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https://mathematica.stackexchange.com/questions/176926/plotting-histograms-with-bin-count-multiplied-by-some-factor	# Plotting histograms with bin count multiplied by some factor Consider, for example Histogram[RandomVariate[NormalDistribution[0, 0.6], 1000]] which gives me, Now, is there a way to change the bin counts, as in, I want to multiply some factor to the bin counts (lets say "0.5"). meaning there should be only about 70 events(y-axis) at -0.5 (x-axis) instead of about 140 events. how do I reflect this on this histogram? --Thanks much! The function fh in Histogram[data, bspec, fh] is applied to two arguments: a list of bins {{b1, b2}, {b2, b3}, ...}, and a corresponding list of counts {c1, c2, ...}. The function should return a list of heights to be used for each of the ci]. So you just need to use a height function to scale the list of counts: Histogram[data, Automatic, .5 #2 &] • Perfect! just what I was looking for! Thank you so so much! – nick Jul 8 '18 at 20:14 An alternative approach is to use the option ChartElementFunction with a function that modifies the built-in functions to change the appropriate element of rectangle coordinates: SeedRandom[1] data = RandomVariate[NormalDistribution[0, 0.6], 1000]; Histogram[data, {-2, 2, .2}, ChartElementFunction -> (ChartElementData["Rectangle"][{{1, 1}, {1, .5}} #, ##2] &)] The advantages of this approach are (1) you can use other built-in height functions such as "PDF" without additional work, (2) you can have multiple datasets each with its own scaling (as long as you don't need them "Stacked"), and (3) you can use the tooltips to see the unscaled values of bar heights. The simple method above works only for the default setting ("Bottom") for the option BarOrigin. The following function makes this approach work for non-default BarOrigin settings: ClearAll[ceF] ceF[sc_ : 1][cedf_ : "Rectangle", o : OptionsPattern[]] := Module[{origin = ChartingChartStyleInformation["BarOrigin"], box = #}, Switch[origin, Bottom, box[[2, 2]] = sc box[[2, 2]], Top, box[[2, 1]] = sc box[[2, 1]], Left, box[[1, 2]] = sc box[[1, 2]], Right, box[[1, 1]] = sc box[[1, 1]]]; ChartElementDataFunction[cedf, o][box, ##2]] &; Examples: Histogram[data, {-2, 2, .2}, "PDF", ImageSize -> 300, BarOrigin -> Bottom, ChartElementFunction -> ceF[.5][] ] Histogram[data, {-2, 2, .2}, ColorFunction -> "Rainbow", ImageSize -> 200, BarOrigin -> #, ChartElementFunction -> ceF[.5][]] & /@ {Bottom, Top, Left, Right} // Grid[Partition[#, 2], Dividers -> All] & Multiple datasets each with its own scaling: Histogram[{data, data}, {-2, 2, .2}, ChartElementFunction -> {ceF[.5][], ceF[.8][]}] `	2019-11-14T07:23:59	{ "domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/176926/plotting-histograms-with-bin-count-multiplied-by-some-factor", "openwebmath_score": 0.2726743817329407, "openwebmath_perplexity": 6517.680523303301, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9572778012346834, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8399890482214901 }
http://math.stackexchange.com/questions/174108/difference-between-space-and-algebraic-structure	# Difference between “space” and “algebraic structure” What is the difference between a "space" and an "algebraic structure"? For example, metric spaces and vector spaces are both spaces and algebraic structures. Is a group a space? Is a manifold a space or an algebraic structure, both or neither? - A metric space has no inherent algebraic structure. – Brian M. Scott Jul 23 '12 at 7:00 As I understand it an algebraic structure with at least one operation defined on it. Isn't a metric space an algebraic structure where that operation is "distance", with the regular axioms? – Aqwis Jul 23 '12 at 7:08 The metric isn’t in general an operation on the space $X$: it doesn’t take values in $X$. – Brian M. Scott Jul 23 '12 at 7:10 Guess that the terminology is not too canonical, but I'd say that an "algebraic structure" on a set $X$ is the datum of one or more possibly interacting operations on $X$. By a "space" $X$ we mean a set with some extra-structure of geometric type, such as a metric or a topology. – Andrea Mori Jul 23 '12 at 8:06 If we take "algebraic structure" to be a synonym for "algebra" (in the sense of universal algebra), then an algebraic structure is a set $X$, together with a family of operations on $X$. Recall that given a set $X$, an "operation" on $X$ is a function $X^{\alpha}\to X$, where $\alpha$ is an ordinal. Such a function is called an $\alpha$-ary operation; when $\alpha$ is a natural number, the operation is said to be "finitary" (takes only finitely many arguments). Sometimes, algebraic structures are further enriched with (i) "partial operations" (functions defined on a subset $A\subseteq X^{\alpha}$ rather than all of $X^{\alpha}$), or (ii) $\beta$-ary relations (subsetes of $A^{\beta}$). We can also impose identities (requires that the operations/relations satisfy certain properties such as commutativity, etc). In this sense, vector spaces, groups, rings, fields, etc. are all (enriched) "algebras"; metric spaces are not. "Space" is a bit fuzzier; I would not put "vector spaces" in the class, restricting it rather to things like topological spaces, manifolds, metric spaces, normed spaces, etc. Now, one should realize that you this does not have to be a dichotomy: you can have structures that include both kinds of data: a topological group is both an algebraic structure (a group) and a space (topological space), in a way that makes both structures interact with one another "nicely". Normed vector spaces are both algebraic structures (vector spaces), and "spaces" (normed spaces, hence metric, hence topological), where, again, we ask that the two structures interact nicely. In fact, there is a lot of interesting stuff that can be obtained by having the two kinds of structures and "playing them off against one another." For example, Stone Duality and Priestley Duality exploit this kind of "structured topological space" (a topological space that also has operations, partial operations, and relations that interact well with the topology). -	2013-05-22T23:38:49	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/174108/difference-between-space-and-algebraic-structure", "openwebmath_score": 0.9365767240524292, "openwebmath_perplexity": 470.06294377812657, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.957277806109987, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8399890432978551 }
https://math.stackexchange.com/questions/552854/volume-of-a-sphere-with-three-holes-drilled-in-it	# Volume of a sphere with three holes drilled in it. Suppose that the sphere $x^2+y^2+z^2=9$ has three holes of radius $1$ drilled through it. One down the $z$-axis, one along the $x$-axis, and one along the $y$-axis. What is the volume of the resulting solid? I can do it for two holes but I'm stuck on three. • What makes it more difficult with three holes? – M.B. Nov 5 '13 at 12:58 The 3 holes such drilled form an intersection of 3 cylinders in the center, plus 6 cylinder/cap pairs. I will treat each problem in turn. Here is a picture of the situation: Intersection of 3 cylinders The problem is to find the volume of three orthogonal, intersecting cylinders: \begin{align}x^2+y^2&=1\\x^2+z^2&=1\\ y^2+z^2&=1\end{align} The intersection region is pictured below: We note that there are two ways to bound the volume over $x$: \begin{align}\|x\| &\le \sqrt{1-y^2}\\ \|x\| &\le \sqrt{1-z^2}\end{align} Since we are computing the volume of the interior of the region defined by these bounds, it stands to reason that $\|x\|$ must be bounded by the smaller of these two bounds: $$\|x\| \le \min{\left(\sqrt{1-y^2},\sqrt{1-z^2}\right)}$$ so that the volume integral takes the form $$\int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: \int_{-m(y,z)}^{m(y,z)} dx = 2 \int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: m(y,z)$$ Below is a representation of the integration region for this integral: The reason for the lines is because $\sqrt{1-y^2} \lt \sqrt{1-z^2}$ according to whether $\|y\| \gt \|z\|$. The integral is then symmetric over the regions bounded by the sloped lines; thus, we need only consider one such region and the others will yield the same result. Let's then consider the region surrounding the positive $y$ (horizontal) axis in the above figure. In this case, $m(y,z) = \sqrt{1-y^2}$; when we use polar coordinates, the integral becomes \begin{align}8 \int_{-\pi/4}^{\pi/4} d\phi \: \int_0^1 d\rho \, \rho \sqrt{1-\rho^2 \cos^2{\phi}} &= 4 \frac{2}{3} \int_{-\pi/4}^{\pi/4} d\phi \: \left( 1- \left\|\sin^3{\phi}\right\|\right) \sec^2{\phi}\\ &= \frac{16}{3} - \frac{16}{3} \int_0^{\pi/4} d\phi \: \sin^3{\phi} \, \sec^2{\phi}\\ &= 8 \left (2 - \sqrt{2}\right ) \end{align} Cylinder and cap The cross-sectional geometry here corresponds to a rectangle of width $2$ inscribed symmetrically about a diameter of a circle of radius $3$, with the cap corresponding to the resulting circular segment. The distance from center of circle to short edge of rectangle is $\sqrt{3^2-1^2}=2 \sqrt{2}$, so the height of the cylinder outside the intersection above is $2 \sqrt{2}-1$. The volume of a cylinder is thus $\pi \cdot 1^2 \cdot (2 \sqrt{2}-1) = (2 \sqrt{2}-1) \pi$. The volume of the cap is the difference between the volume of the spherical sector subtended by the solid angle defined by the hole and the corresponding cone. The volume of the spherical sector is given by $\frac13 (3)^3 \Omega$, where $\Omega$ is the solid angle. We find $\Omega$ by integrating over angle in spherical coordinates: $$\Omega = \int_0^{\theta_0} d\theta \, \sin{\theta} \, \int_0^{2 \pi} d\phi = 2 \pi (1-\cos{\theta_0})$$ where $\sin{\theta_0} = \frac13 \implies \cos{\theta_0} = 2 \sqrt{2}/3$. Thus the volume of a sector is $6\pi (3-2 \sqrt{2})$. The volume of a cone is $\frac13 \pi (1^2) 2 \sqrt{2} = 2 \sqrt{2} \pi/3$. Thus, the volume of a cap is $$6\pi (3-2 \sqrt{2}) - \frac{2 \sqrt{2} \pi}{3} = \left (18 - \frac{38 \sqrt{2}}{3}\right )\pi$$ Putting it all together The volume of the holes is $6$ times the sum of the volumes of the cylinder and cap, plus the volume of the intersection: $$V_{\text{holes}} = 6 \pi (2 \sqrt{2}-1) + \left (108- 76 \sqrt{2}\right )\pi + 8 (2-\sqrt{2})= (102-64\sqrt{2})\pi + 8 (2-\sqrt{2})$$ The volume left over after drilling is therefore the volume of the sphere minus the volume of the holes, or $$V = \frac{4 \pi}{3} (3)^3 - V_{\text{holes}} = (64 \sqrt{2}-66) \pi - 8 (2-\sqrt{2}) \approx 72.31$$ compared with the original volume of the sphere $36 \pi \approx 113.1$. • you should know that the intersection area is not like a cube. i think you have missed to compute the additional volume to find the total volume of holes. help me if i am wrong. – Bhaskara-III Sep 4 '16 at 14:49 • @Bhaskara-III When did I say it was like a cube? My diagram clearly illustrates what the volume is. – Ron Gordon Sep 4 '16 at 15:02 • you have considered 6 cylinders each of length $2\sqrt 2-1$ outside the intersection region which have flat circular ends of radius $1$. actually, you can't separate them with flat faces, in reality each cylinder has one end as flat circle of radius $1$ while other as non planar complicated shape i.e. not as a flat circle. the volume of each cylinder is not equal to $\color{red}{\pi \cdot1^2\cdot (2\sqrt 2-1)}$ that's why your answer has error in the value. – Bhaskara-III Sep 8 '16 at 7:02 • The figure was generated in Mathematica and is based on no assumption. I dispute your assertion. If you are going to stick with it, then why don't you give us an idea of what this "complicated surface" looks like. – Ron Gordon Sep 8 '16 at 13:19 • you may have a close look on my picture added in the answer which shows that the common region of three cylinders does not have any flat face. – Bhaskara-III Sep 11 '16 at 10:35 It is just a hint to visualize the region of intersection inside the sphere of radius 3. here is picture showing that the region of intersection of three cylinders ,each having radius 1, doesn't have any flat face. volume of intersection region has been calculated in the answer as $8(2-\sqrt 2)$. Now, how to calculate the volume of remaining region of holes? in the answer, it has been assumed that the faces of intersection region are flat. i don't know why. please someone clarify my doubt.	2019-11-14T04:15:02	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/552854/volume-of-a-sphere-with-three-holes-drilled-in-it", "openwebmath_score": 0.951598048210144, "openwebmath_perplexity": 338.96203726818834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990587412635403, "lm_q2_score": 0.847967764140929, "lm_q1q2_score": 0.8399861934785905 }
http://mathhelpforum.com/algebra/134295-range-function.html	# Thread: The range of this function? 1. ## The range of this function? The function is f(x)= [6/(x+2)]-3 I wanted to find the range, also could you please post your working as I want to understand how you came to your answer, thanks. Regards Ossy 2. $f(x)= \frac{6}{x+2}-3$ Where is $f(x)$ undefined? Naturally we look for such things as division by $0.$ When will we have division by $0?$ Clearly when $x+2 = 0,$ $i.e.,$ when $x= -2.$ So $f(x)$ take on all values except when $x = -2.$ What is your range? 3. Originally Posted by osmosis786 The function is f(x)= [6/(x+2)]-3 I wanted to find the range, also could you please post your working as I want to understand how you came to your answer, thanks. Regards Ossy one method of determining the range of a function is to determine the domain of the function's inverse. of course, this depends on whether the inverse exists and is relatively easy to find. $y = \frac{6}{x+2} - 3$ swap variables ... $x = \frac{6}{y+2} - 3$ solve for $y$ ... $x+3 = \frac{6}{y+2}$ $y+2 = \frac{6}{x+3}$ $y = \frac{6}{x+3} - 2$ this is the inverse function ... note that its domain is all real values $x$ such that $x \ne -3$. so ... the range of the original function is all real $y$ , $y \ne -3$. 4. Hello, osmosis786! When possible, I like to sketch a graph. Find the range of: . $f(x)\:=\:\frac{6}{x+2} -3$ The graph of . $y \,=\,\frac{1}{x}$ .looks like this: Code: \| \|* \| \| * \| * \| * \| * --------------+---------------- * \| * \| * \| * \| \| \| \| Domain: . $x\,\neq\,0$ . Range: . $y\,\neq\,0$ The graph of . $y \:=\:\frac{1}{x+2}$ . . is the previous graph moved two units to the left. Code: : \| : \| : \| : * \| : \| : \| : \| * --------------+---+---------------- * -2 \| * : \| * : \| * : \| : \| : \| : \| Domain: . $x\,\neq\,-2$ . Range: . $y\,\neq\,0$ The graph of . $y \:=\:\frac{6}{x+2}$ . rises more "steeply" . . and "flattens slower", but has the same basic shape. The graph of . $y \:=\:\frac{6}{x+2} - 3$ . is the previous graph lowered 3 units. Code: : \| : \| : \| : * \| --------------+--\|-------------- : \| : \| * - - - - - - o - \| - - - - - - * (-2,-3)\| * : \| * : \| * : \| : \| *: \| : \| Domain: . $x\,\neq\,-2$ . Range: . $y \,\neq\,-3$	2016-08-27T10:15:11	{ "domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/134295-range-function.html", "openwebmath_score": 0.8695751428604126, "openwebmath_perplexity": 2220.7708103413433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874098566368, "lm_q2_score": 0.8479677602988601, "lm_q1q2_score": 0.8399861873163813 }
https://math.stackexchange.com/questions/3760040/number-of-non-decreasing-sequence-a-i-such-that-every-a-i-geq-i	# Number of non-decreasing sequence $\{a_i\}$ such that every $a_i \geq i$ Find the number of non-decreasing sequences $$a_1, a_2, a_3, a_4, a_5$$ such that $$a_i \geq 1$$, $$a_5 \leq 20$$ and $$a_i \geq i$$; ## My attempt I tried to use the Inclusion-Exclusion principle, the number of non-decreasing sequences $$a_1, a_2, a_3, a_4, a_5$$ such that $$a_i \geq 1$$,$$a_5 \leq 20$$ is $${24\choose5}$$, However, I am having trouble counting the number of such sequences so that there is some $$a_i \lt i$$. I tried to seperate them into $$4$$ cases $$\{a_1, 1, a_3, a_4, a_5\}$$, $$\{a_1, a_2, 2, a_4, a_5\}$$, $$\{a_1, a_2, a_3, 3, a_5\}$$, $$\{a_1, a_2, a_3, a_4, 4\}$$, but there are lots of overlapping cases and I do not want to handle them. ## Question • Should I use Inclusion-Exclusion principle here? If yes, is there any smarter way than mine? • What is the most efficient way to find the answer? • Hint: Consider all sequences of distint numbers and order them-they automatically fulfill the requirements, you have $\binom{20,5}$ possibilities, now consider the cases of multiple equals seperately. – IMOPUTFIE Jul 17 '20 at 11:31 • @IMOPUTFIE Thanks for the hints, but I have another question, consider the case of the form $\{a,a,b,c,d\}$, I should use inclusion-exclusion right? Because the only case for this to be invalid is $\{1,1,a,b,c\}$ – Learning Mathematics Jul 17 '20 at 11:48 • @IMOPUTFIE Can you elaborate? While I understand Brain's approach, I would really want to know how do you consider the cases of multiple in an organised way – Learning Mathematics Jul 18 '20 at 11:36 I found it easiest to convert it to a problem in counting paths on the integer lattice in the plane: it can be solved using the reflection method, one of the standard ways to show that $$C_n=\frac1{n+1}\binom{2n}n$$, where $$C_n$$ is the $$n$$-th Catalan number. Suppose that $$\langle a_1,\ldots,a_5\rangle$$ is such a sequence. We can interpret it as directions for a walk on the integer lattice in the plane, starting at the origin: we first take $$a_1$$ steps north to $$\langle 0,a_1\rangle$$, then one step east to $$\langle 1,a_1\rangle$$, then $$a_2-a_1$$ steps north to $$\langle 1,a_2\rangle$$ and one step east to $$\langle 2,a_2\rangle$$, and so on, finishing by taking $$20-a_5$$ steps north from $$\langle 5,a_5\rangle$$ to $$\langle 5,20\rangle$$; the requirement that each $$a_k\ge k$$ is then the requirement that this path never drop below the diagonal $$y=x$$. Moreover, each NE path (i.e., a path using only steps to the north and to the east) from $$\langle 0,0\rangle$$ to $$\langle 5,20\rangle$$ that never drops below the diagonal corresponds to a unique sequence $$\langle a_1,\ldots,a_5\rangle$$ satisfying the conditions of the problem, so our problem reduces to counting such paths. Suppose that a path first drops below the diagonal at $$\langle k,k-1\rangle$$; after that point it must take $$5-k$$ steps to the east and $$21-k$$ to the north. If we reflect it in the diagonal, we get a path starting at $$\langle k,k-1\rangle$$ and taking $$21-k$$ steps to the east and $$5-k$$ steps north and thus ends at $$\langle 21,4\rangle$$. Conversely, any NE path from $$\langle 0,0\rangle$$ to $$\langle 21,4\rangle$$ must stay on or above the diagonal until it hits a point of the form $$\langle k,k-1\rangle$$, and reflecting the remainder of the path in the diagonal gives us a path from $$\langle 0,0\rangle$$ to $$\langle 5,20\rangle$$ that first drops below the diagonal at $$\langle k,k-1\rangle$$. There are clearly $$\binom{25}5$$ NE paths from $$\langle 0,0\rangle$$ to $$\langle 5,20\rangle$$. There is a bijection between those that drop below the diagonal and NE paths from $$\langle 0,0\rangle$$ to $$\langle 21,4\rangle$$, and there are $$\binom{25}4$$ of those, so there are $$\binom{25}5-\binom{25}4=53130-12650=40480$$ NE paths from $$\langle 0,0\rangle$$ to $$\langle 5,20\rangle$$ that do not drop below the diagonal. More generally, the number of non-decreasing sequences $$a_1,\ldots,a_n$$ such that $$a_1\ge 1$$, $$a_k\ge k$$ for $$k=1\ldots,n$$, and $$a_n\le m$$ is $$\binom{n+m}n-\binom{n+m}{n-1}=\binom{n+m}n-\frac{n}{m+1}\binom{n+m}n=\frac{m+1-n}{m+1}\binom{n+m}n\;.$$ When $$m=n$$ this reduces to $$C_n=\frac1{n+1}\binom{2n}n$$.	2021-08-01T18:34:30	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3760040/number-of-non-decreasing-sequence-a-i-such-that-every-a-i-geq-i", "openwebmath_score": 0.7629914283752441, "openwebmath_perplexity": 117.44973990160902, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874115238967, "lm_q2_score": 0.8479677583778258, "lm_q1q2_score": 0.8399861868272115 }
http://math.stackexchange.com/questions/454583/volume-of-cube-section-above-intersection-with-plane	# Volume of cube section above intersection with plane Suppose we have a unit cube (side=1) and a plane with equation $x+y+z=\alpha$. I'd like to compute the volume of the region that results once the plane sections the cube (above the plane). There are three cases to analyze, and I can't quite visualize one of them. Case 1: $0 \le \alpha < 1$ In this case, the section looks like a triangle, and the volume of interest is 1 minus the volume of the lower left tetrahedron, i.e., $$V = 1 - \int_0^\alpha \int_0^{\alpha-x} \int_0^{\alpha-x-y} dz\,dy\,dx = 1 - \frac{\alpha^3}{6}.$$ Case 3: $2 < \alpha \le 3$. Here, the section is again a triangle, and the volume of interest is the upper right tetrahedron, i.e., $$V = \int_{\alpha-2}^1 \int_{\alpha-x-1}^1 \int_{\alpha-x-y}^1 dz\,dy\,dx = \frac{(3-\alpha)^3}{6}.$$ Case 2: $1 \le \alpha \le 2$. This is where I'm sort of stuck. The section is a hexagon, with one of the inequalities being $\alpha-x-y \le z \le 1$, hence the innermost integral should be $\int_{\alpha-x-y}^1 dz$. The projection of the hexagon slice onto the $xy$-plane is described by $y \ge \alpha-1-x$ and $y \le \alpha-x$. Hence, the area of the hexagon projection is $$A = \int_0^{\alpha-1} \int_{\alpha-x-1}^1 dy\,dx + \int_{\alpha-1}^1 \int_0^{\alpha-x} dy\,dx$$ Question: When I move from $A$ to $V$ am I allowed to distribute the innermost integral between the summing terms, i.e. is it correct to write $$V = \int_0^{\alpha-1} \int_{\alpha-x-1}^1 \int_{\alpha-x-y}^1 dz\,dy\,dx + \int_{\alpha-1}^1 \int_0^{\alpha-x} \int_{\alpha-x-y}^1 dz\,dy\,dx \quad ??$$ If not, what's the approach? Note that there's a neat connection between this problem and figuring out the CDF of a sum of a random variable that has triangular distribution with support on $[0,2]$ and a random variable with uniform distribution on $[0,1]$ (assuming independence). Hence, I know what the answer should be for Case 2 because I worked out the convolution, but I just want to figure out the answer geometrically as well. - Are you certain that it's a pentagon and not a hexagon? – abiessu Jul 29 '13 at 4:52 I think it's a pentagon, because the bottom of the cube is not touched by the slice. I may be wrong. – baudolino Jul 29 '13 at 4:54 If I recall the slicing correctly, you get triangles hexagons triangles when you view the cross-section of a cube as it passes from corner to corner through a surface... Is this a different interpretation from what you are describing? – abiessu Jul 29 '13 at 4:58 I don't think you can get a pentagon by slicing a cube (though I can't prove this). – bubba Jul 29 '13 at 5:39 You're right about it being a hexagon after all (because the plane rests on $xy$ as well). However, in general, getting a pentagon is also possible (the section looks like a baseball diamond). Just make the cut starting at some height $z$ above the base and go at an angle so that you exit the cube on the top facet, but beyond that square's diagonal. – baudolino Jul 30 '13 at 0:24 For $1 \le \alpha \le 2$, it is much easier to visualize the integral by subtracting instead of adding pieces of the volume. As shown in the picture below, when $1 \le \alpha \le 2$, the volume of cube section below intersection of the plane $x + y + z = \alpha$ is the difference of the volume of one big tetrahedron with width/height/depth $= \alpha$ with three smaller ones with width/height/depth $= \alpha-1$. So the volume above the intersection becomes $$1 - \left( \frac16 \alpha^3 - 3 ( \frac16 (\alpha-1)^3 )\right) = 1 - \frac16 \alpha^3 + \frac12 (\alpha-1)^3$$ Update About the question whether this argument can be extended to higher dimension, the answer is yes. Let's look at the 3-dimension $2 \le \alpha \le 3$ case first. As one increases $\alpha$ beyond $2$, the three tetrahedron in first figure start overlap. As shown in second figure, the intersection of the three tetrahedra are now three even smaller tetrahedra of width/height/depth = $\alpha -2$. Previous way to compute the "volume" of cube section below the plane $x + y + z = \alpha$ now subtract too much from this three even smaller tetrahedron. One need to add them back. As a result, the volume above the plane becomes: \begin{align}&1 - \left( \frac16 \alpha^3 - 3(\frac16 (\alpha-1)^3 + 3(\frac16 (\alpha-2)^3 \right)\\ = & 1 - \frac16 \alpha^3 + \frac12 (\alpha-1)^3 - \frac12 (\alpha-2)^3\\ = & \frac{(3-\alpha)^3}{6} \end{align} Let us switch to the $k$-dimension case. To compute the "volume" of the hypercube section above the hyperplane $x_1 + \ldots + x_k = \alpha$, the first step is to subtract the volume of a $k$-simplex of size $\alpha$ from 1. 1. if $\alpha \le 1$, we are done. 2. if $\alpha > 1$, we over subtract the volume of $\binom{k}{1}$ simplices of size $\alpha-1$ and need to add them back. 3. if $\alpha > 2$, the $\binom{k}{1}$ simplices of size $\alpha-1$ in step 2 intersect and the intersection is a union of $\binom{k}{2}$ $\;k$-simplices of size $\alpha-2$. This means in step 2, we have added back too much and need to subtract the volume again. Repeat these arguments and notice in the middle of the process, we need to either add or subtract the volumes of $\binom{k}{i}$ $\;k$-simplices of size $\alpha-i$. The "volume" of interest finally becomes: $$1 -\sum_{i=0}^{\lfloor \alpha \rfloor} (-1)^i \binom{k}{i} \frac{(\alpha-i)^k}{k!}$$ - This is a great way to think about it. Am I correct in asserting that one can expand this argument so that on a $[0,1]^k$ hypercube the "volume" of interest should be given by $$V = 1- \dfrac{1}{k!}\sum_{i=0}^{\lfloor \alpha \rfloor} {(-1)^i \binom{k}{i} (\alpha-i)^k} \quad ?$$ – baudolino Jul 30 '13 at 22:16 +1. Good idea, and nice picture. – bubba Jul 31 '13 at 1:45 Put $\alpha:=1+\beta$ with $0<\beta<1$. Then the plane $x+y+z=1+\beta$ cuts the top face of the cube in the segment connecting $(0,\beta)$ with $(\beta,0)$ at $z$-level $1$, and cuts the bottom face of the cube in the segment connecting $(\beta,1)$ with $(1,\beta)$ at $z$-level $0$. Consider the vertical stalk through a fixed point $(x,y)$ on the bottom face $Q$ of the cube. It intersects your polyhedron $P$ in a vertical segment whose length $\ell(x,y)$ is given by $$\ell(x,y)=\cases{0&(x+y<\beta) \cr x+y-\beta\quad&(\beta\leq x+y\leq \beta+1) \cr 1&(x+y>\beta+1)\cr}\ .$$ The volume of $P$ is then given by \eqalign{{\rm vol}(P)&=\int\nolimits_Q \ell(x,y)\ {\rm d}(x,y)\cr &=\int_0^\beta\int_{\beta-x}^1 (x+y-\beta)\ dy\ dx+\int_\beta^1\int_0^{1+\beta-x}(x+y-\beta)\ dy\ dx +{1\over2}(1-\beta)^2\ .\cr} - This can be interpreted as trying to find $1$ minus the CDF of the sum of three uniformly distributed random variables in $[0, 1]$. The PDF is the quadratic B-spline given by $$f(x)=\begin{cases} \frac{x^2}{2} & x\in[0,1] \\ \frac{x^2}{2} - \frac{3(x-1)^2}{2} & x\in[1,2]\\ \frac{x^2}{2} - \frac{3(x-1)^2}{2} +\frac{3(x-2)^2}{2} & x\in[2,3] \end{cases}$$ The CDF follows from this by integration. - Box splines, perhaps? – bubba Jul 29 '13 at 10:49 Thank you, I have mentioned this solution approach in my question. I want to derive the answer in another fashion, using geometrical intuition. – baudolino Jul 30 '13 at 0:25 The object created by slicing is a polyhedron, obviously. Then, the volume of a polyhedron can be computed conveniently by decomposing it into tetrahedra and adding up the (signed) volumes of these tetrahedra. You can find an explanation of the technique in this document. There is also a paper by Michael Kallay on the same subject that gives slightly simpler formulae. Ask again if this is of interest and you can't find it. All of this is still computation of volume intergals, really. But, in the case of tetrahedra, the values of the integrals can be expressed by simple closed-form formulae, so the integration is somewhat hidden. You'll still need to consider several cases (depending on where/how the plane slices the cube). Actually, I think there are three cases, depending on whether the planar slice has three, five, or six sides. These are shown in the picture below: The blue and brown slices are 3-sided, the red and green ones are 5-sided, and the pink one is 6-sided. - I had the same problem and I came up with the following very simple (and very implicit) formula (I'm unfamiliar with stackexchange's maths notation, forgive me for posting Mathematica code): volPart[q___, r_] := Module[{v, d, t, h}, v = Select[Chop[{q}], (# =!= 0) &]; d = Length[v]; t = Tuples[{-1, 1}, d]; h = (t/2).v + r; (HeavisideTheta[h] h^d).(Times @@@ t)/(d! Times @@ v) ] The return value of volPart[q1, q2, ..., r] should be a volume of a part of the unit hypercube centered at $\{0, 0, ...\}$ above the hyper-plane $x1 q1 + x2 q2 + ... + r > 0$. By appropriate scaling and offsetting the arguments, the formula could be generalised to other cases. For zero valued numerical coefficients in linear form the formula effectively reduces to lower dimension case. For symbolic coefficients approaching zero, the limit must be taken, though. The basic idea is that the volume above the hyper-plane for fixed arguments is a linear combination of vertices' "heights" above (but only these above) the hyper-plane raised to the power of effective dimension. Note the alternating sign for each connected vertex (Times @@@ t). In your particular case this reduces to: $$\frac{\alpha^3{\theta}(-\alpha)+3(1-\alpha)^3{\theta}(1-\alpha)-3(2-\alpha)^3{\theta}(2-\alpha)+(3-\alpha)^3{\theta}(3-\alpha)}{6}$$ By expanding $\theta$, this could also be rewritten as: $$\begin{cases} 1 & \alpha \le 0\\ 1-\frac{\alpha^3}{6} & 0 < \alpha \le 1 \\ \frac{\alpha^3}{3}-\frac{3 \alpha^2}{2}+\frac{3 \alpha}{2}+\frac{1}{2} & 1 < \alpha \le 2 \\ -\frac{(\alpha-3)^3}{6} & 2 < \alpha \le 3 \\ 0 & 3 < \alpha\end{cases}$$ -	2016-05-24T23:47:55	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/454583/volume-of-cube-section-above-intersection-with-plane", "openwebmath_score": 0.8820827007293701, "openwebmath_perplexity": 307.7834667011739, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534392852384, "lm_q2_score": 0.8558511451289037, "lm_q1q2_score": 0.8399780499029723 }
https://math.stackexchange.com/questions/2180685/find-matrix-p-such-that-p-1ap-b	# Find matrix $P$ such that $P^{-1}AP=B$ Given $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & -3 \\ 1 & 3 & 2 \end{bmatrix}$$ $$B= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & -3 \\ 0 & 3 & 2 \end{bmatrix}$$ find $P$ such that $P^{-1} A P = B$. Firstly I said that $AP=PB$ Solved the 9 equations in 9 unknowns. and got that: $$P= \left( \begin{array}{ccc} -10x & 0 & 0 \\ 3x & y & z \\ x & -z & y \end{array} \right)$$ Then I used computer to find $P^{-1}$ in terms of those unknowns and plugged it back in to $P^{-1}AP=B$ Compared the coefficients and i end up with $B= \left( \begin{array}{ccc} 1 & 0 & 0 \\ -z/10x & 2 & -3 \\ 1-y/10x & 3 & 2 \end{array} \right)$ Set $x=1$, $y=10$, $z=0$ and indeed $P^{-1}AP=B$ The doubts I am having is the fact that P is not unique. I could set x,y,z to different numbers. Can anyone explain this? Thank you. • Left multiply both sides by $P$? Is there a better method? – Scavenger23 Mar 10 '17 at 15:16 • That is very true. Thank you. I believe it won't change anything no matter which way i do it? – Scavenger23 Mar 10 '17 at 15:17 • Suppose that $P$ solves your problem, then $\alpha P$ will also solve it for any $\alpha \neq 0$ so the solution is never unique. – Surb Mar 10 '17 at 16:14 Using SymPy: >>> from sympy import * >>> X = MatrixSymbol('X',3,3) >>> A = Matrix([[ 1, 0, 0], [ 0, 2,-3], [ 1, 3, 2]]) >>> B = Matrix([[ 1, 0, 0], [ 0, 2,-3], [ 0, 3, 2]]) >>> simplify(Matrix(AX - XB)) Matrix([ [ 0, -X[0, 1] - 3X[0, 2], 3X[0, 1] - X[0, 2]], [ X[1, 0] - 3X[2, 0], -3(X[1, 2] + X[2, 1]), 3(X[1, 1] - X[2, 2])], [X[0, 0] + 3X[1, 0] + X[2, 0], X[0, 1] + 3X[1, 1] - 3X[2, 2], X[0, 2] + 3X[1, 2] + 3X[2, 1]]]) If $\rm A X - X B = O_3$, we have an underdetermined homogeneous system of $3^2 - 1$ linear equations in $3^2$ unknowns. Thus, we have at least $1$ degree of freedom. We vectorize the matrix equation $$\left( \mathrm I_3 \otimes \mathrm A - \mathrm B^{\top} \otimes \mathrm I_3 \right) \mbox{vec} (\mathrm X) = 0_9$$ and let $\mathrm M := \mathrm I_3 \otimes \mathrm A - \mathrm B^{\top} \otimes \mathrm I_3$. Using SymPy: >>> I3 = Identity(3) >>> O3 = ZeroMatrix(3,3) >>> M = BlockMatrix([[A - I3, O3, O3], [ O3, A - 2I3, -3I3], [ O3, 3I3, A - 2I3]]) >>> Matrix(M) Matrix([ [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, -3, 0, 0, 0, 0, 0, 0], [1, 3, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, -1, 0, 0, -3, 0, 0], [0, 0, 0, 0, 0, -3, 0, -3, 0], [0, 0, 0, 1, 3, 0, 0, 0, -3], [0, 0, 0, 3, 0, 0, -1, 0, 0], [0, 0, 0, 0, 3, 0, 0, 0, -3], [0, 0, 0, 0, 0, 3, 1, 3, 0]]) >>> Matrix(M).rank() 6 Hence, we have $3^2 - 6 = 3$ degrees of freedom. • Yes I know realised that the answer I posted only works for PAP^-1 P^-1AP allows you to set all three unknowns, as long as resulting P is invertible. – Scavenger23 Mar 10 '17 at 20:04 The computation is correct, and we cannot expect that $P$ is unique in general. A somewhat extreme case is $A=B=0$, where any invertible $P$ will do.	2019-04-20T12:35:06	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2180685/find-matrix-p-such-that-p-1ap-b", "openwebmath_score": 0.6917913556098938, "openwebmath_perplexity": 111.13684861792456, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534392852383, "lm_q2_score": 0.855851143290548, "lm_q1q2_score": 0.8399780480987117 }
http://math.stackexchange.com/questions/174828/a-probably-trivial-induction-problem-sum-2nk-2-lt1/174857	# A (probably trivial) induction problem: $\sum_2^nk^{-2}\lt1$ So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither. - Why do you think that this problem can be solved using induction? – joriki Jul 24 '12 at 22:08 @joriki the person who gave it to me said it could? So, ethos I guess... – Paquito Jul 24 '12 at 22:10 what about comparing it with $ln2$ – Theorem Jul 24 '12 at 22:17 As often happens with induction proofs, the easiest approach to proving this statement (which doesn't seem inducable at all - after all, how does knowing the sum for $n$ is less than $1$ tell you anything about the sum for $n+1$?) via induction is to transform it into a stronger one: $$\mathrm{For\ all\ } n\geq2, \frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2} \lt 1-\frac{1}{n}.$$ Now, the answer becomes a matter of simple algebra: $$\sum_{i=2}^{n+1} \frac{1}{i^2} = \sum_{i=2}^{n} \frac{1}{i^2} +\frac{1}{(n+1)^2}\lt 1-\frac{1}{n}+\frac{1}{(n+1)^2}\lt 1-\frac{1}{n}+\frac{1}{n(n+1)} = 1-\frac{1}{n+1}.$$ - Awesome! Many thanks...Might I ask what motivated the choice of subtracting 1/n? – Paquito Jul 24 '12 at 22:21 @Paquito To a certain extent, intuition - since the difference between consecutive terms of the form $1/n$ is on the order of $1/n^2$, I could see that adding a quadratic term to a $1/n$-sized 'gap' between the sum and 1 would give a $1/n+1$-sized gap; from there it was just a quick dig to see whether $1-1/n$ itself would work or whether I would need to do something like $1-1/(n+1)$. – Steven Stadnicki Jul 24 '12 at 22:26 That's beautiful, Steven! – Bruno Joyal Jul 24 '12 at 22:40 Of course, this is also the "integral test" estimate. – Robert Israel Jul 24 '12 at 23:53 @Paquito What Steven did is a classical example of inventor's paradox. Similarly proving $\frac{(2n-1)!!}{(2n)!!} < \frac{1}{\sqrt{n}}$ does not give in to the method of induction, while a stronger inequality $\frac{(2n-1)!!}{(2n)!!} < \frac{1}{\sqrt{n+1}}$ readily does. The explanation for why this works, is that the assumptions of the induction step are stronger in the latter case, allowing to draw stronger conclusion. – Sasha Jul 25 '12 at 1:22 Another proof is by comparison: note that $${1 \over k^2} < {1 \over (k-1)k}$$ for all integers $k \ge 2$. Therefore $${1 \over 2^2} + {1 \over 3^2} + \cdots + {1 \over n^2} < {1 \over 1 \times 2} + {1 \over 2 \times 3} + \cdots + {1 \over (n-1) \times n}$$ and now you need to find the sum on the right-hand side. But you can actually write $${1 \over (k-1)k} = {1 \over k-1} - {1 \over k}$$ (this is just the usual partial fraction decomposition) and therefore $${1 \over 1 \times 2} + {1 \over 2 \times 3} + \cdots + {1 \over (n-1) \times n} = \left( {1 \over 1} - {1 \over 2} \right) + \left( {1 \over 2} - {1 \over 3} \right) + \cdots + \left( {1 \over n-1} - {1 \over n} \right)$$ and the right-hand side what's called a telescoping sum'' -- that is, the pairs of terms $-1/2$ and $+1/2$, $-1/3$ and $+1/3$, and so on cancel. So the right-hand side is $1 - 1/n$, which is less than 1. This came to mind pretty much immediately for me, because I happened to know that $\sum_{k \ge 2}^\infty 1/(k(k-1)) = 1$, but if you didn't know that ahead of time it would be a bit tricky to discover. - Yet another approach : Let us first analyze the sum till infinity. Let $$S= \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$ $$\Rightarrow S=(\frac{1}{2^2}+\frac{1}{4^2}+ \cdots\infty) +(\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty )$$$$\Rightarrow S= \frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\infty)+ S'$$ Where $$S'=\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty$$ $$\Rightarrow S=\frac{1}{4}(1+S)+S'$$ $$\Rightarrow 3S=4S'+1........ Eqn(1)$$ Now examine the following inequality $$(\frac{1}{2^2}-\frac{1}{3^2})+(\frac{1}{4^2}-\frac{1}{5^2})+ \cdots \infty > 0$$ $$\Rightarrow \frac{1}{2^2}+\frac{1}{4^2}+\cdots > \frac{1}{3^2}+\frac{1}{5^2}+ \cdots$$ $$\Rightarrow \frac{1}{4}(1+\frac{1}{2^2}+\cdots) >\frac{1}{3^2}+\frac{1}{5^2}+ \cdots$$ $$\Rightarrow \frac{1}{4}(1+S)> S'$$ $$\Rightarrow (1+S)> 4S'......Eqn(2)$$ From Equation 1 and 2 we get $$1+S> 3S-1$$ $$\Rightarrow 1> S$$ Which shows $$1> \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$ -	2015-07-06T03:31:17	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/174828/a-probably-trivial-induction-problem-sum-2nk-2-lt1/174857", "openwebmath_score": 0.8994263410568237, "openwebmath_perplexity": 298.05780074192336, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534316905262, "lm_q2_score": 0.8558511488056151, "lm_q1q2_score": 0.8399780470115501 }
http://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3	# Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$ One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor? $$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$ - Closely related is math.stackexchange.com/questions/134051 – Dave L. Renfro Jun 13 '12 at 18:51 The statement $\dfrac{\tan(x)-x}{x^3} \to c$ as $x \to 0$ is equivalent to $\tan(x) = x + c x^3 + o(x^3)$ as $x \to 0$, so this is a statement about a Taylor polynomial of $\tan(x)$, and I'm not sure what would count as doing that "without Taylor". However, one thing you could do is start from $$\sin(x) = x + o(x)$$ integrate to get $$\cos(x) = 1 - x^2/2 + o(x^2)$$ then $$\sec(x) = \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$$ $$\sec^2(x) = \left(1 + x^2/2 + o(x^2)\right)^2 = 1 + x^2 + o(x^2)$$ and integrate again to get $$\tan(x) = x + x^3/3 + o(x^3)$$ - thanks. Your approach is nice and interesting. Initially my first thought was to find some geometrical approch, but it doesn't seem that easy.I'm still trying to find a link between this limit and the intersection point of medians in a triangle. – Chris's sis Jun 13 '12 at 19:48 Here's a way which avoids derivatives and integrals. Assume that we know that $\frac{\sin x}{x} \to 1$ as $x \to 0$. Then we also know that $\frac{1-\cos x}{x^2} = \frac12 \left( \frac{\sin(x/2)}{x/2} \right)^2 \to \frac12$. Now, $$\frac{\tan x - x}{x^3} = \frac{1}{\cos x} \left( \frac{\sin x - x}{x^3} + \frac{1-\cos x}{x^2} \right),$$ so we are done if we can compute $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac16$. The reason that I rewrote it like this is that I was asked by a colleague about ten years ago whether that limit could be done in an elementary way. :-) I came up with the following: Let $$f(x) = \frac{x - \sin x}{x^3} = \frac{1 - \frac{\sin x}{x}}{x^2}.$$ (Here I've changed the sign so that the limit will be positive.) Since $f$ is an even function, it's enough to consider $x>0$. Fix a positive integer $n$. To begin with, we have $$x = 2^n \frac{x}{2^n} > 2^n \sin \frac{x}{2^n}.$$ (I'm assuming that we also know that $0 < \sin x < x < \tan x$ for $0 < x < \pi/2$.) Multiply this inequality by $\prod_{k=1}^n \cos\frac{x}{2^k}$ and use the double angle formula repeatedly, as follows (illustrated for the case $n=3$): $$\begin{split} x \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2} & > 2^3 \sin\frac{x}{8} \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2} \\ & = 2^2 \sin\frac{x}{4} \cos\frac{x}{4} \cos\frac{x}{2} \\ & = 2^1 \sin\frac{x}{2} \cos\frac{x}{2} \\ & = \sin x. \end{split}$$ This implies (again for $n=3$, but the general pattern is hopefully clear) $$\begin{split} 1 - \frac{\sin x}{x} & > 1 - \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2} \\ & = \left( 1 - \cos\frac{x}{8} \right) + \cos\frac{x}{8} \left( 1 - \cos\frac{x}{4} \right) + \cos\frac{x}{8} \cos\frac{x}{4} \left( 1 - \cos\frac{x}{2} \right). \end{split}$$ We know that $\frac{1 - \cos(x/2^k)}{x^2} = \frac{1 - \cos(x/2^k)}{2^{2k} (x/2^k)^2} \to \frac{1}{2^{2k+1}}$, so after dividing this inequality by $x^2$ we find in the limit (for general $n$) that $$\liminf_{x \to 0^+} f(x) \ge \sum_{k=1}^n \frac{1}{2^{2k+1}} = \frac16 \left( 1 - \frac{1}{4^n} \right).$$ This holds for every $n$, hence $$\liminf_{x \to 0^+} f(x) \ge \frac16.$$ The other direction is similar. Start with $$x = 2^n \frac{x}{2^n} < 2^n \tan\frac{x}{2^n} = 2^n \frac{\sin(x/2^n)}{\cos(x/2^n)}.$$ This leads to $$\begin{split} 1 - \frac{\sin x}{x} & < 1 - \cos\frac{x}{2^n} \cdot (\text{same product of cosines as above}) \\ & = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (1 - (\text{that product})) \\ & = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (\text{same expression as above}). \end{split}$$ Divide by $x^2$ and let $x \to 0^+$: $$\limsup_{x \to 0^+} f(x) \le \frac{1}{2^{2n+1}} + \frac16 \left( 1 - \frac{1}{4^n} \right).$$ Let $n \to \infty$: $$\limsup_{x \to 0^+} f(x) \le \frac16.$$ It follows that $\lim_{x \to 0^+} f(x) = \frac16$, and therefore by symmetry $\lim_{x \to 0} f(x) = \frac16$,which is what we wanted to show. - :hehe, this is an amazingly nice answer. Thanks! A lot of stuff in this answer! :) – Chris's sis Jun 14 '12 at 8:04 btw, the whole proof is very clear. – Chris's sis Jun 14 '12 at 8:12 @Chris: Thank you! – Hans Lundmark Jun 14 '12 at 10:34 @HansLundmark This is neat. +1 – user17762 Oct 20 '12 at 17:16 There's the Cauchy or extended mean value theorem which says that ${\displaystyle {f(x) - f(y) \over g(x) - g(y)} = {f'(c) \over g'(c)}}$ for some $c$ between $x$ and $y$. You can apply it here with $f(x) = \tan(x) - x$ and $g(x) = x^3$, and you get that for some $0 < c < x$ you have $$\frac{\tan(x) - x}{x^3} = {\sec^2(c) - 1 \over 3c^2}$$ $$= {\tan^2(c) \over 3c^2}$$ $$= {1 \over 3}{1 \over \cos^2(c)}{\sin^2(c) \over c^2}$$ Now take limits as $x$ goes to zero; $c$ goes to zero and the limit is $1/3$. This is all somewhat tongue in cheek of course, since you can get L'Hopital pretty quickly from the extended mean value theorem, but it does satisfy your request to not use it or Taylor polynomials :) - that's right. Thanks. I was thinking of some geometrical approach. – Chris's sis Jun 13 '12 at 19:44 Encouraged by Hans Lundmark's answer, I'm posting my own solution without derivatives and integrals. The triple-angle formula for $\tan$ is $$\tan 3\theta = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.$$ Suppose $\lim_{x\to0}(\tan x-x)/x^3 = c$. Letting $x = 3\theta$, we then have \begin{align} c &= \lim_{x\to0} \frac{\tan x-x}{x^3} \\ &= \lim_{\theta\to0} \frac{\tan 3\theta-3\theta}{27\theta^3} \\ &= \lim_{\theta\to0} \frac{3\tan\theta - \tan^3\theta-3\theta+9\theta\tan^2\theta}{27\theta^3(1-3\tan^2\theta)} \end{align} We can get rid of $1/(1-3\tan^2\theta)$ because its limit is $1$. Next we start pulling out terms and find \begin{align} c &= \lim_{\theta\to0}\frac{3\tan\theta-3\theta}{27\theta^3} - \lim_{\theta\to0}\frac{\tan^3\theta}{27\theta^3} + \lim_{\theta\to0}\frac{9\theta\tan^2\theta}{27\theta^3} \\ &= \frac19c - \frac1{27} + \frac{1}{3}, \end{align} because $\lim_{\theta\to0}(\tan\theta)/\theta = 1$. So $8c/9 = 8/27$, or $c = 1/3$. - Nice. Similar method to mine! – user17762 Jun 14 '12 at 7:07 Ah, much simpler than my answer! Except that you use heavy machinery like the triple angle formula and I only used the double angle formula. ;-) – Hans Lundmark Jun 14 '12 at 7:59 @Rahul Narain: nice answer! Thanks. – Chris's sis Jun 14 '12 at 8:08 @Hans: Yes, I pulled out the big guns for this one! :) Actually, I just didn't realize that the double angle formula works fine with this approach -- it becomes equivalent to Marvis' answer. – Rahul Jun 14 '12 at 9:21 There's a problem with the existence of the limit, which has to be shown before this argument is complete. See my comment to Marvis's answer. – Hans Lundmark Jun 14 '12 at 14:10 Here is a different approach. Let $$L = \lim_{x \to 0} \dfrac{\tan(x) - x}{x^3}$$ Replacing $x$ by $2y$, we get that \begin{align} L & = \lim_{y \to 0} \dfrac{\tan(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2 \tan(y) + 2 \tan(y) - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \dfrac{\dfrac{2 \tan^3(y)}{1 - \tan^2(y)} + 2 \tan(y) - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} + \dfrac{2 \tan(y) - 2y}{8y^3} \right)\\ & = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} \right) + \lim_{y \to 0} \left(\dfrac{2 \tan(y) - 2y}{8y^3} \right)\\ & = \dfrac14 \lim_{y \to 0} \left(\dfrac{\tan^3(y)}{y^3} \dfrac1{1 - \tan^2(y)} \right) + \dfrac14 \lim_{y \to 0} \left(\dfrac{\tan(y) - y}{y^3} \right)\\ & = \dfrac14 + \dfrac{L}4 \end{align} Hence, $$\dfrac{3L}{4} = \dfrac14 \implies L = \dfrac13$$ EDIT In Hans Lundmark answer, evaluating the desired limit boils down to evaluating $$S=\lim_{x \to 0} \dfrac{\sin(x)-x}{x^3}$$ The same idea as above can be used to evaluate $S$ as well. Replacing $x$ by $2y$, we get that \begin{align} S & = \lim_{y \to 0} \dfrac{\sin(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2y}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2 \sin(y) + 2 \sin(y) - 2y}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 \sin(y) - 2y}{8y^3} + \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y)-2 \sin(y)}{8y^3}\\ & = \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) - y}{y^3} - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) (1 - \cos(y))}{y^3}\\ & = \dfrac{S}4 - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) 2 \sin^2(y/2)}{y^3}\\ & = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \lim_{y \to 0} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 - \dfrac18\\ \dfrac{3S}4 & = - \dfrac18\\ S & = - \dfrac16 \end{align} - Huh, I had assumed that to get $\tan$ to third order one would have to use the triple angle formula. I guess I was mistaken. – Rahul Jun 14 '12 at 7:11 @Marvis: very elegant! Thanks! – Chris's sis Jun 14 '12 at 8:05 @Marvis: this proof also works for showing that $\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^2} = 0$. – Chris's sis Jun 14 '12 at 8:16 @Chris Yes. This is powerful neat little trick. I learnt this trick in my last year at high school and yes you can show that $\lim_{x \to 0} \dfrac{\tan(x)-x}{x^2} =0$ using this trick as well. – user17762 Jun 14 '12 at 8:23 @Marvis: i've just noticed now that it works great and very simple by only using: $\sin(x)<x<\tan(x),\space 0< x <\frac{\pi}{2}.$. It's nice to have a lot of tackling ways for each problem. :-) – Chris's sis Jun 14 '12 at 10:16	2014-04-17T09:50:52	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3", "openwebmath_score": 0.9990609288215637, "openwebmath_perplexity": 598.8449403333569, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534327754852, "lm_q2_score": 0.8558511469672594, "lm_q1q2_score": 0.839978046135853 }
https://math.stackexchange.com/questions/703260/using-lagrange-form-of-the-remainder-with-cosh	# Using Lagrange form of the remainder with cosh I am trying to find "$\cosh 4$ using the sixth partial sum ($n=5$) of the Maclaurin series" for the function. I am also trying to use "the Lagrange form of the remainder to estimate the number of decimal places to which the partial sum" is accurate. For the first part, I am getting $\displaystyle\sum_{n=0}^5 \frac{4^{2n}}{(2n)!} \approx 27.2699$, which agrees with the provided answer. However, I am having trouble estimating the error. Here is what I have done so far. The Lagrange form of the remainder is $\displaystyle R_n = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$, where here $a=0$ (Maclaurin), $x=4$ (the value for which we want $\cosh$), and $0=a\leq c\leq x=4$. To find an upper bound for $f^{(n+1)}(c)$, since all derivatives of $\cosh$ are less than or equal to $\cosh$, I see that using $\cosh$ will work, and I also see that $4$ will yield the highest value, so $f^{(n+1)}(c) \leq \cosh 4$. Since we are assuming that we don't know what $e$ is, we have $$f^{(n+1)}(c) \leq \cosh 4 = \frac{e^4+e^{-4}}{2} < \frac{3^4+2^{-4}}{2} < 41.$$ Using the Lagrange form of the remainder, I get $\displaystyle R_n < \frac{41\cdot 4^{5+1}}{(5+1)!}\approx 233$, which is much greater than the error provided in the answer. It seems that the solution is using $2(n+1)$ instead of $n+1$ in the denominator of the Lagrange form (with the factorial), and likewise in the exponent for $x$ (for using these yields the provided answer). However, using this seems rather arbitrary as it does not appear in the general formula. How do I reconcile the general form of the remainder with the provided answer? I would not agree that $\sum_0^5 \frac{4^{2i}}{(2i)!}$ is the sixth partial sum. The "zero terms" should be counted. The remainder, in the Lagrange sense, is the same as the one we get if we take the terms up to and including the term in $x^{11}$. (Yes, it is $0$). Thus if you are going to use the Lagrange remainder formula as quoted in Wikipedia, for the "$k$" in that formula, you need to use $11$. I am going to elaborate on André’s answer after having thought about the problem some more. If we let $T_n(x)$ be the $n$th-degree Taylor polynomial of $\cosh$ at $0$ (since Maclaurin), and $R_n(x)$ be the remainder of the Taylor series, then we can write $$\cosh x = T_n(x) + R_n(x).$$ We can also write $$\cosh x = 1+ 0 + \frac{x^2}{2!} + 0 + \frac{x^4}{4!} + \cdots + \frac{x^8}{8!} + 0 + \frac{x^{10}}{10!} + 0+ \cdots = \sum_{i=0}^\infty \frac{x^{2i}}{(2i)!}.$$ Now if we want to use the partial sums of $\cosh$, we notice that we don't get $\sum_{i=0}^n \frac{x^{2i}}{(2i)!} = T_n(x)$, but rather $\sum_{i=0}^n \frac{x^{2i}}{(2i)!} = T_{2n}(x)$. In the original problem, when it asked to find the “sixth partial sum ($n=5$)”, the “$n$” under consideration was the one associated with the partial sum of $\cosh$, not the one in the Taylor polynomial. In other words, the problem wanted one to evaluate $\left .\sum_{i=0}^5 \frac{x^{2i}}{(2i)!}\right\|_{x=4}$. However, for this sum, the associated Taylor polynomial is actually $T_{2\cdot5}(4)=T_{10}(4) = T_{11}(4)$ (the last equality just happens to work since every odd-powered term is $0$, and will help with getting the error down even smaller). Therefore we have $\cosh 4 = T_{11}(4) + R_{11}(4)$. To get the error, we want \begin{align} R_{11}(4) &= \frac{\cosh^{11+1}(c)\cdot 4^{11+1}}{(11+1)!}\\ &< \frac{41\cdot 4^{12}}{12!} \\ &< 2,\end{align} which agrees with the provided answer.	2019-09-21T15:03:45	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/703260/using-lagrange-form-of-the-remainder-with-cosh", "openwebmath_score": 0.9882498979568481, "openwebmath_perplexity": 98.69650134751436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534327754852, "lm_q2_score": 0.8558511469672594, "lm_q1q2_score": 0.839978046135853 }
https://math.stackexchange.com/questions/1055638/solving-3x-equiv-4-pmod-7	# Solving $3x\equiv 4\pmod 7$ I'm trying to learn about linear congruences of the form ax = b(mod m). In my book, it's written that if $\gcd(a, m) = 1$ then there must exist an integer $a'$ which is an inverse of $a \pmod{m}$. I'm trying to solve this example: $$3x \equiv 4 \pmod 7$$ First I noticed $\gcd(3, 7) = 1$. Therefore, there must exist an integer which is the multiplicative inverse of $3 \pmod 7$. According to Bezout's Theorem, if $\gcd(a, m) = 1$ then there are integers $s$ and $t$ such that $sa+tm=1$ where $s$ is the multiplicative inverse of $a\pmod{m}$. Using that theorem: \begin{align}7 = 3\cdot2 +1\\7 - 3\cdot2 = 1 \\-2\cdot3 + 7 = 1\end{align} $s=-2$ in the above equation so $-2$ is the inverse of $3 \pmod{7}$. The book says that the next step to solve $3x \equiv 4 \pmod{7}$ is to multiply $-2$ on both sides. By doing that I get: \begin{align}-2\cdot3x \equiv -2\cdot4 \pmod 7\\-6x\equiv -8 \pmod 7\end{align} What should I do after that? I am working on this problem for hours. Thanks :) \begin{align} 3x\equiv4\pmod{7} & (\text{Original equation})\\3x\equiv -3\pmod{7} &(\text{Replaced 4 with -3(by subtracting 7)})\\x\equiv-1\pmod{7}& (\text{Divide each side by 3})\\ x\equiv6\pmod{7} &(\text{replaced -1 with 6 (by adding 7))} \end{align} P.S.- The reason you can add or subtract $7$ is one of the properties of $\pmod{7}$. You can add or subtract multiples of $7$ to the number in front of the $mod$ without effecting the equation. • thanks for replying. Your answer is a lot simpler. Will this subtracting/adding always work out the problem? – Tehmas Dec 7 '14 at 13:32 • yes you can always add multiples of $7$ (in this particular example) since $7\equiv0\pmod{7}$ so what you're doing is really just adding $0$ to the equation. More generally $a\equiv0\pmod{a}$ (just to make sure I don't confuse you and make you add $7$ to all mod equations lol – Fmonkey2001 Dec 7 '14 at 13:36 • Ahan. But you didn't even use the inverse(-2). Was it a useless step? @_@ – Tehmas Dec 7 '14 at 13:39 • It wasn't useless, it was just a different method to solving the same problem. You'd get the same answer if you did it your way. Just like @N. F. Taussig did you could always substitute the answer of $6$ back into the original equation to see that $x=6 \rightarrow 18\equiv4\pmod{7}$ – Fmonkey2001 Dec 7 '14 at 13:48 • Thanks Fmonkey2001. :) Your method is truly simpler and thanks for replying to other questions as well. – Tehmas Dec 7 '14 at 14:43 You have arrived at $$-6x=-8\pmod{7}.$$ Now: $$-6x=-8\pmod{7} \underbrace{\iff}_{\mathrm{add}\: 7x=0\pmod{7}} 7x-6x=-8\pmod{7}\\ \iff x=-8\pmod{7}\underbrace{=}_{\mathrm{add}\: 14=0\pmod{7}}(2\cdot 7-8)\pmod{7}=6\pmod{7}.$$ You obtained $$-2 \cdot 3x \equiv -8 \pmod{7}$$ Simplifying yields $$-6x \equiv -8 \pmod{7}$$ Observe that $-6 \equiv 1 \pmod{7}$ and that $-8 \equiv 6 \pmod{7}$. Thus, we obtain $$x \equiv 6 \pmod{7}$$ Check: If $x \equiv 6 \pmod{7}$, then $3x \equiv 3 \cdot 6 \equiv 18 \equiv 4 \pmod{7}$. • Taussig, thank you for replying. I don't understand this: "Observe that −6≡1(mod7) and that −8≡6(mod7)." – Tehmas Dec 7 '14 at 13:43 • $-6 \equiv 1 \pmod{7}$ since both $-6$ and $1$ have remainder $1$ when divided by $7$. $-6 = -1 \cdot 7 + 1$, so $-6 \equiv 1 \pmod{7}$. Note that when you add $7$ to $-6$ you obtain $1$. Similarly, $-8 = -2 \cdot 7 + 6$, so $-8 \equiv 6 \pmod{7}$. – N. F. Taussig Dec 7 '14 at 19:27 $$3x\equiv4\pmod{7}\\-6x\equiv -8\pmod{7}\\-6x\equiv-1-7\\-6x\equiv-1\pmod{7}\\(7-6)x\equiv-1\equiv6\pmod{7}$$ • kingW3, thank you for answering. Can you please elaborate step 3? – Tehmas Dec 7 '14 at 13:15 • @Tehmas: $8 \equiv 1\pmod 7$. – user 170039 Dec 7 '14 at 13:16 • @Tehmas Edited,you can also look the comment below you – kingW3 Dec 7 '14 at 13:25 • I apologize for repeatedly asking about step 3 but what did you do there now? −6x≡−1−7 – Tehmas Dec 7 '14 at 13:47 • This was two steps in one. The first thing he did was to add $7$ to $-1$ to get $6$. The second part was a little bit harder to see. He added $7x$ to the left side since any multiples of $7$ are $0\pmod{7}$ Then after he did that he factored out the $x$to get $(7-6)x$ then he simplified that to just get $x$ – Fmonkey2001 Dec 7 '14 at 14:20	2019-10-16T02:08:58	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1055638/solving-3x-equiv-4-pmod-7", "openwebmath_score": 0.7894480228424072, "openwebmath_perplexity": 506.39700833767273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534322330057, "lm_q2_score": 0.8558511414521923, "lm_q1q2_score": 0.8399780402587899 }
https://math.stackexchange.com/questions/1672931/how-can-we-show-that-all-the-roots-of-some-irreducible-polynomial-are-not-of-alg	How can we show that all the roots of some irreducible polynomial are not of algebraically equal status? In studying Galois theory, I found that all roots of some irreducible polynomial are not of algebraically equal status, because the Galois group of some irreducible polynomial may not be full symmetric group $S_n$. But I am searching a concrete example elucidating that all roots are not algebraically equal and should be distinct in an algebraic way. One of my attempt is like this. Let $p(t)$ be an irreducible separable polynomial over $F$ and $E$ its spliting field over $F$. Let $\alpha_1,\alpha_2,\cdots,\alpha_n$ are all roots of $p(t)$. Then $E=F(\alpha_1,\alpha_2,\cdots,\alpha_n)$ and we consider the tower of fields $F\le F(\alpha_1) \le F(\alpha_1,\alpha_2)\le \cdots \le F(\alpha_1,\alpha_2,\cdots,\alpha_n)$. I guess that the $\deg (irr(\alpha_i,F(\alpha_j)))$ may differ depending on the choice of $\alpha_i$ and $\alpha_j$. If this is true, I can suggest this to support my claim that all roots are not algebraically equal. But I am not able to find an apt example supporting my guess. Do you know some example verifying my guess? Or if you have any idea which helps convince that all roots are not algebraically equal, please share with me. Thanks for reading my question and any comment will be appreciated! • What does "of algebraically equal status" mean? – Eric Wofsey Feb 26 '16 at 8:30 • I think I better phrasing would be to just say that you are looking for a concrete "reason" that the Galois group cannot be all of $S_n$ in some example. – Eric Wofsey Feb 26 '16 at 8:52 • There is no clear definition and it depends on as you feel it. I don't know how to express it. But I have some feeling that they are not equal algebraically. For example, consider irreducible polynomial $p(t)=t^4-2\in \mathbb{Q}[t]$. Let $\alpha=2^{\frac{1}{4}}$, $\beta=2^{\frac{1}{4}}i$, $\gamma=-2^{\frac{1}{4}}$. Then $\alpha$, $\beta$, $\gamma$ are all roots of $p(t)$, but their reciprocal relation over $\mathbb{Q}$ is a bit different. For example, $\alpha+\gamma=0$ whereas $(\frac{\alpha}{\beta})^2+1=0$. – user29422 Feb 26 '16 at 8:53 • The distinctive status of each roots I think is the whole relationship of each roots of some irreducible polynomial with other elements in $\bar{\mathbb{Q}}$ over $\mathbb{Q}$. – user29422 Feb 26 '16 at 8:59 Sure, what you're asking for can happen. For a simple example, take $F=\mathbb{Q}$ and $p(t)=t^4-2$. The roots are $\alpha_1=\sqrt[4]{2}$, $\alpha_2=-\sqrt[4]{2}$, $\alpha_3=i\sqrt[4]{2}$, and $\alpha_4=-i\sqrt[4]{2}$. Note then that $\alpha_2\in F(\alpha_1)$ (so its minimal polynomial over $F(\alpha_1)$ has degree $1$), while $\alpha_3\not\in F(\alpha_1)$ (and its minimal polynomial over $F(\alpha_1)$ has degree $2$). However, I would object somewhat to your phrasing that this means the roots are "algebraically distinct". It is always true that the Galois group acts transitively on the roots (as long as $p$ is irreducible): that is, for any $i$ and $j$, there is an automorphism of $E$ over $F$ that maps $\alpha_i$ to $\alpha_j$. So you can't really distinguish $\alpha_i$ from $\alpha_j$ (at least, from the perspective of $F$). All you can really say is that you can distinguish certain subsets of the roots from other subsets of the roots of the same size. For instance, in the example above, the set $\{\alpha_1,\alpha_2\}$ is in a strong sense distinguishable from $\{\alpha_1,\alpha_3\}$ over $F$. • +1 Nice observation. From a group-theoretic point of view, I think the idea is that the Galois group is the dihedral group $G$ of order $8$, which acts transitively, but imprimitively on the roots. If ones sees $G$ as acting on the vertices of a square, a block system is given by the diagonals $\{ \alpha_{1}, \alpha_{2}\}$ and $\{ \alpha_{3}, \alpha_{4}\}$. – Andreas Caranti Feb 26 '16 at 9:09	2019-07-16T05:58:54	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1672931/how-can-we-show-that-all-the-roots-of-some-irreducible-polynomial-are-not-of-alg", "openwebmath_score": 0.9033700823783875, "openwebmath_perplexity": 131.5978809307098, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363758493942, "lm_q2_score": 0.8519528076067262, "lm_q1q2_score": 0.8399712635264919 }
http://www.real-statistics.com/tests-normality-and-symmetry/analysis-skewness-kurtosis/	# Analysis of Skewness and Kurtosis Since the skewness and kurtosis of the normal distribution are zero, values for these two parameters should be close to zero for data to follow a normal distribution. • A rough measure of the standard error of the skewness is $\sqrt{6/n}$ where n is the sample size. • A rough measure of the standard error of the kurtosis is $\sqrt{24/n}$ where n is the sample size. If the absolute value of the skewness for the data is more than twice the standard error this indicates that the data are not symmetric, and therefore not normal. Similarly if the absolute value of the kurtosis for the data is more than twice the standard error this is also an indication that the data are not normal. Example 1: Use the skewness and kurtosis statistics to gain more evidence as to whether the data in Example 1 of Graphical Tests for Normality and Symmetry is normally distributed. As we can see from Figure 4 of Graphical Tests for Normality and Symmetry (cells D13 and D14), the skewness for the data in Example 1 is .23 and the kurtosis is -1.53. The standard error for the skewness is .55 (cell D16) the standard error for the kurtosis is 1.10 (cell D17). Both statistics are within two standard errors, which suggest that the data is likely to be relatively normally distributed. Observation: Related to the above properties is the Jarque-Barre (JB) test for normality which tests the null hypothesis that data from a sample of size n with skewness skew and kurtosis kurt For Example 1 based on using the functions SKEW and KURT to calculate the sample skewness and kurtosis values. Since CHISQ.DIST.RT(2.13, 2) = .345 > .05, based on the JB test, we conclude there isn’t sufficient evidence to rule out the data coming from a normal population. The JB test can also be calculated using the SKEWP (or SKEW.P) and KURTP functions to obtain the population values of skewness and kurtosis. In this case, we obtain Since CHISQ.DIST.RT(1.93, 2) = .382 > .05, once again we conclude there isn’t sufficient evidence to rule out the data coming from a normal population. Real Statistics Functions: The Real Statistics Resource Pack contains the following functions. JARQUE(R1, pop) = the Jarque-Barre test statistic JB for the data in the range R1 JBTEST(R1, pop) = p-value of the Jarque-Barre test on the data in R1 If pop = TRUE (default), the population version of the test is used; otherwise the sample version of the test is used. Any empty cells or cells containing non-numeric data are ignored. For Example 1, we see that JARQUE(A4:A23) = 1.93 and JBTEST(A4:A23) = .382. Similarly, JARQUE(A4:A23, FALSE) = 2.13 and JBTEST(A4:A23, FALSE) = .345. Observation: See D’Agostino-Pearson Test for another more accurate test for normality which is based on the skewness and kurtosis of sample data. ### 21 Responses to Analysis of Skewness and Kurtosis 1. soharb says: I think there is some thing wrong with this formula for example for this series 26.83946269 26.95131935 8.371060164 10.40495872 18.38858378 20.12905135 24.2843167 1.76670796 20.19191695 41.06557085 16.09877032 13.34390071 0.426210193 28.31166689 11.89051087 109.3641761 25.50859431 61.26802436 32.5178008 66.58119511 41.27546773 14.67351611 2.048435245 28.01590722 44.93746991 the JARQUE(R1)=38.28239095 but if we use an array formula like this: =COUNT(A2:A26)(((((SUM((A2:A26-AVERAGE(A2:A26))^3)/COUNT(A2:A26))/((SUM((A2:A26-AVERAGE(A2:A26))^2)/COUNT(A2:A26))^1.5))^2)/6)+((((SUM((A2:A26-AVERAGE(A2:A26))^4)/COUNT(A2:A26))/((SUM((A2:A26-AVERAGE(A2:A26))^2)/COUNT(A2:A26))^2)-3))^2)/24) + CTRL + SHIFT + ENTER • Charles says: I am using the following Excel formula =COUNT(A2:A26)(SKEW(A2:A26)^2/6+KURT(A2:A26)^2/24) Charles • soharb says: Then there is some thing wrong (bug) in excel formula, since I calculated the SKEW, KURT and JB with “EViews 9.5” and my array formula turn up to be the correct answer! • Charles says: What value did you get for SKEW and KURT_ Charles • soharb says: EViews 9.5: SKEW= 1.769081 KURT= 3.620125 JB= 26.69155 Excel regular formula: =SKEW(A2:A26) = 1.884063081 =SKEW.P(A2:A26) =1.769080723 =KURT(A2:A26) = 4.748928357 Note: there is no KURT.P!!! Excel array formula: for SKEW =((SUM((A2:A26-AVERAGE(A2:A26))^3)/COUNT(A2:A26))/((SUM((A2:A26-AVERAGE(A2:A26))^2)/COUNT(A2:A26))^1.5)) + CTRL + SHIFT + ENTER =1.769080723 for KURT =((SUM((A2:A26-AVERAGE(A2:A26))^4)/COUNT(A2:A26))/((SUM((A2:A26-AVERAGE(A2:A26))^2)/COUNT(A2:A26))^2)-3) + CTRL + SHIFT + ENTER =3.620124598 • Charles says: Soharb, Thanks for sending me this information. It looks like if we use the population values of skewness and kurtosis then we get the result that you have seen from EViews. In particular, the Real Statistics Resource Pack has functions SKEWP and KURTP. If these functions are used then the formula =COUNT(A2:A26)*(SKEWP(A2:A26)^2/6+KURT(A2:A26)^2/24) yields the result 26.69155. Thanks for bringing this up. I will revise the JARQUE and JBTEST functions in the next release of the software. Charles 2. Jpso says: Hi and congrats for the great initiative. When you refer to Kurtosis, you mean the Excess kurtosis (i.e. kurt-3) or the outright kurtosis? For example when I perform the “D’Agostino-Pearson Test” as described in the relevant section (i.e. using outright kurtosis) I get results suggesting rejection of the null hypothesis, even if I use Kurt=3, Skew=0, which is the ND standards stats. Thank you. • Charles says: Jpso, I am using excess kurtosis (as does Excel). Charles 3. david oluyole ajekigbe says: thank you very much for this information. i have gained a lot from it. it will be appreciated if you can please attend to the question of zohreh of february 28, 2016 @ 9.31pm . i also will like to name of the person for reference. thank you . david • Charles says: David, As I wrote in response to that comment “We often use alpha = .05 as the significance level for statistical tests. The critical value for a two tailed test of normal distribution with alpha = .05 is NORMSINV(1-.05/2) = 1.96, which is approximately 2 standard deviations (i.e. standard errors) from the mean. This is source of the rule of thumb that you are referring to. The Jarque-Barre and D’Agostino-Pearson tests for normality are more rigorous versions of this rule of thumb.” Thus, it is difficult to attribute this rule of thumb to one person, since this goes back to the beginning of statistics, or at least the use of the value 1.96. You will find this value of 1.96 in any elementary book on statistics. Charles 4. Denny Yu says: Thank you very much! The Real Statistics Functions are really of great help. However, I came across a problem that JBTEST, as well as DPTEST, doesn’t allow ranges expressed in array form. For example, the expression: =jbtest(IF(INDIRECT(“G”&6):INDIRECT(“G”&10)0,INDIRECT(“AE”&6):INDIRECT(“AE”&10))) cannot be recognized by Excel and the result is #VALUE!. By comparing with another expression: =jbtest(INDIRECT(“AE”&6):INDIRECT(“AE”&10)) in Evaluating Fomula, I found that JBTEST can only read data with form of “Am:Bn”, not expressed in a set of data like “0.1, 0.2, …”. Is there any solution to it? I have to deal with ranges within which there are certain values that should not be included in the test. Thank you again! • Charles says: Denny, The current implementation of these functions supports only arrays which are ranges. I have just changed this so that they should support any arrays. I will include these changes in the next release of the software. I hope to issue this release in the next few days. Charles • Denny Yu says: I’m really looking forward to it. 5. Zohreh says: Salaam May you please cite the reference for “If the absolute value of the skewness for the data is more than twice the standard error this indicates that the data are not symmetric, and therefore not normal”. I need it. Thanks. • Charles says: We often use alpha = .05 as the significance level for statistical tests. The critical value for a two tailed test of normal distribution with alpha = .05 is NORMSINV(1-.05/2) = 1.96, which is approximately 2 standard deviations (i.e. standard errors) from the mean. This is source of the rule of thumb that you are referring to. The Jarque-Barre and D’Agostino-Pearson tests for normality are more rigorous versions of this rule of thumb. Charles • Zohreh says: Thanks for replying. I’ve heard that one way to check normality is to divide skewness by standard error, if the results falls between the range +-1.96, then normality will be satisfies. Using this formula my data was proved to be not normal. I used another formula to which you referred “If the absolute value of the skewness for the data is more than twice the standard error this indicates that the data are not symmetric, and therefore not normal”, then my data revealed to be normal. As I want to use the latter procedure in my study I need to cite the name of the person whose opinion I will use. By reference I meant based on whose opinion “If the absolute value of the skewness for the data is … Will you please provide the name of the person? Many thanks… 6. Rajesh says: Data distribution free how to apply 2 way anova • Charles says: Sorry, but I don’t understand your question. Charles 7. Colin says: Sir What are the rough measure of the standard error of the skewness and kurtosis ? I cannot see the pictures. • Charles says: Colin, A rough measure of the standard error of the skewness is \sqrt{6/n} where n is the sample size. A rough measure of the standard error of the kurtosis is \sqrt{24/n} where n is the sample size. Charles 8. My brother recommended I would possibly like this web site. He was once entirely right. This post actually made my day. You cann’t imagine just how much time I had spent for this info! Thank you!	2016-09-26T01:55:24	{ "domain": "real-statistics.com", "url": "http://www.real-statistics.com/tests-normality-and-symmetry/analysis-skewness-kurtosis/", "openwebmath_score": 0.7346479296684265, "openwebmath_perplexity": 1028.7116975059869, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363708905831, "lm_q2_score": 0.851952809486198, "lm_q1q2_score": 0.8399712611548583 }
https://www.physicsforums.com/threads/limit-proof.793144/	# Limit proof 1. Jan 19, 2015 ### ciubba 1. The problem statement, all variables and given/known data Prove $$lim_{x->4}\frac{x-4}{\sqrt{x}-2}=4$$ 2. Relevant equations Epsilon\delta definition 3. The attempt at a solution I can see that a direct evaluation at 4 leads to an indeterminate form, so: $$\frac{x-4}{\sqrt{x}-2}*\frac{\sqrt{x}+2}{\sqrt{x}+2} \; \mbox{simplifies to} \; \sqrt{x}+2 \; \mbox{when} \; x \ne 4$$ Via epsilon\delta definition, $$\|\sqrt{x}+2-4\|<\epsilon \; \mbox{whenever} \; \|x-4\|<\delta$$ Expanding the right side, $$-\delta<x-4<\delta$$ $$-\delta+4<x<\delta+4$$ $$\sqrt{-\delta+4}<\sqrt{x}<\sqrt{\delta+4} \; \mbox{when} \; \delta \leq 4$$ $$\sqrt{-\delta+4}-2<\sqrt{x}-2<\sqrt{\delta+4}-2$$ Thus, $$\delta=min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}$$ Given this definition of delta, it is elementary to work backwards towards $$\|\sqrt{x}+2\|<\delta$$ Have I made any errors in my proof? 2. Jan 19, 2015 ### perplexabot Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier. 3. Jan 19, 2015 ### pasmith What is $\delta$ when $\epsilon = 10$? What about when $\epsilon = 10^{-5}$? For every $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < \|x - 4\| < \delta$ then $\|\frac{x - 4}{\sqrt{x} - 2} - 4\| < \epsilon$. You need to relate your $\delta$ to the $\epsilon$. 4. Jan 19, 2015 ### Ray Vickson He could have used l'Hospital's rule (yes, it really is spelled Hospital) but his way is easier. Alternatively, he could have set $\sqrt{x} = y$ and then evaluated $$\lim_{y \to 2} \frac{y^2 - 4}{y-2},$$ because in this form l'Hospital's rule is much nicer to use. 5. Jan 19, 2015 ### ciubba Ah, I wish I was allowed to use that technique, but our prof. is adamant on teaching delta ep. definitions first. Oops, I mistyped. I meant to say "Given this definition of delta, it is elementary to work backwards towards $$\|\sqrt{x}+2\mathbf{-4}\|<\boldsymbol{\epsilon}$$ My main question is whether or not it is correct to say that, for any epsilon, delta is $$min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}$$ Ah, I didn't think to do a u-sub. I'll definitely try that next time!	2017-10-22T08:44:06	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/limit-proof.793144/", "openwebmath_score": 0.8043165802955627, "openwebmath_perplexity": 970.9712102879457, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363737832231, "lm_q2_score": 0.8519528000888386, "lm_q1q2_score": 0.8399712543540526 }
http://farmaciacoverciano.it/dhti/when-two-dice-are-rolled-find-the-probability-of-getting-a-sum-of-5-or-6.html	# When Two Dice Are Rolled Find The Probability Of Getting A Sum Of 5 Or 6 With two dice, there are 6 x 6 = 36 possible outcomes. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. Number of outcomes of the experiment that are favorable to the event that a sum of two events is 6. The logic is there are six sides to each die, so for each number on one You did the math for the probability of rolling a dice twice and getting a multiple of 3 on both rolls. Think about a dice. a sum less than 4 or greater than 9 d. 1 in 6 x 1 in 6 = 1 in 36. Probability that sum is neither 7 or 11 = 1 - Probability that the sum is 7 and 11. Find the variance and standard deviation of X. [3 Marks) 0. Dependent Event - An event whose probability of occurring is influenced by (i. In the example you gave, I find it much easier to start by calculating the probability of NOT rolling a 5 across multilple throws, because these probabilities can be just multiplied together. Two fair dice are rolled and the sum of the points is noted. What is the. Question: Question 1 If Two Dice Are Rolled One Time, Find The Probability Of Getting A Sum Greater Than 6 And Less Than 12. Texas A&M University. Noticing these patterns can make counting much easier. 4 And P (B) = 0. When two dice are rolled, we get 36 possible outcome like (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) …………. "If you roll a dice three times, what is the probability of rolling a 6 at least once?" The correct answer is 91/216. Best Answer. For example if n. There are 36 different combinations that can be rolled using 2 die. The probability of the union of two mutually exclusive events — P(one OR five) — is the sum of their probabilites, ie, P(one OR five) = 1/6 + 1/6 = 2/6 = 1/3. Rolling Dice. Expected Value of a. Then P(A) = 4 52. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. Two fair dice are rolled and the sum of the points is noted. For example: 1 roll: 5/6 (83. of ways are - 1 , 1 1 , 2 2 , 1 1 , 4 4 , 1 1 , 6. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. and only one way to roll a 12 (6-6, or boxcars). EXPERIMENTAL PROBABILITIES Simulate rolling two dice 120 times. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. Two dice are tossed. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. Sample space S = {H,T} and n(s) = 2. That intuition is wrong. When you roll a pair of dice there are 36 possible outcomes. To find the probability we use the mutually exclusive probability formula P(A) + P(B). A sum less than or equal to 4. Isn’t that kind of cool?. Two dice are tossed. So the probability of getting a sum of 4 is 3/36 or 1/12. 10 5 13 ! Find the probability distribution. So 1/36 is part of the. Sum of Two Dice. hi Dakotah :) A number cube is rolled 20 times and lands on 1 two times and on 5 four times. Rolling two dice. Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. The fundamental counting principle tells us there are 66=36 ways to roll two dice, all of them equally likely if the dice are fair. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and 1. Let B be the event - The sum of the top faces of the 3 dice >= 5. What is the. Rolling Two Dice If two dice are rolled one time, find the probability of getting these results. [3 Marks) 0. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. Total possible outcomes = 36. (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) since each 6 numbers. (d) an even number appears on the black dice or the sum of the numbers on the two dice is 7. (it's easier to count the 6 non-red ones and subtract from 36 to get 30). When two dice are thrown, find the probability of getting a number always greater than 4 on the second die. What is the probability that the sum of two rolled dice will equal a prime number? We find this number by multiplying 6 x 6. the probability that the sum is 6 given that at least one of the numbers is less than 3. To find the probability determine the number of successful outcomes divided by the number of possible outcomes overall. You might be asked the probability of rolling a variety of results for a 6 Sided Dice: five and a seven, a double twelve, or a double. But this time, the dice aren't fair: For each die, a 1 is twice as likely to be rolled as a 2, a 2 is twice as likely to be rolled as a 3, , and a 5 is twice as likely to be rolled as a 6 (in other words, each number is twice as likely as the number that follows it). Hi I wrote code for Java program to simulate the rolling of 2 dice. Sum up all of the Power on the Groups in the uncontrolled area. The game is designed as such that you throw a pair of dice and get money back or loose money. Roll 4 6-sided dice, keep the highest 3 and sum them: sum largest 3 4d6. As the chart shows the closer the total is to 7 the greater is the probability of it being thrown. two-dice-are-rolled-simultaneously-find-the-probability-of-getting-a-total-of-9SingleChoice5b5cc7d1e4d2b419777512684. of favorable. When you roll a pair of dice there are 36 possible outcomes. a sum that is divisible by 4 e. Brian Veitch 101,870 views. $\endgroup$ – Squirtle Aug 5 '13 at 19:31. What is the probability that the sum of two of the faces rolled equals the value of the other rolled face?. You might roll a single 6, which means one of the dice is showing 6 and the others are all showing 1-5, and there are 6 different ways to choose which die is showing 6. What is the. We want sum to be greater than 16, So, sum could be either 17 or 18. two-dice-are-rolled-simultaneously-find-the-probability-of-getting-a-total-of-9SingleChoice5b5cc7d1e4d2b419777512684. Find the Probability of Getting : (Ii)Sum Divisible by 5 Concept: Simple Problems on Single Events. Number of outcomes of the experiment that are favorable to the event that a sum of two events is 6. Here, the sample space is given when two dice are rolled. Since there are a total of 6 x 6 = 36 outcomes, then the probability that the sum will be greater than 10 is 3/36 = 1/12. It compiles alright but I am not getting the output. The probability of not rolling a sum of six with two fair dice is 1 minus the probability of rolling a sum of six. The sum is 2 /9. Thus, the probability of two odd numbers (no even numbers) is (1/2)(1/2) = 1/4. What is the probability of getting a flush in a. Get a free answer to a quick problem. from Rosemount. Solution Two Dice (Each Bearing Numbers 1 to 6) Are Rolled Together. Roll 3d6 six times, then pick the best result. What is the probability that the sum of the two tosses is 4?. Okay, so basically it says someone rolls two dice and its asking about the probability of rolling certain sums of the two dice. As such, the probability of both dice (dice 1 and Dice 2) rolling a 1 is 1/36, calculated as 1/6 x 1/6. Of these, 6 have a sum less than five, 1+1, 1+2, 1+3, 2+1, 2+2, and 3+1. There are 4 combinations that have the sum of 9. 33 Question 3 Let A And B Be Two Independent Event, Such That P (A) = 0. Chapter 13 Probability Solutions covers multiple exercises. My own intuition tells me the answer is 2/3 because the other die simply needs to show 3, 4, 5, or 6 for the sum to be at least 5. Sum up all of the Power on the Groups in the uncontrolled area. The probability of one dice not being a particular number is 5/6. The events "getting sum less than 8" and. If we discard the 6 rolls that gave the same numbers, then the odds of getting a six is. What is the distribution of the sum? 30. (d) A sum that is divisible by 4. The other two singletons can be among the other five. Current Stock: Quantity: Decrease Quantity: 1 Increase Quantity: Add to Wish List Description 1. Which of the pairs of events below is dependent? ____&lowbar. ECEN 303 - Fall 2011. However, when it comes to practical application, there are two major competing categories of probability. How likely is it to choose a random number between 10 and 100 that is a multiple of 9? 6. What is the expected value when we roll a fair die? There are six possible outcomes: 1, 2, 3, 4, 5, 6. TE Thaddeus Moss was signed after going undrafted. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. Find the probability of the lost card being a diamond. Independent probabilities are calculated using: Probability of both = Probability of outcome one × Probability of outcome two So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0. Therefore, the probability of rolling a 6 is 36 5. So we just need to work out the probability of rolling a 7, then take half of what's left. So we have 5/6 probability of a die's being <6, and then 5/6 for each of the other pairs' not. Dice roll probability: 6 Sided Dice Example. To find the probability we use the mutually exclusive probability formula P(A) + P(B). Two dice are rolled. Here, the sample space is given when two dice are rolled. A card from a pack of 5 2 cards is lost. The probability of not rolling a sum of six with two fair dice is 1 minus the probability of rolling a sum of six. Example 7: A die is rolled, find the probability of getting a 3. It’s very common to find questions about dice rolling in probability and statistics. (i) To get the sum of numbers 4 or 5 favourable outcomes are: (1, 3) ,(3, 1) , (2,2). Two fair dice are rolled and the sum of the points is noted. What is the probability of getting a straight by a single throw of 5. You ask for P(A\|B). A single die is rolled twice. The other two singletons can be among the other five. For four six-sided dice, the most common roll is 14, with probability 73/648; and the least common rolls are 4 and 24, both with probability 1/1296. They were both great football fans and decided to introduce this game to the workers of the factory. Two fair dice are rolled and the sum of the points is noted. There are 62=36 possible outcomes when a pair of dice are rolled. a sum less than 13. That intuition is wrong. Example 8: A die is rolled, find the probability of getting an even number. When two dice are thrown together total possible outcomes = 6 X 6 = 36 Favourable outcomes when both dice have number more than 3 are (4, 4), (4, 5),(4, 6), (5, 4), (5, 5). Sums of two independent Binomial random variables. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. This is a good introduction to probability, since you can see which combinations are more. Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die. Throwing Dice More Than Once. 444%) probability of NOT rolling a 5. Find the probability of the lost card being a diamond. When two 6 sided dice is tossed, we get a pair of outcomes. When two six-sided dice are tossed, there are 36 possible outcomes as shown. Two dice are thrown simultaneously. Find the expected number of games that are played when. There are 36 permutations of two dice. The probability of choosing a green marble from the jar. (i) Prime numbers = 2, 3 and 5 Favourable number of events = 3 Probability that it will be a prime. Because there are 36 possibilities in all, and the sum of their probabilities must equal. The probability of getting an outcome of "head-head" is 1 out of 4 outcomes, or, in numerical terms, 1/4, 0. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. (a) Find the conditional probability of obtaining a sum greater than 9, given that Given that the two numbers appearing on throwing two dice are different. The term "snake eyes" is the outcome of rolling the dice and getting only one pip on each die. Calulate the probability of getting a double or treble on a dartboard. When you roll a pair of dice there are 36 possible outcomes. dice = 3 , we’re rolling three four-sided dice, i. The probability of getting a sum of 5 when rolling two dice is 4/36 = 1/9 because there are 4 ways to get a five and there are 36 ways to roll the dice (Fundamental Counting Principle - 6 ways to roll the. Using an organized list, table, tree diagram, or method of your choosing, develop a list of all 16 possible outcomes (for example, Die #1 = 1 and Die #2 = 2 for a difference of 1; Die #1 = 1 and Die #2 = 4 for a difference of 3; and so on). If the two dice are fair and independent , each possibility (a,b) is equally likely. Compute the total probability of getting a 4 or a 9: 3/36 + 4/36 = (3 + 4)/36. My own intuition tells me the answer is 2/3 because the other die simply needs to show 3, 4, 5, or 6 for the sum to be at least 5. The probabilities of rolling several numbers using two dice. > Consider this matrix for two dice roll game $\textrm{Total outcomes with the sum}$ $3 = 2$ \textrm{Total outcomes with the sum}[/math. For three six-sided dice, the most common rolls are 10 and 11, both with probability 1/8; and the least common rolls are 3 and 18, both with probability 1/216. We start with writing a table to Discrete = This means that if I pick any two consecutive outcomes. Find each experimental probability. What is the probability of getting a number other than 6?. Notice how for two or more dice the number of combinations equals the sum of combinations one column to the left, starting from one row higher to seven rows higher. When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. California would be the lead Group, as it has two opposed alignments. The probability of throwing any given total is the number of ways to throw that total divided by the total number of combinations (36). I hope this helps, Harley Go to Math Central. Here we consider two events: A - (finding a sum of 8) & B (getting at least one 4) A : Probability of A is 5/36 because Now, the probability of either of the incidents happening is P(AUB)=P(A)+P(B)-P(AnnB) i. Major changes in Python environment : . There are 63 = 18 ways to get two numbers of the same parity (the first can be any of the 6 numbers, and the second has to be 3 of the possible 6 which have the same parity), giving a total of 18 ways to get an even sum out of a possible of 66 = 36 outcomes (we don't have to consider if the first number is even or odd since there are an equal. Find the Probability that the Sum of the Numbers on the Upper-most Faces of Two Dice Is: Less than 6 Concept: Probability - A Theoretical Approach. P(2 twos and 2 ones and the other two different ) + P(2 sixes and 2 ones and 2 of some other number). Find the probability that the student sold 11-15 shirts or less than 6 T's No of T-shirts No of Club Members 0 1 1-5 15 6-10 13 11-15 3 16-20 6 20+ 1. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; (b) X is the value on the first die and Y is the larger of the two values; (c) X is the smallest and Y is the largest value obtained on the dice. A standard deck of cards has 12 face. Experimental Probability: Experiment with probability using a fixed size section spinner, a variable section spinner, two regular 6-sided dice or customized dice. Try the following: 1. Find the probability of: getting a number greater than 3 on each die. There are a total of 36 different rolls with two dice, with any sum from 2 to 12 possible. Probability = 1 ÷ 36 = 0. Three fair, n-sided dice are rolled. In 1887, two other cotton industrialists from Lancashire, Clement and Harry Charnock, moved to work at a cotton factory in Orekhovo-Zuevo, near Moscow. Get an answer for 'When two dice are thrown what is the probability that the sum is 8. (f) A sum less than 13. P(sum divisible by 5)= 7/36. Suppose we have 3 unbiased coins and we have to find the probability of getting at least 2 heads, so there are 23 = 8 ways to toss these coins, i. dice = 3 , we’re rolling three four-sided dice, i. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. Question: Question 1 If Two Dice Are Rolled One Time, Find The Probability Of Getting A Sum Greater Than 6 And Less Than 12. Rolling more dice. of the event of rolling a. Highest Possible Sum Using 4: 4 + 6 = 10. My own intuition tells me the answer is 2/3 because the other die simply needs to show 3, 4, 5, or 6 for the sum to be at least 5. Therefore the number of possible outcomes will be 66 = 36. What is the conditional probability that at least one lands on 6 given that the dice land Probability of a woman being a smoker given she has ectopic pregnancy • E: ectopic pregnancy Did you find mistakes in interface or texts? Or do you know how to improveStudyLib UI?. Question 4: Two dice are rolled, find the probability that the sum is a) equal to 1 b) equal to 4 c) less than 13 Solution to Question 4: a) The sample space S of two dice is shown below. Algebra -> Probability-and-statistics -> SOLUTION: Two dice are rolled. (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) since each 6 numbers. Find the probability that a 5 will occur first. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. When two are rolled, simultaneously, find the probability that the sum of the numbers on them is at least 9. Let X denote the sum of the number of dots on the top faces. When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i. , HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. The table marks with X all the possible combinations of throws. Example 7: A die is rolled, find the probability of getting a 3. (ii) 7 , 8 or 9. The events "getting sum less than 8" and. If you roll a die will obtain 1, 2, 3, 4, 5 or 6? Probability measures and quantifies "how likely" an Let us define event E as the set of possible outcomes where the sum of the numbers on the faces of the two dice is equal to four. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. (i) Prime numbers = 2, 3 and 5 Favourable number of events = 3 Probability that it will be a prime. The probability of getting less than 8 is the sum of the probabilities of 2-7:. (1, 1, 1) = 1+1+1=3. Suppose that each game played is, independently, won by team A with probability p. Let’s say you need the probability of rolling a 5 and a 4. Independent probabilities are calculated using: Probability of both = Probability of outcome one × Probability of outcome two So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0. Subject: Re: Probability: Two six-sided dice, rolling two numbers in order. two dice are rolled find the probability of getting a 5 on either dice or the sum of both dice is 5. Major changes in Python environment : . 5 ways to get a sum of 6. What is the probability that the sum of the two dice will not be a 6? 31/36. SOLUTION: Two dice are rolled. Thus, the probability of two odd numbers (no even numbers) is (1/2)(1/2) = 1/4. Total number of outcomes = 66 = 36, Each die can take a number from 1 to 6 i. A standard deck of cards has 12 face. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. Get an answer for 'When two dice are thrown what is the probability that the sum is 8. The probability of rolling an even number on 1 die is 3/6. The probability of two dice not being a that number is 5/6 x 5/6 = 25/36. Let B be the event - The sum of the top faces of the 3 dice >= 5. from Rosemount. Find maximum subset sum formed by partitioning any subset of array into 2 partitions with equal sum. The ways to get a 4 are: 1+3, 2+2, 3+1 = 3 ways. There are 6 6 possible outcomes. This is the aptitude questions and answers section on "Probability" with explanation for various interview, competitive examination and entrance test. Get a free answer to a quick problem. [3 Marks) 0. Let X1 and X2 be the outcomes, and let S2 = X1 + X2 be the sum of these outcomes. In a tabular format it should show the frequency of rolling 2,3,4,5,6,7,8,9,10,11,12 and also the percentage. Throwing a 10 yields 0. With dice there is: 1 way to get a sum of 2. Roll two dice. What is the probability that the sum of the two dice will not be a 6? 31/36. When rolling one die, the probability of getting a 4 is 1 in 6, or 0. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. Probability - Quantitative Aptitude objective type questions with answers & explanation (MCQs) for job Now a shirt is picked from second box. For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. Of these, 6 have a sum less than five, 1+1, 1+2, 1+3, 2+1, 2+2, and 3+1. Find the Mean of the Roll z column; OK; Repeat process except find the Standard Deviation of the Roll z column; By hand (with a calculator) square the standard deviation to get the variance. If a pair of dice are rolled 5 times, what is the probability of getting a sum of 5 every time?. Isn’t that kind of cool?. Each dice has six combinations which are independent. (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) since each 6 numbers. the black die resulted in a 5. (i) Prime numbers = 2, 3 and 5 Favourable number of events = 3 Probability that it will be a prime. We have to find what is the probability that the sum of numbers rolled is either 5 or 12 We know that, probability of an event = Now, total outcomes for two dices = 6 for 1st dice x 6 for 2nd dice = 6 x 6 = 36. Draw a card. a sum that is divisible by 4 e. Let's write events!! Let A be the event - The sum of the top faces of 3 dice > 8. That intuition is wrong. When two dice are thrown, find the probability of getting a number always greater than 4 on the second die. With dice there is: 1 way to get a sum of 2. Find the probability that a 5 will occur first. 1 die, 2 dice. When two dice are rolled, total no. The probability of getting a sum of 5 when rolling two dice is 4/36 = 1/9 because there are 4 ways to get a five and there are 36 ways to roll the dice (Fundamental Counting Principle - 6 ways to roll the. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; In this first case, we have p X,Y ( x,y ) 2 3 4 5 6 d Suppose you dont know the probability p of getting a ticket but you got 5. Rolling Dice. Find the Mean of the Roll z column; OK; Repeat process except find the Standard Deviation of the Roll z column; By hand (with a calculator) square the standard deviation to get the variance. For example, the event "the sum of the faces showing on the two dice equals six" consists of the five The probability of an event is defined to be the ratio of the number of cases favourable to the. Four fair, 6-sided dice are rolled. In the experiment of rolling two dice think of one as red and the other as green and list the possible Here, for example, the (3,5) in third row and fifth column means a 3 was rolled on the red die and a 5 Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6. This probability of both dice rolling a 2 or 3 or 4 or 5 or 6 is also 1/36. To find the probability. A Recursion Formula for the Probability Distribution of the Sum of k Dice In this section we derive a recursion formula for the probability distribution ofthe sum of j dice, using the probability distribution ofthe sum of 7 -1 dice. 333%) probability of NOT rolling a 5 2 rolls: (5/6) x (5/6) (69. With every new roll the probability the next four rolls will be all double sixes is (1/36)4 = 1 in 1679616. Event Die 1 Die 2 Sum 1 2 3 4 5 6 7 8 9. The other die roll could be a 1, 2, 3, 4, 5, or 6. 2) a sum of 6 or 7 or 8 b) doubles or a sum of 4 or 6 c) a sum greater than 9 or less than 4, Please help me. Rolling Dice. The total of points is 21 and the actual corresponding dice roll (we have to sum 1 pre-assigned point to each die) would be {2,7,1,5,1,1,4,2,7,1}, with sum 31 but with two outlaw dice. (f) A sum less than 13. A pair of dice, two different colors (for example, red and blue) A piece of paper; Some M&M’s or another little treat; What You Do: Tell your child that he's going to learn all about probability using nothing but 2 dice. Two 6-sided dice are rolled. a sum less than 13. Dependent Event - An event whose probability of occurring is influenced by (i. two dice are rolled find the probability of getting a 5 on either dice or the sum of both dice is 5. Is this solution Helpfull? Yes (28) \| No (6). The probability distribution of a discrete random variable X is a listing of each possible value x taken by X. What is the. A single die is rolled twice. I recently got asked how to find the probability of rolling a sum of 12 with two dice. Therefore, in this example, we could write: p1 = p2 = p3 = p4 = p5 = p6 = where p1 ≡ probability of rolling a 1, p2 ≡ probability of rolling a 2, etc. 16&comma. Find the probability of correctly answering the first 2 questions on a multiple choice test if random guesses are made and each question has 5 possible answers. What is the. Throwing a 6,5,4,3,2 or 1 deducts 0. sum that takes two arguments: n. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; (b) X is the value on the first die and Y is the larger of the two values; (c) X is the smallest and Y is the largest value obtained on the dice. > Consider this matrix for two dice roll game [math]\textrm{Total outcomes with the sum} $3 = 2$ $\textrm{Total outcomes with the sum}[/math. The probability of not rolling a sum of six with two fair dice is 1 minus the probability of rolling a sum of six. 4d10 are enough to sample uniformly from between 1 and 10,000), but it becomes increasingly tedious to generate larger numbers. the probability of the sum being: 2 is 1/36 3 is 2/36 4 is 3/36 5 is 4/36 6 is 5/36 7 is 6/36 8 is 5/36 9 is 4/36 10 is 3/36 11 is 2/36 12 is 1/36 It then asks: P(the sum of the two dice equals 2) P(the sum of the two. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. The logic is there are six sides to each die, so for each number on one You did the math for the probability of rolling a dice twice and getting a multiple of 3 on both rolls. We want sum to be greater than 16, So, sum could be either 17 or 18. Online binomial probability calculator using the Binomial Probability Function and the Binomial Entering 0. If I roll two dice, does my probability of rolling a six on one of them increase, or does it stay at 1/6? Mike R. the probability that the sum is 6 given that at least one of the numbers is less than 3. Let B be the event - The sum of the top faces of the 3 dice >= 5. 7) F Two dice are rolled. There are 36 different combinations that can be rolled using 2 die. > Consider this matrix for two dice roll game [math]\textrm{Total outcomes with the sum}$ $3 = 2$ $\textrm{Total outcomes with the sum}[/math. A pair of dice is rolled until either the two numbers on the dice agree or the diﬀerence of the two numbers on the dice is 1 (such as a 4 and a 5, or a 2 and a 1). The probability of either of the incidents happening is 5/12. two dice are rolled find the probability of getting a 5 on either dice or the sum of both dice is 5. 8: 5/369: 4/3610: 3/3611: 2/3612: 1/36. Question 4: Two dice are rolled, find the probability that the sum is a) equal to 1 b) equal to 4 c) less than 13 Solution to Question 4: a) The sample space S of two dice is shown below. A standard deck of cards has 12 face. Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die. Hamilton sat just 1. Dice and Dice Games. Number of outcomes of the experiment that are favorable to the event that a sum of two events is 6. Major changes in Python environment : . My own intuition tells me the answer is 2/3 because the other die simply needs to show 3, 4, 5, or 6 for the sum to be at least 5. Two fair dice are rolled and the sum of the points is noted. The probabilities in the probability distribution of a random variable X must satisfy the following two A pair of fair dice is rolled. Since there are 6 \times 6 = 36 total dice rolls and 1/3 of those are a multiple of three, the number which are divisible by three is (1/3)(36) = \boxed{12}. Find probability nobody gets own hat. equals to prime number when we add two rolled dice from 1 to 6 So no. Find the probability of getting a sum of 6. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. Memorizing the making of the above picture makes the. If two dice are rolled one time, find the probability of getting these results. A sum less than or equal to 4. You ask for P(A\|B). There are 6 6 possible outcomes. Is this unusual? On average, it will occur about 1 in 12 times. Let B be the event - The sum of the top faces of the 3 dice >= 5. Rolling more dice. dice tells how many dice we roll. To support your homeschooling, we're including unlimited answers with your free account for the time being. Assuming that the dice are unbiased or not " loaded". The probability of the two dice totaling an even number is 1/2. It is assume each die is fair and 6-sided. Now, favourable outcomes = sum. In this skilltest, we tested our. 4d6, drop lowest, reroll if max < 14 or reroll if the sum of the modifiers is < 1. When rolling one die, the probability of getting a 4 is 1 in 6, or 0. 7) F Two dice are rolled. 6 outcomes on one die X 6 outcomes on other die = 36 outcomes. So we just need to work out the probability of rolling a 7, then take half of what's left. Let B be the event - The sum of the top faces of the 3 dice >= 5. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice. (i) Prime numbers = 2, 3 and 5 Favourable number of events = 3 Probability that it will be a prime. Find the probability that a 5 will occur first. Explanation of the fundamental concepts of probability distributions. For example: 1 roll: 5/6 (83. Find the probability of getting a sum of 6 when rolling a pair of dice. (i) To get the sum of numbers 4 or 5 favourable outcomes are: (1, 3) ,(3, 1) , (2,2). A Collection of Dice Problems Matthew M. So the probability of a sum of at least 5 is 30 out of 36, which gives us the fraction which reduces to. When two balanced dice are rolled, 36 equally likely outcomes. Then P(A) = 4/36 and P(B) = 6/36. The probability that it is a double with a sum of 11 is zero (0) When Two Balanced Dice Are Rolled, There Are 36 Possible. Is that unusual enough? We have to be careful when we characterize an event as unusual. 1/18 5/36 1/6 1/9. Speech recognition, image recognition, finding patterns in a dataset, object classification in photographs, character text generation, self-driving cars and many more are just a few examples. What is the probability of getting a 5 after rolling a single 6-sided die? 1/6 or 16. What is the probability of getting a number other than 6?. You ask for P(A\|B). Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. What is the probability of getting a flush in a. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment. ECEN 303 - Fall 2011. Suppose we have 3 unbiased coins and we have to find the probability of getting at least 2 heads, so there are 23 = 8 ways to toss these coins, i. hi Dakotah :) A number cube is rolled 20 times and lands on 1 two times and on 5 four times. Independent probabilities are calculated using: Probability of both = Probability of outcome one × Probability of outcome two So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0. Find the probability of getting a sum of 7. Rolling Dice. Obviously with two dice you can't get less than 2 or more than 12, so the only squares are 4 and 9. No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. Example 8: A die is rolled, find the probability of getting an even number. A Recursion Formula for the Probability Distribution of the Sum of k Dice In this section we derive a recursion formula for the probability distribution ofthe sum of j dice, using the probability distribution ofthe sum of 7 -1 dice. What Is The Probability That The Sum Of 8 Does Not Occur?. Dice roll probability: 6 Sided Dice Example. We start with writing a table to Discrete = This means that if I pick any two consecutive outcomes. The sum is 2 /9. If we assume the die is perfectly balanced, the probability of any particular outcome (say, rolling a ‘3’) is 1 out of 6. asked by Jacqueline on August 28, 2015; math. Solution Two Different Dice Are Thrown at the Same Time. Calculate the is the conditional probability that the Finding P (E): The probability of getting 4 atleast once is. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. Two rolls are independent and identically distributed, with probability of rolling a particular number being 1/6. Is this unusual? On average, it will occur about 1 in 12 times. Let's write events!! Let A be the event - The sum of the top faces of 3 dice > 8. The probability of appearance of any of two incompatible events is equal to the sum of the The conditional probability of an event B with the condition that an event A has already happened is. a sum that is divisible by 4 e. Two tetrahedral dice (four-sided dice) are thrown. A dice is thrown, cases 1,2,3,4,5,6 form an exhaustive set of events. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). Experimental Probability: Experiment with probability using a fixed size section spinner, a variable section spinner, two regular 6-sided dice or customized dice. It's somehow different than previously because only a part of the whole set has to match the conditions. It is assume each die is fair and 6-sided. Roll two dice. Memorizing the making of the above picture makes the. Of these, five sum to six, 1+5, 2+4, 3+3, 4+2, and 5+1. Two different coins are tossed randomly. Is that unusual enough? We have to be careful when we characterize an event as unusual. Of these, five sum to six, 1+5, 2+4, 3+3, 4+2, and 5+1. Find the probability of getting two numbers whose sum is greater than 10. Two rolls are independent and identically distributed, with probability of rolling a particular number being 1/6. 5 or 1/2 in the calculator and 100 for the number of trials and 50 for "Number of events" we get Example 2: Dice rolling. > Consider this matrix for two dice roll game [math]\textrm{Total outcomes with the sum}$ $3 = 2$ [math]\textrm{Total outcomes with the sum}[/math. (e) A sum of 14. [3 Marks) 0. Compute the total probability of getting a 4 or a 9: 3/36 + 4/36 = (3 + 4)/36. The probability of rolling a six on a single roll of a die is 1/6 because there is only 1 way to roll a six out of 6 ways it could be rolled. The pair can be any one of 6 numbers. The probability of either of the incidents happening is 5/12. Texas A&M University. Calulate the probability of getting a double or treble on a dartboard. Find the expected number of times one needs to roll a dice before getting 4 sixes. Therefore, x can be any number from. It compiles alright but I am not getting the output. That takes care of the winning or losing probabilities for the naturals (7,11) and the craps (2,3,12) outcomes. Is this unusual? On average, it will occur about 1 in 12 times. a sum of 14 f. Roll each attribute in order - do not assign numbers to stats as you see fit. Find the probability of the lost card being a diamond. "If you roll a dice three times, what is the probability of rolling a 6 at least once?" The correct answer is 91/216. A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. The proability of getting neither is equal to the probability of getting anything other than 7 or 8. Thus, the probability of two odd numbers (no even numbers) is (1/2)(1/2) = 1/4. Let A be event of rolling a 5 and B of rolling a 7. Total possible outcomes = 36. If one of the dice shows 1 to 4, the sum will not be greater than 10. 2 ways to get a sum of 3. You may get a side with 1, 2, 3, 4, 5, or 6 dots. 16667, to turn up when rolled, if the die (D) is unbiased. If I roll two dice, does my probability of rolling a six on one of them increase, or does it stay at 1/6? Mike R. the probability that the sum is 6 given that at least one of the numbers is less than 3. The ways to get a 9 are: 3+6, 4+5, 5+4, 6+3 = 4 ways. Let E denote the event that the number landing uppermost on the first die is a 3, and let F denote the event that the sum of Which pair has equally likely outcomes? Check the two choices below which have equal probabilities of success. 33 Question 3 Let A And B Be Two Independent Event, Such That P (A) = 0. The number of possible outcomes in E is 1 and the number of possible outcomes in S is 6. The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). The sum of the two dice you rolled is. Of these, 6 have a sum less than five, 1+1, 1+2, 1+3, 2+1, 2+2, and 3+1. The probability of rolling any number twice in a row is 1/6, because there are six ways to roll a specific number twice in a row (6 x 1/36). Independent probabilities are calculated using: Probability of both = Probability of outcome one × Probability of outcome two So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0. 7) F Two dice are rolled. There are 36 permutations of rolling two dice. The odds of rolling two dice and the sum being greater than 9 are 6 to 30. Find the probability of getting two numbers whose sum is greater than 10. When two six-sided dice are tossed, there are 36 possible outcomes as shown. The game is designed as such that you throw a pair of dice and get money back or loose money. In this way, the difference value for any roll of the two dice will always be positive or 0. Two different dice are thrown together. So the probability of not getting 7 or 11 is 7/9. Sum of dices when three dices are rolled together If 1 appears on the first dice, 1 on the second dice and 1 on the third dice. When two dice are rolled, we get 36 possible outcome like (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) …………. Throwing Dice More Than Once. If a pair of dice are rolled 5 times, what is the probability of getting a sum of 5 every time?. Example: Roll two 6-sided dice 0. That takes care of the winning or losing probabilities for the naturals (7,11) and the craps (2,3,12) outcomes. 78% If you need to get the probability of acquiring two different numbers when you roll a pair of dice, the calculation becomes a bit different. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment. Assuming that the dice are unbiased or not " loaded". From the remaining cards of the pack, two cards are drawn and are found to be diamonds. To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18. Here are a few examples that show off Troll's dice roll language: Roll 3 6-sided dice and sum them: sum 3d6. The probability that the first die rolls 3 and the second die rolls 1 is also 1/36. Roll each attribute in order - do not assign numbers to stats as you see fit. when two dice are rolled, find the probability of getting: a. The total of points is 21 and the actual corresponding dice roll (we have to sum 1 pre-assigned point to each die) would be {2,7,1,5,1,1,4,2,7,1}, with sum 31 but with two outlaw dice. Suppose that the first die we roll comes up as a 1. Find the probability of getting a sum of 6 when rolling a pair of dice. What is the probability of rolling a 6 with a pair of standard dice? There are five ways to roll a 6: (1,5)(2,4)(3,3)(4,2), and (5,1). That takes care of the winning or losing probabilities for the naturals (7,11) and the craps (2,3,12) outcomes. It is a relatively standard problem to calculate the probability of the sum obtained by rolling two dice. Throwing a 6,5,4,3,2 or 1 deducts 0. What is the. As the chart shows the closer the total is to 7 the greater is the probability of it being thrown. Rolling two dice. Here, the sample space is given when two dice are rolled. What is the probabilities of getting at least a 1 OR a 5 with 1 die, 2 dice, 3 dice, etc. a sum less than 4 or greater than 9 d. Calulate the probability of getting a double or treble on a dartboard. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction. 6 outcomes on one die X 6 outcomes on other die = 36 outcomes. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. 5 ways to get a sum of 6. A black and a red dice are rolled Let us take first numbers to have been appeared on the black die and the second numbers on rolled. 4d10 are enough to sample uniformly from between 1 and 10,000), but it becomes increasingly tedious to generate larger numbers. The probability of getting less than 8 is the sum of the probabilities of 2-7:. What Is The Probability That The Sum Of 8 Does Not Occur?. Two fair dice are rolled. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and 1. Question: Question 1 If Two Dice Are Rolled One Time, Find The Probability Of Getting A Sum Greater Than 6 And Less Than 12. This resulted in the first professional. Roll two dice. Rolling more dice. Find the probability that a 5 occurs first. Here, the sample space is given when two dice are rolled. Let B be the event - The sum of the top faces of the 3 dice >= 5. Question 1033885: Three dice are tossed. Since there are $6 \times 6 = 36$ total dice rolls and $1/3$ of those are a multiple of three, the number which are divisible by three is $(1/3)(36) = \boxed{12}$. We’ll look at two approaches to finding the likely outcomes in kdb/q: Method 1 – Enumeration of all possibilities. the probability that the sum is 6 given that at least one of the numbers is less than 3. A card from a pack of 5 2 cards is lost. Two fair dice are rolled and the sum of the points is noted. There are 63 = 18 ways to get two numbers of the same parity (the first can be any of the 6 numbers, and the second has to be 3 of the possible 6 which have the same parity), giving a total of 18 ways to get an even sum out of a possible of 66 = 36 outcomes (we don't have to consider if the first number is even or odd since there are an equal. If a fair dice is thrown 10 times, what is the probability of throwing at. Remind him that there are 6 options on both sides. Sum of dices when three dices are rolled together If 1 appears on the first dice, 1 on the second dice and 1 on the third dice. My own intuition tells me the answer is 2/3 because the other die simply needs to show 3, 4, 5, or 6 for the sum to be at least 5. When two dice are thrown together total possible outcomes = 6 X 6 = 36 Favourable outcomes when both dice have number more than 3 are (4, 4), (4, 5),(4, 6), (5, 4), (5, 5). When two dice are rolled, the probability of getting an even number on at least one die is 3/4. The probability of one dice not being a particular number is 5/6. (a) Find the conditional probability of obtaining a sum greater than 9, given that Given that the two numbers appearing on throwing two dice are different. (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) since each 6 numbers. When two six-sided dice are tossed, there are 36 possible outcomes as shown. If a pair of dice are rolled 5 times, what is the probability of getting a sum of 5 every time?. 3) Drawing a card from a regular deck of 52 playing cards has 52 possible outcomes. The probability of throwing any given total is the number of ways to throw that total divided by the total number of combinations (36). asked by yorkie16 on April 25, 2009; arithmetic. find the probability of obtaining Of these the sum of 5 is only possible with 1+4 and 2+3. 33 Question 3 Let A And B Be Two Independent Event, Such That P (A) = 0. That intuition is wrong. "getting a sum of odd number" would be mutually exclusive (disjoint). You absolutely need to count $2+5$ AND $5+2$, because the die act independently of each other its not like if one possibility exists the other doesn't. For example: 1 roll: 5/6 (83. The sum of the two numbers rolled are shown below:. A black and a red dice are rolled Let us take first numbers to have been appeared on the black die and the second numbers on rolled. What is the probability that the sum of two of the faces rolled equals the value of the other rolled face?. Dice and Dice Games. outcomes when two dice are tossed. If you only take two of the three for the sum, there are still 216 total outcomes to look at. If you roll a die will obtain 1, 2, 3, 4, 5 or 6? Probability measures and quantifies "how likely" an Let us define event E as the set of possible outcomes where the sum of the numbers on the faces of the two dice is equal to four. Use separate lists for the results of each die, and a third list for the sum. Probability Chapter 13 of Class 12 is one of the most important topics in Maths CBSE Board Exams. Find probability nobody gets own hat. (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) since each 6 numbers. As the chart shows the closer the total is to 7 the greater is the probability of it being thrown. A single die is rolled twice. The probability of Dice 2 rolling a 1 is also 1/6. You may get a side with 1, 2, 3, 4, 5, or 6 dots. So there are 6*combin (5,2)=60 combinations already. As such, the probability of both dice (dice 1 and Dice 2) rolling a 1 is 1/36, calculated as 1/6 x 1/6. Find each experimental probability. the probability that the sum is 6 given that at least one of the numbers is less than 3. Hence, the combination (1,3) is Below you can check our random "roll of dice" generator. 9,10) things get slightly more complicated. What is the probability that the sum of two rolled dice will equal a prime number? We find this number by multiplying 6 x 6. What is the probability that the sum of the two tosses is 4?. Two fair dice are rolled and the sum of the points is noted. 2) a sum of 6 or 7 or 8 b) doubles or a sum of 4 or 6 c) a sum greater than 9 or less than 4, Please help me. , HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. The numbers for the games so far are listed below. Question: Question 1 If Two Dice Are Rolled One Time, Find The Probability Of Getting A Sum Greater Than 6 And Less Than 12. Find the joint probability mass function of X and Y when. If one of the dice shows 1 to 4, the sum will not be greater than 10. The probability of getting each of the dice rolls are: 2: 1/36. Find the probability distribution for the ‘sum of two dice’. Memorizing the making of the above picture makes the. Let X denote the sum of the number of dots on the top faces. 9 Prove that you cannot load two dice in such a way that the probabilities for any sum from 2 to To get a better understanding of this important result, we will look at some examples. What is the probability that the sum of two rolled dice is less than or equal to 9? I got 1/5 because its greater? help. Then, we subtract 5, 15 and 30 from 100, which gives us 100 - 5 - 15 - 30 = 50. Single die roll probability tables. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. Major changes in Python environment : . a sum less than 4 or greater than 9 d. Let us understand the sample space of rolling two dice. Suppose that the first die we roll comes up as a 1. Explanation of the fundamental concepts of probability distributions. A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. Find the probabilities of rolling different sums. Hence the probability of getting a 3 is P(E) = 1 / 6. Get an answer for 'When two dice are thrown what is the probability that the sum is 8. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. Write an even more general version of the function two. That intuition is wrong. Step by step we: Generate the possible outcomes for one die. The probability of getting a sum of 5 is 1 /9 and the probability of getting a sum of 9 is 1 /9. Some of the probabilities are easy to find. Type it in the session window. Two dice are rolled. For the "point" outcomes (4,5,6,8,. Get a free answer to a quick problem. What is the probability that the sum of the two dice will not be a 6? 31/36. For three six-sided dice, the most common rolls are 10 and 11, both with probability 1/8; and the least common rolls are 3 and 18, both with probability 1/216. lis03pfxr3, ujocjh1y64, s12rmgtrh8c7r, kym10de4b5e633r, qme1yq56uk2x, ymlickuxmk8, apzqcwbno2to, slmj4cak4a, 0xy0iog05ybhmfd, gdfmdbew5q1, 96imk2cxbvbdol, l8xudpv0j1, nwazaap0onbm, 6b1rvwwrbg25g, dhhhfhlqcxj7t01, py5nq9oz5dw7o, 8iqcxkyvtn7wh1, 5q2qvtn5knmihv, z6zloxs9s19wn, 6lykvprhlrbnu8w, o1wgudqihtvg, tlh2mmq5umx, 6iuew17hsvdmp, ztt05u5nsncl, fobtt2lkbg9v, ygjk3lvdrt3qt, na18jcn35u	2020-06-06T16:19:09	{ "domain": "farmaciacoverciano.it", "url": "http://farmaciacoverciano.it/dhti/when-two-dice-are-rolled-find-the-probability-of-getting-a-sum-of-5-or-6.html", "openwebmath_score": 0.6622301936149597, "openwebmath_perplexity": 208.79297532556296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9859363750229257, "lm_q2_score": 0.8519527963298946, "lm_q1q2_score": 0.8399712517041412 }
https://stats.stackexchange.com/questions/176702/how-to-calculate-the-expected-value-of-a-standard-normal-distribution	# How to calculate the expected value of a standard normal distribution? I would like to learn how to calculate the expected value of a continuous random variable. It appears that the expected value is $$E[X] = \int_{-\infty}^{\infty} xf(x)\mathrm{d}x$$ where $f(x)$ is the probability density function of $X$. Suppose the probability density function of $X$ is $$f(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}$$ which is the density of the standard normal distribution. So, I would first plug in the PDF and get $$E[X] = \int_{-\infty}^{\infty} x\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}\mathrm{d}x$$ which is a rather messy looking equation. The constant $\displaystyle\frac{1}{\sqrt{2\pi}}$ can be moved outside the integral, giving $$E[X] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} xe^{\frac{-x^{2}}{2}}\mathrm{d}x.$$ I get stuck here. How do I calculate integral? Am I doing this correctly this far? Is the simplest way to get the expected value? • your question title is misleading. You are in fact trying to calculate the expected value of a standard normal random variable. You can also calculate the expected value of a function of a RV. I would rather put in the title: "How to calculate the expected value of a standard normal distribution." Or "How to calculate the expected value of a continuous random variable." – Gumeo Oct 13 '15 at 9:05 • @GuðmundurEinarsson corrected. – mmh Oct 13 '15 at 11:07 • "I get stuck here. How do I calculate integral?" Find the derivative of $-e^{-\frac{x^2}{2}}$. (No, I am not being facetious and suggesting needless busywork to you; I am deadly serious; Just Do It!). Then stare very hard at the derivative you have found. – Dilip Sarwate Dec 3 '15 at 5:19 You are almost there, follow your last step: $$E[X] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} xe^{\displaystyle\frac{-x^{2}}{2}}\mathrm{d}x\\=-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-x^2/2}d(-\frac{x^2}{2})\\=-\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\mid_{-\infty}^{\infty}\\=0$$. Or you can directly use the fact that $xe^{-x^2/2}$ is an odd function and the limits of the integral are symmetry. • The symmetry argument only works if both halves are themselves convergent. – Glen_b Oct 13 '15 at 10:50 • Could you explain what happens on the second row? – mmh Oct 13 '15 at 11:12 • Glen's comment is correct if it is not convergent then change-of-variables will not work – Deep North Oct 13 '15 at 11:23 • The second row is equal to the first row since $d(-\frac{x^2}{2})=-xdx$ also note the negative sign at the beginning. Then you can think of change of variable for integration, then you change it back since the limits did not change. Or you can use integrate by parts. And remember $\int_{a}^{b}e^y dy=e^y\mid_{a}^{b}$ – Deep North Oct 13 '15 at 11:37 • To use symmetry to get the mean you need to know that $\int_0^\infty xf(x) dx$ converges - it does for this case, but more generally you can't assume it. For example, the symmetry argument would say that the mean of the standard Cauchy is 0, but it doesn't have one. – Glen_b Dec 3 '15 at 1:23 Since you want to learn methods for computing expectations, and you wish to know some simple ways, you will enjoy using the moment generating function (mgf) $$\phi(t) = E[e^{tX}].$$ The method works especially well when the distribution function or its density are given as exponentials themselves. In this case, you don't actually have to do any integration after you observe $$t^2/2 -\left(x - t\right)^2/2 = t^2/2 + (-x^2/2 + tx - t^2/2) = -x^2/2 + tx,$$ because, writing the standard normal density function at $x$ as $C e^{-x^2/2}$ (for a constant $C$ whose value you will not need to know), this permits you to rewrite its mgf as $$\phi(t) = C\int_\mathbb{R} e^{tx} e^{-x^2/2} dx = C\int_\mathbb{R} e^{-x^2/2 + tx} dx = e^{t^2/2}C\int_\mathbb{R} e^{-(x-t)^2/2} dx .$$ On the right hand side, following the $e^{t^2/2}$ term, you will recognize the integral of the total probability of a Normal distribution with mean $t$ and unit variance, which therefore is $1$. Consequently $$\phi(t) = e^{t^2/2}.$$ Because the Normal density gets small at large values so rapidly, there are no convergence issues regardless of the value of $t$. $\phi$ is recognizably analytic at $0$, meaning it equals its MacLaurin series $$\phi(t) = e^{t^2/2} = 1 + (t^2/2) + \frac{1}{2} \left(t^2/2\right)^2 + \cdots + \frac{1}{k!}\left(t^2/2\right)^k + \cdots.$$ However, since $e^{tX}$ converges absolutely for all values of $tX$, we also may write $$E[e^{tX}] = E\left[1 + tX + \frac{1}{2}(tX)^2 + \cdots + \frac{1}{n!}(tX)^n + \cdots\right] \\ = 1 + E[X]t + \frac{1}{2}E[X^2]t^2 + \cdots + \frac{1}{n!}E[X^n]t^n + \cdots.$$ Two convergent power series can be equal only if they are equal term by term, whence (comparing the terms involving $t^{2k} = t^n$) $$\frac{1}{(2k)!}E[X^{2k}]t^{2k} = \frac{1}{k!}(t^2/2)^k = \frac{1}{2^kk!} t^{2k},$$ implying $$E[X^{2k}] = \frac{(2k)!}{2^kk!},\ k = 0, 1, 2, \ldots$$ (and all expectations of odd powers of $X$ are zero). For practically no effort you have obtained the expectations of all positive integral powers of $X$ at once. Variations of this technique can work just as nicely in some cases, such as $E[1/(1-tX)] = E[1 + tX + (tX)^2 + \cdots + (tX)^n + \cdots]$, provided the range of $X$ is suitably limited. The mgf (and its close relative the characteristic function $E[e^{itX}]$) are so generally useful, though, that you will find them given in tables of distributional properties, such as in the Wikipedia entry on the Normal distribution.	2019-10-23T10:17:22	{ "domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/176702/how-to-calculate-the-expected-value-of-a-standard-normal-distribution", "openwebmath_score": 0.896049439907074, "openwebmath_perplexity": 206.88384891563888, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363696508799, "lm_q2_score": 0.8519527982093666, "lm_q1q2_score": 0.8399712489804516 }
https://math.stackexchange.com/questions/2443864/solve-the-radical-equation-x-sqrtx25-2x1-sqrt4x24x6-0	# Solve the radical equation $x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$ Solve the following equation: $$x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$$ I wanted to solve this equation. First I tried to change the equations under the roots to the complete square to simplify them out, but it just became more complicated. Can someone help me with this equation, please? • Solving it by hand? You could multiply both sides by $x\sqrt{x^2+5}\color{red}{-}(2x+1)\sqrt{4x^2+4x+6}$. This will arrive at $-15x^4-32x^3-39x^2-28x-6=0$ which you can attempt to factor (trial and error or using general method for quartics, but that would be incredibly messy). You'll arrive at $-(x+1)(3x+1)(5x^2+4x+6)$ which yields four roots, one of which was introduced with our first changes when we multiplied by the conjugate, the second is our actual solution to the original problem, and the remaining two are complex. – JMoravitz Sep 25 '17 at 0:39 • But how could we factor it? Even the trial and error is very hard to approach and to guess the roots. – Nariman Zendehrooh Sep 25 '17 at 3:38 • @Nariman Zendehrooh:I edited my answer to show how to use the substitution trick to get a factored form. – quasi Sep 25 '17 at 3:42 Assuming you want real roots, you can use this trick . . . Let $f\colon \mathbb{R} \to \mathbb{R}$ be given by $f(t) = t\sqrt{t^2+5}$. Then $f$ is an odd function. Also, $f$ is strictly increasing, hence $f$ is one-to-one. Then, letting $u=2x+1$, \begin{align} &x\sqrt{x^2+5}+(2x+1)\sqrt{4x^2 + 4x + 6}=0 \qquad\qquad\qquad\qquad\;\; \\[4pt] \iff\;&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt] \iff\;&f(x) + f(u)=0\\[4pt] \iff\;&f(x) = -f(u)\\[4pt] \iff\;&f(x) = f(-u)\qquad\text{[since $f$ is odd]}\\[4pt] \iff\;&x = -u\qquad\qquad\;\;\text{[since $f$ is one-to-one]}\\[4pt] \iff\;&x = -(2x+1)\\[4pt] \iff\;&3x+1 = 0\\[4pt] \iff\;&x = -{\small{\frac{1}{3}}}\\[4pt] \end{align} As an alternative, using the same trick, you can get a factored form: \begin{align} &x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt] \implies\;&x\sqrt{x^2+5}=-u\sqrt{u^2 + 5}\\[4pt] \implies\;&x^2(x^2+5)=u^2(u^2 + 5)\\[4pt] \implies\;&x^4+5x^2=u^4+5u^2\\[4pt] \implies\;&(u^4-x^4)+5(u^2-x^2)=0\\[4pt] \implies\;&(u^2-x^2)(u^2+x^2)+5(u^2-x^2)=0\\[4pt] \implies\;&(u^2-x^2)(u^2+x^2+5)=0\\[4pt] \implies\;&(u-x)(u+x)(u^2+x^2+5)=0\\[4pt] \implies\;&(x+1)(3x+1)(5x^2+4x+6)=0\qquad\text{[replacing $u$ by $2x+1$]} \end{align} The two candidate real roots, $x=-1,\;x=-{\large{\frac{1}{3}}}\;$need to be verified against the original equation since, when squaring both sides, extraneous real roots were potentially introduced. In this case, as it turns out, the candidate root $x=-{\large{\frac{1}{3}}}$ is ok, but the candidate root $x=-1$ fails, so is not an actual root. • Wow! That was a great trick to simplify and get the solution. But I wondered how it came to your mind in the first place that you have to define f(t). Is there any clue? – Nariman Zendehrooh Sep 25 '17 at 3:44 • I recognized that $$4x^2+4x+6 = (4x^2+4x+1) + 5 = (2x+1)^2 + 5$$ and from there, defining $f(t) = t\sqrt{t^2 + 5}$ was a natural move. – quasi Sep 25 '17 at 3:47	2020-04-07T08:16:13	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2443864/solve-the-radical-equation-x-sqrtx25-2x1-sqrt4x24x6-0", "openwebmath_score": 0.9582511782646179, "openwebmath_perplexity": 603.8355338098116, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363713038174, "lm_q2_score": 0.8519527944504227, "lm_q1q2_score": 0.8399712466825968 }
https://gmatclub.com/forum/the-probability-that-event-a-occurs-is-0-4-and-the-probabil-106950.html	GMAT Question of the Day: Daily via email \| Daily via Instagram New to GMAT Club? Watch this Video It is currently 25 May 2020, 07:45 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # The probability that event A occurs is 0.4, and the probabil Author Message TAGS: ### Hide Tags Manager Status: Do and Die!! Joined: 15 Sep 2010 Posts: 237 The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags Updated on: 28 Mar 2018, 03:13 6 1 31 00:00 Difficulty: 25% (medium) Question Stats: 72% (01:36) correct 28% (02:07) wrong based on 538 sessions ### HideShow timer Statistics The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 _________________ I'm the Dumbest of All !! Originally posted by shrive555 on 26 Dec 2010, 17:47. Last edited by abhimahna on 28 Mar 2018, 03:13, edited 1 time in total. Edited the question Retired Moderator Joined: 03 Aug 2010 Posts: 156 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 26 Dec 2010, 20:42 2 3 P(A or B) = P (A) + P(B) - p(a n b) 0.6= 0.4 + P(B) - 0.25 P(B) = 0.45 ##### General Discussion Manager Joined: 31 May 2010 Posts: 57 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 26 Dec 2010, 22:30 1 P(AandB) = pA + pB - p(AintersectionB) 0.6= 0.4 + p(B) - 0.25 = 0.45 Manager Status: UC Berkeley 2012 Joined: 03 Jul 2010 Posts: 142 Location: United States Concentration: Economics, Finance GPA: 4 WE: Consulting (Investment Banking) Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 27 Dec 2010, 02:29 1 you can state both equations only if they are independent from each other ... Manager Joined: 30 May 2013 Posts: 143 Location: India Concentration: Entrepreneurship, General Management GPA: 3.82 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 10 Aug 2013, 04:33 hirendhanak wrote: P(A or B) = P (A) + P(B) - p(a n b) 0.6= 0.4 + P(B) - 0.25 P(B) = 0.45 Hi. Can u tell me wat is P(AandB)???? regards, Rrsnathan. Director Status: Far, far away! Joined: 02 Sep 2012 Posts: 998 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 10 Aug 2013, 04:59 1 rrsnathan wrote: hirendhanak wrote: P(A or B) = P (A) + P(B) - p(a n b) 0.6= 0.4 + P(B) - 0.25 P(B) = 0.45 Hi. Can u tell me wat is P(AandB)???? regards, Rrsnathan. P(A and B)= probability both events (A,B) occur= P(A)P(B). Senior Manager Joined: 10 Jul 2013 Posts: 274 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 10 Aug 2013, 13:10 shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? 0.05 0.15 0.45 0.50 0.55 P(A) = .40 P(A and B) = P(A) P(B) = .25 ----1 P(A or B) = P (A) + P(B) = .60 -----2 Both 1 & 2 give different results. 1 => P(B) = $$.25/P(A)$$ = 0.62 2=> P(B) = .60 - .40 = .20 whats going wrong ...i don;t know The only correction you need is to minus... P(A or B) = P (A) + P(B) - P(A) * P(B) = or, 0.60 = 0.4 +p(B) - 0.25 so, p(B) = 0.45 (C) we all make mistake....don't worry about it... Director Joined: 12 Nov 2016 Posts: 666 Location: United States Schools: Yale '18 GMAT 1: 650 Q43 V37 GRE 1: Q157 V158 GPA: 2.66 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 12 Oct 2017, 19:30 shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 P(A) = .40 P(A and B) = P(A) * P(B) = .25 ----1 P(A or B) = P (A) + P(B) = .60 -----2 Both 1 & 2 give different results. 1 => P(B) = $$.25/P(A)$$ = 0.62 2=> P(B) = .60 - .40 = .20 whats going wrong ...i don;t know If A and B are overlapping events then P (A or B) = P(A) + P(B) - P(A and B) C Director Joined: 12 Nov 2016 Posts: 666 Location: United States Schools: Yale '18 GMAT 1: 650 Q43 V37 GRE 1: Q157 V158 GPA: 2.66 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 12 Oct 2017, 19:36 shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 P(A) = .40 P(A and B) = P(A) * P(B) = .25 ----1 P(A or B) = P (A) + P(B) = .60 -----2 Both 1 & 2 give different results. 1 => P(B) = $$.25/P(A)$$ = 0.62 2=> P(B) = .60 - .40 = .20 whats going wrong ...i don;t know Another way of looking at this is that both events are overlapping events- A and B can occur at the same time - this is basically the same as Total= A + B -both --> is the same thing as saying P(A or B)= P(A) + P(B) -P(A and B) This is one application of the overlapping set formula to probability Director Joined: 24 Oct 2016 Posts: 664 GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags Updated on: 08 Jun 2018, 12:00 Nunuboy1994 wrote: shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 P(A) = .40 P(A and B) = P(A) * P(B) = .25 ----1 P(A or B) = P (A) + P(B) = .60 -----2 Both 1 & 2 give different results. 1 => P(B) = $$.25/P(A)$$ = 0.62 2=> P(B) = .60 - .40 = .20 whats going wrong ...i don;t know Another way of looking at this is that both events are overlapping events- A and B can occur at the same time - this is basically the same as Total= A + B -both --> is the same thing as saying P(A or B)= P(A) + P(B) -P(A and B) This is one application of the overlapping set formula to probability I have a similar question as shrive555. Not really convinced with Nunuboy1994's explanation. shrive555 When events are mutually exhaustive, then P(A or B) = P(A) + P(B). So we can't use that formula (i.e. your point 2) since the Q tells us that both events can happen together. Hey abhimahna Here's my question: How do we know that the events are not independent? For independent events, we know that P(A and B) = P(A) * P(B), which would give a different answer. Originally posted by dabaobao on 04 Jun 2018, 17:15. Last edited by dabaobao on 08 Jun 2018, 12:00, edited 1 time in total. Senior PS Moderator Joined: 26 Feb 2016 Posts: 3253 Location: India GPA: 3.12 The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 04 Jun 2018, 22:48 1 1 Alternate approach P(Total) = 1 \| P(Event A) = 0.4 \| P(Both) = 0.25 (from question stem) P(Neither) = 1 - P(Either event A or event B) = 1 - 0.6 = 0.4 P(Total) = P(Event A) + P(Event B) - P(Both) + P(Neither) Substituting values, $$1 = 0.4 + P(Event B) - 0.25 + 0.4$$ -> $$1 = 0.8 - 0.25 + P(Event B)$$ -> $$P(Event B) = 1 - 0.55$$ = 0.45(Option C) _________________ You've got what it takes, but it will take everything you've got Director Joined: 24 Oct 2016 Posts: 664 GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags Updated on: 15 Jun 2018, 10:30 1 pushpitkc wrote: Alternate approach P(Total) = 1 \| P(Event A) = 0.4 \| P(Both) = 0.25 (from question stem) P(Neither) = 1 - P(Either event A or event B) = 1 - 0.6 = 0.4 P(Total) = P(Event A) + P(Event B) - P(Both) + P(Neither) Substituting values, $$1 = 0.4 + P(Event B) - 0.25 + 0.4$$ -> $$1 = 0.8 - 0.25 + P(Event B)$$ -> $$P(Event B) = 1 - 0.55$$ = 0.45(Option C) Thanks pushpitkc for that alternate approach. We could do that if we know that we have overlapping sets. Hey abhimahna, any idea what's the relationship between overlapping and independent sets? Why can't we use the independent events formula here, P(A and B) = P(A) * P(B), which would give a different answer. Originally posted by dabaobao on 05 Jun 2018, 04:00. Last edited by dabaobao on 15 Jun 2018, 10:30, edited 1 time in total. Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 10547 Location: United States (CA) Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 06 Jun 2018, 15:42 shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 We can use the formula: P(A or B) = P(A) + P(B) - P(A and B) So we have: 0.6 = 0.4 + P(B) - 0.25 0.6 = 0.15 + P(B) 0.45 = P(B) _________________ # Scott Woodbury-Stewart Founder and CEO [email protected] 202 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Board of Directors Status: Emory Goizueta Alum Joined: 18 Jul 2015 Posts: 3586 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 15 Jun 2018, 10:44 1 dabaobao wrote: Hey abhimahna, any idea what's the relationship between overlapping and independent sets? Why can't we use the independent events formula here, P(A and B) = P(A) * P(B), which would give a different answer. Hey dabaobao , When two events are said to be independent of each other, what this means is that the probability that one event occurs in no way affects the probability of the other event occurring. Example - Say you rolled a die and flipped a coin. Here, no such case is happening. We have two events A and B as well as we have an overlap between them as well Both A and B. Hence, we will use the formula mentioned above. Also, please note that I would not consider A and B independent events unless explicitly stated in the question. Does that make sense? _________________ My LinkedIn abhimahna. \| My GMAT Story: From V21 to V40 \| My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub \| Importance of an Error Log! Verbal Resources: All SC Resources at one place \| All CR Resources at one place GMAT Club Inbuilt Error Log Functionality - View More \| Best Reply Functionality on GMAT Club! New Visa Forum - Ask all your Visa Related Questions - here \| Have OPT questions? - Post them here. Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free Check our new About Us Page here. \| Manager Joined: 07 Feb 2017 Posts: 167 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 15 Jun 2018, 11:18 0.6 = 0.4 + PB - 0.25 PB = 0.45 Senior Manager Joined: 17 Mar 2014 Posts: 451 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 21 Jul 2018, 07:28 3 shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 Using Manhattan book formula Attachments WhatsApp Image 2018-07-21 at 11.27.06.jpeg [ 86.14 KiB \| Viewed 24306 times ] Intern Joined: 22 Dec 2019 Posts: 2 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 22 Dec 2019, 22:13 Please forgive my ignorance, but why P(B) not equal to 0.625?? as P(A) = .40 P(A and B) = P(A) * P(B) = .25 P(B) = .25/P(A) = 0.625 Intern Joined: 22 Dec 2019 Posts: 2 Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 22 Dec 2019, 22:18 pushpitkc wrote: Alternate approach P(Total) = 1 \| P(Event A) = 0.4 \| P(Both) = 0.25 (from question stem) P(Neither) = 1 - P(Either event A or event B) = 1 - 0.6 = 0.4 P(Total) = P(Event A) + P(Event B) - P(Both) + P(Neither) Substituting values, $$1 = 0.4 + P(Event B) - 0.25 + 0.4$$ -> $$1 = 0.8 - 0.25 + P(Event B)$$ -> $$P(Event B) = 1 - 0.55$$ = 0.45(Option C) Please forgive my ignorance, but why P(B) not equal to 0.625?? as P(A) = .40 P(A and B) = P(A) * P(B) = .25 P(B) = .25/P(A) = 0.625 Secondly, can you please explain what do you mean by "1 \| P(Event A) = 0.4 \| P(Both)" =0.25? I am not familiar with why "total 1" sign "\|" P(A)=0.4 sign "\|" P(Both)? Senior Manager Status: So far only Dreams i have!! Joined: 05 Jan 2015 Posts: 305 WE: Consulting (Consulting) Re: The probability that event A occurs is 0.4, and the probabil [#permalink] ### Show Tags 22 Dec 2019, 23:01 gaoyuskr wrote: pushpitkc wrote: Alternate approach P(Total) = 1 \| P(Event A) = 0.4 \| P(Both) = 0.25 (from question stem) P(Neither) = 1 - P(Either event A or event B) = 1 - 0.6 = 0.4 P(Total) = P(Event A) + P(Event B) - P(Both) + P(Neither) Substituting values, $$1 = 0.4 + P(Event B) - 0.25 + 0.4$$ -> $$1 = 0.8 - 0.25 + P(Event B)$$ -> $$P(Event B) = 1 - 0.55$$ = 0.45(Option C) Please forgive my ignorance, but why P(B) not equal to 0.625?? as P(A) = .40 P(A and B) = P(A) * P(B) = .25 P(B) = .25/P(A) = 0.625 Secondly, can you please explain what do you mean by "1 \| P(Event A) = 0.4 \| P(Both)" =0.25? I am not familiar with why "total 1" sign "\|" P(A)=0.4 sign "\|" P(Both)? Hi gaoyuskr Probability formula: Event (A OR B) or P (A U B) where P (A ∩ B) = P(A and B) P (A U B) = P (A) + P (B) – P (A ∩ B) Using values: .6 = .4 + P(B) - .25 P(B) = .45 Option C! Hope it is clear. _________________ Unable to Give Up! Re: The probability that event A occurs is 0.4, and the probabil [#permalink] 22 Dec 2019, 23:01	2020-05-25T15:45:16	{ "domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-probability-that-event-a-occurs-is-0-4-and-the-probabil-106950.html", "openwebmath_score": 0.6984884738922119, "openwebmath_perplexity": 3070.527046349151, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9518632343454895, "lm_q2_score": 0.8824278680004706, "lm_q1q2_score": 0.8399506445115226 }
http://calculus123.com/wiki/Student%27s_guide_to_proof_writing	This site is devoted to mathematics and its applications. Created and run by Peter Saveliev. # Student's guide to proof writing These are 10 rules (or tests, or clues) that you can use to check if there may be problems with what you are about to submit... Once upon a time a lecture was being given: Pythagorean Theorem: $a^2+b^2=c^2$ for the sides of a right triangle... Homework assignment: Prove the theorem. . . Law of Cosines: $a^2+b^2-2ab\cos \alpha=c^2$ for the sides of a triangle with angle $\alpha$... The week after this was submitted: Homework solution: To prove the Pythagorean Theorem, take the Law of Cosines and set $\alpha = 90$ degrees, done. What is wrong with this picture? The argument is circular as the proof of the Law of Cosines is (usually) based on the Pythagorean Theorem. That's why the lectures that follow the assignment should be off-limits. Rule 1: If you use in your proof results presented in lectures after the homework was assigned, this is very likely not what's expected from you. To check all the lectures may be impossible, and the purpose of the rule is to eliminate any chance of circular reasoning. The version of this rule for the professor to follow: Don't assign homework until all the necessary material has been covered. The example provides a rationale for the following: Rule 2: If your proof is "The statement is just a particular case of this theorem", that's probably not what's expected from you. Be especially careful if this theorem comes from outside the course. Simply put, a very short proof generally is a bad sign. This rule is meant to prevent you from being tempted by an easy way out. A proof from scratch -- based on the concepts being discussed in the class you are taking -- is expected just about every single time. To summarize the two rules: the proof should be as localized as possible, but only on one side (think $(a-\epsilon,a]$). Rule 3: If your proof doesn't provide the definition or quote a theorem for each concept used, it is probably flawed. The danger is that the proof is superficial and hand-wavy. A proof that looks like an essay can probably use a lot more structure. Rule 4: If your proof is long but has no lemmas, it is likely to be flawed. It is certainly not very readable. It is not readable for the professor but for you too -- that's why there is a chance it may have problems that you've overlooked. Note: In just about any undergraduate or graduate course a one-page proof in TeX (with no illustrations) is long. The rule doesn't mean to proclaim that long is bad. Just the opposite is true: one shouldn't keep any part of the proof for himself. Rule 5: If you have an “easy” part of the proof in your head but don't put it on paper, this part is likely to be challenged. Rule 6: If you don't introduce all or most of the objects in the proof by giving them letter names, your proof may be flawed. The danger is again that the proof is superficial and hand-wavy. Always start with "Let $x$ be..." Note: How to choose good notation is a subject of a separate discussion. Rule 7: If you introduce an object in your proof and then never use it, this is a problem and might be a sign of other problems. That would leave a very bad impression... Rule 8: If you don't use all of the conditions of the theorem you are to prove, your proof is very likely to be flawed. You simply don't see a lot of theorems with redundant conditions. Your drawings may fool you. In fact, draw a triangle... Now, take a closer look at it; you are likely to discover that your triangle is either a right triangle or an isosceles. A proof based on this picture could easily be flawed. Rule 9: If removing all pictures from your proof makes it incomplete or just hard to follow, it is probably not rigorous enough. Let illustrations illustrate... And finally, just because your professor or your textbook violate, as they often do, some or all of these rules, don't assume that you are off the hook. Rule 10: If you write your proof as just a variation of a proof taken from a lecture or a book, it is likely that higher standards will be applied to yours.	2019-07-21T01:04:34	{ "domain": "calculus123.com", "url": "http://calculus123.com/wiki/Student%27s_guide_to_proof_writing", "openwebmath_score": 0.6310448050498962, "openwebmath_perplexity": 440.40438925839203, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9518632288833652, "lm_q2_score": 0.882427872638409, "lm_q1q2_score": 0.839950644106275 }
https://mathematica.stackexchange.com/questions/126567/coding-a-simulation-of-multi-species-population-dynamics	# Coding a simulation of multi-species population dynamics I want to code several steps in a recursion equation (in my case, migration, followed by selection, followed by mating, etc in a population). I have seen this done using a lot of copying and pasting and Do loops. I am hoping someone has a better suggestion that makes use of Mathematica's recursion equation construction and memoization. In step 1 I was able to code a pair of coupled recursion equations. Step 1: Frequency of genotype i in population 1 (x1) and population 2 (x2) at time t depends on migration rate between the two populations and the frequency of genotype i in each at time t - 1. parameters = {m->0.01} x1[i_, t_] := x1[i, t] = (1 - m) x1[i, t - 1] + m x2[i, t - 1] /. parameters x2[i_, t_] := x2[i, t] = (1 - m) x2[i, t - 1] + m x1[i, t - 1] /. parameters x1[1,0] := 1 x2[1,0] := 0 ListPlot[Table[{{t, x1[1, t]}, {t, x2[1, t]}}, {t, 0, 1000}], PlotRange -> {0, 1}] Now I want to add a step to my recursion equations so that selection also occurs in the same generation. Step 2 depends on x1[i, t] and x2[i, t] (as they are currently written) but I would not consider it time t + 1. Not a reproducible example below because I don't know how to code this part Step 2: Selection in population 1 x1Selection[1,t] = ((1 + s1) x1[1,t])/(1 + s1 x1[1,t]) x2Selection[1,t] = x2[1,t] Step 3: Not shown Step 4: Not shown So I want to go through these steps and get the frequency of genotype i at time t, after migration, selection, etc have happened. Then repeat step 1 using x1Final[1, t] and x2Final[1, t] in place of x1[1, t - 1] and x2[1, t - 1], respectively. Is there an elegant way to code a multistep recursion equation without copying and pasting the output of the previous step into the next step (because this gets considerably uglier in steps 3 and 4)? That is, without writing something like: parameters = {m->0.01, s1->0.05} x1[i_, t_] := x1[i, t] = ((1 + s1) ((1 - m) x1[i, t - 1] + m x2[i, t - 1]))/(1 + s1 ((1 - m) x1[i, t - 1] + m x2[i, t - 1])) /. parameters x2[i_, t_] := x2[i, t] = (1 - m) x2[i, t - 1] + m x1[i, t - 1] /. parameters x1[1,0] := 1 x2[1,0] := 0 ListPlot[Table[{{t, x1[1, t]}, {t, x2[1, t]}}, {t, 0, 1000}], PlotRange -> {0, 1}] • You can use Composition to combine the different steps for your Genetic Algorithm and then use NestList to recursively simulate new generations. – Thies Heidecke Sep 17 '16 at 23:56 • I suggest omitting any attempt to memoise until you have fully debugged your code (and only implement it if essential for efficiency). – mikado Sep 18 '16 at 7:46 • @ThiesHeidecke I am reading through the documentation and having trouble understanding how to set up the function definitions to use with Composition. Would it be like x1Migration[i, t] = (1 - m) x1[i, t] + m x2[i, t] and then x1Selection[x1Migration[i,t]]:= ((1 + s1) x1Migration[i,t])/(1 + s1 x1Migration[i,t]) and then Composition[x1, x1Selection, x1Migration][i,t]? – biologyUser Sep 18 '16 at 18:13 I would drop the whole idea of using recursion equations and simply write a function that expresses how the next generation depends on the current one. Like so. With[{m = .01, s1 = .01}, nextGen[{x1_, x2_}] := Module[{x1tmp, x2tmp}, x1tmp = (1 - m) x1 + m x2; x1tmp = (1 + s1) x1tmp/(1 + s1 x1tmp); x2tmp = (1 - m) x2 + m x1; {x1tmp, x2tmp}]] The above function is easy to extend with further algorithmic steps (e.g., your steps 3 and 4, whatever they will be) and to the addition of species (e.g., {x1, x2, x3}) should that be needed. It also evaluates quite quickly. With nextGen, the history over 1000 generations can be generated with populations = Transpose @ NestList[nextGen, {1, 0}, 1000]; ListPlot[populations, PlotRange -> {0, 1}] ### Update In response to the OP's comment, I give a vectorized version of nextGen. With[{m = .01, s = {.0024, -.002}}, nextGen[pop_List] := Block[{sp}, sp[1] = (1 - m) pop[[1]] + m pop[[2]]; sp[2] = (1 - m) pop[[2]] + m pop[[1]]; sp[1] = (1 + s[[1]]) sp[1]/(1 + s[[1]] sp[1]); sp[2] = (1 + s[[2]]) sp[2]/(1 + s[[2]] sp[2]); Table[sp[i], {i, Length[pop]}]]] Nothing changes in the way the simulation is run. populations = Transpose @ NestList[nextGen, {1, 0}, 1000]; ListPlot[populations, PlotRange -> {0, 1}] The result is different because the 2nd species also undergoes selection and the selection parameters have been given different values (just to produce an different plot). • Is there a way to do it without giving all the different species different names? I'll need to keep track of eight total. I was hoping to index them by population (k) and genes (pi and ti) to reduce the chance of making errors later. e.g. migration: xm[k_, pi_, ti_] := (1 - m) x[k, pi, ti] + m x[3 - k, pi, ti] /.parameters followed by selection: xs[k_,pi_,ti_] := ((1 + KroneckerDelta[ti, k] s[k]) xm[k, pi, ti])/(1 + s[k] (Sum[xm[k, pi, k], {pi, 1, 2}])) – biologyUser Sep 18 '16 at 17:26 • @user43111. What you bring up in your comment looks like a different question than the one you asked above. I don't like bait and switch. You should edit your question so it expresses your full, actual problem. Do not dribble it out a bit at a time -- especially not in a comment to an answer. – m_goldberg Sep 18 '16 at 20:11 • @biologyUser I'm a bit confused about what exactly you're trying to model and whether the equations make sense. Are you using the words population and species interchangeably, or are you envisioning multiple interacting species (or genotypes?) in a multi-patch system? – Chris K Sep 20 '16 at 13:19	2019-07-16T21:30:15	{ "domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/126567/coding-a-simulation-of-multi-species-population-dynamics", "openwebmath_score": 0.2148144245147705, "openwebmath_perplexity": 2683.7716337328075, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9609517050371972, "lm_q2_score": 0.8740772482857833, "lm_q1q2_score": 0.8399460220744449 }
https://www.jiskha.com/display.cgi?id=1245121807	# irrational numbers?! posted by Sanchia what are rational and irrational numbers?? i know that irrational numbers are numbers that cannot be expressed as a fraction, but i'm still confused. there is this question: which of the following are irrational numbers: √2, √8, 22/7, pi, 2√3 i know that 22/7 is rational, pi is irrational, but what about those square roots? 1. drwls Square roots of numbers that are not the squares of integers are all irrational. There is a way to prove that but I forgot the details. You have to assume that a fraction works and then prove that the assumption leads to a contradiction Any number that does not meet the definition of being rational is irrational. 2. Reiny - √2 is either rational or irrational. - Assume that √ is rational so that √2 = a/b, with a/b in lowest terms - Square both sides to get 2 = a^2/b^2 then a^2 = 2b^2 - the right side of this equation is clearly an even number, since anything multiplied by 2 is even - so a^2 must be even. We also know that if we square an odd number the result is odd, and if we square an even number the result is even so a must be even so a could be written as 2k - rewriting our equation as 2b^2= (2k)^2 2b^2 = 4k^2 b^2 = 2k^2 by the same argument as above 2k^2 is even , so b has to be even which means a and b are both even, therefore a/b is not in lowest terms BUT that contradicts my assumption that a/b was a fraction in lowest terms so √2 = a/b is a false statement therefore √2 cannot be rational, and must then be irrational the same argument could be used for √3 and all other square roots 3. brandy true or false does this number represent a rational number 0.20200200020000200000 4. Amber Is √4 Irrational? 5. Cereal.... is life. -Life Cereal is this seriouly the answer for the LEAP test feecepum you're not going to get ALL 50 answers -_- ## Similar Questions 1. ### Math What are irrational, rational, and natural numbers? 2. ### Algebra 28. Use the Distributive Property to simplify x(4x^2 + x + 4) Is it 4x^3 + x^2 + 4x? 3. ### math 1. Which of the following numbers is an example of an integer? 4. ### Math 1a) Prove that there exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 1b) use similar idea to prove that there exists a rational number a and an irrational … There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b so … 6. ### THE HOMEWORK IS DUE TODAY- HELP! HELP! PLEASE There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b so … 7. ### math There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b so … 8. ### math Which statement is true? A. All irrational numbers are also rational numbers. B. Half of the irrational numbers are also rational numbers. C. One-third of the irrational numbers are also rational numbers D. Irrational numbers cannot 9. ### Math Check Pleas Thank You 1. Which of the following numbers is an example of an integer? 10. ### math Which statements are true for irrational numbers written in decimal form? More Similar Questions	2018-06-19T14:37:07	{ "domain": "jiskha.com", "url": "https://www.jiskha.com/display.cgi?id=1245121807", "openwebmath_score": 0.8192746043205261, "openwebmath_perplexity": 1142.8662794617562, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9609517072737737, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8399460066926067 }
https://math.stackexchange.com/questions/735900/limits-to-infinity	# Limits to infinity? As a part of homework, I was asked What does $\lim_{x\to a} f(x)=\infty$ mean? In an earlier calculus class I was taught that in order for $L=\lim_{x\to a}f(x)$ to exist, we need that $L=\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$. In an explicit(but extremely non-rigorous) example the teacher explained with a graphic in which the two two sides of a vertical asymptote tended to $\infty$, the limit didn't exist because we couldn't equate both lateral limits, since we cannot compare two infinities. Is this right? If so, I would answer "It doesn't make sense, since we would need that both lateral limits at $a$...equated to the same infinity?" If it is not right and that does make some sense, may you clarify this point to me about that definition? • In it's current form, it is meaningless. The limit of what? As for the definition, one can make sense of it as follows: $\forall M \in \mathbb{R}$, there exists $\delta > 0$ such that $0 < \|x-a\| < \delta$ (note my initial error, it should be a deleted neighbourhood) implies $f(x) > M$. The one-sided limit on the right would be $0 < x - a < \delta$; the one on the left would be $0 < a - x < \delta$. The two-sided limit only makes sense if they are equal. – Chris K Apr 2 '14 at 0:19 • @ChrisK Sorry, I was kind of absentminded when I wrote this, the post is edited now. Using that $\epsilon-\delta$ definition, it is clear that both limits tend to $\infty$. But is it common to write that the limit is $\infty$, even if we can't equate both lateral lmimits? – chubakueno Apr 2 '14 at 0:29 • If both limits aren't equal, we can only make sense of one-sided limits. We then say the two-sided limit does not exist. – Chris K Apr 2 '14 at 1:23 • @chubakueno If it helps, think of $\infty$ as notation for "grows without bound", rather than representing some actual value. – augurar Apr 3 '14 at 3:39 $\lim_{x\to a}f(x)\to\infty$ means that as you go closer and closer to $a$, the value of $f(x)$ grows arbitrarily large. Now, if you approach $-\infty$ as you go closer to $a$ from the left (or right), and if you approach $+\infty$ as you go closer to $a$ from the right (or left), then the limit does not exist. The reason is that the limit has two values at the same point. To be mathematically precise, since we know that the left- and right-hand limits are the same if and only if the ordinary limit exists, and that the left- and right-hand limits are not the same, $\lim_{x\to a}f(x)$ is undefined. I believe that this is what you mean by "two sides of a vertical asymptote tended to $∞$." • Yes, that was what I meant. Alas(it seems) there is not an universal convention about this. The epsilon delta arguments can prove that a function can grow without bound towards $+\infty$, but with the comments and answers here I think that $\lim_{x\to a} f(x)=\infty$ is not universally accepted. – chubakueno Apr 2 '14 at 0:50 • @chubakueno Yes, that's true... – user122283 Apr 2 '14 at 0:57 There is still some difference of feeling on how to talk about these. In the traditional discussion of limits, a limit should be a finite number that is "approached arbitrarily closely" by the function (made rigorous by $\ \epsilon - \delta \$ proofs, or the like). So when a function does not approach a finite value as $\ x \ \rightarrow \ a \ ,$ we say the limit "does not exist". But there are various ways this can happen. The oxymoronic phrase "infinite limit" is introduced in some texts to indicate that a function grows "without limit" to the same signed infinity for $\ x \$ approaching $\ a \$ "from both sides". Hence, these authors will write $\ \lim_{x \rightarrow 0} \ \frac{1}{x^2} \ = \ +\infty \ ,$ for example; they would still write $\ \lim_{x \rightarrow 0} \ \frac{1}{x} \$ DNE . • I think you have a typo... $lim_{x->0} \frac{1}{x^2}$ is certainly not $0$! – augurar Apr 2 '14 at 0:24 • Oops, sorry! Thinking too much about the zeroes. Thanks. – colormegone Apr 2 '14 at 0:25 • So It means that it grows without limit to the positive infinity, even though lateral limits are not equal would be the most politically correct answer? – chubakueno Apr 2 '14 at 0:35 • I'm not sure exactly what a "lateral limit" is; your post is the first time I've seen the term. How does it differ from a "one-sided" limit? The issue of "comparing infinities" is being avoided, since the notion of calling something an "infinite limit" is just a convention to summarize a particular situation (and one which is not universally accepted). – colormegone Apr 2 '14 at 0:44 • We are talking about the same thing, it just just my bad use of english :) (that I am currently trying to improve) – chubakueno Apr 2 '14 at 0:47	2019-11-20T02:52:14	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/735900/limits-to-infinity", "openwebmath_score": 0.9388114809989929, "openwebmath_perplexity": 346.0970004662118, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307684643189, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8399339302555958 }
https://math.stackexchange.com/questions/292230/velocity-definite-integral	# Velocity & Definite Integral I am not sure how to approach this problem. Problem Suppose a certain object moves in a straight line with the following velocity, where $v$ is in meters per second and $t$ is in seconds: $v(t) = -2 + t + 3\sin(\pi t)$ Without using your calculator, but instead using properties of definite integrals and facts you know about area, determine the net change in distance of the object from $t = 0$ to time $t = 6$ and find the object's average velocity on this interval. My impression is that the net change in distance would be: $d = \int_{0}^{6} \|v(t)\| dt$ But since we cannot use our calculators I am unsure. We want the net change in displacement from $t=0$ to $t=6$. This is $$\int_0^6 v(t)\,dt,$$ no absolute values. With absolute values, you are computing the total distance travelled, which would be relevant for gasoline consumption, but is not what is asked for. The average velocity in our time interval is $$\frac{1}{6}\int_0^6 v(t)\,dt.$$ As to doing it without calculation, by properties of area, note that $\sin \pi t$ goes through $3$ full cycles. By the picture of sine, the integral of the sine part is $0$. (There is exactly as much area below the axis as above.) As to the integral of the $-2+t$ part, draw the line $y=-2+t$ (the usual $x$-axis is now called the $t$-axis). The integral is (sort of) the area under this curve, and "above" the $t$-axis, except that from $t=0$ to $t=2$ this area has to be viewed as negative. So it is the area of a certain triangle, $t=2$ to $t=6$, minus the area of a certain triangle, $t=0$ to $t=2$. A picture (for the $y=-2+t$ part, forget about the sine part) is essential. • That makes sense, but to confirm: $\Delta d = 6$ m and the average velocity would then be $v_{ave} = \frac{1}{6} (6) = 1$ m/s. – Biff Feb 1 '13 at 18:13 • Yes, the change in displacement is indeed $6$. – André Nicolas Feb 1 '13 at 18:14 I think net change would not have the absolute value; you would use absolute value when you want to know how much total movement there was. So you simply evaluate $$\int_0^6 v(t) \, dt$$ For your function, you would evaluate \begin{align} \int_0^6 (-2 +t + 2 \sin{(\pi t)}\, dt &= -2 \int_0^6 dt + \int_0^6 t \, dt + 2 \int_0^6 \sin{(\pi t)} \, dt \\ &=-2 (6-0) + \frac{1}{2} (6^2 - 0^2) + \frac{2}{\pi} (1-\cos{(6 \pi)}) \\ &= -12 + 18 \\ &= 6 \end{align} That is, the net distance is $6$ m. Note that I used the fact that $\cos{(6 \pi)} = 1$.	2019-09-24T08:46:02	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/292230/velocity-definite-integral", "openwebmath_score": 0.9966103434562683, "openwebmath_perplexity": 288.4295985161888, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307684643189, "lm_q2_score": 0.8633916152464017, "lm_q1q2_score": 0.8399339285458066 }
https://madagas-car.pl/maximilien-robespierre-dqlihta/6351fa-maximum-number-of-edges-in-a-disconnected-graph	Let [FONT=MathJax_Math-italic]k and [/FONT][FONT=MathJax_Math-italic]n - k [/FONT] be the number of vertices in the two pieces. Home Browse by Title Periodicals Discrete Mathematics Vol. mRNA-1273 vaccine: How do you say the “1273” part aloud? So, there is a net gain in the number of edges. Alternate solution a simple connected planar graph G with 10 vertices and 25 edges have 17 faces, Maximum set of edges or vertices that doesn't disconnect graph. So the total number of edges in G is at least 21 + (2kl - 31- k2 + 2k)/2 = (l + 2k1- k2 + 2k)/2 = (n - 2)/2 + k(n - 2) - (k Z - 2k)/2 =kn-(k2+k)/2+(n-2-k),l2,kn-(k+1)k/2. The last remaining question is how many vertices are in each component. Consider a graph of only 1 vertex and no edges. Then, each vertex in the first piece has degree at most $k-1$, therefore the number of edges in the first component is at most $\frac{k(k-1)}{2}$, while the number of edges in the second component is at most $\frac{(n-k)(n-k-1)}{2}$. Even if it has more than 2 components, you can think about it as having 2 "pieces", not necessarily connected. Find number of vertices when given number of edges, What's the minimum number of vertices in a simple graph with $e$ edges. You can also prove that you only get equality for $k=1$ or $k=n-1$. How can there be a custom which creates Nosar? A graph or multigraph is k-edge-connected if it cannot be disconnected by deleting fewer than k edges. Beethoven Piano Concerto No. Let G be a graph with n vertices. Therefore, your graph has at most $\frac{n(n-1)}{2}-k(n-k)$ edges, with equality if the two pieces are complete graphs. formalizes this argument). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Proof. 3: Last notes played by piano or not? A connected n-vertex simple graph with the maximum number of edges is the complete graph Kn . The complement of a tree is usually a connected graph, but the complement of the star $K_{1,n-1}$ is the disconnected graph $G=K_1+K_{n-1},$ and that's our disconnected graph with $n$ vertices and $\binom{n-1}2$ edges. Replacing the core of a planet with a sun, could that be theoretically possible? Therefore our disconnected graph will have only two partions because as number of partition increases number of edges will decrease. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. How to derive it using the handshake theorem? 1-3 Maximum number of edges in a critically k-connected graph article Maximum number of edges in a critically k-connected graph Request PDF \| Maximum number of edges in a critically k-connected graph \| A k-connected graph G is said to be critically k-connected if G−v is not k-connected for any v∈V(G). Now assume that First partition has x vertices and second partition has (n-x) vertices. Please use Mathjax for better impact and readability, The maximum no. In a simple undirected graph with n vertices what is maximum no of edges that you can have keeping the graph disconnected? Maximum number edges to make Acyclic Undirected/Directed Graph Dijkstra’s – Shortest Path Algorithm (SPT) - Adjacency Matrix - Java Implementation Categories Graphs , Intermediate , Software Development Engineer (SDE) , Software Engineer Tags Intermediate Leave a comment Post navigation Maximum number of edges in connected graphs with a given domination number Am I allowed to call the arbiter on my opponent's turn? Thanks for contributing an answer to Mathematics Stack Exchange! deleted , so the number of edges decreases . Suppose we have been provided with an undirected graph that has been represented as an adjacency list, where graph[i] represents node i's neighbor nodes. What is the minimum number of edges G could have and still be connected? It only takes a minute to sign up. 24 21 25 16. Now, according to Handshaking Lemma, the total number of edges in a connected component of an undirected graph is equal to half of the total sum of the degrees of all of its vertices. What is the maximum number of edges in a bipartite graph having 10 vertices? Why does "nslookup -type=mx YAHOO.COMYAHOO.COMOO.COM" return a valid mail exchanger? Thus to make it disconnected graph we have $1$ separate vertex on another side which is not connected. It has n(n-1)/2 edges . The maximum number of edges with n=3 vertices −. 1)(n ? What is the maximum number of edges possible in this graph? Maximum number of edges in a simple graph? Now if a graph is not connected, it has at least two connected components. Hence, every n-vertex graph with fewer than n 1 edges has at least two components and is disconnected. It is minimally k -edge-connected if it loses this property when any edges are deleted. If we divide Kn into two or more coplete graphs then some edges are. So the maximum edges in this case will be $\dfrac{(n-k)(n-k+1)}{2}$. Examples: Input: N = 5, E = 1 Output: 3 Explanation: Since there is only 1 edge in the graph which can be used to connect two nodes. How many connected graphs over V vertices and E edges? This can be proved by using the above formulae. I think that the smallest is (N-1)K. The biggest one is NK. Use MathJax to format equations. Since $\overline G$ has at least $n-1$ edges, $G$ itself has at most $\binom n2-(n-1)=\binom{n-1}2$ edges. Origin of “Good books are the warehouses of ideas”, attributed to H. G. Wells on commemorative £2 coin? Is it normal to need to replace my brakes every few months? This is because instead of counting edges, you can count all the possible pairs of vertices that could be its endpoints. Can I print plastic blank space fillers for my service panel? To finish the problem, just prove that for $1 \leq k \leq k-1$ we have Since we have to find a disconnected graph with maximum number of edges with n vertices. By Lemma 9, every graph with n vertices and k edges has at least n k components. According to this paper, Case 3(b): t , 2. The contrapositive of this is that every connected n-vertex graph has at least n 1 edges. Let's assume $n\ge2$ so that the question makes sense; there is no disconnected graph on one vertex. How many edges to be removed to always guarantee disconnected graph? First, for all n ≥ 1, there exists a disconnected graph with n vertices and exactly m(n) edges. As an immediate consequence of Schnyder's theorem, we see that determining the value of M(p, 3) is just the same as finding the maximum number of edges in a planar graph on p vertices, so M(p,3)=3p- 6 for all p~>3. That's the same as the maximum number of [unique] handshakes among $n$ people. If the edge is removed, the graph becomes disconnected… $\endgroup$ – Jon Noel Jun 25 '17 at 16:53 6-20. Proof. Graphs with bounded chromatic number can be drawn on the three-dimensional grid with O(n 2 ) volume, as shown by Pach et al. 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References or personal experience, we introduce the following concept: Def to stop throwing food once 's... Count all the connected components that you can count all the possible pairs of vertices the!	2021-06-15T03:53:30	{ "domain": "madagas-car.pl", "url": "https://madagas-car.pl/maximilien-robespierre-dqlihta/6351fa-maximum-number-of-edges-in-a-disconnected-graph", "openwebmath_score": 0.35139915347099304, "openwebmath_perplexity": 705.4726990497503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274402, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8399339254567376 }
https://math.stackexchange.com/questions/3385939/constructing-a-function-based-on-a-real-world-scenario	# Constructing a function based on a real-world scenario A random thought came into my head today when I was in the subway: Suppose we have a train in a subway where the stations are evenly spaced in a straight line. The train accelerates for some amount of time, moves with a constant speed for some amount of time, and starts decelerating until it reaches the next station such that the same amount of time spent accelerating and decelerating are the same, and the magnitudes of both are the same. Create a possible function that illustrates this scenario. I decided that I would try to use some combination of a sinusoidal function and a linear function, since the train would try to do the same thing during certain periods of time, and the train is moving only in one direction. I hopped onto Desmos and played around a little bit, and I was able to create a function that gets me somewhat close to what I want. $$d_1(t)=t-\frac1{2\pi}\sin 2\pi t$$ I chose the constant multiple $$\frac1{2\pi}$$ to simplify the distance between each station to $$1$$, and the multiplier $$2\pi$$ for $$t$$ to suggest that it takes $$1$$ unit of time to get from one station to the next. Choosing the sinusoidal function means that there will be no length of time where the train is moving at a constant speed, and the acceleration and deceleration will be exactly opposite, separated by the point of inflection halfway between the stations. The problem: trains usually stop at every station. I want to, arbitrarily at first, modify the function so that the train waits for the same amount of time as it takes to travel from one station to another. Eventually, I want to be able to construct a function that can describe the waiting time in some proportion to the traveling time. I came up with the function $$d_2(t)=\begin{cases}t-\frac{\lfloor t\rfloor}2-\frac1{2\pi}\sin 2\pi t,& \lfloor t\rfloor \textrm{ is even}\\ \frac{\lceil t\rceil}2,& \lfloor t\rfloor \textrm{ is odd} \end{cases}$$ This gets me something closer to what I want. My questions: 1. What alternative ways of representing the scenario are there? For example, would it be possible to construct a piecewise function based only on polynomials that meets the criteria above? 2. Is there a way to represent this particular scenario without use of a piecewise function? 3. Suppose there must be a stretch of time where the train is moving at a constant speed between stations. How can I factor that in to the equation for the function? 4. Suppose the train makes the return trip in a similar manner, and it makes some number of return trips daily. Does this suggest that it can be written as a periodic function? If so, what would be the equation for that function? • I know you tagged this as precalculus, but if you start by just considering the acceleration, you can then anti-differentiate to get velocity and anti-differentiate again to get distance. From the description you gave, you would end up with a piecewise function of quadratics. Oct 8 '19 at 20:25 • Yes, I see where you are coming from, but I want to suppose that the acceleration may or may not be constant -- having the same sign is sufficient (e.g. having an acceleration $x>0$ in one interval, then having the deceleration (or negative acceleration) be the same magnitude, but different sign). Thanks for input; I will remove that tag since concavity of a function is based on knowledge of calculus. Oct 8 '19 at 20:42 • You can try to make formal your conditions as differential equations, solving which will give you an infinite family of possible solutions. Oct 8 '19 at 20:52 • Particularly question 4, because of the periodicity aspect, made me think of Fourier Series: en.wikipedia.org/wiki/Fourier_series Oct 8 '19 at 21:40 • If you want your function to be constant in some sections, than it cannot be analytic. In your situation this translates to the fact that you need a piecewise definition. Oct 9 '19 at 7:05 Just collecting various comments all in one place, and filling in some holes / details. First, a function $$d(t)$$ is simply a mapping from values in the domain (in your case: time) to values in the range (in your case: distance). Whether you can write that function as a "nice" formula of "well-known" expressions, is kinda irrelevant. I should add though, the confusion between (abstract) functions and (nice) formulas is historic, so if you have the same confusion, you're not alone. :) In the context of $$d(t)$$ being the motion of a train, there are reasonable assumptions to be adopted, e.g. $$d$$ should be continuous (the train does not "teleport"), differentiable (velocity is well defined) and maybe twice differentiable (acceleration is well defined). However, even after adopting these, there is nothing special about sinusoids. (In the context of a moving pendulum, or (idealized) planetary rotation, maybe sinusoids would be natural, but a moving train...?) Q1: As answered by @79037662 it is surely possible to model the train motion $$d(t)$$ as piecewise polynomial - and in fact for a train this would probably be more natural than modeling by sinusoids. Assuming no friction (ha!) a zero acceleration would give constant velocity $$v(t)$$ and linear $$d(t)$$, a constant non-zero acceleration would give linear velocity $$v(t)$$ and quadratic $$d(t)$$, a linear acceleration would give quadratic $$v(t)$$ and cubic $$d(t)$$, etc. If you have pieces of acceleration you will have pieces of polynomial $$d(t)$$ and once you have pieces of polynomials you simply have to be careful to "glue" them together. The piecewise formula might be tedious to write out but there is nothing magical going on. Q2: As answered by @quarague it is not possible for your function to have a constant stretch and also be analytic, and "analytic" very roughly translates to something with a nice formula (and probably includes most "familiar" functions you have in mind: polynomials, sinusoids, exponentials, etc). This is actually a non-trivial result. This applies to both distance (constant distance implying stopping at a station) and velocity: If you have constant stretches and non-constant stretches (in either distance or velocity), then you will have to write your function as piecewise. Q3: Constant speed simply means that part of $$d(t)$$ is linear, e.g. $$d(t) = v t + C$$. Here $$v$$ is the constant velocity. You get this $$d(t)$$ by integrating $$v$$ over $$t$$. Q4: As answered by @AdamRubinson Fourier Series would be a way to go, but for the purpose of writing out the function it's probably overkill. Suppose $$d^(t)$$ for $$t \in [0,1]$$ describes what happens to the train in the first hour, and $$d^(0)=d^(1)$$ s.t. the train makes a complete round trip. Further suppose the train does this every hour, then you can simply write the overall $$d(t)$$ as $$d(t) = d^(t - \lfloor t \rfloor)$$ Here $$\lfloor t \rfloor$$ is the standard notation for the "floor" function which denotes the greatest integer $$\le t$$, so e.g. if $$t=13.7$$ then $$\lfloor t \rfloor = 13$$ and $$t - \lfloor t \rfloor = 0.7$$ gives the fraction part of $$t$$. Note that while "floor" is a "standard" function, it is piecewise. Which in a roundabout way brings us back to the your Q2: you can always describe a piecewise function in a non-piecewise way... if you define a new name for it! :) Hope this helps (as opposed to confuses even further!) • Thanks for a detailed response! Got a final question for you: suppose I wanted to create a polynomial function for my above $d_1(t)$ on, say, $[0,m]$. Is there a way to ensure that the "bumps" that I have are all uniform (like, $(n,n+1)$ acts the same as $(n-1,n)$ except one unit up)? It is (in addition to my naivety or my lack of training) why I tried to think about it using a periodic function in the first place. Oct 9 '19 at 16:12 • Say $x$ is a variable $\in (0,1)$, then you seem to be saying you want $f(n+x) = f(n-1 + x) + 1$. The only polynomial function that does that would be linear; quadratics, cubics etc all never do that. So if $f$ is polynomial but not linear, then you have to specify $f$ as piecewise polynomial, e.g. maybe use my idea in Q4 or something similar. Oct 9 '19 at 17:18	2022-01-18T10:53:51	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3385939/constructing-a-function-based-on-a-real-world-scenario", "openwebmath_score": 0.7681986689567566, "openwebmath_perplexity": 258.9404844407216, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830768464319, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8399339199968598 }
https://math.stackexchange.com/questions/4375095/width-of-this-shape	"Width" of this shape I know the length of the arc A, the sides L, the bottom W and the maximum height H. I would like to calculate the maximum width. What would I derive a formula for it? Edits: The whole shape is symmetrical on both sides of H. There is no guarantee that L are radii of the circle that would be formed from A. Johan Biemans, who posted below, first posed the question to me. • Are $W$, $H$ and $A$ in the center? Feb 6 at 0:22 • Yes they are. Thanks for checking. Feb 6 at 0:28 • If you can get the radius $r$, then use the formular to find chord length: $2 × \sqrt(r^2− d^2)$ where $d$ is perpendicular distance from the centre – Abel Feb 6 at 0:31 • Is $A$ a circular arc? Feb 7 at 16:39 • There seems to be a lot of confusion about your diagram. So first of all, is the figure symmetric about H? And second, are the lines L normal to the arc A, that is, are the lines L along radii of A? Feb 8 at 15:58 I will begin this by saying that the following is only true if $$L\parallel L'$$ for each pair of legs, $$L+L'=R$$ and if $$A$$ is a circular arc. I've added a bit to your diagram, as you can see below. For the sake of clarity, I have extended the sides to meet at a point. Additionally, I have notated the distance from the point to the middle of the arc as $$R$$ and the distance from the top of the arc to the midpoint of $$D$$ as $$P$$. From here, we have a few basic geometric theorems we can use. Namely, if we allow $$\theta$$ to be the angle between the two $$L'$$ legs of our newly formed traingle, • $$A=R\theta$$ • $$D=2R\sin{(\frac{A}{2R})}$$ • $$P=R-\sqrt{R^2-(\frac{D}{2})^2}=R(1-\sqrt{1-(\frac{D}{2R})^2})$$ To start, we must make our first, most useful substitution: $$R=\frac{A}{\theta}$$. Swapping this out in our definition of $$D$$ and our formula for $$P$$, we get • $$D=\frac{2A}{\theta}\sin{(\frac{\theta}{2})}$$ • $$P=\frac{A}{\theta}(1-\sqrt{1-(\frac{D\theta}{2A})^2})$$ Plugging our new formula for $$D$$ into our one for $$P$$, we get • $$P=\frac{A}{\theta}(1-\sqrt{1-\sin^2{(\frac{\theta}{2})}})$$ Here's where everything comes into play. If, as we said, $$L\parallel L'$$, then the triangle with sides $$L', L',$$ and $$W$$ will be similar to the triangle with sides $$L+L', L+L',$$ and $$D$$. If these two triangles are similar, then the following relation will also be true: • $$\frac{R-H}{R-P} = \frac{W}{D}$$ Expanding this out, we get • $$\frac{\frac{A}{\theta}-H}{\frac{A}{\theta}-\frac{A}{\theta}(1-\sqrt{1-\sin^2{(\frac{\theta}{2})}})} = \frac{W}{\frac{2A}{\theta}\sin{(\frac{\theta}{2})}}$$ This can be massively simplified if we multiply both sides by $$\frac{\theta}{\theta}$$. Doing so yields us • $$\frac{A-H\theta}{A\sqrt{1-\sin^2{(\frac{\theta}{2})}}} = \frac{W\theta}{2A\sin{(\frac{\theta}{2})}}$$ We can rearrange this to give us • $$\frac{A-H\theta}{W\theta}=\frac{\sqrt{1-\sin^2 (\frac{\theta}{2})}}{2\sin{(\frac{\theta}{2})}}$$ And now, squaring each side, we see • $$(\frac{A-H\theta}{W\theta})^2 = \frac{1-\sin^2 (\frac{\theta}{2})}{4\sin^2{(\frac{\theta}{2})}}$$ It is at this point I must concede that I don't know how you would solve for $$\theta$$ here, or if, in fact, you can solve for $$\theta$$. However, if you can, then the solution follows very naturally. If $$\theta = f(A,H,W)$$, then very simply, the width of the shape will be $$D=\frac{2A}{f(A,H,W)}\sin(\frac{f(A,H,W)}{2})$$. I understand that this does not solve the problem in general and only solves it if a certain requirement is met, but at the very least, I hope that this can help you get on the right path. • Thank you! I was hoping for a general solution but I'll study yours and learn some things for sure. Feb 8 at 6:42 • An update after thinking on it some more - that last equality is actually equal to $(\frac{A-H\theta}{W\theta})^2 = \frac{1}{4}\cot^2{(\frac{\theta}{2})}$, which further simplifies down to $\frac{A-H\theta}{W\theta} = \frac{1}{2}\cot{(\frac{\theta}{2})}$. This looks easier to solve, but I'm still not sure how to approach it. Feb 8 at 13:23 I am making some assumptions about your diagram with this solution. First of all, that it is symmetric about $$H$$ and second that the lines $$L$$ are normal to the arc $$A$$, that is, they are radial lines. Here are the basic relations you need: The arc $$s=A=R\theta$$, where $$R$$ is the radius of the arc and $$\theta$$ is the angle of the sector (both as yet unknown). The unknown chord $$c=2R\sin(\theta/2)$$. Extending the lines $$L$$ so that they intersect at the origin for the arc, we'll have two similar triangle for which we can say $$\frac cW=\frac{R}{R-L}$$ Substituting the above relations we can find the following equation for $$\theta$$ in terms of known quantities, to wit $$2R\sin\frac{\theta}{2}=\frac{RW}{R-L}\\ 2\sin\frac{\theta}{2}\big(\frac{s}{\theta}-L\big)=W$$ You will have to solve for $$\theta$$ by iteration, then get $$R=A/\theta$$ and $$c=2R\sin(\theta/2)=2A\sin(\theta/2)/\theta$$. I can't post replies to comments because I am a new user, but in the diagram in the OP: • Line H is perpendicular to W • Line H bisects W, Unknown and A • Line W and Unknown are parallel • Arc A is a circular arc • The L lines are not along the Radii of A, but they are symmetrical. (The shape is mirrored along H) Diagram Here is a working version of the calculator if anyone is interested: https://jbiemans77.github.io/CurvedShapeCalculator/CurvedShapeCalculator/index.html And the source code: https://github.com/jbiemans77/CurvedShapeCalculator My solution was a little less orthadox but I thought I would share to see if it could spark someone else. I couldn't find a way to make a forumla that worked, so instead, I used a trial and error approach. Unfortunately, because L is not on a radius line of A, I can't use the solutions above. I created a triangle on the corner and made a guess at the angle (starting at 45). I then used the information I had to calculate the unknown and the arc length. I compared my arc length to the A value I was looking for, and then kept adjusting the angle until I got a matching arc length. Once I did, I found the unknown. It usually took the program between 10-20 attempts before it could narrow it down to the correct answer. I can share the code if anyone wants. The basic math though was: //Use the triangle to find the height of the arc and length of the unknown P (Height Of Arc) = H - (L * (Sin(90-aA) / Sin(90) )) U (Cord Length - Unknown) = W + (L * (Sin(aA) / Sin(90) )) * 2 //Use the Cord length and height of arc to find the radius and the arclength RA = (((U * U) / (P * 4)) + P) / 2 A = ((Asin(U / (RA * 2))) * 2) * RA ( I tried to simplify it out to make it clean. The calculation part of the code is:) float topAngle = angleToTry; float bottomAngle = 90-topAngle; float topAngleRadians = ConvertDegreeToRadians(topAngle); float hypotenuseC = lengthOfSides; float heightOfArc = heightOfShape-heightOfTriangle; float cordLength = lengthOfBottom + baseOfTriangle + baseOfTriangle; finalWidthOfPiece = cordLength; float radiusOfArc = (((cordLengthcordLength) / (heightOfArc4)) + heightOfArc) / 2; float arcLength = ((Mathf.Asin(cordLength / (radiusOfArc * 2))) * 2) * radiusOfArc; Now that I am advised that the $$L$$'s are not radial to the arc, I'll take a different view. This is not a solution, per se, but is too long for a comment. So, the unknown chord, $$c$$, can be expressed in terms of both the arc and the trapezoid. Let's define $$\alpha$$ as the obtuse angle ($$LW$$) of the trapezoid, $$R$$ as the radius of the arc, $$h$$ as the height of circular segment, and $$\theta$$ as the angle of the circular sector. Now, for the circular arc, $$c=2R\sin\frac{\theta}{2}=2h\tan\frac{\theta}{2}$$ and for the trapezoid, $$c=W+2L\cos(\pi-\alpha)=W-2L\cos\alpha,\quad \alpha\in (\pi/2,\pi)$$ We can express the segment height in terms of $$H,L,\alpha$$, $$h=H-L\sin(\pi-\alpha)=H-L\sin\alpha$$ Combining the above, we find $$\tan\frac{\theta}{2}=\frac{W-2L\cos\alpha}{2(H-L\sin\alpha)}$$ This is as far as I can go without specific values for $$L,W,H$$, but it suggests that there are multiple possible solutions. Edit: Without the assumption that the $$L$$ segments are along the radius of the arc, the given information is not enough to identify a unique solution to the problem. See an example in the figure below. In this example, for ease of verification, the angles of the arcs in the large and small green circles are $$\frac{\pi}{4}$$ and $$\frac{\pi}{2}$$, respectively, and the proportion of their radii is $$\frac21$$. Therefore it is easy to see that the two green arcs are of equal length. The radii of the dashed blue circles are $$L$$. It can be seen that there are an infinite number of solutions to this problem. On the other hand, with the assumption that the $$L$$ segments are along the radius of the arc (thank you @DavidK and OP) the problem can be solved as described in my pre-edit answer: Extend the sides to meet at point $$O$$. This is the center of the circle that includes your arc. Let the radius of this circle be $$R$$ and the angle between the two radii be $$\alpha$$. Then $$\sin \frac{\alpha}{2} = \frac{\frac W2}{R-L}$$ $$\cos \frac{\alpha}{2} = \frac{R-H}{R-L}$$ (Why?) From here solve the equation $$\sin^2\frac{\alpha}{2} + \cos^2\frac{\alpha}{2} = 1$$ to find $$R$$. Can you take it from here? • You assume that the segments labeled $L$ lie along radii of the arc. I see nothing in the problem statement to guarantee that this is true. Feb 7 at 4:22 • @DavidK Good observation, thanks. Feb 7 at 19:32 • There is a comment under the problem about $W$, $H$ and $A$ being in the center, which was confirmed by the OP. Reading that, I jumped to the assumption that the orange segment is the perpendicular bissector of segment $W$. Feb 7 at 19:41 • Actually I'm fairly sure that the orange segment is the perpendicular bisector of the bottom edge. But the question is the angle between the side edges and the arc. Suppose you have $A,L,W,H$ so that the $L$ segments are radii of the arc. Reduce $A.$ The figure still exists, but the edges $L$ now end inside the old figure and the arc has a smaller radius. The center of the arc has moved upward but the intersection of the $L$ lines (when extended) has moved downward, so those segments can no longer lie on radii. Feb 8 at 2:46 • It's true that I was hoping for a solution that does not constrain the angle where the arc meets the sides at all. But I'm learning from these solutions! Feb 8 at 6:44 Cye Waldman got me thinking about this in a slightly different way. I was always focused on solving one shape before the other, but I should have been thinking of both shapes at the same time. The following mess works to solve the arc length given $$W$$, $$L$$, $$H$$ and the top angle of the triangle (I labeled it $$a_B$$ on my previous response.) However, in this case, we don't have $$a$$ (the angle), but we do have the other side $$A$$ (the arc length). Is there any way to isolate a in the equation below to solve for it given $$L$$, $$W$$, $$H$$ and $$A$$? Or is it way too much of a mess to do so? $$A = \sin^{-1}{\!\left(\frac{W + L\sin{\!(a)} + L\sin{\!(a)}}{\left(\frac{(W + L\sin{\!(a)} + L\sin{\!(a)})(W + L\sin{\!(a)} + L\sin{\!(a)})}{4(H - L\sin{\!(90^{\circ}-a)})} + H - L\sin{\!(90^{\circ}-a)}\right) / 2 \cdot 2}\right)} \cdot 2\left(\frac{(W + L\sin{\!(a)} + L\sin{\!(a)})(W + L\sin{\!(a)} + L\sin{\!(a)})}{4(H - L\sin{\!(90^{\circ}-a)})} + (H - L\sin{\!(90^{\circ}-a)})\right) / 2$$ I was able to clean it up a bit with the help of a math site: $$A = \sin^{-1}{\!\left(\frac{W + 2L\sin{\!(a_B)}}{H + \frac{(W + 2L\sin{\!(a_B)})^2}{4H + 4L\sin{\!(a_B - 90^{\circ})}} + L\sin{\!(a_B - 90^{\circ})}}\cdot\left(H + \frac{(W + 2L\sin{\!(a_B)})^2}{4H + 4L\sin{\!(a_B - 90^{\circ})}} + L\sin{\!(a_B - 90^{\circ})}\right)\right)}$$ • Please review the way I've used MathJax to format your question and use similar techniques in the future. Feb 15 at 1:36 • Thank you, I couldn't figure out how to display it like that. I will try to learn more before future posts. Feb 15 at 13:15 • @JacobManaker, do you have access to the initial equation I posted still ? I think there might be a formatting error in the MathJax version. Thank you. Feb 25 at 14:47 • If you click on "edited Feb 15 at 1:35" it will show your original version and the changes I made. (That's what I meant by "review"; my comment may have seemed a little critical of you if you didn't know that, for which I apologize.) Feb 25 at 20:44 • Thank you, I didn't know how to find that information. It turned out that my "cleaned up" version, had some errors in it, and when you fixed the formatting it made that error much more apparent. I just trusted the software that I ran it through without first double checking that it simplified it correctly. Also don't worry about the comment, I am glad that it was critical because I do need to learn. It was constructive criticism and it was also done with the corrections (not just telling me to fix it), so as far as I am concerned it was well done. Feb 28 at 15:34	2022-09-25T21:23:46	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4375095/width-of-this-shape", "openwebmath_score": 0.8001608848571777, "openwebmath_perplexity": 287.77765260552144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307716151472, "lm_q2_score": 0.8633916029436189, "lm_q1q2_score": 0.8399339192976796 }
https://math.stackexchange.com/questions/2091094/prove-if-f-is-positive-and-increasing-on-a-b-then-l-n-%E2%89%A4-a-%E2%89%A4-r-n-riemann	Prove if f is positive and increasing on $[a, b]$ then $L_n ≤ A ≤ R_n.$ (riemann sum) Prove if f is positive and increasing on $[a, b]$ then for all $n\ge 0$ we have $L_n \le A \le R_n$. (Riemann sum) Let $A$ denote the actual area. Let $L_n$ denote the left Riemann sum. Let $R_n$ denote the right Riemann sum. So far what I did: $$\int_{a}^{b} f(a) dx \leq \int_{a}^{b} f(x) dx \leq \int_{a}^{b} f(b)$$ But after this I am stuck on the proof? 2 Answers Just write the definitions $$L_n = \sum_{i=0}^{n-1}f(x_i){b-a\over n}\le \sup_m L_m = A= \inf_m R_m \le \sum_{i=1}^n f(x_i){b-a\over n}$$ here the $\sup=\limsup$ by monotonicity of $f$. and if you're not familiar with the $\sup$ notation, then try this $$L_n =\sum_{i=0}^{n-1}f(x_i){b-a\over n}=\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}}f(x_i)\,dx \le \sum_{i=1}^{n-1}\int_{x_i}^{x_{i+1}}f(x)\,dx$$ $$= \int_a^b f(x)\,dx$$ $$=\sum_{i=1}^n\int_{x_{i-1}}^{x_i}f(x)\le \sum_{i=1}^n\int_{x_{i-1}}^{x_i}f(x_i)\,dx =\sum_{i=1}^nf(x_i){b-a\over n}=R_n$$ • Does the downvoter want to explain what he/she thinks is wrong with my answer? – Adam Hughes Jan 10 '17 at 0:22 • Is the first claim clearly true? I can see that $L_n \le \sup_m L_m$, and for an integrable $f$, $\limsup_m L_m = A$, but I think that going from $\sup_m L_m$ to $\limsup_m L_m$ possibly requires a little more justification. (same comment w.r.t. the upper bound). – πr8 Jan 10 '17 at 0:47 • @πr8 monotonicity means $\limsup = \sup$. I've edited this detail in. – Adam Hughes Jan 10 '17 at 0:51 • The function is monotone, but is it clear that the upper/lower sums are monotone as a result of this? Will change to an upvote in any case - was hasty to downvote, and the remainder of your answer is exemplary. – πr8 Jan 10 '17 at 1:01 For partition $x_0<x_1<\cdots<x_n$ Left Riemann sum: $= \sum_\limits{i=1}^{n} f(x_{i-1})(x_{i}-x_{i-1})$ Right Riemann sum: $= \sum_\limits{i=1}^{n} f(x_i)(x_i-x_{i-1})$	2019-06-20T07:03:03	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2091094/prove-if-f-is-positive-and-increasing-on-a-b-then-l-n-%E2%89%A4-a-%E2%89%A4-r-n-riemann", "openwebmath_score": 0.9433050751686096, "openwebmath_perplexity": 555.6926573220067, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307676766119, "lm_q2_score": 0.8633916047011595, "lm_q1q2_score": 0.8399339176069709 }
https://math.stackexchange.com/questions/2048469/how-to-prove-that-a-function-is-affine/2048484	# How to prove that a function is affine? I am trying to understand the concept of affinity of functions. First, I thought that every affine function has to be a linear function, too, because my teacher's notes define linear and affine functions as follows: $$T(\sum_{i=0}^n \alpha_iu_i) = \sum_{i=0}^n\alpha_iT(u_i)$$ is a linear function. An affine function is defined as $T(\sum_{i=0}^n \alpha_iu_i)$ with $\sum_{i=0}^n \alpha_i = 1$ and the above condition of a linear function. Then, I found the example of $f(x) = 2x + 3$ which is an affine function but not linear which is pretty confusing to me (I understand why it is not linear, but have no clue as to why it is affine according to the definitions). I also have to solve a problem such as: $$T: \mathbb{R} \to \mathbb{R}, T(x,y,z) := (x − z + 1, y - 5, z - y, 2)$$ but I really have no idea how to proof if it's affine or not. There are no $\alpha$ and that function is not linear, so I am kind of stuck here. I appreciate any sort of help, like links to websites or anything that helps me to understand this because I have no strategy to solve this problem. • Every linear function is affine but not the other way around. Think of an affine function as a linear function plus a constant. – copper.hat Dec 7 '16 at 18:53 A function is affine iff $$T(\lambda x + (1-\lambda) y) = \lambda T(x) + (1-\lambda) T(y)$$ for all $$x,y$$ and $$\lambda \in \mathbb{R}$$. It is straightforward to show that the above definition is the same as the one in the question. You can use this to answer your question by checking that $$T(\lambda (x_1,y_1,z_1) + (1-\lambda)(x_2,y_2,z_2)) = \lambda T((x_1,y_1,z_1)) + (1-\lambda)T((x_2,y_2,z_2))$$ for any scalar $$\lambda$$. Also, note that if $$T$$ is affine and we let $$L(x) = T(x)-T(0)$$, then $$L(\lambda x) = T(\lambda x) - T(0) = T (\lambda x + (1-\lambda) 0) - T(0) =\lambda T(x)+(1-\lambda) T(0) - T(0) = \lambda L(x)$$, and $$L(x+y) = 2 L({1 \over 2} (x+y))= 2 T({1 \over 2} (x+y)) - 2T(0)= 2( {1 \over 2}(T(x)+T(y)))-2 T(0) = L(x)+L(y)$$. Hence $$L$$ is linear, and we can write $$T(x) = T(0)+L(x)$$. It is easy to check that if $$L$$ is linear and $$c$$ is a constant vector, then $$T(x) = L(x)+ c$$ is affine. So we see that $$T$$ is affine iff it can be written as a constant plus a linear function. In your case, we see that we can write $$T((x,y,z)) = A (x,y,z)^T + b$$ for some matrix $$A$$ and some vector $$b$$. • Ok, I think I am getting the hang of it. But just to be clear: the function is not linear so this approach would pretty much end up with the same result, right? If multiplying with a scalar proofs that it is not a linear function, following this approach, it would also proof that it is not affine, right? – Urmir Dec 7 '16 at 19:05 • The function you have is not linear (it is affine). The above means you just need to check for two $u$s, not $n$ of them... – copper.hat Dec 7 '16 at 19:06 • So I randomly choose six values plus an $\lambda$ and check if the equation holds for these values? – Urmir Dec 7 '16 at 19:16 • No, there's no randomness. Compute the three expressions $T(\lambda (x_1,y_1,z_1) + (1-\lambda)(x_2,y_2,z_2))$, $T((x_1,y_1,z_1))$, and $T((x_2,y_2,z_2))$ and check that the above equality holds. – copper.hat Dec 7 '16 at 19:18 • So what I did is $$T(\alpha (x_1,y_1,z_1) + (1-\alpha)(x_2,y_2,z_2)) = ((\alpha x_1 + (1-\alpha)(x_2))-(\alpha y_1 + (1-\alpha)y_2)+2, ...$$ and so on for the equation in my question above. I ended up with $\alpha(x_1-x2+2, ...)$ and $(1-\alpha)(y_1-y_2+2, ...)$. Is that path to the solution? – Urmir Dec 7 '16 at 19:46 Every affine function is a linear function "plus a constant vector". That is, for every affine function $A(x)$, there is a vector $v$ and a linear function $T(x)$ such that $A(x)=T(x)+v$. What is the difference between linear and affine function	2021-03-07T19:15:34	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2048469/how-to-prove-that-a-function-is-affine/2048484", "openwebmath_score": 0.9064493179321289, "openwebmath_perplexity": 85.36728471617997, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307668889048, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.839933916926871 }
https://math.stackexchange.com/questions/2501158/prove-that-there-is-only-a-single-point-of-minimum-distance-for-n3-points	# Prove that there is only a single point of minimum distance for $N>3$ points We want to mimimize the sum of distances from $n$ distinct points. Prove that there exists only one such point for $n>3$ if all the $n$ points lie on a single plane (and not on a single line) The problem seems quite tough, but might possess an elegant solution. I tried creating two PDE's (partial differential equations) for $x$ and $y$ coordinates and couldn't find anything fruitful. Might be something towards vectors and their sum. • Would you care to explain the statement ''Prove that there exists only one such point for n>3 on a single plane.'' It is not clear to me what you mean. You could minimize the sum of those distances, but the unclear part is ''on a single plane'' – John D Nov 2 '17 at 10:38 • @Magnusseen i meant all points lie on a single plane – avz2611 Nov 2 '17 at 10:40 • @JohnWatson Partial Differential Equations – avz2611 Nov 2 '17 at 10:40 • I think some confusion is created by the wording, hence the questions. But as I understand it, you are given a fixed number of points, say $k$, on a plane and you seek to show that there exists a unique point on the plane, different from these $k$ points, such that the distance of it from all these $k$ points is minimum. – MathematicianByMistake Nov 2 '17 at 10:46 • Glad to clarify. I suggest you edit because it seems-to me at least-as a very nice problem. +1 – MathematicianByMistake Nov 2 '17 at 10:49 lemma:suppose $A,B,C$ are three distinic points on plane , and $M$ is the middle point of $BC$, then $AM \le {{AB+AC} \over 2}$ proof: if three points $A,B,C$ lie on a line it is clear the condtion is true, now consider the case which three point create a triangle , suppose $A_1$ be the symmetry point of $A$ into $M$ then quadrilateral $ABA_1C$ diameters , cuting each other in to half , so it is a parallelogram.please note that $AA_1=2AM$ and $AB=CA_1$ now accordin to the triangle inequality in the $AA_1C$ we have $AA_1 <AC+CA_1$ thus $2AM<AB+AC$ and it is proved. now lets solve problem using this lemma: suppose ${A_1,A_2,A_3,....,A_n}$ are $n$ distinic point and suppose there exist two point that has minimumthe sum of distances from $n$ distinct points, call them $X$ and $Y$. call the middle point of $X$ and $Y$ the $M$,now we are using above lemma in $A_1XY$ and $A_2XY$ and ... $A_nXY$ (please note that even if three point lies on a line then lemma is true): $A_1M \le {{A_1X+A_1Y}\over 2}$ $A_2M \le {{A_2X+A_2Y}\over 2}$ ..... $A_nM \le {{A_nX+A_nY}\over 2}$ add all of them ,$A_1M+A_2M+...+A_nM \le { {A_1X+...+A_nX+A_1Y+...+A_nY} \over 2}$ but since all the points is not on a single line so at least one of the inequalitys is strict inequality thus $A_1M+A_2M+...+A_nM < { {A_1X+...+A_nX+A_1Y+...+A_nY} \over 2}$, and since $X$ and $Y$ has minimum sum,call it H, so $A_1M+A_2M+...+A_nM < H={2H\over 2}={ {A_1X+...+A_nX+A_1Y+...+A_nY} \over 2}$ so the point $M$ is the desired point which is contradiction.(note you should add all the point is not on a single line other wisee there exist two points) • Oh my goodness i forgot to mention that thanks for mentioning this! – avz2611 Nov 2 '17 at 11:39 • i really like this solution, clean and simple to understand +1! – avz2611 Nov 2 '17 at 14:33 Hints & References You are seeking to prove that the Geometric median is unique. This holds true only if the points are not colinear, an assumption that should be added at the description. This paper-"The multivariate L1-median and associated data depth"-presents a generalized approach. This answer on-"How to find out Geometric Median" on stackoverflow provides an algorithm for finding such a point. Also this post-"Geometric median (or Fermat-Weber problem), including continuous case"-here on MSE might interest you- Here is your sum $$S(r) = \sum_{k=1}^n \lVert r - r_k \rVert$$ where $r, r_k \in \mathbb{R}^N$. The $i$-th component of the $r$-gradient of $S$ in the Euclidean norm is \begin{align} \partial_i S(r) &= \partial_i \sum_{k=1}^n \lVert r - r_k \rVert \\ &= \partial_i \sum_{k=1}^n \sqrt{\sum_{j=1}^N \left(x_j - x_{k,j} \right)^2} \\ &= \sum_{k=1}^n \frac{1}{2}\lVert r - r_k \rVert^{-1} \sum_{j=1}^N 2(x_j-x_{j,k})\delta_{ij} \\ &= \sum_{k=1}^n \frac{x_i-x_{k,i}}{\lVert r - r_k \rVert} \end{align} So critical points $r$ of $S$ (which are candidates for extrema) fulfill the condition $$\sum_{k=1}^n \frac{r-r_k}{\lVert r - r_k \rVert} = 0$$ This seems to be the characterization of the geometric median (Wikipedia article). The Wikipedia article links to Vardi & Zhang (2000) for a uniqueness proof, which was already cited by MathematicianByMistake. • I dont get it why does this prove uniqueness – avz2611 Nov 2 '17 at 11:36 • It does not. It just provides the starting point, why one lands at the geometric median – mvw Nov 2 '17 at 13:04 Suppose the minimum sum occurs at two points, $P$ and $Q$. Let $M$ be the midpoint of the line segment $PQ$. For all points not on the line $PQ$, the distance to $M$ is strictly less than the average of the distance to $P$ and $Q$. Since it's assumed that not all $n$ points are on a single line, the sum of their distances to $M$ is strictly less than the average of their sums to $P$ and $Q$ -- which is to say, strictly less than the minimum sum. That's a contradiction, hence the mimimum sum occurs at a unique point. Remark: As Nick Pavlov points out, this answer essentially duplicates Dexpectra's. That answer contains a nice proof of the key observation, that the distance to $M$ is less than the average of the distances to $P$ and $Q$ (by viewing $M$ as the center of a parallelogram with $P$ and $Q$ as oppositve vertices). • this might be worded more succinctly than Dexpectra's answer, but is otherwise exactly the same; perhaps we need a [duplicate] flag for answers now, too. – Nick Pavlov Nov 2 '17 at 12:28 • @NickPavlov, I have to admit, I glanced at Dexpectra's answer but didn't read it. – Barry Cipra Nov 2 '17 at 12:31	2019-07-20T09:25:26	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2501158/prove-that-there-is-only-a-single-point-of-minimum-distance-for-n3-points", "openwebmath_score": 0.9601722359657288, "openwebmath_perplexity": 438.87194875546265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307668889047, "lm_q2_score": 0.8633916011860785, "lm_q1q2_score": 0.8399339135072921 }
http://ucpd.aoru.pw/p-series-calculus.html	# P Series Calculus In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. 8 Infinite Series 8. The course emphasizes not just getting answers, but asking the question "why is this true?". 8 Taylor and Maclaurin Series Chapter 8 Review Chapter 8 Group Research. n th-Term Test for Divergence If the sequence {a n} does not converge to zero, then the series a n diverges. ANALYTIC GEOMETRY & CALCULUS II Textbook or ebook: thCalculus – 10 Edition by Larson and Edwards, Cengage Learning Publisher. Alternating Series Test: If the series has the form 1 n a n, then the series converges if 0 aa nn 1 (decreasing terms) for all n, for some n, and lim 0 n n b. 8 where Cis the number of correct responses and Iis the number of incorrect responses. (These courses have been audited and approved by the College Board to carry the AP designation and have been acknowledged to provide students with a college. Math 132 - Calculus II - Course Outline (Spring, Fundamental Theorem of Calculus Part I. Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{n^2}{n^4}$ is convergent or divergent. 17Calculus - You CAN ace calculus. If p > 1 then the sum of the p-series is ζ(p), i. We had a guess for pi: somewhere between 2. 4 The Comparison Tests 9. Calculus I and II). Calculus X is the most effective study tool for calculus at the AP or intro college level. Power Series i. The worked-out solutions give students a way to check their answers, ensure that they took the correct steps to arrive at an answer, and help them understand how to solve even the toughest problems. Herep is the parameter. You will need to find one of your fellow class mates to see if there is something in these notes that wasn't covered in class. Course Description AP Calculus has two levels, Calculus AB and Calculus BC. 5 The Ratio Test and the Root Test 8. One of them contains the terms of the series, represented by arrows. Free series convergence calculator - test infinite series for convergence step-by-step. Limit comparison test with a p-series calculus 2? Use the limit comparison test with a p-series to determine whether the following series are convergent or divergent. 6 Alternating Series; Absolute and Conditional Convergence 8. You may want to review that material before trying these problems. The 1st Fundamental Theorem of Calculus is an extremely important theorem that allows us to find the area under a curve over an interval. Successful students will earn 3 credit hours. We will now look at some examples of specifically applying the p-Series test. Se você está vendo esta mensagem, significa que estamos tendo problemas para carregar recursos externos em nosso website. A 15 100 1 1,lim nn54 n 5 5 ss →∞ ⎛⎞+n ==⎜⎟ ⋅1= ⎝⎠+n C I. Distinguishing features of the course include: 1) the introduction and use of Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus;. Again, Excel can be used to demonstrate this behavior. Math 112 Worksheet 4: Series Convergence Testing I Instructions: Determine whether the following series converge or diverge. This calculator will find the sum of arithmetic, geometric, power, infinite, and binomial series, as well as the partial sum. We also use the fact that sin T / T approaches 1 when T approaches 0. It is part of the unit on Infinite Series, (Chapter 9 in most of the Larson textbooks. In addition, L'Hôpital's rule, improper integrals, and using partial fractions to integrate rational functions are commonly. The following exercises test your understanding of infinite sequences and series. Evaluating Limits Analytically. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. Now is the time to redefine your true self using Slader's free Stewart Calculus answers. p-series Series of the form X1 np, where pis a constant power, are called p-series. instructions accordingly as quizzes and tests in Honors AP Calculus AB will have two parts: no calculator allowed and graphing calculator permitted. So we can say the p-Series converges if, and only if, this integral right over here converges. It is suitable for a one-semester course, normally known as "Vector Calculus", "Multivariable Calculus", or simply "Calculus III". Recall the p-Test: Regardless of the value of the number p, the improper integral is always divergent. 3 Direct Comparison, Limit Comparison, & Ratio Test. Note: The harmonic series is a p-series with p =1. 11 Suppose f is a continuous, positive, decreasing function for x z: 1 and let. The p-series test determines the nature of convergence of a p-series as follows: The p-series converges if and diverges if. Each question is followed by a "Hint" (usually a quick indication of the most efficient way to work the problem), the "Answer only" (what it sounds like), and finally a "Full solution" (showing all the steps required to get to the right answer). 1 n=1 ( 1) n b. So x = 6 must be included in the interval of convergence. Solved Example: Consider the series. Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{n^2}{n^4}$ is convergent or divergent. AP CALCULUS BC Section Number: 9. When p 1, the series is the divergent Harmonic series. Marginal cost, marginal revenue, and marginal profit. 3 Direct Comparison, Limit Comparison, & Ratio Test. Binomial series ( ) 1 2 2 3 3( ) ( )( ) 1. A series, which is not a list of terms like a sequence, is the sum of the terms in a sequence. So in this particular case, our p, for this p-series, is equal to five. (Reprinted in Selected Papers on Calculus, The Mathematical Association of America, 1968, 353-354. p a p a p a a a a p n c n n c n n n n n n n c n No Conclusion, when diverges, when converges, when if then n n 8. Alternating Series remainder For a convergent alternating series, the absolute value of the remainder in approximating the sum with the first n partial sums is less than or equal to the value of the first neglected term. Parallel Cross Sections. However their convergence or divergence depends on the denominator's exponent, p. This section introduced us to series and defined a few special types of series whose convergence properties are well known: we know when a $$p$$-series or a geometric series converges or diverges. Problems to look at. The latest versions may be found by. (Reprinted in Selected Papers on Calculus, The Mathematical Association of America, 1968, 353–354. Then generalize to define an entire class of series called p-series, and prove a theorem showing when they converge. X1 n=1 n3 n4 1 2. CONICS, PARAMETRIC EQUATIONS, AND POLAR COORDINATES. In this part of the exercise we will see how to use information about P 1 k 2 to determine information about P 1 k 2+k. We know that this series is convergent because it is a p-series. 3 # 1,17,21,25,35,59 Vocabulary: p-series Study Questions: When should you use the integral test to check the converge or divergence of a series? Prove the p-series test using the integral test?. ) [3] Rick Kreminski, Using Simpson’s Rule to Approximate Sums of Inﬁnite Series, The College. 3 Day 1 - Packet 1, 7, 14, 17, 28; 10. ) Summarizing the above work, we know that 4 is not included, but 6 is. Let's make our guess better. View Notes - Section 9. The series ($\sum \frac{1}{n}$), known as the harmonic series, diverges by to the p-series test. Recall the p-Test: Regardless of the value of the number p, the improper integral is always divergent. Diverges by limit comparison with harmonic series. MATH 2414 - Calculus II Summary of Series Facts Geometric Series diverges. Late transcendentals and multivariable versions are also available. So this would be the first term, in this p-Series, this would just be an area of one. 10· 9· 8· 7. Theorem (Monotonic Sequence Theorem). Close with the sum of the harmonic series, the fascinating Euler-Mascheroni constant, which is not known to be rational or irrational. Strategy for Testing Series: Solutions 1. Proof - Convergence of a p-Series Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Arithmetic and Geometric Series Definitions: If a = 0 the series is often called a Maclaurin series. We know that the series P 1 k 2 is a p-series with p = 2 > 1 and so P 1 k 2 converges. When p= 1, the p-series is the harmonic series which we know diverges. Each question is followed by a "Hint" (usually a quick indication of the most efficient way to work the problem), the "Answer only" (what it sounds like), and finally a "Full solution" (showing all the steps required to get to the right answer). To be honest, many students will never see series outside of their calculus class. (In particular, if p > 1, then the graph is concave up, such as the parabola y = x2. Convergence of Infinite Series. 3 The Integral Test; p-series 8. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. AP Calculus BC Course Syllabus Resources Larson, Ron, Robert P. X1 k=2 1 4k2 The series is a constant multiple of a p-series with p= 2 >1, so the series converges. 2 Linear Models and Rates of Change [11] P. p-series Test File. December 17, 2016 Final Exam Math 162 (Calculus IIA) and the original integral is Z arcsinxdx = xarcsinx+ p 1 x2 +C: 4. Calculus Test One Section One Multiple-Choice No Calculators Time—30 minutes Number of Questions—15. This calculus course covers differentiation and integration of functions of one variable, and concludes with a brief discussion of infinite series. PERMUTATIONS AND COMBINATIONS. Download free-response questions from past exams along with scoring guidelines, sample responses from exam takers, and scoring distributions. Problem Solving. Calculus II Calculators; Math Problem Solver (all calculators) Series and Sum Calculator. nth-Term Test for Divergence. Below is a checklist through which one can run a given series to try to determine its convergence classification. Series is one of those topics that many students don’t find all that useful. Who Am I 4. Moreover, we have. In other words, series is an important topic even if you won’t ever see any of the applications. Combination problems. X1 n=1 n3 n4 1 2. * Series of constants + Motivating examples, including decimal expansion + Geometric series with applications + The harmonic series. The course is intended to be challenging and demanding. (b) n-th Term Test for Divergence: If. I Leave out the theory and all the wind. 17Calculus - You CAN ace calculus. Chapter P Preparation for Calculus [1] P. AP CALCULUS AB/BC I. In an Arithmetic Sequence the difference between one term and the next is a But a sum of an infinite sequence it is called a "Series". A p-series can be either divergent or convergent, depending on its value. Recall the p-Test: Regardless of the value of the number p, the improper integral is always divergent. The course teaches all topics associated with Functions, Graphs, Limits, Derivatives, Integrals, Polynomial Approximations, Series, Parametric and Polar Equations. Moreover, we have. Thisis better example to do the Converges. Sigma notation, divergent series, convergent series. Equal Opportunity Notice The Issaquah School District complies with all applicable federal and state rules and regulations and does not discriminate on the basis of sex, race, creed, religion, color, national origin, age, honorably discharged veteran or military status, sexual orientation including gender expression or identity, the presence of any sensory, mental or physical disability, or. Wolfram Demonstrations Project. Since the harmonic series diverges, these. We also use the fact that sin T / T approaches 1 when T approaches 0. Theorem (Monotonic Sequence Theorem). AP Calculus Questions Similar to AB Exams Free AB Calculus Test Practice Questions with Answers - Sample 1. We will now look at some examples of specifically applying the p-Series test. In this fifth part--part five of five--we cover a calculus for sequences, numerical methods, series and convergence tests, power and Taylor series, and conclude the course with a final exam. When p= 1, the p-series is the harmonic series which we know diverges. By using higher derivatives, the idea of a tangent line can be extended to the idea of polynomials of higher degree which are "tangent" in some sense to a given curve. A power series is an infinite series. AP Calculus BC - Sequences and Series Chapter 11- AP Exam Problems solutions 1. You Can Turn Your Calculus Grade Around. Therefore, our series is absolutely convergent (and therefore convergent). Section 11. The course teaches all topics associated with Functions, Graphs, Limits, Derivatives, Integrals, Polynomial Approximations, Series, Parametric and Polar Equations. ANALYTIC GEOMETRY & CALCULUS II Textbook or ebook: thCalculus – 10 Edition by Larson and Edwards, Cengage Learning Publisher. Integral P Series Homework Link Follow the above link for file with integral test and P series test problems. An unanswered question earns. Side note: most of the BC exam is AB, so if your AB knowledge is good don't worry too much, just learn the new thing. 1 - Area Between Curves. Processing is an open source programming language and environment for people who want to create images, animations, and interactions. The sample tests are just to give you an idea of the a general idea of the topics covered, the level of difficulty, how questions may be worded and, if solutions are provided, what is the acceptable level of detail required in the solutions. Test and Worksheet Generators for Math Teachers. A Series Test Gauntlet. MATH 2414 - Calculus II Summary of Series Facts Geometric Series diverges. Here is another type of series that may converge or diverge dependent upon the value of a parameter. Proof of the method If P(1) is OK, then we can use this to deduce that P(2) is true and then use this to show that P(3) is true and so on. This is a challenging sub-section of algebra that requires the solver to look for patterns in a series of fractions and use lots of logical thinking. Free series convergence calculator - test infinite series for convergence step-by-step. At the end of the PayPal checkout, you will be sent an email containing your key and download instructions. Is the series similar to a $$p$$-series or a geometric series? If so, try the Comparison Test. Note as well that there really isn't one set of guidelines that will always work and so you always need to be flexible in following this set of guidelines. com: Calculus with Analytic Geometry, Alternate (9780395889022) by Ron Larson; Robert P. com Functions READ: Linear Functions Find Slope Find y=mx+b All-in-one-Function Explorer Evaluate Function Find Domain of f(x) Find Range of f(x) Intersection of 2 Functions Composition of 2 Functions f(g(x)) Do the Quadratic Equation Complete the Square. Now you might immediately recognize this as a p-series, and a p-series has the general form of the sum, going from n equals one to infinity, of one over n to the p, where p is a positive value. 2 Introduction to Infinite Series; Geometric Series 8. Its sum is nite for p>1 and is in nite for p 1. Does the celebrated harmonic series diverge or converge? Discover a proof using the integral test. 32 min 3 Examples. If p=1, we call the resulting series the harmonic series: By the above theorem, the harmonic series does not converge. Enter CONV for convergent, DIV for divergent , and the value of p. As a known series, only a handful are used as often in comparisons. (d) If you want to show a series converges, compare it to a larger series that also converges. For the the series in form of 1/nᴾ, the easiest way to determine its convergence is using the p-series test:. @radicalnumber @desmos used for exploring vertex form of a quadratic, students liked seeing how a,h, and k values effect the graph. Also available are scanned solutions to problems in differential, integral and multi-variable calculus and series. So we are using the p series calculus function. 6 Alternating Series and Absolute Convergence 9. Useful Materials and Links: I'll probably post small notes about particular lectures here if need be. CONICS, PARAMETRIC EQUATIONS, AND POLAR COORDINATES. 4 Absolute and Conditional Convergence; Alternating Series Jiwen He 1 Convergence Tests Basic Series that Converge or Diverge Basic Series that Converge Geometric series: X xk, if \|x\| < 1 p-series: X 1 kp, if p > 1 Basic Series that Diverge Any series X a k for which lim k→∞ a k 6= 0 p-series: X 1 kp, if p ≤ 1. Red and Black 7. Parametric Equations and Calculus. 1 Sequences and Their Limits 8. First we'll prove P(1); this is called "anchoring the induction". On each upward bounce the ball returns to 3 1 of it previous height. Recall the p-Test: Regardless of the value of the number p, the improper integral is always divergent. Most series that we encounter are not one of these types, but we are still interested in knowing whether or not they converge. Sigma notation, divergent series, convergent series. Comparison Tests (19 minutes, SV3 » 52 MB, H. But since 1/n^(3/2) is a converging p-series, the series in question converges also. To distinguish between them we specify what their improper point is. The p-Integrals. 4 Integral Test and Alternating Series Test. Series Convergence & Divergence This page of videos will cover most of your series needs, so that's why it's so long. P an = s YES P an Diverges NO TAYLOR SERIES Does an = f(n)(a) n! (x −a) n? NO YES Is x in interval of convergence? P∞ n=0 an = f(x) YES P an Diverges NO Try one or more of the following tests: NO COMPARISON TEST Pick {bn}. Distinguishing features of the course include: 1) the introduction and use of Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus;. CONICS, PARAMETRIC EQUATIONS, AND POLAR COORDINATES. Free math problem solver answers your calculus homework questions with step-by-step explanations. APﬁ Calculus BC 2002 Free-Response Questions These materials were produced by Educational Testing Service ﬁ (ETS ), which develops and administers the examinations of the Advanced Placement Program for the College Board. Learn ap calculus series with free interactive flashcards. Free math lessons and math homework help from basic math to algebra, geometry and beyond. The unique pneumatic toggle clamp mechanism can develop up to 60 tons of clamp force with regular shop air. I hope this helps!. A ball is dropped from a height of 9. 1 n=1 ( 1) n b. Three big theorems are found in this chapter: 1st Fundamental Theorem of Calculus, 2nd Fundamental Theorem of Calculus, and the Mean Value Theorem for Integrals. Geometric Series 1. However their convergence or divergence depends on the denominator's exponent, p. Serioes of this type are called p-series. 5 Notes for AP Calculus class (I guess college calculus, too). Preface: The goal of this text is to help students learn to use calculus intelligently for solving a wide variety of mathematical and physical problems. Convergence of Power Series Lecture Notes Consider a power series, say 0 B œ " B B B B âa b # $%. On each upward bounce the ball returns to 3 1 of it previous height. 3 LECTURE NOTES Topics: The Integral Test and p-Series - The. CONICS, PARAMETRIC EQUATIONS, AND POLAR COORDINATES. Distinguishing features of the course include: 1) the introduction and use of Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus;. 5 The Root and Ratio Tests 9. If the series has a finite number of terms, it is a simple matter to find the sum of the series by adding the terms. sum_(k=1)^oo (1 / k^p) Please express your views of this topic Dot Product Properties by commenting on blog. Sequences in Calculus. Series can expand about the point x = ∞. Their basic terminology is summarized in the table below. HD videos covering everything you need to know in Calculus I, II & III + Linear Algebra and Differential Equations; Now Offering ALGEBRA, PRECALCULUS, TRIG, & MATH. Look closely at the preceding example. Does the celebrated harmonic series diverge or converge? Discover a proof using the integral test. 8 where Cis the number of correct responses and Iis the number of incorrect responses. is convergent if p > 1 and divergent otherwise. Overview of Sequences and Series with terms and definitions. a) Using Maple, graph the first twenty terms of the series. A Series Test Gauntlet. As a study aid, I have listed below the major. Limits An Introduction to Limits Epsilon-Delta Definition of the Limit Evaluating Limits Numerically Understanding Limits Graphically Evaluating Limits Analytically Continuity Continuity at a Point Properties of Continuity Continuity on an Open/Closed Interval Intermediate Value Theorem Limits Involving Infinity Infinite Limits Vertical Asymptotes. It is suitable for a one-semester course, normally known as "Vector Calculus", "Multivariable Calculus", or simply "Calculus III". Using the p-Series test determine if the series$\sum_{n=1}^{\infty} \frac{n^2}{n^4}$is convergent or divergent. The absolute values of the series is a divergent p-series with. AP Calculus BC Course Syllabus Resources Larson, Ron, Robert P. Heath and Company, 5th edition 1994 – student issued textbook Hughes-Hallett, Deborah, et al. PCHS AP CALCULUS. First we'll prove P(1); this is called "anchoring the induction". Teach yourself calculus. AP CALCULUS AB/BC I. Expression: Variable. Chapter 8: Inﬁnite Sequences and Series, Calculus (3e) by Stewart Important theorems and convergence tests (compiled by Niels Joaquin) Theorem (Squeeze Theorem for Sequences). If this was your ID you would only type in BD92F455. It takes the following form: Here's a common example of a p-series, when p = 2: Here are a few other examples of p-series: Remember not to confuse p-series with geometric series. We will now look at some examples of specifically applying the p-Series test. The series includes High School Chemistry, AP Chemistry, General Chemistry, Organic Chemistry and Biochemistry. p-series Test File. For examples, look at the sample questions released by SOA. p-Series Convergence The p-series is given by 1/n p = 1/1 p + 1/2 p + 1/3 p + where p > 0 by definition. While taking the Advanced Placement (AP) Calculus BC exam is not. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. To achieve this goal, students will gain a thorough understanding of the topics covered in the course outline (typical Calculus 2 course in college). The sum of all combinations. If you continue browsing the site, you agree to the use of cookies on this website. When p = 1, the p-series is the harmonic series, which diverges. 2 Geometric Series, p-Series, and nth Term Test. Factor x inside the square root and use the fact that sqrt (x). Calculus: Development of Major Content Strands PDF If you have adopted the CPM curriculum and do not have a teacher edition, please contact our Business Office at (209) 745-2055 for information to obtain a copy. ∫f(x) dx Calculus alert! Calculus is a branch of mathematics that originated with scientific questions concerning rates of change. Free math problem solver answers your calculus homework questions with step-by-step explanations. We’re currently working with$ p $-series in my Calculus class, and I’ve fallen for the apparently common misconception that the infinite sum$ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{p}} $, where$ p = 1 $, should converge, whereas in reality, it diverges. This guide offers step-by-step solutions for all odd-numbered text exercises in Calculus of a Single Variable 11e (Chapters P-11 of Calculus 11e). Problem Solving. Power Series. Alternating Series remainder For a convergent alternating series, the absolute value of the remainder in approximating the sum with the first n partial sums is less than or equal to the value of the first neglected term. Saleem Watson, who received his doctorate degree under Stewart's instruction, and Daniel Clegg, a former colleague of Stewart's, will author the revised series, which has been used by more than 8 million students over the last fifteen years. We explain calculus and give you hundreds of practice problems, all with complete, worked out, step-by-step solutions. AP Calculus Questions Similar to BC Exams. [Note: the list is not definite; you may learn all or some of the things mentioned, or you may learn other topics not listed her. The 1st Fundamental Theorem of Calculus is an extremely important theorem that allows us to find the area under a curve over an interval. The first term in the series is a, and the last one is a+(n-1)d, so we can say the sum of the series is the first term plus the last term multiplied by the number of terms divided by 2. 58 The AP CALCULUS PROBLEM BOOK A. You will need to find one of your fellow class mates to see if there is something in these notes that wasn't covered in class. Since the harmonic series diverges, these. (These courses have been audited and approved by the College Board to carry the AP designation and have been acknowledged to provide students with a college. 3 Problem 2E. AP Calculus. A Series Test Gauntlet. Calculus For Dummies, 2nd Edition makes calculus manageable—even if you're one of the many students who sweat at the thought of it. Calculus is not explicitly tested on P (one of the changes from 1 to P). AP Calculus BC Review — Chapter 12 (Sequences and Series), Part One Things to Know and Be Able to Do ¾ Understand the distinction between sequences and series ¾ Understand the meaning of convergence (absolute and conditional), divergence, boundedness and how to test for each. Lecture 25/26 : Integral Test for p-series and The Comparison test In this section, we show how to use the integral test to decide whether a series of the form X1 n=a 1 np (where a 1) converges or diverges by comparing it to an improper integral. Webassign Answers. After taking this course you'll be able to work with functions in a variety of ways, and be able to use derivatives to solve a variety of problems, which is math-speak for having the skills to build the future of technology. Index for Calculus Math terminology from differential and integral calculus for functions of a single variable. Alexandru Cibotarica at Ivy Tech Community College - StudyBlue. Proof - Convergence of a p-Series Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Video transcript. Convergence of Infinite Series. Learners in this course can earn a certificate in the series by signing up for Coursera's verified certificate program and passing the series' final exam. Conics and Calculus. This book covers calculus in two and three variables. Series Calculator computes sum of a series over the given interval. Hostetler; Bruce H. 264 » 20 MB). There is no general method of determining the test you should use to check the convergence of a series. Topic 9: Calculus Option Series Part 1 A series consists of C v+C w+C x…qK S=z T St_ qK s=z S T st_ It is denoted by: For a total sum For a partial sum The Divergence Test states:. Awesome, right? Intro to Convergence & Divergence (free) As you probably noticed, there are a LOT of videos in …. BC Calculus - "Fun with P Series" For the following P series: ∑ n=1 ∞ 1 n1. Find more Mathematics widgets in Wolfram\|Alpha. Lecture 25/26 : Integral Test for p-series and The Comparison test In this section, we show how to use the integral test to decide whether a series of the form X1 n=a 1 np (where a 1) converges or diverges by comparing it to an improper integral. Fundamental Theorem of Calculus; Simpson's Rule & Trapezoid Rule; Improper Integrals; Sequences; Series. Alternating Series remainder For a convergent alternating series, the absolute value of the remainder in approximating the sum with the first n partial sums is less than or equal to the value of the first neglected term. Processing is an open source programming language and environment for people who want to create images, animations, and interactions. Factor x inside the square root and use the fact that sqrt (x). calculus to graduate-level classes in algebra and numerical analysis. We’re currently working with$ p $-series in my Calculus class, and I’ve fallen for the apparently common misconception that the infinite sum$ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{p}} $, where$ p = 1 \$, should converge, whereas in reality, it diverges. Problem Solving. In general, a p-series follows the following form: p-series are useful because of the following theorem: The p-series is convergent if p > 1 and divergent otherwise. Series Formulas 1. Integral Test Finally, let's talk about a convergence test that may help out in certain specialized situations. 1 Convergence of Series - Free download as Powerpoint Presentation (. An important type of series is called the p-series. Excerpts from "How To Ace Calculus". Most of the applications are beyond the scope of most Calculus courses and tend to occur in classes that many students don’t take. Calculus Test Two Section One Multiple-Choice Calculators Allowed Time—45 minutes Number of Questions—15 The scoring for this section is determined by the formula [C −(0. 1 n=1 +1 n, where n > 0 for all n, is called an alternating series, because the terms alternate between positive and negative values. I A series of the form P. December 17, 2016 Final Exam Math 162 (Calculus IIA) and the original integral is Z arcsinxdx = xarcsinx+ p 1 x2 +C: 4. The course emphasizes not just getting answers, but asking the question "why is this true?". 10 Convergence of Taylor Series 10. Two graphs are included. Side note: most of the BC exam is AB, so if your AB knowledge is good don't worry too much, just learn the new thing. 6 Ratio Test and Root Tests notes by Tim Pilachowski The geometric series r cr cr m n n m − ∑ = ∞ = 1 if and only if r <1. Differentiation rules 3. Wolfram Demonstrations Project. Chapter 11 Infinite Sequences and Series Test 4 [James Stewart Calculus 8E] - Mathematics 212 with Dr. Unfortunately some improper integrals fails to fall under the scope of these tests but we will not deal with them here. 1 - Area Between Curves. questions Free-response: 3 questions (partial) Personal Progress Check 2. However, the geometric series is an exception. Excerpts from "How To Ace Calculus". Thinkwell's Calculus course covers both Calculus I and Calculus II, each of which is a one-semester course in college. Typically these tests are used to determine convergence of series that are similar to geometric series or p-series. It is part of the unit on Infinite Series, (Chapter 9 in most of the Larson textbooks. Listed #'s correspond to class lessons and topics. (a) Consider the series X 1 k 2 and X 1 k 2 + k. The p-series test determines the nature of convergence of a p-series as follows: The p-series converges if and diverges if. This is useful for analysis when the sum of a series online must be presented and found as a solution. The Confrontation 3. Calculus Test One Section One Multiple-Choice No Calculators Time—30 minutes Number of Questions—15. (In particular, if p > 1, then the graph is concave up, such as the parabola y = x2.	2020-04-04T05:26:23	{ "domain": "aoru.pw", "url": "http://ucpd.aoru.pw/p-series-calculus.html", "openwebmath_score": 0.7618722915649414, "openwebmath_perplexity": 850.8486311592696, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9928785729400543, "lm_q2_score": 0.8459424353665381, "lm_q1q2_score": 0.8399181180161626 }
https://nhigham.com/2020/11/24/what-is-a-nonnormal-matrix/	# What Is a (Non)normal Matrix? An $n\times n$ matrix is normal if $A^A = AA^$, that is, if $A$ commutes with its conjugate transpose. Although the definition is simple to state, its significance is not immediately obvious. The definition says that the inner product of the $i$th and $j$th columns equals the inner product of the $i$th and $j$th rows for all $i$ and $j$. For $i = j$ this means that the $i$th row and the $i$th column have the same $2$-norm for all $i$. This fact can easily be used to show that a normal triangular matrix must be diagonal. It then follows from the Schur decomposition that $A$ is normal if it is unitarily diagonalizable: $A = QDQ^$ for some unitary $Q$ and diagonal $D$, where $D$ contains the eigenvalues of $A$ on the diagonal. Thus the normal matrices are those with a complete set of orthonormal eigenvectors. For a general diagonalizable matrix, $A = XDX^{-1}$, the condition number $\kappa(X) = \\|X\| \\|X^{-1}\\|$ can be arbitrarily large, but for a normal matrix $X$ can be taken to have 2-norm condition number $1$. This property makes normal matrices well-behaved for numerical computation. Many equivalent conditions to $A$ being normal are known: seventy are given by Grone et al. (1987) and a further nineteen are given by Elsner and Ikramov (1998). The normal matrices include the classes of matrix given in this table: Real Complex Diagonal Diagonal Symmetric Hermitian Skew-symmetric Skew-Hermitian Orthogonal Unitary Circulant Circulant Circulant matrices are $n\times n$ Toeplitz matrices in which the diagonals wrap around: $\notag \begin{bmatrix} c_1 & c_n & \dots & c_2 \\ c_2 & c_1 & \dots & \vdots \\ \vdots & \ddots & \ddots & c_n \\ c_n & \dots & c_2 & c_1 \\ \end{bmatrix}.$ They are diagonalized by a unitary matrix known as the discrete Fourier transform matrix, which has $(r,s)$ element $\exp( -2\pi \mathrm{i} (r-1)(s-1) / n )$. A normal matrix is not necessarily of the form given in the table, even for $n = 2$. Indeed, a $2\times 2$ normal matrix must have one of the forms $\notag \left[\begin{array}{@{\mskip2mu}rr@{\mskip2mu}} a & b\\ b & c \end{array}\right], \quad \left[\begin{array}{@{}rr@{\mskip2mu}} a & b\\ -b & a \end{array}\right].$ The first matrix is symmetric. The second matrix is of the form $aI + bJ$, where $J = \bigl[\begin{smallmatrix}\!\phantom{-}0 & 1\\\!-1 & 0 \end{smallmatrix}\bigr]$ is skew-symmetric and satisfies $J^2 = -I$, and it has eigenvalues $a \pm \mathrm{i}b$. It is natural to ask what the commutator $C = AA^- A^A$ can look like when $A$ is not normal. One immediate observation is that $C$ has zero trace, so its eigenvalues sum to zero, implying that $C$ is an indefinite Hermitian matrix if it is not zero. Since an indefinite matrix has at least two different nonzero eigenvalues, $C$ cannot be of rank $1$. In the polar decomposition $A = UH$, where $U$ is unitary and $H$ is Hermitian positive semidefinite, the polar factors commute if and only if $A$ is normal. The field of values, also known as the numerical range, is defined for $A\in\mathbb{C}^{n\times n}$ by $F(A) = \biggl\{\, \displaystyle\frac{z^Az}{z^z} : 0 \ne z \in \mathbb{C}^n \, \biggr\}.$ The set $F(A)$ is compact and convex (a nontrivial property proved by Toeplitz and Hausdorff), and it contains all the eigenvalues of $A$. Normal matrices have the property that the field of values is the convex hull of the eigenvalues. The next figure illustrates two fields of values, with the eigenvalues plotted as dots. The one on the left is for the nonnormal matrix gallery('smoke',16) and that on the right is for the circulant matrix gallery('circul',x) with x constructed as x = randn(16,1); x = x/norm(x). ## Measures of Nonnormality How can we measure the degree of nonnormality of a matrix? Let $A$ have the Schur decomposition $A = QTQ^$, where $Q$ is unitary and $T$ is upper triangular, and write $T = D+M$, where $D = \mathrm{diag}(\lambda_i)$ is diagonal with the eigenvalues of $A$ on its diagonal and $M$ is strictly upper triangular. If $A$ is normal then $M$ is zero, so $\\|M\\|_F$ is a natural measure of how far $A$ is from being normal. While $M$ depends on $Q$ (which is not unique), its Frobenius norm does not, since $\\|A\\|_F^2 = \\|T\\|_F^2 = \\|D\\|_F^2 + \\|M\\|_F^2$. Accordingly, Henrici defined the departure from normality by $\notag \nu(A) = \biggl( \\|A\\|_F^2 - \displaystyle\sum_{j=1}^n \|\lambda_j\|^2 \biggr)^{1/2}.$ Henrici (1962) derived an upper bound for $\nu(A)$ and Elsner and Paardekooper (1987) derived a lower bound, both based on the commutator: $\notag \displaystyle\frac{\\|A^A-AA^\\|_F}{4\\|A\\|_2} \le \nu(A) \le \Bigl( \displaystyle\frac{n^3-n}{12} \Bigr)^{1/4} \\|A^A-AA^\\|_F^{1/2}.$ The distance to normality is $\notag d(A) = \min \bigl\{\, \\|E\\|_F: ~A+E \in \mathbb{C}^{n\times n}~~\mathrm{is~ normal} \,\bigr\}.$ This quantity can be computed by an algorithm of Ruhe (1987). It is trivially bounded above by $\nu(A)$ and is also bounded below by a multiple of it (László, 1994): $\notag \displaystyle\frac{\nu(A)}{n^{1/2}} \le d(A) \le \nu(A).$ Normal matrices are a particular class of diagonalizable matrices. For diagonalizable matrices various bounds are available that depend on the condition number of a diagonalizing transformation. Since such a transformation is not unique, we take a diagonalization $A = XDX^{-1}$, $D = \mathrm{diag}(\lambda_i)$, with $X$ having minimal 2-norm condition number: $\kappa_2(X) = \min\{\, \kappa_2(Y): A = YDY^{-1}, ~D~\mathrm{diagonal} \,\}.$ Here are some examples of such bounds. We denote by $\rho(A)$ the spectral radius of $A$, the largest absolute value of any eigenvalue of $A$. • By taking norms in the eigenvalue-eigenvector equation $Ax = \lambda x$ we obtain $\rho(A) \le \\|A\\|_2$. Taking norms in $A = XDX^{-1}$ gives $\\|A\\|_2 \le \kappa_2(X) \\|D\\|_2 = \kappa_2(X)\rho(A)$. Hence $\notag \displaystyle\frac{\\|A\\|_2}{\kappa_2(X)} \le \rho(A) \le \\|A\\|_2.$ • If $A$ has singular values $\sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_n$ and its eigenvalues are ordered $\|\lambda_1\| \ge \|\lambda_2\| \ge \cdots \ge \|\lambda_n\|$, then (Ruhe, 1975) $\notag \displaystyle\frac{\sigma_i(A)}{\kappa_2(X)} \le \|\lambda_i(A)\| \le \kappa_2(X) \sigma_i(A), \quad i = 1\colon n.$ Note that for $i=1$ the previous upper bound is sharper. • For any real $p > 0$, $\notag \displaystyle\frac{\rho(A)^p}{\kappa_2(X)} \le \\|A^p\\|_2 \le \kappa_2(X) \rho(A)^p.$ • For any function $f$ defined on the spectrum of $A$, $\notag \displaystyle\frac{\max_i\|f(\lambda_i)\|}{\kappa_2(X)} \le \\|f(A)\\|_2 \le \max_i\|f(\lambda_i)\|.$ For normal $A$ we can take $X$ unitary and so all these bounds are equalities. The condition number $\kappa_2(X)$ can therefore be regarded as another measure of non-normality, as quantified by these bounds. ## References This is a minimal set of references, which contain further useful references within.	2021-10-17T22:13:39	{ "domain": "nhigham.com", "url": "https://nhigham.com/2020/11/24/what-is-a-nonnormal-matrix/", "openwebmath_score": 0.9715813398361206, "openwebmath_perplexity": 166.27571390853623, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9928785723078832, "lm_q2_score": 0.8459424314825853, "lm_q1q2_score": 0.8399181136250886 }
https://math.stackexchange.com/questions/3060767/why-is-frac15-sqrt4125-sqrt45-15-sqrt5-and-not-15-sqrt425	# Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$? I have an expression I am to simplify: $$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$ I arrived at $$15\sqrt[4]{25}$$. My textbook tells me that the answer is in fact $$15\sqrt{5}$$. Here is my thought process: $$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$ = $$\frac{15\sqrt[4]{5}\sqrt[4]{25}}{\sqrt[4]{5}}$$ = (cancel out $$\sqrt[4]{5}$$ present in both numerator and denominator) leaving: $$15\sqrt[4]{25}$$ Where did I go wrong and how can I arrive at $$15\sqrt{5}$$? • The fourth root is the square root of the square root ... Jan 3 '19 at 16:54 $$\frac{\sqrt[4] {125}}{\sqrt[4] 5} = \frac{\sqrt[4] {5^3}}{\sqrt[4] 5} = \sqrt[4]{5^2} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt 5$$ Even quicker, $$\sqrt[4]{25}$$ means $$\sqrt{\sqrt{25}}$$, which becomes $$\sqrt{5}$$. Write $$\sqrt[4]{\frac{125}{5}}=\sqrt[4]{25}=\sqrt{5}$$ It turns out that $$\sqrt[4]{25}=\sqrt{5}$$. This is because $$25=5^2$$, so that $$\sqrt[4]{25}=\sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=\sqrt{5}.$$ So, you are correct, as is the book. • Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me. Jan 3 '19 at 17:09	2021-09-21T04:49:31	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3060767/why-is-frac15-sqrt4125-sqrt45-15-sqrt5-and-not-15-sqrt425", "openwebmath_score": 0.9452316761016846, "openwebmath_perplexity": 210.3992910117195, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226317625857, "lm_q2_score": 0.8596637541053281, "lm_q1q2_score": 0.8399109434668921 }
https://socratic.org/questions/express-2-00-as-a-percentage-of-8-00	# Express $2.00 as a percentage of$8.00 ? ##### 2 Answers Write your answer here... Start with a one sentence answer Then teach the underlying concepts Don't copy without citing sources preview ? #### Answer Write a one sentence answer... #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer Describe your changes (optional) 200 1 Nam D. Share Mar 9, 2018 #### Answer: 25% #### Explanation: Given: ($2)/($8) We have: (color(red)cancelcolor(black)$2)/(color(red)cancelcolor(black)$8)=1/4 Now, we convert $\frac{1}{4}$ to a percent. We got: 1/4100%=25% :.($2)/($8)=25% Was this helpful? Let the contributor know! Write your answer here... Start with a one sentence answer Then teach the underlying concepts Don't copy without citing sources preview ? #### Answer Write a one sentence answer... #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer Describe your changes (optional) 200 1 Mar 9, 2018 25% #### Explanation: First let's think about a normal percentage as a ratio. If I wanted to think about 4% as a ratio I would say well... That's 4 parts of 100% or $\frac{4}{100}$ and from there I could simplify the fraction into $\frac{1}{25}$ and now I have a mathematical representation of 4%. Anything that I multiply by $\frac{1}{25}$ will give me 4% of that thing. "How does this help me?" you say I'm very glad you asked Let's think about $8$ as our 100% and we want to find 2 parts of 8, so using what we just learned we say that "2 parts of 8 is the same as $\frac{2}{8}$." To turn $\frac{2}{8}$ into a percentage we just need the denominator to equal 100 since earlier we said that 8 is our 100% We could use an equation for this, so let's use an equation for this :D We know that $8 \cdot \left(\text{some number}\right) = 100$ so; $8 x = 100$ $\frac{1}{8} \cdot 8 x = 100 \cdot \frac{1}{8} = \frac{100}{8} = \frac{25}{2} = 12.5$ Now we know that if we multiply the denominator by 12.5 we will get 100 and if we multiply the numerator by 12.5 we will get a ratio with 100 as the denominator, so our percentage must be the numerator! 2/812.5/12.5 = 25/100 = 25% Was this helpful? Let the contributor know! ##### Just asked! See more • 9 minutes ago • 9 minutes ago • 11 minutes ago • 11 minutes ago • 55 seconds ago • 2 minutes ago • 2 minutes ago • 4 minutes ago • 7 minutes ago • 8 minutes ago • 9 minutes ago • 9 minutes ago • 11 minutes ago • 11 minutes ago	2018-03-24T17:58:57	{ "domain": "socratic.org", "url": "https://socratic.org/questions/express-2-00-as-a-percentage-of-8-00", "openwebmath_score": 0.9287996888160706, "openwebmath_perplexity": 3098.363820310095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226300899745, "lm_q2_score": 0.8596637541053281, "lm_q1q2_score": 0.8399109420290087 }
https://math.stackexchange.com/questions/1876240/proving-or-disproving-existence-of-a-continuous-function	# Proving or Disproving Existence of a Continuous Function I'm studying for an exam in topology; this is a question from a previous exam several years ago -- it's not being graded, I just want to know how to handle it. I'm more concerned here with learning the thought process involved in answering the following question(s) than I am with just being given the actual answers. If someone can provide a little guidance or maybe a hint or two, that would be great! The question is this: Consider the following result, which you can assume to be true without a proof: () For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto the unit interval $I=[0,1]$. Which of the following variations of () is true? Justify your answers. (a) For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto $\mathbb{R}$. (b) For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto the unit square $I^2=[0,1]^2$. (c) For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto the Cantor ternary set $C$. The work so far... The definition of continuity that I'm working with is this: A function $f:X \rightarrow Y$ is continuous if for every set V open in Y, the set $f^{-1}(V)$ is open in $X$. This tells me that, in some sense, the topology on the domain must have more (or an equal number of) open sets than the topology on the range space. I'm also aware that several properties are preserved under continuous maps: compactness, convergence of sequences, connectedness, path connectedness, the Lindelof property, and separability. Statement (a) then should be false. $F$ is compact, as it is either a closed interval or a union of closed intervals in $\mathbb{R}$. The image of $F$ under any continuous function would also have to be compact. However, $\mathbb{R}$ is not compact. Thus such a continuous function does not exist. I'm inclined to think that (b) is true, mainly because I can't come up with a reason why it wouldn't be. $I^2$ is not path connected, but $F$ isn't necessarily either. $I^2$ is connected and compact, but I don't think that really gives me anything to rule out the existence of the desired function. Now, if I want to show that such a function does exist, the only convincing argument I can come up with is to define one, but I'm not really sure how to do that here. I think (c) is false, mostly based on intuition. I think it's going to come down to showing that the cardinality of the Cantor set is too big for a surjective function to exist, but I wouldn't put money on it. Any help at all would be fantastic, but bonus points for putting a lot of detail into your thought process! I really want to learn how to handle this type of problem on my own, not just have an answer to this particular problem. • (a) is false, but your reasoning is not correct, for example you could take $F$ to be the reals in which case the identity is a suitable map. You can just produce a counterexample by taking $F$ to be compact. Jul 30 '16 at 19:04 You are right about (a) (although you should be careful: a finite union of closed intervals is compact, but an infinite union need not be - consider $[0, 1]\cup [2, 3]\cup [4, 5]\cup . . .$). EDIT: Actually I misread it - your reasoning for (a) is wrong. How do you know $F$ is compact? Rather, the point is that there exists an uncountable closed set which is compact, and such a set is a counterexample to (a). Your guess is right for part (b), but your reason is wrong (note that both $I^2$ and $I$ are path connected). HINT: it would be enough to find a continuous map from $I$ onto $I^2$ - do you see why, and have you heard of the Peano curve? Your guess for (c) is right, but again, for the wrong reason: the Cantor set and $I$ have the same cardinality. Instead, consider (similarly to (a)): what can you say about the image of a connected set under a continuous map? • @Keith Your second sentence ("In order . . .") is not true: the set I describe above is closed but not compact, and not a finite union of closed intervals. But this doesn't effect your reasoning for (a): all you need is a single uncountable closed compact set, and $I$ works. Re: the Peano curve, you don't need to define it differently in topology - the point is just that it's a continuous surjective map from $I$ to $I^2$. Jul 30 '16 at 18:56 • For (a), I see what you're saying. For (b), I'm not sure I see why it's enough to find a continuous map from $I$ onto $I^2$. Is it true in general that all closed subsets of $\mathbb{R}$ are homeomorphic with one another (and is the analogous statement true for arbitrary $\mathbb{R}^n$)? Or is there some deeper reason I don't see? Jul 30 '16 at 23:12 • @Keith Suppose I find a continuous map from $I$ onto $I^2$. The question allows you to assume already that there is a continuous map from any uncountable closed $F$ onto $I$. Do you see how to combine these? (And it is very false that any two closed subsets of $\mathbb{R}$ are homeomorphic - consider e.g. the closed sets $\emptyset$, $\{0\}$, $\{0, 1\}$, $[0, 1]$, $[0, 1]\cup [2, 3]$, $\mathbb{R}$.) Jul 30 '16 at 23:17	2021-09-18T19:22:42	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1876240/proving-or-disproving-existence-of-a-continuous-function", "openwebmath_score": 0.8657563328742981, "openwebmath_perplexity": 78.85529610782326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226307590189, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8399109408477629 }
http://math.stackexchange.com/questions/317235/covariance-and-variance-of-a-poisson-r-v	# Covariance and variance of a Poisson r.v. Given a Poisson process $N(t),t\geq 0$ with rate $\lambda$ and another r.v. $T$ independent of $N(t)$ with mean $\mu$ and variance $\sigma^2$, I would like to compute the following quantities: $$\mathbb{Cov}(T,N(T)) \ \ \mbox{ and } \ \ \mathbb{Var}(N(T))$$ My guess is respectively: $\lambda \mu + \lambda \sigma^2$ and $\sigma^2\lambda$. But I am not sure it is correct nor how to justify some steps. Anyone knows? Thank you very much! - For the variance, use the conditional variance formula: $$\text{var}(N(T)) = E[\text{var}(N(T)\mid T] + \text{var}(E[N(T)\mid T).$$ This will show why your guessed answer is not correct (it is missing one term from the displayed expression above). – Dilip Sarwate Feb 28 '13 at 21:48 Since $$\mathbb{Cov}\left(T, N(T)\right) = \mathbb{E}\left(T N(T)\right) - \mathbb{E}(T) \mathbb{E}\left(N(T)\right) = \mathbb{E}\left(T \, \mathbb{E}\left(N(T)\mid T\right)\right) - \mathbb{E}(T) \, \mathbb{E}\left(\mathbb{E}\left(N(T)\mid T\right)\right)$$ But $\mathbb{E}\left(N(T) \mid T\right) = \lambda T$, thus $$\mathbb{Cov}\left(T, N(T)\right) = \mathbb{E}\left(\lambda T^2\right) - \lambda \mathrm{E}(T)^2 = \lambda \mathbb{Var}(T)$$ Similarly: $$\begin{eqnarray} \mathbb{Var}\left(N(T)\right) &=& \mathbb{E}\left( \mathbb{E}\left(N(T)^2\|T\right) \right) - \mathbb{E}\left(\mathbb{E}\left( N(T) \mid T\right)\right)^2 \\ &=& \mathbb{E}\left( \lambda^2 T^2 + \lambda T\right) - \mathbb{E}\left(\lambda T\right)^2 \\ &=& \lambda^2 \mathbb{Var}(T) + \lambda \mathbb{E}(T) \end{eqnarray}$$ Thanks a lot! Yes I see it, this is exactly what I did but something seemed strange to me.... Since we compose $N(T)$, it could be that we know the random variable $T$ but not $N$, what I mean is that $N(T) = N(T(\omega))(\tilde{\omega})$ where $\omega \in \Omega$, $\tilde{\omega} \in \tilde{\Omega}$ are the sample spaces of each variable. So when averaging out one, we should obtain a random variable? shouldn't we? – Dan Feb 28 '13 at 21:53 Yes, I think you are correct. The conditioning by $T$ indicates that we average over $\tilde\omega$ while keeping $\omega$ fixed, and then average over $\omega$. – Sasha Feb 28 '13 at 22:03	2015-07-02T16:40:16	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/317235/covariance-and-variance-of-a-poisson-r-v", "openwebmath_score": 0.9581418633460999, "openwebmath_perplexity": 203.15259132298777, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226300899744, "lm_q2_score": 0.8596637487122111, "lm_q1q2_score": 0.8399109367598113 }
https://math.stackexchange.com/questions/830413/calculating-the-arc-length-of-a-circle-segment	# Calculating the arc length of a circle segment I would like to calculate the arc length of a circle segment, i.e. I know the start coordinates (x/y) of the circle segment, the end coordinates (x/y) and the x and y distances from the starting point to the center point of the circle segment. I know that I can calculate the circumference with 2 * radius * PI. Consequently, I would have to calculate the radius and the angle of the circle segment via Pythagorean theorem and sin and cos. My question: Is there a simple formula where I just have to put in start-coordinates, end-coordinates and the circle origin point coordinates? Thanks. enne • I have never seen the phrase "bow length" used before. What does it mean? – Gerry Myerson Jun 11 '14 at 12:34 • I mean the circle length, I replaced the term "bow" with "circle". – enne87 Jun 11 '14 at 12:40 • Your question is still not very clear. If you have translated this, please provide the original text also. – M. Vinay Jun 11 '14 at 12:41 • @GerryMyerson, I think OP wants to find the length of the arc between the two points given. – Vikram Jun 11 '14 at 12:42 • @GerryMyerson Arc length. Could be a word-for-word translation from the German Bogenlänge. – Daniel Fischer Jun 11 '14 at 12:42 Use $(x-h)^2+(y-k)^2=r^2$...(eq I), where $(h,k)$ is the center and $r$ is the radius. Your third point can be calculated as $x_3=x_1+$(horizontal displacement) and $y_3=y_1+$(vertical displacement), you have mentioned that you know x and y distance. the x and y distances from the starting point to the center point of the circle segment. Now, in eq (I), put $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$ to get three equations, solve these three equations simultaneously to get $(h,k)$. Next, use $(h,k)$ and any of the three points we know to get $r$ with the help of eq (I). Now shift the circle so that it's origin coincides with $(0,0)$, this will help you find the angle $\theta$ easily. Now, find $\theta$ and the length of the arc=$r\theta$ EDIT:Consider a circle with 3 points on it. $x_1\equiv(2+2\sqrt{3},0) \hspace{12 pt}x_2\equiv(6,2)\hspace{12 pt}x_3\equiv(2+2\sqrt{3},4)$ If you use eq (I) you will get following 3 eqns: $29.85+h^2-10.91h+k^2=r^2$...(eqn 1) $45.85+h^2-10.91h+k^2-8k=r^2$...(eqn 2) $40+h^2-12h+k^2-4k=r^2$...(eqn 3) subtract eqn 3 from eqn 1 to get one eqn subtract eqn 3 from eqn 2 to get second eqn solve these last two eqn and you will get $(h,k)\equiv(2,2)$, the center of the circle, substitute $(h,k)$ in eqn 3 (or 1 or 2), you will get $r=4$ No need to do transform the center of the circle to the origin, we know the radius, so find what angle points $x_1$ and$x_3$ makes with the horizontal line using trigonometry ($\sin \theta=2/4=1/2 \therefore \theta=\pi/6$) Length of the arc=$r\theta=4*\large\frac{\pi}{6}=\large \frac{2\pi}{3}$ • I don't know why you are give 3 points on the circle, only two are enough to find the center,(2 unknowns, 2 equations) – Vikram Jun 11 '14 at 13:36 • replace the word "origin" in second last line with "center" – Vikram Jun 11 '14 at 13:47 • Sorry, probably I'm too stupid, but can you please give me an example how this works? – enne87 Jun 12 '14 at 10:03 • Hey, cool thing, thanks :D – enne87 Jun 12 '14 at 15:32 • @enne87, you’re welcome – Vikram Jun 12 '14 at 16:01 You can derive a simple formula using the law of cosines. In fact, while all the planar geometry is helpful for visualization, there's really no need for most of it. You have 3 points: your arc start and stop points, which I'll call $A$ and $B$, and your circle center, $C$. The angle for the arc you're wanting to measure, I'll call it $\theta$, is the angle of the triangle $ABC$ at point $C$. Because $C$ is the center of the circle that $A$ and $B$ are on, the triangle sides $AC$ and $BC$ are equal to your circle's radius, $r$. We'll call the length of $AB$, the remaining side, $d$ (see picture). (By the way, if $A=(x_1,y_1)$ and $B=(x_2,y_2)$, then $d=\sqrt {(x_1-x_2)^2 + (y_1-y_2)^2}$.) According to the law of cosines, $\cos (\theta )={{r^2+r^2-d^2}\over {2rr}}=1- {{d^2}\over {2r^2}}$. So all you need is the distance between the end points of your arc and the radius of the circle to compute the angle, $\theta = \arccos (1- {{d^2}\over {2r^2}})$ Lastly, the length is calculated - $Length = r\theta$ Where $\theta$ is expressed in radians. • Hi Zimul8r, thanks for your help. I just tried your approach with a circle segment where θ = 90° and r = 4. This means this is an exact quarter circle segment. So d = sqrt(4²+4²) = 5.6568; Consequently, cos(θ) = (1-5.6568²) / 32 = -0.9687. arccos(-0.9687) = 14.373rad ?? What am I doing wrong here? – enne87 Jun 12 '14 at 9:40 • Ah sorry, now it works.Thanks for the great approach. – enne87 Jun 12 '14 at 15:32	2019-06-16T06:34:29	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/830413/calculating-the-arc-length-of-a-circle-segment", "openwebmath_score": 0.7579015493392944, "openwebmath_perplexity": 390.4446218473956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226341042414, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.8399109366979328 }
https://mathoverflow.net/questions/217612/primorial-puzzlement	# primorial puzzlement Let $x_n$ be the smallest positive integer which is not a quadratic residue modulo any of the first $n$ odd primes. The question is: is there any bound on how quickly $x_n$ grows as a function of $n?$ Now, since the probability of being a residue modulo any given prime is very close to $1/2,$ by an independence heuristic, $x_n$ should grow roughly like $2^n.$ The problem is that independence clearly does not hold here: if you ask the "opposite" question: how big is $y_n,$ the smallest integer which is a square modulo all of the first $n$ odd primes, the answer is $y_n=1.$ When in doubt, experiment, and computing $\log x_n$ for the $n=2, \dotsc, 20$ gives the graph: In case you wonder, the slope is around $0.8$ (so a bit bigger than $\log 2,$ but $20$ is pretty small, so $x_n \sim 2^n$ seems like a reasonable guess. Does anyone have any nontrivial bounds, even mod "standard conjectures"? EDIT Noam Elkies points out that a few more terms have been computed (I have gotten up to 24 in the meantime, but 27 is even better :)) The slope of the log is down to $0.76,$ so the conjecture that $\log(x_n) \sim n \log 2$ seems quite reasonable. Further Edit In fact, if you ponder Noam's comment, you will note that the heuristic will give something like $\log n 2^n$ for the conjectured growth rate... More Values Using some of joro's ideas, and a little more, we can push the list of values to 40. {2, 2, 17, 17, 83, 167, 227, 398, 398, 5297, 64382, 69647, 116387, 214037, 214037, 430022, 5472953, 5472953, 8062073, 8062073, 8062073, 41941577, 86374763, 163520117, 163520117, 231912722, 231912722, 231912722, 545559467, 1728061733, 2832363203, 5638787822, 5638787822, 6154772762, 6154772762, 7012246247, 7012246247, 7012246247, 6571091781638, 7218195919667} • A few more terms, up to $x_{29} = 1728061733$: oeis.org/A206095 Sep 6 '15 at 22:24 • One might expect the slope to start a bit above $\log 2$, because the probability of $(x/p) = -1$ for small $x$ is $(p-1) \, / \, (2p)$, which is a bit less than $1/2$, and noticeably less for $p$ small. Sep 6 '15 at 22:27 • @NoamD.Elkies Thanks! I incorporated this into the latest edit. Sadly, oeis seems to make no further statement... Sep 6 '15 at 22:32 • @NoamD.Elkies True, but that would probably affect the intercept more than the slope. If you don't control for the intercept, it is, indeed, negative, but the slope than becomes a bit higher. Again, the numbers (be they 20 or 27) are small enough that I would not read too much into them... Sep 6 '15 at 22:43 • I think the standard conjecture is that the $n$th prime satisfies $p_n\sim n\log n\sim (\log x_n)(\log\log x_n)/\log 2$, simply transferring from pseudosquare guesses. Which is the same as what you write. Sep 7 '15 at 15:12 I'll try to unify all the previous answers (of GH from MO, Will Sawin and Hurkyl) and also indicate unconditional results on this problem. It turns out that one can get a surprisingly decent unconditional upper bound here, which is something I hadn't appreciated fully previously. As noted in Hurkyl's answer above, a unifying problem is the following: Given a non-trivial subgroup $H$ of $({\Bbb Z}/q{\Bbb Z})^$ with index $h$, bound (i) the least prime $p$ lying in a given coset of $H$, or (ii) the least integer $n$ lying in a given coset of $H$. Obviously the second problem is easier than the first, and as noted in Hurkyl's answer Theorem 1.4 of Lamzouri, Li and Soundararajan gives a good bound on GRH for the first problem: the least prime is essentially bounded by $((h-1) \log q)^2$ (see the precise form) and this result unifies both the least quadratic non-residue problem and also the least prime in an arithmetic progression problem. Unconditionally, for problem (i), by Linnik's theorem on the least prime in an arithmetic progression we always have a bound of $\ll q^{L}$, for an absolute constant $L$, and the current record here is that $L=5$ is permissible by Xylouris. When $h$ is small, then a somewhat better bound can be given: for $h=2$ (the case of the least prime that splits, or the least prime that is inert in a quadratic field) this is classical and well known, the case of split primes being harder and the bound here is $q^{\frac 14+\epsilon}$ (using Burgess for character sums and Siegel's ineffective bound for $L(1,\chi)$). For larger, but still bounded $h$, the matter was treated by Elliott in the context of the least prime $h$-th power residue, and he obtained the bound $q^{(h-1)/4+\epsilon}$. Of course for large $h$ (i.e. $h\ge 22$) this becomes inferior to Linnik's theorem. A recent paper of Pollack addresses a somewhat more general version of this question, and obtains the bound $\ll q^{(h-1)/4+\epsilon}$. Now for unconditional results towards the easier problem (ii). Assume that $h \le q^{\epsilon}$ and express the condition of belonging to a coset of $H$ in terms of the characters that are trivial on $H$ (see the paper of Lamzouri, Li and Soundararajan for details). Then use the Burgess bounds on the character sums that arise. This argument gives that the least integer $n$ in any given coset is $\ll q^{\frac 14+\epsilon}$. This is a (modest) improvement over the trivial bound of $q$. For a general modulus $q$ the Burgess bound gives the best known result on this problem. However, for moduli $q$ that are very smooth (and this will be important for the present problem) one can obtain a better result by a $q$-van der Corput argument due to Graham and Ringrose (following Heath-Brown). From the Graham-Ringrose argument (see either the original paper, or Corollary 12.14 of Iwaniec and Kowalski, or estimate (4.1) and Lemma 4.2 from Granville and Soundararajan), one can show that if all the prime factors of $q$ are bounded by $q^{o(1)}$ then the least integer in any prescribed coset is $\ll q^{\epsilon}$ for any fixed $\epsilon >0$. Naturally there is a connection between how small $\epsilon$ can be and the smoothness of $q$, and one can find uniform bounds in the results referenced above. Returning to the problem at hand, if $p_1, \ldots, p_n$ are the first $n$ odd primes, and we set $q=p_1\cdots p_n = n^{(1+o(1))n}$ and $H$ to be the subgroup of squares mod $q$ (which has index $2^n$) then on GRH we can find a prime $p \ll 4^n (n\log n)^2$ lying in any given coset of $H$. That is given any choice of signs $\epsilon_j = \pm 1$, there is a prime $p\ll 4^n (n\log n)^2$ with $(\frac{p}{p_j}) = \epsilon_j$ (in particular we can take all $\epsilon_j=-1$ as asked for in the problem). This was noted in Hurkyl's answer. Note now that unconditionally by the Burgess estimate we may find a number $N \ll n^{n(1/4+\epsilon)}$ with $(\frac{N}{p_j}) =\epsilon_j$. But note that we are dealing now with a modulus $q$ of size $n^n$ all of whose prime factors are of size $\ll n\log n$; in other words, $q$ is very smooth indeed and Graham-Ringrose applies and gives an unconditional upper bound of $n^{\epsilon n}$. In fact, by keeping track of uniformity, one gets an unconditional upper bound of $n^{c n/\log \log n}$ for some positive constant $c$. Now consider lower bounds in the problem. Here we must make use of the fact that we are looking for the smallest number $N$ with $(\frac{N}{p_j})=-1$ for all $j\le n$ -- for a random choice of signs $\epsilon_j$ there may of course be a very small number $N$ (just pick the number $N$ and see what the signs are!). Note that $N$ or $4N$ would be a discriminant, and we are asking for a quadratic character $\pmod {N}$ or $\pmod {4N}$ taking value $-1$ on the first $n$ small primes. This fits our framework by taking $q=4N$, and $H$ to be the subgroup of index $2$ on which this character (say $\chi$) is trivial. On GRH we know that there is an odd prime $\ell \le (1+o(1)) (\log N)^2$ with $\chi(\ell) =1$, and so we must have $(\log N)^2 \ge (1+o(1)) n\log n$ or $$N\ge \exp((1+o(1)) \sqrt{n\log n}).$$ This is the conditional bound mentioned in Will Sawin's answer, and what I indicated in my comment there. Unconditionally, the Burgess bound and Siegel's theorem argument (for problem (i)) gives that there is a prime $\ell \le N^{\frac 14+\epsilon}$ with $\chi(\ell)=1$, and so $N\ge n^{4+o(1)}$. To summarize, on GRH we know that the answer lies between $$\exp((1+o(1)) \sqrt{n\log n} ) \text{ and } \ll 4^n (n\log n)^2$$ and unconditionally we know that it is between $$n^{4+o(1)} \text{ and } \ll n^{cn/\log \log n}.$$ The probabilistic conjecture that the answer should be somewhere near $2^n$ seems reasonable. I think the standard approach to this problem is to estimate the sum $$S(x) := \sum_{m=1}^x\prod_{k=1}^n\left(1-\left(\frac{m}{p_k}\right)\right),$$ where $p_k$ is the $k$-th odd prime. Indeed, we are looking for the smallest $x$ such that $S(x)>0$. Factoring out, we get $$S(x) = \sum_{d\mid p_1\dots p_n}(-1)^{\omega(d)}\sum_{m=1}^x \left(\frac{m}{d}\right).$$ The inner sum equals $x$ for $d=1$, while it is much smaller for $d>1$ due to the oscillating nature of the Jacobi symbol. We can quantify this under GRH using an explicit bound of Chandee (Explicit upper bounds for $L$-functions on the critical line, Proceedings of the AMS), and we infer $$S(x)=x+O_\epsilon\bigl(2^n e^{(3/8+\epsilon)n}\sqrt{x}\bigr).$$ In particular, under GRH, we see that $x_n$ grows at most exponentially, namely $$x_n \ll_\epsilon 4^n e^{(3/4+\epsilon)n}.$$ Added. We can improve the above argument by restricting $m$ to prime values exceeding $p_n$ and adding the weights $\Lambda(m)$. We get, by Theorem 5.15 in Iwaniec-Kowalski: Analytic number theory, $x_n\ll 4^n n^5$ (under GRH). • Sorry, what does the second last formula mean? There must be a typo of some sort... Sep 7 '15 at 0:08 • @IgorRivin: Sorry, "$S(x)=$" was missing, I added this. Sep 7 '15 at 0:11 • Can we also use GRH to give a lower bound by upper bounding the sum $\sum_{k=1}^{p_{n+1}-1} \mu(k) \left( \frac{ x } {k} \right)$ for $x$ small, using the fact, from quadratic reciprocity, that $\left( \frac{ x } {k} \right)$ is a Dirichlet character? Sep 7 '15 at 2:48 • @WillSawin: Good idea. I think this way we can get $\log x_n\gg(\log n)(\log\log n)$ or, using more optimistic conjectural upper bounds for $L$-functions perhaps $\log x_n\gg(\log n)^{2-o(1)}$. Sep 7 '15 at 7:39 • The OP's question (with his intent that none of the first $n$ odd primes divide $x_n$) is equivalent to asking for the smallest $x$ such that the first $n$ odd primes are all inert in $\mathbb{Q}(\sqrt{x})$. If you have an upper bound on the smallest odd split prime in terms of the discriminant, then you get a lower bound on $x_n$. – user13113 Sep 7 '15 at 22:06 I think one can get a lower bound more like $x_n \geq e^{ c \sqrt{n \log n}}$, or equivalent $p_{n+1} = O ( (\log x_n)^2)$, conditional on GRH. One simply reverses the sign in the argument giving the upper bound on the least quadratic nonresidue to obtain an upper bound on the first prime at which a quadratic Dirichlet character takes the value 1. This argument is due to Ankeny, in The Least Quadratic Non-Residue. Apply the same argument from Theorem 2 to the Dirichlet character $\chi(d) = \left( \frac{x_n}{d} \right)$. The condition that $\chi(p) = 1$ for small $p$ becomes $\chi(p) = -1$ for small $p$. The same argument goes through with the sign reversed and gives the same bound. • See also Theorem 1.4 of arxiv.org/pdf/1309.3595v1.pdf for a recent explicit version. Also unconditionally, Linnik and Vinogradov knew that the least prime that splits completely in a quadratic field is $\le \|D\|^{1/4+o(1)}$ (see Pollack ams.org/journals/proc/2014-142-06/S0002-9939-2014-12199-X/… ) ; this follows from the Burgess bounds and Siegel's theorem. Sep 7 '15 at 14:20 • @Lucia So you're saying that unconditionally, in Igor's notation, we have $x_n > n^{4 - o(1)}$? Sep 7 '15 at 14:23 • Yes, that's right. Sep 7 '15 at 14:23 • The result of Linnik-Vinogradov is ineffective. This is a substantial difference between the case of $\chi(p)=-1$ versus $\chi(p)=1$ for all the small primes. But under GRH it should not matter (maybe only a log factor somewhere). Sep 7 '15 at 15:06 • Your observation is very nice, by the way. Sep 7 '15 at 19:41 Very partial answer for computational approach. I get $x_{30}=2832363203,x_{31}=x_{32}=5638787822,x_{33}=6154772762$ in relatively short time and other results agree with OEIS. The performance likely can be greatly improved by working in something faster than sage. The main idea is to work in many arithmetic progression with large step. Set $M$ the product of the $k$ odd primes. Build increasing sequence $0 \le b_i \le M-1$ s.t. $b_i$ is quadratic non-residue modulo the first $k$ odd primes. Experimentally $\|\{b_i\}\|$ is significantly smaller than $M$. Work in the arithmetic progressions $Mj+b_i$ in increasing order. Sample sage implementation def igorprimes1(n): """ build M, b_i """ pri=[] res=[] for k in [ 2 .. n+1]: p=nth_prime(k) pri += [p] for k in xrange(prod(pri)): for p in pri: if kronecker(k,p) != -1: break res += [k] return res,prod(pri) def igorprimes2(res,M,n): """ computes x_n = A206095(n) """ pri=[nth_prime(k) for k in [ 2 .. n+1] ] r=0 while True: M1=Mr for a in res: M2=M1 + a for p in pri: if kronecker(M2%p,p) != -1: break return M2 r += 1 • Actually, my Sage complains when I run igorprimes1(20) that there is some number bigger than a C int. (my original computations were done in a completely brute force manner in Mathematica) Sep 7 '15 at 13:51 • @IgorRivin This is normal, $igorprimes1$ doesn't expect such large values (it does loop to 21#/2). Try: tt=igorprimes1(7);igorprimes2(tt[0],tt[1],33) – joro Sep 7 '15 at 14:14 • Ah, I was guessing it was something like that! Sep 7 '15 at 14:23 • You sieve first, then test, but you don't sieve all the way. Sep 7 '15 at 14:33 • @IgorRivin Something like this -- there are two stages as explained. First find M,b_i, then look for solutions in the arithmetic progressions M j+ b_i. As the code is currently written, it may lose small solutions if the first stage is sufficiently large, but this is easy to fix. – joro Sep 7 '15 at 14:55 Lucia's comment above mentions theorem 1.4 from CONDITIONAL BOUNDS FOR THE LEAST QUADRATIC NON-RESIDUE AND RELATED PROBLEMS : assuming the GRH, Let $q \geq 20000$, $H$ be a proper subgroup of $G = (\mathbb{Z} / q \mathbb{Z})^\times$, $h = [G : H]$, and $p$ be the smallest prime in a given coset of $H$. Then either $p \leq 10^9$ or $$p \leq \left( (h-1) \log q + 3(h+1) + \frac{5}{2} (\log \log q)^2 \right)^2$$ If we let $q = \prod_{k=1}^n p_k$ and $H$ be the subgroup of quadratic residues, then the OP asks for the smallest positive integer in a particular coset of $H$. This is bounded above by the smallest prime in the coset, which can be bounded by the theorem. In particular, we have $h = 2^n$ and $\log q = (1 + o(1)) n \log n$, and consequently we infer $$x_n = O\left( 4^n n^2 (\log n)^2 \right)$$ This is a slight improvement over the answer above. • Cool! By the way, if you look at the numbers (well, let your computer look at them...) about 1/2 are primes. I am not sure if this trend continues... Sep 11 '15 at 12:14 • Good observation! Sep 11 '15 at 12:49 • By the way, none of the answers mention any nontrivial unconditional upper bound. Is there really nothing one can say? Sep 11 '15 at 12:55 On the possible precision of an asymptotic estimate... At "random" we expect $x_{n+1} = x_n$ with probability 1/2. For any particular size $B$, we expect there to be some $n \sim B$ with $x_n = x_{n + \lfloor \log_2 B \rfloor}$. Supposing that we estimate $x_n \sim 2^n \log n$, if $n$ is as above we have $$2^n \log n \sim 2^{n + \log_2 n} \log(n + \log_2 n)$$ Consequently, if this is the right estimate, it is unlikely that it can be stated more precisely than $$x_n = 2^{n + O(\log n)}$$ If we treat a quadratic residue as a Bernoulli random trial with probability 1/2, then we can model $$x_n \approx G_n \log n$$ Where $G_n$ is a geometric random variable with parameter $1/2^n$, and $\log n$ estimates the effect of having to be relatively prime to the first $n$ odd primes. Recall that the cumulative distribution for a geometric random variable is given by $$P(G_n > m) = (1 - 1/2^n)^m \approx \exp(-m/2^n)$$ We have $$P(G_n < a \log n \, 2^n) \approx 1 - n^{-a}$$ If we blithely assume the $G_n$ are independent random variables, then we have $$P(\forall n: G_n < a \log n\, 2^n) = \prod_n (1 - n^{-a})$$ which diverges to zero if $a=1$. Therefore: • $\forall n: x_n < 2^n (\log n)^2$ happens with probability zero • $x_n = O(2^n (\log n)^2)$ with probability 1 Similarly, $$P(G_n > c 2^n / n^a) \approx \exp(-c/n^a)$$ $$P(\forall n: G_n > 2^n / n^a) \approx \exp\left (-\sum_{n=1}^{\infty} 1/n^a \right)$$ and therefore • $x_n = \Omega(2^n (\log n) / n)$ happens with probability zero • $x_n = \Omega(2^n (\log n) / n^{1+\epsilon})$ with positive probability A more accurate model which makes the $G_n$ truly independent would be $x_n \approx x_{n-1} + B_n G_n \log n$ where the $B_n$ are independent Bernoulli random variables with parameter $1/2$. I'm pretty sure this doesn't affect the asymptotics above, but I have not verified it. This implies that the estimate $x_n = 2^{n + O(\log n)}$ given in the initial section should be expected to happen "at random" with probability 1, and we can't expect to do better in the sense that $x_n = 2^{n + o(\log n)}$ should happen with probability 0. • Certainly, the experiments indicate that there is a lot of jumping around, so I am sure you are right. Do you think that even this level of precision is possible? Sep 10 '15 at 7:10 • See the last two data points, just added :) Sep 10 '15 at 7:12 • I'm not sure. My main tool for conjecturing these things is to model everything as random variables, but I haven't fully worked out the right model, and I feel this example is outside of the domain where I have confidence the approach I want to use will give reasonable results anyways. – user13113 Sep 10 '15 at 7:47 • ... but my impression is that this approach leads to $E(x_n) \approx 2^n \log n$ and $Var(x_n) \approx 4^n (\log n)^2$, which would suggest that you can't do better than $O(2^n \log n)$ (and I haven't worked out the probability that such an expression holds for all of the random variables!). But my vague impression is that in these sorts of problems, the actual distributions tend not to vary to such great extremes as might be predicted by a simple random model. – user13113 Sep 10 '15 at 8:08 • The model I used, incidentally, is that $x_n = x_{n-1} + G_n B_n \log n$, where the $G_n$ and $B_n$ are independent, $B_n$ is a bernoulli variable with parameter 1/2 (to reflect the possibility that $x_n = x_{n-1}$), and $G_n$ is geometric with parameter $1/2^n$ (to reflect the need to do more random trials until you find all non-residues). (and the $\log n$ factor represents the need to be relatively prime to the first $n$ odd primes) – user13113 Sep 10 '15 at 8:13	2021-12-03T04:49:08	{ "domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/217612/primorial-puzzlement", "openwebmath_score": 0.9119747281074524, "openwebmath_perplexity": 247.02325675948828, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022625406662, "lm_q2_score": 0.8596637505099168, "lm_q1q2_score": 0.8399109344901365 }
https://math.stackexchange.com/questions/1840037/number-of-positive-unequal-integer-solutions-of-xyzw-20	# Number of positive unequal integer solutions of $x+y+z+w=20$ What is the number of positive different integer solutions of $x+y+z+w=20$, where $x,y,z,w$ are all different and positive? It would be nice if coding is not used. I am given the answer $552$. HINT: Each solution of the desired type is a permutation of a solution in which $x<y<z<w$, and every permutation of such an increasing solution is of the desired type. Each increasing solution has $4!=24$ permutations, so you need only count the increasing solutions and multiply by $24$. Added: This can be done fairly easily by brute force. For other methods see this question. To see what Marc van Leeuwen is doing in his answer, note that \begin{align} \frac1{1-x}&=1+x+x^2+x^3+\ldots=\sum_{n\ge 0}x^n\\ \frac1{1-x^2}&=1+x^2+x^4+x^6+\ldots=\sum_{n\ge 0}x^{2n}\\ \frac1{1-x^3}&=1+x^3+x^6+x^9+\ldots=\sum_{n\ge 0}x^{3n}\\ \frac1{1-x^4}&=1+x^4+x^8+x^{12}+\ldots=\sum_{n\ge 0}x^{4n} \end{align} If you take the product of these, each term in that product (before you collect like terms) will be of the form $$x^{k_1}x^{2k_2}x^{3k_3}x^{4k_4}=x^{k_1+2k_2+3k_3+4k_4}\tag{1}$$ for some non-negative integers $k_1,k_2,k_3$, and $k_4$. When you combine like powers of $x$, the coefficient of $x^{10}$ will be the number of terms like $(1)$ in which $k_1+2k_2+3k_3+4k_4=10$. Finally, there is a bijection between such $4$-tuples $\langle k_1,k_2,k_3,k_4\rangle$ and partitions of $10$ into parts of sizes $1,2,3$, or $4$: $k_i$ is the number of parts of size $i$. And as Marc had already briefly explained, the number of such partitions is equal to the number that we want. You may find it helpful to look at this Wikipedia article; when Marc talks about Young diagrams, he’s talking about what the article calls Ferrers diagrams. • Is it by taking $x1$=$x$-1, $y1$=$y$-2, and so on so u get $x1+y1+z1+w1$=10 so the no of solutions is 23?? – MathMan Jun 26 '16 at 9:40 • @Harish: Yes, that’s essentially the method used by Marc van Leeuwen in his answer to the question to which I linked in the addition to my answer. – Brian M. Scott Jun 26 '16 at 10:02 • Yes i understood most of that but why is it 1/1−X×1/1−X2×1/1−X3×1/1−X4?? I did not get that part and i couldn't ask him because of my lack of reputations – MathMan Jun 26 '16 at 10:06 • @Harish: Have you done any work yet with generating functions? – Brian M. Scott Jun 26 '16 at 10:09 • @Harish: That makes it pretty difficult to explain what he was doing there. Let me see if I can add something to my answer to give you at least the general idea. – Brian M. Scott Jun 26 '16 at 10:12 By inclusion-exclusion over the distinctness conditions: There are $\binom{20-1}3=\binom{19}3$ solutions disregarding the conditions. There are $\binom42$ ways to choose one pair that fails to be distinct, with common value $k$, and $19-2k$ options for the remaining two values, for a total of \begin{align} \sum_{k=1}^9(19-2k)&=9\cdot19-2\cdot\frac{9(9+1)}2\\ &=81\;. \end{align} There are $3$ ways to choose two non-overlapping pairs that fail to be distinct, and $\binom{10-1}1=9$ options for their values, and $12$ ways to choose two overlapping pairs that fail to be distinct, so that three values are equal, and then $6$ different options for the values. There are $\binom{\binom42}3=\binom63$ ways to choose three pairs, of which $4$ constrain only $3$ values to be equal, leading to $6$ options as above, and the remaining $\binom63-4$ constrain all $4$ values to be equal, leaving only $1$ option. For $k=4,5,6$, there are $\binom{\binom42}k=\binom6k$ ways to choose $k$ pairs, all of which constrain all $4$ values to be equal and thus leave only $1$ option. Thus by inclusion-exclusion the total count is $$\binom{19}3-\binom42\cdot81+3\cdot9+12\cdot6-4\cdot6-\left(\binom63-4\right)+\binom64-\binom65+\binom66=552\;.$$ Just for comparison with Brian Scott's solution, here is the longhand count solution. As is observed in several places above, it is sufficient to find the number of solutions with $x<y<z<w$ and multiply by 24, because each such solution can be permuted to give 24 solutions to the original question. Next note that $1+2+3+4=10$ is the smallest total for 4 unequal positive integers, so we are not going to get many solutions. Now $4+5+6+7>20$, so $x=1,2$ or 3. Since $3+4+5+6=18$ the only possible solutions for $x=3$ are $(3,4,6,7),(3,4,5,8)$. Similarly for $x=2$ we must have $y=3,4$ or 5. $y=5$ gives only $(2,5,6,7)$. $y=4$ gives only $(2,4,6,8),(2,4,5,9)$. $y=3$ gives only $(2,3,4,11),(2,3,5,10),(2,3,6,9),(2,3,7,8)$. Finally in a similar way $x=1$ implies $y=2,3,4,5$ and gives solutions $(1,2,3,4),(1,2,4,13),(1,2,5,12),(1,2,6,11),(1,2,7,10),(1,2,8,9)$ and $(1,3,4,12),(1,3,5,11),(1,3,6,10),(1,3,7,9)$ and $(1,4,5,10),(1,4,6,9),(1,4,7,8)$, and $(1,5,6,8)$. Total 23, hence 552 to question. --------- shorter version ------ I have spelt this out in full detail, but note that since only the count is required setting out all those cases with $z+w=k$ is not really needed. It is enough to say something like: $x=3$ implies $y=4$ and 2 solutions $x=2$ implies: $y=3$, 4 solutions; or $y=4$, 2 solutions; or $y=5$, 1 solution $x=1$ implies: $y=2$, 6 solutions; or $y=3$, 4 solutions; or $y=4$, 3 solutions; or $y=5$, 1 solution. Total 23 solutions. Times 4! gives 552 to the original question. ----------- comment ---------- Teachers often give simple questions for you to practice methods like generating functions. But simple questions can often be answered faster by more elementary methods. Often fastest is best! Here is the best Solution I found when I was preparing for IIT-JEE 2000-2003. suppose equation is $x + y +z + t = N$ find $k= [(N-2n-4)/2]$, where n is the least integer present is the solution. for example to find number of solutions in which least number present is $1$, put $n=1$. Now draw the following diagram Stop when $1$ comes. Now If $N$ is even add lower row terms along with middle row terms and multiply the sum with $4!$. This is the number of unequal integral solution of above equation having n as the least integer. If $N$ is odd add upper row terms along with middle row terms and multiply the sum with $4!$. Repeat the process for other possible values of $N$. In above case $n$ can have values $0, 1, 2$, & $3$, if we are interested in finding non-negative unequal integral solutions, because $4 + 5 + 6 + 7 > 20$ for $n=0$, there are $24\times 4!$ solution for $n=1$, there are $14\times 4!$ solutions for $n=2$, there are $7\times 4!$ solutions and for $n=3$, there are $2\times 4!$ solutions. Total number of non-negative solutions can be obtained by adding all$= 1128$ If you are interested in only positive unequal integral solutions drop the case of n=0, and you get $552$ If you wish to obtain solution for $x, y, z, t > 1$, drop the case $n=0$ and $n=1$ both, and you get $216$. and so on.......	2019-12-12T03:41:56	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1840037/number-of-positive-unequal-integer-solutions-of-xyzw-20", "openwebmath_score": 0.818031370639801, "openwebmath_perplexity": 282.2537124383555, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226327661525, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.839910933791227 }
http://math.stackexchange.com/questions/50671/why-isnt-there-a-good-product-formula-for-antiderivatives	# Why isn't there a good product formula for antiderivatives? Computing antiderivatives is more challenging than computing derivatives, in part due to the lack of a product formula''; namely, while $(fg)'$ can be expressed in terms of $f,f',g,g'$, there seems to be no way to express $\int fg$ in terms of $f, \int f, g \int g$ and related quantities. Is there any intuitive reason or heuristic explanation for why no such formula exists? I'm looking for a non-rigorous explanation which can be understood by first-year calculus students. - See also mathoverflow.net/questions/66377/… . – Qiaochu Yuan Jul 10 '11 at 18:52 Maybe use the notion of "elementary function". There are elementary functions $A,B$ such that both $\int A$ and $\int B$ are elementary, but $\int(AB)$ is not elementary. So there can be no simple "product rule" for integration, since any simple combinations of elementary functions is again elementary. – GEdgar Jul 10 '11 at 20:26 Because there isn't a good product formula for derivatives? – André Nicolas Jul 10 '11 at 20:30 @GEdgar - thats an excellent suggestion. Because I'm looking for something I can explain to a first-year calculus student, should I go in this direction, I would need an elementary function and a non-rigorous explanation, understandable by such a student, for why its antiderivative is not elementary. Does something like that exist at this level? – robinson Jul 10 '11 at 20:38 @user6312 - I suppose I neglected to say what, exactly, I mean by "good." I would be happy with any expression for $\int fg$ in terms of $\int f, f, f', f'', \ldots$ and $\int g, g, g', g'', \ldots$, since such an expression would turn the computation of antiderivatives of products into simple mechanics. – robinson Jul 10 '11 at 20:40 Expanding on the idea of @Bill Dubuque, let us hypothesize that we have a differentiable function $p(x_1,x_2,x_3,x_4)$ such that $\int fg=p(f,g,\int f,\int g)$ for all differentiable functions $f$ and $g$. (Assuming $f$ and $g$ are merely continuous is not good enough, since then the right-hand side may not be differentiable, whereas the left-hand side must be.) More precisely, we assume that whenever $F$ and $G$ are twice differentiable, we have $$\frac d{dx}p(F'(x),G'(x),F(x),G(x)) = F'(x)G'(x).$$ Let us take $F(x)=ux^2+ax+c$ and $G(x)=vx^2+bx+d$. Using the multi-variable chain rule, we get \begin{align} ab &= F'(0)G'(0) = \frac d{dx}p(F'(x),G'(x),F(x),G(x))\bigg\|_{x=0}\\ &= 2up_1(a,b,c,d) + 2vp_2(a,b,c,d) + ap_3(a,b,c,d) + bp_4(a,b,c,d). \end{align} (Here, $p_j=\partial p/\partial x_j$.) Since the left-hand side does not depend on $u$ and $v$, we must have $p_1=p_2=0$, for all $a,b,c,d$. But this implies that the function $p$ depends only on $x_3$ and $x_4$. So the above reduces to $$ab = ap_3(c,d) + bp_4(c,d).$$ Taking $a=1$ and $b=0$ gives $p_3=0$, for all $c,d$. Taking $a=0$ and $b=1$ gives $p_4=0$, for all $c,d$. Finally, then, we have that the function $p$ is a constant function. Since a constant function cannot satisfy our hypothesis, this shows that there can be no such function. - HINT $\$ If there were a product rule $\rm\:\int f\:g\ =\ p(f,g,\int f,\int g)\:$ for some polynomial $\rm\:p\:$ then for $\rm\:f = x,\ g = 1/x^2\:$ this would imply that $\rm\: log(x)\:$ is a rational function. - Thanks, this is exactly the sort of thing I'm looking for. Now suppose the student to whom I'm explaining this were to object that perhaps we should look for a non-polynomial $p$? I do see that this begins to look rather unlikely, but is there a simple argument to shut this possibility down? – robinson Jul 10 '11 at 21:32 I'd ask what they mean by "non-polynomial", because if they mean some rational function, then I believe the same argument works. If they mean any arbitrary continuous function or something, then it will be a bit trickier I'd guess, but I'm not sure I'd consider that a "rule" anymore. – Matt Jul 10 '11 at 22:02 +1 This is a very good answer. Although, I wonder, is there a intuitive explanation for why we shouldn't expect $\log(x)$ to be a rational function? (Besides a formal proof that it is not, that is.) – Zev Chonoles Jul 10 '11 at 23:23 Here's another one: If there were a product rule for antiderivatives then you could integrate $$e^{-x^2/2}=\bigl({1\over x}\bigr)\cdot\bigl(x e^{-x^2/2}\bigr)$$ in elementary terms. – Christian Blatter Jul 11 '11 at 8:07 Differentiating is somewhat like taking the difference of two things (that are close together). Integrating is somewhat like taking the sum of many things (that may be fluctuating). Suppose $f$ and $g$ are functions that we "know" how to differentiate and integrate. That's somewhat analogous to saying we have expressions for $f_2-f_1$ $g_2-g_1$ $f_1+\cdots+f_n$ $g_1+\cdots+g_n$ Wanting to differentiate or integrate the product $fg$ is somewhat analogous to wanting formulas for $f_2g_2-f_1g_1$ or for $f_1g_1+\cdots+f_ng_n$. With a difference of two things, you can do algebraic "tricks" such as $f_2g_2-f_1g_1=f_2g_2-f_1g_2+f_1g_2-f_1g_1$. It's not so easy to find an algebraic "trick" for expressing $f_1g_1+\cdots+f_ng_n$ in terms of $f_1+\cdots+f_n$ and $g_1+\cdots+g_n$. - Yes, this is more or less how I think about it. Another way to phrase the same observation is that differentiation is local, while integration is global. – Qiaochu Yuan Jul 10 '11 at 22:14 Maybe use the notion of "elementary function". There are elementary functions $A,B$ such that both $\int A$ and $\int B$ are elementary, but $\int(AB)$ is not elementary. So there can be no simple "product rule" for integration, since any simple combinations of elementary functions is again elementary. Example: $A(x) = \cos x$, $B(x) = \frac{1}{x}$. Then $\int \cos x\,dx = \sin x + C$ and $\int \frac{1}{x}\,dx = \ln x +C$ are elementary, but $\int \frac{\cos x}{x}\,dx$ is not. - Ok, but is there any simple way to convince a first year calculus student that $\int cos(x)/x$ is not elementary? That is the intended audience. – robinson Jul 10 '11 at 21:27 Arguably integration by parts is the anti-derivative form of the product rule; in fact it is derived from the differentiation (i.e. the usual) product rule. - Sure, but it expresses the antiderivative of one product in terms of the antiderivative of another product. I'm wondering if there is an intuitive reason we can't do even better and get rid of the products entirely. – robinson Jul 10 '11 at 18:49 I'm not sure I have a good explanation beyond "that's just how it works" :) Hopefully someone else will be able to give you a more satisfying, intuitive answer. – Zev Chonoles Jul 10 '11 at 19:01	2015-08-03T15:22:09	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/50671/why-isnt-there-a-good-product-formula-for-antiderivatives", "openwebmath_score": 0.9527915716171265, "openwebmath_perplexity": 191.5125882013445, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022632097108, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.8399109332160738 }
https://math.stackexchange.com/questions/4206782/finding-lambda-for-which-alpha-i-is-a-root-of-z2-lambda-z-4	# Finding $\lambda$ for which $\alpha=i$ is a root of $z^2+\lambda z-4$ (Note: $$i$$ refers to the imaginary unit) I am trying to find the values of $$\lambda$$ for which $$\alpha =i$$ is a root of the quadratic: $$z^2+\lambda z-4$$ Using the quadratic formula I got: $$z=\frac{-\lambda \pm \sqrt{\lambda^2+16}}{2}$$ (Note: I believe that $$\alpha$$ refers to the root that is equal to $$\frac{-b \textbf{+} \sqrt{b^2-4ac}}{2a}$$, but I could be mistaken) Therefore, to find the value of $$\lambda$$ for which $$\alpha =i$$, I have to solve the equation: $$i=\frac{-\lambda + \sqrt{\lambda^2+16}}{2}$$ Here is an outline of the steps I took: Multiply both sides by $$2$$ and add $$\lambda$$ to both sides: $$2i+\lambda=\sqrt{\lambda^2 +16}$$ Square both sides: $$4i^2 +4\lambda i +\lambda^2 =\lambda^2 +16$$ Subtract $$\lambda^2$$ from both sides then solve for $$\lambda$$: $$\lambda=-5i$$ However, $$\lambda=-5i$$ does not solve the equation ($$i \neq 4i$$ ), which is very confusing. Where did I go wrong? P.S: I did find that $$\lambda=-5i$$ was a solution to the other root, $$i=\frac{-\lambda - \sqrt{\lambda^2+16}}{2}$$. • if $i$ is a root doesn't it mean that $i^2+i\lambda -4=0$,$\lambda=-5i$ Jul 25, 2021 at 11:29 • Yes, that's part of the problem. Then why doesn't $i=\frac{-\lambda + \sqrt{\lambda^2+16}}{2}$ work when you substitute $\lambda=-5i$? Jul 25, 2021 at 11:38 • It does? $$i=\frac{-\lambda - \sqrt{\lambda^2+16}}{2}=i=\frac{5i - \sqrt{-25+16}}{2}=\frac{2i}{2}=i$$ It's just one of the two roots Jul 25, 2021 at 11:49 • Yes, I mentioned that at the end of my post. What I don't understand is why $i \neq \frac{-(5i)+ \sqrt{(-5i)^2+16}}{2}=4i$ Jul 25, 2021 at 11:52 • @TookieWookie: Sorry if this doesn't address your specific confusion, but the simplest way to solve this is as follows: if $i$ is a root of the quadratic $z^2+\lambda z-4$, then $$i^2+\lambda i-4=0\iff \lambda=-5i \, .$$ – Joe Jul 25, 2021 at 18:20 Each non-zero real or complex number $$w$$ has exactly two square roots $$r_1, r_2$$ which differ by sign ($$r_2 = - r_1$$). If $$w$$ is a positive real number, then both $$r_1, r_2$$ are real and the standard convention is to write $$\sqrt w$$ for the positive square root. In all other cases there is no universally accepted interpretation of $$\sqrt w$$. If $$w$$ is a negative real number, then one may of course use the definition $$\sqrt w = i \sqrt {\lvert w \rvert}$$ (as you do in your calculation). This is a good and reasonable choice for $$\sqrt w$$, but it is only one of two possible choices. For $$w \notin \mathbb R$$ it is even more arbitrary to define a unique value of $$\sqrt w$$. You corrrectly state that the quadratic equation $$z^2 + \lambda z - 4 = 0 \tag{1}$$ has the two solutions $$z=\frac{-\lambda \pm \sqrt{\lambda^2+16}}{2} \tag{2}$$ As explained above, this formula is based on a specific interpretation of the expression $$\sqrt{\lambda^2+16}$$; it involves a convention concerning the symbol $$\sqrt{\phantom x }$$. Perhaps it would be a more neutral way to say that the solutions of $$(1)$$ have the form $$z=\frac{-\lambda + r}{2} \tag{3}$$ where $$r$$ is any square root of $$\lambda^2+16$$, i.e. any solution of $$r^2 = \lambda^2+16 . \tag{4}$$ If you want to have $$i$$ as a solution of $$(1)$$, you have to determine $$\lambda \in \mathbb C$$ and $$r$$ with $$r^2 = \lambda^2+16$$ such that $$i=\frac{-\lambda + r}{2} \tag{5}$$ You write right from the start $$r = \sqrt{\lambda^2+16}$$, but this is inadequate if you use a fixed convention for the symbol $$\sqrt{\phantom x }$$. In fact, you must be aware that only one of the two solutions of $$(4)$$ will satisfy $$(5)$$ and the other produces a value $$\ne i$$. Can you be sure that a fixed square root convention produces the appropriate value for $$r$$? No, you can't. Your calculation correctly gives $$\lambda = -5i$$ and thus $$\lambda^2+16 = -9$$. The above convention for square roots of negative real numbers produces then $$r = \sqrt{-9} = 3i$$ which does not satisfy $$(5)$$. The other solution of $$(4)$$ is $$r = -3i$$ and this is the adequate one which satisfies $$(5)$$. Update: A simpler approach is this. You want to find $$\lambda$$ such that $$i$$ is a root of $$(1)$$. Let $$\rho$$ be the second root of $$(1)$$. Then $$(z-i)(z-\rho) = z^2 + \lambda z - 4$$ which gives $$i\rho = -4$$ and $$-(i + \rho) = \lambda$$. The first equation implies $$\rho = 4i$$ and the second implies then $$\lambda = -5i$$. • Thank you! I understand now... except for the bit after "Update: " How do you go from $(z-i)(z-\rho) = z^2 + \lambda z - 4$ to $i\rho = -4$ and $-(i + \rho) = \lambda$? Jul 26, 2021 at 8:20 • @TookieWookie We have $(z-i)(z-\rho) = z^2 - (i +\rho)z + i \rho$. Now compare the coefficients. Recall Vieta's formulas for quadratic equations! Jul 26, 2021 at 8:28 If a quadratic equation is given with certain coefficients its roots can be found. Else if the roots are given we can find out the quadratic equation that contains/defines these roots. The only quadratic equation which has roots $$\pm \alpha = \pm i\;$$ is: $$z^2+1=0\text { or } z^2+\alpha^2=0$$ where the complex conjugate has to be included. It is not clear how the given equation can be imagined or thought of to be able to accommodate these roots. To me appears such adjustment is not possible. • If we want both $\pm i$ to be roots, your argument is correct. But the OP only requires that $i$ is a root. Jul 25, 2021 at 23:12 • A quadratic equation with real coefficients can have a single isolated stand alone imaginary root? The totality needs to be considered in an answer even if OP asks for an impossible scenario ... or so I assume. Jul 25, 2021 at 23:59 • You are right again if the coefficients of the quadratic equation are required to be real. But the OP does not assume this. Jul 26, 2021 at 6:55 • ok thanks, .. your last edit clarifies it well. Jul 26, 2021 at 11:51 It isn't often discussed in textbooks (at least those I've seen) that polynomial and synthetic division also work for complex numbers. If we already know one of the zeroes of the polynomial $$\ z^2 + \lambda z - 4 \$$ is $$\ i \ \ ,$$ we may also carry out the calculation We must then have $$\ -5 + \lambda·i \ = \ 0 \ \Rightarrow \ \lambda \ = \ \frac{5}{i} \ = \ -5i \ \ .$$ The "reduced" polynomial is then $$\ z + (\lambda + i) \ = \ z + [-5i] + i \ = \ z - 4i \ \ ,$$ confirming that the second zero is $$\ z = 4i \ \$$ and that the quadratic polynomial is $$\ z^2 - 5i·z - 4 \ \ .$$ A small extension (which is not difficult to show) of the theorem that a quadratic polynomial with real coefficients has complex conjugate zeroes is that a quadratic polynomial has a real constant term and imaginary linear coefficent if and only if it has two purely imaginary zeroes.	2022-05-26T05:44:36	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4206782/finding-lambda-for-which-alpha-i-is-a-root-of-z2-lambda-z-4", "openwebmath_score": 0.9237287640571594, "openwebmath_perplexity": 131.8115896204816, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226314280636, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.8399109326409205 }