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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.
| 2026-01-26T11:29:09.672176
|
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.
| 2026-01-26T11:29:10.897038
|
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
| 2026-01-26T11:29:12.138434
|
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.
| 2026-01-26T11:29:13.460058
|
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 $2$.
| 2026-01-26T11:29:15.052144
|
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]
| 2026-01-26T11:29:16.459675
|
|
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!
| 2026-01-26T11:29:17.804728
|
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.
| 2026-01-26T11:29:19.320024
|
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.
| 2026-01-26T11:29:21.065070
|
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.
| 2026-01-26T11:29:22.546435
|
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.
| 2026-01-26T11:29:24.183639
|
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!
| 2026-01-26T11:29:25.878151
|
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]
| 2026-01-26T11:29:27.339132
|
|
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.
| 2026-01-26T11:29:28.865987
|
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.
| 2026-01-26T11:29:30.384720
|
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.
| 2026-01-26T11:29:32.045700
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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 $$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$
| 2026-01-26T11:29:33.677351
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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!
| 2026-01-26T11:29:35.209337
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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 $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$
| 2026-01-26T11:29:36.800032
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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.
| 2026-01-26T11:29:38.544477
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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.
| 2026-01-26T11:29:40.041236
|
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
|
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
|
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!
| 2026-01-26T11:29:41.614944
|
199,676
|
What are imaginary numbers?
|
At school, I really struggled to understand the concept of imaginary numbers. My teacher told us that an imaginary number is a number that has something to do with the square root of $-1$. When I tried to calculate the square root of $-1$ on my calculator, it gave me an error. To this day I still do not understand imaginary numbers. It makes no sense to me at all. Is there someone here who totally gets it and can explain it? Why is the concept even useful?
| 526
|
complex-numbers, definition
|
https://math.stackexchange.com/questions/199676/what-are-imaginary-numbers
|
Let's go through some questions in order and see where it takes us. [Or skip to the bit about complex numbers below if you can't be bothered.] What are natural numbers? It took quite some evolution, but humans are blessed by their ability to notice that there is a similarity between the situations of having three apples in your hand and having three eggs in your hand. Or, indeed, three twigs or three babies or three spots. Or even three knocks at the door. And we generalise all of these situations by calling it 'three'; same goes for the other natural numbers. This is not the construction we usually take in maths, but it's how we learn what numbers are. Natural numbers are what allow us to count a finite collection of things. We call this set of numbers $\mathbb{N}$. What are integers? Once we've learnt how to measure quantity, it doesn't take us long before we need to measure change, or relative quantity. If I'm holding three apples and you take away two, I now have 'two fewer' apples than I had before; but if you gave me two apples I'd have 'two more'. We want to measure these changes on the same scale (rather than the separate scales of 'more' and 'less'), and we do this by introducing negative natural numbers: the net increase in apples is $-2$. We get the integers from the naturals by allowing ourselves to take numbers away: $\mathbb{Z}$ is the closure of $\mathbb{N}$ under the operation $-$. What are rational numbers? My friend and I are pretty hungry at this point but since you came along and stole two of my apples I only have one left. Out of mutual respect we decide we should each have the same quantity of apple, and so we cut it down the middle. We call the quantity of apple we each get 'a half', or $\frac{1}{2}$. The net change in apple after I give my friend his half is $-\frac{1}{2}$. We get the rationals from the integers by allowing ourselves to divide integers by positive integers [or, equivalently, by nonzero integers]: $\mathbb{Q}$ is (sort of) the closure of $\mathbb{Z}$ under the operation $\div$. What are real numbers? I find some more apples and put them in a pie, which I cook in a circular dish. One of my friends decides to get smart, and asks for a slice of the pie whose curved edge has the same length as its straight edges (i.e. arc length of the circular segment is equal to its radius). I decide to honour his request, and using our newfangled rational numbers I try to work out how many such slices I could cut. But I can't quite get there: it's somewhere between $6$ and $7$; somewhere between $\frac{43}{7}$ and $\frac{44}{7}$; somewhere between $\frac{709}{113}$ and $\frac{710}{113}$; and so on, but no matter how accurate I try and make the fractions, I never quite get there. So I decide to call this number $2\pi$ (or $\tau$?) and move on with my life. The reals turn the rationals into a continuum, filling the holes which can be approximated to arbitrary degrees of accuracy but never actually reached: $\mathbb{R}$ is the completion of $\mathbb{Q}$. What are complex numbers? [Finally!] Our real numbers prove to be quite useful. If I want to make a pie which is twice as big as my last one but still circular then I'll use a dish whose radius is $\sqrt{2}$ times bigger. If I decide this isn't enough and I want to make it thrice as big again then I'll use a dish whose radius is $\sqrt{3}$ times as big as the last. But it turns out that to get this dish I could have made the original one thrice as big and then that one twice as big; the order in which I increase the size of the dish has no effect on what I end up with. And I could have done it in one go, making it six times as big by using a dish whose radius is $\sqrt{6}$ times as big. This leads to my discovery of the fact that multiplication corresponds to scaling $-$ they obey the same rules. (Multiplication by negative numbers responds to scaling and then flipping.) But I can also spin a pie around. Rotating it by one angle and then another has the same effect as rotating it by the second angle and then the first $-$ the order in which I carry out the rotations has no effect on what I end up with, just like with scaling. Does this mean we can model rotation with some kind of multiplication, where multiplication of these new numbers corresponds to addition of the angles? If I could, then I'd be able to rotate a point on the pie by performing a sequence of multiplications. I notice that if I rotate my pie by $90^{\circ}$ four times then it ends up how it was, so I'll declare this $90^{\circ}$ rotation to be multiplication by '$i$' and see what happens. We've seen that $i^4=1$, and with our funky real numbers we know that $i^4=(i^2)^2$ and so $i^2 = \pm 1$. But $i^2 \ne 1$ since rotating twice doesn't leave the pie how it was $-$ it's facing the wrong way; so in fact $i^2=-1$. This then also obeys the rules for multiplication by negative real numbers. Upon further experimentation with spinning pies around we discover that defining $i$ in this way leads to numbers (formed by adding and multiplying real numbers with this new '$i$' beast) which, under multiplication, do indeed correspond to combined scalings and rotations in a 'number plane', which contains our previously held 'number line'. What's more, they can be multiplied, divided and rooted as we please. It then has the fun consequence that any polynomial with coefficients of this kind has as many roots as its degree; what fun! The complex numbers allow us to consider scalings and rotations as two instances of the same thing; and by ensuring that negative reals have square roots, we get something where every (non-constant) polynomial equation can be solved: $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$. [Final edit ever: It occurs to me that I never mentioned anything to do with anything 'imaginary', since I presumed that Sachin really wanted to know about the complex numbers as a whole. But for the sake of completeness: the imaginary numbers are precisely the real multiples of $i$ $-$ you scale the pie and rotate it by $90^{\circ}$ in either direction. They are the rotations/scalings which, when performed twice, leave the pie facing backwards; that is, they are the numbers which square to give negative real numbers.] What next? I've been asked in the comments to mention quaternions and octonions. These go (even further) beyond what the question is asking, so I won't dwell on them, but the idea is: my friends and I are actually aliens from a multi-dimensional world and simply aren't satisfied with a measly $2$-dimensional number system. By extending the principles from our so-called complex numbers we get systems which include copies of $\mathbb{C}$ and act in many ways like numbers, but now (unless we restrict ourselves to one of the copies of $\mathbb{C}$) the order in which we carry out our weird multi-dimensional symmetries does matter. But, with them, we can do lots of science. I have also completely omitted any mention of ordinal numbers, because they fork off in a different direction straight after the naturals. We get some very exciting stuff out of these, but we don't find $\mathbb{C}$ because it doesn't have any natural order relation on it. Historical note The above succession of stages is not a historical account of how numbers of different types are discovered. I don't claim to know an awful lot about the history of mathematics, but I know enough to know that the concept of a number evolved in different ways in different cultures, likely due to practical implications. In particular, it is very unlikely that complex numbers were devised geometrically as rotations-and-scalings $-$ the needs of the time were algebraic and people were throwing away (perfectly valid) equations because they didn't think $\sqrt{-1}$ could exist. Their geometric properties were discovered soon after. However, this is roughly the sequence in which these number sets are (usually) constructed in ZF set theory and we have a nice sequence of inclusions $$1 \hookrightarrow \mathbb{N} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow \mathbb{R} \hookrightarrow \mathbb{C}$$ Stuff to read The other answers to this question give very insightful ways of getting $\mathbb{C}$ from $\mathbb{R}$ in different ways, and discussing how and why complex numbers are useful $-$ there's only so much use to spinning pies around. A Visual, Intuitive Guide to Imaginary Numbers $-$ thanks go to Joe, in the comments, for pointing this out to me. Some older questions, e.g. here and here, have some brilliant answers. I'd be glad to know of more such resources; feel free to post any in the comments.
| 828
| true
|
Q: What are imaginary numbers?
At school, I really struggled to understand the concept of imaginary numbers. My teacher told us that an imaginary number is a number that has something to do with the square root of $-1$. When I tried to calculate the square root of $-1$ on my calculator, it gave me an error. To this day I still do not understand imaginary numbers. It makes no sense to me at all. Is there someone here who totally gets it and can explain it? Why is the concept even useful?
A: Let's go through some questions in order and see where it takes us. [Or skip to the bit about complex numbers below if you can't be bothered.] What are natural numbers? It took quite some evolution, but humans are blessed by their ability to notice that there is a similarity between the situations of having three apples in your hand and having three eggs in your hand. Or, indeed, three twigs or three babies or three spots. Or even three knocks at the door. And we generalise all of these situations by calling it 'three'; same goes for the other natural numbers. This is not the construction we usually take in maths, but it's how we learn what numbers are. Natural numbers are what allow us to count a finite collection of things. We call this set of numbers $\mathbb{N}$. What are integers? Once we've learnt how to measure quantity, it doesn't take us long before we need to measure change, or relative quantity. If I'm holding three apples and you take away two, I now have 'two fewer' apples than I had before; but if you gave me two apples I'd have 'two more'. We want to measure these changes on the same scale (rather than the separate scales of 'more' and 'less'), and we do this by introducing negative natural numbers: the net increase in apples is $-2$. We get the integers from the naturals by allowing ourselves to take numbers away: $\mathbb{Z}$ is the closure of $\mathbb{N}$ under the operation $-$. What are rational numbers? My friend and I are pretty hungry at this point but since you came along and stole two of my apples I only have one left. Out of mutual respect we decide we should each have the same quantity of apple, and so we cut it down the middle. We call the quantity of apple we each get 'a half', or $\frac{1}{2}$. The net change in apple after I give my friend his half is $-\frac{1}{2}$. We get the rationals from the integers by allowing ourselves to divide integers by positive integers [or, equivalently, by nonzero integers]: $\mathbb{Q}$ is (sort of) the closure of $\mathbb{Z}$ under the operation $\div$. What are real numbers? I find some more apples and put them in a pie, which I cook in a circular dish. One of my friends decides to get smart, and asks for a slice of the pie whose curved edge has the same length as its straight edges (i.e. arc length of the circular segment is equal to its radius). I decide to honour his request, and using our newfangled rational numbers I try to work out how many such slices I could cut. But I can't quite get there: it's somewhere between $6$ and $7$; somewhere between $\frac{43}{7}$ and $\frac{44}{7}$; somewhere between $\frac{709}{113}$ and $\frac{710}{113}$; and so on, but no matter how accurate I try and make the fractions, I never quite get there. So I decide to call this number $2\pi$ (or $\tau$?) and move on with my life. The reals turn the rationals into a continuum, filling the holes which can be approximated to arbitrary degrees of accuracy but never actually reached: $\mathbb{R}$ is the completion of $\mathbb{Q}$. What are complex numbers? [Finally!] Our real numbers prove to be quite useful. If I want to make a pie which is twice as big as my last one but still circular then I'll use a dish whose radius is $\sqrt{2}$ times bigger. If I decide this isn't enough and I want to make it thrice as big again then I'll use a dish whose radius is $\sqrt{3}$ times as big as the last. But it turns out that to get this dish I could have made the original one thrice as big and then that one twice as big; the order in which I increase the size of the dish has no effect on what I end up with. And I could have done it in one go, making it six times as big by using a dish whose radius is $\sqrt{6}$ times as big. This leads to my discovery of the fact that multiplication corresponds to scaling $-$ they obey the same rules. (Multiplication by negative numbers responds to scaling and then flipping.) But I can also spin a pie around. Rotating it by one angle and then another has the same effect as rotating it by the second angle and then the first $-$ the order in which I carry out the rotations has no effect on what I end up with, just like with scaling. Does this mean we can model rotation with some kind of multiplication, where multiplication of these new numbers corresponds to addition of the angles? If I could, then I'd be able to rotate a point on the pie by performing a sequence of multiplications. I notice that if I rotate my pie by $90^{\circ}$ four times then it ends up how it was, so I'll declare this $90^{\circ}$ rotation to be multiplication by '$i$' and see what happens. We've seen that $i^4=1$, and with our funky real numbers we know that $i^4=(i^2)^2$ and so $i^2 = \pm 1$. But $i^2 \ne 1$ since rotating twice doesn't leave the pie how it was $-$ it's facing the wrong way; so in fact $i^2=-1$. This then also obeys the rules for multiplication by negative real numbers. Upon further experimentation with spinning pies around we discover that defining $i$ in this way leads to numbers (formed by adding and multiplying real numbers with this new '$i$' beast) which, under multiplication, do indeed correspond to combined scalings and rotations in a 'number plane', which contains our previously held 'number line'. What's more, they can be multiplied, divided and rooted as we please. It then has the fun consequence that any polynomial with coefficients of this kind has as many roots as its degree; what fun! The complex numbers allow us to consider scalings and rotations as two instances of the same thing; and by ensuring that negative reals have square roots, we get something where every (non-constant) polynomial equation can be solved: $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$. [Final edit ever: It occurs to me that I never mentioned anything to do with anything 'imaginary', since I presumed that Sachin really wanted to know about the complex numbers as a whole. But for the sake of completeness: the imaginary numbers are precisely the real multiples of $i$ $-$ you scale the pie and rotate it by $90^{\circ}$ in either direction. They are the rotations/scalings which, when performed twice, leave the pie facing backwards; that is, they are the numbers which square to give negative real numbers.] What next? I've been asked in the comments to mention quaternions and octonions. These go (even further) beyond what the question is asking, so I won't dwell on them, but the idea is: my friends and I are actually aliens from a multi-dimensional world and simply aren't satisfied with a measly $2$-dimensional number system. By extending the principles from our so-called complex numbers we get systems which include copies of $\mathbb{C}$ and act in many ways like numbers, but now (unless we restrict ourselves to one of the copies of $\mathbb{C}$) the order in which we carry out our weird multi-dimensional symmetries does matter. But, with them, we can do lots of science. I have also completely omitted any mention of ordinal numbers, because they fork off in a different direction straight after the naturals. We get some very exciting stuff out of these, but we don't find $\mathbb{C}$ because it doesn't have any natural order relation on it. Historical note The above succession of stages is not a historical account of how numbers of different types are discovered. I don't claim to know an awful lot about the history of mathematics, but I know enough to know that the concept of a number evolved in different ways in different cultures, likely due to practical implications. In particular, it is very unlikely that complex numbers were devised geometrically as rotations-and-scalings $-$ the needs of the time were algebraic and people were throwing away (perfectly valid) equations because they didn't think $\sqrt{-1}$ could exist. Their geometric properties were discovered soon after. However, this is roughly the sequence in which these number sets are (usually) constructed in ZF set theory and we have a nice sequence of inclusions $$1 \hookrightarrow \mathbb{N} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow \mathbb{R} \hookrightarrow \mathbb{C}$$ Stuff to read The other answers to this question give very insightful ways of getting $\mathbb{C}$ from $\mathbb{R}$ in different ways, and discussing how and why complex numbers are useful $-$ there's only so much use to spinning pies around. A Visual, Intuitive Guide to Imaginary Numbers $-$ thanks go to Joe, in the comments, for pointing this out to me. Some older questions, e.g. here and here, have some brilliant answers. I'd be glad to know of more such resources; feel free to post any in the comments.
| 2026-01-26T11:29:43.181514
|
3,869
|
What is the intuitive relationship between SVD and PCA?
|
Singular value decomposition (SVD) and principal component analysis (PCA) are two eigenvalue methods used to reduce a high-dimensional data set into fewer dimensions while retaining important information. Online articles say that these methods are 'related' but never specify the exact relation. What is the intuitive relationship between PCA and SVD? As PCA uses the SVD in its calculation, clearly there is some 'extra' analysis done. What does PCA 'pay attention' to differently than the SVD? What kinds of relationships do each method utilize more in their calculations? Is one method 'blind' to a certain type of data that the other is not?
| 466
|
linear-algebra, matrices, statistics, svd, principal-component-analysis
|
https://math.stackexchange.com/questions/3869/what-is-the-intuitive-relationship-between-svd-and-pca
|
(I assume for the purposes of this answer that the data has been preprocessed to have zero mean.) Simply put, the PCA viewpoint requires that one compute the eigenvalues and eigenvectors of the covariance matrix, which is the product $\frac{1}{n-1}\mathbf X\mathbf X^\top$, where $\mathbf X$ is the data matrix. Since the covariance matrix is symmetric, the matrix is diagonalizable, and the eigenvectors can be normalized such that they are orthonormal: $\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf W\mathbf D\mathbf W^\top$ On the other hand, applying SVD to the data matrix $\mathbf X$ as follows: $\mathbf X=\mathbf U\mathbf \Sigma\mathbf V^\top$ and attempting to construct the covariance matrix from this decomposition gives $$ \frac{1}{n-1}\mathbf X\mathbf X^\top =\frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf U\mathbf \Sigma\mathbf V^\top)^\top = \frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf V\mathbf \Sigma\mathbf U^\top) $$ and since $\mathbf V$ is an orthogonal matrix ($\mathbf V^\top \mathbf V=\mathbf I$), $\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf U\mathbf \Sigma^2 \mathbf U^\top$ and the correspondence is easily seen (the square roots of the eigenvalues of $\mathbf X\mathbf X^\top$ are the singular values of $\mathbf X$, etc.) In fact, using the SVD to perform PCA makes much better sense numerically than forming the covariance matrix to begin with, since the formation of $\mathbf X\mathbf X^\top$ can cause loss of precision. This is detailed in books on numerical linear algebra, but I'll leave you with an example of a matrix that can be stable SVD'd, but forming $\mathbf X\mathbf X^\top$ can be disastrous, the Läuchli matrix: $\begin{pmatrix}1&1&1\\ \epsilon&0&0\\0&\epsilon&0\\0&0&\epsilon\end{pmatrix}^\top,$ where $\epsilon$ is a tiny number.
| 388
| true
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Q: What is the intuitive relationship between SVD and PCA?
Singular value decomposition (SVD) and principal component analysis (PCA) are two eigenvalue methods used to reduce a high-dimensional data set into fewer dimensions while retaining important information. Online articles say that these methods are 'related' but never specify the exact relation. What is the intuitive relationship between PCA and SVD? As PCA uses the SVD in its calculation, clearly there is some 'extra' analysis done. What does PCA 'pay attention' to differently than the SVD? What kinds of relationships do each method utilize more in their calculations? Is one method 'blind' to a certain type of data that the other is not?
A: (I assume for the purposes of this answer that the data has been preprocessed to have zero mean.) Simply put, the PCA viewpoint requires that one compute the eigenvalues and eigenvectors of the covariance matrix, which is the product $\frac{1}{n-1}\mathbf X\mathbf X^\top$, where $\mathbf X$ is the data matrix. Since the covariance matrix is symmetric, the matrix is diagonalizable, and the eigenvectors can be normalized such that they are orthonormal: $\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf W\mathbf D\mathbf W^\top$ On the other hand, applying SVD to the data matrix $\mathbf X$ as follows: $\mathbf X=\mathbf U\mathbf \Sigma\mathbf V^\top$ and attempting to construct the covariance matrix from this decomposition gives $$ \frac{1}{n-1}\mathbf X\mathbf X^\top =\frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf U\mathbf \Sigma\mathbf V^\top)^\top = \frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf V\mathbf \Sigma\mathbf U^\top) $$ and since $\mathbf V$ is an orthogonal matrix ($\mathbf V^\top \mathbf V=\mathbf I$), $\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf U\mathbf \Sigma^2 \mathbf U^\top$ and the correspondence is easily seen (the square roots of the eigenvalues of $\mathbf X\mathbf X^\top$ are the singular values of $\mathbf X$, etc.) In fact, using the SVD to perform PCA makes much better sense numerically than forming the covariance matrix to begin with, since the formation of $\mathbf X\mathbf X^\top$ can cause loss of precision. This is detailed in books on numerical linear algebra, but I'll leave you with an example of a matrix that can be stable SVD'd, but forming $\mathbf X\mathbf X^\top$ can be disastrous, the Läuchli matrix: $\begin{pmatrix}1&1&1\\ \epsilon&0&0\\0&\epsilon&0\\0&0&\epsilon\end{pmatrix}^\top,$ where $\epsilon$ is a tiny number.
| 2026-01-26T11:29:44.598509
|
39,802
|
To sum $1+2+3+\cdots$ to $-\frac1{12}$
|
$$\sum_{n=1}^\infty\frac1{n^s}$$ only converges to $\zeta(s)$ if $\text{Re}(s)>1$. Why should analytically continuing to $\zeta(-1)$ give the right answer?
| 466
|
analysis, complex-analysis, riemann-zeta, divergent-series
|
https://math.stackexchange.com/questions/39802/to-sum-123-cdots-to-frac112
|
there are many ways to see that your result is the right one. What does the right one mean? It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$. It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook. Exponential regulator method Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes $$ S = \sum_{n=1}^{\infty} n e^{-\epsilon n} $$ Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $\epsilon\to 0$ limit of the formula for $S$. In physics, $\epsilon$ would be viewed as a kind of "minimum distance" that can be resolved. The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself) $$ S = \frac{e^\epsilon}{(e^\epsilon-1)^2} $$ We will only need some Laurent expansion around $\epsilon = 0$. $$ S = \frac{1+\epsilon+\epsilon^2/2 + O(\epsilon^3)}{(\epsilon+\epsilon^2/2+\epsilon^3/6+O(\epsilon^4))^2} $$ We have $$ S = \frac{1}{\epsilon^2} \frac{1+\epsilon+\epsilon^2/2+O(\epsilon^3)}{(1+\epsilon/2+\epsilon^2/6+O(\epsilon^3))^2} $$ You see that the $1/\epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command 1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}] and the result is $$ \frac{1}{\epsilon^2} - \frac{1}{12} + \frac{\epsilon^2}{240} + O(\epsilon^4) $$ In the $\epsilon\to 0$ limit we were interested in, the $\epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/\epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum. In physics applications, $\epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $\exp(-(\epsilon+\epsilon^2)n)$. There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function. Euler's method Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too. We will work with two sums, $$ S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots $$ Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion $$ \frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + \dots $$ Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now: $$ T = (1+2+3+\dots) - 2\times (2+4+6+\dots) = (1+2+3+\dots) (1 - 4) = -3S$$ so $S=-T/3=-1/12$. A zeta-function calculation A somewhat unusual calculation of $\zeta(-1)=-1/12$ of mine may be found in the Pictures of Yellows Roses, a Czech student journal. The website no longer works, although a working snapshot of the original website is still available through the WebArchive (see this link). A 2014 English text with the same evaluation at the end can be found at The Reference Frame. The comments were in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.
| 248
| false
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Q: To sum $1+2+3+\cdots$ to $-\frac1{12}$
$$\sum_{n=1}^\infty\frac1{n^s}$$ only converges to $\zeta(s)$ if $\text{Re}(s)>1$. Why should analytically continuing to $\zeta(-1)$ give the right answer?
A: there are many ways to see that your result is the right one. What does the right one mean? It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$. It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook. Exponential regulator method Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes $$ S = \sum_{n=1}^{\infty} n e^{-\epsilon n} $$ Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $\epsilon\to 0$ limit of the formula for $S$. In physics, $\epsilon$ would be viewed as a kind of "minimum distance" that can be resolved. The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself) $$ S = \frac{e^\epsilon}{(e^\epsilon-1)^2} $$ We will only need some Laurent expansion around $\epsilon = 0$. $$ S = \frac{1+\epsilon+\epsilon^2/2 + O(\epsilon^3)}{(\epsilon+\epsilon^2/2+\epsilon^3/6+O(\epsilon^4))^2} $$ We have $$ S = \frac{1}{\epsilon^2} \frac{1+\epsilon+\epsilon^2/2+O(\epsilon^3)}{(1+\epsilon/2+\epsilon^2/6+O(\epsilon^3))^2} $$ You see that the $1/\epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command 1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}] and the result is $$ \frac{1}{\epsilon^2} - \frac{1}{12} + \frac{\epsilon^2}{240} + O(\epsilon^4) $$ In the $\epsilon\to 0$ limit we were interested in, the $\epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/\epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum. In physics applications, $\epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $\exp(-(\epsilon+\epsilon^2)n)$. There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function. Euler's method Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too. We will work with two sums, $$ S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots $$ Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion $$ \frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + \dots $$ Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now: $$ T = (1+2+3+\dots) - 2\times (2+4+6+\dots) = (1+2+3+\dots) (1 - 4) = -3S$$ so $S=-T/3=-1/12$. A zeta-function calculation A somewhat unusual calculation of $\zeta(-1)=-1/12$ of mine may be found in the Pictures of Yellows Roses, a Czech student journal. The website no longer works, although a working snapshot of the original website is still available through the WebArchive (see this link). A 2014 English text with the same evaluation at the end can be found at The Reference Frame. The comments were in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.
| 2026-01-26T11:29:46.564273
|
54,506
|
Is this Batman equation for real?
|
HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real? Batman Equation in text form: \begin{align} &\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\ &\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\ &\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\ &\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\ &\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0 \end{align}
| 466
|
geometry, algebra-precalculus, graphing-functions, plane-curves
|
https://math.stackexchange.com/questions/54506/is-this-batman-equation-for-real
|
As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this: So the curve $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $|x|>3$ and $y > -3\sqrt{33}/7$: That's the first factor. The second factor is quite ingeniously done. The curve $\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like: This is got by adding $y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected: and $y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$: The third factor $9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines y = 9 - 8|x|: truncated to the region $0.75 Similarly, the fourth factor $3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines $y = 3|x| + 0.75$: truncated to the region $0.5 The fifth factor $2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line $y = 2.25$ truncated to $-0.5 Finally, $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like: so the sixth factor $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving: (the software, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)
| 1,075
| true
|
Q: Is this Batman equation for real?
HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real? Batman Equation in text form: \begin{align} &\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\ &\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\ &\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\ &\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\ &\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0 \end{align}
A: As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this: So the curve $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $|x|>3$ and $y > -3\sqrt{33}/7$: That's the first factor. The second factor is quite ingeniously done. The curve $\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like: This is got by adding $y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected: and $y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$: The third factor $9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines y = 9 - 8|x|: truncated to the region $0.75 Similarly, the fourth factor $3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines $y = 3|x| + 0.75$: truncated to the region $0.5 The fifth factor $2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line $y = 2.25$ truncated to $-0.5 Finally, $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like: so the sixth factor $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving: (the software, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)
| 2026-01-26T11:29:48.230889
|
178,940
|
Proofs that every mathematician should know.
|
There are mathematical proofs that have that "wow" factor in being elegant, simplifying one's view of mathematics, lifting one's perception into the light of knowledge, etc. So I'd like to know what mathematical proofs you've come across that you think other mathematicians should know, and why.
| 460
|
soft-question, big-list
|
https://math.stackexchange.com/questions/178940/proofs-that-every-mathematician-should-know
|
Here is my favourite "wow" proof . Theorem There exist two positive irrational numbers $s,t$ such that $s^t$ is rational. Proof If $\sqrt2^\sqrt 2$ is rational, we may take $s=t=\sqrt 2$ . If $\sqrt 2^\sqrt 2$ is irrational , we may take $s=\sqrt 2^\sqrt 2$ and $t=\sqrt 2$ since $(\sqrt 2^\sqrt 2)^\sqrt 2=(\sqrt 2)^ 2=2$.
| 353
| false
|
Q: Proofs that every mathematician should know.
There are mathematical proofs that have that "wow" factor in being elegant, simplifying one's view of mathematics, lifting one's perception into the light of knowledge, etc. So I'd like to know what mathematical proofs you've come across that you think other mathematicians should know, and why.
A: Here is my favourite "wow" proof . Theorem There exist two positive irrational numbers $s,t$ such that $s^t$ is rational. Proof If $\sqrt2^\sqrt 2$ is rational, we may take $s=t=\sqrt 2$ . If $\sqrt 2^\sqrt 2$ is irrational , we may take $s=\sqrt 2^\sqrt 2$ and $t=\sqrt 2$ since $(\sqrt 2^\sqrt 2)^\sqrt 2=(\sqrt 2)^ 2=2$.
| 2026-01-26T11:29:49.943020
|
1,002
|
Fourier transform for dummies
|
What is the Fourier transform? What does it do? Why is it useful (in math, in engineering, physics, etc)? This question is based on Kevin Lin's question, which didn't quite fit in MathOverflow. Answers at any level of sophistication are welcome.
| 457
|
soft-question, fourier-analysis, fourier-transform
|
https://math.stackexchange.com/questions/1002/fourier-transform-for-dummies
|
The ancient Greeks had a theory that the sun, the moon, and the planets move around the Earth in circles. This was soon shown to be wrong. The problem was that if you watch the planets carefully, sometimes they move backwards in the sky. So Ptolemy came up with a new idea - the planets move around in one big circle, but then move around a little circle at the same time. Think of holding out a long stick and spinning around, and at the same time on the end of the stick there's a wheel that's spinning. The planet moves like a point on the edge of the wheel. Well, once they started watching really closely, they realized that even this didn't work, so they put circles on circles on circles... Eventually, they had a map of the solar system that looked like this: This "epicycles" idea turns out to be a bad theory. One reason it's bad is that we know now that planets orbit in ellipses around the sun. (The ellipses are not perfect because they're perturbed by the influence of other gravitating bodies, and by relativistic effects.) But it's wrong for an even worse reason that that, as illustrated in this wonderful youtube video. In the video, by adding up enough circles, they made a planet trace out Homer Simpson's face. It turns out we can make any orbit at all by adding up enough circles, as long as we get to vary their size and speeds. So the epicycle theory of planetary orbits is a bad one not because it's wrong, but because it doesn't say anything at all about orbits. Claiming "planets move around in epicycles" is mathematically equivalent to saying "planets move around in two dimensions". Well, that's not saying nothing, but it's not saying much, either! A simple mathematical way to represent "moving around in a circle" is to say that positions in a plane are represented by complex numbers, so a point moving in the plane is represented by a complex function of time. In that case, moving on a circle with radius $R$ and angular frequency $\omega$ is represented by the position $$z(t) = Re^{i\omega t}$$ If you move around on two circles, one at the end of the other, your position is $$z(t) = R_1e^{i\omega_1 t} + R_2 e^{i\omega_2 t}$$ We can then imagine three, four, or infinitely-many such circles being added. If we allow the circles to have every possible angular frequency, we can now write $$z(t) = \int_{-\infty}^{\infty}R(\omega) e^{i\omega t} \mathrm{d}\omega.$$ The function $R(\omega)$ is the Fourier transform of $z(t)$. If you start by tracing any time-dependent path you want through two-dimensions, your path can be perfectly-emulated by infinitely many circles of different frequencies, all added up, and the radii of those circles is the Fourier transform of your path. Caveat: we must allow the circles to have complex radii. This isn't weird, though. It's the same thing as saying the circles have real radii, but they do not all have to start at the same place. At time zero, you can start however far you want around each circle. If your path closes on itself, as it does in the video, the Fourier transform turns out to simplify to a Fourier series. Most frequencies are no longer necessary, and we can write $$z(t) = \sum_{k=-\infty}^\infty c_k e^{ik \omega_0 t}$$ where $\omega_0$ is the angular frequency associated with the entire thing repeating - the frequency of the slowest circle. The only circles we need are the slowest circle, then one twice as fast as that, then one three times as fast as the slowest one, etc. There are still infinitely-many circles if you want to reproduce a repeating path perfectly, but they are countably-infinite now. If you take the first twenty or so and drop the rest, you should get close to your desired answer. In this way, you can use Fourier analysis to create your own epicycle video of your favorite cartoon character. That's what Fourier analysis says. The questions that remain are how to do it, what it's for, and why it works. I think I will mostly leave those alone. How to do it - how to find $R(\omega)$ given $z(t)$ is found in any introductory treatment, and is fairly intuitive if you understand orthogonality. Why it works is a rather deep question. It's a consequence of the spectral theorem. What it's for has a huge range. It's useful in analyzing the response of linear physical systems to an external input, such as an electrical circuit responding to the signal it picks up with an antenna or a mass on a spring responding to being pushed. It's useful in optics; the interference pattern from light scattering from a diffraction grating is the Fourier transform of the grating, and the image of a source at the focus of a lens is its Fourier transform. It's useful in spectroscopy, and in the analysis of any sort of wave phenomena. It converts between position and momentum representations of a wavefunction in quantum mechanics. Check out this question on physics.stackexchange for more detailed examples. Fourier techniques are useful in signal analysis, image processing, and other digital applications. Finally, they are of course useful mathematically, as many other posts here describe.
| 452
| false
|
Q: Fourier transform for dummies
What is the Fourier transform? What does it do? Why is it useful (in math, in engineering, physics, etc)? This question is based on Kevin Lin's question, which didn't quite fit in MathOverflow. Answers at any level of sophistication are welcome.
A: The ancient Greeks had a theory that the sun, the moon, and the planets move around the Earth in circles. This was soon shown to be wrong. The problem was that if you watch the planets carefully, sometimes they move backwards in the sky. So Ptolemy came up with a new idea - the planets move around in one big circle, but then move around a little circle at the same time. Think of holding out a long stick and spinning around, and at the same time on the end of the stick there's a wheel that's spinning. The planet moves like a point on the edge of the wheel. Well, once they started watching really closely, they realized that even this didn't work, so they put circles on circles on circles... Eventually, they had a map of the solar system that looked like this: This "epicycles" idea turns out to be a bad theory. One reason it's bad is that we know now that planets orbit in ellipses around the sun. (The ellipses are not perfect because they're perturbed by the influence of other gravitating bodies, and by relativistic effects.) But it's wrong for an even worse reason that that, as illustrated in this wonderful youtube video. In the video, by adding up enough circles, they made a planet trace out Homer Simpson's face. It turns out we can make any orbit at all by adding up enough circles, as long as we get to vary their size and speeds. So the epicycle theory of planetary orbits is a bad one not because it's wrong, but because it doesn't say anything at all about orbits. Claiming "planets move around in epicycles" is mathematically equivalent to saying "planets move around in two dimensions". Well, that's not saying nothing, but it's not saying much, either! A simple mathematical way to represent "moving around in a circle" is to say that positions in a plane are represented by complex numbers, so a point moving in the plane is represented by a complex function of time. In that case, moving on a circle with radius $R$ and angular frequency $\omega$ is represented by the position $$z(t) = Re^{i\omega t}$$ If you move around on two circles, one at the end of the other, your position is $$z(t) = R_1e^{i\omega_1 t} + R_2 e^{i\omega_2 t}$$ We can then imagine three, four, or infinitely-many such circles being added. If we allow the circles to have every possible angular frequency, we can now write $$z(t) = \int_{-\infty}^{\infty}R(\omega) e^{i\omega t} \mathrm{d}\omega.$$ The function $R(\omega)$ is the Fourier transform of $z(t)$. If you start by tracing any time-dependent path you want through two-dimensions, your path can be perfectly-emulated by infinitely many circles of different frequencies, all added up, and the radii of those circles is the Fourier transform of your path. Caveat: we must allow the circles to have complex radii. This isn't weird, though. It's the same thing as saying the circles have real radii, but they do not all have to start at the same place. At time zero, you can start however far you want around each circle. If your path closes on itself, as it does in the video, the Fourier transform turns out to simplify to a Fourier series. Most frequencies are no longer necessary, and we can write $$z(t) = \sum_{k=-\infty}^\infty c_k e^{ik \omega_0 t}$$ where $\omega_0$ is the angular frequency associated with the entire thing repeating - the frequency of the slowest circle. The only circles we need are the slowest circle, then one twice as fast as that, then one three times as fast as the slowest one, etc. There are still infinitely-many circles if you want to reproduce a repeating path perfectly, but they are countably-infinite now. If you take the first twenty or so and drop the rest, you should get close to your desired answer. In this way, you can use Fourier analysis to create your own epicycle video of your favorite cartoon character. That's what Fourier analysis says. The questions that remain are how to do it, what it's for, and why it works. I think I will mostly leave those alone. How to do it - how to find $R(\omega)$ given $z(t)$ is found in any introductory treatment, and is fairly intuitive if you understand orthogonality. Why it works is a rather deep question. It's a consequence of the spectral theorem. What it's for has a huge range. It's useful in analyzing the response of linear physical systems to an external input, such as an electrical circuit responding to the signal it picks up with an antenna or a mass on a spring responding to being pushed. It's useful in optics; the interference pattern from light scattering from a diffraction grating is the Fourier transform of the grating, and the image of a source at the focus of a lens is its Fourier transform. It's useful in spectroscopy, and in the analysis of any sort of wave phenomena. It converts between position and momentum representations of a wavefunction in quantum mechanics. Check out this question on physics.stackexchange for more detailed examples. Fourier techniques are useful in signal analysis, image processing, and other digital applications. Finally, they are of course useful mathematically, as many other posts here describe.
| 2026-01-26T11:29:51.529036
|
30,732
|
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?
|
How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method. In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
| 457
|
sequences-and-series, convergence-divergence, power-series, geometric-series, faq
|
https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn
|
No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$ Notice that \begin{align*} S_{m}-rS_{m} & = -mr^{m+1}+\sum_{n=1}^{m}r^{n}\\ & = -mr^{m+1}+\frac{r-r^{m+1}}{1-r} \\ & =\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}. \end{align*} Hence $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$ This equality holds for any $r$, but in your case we have $r=\frac{1}{3}$ and a factor of $\frac{2}{3}$ in front of the sum. That is \begin{align*} \sum_{n=1}^{\infty}\frac{2n}{3^{n+1}} & = \frac{2}{3}\lim_{m\rightarrow\infty}\frac{m\left(\frac{1}{3}\right)^{m+2}-(m+1)\left(\frac{1}{3}\right)^{m+1}+\left(\frac{1}{3}\right)}{\left(1-\left(\frac{1}{3}\right)\right)^{2}} \\ & =\frac{2}{3}\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^{2}} \\ & =\frac{1}{2}. \end{align*} Added note: We can define $$S_m^k(r) = \sum_{n=1}^m n^k r^n.$$ Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$. We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$. This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for. This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_m^k(r)$ in general for any $k$? It turns out we can, and the formula is similar to the formula for $\sum_{n=1}^m n^k$, and involves the Bernoulli numbers. In particular, the denominator is $(1-r)^{k+1}$.
| 405
| true
|
Q: How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?
How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method. In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
A: No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$ Notice that \begin{align*} S_{m}-rS_{m} & = -mr^{m+1}+\sum_{n=1}^{m}r^{n}\\ & = -mr^{m+1}+\frac{r-r^{m+1}}{1-r} \\ & =\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}. \end{align*} Hence $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$ This equality holds for any $r$, but in your case we have $r=\frac{1}{3}$ and a factor of $\frac{2}{3}$ in front of the sum. That is \begin{align*} \sum_{n=1}^{\infty}\frac{2n}{3^{n+1}} & = \frac{2}{3}\lim_{m\rightarrow\infty}\frac{m\left(\frac{1}{3}\right)^{m+2}-(m+1)\left(\frac{1}{3}\right)^{m+1}+\left(\frac{1}{3}\right)}{\left(1-\left(\frac{1}{3}\right)\right)^{2}} \\ & =\frac{2}{3}\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^{2}} \\ & =\frac{1}{2}. \end{align*} Added note: We can define $$S_m^k(r) = \sum_{n=1}^m n^k r^n.$$ Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$. We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$. This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for. This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_m^k(r)$ in general for any $k$? It turns out we can, and the formula is similar to the formula for $\sum_{n=1}^m n^k$, and involves the Bernoulli numbers. In particular, the denominator is $(1-r)^{k+1}$.
| 2026-01-26T11:29:53.076805
|
617,625
|
On "familiarity" (or How to avoid "going down the Math Rabbit Hole"?)
|
Anyone trying to learn mathematics on his/her own has had the experience of "going down the Math Rabbit Hole." For example, suppose you come across the novel term vector space, and want to learn more about it. You look up various definitions, and they all refer to something called a field. So now you're off to learn what a field is, but it's the same story all over again: all the definitions you find refer to something called a group. Off to learn about what a group is. Ad infinitum. That's what I'm calling here "to go down the Math Rabbit Hole." Upon first encountering the situation described above one may think: "well, if that's what it takes to learn about vector spaces, then I'll have to toughen up, and do it." I picked this particular example, however, because I'm sure that the course of action it envisions is one that is not just arduous: it is in fact utterly misguided. I can say so with some confidence, for this particular case, thanks to some serendipitous personal experience. It turns out that, luckily for me, some kind calculus professor in college gave me the tip to take a course in linear algebra (something that I would have never thought of on my own), and therefore I had the luxury of learning about vector spaces without having to venture into the dreaded MRH. I did well in this class, and got a good intuitive grasp of vector spaces, but even after I had studied for my final exams (let alone the first day of class), I couldn't have said what a field was. Therefore, from my experience, and that of pretty much all my fellow students in that class, I know that one does not need to know a whole lot about fields to get the hang of vector spaces. All one needs is a familiarity with some field (say $\mathbb{R}$). Now, it's hard to pin down more precisely what this familiarity amounts to. The only thing that I can say about it is that it is a state somewhere between, and quite distinct from, (a) the state right after reading and understanding the definition of whatever it is one wants to learn about (say, "vector spaces"), and (b) the state right after acing a graduate-level pure math course in that topic. Even harder than defining this familiarity is coming up with an efficient way to attain it... I'd like to ask all the math autodidacts reading this: how do you avoid falling into the Math Rabbit Hole? And more specifically, how do you efficiently attain enough familiarity with pre-requisite concepts to move on to the topics that you want to learn about? PS: John von Neumann allegedly once said "Young man, in mathematics you don't understand things. You just get used to them." I think that this "getting used to things" is much of what I'm calling familiarity above. The problem of learning mathematics efficiently then becomes the problem of "getting used to things" quickly. EDIT: Several answers and comments have suggested to use textbooks rather than, say, Wikipedia, to learn math. But textbooks usually have the same problem. There are exceptions, such as Gilbert Strang's books, which generally avoid technicalities and instead focus on the big picture. They are indeed ideal introductions to a subject, but they are exceedingly rare. For example, as I already mentioned in one comment, I've been looking for an intro book on homotopy theory that focuses on the big picture, to no avail; all the books I've found bristle with technicalities from the get go: Hausdorff this, locally compact that, yadda yadda... I'm sure that when one mathematician asks another for an introduction to some branch of math, the latter does not start spewing all these formal technicalities, but instead gives a big-picture account, based on simple examples. I wish authors of mathematics books sometimes wrote books in such an informal vein. Note that I'm not talking here about books written for math-phobes (in fact I detest it when a math book adopts a condescending "for-dummies", "let's-not-fry-our-little-brains-now" tone). Informal does not mean "dumbed down". There's a huge gap in the mathematics literature (at least in English), and I can't figure out why. (BTW, I'm glad that MJD brought up Strang's Linear Algebra book, because it's a concrete example that shows it's not impossible to write a successful math textbook that stays on the big picture, and doesn't fuss over technicalities. It goes without saying that I'm not advocating that all math books be written this way. Attention to such technical details, precision, and rigor are all essential to doing mathematics, but they can easily overwhelm an introductory exposition.)
| 453
|
soft-question, self-learning
|
https://math.stackexchange.com/questions/617625/on-familiarity-or-how-to-avoid-going-down-the-math-rabbit-hole
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Your example makes me think of graphs. Imagine some nice, helpful fellow came along, and made a big graph of every math concept ever, where each concept is one node and related concepts are connected by edges. Now you can take a copy of this graph, and color every node green based on whether you "know" that concept (unknowns can be grey). How to define "know"? In this case, when somebody mentions that concept while talking about something, do you immediately feel confused and get the urge to look the concept up? If no, then you know it (funnily enough, you may be deluding yourself into thinking you know something that you completely misunderstand, and it would be classed as "knowing" based on this rule - but that's fine and I'll explain why in a bit). For purposes of determining whether you "know" it, try to assume that the particular thing the person is talking about isn't some intricate argument that hinges on obscure details of the concept or bizarre interpretations - it's just mentioned matter-of-factly, as a tangential remark. When you are studying a topic, you are basically picking one grey node and trying to color it green. But you may discover that to do this, you must color some adjacent grey nodes first. So the moment you discover a prerequisite node, you go to color it right away, and put your original topic on hold. But this node also has prerequisites, so you put it on hold, and... What you are doing is known as a depth first search. It's natural for it to feel like a rabbit hole - you are trying to go as deep as possible. The hope is that sooner or later you will run into a wall of greens, which is when your long, arduous search will have born fruit, and you will get to feel that unique rush of climbing back up the stack with your little jewel of recursion terminating return value. Then you get back to coloring your original node and find out about the other prerequisite, so now you can do it all over again. DFS is suited for some applications, but it is bad for others. If your goal is to color the whole graph (ie. learn all of math), any strategy will have you visit the same number of nodes, so it doesn't matter as much. But if you are not seriously attempting to learn everything right now, DFS is not the best choice. So, the solution to your problem is straightforward - use a more appropriate search algorithm! Immediately obvious is breadth-first search. This means, when reading an article (or page, or book chapter), don't rush off to look up every new term as soon as you see it. Circle it or make a note of it on a separate paper, but force yourself to finish your text even if its completely incomprehensible to you without knowing the new term. You will now have a list of prerequisite nodes, and can deal with them in a more organized manner. Compared to your DFS, this already makes it much easier to avoid straying too far from your original area of interest. It also has another benefit which is not common in actual graph problems: Often in math, and in general, understanding is cooperative. If you have a concept A which has prerequisite concept B and C, you may find that B is very difficult to understand (it leads down a deep rabbit hole), but only if you don't yet know the very easy topic C, which if you do, make B very easy to "get" because you quickly figure out the salient and relevant points (or it may be turn out that knowing either B or C is sufficient to learn A). In this case, you really don't want to have a learning strategy which will not make sure you do C before B! BFS not only allows you to exploit cooperativities, but it also allows you to manage your time better. After your first pass, let's say you ended up with a list of 30 topics you need to learn first. They won't all be equally hard. Maybe 10 will take you 5 minutes of skimming wikipedia to figure out. Maybe another 10 are so simple, that the first Google Image diagram explains everything. Then there will be 1 or 2 which will take days or even months of work. You don't want to get tripped up on the big ones while you have the small ones to take care of. After all, it may turn out that the big topic is not essential, but the small topic is. If that's the case, you would feel very silly if you tried to tackle the big topic first! But if the small one proves useless, you haven't really lost much energy or time. Once you're doing BFS, you might as well benefit from the other, very nice and clever twists on it, such as Dijkstra or A*. When you have the list of topics, can you order them by how promising they seem? Chances are you can, and chances are, your intuition will be right. Another thing to do - since ultimately, your aim is to link up with some green nodes, why not try to prioritize topics which seem like they would be getting closer to things you do know? The beauty of A* is that these heuristics don't even have to be very correct - even "wrong" or "unrealistic" heuristics may end up making your search faster.
| 221
| true
|
Q: On "familiarity" (or How to avoid "going down the Math Rabbit Hole"?)
Anyone trying to learn mathematics on his/her own has had the experience of "going down the Math Rabbit Hole." For example, suppose you come across the novel term vector space, and want to learn more about it. You look up various definitions, and they all refer to something called a field. So now you're off to learn what a field is, but it's the same story all over again: all the definitions you find refer to something called a group. Off to learn about what a group is. Ad infinitum. That's what I'm calling here "to go down the Math Rabbit Hole." Upon first encountering the situation described above one may think: "well, if that's what it takes to learn about vector spaces, then I'll have to toughen up, and do it." I picked this particular example, however, because I'm sure that the course of action it envisions is one that is not just arduous: it is in fact utterly misguided. I can say so with some confidence, for this particular case, thanks to some serendipitous personal experience. It turns out that, luckily for me, some kind calculus professor in college gave me the tip to take a course in linear algebra (something that I would have never thought of on my own), and therefore I had the luxury of learning about vector spaces without having to venture into the dreaded MRH. I did well in this class, and got a good intuitive grasp of vector spaces, but even after I had studied for my final exams (let alone the first day of class), I couldn't have said what a field was. Therefore, from my experience, and that of pretty much all my fellow students in that class, I know that one does not need to know a whole lot about fields to get the hang of vector spaces. All one needs is a familiarity with some field (say $\mathbb{R}$). Now, it's hard to pin down more precisely what this familiarity amounts to. The only thing that I can say about it is that it is a state somewhere between, and quite distinct from, (a) the state right after reading and understanding the definition of whatever it is one wants to learn about (say, "vector spaces"), and (b) the state right after acing a graduate-level pure math course in that topic. Even harder than defining this familiarity is coming up with an efficient way to attain it... I'd like to ask all the math autodidacts reading this: how do you avoid falling into the Math Rabbit Hole? And more specifically, how do you efficiently attain enough familiarity with pre-requisite concepts to move on to the topics that you want to learn about? PS: John von Neumann allegedly once said "Young man, in mathematics you don't understand things. You just get used to them." I think that this "getting used to things" is much of what I'm calling familiarity above. The problem of learning mathematics efficiently then becomes the problem of "getting used to things" quickly. EDIT: Several answers and comments have suggested to use textbooks rather than, say, Wikipedia, to learn math. But textbooks usually have the same problem. There are exceptions, such as Gilbert Strang's books, which generally avoid technicalities and instead focus on the big picture. They are indeed ideal introductions to a subject, but they are exceedingly rare. For example, as I already mentioned in one comment, I've been looking for an intro book on homotopy theory that focuses on the big picture, to no avail; all the books I've found bristle with technicalities from the get go: Hausdorff this, locally compact that, yadda yadda... I'm sure that when one mathematician asks another for an introduction to some branch of math, the latter does not start spewing all these formal technicalities, but instead gives a big-picture account, based on simple examples. I wish authors of mathematics books sometimes wrote books in such an informal vein. Note that I'm not talking here about books written for math-phobes (in fact I detest it when a math book adopts a condescending "for-dummies", "let's-not-fry-our-little-brains-now" tone). Informal does not mean "dumbed down". There's a huge gap in the mathematics literature (at least in English), and I can't figure out why. (BTW, I'm glad that MJD brought up Strang's Linear Algebra book, because it's a concrete example that shows it's not impossible to write a successful math textbook that stays on the big picture, and doesn't fuss over technicalities. It goes without saying that I'm not advocating that all math books be written this way. Attention to such technical details, precision, and rigor are all essential to doing mathematics, but they can easily overwhelm an introductory exposition.)
A: Your example makes me think of graphs. Imagine some nice, helpful fellow came along, and made a big graph of every math concept ever, where each concept is one node and related concepts are connected by edges. Now you can take a copy of this graph, and color every node green based on whether you "know" that concept (unknowns can be grey). How to define "know"? In this case, when somebody mentions that concept while talking about something, do you immediately feel confused and get the urge to look the concept up? If no, then you know it (funnily enough, you may be deluding yourself into thinking you know something that you completely misunderstand, and it would be classed as "knowing" based on this rule - but that's fine and I'll explain why in a bit). For purposes of determining whether you "know" it, try to assume that the particular thing the person is talking about isn't some intricate argument that hinges on obscure details of the concept or bizarre interpretations - it's just mentioned matter-of-factly, as a tangential remark. When you are studying a topic, you are basically picking one grey node and trying to color it green. But you may discover that to do this, you must color some adjacent grey nodes first. So the moment you discover a prerequisite node, you go to color it right away, and put your original topic on hold. But this node also has prerequisites, so you put it on hold, and... What you are doing is known as a depth first search. It's natural for it to feel like a rabbit hole - you are trying to go as deep as possible. The hope is that sooner or later you will run into a wall of greens, which is when your long, arduous search will have born fruit, and you will get to feel that unique rush of climbing back up the stack with your little jewel of recursion terminating return value. Then you get back to coloring your original node and find out about the other prerequisite, so now you can do it all over again. DFS is suited for some applications, but it is bad for others. If your goal is to color the whole graph (ie. learn all of math), any strategy will have you visit the same number of nodes, so it doesn't matter as much. But if you are not seriously attempting to learn everything right now, DFS is not the best choice. So, the solution to your problem is straightforward - use a more appropriate search algorithm! Immediately obvious is breadth-first search. This means, when reading an article (or page, or book chapter), don't rush off to look up every new term as soon as you see it. Circle it or make a note of it on a separate paper, but force yourself to finish your text even if its completely incomprehensible to you without knowing the new term. You will now have a list of prerequisite nodes, and can deal with them in a more organized manner. Compared to your DFS, this already makes it much easier to avoid straying too far from your original area of interest. It also has another benefit which is not common in actual graph problems: Often in math, and in general, understanding is cooperative. If you have a concept A which has prerequisite concept B and C, you may find that B is very difficult to understand (it leads down a deep rabbit hole), but only if you don't yet know the very easy topic C, which if you do, make B very easy to "get" because you quickly figure out the salient and relevant points (or it may be turn out that knowing either B or C is sufficient to learn A). In this case, you really don't want to have a learning strategy which will not make sure you do C before B! BFS not only allows you to exploit cooperativities, but it also allows you to manage your time better. After your first pass, let's say you ended up with a list of 30 topics you need to learn first. They won't all be equally hard. Maybe 10 will take you 5 minutes of skimming wikipedia to figure out. Maybe another 10 are so simple, that the first Google Image diagram explains everything. Then there will be 1 or 2 which will take days or even months of work. You don't want to get tripped up on the big ones while you have the small ones to take care of. After all, it may turn out that the big topic is not essential, but the small topic is. If that's the case, you would feel very silly if you tried to tackle the big topic first! But if the small one proves useless, you haven't really lost much energy or time. Once you're doing BFS, you might as well benefit from the other, very nice and clever twists on it, such as Dijkstra or A*. When you have the list of topics, can you order them by how promising they seem? Chances are you can, and chances are, your intuition will be right. Another thing to do - since ultimately, your aim is to link up with some green nodes, why not try to prioritize topics which seem like they would be getting closer to things you do know? The beauty of A* is that these heuristics don't even have to be very correct - even "wrong" or "unrealistic" heuristics may end up making your search faster.
| 2026-01-26T11:29:54.856201
|
406,099
|
My son's Sum of Some is beautiful! But what is the proof or explanation?
|
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation: $$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$ Why is this so? What is the proof or explanation? Is it true for any $n$? His own presentation of his finding: Every one of these numbers is two times the number before it. $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$. I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that. If I then skip three ones and read the fourth, that one equals my sum!
| 447
|
sequences-and-series
|
https://math.stackexchange.com/questions/406099/my-sons-sum-of-some-is-beautiful-but-what-is-the-proof-or-explanation
|
Factor out the $2^n$ and you get: $2^n (100+20+8) = 2^n 128 = 2^{n+7}$ since $2^7 = 128$
| 459
| false
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Q: My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation: $$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$ Why is this so? What is the proof or explanation? Is it true for any $n$? His own presentation of his finding: Every one of these numbers is two times the number before it. $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$. I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that. If I then skip three ones and read the fourth, that one equals my sum!
A: Factor out the $2^n$ and you get: $2^n (100+20+8) = 2^n 128 = 2^{n+7}$ since $2^7 = 128$
| 2026-01-26T11:29:56.395250
|
3,852
|
If $AB = I$ then $BA = I$
|
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$. I do not understand anything more than the following. Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix. If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem? P.S.: Please avoid using the transpose and/or inverse of a matrix.
| 409
|
linear-algebra, matrices, inverse
|
https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i
|
Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence. Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices). Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$. Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.
| 326
| true
|
Q: If $AB = I$ then $BA = I$
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$. I do not understand anything more than the following. Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix. If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem? P.S.: Please avoid using the transpose and/or inverse of a matrix.
A: Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence. Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices). Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$. Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.
| 2026-01-26T11:29:57.886417
|
630,339
|
Pedagogy: How to cure students of the "law of universal linearity"?
|
One of the commonest mistakes made by students, appearing at every level of maths education up to about early undergraduate, is the so-called “Law of Universal Linearity”: $$ \frac{1}{a+b} \mathrel{\text{“=”}} \frac{1}{a} + \frac{1}{b} $$ $$ 2^{-3} \mathrel{\text{“=”}} -2^3 $$ $$ \sin (5x + 3y) \mathrel{\text{“=”}} \sin 5x + \sin 3y$$ and so on. Slightly more precisely, I’d call it the tendency to commute or distribute operations through each other. They don't notice that they’re doing anything, except for operations where they’ve specifically learned not to do so. Does anyone have a good cure for this — a particularly clear and memorable explanation that will stick with students? I’ve tried explaining it several ways, but never found an approach that I was really happy with, from a pedagogical point of view.
| 407
|
algebra-precalculus, education
|
https://math.stackexchange.com/questions/630339/pedagogy-how-to-cure-students-of-the-law-of-universal-linearity
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I think this is a symptom of how students are taught basic algebra. Rather than being told explicit axioms like $a(x+y)= ax+ay$ and theorems like $(x+y)/a = x/a+y/a,$ students are bombarded with examples of how these axioms/theorems are used, without ever being explicitly told: hey, here's a new rule you're allowed to use from now on. So they just kind of wing it. They learn to guess. So the solution, really, is to teach the material properly. Make it clear that $a(x+y)=ax+ay$ is a truth (perhaps derive it from a geometric argument). Then make it clear how to use such truths: for example, we can deduce that $3 \times (5+1) = (3 \times 5) + (3 \times 1)$. We can also deduce that $x(x^2+1) = xx^2 + x 1$. Then make it clear how to use those truths. For example, if we have an expression possessing $x(x^2+1)$ as a subexpression, we're allowed to replace this subexpression by $x x^2 + x 1.$ The new expression obtained in this way is guaranteed to equal the original, because we replaced a subexpression with an equal subexpression. Perhaps have a cheat-sheet online, of all the truths students are allowed to use so far, which is updated with more truths as the class progresses. I think that, if you teach in this way, students will learn to trust that if a rule (truth, whatever) hasn't been explicitly written down, then its either false, or at the very least, not strictly necessary to solve the problems at hand. This should cure most instances of universal linearity.
| 182
| false
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Q: Pedagogy: How to cure students of the "law of universal linearity"?
One of the commonest mistakes made by students, appearing at every level of maths education up to about early undergraduate, is the so-called “Law of Universal Linearity”: $$ \frac{1}{a+b} \mathrel{\text{“=”}} \frac{1}{a} + \frac{1}{b} $$ $$ 2^{-3} \mathrel{\text{“=”}} -2^3 $$ $$ \sin (5x + 3y) \mathrel{\text{“=”}} \sin 5x + \sin 3y$$ and so on. Slightly more precisely, I’d call it the tendency to commute or distribute operations through each other. They don't notice that they’re doing anything, except for operations where they’ve specifically learned not to do so. Does anyone have a good cure for this — a particularly clear and memorable explanation that will stick with students? I’ve tried explaining it several ways, but never found an approach that I was really happy with, from a pedagogical point of view.
A: I think this is a symptom of how students are taught basic algebra. Rather than being told explicit axioms like $a(x+y)= ax+ay$ and theorems like $(x+y)/a = x/a+y/a,$ students are bombarded with examples of how these axioms/theorems are used, without ever being explicitly told: hey, here's a new rule you're allowed to use from now on. So they just kind of wing it. They learn to guess. So the solution, really, is to teach the material properly. Make it clear that $a(x+y)=ax+ay$ is a truth (perhaps derive it from a geometric argument). Then make it clear how to use such truths: for example, we can deduce that $3 \times (5+1) = (3 \times 5) + (3 \times 1)$. We can also deduce that $x(x^2+1) = xx^2 + x 1$. Then make it clear how to use those truths. For example, if we have an expression possessing $x(x^2+1)$ as a subexpression, we're allowed to replace this subexpression by $x x^2 + x 1.$ The new expression obtained in this way is guaranteed to equal the original, because we replaced a subexpression with an equal subexpression. Perhaps have a cheat-sheet online, of all the truths students are allowed to use so far, which is updated with more truths as the class progresses. I think that, if you teach in this way, students will learn to trust that if a rule (truth, whatever) hasn't been explicitly written down, then its either false, or at the very least, not strictly necessary to solve the problems at hand. This should cure most instances of universal linearity.
| 2026-01-26T11:29:59.429272
|
820,686
|
'Obvious' theorems that are actually false
|
It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true". Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem). But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false (i.e. counter-intuitively-false theorems)? The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$ ). I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$. I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself. Thanks!
| 394
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soft-question, big-list
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https://math.stackexchange.com/questions/820686/obvious-theorems-that-are-actually-false
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Theorem (false): One can arbitrarily rearrange the terms in a convergent series without changing its value.
| 285
| true
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Q: 'Obvious' theorems that are actually false
It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true". Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem). But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false (i.e. counter-intuitively-false theorems)? The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$ ). I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$. I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself. Thanks!
A: Theorem (false): One can arbitrarily rearrange the terms in a convergent series without changing its value.
| 2026-01-26T11:30:00.885855
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243,770
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Can every proof by contradiction also be shown without contradiction?
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Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction? As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians. Is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.
| 389
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logic, proof-writing, propositional-calculus, proof-theory
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https://math.stackexchange.com/questions/243770/can-every-proof-by-contradiction-also-be-shown-without-contradiction
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To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $\bot$ represent an identically false proposition. Then $\lnot A$, the negation of $A$, is equivalent to $A \to \bot$, and we take the latter to be the definition of the former in terms of $\bot$. There are two key logical principles that express different parts of what we call "proof by contradiction": The principle of explosion: for any statement $A$, we can take "$\bot$ implies $A$" as an axiom. This is also called ex falso quodlibet. The law of the excluded middle: for any statement $A$, we can take "$A$ or $\lnot A$" as an axiom. In proof theory, there are three well known systems: Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose. Intuitionistic logic includes minimal logic and the principle of explosion Classical logic includes intuitionistic logic and the law of the excluded middle It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly. The scheme "If $A$ implies a contradiction, then $\lnot A$ must hold" is true even in intuitionistic logic, because $\lnot A$ is just an abbreviation for $A \to \bot$, and so that scheme just says "if $A \to \bot$ then $A \to \bot$". But in intuitionistic logic, if we prove $\lnot A \to \bot$, this only shows that $\lnot \lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\lnot \lnot A$ implies $A$, which means that in classical logic if we can prove $\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.
| 309
| true
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Q: Can every proof by contradiction also be shown without contradiction?
Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction? As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians. Is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.
A: To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $\bot$ represent an identically false proposition. Then $\lnot A$, the negation of $A$, is equivalent to $A \to \bot$, and we take the latter to be the definition of the former in terms of $\bot$. There are two key logical principles that express different parts of what we call "proof by contradiction": The principle of explosion: for any statement $A$, we can take "$\bot$ implies $A$" as an axiom. This is also called ex falso quodlibet. The law of the excluded middle: for any statement $A$, we can take "$A$ or $\lnot A$" as an axiom. In proof theory, there are three well known systems: Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose. Intuitionistic logic includes minimal logic and the principle of explosion Classical logic includes intuitionistic logic and the law of the excluded middle It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly. The scheme "If $A$ implies a contradiction, then $\lnot A$ must hold" is true even in intuitionistic logic, because $\lnot A$ is just an abbreviation for $A \to \bot$, and so that scheme just says "if $A \to \bot$ then $A \to \bot$". But in intuitionistic logic, if we prove $\lnot A \to \bot$, this only shows that $\lnot \lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\lnot \lnot A$ implies $A$, which means that in classical logic if we can prove $\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.
| 2026-01-26T11:30:02.606322
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290,435
|
Find five positive integers whose reciprocals sum to $1$
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Find a positive integer solution $(x,y,z,a,b)$ for which $$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$ Is your answer the only solution? If so, show why. I was surprised that a teacher would assign this kind of problem to a 5th grade child. (I'm a college student tutor) This girl goes to a private school in a wealthy neighborhood. Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $ x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering "unit fractions" in her class.
| 381
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algebra-precalculus, egyptian-fractions
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https://math.stackexchange.com/questions/290435/find-five-positive-integers-whose-reciprocals-sum-to-1
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The perfect number $28=1+2+4+7+14$ provides a solution: $$\frac1{28}+\frac1{14}+\frac17+\frac14+\frac12=\frac{1+2+4+7+14}{28}=1\;.$$ If they’ve been doing unit (or ‘Egyptian’) fractions, I’d expect some to see that since $\frac16+\frac13=\frac12$, $$\frac16+\frac16+\frac16+\frac16+\frac13=1$$ is a solution, though not a much more interesting one than the trivial solution. The choice of letters might well suggest the solution $$\frac16+\frac16+\frac16+\frac14+\frac14\;.$$ A little playing around would show that $\frac14+\frac15=\frac9{20}$, which differs from $\frac12$ by just $\frac1{20}$; that yields the solution $$\frac1{20}+\frac15+\frac14+\frac14+\frac14\;.$$ If I were the teacher, I’d hope that some kids would realize that since the average of the fractions is $\frac15$, in any non-trivial solution at least one denominator must be less than $5$, and at least one must be greater than $5$. Say that $x\le y\le z\le a\le b$. Clearly $x\ge 2$, so let’s try $x=2$. Then we need to solve $$\frac1y+\frac1z+\frac1a+\frac1b=\frac12\;.$$ Now $y\ge 3$. Suppose that $y=3$; then $$\frac1z+\frac1a+\frac1b=\frac16\;.$$ Now $1,2$, and $3$ all divide $36$, and $\frac16=\frac6{36}$, so we can write $$\frac1{36}+\frac1{18}+\frac1{12}=\frac{1+2+3}{36}=\frac6{36}=\frac16\;,$$ and we get another ‘nice’ solution, $$\frac12+\frac13+\frac1{12}+\frac1{18}+\frac1{36}\;.$$
| 435
| true
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Q: Find five positive integers whose reciprocals sum to $1$
Find a positive integer solution $(x,y,z,a,b)$ for which $$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$ Is your answer the only solution? If so, show why. I was surprised that a teacher would assign this kind of problem to a 5th grade child. (I'm a college student tutor) This girl goes to a private school in a wealthy neighborhood. Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $ x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering "unit fractions" in her class.
A: The perfect number $28=1+2+4+7+14$ provides a solution: $$\frac1{28}+\frac1{14}+\frac17+\frac14+\frac12=\frac{1+2+4+7+14}{28}=1\;.$$ If they’ve been doing unit (or ‘Egyptian’) fractions, I’d expect some to see that since $\frac16+\frac13=\frac12$, $$\frac16+\frac16+\frac16+\frac16+\frac13=1$$ is a solution, though not a much more interesting one than the trivial solution. The choice of letters might well suggest the solution $$\frac16+\frac16+\frac16+\frac14+\frac14\;.$$ A little playing around would show that $\frac14+\frac15=\frac9{20}$, which differs from $\frac12$ by just $\frac1{20}$; that yields the solution $$\frac1{20}+\frac15+\frac14+\frac14+\frac14\;.$$ If I were the teacher, I’d hope that some kids would realize that since the average of the fractions is $\frac15$, in any non-trivial solution at least one denominator must be less than $5$, and at least one must be greater than $5$. Say that $x\le y\le z\le a\le b$. Clearly $x\ge 2$, so let’s try $x=2$. Then we need to solve $$\frac1y+\frac1z+\frac1a+\frac1b=\frac12\;.$$ Now $y\ge 3$. Suppose that $y=3$; then $$\frac1z+\frac1a+\frac1b=\frac16\;.$$ Now $1,2$, and $3$ all divide $36$, and $\frac16=\frac6{36}$, so we can write $$\frac1{36}+\frac1{18}+\frac1{12}=\frac{1+2+3}{36}=\frac6{36}=\frac16\;,$$ and we get another ‘nice’ solution, $$\frac12+\frac13+\frac1{12}+\frac1{18}+\frac1{36}\;.$$
| 2026-01-26T11:30:04.086086
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505,367
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Collection of surprising identities and equations.
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What are some surprising equations/identities that you have seen, which you would not have expected? This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc. I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$. Please write a single identity (or group of identities) in each answer. I found this list of Funny identities, in which there is some overlap.
| 380
|
soft-question, big-list
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https://math.stackexchange.com/questions/505367/collection-of-surprising-identities-and-equations
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This one by Ramanujan gives me the goosebumps: $$ \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}. $$ P.S. Just to make this more intriguing, define the fundamental unit $U_{29} = \frac{5+\sqrt{29}}{2}$ and fundamental solutions to Pell equations, $$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$ $$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$ $$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$ then we can see those integers all over the formula as, $$\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k} = \frac{1}{\pi} $$ Nice, eh?
| 268
| false
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Q: Collection of surprising identities and equations.
What are some surprising equations/identities that you have seen, which you would not have expected? This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc. I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$. Please write a single identity (or group of identities) in each answer. I found this list of Funny identities, in which there is some overlap.
A: This one by Ramanujan gives me the goosebumps: $$ \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}. $$ P.S. Just to make this more intriguing, define the fundamental unit $U_{29} = \frac{5+\sqrt{29}}{2}$ and fundamental solutions to Pell equations, $$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$ $$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$ $$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$ then we can see those integers all over the formula as, $$\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k} = \frac{1}{\pi} $$ Nice, eh?
| 2026-01-26T11:30:05.677455
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11,150
|
Zero to the zero power – is $0^0=1$?
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Could someone provide me with a good explanation of why $0^0=1$? My train of thought: $$x>0\\ 0^x=0^{x-0}=\frac{0^x}{0^0}$$ so $$0^0=\frac{0^x}{0^x}=\,?$$ Possible answers: $0^0\cdot0^x=1\cdot0^0$, so $0^0=1$ $0^0=\frac{0^x}{0^x}=\frac00$, which is undefined PS. I've read the explanation on mathforum.org, but it isn't clear to me.
| 379
|
algebra-precalculus, exponentiation, indeterminate-forms, faq
|
https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1
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In general, there is no good answer as to what $0^0$ "should" be, so it is usually left undefined. Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$? Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined. However, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote "size" (cardinalities), then the "$A^B$" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period. Added 2: the same holds in Discrete Mathematics, when we are mostly interested in "counting" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as "maps from $\{1,2,\ldots,m\}$ to $\\{1,2,\ldots,n\\}$" when interpreted appropriately, so it is again the same thing as in set theory). So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$). Your "train of thoughts" don't really work: If $x\neq 0$, then $0^x$ means "the number of ways to make $x$ choices from $0$ possibilities". This number is $0$. So for any number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ "should" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ is the number of ways of making no selections from no choices. Coda. In the end, it is a matter of definition and utility. In Calculus and algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions. We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that "useful and natural" way should be, so we define it that way.
| 364
| true
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Q: Zero to the zero power – is $0^0=1$?
Could someone provide me with a good explanation of why $0^0=1$? My train of thought: $$x>0\\ 0^x=0^{x-0}=\frac{0^x}{0^0}$$ so $$0^0=\frac{0^x}{0^x}=\,?$$ Possible answers: $0^0\cdot0^x=1\cdot0^0$, so $0^0=1$ $0^0=\frac{0^x}{0^x}=\frac00$, which is undefined PS. I've read the explanation on mathforum.org, but it isn't clear to me.
A: In general, there is no good answer as to what $0^0$ "should" be, so it is usually left undefined. Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$? Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined. However, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote "size" (cardinalities), then the "$A^B$" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period. Added 2: the same holds in Discrete Mathematics, when we are mostly interested in "counting" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as "maps from $\{1,2,\ldots,m\}$ to $\\{1,2,\ldots,n\\}$" when interpreted appropriately, so it is again the same thing as in set theory). So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$). Your "train of thoughts" don't really work: If $x\neq 0$, then $0^x$ means "the number of ways to make $x$ choices from $0$ possibilities". This number is $0$. So for any number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ "should" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ is the number of ways of making no selections from no choices. Coda. In the end, it is a matter of definition and utility. In Calculus and algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions. We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that "useful and natural" way should be, so we define it that way.
| 2026-01-26T11:30:07.552890
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155
|
How can you prove that a function has no closed form integral?
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In the past, I've come across statements along the lines of "function $f(x)$ has no closed form integral", which I assume means that there is no combination of the operations: addition/subtraction multiplication/division raising to powers and roots trigonometric functions exponential functions logarithmic functions which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example. What sort of techniques are used to prove statements like this? What is this branch of mathematics called? Merged with "How to prove that some functions don't have a primitive" by Ismael: Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions. I wonder how we can prove that kind of assertion?
| 373
|
real-analysis, calculus, integration, faq, differential-algebra
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https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral
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It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions $f$ and $g$, $g$ non-constant, the antiderivative of $$f(x)\exp(g(x)) \, \mathrm dx$$ can be expressed in terms of elementary functions if and only if there exists some rational function $h$ such that it is a solution of $$f = h' + hg'$$ $e^{x^2}$ is another classic example of such a function with no elementary antiderivative. I don't know how much math you've had, but some of this paper might be comprehensible in its broad strokes: https://ksda.ccny.cuny.edu/PostedPapers/liouv06.pdf Liouville's original paper: Liouville, J. "Suite du Mémoire sur la classification des Transcendantes, et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients." J. Math. Pure Appl. 3, 523-546, 1838. Michael Spivak's book on Calculus also has a section with a discussion of this.
| 169
| true
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Q: How can you prove that a function has no closed form integral?
In the past, I've come across statements along the lines of "function $f(x)$ has no closed form integral", which I assume means that there is no combination of the operations: addition/subtraction multiplication/division raising to powers and roots trigonometric functions exponential functions logarithmic functions which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example. What sort of techniques are used to prove statements like this? What is this branch of mathematics called? Merged with "How to prove that some functions don't have a primitive" by Ismael: Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions. I wonder how we can prove that kind of assertion?
A: It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions $f$ and $g$, $g$ non-constant, the antiderivative of $$f(x)\exp(g(x)) \, \mathrm dx$$ can be expressed in terms of elementary functions if and only if there exists some rational function $h$ such that it is a solution of $$f = h' + hg'$$ $e^{x^2}$ is another classic example of such a function with no elementary antiderivative. I don't know how much math you've had, but some of this paper might be comprehensible in its broad strokes: https://ksda.ccny.cuny.edu/PostedPapers/liouv06.pdf Liouville's original paper: Liouville, J. "Suite du Mémoire sur la classification des Transcendantes, et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients." J. Math. Pure Appl. 3, 523-546, 1838. Michael Spivak's book on Calculus also has a section with a discussion of this.
| 2026-01-26T11:30:09.361707
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23,312
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What is the importance of eigenvalues/eigenvectors?
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What is the importance of eigenvalues/eigenvectors?
| 372
|
linear-algebra, matrices, eigenvalues-eigenvectors, soft-question
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https://math.stackexchange.com/questions/23312/what-is-the-importance-of-eigenvalues-eigenvectors
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Short Answer Eigenvectors make understanding linear transformations easy. They are the "axes" (directions) along which a linear transformation acts simply by "stretching/compressing" and/or "flipping"; eigenvalues give you the factors by which this compression occurs. The more directions you have along which you understand the behavior of a linear transformation, the easier it is to understand the linear transformation; so you want to have as many linearly independent eigenvectors as possible associated to a single linear transformation. Slightly Longer Answer There are a lot of problems that can be modeled with linear transformations, and the eigenvectors give very simply solutions. For example, consider the system of linear differential equations \begin{align*} \frac{dx}{dt} &= ax + by\\\ \frac{dy}{dt} &= cx + dy. \end{align*} This kind of system arises when you describe, for example, the growth of population of two species that affect one another. For example, you might have that species $x$ is a predator on species $y$; the more $x$ you have, the fewer $y$ will be around to reproduce; but the fewer $y$ that are around, the less food there is for $x$, so fewer $x$s will reproduce; but then fewer $x$s are around so that takes pressure off $y$, which increases; but then there is more food for $x$, so $x$ increases; and so on and so forth. It also arises when you have certain physical phenomena, such a particle on a moving fluid, where the velocity vector depends on the position along the fluid. Solving this system directly is complicated. But suppose that you could do a change of variable so that instead of working with $x$ and $y$, you could work with $z$ and $w$ (which depend linearly on $x$ and also $y$; that is, $z=\alpha x+\beta y$ for some constants $\alpha$ and $\beta$, and $w=\gamma x + \delta y$, for some constants $\gamma$ and $\delta$) and the system transformed into something like \begin{align*} \frac{dz}{dt} &= \kappa z\\\ \frac{dw}{dt} &= \lambda w \end{align*} that is, you can "decouple" the system, so that now you are dealing with two independent functions. Then solving this problem becomes rather easy: $z=Ae^{\kappa t}$, and $w=Be^{\lambda t}$. Then you can use the formulas for $z$ and $w$ to find expressions for $x$ and $y$.. Can this be done? Well, it amounts precisely to finding two linearly independent eigenvectors for the matrix $\left(\begin{array}{cc}a & b\\c & d\end{array}\right)$! $z$ and $w$ correspond to the eigenvectors, and $\kappa$ and $\lambda$ to the eigenvalues. By taking an expression that "mixes" $x$ and $y$, and "decoupling it" into one that acts independently on two different functions, the problem becomes a lot easier. That is the essence of what one hopes to do with the eigenvectors and eigenvalues: "decouple" the ways in which the linear transformation acts into a number of independent actions along separate "directions", that can be dealt with independently. A lot of problems come down to figuring out these "lines of independent action", and understanding them can really help you figure out what the matrix/linear transformation is "really" doing.
| 385
| true
|
Q: What is the importance of eigenvalues/eigenvectors?
What is the importance of eigenvalues/eigenvectors?
A: Short Answer Eigenvectors make understanding linear transformations easy. They are the "axes" (directions) along which a linear transformation acts simply by "stretching/compressing" and/or "flipping"; eigenvalues give you the factors by which this compression occurs. The more directions you have along which you understand the behavior of a linear transformation, the easier it is to understand the linear transformation; so you want to have as many linearly independent eigenvectors as possible associated to a single linear transformation. Slightly Longer Answer There are a lot of problems that can be modeled with linear transformations, and the eigenvectors give very simply solutions. For example, consider the system of linear differential equations \begin{align*} \frac{dx}{dt} &= ax + by\\\ \frac{dy}{dt} &= cx + dy. \end{align*} This kind of system arises when you describe, for example, the growth of population of two species that affect one another. For example, you might have that species $x$ is a predator on species $y$; the more $x$ you have, the fewer $y$ will be around to reproduce; but the fewer $y$ that are around, the less food there is for $x$, so fewer $x$s will reproduce; but then fewer $x$s are around so that takes pressure off $y$, which increases; but then there is more food for $x$, so $x$ increases; and so on and so forth. It also arises when you have certain physical phenomena, such a particle on a moving fluid, where the velocity vector depends on the position along the fluid. Solving this system directly is complicated. But suppose that you could do a change of variable so that instead of working with $x$ and $y$, you could work with $z$ and $w$ (which depend linearly on $x$ and also $y$; that is, $z=\alpha x+\beta y$ for some constants $\alpha$ and $\beta$, and $w=\gamma x + \delta y$, for some constants $\gamma$ and $\delta$) and the system transformed into something like \begin{align*} \frac{dz}{dt} &= \kappa z\\\ \frac{dw}{dt} &= \lambda w \end{align*} that is, you can "decouple" the system, so that now you are dealing with two independent functions. Then solving this problem becomes rather easy: $z=Ae^{\kappa t}$, and $w=Be^{\lambda t}$. Then you can use the formulas for $z$ and $w$ to find expressions for $x$ and $y$.. Can this be done? Well, it amounts precisely to finding two linearly independent eigenvectors for the matrix $\left(\begin{array}{cc}a & b\\c & d\end{array}\right)$! $z$ and $w$ correspond to the eigenvectors, and $\kappa$ and $\lambda$ to the eigenvalues. By taking an expression that "mixes" $x$ and $y$, and "decoupling it" into one that acts independently on two different functions, the problem becomes a lot easier. That is the essence of what one hopes to do with the eigenvectors and eigenvalues: "decouple" the ways in which the linear transformation acts into a number of independent actions along separate "directions", that can be dealt with independently. A lot of problems come down to figuring out these "lines of independent action", and understanding them can really help you figure out what the matrix/linear transformation is "really" doing.
| 2026-01-26T11:30:11.244342
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11
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Is it true that $0.999999999\ldots=1$?
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I'm told by smart people that $$0.999999999\ldots=1$$ and I believe them, but is there a proof that explains why this is?
| 371
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real-analysis, algebra-precalculus, real-numbers, big-list, decimal-expansion
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https://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-ldots-1
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What does it mean when you refer to $.99999\ldots$? Symbols don't mean anything in particular until you've defined what you mean by them. In this case the definition is that you are taking the limit of $.9$, $.99$, $.999$, $.9999$, etc. What does it mean to say that limit is $1$? Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$. But certainly whatever number you choose your number is bigger than $10^{-k}$ for some $k$. So I can just pick my point to be the $k$th spot in the sequence. A more intuitive way of explaining the above argument is that the reason $.99999\ldots = 1$ is that their difference is zero. So let's subtract $1.0000\ldots -.99999\ldots = .00000\ldots = 0$. That is, $1.0 -.9 = .1$ $1.00-.99 = .01$ $1.000-.999=.001$, $\ldots$ $1.000\ldots -.99999\ldots = .000\ldots = 0$
| 445
| true
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Q: Is it true that $0.999999999\ldots=1$?
I'm told by smart people that $$0.999999999\ldots=1$$ and I believe them, but is there a proof that explains why this is?
A: What does it mean when you refer to $.99999\ldots$? Symbols don't mean anything in particular until you've defined what you mean by them. In this case the definition is that you are taking the limit of $.9$, $.99$, $.999$, $.9999$, etc. What does it mean to say that limit is $1$? Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$. But certainly whatever number you choose your number is bigger than $10^{-k}$ for some $k$. So I can just pick my point to be the $k$th spot in the sequence. A more intuitive way of explaining the above argument is that the reason $.99999\ldots = 1$ is that their difference is zero. So let's subtract $1.0000\ldots -.99999\ldots = .00000\ldots = 0$. That is, $1.0 -.9 = .1$ $1.00-.99 = .01$ $1.000-.999=.001$, $\ldots$ $1.000\ldots -.99999\ldots = .000\ldots = 0$
| 2026-01-26T11:30:13.085506
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1,633,704
|
Calculating the length of the paper on a toilet paper roll
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Fun with Math time. My mom gave me a roll of toilet paper to put it in the bathroom, and looking at it I immediately wondered about this: is it possible, through very simple math, to calculate (with small error) the total paper length of a toilet roll? Writing down some math, I came to this study, which I share with you because there are some questions I have in mind, and because as someone rightly said: for every problem there always are at least 3 solutions. I started by outlining the problem in a geometrical way, namely looking only at the essential: the roll from above, identifying the salient parameters: Parameters $r = $ radius of internal circle, namely the paper tube circle; $R = $ radius of the whole paper roll; $b = R - r = $ "partial" radius, namely the difference of two radii as stated. First Point I treated the whole problem in the discrete way. [See the end of this question for more details about what does it mean] Calculation In a discrete way, the problem asks for the total length of the rolled paper, so the easiest way is to treat the problem by thinking about the length as the sum of the whole circumferences starting by radius $r$ and ending with radius $R$. But how many circumferences are there? Here is one of the main points, and then I thought about introducing a new essential parameter, namely the thickness of a single sheet. Notice that it's important to have to do with measurable quantities. Calling $h$ the thickness of a single sheet, and knowing $b$ we can give an estimate of how many sheets $N$ are rolled: $$N = \frac{R - r}{h} = \frac{b}{h}$$ Having to compute a sum, the total length $L$ is then: $$L = 2\pi r + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi R$$ or better: $$L = 2\pi (r + 0h) + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi (r + Nh)$$ In which obviously $2\pi (r + 0h) = 2\pi r$ and $2\pi(r + Nh) = 2\pi R$. Writing it as a sum (and calculating it) we get: $$ \begin{align} L = \sum_{k = 0}^N\ 2\pi(r + kh) & = 2\pi r + 2\pi R + \sum_{k = 1}^{N-1}\ 2\pi(r + kh) \\\\ & = 2\pi r + 2\pi R + 2\pi \sum_{k = 1}^{N-1} r + 2\pi h \sum_{k = 1}^{N-1} k \\\\ & = 2\pi r + 2\pi R + 2\pi r(N-1) + 2\pi h\left(\frac{1}{2}N(N-1)\right) \\\\ & = 2\pi r N + 2\pi R + \pi hN^2 - \pi h N \end{align} $$ Using now: $N = \frac{b}{h}$; $R = b - a$ and $a = R - b$ (because $R$ is easily measurable), we arrive after little algebra to $$\boxed{L = 4\pi b + 2\pi R\left(\frac{b}{h} - 1\right) - \pi b\left(1 + \frac{b}{h}\right)}$$ Small Example: $h = 0.1$ mm; $R = 75$ mm; $b = 50$ mm thence $L = 157$ meters which might fit. Final Questions: 1) Could it be a good approximation? 2) What about the $\gamma$ factor? Namely the paper compression factor? 3) Could exist a similar calculation via integration over a spiral path? Because actually it's what it is: a spiral. Thank you so much for the time spent for this maybe tedious maybe boring maybe funny question!
| 365
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calculus, integration, summation, recreational-mathematics
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https://math.stackexchange.com/questions/1633704/calculating-the-length-of-the-paper-on-a-toilet-paper-roll
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The assumption that the layers are all cylindrical is a good first approximation. The assumption that the layers form a logarithmic spiral is not a good assumption at all, because it supposes that the thickness of the paper at any point is proportional to its distance from the center. This seems to me to be quite absurd. An alternative assumption is that the layers form an Archimedean spiral. This is slightly more realistic, since it says the paper has a uniform thickness from beginning to end. But this assumption is not a much more realistic than the assumption that all layers are cylindrical; in fact, in some ways it is less realistic. Here's how a sheet of thickness $h$ actually wraps around a cylinder. First, we glue one side of the sheet (near the end of the sheet) to the surface of the cylinder. Then we start rotating the cylinder. As the cylinder rotates, it pulls the outstretched sheet around itself. Near the end of the first full rotation of the cylinder, the wrapping looks like this: Notice that the sheet lies directly on the surface of the cylinder, that is, this part of the wrapped sheet is cylindrical. At some angle of rotation, the glued end of the sheet hits the part of the sheet that is being wrapped. The point where the sheet is tangent to the cylinder at that time is the last point of contact with the cylinder; the sheet goes straight from that point to the point of contact with the glued end, and then proceeds to wrap in a cylindrical shape around the first layer of the wrapped sheet, like this: As we continue rotating the cylinder, it takes up more and more layers of the sheet, each layer consisting of a cylindrical section going most of the way around the roll, followed by a flat section that joins this layer to the next layer. We end up with something like this: Notice that I cut the sheet just at the point where it was about to enter another straight section. I claim (without proof) that this produces a local maximum in the ratio of the length of the wrapped sheet of paper to the greatest thickness of paper around the inner cylinder. The next local maximum (I claim) will occur at the corresponding point of the next wrap of the sheet. The question now is what the thickness of each layer is. The inner surface of the cylindrical portion of each layer of the wrapped sheet has less area than the outer surface, but the portion of the original (unwrapped) sheet that was wound onto the roll to make this layer had equal area on both sides. So either the inner surface was somehow compressed, or the outer surface was stretched, or both. I think the most realistic assumption is that both compression and stretching occurred. In reality, I would guess that the inner surface is compressed more than the outer surface is stretched, but I do not know what the most likely ratio of compression to stretching would be. It is simpler to assume that the two effects are equal. The length of the sheet used to make any part of one layer of the roll is therefore equal to the length of the surface midway between the inner and outer surfaces of that layer. For example, to wrap the first layer halfway around the central cylinder of radius $r$, we use a length $\pi\left(r + \frac h2\right)$ of the sheet of paper. The reason this particularly simplifies our calculations is that the length of paper used in any part of the roll is simply the area of the cross-section of that part of the roll divided by the thickness of the paper. The entire roll has inner radius $r$ and outer radius $R = r + nh$, where $n$ is the maximum number of layers at any point around the central cylinder. (In the figure, $n = 5$.) The blue lines are sides of a right triangle whose vertices are the center of the inner cylinder and the points where the first layer last touches the inner cylinder and first touches its own end. This triangle has hypotenuse $r + h$ and one leg is $r$, so the other leg (which is the length of the straight portion of the sheet) is $$ \sqrt{(r + h)^2 - r^2} = \sqrt{(2r + h)h}.$$ Each straight portion of each layer is connected to the next layer of paper by wrapping around either the point of contact with the glued end of the sheet (the first time) or around the shape made by wrapping the previous layer around this part of the layer below; this forms a segment of a cylinder between the red lines with center at the point of contact with the glued end. The angle between the red lines is the same as the angle of the blue triangle at the center of the cylinder, namely $$ \alpha = \arccos \frac{r}{r+h}.$$ Now let's add up all parts of the roll. We have an almost-complete hollow cylinder with inner radius $r$ and outer radius $R$, missing only a segment of angle $\alpha$. The cross-sectional area of this is $$ A_1 = \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2).$$ We have a rectangular prism whose cross-sectional area is the product of two of its sides, $$ A_2 = (R - r - h) \sqrt{(2r + h)h}.$$ Finally, we have a segment of a cylinder of radius $R - r - h$ (between the red lines) whose cross-sectional area is $$ A_3 = \frac{\alpha}{2} (R - r - h)^2.$$ Adding this up and dividing by $h$, the total length of the sheet comes to \begin{align} L &= \frac1h (A_1+A_2+A_3)\\ &= \frac1h \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2) + \frac1h (R - r - h) \sqrt{(2r + h)h} + \frac{\alpha}{2h} (R - r - h)^2. \end{align} For $n$ layers on a roll, using the formula $R = r + nh$, we have $R - r = nh$, $R + r = 2r + nh$, $R^2 - r^2 = (R+r)(R-r) = (2r + nh)nh$, and $R - r - h = (n - 1)h$. The length then is \begin{align} L &= \left(\pi - \frac{\alpha}{2} \right) (2r + nh)n + (n - 1) \sqrt{(2r + h)h} + \frac{\alpha h}{2} (n - 1)^2\\ &= 2n\pi r + n^2\pi h + (n-1) \sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}\\ &= n (R + r) \pi + (n-1) \sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}. \end{align} One notable difference between this estimate and some others (including the original) is that I assume there can be at most $(R-r)/h$ layers of paper over any part of the central cylinder, not $1 + (R-r)/h$ layers. The total length is the number of layers times $2\pi$ times the average radius, $(R + r)/2$, adjusted by the amount that is missing in the section of the roll that is only $n - 1$ sheets thick. Things are not too much worse if we assume a different but uniform ratio of inner-compression to outer-stretching, provided that we keep the same paper thickness regardless of curvature; we just have to make an adjustment to the inner and outer radii of any cylindrical segment of the roll, which I think I'll leave as "an exercise for the reader." But this involves a change in volume of the sheet of paper. If we also keep the volume constant, we find that the sheet gets thicker or thinner depending on the ratio of stretch to compression and the curvature of the sheet. With constant volume, the length of paper in the main part of the roll (everywhere where we get the the full number of layers) is the same as in the estimate above, but the total length of the parts of the sheet that connect one layer to the next might change slightly. Update: Per request, here are the results of applying the formula above to the input values given as an example in the question: $h=0.1$, $R=75$, and $r=25$ (inferred from $R-r=b=50$), all measured in millimeters. Since $n = (R-r)/h$, we have $n = 500$. For a first approximation of the total length of paper, let's consider just the first term of the formula. This gives us $$ L_1 = n (R + r) \pi = 500 \cdot 100 \pi \approx 157079.63267949, $$ or about $157$ meters, the same as in the example in the question. The remaining two terms yield \begin{align} L - L_1 &= (n-1)\sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos\frac{r}{r+h} \\ &= 499\sqrt{50.1 \cdot 0.1} - (500(25.1) - 0.05)\arccos\frac{25}{25.1} \\ &\approx -3.72246774. \end{align} This is a very small correction, less than $2.4\times 10^{-5} L_1$. In reality (as opposed to my idealized model of constant-thickness constant-volume toilet paper), this "correction" is surely insignificant compared to the uncertainties of estimating the average thickness of the paper in each layer of a roll (not to mention any non-uniformity in how it is rolled by the manufacturing machinery). We can also compare $\lvert L - L_1 \rvert$ to the amount of paper that would be missing if the paper in the "flat" segment of the roll were instead $n - 1$ layers following the curve of the rest of the paper. The angle $\alpha$ is about $0.089294$ radians (about $5.1162$ degrees), so if the missing layer were the innermost layer, its length would be $25.05 \alpha \approx 2.24$, and if it were the outermost layer it would be $74.95 \alpha \approx 6.69$ (in millimeters). Just for amusement, I also tried expanding $L - L_1$ as a power series around $h = 0$ (with a little help from Wolfram Alpha). (To make $L - L_1$ a function of one variable $h$ with constants $R$ and $r$, make the substitution $n = (R - r)/h$.) This turns out to be a series of powers of $\sqrt h$ whose leading term is $$ -\frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h. $$ Plugging in the values from the example, this evaluates to approximately $-3.7267799625$. If you really wanted the length of the idealized toilet roll to the nearest millimeter, but could tolerate an error of a few $\mu\mathrm m$ (for typical dimensions of a toilet roll), a suitable approximation would be $$ L \approx \frac{\pi (R^2 - r^2)}{h} - \frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h. $$
| 275
| true
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Q: Calculating the length of the paper on a toilet paper roll
Fun with Math time. My mom gave me a roll of toilet paper to put it in the bathroom, and looking at it I immediately wondered about this: is it possible, through very simple math, to calculate (with small error) the total paper length of a toilet roll? Writing down some math, I came to this study, which I share with you because there are some questions I have in mind, and because as someone rightly said: for every problem there always are at least 3 solutions. I started by outlining the problem in a geometrical way, namely looking only at the essential: the roll from above, identifying the salient parameters: Parameters $r = $ radius of internal circle, namely the paper tube circle; $R = $ radius of the whole paper roll; $b = R - r = $ "partial" radius, namely the difference of two radii as stated. First Point I treated the whole problem in the discrete way. [See the end of this question for more details about what does it mean] Calculation In a discrete way, the problem asks for the total length of the rolled paper, so the easiest way is to treat the problem by thinking about the length as the sum of the whole circumferences starting by radius $r$ and ending with radius $R$. But how many circumferences are there? Here is one of the main points, and then I thought about introducing a new essential parameter, namely the thickness of a single sheet. Notice that it's important to have to do with measurable quantities. Calling $h$ the thickness of a single sheet, and knowing $b$ we can give an estimate of how many sheets $N$ are rolled: $$N = \frac{R - r}{h} = \frac{b}{h}$$ Having to compute a sum, the total length $L$ is then: $$L = 2\pi r + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi R$$ or better: $$L = 2\pi (r + 0h) + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi (r + Nh)$$ In which obviously $2\pi (r + 0h) = 2\pi r$ and $2\pi(r + Nh) = 2\pi R$. Writing it as a sum (and calculating it) we get: $$ \begin{align} L = \sum_{k = 0}^N\ 2\pi(r + kh) & = 2\pi r + 2\pi R + \sum_{k = 1}^{N-1}\ 2\pi(r + kh) \\\\ & = 2\pi r + 2\pi R + 2\pi \sum_{k = 1}^{N-1} r + 2\pi h \sum_{k = 1}^{N-1} k \\\\ & = 2\pi r + 2\pi R + 2\pi r(N-1) + 2\pi h\left(\frac{1}{2}N(N-1)\right) \\\\ & = 2\pi r N + 2\pi R + \pi hN^2 - \pi h N \end{align} $$ Using now: $N = \frac{b}{h}$; $R = b - a$ and $a = R - b$ (because $R$ is easily measurable), we arrive after little algebra to $$\boxed{L = 4\pi b + 2\pi R\left(\frac{b}{h} - 1\right) - \pi b\left(1 + \frac{b}{h}\right)}$$ Small Example: $h = 0.1$ mm; $R = 75$ mm; $b = 50$ mm thence $L = 157$ meters which might fit. Final Questions: 1) Could it be a good approximation? 2) What about the $\gamma$ factor? Namely the paper compression factor? 3) Could exist a similar calculation via integration over a spiral path? Because actually it's what it is: a spiral. Thank you so much for the time spent for this maybe tedious maybe boring maybe funny question!
A: The assumption that the layers are all cylindrical is a good first approximation. The assumption that the layers form a logarithmic spiral is not a good assumption at all, because it supposes that the thickness of the paper at any point is proportional to its distance from the center. This seems to me to be quite absurd. An alternative assumption is that the layers form an Archimedean spiral. This is slightly more realistic, since it says the paper has a uniform thickness from beginning to end. But this assumption is not a much more realistic than the assumption that all layers are cylindrical; in fact, in some ways it is less realistic. Here's how a sheet of thickness $h$ actually wraps around a cylinder. First, we glue one side of the sheet (near the end of the sheet) to the surface of the cylinder. Then we start rotating the cylinder. As the cylinder rotates, it pulls the outstretched sheet around itself. Near the end of the first full rotation of the cylinder, the wrapping looks like this: Notice that the sheet lies directly on the surface of the cylinder, that is, this part of the wrapped sheet is cylindrical. At some angle of rotation, the glued end of the sheet hits the part of the sheet that is being wrapped. The point where the sheet is tangent to the cylinder at that time is the last point of contact with the cylinder; the sheet goes straight from that point to the point of contact with the glued end, and then proceeds to wrap in a cylindrical shape around the first layer of the wrapped sheet, like this: As we continue rotating the cylinder, it takes up more and more layers of the sheet, each layer consisting of a cylindrical section going most of the way around the roll, followed by a flat section that joins this layer to the next layer. We end up with something like this: Notice that I cut the sheet just at the point where it was about to enter another straight section. I claim (without proof) that this produces a local maximum in the ratio of the length of the wrapped sheet of paper to the greatest thickness of paper around the inner cylinder. The next local maximum (I claim) will occur at the corresponding point of the next wrap of the sheet. The question now is what the thickness of each layer is. The inner surface of the cylindrical portion of each layer of the wrapped sheet has less area than the outer surface, but the portion of the original (unwrapped) sheet that was wound onto the roll to make this layer had equal area on both sides. So either the inner surface was somehow compressed, or the outer surface was stretched, or both. I think the most realistic assumption is that both compression and stretching occurred. In reality, I would guess that the inner surface is compressed more than the outer surface is stretched, but I do not know what the most likely ratio of compression to stretching would be. It is simpler to assume that the two effects are equal. The length of the sheet used to make any part of one layer of the roll is therefore equal to the length of the surface midway between the inner and outer surfaces of that layer. For example, to wrap the first layer halfway around the central cylinder of radius $r$, we use a length $\pi\left(r + \frac h2\right)$ of the sheet of paper. The reason this particularly simplifies our calculations is that the length of paper used in any part of the roll is simply the area of the cross-section of that part of the roll divided by the thickness of the paper. The entire roll has inner radius $r$ and outer radius $R = r + nh$, where $n$ is the maximum number of layers at any point around the central cylinder. (In the figure, $n = 5$.) The blue lines are sides of a right triangle whose vertices are the center of the inner cylinder and the points where the first layer last touches the inner cylinder and first touches its own end. This triangle has hypotenuse $r + h$ and one leg is $r$, so the other leg (which is the length of the straight portion of the sheet) is $$ \sqrt{(r + h)^2 - r^2} = \sqrt{(2r + h)h}.$$ Each straight portion of each layer is connected to the next layer of paper by wrapping around either the point of contact with the glued end of the sheet (the first time) or around the shape made by wrapping the previous layer around this part of the layer below; this forms a segment of a cylinder between the red lines with center at the point of contact with the glued end. The angle between the red lines is the same as the angle of the blue triangle at the center of the cylinder, namely $$ \alpha = \arccos \frac{r}{r+h}.$$ Now let's add up all parts of the roll. We have an almost-complete hollow cylinder with inner radius $r$ and outer radius $R$, missing only a segment of angle $\alpha$. The cross-sectional area of this is $$ A_1 = \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2).$$ We have a rectangular prism whose cross-sectional area is the product of two of its sides, $$ A_2 = (R - r - h) \sqrt{(2r + h)h}.$$ Finally, we have a segment of a cylinder of radius $R - r - h$ (between the red lines) whose cross-sectional area is $$ A_3 = \frac{\alpha}{2} (R - r - h)^2.$$ Adding this up and dividing by $h$, the total length of the sheet comes to \begin{align} L &= \frac1h (A_1+A_2+A_3)\\ &= \frac1h \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2) + \frac1h (R - r - h) \sqrt{(2r + h)h} + \frac{\alpha}{2h} (R - r - h)^2. \end{align} For $n$ layers on a roll, using the formula $R = r + nh$, we have $R - r = nh$, $R + r = 2r + nh$, $R^2 - r^2 = (R+r)(R-r) = (2r + nh)nh$, and $R - r - h = (n - 1)h$. The length then is \begin{align} L &= \left(\pi - \frac{\alpha}{2} \right) (2r + nh)n + (n - 1) \sqrt{(2r + h)h} + \frac{\alpha h}{2} (n - 1)^2\\ &= 2n\pi r + n^2\pi h + (n-1) \sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}\\ &= n (R + r) \pi + (n-1) \sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}. \end{align} One notable difference between this estimate and some others (including the original) is that I assume there can be at most $(R-r)/h$ layers of paper over any part of the central cylinder, not $1 + (R-r)/h$ layers. The total length is the number of layers times $2\pi$ times the average radius, $(R + r)/2$, adjusted by the amount that is missing in the section of the roll that is only $n - 1$ sheets thick. Things are not too much worse if we assume a different but uniform ratio of inner-compression to outer-stretching, provided that we keep the same paper thickness regardless of curvature; we just have to make an adjustment to the inner and outer radii of any cylindrical segment of the roll, which I think I'll leave as "an exercise for the reader." But this involves a change in volume of the sheet of paper. If we also keep the volume constant, we find that the sheet gets thicker or thinner depending on the ratio of stretch to compression and the curvature of the sheet. With constant volume, the length of paper in the main part of the roll (everywhere where we get the the full number of layers) is the same as in the estimate above, but the total length of the parts of the sheet that connect one layer to the next might change slightly. Update: Per request, here are the results of applying the formula above to the input values given as an example in the question: $h=0.1$, $R=75$, and $r=25$ (inferred from $R-r=b=50$), all measured in millimeters. Since $n = (R-r)/h$, we have $n = 500$. For a first approximation of the total length of paper, let's consider just the first term of the formula. This gives us $$ L_1 = n (R + r) \pi = 500 \cdot 100 \pi \approx 157079.63267949, $$ or about $157$ meters, the same as in the example in the question. The remaining two terms yield \begin{align} L - L_1 &= (n-1)\sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos\frac{r}{r+h} \\ &= 499\sqrt{50.1 \cdot 0.1} - (500(25.1) - 0.05)\arccos\frac{25}{25.1} \\ &\approx -3.72246774. \end{align} This is a very small correction, less than $2.4\times 10^{-5} L_1$. In reality (as opposed to my idealized model of constant-thickness constant-volume toilet paper), this "correction" is surely insignificant compared to the uncertainties of estimating the average thickness of the paper in each layer of a roll (not to mention any non-uniformity in how it is rolled by the manufacturing machinery). We can also compare $\lvert L - L_1 \rvert$ to the amount of paper that would be missing if the paper in the "flat" segment of the roll were instead $n - 1$ layers following the curve of the rest of the paper. The angle $\alpha$ is about $0.089294$ radians (about $5.1162$ degrees), so if the missing layer were the innermost layer, its length would be $25.05 \alpha \approx 2.24$, and if it were the outermost layer it would be $74.95 \alpha \approx 6.69$ (in millimeters). Just for amusement, I also tried expanding $L - L_1$ as a power series around $h = 0$ (with a little help from Wolfram Alpha). (To make $L - L_1$ a function of one variable $h$ with constants $R$ and $r$, make the substitution $n = (R - r)/h$.) This turns out to be a series of powers of $\sqrt h$ whose leading term is $$ -\frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h. $$ Plugging in the values from the example, this evaluates to approximately $-3.7267799625$. If you really wanted the length of the idealized toilet roll to the nearest millimeter, but could tolerate an error of a few $\mu\mathrm m$ (for typical dimensions of a toilet roll), a suitable approximation would be $$ L \approx \frac{\pi (R^2 - r^2)}{h} - \frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h. $$
| 2026-01-26T11:30:14.733982
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259,584
|
Why don't we define "imaginary" numbers for every "impossibility"?
|
Before, the concept of imaginary numbers, the number $i = \sqrt{-1}$ was shown to have no solution among the numbers that we had. So we declared $i$ to be a new type of number. How come we don't do the same for other "impossible" equations, such as $x = x + 1$, or $x = 1/0$? Edit: OK, a lot of people have said that a number $x$ such that $x = x + 1$ would break the rule that $0 \neq 1$. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining $-1$ to be the number such that $-1 + 1 = 0$. Note that this breaks the "rule" that "if $x \leq y$, then $ax \leq ay$", which was true for all $a, x, y$ before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing.
| 361
|
soft-question
|
https://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility
|
Here's one key difference between the cases. Suppose we add to the reals an element $i$ such that $i^2 = -1$, and then include everything else you can get from $i$ by applying addition and multiplication, while still preserving the usual rules of addition and multiplication. Expanding the reals to the complex numbers in this way does not enable us to prove new equations among the original reals that are inconsistent with previously established equations. Suppose by contrast we add to the reals a new element $k$ postulated to be such that $k + 1 = k$ and then also add every further element you can get by applying addition and multiplication to the reals and this new element $k$. Then we have, for example, $k + 1 + 1 = k + 1$. Hence -- assuming that old and new elements together still obey the usual rules of arithmetic -- we can cheerfully subtract $k$ from each side to "prove" $2 = 1$. Ooops! Adding the postulated element $k$ enables us to prove new equations flatly inconsistent what we already know. Very bad news! Now, we can in fact add an element like $k$ consistently if we are prepared to alter the usual rules of addition. That is to say, if we not only add new elements but also change the rules of arithmetic at the same time, then we can stay safe. This is, for example, exactly what happens when we augment the finite ordinals with infinite ordinals. We get a consistent theory at the cost e.g. of having cases such as $\omega + 1 \neq 1 + \omega$ and $1 + 1 + \omega = 1 + \omega$.
| 194
| true
|
Q: Why don't we define "imaginary" numbers for every "impossibility"?
Before, the concept of imaginary numbers, the number $i = \sqrt{-1}$ was shown to have no solution among the numbers that we had. So we declared $i$ to be a new type of number. How come we don't do the same for other "impossible" equations, such as $x = x + 1$, or $x = 1/0$? Edit: OK, a lot of people have said that a number $x$ such that $x = x + 1$ would break the rule that $0 \neq 1$. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining $-1$ to be the number such that $-1 + 1 = 0$. Note that this breaks the "rule" that "if $x \leq y$, then $ax \leq ay$", which was true for all $a, x, y$ before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing.
A: Here's one key difference between the cases. Suppose we add to the reals an element $i$ such that $i^2 = -1$, and then include everything else you can get from $i$ by applying addition and multiplication, while still preserving the usual rules of addition and multiplication. Expanding the reals to the complex numbers in this way does not enable us to prove new equations among the original reals that are inconsistent with previously established equations. Suppose by contrast we add to the reals a new element $k$ postulated to be such that $k + 1 = k$ and then also add every further element you can get by applying addition and multiplication to the reals and this new element $k$. Then we have, for example, $k + 1 + 1 = k + 1$. Hence -- assuming that old and new elements together still obey the usual rules of arithmetic -- we can cheerfully subtract $k$ from each side to "prove" $2 = 1$. Ooops! Adding the postulated element $k$ enables us to prove new equations flatly inconsistent what we already know. Very bad news! Now, we can in fact add an element like $k$ consistently if we are prepared to alter the usual rules of addition. That is to say, if we not only add new elements but also change the rules of arithmetic at the same time, then we can stay safe. This is, for example, exactly what happens when we augment the finite ordinals with infinite ordinals. We get a consistent theory at the cost e.g. of having cases such as $\omega + 1 \neq 1 + \omega$ and $1 + 1 + \omega = 1 + \omega$.
| 2026-01-26T11:30:16.330727
|
258,736
|
Limit of sequence of growing matrices
|
Let $$ H=\left(\begin{array}{cccc} 0 & 1/2 & 0 & 1/2 \\ 1/2 & 0 & 1/2 & 0 \\ 1/2 & 0 & 0 & 1/2\\ 0 & 1/2 & 1/2 & 0 \end{array}\right), $$ $K_1=\left(\begin{array}{c}1 \\ 0\end{array}\right)$ and consider the sequence of matrices defined by $$ K_L = \underset{2^{L}\times 2^{L}}{\underbrace{\left[H\otimes I_{2^{L-2}}\right]}}\underset{2^{L}\times 2^{L-1}}{\underbrace{\left[I_2 \otimes K_{L-1}\right]}}\in\mathbb{R}^{2^L\times 2^{L-1}}, $$ where $\otimes$ denotes the Kronecker product, and $I_n$ is the $n\times n$ identity matrix. I am interested in the limiting behaviour of the singular values of $K_L$ as $L$ tends to infinity. Some calculations indicate that the $2^L\times 2^{L-1}$-matrix $K_L$ has $L$ non-zero singular values and that the empirical distribution of those nonzero singular values converges to some limit. Can this limit be described in terms of the matrix $H$? I am wondering if it is possible to use some kind of fixed-point theorem to characterise the limit (in any sense) $\lim_{L\to\infty}K_L$ as an operator on some sequence space. Edit: I did some more experiments and it seems that the limiting behaviour of the singular values of $K_L$ does not only depend on the matrix $H$, but also on the initial value $K_1$. To illustrate this, let $K_1(\alpha)=\left(\begin{array}{c}1 \\ \alpha\end{array}\right)$ and consider the sequence $$ K_L(\alpha) = \left[H\otimes I_{2^{L-2}}\right]\left[I_2 \otimes K_{L-1}(\alpha)\right]. $$ The largest singular value of $K_{10}(\alpha)$ is depicted in the following figure. (The graph looks essentially the same for all $L\geq 4$ instead of $L=10$.) $K_{10}(\alpha)$" /> The minimum is approximately $(-.2936,0.7696)$. This makes it unlikely for fixed-point arguments to work in this setting. I, therefore, modify my question and ask if the limiting behaviour of the singular values of $K_L$ (or $K_L(\alpha)$) can be characterised directly in terms of $H$ and the initial value $K_1$ (or $K_1(\alpha)$). Edit 2 (March 2015): As the question is still receiving attention, let me add that I came up with a conjecture for the asymptotic behaviour of the singular values of $K_L(\alpha),$ as detailed in this MO post.
| 350
|
linear-algebra, matrices, convergence-divergence, operator-theory
|
https://math.stackexchange.com/questions/258736/limit-of-sequence-of-growing-matrices
| 0
| false
|
Q: Limit of sequence of growing matrices
Let $$ H=\left(\begin{array}{cccc} 0 & 1/2 & 0 & 1/2 \\ 1/2 & 0 & 1/2 & 0 \\ 1/2 & 0 & 0 & 1/2\\ 0 & 1/2 & 1/2 & 0 \end{array}\right), $$ $K_1=\left(\begin{array}{c}1 \\ 0\end{array}\right)$ and consider the sequence of matrices defined by $$ K_L = \underset{2^{L}\times 2^{L}}{\underbrace{\left[H\otimes I_{2^{L-2}}\right]}}\underset{2^{L}\times 2^{L-1}}{\underbrace{\left[I_2 \otimes K_{L-1}\right]}}\in\mathbb{R}^{2^L\times 2^{L-1}}, $$ where $\otimes$ denotes the Kronecker product, and $I_n$ is the $n\times n$ identity matrix. I am interested in the limiting behaviour of the singular values of $K_L$ as $L$ tends to infinity. Some calculations indicate that the $2^L\times 2^{L-1}$-matrix $K_L$ has $L$ non-zero singular values and that the empirical distribution of those nonzero singular values converges to some limit. Can this limit be described in terms of the matrix $H$? I am wondering if it is possible to use some kind of fixed-point theorem to characterise the limit (in any sense) $\lim_{L\to\infty}K_L$ as an operator on some sequence space. Edit: I did some more experiments and it seems that the limiting behaviour of the singular values of $K_L$ does not only depend on the matrix $H$, but also on the initial value $K_1$. To illustrate this, let $K_1(\alpha)=\left(\begin{array}{c}1 \\ \alpha\end{array}\right)$ and consider the sequence $$ K_L(\alpha) = \left[H\otimes I_{2^{L-2}}\right]\left[I_2 \otimes K_{L-1}(\alpha)\right]. $$ The largest singular value of $K_{10}(\alpha)$ is depicted in the following figure. (The graph looks essentially the same for all $L\geq 4$ instead of $L=10$.) $K_{10}(\alpha)$" /> The minimum is approximately $(-.2936,0.7696)$. This makes it unlikely for fixed-point arguments to work in this setting. I, therefore, modify my question and ask if the limiting behaviour of the singular values of $K_L$ (or $K_L(\alpha)$) can be characterised directly in terms of $H$ and the initial value $K_1$ (or $K_1(\alpha)$). Edit 2 (March 2015): As the question is still receiving attention, let me add that I came up with a conjecture for the asymptotic behaviour of the singular values of $K_L(\alpha),$ as detailed in this MO post.
A: [No answer available]
| 2026-01-26T11:30:17.999085
|
|
250
|
A challenge by R. P. Feynman: give counter-intuitive theorems that can be translated into everyday language
|
The following is a quote from Surely you're joking, Mr. Feynman. The question is: are there any interesting theorems that you think would be a good example to tell Richard Feynman, as an answer to his challenge? Theorems should be totally counter-intuitive, and be easily translatable to everyday language. (Apparently the Banach-Tarski paradox was not a good example.) Then I got an idea. I challenged them: "I bet there isn't a single theorem that you can tell me - what the assumptions are and what the theorem is in terms I can understand - where I can't tell you right away whether it's true or false." It often went like this: They would explain to me, "You've got an orange, OK? Now you cut the orange into a finite number of pieces, put it back together, and it's as big as the sun. True or false?" "No holes." "Impossible! "Ha! Everybody gather around! It's So-and-so's theorem of immeasurable measure!" Just when they think they've got me, I remind them, "But you said an orange! You can't cut the orange peel any thinner than the atoms." "But we have the condition of continuity: We can keep on cutting!" "No, you said an orange, so I assumed that you meant a real orange." So I always won. If I guessed it right, great. If I guessed it wrong, there was always something I could find in their simplification that they left out.
| 348
|
soft-question, big-list, examples-counterexamples
|
https://math.stackexchange.com/questions/250/a-challenge-by-r-p-feynman-give-counter-intuitive-theorems-that-can-be-transl
|
Every simple closed curve that you can draw by hand will pass through the corners of some square. The question was asked by Toeplitz in 1911, and has only been partially answered in 1989 by Stromquist. As of now, the answer is only known to be positive, for the curves that can be drawn by hand. (i.e. the curves that are piecewise the graph of a continuous function) I find the result beyond my intuition. For details, see http://www.webpages.uidaho.edu/~markn/squares/ (the figure is also borrowed from this site)
| 323
| false
|
Q: A challenge by R. P. Feynman: give counter-intuitive theorems that can be translated into everyday language
The following is a quote from Surely you're joking, Mr. Feynman. The question is: are there any interesting theorems that you think would be a good example to tell Richard Feynman, as an answer to his challenge? Theorems should be totally counter-intuitive, and be easily translatable to everyday language. (Apparently the Banach-Tarski paradox was not a good example.) Then I got an idea. I challenged them: "I bet there isn't a single theorem that you can tell me - what the assumptions are and what the theorem is in terms I can understand - where I can't tell you right away whether it's true or false." It often went like this: They would explain to me, "You've got an orange, OK? Now you cut the orange into a finite number of pieces, put it back together, and it's as big as the sun. True or false?" "No holes." "Impossible! "Ha! Everybody gather around! It's So-and-so's theorem of immeasurable measure!" Just when they think they've got me, I remind them, "But you said an orange! You can't cut the orange peel any thinner than the atoms." "But we have the condition of continuity: We can keep on cutting!" "No, you said an orange, so I assumed that you meant a real orange." So I always won. If I guessed it right, great. If I guessed it wrong, there was always something I could find in their simplification that they left out.
A: Every simple closed curve that you can draw by hand will pass through the corners of some square. The question was asked by Toeplitz in 1911, and has only been partially answered in 1989 by Stromquist. As of now, the answer is only known to be positive, for the curves that can be drawn by hand. (i.e. the curves that are piecewise the graph of a continuous function) I find the result beyond my intuition. For details, see http://www.webpages.uidaho.edu/~markn/squares/ (the figure is also borrowed from this site)
| 2026-01-26T11:30:19.590326
|
362,446
|
Nice examples of groups which are not obviously groups
|
I am searching for some groups, where it is not so obvious that they are groups. In the lecture's script there are only examples like $\mathbb{Z}$ under addition and other things like that. I don't think that these examples are helpful to understand the real properties of a group, when only looking to such trivial examples. I am searching for some more exotic examples, like the power set of a set together with the symmetric difference, or an elliptic curve with its group law.
| 348
|
abstract-algebra, group-theory, examples-counterexamples, big-list
|
https://math.stackexchange.com/questions/362446/nice-examples-of-groups-which-are-not-obviously-groups
|
Homological algebra. Let $A,B$ be abelian groups (or more generally objects of an abelian category) and consider the set of isomorphism classes of abelian groups $C$ together with an exact sequence $0 \to B \to C \to A \to 0$ (extensions of $A$ by $B$). It turns out that this set has a canonical group structure (isn't that surprising?!), namely the Baer sum, and that this group is isomorphic to $\mathrm{Ext}^1(A,B)$. This is also quite helpful to classify extensions for specific $A$ and $B$, since $\mathrm{Ext}$ has two long exact sequences. For details, see Weibel's book on homological algebra, Chapter 3. Similarily many obstructions in deformation theories are encoded in certain abelian groups. Combinatorial game theory. A two-person game is called combinatorial if no chance is involved and the ending condition holds, so that in each case one of the two players wins. Each player has a set of possible moves, each one resulting in a new game. There is a notion of equivalent combinatorial games. It turns out that the equivalence classes of combinatorial games can be made into a (large) group. The zero game $0$ is the game where no moves are available. A move in the sum $G+H$ of two games $G,H$ is just a move in exactly one of $G$ or $H$. The inverse $-G$ of a game $G$ is the one where the possibles moves for the two players are swapped. The equation $G+(-G)=0$ requires a proof. An important subgroup is the class of impartial games, where the same moves are available for both players (or equivalently $G=-G$). This extra structure already suffices to solve many basic combinatorial games, such as Nim. In fact, one the first results in combinatorial game theory is that the (large) group of impartial combinatorial games is isomorphic to the ordinal numbers $\mathbf{On}$ with a certain group law $\oplus$, called the Nim-sum (different from the usual ordinal addition). This identification is given by the nimber. This makes it possible to reduce complicated games to simpler ones, in fact in theory to a trivial one-pile Nim game. Even the restriction to finite ordinal numbers gives an interesting group law on the set of natural numbers $\mathbb{N}$ (see Jyrki's answer). All this can be found in the fantastic book Winning Ways ... by Conway, Berlekamp, Guy, and in Conway's On Numbers and Games. A more formal introduction can be found in this paper by Schleicher, Stoll. There you also learn that (certain) combinatorial games actually constitute a (large) totally ordered field, containing the real numbers as well as the ordinal numbers. You couldn't have guessed this rich structure from their definition, right? Algebraic topology. If $X$ is a based space, the set of homotopy classes of pointed maps $S^n \to X$ has a group structure; this is the $n$th homotopy group $\pi_n(X)$ of $X$. For $n=1$ the group structure is quite obvious, since we can compose paths and go paths backwards. But at first sight it is not obvious that we can do something like that in higher dimensions. Essentially this comes down to the cogroup structure of $S^n$. There is a nice geometric proof that $\pi_n(X)$ is abelian for $n>1$.
| 184
| false
|
Q: Nice examples of groups which are not obviously groups
I am searching for some groups, where it is not so obvious that they are groups. In the lecture's script there are only examples like $\mathbb{Z}$ under addition and other things like that. I don't think that these examples are helpful to understand the real properties of a group, when only looking to such trivial examples. I am searching for some more exotic examples, like the power set of a set together with the symmetric difference, or an elliptic curve with its group law.
A: Homological algebra. Let $A,B$ be abelian groups (or more generally objects of an abelian category) and consider the set of isomorphism classes of abelian groups $C$ together with an exact sequence $0 \to B \to C \to A \to 0$ (extensions of $A$ by $B$). It turns out that this set has a canonical group structure (isn't that surprising?!), namely the Baer sum, and that this group is isomorphic to $\mathrm{Ext}^1(A,B)$. This is also quite helpful to classify extensions for specific $A$ and $B$, since $\mathrm{Ext}$ has two long exact sequences. For details, see Weibel's book on homological algebra, Chapter 3. Similarily many obstructions in deformation theories are encoded in certain abelian groups. Combinatorial game theory. A two-person game is called combinatorial if no chance is involved and the ending condition holds, so that in each case one of the two players wins. Each player has a set of possible moves, each one resulting in a new game. There is a notion of equivalent combinatorial games. It turns out that the equivalence classes of combinatorial games can be made into a (large) group. The zero game $0$ is the game where no moves are available. A move in the sum $G+H$ of two games $G,H$ is just a move in exactly one of $G$ or $H$. The inverse $-G$ of a game $G$ is the one where the possibles moves for the two players are swapped. The equation $G+(-G)=0$ requires a proof. An important subgroup is the class of impartial games, where the same moves are available for both players (or equivalently $G=-G$). This extra structure already suffices to solve many basic combinatorial games, such as Nim. In fact, one the first results in combinatorial game theory is that the (large) group of impartial combinatorial games is isomorphic to the ordinal numbers $\mathbf{On}$ with a certain group law $\oplus$, called the Nim-sum (different from the usual ordinal addition). This identification is given by the nimber. This makes it possible to reduce complicated games to simpler ones, in fact in theory to a trivial one-pile Nim game. Even the restriction to finite ordinal numbers gives an interesting group law on the set of natural numbers $\mathbb{N}$ (see Jyrki's answer). All this can be found in the fantastic book Winning Ways ... by Conway, Berlekamp, Guy, and in Conway's On Numbers and Games. A more formal introduction can be found in this paper by Schleicher, Stoll. There you also learn that (certain) combinatorial games actually constitute a (large) totally ordered field, containing the real numbers as well as the ordinal numbers. You couldn't have guessed this rich structure from their definition, right? Algebraic topology. If $X$ is a based space, the set of homotopy classes of pointed maps $S^n \to X$ has a group structure; this is the $n$th homotopy group $\pi_n(X)$ of $X$. For $n=1$ the group structure is quite obvious, since we can compose paths and go paths backwards. But at first sight it is not obvious that we can do something like that in higher dimensions. Essentially this comes down to the cogroup structure of $S^n$. There is a nice geometric proof that $\pi_n(X)$ is abelian for $n>1$.
| 2026-01-26T11:30:21.004596
|
318,299
|
Any open subset of $\Bbb R$ is a countable union of disjoint open intervals
|
Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals. This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.
| 342
|
real-analysis, general-topology, big-list, faq
|
https://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-countable-union-of-disjoint-open-intervals
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Here’s one to get things started. Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (The term interval here includes unbounded intervals, i.e., rays.) Let $\mathscr{I}$ be the set of $\sim$-classes. Clearly $U=\bigcup_{I \in \mathscr{I}} I$. For each $I\in\mathscr{I}$ choose a rational $q_I\in I$; the map $\mathscr{I}\to\Bbb Q:I\mapsto q_I$ is injective, so $\mathscr{I}$ is countable. A variant of the same basic idea is to let $\mathscr{I}$ be the set of open intervals that are subsets of $U$. For $I,J\in\mathscr{I}$ define $I\sim J$ iff there are $I_0=I,I_1,\dots,I_n=J\in\mathscr{I}$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$. Then $\sim$ is an equivalence relation on $\mathscr{I}$. For $I\in\mathscr{I}$ let $[I]$ be the $\sim$-class of $I$. Then $\left\{\bigcup[I]:I\in\mathscr{I}\right\}$ is a decomposition of $U$ into pairwise disjoint open intervals. Both of these arguments generalize to any LOTS (= Linearly Ordered Topological Space), i.e., any linearly ordered set $\langle X,\le\rangle$ with the topology generated by the subbase of open rays $(\leftarrow,x)$ and $(x,\to)$: if $U$ is a non-empty open subset of $X$, then $U$ is the union of a family of pairwise disjoint open order-convex sets. (A set $C\subseteq X$ is order-convex if whenever $u,v\in C$ and $u, then $x\in C$. Intervals and rays are always order-convex, and the converse is true if the order is complete.) It does take a little work to verify that these sets are open. Alternatively, $U$ is the union of a family of pairwise disjoint intervals, where the intervals are allowed to have endpoints in the completion of the linear order. In general the family need not be countable, of course.
| 234
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Q: Any open subset of $\Bbb R$ is a countable union of disjoint open intervals
Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals. This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.
A: Here’s one to get things started. Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (The term interval here includes unbounded intervals, i.e., rays.) Let $\mathscr{I}$ be the set of $\sim$-classes. Clearly $U=\bigcup_{I \in \mathscr{I}} I$. For each $I\in\mathscr{I}$ choose a rational $q_I\in I$; the map $\mathscr{I}\to\Bbb Q:I\mapsto q_I$ is injective, so $\mathscr{I}$ is countable. A variant of the same basic idea is to let $\mathscr{I}$ be the set of open intervals that are subsets of $U$. For $I,J\in\mathscr{I}$ define $I\sim J$ iff there are $I_0=I,I_1,\dots,I_n=J\in\mathscr{I}$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$. Then $\sim$ is an equivalence relation on $\mathscr{I}$. For $I\in\mathscr{I}$ let $[I]$ be the $\sim$-class of $I$. Then $\left\{\bigcup[I]:I\in\mathscr{I}\right\}$ is a decomposition of $U$ into pairwise disjoint open intervals. Both of these arguments generalize to any LOTS (= Linearly Ordered Topological Space), i.e., any linearly ordered set $\langle X,\le\rangle$ with the topology generated by the subbase of open rays $(\leftarrow,x)$ and $(x,\to)$: if $U$ is a non-empty open subset of $X$, then $U$ is the union of a family of pairwise disjoint open order-convex sets. (A set $C\subseteq X$ is order-convex if whenever $u,v\in C$ and $u, then $x\in C$. Intervals and rays are always order-convex, and the converse is true if the order is complete.) It does take a little work to verify that these sets are open. Alternatively, $U$ is the union of a family of pairwise disjoint intervals, where the intervals are allowed to have endpoints in the completion of the linear order. In general the family need not be countable, of course.
| 2026-01-26T11:30:22.676209
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3,444
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Intuition for the definition of the Gamma function?
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In these notes by Terence Tao is a proof of Stirling's formula. I really like most of it, but at a crucial step he uses the integral identity $$n! = \int_{0}^{\infty} t^n e^{-t} dt$$ , coming from the Gamma function. I have a mathematical confession to make: I have never "grokked" this identity. Why should I expect the integral on the right to give me the number of elements in the symmetric group on $n$ letters? (It's not that I don't know how to prove it. It's quite fun to prove; my favorite proof observes that it is equivalent to the integral identity $\int_{0}^{\infty} e^{(x-1)t} dt = \frac{1}{1 - x}$. But if someone were to ask me, "Yes, but why, really?" I would have no idea what to say.) So what are more intuitive ways of thinking about this identity? Is there a probabilistic interpretation? What kind of random variable has probability density function $\frac{t^n}{n!} e^{-t}$? (What does this all have to do with Tate's thesis?) As a rough measure of what I'm looking for, your answer should make it obvious that $t^n e^{-t}$ attains its maximum at $t = n$. Edit: The kind of explanation I'm looking for, as I described in the comments, is similar to this explanation of the beta integral.
| 338
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probability-theory, special-functions, intuition, gamma-function
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https://math.stackexchange.com/questions/3444/intuition-for-the-definition-of-the-gamma-function
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I haven't quite got this straight yet, but I think one way to go is to think about choosing points at random from the positive reals. This answer is going to be rather longer than it really needs to be, because I'm thinking about this in a few (closely related) ways, which probably aren't all necessary, and you can decide to reject the uninteresting parts and keep anything of value. Very roughly, the idea is that if you "randomly" choose points from the positive reals and arrange them in increasing order, then the probability that the $(n+1)^\text{th}$ point is in a small interval $(t,t+dt)$ is a product of probabilities of independent events, $n$ factors of $t$ for choosing $n$ points in the interval $[0,t]$, one factor of $e^{-t}$ as all the other points are in $[t,\infty)$, one factor of $dt$ for choosing the point in $(t,t+dt)$, and a denominator of $n!$ coming from the reordering. At least, as an exercise in making a simple problem much harder, here it goes... I'll start with a bit of theory before trying to describe intuitively why the probability density $\dfrac{t^n}{n!}e^{-t}$ pops out. We can look at the homogeneous Poisson process (with rate parameter $1$). One way to think of this is to take a sequence on independent exponentially distributed random variables with rate parameter $1$, $S_1,S_2,\ldots$, and set $T_n=S_1+\cdots+S_n$. As has been commented on already, $T_{n+1}$ has the probability density function $\dfrac{t^n}{n!}e^{-t}$. I'm going to avoid proving this immediately though, as it would just reduce to manipulating some integrals. Then, the Poisson process $X(t)$ counts the number of times $T_i$ lying in the interval $[0,t]$. We can also look at Poisson point processes (aka, Poisson random measures, but that Wikipedia page is very poor). This is just makes rigorous the idea of randomly choosing unordered sets of points from a sigma-finite measure space $(E,\mathcal{E},\mu)$. Technically, it can be defined as a set of nonnegative integer-valued random variables $\{N(A)\colon A\in\mathcal{E}\}$ counting the number of points chosen from each subset A, such that $N(A)$ has the Poisson distribution of rate $\mu(A)$ and $N(A_1),N(A_2),\ldots$ are independent for pairwise disjoint sets $A_1,A_2,\ldots$. By definition, this satisfies $$ \begin{array}{}\mathbb{P}(N(A)=n)=\dfrac{\mu(A)^n}{n!}e^{-\mu(A)}.&&(1)\end{array} $$ The points $T_1,T_2,\ldots$ above defining the homogeneous Poisson process also define a Poisson random measure with respect to the Lebesgue measure $(\mathbb{R}\_+,{\cal B},\lambda)$. Once you forget about the order in which they were defined and just regard them as a random set that is, which I think is the source of the $n!$. If you think about the probability of $T_{n+1}$ being in a small interval $(t,t+\delta t)$ then this is just the same as having $N([0,t])=n$ and $N((t,t+\delta t))=1$, which has probability $\dfrac{t^n}{n!}e^{-t}\delta t$. So, how can we choose points at random so that each small set $\delta A$ has probability $\mu(\delta A)$ of containing a point, and why does $(1)$ pop out? I'm imagining a hopeless darts player randomly throwing darts about and, purely by luck, hitting the board with some of them. Consider throwing a very large number $N\gg1$ of darts, independently, so that each one only has probability $\mu(A)/N$ of hitting the set, and is distributed according to the probability distribution $\mu/\mu(A)$. This is consistent, at least, if you think about the probability of hitting a subset $B\subseteq A$. The probability of missing with all of them is $(1-\mu(A)/N)^N=e^{-\mu(A)}$. This is a multiplicative function due to independence of the number hitting disjoint sets. To get the probability of one dart hitting the set, multiply by $\mu(A)$ (one factor of $\mu(A)/N$ for each individual dart, multiplied by $N$ because there are $N$ of them). For $n$ darts, we multiply by $\mu(A)$ $n$ times, for picking $n$ darts to hit, then divide by $n!$ because we have over-counted the subsets of size $n$ by this factor (due to counting all $n!$ ways of ordering them). This gives $(1)$. I think this argument can probably be cleaned up a bit. Getting back to choosing points randomly on the positive reals, this gives a probability of $\dfrac{t^n}{n!}e^{-t}dt$ of picking $n$ in the interval $[0,t]$ and one in $(t,t+dt)$. If we sort them in order as $T_1\lt T_2\lt\cdots$ then $\mathbb{P}(T_1\gt t)=e^{-t}$, so it is exponentially distributed. Conditional on this, $T_2,T_3,\ldots$ are chosen randomly from $[T_1,\infty)$, so we see that the differences $T_{i+1}-T_{i}$ are independent and identically distributed. Why is $\dfrac{t^n}{n!}e^{-t}$ maximized at $t=n$? I'm not sure why the mode should be a simple property of a distribution. It doesn't even exist except for unimodal distributions. As $T_{n+1}$ is the sum of $n+1$ IID random variables of mean one, the law of large numbers suggests that it should be peaked approximately around $n$. The central limit theorem goes further, and gives $\dfrac{t^n}{n!}e^{-t}\approx\dfrac{1}{\sqrt{2\pi n}}e^{-(t-n)^2/{2n}}$. Stirling's formula is just this evaluated at $t=n$. What's this to do with Tate's thesis? I don't know, and I haven't read it (but intend to), but have a vague idea of what it's about. If there is anything to do with it, maybe it is something to do with the fact that we are relating the sums of independent random variables $S_1+\cdots+S_n$ distributed with respect to the Haar measure on the multiplicative group $\mathbb{R}_+$ (edit: oops, that's not true, the multiplicative Haar measure has cumulative distribution given by $\log$, not $\exp$) with randomly chosen sets according to the Haar measure on the additive group $\mathbb{R}$.
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Q: Intuition for the definition of the Gamma function?
In these notes by Terence Tao is a proof of Stirling's formula. I really like most of it, but at a crucial step he uses the integral identity $$n! = \int_{0}^{\infty} t^n e^{-t} dt$$ , coming from the Gamma function. I have a mathematical confession to make: I have never "grokked" this identity. Why should I expect the integral on the right to give me the number of elements in the symmetric group on $n$ letters? (It's not that I don't know how to prove it. It's quite fun to prove; my favorite proof observes that it is equivalent to the integral identity $\int_{0}^{\infty} e^{(x-1)t} dt = \frac{1}{1 - x}$. But if someone were to ask me, "Yes, but why, really?" I would have no idea what to say.) So what are more intuitive ways of thinking about this identity? Is there a probabilistic interpretation? What kind of random variable has probability density function $\frac{t^n}{n!} e^{-t}$? (What does this all have to do with Tate's thesis?) As a rough measure of what I'm looking for, your answer should make it obvious that $t^n e^{-t}$ attains its maximum at $t = n$. Edit: The kind of explanation I'm looking for, as I described in the comments, is similar to this explanation of the beta integral.
A: I haven't quite got this straight yet, but I think one way to go is to think about choosing points at random from the positive reals. This answer is going to be rather longer than it really needs to be, because I'm thinking about this in a few (closely related) ways, which probably aren't all necessary, and you can decide to reject the uninteresting parts and keep anything of value. Very roughly, the idea is that if you "randomly" choose points from the positive reals and arrange them in increasing order, then the probability that the $(n+1)^\text{th}$ point is in a small interval $(t,t+dt)$ is a product of probabilities of independent events, $n$ factors of $t$ for choosing $n$ points in the interval $[0,t]$, one factor of $e^{-t}$ as all the other points are in $[t,\infty)$, one factor of $dt$ for choosing the point in $(t,t+dt)$, and a denominator of $n!$ coming from the reordering. At least, as an exercise in making a simple problem much harder, here it goes... I'll start with a bit of theory before trying to describe intuitively why the probability density $\dfrac{t^n}{n!}e^{-t}$ pops out. We can look at the homogeneous Poisson process (with rate parameter $1$). One way to think of this is to take a sequence on independent exponentially distributed random variables with rate parameter $1$, $S_1,S_2,\ldots$, and set $T_n=S_1+\cdots+S_n$. As has been commented on already, $T_{n+1}$ has the probability density function $\dfrac{t^n}{n!}e^{-t}$. I'm going to avoid proving this immediately though, as it would just reduce to manipulating some integrals. Then, the Poisson process $X(t)$ counts the number of times $T_i$ lying in the interval $[0,t]$. We can also look at Poisson point processes (aka, Poisson random measures, but that Wikipedia page is very poor). This is just makes rigorous the idea of randomly choosing unordered sets of points from a sigma-finite measure space $(E,\mathcal{E},\mu)$. Technically, it can be defined as a set of nonnegative integer-valued random variables $\{N(A)\colon A\in\mathcal{E}\}$ counting the number of points chosen from each subset A, such that $N(A)$ has the Poisson distribution of rate $\mu(A)$ and $N(A_1),N(A_2),\ldots$ are independent for pairwise disjoint sets $A_1,A_2,\ldots$. By definition, this satisfies $$ \begin{array}{}\mathbb{P}(N(A)=n)=\dfrac{\mu(A)^n}{n!}e^{-\mu(A)}.&&(1)\end{array} $$ The points $T_1,T_2,\ldots$ above defining the homogeneous Poisson process also define a Poisson random measure with respect to the Lebesgue measure $(\mathbb{R}\_+,{\cal B},\lambda)$. Once you forget about the order in which they were defined and just regard them as a random set that is, which I think is the source of the $n!$. If you think about the probability of $T_{n+1}$ being in a small interval $(t,t+\delta t)$ then this is just the same as having $N([0,t])=n$ and $N((t,t+\delta t))=1$, which has probability $\dfrac{t^n}{n!}e^{-t}\delta t$. So, how can we choose points at random so that each small set $\delta A$ has probability $\mu(\delta A)$ of containing a point, and why does $(1)$ pop out? I'm imagining a hopeless darts player randomly throwing darts about and, purely by luck, hitting the board with some of them. Consider throwing a very large number $N\gg1$ of darts, independently, so that each one only has probability $\mu(A)/N$ of hitting the set, and is distributed according to the probability distribution $\mu/\mu(A)$. This is consistent, at least, if you think about the probability of hitting a subset $B\subseteq A$. The probability of missing with all of them is $(1-\mu(A)/N)^N=e^{-\mu(A)}$. This is a multiplicative function due to independence of the number hitting disjoint sets. To get the probability of one dart hitting the set, multiply by $\mu(A)$ (one factor of $\mu(A)/N$ for each individual dart, multiplied by $N$ because there are $N$ of them). For $n$ darts, we multiply by $\mu(A)$ $n$ times, for picking $n$ darts to hit, then divide by $n!$ because we have over-counted the subsets of size $n$ by this factor (due to counting all $n!$ ways of ordering them). This gives $(1)$. I think this argument can probably be cleaned up a bit. Getting back to choosing points randomly on the positive reals, this gives a probability of $\dfrac{t^n}{n!}e^{-t}dt$ of picking $n$ in the interval $[0,t]$ and one in $(t,t+dt)$. If we sort them in order as $T_1\lt T_2\lt\cdots$ then $\mathbb{P}(T_1\gt t)=e^{-t}$, so it is exponentially distributed. Conditional on this, $T_2,T_3,\ldots$ are chosen randomly from $[T_1,\infty)$, so we see that the differences $T_{i+1}-T_{i}$ are independent and identically distributed. Why is $\dfrac{t^n}{n!}e^{-t}$ maximized at $t=n$? I'm not sure why the mode should be a simple property of a distribution. It doesn't even exist except for unimodal distributions. As $T_{n+1}$ is the sum of $n+1$ IID random variables of mean one, the law of large numbers suggests that it should be peaked approximately around $n$. The central limit theorem goes further, and gives $\dfrac{t^n}{n!}e^{-t}\approx\dfrac{1}{\sqrt{2\pi n}}e^{-(t-n)^2/{2n}}$. Stirling's formula is just this evaluated at $t=n$. What's this to do with Tate's thesis? I don't know, and I haven't read it (but intend to), but have a vague idea of what it's about. If there is anything to do with it, maybe it is something to do with the fact that we are relating the sums of independent random variables $S_1+\cdots+S_n$ distributed with respect to the Haar measure on the multiplicative group $\mathbb{R}_+$ (edit: oops, that's not true, the multiplicative Haar measure has cumulative distribution given by $\log$, not $\exp$) with randomly chosen sets according to the Haar measure on the additive group $\mathbb{R}$.
| 2026-01-26T11:30:24.346163
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942,263
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Really advanced techniques of integration (definite or indefinite)
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Okay, so everyone knows the usual methods of solving integrals, namely u-substitution, integration by parts, partial fractions, trig substitutions, and reduction formulas. But what else is there? Every time I search for "Advanced Techniques of Symbolic Integration" or "Super Advanced Integration Techniques", I get the same results which end up only talking about the methods mentioned above. Are there any super obscure and interesting techniques for solving integrals? As an example of something that might be obscure, the formula for "general integration by parts " for $n$ functions $f_j, \ j = 1,\cdots,n$ is given by $$ \int{f_1'(x)\prod_{j=2}^n{f_j(x)}dx} = \prod_{i=1}^n{f_i(x)} - \sum_{i=2}^n{\int{f_i'(x)\prod_{\substack{j=1 \\ j \neq i}}^n{f_j(x)}dx}} $$ which is not necessarily useful nor difficult to derive, but is interesting nonetheless. So out of curiosity, are there any crazy unknown symbolic integration techniques?
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integration, soft-question, definite-integrals, indefinite-integrals, big-list
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https://math.stackexchange.com/questions/942263/really-advanced-techniques-of-integration-definite-or-indefinite
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Here are a few. The first one is included because it's not very well known and is not general, though the ones that follow are very general and very useful. A great but not very well known way to find the primitive of $f^{-1}$ in terms of the primitive of $f$, $F$, is (very easy to prove: just differentiate both sides and use the chain rule): $$ \int f^{-1}(x)\, dx = x \cdot f^{-1}(x)-(F \circ f^{-1})(x)+C. $$ Examples: $$ \begin{aligned} \displaystyle \int \arcsin(x)\, dx &= x \cdot \arcsin(x)- (-\cos\circ \arcsin)(x)+C \\ &=x \cdot \arcsin(x)+\sqrt{1-x^2}+C. \end{aligned} $$ $$ \begin{aligned} \int \log(x)\, dx &= x \cdot \log(x)-(\exp \circ \log)(x) + C \\ &= x \cdot \left( \log(x)-1 \right) + C. \end{aligned} $$ This one is more well known, and extremely powerful, it's called differentiating under the integral sign. It requires ingenuity most of the time to know when to apply, and how to apply it, but that only makes it more interesting. The technique uses the simple fact that $$ \frac{\mathrm d}{\mathrm d x} \int_a^b f \left({x, y}\right) \mathrm d y = \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right) \mathrm d y. $$ Example: We want to calculate the integral $\int_{0}^{\infty} \frac{\sin(x)}{x} dx$. To do that, we unintuitively consider the more complicated integral $\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx$ instead. Let $$ I(t)=\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx,$$ then $$ I'(t)=-\int_{0}^{\infty} e^{-tx} \sin(x) dx=\frac{e^{-t x} (t \sin (x)+\cos (x))}{t^2+1}\bigg|_0^{\infty}=\frac{-1}{1+t^2}.$$ Since both $I(t)$ and $-\arctan(t)$ are primitives of $\frac{-1}{1+t^2}$, they must differ only by a constant, so that $I(t)+\arctan(t)=C$. Let $t\to \infty$, then $I(t) \to 0$ and $-\arctan(t) \to -\pi/2$, and hence $C=\pi/2$, and $I(t)=\frac{\pi}{2}-\arctan(t)$. Finally, $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx = I(0) = \frac{\pi}{2}-\arctan(0) = \boxed{\frac{\pi}{2}}. $$ This one is probably the most commonly used "advanced integration technique", and for good reasons. It's referred to as the "residue theorem" and it states that if $\gamma$ is a counterclockwise simple closed curve, then $\displaystyle \int_\gamma f(z) dz = 2\pi i \sum_{k=1}^n \operatorname{Res} ( f, a_k )$ . It will be difficult for you to understand this one without knowledge in complex analysis, but you can get the gist of it with the wiki article. Example: We want to compute $\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx$. The poles of our function $f(z)=\frac{x^2}{1+x^4}$ in the upper half plane are $a_1=e^{i \frac{\pi}{4}}$ and $a_2=e^{i \frac{3\pi}{4}}$. The residues of our function at those points are $$\operatorname{Res}(f,a_1)=\lim_{z\to a_1} (z-a_1)f(z)=\frac{e^{i \frac{-\pi}{4}}}{4},$$ and $$\operatorname{Res}(f,a_2)=\lim_{z\to a_2} (z-a_2)f(z)=\frac{e^{i \frac{-3\pi}{4}}}{4}.$$ Let $\gamma$ be the closed path around the boundary of the semicircle of radius $R>1$ on the upper half plane, traversed in the counter-clockwise direction. Then the residue theorem gives us ${1 \over 2\pi i} \int_\gamma f(z)\,dz=\operatorname{Res}(f,a_1)+\operatorname{Res}(f,a_2)={1 \over 4}\left({1-i \over \sqrt{2}}+{-1-i \over \sqrt{2}}\right)={-i \over 2 \sqrt{2}}$ and $ \int_\gamma f(z)\,dz= {\pi \over \sqrt{2}}$. Now, by the definition of $\gamma$, we have: $$\int_\gamma f(z)\,dz = \int_{-R}^R \frac{x^2}{1+x^4} dx + \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz = {\pi \over \sqrt{2}}.$$ For the integral on the semicircle $$ \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz, $$ we have $$ \begin{aligned} \left| \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \right| &\leq \int_0^\pi \left| {i (R e^{it})^3 \over 1+(R e^{it})^4} \right| dz \\ &\leq \int_0^\pi {R^3 \over R^4-1} dz={\pi R^3 \over R^4-1}. \end{aligned} $$ Hence, as $R\to \infty$, we have ${\pi R^3 \over R^4-1} \to 0$, and hence $\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \to 0$. Finally, $$ \begin{aligned} \int_{-\infty}^\infty \frac{x^2}{1+x^4} dx &= \lim_{R\to \infty} \int_{-R}^R \frac{x^2}{1+x^4} dx \\ &= \lim_{R\to \infty} {\pi \over \sqrt{2}}-\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz =\boxed{{\pi \over \sqrt{2}}}. \end{aligned} $$ My final "technique" is the use of the mean value property for complex analytic functions, or Cauchy's integral formula in other words: $$ \begin{aligned} f(a) &= \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-a}\, dz \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} f\left(a+e^{ix}\right) dx. \end{aligned} $$ Example: We want to compute the very messy looking integral $\int_0^{2\pi} \cos (\cos (x)+1) \cosh (\sin (x)) dx$. We first notice that $$ \begin{aligned} &\hphantom{=} \cos [\cos (x)+1] \cosh [\sin (x)] \\ &=\Re\left\{ \cos [\cos (x)+1] \cosh [\sin (x)] -i\sin [\cos (x)+1] \sinh [\sin (x)] \right\} \\ &= \Re \left[ \cos \left( 1+e^{i x} \right) \right]. \end{aligned} $$ Then, we have $$ \begin{aligned} \int_0^{2\pi} \cos [\cos (x)+1] \cosh [\sin (x)] dx &= \int_0^{2\pi} \Re \left[ \cos \left( 1+e^{i x} \right) \right] dx \\ &= \Re \left[ \int_0^{2\pi} \cos \left( 1+e^{i x} \right) dx \right] \\ &= \Re \left( \cos(1) \cdot 2 \pi \right)= \boxed{2 \pi \cos(1)}. \end{aligned} $$
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Q: Really advanced techniques of integration (definite or indefinite)
Okay, so everyone knows the usual methods of solving integrals, namely u-substitution, integration by parts, partial fractions, trig substitutions, and reduction formulas. But what else is there? Every time I search for "Advanced Techniques of Symbolic Integration" or "Super Advanced Integration Techniques", I get the same results which end up only talking about the methods mentioned above. Are there any super obscure and interesting techniques for solving integrals? As an example of something that might be obscure, the formula for "general integration by parts " for $n$ functions $f_j, \ j = 1,\cdots,n$ is given by $$ \int{f_1'(x)\prod_{j=2}^n{f_j(x)}dx} = \prod_{i=1}^n{f_i(x)} - \sum_{i=2}^n{\int{f_i'(x)\prod_{\substack{j=1 \\ j \neq i}}^n{f_j(x)}dx}} $$ which is not necessarily useful nor difficult to derive, but is interesting nonetheless. So out of curiosity, are there any crazy unknown symbolic integration techniques?
A: Here are a few. The first one is included because it's not very well known and is not general, though the ones that follow are very general and very useful. A great but not very well known way to find the primitive of $f^{-1}$ in terms of the primitive of $f$, $F$, is (very easy to prove: just differentiate both sides and use the chain rule): $$ \int f^{-1}(x)\, dx = x \cdot f^{-1}(x)-(F \circ f^{-1})(x)+C. $$ Examples: $$ \begin{aligned} \displaystyle \int \arcsin(x)\, dx &= x \cdot \arcsin(x)- (-\cos\circ \arcsin)(x)+C \\ &=x \cdot \arcsin(x)+\sqrt{1-x^2}+C. \end{aligned} $$ $$ \begin{aligned} \int \log(x)\, dx &= x \cdot \log(x)-(\exp \circ \log)(x) + C \\ &= x \cdot \left( \log(x)-1 \right) + C. \end{aligned} $$ This one is more well known, and extremely powerful, it's called differentiating under the integral sign. It requires ingenuity most of the time to know when to apply, and how to apply it, but that only makes it more interesting. The technique uses the simple fact that $$ \frac{\mathrm d}{\mathrm d x} \int_a^b f \left({x, y}\right) \mathrm d y = \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right) \mathrm d y. $$ Example: We want to calculate the integral $\int_{0}^{\infty} \frac{\sin(x)}{x} dx$. To do that, we unintuitively consider the more complicated integral $\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx$ instead. Let $$ I(t)=\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx,$$ then $$ I'(t)=-\int_{0}^{\infty} e^{-tx} \sin(x) dx=\frac{e^{-t x} (t \sin (x)+\cos (x))}{t^2+1}\bigg|_0^{\infty}=\frac{-1}{1+t^2}.$$ Since both $I(t)$ and $-\arctan(t)$ are primitives of $\frac{-1}{1+t^2}$, they must differ only by a constant, so that $I(t)+\arctan(t)=C$. Let $t\to \infty$, then $I(t) \to 0$ and $-\arctan(t) \to -\pi/2$, and hence $C=\pi/2$, and $I(t)=\frac{\pi}{2}-\arctan(t)$. Finally, $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx = I(0) = \frac{\pi}{2}-\arctan(0) = \boxed{\frac{\pi}{2}}. $$ This one is probably the most commonly used "advanced integration technique", and for good reasons. It's referred to as the "residue theorem" and it states that if $\gamma$ is a counterclockwise simple closed curve, then $\displaystyle \int_\gamma f(z) dz = 2\pi i \sum_{k=1}^n \operatorname{Res} ( f, a_k )$ . It will be difficult for you to understand this one without knowledge in complex analysis, but you can get the gist of it with the wiki article. Example: We want to compute $\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx$. The poles of our function $f(z)=\frac{x^2}{1+x^4}$ in the upper half plane are $a_1=e^{i \frac{\pi}{4}}$ and $a_2=e^{i \frac{3\pi}{4}}$. The residues of our function at those points are $$\operatorname{Res}(f,a_1)=\lim_{z\to a_1} (z-a_1)f(z)=\frac{e^{i \frac{-\pi}{4}}}{4},$$ and $$\operatorname{Res}(f,a_2)=\lim_{z\to a_2} (z-a_2)f(z)=\frac{e^{i \frac{-3\pi}{4}}}{4}.$$ Let $\gamma$ be the closed path around the boundary of the semicircle of radius $R>1$ on the upper half plane, traversed in the counter-clockwise direction. Then the residue theorem gives us ${1 \over 2\pi i} \int_\gamma f(z)\,dz=\operatorname{Res}(f,a_1)+\operatorname{Res}(f,a_2)={1 \over 4}\left({1-i \over \sqrt{2}}+{-1-i \over \sqrt{2}}\right)={-i \over 2 \sqrt{2}}$ and $ \int_\gamma f(z)\,dz= {\pi \over \sqrt{2}}$. Now, by the definition of $\gamma$, we have: $$\int_\gamma f(z)\,dz = \int_{-R}^R \frac{x^2}{1+x^4} dx + \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz = {\pi \over \sqrt{2}}.$$ For the integral on the semicircle $$ \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz, $$ we have $$ \begin{aligned} \left| \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \right| &\leq \int_0^\pi \left| {i (R e^{it})^3 \over 1+(R e^{it})^4} \right| dz \\ &\leq \int_0^\pi {R^3 \over R^4-1} dz={\pi R^3 \over R^4-1}. \end{aligned} $$ Hence, as $R\to \infty$, we have ${\pi R^3 \over R^4-1} \to 0$, and hence $\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \to 0$. Finally, $$ \begin{aligned} \int_{-\infty}^\infty \frac{x^2}{1+x^4} dx &= \lim_{R\to \infty} \int_{-R}^R \frac{x^2}{1+x^4} dx \\ &= \lim_{R\to \infty} {\pi \over \sqrt{2}}-\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz =\boxed{{\pi \over \sqrt{2}}}. \end{aligned} $$ My final "technique" is the use of the mean value property for complex analytic functions, or Cauchy's integral formula in other words: $$ \begin{aligned} f(a) &= \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-a}\, dz \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} f\left(a+e^{ix}\right) dx. \end{aligned} $$ Example: We want to compute the very messy looking integral $\int_0^{2\pi} \cos (\cos (x)+1) \cosh (\sin (x)) dx$. We first notice that $$ \begin{aligned} &\hphantom{=} \cos [\cos (x)+1] \cosh [\sin (x)] \\ &=\Re\left\{ \cos [\cos (x)+1] \cosh [\sin (x)] -i\sin [\cos (x)+1] \sinh [\sin (x)] \right\} \\ &= \Re \left[ \cos \left( 1+e^{i x} \right) \right]. \end{aligned} $$ Then, we have $$ \begin{aligned} \int_0^{2\pi} \cos [\cos (x)+1] \cosh [\sin (x)] dx &= \int_0^{2\pi} \Re \left[ \cos \left( 1+e^{i x} \right) \right] dx \\ &= \Re \left[ \int_0^{2\pi} \cos \left( 1+e^{i x} \right) dx \right] \\ &= \Re \left( \cos(1) \cdot 2 \pi \right)= \boxed{2 \pi \cos(1)}. \end{aligned} $$
| 2026-01-26T11:30:25.883317
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158,219
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Is a matrix multiplied with its transpose something special?
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In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together. Is $A A^\mathrm T$ something special for any matrix $A$?
| 332
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matrices
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https://math.stackexchange.com/questions/158219/is-a-matrix-multiplied-with-its-transpose-something-special
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The main thing is presumably that $AA^T$ is symmetric. Indeed $(AA^T)^T=(A^T)^TA^T=AA^T$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real. Moreover if $A$ is invertible, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$ Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors. Last but not least if one is interested in how much the linear map represented by $A$ changes the norm of a vector one can compute $$\sqrt{\left}=\sqrt{\left}$$ which simplifies for eigenvectors $x$ to the eigenvalue $\lambda$ to $$\sqrt{\left}=\sqrt \lambda\sqrt{\left},$$ The determinant is just the product of these eigenvalues.
| 219
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Q: Is a matrix multiplied with its transpose something special?
In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together. Is $A A^\mathrm T$ something special for any matrix $A$?
A: The main thing is presumably that $AA^T$ is symmetric. Indeed $(AA^T)^T=(A^T)^TA^T=AA^T$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real. Moreover if $A$ is invertible, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$ Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors. Last but not least if one is interested in how much the linear map represented by $A$ changes the norm of a vector one can compute $$\sqrt{\left}=\sqrt{\left}$$ which simplifies for eigenvectors $x$ to the eigenvalue $\lambda$ to $$\sqrt{\left}=\sqrt \lambda\sqrt{\left},$$ The determinant is just the product of these eigenvalues.
| 2026-01-26T11:30:27.460757
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21,792
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Norms Induced by Inner Products and the Parallelogram Law
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Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$. It's not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality $$ 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2 $$ holds for all pairs $u, v \in V$. I'm having difficulty with the converse. Assuming the parallelogram identity, I'm able to convince myself that the inner product should be $$ \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 - \lVert u - v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u\rVert^2 - \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u - v\rVert^2}{4} $$ I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$. How would one go about proving this?
| 331
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linear-algebra, functional-analysis, normed-spaces, inner-products
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https://math.stackexchange.com/questions/21792/norms-induced-by-inner-products-and-the-parallelogram-law
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Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit easier to prove. So, assume that the norm $\|\cdot\|$ satisfies the parallelogram law $$2 \Vert x \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y \Vert^2 + \Vert x - y \Vert^2$$ for all $x,y \in V$ and put $$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right).$$ We're dealing with real vector spaces and defer the treatment of the complex case to Step 4 below. Step 0. $\langle x, y \rangle = \langle y, x\rangle$ and $\Vert x \Vert = \sqrt{\langle x, x\rangle}$. Obvious. Step 1. The function $(x,y) \mapsto \langle x,y \rangle$ is continuous with respect to $\Vert \cdot \Vert$. Continuity with respect to the norm $\Vert \cdot\Vert$ follows from the fact that addition and negation are $\Vert \cdot \Vert$-continuous, that the norm itself is continuous and that sums and compositions of continuous functions are continuous. Remark. This continuity property of the (putative) scalar product will only be used at the very end of step 3. Until then the solution consists of purely algebraic steps. Step 2. We have $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z\rangle$. By the parallelogram law we have $$2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y + z \Vert^2 + \Vert x - y + z\Vert^2 .$$ This gives $$\begin{align*} \Vert x + y + z \Vert^2 & = 2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 - \Vert x - y + z \Vert^2 \\ & = 2\Vert y + z \Vert^2 + 2\Vert x \Vert^2 - \Vert y - x + z \Vert^2 \end{align*}$$ where the second formula follows from the first by exchanging $x$ and $y$. Since $A = B$ and $A = C$ imply $A = \frac{1}{2} (B + C)$ we get $$\Vert x + y + z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x + z \Vert^2 + \Vert y + z \Vert^2 - \frac{1}{2}\Vert x - y + z \Vert^2 - \frac{1}{2}\Vert y - x + z \Vert^2.$$ Replacing $z$ by $-z$ in the last equation gives $$\Vert x + y - z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x - z \Vert^2 + \Vert y - z \Vert^2 - \frac{1}{2}\Vert x - y - z \Vert^2 - \frac{1}{2}\Vert y - x - z \Vert^2.$$ Applying $\Vert w \Vert = \Vert - w\Vert$ to the two negative terms in the last equation we get $$\begin{align*}\langle x + y, z \rangle & = \frac{1}{4}\left(\Vert x + y + z \Vert^2 - \Vert x + y - z \Vert^2\right) \\ & = \frac{1}{4}\left(\Vert x + z \Vert^2 - \Vert x - z \Vert^2\right) + \frac{1}{4}\left(\Vert y + z \Vert^2 - \Vert y - z \Vert^2\right) \\ & = \langle x, z \rangle + \langle y, z \rangle \end{align*}$$ as desired. Step 3. $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{R}$. This clearly holds for $\lambda = -1$ and by step 2 and induction we have $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{N}$, thus for all $\lambda \in \mathbb{Z}$. If $\lambda = \frac{p}{q}$ with $p,q \in \mathbb{Z}, q \neq 0$ we get with $x' = \dfrac{x}{q}$ that $$q \langle \lambda x, y \rangle = q\langle p x', y \rangle = p \langle q x', y \rangle = p\langle x,y \rangle,$$ so dividing this by $q$ gives $$\langle \lambda x , y \rangle = \lambda \langle x, y \rangle \qquad\text{for all } \lambda \in \mathbb{Q}.$$ We have just seen that for fixed $x,y$ the continuous function $\displaystyle t \mapsto \frac{1}{t} \langle t x,y \rangle$ defined on $\mathbb{R} \smallsetminus \{0\}$ is equal to $\langle x,y \rangle$ for all $t \in \mathbb{Q} \smallsetminus \{0\}$, thus equality holds for all $t \in \mathbb{R} \smallsetminus \{0\}$. The case $\lambda = 0$ being trivial, we're done. Step 4. The complex case. Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, observe that $\langle ix,y \rangle = i \langle x, y \rangle$ and $\langle x, y \rangle = \overline{\langle y, x \rangle}$ and apply the case of real scalars twice (to the real and imaginary parts of $\langle \cdot, \cdot \rangle$). Addendum. In fact we can weaken requirements of Jordan von Neumann theorem to $$ 2\Vert x\Vert^2+2\Vert y\Vert^2\leq\Vert x+y\Vert^2+\Vert x-y\Vert^2 $$ Indeed after substitution $x\to\frac{1}{2}(x+y)$, $y\to\frac{1}{2}(x-y)$ and simplifications we get $$ \Vert x+y\Vert^2+\Vert x-y\Vert^2\leq 2\Vert x\Vert^2+2\Vert y\Vert^2 $$ which together with previous inequality gives the equality.
| 257
| false
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Q: Norms Induced by Inner Products and the Parallelogram Law
Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$. It's not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality $$ 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2 $$ holds for all pairs $u, v \in V$. I'm having difficulty with the converse. Assuming the parallelogram identity, I'm able to convince myself that the inner product should be $$ \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 - \lVert u - v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u\rVert^2 - \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u - v\rVert^2}{4} $$ I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$. How would one go about proving this?
A: Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit easier to prove. So, assume that the norm $\|\cdot\|$ satisfies the parallelogram law $$2 \Vert x \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y \Vert^2 + \Vert x - y \Vert^2$$ for all $x,y \in V$ and put $$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right).$$ We're dealing with real vector spaces and defer the treatment of the complex case to Step 4 below. Step 0. $\langle x, y \rangle = \langle y, x\rangle$ and $\Vert x \Vert = \sqrt{\langle x, x\rangle}$. Obvious. Step 1. The function $(x,y) \mapsto \langle x,y \rangle$ is continuous with respect to $\Vert \cdot \Vert$. Continuity with respect to the norm $\Vert \cdot\Vert$ follows from the fact that addition and negation are $\Vert \cdot \Vert$-continuous, that the norm itself is continuous and that sums and compositions of continuous functions are continuous. Remark. This continuity property of the (putative) scalar product will only be used at the very end of step 3. Until then the solution consists of purely algebraic steps. Step 2. We have $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z\rangle$. By the parallelogram law we have $$2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y + z \Vert^2 + \Vert x - y + z\Vert^2 .$$ This gives $$\begin{align*} \Vert x + y + z \Vert^2 & = 2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 - \Vert x - y + z \Vert^2 \\ & = 2\Vert y + z \Vert^2 + 2\Vert x \Vert^2 - \Vert y - x + z \Vert^2 \end{align*}$$ where the second formula follows from the first by exchanging $x$ and $y$. Since $A = B$ and $A = C$ imply $A = \frac{1}{2} (B + C)$ we get $$\Vert x + y + z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x + z \Vert^2 + \Vert y + z \Vert^2 - \frac{1}{2}\Vert x - y + z \Vert^2 - \frac{1}{2}\Vert y - x + z \Vert^2.$$ Replacing $z$ by $-z$ in the last equation gives $$\Vert x + y - z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x - z \Vert^2 + \Vert y - z \Vert^2 - \frac{1}{2}\Vert x - y - z \Vert^2 - \frac{1}{2}\Vert y - x - z \Vert^2.$$ Applying $\Vert w \Vert = \Vert - w\Vert$ to the two negative terms in the last equation we get $$\begin{align*}\langle x + y, z \rangle & = \frac{1}{4}\left(\Vert x + y + z \Vert^2 - \Vert x + y - z \Vert^2\right) \\ & = \frac{1}{4}\left(\Vert x + z \Vert^2 - \Vert x - z \Vert^2\right) + \frac{1}{4}\left(\Vert y + z \Vert^2 - \Vert y - z \Vert^2\right) \\ & = \langle x, z \rangle + \langle y, z \rangle \end{align*}$$ as desired. Step 3. $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{R}$. This clearly holds for $\lambda = -1$ and by step 2 and induction we have $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{N}$, thus for all $\lambda \in \mathbb{Z}$. If $\lambda = \frac{p}{q}$ with $p,q \in \mathbb{Z}, q \neq 0$ we get with $x' = \dfrac{x}{q}$ that $$q \langle \lambda x, y \rangle = q\langle p x', y \rangle = p \langle q x', y \rangle = p\langle x,y \rangle,$$ so dividing this by $q$ gives $$\langle \lambda x , y \rangle = \lambda \langle x, y \rangle \qquad\text{for all } \lambda \in \mathbb{Q}.$$ We have just seen that for fixed $x,y$ the continuous function $\displaystyle t \mapsto \frac{1}{t} \langle t x,y \rangle$ defined on $\mathbb{R} \smallsetminus \{0\}$ is equal to $\langle x,y \rangle$ for all $t \in \mathbb{Q} \smallsetminus \{0\}$, thus equality holds for all $t \in \mathbb{R} \smallsetminus \{0\}$. The case $\lambda = 0$ being trivial, we're done. Step 4. The complex case. Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, observe that $\langle ix,y \rangle = i \langle x, y \rangle$ and $\langle x, y \rangle = \overline{\langle y, x \rangle}$ and apply the case of real scalars twice (to the real and imaginary parts of $\langle \cdot, \cdot \rangle$). Addendum. In fact we can weaken requirements of Jordan von Neumann theorem to $$ 2\Vert x\Vert^2+2\Vert y\Vert^2\leq\Vert x+y\Vert^2+\Vert x-y\Vert^2 $$ Indeed after substitution $x\to\frac{1}{2}(x+y)$, $y\to\frac{1}{2}(x-y)$ and simplifications we get $$ \Vert x+y\Vert^2+\Vert x-y\Vert^2\leq 2\Vert x\Vert^2+2\Vert y\Vert^2 $$ which together with previous inequality gives the equality.
| 2026-01-26T11:30:28.807365
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1,594,740
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V.I. Arnold says Russian students can't solve this problem, but American students can -- why?
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In a book of word problems by V.I Arnold, the following appears: The hypotenuse of a right-angled triangle (in a standard American examination) is $10$ inches, the altitude dropped onto it is 6 inches. Find the area of the triangle. American school students had been coping successfully with this problem for over a decade. But then Russian school students arrived from Moscow, and none of them was able to solve it as had their American peers (giving $30$ square inches as the answer). Why? Here's the book. I assume the answer is some joke at the expense of the Americans, but I don't get it. Possibly a joke about inches? Anyone?
| 325
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geometry, triangles, area, puzzle, word-problem
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https://math.stackexchange.com/questions/1594740/v-i-arnold-says-russian-students-cant-solve-this-problem-but-american-student
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There is no such right triangle. The maximum possible altitude is half the hypotenuse (inscribe the triangle into a circle to see this), which here is $5$ inches. You would only get $30$ square inches if you tried to compute the area without checking whether the triangle actually exists.
| 305
| true
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Q: V.I. Arnold says Russian students can't solve this problem, but American students can -- why?
In a book of word problems by V.I Arnold, the following appears: The hypotenuse of a right-angled triangle (in a standard American examination) is $10$ inches, the altitude dropped onto it is 6 inches. Find the area of the triangle. American school students had been coping successfully with this problem for over a decade. But then Russian school students arrived from Moscow, and none of them was able to solve it as had their American peers (giving $30$ square inches as the answer). Why? Here's the book. I assume the answer is some joke at the expense of the Americans, but I don't get it. Possibly a joke about inches? Anyone?
A: There is no such right triangle. The maximum possible altitude is half the hypotenuse (inscribe the triangle into a circle to see this), which here is $5$ inches. You would only get $30$ square inches if you tried to compute the area without checking whether the triangle actually exists.
| 2026-01-26T11:30:30.391358
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76,491
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Multiple-choice question about the probability of a random answer to itself being correct
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I found this math "problem" on the internet, and I'm wondering if it has an answer: Question: If you choose an answer to this question at random, what is the probability that you will be correct? a. $25\%$ b. $50\%$ c. $0\%$ d. $25\%$ Does this question have a correct answer?
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probability
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https://math.stackexchange.com/questions/76491/multiple-choice-question-about-the-probability-of-a-random-answer-to-itself-bein
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No, it is not meaningful. 25% is correct iff 50% is correct, and 50% is correct iff 25% is correct, so it can be neither of those two (because if both are correct, the only correct answer could be 75% which is not even an option). But it cannot be 0% either, because then the correct answer would be 25%. So none of the answers are correct, so the answer must be 0%. But then it is 25%. And so forth. It's a multiple-choice variant (with bells and whistles) of the classical liar paradox, which asks whether the statement This statement is false. is true or false. There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value. Edited much later to add: There's a variant of this puzzle that's very popular on the internet at the moment, in which answer option (c) is 60% rather than 0%. In this variant it is at least internally consistent to claim that all of the answers are wrong, and so the possibility of getting a right one by choosing randomly is 0%. Whether this actually resolves the variant puzzle is more a matter of taste and temperament than an objective mathematical question. It is not in general true for self-referencing questions that simply being internally consistent is enough for an answer to be unambiguously right; otherwise the question Is the correct answer to this question "yes"? would have two different "right" answers, because "yes" and "no" are both internally consistent. In the 60% variant of the puzzle it is happens that the only internally consistent answer is "0%", but even so one might, as a matter of caution, still deny that such reasoning by elimination is valid for self-referential statements at all. If one adopts this stance, one would still consider the 60% variant meaningless. One rationale for taking this strict position would be that we don't want to accept reasoning by elimination on True or false? The Great Pumpkin exists. Both of these statements are false. where the only internally consistent resolution is that the first statement is true and the second one is false. However, it appears to be unsound to conclude that the Great Pumpkin exists on the basis simply that the puzzle was posed. On the other hand, it is difficult to argue that there is no possible principle that will cordon off the Great Pumpkin example as meaningless while still allowing the 60% variant to be meaningful. In the end, though, these things are more matters of taste and philosophy than they are mathematics. In mathematics we generally prefer to play it safe and completely refuse to work with explicitly self-referential statements. This avoids the risk of paradox, and does not seem to hinder mathematical arguments about the things mathematicians are ordinarily interested in. So whatever one decides to do with the question-about-itself, what one does is not really mathematics.
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Q: Multiple-choice question about the probability of a random answer to itself being correct
I found this math "problem" on the internet, and I'm wondering if it has an answer: Question: If you choose an answer to this question at random, what is the probability that you will be correct? a. $25\%$ b. $50\%$ c. $0\%$ d. $25\%$ Does this question have a correct answer?
A: No, it is not meaningful. 25% is correct iff 50% is correct, and 50% is correct iff 25% is correct, so it can be neither of those two (because if both are correct, the only correct answer could be 75% which is not even an option). But it cannot be 0% either, because then the correct answer would be 25%. So none of the answers are correct, so the answer must be 0%. But then it is 25%. And so forth. It's a multiple-choice variant (with bells and whistles) of the classical liar paradox, which asks whether the statement This statement is false. is true or false. There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value. Edited much later to add: There's a variant of this puzzle that's very popular on the internet at the moment, in which answer option (c) is 60% rather than 0%. In this variant it is at least internally consistent to claim that all of the answers are wrong, and so the possibility of getting a right one by choosing randomly is 0%. Whether this actually resolves the variant puzzle is more a matter of taste and temperament than an objective mathematical question. It is not in general true for self-referencing questions that simply being internally consistent is enough for an answer to be unambiguously right; otherwise the question Is the correct answer to this question "yes"? would have two different "right" answers, because "yes" and "no" are both internally consistent. In the 60% variant of the puzzle it is happens that the only internally consistent answer is "0%", but even so one might, as a matter of caution, still deny that such reasoning by elimination is valid for self-referential statements at all. If one adopts this stance, one would still consider the 60% variant meaningless. One rationale for taking this strict position would be that we don't want to accept reasoning by elimination on True or false? The Great Pumpkin exists. Both of these statements are false. where the only internally consistent resolution is that the first statement is true and the second one is false. However, it appears to be unsound to conclude that the Great Pumpkin exists on the basis simply that the puzzle was posed. On the other hand, it is difficult to argue that there is no possible principle that will cordon off the Great Pumpkin example as meaningless while still allowing the 60% variant to be meaningful. In the end, though, these things are more matters of taste and philosophy than they are mathematics. In mathematics we generally prefer to play it safe and completely refuse to work with explicitly self-referential statements. This avoids the risk of paradox, and does not seem to hinder mathematical arguments about the things mathematicians are ordinarily interested in. So whatever one decides to do with the question-about-itself, what one does is not really mathematics.
| 2026-01-26T11:30:31.916850
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112,067
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How discontinuous can a derivative be?
|
There is a well-known result in elementary analysis due to Darboux which says if $f$ is a differentiable function then $f'$ satisfies the intermediate value property. To my knowledge, not many "highly" discontinuous Darboux functions are known--the only one I am aware of being the Conway base 13 function--and few (none?) of these are derivatives of differentiable functions. In fact they generally cannot be since an application of Baire's theorem gives that the set of continuity points of the derivative is dense $G_\delta$. Is it known how sharp that last result is? Are there known Darboux functions which are derivatives and are discontinuous on "large" sets in some appropriate sense?
| 316
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real-analysis, derivatives, examples-counterexamples
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https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be
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What follows is taken (mostly) from more extensive discussions in the following sci.math posts: http://groups.google.com/group/sci.math/msg/814be41b1ea8c024 [23 January 2000] http://groups.google.com/group/sci.math/msg/3ea26975d010711f [6 November 2006] http://groups.google.com/group/sci.math/msg/05dbc0ee4c69898e [20 December 2006] Note: The term interval is restricted to nondegenerate intervals (i.e. intervals containing more than one point). The continuity set of a derivative on an open interval $J$ is dense in $J.$ In fact, the continuity set has cardinality $c$ in every subinterval of $J.$ On the other hand, the discontinuity set $D$ of a derivative can have the following properties: $D$ can be dense in $\mathbb R$. $D$ can have cardinality $c$ in every interval. $D$ can have positive measure. (Hence, the function can fail to be Riemann integrable.) $D$ can have positive measure in every interval. $D$ can have full measure in every interval (i.e. measure zero complement). $D$ can have a Hausdorff dimension zero complement. $D$ can have an $h$-Hausdorff measure zero complement for any specified Hausdorff measure function $h.$ More precisely, a subset $D$ of $\mathbb R$ can be the discontinuity set for some derivative if and only if $D$ is an $F_{\sigma}$ first category (i.e. an $F_{\sigma}$ meager) subset of $\mathbb R.$ This characterization of the discontinuity set of a derivative can be found in the following references: Benedetto [1] (Chapter 1.3.2, Proposition, 1.10, p. 30); Bruckner [2] (Chapter 3, Section 2, Theorem 2.1, p. 34); Bruckner/Leonard [3] (Theorem at bottom of p. 27); Goffman [5] (Chapter 9, Exercise 2.3, p. 120 states the result); Klippert/Williams [7]. Regarding this characterization of the discontinuity set of a derivative, Bruckner and Leonard [3] (bottom of p. 27) wrote the following in 1966: Although we imagine that this theorem is known, we have been unable to find a reference. I have found the result stated in Goffman's 1953 text [5], but nowhere else prior to 1966 (including Goffman's Ph.D. Dissertation). Interestingly, in a certain sense most derivatives have the property that $D$ is large in all of the ways listed above (#1 through #7). In 1977 Cliff Weil [8] published a proof that, in the space of derivatives with the sup norm, all but a first category set of such functions are discontinuous almost everywhere (in the sense of Lebesgue measure). When Weil's result is paired with the fact that derivatives (being Baire $1$ functions) are continuous almost everywhere in the sense of Baire category, we get the following: (A) Every derivative is continuous at the Baire-typical point. (B) The Baire-typical derivative is not continuous at the Lebesgue-typical point. Note that Weil's result is stronger than simply saying that the Baire-typical derivative fails to be Riemann integrable (i.e. $D$ has positive Lebesgue measure), or even stronger than saying that the Baire-typical derivative fails to be Riemann integrable on every interval. Note also that, for each of these Baire-typical derivatives, $\{D, \; {\mathbb R} - D\}$ gives a partition of $\mathbb R$ into a first category set and a Lebesgue measure zero set. In 1984 Bruckner/Petruska [4] (Theorem 2.4) strengthened Weil's result by proving the following: Given any finite Borel measure $\mu,$ the Baire-typical derivative is such that the set $D$ is the complement of a set that has $\mu$-measure zero. In 1993 Kirchheim [5] strengthened Weil's result by proving the following: Given any Hausdorff measure function $h,$ the Baire-typical derivative is such that the set $D$ is the complement of a set that has Hausdorff $h$-measure zero. [1] John J. Benedetto, Real Variable and Integration With Historical Notes, Mathematische Leitfäden. Stuttgart: B. G. Teubne, 1976, 278 pages. [MR 58 #28328; Zbl 336.26001] [2] Andrew M. Bruckner, Differentiation of Real Functions, 2nd edition, CRM Monograph Series #5, American Mathematical Society, 1994, xii + 195 pages. [The first edition was published in 1978 as Springer-Verlag's Lecture Notes in Mathematics #659. The second edition is essentially unchanged from the first edition with the exception of a new chapter on recent developments (23 pages) and 94 additional bibliographic items.] [MR 94m:26001; Zbl 796.26001] [3] Andrew M. Bruckner and John L. Leonard, Derivatives, American Mathematical Monthly 73 #4 (April 1966) [Part II: Papers in Analysis, Herbert Ellsworth Slaught Memorial Papers #11], 24-56. [MR 33 #5797; Zbl 138.27805] [4] Andrew M. Bruckner and György Petruska, Some typical results on bounded Baire $1$ functions, Acta Mathematica Hungarica 43 (1984), 325-333. [MR 85h:26004; Zbl 542.26004] [5] Casper Goffman, Real Functions, Prindle, Weber & Schmidt, 1953/1967, x + 261 pages. [MR 14,855e; Zbl 53.22502] [6] Bernd Kirchheim, Some further typical results on bounded Baire one functions, Acta Mathematica Hungarica 62 (1993), 119-129. [94k:26008; Zbl 786.26002] [7] John Clayton Klippert and Geoffrey Williams, On the existence of a derivative continuous on a $G_{\delta}$, International Journal of Mathematical Education in Science and Technology 35 (2004), 91-99. [8] Clifford Weil, The space of bounded derivatives, Real Analysis Exchange 3 (1977-78), 38-41. [Zbl 377.26005]
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Q: How discontinuous can a derivative be?
There is a well-known result in elementary analysis due to Darboux which says if $f$ is a differentiable function then $f'$ satisfies the intermediate value property. To my knowledge, not many "highly" discontinuous Darboux functions are known--the only one I am aware of being the Conway base 13 function--and few (none?) of these are derivatives of differentiable functions. In fact they generally cannot be since an application of Baire's theorem gives that the set of continuity points of the derivative is dense $G_\delta$. Is it known how sharp that last result is? Are there known Darboux functions which are derivatives and are discontinuous on "large" sets in some appropriate sense?
A: What follows is taken (mostly) from more extensive discussions in the following sci.math posts: http://groups.google.com/group/sci.math/msg/814be41b1ea8c024 [23 January 2000] http://groups.google.com/group/sci.math/msg/3ea26975d010711f [6 November 2006] http://groups.google.com/group/sci.math/msg/05dbc0ee4c69898e [20 December 2006] Note: The term interval is restricted to nondegenerate intervals (i.e. intervals containing more than one point). The continuity set of a derivative on an open interval $J$ is dense in $J.$ In fact, the continuity set has cardinality $c$ in every subinterval of $J.$ On the other hand, the discontinuity set $D$ of a derivative can have the following properties: $D$ can be dense in $\mathbb R$. $D$ can have cardinality $c$ in every interval. $D$ can have positive measure. (Hence, the function can fail to be Riemann integrable.) $D$ can have positive measure in every interval. $D$ can have full measure in every interval (i.e. measure zero complement). $D$ can have a Hausdorff dimension zero complement. $D$ can have an $h$-Hausdorff measure zero complement for any specified Hausdorff measure function $h.$ More precisely, a subset $D$ of $\mathbb R$ can be the discontinuity set for some derivative if and only if $D$ is an $F_{\sigma}$ first category (i.e. an $F_{\sigma}$ meager) subset of $\mathbb R.$ This characterization of the discontinuity set of a derivative can be found in the following references: Benedetto [1] (Chapter 1.3.2, Proposition, 1.10, p. 30); Bruckner [2] (Chapter 3, Section 2, Theorem 2.1, p. 34); Bruckner/Leonard [3] (Theorem at bottom of p. 27); Goffman [5] (Chapter 9, Exercise 2.3, p. 120 states the result); Klippert/Williams [7]. Regarding this characterization of the discontinuity set of a derivative, Bruckner and Leonard [3] (bottom of p. 27) wrote the following in 1966: Although we imagine that this theorem is known, we have been unable to find a reference. I have found the result stated in Goffman's 1953 text [5], but nowhere else prior to 1966 (including Goffman's Ph.D. Dissertation). Interestingly, in a certain sense most derivatives have the property that $D$ is large in all of the ways listed above (#1 through #7). In 1977 Cliff Weil [8] published a proof that, in the space of derivatives with the sup norm, all but a first category set of such functions are discontinuous almost everywhere (in the sense of Lebesgue measure). When Weil's result is paired with the fact that derivatives (being Baire $1$ functions) are continuous almost everywhere in the sense of Baire category, we get the following: (A) Every derivative is continuous at the Baire-typical point. (B) The Baire-typical derivative is not continuous at the Lebesgue-typical point. Note that Weil's result is stronger than simply saying that the Baire-typical derivative fails to be Riemann integrable (i.e. $D$ has positive Lebesgue measure), or even stronger than saying that the Baire-typical derivative fails to be Riemann integrable on every interval. Note also that, for each of these Baire-typical derivatives, $\{D, \; {\mathbb R} - D\}$ gives a partition of $\mathbb R$ into a first category set and a Lebesgue measure zero set. In 1984 Bruckner/Petruska [4] (Theorem 2.4) strengthened Weil's result by proving the following: Given any finite Borel measure $\mu,$ the Baire-typical derivative is such that the set $D$ is the complement of a set that has $\mu$-measure zero. In 1993 Kirchheim [5] strengthened Weil's result by proving the following: Given any Hausdorff measure function $h,$ the Baire-typical derivative is such that the set $D$ is the complement of a set that has Hausdorff $h$-measure zero. [1] John J. Benedetto, Real Variable and Integration With Historical Notes, Mathematische Leitfäden. Stuttgart: B. G. Teubne, 1976, 278 pages. [MR 58 #28328; Zbl 336.26001] [2] Andrew M. Bruckner, Differentiation of Real Functions, 2nd edition, CRM Monograph Series #5, American Mathematical Society, 1994, xii + 195 pages. [The first edition was published in 1978 as Springer-Verlag's Lecture Notes in Mathematics #659. The second edition is essentially unchanged from the first edition with the exception of a new chapter on recent developments (23 pages) and 94 additional bibliographic items.] [MR 94m:26001; Zbl 796.26001] [3] Andrew M. Bruckner and John L. Leonard, Derivatives, American Mathematical Monthly 73 #4 (April 1966) [Part II: Papers in Analysis, Herbert Ellsworth Slaught Memorial Papers #11], 24-56. [MR 33 #5797; Zbl 138.27805] [4] Andrew M. Bruckner and György Petruska, Some typical results on bounded Baire $1$ functions, Acta Mathematica Hungarica 43 (1984), 325-333. [MR 85h:26004; Zbl 542.26004] [5] Casper Goffman, Real Functions, Prindle, Weber & Schmidt, 1953/1967, x + 261 pages. [MR 14,855e; Zbl 53.22502] [6] Bernd Kirchheim, Some further typical results on bounded Baire one functions, Acta Mathematica Hungarica 62 (1993), 119-129. [94k:26008; Zbl 786.26002] [7] John Clayton Klippert and Geoffrey Williams, On the existence of a derivative continuous on a $G_{\delta}$, International Journal of Mathematical Education in Science and Technology 35 (2004), 91-99. [8] Clifford Weil, The space of bounded derivatives, Real Analysis Exchange 3 (1977-78), 38-41. [Zbl 377.26005]
| 2026-01-26T11:30:33.444844
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139,503
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In the history of mathematics, has there ever been a mistake?
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I was just wondering whether or not there have been mistakes in mathematics. Not a conjecture that ended up being false, but a theorem which had a proof that was accepted for a nontrivial amount of time before someone found a hole in the argument. Does this happen anymore now that we have computers? I imagine not. But it seems totally possible that this could have happened back in the Enlightenment. Feel free to interpret this how you wish!
| 315
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math-history, big-list, fake-proofs
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https://math.stackexchange.com/questions/139503/in-the-history-of-mathematics-has-there-ever-been-a-mistake
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In 1933, Kurt Gödel showed that the class called $\lbrack\exists^*\forall^2\exists^*, {\mathrm{all}}, (0)\rbrack$ was decidable. These are the formulas that begin with $\exists a\exists b\ldots \exists m\forall n\forall p\exists q\ldots\exists z$, with exactly two $\forall$ quantifiers, with no intervening $\exists$s. These formulas may contain arbitrary relations amongst the variables, but no functions or constants, and no equality symbol. Gödel showed that there is a method which takes any formula in this form and decides whether it is satisfiable. (If there are three $\forall$s in a row, or an $\exists$ between the $\forall$s, there is no such method.) In the final sentence of the same paper, Gödel added: In conclusion, I would still like to remark that Theorem I can also be proved, by the same method, for formulas that contain the identity sign. Mathematicians took Gödel's word for it, and proved results derived from this one, until the mid-1960s, when Stål Aanderaa realized that Gödel had been mistaken, and the argument Gödel used would not work. In 1983, Warren Goldfarb showed that not only was Gödel's argument invalid, but his claimed result was actually false, and the larger class was not decidable. Gödel's original 1933 paper is Zum Entscheidungsproblem des logischen Funktionenkalküls (On the decision problem for the functional calculus of logic) which can be found on pages 306–327 of volume I of his Collected Works. (Oxford University Press, 1986.) There is an introductory note by Goldfarb on pages 226–231, of which pages 229–231 address Gödel's error specifically.
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Q: In the history of mathematics, has there ever been a mistake?
I was just wondering whether or not there have been mistakes in mathematics. Not a conjecture that ended up being false, but a theorem which had a proof that was accepted for a nontrivial amount of time before someone found a hole in the argument. Does this happen anymore now that we have computers? I imagine not. But it seems totally possible that this could have happened back in the Enlightenment. Feel free to interpret this how you wish!
A: In 1933, Kurt Gödel showed that the class called $\lbrack\exists^*\forall^2\exists^*, {\mathrm{all}}, (0)\rbrack$ was decidable. These are the formulas that begin with $\exists a\exists b\ldots \exists m\forall n\forall p\exists q\ldots\exists z$, with exactly two $\forall$ quantifiers, with no intervening $\exists$s. These formulas may contain arbitrary relations amongst the variables, but no functions or constants, and no equality symbol. Gödel showed that there is a method which takes any formula in this form and decides whether it is satisfiable. (If there are three $\forall$s in a row, or an $\exists$ between the $\forall$s, there is no such method.) In the final sentence of the same paper, Gödel added: In conclusion, I would still like to remark that Theorem I can also be proved, by the same method, for formulas that contain the identity sign. Mathematicians took Gödel's word for it, and proved results derived from this one, until the mid-1960s, when Stål Aanderaa realized that Gödel had been mistaken, and the argument Gödel used would not work. In 1983, Warren Goldfarb showed that not only was Gödel's argument invalid, but his claimed result was actually false, and the larger class was not decidable. Gödel's original 1933 paper is Zum Entscheidungsproblem des logischen Funktionenkalküls (On the decision problem for the functional calculus of logic) which can be found on pages 306–327 of volume I of his Collected Works. (Oxford University Press, 1986.) There is an introductory note by Goldfarb on pages 226–231, of which pages 229–231 address Gödel's error specifically.
| 2026-01-26T11:30:34.870590
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223,235
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Please explain the intuition behind the dual problem in optimization.
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I've studied convex optimization pretty carefully, but don't feel that I have yet "grokked" the dual problem. Here are some questions I would like to understand more deeply/clearly/simply: How would somebody think of the dual problem? What thought process would lead someone to consider the dual problem and to recognize that it's valuable/interesting? In the case of a convex optimization problem, is there any obvious reason to expect that strong duality should (usually) hold? It often happens that the dual of the dual problem is the primal problem. However, this seems like a complete surprise to me. Is there any intuitive reason to expect that this should happen? Does the use of the word "dual" or "duality" in optimization have anything to do with the dual space in linear algebra? Or are they just different concepts that go by the same name. What about the use of the word "dual" in projective geometry — is there a connection there? You can define the dual problem and prove theorems about strong duality without ever mentioning the Fenchel conjugate. For example, Boyd and Vandenberghe prove a strong duality theorem without mentioning the Fenchel conjugate in their proof. And yet, people often talk as if the Fenchel conjugate is somehow the "essence" of duality, and make it sound as if the whole theory of duality is based on the Fenchel conjugate. Why is the Fenchel conjugate considered to have such fundamental importance? Note: I will now describe my current level of understanding of the intuition behind the dual problem. Please tell me if you think I might be missing any basic insights. I have read the excellent notes about convex optimization by Guilherme Freitas, and in particular the part about "penalty intuition". When we are trying to solve \begin{align*} \text{minimize} &\quad f(x) \\ \text{such that} & \quad h(x) \leq 0 \end{align*} one might try to eliminate the constraints by introducing a penalty when constraints are violated. This gives us the new unconstrained problem \begin{equation} \text{minimize} \quad f(x) + \langle \lambda ,h(x) \rangle \end{equation} where $\lambda \geq 0$. It's not hard to see that for a given $\lambda \geq 0$, the optimal value of this unconstrained problem is less than or equal to the optimal value for the constrained problem. This gives us a new problem — find $\lambda$ so that the optimal value for the unconstrained problem is as large as possible. That is one way to imagine how somebody might have thought of the dual problem. Is this the best intuition for where the dual problem comes from? Another viewpoint: the KKT conditions can be derived using what Freitas calls the "geometric intuition". Then, if we knew the value of the multipliers $\lambda$, it would be (often) much easier to find $x$. So, a new problem is to find $\lambda$. And if we can somehow recognize that $\lambda$ is a maximizer for the dual problem, then this suggests that we might try solving the dual problem. Please explain or give references to any intuition that you think I might find interesting, even if it's not directly related to what I asked.
| 313
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optimization, convex-optimization, intuition, lagrange-multiplier, karush-kuhn-tucker
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https://math.stackexchange.com/questions/223235/please-explain-the-intuition-behind-the-dual-problem-in-optimization
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Here's what's really going on with the dual problem. (This is my attempt to answer my own question, over a year after originally asking it.) (A very nice presentation of this material is given in Ekeland and Temam. These ideas are also in Rockafellar.) Let $V$ be a finite dimensional normed vector space over $\mathbb R$. (Working in an inner product space or just in $\mathbb R^N$ risks concealing the fundamental role that the dual space plays in duality in convex optimization.) The basic idea behind duality in convex analysis is to think of a convex set in terms of its supporting hyperplanes. (A closed convex set $\Omega$ can be "recovered" from its supporting hyperplanes by taking the intersection of all closed half spaces containing $\Omega$. The set of all supporting hyperplanes to $\Omega$ is sort of a "dual representation" of $\Omega$.) For a convex function $f$ (whose epigraph is a convex set), this strategy leads us to think about $f$ in terms of affine functions $\langle m^*, x \rangle - \alpha$ which are majorized by $f$. (Here $m^* \in V^*$ and we are using the notation $\langle m^*, x \rangle = m^*(x)$.) For a given slope $m^* \in V^*$, we only need to consider the "best" choice of $\alpha$ -- the other affine minorants with slope $m^*$ can be disregarded. \begin{align*} & f(x) \geq \langle m^*,x \rangle - \alpha \quad \forall x \in V \\ \iff & \alpha \geq \langle m^*, x \rangle - f(x) \quad \forall x \in V \\ \iff & \alpha \geq \sup_{x \in V} \quad \langle m^*, x \rangle - f(x) \end{align*} so the best choice of $\alpha$ is \begin{equation} f^*(m^*) = \sup_{x \in V} \quad \langle m^*, x \rangle - f(x). \end{equation} If this supremum is finite, then $\langle m^*,x \rangle - f^*(m^*)$ is the best affine minorant of $f$ with slope $m^*$. If $f^*(m^*) = \infty$, then there is no affine minorant of $f$ with slope $m^*$. The function $f^*$ is called the "conjugate" of $f$. The definition and basic facts about $f^*$ are all highly intuitive. For example, if $f$ is a proper closed convex function then $f$ can be recovered from $f^*$, because any closed convex set (in this case the epigraph of $f$) is the intersection of all the closed half spaces containing it. (I still think the fact that the "inversion formula" $f = f^{**}$ is so simple is a surprising and mathematically beautiful fact, but not hard to derive or prove with this intuition.) Because $f^*$ is defined on the dual space, we see already the fundamental role played by the dual space in duality in convex optimization. Given an optimization problem, we don't obtain a dual problem until we specify how to perturb the optimization problem. This is why equivalent formulations of an optimization problem can lead to different dual problems. By reformulating it we have in fact specified a different way to perturb it. As is typical in math, the ideas become clear when we work at an appropriate level of generality. Assume that our optimization problem is \begin{equation*} \operatorname*{minimize}_{x} \quad \phi(x,0). \end{equation*} Here $\phi:X \times Y \to \bar{\mathbb R}$ is convex. Standard convex optimization problems can be written in this form with an appropriate choice of $\phi$. The perturbed problems are \begin{equation*} \operatorname*{minimize}_{x} \quad \phi(x,y) \end{equation*} for nonzero values of $y \in Y$. Let $h(y) = \inf_x \phi(x,y)$. Our optimization problem is simply to evaluate $h(0)$. From our knowledge of conjugate functions, we know that \begin{equation*} h(0) \geq h^{**}(0) \end{equation*} and that typically we have equality. For example, if $h$ is subdifferentiable at $0$ (which is typical for a convex function) then $h(0) = h^{**}(0)$. The dual problem is simply to evaluate $h^{**}(0)$. In other words, the dual problem is: \begin{equation*} \operatorname*{maximize}_{y^* \in Y^*} \quad - h^*(y^*). \end{equation*} We see again the fundamental role that the dual space plays here. It is enlightening to express the dual problem in terms of $\phi$. It's easy to show that the dual problem is \begin{equation*} \operatorname*{maximize}_{y^* \in Y^*} \quad - \phi^*(0,y^*). \end{equation*} So the primal problem is \begin{equation*} \operatorname*{minimize}_{x \in X} \quad \phi(x,0) \end{equation*} and the dual problem (slightly restated) is \begin{equation*} \operatorname*{minimize}_{y^* \in Y^*} \quad \phi^*(0,y^*). \end{equation*} The similarity between these two problems is mathematically beautiful, and we can see that if we perturb the dual problem in the obvious way, then the dual of the dual problem will be the primal problem (assuming $\phi = \phi^{**}$). The natural isomorphism between $V$ and $V^{**}$ is of fundamental importance here. The key facts about the dual problem -- strong duality, the optimality conditions, and the sensitivity interpretation of the optimal dual variables -- all become intuitively clear and even "obvious" from this viewpoint. An optimization problem in the form \begin{align*} \operatorname*{minimize}_x & \quad f(x) \\ \text{subject to} & \quad g(x) \leq 0, \end{align*} can be perturbed as follows: \begin{align*} \operatorname*{minimize}_x & \quad f(x) \\ \text{subject to} & \quad g(x) + y \leq 0. \end{align*} This perturbed problem has the form given above with \begin{equation*} \phi(x,y) = \begin{cases} f(x) \quad \text{if } g(x) + y \leq 0 \\ \infty \quad \text{otherwise}. \end{cases} \end{equation*} To find the dual problem, we need to evaluate $-\phi^*(0,y^*)$, which is a relatively straightforward calculation. \begin{align*} -\phi^*(0,y^*) &= -\sup_{g(x) + y \leq 0} \quad \langle y^*,y \rangle - f(x) \\ &= -\sup_{\substack{ x \\ q \geq 0 }} \quad \langle y^*, -g(x) - q \rangle - f(x) \\ &= \inf_{\substack{ x \\ q \geq 0 }} \quad f(x) + \langle y^*, g(x) \rangle + \langle y^*, q \rangle. \end{align*} We can minimize first with respect to $q$, and we will get $-\infty$ unless $\langle y^*, q \rangle \geq 0$ for all $q \geq 0$. In other words, we will get $-\infty$ unless $y^* \geq 0$. The dual function is \begin{equation*} -\phi^*(0,y^*) = \begin{cases} \inf_x \quad f(x) + \langle y^*, g(x) \rangle \quad \text{if } y^* \geq 0 \\ -\infty \quad \text{otherwise}. \end{cases} \end{equation*} This is the expected result.
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Q: Please explain the intuition behind the dual problem in optimization.
I've studied convex optimization pretty carefully, but don't feel that I have yet "grokked" the dual problem. Here are some questions I would like to understand more deeply/clearly/simply: How would somebody think of the dual problem? What thought process would lead someone to consider the dual problem and to recognize that it's valuable/interesting? In the case of a convex optimization problem, is there any obvious reason to expect that strong duality should (usually) hold? It often happens that the dual of the dual problem is the primal problem. However, this seems like a complete surprise to me. Is there any intuitive reason to expect that this should happen? Does the use of the word "dual" or "duality" in optimization have anything to do with the dual space in linear algebra? Or are they just different concepts that go by the same name. What about the use of the word "dual" in projective geometry — is there a connection there? You can define the dual problem and prove theorems about strong duality without ever mentioning the Fenchel conjugate. For example, Boyd and Vandenberghe prove a strong duality theorem without mentioning the Fenchel conjugate in their proof. And yet, people often talk as if the Fenchel conjugate is somehow the "essence" of duality, and make it sound as if the whole theory of duality is based on the Fenchel conjugate. Why is the Fenchel conjugate considered to have such fundamental importance? Note: I will now describe my current level of understanding of the intuition behind the dual problem. Please tell me if you think I might be missing any basic insights. I have read the excellent notes about convex optimization by Guilherme Freitas, and in particular the part about "penalty intuition". When we are trying to solve \begin{align*} \text{minimize} &\quad f(x) \\ \text{such that} & \quad h(x) \leq 0 \end{align*} one might try to eliminate the constraints by introducing a penalty when constraints are violated. This gives us the new unconstrained problem \begin{equation} \text{minimize} \quad f(x) + \langle \lambda ,h(x) \rangle \end{equation} where $\lambda \geq 0$. It's not hard to see that for a given $\lambda \geq 0$, the optimal value of this unconstrained problem is less than or equal to the optimal value for the constrained problem. This gives us a new problem — find $\lambda$ so that the optimal value for the unconstrained problem is as large as possible. That is one way to imagine how somebody might have thought of the dual problem. Is this the best intuition for where the dual problem comes from? Another viewpoint: the KKT conditions can be derived using what Freitas calls the "geometric intuition". Then, if we knew the value of the multipliers $\lambda$, it would be (often) much easier to find $x$. So, a new problem is to find $\lambda$. And if we can somehow recognize that $\lambda$ is a maximizer for the dual problem, then this suggests that we might try solving the dual problem. Please explain or give references to any intuition that you think I might find interesting, even if it's not directly related to what I asked.
A: Here's what's really going on with the dual problem. (This is my attempt to answer my own question, over a year after originally asking it.) (A very nice presentation of this material is given in Ekeland and Temam. These ideas are also in Rockafellar.) Let $V$ be a finite dimensional normed vector space over $\mathbb R$. (Working in an inner product space or just in $\mathbb R^N$ risks concealing the fundamental role that the dual space plays in duality in convex optimization.) The basic idea behind duality in convex analysis is to think of a convex set in terms of its supporting hyperplanes. (A closed convex set $\Omega$ can be "recovered" from its supporting hyperplanes by taking the intersection of all closed half spaces containing $\Omega$. The set of all supporting hyperplanes to $\Omega$ is sort of a "dual representation" of $\Omega$.) For a convex function $f$ (whose epigraph is a convex set), this strategy leads us to think about $f$ in terms of affine functions $\langle m^*, x \rangle - \alpha$ which are majorized by $f$. (Here $m^* \in V^*$ and we are using the notation $\langle m^*, x \rangle = m^*(x)$.) For a given slope $m^* \in V^*$, we only need to consider the "best" choice of $\alpha$ -- the other affine minorants with slope $m^*$ can be disregarded. \begin{align*} & f(x) \geq \langle m^*,x \rangle - \alpha \quad \forall x \in V \\ \iff & \alpha \geq \langle m^*, x \rangle - f(x) \quad \forall x \in V \\ \iff & \alpha \geq \sup_{x \in V} \quad \langle m^*, x \rangle - f(x) \end{align*} so the best choice of $\alpha$ is \begin{equation} f^*(m^*) = \sup_{x \in V} \quad \langle m^*, x \rangle - f(x). \end{equation} If this supremum is finite, then $\langle m^*,x \rangle - f^*(m^*)$ is the best affine minorant of $f$ with slope $m^*$. If $f^*(m^*) = \infty$, then there is no affine minorant of $f$ with slope $m^*$. The function $f^*$ is called the "conjugate" of $f$. The definition and basic facts about $f^*$ are all highly intuitive. For example, if $f$ is a proper closed convex function then $f$ can be recovered from $f^*$, because any closed convex set (in this case the epigraph of $f$) is the intersection of all the closed half spaces containing it. (I still think the fact that the "inversion formula" $f = f^{**}$ is so simple is a surprising and mathematically beautiful fact, but not hard to derive or prove with this intuition.) Because $f^*$ is defined on the dual space, we see already the fundamental role played by the dual space in duality in convex optimization. Given an optimization problem, we don't obtain a dual problem until we specify how to perturb the optimization problem. This is why equivalent formulations of an optimization problem can lead to different dual problems. By reformulating it we have in fact specified a different way to perturb it. As is typical in math, the ideas become clear when we work at an appropriate level of generality. Assume that our optimization problem is \begin{equation*} \operatorname*{minimize}_{x} \quad \phi(x,0). \end{equation*} Here $\phi:X \times Y \to \bar{\mathbb R}$ is convex. Standard convex optimization problems can be written in this form with an appropriate choice of $\phi$. The perturbed problems are \begin{equation*} \operatorname*{minimize}_{x} \quad \phi(x,y) \end{equation*} for nonzero values of $y \in Y$. Let $h(y) = \inf_x \phi(x,y)$. Our optimization problem is simply to evaluate $h(0)$. From our knowledge of conjugate functions, we know that \begin{equation*} h(0) \geq h^{**}(0) \end{equation*} and that typically we have equality. For example, if $h$ is subdifferentiable at $0$ (which is typical for a convex function) then $h(0) = h^{**}(0)$. The dual problem is simply to evaluate $h^{**}(0)$. In other words, the dual problem is: \begin{equation*} \operatorname*{maximize}_{y^* \in Y^*} \quad - h^*(y^*). \end{equation*} We see again the fundamental role that the dual space plays here. It is enlightening to express the dual problem in terms of $\phi$. It's easy to show that the dual problem is \begin{equation*} \operatorname*{maximize}_{y^* \in Y^*} \quad - \phi^*(0,y^*). \end{equation*} So the primal problem is \begin{equation*} \operatorname*{minimize}_{x \in X} \quad \phi(x,0) \end{equation*} and the dual problem (slightly restated) is \begin{equation*} \operatorname*{minimize}_{y^* \in Y^*} \quad \phi^*(0,y^*). \end{equation*} The similarity between these two problems is mathematically beautiful, and we can see that if we perturb the dual problem in the obvious way, then the dual of the dual problem will be the primal problem (assuming $\phi = \phi^{**}$). The natural isomorphism between $V$ and $V^{**}$ is of fundamental importance here. The key facts about the dual problem -- strong duality, the optimality conditions, and the sensitivity interpretation of the optimal dual variables -- all become intuitively clear and even "obvious" from this viewpoint. An optimization problem in the form \begin{align*} \operatorname*{minimize}_x & \quad f(x) \\ \text{subject to} & \quad g(x) \leq 0, \end{align*} can be perturbed as follows: \begin{align*} \operatorname*{minimize}_x & \quad f(x) \\ \text{subject to} & \quad g(x) + y \leq 0. \end{align*} This perturbed problem has the form given above with \begin{equation*} \phi(x,y) = \begin{cases} f(x) \quad \text{if } g(x) + y \leq 0 \\ \infty \quad \text{otherwise}. \end{cases} \end{equation*} To find the dual problem, we need to evaluate $-\phi^*(0,y^*)$, which is a relatively straightforward calculation. \begin{align*} -\phi^*(0,y^*) &= -\sup_{g(x) + y \leq 0} \quad \langle y^*,y \rangle - f(x) \\ &= -\sup_{\substack{ x \\ q \geq 0 }} \quad \langle y^*, -g(x) - q \rangle - f(x) \\ &= \inf_{\substack{ x \\ q \geq 0 }} \quad f(x) + \langle y^*, g(x) \rangle + \langle y^*, q \rangle. \end{align*} We can minimize first with respect to $q$, and we will get $-\infty$ unless $\langle y^*, q \rangle \geq 0$ for all $q \geq 0$. In other words, we will get $-\infty$ unless $y^* \geq 0$. The dual function is \begin{equation*} -\phi^*(0,y^*) = \begin{cases} \inf_x \quad f(x) + \langle y^*, g(x) \rangle \quad \text{if } y^* \geq 0 \\ -\infty \quad \text{otherwise}. \end{cases} \end{equation*} This is the expected result.
| 2026-01-26T11:30:36.533073
|
514
|
Conjectures that have been disproved with extremely large counterexamples?
|
I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture. I'm sure that everyone here is familiar with it; it describes an operation on a natural number – $n/2$ if it is even, $3n+1$ if it is odd. The conjecture states that if this operation is repeated, all numbers will eventually wind up at $1$ (or rather, in an infinite loop of $1-4-2-1-4-2-1$). I fired up Python and ran a quick test on this for all numbers up to $5.76 \times 10^{18}$ (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at $1$. Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.) I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?" To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!" And he said, "It is my conjecture that there are none! (and if any, they are rare)". Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?
| 312
|
big-list, conjectures, big-numbers
|
https://math.stackexchange.com/questions/514/conjectures-that-have-been-disproved-with-extremely-large-counterexamples
|
My favorite example, which I'm surprised hasn't been posted yet, is the conjecture: $n^{17}+9 \text{ and } (n+1)^{17}+9 \text{ are relatively prime}$ The first counterexample is $n=8424432925592889329288197322308900672459420460792433$
| 178
| false
|
Q: Conjectures that have been disproved with extremely large counterexamples?
I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture. I'm sure that everyone here is familiar with it; it describes an operation on a natural number – $n/2$ if it is even, $3n+1$ if it is odd. The conjecture states that if this operation is repeated, all numbers will eventually wind up at $1$ (or rather, in an infinite loop of $1-4-2-1-4-2-1$). I fired up Python and ran a quick test on this for all numbers up to $5.76 \times 10^{18}$ (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at $1$. Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.) I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?" To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!" And he said, "It is my conjecture that there are none! (and if any, they are rare)". Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?
A: My favorite example, which I'm surprised hasn't been posted yet, is the conjecture: $n^{17}+9 \text{ and } (n+1)^{17}+9 \text{ are relatively prime}$ The first counterexample is $n=8424432925592889329288197322308900672459420460792433$
| 2026-01-26T11:30:38.273872
|
96,331
|
In Russian roulette, is it best to go first?
|
Assume that we are playing a game of Russian roulette (6 chambers) and that there is no shuffling after the shot is fired. I was wondering if you have an advantage in going first? If so, how big of an advantage? I was just debating this with friends, and I wouldn't know what probability to use to prove it. I'm thinking binomial distribution or something like that. If $n=2$, then there's no advantage. Just $50/50$ if the person survives or dies. If $n=3$, then maybe the other guy has an advantage. The person who goes second should have an advantage. Or maybe I'm wrong.
| 308
|
probability
|
https://math.stackexchange.com/questions/96331/in-russian-roulette-is-it-best-to-go-first
|
For a $2$ Player Game, it's obvious that player one will play, and $\frac16$ chance of losing. Player $2$, has a $\frac16$ chance of winning on turn one, so there is a $\frac56$ chance he will have to take his turn. (I've intentionally left fractions without reducing them as it's clearer where the numbers came from) Player 1 - $\frac66$ (Chance Turn $1$ happening) $\times \ \frac16$ (chance of dying) = $\frac16$ Player 2 - $\frac56$ (Chance Turn $2$ happening) $\times \ \frac15$ (chance of dying) = $\frac16$ Player 1 - $\frac46$ (Chance Turn $3$ happening) $\times \ \frac14$ (chance of dying) = $\frac16$ Player 2 - $\frac36$ (Chance Turn $4$ happening) $\times \ \frac13$ (chance of dying) = $\frac16$ Player 1 - $\frac26$ (Chance Turn $5$ happening) $\times \ \frac12$ (chance of dying) = $\frac16$ Player 2 - $\frac16$ (Chance Turn $6$ happening) $\times \ \frac11$ (chance of dying) = $\frac16$ So the two player game is fair without shuffling. Similarly, the $3$ and $6$ player versions are fair. It's the $4$ and $5$ player versions where you want to go last, in hopes that the bullets will run out before your second turn. For a for $4$ player game, it's: P1 - $\frac26$, P2 - $\frac26$, P3 - $\frac16$, P4 - $\frac16$ Now, the idea in a $2$ player game is that it is best to be player $2$, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player $1$ (or your captor), thus winning, changing your total odds of losing to P1 - $\frac36$, P2 - $\frac26$, Captor - $\frac16$
| 359
| false
|
Q: In Russian roulette, is it best to go first?
Assume that we are playing a game of Russian roulette (6 chambers) and that there is no shuffling after the shot is fired. I was wondering if you have an advantage in going first? If so, how big of an advantage? I was just debating this with friends, and I wouldn't know what probability to use to prove it. I'm thinking binomial distribution or something like that. If $n=2$, then there's no advantage. Just $50/50$ if the person survives or dies. If $n=3$, then maybe the other guy has an advantage. The person who goes second should have an advantage. Or maybe I'm wrong.
A: For a $2$ Player Game, it's obvious that player one will play, and $\frac16$ chance of losing. Player $2$, has a $\frac16$ chance of winning on turn one, so there is a $\frac56$ chance he will have to take his turn. (I've intentionally left fractions without reducing them as it's clearer where the numbers came from) Player 1 - $\frac66$ (Chance Turn $1$ happening) $\times \ \frac16$ (chance of dying) = $\frac16$ Player 2 - $\frac56$ (Chance Turn $2$ happening) $\times \ \frac15$ (chance of dying) = $\frac16$ Player 1 - $\frac46$ (Chance Turn $3$ happening) $\times \ \frac14$ (chance of dying) = $\frac16$ Player 2 - $\frac36$ (Chance Turn $4$ happening) $\times \ \frac13$ (chance of dying) = $\frac16$ Player 1 - $\frac26$ (Chance Turn $5$ happening) $\times \ \frac12$ (chance of dying) = $\frac16$ Player 2 - $\frac16$ (Chance Turn $6$ happening) $\times \ \frac11$ (chance of dying) = $\frac16$ So the two player game is fair without shuffling. Similarly, the $3$ and $6$ player versions are fair. It's the $4$ and $5$ player versions where you want to go last, in hopes that the bullets will run out before your second turn. For a for $4$ player game, it's: P1 - $\frac26$, P2 - $\frac26$, P3 - $\frac16$, P4 - $\frac16$ Now, the idea in a $2$ player game is that it is best to be player $2$, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player $1$ (or your captor), thus winning, changing your total odds of losing to P1 - $\frac36$, P2 - $\frac26$, Captor - $\frac16$
| 2026-01-26T11:30:40.004225
|
49,229
|
Why can ALL quadratic equations be solved by the quadratic formula?
|
In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't. Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?
| 307
|
algebra-precalculus, polynomials, quadratics
|
https://math.stackexchange.com/questions/49229/why-can-all-quadratic-equations-be-solved-by-the-quadratic-formula
|
I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula. Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.) We want to solve the equation $$ax^2+bx+c=0,$$ where $a \ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset. Instead, multiply both sides by $4a$. We obtain the equivalent equation $$4a^2x^2 +4abx+4ac=0.\tag{1}$$ Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely, $$4a^2x^2+4abx=(2ax+b)^2-b^2.$$ So our equation can be rewritten as $$(2ax+b)^2 -b^2+4ac=0 \tag{2}$$ or equivalently $$(2ax+b)^2=b^2-4ac. \tag{3}$$ Now it's all over. We find that $$2ax+b=\pm\sqrt{b^2-4ac} \tag{4}$$ and therefore $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \tag{5}$$ No fractions until the very end! Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form. Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.
| 547
| false
|
Q: Why can ALL quadratic equations be solved by the quadratic formula?
In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't. Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?
A: I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula. Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.) We want to solve the equation $$ax^2+bx+c=0,$$ where $a \ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset. Instead, multiply both sides by $4a$. We obtain the equivalent equation $$4a^2x^2 +4abx+4ac=0.\tag{1}$$ Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely, $$4a^2x^2+4abx=(2ax+b)^2-b^2.$$ So our equation can be rewritten as $$(2ax+b)^2 -b^2+4ac=0 \tag{2}$$ or equivalently $$(2ax+b)^2=b^2-4ac. \tag{3}$$ Now it's all over. We find that $$2ax+b=\pm\sqrt{b^2-4ac} \tag{4}$$ and therefore $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \tag{5}$$ No fractions until the very end! Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form. Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.
| 2026-01-26T11:30:41.614650
|
513,239
|
One question to know if the number is 1, 2 or 3
|
I've recently heard a riddle, which looks quite simple, but I can't solve it. A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "Yes", "No", or "I don't know," and after the girl answers it, he knows what the number is. What is the question? Note that the girl is professional in maths and knows EVERYTHING about these three numbers. EDIT: The person who told me this just said the correct answer is: "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"
| 307
|
puzzle
|
https://math.stackexchange.com/questions/513239/one-question-to-know-if-the-number-is-1-2-or-3
|
"I am thinking of a number which is either 0 or 1. Is the sum of our numbers greater than 2?"
| 280
| true
|
Q: One question to know if the number is 1, 2 or 3
I've recently heard a riddle, which looks quite simple, but I can't solve it. A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "Yes", "No", or "I don't know," and after the girl answers it, he knows what the number is. What is the question? Note that the girl is professional in maths and knows EVERYTHING about these three numbers. EDIT: The person who told me this just said the correct answer is: "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"
A: "I am thinking of a number which is either 0 or 1. Is the sum of our numbers greater than 2?"
| 2026-01-26T11:30:43.075672
|
260,656
|
Can't argue with success? Looking for "bad math" that "gets away with it"
|
I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare"). One example would be "cancelling" the 6s in $$\frac{64}{16}$$ Another one would be something like $$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;$$ Yet another one would be $$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;$$ Note that I am specifically not interested in mathematical fallacies (aka spurious proofs). Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for are cases where one arrives at valid ends by (shockingly) wrong means.
| 306
|
examples-counterexamples, recreational-mathematics, big-list
|
https://math.stackexchange.com/questions/260656/cant-argue-with-success-looking-for-bad-math-that-gets-away-with-it
|
I was quite amused when a student produced the following when cancelling a fraction: $$\frac{x^2-y^2}{x-y}$$ He began by "cancelling" the $x$ and the $y$ on top and bottom, to get: $$\frac{x-y}{-}$$ and then concluded that "two negatives make a positive", so the final answer has to be $x+y$.
| 454
| false
|
Q: Can't argue with success? Looking for "bad math" that "gets away with it"
I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare"). One example would be "cancelling" the 6s in $$\frac{64}{16}$$ Another one would be something like $$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;$$ Yet another one would be $$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;$$ Note that I am specifically not interested in mathematical fallacies (aka spurious proofs). Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for are cases where one arrives at valid ends by (shockingly) wrong means.
A: I was quite amused when a student produced the following when cancelling a fraction: $$\frac{x^2-y^2}{x-y}$$ He began by "cancelling" the $x$ and the $y$ on top and bottom, to get: $$\frac{x-y}{-}$$ and then concluded that "two negatives make a positive", so the final answer has to be $x+y$.
| 2026-01-26T11:30:44.726394
|
1,833,459
|
Math without pencil and paper
|
For someone who is physically unable to use a pencil and paper, what would be the best way to do math? In my case, I have only a little movement in my fingers. I can move a computer mouse and press the left button. Currently I do very little math and when I do I use MS Office but this is a very slow way of doing it and will break my thinking process. EDIT: Thank you for the answers. I have a lot to try. I won't accept an answer since it's dependent on the disability but thank you all.
| 305
|
math-software
|
https://math.stackexchange.com/questions/1833459/math-without-pencil-and-paper
|
(Background: I have a chronic pain condition, and it is extremely painful for me to do any sort of repetitive fine motor activity, including writing and typing. I earned an undergraduate degree in math with quite little handwriting, and I'm currently a graduate student.) Unfortunately, I've never been able to find any one solution or approach to replace the work that many people do with pencil and paper, but there are lots of partial solutions that work for different situations and types of mathematics. Here are some of the ones that I use and which, based on your description, might be applicable. Getting really proficient at mental math. I think you've already done a lot of this, so all I'm going to say is that I can do a lot more math in my head than I would've believed was possible pre-disability. It just takes a lot of practice. Of course, it doesn't work for everything; some calculations just require remembering too many things at once. Speech-to-text. If you have clear speech, then a speech-to-text system is potentially very helpful. (I use Dragon NaturallySpeaking.) Maybe you already use one for general writing and computer use. If so, then you have almost certainly noticed that they are not designed for mathematics (or, for that matter, computer science), so some additional software is necessary in order to do math. I use a system based on NatLaTeX to dictate all of my formal mathematics, including anything in that I'm going to turn in for my coursework. Basically, NatLaTeX defines a speakable form of many common LaTeX commands, including everything you need for most mathematical expressions. Using a custom vocabulary in Dragon, I can dictate a plain text file containing this NatLaTeX source. I then use in scripts from the NatLaTeX project to transform my dictated text into actual LaTeX source, which I can then compile into nicely typeset mathematics using a standard LaTeX compiler. (Actually, I use a batch file to automate the process.) Just as a note, I have made several modifications to NatLaTeX in order to optimize it for mathematics (the original author was a physicist) and to adjust for changes in LaTeX. Feel free to contact me if you want a copy of the modified scripts. I do eventually intend to post them somewhere online, but I need to spend some time updating the documentation first (and that's really hard to justify spending time on it while I'm preparing for comprehensive exams!). The big advantage to a dictation system like this is that you get to go through the process of formally presenting your work, which is really helpful for checking understanding and practicing proof writing. Disadvantages include a steep learning curve and not being able to see your work (typeset, at least) in real time. You also have very limited choices in text editors because Dragon will only "play nice" with a couple of them. Programming languages and computer algebra systems. When you actually need to do calculations, plot functions, etc., it is hard to beat a good computer algebra system. There are lots of choices out there, and I think the choice of which one to use ultimately comes down to personal preference and perhaps compatibility with whatever assistive technology you use. Of course, you still face the problem of how to handle getting input to that computer algebra system. Typing with one finger on an on-screen keyboard sounds rather slow and tedious. Here are a few alternatives you might want to look into. If you have head and neck movement, a head pointer is one option. I sometimes use one that actually marketed as a gaming device called TrackIR. (Gaming peripherals are much less expensive than assistive technology peripherals!) You can use this to type on an on-screen keyboard or to interact with the input panels found in many computer algebra systems. For mouse click, you could use a switch with your finger or a dwell/hover click. Eye tracking is a technology that has recently become much more affordable due to relatively new consumer-level devices marketed for gaming applications. Just a couple months ago, I got an Tobii EyeX eye tracker, and it has been great! I use it with a simple mouse emulation script and Dasher to write code in Sage. Dasher is really cool, by the way, and quite good for text entry with low-precision input devices like eye trackers. It's also much faster than most on-screen keyboards. Just as a warning, the EyeX is not intended as an assistive device so you do have to do a some software configuration in order to make it do what you want it to do. But you're studying computer science, so I don't think that should be a problem for you. (FreePIE, OptiKey, and Dasher are all good pieces of free software to look up and consider for use with the EyeX.) The big advantage to computer algebra systems is that you can do all of the numerical calculations and tedious symbolic manipulations without needing to ever write down a single number. The disadvantage is that hiding all of the details can sometimes hinder your conceptual understanding. Those are the main strategies I use. I hope something here can help you out, too. EDIT: There has been a request for some examples of what NatLaTex can do, so here are a couple of different examples pulled files on my hard drive. Discrete Math Example: NatLaTeX Input (dictated with Dragon) Given a poset "(P, precedes)", a collection of linear extensions "{calligraphy R } equals left curly brace precedes sub one, precedes sub two, low dots, precedes sub k right curly brace" is called a ``realizer'' of "P" if "precedes equals intersection of sub {i equals one } to the k precedes sub k", where each relation "precedes sub i" is interpreted as a set of ordered pairs and "intersection of" is set intersection. Equivalently, "{calligraphy R }" is a realizer of "P" if, for all "p, q in P", "p precedes q" if and only if "p precedes sub i q" for all "one less than or equal to i less than or equal to k". LaTeX Output Given a poset \( ( P , \prec )\), a collection of linear extensions \( { \mathcal R } = \{ \prec_1 , \prec_2 , \ldots , \prec_k \}\) is called a ``realizer'' of \( P\) if \( \prec = \bigcap_{ i = 1 }^k \prec_k\), where each relation \( \prec_i\) is interpreted as a set of ordered pairs and \( \bigcap\) is set intersection. Equivalently, \( { \mathcal R }\) is a realizer of \( P\) if, for all \( p , q \in P\), \( p \prec q\) if and only if \( p \prec_i q\) for all \( 1 \leq i \leq k\). Analysis Example: NatLaTeX Input (dictated with Dragon) begin theorem [Monotone Convergence Theorem] Let "left curly brace f sub n right curly brace sub {n equals one } to the infinity" be a sequence of nonnegative measurable functions with "f sub one less than or equal to f sub two less than or equal to low dots less than or equal to f sub n less than or equal to f sub {n +1 } less than or equal to low dots" and "limit of sub n f sub n equals f" (pointwise). Then, "f" is measurable and @begin{equation} limit of sub {n right arrow infinity } integral f sub n d Greek mu equals integral limit of sub {n right arrow infinity } f sub n d Greek mu equals integral f d Greek mu @end{equation} end theorem LaTeX Output \begin{theorem}[Monotone Convergence Theorem] Let \( \{ f_n \}_{ n = 1 }^\infty\) be a sequence of nonnegative measurable functions with \( f_1 \leq f_2 \leq \ldots \leq f_n \leq f_{ n + 1 } \leq \ldots\) and \( \lim_n f_n = f \) (pointwise). Then, \( f \) is measurable and \begin{equation} \lim_{ n \rightarrow \infty } \int f_n d \mu = \int \lim_{ n \rightarrow \infty } f_n d \mu = \int f d \mu \end{equation} \end{theorem}
| 221
| false
|
Q: Math without pencil and paper
For someone who is physically unable to use a pencil and paper, what would be the best way to do math? In my case, I have only a little movement in my fingers. I can move a computer mouse and press the left button. Currently I do very little math and when I do I use MS Office but this is a very slow way of doing it and will break my thinking process. EDIT: Thank you for the answers. I have a lot to try. I won't accept an answer since it's dependent on the disability but thank you all.
A: (Background: I have a chronic pain condition, and it is extremely painful for me to do any sort of repetitive fine motor activity, including writing and typing. I earned an undergraduate degree in math with quite little handwriting, and I'm currently a graduate student.) Unfortunately, I've never been able to find any one solution or approach to replace the work that many people do with pencil and paper, but there are lots of partial solutions that work for different situations and types of mathematics. Here are some of the ones that I use and which, based on your description, might be applicable. Getting really proficient at mental math. I think you've already done a lot of this, so all I'm going to say is that I can do a lot more math in my head than I would've believed was possible pre-disability. It just takes a lot of practice. Of course, it doesn't work for everything; some calculations just require remembering too many things at once. Speech-to-text. If you have clear speech, then a speech-to-text system is potentially very helpful. (I use Dragon NaturallySpeaking.) Maybe you already use one for general writing and computer use. If so, then you have almost certainly noticed that they are not designed for mathematics (or, for that matter, computer science), so some additional software is necessary in order to do math. I use a system based on NatLaTeX to dictate all of my formal mathematics, including anything in that I'm going to turn in for my coursework. Basically, NatLaTeX defines a speakable form of many common LaTeX commands, including everything you need for most mathematical expressions. Using a custom vocabulary in Dragon, I can dictate a plain text file containing this NatLaTeX source. I then use in scripts from the NatLaTeX project to transform my dictated text into actual LaTeX source, which I can then compile into nicely typeset mathematics using a standard LaTeX compiler. (Actually, I use a batch file to automate the process.) Just as a note, I have made several modifications to NatLaTeX in order to optimize it for mathematics (the original author was a physicist) and to adjust for changes in LaTeX. Feel free to contact me if you want a copy of the modified scripts. I do eventually intend to post them somewhere online, but I need to spend some time updating the documentation first (and that's really hard to justify spending time on it while I'm preparing for comprehensive exams!). The big advantage to a dictation system like this is that you get to go through the process of formally presenting your work, which is really helpful for checking understanding and practicing proof writing. Disadvantages include a steep learning curve and not being able to see your work (typeset, at least) in real time. You also have very limited choices in text editors because Dragon will only "play nice" with a couple of them. Programming languages and computer algebra systems. When you actually need to do calculations, plot functions, etc., it is hard to beat a good computer algebra system. There are lots of choices out there, and I think the choice of which one to use ultimately comes down to personal preference and perhaps compatibility with whatever assistive technology you use. Of course, you still face the problem of how to handle getting input to that computer algebra system. Typing with one finger on an on-screen keyboard sounds rather slow and tedious. Here are a few alternatives you might want to look into. If you have head and neck movement, a head pointer is one option. I sometimes use one that actually marketed as a gaming device called TrackIR. (Gaming peripherals are much less expensive than assistive technology peripherals!) You can use this to type on an on-screen keyboard or to interact with the input panels found in many computer algebra systems. For mouse click, you could use a switch with your finger or a dwell/hover click. Eye tracking is a technology that has recently become much more affordable due to relatively new consumer-level devices marketed for gaming applications. Just a couple months ago, I got an Tobii EyeX eye tracker, and it has been great! I use it with a simple mouse emulation script and Dasher to write code in Sage. Dasher is really cool, by the way, and quite good for text entry with low-precision input devices like eye trackers. It's also much faster than most on-screen keyboards. Just as a warning, the EyeX is not intended as an assistive device so you do have to do a some software configuration in order to make it do what you want it to do. But you're studying computer science, so I don't think that should be a problem for you. (FreePIE, OptiKey, and Dasher are all good pieces of free software to look up and consider for use with the EyeX.) The big advantage to computer algebra systems is that you can do all of the numerical calculations and tedious symbolic manipulations without needing to ever write down a single number. The disadvantage is that hiding all of the details can sometimes hinder your conceptual understanding. Those are the main strategies I use. I hope something here can help you out, too. EDIT: There has been a request for some examples of what NatLaTex can do, so here are a couple of different examples pulled files on my hard drive. Discrete Math Example: NatLaTeX Input (dictated with Dragon) Given a poset "(P, precedes)", a collection of linear extensions "{calligraphy R } equals left curly brace precedes sub one, precedes sub two, low dots, precedes sub k right curly brace" is called a ``realizer'' of "P" if "precedes equals intersection of sub {i equals one } to the k precedes sub k", where each relation "precedes sub i" is interpreted as a set of ordered pairs and "intersection of" is set intersection. Equivalently, "{calligraphy R }" is a realizer of "P" if, for all "p, q in P", "p precedes q" if and only if "p precedes sub i q" for all "one less than or equal to i less than or equal to k". LaTeX Output Given a poset \( ( P , \prec )\), a collection of linear extensions \( { \mathcal R } = \{ \prec_1 , \prec_2 , \ldots , \prec_k \}\) is called a ``realizer'' of \( P\) if \( \prec = \bigcap_{ i = 1 }^k \prec_k\), where each relation \( \prec_i\) is interpreted as a set of ordered pairs and \( \bigcap\) is set intersection. Equivalently, \( { \mathcal R }\) is a realizer of \( P\) if, for all \( p , q \in P\), \( p \prec q\) if and only if \( p \prec_i q\) for all \( 1 \leq i \leq k\). Analysis Example: NatLaTeX Input (dictated with Dragon) begin theorem [Monotone Convergence Theorem] Let "left curly brace f sub n right curly brace sub {n equals one } to the infinity" be a sequence of nonnegative measurable functions with "f sub one less than or equal to f sub two less than or equal to low dots less than or equal to f sub n less than or equal to f sub {n +1 } less than or equal to low dots" and "limit of sub n f sub n equals f" (pointwise). Then, "f" is measurable and @begin{equation} limit of sub {n right arrow infinity } integral f sub n d Greek mu equals integral limit of sub {n right arrow infinity } f sub n d Greek mu equals integral f d Greek mu @end{equation} end theorem LaTeX Output \begin{theorem}[Monotone Convergence Theorem] Let \( \{ f_n \}_{ n = 1 }^\infty\) be a sequence of nonnegative measurable functions with \( f_1 \leq f_2 \leq \ldots \leq f_n \leq f_{ n + 1 } \leq \ldots\) and \( \lim_n f_n = f \) (pointwise). Then, \( f \) is measurable and \begin{equation} \lim_{ n \rightarrow \infty } \int f_n d \mu = \int \lim_{ n \rightarrow \infty } f_n d \mu = \int f d \mu \end{equation} \end{theorem}
| 2026-01-26T11:30:46.405361
|
24,241
|
Why do mathematicians use single-letter variables?
|
I have much more experience programming than I do with advanced mathematics, so perhaps this is just a comfort thing with me, but I often get frustrated when I try to follow mathematical notation. Specifically, I get frustrated trying to keep track of what each variable signifies. As a programmer, this would be completely unacceptable no matter how many comments you added explaining it: float A(float P, float r, float n, float t) { return P * pow(1 + r / n, n * t); } Yet a mathematician would have no problem with this: $A = P\ \left(1+\dfrac{r}{n}\right)^{nt}$ where $A$ = final amount $P$ = principal amount (initial investment) $r$ = annual nominal interest rate (as a decimal) $n$ = number of times the interest is compounded per year $t$ = number of years So why don't I ever see the following? $\text{final_amount} = \text{principal}\; \left(1+\dfrac{\text{interest_rate}}{\text{periods_per_yr}}\right)^{\text{periods_per_yr}\cdot\text{years}}$
| 301
|
soft-question, notation, math-history
|
https://math.stackexchange.com/questions/24241/why-do-mathematicians-use-single-letter-variables
|
I think one reason is that often one does not want to remember what the variable names really represent. As an example, when we choose to talk about the matrix $(a_{ij})$ instead of the matrix $(\mathrm{TransitionProbability}_{ij})$, this expresses the important fact that once we have formulated our problem in terms of matrices, it is perfectly safe to forget where the problem came from originally -- in fact, remembering what the matrix "really" describes might only be unnecessary psychological baggage that prevents us from applying all linear-algebraic tools at our disposal. (As an aside, have you ever seen code written by a mathematician? It very often looks exactly like your first example.)
| 240
| false
|
Q: Why do mathematicians use single-letter variables?
I have much more experience programming than I do with advanced mathematics, so perhaps this is just a comfort thing with me, but I often get frustrated when I try to follow mathematical notation. Specifically, I get frustrated trying to keep track of what each variable signifies. As a programmer, this would be completely unacceptable no matter how many comments you added explaining it: float A(float P, float r, float n, float t) { return P * pow(1 + r / n, n * t); } Yet a mathematician would have no problem with this: $A = P\ \left(1+\dfrac{r}{n}\right)^{nt}$ where $A$ = final amount $P$ = principal amount (initial investment) $r$ = annual nominal interest rate (as a decimal) $n$ = number of times the interest is compounded per year $t$ = number of years So why don't I ever see the following? $\text{final_amount} = \text{principal}\; \left(1+\dfrac{\text{interest_rate}}{\text{periods_per_yr}}\right)^{\text{periods_per_yr}\cdot\text{years}}$
A: I think one reason is that often one does not want to remember what the variable names really represent. As an example, when we choose to talk about the matrix $(a_{ij})$ instead of the matrix $(\mathrm{TransitionProbability}_{ij})$, this expresses the important fact that once we have formulated our problem in terms of matrices, it is perfectly safe to forget where the problem came from originally -- in fact, remembering what the matrix "really" describes might only be unnecessary psychological baggage that prevents us from applying all linear-algebraic tools at our disposal. (As an aside, have you ever seen code written by a mathematician? It very often looks exactly like your first example.)
| 2026-01-26T11:30:48.380895
|
2,667,980
|
Why does this innovative method of subtraction from a third grader always work?
|
My daughter is in year $3$ and she is now working on subtraction up to $1000.$ She came up with a way of solving her simple sums that we (her parents) and her teachers can't understand. Here is an example: $61-17$ Instead of borrowing, making it $50+11-17,$ and then doing what she was told in school $11-7=4,$ $50-10=40 \Longrightarrow 40+4=44,$ she does the following: Units of the subtrahend minus units of the minuend $=7-1=6$ Then tens of the minuend minus tens of the subtrahend $=60-10=50$ Finally she subtracts the first result from the second $=50-6=44$ As it is against the first rule children learn in school regarding subtraction (subtrahend minus minuend, as they cannot invert the numbers in subtraction as they can in addition), how is it possible that this method always works? I have a medical background and am baffled with this… Could someone explain it to me please? Her teachers are not keen on accepting this way when it comes to marking her exams.
| 301
|
arithmetic
|
https://math.stackexchange.com/questions/2667980/why-does-this-innovative-method-of-subtraction-from-a-third-grader-always-work
|
So she is doing \begin{align*} 61-17=(60+1)-(10+7)&=(60-10)-(7-1)\\ & = 50-6\\ & =44 \end{align*} She manage to have positive results on each power of ten group up to a multiplication by $\pm 1$ and sums at the end the pieces ; this is kind of smart :) Conclusion : If she is comfortable with this system, let her do...
| 291
| false
|
Q: Why does this innovative method of subtraction from a third grader always work?
My daughter is in year $3$ and she is now working on subtraction up to $1000.$ She came up with a way of solving her simple sums that we (her parents) and her teachers can't understand. Here is an example: $61-17$ Instead of borrowing, making it $50+11-17,$ and then doing what she was told in school $11-7=4,$ $50-10=40 \Longrightarrow 40+4=44,$ she does the following: Units of the subtrahend minus units of the minuend $=7-1=6$ Then tens of the minuend minus tens of the subtrahend $=60-10=50$ Finally she subtracts the first result from the second $=50-6=44$ As it is against the first rule children learn in school regarding subtraction (subtrahend minus minuend, as they cannot invert the numbers in subtraction as they can in addition), how is it possible that this method always works? I have a medical background and am baffled with this… Could someone explain it to me please? Her teachers are not keen on accepting this way when it comes to marking her exams.
A: So she is doing \begin{align*} 61-17=(60+1)-(10+7)&=(60-10)-(7-1)\\ & = 50-6\\ & =44 \end{align*} She manage to have positive results on each power of ten group up to a multiplication by $\pm 1$ and sums at the end the pieces ; this is kind of smart :) Conclusion : If she is comfortable with this system, let her do...
| 2026-01-26T11:30:50.049865
|
8,814
|
Funny identities
|
Here is a funny exercise $$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$ (If you prove it don't publish it here please). Do you have similar examples?
| 300
|
big-list
|
https://math.stackexchange.com/questions/8814/funny-identities
|
$$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$
| 247
| true
|
Q: Funny identities
Here is a funny exercise $$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$ (If you prove it don't publish it here please). Do you have similar examples?
A: $$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$
| 2026-01-26T11:30:51.895448
|
166,869
|
Is '$10$' a magical number or I am missing something?
|
It's a hilarious witty joke that points out how every base is '$10$' in its base. Like, \begin{align} 2 &= 10\ \text{(base 2)} \\ 8 &= 10\ \text{(base 8)} \end{align} My question is if whoever invented the decimal system had chosen $9$ numbers or $11$, or whatever, would this still be applicable? I am confused - Is $10$ a special number which we had chosen several centuries ago or am I missing a point?
| 296
|
notation, number-systems
|
https://math.stackexchange.com/questions/166869/is-10-a-magical-number-or-i-am-missing-something
|
Short answer: your confusion about whether ten is special may come from reading aloud "Every base is base 10" as "Every base is base ten" — this is wrong; not every base is base ten, only base ten is base ten. It is a joke that works better in writing. If you want to read it aloud, you should read it as "Every base is base one-zero". You must distinguish between numbers and representations. A pile of rocks has some number of rocks; this number does not depend on what base you use. A representation is a string of symbols, like "10", and depends on the base. There are "four" rocks in the cartoon, whatever the base may be. (Well, the word "four" may vary with language, but the number is the same.) But the representation of this number "four" may be "4" or "10" or "11" or "100" depending on what base is used. The number "ten" — the number of dots in ".........." — is not mathematically special. In different bases it has different representations: in base ten it is "10", in base six it is "14", etc. The representation "10" (one-zero) is special: whatever your base is, this representation denotes that number. For base $b$, the representation "10" means $1\times b + 0 = b$. When we consider the base ten that we normally use, then "ten" is by definition the base for this particular representation, so it is in that sense "special" for this representation. But this is only an artefact of the base ten representation. If we were using the base six representation, then the representation "10" would correspond to the number six, so six would be special in that sense, for that representation.
| 256
| true
|
Q: Is '$10$' a magical number or I am missing something?
It's a hilarious witty joke that points out how every base is '$10$' in its base. Like, \begin{align} 2 &= 10\ \text{(base 2)} \\ 8 &= 10\ \text{(base 8)} \end{align} My question is if whoever invented the decimal system had chosen $9$ numbers or $11$, or whatever, would this still be applicable? I am confused - Is $10$ a special number which we had chosen several centuries ago or am I missing a point?
A: Short answer: your confusion about whether ten is special may come from reading aloud "Every base is base 10" as "Every base is base ten" — this is wrong; not every base is base ten, only base ten is base ten. It is a joke that works better in writing. If you want to read it aloud, you should read it as "Every base is base one-zero". You must distinguish between numbers and representations. A pile of rocks has some number of rocks; this number does not depend on what base you use. A representation is a string of symbols, like "10", and depends on the base. There are "four" rocks in the cartoon, whatever the base may be. (Well, the word "four" may vary with language, but the number is the same.) But the representation of this number "four" may be "4" or "10" or "11" or "100" depending on what base is used. The number "ten" — the number of dots in ".........." — is not mathematically special. In different bases it has different representations: in base ten it is "10", in base six it is "14", etc. The representation "10" (one-zero) is special: whatever your base is, this representation denotes that number. For base $b$, the representation "10" means $1\times b + 0 = b$. When we consider the base ten that we normally use, then "ten" is by definition the base for this particular representation, so it is in that sense "special" for this representation. But this is only an artefact of the base ten representation. If we were using the base six representation, then the representation "10" would correspond to the number six, so six would be special in that sense, for that representation.
| 2026-01-26T11:30:53.600410
|
384,861
|
Is mathematics one big tautology?
|
Is mathematics one big tautology? Let me put the question in clearer terms: Mathematics is a deductive system; it works by starting with arbitrary axioms, and deriving therefrom "new" properties through the process of deduction. As such, it would seem that we are simply creating a string of equivalences; each property can be traced back logically to the axioms. It must be that way, that's how deductive systems work! If that be the case, then in what sense are we introducing novel or new ideas? It would seem that everything is simply equivalent to the fundamental set of axioms that we choose to start with. Is there a precise step in mathematical derivation that we can isolate as going beyond pure logic? If so, then how does this fit in with the fact that mathematics is deductive? Must we change our view of mathematics as purely deductive? And if not, then is there a way to reconcile the feeling of creativity in mathematics with the fact that it boils down to pure logic? I'm trying to figure out the true nature of what's going on here.
| 295
|
soft-question, philosophy, foundations
|
https://math.stackexchange.com/questions/384861/is-mathematics-one-big-tautology
|
Disclaimer: different people view this differently. I side with Lakatos: Logic is a tool. Proofs are a way to verify one's intuition (and in many cases to improve one's intuition) and it is a tool to check the consistency of theories in a process of refining the axioms. The fact that every proof boils down to a tautology is true but irrelevant to mathematics. Here is an isomorphic question to the question you posed: A painting is just blobs of paint of different colour on canvas. So, are we to deduce from this fact that the art of painting is reduced to just placing paint on canvas? Technically, the answer is yes. But the painter does much more than that. In fact, it is clear that while the painter must possess quite a large amount of skill in placing paint on canvas, this skill is the least relevant (while absolutely necessary) for the creative process of painting. So it is with mathematics. Being able to prove is essential, but is the least relevant skill for doing mathematics. In mathematics we don't deduce things from axioms. Rather we try to capture a certain idea by introducing axioms, check which theorems follow from the axioms and compare these results against the idea we are trying to capture. If the results agree we are happy. If the results disagree, we change the axioms. The ideas we try to capture transcend the deductive system. The deductive system is there to help us find consequences from the axioms, but it does not tell us how to gauge the validity of results against the idea we try to capture, nor how to adjust the axioms. This is my personal point of view of what mathematics is (or at least what a sizable portion of it is). It is very close to what physics is. Physics is not just some theories about matter and its interactions with stuff. Rather it is trying to model reality. So does mathematics, it's just not entirely clear which reality it is trying to model.
| 281
| true
|
Q: Is mathematics one big tautology?
Is mathematics one big tautology? Let me put the question in clearer terms: Mathematics is a deductive system; it works by starting with arbitrary axioms, and deriving therefrom "new" properties through the process of deduction. As such, it would seem that we are simply creating a string of equivalences; each property can be traced back logically to the axioms. It must be that way, that's how deductive systems work! If that be the case, then in what sense are we introducing novel or new ideas? It would seem that everything is simply equivalent to the fundamental set of axioms that we choose to start with. Is there a precise step in mathematical derivation that we can isolate as going beyond pure logic? If so, then how does this fit in with the fact that mathematics is deductive? Must we change our view of mathematics as purely deductive? And if not, then is there a way to reconcile the feeling of creativity in mathematics with the fact that it boils down to pure logic? I'm trying to figure out the true nature of what's going on here.
A: Disclaimer: different people view this differently. I side with Lakatos: Logic is a tool. Proofs are a way to verify one's intuition (and in many cases to improve one's intuition) and it is a tool to check the consistency of theories in a process of refining the axioms. The fact that every proof boils down to a tautology is true but irrelevant to mathematics. Here is an isomorphic question to the question you posed: A painting is just blobs of paint of different colour on canvas. So, are we to deduce from this fact that the art of painting is reduced to just placing paint on canvas? Technically, the answer is yes. But the painter does much more than that. In fact, it is clear that while the painter must possess quite a large amount of skill in placing paint on canvas, this skill is the least relevant (while absolutely necessary) for the creative process of painting. So it is with mathematics. Being able to prove is essential, but is the least relevant skill for doing mathematics. In mathematics we don't deduce things from axioms. Rather we try to capture a certain idea by introducing axioms, check which theorems follow from the axioms and compare these results against the idea we are trying to capture. If the results agree we are happy. If the results disagree, we change the axioms. The ideas we try to capture transcend the deductive system. The deductive system is there to help us find consequences from the axioms, but it does not tell us how to gauge the validity of results against the idea we try to capture, nor how to adjust the axioms. This is my personal point of view of what mathematics is (or at least what a sizable portion of it is). It is very close to what physics is. Physics is not just some theories about matter and its interactions with stuff. Rather it is trying to model reality. So does mathematics, it's just not entirely clear which reality it is trying to model.
| 2026-01-26T11:30:55.248736
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2,072,308
|
Help with a prime number spiral which turns 90 degrees at each prime
|
I awoke with the following puzzle that I would like to investigate, but the answer may require some programming (it may not either). I have asked on the meta site and believe the question to be suitable and hopefully interesting for the community. I will try to explain the puzzle as best I can then detail the questions I am interested in after. Imagine squared paper. In one square write the number $1.$ Continue to write numbers from left to right (as normal) until you reach a prime. The next number after a prime should be written in the square located $90$ degrees clockwise to the last. You then continue writing numbers in that direction. This procedure should be continued indefinitely. Here is a sample of the grid: $$\begin{array}{} 7&8&9&10&11&40&41 \\6&1&2&&12&&42\\5&4&3&14&13&44&43\\&&34&&26\\&&33&&27\\&&32&&28\\&&31&30&29\end{array}$$ Note that the square containing 3 also contains 15 (I couldn't put it in without confusing the diagram. In fact some squares contain multiple entries. I would have liked to see an expanded version of the diagram. I originally thought of shading squares that contain at least one number. Questions Does the square surrounded by $2,3,9,10,11,12,13,14$ ever get shaded? If so, will the whole grid ever be shaded? Is there a maximum number of times a square can be visited? I have got to 4 times but it is easy to make mistakes by hand. Are there any repeated patterns in the gaps? I have other ideas but this is enough for now as I have genuinely no idea how easy or difficult this problem is. Please forgive me for not taking it any further as it is so easy to make mistakes. I hope this is interesting for the community and look forwards to any results. Thanks. Any questions I'll do my best to clarify. Side note: I observed that initially at least the pattern likes to cling to itself but I suspect it doesn't later on.
| 291
|
prime-numbers, visualization, pattern-recognition
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https://math.stackexchange.com/questions/2072308/help-with-a-prime-number-spiral-which-turns-90-degrees-at-each-prime
|
Just for visual amusement, here are more pictures. In all cases, initial point is a large red dot. Primes up to $10^5$: Primes up to $10^6$: Primes up to $10^6$ starting gaps of length $>6$: Primes up to $10^7$ starting gaps of length $>10$: Primes up to $10^8$ starting gaps of length $>60$: For anyone interested, all the images were generated using Sage and variations of the following code: d = 1 p = 0 M = [] prim = prime_range(10^8) diff = [] for i in range(len(prim)-1): diff.append(prim[i+1]-prim[i]) for k in diff: if k>60: M.append(p) d = -d*I p = p+k*d save(list_plot(M,aspect_ratio = 1,axes = false,pointsize = 1,figsize = 20)+point((0,0),color = 'red'),'8s.png')
| 213
| false
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Q: Help with a prime number spiral which turns 90 degrees at each prime
I awoke with the following puzzle that I would like to investigate, but the answer may require some programming (it may not either). I have asked on the meta site and believe the question to be suitable and hopefully interesting for the community. I will try to explain the puzzle as best I can then detail the questions I am interested in after. Imagine squared paper. In one square write the number $1.$ Continue to write numbers from left to right (as normal) until you reach a prime. The next number after a prime should be written in the square located $90$ degrees clockwise to the last. You then continue writing numbers in that direction. This procedure should be continued indefinitely. Here is a sample of the grid: $$\begin{array}{} 7&8&9&10&11&40&41 \\6&1&2&&12&&42\\5&4&3&14&13&44&43\\&&34&&26\\&&33&&27\\&&32&&28\\&&31&30&29\end{array}$$ Note that the square containing 3 also contains 15 (I couldn't put it in without confusing the diagram. In fact some squares contain multiple entries. I would have liked to see an expanded version of the diagram. I originally thought of shading squares that contain at least one number. Questions Does the square surrounded by $2,3,9,10,11,12,13,14$ ever get shaded? If so, will the whole grid ever be shaded? Is there a maximum number of times a square can be visited? I have got to 4 times but it is easy to make mistakes by hand. Are there any repeated patterns in the gaps? I have other ideas but this is enough for now as I have genuinely no idea how easy or difficult this problem is. Please forgive me for not taking it any further as it is so easy to make mistakes. I hope this is interesting for the community and look forwards to any results. Thanks. Any questions I'll do my best to clarify. Side note: I observed that initially at least the pattern likes to cling to itself but I suspect it doesn't later on.
A: Just for visual amusement, here are more pictures. In all cases, initial point is a large red dot. Primes up to $10^5$: Primes up to $10^6$: Primes up to $10^6$ starting gaps of length $>6$: Primes up to $10^7$ starting gaps of length $>10$: Primes up to $10^8$ starting gaps of length $>60$: For anyone interested, all the images were generated using Sage and variations of the following code: d = 1 p = 0 M = [] prim = prime_range(10^8) diff = [] for i in range(len(prim)-1): diff.append(prim[i+1]-prim[i]) for k in diff: if k>60: M.append(p) d = -d*I p = p+k*d save(list_plot(M,aspect_ratio = 1,axes = false,pointsize = 1,figsize = 20)+point((0,0),color = 'red'),'8s.png')
| 2026-01-26T11:30:56.910664
|
38,517
|
In (relatively) simple words: What is an inverse limit?
|
I am a set theorist in my orientation, and while I did take a few courses that brushed upon categorical and algebraic constructions, one has always eluded me. The inverse limit. I tried to ask one of the guys in my office, and despite a very shady explanation he ended up muttering that "you usually take an already known construction." The Wikipedia article presents two approaches, the algebraic and the categorical. While the categorical is extremely vague for me, the algebraic one is too general and the intuition remains hidden underneath the text in a place I cannot find it. Since I am not too familiar with categories, the explanation most people would try to give me which is categorical in nature seems to confuse me - as I keep asking this question over and over every now and then. Could anyone explain to me in non-categorical terms what is the idea behind an inverse limit? (I am roughly familiar with its friend "direct limit", if that helps) (While editing, I can say that the answers given so far are very interesting, and I have read them thoroughly, although I need to give it quite some thinking before I can comment on all of them right now.)
| 290
|
category-theory, intuition, universal-algebra, limits-colimits
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https://math.stackexchange.com/questions/38517/in-relatively-simple-words-what-is-an-inverse-limit
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I like George Bergman's explanation (beginning in section 7.4 of his Invitation to General Algebra and Universal Constructions). We start with a motivating example. Suppose you are interested in solving $x^2=-1$ in $\mathbb{Z}$. Of course, there are no solutions, but let's ignore that annoying reality for a moment. We use the notation $\mathbb{Z}_n$ for $\mathbb Z / n \mathbb Z$. The equation has a solution in the ring $\mathbb{Z}_5$ (in fact, two: both $2$ and $3$, which are the same up to sign). So we want to find a solution to $x^2=-1$ in $\mathbb{Z}$ which satisfies $x\equiv 2 \pmod{5}$. An integer that is congruent to $2$ modulo $5$ is of the form $5y+2$, so we can rewrite our original equation as $(5y+2)^2 = -1$, and expand to get $25y^2 + 20y = -5$. That means $20y\equiv -5\pmod{25}$, or $4y\equiv -1\pmod{5}$, which has the unique solution $y\equiv 1\pmod{5}$. Substituting back we determine $x$ modulo $25$: $$x = 5y+2 \equiv 5\cdot 1 + 2 = 7 \pmod{25}.$$ Continue this way: putting $x=25z+7$ into $x^2=-1$ we conclude $z\equiv 2 \pmod{5}$, so $x\equiv 57\pmod{125}$. Using Hensel's Lemma, we can continue this indefinitely. What we deduce is that there is a sequence of residues, $$x_1\in\mathbb{Z}_5,\quad x_2\in\mathbb{Z}_{25},\quad \ldots, x_{i}\in\mathbb{Z}_{5^i},\ldots$$ each of which satisfies $x^2=-1$ in the appropriate ring, and which are "consistent", in the sense that each $x_{i+1}$ is a lifting of $x_i$ under the natural homomorphisms $$\cdots \stackrel{f_{i+1}}{\longrightarrow} \mathbb{Z}_{5^{i+1}} \stackrel{f_i}{\longrightarrow} \mathbb{Z}_{5^i} \stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow} \mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow} \mathbb{Z}_5.$$ Take the set of all strings $(\ldots,x_i,\ldots,x_2,x_1)$ such that $x_i\in\mathbb{Z}_{5^i}$ and $f_i(x_{i+1}) = x_i$, $i=1,2,\ldots$. This is a ring under componentwise operations. What we did above shows that in this ring, you do have a square root of $-1$. Added. Bergman here inserts the quote, "If the fool will persist in his folly, he will become wise." We obtained the sequence by stubbornly looking for a solution to an equation that has no solution, by looking at putative approximations, first modulo 5, then modulo 25, then modulo 125, etc. We foolishly kept going even though there was no solution to be found. In the end, we get a "full description" of what that object must look like; since we don't have a ready-made object that satisfies this condition, then we simply take this "full description" and use that description as if it were an object itself. By insisting in our folly of looking for a solution, we have become wise by introducing an entirely new object that is a solution. This is much along the lines of taking a Cauchy sequence of rationals, which "describes" a limit point, and using the entire Cauchy sequence to represent this limit point, even if that limit point does not exist in our original set. This ring is the $5$-adic integers; since an integer is completely determined by its remainders modulo the powers of $5$, this ring contains an isomorphic copy of $\mathbb{Z}$. Essentially, we are taking successive approximations to a putative answer to the original equation, by first solving it modulo $5$, then solving it modulo $25$ in a way that is consistent with our solution modulo $5$; then solving it modulo $125$ in a way that is consistent with out solution modulo $25$, etc. The ring of $5$-adic integers projects onto each $\mathbb{Z}_{5^i}$ via the projections; because the elements of the $5$-adic integers are consistent sequences, these projections commute with our original maps $f_i$. So the projections are compatible with the $f_i$ in the sense that for all $i$, $f_i\circ\pi_{i+1} = \pi_{i}$, where $\pi_k$ is the projection onto the $k$th coordinate from the $5$-adics. Moreover, the ring of $5$-adic integers is universal for this property: given any ring $R$ with homomorphisms $r_i\colon R\to\mathbb{Z}_{5^i}$ such that $f_i\circ r_{i+1} = r_i$, for any $a\in R$ the tuple of images $(\ldots, r_i(a),\ldots, r_2(a),r_1(a))$ defines an element in the $5$-adics. The $5$-adics are the inverse limit of the system of maps $$\cdots\stackrel{f_{i+1}}{\longrightarrow}\mathbb{Z}_{5^{i+1}}\stackrel{f_i}{\longrightarrow}\mathbb{Z}_{5^i}\stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow}\mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow}\mathbb{Z}_5.$$ So the elements of the inverse limit are "consistent sequences" of partial approximations, and the inverse limit is a way of taking all these "partial approximations" and combine them into a "target object." More generally, assume that you have a system of, say, rings, $\{R_i\}$, indexed by an directed set $(I,\leq)$ (so that for all $i,j\in I$ there exists $k\in I$ such that $i,j\leq k$), and a system of maps $f_{rs}\colon R_s\to R_r$ whenever $r\leq s$ which are "consistent" (if $r\leq s\leq t$, then $f_{rs}\circ f_{st} = f_{rt}$), and let's assume that the $f_{rs}$ are surjective, as they were in the example of the $5$-adics. Then you can think of the $R_i$ as being "successive approximations" (with a higher indexed $R_i$ as being a "finer" or "better" approximation than the lower indexed one). The directedness of the index set guarantees that given any two approximations, even if they are not directly comparable to one another, you can combine them into an approximation which is finer (better) than each of them (if $i,j$ are incomparable, then find a $k$ with $i,j\leq k$). The inverse limit is a way to combine all of these approximations into an object in a consistent manner. If you imagine your maps as going right to left, you have a branching tree that is getting "thinner" as you move left, and the inverse limit is the combination of all branches occurring "at infinity". Added. The example of the $p$-adic integers may be a bit misleading because our directed set is totally ordered and all maps are surjective. In the more general case, you can think of every chain in the directed set as a "line of approximation"; the directed property ensures that any finite number of "lines of approximation" will meet in "finite time", but you may need to go all the way to "infinity" to really put all the lines of approximation together. The inverse limit takes care of this. If the directed set has no maximal elements, but the structure maps are not surjective, it turns out that no element that is not in the image will matter; essentially, that element never shows up in a net of "successive approximations", so it never forms part of a "consistent system of approximations" (which is what the elements of the inverse limit are).
| 287
| true
|
Q: In (relatively) simple words: What is an inverse limit?
I am a set theorist in my orientation, and while I did take a few courses that brushed upon categorical and algebraic constructions, one has always eluded me. The inverse limit. I tried to ask one of the guys in my office, and despite a very shady explanation he ended up muttering that "you usually take an already known construction." The Wikipedia article presents two approaches, the algebraic and the categorical. While the categorical is extremely vague for me, the algebraic one is too general and the intuition remains hidden underneath the text in a place I cannot find it. Since I am not too familiar with categories, the explanation most people would try to give me which is categorical in nature seems to confuse me - as I keep asking this question over and over every now and then. Could anyone explain to me in non-categorical terms what is the idea behind an inverse limit? (I am roughly familiar with its friend "direct limit", if that helps) (While editing, I can say that the answers given so far are very interesting, and I have read them thoroughly, although I need to give it quite some thinking before I can comment on all of them right now.)
A: I like George Bergman's explanation (beginning in section 7.4 of his Invitation to General Algebra and Universal Constructions). We start with a motivating example. Suppose you are interested in solving $x^2=-1$ in $\mathbb{Z}$. Of course, there are no solutions, but let's ignore that annoying reality for a moment. We use the notation $\mathbb{Z}_n$ for $\mathbb Z / n \mathbb Z$. The equation has a solution in the ring $\mathbb{Z}_5$ (in fact, two: both $2$ and $3$, which are the same up to sign). So we want to find a solution to $x^2=-1$ in $\mathbb{Z}$ which satisfies $x\equiv 2 \pmod{5}$. An integer that is congruent to $2$ modulo $5$ is of the form $5y+2$, so we can rewrite our original equation as $(5y+2)^2 = -1$, and expand to get $25y^2 + 20y = -5$. That means $20y\equiv -5\pmod{25}$, or $4y\equiv -1\pmod{5}$, which has the unique solution $y\equiv 1\pmod{5}$. Substituting back we determine $x$ modulo $25$: $$x = 5y+2 \equiv 5\cdot 1 + 2 = 7 \pmod{25}.$$ Continue this way: putting $x=25z+7$ into $x^2=-1$ we conclude $z\equiv 2 \pmod{5}$, so $x\equiv 57\pmod{125}$. Using Hensel's Lemma, we can continue this indefinitely. What we deduce is that there is a sequence of residues, $$x_1\in\mathbb{Z}_5,\quad x_2\in\mathbb{Z}_{25},\quad \ldots, x_{i}\in\mathbb{Z}_{5^i},\ldots$$ each of which satisfies $x^2=-1$ in the appropriate ring, and which are "consistent", in the sense that each $x_{i+1}$ is a lifting of $x_i$ under the natural homomorphisms $$\cdots \stackrel{f_{i+1}}{\longrightarrow} \mathbb{Z}_{5^{i+1}} \stackrel{f_i}{\longrightarrow} \mathbb{Z}_{5^i} \stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow} \mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow} \mathbb{Z}_5.$$ Take the set of all strings $(\ldots,x_i,\ldots,x_2,x_1)$ such that $x_i\in\mathbb{Z}_{5^i}$ and $f_i(x_{i+1}) = x_i$, $i=1,2,\ldots$. This is a ring under componentwise operations. What we did above shows that in this ring, you do have a square root of $-1$. Added. Bergman here inserts the quote, "If the fool will persist in his folly, he will become wise." We obtained the sequence by stubbornly looking for a solution to an equation that has no solution, by looking at putative approximations, first modulo 5, then modulo 25, then modulo 125, etc. We foolishly kept going even though there was no solution to be found. In the end, we get a "full description" of what that object must look like; since we don't have a ready-made object that satisfies this condition, then we simply take this "full description" and use that description as if it were an object itself. By insisting in our folly of looking for a solution, we have become wise by introducing an entirely new object that is a solution. This is much along the lines of taking a Cauchy sequence of rationals, which "describes" a limit point, and using the entire Cauchy sequence to represent this limit point, even if that limit point does not exist in our original set. This ring is the $5$-adic integers; since an integer is completely determined by its remainders modulo the powers of $5$, this ring contains an isomorphic copy of $\mathbb{Z}$. Essentially, we are taking successive approximations to a putative answer to the original equation, by first solving it modulo $5$, then solving it modulo $25$ in a way that is consistent with our solution modulo $5$; then solving it modulo $125$ in a way that is consistent with out solution modulo $25$, etc. The ring of $5$-adic integers projects onto each $\mathbb{Z}_{5^i}$ via the projections; because the elements of the $5$-adic integers are consistent sequences, these projections commute with our original maps $f_i$. So the projections are compatible with the $f_i$ in the sense that for all $i$, $f_i\circ\pi_{i+1} = \pi_{i}$, where $\pi_k$ is the projection onto the $k$th coordinate from the $5$-adics. Moreover, the ring of $5$-adic integers is universal for this property: given any ring $R$ with homomorphisms $r_i\colon R\to\mathbb{Z}_{5^i}$ such that $f_i\circ r_{i+1} = r_i$, for any $a\in R$ the tuple of images $(\ldots, r_i(a),\ldots, r_2(a),r_1(a))$ defines an element in the $5$-adics. The $5$-adics are the inverse limit of the system of maps $$\cdots\stackrel{f_{i+1}}{\longrightarrow}\mathbb{Z}_{5^{i+1}}\stackrel{f_i}{\longrightarrow}\mathbb{Z}_{5^i}\stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow}\mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow}\mathbb{Z}_5.$$ So the elements of the inverse limit are "consistent sequences" of partial approximations, and the inverse limit is a way of taking all these "partial approximations" and combine them into a "target object." More generally, assume that you have a system of, say, rings, $\{R_i\}$, indexed by an directed set $(I,\leq)$ (so that for all $i,j\in I$ there exists $k\in I$ such that $i,j\leq k$), and a system of maps $f_{rs}\colon R_s\to R_r$ whenever $r\leq s$ which are "consistent" (if $r\leq s\leq t$, then $f_{rs}\circ f_{st} = f_{rt}$), and let's assume that the $f_{rs}$ are surjective, as they were in the example of the $5$-adics. Then you can think of the $R_i$ as being "successive approximations" (with a higher indexed $R_i$ as being a "finer" or "better" approximation than the lower indexed one). The directedness of the index set guarantees that given any two approximations, even if they are not directly comparable to one another, you can combine them into an approximation which is finer (better) than each of them (if $i,j$ are incomparable, then find a $k$ with $i,j\leq k$). The inverse limit is a way to combine all of these approximations into an object in a consistent manner. If you imagine your maps as going right to left, you have a branching tree that is getting "thinner" as you move left, and the inverse limit is the combination of all branches occurring "at infinity". Added. The example of the $p$-adic integers may be a bit misleading because our directed set is totally ordered and all maps are surjective. In the more general case, you can think of every chain in the directed set as a "line of approximation"; the directed property ensures that any finite number of "lines of approximation" will meet in "finite time", but you may need to go all the way to "infinity" to really put all the lines of approximation together. The inverse limit takes care of this. If the directed set has no maximal elements, but the structure maps are not surjective, it turns out that no element that is not in the image will matter; essentially, that element never shows up in a net of "successive approximations", so it never forms part of a "consistent system of approximations" (which is what the elements of the inverse limit are).
| 2026-01-26T11:30:58.754849
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671,076
|
Is $7$ the only prime followed by a cube?
|
I discovered this site which claims that "$7$ is the only prime followed by a cube". I find this statement rather surprising. Is this true? Where might I find a proof that shows this? In my searching, I found this question, which is similar but the answers seem focused on squares next to cubes. Any ideas?
| 289
|
prime-numbers
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https://math.stackexchange.com/questions/671076/is-7-the-only-prime-followed-by-a-cube
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This is certainly true. Suppose $n^3 - 1$ is prime, for some $n$. We get that $n^3-1 = (n-1)(n^2 + n + 1)$ and so we have that $n-1$ divides $n^3 - 1$. If $n-1>1$ then we're done, as we have a contradiction to $n^3 - 1$ being prime.
| 300
| true
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Q: Is $7$ the only prime followed by a cube?
I discovered this site which claims that "$7$ is the only prime followed by a cube". I find this statement rather surprising. Is this true? Where might I find a proof that shows this? In my searching, I found this question, which is similar but the answers seem focused on squares next to cubes. Any ideas?
A: This is certainly true. Suppose $n^3 - 1$ is prime, for some $n$. We get that $n^3-1 = (n-1)(n^2 + n + 1)$ and so we have that $n-1$ divides $n^3 - 1$. If $n-1>1$ then we're done, as we have a contradiction to $n^3 - 1$ being prime.
| 2026-01-26T11:31:00.392767
|
976,462
|
A 1400 years old approximation to the sine function by Mahabhaskariya of Bhaskara I
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The approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. I wondered how much this could be improved using our computers and so I tried (very immodestly) to see if we could do better using $$\sin(x) \simeq \frac{a (\pi -x) x}{5 \pi ^2-b (\pi -x) x}$$ I so computed $$\Phi(a,b)=\int_0^{\pi} \left(\sin (x)-\frac{a (\pi -x) x}{5 \pi ^2-b (\pi -x)x}\right)^2 dx$$ the analytical expression of which not being added to the post. Settings the derivatives equal to $0$ and solving for $a$ and $b$, I arrived to $a=15.9815,b=4.03344$ so close to the original approximation ! What is interesting is to compare the values of $\Phi$ : $2.98 \times 10^{-6}$ only decreased to $2.17 \times 10^{-6}$. Then, no improvement and loss of attractive coefficients. Now, since this is a matter of etiquette on this site, I ask a simple question: with all the tools and machines we have in our hands, could any of our community propose something as simple (or almost) for basic trigonometric functions ? In the discussions, I mentioned one I made (it is probable that I reinvented the wheel) in the same spirit $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$ which is amazing too !
| 287
|
trigonometry, approximation, math-history
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https://math.stackexchange.com/questions/976462/a-1400-years-old-approximation-to-the-sine-function-by-mahabhaskariya-of-bhaskar
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One simple way to derive this is to come up with a parabola approximation. Just getting the roots correct we have $$f(x)=x(\pi-x)$$ Then, we need to scale it (to get the heights correct). And we are gonna do that by dividing by another parabola $p(x)$ $$f(x)=\frac{x(\pi-x)}{p(x)}$$ Let's fix this at three points (thus defining a parabola). Easy rational points would be when $\sin$ is $1/2$ or $1$. So we fix it at $x=\pi/6,\pi/2,5\pi/6$. We want $$f(\pi/6)=f(5\pi/6)=1/2=\frac{5\pi^2/36}{p(\pi/6)}=\frac{5\pi^2/36}{p(5\pi/6)}$$ And we conclude that $p(\pi/6)=p(5\pi/6)=5\pi^2/18$ We do the same at $x=\pi/2$ to conclude that $p(\pi/2)=\pi^2/4$. The only parabola through those points is $$p(x)=\frac{1}{16}(5\pi^2-4x(\pi-x))$$ And thus we have the original approximation. In the spirit of answering the question: This method could be applied for most trig functions on some small symmetric bound.
| 156
| false
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Q: A 1400 years old approximation to the sine function by Mahabhaskariya of Bhaskara I
The approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. I wondered how much this could be improved using our computers and so I tried (very immodestly) to see if we could do better using $$\sin(x) \simeq \frac{a (\pi -x) x}{5 \pi ^2-b (\pi -x) x}$$ I so computed $$\Phi(a,b)=\int_0^{\pi} \left(\sin (x)-\frac{a (\pi -x) x}{5 \pi ^2-b (\pi -x)x}\right)^2 dx$$ the analytical expression of which not being added to the post. Settings the derivatives equal to $0$ and solving for $a$ and $b$, I arrived to $a=15.9815,b=4.03344$ so close to the original approximation ! What is interesting is to compare the values of $\Phi$ : $2.98 \times 10^{-6}$ only decreased to $2.17 \times 10^{-6}$. Then, no improvement and loss of attractive coefficients. Now, since this is a matter of etiquette on this site, I ask a simple question: with all the tools and machines we have in our hands, could any of our community propose something as simple (or almost) for basic trigonometric functions ? In the discussions, I mentioned one I made (it is probable that I reinvented the wheel) in the same spirit $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$ which is amazing too !
A: One simple way to derive this is to come up with a parabola approximation. Just getting the roots correct we have $$f(x)=x(\pi-x)$$ Then, we need to scale it (to get the heights correct). And we are gonna do that by dividing by another parabola $p(x)$ $$f(x)=\frac{x(\pi-x)}{p(x)}$$ Let's fix this at three points (thus defining a parabola). Easy rational points would be when $\sin$ is $1/2$ or $1$. So we fix it at $x=\pi/6,\pi/2,5\pi/6$. We want $$f(\pi/6)=f(5\pi/6)=1/2=\frac{5\pi^2/36}{p(\pi/6)}=\frac{5\pi^2/36}{p(5\pi/6)}$$ And we conclude that $p(\pi/6)=p(5\pi/6)=5\pi^2/18$ We do the same at $x=\pi/2$ to conclude that $p(\pi/2)=\pi^2/4$. The only parabola through those points is $$p(x)=\frac{1}{16}(5\pi^2-4x(\pi-x))$$ And thus we have the original approximation. In the spirit of answering the question: This method could be applied for most trig functions on some small symmetric bound.
| 2026-01-26T11:31:02.132135
|
5,248
|
Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?
|
A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int\limits_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$ Well, can anyone prove this without using Residue theory? I actually thought of using the series representation of $\sin x$: $$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.
| 286
|
calculus, complex-analysis, analysis, definite-integrals, improper-integrals
|
https://math.stackexchange.com/questions/5248/evaluating-the-integral-int-0-infty-frac-sin-x-x-mathrm-dx-frac-pi
|
I believe this can also be solved using double integrals. It is possible (if I remember correctly) to justify switching the order of integration to give the equality: $$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Notice that $$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$ This leads us to $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Now the right hand side can be found easily, using integration by parts. $$\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\ &= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}. \end{align*}$$ Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$ Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$
| 277
| false
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Q: Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?
A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int\limits_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$ Well, can anyone prove this without using Residue theory? I actually thought of using the series representation of $\sin x$: $$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.
A: I believe this can also be solved using double integrals. It is possible (if I remember correctly) to justify switching the order of integration to give the equality: $$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Notice that $$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$ This leads us to $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Now the right hand side can be found easily, using integration by parts. $$\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\ &= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}. \end{align*}$$ Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$ Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$
| 2026-01-26T11:31:03.751545
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860,294
|
How does one prove the determinant inequality $\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$?
|
Reposted on MathOverflow Let $\,A,B,C\in M_{n}(\mathbb C)\,$ be Hermitian and positive definite matrices such that $A+B+C=I_{n}$, where $I_{n}$ is the identity matrix. Show that $$\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n \det \left(A^2+B^2+C^2\right)$$ This problem is a test question from China (xixi). It is said one can use the equation $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$ but I can't use this to prove it. Can you help me?
| 286
|
linear-algebra, matrices, inequality, determinant, hermitian-matrices
|
https://math.stackexchange.com/questions/860294/how-does-one-prove-the-determinant-inequality-det-left6a3b3c3i-n-ri
|
Here is a partial and positive result, valid around the "triple point" $A=B=C= \frac13\mathbb 1$. Let $A,B,C\in M_n(\mathbb C)$ be Hermitian satisfying $A+B+C=\mathbb 1$, and additionally assume that $$\|A-\tfrac13\mathbb 1\|\,,\,\|B-\tfrac13\mathbb 1\|\,,\, \|C-\tfrac13\mathbb 1\|\:\leqslant\:\tfrac16\tag{1}$$ in the spectral or operator norm. (In particular, $A,B,C$ are positive-definite.) Then we have $$6\left(A^3+B^3+C^3\right)+\mathbb 1\:\geqslant\: 5\left(A^2+B^2+C^2\right)\,.\tag{2}$$ Proof: Let $A_0=A-\frac13\mathbb 1$ a.s.o., then $A_0+B_0+C_0=0$, or $\,\sum_\text{cyc}A_0 =0\,$ in notational short form. Consider the Sum of squares $$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}\big(A_0^2 + \tfrac23 A_0+ \tfrac19\mathbb 1\big) \:=\: \sum_\text{cyc}A_0^2 \:+\: \tfrac13\mathbb 1$$ Sum of cubes $$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3 \:=\: \sum_\text{cyc}\big(A_0^3 + 3A_0^2\cdot\tfrac13 + 3A_0\cdot\tfrac1{3^2} + \tfrac1{3^3}\mathbb 1\big) \\ \;=\: \sum_\text{cyc}A_0^3 \:+\: \sum_\text{cyc}A_0^2 \:+\: \tfrac19\mathbb 1$$ to obtain $$6\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3+\mathbb 1 \;-\; 5\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}A_0^2\,(\mathbb 1 + 6A_0) \:\geqslant\: 0$$ where positivity is due to each summand being a product of commuting positive-semidefinite matrices. $\quad\blacktriangle$ Two years later observation: In order to conclude $(2)$ the additional assumptions $(1)$ may be weakened a fair way off to $$\tfrac16\mathbb 1\:\leqslant\: A,B,C\tag{3}$$ or equivalently, assuming the smallest eigenvalue of each matrix $A,B,C\,$ to be at least $\tfrac16$. Proof: Consider the very last summand in the preceding proof. Revert notation from $A_0$ to $A$ and use the same argument, this time based on $(3)$, to obtain $$\sum_\text{cyc}\big(A-\tfrac13\mathbb 1\big)^2\,(6A -\mathbb 1)\:\geqslant\: 0\,.\qquad\qquad\blacktriangle$$
| 12
| false
|
Q: How does one prove the determinant inequality $\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$?
Reposted on MathOverflow Let $\,A,B,C\in M_{n}(\mathbb C)\,$ be Hermitian and positive definite matrices such that $A+B+C=I_{n}$, where $I_{n}$ is the identity matrix. Show that $$\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n \det \left(A^2+B^2+C^2\right)$$ This problem is a test question from China (xixi). It is said one can use the equation $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$ but I can't use this to prove it. Can you help me?
A: Here is a partial and positive result, valid around the "triple point" $A=B=C= \frac13\mathbb 1$. Let $A,B,C\in M_n(\mathbb C)$ be Hermitian satisfying $A+B+C=\mathbb 1$, and additionally assume that $$\|A-\tfrac13\mathbb 1\|\,,\,\|B-\tfrac13\mathbb 1\|\,,\, \|C-\tfrac13\mathbb 1\|\:\leqslant\:\tfrac16\tag{1}$$ in the spectral or operator norm. (In particular, $A,B,C$ are positive-definite.) Then we have $$6\left(A^3+B^3+C^3\right)+\mathbb 1\:\geqslant\: 5\left(A^2+B^2+C^2\right)\,.\tag{2}$$ Proof: Let $A_0=A-\frac13\mathbb 1$ a.s.o., then $A_0+B_0+C_0=0$, or $\,\sum_\text{cyc}A_0 =0\,$ in notational short form. Consider the Sum of squares $$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}\big(A_0^2 + \tfrac23 A_0+ \tfrac19\mathbb 1\big) \:=\: \sum_\text{cyc}A_0^2 \:+\: \tfrac13\mathbb 1$$ Sum of cubes $$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3 \:=\: \sum_\text{cyc}\big(A_0^3 + 3A_0^2\cdot\tfrac13 + 3A_0\cdot\tfrac1{3^2} + \tfrac1{3^3}\mathbb 1\big) \\ \;=\: \sum_\text{cyc}A_0^3 \:+\: \sum_\text{cyc}A_0^2 \:+\: \tfrac19\mathbb 1$$ to obtain $$6\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3+\mathbb 1 \;-\; 5\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}A_0^2\,(\mathbb 1 + 6A_0) \:\geqslant\: 0$$ where positivity is due to each summand being a product of commuting positive-semidefinite matrices. $\quad\blacktriangle$ Two years later observation: In order to conclude $(2)$ the additional assumptions $(1)$ may be weakened a fair way off to $$\tfrac16\mathbb 1\:\leqslant\: A,B,C\tag{3}$$ or equivalently, assuming the smallest eigenvalue of each matrix $A,B,C\,$ to be at least $\tfrac16$. Proof: Consider the very last summand in the preceding proof. Revert notation from $A_0$ to $A$ and use the same argument, this time based on $(3)$, to obtain $$\sum_\text{cyc}\big(A-\tfrac13\mathbb 1\big)^2\,(6A -\mathbb 1)\:\geqslant\: 0\,.\qquad\qquad\blacktriangle$$
| 2026-01-26T11:31:05.511769
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80,246
|
The Mathematics of Tetris
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I am a big fan of the old-school games and I once noticed that there is a sort of parity associated to one and only one Tetris piece, the $\color{purple}{\text{T}}$ piece. This parity is found with no other piece in the game. Background: The Tetris playing field has width $10$. Rotation is allowed, so there are then exactly $7$ unique pieces, each of which is composed of $4$ blocks. For convenience, we can name each piece by a letter. See this Wikipedia page for the Image ($\color{cyan}{\text{I}}$ is for the stick piece, $\color{goldenrod}{\text{O}}$ for the square, and $\color{green}{\text{S}},\color{purple}{\text{T}},\color{red}{\text{Z}},\color{orange}{\text{L}},\color{blue}{\text{J}}$ are the others) There are $2$ sets of $2$ pieces which are mirrors of each other, namely $\color{orange}{\text{L}}, \color{blue}{\text{J}}$ and $\color{green}{\text{S}},\color{red}{\text{Z}}$ whereas the other three are symmetric $\color{cyan}{\text{I}},\color{goldenrod}{\text{O}}, \color{purple}{\text{T}}$ Language: If a row is completely full, that row disappears. We call it a perfect clear if no blocks remain in the playing field. Since the blocks are size 4, and the playing field has width $10$, the number of blocks for a perfect clear must always be a multiple of $5$. My Question: I noticed while playing that the $\color{purple}{\text{T}}$ piece is particularly special. It seems that it has some sort of parity which no other piece has. Specifically: Conjecture: If we have played some number of pieces, and we have a perfect clear, then the number of $\color{purple}{\text{T}}$ pieces used must be even. Moreover, the $\color{purple}{\text{T}}$ piece is the only piece with this property. I have verified the second part; all of the other pieces can give a perfect clear with either an odd or an even number used. However, I am not sure how to prove the first part. I think that assigning some kind of invariant to the pieces must be the right way to go, but I am not sure. Thank you,
| 284
|
recreational-mathematics, problem-solving, packing-problem, parity, invariance
|
https://math.stackexchange.com/questions/80246/the-mathematics-of-tetris
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My colleague, Ido Segev, pointed out that there is a problem with most of the elegant proofs here - Tetris is not just a problem of tiling a rectangle. Below is his proof that the conjecture is, in fact, false.
| 206
| true
|
Q: The Mathematics of Tetris
I am a big fan of the old-school games and I once noticed that there is a sort of parity associated to one and only one Tetris piece, the $\color{purple}{\text{T}}$ piece. This parity is found with no other piece in the game. Background: The Tetris playing field has width $10$. Rotation is allowed, so there are then exactly $7$ unique pieces, each of which is composed of $4$ blocks. For convenience, we can name each piece by a letter. See this Wikipedia page for the Image ($\color{cyan}{\text{I}}$ is for the stick piece, $\color{goldenrod}{\text{O}}$ for the square, and $\color{green}{\text{S}},\color{purple}{\text{T}},\color{red}{\text{Z}},\color{orange}{\text{L}},\color{blue}{\text{J}}$ are the others) There are $2$ sets of $2$ pieces which are mirrors of each other, namely $\color{orange}{\text{L}}, \color{blue}{\text{J}}$ and $\color{green}{\text{S}},\color{red}{\text{Z}}$ whereas the other three are symmetric $\color{cyan}{\text{I}},\color{goldenrod}{\text{O}}, \color{purple}{\text{T}}$ Language: If a row is completely full, that row disappears. We call it a perfect clear if no blocks remain in the playing field. Since the blocks are size 4, and the playing field has width $10$, the number of blocks for a perfect clear must always be a multiple of $5$. My Question: I noticed while playing that the $\color{purple}{\text{T}}$ piece is particularly special. It seems that it has some sort of parity which no other piece has. Specifically: Conjecture: If we have played some number of pieces, and we have a perfect clear, then the number of $\color{purple}{\text{T}}$ pieces used must be even. Moreover, the $\color{purple}{\text{T}}$ piece is the only piece with this property. I have verified the second part; all of the other pieces can give a perfect clear with either an odd or an even number used. However, I am not sure how to prove the first part. I think that assigning some kind of invariant to the pieces must be the right way to go, but I am not sure. Thank you,
A: My colleague, Ido Segev, pointed out that there is a problem with most of the elegant proofs here - Tetris is not just a problem of tiling a rectangle. Below is his proof that the conjecture is, in fact, false.
| 2026-01-26T11:31:07.154301
|
364,452
|
Evaluate $\int_{0}^{\frac{\pi}2}\frac1{(1+x^2)(1+\tan x)}\,\Bbb dx$
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Evaluate the following integral $$ \tag1\int_{0}^{\frac{\pi}{2}}\frac1{(1+x^2)(1+\tan x)}\,\Bbb dx $$ My Attempt: Letting $x=\frac{\pi}{2}-x$ and using the property that $$ \int_{0}^{a}f(x)\,\Bbb dx = \int_{0}^{a}f(a-x)\,\Bbb dx $$ we obtain $$ \tag2\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{\left(1+\left(\frac{\pi}{2}-x\right)^2\right)(1+\tan x)}\,\Bbb dx $$ Now, add equation $(1)$ and $(2)$. After that I do not understand how I can proceed further.
| 284
|
calculus, real-analysis, integration, definite-integrals, improper-integrals
|
https://math.stackexchange.com/questions/364452/evaluate-int-0-frac-pi2-frac11x21-tan-x-bbb-dx
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Here is an approach. We give some preliminary results. The poly-Hurwitz zeta function The poly-Hurwitz zeta function may initially be defined by the series $$ \begin{align} \displaystyle \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re a >-1, \, \Re b >-1, \, \Re s>0. \tag1 \end{align} $$ This special function is a natural extension of the Hurwitz zeta function initially defined as $$ \zeta(s,a)=\sum_{n=0}^{\infty} \frac{1}{(n+a)^s}, \quad \Re a>0, \Re s>1, \tag2 $$ which is a natural extension itself of the Riemann zeta function initially defined as $$ \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}, \quad \Re s>1. \tag3 $$ The poly-Hurwitz function appears in different places with different notations, one may find it here: [Masri, p. 2 and p. 15 (2004)], [Murty, p. 17 (2006)], [Sinha, p. 45 (2002)]. In this answer we are dealing with a simplified version of a general poly-Hurwitz function. The series in $(1)$ converges absolutely for $\displaystyle \Re s>0$. Moreover, the convergence of the series is uniform on every half-plane $$\displaystyle H_{\delta}=\left\{s \in \mathbb{C}, \Re s \geq \delta \right\}, \, \delta \in \mathbb{R},\, \delta>0,$$ therefore the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot \mid a,b)$ is analytic on the half-plane $\displaystyle \Re s>0$. Let $a$, $b$ and $s$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re s >0$. One may observe that $$ \begin{align} \zeta(s\mid a,b) & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}\\ & = \sum_{n=1}^{+\infty} \frac{(n+b)+(a-b)}{(n+a)^{s+1}(n+b)}\\ & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}}+(a-b)\sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}(n+b)} \tag4 \end{align} $$ giving the functional identity $$ \begin{align} \zeta(s \mid a,b) = \zeta(s+1,a+1) +(a-b) \zeta(s+1 \mid a,b) \tag5 \end{align} $$ where $\displaystyle \zeta(\cdot,\cdot)$ is the standard Hurwitz zeta function. From $(5)$, we obtain by induction, for $n=1,2,3,\ldots $, $$ \begin{align} \zeta(s\mid a,b) = \sum_{k=1}^{n}(a-b)^{k-1}\zeta(s+k,a+1) +(a-b)^n\zeta(s+n \mid a,b). \tag6 \end{align} $$ We use $(6)$ to extend $\displaystyle \zeta(\cdot \mid a,b)$ to a meromorphic function on each open set $\Re s>-n $, $n\geq 1$. Since the Hurwitz zeta function is analytic on the whole complex plane except for a simple pole at $1$ with residue $1$, then from $(6)$ the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot\mid a,b)$ is analytic on the whole complex plane except for a simple pole at $0$ with residue $1$. The poly-Stieltjes constants In 1885 Stieltjes has found that the Laurent series expansion around $1$ of the Riemann zeta function $$ \zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag7 $$ is such that the coefficients of the regular part of the expansion are given by $$ \begin{align} \gamma_k& = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right). \end{align} \tag8 $$ Euler was the first to define a constant of this form (1734) $$ \begin{align} \gamma & = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots. \end{align} $$ The constants $\displaystyle \gamma_k$ are called the Stieltjes constants and due to the fact that $\displaystyle \gamma_0=\gamma$ they are sometimes called the generalized Euler's constants. Similarly, Wilton (1927) and Berndt (1972) established that the Laurent series expansion in the neighbourhood of $1$ of the Hurwitz zeta function $$ \begin{align} \zeta(1+s,a) = \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a>0, \,s\neq 0, \tag{9} \end{align} $$ is such that the coefficients of the regular part of the expansion are given by $$ \begin{align} \gamma_k(a)& = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a>0, \end{align} \tag{10} $$ with $\displaystyle \gamma_{0}(a)=-\psi(a)=-\Gamma'(a)/\Gamma(a)$. The coefficients $\gamma_k(a)$ are called the generalized Stieltjes constants. We have seen from $(6)$ that the poly-Hurwitz zeta function admits a Laurent series expansion around $0$. Let's denote by $\displaystyle\gamma_k(a,b)$ the coefficients of the regular part of $\displaystyle \zeta(\cdot\mid a,b)$ around $0$. I will call these coefficients the poly-Stieltjes constants. Do we have an analog of $(10)$ for $\displaystyle\gamma_k(a,b)$? The following result is new. Theorem 1. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. Consider The poly-Hurwitz zeta function $$ \begin{align} \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re s>0. \tag{11} \end{align} $$ Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$, $$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag{12} $$ where the poly-Stieltjes constants $\displaystyle \gamma_k(a,b)$ are given by $$ \begin{align} \gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) \end{align} \tag{13} $$ with $$ \gamma_{0}(a,b)=-\psi(b+1)=-\Gamma'(b+1)/\Gamma(b+1). \tag{14}$$ Proof. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. We first assume $\Re s>0$. Observing that, for each $n \geq 1$, $$ \left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}Abram. & Steg. p. 258 6.3.16): $$ \begin{align} -\psi(b+1) &= \gamma - \sum_{n=1}^{\infty} \left( \frac1n - \frac1{b+n} \right) \\ &=\lim_{N\to+\infty}\left(\gamma - \sum_{n=1}^N\left( \frac1n - \frac1{b+n} \right)\right)\\ &=\lim_{N\to+\infty}\left(\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)-\left(\sum_{n=1}^N\frac1n-\ln N-\gamma \right)\right)\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)\\\\ &=\gamma_0(a,b) \end{align} $$ using $(13)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$ One of the consequences of Theorem 1 is the new possibility to express some series in terms of the poly-Stieltjes constants. Theorem 2. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$. Then $$ \begin{align} (b-a)\sum_{n=1}^{+\infty} \frac{\log (n+c)}{(n+a)(n+b)}=\gamma_1(c,a)-\gamma_1(c,b), \tag{15} \end{align} $$ similarly $$ \begin{align} \sum_{n=1}^{+\infty} \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)=\gamma_1(a,b)-\gamma_1(c,b), \tag{16} \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$ Proof. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$. We have $$ (b-a)\frac{\log (n+c)}{(n+a)(n+b)}=\frac{\log (n+c)}{n+a}-\frac{\log (n+c)}{n+b} $$ giving, for $N\geq1$, $$ \begin{align} (b-a)&\sum_{n=1}^N \frac{\log (n+c)}{(n+a)(n+b)}=\\\\ & \left(\sum_{n=1}^N\frac{\log (n+c)}{n+a}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{17} \end{align} $$ letting $N \to \infty$ and using $(13)$ gives $(15)$. We have, for $N\geq1$, $$ \begin{align} &\sum_{n=1}^N \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)\\\\ &= \left(\sum_{n=1}^N\frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{18} \end{align} $$ letting $N \to \infty$ and using $(13)$ gives $(16)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$ Juantheron's integral Let’s first give a numerical evaluation of Juantheron’s integral. I would like to thank Jonathan Borwein and David H. Bailey who obtained the result below to $1000$ digits, in just 3.9 seconds run time, using David's new MPFUN-MPFR software, along with the tanh-sinh quadrature program included with the MPFUN-MPFR package. They also tried the integral with David's MPFUN-Fort package, which has a completely different underlying multiprecision system, and they obtained the same result below. Finally, they computed the integral with Mathematica $11.0$; it agreed with the result below, although it required about 10 times longer to run. Proposition 1. We have $$ \begin{align} \int_{0}^{\Large\frac{\pi}2}\!&\frac1{(1+x^2)(1+\tan x)}\mathrm dx\\\\= 0.&59738180945180348461311323509087376430643859042555\\ &67307703207161550311033249824121789098990404474443\\ &73300942847961727020952797366230453350097928752529\\ &62099371263365268445580755896768905606293308536674\\ &89639352215352393870280616186538538722285601087082\\ &81730013060929540132583577240799018025603130403772\\ &83596189879605956759516344861849456740112012597646\\ &30195536341071109827787231788650530475635336662512\\ &50757672078320586388500276160658476344052492489409\\ &64026178233152015087197531148322444147655936720008 \tag{19}\\ &40650450631581050321100329502169853063154902765446\\ &58804861176982696627707544105655815406116180984371\\ &54148587721902800400109013880620460529382772599713\\ &06874977209651994186527207589425408866256042399213\\ &80515694164361264997143539392018681691584285790381\\ &65536517701019826846772718498479534803417547866296\\ &23842162877309354675086691711521468623807334908897\\ &71491673168051054009130049879837629516862688171756\\ &13790927986073268994254629238035029442300668334396\\ &901581838911515359223628586133156893962372426055\cdots \end{align} $$ David H. Bailey confirmed that Mathematica $11.0$, in spite of the great numerical precision, could not find a closed-form of the integral. The next result proves that the OP integral admits a closed form in terms of the poly-Stieltjes constants. Proposition 2. We have $$ \begin{align} \int_{0}^{\Large\frac{\pi}2}\!\!\frac1{(1+x^2)(1+\tan x)}\mathrm dx &=\frac{(e^2+1)^2}{2(e^4+1)}\arctan \! \frac{\pi}{2}-\frac{e^4-1}{4(e^4+1)}\log\left(1+\frac{\pi^2}{4}\right)\\\\ &+\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{20}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right) \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$ Proof. We proceed in three steps. Step 1. One may write $$ \require{cancel} \begin{align} &\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\mathrm dx \\ &=\int_{0}^{\Large\frac{\pi}{2}}\frac{\cos x}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\ &=\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{(\cos x+\sin x)+(\cos x-\sin x)}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\ &=\frac12\int_{0}^{\Large\frac{\pi}{2}}\!\frac{1}{1+x^2}\mathrm dx+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)}\frac{(\cos x-\sin x)}{(\cos x+\sin x)}\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{1+x^2}\tan (x-\pi/4)\:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{-\Large\frac{\pi}{4}}^{\Large\frac{\pi}4}\frac{1}{1+(x+\pi/4)^2}\tan x \:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_0^{\Large\frac{\pi}4}\left(\frac1{1+(x+\pi/4)^2}-\frac1{1+(x-\pi/4)^2}\right)\tan x \:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}-\frac{\Im}2 \!\int_{0}^{\Large\frac{\pi}{4}}\!\!\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx \tag{21} \end{align} $$ Let’s evaluate the latter integral. Step 2. One may recall that the tangent function, as a meromorphic function, can be expressed as an infinite sum of rational functions: $$ \tan x = \sum_{n=0}^{+\infty} \frac{2x}{\pi^2 (n+1/2)^2-x^2}, \quad x \neq \pm \pi/2, \pm 3\pi/2,\pm 5\pi/2,\ldots. \tag{22} $$ We have the inequality $$ \sup_{x \in [0,\pi/4]}\left|\frac{2x}{\pi^2 (n+1/2)^2-x^2}\right|\leq \frac1{(n+1/2)^2}, \quad n=0,1,2,\ldots, \tag{23} $$ the convergence in $(22)$ is then uniform on $[0,\pi/4]$. Thus, plugging $(22)$ into $(21)$, we are allowed to integrate $(21)$ termwise. Each term, via a partial fraction decomposition, is evaluated to obtain $$ \begin{align} \int_{0}^{\Large\frac{\pi}4}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right)\frac{2x}{\pi^2 (n+1/2)^2-x^2}\:\mathrm dx\\ &=\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\ &+\frac1{\pi}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\ &+\frac1{\pi}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right) \end{align} \tag{24} $$ where for the sake of convenience we have set $\tau:=\pi/4+i$. Step 3. We sum $(24)$ from $n=0$ to $\infty$ obtaining $$ \begin{align} \int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\ &=\sum_{n=0}^{\infty}\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\ &+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\ &+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right), \tag{25} \end{align} $$ then, singling out the first terms in the two last series and using Theorem $2$ $(16)$, we get $$ \begin{align} \int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\ &=\tan \tau \log \left( \frac{4\tau-\pi}{4\tau+\pi}\right) +\frac{4\pi}{\pi^2 -4\tau^2}\log 3 \\&+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 +\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 +\frac{\tau}{\pi}\!\right)\tag{26}\\ &+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 -\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 -\frac{\tau}{\pi}\!\right) \end{align} $$ and the substitution $\tau=\pi/4+i$ gives the desired result. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$ Achille Hui's conjecture is true. Achille Hui has announced in the comments that the OP integral is equal to $$ \begin{align} \frac{\arctan(\frac{\pi}{2}) - t\log\sqrt{1+\frac{\pi^2}{4}}}{1+t^2}+\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } \tag{27} \end{align} $$ with $\displaystyle t := \tanh(1)$. The first term in $(27)$, with a little algebra is seen to be equal to the sum of the first two terms on the right hand side of $(20)$. We establish the veracity of the conjecture using Proposition $2$ and using the next result. Proposition 3. We have $$ \begin{align} &\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }\\\\ &=\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{28}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right) \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).$$ Proof. Observe that the first term of the series on the left hand side of $(28)$, given by $n=0$, is just equal to $$ \frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)}. $$ By a partial fraction decomposition, one may check that $$ \begin{align} \frac{\pi^2}4 &\frac{ (2n+1) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } =\frac{\Im}{2\pi}\left(\!\frac1{n+\frac34+\frac{i}{\pi}}-\frac1{n+\frac14+\frac{i}{\pi}}\!\right) \tag{29} \end{align} $$ then, multiplying $(29)$ by $\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right)\right)$ and summing from $n=1$ to $\infty$ we get, using Theorem $2$ $(16)$, the result $(28)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$
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Q: Evaluate $\int_{0}^{\frac{\pi}2}\frac1{(1+x^2)(1+\tan x)}\,\Bbb dx$
Evaluate the following integral $$ \tag1\int_{0}^{\frac{\pi}{2}}\frac1{(1+x^2)(1+\tan x)}\,\Bbb dx $$ My Attempt: Letting $x=\frac{\pi}{2}-x$ and using the property that $$ \int_{0}^{a}f(x)\,\Bbb dx = \int_{0}^{a}f(a-x)\,\Bbb dx $$ we obtain $$ \tag2\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{\left(1+\left(\frac{\pi}{2}-x\right)^2\right)(1+\tan x)}\,\Bbb dx $$ Now, add equation $(1)$ and $(2)$. After that I do not understand how I can proceed further.
A: Here is an approach. We give some preliminary results. The poly-Hurwitz zeta function The poly-Hurwitz zeta function may initially be defined by the series $$ \begin{align} \displaystyle \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re a >-1, \, \Re b >-1, \, \Re s>0. \tag1 \end{align} $$ This special function is a natural extension of the Hurwitz zeta function initially defined as $$ \zeta(s,a)=\sum_{n=0}^{\infty} \frac{1}{(n+a)^s}, \quad \Re a>0, \Re s>1, \tag2 $$ which is a natural extension itself of the Riemann zeta function initially defined as $$ \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}, \quad \Re s>1. \tag3 $$ The poly-Hurwitz function appears in different places with different notations, one may find it here: [Masri, p. 2 and p. 15 (2004)], [Murty, p. 17 (2006)], [Sinha, p. 45 (2002)]. In this answer we are dealing with a simplified version of a general poly-Hurwitz function. The series in $(1)$ converges absolutely for $\displaystyle \Re s>0$. Moreover, the convergence of the series is uniform on every half-plane $$\displaystyle H_{\delta}=\left\{s \in \mathbb{C}, \Re s \geq \delta \right\}, \, \delta \in \mathbb{R},\, \delta>0,$$ therefore the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot \mid a,b)$ is analytic on the half-plane $\displaystyle \Re s>0$. Let $a$, $b$ and $s$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re s >0$. One may observe that $$ \begin{align} \zeta(s\mid a,b) & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}\\ & = \sum_{n=1}^{+\infty} \frac{(n+b)+(a-b)}{(n+a)^{s+1}(n+b)}\\ & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}}+(a-b)\sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}(n+b)} \tag4 \end{align} $$ giving the functional identity $$ \begin{align} \zeta(s \mid a,b) = \zeta(s+1,a+1) +(a-b) \zeta(s+1 \mid a,b) \tag5 \end{align} $$ where $\displaystyle \zeta(\cdot,\cdot)$ is the standard Hurwitz zeta function. From $(5)$, we obtain by induction, for $n=1,2,3,\ldots $, $$ \begin{align} \zeta(s\mid a,b) = \sum_{k=1}^{n}(a-b)^{k-1}\zeta(s+k,a+1) +(a-b)^n\zeta(s+n \mid a,b). \tag6 \end{align} $$ We use $(6)$ to extend $\displaystyle \zeta(\cdot \mid a,b)$ to a meromorphic function on each open set $\Re s>-n $, $n\geq 1$. Since the Hurwitz zeta function is analytic on the whole complex plane except for a simple pole at $1$ with residue $1$, then from $(6)$ the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot\mid a,b)$ is analytic on the whole complex plane except for a simple pole at $0$ with residue $1$. The poly-Stieltjes constants In 1885 Stieltjes has found that the Laurent series expansion around $1$ of the Riemann zeta function $$ \zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag7 $$ is such that the coefficients of the regular part of the expansion are given by $$ \begin{align} \gamma_k& = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right). \end{align} \tag8 $$ Euler was the first to define a constant of this form (1734) $$ \begin{align} \gamma & = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots. \end{align} $$ The constants $\displaystyle \gamma_k$ are called the Stieltjes constants and due to the fact that $\displaystyle \gamma_0=\gamma$ they are sometimes called the generalized Euler's constants. Similarly, Wilton (1927) and Berndt (1972) established that the Laurent series expansion in the neighbourhood of $1$ of the Hurwitz zeta function $$ \begin{align} \zeta(1+s,a) = \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a>0, \,s\neq 0, \tag{9} \end{align} $$ is such that the coefficients of the regular part of the expansion are given by $$ \begin{align} \gamma_k(a)& = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a>0, \end{align} \tag{10} $$ with $\displaystyle \gamma_{0}(a)=-\psi(a)=-\Gamma'(a)/\Gamma(a)$. The coefficients $\gamma_k(a)$ are called the generalized Stieltjes constants. We have seen from $(6)$ that the poly-Hurwitz zeta function admits a Laurent series expansion around $0$. Let's denote by $\displaystyle\gamma_k(a,b)$ the coefficients of the regular part of $\displaystyle \zeta(\cdot\mid a,b)$ around $0$. I will call these coefficients the poly-Stieltjes constants. Do we have an analog of $(10)$ for $\displaystyle\gamma_k(a,b)$? The following result is new. Theorem 1. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. Consider The poly-Hurwitz zeta function $$ \begin{align} \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re s>0. \tag{11} \end{align} $$ Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$, $$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag{12} $$ where the poly-Stieltjes constants $\displaystyle \gamma_k(a,b)$ are given by $$ \begin{align} \gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) \end{align} \tag{13} $$ with $$ \gamma_{0}(a,b)=-\psi(b+1)=-\Gamma'(b+1)/\Gamma(b+1). \tag{14}$$ Proof. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. We first assume $\Re s>0$. Observing that, for each $n \geq 1$, $$ \left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}Abram. & Steg. p. 258 6.3.16): $$ \begin{align} -\psi(b+1) &= \gamma - \sum_{n=1}^{\infty} \left( \frac1n - \frac1{b+n} \right) \\ &=\lim_{N\to+\infty}\left(\gamma - \sum_{n=1}^N\left( \frac1n - \frac1{b+n} \right)\right)\\ &=\lim_{N\to+\infty}\left(\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)-\left(\sum_{n=1}^N\frac1n-\ln N-\gamma \right)\right)\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)\\\\ &=\gamma_0(a,b) \end{align} $$ using $(13)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$ One of the consequences of Theorem 1 is the new possibility to express some series in terms of the poly-Stieltjes constants. Theorem 2. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$. Then $$ \begin{align} (b-a)\sum_{n=1}^{+\infty} \frac{\log (n+c)}{(n+a)(n+b)}=\gamma_1(c,a)-\gamma_1(c,b), \tag{15} \end{align} $$ similarly $$ \begin{align} \sum_{n=1}^{+\infty} \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)=\gamma_1(a,b)-\gamma_1(c,b), \tag{16} \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$ Proof. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$. We have $$ (b-a)\frac{\log (n+c)}{(n+a)(n+b)}=\frac{\log (n+c)}{n+a}-\frac{\log (n+c)}{n+b} $$ giving, for $N\geq1$, $$ \begin{align} (b-a)&\sum_{n=1}^N \frac{\log (n+c)}{(n+a)(n+b)}=\\\\ & \left(\sum_{n=1}^N\frac{\log (n+c)}{n+a}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{17} \end{align} $$ letting $N \to \infty$ and using $(13)$ gives $(15)$. We have, for $N\geq1$, $$ \begin{align} &\sum_{n=1}^N \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)\\\\ &= \left(\sum_{n=1}^N\frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{18} \end{align} $$ letting $N \to \infty$ and using $(13)$ gives $(16)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$ Juantheron's integral Let’s first give a numerical evaluation of Juantheron’s integral. I would like to thank Jonathan Borwein and David H. Bailey who obtained the result below to $1000$ digits, in just 3.9 seconds run time, using David's new MPFUN-MPFR software, along with the tanh-sinh quadrature program included with the MPFUN-MPFR package. They also tried the integral with David's MPFUN-Fort package, which has a completely different underlying multiprecision system, and they obtained the same result below. Finally, they computed the integral with Mathematica $11.0$; it agreed with the result below, although it required about 10 times longer to run. Proposition 1. We have $$ \begin{align} \int_{0}^{\Large\frac{\pi}2}\!&\frac1{(1+x^2)(1+\tan x)}\mathrm dx\\\\= 0.&59738180945180348461311323509087376430643859042555\\ &67307703207161550311033249824121789098990404474443\\ &73300942847961727020952797366230453350097928752529\\ &62099371263365268445580755896768905606293308536674\\ &89639352215352393870280616186538538722285601087082\\ &81730013060929540132583577240799018025603130403772\\ &83596189879605956759516344861849456740112012597646\\ &30195536341071109827787231788650530475635336662512\\ &50757672078320586388500276160658476344052492489409\\ &64026178233152015087197531148322444147655936720008 \tag{19}\\ &40650450631581050321100329502169853063154902765446\\ &58804861176982696627707544105655815406116180984371\\ &54148587721902800400109013880620460529382772599713\\ &06874977209651994186527207589425408866256042399213\\ &80515694164361264997143539392018681691584285790381\\ &65536517701019826846772718498479534803417547866296\\ &23842162877309354675086691711521468623807334908897\\ &71491673168051054009130049879837629516862688171756\\ &13790927986073268994254629238035029442300668334396\\ &901581838911515359223628586133156893962372426055\cdots \end{align} $$ David H. Bailey confirmed that Mathematica $11.0$, in spite of the great numerical precision, could not find a closed-form of the integral. The next result proves that the OP integral admits a closed form in terms of the poly-Stieltjes constants. Proposition 2. We have $$ \begin{align} \int_{0}^{\Large\frac{\pi}2}\!\!\frac1{(1+x^2)(1+\tan x)}\mathrm dx &=\frac{(e^2+1)^2}{2(e^4+1)}\arctan \! \frac{\pi}{2}-\frac{e^4-1}{4(e^4+1)}\log\left(1+\frac{\pi^2}{4}\right)\\\\ &+\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{20}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right) \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$ Proof. We proceed in three steps. Step 1. One may write $$ \require{cancel} \begin{align} &\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\mathrm dx \\ &=\int_{0}^{\Large\frac{\pi}{2}}\frac{\cos x}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\ &=\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{(\cos x+\sin x)+(\cos x-\sin x)}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\ &=\frac12\int_{0}^{\Large\frac{\pi}{2}}\!\frac{1}{1+x^2}\mathrm dx+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)}\frac{(\cos x-\sin x)}{(\cos x+\sin x)}\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{1+x^2}\tan (x-\pi/4)\:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{-\Large\frac{\pi}{4}}^{\Large\frac{\pi}4}\frac{1}{1+(x+\pi/4)^2}\tan x \:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_0^{\Large\frac{\pi}4}\left(\frac1{1+(x+\pi/4)^2}-\frac1{1+(x-\pi/4)^2}\right)\tan x \:\mathrm dx\\ &=\frac12 \arctan\! \frac{\pi}{2}-\frac{\Im}2 \!\int_{0}^{\Large\frac{\pi}{4}}\!\!\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx \tag{21} \end{align} $$ Let’s evaluate the latter integral. Step 2. One may recall that the tangent function, as a meromorphic function, can be expressed as an infinite sum of rational functions: $$ \tan x = \sum_{n=0}^{+\infty} \frac{2x}{\pi^2 (n+1/2)^2-x^2}, \quad x \neq \pm \pi/2, \pm 3\pi/2,\pm 5\pi/2,\ldots. \tag{22} $$ We have the inequality $$ \sup_{x \in [0,\pi/4]}\left|\frac{2x}{\pi^2 (n+1/2)^2-x^2}\right|\leq \frac1{(n+1/2)^2}, \quad n=0,1,2,\ldots, \tag{23} $$ the convergence in $(22)$ is then uniform on $[0,\pi/4]$. Thus, plugging $(22)$ into $(21)$, we are allowed to integrate $(21)$ termwise. Each term, via a partial fraction decomposition, is evaluated to obtain $$ \begin{align} \int_{0}^{\Large\frac{\pi}4}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right)\frac{2x}{\pi^2 (n+1/2)^2-x^2}\:\mathrm dx\\ &=\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\ &+\frac1{\pi}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\ &+\frac1{\pi}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right) \end{align} \tag{24} $$ where for the sake of convenience we have set $\tau:=\pi/4+i$. Step 3. We sum $(24)$ from $n=0$ to $\infty$ obtaining $$ \begin{align} \int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\ &=\sum_{n=0}^{\infty}\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\ &+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\ &+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right), \tag{25} \end{align} $$ then, singling out the first terms in the two last series and using Theorem $2$ $(16)$, we get $$ \begin{align} \int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\ &=\tan \tau \log \left( \frac{4\tau-\pi}{4\tau+\pi}\right) +\frac{4\pi}{\pi^2 -4\tau^2}\log 3 \\&+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 +\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 +\frac{\tau}{\pi}\!\right)\tag{26}\\ &+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 -\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 -\frac{\tau}{\pi}\!\right) \end{align} $$ and the substitution $\tau=\pi/4+i$ gives the desired result. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$ Achille Hui's conjecture is true. Achille Hui has announced in the comments that the OP integral is equal to $$ \begin{align} \frac{\arctan(\frac{\pi}{2}) - t\log\sqrt{1+\frac{\pi^2}{4}}}{1+t^2}+\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } \tag{27} \end{align} $$ with $\displaystyle t := \tanh(1)$. The first term in $(27)$, with a little algebra is seen to be equal to the sum of the first two terms on the right hand side of $(20)$. We establish the veracity of the conjecture using Proposition $2$ and using the next result. Proposition 3. We have $$ \begin{align} &\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }\\\\ &=\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{28}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right) \end{align} $$ with the poly-Stieltjes constant $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).$$ Proof. Observe that the first term of the series on the left hand side of $(28)$, given by $n=0$, is just equal to $$ \frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)}. $$ By a partial fraction decomposition, one may check that $$ \begin{align} \frac{\pi^2}4 &\frac{ (2n+1) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } =\frac{\Im}{2\pi}\left(\!\frac1{n+\frac34+\frac{i}{\pi}}-\frac1{n+\frac14+\frac{i}{\pi}}\!\right) \tag{29} \end{align} $$ then, multiplying $(29)$ by $\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right)\right)$ and summing from $n=1$ to $\infty$ we get, using Theorem $2$ $(16)$, the result $(28)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$
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23,902
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What is the practical difference between a differential and a derivative?
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I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative: So, a derivative is the rate of change of a function with respect to changes in its variable, this much I get. Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives? And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest. Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle. I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it. Any assistance?
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calculus
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https://math.stackexchange.com/questions/23902/what-is-the-practical-difference-between-a-differential-and-a-derivative
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Originally, "differentials" and "derivatives" were intimately connected, with derivative being defined as the ratio of the differential of the function by the differential of the variable (see my previous discussion on the Leibnitz notation for the derivative). Differentials were simply "infinitesimal changes" in whatever, and the derivative of $y$ with respect to $x$ was the ratio of the infinitesimal change in $y$ relative to the infinitesimal change in $x$. For integrals, "differentials" came in because, in Leibnitz's way of thinking about them, integrals were the sums of infinitely many infinitesimally thin rectangles that lay below the graph of the function. Each rectangle would have height $y$ and base $dx$ (the infinitesimal change in $x$), so the area of the rectangle would be $y\,dx$ (height times base), and we would add them all up as $S\; y\,dx$ to get the total area (the integral sign was originally an elongated $S$, for "summa", or sum). Infinitesimals, however, cause all sorts of headaches and problems. A lot of the reasoning about infinitesimals was, well, let's say not entirely rigorous (or logical); some differentials were dismissed as "utterly inconsequential", while others were taken into account. For example, the product rule would be argued by saying that the change in $fg$ is given by $$(f+df)(g+dg) -fg = fdg + gdf + df\,dg,$$ and then ignoring $df\,dg$ as inconsequential, since it was made up of the product of two infinitesimals; but if infinitesimals that are really small can be ignored, why do we not ignore the infinitesimal change $dg$ in the first factor? Well, you can wave your hands a lot of huff and puff, but in the end the argument essentially broke down into nonsense, or the problem was ignored because things worked out regardless (most of the time, anyway). Anyway, there was a need of a more solid understanding of just what derivatives and differentials actually are so that we can really reason about them; that's where limits came in. Derivatives are no longer ratios, instead they are limits. Integrals are no longer infinite sums of infinitesimally thin rectangles, now they are limits of Riemann sums (each of which is finite and there are no infinitesimals around), etc. The notation is left over, though, because it is very useful notation and is very suggestive. In the integral case, for instance, the "dx" is no longer really a quantity or function being multiplied: it's best to think of it as the "closing parenthesis" that goes with the "opening parenthesis" of the integral (that is, you are integrating whatever is between the $\int$ and the $dx$, just like when you have $2(84+3)$, you are multiplying by $2$ whatever is between the $($ and the $)$ ). But it is very useful, because for example it helps you keep track of what changes need to be made when you do a change of variable. One can justify the change of variable without appealing at all to "differentials" (whatever they may be), but the notation just leads you through the necessary changes, so we treat them as if they were actual functions being multiplied by the integrand because they help keep us on the right track and keep us honest. But here is an ill-kept secret: we mathematicians tend to be lazy. If we've already come up with a valid argument for situation A, we don't want to have to come up with a new valid argument for situation B if we can just explain how to get from B to A, even if solving B directly would be easier than solving A (old joke: a mathematician and an engineer are subjects of a psychology experiment; first they are shown into a room where there is an empty bucket, a trashcan, and a faucet. The trashcan is on fire. Each of them first fills the bucket with water from the faucet, then dumps it on the trashcan and extinguishes the flames. Then the engineer is shown to another room, where there is again a faucet, a trashcan on fire, and a bucket, but this time the bucket is already filled with water; the engineer takes the bucket, empties it on the trashcan and puts out the fire. The mathematican, later, comes in, sees the situation, takes the bucket, and empties it on the floor, and then says "which reduces it to a previously solved problem.") Where were we? Ah, yes. Having to translate all those informal manipulations that work so well and treat $dx$ and $dy$ as objects in and of themselves, into formal justifications that don't treat them that way is a real pain. It can be done, but it's a real pain. Instead, we want to come up with a way of justifying all those manipulations that will be valid always. One way of doing it is by actually giving them a meaning in terms of the new notions of derivatives. And that is what is done. Basically, we want the "differential" of $y$ to be the infinitesimal change in $y$; this change will be closely approximated to the change along the tangent to $y$; the tangent has slope $y'(a)$. But because we don't have infinitesimals, we have to say how much we've changed the argument. So we define "the differential in $y$ at $a$ when $x$ changes by $\Delta x$", $d(y,\Delta x)(a)$, as $d(y,\Delta x)(a) = y'(a)\Delta x$. This is exactly the change along the tangent, rather than along the graph of the function. If you take the limit of $d(y,\Delta x)$ over $\Delta x$ as $\Delta x\to 0$, you just get $y'$. But we tend to think of the limit of $\Delta x\to 0$ as being $dx$, so abuse of notation leads to "$dy = \frac{dy}{dx}\,dx$"; this is suggestive, but not quite true literally; instead, one then can show that arguments that treat differentials as functions tend to give the right answer under mild assumptions. Note that under this definition, you get $d(x,\Delta x) = 1\Delta x$, leading to $dx = dx$. Also, notice an interesting reversal: originally, differentials came first, and they were used to define the derivative as a ratio. Today, derivatives come first (defined as limits), and differentials are defined in terms of the derivatives. What is the practical difference, though? You'll probably be disappointed to hear "not much". Except one thing: when your functions represent actual quantities, rather than just formal manipulation of symbols, the derivative and the differential measure different things. The derivative measures a rate of change, while the differential measures the change itself. So the units of measurement are different: for example, if $y$ is distance and $x$ is time, then $\frac{dy}{dx}$ is measured in distance over time, i.e., velocity. But the differential $dy$ is measured in units of distance, because it represents the change in distance (and the difference/change between two distances is still a distance, not a velocity any more). Why is it useful to have the distinction? Because sometimes you want to know how something is changing, and sometimes you want to know how much something changed. It's all nice and good to know the rate of inflation (change in prices over time), but you might sometimes want to know how much more the loaf of bread is now (rather than the rate at which the price is changing). And because being able to manipulate derivatives as if they were quotients can be very useful when dealing with integrals, differential equations, etc, and differentials give us a way of making sure that these manipulations don't lead us astray (as they sometimes did in the days of infinitesimals). I'm not sure if that answers your question or at least gives an indication of where the answers lie. I hope it does. Added. I see Qiaochu has pointed out that the distinction becomes much clearer once you go to higher dimensions/multivariable calculus, so the above may all be a waste. Still... Added. As Qiaochu points out (and I mentioned in passing elsewhere), there are ways in which one can give formal definitions and meanings to infinitesimals, in which case we can define differentials as "infinitesimal changes" or "changes along infinitesimal differences"; and then use them to define derivatives as integrals just like Leibnitz did. The standard example of being able to do this is Robinson's non-standard analysis Or if one is willing to forgo looking at all kinds of functions and only at some restricted type of functions, then you can also give infinitesimals, differentials, and derivatives substance/meaning which is much closer to their original conception.
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Q: What is the practical difference between a differential and a derivative?
I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative: So, a derivative is the rate of change of a function with respect to changes in its variable, this much I get. Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives? And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest. Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle. I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it. Any assistance?
A: Originally, "differentials" and "derivatives" were intimately connected, with derivative being defined as the ratio of the differential of the function by the differential of the variable (see my previous discussion on the Leibnitz notation for the derivative). Differentials were simply "infinitesimal changes" in whatever, and the derivative of $y$ with respect to $x$ was the ratio of the infinitesimal change in $y$ relative to the infinitesimal change in $x$. For integrals, "differentials" came in because, in Leibnitz's way of thinking about them, integrals were the sums of infinitely many infinitesimally thin rectangles that lay below the graph of the function. Each rectangle would have height $y$ and base $dx$ (the infinitesimal change in $x$), so the area of the rectangle would be $y\,dx$ (height times base), and we would add them all up as $S\; y\,dx$ to get the total area (the integral sign was originally an elongated $S$, for "summa", or sum). Infinitesimals, however, cause all sorts of headaches and problems. A lot of the reasoning about infinitesimals was, well, let's say not entirely rigorous (or logical); some differentials were dismissed as "utterly inconsequential", while others were taken into account. For example, the product rule would be argued by saying that the change in $fg$ is given by $$(f+df)(g+dg) -fg = fdg + gdf + df\,dg,$$ and then ignoring $df\,dg$ as inconsequential, since it was made up of the product of two infinitesimals; but if infinitesimals that are really small can be ignored, why do we not ignore the infinitesimal change $dg$ in the first factor? Well, you can wave your hands a lot of huff and puff, but in the end the argument essentially broke down into nonsense, or the problem was ignored because things worked out regardless (most of the time, anyway). Anyway, there was a need of a more solid understanding of just what derivatives and differentials actually are so that we can really reason about them; that's where limits came in. Derivatives are no longer ratios, instead they are limits. Integrals are no longer infinite sums of infinitesimally thin rectangles, now they are limits of Riemann sums (each of which is finite and there are no infinitesimals around), etc. The notation is left over, though, because it is very useful notation and is very suggestive. In the integral case, for instance, the "dx" is no longer really a quantity or function being multiplied: it's best to think of it as the "closing parenthesis" that goes with the "opening parenthesis" of the integral (that is, you are integrating whatever is between the $\int$ and the $dx$, just like when you have $2(84+3)$, you are multiplying by $2$ whatever is between the $($ and the $)$ ). But it is very useful, because for example it helps you keep track of what changes need to be made when you do a change of variable. One can justify the change of variable without appealing at all to "differentials" (whatever they may be), but the notation just leads you through the necessary changes, so we treat them as if they were actual functions being multiplied by the integrand because they help keep us on the right track and keep us honest. But here is an ill-kept secret: we mathematicians tend to be lazy. If we've already come up with a valid argument for situation A, we don't want to have to come up with a new valid argument for situation B if we can just explain how to get from B to A, even if solving B directly would be easier than solving A (old joke: a mathematician and an engineer are subjects of a psychology experiment; first they are shown into a room where there is an empty bucket, a trashcan, and a faucet. The trashcan is on fire. Each of them first fills the bucket with water from the faucet, then dumps it on the trashcan and extinguishes the flames. Then the engineer is shown to another room, where there is again a faucet, a trashcan on fire, and a bucket, but this time the bucket is already filled with water; the engineer takes the bucket, empties it on the trashcan and puts out the fire. The mathematican, later, comes in, sees the situation, takes the bucket, and empties it on the floor, and then says "which reduces it to a previously solved problem.") Where were we? Ah, yes. Having to translate all those informal manipulations that work so well and treat $dx$ and $dy$ as objects in and of themselves, into formal justifications that don't treat them that way is a real pain. It can be done, but it's a real pain. Instead, we want to come up with a way of justifying all those manipulations that will be valid always. One way of doing it is by actually giving them a meaning in terms of the new notions of derivatives. And that is what is done. Basically, we want the "differential" of $y$ to be the infinitesimal change in $y$; this change will be closely approximated to the change along the tangent to $y$; the tangent has slope $y'(a)$. But because we don't have infinitesimals, we have to say how much we've changed the argument. So we define "the differential in $y$ at $a$ when $x$ changes by $\Delta x$", $d(y,\Delta x)(a)$, as $d(y,\Delta x)(a) = y'(a)\Delta x$. This is exactly the change along the tangent, rather than along the graph of the function. If you take the limit of $d(y,\Delta x)$ over $\Delta x$ as $\Delta x\to 0$, you just get $y'$. But we tend to think of the limit of $\Delta x\to 0$ as being $dx$, so abuse of notation leads to "$dy = \frac{dy}{dx}\,dx$"; this is suggestive, but not quite true literally; instead, one then can show that arguments that treat differentials as functions tend to give the right answer under mild assumptions. Note that under this definition, you get $d(x,\Delta x) = 1\Delta x$, leading to $dx = dx$. Also, notice an interesting reversal: originally, differentials came first, and they were used to define the derivative as a ratio. Today, derivatives come first (defined as limits), and differentials are defined in terms of the derivatives. What is the practical difference, though? You'll probably be disappointed to hear "not much". Except one thing: when your functions represent actual quantities, rather than just formal manipulation of symbols, the derivative and the differential measure different things. The derivative measures a rate of change, while the differential measures the change itself. So the units of measurement are different: for example, if $y$ is distance and $x$ is time, then $\frac{dy}{dx}$ is measured in distance over time, i.e., velocity. But the differential $dy$ is measured in units of distance, because it represents the change in distance (and the difference/change between two distances is still a distance, not a velocity any more). Why is it useful to have the distinction? Because sometimes you want to know how something is changing, and sometimes you want to know how much something changed. It's all nice and good to know the rate of inflation (change in prices over time), but you might sometimes want to know how much more the loaf of bread is now (rather than the rate at which the price is changing). And because being able to manipulate derivatives as if they were quotients can be very useful when dealing with integrals, differential equations, etc, and differentials give us a way of making sure that these manipulations don't lead us astray (as they sometimes did in the days of infinitesimals). I'm not sure if that answers your question or at least gives an indication of where the answers lie. I hope it does. Added. I see Qiaochu has pointed out that the distinction becomes much clearer once you go to higher dimensions/multivariable calculus, so the above may all be a waste. Still... Added. As Qiaochu points out (and I mentioned in passing elsewhere), there are ways in which one can give formal definitions and meanings to infinitesimals, in which case we can define differentials as "infinitesimal changes" or "changes along infinitesimal differences"; and then use them to define derivatives as integrals just like Leibnitz did. The standard example of being able to do this is Robinson's non-standard analysis Or if one is willing to forgo looking at all kinds of functions and only at some restricted type of functions, then you can also give infinitesimals, differentials, and derivatives substance/meaning which is much closer to their original conception.
| 2026-01-26T11:31:10.530341
|
885,879
|
Meaning of Rays in Polar Plot of Prime Numbers
|
I recently began experimenting with gnuplot and I quickly made an interesting discovery. I plotted all of the prime numbers beneath 1 million in polar coordinates such that for every prime $p$, $(r,\theta) = (p,p)$. I was not expecting anything in particular, I was simply trying it out. The results are fascinating. When looking at the primes beneath 30000, a spiral pattern can be seen: For comparison, here is the same graph with the multiples of 3 and 7 superimposed on it. Primes are in yellow, multiples of 3 and 7 in green and red respectively. What is really interesting to me, though, is the behavior when the range is increased. Multiples of a given number appear to spiral out in the same pattern into infinity, but the primes begin to form rays in groups of 3 or 4. See below: Compared to multiples of 3 and 7 again: Now, I must admit that I am very much a novice mathematician with little experience beyond trigonometry. I am just going into Calculus and Discrete Mathematics this upcoming fall. I know that there is something called the Prime Number theorem - are these patterns related to it? Are these rays the same phenomenon as the diagonal lines found in Ulam Spirals? EDIT: In response to Greg Martin's explanation, I decided to add a couple more graphs. To see why they are relevant, read his answer. $(r,\theta)=(n,n), n \in \mathbb{N}$
| 279
|
prime-numbers, polar-coordinates
|
https://math.stackexchange.com/questions/885879/meaning-of-rays-in-polar-plot-of-prime-numbers
|
What we're seeing is arithmetic progressions (not prime-producing polynomials) of primes, combined with a classical phenomenon about rational approximations. When the integers (or any subset of them) are represented by the polar points $(n,n)$, of course integers that are close to a multiple of $2\pi$ apart from each other will lie close to the same central ray. Figuring out when integers are close to a multiple of $2\pi$ apart is a perfect job for continued fractions. The continued fraction of $2\pi$ is $\langle 6; 3,1,1,7,2,146,\dots\rangle$, giving the convergents $$ \left\{6,\frac{19}{3},\frac{25}{4},\frac{44}{7},\frac{333}{53},\frac{710}{113},\frac{103993}{16551},\dots\right\}, $$ which are the rational approximations of $2\pi$ that will dominate the picture on different scales. For example, if you plot the polar points $(n,n)$ for $1\le n\le 25000$, you will notice the points aligning themselves into $44$ spirals: jumping ahead from $n$ to $n+44$ is almost the same as going around the circle $7$ times (note the convergent $\frac{44}7$ showing up); moving from $n$ to $n+1$ jumps ahead $7$ spirals. Each spiral corresponds to an arithmetic progression $a\pmod{44}$; going from one spiral to the next one counterclockwise corresponding to changing the arithmetic progression from $a\pmod{44}$ to $a+19\pmod{44}$ (note that $19\equiv7^{-1}\pmod{44}$). If instead you plot only the primes $(p,p)$, you will get reasonable representation in the $\phi(44)=20$ spirals corresponding to arithmetic progressions $a\pmod{44}$ where $\gcd(a,44)=1$, and no primes in the other $24$ spirals. That's what we're seeing in the top two pictures. As the scale moves farther out, these particular spirals become more tightly wound and harder to see (go from the 1st picture to the 5th, then the 3rd, then the 4th), and the next convergent takes over. In this case, the convergent $\frac{710}{113}$ is an extremely good rational approximation to $2\pi$ (as we know from the large partial quotient $146$). Therefore the integer points $(n,n)$ will group themselves into $710$ spirals, but these spirals are so close to straight lines at the beginning that they almost don't look like spirals, and persist for a large interval of possible scales. Each ray thus corresponds to an arithmetic progression $a\pmod{710}$. When we plot only prime points $(p,p)$ (the 4th picture is best here), we will only see the $\phi(710)=280$ arithmetic progressions $a\pmod{710}$ where $\gcd(a,710)=1$. The fact that the visible rays are mostly grouped in fours is a consequence of the fact that $5\mid710$ and so every fifth ray doesn't contain primes. Really, though, we are seeing four out of every ten rays rather than four out of every five; the arithmetic progressions $a\pmod{710}$ with $a$ even have no primes at all and are thus invisible. There are four exceptional groups containing only three rays instead of four; these correspond to the four arithmetic progressions $a\pmod{710}$ where $a$ is a multiple of $71$ but not a multiple of $2$ or $5$.
| 200
| true
|
Q: Meaning of Rays in Polar Plot of Prime Numbers
I recently began experimenting with gnuplot and I quickly made an interesting discovery. I plotted all of the prime numbers beneath 1 million in polar coordinates such that for every prime $p$, $(r,\theta) = (p,p)$. I was not expecting anything in particular, I was simply trying it out. The results are fascinating. When looking at the primes beneath 30000, a spiral pattern can be seen: For comparison, here is the same graph with the multiples of 3 and 7 superimposed on it. Primes are in yellow, multiples of 3 and 7 in green and red respectively. What is really interesting to me, though, is the behavior when the range is increased. Multiples of a given number appear to spiral out in the same pattern into infinity, but the primes begin to form rays in groups of 3 or 4. See below: Compared to multiples of 3 and 7 again: Now, I must admit that I am very much a novice mathematician with little experience beyond trigonometry. I am just going into Calculus and Discrete Mathematics this upcoming fall. I know that there is something called the Prime Number theorem - are these patterns related to it? Are these rays the same phenomenon as the diagonal lines found in Ulam Spirals? EDIT: In response to Greg Martin's explanation, I decided to add a couple more graphs. To see why they are relevant, read his answer. $(r,\theta)=(n,n), n \in \mathbb{N}$
A: What we're seeing is arithmetic progressions (not prime-producing polynomials) of primes, combined with a classical phenomenon about rational approximations. When the integers (or any subset of them) are represented by the polar points $(n,n)$, of course integers that are close to a multiple of $2\pi$ apart from each other will lie close to the same central ray. Figuring out when integers are close to a multiple of $2\pi$ apart is a perfect job for continued fractions. The continued fraction of $2\pi$ is $\langle 6; 3,1,1,7,2,146,\dots\rangle$, giving the convergents $$ \left\{6,\frac{19}{3},\frac{25}{4},\frac{44}{7},\frac{333}{53},\frac{710}{113},\frac{103993}{16551},\dots\right\}, $$ which are the rational approximations of $2\pi$ that will dominate the picture on different scales. For example, if you plot the polar points $(n,n)$ for $1\le n\le 25000$, you will notice the points aligning themselves into $44$ spirals: jumping ahead from $n$ to $n+44$ is almost the same as going around the circle $7$ times (note the convergent $\frac{44}7$ showing up); moving from $n$ to $n+1$ jumps ahead $7$ spirals. Each spiral corresponds to an arithmetic progression $a\pmod{44}$; going from one spiral to the next one counterclockwise corresponding to changing the arithmetic progression from $a\pmod{44}$ to $a+19\pmod{44}$ (note that $19\equiv7^{-1}\pmod{44}$). If instead you plot only the primes $(p,p)$, you will get reasonable representation in the $\phi(44)=20$ spirals corresponding to arithmetic progressions $a\pmod{44}$ where $\gcd(a,44)=1$, and no primes in the other $24$ spirals. That's what we're seeing in the top two pictures. As the scale moves farther out, these particular spirals become more tightly wound and harder to see (go from the 1st picture to the 5th, then the 3rd, then the 4th), and the next convergent takes over. In this case, the convergent $\frac{710}{113}$ is an extremely good rational approximation to $2\pi$ (as we know from the large partial quotient $146$). Therefore the integer points $(n,n)$ will group themselves into $710$ spirals, but these spirals are so close to straight lines at the beginning that they almost don't look like spirals, and persist for a large interval of possible scales. Each ray thus corresponds to an arithmetic progression $a\pmod{710}$. When we plot only prime points $(p,p)$ (the 4th picture is best here), we will only see the $\phi(710)=280$ arithmetic progressions $a\pmod{710}$ where $\gcd(a,710)=1$. The fact that the visible rays are mostly grouped in fours is a consequence of the fact that $5\mid710$ and so every fifth ray doesn't contain primes. Really, though, we are seeing four out of every ten rays rather than four out of every five; the arithmetic progressions $a\pmod{710}$ with $a$ even have no primes at all and are thus invisible. There are four exceptional groups containing only three rays instead of four; these correspond to the four arithmetic progressions $a\pmod{710}$ where $a$ is a multiple of $71$ but not a multiple of $2$ or $5$.
| 2026-01-26T11:31:12.135441
|
279,079
|
How to read a book in mathematics?
|
How is it that you read a mathematics book? Do you keep a notebook of definitions? What about theorems? Do you do all the exercises? Focus on or ignore the proofs? I have been reading Munkres, Artin, Halmos, etc. but I get a bit lost usually around the middle. Also, about how fast should you be reading it? Any advice is wanted, I just reached the upper division level.
| 278
|
soft-question
|
https://math.stackexchange.com/questions/279079/how-to-read-a-book-in-mathematics
|
This method has worked well for me (but what works well for one person won't necessarily work well for everyone). I take it in several passes: Read 0: Don't read the book, read the Wikipedia article or ask a friend what the subject is about. Learn about the big questions asked in the subject, and the basics of the theorems that answer them. Often the most important ideas are those that can be stated concisely, so you should be able to remember them once you are engaging the book. Read 1: Let your eyes jump from definition to lemma to theorem without reading the proofs in between unless something grabs your attention or bothers you. If the book has exercises, see if you can do the first one of each chapter or section as you go. Read 2: Read the book but this time read the proofs. But don't worry if you don't get all the details. If some logical jump doesn't make complete sense, feel free to ignore it at your discretion as long as you understand the overall flow of reasoning. Read 3: Read through the lens of a skeptic. Work through all of the proofs with a fine toothed comb, and ask yourself every question you think of. You should never have to ask yourself "why" you are proving what you are proving at this point, but you have a chance to get the details down. This approach is well suited to many math textbooks, which seem to be written to read well to people who already understand the subject. Most of the "classic" textbooks are labeled as such because they are comprehensive or well organized, not because they present challenging abstract ideas well to the uninitiated. (Steps 1-3 are based on a three step heuristic method for writing proofs: convince yourself, convince a friend, convince a skeptic)
| 241
| true
|
Q: How to read a book in mathematics?
How is it that you read a mathematics book? Do you keep a notebook of definitions? What about theorems? Do you do all the exercises? Focus on or ignore the proofs? I have been reading Munkres, Artin, Halmos, etc. but I get a bit lost usually around the middle. Also, about how fast should you be reading it? Any advice is wanted, I just reached the upper division level.
A: This method has worked well for me (but what works well for one person won't necessarily work well for everyone). I take it in several passes: Read 0: Don't read the book, read the Wikipedia article or ask a friend what the subject is about. Learn about the big questions asked in the subject, and the basics of the theorems that answer them. Often the most important ideas are those that can be stated concisely, so you should be able to remember them once you are engaging the book. Read 1: Let your eyes jump from definition to lemma to theorem without reading the proofs in between unless something grabs your attention or bothers you. If the book has exercises, see if you can do the first one of each chapter or section as you go. Read 2: Read the book but this time read the proofs. But don't worry if you don't get all the details. If some logical jump doesn't make complete sense, feel free to ignore it at your discretion as long as you understand the overall flow of reasoning. Read 3: Read through the lens of a skeptic. Work through all of the proofs with a fine toothed comb, and ask yourself every question you think of. You should never have to ask yourself "why" you are proving what you are proving at this point, but you have a chance to get the details down. This approach is well suited to many math textbooks, which seem to be written to read well to people who already understand the subject. Most of the "classic" textbooks are labeled as such because they are comprehensive or well organized, not because they present challenging abstract ideas well to the uninitiated. (Steps 1-3 are based on a three step heuristic method for writing proofs: convince yourself, convince a friend, convince a skeptic)
| 2026-01-26T11:31:13.693919
|
17,152
|
Given an infinite number of monkeys and an infinite amount of time, would one of them write Hamlet?
|
Of course, we've all heard the colloquialism "If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet." I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact, which got me thinking... Is this really true? Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite. Therefore, it isn't necessarily given that the monkeys would write Hamlet. Could someone who's better at this kind of math than me tell me if this is correct? Or is there more to it than I'm thinking?
| 277
|
probability, infinity
|
https://math.stackexchange.com/questions/17152/given-an-infinite-number-of-monkeys-and-an-infinite-amount-of-time-would-one-of
|
I found online the claim (which we may as well accept for this purpose) that there are $32241$ words in Hamlet. Figuring $5$ characters and one space per word, this is $193446$ characters. If the character set is $60$ including capitals and punctuation, a random string of $193446$ characters has a chance of $1$ in $60^{193446}$ (roughly $1$ in $10^{344000}$) of being Hamlet. While very small, this is greater than zero. So if you try enough times, and infinity times is certainly enough, you will probably produce Hamlet. But don't hold your breath. It doesn't even take an infinite number of monkeys or an infinite number of tries. Only a product of $10^{344001}$ makes it very likely. True, this is a very large number, but most numbers are larger.
| 336
| false
|
Q: Given an infinite number of monkeys and an infinite amount of time, would one of them write Hamlet?
Of course, we've all heard the colloquialism "If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet." I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact, which got me thinking... Is this really true? Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite. Therefore, it isn't necessarily given that the monkeys would write Hamlet. Could someone who's better at this kind of math than me tell me if this is correct? Or is there more to it than I'm thinking?
A: I found online the claim (which we may as well accept for this purpose) that there are $32241$ words in Hamlet. Figuring $5$ characters and one space per word, this is $193446$ characters. If the character set is $60$ including capitals and punctuation, a random string of $193446$ characters has a chance of $1$ in $60^{193446}$ (roughly $1$ in $10^{344000}$) of being Hamlet. While very small, this is greater than zero. So if you try enough times, and infinity times is certainly enough, you will probably produce Hamlet. But don't hold your breath. It doesn't even take an infinite number of monkeys or an infinite number of tries. Only a product of $10^{344001}$ makes it very likely. True, this is a very large number, but most numbers are larger.
| 2026-01-26T11:31:15.440578
|
160,248
|
Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
|
I'm supposed to calculate: $$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.
| 277
|
calculus, real-analysis, sequences-and-series, limits
|
https://math.stackexchange.com/questions/160248/evaluating-lim-limits-n-to-infty-e-n-sum-limits-k-0n-fracnkk
|
The probabilistic way: This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$. By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$. Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED. The analytical way, completing your try: Hence, I know that what I need to do is to find $\lim\limits_{n\to\infty}I_n$, where $$ I_n=\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt.$$ To begin with, let $u(t)=(1-t)e^t$, then $I_n=\dfrac{e^{-n}n^n}{n!}nJ_n$ with $$ J_n=\int_{0}^1 u(t)^n\mathrm dt. $$ Now, $u(t)\leqslant\mathrm e^{-t^2/2}$ hence $$ J_n\leqslant\int_0^1\mathrm e^{-nt^2/2}\mathrm dt\leqslant\int_0^\infty\mathrm e^{-nt^2/2}\mathrm dt=\sqrt{\frac{\pi}{2n}}. $$ Likewise, the function $t\mapsto u(t)\mathrm e^{t^2/2}$ is decreasing on $t\geqslant0$ hence $u(t)\geqslant c_n\mathrm e^{-t^2/2}$ on $t\leqslant1/n^{1/4}$, with $c_n=u(1/n^{1/4})\mathrm e^{-1/(2\sqrt{n})}$, hence $$ J_n\geqslant c_n\int_0^{1/n^{1/4}}\mathrm e^{-nt^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\int_0^{n^{1/4}}\mathrm e^{-t^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\sqrt{\frac{\pi}{2}}(1+o(1)). $$ Since $c_n\to1$, all this proves that $\sqrt{n}J_n\to\sqrt{\frac\pi2}$. Stirling formula shows that the prefactor $\frac{e^{-n}n^n}{n!}$ is equivalent to $\frac1{\sqrt{2\pi n}}$. Regrouping everything, one sees that $I_n\sim\frac1{\sqrt{2\pi n}}n\sqrt{\frac\pi{2n}}=\frac12$. Moral: The probabilistic way is shorter, easier, more illuminating, and more fun. Caveat: My advice in these matters is, clearly, horribly biased.
| 200
| false
|
Q: Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
I'm supposed to calculate: $$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.
A: The probabilistic way: This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$. By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$. Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED. The analytical way, completing your try: Hence, I know that what I need to do is to find $\lim\limits_{n\to\infty}I_n$, where $$ I_n=\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt.$$ To begin with, let $u(t)=(1-t)e^t$, then $I_n=\dfrac{e^{-n}n^n}{n!}nJ_n$ with $$ J_n=\int_{0}^1 u(t)^n\mathrm dt. $$ Now, $u(t)\leqslant\mathrm e^{-t^2/2}$ hence $$ J_n\leqslant\int_0^1\mathrm e^{-nt^2/2}\mathrm dt\leqslant\int_0^\infty\mathrm e^{-nt^2/2}\mathrm dt=\sqrt{\frac{\pi}{2n}}. $$ Likewise, the function $t\mapsto u(t)\mathrm e^{t^2/2}$ is decreasing on $t\geqslant0$ hence $u(t)\geqslant c_n\mathrm e^{-t^2/2}$ on $t\leqslant1/n^{1/4}$, with $c_n=u(1/n^{1/4})\mathrm e^{-1/(2\sqrt{n})}$, hence $$ J_n\geqslant c_n\int_0^{1/n^{1/4}}\mathrm e^{-nt^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\int_0^{n^{1/4}}\mathrm e^{-t^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\sqrt{\frac{\pi}{2}}(1+o(1)). $$ Since $c_n\to1$, all this proves that $\sqrt{n}J_n\to\sqrt{\frac\pi2}$. Stirling formula shows that the prefactor $\frac{e^{-n}n^n}{n!}$ is equivalent to $\frac1{\sqrt{2\pi n}}$. Regrouping everything, one sees that $I_n\sim\frac1{\sqrt{2\pi n}}n\sqrt{\frac\pi{2n}}=\frac12$. Moral: The probabilistic way is shorter, easier, more illuminating, and more fun. Caveat: My advice in these matters is, clearly, horribly biased.
| 2026-01-26T11:31:17.000320
|
78,575
|
Derivative of sigmoid function $\sigma (x) = \frac{1}{1+e^{-x}}$
|
In my AI textbook there is this paragraph, without any explanation. The sigmoid function is defined as follows $$\sigma (x) = \frac{1}{1+e^{-x}}.$$ This function is easy to differentiate because $$\frac{d\sigma (x)}{d(x)} = \sigma (x)\cdot (1-\sigma(x)).$$ It has been a long time since I've taken differential equations, so could anyone tell me how they got from the first equation to the second?
| 276
|
calculus, derivatives
|
https://math.stackexchange.com/questions/78575/derivative-of-sigmoid-function-sigma-x-frac11e-x
|
Let's denote the sigmoid function as $\sigma(x) = \dfrac{1}{1 + e^{-x}}$. The derivative of the sigmoid is $\dfrac{d}{dx}\sigma(x) = \sigma(x)(1 - \sigma(x))$. Here's a detailed derivation: $$ \begin{align} \dfrac{d}{dx} \sigma(x) &= \dfrac{d}{dx} \left[ \dfrac{1}{1 + e^{-x}} \right] \\ &= \dfrac{d}{dx} \left( 1 + \mathrm{e}^{-x} \right)^{-1} \\ &= -(1 + e^{-x})^{-2}(-e^{-x}) \\ &= \dfrac{e^{-x}}{\left(1 + e^{-x}\right)^2} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{e^{-x}}{1 + e^{-x}} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{(1 + e^{-x}) - 1}{1 + e^{-x}} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \left( \dfrac{1 + e^{-x}}{1 + e^{-x}} - \dfrac{1}{1 + e^{-x}} \right) \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \left( 1 - \dfrac{1}{1 + e^{-x}} \right) \\ &= \sigma(x) \cdot (1 - \sigma(x)) \end{align} $$
| 460
| true
|
Q: Derivative of sigmoid function $\sigma (x) = \frac{1}{1+e^{-x}}$
In my AI textbook there is this paragraph, without any explanation. The sigmoid function is defined as follows $$\sigma (x) = \frac{1}{1+e^{-x}}.$$ This function is easy to differentiate because $$\frac{d\sigma (x)}{d(x)} = \sigma (x)\cdot (1-\sigma(x)).$$ It has been a long time since I've taken differential equations, so could anyone tell me how they got from the first equation to the second?
A: Let's denote the sigmoid function as $\sigma(x) = \dfrac{1}{1 + e^{-x}}$. The derivative of the sigmoid is $\dfrac{d}{dx}\sigma(x) = \sigma(x)(1 - \sigma(x))$. Here's a detailed derivation: $$ \begin{align} \dfrac{d}{dx} \sigma(x) &= \dfrac{d}{dx} \left[ \dfrac{1}{1 + e^{-x}} \right] \\ &= \dfrac{d}{dx} \left( 1 + \mathrm{e}^{-x} \right)^{-1} \\ &= -(1 + e^{-x})^{-2}(-e^{-x}) \\ &= \dfrac{e^{-x}}{\left(1 + e^{-x}\right)^2} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{e^{-x}}{1 + e^{-x}} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{(1 + e^{-x}) - 1}{1 + e^{-x}} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \left( \dfrac{1 + e^{-x}}{1 + e^{-x}} - \dfrac{1}{1 + e^{-x}} \right) \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \left( 1 - \dfrac{1}{1 + e^{-x}} \right) \\ &= \sigma(x) \cdot (1 - \sigma(x)) \end{align} $$
| 2026-01-26T11:31:18.686987
|
165,696
|
Your favourite application of the Baire Category Theorem
|
I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power. Here's the theorem (with proof) and two applications: (Baire) A non-empty complete metric space $X$ is not a countable union of nowhere dense sets. Proof: Let $X = \bigcup U_i$ where $\mathring{\overline{U_i}} = \varnothing$. We construct a Cauchy sequence as follows: Let $x_1$ be any point in $(\overline{U_1})^c$. We can find such a point because $(\overline{U_1})^c \subset X$ and $X$ contains at least one non-empty open set (if nothing else, itself) but $\mathring{\overline{U_1}} = \varnothing$ which is the same as saying that $\overline{U_1}$ does not contain any open sets hence the open set contained in $X$ is contained in $\overline{U_1}^c$. Hence we can pick $x_1$ and $\varepsilon_1 > 0$ such that $B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c$. Next we make a similar observation about $U_2$ so that we can find $x_2$ and $\varepsilon_2 > 0$ such that $B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})$. We repeat this process to get a sequence of balls such that $B_{k+1} \subset B_k$ and a sequence $(x_k)$ that is Cauchy. By completeness of $X$, $\lim x_k =: x$ is in $X$. But $x$ is in $B_k$ for every $k$ hence not in any of the $U_i$ and hence not in $\bigcup U_i = X$. Contradiction. $\Box$ Here is one application (taken from here): Claim: $[0,1]$ contains uncountably many elements. Proof: Assume that it contains countably many. Then $[0,1] = \bigcup_{x \in [0,1]} \{x\}$ and since $\{x\}$ are nowhere dense sets, $X$ is a countable union of nowhere dense sets. But $[0,1]$ is complete, so we have a contradiction. Hence $X$ has to be uncountable. And here is another one (taken from here): Claim: The linear space of all polynomials in one variable is not a Banach space in any norm. Proof: "The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem."
| 275
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general-topology, functional-analysis, big-list, baire-category
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https://math.stackexchange.com/questions/165696/your-favourite-application-of-the-baire-category-theorem
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If $P$ is an infinitely differentiable function such that for each $x$, there is an $n$ with $P^{(n)}(x)=0$, then $P$ is a polynomial. (Note $n$ depends on $x$.) See the discussion in Math Overflow.
| 79
| false
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Q: Your favourite application of the Baire Category Theorem
I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power. Here's the theorem (with proof) and two applications: (Baire) A non-empty complete metric space $X$ is not a countable union of nowhere dense sets. Proof: Let $X = \bigcup U_i$ where $\mathring{\overline{U_i}} = \varnothing$. We construct a Cauchy sequence as follows: Let $x_1$ be any point in $(\overline{U_1})^c$. We can find such a point because $(\overline{U_1})^c \subset X$ and $X$ contains at least one non-empty open set (if nothing else, itself) but $\mathring{\overline{U_1}} = \varnothing$ which is the same as saying that $\overline{U_1}$ does not contain any open sets hence the open set contained in $X$ is contained in $\overline{U_1}^c$. Hence we can pick $x_1$ and $\varepsilon_1 > 0$ such that $B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c$. Next we make a similar observation about $U_2$ so that we can find $x_2$ and $\varepsilon_2 > 0$ such that $B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})$. We repeat this process to get a sequence of balls such that $B_{k+1} \subset B_k$ and a sequence $(x_k)$ that is Cauchy. By completeness of $X$, $\lim x_k =: x$ is in $X$. But $x$ is in $B_k$ for every $k$ hence not in any of the $U_i$ and hence not in $\bigcup U_i = X$. Contradiction. $\Box$ Here is one application (taken from here): Claim: $[0,1]$ contains uncountably many elements. Proof: Assume that it contains countably many. Then $[0,1] = \bigcup_{x \in [0,1]} \{x\}$ and since $\{x\}$ are nowhere dense sets, $X$ is a countable union of nowhere dense sets. But $[0,1]$ is complete, so we have a contradiction. Hence $X$ has to be uncountable. And here is another one (taken from here): Claim: The linear space of all polynomials in one variable is not a Banach space in any norm. Proof: "The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem."
A: If $P$ is an infinitely differentiable function such that for each $x$, there is an $n$ with $P^{(n)}(x)=0$, then $P$ is a polynomial. (Note $n$ depends on $x$.) See the discussion in Math Overflow.
| 2026-01-26T11:31:20.774269
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264,371
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Fun but serious mathematics books to gift advanced undergraduates.
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I am looking for fun, interesting mathematics textbooks which would make good studious holiday gifts for advanced mathematics undergraduates or beginning graduate students. They should be serious but also readable. In particular, I am looking for readable books on more obscure topics not covered in a standard undergraduate curriculum which students may not have previously heard of or thought to study. Some examples of suggestions I've liked so far: On Numbers and Games, by John Conway. Groups, Graphs and Trees: An Introduction to the Geometry of Infinite Groups, by John Meier. Ramsey Theory on the Integers, by Bruce Landman. I am not looking for pop math books, Gödel, Escher, Bach, or anything of that nature. I am also not looking for books on 'core' subjects unless the content is restricted to a subdiscipline which is not commonly studied by undergrads (e.g., Finite Group Theory by Isaacs would be good, but Abstract Algebra by Dummit and Foote would not).
| 275
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reference-request, soft-question, big-list
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https://math.stackexchange.com/questions/264371/fun-but-serious-mathematics-books-to-gift-advanced-undergraduates
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Check into Generatingfunctionology by Herbert Wilf. From the linked (author's) site, the second edition is available for downloading as a pdf. There is also a link to the third edition, available for purchase. It's a very helpful, useful, readable, fun, (and short!) book that a student could conceivably cover over winter break. Another promising book by John Conway (et. al.) is The Symmetries of Things, which may very well be of interest to students. One additional suggestion, as it is a classic well worth being placed on any serious student's bookshelf: How to Solve It by George Polya.
| 79
| false
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Q: Fun but serious mathematics books to gift advanced undergraduates.
I am looking for fun, interesting mathematics textbooks which would make good studious holiday gifts for advanced mathematics undergraduates or beginning graduate students. They should be serious but also readable. In particular, I am looking for readable books on more obscure topics not covered in a standard undergraduate curriculum which students may not have previously heard of or thought to study. Some examples of suggestions I've liked so far: On Numbers and Games, by John Conway. Groups, Graphs and Trees: An Introduction to the Geometry of Infinite Groups, by John Meier. Ramsey Theory on the Integers, by Bruce Landman. I am not looking for pop math books, Gödel, Escher, Bach, or anything of that nature. I am also not looking for books on 'core' subjects unless the content is restricted to a subdiscipline which is not commonly studied by undergrads (e.g., Finite Group Theory by Isaacs would be good, but Abstract Algebra by Dummit and Foote would not).
A: Check into Generatingfunctionology by Herbert Wilf. From the linked (author's) site, the second edition is available for downloading as a pdf. There is also a link to the third edition, available for purchase. It's a very helpful, useful, readable, fun, (and short!) book that a student could conceivably cover over winter break. Another promising book by John Conway (et. al.) is The Symmetries of Things, which may very well be of interest to students. One additional suggestion, as it is a classic well worth being placed on any serious student's bookshelf: How to Solve It by George Polya.
| 2026-01-26T11:31:22.577252
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82,467
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Eigenvectors of real symmetric matrices are orthogonal
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Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.
| 272
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linear-algebra, matrices, reference-request, eigenvalues-eigenvectors, symmetric-matrices
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https://math.stackexchange.com/questions/82467/eigenvectors-of-real-symmetric-matrices-are-orthogonal
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For any real matrix $A$ and any vectors $\mathbf{x}$ and $\mathbf{y}$, we have $$\langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle.$$ Now assume that $A$ is symmetric, and $\mathbf{x}$ and $\mathbf{y}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda$ and $\mu$. Then $$\lambda\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda\mathbf{x},\mathbf{y}\rangle = \langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle = \langle\mathbf{x},A\mathbf{y}\rangle = \langle\mathbf{x},\mu\mathbf{y}\rangle = \mu\langle\mathbf{x},\mathbf{y}\rangle.$$ Therefore, $(\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle = 0$. Since $\lambda-\mu\neq 0$, then $\langle\mathbf{x},\mathbf{y}\rangle = 0$, i.e., $\mathbf{x}\perp\mathbf{y}$. Now find an orthonormal basis for each eigenspace; since the eigenspaces are mutually orthogonal, these vectors together give an orthonormal subset of $\mathbb{R}^n$. Finally, since symmetric matrices are diagonalizable, this set will be a basis (just count dimensions). The result you want now follows.
| 371
| true
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Q: Eigenvectors of real symmetric matrices are orthogonal
Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.
A: For any real matrix $A$ and any vectors $\mathbf{x}$ and $\mathbf{y}$, we have $$\langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle.$$ Now assume that $A$ is symmetric, and $\mathbf{x}$ and $\mathbf{y}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda$ and $\mu$. Then $$\lambda\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda\mathbf{x},\mathbf{y}\rangle = \langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle = \langle\mathbf{x},A\mathbf{y}\rangle = \langle\mathbf{x},\mu\mathbf{y}\rangle = \mu\langle\mathbf{x},\mathbf{y}\rangle.$$ Therefore, $(\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle = 0$. Since $\lambda-\mu\neq 0$, then $\langle\mathbf{x},\mathbf{y}\rangle = 0$, i.e., $\mathbf{x}\perp\mathbf{y}$. Now find an orthonormal basis for each eigenspace; since the eigenspaces are mutually orthogonal, these vectors together give an orthonormal subset of $\mathbb{R}^n$. Finally, since symmetric matrices are diagonalizable, this set will be a basis (just count dimensions). The result you want now follows.
| 2026-01-26T11:31:24.172790
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3,749
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Why do we care about dual spaces?
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When I first took linear algebra, we never learned about dual spaces. Today in lecture we discussed them and I understand what they are, but I don't really understand why we want to study them within linear algebra. I was wondering if anyone knew a nice intuitive motivation for the study of dual spaces and whether or not they "show up" as often as other concepts in linear algebra? Is their usefulness something that just becomes more apparent as you learn more math and see them arise in different settings? Edit I understand that dual spaces show up in functional analysis and multilinear algebra, but I still don't really understand the intuition/motivation behind their definition in the standard topics covered in a linear algebra course. (Hopefully, this clarifies my question)
| 271
|
linear-algebra, soft-question, intuition, dual-spaces
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https://math.stackexchange.com/questions/3749/why-do-we-care-about-dual-spaces
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Let $V$ be a vector space (over any field, but we can take it to be $\mathbb R$ if you like, and for concreteness I will take the field to be $\mathbb R$ from now on; everything is just as interesting in that case). Certainly one of the interesting concepts in linear algebra is that of a hyperplane in $V$. For example, if $V = \mathbb R^n$, then a hyperplane is just the solution set to an equation of the form $$a_1 x_1 + \cdots + a_n x_n = b,$$ for some $a_i$ not all zero and some $b$. Recall that solving such equations (or simultaneous sets of such equations) is one of the basic motivations for developing linear algebra. Now remember that when a vector space is not given to you as $\mathbb R^n$, it doesn't normally have a canonical basis, so we don't have a canonical way to write its elements down via coordinates, and so we can't describe hyperplanes by explicit equations like above. (Or better, we can, but only after choosing coordinates, and this is not canonical.) How can we canonically describe hyperplanes in $V$? For this we need a conceptual interpretation of the above equation. And here linear functionals come to the rescue. More precisely, the map $$\begin{align*} \ell: \mathbb{R}^n &\rightarrow \mathbb{R} \\ (x_1,\ldots,x_n) &\mapsto a_1 x_1 + \cdots + a_n x_n \end{align*}$$ is a linear functional on $\mathbb R^n$, and so the above equation for the hyperplane can be written as $$\ell(v) = b,$$ where $v = (x_1,\ldots,x_n).$ More generally, if $V$ is any vector space, and $\ell: V \to \mathbb R$ is any non-zero linear functional (i.e. non-zero element of the dual space), then for any $b \in \mathbb R,$ the set $$\{v \, | \, \ell(v) = b\}$$ is a hyperplane in $V$, and all hyperplanes in $V$ arise this way. So this gives a reasonable justification for introducing the elements of the dual space to $V$; they generalize the notion of linear equation in several variables from the case of $\mathbb R^n$ to the case of an arbitrary vector space. Now you might ask: why do we make them a vector space themselves? Why do we want to add them to one another, or multiply them by scalars? There are lots of reasons for this; here is one: Remember how important it is, when you solve systems of linear equations, to add equations together, or to multiply them by scalars (here I am referring to all the steps you typically make when performing Gaussian elimination on a collection of simultaneous linear equations)? Well, under the dictionary above between linear equations and linear functionals, these processes correspond precisely to adding together linear functionals, or multiplying them by scalars. If you ponder this for a bit, you can hopefully convince yourself that making the set of linear functionals a vector space is a pretty natural thing to do. Summary: just as concrete vectors $(x_1,\ldots,x_n) \in \mathbb R^n$ are naturally generalized to elements of vector spaces, concrete linear expressions $a_1 x_1 + \ldots + a_n x_n$ in $x_1,\ldots, x_n$ are naturally generalized to linear functionals.
| 273
| true
|
Q: Why do we care about dual spaces?
When I first took linear algebra, we never learned about dual spaces. Today in lecture we discussed them and I understand what they are, but I don't really understand why we want to study them within linear algebra. I was wondering if anyone knew a nice intuitive motivation for the study of dual spaces and whether or not they "show up" as often as other concepts in linear algebra? Is their usefulness something that just becomes more apparent as you learn more math and see them arise in different settings? Edit I understand that dual spaces show up in functional analysis and multilinear algebra, but I still don't really understand the intuition/motivation behind their definition in the standard topics covered in a linear algebra course. (Hopefully, this clarifies my question)
A: Let $V$ be a vector space (over any field, but we can take it to be $\mathbb R$ if you like, and for concreteness I will take the field to be $\mathbb R$ from now on; everything is just as interesting in that case). Certainly one of the interesting concepts in linear algebra is that of a hyperplane in $V$. For example, if $V = \mathbb R^n$, then a hyperplane is just the solution set to an equation of the form $$a_1 x_1 + \cdots + a_n x_n = b,$$ for some $a_i$ not all zero and some $b$. Recall that solving such equations (or simultaneous sets of such equations) is one of the basic motivations for developing linear algebra. Now remember that when a vector space is not given to you as $\mathbb R^n$, it doesn't normally have a canonical basis, so we don't have a canonical way to write its elements down via coordinates, and so we can't describe hyperplanes by explicit equations like above. (Or better, we can, but only after choosing coordinates, and this is not canonical.) How can we canonically describe hyperplanes in $V$? For this we need a conceptual interpretation of the above equation. And here linear functionals come to the rescue. More precisely, the map $$\begin{align*} \ell: \mathbb{R}^n &\rightarrow \mathbb{R} \\ (x_1,\ldots,x_n) &\mapsto a_1 x_1 + \cdots + a_n x_n \end{align*}$$ is a linear functional on $\mathbb R^n$, and so the above equation for the hyperplane can be written as $$\ell(v) = b,$$ where $v = (x_1,\ldots,x_n).$ More generally, if $V$ is any vector space, and $\ell: V \to \mathbb R$ is any non-zero linear functional (i.e. non-zero element of the dual space), then for any $b \in \mathbb R,$ the set $$\{v \, | \, \ell(v) = b\}$$ is a hyperplane in $V$, and all hyperplanes in $V$ arise this way. So this gives a reasonable justification for introducing the elements of the dual space to $V$; they generalize the notion of linear equation in several variables from the case of $\mathbb R^n$ to the case of an arbitrary vector space. Now you might ask: why do we make them a vector space themselves? Why do we want to add them to one another, or multiply them by scalars? There are lots of reasons for this; here is one: Remember how important it is, when you solve systems of linear equations, to add equations together, or to multiply them by scalars (here I am referring to all the steps you typically make when performing Gaussian elimination on a collection of simultaneous linear equations)? Well, under the dictionary above between linear equations and linear functionals, these processes correspond precisely to adding together linear functionals, or multiplying them by scalars. If you ponder this for a bit, you can hopefully convince yourself that making the set of linear functionals a vector space is a pretty natural thing to do. Summary: just as concrete vectors $(x_1,\ldots,x_n) \in \mathbb R^n$ are naturally generalized to elements of vector spaces, concrete linear expressions $a_1 x_1 + \ldots + a_n x_n$ in $x_1,\ldots, x_n$ are naturally generalized to linear functionals.
| 2026-01-26T11:31:25.688221
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275,310
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What is the difference between linear and affine function?
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I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated.
| 271
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linear-algebra, affine-geometry
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https://math.stackexchange.com/questions/275310/what-is-the-difference-between-linear-and-affine-function
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A linear function fixes the origin, whereas an affine function need not do so. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else. Linear functions between vector spaces preserve the vector space structure (so in particular they must fix the origin). While affine functions don't preserve the origin, they do preserve some of the other geometry of the space, such as the collection of straight lines. If you choose bases for vector spaces $V$ and $W$ of dimensions $m$ and $n$ respectively, and consider functions $f\colon V\to W$, then $f$ is linear if $f(v)=Av$ for some $n\times m$ matrix $A$ and $f$ is affine if $f(v)=Av+b$ for some matrix $A$ and vector $b$, where coordinate representations are used with respect to the bases chosen.
| 245
| false
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Q: What is the difference between linear and affine function?
I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated.
A: A linear function fixes the origin, whereas an affine function need not do so. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else. Linear functions between vector spaces preserve the vector space structure (so in particular they must fix the origin). While affine functions don't preserve the origin, they do preserve some of the other geometry of the space, such as the collection of straight lines. If you choose bases for vector spaces $V$ and $W$ of dimensions $m$ and $n$ respectively, and consider functions $f\colon V\to W$, then $f$ is linear if $f(v)=Av$ for some $n\times m$ matrix $A$ and $f$ is affine if $f(v)=Av+b$ for some matrix $A$ and vector $b$, where coordinate representations are used with respect to the bases chosen.
| 2026-01-26T11:31:27.404183
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237,002
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Too old to start math
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I'm sorry if this question goes against the meta for posting questions - I attached all the "beware, this is a soft-question" tags I could. This is a question I've been asking myself now for some time. In most areas, there's a "cut off age" to be good at something. For example, you're not going to make the NHL if you start playing hockey at 20. It just won't happen. So my question then, how late is too late to start studying math and make a career out of it? By "start studying math" I mean, to really try to understand and comprehend the material (as opposed to just being able to do well in a formal, intuitional environment). I don't mean this from a "do what you love, its not too late" motivational perspective. I mean this from a purely biological perspective; at approximately what age has your brain's capacity to learn effectively and be influenced by your learning stop? When does the biological clock for learning new math run out? My reasoning for asking this question is (for those who care): I love math. Really I do. But , having spent the first 21 years of my life in sports/video games/obtaining a degree in a scientific field which I care nothing of/etc, despite all my best attempts at trying to learn math, am I just too late starting to ever actually be good enough at it to make it a career? I've almost completed my second degree (in Math), but find that in many cases, despite how I look at a problem, I lack the intuition to comprehend it. I'm going to single him out (sorry), only as an example, but Qiaochu Yuan is my age. Note 1: If this question isn't a suitable post, I won't be offended at all if you vote to close - I know this question borders what's acceptable to ask. Note 2: Thanks to everyone for reading and taking the time for the great responses. Really appreciate it!
| 270
|
soft-question, self-learning, advice, learning
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https://math.stackexchange.com/questions/237002/too-old-to-start-math
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Karl Weierstrass was in his 40's when he got his PHD. There are a dozen other counterexamples, a number fairly recent. A good set of examples can be found in the thread on MO here.This myth of "science is a game for the young" is one of the falsest and most destructive canards in modern society. Don't listen to it. You only get one life and when it's over, that's it. When you're dead a hundred million years, you'll be dead the tiniest most infinitesimal fraction of all the time you'll ever be dead. So stop listening to career advice from teenagers, grab a calculus book and get to work. That's my advice.
| 252
| true
|
Q: Too old to start math
I'm sorry if this question goes against the meta for posting questions - I attached all the "beware, this is a soft-question" tags I could. This is a question I've been asking myself now for some time. In most areas, there's a "cut off age" to be good at something. For example, you're not going to make the NHL if you start playing hockey at 20. It just won't happen. So my question then, how late is too late to start studying math and make a career out of it? By "start studying math" I mean, to really try to understand and comprehend the material (as opposed to just being able to do well in a formal, intuitional environment). I don't mean this from a "do what you love, its not too late" motivational perspective. I mean this from a purely biological perspective; at approximately what age has your brain's capacity to learn effectively and be influenced by your learning stop? When does the biological clock for learning new math run out? My reasoning for asking this question is (for those who care): I love math. Really I do. But , having spent the first 21 years of my life in sports/video games/obtaining a degree in a scientific field which I care nothing of/etc, despite all my best attempts at trying to learn math, am I just too late starting to ever actually be good enough at it to make it a career? I've almost completed my second degree (in Math), but find that in many cases, despite how I look at a problem, I lack the intuition to comprehend it. I'm going to single him out (sorry), only as an example, but Qiaochu Yuan is my age. Note 1: If this question isn't a suitable post, I won't be offended at all if you vote to close - I know this question borders what's acceptable to ask. Note 2: Thanks to everyone for reading and taking the time for the great responses. Really appreciate it!
A: Karl Weierstrass was in his 40's when he got his PHD. There are a dozen other counterexamples, a number fairly recent. A good set of examples can be found in the thread on MO here.This myth of "science is a game for the young" is one of the falsest and most destructive canards in modern society. Don't listen to it. You only get one life and when it's over, that's it. When you're dead a hundred million years, you'll be dead the tiniest most infinitesimal fraction of all the time you'll ever be dead. So stop listening to career advice from teenagers, grab a calculus book and get to work. That's my advice.
| 2026-01-26T11:31:28.965350
|
2,855,975
|
What is the maximum volume that can be contained by a sheet of paper?
|
I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty nontrivial, I believe) question: Q: By properly folding a common $210mm\times 297mm$ sheet of paper, what is the maximum amount of water such a sheet is able to contain? The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like? Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $. As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.
| 270
|
geometry, optimization, recreational-mathematics, volume
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https://math.stackexchange.com/questions/2855975/what-is-the-maximum-volume-that-can-be-contained-by-a-sheet-of-paper
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This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem. First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions $$\phi:P\subset\mathbb{R}^2\to\mathbb{R}^3,\qquad \|d\phi^Td\phi\|_2 \leq 1$$ of the piece of paper $P$ into $\mathbb{R}^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $\phi$ sends the paper boundary $\partial P$ to a curve $\gamma$ in the plane. We can thus partition your problem into two pieces: (I) given a fixed curve $\gamma$, what is the volume of the volume-maximizing surface $M_{\gamma}$ with $\phi(\partial P) = \gamma$? (II) Can we characterize $\gamma$ for which $M_{\gamma}$ has maximum volume? Let's consider the case where $\gamma$ is given. We can partition $M_{\gamma}$ into 1) regions of pure tension, where $d\phi^Td\phi = I$; in these regions $M_{\gamma}$ is, by definition, developable; 2) regions where one direction is in tension and one in compression, $\|d\phi^Td\phi\|_2 = 1$ but $\det d\phi^Td\phi We need not consider $\|d\phi^Td\phi\|_2 Let us look at the regions of type (2). We can trace on these regions a family of curves $\tau$ along which $\phi$ is an isometry. Since $M_{\gamma}$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_{\gamma}$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $\tau$. In other words, for some stress $\sigma$ constant along each $\tau$, at all points $\tau(s)$ along $\tau$ we have $$\hat{n} = \sigma \tau''(s)$$ where $\hat{n}$ the surface normal; it follows that (1) the $\tau$ follow geodesics on $M_{\gamma}$, (2) each $\tau$ has constant curvature. The only thing I can say about problem (II) is that for the optimal $\gamma$, the surface $M_\gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $\approx 1.236$ liters, less than Joriki's solution). I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $\gamma$ and $M_{\gamma}$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters). EDIT 2: A sketch of the orientation of $\tau$ on the surface:
| 142
| true
|
Q: What is the maximum volume that can be contained by a sheet of paper?
I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty nontrivial, I believe) question: Q: By properly folding a common $210mm\times 297mm$ sheet of paper, what is the maximum amount of water such a sheet is able to contain? The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like? Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $. As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.
A: This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem. First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions $$\phi:P\subset\mathbb{R}^2\to\mathbb{R}^3,\qquad \|d\phi^Td\phi\|_2 \leq 1$$ of the piece of paper $P$ into $\mathbb{R}^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $\phi$ sends the paper boundary $\partial P$ to a curve $\gamma$ in the plane. We can thus partition your problem into two pieces: (I) given a fixed curve $\gamma$, what is the volume of the volume-maximizing surface $M_{\gamma}$ with $\phi(\partial P) = \gamma$? (II) Can we characterize $\gamma$ for which $M_{\gamma}$ has maximum volume? Let's consider the case where $\gamma$ is given. We can partition $M_{\gamma}$ into 1) regions of pure tension, where $d\phi^Td\phi = I$; in these regions $M_{\gamma}$ is, by definition, developable; 2) regions where one direction is in tension and one in compression, $\|d\phi^Td\phi\|_2 = 1$ but $\det d\phi^Td\phi We need not consider $\|d\phi^Td\phi\|_2 Let us look at the regions of type (2). We can trace on these regions a family of curves $\tau$ along which $\phi$ is an isometry. Since $M_{\gamma}$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_{\gamma}$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $\tau$. In other words, for some stress $\sigma$ constant along each $\tau$, at all points $\tau(s)$ along $\tau$ we have $$\hat{n} = \sigma \tau''(s)$$ where $\hat{n}$ the surface normal; it follows that (1) the $\tau$ follow geodesics on $M_{\gamma}$, (2) each $\tau$ has constant curvature. The only thing I can say about problem (II) is that for the optimal $\gamma$, the surface $M_\gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $\approx 1.236$ liters, less than Joriki's solution). I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $\gamma$ and $M_{\gamma}$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters). EDIT 2: A sketch of the orientation of $\tau$ on the surface:
| 2026-01-26T11:31:30.700501
|
61,497
|
Why are rings called rings?
|
I've done some search in Internet and other sources about this question. Why the name ring to this particular object? Just curiosity. Thanks.
| 269
|
abstract-algebra, ring-theory, terminology, math-history
|
https://math.stackexchange.com/questions/61497/why-are-rings-called-rings
|
The name "ring" is derived from Hilbert's term "Zahlring" (number ring), introduced in his Zahlbericht for certain rings of algebraic integers. As for why Hilbert chose the name "ring", I recall reading speculations that it may have to do with cyclical (ring-shaped) behavior of powers of algebraic integers. Namely, if $\:\alpha\:$ is an algebraic integer of degree $\rm\:n\:$ then $\:\alpha^n\:$ is a $\rm\:\mathbb Z$-linear combination of lower powers of $\rm\:\alpha\:,\:$ thus so too are all higher powers of $\rm\:\alpha\:.\:$ Hence all powers cycle back onto $\rm\:1,\:\alpha,\:,\ldots,\alpha^{n-1}\:,\:$ i.e. $\rm\:\mathbb Z[\alpha]\:$ is a finitely generated $\:\mathbb Z$-module. Possibly also the motivation for the name had to do more specifically with rings of cyclotomic integers. However, as plausible as that may seem, I don't recall the existence of any historical documents that provide solid evidence in support of such speculations. Beware that one has to be very careful when reading such older literature. Some authors mistakenly read modern notions into terms which have no such denotation in their original usage. To provide some context I recommend reading Lemmermeyer and Schappacher's Introduction to the English Edition of Hilbert’s Zahlbericht. Below is a pertinent excerpt. Below is an excerpt from Leo Corry's Modern algebra and the rise of mathematical structures, p. 149. Below are a couple typical examples of said speculative etymology of the term "ring" via the "circling back" nature of integral dependence, from Harvey Cohn's Advanced Number Theory, p. 49. $\quad$The designation of the letter $\mathfrak D$ for the integral domain has some historical importance going back to Gauss's work on quadratic forms. Gauss $\left(1800\right)$ noted that for certain quadratic forms $Ax^2+Bxy+Cy^2$ the discriminant need not be square-free, although $A$, $B$, $C$ are relatively prime. For example, $x^2-45y^2$ has $D=4\cdot45$. The $4$ was ignored for the reason that $4|D$ necessarily by virtue of Gauss's requirement that $B$ be even, but the factor of $3^2$ in $D$ caused Gauss to refer to the form as one of "order $3$." Eventually, the forms corresponding to a value of $D$ were called an "order" (Ordnung). Dedekind retained this word for what is here called an "integral domain." $\quad$The term "ring" is a contraction of "Zahlring" introduced by Hilbert $\left(1892\right)$ to denote (in our present context) the ring generated by the rational integers and a quadratic integer $\eta$ defined by $$\eta^2+B\eta+C=0.$$ It would seem that module $\left[1,\eta\right]$ is called a Zahlring because $\eta^2$ equals $-B\eta-C$ "circling directly back" to an element of $\left[1,\eta\right]$ . This word has been maintained today. Incidentally, every Zahlring is an integral domain and the converse is true for quadratic fields. and from Rotman's Advanced Modern Algebra, p. 81.
| 256
| true
|
Q: Why are rings called rings?
I've done some search in Internet and other sources about this question. Why the name ring to this particular object? Just curiosity. Thanks.
A: The name "ring" is derived from Hilbert's term "Zahlring" (number ring), introduced in his Zahlbericht for certain rings of algebraic integers. As for why Hilbert chose the name "ring", I recall reading speculations that it may have to do with cyclical (ring-shaped) behavior of powers of algebraic integers. Namely, if $\:\alpha\:$ is an algebraic integer of degree $\rm\:n\:$ then $\:\alpha^n\:$ is a $\rm\:\mathbb Z$-linear combination of lower powers of $\rm\:\alpha\:,\:$ thus so too are all higher powers of $\rm\:\alpha\:.\:$ Hence all powers cycle back onto $\rm\:1,\:\alpha,\:,\ldots,\alpha^{n-1}\:,\:$ i.e. $\rm\:\mathbb Z[\alpha]\:$ is a finitely generated $\:\mathbb Z$-module. Possibly also the motivation for the name had to do more specifically with rings of cyclotomic integers. However, as plausible as that may seem, I don't recall the existence of any historical documents that provide solid evidence in support of such speculations. Beware that one has to be very careful when reading such older literature. Some authors mistakenly read modern notions into terms which have no such denotation in their original usage. To provide some context I recommend reading Lemmermeyer and Schappacher's Introduction to the English Edition of Hilbert’s Zahlbericht. Below is a pertinent excerpt. Below is an excerpt from Leo Corry's Modern algebra and the rise of mathematical structures, p. 149. Below are a couple typical examples of said speculative etymology of the term "ring" via the "circling back" nature of integral dependence, from Harvey Cohn's Advanced Number Theory, p. 49. $\quad$The designation of the letter $\mathfrak D$ for the integral domain has some historical importance going back to Gauss's work on quadratic forms. Gauss $\left(1800\right)$ noted that for certain quadratic forms $Ax^2+Bxy+Cy^2$ the discriminant need not be square-free, although $A$, $B$, $C$ are relatively prime. For example, $x^2-45y^2$ has $D=4\cdot45$. The $4$ was ignored for the reason that $4|D$ necessarily by virtue of Gauss's requirement that $B$ be even, but the factor of $3^2$ in $D$ caused Gauss to refer to the form as one of "order $3$." Eventually, the forms corresponding to a value of $D$ were called an "order" (Ordnung). Dedekind retained this word for what is here called an "integral domain." $\quad$The term "ring" is a contraction of "Zahlring" introduced by Hilbert $\left(1892\right)$ to denote (in our present context) the ring generated by the rational integers and a quadratic integer $\eta$ defined by $$\eta^2+B\eta+C=0.$$ It would seem that module $\left[1,\eta\right]$ is called a Zahlring because $\eta^2$ equals $-B\eta-C$ "circling directly back" to an element of $\left[1,\eta\right]$ . This word has been maintained today. Incidentally, every Zahlring is an integral domain and the converse is true for quadratic fields. and from Rotman's Advanced Modern Algebra, p. 81.
| 2026-01-26T11:31:32.450708
|
2,821,112
|
"Integral milking": working backward to construct nontrivial integrals
|
I begin this post with a plea: please don't be too harsh with this post for being off topic or vague. It's a question about something I find myself doing as a mathematician, and wonder whether others do it as well. It is a soft question about recreational mathematics - in reality, I'm shooting for more of a conversation. I know that a lot of users on this site (e.g. Cleo, Jack D'Aurizio, and so on) are really good at figuring out crafty ways of solving recreational definite integrals, like $$\int_{\pi/2}^{\pi} \frac{x\sin(x)}{5-4\cos(x)} \, dx$$ or $$\int_0^\infty \bigg(\frac{x-1}{\ln^2(x)}-\frac{1}{\ln(x)}\bigg)\frac{dx}{x^2+1}$$ When questions like this pop up on MSE, the OP provides an integral to evaluate, and the answerers can evaluate it using awesome tricks including (but certainly not limited to): Clever substitution Exploitation of symmetry in the integrand Integration by parts Expanding the integrand as a series Differentiating a well-know integral-defined function, like the Gamma or Beta functions Taking Laplace and Inverse Laplace transforms But when I play around with integrals on my own, I don't always have a particular problem to work on. Instead, I start with a known integral, like $$\int_0^\pi \cos(mx)\cos(nx) \, dx=\frac{\pi}{2}\delta_{mn},\space\space \forall m,n\in \mathbb Z^+$$ and "milk" it, for lack of a better word, to see how many other obscure, rare, or aesthetically pleasing integrals I can derive from it using some of the above techniques. For example, using the above integral, one might divide both sides by $m$, getting $$\int_0^\pi \frac{\cos(mx)}{m}\cos(nx) \, dx=\frac{\pi}{2m}\delta_{mn},\space\space \forall m,n,k\in \mathbb Z^+$$ Then, summing both sides from $m=1$ to $\infty$, and exploiting a well-known Fourier Series, obtain $$\int_0^\pi \cos(nx)\ln(2-2\cos(x)) \, dx=-\frac{\pi}{n},\space\space \forall n\in \mathbb Z^+$$ or, after a bit of algebra, the aesthetically pleasing result $$\int_0^{\pi/2} \cos(2nx)\ln(\sin(x)) \, dx=-\frac{\pi}{4n},\space\space \forall n\in \mathbb Z^+$$ After pulling a trick like this, I look through all of my notebooks and integral tables for other known integrals on which I can get away with the same trick, just to see what integrals I can "milk" out of them in the same way. This is just an example - even using the same starting integral, countless others can be obtained by using other Fourier Series, Power Series, integral identities, etc. For example, some integrals derived from the very same starting integral include $$\int_0^\pi \frac{\cos(nx)}{q-\cos(x)} \, dx=\frac{\pi(q-\sqrt{q^2-1})^{n+1}}{1-q^2+q\sqrt{q^2-1}}$$ $$\int_0^\pi \frac{dx}{(1+a^2-2a\cos(x))(1+b^2-2b\cos(mx))}=\frac{\pi(1+a^m b)}{(1-a^2)(1-b^2)(1-a^m b)}$$ and the astounding identity $$\int_0^{\pi/2}\ln{\lvert\sin(mx)\rvert}\cdot \ln{\lvert\sin(nx)\rvert}\, dx=\frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}$$ Everyone seems to be curious about the proof of this last identity. A proof can be found in my answer here. I just pick a starting integral, and using every technique I know as many times as possible, try to come up with the most exotic integrals as I can, rather than picking a specific integral and trying to solve it. Of course, integrals generated this way would be poor (or at least extremely difficult) candidates for contest problems or puzzles to evaluate given the integral, since they are derived "backwards," and determining the derivation given the integral is likely much harder than pursuing the vague goal of a "nice-looking integral" with no objective objective (ha ha). QUESTION: Do you (residents of MSE who regularly answer/pose recreational definite integral questions) do this same activity, in which you try to generate, rather than solve, cool integrals? If so, what are some integrals you have come up with in this way? What strategies do you use? Does anyone care to opine on the value (or perhaps lack of value) of seeking integrals in this way? Cheers!
| 269
|
integration, definite-integrals, soft-question, recreational-mathematics, big-list
|
https://math.stackexchange.com/questions/2821112/integral-milking-working-backward-to-construct-nontrivial-integrals
|
Yes, definitely. For example, I found that $$ m\int_0^{\infty} y^{\alpha} e^{-y}(1-e^{-y})^{m-1} \, dy = \Gamma(\alpha+1) \sum_{k \geq 1} (-1)^{k-1} \binom{m}{k} \frac{1}{k^{\alpha}} $$ (and related results for particular values of $\alpha$) while mucking about with some integrals. Months later, I was reading a paper about a particular regularisation scheme (loop regularisation) useful in particle physics, and was rather surprised to recognise the sum on the right! I was then able to use the integral to prove that such sums have a particular asymptotic that was required for the theory to actually work as intended, which the original author had verified numerically but not proved. The resulting paper's on arXiv here. Never let it be said that mucking about with integrals is a pointless pursuit!
| 170
| false
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Q: "Integral milking": working backward to construct nontrivial integrals
I begin this post with a plea: please don't be too harsh with this post for being off topic or vague. It's a question about something I find myself doing as a mathematician, and wonder whether others do it as well. It is a soft question about recreational mathematics - in reality, I'm shooting for more of a conversation. I know that a lot of users on this site (e.g. Cleo, Jack D'Aurizio, and so on) are really good at figuring out crafty ways of solving recreational definite integrals, like $$\int_{\pi/2}^{\pi} \frac{x\sin(x)}{5-4\cos(x)} \, dx$$ or $$\int_0^\infty \bigg(\frac{x-1}{\ln^2(x)}-\frac{1}{\ln(x)}\bigg)\frac{dx}{x^2+1}$$ When questions like this pop up on MSE, the OP provides an integral to evaluate, and the answerers can evaluate it using awesome tricks including (but certainly not limited to): Clever substitution Exploitation of symmetry in the integrand Integration by parts Expanding the integrand as a series Differentiating a well-know integral-defined function, like the Gamma or Beta functions Taking Laplace and Inverse Laplace transforms But when I play around with integrals on my own, I don't always have a particular problem to work on. Instead, I start with a known integral, like $$\int_0^\pi \cos(mx)\cos(nx) \, dx=\frac{\pi}{2}\delta_{mn},\space\space \forall m,n\in \mathbb Z^+$$ and "milk" it, for lack of a better word, to see how many other obscure, rare, or aesthetically pleasing integrals I can derive from it using some of the above techniques. For example, using the above integral, one might divide both sides by $m$, getting $$\int_0^\pi \frac{\cos(mx)}{m}\cos(nx) \, dx=\frac{\pi}{2m}\delta_{mn},\space\space \forall m,n,k\in \mathbb Z^+$$ Then, summing both sides from $m=1$ to $\infty$, and exploiting a well-known Fourier Series, obtain $$\int_0^\pi \cos(nx)\ln(2-2\cos(x)) \, dx=-\frac{\pi}{n},\space\space \forall n\in \mathbb Z^+$$ or, after a bit of algebra, the aesthetically pleasing result $$\int_0^{\pi/2} \cos(2nx)\ln(\sin(x)) \, dx=-\frac{\pi}{4n},\space\space \forall n\in \mathbb Z^+$$ After pulling a trick like this, I look through all of my notebooks and integral tables for other known integrals on which I can get away with the same trick, just to see what integrals I can "milk" out of them in the same way. This is just an example - even using the same starting integral, countless others can be obtained by using other Fourier Series, Power Series, integral identities, etc. For example, some integrals derived from the very same starting integral include $$\int_0^\pi \frac{\cos(nx)}{q-\cos(x)} \, dx=\frac{\pi(q-\sqrt{q^2-1})^{n+1}}{1-q^2+q\sqrt{q^2-1}}$$ $$\int_0^\pi \frac{dx}{(1+a^2-2a\cos(x))(1+b^2-2b\cos(mx))}=\frac{\pi(1+a^m b)}{(1-a^2)(1-b^2)(1-a^m b)}$$ and the astounding identity $$\int_0^{\pi/2}\ln{\lvert\sin(mx)\rvert}\cdot \ln{\lvert\sin(nx)\rvert}\, dx=\frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}$$ Everyone seems to be curious about the proof of this last identity. A proof can be found in my answer here. I just pick a starting integral, and using every technique I know as many times as possible, try to come up with the most exotic integrals as I can, rather than picking a specific integral and trying to solve it. Of course, integrals generated this way would be poor (or at least extremely difficult) candidates for contest problems or puzzles to evaluate given the integral, since they are derived "backwards," and determining the derivation given the integral is likely much harder than pursuing the vague goal of a "nice-looking integral" with no objective objective (ha ha). QUESTION: Do you (residents of MSE who regularly answer/pose recreational definite integral questions) do this same activity, in which you try to generate, rather than solve, cool integrals? If so, what are some integrals you have come up with in this way? What strategies do you use? Does anyone care to opine on the value (or perhaps lack of value) of seeking integrals in this way? Cheers!
A: Yes, definitely. For example, I found that $$ m\int_0^{\infty} y^{\alpha} e^{-y}(1-e^{-y})^{m-1} \, dy = \Gamma(\alpha+1) \sum_{k \geq 1} (-1)^{k-1} \binom{m}{k} \frac{1}{k^{\alpha}} $$ (and related results for particular values of $\alpha$) while mucking about with some integrals. Months later, I was reading a paper about a particular regularisation scheme (loop regularisation) useful in particle physics, and was rather surprised to recognise the sum on the right! I was then able to use the integral to prove that such sums have a particular asymptotic that was required for the theory to actually work as intended, which the original author had verified numerically but not proved. The resulting paper's on arXiv here. Never let it be said that mucking about with integrals is a pointless pursuit!
| 2026-01-26T11:31:34.676562
|
2,217,248
|
"Which answer in this list is the correct answer to this question?"
|
I received this question from my mathematics professor as a leisure-time logic quiz, and although I thought I answered it right, he denied. Can someone explain the reasoning behind the correct solution? Which answer in this list is the correct answer to this question? All of the below. None of the below. All of the above. One of the above. None of the above. None of the above. I thought: $2$ and $3$ contradict so $1$ cannot be true. $2$ denies $3$ but $3$ affirms $2,$ so $3$ cannot be true $2$ denies $4,$ but as $1$ and $3$ are proven to be false, $4$ cannot be true. $6$ denies $5$ but not vice versa, so $5$ cannot be true. at this point only $2$ and $6$ are left to be considered. I thought choosing $2$ would not deny $1$ (and it can't be all of the below and none of the below) hence I thought the answer is $6.$ I don't know the correct answer to the question. Thanks!
| 268
|
logic, soft-question, puzzle
|
https://math.stackexchange.com/questions/2217248/which-answer-in-this-list-is-the-correct-answer-to-this-question
|
// gcc ImpredictivePropositionalLogic1.c -o ImpredictivePropositionalLogic1.exe -std=c99 -Wall -O3 /* Which answer in this list is the correct answer to this question? (a) All of the below. (b) None of the below. (c) All of the above. (d) One of the above. (e) None of the above. (f) None of the above. */ #include #define iff(x, y) ((x)==(y)) int main() { printf("a b c d e f\n"); for (int a = 0; a This outputs: a b c d e f 0 0 0 0 1 0 The main point I'd like to get across is that you cannot assume at the outset that there is only 1 satisfying assignment. For example consider the question: Which of the following is true? (a) both of these (b) both of these You might be tempted to say that both (a) and (b) are true. But it is also consistent that both (a) and (b) are false. The tendency to assume singularity from definitions isn't correct when the definitions are impredictive.
| 230
| false
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Q: "Which answer in this list is the correct answer to this question?"
I received this question from my mathematics professor as a leisure-time logic quiz, and although I thought I answered it right, he denied. Can someone explain the reasoning behind the correct solution? Which answer in this list is the correct answer to this question? All of the below. None of the below. All of the above. One of the above. None of the above. None of the above. I thought: $2$ and $3$ contradict so $1$ cannot be true. $2$ denies $3$ but $3$ affirms $2,$ so $3$ cannot be true $2$ denies $4,$ but as $1$ and $3$ are proven to be false, $4$ cannot be true. $6$ denies $5$ but not vice versa, so $5$ cannot be true. at this point only $2$ and $6$ are left to be considered. I thought choosing $2$ would not deny $1$ (and it can't be all of the below and none of the below) hence I thought the answer is $6.$ I don't know the correct answer to the question. Thanks!
A: // gcc ImpredictivePropositionalLogic1.c -o ImpredictivePropositionalLogic1.exe -std=c99 -Wall -O3 /* Which answer in this list is the correct answer to this question? (a) All of the below. (b) None of the below. (c) All of the above. (d) One of the above. (e) None of the above. (f) None of the above. */ #include #define iff(x, y) ((x)==(y)) int main() { printf("a b c d e f\n"); for (int a = 0; a This outputs: a b c d e f 0 0 0 0 1 0 The main point I'd like to get across is that you cannot assume at the outset that there is only 1 satisfying assignment. For example consider the question: Which of the following is true? (a) both of these (b) both of these You might be tempted to say that both (a) and (b) are true. But it is also consistent that both (a) and (b) are false. The tendency to assume singularity from definitions isn't correct when the definitions are impredictive.
| 2026-01-26T11:31:36.502082
|
94,827
|
What books must every math undergraduate read?
|
I'm still a student, but the same books keep getting named by my tutors (Rudin, Royden). I've read Baby Rudin and begun Royden though I'm unsure if there are other books that I "should" be working on if I want to study beyond Masters. I'm not there yet as I'm on a four year course and had a gap year between Years 3 and 4. Please recommend for Algebra, Linear Algebra and Categories - Analysis, Set Theory, Measure theory (an area I have seen too little books dedicated for). E.g. Spivak is very good for self learning basic real analysis, but Rudin really cuts to the heart.
| 264
|
reference-request, soft-question, big-list, book-recommendation
|
https://math.stackexchange.com/questions/94827/what-books-must-every-math-undergraduate-read
|
EDIT: I now think that this list is long enough that I shall be maintaining it over time--updating it whenever I use a new book/learn a new subject. While every suggestion below should be taken with a grain of salt--I will say that I spend a huge amount of time sifting through books to find the ones that conform best to my (and hopefully your!) learning style. Here is my two cents (for whatever that's worth). I tried to include all the topics I could imagine you could want to know at this point. I hope I picked the right level of difficult. Feel absolutely free to ask my specific opinion about any book. Basic Analysis: Rudin--Apostol Measure Theory: Royden (only if you get the newest fourth edition)--Folland General Algebra: D&F--Rotman--Lang--Grillet Finite Group Theory: Isaacs-- Kurzweil General Group Theory: Robinson--Rotman Ring Theory: T.Y. Lam-- times two Commutative Algebra: Eisenbud--A&M--Reid Homological Algebra: Weibel--Rotman--Vermani Category Theory: Mac Lane--Adamek et. al--Berrick et. al--Awodey--Mitchell Linear Algebra: Roman--Hoffman and Kunze--Golan Field Theory: Morandi--Roman Complex Analysis: Ahlfors--Cartan--Freitag Riemann Surfaces: Varolin(great first read, can be a little sloppy though)--Freitag(overall great book for a second course in complex analysis!)--Forster(a little more old school, and with a slightly more algebraic bend then a differential geometric one)--Donaldson SCV: Gunning et. al--Ebeling Point-set Topology: Munkres--Steen et. al--Kelley Differential Topology: Pollack et. al--Milnor--Lee Algebraic Topology: Bredon--May-- Bott and Tu (great, great book)--Rotman--Massey--Tom Dieck Differential Geometry: Do Carmo--Spivak--Jost--Lee Representation Theory of Finite Groups: Serre--Steinberg--Liebeck--Isaacs General Representation Theory: Fulton and Harris--Humphreys--Hall Representation Theory of Compact Groups: Tom Dieck et. al--Sepanski (Linear) Algebraic Groups: Springer--Humphreys "Elementary" Number Theory: Niven et. al--Ireland et. al Algebraic Number Theory: Ash--Lorenzini--Neukirch--Marcus--Washington Fourier Analysis--Katznelson Modular Forms: Diamond and Shurman--Stein Local Fields: Lorenz and Levy--Read chapters 23,24,25. This is by far my favorite quick reference, as well as "learning text" for the basics of local fields one needs to break into other topics (e.g. class field theory). Serre--This is the classic book. It is definitely low on the readability side, especially notationally. It also has a tendency to consider things in more generality than is needed at a first go. This isn't bad, but is not good if you're trying to "brush up" or quickly learn local fields for another subject. Fesenko et. al--A balance between 1. and 2. Definitely more readable than 2., but more comprehensive than 1. If you are wondering whether or not so-and-so needs Henselian, this is the place I'd check. Iwasawa--A great place to learn the bare-bones of what one might need to learn class field theory. I am referencing, in particular, the first three chapters. If you are dead-set on JUST learning what you need to, this is a pretty good reference, but if you're likely to wonder about why so-and-so theorem is true, or get a broader understanding of the basics of local fields, I recommend 1. Class Field Theory: Lorenz and Levy--Read chapters 28-32, second only to Iwasawa, but with a different flavor (cohomological vs. formal group laws) Tate and Artin--The classic book. A little less readable then any of the alternatives here. Childress--Focused mostly on the global theory opposed to the local. Actually deduces local at the end as a result of global. Thus, very old school. Iwasawa (read the rest of it!) Milne--Where I first started learning it. Very good, but definitely roughly hewn. A lot of details are left out, and he sometimes forgets to tell you where you are going. Metric Groups: Markley Algebraic Geometry: Reid--Shafarevich--Hartshorne--Griffiths and Harris--Mumford
| 233
| true
|
Q: What books must every math undergraduate read?
I'm still a student, but the same books keep getting named by my tutors (Rudin, Royden). I've read Baby Rudin and begun Royden though I'm unsure if there are other books that I "should" be working on if I want to study beyond Masters. I'm not there yet as I'm on a four year course and had a gap year between Years 3 and 4. Please recommend for Algebra, Linear Algebra and Categories - Analysis, Set Theory, Measure theory (an area I have seen too little books dedicated for). E.g. Spivak is very good for self learning basic real analysis, but Rudin really cuts to the heart.
A: EDIT: I now think that this list is long enough that I shall be maintaining it over time--updating it whenever I use a new book/learn a new subject. While every suggestion below should be taken with a grain of salt--I will say that I spend a huge amount of time sifting through books to find the ones that conform best to my (and hopefully your!) learning style. Here is my two cents (for whatever that's worth). I tried to include all the topics I could imagine you could want to know at this point. I hope I picked the right level of difficult. Feel absolutely free to ask my specific opinion about any book. Basic Analysis: Rudin--Apostol Measure Theory: Royden (only if you get the newest fourth edition)--Folland General Algebra: D&F--Rotman--Lang--Grillet Finite Group Theory: Isaacs-- Kurzweil General Group Theory: Robinson--Rotman Ring Theory: T.Y. Lam-- times two Commutative Algebra: Eisenbud--A&M--Reid Homological Algebra: Weibel--Rotman--Vermani Category Theory: Mac Lane--Adamek et. al--Berrick et. al--Awodey--Mitchell Linear Algebra: Roman--Hoffman and Kunze--Golan Field Theory: Morandi--Roman Complex Analysis: Ahlfors--Cartan--Freitag Riemann Surfaces: Varolin(great first read, can be a little sloppy though)--Freitag(overall great book for a second course in complex analysis!)--Forster(a little more old school, and with a slightly more algebraic bend then a differential geometric one)--Donaldson SCV: Gunning et. al--Ebeling Point-set Topology: Munkres--Steen et. al--Kelley Differential Topology: Pollack et. al--Milnor--Lee Algebraic Topology: Bredon--May-- Bott and Tu (great, great book)--Rotman--Massey--Tom Dieck Differential Geometry: Do Carmo--Spivak--Jost--Lee Representation Theory of Finite Groups: Serre--Steinberg--Liebeck--Isaacs General Representation Theory: Fulton and Harris--Humphreys--Hall Representation Theory of Compact Groups: Tom Dieck et. al--Sepanski (Linear) Algebraic Groups: Springer--Humphreys "Elementary" Number Theory: Niven et. al--Ireland et. al Algebraic Number Theory: Ash--Lorenzini--Neukirch--Marcus--Washington Fourier Analysis--Katznelson Modular Forms: Diamond and Shurman--Stein Local Fields: Lorenz and Levy--Read chapters 23,24,25. This is by far my favorite quick reference, as well as "learning text" for the basics of local fields one needs to break into other topics (e.g. class field theory). Serre--This is the classic book. It is definitely low on the readability side, especially notationally. It also has a tendency to consider things in more generality than is needed at a first go. This isn't bad, but is not good if you're trying to "brush up" or quickly learn local fields for another subject. Fesenko et. al--A balance between 1. and 2. Definitely more readable than 2., but more comprehensive than 1. If you are wondering whether or not so-and-so needs Henselian, this is the place I'd check. Iwasawa--A great place to learn the bare-bones of what one might need to learn class field theory. I am referencing, in particular, the first three chapters. If you are dead-set on JUST learning what you need to, this is a pretty good reference, but if you're likely to wonder about why so-and-so theorem is true, or get a broader understanding of the basics of local fields, I recommend 1. Class Field Theory: Lorenz and Levy--Read chapters 28-32, second only to Iwasawa, but with a different flavor (cohomological vs. formal group laws) Tate and Artin--The classic book. A little less readable then any of the alternatives here. Childress--Focused mostly on the global theory opposed to the local. Actually deduces local at the end as a result of global. Thus, very old school. Iwasawa (read the rest of it!) Milne--Where I first started learning it. Very good, but definitely roughly hewn. A lot of details are left out, and he sometimes forgets to tell you where you are going. Metric Groups: Markley Algebraic Geometry: Reid--Shafarevich--Hartshorne--Griffiths and Harris--Mumford
| 2026-01-26T11:31:38.179132
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2,746
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Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?
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If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer. If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$. Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.
| 261
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sequences-and-series, number-theory, harmonic-numbers
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https://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-sum-limits-k-1n-frac1k-is-never-an-int
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Hint $ $ There is a $\rm\color{darkorange}{unique}$ denominator $\rm\,\color{#0a0} {2^K}$ having maximal power of $\:\!2,\,$ so scaling by $\rm\,\color{#c00}{2^{K-1}}$ we deduce a contradiction $\large \rm\, \frac{1}2 = \frac{c}d \,$ with odd $\rm\,d \:$ (vs. $\,\rm d = 2c),\,$ e.g. $$\begin{eqnarray} & &\rm\ \ \ \ \color{0a0}{m} &=&\ \ 1 &+& \frac{1}{2} &+& \frac{1}{3} &+&\, \color{#0a0}{\frac{1}{4}} &+& \frac{1}{5} &+& \frac{1}{6} &+& \frac{1}{7} \\ &\Rightarrow\ &\rm\ \ \color{#c00}{2}\:\!m &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &+&\, \color{#0a0}{\frac{1}{2}} &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M}\\ &\Rightarrow\ & -\color{#0a0}{\frac{1}{2}}\ \ &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &-&\rm \color{#c00}{2}\:\!m &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M} \end{eqnarray}$$ All denom's in the prior fractions are odd so they sum to fraction with odd denom $\rm\,d\, |\, 3\cdot 5\cdot 7$. Note $ $ Said $\rm\color{darkorange}{uniqueness}$ has easy proof: if $\rm\:j\:\! 2^K$ is in the interval $\rm\,[1,n]\,$ then so too is $\,\rm \color{#0a0}{2^K}\! \le\, j\:\!2^K.\,$ But if $\,\rm j\ge 2\,$ then the interval contains $\rm\,2^{K+1}\!= 2\cdot\! 2^K\! \le j\:\!2^K,\,$ contra maximality of $\,\rm K$. The argument is more naturally expressed using valuation theory, but I purposely avoided that because Anton requested an "elementary" solution. The above proof can easily be made comprehensible to a high-school student. Generally we can similarly prove that a sum of fractions is nonintegral if the highest power of a prime $\,p\,$ in any denominator occurs in $\rm\color{darkorange}{exactly\ one}$ denominator, e.g. see the Remark here where I explain how it occurs in a trickier multiplicative form (from a contest problem). In valuation theory, this is a special case of a basic result on the valuation of a sum (sometimes called the "dominance lemma" or similar). Another common application occurs when the sum of fractions arises from the evaluation of a polynomial, e.g. see here and its comment.
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| true
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Q: Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?
If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer. If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$. Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.
A: Hint $ $ There is a $\rm\color{darkorange}{unique}$ denominator $\rm\,\color{#0a0} {2^K}$ having maximal power of $\:\!2,\,$ so scaling by $\rm\,\color{#c00}{2^{K-1}}$ we deduce a contradiction $\large \rm\, \frac{1}2 = \frac{c}d \,$ with odd $\rm\,d \:$ (vs. $\,\rm d = 2c),\,$ e.g. $$\begin{eqnarray} & &\rm\ \ \ \ \color{0a0}{m} &=&\ \ 1 &+& \frac{1}{2} &+& \frac{1}{3} &+&\, \color{#0a0}{\frac{1}{4}} &+& \frac{1}{5} &+& \frac{1}{6} &+& \frac{1}{7} \\ &\Rightarrow\ &\rm\ \ \color{#c00}{2}\:\!m &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &+&\, \color{#0a0}{\frac{1}{2}} &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M}\\ &\Rightarrow\ & -\color{#0a0}{\frac{1}{2}}\ \ &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &-&\rm \color{#c00}{2}\:\!m &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M} \end{eqnarray}$$ All denom's in the prior fractions are odd so they sum to fraction with odd denom $\rm\,d\, |\, 3\cdot 5\cdot 7$. Note $ $ Said $\rm\color{darkorange}{uniqueness}$ has easy proof: if $\rm\:j\:\! 2^K$ is in the interval $\rm\,[1,n]\,$ then so too is $\,\rm \color{#0a0}{2^K}\! \le\, j\:\!2^K.\,$ But if $\,\rm j\ge 2\,$ then the interval contains $\rm\,2^{K+1}\!= 2\cdot\! 2^K\! \le j\:\!2^K,\,$ contra maximality of $\,\rm K$. The argument is more naturally expressed using valuation theory, but I purposely avoided that because Anton requested an "elementary" solution. The above proof can easily be made comprehensible to a high-school student. Generally we can similarly prove that a sum of fractions is nonintegral if the highest power of a prime $\,p\,$ in any denominator occurs in $\rm\color{darkorange}{exactly\ one}$ denominator, e.g. see the Remark here where I explain how it occurs in a trickier multiplicative form (from a contest problem). In valuation theory, this is a special case of a basic result on the valuation of a sum (sometimes called the "dominance lemma" or similar). Another common application occurs when the sum of fractions arises from the evaluation of a polynomial, e.g. see here and its comment.
| 2026-01-26T11:31:39.820446
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255,063
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Why study Algebraic Geometry?
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I'm going to start self-stydying algebraic geometry very soon. So, my question is why do mathematicians study algebraic geometry? What are the types of problems in which algebraic geometers are interested in? And what are some of the most beautiful theorems in algebraic geometry?
| 259
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algebraic-geometry, soft-question
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https://math.stackexchange.com/questions/255063/why-study-algebraic-geometry
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NEW ADDITION: a big list of freely available online courses on algebraic geometry, from introduction to advanced topics, has been compiled in this other answer. And a digression on motivation for studying the subject along with a self-learning guide of books is in this new answer. There are other similar questions, above all asking for references for self-studying, whose answers may be helpful: (Undergraduate) Algebraic Geometry Textbook Recomendations. Reference for Algebraic Geometry. Best Algebraic Geometry text book? (other than Hartshorne). My personal recommendation is that you start and get your motivation in the following freely available notes. They are extremely instructive, from the very basics of complex algebraic curves up to schemes and intersection theory with Grothendieck-Riemann-Roch, and prove of some of the theorems I mention below. They are excellent for self-study mixing rigor with many pictures! (sadly, something quite unusual among AG references): Matt Kerr - Lecture Notes Algebraic Geometry III/IV, Washington University in St. Louis. Andreas Gathmann - Class Notes: Algebraic Geometry, University of Kaiserslautern. For a powerful, long and abstract course, suitable for self-study, these notes have become famous: Ravi Vakil - Foundations of Algebraic Geometry, Stanford University. Also, there are many wonderful lecture videos for complete courses on elementary algebraic geometry, algebraic surfaces and beyond, by the one old master: Miles Reid - Lecture Courses on Video (WCU project at Sogang University), where you can really start at a slow pace (following his undergraduate textbook) to get up to the surface classification theorem. Now, Algebraic Geometry is one of the oldest, deepest, broadest and most active subjects in Mathematics with connections to almost all other branches in either a very direct or subtle way. The main motivation started with Pierre de Fermat and René Descartes who realized that to study geometry one could work with algebraic equations instead of drawings and pictures (which is now fundamental to work with higher dimensional objects, since intuition fails there). The most basic equations one could imagine to start studying were polynomials on the coordinates of your plane or space, or in a number field in general, as they are the most basic constructions from the elementary arithmetic operations. Equations of first order, i.e. linear polynomials, are the straight lines, planes, linear subspaces and hyperplanes. Equations of second order turned out to comprise all the classical conic sections; in fact the conics classification in the affine, Euclidean and projective cases (over the real and complex numbers) is the first actual algebraic geometry problem that every student is introduced to: the classification of all possible canonical forms of polynomials of degree 2 (either under affine transformations or isometries in variables $(x,y)$, or projective transformations in homogeneous variables $[x:y:z]$). Thus the basic plane curves over the real numbers can be studied by the algebraic properties of polynomials. Working over the complex numbers is actually more natural, as it is the algebraic closure of the reals and so it simplifies a lot the study tying together the whole subject, thanks to elementary things like the fundamental theorem of algebra and the Hilbert Nullstellensatz. Besides, working within projective varieties, enlarging our ambient space with the points at infinity, also helps since then we are dealing with topologically compact objects and pathological results disappear, e.g. all curves intersect at least at a point, giving the beautiful Bézout's theorem. From a purely practical point of view, one has to realize that all other analytic non-polynomial functions can be approximated by polynomials (e.g. by truncating the series), which is actually what calculators and computers do when computing trigonometric functions for example. So when any software plots a transcendental surface (or manifold), it is actually displaying a polynomial approximation (an algebraic variety). So the study of algebraic geometry in the applied and computational sense is fundamental for the rest of geometry. From a pure mathematics perspective, the case of projective complex algebraic geometry is of central importance. This is because of several results, like Lefschetz's principle by which doing (algebraic) geometry over an algebraically closed field of characteristic $0$ is essentially equivalent to doing it over the complex numbers; furthermore, Chow's theorem guarantees that all projective complex manifolds are actually algebraic, meaning that differential geometry deals with the same objects as algebraic geometry in that case, i.e. complex projective manifolds are given by the zero locus of a finite number of homogeneous polynomials! This was strengthened by Jean-Pierre Serre's GAGA theorems, which unified and equated the study of analytic geometry with algebraic geometry in a very general setting. Besides, in the case of projective complex algebraic curves one is actually working with compact orientable real surfaces (since these always admit a holomorphic structure), therefore unifying the theory of compact Riemann surfaces of complex analysis with the differential geometry of real surfaces, the algebraic topology of 2-manifolds and the algebraic geometry of algebraic curves! Here one finds wonderful relations and deep results like all the consequences of the concept of degree, index and curvature, linking together the milestone theorems of Gauß-Bonnet, Poincaré-Hopf and Riemann-Roch theorem! In fact the principal classification of algebraic curves is given in terms of their genus which is an invariant proved to be the same in the different perspectives: the topological genus of number of doughnut holes, the arithmetic genus of the Hilbert polynomial of the algebraic curve and the geometric genus as the number of independent holomorphic differential 2-forms over the Riemann surface. Analogously, the study of real 4-manifolds in differential geometry and differential topology is of central importance in mathematics per se but also in theoretical and mathematical physics, for example in gauge theory, so the study of complex algebraic surfaces gives results and provides tools. The full birational classification of algebraic surfaces was worked out decades ago in the Kodaira-Enriques theorem and served as a starting point to Mori's minimal model program to birationally classify all higher-dimensional (projective) complex algebraic varieties. A fundamental difference with other types of geometry is the presence of singularities, which play a very important role in algebraic geometry as many of the obstacles are due to them, but the fundamental Hironaka's resolution theorem guarantees that, at least in characteristic zero, varieties always have a smooth birational model. Also the construction and study of moduli spaces of types of geometric objects is a very important topic (e.g. Deligne-Mumford construction), since the space of all such objects is often an algebraic-geometric object itself. There are also many interesting problems and results in enumerative geometry and intersection theory, starting from the classic and amazing Cayley-Salmon theorem that all smooth cubic surfaces defined over an algebraic closed field contain exactly 27 straight lines, the Thom-Porteus formula for degeneracy loci, Schubert calculus up to modern quantum cohomology with Kontsevich's and ELSV formulas; Torelli's theorem on the reconstruction of algebraic curves from their Jacobian variety, and finally the cornerstone (Grothendieck)-Hirzebruch-Riemann-Roch theorem computing the number of independent global sections of vector bundles, actually their Euler-Poincaré characteristics, by the intersection numbers of generic zero loci of characteristic classes over the variety. Besides all this, since the foundational immense work of Alexandre Grothendieck, the subject has got very solid and abstract foundations so powerful to fuse algebraic geometry with number theory, as many were hoping before. Thus, the abstract algebraic geometry of sheaves and schemes plays nowadays a fundamental role in algebraic number theory disguised as arithmetic geometry. Wondeful results in Diophantine geometry like Faltings theorem and Mordell-Weil theorem made use of all these advances, along with the famous proof of Wiles of Fermat's last theorem. The development of abstract algebraic geometry was more or less motivated to solve the remarkable Weil conjectures relating the number of solutions of polynomials over finite number fields to the geometry of the complex variety defined by the same polynomials. For this, tremendous machinery was worked out, like étale cohomology. Also, trying to apply complex geometry constructions to arithmetic has led to Arakelov geometry and the arithmetic Grothendieck-Riemann-Roch among other results. Related to arithmetic geometry, thanks to schemes, there has emerged a new subject of arithmetic topology, where properties of the prime numbers and algebraic number theory have relationships and dualities with the theory of knots, links and 3-dimensional manifolds! This is a very mysterious and interesting new topic, since knots and links also appear in theoretical physics (e.g. topological quantum field theories). Also, anabelian geometry interestingly has led the way to studies on the relationships between the topological fundamental group of algebraic varieties and the Galois groups of arithmetic number field extensions. So, mathematicians study algebraic geometry because it is at the core of many subjects, serving as a bridge between seemingly different disciplines: from geometry and topology to complex analysis and number theory. Since in the end, any mathematical subject works within specified algebras, studying the geometry those algebras define is a useful tool and interesting endeavor in itself. In fact, the requirement of being commutative algebras has been dropped since the work of Alain Connes and the whole 'new' subject of noncommmutative geometry has flourished, in analytic and algebraic styles, to try to complete the geometrization of mathematics. On the other hand it attempts to give a quantum counterpart to classical geometries, something of extreme interest in fundamental physics (complex algebraic geometry and noncommutative geometry appear almost necessarily in one way or another in any attempt to unify the fundamental forces with gravity, i.e. quantum field theory with general relativity; even abstract and categorical algebraic geometry play a role in topics like homological mirror symmetry and quantum cohomology, which originated in physics). Therefore, the kind of problems mathematicians try to solve in algebraic geometry are related to much of everything else, mostly: anything related to the classification (as fine as possible) of algebraic varieties (and schemes, maybe someday), their invariants, singularities, deformations and moduli spaces, intersections, their topology and differential geometry, and framing arithmetic problems in terms of geometry. There are many interesting open problems: Birational minimal model program for all varieties, Hodge conjecture, Jacobian conjecture, Hartshorne's conjecture, General Griffiths conjecture, Fujita's conjecture, Linearization and cancelation conjectures, Coolidge-Nagata conjecture, Resolution of singularities in nonzero characteristic, Grothendieck's standard conjectures on algebraic cycles, Grothendieck's anabelian section conjecture, Classification of vector bundles over projective spaces, Unirationality of moduli spaces of curves, Unirationality of rationally connected varieties, Full rigorous formalization of mirror symmetry and quantum cohomology, Full theory of a universal cohomology and mixed motives (e.g. Voevodsky vanishing conjecture). In my personal case, I started as a theoretical physicists but switched completely to pure mathematics because of algebraic geometry, and I also began by self-learning. It is a very deep subject with connections to almost everything else, once one has learned enough to realize that. It is also a very demanding field because of the tremendous background one has to master, in commutative and homological algebra for example, before being able to get to the most modern and interesting results. The effort nevertheless pays off! In fact, the route through commutative algebra actually paves the way not only to algebraic geometry but to algebraic number theory and arithmetic geometry. I had a strong background in differential geometry so I arrived at algebraic geometry through complex (Kähler) geometry, and ended up fascinated by even the most abstract incarnations of it. "Algebraic geometry seems to have acquired the reputation of being esoteric, exclusive, and very abstract, with adherents who are secretly plotting to take over all the rest of mathematics. In one respect this last point is accurate..." - David Mumford. So the question could be instead "why not study algebraic geometry!?" I hope this answer motivates you enough to dive into this deep ocean of the mathematical world and to corroborate it yourself. Best luck!
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Q: Why study Algebraic Geometry?
I'm going to start self-stydying algebraic geometry very soon. So, my question is why do mathematicians study algebraic geometry? What are the types of problems in which algebraic geometers are interested in? And what are some of the most beautiful theorems in algebraic geometry?
A: NEW ADDITION: a big list of freely available online courses on algebraic geometry, from introduction to advanced topics, has been compiled in this other answer. And a digression on motivation for studying the subject along with a self-learning guide of books is in this new answer. There are other similar questions, above all asking for references for self-studying, whose answers may be helpful: (Undergraduate) Algebraic Geometry Textbook Recomendations. Reference for Algebraic Geometry. Best Algebraic Geometry text book? (other than Hartshorne). My personal recommendation is that you start and get your motivation in the following freely available notes. They are extremely instructive, from the very basics of complex algebraic curves up to schemes and intersection theory with Grothendieck-Riemann-Roch, and prove of some of the theorems I mention below. They are excellent for self-study mixing rigor with many pictures! (sadly, something quite unusual among AG references): Matt Kerr - Lecture Notes Algebraic Geometry III/IV, Washington University in St. Louis. Andreas Gathmann - Class Notes: Algebraic Geometry, University of Kaiserslautern. For a powerful, long and abstract course, suitable for self-study, these notes have become famous: Ravi Vakil - Foundations of Algebraic Geometry, Stanford University. Also, there are many wonderful lecture videos for complete courses on elementary algebraic geometry, algebraic surfaces and beyond, by the one old master: Miles Reid - Lecture Courses on Video (WCU project at Sogang University), where you can really start at a slow pace (following his undergraduate textbook) to get up to the surface classification theorem. Now, Algebraic Geometry is one of the oldest, deepest, broadest and most active subjects in Mathematics with connections to almost all other branches in either a very direct or subtle way. The main motivation started with Pierre de Fermat and René Descartes who realized that to study geometry one could work with algebraic equations instead of drawings and pictures (which is now fundamental to work with higher dimensional objects, since intuition fails there). The most basic equations one could imagine to start studying were polynomials on the coordinates of your plane or space, or in a number field in general, as they are the most basic constructions from the elementary arithmetic operations. Equations of first order, i.e. linear polynomials, are the straight lines, planes, linear subspaces and hyperplanes. Equations of second order turned out to comprise all the classical conic sections; in fact the conics classification in the affine, Euclidean and projective cases (over the real and complex numbers) is the first actual algebraic geometry problem that every student is introduced to: the classification of all possible canonical forms of polynomials of degree 2 (either under affine transformations or isometries in variables $(x,y)$, or projective transformations in homogeneous variables $[x:y:z]$). Thus the basic plane curves over the real numbers can be studied by the algebraic properties of polynomials. Working over the complex numbers is actually more natural, as it is the algebraic closure of the reals and so it simplifies a lot the study tying together the whole subject, thanks to elementary things like the fundamental theorem of algebra and the Hilbert Nullstellensatz. Besides, working within projective varieties, enlarging our ambient space with the points at infinity, also helps since then we are dealing with topologically compact objects and pathological results disappear, e.g. all curves intersect at least at a point, giving the beautiful Bézout's theorem. From a purely practical point of view, one has to realize that all other analytic non-polynomial functions can be approximated by polynomials (e.g. by truncating the series), which is actually what calculators and computers do when computing trigonometric functions for example. So when any software plots a transcendental surface (or manifold), it is actually displaying a polynomial approximation (an algebraic variety). So the study of algebraic geometry in the applied and computational sense is fundamental for the rest of geometry. From a pure mathematics perspective, the case of projective complex algebraic geometry is of central importance. This is because of several results, like Lefschetz's principle by which doing (algebraic) geometry over an algebraically closed field of characteristic $0$ is essentially equivalent to doing it over the complex numbers; furthermore, Chow's theorem guarantees that all projective complex manifolds are actually algebraic, meaning that differential geometry deals with the same objects as algebraic geometry in that case, i.e. complex projective manifolds are given by the zero locus of a finite number of homogeneous polynomials! This was strengthened by Jean-Pierre Serre's GAGA theorems, which unified and equated the study of analytic geometry with algebraic geometry in a very general setting. Besides, in the case of projective complex algebraic curves one is actually working with compact orientable real surfaces (since these always admit a holomorphic structure), therefore unifying the theory of compact Riemann surfaces of complex analysis with the differential geometry of real surfaces, the algebraic topology of 2-manifolds and the algebraic geometry of algebraic curves! Here one finds wonderful relations and deep results like all the consequences of the concept of degree, index and curvature, linking together the milestone theorems of Gauß-Bonnet, Poincaré-Hopf and Riemann-Roch theorem! In fact the principal classification of algebraic curves is given in terms of their genus which is an invariant proved to be the same in the different perspectives: the topological genus of number of doughnut holes, the arithmetic genus of the Hilbert polynomial of the algebraic curve and the geometric genus as the number of independent holomorphic differential 2-forms over the Riemann surface. Analogously, the study of real 4-manifolds in differential geometry and differential topology is of central importance in mathematics per se but also in theoretical and mathematical physics, for example in gauge theory, so the study of complex algebraic surfaces gives results and provides tools. The full birational classification of algebraic surfaces was worked out decades ago in the Kodaira-Enriques theorem and served as a starting point to Mori's minimal model program to birationally classify all higher-dimensional (projective) complex algebraic varieties. A fundamental difference with other types of geometry is the presence of singularities, which play a very important role in algebraic geometry as many of the obstacles are due to them, but the fundamental Hironaka's resolution theorem guarantees that, at least in characteristic zero, varieties always have a smooth birational model. Also the construction and study of moduli spaces of types of geometric objects is a very important topic (e.g. Deligne-Mumford construction), since the space of all such objects is often an algebraic-geometric object itself. There are also many interesting problems and results in enumerative geometry and intersection theory, starting from the classic and amazing Cayley-Salmon theorem that all smooth cubic surfaces defined over an algebraic closed field contain exactly 27 straight lines, the Thom-Porteus formula for degeneracy loci, Schubert calculus up to modern quantum cohomology with Kontsevich's and ELSV formulas; Torelli's theorem on the reconstruction of algebraic curves from their Jacobian variety, and finally the cornerstone (Grothendieck)-Hirzebruch-Riemann-Roch theorem computing the number of independent global sections of vector bundles, actually their Euler-Poincaré characteristics, by the intersection numbers of generic zero loci of characteristic classes over the variety. Besides all this, since the foundational immense work of Alexandre Grothendieck, the subject has got very solid and abstract foundations so powerful to fuse algebraic geometry with number theory, as many were hoping before. Thus, the abstract algebraic geometry of sheaves and schemes plays nowadays a fundamental role in algebraic number theory disguised as arithmetic geometry. Wondeful results in Diophantine geometry like Faltings theorem and Mordell-Weil theorem made use of all these advances, along with the famous proof of Wiles of Fermat's last theorem. The development of abstract algebraic geometry was more or less motivated to solve the remarkable Weil conjectures relating the number of solutions of polynomials over finite number fields to the geometry of the complex variety defined by the same polynomials. For this, tremendous machinery was worked out, like étale cohomology. Also, trying to apply complex geometry constructions to arithmetic has led to Arakelov geometry and the arithmetic Grothendieck-Riemann-Roch among other results. Related to arithmetic geometry, thanks to schemes, there has emerged a new subject of arithmetic topology, where properties of the prime numbers and algebraic number theory have relationships and dualities with the theory of knots, links and 3-dimensional manifolds! This is a very mysterious and interesting new topic, since knots and links also appear in theoretical physics (e.g. topological quantum field theories). Also, anabelian geometry interestingly has led the way to studies on the relationships between the topological fundamental group of algebraic varieties and the Galois groups of arithmetic number field extensions. So, mathematicians study algebraic geometry because it is at the core of many subjects, serving as a bridge between seemingly different disciplines: from geometry and topology to complex analysis and number theory. Since in the end, any mathematical subject works within specified algebras, studying the geometry those algebras define is a useful tool and interesting endeavor in itself. In fact, the requirement of being commutative algebras has been dropped since the work of Alain Connes and the whole 'new' subject of noncommmutative geometry has flourished, in analytic and algebraic styles, to try to complete the geometrization of mathematics. On the other hand it attempts to give a quantum counterpart to classical geometries, something of extreme interest in fundamental physics (complex algebraic geometry and noncommutative geometry appear almost necessarily in one way or another in any attempt to unify the fundamental forces with gravity, i.e. quantum field theory with general relativity; even abstract and categorical algebraic geometry play a role in topics like homological mirror symmetry and quantum cohomology, which originated in physics). Therefore, the kind of problems mathematicians try to solve in algebraic geometry are related to much of everything else, mostly: anything related to the classification (as fine as possible) of algebraic varieties (and schemes, maybe someday), their invariants, singularities, deformations and moduli spaces, intersections, their topology and differential geometry, and framing arithmetic problems in terms of geometry. There are many interesting open problems: Birational minimal model program for all varieties, Hodge conjecture, Jacobian conjecture, Hartshorne's conjecture, General Griffiths conjecture, Fujita's conjecture, Linearization and cancelation conjectures, Coolidge-Nagata conjecture, Resolution of singularities in nonzero characteristic, Grothendieck's standard conjectures on algebraic cycles, Grothendieck's anabelian section conjecture, Classification of vector bundles over projective spaces, Unirationality of moduli spaces of curves, Unirationality of rationally connected varieties, Full rigorous formalization of mirror symmetry and quantum cohomology, Full theory of a universal cohomology and mixed motives (e.g. Voevodsky vanishing conjecture). In my personal case, I started as a theoretical physicists but switched completely to pure mathematics because of algebraic geometry, and I also began by self-learning. It is a very deep subject with connections to almost everything else, once one has learned enough to realize that. It is also a very demanding field because of the tremendous background one has to master, in commutative and homological algebra for example, before being able to get to the most modern and interesting results. The effort nevertheless pays off! In fact, the route through commutative algebra actually paves the way not only to algebraic geometry but to algebraic number theory and arithmetic geometry. I had a strong background in differential geometry so I arrived at algebraic geometry through complex (Kähler) geometry, and ended up fascinated by even the most abstract incarnations of it. "Algebraic geometry seems to have acquired the reputation of being esoteric, exclusive, and very abstract, with adherents who are secretly plotting to take over all the rest of mathematics. In one respect this last point is accurate..." - David Mumford. So the question could be instead "why not study algebraic geometry!?" I hope this answer motivates you enough to dive into this deep ocean of the mathematical world and to corroborate it yourself. Best luck!
| 2026-01-26T11:31:41.456402
|
348,198
|
Best Fake Proofs? (A M.SE April Fools Day collection)
|
In honor of April Fools Day $2013$, I'd like this question to collect the best, most convincing fake proofs of impossibilities you have seen. I've posted one as an answer below. I'm also thinking of a geometric one where the "trick" is that it's very easy to draw the diagram wrong and have two lines intersect in the wrong place (or intersect when they shouldn't). If someone could find and link this, I would appreciate it very much.
| 258
|
soft-question, big-list, fake-proofs
|
https://math.stackexchange.com/questions/348198/best-fake-proofs-a-m-se-april-fools-day-collection
|
$$x^2=\underbrace{x+x+\cdots+x}_{(x\text{ times})}$$ $$\frac{d}{dx}x^2=\frac{d}{dx}[\underbrace{x+x+\cdots+x}_{(x\text{ times})}]$$ $$2x=1+1+\cdots+1=x$$ $$2=1$$
| 208
| false
|
Q: Best Fake Proofs? (A M.SE April Fools Day collection)
In honor of April Fools Day $2013$, I'd like this question to collect the best, most convincing fake proofs of impossibilities you have seen. I've posted one as an answer below. I'm also thinking of a geometric one where the "trick" is that it's very easy to draw the diagram wrong and have two lines intersect in the wrong place (or intersect when they shouldn't). If someone could find and link this, I would appreciate it very much.
A: $$x^2=\underbrace{x+x+\cdots+x}_{(x\text{ times})}$$ $$\frac{d}{dx}x^2=\frac{d}{dx}[\underbrace{x+x+\cdots+x}_{(x\text{ times})}]$$ $$2x=1+1+\cdots+1=x$$ $$2=1$$
| 2026-01-26T11:31:43.037539
|
2,694
|
What is the importance of the Collatz conjecture?
|
I have been fascinated by the Collatz problem since I first heard about it in high school. Take any natural number $n$. If $n$ is even, divide it by $2$ to get $n / 2$, if $n$ is odd multiply it by $3$ and add $1$ to obtain $3n + 1$. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach $1$. [...] Paul Erdős said about the Collatz conjecture: "Mathematics is not yet ready for such problems." He offered $500 USD for its solution. QUESTIONS: How important do you consider the answer to this question to be? Why? Would you speculate on what might have possessed Paul Erdős to make such an offer? EDIT: Is there any reason to think that a proof of the Collatz Conjecture would be complex (like the FLT) rather than simple (like PRIMES is in P)? And can this characterization of FLT vs. PRIMES is in P be made more specific than a bit-length comparison?
| 257
|
number-theory, elementary-number-theory, prime-numbers, soft-question, collatz-conjecture
|
https://math.stackexchange.com/questions/2694/what-is-the-importance-of-the-collatz-conjecture
|
Most of the answers so far have been along the general lines of 'Why hard problems are important', rather than 'Why the Collatz conjecture is important'; I will try to address the latter. I think the basic question being touched on is: In what ways does the prime factorization of $a$ affect the prime factorization of $a+1$? Of course, one can always multiply out the prime factorization, add one, and then factor again, but this throws away the information of the prime factorization of $a$. Note that this question is also meaningful in other UFDs, like $\mathbb{C}[x]$. It seems very hard to come up with answers to this question that don't fall under the heading of 'immediate', such as distinct primes in each factorization. This seems to be in part because a small change in the prime factorization for $a$ (multiplication by a prime, say) can have a huge change in the prime factorization for $a+1$ (totally distinct prime support perhaps). Therefore, it is tempting to regard the act of adding 1 as an essentially-random shuffling of the prime factorization. The most striking thing about the Collatz conjecture is that it seems to be making a deep statement about a subtle relation between the prime factorizations of $a$ and $a+1$. Note that the Collatz iteration consists of three steps; two of which are 'small' in terms of the prime factorization, and the other of which is adding one: multiplying by 3 has a small effect on the factorization. adding 1 has a (possibly) huge effect on the factorization. factoring out a power of 2 has a small effect on the factorization (in that it doesn't change the other prime powers in the factorization). So, the Collatz conjecture seems to say that there is some sort of abstract quantity like 'energy' which cannot be arbitrarily increased by adding 1. That is, no matter where you start, and no matter where this weird prime-shuffling action of adding 1 takes you, eventually the act of pulling out 2s takes enough energy out of the system so that you reach 1. I think it is for reasons like this that mathematicians suspect that a solution of the Collatz conjecture will open new horizons and develop new and important techniques in number theory.
| 383
| true
|
Q: What is the importance of the Collatz conjecture?
I have been fascinated by the Collatz problem since I first heard about it in high school. Take any natural number $n$. If $n$ is even, divide it by $2$ to get $n / 2$, if $n$ is odd multiply it by $3$ and add $1$ to obtain $3n + 1$. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach $1$. [...] Paul Erdős said about the Collatz conjecture: "Mathematics is not yet ready for such problems." He offered $500 USD for its solution. QUESTIONS: How important do you consider the answer to this question to be? Why? Would you speculate on what might have possessed Paul Erdős to make such an offer? EDIT: Is there any reason to think that a proof of the Collatz Conjecture would be complex (like the FLT) rather than simple (like PRIMES is in P)? And can this characterization of FLT vs. PRIMES is in P be made more specific than a bit-length comparison?
A: Most of the answers so far have been along the general lines of 'Why hard problems are important', rather than 'Why the Collatz conjecture is important'; I will try to address the latter. I think the basic question being touched on is: In what ways does the prime factorization of $a$ affect the prime factorization of $a+1$? Of course, one can always multiply out the prime factorization, add one, and then factor again, but this throws away the information of the prime factorization of $a$. Note that this question is also meaningful in other UFDs, like $\mathbb{C}[x]$. It seems very hard to come up with answers to this question that don't fall under the heading of 'immediate', such as distinct primes in each factorization. This seems to be in part because a small change in the prime factorization for $a$ (multiplication by a prime, say) can have a huge change in the prime factorization for $a+1$ (totally distinct prime support perhaps). Therefore, it is tempting to regard the act of adding 1 as an essentially-random shuffling of the prime factorization. The most striking thing about the Collatz conjecture is that it seems to be making a deep statement about a subtle relation between the prime factorizations of $a$ and $a+1$. Note that the Collatz iteration consists of three steps; two of which are 'small' in terms of the prime factorization, and the other of which is adding one: multiplying by 3 has a small effect on the factorization. adding 1 has a (possibly) huge effect on the factorization. factoring out a power of 2 has a small effect on the factorization (in that it doesn't change the other prime powers in the factorization). So, the Collatz conjecture seems to say that there is some sort of abstract quantity like 'energy' which cannot be arbitrarily increased by adding 1. That is, no matter where you start, and no matter where this weird prime-shuffling action of adding 1 takes you, eventually the act of pulling out 2s takes enough energy out of the system so that you reach 1. I think it is for reasons like this that mathematicians suspect that a solution of the Collatz conjecture will open new horizons and develop new and important techniques in number theory.
| 2026-01-26T11:31:44.799405
|
253,910
|
The Integral that Stumped Feynman?
|
In "Surely You're Joking, Mr. Feynman!," Nobel-prize winning Physicist Richard Feynman said that he challenged his colleagues to give him an integral that they could evaluate with only complex methods that he could not do with real methods: One time I boasted, "I can do by other methods any integral anybody else needs contour integration to do." So Paul [Olum] puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration! He was always deflating me like that. He was a very smart fellow. Does anyone happen to know what this integral was?
| 257
|
real-analysis, complex-analysis, reference-request, integration, contour-integration
|
https://math.stackexchange.com/questions/253910/the-integral-that-stumped-feynman
|
I doubt that we will ever know the exact integral that vexed Feynman. Here is something similar to what he describes. Suppose $f(z)$ is an analytic function on the unit disk. Then, by Cauchy's integral formula, $$\oint_\gamma \frac{f(z)}{z}dz = 2\pi i f(0),$$ where $\gamma$ traces out the unit circle in a counterclockwise manner. Let $z=e^{i\phi}$. Then $\int_0^{2\pi}f(e^{i\phi}) d\phi = 2\pi f(0).$ Taking the real part of each side we find $$\begin{equation*} \int_0^{2\pi} \mathrm{Re}(f(e^{i\phi}))d\phi = 2\pi \mathrm{Re}(f(0)).\tag{1} \end{equation*}$$ (We could just as well take the imaginary part.) Clearly we can build some terrible integrals by choosing $f$ appropriately. Example 1. Let $\displaystyle f(z) = \exp\frac{2+z}{3+z}$. This is a mild choice compared to what could be done ... In any case, $f$ is analytic on the disk. Applying (1), and after some manipulations of the integrand, we find $$\int_0^{2\pi} \exp\left(\frac{7+5 \cos\phi}{10+6\cos\phi}\right) \cos \left( \frac{\sin\phi}{10+6 \cos\phi} \right) d\phi = 2\pi e^{2/3}.$$ Example 2. Let $\displaystyle f(z) = \exp \exp \frac{2+z}{3+z}$. Then \begin{align*}\int_0^{2\pi} & \exp\left( \exp\left( \frac{7+5 \cos \phi}{10+6 \cos \phi} \right) \cos\left( \frac{\sin \phi}{10+6 \cos \phi} \right) \right) \\ & \times\cos\left( \exp\left( \frac{7+5 \cos \phi}{10+6 \cos \phi} \right) \sin\left( \frac{\sin \phi}{10+6 \cos \phi} \right) \right) = 2\pi e^{e^{2/3}}. \end{align*}
| 83
| false
|
Q: The Integral that Stumped Feynman?
In "Surely You're Joking, Mr. Feynman!," Nobel-prize winning Physicist Richard Feynman said that he challenged his colleagues to give him an integral that they could evaluate with only complex methods that he could not do with real methods: One time I boasted, "I can do by other methods any integral anybody else needs contour integration to do." So Paul [Olum] puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration! He was always deflating me like that. He was a very smart fellow. Does anyone happen to know what this integral was?
A: I doubt that we will ever know the exact integral that vexed Feynman. Here is something similar to what he describes. Suppose $f(z)$ is an analytic function on the unit disk. Then, by Cauchy's integral formula, $$\oint_\gamma \frac{f(z)}{z}dz = 2\pi i f(0),$$ where $\gamma$ traces out the unit circle in a counterclockwise manner. Let $z=e^{i\phi}$. Then $\int_0^{2\pi}f(e^{i\phi}) d\phi = 2\pi f(0).$ Taking the real part of each side we find $$\begin{equation*} \int_0^{2\pi} \mathrm{Re}(f(e^{i\phi}))d\phi = 2\pi \mathrm{Re}(f(0)).\tag{1} \end{equation*}$$ (We could just as well take the imaginary part.) Clearly we can build some terrible integrals by choosing $f$ appropriately. Example 1. Let $\displaystyle f(z) = \exp\frac{2+z}{3+z}$. This is a mild choice compared to what could be done ... In any case, $f$ is analytic on the disk. Applying (1), and after some manipulations of the integrand, we find $$\int_0^{2\pi} \exp\left(\frac{7+5 \cos\phi}{10+6\cos\phi}\right) \cos \left( \frac{\sin\phi}{10+6 \cos\phi} \right) d\phi = 2\pi e^{2/3}.$$ Example 2. Let $\displaystyle f(z) = \exp \exp \frac{2+z}{3+z}$. Then \begin{align*}\int_0^{2\pi} & \exp\left( \exp\left( \frac{7+5 \cos \phi}{10+6 \cos \phi} \right) \cos\left( \frac{\sin \phi}{10+6 \cos \phi} \right) \right) \\ & \times\cos\left( \exp\left( \frac{7+5 \cos \phi}{10+6 \cos \phi} \right) \sin\left( \frac{\sin \phi}{10+6 \cos \phi} \right) \right) = 2\pi e^{e^{2/3}}. \end{align*}
| 2026-01-26T11:31:46.779113
|
209,241
|
Exterior Derivative vs. Covariant Derivative vs. Lie Derivative
|
In differential geometry, there are several notions of differentiation, namely: Exterior Derivative, $d$ Covariant Derivative/Connection, $\nabla$ Lie Derivative, $\mathcal{L}$. I have listed them in order of appearance in my education/in descending order of my understanding of them. Note, there may be others that I am yet to encounter. Conceptually, I am not sure how these three notions fit together. Looking at their definitions, I can see that there is even some overlap between the collection of objects they can each act on. I am trying to get my head around why there are (at least) three different notions of differentiation. I suppose my confusion can be summarised by the following question. What does each one do that the other two can't? I don't just mean which objects can they act on that the other two can't, I would like a deeper explanation (if it exists, which I believe it does). In terms of their geometric intuition/interpretation, does it make sense that we need these different notions? Note, I have put the reference request tag on this question because I would be interested to find some resources which have a discussion of these notions concurrently, as opposed to being presented as individual concepts.
| 256
|
reference-request, differential-geometry, differential-forms, exterior-algebra, big-picture
|
https://math.stackexchange.com/questions/209241/exterior-derivative-vs-covariant-derivative-vs-lie-derivative
|
Short answer: the exterior derivative acts on differential forms; the Lie derivative acts on any tensors and some other geometric objects (they have to be natural, e.g. a connection, see the paper of P. Petersen below); both the exterior and the Lie derivatives don't require any additional geometric structure: they rely on the differential structure of the manifold; the covariant derivative needs a choice of connection which sometimes (e.g. in a presence of a semi-Riemannian metric) can be made canonically; there are relationships between these derivatives. For a longer answer I would suggest the following selection of papers T. J. Willmore, The definition of Lie derivative R. Palais, A definition of the exterior derivative in terms of Lie derivatives P. Petersen, The Ricci and Bianchi Identities Of course, there is a lot more to say. Edit. I decided to extend my answer as I believe that there are some essential points which have not been discussed yet. An encyclopedic reference that treats all these derivatives concurrently at a modern level of generality is I.Kolar, P.W. Michor, J. Slovak, Natural Operations in Differential Geometry (Springer 1993), freely available online here. I would not even dare to summarize this resource since it has an abysmal deepness and all-round completeness, and indeed covers all the parts of the original question. Moreover, I believe that the bibliography list of this book contains almost any further relevant reference. As it has been already mentioned by many in this discussion, these operations are intimately related. It cannot be overemphasized that the most important feature that they all share is naturality (they commute with pullback, and this, in particular, makes them coordinate-free). See KMS cited above and its bibliography, and specifically the following references may be useful: R. Palais, Natural Operations on Differential Forms, e.g. here or here. C.L. Terng, Natural Vector Bundles and Natural Differential Operators, e.g. here It turns out that their naturality forces them to be unique if we impose on them some basic properties, such as $d \circ d = 0$ for the exterior derivative. One way to prove that and further references could be found in: D. Krupka, V. Mikolasova, On the uniqueness of some differential invariants: $d$, $[,]$, $\nabla $, see here. Also it is interesting that the Bianchi identities for the connection follow from the naturality and the property $d \circ d = 0$ for the exterior derivative, see Ph. Delanoe, On Bianchi identities, e.g. here. The reference list that I produce here is too far from being complete in any sense. I only would add one classical treatment that I personally used to comprehend some of the fundamental notions related to Lie derivatives (in particular, the Lie derivative of a connection!): K. Yano, The Theory Of Lie Derivatives And Its Applications, freely available here Indeed, my comments are speculative and sparse. I wish if this question were answered by someone like P. Michor, to be honest :-)
| 98
| true
|
Q: Exterior Derivative vs. Covariant Derivative vs. Lie Derivative
In differential geometry, there are several notions of differentiation, namely: Exterior Derivative, $d$ Covariant Derivative/Connection, $\nabla$ Lie Derivative, $\mathcal{L}$. I have listed them in order of appearance in my education/in descending order of my understanding of them. Note, there may be others that I am yet to encounter. Conceptually, I am not sure how these three notions fit together. Looking at their definitions, I can see that there is even some overlap between the collection of objects they can each act on. I am trying to get my head around why there are (at least) three different notions of differentiation. I suppose my confusion can be summarised by the following question. What does each one do that the other two can't? I don't just mean which objects can they act on that the other two can't, I would like a deeper explanation (if it exists, which I believe it does). In terms of their geometric intuition/interpretation, does it make sense that we need these different notions? Note, I have put the reference request tag on this question because I would be interested to find some resources which have a discussion of these notions concurrently, as opposed to being presented as individual concepts.
A: Short answer: the exterior derivative acts on differential forms; the Lie derivative acts on any tensors and some other geometric objects (they have to be natural, e.g. a connection, see the paper of P. Petersen below); both the exterior and the Lie derivatives don't require any additional geometric structure: they rely on the differential structure of the manifold; the covariant derivative needs a choice of connection which sometimes (e.g. in a presence of a semi-Riemannian metric) can be made canonically; there are relationships between these derivatives. For a longer answer I would suggest the following selection of papers T. J. Willmore, The definition of Lie derivative R. Palais, A definition of the exterior derivative in terms of Lie derivatives P. Petersen, The Ricci and Bianchi Identities Of course, there is a lot more to say. Edit. I decided to extend my answer as I believe that there are some essential points which have not been discussed yet. An encyclopedic reference that treats all these derivatives concurrently at a modern level of generality is I.Kolar, P.W. Michor, J. Slovak, Natural Operations in Differential Geometry (Springer 1993), freely available online here. I would not even dare to summarize this resource since it has an abysmal deepness and all-round completeness, and indeed covers all the parts of the original question. Moreover, I believe that the bibliography list of this book contains almost any further relevant reference. As it has been already mentioned by many in this discussion, these operations are intimately related. It cannot be overemphasized that the most important feature that they all share is naturality (they commute with pullback, and this, in particular, makes them coordinate-free). See KMS cited above and its bibliography, and specifically the following references may be useful: R. Palais, Natural Operations on Differential Forms, e.g. here or here. C.L. Terng, Natural Vector Bundles and Natural Differential Operators, e.g. here It turns out that their naturality forces them to be unique if we impose on them some basic properties, such as $d \circ d = 0$ for the exterior derivative. One way to prove that and further references could be found in: D. Krupka, V. Mikolasova, On the uniqueness of some differential invariants: $d$, $[,]$, $\nabla $, see here. Also it is interesting that the Bianchi identities for the connection follow from the naturality and the property $d \circ d = 0$ for the exterior derivative, see Ph. Delanoe, On Bianchi identities, e.g. here. The reference list that I produce here is too far from being complete in any sense. I only would add one classical treatment that I personally used to comprehend some of the fundamental notions related to Lie derivatives (in particular, the Lie derivative of a connection!): K. Yano, The Theory Of Lie Derivatives And Its Applications, freely available here Indeed, my comments are speculative and sparse. I wish if this question were answered by someone like P. Michor, to be honest :-)
| 2026-01-26T11:31:48.418105
|
1,188,845
|
Does the square or the circle have the greater perimeter? A surprisingly hard problem for high schoolers
|
An exam for high school students had the following problem: Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and $C$ as shown on the diagram. Which of the geometric figures has the greater perimeter, the square or the circle? Of course, there are some ways to solve this problem. One method is as follows: assume the side lengths of the square is $1$, put everything somewhere on a Cartesian coordinate system, find the midpoint of the circle using the coordinates of $E$, $B$ and $C$, then find the radius of the circle, and finally use the radius to calculate the circle's circumference and compare it to the perimeter of the square. The problem with that method is that ostensibly this problem is supposed to be very simple; it shouldn't require the student to know the formula for the midpoint of a circle given three coordinates. Therefore the question here is: does there exist a simple way to solve the problem without knowing any complicated geometric formulas?
| 256
|
geometry, circles, quadrilateral
|
https://math.stackexchange.com/questions/1188845/does-the-square-or-the-circle-have-the-greater-perimeter-a-surprisingly-hard-pr
|
Perhaps the examiner intended the students to notice the square is determined by a $(3, 4, 5)$ triangle, because $3 + 5 = 4 + 4$ (!): Consequently, as several others have noted, $$ \frac{\text{perimeter of the circle}}{\text{perimeter of the square}} = \frac{5 \cdot 2\pi}{4 \cdot 8} = \frac{\pi}{3.2} For an approach less dependent on inspiration, taking the origin of the coordinate system at the center of the circle seems easier than placing the origin at the center of the square. Without loss of generality, assume the circle has unit radius: Equating the lengths of the horizontal and vertical sides of the square in this diagram, we read off $$ x + 1 = 2y\quad\text{(or $x = 2y - 1$).} $$ Invoking the Pythagorean theorem and substituting the preceding line, \begin{align*} 0 &= x^{2} + y^{2} - 1 \\ &= (2y - 1)^{2} + y^{2} - 1 \\ &= 5y^{2} - 4y \\ &= y(5y - 4). \end{align*} Clearly $y \neq 0$, so $y = 4/5$, $x = 3/5$, and we notice the Examiner's Favorite Triangle.
| 224
| true
|
Q: Does the square or the circle have the greater perimeter? A surprisingly hard problem for high schoolers
An exam for high school students had the following problem: Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and $C$ as shown on the diagram. Which of the geometric figures has the greater perimeter, the square or the circle? Of course, there are some ways to solve this problem. One method is as follows: assume the side lengths of the square is $1$, put everything somewhere on a Cartesian coordinate system, find the midpoint of the circle using the coordinates of $E$, $B$ and $C$, then find the radius of the circle, and finally use the radius to calculate the circle's circumference and compare it to the perimeter of the square. The problem with that method is that ostensibly this problem is supposed to be very simple; it shouldn't require the student to know the formula for the midpoint of a circle given three coordinates. Therefore the question here is: does there exist a simple way to solve the problem without knowing any complicated geometric formulas?
A: Perhaps the examiner intended the students to notice the square is determined by a $(3, 4, 5)$ triangle, because $3 + 5 = 4 + 4$ (!): Consequently, as several others have noted, $$ \frac{\text{perimeter of the circle}}{\text{perimeter of the square}} = \frac{5 \cdot 2\pi}{4 \cdot 8} = \frac{\pi}{3.2} For an approach less dependent on inspiration, taking the origin of the coordinate system at the center of the circle seems easier than placing the origin at the center of the square. Without loss of generality, assume the circle has unit radius: Equating the lengths of the horizontal and vertical sides of the square in this diagram, we read off $$ x + 1 = 2y\quad\text{(or $x = 2y - 1$).} $$ Invoking the Pythagorean theorem and substituting the preceding line, \begin{align*} 0 &= x^{2} + y^{2} - 1 \\ &= (2y - 1)^{2} + y^{2} - 1 \\ &= 5y^{2} - 4y \\ &= y(5y - 4). \end{align*} Clearly $y \neq 0$, so $y = 4/5$, $x = 3/5$, and we notice the Examiner's Favorite Triangle.
| 2026-01-26T11:31:50.058507
|
546,155
|
Proof that the trace of a matrix is the sum of its eigenvalues
|
I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.
| 255
|
linear-algebra, matrices, eigenvalues-eigenvectors, trace
|
https://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues
|
These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by $$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$ On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients, we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.
| 339
| true
|
Q: Proof that the trace of a matrix is the sum of its eigenvalues
I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.
A: These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by $$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$ On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients, we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.
| 2026-01-26T11:31:51.797928
|
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