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timestamp[ns]date 2026-01-26 11:29:09
2026-01-26 11:31:51
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|---|---|---|---|---|---|---|---|---|---|---|
733,754
|
Visually stunning math concepts which are easy to explain
|
Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time. Do you know of any other concepts like these?
| 1,669
|
soft-question, recreational-mathematics, education, big-list, visualization
|
https://math.stackexchange.com/questions/733754/visually-stunning-math-concepts-which-are-easy-to-explain
|
I think if you look at this animation and think about it long enough, you'll understand: Why circles and right-angle triangles and angles are all related. Why sine is "opposite over hypotenuse" and so on. Why cosine is simply sine but offset by $\frac{\pi}{2}$ radians.
| 1,171
| false
|
Q: Visually stunning math concepts which are easy to explain
Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time. Do you know of any other concepts like these?
A: I think if you look at this animation and think about it long enough, you'll understand: Why circles and right-angle triangles and angles are all related. Why sine is "opposite over hypotenuse" and so on. Why cosine is simply sine but offset by $\frac{\pi}{2}$ radians.
| 2026-01-26T11:29:09.672000
|
21,199
|
Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?
|
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $\frac{\textrm{d}y}{\textrm{d}x}$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $\textrm{d}y = f'(x)\textrm{d}x$ we are able to plug in values for $\textrm{d}x$ and calculate a $\textrm{d}y$ (differential). Then if we rearrange we get $\frac{\textrm{d}y}{\textrm{d}x}$ which could be seen as a ratio. I wonder if the author say this because $\mbox{d}x$ is an independent variable, and $\textrm{d}y$ is a dependent variable, for $\frac{\textrm{d}y}{\textrm{d}x}$ to be a ratio both variables need to be independent.. maybe?
| 1,332
|
calculus, analysis, math-history, nonstandard-analysis
|
https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio
|
Historically, when Leibniz conceived of the notation, $\frac{dy}{dx}$ was supposed to be a quotient: it was the quotient of the "infinitesimal change in $y$ produced by the change in $x$" divided by the "infinitesimal change in $x$". However, the formulation of calculus with infinitesimals in the usual setting of the real numbers leads to a lot of problems. For one thing, infinitesimals can't exist in the usual setting of real numbers! Because the real numbers satisfy an important property, called the Archimedean Property: given any positive real number $\epsilon\gt 0$, no matter how small, and given any positive real number $M\gt 0$, no matter how big, there exists a natural number $n$ such that $n\epsilon\gt M$. But an "infinitesimal" $\xi$ is supposed to be so small that no matter how many times you add it to itself, it never gets to $1$, contradicting the Archimedean Property. Other problems: Leibniz defined the tangent to the graph of $y=f(x)$ at $x=a$ by saying "Take the point $(a,f(a))$; then add an infinitesimal amount to $a$, $a+dx$, and take the point $(a+dx,f(a+dx))$, and draw the line through those two points." But if they are two different points on the graph, then it's not a tangent, and if it's just one point, then you can't define the line because you just have one point. That's just two of the problems with infinitesimals. (See below where it says "However...", though.) So Calculus was essentially rewritten from the ground up in the following 200 years to avoid these problems, and you are seeing the results of that rewriting (that's where limits came from, for instance). Because of that rewriting, the derivative is no longer a quotient, now it's a limit: $$\lim_{h\to0 }\frac{f(x+h)-f(x)}{h}.$$ And because we cannot express this limit-of-a-quotient as a-quotient-of-the-limits (both numerator and denominator go to zero), then the derivative is not a quotient. However, Leibniz's notation is very suggestive and very useful; even though derivatives are not really quotients, in many ways they behave as if they were quotients. So we have the Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du}\;\frac{du}{dx}$$ which looks very natural if you think of the derivatives as "fractions". You have the Inverse Function theorem, which tells you that $$\frac{dx}{dy} = \frac{1}{\quad\frac{dy}{dx}\quad},$$ which is again almost "obvious" if you think of the derivatives as fractions. So, because the notation is so nice and so suggestive, we keep the notation even though the notation no longer represents an actual quotient, it now represents a single limit. In fact, Leibniz's notation is so good, so superior to the prime notation and to Newton's notation, that England fell behind all of Europe for centuries in mathematics and science because, due to the fight between Newton's and Leibniz's camp over who had invented Calculus and who stole it from whom (consensus is that they each discovered it independently), England's scientific establishment decided to ignore what was being done in Europe with Leibniz notation and stuck to Newton's... and got stuck in the mud in large part because of it. (Differentials are part of this same issue: originally, $dy$ and $dx$ really did mean the same thing as those symbols do in $\frac{dy}{dx}$, but that leads to all sorts of logical problems, so they no longer mean the same thing, even though they behave as if they did.) So, even though we write $\frac{dy}{dx}$ as if it were a fraction, and many computations look like we are working with it like a fraction, it isn't really a fraction (it just plays one on television). However... There is a way of getting around the logical difficulties with infinitesimals; this is called nonstandard analysis. It's pretty difficult to explain how one sets it up, but you can think of it as creating two classes of real numbers: the ones you are familiar with, that satisfy things like the Archimedean Property, the Supremum Property, and so on, and then you add another, separate class of real numbers that includes infinitesimals and a bunch of other things. If you do that, then you can, if you are careful, define derivatives exactly like Leibniz, in terms of infinitesimals and actual quotients; if you do that, then all the rules of Calculus that make use of $\frac{dy}{dx}$ as if it were a fraction are justified because, in that setting, it is a fraction. Still, one has to be careful because you have to keep infinitesimals and regular real numbers separate and not let them get confused, or you can run into some serious problems.
| 1,605
| true
|
Q: Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $\frac{\textrm{d}y}{\textrm{d}x}$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $\textrm{d}y = f'(x)\textrm{d}x$ we are able to plug in values for $\textrm{d}x$ and calculate a $\textrm{d}y$ (differential). Then if we rearrange we get $\frac{\textrm{d}y}{\textrm{d}x}$ which could be seen as a ratio. I wonder if the author say this because $\mbox{d}x$ is an independent variable, and $\textrm{d}y$ is a dependent variable, for $\frac{\textrm{d}y}{\textrm{d}x}$ to be a ratio both variables need to be independent.. maybe?
A: Historically, when Leibniz conceived of the notation, $\frac{dy}{dx}$ was supposed to be a quotient: it was the quotient of the "infinitesimal change in $y$ produced by the change in $x$" divided by the "infinitesimal change in $x$". However, the formulation of calculus with infinitesimals in the usual setting of the real numbers leads to a lot of problems. For one thing, infinitesimals can't exist in the usual setting of real numbers! Because the real numbers satisfy an important property, called the Archimedean Property: given any positive real number $\epsilon\gt 0$, no matter how small, and given any positive real number $M\gt 0$, no matter how big, there exists a natural number $n$ such that $n\epsilon\gt M$. But an "infinitesimal" $\xi$ is supposed to be so small that no matter how many times you add it to itself, it never gets to $1$, contradicting the Archimedean Property. Other problems: Leibniz defined the tangent to the graph of $y=f(x)$ at $x=a$ by saying "Take the point $(a,f(a))$; then add an infinitesimal amount to $a$, $a+dx$, and take the point $(a+dx,f(a+dx))$, and draw the line through those two points." But if they are two different points on the graph, then it's not a tangent, and if it's just one point, then you can't define the line because you just have one point. That's just two of the problems with infinitesimals. (See below where it says "However...", though.) So Calculus was essentially rewritten from the ground up in the following 200 years to avoid these problems, and you are seeing the results of that rewriting (that's where limits came from, for instance). Because of that rewriting, the derivative is no longer a quotient, now it's a limit: $$\lim_{h\to0 }\frac{f(x+h)-f(x)}{h}.$$ And because we cannot express this limit-of-a-quotient as a-quotient-of-the-limits (both numerator and denominator go to zero), then the derivative is not a quotient. However, Leibniz's notation is very suggestive and very useful; even though derivatives are not really quotients, in many ways they behave as if they were quotients. So we have the Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du}\;\frac{du}{dx}$$ which looks very natural if you think of the derivatives as "fractions". You have the Inverse Function theorem, which tells you that $$\frac{dx}{dy} = \frac{1}{\quad\frac{dy}{dx}\quad},$$ which is again almost "obvious" if you think of the derivatives as fractions. So, because the notation is so nice and so suggestive, we keep the notation even though the notation no longer represents an actual quotient, it now represents a single limit. In fact, Leibniz's notation is so good, so superior to the prime notation and to Newton's notation, that England fell behind all of Europe for centuries in mathematics and science because, due to the fight between Newton's and Leibniz's camp over who had invented Calculus and who stole it from whom (consensus is that they each discovered it independently), England's scientific establishment decided to ignore what was being done in Europe with Leibniz notation and stuck to Newton's... and got stuck in the mud in large part because of it. (Differentials are part of this same issue: originally, $dy$ and $dx$ really did mean the same thing as those symbols do in $\frac{dy}{dx}$, but that leads to all sorts of logical problems, so they no longer mean the same thing, even though they behave as if they did.) So, even though we write $\frac{dy}{dx}$ as if it were a fraction, and many computations look like we are working with it like a fraction, it isn't really a fraction (it just plays one on television). However... There is a way of getting around the logical difficulties with infinitesimals; this is called nonstandard analysis. It's pretty difficult to explain how one sets it up, but you can think of it as creating two classes of real numbers: the ones you are familiar with, that satisfy things like the Archimedean Property, the Supremum Property, and so on, and then you add another, separate class of real numbers that includes infinitesimals and a bunch of other things. If you do that, then you can, if you are careful, define derivatives exactly like Leibniz, in terms of infinitesimals and actual quotients; if you do that, then all the rules of Calculus that make use of $\frac{dy}{dx}$ as if it were a fraction are justified because, in that setting, it is a fraction. Still, one has to be careful because you have to keep infinitesimals and regular real numbers separate and not let them get confused, or you can run into some serious problems.
| 2026-01-26T11:29:10.897000
|
379,927
|
If it took 10 minutes to saw a board into 2 pieces, how long will it take to saw another into 3 pieces?
|
So this is supposed to be really simple, and it's taken from the following picture: Text-only: It took Marie $10$ minutes to saw a board into $2$ pieces. If she works just as fast, how long will it take for her to saw another board into $3$ pieces? I don't understand what's wrong with this question. I think the student answered the question wrong, yet my friend insists the student got the question right. I feel like I'm missing something critical here. What am I getting wrong here?
| 1,102
|
arithmetic, word-problem
|
https://math.stackexchange.com/questions/379927/if-it-took-10-minutes-to-saw-a-board-into-2-pieces-how-long-will-it-take-to-saw
|
Haha! The student probably has a more reasonable interpretation of the question. Of course, cutting one thing into two pieces requires only one cut! Cutting something into three pieces requires two cuts! ------------------------------- 0 cuts/1 piece/0 minutes ---------------|--------------- 1 cut/2 pieces/10 minutes ---------|-----------|--------- 2 cuts/3 pieces/20 minutes This is a variation of the "fence post" problem: how many posts do you need to build a 100 foot long fence with 10 foot sections between the posts? Answer: 11 You have to draw the problem to get it...See below, and count the posts! |-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| 0-----10----20----30----40----50----60----70----80----90---100
| 922
| true
|
Q: If it took 10 minutes to saw a board into 2 pieces, how long will it take to saw another into 3 pieces?
So this is supposed to be really simple, and it's taken from the following picture: Text-only: It took Marie $10$ minutes to saw a board into $2$ pieces. If she works just as fast, how long will it take for her to saw another board into $3$ pieces? I don't understand what's wrong with this question. I think the student answered the question wrong, yet my friend insists the student got the question right. I feel like I'm missing something critical here. What am I getting wrong here?
A: Haha! The student probably has a more reasonable interpretation of the question. Of course, cutting one thing into two pieces requires only one cut! Cutting something into three pieces requires two cuts! ------------------------------- 0 cuts/1 piece/0 minutes ---------------|--------------- 1 cut/2 pieces/10 minutes ---------|-----------|--------- 2 cuts/3 pieces/20 minutes This is a variation of the "fence post" problem: how many posts do you need to build a 100 foot long fence with 10 foot sections between the posts? Answer: 11 You have to draw the problem to get it...See below, and count the posts! |-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| 0-----10----20----30----40----50----60----70----80----90---100
| 2026-01-26T11:29:12.138000
|
71,874
|
Can I use my powers for good?
|
I hesitate to ask this question, but I read a lot of the career advice from MathOverflow and math.stackexchange, and I couldn't find anything similar. Four years after the PhD, I am pretty sure that I am going to leave academia soon. I do enjoy teaching and research, but the alpha-maleness, massive egos and pressure to publish are really unappealing to me, and I have never felt that I had the mathematical power to prove interesting results. However, I am really having trouble thinking of anything else to do. Most people seem to think that the main careers open to mathematicians are in banking and finance. I really want to work in some field where I can use mathematics, but it is also important to me to feel like I am contributing something positive or at least not actively doing harm. For this reason, financial speculation is very unappealing to me, although I do find the underlying mathematics quite fascinating. Here is my question: what careers which make a positive contribution to society might be open to academic mathematicians who want to change careers?
| 920
|
soft-question, career-development
|
https://math.stackexchange.com/questions/71874/can-i-use-my-powers-for-good
|
If you are in the US, there are several thousand institutions of higher learning, and at many of them there is very little "pressure to publish". At others, the "pressure to publish" can be met by publishing a textbook or some work of scholarship that does not require proofs of interesting (original) results. High schools also need qualified Mathematics teachers. Consider staying in academia, just moving to a different part of it, as an option for using your powers to do good. I suspect, but cannot be sure, that much of what I've written applies outside the US as well.
| 313
| false
|
Q: Can I use my powers for good?
I hesitate to ask this question, but I read a lot of the career advice from MathOverflow and math.stackexchange, and I couldn't find anything similar. Four years after the PhD, I am pretty sure that I am going to leave academia soon. I do enjoy teaching and research, but the alpha-maleness, massive egos and pressure to publish are really unappealing to me, and I have never felt that I had the mathematical power to prove interesting results. However, I am really having trouble thinking of anything else to do. Most people seem to think that the main careers open to mathematicians are in banking and finance. I really want to work in some field where I can use mathematics, but it is also important to me to feel like I am contributing something positive or at least not actively doing harm. For this reason, financial speculation is very unappealing to me, although I do find the underlying mathematics quite fascinating. Here is my question: what careers which make a positive contribution to society might be open to academic mathematicians who want to change careers?
A: If you are in the US, there are several thousand institutions of higher learning, and at many of them there is very little "pressure to publish". At others, the "pressure to publish" can be met by publishing a textbook or some work of scholarship that does not require proofs of interesting (original) results. High schools also need qualified Mathematics teachers. Consider staying in academia, just moving to a different part of it, as an option for using your powers to do good. I suspect, but cannot be sure, that much of what I've written applies outside the US as well.
| 2026-01-26T11:29:13.460000
|
12,906
|
The staircase paradox, or why $\pi\ne4$
|
What is wrong with this proof? Is $\pi=4?$
| 915
|
geometry, analysis, convergence-divergence, pi, fake-proofs
|
https://math.stackexchange.com/questions/12906/the-staircase-paradox-or-why-pi-ne4
|
This question is usually posed as the length of the diagonal of a unit square. You start going from one corner to the opposite one following the perimeter and observe the length is $2$, then take shorter and shorter stair-steps and the length is $2$ but your path approaches the diagonal. So $\sqrt{2}=2$. In both cases, you are approaching the area but not the path length. You can make this more rigorous by breaking into increments and following the proof of the Riemann sum. The difference in area between the two curves goes nicely to zero, but the difference in arc length stays constant. Edit: making the square more explicit. Imagine dividing the diagonal into $n$ segments and a stairstep approximation. Each triangle is $(\frac{1}{n},\frac{1}{n},\frac{\sqrt{2}}{n})$. So the area between the stairsteps and the diagonal is $n \frac{1}{2n^2}$ which converges to $0$. The path length is $n \frac{2}{n}$, which converges even more nicely to $2$.
| 574
| true
|
Q: The staircase paradox, or why $\pi\ne4$
What is wrong with this proof? Is $\pi=4?$
A: This question is usually posed as the length of the diagonal of a unit square. You start going from one corner to the opposite one following the perimeter and observe the length is $2$, then take shorter and shorter stair-steps and the length is $2$ but your path approaches the diagonal. So $\sqrt{2}=2$. In both cases, you are approaching the area but not the path length. You can make this more rigorous by breaking into increments and following the proof of the Riemann sum. The difference in area between the two curves goes nicely to zero, but the difference in arc length stays constant. Edit: making the square more explicit. Imagine dividing the diagonal into $n$ segments and a stairstep approximation. Each triangle is $(\frac{1}{n},\frac{1}{n},\frac{\sqrt{2}}{n})$. So the area between the stairsteps and the diagonal is $n \frac{1}{2n^2}$ which converges to $0$. The path length is $n \frac{2}{n}$, which converges even more nicely to $2$.
| 2026-01-26T11:29:15.052000
|
358,423
|
A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?
|
Background. If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof of this fact I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem. It is worth mentioning that Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$. T. Coquand and H. Lombardi have found a surprisingly elementary, first-order characterization of the Krull dimension that does not use prime ideals at all. T. Coquand, H. Lombardi, A Short Proof for the Krull Dimension of a Polynomial Ring, The American Mathematical Monthly, Vol. 112, No. 9 (Nov., 2005), pp. 826-829 (4 pages) You can read the article here. Here is a summary. For $x \in R$ let $R_{\{x\}}$ denote the localization of $R$ at the multiplicative subset $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have $$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \label{1}\tag{$\ast$}$$ It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that $$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$ You can use this to define the Krull dimension. This new characterization of $\dim(R) \leq k$ can be seen as a statement in first-order logic (whereas the usual definition with prime ideals uses second-order logic): Use two sorts $N,R$, the usual ring operations for $R$, but also the "mixed" operation $N \times R \to R$, $(n,x) \mapsto x^n$. Every ring becomes a model of this language. A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory. Question. Can we use the characterization \eqref{1} of the Krull dimension by Coquand-Lombardi to prove the formula $$\dim(R[T])=\dim(R)+1$$ for Noetherian commutative rings $R$? Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$. I suspect that this can only work if we find a first-order property of commutative rings (with powers) which is satisfied in particular by Noetherian rings and prove the formula for these rings. Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization. If you post an answer that doesn't mention the characterization, then it's not an answer to my question and therefore offtopic. As of writing this, 20 answers have been posted, all of which have been deleted.
| 891
|
ring-theory, commutative-algebra, noetherian, krull-dimension, dimension-theory-algebra
|
https://math.stackexchange.com/questions/358423/a-proof-of-dimrt-dimr1-without-prime-ideals
| 0
| false
|
Q: A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?
Background. If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof of this fact I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem. It is worth mentioning that Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$. T. Coquand and H. Lombardi have found a surprisingly elementary, first-order characterization of the Krull dimension that does not use prime ideals at all. T. Coquand, H. Lombardi, A Short Proof for the Krull Dimension of a Polynomial Ring, The American Mathematical Monthly, Vol. 112, No. 9 (Nov., 2005), pp. 826-829 (4 pages) You can read the article here. Here is a summary. For $x \in R$ let $R_{\{x\}}$ denote the localization of $R$ at the multiplicative subset $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have $$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \label{1}\tag{$\ast$}$$ It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that $$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$ You can use this to define the Krull dimension. This new characterization of $\dim(R) \leq k$ can be seen as a statement in first-order logic (whereas the usual definition with prime ideals uses second-order logic): Use two sorts $N,R$, the usual ring operations for $R$, but also the "mixed" operation $N \times R \to R$, $(n,x) \mapsto x^n$. Every ring becomes a model of this language. A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory. Question. Can we use the characterization \eqref{1} of the Krull dimension by Coquand-Lombardi to prove the formula $$\dim(R[T])=\dim(R)+1$$ for Noetherian commutative rings $R$? Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$. I suspect that this can only work if we find a first-order property of commutative rings (with powers) which is satisfied in particular by Noetherian rings and prove the formula for these rings. Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization. If you post an answer that doesn't mention the characterization, then it's not an answer to my question and therefore offtopic. As of writing this, 20 answers have been posted, all of which have been deleted.
A: [No answer available]
| 2026-01-26T11:29:16.459000
|
|
8,337
|
Different ways to prove $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$ (the Basel problem)
|
As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$ However, Euler was Euler and he gave other proofs. I believe many of you know some nice proofs of this, can you please share it with us?
| 889
|
sequences-and-series, fourier-analysis, big-list, faq, euler-sums
|
https://math.stackexchange.com/questions/8337/different-ways-to-prove-sum-k-1-infty-frac1k2-frac-pi26-the-b
|
OK, here's my favorite. I thought of this after reading a proof from the book "Proofs from the book" by Aigner & Ziegler, but later I found more or less the same proof as mine in a paper published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9 (EDIT: ...which is actually the proof that I read in Aigner & Ziegler). When $0 Although $S_n$ looks like a complicated sum, it can actually be computed fairly easily. To begin with, $$\frac{1}{\sin^2 x} + \frac{1}{\sin^2 (\frac{\pi}{2}-x)} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x \cdot \sin^2 x} = \frac{4}{\sin^2 2x}.$$ Therefore, if we pair up the terms in the sum $S_n$ except the midpoint $\pi/4$ (take the point $x_k$ in the left half of the interval $(0,\pi/2)$ together with the point $\pi/2-x_k$ in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into $2^{n-1}$ parts. And the midpoint $\pi/4$ contributes with $1/\sin^2(\pi/4)=2$ to the sum. In short, $$S_n = 4 S_{n-1} + 2.$$ Since $S_1=2$, the solution of this recurrence is $$S_n = \frac{2(4^n-1)}{3}.$$ (For example like this: the particular (constant) solution $(S_p)_n = -2/3$ plus the general solution to the homogeneous equation $(S_h)_n = A \cdot 4^n$, with the constant $A$ determined by the initial condition $S_1=(S_p)_1+(S_h)_1=2$.) We now have $$ \frac{2(4^n-1)}{3} - (2^n-1) \leq \frac{4^{n+1}}{\pi^2} \sum_{k=1}^{2^n-1} \frac{1}{k^2} \leq \frac{2(4^n-1)}{3}.$$ Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. This squeezes the partial sums between two sequences both tending to $\pi^2/6$. Voilà!
| 402
| true
|
Q: Different ways to prove $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$ (the Basel problem)
As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$ However, Euler was Euler and he gave other proofs. I believe many of you know some nice proofs of this, can you please share it with us?
A: OK, here's my favorite. I thought of this after reading a proof from the book "Proofs from the book" by Aigner & Ziegler, but later I found more or less the same proof as mine in a paper published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9 (EDIT: ...which is actually the proof that I read in Aigner & Ziegler). When $0 Although $S_n$ looks like a complicated sum, it can actually be computed fairly easily. To begin with, $$\frac{1}{\sin^2 x} + \frac{1}{\sin^2 (\frac{\pi}{2}-x)} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x \cdot \sin^2 x} = \frac{4}{\sin^2 2x}.$$ Therefore, if we pair up the terms in the sum $S_n$ except the midpoint $\pi/4$ (take the point $x_k$ in the left half of the interval $(0,\pi/2)$ together with the point $\pi/2-x_k$ in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into $2^{n-1}$ parts. And the midpoint $\pi/4$ contributes with $1/\sin^2(\pi/4)=2$ to the sum. In short, $$S_n = 4 S_{n-1} + 2.$$ Since $S_1=2$, the solution of this recurrence is $$S_n = \frac{2(4^n-1)}{3}.$$ (For example like this: the particular (constant) solution $(S_p)_n = -2/3$ plus the general solution to the homogeneous equation $(S_h)_n = A \cdot 4^n$, with the constant $A$ determined by the initial condition $S_1=(S_p)_1+(S_h)_1=2$.) We now have $$ \frac{2(4^n-1)}{3} - (2^n-1) \leq \frac{4^{n+1}}{\pi^2} \sum_{k=1}^{2^n-1} \frac{1}{k^2} \leq \frac{2(4^n-1)}{3}.$$ Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. This squeezes the partial sums between two sequences both tending to $\pi^2/6$. Voilà!
| 2026-01-26T11:29:17.804000
|
44,704
|
How to study math to really understand it and have a healthy lifestyle with free time?
|
Here's my issue I faced; I worked really hard studying Math, so because of that, I started to realised that I understand things better. However, that comes at a big cost: In the last few years, I had practically zero physical exercise, I've gained $30$ kg, I've spent countless hours studying at night, constantly had sleep deprivation, lost my social life, and developed health problems. My grades are quite good, but I feel as though I'm wasting my life. I love mathematics when it's done my way, but that's hardly ever. I would very much like my career to be centered around mathematics (topology, algebra or something similar). I want to really understand things and I want the proofs to be done in a (reasonably) rigorous way. Before, I've been accused of being a formalist but I don't consider myself one at all. However, I admit that I am a perfectionist. For comparison, the answers of Theo, Arturo, Jim Belk, Mariano, etc. are absolutely rigorous enough for me. From my experience, $80$% or more mathematics in our school is done in a sketchy, "Hmm, probably true" kind of way (just like reading cooking recipes), which bugs the hell out of me. Most classmates adapt to it but, for some reason, I can't. I don't understand things unless I understand them (almost) completely. They learnt "how one should do things", but less often do they ask themselves WHY is this correct. I have two friend physicists, who have the exact same problem. One is at the doctorate level, constantly frustrated, while the other abandoned physics altogether after getting a diploma. Apart from one $8$, he had a perfect record, all are $10$s. He mentioned that he doesn't feel he understands physics well enough. From my experience, ALL his classmates understand less than he does, they just go with the flow and accept certain statements as true. Did you manage to study everything on time, AND sufficiently rigorous, that you were able to understand it?** ADDITIONS: Frequently, I tend to be the only one who find serious issues in the proofs, the formulations of theorems, and the worked out exercises at classes. Either everyone else understands everything, most or doesn't understand and doesn't care the possible issues. Often, do I find holes in the proofs and that hypotheses are missing in the theorem. When I present them to the professor, he says that I'm right, and mentioned I'm very precise. How is this precise, when the theorem doesn't hold in its current state? Are we even supposed to understand proofs? Are the proofs actually really just sketches? How on earth is one then supposed to be able to discover mathematical truths? Is the study of Mathematics just one big joke and you're not supposed to take it too seriously? NOTE: I have a bunch of sports I like and used to do. Furthermore, I had a perfectly good social life before, so you don't need to give advice regarding that. I don't socialize and do sport because digesting proofs and trying to understand the ideas behind it all eats up all my time. If I go hiking, it will take away $2$ days, one to actually walk + one to rest and regenerate. If I go train MMA, I won't be focused for the whole day. I can't just switch from boxing to diagram chasing in a moment. Also, I can't just study for half an hour. The way I study is: I open the book, search up what I already know but forgot from the previous day, and then go from theorem to theorem, from proof to proof, correcting mistakes, adding clarifications, etc. etc. To add on, I have a bad habit of having difficulty starting things. However when I do start, I start 'my engine', and I have difficulty stopping, especially if it's going good. That's why I unintentionally spend an hour or two before studying just doing the most irrelevant stuff, just to avoid study. This happens especially when I had more math than I can shove down my throat which I have, for mental preparations to begin studying. But, as my engine really starts and studying goes well (proven a lot, understood a lot), it's hard for me to stop, so I often stay late at night, up to 4 a.m., 5 a.m. & 6 a.m. When the day of the exam arrives, I don't go to sleep at all, and the night and day are reversed. I go to sleep at 13h and wake at 21h... I know it's not good but I can't seem to break this habit. If I'm useless through the whole day, I feel a need (guilty conscience) to do at least something useful before I go to sleep. I know this isn't supposed to happen if one loves mathematics. However, when it's 'forced upon you' what and how much and in what amount of time you have to study, you start being put off by math. Mathematics stops being enjoyment/fun and becomes hard work that just needs to be done.
| 867
|
soft-question, advice
|
https://math.stackexchange.com/questions/44704/how-to-study-math-to-really-understand-it-and-have-a-healthy-lifestyle-with-free
|
In my view the central question that you should ask yourself is what is the end goal of your studies. As an example, American college life as depicted in film is hedonistic and certainly not centered on actual studies. Your example is the complete opposite - you describe yourself as an ascetic devoted to scholarship. Many people consider it important to lead a balanced life. If such a person were confronted with your situation, they might look for some compromise, for example investing fewer time on studies in return for lower grades. If things don't work out, they might consider opting out of the entire enterprise. Your viewpoint might be different - for you the most important dimension is intellectual growth, and you are ready to sacrifice all for its sake. It has been mentioned in another answer that leading a healthy lifestyle might contribute to your studies. People tend to "burn out" if they work too hard. I have known such people, and they had to periodically "cool off" in some far-off place. On the contrary, non-curricular activities can be invigorating and refreshing. Another, similar aspect is that of "being busy". Some people find that by multitasking they become more productive in each of their individual "fronts". But that style of life is not for every one. Returning to my original point, what do you expect to accomplish by being successful in school? Are you aiming at an academic career? Professional career? In North America higher education has become a rite of passage, which many graduates find very problematic for the cost it incurs. For them the issue is often economical - education is expensive in North America. You might find out that having completed your studies, you must turn your life to some very different track. You may come to realize that you have wasted some best years of your life by studying hard to the exclusion of everything else, an effort which would eventually lead you nowhere. This is the worst-case scenario. More concretely, I suggest that you plan ahead and consider whether the cost is worth it. That requires both an earnest assessment of your own worth, and some speculation of the future job market. You should also estimate how important you are going to consider these present studies in your future - both from the economical and the "cultural" perspective. This all might sound discouraging, but your situation as you describe it is quite miserable. Not only are you not satisfied with it, but it also looks problematic for an outside observer. However, I suspect that you're exaggerating, viewing the situation from a romantic, heroic perspective. It's best therefore to talk to people who know you personally. Even better, talk to people who're older than you and in the next stage of "life". They have a wider perspective on your situation, which they of their acquaintances have just still vividly recall. However, even their recommendations must be taken with a grain of salt, since their present worries are only part of the larger picture, the all-encompassing "life". Finally, a few words more pertinent to the subject at hand. First, learning strategy. I think the best way to learn is to solve challenging exercises. The advice given here, trying to "reconstruct" the textbook before reading it, seems very time consuming, and in my view, concentrating the effort at the wrong place The same goes for memorizing theorems - sometimes one can only really "understand" the proof of a theorem by studying a more advanced topic. Even the researcher who originally came out with the proof probably didn't "really" understand it until a larger perspective was developed. Memorizing theorems is not your choice but rather a necessity. I always disliked regurgitation and it is regrettable that this is forced unto you. I'm glad that my school would instead give us actual problems to solve - that's much closer to research anyway. Since you have to go through this lamentable process, try to come up with a method of memorization which has other benefits as well - perhaps aim at a better understanding of "what is going on" rather than the actual steps themselves. This is an important skill. Second, one of the answers suggests trying to deduce as many theorems as possible as the "mathematical" thing that ought to be done after seeing a definition. I would suggest rather the opposite - first find out what the definition entails, and then try to understand why the concept was defined in the first place, and why in that particular way. It is common in mathematics to start studying a subject with a long list of "important definitions", which have no import at all at that stage. You will have understood the subject when you can explain where these definitions are coming from, what objects they describe; and when you can "feel" these objects intuitively. This is a far cry from being able to deduce some facts that follow more-or-less directly from the definitions.
| 289
| true
|
Q: How to study math to really understand it and have a healthy lifestyle with free time?
Here's my issue I faced; I worked really hard studying Math, so because of that, I started to realised that I understand things better. However, that comes at a big cost: In the last few years, I had practically zero physical exercise, I've gained $30$ kg, I've spent countless hours studying at night, constantly had sleep deprivation, lost my social life, and developed health problems. My grades are quite good, but I feel as though I'm wasting my life. I love mathematics when it's done my way, but that's hardly ever. I would very much like my career to be centered around mathematics (topology, algebra or something similar). I want to really understand things and I want the proofs to be done in a (reasonably) rigorous way. Before, I've been accused of being a formalist but I don't consider myself one at all. However, I admit that I am a perfectionist. For comparison, the answers of Theo, Arturo, Jim Belk, Mariano, etc. are absolutely rigorous enough for me. From my experience, $80$% or more mathematics in our school is done in a sketchy, "Hmm, probably true" kind of way (just like reading cooking recipes), which bugs the hell out of me. Most classmates adapt to it but, for some reason, I can't. I don't understand things unless I understand them (almost) completely. They learnt "how one should do things", but less often do they ask themselves WHY is this correct. I have two friend physicists, who have the exact same problem. One is at the doctorate level, constantly frustrated, while the other abandoned physics altogether after getting a diploma. Apart from one $8$, he had a perfect record, all are $10$s. He mentioned that he doesn't feel he understands physics well enough. From my experience, ALL his classmates understand less than he does, they just go with the flow and accept certain statements as true. Did you manage to study everything on time, AND sufficiently rigorous, that you were able to understand it?** ADDITIONS: Frequently, I tend to be the only one who find serious issues in the proofs, the formulations of theorems, and the worked out exercises at classes. Either everyone else understands everything, most or doesn't understand and doesn't care the possible issues. Often, do I find holes in the proofs and that hypotheses are missing in the theorem. When I present them to the professor, he says that I'm right, and mentioned I'm very precise. How is this precise, when the theorem doesn't hold in its current state? Are we even supposed to understand proofs? Are the proofs actually really just sketches? How on earth is one then supposed to be able to discover mathematical truths? Is the study of Mathematics just one big joke and you're not supposed to take it too seriously? NOTE: I have a bunch of sports I like and used to do. Furthermore, I had a perfectly good social life before, so you don't need to give advice regarding that. I don't socialize and do sport because digesting proofs and trying to understand the ideas behind it all eats up all my time. If I go hiking, it will take away $2$ days, one to actually walk + one to rest and regenerate. If I go train MMA, I won't be focused for the whole day. I can't just switch from boxing to diagram chasing in a moment. Also, I can't just study for half an hour. The way I study is: I open the book, search up what I already know but forgot from the previous day, and then go from theorem to theorem, from proof to proof, correcting mistakes, adding clarifications, etc. etc. To add on, I have a bad habit of having difficulty starting things. However when I do start, I start 'my engine', and I have difficulty stopping, especially if it's going good. That's why I unintentionally spend an hour or two before studying just doing the most irrelevant stuff, just to avoid study. This happens especially when I had more math than I can shove down my throat which I have, for mental preparations to begin studying. But, as my engine really starts and studying goes well (proven a lot, understood a lot), it's hard for me to stop, so I often stay late at night, up to 4 a.m., 5 a.m. & 6 a.m. When the day of the exam arrives, I don't go to sleep at all, and the night and day are reversed. I go to sleep at 13h and wake at 21h... I know it's not good but I can't seem to break this habit. If I'm useless through the whole day, I feel a need (guilty conscience) to do at least something useful before I go to sleep. I know this isn't supposed to happen if one loves mathematics. However, when it's 'forced upon you' what and how much and in what amount of time you have to study, you start being put off by math. Mathematics stops being enjoyment/fun and becomes hard work that just needs to be done.
A: In my view the central question that you should ask yourself is what is the end goal of your studies. As an example, American college life as depicted in film is hedonistic and certainly not centered on actual studies. Your example is the complete opposite - you describe yourself as an ascetic devoted to scholarship. Many people consider it important to lead a balanced life. If such a person were confronted with your situation, they might look for some compromise, for example investing fewer time on studies in return for lower grades. If things don't work out, they might consider opting out of the entire enterprise. Your viewpoint might be different - for you the most important dimension is intellectual growth, and you are ready to sacrifice all for its sake. It has been mentioned in another answer that leading a healthy lifestyle might contribute to your studies. People tend to "burn out" if they work too hard. I have known such people, and they had to periodically "cool off" in some far-off place. On the contrary, non-curricular activities can be invigorating and refreshing. Another, similar aspect is that of "being busy". Some people find that by multitasking they become more productive in each of their individual "fronts". But that style of life is not for every one. Returning to my original point, what do you expect to accomplish by being successful in school? Are you aiming at an academic career? Professional career? In North America higher education has become a rite of passage, which many graduates find very problematic for the cost it incurs. For them the issue is often economical - education is expensive in North America. You might find out that having completed your studies, you must turn your life to some very different track. You may come to realize that you have wasted some best years of your life by studying hard to the exclusion of everything else, an effort which would eventually lead you nowhere. This is the worst-case scenario. More concretely, I suggest that you plan ahead and consider whether the cost is worth it. That requires both an earnest assessment of your own worth, and some speculation of the future job market. You should also estimate how important you are going to consider these present studies in your future - both from the economical and the "cultural" perspective. This all might sound discouraging, but your situation as you describe it is quite miserable. Not only are you not satisfied with it, but it also looks problematic for an outside observer. However, I suspect that you're exaggerating, viewing the situation from a romantic, heroic perspective. It's best therefore to talk to people who know you personally. Even better, talk to people who're older than you and in the next stage of "life". They have a wider perspective on your situation, which they of their acquaintances have just still vividly recall. However, even their recommendations must be taken with a grain of salt, since their present worries are only part of the larger picture, the all-encompassing "life". Finally, a few words more pertinent to the subject at hand. First, learning strategy. I think the best way to learn is to solve challenging exercises. The advice given here, trying to "reconstruct" the textbook before reading it, seems very time consuming, and in my view, concentrating the effort at the wrong place The same goes for memorizing theorems - sometimes one can only really "understand" the proof of a theorem by studying a more advanced topic. Even the researcher who originally came out with the proof probably didn't "really" understand it until a larger perspective was developed. Memorizing theorems is not your choice but rather a necessity. I always disliked regurgitation and it is regrettable that this is forced unto you. I'm glad that my school would instead give us actual problems to solve - that's much closer to research anyway. Since you have to go through this lamentable process, try to come up with a method of memorization which has other benefits as well - perhaps aim at a better understanding of "what is going on" rather than the actual steps themselves. This is an important skill. Second, one of the answers suggests trying to deduce as many theorems as possible as the "mathematical" thing that ought to be done after seeing a definition. I would suggest rather the opposite - first find out what the definition entails, and then try to understand why the concept was defined in the first place, and why in that particular way. It is common in mathematics to start studying a subject with a long list of "important definitions", which have no import at all at that stage. You will have understood the subject when you can explain where these definitions are coming from, what objects they describe; and when you can "feel" these objects intuitively. This is a far cry from being able to deduce some facts that follow more-or-less directly from the definitions.
| 2026-01-26T11:29:19.320000
|
668
|
What's an intuitive way to think about the determinant?
|
In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?
| 849
|
linear-algebra, matrices, determinant, intuition
|
https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant
|
Your trouble with determinants is pretty common. They’re a hard thing to teach well, too, for two main reasons that I can see: the formulas you learn for computing them are messy and complicated, and there’s no “natural” way to interpret the value of the determinant, the way it’s easy to interpret the derivatives you do in calculus at first as the slope of the tangent line. It’s hard to believe things like the invertibility condition you’ve stated when it’s not even clear what the numbers mean and where they come from. Rather than show that the many usual definitions are all the same by comparing them to each other, I’m going to state some general properties of the determinant that I claim are enough to specify uniquely what number you should get when you put in a given matrix. Then it’s not too bad to check that all of the definitions for determinant that you’ve seen satisfy those properties I’ll state. The first thing to think about if you want an “abstract” definition of the determinant to unify all those others is that it’s not an array of numbers with bars on the side. What we’re really looking for is a function that takes N vectors (the N columns of the matrix) and returns a number. Let’s assume we’re working with real numbers for now. Remember how those operations you mentioned change the value of the determinant? Switching two rows or columns changes the sign. Multiplying one row by a constant multiplies the whole determinant by that constant. The general fact that number two draws from: the determinant is linear in each row. That is, if you think of it as a function $\det: \mathbb{R}^{n^2} \rightarrow \mathbb{R}$, then $$ \det(a \vec v_1 +b \vec w_1 , \vec v_2 ,\ldots,\vec v_n ) = a \det(\vec v_1,\vec v_2,\ldots,\vec v_n) + b \det(\vec w_1, \vec v_2, \ldots,\vec v_n),$$ and the corresponding condition in each other slot. The determinant of the identity matrix $I$ is $1$. I claim that these facts are enough to define a unique function that takes in N vectors (each of length N) and returns a real number, the determinant of the matrix given by those vectors. I won’t prove that, but I’ll show you how it helps with some other interpretations of the determinant. In particular, there’s a nice geometric way to think of a determinant. Consider the unit cube in N dimensional space: the set of N vectors of length 1 with coordinates 0 or 1 in each spot. The determinant of the linear transformation (matrix) T is the signed volume of the region gotten by applying T to the unit cube. (Don’t worry too much if you don’t know what the “signed” part means, for now). How does that follow from our abstract definition? Well, if you apply the identity to the unit cube, you get back the unit cube. And the volume of the unit cube is 1. If you stretch the cube by a constant factor in one direction only, the new volume is that constant. And if you stack two blocks together aligned on the same direction, their combined volume is the sum of their volumes: this all shows that the signed volume we have is linear in each coordinate when considered as a function of the input vectors. Finally, when you switch two of the vectors that define the unit cube, you flip the orientation. (Again, this is something to come back to later if you don’t know what that means). So there are ways to think about the determinant that aren’t symbol-pushing. If you’ve studied multivariable calculus, you could think about, with this geometric definition of determinant, why determinants (the Jacobian) pop up when we change coordinates doing integration. Hint: a derivative is a linear approximation of the associated function, and consider a “differential volume element” in your starting coordinate system. It’s not too much work to check that the area of the parallelogram formed by vectors $(a,b)$ and $(c,d)$ is $\Big|{}^{a\;b}_{c\;d}\Big|$ either: you might try that to get a sense for things.
| 518
| true
|
Q: What's an intuitive way to think about the determinant?
In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?
A: Your trouble with determinants is pretty common. They’re a hard thing to teach well, too, for two main reasons that I can see: the formulas you learn for computing them are messy and complicated, and there’s no “natural” way to interpret the value of the determinant, the way it’s easy to interpret the derivatives you do in calculus at first as the slope of the tangent line. It’s hard to believe things like the invertibility condition you’ve stated when it’s not even clear what the numbers mean and where they come from. Rather than show that the many usual definitions are all the same by comparing them to each other, I’m going to state some general properties of the determinant that I claim are enough to specify uniquely what number you should get when you put in a given matrix. Then it’s not too bad to check that all of the definitions for determinant that you’ve seen satisfy those properties I’ll state. The first thing to think about if you want an “abstract” definition of the determinant to unify all those others is that it’s not an array of numbers with bars on the side. What we’re really looking for is a function that takes N vectors (the N columns of the matrix) and returns a number. Let’s assume we’re working with real numbers for now. Remember how those operations you mentioned change the value of the determinant? Switching two rows or columns changes the sign. Multiplying one row by a constant multiplies the whole determinant by that constant. The general fact that number two draws from: the determinant is linear in each row. That is, if you think of it as a function $\det: \mathbb{R}^{n^2} \rightarrow \mathbb{R}$, then $$ \det(a \vec v_1 +b \vec w_1 , \vec v_2 ,\ldots,\vec v_n ) = a \det(\vec v_1,\vec v_2,\ldots,\vec v_n) + b \det(\vec w_1, \vec v_2, \ldots,\vec v_n),$$ and the corresponding condition in each other slot. The determinant of the identity matrix $I$ is $1$. I claim that these facts are enough to define a unique function that takes in N vectors (each of length N) and returns a real number, the determinant of the matrix given by those vectors. I won’t prove that, but I’ll show you how it helps with some other interpretations of the determinant. In particular, there’s a nice geometric way to think of a determinant. Consider the unit cube in N dimensional space: the set of N vectors of length 1 with coordinates 0 or 1 in each spot. The determinant of the linear transformation (matrix) T is the signed volume of the region gotten by applying T to the unit cube. (Don’t worry too much if you don’t know what the “signed” part means, for now). How does that follow from our abstract definition? Well, if you apply the identity to the unit cube, you get back the unit cube. And the volume of the unit cube is 1. If you stretch the cube by a constant factor in one direction only, the new volume is that constant. And if you stack two blocks together aligned on the same direction, their combined volume is the sum of their volumes: this all shows that the signed volume we have is linear in each coordinate when considered as a function of the input vectors. Finally, when you switch two of the vectors that define the unit cube, you flip the orientation. (Again, this is something to come back to later if you don’t know what that means). So there are ways to think about the determinant that aren’t symbol-pushing. If you’ve studied multivariable calculus, you could think about, with this geometric definition of determinant, why determinants (the Jacobian) pop up when we change coordinates doing integration. Hint: a derivative is a linear approximation of the associated function, and consider a “differential volume element” in your starting coordinate system. It’s not too much work to check that the area of the parallelogram formed by vectors $(a,b)$ and $(c,d)$ is $\Big|{}^{a\;b}_{c\;d}\Big|$ either: you might try that to get a sense for things.
| 2026-01-26T11:29:21.065000
|
216,343
|
Does $\pi$ contain all possible number combinations?
|
$\pi$ Pi Pi is an infinite, nonrepeating $($sic$)$ decimal - meaning that every possible number combination exists somewhere in pi. Converted into ASCII text, somewhere in that infinite string of digits is the name of every person you will ever love, the date, time and manner of your death, and the answers to all the great questions of the universe. Is this true? Does it make any sense ?
| 810
|
elementary-number-theory, irrational-numbers, pi
|
https://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations
|
It is not true that an infinite, non-repeating decimal must contain ‘every possible number combination’. The decimal $0.011000111100000111111\dots$ is an easy counterexample. However, if the decimal expansion of $\pi$ contains every possible finite string of digits, which seems quite likely, then the rest of the statement is indeed correct. Of course, in that case it also contains numerical equivalents of every book that will never be written, among other things.
| 1,020
| true
|
Q: Does $\pi$ contain all possible number combinations?
$\pi$ Pi Pi is an infinite, nonrepeating $($sic$)$ decimal - meaning that every possible number combination exists somewhere in pi. Converted into ASCII text, somewhere in that infinite string of digits is the name of every person you will ever love, the date, time and manner of your death, and the answers to all the great questions of the universe. Is this true? Does it make any sense ?
A: It is not true that an infinite, non-repeating decimal must contain ‘every possible number combination’. The decimal $0.011000111100000111111\dots$ is an easy counterexample. However, if the decimal expansion of $\pi$ contains every possible finite string of digits, which seems quite likely, then the rest of the statement is indeed correct. Of course, in that case it also contains numerical equivalents of every book that will never be written, among other things.
| 2026-01-26T11:29:22.546000
|
637,728
|
Splitting a sandwich and not feeling deceived
|
This is a problem that has haunted me for more than a decade. Not all the time - but from time to time, and always on windy or rainy days, it suddenly reappears in my mind, stares at me for half an hour to an hour, and then just grins at me, and whispers whole day: "You will never solve me..." Please save me from this torturer. Here it is: Let's say there are two people and a sandwich. They want to share the sandwich, but they don't trust each other. However, they found the way how both of them will have a lunch without feeling deceived: One of them will cut the sandwich in two halves, and another will choose which half will be his. Fair, right? The problem is: Is there such mechanism for three people and a sandwich? EDIT: This was roller-coaster for me. Now, it turns out that there are at least two books devoted exclusively on this problem and its variations: Fair Division Cake Cutting Algorithms Yesterday, I was in a coffee shop in a small company. We ordered coffee and some chocolate cakes. As I was cutting my cake for my first bite, I felt sweat on my forehead. I thought, 'What if some of my buddies just interrupt me and say: Stop! You are not cutting the cake in a fair manner!' My hands started shaking in fear of that. But, no, nothing happened, fortunately.
| 697
|
game-theory, fair-division
|
https://math.stackexchange.com/questions/637728/splitting-a-sandwich-and-not-feeling-deceived
|
For more than two, the moving knife is a nice solution. Somebody takes a knife and moves it slowly across the sandwich. Any player may say "cut". At that moment, the sandwich is cut and the piece given to the one who said "cut". As he has said that is an acceptable piece, he believes he has at least $\frac 1n$ of the sandwich. The rest have asserted (by not saying "cut") that is it at most $\frac 1n$ of the sandwich, so the average available is now at least their share. Recurse.
| 288
| false
|
Q: Splitting a sandwich and not feeling deceived
This is a problem that has haunted me for more than a decade. Not all the time - but from time to time, and always on windy or rainy days, it suddenly reappears in my mind, stares at me for half an hour to an hour, and then just grins at me, and whispers whole day: "You will never solve me..." Please save me from this torturer. Here it is: Let's say there are two people and a sandwich. They want to share the sandwich, but they don't trust each other. However, they found the way how both of them will have a lunch without feeling deceived: One of them will cut the sandwich in two halves, and another will choose which half will be his. Fair, right? The problem is: Is there such mechanism for three people and a sandwich? EDIT: This was roller-coaster for me. Now, it turns out that there are at least two books devoted exclusively on this problem and its variations: Fair Division Cake Cutting Algorithms Yesterday, I was in a coffee shop in a small company. We ordered coffee and some chocolate cakes. As I was cutting my cake for my first bite, I felt sweat on my forehead. I thought, 'What if some of my buddies just interrupt me and say: Stop! You are not cutting the cake in a fair manner!' My hands started shaking in fear of that. But, no, nothing happened, fortunately.
A: For more than two, the moving knife is a nice solution. Somebody takes a knife and moves it slowly across the sandwich. Any player may say "cut". At that moment, the sandwich is cut and the piece given to the one who said "cut". As he has said that is an acceptable piece, he believes he has at least $\frac 1n$ of the sandwich. The rest have asserted (by not saying "cut") that is it at most $\frac 1n$ of the sandwich, so the average available is now at least their share. Recurse.
| 2026-01-26T11:29:24.183000
|
323,334
|
What was the first bit of mathematics that made you realize that math is beautiful? (For children's book)
|
I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education. I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers. However, I myself am mathematically unsophisticated. What was the first bit of mathematics that made you realize that math is beautiful? For the purposes of this children's book, accessible answers would be appreciated.
| 660
|
soft-question, education, big-list
|
https://math.stackexchange.com/questions/323334/what-was-the-first-bit-of-mathematics-that-made-you-realize-that-math-is-beautif
|
This wasn't the first, but it's definitely awesome: This is a proof of the Pythagorean theorem, and it uses no words!
| 315
| false
|
Q: What was the first bit of mathematics that made you realize that math is beautiful? (For children's book)
I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education. I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers. However, I myself am mathematically unsophisticated. What was the first bit of mathematics that made you realize that math is beautiful? For the purposes of this children's book, accessible answers would be appreciated.
A: This wasn't the first, but it's definitely awesome: This is a proof of the Pythagorean theorem, and it uses no words!
| 2026-01-26T11:29:25.878000
|
952,466
|
Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
|
Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ is connected. Question: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$? Full disclosure: I've now crossposted the question on MO. Various remarks If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = \sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not. With the bijection, it holds true in $n = 1$. But this is using the order structure of $\mathbb{R}$: a bijection that preserves connectedness on $\mathbb{R}$ must be monotone. As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.) Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open? Some properties of $\mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $\mathbb{R}^n$ with the standard topology to itself, self-maps of some arbitrary topological space, it is easy to make the answer go either way. For example, if the topological space $(X,\tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by the cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected. On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = \mathbb{Z}$ be equipped with the topology generated by: $$ \{\mathbb{N}\} \cup \{ \{k\} \mid k \in \mathbb{Z} \setminus \mathbb{N} \} $$ Then the map $k \mapsto k+\ell$ for any $\ell > 0$ maps connected sets to connected sets, but its inverse $k\mapsto k-\ell$ can map connected sets to disconnected ones. Working a bit harder one can construct in similar vein examples which are Hausdorff.
| 656
|
general-topology, metric-spaces, examples-counterexamples, connectedness
|
https://math.stackexchange.com/questions/952466/is-there-a-bijection-of-mathbbrn-with-itself-such-that-the-forward-map-is
| 0
| false
|
Q: Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ is connected. Question: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$? Full disclosure: I've now crossposted the question on MO. Various remarks If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = \sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not. With the bijection, it holds true in $n = 1$. But this is using the order structure of $\mathbb{R}$: a bijection that preserves connectedness on $\mathbb{R}$ must be monotone. As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.) Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open? Some properties of $\mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $\mathbb{R}^n$ with the standard topology to itself, self-maps of some arbitrary topological space, it is easy to make the answer go either way. For example, if the topological space $(X,\tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by the cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected. On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = \mathbb{Z}$ be equipped with the topology generated by: $$ \{\mathbb{N}\} \cup \{ \{k\} \mid k \in \mathbb{Z} \setminus \mathbb{N} \} $$ Then the map $k \mapsto k+\ell$ for any $\ell > 0$ maps connected sets to connected sets, but its inverse $k\mapsto k-\ell$ can map connected sets to disconnected ones. Working a bit harder one can construct in similar vein examples which are Hausdorff.
A: [No answer available]
| 2026-01-26T11:29:27.339000
|
|
1,681,993
|
Why is $1 - \frac{1}{1 - \frac{1}{1 - \ldots}}$ not real?
|
So we all know that the continued fraction containing all $1$s... $$ x = 1 + \frac{1}{1 + \frac{1}{1 + \ldots}}. $$ yields the golden ratio $x = \phi$, which can easily be proven by rewriting it as $x = 1 + \dfrac{1}{x}$, solving the resulting quadratic equation and assuming that a continued fraction that only contains additions will give a positive number. Now, a friend asked me what would happen if we replaced all additions with subtractions: $$ x = 1 - \frac{1}{1 - \frac{1}{1 - \ldots}}. $$ I thought "oh cool, I know how to solve this...": \begin{align} x &= 1 - \frac{1}{x} \\ x^2 - x + 1 &= 0. \end{align} And voila, I get... $$ x \in \{e^{i\pi/3}, e^{-i\pi/3} \} .$$ Ummm... why does a continued fraction containing only $1$s, subtraction and division result in one of two complex (as opposed to real) numbers? (I have a feeling this is something like the $\sum_i (-1)^i$ thing, that the infinite continued fraction isn't well-defined unless we can express it as the limit of a converging series, because the truncated fractions $1 - \frac{1}{1-1}$ etc. aren't well-defined, but I thought I'd ask for a well-founded answer. Even if this is the case, do the two complex numbers have any "meaning"?)
| 651
|
complex-numbers, recursion, continued-fractions
|
https://math.stackexchange.com/questions/1681993/why-is-1-frac11-frac11-ldots-not-real
|
You're attempting to take a limit. $$x_{n+1} = 1-\frac{1}{x_n}$$ This recurrence actually never converges, from any real starting point. Indeed, $$x_2 = 1-\frac{1}{x_1}; \\ x_3 = 1-\frac{1}{1-1/x_1} = 1-\frac{x_1}{x_1-1} = \frac{1}{1-x_1}; \\ x_4 = x_1$$ So the sequence is periodic with period 3. Therefore it converges if and only if it is constant; but the only way it could be constant is, as you say, if $x_1$ is one of the two complex numbers you found. Therefore, what you have is actually basically a proof by contradiction that the sequence doesn't converge when you consider it over the reals. However, you have found exactly the two values for which the iteration does converge; that is their significance. Alternatively viewed, the map $$z \mapsto 1-\frac{1}{z}$$ is a certain transformation of the complex plane, which has precisely two fixed points. You might find it an interesting exercise to work out what that map does to the complex plane, and examine in particular what it does to points on the real line.
| 558
| true
|
Q: Why is $1 - \frac{1}{1 - \frac{1}{1 - \ldots}}$ not real?
So we all know that the continued fraction containing all $1$s... $$ x = 1 + \frac{1}{1 + \frac{1}{1 + \ldots}}. $$ yields the golden ratio $x = \phi$, which can easily be proven by rewriting it as $x = 1 + \dfrac{1}{x}$, solving the resulting quadratic equation and assuming that a continued fraction that only contains additions will give a positive number. Now, a friend asked me what would happen if we replaced all additions with subtractions: $$ x = 1 - \frac{1}{1 - \frac{1}{1 - \ldots}}. $$ I thought "oh cool, I know how to solve this...": \begin{align} x &= 1 - \frac{1}{x} \\ x^2 - x + 1 &= 0. \end{align} And voila, I get... $$ x \in \{e^{i\pi/3}, e^{-i\pi/3} \} .$$ Ummm... why does a continued fraction containing only $1$s, subtraction and division result in one of two complex (as opposed to real) numbers? (I have a feeling this is something like the $\sum_i (-1)^i$ thing, that the infinite continued fraction isn't well-defined unless we can express it as the limit of a converging series, because the truncated fractions $1 - \frac{1}{1-1}$ etc. aren't well-defined, but I thought I'd ask for a well-founded answer. Even if this is the case, do the two complex numbers have any "meaning"?)
A: You're attempting to take a limit. $$x_{n+1} = 1-\frac{1}{x_n}$$ This recurrence actually never converges, from any real starting point. Indeed, $$x_2 = 1-\frac{1}{x_1}; \\ x_3 = 1-\frac{1}{1-1/x_1} = 1-\frac{x_1}{x_1-1} = \frac{1}{1-x_1}; \\ x_4 = x_1$$ So the sequence is periodic with period 3. Therefore it converges if and only if it is constant; but the only way it could be constant is, as you say, if $x_1$ is one of the two complex numbers you found. Therefore, what you have is actually basically a proof by contradiction that the sequence doesn't converge when you consider it over the reals. However, you have found exactly the two values for which the iteration does converge; that is their significance. Alternatively viewed, the map $$z \mapsto 1-\frac{1}{z}$$ is a certain transformation of the complex plane, which has precisely two fixed points. You might find it an interesting exercise to work out what that map does to the complex plane, and examine in particular what it does to points on the real line.
| 2026-01-26T11:29:28.865000
|
111,440
|
Examples of patterns that eventually fail
|
Often, when I try to describe mathematics to the layman, I find myself struggling to convince them of the importance and consequence of "proof". I receive responses like: "surely if Collatz is true up to $20×2^{58}$, then it must always be true?"; and "the sequence of number of edges on a complete graph starts $0,1,3,6,10$, so the next term must be 15 etc." Granted, this second statement is less logically unsound than the first since it's not difficult to see the reason why the sequence must continue as such; nevertheless, the statement was made on a premise that boils down to "interesting patterns must always continue". I try to counter this logic by creating a ridiculous argument like "the numbers $1,2,3,4,5$ are less than $100$, so surely all numbers are", but this usually fails to be convincing. So, are there any examples of non-trivial patterns that appear to be true for a large number of small cases, but then fail for some larger case? A good answer to this question should: be one which could be explained to the layman without having to subject them to a 24 lecture course of background material, and have as a minimal counterexample a case which cannot (feasibly) be checked without the use of a computer. I believe conditions 1. and 2. make my question specific enough to have in some sense a "right" (or at least a "not wrong") answer; but I'd be happy to clarify if this is not the case. I suppose I'm expecting an answer to come from number theory, but can see that areas like graph theory, combinatorics more generally and set theory could potentially offer suitable answers.
| 643
|
big-list, examples-counterexamples
|
https://math.stackexchange.com/questions/111440/examples-of-patterns-that-eventually-fail
|
I'll translate an entry in the blog Gaussianos ("Gaussians") about Polya's conjecture, titled: A BELIEF IS NOT A PROOF. We'll say a number is of even kind if in its prime factorization, an even number of primes appear. For example $6 = 2\cdot 3$ is a number of even kind. And we'll say a number is of odd kind if the number of primes in its factorization is odd. For example, $18 = 2·3·3$ is of odd kind. ($1$ is considered of even kind). Let $n$ be any natural number. We'll consider the following numbers: $E(n) =$ number of positive integers less or equal to $n$ that are of even kind. $O(n) =$ number of positive integers less or equal to $n$ that are of odd kind. Let's consider $n=7$. In this case $O(7) = 4$ (number 2, 3, 5 and 7 itself) and $E(7) = 3$ ( 1, 4 and 6). So $O(7) >E(7)$. For $n = 6$: $O(6) = 3$ and $E(6) = 3$. Thus $O(6) = E(6)$. In 1919 George Polya proposed the following result, know as Polya's Conjecture: For all $n > 2$, $O(n)$ is greater than or equal to $E(n)$. Polya had checked this for $n In 1962, Lehman found an explicit counterexample: for $n = 906180359$, we have $O(n) = E(n) – 1$, so: $$O(906180359) By an exhaustive search, the smallest counterexample is $n = 906150257$, found by Tanaka in 1980. Thus Polya's Conjecture is false. What do we learn from this? Well, it is simple: unfortunately in mathematics we cannot trust intuition or what happens for a finite number of cases, no matter how large the number is. Until the result is proved for the general case, we have no certainty that it is true.
| 436
| false
|
Q: Examples of patterns that eventually fail
Often, when I try to describe mathematics to the layman, I find myself struggling to convince them of the importance and consequence of "proof". I receive responses like: "surely if Collatz is true up to $20×2^{58}$, then it must always be true?"; and "the sequence of number of edges on a complete graph starts $0,1,3,6,10$, so the next term must be 15 etc." Granted, this second statement is less logically unsound than the first since it's not difficult to see the reason why the sequence must continue as such; nevertheless, the statement was made on a premise that boils down to "interesting patterns must always continue". I try to counter this logic by creating a ridiculous argument like "the numbers $1,2,3,4,5$ are less than $100$, so surely all numbers are", but this usually fails to be convincing. So, are there any examples of non-trivial patterns that appear to be true for a large number of small cases, but then fail for some larger case? A good answer to this question should: be one which could be explained to the layman without having to subject them to a 24 lecture course of background material, and have as a minimal counterexample a case which cannot (feasibly) be checked without the use of a computer. I believe conditions 1. and 2. make my question specific enough to have in some sense a "right" (or at least a "not wrong") answer; but I'd be happy to clarify if this is not the case. I suppose I'm expecting an answer to come from number theory, but can see that areas like graph theory, combinatorics more generally and set theory could potentially offer suitable answers.
A: I'll translate an entry in the blog Gaussianos ("Gaussians") about Polya's conjecture, titled: A BELIEF IS NOT A PROOF. We'll say a number is of even kind if in its prime factorization, an even number of primes appear. For example $6 = 2\cdot 3$ is a number of even kind. And we'll say a number is of odd kind if the number of primes in its factorization is odd. For example, $18 = 2·3·3$ is of odd kind. ($1$ is considered of even kind). Let $n$ be any natural number. We'll consider the following numbers: $E(n) =$ number of positive integers less or equal to $n$ that are of even kind. $O(n) =$ number of positive integers less or equal to $n$ that are of odd kind. Let's consider $n=7$. In this case $O(7) = 4$ (number 2, 3, 5 and 7 itself) and $E(7) = 3$ ( 1, 4 and 6). So $O(7) >E(7)$. For $n = 6$: $O(6) = 3$ and $E(6) = 3$. Thus $O(6) = E(6)$. In 1919 George Polya proposed the following result, know as Polya's Conjecture: For all $n > 2$, $O(n)$ is greater than or equal to $E(n)$. Polya had checked this for $n In 1962, Lehman found an explicit counterexample: for $n = 906180359$, we have $O(n) = E(n) – 1$, so: $$O(906180359) By an exhaustive search, the smallest counterexample is $n = 906150257$, found by Tanaka in 1980. Thus Polya's Conjecture is false. What do we learn from this? Well, it is simple: unfortunately in mathematics we cannot trust intuition or what happens for a finite number of cases, no matter how large the number is. Until the result is proved for the general case, we have no certainty that it is true.
| 2026-01-26T11:29:30.384000
|
2,755
|
Why can you turn clothing right-side-out?
|
My nephew was folding laundry, and turning the occasional shirt right-side-out. I showed him a "trick" where I turned it right-side-out by pulling the whole thing through a sleeve instead of the bottom or collar of the shirt. He thought it was really cool (kids are easily amused, and so am I). So he learned that you can turn a shirt or pants right-side-out by pulling the material through any hole, not just certain ones. I told him that even if there was a rip in the shirt, you could use that to turn it inside-out or right-side-out, and he was fascinated by this and asked "why?" I don't really know the answer to this. Why is this the case? What if the sleeves of a long-sleeve shirt were sewn together at the cuff, creating a continuous tube from one sleeve to the other? Would you still be able to turn it right-side-out? Why? What properties must a garment have so that it can be turned inside-out and right-side-out? Sorry if this is a silly question, but I've always wondered. I wouldn't even know what to google for, so that is why I am asking here. If you know the answer to this, could you please put it into layman's terms? Update: Wow, I really appreciate all the participation. This is a really pleasant community and I have learned a lot here. It seems that the answer is that you need at least one puncture in the garment through which to push or pull the fabric. It appears that you can have certain handles, although it's not usually practical with clothing due to necessary stretching. Accepted (a while ago actually -- sorry for not updating sooner) Dan's answer because among the answers that I understand, it is the highest ranked by this community.
| 640
|
general-topology, algebraic-topology
|
https://math.stackexchange.com/questions/2755/why-can-you-turn-clothing-right-side-out
|
First, a warning. I suspect this response is likely not going to be immediately comprehensible. There is a formal set-up for your question, there are tools available to understand what's going on. They're not particularly light tools, but they exist and they're worthy of being mentioned. Before I write down the main theorem, let me set-up some terminology. The tools belong to a subject called manifold theory and algebraic topology. The names of the tools I'm going to use are called things like: the isotopy extension theorem, fibre-bundles, fibrations and homotopy-groups. You have a surface $\Sigma$, it's your shirt or whatever else you're interested in, some surface in 3-dimensional space. Surfaces have automorphism groups, let me call it $\operatorname{Aut}(\Sigma)$. These are, say, all the self-homeomorphisms or diffeomorphisms of the surface. And surfaces can sit in space. A way of putting a surface in space is called an embedding. Let's call all the embeddings of the surface $\operatorname{Emb}(\Sigma, \mathbb R^3)$. $\operatorname{Emb}(\Sigma, \mathbb R^3)$ is a set, but in the subject of topology these sets have a natural topology as well. We think of them as a space where "nearby" embeddings are almost the same, except for maybe a little wiggle here or there. The topology on the set of embeddings is called the compact-open topology (see Wikipedia, for details on most of these definitions). Okay, so now there's some formal nonsense. Look at the quotient space $\operatorname{Emb}(\Sigma, \mathbb R^3)/\operatorname{Aut}(\Sigma)$. You can think of this as all ways $\Sigma$ can sit in space, but without any labelling -- the surface has no parametrization. So it's the space of all subspaces of $\mathbb R^3$ that just happen to be homeomorphic to your surface. Richard Palais has a really nice theorem that puts this all into a pleasant context. The preamble is we need to think of everything as living in the world of smooth manifolds -- smooth embeddings, $\operatorname{Aut}(\Sigma)$ is the diffeomorphism group of the surface, etc. There are two locally-trivial fibre bundles (or something more easy to prove -- Serre fibrations), this is the "global" isotopy-extension theorem: $$\operatorname{Diff}(\mathbb R^3, \Sigma) \to \operatorname{Diff}(\mathbb R^3) \to \operatorname{Emb}(\Sigma, \mathbb R^3)/\operatorname{Aut}(\Sigma)$$ $$\operatorname{Diff}(\mathbb R^3 \operatorname{fix} \Sigma) \to \operatorname{Diff}(\mathbb R^3, \Sigma) \to \operatorname{Aut}(\Sigma)$$ here $\operatorname{Diff}(\mathbb R^3)$ indicates diffeomorphisms of $\mathbb R^3$ that are the identity outside of a sufficiently large ball, say. So the Palais theorem, together with the homotopy long exact sequence of a fibration, is giving you a language that allows you to translate between automorphisms of your surface, and motions of the surface in space. It's a theorem of Jean Cerf's that $\operatorname{Diff}(\mathbb R^3)$ is connected. A little diagram chase says that an automorphism of a surface can be realized by a motion of that surface in 3-space if and only if that automorphism of the surface extends to an automorphism of 3-space. For closed surfaces, the Jordan-Brouwer separation theorem gives you an obstruction to turning your surface inside-out. But for non-closed surfaces you're out of tools. To figure out if you can realize an automorphism as a motion, you literally have to try to extend it "by hands". This is a very general phenomena -- you have one manifold sitting in another, but rarely does an automorphism of the submanifold extend to the ambient manifold. You see this phenomena happening in various other branches of mathematics as well -- an automorphism of a subgroup does not always extend to the ambient group, etc. So you try your luck and try to build the extension yourself. In some vague sense that's a formal analogy between the visceral mystery of turning the surface inside-out and a kind of formalized mathematical problem, but of a fundamentally analogous feel. We're looking for automorphisms that reverse orientation. For an arbitrary surface with boundary in 3-space, it's not clear if you can turn the surface inside out. This is because the surface might be knotted. Unknotted surfaces are examples like your t-shirt. Let's try to cook up something that can't be turned inside-out. The automorphism group of a 3-times punctured sphere has 12 path-components (12 elements up to isotopy). There are 6 elements that preserve orientation, and 6 that reverse. In particular the orientation-reversing automorphisms reverse the orientation of all the boundary circles. So if you could come up with a knotted pair-of-pants (3-times punctured surface) so that its boundary circles did not admit a symmetry that reversed the orientations of all three circles simultaneously, you'd be done. Maybe this doesn't seem like a reduction to you, but it is. For example, there are things called non-invertible knots: http://en.wikipedia.org/wiki/Invertible_knot So how do we cook up a knotted pair-of-pants from that? Here's the idea. The non-invertible knot in the link above is sometimes called $8_{17}$. Here is another picture of it: http://katlas.org/wiki/8_17 Here is a variant on that. Interpret this image as a ribbon of paper that has three boundary circles. One boundary circle is unknotted. One is $8_{17}$. The other is some other knot. It turns out that other knot isn't trivial, nor is it $8_{17}$. So why can't this knotted pair of pants be turned inside-out? Well, the three knots are distinct, and $8_{17}$ can't be reversed. The reason why I know the other knot isn't $8_{17}$? It's a hyperbolic knot and it has a different ($4.40083...$) hyperbolic volume than $8_{17}$ ($10.9859...$). FYI: in some sense this is one of the simplest surfaces with non-trivial boundary that can't be turned inside-out. All discs can be turned inside-out. Similarly, all annuli (regardless of how they're knotted) can be turned inside-out. So for genus zero surfaces, 3 boundary components is the least you can have if you're looking for a surface that can't be turned inside-out. edited to correct for Jason's comment. comment added later: I suggest if you purchase a garment of this form you return it to the manufacturer.
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Q: Why can you turn clothing right-side-out?
My nephew was folding laundry, and turning the occasional shirt right-side-out. I showed him a "trick" where I turned it right-side-out by pulling the whole thing through a sleeve instead of the bottom or collar of the shirt. He thought it was really cool (kids are easily amused, and so am I). So he learned that you can turn a shirt or pants right-side-out by pulling the material through any hole, not just certain ones. I told him that even if there was a rip in the shirt, you could use that to turn it inside-out or right-side-out, and he was fascinated by this and asked "why?" I don't really know the answer to this. Why is this the case? What if the sleeves of a long-sleeve shirt were sewn together at the cuff, creating a continuous tube from one sleeve to the other? Would you still be able to turn it right-side-out? Why? What properties must a garment have so that it can be turned inside-out and right-side-out? Sorry if this is a silly question, but I've always wondered. I wouldn't even know what to google for, so that is why I am asking here. If you know the answer to this, could you please put it into layman's terms? Update: Wow, I really appreciate all the participation. This is a really pleasant community and I have learned a lot here. It seems that the answer is that you need at least one puncture in the garment through which to push or pull the fabric. It appears that you can have certain handles, although it's not usually practical with clothing due to necessary stretching. Accepted (a while ago actually -- sorry for not updating sooner) Dan's answer because among the answers that I understand, it is the highest ranked by this community.
A: First, a warning. I suspect this response is likely not going to be immediately comprehensible. There is a formal set-up for your question, there are tools available to understand what's going on. They're not particularly light tools, but they exist and they're worthy of being mentioned. Before I write down the main theorem, let me set-up some terminology. The tools belong to a subject called manifold theory and algebraic topology. The names of the tools I'm going to use are called things like: the isotopy extension theorem, fibre-bundles, fibrations and homotopy-groups. You have a surface $\Sigma$, it's your shirt or whatever else you're interested in, some surface in 3-dimensional space. Surfaces have automorphism groups, let me call it $\operatorname{Aut}(\Sigma)$. These are, say, all the self-homeomorphisms or diffeomorphisms of the surface. And surfaces can sit in space. A way of putting a surface in space is called an embedding. Let's call all the embeddings of the surface $\operatorname{Emb}(\Sigma, \mathbb R^3)$. $\operatorname{Emb}(\Sigma, \mathbb R^3)$ is a set, but in the subject of topology these sets have a natural topology as well. We think of them as a space where "nearby" embeddings are almost the same, except for maybe a little wiggle here or there. The topology on the set of embeddings is called the compact-open topology (see Wikipedia, for details on most of these definitions). Okay, so now there's some formal nonsense. Look at the quotient space $\operatorname{Emb}(\Sigma, \mathbb R^3)/\operatorname{Aut}(\Sigma)$. You can think of this as all ways $\Sigma$ can sit in space, but without any labelling -- the surface has no parametrization. So it's the space of all subspaces of $\mathbb R^3$ that just happen to be homeomorphic to your surface. Richard Palais has a really nice theorem that puts this all into a pleasant context. The preamble is we need to think of everything as living in the world of smooth manifolds -- smooth embeddings, $\operatorname{Aut}(\Sigma)$ is the diffeomorphism group of the surface, etc. There are two locally-trivial fibre bundles (or something more easy to prove -- Serre fibrations), this is the "global" isotopy-extension theorem: $$\operatorname{Diff}(\mathbb R^3, \Sigma) \to \operatorname{Diff}(\mathbb R^3) \to \operatorname{Emb}(\Sigma, \mathbb R^3)/\operatorname{Aut}(\Sigma)$$ $$\operatorname{Diff}(\mathbb R^3 \operatorname{fix} \Sigma) \to \operatorname{Diff}(\mathbb R^3, \Sigma) \to \operatorname{Aut}(\Sigma)$$ here $\operatorname{Diff}(\mathbb R^3)$ indicates diffeomorphisms of $\mathbb R^3$ that are the identity outside of a sufficiently large ball, say. So the Palais theorem, together with the homotopy long exact sequence of a fibration, is giving you a language that allows you to translate between automorphisms of your surface, and motions of the surface in space. It's a theorem of Jean Cerf's that $\operatorname{Diff}(\mathbb R^3)$ is connected. A little diagram chase says that an automorphism of a surface can be realized by a motion of that surface in 3-space if and only if that automorphism of the surface extends to an automorphism of 3-space. For closed surfaces, the Jordan-Brouwer separation theorem gives you an obstruction to turning your surface inside-out. But for non-closed surfaces you're out of tools. To figure out if you can realize an automorphism as a motion, you literally have to try to extend it "by hands". This is a very general phenomena -- you have one manifold sitting in another, but rarely does an automorphism of the submanifold extend to the ambient manifold. You see this phenomena happening in various other branches of mathematics as well -- an automorphism of a subgroup does not always extend to the ambient group, etc. So you try your luck and try to build the extension yourself. In some vague sense that's a formal analogy between the visceral mystery of turning the surface inside-out and a kind of formalized mathematical problem, but of a fundamentally analogous feel. We're looking for automorphisms that reverse orientation. For an arbitrary surface with boundary in 3-space, it's not clear if you can turn the surface inside out. This is because the surface might be knotted. Unknotted surfaces are examples like your t-shirt. Let's try to cook up something that can't be turned inside-out. The automorphism group of a 3-times punctured sphere has 12 path-components (12 elements up to isotopy). There are 6 elements that preserve orientation, and 6 that reverse. In particular the orientation-reversing automorphisms reverse the orientation of all the boundary circles. So if you could come up with a knotted pair-of-pants (3-times punctured surface) so that its boundary circles did not admit a symmetry that reversed the orientations of all three circles simultaneously, you'd be done. Maybe this doesn't seem like a reduction to you, but it is. For example, there are things called non-invertible knots: http://en.wikipedia.org/wiki/Invertible_knot So how do we cook up a knotted pair-of-pants from that? Here's the idea. The non-invertible knot in the link above is sometimes called $8_{17}$. Here is another picture of it: http://katlas.org/wiki/8_17 Here is a variant on that. Interpret this image as a ribbon of paper that has three boundary circles. One boundary circle is unknotted. One is $8_{17}$. The other is some other knot. It turns out that other knot isn't trivial, nor is it $8_{17}$. So why can't this knotted pair of pants be turned inside-out? Well, the three knots are distinct, and $8_{17}$ can't be reversed. The reason why I know the other knot isn't $8_{17}$? It's a hyperbolic knot and it has a different ($4.40083...$) hyperbolic volume than $8_{17}$ ($10.9859...$). FYI: in some sense this is one of the simplest surfaces with non-trivial boundary that can't be turned inside-out. All discs can be turned inside-out. Similarly, all annuli (regardless of how they're knotted) can be turned inside-out. So for genus zero surfaces, 3 boundary components is the least you can have if you're looking for a surface that can't be turned inside-out. edited to correct for Jason's comment. comment added later: I suggest if you purchase a garment of this form you return it to the manufacturer.
| 2026-01-26T11:29:32.045000
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562,694
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Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \mathrm dx$
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I need help with this integral: $$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$ The integrand graph looks like this: $\hspace{1in}$ The approximate numeric value of the integral: $$I\approx8.372211626601275661625747121...$$ Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in WolframAlpha and ISC+ did not return plausible closed form candidates either. But I still hope there might be a closed form for it. I am also interested in cases when only numerator or only denominator is present under the logarithm.
| 632
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calculus, integration, definite-integrals, contour-integration, closed-form
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https://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1
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I will transform the integral via a substitution, break it up into two pieces and recombine, perform an integration by parts, and perform another substitution to get an integral to which I know a closed form exists. From there, I use a method I know to attack the integral, but in an unusual way because of the 8th degree polynomial in the denominator of the integrand. First sub $t=(1-x)/(1+x)$, $dt=-2/(1+x)^2 dx$ to get $$2 \int_0^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} $$ Now use the symmetry from the map $t \mapsto 1/t$. Break the integral up into two as follows: \begin{align} & 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \end{align} Sub $t=u^2$ to get $$4 \int_0^{1} \frac{du}{1-u^2} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}$$ Integrate by parts: $$\left [2 \log{\left (\frac{1+u}{1-u} \right )} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}\right ]_0^1 \\- 32 \int_0^1 du \frac{\left(u^5-6 u^3+u\right)}{\left(u^4-2 u^2+5\right) \left(5 u^4-2 u^2+1\right)} \log{\left (\frac{1+u}{1-u} \right )}$$ One last sub: $u=(v-1)/(v+1)$ $du=2/(v+1)^2 dv$, and finally get $$8 \int_0^{\infty} dv \frac{(v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v}$$ With this form, we may finally conclude that a closed form exists and apply the residue theorem to obtain it. To wit, consider the following contour integral: $$\oint_C dz \frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} \log^2{z}$$ where $C$ is a keyhole contour about the positive real axis. This contour integral is equal to (I omit the steps where I show the integral vanishes about the circular arcs) $$-i 4 \pi \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1}$$ It should be noted that the second integral vanishes; this may be easily seen by exploiting the symmetry about $v \mapsto 1/v$. On the other hand, the contour integral is $i 2 \pi$ times the sum of the residues about the poles of the integrand. In general, this requires us to find the zeroes of the eight degree polynomial, which may not be possible analytically. Here, on the other hand, we have many symmetries to exploit, e.g., if $a$ is a root, then $1/a$ is a root, $-a$ is a root, and $\bar{a}$ is a root. For example, we may deduce that $$z^8+4 z^6+70z^4+4 z^2+1 = (z^4+4 z^3+10 z^2+4 z+1) (z^4-4 z^3+10 z^2-4 z+1)$$ which exploits the $a \mapsto -a$ symmetry. Now write $$z^4+4 z^3+10 z^2+4 z+1 = (z-a)(z-\bar{a})\left (z-\frac{1}{a}\right )\left (z-\frac{1}{\bar{a}}\right )$$ Write $a=r e^{i \theta}$ and get the following equations: $$\left ( r+\frac{1}{r}\right ) \cos{\theta}=-2$$ $$\left (r^2+\frac{1}{r^2}\right) + 4 \cos^2{\theta}=10$$ From these equations, one may deduce that a solution is $r=\phi+\sqrt{\phi}$ and $\cos{\theta}=1/\phi$, where $\phi=(1+\sqrt{5})/2$ is the golden ratio. Thus the poles take the form $$z_k = \pm \left (\phi\pm\sqrt{\phi}\right) e^{\pm i \arctan{\sqrt{\phi}}}$$ Now we have to find the residues of the integrand at these 8 poles. We can break this task up by computing: $$\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1) \log^2{z}}{z^8+4 z^6+70z^4+4 z^2+1}\right ]=\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] \log^2{z_k}$$ Here things got very messy, but the result is rather unbelievably simple: $$\operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] = \text{sgn}[\cos{(\arg{z_k})}]$$ EDIT Actually, this is a very simple computation. Inspired by @sos440, one may express the rational function of $z$ in a very simple form: $$\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} = -\left [\frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right ]$$ where $$p(z)=z^4+4 z^3+10 z^2+4 z+1$$ The residue of this function at the poles are then easily seen to be $\pm 1$ according to whether the pole is a zero of $p(z)$ or $p(-z)$. END EDIT That is, if the pole has a positive real part, the residue of the fraction is $+1$; if it has a negative real part, the residue is $-1$. Now consider the log piece. Expanding the square, we get 3 terms: $$\log^2{|z_k|} - (\arg{z_k})^2 + i 2 \log{|z_k|} \arg{z_k}$$ Summing over the residues, we find that because of the $\pm1$ contributions above, that the first and third terms sum to zero. This leaves the second term. For this, it is crucial that we get the arguments right, as $\arg{z_k} \in [0,2 \pi)$. Thus, we have $$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$
| 1,075
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Q: Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \mathrm dx$
I need help with this integral: $$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$ The integrand graph looks like this: $\hspace{1in}$ The approximate numeric value of the integral: $$I\approx8.372211626601275661625747121...$$ Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in WolframAlpha and ISC+ did not return plausible closed form candidates either. But I still hope there might be a closed form for it. I am also interested in cases when only numerator or only denominator is present under the logarithm.
A: I will transform the integral via a substitution, break it up into two pieces and recombine, perform an integration by parts, and perform another substitution to get an integral to which I know a closed form exists. From there, I use a method I know to attack the integral, but in an unusual way because of the 8th degree polynomial in the denominator of the integrand. First sub $t=(1-x)/(1+x)$, $dt=-2/(1+x)^2 dx$ to get $$2 \int_0^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} $$ Now use the symmetry from the map $t \mapsto 1/t$. Break the integral up into two as follows: \begin{align} & 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \end{align} Sub $t=u^2$ to get $$4 \int_0^{1} \frac{du}{1-u^2} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}$$ Integrate by parts: $$\left [2 \log{\left (\frac{1+u}{1-u} \right )} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}\right ]_0^1 \\- 32 \int_0^1 du \frac{\left(u^5-6 u^3+u\right)}{\left(u^4-2 u^2+5\right) \left(5 u^4-2 u^2+1\right)} \log{\left (\frac{1+u}{1-u} \right )}$$ One last sub: $u=(v-1)/(v+1)$ $du=2/(v+1)^2 dv$, and finally get $$8 \int_0^{\infty} dv \frac{(v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v}$$ With this form, we may finally conclude that a closed form exists and apply the residue theorem to obtain it. To wit, consider the following contour integral: $$\oint_C dz \frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} \log^2{z}$$ where $C$ is a keyhole contour about the positive real axis. This contour integral is equal to (I omit the steps where I show the integral vanishes about the circular arcs) $$-i 4 \pi \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1}$$ It should be noted that the second integral vanishes; this may be easily seen by exploiting the symmetry about $v \mapsto 1/v$. On the other hand, the contour integral is $i 2 \pi$ times the sum of the residues about the poles of the integrand. In general, this requires us to find the zeroes of the eight degree polynomial, which may not be possible analytically. Here, on the other hand, we have many symmetries to exploit, e.g., if $a$ is a root, then $1/a$ is a root, $-a$ is a root, and $\bar{a}$ is a root. For example, we may deduce that $$z^8+4 z^6+70z^4+4 z^2+1 = (z^4+4 z^3+10 z^2+4 z+1) (z^4-4 z^3+10 z^2-4 z+1)$$ which exploits the $a \mapsto -a$ symmetry. Now write $$z^4+4 z^3+10 z^2+4 z+1 = (z-a)(z-\bar{a})\left (z-\frac{1}{a}\right )\left (z-\frac{1}{\bar{a}}\right )$$ Write $a=r e^{i \theta}$ and get the following equations: $$\left ( r+\frac{1}{r}\right ) \cos{\theta}=-2$$ $$\left (r^2+\frac{1}{r^2}\right) + 4 \cos^2{\theta}=10$$ From these equations, one may deduce that a solution is $r=\phi+\sqrt{\phi}$ and $\cos{\theta}=1/\phi$, where $\phi=(1+\sqrt{5})/2$ is the golden ratio. Thus the poles take the form $$z_k = \pm \left (\phi\pm\sqrt{\phi}\right) e^{\pm i \arctan{\sqrt{\phi}}}$$ Now we have to find the residues of the integrand at these 8 poles. We can break this task up by computing: $$\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1) \log^2{z}}{z^8+4 z^6+70z^4+4 z^2+1}\right ]=\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] \log^2{z_k}$$ Here things got very messy, but the result is rather unbelievably simple: $$\operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] = \text{sgn}[\cos{(\arg{z_k})}]$$ EDIT Actually, this is a very simple computation. Inspired by @sos440, one may express the rational function of $z$ in a very simple form: $$\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} = -\left [\frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right ]$$ where $$p(z)=z^4+4 z^3+10 z^2+4 z+1$$ The residue of this function at the poles are then easily seen to be $\pm 1$ according to whether the pole is a zero of $p(z)$ or $p(-z)$. END EDIT That is, if the pole has a positive real part, the residue of the fraction is $+1$; if it has a negative real part, the residue is $-1$. Now consider the log piece. Expanding the square, we get 3 terms: $$\log^2{|z_k|} - (\arg{z_k})^2 + i 2 \log{|z_k|} \arg{z_k}$$ Summing over the residues, we find that because of the $\pm1$ contributions above, that the first and third terms sum to zero. This leaves the second term. For this, it is crucial that we get the arguments right, as $\arg{z_k} \in [0,2 \pi)$. Thus, we have $$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$
| 2026-01-26T11:29:33.677000
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11,669
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Mathematical difference between white and black notes in a piano
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The division of the chromatic scale in $7$ natural notes (white keys in a piano) and $5$ accidental ones (black) seems a bit arbitrary to me. Apparently, adjacent notes in a piano (including white or black) are always separated by a semitone. Why the distinction, then? Why not just have scales with $12$ notes? (apparently there's a musical scale called Swara that does just that) I've asked several musician friends, but they lack the math skills to give me a valid answer. "Notes are like that because they are like that." I need some mathematician with musical knowledge (or a musician with mathematical knowledge) to help me out with this. Mathematically, is there any difference between white and black notes, or do we make the distinction for historical reasons only?
| 601
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music-theory
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https://math.stackexchange.com/questions/11669/mathematical-difference-between-white-and-black-notes-in-a-piano
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The first thing you have to understand is that notes are not uniquely defined. Everything depends on what tuning you use. I'll assume we're talking about equal temperament here. In equal temperament, a half-step is the same as a frequency ratio of $\sqrt[12]{2}$; that way, twelve half-steps makes up an octave. Why twelve? At the end of the day, what we want out of our musical frequencies are nice ratios of small integers. For example, a perfect fifth is supposed to correspond to a frequency ratio of $3 : 2$, or $1.5 : 1$, but in equal temperament it doesn't; instead, it corresponds to a ratio of $2^{ \frac{7}{12} } : 1 \approx 1.498 : 1$. As you can see, this is not a fifth; however, it is quite close. Similarly, a perfect fourth is supposed to correspond to a frequency ratio of $4 : 3$, or $1.333... : 1$, but in equal temperament it corresponds to a ratio of $2^{ \frac{5}{12} } : 1 \approx 1.335 : 1$. Again, this is not a perfect fourth, but is quite close. And so on. What's going on here is a massively convenient mathematical coincidence: several of the powers of $\sqrt[12]{2}$ happen to be good approximations to ratios of small integers, and there are enough of these to play Western music. Here's how this coincidence works. You get the white keys from $C$ using (part of) the circle of fifths. Start with $C$ and go up a fifth to get $G$, then $D$, then $A$, then $E$, then $B$. Then go down a fifth to get $F$. These are the "neighbors" of $C$ in the circle of fifths. You get the black keys from here using the rest of the circle of fifths. After you've gone up a "perfect" perfect fifth twelve times, you get a frequency ratio of $3^{12} : 2^{12} \approx 129.7 : 1$. This happens to be rather close to $2^7 : 1$, or seven octaves! And if we replace $3 : 2$ by $2^{ \frac{7}{12} } : 1$, then we get exactly seven octaves. In other words, the reason you can afford to identify these intervals is because $3^{12}$ happens to be rather close to $2^{19}$. Said another way, $$\log_2 3 \approx \frac{19}{12}$$ happens to be a good rational approximation, and this is the main basis of equal temperament. (The other main coincidence here is that $\log_2 \frac{5}{4} \approx \frac{4}{12}$; this is what allows us to squeeze major thirds into equal temperament as well.) It is a fundamental fact of mathematics that $\log_2 3$ is irrational, so it is impossible for any kind of equal temperament to have "perfect" perfect fifths regardless of how many notes you use. However, you can write down good rational approximations by looking at the continued fraction of $\log_2 3$ and writing down convergents, and these will correspond to equal-tempered scales with more notes. Of course, you can use other types of temperament, such as well temperament; if you stick to $12$ notes (which not everybody does!), you will be forced to make some intervals sound better and some intervals sound worse. In particular, if you don't use equal temperament then different keys sound different. This is a major reason many Western composers composed in different keys; during their time, this actually made a difference. As a result when you're playing certain sufficiently old pieces you aren't actually playing them as they were intended to be heard - you're using the wrong tuning. Edit: I suppose it is also good to say something about why we care about frequency ratios which are ratios of small integers. This has to do with the physics of sound, and I'm not particularly knowledgeable here, but this is my understanding of the situation. You probably know that sound is a wave. More precisely, sound is a longitudinal wave carried by air molecules. You might think that there is a simple equation for the sound created by a single note, perhaps $\sin 2\pi f t$ if the corresponding tone has frequency $f$. Actually this only occurs for tones which are produced electronically; any tone you produce in nature carries with it overtones and has a Fourier series $$\sum \left( a_n \sin 2 \pi n f t + b_n \cos 2 \pi n f t \right)$$ where the coefficients $a_n, b_n$ determine the timbre of the sound; this is why different instruments sound different even when they play the same notes, and has to do with the physics of vibration, which I don't understand too well. So any tone which you hear at frequency $f$ almost certainly also has components at frequency $2f, 3f, 4f, ...$. If you play two notes of frequencies $f, f'$ together, then the resulting sound corresponds to what you get when you add their Fourier series. Now it's not hard to see that if $\frac{f}{f'}$ is a ratio of small integers, then many (but not all) of the overtones will match in frequency with each other; the result sounds a more complex note with certain overtones. Otherwise, you get dissonance as you hear both types of overtones simultaneously and their frequencies will be similar, but not similar enough. Edit: You should probably check out David Benson's "Music: A Mathematical Offering", the book Rahul Narain recommended in the comments for the full story. There was a lot I didn't know, and I'm only in the introduction!
| 562
| true
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Q: Mathematical difference between white and black notes in a piano
The division of the chromatic scale in $7$ natural notes (white keys in a piano) and $5$ accidental ones (black) seems a bit arbitrary to me. Apparently, adjacent notes in a piano (including white or black) are always separated by a semitone. Why the distinction, then? Why not just have scales with $12$ notes? (apparently there's a musical scale called Swara that does just that) I've asked several musician friends, but they lack the math skills to give me a valid answer. "Notes are like that because they are like that." I need some mathematician with musical knowledge (or a musician with mathematical knowledge) to help me out with this. Mathematically, is there any difference between white and black notes, or do we make the distinction for historical reasons only?
A: The first thing you have to understand is that notes are not uniquely defined. Everything depends on what tuning you use. I'll assume we're talking about equal temperament here. In equal temperament, a half-step is the same as a frequency ratio of $\sqrt[12]{2}$; that way, twelve half-steps makes up an octave. Why twelve? At the end of the day, what we want out of our musical frequencies are nice ratios of small integers. For example, a perfect fifth is supposed to correspond to a frequency ratio of $3 : 2$, or $1.5 : 1$, but in equal temperament it doesn't; instead, it corresponds to a ratio of $2^{ \frac{7}{12} } : 1 \approx 1.498 : 1$. As you can see, this is not a fifth; however, it is quite close. Similarly, a perfect fourth is supposed to correspond to a frequency ratio of $4 : 3$, or $1.333... : 1$, but in equal temperament it corresponds to a ratio of $2^{ \frac{5}{12} } : 1 \approx 1.335 : 1$. Again, this is not a perfect fourth, but is quite close. And so on. What's going on here is a massively convenient mathematical coincidence: several of the powers of $\sqrt[12]{2}$ happen to be good approximations to ratios of small integers, and there are enough of these to play Western music. Here's how this coincidence works. You get the white keys from $C$ using (part of) the circle of fifths. Start with $C$ and go up a fifth to get $G$, then $D$, then $A$, then $E$, then $B$. Then go down a fifth to get $F$. These are the "neighbors" of $C$ in the circle of fifths. You get the black keys from here using the rest of the circle of fifths. After you've gone up a "perfect" perfect fifth twelve times, you get a frequency ratio of $3^{12} : 2^{12} \approx 129.7 : 1$. This happens to be rather close to $2^7 : 1$, or seven octaves! And if we replace $3 : 2$ by $2^{ \frac{7}{12} } : 1$, then we get exactly seven octaves. In other words, the reason you can afford to identify these intervals is because $3^{12}$ happens to be rather close to $2^{19}$. Said another way, $$\log_2 3 \approx \frac{19}{12}$$ happens to be a good rational approximation, and this is the main basis of equal temperament. (The other main coincidence here is that $\log_2 \frac{5}{4} \approx \frac{4}{12}$; this is what allows us to squeeze major thirds into equal temperament as well.) It is a fundamental fact of mathematics that $\log_2 3$ is irrational, so it is impossible for any kind of equal temperament to have "perfect" perfect fifths regardless of how many notes you use. However, you can write down good rational approximations by looking at the continued fraction of $\log_2 3$ and writing down convergents, and these will correspond to equal-tempered scales with more notes. Of course, you can use other types of temperament, such as well temperament; if you stick to $12$ notes (which not everybody does!), you will be forced to make some intervals sound better and some intervals sound worse. In particular, if you don't use equal temperament then different keys sound different. This is a major reason many Western composers composed in different keys; during their time, this actually made a difference. As a result when you're playing certain sufficiently old pieces you aren't actually playing them as they were intended to be heard - you're using the wrong tuning. Edit: I suppose it is also good to say something about why we care about frequency ratios which are ratios of small integers. This has to do with the physics of sound, and I'm not particularly knowledgeable here, but this is my understanding of the situation. You probably know that sound is a wave. More precisely, sound is a longitudinal wave carried by air molecules. You might think that there is a simple equation for the sound created by a single note, perhaps $\sin 2\pi f t$ if the corresponding tone has frequency $f$. Actually this only occurs for tones which are produced electronically; any tone you produce in nature carries with it overtones and has a Fourier series $$\sum \left( a_n \sin 2 \pi n f t + b_n \cos 2 \pi n f t \right)$$ where the coefficients $a_n, b_n$ determine the timbre of the sound; this is why different instruments sound different even when they play the same notes, and has to do with the physics of vibration, which I don't understand too well. So any tone which you hear at frequency $f$ almost certainly also has components at frequency $2f, 3f, 4f, ...$. If you play two notes of frequencies $f, f'$ together, then the resulting sound corresponds to what you get when you add their Fourier series. Now it's not hard to see that if $\frac{f}{f'}$ is a ratio of small integers, then many (but not all) of the overtones will match in frequency with each other; the result sounds a more complex note with certain overtones. Otherwise, you get dissonance as you hear both types of overtones simultaneously and their frequencies will be similar, but not similar enough. Edit: You should probably check out David Benson's "Music: A Mathematical Offering", the book Rahul Narain recommended in the comments for the full story. There was a lot I didn't know, and I'm only in the introduction!
| 2026-01-26T11:29:35.209000
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75,130
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How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
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How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution. This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.
| 567
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calculus, limits, trigonometry, limits-without-lhopital
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https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1
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The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$ \cos(x)\le\frac{\sin(x)}{x}\le1\tag{2} $$ Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3} $$ Also, dividing $(2)$ by $\cos(x)$, we get that $$ 1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4} $$ Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$
| 671
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Q: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution. This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.
A: The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$ \cos(x)\le\frac{\sin(x)}{x}\le1\tag{2} $$ Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3} $$ Also, dividing $(2)$ by $\cos(x)$, we get that $$ 1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4} $$ Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$
| 2026-01-26T11:29:36.800000
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154
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Do complex numbers really exist?
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Complex numbers involve the square root of negative one, and most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning. What's the best way to explain to a non-mathematician that complex numbers are necessary and meaningful, in the same way that real numbers are? This is not a Platonic question about the reality of mathematics, or whether abstractions are as real as physical entities, but an attempt to bridge a comprehension gap that many people experience when encountering complex numbers for the first time. The wording, although provocative, is deliberately designed to match the way that many people actually ask this question.
| 546
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soft-question, complex-numbers, education, philosophy
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https://math.stackexchange.com/questions/154/do-complex-numbers-really-exist
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There are a few good answers to this question, depending on the audience. I've used all of these on occasion. A way to solve polynomials We came up with equations like $x - 5 = 0$, what is $x$?, and the naturals solved them (easily). Then we asked, "wait, what about $x + 5 = 0$?" So we invented negative numbers. Then we asked "wait, what about $2x = 1$?" So we invented rational numbers. Then we asked "wait, what about $x^2 = 2$?" so we invented irrational numbers. Finally, we asked, "wait, what about $x^2 = -1$?" This is the only question that was left, so we decided to invent the "imaginary" numbers to solve it. All the other numbers, at some point, didn't exist and didn't seem "real", but now they're fine. Now that we have imaginary numbers, we can solve every polynomial, so it makes sense that that's the last place to stop. Pairs of numbers This explanation goes the route of redefinition. Tell the listener to forget everything he or she knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good "realistic" use of pairs of numbers (many exist). Then, show that in this system, $(0,1) * (0,1) = (-1,0)$, in other words, we've defined a new system, under which it makes sense to say that $\sqrt{-1} = i$, when $i=(0,1)$. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there is an answer to $\sqrt{-1}$. The historical explanation Explain the history of the imaginary numbers. Showing that mathematicians also fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end. I'm a little rusty, but I think there were certain equations that kept having parts of them which used $\sqrt{-1}$, and the mathematicians kept throwing out the equations since there is no such thing. Then, one mathematician decided to just "roll with it", and kept working, and found out that all those square roots cancelled each other out. Amazingly, the answer that was left was the correct answer (he was working on finding roots of polynomials, I think). Which lead him to think that there was a valid reason to use $\sqrt{-1}$, even if it took a long time to understand it.
| 378
| true
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Q: Do complex numbers really exist?
Complex numbers involve the square root of negative one, and most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning. What's the best way to explain to a non-mathematician that complex numbers are necessary and meaningful, in the same way that real numbers are? This is not a Platonic question about the reality of mathematics, or whether abstractions are as real as physical entities, but an attempt to bridge a comprehension gap that many people experience when encountering complex numbers for the first time. The wording, although provocative, is deliberately designed to match the way that many people actually ask this question.
A: There are a few good answers to this question, depending on the audience. I've used all of these on occasion. A way to solve polynomials We came up with equations like $x - 5 = 0$, what is $x$?, and the naturals solved them (easily). Then we asked, "wait, what about $x + 5 = 0$?" So we invented negative numbers. Then we asked "wait, what about $2x = 1$?" So we invented rational numbers. Then we asked "wait, what about $x^2 = 2$?" so we invented irrational numbers. Finally, we asked, "wait, what about $x^2 = -1$?" This is the only question that was left, so we decided to invent the "imaginary" numbers to solve it. All the other numbers, at some point, didn't exist and didn't seem "real", but now they're fine. Now that we have imaginary numbers, we can solve every polynomial, so it makes sense that that's the last place to stop. Pairs of numbers This explanation goes the route of redefinition. Tell the listener to forget everything he or she knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good "realistic" use of pairs of numbers (many exist). Then, show that in this system, $(0,1) * (0,1) = (-1,0)$, in other words, we've defined a new system, under which it makes sense to say that $\sqrt{-1} = i$, when $i=(0,1)$. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there is an answer to $\sqrt{-1}$. The historical explanation Explain the history of the imaginary numbers. Showing that mathematicians also fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end. I'm a little rusty, but I think there were certain equations that kept having parts of them which used $\sqrt{-1}$, and the mathematicians kept throwing out the equations since there is no such thing. Then, one mathematician decided to just "roll with it", and kept working, and found out that all those square roots cancelled each other out. Amazingly, the answer that was left was the correct answer (he was working on finding roots of polynomials, I think). Which lead him to think that there was a valid reason to use $\sqrt{-1}$, even if it took a long time to understand it.
| 2026-01-26T11:29:38.544000
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302,023
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Best Sets of Lecture Notes and Articles
|
Let me start by apologizing if there is another thread on math.se that subsumes this. I was updating my answer to the question here during which I made the claim that "I spend a lot of time sifting through books to find [the best source]". It strikes me now that while I love books (I really do), I often find that I learn best from sets of lecture notes and short articles. There are three particular reasons that make me feel this way. $1.$ Lecture notes and articles often times take on a very delightful informal approach. They generally take time to bring to the reader's attention some interesting side fact that would normally be left out of a standard textbook (lest it be too big). Lecture notes and articles are where one generally picks up on historical context, overarching themes (the "birds eye view"), and neat interrelations between subjects. $2.$ It is the informality that often allows writers of lecture notes or expository articles to mention some "trivial fact" that every textbook leaves out. Whenever I have one of those moments where a definition just doesn't make sense, or a theorem just doesn't seem right it's invariably a set of lecture notes that sets everything straight for me. People tend to be more honest in lecture notes, to admit that a certain definition or idea confused them when they first learned it, and to take the time to help you understand what finally enabled them to make the jump. $3.$ Often times books are very outdated. It takes a long time to write a book, to polish it to the point where it is ready for publication. Notes often times are closer to the heart of research, closer to how things are learned in the modern sense. It is because of reasons like this that I find myself more and more carrying around a big thick manila folder full of stapled together articles and why I keep making trips to Staples to get the latest set of notes bound. So, if anyone knows of any set of lecture notes, or any expository articles that fit the above criteria, please do share! I'll start: People/Places who have a huge array of fantastic notes: K Conrad Pete L Clark Milne Stein Igusa Hatcher Andrew Baker (Contributed by Andrew) Garrett (Contributed by Andrew) Frederique (Contributed by Mohan) Ash B Conrad Matthew Emerton (not technically notes, but easily one of the best reads out there). Geraschenko A collection of the "What is..." articles in the Notices Brian Osserman ALGANT Masters Theses (an absolutely stupendous collection of masters theses in various aspects of algebraic geometry/algebraic number theory). The Stacks Project (an open source 'textbook' with the goal in mind to have a completely self-contained exposition of the theory of stacks. Because such a huge amount of background is required, it contains detailed articles about commutative algebra, homological algebra, set theory, topology, category theory, sheaf theory, algebraic geometry, etc.). Harvard undergraduate theses (an excellent collection of the mathematics undergraduate theses completed in the last few years at Harvard). Bas Edixhoven (this is a list of notes from talks that Edixhoven has given over the years). Model Theory: The Model Theory of Fields-Marker Number Theory: Algebraic Number Theory-Conrad Algebraic Number Theory-Weston Class Field Theory-Lemmermeyer Compilation of Notes from Things of Interest to Number Theorists Elliptic Modular Forms-Don Zagier Modular Forms-Martin What is a Reciprocity Law?-Wyman Class Field Theory Summarized-Garbanati Three Lectures About the Arithmetic of Elliptic Curves-Mazur Congruences Between Modular Forms-Calegari Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture-Rubin Simple Proof of Kronecker Weber-Ordulu Tate's Thesis-Binder Introduction to Tate's Thesis-Leahy [A Summary of CM Theory of Elliptic Curves-Getz] An Elementary Introduction to the Langland's Program-Gelbart $p$-adic Analysis Compared to Real Analysis-Katok (Contributed by Andrew; no longer on-line - but here is a snapshot from the Wayback Machine) Representation of $p$-adic Groups-Vinroot Counting Special Points: Logic, Diophantine Geometry, and Transcendence Theory-Scanlon Algebraic Number Theory-Holden The Theory of Witt Vectors-Rabinoff Complex Geometry: Complex Analytic and Differential Geometry-Demailly Weighted $L^2$ Estimes for the $\bar{\partial}$ Operator on a Complex Manifold Demailly Uniformization Theorem-Chan Analytic Vector Bundles-Andrew (These notes are truly amazing) Complex Manifolds-Koppensteiner Kahler Geometry and Hodge Theory-Biquard and Horing Kahler Geometry-Speyer Differential Topology/Geometry: Differential Topology-Dundas Spaces and Questions-Gromov Introduction to Cobordism-Weston The Local Structure of Smooth Maps of Manifolds-Bloom Groups Acting on the Circle-Ghys Lie Groups-Ban (comes with accompanying lecture videos) Very Basic Lie Theory-Howe Differential Geometry of Curves and Surfaces-Shifrin (Contributed by Andrew) A Visual Introduction to Riemannian Curvatures and Some Discrete Generlizations-Ollivier Algebra: Geometric Group Theory-Bowditch Categories and Homological Algebra-Schapira Category Theory-Leinster (Contributed by Bruno Stonek) Category Theory-Chen (Contributed by Bruno Stonek) Commutative Algebra-Altman and Klein (Contributed by Andrew) Finite Group Representation Theory-Bartel (Contributed by Mohan) Representation Theory-Etingof Commutative Algebra-Haines Geometric Commutative Algebra-Arrondo Examples in Category Theory-Calugereanu and Purdea Topology Homotopy Theories and Model Categories-Dwyer and Spalinski (Contributed by Elden Elmanto) Algebraic Geometry: Foundations of Algebraic Geometry-Vakil Analytic Techniques in Algebraic Geometry-Demailly Algebraic Geometry-Gathmann (Contributed by Mohan) Oda and Mumford's Algebraic Geometry Notes (Pt. II) Galois Theory for Schemes-Lenstra Rational Points on Varieties-Poonen Teaching Schemes-Mazur NOTE: This may come in handy for those who, like me, don't like a metric ton of PDFs associated to a single document: https://www.pdfmerge.com/
| 538
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self-learning, big-list, learning, online-resources
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https://math.stackexchange.com/questions/302023/best-sets-of-lecture-notes-and-articles
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In no particular order: Algebraic number theory notes by Sharifi: http://math.arizona.edu/~sharifi/algnum.pdf Dalawat's first course in local arithmetic: http://arxiv.org/abs/0903.2615 Intro to top grps: http://www.mat.ucm.es/imi/documents/20062007_Dikran.pdf Representation theory resources: http://www.math.columbia.edu/~khovanov/resources/ Classical invariant theory: http://jones.math.unibas.ch/~kraft/Papers/KP-Primer.pdf CRing project: http://people.fas.harvard.edu/~amathew/CRing.pdf - The notes are huge & has many authors - including MSE's Zev, Akhil (no longer active) & Darij. Check the ToC. Partitions bijections, a survey: http://www.math.ucla.edu/~pak/papers/psurvey.pdf Hidden subgroup problem (review, open stuff): http://arxiv.org/abs/quant-ph/0411037 Spirit of moonshine: http://www.math.harvard.edu/theses/senior/booher/booher.pdf Vertex operator algebras and modular forms: http://arxiv.org/abs/0909.4460 Categorified algebra & quantum mechanics: http://arxiv.org/abs/math/0601458 Exponential sums over finite fields: http://www.math.ethz.ch/~kowalski/exp-sums.pdf Gauss sums: http://math.mit.edu/~brubaker/houghexpsums05.pdf Adeles over $\Bbb Q$: https://www.maths.nottingham.ac.uk/personal/ibf/text/gl1.pdf, followed by automo reps over GL(1,A) https://www.maths.nottingham.ac.uk/personal/ibf/text/gl2.pdf Invariant thry: http://www.win.tue.nl/~jdraisma/teaching/invtheory0910/lecturenotes11.pdf Species: http://www.newton.ac.uk/programmes/CSM/Abstract3/Species_intro.pdf FLT: http://www.math.mcgill.ca/darmon/pub/Articles/Expository/05.DDT/paper.pdf Categorical concepts: http://www.math.harvard.edu/~eriehl/266x/survey.pdf Groups, Rings, Fields (Lenstra): http://websites.math.leidenuniv.nl/algebra/topics.pdf, which is part of algebra notes: http://websites.math.leidenuniv.nl/algebra/ If we're going to mention Hatcher (famous to me for the algebraic topology notes), we might as well also mention a few other books that are online, like Algebra chapter 0, Stanley's insane first volume of Enumerative Combinatorics (which reminds me: generatingfunctionology). Also I don't see topology without tears mentioned. The sheer number of books and notes on differential geometry and lie theory is mind-boggling, so I'll have to update later with the juicier ones. Let's not forget the AMS notes online back through 1995 - they're very nice reading as well.
| 72
| false
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Q: Best Sets of Lecture Notes and Articles
Let me start by apologizing if there is another thread on math.se that subsumes this. I was updating my answer to the question here during which I made the claim that "I spend a lot of time sifting through books to find [the best source]". It strikes me now that while I love books (I really do), I often find that I learn best from sets of lecture notes and short articles. There are three particular reasons that make me feel this way. $1.$ Lecture notes and articles often times take on a very delightful informal approach. They generally take time to bring to the reader's attention some interesting side fact that would normally be left out of a standard textbook (lest it be too big). Lecture notes and articles are where one generally picks up on historical context, overarching themes (the "birds eye view"), and neat interrelations between subjects. $2.$ It is the informality that often allows writers of lecture notes or expository articles to mention some "trivial fact" that every textbook leaves out. Whenever I have one of those moments where a definition just doesn't make sense, or a theorem just doesn't seem right it's invariably a set of lecture notes that sets everything straight for me. People tend to be more honest in lecture notes, to admit that a certain definition or idea confused them when they first learned it, and to take the time to help you understand what finally enabled them to make the jump. $3.$ Often times books are very outdated. It takes a long time to write a book, to polish it to the point where it is ready for publication. Notes often times are closer to the heart of research, closer to how things are learned in the modern sense. It is because of reasons like this that I find myself more and more carrying around a big thick manila folder full of stapled together articles and why I keep making trips to Staples to get the latest set of notes bound. So, if anyone knows of any set of lecture notes, or any expository articles that fit the above criteria, please do share! I'll start: People/Places who have a huge array of fantastic notes: K Conrad Pete L Clark Milne Stein Igusa Hatcher Andrew Baker (Contributed by Andrew) Garrett (Contributed by Andrew) Frederique (Contributed by Mohan) Ash B Conrad Matthew Emerton (not technically notes, but easily one of the best reads out there). Geraschenko A collection of the "What is..." articles in the Notices Brian Osserman ALGANT Masters Theses (an absolutely stupendous collection of masters theses in various aspects of algebraic geometry/algebraic number theory). The Stacks Project (an open source 'textbook' with the goal in mind to have a completely self-contained exposition of the theory of stacks. Because such a huge amount of background is required, it contains detailed articles about commutative algebra, homological algebra, set theory, topology, category theory, sheaf theory, algebraic geometry, etc.). Harvard undergraduate theses (an excellent collection of the mathematics undergraduate theses completed in the last few years at Harvard). Bas Edixhoven (this is a list of notes from talks that Edixhoven has given over the years). Model Theory: The Model Theory of Fields-Marker Number Theory: Algebraic Number Theory-Conrad Algebraic Number Theory-Weston Class Field Theory-Lemmermeyer Compilation of Notes from Things of Interest to Number Theorists Elliptic Modular Forms-Don Zagier Modular Forms-Martin What is a Reciprocity Law?-Wyman Class Field Theory Summarized-Garbanati Three Lectures About the Arithmetic of Elliptic Curves-Mazur Congruences Between Modular Forms-Calegari Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture-Rubin Simple Proof of Kronecker Weber-Ordulu Tate's Thesis-Binder Introduction to Tate's Thesis-Leahy [A Summary of CM Theory of Elliptic Curves-Getz] An Elementary Introduction to the Langland's Program-Gelbart $p$-adic Analysis Compared to Real Analysis-Katok (Contributed by Andrew; no longer on-line - but here is a snapshot from the Wayback Machine) Representation of $p$-adic Groups-Vinroot Counting Special Points: Logic, Diophantine Geometry, and Transcendence Theory-Scanlon Algebraic Number Theory-Holden The Theory of Witt Vectors-Rabinoff Complex Geometry: Complex Analytic and Differential Geometry-Demailly Weighted $L^2$ Estimes for the $\bar{\partial}$ Operator on a Complex Manifold Demailly Uniformization Theorem-Chan Analytic Vector Bundles-Andrew (These notes are truly amazing) Complex Manifolds-Koppensteiner Kahler Geometry and Hodge Theory-Biquard and Horing Kahler Geometry-Speyer Differential Topology/Geometry: Differential Topology-Dundas Spaces and Questions-Gromov Introduction to Cobordism-Weston The Local Structure of Smooth Maps of Manifolds-Bloom Groups Acting on the Circle-Ghys Lie Groups-Ban (comes with accompanying lecture videos) Very Basic Lie Theory-Howe Differential Geometry of Curves and Surfaces-Shifrin (Contributed by Andrew) A Visual Introduction to Riemannian Curvatures and Some Discrete Generlizations-Ollivier Algebra: Geometric Group Theory-Bowditch Categories and Homological Algebra-Schapira Category Theory-Leinster (Contributed by Bruno Stonek) Category Theory-Chen (Contributed by Bruno Stonek) Commutative Algebra-Altman and Klein (Contributed by Andrew) Finite Group Representation Theory-Bartel (Contributed by Mohan) Representation Theory-Etingof Commutative Algebra-Haines Geometric Commutative Algebra-Arrondo Examples in Category Theory-Calugereanu and Purdea Topology Homotopy Theories and Model Categories-Dwyer and Spalinski (Contributed by Elden Elmanto) Algebraic Geometry: Foundations of Algebraic Geometry-Vakil Analytic Techniques in Algebraic Geometry-Demailly Algebraic Geometry-Gathmann (Contributed by Mohan) Oda and Mumford's Algebraic Geometry Notes (Pt. II) Galois Theory for Schemes-Lenstra Rational Points on Varieties-Poonen Teaching Schemes-Mazur NOTE: This may come in handy for those who, like me, don't like a metric ton of PDFs associated to a single document: https://www.pdfmerge.com/
A: In no particular order: Algebraic number theory notes by Sharifi: http://math.arizona.edu/~sharifi/algnum.pdf Dalawat's first course in local arithmetic: http://arxiv.org/abs/0903.2615 Intro to top grps: http://www.mat.ucm.es/imi/documents/20062007_Dikran.pdf Representation theory resources: http://www.math.columbia.edu/~khovanov/resources/ Classical invariant theory: http://jones.math.unibas.ch/~kraft/Papers/KP-Primer.pdf CRing project: http://people.fas.harvard.edu/~amathew/CRing.pdf - The notes are huge & has many authors - including MSE's Zev, Akhil (no longer active) & Darij. Check the ToC. Partitions bijections, a survey: http://www.math.ucla.edu/~pak/papers/psurvey.pdf Hidden subgroup problem (review, open stuff): http://arxiv.org/abs/quant-ph/0411037 Spirit of moonshine: http://www.math.harvard.edu/theses/senior/booher/booher.pdf Vertex operator algebras and modular forms: http://arxiv.org/abs/0909.4460 Categorified algebra & quantum mechanics: http://arxiv.org/abs/math/0601458 Exponential sums over finite fields: http://www.math.ethz.ch/~kowalski/exp-sums.pdf Gauss sums: http://math.mit.edu/~brubaker/houghexpsums05.pdf Adeles over $\Bbb Q$: https://www.maths.nottingham.ac.uk/personal/ibf/text/gl1.pdf, followed by automo reps over GL(1,A) https://www.maths.nottingham.ac.uk/personal/ibf/text/gl2.pdf Invariant thry: http://www.win.tue.nl/~jdraisma/teaching/invtheory0910/lecturenotes11.pdf Species: http://www.newton.ac.uk/programmes/CSM/Abstract3/Species_intro.pdf FLT: http://www.math.mcgill.ca/darmon/pub/Articles/Expository/05.DDT/paper.pdf Categorical concepts: http://www.math.harvard.edu/~eriehl/266x/survey.pdf Groups, Rings, Fields (Lenstra): http://websites.math.leidenuniv.nl/algebra/topics.pdf, which is part of algebra notes: http://websites.math.leidenuniv.nl/algebra/ If we're going to mention Hatcher (famous to me for the algebraic topology notes), we might as well also mention a few other books that are online, like Algebra chapter 0, Stanley's insane first volume of Enumerative Combinatorics (which reminds me: generatingfunctionology). Also I don't see topology without tears mentioned. The sheer number of books and notes on differential geometry and lie theory is mind-boggling, so I'll have to update later with the juicier ones. Let's not forget the AMS notes online back through 1995 - they're very nice reading as well.
| 2026-01-26T11:29:40.041000
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206,890
|
"The Egg:" Bizarre behavior of the roots of a family of polynomials.
|
In this MO post, I ran into the following family of polynomials: $$f_n(x)=\sum_{m=0}^{n}\prod_{k=0}^{m-1}\frac{x^n-x^k}{x^m-x^k}.$$ In the context of the post, $x$ was a prime number, and $f_n(x)$ counted the number of subspaces of an $n$-dimensional vector space over $GF(x)$ (which I was using to determine the number of subgroups of an elementary abelian group $E_{x^n}$). Anyway, while I was investigating asymptotic behavior of $f_n(x)$ in Mathematica, I got sidetracked and (just for fun) looked at the set of complex roots when I set $f_n(x)=0$. For $n=24$, the plot looked like this: (The real and imaginary axes are from $-1$ to $1$.) Surprised by the unusual symmetry of the solutions, I made the same plot for a few more values of $n$. Note the clearly defined "tails" (on the left when even, top and bottom when odd) and "cusps" (both sides). You can see that after approximately $n=60$, the "circle" of solutions starts to expand into a band of solutions with a defined outline. To fully absorb the weirdness of this, I animated the solutions from $n=2$ to $n=112$. The following is the result: Pretty weird, right!? Anyhow, here are my questions: First, has anybody ever seen anything at all like this before? What's up with those "tails?" They seem to occur only on even $n$, and they are surely distinguishable from the rest of the solutions. Look how the "enclosed" solutions rotate as $n$ increases. Why does this happen? [Explained in edits.] Anybody have any idea what happens to the solution set as $n\rightarrow \infty$? Thanks to @WillSawin, we now know that all the roots are contained in an annulus that converges to the unit circle, which is fantastic. So, the final step in understanding the limit of the solution sets is figuring out what happens on the unit circle. We can see from the animation that there are many gaps, particularly around certain roots of unity; however, they do appear to be closing. The natural question is, which points on the unit circle "are roots in the limit"? In other words, what are the accumulation points of $\{z\left|z\right|^{-1}:z\in\mathbb{C}\text{ and }f_n(z)=0\}$? Is the set of accumulation points dense? @NoahSnyder's heuristic of considering these as a random family of polynomials suggests it should be- at least, almost surely. These are polynomials in $\mathbb{Z}[x]$. Can anybody think of a way to rewrite the formula (perhaps recursively?) for the simplified polynomial, with no denominator? If so, we could use the new formula to prove the series converges to a function on the unit disc, as well as cut computation time in half. [See edits for progress.] Does anybody know a numerical method specifically for finding roots of high degree polynomials? Or any other way to efficiently compute solution sets for high $n$? [Thanks @Hooked!] Thanks everyone. This may not turn out to be particularly mathematically profound, but it sure is neat. EDIT: Thanks to suggestions in the comments, I cranked up the working precision to maximum and recalculated the animation. As Hurkyl and mercio suspected, the rotation was indeed a software artifact, and in fact evidently so was the thickening of the solution set. The new animation looks like this: So, that solves one mystery: the rotation and inflation were caused by tiny roundoff errors in the computation. With the image clearer, however, I see the behavior of the cusps more clearly. Is there an explanation for the gradual accumulation of "cusps" around the roots of unity? (Especially 1.) EDIT: Here is an animation $Arg(f_n)$ up to $n=30$. I think we can see from this that $f_n$ should converge to some function on the unit disk as $n\rightarrow \infty$. I'd love to include higher $n$, but this was already rather computationally exhausting. Now, I've been tinkering and I may be onto something with respect to point $5$ (i.e. seeking a better formula for $f_n(x)$). The folowing claims aren't proven yet, but I've checked each up to $n=100$, and they seem inductively consistent. Here denote $\displaystyle f_n(x)=\sum_{m}a_{n,m}x^m$, so that $a_{n,m}\in \mathbb{Z}$ are the coefficients in the simplified expansion of $f_n(x)$. First, I found $\text{deg}(f_n)=\text{deg}(f_{n-1})+\lfloor \frac{n}{2} \rfloor$. The solution to this recurrence relation is $$\text{deg}(f_n)=\frac{1}{2}\left({\left\lceil\frac{1-n}{2}\right\rceil}^2 -\left\lceil\frac{1-n}{2}\right\rceil+{\left\lfloor \frac{n}{2} \right\rfloor}^2 + \left\lfloor \frac{n}{2} \right\rfloor\right)=\left\lceil\frac{n^2}{4}\right\rceil.$$ If $f_n(x)$ has $r$ more coefficients than $f_{n-1}(x)$, the leading $r$ coefficients are the same as the leading $r$ coefficients of $f_{n-2}(x)$, pairwise. When $n>m$, $a_{n,m}=a_{n-1,m}+\rho(m)$, where $\rho(m)$ is the number of integer partitions of $m$. (This comes from observation, but I bet an actual proof could follow from some of the formulas here.) For $n\leq m$ the $\rho(m)$ formula first fails at $n=m=6$, and not before for some reason. There is probably a simple correction term I'm not seeing - and whatever that term is, I bet it's what's causing those cusps. Anyhow, with this, we can make almost make a recursive relation for $a_{n,m}$, $$a_{n,m}= \left\{ \begin{array}{ll} a_{n-2,m+\left\lceil\frac{n-2}{2}\right\rceil^2-\left\lceil\frac{n}{2}\right\rceil^2} & : \text{deg}(f_{n-1}) m \\ ? & : m \leq \text{deg}(f_{n-1}) \text{ and } n \leq m \end{array} \right. $$ but I can't figure out the last part yet. EDIT: Someone pointed out to me that if we write $\lim_{n\rightarrow\infty}f_n(x)=\sum_{m=0}^\infty b_{m} x^m$, then it appears that $f_n(x)=\sum_{m=0}^n b_m x^m + O(x^{n+1})$. The $b_m$ there seem to me to be relatively well approximated by the $\rho(m)$ formula, considering the correction term only applies for a finite number of recursions. So, if we have the coefficients up to an order of $O(x^{n+1})$, we can at least prove the polynomials converge on the open unit disk, which the $Arg$ animation suggests is true. (To be precise, it looks like $f_{2n}$ and $f_{2n+1}$ may have different limit functions, but I suspect the coefficients of both sequences will come from the same recursive formula.) With this in mind, I put a bounty up for the correction term, since from that all the behavior will probably be explained. EDIT: The limit function proposed by Gottfriend and Aleks has the formal expression $$\lim_{n\rightarrow \infty}f_n(x)=1+\prod_{m=1}^\infty \frac{1}{1-x^m}.$$ I made an $Arg$ plot of $1+\prod_{m=1}^r \frac{1}{1-x^m}$ for up to $r=24$ to see if I could figure out what that ought to ultimately end up looking like, and came up with this: Purely based off the plots, it seems not entirely unlikely that $f_n(x)$ is going to the same place this is, at least inside the unit disc. Now the question is, how do we determine the solution set at the limit? I speculate that the unit circle may become a dense combination of zeroes and singularities, with fractal-like concentric "circles of singularity" around the roots of unity... :)
| 529
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abstract-algebra, complex-analysis, algebraic-geometry, numerical-methods, recreational-mathematics
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https://math.stackexchange.com/questions/206890/the-egg-bizarre-behavior-of-the-roots-of-a-family-of-polynomials
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First, has anybody ever seen anything at all like this before? Yes, and in fact the interesting patterns that arise here are more than just a mathematical curiosity, they can be interpreted to have a physical context. Statistical Mechanics In a simple spin system, say the Ising model, a discrete set of points are arranged on a grid. In physics, we like to define the energy of the system by the Hamiltonian, which gives the energy of any particular microstate. In this system, if the spins are aligned they form a bond. This favorable and the energy is negative. If they are misaligned, the energy is positive. Let's consider a simple system of two points, adjacent to each other. Furthermore, let each site point up (1) or down (-1). For an Ising-like system we would write the Hamiltonian as: $$ H = - \sum_{ij} J \sigma_i \sigma_j $$ where $\sigma_i$ is the spin of the $i$th point and the summation runs over all pairs of adjacent sites. $J$ is the strength of the bond (which we can set to one for our example). In our simple system we have only four possible states: 0 - 0 H = -J 1 - 0 H = 0 0 - 1 H = 0 1 - 1 H = -J Now we can write the partition function $\mathcal{Z}$, a term which encompasses all information of the Hamiltonian from the perspective of statistical mechanics: $$ \mathcal{Z} = \sum_s \exp (H(s)/kT) $$ Here the summation runs over all possible (micro)states of the system. The partition function is really useful as it is related to the free energy $A = -kT \ln{\mathcal{Z} }$. When the partition function goes to zero, the free energy explodes and this signifies a phase change - a physically interesting event. What about our simple system? $$ \mathcal{Z} = 2 \exp({\beta J}) + 2 = 2x + 2 $$ You'll notice that I changed $x=\exp({\beta J})$ to make things a little neater. You may also notice that $\mathcal{Z}$ looks like polynomial. Which means if we want to find the interesting events in the system we find the zeros of the partition function $\mathcal{Z}=0$. This zero will correspond to a particular temperature $T$. In this case the only temperature we get is a complex one ... Complex Temperatures? Before you discount the idea that a temperature not on the real number line is impossible (and that $Tmuch like the pattern you've shown above. For a finite spin system, you'll never find a zero on the real axis, however... Anybody have any idea what happens to the solution set as n→∞? At the thermodynamic limit (which corresponds to an infinite number of sites) the points become dense on the plane. At this limit the points can touch the real axis (corresponding to a phase change in the system). For example, in the 2D Ising model the points do touch the real axis (and make a beautiful circle on the complex plane) where the system undergoes a phase transition from ordered to disordered. Prior work The study of these zeros (from a physics perspective) is fascinating and started with the seminal papers by Yang and Lee: Yang, C. N.; Lee, T. D. (1952), "Statistical Theory of Equations of State and Phase Transitions. I. Theory of Condensation", Physical Review 87: 404–409, doi:10.1103/PhysRev.87.404 Lee, T. D.; Yang, C. N. (1952), "Statistical Theory of Equations of State and Phase Transitions. II. Lattice Gas and Ising Model", Physical Review 87: 410–419, doi:10.1103/PhysRev.87.410 Which are surprisingly accessible. For a good time, search for images of Yang-Lee zeros. In addition you can extend the fugacity to the complex plane, these are called the Fisher zeros and make even more complex patterns!
| 157
| false
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Q: "The Egg:" Bizarre behavior of the roots of a family of polynomials.
In this MO post, I ran into the following family of polynomials: $$f_n(x)=\sum_{m=0}^{n}\prod_{k=0}^{m-1}\frac{x^n-x^k}{x^m-x^k}.$$ In the context of the post, $x$ was a prime number, and $f_n(x)$ counted the number of subspaces of an $n$-dimensional vector space over $GF(x)$ (which I was using to determine the number of subgroups of an elementary abelian group $E_{x^n}$). Anyway, while I was investigating asymptotic behavior of $f_n(x)$ in Mathematica, I got sidetracked and (just for fun) looked at the set of complex roots when I set $f_n(x)=0$. For $n=24$, the plot looked like this: (The real and imaginary axes are from $-1$ to $1$.) Surprised by the unusual symmetry of the solutions, I made the same plot for a few more values of $n$. Note the clearly defined "tails" (on the left when even, top and bottom when odd) and "cusps" (both sides). You can see that after approximately $n=60$, the "circle" of solutions starts to expand into a band of solutions with a defined outline. To fully absorb the weirdness of this, I animated the solutions from $n=2$ to $n=112$. The following is the result: Pretty weird, right!? Anyhow, here are my questions: First, has anybody ever seen anything at all like this before? What's up with those "tails?" They seem to occur only on even $n$, and they are surely distinguishable from the rest of the solutions. Look how the "enclosed" solutions rotate as $n$ increases. Why does this happen? [Explained in edits.] Anybody have any idea what happens to the solution set as $n\rightarrow \infty$? Thanks to @WillSawin, we now know that all the roots are contained in an annulus that converges to the unit circle, which is fantastic. So, the final step in understanding the limit of the solution sets is figuring out what happens on the unit circle. We can see from the animation that there are many gaps, particularly around certain roots of unity; however, they do appear to be closing. The natural question is, which points on the unit circle "are roots in the limit"? In other words, what are the accumulation points of $\{z\left|z\right|^{-1}:z\in\mathbb{C}\text{ and }f_n(z)=0\}$? Is the set of accumulation points dense? @NoahSnyder's heuristic of considering these as a random family of polynomials suggests it should be- at least, almost surely. These are polynomials in $\mathbb{Z}[x]$. Can anybody think of a way to rewrite the formula (perhaps recursively?) for the simplified polynomial, with no denominator? If so, we could use the new formula to prove the series converges to a function on the unit disc, as well as cut computation time in half. [See edits for progress.] Does anybody know a numerical method specifically for finding roots of high degree polynomials? Or any other way to efficiently compute solution sets for high $n$? [Thanks @Hooked!] Thanks everyone. This may not turn out to be particularly mathematically profound, but it sure is neat. EDIT: Thanks to suggestions in the comments, I cranked up the working precision to maximum and recalculated the animation. As Hurkyl and mercio suspected, the rotation was indeed a software artifact, and in fact evidently so was the thickening of the solution set. The new animation looks like this: So, that solves one mystery: the rotation and inflation were caused by tiny roundoff errors in the computation. With the image clearer, however, I see the behavior of the cusps more clearly. Is there an explanation for the gradual accumulation of "cusps" around the roots of unity? (Especially 1.) EDIT: Here is an animation $Arg(f_n)$ up to $n=30$. I think we can see from this that $f_n$ should converge to some function on the unit disk as $n\rightarrow \infty$. I'd love to include higher $n$, but this was already rather computationally exhausting. Now, I've been tinkering and I may be onto something with respect to point $5$ (i.e. seeking a better formula for $f_n(x)$). The folowing claims aren't proven yet, but I've checked each up to $n=100$, and they seem inductively consistent. Here denote $\displaystyle f_n(x)=\sum_{m}a_{n,m}x^m$, so that $a_{n,m}\in \mathbb{Z}$ are the coefficients in the simplified expansion of $f_n(x)$. First, I found $\text{deg}(f_n)=\text{deg}(f_{n-1})+\lfloor \frac{n}{2} \rfloor$. The solution to this recurrence relation is $$\text{deg}(f_n)=\frac{1}{2}\left({\left\lceil\frac{1-n}{2}\right\rceil}^2 -\left\lceil\frac{1-n}{2}\right\rceil+{\left\lfloor \frac{n}{2} \right\rfloor}^2 + \left\lfloor \frac{n}{2} \right\rfloor\right)=\left\lceil\frac{n^2}{4}\right\rceil.$$ If $f_n(x)$ has $r$ more coefficients than $f_{n-1}(x)$, the leading $r$ coefficients are the same as the leading $r$ coefficients of $f_{n-2}(x)$, pairwise. When $n>m$, $a_{n,m}=a_{n-1,m}+\rho(m)$, where $\rho(m)$ is the number of integer partitions of $m$. (This comes from observation, but I bet an actual proof could follow from some of the formulas here.) For $n\leq m$ the $\rho(m)$ formula first fails at $n=m=6$, and not before for some reason. There is probably a simple correction term I'm not seeing - and whatever that term is, I bet it's what's causing those cusps. Anyhow, with this, we can make almost make a recursive relation for $a_{n,m}$, $$a_{n,m}= \left\{ \begin{array}{ll} a_{n-2,m+\left\lceil\frac{n-2}{2}\right\rceil^2-\left\lceil\frac{n}{2}\right\rceil^2} & : \text{deg}(f_{n-1}) m \\ ? & : m \leq \text{deg}(f_{n-1}) \text{ and } n \leq m \end{array} \right. $$ but I can't figure out the last part yet. EDIT: Someone pointed out to me that if we write $\lim_{n\rightarrow\infty}f_n(x)=\sum_{m=0}^\infty b_{m} x^m$, then it appears that $f_n(x)=\sum_{m=0}^n b_m x^m + O(x^{n+1})$. The $b_m$ there seem to me to be relatively well approximated by the $\rho(m)$ formula, considering the correction term only applies for a finite number of recursions. So, if we have the coefficients up to an order of $O(x^{n+1})$, we can at least prove the polynomials converge on the open unit disk, which the $Arg$ animation suggests is true. (To be precise, it looks like $f_{2n}$ and $f_{2n+1}$ may have different limit functions, but I suspect the coefficients of both sequences will come from the same recursive formula.) With this in mind, I put a bounty up for the correction term, since from that all the behavior will probably be explained. EDIT: The limit function proposed by Gottfriend and Aleks has the formal expression $$\lim_{n\rightarrow \infty}f_n(x)=1+\prod_{m=1}^\infty \frac{1}{1-x^m}.$$ I made an $Arg$ plot of $1+\prod_{m=1}^r \frac{1}{1-x^m}$ for up to $r=24$ to see if I could figure out what that ought to ultimately end up looking like, and came up with this: Purely based off the plots, it seems not entirely unlikely that $f_n(x)$ is going to the same place this is, at least inside the unit disc. Now the question is, how do we determine the solution set at the limit? I speculate that the unit circle may become a dense combination of zeroes and singularities, with fractal-like concentric "circles of singularity" around the roots of unity... :)
A: First, has anybody ever seen anything at all like this before? Yes, and in fact the interesting patterns that arise here are more than just a mathematical curiosity, they can be interpreted to have a physical context. Statistical Mechanics In a simple spin system, say the Ising model, a discrete set of points are arranged on a grid. In physics, we like to define the energy of the system by the Hamiltonian, which gives the energy of any particular microstate. In this system, if the spins are aligned they form a bond. This favorable and the energy is negative. If they are misaligned, the energy is positive. Let's consider a simple system of two points, adjacent to each other. Furthermore, let each site point up (1) or down (-1). For an Ising-like system we would write the Hamiltonian as: $$ H = - \sum_{ij} J \sigma_i \sigma_j $$ where $\sigma_i$ is the spin of the $i$th point and the summation runs over all pairs of adjacent sites. $J$ is the strength of the bond (which we can set to one for our example). In our simple system we have only four possible states: 0 - 0 H = -J 1 - 0 H = 0 0 - 1 H = 0 1 - 1 H = -J Now we can write the partition function $\mathcal{Z}$, a term which encompasses all information of the Hamiltonian from the perspective of statistical mechanics: $$ \mathcal{Z} = \sum_s \exp (H(s)/kT) $$ Here the summation runs over all possible (micro)states of the system. The partition function is really useful as it is related to the free energy $A = -kT \ln{\mathcal{Z} }$. When the partition function goes to zero, the free energy explodes and this signifies a phase change - a physically interesting event. What about our simple system? $$ \mathcal{Z} = 2 \exp({\beta J}) + 2 = 2x + 2 $$ You'll notice that I changed $x=\exp({\beta J})$ to make things a little neater. You may also notice that $\mathcal{Z}$ looks like polynomial. Which means if we want to find the interesting events in the system we find the zeros of the partition function $\mathcal{Z}=0$. This zero will correspond to a particular temperature $T$. In this case the only temperature we get is a complex one ... Complex Temperatures? Before you discount the idea that a temperature not on the real number line is impossible (and that $Tmuch like the pattern you've shown above. For a finite spin system, you'll never find a zero on the real axis, however... Anybody have any idea what happens to the solution set as n→∞? At the thermodynamic limit (which corresponds to an infinite number of sites) the points become dense on the plane. At this limit the points can touch the real axis (corresponding to a phase change in the system). For example, in the 2D Ising model the points do touch the real axis (and make a beautiful circle on the complex plane) where the system undergoes a phase transition from ordered to disordered. Prior work The study of these zeros (from a physics perspective) is fascinating and started with the seminal papers by Yang and Lee: Yang, C. N.; Lee, T. D. (1952), "Statistical Theory of Equations of State and Phase Transitions. I. Theory of Condensation", Physical Review 87: 404–409, doi:10.1103/PhysRev.87.404 Lee, T. D.; Yang, C. N. (1952), "Statistical Theory of Equations of State and Phase Transitions. II. Lattice Gas and Ising Model", Physical Review 87: 410–419, doi:10.1103/PhysRev.87.410 Which are surprisingly accessible. For a good time, search for images of Yang-Lee zeros. In addition you can extend the fugacity to the complex plane, these are called the Fisher zeros and make even more complex patterns!
| 2026-01-26T11:29:41.614000
|
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- Formats: CSV and JSON
- Typical fields:
question_id(int)question_title(string)question_body(string, cleaned of HTML)question_score(int)question_tags(string, comma-separated)question_url(string)answer_body(string, cleaned of HTML; empty if no answer)answer_score(int)answer_accepted(bool)combined_text(string, "Q:\n\n<body>\n\nA: <answer or placeholder>")</li> <li><code>scraped_at</code> (ISO 8601 timestamp)</li> </ul> </li> </ul> <p>Example CSV header:</p> <pre><code>question_id,question_title,question_body,question_score,question_tags,question_url,answer_body,answer_score,answer_accepted,combined_text,scraped_at </code></pre> <h2 class="relative group flex items-center"> <a id="license-and-attribution" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#license-and-attribution" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> License and attribution </span> </h2> <p>Content in this dataset is user-contributed content from Stack Exchange and is licensed under the Creative Commons Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.</p> <p>When using or redistributing the dataset, include attribution. A suggested attribution statement:</p> <p>"Data from Mathematics Stack Exchange (math.stackexchange.com). Licensed under CC BY-SA 4.0."</p> <p>Also include a link to the original question pages (the <code>question_url</code> field) when practical.</p> <h2 class="relative group flex items-center"> <a id="how-the-dataset-was-generated" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#how-the-dataset-was-generated" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> How the dataset was generated </span> </h2> <p>The dataset was produced by a script that uses the official Stack Exchange API:</p> <ul> <li>Requests top-voted mathematics questions (configurable count)</li> <li>For each question, fetches the top answer (if any)</li> <li>Cleans HTML from bodies (HTML entities unescaped and tags removed), collapses whitespace</li> <li>Produces both CSV and JSON exports</li> <li>Respects rate limits and inserts polite delays</li> </ul> <p>Notes:</p> <ul> <li>The generator script is not included in this repository. If you want to include it, add the script file (for example <code>scripts/generate_math_se_dataset.py</code>) and reference it here.</li> <li>For reproducibility, include the script, the exact arguments used, and the timestamp of generation.</li> </ul> <h2 class="relative group flex items-center"> <a id="regenerating-the-dataset" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#regenerating-the-dataset" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Regenerating the dataset </span> </h2> <p>Prerequisites:</p> <ul> <li>Python 3.8+ (or compatible)</li> <li><code>requests</code> (for API calls)</li> <li>Optional: a Stack Exchange API key (recommended for higher rate limits)</li> </ul> <p>Important considerations:</p> <ul> <li>Respect Stack Exchange API rate limits; use an API key if you will fetch many pages.</li> <li>The script should include backoff/retry and delays between requests; do not parallelize requests without careful rate-limit handling.</li> <li>If exporting large datasets, consider paging and storing intermediate results.</li> </ul> <h2 class="relative group flex items-center"> <a id="data-quality--known-limitations" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#data-quality--known-limitations" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Data quality & known limitations </span> </h2> <ul> <li>Bodies are cleaned of HTML but some formatting (math markup, code blocks, LaTeX) may be simplified or lost; consider preserving LaTeX if you need formula fidelity.</li> <li>Only the top answer (by score) is captured — accepted vs. highest-scored may differ; both are recorded if available.</li> <li>Potential biases:<ul> <li>Questions with higher votes and English-language questions are overrepresented if using a top-voted query.</li> <li>Highly specialized or duplicate content may appear.</li> </ul> </li> <li>Empty <code>answer_body</code> means no answer was available at scraping time.</li> <li>Some fields may contain non-ASCII characters; files are exported with UTF-8 encoding.</li> </ul> <h2 class="relative group flex items-center"> <a id="suggested-preprocessing-for-ml" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#suggested-preprocessing-for-ml" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Suggested preprocessing for ML </span> </h2> <ul> <li>Deduplicate entries by <code>question_id</code>.</li> <li>Normalize or preserve LaTeX as needed (consider using a LaTeX-aware tokenizer when training models that must understand formulas).</li> <li>Truncate or chunk very long texts (both question and answer) when fitting into model context windows.</li> <li>Optionally filter by tags for domain-specific datasets (e.g., <code>algebra</code>, <code>calculus</code>, <code>probability</code>).</li> </ul> <h2 class="relative group flex items-center"> <a id="use-cases" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#use-cases" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Use cases </span> </h2> <ul> <li>Training and evaluating question-answering systems and math tutoring chatbots</li> <li>Fine-tuning language models on math Q&A patterns</li> <li>Research on problem-solving strategies and reasoning in mathematics</li> <li>Building retrieval-augmented systems with real Q&A examples</li> </ul> <h2 class="relative group flex items-center"> <a id="privacy-ethics-and-policy" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#privacy-ethics-and-policy" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Privacy, ethics, and policy </span> </h2> <ul> <li>The dataset contains user-contributed content under CC BY-SA; comply with the license and provide attribution.</li> <li>Do not claim the content as your own; include proper attribution and linkbacks where practical.</li> <li>Avoid publishing personally identifiable information beyond what is already publicly available on Stack Exchange.</li> </ul> <h2 class="relative group flex items-center"> <a id="citation" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#citation" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Citation </span> </h2> <p>If you use this dataset in publications, we suggest citing it as:</p> <p>Furkan Nar or <a href="https://github.com/TheOfficialFurkanNar/StackMathematics" rel="nofollow">https://github.com/TheOfficialFurkanNar/StackMathematics</a>. (2026). Mathematics Stack Exchange Q&A Dataset. Retrieved from <a href="https://github.com/TheOfficialFurkanNar/StackMathematics" rel="nofollow">https://github.com/TheOfficialFurkanNar/StackMathematics</a> (data originally from mathematics.stackexchange.com; licensed CC BY-SA 4.0)</p> <h2 class="relative group flex items-center"> <a id="contact" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#contact" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Contact </span> </h2> <p>Contact via: <a href="mailto:furkannar168@hotmail.com" rel="nofollow">furkannar168@hotmail.com</a> For questions about this dataset or generation script, open an issue in this repository or contact the maintainer.</p> <h2 class="relative group flex items-center"> <a id="community--support" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" href="#community--support" > <span class="header-link"><svg class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 256 256"><path d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z" fill="currentColor"></path></svg></span> </a> <span> Community & Support </span> </h2> <p>Join the community to ask questions, discuss the dataset, and get help:</p> <ul> <li>Discord: <a href="https://discord.gg/dcjh8S5H" rel="nofollow">https://discord.gg/dcjh8S5H</a></li> <li></li> </ul>
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