difference between nested interval theorem and cantor intersection theorem What is difference between the nested interval theorem and Cantor intersection theorem? In $\mathbb{R}$, are they the same? When are they different?
Intuitively, I feel they are same for $\mathbb{R^n}$. But what happens in a general space?

Nested Interval Theorem
  Consider a family of closed and bounded real intervals, $I_1 = [a_1, b_1], I_2 = [a_2, b_2], \ldots$.
  If $a_n \leq a_{n+1}$ and $b_{n+1} \leq b_n$ for all $n$ then there is an $x$ which is in every $I_n$, that is, $\displaystyle\bigcap_{n=1}^{\infty} I_n \neq \emptyset$.

  Cantor's Intersection Theorem
  Let S be a complete metric space. Consider a decreasing, nested sequence $\{C_k\}$ of nonempty closed and bounded sets in $S$. Then, $\displaystyle\bigcap_{k=1}^{\infty} C_k \neq \emptyset$.

It seems to me that Cantor's Intersection Theorem generalizes the Nested Interval Theorem to arbitrary complete spaces.

To see why the completeness hypothesis is necessary, consider $\{C_k\}={[0,\frac{1}{k}]}$. If $S=\mathbb{R}$, then $\displaystyle\bigcap_{k=1}^{\infty} C_k = \{0\}$. If $S=\mathbb{R}-\{0\}$, then $\displaystyle\bigcap_{k=1}^{\infty} C_k = \emptyset$.


All possible combinations of x letters (what is this called in mathematics) Firstly, thank you for looking at my question.

I would like to know what this kind of problem is called in mathematics:

Given a set of letters, find all possible 'words' you can make with those letters. For example for 'abc' the solution would be:

a, b, c, ab, ac, abc, ba, bac, bca, ca, cab, cba

Some background, I am writing a computer program to play Scrabble and need to generate all possible words given from a set of letters. I'm researching algorithms for this problem but couldn't quite figure out what the general name is for this type of problem. I'm curious to find out so I thought I would ask.

I thought this was a type of permutation problem but reading up on Permutations I see that the length of the result is set, not variable. And it's not a Combination since the order matters.

The problem is about permutations of k-order. You can read about this kind of problems in the book "Discrete Mathematics and Its Applications" (Kenneth Rosen)

Regards!.

All possible outcomes of a probability problem are called the "sample space". Is that what you were looking for?


Can we prove the existence of a gcd in $\mathbb Z$ without using division with remainder? For $a,b\in\mathbb Z$ not both $0$, we say $d$ is a gcd of $a$ and $b$ if $d$ is a common divisor of $a$ and $b$ and if every common divisor of $a$ and $b$ divides $d$. With this definition, can we prove the existence of a gcd of two integers (not both zero) without using

Division with remainder (that is, without proving that $\mathbb Z$ is Euclidean first)

or anything that relies on this, such as

Euclid's lemma: $p\mid ab\implies p\mid a\vee p\mid b$
Uniqueness of prime factorisation (which already relies on Euclid's lemma)?

I am aware that "without using ..." is quite vague, so I'll reformulate my question as follows:

  Is any proof known of the existence of the gcd which does not rely on the above?

One can eliminate  explicit use of the division algorithm by replacing its use by its inductive essence (i.e. rewriting the proof in mathematical "assembly language"). Below is one simple way.

The set $\rm\:S\:$ of all integers of the form $\rm\:a\:x + b\:y,\,\ x,y\in \mathbb Z,\:$ is closed under subtraction so, by the Lemma below, every positive $\rm\:n\in S\:$ is divisible by $\rm\:d = $ least positive $\rm\in S,\:$ so $\rm\:a,b\in S\:$ $\Rightarrow$ $\rm\:d\:|\:a,b.\:$ So $\rm\:d\:$ is a common divisor of $\rm\:a,b,\:$ necessarily greatest: $\rm\:c\:|\:a,b\:$ $\Rightarrow$ $\rm\:c\:|\:d = a\,x_1\!+\! b\,y_1\,$ ($\Rightarrow$ $\rm\:c\le d).$

Lemma $\ \ $ Let $\,\rm S\ne\emptyset \,$ be a set of integers $>0$ closed under subtraction $> 0,\,$ i.e. for all $\rm\,n,m\in S, \,$ $\rm\ n > m\ \Rightarrow\  n-m\, \in\, S.\,$  Then every element of $\rm\,S\,$ is a multiple of the least element $\rm\:\ell = \min\, S.$

Proof $\ $ If not there is a least nonmultiple $\rm\,n\in S,\,$ contra  $\rm\,n-\ell \in S\,$ is a nonmultiple of $\rm\,\ell.$

Remark $\ $ Using the same idea one can give "elementary" proofs of other theorems of number theory, e.g. the  proof of FTA = Fundamental Theorem of Arithmetic due to Zermelo (and others).

However, this is exactly the opposite of what one desires from a pedagogical standpoint. The innate ideal theoretic structure that underlies many of these results should be highlighted - not eliminated (for example, see posts on denominator ideals and order ideals)

I have a unique method to prove the existence of gcd

Since Euclidean Algorithm provides gcd,so proving that Euclidean Algorithm terminates might help us to prove that gcd exists.

So let's start.... Consider "a", "b" such that "a", "b"$\in$ $Z^+$

We know that Euclidean Algorithm ends when remainder becomes "0" ---(1)

Let us assume that Euclidean Algorithm never ends, So according to statement (1) none of remainder should be zero.

Applying euclidean algorithm to "a","b"....

a=bq1+r1, 0< r1 < b

b=r1q2+r2, 0< r2 < r1

r1=r2q3+r3, 0< r3 < r2

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

So continuing the process will give us b>r1>r2>r3>r4>r5>r6..... ∞ .Since b,r1,r2,r3,r4,r5...∞ are positive integers

So we got a infinite sequence of strictly decreasing positive integers ...

But according to statement of infinite descent which is a consequence of well ordering Principle, no such sequence exists.

So we are left with a contradiction!!

Hence Euclidean Algorithm terminates.

Since Euclidean Algorithm provides gcd

So gcd also exists!!!!

Hope it helps


Why do imaginary numbers work (somewhat philosophical question)? Asking as a layman, I've always puzzled over imaginary numbers and how they can be used to solve problems involving real numbers or quantities only (e.g. contour integration methods or Fourier analysis obtaining the frequency domain).  It seems strange how using an unreal number conjoined with the normal rules of arithmetic can sometimes produce results which are difficult or impossible using real variables/numbers alone.  I don't know of any imaginary units of any physical quantity in the real world.

Could someone direct me to a reference (perhaps a philosophy of mathematics text) which explains how these complex analysis methods work?  Maybe all these methods somehow  concisely represent or parse certain lengthy or complicated real-number operations 'behind the scenes'?  Another useful reference would be one which shows how one can obtain with real numbers only, any result obtainable using methods involving imaginary numbers.

When you are a little kid you learn to count and add whole numbers. In first grade, a problem like $3-5$ is said to have no answer, because you can't take $5$ apples from a pile of $3$. This makes total sense and you're ok with that. A little later you learn about negative numbers and at first you're like "whoa" but before long negative numbers are every bit as legitimate as positive ones.

Then you learn about multiplication and division and at first $13\div5$ doesn't have an answer because $5$ "doesn't go into" $13$. This makes perfect sense and you're ok with that. Then you start saying it's $2$ with remainder $3$ and finally you're told of fractions and $13\div5$ is a perfectly fine number called $2 \frac{3}{5}$ or $\frac{13}{5}$. It's a little strange at first but at some point you wonder how you could've ever gotten along without these fractional numbers.

Sometime later you run into problems that seem like they should have an answer like what's the length across the diagonal of a square but no fraction fits the bill. You're in good company as this bothered some pretty smart people over the ages. But we overcome the problem and add new numbers like $\sqrt{2}$ into the mix. It's a number whose square is $2$. This makes perfect sense and you're ok with that. These numbers solve equations like $x^2=2$ or any other equation you can make with integers and all the usual operations of arithmetic.

The $\sqrt{2}$ thing takes a little getting used to at first because it's 1.41421356237... and on and on infinitely and randomly and for the first time you're questioning the "reality" of such numbers. Yet surely the length of the diagonal is real so you accept it.

Around the same time you learn about $\pi$, another number with an infinite decimal, which is slightly more mysterious because it's got a greek name. Probably you're not told it doesn't solve any equation with integers and arithmetic operations but if you study math you find out it really is a bit weirder, and that weirdness has a name, transcendental. But you accept all these transcendentals like $\pi$ and $\sin 1$ and $\log 2$ because they seem to have a value somewhere on the number line.

At this point you think you're done because what else could there be? The whole line is accounted for. Equations like $x^2+1=0$ have no solution but you're ok with that.

You're introduced to complex numbers which solve equations like $x^2+1=0$ but they are presented as some kind of trick or device. Turns out they have all kinds of uses but they're usually not considered part of reality the way all the other numbers are. Why not?

It's often brought up that you don't need complex numbers, that you could do all the same calculations and analysis using real numbers. But that's like saying rational aren't necessary. Anything you can do with rational numbers you can do by carrying around pairs of integers.

To me complex numbers are as "real" as all the others, only more so. It just takes a little longer to accept them than it did for fractions, say. Fractions made you stretch your imagination a little. Complex numbers stretch your imagination more. That's part of makes them beautiful.

There may be dissenting comments on this but there is a sense in which complex numbers (and quaternions and octonions) are built into ordinary geometry and are not something that is just tacked on. This is done through Grassmann algebra, which may be the best axiomatic description of geometry. It is a graded algebra with scalars, points, vectors, bivectors, etc. and various types of products: exterior, regressive, interior, Clifford along with a complement.

Just to give a taste of this, here is an interior product of a bivector and a vector (both are basis vectors of the Grassmann algebra of the plane). The product is then simplified using the Euclidean metric and built-in rules of the algebra. The interior product is symbolized by a CircleMinus, which looks a little like a capital theta.

The result is a 90 degree counterclockwise rotation, so the bivector acts just as i does in the imaginary plane. Here it is applied four times to rotate around the plane.

The point is that these elements and products were designed for a complete algebra for ordinary geometry and yet they have built in operations that act like complex arithmetic. This makes complex numbers seem less artificial. It is possible to add generalized Grassmann products, hypercomplex products and then more formally build complex, quaternion and octonion algebras.

Some taste for this can be obtained in Grassmann Algebra Volume 1 by John Browne. (But the generalized Grassmann products will appear in Volume 2.) or books such as Geometric Algebra for Physicists by Chris Doran & Anthony Lasenby.


Proof that a continuous function on a compact set is uniformly continuous, compact set not containing infimum I'm reviewing the proof of the following statement (Baby Rudin 4.19): Let $f$ be a continuous mapping of a compact metric space $X$ into a metric space $Y$. Then $f$ is uniformly continuous on $X$.
The general idea of the proof makes sense. For a fixed $\varepsilon>0$, we look at all the $ \phi(p)$ (which would usually be denoted as $\delta$), and pick one that is small enough to work for all $p\in X$. We do this by finding a finite subcover of the open cover formed by the collection of $J(p)=N_{1/2\phi(p}(p)$, call it [the finite subcover] $\{J(p_1),\ldots,J(p_n)\} $. We then let $$ \delta=\frac{1}{2}\min\{\phi(p_1),\ldots,\phi(p_2)\}.$$ Cool solution, because we cannot take $$\delta=\frac{1}{2}\inf\limits_{p\in X}\{\phi(p)\}. $$ My question is, why can't we take the infimum over all of $X$ even when $X$ is compact? Rudin writes "the inf of an infinite set of positive numbers may very well be $0$", but I cannot think of any example where this is the case for a compact $X$. Surely if $X\subset\mathbb{R}$, then $\inf_{p\in X}\{\phi(p)\}>0$, as $X$ is closed and bounded, and a closed and bounded set in $\mathbb{R}$ contains its infimum.
This makes me think that any compact $X$ which is an infinite collection of positive numbers such that $\inf X=0$ must somehow lack the least-upper-bound/greatest-lower-bound property, as this is what ensures a closed and bounded set in $\mathbb{R}$ contains it infimum (Baby Rudin 2.28), and any compact $X$ is closed and bounded regardless of the metric space. So perhaps we cannot take $\inf_{p\in X}\{\phi(p)\}$ as it is not defined?
I would love anyone's advice, insight, or examples!

The relevant set here is not $X$, but the range of $\phi(p)$. Is this compact? In fact it is, because $\phi$ is a continuous function of $p$, and the continuous image of a compact set is compact. So you could prove it that way. It requires two facts: that the continuous image of a compact set is compact; and that the infimum of a compact set of strictly positive numbers is strictly positive.
But Rudin's proof just uses the definition of compactness, in a natural way. So it is simpler and more straightforward.


Expected time to roll all 1 through 6 on a die What is the average number of times it would it take to roll a fair 6-sided die and get all numbers on the die?  The order in which the numbers appear does not matter.

I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation.  We were given the answer $(1-(\frac56)^n)^6 = .5$ or $n = 12.152$

Can someone please explain this to me, possibly with a link to a general topic?

The time until the first result appears is $1$. After that, the random time until a second (different) result appears is geometrically distributed with parameter of success $5/6$, hence with mean $6/5$ (recall that the mean of a geometrically distributed random variable is the inverse of its parameter). After that, the random time until a third (different) result appears is geometrically distributed with parameter of success $4/6$, hence with mean $6/4$. And so on, until the random time of appearance of the last and sixth result, which is geometrically distributed with parameter of success $1/6$, hence with mean $6/1$. This shows that the mean total time to get all six results is
$$\sum_{k=1}^6\frac6k=\frac{147}{10}=14.7.$$

Edit: This is called the coupon collector problem. For a fair $n$-sided die, the expected number of attempts needed to get all $n$ values is
$$n\sum_{k=1}^n\frac1k,$$ which, for large $n$, is approximately $n\log n$. This stands in contrast with the mean time needed to complete only some proportion $cn$ of the whole collection, for some fixed $c$ in $(0,1)$, which, for large $n$, is approximately $-\log(1-c)n$. One sees that most of the $n\log n$ steps needed to complete the full collection are actually spent completing the last one per cent, or the last one per thousand, or the last whatever percentage of the collection.

Associate a success with each number appearing that has not appeared before. Let $X_i$ be the number of trials between the $i^{th}$ success and the $(i + 1)^{st}$ success.

Let $X$ be the random variable representing the total number of trials required for the required event, and $E[X]$ be the required expected value.

Then by linearity of Expectation, we have $E[X] = 1 + \sum_{i=1}^{5}E[X_i]$.

To calculate $E[X_i]$, consider the following,
after receiving $i−1$ different numbers i.e after $i −1$ successes, each subsequent trial has probability $(6 − i)/6$ of getting a number that has not been appeared before.

Therefore,the random variable $X_i$ is geometric with parameter $p_i = (6−i)/6$, therefore $E[X_i] = 1/p_i = 6/(6-i)$.

It follows that $E[X] = 1 + 6\sum_{i=1}^{5}1/i$.

Hence $E[X]=14.7$.


Graph contains triangle Prove that if a simple graph of order n has more than n^2/4 edges then it contains a triangle.

I know Martels theorem states the opposite condition for a triangle free graph but I'm not sure how to prove this condition.

Let $G$ be a simple graph of order $n$. Then Mantel's Theorem states that:

  If $G$ is triangle-free, then $G$ has at most $n^2/4$ edges.

But by taking the contrapositive of this implication, we get exactly what we want:

  If $G$ has more than $n^2/4$ edges, then $G$ contains a triangle.

Let $G$ be a graph of order $n$ and size $m$. Assume that $G$ is $K_3$-free. Then by Mantel's Theorem we know that $m\leq {n^2\over 4}$. However, this is a contradiction since we are given that $m> {n^2\over 4}$. Thus if $G$ has order $n$ and $m>{n^2\over 4}$, then $G$ contains a triangle.


Find the values of k for which the equation $x^3-9x^2 +24x +k =0$ may have multiple roots and solve the equation in each case. This question is from Theory of Equations. Help please.

A polynomial $\;f(x)\;$  over a field has a multiple root $\;\alpha\;$ iff $\;f(\alpha)=f'(\alpha)=0\;$. Now, here we have

$$f(x)=x^3-9x^2+24x+k\implies f'(x)=3x^2-18x+24=0\iff$$

$$x^2-6x+8=(x-2)(x-4)=0\iff x=2,4$$

So it must either $\;x=2,\,4\;$ is a multiple root. Now check for what values of $\;k\;$ this happens.

One way to solve the problem is to find the roots of the derivative of the polynomial $p(x)$, because if $p(x)$ has multiple roots, then $p'(x)$ will have the same roots. To see this, say $a$ is a root of $p(x)$ with multiplicity $n$, then we can write $p(x)=(x-a)^ng(x)$ where $g(a)\not=0.$ Then $p'(x)=n(x-a)^{n-1}g(x)+(x-a)^ng'(x)$, i.e. $a$ is also a root of $p'(x)$.

So to solve your problem, you can take the derivative of your polynomial and then find the roots and then put those roots back to the original polynomial to calculate the values of $k.$


Am I missing something in the answer to Spivak's Calculus, problem 5-39(vi)? Am I missing something in the answer to Spivak's Calculus (4E), problem 5-39(vi)?

Earlier, problem 5-39(v)(p. 113) has established that $$\mathop{\lim}\limits_{x \to \infty}\sqrt{x^{2}+2x}-x=1$$ and then problem 5-39(vi) asks for $$\mathop{\lim}\limits_{x \to \infty}x\left(\sqrt{x+2}-\sqrt{x}\right).$$

Spivak's key (Combined Answer Book, p.78) determines this answer by going through several algebraic manipulations; but for $x>0$, $$x\left(\sqrt{x+2}-\sqrt{x}\right)=\sqrt{x}\left(\sqrt{x^{2}+2x}-x\right)$$ so that $$\mathop{\lim}\limits_{x \to \infty}x\left(\sqrt{x+2}-\sqrt{x}\right)=\left(\mathop{\lim}\limits_{x \to \infty}\sqrt{x}\right)\cdot1$$ can be determined by simply substituting the limit found in the preceding problem. Does taking this "shortcut" miss some required steps?

I wouldn't ask, except that where Spivak can refer to the answer to a previous problem to save work and space, he nearly always does. When he goes to the trouble of working out all the steps it is usually to illustrate something that would otherwise have been missed. But it seems that here the thing to observe is that the previously found limit exists and can be used.

I see nothing wrong with your argument. The only possibly delicate point is that $\lim\limits_{x\to\infty}\sqrt{x}=\infty$. Since $\infty\cdot1$ is not an indeterminate form, everything seems copacetic.

$\lim_{x \to \infty}(\sqrt{x+2}-\sqrt{x})\frac{(\sqrt{x+2}+\sqrt{x})}{(\sqrt{x+2}+\sqrt{x})}x=$

$=\lim_{x \to \infty}\frac{x+2-x}{\sqrt{x+2}+\sqrt{x}}x=\lim_{x \to \infty}\frac{2x}{\sqrt{x+2}+\sqrt{x}}=$

$=\lim_{x \to \infty}\frac{2x/x}{\sqrt{x+2}/x+\sqrt{x}/x}$$=\lim_{x \to \infty}\dfrac{2}{\sqrt{\frac{1}{x}+\frac{2}{x^2}}+\sqrt{\frac{1}{x}}}=\frac{2}{0}=\infty$


How do I sell out with abstract algebra? My plan as an undergraduate was unequivocally to be a pure mathematician, working as an algebraist as a bigshot professor at a bigshot university.  I'm graduating this month, and I didn't get into where I expected to get into.  My letters were great and I'm published, but my GRE was bad and my grades were good but not perfect.  My current plan, I guess, is to start a PhD program at my backup school, then reapply to the better schools next year.

Reality is starting to hit, though, and I'm starting to think about "selling out."   I would still love to work in algebra, but I'm not as in love with the Ivory Tower as I was a few years ago, and I don't want to give up my entire life for it.  If the institution isn't going to let me do what I wanted to do, or if I'll never be as talented as I wanted to be, it isn't worth the sacrifice.  In other words, I'd rather be a well paid applied mathematician in industry than a poor, mediocre pure mathematician at a low end university.

The problem is it seems that most of the applied jobs out there are all about analysis / continuous mathematics, and I am firmly in the algebra / discrete camp. I really do not want to spend my life solving fluid flow PDEs. I always hear about cryptography as an "applied algebra" job, but I'm not particularly crazy about working for NSA or a telecom (plus crypto can't be the only option).

I read some of the answers from Can I use my powers for good? but it's not clear to me which of these suggestions value algebraic thinking.  Many seem very quantitative, rather than structural - is it possible to avoid this in industry?  Also, I have a lot of debt from a long undergraduate career across several majors, so "how much" is unfortunately also a concern.  I don't want to sell out cheap.

  Are there applied math jobs in industry which focus on structural mathematics reminiscent of abstract algebra, earn an appreciably high salary, and aren't cryptography?

and

  How would one best go about pursuing these jobs starting as a recent graduate / first year graduate student?

First, let me say that I'm impressed by your maturity and wisdom. It's not easy to recognize your own limitations, accept them, and adapt. Most people have to learn the hard way, by living through a few decades of struggle and frustration. Some people actually enjoy struggle and frustration, though. Your choice.

I have been an "industrial" mathematician for 40 years (I "sold out" long ago). I work in the software industry. I don't have a Ph.D, and I don't write research papers (not very often, anyway). I don't spend a great deal of my time doing mathematics, and almost no time at all doing original mathematics, but I do write "mathematician" on my tax return every year. My work is interesting (to me), and I've made quite a lot of money.

From the suggestions below, it's clear that some people judge the fabric of a profession by reading its research literature. This is a hopelessly misleading approach. What happens in day-to-day work in any industry is very far removed from what you read in research papers. If you want to know what it's like working in industry, you should ask people who work in industry. And this is not a very good place to do that. Most of the people who hang around here are university faculty, grad students, and (recently) kids trying to get someone to do their homework for them. If you want to know what software developers do, for example, ask at StackOverflow.

I'll repeat some of the advice from others. Learn some computer science. Learn about basic algorithms, and get good at programming in some mainstream language like C/C++ or Java (not Haskell or OCaml). It's not that difficult, and it's great fun when your code works.

Accept that no-one is likely to pay you to do original research (except on a very small scale as part of a larger project). Especially not mathematical research. People in industry are expected to create working saleable products/processes/systems with a high degree of predictability. Research is too risky. If it were less risky, and the results were more certain, then it wouldn't be research.

Think about what it means to "sell out". One definition says that selling out is doing what society (and your employer) want you to do, rather than what you want to do. But society (or your employer) will only be willing to pay you if your work is valuable to them. So, in some sense, selling out is inevitable unless you're going to be a hermit poet or you're independently wealthy. The best you can hope for is that your work is interesting and fulfilling (in addition to being valuable), and that you don't have to do anything that you find morally distasteful. If you think that making money is distasteful, stick to academia.

To answer your question, I'm not personally aware of any places in industry where significant numbers of people spend time pondering the workings of abstract algebraic constructs. I don't say that they don't exist -- just that I'm not personally aware of them. My mathematical work mostly involves differential geometry (in 2D and 3D space), approximation of functions, numerical methods (root finding, minimization, etc.), very simple linear algebra, and occasionally a bit of algebraic geometry. I very rarely do any original mathematics. I typically use software packages written by other people, and I only need to know enough mathematics to understand the limitations of these packages and their applicability to my problems. If you want to work on the development of the mathematical software tools used by people like me, check out companies like Wolfram, MathWorks, MapleSoft, Rogue Wave, NAG. But be aware that these are (mostly) fairly small companies and they don't employ very many people. And they won't hire you unless you have good programming skills.

I mostly work with manufacturing companies -- people who design cars, airplanes, consumer electronics gadgets and so on. Think about what those companies are trying to do -- they want to create more attractive products, more quickly, with lower costs. How can you (and your expertise) help them do that? Contemplate this until you identify some place where you can imagine that you might fit in and be happy. Or, pick some other industry and go through the same sort of reasoning. The key is to find some place where your skills can add value.

Stop thinking of your work as your life. You'll still be the same person, regardless of whether you're winning Fields Medals or hacking code. Your children will love you just as much either way.

NSA is not the only place for cryptographers. There are various research labs like HP   , Microsoft Research  labs where cryptographers and number theorists are hired.

Also, algebraic coding theory is another area where you can apply your skills of abstract algebra.


Finding limit without using L'Hopital rule How to prove $$\lim_{x \to \infty} \frac {\ln(x)}{x}=0$$ without using L'Hospital Rule. Just by using some basis limit properties.

For sufficiently large $x$:$0\le \ln (x)\le \sqrt x$. Now divide by $x$ and squeeze.

The Taylor series for $\ln(x)$ in $-1\le x \le 1$ is given by :

You can substitute this and then apply the basic limits results to get the answer.


A summation with binomial coefficients Evaluate

There seemed to be some problem with stackexchange's math rendering but Ian corrected whatever error was there in the expression.Thanks

$$5050 \frac {\left( \sum _{r=0}^{100} \frac {{100\choose r}}{50r+1}\cdot (-1)^r\right) - 1}{\left( \sum _{r=0}^{101}\frac{{101\choose r}}{50r+1} \cdot (-1)^r\right) - 1}$$

The original definite integral that led to this was

$$5050\frac{\int_0^1(1-x^{50})^{100} dx}{\int_0^1(1-x^{50})^{101} dx} $$

I used binomail to arrive at the above expression.

I already know a technique by definite integration but I want one by sequence and series.

   The answer is 5051

Use of sequence and series is not suggested; and a possible way is outlined, which is very long and useless, this is due to the fact of bounded multiplication of r with 50 in denominator, if the r would have been free, it would be easy to use this method to caluculate:

Brute Force

$$\frac{100\choose r}{50r+1}=1-50.\frac{r.{100\choose r}}{50r+1}$$
Use: $$r{100\choose r}=100.{99\choose{r-1}}$$
Then
$$\frac{{99\choose{r-1}}}{50r+1}=\left[1-50.\frac{r.{99\choose {r-1}}}{50r+1}\right]$$
Again use:
$$r{99\choose {r-1}}=99.{98\choose{r-2}}\text{ after }\frac{r.{99\choose {r-1}}}{50r+1}=\frac{(r-1).{99\choose {r-1}}}{50r+1}+\frac{{99\choose {r-1}}}{50r+1}$$
and so on...

The way you are mentioning- definite integration of $\beta$-function (though you may not be realising it is the $\beta$-function)- is the best method in my opinion.

You should use these(preferred but not necessary) while summing binomial coefficients:
$$r{n\choose r}=n{{n-1}\choose{r-1}};\frac1r{n\choose r}=\frac1{n+1}{{n+1}\choose{r+1}}$$
Secondly, use partial fractions or partial rationals.


$X_n \rightarrow_{a.s.} X$ versus $X_n=X$ wpa 1 Consider a sequence of real-valued random variables $\{X_n\}_{\forall n \in \mathbb{N}}$ and a real-valued random variable $X$

All r.v. are defined on the probability space $
(\Omega, \mathcal{F}, P)$

Could explain what is the relation (equivalent, one implies the other, etc) between

$$
(1) \hspace{1cm}X_n \rightarrow_{a.s.} X \text{ as $n\rightarrow \infty$}
$$

and
$$
(2) \hspace{1cm}X_n =X \text{ with probability approaching 1 as $n\rightarrow \infty$}
$$

Some considerations:

Using the definitions,

(1) $P(\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty}X_n(\omega)=X(\omega))=1$

(2) $\lim_{n\rightarrow \infty} P(\omega \in \Omega \text{ s.t. } X_n(\omega)=X(\omega))=1$

So (1) has the limit inside, (2) has the limit outside.

(2) may look very similar to $X_n\rightarrow_pX$, where $X_n\rightarrow_pX$ means that $\forall \epsilon>0$ $\lim_{n\rightarrow \infty} P(\omega \in \Omega \text{ s.t. } X_n-X\leq \epsilon)=1$

What can we deduce from here?

In fact, $(2)$ is stronger than convergence in probability.  (Assume $(2)$ and take $\epsilon > 0$.  Then $\{X_n = X\} \subseteq \{|X_n - X| < \epsilon\}$.)

Under the given conditions, there are no implications between these statements.  Take $(\Omega, \mathcal{F}, P) = ([0,1], \mathcal{B}([0,1]), \lambda)$. i.e. the Borel sets on $[0,1]$ equipped with the Lebesgue measure.

when both statements are true: simply take $X_n = c$, i.e. be a constant sequence of constant random variable.
$(1)$ holds but $(2)$ does not: adapt the textbook example for pointwise-convergence.  Take $X_n(\omega) = \omega^n$ and $X \equiv 0$.

$X_n \overset{a.s.}{\to} X$
$\forall n \in \Bbb{N}, P(X_n = X) = P(\{0\}) = 0$

$(2)$ holds but $(1)$ does not: adapt the classic "sliding bump" functions for showing that convergence in measure doesn't implies convergence almost everywhere.  Take $Y_{m,n} = 1_{\left[\frac{m-1}{n}, \frac mn \right]}$ for $m \in \{1,\dots,n\}$ and $n \in \Bbb{N}$.  i.e. $Y_{m,n}$ is a "bump $\left[\frac{m-1}{n}, \frac mn \right]$ of width $\frac1n$ sliding from left to right as $m$ runs from $1$ to $n$".  Enumerate $(Y_{m,n})$ as $\{Y_{1,1},Y_{2,1},Y_{2,2}, \dots\}$ and denote this sequence of random variables as $X_k$.

$X_k$ oscillates almost surely as every point on $[0,1]$ is "visited by infinitely many bumps".  For all $\omega \in [0,1]$, choose $n \in \Bbb{N}$.  As the "bumps" $\left[\frac{m-1}{n}, \frac mn \right]$ cover the whole space $[0,1]$, there exists an $m$ so that $Y_{m,n}(\omega) = 1$.  Therefore, $Y_{m,n}(\omega) = 1$ infinitely often.  Idem for $Y_{m,n}(\omega) = 0$, so $(1)$ doesn't hold.
$P(Y_{m,n} = 0) = \dfrac{n-1}{n} \xrightarrow[n\to\infty]{} 1$, so $(2)$ follows.

Remarks: $(2) \implies (1)$ is "partially true", in the sense that $(2)$ implies convergence in probability implies convergence of a subsequence almost surely.

TL;DR Not a full answer but no inclusion for similar reasons to Is it correct to say that ($\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$?

I like to this of this in terms of Fatou's lemma and important inequalities:

Fatou's Lemma:

$$
\mathbb P\left(\liminf A_n\right) \leq \liminf\mathbb P\left(A_n\right)\leq \limsup\mathbb P\left(A_n\right) \leq \mathbb P\left(\limsup A_n\right).
$$

Important inequalities:

Williams - Probability with Martingales

Deduced similarly:

(iii) If $\liminf x_n > z$, then

$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n < z $, then

$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)

$(2)$ is read as $\lim P(X_n = X) = 1$. By Fatou's Lemma, we have equality to 1 for the middle and RHS equality but there's no guarantee that the LHS equality equals 1 too. Even if LHS equals 1, $P(\lim[X_n=X])=1$ doesn't imply $P([\lim X_n]=X)=1$. Besides constructing counterexamples, you can also try to change equalities to inequalities with $\varepsilon$ and try, to no avail, to use the important inequalities to prove the implication.


In how many ways can you split six persons in two groups? In how many ways can you split six persons in two groups?

I think that I should use the binomial coefficient to calculate this but I dont know how.

If the two groups has to have equal size, then each group should have three persons. Then the calculation would be:

$${{6}\choose{3}}/2 = 10$$

But since the groups now can have different sizes I dont really know. I would guess that you have to add up multiple binomial coefficients, something like this:

$$ {{6}\choose{1}} + {{6}\choose{2}} + {{6}\choose{3}}=41$$

But the answer should be 31 so that does not seems to be quite right. I would like some suggestions on this!

I assume the two groups must be non-empty!  It helps to know that Alicia is one of the people.

Joining Alicia will be any subset of the remaining people, except for the full set. So there are $2^5-1$ ways to choose the people who will be in Alicia's group.

Total number of combinations of 3 from 6 = 6C3 = 20
But this answer needs to be divided by two as it asks for number of groups so each Combo of three cancels out its opposite pair. 20/2 = 10.

Number of uneven groups = number of 1 and 5s and number of 2 and 4s
= 5C1 + 4C2 = 21.  Again splits of 5 and 1 and 4 and 2 are not considered as they are simply mirrors of the 1-5, 2-4 combinations.


Do you need true randomness to beat the two-envelope game? A well-known (non-)paradox in probability involves a two-envelope game played between two players, $A$ and $B$:

$A$ selects two distinct (real) numbers, $x$ and $y$, writing each one down on a card and sealing each card in an envelope, then presenting the two envelopes to $B$.
$B$ chooses one of the envelopes and looks at the card inside.  They then guess whether the number on the card they've chosen is the larger or smaller of the two numbers.

The 'paradox' here is that regardless of $A$'s scheme for choosing numbers — and even if $A$ knows $B$'s strategy in advance — there's a strategy for $B$ that will achieve a better-than-even success rate in the long run: choose an envelope at random, then map the number in it onto the interval $(0,1)$ using some (arbitrary) monotonic function.  Choose a random deviate $U\in(0,1)$, and then guess 'higher' or 'lower' according to whether (the mapping of) the number looked at is higher or lower than the generated random deviate.

I'll skip the analysis of this strategy here (see Do better than chance or Who discovered this number-guessing paradox? for more details), but note that it explicitly relies on having a source of random deviates. My question is whether this is necessary for $B$ to have the advantage.  More specifically, consider the following variant of the game:

B chooses computable functions $f():\mathbb{N}\mapsto\{0,1\}$ and $g():\mathbb{N}\mapsto\mathbb{Q}\cap(0,1)$.  Note that $A$ knows nothing about these functions, other than that they are computable.
For each integer $n$, in turn:

$A$ selects two distinct real numbers $x,y\in(0,1)$, writing each down on a card and presenting them in sealed envelopes.  (I'm restricting the numbers here to eliminate the mathematically-moot mapping step.)
$B$ computes $f(n)$; if $f(n)=0$ then $B$ chooses $x$, and if $f(n)=1$ then $B$ chooses $y$.  Call $B$'s chosen number $z$.
$B$ computes $g(n)$; if $g(n)\leq z$ then $B$ guesses 'higher', otherwise $B$ guesses 'lower'.

Note that this is essentially $B$ following the strategy in the usual version of the game, except that $B$ is following a computable strategy rather than a purely randomized one.

  Can $A$ win this game in the long run?

Since $A$ doesn't know what computable strategy $B$ is following, they'll clearly have to take some dovetailing approach; instinctively it feels like randomness is inherent in $B$'s ability to win with the usual strategy and that $A$, with the knowledge that $B$'s strategy is actually computable, should be able to 'game the system' and win.  Unfortunately, I can't see a clear proof here (and I wouldn't be entirely shocked to learn that I'm wrong).  Is anything known about this problem?

EDIT: to clarify, I should point out that unlike $B$, the strategy that $A$ follows does not have to be computable; $A$ can, for instance, take advantage of an oracle that enumerates the total computable functions.  For example, this ensures that $A$ can guarantee they'll win at least once: enumerate all possible pairs $\langle f_n(), g_n()\rangle$ of recursive functions and in round $i$ behave as though $B$'s selections for the round will be $f_i(i)$ and $g_i(i)$ (by choosing values that will win given that these are $B$'s selections).  Also, I suspect $f()$ is actually superfluous and that we can ask the analogous question for the strategy where $B$ always guesses $x$, but if the presence or absence of $f()$ does matter then it'd be interesting to know that too.

Your question is too ambiguous. I assume that you are asking for a strategy for A that guarantees winning in the long run no matter what strategy B chooses that is unknown to A, but even in that case it is not clear what is considered as winning. In this answer I assume that you want the limit infimum of the ratio of wins to games to be more than half.

(1) A plays the same against any B : A cannot guarantee to win!

If A's strategy does not depend on the game history, then A certainly has no winning strategy because for any strategy for A that wins against a strategy for B, the opposite strategy for B (pick the same way but guess the opposite) wins against that strategy for A. This applies even if A's strategy is not computable. However, B may have a better strategy if A is restricted, as we shall see below.

(2) A and B plays computably : Neither can guarantee to win

If A has a computable strategy, we can trivially construct a strategy for B so that B will always win A, simply by simulating A's strategy for each move and playing against it. Similarly if A knows B's strategy A can always wins. Thus there is no single strategy that either side can use to guarantee a win, just like in Scissors-Paper-Stone. Note that this applies even if the opponent does not remember the game history, since the counter-example strategy does not need to.

(3) A can use the halting oracle and B plays computably: A wins!

For the $k$-th move, A runs all halting computable functions on the entire game history as input. If one of them outputs a sequence of length $k$ that matches all the previous $(k-1)$ moves by B, then A plays against that sequence for the current move. As before, if more than one halting computable function does, A takes the one with the smallest description. If B is playing by some computable function, then clearly A would eventually settle on an equivalent function with smallest description (guaranteed to exist by induction) because eventually no function with smaller description would satisfy the criterion. From that point onwards A would win every game!

(3') B can use the halting oracle and A plays computably: B wins!

Same argument as in (3).

Fonctions $f$ and $g$ are total recursive functions. For player $B$ to be able to game the system, the main problem is that the set of total recursive functions is not recursively enumerable. Hence he can't (as it is the only way) enumerate them until he finds some functions that match the values already given, knowing that ultimately this process will give him the real functions $f$ and $g$ used by payer $A$.

But in practice, player $B$ can observe the time used by player $A$ to give his answers, and then guess the time (or more precisely, the number of computations steps) needed to compute $f$ and $g$. Knowing that he can find the smaller program that match both answers of $f$ and $g$ already known in the (estimated) computation time. The process will ultimately converges to the real $f$ and $g$ and leads $B$ to ultimate victory.


Give an example of an operator on a finite dimensional vector space. Give an example of an operator on a finite-dimensional real vector space such that 0 is the only eigencvalue of T but T is not nilpotent.

I've been stuck on this problem for a while, the main example that comes to mind is the shift operator because the only eigenvalue of it is $\lambda = 0$ such that $T(z_1, \dots z_n) = (0, z_1, \dots, z_{n-1})$

However this is obviously nilpotent

So instead I tried an operator that rotates like this

$T(z_1, \dots z_n) = (z_n, z_1, \dots, z_{n-1})$

or flips them so that $T(z_1, \dots z_n) = (z_n, \dots, z_1)$

But neither of these have an eigenvalue of $lambda = 0$ because $z_1 \dots z_n$ would all have to be zero.

The differentiation map is also nilpotent.

Can anyone help me?

The result isn't true if $0$ is the only eigenvalue of $T$. This is because, then the characteristic equation for $T$ will be $x^n=0$, and by Cayley-Hamilton theorem, $T^n=0$ i.e. $T$ is nilpotent. I guess the question means that $T$ has only one real eigenvalue- which is $0$. The other eigenvalues are complex.

If that's the case then try to find a $3$x$3$ matrix which is similar to a diagonal matrix with diagonals $0,i,-i$. (This is easy).

Take $V=\Bbb R^n$ ($n\geq 3$) real vector space , and $(e_1,\ldots, e_n)$ a canonical basis, and $T\in \mathcal(L)(V)$ defined by: $T(e_1)=e_2$ , $T(e_2)=-e_1$, and $T(e_k)=0$ for $k\geq 3$. $0$ is the only eigenvalue of $T$ and $T$ is not nilpotent.


Given the points $A,B,C,D$ in a straight line $m$ and $A,E,F,G$ in a straight line $n$, how many triangles can be formed with these points? Given the points $A,B,C,D$ in a line $m$ and $A,E,F,G$ in a straight line $n$, how many triangles can be formed with these points?

I've done the following:

I've used the following heuristic: Chose one point on $m$, then another point on $m$ which is different of the previous chosen point. This yields $4*3$, now choose a new point on $n$, this gives me $4*3*3=36$, now choose one point on $n$ and another point on $n$ different of the last point chosen and different of $A$. This gives me $3*2*3=12$ and hence $36+12=48$.

I've also tried to do it in the following way:

$${\text{#points in m} \choose 2}*\text{#points in n (w/o A)}+ {\text{#points in n (w/o A)} \choose 2}*\text{#points in m (w/o A)}$$

Then:

$${4 \choose 2}*3 + {3 \choose 2}*3=27$$

But none of them is right. I may be missing some triangles but I cant see which ones are these.

Outline: First let us count the triangles that have $A$ as a vertex. We need to choose a point other than $A$ on each line.

Next we count the triangles that don't have $A$ as a vertex. We pick $2$ points other than $A$ on one line, and $1$ on the other, or the other way around.

Here is an example,
If you choose BEGC: one line from BE, one from GC, and one from EG--> you have one more triangle. Do the same  for the rest and I think you may get the result.


$K$ is a limit point of $(x_{n})_{n = 0}^\infty \Rightarrow$ There is a subsequence of $(x_{n})_{n = 0}^\infty$ that converges to $K$. I have to prove this exercise for my math study. It's form Terence Tao's Analysis I:

Assume $(x_{n})_{n = 0}^\infty$ is a sequence of real numbers, and $K \in \mathbb{R}$.
I have to prove the following:

$K$ is a limit point of $(x_{n})_{n = 0}^\infty \Rightarrow$ There is a subsequence of $(x_{n})_{n = 0}^\infty$ that converges to K.

This is my proof:

$K$ is a limit point of $(x_{n})_{n = 0}^\infty$
$\Rightarrow$ $\forall$ $N \in \mathbb{N}$, $\forall$ $\epsilon > 0 \in \mathbb{R}$ there is a $n \geqslant N $ such that $|x_{n} - K| < \epsilon$
$\Rightarrow$ Now take the subsequence of all $x_{n}$ with $n \geqslant N$ for wich $|x_{n} - K| < \epsilon$ and call this sequence $(a_{n})_{n = 0}^\infty$
$\Rightarrow$ $(a_{n})_{n = 0}^\infty$ is a subsequence of $(x_{n})_{n = 0}^\infty$ and it converges to K
$\Rightarrow$ There is a subsequence of $(x_{n})_{n = 0}^\infty$ that converges to K

Is this correct? and if not, how should I do it?

Thanks in advance!

Here there are two cases.

Case 1: Suppose $\{x_n|n\in \mathbb{N}\}$ is finite. Then there exists a sequence $(n_k)$ of positive integers such that $n_1<n_2\dots<n_k<\dots$ and $x_{n_k}=x_{n_1}=K$ for each $k\in \mathbb{N}$. Then $(x_{n_k})$ is convergent and $\lim\limits_{k\to \infty}x_{n_k}=K$.

Case 2: Now suppose $\{x_n|n\in \mathbb{N}\}$ is infinite. Since $K$ is a limit point of $(x_n)$, for each $ \epsilon >0$, $\left( B_{\epsilon}(K)\setminus \{K\}\right) \cap \{x_n|n\in \mathbb{N}\}\ne \varnothing $ where $B_{\epsilon}(K)$ is the open ball with center $K$ and radius $\epsilon$.

Let $n_1=\operatorname{least element of }\{n\in \mathbb{N}|x_n\in B_{\epsilon}(K)\setminus \{K\}\}$ and $ r_1=\min \{1/2,|x_{n_1}-K|\} $.

Let $n_2=\operatorname{least element of }\{n>n_1|x_n\in B_{r_1}(K)\setminus \{K\}\}$ and $ r_2=\min \{1/3,|x_{n_2}-K|\} $.

Let $n_3=\operatorname{least element of }\{n>n_2|x_n\in B_{r_2}(K)\setminus \{K\}\}$ and $ r_3=\min \{1/4,|x_{n_3}-K|\} $.

$\vdots$

Proceeding in this manner obtain a sequence of integers $(n_k)$ such that

$n_k=\operatorname{least element of}\{n>n_{k-1}|x_n\in B_{r_{k-1}}(K)\setminus \{K\}\}$ where $ r_{k-1}=\min \{1/k,|x_{n_{k-1}}-K|\} $.

Thus $(x_{n_k})$ is a sub sequence of $(x_n)$ such that $|x_{n_k}-K|<\frac{1}{k}$ for each $k\in \mathbb{N}$.

So $\lim\limits_{k\to \infty}x_{n_k}=K$. That is $(x_{n_k})$ converges to $K$.

Let $\varepsilon > 0.$ For every $N \in \mathbb{N},$ let $n_{N}$ be an integer such that
$$|x_{n_{N}} - K| < \varepsilon.$$
Then the sequence $(x_{n_{N}})$ is a subsequence of $(x_{n})$ and $\to K.$


How determine all functions $f: R \mapsto R$ differentiable, are not identically zero and such that $f(x)= x f^{\prime}(\frac{x}{\sqrt{3}})$? How determine all functions $f: R \mapsto R$ differentiable, are not identically zero and such that $f(x)= x f^{\prime}(\frac{x}{\sqrt{3}})$ for $x \in R$?

One class of solutions is $f(x) = a x + b x^3$ for constants $a,b$.  These are all the solutions that are analytic.  There are also non-analytic solutions such as
$$ f(x) = \cases{ a x + b_1 x^3 & for $x \ge 0$\cr
                      a x + b_2 x^3 & for $x < 0$} $$
Also, if $r = a + b i$ is any of the (I would guess infinitely many) complex zeros of
the function $g(z) = 3^{z/2} - \sqrt{3} z$ with $a = \text{Re}(r) \ge 1$,
$ |x|^a \cos(b \log |x|)$ and $ |x|^a \sin(b \log |x|)$ are solutions (defined to be $0$ at $x=0$).  Again you can "mix and match" the solutions for positive and negative $x$, and take linear combinations.


How can I tell if one polynomial is greater than another over an interval? $y_1=ax^2 + bx + c$

$y_2=dx^2 + ex + f$

Question: what is the proper way for me to test that $y_1 > y_2$ over an interval (say 0 to 10, inclusive) for all numbers in that interval?

I assume it has something to do with testing the integral of y1 - y2, but not sure.

=== Added some context ===

I'm working on a continuous game theory program written in Python, and am working on the concept of strict domination. $y1$ and $y2$ are meant to be the utility functions for a given agent. $y1$ strictly dominates $y2$ if, over the range of other player actions in question (i.e. the interval of $x$), $y1>y2$.

Solve $ y_{1} = y_{2} $ and then use the solutions (if there are any) to construct the sign diagram of $ y_{1} - y_{2} $ over the interval.

Having $ \displaystyle \int_{a}^{b} (y_{1} - y_{2}) \,d{x} > 0 $ is a necessary condition for $ y_{1} > y_{2} $ over the interval $ [a,b] $, but definitely not sufficient.

One method is to check the first point, so in this case, check $y_1(0)\geq y_2(0)$. Then consider the derivatives and verify that $y_1'(x)\geq y_2'(x)$ for all $x\in[0,10]$.


Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui  a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$

Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}
\implies a_{n+1} = \frac{4a_n} {2a_n - 3}
\implies a_{n+1} = \frac{\frac{4a_n} {a_n} } {\frac{2a_n}{a_n} - \frac{3} {a_n} }
\implies  a_{n+1} = \frac{4} {2 - \frac{3} {a_n} }
\implies \frac{1 } {a_{n+1}} = \frac{2 - \frac{3} {a_n} } {4}
\implies \frac{1 } {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}$$
Then go to where????

If you have proved that the sequence has a limit $l$, then $l$ can be calculated by the following:

$$\lim_{n\to \infty}2a_{n+1}a_n = \lim_{n\to \infty}(4a_n + 3a_{n+1}) $$

and hence

$$2l^2=4l+3l$$


How to do division of two numbers which are already under modulo 'm'? How to do division for the following example?

Case 1 : Without modulo

n1 = 40, n2 = 8

Quotient = n1/n2 = 5

Case 2 : With modulo

m = 6

n1 = n1 mod m = 4 (AND)
n2 = n2 mod m = 2

Quotient = 4 / 2 = 2

Now,In case 1, Quotient = 5, but in case 2, Quotient = 2. How to do division on numbers such that both numbers are already under modulo 'm'?

This only works if $m$ is a prime and thus the group of residues is a field. If $m$ is not prime, the corresponding congruences modulo $m$ don't have unique solutions.

In your case, the “quotient” $x\equiv4/2$ corresponds to the congruence $2x\equiv4$. Modulo $6$, this congruence has two different solutions, $x=2$ and $x=5$. More generally, if $m$ is not prime and $a$ in $ax\equiv b$ is not coprime with $m$, then the solution is only fixed up to multiples of $m/\gcd(a,m)$.


How can I improve my problem solving/critical thinking skills and learn higher math? I'm a rising sophomore in high school. So far, I've taken Algebra One, Two, and Geometry in school. I want to learn higher math such as precalculus/trigonometry, calculus, linear algebra, and more, so I can go into topics such as cryptography, advanced computer science, and possibly take the AMC and other olympiad tests (I'm not too interested in that).

The only problem, though, is that my abilities in problem solving and other stuff in math aren't that good. I do pretty well in my classes (high As) but that doesn't mean anything. The U.S. system doesn't seem too good in actually teaching math.

For example, I can do whatever is on my homework or tests. But, if I'm given a more difficult problem than usual concerning a topic I learned (say logarithms or something), I can't solve it.

I feel like this is going to be a hindrance to me learning higher math, doing well in more difficult subjects like calculus and linear algebra, doing well on olympiad tests, and going into math-heavy fields like computer science and cryptography.

So, how can I change all of this and improve my skills? Are there any books that teach problem-solving, mathematical thinking, and higher math (or something like precalculus)? Again, I want to better these skills so I can do well not only in math, but other fields.

Any help is really appreciated.

I'm going to take a different approach. Yes, you should buy the Polya books, I also recommend looking at learning how to learn on Edx for an interesting take on learning techniques. But do something else as well: watch the Khan academy videos on trig, then the first couple of MIT OCW calc videos. Then take Robert Ghrists Calculus course on Coursera and take the A.P. Calculus exam (I actually did this in one year, and  it wasn't very hard-not because I'm so smart, I know from experience that I am at best mediocre in a real math class. It's just A.P. Calculus doesn't take a ton of real math skill). Then, for the final step, see if you can take classes at a local Univ. in real math. Their is no way to learn math like learning from actual mathematicians, this will get you college credit, and it will look good applying to college.

Agreed on the US school system... the best way to fix it is to forget everything they're trying to teach you! Okay maybe don't forget everything, but in my experience US math classes often try to teach you how to solve problems the wrong way, which is why many students end up hating math. Math isn't about following a series of steps towards a defined problem, computers do that. It's about creatively finding those steps.

My best advice is to do your best to forget about memorizing methods, and focus on understanding why those methods work. I'll always remember my Calculus 1 final exam where I derived the equation for arc length in the margins because I forgot it. My professor thought I wanted extra credit.

It's all well and good to be able to spit out that the area of a triangle is (1/2)bh, but it is much, much more helpful to know that a triangle is just half of a rectangle.


Using factorisation, $A=PJP^-1$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.

$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{-1} & \mathbf{1} \end{bmatrix}. $$

Not sure how to finish this problem as the matrix $J$ is not diagonal, but $J^k$ can still be found somehow. How exactly is $J^k$ obtained in this case?

You have basically put your matrix $A$ in Jordan normal form.
This means that you can split your matrix $J$ as follows:
$$J = D + N$$
where $D$ is diagonal and $N$ is nilpotent, i.e. there exists an a positive integer $n$ such that $N^n = 0$.

In your case
$$J = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix}
=
\underbrace{
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}}_{=D}
+
\underbrace{
\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}}_{=N}
$$
and in your case you have $N^2 = 0$.

The basic idea is that when you expand your power $J^k = (D+N)^k$ using the binomial theorem, terms with $N^i$ with $i \geq n$ will disappear, since $N^n = 0$. In your case:
$$(D+N)^k = {k \choose 0} D^k + {k \choose 1} D^{k-1}N + \dots$$
where the remaining terms will be zero, e.g. the next term will be
$${k \choose 2}D^{k-2} \underbrace{N^2}_{=0} = 0.$$

More information can be found in this math.SE post.

Hint: It is related to the Fibonacci numbers.


Equivalent definition of metric compatibility for a connection: what does $\nabla g$ mean? So the definition I know for metric compatibility is:

$$Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ),$$

which make sense, as $g(Y,Z)$ is a smooth function from the manifold to reals and we think of $X$ as a derivation. So now I read this apparent equivalent definition that says $\nabla g=0$. Can someone explain what this means? How can I do $\nabla$ of $g$ I thought $g_p$ is an element of $T_p^*M\otimes T_p^*M$ at every point. Furthermore after you explain the meaning of this can you show me that these two definitions are indeed equivalent?

It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction.

Given a connection

\begin{align*}
\nabla : \Gamma(TM) \times \Gamma(TM) &\to \Gamma(TM)\\
(X, Y) &\mapsto \nabla_XY
\end{align*}

on $TM$, there is an associated connection (which I will also denote $\nabla$) on $T^*M$ given by

\begin{align*}
\nabla : \Gamma(TM) \times \Gamma(T^*M) &\to \Gamma(T^*M)\\
(X, \alpha) &\mapsto \nabla_X\alpha
\end{align*}

where $(\nabla_X\alpha)(Y) := X(\alpha(Y)) - \alpha(\nabla_XY)$. With this definition, together with the definition $\nabla_Xf = Xf$ for a smooth function $f$, we see that the following identity holds:

$$\nabla_X(\alpha(Y)) = (\nabla_X\alpha)(Y) + \alpha(\nabla_XY).$$

More generally, given a $(p, q)$-tensor $T$, we implicitly define the covariant derivative $\nabla_XT$, which is again a $(p, q)$-tensor, by the following equation:

\begin{align*}
\nabla_X(T(Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q)) =&\ (\nabla_XT)(Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q)\\
&+ \sum_{i=1}^pT(Y_1, \dots, \nabla_XY_i, \dots, Y_p, \alpha_1, \dots, \alpha_q)\\
&+ \sum_{j=1}^qT(Y_1, \dots, Y_p, \alpha_1, \dots, \nabla_X\alpha_j, \dots, \alpha_q).\qquad (\ast)
\end{align*}

One could instead consider the covariant derivative of $T$ as a $(p+1, q)$-tensor $\nabla T$ given by

$$(\nabla T)(X, Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q) := (\nabla_XT)(Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q).$$

Now, $g$ is a $(2, 0)$-tensor. So if $Y$ and $Z$ are vector fields, $g(Y, Z)$ is a smooth function and hence

$$\nabla_X(g(Y, Z)) = X(g(Y, Z)).$$

On the other hand, by $(\ast)$,

$$\nabla_X(g(Y, Z)) = (\nabla_Xg)(Y, Z) + g(\nabla_XY, Z) + g(Y, \nabla_XZ).$$

Using these two equations, we see that

$$(\nabla g)(X, Y, Z) = (\nabla_X g)(Y, Z) = X(g(Y, Z)) - g(\nabla_XY, Z) - g(Y, \nabla_XZ).$$

So we see that $\nabla$ is compatible with the metric $g$ if and only if $(\nabla g)(X, Y, Z) = 0$ for all vector fields $X, Y, Z$ (i.e. $\nabla g = 0$).

It means the same thing, $\nabla g=0$ is equivalent to saying that $g$ is parallel relatively to the connection $\nabla$ which is equivalent to saying that $Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_X,Z)$.

It is the answer, the covariant derivative of the $n$ tensor $T$ is defined by $\nabla_XT(X_1,..,X_n)=X.T(X_1,..,X_n)-\sum_i(X_1,..,\nabla_XX_i,..,X_n)$


$∀x(P(x)∨¬P(x))$ fitch proof without premises? How should I go about solving this? Am I able to solve this with contradiction?
I tried starting with $¬∀x(P(x)∨¬P(x))$, but I don't know where to go with it.
Some help would be nice, thank you

Yes, you can prove this by contradiction, but not assume $¬∀x(P(x)∨¬P(x)),$ instead we will assume $¬(P(a)∨¬P(a))$ so we can use $\forall$ Intro :$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\fitch{1.}{\fitch{2.~~\boxed{a}}{\fitch{3.~~\neg(P(a)\lor\neg P(a))}{\fitch{4.~~P(a)}{5.~~P(a)\lor\neg P(a)\hspace{10ex}\lor\text{ Intro }4\\6.~~\bot\hspace{22.9ex}\bot\text{ Intro }3,5}\\7.~~\neg P(a)\hspace{22ex}\neg\text{ Intro }4-6\\8.P(a)\lor\neg P(a)\hspace{14.5ex}\lor\text{ Intro }7\\9.~~\bot\hspace{26.2ex}\bot\text{ Intro }3,8}\\10.~~P(a)\lor\neg P(a)\hspace{16.9ex}\neg\text{ Intro }3-9}\\11.~~\forall x(P(x)\lor\neg P(x))\hspace{16ex}\forall\text{ Intro }2-10}$$


Proving there are infinitely many rational numbers in $[x,y]$ Prove if $x$ and $y$ are real numbers with $x \lt y$, then there are infinitely many rational numbers in the interval $[x,y]$.

What I got so far:

Let $x,y \in \Bbb R$ with $x \lt y$

Let $S = [x,y]$

By the density of $\Bbb Q$ in $\Bbb R$, $\exists r \in \Bbb Q$ such that $x \lt r \lt y$ where $r \in S$.

This is where I got stuck.

If you have only $n$ finitely many rational numbers between $x$ and $y$, put them in order, $x, r_1, r_2, \ldots r_n, y$.  Can you now prove that there were more than $n$ rational numbers between $x$ and $y$?  If so, you have contradicted your original assumption that there were only $n$ such numbers, which means no such $n$ can exists and there must be infinitely many rational numbers between $x$ and $y$.

Consider the interval $[x,y]$.  Find two rational numbers $q_{min}$ and $q_{max}$ such that $x < q_{min} < q_{max} < y$.  Let $r=q_{min}-q_{max}$.  Now make a new number: $q_1 = q_{min} + r/2$.  Let $r_1=q_{max}-q_1$.  Now make a new number: $q_2=q_1+r_1/2$.  Let $r_2=q_{max}-q_2$.  Follow the pattern.  Each of these numbers is a rational number; each number is unique.  The set constructed is countably infinite in size.


Question about a linear operator and its matrix being normal and self-adjoint So I have a linear operator $T$, an orthonormal basis $\epsilon$, and I have calculated its matrix, $A=[T]_\epsilon$. I know that the matrix is normal, but not self adjoint. I found this by calculating $A^*$, and then showing that $A^*A=AA^*$, and that $A\neq A^*$. However, using I am not quite sure how I would show if $T$ is normal or self-adjoint, without calculating $T^*$. Is it possible, using the information I have about $A$?

Since $\epsilon$ is an orthonormal basis, you have

$$ [T \circ T^{*}]_{\epsilon} = [T]_{\epsilon} [T^{*}]_{\epsilon} = [T]_{\epsilon} ([T]_{\epsilon})^{*} = AA^{*} = A^{*}A = [T^{*} \circ T]_{\epsilon} $$

which implies that $T \circ T^{*} = T^{*} \circ T$ and so $T$ is normal. More generally, this argument shows that an operator $T$ is normal / unitary / Hermitian / etc if and only if the matrix representing $T$ with respect to some (or any) orthonormal basis is normal / unitary / Hermitian / etc.

An operator on a finite-dimensional vector space is normal iff it can be diagonalized with orthogonal eigenvectors.


Proving a result of $\text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)$ Show that for $0<\theta<2\pi$:
  $$\text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\frac{1}{2}+\frac{\text{sin}\left(n+\frac{1}{2}\right)\theta}{2\text{sin}\frac{\theta}{2}}$$

I am stuck up to this point
\begin{align}
\text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)&=\frac{1-\text{cos}\theta-\text{cos}(n\theta+\theta)+\text{cos}\theta\text{cos}(n\theta+\theta)}{2-2\text{cos}\theta} \\
&=\frac{1-\text{cos}\theta}{2-2\text{cos}\theta}+\frac{\text{cos}(n\theta+\theta)(\text{cos}\theta-1)}{2-2\text{cos}\theta} \\
&=\frac{1}{2}-\frac{\text{cos}(n\theta+\theta)}{2}
\end{align}

I hint would be very helpful.

HINT

It seems useful refer to   Lagrange's trigonometric identities

$$ \sum_{n=1}^N \cos (n\theta)  = -\frac{1}{2}+\frac{\sin\left(\left(N+\frac{1}{2}\right)\theta\right)}{2\sin\left(\frac{\theta}{2}\right)}$$

observing that for $\theta \neq 2k\pi$

$$\text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\text{Re}(1+e^{i\theta}+\ldots+ e^{in\theta})$$


Question about the polynomial remainder theorem Given that $f(x) = x^3 - x^2 - ax - b$ has a factor $x - 3$ but leaves a remainder of $13x - 11$ when divided by $x + 4$, find $a$ and $b$.

I get that in order for $x-3$ to be a factor then according to the remainder theorem, to find remainder of $f(x)$ divided by $x-a$, calculate $f(a)$ which in this case $f(3)=0$.

So, $f(3)= 18-3a-b=0$

$3a+b=18$ is the first equation found.

It starts to get tricky when dividing $f(x)$ by $x+4$ and finding values of $a$ and $b$ such that the remainder = $13x-11$ to find our second equation according to the answer , $a=5$ , $b=3$

Could someone please help me with this. This is supposed to be grade 10 math and i've got a degree in mathematics yet I still can't get it.


Smallest $k$ Such that $13 + 4 \cdot k \cdot p^2$ is a Perfect Odd Square Given a prime number $p$, I am looking to find the smallest positive integer $k$ such that the following equation $$13 + 4 \cdot k \cdot p^2$$ produces a perfect odd square. All variables are integers. For example, for the prime $43$, $k = 3$. For $p=103$ , it turns out that $k = 1391$. A computer program can solve this for small prime numbers. It is easy to prove that $k$ has to be odd too, which improves the search. But for larger primes, say $p>10^4$, the naive approach of incrementing $k$ untill a suitable value is found just takes a long time.
It is important to mention that not all primes have any solution at all. For those which do have a solution, I am interested in an efficient way for finding it.
Is there any other approach to tackle this? Perhaps one that relates to number theory? Or any other field really which may prove useful.

There is one major optimization screaming at me here.
Check each square sequentially for whether or not it is the “odd square” that the formula equals. This will be faster because $n^2$ (for odd $n$) grows faster than the current linear formula dependent on $k$.
Of course you would start with the first square greater than $13 + 4p^2$ since any lower square is impossible.
This method will be faster when $\frac {n^2}{4p^2} > n - \sqrt{4p^2} = n - 2p$.
I do not know whether or not this equation ever works out to being true. However, for sufficiently large $p$ I strongly suspect that iterating through the squares will be faster.
One may note that my formula assumes that every multiple of $p^2$ needs to be as well as every $n^2$. This cancels out as I would divide both sides by $2$. Therefore it is irrelevant.
EDIT:
I thought about this a bit more. For sufficiently small $k$ iterating through squares will be slower (because the growth rate of sequential squares will be smaller than the growth of sequential multiples of $4p^2$). Once $k > 2p^2 - 1$ the growth of sequential squares outpaces the linear growth of your formula. Therefore you should add something in your code to start counting by squares once you reach $k = 2p^2 - 2$. The value of $n$ to start iterating squares would then be $n = 2p^2 - 1$.
This should be about as fast you can get (assuming $k$ exists) other than iterating through odd values of $k$ and $n$.


Number is equal to nth power of summation of its digits I am working on problem from code wars. Following is the question

The number $81$ has a special property, a certain power of the sum of its digits is equal to $81$ (nine squared). Eighty one ($81$), is the first number in having this property (not considering numbers of one digit). The next one, is $512$. Let's see both cases with the details
$8 + 1 = 9$ and $9^2 = 81$
$512 = 5 + 1 + 2 = 8$ and $8^3 = 512$
We need to make a function, power_sumDigTerm(), that receives a number n and may output the n-th term of this sequence of numbers. The cases we presented above means that
power_sumDigTerm($1$) == $81$
power_sumDigTerm($2$) == $512$
Happy coding!

Only possible way I can think to solve this problem is by using brute force approach. But it is too time consuming Is there any mathematical theory I can apply to solve this problem.

Denis,

try to calculate all powers (2 to 180) ^ (2 to 63), sort this and remove duplicates (3^4 == 9^2).

Now you can go in the list of powers and check if powering the sum of the digits will be the original number. I.e: use your brute force only for this items.

I finished the challenge with total calculation time of 52ms from server.
But to achieve this I reduced the range to fit the problem.


Tangent even derivative How to show that $\tan^{(2n)}(0)=0$ for each $n \in \mathbb{N}$ ? I can't quite see the inductive step (or maybe there's another way to do it).

Prove by induction that every even derivative of tangent is a linear combinaion of odd powers of tangent.

You can use the Taylor series for $\tan{x}$. This is $$\tan{x} = \sum_{k=1}^{\infty} \frac{U_{2n+1}x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + O(x^5)$$


Integration by parts of $\sin(x)e^x$ I'm studying analysis and I have a problem with the integral calculation of this function:

$$\int \sin(x)e^x\mathrm dx$$

I start by using integration by parts and I obtained this:

$$\int \sin(x)e^x\mathrm dx = \sin(x)e^x-\int \cos(x)e^x\mathrm dx$$

But now I don't know how to continue, will I enter in a infinite loop if I repeat integration by parts?

I do it one more time like the comment above said and I have:

$$\int \sin(x)e^x\mathrm dx = \sin(x)e^x-\cos(x)e^x-\int -\sin(x)e^x\mathrm dx$$

And if I move to the left I obtain:

$$\int \sin(x)e^x\mathrm dx+\int -\sin(x)e^x\mathrm dx = \sin(x)e^x-\cos(x)e^x$$

Right?

Another way could be to consider $$I=\int \sin(x)\,e^x\,dx$$ $$J=\int \cos(x)\,e^x\,dx$$ Then $$J+i I=\int e^{ix}\,e^x\,dx=\int e^{(1+i)x}\,dx=\frac { e^{(1+i)x}} {1+i}=\left(\frac{1}{2}-\frac{i}{2}\right) e^{(1+i) x}$$ Expand the last term to get $$J+i I=\frac{1}{2} e^x \sin (x)+\frac{1}{2} e^x \cos (x)+i \left(\frac{1}{2} e^x \sin
   (x)-\frac{1}{2} e^x \cos (x)\right)$$

Doing the same for $$K=\int \sin(ax)\,e^{bx}\,dx$$ $$L=\int \cos(ax)\,e^{bx}\,dx$$ $$L+iK=\int e^{iax}e^{bx}\,dx=\int e^{(b+ia)x}dx=\frac {e^{(b+ia)x}}{b+ia}=\frac {b-ia}{a^2+b^2}{e^{(b+ia)x}}$$ and expanding again $$L+iK=\frac{a e^{b x} \sin (a x)}{a^2+b^2}+\frac{b e^{b x} \cos (a x)}{a^2+b^2}+i
   \left(\frac{b e^{b x} \sin (a x)}{a^2+b^2}-\frac{a e^{b x} \cos (a
   x)}{a^2+b^2}\right)$$

Let $\int\sin xe^x = I$.

$I = \sin xe^x - \cos xe^x -I$

$2I = \sin xe^x - \sin xe^x$

$I = (\sin xe^x - \cos xe^x)/2.$


How many numbers are in the Fibonacci sequence below a given threshold? Assuming I'm asked to generate Fibonacci numbers up to N, how many numbers will I generate? I'm looking for the count of Fibonacci numbers up to N, not the Nth number.

So, as an example, if I generate Fibonacci numbers up to 25, I will generate:

1, 1, 2, 3, 5, 8, 13, 21
that's 8 numbers

How do I calculate this mathematically for an arbitrary "n"?

As mentioned in the comments, Binet's formula,

$$F_n=\frac1{\sqrt{5}}(\phi^n-(-\phi)^{-n})$$

where $\phi=\frac12(1+\sqrt 5)$ is the golden ratio, is a closed-form expression for the Fibonacci numbers. See this related question for a few proofs.

As for counting how many Fibonacci numbers are there that are less than or equal to a given number $n$, one can derive an estimate from Binet's formula. The second term in the formula can be ignored for large enough $n$, so

$$F_n\approx\frac{\phi^n}{\sqrt{5}}$$

Solving for $n$ here gives

$$n=\frac{\log\,F_n+\frac12\log\,5}{\log\,\phi}$$

Taking the floor of that gives a reasonable estimate; that is, the expression

$$\left\lfloor\frac{\log\,n+\frac12\log\,5}{\log\,\phi}\right\rfloor$$

can be used to estimate the number of Fibonacci numbers $\le n$. This is inaccurate for small $n$, but does better for large $n$.

It turns out that by adding a fudge term of $\frac12$ to $n$, the false positives of the previous formula disappear. (Well, at least in the range I tested.) Thus,

$$\left\lfloor\frac{\log\left(n+\frac12\right)+\frac12\log\,5}{\log\,\phi}\right\rfloor$$

gives better results.

I don't think you will be able to find a closed-form formula for the number of Fibonacci numbers less than $N$ - if you had chosen $N = 26$ instead of $N = 25$, you would have gotten the same answer, but if you had set $N = 34$, the answer would have suddenly jumped to $9$. As has been suggested before, you can use Binet's formula:

nth Fibonacci number $F_n= (\phi^n - (-\phi)^{-n})/(\sqrt 5)$ where $\phi = (1 + (\sqrt 5))/2$

Simply plug in values of $n$ until $F_n > N$; then the number you are looking for is $n-1$.


Proving that $\langle x, y \rangle$ is a not principal ideal of $\mathbb{Q}[x, y]$ My idea was to prove that $x, y$ aren't elements of this ideal, the way I thought about it is that being part of this ideal forces $x$ to be written in the form of a product between $2$ polynomials in $\mathbb{Q}[x, y]$, but $1$ of those polynomials will have to be $x$ and the other will have to be $1$, the unit, which isn't possible... I don't know how to formulate this correctly.
Could someone please help me with this question?

Your idea is good. In fact, assuming the ideal is generated by some polynomial $P$, we deduce the following :

The degree in $y$ of $P$, when $P$ is seen as an element of $\mathbb{Q}[x][y]$, has to be $0$, because $x$ is in the ideal generated by $P$. Can you explain why ?

Note that $A:=\mathbb{Q}[x]$ is a domain, so we can work with polynomials with coefficients in $A$ without much trouble.

Similarly, the degree in $x$ of $P$ has to be $0$.

So $P$ is constant.
So, the ideal generated by $P$ is either the whole ring, or the $0$ ideal. In both cases, it is not the ideal $(x,y)$. By contradiction, the result is proved.

Suppose $x=f(x,y)$ and $y=f(x,y)$ for some polynomial $f(x,y)$.
In view of $x=f(x,y)$, write $f(x,y)$ as a polynomial in $x$ with coefficients given by polynomials in $y$ such as in $x^2(y+1) + x(y-1) + 1$.
Then $x= x^n f_n(y) +\ldots +xf_1(y) + f_0$. By comparing degrees, $n=1$ and then $f_1(y)=1$ by the shorting rule, i.e., $f(x,y)=x$. Thus there is no other representation of the polynomial $x$.
Similar for $y=f(x,y)$.


Integral $\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^b}$ For Quantum Mechanics I need to normalize two wavefunctions and I haven't been able to figure out the integrals. The integrals involve a normalization constant, and 2 flexible parameters since these wavefunctions are then passed to the variation principle to solve the upper bound to the groundstate energy. The Quantum behind the math isn't important in what I am trying to solve

The following are the integrals I cannot figure out, find in an integral table, or use wolframalpha to determine:

Note: 'A' is the normalization constant that can be Complex; what I need to solve for. 'a'and 'b' are the flexible parameters

$$
1=|A|^2\int_{-\infty}^{\infty} (a^2+x^2)^{-b}dx
$$
$$
1=4\pi |A|^2\int_{0}^{\infty} (a+r)^{-2b}r^2dr
$$

So in both cases I need to solve for A (or A squared) in terms of the flexible parameters 'a'and 'b'.

For the first normalization constant
$$
|A|^2\int_{-\infty}^{\infty} (a^2+x^2)^{-b}dx=1,
$$
note the integrand is even so we can write
$$
|A|^2\int_0^\infty \frac{1}{(x^2+a^2)^{b}}dx=\frac{1}{2}.
$$
We can define this integral as
$$
I_1\equiv \int_0^\infty \frac{1}{(x^2+a^2)^{b}}dx, \quad \Re(a^2)>0, \Re(b)>\frac{1}{2}.
$$
Note the constraints placed on $a,b$, otherwise your integral isn't convergent.  Make sure you understand this from a physical standpoint as well as this is quantum mechanics. This integral can be calculated by first considering
$$
I_{1,2}(a)\equiv\int_0^\infty \frac{1}{x^2+a^2}dx,
$$
now differentiate this once w.r.t a to obtain
$$
I'_{1,2}(a)=\int_0^\infty \frac{\partial}{\partial a}\frac{1}{x^2+a^2}dx=-2a\int_0^\infty \frac{dx}{(x^2+a^2)^2}.
$$
If you generalize to differentiating n times, you can see the pattern easily.  We obtain
$$
I^{(n)}_{1,2}(a)=(-1)^{b-1}\dfrac{(2b-3)!!}{2^{b}a^{2b-1}}\cdot\pi.
$$
We can integrate this result to obtain $I_1$ which is given by
$$
I_1=\int_0^\infty \frac{1}{(x^2+a^2)^{b}}dx=\frac{1}{2\Gamma(b)} \sqrt{\frac{\pi}{a^2}} a^{2-2b}\Gamma\big(b-\frac{1}{2}\big), \quad \Re(a^2)>0, \Re(b)>\frac{1}{2}.
$$
Thus your first normalization integral gives us
$$
|A|^2\int_{0}^{\infty} (a^2+x^2)^{-b}dx=\frac{1}{2}=|A|^2\frac{1}{2\Gamma(b)} \sqrt{\frac{\pi}{a^2}} a^{2-2b}\Gamma\big(b-\frac{1}{2}\big)=\frac{1}{2},
$$
you can solve for A.  I am still working on your second normalization integral.

Your second normalization integral given by
$$
1=4\pi |A|^2\int_{-\infty}^{\infty} (a+r)^{-2b}r^2dr\equiv 4\pi |A|^2 I_2.
$$
We now calculate $I_2$ which is given by
$$
I_2=\int_{-\infty}^{\infty} \frac{r^2}{(a+r)^{2b}}dr=0.
$$
However this integral evaluates to 0 because of the lower limit of integration.  I think you meant to write
$$
I_2=\int_{0}^{\infty} \frac{r^2}{(a+r)^{2b}}dr=\frac{a^{3-2b}}{4b^3-12b^2+11b-3}, \Re(a)>0, 2\Re(b)>3, \arg(a^{-1})\leq \pi.
$$
The integral should be from $0$ to $\infty$ because the radius is always positive in physics.  Thus your second normalization constant can be calculated by
$$
1=4\pi |A|^2\cdot \frac{a^{3-2b}}{4b^3-12b^2+11b-3},
$$
solve for A.  Let me know what you think, and if you need more help.  Also, clarify with me that you agree you posted the second normalization integral incorrectly because of the lower bound you wrote as $-\infty$.

UPDATE: Gamma Functions in elementary quantum mechanics $\Gamma(n)=(n-1)!$.

Also, in regards to Hydrogen
$$
\int_0^\infty x^n e^{-x} dx=n!=\Gamma(n+1), \quad \Re(n) >-1
$$
this is what $\big<r^n\big>$ is equal to if we are discussing the ground state of Hydrogen $\psi \propto e^{-r/a}$. Common thing to do is use a Gaussian $\psi \propto e^{-r^2/a^2}$ and use the variational principle to estimate the Hydrogen ground state energy. You will see the difference.  Another quantity that comes up often is $\big<x^{2p}\big>$ for the quantum harmonic oscillator.  We need integrals of the form (ground state of oscillator)
$$
\int_{-\infty}^\infty e^{-x^2}x^{2p}dx=\frac{1}{2}\big(1+(-1)^{2p}\big) \Gamma\big(\frac{1}{2}+p\big), \quad \Re(p) > -\frac{1}{2}.
$$
Put p=1, and see what you get by hand (think about it physically) and check it with this formula!

$\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{{\cal I}\pars{\mu,\nu} \equiv
     \int_{0}^{\infty}\pars{1 + \mu x^{2}}^{-\nu}\,\dd x\,,\quad\mu > 0,\
     \nu > {3 \over 4}
     \quad\mbox{and}\quad{\cal I}\pars{\mu,\nu} =
     \mu^{-1/2}\int_{0}^{\infty}\pars{1 + x^{2}}^{-\nu}\,\dd x.\ \mbox{Also},
     \partiald{{\cal I}\pars{\mu,\nu}}{\mu}=-\nu\int_{0}^{\infty}\pars{1 + \mu x^{2}}^{-\nu - 1}x^{2}\,\dd x}$

  With $\ds{t \equiv {1 \over 1 + x^{2}}\quad\imp\quad x = \pars{{1 \over t} - 1}^{1/2}}$:
  \begin{align}
\color{#00f}{\large{\cal I}\pars{\mu,\nu}}&=
\mu^{-1/2}\int_{0}^{1}
t^{\nu}\,\half\,\pars{{1 \over t} - 1}^{-1/2}\pars{-\,{1 \over t^{2}}}\,\dd t
=\half\,\mu^{-1/2}\int_{0}^{1}t^{\nu - 3/2}\pars{1 - t}^{-1/2}\,\dd t
\\[3mm]&=\color{#00f}{\large\half\,\mu^{-1/2}\
\overbrace{{\rm B}\pars{\nu - \half,\half}}^{\mbox{Beta Function}}}
\end{align}

\begin{align}
\color{#00f}{\large\int_{-\infty}^{\infty}\pars{a^2 + x^{2}}^{-b}\,\dd x}
&={2 \over \verts{a}^{2b}}\int_{0}^{\infty}
\pars{1 + {1 \over a^{2}}\,x^{2}}^{-b}\,\dd x
={2 \over \verts{a}^{2b}}\,\half\,\verts{a}{\rm B}\pars{b - \half,\half}
\\[3mm]&=\color{#00f}{\large{1 \over \verts{a}^{2b - 1}}\,{\rm B}\pars{b - \half,\half}}
\end{align}

  \begin{align}
\color{#00f}{\large\int_{-\infty}^{\infty}\pars{a^2 + x^{2}}^{-2b}\,x^{2}\dd x}
&={2 \over \verts{a}^{2b}}\int_{0}^{\infty}
\pars{1 + {1 \over a^{2}}\,x^{2}}^{-2b}\,x^{2}\dd x
\\[3mm]&=\left.{2 \over \verts{a}^{2b}}\pars{-\,{1 \over \nu}}\,
\partiald{{\cal I}\pars{\mu,\nu}}{\mu}
\right\vert_{\mu = {1 \over a^{2}},\nu = 2b - 1}
\\[3mm]&={2 \over \verts{a}^{2b}}\pars{-\,{1 \over 2b - 1}}
\pars{-\,{1 \over 4}\,\verts{a}^{3}}\,{\rm B}\pars{2b - {3 \over 2},\half}
\\[3mm]&=\color{#00f}{\large%
\half\,{1 \over \verts{a}^{2b - 3}}\,{1 \over 2b - 1}\,
{\rm B}\pars{2b - {3 \over 2},\half}}
\end{align}


flatness and isomorphism Given: domains $A$ and $B$ such that $A\subseteq B$ and such that their fraction fields are equal. Also $A$ is a unique factorization domain and if $a \in A$ is a unit in $B$, it is a unit in $A$. I want to show that if $A \subseteq B$ is flat then $B=A$. I think I need to use the fact that every element in $B$ can be written as $\frac{a_1}{a_2}$ with $a_1,a_2 \in A$. Does anyone have an idea?

The assumption $A \cap B^* = A^*$ implies that $\mathfrak{p}B \neq B$ for all prime ideals $\mathfrak{p} \subseteq A$, i.e. that $B$ is faithfully flat over $A$. Now use SE/482908.


How can you find $m$ in $mx^2+(m-3)x+1=0 $ so that there is only one solution How can you find $m$ in $$mx^2+(m-3)x+1=0 $$ so that there is only one solution.

I tried to solve it by quadratic equation but I end up with two solutions.

So I want it know that is there a way so that I'll only get one solution the end.

Thanks.

A quadratic equation always has $2$ solutions, by the Fundamental Theorem of Algebra.

So if it has one distinct solution, then the two roots must be same.

Hence the discriminant must be equated to $0$.

Therefore, $$(m-3)^2-4\cdot m\cdot 1=0$$
$$m^2-10m+9=0$$
$$(m-9)(m-1)=0$$

So required value of $m$ is $1$ and $9$.

If we want only one solution, we need the discriminant of the quadratic equation to equal $0$. That is, in $$x = {3 - m \pm \sqrt{(m-3)^2 - 4m} \over 2m},$$ we need the stuff under the radical $(m-3)^2 - 4m = 0$. So $$m^2 - 6m + 9 - 4m = m^2 - 10m + 9 = (m-9)(m-1)=0.$$ So we get two answers, $m = 9$ and $m=1$. These are the only two solutions because if this is the case, then $$x = {3-9\pm \sqrt{6^2-4(9)} \over 2(9)} = {-6 \pm 0 \over 18} = -{1 \over 3}$$ and $$x = {2 - 1 \pm \sqrt{(-2)^2-4} \over 2} = {1 \pm 0 \over 2} = {1 \over 2}$$

Notice how each case of $m=9$ and $m=1$ respectively yields only one solution.


Is this alternative proof of Theorem 3.7 ("Baby" Rudin, Ch. 3) correct and, if so, well written? Rudin, in his Principles of Mathematical Analysis, proves the following theorem:

  The subsequential limits of a sequence $\{p_n\}$ in a metric space $X$ form a closed subset of $X$.

I've tried to come up with a different proof, but I have a few doubts about its correctness:

Let $E$ be the set of the subsequential limits of $\{p_n\}$ and $q$ a limit point of $E$. We have to prove that $q\in E$.

Since $q$ is a limit point of $E$, there exists a sequence $\{a_n\}$ in $E$ that converges to $q$ (this is a theorem that Rudin proves earlier in the book). Let $V_n$ be the open ball of center $a_n$ and radius $d(q,a_n)$.

Since there is a subsequence of $\{p_n\}$ which converges to $a_n$, $V_n \cap p(\mathbb{N})$, where $p(\mathbb{N})$ is the range of the sequence $\{p_n\}$, is nonempty for every $n\in \mathbb{N}$; that is, $V_n$ always contains at least one term of the sequence.

Therefore, by the axiom of countable choiche, there exists a sequence $\{q_n\}$ such that $q_n \in V_n\cap p(\mathbb{N})$ for every $n$. This happens to be a subsequence of $\{p_n\}$.

Now it is left to prove that $q_n$ converges to $q$: let $\epsilon >0$ be given; then there is $N$ such that $d(q,a_n)<\epsilon$ if $n>N$. We also find that $d(a_n,q_n)<d(q,a_n)$; thus, by the triangle inequality, $d(q,q_n)\leq d(a_n,q_n)+d(q,a_n)<2\epsilon$. Hence the thesis.

I realize that it's definitely more complicated that the one offered by the author, but is it sound nonetheless? Is it written in a fashion that would be accepted in an exam?

Here is how to prove it without the axiom of choice:...[0]:.. Let $q$ be a limit point of $E$...[1]:Observe that for any member of $E$ there is a strictly increasing $f : N \to N$ such that  $(p_{f(n)})_{n \in N}$ converges to it...[2]:.. For all $r>0$, the set  $S(r)= \{ n \in N : p_n \in B(q,r) \}$ is infinite because...[2a]:.. B(q,r) contains some $q^* \in E$ and,by [1], $q^*$is the limit of a sequence $(p_{f(n) : n \in N})$ for some strictly increasing $f:N \to N$, and ...[2b]:.. There exists, by the Triangle Inequality, some $s>0$ for which $B(q^*,s) \subset B(q,r)$, so... [2c}:..The infinite set $\{ n \in N : p_{f(n)} \in C \}$ is a subset of $S(r)$....[3]:..Now define $g:N \to N $ recursively by letting $g(0)= \min \{ k: p_k \in B(q,1) \}$ and $g(n+1)= \min \{ k>g(n) : p_k \in B(q,2^{-n-1} \}$. Then $g$ exists by [2] and is strictly increasing and $(p_{g(n)})_{n \in N}$ converges to $q$.


Evaluate $\lim_{x \rightarrow \infty}\left( \sqrt[3]{x^3+1} -\sqrt[4]{x^4+1}\right)$ Could you help to evaluate $$\lim_{x \rightarrow \infty}\left( \sqrt[3]{x^3+1} -\sqrt[4]{x^4+1}\right)?$$
I tried all methods I know of to no avail

That's equivalent to $$\lim_{x \rightarrow 0} \frac{(x^3+1)^{1/3} -(x^4+1)^{1/4}}{x},$$ to which you may apply L'hopital.


Can someone kindly explain idea behind the sum of projections? Here is my question, which is part of the beginning of the chi-squared decomposition theorem proof :
Suppose that $P_1, ....P_k $ are symmetric projection matices with sum the identity:

$I = P_1 + P_2 + ... + P_k $

Then squaring,

$ I = I^2 = \sum_i P_i^2 + \sum_{i<j} P_i P_j $

My question is why after squaring, it yields the term $\sum_{i <j} P_i P_j$ , how is it derived ???

Actually, under your conditions, $$I=I^2=\sum_{_i}P_i + 2\sum_{i<j}P_i P_j$$
because $P_i^2 = P_i$ (all projection matrices are idempotent) and because $P_i$ , $P_j$ are symmetric with the same dimensions, so $P_i P_j + P_j P_i = 2 P_i P_j$.

This is simple algebra $(P_1+...+P_n)^2=(P_1+...+P_n)(P_1+...+P_n)$.


When is the image of a null set a null set? I came upon this question here which contains the following statement:

  It is easy to prove that if $A \subset \mathbb{R}$ is null (has
  measure zero) and $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lipschitz
  then $f(A)$ is null. You can generalize this to $\mathbb{R}^n$ without
  difficulty.

I believe that Lipschitz here refers to Lipschitz continuity.

When I saw the statement it seemed to me that Lipschitz continuity was too strong. For example, Lipschitz continuous implies continuous. Is it conceivable that a continuous (read "nice") function maps a measurable set to a non measurable set?

It seems to me that this would be the only case for which the image can have non zero measure since it seems intuitively imperative that any image cannot have larger measure than the original set.

  Please could someone enlighten me on the minimal condition for which
  $f: \mathbb R^n \to \mathbb R^m$ maps null sets to null sets? And if
  possible point out any mistakes in my thoughts above, I would greatly
  appreciate it.

Note: I expect the answer to be that $f$ should be measurable or some similarly weak condition.

No, there is a continuous function from the unit interval that sends the Cantor set onto the unit interval. Since the Cantor set is null, the contradicts your conjecture.

See https://en.m.wikipedia.org/wiki/Cantor_function

Since all continuous functions are measurable, your guess after the question is also wrong.

Absolute continuity is sufficient, but be clear- there are plenty of discontinuous functions that have this property. It will be unlikely that you'll find a minimal condition in any strict notion of "minimal." And it's not entirely clear how to extend the definition of absolute continuity to multiple dimensions.

Do you know Peano curve ?

It maps $[0,1]$ to an area in $\mathbb{R}^2$ with positive 2 dimensional Lebesgue measure


What does it mean to integrate with respect to the distribution function? If $f(x)$ is a density function and $F(x)$ is a distribution function of a random variable $X$ then I understand that the expectation of x is often written as:

$$E(X) = \int x f(x) dx$$

where the bounds of integration are implicitly $-\infty$ and $\infty$. The idea of multiplying x by the probability of x and summing makes sense in the discrete case, and it's easy to see how it generalises to the continuous case.  However, in Larry Wasserman's book All of Statistics he writes the expectation as follows:

$$E(X) = \int x dF(x)$$

I guess my calculus is a bit rusty, in that I'm not that familiar with the idea of integrating over functions of $x$ rather than just $x$.

What does it mean to integrate over the distribution function?
Is there an analogous process to repeated summing in the discrete case?
Is there a visual analogy?

UPDATE:
I just found the following extract from Wasserman's book (p.47):

  The notation $\int x d F(x)$ deserves some comment. We use it merely
  as a convenient unifying notation so that we don't have to write
  $\sum_x x f(x)$ for discrete random variables and $\int x f(x) dx$ for
  continuous random variables, but you should be aware that $\int x d F(x)$  has a precise meaning that is discussed in a real analysis
  course.

Thus, I would be interested in any insights that could be shared about what is the precise meaning that would be discussed in a real analysis course?

There are many definitions of the integral, including the Riemann integral, the Riemann-Stieltjes integral (which generalizes and expands upon the Riemann integral), and the Lebesgue integral (which is even more general.)   If you're using the Riemann integral, then you can only integrate with respect to a variable (e.g. $x$), and the notation $dF(x)$ isn't defined.

The Riemann-Stieltjes integral generalizes the concept of the Riemann integral and allows for integration with respect to a cumulative distribution function that isn't continuous.

The notation $\int_{a}^{b} g(x)dF(x)$ is roughly equivalent of $\int_{a}^{b} g(x) f(x) dx$ when $f(x)=F'(x)$.  However, if $F(x)$ is a function that isn't differentiable at all points, then you simply can't evaluate $\int_{a}^{b} g(x) f(x) dx$, since $f(x)=F'(x)$ isn't defined.

In probability theory, this situation occurs whenever you have a random variable with a discontinuous cumulative distribution function.  For example, suppose $X$ is $0$ with probability $\frac{1}{2}$ and $1$ with probability $\frac{1}{2}$.  Then

$$
\begin{align}
F(x) &= 0 & x &< 0 \\
F(x) &= 1/2 & 0 &\leq x < 1  \\
F(x) &= 1 & x &\geq 1 \\
\end{align}
$$

Clearly, $F(x)$ doesn't have a derivative at $x=0$ or $x=1$, so there isn't a probability density function $f(x)$ at those points.

Now, suppose that we want to evaluate $E[X^3]$.  This can be written, using the Riemann-Stieltjes integral, as

$$E[X^3]=\int_{-\infty}^{\infty} x^3 dF(x).$$

Note that because there isn't a probability density function $f(x)$, we can't write this as

$$E[X^{3}]=\int_{-\infty}^{\infty} x^3 f(x) dx.$$

However, we can use the fact that this random variable is discrete to evaluate the expected value as:

$$E[X^{3}]=(0)^{3}(1/2)+(1)^{3}(1/2)=1/2$$

So, the short answer to your question is that you need to study alternative definitions of the integral, including the Riemann and Riemann-Stieltjes integrals.

The definition of dF(x) is f(x). F(x) is the cumulative distribution function (AKA the CDF). f(x) is the probability density function (PDF). See for yourself:
http://mathworld.wolfram.com/DistributionFunction.html


what is the interval of integration of $\int^a_b \frac {dx}{dt}dt $ for $\int^a_b dt$ the interval of integration is [a,b] so how about
$$\int^a_b \frac {dx}{dt}dt = \int^a_b dx $$
Does the interval of integral still [a,b] or it's changed. If it's, in what way it's changed?

There seem to be multiple issues:

If $a < b$, then $\displaystyle \int_{a}^{b} \dots$ denotes an integral over $[a, b]$. (The left-hand endpoint is the lower limit.)
If $b < a$, then $\displaystyle \int_{a}^{b} \dots = -\int_{b}^{a} \dots$, and the interval of integration would be $[b, a]$.
If $x$ is a continuously-differentiable function on $[a, b]$, the "substitution rule" or "change of variables formula" reads
$$
\int_{a}^{b} \frac{dx}{dt}\, dt = \int_{x(a)}^{x(b)} dx.
$$
(By the fundamental theorem, each side is equal to $x(b) - x(a)$.) Thus, the interval of integration on the right is $\bigl[x(a), x(b)\bigr]$ if $x(a) < x(b)$, or is $\bigl[x(b), x(a)\bigr]$ if $x(b) < x(a)$.

The differentials $dt$ cancel out. So you are referring to variable $x$ only and its interval is $(a,b)$ and its integral is $(a-b)$.


How to find the indefinite integral of sin(sin(x))dx? (I got this function by mistake, when I miswrote other function. Now I'm curious how to find the antiderivative of what I miswrote)

I have no a clue how to calculate it and neither does Wolfram Alpha or any other site that I tried. Trig formulas from school course don't seem to be useful too.


Explain why this homotopy works? http://www.math.cornell.edu/~hatcher/AT/AT.pdf

It's page 351, but I'm having trouble with the line highlighted.

The highlighted part makes no sense to me. I can see why $f_0=f$ as you would do this $f_0=(1-0)f+0g=f$, but then how do you get $f_1|K_1=g|K_1$. I was thinking that $f_1|K_1=(1-\varphi)f+\varphi g$?

So how do you get that to be equal to g. I know g and f agree on vertices. But, I don't see how you get it.  The before this bit is fine, but then this bit confuses me.

Since $\phi$ is linear on simplices and constant on vertices in $K_1$, we must have $\phi$ constant (and equal to 1) on all of $K_1$, which is what you want.

Without answering this question on Hatcher's proof, I mention that the proof of the Cellular Approximation Theorem in Section 7.6 of my book "Topology and groupoids" uses no simplicial subdivisions, but only cubical ones, and you can find the same proof in older editions of this book, with different titles, which may be in your library.  I took it  long ago from handwritten notes of J.F. Adams. (The proof given restricts to the case of finite cell complexes, but the topology in the general case is arranged so that the same proof goes through!) Here is the key step.

The following statements are true for each $n \geqslant 1$.

$\alpha(n)$  Any map $S^r \to S^n$ with $r &lt n$ is
inessential.

$\beta(n)$ Any map $S^r \to S^n$ with $r &lt n$ extends
over $E^{r+1}$.

$\gamma(n)$  Let $B$ be path-connected and let $Q$ be formed by attaching a
finite number of $n$-cells to $B$. Then any map
$$ (E^r,S^{r-1}) \to (Q,B)$$
with $r&ltn$ is deformable into $B$.

The proof is by induction by means of the implications
$$ \gamma(n) \Rightarrow \alpha(n)
\Leftrightarrow \beta(n) \Rightarrow \gamma(n+1)$$
the only difficult step being the proof of $\beta(n) \Rightarrow
\gamma(n+1)$.


Is $n=26$ the only integer satisfying $n-1$ being a square and $n+1$ being a cube? I guess it is possible to solve it by ring theory, but I would prefer an elementary method instead. I can only figure out that $n$ is an even integer, otherwise there will be contradiction w.r.t. modulo $4$. I have also tried to modulo several other integers(3, 5, 8, 9, 10) but haven't got any contradiction yet. Any discussion is welcomed!

I wont give you the whole solution but try fuctoring n-1 and n+1, and compare them for some specific modulus
Alse you might want to consider quadratic reciprocoty


When does sum of two units give a unit? Let $A$ be a ring and $u,v \in A^\times$. When do we have that $u + v \in A^\times$?

I think that A is needed to be an integral domain. For example consider $\mathbb{Z/6}$. Both $1$ and $5$ is a unit but their sum $1+5=0$ is not a unit.

I do not think there is such a condition. For any unital ring such that $1 \neq 0,$ $1$ and $-1$ are units, but $1+(-1)=0,$ which is not a unit.

In any field, the sum of two units is a unit, unless they're additive inverses.  For a non-field example, let $\zeta$ be a primitive sixth-root of unity and let $A$ be the ring of integers in $\mathbb{Q}[\zeta]$. Then note that $\zeta^2+1 = \zeta$.  (Because $0 = \zeta^6 - 1 = (\zeta^3-1)(\zeta^3+1) = (\zeta-1)(\zeta^2+\zeta+1)(\zeta+1)(\zeta^2-\zeta+1)$ and we can set this last factor equal to $0$.)


Prove that $ 3x^{10} - y^{10} = 1991 $ has no integral solutions. Prove that $ 3x^{10} - y^{10} = 1991 $ has no integral solutions.

Attempt:

$ 11|1991 $
Therefore, for the equation to have a solution:
$11| 3x^{10} - y^{10} $
If I can prove that 11 does not divide $ 3x^{10} - y^{10} $ for any integral x and y, then the problem will be solved.
I am unable to prove that.
Is my method right? If yes , please help me with where i am stuck. If no, please provide an alternative solution.

For any integer $x$ we have $x^{10}\equiv\{0,1\}\pmod{11}$ due to Fermat's little theorem. Since $11\mid 1991$, assuming that $3x^{10}-y^{10}=1991$ we have that both $x$ and $y$ have to be multiples of $11$, but in such a case $3x^{10}-y^{10}$ is a multiple of $11^{10}$, while $1991$ is not.

First show by putting x=2 and y=1 that has no solution and then write the comment that on taking values greater than 2 the difference becomes greater than 1991.You can show this by taking a function f(x) =3x^10-(x-1) ^10.And show that f(x) is an increasing function.
Hope this help you.


Range for the function $f(x) = 3x + 2$ with domain $x > 0$ The function below is defined for continuous domains

Sketch the graph and state the range of the function

Question: $f(x) = 3x + 2$ for the domain $\{x \in \mathbb{R} : x > 0 \}$

The straight line cuts the $y$-axis at $(2, 0)$ but since $x >0$ why is the answer for the range still $f(x) > 2$? Shouldn't you be using $x = 0.1$ or $x = 1$ to find the $y$-intercept since the domain originally stated $x > 0$ (hence $x \not= 0$)

Thanks.

Edit: I have just noticed that the range is $f(x) > 2$ and not $f(x) \ge 2$ which solves everything.

$x>0\Rightarrow 3x+2>2\Rightarrow f(x)\in (2,\infty)$

Yeah... But we first drew the graph of function over $R$ and then smartly erased the part not needed. Your case will yield the same answer.

Your Ad Here answers the range doubt.


Show $Ax = e_{m+1}$ is consistent. $A = \binom{C}{a^{T}}$. $C$ is $mxn$ matrix. $a$ is $n$-vector that is lin. indep. of rows of $A$. $C$ is an $mxn$ matrix of any shape, shape, rank, etc.  $a$ is a $n$-vector that is linearly independent of the rows of $C$. Let $A = \binom{C}{a^{T}}$.  Let $e_{m+1}$ denote the $m+1$ Identity vector.  Show that $Ax=e_{m+1}$ is consistent.

I don't know where to start.

Note that $Ax=e_{m+1}$ means that you should find an $x$ with $Cx=0$ and $a^Tx=1$. Because you can scale $x$ appropriately if necessary, it suffices to find $x\in \operatorname{ker}(C)$ with $x\notin \operatorname{ker}{a^T}$. This is possible because $\operatorname{ker}(C)\subseteq \operatorname{ker}{a^T}$ implies that $a^T$ is linearly dependent to the rows of $C$, see e.g. this math.stackexchange question with $L_i$ taken to be the functionals given by multiplying with the rows of $C$ and $L$ is taken to be the functional given by multiplying with $a^T$.

This is going to be hard to prove because it is false.

Consider
$$
C = \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 1 & 1
\end{array}\right) \\
a^T = ( -1, 0, 1 )
$$
which is not linearly dependent on the rows of $C$.

Then
$$
A = \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 1 & 1 \\ -1 & 0 & 1
\end{array}\right)
$$ and $ Ax = e_3$ is
$$
2x_1 + 2x_2 + 2x_3 = 1 \\x_1 + x_2 + x_3 = 1 \\ -x_1 + x_3 = 1
$$
and those first two equations are inconsistent with each other.