \frac{\beta}{m} v^2 = g$ $\alpha, \beta$ are constants.
I know this is a Riccati equation, and I've tried to find a particular solution, with no success.
Also, $v = v(t)$, but $t$ is not part of the equation explicitly, so maybe that could help somehow?
I don't know what else to try, I'd appreciate a push in the right direction.

Darsen is of course correct for the case where $\dot v =0 \; .$ That being said, my first observation is that this equation is integrable. Observe that $$\dot v  = g - \frac{\alpha}{m}v - \frac{\beta}{m} v^2$$
$$\text{implies}$$
$$\frac{\dot v}{g - \frac{\alpha}{m}v - \frac{\beta}{m} v^2}  = 1 $$
whenever $\dot v \neq 0$ thus we can find an expression for $v$ using
$$\int\frac{dv}{g - \frac{\alpha}{m}v - \frac{\beta}{m} v^2}  = \int dt \; .$$
Use some good integral tables and you will find a solution where $v$ is easy to isolate. Let me know if you need any more guidance.
Edit:
pujfei is also correct in his answer and you'll find that the integral above is a flavor of the $\arctan $ function.

Result
Given the original equation:
$$
v'(t)+\frac{\alpha  }{m}v(t)+\frac{ \beta }{m}v(t)^2=g
$$
The general solution is:
\begin{equation}
v(t)= \frac{-\alpha +\sqrt{-\alpha ^2-4 \beta  g m} \tan \left(\frac{1}{2} \left(-\frac{t \sqrt{-\alpha ^2-4 \beta  g m}}{m}-c_1 \sqrt{-\alpha ^2-4 \beta  g m}\right)\right)}{2 \beta }
\end{equation}
Or:
$$
\begin{align*}
v(t)&=\frac{-\alpha+\sqrt{-\alpha^2-4bgm}\tan{\big[\frac{1}{2}\big(-\frac{\sqrt{-\alpha^2-4{\beta}gm}}{m}{\cdot}t-\sqrt{-\alpha^2-4{\beta}gm}{\cdot}c_1\big)\big]}}{2{\beta}}\\
&=\frac{1}{2{\beta}}\bigg(-\alpha+\sqrt{-\alpha^2-4{\beta}gm}\tan{\big[\frac{1}{2}\big(-c_1\sqrt{-\alpha^2-4{\beta}gm}-\frac{\sqrt{-\alpha^2-4{\beta}gm}}{m}{\cdot}t\big)\big]}\bigg)\\
&=\frac{1}{2{\beta}}\bigg(-\alpha+\sqrt{\Delta}\tan{\big[\frac{1}{2}\big(-c_1\sqrt{\Delta}-\frac{\sqrt{\Delta}}{m}{\cdot}t\big)\big]}\bigg)
\end{align*}
$$
In which the discriminant for this kind of differential equations is apparently:
$$
\Delta=-\alpha^2-4{\beta}gm
$$
Proof
The solution is achieved by applying the following transformation to a general form of Riccati Equation.
A Riccati Equation is an equation that looks like this:
$$y'(x)=q_0(x)+q_1(x)y(x)+q_2(x)y^2(x),q_2(x)\neq0\tag{1}$$
It can always be reduced to a second order linear ordinary differential equation (ODE).
With:
$$y'=q_0(x)+q_1(x)y+q_2(x)y^2\tag{2}$$
Let $v=yq_2$, we'll have:
$$v'=v^2+R(x)v+S(x)\tag{3}$$
where
$$
\left\{
\begin{align}
S&=q_0q_2\\
R&=q_1+\frac{q^{'}_2}{q_2}
\end{align}\tag{4}
\right.
$$
To continue, substitute:
$$v=-\frac{u'}{u}\tag{5}$$
We'll achieve a linear 2nd order ODE:
$$u''-Ru'+Su=0\tag{6}$$
Whose general solution is:
$$
\begin{equation}
u(x)= c_1 e^{\frac{1}{2} x \left(R-\sqrt{R^2-4 S}\right)}+c_2 e^{\frac{1}{2} x \left(R+\sqrt{R^2-4 S}\right)}
\end{equation}\tag{7}
$$
Thus we have:
$$
\begin{align}
y&=-\frac{u'}{q_2u}\\
&=\frac{1}{2q_2}\left(-R+\sqrt{R^2-4 S} \left(-1+\frac{2 c_1}{c_2 e^{x \sqrt{R^2-4 S}}+c_1}\right)\right)
\end{align}\tag{8}
$$
For the instance in this question:
$$
\left\{
\begin{align}
q_0&=g\\
q_1&=-\frac{\alpha}{m}\\
q_2&=-\frac{\beta}{m}\\
R&=q_1=-\frac{\alpha}{m}\\
S&=q_2q_0=-\frac{g\beta}{m}\\
R^2-4S&=\frac{\alpha^2+4{\beta}mg}{m^2}
\end{align}\tag{9}
\right.
$$


Ways to study mathematics while commuting I spend approximately 3 to 4 hours on public transport everyday. I try to maximize the usage of this time by checking email etc on my phone.
Are there any tips to study mathematics while commuting?
Thanks for sharing! Just want to make full use of the time spent commuting!

When going to uni I would spend 5 hours on a train every day. I would use that time to solve lots of problems, individual ones, such as the ones found at the end of each chapter in the book. Studying concepts while commuting is too hard as the train is too distracting an environment. So I would just do problems after problems on the train, and save the studying of the concepts themselves for another time.

IF you want to apply mathematics, e.g. increase your command of algebra: use khan academy. THe courses for math a re simply great: https://www.khanacademy.org/.
I am doing this right now for my daily 2 hour commute. THe caveat: you need internet access, at least sporadically.


How many numbers with all different digits are between $1000$ and $4600$? To answer this question my thought was to solve the problem by parts.
So, first I counted how many numbers with all different digits there are less than $4000$:

For the first position (left to right) there are $4$ possibilities $(0,1,2,3)$. For the second position there are $9$, for the third they are $8$ and for the last we have $7$. Multiplied all  possibilities, $4 \cdot9\cdot8\cdot7=2016$.

Then I counted how many numbers with all different digits there are less than $1000$. In this case one have at most $3$ positions. For the first position there are $10$ possibilities $(0,1,2,...,8,9)$, on the second position there are $9$ possibilities and the last position have $8$ possibilities. So, $10 \cdot9\cdot8=720$

Making the difference, $2016-720$, we get $1296$. Now I will counted the numbers with all different digits between $4000$ and $4600$:

In this part I will counted for less than $4600$, until $4599$. For the first position there is only one possibility. For the second position there are $5$, $(0,1,2,3,5)$. For the third position there are $8$ and the last position have $7$ possibilities. So $1 \cdot5\cdot8\cdot7=280$.

Finally there are $1296+280=1576$ numbers with all different digits between $1000$ and $4600$. Is my thought correct? Thanks

We consider two cases:

The positive integer is at least $1000$ but less than $4000$.  We have three choices for the thousands digit ($1$, $2$, or $3$), nine choices for the hundreds digit (since we must exclude the thousands digit), eight choices for the tens digit (since we must exclude both the thousands digit and the hundreds digit), and seven choices for the units digit (since we must exclude the thousands digit, the hundreds digit, and the tens digit), giving a total of $3 \cdot 9 \cdot 8 \cdot 7 = 1512$ four digit positive integers with distinct digits less than $4000$.
The number is at least $4000$ and at most $4600$.  As you determined, there is one choice for the thousands digit (namely, $4$), five choices for the hundreds digit ($0$, $1$, $2$, $3$, $5$), eight choices for the tens digit, and seven choices for the units digit, which yields $1 \cdot 5 \cdot 8 \cdot 7 = 280$ four digit positive integers with distinct digits between $4000$ and $4600$ inclusive.

Thus, there are $1512 + 280 = 1792$ positive integers $n$ satisfying the inequalities $1000 \leq n \leq 4600$ with distinct digits.

Initially consider
4 possibilities for the thousands;
6 possibilities for the hundreds as there are 7 digits 0 to 6;
8 possibilities for the tens;
7 possibilities for the units;

= 1,344 with a few numbers > 4600

The incorrect possibilities occur when we have 4 in the thousands and 6 in the hundreds. This will occur 8 x 7 times ie 56 times.

So the total number of distinct numbers is 1,344 – 56 = 1,288


How to show linear independency of non-zero vectors of a 3x3 matrix? If $C$ is a $3×3$ matrix and if the non-zero vectors $u$ and $v\in \Bbb R^3$ are such that,

$Cu=2u$ and $Cv=−5v$

How can you show that $u$ and $v$ are linearly independent?

Suppose it is dependent. Then $$u=kv$$ where $k$ is non zero. Then $$Cu=kCv \implies 2u=-5kv$$ Now multiply $2$ on both sides of $u=kv$. Then we have $$2u=2kv$$

Subtract the two equations, we get $$0=7kv$$ and so $k=0$, a contradiction!

Suppose u and v are not independent. That is, suppose there exist numbers, A and B, not both 0 such that Au+ Bv= 0. Then C(Au+ Bv)= ACu+ BCv= A(2u)+ B(-5v)= 2Au- 5Bv= 2(Au+ Bv)- 3Bv= -3Bv= 0.  It follows that v= 0, a contradiction.


Compute $\int_1^e \frac{dx}{x(x+(\ln x)^2)}$ My friend asked me how to integrate the following:
$$\int_1^e \frac{dx}{x(x+(\ln x)^2)}$$

How am I going to solve this?Any help is greatly appreciated.
Thanks.

Assume $u=\ln x$, then
\begin{align}
I:=\int_1^e \frac{dx}{x(x+(\ln x)^2)}
=\int_0^1 \frac{e^u du}{e^u(e^u+u^2)}
&=\int_0^1 \frac{ du}{e^u+u^2}
\\
&=\int_0^1 \frac{e^{-u} du}{1+u^2e^{-u}}
\end{align}
Using geometric series $$\frac{1}{1+u^2e^{-u}}=\sum_{n=0}^{\infty}{(-1)^nu^{2n} e^{-nu} }$$ term by term integration, we have
\begin{align}
I=\sum(-1)^n (\int_0^1 {  u^{2n} e^{-(n+1)u}du}=J_n)
\end{align}
Integrating by parts we get (from tables) $$J_n=-e^{-(n+1) u}\sum_{k=0}^{2n}\frac{(2n)!}{(2n-k)!}(n+1)^{-(k+1)}u^{2n-k}$$


fit sine wave to data I have some astronomical data which I know has a sinusoidal component to it of the form

y = Asin(Ωt+Φ). The period of the sinusoid is equal to a sidereal day. So I know the frequency Ω. So, I just need a way to find out optimal values for A and Φ.

Now, since the period of the sinusoid is so huge and we can track a source only to an extent of about  6 hours, I never get data that would have a full cycle of a sine wave in it. most of the cases I might have only 25 percent of it, at the maximum. In this situation where I don't even have a full cycle, what is the best way to estimate the value of the amplitude and phase of the sinusoid?

Write $y(t) = A\sin(\Omega t)\cos(\phi) + A\cos(\Omega t)\sin(\phi)$. Let $A_1 = A\cos(\phi)$, $A_2 = A\sin(\phi)$. Define the variables $w(t) = \sin(\Omega t)$ and $z(t) = \cos(\Omega t)$. Now you can write
$$y(t) = \begin{bmatrix}w(t) & z(t)\end{bmatrix}\begin{bmatrix}A_1 \\ A_2\end{bmatrix}$$

The rest is linear regression. To be more clear you can write your datapoints as
$$\begin{bmatrix}y_1 \\ \vdots \\ y_n\end{bmatrix} = \begin{bmatrix}w_1 & z_1\\ \vdots & \vdots\\ w_n & z_n \end{bmatrix}\begin{bmatrix}A_1 \\ A_2\end{bmatrix}$$
Using the matrix notation
$$Y = XB$$
You want to solve the following minimization problem
$$  \min_{B} \lVert Y - XB \rVert  = \min_{B} (Y-XB)^\top (Y-XB) $$

If you differentiate the cost function with respect to $B$ and set it to $0$, you will see that $$B = (X^\top X)^{-1}X^\top Y$$.

Having $B$ you can solve for $A$ and $\phi$.

Two samples with $90^\circ$ difference method for $y=A\sin(\omega t+\pi)$ and $T$ as the period

first sample : $y_1=A\sin(\omega t+\pi)$
second sample : $y_2=A\sin\left(\omega\left[t+\frac T4\right]+\pi\right)=A\cos(\omega t+\pi)$
so $A^2=y_1^2+y_2^2$

B.R
s.mosaviasl