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2,633,847 |
It often happens in mathematics that the answer to a problem is "known" long before anybody knows how to prove it. (Some examples of contemporary interest are among the Millennium Prize problems: E.g. Yang-Mills existence is widely believed to be true based on ideas from physics, and the Riemann hypothesis is widely believed to be true because it would be an awful shame if it wasn't. Another good example is Schramm–Loewner evolution , where again the answer was anticipated by ideas from physics.) More rare are the instances where an abstract mathematical "idea" floats around for many years before even a rigorous definition or interpretation can be developed to describe the idea. An example of this is umbral calculus , where a mysterious technique for proving properties of certain sequences existed for over a century before anybody understood why the technique worked, in a rigorous way. I find these instances of mathematical ideas without rigorous interpretation fascinating, because they seem to often lead to the development of radically new branches of mathematics $^1$ . What are further examples of this type? I am mainly interested in historical examples, but contemporary ones (i.e. ideas which have yet to be rigorously formulated) are also welcome. Footnote: I have some specific examples in mind that I will share as an answer, if nobody else does.
|
The notion of probability has been in use since the middle ages or maybe before. But it took quite a while to formalize the probability theory and giving it a rigorous basis in the midst of 20th century. According to wikipedia: There have been at least two successful attempts to formalize probability, namely the Kolmogorov formulation and the Cox formulation . In Kolmogorov's formulation, sets are interpreted as events and probability itself as a measure on a class of sets. In Cox's theorem, probability is taken as a primitive (that is, not further analyzed) and the emphasis is on constructing a consistent assignment of probability values to propositions. In both cases, the laws of probability are the same, except for technical details. There are other methods for quantifying uncertainty, such as the Dempster–Shafer theory or possibility theory, but those are essentially different and not compatible with the laws of probability as usually understood.
|
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|
2,633,858 |
What is the dimension of the subspace $W=\{ F(x) \mid (D^2+1)F(x)=0 \wedge F(x)=0\}$ I solved this and found $F(x)=a \cos(x)+b \sin(x)$, but due to the second condition I think it is the null space. Am I right or wrong?
|
The notion of probability has been in use since the middle ages or maybe before. But it took quite a while to formalize the probability theory and giving it a rigorous basis in the midst of 20th century. According to wikipedia: There have been at least two successful attempts to formalize probability, namely the Kolmogorov formulation and the Cox formulation . In Kolmogorov's formulation, sets are interpreted as events and probability itself as a measure on a class of sets. In Cox's theorem, probability is taken as a primitive (that is, not further analyzed) and the emphasis is on constructing a consistent assignment of probability values to propositions. In both cases, the laws of probability are the same, except for technical details. There are other methods for quantifying uncertainty, such as the Dempster–Shafer theory or possibility theory, but those are essentially different and not compatible with the laws of probability as usually understood.
|
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|
2,635,850 |
A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9).
|
Not necessarily. Look at the following potential $A%$ defined on some region: The associated vector field $F=\mathrm{grad}(A)$ looks like this: Since it is a gradient, it has $\mathrm{curl}(F)=0$. But we can complete it into the following still curl-free vector field: This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero. This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $\Bbb R^3$, then you are safe: $\Bbb R^3$ is simply connected and every curl-free vector field is conservative.
|
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|
2,635,994 |
Intuitively, it seems like the area of a square should always be greater than the length of one of its sides because you can "fit" one of its sides in the space of its area, and still have room left over. However when the length of a side, $s$ , is less than $1$ , then the area $s^2 < s$ , which doesn't make sense to me for the reason above.
|
I think your intuition is failing you because you are trying to compare a 1-dimensional object (the length of a side) with a 2-dimensional object (the area of the interior of the square). You can fit loads of segments into a square of any size -- infinitely many, in fact! That comparison doesn't really mean anything. On the other hand, here's a comparison that does make sense: Set a square of side length $s$ side-by-side with a rectangle whose sides are $s \times 1$. Now you are comparing area to area. The rectangle's area will fit inside the square if and only if $s>1$.
|
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|
2,636,777 |
Why are some integrals that cannot be integrated in elementary terms defined and given names, while others aren’t? Based on what criteria are they chosen? Applicability to real life? And what is the point if we cannot solve them? For example: $$\int\frac{\sin(x)}{x}\,dx=\text{Si}(x), \quad -\int_{-x}^\infty \frac{e^{-t}}{t}\,dt=\text{Ei}(x), \quad \int \cos\left(x^2\right)\,dx = \sqrt{\frac\pi2} \text{C}\left( \sqrt{\frac2\pi}x\right)$$
while others like
$$\int x^x \,dx$$
are not defined.
|
I can define one right now. I hereby declare that
$$
\mathrm{Xi}(x) := \int_0^xt^tdt
$$
Now, the only thing that remains is to see whether other mathematicians pick it up and start using it . As with any other mathematical notation, there isn't really a committee somewhere who sits and weighs criteria to decide what mathematical notation is correct and should be accepted. What decides whether a piece of notation gets accepted and conventional is simply whether enough other mathematicians find it useful and start using it.
|
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|
2,637,587 |
I'm having a trouble understanding the concept. Can you give me a math example where $P\Rightarrow Q$ Is true but $P\Leftrightarrow Q$ is false? Thank you.
|
For real numbers: $$x > 1$$ implies that
$$
x^2 > 1
$$ But $x^2 > 1$ does not imply that $x > 1$. For instance, $(-2)^2 = 4 > 1$, but $-2$ is not greater than $1$.
|
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|
2,637,690 |
If all sides: $a, b, c, d$ are known, is there a formula that can calculate the area of a trapezoid? I know this formula for calculating the area of a trapezoid from its two bases and its height: $$S=\frac {a+b}{2}×h$$ And I know a well-known formula for finding the area of a triangle, called Heron's formula: $$S=\sqrt {p(p-a)(p-b)(p-c)}$$ $$p=\frac{a+b+c}{2}$$ But I could not a formula for finding the area of a trapezoid in the books.
|
This problem is more subtle than some of the other answers here let on. A great deal hinges on whether "trapezoid" is defined inclusively (i.e. as a quadrilateral with at least one pair of parallel sides) or exclusively (i.e. as a quadrilateral with exactly one pair of parallel sides). The former definition is widely considered more mathematically sophisticated, but the latter definition is more traditional, is still extensively used in K-12 education in the United States, and has some advantages. As the other responses have pointed out, if one defines "trapezoid" inclusively, then any parallelogram is automatically a trapezoid, and as the side-lengths of a parallelogram do not determine its area, it is not possible (even conceptually) that there could be a formula for the area of a trapezoid in terms of its side lengths. However, if "trapezoid" is defined exclusively , then things are quite different. Consider a trapezoid with parallel bases of length $a$ and $b$ with $b>a$. Let $\theta$ and $\phi$ respectively denote the angles formed by the legs $c$ and $d$ with the base $b$. Then we have the following relationships:
$$c\cos\theta + d\cos\phi = b-a$$
$$c\sin\theta = d\sin\phi$$
These conditions uniquely determine $\theta$ and $\phi$, and therefore among non-parallelogram trapezoids, choosing the lengths of the parallel sides and the lengths of the bases uniquely determines the figure . In particular we would have $$\cos\theta = \frac{(b-a)^2+c^2-d^2}{2c(b-a)}$$. The height of the trapezoid would then be $h=c\sin\theta$ (or if you prefer $h=d\sin\phi$, which is equal to it), so the area of the trapezoid can (in principal) be computed. If you really want to carry it out, you would have $$\sin\theta = \sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$
so the area would be
$$A=\frac{a+b}{2}c\sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$
I am not sure if there is a simpler expression, however.
|
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|
2,639,243 |
For simplicity's sake, let the Earth be a perfect sphere. Imagine you are drawing an equilateral triangle over its surface. How long should its sides be, for the sum of its angles to be 180.1 degrees?
|
The area of a spherical triangle is exactly $ER^2$ where $E$ is the angular excess. In our case $E=\frac{\pi}{1800}$, so the given triangle covers $\frac{1}{7200}$ of Earth's surface. Assuming $R=1$, l'Huilier's formula (page 184 of my notes) relates the angular excess / the area to the semiperimeter / the side lengths through
$$ \tan\frac{E}{4}=\sqrt{\tan\frac{s}{2}\tan\frac{s-a}{2}\tan\frac{s-b}{2}\tan\frac{s-c}{2}} $$
and in our case we have $a=b=c=\ell$ and $s=\frac{3}{2}\ell$, so $$ E = \frac{\pi}{1800} = 4\arctan\sqrt{\tan\frac{3\ell}{4}\tan^3\frac{\ell}{4}} $$
and by solving
$$ \tan\frac{3\ell}{4}\tan^3\frac{\ell}{4}=\tan^2\frac{\pi}{7200} $$
we get that $\ell$ is approximately $6.347\%$ of the radius $R$. For Earth, $R =6\ 371$ km, and $\ell = 404.377$ km.
|
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|
2,640,860 |
If we really don't know which is bigger if $ i $ is greater or $ 2i $ or so on then why do we plot $ i $ first then $ 2i $ and so on, on the imaginary axis of the Argand plane? My teacher said that imaginary numbers are just points and all are dimensionless so they are incomparable and the distance really doesn't matter. I want to get this more clear
|
There is no way to order complex numbers, in a way that preserves operations in a sensible way. The precise term is ordered field ; among their properties, $x^2
\ge 0$ for every $x$. Since $i^2=-1$, we would need $-1\ge 0$, which is impossible. However, if you want to measure distance , you can do that with a norm . In the complex numbers, this is calculated as $|a+bi|=\sqrt{a^2+b^2}$. Hence, it is correct to say that $2i$ is twice as far from the origin as $i$ is.
|
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|
2,643,036 |
I have a rectangle made up of 30 identical squares (5 tall and 6 wide). By only drawing two lines on the rectangle, split the rectangle into 4 parts where the areas are in the ratio 1:2:3:4. How would one go around doing this? I tried for a solid hour or two; but to no avail. Disclaimer: This question is NOT created by me.
|
First, divide the rectangle equally on its long side, as suggested in the comments. In the left rectangle, mark the point that is $\frac45$ of the way up and centred horizontally; do the same for the right rectangle, but mark the point only $\frac35$ of the way up. Then draw the line connecting the two marked points. Since this line crosses the left and right edges of the whole rectangle, the four small triangles shown above (bounded between the purple and darker grey lines) have equal area, so the left rectangle is divided 1:4 and the right rectangle divided 2:3. Overall, the rectangle has been divided into four parts with ratio 1:2:3:4. This solution does not require moving the pieces as in Donald Splutterwit's answer.
|
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|
2,643,121 |
This is a question from Math Olympiad. If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$. I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.
|
To my surprise, this problem can be solved using geometry. Identify the Euclidean plane with complex plane $\mathbb{C}$.
and let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity. Consider triangle $\triangle ABC$ with vertices at $A = x$, $B = y\omega$ and $C = z\omega^2$. The sides of the triangle equal to $$\begin{align}
a^2 = BC^2 &= | y\omega - z\omega^2|^2 = y^2 + yz+ z^2 = 1\\
b^2 = AC^2 &= | x - z\omega^2|^2 = x^2 + xz + z^2 = 4\\
c^2 = AB^2 &= | x - y\omega|^2 = x^2+xy+y^2 = 3
\end{align}
$$
Since $a^2 + c^2 = b^2$, this is a right angled triangle with area $\mathcal{A} = \frac12 ac
=\frac{\sqrt{3}}{2}$. An alternate way to compute the area is cut the triangle into 3 pieces along the line $OA$, $OB$ and $OC$. This gives us $$\mathcal{A}= \frac12 (xy + yz + xz)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4} (xy+yz+xz)$$
Combine these two results, we find:
$$xy+yz+xz = 2$$ Update For a pure algebraic answer, one can substitute above expression of $a^2,b^2,c^2$ into Heron's formula for area of triangle, $$\mathcal{A} = \frac14\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$
simplify and obtain following algebraic identity $$\begin{array}{rl}
3(xy+yz+zx)^2 =& \phantom{+0}((x^2+xz+y^2) + (y^2+yz+z^2) + (z^2+xz+x^2))^2\\
&-2((x^2+xz+y^2)^2 + (y^2+yz+z^2)^2 + (z^2+xz+x^2)^2)
\end{array}
$$
Using given values of $a, b, c$, we find
$$3(xy+yz+xz)^2 = (3+4+1)^2 - 2(3^2+4^2+1^2) = 64 -2(26) = 12 = 3(2)^2$$
This leads to the same conclusion $xy+yz+xz = 2$ as before.
|
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|
2,644,700 |
This is a very speculative/soft question; please keep this in mind when reading it. Here "higher" means "greater than 3". What I am wondering about is what new geometrical phenomena are there in higher dimensions. When I say new I mean phenomena which are counterintuitive or not analogous to their lower dimensional counterparts. A good example could be hypersphere packing . My main (and sad) impression is that almost all phenomena in higher dimensions could be thought intuitively by dimensional analogy. See for example, this link : What this implies (for me) is the boring consequence that there is no new conceptual richness in higher dimensional geometry beyond the fact than the numbers are larger (for example my field of study is string compactifications and though, at first sight, it could sound spectacular to use orientifolding which set a loci of fixed points which are O3 and O7 planes; the reasoning is pretty much the same as in lower dimensions...) However the question of higher dimensional geometry is very related (for me) to the idea of beauty and complexity: these projections to 2-D of higher dimensional objects totally amazes me (for example this orthonormal projection of a 12-cube ) and makes me think there must be interesting higher dimensional phenomena... I would thank anyone who could give me examples of beautiful ideas implying “visualization” of higher dimensional geometry…
|
In high dimensions, almost all of the volume of a ball sits at its surface. More exactly, if $V_d(r)$ is the volume of the $d$-dimensional ball with radius $r$, then for any $\epsilon>0$, no matter how small, you have
$$\lim_{d\to\infty} \frac{V_d(1-\epsilon)}{V_d(1)} = 0$$
Algebraically that's obvious, but geometrically I consider it highly surprising. Edit: Another surprising fact: In 4D and above, you can have a flat torus, that is, a torus without any intrinsic curvature (like a cylinder in 3D). Even more: You can draw such a torus (not an image of it, the flat torus itself) on the surface of a hyperball (that is, a hypersphere). Indeed, the three-dimensional hypersphere (surface of the four-dimensional hyperball) can be almost completely partitioned into such tori, with two circles remaining in two completely orthogonal planes (thanks to anon in the comments for reminding me of those two leftover circles). Note that the circles could be considered degenerate tori, as the flat tori continuously transform into them (in much the same way as the circles of latitude on the 2-sphere transform into a point at the poles).
|
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|
2,644,706 |
So far,
I'ave used the squeeze theorem with functions $\frac1n$ and $-\frac1n$, and so got the limit $0$, but the answer is supposedly infinity... which makes little sense to me.
|
In high dimensions, almost all of the volume of a ball sits at its surface. More exactly, if $V_d(r)$ is the volume of the $d$-dimensional ball with radius $r$, then for any $\epsilon>0$, no matter how small, you have
$$\lim_{d\to\infty} \frac{V_d(1-\epsilon)}{V_d(1)} = 0$$
Algebraically that's obvious, but geometrically I consider it highly surprising. Edit: Another surprising fact: In 4D and above, you can have a flat torus, that is, a torus without any intrinsic curvature (like a cylinder in 3D). Even more: You can draw such a torus (not an image of it, the flat torus itself) on the surface of a hyperball (that is, a hypersphere). Indeed, the three-dimensional hypersphere (surface of the four-dimensional hyperball) can be almost completely partitioned into such tori, with two circles remaining in two completely orthogonal planes (thanks to anon in the comments for reminding me of those two leftover circles). Note that the circles could be considered degenerate tori, as the flat tori continuously transform into them (in much the same way as the circles of latitude on the 2-sphere transform into a point at the poles).
|
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|
2,647,300 |
The website of my university posted a riddle that goes something like this: Riddle There are three men named 1,2 and 3 and each one has two colored dots on his forehead. Possible colors are black and red . No color is used more than four times . The men can see the colors on the other mens head, but not the ones on their own. A game-master asks the men in the order 1,2,3,1,2,3,... whether they know the colors of their dots. If one says "no", he proceeds with the next man. If one says "yes", he has to state his guess. If he is correct, all men win the game, if he is wrong then all men lose. The men were not given any time to make up a strategy. As it turnes out the men answer "no", "no", "no", "no", "yes". What is the color on the head of man 2? I was going to solve this riddle systematically by trying to reason what color combinations make sense and what information each man can deduce from the others saying "no". However, I came to a much more direct solution, that completely ignores most of the riddle and only works by the assumption that the riddle is solvable . Solution : Man 2 must have a red and a black dot on his forehead. Reason : I (from the meta perspective) only have the information about the answer-pattern and no actual colors. Since the game is symmetric w.r.t. the colors red and black, the game would have the same answer-pattern if we switch the colors. This means, any reasoning that explains that man 2 has e.g. two red dots, will also work to reason that he has two black dots. If any of these two would be the solution, I would not be able to conclusively find it. Hence the solution must be "one red and one black dot". Note that I cannot answer how man 2 came to his conclusion, but this was not part of the question. For me, this kind of solving the riddle was very interesting and I wondered if there are more such riddles out there which (intentionally or unintentionally) can be solved by such meta-assumptions. Question : Are there more such riddles which have a surprisingly easy solution by meta-assumptions like e.g. "the riddle has a solution" or "the riddle is solvable with reasonable effort" etc. Update Because it was asked in the comments, and maybe needs a clarification, here is what I am looking for in other words: I ask for riddles or math-problems which are formulated in a way that suggests some specific meta-assumption (e.g. the riddle must have a solution which I as a reader can find). the meta-information is not obviously presented to the reader as something that should be used to solve the riddle (preferably might not even be needed). the meta-information turns out to be unexpectedly helpful. I am not specifically asking about the assumptions of "existence of a unique solution", but also other meta-assumptions, like e.g. the context of the question, the number of answers (if it is multiple-choice), the expected time-frame of solution, maybe even the color of the paper the question was asked on, etc. Some type of examples which came up a lot (and which I liked) are riddles/problems where the meta-information was the absence of some information . The absence implied that the solution will probably not depend on this missing information and hence we can choose a very simple instance of the problem. One other example that came to my mind (but it does not fit so perfectly) is the graph-problem presented in this video of 3Blue1Brown. It can (and must) be solved by the meta-information that it is presented on a cup and not on a piece of paper. Unfortunatly the riddle can only be solved by recognizing that it is on a cup. I would prefer that the meta-way is not the intended solution.
|
Here's a riddle that fits right into this category. Hope you like it: Begin with a sphere. Drill a circularly-cylindrical hole directly through the center of the sphere [i.e the result is still a 'solid of revolution'], and remove the drilled material. Measure the height of the hole from one circular edge through the body of the sphere to the other circular edge; the measured height is 10 centimeters. What is the volume of the remaining solid?
|
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2,647,304 |
Standing to definitions : A fact is a sentence that states that a relation holds between individuals. A rule is a relation that holds between individuals provided that some other relations hold. According to this, why it is not possible to
construct contradictory descriptions using facts and rules ? For example, i can use as facts the fact that a = b and a = not(b) and i have a contradiction....
|
Here's a riddle that fits right into this category. Hope you like it: Begin with a sphere. Drill a circularly-cylindrical hole directly through the center of the sphere [i.e the result is still a 'solid of revolution'], and remove the drilled material. Measure the height of the hole from one circular edge through the body of the sphere to the other circular edge; the measured height is 10 centimeters. What is the volume of the remaining solid?
|
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|
2,652,605 |
I was watching this Japanese game show and came across this question: The contestants were told that each small cube is 2cm on its side and were asked to find the volume of the above figure. The answer was 3080 $cm^3$. While I was counting the number of cubes for the first row, one of the contestants was able to answer this within a few seconds. I'm curious about how he did it. I assumed the figure was constructed in some sort of pattern and was hoping someone could shed some light on this. (The game show didn't explain how to solve this unfortunately...)
|
I looked at the horizontal layers. Top layer has seven, and each layer below shows seven more. So the number of cubes is
$$
7+14+\cdots+70=\frac{77}2\cdot10=385\,.
$$
|
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|
2,652,608 |
I wrote it as the following sum: $$1 + \sum_{k=1}^n (3k - 2)$$ Which I solved for and got the following formula: $$\frac{3n^2 - n + 2}2$$ But this seems wrong to me because the base case seems incorrect to me. Any help with this?
|
I looked at the horizontal layers. Top layer has seven, and each layer below shows seven more. So the number of cubes is
$$
7+14+\cdots+70=\frac{77}2\cdot10=385\,.
$$
|
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|
2,653,127 |
This one I gave my students today, nobody solve it. Is a following function periodic $f:\mathbb{R}\to\mathbb{R}$ $$f(x) = \cos (x) +\cos(x^2)$$ If someone is interested I can show a solution later.
|
Without using the derivative, the equation $f(x) = f(0)$ has only one solution. Indeed, if $\cos(x)+\cos(x^2)=2$ then $\cos(x) = 1 = \cos(x^2)$ so there exists $p,q \in \mathbb{Z}$ such that $x = 2p \pi$ and $x^2 = 2q \pi$ , so $\pi = \frac{q}{2 p^2}$ . But $\pi$ is not rational ; absurd. This may be overkill, but at least the same reasoning can prove the following : given any $\beta$ -periodic function $g$ with $\beta \in \mathbb{R} \backslash \mathbb{Q}$ and such that $g(x) \neq g(0)$ for $x \in ]0,\beta[$ , the function $x \mapsto g(x)+g(x^2)$ is not periodic. For instance, this is also true if $g$ is the Weierstrass function (if $b\notin 2\mathbb{Z}$ , with the definition used by wiki).
|
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|
2,656,519 |
Let $R$ be a ring without a unit element and $R'$ be a (non trivial) ring with a unit element. Can there be an onto homomorphism from $R$ to $R'$? Some observations: There cannot be an isomorphism , because then the element of $R$ which maps to $1$ in $R'$ must be a unit element in $R$. Also, we cannot have an onto homomorphism from a ring with unity to a ring without unity as $f(1)$ is going to be a unit element. Edit: I am defining a ring homomorphism as a function $ \phi: R \rightarrow R'$ such that for all $a,b$ in $R$,
$$\phi(a+b) = \phi(a) + \phi(b) $$
$$\phi(ab) = \phi(a)\phi(b)$$
|
Let $R$ be the set of even integers. This is a ring without unity. Let $R'$ be the ring of integers modulo $3$. This is a ring with unity. Let $\phi:R\to R'$ be the "reduction modulo $3$" map. This is a surjective homomorphism of rings.
|
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2,656,531 |
Suppose $B=A-A^T$ and I know $B\in\mathbb R^{n\times n}$. What is a simple way to get $A$? And what if I have the constraint that $A_{ij}\ge 0$ $\forall i,j$? Clarifications: $B$ is skew symmetric $a_{ii}$ (the diagonal elements of $A$) are zero
|
Let $R$ be the set of even integers. This is a ring without unity. Let $R'$ be the ring of integers modulo $3$. This is a ring with unity. Let $\phi:R\to R'$ be the "reduction modulo $3$" map. This is a surjective homomorphism of rings.
|
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|
2,656,551 |
I am trying to solve the following problem: Suppose $X$ is a continuous random variable with density $f$ over $[0,1]$. Then show that $(\int_0^1xf(x)dx)(\int_0^1x^3f(x)dx)>(\int_0^1x^2f(x)dx)^2$. I am trying to apply Fubini's theorem, but am getting equality for the two sides. What mistake am I making? Thanks in advance
|
Let $R$ be the set of even integers. This is a ring without unity. Let $R'$ be the ring of integers modulo $3$. This is a ring with unity. Let $\phi:R\to R'$ be the "reduction modulo $3$" map. This is a surjective homomorphism of rings.
|
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|
2,656,690 |
When I read Pinter's A Book of Abstract Algebra , Exercise 7 on page 25 asks whether the operation
$$x*y=\frac{xy}{x+y+1}$$
(defined on the positive real numbers) is associative. At first I considered this to be false, because the expression is so complicated. But when I worked out $(x*y)*z$ and $x*(y*z)$, I found both to be $$\frac{xyz}{xy+yz+zx+x+y+z+1}!$$ Commutativity is easy to see. But associativity can be so counter-intuitive! Can you see this operation is associative without working it out? Are there tricks to do this?
|
A common way to build weird-looking associative operations is to start from a known one, such as multiplication, say on the real numbers or some subset of them, and then to transform it through some bijection $\alpha$, by defining
$$x\ast y=\alpha^{-1}(\alpha(x)\cdot\alpha(y)).$$
Indeed this is equivalent to $\alpha(x\ast y)=\alpha(x)\cdot \alpha(y)$ (so that $\alpha$ is actually an isomorphism ), and it is then easy to check associativity by noticing that
\begin{align*}\alpha(x\ast (y\ast z)) & =\alpha(x)\cdot \alpha(y\ast z) = \alpha(x)\cdot(\alpha(y)\cdot \alpha(z))\\ & =(\alpha(x)\cdot \alpha(y))\cdot \alpha(z) = \alpha(x\ast y)\cdot \alpha(z)\\ & =\alpha((x\ast y)\ast z),\end{align*}
which implies that $x\ast (y\ast z)=(x\ast y)\ast z$ since $\alpha$ is bijective. Other properties, such as commutativity or existence of neutral or inverses, can be done in the same way, depending on the cases. In this case, we can see that
$$\frac{1}{x\ast y}=\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$$so that
$$1+\frac{1}{x\ast y}=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\left(1+\frac{1}{x}\right)\cdot\left(1+\frac{1}{y}\right),$$
so if you define $\alpha(x)=1+\frac{1}{x}$, you can check that it defines a bijection $(0,+\infty)\to (1,+\infty)$, and $\ast$ is just a transformation of the multiplication on $(1,+\infty)$, which explain why it is associative. In fact you can also see right away that it must also be commutative, but that it can't have a neutral element (otherwise $(1,+\infty)$ would have one).
|
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|
2,657,076 |
We have the unit interval $[0,1]$ and we want to find the probability of picking two random numbers $a,b$ from that interval with $|a-b|>0.5$. Must I investigate $[0,1]×[0,1]$? I don't have the faintest idea of how to solve this. The problem is that $[0,1]$ has infinite numbers to pick from… so how to calculate a probability with infinitely many items in the sample space? I would be really happy if somebody shed a light on this.
|
Quick, draw a diagram! Since the two random variables are independent and uniformly distributed, the answer is clearly $\frac14$.
|
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|
2,659,008 |
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$ My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$ So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$ Could some help me to solve it, Thanks in Advanced
|
Let $\xi=\exp\left(\frac{2\pi i}{40}\right)$. The given product equals $$ \frac{1}{16}\left(1+\xi+\xi^{-1}\right)\left(1+\xi^3+\xi^{-3}\right)\left(1+\xi^9+\xi^{-9}\right)\left(1+\xi^{27}+\xi^{-27}\right)$$
or
$$ \frac{1}{16\xi\xi^3\xi^9\xi^{27}}\cdot\frac{\xi^3-1}{\xi-1}\cdot\frac{\xi^9-1}{\xi^3-1}\cdot\frac{\xi^{27}-1}{\xi^9-1}\cdot\frac{\xi^{81}-1}{\xi^{27}-1}$$
or (by telescopic property and the fact that $\xi^{81}=\xi$)
$$ \frac{1}{16 \xi^{1+3+9+27}}=\frac{1}{16\xi^{40}}=\color{red}{\frac{1}{16}}.$$
|
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|
2,659,047 |
Question $1$. $$T\begin{pmatrix}a & b\\\ c & d\end{pmatrix}=(d+b)+(b+2c+d)x+(3c+3d)x^2+4dx^3$$ What is the kernel and nullity of $T$? Is it $1-1$? What is the image and rank of $T$? Is it onto. Answer for $Q1$: $b=c=d=0$. Nullity$=0$ and its $1-1$. The image is $\{1,x,x^2,x^3\}$ and rank$=4$. Its onto since $T$ is $1-1$. Question $2. T(a_1,a_2,a_3)= (a_2,a_3,a_1)$. Is it $1-1$ and onto? Answer for $Q2$: $(a_1, a_2, a_3)=0$ so nullity$=0$ and its $1-1$. $R(T)= \mathbb{R}^3$ and its onto since $T$ is $1-1$.
|
Let $\xi=\exp\left(\frac{2\pi i}{40}\right)$. The given product equals $$ \frac{1}{16}\left(1+\xi+\xi^{-1}\right)\left(1+\xi^3+\xi^{-3}\right)\left(1+\xi^9+\xi^{-9}\right)\left(1+\xi^{27}+\xi^{-27}\right)$$
or
$$ \frac{1}{16\xi\xi^3\xi^9\xi^{27}}\cdot\frac{\xi^3-1}{\xi-1}\cdot\frac{\xi^9-1}{\xi^3-1}\cdot\frac{\xi^{27}-1}{\xi^9-1}\cdot\frac{\xi^{81}-1}{\xi^{27}-1}$$
or (by telescopic property and the fact that $\xi^{81}=\xi$)
$$ \frac{1}{16 \xi^{1+3+9+27}}=\frac{1}{16\xi^{40}}=\color{red}{\frac{1}{16}}.$$
|
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|
2,660,247 |
I have this question from my exam where it's asked to find the sum: $$S=\sum_{k=1}^n \frac{1}{(1-r_k)^2}$$
where $r_k$ are the roots of
$$f(x)=x^n-2x+2\quad,n\ge3$$
I recalled this relation
$$\frac{f'(x)}{f(x)}=\sum_{k=1}^n \frac{1}{x-r_k}$$
and quickly realised that
$$\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$$
and using derivative would easily produce my answer. Indeed
$$\frac{d}{dx}\left( \frac{f'(x)}{f(x)} \right)=\frac{f''(x)f(x)-(f'(x))^2}{f(x)^2}=-\sum_{k=1}^n \frac{1}{(x-r_k)^2}$$
Evaluating at $x=1$ gives
$$S=\frac{(n-2)^2-n(n-1)}{1^2}=4-3n$$
However after the exam I realised that I have no ideea why
$$\frac{f'(x)}{f(x)}=\sum_{k=1}^n \frac{1}{x-r_k}$$
and I just memorized it, is there a nice way to show it?
And of course maybe a more elegant solution to this exam question?
|
$$f(x)=(x-r_1)(x-r_2)\cdots(x-r_n)$$ Using logarithm, $$\ln (f(x))=\ln (x-r_1)+\ln(x-r_2)+\cdots +\ln(x-r_n)$$ Taking derivative, $$\frac{f'(x)}{f(x)}= \frac{1}{(x-r_1)}+\frac{1}{(x-r_2)} +\cdots +\frac{1}{(x-r_n)}$$ Done!
|
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|
2,660,361 |
I don't understand the formula at all: $$e^{ix} = \cos(x) + i \sin(x)$$ I've tried reading all sorts of webpages and answers on the subject but it's just not clicking with me. I don't understand how we can define things when these are already known quantities. We've already got an exact definition for $e$ as $\lim_{n \to 0} (1+\frac{1}{n})^n$, and we've got $i = \sqrt{-1}$, and we've got $\cos$ and $\sin$ as the $x$ and $y$ coordinates of where right triangles meet the unit circle. So I don't understand how we get from those figures to Euler's formula. Then again I also don't understand why complex numbers are of the form $a + bi$ so that might have something to do with it. I have a hard time separating what identities are "emergent" from the previous mathematics and which pieces are "by definition" and why they are defined that way. I don't understand why we start talking about "rotations" as if it's obvious that's what's happening. I don't even know why $e$ is involved in any of this to begin with. Is $e^{ix}$ simply being written in the slightly rearranged complex form $a + ib$ where $a = \cos(x)$ and $b = \sin(x)$? Is this to be interpreted as $\cos(x)$ being the $x$-coordinate and somehow $i \sin(x)$ is the $y$ coordinate?
|
Euler’s identity $$e^{ix}=\cos x+i\sin x$$ is related to the following geometric interpretation and can be proved by series $$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$
|
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|
2,660,364 |
Let $f:[a,b] \rightarrow \mathbb R$ be a continuous function with $f(a)=f(b)$ such that there exists in each point $x\in (a,b)$ the right derivative $f'_{+}(x)$. I want to show that there exist points $c_1,c_2 \in (a,b)$ such that $f'_{+}(c_1) \leq 0$ and $f'_{+}(c_2) \geq 0$.
(It is stated
in Remark to the Rolle's theorem in Course of mathematical analysis by L. Schwarz, chap.3, $\S2$.) This is obvious if $f$ is a constant function. Let us assume that $f$ has a positive value. Then
the point $c_1$ we find exactly as in the proof of the classic Rolle's theorem: let $f$ takes the maximum value in $c_1,$ then $f_{+}'(c_1)=\lim_{x\rightarrow c_1^+} \frac{f(x)-f(c_1)}{x-c_1} \leq 0$. If $\min f>0$, let $f$ takes the minimum value in $c_2$. Then $f_{+}'(c_2)=\lim_{x\rightarrow c_1^+} \frac{f(x)-f(c_1)}{x-c_1} \geq 0$. My question is: How to find $c_2$ in general?
|
Euler’s identity $$e^{ix}=\cos x+i\sin x$$ is related to the following geometric interpretation and can be proved by series $$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$
|
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|
2,660,379 |
In my research I have come across a linear, second-order ordinary differential equation that looks much like Bessel's equation except for one small difference:
\begin{equation}
z^2\frac{\partial^2}{\partial z^2}R_m(z) + z\frac{\partial}{\partial z}R_m(z) + \left(z^2+Wz-m^2\right)R_m(z) = 0,
\end{equation}
where $m$ and $W$ are constants. If $W\rightarrow0$, then Bessel's equation is recovered. I believe this equation may be solved as a series, but its similarity to Bessel's equation makes me wonder if there is another way (maybe through some clever substitution) that makes use of well known functions. I don't especially want to re-invent the wheel and I have tried to look up this equation, but I have not been able to find anything. Does anyone know if this equation has previously established solutions?
|
Euler’s identity $$e^{ix}=\cos x+i\sin x$$ is related to the following geometric interpretation and can be proved by series $$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$
|
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|
2,660,385 |
I have an array {1, 2, 3, 4, 5} and I extract 4 random numbers. I want to find out the probability that at least 3 numbers are identical. I concluded that it's the probability of having exactly 3 identical numbers added to the probability of having exactly 4 identical numbers. The probability of having exactly 3 numbers is: $$\binom{4}{3}*p^{3}*(1-p)^{4-3}$$ Now, my issue is, who is p? As far as I understand, p is: $$\frac{1}{5}$$ Since it represents the probability of extracting any random number, but I'm not sure.
|
Euler’s identity $$e^{ix}=\cos x+i\sin x$$ is related to the following geometric interpretation and can be proved by series $$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$
|
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|
2,660,386 |
Let $f$ be measurable in $\mathbb{R}$ and $\int_\mathbb{R} f dx =1$ Show that for any $r\in(0,1)$, there is a Lebesgue measurable set $E$ such that $\int_Ef dx =r$. I have tried the interval $(-R,R)$. If I can show that $R \mapsto \int_{-R}^R f dx$ is continuous and I can use the intermediate value theorem to show that there is a $R$ such that $\int_{-R}^R f dx=r$. If I consider $|\int f(\chi_{(-R_1,R_1)}-\chi_{(-R_2,R_2)}) dx|$, I do not know to show the continuous since I think that $\int_{-R}^R fdx$ may not be finite for some $R$ or $\int |f|dx$ may not be finite. Thank you so much!
|
Euler’s identity $$e^{ix}=\cos x+i\sin x$$ is related to the following geometric interpretation and can be proved by series $$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$
|
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|
2,661,443 |
For the equation $2^x = 7$ The textbook says to use log base ten to solve it like this $\log 2^x = \log 7$. I then re-arrange it so that it reads $x \log 2 = \log 7$ then divide the RHS by $\log 2$ to isolate the $x$. I understand this part. I can alternatively solve it in an easier way by simply using $\log_2 7$ on my calculator. Using both methods the answer comes to the same which is $2.807$ My question is twofold: Why would the textbook suggest to use log base ten rather than simply using log base two? I can see how using log base ten and the suggested method in the textbook makes me arrive at the same answer but I don't understand WHY this is so. How does base ten play a factor in the whole scope of things. Thank you
|
This works in any base because $\log_b (a^c) = c \log_b(a)$ The practical reason for using base $10$ was a little old fashioned: it allowed the use of tables of logarithms instead of a calculator and reducing these calculations to addition and subtraction. Calculators then provided a $\log_{10}$ function as a carry-over from tables and just called the button log My 1953 elementary statistical tables have logarithms of numbers from $1$ to $10$ and anti-logarithms of numbers from $0$ to $1$. Since these are logarithms base $10$, I can easily deal with all numbers because $\log(a \times 10^n)=n +\log(a)$ Here I want $\dfrac{\log 7}{\log 2}$. My tables tell me $\log 7\approx 0.8451_6$ (with the ${\,}_6$ helping interpolation) and $\log 2\approx 0.3010_{22}$. So that leaves me with trying to calculate $\dfrac{0.8451}{0.3010}$. I cannot be bothered to do long division, so instead I try to calculate $\text{antilog}\,({\log 0.8451 -\log 0.3010})$. My tables tell me $\log 8.45 \approx 0.9269_5$ so I write $\log 0.8451 \approx \bar{1}.9269$ (with the $\bar{1}$ because I wanted $\log (8.451\times 10^{-1})= -1+\log 8.451$). Similarly it tells me $\log 3.01 \approx 0.4786_{14}$ so I write $\log 0.3010 \approx \bar{1}.4786$ I now calculate by hand $\bar{1}.9269 - \bar{1}.4786 = 0.4483$. My tables tell me $\text{antilog}\,0.448 \approx 2.805_7$ and I interpolate the final $3$ using the ${\,}_7$ to give a final approximate answer of $2.807$. Which is what you got with some clever silicon
|
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|
2,662,005 |
Suppose $G$ and $H$ are groups and we have an equivalence of categories between $G\textrm{-}\mathbf{Set}$ and $H\textrm{-}\mathbf{Set}$. (One can think of this as a form of "nonlinear Morita equivalence".) What can be said about $G$ and $H$? I suspect that $G$ and $H$ have to be isomorphic, but I can't prove it. If this is too hard in general, I'm also interested in the case that $G$ and $H$ are assumed to be finite. Here are some things I've tried: Since $G$-modules are the abelian group objects in $G\textrm{-}\mathbf{Set}$, we get a Morita equivalence between $\Bbb{Z}[G]$ and $\Bbb{Z}[H]$. So if we tensor with any field $k$, we get a Morita equivalence between $k[G]$ and $k[H]$. Assume for a moment that $G$ and $H$ are finite. If we take $k = \Bbb{C}$, then this means that $G$ and $H$ have the same number of irreducible representations, so the same number of conjugacy classes. If we take $k=\Bbb{R}, \Bbb{Q}, \overline{\Bbb{F}_p}$, we also get the same number of real, rational and $p$-regular conjugacy classes. While this approach gives some common properties between $G$ and $H$, it's not possible to conclude that $G$ and $H$ are isomorphic, because there are known examples of non-isomorphic finite groups with isomorphic integral group algebras. (See here ) This means that we have to use some of the "nonlinear" information from the category $G\textrm{-}\mathbf{Set}$. Another thing I considered is that the automorphism group of the forgetful functor $G\textrm{-}\mathbf{Set} \to \mathbf{Set}$ is isomorphic to $G$, by a simple Yoneda-argument, so if we could somehow reconstruct the forgetful functor just from the category $G\textrm{-}\mathbf{Set}$, this would show that $G$ and $H$ must be isomorphic. I haven't been able to do this, but I reconstructed some other functors: The terminal object in $G\textrm{-}\mathbf{Set}$ is a one-point set with a trivial action, denote this $G$-set by $\{*\}$, we have a natural bijection $\operatorname{Hom}_{G\textrm{-}\mathbf{Set}}(\{*\},X) \cong X^G$, where $X^G$ denotes the set of fixed points under the action of $G$.
So we can reconstruct the fixed point functor $G\textrm{-}\mathbf{Set} \to \mathbf{Set}$. The left adjoint of that functor is the functor $\mathbf{Set} \to G\textrm{-}\mathbf{Set}$ which gives each set a trivial $G$-action. Denote $X$ with a trivial $G$-action by $X_{triv}$. We have $\operatorname{Hom}_{G\textrm{-}\mathbf{Set}}(X,Y_{triv}) \cong \operatorname{Hom}_{\mathbf{Set}}(X/G,Y)$, so the functor which sends each $G$-set to the orbit space $X/G$ is left adjoint to the functor which gives each set a trivial $G$-action, so we can reconstruct the functor $X \mapsto X/G$ from the category $G\textrm{-}\mathbf{Set}$. Not sure if that's helpful. Maybe it's even possible that $G$ and $H$ don't have to be isomorphic? I'm looking either for a counterexample or a proof.
|
The empty $G$-set is the initial object. The coproduct of $G$-sets is the disjoint union. Call a $G$-set “indecomposable” if it is not the coproduct of two non-empty $G$-sets, so an indecomposable $G$-set is just a transitive one. An epimorphism of $G$-sets is just a surjective map of $G$-sets. Up to isomorphism, there is a unique transitive $G$-set with an epimorphism to every other transitive $G$-set, namely the regular $G$-set, whose automorphism group is $G$. So $G$ can be recovered from the category of $G$-sets.
|
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|
2,662,007 |
I'd like to show that the group given by presentation $G=\langle a,b \ | \ a^2b^2 \rangle$ is not isomorphic to any of $\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, or $\mathbb{Z} \times \mathbb{Z}$. For $G \ncong \mathbb{Z}$, I note that $G$ is not free because it has a relator in its presentation. For $G \ncong \mathbb{Z}/2\mathbb{Z}$, I claim that $G$ cannot be of order 2, since the subgroup $\langle a \rangle$ in $G$ is infinite: if I write $a^n=1$ for some $n \in \mathbb{Z}\ \backslash\{0\}$ then $a^n$ has to be obtained by a finite number of composition of self-inversions, or self-multiplications, or conjugations on the relator $a^2b^2$. All of these operations preserve the presence of $b$, so we can never get to $a^n$. For $G \ncong \mathbb{Z} \times \mathbb{Z}$ I argue that $G$ is not abelian. The abelianization of $G$ is $G_{ab}=\langle a,b \ | \ a^2b^2, aba^{-1}b^{-1} \rangle$. How do I show $G_{ab}$ is not trivial? I'm new to presentations and am completely unsure if the above reasonings are allowed or not. Any help appreciated.
|
The empty $G$-set is the initial object. The coproduct of $G$-sets is the disjoint union. Call a $G$-set “indecomposable” if it is not the coproduct of two non-empty $G$-sets, so an indecomposable $G$-set is just a transitive one. An epimorphism of $G$-sets is just a surjective map of $G$-sets. Up to isomorphism, there is a unique transitive $G$-set with an epimorphism to every other transitive $G$-set, namely the regular $G$-set, whose automorphism group is $G$. So $G$ can be recovered from the category of $G$-sets.
|
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|
2,662,174 |
To my understanding, the Taylor series is a type of power series that provides an approximation of a function at some particular point $x=a$. But under what circumstances is this approximation perfect, and under what circumstances is it "off" even at infinity? I realize is a little hazy so I'll rephrase: By "perfect" I refer to how a regular limit doesn't ever actually reach something but instead provides a sort of error term that you can make as small as you want, so for all practical purposes we treat it as zero error. Whereas for something that is an imperfect approximation maybe that arbitrarily small error piece doesn't exist, or maybe the function is only correct for that particular point and nowhere else, etc. So maybe what I am asking is when the Taylor series provides an equivalent representation of the function over all $x$ in $f$'s domain, and when does it not? And when it doesn't, how do we even know?
|
Limits are exact You have a misunderstanding about limits! A limit, when it exists is just a value. An exact value. It doesn't make sense to talk about the limit reaching some value, or there being some error. $\lim_{x \to 1} x^2$ is just number, and that number is exactly one. What you are describing — these ideas about "reaching" a value with some "error" — are descriptions of the behavior of the expression $x^2$ as $x \to 1$. Among the features of this behavior is that $x^2$ is "reaching" one. By its very definition, the limit is the exact value that its expression is "reaching". $x^2$ may be "approximately" one, but $\lim_{x \to 1} x^2$ is exactly one. Taylor polynomials In this light, nearly everything you've said in your post is not about Taylor series , but instead about Taylor polynomials . When a Taylor series exists, the Taylor polynomial is given simply by truncating the series to finitely many terms. (Taylor polynomials can exist in situations where Taylor series don't) In general, the definition of the $n$-th order Taylor polynomial for an $n$-times differentiable function is the sum $$ \sum_{k=0}^n f^{(k)}(x) \frac{x^k}{k!} $$ Taylor polynomials, generally, are not exactly equal to the original function. The only time that happens is when the original function is a polynonial of degree less than or equal to $n$. The sequence of Taylor polynomials, as $n \to \infty$, may converge to something. The Taylor series is exactly the value that the Taylor polynomials converge to. The error in the approximation of a function by a Taylor polynomial is something people study. One often speaks of the "remainder term" or the "Taylor remainder", which is precisely the error term. There are a number of theorems that put constraints on how big the error term can be. Taylor series can have errors! Despite all of the above, one of the big surprises of real analysis is that a function might not be equal to its Taylor series! There is a notorious example: $$ f(x) = \begin{cases} 0 & x = 0 \\ \exp(-1/x^2) & x \neq 0 \end{cases} $$ you can prove that $f$ is infinitely differentiable everywhere. However, all of its derivatives have the property that $f^{(k)}(0) = 0$, so its Taylor series around zero is simply the zero function. However, we define A function $f$ is analytic at a point $a$ if there is an interval around $a$ on which $f$ is (exactly) equal to its Taylor series. "Most" functions mathematicians actually work with are analytic functions (e.g. all of the trigonometric functions are analytic on their domain), or analytic except for obvious exceptions (e.g. $|x|$ is not analytic at zero, but it is analytic everywhere else).
|
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|
2,665,752 |
We repeat the phrase "approach zero" regularly, but what exactly does it mean to say "$\delta t$ approaches zero"? P.S. If this a stupid question, then please forgive me.
|
Your question isn't stupid, it's the heart of calculus. An introductory step from algebra to calculus is in the context of slope. Algebra allows us to find an average slope, while calculus allows us to find the instantaneous slope. In other words, algebra gives us the slope of a line, while calculus gives us the slope of a point. Slope of a point? Yes, but let's stay with lines for now. The slope of a line is given by the following function. We divide the change in $y$ by the change in $x$: $$m=\frac{y_2 - y_1}{x_2 - x_1}$$ However, if the line is curved, say $f(x)=x^2$, then each point on that line has a different slope. If we are asked to find the slope when $x=5$, then we can approximate it by finding the average slope between $x=4$ and $x=6$: $$m=\frac{6^2-4^2}{6-4}$$ But this isn't the correct answer. If we wanted to get closer to the correct answer, we would choose values that are closer to 5: $$m=\frac{5.1^2-4.9^2}{5.1-4.9}$$ The pattern to notice is that the more accurate our answer becomes, the smaller the difference between $x_2$ and $x_1$. In fact, the correct answer will be found when the difference is zero. However, when we go to write this down, we have a problem: $$m=\frac{0}{0}$$ Specifically, we can't divide by zero. We've gone as far as algebra can take us, and we need a new way to talk about math. We need calculus. In algebra, we saw that we get closer and closer to the correct answer. In calculus, this is called the "limit". We get closer and closer to the limit as the divisor gets closer and closer to zero. The divisor "approaches zero". Finally, we have "What is the limit as x approaches zero?"
|
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|
2,665,782 |
I'm currently working with infinite series for my calculus class, and I'm wondering whether we always (in theory) can establish whether a series is divergent or convergent? Of course, it might be computationally hard, but is there a class of series where we simply lack the tools to determine whether the series converges or diverges?
|
I'm not sure whether this is what you want or not. Consider the series $\displaystyle\sum_{n=1}^\infty a_n$ , with $$a_n=\begin{cases}1&\text{ if }2^n-1\text{ is prime}\\0&\text{ otherwise.}\end{cases}$$ It is not known whether it converges or not.
|
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|
2,665,825 |
I am currently stuck at this question and have no idea how to solve.
I just started out learning linear and I'm really weak in this field. Justify, without evaluating, that the determinant of the following matrix is zero Here's the matrix A: $$\begin{pmatrix}
1 & 0 & 2 & 4\\
-2 & 3 & 8 & 6\\
-1 & 3 & 10 & 10\\
6 & 6 & -3 & 7\\
\end{pmatrix}$$ I searched online but couldn't find something similar.
What I found though was that if it was skew-symmetric ($A^t= -A$)
then the determinant could directly be said to be equal to zero.
But in this case it didn't work with me. Thank you.
|
The third row is the sum of the first and second rows. The rows are not linearly independent, so the determinant is zero.
|
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|
2,665,897 |
How to represent 0 as rational number? $0/0$ is not legitimate, $0/\text{const}$ should be good enough, but what is the right value of const? $0/1$ works for a lot of computational cases, but only on intuitive.
|
There is no single right value. The rational number $0$ can be represented as the quotient of an integer by a non-zero integer in infinitely many ways: $0=\frac 0d$ ($d\in\mathbb{Z}\setminus\{0\}$). Choosing $d=1$ is a natural choice: every rational number can be represented in one and only one way as $\frac nd$ with $n\in\mathbb Z$, $d\in\mathbb N$ and $\gcd(n,d)=1$. In the case of $0$, that representation is $\frac01$.
|
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|
2,665,915 |
A straight line is in $1$ dimension, A square in $2$, but what about a curved line? A straight line looks the same no matter how you look at it but a square doesn't. It looks like a straight line if looked sideways. But a curved line, if looked sideways it looks a shorter straight line. So which dimension does it belong to? Thanks.
|
There is no single right value. The rational number $0$ can be represented as the quotient of an integer by a non-zero integer in infinitely many ways: $0=\frac 0d$ ($d\in\mathbb{Z}\setminus\{0\}$). Choosing $d=1$ is a natural choice: every rational number can be represented in one and only one way as $\frac nd$ with $n\in\mathbb Z$, $d\in\mathbb N$ and $\gcd(n,d)=1$. In the case of $0$, that representation is $\frac01$.
|
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|
2,666,834 |
I know count is simply the amount of times the observations occur in a single bin width, but what is density?
|
Illustrations: Suppose $X_1, X_2, \dots, X_{100}$ is a random sample of size $n$ from a normal
distribution with mean $\mu=100$ and standard deviation $\sigma=12.$ Also, we have bins (intervals) of equal width, which we use to make a histogram. The vertical scale of a 'frequency histogram' shows the number of observations
in each bin. Optionally, we can also put numerical labels atop each bar that show
how many individuals it represents. The vertical scale of a 'density histogram' shows units that make the total
area of all the bars add to $1.$ This makes it possible to show the density
curve of the population using the same vertical scale. From above, we know that the tallest bar has 30 observations, so this bar accounts for relative frequency $\frac{30}{100} = 0.3$ of the observations. The width of this bar is $10.$ So
its density is $0.03$ and its area is $0.03(10) = 0.3.$ The
density curve of the distribution $\mathsf{Norm}(100, 15)$ is also shown superimposed on the histogram. The area beneath this density curve is also $1.$ (By definition, the area beneath a density function is always $1.)$ Optionally, I have added tick marks below the histogram to show the locations of the individual observations. Definitions: If the frequency of the $i$ th bar is $f_i,$ then its relative
frequency is $r_i = f_i/n,$ where $n$ is the sample size. Its density is $d_i = r_i/w_i,$ where $w_i$ is its width. Ordinarily, you should make a density
histogram only if each bar has the same width. Notes: (1) Another type of histogram (that you did not ask about) would be a 'relative frequency' histogram with
relative frequencies (not densities) on the vertical scale. (2) The sample mean of the data shown is $\bar X =102.98$ and the sample standard deviation is $S = 15.37.$ (3) These histograms were made using R statistical software.
|
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|
2,667,980 |
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.
|
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...
|
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|
2,668,264 |
We define a $2\times 2$ Givens rotation matrix as: $${\bf G}(\theta) = \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) &\cos(\theta) \end{bmatrix}.$$ On the other hand, we define a $2\times 2$ hyperbolic rotation matrix as: $${\bf H}(y)=\begin{bmatrix}
\cosh( y) & \sinh( y) \\
\sinh( y) &\cosh( y) \end{bmatrix}.$$ I don't see why do we qualify matrix ${\bf H}$ as a rotation! Suppose we take a 2-D vector $x=[-3, 1]^T$ and we transform it using ${\bf G}(\theta), \theta = 0,\dots, \pi/2$, and ${\bf H}, y = -2,\dots, 2.5$. See below for the result. For me Givens rotation does clearly rotate the initial point around the point $[0,0]^T$ but for the hyperbolic rotation, we see a bending but not a rotation, at least not around a fixed point (I checked for other points and its the same behavior with different bending angles). am I missing something?
|
You can say that ${\bf H}(y)$ is a rotation, not for the usual inner product $$\langle {\bf x}, {\bf y}\rangle_E = x_1y_1 + x_2y_2,$$but for the Lorentz-Minkowski product $$\langle {\bf x}, {\bf y}\rangle_L = x_1y_1 - x_2y_2.$$ In the same way that orthogonal transformations are linear maps preserving $\langle\cdot,\cdot\rangle_E$, we call the linear maps preserving $\langle \cdot,\cdot\rangle_L$ Lorentz transformations . The point is that ${\bf H}(y)$ is not an orthogonal map, but a Lorentz transformation. You can see it as a "rotation" moving points along hyperbolas $xy = {\rm constant}.$ In $\Bbb R^n$, consider $$\langle {\bf x},{\bf y}\rangle_L = x_1y_1+\cdots+x_{n-1}y_{n-1} - x_ny_n.$$Call ${\bf x} \neq {\bf 0}$ spacelike , timelike or lightlike if $\langle {\bf x}, {\bf x}\rangle_L$ is positive, negative, or zero. In the Euclidean case, we call the elements of ${\rm SO}(n,\Bbb R)$ (orthogonal maps with unit determinant) rotations (with respect to the Euclidean inner product). You can mimic that and say that Lorentz transformations with unit determinant are rotations with respect to the Lorentz-Minkowski product. You must be careful in odd dimensions, though. For example, in $n=3$ Lorentz transformations with unit determinant always have an eigenvector. The rotation will be called hyperbolic (resp. elliptic, parabolic) if said eigenvector is spacelike (resp. timelike, lightlike). In your case we consider ${\bf H}(y)$ a hyperbolic rotation seeing the plane $\Bbb R^2$ as the $xz$ (or $yz$) plane in $\Bbb R^3$, so that the eigenvector $(1,0,0)$ of $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cosh y & \sinh y \\ 0 & \sinh y & \cosh y\end{pmatrix}$$is spacelike.
|
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|
2,670,344 |
A friend of mine has set me the challenge of finding an example of the following: Is there a question, that everyone (both mathematicians and non-mathematicians) can understand, that most mathematicians would answer correctly, instantly, but that most non-mathematician would struggle to solve/take longer to arrive at a solution? When I say mathematician, I mean a person who has studied maths at degree level. The key feature of such a question is that it should be understandable to an average person. I realise, of course, that this is a subjective question - what does 'most mathematicians' mean, what would 'most mathematicians' be able to answer? Nonetheless, I would be interested to hear peoples' opinions and ideas: Can you think of a question, that in your opinion, is an example of the above? An example of such a question, I think, would provide a good way of explaining to people how mathematicians think. It could also be a good teaching tool (i.e to show how mathematicians approach problem solving). My friend suggested the following question: Does there exist a completed Sudoku grid with top row $1,2,3,4,5,6,7,8,9$? I won't give the answer (so that you can see for yourself if it works!). When we asked this to fellow maths researchers, nearly all were able to give the correct answer immediately. When I asked the undergraduates that I teach, most of them (but not all) could answer correctly and pretty quickly. I like this question but I'm sure there's a better one. Thanks! Edit: As for the Sudoku question, most of the researchers I asked answered in 10 seconds with the solution: Yes. You can relabel any Sudoku (i.e swaps sets of numbers - change all $1$s for $2$s for example) and still have a valid Sudoku solution. So, in a sense all Sudoku are equivalent to a Sudoku with one to nine in the first row.
|
A good example is the Bridges of Königsberg puzzle. An important city in 18 th century Prussia was the city of Königsberg (modern day Kaliningrad, a Russian enclave) which had seven bridges. The residents played a game: try to cross every bridge precisely once. No one could solve this puzzle for a long time. Euler proved this was impossible. Every mathematician knows the solution and how to solve similar problems, although this is by training rather than their own mental guile :-)
|
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|
2,670,781 |
Is it possible for a parallelogram to have whole-number lengths for all four sides and both diagonals? One idea I had was to arrange four identical right triangles such that the right angles are adjacent. For example if we take four triangles that are all 3-4-5 right triangles, and arrange them so that their legs form a cross with two arms that are 3 units and two arms that are 4 units, their hypotenuses will form a parallelogram in which all sides are 5 units – that is, a rhombus. A second idea was to take two pairs of isosceles triangles with all legs the same length, where one vertex angle is supplementary to another, and arrange them so that their vertex angles are adjacent. For example we could take two triangles with two sides of 6 units that form a 25° angle, and another two triangles with two sides of 6 units that form a 155° angle, and arrange them to form a quadrilateral with 12-unit diagonals – that is, a rectangle. That's all I can imagine. My hunch is that a parallelogram can have whole-number lengths for all four sides and both diagonals only if it is either a rhombus or a rectangle. Is that right? If so, can this limitation be explained elegantly?
|
Your hunch is wrong. A non-rectangular non-rhomboid parallelogram with integer side and diagonal lengths exists: Suppose that the parallelogram is $ABCD$ with $AB=a$ , $AD=b$ and $BD=c$ . By the law of cosines: $$\cos\angle DAB=\frac{a^2+b^2-c^2}{2ab}$$ Since $\angle CDA=\pi-\angle DAB$ , $\cos\angle CDA=-\cos\angle DAB$ . Then $$AC^2=d^2=a^2+b^2-2ab\cos\angle CDA=a^2+b^2+2ab\cdot\frac{a^2+b^2-c^2}{2ab}=2a^2+2b^2-c^2$$ $$\color{red}{c^2+d^2=2(a^2+b^2)}$$ If $c=d$ we have a rectangle; if $a=b$ we have a rhombus. Thus, we look for a number that is a sum of two unequal squares $a$ and $b$ whose double is also a sum of two unequal squares $c$ and $d$ , with triangle inequalities satisfied to ensure the parallelogram is non-degenerate: $|a-b|<c,d<a+b$ . Fixing $a$ and $b$ , a trivial choice is $c=a-b$ and $d=a+b$ because $2(a^2+b^2)=(a-b)^2+(a+b)^2$ . However, these assignments do not satisfy the triangle inequalities, making numbers that have multiple representations as sums of two squares valuable for this problem. I used OEIS A025426 to find them; the first number I saw was $145=9^2+8^2=12^2+1^2$ , whose double is $290=13^2+11^2=17^2+1^2$ . The first representations listed here allowed me to quickly construct the parallelogram above, although it is not the smallest: there is a parallelogram with sides 4 and 7, diagonals 7 and 9. The numbers being squared in the above equations relate to the above parallelogram as follows. The lengths of its sides are 9 and 8, and of its diagonals 13 and 11. If its left side is extended upwards along the same straight line by 8 (so that it is now 16), the resulting quadrilateral has sides 16, 9, 8 and 11, and its diagonals are 17 and 13. Here is a way to efficiently generate infinitely many such integer parallelograms. Let $r$ and $s$ be two coprime integers with $r>s>0$ and $(r,s)\ne(3,1)$ . The product $(2^2+1^2)(r^2+s^2)$ can be written as a sum of two squares in two different ways (the Brahmagupta–Fibonacci identity ): $$(2^2+1^2)(r^2+s^2)=(2r+s)^2+(2s-r)^2=\color{blue}{(2s+r)^2+(2r-s)^2}$$ From these two representations, we can write twice the product as two different sums of two squares too: $$2(2^2+1^2)(r^2+s^2)=\color{blue}{(2(r+s)-(r-s))^2+(2(r-s)+(r+s))^2}
=(2(r-s)-(r+s))^2+(2(r+s)^2+(r-s))^2$$ To satisfy the triangle inequalities we choose $$\color{blue}{a=2r-s\qquad b=2s+r\qquad c=2(r+s)-(r-s)\qquad d=2(r-s)+(r+s)}$$ These are guaranteed to form a non-rectangular non-rhomboid integer parallelogram with the given restrictions on $r$ and $s$ . The one pictured at the top of this answer corresponds to $(r,s)=(5,2)$ and the smallest instance (the one with side lengths 4 and 7) corresponds to $(r,s)=(3,2)$ .
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|
2,670,785 |
For an assignment I need to solve a $4\times 3$ matrix (?) with the Gaussian elimination method, as well as finding the coefficients-matrix (2. language) and the determinant of the matrix.
I am having issues setting up the coefficients-matrix, is this correct? Or should I include the zeroes since there are missing values in each row? A = The coeff-matrix $$x_1-0.25x_2-0.25x_3=50$$
$$-0.25x_1+x_2-0.25x_4=50$$
$$-0.25x_1+x_3-0.25x_4=25$$
$$-0.25x_2-0.25x_3+x_4=25$$ $$A= \begin{pmatrix}
1 & -0.25 & -0.25 \\
-0.25 & 1 & -0.25 \\
-0.25 & 1 & -0.25 \\
-0.25 & -0.25 & 1 \\
\end{pmatrix}$$ If anyone could clarify how I should think around this one I would be grateful. Jonas
|
Your hunch is wrong. A non-rectangular non-rhomboid parallelogram with integer side and diagonal lengths exists: Suppose that the parallelogram is $ABCD$ with $AB=a$ , $AD=b$ and $BD=c$ . By the law of cosines: $$\cos\angle DAB=\frac{a^2+b^2-c^2}{2ab}$$ Since $\angle CDA=\pi-\angle DAB$ , $\cos\angle CDA=-\cos\angle DAB$ . Then $$AC^2=d^2=a^2+b^2-2ab\cos\angle CDA=a^2+b^2+2ab\cdot\frac{a^2+b^2-c^2}{2ab}=2a^2+2b^2-c^2$$ $$\color{red}{c^2+d^2=2(a^2+b^2)}$$ If $c=d$ we have a rectangle; if $a=b$ we have a rhombus. Thus, we look for a number that is a sum of two unequal squares $a$ and $b$ whose double is also a sum of two unequal squares $c$ and $d$ , with triangle inequalities satisfied to ensure the parallelogram is non-degenerate: $|a-b|<c,d<a+b$ . Fixing $a$ and $b$ , a trivial choice is $c=a-b$ and $d=a+b$ because $2(a^2+b^2)=(a-b)^2+(a+b)^2$ . However, these assignments do not satisfy the triangle inequalities, making numbers that have multiple representations as sums of two squares valuable for this problem. I used OEIS A025426 to find them; the first number I saw was $145=9^2+8^2=12^2+1^2$ , whose double is $290=13^2+11^2=17^2+1^2$ . The first representations listed here allowed me to quickly construct the parallelogram above, although it is not the smallest: there is a parallelogram with sides 4 and 7, diagonals 7 and 9. The numbers being squared in the above equations relate to the above parallelogram as follows. The lengths of its sides are 9 and 8, and of its diagonals 13 and 11. If its left side is extended upwards along the same straight line by 8 (so that it is now 16), the resulting quadrilateral has sides 16, 9, 8 and 11, and its diagonals are 17 and 13. Here is a way to efficiently generate infinitely many such integer parallelograms. Let $r$ and $s$ be two coprime integers with $r>s>0$ and $(r,s)\ne(3,1)$ . The product $(2^2+1^2)(r^2+s^2)$ can be written as a sum of two squares in two different ways (the Brahmagupta–Fibonacci identity ): $$(2^2+1^2)(r^2+s^2)=(2r+s)^2+(2s-r)^2=\color{blue}{(2s+r)^2+(2r-s)^2}$$ From these two representations, we can write twice the product as two different sums of two squares too: $$2(2^2+1^2)(r^2+s^2)=\color{blue}{(2(r+s)-(r-s))^2+(2(r-s)+(r+s))^2}
=(2(r-s)-(r+s))^2+(2(r+s)^2+(r-s))^2$$ To satisfy the triangle inequalities we choose $$\color{blue}{a=2r-s\qquad b=2s+r\qquad c=2(r+s)-(r-s)\qquad d=2(r-s)+(r+s)}$$ These are guaranteed to form a non-rectangular non-rhomboid integer parallelogram with the given restrictions on $r$ and $s$ . The one pictured at the top of this answer corresponds to $(r,s)=(5,2)$ and the smallest instance (the one with side lengths 4 and 7) corresponds to $(r,s)=(3,2)$ .
|
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|
2,671,327 |
Suppose that there are $10$ people at a party whose (integer) ages range from $0$ to $100$. Show that there are two distinct, but not necessarily disjoint, subset of people that have exactly the same total age. So my thoughts so far are that you could split the ages $0-100$ $(101$ numbers$)$ into $11$ groups (pigeons) and then categorize these groups into the $10$ people (holes). However this seems backwards and not sound to me. Is there a different way to think about this problem? Any help is appreciated.
|
The sum of ages cannot exceed $100 \times 10=1000$ $(1001$ cages, including $0 )$. The total ways to chose subsets of people are $2^{10}=1024$ $(1024$ pigeons$)$. Can you see that coming?
|
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|
2,671,760 |
$$\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$$ $$= \frac{1-0-(1-0)^3}{1-(1-0)^3}$$ $$=\frac{0}{0}$$ $$\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$$ $$=\lim_{p \rightarrow 0}\frac{p^2-3p+2}{p^2-3p+3}$$ $$=\frac{0^2-3(0)+2}{0^2-3(0)+3}$$ $$=\frac{2}{3}$$ Edit 1: $0\le p \le1$
|
You conveniently omitted the crucial step $$\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$$ $$=\color{red}{\lim_{p \rightarrow 0}\frac{p^3-3p^2+2p}{p^3-3p^2+3p}}$$ $$=\color{red}{\lim_{p \rightarrow 0}\frac pp} \cdot\lim_{p \rightarrow 0}\frac{p^2-3p+2}{p^2-3p+3}$$ $$=\frac{0^2-3(0)+2}{0^2-3(0)+3}$$ $$=\frac{2}{3}$$ The expression in red is clearly of the form $\dfrac 00$ too but will not be any more after you divide the numerator and the denominator by $p$. Functions that behave like that are said to have removable singularities.
|
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|
2,671,765 |
The area of the front, top and side of a right rectangular prism are $36 cm^2, 40cm^2$ and $45cm^2$ respectively. Determine the exact length of the diagonal of the prism. So I know that the diagonal, if the length, width and height were each $a, b, c$, would be $\sqrt{a^2+b^2+c^2}$. Also, $a^2+b^2+c^2 \ge ab+bc+ca$. WLOG, let $ab = 36, bc = 40, ca = 45$. Then $ab+bc+ca = 121$, and the diagonal will be at least $11$. I'm quite inclined to believe this is the answer, but I'm having a problem making sure. For example, I could calculate that $abc = 180\sqrt{2}$, but that doesn't seem to help me... Any help would be really appreciated!
|
You conveniently omitted the crucial step $$\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$$ $$=\color{red}{\lim_{p \rightarrow 0}\frac{p^3-3p^2+2p}{p^3-3p^2+3p}}$$ $$=\color{red}{\lim_{p \rightarrow 0}\frac pp} \cdot\lim_{p \rightarrow 0}\frac{p^2-3p+2}{p^2-3p+3}$$ $$=\frac{0^2-3(0)+2}{0^2-3(0)+3}$$ $$=\frac{2}{3}$$ The expression in red is clearly of the form $\dfrac 00$ too but will not be any more after you divide the numerator and the denominator by $p$. Functions that behave like that are said to have removable singularities.
|
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|
2,672,360 |
I'm thinking in a function and if it's possible to solve that. I have been playing with the cube using the following move: $R U L' U'.$ I notice that the cube solves itself with a certain number of moves: 28 moves to $2\times2\times2$ and 112 to $3\times3\times3.$ (If the cube already is solved). Then I'm trying to create a formula to calculate the number of moves for the another cubes like $4\times4\times4, 5\times5\times5, 6\times6\times6\dots$ Since: $x \rightarrow y$ $1 \rightarrow 0$ $2 \rightarrow 28$ $3 \rightarrow 112$ $4 \rightarrow z$ Where x is the number of the cube $(2 = 2\times2\times2, 3 = 3\times3\times3 \dots)$ and y is the amount of moves, I came up with two formulas: 28*$((x-1)^{(x-1)})$ and 28*$(x-1)^2$ . Thus, the value for $z$ could be 252 or 756 , My questions are: Are any of these formulas correct? If so, which one? Can be my reasoning corret about the formulas? If I'm wrong, answer me why!
|
The answer for a $4 \times 4 \times 4$, $5\times 5\times 5$, and so on cube will continue to be $112$. Here's why. Imagine that you take an $n \times n \times n$ cube and, on each face, glue the pieces in the middle $(n-2) \times (n-2)$ square together, so they cannot be separated. Similarly, glue the middle $n-2$ pieces along each edge together. What you have is no longer an $n \times n \times n$ cube, but a $3 \times 3 \times 3$ cube in which some of the pieces are much larger in size. But it still operates exactly like a $3 \times 3 \times 3$ cube and affords exactly the same twists. Your repeated RULU' move is acting on the $n \times n \times n$ cube in a way that gluing the pieces as above doesn't forbid. So whatever happens on the $n \times n \times n$ cube for $n>3$ is exactly the same as what happens for the $3 \times 3 \times 3$ cube. The only reason that the $2 \times 2 \times 2$ cube is an exception is that here, there are no "edge" pieces at all. So if a sequence of moves in the $3 \times 3 \times 3$ cube scrambles (or flips) the edges but leaves the corners fixed, then on the $2 \times 2 \times 2$ cube, it does nothing, and this exactly describes what $28$ iterations of RULU' do.
|
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|
2,672,376 |
I have this function: $$f(x)=\frac{x^{3}+3(\sqrt{n}-2)x^{2}-24\sqrt{n}x-2}{n^{2}}$$ I have to find the critical point of this funcion depending on $n\ge1\in \mathbb N$. So I have computed the first derivative $$f'(x)=\frac{3x^{2}+6(\sqrt{n}-2)x-24 \sqrt{n}}{n^{2}}$$ So the problem is, since in this case $n^{2}\ge1$ is always positive and $\neq 0$, studying: $$3x^{2}+6(\sqrt{n}-2)x-24 \sqrt{n}=0$$ I've tried and tried to solve this equation without finding the correct answer. Could someone help me? Thank you :).
|
The answer for a $4 \times 4 \times 4$, $5\times 5\times 5$, and so on cube will continue to be $112$. Here's why. Imagine that you take an $n \times n \times n$ cube and, on each face, glue the pieces in the middle $(n-2) \times (n-2)$ square together, so they cannot be separated. Similarly, glue the middle $n-2$ pieces along each edge together. What you have is no longer an $n \times n \times n$ cube, but a $3 \times 3 \times 3$ cube in which some of the pieces are much larger in size. But it still operates exactly like a $3 \times 3 \times 3$ cube and affords exactly the same twists. Your repeated RULU' move is acting on the $n \times n \times n$ cube in a way that gluing the pieces as above doesn't forbid. So whatever happens on the $n \times n \times n$ cube for $n>3$ is exactly the same as what happens for the $3 \times 3 \times 3$ cube. The only reason that the $2 \times 2 \times 2$ cube is an exception is that here, there are no "edge" pieces at all. So if a sequence of moves in the $3 \times 3 \times 3$ cube scrambles (or flips) the edges but leaves the corners fixed, then on the $2 \times 2 \times 2$ cube, it does nothing, and this exactly describes what $28$ iterations of RULU' do.
|
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|
2,672,615 |
Here's a cute question I came up with. Start with a circle, and then choose three points $a$ , $b$ , and $c$ on the circle, and proceed as follows: Draw the triangle inside the circle with vertices $a$ , $b$ , and $c$ Draw the inscribed circle of that triangle, which is tangent to each of the three sides of triangle, and now label these three points of tangency as $a$ , $b$ , and $c$ (so we're updating which points we're calling points $a$ , $b$ , and $c$ ). Repeat from Step 1. This construction gives us a sequence of inscribed circles and triangles that telescope down to a point, a limit point , which is determined only by the initial choice of the points $a$ , $b$ , and $c$ . Which points in the interior of the circle are limit points of this construction? Also, if anyone has ideas for more interesting variations of this question, I'd like to hear them.
|
Without loss of generality assume that the circle has radius $1$, and that the triple $(a,b,c)$ is positively oriented (i.e., they go counterclockwise around the circle). As the problem is circularly symmetric, it suffices to show that we can choose $a,b,c$ so that the limit point has any specified distance from its center. For any $\epsilon > 0$, we can choose $a, b, c$ such that the triangle $abc$ does not intersect the circle of radius $1-\epsilon$. Since the limit point must lie inside the triangle $abc$, it follows that the limit point can come arbitrarily close to the boundary. On the other hand, if the triangle $abc$ is equilateral, the limit point is clearly the center of the circle (as it is fixed under a rotation by $\frac{2 \pi}{3}$ about the center). So, if we can show that the map from $(a,b,c)$ to the limit point is continuous, it will follow that the limit point can have any distance less than $1$ from the center. The set of all possible distances will be a continuous image in $\Bbb{R}$ of the connected set of all possible oriented triples $(a,b,c)$, so it must be an interval; we have showed that this interval both contains $0$ and comes arbitrarily close to $1$.
|
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|
2,672,742 |
It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} $ The answer might be $-1$, but I'm not sure as $i$ is not a real number. Can anyone help?
|
\begin{eqnarray*}
x= a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { \cdots } } } } } } \\
x=a^{ 1+1/2+1/4+1/8+\cdots} \\
x=a^2
\end{eqnarray*}
So it would seem that
\begin{eqnarray*}
i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }=\color{red}{-1}.
\end{eqnarray*}
|
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|
2,674,934 |
Say I have three numbers, $a,b,c\in\mathbb C$ . I know that if $a$ were complex, for $abc$ to be real, $bc=\overline a$ . Is it possible for $b,c$ to both be complex, or is it only possible for one to be, the other being a scalar?
|
For example $z^3=1$, where $z\neq1.$ Id est, $$a=b=c=-\frac{1}{2}+\frac{\sqrt3}{2}i.$$
|
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|
2,675,852 |
E and F are independent iff $\frac{p(E \cap F)}{p(F)}=p(E)$. Also, E and F are independent iff $p(E | F)=p(E)$ Why do I get nonsense if I combine the two? I'd get: E and F are independent iff $p(E | F) = \frac{p(E \cap F)}{p(F)}$ Why doesn't the iff carry over when I combine the two equations?
|
Let $a=\frac{p(E \cap F)}{p(F)}$, $b=p(E | F)$, and $c=p(E)$. You know that the equations $a=c$ and $b=c$ are equivalent to each other (since they are both equivalent to "$E$ and $F$ are independent"). That is, whenever $a=c$ is true, $b=c$ is also true, and conversely. There is no reason for $a=b$ to be equivalent to these two statements, though. If $a=c$ is true, then $b=c$ is also true, and so $a=b$ is true as well. So we have an implication in one direction. But the other direction doesn't work: if we know $a=b$, that doesn't tell us anything about whether $a=c$ or $b=c$ are true. It's entirely possible that $a$ and $b$ are equal to each other, but not to $c$.
|
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|
2,676,994 |
I am referring to two Peano axioms: If $x = y$ then $y = x$ (symmetry) If $x = y$ and $y = z$ then $x = z$ (transitivity) What I don't understand is why we need the transitive axiom. Isn't it already implied by the symmetry axiom? Is it more of a convenience? Can we prove that transitivity is a required axiom? What happens if we were to remove it? Or should I think of this more like "symmetry just says we can physically flip the order of the equalities, and transitivity lets us swap things around as long as they're all equal to each other"?
|
Consider the following relation: $x$ opposes $y$ if and only if $x = -y$. Can you see that the "opposes" relation is symmetrical but not transitive?
|
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|
2,677,018 |
I'm solving this inequality: $$x^4+5x^2 \geq 36$$ I have advanced to this point: $$ x^4+5x^2-36\geq 0$$ I established that $$ u= x^2$$ Therefore $$ x^4+5x^2-36=(u-4)(u+9)$$
$$ (u-4)= (x+2)(x-2)$$
$$ (u+9)= (x^2+9)$$
$$(x+2)(x-2)(x^2+9)\geq 0$$
$$ x\leq -2 \ x\geq 2$$
As you can see I have already solved the inequality, but I need to explain why all the values of $(x^2+9)$ are positive without using square root.
|
Consider the following relation: $x$ opposes $y$ if and only if $x = -y$. Can you see that the "opposes" relation is symmetrical but not transitive?
|
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|
2,677,200 |
This question is mostly from pure curiosity. We know that any formal system cannot completely pin down the natural numbers. So regardless of whether we're reasoning in PA or ZFC or something else, there will be nonstandard models of the natural numbers, which admit the existence of additional integers, larger than all the finite ones. Suppose that for some particular Turing machine $Z$, I have proven that $Z$ halts, but that it does so only after some ridiculously huge number of steps $N$, such as $A(A(A(10)))$, where $A$ is the Ackermann sequence. My question is, in a case like this, how can I know for sure that $N$ is a standard natural number and not a nonstandard one? Of course, in principle I could just simulate the Turing machine until it halts, at which point I would know the value of $N$ and could be sure it is a standard natural number. But in practice I can't do that, because the universe would come to an end long before I was finished. (Let's suppose, unless this is impossible, that there is no way around this for this particular Turing machine; that is, any proof of the exact value of $N$ has a length comparable to $N$.) If $N$ does turn out to be a nonstandard number then the Turing machine does't halt after all, since when simulating it we would have to count up through every single standard natural number before reaching $N$. This would seem to put us in a tricky situation, because we've proven that some $N$ exists with a particular property, but unless we can say for sure that $N$ is a standard natural number then we haven't actually proven the Turing machine halts at all! My question is simply whether it's possible for this situation to occur, or if it's not, why not? I appreciate that the answer to this might depend on the nature of the proof that $Z$ halts, which I haven't specified. If this is the case, which kinds of proof are susceptible to this issue, and which are not?
|
[I will take for granted in this answer that the standard integers "exist" in some Platonic sense, since otherwise it's not clear to me that your question is even meaningful.] You're thinking about this all wrong. Do you believe the axioms of PA are true for the standard integers? Then you should also believe anything you prove from PA is also true for the standard integers. In particular, if you prove that there exists some integer with some property, that existence statement is true in the standard integers. To put it another way, anything you prove from your axioms is true in any model of the axioms, standard or nonstandard. So the existence of nonstandard models is totally irrelevant. All that is relevant is whether the standard model exists (in other words, whether your axioms are true for the standard integers). Now, I should point out that this notion is a lot slipperier for something like ZFC than for something like PA. From a philosophical standpoint, the idea that there actually exists a Platonic "standard set-theoretic universe" that ZFC is correctly describing is a lot less coherent than the corresponding statement for integers. For all we know, ZFC may actually be inconsistent and so it proves all sorts of false statements about the integers. Or maybe it is consistent, but it still proves false statements about the integers (because it only has nonstandard models). But if you do believe that the ZFC axioms are true in their intended interpretation, then you should believe that any consequences of them are also true (including consequences about the integers).
|
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|
2,677,935 |
One thing that has bothered me so far while learning about the Peano axioms is that the use of parentheses just comes out of nowhere and we automatically assume they are true in how they work. For example the proof that $(a+b)+c = a+(b+c)$. Given some base case $(a+b)+0 = a+b = a+(b+0)$ we have for an inductive step: $(a+b)+S(c) = S((a+b)+c) = S(a+(b+c)) = a + S(b+c) = a + (b+S(c))$ But for some reason this bothers me because we haven't really explained yet at all how the parentheses are meant to be treated in the first place. Do we need an additional axiom or definition for this? Is this mentioned anywhere? Or do we just sort of take them for granted?
|
In formal language theory (most relevantly, context-free languages), there is the notion of an abstract syntax tree . A decent chunk of formal language theory is figuring out how to turns flat, linear strings of symbols into this more structured representation. This more structured, tree-like representation is generally what we are thinking of when think about well-formed formulas or terms. For example, when we consider doing induction over all formulas, it is a structural induction over abstract syntax trees, not strings. For the relatively simple and regular languages often used for simple formal logics, it is relatively easy to describe the syntax with a context-free grammar and/or describe an algorithm that will take a string and parse out the abstract syntax tree if possible. For something like $S(a+(b+c))$ this would produce a tree like: S
|
+
/ \
a +
/ \
b c Of course, if I wanted to represent this tree in linear text, i.e. I wanted to serialize it, I would need to pick some textual representation, and in this case the original expression, S(a+(b+c)) , would be one possible representation. Another would be Polish notation, S+a+bc . Another would be S-expressions, (S (+ a (+ b c))) . Another would be something like standard notation but eschewing infix operators, S(+(a,+(b,c))) . Ultimately, what I would strongly recommend is that you think of the abstract syntax trees as primary and the pieces of text you see in books as just serializations of these trees forced upon us by practical limitations. The need to concern ourselves with parsing at all in logic is, to a large extent, just an artifact of the medium we're using. From the perspective of abstract syntax trees, associativity looks like: + +
/ \ / \
+ c = a +
/ \ / \
a b b c Humans are rather good at parsing languages, so it's not generally a barrier to understanding. It is obviously unnecessary to understand context-free grammars to confidently and correctly read most formal notation. If you are interested in parsing, by all means read up on it, but parsing isn't too important to logic so I would not get hung up on it. (Formal languages and formal logic do have a decent overlap in concepts and techniques though.) From a philosophical perspective, worrying about parsing formal languages (an impression some books give) while reading sentences in a massively more complicated natural language is a touch silly.
|
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|
2,678,805 |
I learned calculus for 2 years, but still don't understand the definition of $\ln(x)$ $$\ln(x) = \int_1^x \frac{\mathrm d t}{t}$$ I can't make sense of this definition. How can people find it? Do you have any intuition?
|
We want a function that changes multiplication into addition. That is, we want $$f(xy) = f(x) + f(y).\tag 1 $$ Substituting $y=1,$ we get $f(x) = f(x) + f(1),$ so we know that $f(1) = 0.$ Now, let's suppose that $f$ is differentiable. After all, we want to find as nice a function as possible. Let's hold $y$ constant for the moment, and differentiating $(1)$ gives $$yf'(xy) = f'(x) \implies \frac{f'(xy)}{f'(x)}=\frac{1}{y}$$ Now it's not hard to guess that $f'(x) = 1/x$ fills the bill, and together with $f(1)=0,$ the fundamental theorem of calculus gives us the definition.
|
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|
2,679,304 |
I understand that the main difference between a subspace and a hyperplane is that the subspace must go through the origin. Why does need to happen? In other words, why does a subspace always have to go through the origin? What restricts it from doing otherwise?
|
A subspace is a vector space , then it must satisfy all axioms for a vector space, including the existence of a zero vector.
|
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|
2,681,160 |
The Koch snowflake has finite area, but infinite perimeter, right? So if we make this snowflake have some thickness (like a cake or something), then it appears that you can fill it with paint like this ($\text{finite area} \times \text{thickness} =\text{finite volume}$), but on the other hand, you can't paint its sides ($\text{infinite perimeter}\times\text{thickness}=\text{infinite surface area}$)... Is there something I am getting wrong? Is there a clever trick to it, or am I just misunderstanding something? My teacher (I'm in middle school) told me that this is some higher-level maths I won't understand now.
|
What you've found is a fundamental property (or perhaps it would be better to say lack of property) of arc length in two dimensions — or going up one dimension, surface area in three dimensions: it's not 'continuous' with respect to small changes in shape. You don't even need a Koch snowflake or similar fractal curve to see this; give me any shape — for instance, a circle — and give me an arbitrarily small box that its boundary passes through, and I can give you a shape that looks exactly the same outside the box but that has arbitrarily large arc length. For an example of this, imagine starting with the curve $y=\sin^2(\pi x)$ between $x=0$ and $x=1$. Then the exact arc length of this curve is hard to compute, but we can put a lower bound on it: the curve goes from $y=0$ to $y=1$, then from $y=1$ to $y=0$ again, so it must have total length at least $2$. $(^*)$ But now consider the curve $y=\sin^2(2\pi x)$ between $x=0$ and $x=1$. This is two copies of the sine curve placed next to each other, and its $y$ values take the 'path' mentioned above twice, so it must have total length at least $4$. And more generally, $y=\sin^2(n\pi x)$ has arc length at least $2n$ between $x=0$ and $x=1$. But the curve is always within the rectangle $0\leq x\leq 1\pi$, $0\leq y\leq 1$. So imagine that you tell me you want to get a continuous curve with minimum arc length $\ell$ that fits into a box of size (side length) $d$. Then I know that if I create a curve with arc length $\ell/d$ within a $1\times 1$ box, I can scale it by a factor of $d$ and get a curve of arc length $\ell$ within the $d\times d$ box. But I also know that I can create a curve with arc length at least $\ell/d$ in the $1\times 1$ box by using the curve $y=\sin^2(2\pi\ell x/d)$ (for $0\leq x\leq1$). So I can just "paste" this curve in wherever you tell me; I've only modified things within the $d\times d$ box, but I've made the arclength arbitrarily large. As for what this has to do with 'painting' the snowflake, consider the thickness of the paint you're using to try and paint the sides of it. If you want your paint layer to have a finite thickness $\tau$, then we're not necessarily talking about the Koch snowflake itself, but just a curve that lies within a distance $\tau$ of it everywhere; since the paint 'hides' anything smaller than $\tau$, we can't tell the difference between the two anyway. But for any $\tau$ there are curves that stay within distance $\tau$ of the snowflake but have finite length! So we only need a finite amount of paint to do the painting here. And as your paint gets thinner and thinner, the length that you need to cover will increase — but that's all right, because your paint will spread farther and farther. In the limit, you're talking about 'painting' the infinite length of the snowflake itself — but now your paint layer is infinitely thin, so it shouldn't be too surprising that you can make it stretch as far as you need to. (This is very much analagous to taking a one-square-unit 'bucket' of 2d paint and noting that you can paint a $1$ unit $\times 1$ unit square with it, or a $2$ unit $\times 1/2$ unit rectangle, or a $1000$ unit $\times 0.001$ unit rectangle, or...) Meanwhile, I can't do the same thing with area (or in the 3d case, volume); it's clear to see that any modifications that are made within a $d\times d$ box can only change the area of the figure (one way or the other) by at most $d^2$. $(^*)$ So if you tell me that you want to increase the area bounded by your shape by a thousand square units but only give me a $1/10\times 1/10$ unit box to do it in, I can fairly tell you that that's impossible. Incidnetally, this same lack of continuity for arc length is at play in another very similar 'paradox': The staircase paradox, or why $\pi\ne4$ . The fundamental explanation is the same: the fact that two curves are arbitrarily close to each other along their entire spans tells you nothing about how close their arc lengths are. $(^*)$ Note that I haven't given you proofs of these assertions! You should be skeptical here, because you've already seen that behavior doesn't necessarily match what you might expect where these things are concerned. Fortunately, these assertions are true, but proving them would require going pretty far afield.
|
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|
2,681,545 |
I am studying for my GED. I was homeschooled, so most of the math topics covered are fairly easy, but there's one which I never went over, and which is giving me some trouble. I understand how to do the problems. The math works out. The trouble is that I don't understand why the process to get the solution is different. I was hoping someone here could explain it in terms I can understand. I'm dealing with what my study guide terms 'counting' questions. They deal with the number of possible choices in a given scenario. Here's a typical question: A DJ has enough time to play four songs. She has seven different songs to choose from. How many different orderings of songs can she choose? This is fairly straightforward. The work looks like $(7)(6)(5)(4)$. I understand why this works. She has seven choices first, then six, and so on. It's the next type of question which I can't figure out: A DJ is choosing four new records for his collection. He has seven available choices. How many different groups of records can he choose? This one starts out like the first one, but there's an additional step. You divide thusly: $\frac{(7)(6)(5)(4)}{(4)(3)(2)(1)}$. I know where the numbers are coming from, and how the math works. But I would like to know why. If I understand the principle, the problem will be much easier. The book explains that this is because 'order does not matter'. That doesn't make any sense, as clearly order doesn't matter in the first problem either. There's no further explanation. I need to understand why there is an additional step, so that I will be able to spot the differences between the two problems on the test. Note: I'm not asking for how to solve the problem, or how the math works. I'm asking why two problems, which, to me, look identical, are solved differently.
|
Ordering does matter in the first question. It even says so - "how many different orderings of songs can she choose?". So in the first one, playing "Twinkle Twinkle Little Star" before "Happy Birthday To You" is a different result to playing "Happy Birthday" first. In the second one, the DJ is choosing groups of records, i.e. he's just pulling them out in any order. So in that case, the set {"Toto I", "Toto II", "Toto III", "Toto IV"} is considered the same set as {"Toto II", "Toto IV", "Toto I", "Toto III"}. To switch between the two, you can start with the first case (pick 4 things, in order, from a set of 7), then having listed all of those sets, you note that they can be grouped together into equal-sized sets of redundant orderings. For example, let's say we're just picking 2 options out of {A, B, C, D}. So if order does matter, then the options are: AB AC AD BA BC BD CA CB CD DA DB DC But you can then group those together into options that are re-orderings of each other, i.e. AB BA AC CA AD DA BC CB BD DB CD DC So you picked $4 \times 3$, but then grouped them into sets of $2 \times 1$, giving you $\frac{(4)(3)}{(2)(1)}$ different options in the end.
|
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|
2,683,368 |
If you roll two fair six-sided dice, what is the probability that the sum is $4$ or higher? The answer is $\frac{33}{36}$ or $\frac{11}{12}$. I understand how to arrive at this answer. What I don't understand is why the answer isn't $\frac{9}{11}$? When summing the results after rolling two fair six sided dice, there are $11$ equally possible outcomes: $2$ through $12$. Two of these outcomes are below four, meaning $9$ are greater than or equal to four which is how I arrived at $\frac{9}{11}$. Can someone help explain why that is wrong?
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It is wrong because it is not $11$ equally possible outcome. There is exactly $1$ way to get the sum to be $2$. ($1+1=2$) but there is more than one way to get $3$. ($1+2=3, 2+1=3$)
|
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|
2,685,634 |
I'm reading the text Category theory in context from Emily Riehl, and having trouble to find an example asked on exercise $1.3.viii.$: prove that functors not need to reflect isomorphisms, i.e., find a functor $F:C\rightarrow D$ and a morphism $f$ in $C$ so that $Ff$ is an isomorphism in $D$ but is not an isomorphism in $C$. I know that a non conservative functor from $Top$ to $Set$ might work but can't find the adequate morphism. Any suggestion is apreciated.
|
A very generic counterexample: Take for $\mathcal{D}$ the category with one object $O$ and one morphism $f$. There is a functor $F$ from any category to $\mathcal{D}$ that sends every object to $O$ and any morphism to $f$. Since $f$ is an isomorphism this generates a counterexample for every category which contains at least one morphism which is not an isomorphism.
|
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|
2,686,606 |
It should be simple, but I'm having trouble. The three points are
$$A(1,-2,1)\qquad B(4,-2,-2)\qquad C(4,1,4)$$
The plane I get is
$$x+2y+z+6=0$$
but it obviously does not pass through the three points $A,B,C$.
|
Here's one way to get the requisite plane: Get two different vectors which are in the plane, such as $B-A=(3,0,-3)$ and $C-A=(3,3,3)$. Compute the cross product of the two obtained vectors: $(B-A)×(C-A)=(9,-18,9)$. This is the normal vector of the plane, so we can divide it by 9 and get $(1,-2,1)$. The equation of the plane is thus $x-2y+z+k=0$. To get $k$, substitute any point and solve; we get $k=-6$. The final equation of the plane is $x-2y+z-6=0$.
|
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|
2,686,638 |
I have the following three equations:
\begin{cases}
v_{1f}\cos(37^\circ)+v_{2f}\cos(\theta) & = 3.5 \times 10^5 \\
v_{1f}\sin(37^\circ)-v_{2f}\sin(\theta) & = 0\\
v_{1f}^2+v_{2f}^2 & =(3.5 \times 10^5)^2
\end{cases}
And I want to solve for $v_{1f}$, $v_{2f}$, and $\theta$. This is a system of three equations but it doesn't seem solvable and I've tried everything I know to solve it. For example, nothing can cancel with each other like you would in an easy system, and I've tried using the 3rd equation to solve for $v_{1f}$ or $v_{2f}$ but it still does not come out correctly. I do know the answers, just not how to get them. Here they are:
\begin{cases}
v_{1f}=2.8 \times 10^5 \\
v_{2f}=2.11 \times 10^3 \\
\theta=53^\circ \\
\end{cases}
Am I missing some information needed for solving this? I really appreciate any help with this question. Sorry that I could not show more of my work but I'm stuck and showed what I know so far. Thank you. Also, if someone sees that it isn't solvable that would help as well. EDIT: I messed up typing the 1st equation, fixed now.
|
a) the geometric solution Let's change the symbols so as to keep notation cleaner, and rewrite your input as
$$ \bbox[lightyellow] {
\left\{ \matrix{
a\cos \alpha + b\cos \theta = A \hfill \cr
a\sin \alpha = b\sin \theta \hfill \cr
a^{\,2} + b^{\,2} = A^{\,2} \hfill \cr} \right.
} \tag{1}$$ This corresponds to the right triangle shown in this sketch and the solution is quite simple noting that $\theta = 90^\circ -\alpha$, and applying the sine law. Note that a (limit) solution is also given by
$$ \theta=0, \; a=0, \; b=A$$ b) the algebraic solution The last equation in (1) suggests that we may introduce an additional angle $\phi$ and write
$$
\left\{ \matrix{
a = A\cos \varphi \hfill \cr
b = A\sin \varphi \hfill \cr
A\cos \varphi \cos \alpha + A\sin \varphi \cos \theta = A \hfill \cr
A\cos \varphi \sin \alpha - A\sin \varphi \sin \theta = 0 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{
a = A\cos \varphi \hfill \cr
b = A\sin \varphi \hfill \cr
\cos \varphi \cos \alpha + \sin \varphi \cos \theta = 1 \hfill \cr
\cos \varphi \sin \alpha - \sin \varphi \sin \theta = 0 \hfill \cr} \right.
$$ The last two equations written in matrix terms become
$$
\left( {\matrix{
{\cos \alpha } & {\cos \theta } \cr
{\sin \alpha } & { - \sin \theta } \cr
} } \right)\left( {\matrix{
{\cos \varphi } \cr
{\sin \varphi } \cr
} } \right) = \left( {\matrix{
1 \cr
0 \cr
} } \right)\quad \Rightarrow \quad {\bf M}_{\,\theta } \,{\bf v}_{\,\varphi } = {\bf u}
$$
where the matrix determinant is
$$
\left| {{\bf M}_{\,\theta } } \right| = - \sin \left( {\alpha + \theta } \right)
$$
so that for $\alpha +\theta \ne n \pi$ the matrix is invertible, giving
$$
{\bf M}_{\,\theta } ^{\, - \,{\bf 1}} = {1 \over {\sin \left( {\alpha + \theta } \right)}}\left( {\matrix{
{\sin \theta } & {\cos \theta } \cr
{\sin \alpha } & { - \cos \alpha } \cr
} } \right)
$$
and we get that
$$
{\bf v}_{\,\varphi } = {\bf M}_{\,\theta } ^{\, - \,{\bf 1}} {\bf u}\quad \Rightarrow \quad \left( {\matrix{
{\cos \varphi } \cr
{\sin \varphi } \cr
} } \right) = {1 \over {\sin \left( {\alpha + \theta } \right)}}\left( {\matrix{
{\sin \theta } \cr
{\sin \alpha } \cr
} } \right)
$$ However we shall ad the condition that the modulus of ${\bf v}_{\,\varphi }$ be unitary, i.e.
$$
1 = \left| {{\bf v}_{\,\varphi } } \right|^{\,2} = \overline {\bf u} \,\overline {{\bf M}_{\,\theta } } ^{\, - \,{\bf 1}} {\bf M}_{\,\theta } ^{\, - \,{\bf 1}} {\bf u}\quad \Rightarrow \quad \sin ^{\,2} \theta + \sin ^{\,2} \alpha = \sin ^{\,2} \left( {\alpha + \theta } \right)
$$
which develops to provide
$$
\eqalign{
& \sin ^{\,2} \theta + \sin ^{\,2} \alpha = \sin ^{\,2} \alpha \cos ^{\,2} \theta + \cos ^{\,2} \alpha \sin ^{\,2} \theta + 2\sin \alpha \cos \alpha \cos \theta \sin \theta \cr
& \sin ^{\,2} \alpha \sin ^{\,2} \theta + \sin ^{\,2} \alpha \sin ^{\,2} \theta = 2\sin \alpha \cos \alpha \cos \theta \sin \theta \cr
& \sin \alpha \sin ^{\,2} \theta = \cos \alpha \cos \theta \sin \theta \cr
& \left( {\sin \alpha \sin \theta - \cos \alpha \cos \theta } \right)\sin \theta = 0 \cr
& \sin \theta = 0\; \vee \;\cos \left( {\alpha + \theta } \right) = 0 \cr}
$$ Finally, resuming the various steps we conclude that
$$ \bbox[lightyellow] {
\left\{ \matrix{
\theta \in \left\{ {\pi /2 + k\pi - \alpha ,\;k\pi } \right\} \hfill \cr
\varphi = \arcsin \left( {\sin \alpha /\sin \left( {\alpha + \theta } \right)} \right) \hfill \cr
a = A\cos \varphi \hfill \cr
b = A\sin \varphi \hfill \cr} \right.
} \tag{2}$$ We can see that the algebraic method add some solutions wrt
the geometric method , which actually reduce to those implying negative values for $b$ e.g. $\theta=\pi, \; a=0, \; b=-A$
|
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|
2,687,297 |
It's well known that vacuous truths are a concept, i.e. an implication being true even if the premise is false. What would be the problem with simply redefining this to be evaluated to false? Would we still be able to make systems work with this definition or would it lead to a problem somewhere? Why must it be the case that false -> false is true and false -> true is true?
|
Notice that 3=5 is false. but if 3=5 we can prove 8=8 which is true. $$ 3=5$$ therefore $$ 5=3$$ Add both sides, $$8=8$$ We can also prove that $$ 8=10$$ which is false. $$ 3=5$$ Add $5$ to both sides, we get $$8=10$$ The point is that if we assume a false assumption, then we can claim whatever we like. That means " False $\implies$ False " is true. And " False $\implies$ True " is true.
|
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|
2,689,332 |
I was bored in class one day and wondered to myself if there were any quadratics $x^2+ax+b$ such that $a$ and $b$ are the zeros. I found two: $x^2+x-2,$ and $x^2 -{1\over2}x -{1\over2}$. The comments suggested $x^2+0x+0$, though this seems trivial. I wonder if this applies to other degree polynomials. Clearly it never works for a linear, except for $x+0=0$, as if $x+a=0$, $x=-a$, not $a$. What about cubics, quadratics, or even higher powers? In general, $x^n$ works. What about non-trivial solutions?
|
For the quadratic case $x^2+ax+b=0$, by Viète's formulas we have
$$-a=a+b\qquad b=ab$$
The first formula implies $b=-2a$. Substituting this into the second equation gives $-2a=-2a^2$ or $a=a^2$, so $a=0,1$. These give corresponding $b$ values of 0 and $-2$ respectively, so the unique non-trivial quadratic "auto-solving" polynomial is $x^2+x-2=0$. For the cubic case $x^3+ax^2+bx+c$, the same formulas give
$$-a=a+b+c\qquad b=ab+bc+ca\qquad-c=abc$$
If $c=0$ the situation is just the quadratic case with another zero root. In general, if $p(x)$ is auto-solving, so is $p(x)x^k$ for any $k>0$, since only extra zeros are introduced to both the coefficient and root list. Thus we have for example $x^3+x^2-2x$, but these are trivial. There is a nice non-trivial auto-solving cubic polynomial, $x^3+x^2-x-1$, but also a very ugly one given by
$$a=0.56519771\dots$$
$$b=-1.76929235\dots$$
$$c=0.63889691\dots$$
(It turns out that these values are in the OEIS: A273065 , $-$ A273066 , A273067 .) And this is just the all-real solution! There are two solutions with complex coefficients. One is given by
$$a={-0.78259885\dots}-0.52171371\dots i$$
$$b={0.88464617\dots}-0.58974280\dots i$$
$$c={0.68055154\dots}+1.63317024\dots i$$
and the other is obtained by taking the complex conjugates of the values above. For quartics, there is only one non-trivial real auto-solving polynomial, $x^4+x^3+ax^2+bx+c$ where
$$a=-1.75487766\dots$$
$$b=-0.56984029\dots$$
$$c=0.32471795\dots$$
Even more startlingly, there are no auto-solving real polynomials of degree five or higher . This and the uniqueness of the quartic example were proved by Paul R. Stein in " On Polynomial Equations with Coefficients Equal to Their Roots " (March 1966), American Mathematical Monthly 73 (3), pp. 272-274.
|
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|
2,693,706 |
I understand that differentiation of a function ($\mathbb{R} \rightarrow \mathbb{R} $) at a point is the rate of change in the output for a slight nudge in the input. And this rate of change could be negative or positive. There is no concept of direction for the single-variable function as obvious. Now, my doubt is in the case of the multivariate function ($\mathbb{R}^n \rightarrow \mathbb{R}$) where differentiation is a gradient. And this gradient representing partial differentiation w.r.t. to each basis becomes a direction. This direction is a direction of ascent but not descent, why?. Why it is a direction is of ascent. My question is not at all related to steepest ascent , about which one can find many answers on this forum and read elaborately at this link . An intuitive explanation would be preferable than mathematical at this link .
|
The comments persuaded me to reformulate my answer. For the original (still correct, but sub-optimal) version, see below. The gradient is defined in a completely natural way. There is no completely mathematical reason, why it can be said to point to the steepest ascent. It has more to do with some more or less arbitrary choices being made in several definitions, which break this symmetry. Observation . The concept "gradient points in direction of ascent" also works for single-valued functions $\Bbb R\to\Bbb R$. There is indeed a concept of direction in $\Bbb R$: right and left . A positive derivative is a vector (the gradient) pointing to the right (in the direction of ascent), a negative derivative is a vector pointing to the left (in this case, also the direction of ascent, because the function is decreasing to the right). So since the same observation also applies in 1D, we should start looking for an explanation here. Note : I am going to use the terms "right" and "left" for the direction "positive" and "negative" on the number-line, because this is the standard orientation of the number-line. This is also a symmetry break, but only a notational one. It does not effect the mathematics in any way if we flip these directions. Thanks to the comments, some few definitions could be localized as the rootcause of the broken symmetry. If you are standing on a mountain side, there is no meaning in asking whether it is up-hill or down-hill. This question only makes sense if you define a direction with respect to which we should judge the slope. The same goes for single-valued functions. It has been standard to call a function increasing if it's function graph is uphill to the right . This involves two arbitrary choices: The $y$-axis is pointing upwards, hence increasing function values are seen as going up . This is the most obvious arbitrary choice. Many applications do it the other way around, e.g. line-numbers in text are increasing from top to bottom, and pixels on a screen are usually adressed with an downwards-increasing $y$-axis. The kind of slope is judged w.r.t. to the "arbitrary" direction "right". Why not left? It seems natural, but is not forcing. There might be another arbitrary choice: a positive derivative indicates that the function is increasing . We could have defined it the other way around. Anyways, flipping any single of these definitions will change the gradient from pointing upwards to pointing downwards. Note . Yes I know, "increasing" is formally defined as $x\le y\implies f(x)\le f(y)$, but also this definition is motivated by the visualization of an increasing function graph to the right. No one would use it if the $y$-axis was pointing downwards. Conclusion : The reason for the gradient pointing to the steepest ascent is based in our somewhat biased definitions. This is especially evident in the 1D-case. The derivative is defined in such a way so that it has a positive value (the gradient points to the right) if the function increases. A function is called increasing if its function graph is going uphill to the right. You see how these arbitrary definitions combine to "gradient is pointing uphill". ORIGINAL Because we have a somehow biased definition of differentiation. Let me explain. As noted in a comment, this "gradient pointing in direction of ascent" also works for single-valued functions $\Bbb R\to\Bbb R$. There is indeed a concept of direction in $\Bbb R$: left and right. A positive derivative is a vector (the gradient) pointing to the right (in the direction of ascent), a negative derivative is a vector pointing to the left (also in the direction of ascent, because the function is decreasing). So since the same observation happens in $1$D, we should probably start there looking for an answer. It all happens because the definition of derivative is biased in some sense. Someone once decided that a function is considered increasing if its value gets bigger to the right . You see the broken symmetry? Why to the right, why not to the left? So once one decided that the derivative is positive $-$ the gradient points to the right $-$ when the function grows to the right. Here we have it. Whoever defined it, directly coupled the terms "direction of gradient" and "direction of acsent". Would he had decided to define a function as increasing if its value grows to the left (unnatural, considering our left-to-right reading direction), then the gradient would point to the steepest decent. Note : In this answer I assumed that the number line is oriented with the positive number on the right. This is standard, but another symmetry break (but only a notational one, without impact on the mathematics). You can substitute all left/right above by negative/positive if you want to be indepedent of this broken symmetry.
|
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|
2,694,582 |
Everyone learns about the two "special" right triangles at some point in their math education—the $45-45-90$ and $30-60-90$ triangles—for which we can calculate exact trig function outputs. But are there others? To be specific, are there any values of $y$ and $x$ such that: $y=\sin(x)$; $x$ (in degrees) is not an integer multiple of $30$ or $45$; $x$ and $y$ can both be written as radical expressions? By radical expression, I mean any finite formula involving only integers, addition/subtraction, multiplication/division, and $n$th roots. [Note that I require $x$ also be a radical expression so that we can't simply say "$\arcsin(1/3)$" or something like that as a possible value of $x$, which would make the question trivial.] If yes, are they all known and is there a straightforward way to generate them? If no, what's the proof?
|
There is
$$\cos\frac{\pi}5=\frac{\sqrt5+1}4$$
and similar for cosines and sines of multiples of this. Gauss proved that
one can find expressions for $\cos \pi/p$ involving iterated square roots
where $p$ is prime if and only if $p$ is a Fermat prime (of form $2^{2^k}+1$), so for $p=2$, $3$, $5$, $17$, $257$ and $65537$ (but to date no others are known).
|
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|
2,696,344 |
I should like to evaluate $\log_2{256!}$ or other large numbers to find 'bits' of information. For example, I'd need three bits of information to represent the seven days of the week since $\lceil \log_2{7}\rceil = 3$, but my calculator returns an error for large numbers.
|
If it's about factorials, you can use Stirling's approximation: $$\ln(N!) \approx N\ln(N) - N$$ Due to the fact that $$N! \approx \sqrt{2\pi N}\ N^N\ e^{-N}$$ Error Bound Writing the "whole" Stirling series as $$\ln(n!)\approx n\ln(n)−n+\frac{1}{2}\ln(2\pi n)+\frac{1}{12n} −\frac{1}{360n^3}+\frac{1}{1260n^5}+\ldots $$ it is known that the error in truncating the series is always the opposite sign and at most the same magnitude as the first omitted term. Due to Robbins, we can bound: $$\sqrt{2\pi }n^{n+1/2}e^{-n} e^{\frac{1}{12n+1}} < n! < \sqrt{2\pi }n^{n+1/2} e^{−n} e^{1/12n}$$ More on Stirling Series in Base $2$ Let's develop the question of Stirling series when we have a base $2$ for example. The above approximation has to be read this way: $$log_2(N!) \approx \log_2(\sqrt{2\pi N} N^N\ e^{-N})$$ Due to the fact that we have a non-natural log, it becomes $$\log_2(N!) \approx \frac{1}{2}\log_2(2\pi N) + N\log_2(N) - N\log_2(e)$$ Hence one has to be very careful with the last term which is not $N$ anymore, but $N\log_2(e)$. That being said one can proceed with the rest of Stirling series. See the comments for numerical results. Beauty Report $$\color{red}{256\log_2(256) - 256\log_2(e) + \frac{1}{2}\log_2(2\pi\cdot 256) = 1683.9958175971615}$$ a very good accord with numerical evaluation (for example W. Mathematica) which gives $\log_2(256!) = 1683.9962872242145$.
|
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|
2,696,959 |
In proofs, are "for each" and "for any" synonyms? Or some context is usually required to determine this?
|
Compare A1. If there’s a simple solution for each of the problems, the test is too easy. A2. If there’s a simple solution for any of the problems, the test is too easy. These are not equivalent. The first might be true without the second being true. Compare also the plainly different B1. There isn’t a simple solution for each of the problems. B2. There isn’t a simple solution for any of the problems. So it is important when formally regimenting English into the language of logic to note that, while many standalone or wide scope uses of ‘for any’ and ‘for each’ are equivalent, they do embed differently inside other logical operators. [And before you rush to making synonymy claims at least for unembedded uses, it is worth remarking that there remain complicated differences between 'any' and 'each' even here. For example, 'any' can take plurals as well as singulars, so we can have both 'for any man' and 'for any men' (and these are different -- a table may be too heavy for any man to lift, but not too heavy for any men to lift); but we can't have 'for each men'. And 'any' can take mass nouns, while 'each' can't (so compare, ‘for any ice that doesn’t shift, try salt’ vs the ungrammatical ‘for each ice that doesn’t shift, try salt’). And so it goes. There's a good reason why we introduce formal quantifiers to avoid the vagaries of English usage.]
|
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|
2,696,963 |
I was trying to prove Sobolev inequality and estimate the value of the constant for dimensions greater (or equal) to 3 and these integrals came up
$$I_\pm=\int_0^\infty \frac{r^{n\pm1}}{(b^2+r^2)^n} dr$$
I tried substitutions $r=b\tan t$, $r=b\sinh t$ and even Cauchy formula for integrals... but I didn't get any results. Any hint?
|
Compare A1. If there’s a simple solution for each of the problems, the test is too easy. A2. If there’s a simple solution for any of the problems, the test is too easy. These are not equivalent. The first might be true without the second being true. Compare also the plainly different B1. There isn’t a simple solution for each of the problems. B2. There isn’t a simple solution for any of the problems. So it is important when formally regimenting English into the language of logic to note that, while many standalone or wide scope uses of ‘for any’ and ‘for each’ are equivalent, they do embed differently inside other logical operators. [And before you rush to making synonymy claims at least for unembedded uses, it is worth remarking that there remain complicated differences between 'any' and 'each' even here. For example, 'any' can take plurals as well as singulars, so we can have both 'for any man' and 'for any men' (and these are different -- a table may be too heavy for any man to lift, but not too heavy for any men to lift); but we can't have 'for each men'. And 'any' can take mass nouns, while 'each' can't (so compare, ‘for any ice that doesn’t shift, try salt’ vs the ungrammatical ‘for each ice that doesn’t shift, try salt’). And so it goes. There's a good reason why we introduce formal quantifiers to avoid the vagaries of English usage.]
|
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|
2,697,936 |
Does there exist a set containing infinite elements, whose elements themselves are sets containing infinite elements? I think the answer is no, there is a famous paradox for it but I'm forgetting.
|
There are many sets with this property. One example is $A=\{S\subseteq\mathbb{N}\mid |S|=\infty\}$, i.e. the set of infinite subsets of the naturals.
|
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|
2,698,554 |
This is a grade school problem! Consider the following figure: It is very easy to show that the red area and the blue area equal. I can demonstrate this based on my knowledge related to the computation of the surface areas of circular sectors and triangles. Both areas equal $2\left(\frac{r^2\pi}4-\frac{r^2}2\right)$ where $r$ is the radius of the smaller circles. But, how am I going to show the same if I forget, for good, the formula providing the area of a triangle? I am not able to get rid of my thought process using triangles.
|
See the image with blue parts shifted: The single blue figure is the same part of its small square as the red plus both blue of the big square, hence areas
$$\frac{2\cdot blue + red}{blue}=\frac{big\ square}{small\ square}=4$$
so
$$2\cdot blue + red = 4\cdot blue$$
hence
$$red = 2\cdot blue$$
Q.E.D. Without the formula for the area of a triangle, and without the formula for the area of a circle...
|
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|
2,698,555 |
Question Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$ My attempt Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt{x^2+x+3}$$
$$(d+x)^2=(\sqrt{x^2+x+3})^2$$
$$d^2 + x^2 + 2dx =x^2+x+3$$
$$d^2 +2dx = x +3$$
$$x = \frac{3-d^2}{2d-1}$$ $$d \neq \frac{1}{2}$$ So $x$ will be rational as long as $d \neq \frac{1}{2}$. Now
$$ y = \sqrt{x^2+x+3}$$
$$ y = \sqrt{(\frac{3-d^2}{2d-1})^2 + \frac{3-d^2}{2d-1} + 3}$$
$$ y = \sqrt{\frac{(3-d^2)^2}{(2d-1)^2} + \frac{(3-d^2)(2d-1)}{(2d-1)^2} + 3\frac{(2d-1)^2}{(2d-1)^2}}$$
$$ y = \sqrt{\frac{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}{(2d-1)^2}} $$
$$ y = \frac{\sqrt{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}}{(2d-1)}$$
$$ y = \frac{\sqrt{d^4-2d^3+7d^2-6d+9}}{(2d-1)}$$ I know that again $d \neq \frac{1}{2}$ but I don't know what to do with the numerator. Help
|
See the image with blue parts shifted: The single blue figure is the same part of its small square as the red plus both blue of the big square, hence areas
$$\frac{2\cdot blue + red}{blue}=\frac{big\ square}{small\ square}=4$$
so
$$2\cdot blue + red = 4\cdot blue$$
hence
$$red = 2\cdot blue$$
Q.E.D. Without the formula for the area of a triangle, and without the formula for the area of a circle...
|
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|
2,699,383 |
I had a simple thing to compute with a calculator:
$$\sin\left(2\cos^{-1}\left(\frac{15}{17}\right)\right)$$
I got the decimal answer of about $0.83044983$, but when I typed it in WolframAlpha, it also gave an exact answer of $\frac{240}{289}$. How in the world would one get an exact answer here?
|
Here are three relevant formulas: $\sin 2x = 2 \sin x \cos x$. $\cos \cos^{-1} x = x.$ $\sin \cos^{-1} x = \sqrt{1 - x^2}$ after drawing an appropriate right triangle. Combining these three to get the desired conclusion is left to the interested reader.
|
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|
2,699,410 |
So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*} I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc}
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\end{array} \right),\end{align*}
and \begin{align*}P^3=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*} In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix. I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?
|
This is a periodic Markov chain (with period $2$). Otherwise, there's not much that's terribly unusual about it. User "Iwillnotexist Idonotexist" raised an important point in the comments: Well, something that can be noted for periodic Markov chains is that
by definition they cannot be ergodic, another very important property
of MCs that you may encounter soon. Roughly speaking, an ergodic MC
that runs long enough "forgets" everything about its initial state. If
the MC is periodic, then clearly you must remember some information
about the contents of the initial state, because you're stuck in a
loop of states that you keep on coming back to and aren't forgetting.
|
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|
2,699,450 |
If $P$ is a polynomial with $P(3)=10$ and $P(1)=1$ , then why can't all the coefficients of $P$ be integers? This question was deleted for not enough details half a year ago, therefore I'm providing them. In this question specifically, I'm asking for your help to solve this problem in 8th grader way, because I'm sure that this question might help other students to understand polynomials much better, without any higher, university-level knowledge (some provided answers there are very elegant and understandable even for a 6th grader). To mention more, right now it would be quite hypocritical to say that 'I tried < insert any theorem > but got stuck, hence I'm asking for your help'. So I'm not saying it now, instead of that just simply asking for you to undelete this question for the reasons I mentioned above. Thank you!
|
If $p(x)$ is a polynomial with integer coeficients then for all integers $a,b$ you have $$a-b\mid p(a)-p(b)$$ In particular you have $$3-1\mid p(3)-p(1) = 9$$ A contradiction.
|
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|
2,699,466 |
Given: $f(n) = n ^ {ln (n)}$ $g(n) = ( log_2n )^n$ Is $f(n) \in O(g(n))$, $g(n) \in O(f(n))$, or neither? Why?
|
If $p(x)$ is a polynomial with integer coeficients then for all integers $a,b$ you have $$a-b\mid p(a)-p(b)$$ In particular you have $$3-1\mid p(3)-p(1) = 9$$ A contradiction.
|
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|
2,701,138 |
My daughter in grade 3 is learning about telling time at her school. She eagerly showed me this method she has discovered on her own to tell the minutes part of the time on an analogue clock. I wasn't sure at first because I have never heard about it before but it works really well. Here is her method: Look at the minute hand and see what number it is pointing to, let's say it's $3$. Add a zero at the end to make it $30$. Halve it to get the minutes, so $3$ becomes $30$ and halving it gives $15$, $6$ becomes $60$ and halving it gives $30$ and so on. It works well but I am not sure why. What mathematically justifies the method used?
|
Each number on the clock face is worth five minutes. One good way to multiply by $5$ is to first multiply by $10$, and then divide by $2$. This works because $5=10\div 2$.
|
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|
2,701,182 |
I must once again resort to the advice of this great community. As I was reading about the pigeonhole principle something about its proof struck me as odd. Allow me to explain: After reading the "The Foundations: Logic and Proofs" chapter in Rosen's "Discrete mathematics and its applications" book I was left with the feeling/notion that I can (and I ought to) describe all my proof's statements in symbols. Yet, as you can see in the pigeonhole principle's proof: We use a proof by contraposition. Suppose none of the k boxes has
more than one object. Then the total number of objects would be at
most k. This contradicts the statement that we have k + 1 objects. Without a doubt it has more English words than symbols. Yet the proof is actually without flaws. I struggle with the fact that I can't convert it into the A -> B format (I hope you understand what I'm trying to convey). However, is there a way to symbolically represent what it states (as I'm trying to put it in my mind)? Or the only way to argue for this proof, is by using words? And if so, is there any guide or principle that should tell us when to use words or symbols in our proofs?
|
Your plan is exactly backwards. All proofs should be readable as English prose, i.e. sentences arranged into paragraphs. Symbols may be used as needed, but they need to be human-readable. If you've defined enough symbols, you can write parts of the proof entirely in symbols, provided that they can be parsed back into English. For example, $$\forall\ x\in\mathbb{R}, \exists\ y\in \mathbb{Z}: x\ge y$$
reads as "For all real numbers $x$, there is an integer $y$, such that $x$ is greater than or equal to $y$."
|
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|
2,704,941 |
It's been more than once I've found this expression "induced by", in a sentence of the form "$X$ is induced by $Y$, in mathematics and computer science. I usually associate "induced by" with "generated by". However, I am not always confident regarding its meaning. For example, in the following sentence If a planar subdivision is induced by $n$ line segments... What's the precise meaning of "induced by", in general, and in the sentence above?
|
First, " induce " is a perfectly cromulent English word. The second definition that Google gives is relevant here: bring about or give rise to. In basic vernacular English, it is reasonable to say that "$A$ induces $B$" when $A$ causes $B$, though I think that there is a connotation of indirectness (i.e. there might not be that $A$ directly causes $B$, but $A$ creates the conditions for $B$). In mathematics, this is the definition that is generally meant. When we say that "$A$ induces $B$," we typically mean that $A$ gives rise to $B$, typically in some canonical manner. For example (in an area with which I am more familiar), we often say that a "metric induces a topology". What this means is the following: if $(X,d)$ is a metric space, then the open balls , i.e. the collection
$$
\mathscr{B} := \{ B(x,r) : x\in X, r> 0 \},
$$
where $B(x,r) := \{ y \in X : d(x,y) < r \}$, forms a basis for a topology on $X$. The topology generated by this basis is the topology induced by the metric. That is, the metric gives rise to this topology. After a bit of Googling, a "planar subdivision induced by a set of $n$ line segments" seems to make sense in a similar way. Near as I can tell, a planar subdivision is a partition of the plane, i.e. a division of the plane into a collection of mutually disjoint sets whose union is the plane. A partition has more structure than just a collection of line segments, but a collection of line segments can give rise to a partition in a canonical manner. It is therefore appropriate to say that such a partition is induced by a collection of line segments.
|
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|
2,706,640 |
If I have a specific set of some elements like: $$X = \{1,2,3,4,5,6\}$$ How can one make an expression for the number of elements of the set $X$?
|
You're looking for the cardinality of a set, which is usually denoted by vertical bars: $$| X |$$
|
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|
2,706,648 |
Let $$ \begin{cases} \sqrt{x} = y -1 \\ \sqrt{y} = 11 - x\end{cases}$$ Solve $x$ and $y$ in real numbers. After a long calculation, I get to this equation: $(y−4)(y^3−16y−25)=0$ . I found $x = 9$ and $y = 4$ are the only answers. What's the other methods for solving that maybe using derivation or inequalities ?
|
You're looking for the cardinality of a set, which is usually denoted by vertical bars: $$| X |$$
|
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|
2,707,046 |
Consider the class of rational functions that are the result of dividing one linear function by another: $$\frac{a + bx}{c + dx}$$ One can easily compute that, for $\displaystyle x \neq \frac cd$
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{a + bx}{c + dx}\right) = \frac{bc - ad}{(c+dx)^2} \lessgtr 0 \text{ as } ad - bc \gtrless 0$$
Thus, we can easily check whether such a rational function is increasing or decreasing (on any connected interval in its domain) by checking the determinant of a corresponding matrix \begin{pmatrix}a & b \\ c & d\end{pmatrix} This made me wonder whether there is some known deeper principle that is behind this connection between linear algebra and rational functions (seemingly distant topics), or is this probably just a coincidence?
|
I'll put it more simply. If the determinant is zero, then the linear functions $ax+b$ and $cx+d$ are rendered linearly dependent. For a pair of monovariant functions this forces the ratio between them to be a constant. The zero determinant condition is thereby a natural boundary between increasing and decreasing functions.
|
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|
2,708,006 |
What is the maximum number of intersection points between a quadrilateral and a pentagon, both non-intersecting? I believe the maximum is 16 as shown below, but I have no idea how to prove this. Any help or pointers would be greatly appreciated. (Or a diagram with more than 16 intersection points.)
|
$16$ is the maximum if one proves that a line cannot meet the boundary of a $(2n+1)$-gon at more than $2n$ points. And this is a consequence of the Jordan curve theorem : each time we cross the boundary of a polygon we go from the exterior to the interior or the opposite. If we start at the exterior and we end at the exterior, we have an even number of crossings.
|
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|
2,710,309 |
I have to prove that
$$ \lim_{x\to\infty}\frac{\sqrt x\cos(x-x^2)}{x+1} = 0. $$ I tried squaring both the denominator and numerator to get rid of $\sqrt{x}$ but then $\cos$ becomes $\cos^2$ and I do not know how to solve it/simplify it further. I have also tried to use the squeeze theorem with $a_n = 0$, and $$ b_n =\frac{\sqrt x\cos(x-x^2)}{x+1}, $$ but to no avail. I could not find another function that is greater than $b_n$ that has a limit of $0$.
|
Note that $$-\frac{\sqrt x}{x+1}\le \frac{\sqrt x\cos(x-x^2)}{x+1}\le\frac{\sqrt x}{x+1}$$ and $$\frac{\sqrt x}{x+1}\to 0$$
|
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|
2,710,328 |
First up, this question differs from the other ones on this site as I would like to know the isolated meaning of nabla if that makes sense. Meanwhile, other questions might ask what it means in relation to something else. This might be a very stupid question; it's hard to tell when I struggle to understand what it indicates, and thus this might seem very idiotic for a person fully knowledgeable about its meaning. Now from the document How do the $\nabla x$ and $\nabla \cdot$ notations work? : Currently I interpret the nabla symbol as a way of turning something into a vector. Is my understanding correct? Anyway, what is the meaning of it, and why is it used? (Please try and describe it as simple as possible.) Less important In case there should exist multiple meanings of this symbol, this is the context:
I stumbled upon this symbol when researching neural networks (C denotes the cost function): " -∇C(...)= [*this is a vector of weights and biases*] " ( source )
|
We may think of $ \nabla $ as an operator ( del operator ) in the following sense. It takes a function $f$ and turns it into a vector $\nabla f$ . $\nabla f= \left\langle \frac {\partial f}{\partial x},\frac {\partial f}{\partial y}, \frac {\partial f}{\partial z} \right\rangle $ is called the gradient vector. The gradient vector points to the direction at which your function increases most rapidly. For example if $$ f(x,y,z)= x+3y^2 -10z$$ Then $$ \nabla f (x,y,z)= \langle 1,6y,-10\rangle $$ and if there is a point given, say $(1,3,5)$, we can evaluate $ \nabla f (1,3,5)= \langle 1,18,-10\rangle.$ This vector points at the direction of maximum increase of our function at $(1,3,5).$
|
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|
2,710,728 |
In this Information Security question , we discuss whether or not a $100$ character secret randomly-generated username is equivalent to a $50$ character secret randomly-generated username plus a $50$ character secret randomly-generated password. This answer [now deleted] claims that there is a mathematical difference. It claims: If we assume that the user id can be kept private and is choosen randomly, it would allow for more combinations. If we make an example with a base of $62$ possible characters to choose from $(a..z, A..Z, 0-1)$ , we get: $62^{100} = 10^{179}$ combinations [versus] $62^{50} + 62^{50} = 80^{89}$ combinations Is this correct? It seems erroneous to me; requiring two $50$ character items would be the same number of combinations as requiring one $100$ character item. If I'm mistaken, can you help me understand my error?
|
You are correct, they are equivalent. The answer you quote adds $62^{50}$ and $62^{50}$ instead of multiplying these numbers together. One hundred characters is one hundred characters. If it helps, think about what happens with smaller number of characters. (Take an extreme example, like only allowing the letters "a", "b", and "c" in the username and password, and comparing the number of one character usernames with one character passwords to the number of two character usernames.) How would you write down a hundred character username? You'd write down a hundred characters. How would you write down a fifty character username together with a fifty character password? You'd write down a hundred characters. If you came accross a hundred characters somewhere, how would you know which process was used to create the string? You can translate any string of one hundred characters into either a username/password pair, or into just a username, and this translation is a bijection in each case.
|
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|
2,711,691 |
I have a vector $v \in \mathbb{R}^n$ and I am searching for a mathematical notation for a function that returns $1$ if at least one of the vector's elements is $0$ and $0$ otherwise. The analogue function for a set $S$ would be:
$$
f = \begin{cases} 1 & \text{if } 0 \in S \\ 0 & \text{if } 0 \notin S
\end{cases}
$$
However I am struggling to find a similar notation for a vector. In particular, how do I write: "If any element in $v$ is $0$"?
|
It's best to write it in words:
$$
f(v)
=
\begin{cases}
1&\text{when at least one component of $v$ is zero}\\
0&\text{otherwise}.
\end{cases}
$$
This is perfectly valid mathematical notation.
Using symbols doesn't make it better than using words.
With words you can say directly what you want to say instead of taking an unnecessary detour. If someone insists on using symbols (in order for the function to be implementable on a machine), you can replace "when at least one component of $v$ is zero" by "$0\in\{v_1,\dots,v_n\}$".
That's still readable, but I think the version in words is best for communicating with people.
|
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|
2,711,754 |
I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational.
Thank you for your answers.
|
Yes, it can. Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and
$$
b_n = \frac{a_n}{a_{n-1}}
$$
for $n > 1$. We then have that
$$
b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.
$$ We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.
|
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|
2,713,500 |
Examples $$\sec(x) = \dfrac{1}{\cos(x)}$$
$$\cot(x) = \dfrac{1}{\tan(x)}$$ There are many more out there, but why do we need definitions that can be written with just $\sin , \cos ,\tan $ etc. in maths? Why can't they just be written as their expanded form? Most trigonometric functions can be written with just $\sin \cos$ and $\tan$. Why do we need so many? Additionally, I mean all definitions.
|
There are a lot of trigonometric functions which are defined geometrically, which we rarely use anymore. Many of these are summarized by this image: These all have their uses in particular circumstances. For example, the half versed sine (or haversine) is useful for determining the great circle distance between points , which is incredibly useful if you are trying to navigate. We don't need the haversin, but it is useful, and reduces notation a bit in at least one specific context. The other trig functions are similar—personally, I would rather write
$$ \frac{\mathrm{d}}{\mathrm{d}t} \tan(t) = \sec(t)^2 $$
than
$$ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\sin(t)}{\cos(t)} = \frac{1}{\cos(t)^2}. $$ EDIT: This answer was written when the question seemed to be asking about the "necessity" defining secant and cotangent functions. It seems that the original questioner had a much more general question in mind, i.e. why do we need any definitions at all? The only possible response that that, I think, is because mathematics would be impossible without "definitions." Working under the assumption that the original questioner is in earnest, a partial answer is as follows: A huge part of mathematics is the language we use in order to communicate mathematical ideas. We could, I suppose, never define anything beyond the basic axioms, but then we could never get anything done, and would have no hope of ever communicating our ideas to others. If we don't define a derivative, how do we describe the the motion of a planet? It would be cripplingly inconvenient if we could never write $3$, and always had to write $\{ \{\}, \{\{\}, \{\{\}\} \}, \{\{\}, \{\{\}, \{\{\}\}\} \}$. Not only is that quite hard to read (do you really want to check that I got all of my commas and braces right?), it is horribly inefficient. And this is just to describe a relatively small natural number. It only gets worse from here! The point is that definitions allow us to encapsulate complicated ideas into a short collection of symbols (i.e. words) that allow us to make further deductions. Definitions are at the very heart of mathematics. We can do nothing without them.
|
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|
2,715,752 |
How did mathematicians prior to the coming of calculus derive the area of the circle from scratch, without the use of calculus? The area, $A$, of a circle is $\pi r^2$. Given radius $r$, diameter $d$ and circumference $c$, by definition, $\pi := \frac cd$.
|
There's an interesting method using which you can approximately find the area. Split up the circle into many small sectors, and arrange them as a parallelogram as shown in the image (from wikipedia) The higher the number of sectors you take, the more it tends to a parallelogram, with one of it’s height the radius $r$ , and the other side half the circumference $\pi r$ . Thus, its area tends to $\pi r \cdot r = \pi r^2$
|
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|
2,715,753 |
I want to maximize the following function
$$f(x,y)=\frac{(x+y)^2}{x^2+y^2}$$
over $\mathbb{R}$.
Equating its gradient to zero gives
$$\nabla f(x,y)=0\Rightarrow x=y$$ Then, I used Wolfram to compute its Hessian $H$, here . Substituting $x=y$ into $H$, I saw that $H_{11}=-\frac{1}{x^2}<0$ and that the determinant of $H$ is zero (see previous link). Hence, according to here , the test is inconclusive and the point might be minimum, maximum or saddle point. But again according to Wolfram it is actually a maximum and equals 2 (see here ). How can I show this?
|
There's an interesting method using which you can approximately find the area. Split up the circle into many small sectors, and arrange them as a parallelogram as shown in the image (from wikipedia) The higher the number of sectors you take, the more it tends to a parallelogram, with one of it’s height the radius $r$ , and the other side half the circumference $\pi r$ . Thus, its area tends to $\pi r \cdot r = \pi r^2$
|
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|
2,715,755 |
Let $f$ be a continuous function such that, for all $x \in \mathbb{R}$, it is true that
$$f(x)= \int_0^x e^t f(t)dt$$
Prove that there exists $c \in \mathbb{R}$ such that $f(x)=ce^{(e^x-1)}$. I am not quite sure how to proceed with this one. Maybe using logarithm properties, but just do not know.
|
There's an interesting method using which you can approximately find the area. Split up the circle into many small sectors, and arrange them as a parallelogram as shown in the image (from wikipedia) The higher the number of sectors you take, the more it tends to a parallelogram, with one of it’s height the radius $r$ , and the other side half the circumference $\pi r$ . Thus, its area tends to $\pi r \cdot r = \pi r^2$
|
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|
2,717,251 |
Edit: I've added an answer myself, based on the other answers and comments. Here is a very very informal "proof" (sketch) that Gödel's theorem is wrong (or at least that the idea of the proof is wrong) : Roughly, the proof of Gödel's theorem is as follows: For any decidable and consistent set of axioms $\Phi$ that contain (or imply) the first order Peano's axioms in first order language, we can construct a Gödel sentence $G^\Phi$, such that neither $\Phi\vdash G^\Phi$ nor $\Phi\vdash \neg G^\Phi$, but where we know from an argument in the meta-language that $G^\Phi$ is true. For any such $\Phi$, we will therefore have a counterexample to the completeness of the theory $Th(\Phi)$. Therefore we know that no such $\Phi$ can be complete (where complete means that all first order statements can be proven or disproven from it). Here is a failed proposal to "circumvent" Gödel's theorem that I have already heard someone make: Just define $\Phi_1=\Phi\cup \{G^\Phi\}$, and since we know that $G^\Phi$ is true in arithmetic, we know that $\Phi_1$ is consistent. The problem of course is: We can now formulate a new Gödel sentence $G^{\Phi_1}$, which cannot be proven from in $\Phi_1$, even though it is true in standard arithmetic. Now here is my proposal: Rather than trying to add individual Gödel sentences to the set of axioms, we simply take the enumeration procedure, such that it enumerates $\phi_i \in \Phi$ for the original set of axioms $\Phi$, and also enumerates all successive Gödel sentences $G^{\Phi}, G^{\Phi_1}, G^{\Phi_2},...$. This is possible, since $\Phi$ is decidable, and decidable sets of finite strings are enumerable, so we can enumerate them successively, as $\phi_1, \phi_2, \phi_3$, where $\phi_1$ is the first statement of the enumeration of $\Phi$, and $\phi_2 = G^\Phi$, and $\phi_3$ is the second statement of the enumeration of $\Phi$, etc... We can then define the set of axioms $\Phi_\infty = \{\phi_1, \phi_2, ...\}$. This will also have a Gödel sentence $G^{\Phi_\infty}$. But what we can simply do, is add this to the enumeration procedure as well. And then the next one, and the next, and so forth. We take this process to infinity, just as we did for $\Phi$, and just keep going. Every time a Gödel sentence pops up, we simply add it to the enumeration. Now note that: since the set of first order sentences is countable, the set of Gödel sentences is countable as well (since it is a subset of the set of first order sentences). Therefore we can in this procedure described above enumerate all possible Gödel sentences. The resulting set of sentences forms an enumerable and consistent set of sentences $\Psi$ that contains the original $\Phi$, and additionally
contains the Gödel sentences of all possible sets of axioms $\Phi_x$. Therefore The Gödel sentence of $\Psi$ must be in $\Psi$ itself. Moreover, we can then create a "decidable version" of $\Psi$, by defining $\Psi^*=\{\psi_1, \psi_1 \land \psi_2, \psi_1 \land \psi_2\land \psi_3, ... \}$, for all $\psi_1, \psi_2,... \in \Psi$. We therefore have a consistent and decidable set of first order sentences that are true in standard arithmetic, contain Peano's axioms, and bypass Gödel's proof of incompleteness. This is obviously a contradiction with Gödel's theorem. So where is my "proof sketch" wrong?
|
There are at least two problems here. First, when you say "we take this process to infinity and just keep going", that is a very informal description, and without spending some work on making it more concrete you have no good reason to expect it can actually be made to work. Fortunately, such work has in fact been done, and the standard way of making it concrete is to speak about process steps indexed by transfinite ordinal numbers , which I'm going to suppose is what you are proposing. Then, however, a real problem arises: Gödel's procedure only works when the original theory is recursively axiomatized , that is, it is computable whether a given proposed sentence is one of its axioms or not. In order to do this for one of your intermediate theories, you need to be able to algorithmically describe the process that produced the theory. And there is (vividly speaking, but can be made precise) so far to infinity that there are not enough Turing machines to describe the structure of each of the theories you encounter along the way. So at some point you're going to approach a point where the process that produced the axioms you already have is so complex that the combined theory is not recursively axiomatized anymore, and then the incompleteness theorem stops working. A second, comparatively minor, problem comes later in your argument: since the set of first order sentences is countable, the set of Gödel sentences is countable as well (since it is a subset of the set of first order sentences). Therefore we can in this procedure described above enumerate all possible Gödel sentences. This argument seems to be the form: "There are a countable infinity of foos in total; here we have a set of countably-infinite many foo; therefore the set contains all of them", which is not valid -- consider e.g. the situation where a "foo" is a natural number and the set in question contains all the perfect squares. (Note also that you don't seem to have defined what a "possible Gödel sentence" means, which you really ought to do before claiming that you have all of them).
|
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|
2,717,456 |
My question (more of a hypothesis really) is basically this: If a function $f(x)$ is defined such that $f'(x)$ is not constant and never the same for any 2 values of $x$. Then there do not exist positive integers $a,b,c$ and $a\le b\le c$ such that, $$\int_0^{a} f(x)dx = \int_b^{c} f(x)dx\tag1$$ Taking $f(x)=x^n$ then for $n \gt 1$, $(1)$ becomes Fermat's Last Theorem, while for $n=1$, $(1)$ becomes $a^2 + b^2 = c^2$ Maybe this has something to do with the "curviness" (I am 16 so please forgive me for non-technical language) of the graphs of $x^n$ since for $n=1$ the graph is linear and for $n>1$ the graph is curvy. My school teachers are not talking to me about this because "it is out of syllabus", so I thought maybe this community could help.
|
Interesting. But not true. If you define $f(x)=\dfrac1{x+1}$ , then $$\int_0^1f(x)\,\mathrm dx=\int_1^3f(x)\,\mathrm dx,$$ since $$\int_\alpha^\beta f(x)\,\mathrm dx=\log\left(\frac{\beta+1}{\alpha+1}\right).$$
|
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|
2,718,551 |
I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine . Find all integer solutions of $$\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$$ Initially it looked like a typical quadratic problem, however I hit a dead end each time I solve it. My methodology is as follows,
$$\frac{n^2+mn-1}{mn^2}=\frac{3}{4}$$
$$\implies 4n^2+4mn-4 = 3mn^2$$
$$\implies (4-3m)n^2+4m \cdot n-4=0$$
I used the quadratic formula , and got,
$$n = \frac{-4m \pm \sqrt{(4m)^2-4\cdot(4-3m)\cdot(-4)}}{2(4-3m)}$$
I do the usual algebraic manipulations and drop at,
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$ I am unsure how I go ahead from this. Some help would be much appreciated. Cheers!
|
HINT: solving for $m$ is better, we do not need a quadratic equation:
$$m=\frac{4(n^2-1)}{3n^2-4n}$$
We get $$m=3,n=2$$
|
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