I know there's gaming stackexchange for gaming questions, but I believe this is purely maths related.

I'll try to avoid using game jargon and keep it simple. I'm collecting keys in game, each key taking a fair bit of time to obtain. They open a chest, which generates one random reward from a predeterimed table. On the loot table, there are five different armor pieces, each with a chance of 1/1000 to obtain. The catch is, because of game's inventory limitations, I can't open the chest every time I get a key and just get the five armor pieces that way - I have to do big openings of multiple keys at once. If I don't get a full set, but, for example, 4 out of 5 pieces, it would limit the speed of obtaining future keys.

Which is why I would like to count an optimal amount of keys to have to get the full set at once. Is there a set method or a formula for similar problems?

If the above explaination is too convoluted, I can try to further simplify it if needed.

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     A fascinating article about poet Jericho Brown (by Allison Glock in Garden and Gun magazine) reminded me of the vital role of line-arrangement in creating a poem.  (Emory University professor Brown has won the Pulitzer Prize in poetry for his collection The Tradition  (Copper Canyon Press, 2019)).
      Glock's article, "Jericho Rising," tells of various factors that have influenced Brown's poetry and describes his process of arranging lines, typed on separate strips of paper, into poems.  Three of the lines shown in the article are:

       What is the history of the wound? 
       And a grief so thick you could touch it.
Read more »

from Intersections -- Poetry with Mathematics

I could not find this problem anywhere else, but maybe I am just using the wrong keywords (in that case, I'm sorry).

Considering I have two bags, each one with $N$ elements inside, I need to pair each element of the first bag to an element in the second bag (therefore creating $N$ pairs). The question is how many ways there are to pair these elements.

For example, let's consider two bags with $N=5$:

Bag_1 = [A, B, C, D, E]

Bag_2 = [1, 2, 3, 4, 5]

Creating pairs [A1, B2, C3, D4, E5] (one possibility) or [A2, B1, C3, D4, E5] (two possibilities) and so on...

There are $5! = 120$ ways of pairing these elements.

For the case of one bag containing $N$ different elements I could generalize the answer to be:

$\frac{N!}{\prod{K_i!}}$

Where $K_i$ is the amount of each element present in the second bag (e.g. for [1,1,2,2,3] the denominator would be $2!2!1!$).

But when both "bags" have repeated elements, I could not create a rule for describing how many different ways there were to pair the elements.

This is a simplification of the problem, what I actually have to solve involves $N$ bags (creating not pairs, but groups of $N$ elements). But any light on the problem is appreciated.

Thanks for your attention!

Edit: I also need to mention that elements in both bags could be the same (e.g. both could be letters) and order does not matter in any way (e.g 'AB' is the same as 'BA').

from Hot Weekly Questions - Mathematics Stack Exchange

Prove that there are infinitely many natural $n$ such that $n|2^n+3^n+6^n$

I guess some sort of hensel's lifting method should work but I have no progress,so pls help

from Hot Weekly Questions - Mathematics Stack Exchange

Similarly to "center of a group", also "inner automorphism" has a topological sounding. But, though for the former I could find some possible explanation in this site, for the latter I couldn't. So:

Why are inner automorphisms named this way?

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We needed more toilet paper at home but there wasn't any at the store.

So, I ordered this online.  When it came I was surprised.  Read on ...

The activity: 16-Rolls-Toilet-Paper.pdf

CCSS: 5.MD.C, 6.G.A, 6.RP.A, 7.RP.A, 7.G

For members we have an editable Word docx and solutions.

16-Rolls-Toilet-Paper.docx    Bamboo-Toilet-Paper-solution.pdf

from Yummy Math

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Wikipedia's article on parametric statistical models (https://en.wikipedia.org/wiki/Parametric_model) mentions that you could parameterize all probability distributions with a one-dimensional real parameter, since the set of all probability measures & $\mathbb{R}$ share the same cardinality.

This fact is mentioned in the cited text (Bickel et al, Efficient and Adaptive Estimation for Semiparametric Models), but not proved or elaborated on.

This is pretty neat to me. (If I'd been forced to guess, I would have guessed the set of possible probability distributions to be bigger, since pdfs are functions $\mathbb{R}\rightarrow\mathbb{R}$, and we're counting probability distributions that don't have a density, too. It's got to be countable additivity constraining the number of possible distributions, but how?)

Where could I go to find a proof of this, or is it straightforward enough to outline in an answer here? Does its proof depend on AC or the continuum hypothesis? We need some kind of condition on the cardinality of the sample space that neither Wikipedia or Bickel mention, right (if it's too big, then the number of degenerate probability distributions is too big)?

from Hot Weekly Questions - Mathematics Stack Exchange

I am reading a research paper and am stuck at a point where the author uses angular integrals. I don't have any idea about it and would like help. The angular integral is:

$$I_k (y)=\int_0^\pi {\sin^2(\theta)\cos^{2k}(\theta) \over \beta^2(y)-\cos^2 (\theta)} d\theta$$ The author directly give results to this integral without proof. The results are: $$I_0 (y)=\pi \left(1-\sqrt{1-{1\over\beta^2 (y)} } \right),$$

$$I_1 (y)={-\pi \over 2}+\beta^2 (y)I_0 (y),$$

I don't know how he got these results. Can anyone please help me?

from Hot Weekly Questions - Mathematics Stack Exchange

$\mathbf{Question:}$ Show that $|b-a|\geq|\cos a-\cos b|$ for all real numbers a and b.

$\mathbf{My\ attempt:}$

The Mean Value Theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists $c \in (a,b)$ such that $f'(c)=\frac {f(b)-f(a)}{b-a}.$

Using the MVT where $f(c)=\cos c:$

$$ f'(c)=\frac {f(b)-f(a)}{b-a} $$

$$ f'(c)(b-a)=f(b)-f(a) $$

$$ (-\sin c)(b-a)=\cos(b)-\cos(a) $$

$$ (\sin c)(b-a)=\cos(a)-\cos(b) $$

Taking the absolute value of both sides:

$$ |\sin c||b-a|= |\cos(a)-\cos(b)| $$

Because $|\sin c |\leq 1,\,$ we can bound $|\sin c|$ by $1$ $$ 1 \cdot|b-a| \geq |\cos(a)-\cos(b)| $$

Thus $|b-a|\geq|\cos a-\cos b|$ for all real numbers $a$ and $b$

from Hot Weekly Questions - Mathematics Stack Exchange

Brouwer's Fixed Point Theorem (BFPT) is not provable in Bishop-style constructive mathematics (BISH). For quick orientation, BISH is obtained from classical mathematics by removing the Law of Excluded Middle (LEM) and replacing the full Axiom of Choice by the Axiom of Dependent Choice.

Standard classical proofs of BFPT either rely on LEM or the Bolzano-Weierstrass Theorem, which is equivalent to the Limited Principle of Omniscience, a constructively inadmissible weakening of LEM. BFPT itself is equivalent to Weak König's Lemma in BISH, and thus to the Lesser Limited Principle of Omniscience, another constructively inadmissible weakening of LEM.

However, the following approximate version of BFPT is known to be provable in BISH.

Let $f$ be a uniformly continuous function from $\Delta^n$ into itself. Then for each $\varepsilon>0$ there exists $x\in\Delta^n$ such that $d(f(x),x)<\varepsilon$.

Here's my proof attempt:

The proof in BISH proceeds via Sperner's Lemma. Recall that $$\Delta^n=\bigg\{(x_0,...,x_n)\mid\sum_{i=0}^{n}\leq1\bigg\}.$$

Consider a uniformly continuous function $f$ from $\Delta^n$ into itself. Fix $m,k\in\mathbb{N}$ such that $2^{-k}+2^{-k-2}+2^{-k-3}<2^{-m}$ and $$d(x,y)<2^{-k}+2^{-k-2}\Longrightarrow d(f(x),f(y))<2^{-m},$$ where $d$ is the Euclidian metric.

Consider a simplical subdivision $\mathcal{S}$ of $\Delta^2$ with grid size $2^{-k}$. Specify a labeling as follows. A vertex $x$ of a simplex in $\mathcal{S}$ gets assigned the label $i$ if, and only if, $f(x)_i<x_i$ or $f(x)_i<x_i+2^{-k-3}$. This is constructively well-defined by the Approximate Splitting Principle and constitutes a feasible labeling in the sense of Sperner's Lemma. Thus, $\mathcal{S}$ contains a fully labeled subsimplex $S$. We must now show that for a vertex $s$ of $S$, $d(f(s),s)<(n-1)2^{-m}$ holds (cf. Scarf, 1967; van Dalen, 2011).

This is where I got stuck.

Suppose $s$ is labeled by $1$. I just can't figure out how to infer $d(f(s),s)<(n-1)2^{-m}$ from the fact that $f(s)_1<s_1+<2^{-k-3}$ and uniform continuity of $f$ alone.

I'm sure the argument is fairly easy, but I just can't see it. A "visual" argument for $\Delta^2$ is in the paper by van Dalen referenced above. While it helps me see that the conclusion is true for $\Delta^1$ and $\Delta^2$, I am still unable to derive it formally, let alone generalize it to $\Delta^n$.

Any help would be greatly appreciated.

Thank you in advance!

PS: I am not a mathematician, so bear with me.

from Hot Weekly Questions - Mathematics Stack Exchange

Find the largest integer $n$ such that $n$ is divisible by all positive integers less than $\sqrt[3]{n}$.

420 satisfies the condition since $7<$ $\sqrt[3]{420}<8$ and $420=\operatorname{lcm}\{1,2,3,4,5,6,7\}$

Suppose $n>420$ is an integer such that every positive integer less than $\sqrt[3]{n}$ divides $n .$

Then $\sqrt[3]{n}>7,$ so $420=\operatorname{lcm}(1,2,3,4,5,6,7)$ divides $n$ thus $n \geq 840$ and $\sqrt[3]{n}>9 .$

Thus $2520=\operatorname{lcm}(1,2, \ldots, 9)$ divides $n$ and $\sqrt[3]{n}>13$

now this pattern looks continues,but i am not able to prove that this pattern always continues..

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First question on MSE! I'd appreciate hints, theorem suggestions, or method suggestions regarding the question in the title or below. Please avoid full solutions. I'm studying for an exam coming up and got stuck on this question:

Problem Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a measurable function. Prove that $f(x)$ and $\frac{1}{f(1/x)}$ cannot both be Lebesgue integrable.

I've taken courses based on and read from Royden & Fitzpatrick if that helps with suggestions.

My attempts so far have focused on trying to find contradictions assuming $f$ is integrable: i.e. $\int_{\mathbb{R}} |f| < \infty$ and defining $S_0 := \{x \in \mathbb{R} | f(x) = 0 \}$. I'm thinking that something is happening with zeros and infinities that destroys the measurability of the alternative function.

Thanks in advance!

from Hot Weekly Questions - Mathematics Stack Exchange

So I've been thinking, can someone explain mathematically if I started with a square of paper dimensions 1x1, what the smallest single side dimension could be reached after one fold. My gut instinct is that after only one fold there's no way to make the paper less than half the height/width. When I say single side dimension, imagine drawing a bounding box around the folded paper and taking the smaller of the two side dimensions.

from Hot Weekly Questions - Mathematics Stack Exchange

$\textbf{Question:} \\ $ Find all functions $f : N \to N$ for which the expression $af(a) + bf(b) + 2ab$ is a perfect square for all $a,b \in \mathbb N$

$\textbf{My progress:}$

1.$ p \mid f(np) $ for all primes p and positive integer n.

2.$f(1)$ is a quadratic residue modulo infinitely many primes which (probably) implies $f(1)$ itself is a perfect square

3.all primes other than 2 can occur only once in the prime factor decomposition of $f(1)$

4.2 and 3 together gives $f(1)=2^k$. for even k. Then using $2*f(1)+2$ being a perfect square we get $f(1)=1$

5.And I also noticed $f(a)=a$ works.

Here N means only the positive whole numbers.

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Motivation

NaN as a value exists in IEEE floating-point numbers. Since every operation involving NaN has NaN as the outcome, IEEE floating-point numbers are not fields. I want to define a new algebraic structure so NaN could be encapsulated.

Definitions

A field $(F,+,\times)$ is NaNable if there exists a binary operation $\sim$ such that $(F \cup \{\text{NaN}\}, \sim, +)$ is a field, where $\text{NaN}$ is the identity element of $\sim$.

$(F \cup \{\text{NaN}\}, \sim, +,\times)$ is a NaNification of $(F,+,\times)$, and is a NaN-field.

Examples

Examples of NaNable finite fields include $\mathbb{Z}_2$, $\mathbb{Z}_3$, $\mathbb{Z}_7$, and $\mathbb{Z}_{31}$. The operation tables of NaNification of $\mathbb{Z}_3$ is:

$$ \begin{array}{c|cccc} \sim & \text{NaN} & 0 & 1 & 2 \\ \hline \text{NaN} & \text{NaN} & 0 & 1 & 2 \\ 0 & 0 & \text{NaN} & 2 & 1 \\ 1 & 1 & 2 & \text{NaN} & 0 \\ 2 & 2 & 1 & 0 & \text{NaN} \\ \end{array} $$

$$ \begin{array}{c|cccc} + & \text{NaN} & 0 & 1 & 2 \\ \hline \text{NaN} & \text{NaN} & \text{NaN} & \text{NaN} & \text{NaN} \\ 0 & \text{NaN} & 0 & 1 & 2 \\ 1 & \text{NaN} & 1 & 2 & 0 \\ 2 & \text{NaN} & 2 & 0 & 1 \\ \end{array} $$ $$ \begin{array}{c|ccc} \times & 0 & 1 & 2 \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 \\ 2 & 0 & 2 & 1 \\ \end{array} $$

Note that multiplication involving $\text{NaN}$ is not defined yet.

Questions

1. Is there a notable example of NaNable infinite field?

2. Is there a dedicated name for $\sim$? Not as a symbol, but as an operation?

3. It turns out that if every multiplication involving $\text{NaN}$ is defined as $\text{NaN}$, distributivity is satisfied in the NaN-field. Does this mean I should define it accordingly?

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I opened a stream that started an hour ago. Not wanting to miss anything, I started from the beginning and set it to 1.5x speed. How long will it take for me to catch up?

I know that it will take 40 minutes to watch the hour that I missed ($ \frac {60}{1.5}=40$) but during that time, the stream has generated another 40 minutes that I need to watch. This tells me I need to do some calculus, but it's been a decade since I took that course. Can someone help me come up with an equation?

from Hot Weekly Questions - Mathematics Stack Exchange

It's known that the asymptotes of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is given by $y=\pm\frac{b}{a}x$ if $a>b$.

I tried to find a proof of the fact that why the equations of these asymptotes are like that,however the only reference (Thomas calculus book) that I found explained that the two asymptotes are derived by letting $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$.

As usual the book does not give a reason for that , (Well it's not surprising to see that the author of these elementary books always try to fill their book with every subjects consisting in mathematical without any kind of logical reason.these days publishing an elementary mathematics book is easy-peasy and the responsibility is completely lost ,as Galois says: "Unfortunately what is little recognized is that the most worthwhile scientific books are those in which the author clearly indicates what he does not know; for an author most hurts his readers by concealing difficulties.")

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It would be highly appreciated if someone prove why the equation of the asymptotes have such form.

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Algebraic Identities - (a + b)^n and (a - b)^n Calculators - Example Problems

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With everything that has happened lately, I haven't had too much time to blog!

But, now that things are settling down a bit, I wanted to tell you about some Geometry distance learning activities I have been making.

These activities are great for giving  your students {and you!} a break from the videos and the worksheets.  Students think these activities are fun and you get data :)



1) Use Drag and Drop Activities on Google Slides

I have made several of these activities and I like them because they are self-checking.  You can insert them into learning management systems such as Schoology and students can use them right there.  Here is a really small video of one of my activities.




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Welcome to the twenty-fifth match in this year’s Big Math-Off. Take a look at the two interesting bits of maths below, and vote for your favourite.

You can still submit pitches, and anyone can enter: instructions are in the announcement post.

Here are today’s two pitches.

Nikki Rohlfing – Mathematical card trick

Nikki Rohlfing is a heavy metal roadie turned maths teacher. On twitter he is @heavymetalmaths.

Pupil: “Sir, how did you do that?”

Me: “Magic!”

Yes, that’s how helpful I can be when teaching maths….I probably just ‘multiplied by a clever number 1’, that seems to do the trick for a lot of secondary school maths!

But how about some actual maths/magic combo: here is a trick I learnt a loooong time ago, yet I keep coming back to for its mathematical magic.

This card trick fascinated me as a child, and I remember one week in particular during the summer holidays at my grandad’s house where I investigated a bit more: can I find a way to make it work with more/less cards? Or more iterations of the separating of stacks? etc. I was counting cards for days but got nothing conclusive.

Over the years, I kept thinking about it: part of me was bothered that I couldn’t work it out, but part of me also enjoyed the fact that it was still magical to me (not the actual trick, but the ‘magic’ of the maths behind it!). So here I am, over 20 years later, having another go at cracking the code, and although my maths is frustratingly not up to scratch, I hope to provide some insight at least, and maybe awaken some curiosity!

So to start with, let’s opt for an exhaustive method to confirm it works, that the method shown in the video does indeed always return the required card into the 3^rd place from the top.

I’ll start by labelling the cards 1-8, show the two stacks, and branch off the “yes”/”no” answers to show all possible outcomes.

With each “yes/no” decision, you are creating $2\times2\times2=8$ possible outcomes, which is good, as that relates to each of the 8 cards. What was not obvious to me before setting out the diagram was that if going “yes, yes, yes”, the cards are back in the exact order they started in – magic! And looking at the numbers in the stacks, then there’s quite a few patterns to see, like in the second to last, the numbers 1-4 and 5-8 are already grouped together again no matter what route.

But why does putting the stack with the card in on top result in it being always in third place of the final stack?

After not unlocking much myself in the limited time I have, I had a little internet search, and found a Matt Parker video on a very similar (and arguably better) trick. You may want to watch before reading on, as I’ll refer to his method.

The main difference is that Matt’s trick has more cards, and they get divided into 3 stacks, and the similarity is doing the iterative stacking process 3 times. Matt’s trick is based in ternary maths – is this due to doing the process 3 times or because he has 3 stacks? Given his numbering of 0,1,2 to help order the three stacks, I am sensing it’s the latter. This would lead to mine being in binary, which also links in with the $2^3$ cards/choices perhaps.

Matt’s trick is based upon calculating one less than the participants number in ternary form. Given the three stages, this requires 3 digits per number, which then shows why the “one less” is key, as for example, I would need to go from 0 to 7 rather than 1 to 8, so the 3^rd card in the pack would actually be number “2” when counting from 0. Another reason for using 0-7 is that 8 in binary would be four digits, which wouldn’t work. So looking at how the card always lands in 3^rd position, I should perhaps look at the number 2 in binary and try and apply Matt’s method:

\begin{array}{ccc}\\2^2&2^1&2^0\\0&1&0\end{array}

And again, following Matt’s design, I would place the pack with the participants card first on top if I see a 0, and on the bottom if a see a 1, but reading the binary number from right to left. If you have a pack of cards (or by all means try mentally), you might want to think about where the card ends up before clicking on the reveal below.

Spoiler: [+ show][- hide]

Well, the sharp-minded amongst you will have known this wouldn’t work, as to get it into 3^rd place, I place the stack with the card in always on top. So using the binary method, this would be 000. Hmm, so perhaps I’m computing it wrongly. But, out of interest, what about other numbers? Well, for example, trying “3” (011), gets the card into second place each time, and as you process the binary numbers 0 to 7 you do indeed get all the different positions.

The participants card will move to:

Decimal number 0, binary: 000 $\rightarrow$ 3^rd place

Decimal number 1, binary: 001 $\rightarrow$ 4^th place

Decimal number 2, binary: 010 $\rightarrow$ 1^st place

Decimal number 3, binary: 011 $\rightarrow$ 2^nd place

Decimal number 4, binary: 100 $\rightarrow$ 7^th place

Decimal number 5, binary: 101 $\rightarrow$ 8^th place

Decimal number 6, binary: 110 $\rightarrow$ 5^th place

Decimal number 7, binary: 111 $\rightarrow$ 6^th place

Yet more number patterns, with an odd/even twist. What is also noticeable is how these 0s and 1s link back to the “yes/no” options in my original diagram (although in reverse, and slightly different thinking, but the same underlying principle!).

So looking at the aforementioned 011, means the stack with the participants card in goes (reading right to left) “bottom, bottom, top”, which translates to “no,no,yes” (left to right) in my original notation. It takes the participants card to place 2 in the final stack, which happens to be starting card number 3. In fact, from my diagram, you can see how each binary number 000 to 111 takes card 3, and places it into one of the 8 different places in the pack. This feels like my original trick backwards somehow!

So what has this actually all achieved? Well, two different approaches, working in different directions, using binary and decimal, as well as (perhaps not sensibly) labelling the cards 1-8 when dealing with numbers 0-7….. it has definitely achieved my brain getting muddled a few times!

But one clear break-through is that I now no longer have to rely on revealing the 3^rd card, and can in fact have the participants card pop up anywhere I like! I like this as a possible way to get their card on top, or bottom of the stack, which are key positions for any card magic and could allow an expansion to the routine. The changing of position also allows for greater repetition of the trick without people realising that you just choose the 3^rd card each time.

However, going back to my childhood musings of can this work with more/less cards or iterations, then perhaps I’ve made less progress. Matt’s version shows that at least one more way is possible. But how many different ways could there be? The fact that when you follow my diagram with “yes, yes, yes”, the cards are back in their starting order might be helpful. However, for 3 stacks, like with Matt, I’m not (yet!) sure which combination is required to do the same, given that’s it’s “0/1/2” rather than just “yes/no”. If nothing else, I hope you have a go at doing the card trick with someone; and by all means get in touch if you discover anything I missed!

Pat Ashforth – Some thoughts about becoming a mathematician

Pat Ashforth explores maths through knitting and crochet at woollythoughts.com.

Reading all the pitches in the The Big Lock-Down Math-Off made me start thinking about what influences turn a person into a mathematician or, more specifically, into a particular kind of mathematician.  We all know that there is no dividing line between mathematicians and non-mathematicians. Everyone has some kind of mathematical thinking whether they regard it as maths or not.

I don’t remember much about my maths education in my early life. My father, who was in no way a gambler, taught me to play pontoon when I was old enough to hold the cards and this could have contributed to being at ease with numbers. At the age of three I was one of those awful children who did not ‘rest’ in the afternoon (or sleep much at night either).  I went to a nursery where the other children did rest so, each afternoon, I was sent to the end of the road where I could see the town clock and told to come back at a particular time. I’m not sure I could tell the time properly but I was able to come back and demonstrate, with outstretched arms, the time on the clock. It wouldn’t happen these days.

In Junior School I loved sums, which were more complicated than they are today because we worked in pounds, shillings and pence, and pounds and ounces, and feet and inches. Using several different number bases was just what we did.  There were other lovely things like hundredweights and tons, and pints and gallons.

At the age of eleven I went to the very new local Grammar School. Maths was a revelation and a very different animal from what we have today. It was split into three separate subjects, each with a different teacher – Arithmetic, Algebra and Geometry. I doubt if I had even heard these words before. I remember very little about arithmetic and algebra but geometry had a massive impact. Mr Parkes was terrifying. He was ex-army with the loudest voice imaginable. He always wore a gown. Most Grammar School teachers did in those days. He marched down the corridors and it billowed around him. He had metal tips on the heels of his shoes. That wasn’t unusual either but he seemed to be louder than anyone else. You always knew when he was coming and could get out of the way.

The very first lesson has stuck with me forever. We all had the obligatory geometry set but this didn’t fulfil all of his requirements. Pencils! Two 2H pencils were essential. It was a punishable offence not to have them. On reflection, it makes perfectly good sense. It is not possible to draw a thick line with a 2H pencil and he frequently reminded us that a line, theoretically, has no thickness. The second pencil had to be cut in half. One half was to be used in compasses. We all know how frustrating it is to draw a circle with a bump in it where the pencil hit your fingers. It was all terrifying but it was new and it was fun. To this day every drawing I do has incredibly faint lines and I still feel guilty if I don’t use a 2H pencil to do it.

I loved Maths but I also loved crafts. I wasn’t allowed to do any practical subjects. I was in the stream that had to do Latin instead. I resented that at the time. It was assumed that only boys would do Maths and that they would all be scientists. I was not a scientist. At A Level I wanted to do Maths and English plus something else. This didn’t fit with the school’s Arts/Science paths. To their credit, they rejigged the sixth form teaching to make it possible. Once it became an option I wasn’t the only girl doing Maths. I’m sure my future would have been very different if I had been forced down the science route.

Over time, more by chance than intention I found my own weird way of combining maths and art. I am now very thankful that I didn’t do anything arty in school. I was not directed by other people’s ideas.  I was definitely a geometer and I lay that at the feet of Mr Parkes. You can never know the long term effects of the little things you say and do with students.

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So, which bit of maths made you say “Aha!” the loudest? Vote:

Note: There is a poll embedded within this post, please visit the site to participate in this post's poll.

The poll closes at 9am BST on Monday the 1st, when the next match starts.

If you’ve been inspired to share your own bit of maths, look at the announcement post for how to send it in. The Big Lockdown Math-Off will keep running until we run out of pitches or we’re allowed outside again, whichever comes first.

from The Aperiodical https://ift.tt/2ZQojXS

In Steen and Seebach's "Counterexamples in Topology", we see the definition of the Long Line (counterexample 45).

"The long line $L$ is constructed from the ordinal space $[0, \Omega)$ (where $\Omega$ is the least uncountable ordinal) by placing between each ordinal $\alpha$ and its successor $\alpha + 1$ a copy of the unit interval $I = (0,1)$. $L$ is then linearly ordered, and we give it the order topology."

Having given this a bit of thought, I need clarifying the following.

Are the ordinals $0, 1, 2, \ldots, \alpha, \alpha + 1, \ldots$ part of the space, or is $L$ just $\Omega$ instances of $(0,1)$ concatenated? If the latter, then it appears there may be a homeomorphism between $L$ and $[0,\Omega) \times (0,1)$ under the lexicographic ordering. If the former, then it is very much less simple.

So is $L$ like: $0, (0,1), 1, (0,1), 2, (0,1), \ldots, (0,1), \alpha, (0,1), \alpha + 1, (0,1), \ldots, (0,1), \Omega-1, (0,1)$

or is it like:

$(0,1), (0,1), (0,1), \ldots, (0,1), (0,1), (0,1), \ldots, (0,1), (0,1)$

with $\Omega$ instances of $(0,1)$?

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Find all sequences that has $\sum_{i=1}^\infty a_i$ converges, where $a_i = \sum_{k=i+1}^\infty a_k^2$.

My intuition is that the only sequence of this form is the zero sequence.

Here's what I have so far: $a_n - a_{n+1} = a_{n+1}^2 \implies a_{n+1} = \sqrt{a_n + \frac{1}{4}} - \frac{1}{2}$, but it doesn't seem to lead me anywhere.

Another line of thought is that if $a_i = 0$ for some $i$, it means that $\sum_{k=i+1}^\infty a_k^2=0$, which means that $a_k = 0$ for $k > i$. This will also mean $a_{i-1} = 0, a_{i-2} = 0...$, making the whole sequence the zero sequence.

It means that $a_i >0 $ for all $i$, yet $\lim a_i = 0$.

The last line I've tried is $a_1 = a_2^2 + a_3^2 + a_4^2 + ..., a_2 = a_3^2 + a_4^2 + ....$, so $\sum_{i=1}^\infty a_i = a_2^2 + a_3^2 + a_4^2 + ... +a_3^2 + a_4^2 + .... = a_2^2 + 2a_3^2 + 3a_4^2 = \sum_{i=2}^\infty (i-1)a_i^2$, which implies a stronger condition of having $ia_i^2 \to 0$. I'm hoping to get a contradiction but it doesn't seem to work.

Python seems to suggest that $(a_n) \approx \frac{1}{n}$ for large $n$.

Any hints?

from Hot Weekly Questions - Mathematics Stack Exchange

Let $p,\mu_c \in [0,1]$. Let $(X_i)_{i \in \mathbb{N}}$ be a sequence of i.i.d random variables, each following a Bernoulli distribution $\mathcal{B}(p)$. For $n \in \mathbb{N}$, let $ \mu_n = \frac{1}{n+1} \sum_{i=0}^n X_i $ and $A_n$ the event: "$\mu_n \leq \mu_c$". I am looking for a way to compute $$ \mathbb{P} \left(\bigcap_{n \in \mathbb{N}} A_n \right), $$ that is, the probability that the mean of the sequence never exceeds the critical value $\mu_c$.

You can assume $\mu_c > p$, since this probability is $0$ when $\mu_c \leq p$.

I first wanted to compute it using a martingale, but the optional stopping theorem does not seem to help. I don't know if computing it "by hand" (combinatorially) is possible, but it would be quite unsatisfying anyway. Eventually, it may be somehow linked to this post, but I don't understand Brownian motion well enough to use it correctly.

Thank you for any advice!

Edit: When $p = \frac{1}{2}$ and $\mu_c = \frac{3}{4}$, simulations give $$\mathbb{P} \left(\bigcap_{n \leq 1000} A_n \right) \approx 0.456.$$ I think that this is a very good approximation of $\mathbb{P} \left(\bigcap_{n \in \mathbb{N}} A_n \right)$ (going up to 2000 does not affect the result). This fact may be used to check any suggested answer. I find it surprisingly high, since $\mathbb{P}(A_0)$ alone is equal to $\frac{1}{2}$.

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I found one rather interesting but intractable topology problem.

Prove that a torus triangulation cannot have degrees of vertices $5, 7, 6, 6, 6, 6, \ldots$

Despite various attempts to contract the graph or reduce it to irreducible triangulation, nothing came of it.

What can you recommend?

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Congratulations to Michelle Grant for passing her PhD viva today (Friday 29th May)! Michelle’s thesis is entitled “Data assimilation, tipping events and the importance of underlying dynamics“. The External Examiner was Dr Amos Lawless (Reading) and the Internal Examiner was Naratip Santitissadeekorn and the Chair of the viva was Philip Aston. Michelle was supervised by Anne Skeldon and Ian Roulstone.

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I am trying to find the solution to the equation- $$\Gamma(z)=i$$ I have tried doing it the following way- LHS is- $$\displaystyle \int_{0}^{\infty}t^ze^{-t}\ dt$$ Taking $z=a+ib$, we get- $$\displaystyle \int_{0}^{\infty}t^{a+ib}e^{-t}\ dt$$ or $$\displaystyle \int_{0}^{\infty}t^{a}t^{ib}e^{-t}\ dt$$ Using Euler's formula- $e^{i\theta}=\text{cos}\ \theta +\ i\ \text{sin}\ \theta$, we have- $$\displaystyle \int_{0}^{\infty}t^{a}(\text{cos}\ (b\ \text{ln}t) +\ i\ \text{sin}\ (b\ \text{ln}t)) e^{-t}\ dt$$ After this point, I am not being able to solve this integral. I have tried graphing it on desmos, but it doesn't seem like this question has any solution. How can I approach further in this?

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Source: https://xkcd.com/1856/

from Mean Green Math

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8th Grade Math Worksheet Common Core

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In a model category $C$ and $X \in C$ we can define $\Sigma X$ as the homotopy pushout of $$* \leftarrow X \rightarrow * $$ letting $*$ denote the terminal object.

The way I understand it is that if $C = \text{Ch}_+(R-\text{Mod})$ with the projective model structure, if $D$ is a chain complex, $\Sigma D = D[-1]$ where $D[-1]_n = D_{n-1}$. I believe this is true because of the case when $D = C_*(X;R)$, where $X$ is some topological space.

Consider the map

$$C_*(X;R) \rightarrow C_*(C X;R)$$

where $CX$ is the cone over $X$ and note that it is a cofibrant replacement for $C_*(X) \rightarrow *$ in this case and we can thus deduce from some basic algebraic topology that $\Sigma C_*(X;R) \approx C_*(\Sigma X;R)$ which is weakly equivalent as a chain complex to $C_*(X;R)[-1]$.

First of all, is my intuition for this correct? Is $\Sigma D = D[-1]$ for general $D$? I have no clue how to generalize this argument to general chain complexes $D$ so any help would be greatly appreciated.

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How to Find the Unit Digit in the Product Expression - Concept - Examples with step by step solution

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How to Find the Unit Digit in the Product - Concept - Examples with step by step explanation

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Let $\{e_1,e_2,e_4\}$ be an orthonormal basis for a complex unitary space $V$. Let's define the vectors: $f_j=e_j-\frac14\sum\limits_{i=1}^4e_i, j\in\{1,2,3,4\}$. Let $A\in\mathcal L(V), Ax:=\sum\limits_{j=1}^4\langle x,f_j\rangle f_j$.

Show $A$ is Hermitian and find the orthonormal basis for $V$ in which $A$ is diagonalizable.

Note: typo corrected.

---

My attempt:

Let's compute $f_1,f_2,f_3,f_4$ first.

$\begin{aligned}f_j=e_j-\frac14\sum\limits_{i=1}^4, e_i\implies&f_1=\frac34e_1-\frac14(e_2+e_3+e_4)\\&f_2=\frac34e_2-\frac14(e_1+e_3+e_4)\\&f_3=\frac34e_3-\frac14(e_1+e_2+e_4)\\&f_4=\frac34e_4-\frac14(e_1+e_2+e_3)\end{aligned}$

$\begin{aligned}Ae_i&=\sum\limits_{j=1}^4\langle e_i,f_j\rangle f_j\implies Ae_1=\left\langle e_1,\frac34e_1-\frac14(e_2+e_3+e_4)\right\rangle f_1+\left\langle e_1,\frac34e_2-\frac14(e_1+e_3+e_4)\right\rangle f_2+\left\langle e_1,\frac34e_3-\frac14(e_1+e_2+e_4)\right\rangle f_3+\left\langle e_1,\frac34e_4-\frac14(e_1+e_2+e_3)\right\rangle f_4=\frac34f_1-\frac14(f_2+f_3+f_4)\end{aligned}$

$\ Ae_2=\frac34f_2-\frac14(f_1+f_3+f_4)\\Ae_3=\frac34f_3-\frac14(f_1+f_2+f_4)\\Ae_4=\frac34f_4-\frac14(f_1+f_2+f_3)$

Then, $$[A]_e^f=\begin{bmatrix}\frac34&-\frac14&-\frac14&-\frac14\\-\frac14&\frac34&-\frac14&-\frac14\\-\frac14&-\frac14&\frac34&-\frac14\\-\frac14&-\frac14&-\frac14&\frac34\end{bmatrix}$$

About the matrix representation of $A\in\mathcal L(V)$:

$A\in M_n(\Bbb R)\ \&\ A=A^\tau\ \implies A=A^*\iff A\ \text{is normal}\implies A\text{ is diagonalizable in some orthonormal basis}$ $\{a_1,a_2,a_3,a_4\}$

Let's find the eigenvalues and the corresponding eigenspaces. Using the formula derived here. According to the notation I used in the thread, $a_j=\frac34-\lambda\ \forall j\in\{1,2,3,4\}$ and $x=-\frac14$.

$$\det(A-\lambda I)=\begin{vmatrix}\frac34-\lambda&-\frac14&-\frac14&-\frac14\\-\frac14&\frac34-\lambda&-\frac14&-\frac14\\-\frac14&-\frac14&\frac34-\lambda&-\frac14\\-\frac14&-\frac14&-\frac14&\frac34-\lambda\end{vmatrix}=\left(\frac34-\lambda+\frac14\right)^4\left(1-\frac14\cdot 4\cdot\frac1{\frac34-\lambda+\frac14}\right)=-\lambda(1-\lambda)^3=\lambda(\lambda-1)(1-\lambda)^2\implies\sigma(A)=\{0,1\}$$ Now, $E_A(0)=\ker(A)$: $$\begin{bmatrix}\frac34&-\frac14&-\frac14&-\frac14\\-\frac14&\frac34&-\frac14&-\frac14\\-\frac14&-\frac14&\frac34&-\frac14\\-\frac14&-\frac14&-\frac14&\frac34\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3\\1&1&-3&1\\1&-3&1&1\\-3&1&1&1\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3\\0&0&-4&4\\0&-4&0&4\\0&4&4&-8\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3\\0&0&-1&1\\0&-1&0&1\\0&0&0&0\end{bmatrix}\sim\begin{bmatrix}1&0&0&-1\\0&-1&0&1\\0&0&-1&1\\0&0&0&0\end{bmatrix}$$ $$\implies E_A(0)=\operatorname{span}\left\{\underbrace{\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}}_{v_1},\underbrace{\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}}_{v_2},\underbrace{\begin{bmatrix}0\\0\\-1\\1\end{bmatrix}}_{v_3}\right\}$$ $E_A(1)=\ker(A-I)$: $$\begin{bmatrix}-\frac14&-\frac14&-\frac14&-\frac14\\-\frac14&-\frac14&-\frac14&-\frac14\\-\frac14&-\frac14&-\frac14&-\frac14\\-\frac14&-\frac14&-\frac14&-\frac14\end{bmatrix}\sim\begin{bmatrix}1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$ $$\implies E_A(1)=\operatorname{span}\left\{\underbrace{\begin{bmatrix}1\\1\\1\\1\end{bmatrix}}_{v_4}\right\}$$

Let's apply Gramm-Schmidt to the obtained basis for $V$: $$a_1=\frac1{\|v_1\|}v_1=\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}=b_1$$ $$\begin{aligned}b_2&=v_2-\langle v_2,a_1\rangle a_1\\&=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}0\\-1\\0\\1\end{bmatrix},\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\\&=\frac32\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\end{aligned}$$ $$a_2=\frac1{\|b_2\|}b_2=\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$$ $$\begin{aligned}b_3&=v_3-\langle v_3,a_1\rangle a_1-\langle v_3,a_2\rangle a_2\\&=\begin{bmatrix}0\\0\\-1\\1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}0\\0\\-1\\1\end{bmatrix},\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}0\\0\\-1\\1\end{bmatrix},\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\\&=\begin{bmatrix}0+\frac12\\0+\frac12\\-1\\1-\frac12-\frac12\end{bmatrix}=\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}\end{aligned}$$ $$a_3=\frac1{\|b_3\|}b_3=\frac{\sqrt{6}}3\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}$$ $$\begin{aligned}b_4&=v_4-\langle v_4,a_1\rangle a_1-\langle v_4,a_2\rangle a_2-\langle v_4,a_3\rangle a_3\\&=\begin{bmatrix}1\\1\\1\\1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}-\frac{\sqrt{6}}3\left\langle\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}\right\rangle\frac{\sqrt{6}}3\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}\\&=\begin{bmatrix}1\\1\\1\\1\end{bmatrix}\end{aligned}$$ $$a_4=\frac1{\|b_4\|}b_4=\frac12\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$$

Therefore, a Hermitian operator $A$ is diagonalizable in the orthonormal basis: $$\{a_1,a_2,a_3,a_4\}=\left\{\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix},\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix},\frac{\sqrt{6}}3\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix},\frac12\begin{bmatrix}1\\1\\1\\1\end{bmatrix}\right\}$$

May I ask if this is correct? If so, how could I improve my approach?

Thank you in advance!

from Hot Weekly Questions - Mathematics Stack Exchange

This question was inspired from this problem.

For this problem, the radius of a square will be the distance from its centre to any of its four vertices. A lattice point is a point $(a, b)$ in the plane where $a$ and $b$ are both integers.

Find a function $f$ on the positive real numbers so that $f(r)$ is the largest possible number of lattice points inside any square of radius $r$ centred at the origin. A point on the perimeter is considered to be inside the square.

I misread the problem linked above and was trying to solve this question. I tried writing down the values of $f(r)$ depending on the value of $r$. For example, when $1 \le r <\sqrt 2, f(r)=5$ and when $\sqrt 2 \le r < 2, f(r)=9$. Next, I tried to focus only on the lower bounds, where the square's corners fall on boundary points. I could see that $r$ was of the form $\sqrt {a^2+b^2}$ so maybe I could try a function involving a lower bound. When the square's vertices fall on lattice points, we can use Pick's theorem to calculate the area. Here, $i$ means the number of interior points and $b$ means the number of boundary points.

According to Pick's theorem,
$A=\frac 12b + i-1$
However, as it is a square with diagonal $2r$,
$A=2r^2$
$2r^2=\frac 12b + i-1$

We know that $f(r)=b+i$ when $r$ is such that the square has lattice points as vertices. So:

$f(r)=4r^2+1-i$ or $f(r)=4r^2+b+1$

I could not progress any further. I thought I could form a relationship between the current square and the next largest square because $f\left(r_{\text{Current Square}}\right)=4r^2+1-f\left(r_{\text{Next Largest Square}}\right)$ but I am finding it difficult to define how we can get from a square to the next largest/smallest one. How should I progress? Is it even possible to find $f(r)$?

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Rota writes in his TEN LESSONS I WISH I HAD LEARNED BEFORE I STARTED TEACHING DIFFERENTIAL EQUATIONS:

"It states that every differential polynomial in the two solutions of a second order linear differential equation which is independent of the choice of a basis of solutions equals a polynomial in the Wronskian and in the coefficients of the differential equation (this is the differential equations analogue of the fundamental theorem on symmetric functions, but keep it quiet)."

And:

"Worse, no one realizes that changes of variables are not just a trick; they are a coherent theory (it is the differential analogue of classical invariant theory, but let it pass)."

And:

"For second order linear differential equations, formulas for changes of dependent and independent variables are known, but such formulas are not to be found in any book written in this century, even though they are of the utmost usefulness. Liouville discovered a differential polynomial in the coefficients of a second order linear differential equation which he called the invariant. He proved that two linear second order differential equations can be transformed into each other by changes of variables if and only if they have the same invariant. This theorem is not to be found in any text. It was stated as an exercise in the first edition of my book, but my coauthor insisted that it be omitted from later editions."

Where can I learn more about this?

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Check out this Problem of the Week.
Be sure to let us know how you solved it in the comments below or on social media!

Solution below.



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Developing Our MathMINDs is a series of conversations and resources about math that is intended to be a journey of growth with families over several weeks. Each week, MIND's Lead Mathematician and Product Director Brandon Smith and Content Development Manager Nina Wu will be talking about the adjustments families are making to learning at home, and the opportunities this situation provides for changing our relationship with math.

In this week's Developing Our MathMINDs series, we have mutliple ways for families to celebrate math in the world around us.

Math Walk

First up, Nina takes us on a math walk and provides some great examples of mathematical experiences we can have along the way:

 

Make Your Own Mathematical Newspaper

There are so many great stories to tell, and questions to ask, that involve the math in your everyday life. In fact, you could create your own newspaper, and report on what's happening around you, through the lens of math.

In this downloadable example of the "Family Times," intrepid reporter Nina Wu provides the scoop on the news everyone wants to know about. How much screen time are we having each day? How many different kinds of wildlife live in our neighborhood? How many secret ingredients are in grandma's banana bread recipe? 

What are some mathematical stories you could report on in your own family newspaper? Download Nina's edition for some great ideas right here!

Math and Magic

Our friend Jay Flores from Rockwell Automation is back this week as well. Jay's been doing a series called "It's Not Magic, It's Science" on his social channels, and this past week, he created an installment about math, featuring a cameo from our own JiJi!

In part one, Jay shows us a trick that seems like magic:

Ep 4: It’s Not #Magic It’s #Science! - JiJi Magic with @STMath!

Can you guess how we did it? Check back in tomorrow for a video revealing the “magic.” #stmath #jijimath #MATHMINDs pic.twitter.com/zKoLcsmICE

— Jay Flores (@JayFlores2032) May 21, 2020

But in part two, he pulls the curtain back to explain the math behind it, and challenges families with a problem that builds on the original :

Ep 4.2: It’s Not #Magic It’s #Science! - JiJi Magic with @STMath!

Can you figure out the answer when we roll 5 🎲?

Try with your family #athome! #mathchallenge #jijimath pic.twitter.com/GVzQs2dQZM

— Jay Flores (@JayFlores2032) May 22, 2020

Free Math Storybooks

Last week, we talked about the relationship between stories and math that goes well beyond word problems.  We shared two of new storybook prototypes so families could experience the story of mathematics in a new way. This week, we're back with two more!

Each story uses real life or historical situations to explore a mathematical concept in a non-typical way. Feel free to pause the story to act out or complete the activity described in the story.

Disappearing Moon (Kindergarten-1st Grade)

Cubey Cake (2nd Grade+)

Technical notes: On mobile devices, you may need to turn to landscape to see two page view in full screen. On mobile devices, you may need to reload the story to view it properly.

When you finish a story, please consider taking the survey linked at the end. One of our core values at MIND is continuous improvement; so thank you for helping us make these stories even better for future families. Who knows, your feedback could be incorporated in a new iteration of the digital stories or even a full storybook board game like MathMINDs Games: South of the Sahara.

Resources

• Math Cartoons in the Classroom
• Re-imagining Storytelling to Connect Math, Games and History
• Families Can Bake and Learn Together With This Pi Day Recipe
• Developing Our MathMINDs—Week 6: Experience Math Through Story
• Developing Our MathMINDs—Week 5: Math Activities to Get You Moving
• Developing Our MathMINDs—Week 4: How to be a Rockstar Facilitator

• Developing Our MathMINDs—Week 3: Informative Feedback

• Developing Our MathMINDs—Week 2: A Glossary for New Math Teachers

• Developing Our MathMINDs—Week 1: Adjusting to Learning at Home

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