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July 2020

Let $f(x)$ be a differentiable function on $(0, \infty)$ with $\lim_{x\to \infty} f(x) - xf'(x) = L\in \mathbb{R}$. I'm trying to prove or disprove that $\lim_{x\to\infty} f'(x)$ exists as well. Here's what I have so far: if the limit does exist, it must also equal $\lim_{x\to\infty} \frac{f(x)}{x}$ (divide the limit condition by $x$). Then I rewrote the condition as $x^2 \frac{d}{dx} \frac{f(x)}{x} \to L$, and from this I expect (using some $1/x^2$ asymptotic argument) my statement to be true - but I'm not sure how to rigorously proceed with this argument.. Can someone provide some next steps, or a counter example?



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Faraz Masroor

I'm riffing on an old contest training question I jousted with 40 years ago.

The original problem was:

A solid $20\times20\times20$ cube is built out rectangular bricks of dimensions $2\times2\times1$. Prove that it is possible to "push" a line through the cube in such a way that the line is not obstructed by any of the bricks.

Solution: We need $2000$ bricks to build this cube. Imagine that the edges of the cube align with the coordinate axes, and that the cube is in the first octant with one of its vertices at the origin. So there are $19^2$ lines parallel to the $z$-axis going through the cube, each given by the equations $x=a, y=b, a,b\in\{1,2,\ldots,19\}$, lines parametrized by the choice of the pair $(a,b)$. Similarly, there are $19^2$ lines parallel to the $x$ and $y$-axes for a total of $3\cdot19^2$ lines. It turns out that one of these will go through the cube along the cracks between the bricks. The key observation is that each line will be blocked by an even number of bricks (spoiler hidden below in case you want to think about it yourself).

As $2\cdot3\cdot19^2>2000$ it is impossible that all these lines would be blocked by two or more bricks. Therefore at least one of them is unobstructed, proving the claim.

Ok, that was the background story. On with the actual question.

As the size of the cube, call it $n$, grows, the number of bricks increases as $n^3/4$, but the number of those lines, call them integer lines, increases as a quadratic polynomial of $n$ only. Therefore sooner rather than later the above argument fails to work. In fact, this happens already with $n=22$ as $2\cdot3\cdot21^2<22^3/4$. The parameters $a,b$ obviously ranging from $1$ to $n-1$.

Is it possible to build a solid $22\times22\times22$ cube out of $2\times2\times1$ bricks in such a way that all the integer lines are blocked by at least one (hence at least two) bricks? If this is not possible with $n=22$, what is the smallest value of $n$ for which this construction is possible (if one exists)?


Given that the answer to my question is unknown, I will welcome answers explaining a construction for answerer's choice of $n$.



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Jyrki Lahtonen

Hello,

William here.

I have figured out a trick that makes squaring numbers a lot easier, and this involves the simple trick that (hopefully) is known globally. That trick is the ”square any 2digit number that ends with a 5 in head without calculator”. Perhaps you know this trick; but for those who doesn't :

Consider this example: 35²

The (publicly known) trick is just to take the first digit: (3) and plus it by 1. (4)

3 + 1 = 4

Then multiply the first digit (3) with the 4. (3+1)

3*4 = 12

The last step is - to just 'add 25 at the end' (NOT plus!) (just write 25 at end of the answer 12) like so; 12 25

Done! You've now calculated 35² = 1225 in your head - without any calculator!

Now to the thing I figured out - this also works for big numbers, I took (for the sake of this proof of concept) smaller numbers to begin with, and then moved up to really big numbers. And, those big numbers actually required calculator to solve.

Now - you might be asking what the point is with this if you already have access to a calculator? And if the number is so big anyway that you need a calculator to solve it?

The point is:

What the point of the following is NOT:

It is not to solve it without calculator.

What the point of the following IS:

The point of this is however to (hopefully) help break down a big number more easily.

And, to show you a very interesting ”chain” I found, by mistake actually! (And I just 'went with the flow')

So, Here is the ”chain” or what you’d like to call it:

2 digits = +1 at the first 1 digit:

75² =

7+1 = 8

7*8 = 56

56 and add 25 @ end: 5625

3 digits = +1 at the first 2 digits:

925² =

92+1 = 93

93 * 92 = 8556

8556 and add 25 @ end: 855625

4 digits = +1 at the first 3 digits:

8425² =

842 + 1 = 843

843 * 842 = 709806

709806 add 25 @ end:

8425² = 70980625

6 digits = +1 at the first 5 digits:

906835² =

90683+1 = 90684

90684 * 90683 = 8223497172

8223497172 and add 25 @ end:822349717225

906835² = 8.223497172×10¹ ¹

(my calculator showed this answer)

I am currently trying to discover new ways - of doing this with another ’ending number’ that ends with something else than 5, and is larger than 100

Please, this is my first post about mathematics – If you spotted a error, tell me as soon as possible, and I will fix it! (I appreciate it, really!) Also, if you liked this: Tell me your thoughts!

If you have time, send me a review about my site, and , if you know something I could improve; tell me!.

It would be extremely helpful!

References:

https://en.wikipedia.org/wiki/Square_number

As I said above , You may or may not already know this trick(s) but when I first saw these (by testing out my own, just random stuff, and come across this)

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1. Definitions: Unimodularity and cocommutativity
Let $H$ be a Hopf algebra over a field $\mathbb k$.

  • We call $H$ unimodular if the space of left integrals $I_l(H)$ is equal to the space of right integrals $I_r(H)$.
  • We call $H$ cocommutative if $\tau_{H,H} \circ \Delta = \Delta$. Here, $\Delta$ denotes the coproduct of $H$, while $\tau: H \otimes H \rightarrow H \otimes H; v \otimes w \mapsto w \otimes v$ is the twist map.

2. Question

  • In my lecture notes it says that there are cocommutative, non-unimodular Hopf algebras. What would be an example?

  • Apparently, an example is given in Hopf algebras and their action on rings by Susan Montgomery. However, due to the pandemic I am unable to get it from the library. If you have a copy and could write down the relevant section, that would be very much appreciated.

3. My ideas so far

  • The Taft-Hopf algebra $H$ over a field $\mathbb k$ is not an example: If $H$ is commutative (i.e. root of unity $\zeta =1_{\mathbb k}$), then $H$ is unimodular. In this case, it is even isomorphic to the boring group algebra of the zero group. Otherwise, $H$ is not cocommutative (even though it is non-unimodular then). Non-cocommutativity follows easily from the observation that the square of the antipode is not the identity (if $\zeta \neq 1_{\mathbb k} $).

  • Group algebras: As the coproduct of a group algebra is given by the diagonal map any group algebra is cocommutative. However, any group algebra $\mathbb k[G]$ over a finite group $G$ is unimodular, since $$I_l=I_r=\mathbb k \cdot \sum\limits_{g\in G} g$$ What about infinite groups?

  • Regarding the universal enveloping algebra, tensor algebra, symmetric algebra, alternating algebra I am not sure. What can be said here?

  • Maybe the following proposition turns out to be useful: A finite dimensional Hopf algebra $H$ is unimodular iff its distinguished group-like element/modular element $a \in G(H^*)$ is equal to the counit. Here, the modular element $a$ is the unique linear form such that $t\cdot h = t a(h)$ for all $h\in H, t\in I_l(H)$. It exists because $t\cdot h \in I_l(H)$ and $I_l(H)$ is one dimensional. It can be shown to be a morphism of algebras, hence a group-like element in $H^*$.



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M.C.

I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx$$ With no success, i tried to consider the following integrals $$I=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx,J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$ $$I+J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^2\right)}{1+x^2}\:dx=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^4\right)}{1+x^2}\:dx-\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$ I managed to express that $1$st integral into somewhat known euler sums but that $2$nd integral arrived at a sum i didnt know how to evaluate which was $$2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{\left(2k+1\right)^3}$$ And it seems this approach wont go smooth, could i tackle the main integral differently? maybe with an easier approach?



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stefan

Hi r/math,

W.C. Holland - Distinguished research professor

I am going out on a limb here a bit, but I am wondering if any active members of this community had ever crossed paths with my grandfather, Charles Holland. I am sad to say that he has recently passed away, and I am hunting around for any interactions people may have had with him that I can share with the family.

Both his son (my dad) and myself studied math in college, though neither of us have degrees beyond bachelor's, so we never had the opportunity to become fully immersed in what his field was. (or, you know... even understand it on more than a surface level, hah). I have heard kind words said about him from the math community, and there was even an issue of Mathematica Slovaca dedicated to him, which is pretty cool. I am wondering if anybody here studies in the fields he worked in, and if anybody can give a mostly laymen description of what his contributions were.

I knew him as a loving grandfather and family member who inspired my passion for mathematics, and am curious if this post happens to reach anybody who either studied directly with him, or knows his contributions.

His Mathematics Genealogy page

Thanks for reading! He was a pretty remarkable person and I will greatly miss him.

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How do I integrate $$\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx\quad ?$$

The actual integral that I encountered is:

$$\int_{-\infty}^\infty dx \left(\frac{N}{\cosh(\frac{\pi }{c}(x-1))}+\frac{1}{\cosh(\frac{\pi}{c}x)} \right) 2 \tan^{-1}\left(\frac{2x-2}{c} \right)$$ where c is a constant with $$\Re c>0$$ Not sure if these two terms makes it easier.

I was trying to solve just the last term, but I couldn't make any progress. Numerical integration gives $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx= -1.01334 $. Any hint on how to do it analytically?



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Pradip Kattel

This recurring thread will be for general discussion on whatever math-related topics you have been or will be working on over the week/weekend. This can be anything from math-related arts and crafts, what you've been learning in class, books/papers you're reading, to preparing for a conference. All types and levels of mathematics are welcomed!

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The question is in the title, $n>m>2$ are integers. All text below is the context.

Two weeks ago user759001 asked on integer solutions $x>y\ge 2$ of a Diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$ for integers $m,n\geq 2$. The only known solutions are $(x,y;m,n)=(3,2;2,3)$ and $(98,21;2,3)$. User2020201 showed that $m<n$. I conjectured that there are no solutions when $m|n$ and proved the conjecture in particual cases (when $(m,n)$ is $(2,6)$, $(3,19)$, or $(4,12)$. Also I guess I have a proof when $n=2m$), see this answer.

According to [G], Diophantine equations with two variables of degree greater than two have infinitely many (integer) solutions only in very rare cases. In particular, by a special and very complicated method K. Zigel’ (Siegel?) showed the following

Theorem. Let $P(x,y)$ be an irreducible polynomial of two variables with integer coefficients of a total degree greater than two (that is, $P(x,y)$ contains a monomial $ax^ky^s$, where $k+s>2$). (The irreducibility of $P(x,y)$ means that it cannot be represented as a product of two non-constant polynomials with integer coefficients). If an equation $P(x,y)=0$ has infinitely many integer solutions $(x,y)$ then there exist an integer $r$ and integers $a_i$, $b_i$ for each $-r\le i\le r$ such that if in the equation $P(x,y)=0$ we make a substitution $x=\sum_{i=-r}^r a_it^i$ and $y=\sum_{i=-r}^r b_it^i$ then we obtain an identity.

In order to apply this theorem to user759001’s equation for fixed $n>m>2$ we need irreducibility of the polynomial $y^n+y^{n-1}-x^m-x^{m-1}$. It looks plausible and easy to show, but, unfortunately, I am not a specialist in factorization of multivariable polynomials, so I decided to ask MSE community for help. Thanks.

References

[G] Gel’fand A.O. Solutions of equations in integer numbers, 3rd edn., Moscow, Nauka, 1978, in Russian.



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Alex Ravsky

The knight is moved exactly $10$ times. A knight has $8$ possible ways to move once. So I believe there are $8^{10}= 2^{30} \sim 1$ billion permutations. How many in which the knight ends up on the same square?



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cornelius

Hellow fellow mathematicians,
My uni I just transferred to requires students pursuing a B.S. to also choose a minor. Seeing that I have completed Calculus 1-3, I chose mathematics. I did much better in 3 than 2 since I am horrendous at recognizing patterns in sequences/series, but loved the abstract concepts and 3-D visualizations in 3.
I need two more courses to earn my minor. For those who have taken these classes and other higher-level courses, what would you recommend? I don't expect to find classes extremely similar to 3, but looking for classes people enjoyed or found less challenging than others. I'm hoping to stay away from DiffEQ, as I've heard many horror stories and don't necessarily need it for my minor.

I'm open to any suggestions!

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Algebraic General Topology is a certain theory that generalizes general topology (both point-set topology and pointfree topology) in an algebraic way.

There are three books available:

The "Basics" book generalizes together topological spaces, pretopological spaces, closure spaces, (quasi-)proximity spaces, (directed) graphs (=binary relations) into one concept of funcoids, and both uniform spaces and (directed) graphs (=binary relations) are generalized as reloids, that is filters on binary Cartesian products.

That funcoids and reloids are generalizations of both spaces and functions makes them convenient for study of functions on spaces. For example, continuity is defined by simple algebraic formulas rather than by epsilon-delta notation.

The books further considers pointfree and multidimensional generalizations, various products, etc.

The book 3 "Algebra" generalizes it even further. It shows that every space in topology (all above mentioned spaces, frames, locales, and even metric spaces) is an element of an ordered semigroup action (that is an element of a semigroup action conforming to a partial order). Thus the entire general topology is just ordered semigroup.

And finally "Limit of a Discontinuous function" uses the above theory to define a generalization of limit that is applicable to every (even discontinuous) function. This makes obvious to define derivative and integrals of arbitrary functions. Derivative and integrals as defined in the book are still linear operators! The book considers also some basic properties of differential equations with generalized derivatives.

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The Pythagorean Theorem might have been used in antiquity to build the pyramids, dig tunnels through mountains, and predict eclipse durations, it has been said. But maybe the main interest in the theorem was always more theoretical. Euclid’s proof of the Pythagorean Theorem is perhaps best thought of not as establishing the truth of the theorem but as breaking the truth of the theorem apart into its constituent parts to analyse what makes it tick. Euclid’s Elements as a whole can be read in this way, as a project of epistemological analysis.

Transcript

Let’s read Euclid together. Euclid’s Elements, one of the most important and influential works in human history, who wouldn’t want to read that? “Euclid alone has looked on beauty bare,” as the poets say.

Let’s do some episodes on this where we go through Euclid’s Elements Book I. And here’s the first twist: Let’s read it backwards. Well, not quite. But it’s a good idea to start at the end. Book I of the Elements ends with the Pythagorean Theorem and its converse. It’s not a murder mystery, it won’t spoil the fun to know the ending.

I will explain why I think this is a good idea. This has to do with appreciating the refined goals of the Elements. It’s a very subtle work, in ways that are easy to miss. So I will use this idea of starting at the end as a way of highlighting some things to keep in mind in that regard, so that we approach the text with appreciation of these subtleties.

It might be a bit dry to do only that, so I will also mix it up with some lighter things. Some stories related to the Pythagorean Theorem. Did the Egyptians use the Pythagorean Theorem to build the pyraminds, for example? Is that how they got the angles just right? We will discuss that soon. And I will also play a clip of RoboCop.

I will try to do this for the Elements as a whole: a serious discussion of its finer points, as well as some entertaining tangents exploring the many cultural links of the various parts of the Elements.

So here we go. My first goal is to outline the mindset with which we must approach Euclid’s text.

If you’re a young person, you may look at Euclid’s Elements and say: yeah yeah, triangles and stuff, I saw all of that in high school too; our textbook had proofs just like this thing by Euclid; it’s pretty much the same thing. No, no, no. That’s like listening to Mozart and saying: yeah yeah, big deal, music is music.

Forget it. There’s a world of difference. Euclid is on a whole other level of sophistication than some crappy high school textbook. You wouldn’t know it just by looking at the text though. The text looks the same as any other geometry text. Triangle ABC blah blah blah. It’s the same with musical scores, isn’t it? They all look the same when you just glance at the pages. You can’t tell Mozart from some hack.

We must look deeper to appreciate the subtlety and genius of Euclid. The text itself doesn’t spell that out, just as a Mozart quartet doesn’t have a narrator telling you what’s great about it. But great works reward reflection. The more you study Euclid, the more you interrogate the text, the more you puzzle over its oddities, the more you come to appreciate the mastery that went into crafting everything just right. Euclid knew exactly what he was doing. His work is orders of magnitude more sophisticated than other superficially similar works in the same genre.

The exercise of reading backwards is one angle we can use to start getting a handle on this. If we read Euclid from cover to cover, in the order it’s written, we get a strictly “bottom-up” perspective: we start with the most basic things and gradually get to higher and higher levels of sophistication. That’s how mathematics is typically written down. And with good reason. But the way mathematics comes into being is much more bidirectional. Mathematics grows like a tree: as the branches extend, so do the roots. Starting our Euclid adventure with the Pythagorean Theorem is a way of making us think about this.

Of course when we read Euclid’s proof of the Pythagorean Theorem we find that it is based on earlier results. So you might say: Obviously you have to read those first before you can understand this proof. But that’s a bit simplistic. You could also say: Actually you need to look at the Pythagorean Theorem first because only then can you understand what the purpose is of those earlier propositions. From a purely logical perspective you have to read it linearly from start to finish, but to understand the meaning and purpose of these logical constructions you have to take a step back and interrogate the text from other angles as well. For a dogmatic understanding, it is enough to read it linearly, and parse the logical steps like a machine. But for a critical, independent understanding you want to not only verify the logic but also see how one could arrive at such logical constructions organically.

That goes for any formal mathematics text, still to this day. Or maybe even more so today than ever. The definitions and axioms are the starting points of the way mathematics is written, but often they are almost the end product of the actual creative thought process. Only after you have figured out the hard parts of your theory do you know what the starting points need to be. Or at least there’s an interaction, a back-and-forth negotiation between the top and the bottom of the theory. Each is adapted to the other.

So that’s one reason to read Euclid backwards. It’s a reason that applies to any formal mathematical theory, because they all have this element of bidirectionality.

Actually geometry might be among the more unidirectional formal mathematical theories in how it was conceived, because the results of geometry were known in great detail, long before they were formalised. The tree came before the roots, so to speak.

Here’s another way of visualising it. Think of the Pythagorean Theorem as the apex of a pyramid. The proof reveals which lower, more foundational stones it rests on. Those stones in turn rest on other stones, and so on. Something has to be the bedrock that is considered solid enough not to need any further support beneath it. Euclid’s Elements can be read in two directions: as a way of building up a more and more elaborate structure on top of solid foundations, or as a way of reducing advanced results to their basic components. So when we read the proof of the Pythagorean Theorem, one of the perspectives we should use is to think of it as “boiling down” this somewhat advanced result to more basic ones. This will help us appreciate the purpose and achievement of the more fundamental parts of the Elements when we get to those.

Indeed, by the time Euclid wrote the Elements, the theorems themselves—such as the Pythagorean Theorem—had been known for hundreds or even thousands of years. Even proving the theorem wasn’t all that new. There were plenty of proofs. I bet Euclid knew two dozen proofs of the Pythagorean Theorem.

We shouldn’t think of Euclid as saying: Hey guys, I discovered some things about triangles and stuff; check out this book where I explain how I came up with these theorems.

No, no, no. That’s not at all what Euclid is doing. We must understand, when we read the Elements, that we’re way beyond that.

If you just wanted to convince a random person that the Pythagorean Theorem is true, then there are much better proofs than Euclid’s. Simpler ones. More intuitive, based on simple diagrams. If all you want is a psychologically compelling argument that the Pythagorean Theorem is true then there are better options than Euclid.

Euclid knew all of that, and he chose his proof very deliberately. Because it’s the best proof for his purposes. Namely the purpose of carefully analysing how the truth of the Pythagorean Theorem can be broken down into smaller truths. And more generally to do the same thing for all the truths of geometry in a comprehensive and systematic manner.

So the proof of the Pythagorean Theorem isn’t so much about showing that the theorem is true. It’s more about showing what its ultimate foundations are.

Here’s another metaphor for this. Think of a mathematical theorem as a dish that you cook. The Pythagorean Theorem is like a soup, let’s say. You can whip it up very quickly with store-bought ingredients like stock cubes or just microwaving something from a can. But Euclid doesn’t do store-bought. He’s going to do everything from scratch. And I mean really from scratch. If there’s going to be carrots in there, then Euclid is going to grow his own carrots.

In fact you might say that Euclid is not so interested in cooking at all, even though a proof is like a recipe. Euclid is like a cookbook author who doesn’t like cooking and has no interest in feeding anyone.

Instead he’s more like a chemist who is analyzing the molecular composition of foods. His recipes are not meant as a practical cooking guide but as an analysis of what the core ingredients of the dish are if you deconstruct the recipe as far as you possibly can.

Here we have the idea of reading backwards again: Euclid isn’t really interested in making Pythagorean Theorem soup, but in starting with Pythagorean Theorem soup and taking it apart in the lab. Put it on the Bunsen burner. Different ingredients have different boiling points and so on, so you can carefully separate them out again.

There was already plenty of geometry before Euclid. If theorems are food, everyone was already well fed, so to speak. Everyone already had their favourite dishes and neither they nor Euclid were looking to replace the traditional menus. What Euclid is bringing to the table is not new food but a refined theoretical perspective that stands apart from actual cooking.

The idea of reading Euclid backwards is also related to a famous anecdote recorded about Thomas Hobbes, the 17th-century philosopher. Here’s what it says about Hobbes:

“He was 40 years old before he looked on geometry; which happened accidentally. Being in a gentleman’s library, Euclid’s Elements lay open, and ‘twas the [47th Proposition of Elements Book I, the Pythagorean Theorem]. He read the proposition. By God, sayd he, this is impossible! So he reads the Demonstration of it, which referred him back to such a Proposition; which proposition he read. That referred him back to another, which he also read. [And so on], that at last he was demonstratively convinced of that trueth. This made him in love with geometry.”

It is interesting that Hobbes ended up reading Euclid backwards by accident like this. Precisely what I recommended as a deliberate strategy. But he doesn’t seem to have appreciated the point of doing so the way I have described it. Maybe he could just as well have read the book forwards and had the same experience, as far as this anecdote goes.

Hobbes fell “in love with geometry” by reading it backwards, but others had the same experience reading it forwards. Bertrand Russell, another famous philosopher, read Euclid the conventional way, starting at the beginning, and he still found it, as he later said, “as dazzling as first love”: “I had not imagined there was anything so delicious in the world.” Bertrand Russell was eleven at the time, while Hobbes was 40 when he stumbled upon Euclid. They lived almost three centuries apart. So these anecdotes speak to the universality of Euclid’s text: young or old, forwards or backwards, conservative or socialist, in a society of cars or one of horses—the one thing they have in common is the love that Euclid stirred up in them.

That’s all very nice, but it kind of misses the point in terms of what I have tried to argue was the goal of Euclid’s Elements. What Hobbes and Russell fell in love with was the idea of geometrical proof, it seems. Historically, those epiphanies are better associated with a pre-Euclidean period. We discussed Thales before, and there were plenty of others in the centuries between him and Euclid.

So when you read Euclid, by all means, do fall in love. Be seduced like so many others have been. But also keep in mind that these charms are only part of the greatness of Euclid. Euclid’s Elements can be as good a vehicle as any to have that epiphany of the beauty of mathematics. But to Euclid and many of his readers that was old news.

Euclid wanted to do more than that. He didn’t want to just show how cool it is to prove stuff, although that is lovely. More than that, he wanted to explore the very essence of geometrical knowledge. What are its preconditions, and the source of its certainty? Just as a chemist seeks to decompose any substance into the elements of the periodic table, so Euclid sought to find the “periodic table” of geometry, so to speak: he wanted to uncover the ultimate building blocks of this entire branch of knowledge.

Ok, so that’s my lesson one in how to read Euclid. Start at the back and keep in mind this theme of distillation into ultimate foundations.

So I urge you to go read Euclid that way. I’m not going to go through the proof here; you’ll have to follow along in your own copy of the Elements. I recommend my own edition, for which I added illustrations for each step of the proofs. It’s a joy to read, in my opinion. But it’s too visual to translate into this medium, so I’ll leave that to you to pursue.

Now I wanted to take this opportunity to think about the origin of the Pythagorean Theorem. Part of the appeal of reading Euclid’s Elements is how embedded it is many aspects of human culture and history. So in parallel with our reading of Euclid I wanted to bring up such themes as well.

The Pythagorean Theorem has little to do with Pythagoras. It was discovered independently in several cultures, some of them long before Pythagoras. But never mind the name. The more interesting question is: Why were people interested in this theorem? Why would anybody want to calculate a bunch of hypothenuses?

If you look in a modern geometry textbook, you won’t find any good answers. The book will give you the formula and ask you to apply it in all kinds supposedly real-world cases, but they are all fake and transparently ridiculous. How to calculate the diagonal of a field when you know the lengths of the sides: When would you ever use this? Why wouldn’t you just measure the diagonal then if that’s what you want to know?

Ladder problems is another one of those fake classics. The foot of the ladder is so-and-so far from the wall, and the ladder is so-and-so long, will it reach to such-and-such a height, maybe for instance the ladder of a fire truck to save someone from a burning building? Not a very realistic scenario. Wouldn’t you just try it and see if it worked? Wouldn’t that be just as easy as sitting around making calculations? And why would the distance from the wall to the foot of the ladder be some exact given number? And so on.

It doesn’t make sense that people discovered the Pythagorean Theorem because they were wrestling with practical problems like those. They would not have needed mathematics for that. If they wanted to solve those problems they would have used trial and error and direct measurements.

Unfortunately, ancient textbooks are as ridiculous as modern ones in this regard. Here’s an example from a Chinese text from about the time of Euclid. A 10-feet-high stem of bamboo broke in the wind. It broke into two straight prices. One part remains upright, perpendicular to the ground. But the other part, that broke off but is still attached, tipped over and is now touching the ground, 3 feet away from the base of the stem. How high up the stem did the break occur?

You can calculate this with the Pythagorean Theorem, sure enough, but of course there is no way anyone would ever do something so absurd in the real world. Just measure it, if you want to know. You apparently already measure the distance along the ground and the full height somehow, so why couldn’t you just as well measure this thing? Doesn’t make any sense.

Here’s another scenario some have claimed involves the Pythagorean Theorem. On the Greek island of Samos, there’s an ancient tunnel, which was dug in fact right in the lifetime of Pythagoras.

This tunnel is a marvelous thing, a tribute to the engineering skills of the Greeks. It’s still there today. The tunnel is over one kilometer in length through a big mountain. It was dug to supply the capital with fresh water.

Digging the tunnel was certainly a geometrical project. In fact, the walls still have letters on them, like the lettering of a geometrical diagram. Evidently there was a plan of the tunnel in the form of a drawn diagram, with points makes by letters, and then as it was dug these letters were inscribed on the wall to keep track of how the actual tunnel corresponded to the geometrical plan.

This was all the more essential since the tunnel had to be dug from both ends, in order to complete it in half the time. So the diggers had to be coordinated to ensure they met in the middle. A highly non-trivial problem, which the Greek geometers solved flawlessly.

In fact, at some point the plan even had to change because the rock was becoming to porous. So there was a risk that tunnel would collapse. Therefore it was necessary to make a bend in the tunnel that took it more toward the core of the mountain, which had harder rock. The geometers dealt with this flawlessly as well. They added a shallow isosceles triangle to the diagram. So each digging team had started out along straight lines that would have met in the middle, but halfway through both teams were instructed to make a slight turn which was specified with geometrical precision. So the whole tunnel has a kind of V-shaped bend in the middle. But it still worked. The two digging teams met just as the geometers had calculated.

That’s great stuff, but is it the Pythagorean Theorem? Let me play to you a clip from the History Channel documentary series Engineering an Empire, which claims that it is.

“Eupalinos dug tunnels from each side of the mountain, until they met in the middle. To succeed, Eupalinos had to make sure that each tunnel started at the same vertical height., on opposite sides of the mountain. The tunnels also had to match up on a horizontal plane. Otherwise, they would pass each other like ships in the night.”

“By forging a path from the spring to the city, in short perpendicular lines, Eupalinos could measure each small length in order to calculate two sides of a right triangle. With two known sides of the triangle, the hypothenuse became the path of the tunnel through the mountain.”

So according to the History Channel, the plan for the tunnel was based on the Pythagorean Theorem. The History Channel are not even taking into account the alterations of the plans midway through, by the way. They just discuss the problem of making a straight tunnel.

The presenter of this documentary is Peter Weller, who is also the actor who played RoboCop in the 1987 movie. Turns out he’s also a historian.

I must say though that I disagree with RoboCop’s analysis. The tunnel of Samos was great geometry but it wasn’t the Pythagorean Theorem. The way RoboCop puts it in the documentary, it sounds as if the point was to calculate the length of the tunnel. That’s the hypothenuse that RoboCop is talking about in that clip. But of course the real problem is the coordination of the two digging teams, so they won’t miss each other “like ships in the night,” as RoboCop himself said. How is the length of the hypothenuse supposed to be useful for this? Knowing how long the tunnel is supposed to be doesn’t help you determine the direction of digging.

So I don’t think this tunnel stuff is a great example of real-world motivation for the Pythagorean Theorem. We have to keep looking for where ancient man could have had reason to discover or apply this theorem.

Here’s another such scenario. Did the Egyptians use the Pythagorean Theorem to build the pyramids? I’ll play another clip from another documentary series that claims: yes. This is from The Story of Maths, a BBC documentary presented by Marcus du Sautoy.

“The most impressing thing about the pyramids in the mathematical brilliance that went into making them. Including the first inkling of one of the great theorems of the ancient world: Pythagoras’s Theorem. In order to get perfect right-angled corners on their buildings and pyramids, the Egyptians would have used a rope with knots tied in it. At some point, the Egyptians realised that if they took a triangle with sides marked with 3 knots, 4 knots, and 5 knots, it guaranteed them a perfect right angle.”

The theorem involved here is not the Pythagorean Theorem itself, but the converse of it, which is Proposition 48 in Euclid.

In terms of historical evidence, we really don’t know if the Egyptians did this or not. It’s plausible that they knew this but there’s very little documentary evidence from way back then.

Obviously you can’t believe anything just because Marcus du Sautoy said it in a BBC documentary. Marcus du Sautoy is not a historian, he’s just clowning around. But let’s see, if we’re serious about it, does it make any sense?

I used to be skeptical about this, but I have come to think maybe it’s not so bad. I think the standard formulation about a rope with 3+4+5 equally spaced knots on it is a bit silly. Seems very complicated to get the knots just right.

But you don’t really need one triangular rope. Instead you can just use three separate ropes, of lengths 3, 4, and 5. That’s easy to make. Then when you need to make a right triangle you stretch the 3 and 4 ropes along the intended sides, and you check if the 5 rope fits between their endpoints. Then you have the guy holding the end of the 4 rope move a bit this way or that until it lines up perfectly.

I have to admit, if I was building a pyramid I would probably go with this method. Especially because of the scale of the project. The base of the pyramid is enormous. You would use ropes with lengths 3, 4, 5, but not in feet or meters but some bigger unit. Maybe 30 meters, 40 meters, 50 meters. The ratio is all that matters of course. The longer the ropes, the less significant the measurement error becomes. So it’s a pretty good method I think.

Let me read you a quote here from the book Euclid’s Window by Leonard Mlodinow. I thought it was quite funny.

“Picture a windswept, desolate desert, the date, 2580 B.C. The architect had laid out a papyrus with the plans for your structure. His job was easy—square base, triangular faces—and, oh yeah, it has to be 480 feet high and made of solid stone blocks weighing over 2 tons each. You were charged with overseeing completion of structure. Sorry, no laser sight, no fancy surveyor’s instruments at your disposal, just some wood and rope. As many homeowners know, marking the foundation of a building or the perimeter of even a simple patio using only a carpenter’s square and measuring tape is a difficult task. In building this pyramid, just a degree off from true, and thousands of tons of rocks, thousands of person-years later, hundreds of feet in the air, the triangular faces of your pyramid miss, forming not an apex but a sloppy four-pointed spike. The Pharaohs, worshipped as gods, with armies who cut the phalluses off enemy dead just to help them keep count, were not the kind of all-powerful deities you would want to present with a crooked pyramid. Applied Egyptian geometry became a well-developed subject.”

So that’s a quite comical way of putting it, but the point is well taken, I think. Indeed it does make some sense, this whole thing. The historical and societal context, the mathematics available at that time, the need to make exact right angles, the method for doing so using strings and a Pythagorean triple: that is all quite plausible, I would say.

It’s hardly plausible that they would have discovered the Pythagorean Theorem this way, by starting with the problem of making right angles. But it is plausible that may have used knowledge of the 3-4-5 special case of the converse of the Pythagorean Theorem to make right angles.

Here’s another proposal for the possible origins of the Pythagorean Theorem. This proposal is from van der Waerden’s book Geometry and Algebra in Ancient Civilizations. He proposes that the original motivation for the discovery of the Pythagorean Theorem might have been related to eclipses. Namely, calculating the duration of a lunar eclipse.

Indeed, astronomy was important to many ancient peoples. You know the Stonehenge, Maya temples aligned with solstices and so on. People cared a lot about the sky back then.

Eclipses were a big deal. Probably they were often seen as having some kind of theological significance, some sort of omen, and so on. They were also scientifically important, for instance for exact calendar keeping.

So what do eclipses have to do with the Pythagorean Theorem? Mathematically, this is a neat example. Fun to use in a geometry class.

A lunar eclipse occurs when the moon passes through the shadow cast by the earth. The earth’s shadow is about twice the size of the moon, at that distance. So the moon is approaching this dark spot, it enters it, and keeps moving through it, and comes out at the other side. The whole thing takes maybe an hour or two, it differs.

We can predict in advance how long a particular lunar eclipse is going to last. The determining factor is whether the path of the moon goes right through the middle of the earth’s shadow, or cuts across it off center. The moon’s orbit is complicated and it’s different each time. Sometimes it’s coming in a bit high and sometimes a bit low. We can see this by comparing its position to the stars.

So this means that the problem of calculating the duration of an eclipse comes down to calculating the length of a line cutting through a circle, not necessarily through the middle. We assume that the moon’s speed is constant throughout the eclipse. So the duration of the eclipse is determined by how big of a segment of the moon’s path is in the circular shadow cast by the earth.

This indeed becomes a Pythagorean Theorem problem. You can picture it like this. Draw a circle. That’s the shadow cast by the earth. Now draw a line cutting through the circle, but not through the midpoint. That’s the path the moon is moving along. We want to know the length of the segment inside the circle. This is what determines the duration of the eclipse.

Find the midpoint of this segment. Connect it to the center of the circle. This is a known length, because it corresponds to how far off-center the moon was in its approach, which we can determine by comparing its position to the stars. So the distance from the midpoint of the segment to the center of the circle was known before the eclipse began.

Let’s add one more line to the diagram: the line from the center of the circle to the point where the moon’s path entered the circle. That’s of course a radius of the circle, which is known because the size of the earth’s shadow is known.

So now you see why it’s a Pythagorean Theorem problem. The two knowns are two sides of a right-angle triangle, and the sought length is the remaining side.

Could this be how ancient man discovered the Pythagorean Theorem? This hypothesis has one thing going for it, namely that the sought quantity cannot be measured directly in advance of the eclipse. You genuinely need the Pythagorean Theorem to do this. It’s not one of those fake ones where you could just as easily have measured the side you are looking for, instead of measuring the sides you don’t want and then calculating the you do want, as in those fake textbook problems.

Mathematically, that’s all very satisfying. Unfortunately this hypothesis is not very plausible historically. In the Babylonian tradition, mathematics came long before mathematical astronomy. Serious mathematical astronomy such as this, with detailed eclipse calculations and so on, was a preoccupation of the second flowering of ancient Babylonian mathematics. That’s about a thousand years after the first golden age of Babylonian mathematics.

Already the older period had excellent mathematics, including something like the Pythagorean Theorem. One of the most famous old Babylonian clay tablets states the ratio between the side and the diagonal of a square. So it’s essentially a numerical approximation of the square root of 2, in other words. The numerical value the tablet states is very nearly accurate to six decimal places. That’s very accurate indeed. Suppose you used it to compute the diagonal of a square field with a side of a hundred meters. So a football field, basically. Then the Babylonian approximation is off from the exact answer by less than one millimeter.

That’s more than a thousand years before Babylonian priests became obsessed with eclipses for the sake of ensuring the calendric accuracy of their rituals. So the mathematically pleasing hypothesis about the Pythagorean Theorem being discovered to calculate eclipse durations doesn’t really fit the historical record unfortunately.

So what can we conclude from all this? I think it’s safe to say that practical need was never the main driver of mathematics that goes even a bit beyond the basics. The Pythagorean Theorem was discovered because people were fascinated by mathematics for its own sake, not because they needed to calculate stuff. The Chinese didn’t need to know the breaking points of bamboos, the Babylonians didn’t need to know the diagonal of a football field with millimeter accuracy. They were fascinated by the power of mathematical reasoning to discover hidden relationships, and that’s why they explored these things.

This is also how we should read Euclid. The proof of the Pythagorean Theorem is not so much about proving that the theorem is true. It’s more about exploring the basis for this knowledge. Mathematics was always explored for this reason.

Discovering mathematics was like discovering magic. It impresses us as a powerful force that can do incredible things. We want to understand it: How is this possible? What makes this magic tick? It is so unlike anything else we are familiar with, it’s like a portal to a divine realm. We feel a spiritual imperative to understand it.

Already ancient civilisations started along this path, and Euclid does the same. If mathematics is magic, Euclid’s Elements is not a book of spells, but a scientific investigation of how there can be such a thing as magic at all.

Or to use another metaphor, we have to dissect mathematics like an alien corpse to discover the secrets of it mysterious inner workings. The Pythagorean Theorem is the alien: a weird thing that seems to have superhuman powers. Euclid’s proof is not a recipe to give you alien abilities; rather, it is the result of his through dissection of an alien he found in the wild.

So let’s read Euclid this way, as an exploration into the inner mechanisms—the heartbeat—of these strange entities, these superhuman theorems, that have impressed mankind with their seemingly magical and divine aura for many thousands of years.



from Intellectual Mathematics

$\bullet~$Problem: Show that $\cos\bigg(\dfrac{2\pi}{n}\bigg)$ is an algebraic number [where $n$ $\in$ $\mathbb{Z}/\{0\}$].


$\bullet~$ My approach:

Let's consider the following polynomial in $\mathbb{Z}[x]$ in recursive terms. \begin{align*} &T_{0}(x) = 1\\ &T_{1}(x) = x\\ &T_{n + 1}(x) = 2x T_{n}(x) - T_{n-1}(x) \end{align*} $\bullet~$ $\textbf{Claim:}$ The polynomial $T_{n}(x)$ for any $n$ $\in$ $\mathbb{N}$ satisfies the following \begin{align*} T_{n}(\cos(\theta)) = \cos(n\theta) \end{align*} $\bullet~$Proof: We'll use induction on $n$ for this proof.

At first, we easily obtain that for $n = 0$ the given is true.

Now for some $n = k$, we assume that \begin{align*} T_{k}(\cos(\theta)) = \cos(k\theta) \end{align*} Therefore we need to prove for $n = (k + 1)$.

Now from the recursion relation of $T_{n}(x)$ we have \begin{align*} T_{k + 1}(\cos(\theta)) & = 2 \cos(\theta)T_{k}(\cos(\theta)) - T_{k -1}(\cos(\theta))\\ & = 2 \cos(\theta) \cos(k\theta) - \cos((k -1)\theta)\\ & = 2 \cos(\theta) \cos(k\theta) - \cos(k\theta) \cos(\theta) - \sin(k\theta)\sin(\theta)\\ & = \cos((k + 1)\theta) \end{align*} Hence by induction hypothesis, we obtain that our claim is true.

Therefore we have \begin{align*} T_{n}\Bigg(\cos\bigg(\frac{2\pi}{n}\bigg)\Bigg) = \cos(2\pi) = 1 \end{align*} Therefore we just need to consider a polynomial $P(x) = T_{n}(x) - 1.~$ As $T_{n}(x) \in \mathbb{Z}[x]$ it implies $P(x) \in \mathbb{Z}[x]$

Therefore we have $\cos\big(\frac{2\pi}{n}\big)$ is an algebraic number.


Please check the solution and point out the glitches :)



from Hot Weekly Questions - Mathematics Stack Exchange
Ralph Clausen

Good evening!

I have the opportunity to carry out research at my university, along my BSc studies in Computer Science.

This 2 years long research programme entails 1 year of interdisciplary research.

We, the students, have to propose some ideas and then research groups will be formed.

Majors in the programme:

• Computer Science;

• Mathematics;

• Physics&Astronomy;

• Biology;

• Chemistry;

• Molecular Life Sciences;

While I have a ton of ideas(mostly involving mathematics and computer science, some physics as well), I would love to hear your suggestions.

Some of the projects in the past years have led to published papers;

To give a bit more context:

• The groups are composed by 2nd year BSc students;

• I am very interested in machine learning, computational models and mathematical finance.

Thank you for your time.

Sorry for formatting and typos.

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from math https://ift.tt/33aly57 https://ift.tt/eA8V8J

1. Context: The notion of an integral
Let $H$ be a Hopf algebra over a field $\mathbb k$. We call its $\mathbb k$-linear subspace $$ I_l= \{x \in H; h \cdot x=\epsilon(h)x \quad for \>all\>h\in H\} $$ the space of left integrals. In other words, $I_l$ is the space of left invariants for $H$ acting on itself by multiplication. In a similar manner one can define (the space of) right (co)integrals.

Integrals seem to have a wide range of applications. For instance, they appear in a strong "(Hopf algebra) version" of Maschke's theorem, i.e. they are related to the semisimplicity of a Hopf algebra.

2. Question

  • Why are integrals called integrals?
  • Specifically, I think I overheard someone saying that they can be related to the notion of an integral in calculus. How so?


from Hot Weekly Questions - Mathematics Stack Exchange
M.C.

According to it's reviews, it is said to be one of the best books for beginners on discrete mathematics. I would like advice, and any other recommendations, along with insight on this book or other books.

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from math https://ift.tt/2CU0fdG https://ift.tt/eA8V8J

This, things like this is what I searching for. Unfortunately, there are not many channels that post this type of content. And search engines always seem to return crap. I want to solve puzzles that are hard not because of the conceptual level but because of the creativity that went into it. TED-Ed itself redirects to Brilliant.org, but it doesn't seem so interesting. It contains courses on math and science almost at a high-school level which is boring, especially considering that I myself am an advanced mathematics user. And the question "Is Brilliant worth it?" seems to confirm my suspicion that Brilliant doesn't exactly have these types of puzzles.

Can someone suggest books/resources that are worth reading?

P.S. If you think I am wrong in case of brilliant, you are welcome to suggest otherwise as well.

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from math https://ift.tt/314JQee https://ift.tt/eA8V8J

Hi, I volunteer as a math tutor children (usually age 6 to 12). Many of them have learning difficulties, and of course, covid19 doesn’t make the situation better. I always try to break down concepts by using stories/images/drawings. I really want each of them to succeed and gain confidence in mathematics. Any ideas, thoughts, resources for children?

Thank you

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from math https://ift.tt/33bVp6e https://ift.tt/eA8V8J

Can You Learn Math From a Game?

The statement above made 20 years ago would have been met with a lot of skepticism. Now, with over 20 years of history, millions of students across the country are learning problem-driven mathematics in the ST Math program. With the release of the 6th version, the new ST Math has doubled down on its learning science roots to create deeper levels of learning.

Below are some of the key highlights for how we make this learning happen...

Preserving the Magic of Spatial Temporal Math

Leg_Drape_GameFigure 1. The game Leg Drape shown with the addition of the Go Button, hint functionality, and keyboard controls

At the heart of ST Math is the patented spatial-temporal puzzles. While maintaining the zen-like nature of ST Math, there have been a number of enhancements to individual games. Hint functionality allows students and teachers to more precisely see which elements on the screen are interactive. Enhanced animations make the animation sequence easier to follow. All games in the new ST Math support keyboard control, allowing students to use the keyboard in order to play them.

In order to provide an additional opportunity for students to “Think Before They Click,” a Go Button has been added to each game to ensure students have a second chance before they make their final answer. All of these elements are enhancements designed around the spatial-temporal game play that is at the core of ST Math. 

Doubling Down on Informative Feedback 

The center of learning in ST Math lies in the informative feedback. Based on engaging students in the perception-action cycle, this unique approach of displaying visual animations in response to student input is what marks ST Math as one of the pioneers in its field.

STMath_Push_Box_Gif
Figure 2. Using the animation controls to control the animation in the game ‘Push Box’

To build games for the new ST Math, we used a novel development process that puts the animations on a timeline, as opposed to a more traditional approach of implementing animation through code sentences. This feat of engineering allowed us to add animation controls which gives students  direct control over the informative feedback animation.

Since the feedback acts as a visual proof of the mathematics, this turns every game into a visual, mathematics proof simulation of sorts, allowing every student real-time access to this tool. In addition to student use, this feature can be especially powerful for teachers facilitating a student, by allowing them to pause, rewind, and focus on a specific point of the animation to discuss the key mathematics idea. This feature enables a deeper level of interaction with the feedback in ST Math.

Deepening Students' Mathematics Schema

One of the beautiful aspects of ST Math is how the puzzles allow for a range of strategies to be used while solving them. While the animated feedback does encourage a strategy, the nature of the visual model allows students to flexibly think about the mathematics and have the freedom to derive and test their own strategies, much like a scientist performing a science experiment.

However, since most of this thinking is done in the student’s mind, the variety of approaches to solving a puzzle are typically invisible to teachers.

Fruit_Monster_Game
Figure 3. In the 3rd grade multiplication and division game Fruit Monster, the annotation tool is used to show three different strategies that may be used to solve the problem.

The new annotation tool allows students to record their thinking on the screen, bringing students’ strategies to the forefront. This can be used by a student individually, students sharing strategies with each other, or teachers using the tool as a method to discuss the strategy taken by a student.

Since ST Math enables multiple strategies for solving its puzzles, this affords teachers the opportunity to present and discuss multiple strategies used by students, as shown above. This can lend itself well while discussing a puzzle with the class, extending the schemas students have built in ST Math and allowing ST Math to be the foundation for strategy development. 

Boosting Game-Based Learning 

Student_home_screen-STMathFigure 4. Student home screen in the new ST Math

One of the most powerful elements of ST Math is the way it captures deep student engagement.  Much of this is the result of ST Math’s application of game-based learning principles to the student learning experience. From puzzle driven scenarios to traversing through Learning Objectives, ST Math embodies what it truly means to be a game-based learning experience.

The new ST Math takes this experience even deeper. 

Learning_Objective_STMathFigure 5: Visual of a Learning Objective in ST Math shown as a bridge with pillars and boxes for students to play through 

One of the most impactful aspects of game-based learning is the role of story. Rather than present learning scenarios in the abstract, stories provide a narrative students can connect with. Simply adding a character can create this effect, but even better when all the elements weave through the story context.

In the new ST Math, JiJi starts the Journey in the arctic and travels from island to island. Each Learning Objective, embodied as an island, has a spatial-temporal representation of the sequence of levels required to get to the end of  the bridge. Along the bridge are visualizations of the essential learning elements: each pillar represents a game, and each block represents a level in that game.

This stays true to ST Math’s principle of only showing the essential elements of the learning experience and avoiding adding items that add additional cognitive load. 

Enhancing Student Agency over Learning

Student agency has become a huge focus in recent years. However, finding opportunities that enable this while maintaining meaningful learning is not always the easiest to design. In the new ST Math, in addition to the annotation tools and animation controls, we’ve created additional opportunities to put more agency in the hands of the students.  

Student_Data_Log_STMathFigure 6. Student data log showing this weeks and last weeks time and puzzle data, as well as the cumulative numbers shown in the student history

One unique element of increasing student agency is allowing students to set and work toward their own goals. Students now have enhanced data views of the number of puzzles and minutes they have achieved per week, on average, and total.

This provides the opportunity for students to set goals based on their previous performance, and reflect and revise those goals as they progress. For example, students may set weekly goals on the way to yearly goals, which gives them more power to adjust their effort because they can measure progress toward those goals at every session.

The above are just a subset of the many elements designed to enhance and deepen learning in the new ST Math. For additional information, take a look at the other features or experience the games on the ST Math play page yourself!

Further Reading:



from MIND Research Institute Blog https://ift.tt/2CVdFWR
Ki Karou

Let $f\in C^1(\mathbb{R})$, with $f(0)=0$ and $f'(0)=0$. Furthermore, assume that in some neighborhood around $0$, $f$ and $f'$ have no additional zeros, so $f^{-1}(\{0\})=\{0\}=(f')^{-1}(\{0\})$. I want to show that $\lim_{x\to 0} \frac{f(x)}{f'(x)}=0$.

EDIT: According to a comment, this statement might be false. Would it be possible to prove the following, weaker statement: If $\lim_{x\to 0} \frac{f(x)}{f'(x)}=y$, then, $y\in\{0,+\infty,-\infty\}$?

My attempt so far is to write $$ \lim_{x\to 0} \frac{f(x)}{f'(x)} = \lim_{x\to 0} \lim_{h\to 0} \frac{hf(x)}{f(x+h)-f(x)} \stackrel{?}{=} \lim_{h\to 0} \lim_{x\to 0} \frac{hf(x)}{f(x+h)-f(x)} = \lim_{h\to 0} \frac{h\cdot 0}{f(h)-0} = 0. $$ As indicated by the "?" above the "=", I am not sure how to prove that I am allowed to exchange these limits. I tried to apply the Moore-Osgood theorem. If I understand the theorem correctly, it boils down to showing:

  1. For all $h\neq 0$, the limit $\lim_{x\to 0} \frac{hf(x)}{f(x+h)-f(x)}$ exists. This limit is always equal to $0$, by the same calculation as above.
  2. For all $x\neq 0$, the limit $\lim_{h\to 0} \frac{hf(x)}{f(x+h)-f(x)}$ exists. This limit is equal to $\frac{f(x)}{f'(x)}$ and thus exists.
  3. One of the limits converges uniformly, i.e., either the first limit converges uniformly for $h\neq 0$, or the second limit converges uniformly for $x\neq 0$.

Unfortunately, I am stuck at showing uniform convergence of either of the two limits.

I have the following questions:

  1. Is the statement I am trying to prove correct, or do I need further assumptions?
  2. Is my proof strategy correct so far? Is there a simpler way?
  3. Is one of the limits actually uniform? If so, can someone give me a hint on how to show it?


from Hot Weekly Questions - Mathematics Stack Exchange
chafner

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