Boundary conditions for 2D navier stokes equation (incompressible, stationary) I am asked to give the boundary conditions for the following duct flow

We are using a cartesian coordinate system. And $u$ and $v$ are the $x$ and $y$ velocities of the flow. $n_i=(n_{ix},n_{iy})$ is the unit normal vector of the inlet area, $n_0=(n_{ox},n_{oy})$ is the unit normal vector for the outlet area and $n_w$ is the unit normal for the walls.

I would guess that the following boundary conditions are necessary:

Boundary conditions on wall: $(u,v)=(0,0)$ at the wall (no-slip + no-through)
Inlet condition:$(u,v) \cdot (n_{ix},n_{iy})$ is given.
Outlet condition: $(u,v) \cdot (n_{ox},n_{oy})$ is given.

  Are my boundary conditions right? I am a little bit confused because
  our solution says $u\cdot n_{ix} +u\cdot n_{iy}$ (inlet),
  $v\cdot n_{ix} +v\cdot n_{iy}$ (inlet), $u\cdot n_{ox} + u\cdot n_{oy}$
  (outlet) and $v\cdot n_{ox} + v\cdot n_{oy}$ (outlet) need to be
  specified.

One thing we can say for sure is that the boundary conditions as proposed in the question as "our solution" are wrong; I've never seen such a strange application of the inner product anyway.
Another thing that can be said is that the boundary conditions as proposed by the OP are somehow better - that is: inlet and outlet conditions would rather be applicable for irrotational flow, which is a special case of Navier-Stokes - but in general they cannot be deemed correct as well.

With the 2-D Navier-Stokes equations for incompressible & stationary flow, there is an abundance of more or less proper boundary conditions in literature. Some of these are incredibly complicated, so I'd suggest to hunt for the simple ones. More or less by coincidence, I've stumbled upon a decent example for duct flow:

What Are the Navier-Stokes Equations?

The acccompanying picture illustrating the boundary conditions is resemblant to the OP's:

Then the article says: The fluid velocity is specified at the inlet and pressure prescribed at the outlet.
A no-slip boundary condition (i.e., the velocity is set to zero) is specified at the walls.

This means that, at the inlet area, the full velocity vector field must be specified: $(u_i,v_i) = (U,V)$. If there is no reason to assume otherwise, then an uniform velocity field may be imposed.

The pressure condition at the outlet area comes as a slight surprise for me, but I think it's more reasonable than trying to impose a velocity field. The reason is that you will get in trouble while trying to fulfull the global conservation laws for mass and momentum. An extreme example is the outlet velocity field $(u_o,v_o) = (0,0)$ that will suck all mass into nothingness; I can only hope that the CFD code will protest against this, but have the sad experience that too often it will not.If there is no reason to assume otherwise, then an uniform pressure field may be imposed.

Last but not least, the no-slip boundary conditions are commonly assumed to be correct.

If is is assumed in addition that the flow is irrotational, then the no-slip boundary conditions must be replaced by impermeability conditions, of the form $\;(u,v) \cdot (n_{wx},n_{wy}) = 0$ .The boundary conditions as proposed by the OP for the inlet area are entirely correct for such ideal / potential flow, as it is called : $(u,v) \cdot (n_{ix},n_{iy}) = $ given.
Boundary conditions for the outlet area are: velocities parallel to the normal $\,(u,v)\, //\, (n_{ox},n_{oy})$ . Then let the CFD code be good enough to assure that e.g. global mass conservation is guaranteed.

Mr. YouMath
That isnt QUITE what I am saying, but I am grateful that you pushing me to refine my answer. I realize now that the question as posed admits at least three interpretations, as I shall now list, but no interpretation is very satisfying.
That is the reason why I claimed that there was no good answer to the question, but to be convincing it needs some elaboration.

INTERPRETATION ONE
The question is put to a Mathematician (M); Is the problem defined by these equations (E) and these boundary conditions (B) well-posed? It is not, as Han seems to think, to give all B leading to well-posed problems for E, because there could be an unlimited number of them, but it is usual to give a list of common well-posed B (Dirichlet, Neumann, Robin) that we might describe as pre-approved.
I doubt that the question was meant to be so profound.

INTERPRETATION TWO
The question is put to a designer (D).
Among the set of pre-approved B, or perhaps the set of not-yet-approved B, what specific B gives the best representation of the particular question that D wishes to ask, assuming that D knows what that is? (By specific B I mean that, although M may have said "prescribe the normal velocity", there might be some doubt about the actual value to be prescribed. Where do we get it from?) This is subjective, and will almost always involve physical intuition. Best here to take a specific example, one that I know a little about. Suppose that the domain represents the inlet region
of a jet engine. The inflow represents external air drawn in from the outside air. Probably that flow is uniform "at infinity", and its passage into the inlet is essentially inviscid, so D will know its Bernouilli constant and that it is irrotational. That does not determine it completely, although almost surely the inflow will not be uniform. D could guess the velocity components consistent with the information they have, or D could couple their calculations with an external calculation (say of potential flow, as opposed to NS internally)

At outlet, the rest of the universe consists of a jet engine. Obviously D has already decided not to include the details of this. Experience has shown that the most important aspect of the engine is to impose the static pressure at outlet, although M would surely approve of other choices.
But then, what should the imposed static pressure be? The question does not allow for any discussion of these issues, so this cannot be the intended interpretation.

The specfic question here is likely to be what shape of the duct leading from inlet to engine provides the engine with the most uniform inflow? Because that was the condition assumed by the engine designer. There are bonus points for a short duct that will weigh less. Because the flow is subsonic, altering the duct shape will alter the flow at inlet although not at infinity. This is the sense in which you need to know the solution before you know the boundary conditions. You may know the SORT OF B that can be used, but not the actual values. Ways out; do a bigger calculation, iterate, divide the universe into different bits.

INTERPRETATION THREE. The question is put to the code-writer (W). At a boundary of any kind, the numerical procedure used to predict the flow elsewhere will fail, because information about the flow variables is not now available from all directions. We then speak of a boundary procedure. Just as not every method for the interior flow is accurate or stable, so not every boundary procedure is accurate or stable. It will depend on the boundary condition that we wish to impose, and on the interior method. It is the task of the algorithm designer (A) to investigate this, perhaps to publish the results. The code writer (W) must understand the conclusions and make choices that are consistent with them. This should be discussed in the users manual, perhaps to describe limitations or perhaps to point out various options. If it isn't, or if you don't understand the discussion, consult the help desk (H), but of course you do not ask them about interpretation one. A reputable software company will work with you to resolve any difficulties if your particular case is not already covered.


Does a limit exist at a cusp or sharp point Does a limit exist at a sharp curve or point? The limit as $x \to c+$ and $x \to c-$ won't equal at a cusp right? (or would it in some cases? So depending on that, does a limit always or never or sometimes exists at a cusp.

Thanks guys

Yes there exists a limit at a sharp point.
According to the definition of limit. Limit $L$ exists if $$\lim _{x\rightarrow n^+}f\left( x\right) =\lim _{x\rightarrow n^-}f\left( x\right) $$
The function is of course still continuous at the cusp so the limit exists and is evaluated as
$$\lim _{x\rightarrow n^+}f\left( x\right) = L =\lim _{x\rightarrow n^-}f\left( x\right) $$

There is a mathematical way to deal with sharp edges or cusps,
if $f(x)$ is the function with sharp edge at $x=c$ then shift the origin at $(c,0)$ and then obtain fourier series for $f(x)$,
Let $f(x)=a_0+a_1\cos{x}+a_2\cos{2x}...+b_1\sin{x}+b_2\sin{2x}...$
Then, $f'(0)=\sum_{1}^{\infty}nb_n$
If you can evaluate it you could obtain the derivative at the sharp edge.


Confidence interval estimate of population mean HELP Question:

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow.

Develop a 95% confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places.

Data: 6
4
6
8
7
7
6
3
3
8
10
4
8
7
8
7
5
9
5
8
4
3
8
5
5
4
4
4
8
4
5
6
2
5
9
9
8
4
8
9
9
5
9
7
8
3
10
8
9
6

My answer:

Sameple: 50
Mean: 6.34
SD: 2.1629

I got -5.73 and +6.95 so    (0.61,13.29)

but it keeps saying that its the wrong answer. Anyone out there have any ideas?

You are doing the arithmetic wrong. For a 95% confidence interval, you need to find the values which are on $\pm 1.96 \times \sigma$ of the mean. I don't know where your -5.73 and +6.95 values are coming from, the confidence interval will be symmetric about the mean on a normal distribution. Try it again and if you are still struggling maybe take a look at this page: http://onlinestatbook.com/2/estimation/mean.html


If $\log_27=x$ is $x$ irrational or rational? $x$ will definitely be between $2$ and $3$, but why does it have to irrational? Let’s assume that you don’t have the log table and no calculator, then how do you determine if it’s rational or irrational?

It’s probably an obvious answer like $x$ will always be irrational and all that, but I’d still like to know.

Thanks a lot.

Hint $\log_2 (7) =x \Leftrightarrow 2^x=7$. Now, if $x=\frac{m}{n}$ would be rational then
$$2^m=7^n$$

Another approach:

$\log _2 a=x$ is integer if and only if $a=2^n; n∈N$

But $7=2^3-1$, so x is not integer which means it is a fraction or a rational number with denominator greater than unity.


Find the number of generators of the cyclic group $\mathbb{Z}_{p^r}$ Let $p$ be a prime number. Find the number of generators of the cyclic group $\mathbb{Z}_{p^r}$, where $r \in \mathbb{Z} \geq 1$

I'm trying to understand the question and am experimenting with $p=5$ and $r=1,2,3$.

When $r=1$ it generates $\mathbb{Z_5}$, where every non-zero element is a generator of the group.

When $r=2$ it generates $\mathbb{Z_{10}}$. All the elements relatively prime to $10$ are $1,3,7,$ and $9$, also $4$ generators.

When $r=3$ it generates $\mathbb{Z_{15}}$. All of the elements relatively prime to $15$ are $1,2,4,7,8,11,13$, and $14$, which are $8$ generators.

So I'm trying to figure out how to find the number of relatively prime elements for the general group $\mathbb{Z}_{p^r}$

You have the right idea, but remember it's $p^r$, not $pr$.  So for instance, for $p=5$ and $r=2$, you get $\mathbb{Z}_{25}$, not $\mathbb{Z}_{10}$.

This also makes the question easier to answer: you just have to count how many integers between $1$ and $p^r$ are relatively prime to $p^r$.  An integer is relatively prime to $p^r$ iff it is not divisible by $p$ (why?).  To count such integers, you may find it easier to first count the integers between $1$ and $p^r$ that are divisible by $p$.  I'll let you finish from here.

I think the formula to calculate the answer of above question is $p^r-p^{r-1}.$
Please re-check it and if there is counter example then notify me.


Parametrization of $a^2+b^2+c^2=d^2+e^2+f^2$ Is there an existing parametrization of the equation above that is similar to Brahmagupta's identity for $a^2+b^2=c^2+d^2$? I need either a reference to look it up or a hint to solve it. Thanks.

There is a parameterization for every equal sums of squares equation
$$
X_1^2 + \dotsb + X_m^2 = Y_1^2 + \dotsb + Y_n^2
$$
with $n,m$ positive integers and all $X_i,Y_i$ integers. The papers by Barnett and Bradley are my first recommendations.

Above equation (a^2+b^2+c^2)=(d^2+e^2+f^2) has parametric solution.
Refer to Tito piezas on line book "collection of algebraic Identities".
Section sum of squares. The answer is given below;

(a,b,c)=[(p+q),(r+s),(t+u)]  and

(d,e,f)=[(p-q),(r-s),(t-u)]

Condition is (pq+rs+tu)=0

After parametrization of (pq+rs+tu)=0, the parametric solution in
one variable is given below:

(a,b,c)=[(4k^2+12k+1),(3k-8),(7k+28)]    and

(d,e,f)=[(4k^2+12k-13),(k+2),(13k+26)]


Euler's theorem (modular arithmetic) for non-coprime integers I am trying to calculate $10^{130} \bmod 48$ but I need to use Euler's theorem in the process.

I noticed that 48 and 10 are not coprime so I couldn't directly apply Euler's theorem. I tried breaking it down into $5^{130}2^{130} \bmod 48$ and I was sucessfully able to get rid of the 5 using Euler's theorem but now I'm stuck with $2^{130} \bmod 48$. $2^{130}$ is still a large number and unfortunately 2 and 48 are not coprime.

Could someone lead me in the right direction regarding how would I proceed from here?

In this case it pays to try the dumb strategy first: by calculating a few powers of 10, you quickly find that for all $n \geq4:$ $10^n = 16 \bmod 48$. Hence $10^{130} = 16 \bmod 48$.


Prove that if $f'(x)≠0$ on an interval $( a , b)$ on $A$ for a differentiable function $f$, then $f$ is one-to-one on $A$. Prove that if $f'(x)≠0$ on an interval $( a , b)$ on $A$ for a differentiable function $f$, then $f$ is one-to-one on $A$.

I was wondering whether the proof in the attached picture is correct. Could anyone please help me?

Not quite. If $f$ is not one-to-one on $(a,b)$ then there exist two distinct points $x,y \in (a,b)$ satisfying $f(x) = f(y)$. But there exists a point $z$ in between $x$ and $y$ satisfying $$0 = \frac{f(y) - f(x)}{y-x} = f'(z) \not= 0$$ which is the contradiction you need.

There is no reason for $f(a) = f(b)$ to hold.

Your pictured proof just shows that $f$ is injective.  You now have to show it is surjective as well.


Proving the set of irrational numbers in $\mathbb{R}$ has the same cardinality as $\mathbb{R}$ Not going to show my whole proof here, but this logic work?

We are showing |R-Q| = |R| => |I| = |R|
Since I is a subset of R, |I| must be less than or equal to |R|.
So all we have to show is |I| is not less than |R|.

So started by proving R is uncountable using diagonalization and a contradiction.
Then proved I is uncountable using a contradiction and the fact that I union Q = R.

Then concluded that since I is uncountable and R is uncountable, then|I| cannot be < |R| => |I| = |R|.

Does that make sense?

Just show that there are injective functions from one set to another and vice versa. For one, just take the inclusion. For the other send R\Q in (0,1)\Q and Q to (1,2)\Q for example. ( I'm using the fact the the open interval has a bijection with R and that injections preserve cardinality). Then the result follows by the Cantor-Bernstein-Schroeder theorem.

That $\Bbb R\setminus\Bbb Q$and $\Bbb R$ are both uncountable doesn't imply that they have the same cardinality.  There are infinitely many uncountable cardinals, by Cantor's diagonal argument.

Idea: $\Bbb R=(\Bbb R\setminus\Bbb Q)\cup\Bbb Q$.

But, as you noted, $\Bbb R$ is uncountable.

Presumably you know $\Bbb Q$ is countable.

Hence $\Bbb R\setminus\Bbb Q$ must be of the same cardinality as $\Bbb R$.


A Fibonacci conjecture: $\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}$ Given the Fibonacci sequence $F_n$, that is $$F_1=F_2=1,F_{n+2}=F_{n+1}+F_n.$$
I find that
$$
\log_21<\log_32<\log_53<\log_85<\cdots,
$$
i.e. we have $$
\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}, \quad \text{for} \; n\geqslant1\tag{1}
$$
If $n=2k$, $(1)$ is easy to get, but if $n=2k+1$, I have no idea to prove it.
What's more, if we let $F_1=a,F_2=b$ be two positive integers, then $(1)$ still exstis when $n$ is greater than some integer $m$.
I do not konw whether my conjecture is true.


Unbounded measurable set with different inner and outer measures I'm working on providing a counterexample to the claim that

  A unbounded set $A \subset \mathbb{R}$ is Lebesgue measurable if and only if its inner and
  outer measures are equal. Further, if $B$ is an unbounded measurable set that contains $A$,
  then $A$ is measurable if and only if it divides $B$ cleanly.

Let me clarify which definitions I'm using.

  Lebesgue outer measure is
  $$m^*A = \inf\left\{ \sum_k |I_k| : \{I_k\} \text{ is a covering of $A$ by open intervals}\right\}$$

  Lebesgue inner measure is
  $$m_*A = \sup\left\{ m^*C : C\text{ is closed and }C \subset A\right\}$$

  A set $E$ is Lebesgue measurable if the division $E|E^c$ of $\mathbb{R}$ is so "clean"
  that for each "test set" $X \subset \mathbb{R}$, we have
  $$m^*X = m^*(X \cap E)+ m^*(X \cap E^c)$$

So far, I have thought that the best strategy to disprove the claim is to find an example of an unbounded measurable set that has unequal inner and outer measure. Does this seem like the right direction? Are there any example sets I should study?

The counterexample is as follows. First, we define an unmeasurable set: $V \subset [0,1]$ (perhaps something like the Vitali set). Then, defining $R = (2,\infty)$, we can consider our counterexample function $C = V \sqcup R$. We know that $C$ is unmeasurable because if it were,  since $[0,1]$ is measurable, we know that $C \cap [0,1]$ would be measurable. This clearly isn't the case because $C \cap [0,1] =V$.

Now, we know that $m^*C = m_*C = \infty$, because $m^*R = \infty$ and both inner and outer measure are monotonic. Therefore, we have an unbounded, unmeasurable set with equal inner and outer measure, contradicting the claim.

Theorem:
We know that $m_*(E) \leq m^*(E)$, and that equality holds if and only if $E$ is Lebesgue measurable.

Now any non-measurable set will  serve our purpose.


Demonstration congruences Assuming that $m=p_1^{\alpha_1}...p_r^{\alpha_r}$. Show that $$a\equiv b\pmod m\Longleftrightarrow a\equiv b\pmod {p_i^{\alpha_i}},\;i={1,...,r}$$

I always thought very beautiful statements that contain numbers in this way $$x=p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}p_4^{\alpha_4}...p_w^{\alpha_w}$$and to be here studying congruences, came across eats this issue, which unfortunately do not even know where to start or what to do ... While statements like these, I can not understand them very easily so I ask YOU DO PLEASE DETAILED ...

I thank you ..

Lemma 1: if $c$ and $d$ are coprime and both divide $x$, then their product divides $x$.

Proof: on the hypotheses, $x=cr=ds$ for some integers $r,s$, so $c$ divides $ds$. That, with $c$ being coprime to $d$, implies (by a standard result) that $c$ divides $s$, so $s=ct$ for some integer $t$, so $x=cdt$, so $cd$ divides $x$.

Lemma 2: if $c_1,c_2,\dots,c_r$ are pairwise coprime, and all divide $x$, then their product divides $x$.

Proof: by induction on $r$, with Lemma 1 providing the base of the induction.

Theorem: if $p_1,p_2,\dots,p_r$ are distinct primes, and $m=p_1^{u_1}p_2^{u_2}\times\cdots\times p_r^{u_r}$, and $a\equiv b\pmod{p_i^{u_i}}$ for all $i$, then $a\equiv b\pmod m$.

Proof: the numbers $p_i^{u_i}$ are pairwise coprime, so Lemma 2 applies.

Hint: Show that if $(n_1,n_2)=1$, then $n_1\cdot n_2|(a-b)$ if and only if $n_1|(a-b)$ and $n_2|(a-b)$.


Probability of an event with 2 scenarios How to determine probability if we have 2 scenarios.
For example
If 70% of men have black eyes &
60% of Indians have black eyes
What will be the chances of an Indian Man having black eyes?

The given information is insufficient. I will give two examples to show this.

Case I - There are 300 Indians, 120 of them are men who do not have black eyes, and rest of the Indians are not men, but have black eyes. This satisfies the condition that 60% Indians have black eyes. There are 280 other men who are not Indians and have black eyes. Now, 70% of men have black eyes. In this case, no Indian man has black eyes.

Case II - There are 100 Indians, 60 of them men and having black eyes. Remaining 40 Indians are not men and do not have black eyes. Thus, 60% Indians have black eyes. There are 40 other men who are not Indians out of which 10 have black eyes, the remaining 30 do not have black eyes. Thus, 70% of men have black eyes. In this case, all Indian men have black eyes.

So the answer can be anywhere between and including 0% and 100%.


Characters on symmetric groups $S_n$ and alternating groups $A_n$. For an integer $2 \leq n$ ,
let $S_n$ be the permutation on $n$ symbols and $A_n$ be the alternating group.  Also let's denote by $\mathbb{C}^*$, the multiplicative group of non-zero complex number.
Which of the following is correct?

for every integer $2 \leq n$,
there is a non-trivial homomorphism
$φ:S_n \rightarrow \mathbb{C}^*$.
for every integer $2 \leq n$,
there is a unique non-trivial homomorphism
$φ:S_n \rightarrow \mathbb{C}^*$.
for every integer $3 \leq n$,
there is a non-trivial homomorphism
$φ:A_n \rightarrow \mathbb{C}^*$.
for every integer $5 \leq n$,
there is no non-trivial homomorphism
$φ:A_n \rightarrow \mathbb{C}^*$.

my attempt: for $5 \leq n$ $A_n$ is simple, so option $4$ is correct.

Note that every finite subgroup of $C^\times$ is cyclic.

Option 1 is  correct: Take the sign homomorphism (i.e., $\phi(g)=1$ for $g\in A_n$ and $\phi(g)=-1$ otherwise.

Option 2 is also correct as the sign homomorphism is the only non-trivial homomorphism: For $n\ge 5$, this follows because the kernel must contain the simple subgroup $A_n$. For $n\le 4$, by inspection.

Option 3 is correct because it is weaker than option 1.

Option 4 is wrong because it contradicts option 1.


Minimum value of $p^2+q$ Let $p$ and $q$ be real numbers such that $p^2+q>q^2+p$. Then the minimum value of $p^2+q$ is what?

My argument is as follows:

Since $p,q$ are real numbers, it should not matter if we take $p$ as $x$ and $q$ as $y$.
\begin{align*} p^2+q &> q^2+p \\ x^2-y^2 &> x-y \\ (x-y)(x+y-1)&>0\end{align*}
So, we have to maximize $x^2+y$ subject to the conditions that:
$$x-y>0,\quad x+y-1>0$$ or $$x-y<0,\quad x+y-1<0$$
From the graph of thse, it's clear that the minimum value occurs at $(1/2,1/2)$ in region $1$ (labeled on the graph). So, $p^2+q = (1/2)^2+(1/2) = 3/4$.

However this is not the correct answer.
Can anyone suggest where I'm making a mistake?

I'll solve the following problem.

Let $p^2+q\geq q^2+p$. Find a minimal value of $p^2+q$.

Indeed, $p^2+q+\frac{1}{4}=\frac{1}{2}\left(\left(p+\frac{1}{2}\right)^2+\left(q+\frac{1}{2}\right)^2\right)+\frac{1}{2}(p^2+q-q^2-p)\geq0$.

The equality occurs for $p=q=-\frac{1}{2}$.

Id est, the answer is $-\frac{1}{4}$.

It appears that you made an error early on in your proof. You got (x-y)(x+y-1)>0 when you should've gotten x+y-1>0, as the x-y terms cancel each other out.


Continuous bijection from $(0,1)$ to $[0,1]$ Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.

No. If $f:(0,1) \to [0,1]$ were continuous and bijective, there would be a unique point $x \in (0,1)$ such that $f(x) = 1$. However, since $f$ is continuous, the intervals $[x - \varepsilon, x]$ and $[x, x + \varepsilon]$ would be mapped to intervals $[a,1]$ and $[b,1]$, say. By bijectivity we'd have $a, b \lt 1$. Thus every value strictly between $\max{\{a,b\}}$ and $1$ would be assumed at least twice, contradicting bijectivity.

There does not exist  a continuous bijection from (0,1) to [0,1]. Indeed, let $f$ be such a function. Let consider a sequence  $x_n=1-1/n$. Then from the sequence $(f(x_n))$ we can choose a subsequence $(f(x_{n_k}))$  which is convergent. Let denote this limit by $y$. Obviously, $y \in [0,1]$. Since $f^{-1}$ also is continuous, we get $f^{-1}(y)=\lim_{k \to +\infty}f^{-1}(f(x_{n_k}))=\lim_{k \to \infty}x_{n_k}=1$. But $1 \notin (0,1)$.

Remark(Why $f^{-1}$ must be continuous under our assumption?)  By our assumption $f:(0,1)\to [0,1]$ is continuous bijection. Then  $f:(0,1)\to [0,1]$ must be injective and continuous which following invariance of domain (see, http://en.wikipedia.org/wiki/Invariance_of_domain   is  homeomorphism. Hence $f^{-1}: [0,1]\to (0,1)$ is continuous.


Is there an analogue of the fundamental theorem of calculus for probability distributions without a density? Assume that $\mu$ is a probability measure on $\mathbb{R}$, and that we have the values of the integral $$A(s) = \int_0^s x \,d\mu(x) + \int_s^\infty d\mu(x).$$ If $\mu(x) = f(x) \, dx$, we can retrieve $f$, and hence $\mu$ by taking the derivative

$$A'(x) = xf(x) - f(x).$$

If $f$ is differentiable, the second derivative gives $A''(x) = f(x)$ and we now have $\mu$.

What if we don't know that it has a density function, is their any means of effectively retrieving the probability measure?

(Assuming the measure is non-atomic) for $h > 0$ we have
$$ (s-1) \mu([s,s+h)) \le A(s+h) - A(s) = \int_s^{s+h} (x-1)\; d\mu(x) \le (s+h-1) \mu([s,s+h))$$

and thus if $s > 1$,
$$  \frac{A(s+h)-A(s)}{s+h-1} \le  \mu([s,s+h)) \le \frac{A(s+h)-A(s)}{s-1}$$
If $s+h < 1$, the inequalities are reversed.
By taking sufficiently fine partitions, we can obtain arbitrarily precise bounds on the measure of an interval.


Jordan normal form and spectral decomposition In this post by Terence Tao (exercise 28, point vi), he proves the following theorem (called spectral decomposition):

Theorem:
Let $A$ be a complex square matrix. Then $A$ can be written as

$A=\sum P_j(\lambda_j I+N_j)P_j$

Where $P_j$ are matrices such that $P_j P_l=0$, $P_j^2=P_j$ and $P_1+\dots +P_k=I$
and $N_j$ is a nilpoten matrix of order $\text{mult}(\lambda_j)$ ($\lambda_j$ is the $j$-th eigenvalue)

This theorem seems really connected with the Jordan form of a matrix, although I am not able to prove that: It is clear that the J.F. implies the spectral decomposition, while the opposite is not so easy: I tried to prove it on a canonical basis in order to generalize it after but I didn't get anywhere.

Question:

Is the J.F. actually a consequence of the spectral decomposition? If so, how to prove it?

This almost implies the Jordan canonical form. What is missing is the normal form of nilpotent matrices. That is, $N_j$ is similar to a block diagonal matrix with blocks of the form
$$
\begin{bmatrix}0&1&0&\ldots &0\\0&0&1&\ldots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\ldots &1\\0&0&0&\ldots &0\end{bmatrix}.
$$
This change of basis gives a Jordan basis on each generalized eigenspace, which are the images of the $P_j$.


Differential Equation. Here is the entire problem below:

According to a simple physiological model, an athletic adult male needs 20 calories per day per pound of body weight to maintain his weight. If he consumes more or fewer calories than those required to maintain his weight, his weight changes at a rate proportional to the difference between the number of calories consumed and the number needed to maintain his current weight; the constant of proportionality is 1/3500 pounds per calorie. Suppose that a particular person has a constant caloric intake of H calories per day. Let W(t) be the person's weight in pounds at time t (measured in days).

(a) What differential equation has solution W(t)?

So I already solved for this and got that part correct. My answer was: $$\frac{dW}{dt} = \frac{1}{3500}(H-20W)$$
Below is the part that I am unsure of what to do:

(b) If the person starts out weighing 165 pounds and consumes 3500 calories a day,$$\lim_{t\to \infty} W(t) = \text{?}$$

I at first thought to plug in the H(3500) and W(165) values that were given in the part b question into the answer I got for part A. But then I wasn't sure what to do with that answer, or if that is even the way to go. It's probably a simple solution, but I'm not seeing where to go from here. Any help would be appreciated. Thank you.

When the weight stabilizes (at infinity), the derivative cancels out and

$$H-20W_\infty=0.$$

We're told,

the guy needs 20 calories per day per pound of body weight to maintain his weight, and
he consumes 3500 clories a day.

It follows that if he weighs $3500\div20=175$ pounds, then he will stay at 175 pounds. If he weighs any less than 175 pounds (and consumes 3500 calories per day), his weight will increase; if he weighs any more than 175 pounds, his weight will decrease. It follows that, no matter what he weighs, his weight will, in the long run, approach 175 pounds.

Strictly speaking, there's a bit of handwaving going on here, and one ought to solve the differential equation to get a function $W(t)$, and then evaluate $\lim_{t\to\infty}W(t)$ – but at the very least the handwaving argument tells you what answer to expect.

As for how to solve the differential equation, if your teacher really expects you to solve the differential equation, and doesn't supply you with the tools for doing so, you have a case of academic malpractice, and should go to the head of the department to inform her. But first make absolutely sure that you have not, in fact, been given the tools to solve the equation!


How is this case a reflexive relation? My textbook says, if $R$ is a reflexive relation on set $A=\{1,2,3,4\}$ then $R = \{ (x,x) : \forall x \in A\}$ then it proceeds on with $2$ examples.

$$R_1= \{ (1,1),(2,2),(3,3),(4,4) \}$$ Alrighty, this is fine and makes sense. Every $x\in A$ there is an $(x,x)$ type of elements in $R_1$

Now the trouble starts with this example,

$$R_2= \{(1,1),(2,2),(3,3),(4,4),(1,2) \} $$ is also a reflexive relation?

My question :

  In $R_2$ there exists an element $(1,2)$ which is not of the type $(x,x)$ but we clearly said earlier the Relation will contain all $(x,x)$ type elements. Then why is this a reflexive relation?

The relation can contain other elements too. It must contain all the elements of the type $(x,x)$ in order to be reflexive but it doesn't mean that these are the only elements in the relation.

Your $R_1$ is an example of an identity relation. Which is indeed also a reflexive one. But not every reflexive relation is identity.

Reflexive relation has its elements of the $\{x, x\}$ type where $x$ belongs to $A$. When all identity types are contained in the set then the reflexive relation can contain more than those too.
Therefore:
$$R=\{[(x, x) :x \text{ belongs to $A$ }] \text{ or } [\{x, y\} :x \text{ belongs to $A$, $y$ belongs to $A$}]\} $$
This can be the complete definition for reflexive set.


E/F finite extension F'/F any extension. EF' composite field. Prove [EF':F']<=[E:F] As the question says (we assume E and F' are contained in some field so that the composite EF' is well-defined). I've thought of trying to prove this using indiction by writing E=F(a1,...ar) but unfortunately I haven't really been able to get onto the right track and cant seem to quite get to the answer. Do you know of any hints/ways to solve this problem? Thank you for any comments or answers.

EDIT: I have now managed to solve the base case E=F(a).

lets assum that F'=F(x), then EF=EF(x)=E(x), the minimal polynomial of x over F (say f) is a polynomial in E[X], so the minimal polynomial of x over E must devise f, so [EF':E]<=[F':F].


For any closed subset of $\mathbb{R}$ there is a sequence in $\mathbb{R}$ whose sequential limits is equal to the that subset Question: Let $A$ be a closed subset in $\mathbb{R}$. Prove that there exists a sequence $x_n$ in $\mathbb{R}$ whose set of subsequential limits is exactly equal to $A$.

My approach:
I think this sort of makes sense me if we our space was $\mathbb{C}$, however, I'm not even sure if this statement is true or not. I was thinking if it's possible to prove it by starting to say that the set of all subsequential limits of a sequence in a metric space is closed. However, I'm not sure if for any closed subset in $\mathbb{R}$, we can construct such sequence.

I'm going to assume that you know basic things about compactness. And give you two steps to begin with.

If $A$ is a closed and bounded set, then it is compact. Prove, even in general, that a compact metric space is always the closure of a countable set (or in other words, a compact metric space is always countable). You can do that by finding a particular sequence of open covers, refine each to a finite subcover and use that to define the countable set.
If $A$ is closed and unbounded, then it can be written as the countable union of closed and bounded sets. Namely, if $A$ is closed, then it is the union of countably many compact sets. What do you know about the countable unions of countable sets? Now apply the previous case.

Or, prove directly,

If $A\subseteq\Bbb R$ is covered by open sets, then there is a countable subcover. Use similar tricks to the previous two steps and find a countable dense subset. (This property of $\Bbb R$ is called hereditarily Lindelöf.)

Hint: $A$ contains a countable dense subset.


Determine if the function is surjective Let $\\f: \ \mathbb{R} \to \mathbb{R} $ via $ f(x) = \frac {x}{1+x^2}$.

I know this function is not injective, but im not sure how to determine if the function is surjective or not

In order to show that $f$ is surjective, you must show that for any $y\in\Bbb R,$ there is some $x\in\Bbb R$ such that $f(x)=y.$ That is, we must be able to solve $$\frac{x}{1+x^2}=y$$ for $x$ and find that $x\in\Bbb R.$ The quadratic formula will come in handy in your attempts to do so.

A function is surjective if every element has a pre-image. I will show that this function is not surjective by showing $2$ does not have a pre-image.

Indeed, if $2$ has a pre-image, say $x$ then $$\frac{x}{1+x^2}=2$$
Then $x=2+2x^2$. That is $$2x^2-x+2=0$$
As $1-4(2)(2)<0$. Therefore, this equation has no real roots.
Hence, $f$ cannot be surjective.


Show that $\sum_{n=1}^\infty x_n$ diverges for a sequence defined by $x_{n}=h(x_{n+1})$ Let  $h:[-1,1]\to \mathbb{R}$ be a $C^2$ function such that $h(0)=0$, $h^{\prime}(0)=1$ and $h^{\prime}(x)>1$ for all $x>0$. Take $a_{1} \in (0,1)$. Define the sequence $x_{n}$ by $x_{1} = a_{1}$ and $x_{n}=h(x_{n+1})$ . Prove $\sum_{n=1}^\infty x_n=\infty$.
I tried to do this question but I'm having problems whit the divergence of this series. I know that the limit of the sequence is zero, and I have an idea of how the sequence works geometrically.
An idea of how can I prove the divergence would be appreciated.

The sequence $(x_n)$ is strictly decreasing and positive. In order to show that $\sum_{n=1}^\infty x_n$ diverges we need some lower bound for the $x_n$.
Let $K = \max \{ |f''(x)| : -1 \le x \le 1 \}$. From Taylor's theorem we have
$$
 x_n = h(x_{n+1}) = h(0) + h'(0) x_{n+1}  + \frac{h''(c)}{2} x_{n+1}^2
$$
for some $c \in [-1, 1]$, and therefore
$$
 x_n \le x_{n+1} + \frac K2 x_{n+1}^2  = x_{n+1} (1+\frac K2 x_{n+1})\, .
$$
It follows that
$$
\frac{1}{x_{n+1}} - \frac{1}{x_n} \le \frac{1}{x_{n+1}} - \frac{1}{x_{n+1}(1+ K x_{n+1}/2)} = \frac{K}{2+K x_{n+1}} < \frac K2
$$
for all $n$. Adding these inequalities gives
$$
\frac{1}{x_{n}} - \frac{1}{x_1} \le (n-1) \frac K2
$$
or
$$
 x_n \ge \frac{1}{1/x_1 + (n-1) K/2} \, .
$$
By comparison with the harmonic series it follows that $\sum_{n=1}^\infty x_n = \infty$.

the sequence is monotonically increasing:
$\frac{\mathrm{h}(\mathrm{x})}{\mathrm{x}}=\frac{\mathrm{h}(\mathrm{x})-\mathrm{h}(\mathrm{0})}{\mathrm{x}-0}=\mathrm{h}^{\prime}(\mathrm{y})>1$
where $\mathrm{y} \in(0, \mathrm{x})$.
So $h(x)>x$ for all $x>0$,
therefore $x_{n+1}=h\left(x_{n}\right)>x_{n}$.
Consequently $x_{n} \nrightarrow 0$ and $\sum_{1}^{\infty} x_{n}=\infty$


How to find limit of this function: $\lim_{n\to\infty} 0.99\ldots99^{10^n}$? How to find this limit:

$$ \lim_{n\to\infty}  0.99\ldots99^{10^n},$$

$$ \text{where number of 9 is } n. $$

My solution:
$$ \text{if } n\to\infty \text{ then } 0.99\ldots99 \to\ 0.(9) $$
$$ \text{because } 0.(9) = 1 \implies \lim_{n\to\infty}  0.99\ldots99^{10^n}=1^{10^\infty} = 1 $$

But my solution is wrong. Why? How can I correct it?

The limit you're trying to find is

$$ \lim_{n\to\infty} (1-10^{-n})^{10^n} $$

Substitute $u \mapsto 10^n$ and you get

$$ \lim_{u\to\infty} \left(1-\frac 1 u \right)^u = \frac 1 e $$

Your answer is wrong because 1 to the power infinity is not equal to 1 but it has unspecified value just like 0/0


Prove that odd perfect square is congruent to $1$ modulo $8$ How can we prove that every odd perfect square is congruent to $1$ modulo $8$?

An odd perfect square is of the form $(2k+1)^2$.
$$(2k+1)^2=4k^2+4k+1=4(k^2+k)+1$$
Since $k^2+k=k(k+1)$ is always even, $4(k^2+k)$ is always divisible by $8$. Now it follows that every odd square is congruent to $1$ modulo $8$.

In general, for any integers $m,n$ such that $m$ is odd and $n>2$, $m^{2^{n-2}}$ is congruent to $1$ modulo $2^n$. This follows directly from the fact that $U(\mathbb{Z} /2^n\mathbb{Z})\cong \mathbb{Z}_2\oplus \mathbb{Z}_{2^{n-2}} $.

Consider the group
$$(\mathbb{Z}/8\mathbb{Z})^\times = \{[1], [3], [5], [7]\} $$
under multiplication mod 8. Then
$$[1]^2 =[3]^2 = [5]^2 = [7]^2 \equiv 1 \, \,\mathrm{ mod } \, 8.$$
Since every odd integer is in exactly one of these congruence classes, every odd integer squared is 1 mod 8.


Induction Question. Suppose there are n teams in a football league how can i prove this

Let n > 1 be an integer. Suppose there are n teams in a football league and every two teams
have played against each other exactly once with no ties. Prove that it is possible to number
the teams 1 to n such that at the end of the season team i beats team i + 1 for i = 1,...,n − 1.

I proved this by for example giving n 3 or 4, but what exactly this question want ?

Suppose the first $n-1$ have been arranged as asked.  Try to show that it is possible either
1) to slot the nth team between two teams, or
2) that the nth team beat the first of the $n-1$, or
3) that the nth team lost to the last of the $n-1$.

Define the $n$ teams as $T_i, 1\le i\le n$, and the win/lose relation as $\gt$ where $T_i$ beats $T_j$ iff $i\lt j$, so for example $T_2$ beats $T_3$. This means $T_1$ wins $n-1$ games, $T_2$ wins $n-2$ games and so on, with $T_n$ winning zero games. As each team has won a different number of games, your question has been answered.


An application of Fubini’s theorem on Fourier transform Given $f,g\in L^1(\mathbb{R}^n)$ and we denote the Fourier transform of $f$ by $\widehat{f}$.

I want to prove that
$$\int_{\mathbb{R}^n}\widehat{f}(x)g(x)~dx= \int_{\mathbb{R}^n}f(x)\widehat{g}(x)~dx.$$

Here’s my attempt:
\begin{align}
\int_{\mathbb{R}^n}\widehat{f}(x)g(x)~dx=&\int _{\mathbb{R}^n}\left\{\int _{\mathbb{R}^n} f(t)e^{-2\pi it\cdot x}~dt\right\}g(x)dx\\
=&\int _{\mathbb{R}^n}\left\{\int _{\mathbb{R}^n} g(x)f(t)e^{-2\pi it\cdot x}~dt\right\}dx\\
{\color{red}{=}}&\int _{\mathbb{R}^n}\left\{\int_{\mathbb{R}^n} g(x)f(t)e^{-2\pi it\cdot x}dx\right\}~dt\\
=&\int_{\mathbb{R}^n}\left\{\int_{\mathbb{R}^n} g(x)e^{-2\pi it\cdot x}dx\right\}f(t)~dt\\
=&\int_{\mathbb{R}^n}f(t)\widehat{g}(t)~dt\\
=&\int_{\mathbb{R}^n}f(x)\widehat{g}(x)dx.
\end{align}

In the third equation I used Fubini’s theorem. But here’s something that I’m not sure: if we want to apply Fubini’s theorem, then $F(x,t):= g(x)f(t)e^{-2\pi it\cdot x}$ must be $\mathbb{R}^n\times\mathbb{R}^n$ integrable. But I was stuck when trying to prove that $F(x,t)$ is integrable. Could you give me some help? Thanks!

$$\int_{\mathbb{R}^n \times \mathbb{R}^n}|F(x,t)| \lambda(dx,dt)$$
$$=\int_{\mathbb{R}^n \times \mathbb{R}^n}|g(x) f(t)| \lambda(dx,dt)$$
$$= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} |g(x)||f(t)| \lambda(dx) \lambda(dt)$$
$$= \Vert g \Vert_1 \Vert f \Vert_1 < \infty$$

by Fubini for positive functions.


Why does $x$ divided by zero not equal $x$? Why does $x$ divided by zero not equal $x$? After all, $x$ is not being divided by anything.

Think of division as an "inverse" operation to multiplication.

For instance, let us look at $\frac{a}{b}$ means.

Let $x = \frac{a}{b}$. What it means is when $x$ is multiplied by $b$ it gives $a$.

So, if you want to make sense of $\frac{a}{0}$.

Say $\frac{a}{0}$ is $y$.

So, we want $y$ such that $y \times 0$ gives us $a$.

But we know that $y \times 0 = 0$, $\forall y$.

So there is a logical inconsistency here.

Who says X divided by zero does not equal X?  Let's challenge the conventional thinking (wording) and agree that, yes, if you divide something by nothing, there is no division.  Likewise, if you multiply something by nothing, there is no multiplication.  Don't treat zero as though it were a number, as though it were a part of the continuity of infinitely shrinking numbers less than one.  It's not.  It's a limit those numbers will never reach.  Zero is not another point on a line.  It's a gap.  It comes from the Sanskrit word meaning "empty."

Divide that apple on the table by ... nothing at all, and you've performed no operation.  You're left with the apple, whole and undivided.  Seems perfectly logical. You've done nothing to the apple.  You've done something to the operation, which was to abandon it.  So let's go with that and invite Yuan or another contributor to explain more convincingly why zero isn't the same as nothing.

That's going to be a hard sell, don't you think?  Of course, I like to argue negative numbers don't exist because you can't have less than nothing; you can't take two apples away from one.  But that's another topic.


Why are proofs not written as collections of logic symbols but are instead written in sentences? Mathematical proofs are written as sentences and not as collections of logic symbols.
Through logical operations, it is much easier for me to visualize what the symbols are trying to tell us rather than English text filled with grammar. This is my personal opinion, others may have different opinions.
I just asked this question on another website to find out logical mistakes in my work which is entirely done in the language of propositional logic.
Some people suggested to write it down in sentences in English. Is there any kind of tragedy in writing proofs as collections of logic symbols?

You have not translated the pages from Apostol's book into mathematical logic.
What you have done is to transcribe them into your own idiosyncratic shorthand, which may be useful to you but is less than meaningless to anyone else.
Let's start with the use of the symbol $\stackrel{\mathrm{def}}=.$
In normal mathematics, this tells us that the notation on the left is defined to represent the expression on the right in a general way.
For example, when we write
$$ \cosh x \stackrel{\mathrm{def}}= \frac{e^x + e^{-x}}{2}, \tag1$$
it is a definition of the $\cosh$ function.
In a definition of this sort, a symbol such as $x$ is a variable that can be substituted, so Definition $(1)$, above, tells us not only how to interpret
$\cosh x$; it also says how to interpret
$\cosh y,$ $\cosh t,$ $\cosh a,$ or $\cosh b.$
For example, Definition $(1)$ informs us that
$$ \cosh b = \frac{e^b + e^{-b}}{2}.$$
In your notes, you start with the definition
$$ [a, b] \stackrel{\mathrm{def}}= \text{closed interval in $x$-axis}. $$
Now, setting aside the fact that there are four English words on the right-hand side of that definition (what were you saying about using symbols rather than English text?), you have just defined a bracket notation for us,
"[" followed by a variable followed by "." followed by another variable
followed by "]" and you have informed us that this is a closed interval on the $x$-axis. Now it seems strange that your variable names do not occur on the right-hand side of this definition, and in fact this does make the definition relatively useless in strict logic: which closed interval is denoted by $[a,b]$?
But worse still, on the next line we find out that changing the variable names changes the definition to a closed interval on the $y$-axis, not the $x$-axis.
If you actually succeeded in translating the pages to pure logic, along the way you would realize that the labels "$x$-axis" and "$y$-axis" are hints to help you visualize things, not part of the strict logic of the mathematics itself.
You really need only define the closed-interval notation once.
I would say that some of your uses of $\stackrel{\mathrm{def}}=$ are actually logical definitions of symbols and notation. But many are not.
If you have a good definition of the product of two sets, it is not necessary to write out your interpretation of $P_x \times P_y$ as a "definition."
It would already be defined and (logically) unnecessary to write.
By the, way, symbols such as "$\ldots$" do not belong to the notation of mathematical logic; they are (again) merely hints to understanding.
You also seem to tend to use "$=$" to signify "is a" rather than the standard symmetric, transitive, and reflexive notion of equality.
For example:
$$ Ƃ:Q \to \mathbb R = \mathrm{SF} $$
would mean the same thing as
$$ \mathrm{SF} = Ƃ:Q \to \mathbb R $$
if you were writing in the language of mathematical logic;
and the meaning of the line in which it appears would still be ambiguous.
(Is SF a mathematical constant like $\pi$?)
If you actually were writing in mathematical logic you might have defined SF as a predicate, written in the form
$$ \mathrm{SF}(Ƃ:Q \to \mathbb R). $$
Later on that same line, however, you write $Ƃ:Q_{ij} \to \mathbb R,$
contradicting what you wrote earlier.
The domain of $Ƃ$ could be either $Q$ or $Q_{ij},$ but it cannot be both in the same definition. It seems you want to say that the restriction of $Ƃ$ to $Q_{ij}$ is a constant function, but you have neither the logical notation to describe a restriction of a function to a subdomain nor to say that a function is constant.
You end up defining $©_{ij}$ as a synonym for $Ƃ:Q_{ij} \to \mathbb R$ but not saying anything about new what the function does.
Frankly, without using Apostol's text as a Rosetta Stone for your work,  I think it would be very difficult for anyone else to guess what you mean by all your notations.
I see nothing wrong with making your own notes on a passage of text and equations in which you break everything out in a tabular format with displayed equations and no paragraphs of text.
Just don't expect anyone else to read it. It is for your own use in organizing your thoughts, and that is all.
If you really want to write things like this in mathematical logic, there are various computer-aided proof systems in which you can write your definitions and theorems in completely symbolic language and feed them into the software, which will check them for you.
But I don't know if you would actually find this easier to work with than the text in a book like Apostol's.

In the beginning, there where no mathematical symbols just words!
So young mathematicians love to emulate the older mathematicians.
This why we still love to not use much symbols.


A challenging geometry proof? Given: $C$ on $\overline{AB}$ such that $BC=3AC$ and $m\angle B=2m\angle XCB$.
To show: $AX=2AC+BX$

I have verified this result with trigonometry and analytic geometry and double-checked my work with GeoGebra.  But it seems like such an elegant result that there should be a purely geometric proof.  Any ideas?

I was inspired to investigate this diagram trying to solve a different problem here on MSE.  As far as trying to come up with a proof on my own I tried constructing $M,N$ on $\overline{AX}$ such that $AM=AC$ and $NX=BX$ and drawing some isosceles triangles. That might be fruitful (since you'd only have to show $AM=MN$), but nothing leapt out at me quickly.

OP's own answer shows that the key to the solution is to recognize that point $X$ lies on a hyperbola with foci $A$ and $C$ passing through $B$. Here's a "geometric" derivation of that fact.

Let the trisecting points of $\overline{BC}$ be $S$ and $T$. Let the angle bisector at $C$ meet $\overline{BX}$ at $D$, creating isosceles $\triangle BCD$. Let $\overleftrightarrow{DM}$ (with $M$ the midpoint of $\overline{BC}$ be the extended altitude of this triangle, and let $P$ be the projection of $X$ onto this line.

Then we have
$$\left.\begin{align}
\text{Angle Bis. Thm} &\implies \frac{|CX|}{|DX|}=\frac{|BC|}{|BD|} = \frac{2|BM|}{|BD|} \\[4pt]
\triangle DXP\sim\triangle DBM &\implies \frac{|DX|}{|PX|}=\frac{|BD|}{|BM|}
\end{align}\right\}\implies
\frac{|CX|}{|PX|}=\frac{|CX|}{|DX|}\cdot\frac{|DX|}{|PX|}=2$$
Therefore, $\overleftrightarrow{DM}$ is the directrix, and $C$ the focus, of a hyperbola through $X$ with eccentricity $2$.

Since trisection point $T$ divides $\overline{MC}$ in the ratio $1:2$, it must be a vertex of the hyperbola. Moreover, since $|ST|:|SC|=1:2$, it follows that $S$ is the center of the hyperbola. By symmetry across that center, $B$ and $A$ are the other vertex and focus, respectively, and the result follows. $\square$

Unfortunately, when I tried to edit DinosaurEgg's proof, the mods rejected it for "deviating from the original intent of the post".  With all due respect, I disagree with them, but it was a pretty dramatic revision.  So I'll post it here so that people can appreciate the elegance.  DinosaurEgg, if you see this, feel free to edit this into your answer if you like, and I'll delete this answer.

Construct $P$ on $\overline{AB}$ such that $\angle XPB\cong\angle B$. $\triangle XPB$ is obviously isosceles.  Since $\angle XCP=\theta$ and $\angle XPB=2\theta$, $\triangle XCP$ is also isosceles by the External Angle Theorem.  Therefore, $XB=XP=PC$. Dropping the perpendicular from $X$, it should be clear that $\cos B=\frac{PB/2}{XB}$, or $$PB=2XB\cos B$$ Since we know that $BC=3AC$, it follows that $AB=4AC$ and $PB=BC-PC=3AC-XB$.

Combining all of that with the Law of Cosines applied to $\triangle ABX$ yields

$$\begin{eqnarray*}
AX & ^{2}= & XB^{2}+AB^{2}-2AB\cdot XB\cos B\\
 & = & XB^{2}+AB^{2}-AB\cdot PB\\
 & = & XB^{2}+(4AC)^{2}-4AC(3AC-XB)\\
 & = & XB^{2}+16AC^{2}-12AC^{2}+4XB\cdot AC\\
 & = & XB^{2}+4XB\cdot AC+4AC^{2}\\
 & = & (XB+2AC)^{2}
\end{eqnarray*}$$


Compute value of $\pi$ up to 8 digits I am quite lost on how approximate the value of $\pi$ up to 8 digits with a confidence of 99% using Monte Carlo. I think this requires a large number of trials but how can I know how many trials?

I know that a 99% confidence interval is 3 standard deviations away from the mean in a normal distribution. From the central limit theorem the standard deviation of the sample mean (or standard error) is proportional to the standard deviation of the population $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}$

So I have something that relates the size of the sample (i.e. number of trials) with the standard deviation, but then I don't know how to proceed from here. How does the "8 digit precision" comes into play?

UPDATE

Ok I think I am close to understand it. From CLT we have $\displaystyle \sigma_{M} = \frac{\sigma}{\sqrt{N}}$ so in this case $\sigma = \sqrt{p(1-p)}$ therfore $\displaystyle \sigma_{M} = \frac{\sqrt{p(1-p)}}{\sqrt{N}}$

Then from the Bernoulli distribution, $\displaystyle \mu = p = \frac{\pi}{4}$ therefore

$$\sigma_{M}=\frac{\sqrt{\pi(4-\pi)}}{\sqrt{N}}$$ but what would be the value of $\sigma_{M}$? and then I have $\pi$ in the formula but is the thing I am trying to approximate so how does this work? and still missing the role of the 8 digit precision.

I may completely off the mark, but I guess that Monte Carlo here refers to approximating some integral by a random process, and that the integral must be chosen in a way that knowing its value allows us to calculate $\pi$.

Let's guess that the integral we are interested in is
$$
\frac\pi4=\int_0^1\sqrt{1-x^2}\,dx
$$
giving the probability that a random point in the square $[0,1]\times[0,1]$ (so $x$ and $y$ independent and uniformly distributed in $[0,1]$) is also in the unit disk.

Assume that we generate $N$ such points $(x,y)$ and record the number of successes (points in the unit disk) $M$. We know from crude estimates for $\pi$, say $3<\pi<4$, that the success rate $p$ of an individual point is between $3/4$ and $1$. Therefore the standard deviation of $M$ is bounded from above by
$$\sigma(M)=\sqrt{Np(1-p)}<\frac{\sqrt{3N}}4.$$

We approximate
$$
\pi\approx\frac{4M}N.
$$
This has SD $\sigma(\pi)<4\sigma(M)/N=\sqrt{3/N}$.

For 99% confidence we want $3\sigma(\pi)<10^{-8}$ or equivalently $\sqrt{N}>3\sqrt{3}\cdot 10^8$. This suggests that generating $N\approx 27\cdot10^{16}$ random points on the unit square would do the trick :-)

After this gedankenexperiment I quite appreciate Machin's formula. Even the alternating series for $\arctan 1$ beats this.

So this leaves open the possibility that something completely different was wanted?

The usual trick is to use the approximation of 355/113, repeating multiples of 113, until the denominator is 355.  This version of $\pi=355/113$ is the usual implied value when eight digits is given.

Otherwise, the approach is roughly $\sqrt{n}$


Suppose $f:\mathbb{R}^n\to \mathbb{R}^n$ is $C^1$,and $||f(x)-f(y)||\ge||x-y||$, is $f$ surjective? I asked this question a few days ago but haven't get an answer, so I ask again here.

Suppose $f:\mathbb{R}^n\to \mathbb{R}^n$ is $C^1$,and $||f(x)-f(y)||\ge||x-y||$, is $f$ surjective? What if $f$ is just $C^0$?I know it is clearly true when $n=1$, but when $n\ge 1$?

$f$ is clearly injective. Being an injective $C^0$ map $\mathbb{R}^n \to \mathbb{R}^n$, it follows by invariance of domain that  the image is open. The argument on Hagen von Eitzen's answer implies it is also closed. Therefore, the image must be the whole $\mathbb{R}^n$.

The argument due to Hagen is:

  Let $a\in\overline{\operatorname{im}(f)}$. Then a sequence $x_n$ with
  $f(x_n)\to a$ must be Cauchy, hence $x_n\to x$ with $f(x)=a$ (by
  continuity of $f$). We conclude that $\operatorname{im}(f)$ is closed.

Such $f$ is clearly injective and the inverse ${\operatorname{im}(f)}\to\Bbb R^n$ is continuous, hence we have a homeomorphism.

Let $a\in\overline{\operatorname{im}(f)}$.
Then a sequence $x_n$ with $f(x_n)\to a$ must be Cauchy, hence $x_n\to x$ with $f(x)=a$ (by continuity of $f$). We conclude that $\operatorname{im}(f)$ is closed.

Suppose $b\in\partial {\operatorname{im}(f)}$. Then $\operatorname{im}(f)\setminus\{b\}$ is still simply connected, but $\Bbb R^n\setminus\{f^{-1}(b)\}$ is not - contradiction.


how to transform word problem to an equation ?? is there any trick for this? A team f eleven spies were assigned consecutive whole numbers divisible by eleven as their identity code . For telephonic contact they have to use contact code that equals the product of the middle of the identity code and sum of all identity code of the eleven spies . if last of the identity is 209 then the contact code is ________ .

I am totally blank in this question .

What i Have done is :

no. Spies =11,

S= 11s1+ 12s2+13s3+14s4+......21s11 ..

now i am Confusing what to do  ...

You were given that $s_{11} = 209$.  Since the identity codes consist of consecutive multiples of $11$,

$$s_1 = s_{11} - 10 \cdot 11 = 209 - 110 = 99$$

giving the sequence of identity codes $\{99, 110, 121, 132, 143, 154, 165, 176, 187, 198, 209\}$.  Try to complete the problem from there.

Its a old question  i was just solved it for my exams and i came up with the solution. Its a bit tricky question

Spies = 11

{S1,S2,S3,S4,S5,S6,S7,S8,S9,S10,S11}

and their codes will be
{99,110,121,132,143,154,165,176,187,198,209}

Sum of Spies codes = 1694

Product of the middle of the codes = 1*2*3*4*5*6*7*8*9*0 = 0

So the contact code will be 0 + 1694 = 1694


Matrix multiplication: interpreting and understanding the process I have just watched the first half of the 3rd lecture of Gilbert Strang on the open course ware with link:

http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/

It seems that with a matrix multiplication $AB=C$, that the entries as scalars, are formed from the dot product computations of the rows of $A$ with the columns of $B$. Visual interpretations from mechanics of overlpaing forces come to mind immediately because that is the source for the dot product (inner product).

I see the rows of $C$ as being the dot product of the rows of $B$, with the dot product of a particular row of $A$. Similar to the above and it is easy to see this from the individual entries in the matrix $C$ as to which elements change to give which dot products.

For understanding matrix multiplication there is the geometrical interpretation, that the matrix multiplication is a change in the reference system since matrix $B$ can be seen as a transormation operator for rotation, scalling, reflection and skew. It is easy to see this by constructing example $B$ matrices with these effects on $A$. This decomposition is a strong argument and is strongly convincing of its generality. This interpreation is strong but not smooth because I would find smoother an explanation which would be an interpretation begining from the dot product of vectors and using this to explain the process and the interpretation of the results (one which is a bit easier to see without many examples of the putting numbers in and seeing what comes out which students go through).

I can hope that sticking to dot products throughout the explanation and THEN seeing how these can be seen to produce scalings, rotations, and skewings would be better. But, after some simple graphical examples I saw this doesn't work as the order of the columns in matrix $B$ are important and don't show in the graphical representation.

The best explanation I can find is at Yahoo Answers. It is convincing but a bit disappointing (explains why this approach preserves the "composition of linear transformations"; thanks @Arturo Magidin). So the question is: Why does matrix multiplication happen as it does, and are there good practical examples to support it? Preferably not via rotations/scalings/skews (thanks @lhf).

Some comments first. There are several serious confusions in what you write. For example, in the third paragraph, having seen that the entries of $AB$ are obtained by taking the dot product of the corresponding row of $A$ with column of $B$, you write that you view $AB$ as a dot product of rows of $B$ and rows of $A$. It's not.

For another example, you talk about matrix multiplication "happening". Matrices aren't running wild in the hidden jungles of the Amazon, where things "happen" without human beings. Matrix multiplication is defined a certain way, and then the definition is why matrix multiplication is done the way it is done. You may very well ask why matrix multiplication is defined the way it is defined, and whether there are other ways of defining a "multiplication"  on matrices (yes, there are; read further), but that's a completely separate question. "Why does matrix multiplication happen the way it does?" is pretty incoherent on its face.

Another example of confusion is that not every matrix corresponds to a "change in reference system". This is only true, viewed from the correct angle, for invertible matrices.

Standard matrix multiplication. Matrix multiplication is defined the way it is because it corresponds to composition of linear transformations. Though this is valid in extremely great generality, let's focus on linear transformations $T\colon \mathbb{R}^n\to\mathbb{R}^m$. Since linear transformations satisfy $T(\alpha\mathbf{x}+\beta\mathbf{y}) = \alpha T(\mathbf{x})+\beta T(\mathbf{y})$, if you know the value of $T$ at each of $\mathbf{e}_1,\ldots,\mathbf{e}_n$, where $\mathbf{e}^n_i$ is the (column) $n$-vector that has $0$s in each coordinate except the $i$th coordinate where it has a $1$, then you know the value of $T$ at every single vector of $\mathbb{R}^n$.

So in order to describe the value of $T$, I just need to tell you what $T(\mathbf{e}_i)$ is. For example, we can take
$$T(\mathbf{e}_i) = \left(\begin{array}{c}a_{1i}\\a_{2i}\\ \vdots\\ a_{mi}\end{array}\right).$$
Then, since
$$\left(\begin{array}{c}k_1\\k_2\\ \vdots\\k_n\end{array}\right) = k_1\mathbf{e}_1 + \cdots +k_n\mathbf{e}_n,$$ we have
$$T\left(\begin{array}{c}k_1\\k_2\\ \vdots\\ k_n\end{array}\right) = k_1T(\mathbf{e}_1) + \cdots +k_nT(\mathbf{e}_n) = k_1\left(\begin{array}{c}a_{11}\\a_{21}\\ \vdots\\a_{m1}\end{array}\right) + \cdots + k_n\left(\begin{array}{c}a_{1n}\\a_{2n}\\ \vdots\\ a_{mn}\end{array}\right).$$

It is very fruitful, then to keep track of the $a_{ij}$ in some way, and given the expression above, we keep track of them in a matrix, which is just a rectangular array of real numbers. We then think of $T$ as being "given" by the matrix
$$\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{array}\right).$$
If we want to keep track of $T$ this way, then for an arbitrary vector $\mathbf{x} = (x_1,\ldots,x_n)^t$ (the ${}^t$ means "transpose"; turn every rown into a column, every column into a row), then we have that $T(\mathbf{x})$ corresponds to:
$$\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{array}\right) \left(\begin{array}{c}
x_1\\x_2\\ \vdots\\ x_n\end{array}\right) = \left(\begin{array}{c}
a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n\\
a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n\\
\vdots\\
a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n
\end{array}\right).$$

What happens when we have two linear transformations, $T\colon \mathbb{R}^n\to\mathbb{R}^m$ and $S\colon\mathbb{R}^p\to\mathbb{R}^n$? If $T$ corresponds as above to a certain $m\times n$ matrix, then $S$ will likewise correspond to a certain $n\times p$ matrix, say
$$\left(\begin{array}{cccc}
b_{11} & b_{12} & \cdots & b_{1p}\\
b_{21} & b_{22} & \cdots & b_{2p}\\
\vdots & \vdots & \ddots & \vdots\\
b_{n1} & b_{n2} & \cdots & b_{np}
\end{array}\right).$$
What is $T\circ S$? First, it is a linear transformation because composition of linear transformations yields a linear transformation. Second, it goes from $\mathbb{R}^p$ to $\mathbb{R}^m$, so it should correspond to an $m\times p$ matrix. Which matrix? If we let $\mathbf{f}_1,\ldots,\mathbf{f}_p$ be the (column) $p$-vectors given by letting $\mathbf{f}_j$ have $0$s everywhere and a $1$ in the $j$th entry, then the matrix above tells us that
$$S(\mathbf{f}_j) = \left(\begin{array}{c}b_{1j}\\b_{2j}\\ \vdots \\b_{nj}\end{array}\right) = b_{1j}\mathbf{e}_1+\cdots + b_{nj}\mathbf{e}_n.$$

So, what is $T\circ S(\mathbf{f}_j)$? This is what goes in the $j$th column of the matrix that corresponds to $T\circ S$. Evaluating, we have:
\begin{align*}
T\circ S(\mathbf{f}_j) &= T\Bigl( S(\mathbf{f}_j)\Bigr)\\\
&= T\Bigl( b_{1j}\mathbf{e}_1 + \cdots + b_{nj}\mathbf{e}_n\Bigr)\\\
&= b_{1j} T(\mathbf{e}_1) + \cdots + b_{nj}T(\mathbf{e}_n)\\\
&= b_{1j}\left(\begin{array}{c}
a_{11}\\\ a_{21}\\\ \vdots\\\ a_{m1}\end{array}\right) + \cdots + b_{nj}\left(\begin{array}{c} a_{1n}\\a_{2n}\\\ \vdots\\\ a_{mn}\end{array}\right)\\\
&= \left(\begin{array}{c}
a_{11}b_{1j} + a_{12}b_{2j} + \cdots + a_{1n}b_{nj}\\\
a_{21}b_{1j} + a_{22}b_{2j} + \cdots + a_{2n}b_{nj}\\\
\vdots\\\
a_{m1}b_{1j} + a_{m2}b_{2j} + \cdots + a_{mn}b_{nj}
\end{array}\right).
\end{align*}
So if we want to write down the matrix that corresponds to $T\circ S$, then the $(i,j)$th entry will be
$$a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}.$$
So we define the "composition" or product of the matrix of $T$ with the matrix of $S$ to be precisely the matrix of $T\circ S$. We can make this definition without reference to the linear transformations that gave it birth: if the matrix of $T$ is $m\times n$ with entries $a_{ij}$ (let's call it $A$); and the matrix of $S$ is $n\times p$ with entries $b_{rs}$ (let's call it $B$), then the matrix of $T\circ S$ (let's call it $A\circ B$ or $AB$) is $m\times p$ and with entries $c_{k\ell}$, where
$$c_{k\ell} = a_{k1}b_{1\ell} + a_{k2}b_{2\ell} + \cdots + a_{kn}b_{n\ell}$$
by definition. Why? Because then the matrix of the composition of two functions is precisely the product of the matrices of the two functions. We can work with the matrices directly without having to think about the functions.

In point of fact, there is nothing about the dot product which is at play in this definition. It is essentially by happenstance that the $(i,j)$ entry can be obtained as a dot product of something. In fact, the $(i,j)$th entry is obtained as the matrix product of the $1\times n$ matrix consisting of the $i$th row of $A$, with the $n\times 1$ matrix consisting of the $j$th column of $B$. Only if you transpose this column can you try to interpret this as a dot product. (In fact, the modern view is the other way around: we define the dot product of two vectors as a special case of a more general inner product, called the Frobenius inner product, which is defined in terms of matrix multiplication, $\langle\mathbf{x},\mathbf{y}\rangle =\mathrm{trace}(\overline{\mathbf{y}^t}\mathbf{x})$).

And because product of matrices corresponds to composition of linear transformations, all the nice properties that composition of linear functions has will automatically also be true for product of matrices, because products of matrices is nothing more than a book-keeping device for keeping track of the composition of linear transformations. So $(AB)C = A(BC)$, because composition of functions is associative. $A(B+C) = AB + AC$ because composition of linear transformations distributes over sums of linear transformations (sums of matrices are defined entry-by-entry because that agrees precisely with the sum of linear transformations). $A(\alpha B) = \alpha(AB) = (\alpha A)B$, because composition of linear transformations behaves that way with scalar multiplication (products of matrices by scalar are defined the way they are precisely so that they will correspond to the operation with linear transformations).

So we define product of matrices explicitly so that it will match up composition of linear transformations. There really is no deeper hidden reason. It seems a bit incongruous, perhaps, that such a simple reason results in such a complicated formula, but such is life.

Another reason why it is somewhat misguided to try to understand matrix product in terms of dot product is that the matrix product keeps track of all the information lying around about the two compositions, but the dot product loses a lot of information about the two vectors in question. Knowing that $\mathbf{x}\cdot\mathbf{y}=0$ only tells you that $\mathbf{x}$ and $\mathbf{y}$ are perpendicular, it doesn't really tell you anything else. There is a lot of informational loss in the dot product, and trying to explain matrix product in terms of the dot product requires that we "recover" all of this lost information in some way. In practice, it means keeping track of all the original information, which makes trying to shoehorn the dot product into the explanation unnecessary, because you will already have all the information to get the product directly.

Examples that are not just "changes in reference system". Note that any linear transformation corresponds to a matrix. But the only linear transformations that can be thought of as "changes in perspective" are the linear transformations that map $\mathbb{R}^n$ to itself, and which are one-to-one and onto. There are lots of linear transformations that aren't like that. For example, the linear transformation $T$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ defined by
$$T\left(\begin{array}{c}
a\\b\\c\end{array}\right) = \left(\begin{array}{c}b\\2c\end{array}\right)$$
is not a "change in reference system" (because lots of nonzero vectors go to zero, but there is no way to just "change your perspective" and start seeing a nonzero vector as zero) but is a linear transformation nonetheless. The corresponding matrix is $2\times 3$, and is
$$\left(\begin{array}{cc}
0 & 1 & 0\\
0 & 0 & 2
\end{array}\right).$$
Now consider the linear transformation $U\colon\mathbb{R}^2\to\mathbb{R}^2$ given by
$$U\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}3x+2y\\
9x + 6y\end{array}\right).$$
Again, this is not a "change in perspective", because the vector $\binom{2}{-3}$ is mapped to $\binom{0}{0}$. It has a matrix, $2\times 2$, which is
$$\left(\begin{array}{cc}
3 & 2\\
9 & 6
\end{array}\right).$$
So the composition $U\circ T$ has matrix:
$$\left(\begin{array}{cc}
3 & 2\\
9 & 6
\end{array}\right) \left(\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 2
\end{array}\right) = \left(\begin{array}{ccc}
0 & 3 & 4\\
0 & 9 & 12
\end{array}\right),$$
which tells me that
$$U\circ T\left(\begin{array}{c}x\\y\\z\end{array}\right) = \left(\begin{array}{c} 3y + 4z\\ 9y+12z\end{array}\right).$$

Other matrix products. Are there other ways to define the product of two matrices? Sure. There's the Hadamard product, which is the "obvious" thing to try: you can multiply two matrices of the same size (and only of the same size), and you do it entry by entry, just the same way that you add two matrices. This has some nice properties, but it has nothing to do with linear transformations. There's the Kronecker product, which takes an $m\times n$ matrix times a $p\times q$ matrix and gives an $mp\times nq$ matrix. This one is associated to the tensor product of linear transformations. They are defined differently because they are meant to model other operations that one does with matrices or vectors.

I think part of the problem people have with getting used to linear transformations vs. matrices is that they have probably never seen an example of a linear transformation defined without reference to a matrix or a basis. So here is such an example. Let $V$ be the vector space of real polynomials of degree at most $3$, and let $f : V \to V$ be the derivative.

$V$ does not come equipped with a natural choice of basis. You might argue that $\{ 1, x, x^2, x^3 \}$ is natural, but it's only convenient: there's no reason to privilege this basis over $\{ 1, (x+c), (x+c)^2, (x+c)^3 \}$ for any $c \in \mathbb{R}$ (and, depending on what my definitions are, it is literally impossible to do so). More generally, $\{ a_0(x), a_1(x), a_2(x), a_3(x) \}$ is a basis for any collection of polynomials $a_i$ of degree $i$.

$V$ also does not come equipped with a natural choice of dot product, so there's no way to include those in the discussion without making an arbitrary choice. It really is just a vector space equipped with a linear transformation.

Since we want to talk about composition, let's write down a second linear transformation. $g : V \to V$ will send a polynomial $p(x)$ to the polynomial $p(x + 1)$. Note that, once again, I do not need to refer to a basis to define $g$.

Then the abstract composition $gf : V \to V$ is well-defined; it sends a polynomial $p(x)$ to the polynomial $p'(x + 1)$. I don't need to refer to a basis or multiply any matrices to see this; all I am doing is composing two functions.

Now let's do everything in a particular basis to see that we get the same answer using the correct and natural definition of matrix multiplication. We'll use the basis $ \{ 1, x, x^2, x^3 \}$. In this basis $f$ has matrix

$$\left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\\
 0 & 0 & 2 & 0 \\\
 0 & 0 & 0 & 3 \\\
 0 & 0 & 0 & 0 \end{array} \right]$$

and $g$ has matrix

$$\left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\\
 0 & 1 & 2 & 3 \\\
 0 & 0 & 1 & 3 \\\
 0 & 0 & 0 & 1 \end{array} \right].$$

Now I encourage you to go through all the generalities in Arturo's post in this example to verify that $gf$ has the matrix it is supposed to have.


Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$ Here is my attempt:

Let $(x, y)$ be any such point.

Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.

Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$

Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.

Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$

Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?

If you had actually checked each candidate point for equality of distance from the coordinate axes and $(4,2)$, you would have seen that only $(2,2)$ and $(10,10)$ satisfy the conditions of the original question.

(-4,2) (-4,-2) (4,-2)
The original point is 4 units from the y axis and two units from the x axis. i.e distance from each axis. The question is asking to find every point which is 4 units from the y axis AND 2 units from the x axis. There is one such point in each quadrant, which are the three additional points I listed. If you graph it, it should make sense. Those three points are the only ones which will satisfy the condition of distance of 4 from the y axis and distance of 2 from the x axis. It is basically a question about symmetry around each axis.
These represent the remaining three corners of an 8x4 rectangle, centered on the origin.  One point in each quadrant.  I believe most people are overthinking what the problem actually asks for.


distance to the edge of a rounded rectangle I'm trying to figure out a formula to create a rounded rectangle... but I need to do it by finding the distance from $(0, 0)$ to the line/edge of the rectangle given the angle provided as a vector.

Ok.. so here is the formula to plot the rectangle:  $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 1$

In my usage a and b will both be 1. Thus the formula can be simplified into $x^n + y^n = 1$

http://www.wolframalpha.com/input/?i=x%5E4+%2B+y%5E4+%3D+1

My input will be a point in space that can represent a vector.  Say it's (.5,.5).  The angle of this vector is $45^\circ$ and therefore it intersects with the plotted line at a specific distance from $(0, 0)$.  I don't need to know the $x$, $y$ of the plot per se.  I need to know the distance to that point.  If I have the x,y I can easily calculate the distance.  But maybe there is a way to calculate the distance directly.

I know I have: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 1$

and I have $d = \sqrt{x^2 + y^2}$

But alone that isn't enough info for me to figure it out.  I can see a pattern in there though.  OH crap they're the same formula, except that $n=2$, where I want $n=4$ (or $n>4$).  When $d = 1$ the square root of $1 = 1$.

I was able to solve for d like this - but I don't know if this is helping...

  1 = x^n + y^n

  0 = x^n + y^n - 1

  d = sqr(x^2 + y^2)

  0 = sqr(x^2 + y^2) - d

  d = 1 + sqr(x^2 + y^2) - x^n - y^n

But still.. I don't know x and y on the plot.  I'm given x,y in space and have to find the distance to the plotted line where it intersects with that vector.

Maybe rather I need some sort of intersection type formula??  Any ideas?

EDIT ----

Really what I need is to figure out the distance to the line at any given angle.

EDIT 2 --

What about this approach..

Here are the knowns.. I have an x,y out in space.  Therefore I have an angle and a distance for that vector.  I also know the base size of the rounded rectangle.

If it was a circle it's easy cause the radius of a circle is constant.

My answer here might not be very mathematical in terms of equation syntax.. but I'm a programmer, not a mathematician..

It should be possible to calculate the difference between the radius of the circle (known if the object were a circle) and the "radius" at any point on the rounded rectangle.

At an angle of $\theta$ they are the same.  At and angle of 90 they are the same.  But at points between the two they are not.

If the radius of the circle is 1.. then you have $d = 1 + \sqrt{x^2 + y^2} - x^n - y^n$ for the "radius" or distance to the edge of the rounded rectangle.

The difference between the two is delta $d = 1 - x^n - y^n$

So if our desired rectangle is S in size instead of 1 - S is a scale factor over the 1.

If I normalize the vector $(x, y)$ and make it's length 1.  Then use that x, y to figure out the delta d. I can then scale that dd by S and add it to d to get the distance at that angle.

Checking to see if it works in code.

Let the vector be $(a,b)$.  The line it is on is $y=\frac ba x$ (assuming it is not vertical.)  Plug this into your superellipse and get $x^n+\left(\frac ba x\right)^n=1$ or $x^n= \frac 1{1+(\frac ba)^n}=\frac {a^n}{a^n+b^n}, x=\left(\frac {a^n}{a^n+b^n}\right)^{\frac 1n}$ for the intersection point.

For what its worth here is a solution using Qt that finds the intersecting point between a line starting at the center of a rounding rectangle to some place outside of the shape. You have to specify the Rectangle, Radius of the rounded edges and a point B which is outside of the rounded rectangle.

#include <QApplication>
#include <QDebug>
#include <QGraphicsItem>
#include <QGraphicsScene>
#include <QGraphicsView>
#include <QLine>
#include <QPainterPath>
#include <QPoint>
#include <QRect>
#include <limits>
#include <math.h>

#define kDegToRad 0.0174532925

//-----------------------------------------------------------------------------
//-----------------------------------------------------------------------------
bool circleLineIntersectPoint( double aRadius, QPointF aA, QPointF aB, QPointF aC, QPointF &aIntersection)
{
  bool tReturn = true; // false = not intersection point, true == insersected
  aIntersection = QPointF();
  QPointF tD;
  tD.setX(aB.x() - aA.x());
  tD.setY(aB.y() - aA.y());
  if ((tD.x() == 0) && (tD.y() == 0))
  {
    // A and B are the same points, no way to calculate intersection
    tReturn = false;
    return tReturn;
  }

  double tDL = (tD.x() * tD.x() + tD.y() * tD.y());
  double tT = ((aC.x() - aA.x()) * tD.x() + (aC.y() - aA.y()) * tD.y()) / tDL;

  // point on a line nearest to circle center
  QPointF tNearest;
  tNearest.setX(aA.x() + tT * tD.x());
  tNearest.setY(aA.y() + tT * tD.y());

  QLineF tLine(tNearest.x(), tNearest.y(), aC.x(), aC.y());
  double tDist = tLine.length();

  if (tDist == aRadius)
  {
    // line segment touches circle; one intersection point
    aIntersection = tNearest;

    if (tT < 0 || tT > 1)
    {
      // intersection point is not actually within line segment
      tReturn = false;
    }
  }
  else if (tDist < aRadius)
  {
    // two possible intersection points

    double tDT = sqrt(aRadius * aRadius - tDist * tDist) / sqrt(tDL);

    // intersection point nearest to A
//    double tT1 = tT - tDT;
//    QPointF tI1;
//    tI1.setX(aA.x() + tT1 * tD.x());
//    tI1.setY(aA.y() + tT1 * tD.y());
//    if (tT1 < 0 || tT1 > 1)
//    {
//      // intersection point is not actually within line segment
//      tReturn = false;
//    }

//    if(tReturn)  // intersection point farthest from A
//    {
      double tT2 = tT + tDT;
      QPointF tI2;
      tI2.setX(aA.x() + tT2 * tD.x());
      tI2.setY(aA.y() + tT2 * tD.y());
      if (tT2 < 0 || tT2 > 1)
      {
        // intersection point is not actually within line segment
        tReturn = false;
      }
      else
        aIntersection = tI2;
//    }
  }
  else
  {
    // no intersection
    tReturn = false;
  }

  return tReturn;
}

//-----------------------------------------------------------------------------
//-----------------------------------------------------------------------------
int rectangleLineIntersectPoint(QRectF aRect, QPointF aB, QPointF &aPoint)
{
  // Define a line from the center of the rectangle to point B
  QLineF tLine(aRect.center(), aB);

  QPointF tReturn;
  int tType = 0;

  // See if the line intersects the top line of the rectangle
  QLineF tTopLine(aRect.topLeft(), aRect.topRight());
  if(QLineF::BoundedIntersection == tTopLine.intersect(tLine, &aPoint)) tType = 1; // Top Line
  else
  {
    // See if the line intersects the right line of the rectangle
    QLineF tRightLine (aRect.topRight(), aRect.bottomRight());
    if(QLineF::BoundedIntersection == tRightLine.intersect(tLine, &aPoint)) tType = 2; // Right Line
    else
    {
      // See if the line intersects the bottom line of the rectangle
      QLineF tBottomLine(aRect.bottomLeft(), aRect.bottomRight());
      if(QLineF::BoundedIntersection == tBottomLine.intersect(tLine, &aPoint)) tType = 3; // Bottom Line
      else
      {
        // See if the line intersects the left line of the rectangle
        QLineF tLeftLine(aRect.topLeft(), aRect.bottomLeft());
        if(QLineF::BoundedIntersection == tLeftLine.intersect(tLine, &aPoint)) tType = 4; // Left Line
        else
          tType = 0; // This should never happen since the center of the line is the center of the rect
      }
    }
  }
  return tType;
}

//-----------------------------------------------------------------------------
//-----------------------------------------------------------------------------
bool roundedRectLineIntersectionPoint(double aRadius, QRectF aRect, QPointF aB, QPointF &aPoint)
{
  bool tReturn = true; // False = not intersection point

  // Compute the rectangle to line intersection point
  int tType = rectangleLineIntersectPoint(aRect, aB, aPoint);

  //  Based on the type figure out if we need to compute the intersection with
  //  rounded part of the rounded rectangle
  QPointF tC; // Center of the circle that makes up the rounded part
  bool tInRoundedEdge = true;
  switch(tType)
  {
    case 0: tReturn = false; tInRoundedEdge = false; break;

    // Top Line
    case 1: if(aPoint.x() <= aRect.left() + aRadius) // Top Left
              tC = QPointF(aRect.left() + aRadius, aRect.top() + aRadius);
            else
            if(aPoint.x() >= aRect.right() - aRadius) // Top Right
              tC = QPointF(aRect.right() - aRadius, aRect.top() + aRadius);
            else
              tInRoundedEdge = false;
            break;

    // Right Line
    case 2: if(aPoint.y() <= aRect.top() + aRadius) // Top right
              tC = QPointF(aRect.right() - aRadius, aRect.top() + aRadius);
            else
            if(aPoint.y() >= aRect.bottom() - aRadius) // Bottom Right
              tC = QPointF(aRect.right() - aRadius, aRect.bottom() - aRadius);
            else
              tInRoundedEdge = false;
            break;
    // Bottom Line
    case 3: if(aPoint.x() <= aRect.left() + aRadius) // Bottom Left
              tC = QPointF(aRect.left() + aRadius, aRect.bottom() - aRadius);
            else
            if(aPoint.x() >= aRect.right() - aRadius) // Bottom Right
              tC = QPointF(aRect.right() - aRadius, aRect.bottom() - aRadius);
            else
              tInRoundedEdge = false;
            break;
    // Left Line
    case 4: if(aPoint.y() <= aRect.top() + aRadius) // Top Left
              tC = QPointF(aRect.left() + aRadius, aRect.top() + aRadius);
            else
            if(aPoint.y() >= aRect.bottom() - aRadius) // Bottom left
              tC = QPointF(aRect.left() + aRadius, aRect.bottom() - aRadius);
            else
              tInRoundedEdge = false;
            break;
  }

  if(tReturn && tInRoundedEdge)
    tReturn = circleLineIntersectPoint( aRadius, aRect.center(), aB, tC, aPoint);

  return tReturn;
}

//-----------------------------------------------------------------------------
//-----------------------------------------------------------------------------
int main(int argc, char *argv[])
{
  QApplication app(argc, argv);

  QGraphicsView view;
  QGraphicsScene *scene = new QGraphicsScene;
  view.setScene(scene);

  QRectF tRect(-300, -150, 600, 300);
  double tLineLength = 800.0;
  qreal tRadius = 100.0;

  QPainterPath tPath;
//  tPath.addRoundedRect(tRect, tRadius, Qt::AbsoluteSize);
  tPath.addRect(tRect);
  scene->addPath(tPath, QPen(Qt::red), QBrush(Qt::green));

  QPointF tPoint;
  for(double tIndex = 0; tIndex < 360; tIndex+=1)
  {
    double tX = tLineLength * sin(tIndex * kDegToRad);
    double tY = tLineLength * cos(tIndex * kDegToRad);
    roundedRectLineIntersectionPoint(tRadius, tRect, QPointF(tX, tY), tPoint);
    scene->addLine(QLineF(tRect.center(), QPointF(tPoint.x(), tPoint.y())));
  }

  view.show();

  return app.exec();
}


Let $x = a+2b$, $y = b+2c$, and $z = c + 2a$, then $x+y+z = 3(a+b+c) = 9$, and we have that $f''(x) = -\dfrac{2}{9}\cdot x^{-\frac{5}{3}} < 0$ on $(0,9)$ with $f(x) = \sqrt[3]{x}$. Thus $f$ is concave and we have:

$LHS = f(x) + f(y) + f(z) \leq 3f\left(\frac{x+y+z}{3}\right) = 3f(3) = 3\sqrt[3]{3} = RHS$

let $$f(a,b,c)=\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\tag{1}$$
and $$g(a,b,c)=a+b+c-3=0\tag{2}$$

Using Lagarange's Multiplier we have

$$\frac{\frac{\partial f}{\partial a}}{\frac{\partial g}{\partial a}}=
\frac{\frac{\partial f}{\partial b}}{\frac{\partial g}{\partial b}}=
\frac{\frac{\partial f}{\partial c}}{\frac{\partial g}{\partial c}}=k$$
This implies
$$(a+2b)^{-2/3}+2\cdot(c+2a)^{-2/3}=3k \tag{3}$$
$$(b+2c)^{-2/3}+2\cdot(a+2b)^{-2/3}=3k \tag{4}$$
$$(c+2a)^{-2/3}+2\cdot(2+2c)^{-2/3}=3k \tag{5}$$

Solving $(3),(4),(5)$ Simultaneously We get
$$a=b=c$$substituting this in $(2)$ we get,

$$a=b=c=1$$
It can be shown that Maxima occurs at above mentioned values of parameters,
Putting these values in $(1)$ we get

$$f(1,1,1)=\sqrt[3]{3}+\sqrt[3]{3}+\sqrt[3]{3}=3\sqrt[3]{3}$$
Since $f(1,1,1)=3\sqrt[3]{3}$ is Maximum values of $f(a,b,c)$ this implies
$$f(a,b,c)=\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$
$$\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$


How to differentiate $(x+1)^{3/2}$ using the limit definition of the derivative? I want to use the definition of the derivative to find $f'$ of $f(x)=(x+1)^{3/2}$.

I solved it using the chain rule. Would like to try to solve it using the definition of a derivative:

$$
\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.
$$

Don't really know how I should start it. Should I make it $\sqrt{(x+1)^3}$? Or keep the exponent as $3/2$? Other guidance would be appreciated.

You can also use the product rule:
$$\left[(x+1)^{3/2}\right]'=\left[(x+1)\sqrt{x+1}\right]'=$$
$$\sqrt{x+1}\cdot\lim_{h\rightarrow 0} \frac{(x+h+1)-(x+1)}{h}+(x+1)\cdot\lim_{h\rightarrow 0} \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}=$$
$$\sqrt{x+1}+(x+1)\cdot \lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h+1}+\sqrt{x+1})}=$$
$$\sqrt{x+1}+\frac{\sqrt{x+1}}{2}=\frac{3}{2}\sqrt{x+1}.$$


Why the only binary MDS codes are trivial ones? Why the only binary MDS codes are trivial ones?

I have been thinking how to draw a contradiction by assuming the MDS code is not trivial.

Thank you very much!

This is Proposition 9.2 on p. 212 of Elements of Algebraic Coding Theory by L. R. Vermani.

  Definiton 9.2

  We have shown that linear $[n, 1, n]$, $[n, n- 1, 2]$ and $[n, n, 1]$ codes exist
  over any finite field $F$ and these are MDS codes. These are called trivial MDS
  codes.

  Proposition 9.2

  The only binary MDS codes are the trivial codes.

  Proof

  Let $C$ be a binary $[n, k, d]$ MDS code. If $k = 1$, then $C$ is a trivial MDS code and
  so we may suppose that $k > 1$. Let $G$ be a generator matrix of $C$ with the first
  $k$ columns of $G$ forming the identity matrix. If $n > k + 1$, then $C$ has a column,
  say $j$th, of weight less than $k$ and greater than $1$. Suppose that the $i$th entry
  of this column is $0$. Then the first $k$ columns of $G$ except the $i$th together
  with the $j$th column are linearly dependent. This proves that $C$ cannot be an
  MDS code. Hence

  $$k \le n \le k + 1$$

  and $C$ is a trivial MDS code.

The only binary codes that are MDS are the trivial (n, n, 1) universe codes, the (n, n − 1, 2) single-parity-check (SPC) codes, and the (n, 1, n) repetition codes.
http://web.stanford.edu/class/ee392d/Chap8.pdf


Hadamard production for $\sin z$. We know that all zeros of $\sin z$ are $z=k\pi,\,k\in\mathbb Z$, all of which are of order $1$.

So by Hadamard Factorization theorem, we have
$$
\sin z=e^{P(z)}\cdot z\cdot\prod_{n=1}^{\infty}(1-\frac{z^2}{n^2\pi^2}),
$$
where $P(z)$ is a polynomial of degree $\leq1$.

Actually,
$$
\sin z=z\cdot\prod_{n=1}^{\infty}(1-\frac{z^2}{n^2\pi^2}).
$$

So I wonder how to prove $P(z)=0$?

You already know that $P$ has degree at most one: $P(z) = az+b$. Then $\sin(-z) = -\sin(z)$ implies $e^{P(z)} = e^{P(-z)}$, i.e. $1 = e^{P(z)-P(-z)} = e^{2az}$, so that $a=0$  and $P$ is constant.

Then use $\frac{\sin(z)}{z} \to 1$ for $z\to0$ to conclude that $e^{P(z)} = 1$.


Dominant Eigenvalues I'm trying to understand dominant eigenvalues and I found this website that has a explanation of it (Definition 9.2). In the example, the power method is used to find the dominant eigenvector which correspondes to the eigenvalue of 1.

When I calculate the eigenvalues and vectors of the matrix in the example, I got this result:

The first line are the eigenvalues and the second row the eigenvectors.
As you can see, the eigenvector that corresponde to to the eigenvalue of 1 is

{-0.577, -0.577, -0.577}

If I calculate the powers of the matrix, I find that after M^9, it converges as shown in the website

I don't understand what is the difference between the eigenvector that I found that corresponde to to the eigenvalue of 1 and the eigenvector that is found after elevating the matrix many times, and that the website described also as the eigenvector of eigenvalue 1.

What do you mean by "the eigenvector"? For each eigenvalue, there are infinitely many eigenvectors, and you typically get a basis as "the" eigenvectors.

From your post, it looks like the general eigenvector for $\lambda=1$ has the form $(t,t,t)$ and you obtained the one with $t= -0.57735$. But other values of $t$ lead to other eigenvectors, and if I remember right, the power method produces a probability eigenvector, thus in this case $t=\frac{1}{3}$.

P.S. Also note, if you are reffering to the first 3x3 example on that page (it would had been very helpfull if you included the matrix you calculated), you should note that they find the eigenvector for $**A^T**$. In that example, it is trivial to see that $(t,t,t)$ is an eigenvector for $A$, but they don't calculate the eigenvectors of $A$!

Imagine someone has a list of numbers, known (to that person) but kept secret. Now imagine you have the ability to do some limited manipulation of those numbers and extract some information from the person about this set of numbers. You can ask that someone to alter every number in the same way by raising them to a power. Then you can ask that someone to assign each separate number to a specific (to that number) vector. The information then that you can get is the vector sum. If each but one number is smaller than unity in magnitude, they (the small numbers) will in the power exponential become small and their corresponding vector is scaled to a tiny bit. The number that is close or equal to one (or really just the largest in magnitude) will dominate, and the vector corresponding to that number will be apparent.

Now imagine all that again, but the person with the list of numbers is actually a matrix, the list of numbers are its eigenvalues, and the corresponding vectors for those numbers are the eigenvectors. That is what is called the power method.


Prove the open mapping theorem by using maximum modulus principle The open mapping theorem says a non constant analytic function maps open sets to open sets.

The maximum modulus principle says if $f$ a non constant analytic function on an open connected set $D\subset\mathbb{C}$, then $|f|$ does not attain a local maximum on $D$.

It it known that one application of the open mapping theorem is to prove the maximum modulus principle. But what about the other way around? Can we use the maximum modulus principle(possibly plus some other results) to prove the open mapping theorem?

The reason I am interested in this question is because after I see the proof in wiki-pedia, personally I found the idea in this proof somewhat "hidden", it is not that intuitive (at least to me).

Sorry for the late response. Here's a proof of the open mapping theorem assuming the maximum modulus principle.

First, we need the "minimum modulus principle". That is, if $f$ is a non-constant analytic function on an open connected set $D \subset \mathbb{C}$, and $f$ has no zeroes in $D$, then $| f |$  cannot attain a minimum in $D$. The proof follows trivially by applying the maximum modulus principal to the function $1/f$ which is analytic on $D$.

Now suppose $D \subset \mathbb{C}$ is open and connected, and $f$ is a non-constant analytic function on $D$. Let $U \subset D$ be open, and let $w_0 \in f(U)$, say $w_0=f(z_0)$ with $z_0 \in U$. We must show that there is a disc centered at $w_0$ which is contained in $f(U)$.

Choose $t>0$ so that $\overline {D_t(z_0)} \subset U$ and $f(z) \neq w_0$ for any $z \in \overline {D_t(z_0)}$ other than $z_0$. Let $m=\inf \{|f(z)-w_0| : |z-z_0|=t \} > 0 $. Suppose $|w-w_0| < m/3$, and that there is no $z \in U$ such that $f(z)=w$. Then the function $g(z)=f(z)-w$ is analytic, non-constant, and has no zeroes in the open connected set $D_t(z_0)$, so the minimum modulus principle shows that $g$ cannot attain a minimum modulus in $D_t(z_0)$. However, $g$ does attain a minimum modulus in the compact set $ \overline {D_t(z_0)} $, so this minimum modulus must occur on the boundary circle defined by $|z-z_0|=t$. But if $|z-z_0|=t$, then

$|g(z)|=|f(z)-w| \geq |f(z)-w_0| - |w_0-w| \geq 2m/3 $, and

$|g(z_0)| = |w_0-w| < m/3 < 2m/3$.

This gives a contradiction since $z_0$ is obviously in the interior of the disc in question. Therefore $f(z)=w$ for some $z \in D_t(z_0)$, and $D_{m/3}(w) \subset f(U)$, showing that $f(U)$ is open and proving the theorem.

You can't prove the open mapping theorem (OMT) with the maximum modulus principle. Because the maximum modulus principle is not a tool that is suitable for proving the open mapping theorem. The maximum modulus principle is insufficiently sophisticated to understand the topology of the complex plane. It is merely a statement about $|f|$, not about $f$ itself. The maximum modulus principle holds for lots of maps, e.g., the product of $f$ with any unimodular function, which have no reason to be  open. Etc., etc.

I will try to explain the idea of the OMT proof instead. Pick a point $z_0$ in the domain of $f$. Consider the series $f(z)=f(z_0)+a_k(z-z_0)^k+\dots$ where by $k$ is the first index after $0$ such that $a_k\ne 0$. Let $F(z)=f(z_0)+a_k(z-z_0)^k$. The map $F$ is open, because it's just a power of $z$ composed with some shifts. (The power of $z$ can be written out explicitly in polar coordinates, to check that it's open.)

When $z$ is sufficiently close to $z_0$, $|f(z)-F(z)|$ is very small, because it consists of powers of $z-z_0$ higher than $k$. This allows us to use Rouché's theorem to conclude that the values attained by $F$ will also be attained by $f$ (I am omitting the details of estimates here). Since the image of $F$ contains a neighborhood of $f(z_0)$, so does the image of $f$.


Cauchy Sequence $|a_n - a_m| < \epsilon $ Question

  A sequence $\{a_n\}$ of real numbers is said to be a Cauchy sequence of for
  each $\epsilon$ > 0 there exists a number $N > 0$ such that m, $n > N$ implies
  that $|a_n − a_m| <\epsilon$.

  Prove that every convergent sequence is a Cauchy sequence

Attempt

This is my first time hearing what a cauchy sequence is. I have no idea how to even start this. I googled cauchy sequence and I think its when $a_n$ converges to $a_{n+1}$?

Attempt:

WTS: $\exists a_m \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - a_m| < \epsilon$

Let $\epsilon > 0$ be arbitrary

Choose N such that for $n > N$ we have $|a_n - a_m| < \epsilon$

Suppose $n > N$, then

??

Could someone point me to the right direction? Thx.

The sequence $(a_{n})$ is convergent by assumption; let $l := \lim_{n}a_{n}$. Let $\varepsilon > 0$. Then there is some $N$ such that $|a_{n} - l| < \varepsilon/2$ for all $n \geq N$. Note that
$$
|a_{n} - a_{m}| \leq |a_{n}-l| + |l-a_{m}| < \varepsilon/2 + \varepsilon/2 = \varepsilon
$$
for all $n,m \geq N$.

The major tool here is the triangle inequality.

If it is not Cauchy, it is obviously not convergent: for |a_n -a_m| is greater than some epsilon for arbitrarily large n,m. This assures a_n and a_m do not both get within epsilon/2 of some a_infinity.