Is it trivial to say $\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k}$ Is it trivial to say
$$\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k},$$

considering the fact that we know
$$\mathop {\lim }\limits_{n \to \infty } {(1 + {1 \over n})^n} = e?$$

Since $\frac kn = \frac 1 {\frac nk}$ and $n = \frac nk \cdot k$ we have
$$\lim_{n\to \infty} (1+\frac kn)^n = \lim_{n\to\infty} \left( (1 + \frac{1}{\frac nk})^{\frac nk} \right)^k = \left ( \lim_{u\to\infty} (1+\frac 1u)^u \right)^k = e^k$$
If $k<0$, you have
$$(1+\frac 1u)^{-u} = \left((1+\frac 1u)^u\right)^{-1} \to e^{-1} \qquad (u\to\infty)$$

By $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e,n\in\mathbb{N}$, we have
$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e,\quad x\in \mathbb{R}$$


Counting paths in multidimensional grids Let's consider a grid in a multidimensional space. Each grid has 'n' number of lines, spaced one unit apart. A path (or walk) of 'm' steps is between two grid intersections. A step is a single unit movement in any dimension (direction).

I'll cut through the fluff by providing an example :

Let's say we have 2 dimensions and 3 grid lines. The grid intersections (x,y), where x,y ∈ {0,1,2}.

There will be six valid walks starting at (0,0) of length m=2:

(0,0)→(0,1)→(0,0)

(0,0)→(0,1)→(0,2)

(0,0)→(0,1)→(1,1)

(0,0)→(1,0)→(0,0)

(0,0)→(1,0)→(2,0)

(0,0)→(1,0)→(1,1)

Then generalize this for d dimensions, n gridlines, m steps and any starting point.

I'm not interested in knowing what the paths look like, just how many of them there are based on the starting location and the other parameters.

I wrote a python script (iterative and recursive) that kinda generates this (at each intersection, it generates all the possible following steps), but it chokes when the dimensions and steps start to get larger. Since we're supposed to know what's the highest count of paths possible considering all starting locations, this "brute-force" approach is definitely not the way to go I believe.

I was wondering if there was a way to solve this in a simpler manner, involving math to generate the number of final intersections to be visited based on the different parameters, without having to actually generate the paths maybe ?

The number of shortest paths from $(0,0,0)$ to $(a,b,c)$ is
$$\frac{(a+b+c)!}{a!\;b!\;c!}$$
with an obvious extension to higher dimensions.  This is a Multinomial Coefficient.

---Please note---
I'm not well versed in the topic being discussed.
Someone mentioned the same problem to me. This is my approach, which was based mostly on brute force and some chance encounter of adjacency matrices.

For anyone who is well versed in this, is this an approach worth using? Basically, did I get the answer?

I considered the following situation:
If I can create an matrix of for each of the possible coordinates (grid lines) and then list out possible single step (of length 1 unit), taking powers of the matrix following by sum of rows (or columns) of the result will yield possible total number of paths of length(l) taken 1 step at a time.

I have attempted this in R as follows:

library(Matrix)

q <- list()#will house coordinates for a xyzk plane where each point will be enumerated
counter = 1
for (i in 0:10){
  for (j in 0:10){
    for (k in 0:10){
      for (l in 0:10){
        q[[counter]]<- c(i,j,k,l)
        counter = counter +1
      }
    }}}
rm(counter,i,j,k,l)  #view length(q), should be 10^4
length(q)
a <- matrix(ncol=length(q),nrow=length(q),byrow = T)#making space  for adjacency matrix
for(i in 1:length(q)){
  for(j in 1:length(q)){
    if(i==j){
      a[i,j] <- 0#all diagonals are 0
    }else{
      if(sum(abs(q[[j]]-q[[i]]))==1){ #this calculates absolute path length
        #since we are moving 1 step at a time we are only interested in
        #step sizes of 1, so length = 1
        a[i,j] <- 1
      }else{a[i,j] <- 0}
    }
  }
}
rm(i,j,q)
\#save space and speedier calculation by making it a sparse matrix
a_sparse <- as(a,"dgCMatrix")
\#remove large a
rm(a)

\#go 10 steps, each step is a power of the adjacency matrix
mat_power <- function(m, pow){
  v = as(diag(dim(m)[1]),"dgCMatrix")
  for(i in 1:pow){
    v <- v%*%m
  }
  return(v)
}
\#the calculation will take a bit of time
\#note that using dgCMatrix saves space and also expedites calculation
\#do not attempt to calculate powers for such large matrix from large matrix A

steps <- mat_power(a_sparse,10)
sums <- apply(steps, 1, sum)

\# a neat thing to be noted is that this will yield steps of size 10 (taken 1 step at a time) for all possible starting points in that grid line.

So starting form (0,0,0,0) I got a total of 44569728 steps that can be taken of size 10 (taken 1 at a time)


How do I get the square root of a complex number? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?

This is a three parts post. The first part was written by user Did; it provides a formula and some brief comments on it. The second part was written by user Hans Lundmark; it provides a geometric way to understand the formula. The third part was written by user t.b.; it provides some explanatory pictures and some brief comments on them.

(Did) If one is able to compute the square root of every positive real number and the modulus of every complex number, a nice formula for the principal square root $\sqrt{z}$ of $z$ is
$$
\sqrt{z}=\sqrt{r}\frac{z+r}{|z+r|},\quad r=|z|.
$$
Try to prove it and you will see, it works...

The principal square root is the one with a positive real part. The only case when the formula fails is when there is no principal square root, that is, when $z$ is a negative real number.

No sine or cosine is involved, one does not even need to solve second degree polynomials, one just uses squares and square roots. For example, for $z=9+4\mathrm{i}$,
$$
\sqrt{z}=\frac{9+\sqrt{97}+4\mathrm{i}}{\sqrt{2(9+\sqrt{97})}}.
$$

(HL) There's a geometric way of understanding the formula in Did's answer. To find a square root of a given complex number $z$, you first want to find a complex number $w$ which has half the argument of $z$ (since squaring doubles the argument). Compute $r=|z|$ and let $w = z+r$; thus $w$ lies $r$ steps to the right of $z$ in the complex plane. Draw a picture of this, and it should be clear that the points $0$, $z$ and $w$ form an isosceles triangle, from which one sees that the line from $0$ to $w$ bisects the angle between the real axis and the line from $0$ to $z$. In other words, $w$ has half the argument of $z$, as desired. Now it only remains to multiply $w$ by some suitable real constant $c$ so that $|cw|^2 = |z|$; then we will have $(cw)^2=z$ and hence $cw$ is a square root of $z$. Obviously, $c=\pm\sqrt{|z|}/|w|$ will do the trick, so this method only fails when $w$ happens to be zero, i.e., if $z$ is a negative real number.

(t.b.) Following a suggestion of Did, I take the liberty of adding two pictures I originally posted as a separate answer, but it seemed better to have them here:

Here's the picture for $z = 9 + 4i$:

Remark: The construction of the square roots is geometrically exact. That is to say, they were constructed using straightedge and compass only. I decided to hide the construction, as it seems rather obfuscating the intended illustration than adding to it. Nevertheless, I suggest taking a moment and thinking about how you would achieve the geometric construction.

Added (t.b.)

Here's the construction I used: Intersect the circle around $z/2$ through $z$ with the tangent to the unit circle orthogonal to $z$. Then $h^2 = (|z|-1)\cdot 1$ and thus the red circle has radius $\sqrt{|z|}$. It remains to intersect the red circle with the angular bisector of the $x$-axis and $z$ which I constructed using the process Hans described in his part of the post.

The pictures were created using GeoGebra.

ok, Would someone please tell me if I'm wrong about this.

We want $\pm\sqrt{a+bi}.$  $($This kind of square root is not well-defined without the $\text{“}{\pm}\text{''}.)$

If $a+bi = r(\cos\theta+i\sin\theta)$ then $\displaystyle\tan\frac\theta2 = \frac{\sin\theta}{1+\cos\theta} = \frac b {\sqrt{a^2+b^2}+a},$ so
$$
\sin\frac\theta2 = \frac b{\text{something}} \text{ and } \cos\frac\theta 2 = \frac{\sqrt{b^2+a^2}+a}{\text{something}}
$$
where $\text{“something''}$ makes $\sin^2+\cos^2=1,$ so the efficient way to think about this is the tangent half-angle formula. Or is there a more efficient way?


Motivation of the Logarithm Suppose that someone wants to calculate approximately the product of 101,123,958,959,055 and 342,234,234,234,236 without using a computer. Since these numbers are so long, completely carrying out multiplication by hand would take a long time because each digit in one number must be multiplied by every digit in the other number. I understand how a logarithm together with a logarithm table can help approximate this product, but why is this better than approximating it by the product of 101,000,000,000,000 and 342,000,000,000,000 (or some other number of the digits replaced by zeros)?

Logarithms were not motivated by the desire to get exact solutions to problems such as multiplying two numbers with $15$ significant digits each.

With a decent table of logarithms you might have a precision of five digits, so indeed the answer you would get by using logarithms to compute
$101{,}123{,}958{,}959{,}055 \times 342{,}234{,}234{,}234{,}236$
would be no more accurate than if you simply multiplied
$101{,}120{,}000{,}000{,}000 \times 342{,}230{,}000{,}000{,}000.$
It might be possible to go to more digits with a better table and sophisticated interpolation techniques, but not many more digits.

Adding two five digit numbers is still easier to do than multiplying two five-digit numbers if the only tools at your disposal are pencil and paper.

This still might not have been enough motivation to purchase a table of logarithms if you only needed to do such a multiplication once a year. But there are problems that require you to perform many multiplications to get just one answer.
Therefore there was an incentive to develop and publish tables of logarithms and promote their use.

Meanwhile, logarithms also turned out to have some interesting mathematical properties, for example providing a solution for the integral of $\frac1x$
(the only power of $x$ whose integral is not another power of $x$).
I think the main reason we study logarithms today is because of things like that, not because we want to get the product of two $n$-digit numbers.

Direct multiplication of two $n$ digits numbers takes on the order of $n^2$ elementary operations (digits multiplies and additions with carries).

When they were performed with tables of logarithms, full $n$ digits were obtained, using high-precision tables plus interpolation techniques. Then additions were performed with just order $n$ elementary operations, and the total effort was more acceptable.


Point positioning inside rectangle with only two known vertexes I'm working on a js widget, and I've come across a positioning problem, which I can't seem to solve with my limited geometry knowledge or by help of Wikipedia/google.

I have a quadrilateral rectangle, which is positioned at an angle. I know its two opposite vertexes and width/height ratio. And there's a point on it, which coordinates I also know. I need to find how far (in %s of width/height) is that point from rectangle's sides. Is it possible to do so?

Let $(x_0,y_0)$ the coords of the lowest point, and $(x_1,y_1)$ the coords of the highest.  Let also $\alpha$ the angfle formed by the line passing throught $(x_0,y_0)$ parallel to $h$. We can write the equation of this line as:
$$L_h: y-y_0=\tan(\alpha)(x-x_0)$$
Now, we have to find the equation of the line paasing throught $(x_1,y_1)$ and parallel to $w$. We have:
$$L_w: y-y_1=-\frac{1}{\tan(\alpha)}(x-x_1)$$
Let $(X,Y)$ the coords of the fixed point. Now, we have done, because we can use the distance formula. Let $\delta_h$ the distance of the fixed point from the line $L_h$, we have:
$$\delta_h=\frac{|\tan(\alpha)X-Y+y_0-\tan(\alpha)x_0|}{\sqrt{\tan^2(\alpha)+1}}$$
Let $\delta_w$ the distance of the fixed point from the line $L_w$, we have:
$$\delta_w=\frac{|-\frac{1}{\tan(\alpha)}X-Y+\frac{x_1}{\tan(\alpha)}+y_1|}{\sqrt{\left(\frac{1}{\tan(\alpha)}\right)^2+1}}$$


I am confused with my knowledge of polynomials? Is $x^2+cos(x)=0$ a polynomial?

Please help( I know its a naive question)

It is not. If it were, then
$$\cos(x) = x^2 + \cos(x) - x^2$$ whould be a polynomial as well, bus $\cos(x)$ has too many zeros ( infinite). A (non-zero) polynomial can only have a finite number of zeros.

It's not! Only powers of $x$ can appear in a polynomial, no trig functions. Also, technically, a polynomial is an expression not an equality.


Evaluate limit $\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}$ How to evaluate $$\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}?$$
Only idea I can think of is sandwich theorem, but then I get $0^0$.

I went ahead and solved this using Sandwich Theorem but took logarithm first:

Given y = $(1+ sin x)^{1/x}$

Taking logarithm on both sides:

log y = $(1/x) log (1 + sinx)$

We know that $-1< sin x < 1$

0 < $1 + sinx$< 2 (Adding 1 on all sides)

$(log 0)/x < log [1+ sin x]/x < (log 2)/x$  (Taking Log on both sides and dividing by $x$)

Now using LH Rule for computing the $\lim_{x \to \infty}$ $(log 0)/x$ :

We get the limit as 0 after differentiating the numerator and denominator. The numerator being a constant and denominator being differentitaed to 1, gives 0.

Also, $\lim_{x \to \infty}$ $(log 2)/x$ = $0$

Hence $\lim_{x \to \infty}$ $log [1+ sin x]/x$ = $0$ by Sandwich Theorem.

Since $logy$ = $(1/x) log (1 + sinx)$,
from above we have;

$\lim_{x \to \infty}$ $logy$ = $0$

$implies$ $\lim_{x \to \infty}$ $y$ = $e^0$ = $1$

Answer: $\lim_{x \to \infty}$ $(1+ sin x)^{1/x}$ = $1$


The norm $L ^ p$ of $f$ is bigger than the norm of $f$ minus its mean? Let $f \in L^p (\Omega)$ and $\tilde{f}=\frac{1}{|\Omega|}\int_{\Omega}f$, where $\Omega \subset R^N$ is open and bounded, I want to know if the following statement is true or false

$\| f- \tilde{f} \|_p \leq \| f \|_{p}$

I have tried with some functions and I think it is true. If the mean is zero there is nothing to do, in otherwise if $\tilde{f}$ is different to zero, I believe that the function $f-\tilde{f}$ will be closer to the reference axes so its integral will have a lower value. but I have no idea how to prove or refute the statement, if you could give me an idea. I have this

Lemm1. There exist C independent f such that $\| f- \tilde{f} \|_p \leq C \| f \|_{p}$

Proof: $\| f- \tilde{f} \|_p \leq \|f\|_p+\| \tilde{f} \|_p=\|f\|_p+|\tilde{f}| |\Omega|^{\frac{1}{p}}$
$$= \|f\|_p+|\Omega|^{\frac{1}{p}-1}|\int_{\Omega}f| $$
$$\leq \|f\|_p+|\Omega|^{\frac{1}{p}-1}\|f\|_1$$
Since $L^p \subset L^1$ we obtain the statement.

Thank you.
I'm sorry for not giving much in the solution. excuse my english.

Sorry initially I asked the contrary statement this was all my fault

I have this conjecture
$\| f- \tilde{f} \|_p = \inf_{k \in K}\| f +k\|_{p}$ False. See comments
Where K is the set of all constants.

Consider $L^1[0, 1]$ and $f(x) = 1+x$, then we see that
\begin{align}
\tilde f = \|f\|_1=\int^1_0 1+x\ dx = \frac{3}{2}
\end{align}
which means
\begin{align}
\|f-\tilde f\|_1 = \int^1_0 \left|x-\frac{1}{2}\right|\ dx = \frac{1}{4}.
\end{align}
Hence the stated inequality doesn't hold.

However, a better question would be: Does there exists $C>0$ depending only on $\Omega=[0, 1]$ such that
\begin{align}
\|f\|_1 \leq C\|f-\tilde f\|_1?
\end{align}


Why does L'Hopital not apply to $\lim_{x \to \infty} \frac{5x+3}{\sqrt{4x^2-x+7}}$? I am given the following limit:

$$\lim_{x \to \infty} \frac{5x+3}{\sqrt{4x^2-x+7}}$$

By simply dividing all factors by $x$, one can wind up with the result $\frac{5}{2}$. However, this is also a limit of the form $\frac{\infty}{\infty}$, thus one should be able to apply L'Hopital:

$$\lim_{x \to \infty} \frac{5}{\frac{8x-1}{2\sqrt{4x^2-x+7}}} = 0$$

seeing as we are dividing a constant number by something infinitely large. Where is the mistake in my logic here?

EDIT: Found the mistake after writing this question, will update post with correct answer if this is not already sufficiently answered when I'm finished calculating.

The application of L'Hospital's on radical  seems to be never ending.

$$\text{As }\lim_{x \to \infty} \frac{5x+3}{\sqrt{4x^2-x+7}}=\sqrt{\lim_{x \to \infty} \frac{(5x+3)^2}{4x^2-x+7}},$$

Now applying L'Hospital's
$$\lim_{x \to \infty} \frac{(5x+3)^2}{4x^2-x+7}=\lim_{x \to \infty} \frac{2(5x+3)\cdot5}{8x-1}$$ which is again of the form $\displaystyle\frac\infty\infty$

So applying L'Hospital's again, $$ \lim_{x \to \infty} \frac{10(5x+3)}{8x-1}=\lim_{x \to \infty} \frac{10\cdot5}8=\frac{25}4$$


A Proof relating to the Disjunctive normal form Here is a problem from Kenneth Rosen's Discrete Mathematics and its Applications, Section 1.3

  Suppose that a truth table in $n$ propositional variables is specified.
  Show that a compound proposition with this truth table can be formed
  by taking the disjunction of conjunctions of the variables or their
  negations, with one conjunction included for each combination of
  values for which the compound proposition is true. The resulting
  compound proposition is said to be disjunctive normal form.

I have been stuck in this question for quite some time and I would really appreciate a nudge in the right direction.

This is a homework problem and I am not expecting a complete answer. However, what I would like is to get a feel of what a Disjunctive Normal form is. I have been looking it up on the internet but no explanation is a good enough explanation for me. I would really appreciate an explanation.

Thanks a lot!

This example is for explanation of what a Disjunctive Normal form is,
On purpose i don't simplify the resulting formula,
The question is about homework and it should stay homework.

suppose you have the truth table:

P  Q    Value
T  T      T
T  F      T
F  T      F
F  F      T

and you want to transform that into a formula in the Disjunctive Normal form.

First add to the true values the formula that makes that row valid

  P  Q    Value   formula
     T  T      T     $ (P \land Q )$
     T  F      T     $ (P \land \neg Q )$
     F  T      F     ------------------ (this value is false so don't mention formula)
     F  F      T     $ (\neg P \land \neg Q )$

(sorry had some problems with the layout)

and then or them together

$ (P \land Q ) \lor (P \land \neg Q ) \lor (\neg P \land \neg Q ) $

And this formula is in Disjunctive Normal form.

notice this formula can be simplified but that is out of scope for this answer, see
http://en.wikipedia.org/wiki/Disjunctive_normal_form
and
http://en.wikipedia.org/wiki/Karnaugh_map
for that.

Good luck

Show that ¬ and ∧ form a functionally complete collection
of logical operators. [Hint: First use a De Morgan
law to show that p ∨ q is logically equivalent to
¬(¬p ∧ ¬q)


Circle inscribed between quarter circles - proving its center point Square $ABCD$ has a side length of $1$. $BGD$ and $AFC$ are quarter circles, and there is an inscribed circle between the quarter circles. Two points of tangency $F$ and $G$ are labeled.

How can I prove $O$ is the center of the inscribed circle?

I was thinking to show $EFGH$ is a rectangle (or equivalently an inscribed parallelogram). It is easy to see $FG$ and $EH$ are parallel to $AD$ by symmetry, but I cannot justify why $EF$ and $HG$ are parallel to $AB$ (well I thought it was obvious, but upon further thinking it didn't seem so obvious!).

The problem of solving for the radius of the circle has two solutions on Quora. However, I was having issues justifying the steps of the proofs. For example, the first proof seems to assume $O$ is the center of the circle. And the second states the three points of tangency in the circle form an equilateral triangle (which doesn't seem obvious to me).

It is definitely true $O$ is the center, but I'm at a mental block for proving this obvious thing. Any suggestions would be much appreciated, thanks!

Disclosure: I run the YouTube channel MindYourDecisions. I will give credit by linking to this thread in the "sources" for the video/blog post.

$\overline{AG}$ is perpendicular to $\bigcirc O$ (at $G$) and hence $\overline{EG}$ is one of its diameters.  Likewise for $\overline{DF}$.  Two distinct diameters always meet at the center. Q.E.D.

Note:  FareedAbiFarraj's demonstration that $\overline{OF} = \overline{OG}$ does not prove that $O$ is the center:

I had earlier solved this problem and found out the radius of the inscribed circle to be 3/8 of side of the square. If side of the square is taken as "a" then the co-ordinates of centre is (a/2,3a/8). I am taking point A as the origin. We can get the co-ordinates F and G by solving the equations of the inscribed and quarter circles. Then we can prove the colinearity of AOG and DOF.


Solving $\sin 2x + \cos 2x =-1$ $$\sin 2x + \cos 2x =-1$$

How would I go about solving this equation? Some hint on how to start, so I can try to figure it out on my own.
Thank you

$$\sin 2x+\cos 2x=-1$$
$$2\sin x\cos x+\cos^2x-\sin^2x+\sin^2x+\cos^2x=0$$
$$2\sin x\cos x+2\cos^2x=0$$
$$\cos x(\sin x+\cos x)=0$$
$$\cos x=0\Rightarrow x=\frac{\pi}{2}+k\pi,\tan x=-1\Rightarrow x=-\frac{\pi}{4}+k\pi$$

From $\sin 2x + \cos 2x =-1$ we get with $t=2x$:

$1=\sin^2 t+\cos^2 t+2 \sin t \cos t=1+ \sin(2t)$ .

Hence $ \sin(4x)=0$.


Proving the class of countable ordinals is closed under ordinal exponentiation in ZF I managed to prove that given the axiom of choice, the class (or is it a set?) of countable ordinals is closed under exponentiation, since the axiom of choice implies that the countable union of countable sets is countable (although this is not true in ZF).

I managed to prove that $\omega^\omega$ is countable in ZF by comparing it to the set of polynomials in $\omega$ with natural coefficients, but am struggling to extend it to more general ordinals.

Any help would be most welcome. Thank you in advance

You're right that in ZF, the countable union of countable sets need not be countable. However, the countable union of counted sets is always countable: that is, if I have countably many sets $A_i$, and a set of injections $f_i: A_i\rightarrow \omega$, then $\bigcup A_i$ is indeed countable. The weird examples come when I have (in some model of ZF) a countable collection $\{A_i\}$ of countable sets, but I don't have a set of injections of the $A_i$s into $\omega$.

So the goal is to make $\alpha^\beta$ explicitly counted. Specifically:

  Can you find an injection from $\alpha^\beta$ to $\omega$, given injections $f: \alpha\rightarrow\omega$ and $g:\beta\rightarrow\omega$?

HINT: it will be easier to do this if you work with the "explicit" description of ordinal exponentiation - that $\alpha^\beta$ is the set of all maps $\beta\rightarrow\alpha$ which are nonzero at only finitely many values, ordered lexicographically . . .

Note that the "given $f$ and $g$" can't be done away with: it is consistent with ZF that there is no map $H$ from $\{$countable ordinals$\}$ such that $H(\alpha)$ is an injection from $\alpha$ into $\omega$, that is, there's no "canonical" way to count each countable ordinal. (Indeed, such an $H$ existing implies that the union of countably many countable ordinals is countable - that is, that $\omega_1$ is regular!)

The following argument is overkill and probably not helpful to you. However, I thought I'd add it anyways - just for the heck of it.

It's easy to verify that ordinal arithmetic (i.e. addition, multiplication, exponentiation) is absolute between transitive models of $\operatorname{ZF}^{-}$. In particular, for any $\alpha, \beta < \omega_{1}$, we may calculate $\alpha^\beta$ in $L_{\omega_1}$ and obtain the correct value. Since $L_{\omega_1} \cap \operatorname{On}= \omega_1$, this implies that $\alpha^\beta$ is countable.


Find the solution set for $[\sin^{-1}x]>[\cos^{-1}x]$, where $[.]$ is greatest integer function The values of $[\sin^{-1}x] \in [-2,1]$ and $[\cos^{-1}x] \in [0,3]$
How do i solve it from here? I have no idea on how to go further. I am not able to convert the graphs to GIFs, and I can’t do it algebraically either.

Note that
$$\lfloor \sin^{-1}x \rfloor=\begin{cases}1, & \sin 1\le x\le 1 \\ 0,& 0\le x\lt \sin 1 \\ -1,& -\sin 1\le x\lt 0 \\ -2,& -1\le x\lt -\sin 1 \end{cases} $$ and
$$\lfloor \cos^{-1}x \rfloor=\begin{cases} 0, &\cos1\lt x\le 1 \\ \vdots\end{cases} $$ We don't need to worry about the other values, as they will turn out to be $\ge 1$, but $\sin^{-1} x\le 1$. Your inequality will only be true, when $$\lfloor \sin^{-1} x\rfloor =1  \land \lfloor \cos^{-1} x\rfloor =0$$ That is, we need to take the intersection of the range of values for which the two equalities hold. The answer is hence $$[\sin 1,1] \cap (\cos 1, 1]\\=\color{purple}{[\sin 1, 1]} \\ (\because \sin 1\gt \cos 1) $$

Not a complete solution
For $x\le0,[\sin^{-1}x]\le-1$  and $[\cos^{-1}x]>\left[\dfrac\pi2\right]=1$
Clearly we need $[\sin^{-1}x],[\cos^{-1}x]\in [0,1]$
$1=[\sin^{-1}x+\cos^{-1}x]\ge[\sin^{-1}x]+[\cos^{-1}x]>2[\cos^{-1}x]$
$\implies [\cos^{-1}x]<\dfrac12$
But $0\le[\cos^{-1}x]\le3\implies[\cos^{-1}x]=0\implies0\le \cos^{-1}x<1\iff x>\cos1\ \ \ \  (1)$
Again  as $[\cos^{-1}]x\ge0,[\sin^{-1}x] $ must be $>0\iff[\sin^{-1}x]\ge1\iff x\ge\sin1\ \ \ \  (2)$
Now $\sin1-\cos1=\sqrt2\sin\left(1-\dfrac\pi4\right)>0,$ so by $(1),(2), x\ge\sin1$ which is obviously a necessary condition
I am trying to find the sufficient condition.


Integrating $\int_0^ex^{1/x}\ \mathrm dx$ Compute
$$\int_0^ex^{1/x}\;\mathrm dx.$$

There is an analytical anti-derivative found in this answer. How does one compute this?

Using the anti-derivative approach we have

$$\int x^{1/x}\;\mathrm d x=x + \frac{\log^2x}{2}-\sum^\infty_{n=2}\sum^n_{k=0}\frac{\log^{n-k}x\;}{x^{n-1}(n-k)!(n-1)^{k+1}}+C$$

Now there is a problem for this anti-derivative as it gets near $0$ (apparent from the $\log$ s). I do not know how to prove this but if $f(x)=x^{1/x}$ and $\lim_{x\to0}f(x)=0$ can be defined then I want to say that $\lim_{x\to0}F(x)=0$. Assuming my non-rigorous logic and some logic from this question we have $$\int_0^ex^{1/x}\;\mathrm dx=\sum _{n=2}^{\infty } \frac{ \Gamma (n+1,\;n-1)}{(n-1)^{n+1}\;n!}+\frac{1}{2}+e$$

Update: Using Mathematica I computed some numerical integrals to find that $F(0)\approx1.53328$. Therefore we have $$\int_0^ex^{1/x}\;\mathrm dx=\sum _{n=2}^{\infty } \frac{ \Gamma (n+1,\;n-1)}{(n-1)^{n+1}\;n!}+\frac{1}{2}+e-\lim_{x\to0}\int x^{1/x}\;\mathrm dx$$

Here is a way a physicist is approaching the problem.

First, the variable change $x=\frac{1}{y}$ gives:

$$\int_0^ex^{1/x}\;\mathrm dx=\int_{\frac{1}{e}}^{\infty}\frac{y^{-y}}{y^2}dy$$

Now, as a first approximation, we use the following inequality:

$$\int_{\frac{1}{e}}^{\infty}\frac{y^{-y}}{y^2}dy<\int_{\frac{1}{e}}^{\infty}\frac{y^{-\frac{1}{e}}}{y^2}dy=\int_{\frac{1}{e}}^{\infty}y^{-\frac{1}{e}-2}dy=\frac{e^{2+\frac{1}{e}}}{e+1}$$

Finally:

$$0<\int_0^ex^{1/x}\;\mathrm dx<\frac{e^{2+\frac{1}{e}}}{e+1};\;e>0$$

This result can be probably improved.

The answer is approximately 2.66183. I hope this helps. I evaluated it, but I am new to the website and wasn't sure how to write out the calculations.


Asymptotic behavior of $-gTt-gT^2e^{\frac{-t}{T}}$ for small $t$ I want to solve this using Taylor series expansion of $e^{f(x)}$

$$\begin{align}x=-gTt-gT^2e^{\frac{-t}{T}}+gT^2+x_0\end{align}$$
Show that for small values of t $(t\ll T)$, the equation for x is approximately $x\approx{x_0-\frac{1}{2}gt^2}$ , where T is a constant

Please help.

Using Taylor Series You get:

$e^{-t/T} \approx 1 - t/T + 0.5 t^2 / T^2$


Some questions about proof of Theorem 2.43 in Baby Rudin I will include the proof here and highlight the parts that are giving me trouble.

  Theorem $\hspace{5 pt}$ Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.

  Proof $\hspace{5 pt}$ Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots$. We shall construct a sequence $\{V_n\}$ of neighborhoods as follows.

  Let $V_1$ be any neighborhood of $x_1$.

1) ^ Are we using the Axiom of Choice here? How can we have an arbitrary set in a construction?

  If $V_1$ consists of all $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y} - \mathbf{x_1}| < r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y} - \mathbf{x_1}| \leq r$.

2) ^ It makes sense intuitively, but how do we prove this last statement?

  Suppose $V_n$ has been constructed, so that $V_n \cap P$ is not empty. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $x_n \notin \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P$ is not empty. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.

3) ^ I really don't get this whole paragraph much at all. Could someone explain it in a more step-by-step way?

  Put $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact.

4) ^ "closed" comes from it being a closure and "bounded" comes from the definition of neighborhood, correct?

  Since $x_n \notin K_{n+1}$, no point of $P$ lies in $\bigcap_1^\infty K_n$. Since $K_n \subset P$, this implies that $\bigcap_1^\infty K_n$ is empty. But each $K_n$ is nonempty, by (iii), and $K_n \supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.

$1):$ I don't think AC is used here. We just look at all possible neighborhoods of $x_1$ and pick one. After that, the construction can proceed.

$2):$ If $(y_n)$ is a sequence in $V_1$ and $y_n\rightarrow y$, then for all $\epsilon>0$ there is an $y_n$ such that: $|x_1-y|\le |x_1-y_n|+|y_n-y|\le r+\epsilon$. Then we must also have $|x_1-y|\le r$.

$3):$ Since $x_n$ is a limit point of $P$ and $V_n$ a neighborhood of $x_n$, there is a $p\in V_n\cap P$ such that $p\neq x_n$. Since $V_n$ is open, there is an $\epsilon$ such that $B_{\epsilon}(p)\subset V_n$. Put $\delta:= \min \left\{\frac{\epsilon}{2},\frac{|p-x_n|}{2}\right\}$. Then $x_n\not\in \overline{B_{\delta}(p)}$ and $\overline{B_{\delta}(p)} \subset V_n$. So we define $V_{n+1}:=B_{\delta}(p)$. Since $p\in V_{n+1}$, $p$ is a limit point of $P$ and $V_{n+1}$ is open, we have $V_{n+1}\cap P\neq \emptyset$.

$4):$ Yes, the sets are closed and bounded since they are closed balls with radius $r$.

My proof here includes a step-by-step construction of $V_{n+1}$ for step 3:

Proof of Baby Rudin Theorem 2.43

You might find it helpful. Step 3 is what most people (myself included) found puzzling about 2.43 when they first encountered it.


Question about Certain Upper Bounds of the Sum of Divisors Function It has been proven that $$\limsup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n}=e^\gamma$$ where $\sigma(n)$ is the sum of divisors and $\gamma$ is Euler's constant ($0.5772...$). The way I understand it is that for every $\epsilon$, there is some $b$, so that when $n>b$, $(e^\gamma + \epsilon)n \ln\ln n>\sigma(n)$. Is there a way to find such smallest $b$ for every $\epsilon$ to make above inequality true (that is, for any smaller $b$, the inequality wouldn't hold)? If there is not a way to find the exact smallest $b$, is there at least a way to find a "tight" estimate/upper bound (any smaller estimate/upper bound wouldn't work) on such smallest $b$?

We have the following (unconditional) inequality due to Robin:
$$\sigma(n)<e^\gamma n\log\log n+\frac{0.6483 n}{\log\log n}$$
for $n\geq 3$. Therefore, to have the inequality you ask it's enough to have
$$e^\gamma n\log\log n+\frac{0.6483 n}{\log\log n}\leq(e^\gamma+\varepsilon)n\log\log n.$$
Elementary manipulations show that this is equivalent to
$$n\geq e^{e^{\sqrt{0.6483/\varepsilon}}},$$
so we may take the right-hand side as the upper bound for $b$, provided it's at least $3$.

There are better inequalities known nowadays, for instance the following due to Axler (A new upper bound for the sum of divisors function) gives
$$\sigma(n)<e^\gamma n\log\log n+\frac{0.1209 n}{(\log\log n)^2},$$
for $n>2520$, and we can similarly derive an upper bound for $b$:
$$b\leq e^{e^{\sqrt[3]{0.1209/\varepsilon}}}$$
if the right-hand side is at least $2520$.

As for finding the exact value of $b$, this already shows that this is in principle algorithmically possible, since we just have to find the largest $n$ smaller than the given bound for a given $\varepsilon$.

If we admit conjectural bounds, by results of Robin under RH the inequality holds for $\varepsilon=0$ for $n>5040$, and therefore $b\leq 5040$ for all $\varepsilon$, and for all small enough values of $\varepsilon$ this is exact. Therefore under RH, $b=5040$ for small enough $\varepsilon$.

The definition of $\limsup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n}=e^\gamma$ is $$f(n) = \sup_{m \ge n} \frac{\sigma(m)}{m \ln \ln m}, \qquad \lim_{n \to \infty}  f(n) = e^{-\gamma}$$
$\forall \epsilon>0, \exists b,\forall n \ge b,| f(n) -e^{-\gamma}|<\epsilon$ is the definition of $\lim_{n \to \infty}  f(n) = e^{-\gamma}$.
$\lim_{n \to \infty}  f(n) = e^{-\gamma}$ follows from Mertens theorems.
Robin Nicolas Lagarias proved that the RH holds iff $\forall n,f(n) = e^{-\gamma}$.
Thus $f(n)$ is computable iff the RH is (dis)provable. If it is computable then for a fixed $\epsilon$ the smallest $b$ is computable.


Can we solve a simple set of equations in radicals? Suppose  $x,y,z$  are non-zero coprime integers with $x+y \ne z$ and which satisfy the following three equations:

$\operatorname{rad}(x)=\operatorname{rad}(z-y)$
$\operatorname{rad}(y)=\operatorname{rad}(z-x)$
$\operatorname{rad}(z)=\operatorname{rad}(x+y)$

The radical of $n$ is defined as the product of all prime divisors of $n$. Hence $\operatorname{rad}(n)=\prod_{p|n}{p}$. For instance $\operatorname{rad}(20)=2.5=10$. We extend the definition by defining $\operatorname{rad}(0)=0$, $\operatorname{rad}(1)=1$ and $\operatorname{rad}(-n)=\operatorname{rad}(n)$.

There are endless no coprime solutions. For instance choose $x=\operatorname{rad}(2^n \mp 1)$ with $n>1$, and consider $(x,x,\pm2^nx)$.

It is not difficult to check that if $(x,y,z)$ is a solution, also $(-x,-y,-z), (y,x,z),(z,-x,y)$ and $(-y,z,x)$ are solutions.

Beside these symmetries, can we prove that $(5,27,2)$ is the only coprime solution?

I found a general expression for a solution. For the sake of comprehensibility, I reduce the solution through reverse reasoning. As shorthand we write $r_n=\operatorname{rad}(n)$.

I use two -easy to check- observations.
1) Let $n=\prod_i{p_i^{e_i}}$ with $p_i$ prime. Choose $m=\max(e_i)$, then $n$ divides $(r_n)^m$. Hence, for some integer $s$ we have $(r_n)^m=sn$.
2) If $a$ and $b$ are integers with radical $r_a$ and $r_b$, then $s=\frac{r_a}{\gcd(r_a,r_b)}.\frac{r_b}{\gcd(r_a,r_b)}$ is the smallest non-negative integer such that $\operatorname{rad}(sa)=\operatorname{rad}(sb)$.

Suppose $(x,y,z)$ is a solution. Using the first observation, there exist pairs of non-zero integers $(a,b),(c,d),(e,f)$ such that $ax=b(z-y)$,  $cy=d(z-x)$ and  $ez=f(x+y)$. We can assume that the pairs $(a,b),(c,d),(e,f)$ are coprime by removing the common factor.
In matrix form we have:

\begin{equation*}
\begin{pmatrix}  a & b & -b \\ d & c &-d \\ -f & -f  & e \end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
=
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}

Hence, the determinant is zero which implies $e(ac-bd)=f\big(b(c-d)+d(a-b)\big)$. It is straightforward to verify that an eigenvector is equal to:
\begin{pmatrix} b(c-d) \\ d(a-b) \\ ac-bd \end{pmatrix}
Note that $ac-bd \ne 0$ as $ac=bd$ implies $axcy=b(z-y)d(z-x)=acz(z-x-y)+acxy$ thus $z(z-x-y)=0$ which is not possible.

Hence, we conclude that $(x,y,z)=\lambda \big( b(c-d),d(a-b),ac-bd \big)$. Using the second observation, we get an expression for $\lambda$. Define $\lambda=lcm(s_x,s_y,s_z)=lcm(\frac{r_{ab}}{\operatorname{rad}(\gcd(ab,c-d))}, \frac{r_{cd}}{\operatorname{rad}(\gcd(cd,a-b))},$ $\frac{r_{ef}}{\operatorname{rad}\big(\gcd(ac-bd,b(c-d)+d(a-b))\big)})$. We can divide $\lambda$ by $\gcd\big(b(c-d),d(a-b),ac-bd\big)$ to avoid redundancy.

Note that the parameters $e$ and $f$ are completely determined by the parameters $(a,b),(c,d)$. From the relation $e(ac-bd)=f\big(b(c-d)+d(a-b)\big)$ and the fact that $e,f$ are coprime, we conclude that $e=\pm \frac{b(c-d)+d(a-b)}{\gcd \big(ac-bd,b(c-d)+d(a-b) \big)}$ and $f=\pm \frac{ac-bd}{\gcd \big(ac-bd,b(c-d)+d(a-b) \big)}$. The sign of $e,f$ is determined by $\operatorname{sign}(a-b)=\operatorname{sign}(f-e)$.

Hence, we need two pairs of coprime non-zero integers $(a,b),(c,d)$ with $ac-bd \ne 0$ and $bc+ad \ne 2bd$ to construct a solution. However, the above general expression does not help me to prove that $(5,27,2)$ is the only coprime solution.


I need help with constructing a "3D" cone out of paper with precise angles. I just can't seem to wrap my head around it. - I need to make an accurate and precise model of a cone out of paper, with it's angle around the cap/spike equal to $90^° $ - I know it seems obvious but getting the Side Area to stick itself into a 3D model is going very wrong so far for me - even though I used this relationship:$$\theta = 2\pi \sin\frac\alpha2$$ image link

Sorry, my mistake. Corrected answer

$ L =\frac{R}{sin(\alpha/2)}$ . Cut a sector of circle with radius $L$ and angle it makes

$ \theta = 360^o \dfrac{R}{L}$

See the figure.. It explains. Take a rectangular paper with one side $2 \pi R$.

Now from cone figure

$ L =\frac{R}{sin(\alpha/2)}$

Now from paper figure what is $P$?

Simply

$P = \sqrt{L^2-(\pi R)^2}=\sqrt{\frac{R^2}{sin^2(\alpha/2)^2}-(\pi R)^2}$

$P = R \sqrt{cosec^2(\alpha/2)-\pi ^2}$

Draw a perpendicular bisector in paper with length $P$, Complete triangle , cut and glue. :)


Time-based probability question Two adult male baboons are introduced to the same $50$ ft. square cage.  Male A looks at Male B for a total of $5$ hours in the first ($24$ hour) day, and Male B looks at Male A for a total of $3$ hours in the same first day.  If the two baboons behave independently of each other, what is the probability that they will look at each other at the same time (at least once) on that first day?

My thinking: I think this can be solved geometrically, but I am having trouble interpreting it. Can I interpret it to $0 < B < 5$ and $0 < A < 3$ and $A-B > 0$?

Another solution I had for this question is that: Two baboons only look together if A's $5$ hour is happening within the $3$ hours of B looking at A. So the probability would be $\frac{3}{5}\cdot\frac{5}{24}=\frac{1}{8}$? I think it is wrong though.

Decomposing the day into $n$ equal timeslots with $n$ large and assuming each animal chooses randomly uniformly and independently the necessary number of timeslots, one gets a probability of no eye contact approximately equal to $\mathrm e^{-n/10}.$

Let $u=5/24$ and $v=8/24$, then Male A looks at Male B during $a$ timeslots, with $a=un$, and Male B looks at Male A during $b$ timeslots, with $b=vn$. Fix the $a$ timeslots used by Male A. No eye contact happens if Male B uses $b$ distinct timeslots taken from the $n-a$ remaining "free" timeslots. Consider that Male B chooses his $b$ timeslots sequentially. The first timeslot Male B chooses is free with probability $(n-a)/n$. Conditionally on this, the second timeslot Male B chooses is free with probability $(n-a-1)/(n-1)$, and so on until timeslot number $b$, which is free with conditional probability $(n-a-b+1)/(n-b+1)$. Thus, the exact probability of no eye contact is $$p=\prod_{i=0}^{b-1}\frac{n-a-i}{n-i}=\frac{(n-a)!\,(n-b)!}{n!\,(n-a-b)!}.$$ When $u$ and $v$ are fixed and $n\to\infty$ with $a=un$ and $b=vn$, $$\frac{\log p}n\to-\theta=(1-u)\log(1-u)+(1-v)\log(1-v)-(1-u-v)\log(1-u-v),$$ hence one can approximate $p$ as $$p\approx\mathrm e^{-\theta n}.$$
If $u=5/24$ and $v=8/24$, $\theta\approx.09768249$. For a decomposition into hours, $n=24$ hence $p\approx9.6\%$. For a decomposition into minutes, $n=1440$ hence $p\approx10^{-61}$.
For a decomposition into seconds, $n=86,400$ hence $p\approx10^{-3665}$.

Having said that (which solves the problem as interpreted by the OP), my impression is that the real question assumes some continuous periods of 5 hours, respectively 3 hours, in the day, and asks for the probability of no eye contact, that is, that the time intervals corresponding to these periods do not intersect.

This happens if and only if the midpoints $X$ and $Y$ of the intervals are at distance more than $4$, $X$ being uniform on $(2.5,21,5)$ and $Y$ uniform on $(1.5,22.5)$. Equivalently, one can consider $X$ uniform on $(0,19)$ and $Y$ uniform on $(0,21)$. Then $p$ is the normalized sum of the area of the two triangles $X\gt Y+4$ and $Y\gt X+4$, that is, $$p=\frac{\frac1217^2+\frac1215^2}{19\cdot21}=\frac{257}{399}\approx64\%.$$

When I revisit this question:

Let event A be that the male baboon A looking at the male baboon B. P(A)=5/24

Let event B be that the male baboon B looking at the male baboon A. P(B)=3/24

So P(A and B)=5/24*3/24=5/192


Showing identity of ideals in $\mathbb{Q}[x,y]$ Problem: Let $R=\mathbb{Q}[x,y]$ and consider the ideal
  $$I=\{f\in R:f(0,0)=f(1,1)=0\}.$$ Prove that $I=\langle x-y,y-x^2\rangle.$

Since $x-y$ and $y-x^2$ belong to $I$ we have $\langle x-y,y-x^2\rangle\subset I.$
To show the other relation I don't know even how to start.

To start take any element $f$ of $I$ which is of the form $a_0+a_1x+a_2x^2+....+a_nx^n$ where $a_i\in \mathbb Q[y]$ and $f(0,0)=f(1,1)=0$

Again note that $a_i=b_0+b_1 y+b_2y^2+.....+b_ny^n$ where $b_i\in \mathbb Q$.

Apply the two conditions you will that $b_0=0$ from one of them .

See  if you can make use of the other one and make $f$ of the form $f=(x-y)g(x,y)+(y-x^2)h(x);g(x,y),h(x,y)\in \mathbb Q[x,y]$


Can a function $f$ have $L^p$ norms $\Vert f\Vert_p =p$ for all $1\leq p<\infty$? I have tried to show that such a function must be in $L^{\infty}$ and thus it is impossible for such a function to exist since, in that case, $$\infty =\lim_{p\rightarrow \infty} \Vert f\Vert_p=\Vert f\Vert_{\infty}.$$
Basic estimates seem to fail and I can't seem to construct a counterexample.

What if we replace $p$ with something that grows at different rates, like $e^p$ or $\log p$?

There are two cases to consider. Let's first start with $f\in L^p(X)$ where $X$ has finite Lebesgue measure. The $L^p$ spaces are nested. We have that if $f\in L^q(X)$, then $||f||_{L^p}\leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}||f||_{L^q}$

If $||f||_{L^q}=q$ then $\mu(X)^{\frac{1}{p}-\frac{1}{q}}*q$

We want to show that $||f||_{L^p}=p$, if this is true, then $p\leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}*q$. We can rewrite this as $\frac{p}{q}\leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}$. We know $f=0$, which has $L^p$ norm equal $0$ cannot be such a function. For now, lets just take $p=2$ and $q=3$.

We have $\frac{2}{3}\leq \mu(X)^{\frac{1}{6}}$ and so $\frac{2^6}{3^6}\leq \mu(X)$. This means that in order for a function $f$ to have $||f||_{L^2}=2$ and $||f||_{L^3}=3$ we must have the $\frac{2^6}{3^6}\leq \mu(X)$. You can toy around with other $p,q$ values to establish similar bounds.

That said, I am not sure such a function can be constructed. Note that if $||f||_{L^p}=p$, then let's start with $p=1$. As an example, conside $f=exp(2\pi ix)$ this has $L^1([0,1])$ norm of $1$, and even if we insert some scaling factor $aexp(2\pi ix)$, it is impossible to get the desired value for the other $L^p$ norms.

If you are unsure of the algebra, recall you can pull the constant out during integration and so we would need $a^p=p$ and only $a=0,1,$ solve this, and so it is not possible for our function.

I think it is possible to show that no such $f$ can exist, but a clean proof of this evades me at the moment.

The second case is when $\mu(X)=\infty$, and here we don't have nesting of $L^p$ spaces. While I haven't worked it out, I don't think such an $f$ can exist here either.

The magic words are: Riesz-Thorin interpolation Theorem.


A fair 6-sided die is rolled 6 times, what's the probability the outcome has exactly 2 or 3 elements? A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements?

I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! :)

Exactly two numbers appear:

There are $\binom{6}{2}$ ways for two of the six numbers to appear.  On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear.  However, of these $2^6$ sequences, there are two in which the same number is rolled six times.  Thus, the probability of exactly two numbers appearing in six rolls of the die is
$$\frac{\binom{6}{2} \cdot (2^6 - 2)}{6^6}$$

Exactly three numbers appear:

There are $\binom{6}{3}$ ways for three of the six numbers to appear.  On each of the six rolls of the die, there are three possible outcomes, giving $3^6$ sequences containing these three numbers.  However, we have counted the sequences in which not all three of the numbers appear.  There are $\binom{3}{2}$ ways for two of the numbers to appear.  For each such pair, there are $2^6$ sequences, of which two consist of sequences in which one number is rolled all six times.  Thus, there are $\binom{3}{2}(2^6 - 2)$ sequences in which exactly two of the three numbers appear.  The number of sequences in which exactly one of the three numbers appears is $3$.  Hence, the probability that exactly three of the numbers appear is
$$\frac{\binom{6}{3}[3^6 - \binom{3}{2}(2^6 - 2) - 3]}{6^6}$$

My thanks to @bof for clarifying my thinking about the first problem.  Any errors that remain are entirely my responsibility.

Why can't we use the multinomial formula? We want to have a sequence of 2 repeated elements 3 each. This sounds like $\frac{6!}{3!*3!}$, right?
For example we have {1,1,1,2,2,2} or a combination of this. Then we get $\frac{6!}{3!*3!}$ since we have 2 sets of 3 repeated elements and then we divide by the sample space times 6 (there are six possibilities). The result seems to be incorrect using this method. I tried using a 3^3 tree and it seemed ok. Any thoughts?


Prove that $n! > n^5$ I'm trying to prove that $n! > n^5$ for large enough values of n. While it seems obvious that this should be true, I have no idea how to prove it rigorously.

EDIT:

So, looking at the comments, here's what I'm thinking now. What do folks think? Is the following reasonable?

The idea is to take the limit of the ratio of $(n+1)!/n!$ vs $(n+1)^5/n^5$ to show that the former is much larger than the latter, and thus that when n > M, where M is some positive integer, $n!$ must be growing much larger than $n^5$:

Let us assume $\frac{(n+1)!}{n!} > \frac{(n+1)^5}{n^5}$

$$
\lim_{n\to\inf} \frac{(n+1)!}{n!} > \frac{(n+1)^5}{n^5}\\
= \lim_{n\to\inf} n+1 > (\frac{n+1}{n})^5\\
= \lim_{n\to\inf} n+1 > (1 + \frac{1}{n})^5\\
\ldots
=\lim_{n\to\inf} n+1 > (1 + 3/n + 8/n^2 + 10/n^3 + 5/n^4 + 1/n^5)\\
= inf > 1
$$

(Edited [again]: I did not know you could use latex)

For $n \geq 10$, we have
$$n! = n(n-1) \cdots (n-4)(n - 5)! \geq \left(\frac{n}{2}\right)^5 (n - 5)! = n^5\left(\frac{(n -5)!}{32}\right) > n^5.$$
(The bound above is not sharp, but it's sufficient for the problem. Also, the same argument shows that for any $p > 0$, we have $n! > n^p$ for sufficiently large $n$.)

The first derivative of those should be useful in your proof.  Start with n = $10$ for example.

Someone asked me to explain further but I am not a math person so it will be a simple explanation.

For n = $10$ we have $10$! vs. $10^5$ which is $3,628,800$ vs. $100,000$ so it is already larger at the point.  If we increase n to $11$, the $10$! increases by a factor of $11$ but the $10^5$ increases by $1.1^5$ which is only about $1.61$.  If we increase n to $12$ you can see a "breakaway" between them so just prove this breakaway by showing the "discrete 1st derivative" of n! is increasing greater than that of $n^5$ and you got it.


Riemann Integrals converging to $0$ Let's assume that $\{f_n\}$ is a sequence of real,continuous,nonnegative functions on $[0,1]$ such that $$\int_0^1f_n(x)dx\overset{n\to\infty}{\longrightarrow} 0$$
I need to prove, or disprove that there exist points  $x_o\in[0,1]$ such that $\lim\limits_{n\to\infty}f_n(x_0)=0$.

I feel that there must exist such points. So, suppose on the contrary that $\forall x\in[0,1] ,f_n(x)\nrightarrow0$. Taking an $x$ we can find a subsequence $\{f_{k_n}\}$ of $\{f_n\}$ such that $f_{k_n}(x)\geq\varepsilon$ for some $\varepsilon >0$ and $\forall n\in \mathbb N$. But how can this work? I can find an area around $x$ in which $f_{k_n}$ is away from zero , but what about when $n$ gets larger? The problem is that $f_{k_n}(y)$ can still converge to zero for $y\neq x$ in that area, and I am not getting a contradiction since I am working on a subsequence.

Do you think there can be a counterexample?

This is false. To see why not, try constructing a counterexample in the following way (the counterexample I give is not continuous, but you can make it continuous with a modification).

For notational ease, I'll write this as a kind of array of functions, which you can make into a linear sequence in a clear way. Write $f_{0,1}(x) = 1$ identically, and
$$
f_{1,1}(x) = \begin{cases}1 & x \in [0,1/2)\\ 0 & x \in [1/2, 1] \end{cases},~~
f_{1,2}(x) = \begin{cases}0 & x \in [0,1/2)\\ 1 & x \in [1/2, 1] \end{cases}
$$
Generally, for $k > 1, 1 \leq l < 2^k$, we define
$$
f_{k,l}(x) = \begin{cases}1 & x \in [(l-1)2^{-k},l 2^{-k})\\ 0 & \text{else} \end{cases}, ~~ f_{k,2^k}(x) = \begin{cases}1 & x \in [1-2^{-k},1]\\ 0 & \text{else} \end{cases},
$$
For $k$ fixed, varying $l$ has the effect of shifting around the 'bump' of width $2^{-k}$. Writing $g_1 = f_{0,1}, g_2 = f_{1,1}, g_3 = f_{1,2}$ and so on, we have now a sequence of functions $g_n$ for which
$$
\int_0^1 g_n dx \rightarrow 0
$$
Can you see why there is no point $x_0 \in [0,1]$ for which $g_n(x) \rightarrow 0$? Moreover, do you see how to turn this into a counterexample for when the $g_n$ are continuous?

because $f_n: [0,1] \to \mathbb{R}_{\geq0}$, $f_n \to f$ in L1  $\implies f_n \to 0$ in measure.  This however is not enough.  You can come up with $f_n$ s.t. $f_n \notin L_\infty $ over any subset of [0,1] and have it converge to 0 in $L_1$  You can have subsequences with this property though.


On a limit exercise $ \lim_{x \rightarrow 0} \ln \Big( \frac{\sin x e^{x/ (x+1)} - x}{x \sin x} \Big)/ \arctan x $. I want to solve this limit

$$ \lim_{x \rightarrow 0} \ln \Big( \frac{\sin x e^{x/ (x+1)} - x}{x \sin x} \Big)/ \arctan x $$

So this is what I did:

$$\Bigg( \frac{\sin x e^{x/ (x+1)} - x}{x \sin x} \Bigg) = \Bigg( \frac{ (x -1/6x^3 + o(x^3)) (1+ x/(x+1) + (x/2(x+1))^2 + o(x^2)) - x}{x \sin x} \Bigg)$$

$$= \Bigg( \frac{x^2 (x+1) 6 +3x^3 - x^3 (x+1)^2 + o(x^3)}{(x+1)^2 6 x^2(1 -1/6 x^2 + o(x^2)   )} \Bigg) = \Bigg( \frac{1+ 4/3x +o(x)}{(x+1)^2 (1 -1/6 x^2 + o(x^2)   )} \Bigg) $$

So now $$\ln \Big( \frac{\sin x e^{x/ (x+1)} - x}{x \sin x} \Big) = \ln(1+ 4/3x +o(x)) -\ln(x+1)^2 - \ln(1 -1/6 x^2 + o(x^2))   $$

and if I expand the arctan and these logs with Taylor I obtain that the limit is equal to $4/3$ except the limit should be equal to $-2/3$.

When I taylor expand first in the exponent of $e ^{x / (x+1)}$  I get the correct result, but what I am doing above should be correct too right?

Could someone kindly tell me where my mistake is?

Using theorem of arithmetic borders:
$$\lim_{x→0}\frac{\ln\left(\frac{\sin x \exp\left(\frac{x}{x+1}\right)-x}{x\sin x}\right)}{\arctan x}=\lim_{x→0}\frac{\ln\left(\sin x \exp\left(\frac{x}{x+1}\right)-x\right)-\ln\left(x\sin x\right)}{\arctan x}$$
We calculating more simply limit:
$$\lim_{x→0} \frac{1}{\arctan x}=\frac{1}{\arctan 0}=\frac{1}{0}=\infty$$
Now we calculating limit:
$$\lim_{x→0}\frac{\ln\left(\sin x \exp\left(\frac{x}{x+1}\right)-x\right)-\ln\left(x\sin x\right)}{\arctan x}=\left[\frac{0}{0}\right]=\lim_{x→0}\frac{\left(\ln\left(\sin x \exp\left(\frac{x}{x+1}\right)-x\right)-\ln\left(x\sin x\right)\right)^\prime}{\left(\arctan x\right)^\prime}=\lim_{x→0}\frac{\frac{1}{\exp\left(\frac{x}{x+1}\right)\sin x-x}\left(-\frac{\left(\exp\left(\frac{x}{x+1}\right)\sin x-x\right)\csc x}{x^2}+\frac{\csc x\left(\exp\left(\frac{x}{x+1}\right)\left(\frac{1}{x+1}-\frac{x}{\left(x+1\right)^2}\right)\sin x+
\exp\left(\frac{x}{x+1}\right)\cos x-1\right)}{x}-\frac{\left(\exp\left(\frac{x}{x+1}\right)\sin x-x\right)\cot x\csc x}{x}\right)}{\frac{1}{x^2+1}}=\frac{\frac{1}{\exp\left(\frac{0}{0+1}\right)\sin 0-0}\left(-\frac{\left(\exp\left(\frac{0}{0+1}\right)\sin 0-0\right)\csc 0}{0^2}+\frac{\csc 0\left(\exp\left(\frac{0}{0+1}\right)\left(\frac{1}{0+1}-\frac{0}{\left(0+1\right)^2}\right)\sin 0+
\exp\left(\frac{0}{0+1}\right)\cos 0-1\right)}{0}-\frac{\left(\exp\left(\frac{0}{0+1}\right)\sin 0-0\right)\cot 0\csc 0}{0}\right)}{\frac{1}{0^2+1}}=\frac{-\frac{2}{3}}{\frac{1}{1}}=-\frac{2}{3}$$


To express a vector as a linear combination of vectors, do the vectors need to linearly independent? For example, write (-3,4,1) as a linear combination of (1,2,1) (8,1,2) and (4,3,2)

Is this possible considering these vectors are not linearly independent?
I ask this because when constructing these into a matrix and running a reduced row echelon form, I do not have a clear answer, and I figure this is the reason.

Thank you for the help.

In order to determine whether it is possible to write $(-3,4,1)$ as a linear combination of $(1,2,1), (8,1,2),$ and $(4,3,2),$ the standard approach is to put these vectors together as the columns of a matrix, then row reduce. In particular, we have
$$
\left(\begin{array}{cccc} 1 & 8 & 4 & -3\\ 2 & 1 & 3 & 4\\ 1 & 2 & 2 & 1 \end{array}\right)
\leadsto \left(\begin{array}{cccc} 1 & 0 & \frac{4}{3} & \frac{7}{3}\\ 0 & 1 & \frac{1}{3} & -\frac{2}{3}\\ 0 & 0 & 0 & 0 \end{array}\right).
$$
Because the last column of the row-reduced matrix is not a pivot column, it is possible to write $(-3,4,1)$ as a linear combination of the other three vectors.  In particular: we see that in the row-reduced matrix, the last column can be written as $\frac 73$ of the first column added to $- \frac 23$ of the second column.  Because this holds for the row-reduced matrix, it also holds for the original matrix. That is, we have
$$
(-3,4,1) = \frac 73 \cdot (1,2,1) - \frac 23 \cdot (8,1,2).
$$

Not exactly.

In your example, suppose the coeffients are a b and c.
Then we want to solve the equation:
$$
\begin{bmatrix}
1 & 8 & 4\\
2 & 1 & 3\\
1 & 2 & 2
\end{bmatrix}
\begin{bmatrix}
a\\b\\c
\end{bmatrix}=
\begin{bmatrix}
-3 \\4 \\1
\end{bmatrix}
$$
Because the base vectors are dependant, the equation has no solutions that can be expressed as $x=A^{-1}b$.


Revisited$^2$: Why is $(-n)^2$ divergent? How can it be shown rigorously? Why is $(-n)^2$ divergent? How is this proven? I've tried using the $\epsilon$ definition of convergence to come to a contradiction, but I don't know that using the definition is the way to go. I get that $n^2-s\leq \frac{1}{n}$. Not sure where to go from here. A hint would be nice.

Attempt1:

Say $\lim_{n\rightarrow\infty}=(-n)^2=s$. Then given an $\epsilon>0$ we can find an $N\in\mathbb{N}$ so that $\lvert (-n)^2-s\rvert \leq \epsilon$ for every $n\geq N$. But if $n\geq N$, then we must have that
$$\lvert (-n)^2-s\rvert\leq \frac{1}{n}\leq\frac{1}{N},$$
but . . . hmm . . . I don't know that this is a fruitful approach.

Attempt2:

Assuming that the negation of convergence is

  $$\text{$\exists\epsilon\leq 0$ s.t. $\forall N\in\mathbb{N}$ $n>N$ so that $\lvert s_n-s\rvert \geq \epsilon$},$$

then if we let $\epsilon=0$, then $\lvert s_n-s\rvert \geq \epsilon\implies \lvert n^2-s\rvert \geq 0$, which means that $n^2\geq s$ and $s\geq n^2$, so $s=n^2$, which means $\lvert s_n-s\rvert \geq \epsilon$ holds as $0\geq 0$, right?

Easiest to use the negation of the definition of Cauchy sequence:

There exists $\epsilon > 0$ such that for any $N$, there is some $n,m > N$ such that $|x_n - x_m| > \epsilon$.

Take time to parse this statement and it won't be too bad to prove what you want.

Note that $(-n)^2 = (-1)^2 n^2= n^2$. So, proving that $(-n)^2 \to \infty$ is the same as proving that $n^2 \to \infty$, which is an easy and well known proof.


Jordan–Bareiss algorithm Jordan–Bareiss Algorithm is an algorithm that find determinant of $n \times n$ square matrix $M$. Its algorithm is given below:

Initialize the sign $s := 1$ and set $c := m^{(-1)}_{00}
 = 1;$
loop over rows $M_k$ of $M$ for $k ∈ \{1, \dots , n\}$:
a) find the pivot $m_{ik} \neq 0$ with $i ≥ k$;
b) if no such $m_{ik}$ exists, then return $0$ and quit;
c) if $i \neq k$, then swap the rows $M_k$ and $M_i$ and update the sign, setting $s := −s$.
d) replace every row $M_i$ with $i ∈ \{k + 1, \dots , n\}$ by an expression
$\hspace{2.4cm}$ $\frac{(m_{kk}·M_i − m_{ik}·M_k)}{c}$
e) update the divisor $c$ setting $c:= m_{kk}$;
return $\det M = s·m_{nn} $.

Here is my trial using SageMath but its giving wrong answer.
def JordanBarreiss(M):
    assert M.is_square(), "It must be a square matrix"
    n=M.dimensions()[0]
    s=1
    c=1
    for i in [1,..,n-1]:
        for k in [1,..,n-1]:
            if M[i,k] == 0 and i < k:
                return 0
            if i != k:
                M.swap_rows(i,k)
                s=-s
                for j in [k+1,..,n-1]:
                    M[j] = (M[k,k]*M[j] - M[j,k]*M[k])/c
        c = M[k,k]
    return M[n-1,n-1]

A = matrix(QQ, [[5,1,-9,0],[-1,3,-1,1],[8,-2,1,-3],[2,2,1,2]]);
JordanBarreiss(A);

out[110]  0

The expected output M = 5  1  -9   0
                        0  16 -16  5
                        0  0   196 -30
                        0  0   0   475
where M[n-1,n-1] = 475 (det(M))

Please, it will be highly appreciated if anyone could help to fix this error, Python is also allowed. thanks.

Now I get it, thanks all
def JordanBarreiss(M):
    """
     JordanBareiss: Function to compute determinant of nxn square matrix M of size this algorithm

     INPUT:

         - A square matrix M

    OUTPUT:

        - det(M) 'determinant of the given matrix'

    """
    assert M.is_square(), "Only Square matrix allowed"
    n=M.dimensions()[0]
    c=1
    for k in range(n-1):
        for i in range(k+1,n):
            for j in [k+1,..,n-1]:
                M[i,j] = (M[i,j]*M[k,k] - M[i,k]*M[k,j])/c
        c=M[k,k]
    return M[n-1,n-1]

```


put n balls into n boxes Suppose we are given n balls, labeled 1, · · · , n, as well as n boxes, also labeled 1, · · · , n.
We randomly draw one of the n balls and one of the n boxes, and place the ball in the box. This
procedure is repeated n times, until every ball is in a box. Each time the ball is chosen uniformly
from the remaining balls, and the box is chosen uniformly from the remaining boxes.

(a) What is the probability that the ball labeled k ends up in the box labeled k?

(b) What is the expected number of label matches?

(c) What is the variance of the number of label matches?

I think for (a), should it just be 1/n^2? But I think it is not correct, because it is without replacement. Can anyone help me with this? Thanks a lot.

Let $\chi_i$ be the binary random variable representing the event that ball $i$ is placed into box $i$.  That is to say, $\chi_i=\begin{cases}1&\text{if ball}~i~\text{is matched with box}~i\\0&\text{otherwise}\end{cases}$

Question (1) can be rephrased as calculating $Pr(\chi_k=1)$.  As alluded to in the comments we recognize that the arrangement of balls in boxes is precisely a permutation and each position in the permutation is equally likely to be any of the numbers (or equivalently each box is equally likely to have any of the balls).  For a more formal explanation, one can approach this with multiplication principle noting that there are $(n-1)!$ ways to arrange balls into boxes satisfying that box $k$ gets ball $k$.  As a result, the probability we are looking for is $Pr(\chi_k=1)=\frac{1}{n}$.

Letting $X$ represent the total number of boxes with their correct balls (i.e. the number of fixed points in the permutation) we recognize that $X=\chi_1+\chi_2+\dots+\chi_n$

Question (2) asks us to find $E[X]$ which using the above and linearity of expectation gives us $E[X]=E[\chi_1+\chi_2+\dots+\chi_n]=E[\chi_1]+E[\chi_2]+\dots+E[\chi_n]=n\cdot\frac{1}{n}=1$

Question (3) asks us to find $Var[X]$ which we have the option to expand as $E[X^2]-E[X]^2$.  We already found $E[X]=1$ earlier, so all that remains apart from arithmetic is to calculate $E[X^2]$.  Using that $X=\chi_1+\chi_2+\dots+\chi_n$ again we find:

$$\begin{array}{rl}X^2 &= (\chi_1+\chi_2+\dots+\chi_n)(\chi_1+\chi_2+\dots+\chi_n)\\ &= \chi_1\chi_1+\chi_1\chi_2+\chi_1\chi_3+\dots+\chi_n\chi_{n-1}+\chi_n\chi_n\\
&=\sum\limits_{k=1}^n\chi_k^2 + 2\sum\limits_{k=1}^n\sum\limits_{j=k+1}^n\chi_k\chi_j\end{array}$$

We have then $E[X^2]=E\left[\sum\limits_{k=1}^n\chi_k^2 + 2\sum\limits_{k=1}^n\sum\limits_{j=k+1}^n\chi_k\chi_j\right]$ which simplifies quite nicely due to symmetry and properties of binomial random variables.

   $E[X^2]=nE[\chi_1] + 2\binom{n}{2}E[\chi_1\chi_2]$

From here there is one small probability calculation left and then a bit of arithmetic to finish.

   You need to calculate $E[\chi_1\chi_2]$.  Remember that the product of two binary random variables is again a binary random variable and the expected value of a binary random variable is just the probability that the random variable takes the value 1.  That is to say $\chi_1\chi_2=\begin{cases}1&\text{if ball}~1~\text{is in box}~1~\text{and ball}~2~\text{is in box}~2\\0&\text{otherwise}\end{cases}$ and $E[\chi_1\chi_2]=Pr(\chi_1\chi_2=1)$

Continued:

   $Pr(\chi_1\chi_2=1)=Pr(\chi_1=1)\cdot Pr(\chi_2=1\mid \chi_1=1)=\frac{1}{n}\cdot\frac{1}{n-1}$ (note, these random variables are dependent on one another).  We get then $$Var[X]=E[X^2]-E[X]^2=n\cdot\frac{1}{n}+2\binom{n}{2}\frac{1}{n(n-1)}-1=1+1-1=1$$

Update (The original answer was wrong, since I had not read the problem carefully. I then overlooked that boxes, once chosen, are taken out of the game as well.)

The described protocol chooses a permutation $\pi\in{\cal S}_n$ uniformly at random. The probability that for a given $k\in[n]$ we have $\pi(k)=k$ then is ${1\over n}$, by symmetry. The linearity of expectation then enforces that the expected number of correct matches is $n\cdot{1\over n}=1$. – Only the computation of the variance necessitates more work.


Basic solving for fractions Can someone help me with this? I am a beginner to physics and in a question i need no isolate a constant to solve for another one.

A/B=C/D solve for D step by step please.

We start with $\frac{A}{B}=\frac{C}{D}$.

Our goal is to isolate $D$. (Of course since $B$ and $D$ are in the denominator, they are assumed to be non-zero).

First we want to get $D$ into the numerator. We multiply both sides by $D$.

$\frac{AD}{B}=C$

Now we multiply both sides by $B$.

$AD=BC$.

Lastly, we divide both sides by $A$ (once again, this assumes that $A$ is non-zero).

We get that $D=\frac{BC}{A}$


Horizontal tangent line of a parametric curve Suppose $x=t^2,y=t^3$ is a parametric curve. Here's a quote from my textbook:

  The origin, which corresponds to $t=0$, is a singular point of the parametric curve, because $dx/dt=2t,dy/dt=3t^2$ are both zero when $t=0$.

So far so good.

But then they write:

  However, the curve has a horizontal tangent line at the origin, because for all $t\neq 0$:
  $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{2}t$$
  And thus:
  $$\lim_{t\to 0^+} \frac{dy}{dx}=\lim_{t\to 0^-} \frac{dy}{dx}=0$$

It looked a little odd for me. Nevertheless, I decided to use the same argument to show that the parametric curve $x=2\cos t - \cos (2t), y=2\sin t - \sin(2t)$ has a horizontal tangent line at $t=0$, that is at $(1,0)$.

However my professor said that this is wrong ("because the derivative is not zero" - indeed, $\frac{dy}{dx}\Big|_{t=0}$ is undefined - "$0/0$").

So who is right? Is the existence of the limit a sufficient condition for the (horizontal) tangent line to exist, as my textbook says, or not? I'm confused.

Thanks.

Short version of the question: can a parametric curve have a horizontal tangent line at a singular point?

I.e. is $\lim_{t\to 0} \frac{dy}{dx}=0$ a sufficient condition for a horizontal line to exist (at $t=0$)? (even if the derivative $\frac{dy}{dx}\Big |_{t=0}$ itself doesn't exist).

Notice that both your curves are algebraic curves, of equations $x^3 - y^2 = 0$ and $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ respectively.

We say that the line $ax + by + c = 0$ is tangent at the point $(x_0, y_0)$ to the curve given by $F(x,y) = 0$ if and only if $(x_0, y_0)$ is a multiple root of the system $\begin{cases} F(x,y) = 0 \\ ax + by + c = 0 \end{cases}$ (this is the definition that was originally used by algebraic geometers).

In your case, the points are $(0,0)$ in the first case and $(1, 0)$ in the second, and the line to be checked is $y=0$.

Plugging $y = 0$ into $x^3 - y^2 = 0$ gives $x^3 = 0$, which indeed has $x=0$ as a triple root, therefore $y=0$ is tangent to $x^3 - y^2 = 0$ at $(0,0)$.

Plugging $y = 0$ into $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ gives $(x^2-1)^2 - 4(x-1)^2 = 0$, or equivalently $(x-1)^2 ((x+1)^2 - 4) = 0$, or again $(x-1)^3 (x+3) = 0$, which indeed has $x=1$ as a triple root (the root $-3$ being simple, meaning that at the point $(-3, 0)$ the line $y=0$ intersects the curve as a secant), therefore $y=0$ is tangent to $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ at $(1,0)$.

Since the bounty giver is still unsure, let us follow Wikipedia and compute the intersection number in the first case. Wikipedia gives several methods, one of them is to consider the ideal generated by $y^2 - x^3$ and by $y$ in $\Bbb R[[x,y]]$ (the ring of formal series in $x$ and $y$) and compute the dimension of the quotient vector space $\Bbb R [[x,y]] / (x^3-y^2, y)$.

In the quotient space, both $y$ and $x^3 - y^2$ will become $0$, which implies that $x^3$ also becomes $0$. Therefore, the only powers of $x$ and $y$ that survive in the quotient set are $x^0 = y^0 = 1$, and $x$ and $x^2$ - a total of $3$ linearly independent powers, so $3$ will be the intersection number of $y=0$ and $x^3-y^2$ at $(0,0)$. Since everything $\ge 2$ means tangency, this shows that the line $y=0$ is tangent to $x^3-y^2$ at $(0,0)$.

The same thing could be done with what Wikipedia denotes by $I_{(0,0)}$, letting $P = y$ and $Q = x^3 - y^2$. Applying the properties that you see on that page (the numbers above equal signs are the Wikipedia properties that I apply),

$$I_{(0,0)} (y, x^3-y^2) \overset 6 = I_{(0,0)} (y, x^3) \overset 5 = I_{(0,0)}(x,y) + I_{(0,0)}(x,y) + I_{(0,0)}(x,y) \overset 4 = 1 + 1 + 1 = 3 .$$

The same computations could be done for the second example, it is just that they are more tedious. First, let us translate the curve such that the cusp moves from $(1,0)$ to $(0,0)$ (because, for simplicity, Wikipedia's formulae are given only for $(0,0)$). To do this we shall make the change of variable $x = u + 1$, which will lead to $(u^2 +2u +y^2)^2 -4(u^2+y^2)$. Next,

$$I_{(0,0)} ((u^2 +2u +y^2)^2 -4(u^2+y^2), y) \overset 6 = I_{(0,0)} (u^4 + 4u^3, y) = I_{(0,0)} (u^3 (u+4), y) \overset 5 = I_{(0,0)} (u^3, y) + \\ I_{(0,0)} (u+4, y) = I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u+4, y) \overset {3, 4} = 1+1+1+0 = 3 .$$

Again, we obtain an intersection number $\ge 2$, which means tangency.

What others have correctly stated is that the two plane curves given in the problem are not differentiable submanifolds of $\Bbb R^2$ at the singular points under discussion, therefore they do not fit into the framework of diferential geometry, therefore we may not speak of their tangent spaces at the singular points as defined in differential geometry (a proof that an almost identical curve cannot be a smooth submanifold can be found here). Counterintuitively, though, they do have tangent lines! There is no contradiction, the thing is that we use two meanings for "being tangent" that are not synonymous: one comes from differential geometry ("tangent space"), and it doesn't apply here, the other from algebraic geometry ("intersection number"), and it does apply here. In the latter framework, we may speak about "a given line being tangent (or not) to a curve in a point" without speaking of "the tangent space at that curve in that point".

$y=0$ is tangent to those curves at the specified points. Those curves do not have nicely defined tangent spaces at those points.

You are referring to two different things here..

For the semi-cubical parabola curve $y=x^{3/2}$ there is horizontal tangent at the cusp y=0 or t=0.

For cardoid, depending on the cusp contact point chosen, slope of cusp tangent varies, look at all cusps of  epicycloids:

EDIT 1:

Since it is passing through infinite curvature which is rate of change of tangent inclination, the slope is changing very fast and undefined.

It can be also seen at the point $(1,0)$ Changing the multiple angle

$$ x=2\cos t - \cos (n t), y=2\sin t - \sin( n t) $$

has the effect of influencing curvature. It changes from (positive for circle) to negative for looped case, via the present central cardoid case of infinite curvature.

The direction of tangent is vertical, indeterminate and vertical respectively


How to compute $\int_0^\infty \frac{e^{-t}-1}{t^{s+1}}dt$ I came across the following integral in a textbook without explanation. How can I prove it?

$$\int_0^\infty \frac{e^{-t}-1}{t^{s+1}}dt=\Gamma(-s)$$

Here $s\in(0,1)$.


Generate a number with a die that has three 0s and three 1s Let's say I have a regular fair die, but instead of digits 1 through 6, there are three 0s and three 1s.

I can generate a random $n$-digit binary number as follows: if initially I roll a zero (or a series of them), I ignore them; then what is left to do is just roll the die $n$ times. [Thanks to user Alex Meiburg, I have learned that this strategy does generate random numbers, but not uniformly; as he pointed out, generating, e.g., '101' is 3 (!) times more likely than generating '111'. Given this, I would appreciate if someone proposed a strategy lacking this oddity.]

Question: What is the simplest algorithm to generate a random $n$-digit number in base-$b$, $b\neq2$ using this "binary" die?

I have been trying to come up with an answer for a few hours, but I feel that there is none.

This method is probably best explained by an illustrative example. Suppose you want a random four-digit number in base 6: anything from $1000_6$ to $5555_6$ inclusive (for a total of $1295-216+1=1080$ possibilities).

Since $10$ bits would only give you $1024$ possibilities, we need to use at least $11$ bits, so generate $11$ random bits (by rolling the "die" 11 times), and write them as a binary number (like $00000000110_2=6$). If that number is between $0$ and $1079$ inclusive, add $6^3$ to it to get to get a number between $1000_6$ and $5555_6$. If the binary number was between $1080$ and $2047$ inclusive: too bad, try again by generating 11 new random bits.

In a bad case like this where nearly half of the possibilities are too high to use, the expected number of tries is a little under 2, so this method with retries like this isn't so terrible.

Side note: This algorithm/question is very closely related to the one at Generate random integers from random bit sequence.

I'm not sure what your difficulty is.

To generate a $n$-digit binary number.

First roll produces $0$ or $1$: this is the first digit.

Second roll produces $0$ or $1$: this is the second digit.

Keep rolling until you have $n$ digits.

This will give you (in binary) the numbers from $0$ to $2^{n}-1$. Each will be equally likely to occur.

To generate a $n$-digit number for another base $b \ne2$, first find the smallest value $k$ such that $2^k >b$.

For the first digit, produce a $k$-digit binary number $n$. If $n \ge b$ then it is not suitable for use in base $b$, so generate another $k$-digit binary number $n$ until $n <b$. This is the first digit.

For the second digit, produce a $k$-digit binary number $n$. If $n \ge b$ then it is not suitable for use in base $b$, so generate another $k$-digit binary number $n$ until $n <b$. This is the second digit.

Continue until you have a list of $n$ digits.


Riemann integrability implies Darboux integrability I'm trying to prove the following:

Suppose that given $\epsilon>0$, there is a $\delta>0$ such that if $P$ is any partition of $Q$ (a rectangle in $\mathbb{R}^n$)  of mesh less than $\delta$, and if, for each subrectangle $R$ determined by $P$, $\mathbf{x_R}$ is a point of $R$, the condition:
$$\left | \sum_R f(\mathbf{x_R})v(R)-\int_Qf \right |<\epsilon$$
holds (here $v(R)$ is the volume of $R$), then $f$ is Darboux integrable (clearly $f$ is bounded).
The converse is not that difficult, the proof makes use of the Riemann condition (i.e. $U(f,P)-L(f,P)<\epsilon)$. So, if only someone gave me a hint, it would be appreciated.

Let $\epsilon>0$ be given and let $P=\{R_1, \ldots, R_n\}$ be a partition of $Q$ of mesh less than $\delta(\epsilon/2)$. We know that for every $i$ there exists an $x_i \in R_i$ such that $$\sup_{x\in R_i}\{f(x)\}-f(x_i)<\frac{\epsilon}{2v(Q)}$$ Let $S(f,P,\{x_i\})$ be the Riemann sum corresponding to this choice of points.

What can you say about $U(f,P)-S(f,P,\{x_i\})$ and how can you use this to prove that $$|U(f,P)-\int_Qf|<\epsilon$$


Prove that $11,111,111.....$ is not a perfect square I know there are many other same related questions, but as my profile says I don't have a mentor, I just wanted you guys to check my solution.
First I noticed that,
$11$ = $10 + 1$,
$111$ = $10^2+10+1$,
Similarly, $\underbrace{111\cdots111}_{n\,\text{times}}$ = $10^{n-1}+10^{n-2}+...+10+1$,
which  can easily be evaluated as it is a finite geometric series.
$$\sum_{i=0}^{n-1} 10^i=\frac{10^n-1}{9}$$ for $n \ge
2$.
After this, I checked perfect square modulo $4$,
Let $j$ be any integer, then if $j$ is even, then $j=2k \Rightarrow j^2 = 4k^2 \equiv 0 \mod 4$
and if $j$ is odd then, $j = 2k+1 \Rightarrow j^2 = 4k^2+4k+1 \equiv 1 (\mod4)$
Then, Since $\underbrace{111\cdots111}_{n\,\text{times}} = \frac{10^n-1}{9} \Rightarrow (9) (\underbrace{111\cdots111}_{n\,\text{times}}) = {10^n-1}$ $\equiv$ $2^n-1$ $(\mod 4)$
$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv 2^n-1 \mod 4$
Since for $n \ge 2, 2^n$ is always a multiple of $4$. Hence,
$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv -1 \mod 4$
$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv 3\mod 4$
But a perfect square leaves remainder $0$ if it is even and $1$ if it is odd. So, clearly for
$n \ge 2,$  $(\underbrace{111\cdots111}_{n\,\text{times}})$ is not a perfect square.
I know there are other questions related to this but I just wanted to verify if my solution was correct. Thank you and suggestions would be appreciated!

base 3:  102 * 102 = 11111
base 7:   26 *  26 = 1111
These disprove the assertion.
In decimal, any number 11 modulo 100 is 3 mod 4, and is thus not square.


If $R=\{(x,y): x\text{ is wife of } y\}$, then is $R$ transitive? If $R=\{(x,y): x\text{ is wife of } y\}$, determine whether the relation $R$ is transitive or not.

My Try: For Transitivity, If $(a,b) \in R$ and $(b,c)\in R\;,$ Then $(a,c)\in R.$.

Here If $x$ is a wife of $y$, then $y$ is not a wife of $z$. Therefore $x$ is not a wife of $z$.

Means $(x,y)\in R$ and $(y,z)\notin R$, then $(x,z)\notin R$

So $R$ is not a Transitive Relation.

But in Book, It is Transitive Relation.

Plz explain me with example. How this can be possible? Thanks

The best explanation is probably that the possibility of same-sex marriages were not on the author's mind when he/she came up with the exercise.

If we suppose that marriages is always between a man and a woman (and genders are binary, bla bla bla), then the relation is indeed transitive, but vacuously so.

The condition it has to satisfy is that if $x$ is the wife of $y$ and $y$ is the wife of $z$, then $x$ is the wife of $z$. However this is true because the premise (i.e. the part after "if") is impossible to satisfy (under our assumed assumptions): $x$ can only be the wife of $y$ if $y$ is a man, and $y$ can only be the wife of $z$ if $y$ is a woman. Since nothing is both a man and a woman, both of these cannot be true at the same time.

In other words, the only way for a relation not to be transitive is if there is some $x$, $y$, and $z$ such that $xRy$ and $yRz$ and not $xRz$. If there is no such example, the relation is transitive -- no matter why there is no example.

Thus, the relation is transitive for the same reason that the empty relation is transitive: Because there are no triples $(x,y,z)$ such that $xRy$ and $yRz$, the condition for being transitive is not automatically met.

On the other hand, if $R$ includes information about at least one lesbian marriage, then it fails to be transitive: Suppose Alice and Betty are married. Then, in the case $(x,y,z)=(\mathit{Alice},\mathit{Betty},\mathit{Alice})$ we have that $\mathit{Alice}\mathrel{R}\mathit{Betty}$ and $\mathit{Betty}\mathrel{R}\mathit{Alice}$, yet not $\mathit{Alice}\mathrel{R}\mathit{Alice}$, since Alice is not her own wife.

It is transitive, considering that a set (y,z) (implying that y who is wife of x, is husband of z) does not exist.
To prove transitive condition you need to have the set (x,y) and also (y,z).Then you find out whether (y,z) is in the relation to figure out the transitive condition.
And if the condition (y,z) does not exist (like in the above case) you don't need test for transitiveness. It is trasitive by default.


Conditional Probability with Anti Set Ethnic Group    O    A    B    AB
1             .082 .106 .008 .004
2             .135 .141 .018 .006
3             .215 .200 .065 .020

A given population is in one of the four above blood types and one of the three above ethnicities.
Above is given the "joint" probabilities.
Have to calculate probability of a randomly selected individual being from ethnicity 1 given  that their blood type is NOT B.

My Solution:
$P(1|B') = \dfrac{P(1 \cap (O \cup A \cup AB))}{P(O \cup A \cup AB)}$

But this yielded the wrong answer ... Any idea how to approach this?

Question on your answer is given by the Bayes' theorem. I will try to write a complete solution.

Let $B$ denote that given person has a blood type B and denote $C_{i}$ that given person belongs to the i-th ethnicity group. I assume that $P(C_{1}) = P(C_{2}) = P(C_{3}) = \frac{1}{3}$ (which should have been definitely given in the assignment).

According to the Bayes' theorem we have

$P(C_{1}|B) = \frac{P(B|C_{1}) \cdot P(C_{1})}{\sum^{3}_{i=1} P(B|C_{i}) \cdot P(C_{i})}$.

All the terms on the right hand side of the previous equations are given by your table, so you can easily compute the desired probability $P(C_{1}|B)$.


How prove $\lim_{x\to 0}f(x) = 0\iff\lim_{x\to 0}xf(x) = 0 $? Let $f:R \rightarrow R$ such that  $$| f(x+y)-f(x)-f(y) |\le |x-y|,$$ for all $x, y \in R.$ How can I prove that  $$\lim_{x\to 0}f(x) = 0\iff\lim_{x\to 0}xf(x) = 0? $$

$(\implies)$

Straightforward.

$(\impliedby)$

Edit: Following David C. Ullrich's comment, I turned the wlog assumption $f \geqslant 0$ into using $|f(x)|$ instead of $f(x)$ everywhere.

First note that substituting $x = y$ yields $|f(2x) - f(x) - f(x)| \leqslant 0$, so $f(2x) = 2f(x)$ for every $x \in \mathbb{R}$.

Suppose that $\displaystyle \lim_{x \to 0} f(x) \neq 0$. We will show that also $\displaystyle \lim_{x \to 0} x \cdot f(x) \neq 0$.

Exists such $\varepsilon > 0$ that there are arbitrarily small $x \neq 0$ with $|f(x)| \geqslant \varepsilon$. Let $\delta > 0$. There is such $x \neq 0$ that $|x| < \delta^2$ and $|f(x)| \geqslant \varepsilon$. There is a unique $n \in \mathbb{N}$ such that $\frac{\delta}{2} \leqslant |x| \cdot 2^n < \delta$ and then

$$|f(x \cdot 2^n)| = 2^n \cdot |f(x)| \geqslant \frac{\delta}{2 |x|} \cdot \varepsilon > \frac{\delta}{2 \delta^2} \cdot \varepsilon = \frac{\varepsilon}{2 \delta}.$$

Thus $|x \cdot 2^n| < \delta$ and

$$|x \cdot 2^n| \cdot |f(x \cdot 2^n)| \geqslant \frac{\delta}{2} \cdot \frac{\varepsilon}{2 \delta} = \frac{\varepsilon}{4}.$$

From right to left

If $x=y$, we have
$$
|f(2x)-2f(x)|\le |x-x|\Longrightarrow f(2x)=2f(x)\Longrightarrow f(x)=ax,
$$
where $a$ is a constant, and statement is obvious.


For a normal subgroup, the factor group is isomorphic to the range mod the image of a surjective homomorphism. Suppose that $G_{1} \trianglelefteq G$ and $\phi \colon G\rightarrow H$ a surjective homomorphism.  I need to prove (or disprove, but somehow I get the feeling that the theorem is true) that $G/G_{1}\cong H/\phi(G_{1})$.

I've been trying to think how the isomorphism theorems bear on this, which doesn't seem obvious since I don't know the kernel of the mapping so the first theorem seems inappropriate and the other two don't seem to say anything directly about homomorphisms.  If I try to use them anyway, I end up saying this:

From the first homomorphism theorem $G/\ker\phi \cong \phi(G)$.  If I restrict the map to $G_{1}$ then I get $G_{1}/\ker\phi \cong \phi(G_{1})$.  If I'm to use the second homomorphism theorem then I have to pick something to be the "main group" and then two subgroups.  I could choose $G$ with subgroups $G_{1}$ and $\ker \phi$.  Then I'd have $G_{1}/(G_{1}\cap \ker \phi) \cong G_{1}\ker\phi / \ker\phi$.  That doesn't seem to help.  Alternately I could choose the group $H$ but I don't see more than one natural choice for a subgroup of that.

In the third isomorphism theorem I don't see how I can apply it here at all since I don't think I have any reason to believe that $\ker\phi \subset G_{1}$.

[As a further thought, perhaps the theorem is false:  It sounds like it wouldn't be hard to think of an example where $\phi(G_{1})=H$ in which case $H/\phi(G_{1})$ is the trivial group, while it wouldn't necessarily be true that $G/G_{1}$ is.  But this seems to go against the hint that I was given in the problem, that I should use the isomorphism theorems.]

This is not true without some more restrictions on $\phi$. Let $G$ be $\Bbb Z^3$, $G_1$ the subgroup generated by $(1,0,0)$, $H=\Bbb Z$, and $\phi$ projection onto the last coordinate.

$\Bbb Z^3/\Bbb Z \not\approx \Bbb Z/(0)$

If you demand that $\phi^{-1}\circ\phi G_1 = G_1$, then you might be on to something.

https://faculty.etsu.edu/gardnerr/5410/notes/I-5.pdf you might find this one helpful!


Find the derivative of $\frac{2}{x^2}$ directly from definition I have to find the derivative of $\frac{2}{x^2}$ using only the definition. I have tried to do this several times but I always get the wrong answer - could you tell me where I made an error?

$$\frac{d}{dx}[\frac{2}{x^2}] = \lim_{h \to 0} \frac{\frac{2}{(x+h)^2}- \frac{2}{x^2}}{h} = \lim_{h \to 0} \frac{2x^2-2(x+h)^2}{x^2(x+h)^2h} = \lim_{h \to 0} \frac{-4xh -2h^2}{x^4h + 2x^3h^2 + x^2h^3} = \lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4}{x^4}$$
As you see, I keep getting $x^4$ instead of $x^3$ but I simply cannot see where I made a mistake - any suggestions?


Compactness of the closed interval [0,1] In general topology, a topological space is said to be compact, if every one of its open cover has a finite subcover.
However, I cannot see the compactness of the close interval [0,1] from the above definition.
To be a little specific,let us consider the following open cover for [0,1]: $C= \{[0,1/2),(1/3,3/4), (2/3,1]\}$. Now, the open interval (1/3,3/4) itself has at least one open cover (let's call it P) which does not have a finite subcover. We use P to cover the open interval (1/3,3/4). This gives a new open cover C' for the interval [0,1]. It looks like C' does not have an finite subcover, since C' includes P which does not have a finite subcover.

Of course I misunderstood something here. If somebody can catch my error it will be very helpful.

So you get an open cover by retaining $[0,1/2)$ and $(2/3,1]$
but replacing $(1/3,3/4)$ by a bunch $P$ of open sets where no finite
collection covers $(1/3,3/4)$.

You can do this.

But it is still the case that this new covering $C'$ has a finite
subcovering. Don't forget that $[0,1/2)$ and $(2/3,1]$
are still available. If we used both of these, all we have to do
is find a finite subset of $P$ that covers $[1/2,2/3]$. (We don't
need it to cover all of $(1/3,3/4)$.) As $[1/2,2/3]$ is compact,
then there will be such a finite subset.

This kind of thing (compactness of $[0,1]$) is typically proved via an appeal to the Heine-Borel theorem, which says that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded.  The subset $[0,1]$ is closed and bounded and therefore compact.


Is the element $(0,0,0)\in\mathbb{R}^3$ a divisor of zero? (I'm assuming that $\mathbb{R}^3=\mathbb{R}\times\mathbb{R}\times\mathbb{R}$)

In my assignment, I'm told to prove that exactly one of the following can be true for an element $(x,y,z)\in\mathbb{R}^3$

$(x,y,z) = (0,0,0)$
$(x,y,z)$ is invertible
$(x,y,z)$ is a divisor of zero

I'm starting with the first. Given the definition of the operations on the ring $A=\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ (we were given non-standard multiplication), the multiplicative identity is $(1,0,0)$.

There's no way that, given our definition, $(0,0,0)(0,0,0)^{-1}=(1,0,0)$ so it's not invertible.

However, my issue is with showing that it is not a zero divisor. Whenever I use algebra to find each component of another element in the ring $p$, such that $p\cdot(0,0,0)=(0,0,0)\cdot p= (0,0,0)$, this is always true. Any suggestion?

By "zero divisor," your book probably means that $v \in \Bbb{R}^3$ is a zero divisor if and only if there exists some $a \in \Bbb{R}^3$ such that $a, v \neq 0$ and $av=\textbf{0}$. Therefore, $(0, 0, 0)$ is not a zero divisor because $(0, 0, 0) \neq \textbf{0}$ is false.

It is important to note that $(0, 0, 0)$ is intentionally excluded from the definition of zero divisor because some mathematicians think that this is a more useful definition of zero divisor. This way, we can talk about rings without zero divisors called integral domains, where if $av=\textbf{0}$, then either $a=\textbf{0}$ or $v=\textbf{0}$. This would be rings like $\Bbb{R}$ or $\Bbb{R}^3$ where you can't get $\textbf{0}$ by multiplying two non-zero elements.

Thus, because of the definition of zero divisor used in your book, $(0, 0, 0)$ is not a zero divisor.

You can actually have zero divisors depending on how multiplication is defined. If addition is point-wise, then (1,0,1)(0,1,0)=(0,0,0).
You said you were given non-standard multiplication. Was it defined for you?


Prove that: the center of any group is characteristic subgroup . Let $G$ be any group , $Z(G)$ is the center of the group $G$ , prove that :
$\forall \tau \in Aut(G) , \tau [(Z(G)] = Z(G)$

My first trial was to prove that the center of any group is the unique subgroup of its order hence is a characteristic . but i found a counterexample easily which is $D_8$ the Dihedrial group of order $8$ ( i use the notation $D_{2n}$ )

So , any hints ?

Can you show that for any $x\in Z(G)$ and $y\in G$, $\tau(x)\tau(y)=\tau(y)\tau(x)$? This will say $\tau(x)$ commutes with every element of $G$ (since $y$ was arbitrary), thus $\tau(x)\in Z(G)$. This shows containment in one direction, can you show containment in the other direction now?

consider that any Automorphism preserves the order, means order of Z(G) = order of τ(Z(G)).


Prove the following theorem in propositional calculus I have the following Hilbert-style propositional calculus, having the logical connectives $\{\neg,\rightarrow\}$ of arity one and two respectively, and the following axioms:

A1: $A\rightarrow(B\rightarrow A)$

A2: $(A\rightarrow(B\rightarrow C))\rightarrow((A\rightarrow B)\rightarrow(A\rightarrow C))$

A3: $(\neg A\rightarrow \neg B)\rightarrow (B\rightarrow A).$

$A,B,C$ are any well-formed formulas (wffs). The only inference rule is modus ponens. I want to prove the following theorem:

$\vdash (\neg A\rightarrow A)\rightarrow A$ for any wff $A$.

You may use any or all of the following theorems without proof:

$\vdash A\rightarrow A$ for any wff $A$.

$A\rightarrow B, B\rightarrow C\vdash A\rightarrow C$ for any wffs $A,B,C$.

$A\rightarrow (B\rightarrow C), B\vdash A\rightarrow C$ for any wffs $A,B,C$.

$\vdash \neg A\rightarrow (A\rightarrow B)$ for any wffs $A,B$.

Please avoid using the deduction theorem unless absolutely necessary.

Lemma

1) $\vdash \lnot A \to (A \to B)$ --- Th.4

2) $\vdash (\lnot A \to (A \to B)) \to ((\lnot A \to A) \to (\lnot A \to \lnot B))$ --- Ax.2

3) $\vdash (\lnot A \to A) \to (\lnot A \to \lnot B)$ --- from 1) and 2)

4) $\vdash (\lnot A \to \lnot B) \to (B \to A)$ --- Ax.3

  5) $\vdash (\lnot A \to A) \to (B \to A)$ --- from 3) and 4) by Th.2

Proof

1) $\vdash (\lnot A \to A) \to ((\lnot A \to A) \to A)$ --- from Lemma

2) $\vdash ((\lnot A \to A) \to ((\lnot A \to A) \to A)) \to (((\lnot A \to A) \to (\lnot A \to A)) \to ((\lnot A \to A) \to A))$ --- Ax.2

3) $\vdash ((\lnot A \to A) \to (\lnot A \to A)) \to ((\lnot A \to A) \to A)$ --- from 1) and 2)

4) $\vdash (\lnot A \to A) \to (\lnot A \to A)$ --- from Th.1

  5) $\vdash (\lnot A \to A) \to A$ --- from 3) and 4)

I use Polish notation.  Thus, instead I'll translate every wff of the form $\lnot$$\alpha$ into the form N$\alpha$.  Also, I'll translate every wff of the form ($\alpha$$\rightarrow$$\beta$) into the form C$\alpha$$\beta$.  Thus, the axioms can get translated to
A1 CxCyx
A2 CCxCyzCCxyCxz
A3 CCNxNyCxy
I'll also use spacing to help to indicate where a substitution of an axiom or previously proved theorem took place.

1 CxCyx
2 CCxCyzCCxyCxz
3 CCNxNyCyx
4 C CxCyx C z CxCyx from 1
5 CzCxCyx from {4, 1}
6 CC CxCyz C Cxy Cxz CC CxCyz Cxy C CxCyz Cxz from 2
7 CCCxCyzCxyCCxCyzCxz from {6, 2}
8 C CCNxNyCyx C z CCNxNyCyx from 1
9 CzCCNxNyCyx from {8, 3}
10 C CxCCyxz C x C y x from 5
11 CCC x C Cyx z C x Cyx CC x C Cyx z C x z from 7
12 CCxCCyxzCxz from {11, 10}
13 CC z C CNxNy Cyx CC z CNxNy C z Cyx from 2
14 CCzCNxNyCzCyx from {13, 9}
15 CC Nz CC Ny Nz Czy C Nz Czy from 12
16 C Nz CC Ny Nz C z y from 9
17 CNzCzy from {15, 16}
18 CC Nz C z y CC Nz z C Nz y from 2
19 CCNzzCNzy from {18, 17}
20 CC CNzz CN z N y C CNzz C y z from 14
21 CCN z z CN z Ny from 19
22 CCNzzCyz from {20, 21}
23 CC CNxx CC y CNxx x C CNxx x from 12
24 CCN x x C CyCNxx x from 22
25 CCNxxx from {23, 24}


What axiom(s) do I need to prove that every nonempty set of natural numbers has a smallest element? Math people:

The title is the question.  I am not a logician, and to me, it seems self-evident that any nonempty set of natural numbers has a smallest element.  I am reading an analysis book that uses the Completeness Axiom of the real numbers to prove this "fact" (I put "fact" in quotes because a logician might not accept something so obvious as a fact but might require that I assume some axiom).  To me, this really seems like overkill, and that you should be able to prove it using something much weaker.

EDIT: I am not asking specifically about axioms of ZFC.  My question does not even mention ZFC.  I highly doubt that Choice is necessary here and I suspect it may not even be helpful.  I would like (i) confirmation that using the Completeness Axiom of the real numbers to prove that every nonempty set of natural numbers has a least element is massive overkill and (ii) some weaker axioms that give me the same conclusion, the weaker the better.

Most people, logicians included, will indeed agree that intuitively the claim that every non-empty subset of the natural numbers has a least element (which is knows as the well-ordering principle of the naturals) is self-evident. Another such self-evident property of the natural numbers is the principle of induction.

As is often the case with self-evident things, actually proving them can be tricky. It turns out (and it's not hard to prove directly) that the well-ordering of the naturals is equivalent to the principle of induction (which is a second order principle).

Now things become more complicated when one realizes that there exist first order models of arithmetic in which all of the Peano axioms hold as well as all the induction scheme axioms (that is, one axiom per first order formula) in which the second order principle of induction fails. That means that in such models of the naturals the well-ordering principle fails too. So things are not so simple, or self-evident, after all.

I do agree that using the completeness of the reals in order to prove the well-ordering of the naturals is a bit of an over kill as the former is less intuitively clear than the latter. But, it is instructive to see that imposing completeness on the reals (a second order property again) forces the naturals to obey the principle of induction.

Assume that for all natural numbers "a", (a+0)=a.  With both assumptions in place, it follows that (0+0)=0.  No other natural number has the property that (x+x)=x.  Thus, 0 comes as unique.  Addition comes as monotonically increasing.  That is,

For all x, for all y belonging to the set of natural numbers

 If x<y, then (c+x)<(c+y), where c indicates any constant.

 Also, if (c+x)<(c+y), then x<y.

Now suppose that x=0.  By the second statement, it follows that

 if (c+0)<(c+y), then 0<y.

"y" qualifies as an arbitrary natural number, so 0 comes as the least natural number.  That gives us a "hint".

Now consider any arbitrary set of natural numbers.  Define a monotonicity relation for a function $ on that arbitrary set such that

 if (c $ x)<(c $ y), then x<y.  And if x<y, then (c $ x) < (c $ y)

Define another function # such that there exists some element j belonging to the arbitrary set of natural numbers such that for all x, (j # x)=j.  It follows that (j # j)=j.  Now consider the statement

 if (c $ j)<(c $ y), then j<y.

Since y came as arbitrary, j comes as the least element of the set.  Since the set came as arbitrary it holds for all subsets of natural numbers.


Will 0.99999999 eventually become equal to 1? I am currently learning about fractions, and there is something that I am finding it hard to make sense of. When a fraction it added to the right of the decimal point, the number becomes slightly bigger right? So for example if I add enough fractions to the 0.99999999, will this number eventually become equal to 1?

So will this number eventually become equal to 1 (if I added enough 9): 0.9999999999999999999999999999999999999999999999999999999999999999999999...9

Yes and no. When you are adding more $9$'s to the end of the number, you are adding $0.9$, $0.09$, or $9(0.1)^n$, where $n$ is the $n^{th}$ digit. So if we look at it with a geometric series, we find that adding $n \,9's=\sum_{k=1}^{n}  9\frac{1}{10}^k=\sum_{k= 0}^{n-1} \frac{9}{10} \left(\frac{1}{10}\right)^{k}= \dfrac{9}{10}\dfrac{1-\left(\frac{1}{10}\right)^{n-1}}{1-\frac{1}{10}}$ Let's see if this can equal $1$ for any $n$:

$$1=1-\frac{1}{10}^{n-1}\implies 0=-\frac{1}{10}^{n-1}$$This is never true because an exponential is always positive. So no matter how many $9$'s you add, you will never reach $1$. However, if you add infinite $9$'s, we can take the limit as $n\to \infty$ and see that infinite 9's does equal 1.

Here is a kind of proof with handmade animation, Hope it will be useful.

Another one is here too


Is this function differentiable at 0? I would like to know if this function is differentiable at the origin:

$$f(x) =
\left\{
 \begin{array}{cl}
  x+x^2  & \mbox{if } x \in \mathbb{Q}; \\
  x & \mbox{if } x \not\in \mathbb{Q}. \end{array}
\right.$$

Intuitively, I know it is, but I don't know how to prove it.

Any ideas?

Thanks a lot.

$f$ can be either $x + x^2$, or $x$ (depending on the value of $x$), right?

Firstly, you can prove that $f$ is continuous at 0, by noticing that $0 < |f(x)| \le |x| + x^2$

And, you know that these 2 functions ($x + x^2$, and $x$)' derivatives at 0 are both 1. So, in order to prove that the derivative at 0 is indeed 1, you should use Squeeze Theorem, like this:

Right derivative at 0

For $x > 0$, we have the inequality: $x < x + x^2$, so the upper bound for $f$ is $x + x^2$, and the lower bound is of course $x$. So we'll have:

$\begin{align} & x \le f(x) \le x + x^2 \\
\Rightarrow &\dfrac{x}{x} \le \dfrac{f(x)}{x} \le \dfrac{x + x^2}{x} \quad \mbox{since }x > 0\mbox{, the inequality sign doesn't change}\\
\Rightarrow &1 \le \dfrac{f(x) - 0}{x - 0} \le \dfrac{x + x^2}{x}\\
\Rightarrow &1 \le \dfrac{f(x) - f(0)}{x - 0} \le \dfrac{x + x^2}{x}
\end{align}$

...

From here, you can apply Squezze Theorem, it should be easy, let's give it a try.

Left derivative at 0

Since $x \rightarrow 0^-$, we'll still have the inequality: $x < x+x^2$, so the upper bound for $f$ is still $x + x^2$, and the lower bound is of course $x$. So we'll have:

$\begin{align} & x \le f(x) \le x + x^2 \\
&...\end{align}$

Can you take it from here? Remember that when dividing to $x$ in this case, you have to change the signs (from > to <; and from < to >), as $x < 0$.

You need to show that if $h$ is near $0$, then $\displaystyle\frac{f(h)}h$ is near $1$ (which should be the value of $f'(0)$).

But $\displaystyle \frac{f(h)}h$ is either $1$ or $1+h$.

In detail: We have that $f(0)=0$.

We can anticipate that, if it exists, the derivative of $f$ at $0$ is $1$, since $g'(0)=j'(0)=1$ where $g(x)=x$ and $j(x)=x+x^2$. However, to see whether $f'(0)$ exists and what its value is, we simply use the definition of the derivative at $0$:
 $$ \lim_{h\to0}\frac{f(0+h)-f(0)}h=\lim_{h\to0}\frac{f(h)}h. $$
From the definition of $f$, we see that the fraction $\displaystyle \frac{f(h)}h$ equals either $1$ or $1+h$, depending on whether $h$ is irrational or not.

It should be clear now that the limit of this expression is $1$ as $h$ approaches $0$, but we can proceed to verify this from the definition of limit:

We have that $\displaystyle\lim_{h\to0}\frac{f(h)}h=1$ if and only if for every $\epsilon>0$ there is a $\delta>0$ such that if $0<|h-0|<\delta$, then
 $$\left|\frac{f(h)}h-1\right|<\epsilon.$$
That is, given $\epsilon>0$, we must find a $\delta>0$ such that $0<|h|<\delta$ implies that both $|1-1|<\epsilon$ and $|(1+h)-1|<\epsilon$.

The first of these inequalities always holds, so we only need to look at the second one, which simplifies to $|h|<\epsilon$. Now we see that if we take $\delta=\epsilon$, then what we need to verify is that $0<|h|<\epsilon$ implies $|h|<\epsilon$, which is always the case, and we are done: We have shown that $f'(0)$ exists, and equals $1$.

(Of course, we could just as well take as $\delta$ anything that is both positive and less than $\epsilon$.)