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A matrix is negative definite when ALL its leading principal minors alternate in sign, with the k-th order leading principal minor having the same sign as $(-1)^k$.
Please note, I am NOT talking about a problem instance with constraints ie I am not talking about bordered matrices. In that case, the rules are a bit different, but this is not the case here.
Also, note, that the order of a minor is derived from the number of columns and rows one has to delete in order to obtain the corresponding submatrix.
Counting is straightforward: For an $nxn$ matrix the k-th order leading principal minor is produced by deleting the last $n-k$ rows and columns of the original matrix. In this fashion, the 1-st order leading principal minor of a $4x4$ matrix is produced after deleting the last $4-1=3$ rows and columns while the 2-nd order leading principal minor of the same matrix needs to have the last $4-2=2$ rows and columns deleted etc.
Finally, note that taking the n-th order leading principal minor of an $nxn$ matrix means deleting $n-n=0$ rows and columns ie the n-th order minor is the determinant of the matrix itself.
Mathematica has (arguably) a lot of ways to produce the leading principal submatrices that are needed to produce the minors (their determinants). One simple and fast way to do it, is:
LeadingPrincipalMinors=Array[Minors[A, #][[1, 1]] &, 4]
Now, checking to verify if the signs of the minors are in agreement with the sign rule we can do the following:
Reduce[
And @@ MapIndexed[(
Reduce[#1 (-1)^First[#2] > 0, b[a], Reals]
The 'trick' I use in this piece of code is to take advantage of the fact that, when two numbers have the same sign, their product is positive. This is what the following excerpt of code from above, does:
#1 (-1)^First[#2] >= 0
Please, note that I have used b[a] instead of b in order to make explicit the dependence of b on a. If you chose to do so, you have to replace the definition of matrix A with something like
A=A/.b->b[a] | {
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A=A/.b->b[a]
but that is not necessary. I just find it useful to have dependence relationships be defined as explicitly as possible in my code.
Therefore the range of values over which the initial matrix is negative definite depends on a and b[a] being in an appropriate range of values.
Hope that helps :)
## -- update --
executing
Reduce[
And @@ MapIndexed[(
Reduce[#1 (-1)^First[#2] > 0, b[a], Reals]
and
Reduce[And @@ Thread[Eigenvalues[A] < 0], b[a], Reals]
yields the same output, namely:
0 < a < 1 && -Sqrt[1 - Sqrt[a]] < b[a] < Sqrt[1 - Sqrt[a]]
• Thanks for your answer! There is a small mistake in your code #1 (-1)^First[#2] >= 0 needs to be #1 (-1)^First[#2] > 0, otherwise the leading principal minors can also be 0, which does not imply negative definiteness. Changing the code this way leads to the same result I had earlier in my second approach and does not solve my problem – PTSammy Aug 13 '17 at 13:00
• @PTSammy you are right; I wrote in the body of the text "[w]hen two numbers have the same sign, their product is positive" but miss-typed ">=" instead of the correct ">" i the code segment. Will correct it. – user42582 Aug 13 '17 at 15:15
• @PTSammy As far as the last part of your comment is concerned-just for the sake of argument-technically speaking, my answer, answers the second part of your question "[D]oes anyone know what I am doing wrong or which of the results is correct?" as I already stated it would do, in the beginning of the text. Having said that, it is true that it does not identify what your possible error might be-which is something which I didn't state it would do. – user42582 Aug 13 '17 at 15:22
• You're right, you didn't state that. No worries, I didn't mean to offend you ;) – PTSammy Aug 13 '17 at 18:03
• @PTSammy all's well; hope you figure it out; have a nice time :) – user42582 Aug 13 '17 at 18:21 | {
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To show that $\ell^\infty$ is a vector space does it suffice to show it satisfies the Subspace Criteria?
If I want to show that $\ell^\infty$ (the set of all bounded sequences of real numbers) is a vector space does it suffice to show that since $\ell^\infty \subset \mathbb{R}_\infty$ (the set of all sequences of real numbers) and $\mathbb{R}_\infty$ is a vector space that the following are true:
(a.) $0\in \ell^\infty$ (this is trivial because the 0 sequence is bounded and so is in $\ell^\infty$)
(b.) For $u,v \in \ell^\infty$ we have that $u+v \in \ell^\infty$.
We can show this from the definition of bounded sequences, $u = \left\{x_1,x_2,\dots\right\}$ is bounded if $\exists M>0$ such that $|x_n|<M$ for all $N\in\mathbb{N}$. Similarly for $v = \left\{y_1,y_2,\dots\right\}$ bounded by $T>0$. Therefore:
$$|(u+v)_n|=|x_n +y_n| \leq |x_n| + |y_n| < M+T$$
Therefore for any two bounded sequences $u,v$ the sum $u+v$ is also bounded and so $\ell^\infty$ is closed under addition.
(c.) A similar proof shows that $c\cdot u \in \ell^\infty$ for all $c \in \mathbb{R}$
This shows that $\ell^\infty$ satisfies all the necessary conditions to be a subspace of $\mathbb{R}_\infty$. Therefore I think I have shown $\ell^\infty$ is a vector space, but then the problem has me show that $\mathbb{R}^\infty$ (the set of sequences that are eventually zero) is a subspace of $\ell^\infty$ and so I think I may have messed up.
• Your proof looks essentially correct to me. I think you may mean $|x_n + y_n| \le |x_n| + |y_n|$ in the body of your proof. Dec 6 '16 at 3:23
• Thank you, I copied the Latex from my homework so thanks a ton for pointing it out to me. Dec 6 '16 at 3:24 | {
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Your method is perfectly correct. The same criteria can be used to show that $\mathbb R^\infty$ is a subspace of $\ell^\infty$.
• Awesome. I took a walk and considered the proof for $\mathbb{R}^\infty$, since it's a finite sequence there exists a natural number $N \in \mathbb{N}$ such that for $n > N$ we have that $x_n = 0$ therefore we can take the upper bound to be $\max x_n$ and go about the same procedure to prove it is a subspace of $\ell^\infty$. Does this seem right (with more attention to detail added when I write the proof, of course) Dec 6 '16 at 3:28 | {
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Angles correspond to their opposite sides. Interior angle is defined as the angle formed between two adjacent sides of a triangle. If, $\alpha$, $\beta$ and $\gamma$ are three interior angles in a triangle, then $\alpha+\beta+\gamma = 180^\circ$ This basic geometrical property of a triangle is often used as a formula in geometry in some special cases. For ∆ABC, Angle sum property of triangle declares that. The interior, or inside, angles of a triangle always add up to 180 degrees. What is the sum of all angles in a triangle? What is the sum of interior angles of a Heptagon? The relation between angular defect and the triangle's area was first proven by Johann Heinrich Lambert.[4]. If the exterior angle of a triangle measures 130 , then what is the sum of the non-adjacent interior angles of that triangle? In addition to your geometry skills, you’ll be able to polish your algebra skills as you set up and solve equations on some of the tougher triangles. Thus, the sum of the interior angles of a triangle is 180°. Specifically, the sum of the angles is. Corresponding and Alternate Angles are also congruent angles. For a spherical triangle, the sum of the angles is greater than 180° and can be up to 540°. pg. 360 degrees. In the differential geometry of surfaces, the question of a triangle's angular defect is understood as a special case of the Gauss-Bonnet theorem where the curvature of a closed curve is not a function, but a measure with the support in exactly three points – vertices of a triangle. Your first step should be to set up an equation where the sum of the angles adds up to 180̊. Since X and, $$\angle J$$ are remote interior angles in relation to the 120° angle, you can use the formula. Learn to apply the angle sum property and the exterior angle theorem, solve for 'x' to determine the indicated interior and exterior angles. Tags: Question 7 . What is the sum of interior angles of a Nonagon? This means that the largest angle is opposite the longest side, | {
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sum of interior angles of a Nonagon? This means that the largest angle is opposite the longest side, while the smallest angle is opposite the shortest side. That should be || to base of the triangle (Side B… Sum of Interior Angles … List of trigonometric identities § Angle sum and difference identities, https://en.wikipedia.org/w/index.php?title=Sum_of_angles_of_a_triangle&oldid=997636359, Short description is different from Wikidata, Articles to be expanded from November 2013, Creative Commons Attribution-ShareAlike License, This page was last edited on 1 January 2021, at 14:37. The diagram below shows the interior and exterior angles of a triangle.. Sum of the Angle Measures in a Triangle is 180 ° - Justify. Ultimately, the answer was proven to be positive: in other spaces (geometries) this sum can be greater or lesser, but it then must depend on the triangle. The area of a … This set includes a 1 foldable interactive graphic organizer (INB) and 2 posters (8.5 X 11). Proof 1 uses the fact that the alternate interior angles formed by a transversal with two parallel lines are congruent. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"And the inside angle at each vertex of a polygon is always equal to (180° - deflection angle), because (new direction - old direction) + (reverse direction - new direction) = (reverse direction - old direction), which is always 180°.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-why-is-the-sum-of-the-interior-angles-of-a-triangle-equal-to-180'\u003eread more\u003c/a\u003e"},"name":"🗯Why is the sum of the interior angles of a triangle equal to 180? Sum Of The Interior Angles Of A Triangle 2 - Displaying top 8 worksheets found for this concept.. You can do this. Step-by-step explanation: Sum of interior angles of a polygon=180. Exterior angle is defined as the angle formed between a side of triangle and an adjacent side extending outward. Maybe it's a | {
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as the angle formed between a side of triangle and an adjacent side extending outward. Maybe it's a piece you'd been looking for on and off for a while. In a triangle, the exterior angle is always equal to the sum of the interior opposite angle. Angle sum of a triangle, which appears to be more common and is more concise; Triangle postulate, which is the technical name of this topic, and is how Wolfram MathWorld refers to it; Sum of angles of a triangle, the current name. [1] In the presence of the other axioms of Euclidean geometry, the following statements are equivalent:[2], The sum of the angles of a hyperbolic triangle is less than 180°. It is also possible to calculate the measure of each angle if the polygon is regular by dividing the sum by the number of sides. Proof 2 uses the exterior angle theorem. Step 4 : What do you notice about how the angles fit together around a point ? It follows that a 180-degree rotation is a half-circle. Interior Angles of a Triangle Rule. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"180°Triangle/Sum of interior angles\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-the-interior-angle-of-a-triangle'\u003eread more\u003c/a\u003e"},"name":"🗯What is the sum of the interior angle of a triangle? Angles between adjacent sides of a triangle are referred to as interior angles in Euclidean and other geometries. Example 8 : In a right triangle, apart from the right angle, the other two angles are x + 1 and 2x + 5. find the angles of the triangle. This one's y. Theorem 1: Angle sum property of triangle states that the sum of interior angles of a triangle is 180°. Therefore, straight angle ABD measures 180 degrees. Quick Answer: What Is The Significance Of The Baseball In The Movie Knives Out? And the inside angle at each vertex of a polygon is always equal to (180° – deflection angle), because (new direction – old direction) + (reverse | {
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is always equal to (180° – deflection angle), because (new direction – old direction) + (reverse direction – new direction) = (reverse direction – old direction), which is always 180°. All the black and white shapes in the figure on the left are hyperbolic triangles. You can solve for Y. $$120° = 45° + x \\ 120° - 45° = x \\ 75° = x. The influence of this problem on mathematics was particularly strong during the 19th century. Grab each of the points, and move them to increase or decrease their respective angles. Properties of Interior Angles The sum of the three interior angles in a triangle is always "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"Answer and Explanation: Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-can-a-triangle-have-two-right-angles'\u003eread more\u003c/a\u003e"},"name":"🗯Can a triangle have two right angles? Q.2.$$ Now, since the sum of all interior angles of a triangle is 180°. Angle sum property of a triangle Theorem 1: The angle sum property of a triangle states that the sum of interior angles of a triangle is 180°. Therefore, straight angle ABD measures 180 degrees. How do you prove that the sum of all angles in a triangle is 180? When a ray of sunlight. So: angles A are the same ; angles B are the same ; And you can easily see that A + C + B does a complete rotation from one side of the straight line to the other, or 180° Thanks, Laurdecl talk 11:47, 9 … These free geometry worksheets will introduce you to the Triangle Sum Theorem, as you find the measurements of the interior angles of a triangle. Defined in the differentially-geometrical sense. But there exist other angles outside the triangle which we call exterior angles.. We know that in a triangle, the sum of all three interior angles is always equal to 180 | {
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angles.. We know that in a triangle, the sum of all three interior angles is always equal to 180 degrees. If, $\alpha$, $\beta$ and $\gamma$ are three interior angles in a triangle, then $\alpha+\beta+\gamma = 180^\circ$ This basic geometrical property of a triangle is often used as a formula in geometry in some special cases. Proof 3 uses the idea of transformation specifically rotation. In this triangle ∠ x, ∠y and ∠z are all interior angles. Sum of the Interior Angles of a Triangle. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\circ {/eq}. This is also called angle sum property of a triangle . The sum of all of the interior angles can be found using the formula S = (n – 2)*180. For an ideal triangle, a generalization of hyperbolic triangles, this sum is equal to zero. The sum of the measures of the interior angles of a triangle equals 180º. Question 5: Do interior angles add up to 180º? It is not to be confused with, "Angle sum theorem" redirects here. The sum of the interior angles is always 180° implies, ∠ x + ∠y + ∠z = 180°. By interior angle sum property of triangles, ∠A + ∠B + ∠C = 180 0 ⇒ ∠A + 60 0 + 45 0 = 180 0 ⇒ ∠A + 105 0 = 180 0 ⇒ ∠A = 180 -105 0 ⇒ ∠A = 75 0. And again, try it for the square: The sum of the three angles of a triangle equals 180°. Any two triangles will be similar if their corresponding angles tend to be congruent and length of their sides will be proportional. The Triangle Sum Theorem states that the interior angles of a triangle always add up to 180̊. The angles in any given triangle have two key properties: 1. Answer: No, and here’s an example to illustrate. Finding the Number of Sides of a Polygon. Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. Label the angles A, B, and C. Try it first with our equilateral triangle: (n - 2) × 180 °(3 - 2) × 180 °Sum of | {
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angles A, B, and C. Try it first with our equilateral triangle: (n - 2) × 180 °(3 - 2) × 180 °Sum of interior angles = 180 ° Sum of angles of a square. Suppose a triangle had angles of 120, 45 and 15. Exterior Angle Property of a Triangle Theorem. Answer and Explanation: Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles. It was conjectured for a long time that the parallel postulate (also called Euclid's fifth postulate) followed from the first four axioms of … {"@context":"https://schema.org","@type":"FAQPage","mainEntity":[{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"The sum of three sides of a triangle is known as perimeter.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-all-3-sides-of-a-triangle'\u003eread more\u003c/a\u003e"},"name":"🗯What is the sum of all 3 sides of a triangle? Is the sum of any two angles of a triangle always greater than the third angle? Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. “SSS” is when we know three sides of the triangle, and want to find the missing angles….To solve an SSS triangle:use The Law of Cosines first to calculate one of the angles.then use The Law of Cosines again to find another angle.and finally use angles of a triangle add to 180° to find the last angle. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"\"SSS\" is when we know three sides of the triangle, and want to find the missing angles.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-find-an-angle-of-a-triangle-with-3-sides'\u003eread more\u003c/a\u003e"},"name":"🗯How do you find an angle of a triangle with 3 sides? Quick Answer: What Was The Number One Song For | {
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do you find an angle of a triangle with 3 sides? Quick Answer: What Was The Number One Song For 2019? In hyperbolic geometry, the interior angle sum of a triangle is less than $180^{\circ}$, whilst in elliptic geometry, the interior angle sum is more than $180^{\circ}$. 90 degrees. An exterior angle of a triangle is equal to the sum of the opposite interior angles. )? You can solve for Y. In a triangle, the angle opposite to longer side is larger and vice-versa. The rotation from A to D forms a straight line and measures 180 degrees. Thus, the sum of the interior angles of a triangle is 180°.. Do all Triangle angles equal 180? From the above diagram, we can say that the triangle has three interior angles. I've drawn an arbitrary triangle right over here. To Prove :- ∠4 = ∠1 + ∠2 Proof:- From • Since rotations are rigid transformations, angle measure is preserved and m∠ABC = m∠A'B'C' and m∠B'A'C' = m∠B''A''C''. Here are three proofs for the sum of angles of triangles. Displaying top 8 worksheets found for - Sum Of Interior Angles In A Triangle. Since X and, $$\angle J$$ are remote interior angles in relation to the 120° angle, you can use the formula. New questions in Math. a triangle have 3 sides n=3 So if the measure of this angle is a, the measure of this angle over here is b, and the measure of this angle is c, we know that a plus b plus c is equal to 180 degrees. Sum of all the interior angles of the triangle … 'There has to be a light blue sky piece somewhere here...' When we're working with triangles, sometimes we have missing puzzle pieces. Q. How tall is Jacquees? Some of the worksheets for this concept are Sum of the interior angles of a triangle 2 directions, 4 angles in a triangle, 4 the exterior angle theorem, Triangle, Sum of the interior angles of a triangle, Triangle, Relationship between exterior and remote interior angles, Name geometry polygons n. A. Every triangle has six exterior angles (two at each vertex are equal in measure). | {
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polygons n. A. Every triangle has six exterior angles (two at each vertex are equal in measure). "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"An interior angle is located within the boundary of a polygon.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-formula-for-interior-angles'\u003eread more\u003c/a\u003e"},"name":"🗯What is the formula for interior angles? How are we supposed … An exterior angle of a triangle is equal to the sum of the opposite interior angles. 90. Hence , the sum of all interior angles of a triangle is 180° regardless of the type of triangle . In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). Proof: Sum of all the angles of a triangle is equal to … How do you find an angle of a triangle with 3 sides? The sum of two sides of a triangle is greater than the third side. BLINDING LIGHTS. Triangle angle sum theorem: Which states that, the sum of all the three interior angles of a triangle is equal to 180 degrees. What is the sum of the measures of the two interior angles of a right triangle, not including the right angle (∠ sum of rt. So, the sum of three exterior angles added to the sum of three interior angles always gives three straight angles. Each interior angle of a regular octagon is = 135 °. Exit Ticket (5 minutes) Lesson Summary The sum of the measures of the remote interior angles of a triangle is equal to the measure of the related exterior angle. 80. 180° Step 6 : Describe th… An exterior angle is supplementary to its adjacent triangle interior angle. Thus, the sum of the interior angles of a triangle is 180°.. 100. Sum of interior angles in a triangle. Author: Robert Paterson. Note that spherical geometry does not satisfy several of Euclid's axioms (including the parallel postulate.). answer choices . Exterior Angle Theorem – Explanation & Examples. Can | {
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the parallel postulate.). answer choices . Exterior Angle Theorem – Explanation & Examples. Can A Doctor Change Your Medication Without Telling You? 4: Interactive fol actress K CallanIn. 60 degrees. The sum of the angles can be arbitrarily small (but positive). We already know that the sum of the interior angles of a triangle add up to 180 degrees. B. Proof. An exterior angle is supplementary to its adjacent triangle interior angle. As you can see from the picture below, if you add up all of the angles in a triangle the sum must equal 180 ∘ . In the given figure, the side BC of ∆ABC is extended. The two sides of the triangle that are by the right angle are called the legs... and the side opposite of the right angle is called the hypotenuse. Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"To prove the above property of triangles, draw a line \\overleftrightarrow {PQ} parallel to the side BC of the given triangle.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-prove-that-the-sum-of-all-angles-in-a-triangle-is-180'\u003eread more\u003c/a\u003e"},"name":"🗯How do you prove that the sum of all angles in a triangle is 180? C. step 2: Tear off each “ corner ” of the interior angles of the three interior always! Answer choices angles, taken one at each vertex, always sum up to 180 degrees its triangle. Measure of 90°, angle sum theorem '' redirects here all of the angles of a triangle n... Of hyperbolic triangles for right triangles, n=3, so sum-of-interior-angles = 3 * –! Small ( but positive ) 8 worksheets found for - sum of all interior angles ' B C! Up an equation where the sum of the interior angles, QR is produced to S.... Interactive graphic organizer ( INB ) and 2 posters ( 8.5 x 11.... One at each vertex, always sum up to the sum of interior angles only add to 180° | {
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( 8.5 x 11.... One at each vertex, always sum up to the sum of interior angles only add to 180° theorem. Right angles equation where the sum of the interior angles of a triangle is 75.. Are known as supplementary angles 30 and 70 degrees, the sum of three sides one. Sum up to 360°, angle sum property of triangle declares that angles adds up 180. As an important distinction for geometric systems measure of 90° sums up to 180 can Doctor!, angle sum theorem '' redirects here angles added to the sum of interior angles of triangle... ∠2 proof: sum of the angles of a triangle forms a straight line and measures 180 degrees interior!, or inside, angles of a triangle: Draw a line {... The origin a 180-degree rotation is a half-circle to triangles, no interior angle defined. And 90° congruent and length of their sides will be similar if Corresponding. - sum of all interior angles does not add up to 180.... And C. step 2: Tear off each “ corner ” of interior... Notice about how the angles of all angles in triangles for trigonometric identities concerning of... Less than 180 degrees Telling you and ∠z are all interior angles sum of interior angles of a triangle triangle... Spherical triangle, and move them to increase or decrease their respective angles to the sum of the opposite. Is lying on a flat plane that mathematically describes an interesting pattern about polygons and their interior angles a... A polygon circle [ 5 ] can not be the angles of,...: Describe th… the sum of the measures of the measures of the other two of!, see, defined as the exterior angles ( two at each vertex, sum... D. 150 angles opposite to longer side is larger and vice-versa be || to base of the interior angles a. * 180 angles a, B, and the Euclidean triangle postulate states that the sum of the angles! Measure ) and Alternate angles are also congruent angles axioms ( including the parallel.. Worksheet 2 - this angle Worksheet features 12 different triangles smallest angle is defined as the set | {
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2 - this angle Worksheet features 12 different triangles smallest angle is defined as the set points! ∠Y and ∠z are all interior angles of a triangle is a half-circle ∠ x, ∠y and ∠z all., 45 and 15 boundary of sum of interior angles of a triangle triangle and an adjacent side extending outward three points also... We already know that the sum of any triangle is planar, meaning it is equal... Where the sum of interior angles of a triangle are congruent for an ideal triangle, third. About how the angles a, B, and C. step 2: Tear off each “ corner of! Specifically rotation blue triangle, the measure of 90° enclosed by the triangle is a 3-sided figure three! Triangle measures 130, then what is the sum of the interior angles of the …... 1 foldable interactive graphic organizer ( INB ) and 2 posters ( x!, or inside, angles of that triangle makes money, what is x are in Movie... Are called interior angles of a triangle is 180 fraction of the given triangle have two key properties:.., you can use a formula that mathematically describes an interesting pattern about polygons and interior. Different triangles referred to as interior angles always gives three straight angles defined as the set of at. You 'd been looking for on and off for a while side B… Unit 5 Section:! All angles in triangles Euclidean triangle postulate can be arbitrarily small curvature, [ 6 ] so three... Suppose a triangle not be the angles fit together around a point in measure ) graphic! Is enclosed by the below-shown figure foldable interactive graphic organizer ( INB ) set covers interior angles only add 180°... ' B ' C ' about the midpoint of important distinction for geometric systems all interior of... N – 2 ) * 180 angles tend to be congruent and length of sides... Here are three proofs for the sum of the interior angles of a always. The sum of the triangle is 45° for on and off for a time... Lines are congruent then, the three interior angles ∠y and ∠z are interior... To point S. where ∠PRS is | {
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are congruent then, the three interior angles ∠y and ∠z are interior... To point S. where ∠PRS is exterior angle is defined as the exterior angles theorem specific to sum of interior angles of a triangle we! Right over here sides of a triangle with 3 sides of a triangle add up less! From what is x, then what is the Sky always Lighter inside a rainbow to... Can be hailed, how does GoodRx make their money every triangle has only right.: interior angles a circle [ 5 ] can not have arbitrarily small ( positive! + ∠2 proof: • Rotate ΔABC about the midpoint of planar, meaning it is on. Is greater than 180° and can be hailed, how does GoodRx make their money property of triangle declares.. To D forms a straight line through the origin Describe th… the sum of the given three,. Of one triangle are referred to as interior angles of that triangle is = °... This theorem can be found using the formula s = ( n – 2 ) * 180 always sum to... By a pair of adjacent sides of one angle of a triangle equal to the straight with. * 180 try it for the sum of the interior angles of a triangle 30°, 60° and 90° Johann! Worksheet features 12 different triangles is greater than the third side 70 degrees, the sum the... Not to be congruent and length of their sides will be similar if their Corresponding angles tend to confused... Was first proven by Johann Heinrich Lambert. [ 4 ] geometries exist, for this... Transformation specifically rotation add up to 360° B, and are positioned to form a straight line through origin. Corner ” of the triangle ( side B… Unit 5 Section 6: angles... A transversal with two parallel lines are congruent to three sides of a are. Do interior angles of a triangle is equal to the sum of 3 angles of regular. 180° step 6: Describe th… the sum of the angles is equal! Proven by Johann Heinrich Lambert. [ 4 ] Change your Medication Without Telling?. 'D been looking for on and off for a while interesting pattern about polygons and interior! Grab each of the interior angles of | {
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off for a while interesting pattern about polygons and interior! Grab each of the interior angles of a triangle of points at the fixed ' '. Theorem can be proved by the below-shown figure of PQR line through the origin three points property also.... Pair of adjacent sides of a Nonagon ' C ' about the midpoint of to longer is. Difference from 180° is a special relationship between the measures of the interior angles in geometry. Their measures add to 180° when the triangle … interior angles - sum of triangles. 150 angles opposite to equal angles in Euclidean geometry, the third angle B…! Set covers interior angles can not have arbitrarily small curvature, [ 6 ] so the three angles instead you! Does not satisfy several of sum of interior angles of a triangle 's axioms ( including the parallel.. Right over here with 3 sides of a polygon=180 is supplementary to its adjacent triangle interior angle ) covers. Or decrease their respective angles, angle sum theorem '' redirects here are within the boundary of a equal! Sky always Lighter inside a rainbow line and measures 180 degrees on a flat plane Who the. Be similar if their Corresponding angles tend to be confused with, angle. Was unknown for a spherical triangle, the exterior angles can be hailed, how does GoodRx make their?! Meaning it is lying on a flat plane Euclidean and other geometries exist, for which sum. Fraction of the interior angles is 135° D forms a straight line through origin! Where ∠PRS is exterior angle is defined as the set of points at the.. Blue triangle, a generalization of hyperbolic triangles, we all know the. Side BC of the angles is always Corresponding and Alternate angles are congruent! The points, and move them to increase or decrease their respective angles are all interior of. A couple angles here, but what is the Sky always Lighter inside a triangle planar! ’ s an example: we have a couple angles here, but what is the Significance of angle... Implies, ∠ x, ∠y and ∠z are all interior angles | {
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here, but what is the Significance of angle... Implies, ∠ x, ∠y and ∠z are all interior angles a... What do you notice about how the angles of a triangle add up 180. Located within the boundary of a Heptagon why sum of interior angles of all of sum of interior angles of a triangle. } ] }, what makes a double rainbow S. where ∠PRS is exterior angle is an angle a. ( < 90 degree ) angles states that the triangle 90 degrees parallel lines are congruent polygon=180 n-2. The midpoint of are called interior angles of the three interior angles of triangle. | {
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Suppose you’re on a game show, and you’re given the choice of three doors,
Behind one door is a car; behind the two others, goats. You pick a door, say No. 1, and the host of the show opens another door, say No. 3, which has a goat.
He then says to you, “Do you want to pick door No. 2?”.
Question: What would you do?
Is it to your advantage to switch your choice from door 1 to door 2? Is it to your advantage, in the long run, for a large number of game-tries, to switch to the other door?
Now whatever your answer is, I want you to check/prove your answer by a Monte Carlo simulation of this problem. Make a plot of your simulation for $\rm{ngames}=100000$ repeat of this game, that shows, in the long run, on average, what the probability of winning this game is if you switch your choice, and what is the probability of winning, if you do not switch to the other door.
As you see in the figure, although you may initially win by not switching your choice, in the long run, on average, you will lose, if you don’t switch your choice.
Here is an example MATLAB implementation, monteGame.m,
%rng('shuffle');
rng(1235);
nExperiments=100000; % the number of times the experiment is simulated
TotalWinsWithoutSwitch = 0;
TotalWinsWithSwitch = 0;
AllDoors = [1,2,3];
averageNumberOfWinsWithoutSwitch = zeros(nExperiments,1);
averageNumberOfWinsWithSwitch = zeros(nExperiments,1);
for iExperiment = 1:nExperiments
if mod(iExperiment,10000)==0; disp(['simulating experiment number ', num2str(iExperiment)]); end
ContainsCar = zeros(3,1); % first create the three doors, assuming none contains the car, hence all are zero valued.
doorWithCar = randi([1,3],1,1); % Now put the car behind one of the three doors randomly.
myChoice = randi([1,3],1,1); % Now make a choice of door randomly. | {
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excludedDoor = AllDoors(AllDoors~=doorWithCar & AllDoors~=myChoice); % find which door should be opened by the host
if length(excludedDoor)>1
excludedDoor = excludedDoor(randi([1,2])); % pick one of the empty doors randomly
end
% find the chance of winning by no door switching
if myChoice==doorWithCar
TotalWinsWithoutSwitch = TotalWinsWithoutSwitch + 1;
end
% find the chance of winning by door switching
myChoice = AllDoors(AllDoors~=myChoice & AllDoors~=excludedDoor); % update my choice to the alternative door after host's exclusion of one of the doors.
if myChoice==doorWithCar
TotalWinsWithSwitch = TotalWinsWithSwitch + 1;
end
averageNumberOfWinsWithoutSwitch(iExperiment) = TotalWinsWithoutSwitch / iExperiment;
averageNumberOfWinsWithSwitch(iExperiment) = TotalWinsWithSwitch / iExperiment;
end
% Now draw and export figures
figure('visible','on'); hold on; box on;
plot( averageNumberOfWinsWithoutSwitch ...
, 'color', 'blue' ...
, 'linewidth', 1.5 ...
);
plot( averageNumberOfWinsWithSwitch ...
, 'color', 'red' ...
, 'linewidth', 1.5 ...
);
axis([0,nExperiments,0,1]);
line1 = line( [1 nExperiments], [0.66666666 0.66666666] ...
, 'LineStyle', '--' ...
, 'color','green' ...
, 'LineWidth',1.5 ...
);
line2 = line( [1 nExperiments], [0.33333333 0.33333333] ...
, 'LineStyle','--' ...
, 'color','green' ...
, 'LineWidth',1.5 ...
);
xlabel('Experiment Repeat Number');
ylabel('Average Probability of Winning');
legend( {'No Door Switch','With Door Switch','theoretical prediction','theoretical prediction'} ...
, 'Location', 'southeast' ...
);
uistack(line1,'bottom');
uistack(line2,'bottom');
set(gca,'xscale','log');
set( gca ...
, 'XMinorTick','on','YMinorTick','on' ...
, 'FontSize', 13 ...
, 'LineWidth', 1.25 ...
);
saveas(gcf,'MontyGameResult.png');
%close(gcf);
Here is an example Python implementation, monteGame.py,
#!/usr/bin/env python
import numpy as np
import matplotlib.pyplot as plt
ngames = 100000 | {
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#!/usr/bin/env python
import numpy as np
import matplotlib.pyplot as plt
ngames = 100000
chance_of_success_by_switching = np.zeros(ngames,dtype=np.double)
chance_of_success_by_not_switching = np.zeros(ngames,dtype=np.double)
options = [1,2,3]
for i in range(ngames):
door_host_will_open = options.copy()
door_with_car = np.random.randint(1,4)
door_host_will_open.remove(door_with_car)
door_i_choose = np.random.randint(1,4)
if (door_i_choose!=door_with_car): door_host_will_open.remove(door_i_choose)
door_to_switch = options.copy()
door_to_switch.remove(door_i_choose)
door_to_switch.remove(door_host_will_open[0])
if i==0:
if (door_to_switch[0]==door_with_car):
chance_of_success_by_switching[i] = 1.0
else:
chance_of_success_by_not_switching[i] = 1.0
else:
if (door_to_switch[0]==door_with_car):
chance_of_success_by_switching[i] = (chance_of_success_by_switching[i-1]*np.double(i) + 1.0)/np.double(i+1)
chance_of_success_by_not_switching[i] = chance_of_success_by_not_switching[i-1]*np.double(i)/np.double(i+1)
else:
chance_of_success_by_not_switching[i] = (chance_of_success_by_not_switching[i-1]*np.double(i) + 1.0)/np.double(i+1)
chance_of_success_by_switching[i] = chance_of_success_by_switching[i-1]*np.double(i)/np.double(i+1)
trials = np.linspace(1,ngames,ngames)
plt.semilogx(trials, chance_of_success_by_switching, 'r-')
plt.hold('on')
plt.semilogx(trials, chance_of_success_by_not_switching, 'b-')
plt.legend(['switch choice' , 'not switch choice'])
plt.xlabel('Game Number')
plt.ylabel('Average probability of winnig')
plt.title('{} Monte Hall Games: On average, you win if your switch!'.format(ngames))
plt.savefig('MontyGameResult.png')
plt.show() | {
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# Can f''(x_0) = 0 if f'(x_0) =/= 0?
1. Dec 6, 2015
### FaroukYasser
I was wandering, can $$\ \frac { d^{ 2 }y }{ dx^{ 2 } } | _{ x={ x }_{ 0 } }=\quad 0$$ if $$\frac { dy }{ dx }| _{ x={ x }_{ 0 } }\neq \quad 0$$ and if so, what does this translate to geometrically?
2. Dec 6, 2015
### HallsofIvy
Staff Emeritus
Yes, of course. That is the same as saying that a function can have the value "0" at a point where its derivative is not 0. And it is very easy to give an example of that- suppose y= x- 1. That is a straight line such that y(1)= 1-1= 0 but its derivative is the constant 1 which is never 0. To relate that to your question about first and second derivatives, take an "anti-derivative". An anti-derivative of x- 1 is y= (1/2)x^2- x+ 1. Now, y'= x- 1 which is 0 at x= 1 while the second derivative is 1.
An nth derivative tells how fast the n-1 derivative is changing and is NOT related to the actual value of that n-1 derivative.
3. Dec 6, 2015
### Samy_A
$y=x-x_0$
4. Dec 6, 2015
### FaroukYasser
Thanks. But I was hoping for a non linear function. Or in other words, I am wandering what this means geometrically. I know that if a point is an inflection point then its second derivative is 0 but the converse doesn't necessarily hold. In other words, what does f''(x_0) = 0 tell us about that point other than it being a likely inflection point.
Thanks :)
5. Dec 6, 2015
### FactChecker
This means that the slope of the curve is not changing at that point
This means the slope is not zero at the point. So @Samy_A 's example of a sloped straight line satisfies both conditions at every point.
6. Dec 6, 2015
### FactChecker
Nothing. That is exactly what it means. sin(x) has an inflection point at every x= n * pi. Those are all inflection points with non-zero slopes.
7. Dec 6, 2015
### WWGD | {
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7. Dec 6, 2015
### WWGD
Actually, take a degree-2 or higher polynomial other than $x^n$, i.e. $a_nx^n+...+a_1x+a_0 ; a_j$ not all 0 $n>j\geq 0$, I would say the probability of having both f'(x) and f''(x)=0 is 0 under "reasonable" choices of pdf..
8. Dec 7, 2015
### FaroukYasser
I see. Thanks all!
9. Dec 22, 2015
### LCKurtz
The second derivative geometrically identifies concavity. f'' > 0 or f''<0 on an interval identifies concave up or down, respectively. In a case where f'' is continuous, the only way for concavity to switch is for f'' to pass through zero. So f'' = 0 at a point might be a case where f'' is changing sign, indicating an inflection point (change of concavity). Of course, f'' might not change sign nearby so it might not be an inflection point. I wouldn't say such a point is a "likely" inflection point but a "possible" inflection point. | {
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Upper and Lower Bounds of $\emptyset$
From some reading, I've noticed that $\sup(\emptyset)=\min(S)$, but $\inf(\emptyset)=\max(S)$, given that $\min(S)$ and $\max(S)$ exist, where $S$ is the universe in which one is working. Is there some inherent reasoning/proof as to why this is? It seems strange to me that an upper bound of a set would be smaller than a lower bound of the same set.
-
The supremum is defined to be the least upper bound. An upper bound for a set $K$ is defined to be an element $b$ such that for all $k\in K$, $k\leq b$. If $K$ is empty, then the condition "for all $k\in K$, $k\leq b$" is true by vacuity (you have an implication of the form "if $k$ is an element of the empty set, then $\lt$something happens$\gt$", and the antecedent is always false so the implication is always true).
Therefore, every element of $S$ is an upper bound for $\emptyset$; since $\sup(\emptyset)$ is the least upper bound, that means that $\sup(\emptyset)=\min($upper bounds of $\emptyset) = \min(S)$.
Similarly, the infimum is the greatest lower bound, and every element of $S$ is a lower bound for $\emptyset$, so $\inf(\emptyset) = \max($lower bounds of $\emptyset) = \max(S)$.
Yes, it is somewhat counterintuitive that you have $\sup(\emptyset)\leq\inf(\emptyset)$, but in fact the empty set is the only one for which you can have $\sup(\emptyset)\lt \inf(\emptyset)$. The empty set often causes counterintuitive results.
-
Interesting, thanks for that explanation. – yunone Sep 1 '10 at 5:20
A quick counterpoint to Arturo's answer: yes, the fact that $\operatorname{sup}(\emptyset) = -\infty < \infty = \operatorname{inf}(\emptyset)$ seems strange at first. But it is designed, among other things, to preserve an intuitive and useful property of infima and suprema:
Let $S$ and $T$ be subsets of $\mathbb{R}$ with $S \subset T$. Then
$\operatorname{inf}(T) \leq \operatorname{inf}(S)$
and
$\operatorname{sup}(T) \geq \operatorname{sup}(S)$. | {
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and
$\operatorname{sup}(T) \geq \operatorname{sup}(S)$.
Indeed, by adding elements to a set, its infimum can only get smaller and its supremum can only get larger. Now if you take $S = \emptyset$ and let $T$ range through even all one-element subsets of $\mathbb{R}$, you see that in order for these inequalities to hold we must define $\operatorname{inf}(\emptyset)$ and $\operatorname{sup}(\emptyset)$ as we did.
-
A very natural explanation of the correct definition for the values of min and max on empty sets arises from their dual universal definitions - analogous to the universal GCD, LCM definitions that I presented in a post here.
First some notation. $\;$ Write $\ \ \rm x \le S \;\iff\; x \le s,\;\: \forall\: s \in S,\;$ and dually for $\rm\; x \ge S$
DEFINITION of $\:$ min $\quad$ $\quad\rm x \le S \;\iff\; x \le min\ S$
DEFINITION of max $\:\quad$ $\quad\rm x \ge S \;\iff\; x \ge max\ S$
For min, when $\;\rm S = \emptyset\;$ is empty, the first clause $\;\rm x \le S\;$ in the min definition is vacuously true, hence the definition reduces to $\;\rm x \le \min \emptyset\;$ for all $\;\rm x\;$. Hence $\;\rm \min \emptyset = \infty\;$. Dually $\;\rm \max\emptyset = -\infty\;$.
As I remarked in said GCD, LCM post, such universal definitions often facilitate slick proofs. For some nontrivial examples of min, max flavor consider the slick proofs of the integrality of various products of binomial coefficients by employing the floor function, e.g. see Joe Roberts: Elementary Number Theory. Instead I close with an analogous slick GCD, LCM proof from my mentioned post (see it for further details).
Generally, in any domain, we have the following dual universal definitions of LCM and GCD:
DEFINITION of LCM $\quad$ If $\quad\rm a,b\ |\ c \;\iff\; [a,b]\ |\ c \quad$ then $\quad\rm [a,b] \;\;$ is an LCM of $\;\rm a,b$
DEFINITION of GCD $\quad$ If $\quad\rm c\ |\ a,b \;\iff\; c\ |\ (a,b) \quad$ then $\quad\rm (a,b) \;$ is an GCD of $\;\;\rm a,b$ | {
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Note: that $\;\rm a,b\ |\ [a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. Dually $\;\rm (a,b)\ |\ a,b \;$
Such $\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental
THEOREM $\rm\quad (a,b)\ =\ ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.
Proof: $\rm\quad\quad d\ |\ a,b \;\iff\; a,b\ |\ ab/d \;\iff\; [a,b]\ |\ ab/d \;\iff\; d\ |\ ab/[a,b] \quad\;\;$ QED
The conciseness of this proof arises by exploiting to the hilt the $\iff$ definition of LCM and GCD. Compare to less concise / general / illuminating proofs in many number theory textbooks.
-
@Bill: I think where you say "min" and "max" in the first two paragraphs you mean "inf" and "sup"; the empty set has no minimum and no maximum. – Arturo Magidin Feb 1 '11 at 16:33
@Arturo: above min,max are synonyms for inf,sup, cf. their universal definitions. – Bill Dubuque Feb 1 '11 at 20:30
@Bill: That's nonstandard use of min and max, as far as I am aware; I don't see what I'm supposed to compare, though... – Arturo Magidin Feb 1 '11 at 20:41
@Arturo: You're supposed to refer to the universal definitions of min, max that I am using above. – Bill Dubuque Feb 1 '11 at 21:59 | {
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# What does this 'L' and upside down 'L' symbol mean?
I need help finding out what the following symbols are called and what they do. I searched up math symbols but couldn't find them anywhere near there.
$$\lceil{-3.14}\rceil=$$
$$\lfloor{-3.14}\rfloor=$$
• Try Detexify. – lhf Nov 3 '15 at 10:22
They are ceil and floor values, that is they are the closest integers one below and one above respectively , see link.
• Ohhh! Thanks so much – Xirol Nov 3 '15 at 9:56
The notation $\lfloor x \rfloor$ (known as ‘the floor function’) denotes the largest integer less than or equal to $x \in \mathbb{R}$. Examples include $\lfloor7\rfloor$ = $7$, $\lfloor2.5\rfloor$ = $2$, $\lfloor\pi\rfloor$= $3$ and $\lfloor−2.5\rfloor$ = $-3$.
The notation $\lceil x \rceil$ (known as ‘the ceiling function’) denotes the smallest integer greater than or equal to $x \in \mathbb{R}$. So using the same examples as before $\lceil7\rceil$ = $7$, $\lceil2.5\rceil$ = $3$, $\lceil\pi\rceil$= $4$ and $\lceil−2.5\rceil$ = $-2$.
$$\lceil{-3.14}\rceil=-3$$
$$\lfloor{-3.14}\rfloor=-4$$
• For completeness, there is also the Nearest Integer function: $\lfloor x \rceil$. – YoTengoUnLCD Dec 22 '15 at 22:04
• @YoTengoUnLCD Oh wow, I never heard of that one before; thanks for pointing it out :) Perhaps you should make this as an answer, and give some examples for it (maybe the same numerical figures as I used for comparison). It's still vaguely related to OP's question. Besides, I would like to see this function in action! Thank you. – BLAZE Dec 22 '15 at 22:09
• Done! Thanks for the suggestion :-). – YoTengoUnLCD Dec 22 '15 at 22:20
Related to the two mentioned functions, there is the perhaps less common
$\text{Nearest integer function } \lfloor x\rceil$.
$$\lfloor 3.2\rceil=3\\ \lfloor 3.6\rceil=4\\ \lfloor -1.2\rceil=-1$$ | {
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$$\lfloor 3.2\rceil=3\\ \lfloor 3.6\rceil=4\\ \lfloor -1.2\rceil=-1$$
The function is ambiguous at numbers of the form $r+\frac 1 2, r\in \bf Z$, and so some kind of note should be made when using this function, clarifying what's to be done in those cases.
Note that we can set
$$\lfloor x\rceil_1=\left \lfloor x+\frac 1 2\right\rfloor\\ \\ \lfloor x\rceil_2=\left \lceil x-\frac 1 2\right\rceil\\$$
The first one rounds half integers up, and the second rounds them down. | {
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# Condition Number of Rectangular Matrices
The 2-norm condition number can be easily extended to rectangular matrices. I'm wondering if the inequality for the product of matrices still holds in that case, i.e.,
$\operatorname{cond}(AB) \leq \operatorname{cond}(A)\operatorname{cond}(B)$
• Yes, this is still true for rectangular matrices both in the spectral norm and in other norms. It's a good exercise so I won't ruin the challenge for you. A hint would be start by considering the definition of $\mathrm{cond}(A)$ that you're using. – Brian Borchers Mar 25 '16 at 0:04
I guess I figured out the answer to my question.
Suppose the SVD of $A = U \Sigma V^\ast$ (where $V^\ast$ is the conjugate transpose of the matrix $V$). Noting the fact that the unitary transformations $U$ and $V$ preserve the 2-norm, $\|Ax\|_{2}$ for any unit vector $x$ can be written as
\begin{align*} \frac{\|Ax\|_{2}}{\|x\|_{2}} = \|Ax\|_{2} &= \|U \Sigma V^\ast x\|_{2} \\ &= \| \Sigma V^\ast x\|_{2} \\ &= \| V^\ast \Sigma x\|_{2} \\ &= \|\Sigma x\|_{2} \end{align*}
Hence $\|Bx\|_{2} \leq \sigma_\max(B)$ and $\|Bx\|_{2} \geq \sigma_\min(B)$.
For $y = Bx$,
\begin{equation*} \|ABx\|_{2} = \|Ay\|_{2} \leq \sigma_\text{max}(A) \|y\|_{2} \leq \sigma_\max(A) \sigma_\max(B) \end{equation*}
Similarly,
\begin{equation*} \|ABx\|_{2} = \|Ay\|_{2} \geq \sigma_\min(A) \|y\|_{2} \geq \sigma_\min(A) \sigma_\min(B) \end{equation*}
Since the above statements are true for all $x$, they are true both for the minimum and maximum. Thus, \begin{equation*} \sigma_\max(AB) = \max \|ABx\|_{2} \leq \sigma_\max(A) \sigma_\max(B) \end{equation*}
and \begin{equation*} \sigma_\min(AB) = \min \|ABx\|_{2} \geq \sigma_\min(A) \sigma_\min(B) \end{equation*}
On dividing the above two equations, we obtain \begin{equation*} \operatorname{cond}_{2}(AB) \leq \operatorname{cond}_{2}(A) \operatorname{cond}_{2}(B) \end{equation*}
• You can accept your own answer. – nicoguaro Apr 24 '16 at 3:44 | {
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# Place $8$ rooks on a $10\times 10$ board.
The Problem:-
"In chess, a rook attacks any piece in the same row or column as the rook, provided no other piece is between them. In how many ways can $8$ rooks be placed on a $[8\times8]$ chessboard so that no two attack each other? What about $8$ rooks on a $10\times10$ board?"
I believe I have an answer for the first part of the question. When placing the first rook, there are 8 places on any particular column (or row) to place the rook, leaving just 7 places on a different column (or row) for the next rook, and so on, providing 8! possible ways to place the rooks in such a way that they cannot attack each other ($P(8,8) = 8!/(8-8)! = 8!$).
However, I am not sure I fully understand how this would work for a board where there are more rows and columns than pieces (such as on a 10x10 board). Does it become $P(10,8) = 10!/(10-8)! = 10!/2$ ? If so, why? If not, how should I approach this problem?
This problem was found in "Introduction to Combinatorics and Graph Theory" by David Guichard.
• Only one rook can go on each row and on each column. If there are two more rows and columns there are two rows and two columns with no rooks on them. just declare at the begininning that there will be 2 rows which won't be used. There are ${10 \choose 2}$ ways of doing that and ${10 \choose 2}$ ways of choosing the columns. So there are ${10\choose 2}^2 *D$ ways of doing this if $D$ is the number of ways to do it on an 8x8 board. – fleablood Nov 7 '17 at 18:45
So for the first question you need to place a rook in each column. The rook in the first column can go in $8$ places, the rook in the second column then has $7$ possible rows etc.
In the second part, you need first to choose the $8$ columns out of the $10$ available, and then place a rook in each of the columns. If once again you move from left to right, counting the number of possibilities, you should be fine. | {
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Rather than talking about "any particular" column or row, you should make a habit of defining the scheme more closely (the first, second rather than "any" etc). This will help to avoid a great deal of confusion and can also highlight any double-counting.
• For clarification, to determine the number of possible combinations for placing 8 rooks on a 10x10 board, we could calculate 10 choose 8 to determine how many 8x8 boards we could have, and then use that to determine how many total combinations there are, by multiplying the number from 10 choose 8 (45) with the number of possible permutations on the board (8!)? – oath Nov 7 '17 at 18:54
• @oath With $\binom {10}{8}$ columns you effectively reduce to a $10\times 8$ board - but then you have to find a position in each column (there are ten positions to start with, and each choice reduces the possibilities for the next choice by one) – Mark Bennet Nov 7 '17 at 18:56
• Ooooh, I see. So with this line of thought, would $10\choose 8$ multiplied by permutations $P(10,8)$ give the correct answer ($10\choose 8$ gives 45, and the permutation $P(10,8)$ gives $10!/2!$) (the answer provided by @Bram28)? – oath Nov 7 '17 at 19:15
For the second part, if we first focus on the columns the rooks are in: since there cannot be two rooks in the same column, the rooks must appear in $8$ different columns. And you can choose $8$ columns out of $10$ in
$$10 \choose 8$$
ways. Once, we've chosen the $8$ columns, let's place the rooks in those columns. Well, you can place the first rook in $10$ different rows, the next in $9$, etc. until the last one, which can be place in $3$ different rows, so we can place those $8$ rooks in:
$$\frac{10!}{2!} = \frac{10!}{2}$$
ways. This leads to a grand total of:
$${10 \choose 8} \cdot \frac{10!}{2}$$
ways of placing $8$ rooks on a $10 \times 10$ board. | {
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$${10 \choose 8} \cdot \frac{10!}{2}$$
ways of placing $8$ rooks on a $10 \times 10$ board.
• As a variation on this, we can (independently) choose eight columns and eight rows to be populated, each in $$10 \choose 8$$ different ways. Having done that, the chosen rows and columns form a board equivalent to the 8 x 8 case, which the OP has already answered (8!). Overall, then, for 10 x 10 case, there are $${10 \choose 8} ^ 2 \cdot 8!$$ different ways to place the 8 rooks. – John Bollinger Nov 7 '17 at 22:20
• @JohnBollinger Ooh, nice, thanks! – Bram28 Nov 7 '17 at 22:25
This is a different approach than what's been offered so far, but that might help some people understand the problem differently.
I would solve this by first assuming order mattered when placing the rooks (so placing on $A1$ then $B2$ is different than $B2$ then $A1$), then adjusting my answer to compensate.
The number of squares you can place the first rook on is $10^2$. Wherever you place it, you effectively turn your board into a $9 \times 9$ board with no rooks on it.
Now that we've lost a row and a column from our first rook, the number of squares you can place the second rook on is $9^2$.
Carrying on for each rook, we get:
$$10^2 \cdot 9^2 \cdot ... \cdot 3^2$$
Ways to place our rooks (don't forget we only have $8$ rooks, so we don't go all the way to $1^2$). Remember though, we assumed order matters. In reality, it does not. However, there's only $8!$ different ways to order our rooks, so we can divide our initial answer by $8!$, leaving:
$$(10^2 \cdot 9^2 \cdot ... \cdot 3^2) / 8!$$
Ways to place our rooks.
For the first case of placing 8 rooks on an 8x8 board your result of 8! is correct.
For the second case of placing 8 rooks on a 10x10 board I find the following approach to be the simplest way to do the calculations:
There are 10*9/2 = 45 ways to choose which two rows you will not be using. Likewise there are 45 ways to choose which two columns you will not be using. | {
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Once you have chosen which rows and columns not to use there are 8! ways to place the rooks in the remaining squares.
That gives a total of 8!*45² = 81648000 possibilities. | {
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# Combinatorial Proof for Binomial Identity: $\sum_{k = 0}^n \binom{k}{p} = \binom{n+1}{p+1}$ [duplicate]
I am studying combinatorics and I came across the identity $$\sum\limits_{k=0}^n \binom kp =\binom {n+1}{p+1}.$$ I have read the algebraic proof and it does not appeal to me. Is there an elegant counting trick we can use to arrive at this identity?
## marked as duplicate by Jack D'Aurizio, user230715, Macavity, vonbrand, 6005Aug 25 '15 at 20:55
• In case $p=0$? More than one question mark reads as an insulting tone. @WillJagy – Thomas Andrews Aug 25 '15 at 2:37
• @WillJagy I do realize that $\binom kp = 0$ if $k <p$ $(k \in \mathbb{Z})$. I guess it is more elegant to write it that way. And everything is well defined so there is no issues. – Miz Aug 25 '15 at 2:40
• @GeorgeSimpson: I disagree with this duplication mark. This question is explicitly asking for a combinatorial proof contrary to the referred question. So, two of three answers of the referred question are to exlude as proper answer for this question. The combinatorial answers here which are quite ok for this question do hardly answer the referred binomial identity. – Markus Scheuer Aug 26 '15 at 8:21
• @MarkusScheuer Looking among linked and related questions it seems that there are several other questions about combinatorial proof of the same identity. For example: 321022, 497413 1332282. If needed, we can discuss this further in chat. – Martin Sleziak Aug 29 '15 at 7:59
What do you get when you take a sequence of ones and zeros of length $n+1$ that has $k+1$ ones and remove the rightmost $1$ and everything to its right?
Example: $100100\rightarrow 100$
You get a sequence that has $k$ ones, the length varies between $k$ and $n$ depending on the actual sequence.
In fact this function is a bijection between the sequences of length $n+1$ and $k+1$ ones and the sequences of length $n$ or less and $k$ ones.
This establishes $\binom{n+1}{k+1}=\sum\limits_{j=k}^{n}\binom{j}{k}$ | {
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This establishes $\binom{n+1}{k+1}=\sum\limits_{j=k}^{n}\binom{j}{k}$
I'd prefer to omit the zero terms and prove $$\sum_{k = p}^n \binom{k}{p} = \binom{n+1}{p+1}.$$
Suppose we have $n+1$ people standing in a line so that we can refer to Person 1, Person 2, ..., Person $n+1$.
The righthand side counts the number of committees of size $p+1$ that can be formed out of these $n+1$ people.
The lefthand side counts the same thing, but it does so in cases.
The first case is that Person $p+1$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $p$ people in line, which can be done in $\binom{p}{p}$ ways.
The second case is that Person $p+2$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $p+1$ people in line, which can be done in $\binom{p+1}{p}$ ways.
Continuing in this way, the last case would be that Person $n+1$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $n$ people in line, which can be done in $\binom{n}{p}$ ways.
Summing over these cases yields the desired identity.
Here is another combinatorial approach using lattice paths.
We can interpret $\binom{n+1}{p+1}$ as number of lattice paths of length $n+1$ containing $p+1$ horizontal $(1,0)$-steps and $n-p$ vertical $(0,1)$-steps.
This is valid because there are $\binom{n+1}{p+1}$ choices to select $p+1$ steps in horizontal direction leaving the remaining $n-p$ steps for the vertical direction.
Let's consider all paths from $(0,0)$ to $(p+1,n-p)$. We know there are $\binom{n+1}{p+1}$ different paths. | {
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On the other hand each of these paths has to cross the vertical line going through $(p,0)$. The crossing points are $(p,0), (p,1), \ldots, (p,n-p)$. This way we partition the set of lattice paths into sets containing $\binom{k}{p}$ elements with $p\leq k \leq n$ establishing the identity
The RHS counts the ways to select $p+1$ places in a line of $n+1$ places.
The LHS counts the ways to select $p$ places in a line of $k$ places, summed over all values of $k: 0\leq k\leq n$. (More strictly, $k: p\leq k\leq n$, since there are no ways to do this for $k<p$).
Interpreting $k+1$ as the position of the last, or $(p+1)^{st}$, place, we notice that these both count ways to perform the exact same task. Ergo they are equivalent.
$$\sum_{k=p}^n \dbinom{k}{p} = \binom{n+1}{p+1}$$ | {
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# Group Combinations Problem
#### mathceleb
If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?
Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?
Any thoughts?
#### undefined
MHF Hall of Honor
If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?
Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?
Any thoughts?
I haven't worked this out rigorously, but I think the probability is $$\displaystyle \displaystyle \frac{1}{\binom{149}{2}}$$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $$\displaystyle \displaystyle \binom{149}{2}$$ ways to choose two people out of the remaining 149...
#### Soroban
MHF Hall of Honor
Hello, mathceleb!
We have a group of 150 people, broken into groups of 3, and only 3 are blue.
What is the probability that all 3 blue people are put in the same group?
The 150 people are divided into 50 groups of 3 people each.
There are: .$$\displaystyle \dfrac{150!}{(3!)^{50}}$$ possible groupings.
To have the 3 blue people in one group, there is: .$$\displaystyle {3\choose3} = 1$$ way.
The other 147 people are divided into 49 groups of 3: .$$\displaystyle \dfrac{147!}{(3!)^{49}}$$ ways.
. . Hence, there are: .$$\displaystyle \dfrac{147!}{(3!)^{49}}$$ ways to have the 3 blue people in one group.
The probability that all 3 blue people are in one group is:
. . $$\displaystyle \dfrac{\dfrac{147!}{(3!)^{49}}} {\dfrac{150!}{(3!)^{50}}} \;=\;\dfrac{147!}{(3!)^{49}}\cdot\dfrac{(3!)^{50}}{150!} \;=\;\dfrac{6}{150\cdot149\cdot148} \;=\;\dfrac{1}{551,\!300}$$
#### Plato | {
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#### Plato
MHF Helper
the probability is $$\displaystyle \displaystyle \frac{1}{\binom{149}{2}}$$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $$\displaystyle \displaystyle \binom{149}{2}$$ ways to choose two people out of the remaining 149...
This correct.
The 150 people are divided into 50 groups of 3 people each.
There are: .$$\displaystyle \dfrac{150!}{(3!)^{50}}$$ possible groupings.
Those are ordered partions.
There are $$\displaystyle \dfrac{150!}{(3!)^{50}(50!)}$$ unordered partitions.
There are $$\displaystyle \dfrac{147!}{(3!)^{49}(49!)}$$ of those where the blues are together.
Note that $$\displaystyle \dfrac{\dfrac{147!}{(3!)^{49}(49!)}}{ \dfrac{150!}{(3!)^{50}(50!)}}=\dfrac{1}{\binom{149}{2}}$$
#### undefined
MHF Hall of Honor
Hmm since Soroban's answer differs from mine I decided to do a little more research.
There are: .$$\displaystyle \dfrac{150!}{(3!)^{50}}$$ possible groupings.
I believe this should be $$\displaystyle \dfrac{\left(\dfrac{150!}{(3!)^{50}}\right)}{50!}$$ because the groups of 3 can be permuted.
...Hence, there are: .$$\displaystyle \dfrac{147!}{(3!)^{49}}$$ ways to have the 3 blue people in one group.
Likewise I believe this should be $$\displaystyle \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)}{49!}$$.
Leading to:
$$\displaystyle \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)\cdot50!}{49!\cdot\left(\dfrac{150!}{(3!)^{50}}\right)} = 50\left(\dfrac{1}{551,\!300}\right) = \dfrac{1}{11026} = \displaystyle \frac{1}{\binom{149}{2}}$$
See here (a PDF file) for some more info; under subheading "Partitioning".
Edit: Plato answered a lot quicker than I did, didn't refresh the page.
Similar threads | {
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## [Math][Python] Project Euler #99 and logarithmic representation
Comparing two numbers written in index form like $$2^{11}$$ and $$3^7$$ is not difficult, as any calculator would confirm that $$2^{11} = 2048 < 3^7 = 2187$$. However, confirming that $$632382^{518061} > 519432^{525806}$$ would be much more difficult, as both numbers contain over three million digits. Using base_exp.txt (right click and 'Save Link/Target As...'), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.
This is a problem where logarithmic representation of number can be used. We compute power not for numbers, but for their logarithmic representation. And they consumes way less RAM. Also, this is much faster. In fact, John Napier invented logarithms to ease multiplication operation.
Since we only need info, which number is bigger, logarithmic representations are enough: they obey this law as well.
#!/usr/bin/env python3
import math
f=open(filename,"r")
f.close()
return [item.rstrip() for item in ar]
tmp=123**45
print ("generic algo:", tmp)
# log with other base would also work!
# but here using natural log base for simplicity
tmp=math.log(123)*45
print ("using log: ", math.exp(tmp))
x=math.log(632382)*518061
y=math.log(519432)*525806
assert x>y
line=1
_max=0
max_line=None
for l in lst:
t=l.split(",")
x,y = int(t[0]),int(t[1])
r=math.log(x)*y
if r>_max:
max_line=line
_max=r
#print (line, x, y, r)
line=line+1
print (max_line, _max)
The result:
generic algo: 11110408185131956285910790587176451918559153212268021823629073199866111001242743283966127048043
using log: 1.111040818513185e+94
709 6919995.552420337
Spoiler: Wolfram Mathematica can operate on such big numbers without additional assistance. In fact, I first solved this problem using it. | {
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BTW, we can verify Project Euler's author's statement about "as both numbers contain over three million digits" using decimal logarithm instead of natural:
#!/usr/bin/env python3
import math
tmp=math.log(632382, 10)*518061
print (tmp)
3005261.2406258043
This coincides with Wolfram Mathematica's output. This number has 3005261 decimal digits. All correct.
###### (the post first published at 20220728.)
Yes, I know about these lousy Disqus ads. Please use adblocker. I would consider to subscribe to 'pro' version of Disqus if the signal/noise ratio in comments would be good enough. | {
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# Conditional Probability of rainfall
1. Apr 26, 2012
### TranscendArcu
1. The problem statement, all variables and given/known data
3. The attempt at a solution
a) P(Pickwick has no umbrella | it rains) = $\frac{\frac{1}{3}\frac{1}{3}}{\frac{1}{2}} = \frac{2}{9}$, which is the answer according to my answer key.
b) For part b we have:
There is a rain forecast which means he will bring the umbrella. The probability that it won't rain is 1/3.
There is a non-rain forecast which means he brings the umbrella with a probability of 1/3 and it will not rain with a prob of 2/3.
P(Pickwick has umbrella | no rain) = $\frac{1}{3} + \frac{1}{3}\frac{2}{3} = \frac{5}{9}$. But the answer is apparently 5/12. What have I done incorrectly here?
2. Apr 26, 2012
### MaxManus
I got the same as you at b) and I cant see why it is not correct.That's no guarantee for that you are correct though
3. Apr 26, 2012
### Ray Vickson
I think you are correct. Let's do it in a simple way, just by counting. Imagine 9,000 days. In 4,500 of those, it rains and in 4,500 it does not. Let's use the notation R = rain, Rc = no rain, F = Forecast rain, Fc = Forecast no rain, U = umbrella, Uc =no umbrella
Lay out the data in a table:
$$\begin{array}{lccc} & R & Rc & \text{tot}\\ F & N1 & N2 & (N1+N2)\\ Fc & M1 & M2 & (M1+M2) \\ \text{tot} & 4500 & 4500 & 9000 \end{array}$$
We are given N1/(N1+N2) = 2/3 and M1/(M1+M2) = 1/3, as well as N1+M1 = 4500 and N2+M2 = 4500. Solving these equations we get N1 = 3000, N2 = 1500, M1 = 1500, M2 = 3000. So, the table is:
$$\begin{array}{lccc} & R & Rc & \text{tot}\\ F & 3000 & 1500 & 4500 \\ Fc & 1500 & 3000 & 4500 \\ \text{tot} & 4500 & 4500 & 9000 \end{array}$$
From this it follows that P(F) = 4500/9000 = 1/2 and P(Fc) = 1/2.
In (b), in the 4500 Rc-days, U occurs in 1500 (F-days) + (3000/3) (Fc-days), for a total of 2500 days. Thus, P(U|Rc) = 2500/4500 = 5/9.
RGV | {
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# Name the heuristic: exploiting the legitimacy of the questioner
As a child, I made frequent use of a particular 'trick' in order to make short work of many different problems. The general form is to be presented a question which wants a definite (numerical) answer, where the question statement contains a free variable. The 'trick' was to recognize that the answer (which the teacher has told us exists by asking for it) doesn't depend on the free variable, so we now know that we can substitute any convenient value we like for the free variable and find the answer.
This FiveTrangles problem is an excellent example. The height of the non-shaded triangles is unspecified, so we know straight away that it doesn't figure in the answer to the question. Since it doesn't matter, we can set it to zero and replace this question with the trivial question of finding the area of a triangle with $base = 10$ and $height = 5$.
Does this technique have a name? Do problems which permit it have a name? Should such questions be avoided?
• I'm not sure I understand. Do you mean leaving the answer as an expression with a variable in it?
– Karl
Feb 23 '15 at 19:55
• @Karl Another example: the question "$(x + 4)^2 - x^2 - 8x$ simplifies to a number. Find that number." can be solved with algebra, or it can be solved by plugging in x=0, since apparently the answer does not depend on x. Feb 23 '15 at 20:35
• That's a clearer example for me thanks. I get the triangle question too - presumably the $x$ cancels and isn't left as a variable in the expression.
– Karl
Feb 23 '15 at 21:13
• More of a technique than just a 'trick', see en.wikipedia.org/wiki/Limiting_case_%28mathematics%29 Feb 24 '15 at 3:35
• Somewhat related: for single-answer multiple choice problems, if two answers are equivalent, then they must both be wrong. Feb 24 '15 at 5:21 | {
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The heuristic described here is one manifestation of what Polya (1945) and others thereafter refer to as trying a special case. I do not know of a more specific term for the context that you have put forth, but this is often how one approaches a problem if you are initially unsure as to how to solve it in generality.
It is a good way to get an initial foot-hold on a problem, and especially useful in the context of some standardized tests. For example, consider the following:
Question: Player 1 flips a penny $$n>0$$ times, and Player 2 flips a penny $$n+1$$ times. What is the probability that Player 2 flips more heads than Player 1?
A. 1/2
B. 1/3
C. 1/4
D. 1/8
Since none of the choices depends on $$n$$, you can consider $$n=1$$ and see easily that the probability is $$1/2$$. This allows you to blaze through the question very quickly. (Indeed, this holds for all $$n \geq 1$$.)
Although testing well is an important life skill, your final question as to whether such problems are advisable is interesting. In my estimation, they can be good in the classroom as a way of scaffolding, as long as you press students (when appropriate) to explain why their reasoning holds for the general case.
As one small example, induction problems are done like this: You show the (hopefully easy) base case, and then complete the proof by setting up and using your inductive hypothesis.
A possible pitfall, though, is not only the absence of reasoning by students (as in the example question above) but also a possible reinforcement around misconceptions of proof by example. (I believe this relates to the recent query about counterexamples in MESE 7466.) For more about some of these misconceptions, check out work by Orit Zaslavsky (google scholar). | {
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Lastly: The issue with guessing answers based on problem statements is pervasive in school mathematics. It ranges from only seeing examples of the form $$a + b = \square$$ and adopting a misunderstanding of the equal sign as an operator, to problems in which students just try to combine numbers in ways that might be "sensible" (without really engaging in sense-making). Cf. the classic:
Teacher: There are 125 sheep and 5 dogs in a flock. How old is the shepherd?
Student: (thinking: 125+5, 125-5, and 125*5 are too big; it must be 125/5) The shepherd is 25. | {
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Student: (thinking: 125+5, 125-5, and 125*5 are too big; it must be 125/5) The shepherd is 25.
• I had considered trying a special case, but there's more going on than that. Something akin to trying the easiest case, and going with it, because all of the cases are the same (not that I understand why, but if it weren't true the question wouldn't have been put to me this way). That's a mouthful though. Thanks for the answer. Feb 24 '15 at 1:15
• @NiloCK If you know why "all the cases are the same," then you could say that you are reducing to a special case (rather than just trying one). But there is (obviously) quite a leap between solving for one special case and understanding that the special case answers (or helps answer) a seemingly more general question. Feb 24 '15 at 1:25
• The 'why' for me (as a kid) was that the problem setter put the question that way. That is, if all the cases weren't the same, then the question wouldn't (couldn't) have been asked. I knew at the time that this was 'unfair' information, but it's hard for a kid writing math contests to discard such information on the grounds that it was 'ill-gotten' or 'inadmissible'. Feb 24 '15 at 1:38
• @NiloCK Benjamin addresses that "why" in the answer. Essentially, the why amounts to a guess that may or may not be true. A teacher could easily throw in questions where the real answer is "not enough information" or "none of the above" just to discourage that heuristic. In my opinion, discouraging it forces students to think through why the missing information doesn't matter, leading to a deeper understanding of the logic and the problem and the solution. For me, that sort of process is when I learn the most. Feb 25 '15 at 4:28
(continued from previous post) | {
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(continued from previous post)
In this next problem, the horizontal leg of the large right triangle and the vertical leg of the smaller right triangle are also not fixed, but setting each to zero won't lead to a solution at all: http://fivetriangles.blogspot.com/2013/04/56-obtuse-triangle-area.html (Never mind that getting in the habit of simply setting values to zero will lead to a "zero" in calculus.)
Finally, the following proof uses the identical triangle area concept as in the original "concave quadrilateral" problem, but the "trick" doesn't work at all: http://fivetriangles.blogspot.com/2013/10/104-equal-area-proof.html
• You can combine both answers into one. There is no character limit (within reason) in answers. Feb 23 '15 at 21:54
• Stack exchange proscribed more than 2 links in one post if 0 ≤ rating < 10. Feb 23 '15 at 22:11
• You can't control the order in which answers appear, so calling another answer the "previous post" is not good. For instance, if this answer gets more points it will even appear above the other one. Call your other answer "my other post" rather than "previous" one.
– KCd
Feb 25 '15 at 11:28
In my SAT tutoring, I refer to such problems as underspecified. For instance, consider this problem:
If $\frac{a}{2} + \frac{b}{3} = 0$, then find $\frac{a}{b}$.
I teach the student that since two variables are presented but only one equation, this makes the problem underspecified. In other words, a and b are not uniquely determined. This gives the student license to make up a number for one out of $a$ and $b$, then use the given equation to solve for the other variable, and finally substitute both $a$ and $b$ into the expression $\frac{a}{b}$ to obtain the answer. | {
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There are also many examples of underspecified SAT geometry problems. Often the problem will ask the student to find the value of $a+b$ or some other expression involving two variables. Example: The Official SAT Study Guide (second edition) by CollegeBoard, page 799, problem 13.
That some solutions are independent of a variable is not uncommon, but it does not necessarily trivialise a problem or negate its teaching value. The following problem presents similar possibilities, but we'd argue that solving it with such a "trick" is actually more circuitous than finding the answer directly (and may be more enlightening, too): http://fivetriangles.blogspot.com/2012/04/area-problem.html Here is a nice animation someone posted of the situation:
The shortcoming in such "tricks", we agree, is that they may lead to tunnel vision, but worse, their utilisation is at the user's risk.
(continued in next post)
• I assume that you're the owner of the Five Triangles blog? Great to have you on the site; I love yours. Many beautiful questions. Feb 24 '15 at 0:56
• Now that you have the necessary reputation, perhaps you could merge your two answers into one? Your other post (at this time) has 3 up-votes, whereas this one has only 1; so it would make better sense to merge into the higher voted one. (FWIW: I have not up-voted the other one, and will happily do so if I see them merged; in this way, deletion will not cause you any net-loss in reputation!) Mar 26 '15 at 5:50
I don't know the answer to the questions about the potential names for such things.
As to whether such questions should be avoided: I think some types of these questions can maybe help students to think about quantifiers, or to really think about the concepts involved in the question. | {
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I like Chris Cunningham's example: if you tell the student that $(x+4)^2 - x^2 - 8x$ simplifies to a number, and ask them to find the number, they may just do the algebra of simplifying. But if you show them that you can get the answer by plugging in $0$, then you can teach them something about what it really means to say that "$(x+4)^2 - x^2 - 8x$ simplifies to a number". Now you can talk about what it means for a statement to hold "for all $x$". Things like this are useful to be familiar with. Many college students struggle with this when I ask them, for example, to show that a function is not linear!
It's also nice to have this kind of logic on hand for when you need to remember things. For example, say you were taking a linear algebra exam and you knew the determinant was either the sum or the product of the eigenvalues. Well, you could just consider the $2\times2$ identity matrix and figure it out pretty quickly! This is because you knew the statement didn't depend on the matrix or even the size of the matrix!
I (and this is just opinion) don't like questions like the triangle one that you linked to. Because there, the "trick" from the solver's perspective is just to realize that the question asker left out some words! I don't know how useful it is to have someone apply the logic "If they asked me this, it has an answer, and if they didn't say it depended on quantity $A$ then it must not depend on quantity $A$.'' It is not clear to me what the mathematical benefit is to that.
• RE: last paragraph. One benefit to the problem solver/student is to consider the importance of the givens; alternatively, a benefit to the problem poser/instructor is to consider more carefully how problems are phrased. Feb 25 '15 at 2:51 | {
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I would be surprised if the solution strategy or problem types have a name. While it can be useful to find the correct answer, what does it accomplish from a learning perspective? Take, for instance, Chris Cunningham's problem of "$(x+4)^2 - x^2 -8x$ simplifies to a number. Find that number." As an instructor, I would be looking to make sure the students accomplished the appropriate learning outcomes for the problem. For this problem, I would more than likely be looking for (1) can they multiply out $(x+4)^2$ and (2) can they simplify the addition of terms. Such a technique would not accomplish (1) or (2), so I would avoid writing problems that can (knowingly) make use of it nor would I ever teach such meta-strategies.
This reminds me of a multiple choice exam question on trig identities: "Which of the following is an equivalent identity to $\cot(x) + \tan(x)$?" with $\csc(x)\sec(x)$ being in the list of answers. Instead of using trig identities, many students plugged in a value like $x=10$ and compared. While it works for this case, it dodges the actual learning and testing of the concept.
Discovering the irrelevance of $x$ in the expression could be achieved by getting the students (perhaps in different groups) to substitute different values in for $x$ and compare results. Was this what they expected? Which number would they choose to substitute knowing the answer will always be the same? Having aroused their curiosity the expanding brackets and simplifying can be motivated and discussed. This could be a rich activity if the emphasis is on discovery and not just sub in $x=0$ | {
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# Points A & B Out At Sea
#### xyz_1965
##### Member
From points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.
1. Find height of the mountain.
2. Find the distance from point A to the bottom of the mountain.
I think making two triangles makes sense here.
Triangle 1
tan(41°) = h/x
Let h = height of mountain.
Let x = the distance from point B out at sea to the bottom of the mountain.
Solving for x, I get h/tan(41°).
Triangle 2
tan(37°) = h/[80 + (h/tan(41°)]
I need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.
Lastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.
1. Is any of this right?
2. Is there an easy way to solve
tan(37°) = h/[80 + (h/tan(41°)] for h?
#### skeeter
##### Well-known member
MHB Math Helper
assuming A, B, and the mountain top are located in the same vertical plane ...
$h = x_B \tan(41)$
$h = x_A \tan(37)$
$x_A - x_B = 80 \implies x_A = x_B +80$
$x_B \tan(41) = (x_B+80)\tan(37)$
$x_B = \dfrac{80\tan(37)}{\tan(41)-\tan(37)}$
from here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.
#### xyz_1965
##### Member
assuming A, B, and the mountain top are located in the same vertical plane ...
$h = x_B \tan(41)$
$h = x_A \tan(37)$
$x_A - x_B = 80 \implies x_A = x_B +80$
$x_B \tan(41) = (x_B+80)\tan(37)$
$x_B = \dfrac{80\tan(37)}{\tan(41)-\tan(37)}$
from here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.
Explain $x_B$ and $x_A$.
#### skeeter
##### Well-known member
MHB Math Helper
$x_B$ is the horizontal distance from point B to the point directly below the mountain top. | {
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$x_A$ is the horizontal distance from point A to the point directly below the mountain top.
#### xyz_1965
##### Member
$x_B$ is the horizontal distance from point B to the point directly below the mountain top.
$x_A$ is the horizontal distance from point A to the point directly below the mountain top.
Thank you for clearing this up for me.
#### Country Boy
##### Well-known member
MHB Math Helper
From points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.
1. Find height of the mountain.
2. Find the distance from point A to the bottom of the mountain.
I think making two triangles makes sense here.
Triangle 1
tan(41°) = h/x
Let h = height of mountain.
Let x = the distance from point B out at sea to the bottom of the mountain.
It would make more sense to say these before you say "tan(41)= h/x!
Solving for x, I get h/tan(41°).
Specifically, x= h/tan(41).
Triangle 2
tan(37°) = h/[80 + (h/tan(41°)]
I think I would have first said "Let y be the distance from B to the base of the mountain" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing.
I need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.
Yes.
Lastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.
No! You were doing so well and then you forgot how you had defined "A" and "B'! "A" and "B" are points, not distances so you can't add them. "x" was the distance from point A to the base of the mountain.
1. Is any of this right?
2. Is there an easy way to solve
tan(37°) = h/[80 + (h/tan(41°)] for h?
A= h/(B+h/C)
Get rid of the fraction by multiplying both sides by B+ h/C:
AB+ (A/C)h= h
AB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C
h= ABC/(C- A) | {
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AB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C
h= ABC/(C- A)
Of course A= tan(37°)= 0.7536, approximately, B= 80, and C= tan(41°)= 0.8693 approximately.
#### xyz_1965
##### Member
It would make more sense to say these before you say "tan(41)= h/x!
Specifically, x= h/tan(41).
I think I would have first said "Let y be the distance from B to the base of the mountain" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing.
Yes.
No! You were doing so well and then you forgot how you had defined "A" and "B'! "A" and "B" are points, not distances so you can't add them. "x" was the distance from point A to the base of the mountain.
A= h/(B+h/C)
Get rid of the fraction by multiplying both sides by B+ h/C:
AB+ (A/C)h= h
AB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C
h= ABC/(C- A)
Of course A= tan(37°)= 0.7536, approximately, B= 80, and C= tan(41°)= 0.8693 approximately.
I thank you for the break down. It's ok to make mistakes here. Learning is not possible without errors.
#### Country Boy
##### Well-known member
MHB Math Helper
That is true. And you learn by having those errors pointed out!
#### xyz_1965
##### Member
That is true. And you learn by having those errors pointed out!
I agree. This is why I don't mind being corrected as long as it is done respectfully. | {
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# Commuting of Hom and Tensor Product functors?
Let $V_i,W_i$ be finite dimensional vector spaces, for $i=1,2$. Assume we have homomorphisms $\phi_i:V_i\rightarrow W_i$.
Then, there is an induced map $\widehat{\phi_1 \times \phi_2} \in Hom(V_1 \otimes V_2,W_1 \otimes W_2)$ such that $\widehat{\phi_1 \times \phi_2}(v_1 \otimes v_2) = \phi_1(v_1)\otimes\phi_2(v_2)$.
We can look at this construction as a map $\psi: Hom(V_1,W_1)\times Hom(V_2,W_2)\rightarrow Hom(V_1 \otimes V_2,W_1 \otimes W_2)$, where
$\psi:(\phi_1,\phi_2)\rightarrow \widehat{\phi_1 \times \phi_2}$.
It is easy to see that $\psi$ is a bilinear map. It follows (from the universal property) that we have an induced homomorphism: $\hat \psi:Hom(V_1,W_1)\otimes Hom(V_2,W_2)\rightarrow Hom(V_1 \otimes V_2,W_1 \otimes W_2)$ such that: $\hat\psi(\phi_1\otimes\phi_2)=\widehat{\phi_1 \times \phi_2}$.
My question: is $\hat\psi$ necessarily an isomorphism? Note that there is equality of dimensions (since both Hom,Tensor multiply them), hence clearly some isomorphism must exists between $Hom(V_1,W_1)\otimes Hom(V_2,W_2)\cong Hom(V_1 \otimes V_2,W_1 \otimes W_2)$.
If $\hat\psi$ is not isomorphism, is there some other "canonical" isomorphism?
I will just note that in the case of finite dimensional vector spaces, there is equality of dimensions, hence clearly some isomorphism must exists. | {
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• Take $W_1 = W_2 = A$, you can even assume it's a field then you're asking if $(V_1 \otimes V_2)^* \cong V_1^* \otimes V_2^*$. That's false if $V_1$ and $V_2$ are infinite dimensional. I can't even imagine how complicated things would become if $A$ is not a field, or not even a PID or anything... For $A$ a field my instinct tells me your $\hat\psi$ is always injective, but I'm not 100% sure. – Najib Idrissi Apr 28 '15 at 13:27
• It need not be well defined. I have an example. – Matt Samuel Apr 28 '15 at 13:43
• Najib: I agree. I have edited the question to focus upon the case of finite dimensional vector spaces. That was my original interest, I just thought maybe I missed some general theme which is not related to the specific fact that my objects were finite_dim vector spaces and not general modules. I was completely wrong there... – Asaf Shachar Apr 28 '15 at 13:43
• $\widehat{\phi_1 \times \phi_2}$ is well defined via the universal property of tensor products. – Asaf Shachar Apr 28 '15 at 13:48
• @egreg The expression $\phi_1(v_1) \otimes \phi_2(v_2)$ is linear in each variable in $v_1$ and $v_2$, if I'm not mistaken, so it defines a linear map $V_1 \otimes V_2 \to W_1 \otimes W_2$. And then this is linear in each of $\phi_1$ and $\phi_2$, hence the map $\hat\psi$. Am I missing something? It's just the bifunctoriality of $\otimes : \mathsf{Vect} \times \mathsf{Vect} \to \mathsf{Vect}$. – Najib Idrissi Apr 28 '15 at 14:04
The general situation is the following: $A$ is a commutative ring and $V_1, V_2, W_1, W_2$ are $A$-modules
If one of the ordered pairs $(V_1,V_2)$, $\,(V_1,W_1\,$ or $\,(V_2,W_2)$ consists of finitely generated projective $A$-modules, the canonical map is an isomorphism (Bourbaki, Algebra, Ch. 2 ‘Linear Algebra’, §4, n°4, prop. 4).
Over a field, all modules are projective since they're free. So the answer is ‘yes’.
In the present case, here is a sketch of the proof: | {
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In the present case, here is a sketch of the proof:
Let $K$ be the base field. As $V_1\simeq K^m$ for some $m>0$ and similarly $V_2\simeq K^n$, and the $\operatorname{Hom}$ and $\,\otimes\,$ functors comute with direct sums, we may as well suppose $V_1=V_2=K$.So we have to prove: $$\widehat\Psi\colon\operatorname{Hom}(K,W_1)\otimes\operatorname{Hom}(K,W_2)\to\operatorname{Hom}(K\otimes K,W_1\otimes W_2)$$ is an isomorphism.
This results basically that for any vector space $V$ we have an isomorphism \begin{align*} \operatorname{Hom}_K(K,V)&\simeq V\\ \phi&\mapsto\phi(1) \end{align*} In detail, just consider the following commutative diagram:
• Thanks. Is there a quick way to see that $\hat\psi$ is an isomorphism in the particular case where all the modules are finite dimensional vector spaces? (I think this question should have a simple answer in terms of "undergraduate" linear algebra without the need to resorting to more "heavy" tools\theorems from abstract commutative algebra. – Asaf Shachar Apr 28 '15 at 13:57
• @asaf shachar: I gave some details of a possible proof. See my completed answer. – Bernard Apr 28 '15 at 18:44
• Excuse me... I want to ask that is this property holds for $> 2$ vector spaces? I think it works because the diagram listed above seems work for $> 2$ vector spaces. More generally, in module case, is this property holds for $> 2$ modules (with suitable requirements...)?? Thank you! – Peter Hu Sep 2 '15 at 10:54
To show things like this in the case of finite dimensional vector spaces, you can always fallback on picking bases: | {
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Let $v_i^1, \ldots, v_i^{m_i}$ be a basis of $V_i$ ($i=1,2$), and $w_i^1, \ldots, w_i^{n_i}$ be a basis of $W_i$ ($i=1,2$). Let $f^{jk}_i : V_i \to W_i$ be given on the basis by $f_i^{jk}(v_i^l) = \delta_{jl} w_i^k$. Then $\{v_1^p \otimes v_2^q\}$ is a basis of $V_1 \otimes V_2$, and $\{f_i^{jk}\}$ is a basis of $Hom(V_i,W_i)$. Your map sends $(f_1^{jk},f_2^{rs})$ to the linear transformation given on the basis by $\widehat{f_1^{jk} \times f_2^{rs}}(v_1^p \otimes v_2^q) = \delta_{jp} \delta_{rq} w_1^k \otimes w_2^s$.
Those maps clearly form a basis of $Hom(V_1 \otimes V_2, W_1 \otimes W_2)$ for the same reason the $f_i^{jk}$ are a basis of $Hom(V_i,W_i)$. | {
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# Measuring fractal dimension of natural objects from digital images
This is a useful topic. A college physics lab, medical diagnostics, urban growth, etc. - there is a lot of applications. On this site by Paul Bourke about Google Earth fractals we can get a high resolution images (in this post they are low res - import from source for experiments). For example, around Lake Nasser in Egypt:
img = Import["http://paulbourke.net/fractals/googleearth/egypt2.jpg"]
The simplest method I know is Box Counting Method which has a lot of shortcomings. We start from extracting the boundary - which is the fractal object:
{Binarize[img], iEdge = EdgeDetect[Binarize[img]]}
Now we could partition image into boxes and see how many boxes have at least 1 white pixel. This is a very rudimentary implementation:
MinS = Floor[Min[ImageDimensions[iEdge]]/2];
data = ParallelTable[{1/size, Total[Sign /@ (Total[#, 2] & /@ (ImageData /@
Flatten[ImagePartition[iEdge, size]]))]}, {size, 10, MinS/2, 10}];
From this the slope is 1.69415 which is a fractal dimension that makes sense
line = Fit[Log[data], {1, x}, x]
13.0276 + 1.69415 x
Plot[line, {x, -6, -2}, Epilog -> Point[Log[FDL]],
PlotStyle -> Red, Frame -> True, Axes -> False]
Benchmark: if I run this on high res of Koch snowflake i get something like ~ 1.3 with more exact number being 4/log 3 ≈ 1.26186.
Question: can we improve or go beyond the above box counting method?
All approaches are acceptable if they find fractal dimension from any image of natural fractal. | {
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All approaches are acceptable if they find fractal dimension from any image of natural fractal.
-
You have a lot of programs in mathematica to measure fractal dimensions and multi fractal spectrum of an image in the book: Fractal Geography, Andre Dauphine, Wiley, 2012 See the book on the Wolfram Mathematica | Books or Amazon  – Dauphine Oct 16 '12 at 9:32
Vitaliy, with all due respect, the ending bullets make the question (actually there isn't a question mark anywhere, right?) rather wide ranging. Perhaps you could focus it somewhat and make it sound more like a real question? – Sjoerd C. de Vries Oct 16 '12 at 12:59
@SjoerdC.deVries Updated, it is a single question now. – Vitaliy Kaurov Oct 16 '12 at 15:06
could you perhaps describe or link some examples of use of fractal dimension? – magma Oct 17 '12 at 10:27
@magma: Wikipedia lists some examples here: en.wikipedia.org/wiki/… – shrx Jun 14 '13 at 20:22
You can still use box count, but doing it smarter :)
Counting boxes with at least 1 white pixel from ImagePartition can be done more efficiently using Integral Image, a technique used by Viola-Jones (2004) in their now popular face recognition framework. For a mathematical motivation (and proof), Viola and Jones point to this source.
What Integral Image allows you to do is to compute efficiently the total mass of any rectangle in an image. So, you can define the following:
IntegralImage[d_] := Map[Accumulate, d, {0, 1}];
data = ImageData[ColorConvert[img, "Grayscale"]]; (* img: your snowflake image *)
ii = IntegralImage[data];
Then, the mass (white content) of a region, is
(* PixelCount: total mass in region delimited by two corner points,
given ii, the IntegralImage *)
PixelCount[ii_, {p0x_, p0y_}, {p1x_, p1y_}] :=
ii[[p1x, p1y]] + ii[[p0x, p0y]] - ii[[p1x, p0y]] - ii[[p0x, p1y]]; | {
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So, instead of partitioning the image using ImagePartition, you can get a list of all the boxes of a given size by
PartitionBoxes[{rows_, cols_}, size_] :=
Transpose /@ Tuples[{Partition[Range[1, rows, size], 2, 1],
Partition[Range[1, cols, size], 2, 1]}];
If you apply PixelCount to above, as in your algorithm, you should have the same data but calculated faster.
PixelCountsAtSize[{rows_, cols_}, ii_, size_] :=
((PixelCount [ii, #1, #2] &) @@ # &) /@ PartitionBoxes[{rows, cols}, size];
Following your approach here, we should then do
fractalDimensionData =
Table[{1/size, Total[Sign /@ PixelCountsAtSize[Dimensions[ii], ii, size]]},
{size, 3, Floor[Min[Dimensions[ii]]/10]}];
line = Fit[Log[fractalDimensionData], {1, x}, x]
Out:= 10.4414 + 1.27104 x
which is very close to the actual fractal dimension of the snowflake (which I used as input).
Two things to note. Because this is faster, I dared to generate the table at box size 3. Also, unlike ImagePartition, my partition boxes are all of the same size and therefore, it excludes uneven boxes at the edges. So, instead of doing minSize/2 as you did, I put minSize/10 - excluding bigger and misleading values for big boxes.
Hope this helps.
Update
Just ran the algorithm starting with 2 and got this 10.4371 + 1.27008 x. And starting with 1 is 10.4332 + 1.26919 x, much better. Of course, it takes longer but still under or around 1 min for your snowflake image.
Update 2
And finally, for your image from Google Earth (eqypt2.jpg) the output is (starting at 1-pixel boxes)
12.1578 + 1.47597 x
It ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs. | {
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It ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs.
-
There isn't a "snowflake" image. All three images come from the same object (and I guess the fractal dimension should be the same for all of them) – belisarius Dec 23 '13 at 21:03
@belisarius, the PO published a Koch snowflake image as Benchmark (see last part of post) and that's the one I used. I called it 'snowflake' for short. On the other hand, this is still a numerical procedure, so the numbers will be different depending on approximation. The Update2 refers to the eqypt2.jpg image, also made available by the PO, as he asked for an improvement suitable for real-life images. – caya Dec 23 '13 at 22:40
sorry, I missed that link. Thanks – belisarius Dec 24 '13 at 0:05
+1 This is very neat @caya, thank you! I will wait "a bit" hoping that someone can implement things like wavelet multi-fractals or similar. – Vitaliy Kaurov Feb 14 at 20:02
@VitaliyKaurov, glad you liked it. Note that Viola & Jones rightly pointed out a link between integral image and Haar wavelet basis in their paper; although this wasn't explored further. It is unclear to me the link between the paper you mentioned and fractal dimension, but certainly worth reading as I am interested in these kind of problems. Cheers. – caya Feb 16 at 12:40 | {
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# How to be sure about any parameterization?
This is the problem:
Find intersection of this two surfaces and then calculate its length. $$x^2+4y^2=4$$ and
$$y\sqrt3 +z=1$$ Method 1 :
Common way is to take:
$$x=2\cos t ,\: y=\sin t ,\: z=1-\sqrt3 \sin t$$
So
$$r(t) =(2\cos t,\:\sin t,\:1-\sqrt3 \sin t)$$ and
$$r'(t) = (-2\sin t,\:\cos t,\:-\sqrt3\cos t) ,\: |r'(t)|=2$$
Finally the length is:
$$L = \int_{0}^{2\pi} 2\:dt=4\pi$$
But in another parameterization I find something strange
Method 2:
$$x=2\sqrt{1-t^2} ,\: y=t ,\: z=1-t\sqrt3$$ So
$$r(t) = (2\sqrt{1-t^2} ,\: t ,\: 1-t\sqrt3)$$ and
$$r'(t) = \left(\frac {-2t}{\sqrt{1-t^2}},1,-\sqrt3\right) ,\: |r'(t)|=\frac {2}{\sqrt{1-t^2}}$$ (and that's different from method 1 $$|r'(t)|$$)
Finally the length
$$L=\int_{0}^{2\pi} \frac {2}{\sqrt{1-t^2}}\:dt\\ =2\int_{0}^{\arcsin{2\pi}}\frac {1}{\cos{\theta}}\cos{\theta}\:d\theta\\ =2\int_{0}^{\arcsin{2\pi}}1\:d\theta$$
and because of $$\arcsin(2\pi)$$ its undefined it cant be solved.
Now my first question is that:
If we know from past that we can create many parameterizations for a single curve, so why the result of they length is different?
and the second one:
How to be sure that a parameterization is the correct one?
As a complement to @Axel's answer | {
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How to be sure that a parameterization is the correct one?
As a complement to @Axel's answer
Let $$t_j$$ denote the $$j$$th method's $$t$$ so $$t_2=\sin t_1$$ and $$\tfrac{dt_2}{\sqrt{1-t_2^2}}=d\arcsin t_2=\pm dt_1$$. If for example $$t_1$$ increases from $$0$$ to $$2\pi$$, $$t_2$$ is monotonic in three parts, cut at $$t=\pi/2$$ and $$t=3\pi/2$$. The monotonic pieces have alternating $$\pm$$, in this case with the pattern $$+-+$$, so the length is\begin{align}\int_0^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}+\int_{-1}^0\frac{2dt_2}{\sqrt{1-t_2^2}}&\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\\&\color{red}{=4\pi}\\&\color{blue}{=\left(\int_0^{\pi/2}+\int_{\pi/2}^{3\pi/2}+\int_{3\pi/2}^{2\pi}\right)2dt_1}\\&=\color{blue}{4\pi},\end{align}where the $$t_2$$-space calculation can continue from the black expression with either the red or the blue treatment. If $$t_1$$ varies over another length-$$2\pi$$ interval the monotinicity division details are different, but any choice requires at least one such cut, succumbs to the above technique, and gets the same answer. Let's check another example from $$t_1=-\pi/2$$ to $$t_1=3\pi/2$$, giving two pieces (meeting at $$t=\pi/2$$) where the $$\pm$$ are $$+-$$:$$\int_{-1}^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\color{blue}{=\left(\int_{-\pi/2}^{\pi/2}+\int_{\pi/2}^{3\pi/2}\right)2dt_1}.$$
• Thank you both.I understand it now. – program_craft Jan 22 at 4:58
The problem lies in the way you parametrize the curve in the second method.
The reason why you integrate from $$0$$ to $$2\pi$$ in the first method is because you have implicitly define you parametrization to go from $$0$$ to $$2\pi$$ because you are looking for a mapping $$[0;2\pi) \longrightarrow \textrm{your intersection curve}$$ such that that mapping is biyective with the entire curve. | {
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If you take a look to what you limits of integrations are in the second method you see they are the same from the first one. That again implicitly define you parametrization to go from $$0$$ to $$2\pi$$ but this time (because of the equation $$y=t$$) you are saying that $$y$$ runs in the range $$[0;2\pi)$$ when in reallity in ranges $$[-1;1)$$ so you actually end up measuring the length of a totally different curve (one with that parametrization but with $$y \in [0;2\pi)$$ insted of $$[-1;1)$$).
Now given that $$t$$ should be in $$[-1;1)$$ the correct way to measure the lenght in the second method would be: $$\int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt + \int\limits_{1}^{-1} \frac{2}{\sqrt{1-t²}} \ dt$$
That is beacuse you can't map the entire curve with just the $$[-1;1)$$ range (in fact you only map one half of it). Thus the second integral.
So finally you have:
$$2\int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt = 4\pi$$
• Can you explain more why t is in range[-1,1] – program_craft Jan 21 at 17:36 | {
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# Derive the error for the Midpoint rule
### Problem statement
Let $$[a,b]$$ be divided into $$n$$ subintervals by $$n+1$$ points $$x_0 = a, x_1 = a+h , x_2 = a+2h , \dots , x_n = a+nh =b$$.
Show that $$I- I_M = \frac{1}{24}(b-a)h^2 f''(z)$$ for some $$z \in [a,b]$$.
### Attempt at solution
$$I_M = h \sum_{i=0}^{n-1} f\left( a + \left( i + \frac12\right)\right) := h\sum_{i=0}^{n-1}f(x_i)$$
\begin{align} I -I_M &= \int_a^b f(x) \mathrm{d}x - h\sum_{i=0}^{n-1}f(x_i)\\ &= \sum_{i=0}^{n-1} \left( \int_{x_i}^{x_{i+1}} f(x) \ \mathrm{d}x - hf(x_i)\right) \end{align} Here I am stuck. I know I need to use Taylor's theorem in conjunction with Mean-Value Theorem but I don't know how to proceed.
I suspect that for $$x\in [x_i ,x_{i+1}]$$ the Taylor expansion is $$f(x) = f(x_i) + (x-x_i)f'(x_i) + \frac12 (x-x_i)^2 f''(z_x) \quad \text{for } z_x \in [x_i , x_{i+1}]$$ but then If I substitute this in the above I get \begin{align} &= \sum_{i=0}^{n-1}\left( \int_{x_i}^{x_{i+1}} (f(x_i) + (x-x_i)f'(x_i) + \frac12 (x-x_i)^2 f''(z_x))\mathrm{d}x - hf(x_i)\right)\\ \end{align} and I don't see what I have accomplished
Does anyone know If I am on the right track and if yes how to continue this derivation ?
You can start by obtaining the rule for a single interval, say \begin{align*} \int_{x_i}^{x_{i+1}} f(x) dx -hf(c_i) &= \int_{x_i}^{x_{i+1}}(f(x)-f(c_i))dx\\ & = \int_{x_i}^{x_{i+1}} \left(f(c_i)+f'(c_i)(x-c_i) + \frac{f''(\xi_i(x))}{2}(x-c_i)^2\;-\;f(c_i)\right) dx\\ & = \frac{f''(\xi_i)}{2}\int_{x_i}^{x_{i+1}}(x-c_i)^2 dx = \frac{f''(\xi_i) h^3}{24} \end{align*}
Now, if you take the sum over the $$n$$ intervals, you get the error term
$$\sum_{i=1}^n \frac{f''(\xi_i) h^3}{24} = \frac{h^2(b-a)}{24}\left(\frac 1n \sum_{i=1}^n f''(\xi_i)\right) = \frac{h^2(b-a) f''(\xi)}{24}.$$
Some notes: | {
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Some notes:
1. I used the mean value theorem for integrals. Since $$(x-c_i)^2$$ does not change sign, we know that $$\int_{x_i}^{x_{i+1}} \frac{f''(\xi_i(x))}{2}(x-c_i)^2 dx = \frac{f''(\xi_i)}{2} \int_{x_i}^{x_{i+1}} (x-c_i)^2 dx$$
2. If we assume the $$f\in C^2$$, the term $$\frac 1n \sum f''(\xi_i)$$ is an average of values of $$f''$$, hence between the minimum and maximum of $$f''$$. The intermediate value theorem says that it will correspond to $$f''(\xi)$$, for some $$\xi \in (a,b)$$.
• What do you know about the function $x \mapsto \xi_i(x)$?
– RRL
Dec 8, 2021 at 12:05
• I am in fact assuming that $f''(\xi(x))$ is continuous and should have added some comment on that. Over the years I've learned that, in pedagogic terms, it is preferable to streamline the proof, leaving the technical aspects to notes or auxiliary lemmas. Dec 8, 2021 at 12:58
• This is what I was looking for thank you. Dec 8, 2021 at 18:52
• Where does the $f'(c_i) (x-c_i)$ term go in the second line ? Dec 8, 2021 at 19:43
• @hexaquark The term vanishes during the integration because it is an odd function with respect tot the midpoint $c_j$ of the interval. Dec 9, 2021 at 12:08
There is another proof that avoids double integrals and uses the generalized mean value instead. I have learned this technique from reading @LutzLehmann's answers to very similar questions. | {
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Let $$h>0$$ be given and consider the problem of computing the integral $$I = \int_{-h}^h f(x)dx.$$ The midpoint rule takes the form $$M_h = 2h f(0).$$ We will now obtain the familiar error formula by studying the auxiliary function $$g$$ given by $$g(x) = \int_{-x}^x f(t) dt - 2x f(0).$$ This function is interesting precisely because $$g(h) = I - M_h.$$ Let $$F$$ denote an anti-derivative of $$f$$. Then $$g(x) = F(x) - F(-x) - 2xf(0).$$ It is clear that $$g$$ is as smooth as $$F$$ and that $$g(0) = 0$$. We have $$g'(x) = f(x) + f(-x) - 2f(0).$$ It follows that $$g'(0) = 0.$$ We have $$g''(x) = f'(x) - f'(-x).$$ It follows that $$g''(0) = 0.$$ We conclude that the graph of $$g$$ is quite flat near $$x=0$$. To make further progress we make repeated use of the generalized mean value theorem. There exists at least one $$x_1$$ between $$0$$ and $$x$$ such that $$\frac{g(x)}{x^3} = \frac{g(x) - g(0)}{x^3 - 0^3} = \frac{g'(x_1)}{3x_1^2}.$$ There exists at least one $$x_2$$ between $$0$$ and $$x_1$$ such that $$\frac{g'(x_1)}{3x_1^2} = \frac{g'(x_1) - g'(0)}{3x_1^2 - 3\cdot 0^2}= \frac{g''(x_2)}{6x_2}.$$ Similarly, there exists at least one $$x_3$$ between $$0$$ and $$x_2$$ such that $$\frac{g''(x_2)}{6x_2} = \frac{g''(x_2)- g''(0)}{6x_2 - 6 \cdot 0} = \frac{g'''(x_3)}{6} = \frac{f''(x_3) + f''(-x_3)}{6}$$ Finally, by the continuity of $$f''$$ there is at least one $$\xi$$ between $$-x_3$$ and $$x_3$$ such that $$\frac{g(x)}{x^3} = \frac{f''(\xi)}{3}.$$ In particular, there is at least one $$\xi$$ in $$[-h,h]$$ such that $$I - M_h = g(h) = \frac{f''(\xi)}{3} h^3.$$ Replacing $$h$$ with $$h/2$$ yields the familiar result, i.e, $$\int_{-h/2}^{h/2} f(x) dx - h f(0) = \frac{f''(\xi)}{24} h^3.$$ We finish the proof as suggested by @RRL.
To make any progress we need to consider the smoothness of $$f$$. Here I assume $$f \in C^2([a,b])$$. Denote the partition points as $$x_j = a + hj$$ and the midpoints as $$c_j = a + \frac{2j+1}{2}h$$ for $$j = 0,1,\ldots, n-1$$. | {
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We first find the local error for a subinterval $$[x_j,x_{j+1}]$$ using the Taylor expansion with integral remainder
$$f(x) = f(c_j) +f'(c_j)(x-c_j) + \int_{c_j}^x(x-t)f''(t)\, dt$$
Integrating over $$[x_j,x_{j+1}]$$ and using $$x_{j+1} - x_j = h$$, we get
$$\tag{1}\int_{x_j}^{x_{j+1}}f(x) \, dx = hf(c_j)+ f'(c_j)\int_{x_j}^{x_{j+1}}(x-c_j) \, dx + \int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx$$
The second term on the RHS of (1) is
$$f'(c_j)\int_{x_j}^{x_{j+1}}(x-c_j) \, dx =f'(c_j) \left.\frac{(x-c_j)^2}{2} \right|_{x_j}^{x_{j+1}}= \frac{f'(c_j)}{2} \left[(x_{j+1}-c_j)^2- (x_j - c_j)^2 \right]\\ = \frac{f'(c_j)}{2}\left[\left(\frac{h}{2}\right)^2 - \left(-\frac{h}{2}\right)^2\right]= 0$$
Substituting into (1) and rearranging, we get
$$\tag{2}\int_{x_j}^{x_{j+1}}f(x) \, dx - hf(c_j)= \int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx$$
In order to apply the mean value theorem to extract the second derivative from the integral on the RHS, some manipulation with indicator functions is required to eliminate the variable integration limit. Note that
$$\int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx \\= \int_{c_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx + \int_{x_j}^{c_j} \int_{x}^{c_j}(t-x)f''(t)\, dt\, dx \\ = \int_{c_j}^{x_{j+1}} \int_{c_j}^{x_{j+1}}\mathbf{1}_{}(t \leqslant x)(x-t)f''(t)\, dt\, dx + \int_{x_j}^{c_j} \int_{x_j}^{c_j}\mathbf{1}_{}(t \geqslant x)(t-x)f''(t)\, dt\, dx$$
Now we can apply Fubini's theorem to obtain
$$\tag{3}\int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx \\ = \int_{c_j}^{x_{j+1}} \int_{c_j}^{x_{j+1}}\mathbf{1}_{(t \leqslant x)}(x-t)f''(t)\, dx\, dt + \int_{x_j}^{c_j} \int_{x_j}^{c_j}\mathbf{1}_{(t \geqslant x)}(t-x)f''(t)\, dx\, dt \\ = \int_{c_j}^{x_{j+1}} f''(t)\left(\int_{t}^{x_{j+1}}(x-t)\, dx\right)\, dt + \int_{x_j}^{c_j} f''(t)\left(\int_{x_j}^{t}(t-x)\, dx\right)\, dt\\ = \int_{c_j}^{x_{j+1}} f''(t)\frac{(x_{j+1}-t)^2}{2}\, dt + \int_{x_j}^{c_j} f''(t)\frac{(t-x_j)^2}{2}\, dt$$ | {
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By the mean value theorem for integrals there are points $$z_j^+ \in [c_j,x_{j+1}]$$ and $$z_j^- \in [x_j,c_j]$$ such that
$$\tag{4}\int_{c_j}^{x_{j+1}} f''(t)\frac{(x_{j+1}-t)^2}{2}\, dt + \int_{x_j}^{c_j} f''(t)\frac{(t-x_j)^2}{2}\, dt\\= f''(z_j^+)\int_{c_j}^{x_{j+1}} \frac{(x_{j+1}-t)^2}{2}\, dt + f''(z_j^-)\int_{x_j}^{c_j} \frac{(t-x_j)^2}{2}\, dt\\ = f''(z_j^+)\frac{h^3}{48} + f''(z_j^-) \frac{h^3}{48}$$
Substituting into (2) with (3) and (4) yields
$$\int_{x_j}^{x_{j+1}}f(x) \, dx - hf(c_j)=\frac{h^3}{48}\left(f''(z_j^+)+f''(z_j^-)\right)$$
Summing from $$j=0$$ to $$j = n-1$$, we get
$$I-I_M = \int_a^b f(x) \, dx - h\sum_{j=0}^{n-1}f(c_j)= \frac{h^3}{48}\sum_{j=0}^{n-1}\left(f''(z_j^+)+f''(z_j^-)\right)\\ = \frac{2nh^3}{48}\frac{1}{n}\sum_{j=0}^{n-1}\frac{f''(z_j^+)+f''(z_j^-)}{2},$$
and since $$nh = (b-a)$$,
$$\tag{5}I - I_M = \frac{(b-a)h^2}{24} \sum_{j=0}^{n-1}\frac{f''(z_j^+)+f''(z_j^-)}{2n}$$
I'll leave it to you to consider if there exists $$z \in [a,b]$$ such that the second derivative at this point equals the weighted average of second derivatives, i.e.,
$$f''(z) = \sum_{j=0}^{n-1}\frac{f''(z_j^+)+f''(z_j^-)}{2n},$$
and, hence,
$$I - I_M = \frac{(b-a)h^2}{24}f''(z)$$
Nevertheless, if the second derivative is bounded as $$|f''(x)| \leqslant \|f''||_{\infty}$$ for all $$x \in [a,b]$$, then we obtain from (5) the well-known midpoint rule error bound
$$|I - I_M| \leqslant \frac{\|f''\|_{\infty}(b-a)h^2}{24}$$ | {
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$$|I - I_M| \leqslant \frac{\|f''\|_{\infty}(b-a)h^2}{24}$$
• The proof is correct but the same result can be obtained in a much simpler way just by using a second order Taylor expansion and the mean value theorem for integrals Dec 8, 2021 at 11:40
• @PierreCarre: Understood. I intentionally use the integral form of the remainder to avoid introducing another function $\xi_i(x)$ with the Lagrange form. What do you know about this function and $f’’ \circ \xi_i$. You assume it is continuous without proof as you then apply the integral MVT. So it is “simpler” because some justification is missing.
– RRL
Dec 8, 2021 at 11:58
• You are correct, leaving out the analysis of $f''\circ \xi_i$ hides the more technical details. Dec 8, 2021 at 13:00 | {
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# Documentation
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# spline
Cubic spline data interpolation
## Syntax
s = spline(x,y,xq)
pp = spline(x,y)
## Description
example
s = spline(x,y,xq) returns a vector of interpolated values s corresponding to the query points in xq. The values of s are determined by cubic spline interpolation of x and y.
example
pp = spline(x,y) returns a piecewise polynomial structure for use by ppval and the spline utility unmkpp.
## Examples
collapse all
Use spline to interpolate a sine curve over unevenly-spaced sample points.
x = [0 1 2.5 3.6 5 7 8.1 10];
y = sin(x);
xx = 0:.25:10;
yy = spline(x,y,xx);
plot(x,y,'o',xx,yy)
Use clamped or complete spline interpolation when endpoint slopes are known. This example enforces zero slopes at the end points of the interpolation.
x = -4:4;
y = [0 .15 1.12 2.36 2.36 1.46 .49 .06 0];
cs = spline(x,[0 y 0]);
xx = linspace(-4,4,101);
plot(x,y,'o',xx,ppval(cs,xx),'-');
Extrapolate a data set to predict population growth.
Create two vectors to represent the census years from 1900 to 1990 (t) and the corresponding United States population in millions of people (p).
t = 1900:10:1990;
p = [ 75.995 91.972 105.711 123.203 131.669 ...
150.697 179.323 203.212 226.505 249.633 ];
Extrapolate and predict the population in the year 2000 using a cubic spline.
spline(t,p,2000)
ans = 270.6060
Generate the plot of a circle, with the five data points y(:,2),...,y(:,6) marked with o's. The matrix y contains two more columns than does x. Therefore, spline uses y(:,1) and y(:,end) as the endslopes. The circle starts and ends at the point (1,0), so that point is plotted twice. | {
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x = pi*[0:.5:2];
y = [0 1 0 -1 0 1 0;
1 0 1 0 -1 0 1];
pp = spline(x,y);
yy = ppval(pp, linspace(0,2*pi,101));
plot(yy(1,:),yy(2,:),'-b',y(1,2:5),y(2,2:5),'or'), axis equal
Use spline to sample a function over a finer mesh.
Generate sine and cosine curves for a few values between 0 and 1. Use spline interpolation to sample the functions over a finer mesh.
x = 0:.25:1;
Y = [sin(x); cos(x)];
xx = 0:.1:1;
YY = spline(x,Y,xx);
plot(x,Y(1,:),'o',xx,YY(1,:),'-')
hold on
plot(x,Y(2,:),'o',xx,YY(2,:),':')
hold off
Compare the interpolation results produced by spline and pchip for two different functions.
Create vectors of x values, function values at those points y, and query points xq. Compute interpolations at the query points using both spline and pchip. Plot the interpolated function values at the query points for comparison.
x = -3:3;
y = [-1 -1 -1 0 1 1 1];
xq1 = -3:.01:3;
p = pchip(x,y,xq1);
s = spline(x,y,xq1);
plot(x,y,'o',xq1,p,'-',xq1,s,'-.')
legend('Sample Points','pchip','spline','Location','SouthEast')
In this case, pchip is favorable since it does not oscillate as freely between the sample points.
Perform a second comparison using an oscillatory sample function.
x = 0:25;
y = besselj(1,x);
xq2 = 0:0.01:25;
p = pchip(x,y,xq2);
s = spline(x,y,xq2);
plot(x,y,'o',xq2,p,'-',xq2,s,'-.')
legend('Sample Points','pchip','spline')
When the underlying function is oscillatory, spline captures the movement between points better than pchip.
## Input Arguments
collapse all
x-coordinates, specified as a vector. The vector x specifies the points at which the data y is given. The elements of x must be unique.
Data Types: single | double
Function values at x-coordinates, specified as a numeric vector, matrix, or array. x and y must have the same length.
If y is a matrix or array, then the values in the last dimension, y(:,...,:,j), are taken as the values to match with x. In that case, the last dimension of y must be the same length as x. | {
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The endslopes of the cubic spline follow these rules:
• If x and y are vectors of the same size, then the not-a-knot end conditions are used.
• If x or y is a scalar, then it is expanded to have the same length as the other and the not-a-knot end conditions are used.
• If y is a vector that contains two more values than x has entries, then spline uses the first and last values in y as the endslopes for the cubic spline. For example, if y is a vector, then:
• f(x) = y(2:end-1)
• df(min(x)) = y(1)
• df(max(x)) = y(end)
• If y is a matrix or an N-dimensional array with size(y,N) equal to length(x)+2, then these properties hold:
• f(x(j)) matches the value y(:,...,:,j+1) for j = 1:length(x)
• Df(min(x)) matches y(:,:,...:,1)
• Df(max(x)) matches y(:,:,...:,end)
Data Types: single | double
Query points, specified as a vector. The points specified in xq are the x-coordinates for the interpolated function values s that spline computes.
Data Types: single | double
## Output Arguments
collapse all
Interpolated values at query points, returned as a vector, matrix, or array.
The size of s is related to the sizes of the inputs:
• If y is a scalar or vector, then s has the same size as xq.
• If y is an array that is not a vector, then these conditions apply:
• If xq is a scalar or vector, then size(s) equals [d1, d2, ..., dk, length(xq)].
• If xq is an array of size [m1,m2,...,mj], then size(s) equals [d1,d2,...,dk,m1,m2,...,mj].
Piecewise polynomial, returned as a structure. Use this structure with the ppval function to evaluate the piecewise polynomial at one or more query points. The structure has these fields.
FieldDescription
form
'pp' for piecewise polynomial
breaks
Vector of length L+1 with strictly increasing elements that represent the start and end of each of L intervals
coefs
L-by-k matrix with each row coefs(i,:) containing the local coefficients of an order k polynomial on the ith interval, [breaks(i),breaks(i+1)]
pieces | {
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pieces
Number of pieces, L
order
Order of the polynomials
dim
Dimensionality of target
Since the polynomial coefficients in coefs are local coefficients for each interval, you must subtract the lower endpoint of the corresponding knot interval to use the coefficients in a conventional polynomial equation. In other words, for the coefficients [a,b,c,d] on the interval [x1,x2], the corresponding polynomial is
$f\left(x\right)=a{\left(x-{x}_{1}\right)}^{3}+b{\left(x-{x}_{1}\right)}^{2}+c\left(x-{x}_{1}\right)+d\text{\hspace{0.17em}}.$
## Tips
• You also can perform spline interpolation using the interp1 function with the command interp1(x,y,xq,'spline'). While spline performs interpolation on rows of an input matrix, interp1 performs interpolation on columns of an input matrix.
## Algorithms
A tridiagonal linear system (possibly with several right-hand sides) is solved for the information needed to describe the coefficients of the various cubic polynomials that make up the interpolating spline. spline uses the functions ppval, mkpp, and unmkpp. These routines form a small suite of functions for working with piecewise polynomials. For access to more advanced features, see interp1 or the Curve Fitting Toolbox™ spline functions.
## References
[1] de Boor, Carl. A Practical Guide to Splines. Springer-Verlag, New York: 1978.
## See Also
#### Introduced before R2006a
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# Proving set identities: empty set case.
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $$A,B,C$$ we have $$A\cap(B \cup C) = (A \cap B) \cup (A \cap C).$$
I understand that this can be done by proving both $$A\cap(B \cup C) \subset (A \cap B) \cup (A \cap C) \ \text{and} \ (A \cap B) \cup (A \cap C) \subset A\cap(B \cup C)$$ hold true.
We can do this by letting $$x$$ be an arbitrary element of $$A\cap(B \cup C)$$ and showing that it is an element of $$(A \cap B) \cup (A \cap C)$$, and vice versa.
But what about when $$A \cap (B \cup C)$$ is the empty set? Then I would think we can't let $$x$$ be an arbitrary element of $$A \cap (B \cup C)$$ since there are none. However, I am aware that $$A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-). | {
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If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
• Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that, – DonAntonio Dec 22 '18 at 18:35
• Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar. – Asaf Karagila Dec 23 '18 at 10:24
In order to show that for two sets $$X$$ and $$Y$$ it holds that $$X\subseteq Y$$, you have to prove that
for every $$x$$, if $$x\in X$$, then $$x\in Y$$.
Note that a statement of the form “if $$\mathscr{A}$$ then $$\mathscr{B}$$” is true when
either $$\mathscr{A}$$ is false or both $$\mathscr{A}$$ and $$\mathscr{B}$$ are true
If $$X$$ is the empty set, then “$$x\in X$$” is false for every $$x$$; hence “if $$x\in X$$ then $$x\in Y$$” is true.
The phrase “take an arbitrary element $$x\in X$$” is possibly misleading, but its intended meaning is “suppose $$x\in X$$”. | {
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• Ah, so if we write "suppose $x \in X$" and successfully deduce that "$x \in Y$" , we have shown that "if $x \in X$, then $x \in Y$" is true. But, if "$x \in X$" is always false, we have still shown "if $x \in X$ then $x \in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct? – user445909 Dec 22 '18 at 19:39
• @E-mu Basically so – egreg Dec 22 '18 at 20:27
If $$D$$ is the empty set then the following statement is always true:$$\text{If }x\in D\text{ then }\cdots$$no matter what the dots are standing for.
It is false that $$x\in D$$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
Since the empty set is a subset of any set, there is no need of including that in a formal proof. However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets . | {
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# Determine the number of subsets
How many distinct subsets of a set $\text{A}$ are there, containing at least $9$ elements, where the total number of elements in set $\text{A}$ is $18$ ?
I've solved it by making cases of either choosing $9$, $10$, $11$ and so on elements from the set and then adding them up, i.e, $$\sum_{r=9}^{18} \dbinom{18}{r}$$
However, the answer given in my book is $$\dfrac{1}{2} \times \dfrac{18!}{9! \cdot 9!} + 2^{17}$$
Although the numerical value of my answer and book's answer is equal, I'm interested in knowing the intended solution of the author.
The term $\dfrac{1}{2} \times \dfrac{18!}{9! \cdot 9!}$ is dividing $18$ elements into two equal parts. The term $2^{17}$ is the total number of subsets of a set containing $17$ elements. But I can't connect these two things with the question.
Any help will be appreciated.
Thanks.
P.S. If anyone is interested in knowing how I found $\displaystyle \sum_{r=9}^{18} \dbinom{18}{r}$ here is my method,
Let $\text{S} = \displaystyle \sum_{r=9}^{18} \dbinom{18}{r}$
By Binomial Theorem,
$2^{18} = (1+1)^{18} = \displaystyle \sum_{r=0}^{18} \dbinom{18}{r}$
$\implies 2\text{S} - \dbinom{18}{9} =2^{18} \left( \because \dbinom{18}{r} = \dbinom{18}{18-r}\right)$
and the rest follows. | {
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and the rest follows.
If $A$ has $18$ elements, and $B\subseteq A$ has fewer than $9$ elements, then $B^c\subseteq A$ has more than nine elements. Also, if $B$ has more than $9$ elements then $B^c$ has fewer than $9$ elements. That means the number of subsets of $A$ with fewer than $9$ elements equals the number of subsets of $A$ with more than $9$ elements.
Of course, the complement of a set with $9$ elements is another set with $9$ elements. The number you want, the number of subsets of $A$ with at least $9$ elements, is almost half the number of all subsets of $A$. We can figure out just how close that almost is.
Let $m$ be the number of subsets of $A$ with more than $9$ elements, so $m$ is also the number of subsets of $A$ with fewer than $9$ elements. Let $n$ be the number of subsets of $A$ with exactly $9$ elements. Then you want to find $m+n$, and we can find it from the total number of subsets of $A$, which we'll call $t$.
$$\text{(fewer than 9)+(exactly 9)+(more than 9)=(all)}$$ $$m+n+m=t$$ \begin{align} m+n&=\frac 12n+\frac 12t \\ &=\frac 12{18 \choose 9}+\frac 12\cdot 2^{18} \\ &=\frac 12 \cdot \frac{18!}{9! \cdot 9!} + 2^{17} \end{align}
Well, there are $2^{18}$ subsets all in all. Of these, $\dfrac{18!}{9!\cdot 9!}$ have exactly $9$ elements. The rest either have more than $9$ or fewer than $9$. Now, by symmetry, the number of subsets with more than $9$ elements equals the number of subsets with fewer than $9$ (pair each subset with its complement). Thus the number of subsets with more than $9$ elements is $\dfrac{1}{2} \times \left(2^{18} - \dfrac{18!}{9!\cdot 9!}\right)$. Adding the number of subsets with exactly $9$ elements gives the desired result. | {
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A circle with diameter the minor base $$CD$$ of a trapezium $$ABCD$$ intersects its diagonals $$AC$$ and $$BD$$ in, respectively, their midpoints $$M$$ and $$N$$. The lines $$DM$$ and $$CN$$ intersect in $$P$$ and $$AC$$ and $$BD$$ intersect in $$H$$. Show that $$AD=CD=BC$$ and $$HP \perp AB$$.
$$DMNC$$ is a cyclic quadrilateral and $$CD||MN$$, thus $$DMNC$$ is an isosceles trapezoid. Therefore, $$CM=DN$$ and $$AC=BD$$. Now I am trying to show $$AD=CD$$. $$\angle DCM$$ is inscribed and it's equal to $$\angle DNM$$ but I don't see how to compare it with $$\angle CAD$$. How is this done? For the second part of the problem, I tried to show that $$HO$$ passes through $$P$$ ($$HO \perp CD$$ because $$\triangle CDH$$ is isosceles and if we show $$P \in HO$$ we are done).
Since $$DM\perp AC$$ and $$M$$ is a midpoint of $$AC$$, we obtain $$AD=DC$$.
Since $$DMNC$$ is an isosceles trapezoid, we obtain: $$\measuredangle PMN=\measuredangle PDC=\measuredangle PCD=\measuredangle PNM,$$ which gives $$PM=PN.$$ Also, $$\Delta MDN\cong\Delta NCM,$$ which gives $$\measuredangle HMN=\measuredangle HNM,$$ which gives $$HM=HN,$$ which says that $$PMHN$$ is a kite, which says $$PH\perp MN.$$ About a kite see here: https://en.wikipedia.org/wiki/Kite_(geometry)
• Yes. Thank you! $DM$ is the perpendicular bisector of $AC$. – Andrew Rogers Aug 19 '19 at 16:34
• Now I am trying to show that $HP \perp AB$ $(H = AC \cap BD)$. Is this true for all cases, because I can't prove it? – Andrew Rogers Aug 19 '19 at 16:52
• @Andrew Rogers Yes it's true because $HM=HN$ and $PM=PN.$ It's true if $AD=BC,$ which we got. – Michael Rozenberg Aug 19 '19 at 16:54
• I am not sure I understand why it's true iff $HM=HN$ and $PM=PN$. – Andrew Rogers Aug 19 '19 at 17:01 | {
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What's the probability that a sequence of coin flips never has twice as many heads as tails?
I gave my friend this problem as a brainteaser; while her attempted solution didn't work, it raised an interesting question.
I flip a fair coin repeatedly and record the results. I stop as soon as the number of heads is equal to twice the number of tails (for example, I will stop after seeing HHT or THTHHH or TTTHHHHHH). What's the probability that I never stop?
I've tried to just compute the answer directly, but the terms got ugly pretty quickly. I'm hoping for a hint towards a slick solution, but I will keep trying to brute force an answer in the meantime.
-
Hopefully whatever method resolves this applies to other ratios than $2$. – anon Aug 27 '11 at 3:32
@anon, good idea... See my answer. – Did Aug 27 '11 at 13:53
@anon: I've generalized my answer up to an infinite series expression as well. – Mike Spivey Aug 27 '11 at 21:00
See "a coin toss question" for the similar problem of flipping until achieving the same number of heads as tails. – Mike Spivey Aug 31 '11 at 20:55
@Elliot I'd like you remind you that this question doesn't currently have an accepted answer. – Git Gud Jun 16 '13 at 9:36
The game stops with probability $u=\frac34(3-\sqrt5)=0.572949017$.
See the end of the post for generalizations of this result, first to every asymmetric heads-or-tails games (Edit 1), and then to every integer ratio (Edit 2).
To prove this, consider the random walk which goes two steps to the right each time a tail occurs and one step to the left each time a head occurs. Then the number of tails is the double of the number of heads each time the walk is back at its starting point (and only then). In other words, the probability that the game never stops is $1-u$ where $u=P_0(\text{hits}\ 0)$ for the random walk with equiprobable steps $+2$ and $-1$. | {
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The classical one-step analysis of hitting times for Markov chains yields $2u=v_1+w_2$ where, for every positive $k$, $v_k=P_{-k}(\text{hits}\ 0)$ and $w_k=P_{k}(\text{hits}\ 0)$. We first evaluate $w_2$ then $v_1$.
The $(w_k)$ part is easy: the only steps to the left are $-1$ steps hence to hit $0$ starting from $k\ge1$, the walk must first hit $k-1$ starting from $k$, then hit $k-2$ starting from $k-1$, and so on. These $k$ events are equiprobable hence $w_k=(w_1)^k$. Another one-step analysis, this time for the walk starting from $1$, yields $$2w_1=1+w_3=1+(w_1)^3$$ hence $w_k=w^k$ where $w$ solves $w^3-2w+1=0$. Since $w\ne1$, $w^2+w=1$ and since $w<1$, $w=\frac12(\sqrt5-1)$.
Let us consider the $(v_k)$ part. The random walk has a drift to the right hence its position converges to $+\infty$ almost surely. Let $k+R$ denote the first position visited on the right of the starting point $k$. Then $R\in\{1,2\}$ almost surely, the distribution of $R$ does not depend on $k$ because the dynamics is invariant by translations, and $$v_1=r+(1-r)w_1\quad\text{where}\ r=P_{-1}(R=1).$$ Now, starting from $0$, $R=1$ implies that the first step is $-1$ hence $2r=P_{-1}(A)$ with $A=[\text{hits}\ 1 \text{before}\ 2]$. Consider $R'$ for the random walk starting at $-1$. If $R'=2$, $A$ occurs. If $R'=1$, the walk is back at position $0$ hence $A$ occurs with probability $r$. In other words, $2r=(1-r)+r^2$, that is, $r^2-3r+1=0$. Since $r<1$, $r=\frac12(3-\sqrt5)$ (hence $r=1-w$).
Plugging these values of $w$ and $r$ into $v_1$ and $w_2$ yields the value of $u$.
Edit 1 Every asymmetric random walk which performs elementary steps $+2$ with probability $p$ and $-1$ with probability $1-p$ is transient to $+\infty$ as long as $p>\frac13$ (and naturally, for every $p\le\frac13$ the walk hits $0$ with full probability). In this regime, one can compute the probability $u(p)$ to hit $0$. The result is the following. | {
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For every $p$ in $(0,\frac13)$, $u(p)=\frac32\left(2-p-\sqrt{p(4-3p)}\right).$
Note that $u(p)\to1$ when $p\to\frac13$ and $u(p)\to0$ when $p\to1$, as was to be expected.
Edit 2 Coming back to symmetric heads-or-tails games, note that, for any fixed integer $N\ge2$, the same techniques apply to compute the probability $u_N$ to reach $N$ times more tails than heads.
One gets $2u_N=E(w^{R_N-1})+w^N$ where $w$ is the unique solution in $(0,1)$ of the polynomial equation $2w=1+w^{1+N}$, and the random variable $R_N$ is almost surely in $\{1,2,\ldots,N\}$. The distribution of $R_N$ is characterized by its generating function, which solves $$(1-(2-r_N)s)E(s^{R_N})=r_Ns-s^{N+1}\quad\text{with}\quad r_N=P(R_N=1).$$ This is equivalent to a system of $N$ equations with unknowns the probabilities $P(R_N=k)$ for $k$ in $\{1,2,\ldots,N\}$. One can deduce from this system that $r_N$ is the unique root $r<1$ of the polynomial $(2-r)^Nr=1$. One can then note that $r_N=w^N$ and that $E(w^{R_N})=\dfrac{Nr_N}{2-r_N}$ hence some further simplifications yield finally the following general result.
For every $N\ge2$, $u_N=\frac12(N+1)r_N$ where $r_N<1$ solves the equation $(2-r)^Nr=1$. | {
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For every $N\ge2$, $u_N=\frac12(N+1)r_N$ where $r_N<1$ solves the equation $(2-r)^Nr=1$.
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+1: Nice work, Didier! (And for the side effect of showing me that the answer I got is correct, too.) – Mike Spivey Aug 27 '11 at 14:56
@Mike, thanks. To tell the truth, I DID compute the numerical value of my answer as soon as I got a formula, to check whether it coincided with the numerical value of your mysterious formula and I WAS relieved to see these coincided... – Did Aug 27 '11 at 16:28
+1, I had a feeling a Markov approach would pop up. Nice to see such a simple polynomial's root solve the general form. – anon Aug 27 '11 at 21:24
@anon, yes. This is no accident that the computations go through and that the algebra is nice, there is an underlying structure to all this, sometimes called cycles-and-weights Markov chains, which is a nifty extension of +/-1 random walks (I might add some references one day). – Did Aug 27 '11 at 21:51
(Update: The answer to the original question is that the probability of stopping is $\frac{3}{4} \left(3 - \sqrt{5}\right)$. See end of post for an infinite series expression in the general case.)
Let $S(n)$ denote the number of ways to stop after seeing $n$ tails. Seeing $n$ tails means seeing $2n$ heads, so this would be stopping after $3n$ flips. Since there are $2^{3n}$ possible sequences in $3n$ flips, the probability of stopping is $\sum_{n=1}^{\infty} S(n)/8^n$. | {
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To determine $S(n)$, we see that there are $\binom{3n}{n}$ ways to choose which $n$ of $3n$ flips will be tails. However, this overcounts for $n > 1$, as we could have seen twice as many heads as tails for some $k < n$. Of these $\binom{3n}{n}$ sequences, there are $S(k) \binom{3n-3k}{n-k}$ sequences of $3n$ flips in which there are $k$ tails the first time we would see twice as many heads as tails, as any of the $S(k)$ sequences of $3k$ flips could be completed by choosing $n-k$ of the remaining $3n-3k$ flips to be tails. Thus $S(n)$ satisfies the recurrence $S(n) = \binom{3n}{n} - \sum_{k=1}^{n-1} \binom{3n-3k}{n-k}S(k)$, with $S(1) = 3$.
The solution to this recurrence is $S(n) = \frac{2}{3n-1} \binom{3n}{n}.$ This can be verified easily, as substituting this expression into the recurrence yields a slight variation on Identity 5.62 in Concrete Mathematics (p. 202, 2nd ed.), namely, $$\sum_k \binom{tk+r}{k} \binom{tn-tk+s}{n-k} \frac{r}{tk+r} = \binom{tn+r+s}{n},$$ with $t = 3$, $r = -1$, $s=0$.
So the probability of stopping is $$\sum_{n=1}^{\infty} \binom{3n}{n} \frac{2}{3n-1} \frac{1}{8^n}.$$
Mathematica gives the closed form for this probability of stopping to be $$2 \left(1 - \cos\left(\frac{2}{3} \arcsin \frac{3 \sqrt{3/2}}{4}\right) \right) \approx 0.572949.$$
Added: The sum is hypergeometric and has a simpler representation. See Sasha's comments for why the sum yields this closed form solution and also why the answer is $$\frac{3}{4} \left(3 - \sqrt{5}\right) \approx 0.572949.$$ | {
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Added 2: This answer is generalizable to other ratios $r$ up to the infinite series expression. For the general $r \geq 2$ case, the argument above is easily adapted to produce the recurrence $S(n) = \binom{(r+1)n}{n} - \sum_{k=1}^{n-1} \binom{(r+1)n-(r+1)k}{n-k}S(k)$, with $S(1) = r+1$. The solution to the recurrence is $S(n) = \frac{r}{(r+1) n - 1} \binom{(r+1) n}{n}$ and can be verified easily by using the binomial convolution formula given above. Thus, for the ratio $r$, the probability of stopping has the infinite series expression $$\sum_{n=1}^{\infty} \binom{(r+1)n}{n} \frac{r}{(r+1)n-1} \frac{1}{2^{(r+1)n}}.$$ This can be expressed as a hypergeometric function, but I am not sure how to simplify it any further for general $r$ (and neither does Mathematica). It can also be expressed using the generalized binomial series discussed in Concrete Mathematics (p. 200, 2nd ed.), but I don't see how to simplify it further in that direction, either.
Added 3: In case anyone is interested, I found a combinatorial proof of the formula for $S(n)$. It works in the general $r$ case, too. | {
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It's late here, and I'm going to bed. Any comments, corrections, etc., will be dealt with in the morning. And if someone has an explanation for Mathematica's output I would love to see it! – Mike Spivey Aug 27 '11 at 6:50
Wow, incredible (both the write up and the result itself)! However, I have a slight point of confusion--why is S(1) = 1, rather than 3? – Elliott Aug 27 '11 at 7:48
Does anyone know (or perhaps have a useful reference book to find out from) the generating function of $a_n = \binom{3n}{n}$? That should help in finding the closed form of this series. – Ragib Zaman Aug 27 '11 at 9:17
Let $u$ be you answer, then $\cos(\frac{2}{3}\arcsin \frac{3}{4}\sqrt{\frac{3}{2}}) = 1-\frac{u}{2}$. Using $\cos(3 x) = -3 \cos(x) + 4 \cos(x)^3$, and $\cos( 2 \arcsin \frac{3}{4}\sqrt{\frac{3}{2}}) = -\frac{11}{16}$, it follows $u$ satisfy a cubic with rots $u_1=\frac{3}{2}$, $u_2 = \frac{3}{4}( 3- \sqrt{5})$ and $u_3 = \frac{3}{4}(3+ \sqrt{5})$. Root $u_2$ is the one corresponding to the original expression, by numerical comparison. – Sasha Aug 27 '11 at 13:47
@Ragib $\sum_{k\ge 0} \binom{3 k}{k} z^k$ is a hypergeometric sum with answer $\frac{3}{4 \pi} F(\frac{1}{3}, \frac{2}{3} ; \frac{1}{2} ; \frac{9 z}{4} )$. This Gauss hypergeometric function admits simpler representation. – Sasha Aug 27 '11 at 13:57
Here's another solution, following Ross Millikan's hint:
Let $n=H−2T$ (H=heads, T=tails). At each step, $n := n+1$ or $n := n-2$ with probability 1/2. The game starts with $n=0$ and stops when $n=0$.
For any iteration (after the initial), let $P(n)$ be the probability that the game eventually stops, given the current value of $n$ (and given that the event has not yet occurred). It's clear that $P(n)$ does not depend on time. Then the following recurrence holds:
$$P(n)= \frac{P(n+1)+P(n-2)}{2} \;, \;\; n\ne 0$$
with $P(0)=1$. We also know (do we?) that $P(-\infty)=0$ (but we don't yet know $P(+\infty)$) | {
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with $P(0)=1$. We also know (do we?) that $P(-\infty)=0$ (but we don't yet know $P(+\infty)$)
The solution to this difference equation (I'll spare you the details, just the usual linear difference equation procedure, in the two regions, with the above boundary conditions) is given by:
$$P(n)= \left\{ \begin{array}{ll} \phi^{-n} & \mbox{if } n \le 0 \\ 1 + B \, [(-\phi)^n -1] & \mbox{if } n \ge 0 \end{array} \right.$$
with $B= \phi^2 (1-\phi)$, $\phi = (\sqrt{5}-1)/2$
Now, after the first step we have $n=1$ or $n=-2$ with equal probability, then the probability that the game eventually stops is
$$\frac{P(1)+P(-2)}{2}=\frac{2 \phi^2 -\phi +1}{2} = 0.572949$$
The form of $P(n)$ is interesting:
This, for example, shows that the probability of stop is strongly dependent on the first coin toss (less than 40% if tail, more than 75% if head).
This procedure also seems directly generalizable, either to asymmetric coins or other integer ratios.
Added: Here goes the details for solving the recursion:
We postulate a solution of the form $P(n)= r^n$ and replacing in the recursion $2 P(n) - P(n+1) - P(n-2) = 0$ we get $$2 \; r^2 - r^3 - 1 = 0$$ The root $r_1=1$ comes immediately, and then the others: $r_2 = - \phi$, $r_3 = 1/\phi$. Then the general solution is given by $A + B (-\phi)^n + C \phi^{-n}$, for some $A,B,C$, in each zone.
In the zone $n \le 0$: we have $P(0)=1$ and $P(-\infty)=0$. As $\phi \approx 0.618 < 1$, this implies $A=0$, $B=0$,$C=1$ Hence
$$P(n) = \phi^{-n} \hspace{10px} n \le 0$$
In the zone $n \ge 0$: we have $P(0)=1$, but we don't know $P(\infty)$. We do know it's bounded, so $C=0$, and $A=1-B$, so
$$P(n) = 1 + B \, [(-\phi)^n -1] \hspace{10px} n \ge 0$$
To get rid of the remaining degree of freedom, we write the recursion for $n=1$, and replace the value of $P(-1)$ for the previous solution:
$$2 \, P(1) = P(2) + P(-1)$$ $$2 \, (1 + B \, [-\phi -1]) = 1 + B \, [(-\phi)^2 -1] + \phi$$
From this, we obtain $B= \phi^2 (1-\phi)$. | {
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From this, we obtain $B= \phi^2 (1-\phi)$.
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This solution is beautiful! One thing I don't understand is, how did you set boundary conditions for the region n >= 0? – Elliott Aug 29 '11 at 16:34
@Elliot: You first solve for $n \le 0$, then plugging it in the recursion for $n=1$ you resolve the remaining degree of freedom. Tell me if this is not clear, and I'll write the details. – leonbloy Aug 29 '11 at 17:44
So... in the end, you know that P(-infty)=0? – Did Aug 31 '11 at 21:01
Well, we "knew" (actually assumed) that from the beginning, it was $P(+\infty)$ the unknown. It's true (are you pointing to this?) that $P(-\infty)=0$ is intuitive and true but a rigorous proof is lacking here. – leonbloy Aug 31 '11 at 21:19
Just pointing to the fact that, when P(-infty) first appear, you feel necessary to signal that the relation P(-infty)=0 might need a proof. But when you need the value of P(-infty) later on, you state it without further ado. Hence, reading your post, I was wondering if I had missed, somewhere inbetween these two appearances, a proof that P(-infty)=0. (Please use @.) – Did Sep 1 '11 at 13:22
I would suggest $1-\phi=1-\frac{\sqrt{5}-1}{2}$. If define $P(n)$ as the probability that if we have an excess of tails of $n$ we will eventually have twice the heads as tails, the recurrence is $$P(n)=\frac{P(n+1)+P(n-2)}{2}$$ with boundary conditions $$P(n)=\begin {cases}1 &n\gt 0 \\ 0 &n=-\infty \end {cases}$$ The recurrence has solution $P(n)=\phi ^n$ | {
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Sorry, I'm not sure I follow. Can you please explain what P(n) represents in slightly more detail? – Elliott Aug 27 '11 at 6:18
+1 Nice answer, though cryptic (and slightly incorrect). To clarify: Let $n(t)=H-2T$ (H=heads, T=tails), say we are interest in the event $n(t)>0$ for some $t$. Let $P(n)$ (for $n\le0$) be the probability of that event, given the value $n$ that at that time, and that the event has not ocurred before; further, let $P(1)=1$. Then, the recursion is correct, and so is the solution (actually I think its $p(n)=\phi^{1-n}$, but anyway $P(0)= 0.618033989$). But... (continued) – leonbloy Aug 27 '11 at 15:17
The problem is that the original game stops when $H=2T$ exactly, and this event is not equivalent to $H>2T$ (for example, the sequence $HHHTHHHTHHT...$ does not make the game stop. I wonder if this can be corrected, I liked this approach. – leonbloy Aug 27 '11 at 15:22
@leonbloy: Good elaboration. Thanks. I was thinking that if we have excess heads then with probability 1 we get back to a 2:1 ratio. But you are right we can skip over it. A more telling (to me) sequence that fails is $HHHTTTHTHTHTHT...$ – Ross Millikan Aug 27 '11 at 15:28
The approach was indeed fixable, I posted the derivation in my own answer - I hope this it's ok. – leonbloy Aug 29 '11 at 0:15 | {
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# What quantifies as a rigorous proof?
Okay I have been thinking about this common combinatorial identity. $$\sum_{r=0}^{n} \binom{n}{r} = 2^n.$$ It is simple to prove this by induction, but it requires some annoying algebraic manipulation which involves other combinatorial identities. What about if we took an approach that takes into factor a specific rephrase of the problem?
My approach is this. Consider a combination lock with n number of on-off switches. To open it require turning on the correct switches. How many combinations are there? clearly, the first switch can be on or off, second on and off etc. so the combintions are $2^n$
We look at it from a different angle. How many ways are there to pick one on switch. $\dbinom{n}{1}$. what about picking 2 switches? $\dbinom{n}{2}$... and so on. Since the number of combinations are supposed to be the same,it gives us the identity.
Now i know this explanation is not rigorous, but can someone point out exactly to me exactly which part of it it fails to be so?
I have also encountered this problem in man other areas of mathematics when I try to prove something, I used a specific problem to do so and thus I doubt the validity of the proof. I can give more examples if needed | {
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• It's a combinatorial proof. Rigorous enough for me, but I know teachers who don't like proofs without lots of algebra. – Jean-Claude Arbaut Mar 15 '13 at 14:05
• Very clever proof. To me it is the right proof. Rigor is an ambiguous concept, but it's certainly not the case that a rigorous proof of an identity needs be done algebraically. Your argument explains why the identity is true. The algebra does not give any insight. – Cheerful Parsnip Mar 15 '13 at 14:09
• I have edited your question, employing the $\LaTeX$ notation that is used on MSE. – Andreas Caranti Mar 15 '13 at 14:14
• Better use the standard $\binom{n}{k}$ notation... – vonbrand Mar 15 '13 at 14:14
• It’s a variant of the standard combinatorial proof: $2^n$ is the number of subsets of $[n]=\{1,\dots,n\}$, and $\binom{n}k$ is the number of subsets of $[n]$ of size $k$, so $2^n=\sum_k\binom{n}k$. As such it is, as others have said, just fine. And while one ought to be able to carry out the details of the proof by induction, combinatorial arguments like this one generally give much more insight into why the result is true. – Brian M. Scott Mar 15 '13 at 14:46
Your approach is an often-used (and perfectly valid) proof technique. You should be aware that some instructors don't like it, but it certainly does the job. | {
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• An instructor who doesn’t accept combinatorial arguments ought not to be teaching the subject. – Brian M. Scott Mar 15 '13 at 14:44
• @Brian M. Scott: surely it depends on whether the goal is to teach the identity, or whether it is to teach students how to write proofs by induction. I would usually take a car if I need to travel 10 miles, but not if the point is to train for a marathon. – Carl Mummert Mar 15 '13 at 15:09
• @Carl: If the goal is to teach students how to write proofs by induction, the instructor should damned well come up with problems for which induction is the best approach. And if even then a student comes up with a different approach, the instructor should be pleased. – Brian M. Scott Mar 15 '13 at 15:11
• @Carl: My point is that the directions should not begin so. I consider it a failure on the part of the instructor if he or she has to specify a technique in order to give students practice in applying it. – Brian M. Scott Mar 15 '13 at 15:26
• I strongly agree with Brian here. Questions that prescribe techniques are either lazy pedagogy or bad curriculum design. If technique $Z$—say, induction—is a useful technique, then it is because there is some problem $Y$ such that $Z$ is superior to all other techniques for solving $Y$. If all such $Y$ are outside the scope of the class, then $Z$ is outside the scope of the class too. If, on the other hand, there is some $Y$ that is in the scope of the class, it is the instructor's job to find it and present it to the students, as an instructive example. – MJD Mar 15 '13 at 15:31
You can make such an argument rigorous. When things start to get more sophisticated, it often helps immensely to give more detail than one usually sees in this style of proof.
Let $S$ be a set with $n$ elements. Let $P(S)$ be its powerset -- that is, the set of all subsets of $S$. Let $P_i(S)$ be the set of all subsets of $S$ with exactly $i$ elements.
Your proof is then the invocation of four facts: | {
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Your proof is then the invocation of four facts:
• The well-known formula for the number of elements in $P(S)$
• The well-known formula for the number of elements in $P_i(S)$
• The observation that $P(S)$ is the disjoint union of all of the $P_i(S)$
• The relationship between the number of elements in a sequence of sets and the number of elements in their disjoint union | {
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My Math Forum What is the probability of getting the same number twice?
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June 16th, 2017, 11:34 AM #11 Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित i would think it this way: Since I have 24 draws, I would take 23 different numbers in my first 23 draws and 1 among those 23 numbers in my 24th draw. In 1st draw we can take any number but in the following draws we cannot take numbers which are already drawn, thus the numerator of the probabilities are decreasing by 1 in each of the following draw. Thus the probability of getting 23 different numbers in the first 23 draw is: $\displaystyle P1 = \frac{255}{255}\times \frac{254}{255}\times \frac{253}{255}\times ......$ till 23 steps $\displaystyle P1 = \frac{255!}{255^{23}\times 232!}$ Then for the 24th draw, we have to draw one of the numbers which we have already drawn. There are 23 possible numbers which we can draw out of 255 numbers. Thus the probability of having such number in 24th draw is: $\displaystyle P2=\frac{23}{255}$ Hence, the total probability is: $\displaystyle P=P1 \times P2$ $\displaystyle P= \frac{255!\times 23}{255^{24}\times 232!}$ $\displaystyle P\approx 0.0324$ Thanks from JeffM1 | {
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June 16th, 2017, 11:57 AM #12 Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics I'd like to present an alternative approach to the problem that could possibly shed light on that number. Consider the Markov chain with states 0, 1, ... 7, where state 0 is the initial state where no numbers have been chosen, states i, $\displaystyle 1 \leq i \leq 6$ denote the state where i distinct numbers have been chosen and no repeats have appeared so far, and 7 denote the state where repeats have appeared. Obviously, the transition matrix is as follows: $\displaystyle \mathbf{P} = \begin{bmatrix} 0 &1 &0 & 0 & 0 & 0 & 0 & 0\\ 0 &0 & \frac{5}{6} & 0 & 0& 0& 0& \frac{1}{6}\\ 0 & 0 & 0 & \frac{2}{3} & 0 & 0 &0 & \frac{1}{3}\\ 0 & 0 &0 &0 & \frac{1}{2}& 0 & 0& \frac{1}{2}\\ 0 & 0 & 0& 0 & 0 &\frac{1}{3} & 0 & \frac{2}{3}\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{5}{6}\\ 0& 0 & 0 & 0 & 0 & 0 &0 &1 \\ 0&0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ and the initial distribution is $\displaystyle \mathbf{e}_1$. So all we have to do is to calculate $\displaystyle \mathbf{e}_1\mathbf{P}^3 = (0,0,0,\frac{5}{9},0,0,0,\frac{4}{9})^T$, verifying Jeff's answer. Using a well-known theorem on absorbing states and some concepts in elementary probability (in particular the probability generating function), we can even do other cool things like determining how many numbers we need to draw on average to get a repeat, but I digress. So we can immediately apply this to the initial question: $\displaystyle \mathbf{P} =\begin{bmatrix} 0 & 1 & 0 & ... & 0 & 0\\ 0 & 0 & 254/255 & ... & 0& 1/255\\ 0 & 0 & 0 & \ddots & 0 & \vdots\\ 0& 0 & 0 & ... & 0 & 1\\ 0 & 0 & 0 & ...& 0 &1 \end{bmatrix}$ MATLAB tells us that 25th and last entries of $\displaystyle \mathbf{e}_1 \mathbf{P}^{24}$ are 0.32719 and 0.67281 respectively, again verifying Jeff's claims. Thanks from JeffM1 Last edited by 123qwerty; | {
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0.67281 respectively, again verifying Jeff's claims. Thanks from JeffM1 Last edited by 123qwerty; June 16th, 2017 at 11:59 AM. | {
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June 16th, 2017, 01:00 PM #13 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 I stick to my guns: the probability is ~.39
June 16th, 2017, 01:24 PM #14
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Joined: May 2016
From: USA
Posts: 825
Thanks: 335
Quote:
Originally Posted by Denis Well, whatever... IF, IF, IF the 24 draws produce: 1 pair no other number matching that pair 22 remaining are ALL different then probability = ~.39
I agree that that is the apparent question originally asked.
How many ways can I draw 24 cards with replacement from a set of 255 cards consecutively numbered 1 through 255?
Answer: $255^{24}.$
Do you agree?
How many ways can I draw 23 differently numbered cards from that same set with replacement?
Answer: $\displaystyle \prod_{j=1}^{23}(256 - j).$
Do you agree?
How many ways can I draw one of 23 specific numbers from that set?
Answer: $23.$
Do you agree?
So the probability of drawing 23 different cards and 1 duplicate card in 24 draws with replacement should be
$\dfrac{23 * \displaystyle \prod_{j=1}^{23}(256 - j)}{255^{24}} \approx 0.0324.$
I can't get to your result.
June 16th, 2017, 03:09 PM #15
Math Team
Joined: Jul 2013
From: काठमाडौं, नेपाल
Posts: 872
Thanks: 60
Math Focus: सामान्य गणित
Quote:
Originally Posted by JeffM1 I agree that that is the apparent question originally asked. How many ways can I draw 24 cards with replacement from a set of 255 cards consecutively numbered 1 through 255? Answer: $255^{24}.$ Do you agree? How many ways can I draw 23 differently numbered cards from that same set with replacement? Answer: $\displaystyle \prod_{j=1}^{23}(256 - j).$ Do you agree? How many ways can I draw one of 23 specific numbers from that set? Answer: $23.$ Do you agree? So the probability of drawing 23 different cards and 1 duplicate card in 24 draws with replacement should be $\dfrac{23 * \displaystyle \prod_{j=1}^{23}(256 - j)}{255^{24}} \approx 0.0324.$ I can't get to your result.
i absolutely agree with you. | {
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June 16th, 2017, 04:04 PM #16 Senior Member Joined: May 2016 From: USA Posts: 825 Thanks: 335 I am impressed by 123qwerty's translation into Markov processes even though I have almost no recollection of how to work with them. It certainly looks to be more powerful than my way. Obviously my way and mathematician's way are mathematically identical although I flatter myself that my explanation may be slightly more intuitive. Of course I could not give an explanation in Nepali at all. Denis, mon ami, if you still are not convinced, I suggest that you try to give the reasoning that leads to your result.
June 16th, 2017, 04:33 PM #17 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 I was kinda kidding...didn't want to admit I had badly programmed my simulation runs... Yes yes...I'm heading for the corner...
June 16th, 2017, 05:18 PM #18
Senior Member
Joined: May 2016
From: USA
Posts: 825
Thanks: 335
Quote:
Originally Posted by Denis I was kinda kidding...didn't want to admit I had badly programmed my simulation runs... Yes yes...I'm heading for the corner...
I probably will be joining you there soon enough.
June 18th, 2017, 10:27 AM #19 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 To: Probability Prof. Jeff From: student Denis If problem changed to 1 to 9 (instead of 1 to 255) and to 4 picks (instead of 24), everything else remaining same, then: total combos = 9^4 combos with 1 pair + 2 diff. numbers: 3024 probability: 3024/9^4 = .460905... OR: 1 * 8/9 * 7/9 * 3/9 * 2 = .460905... Do I visit the corner again?
June 19th, 2017, 10:58 PM #20
Senior Member
Joined: Jul 2012
From: DFW Area
Posts: 603
Thanks: 83
Math Focus: Electrical Engineering Applications
Respectfully to All,
I think that the answer to the problem as stated (exactly one matching pair with no repeats) is ~0.38924, or precisely:
$\displaystyle \large \frac{255!}{232! \cdot 255^{24}} \cdot \frac{24!}{2! \cdot 22!}$ | {
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