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given diagonal, height, angle between the diagonals and base ( a b ) : * It is true in this case: 4. Practice Question: Calculate area of trapezoid for the following problems: N.B. As a consequence the two legs are also of equal length and it has reflection symmetry. How to find the area of a trapezoid? A trapezoid is a quadrangle with two parallel sides. Customer Voice. Algebra; Geometry; Electrics; Trapezoid definition. The stimulus check calculator computes how much money you are eligible to receive from the coronavirus checks promised by the US government. Calculate the remaining internal angles. Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. α - 30° γ - 125° h – 6 cm a = 4 cm P = 25 cm Now you just follow these steps. Some possible trapezoid shapes are shown below to clarify the concept. Our perimeter calculator supports a lot of the basic shapes and below you can read details about each one, including its perimeter calculation formula. Trapezoid definition. The step by step workout for how to find what is the area and perimeter of a trapezoid. Enter three side lengths and one angle between two of those sides. Step 2: Now click the button “Solve” to get the result. A trapezoidal prism is a solid prism, which has trapezoid cross-sections in one direction and rectangular cross-sections in the other directions. … A trapezium or trapezoid is a 2dimensional shape which has 2 parallel sides and 2 non-parallel sides. Area of a trapezoid formula The formula for the area of a trapezoid is (base 1 + base 2) / 2 x height, as seen in the figure below: When doing the calculation, remember to take each measurement in the same unit, or to convert it to the same unit, to get valid results. Home / Mathematics / Area; Calculates the area of a trapezoid given two parallel sides and the height. Here is an example to understand it better. Omni Calculator is here to change all that - we are | {
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Here is an example to understand it better. Omni Calculator is here to change all that - we are working on a technology that will turn every* calculation-based problem trivial to solve for anyone. Perimeter of a square. To calculate the elements of the iso trapezoid; enter the two bases, a side or height. Prism is often distinguished by the shape of their base polygon. 1. Calculations at an isosceles trapezoid (or isosceles trapezium). This is possible for acute trapezoids or right trapezoids (rectangles). How to find the area of a trapezoid? Its area found to be 40cm 2. read more about us *within reason. All sides 2.Lateral side (leg) and angle at the base 3. Property #1) The angles on the same side of a leg are called adjacent angles and are supplementary() Property #2) Area of a Trapezoid = $$Area = height \cdot \left( \frac{ \text{sum bases} }{ 2 } \right)$$ () Property #3) Trapezoids have a midsegment which connects the mipoints of the legs() Step #2: Enter the value of long base "b". It has 3 fields in which you need to feed value. Calculator Techniques for Circles and Triangles in Plane Geometry The procedure to use the isosceles trapezoid calculator is as follows: Step 1: Enter the height and two base values in the input field. The volume of a trapezoidal prism calculator can find the volume and area of the trapezoidal prism at the same time. Mona Gladys has created this Calculator and 10+ more calculators! Trapezium. Follow Us. It has two parallel sides and the remaining two sides can be of any length, at any angle. This is not saying much about the quadrangle, so it is quite difficult to do calculations. Before jump into the code let’s know what is a trapezoid: A trapezoid is a geometrical figure with four sides in which two sides are parallel to each other. Properties. Choose the number of decimal places and click Calculate. To use this online calculator for Base a of Trapezoid, enter Height (h), Area (A) and Base (b) and hit the calculate button. FAQ. In | {
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Base a of Trapezoid, enter Height (h), Area (A) and Base (b) and hit the calculate button. FAQ. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule—see Trapezoid for more information on terminology) is a technique for approximating the definite integral. Therefore, the length of the other base of the trapezoid is 19 m. Example 5. How to Use the Isosceles Trapezoid Calculator? Formulas, explanations, and graphs for each calculation. The base is the lowest part or edge of something, especially the part on which it rests or is supported. Trapezoid area calculator is simple and easy to use. Our calculator will solve geometrical problems in a few seconds. The 2 parallel sides are called the bases. Using the area of trapezoid calculator: an example. If you only know the side lengths of a regular trapezoid, you can break the trapezoid into simple shapes to find the height and finish your calculation. Anamika Mittal has verified this Calculator and 25+ more calculators! Enter the three side lengths, choose the number of decimal places and click Calculate. An Example of How to Use the Area of a Trapezoid Calculator Sometimes, it’s helpful to have a real-life example to understand the calculator’s inner workings. Base b of Trapezoid calculator uses Base B=2*(Area/Height)-Base A to calculate the Base B, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base. An Example of How to Use the Area of a Trapezoid Calculator Sometimes, it’s helpful to have a real-life example to understand the calculator’s inner workings. Here is an example to understand it better. Anamika Mittal has verified this Calculator and 25+ more calculators! Use the point as decimal separator. By … Area of a Trapezoid Calculator: ... A 4 – sided geometrical figure which has one pair of parallel sides is called a Trapezoid. An isosceles trapezoid is a trapezoid where the | {
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one pair of parallel sides is called a Trapezoid. An isosceles trapezoid is a trapezoid where the base angles have the same measure. Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. It is the space of the internal surface in a figure, it is limited for the perimeter. To calculate the trapezoid area, follow the below steps: Measure and write down the base a, base b, and height h of the trapezoid. Area of a trapezoid Calculator . Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. Other Calculator Techniques. What is its Area? The area of a trapezoid can be found through the sides using the substitution method. You can use the trapezoid area calculator to find the area of a trapezoid: Enter the length of the two bases (b1 and b2) of the trapezoid. Here is a comprehensive set of problems about polygons solved using calculators. Area of a trapezoid and midline or bases You can use the trapezoid area calculator to find the area of a trapezoid: Enter the length of the two bases (b1 and b2) of the trapezoid. : After working out the answer of each of the next questions, click adjacent button of see the correct answer. Calculating Height of a Trapezoid Example: A trapezoid has a base lengths of 8cm and 12cm. Similarly, as γ + δ = 180°, δ = 180° - 125° = 55°. Volume=(1/3)*pi*Height*(Radius 1^2+Radius 2^2+(Radius 1*Radius 2)), Total Surface Area=pi*Radius*(Radius+sqrt(Radius^2+Height^2)), Lateral Surface Area=pi*Radius*sqrt(Radius^2+Height^2), Total Surface Area=2*pi*Radius*(Height+Radius), Area of a Triangle when base and height are given, Area of a Parallelogram when base and height are given, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base and is represented as, The Base b of Trapezoid formula is defined as the twice the product of Area by | {
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is represented as, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base is calculated using. This trapezoid calculator will compute the area of a trapezoid for you. A trapezoid is an interesting four-sided geometric figure. How to calculate the area of the trapezoid in Python. Base B and is denoted by bb symbol. Let's assume that you want to calculate the area of a certain trapezoid. home / calculators / trapezoidal prism volume In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. TRAPEZOID CALCULATOR (A & B) and (B & C) that are supplementary (add to 180°). Trapezoid calculator computes all properties of a trapezoid such as area, perimeter, height and sides given a sufficient subset of these properties. Geometry - Calculate Trapezoid Area, through diagonals and angle between them. An acute trapezoid has two adjacent acute angles on its longer base edge, while an obtuse trapezoid has one acute and one obtuse angle on each base. All sides 2.Lateral side (leg) and angle at the base 3. In Euclidean geometry, a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezium. Base of Trapezoid Calculator. ∫ (). Angles are calculated and displayed in … You can always use our trapezoid calculator above, however, you should be able to calculate the area of trapezoid by yourself. Diagonals, angle between the diagonals and bases or midline 4. How to use the trapezoid calculator Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real numbers and press "calculate" with b being the short base and d being the long base (d > … We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at | {
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h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. α - 30° γ - 125° h – 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = ½(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the … 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base ° = 150° the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to | {
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difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button “ solve ” to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid – sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, δ = 180° 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180°, δ = 180°, =... Especially the part on which it rests or is supported 30° γ - 125° h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as γ + δ = 180° β... Of something, especially the part on which it | {
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structures and relationships, as γ + δ = 180° β... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180° ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180° ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 – sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base | {
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other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base a '' – sided geometrical figure has! Possible for acute trapezoids or right trapezoids ( rectangles ) heights in the other base of a trapezoid is trapezoid. Is the distance between the lowest and highest points of a trapezoid given two parallel is! And you can solve using the same Inputs P = 25 cm Now just... Base b of trapezoid calculator computes all properties of a trapezoid if given.... Between the diagonals and bases or midline 4 angles, and width.! Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the | {
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the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved | {
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polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has. | {
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5 septiembre, 2016 | {
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# Finding the roots, with multiplicity, of a polynomial with compound angle formula
My question is to find the roots, counted with multiplicity, of the polynomial equation
$16x^5-20x^3+5x-1=0$ using the compound angle formula $\sin\left(5\theta\right)=16\sin^5\theta-20\sin^3\theta+5\sin\theta$
So after substituting $x=\sin\theta$, I get to the equation
$\sin(5\theta)=1$
Then I get an infinitude of $\theta$ values, which when I find the sines of these values, all correspond to the distinct solutions $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(-\frac{3\pi}{10}\right)$.
What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.
Also if possible, is there a way to solve this polynomial using the given compound angle formula without the need to find the distinct roots and then determine the ones which repeat?
This is because the solutions to this question say $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$ without any reasoning, which makes me suspect I am unaware of some related theorem.
• Is your first equation correct? Maybe it is $16x^ 5$ instead of $x^5$. Nov 30, 2016 at 14:11
• Technically speaking, $x=\sin \theta$ is not a correct and proper substitution since it makes an assumption that $x \in [-1,1]$. Nov 30, 2016 at 14:49
What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.
If one knows the roots, then it is possible to find their multiplicity by finding whether or not the function's derivative(s) have the same root. See this question for a proof. | {
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• Yes I am aware of that theorem, but how exactly does one figure out if any of those three distinct roots are zeros of the derivative polynomial? Nov 30, 2016 at 15:04
• The derivative polynomial is $80x^4 - 60x^2 + 5$. So you know that $1$ definitely is not a repeated root. Also, you can see that the derivative is even. So it has $2$ positive roots and $2$ negative roots. And you have figured out that the distinct roots of the original equation are definitely $\sin(\frac{\pi}{10})$ and $\sin(\frac{-3\pi}{10})$. Do you see a complete argument? Nov 30, 2016 at 15:13
As already explained, the solutions are: $$x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$$ Because, $$\sin(5\theta)=1 \Rightarrow \theta=\frac{1}{5}\left(\frac{\pi}{2}+2k\pi\right)$$ and setting $k=0,1,2,3,4$ we get all five solutions.
But $\sin\left(\frac{\pi}{10}\right)=\sin\left(\frac{9\pi}{10}\right)$ and $\sin\left(\frac{13\pi}{10}\right)=\sin\left(\frac{17\pi}{10}\right)$.
So the five roots (with multiplicity) are $$1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right)$$.
Let's explain better why if make any other choice of $k$ we will always find an angle congruent to one of those in $S=\{\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10\}$.
We can write $k=5n+r$ with $r \in \{0,1,2,3,4\}$ and $n \in \Bbb Z$. Replacing that at the expression of $\theta$ we get:
$$\theta=\frac{\pi}{10}+\frac{2k\pi}{5}=\frac{\pi}{10}+\frac{2(5n+r)\pi}{5}=\frac{\pi}{10}+2n\pi+\frac{2r\pi}{5} \equiv \frac{\pi}{10}+\frac{2r\pi}{5}$$ | {
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The last equivalence means that the angle $\frac{\pi}{10}+2n\pi$ is congruent to $\frac{\pi}{10}$. That means that both angle stop at the same point on the trigonometric circle and then if we apply any trigonometric function at those angles we will get the same value. The algebric value are different but geometricaly they represent the same point at the circle.
It means that if we want to find the different solution for $\sin\theta$ is enough to take
$$\frac{\pi}{10}+\frac{2r\pi}{5}$$
with $r=0,1,2,3,4$.
• The question seems to be more like: why the specific values of $k=0,1,2,3,4$? Why not $k=-2,-1,1,2,5$? Nov 30, 2016 at 14:50
• When we choose $k=0,1,2,3,4$ we get all different values of $\theta$, wich are $S={\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10}$. If you those any other set for $k$ your new set will be inside $S$. Nov 30, 2016 at 14:58
• What do you mean by "different values of $\theta$"? If we choose different $k$, then obviously the $\theta$'s obtained will be different. And then, how can the new set of $\theta$'s be "inside" the aforementioned set $S$. Nov 30, 2016 at 15:05
• @ArnaldoNascimento could you please clarify what you meant by 'your new set will be inside S'? Nov 30, 2016 at 15:05
• Note that if $k \in \mathbb{Z}, \ \mbox{with}\ k \neq 0,1,2,3$ or $4$ then $k = 5\cdot n + r$ with $r=0,1,2,3\ \mbox{or}\ 4, n \in \mathbb{Z}$. In this case this does not matter because $sin (\frac{1}{5}( \frac{\pi}{2} + 2kpi )) = sin (\frac{1}{5}( \frac{\pi}{2} + 2rpi ))$. Nov 30, 2016 at 15:18 | {
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type/partition - Maple Programming Help
Home : Support : Online Help : Programming : Data Types : Type Checking : Types : type/partition
type/partition
check for an integer partition
Calling Sequence type( expr, 'partition' )
Parameters
expr - anything; any Maple expression
Description
• The call type( expr, 'partition' ) returns true if expr is a partition of a positive integer, and returns false otherwise.
• A partition of a positive integer $n$ is a list [ ${k}_{1},\mathrm{...},{k}_{r}$]. of positive integers ${k}_{i}$, whose sum is equal to $n$, and which is non-decreasing, that is, which satisfies $\mathrm{add}\left({k}_{i},i=1..r\right)=n$ and ${k}_{i}\le {k}_{i+1}$, for $i$ in $1..r-1$.
Examples
> $\mathrm{type}\left(\left\{\left\{1,2\right\},\left\{3,4\right\}\right\},'\mathrm{partition}'\right)$
${\mathrm{false}}$ (1)
> $\mathrm{type}\left(\left[1,2,\frac{2}{3}\right],'\mathrm{partition}'\right)$
${\mathrm{false}}$ (2)
> $\mathrm{type}\left(\left[1,-2,3\right],'\mathrm{partition}'\right)$
${\mathrm{false}}$ (3)
> $\mathrm{type}\left(\left[1,2,3\right],'\mathrm{partition}'\right)$
${\mathrm{true}}$ (4)
> $\mathrm{type}\left(\left[1,2,3,3,4\right],'\mathrm{partition}'\right)$
${\mathrm{true}}$ (5)
> $\mathrm{type}\left(\left[1,2,3,3,2\right],'\mathrm{partition}'\right)$
${\mathrm{false}}$ (6) | {
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# What do curly less than sign mean?
I am reading "Convex Optimization" by Stephen Boyd. He is using a curved greater than and curved less than equal to signs.
$f(x^*) \succcurlyeq \alpha$
or
$f(x*) \preccurlyeq \alpha$
Can someone explain or point me to their meaning.
-
as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly – barlop Feb 9 at 6:43
@barlop If you look at $\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \succcurlyeq, which has curly word in it, not curved. – Ruslan Feb 9 at 7:18
Thanks everyone for answers. As I understand $\succeq$ or $\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \succeq Y, \quad if \quad \| X \| \ge \| Y \|$ where $\| \cdot \|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. – Dinesh K. Feb 9 at 17:53
You should definitely take the tour. This is not a traditional forum! – canaaerus Feb 9 at 18:03
Well I don't know what a traditional forum looks like. :-) – Dinesh K. Feb 9 at 18:20
There's a list of notation in the back of the book. On page 698, $x\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\leq y_i$ for every index $i$.
Edit: The notation is introduced on page 32.
-
Both Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here. | {
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It should be noted that the book does not use $\succeq$, $\preceq$, $\succ$, and $\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for.
The book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.
Let $S$ be a vector space, and let $K\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\alpha K\equiv K$ for all $\alpha>0$; and by pointed, we mean that $K\cap-K=\{0\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \succeq_K y \quad\Longleftrightarrow\quad y \preceq_K x \quad\Longleftrightarrow\quad x - y \in K$$ $$x \succ_K y \quad\Longleftrightarrow\quad y \prec_K x \quad\Longleftrightarrow\quad x - y \in \mathop{\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\in S$, $x \not\succeq_K y$ and $y \not\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\geq$, $\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \succeq_K \alpha y \quad\forall \alpha>0$$ $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \preceq_K \alpha y \quad\forall \alpha<0$$ $$x\succeq_K y, ~ x\preceq_K y \quad\Longrightarrow\quad x=y$$ $$x\succ_K y \quad\Longrightarrow\quad x\not\prec_K y$$ | {
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When the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\succeq$. There are two cases where this is almost always done. First, when $S=\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \succeq_{\mathbb{R}^n_+} y \quad\Longleftrightarrow\quad x_i\geq y_i,~i=1,2,\dots,n$$ Second, when $S$ is the set of symmetric $n\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\mathcal{S}^n_+=\{X\in S\,|\,\lambda_{\text{min}}(X)\geq 0\}$, the inequality is a linear matrix inequality (LMI): $$X \succeq_{\mathcal{S}^n_+} Y \quad\Longleftrightarrow\quad \lambda_{\text{min}}(X-Y)\geq 0$$ In both of these cases, the cone subscript is almost always dropped.
Many texts in convex optimization don't bother with this distinction, and use $\geq$ and $\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\mathbb{R}^n_+$; I think most people rightly assume that $x\geq y$ is considered elementwise when $x,y$ are vectors.
-
Thanks a lot for the detailed answer, and for correcting the symbol as well :-). – Dinesh K. Feb 9 at 16:35
Often these symbols represent partial order relations. The typical "less than" and "greater than" operations both define partial orders on the real numbers. However, there are many other examples of partial orders.
- | {
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# Circular motion question about a car banked
1. Feb 12, 2005
### F|234K
if a curve with a radius of 60 meter is properly banked for a car travelling at 60km/h, what must be the coefficient of static friction for a car not to skid when travelling at 90km/h?
i know that in order to solve the question, one needs to find the angle first, and i found the angle to be 25.3 degrees(not sure to be correct)...
and the answer is 0.39...i just can't seem to get the answer...
2. Feb 12, 2005
### Staff: Mentor
That's correct.
Since you didn't show any work, there is no way to tell where you got stuck. The basic idea is exactly the same as in the no friction case: the only difference is the addition of the friction force on the car ($\mu N$ acting down the incline).
3. Feb 12, 2005
### xanthym
Your calculated road banking angle of 25.3 deg relative to horizontal is correct (also indicated by the previous msg).
The required static friction coefficient for 90 km/h along the same banked road is calculated to be 0.39 in agreement with the provided answer (but apparently not in line with your calculations).
You may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system.
A typical formulation based on the friction horizontal component only is:
$$:(1): \ \ \ \ \frac {v^2} {rg} = Tan(\theta) + \mu_s Cos(\theta)$$
The above equation yields a static friction coefficient of 0.65 for this case. The formulation based on fully coupled horizontal and vertical components is: | {
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$$:(2): \ \ \ \ \frac {v^2} {rg} = \frac {Sin(\theta) + \mu_s Cos(\theta)} {Cos(\theta) - \mu_s Sin(\theta) }$$
This latter formulation yields 0.39 for Coefficient of Static Friction.
The complete system of equations for horizontal and vertical components is given below:
$$:(3): \ \ \ \ \frac{mv^2} {r} = N Sin(\theta) + \mu_s N Cos(\theta)$$
$$:(4): \ \ \ \ 0 = N Cos(\theta) - \mu_s N Sin(\theta) - mg$$
where N is the force Normal to the road surface, (theta) the bank angle, and the other variables defined per usual conventions.
~~
Last edited: Feb 12, 2005
4. Feb 12, 2005
### F|234K
thank you guys very much
"You're may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system."
and yes, i got my answer to be 0.59, thank you xanthym for your detailed and consummate reply which enlightens me on the vertical component of friction. (i only calculated the horizontal part of friction.) | {
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# Is this a valid way to prove this modified harmonic series diverges?
I am trying to find a way to prove that $$\dfrac 11 + \dfrac 12 + \dfrac 13 + \dfrac 14 + \cdots \color{red}{-} \dfrac 18 + \cdots$$
where the pattern repeats every $8$ terms. Knowing about the Riemann Series Theorem, I am a little hesitant about manipulating conditionally convergent series at all. Granted that the harmonic series diverges, is the following a valid way to prove my series diverges?
$$\dfrac 11 + \dfrac 12 + \dfrac 13 + \cdots + \dfrac 17 - \dfrac 18 + \cdots = \sum_{n=1}^{\infty} \dfrac 1n - 2 \cdot \dfrac 18\sum_{n=1}^{\infty} \dfrac 1n$$
$$=\dfrac 34 \sum_{n=1}^{\infty} \dfrac 1n$$
Since the harmonic series diverges, so does $\dfrac 34$ times it.
• I stared at that line many, many times before I finally saw where the $+$ turned into a $-$. – Yakk Oct 21 '16 at 10:55
• @Yakk I suggested an edit to make it more obvious – null Oct 21 '16 at 16:46
• I've made it red. Don't know if there's a better way. – asmeurer Oct 21 '16 at 20:11
• @asmeurer Is it better now? – Ovi Oct 22 '16 at 1:57
• That helps. At first I thought you meant repeat the terms shown, which obviously diverges. – asmeurer Oct 22 '16 at 1:59
## 4 Answers
Your idea is a good one, but, as you suspected, you need to be more careful about this sort of manipulation of conditionally convergent series.
One way to carry out your argument correctly, but with only minor changes, is by looking at partial sums:
Let's write $$a_n=\begin{cases}1,&\text{ if }n\text{ is not a multiple of 8} \\-1,&\text{ if }n\text{ is a multiple of 8},\end{cases},$$ so that your series is $\sum_{n=1}^\infty \frac{a_n}{n}.$
Then for any natural number $N,$ | {
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Then for any natural number $N,$
\begin{align} \sum_{n=1}^{8N}\frac{a_n}{n} &= \sum_{n=1}^{8N}\frac1{n}-2\sum_{n=1}^N \frac1{8n} \\&=\sum_{n=1}^{8N}\frac1{n}-\frac1{4}\sum_{n=1}^N \frac1{n} \\&\ge\sum_{n=1}^{8N}\frac1{n}-\frac1{4}\sum_{n=1}^{8N} \frac1{n}\scriptsize{\quad\text{(because we can only be subtracting *more* positive numbers)}} \\&=\frac3{4}\sum_{n=1}^{8N}\frac1{n}, \end{align}
which approaches $\infty$ as $N$ approaches $\infty,$ since the harmonic series diverges.
• Thank you, I learned something very important; you can manipulate even conditionally convergent series as long as you manipulate them in the context of partial sums. – Ovi Oct 21 '16 at 4:52
• @Ovi -- Yes, because the partial sums are just regular finite sums. But you still need to be careful that your partial sums are entire initial parts of the infinite series. You can't pick and choose which terms to include; all you can do is chop the infinite series off at some finite point. After that, you can apply any normal algebraic operations, because you just have a finite sum. – Mitchell Spector Oct 21 '16 at 4:56
• @ Mitchell Spectator Okay got it – Ovi Oct 21 '16 at 4:57
• And if you want to prove that the conditionally convergent series does converge, you can't cherry-pick which of the partial sums you're looking at (like it happens here), of course. – Henning Makholm Oct 21 '16 at 14:21
• @HenningMakholm Good point. – Mitchell Spector Oct 21 '16 at 15:37
To expand on Mitchell Spector's answer, let's generalize to the problem where every $b$th term is negated.
By the same argument, we get \begin{align} \sum_{n=1}^{bN}\frac{a_n}{n} &= \sum_{n=1}^{bN}\frac1{n}-2\sum_{n=1}^N \frac1{bn} \\&=\sum_{n=1}^{bN}\frac1{n}-\frac2{b}\sum_{n=1}^N \frac1{n} \\&\ge\sum_{n=1}^{bN}\frac1{n}-\frac2{b}\sum_{n=1}^{bN} \frac1{n} \\&=\left(1-\frac2{b}\right)\sum_{n=1}^{bN}\frac1{n} \end{align} | {
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This is nice, since we can see that the divergence proof holds for $b \geq 3$, but fails for $b=2$. This is not strange, since the series for $b=2$ is known to converge to $\log 2$.
This proof isn't valid because of the Riemann series theorem; the original series could still converge even though some rearrangement of it diverges. It's important to focus on the partial sums.
You can argue along the lines that the series $$\sum_{k=0}^{\infty} \frac{1}{8k+1} = 1 + \frac{1}{9} + \frac{1}{17} + \cdots$$ already diverges by comparing its partial sums to the partial sums of the harmonic series $\frac{1}{8} \Big( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \Big).$
Note that \begin{align} &{\tiny\sum_{k=0}^\infty}{\tiny\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}+\frac1{8k+7}-\frac1{8k+8}\right)}\\ &={\tiny\sum_{k=0}^\infty}{\tiny\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}\right)+{\tiny\sum_{k=0}^\infty}\left(\frac1{8k+7}-\frac1{8k+8}\right)}\\ \end{align} For the left hand sum, we have \begin{align} &{\small\sum_{k=0}^\infty\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}\right)}\\ &\ge{\small\frac68\sum_{k=0}^\infty\frac1{k+1}} \end{align} which diverges.
For the right hand sum, we have \begin{align} &{\small\sum_{k=0}^\infty\left(\frac1{8k+7}-\frac1{8k+8}\right)}\\ &\le{\small\frac1{56}+\frac18\sum_{k=1}^\infty\left(\frac1{8k}-\frac1{8k+8}\right)}\\ &=\frac{15}{448} \end{align} A divergent sum plus a convergent sum diverges. | {
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• How did you evaluate $\dfrac 18 \sum_{k=1}^{\infty} \left( \dfrac {1}{8k}-\dfrac {1}{8k+8} \right) = \dfrac {1}{64} \sum_{k=1}^{\infty} \dfrac {1}{k(k+1)}$? I guess its not necessary to calculate the sum because we know it converges anyway, but I'm curious how you got a rational answer, knowing that $\sum_{k=1}^{\infty} \dfrac {1}{k^2} = \dfrac {\pi^2}{6}$ – Ovi Oct 24 '16 at 15:19
• \begin{align} \sum_{k=1}^\infty\frac1{k(k+1)} &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=2}^{n+1}\frac1k\right)\\ &=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)\\ &=1 \end{align} which is a Telescoping Sum. – robjohn Oct 24 '16 at 15:50 | {
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# Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$
Does anyone know how to evaluate the following limit? $$\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$$ The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
Multiply by the conjugate, then reduce: \begin{align*} \lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) &= \lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) \cdot \frac{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\ &= \lim_{x\to\infty} \frac{(x + \sqrt{x + \sqrt{x}}) - x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\ &= \lim_{x\to\infty} \frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \cdot \frac{\dfrac{1}{\sqrt{x}}}{\dfrac{1}{\sqrt{x}}} \\ &= \frac{\sqrt{\lim\limits_{x\to\infty}\dfrac{1}{\sqrt{x}} + 1}}{\sqrt{1 + \lim\limits_{x\to\infty} \dfrac{\sqrt{x + \sqrt{x}}}{x}} + 1} \\ &= \frac{\sqrt{0 + 1}}{\sqrt{1 + 0} + 1} \\ &= \frac{1}{2} \end{align*}
• I like your answer! :) – dmk Jun 15 '14 at 1:46
• thanks i like ur answer to – user157128 Jun 15 '14 at 1:54
• as you did in step 3 to 4 when multiplied by 1/v5 ??? – user157128 Jun 15 '14 at 2:03
$$\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{\frac{3}{2}}}}}+1}$$
I got the above by first multiplying by the conjugate to $\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}$ then dividing both the numerator and denominator by $\frac{1}{\sqrt{x}}$. Now take $x$ to infinity.
• the answer is 1/2, but I tried this and rationalize, but did not reach 0.5 – user157128 Jun 15 '14 at 1:38
• @user157128 what did you get? In the top I get $\sqrt{1 + 0} - 0$. What do you get in the bottom? – DanZimm Jun 15 '14 at 1:40 | {
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\begin{align} \ \lim_{x\rightarrow\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} &= \lim_{x\rightarrow\infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right) \cdot \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \ &= \lim_{x\rightarrow\infty} \frac{x + \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \cdot \frac{1/\sqrt{x}}{1/\sqrt{x}} \\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\frac{1}{x}+\frac{1}{x\sqrt{x}}}+1} \\ \ &= \frac{\sqrt{1}}{\sqrt{1}+1} \\ \ &= \frac{1}{2} \\ \end{align}
Using Taylor expansion:
$$\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$$
$$\sqrt{x + \sqrt{x}+\dfrac{1}{2} + O(\frac{1}{\sqrt{x}})} - \sqrt{x}$$
$$\sqrt{x}+\dfrac{1}{2\sqrt{x}}(\sqrt{x}+\dfrac{1}{2}+ O(\frac{1}{\sqrt{x}})) + O(\frac{1}{\sqrt{x}})- \sqrt{x}$$
$$\dfrac{1}{2} + \dfrac{1}{4\sqrt{x}}$$
Which in the limit case tends to ${\dfrac{1}{2}}$.
The expansion I used was $(a+b)^{1/2} = \sqrt{a}+\frac{b}{2\sqrt{a}} + \cdots$
• It's not clear from what you've written why you get to ignore the error terms in your expansion. – Cheerful Parsnip Jun 15 '14 at 1:50
• @user157107 this seems like a fairly interesting way to go about it... and similar to what I was considering doing. Could you possibly add more details to justify your approach? – Squirtle Jun 15 '14 at 2:13 | {
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# Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other
Find the nonzero vectors $$u,v,w$$ that are perpendicular to the vector $$(1,1,1,1)$$ and to each other.
If I follow algebra, then I get complicated results to solve it as follows:
Let $$u=(u_1,u_2,u_3,u_4), \ v=(v_1,v_2,v_3,v_4) , \ w=(w_1,w_2,w_3,w_4)$$
Then, $$u \cdot (1,1,1,1)=v \cdot (1,1,1,1)=w \cdot (1,1,1,1)=0$$
Also $$u \cdot v=w \cdot u=v \cdot w=0$$.
These gives us
$$u_1+u_2+u_3+u_4=0, \\ v_1+v_2+v_3+v_4=0 , \\ w_1+w_2+w_3+w_4=0, \\ u_1v_1+u_2v_2+u_3v_3+u_4v_4=0, \\ u_1w_1+u_2w_2+u_3v_3+u_4v_4=0, \\ v_1w_1+v_2w_2+v_3w_3+v_4w_4=0.$$
But how to solve for $$u_i, v_i,w_i, \ i=1,2,3,4$$ from here?
Does there exit any other easy method?
Help me out
• Apply Gram-Schmidt to the basis $$((1, 1, 1, 1), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)),$$ or any other basis beginning with $(1, 1, 1, 1)$. – Theo Bendit Sep 27 '18 at 2:37
• Is not in that case the solution will be particular , I mean there can other vectors also which can not be conclude using Gram-schmidt method. – M. A. SARKAR Sep 27 '18 at 2:44
• Yes, it will be particular. If you range over all such bases, then you will obtain all orthonormal bases beginning with $\left(\frac12,\frac12,\frac12,\frac12\right)$, though not uniquely. – Theo Bendit Sep 27 '18 at 2:47
as columns $$\left( \begin{array}{rrrr} 1&-1&-1&-1 \\ 1& 1&-1&-1 \\ 1&0 &2&-1 \\ 1&0&0&3 \end{array} \right)$$ Pattern, done correctly, works in any dimension | {
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$$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right).$$
• That's great! =) – Siong Thye Goh Sep 27 '18 at 3:02
• Excellent method – M. A. SARKAR Sep 27 '18 at 11:13
Consider the Hadamard matrix of which we know that the columns form an orthogonal basis of $$\mathbb{R}^4$$.
$$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1\end{bmatrix}$$
The other columns would give you a solution.
Alternatively, use Gram-Schmidt process.
• But if we consider the Hadamard matrix , then the solution will be particular. There may be other vectors which are perpendicular to $(1,1,1,1)$ except the last 3 column vectors in the Hadamard vectors and its multiples . – M. A. SARKAR Sep 27 '18 at 2:42
• So if we use Gram-Schmidt method then we need a basis – M. A. SARKAR Sep 27 '18 at 2:43
• Theo has given you a basis right? Note that answer is not unique. If you want to describe the set, you have already done so in your post. – Siong Thye Goh Sep 27 '18 at 2:48
• There is a pattern that easily adapts to any dimension... – Will Jagy Sep 27 '18 at 3:01 | {
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# Is it necessary to write limits for a substituted integral?
To solve the following integral, one can use u-substitution: $$\int_2^3 \frac{9}{\sqrt[4]{x-2}} \,dx,$$ With $$u = \sqrt[4]{x-2}$$, our bounds become 0 and 1 respectively. Thus, we end up with the following: $$36\int_0^1{u^2} \,du$$ In the first case, the lower bound is a vertical asymptote so we would have to use limits to find the answer. However, in the second case there's no longer an asymptotal bound - would you still have to write the limits since the original function would've needed limits, or can you just solve this by plugging in the substituted bounds? I know the final answer will be the same either way, but I want to know if it can be considered correct to exclude the limits in the 2nd integral from a technical perspective since the function has been changed. Many thanks in advance!
## 3 Answers
Without too much simplification, the substitution you cite yields
$$\int_2^3\frac9{(x-2)^{1/4}}\,\mathrm dx=36\int_0^1\frac{u^3}u\,\mathrm du$$
which you certainly welcome to treat as an improper integral,
$$36\left(\frac13-\lim_{u\to0^+}\frac{u^3}3\right)$$
but since $$u=0$$ is a removable discontinuity and the limand reduces to $$u^2$$, you may as skip this treatment altogether.
• I'm confused about your last expression. It seems like you want to still have $\int$ present, but the $1-$ seems like you've already taken the integral, but the $\frac{u^3}u$ is from before you've taken the integral... – Teepeemm Feb 20 at 2:05
• Yes, it was wrong as written. I have edited it. – Martin Argerami Feb 20 at 3:51
• Thanks @MartinArgerami ! – user170231 Feb 20 at 14:55
There are certain conditions that must be met for a substitution to be "legal". In most circumstances these conditions are naturally met and so they are not emphasized; you have here one situation in which they are not. | {
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For instance, one set of conditions is given in Anton, Anderson, and Bivens (Calculus, Early Transcendentals, 11th Edition), Theorem 5.9.1:
Theorem 5.9.1. If $$g'(x)$$ is continuous on $$[a,b]$$, and $$f$$ is continuous on an interval containing the values of $$g(x)$$ for $$a\leq x\leq b$$, then $$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.$$
Here you have $$g(x) = \sqrt[4]{x-2}$$, so $$g'(x) = \frac{1}{4}(x-2)^{-3/4}$$... which is not continuous on the interval $$[2,3]$$ that you are working on.
In fact, as you note, the initial integral is improper, which means you aren't really evaluating that integral: you are evaluting a limit, $$\lim_{h\to 2^+}\int_h^3 \frac{9}{\sqrt[4]{x-2}}\,dx.$$ The integral in the limit does satisfy the conditions of the theorem above, so you can make the substitution to get $$\lim_{h\to 2^+}\int_{\sqrt[4]{h-2}}^1 \frac{u^3}{u}\,du = \lim_{a\to 0^+}\int_a^1 u^2\,du,$$ and proceed from there.
• Thank you very much Arturo, your answer was very in-depth and provided a lot of reinforcement for what you were saying. I wish I could select multiple answers to be correct, I thought your inclusion of the cited textbook snippet was really nicely done too. I chose the other answer only because I found it easier to digest/understand at first glance, but after I properly read your comment I understood it more fully. Thank you for your quick and effective contribution :-) – Lord Kanelsnegle Feb 19 at 17:32
Suppose you must. Then we have $$9\lim_{a\to2^+}\int_a^3\frac1{\sqrt[4]{x-2}}\,dx.$$
Let $$u=x-2$$. Then we have $$9\lim_{a\to2^+}\int_{a-2}^1 u^{-1/4}\, du.$$
By the power rule, we have $$9\lim_{a\to2^+} \left.\frac43u^{3/4}\right]_{a-2}^1=9\left[\frac43(1-\lim_{a\to2^+}(a-2)^{3/4})\right]=9\left[\frac43(1-0)\right]=12.$$
Notice that it does not make a difference whether you use $$\lim_{a\to2^+}(a-2)$$ or $$0$$. | {
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Notice that it does not make a difference whether you use $$\lim_{a\to2^+}(a-2)$$ or $$0$$.
• Thanks for the quick and simple response! I found the answer I selected to be more aligned with the question, but I wanted to comment because I appreciate the time you took to write this. Thank you once again. – Lord Kanelsnegle Feb 19 at 17:29 | {
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# Expected Value of flipping through a book
Suppose that a book has $$N$$ pages, and we read through the book as follows. We start from page 0, and if we're on page $$i$$, we randomly flip to a page $$i + 1, i + 2, ..., N$$ with equal probability.
1. What is the expected value of number of flips we need to finish the book?
Intuition tells me that we can, on average, expect to halve the number of pages remaining. This yields $$\log_2(N)$$, but I'm having trouble formalizing it.
1. If $$N = 26$$, what is the probability we flip to page 13 at some point? Assume we start at page 0.
I let $$P_i$$ be the probability we land on page 13 eventually, starting from page $$i$$. Then, $$P_{13} = 1$$, and in general, $$P_{i} = \frac{1}{26 - i}\sum_{k = i + 1}^{13}P_k$$
Evaluating terms like $$P_{12}, P_{11}, P_{10}$$, I see that all of these values are $$\frac{1}{14}$$, including $$P_0$$. Is there a more intuitive reason for such a simple answer?
• Question 1. is unclear ... number of flips we need for what? To read every page in the book? To reach the last page in the book? What does it mean? – Ross Presser Aug 21 at 13:47
• edited for clarity, though it should be implicitly clear that it's often not possible to read every page in the book, and the expected value of flips to read every page is trivial. – user815048 Aug 21 at 20:13
Let's consider the equivalent problem in which we start at page $$n$$ and flip backwards through the book, going to each of the pages $$0, 1, ..., n - 1$$ with equal probability. Let $$E_n$$ be the expected number of flips. Then we have $$E_0 = 0$$ and
$$E_n = 1 + \frac{1}{n} \sum\limits_{i = 0}^{n - 1} E_i$$
Then in particular we have | {
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$$E_n = 1 + \frac{1}{n} \sum\limits_{i = 0}^{n - 1} E_i$$
Then in particular we have
$$$$\begin{split} E_{n + 1} &= 1 + \frac{1}{n + 1} \sum\limits_{i = 0}^{n} E_i \\ &= 1 + \frac{E_n}{n + 1} + \frac{n}{n + 1} \frac{1}{n} \sum\limits_{i = 0}^{n - 1} E_i \\ &= 1 + \frac{E_n}{n + 1} + \frac{n}{n + 1} (1 + \frac{1}{n} \sum\limits_{i = 0}^{n - 1} E_i) - \frac{n}{n + 1} \\ &= 1 - \frac{n}{n + 1} + \frac{1}{n + 1} E_n + \frac{n}{n + 1} E_n \\ &= \frac{1}{n + 1} + E_n \end{split}$$$$
whenever $$n \geq 1$$ (and the identity is easily verified when $$n = 0$$ as well).
Then by induction, we have $$E_n = \sum\limits_{j = 1}^n \frac{1}{j}$$, the $$n$$th harmonic number. This will be asymptotically very close to $$\log_e(n)$$.
Let $$P_n$$ be the expected number of flips in a book with $$n$$ pages. Then $$P_0=0,\ P_1=1$$ and $$P_n=1+\frac1n\sum_{k=0}^{n-1}P_k,\tag1$$ because we have to make one flip, and then we're equally likely to have any number of pages from $$0$$ to $$n-1$$ left to flip through.
We get \begin{align} P_1&=1\\ P_2&=\frac32\\ P_3&=\frac{11}6\\ P_4&=\frac{50}{24}\\ P_5&=\frac{174}{120} \end{align}
The denominators are obviously $$n!$$, so we look for the numerators in OEIS and find A000254, the unsigned Stirling numbers of the first kind.
OESI gives the recurrence $$a_{n+1}=(n+1)a_n+n!$$ for the unsigned Stirling numbers of the first kind, and dividing through by $$(n+1)!$$ we get $$P_{n+1}=P_n+\frac1{n+1}$$ which clearly gives $$P_n=\sum_{k=1}^n\frac1k=H_n,$$ the $$n$$th harmonic number. To complete the problem, we must show that the harmonic numbers satisfy the recurrence $$(1)$$. | {
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Here is how I approached the first part of the problem. Consider a book with exactly $$n$$ pages. Let $$P_1$$ denote the first page you flipped to, and let $$X_n$$ represent the number of pages you flipped until you get to the last page. Note $$P_1$$ is uniformly distributed on the set $$\{1,...,n\}$$ and $$E(X_1)=1$$. Using the total law of expectation we get for $$n\geq2$$ that $$E(X_n)=\sum_{k=1}^{n}E(X_n|P_1=k)P(P_1=k)=\frac{1}{n}\sum_{k=1}^{n}E(X_n|P_1=k)$$
Notice $$E(X_n|P_1=k)=1+E(X_{n-k})$$ and so $$E(X_n)=\frac{1}{n}\sum_{k=1}^{n}\Big[1+E(X_{n-k})\Big]=1+\frac{E(X_0)+\dots+E(X_{n-1})}{n}$$ Replace $$n$$ with $$n+1$$ to get $$E(X_{n+1})=1+\frac{E(X_0)+\dots+E(X_{n})}{n+1}$$ Combining the previous two equations unveils the relation $$(n+1)(E(X_{n+1})-1)=(n+1)E(X_n)-n$$ which is equivalent to saying $$E(X_{n+1})=E(X_n)+\frac{1}{n+1}$$ So finally $$E(X_n)=\sum_{k=1}^{n}\frac{1}{k}$$ | {
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# Math Help - college prob + stat
1. ## college prob + stat
2 Questions:
1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?
2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?
2. Originally Posted by magicpunt
1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?
$P(\mbox{men or woman})=P(\mbox{men})+P(\mbox{woman})$
Let us find,
$P(\mbox{men})$
There are 10 people and 5 are selected there are a total of $_{10}C_5=252$
From 5 men you take 5 men intotal the number of combinations is,
$_5C_5=1$
Thus,
$P(\mbox{men})=\frac{1}{252}$
Similarily ya have,
$P(\mbox{women})=\frac{1}{252}$
Thus, the probability is,
$\frac{1}{252}+\frac{1}{252}=\frac{2}{252}=\frac{1} {126}$
3. 2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?
This is similar to the 'boxes and balls' problems.
There are $5^{10}=9765625$ ways to distribute the people to the 5 floors.
There are $\frac{5!}{2!}=60$ ways that no two people get off on the same floor.
EDIT:
Yes, I misinterpreted the problem. Cerebral flatulence. Soroban's method is solid.
My second part should be $\frac{10!}{2^{5}}$
Therefore, $\frac{5^{10}}{\frac{10!}{2^{5}}}$
This is a version of the classic 'assigning diplomats' problem.
For instance:
How many ways can 10 different diplomats be assigned to 5 different countries if 2 diplomats must be assigned to each country?. | {
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See the similarity with your problem?. Yours is just elevators and people, instead.
This can be done in $\frac{10!}{(2!)^{5}}=113,400$ ways.
The method I used is just another variation on this problem. Keep it, you may need it.
4. Originally Posted by galactus
This is similar to the 'boxes and balls' problems.
There are $5^{10}=9765625$ ways to distribute the people to the 5 floors.
There are $\frac{5!}{2!}=60$ ways that no two people get off on the same floor.
I think you might've misunderstood the problem
The question is: what is P(Exactly 2 people get off at each floor)
5. Hello, magicpunt!
Here's my approach to #2 . . .
2) There are 10 people in an elevator in the basement.
They can get off at any of 5 floors above.
The probability that they get off at each floor is equal
and each person leaving is independant of each other.
What is the probability that 2 people get off at each floor?
Each of the ten people has a choice of five floors.
. . There are: $5^{10}$ways that they can leave the elevator.
If exactly two people get off at each floor,
. . the number of ways is: $\begin{pmatrix}10\\2,2,2,2,2\end{pmatrix} = 113,400$
Therefore, the probabilty is: . $\frac{113,400}{5^{10}}\:=\:\frac{4,536}{390,625}$
6. Originally Posted by magicpunt
2 Questions:
1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?
Another approach to #1
Find the total no. of ways of selecting 5 out of 10= $\frac{10!}{5!5!}=C_5^{10}$
No. of ways in which all are men=1
No. of ways in which all are women=1
Required probability= $\frac{1+1}{C_5^{10}}=\frac{1}{126}$
Malay | {
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# What is the moment of inertia of a Gosper island?
We know that regular hexagons can tile the plane but not in a self-similar fashion. However we can construct a fractal known as a Gosper island, that has the same area as the hexagon but has the property that when surrounded by 6 identical copies produces a similar shape, but with dimensions scaled by a factor of $\sqrt{7}$.
What is the distance between two of the centers? Is it the same as the distance between hexagons of the same area? ie. If I start with a hexagon of area A, then construct a Gosper island and place it next to an identical copy, would the distance still be the same as if they were hexagons? Or does the scaling factor come into play somewhere? Right now I think the answer is $\sqrt{3}/2$, as for the hexagon.
The reason I ask is that I'm trying to calculate the Gosper island's moment of inertia through an axis through its centre of mass and perpendicular to the plane of the island.
If we assume that the moment of inertia is always proportional to the mass, and proportional to the square of a characteristic length scale, then $$I = \gamma Ml^2,$$ where $\gamma$ is a constant, $l$ is the 'diameter' of the island, in a hexagon this would be the distance between two opposite vertices. Shrink the Gosper island by the scaling factor and surround it by six others. This self-similarity technique is super cute, and can be used to calculate the moment of inertia of an equilateral triangle, and can be extended to a square/rectangle quite easily. Fractals, having a high degree of self-similarity, seem amenable to this technique - here I calculate the moment of inertia for a Koch snowflake.
$\hspace{1.3cm}$
Using the principle of superposition, $$I = I_{\text{centre}} + 6I_{\text{edge}},$$ where $$I_{\text{centre}} =\gamma \frac{M}{7}\left(\frac{l}{\sqrt{7}}\right)^2 = \gamma \frac{Ml^2}{49} = \frac{I}{49}.$$ | {
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Now, by the parallel axis theorem $\displaystyle I_{\text{edge}} = I_{\text{COM}} + Md^2$ where $$\displaystyle I_{\text{COM}} = \frac{I}{49}$$ and $\displaystyle d= \frac{\sqrt{3} l}{2}$ (this was one source of error), so $\displaystyle I_{\text{edge}} = \frac{I}{49} + \frac{3Ml^2}{4},$ and \begin{align*} I &= \frac{I}{49} + 6\left(\frac{I}{49}+ \frac{3Ml^2}{4}\right), \\ I & = \frac{I}{7} + \frac{9Ml^2}{2}, \\ \frac{6I}{7} & = \frac{9Ml^2}{2}, \\ I & = \frac{21Ml^2}{4}. \end{align*}
This seems incorrect? It feels wrong, comparing to a disk of radius $l/2$ which has moment of inertia $Ml^2/4$ it seems far too large.
It would also be nice if we could verify our answer numerically or otherwise. Any references are also appreciated. | {
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• The center of mass of the Gosper island is indeed the same as the center of the initial hexagon. This, as well as the moment of inertia can be computed using the technique of self-similar integration as in my answer to this question. I could provide a few more details but I get the feeling this might be an exercise? – Mark McClure May 17 '16 at 13:18
• Nope, just something I thought up after teaching the self-similarity method for finding the MOI of a triangle/square. Thankyou so much for the link! I suppose this can be done for an fractal with enough rotational symmetry to tile the plane? Please provide as many details as you wish, in an answer if you like. – Bennett Gardiner May 18 '16 at 2:05
• By the way, the polar inertia of a disk is not $ml^2/4$ (That is the x or y inertia) but $Ml^2/2$. – Urukann May 20 '16 at 4:40
• Isn't that only if $l$ is the radius though? In my formulation I was trying to use the diagonal of the hexagon, so it made more sense to compare to a disk with $l$ as the diameter. – Bennett Gardiner May 20 '16 at 6:19
• @BennettGardiner: IMO this is a source of confusion (especially if you don't explain it). The standard formula is $I=MR^2$ where $R$ is an equivalent radius (AKA gyration). I found it more effective to work with unit area shapes, so that the mass can be simplified, and $R$ conveys all shape information. – Yves Daoust May 20 '16 at 14:40
I think that your $\frac{\sqrt{3}}{2}$ for the distance between hexagons center is not right... The distance from an hexagon center to an edge is $\frac{r\sqrt{3}}{2}$, thus the distance between two hexagons of radius $\frac{l}{2\sqrt{7}}$ that touch by an edge should be: $$d = l\sqrt{\frac{3}{28}}$$
Now, if we take back your computation: $I = \frac{I}{49} + 6 \left(\frac{I}{49} + \frac{3Ml^2}{28}\right)$
This gives us : $I = \frac{3}{4} M l^2$, which may still not be the right answer.
It would be nice to compare this value to numerical methods, I'll look into it. | {
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It would be nice to compare this value to numerical methods, I'll look into it.
e/ I ran some numerical simulation, unfortunately it does not seem to confirm the above result.
How I did it:
• Starting from the code linked in this page, I generated the all points composing the outer shape of the gosper island.
• Then, I trimmed the generating points to remove duplicates
• I approximate the Gosper island as being star-shaped (This is not perfectly true, at least for $n=5$). Then its inertia is the sum of the inertia of the triangles composed of the origin and two consecutive points. Note that to get the actual inertia, one has to divide by the shape's area (Because when evaluating the triangle's inertia, we consider it has a surfacic mass of 1.). All triangle related formulas are available on Wolfram Alpha.
The results show that indeed the inertia is proportional to $l^2$, but its ratio to the value conjectured above is not 1, but a constant close to $\frac{4}{7}$: $$I_{gosper} \approx 0.568582263418 \frac{3}{4} M l^2$$
Unfortunately, it seems that this error is not related to the star-shaped approximation: I ran another experiment, this time using the Seidel program from the University of North Carolina at Chapel Hill. It allowed me to find a triangulation of the inner area of the Gosper island. Using another (similar) code I could check that the computation for a radius of 1. does yield the same ration between expected inertia (0.75) and the actual inertia (0.42856647032), with a similar ratio of 0.571421960426. Note that this inertia is very close to $\frac{3}{7}$, (best fractional approximation with a denominator inferior to fifteen thousand).
Actually, I had forgotten that the characteristic dimension is not the diameter, but the radius, thus the ratio is 0.142855490107, very close to $\frac{1}{7}$. | {
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Using this method for a Koch snowflake yields pretty correct results: for a snowflake of radius $r \approx 1.44$ (I forgot to scale the step size properly) I get an inertia of $I_{koch} \approx 0.736036125705$ while the one given by $Ml^2/11 \approx 0.7435999999999999$
e/ I found the error: The mass of a "small" Gosper island is not $M$, but $\frac{M}{7}$, thus the missing factor 7. This is due to the fact that we make the assumption of a uniform density Gosper Island, thus its mass is proportional to its area.
We can rewrite our original equation: $$I_{edge} = I_{center} + \frac{M}{7}d^2\\ d = l \sqrt{\frac{3}{28}}$$
Which gives us: $$I = \frac{I}{49} + 6 \left( \frac{I}{49} + \frac{3Ml^2}{7\cdot 28}\right)$$
And finally:
$$I = \frac{3Ml^2}{28}$$ | {
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And finally:
$$I = \frac{3Ml^2}{28}$$
• Bah, forgot to scale the distance $d$. Thanks, this sounds much more reasonable. If you could confirm using numerical methods with some explanation of how you did it, it would put it to rest and I'd definitely accept/award the bounty. – Bennett Gardiner May 19 '16 at 12:42
• I just added some numerical simulation. However, it does not confirm my above conjecture. It may be because my code is quite dirty, but it may also be an error in the reasoning... Cheers – Urukann May 20 '16 at 4:37
• Interesting, thanks for your efforts, looks like we still have some way to go. Can you use the code to check the Koch snowflake case I linked? It code provide a check on the code For the snowflake, $I = Ml^2/11$ where $l$ is the largest 'diameter' of the snowflake (tip to tip) I really felt like this method should work, I wonder where the error lies! – Bennett Gardiner May 20 '16 at 6:22
• Hum, I checked with the Koch snowflake (my code was easier to modify than expected !): I get "correct" results. I noticed that I forgot to use the diameter instead of the radius, so we are off by another factor 4 I'm guessing... – Urukann May 20 '16 at 6:44
• Sorry, this is getting confusing, but no: I calculated the MOI for a radius 1 Gosper Island to be 0.42856647032, that is to say, $I = \frac{3Mr^2}{7} = \frac{3Md^2}{7 \cdot 4}$, meaning that we are missing a factor $\frac{1}{7}$. – Urukann May 20 '16 at 6:58
Yes, the distance between those hexagon-like shapes in a hexagonal lattice is the same as the distance between centres of hexagons of the same area in a hexagonal lattice.
The reason is that "in the long run" the area per mesh must be the same. That is, a sufficiently large circular disk contains both Gosper islands and hexagons in a number approximately equal to area of the disk dividied by area of the shape (with an error in the order of magnitude proportional to the cirumference of the disk). | {
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The area inside the hexagon is made of one whole island and exactly six thirds, by symmetry.
Hence for an island of unit area, the distance between centers is such that
$$3\frac{\sqrt3}2d^2=3,$$
$$d=\sqrt{\frac2{\sqrt3}}.$$
To compute the moment of inertia, we will work with the radius of gyration and for an island of unit area.
$$I=MR^2$$ where $M$ is the mass.
For the assembly in the figure, enlarging by the factor $\sqrt7$ and using the axis theorem,
$$7M7R^2=49I=I+6(I+Md^2),$$
giving
$$R=\frac d{\sqrt7}=\sqrt{\frac{2}{7\sqrt3}}=0.406149258\cdots$$
This compares to the radius of gyration of a unit area disk,
$$R'=\frac1{\sqrt{2\pi}}=0.398942280\cdots$$
and that of a unit area hexagon,
$$R''=\sqrt{\frac{10}{36\sqrt3}}=0.400468569\cdots$$
• Interesting, so to scale this to an island with diameter $l$, would we simply scale this distance by $l$? – Bennett Gardiner May 20 '16 at 6:57
• Or $\sqrt{3/28}l$, like in the other poster's answer? I'm afraid I've confused myself quite a bit with the length scales. – Bennett Gardiner May 20 '16 at 7:00
• @BennettGardiner: I have no idea of the value of the diameter $l$, but you don't need it for the computation of the moment. – Yves Daoust May 20 '16 at 7:03
• I'm still trying to figure out your distance $d$. If that is the radius of gyration, what would the moment of inertia be? – Bennett Gardiner May 20 '16 at 7:32
• My confusion is part of why I asked the question. I am inclined to believe your answer more now that you have posted a comparison with the hexagon of unit area, in my mind they should have been very similar. – Bennett Gardiner May 20 '16 at 14:18 | {
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# Number of strings, when each character must occur even times
I was reading this question on programmers.StackExchange and wanted to try a combinatorics approach to solve the following problem:
Let $S=\{A,B,C,D\}$ be a set of characters and $n$ a positive natural, find the number of strings of length $n$ composed of characters in $S$ such that each character occur even times.
I tried to solve it by myself but I am stuck and I would appreciate a hint to solve this kind of problems. Also, there are similar questions on this site for not quite exactly the same problem, but I didn't find an answer that could help me.
## Current attempt
First, when $n$ is odd, the result must be zero. A string satisfying the desired property would necessarly have an even length, as a permutation of the concatenation of substrings of even lengths (each substring being a character from $S$ repeatead even times).
Now, let's suppose that $n$ is even.
If we have a mapping $c$ from $S$ to $\mathbb{N}$ assigning the number of occurence of each character (e.g. $c_A=0, c_B=2, c_C=4, c_D=0$), then the number of strings $N(c)$ we can produce with this mapping would be a k-permutation of characters, as we choose a string of size $n$ with $c_A$ times $A$, $c_B$ times $B$, etc. :
$$N(c) = \frac{n!}{c_A! c_B! c_C! c_D!}$$
Also, I can evaluate the number of such mappings $c$: this is equivalent to finding the number of strings where characters appear in order with varying occurences. For example, with $n=4$:
$$\begin{array}{c|cccc} & c_A & c_B & c_C &c_D \\ \hline AAAA & 4 & 0 & 0 & 0 \\ AABB & 2 & 2 & 0 & 0 \\ AACC & 2 & 0 & 2 & 0 \\ \dots\\ \end{array}$$
The number of strings of size $n$ with an even number of each character is exactly the same as the number of strings of size $k=n/2$ with single characters (just imagine that AA, BB, CC, DD are characters). So, the number of mappings is the number of combinations of characters from $S$, with repetition, like the following ones with $k=2$: | {
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AA AB AC AD
BB BC BD
CC CD
DD
That is given as a $k$-combination of 4 elements:
$$N_{mapping}(k) = \binom{|S|+k-1}{k} = \binom {3+k}{k}$$
This is where I'am stuck. I wanted to combine this number of mappings with the number of permutations computed by each mapping, previously, but I can't ($N(c)$ depends on actual values of the mapping $c$). Also, I am sure there is a more straightforward approach, but I don't know how to proceed. Thanks for any help you could provide.
• I might miss something, but polynomial coefficient seems to work: $\binom{n}{2k} \cdot \binom{n-2k}{2j} \cdot \binom{n-2k -2j}{2m}$ – Alex Feb 13 '15 at 13:26
• @Alex In your formula, do $k$ $j$ and $m$ stand for half the number of occurences of each character? I understand it as: choose first $2k$ elements from $n$, then $2j$ from $n-2k$, and so on...? Are $k$, $j$ and $m$ supposed to be known? Sorry, I have very little experience with this. – coredump Feb 13 '15 at 13:39
• Yes, but it seems like not every $\binom{2n}{2k}$ is even, e.g. $\binom{6}{4}$ – Alex Feb 13 '15 at 14:56
Generating functions can be applied to this problem. However, since the order of letters in a string is important, exponential generating functions are appropriate (not ordinary ones). So, suppose the alphabet $S$ has $m$ letters and let $a_n$ be the number of strings of length $n$ over the alphabet $S$ with an even number of each letter. Let $g(x)=\sum_{n\ge0}\frac{a_n}{n!}x^n$ be the exponential generating function of $\langle a_n:n\ge0\rangle$. Then, $$g(x)=\left(\sum_{n\textrm{ even}}\frac{x^n}{n!}\right)^m=\left(\frac{e^x+e^{-x}}2\right)^m.$$
In the case of $m=4$, $g(x)=\frac1{16}(e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x})$. Then, $a_n=\left[\frac{x^n}{n!}\right]g(x)$ and if $n$ is even, $a_n=\frac{n!}{16}\left(\frac{2\cdot4^n}{n!}+\frac{8\cdot2^n}{n!}\right)=\frac18(4^n+4\cdot2^n)$. In particular, $a_4=40$. | {
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Note: based on the form of the solution of $a_n$, I suspect there's a direct combinatorial argument (but I don't know it).
• I am very impressed. I should have a look at generating functions. Thanks a lot, I'll study this and try to understand what it all means. – coredump Feb 13 '15 at 21:04
• @coredump: If you want to take a look at generating functions, one excellent place to start is Herb Wilf's free book: math.upenn.edu/~wilf/gfologyLinked2.pdf. After reading his book, I was hooked. – Rus May Feb 13 '15 at 23:41
Let $m=|S|$ be the number of distinct characters, and let $a_n$ be the number of strings in $S^n$ having an even number of each character in $S$. Let
$$g(x)=\sum_{n\ge 0}\frac{a_n}{n!}x^n$$
be the exponential generating function (egf) for $\langle a_n:n\in\Bbb N\rangle$. (The egf is wanted because the order of the characters matters.) The egf for the sequence given by $$a_n=\begin{cases}0,&\text{if }n\text{ is odd}\\1,&\text{if }n\text{ is even}\end{cases}$$ is
$$\sum_{n\ge 0}\frac1{(2n)!}x^{2n}=\frac12(e^x+e^{-x})\;,$$
so
$$g(x)=\left(\sum_{n\ge 0}\frac1{(2n)!}x^{2n}\right)^m=\frac1{2^m}(e^x+e^{-x})^m\;.$$
Now
\begin{align*} (e^x+e^{-x})^m&=\sum_{k=0}^m\binom{m}ke^{kx}e^{-(m-k)x}\\\\ &=\sum_{k=0}^m\binom{m}ke^{(2k-m)x}\\\\ &=\sum_{k=0}^m\binom{m}k\sum_{n\ge 0}\frac{(2k-m)^n}{n!}x^n\\\\ &=\sum_{n\ge 0}\frac1{n!}\left(\sum_{k=0}^m\binom{m}k(2k-m)^n\right)x^n\;, \end{align*}
so
$$a_n=\frac1{2^m}\sum_{k=0}^m\binom{m}k(2k-m)^n\;.\tag{1}$$
Note that $2(m-k)-m=m-2k=-(2k-m)$, so $(1)$ is indeed $0$ when $n$ is odd. When $n$ is even and positive we can rewrite $(1)$ as
$$a_n=\frac1{2^{m-1}}\sum_{k=0}^{\lfloor m/2\rfloor}\binom{m}k(m-2k)^n\;.$$
For instance, if $m=4$, then
\begin{align*} a_{2n}&=\frac18\left(\binom40(4-0)^{2n}+\binom41(4-2)^{2n}+\binom42(4-4)^{2n}\right)\\\\ &=\frac18\left(4^{2n}+4\cdot2^{2n}\right)\\\\ &=\frac18\left(4^{2n}+4^{n+1}\right)\\\\ &=2\cdot4^{n-1}\left(4^{n-1}+1\right)\;. \end{align*} | {
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• May I ask you to explain how you get from the line headed by 'is' to the line headed by 'so'. Do you have a pointer to a book or literature where I could look this up? – Harald Feb 13 '15 at 21:10
• @Harald: This PDF is a concise introduction to exponential generating functions. This PDF has a more general introduction to generating functions. Miklós Bóna, Introduction to Enumerative Combinatorics, is very readable and has everything that you need on the subject. I suspect that the same goes for his A Walk Through Combinatorics, but I’ve not actually seen it. – Brian M. Scott Feb 13 '15 at 21:26 | {
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# Determinant of a 5 × 5 matrix
I have a little problem with a determinant.
Let $A = (a_{ij}) \in \mathbb{R}^{(n, n)}, n \ge 4$ with
$$a_{ij} = \begin{cases} x \quad \mbox{for } \,i = 2, \,\, j \ge 4,\\ d \quad \mbox{for } \,i \ge j, \\ 0 \quad \mbox{else.} \end{cases}$$
So for example, if we choose $n = 5$, the matrix would look like this: $$A = \begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}$$
How can I find the determinant of this matrix?
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A = T \cdot R$. Then I wanted to use $$\det(A) = \det(T \cdot R) = \det(T) \cdot \det(R).$$
to figure out the determinant. This would be something like
$$d^n \cdot \det(R)$$
But is this approach even possible (I don't think so)? Is there any intelligent way of solving this? Thanks in advance.
• What is the determinant for $n=4,5$? Maybe you could conjecture the result and then use induction. Apr 21 '15 at 14:26
• BTW, what is a "rest matrix"? Apr 21 '15 at 14:30
Adding a multiple of one row to another preserves the determinant. Subtract $x/d$ of the last row from the second to get $$\begin{pmatrix} d &0 &0 &0 &0 \\ d-x &d-x &-x &0 &0 \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}$$ and then add $x/d$ of the third row to the second row to get $$\begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &0 &0 \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}.$$ This is lower triangular, so its determinant is the product of its diagonal, which is $d^5$. This all works for the $n$ by $n$ case, so the answer in general is $d^n$. | {
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• "Row operations preserve determinant" - except for multiplying a row by a constant, which multiplies the determinant by the same constant. Apr 21 '15 at 18:13
• @user2357112 Or switching two rows, which multiplies the determinant by $(-1)$. Apr 21 '15 at 18:42
• Two! Two operations. Apr 21 '15 at 18:44
• @user2357112 You're right, I should be more clear. Apr 22 '15 at 7:13
Develop your matrix wrt the first row and get
$$|A|=d\begin{vmatrix}d&0&x&x\\d&d&0&0\\d&d&d&0\\d&d&d&d\end{vmatrix}$$
Develop again wrt the first row but observe that when your pivot points are the $\;x$'s you get determinant zero as there are two identical rows in each case, so we get
$$d^2\begin{vmatrix}d&0&0\\d&d&0\\d&d&d\end{vmatrix}=d^5$$
Try now some inductive argument based on this.
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A=T⋅R$.
That's very much a way to do it. The technique is called LU Decomposition. It produces a lower and upper triangular matrix, allowing trivial determinate calculations. For this reason, you actually only need to find the diagonal elements to get your determinant.
In this case,
$$A = \begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix} = \begin{pmatrix} 1 &0 &0 &0 &0 \\ 1 &1 &0 &0 &0 \\ 1 &1 &1 &0 &0 \\ 1 &1 &1 &1 &0 \\ 1 &1 &1 &1 &1 \\ \end{pmatrix} \cdot \begin{pmatrix} d &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ 0 &0 &d &-x &-x \\ 0 &0 &0 &d &0 \\ 0 &0 &0 &0 &d \\ \end{pmatrix}$$
So
$$\det A = \det L \cdot \det U = 1^5 \cdot d^5$$ | {
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So
$$\det A = \det L \cdot \det U = 1^5 \cdot d^5$$
The first few steps of the method used here (takes longer to texify than to do) are: $$\begin{eqnarray} \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{smallmatrix}\right] &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ 0 &d &d &0 &0 \\ 0 &d &d &d &0 \\ 0 &d &d &d &d \\ \end{smallmatrix}\right] \\ &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &d &0 &x &x \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &0 &0 &0 &0 \\ 0 &0 &0 &0 &0 \\ 0 &0 &d &-x &-x \\ 0 &0 &d &d-x &-x \\ 0 &0 &d &d-x &d-x \\ \end{smallmatrix}\right] \\ &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 & 0\\ 1 & 1\\ 1 & 1\\ 1 & 1\\ 1 & 1\\ \end{smallmatrix}\right] + \cdots \end{eqnarray}$$ | {
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# Project Euler #6 in Java
Project Euler #6:
The sum of the squares of the first ten natural numbers is,
$1^2+2^2+ ... + 10^2 = 385$
The square of the sum of the first ten natural numbers is,
$(1+2+ ... + 10)^2 = 55^2 = 3025$
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 − 385 = 2640$.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Here is my solution:
public class DifferenceFinder {
private static final int MAX = 100;
public static void main(String[] args) {
long time = System.nanoTime();
int result = MAX * (MAX + 1) / 2;
result *= result;
for(int i = 1; i <= MAX; i++) {
result -= i * i;
}
time = System.nanoTime() - time;
System.out.println("Result: " + result
+ "\nTime used to calculate in nanoseconds: " + time);
}
}
It simply does:
$$(1+2+ ... + 100)^2-1^2-2^2- ... -100^2$$
Output:
Result: 25164150
Time used to calculate in nanoseconds: 2231
Questions:
1. Is the simplest solution the most efficient one?
2. Does it smell?
## Method extraction.
Even for simple programs, having multiple responsibilities in a method is poor practice.
Consider simple extractions, like:
private static int squareOfSum(int limit) {
int sum = (limit * (limit + 1)) / 2;
return sum * sum;
}
public static int sumOfSquares(int limit) {
int sum = 0;
for (int i = 1; i <= limit; i++) {
sum += i * i;
}
return sum;
}
int difference = squareOfSum(100) - sumOfSquares(100);
System.out.printf("Difference is %d\n", difference);
By extracting functions, the logic becomes reusable, and discrete. Much better.
Additionally, it allows you to easily change the logic inside the methods to suite your algorithms, and use better algorithms like vnp suggests (+1 to that answer too).
## Timing | {
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## Timing
Your use of the nanos to time your code is misleading. The performance of the code is heavily related to how often the code runs, and, for this example, any performance measurement is unreliable....
• Is the timing really misleading? When something takes ~2 microseconds, is it fair to say that startup costs don't count? – 200_success Jan 22 '15 at 1:24
• @200_success No, what's misleading is suggesting the code takes 2 microseconds when in another context it may take 0.2 microseconds. In addition, there are other things that are not being timed.... but my main concern is using a measure for the timing that is heavily context-dependent – rolfl Jan 22 '15 at 1:26
A sum of first $n$ numbers is $\frac{n(n+1)}{2}$. A sum of squares of first $n$ numbers is $\frac{n(n+1)(2n+1)}{6}$. A difference in question is $\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6} = \frac{(n-2)(n-1)n(n+1)}{12}$.
This solution is likely the most efficient.
• This doesn't seem to work. It gives the result 8165850. – TheCoffeeCup Jan 22 '15 at 3:10
• @MannyMeng verified using the above Summation formulas and it work with much better timings. Just don't compute the difference with command denominator fraction, use the method suggested in the answer above – yadav_vi Jan 22 '15 at 11:38
You can linearize this problem, without losing readability. For example my solution from years back:
int size = 100;
long qos = (long) Math.pow(size * (size + 1) / 2, 2);
long soq = size * (size + 1) * (2 * size + 1) / 6;
System.out.println(qos - soq);
Where qos and soq are known acronyms for me denoting square of sum and sum of squeares.
Note however that my code actually had a potential bug called "Possible Loss of Fraction"
size * (size + 1) / 2
should be
size * (size + 1.0) / 2
In this example
• size * (size + 1) would be of type int
• size * (size + 1.0) would be of type double
In java
• int / int you will get int
• double / int you will get double | {
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In java
• int / int you will get int
• double / int you will get double
In Java you mostly deal with discrete mathematics while equations do not, note this fact as you start solving Euler problems. In addition remember to note that type double is an approximation.
It is also good to use best tool for the task, for example if you know the result will be
You can compute it with wolfram alpha (mathematica/wolfram language) using:
Limit[(n (n - 2) (n - 1) (n + 1))/12, n -> 100]
• In java 1.0 is same as 1d. Where 1f would result in 1.0, but be of type float (equal value, but not same type). – Margus Jan 23 '15 at 10:41 | {
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# Class Number of $\mathbb{Q}(\sqrt[3]{19})$ and Hilbert class field
Finding the class number of $\mathbb{Q}(\sqrt[3]{19})$ is an exercise from Marcus 'Number Field'. This question was uploaded by some other user, but it was removed by now. I have worked on details and have some more questions.
Following are the steps in the exercise:
Let $K=\mathbb{Q}(\alpha)$, $R=\mathcal{O}_K$, and $3R=P^2Q$, where $\alpha=\sqrt[3]{19}$, and $P$, $Q$ are prime ideals in $R$.
(a) Prove that the ideal class group is cyclic, generated by the class containing $P$.
(b) Prove that the number of ideal classes is a multiple of $3$.
(c) Prove that there are either three or six ideal classes.
(d) Prove that there are three ideal classes. (Suggestion: Suppose there were six. Show that none of the ideals $J$ with $||J||\leq 9$ are in the same class with $P^3$. )
After solving this problem, it seems that the calculations from my solution to (a) are enough to conclude that the class number is $3$.
To do (a), we find the Minkowski's bound: $$\frac{3!}{3^3}\frac 4{\pi} \sqrt{1083} \approx 9.3$$ This shows that each ideal class in the ideal class group contains an ideal $J$ with $||J||\leq 9$. Also, each ideal class generating the class group contains a prime ideal lying above $2$, $3$, $5$, $7$.
So, we find prime factorizations of $2R$, $3R$, $5R$, $7R$: $$2R=P_1P_2,\ \ \ 3R=P^2 Q,$$ $$5R= P_3P_4, \ \ \ 7R = P_5,$$ with $P_1=(2, \alpha-1)$, $P=(3, \frac{\alpha^2+\alpha+1}3)$, $P_3=(5,\alpha+1)$.
With a help of SAGE, I found that $$P_1= \frac{\alpha-1}3 P, \ \ \ P_3= \frac{-\alpha+4}3 P, \ \ \ P_5=\frac{7\alpha+14}9 P^3.$$
Then the prime ideals appear above (as factorizations of $2$, $3$, $5$, $7$) belong to one of the three ideal classes of $P$, $P^2$, $I$ (principal).
My questions are
1. Are these calculations enough to conclude that the ideal class group has order $3$?
2. If the answer to 1 is no, then how are the rest of parts solved? | {
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2. If the answer to 1 is no, then how are the rest of parts solved?
3. How do we determine the Hilbert class field of $\mathbb{Q}(\sqrt[3]{19})$?
• And how are you concluding that all the ideal class aren't in fact the principal ideal class? – John M Jul 29 '16 at 2:36
• I have to prove that the ideal class of $P$ is not principal, so the class of $P$ has order $3$. Okay, so part (b) is really necessary to do that. I see. – Sungjin Kim Jul 29 '16 at 2:42
• Except for showing that at least one ideal is nonprincipal, I think you’re OK. Since $P_5=(7)$, principal, and also equals $(z)P^3$ for a number $z$, $P^3$ is principal. $P_2$ has norm $4$, but since $P_1P_2\sim1$, you have $P_2\sim P^2$. $Q$ also hs norm $3$, but it’s equiv to $P$. $P_3$ has norm $5$, so $P_4$ has norm $25$, so it’s not a problem. Seems to me that you’ve covered all bases for the ideal class group to be of order $3$. – Lubin Jul 29 '16 at 2:46
• My comment above may be worthless, but for the Class Field, I’d first look at the cubic subfield of $\Bbb Q(\zeta_{19})$. It’s unramified outside $19$, and you’d have a good chance of showing that its compositum with $K$ is unramified above $19$, over $K$. – Lubin Jul 29 '16 at 3:21
It turns out that the answer to 1 is no as commented by @JohnM. We need to prove that $P$ is not principal. To do this, assume that $P=\beta R$. Then $||P||=||\beta||=3$. So, now the question becomes whether there exists an element $\beta\in R$ with $||\beta||=3$. | {
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Since any element in R can be expressed as $\beta=a+b\alpha+c\alpha^2$ with $a, b, c\in \mathbb{Q}$ and $3a, 3b, 3c \in \mathbb{Z}$, $3a\equiv 3b\equiv 3c \ \mathrm{mod} \ 3$, we see that $$||\beta||=3=a^3+19b^3+19^2c^3-3\cdot 19 abc$$ and $$27\cdot 3 =81= (3a)^3 + 19(3b)^3 + 19^2 (3c)^3 - 3 \cdot 19 (3a)(3b)(3c)\equiv (3a)^3 \ \mathrm{mod} \ 19$$ However, $81 \equiv 5 \ \mathrm{mod} \ 19$ is not a cubic residue modulo $19$ (list of cubic residues mod $19$ is $0, 1, 7, 8, 11, 12, 18$). Therefore, there is no element $\beta\in R$ of norm $3$. This proves that $P$ is not principal.
Then the rest of argument will follow as @Lubin commented. Thus, we needed to solve (b) to conclude, but (c) and (d) are not necessary. This answers my question 2.
To find the Hilbert class field, we use a cubic subfield $K'$ of the cyclotomic field $\mathbb{Q}(\zeta_{19})$ as commented by @Lubin. This method was already discussed by @franzlemmermeyer in the answer to this question.
Let $L$ be composite field of $K'$ and $K=\mathbb{Q}(\alpha)$. Let $\mathcal{P}$ be a prime in $L$. If $\mathcal{P}$ lies over rational prime $p\neq 3, 19$, then $p$ is unramified in both extensions $K$ and $K'$. Thus, the prime $\mathcal{P}\cap K$ lying under $\mathcal{P}$ is unramified in $L$.
Suppose that $\mathcal{P}$ lies over rational prime $3$. Since the rational prime $3$ is inert in $K'$, the residue field degree must be $3$. By the decomposition $3R=P^2Q$, the residue field degree for $P$ and $Q$ over rational prime $3$ must be both $1$. This gives the residue field degree for $\mathcal{P}$ over both $P$ and $Q$ must be $3$, yielding that the ramification indices for $\mathcal{P}$ over both $P$ and $Q$ are $1$. | {
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Now assume that $\mathcal{P}$ is over rational prime $19$. As @franzlemmermeyer answered in the above link, we use Abhyankar's lemma. Since $19$ is totally ramified in both $K$ and $K'$, we have by Abhyankar's lemma that the ramification index for $\mathcal{P}$ over $\mathcal{P}\cap K$ is $1$.
Hence, the extension $L$ over $K$ is unramified over $K$, yielding that $L$ is the Hilbert class field of $K$. This answers my question 3.
Acknowledgement @JohnM, @Lubin, Thank you very much for the helpful comments with great insights. | {
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# ListContourPlot is blank
This is probably very simple. I want to do a simple ListContourPlot.
contourData = Table[ myFunc[x,y], {x,0.1, 2.9}, {y,0.1, 2.9}];
ListContourPlot[contourData]
This works without a problem.
Now, I want to include x and y values so that I can show the values corresponding to specific values of x and y. I modify the Table statement to include x and y values.
contourData = Table[ {x,y, myFunc[x,y]}, {x,0.1, 2.9}, {y,0.1, 2.9}];
ListContourPlot[contourData]
My contour plot is now blank. I'm guessing it is a problem with the Table statement. The maximum minus minimum over this range is about 220. As this is a function of Pythagorean distance, the values vary (so I don't think it is a case of essentially equal values over the ranges of x and y.
What is the correct way to generate data for a ListContourPlot?
• Try ListContourPlot[Flatten[contourData, 1]]. Sep 14 '14 at 19:04
• That was it exactly! Thank you most kindly! Sep 14 '14 at 19:10
As an alternative to @ b.gatessucks excellent comment you might consider DataRange:
ListContourPlot[Table[Sin[x + y], {x, 0, 3, 0.1}, {y, 0, 3, 0.1}], | {
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# calculus
Find the extreme values of the function for the given intervals.
2x^3 – 21x^2 + 72x
a) [0,5]
b) [0,4]
I have both the minimum values for both intervals which is 0. But I am unable to find the maximum. please help!!!
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1. You have probably found two extrema where f'(x)=0.
To identify whether the extremum is a maximum or minimum, use the second derivative test.
If the second derivative is positive, the extremum is a minimum. If the second derivative is negative, the extremum is a maximum.
Here the extrema are at x=3 and x=4.
The second derivative, f"(x)=12x-42,
so for example, f2(3)=-6, so x=3 is a maximum.
Use your calculator to plot the graph of the function to help you solve these problems. After some time, you will be able to visualize the graph from the equation, even without the calculator.
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2. I tried 3 for both (a and b) as a maximum and didn't work....
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3. It is important to know that extreme values include local maxima, local minima and end-points of the given interval.
The question requires the extreme values of the function, so it is the value of the function that counts.
To find extreme values, we calculate the critical points (where f'(0)=0 and where f'(0) does not exist) as well as the value of the function at end-points.
Given
f(x)=2x^3 – 21x^2 + 72x
1. We calculate the points where f'(x)=0, and we have determined that f(3)=81 is a local maximum and f(4)=80 is a local minimum.
2. For [0,4], the values of the function at end-points are:
f(0)=zero, and f(4)=80.
So the list of values to consider are:
(0,zero)
(3,81)
(4,80)
We conclude that the extreme minimum is zero, and the extreme maximum is 81.
3. For [0,5], the values of the function at end-points are f(0)=zero, and f(5)=85.
So the list of values to consider for extreme values are:
(0,zero)
(3,81)
(4,80)
(5,85)
We choose zero as the extreme minimum, and 85 as the extreme maximum.
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# How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$
$$\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$$
WolframAlpha gives a numerical answer of $43.8122$, which appears to be $\sqrt{611\pi}$. And playing with that, it seems that replacing $611$ with $a$ just gives $\sqrt{a\pi}$. My trouble is that the stuff in the exponential always seems to be just a big mess, and I haven't been able to get it into a form I can understand or deal with.
I would greatly appreciate seeing a method for solving this integral.
• I have a strong feeling that you can use Feynman's trick on this, but I am still trying to find the correct one. – UserX Sep 29 '14 at 7:20
• How can you be sure that $\sqrt{a\pi}$ is not just a good approximation ? Could you produce some values with more digits ? – Claude Leibovici Sep 29 '14 at 7:34
• @ClaudeLeibovici I'm not entirely sure if it's exactly $\sqrt{a\pi}$, but for every value of $a$ I've tried, WolframAlpha always gives a number very close to $\sqrt{a\pi}$. I was hoping this fact might help with solving the integral. – user137794 Sep 29 '14 at 7:40
• Working with very high precision, I can confirm that the result is exactly $\sqrt{a\pi}$ what you would get with $$\int_{-\infty}^\infty\exp\left(-\frac{x^2}{a}\right)\ dx=\sqrt{a\pi}$$ – Claude Leibovici Sep 29 '14 at 8:12
Let $\displaystyle\;u(x) = \frac{x^2-13x-1}{x}\;$. As $x$ varies over $\mathbb{R}$, we have
• u(x) increases monotonically from $-\infty$ at $-\infty$ to $+\infty$ at $0^{-}$.
• u(x) increases monotonically from $-\infty$ at $0^{+}$ to $+\infty$ at $+\infty$.
This means as $x$ varies, $u(x)$ covered $(-\infty,\infty)$ twice.
Let $x_1(u) < 0$ and $x_2(u) > 0$ be the two roots of the equation for a given $u$:
$$u = u(x) = \frac{x^2-13x-1}{x} \quad\iff\quad x^2 - (13+u)x - 1 = 0$$ we have $$x_1(u) + x_2(u) = 13 + u \quad\implies\quad \frac{dx_1}{du} + \frac{dx_2}{du} = 1.$$ From this, we find | {
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\begin{align} \int_{-\infty}^\infty e^{-u(x)^2/611} dx &= \left( \int_{-\infty}^{0^{-}} + \int_{0^{+}}^{+\infty}\right) e^{-u(x)^2/611} dx\\ &= \int_{-\infty}^{\infty} e^{-u^2/611}\left(\frac{dx_1}{du} + \frac{dx_2}{du}\right) du\\ &= \int_{-\infty}^{\infty} e^{-u^2/611} du\\ &= \sqrt{611\pi} \end{align}
• This is a new technique for me (+1), I think it can be generalised. If so, what is the condition? Say $u(x)=\dfrac{px^2+qx+r}{sx}$ – Anastasiya-Romanova 秀 Sep 30 '14 at 10:44
• @Anastasiya-Romanova algebraically, $pr < 0$. Geometrically, this ensure the monotonicity of the $u(x)$ and $u$ cover $(-\infty,\infty)$ twice. – achille hui Sep 30 '14 at 12:47
• +1 for its potent to be generalized (to a higher degree, maybe?). – Sangchul Lee Sep 30 '14 at 19:43
• @sos440 the generalization is essentially the "Added" part of orangeskid answer. – achille hui Sep 30 '14 at 19:59
HINT:
For $a$ fixed
$\int_{-\infty}^\infty\exp\left(-\frac{(x^2+sx-b)^2}{a x^2}\right)\ dx$
is constant in $s$ and $b\ge 0$.
$\bf{Added:}$
The function $\frac{x^2 + s x - b}{x} = x - \frac{b}{x} + s$ invariates the Lebesgue measure as @achille hui: showed in his answer.
Let $n \in \mathbb{N}$ $\alpha_1$, $\ldots$, $\alpha_n$ distinct real numbers, $\rho_1$, $\ldots$, $\rho_n$ $>0$ and $\beta \in \mathbb{R}$. The function
$$\phi(x) = x - \sum_{i=1}^n \frac{\rho_i}{x- \alpha_i} -\beta$$
invariates the Lebesgue measure on $\mathbb{R}$.
Lemma: For any $a \in \mathbb{R}$ the equation
\begin{eqnarray*} x- \sum_{i=1}^n \frac{\rho_i}{x- \alpha_i} -\beta = u \end{eqnarray*} has $n+1$ distinct real root with sum $u + \sum_i \alpha_i + \beta$. Use Viete.
Lemma: Let $I$ an interval in $\mathbb{R}$ of length $l$. Then the preimage $\phi^{-1} (I)$ is a union of $n+1$ disjoint intervals of total length $l$.
Consequence: $$\int_{\mathbb{R}} (f\circ \phi)\, d\,\mu = \int_{\mathbb{R}} f\ d\mu$$ | {
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Consequence: $$\int_{\mathbb{R}} (f\circ \phi)\, d\,\mu = \int_{\mathbb{R}} f\ d\mu$$
Composing two rational maps that invariate the measure gets a third one. They will have singularities in general.
For $f(x) = e^{-\frac{x^2}{a}}$ the composition $f (\phi(x))$ is still smooth due to the rapid decay at $\infty$ of $e^{-\frac{x^2}{a}}$.
• That is remarkably simple and straightforward. +1 – Tom-Tom Sep 29 '14 at 9:56
• Would you be able to elaborate on how you concluded this? – user137794 Sep 29 '14 at 11:45
• @user137794: Added more detail. – orangeskid Sep 30 '14 at 6:43
• @Tom-Tom: Thanks! Added more detail. – orangeskid Sep 30 '14 at 16:36
• For $f(x) = e^{-\frac{x^2}{a}}$ the function $f(\phi(x))$ is still smooth. – orangeskid Sep 30 '14 at 20:19
Adding another solution owing to a friend of mine.
Through some algebra, the integral is equivalent to
$$\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx$$
Then using the following identity
$$\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx$$
We have
\begin{align} &\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx\\ =&\int_{-\infty}^\infty \exp\left(-\frac1{611}(x-13)^2\right)\ dx\\ =&\int_{-\infty}^\infty \exp\left(-\frac1{611}x^2\right)\ dx\\ =&\sqrt{611\pi} \end{align}
Using identity in @user137794's answer: $$\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx$$ where the complete proof can be seen here. The problem can be generalised to evaluate
$$\int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx = \sqrt{a\pi}$$
Proof: | {
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$$\int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx = \sqrt{a\pi}$$
Proof:
It's easy to see that $\dfrac{(x^2-bx-1)^2}{ax^2}=\dfrac{1}{a}\left(x-x^{-1}-b\right)^2$, then \begin{align} \int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx &=\int_{-\infty}^\infty \exp\left(-\frac{(x-x^{-1}-b)^2}{a}\right)\ dx\\ &=\int_{-\infty}^\infty \exp\left(-\frac{(x-b)^2}{a}\right)\ dx\\ &=\int_{-\infty}^\infty \exp\left(-\frac{y^2}{a}\right)\ dy\\ &=\sqrt{a}\int_{-\infty}^\infty \exp\left(-z^2\right)\ dz\\ &=\sqrt{a\pi} \end{align} Therefore $$\int_{-\infty}^\infty \exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx = \sqrt{611\pi}$$
• Nice generalization! It's interesting, that $b$ does not matter, on the other hand $-1$ is how important! – user153012 Oct 1 '14 at 12:07 | {
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# Combinatorial proof of a Stirling number identity.
Consider the identity $$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$ where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity $$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$ which essentially states that the number of even subsets of a set is equal to the number of odd subsets.
Now there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?
As an added note, there is a similar identity $$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$ A combinatorial interpretation of this one would also be appreciated.
• May inclusion exclusionn principle will help? – Norbert May 18 '13 at 4:21
Perhaps there is something relating odd ordered partitions with even ordered partitions?
There is indeed. Let's try to construct an involution $T_n$, mapping odd ordered partitions of $n$-element set to even and vice versa: if partition has part $\{n\}$, move $n$ into previous part; otherwise move $n$ into new separate part.
Example: $(\{1,2\},\{\mathbf{5}\},\{3,4\})\leftrightarrow(\{1,2,\mathbf{5}\},\{3,4\})$.
This involution is not defined on partitions of the form $(\{n\},\ldots)$, but for these partitions one can use previous involution $T_{n-1}$ and so on.
Example: $(\{5\},\{4\},\{1,2\},\{\mathbf{3}\})\leftrightarrow(\{5\},\{4\},\{1,2,\mathbf{3}\})$.
In the end only partition without pair will be $(\{n\},\{n-1\},\ldots,\{1\})$. So our (recursively defined) involution gives a bijective proof of $\sum_{\text{k is even}}k!{n \brace k}=\sum_{\text{k is odd}}k!{n \brace k}\pm1$ (cf. 1, 2). | {
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Upd. As for the second identity, the involution $T_n$ is already defined on all cyclically ordered partitions, so $\sum_{\text{k is even}}(k-1)!{n \brace k}=\sum_{\text{k is odd}}(k-1)!{n \brace k}$.
P.S. I can't resist adding that $k!{n \brace k}$ is the number of $(n-k)$-dimensional faces of an $n$-dimensional convex polytope, permutohedron (the convex hull of all vectors formed by permuting the coordinates of the vector $(0,1,2,\ldots,n)$). So $\sum(-1)^{n-k}k!{n \brace k}=1$ since it's the Euler characteristic of a convex polytope.
• much more combinatorial (+1) – robjohn May 23 '13 at 13:30
• Beautiful, thank you. – EuYu May 23 '13 at 18:12
These are not combinatorial interpretations, but they are simple.
The defining equation for Stirling numbers of the second kind is $$\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{x}{k}k!=x^n\tag{1}$$ That is, Stirling numbers of the second kind tell how to write monomials as a combination of falling factorials (or combinatorial polynomials).
Noting that $\displaystyle\binom{-1}{k}=(-1)^k$ and setting $x=-1$ yields \begin{align} \sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}(-1)^kk!= \sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{-1}{k}k!=(-1)^n \end{align}
Since $\displaystyle\binom{x}{k}=\binom{x-1}{k-1}\frac{x}{k}$ and $\begin{Bmatrix}n\\0\end{Bmatrix}=0$ for $n\ge1$, we can rewrite $(1)$ as $$\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{x-1}{k-1}(k-1)!=x^{n-1}\tag{2}$$ Setting $x=0$ yields $$\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}(-1)^{k-1}(k-1)!=0^{n-1}$$ where $0^0=1$ for the case $n=1$.
• Certainly the second part is supposing $n>0$, so that the term for $k=0$ can be dropped. However it fails to accommodate to the fact (even though some wish to deny it) that $0^0=1$. Stated more directly, it fails for $n=1$. – Marc van Leeuwen May 23 '13 at 14:56
• @MarcvanLeeuwen: Good point. I got rid of the exponent when $n=1$ when I shouldn't have. – robjohn May 23 '13 at 15:54 | {
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For the sake of completeness I include a treatment using generating functions. The exponential generating function of the Stirling numbers of the second kind is $$G(z, u) = \exp(u(\exp(z)-1))$$ so that $${n \brace k} = n! [z^n] \frac{(\exp(z) - 1)^k}{k!}.$$ It follows that $$\sum_{k=0}^n (-1)^k k! {n \brace k} = n! [z^n] \sum_{k=0}^n (1-\exp(z))^k = n! [z^n] \sum_{k=0}^\infty (1-\exp(z))^k,$$ where the last equality occurs because the series for $(1-\exp(z))^k$ starts at degree $k.$ But this is just $$n! [z^n] \frac{1}{1-(1-\exp(z))} = n! [z^n] \exp(-z) = (-1)^n,$$ showing the result. | {
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the residuals, that is response minus fitted values if "g" is a ⦠weights. For practical purposes it might be preferable to use a nonlinear least squares approach (e.g., the nls function). For example, if a student had spent 20 hours on an essay, their predicted score would be 160, which doesnât really make sense on a typical 0-100 scale. When present, the objective function is weighted least squares. In literal manner, least square method of regression minimizes the sum of squares of errors that could be made based upon the relevant equation. When the "port" algorithm is used the objective function value printed is half the residual (weighted) sum-of-squares. R-bloggers ... BVLS is implemented in the bvls() function ⦠Each classroom has a least squared mean of 153.5 cm, indicating the mean of classroom B was inflated due to the higher proportion of girls. We want to build a model for using the feature. The object of class "gmm" is a list containing at least: coefficients $$k\times 1$$ vector of coefficients. residuals. object: an object inheriting from class "gls", representing a generalized least squares fitted linear model.. model: a two-sided linear formula object describing the model, with the response on the left of a ~ operator and the terms, separated by + operators, on the right.. model. Moreover, we have studied diagnostic in R which helps in showing graph. If you have any suggestion or feedback, please comment below. Also, we have learned its usage as well as its command. Linear model Background. Now, you are an expert in OLS regression in R with knowledge of every command. And if the data-simulating function does not have the correct form (for example, if the zeroth order term in the denominator is not 1), the fitted curves can be completely wrong. Specifically, I am looking for something that computes intercept and slope. The least squares regression method follows the same cost function as the other methods used to segregate a mixed ⦠Least | {
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method follows the same cost function as the other methods used to segregate a mixed ⦠Least squares method, also called least squares approximation, in statistics, a method for estimating the true value of some quantity based on a consideration of errors in observations or measurements. an optional numeric vector of (fixed) weights. Disadvantages of least-squares regression *As some of you will have noticed, a model such as this has its limitations. The functions 'summary' is used to obtain and print a summary of the results. The most basic way to estimate such parameters is to use a non-linear least squares approach (function nls in R) which basically approximate the non-linear function using a linear one and iteratively try to find the best parameter values . subset. In linear least squares the model contains equations which are linear in the parameters appearing in the parameter vector , so the residuals are given by = â. Note that the following example uses a linear model with the lm function. In non-linear regression the analyst specify a function with a set of parameters to fit to the data. In ordinary least squares (OLS), one seeks ⦠Continue reading â Imagine that one has a data matrix consisting of observations, each with features, as well as a response vector . an optional vector specifying a subset of observations to be used in the fitting process. In the least squares method of data modeling, the objective function, S, =, is minimized, where r is the vector of residuals and W is a weighting matrix. Does R have a function for weighted least squares? We have seen how OLS regression in R using ordinary least squares exist. Changes to the model â see update.formula for details.. data It also compute the J-test of overidentying restriction. List containing at least: coefficients \ ( k\times 1\ ) vector of.... For details.. data we have learned its usage as well as its.... Its limitations for practical purposes it might be preferable to use a | {
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On a recent MathsJam Shout, an Old Chestnut appeared (in this form, due to @jamestanton):
If you’ve not seen it, stop reading here and have a play with it - it’s a classic puzzle for a reason.
Below the line are spoilers.
Counting is hard
The first thing you’d probably try is to draw out the lines and count up the regions.
The first circle has 1 region.
The second has 2 regions.
The third has 4. (I’m spreading these out to try to avoid accidental spoilers, by the way.)
The fourth has 8. A pattern seems to be forming.
The fifth has 16. Aha! It must be…
The sixth has 31. Oh. Let’s count them again: nope, definitely 31 rather than 32. What’s going on?
What is going on?
As you draw more and more points, it gets harder to a: make sure you’re not making your lines coincide and b: count the regions accurately. At MathsJam, we made it to the seventh circle (57 regions) before realising that was about as far as we could go.
So we needed a more methodical approach. The nugget of the solution - for me, at least - was to think about how each line you add to a diagram affects the number of regions - and that’s related to the number of lines it crosses. In fact, adding a line that crosses $n$ other lines adds $n+1$ regions to the circle - which you might like to prove.
So, in adding a fifth point to the fourth circle, we’ll need to add four new lines.
The two lines to the adjacent points each add a single region - they don’t cross any other lines. The lines to the two opposite points each cross two lines, creating three new regions. Overall, that’s 1 + 3 + 3 + 1 new regions, taking the eight regions to 16.
Let’s go a bit more methodically this time
Going from five to six is a bit trickier ((because counting is hard)).
Let’s consider the five existing points, going anticlockwise from our new point (which I’ll imagine is at the bottom of the circle). Each new line will split the polygon into two parts, which I’ll call ‘left’ and ‘right’. | {
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The first line, to the neighbour, crosses no lines, and adds one region.
The second line, to point 2, splits the circle so there is one point to the left and three to the right, all of which are connected - so this line adds $1 \times 3 + 1$ regions.
The third line splits the circle two on each side, so it adds $2 \times 2 + 1$ regions.
The fourth line mirrors the second, and the fifth line mirrors the first, so I might be tempted to write the number of regions added as:
$(0\\times 4 + 1) + (1 \\times 3 + 1) + (2 \\times 2 +1) + (3 \\times 1 + 1) + (4 \\times 0 + 1)=15$
… or better, the new number of regions is $5 + \sum_{i=0}^{4} i(4-i)$
But we can generalise!
Whenever we add a point to a diagram, the process is going to be identical! Each line will divide the existing points into three sets: those to the left, the one being connected, and those to the right. Every point on the left connects to every point on the right exactly once, so the number of lines cut by the new line is always the number of left points multiplied by the number of right points.
So! If we’re moving from $N-1$ points to $N$, we should be adding:
$R\_n = (N-1) + \\sum\_{i=0}^{N-2} i( (N-2) - i)$
new regions.
And we can write that explicitly!
$R\_n = (N-1) + (N-2)\\sum\_{i=0}^{N-2} i - \\sum{i=0}^{N-2} i^2$ $= (N-1) + (N-2)\\frac{(N-2)(N-1)}{2} - \\frac{ (N-2)(N-1)(2N-3)}{6}$ $= \\frac{(N-1)}{6} \\left\[ 6 + 3(N-2)^2 - (N-2)(2N-3) \\right$\] $= \\frac{(N-1)}{6} \\left\[ N^2 - 5N + 12 \\right$\]
Let’s check that with $N=6$, which we’ve already worked out: $R_n = \frac{5}{6}\left[ 36 - 30 + 12 \right] = 15$ new regions.
How about $N=7$? We’re expecting 26: $R_n=\frac{6}{6}\left[ 49 - 35 + 12\right] = 26$. Hooray!
Hang on - that’s just the new regions!
So, the number of new regions each time is $\frac{N-1}{6} \left[ N^2 - 5N + 12 \right]$, and we know when there are no points, there is one region. | {
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So, the number of regions altogether is $R_\sum(M) = 1 + \frac{1}{6}\sum_{N=0}^{M} (N-1)(N^2 - 5N + 12)$.
Expanding that and distributing the sum:
$R_\sum(M) = 1 + \frac{1}{6} \sum_{N=0}^{M} N^3 - \sum_{N=0}^{M} N^2 + \frac{17}{6}\sum_{N=0}^{M} N - \sum_{N=0}^{M} 2$
There follows an awful lot of tedious algebra, which ends up giving $R_\sum(M) = \frac{1}{24}\left(M^4 - 6M^3 + 23M^2 - 18M + 24\right)$
Just to check, we know $R_\sum(6)$ is 31: is that $\frac{1}{24}\left( 6^4 - 6\times6^3 + 23\times 6^2 - 18\times 6 + 24 \right)$? The first two terms cancel, and the rest? It’s $\frac{6}{24} \left( 23 \times 6 - 18 + 4\right)$, or $\frac{1}{4} \times 124 = 31$, as required!
Curiously, it turns out that $R_\sum(10) = 256$, which is also a power of 2 (although half as big as it “ought” to be if it followed the doubling pattern.) | {
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# Combinatorics: how many unique polygons can be drawn using up to p points along a circle?
I am faced with the following problem:
Let p ≥ 3 be a prime number, and mark p points at equal distance on the circumference of a circle like in the pictures below (where p = 7). We can trace polygons by choosing a certain amount of those points as vertices, and connecting them with non-intersecting straight lines.
With three parts:
a) How many different polygons can we form if we consider that a different choice of vertices leads to a different polygon?
This is straightforward: you can have polygons with three vertices, 4 vertices […] up until p vertices, and the amount of polygons you can draw of a given number of vertices k is $$p \choose k$$. Hence the answer is $${p \choose 3} + {p \choose 4} + {...} + {p \choose p-1} + {p \choose p}$$
b) How many different polygons can we form if we count all different rotations of the same polygon as a single polygon?
Well, any given polygon can be rotated up to $$p-1$$ times. Hence each polygon counted from a) is actually being counted 7 times. We adjust for this by dividing through the answer in a) by p, and get the answer as $$\frac{p \choose 3}{p} + \frac{p \choose 4}{p} + {...} + \frac{p \choose p-1}{p} + \frac{p \choose p}{p} = \frac{p \choose 3}{p} + \frac{p \choose 4}{p} + {...} + \frac{p \choose p-2}{p} + 1 + 1$$.
I think all the above makes sense, but I am struggling with the third part:
c) How many different polygons can we form if we count all isometric polygons (all rotations and reflections of the same polygon) as the same ?
And I don't know how to proceed from here. I'm guessing there has to be some pattern to the number of reflections+rotations that any polygon of k vertices has along p points. However, I've tried playing around with small examples and I can't seem to find any generalizable pattern, and I've been stumped for longer than I'd like to admit. Any help would be greatly appreciated. | {
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• Reworded: How many necklaces and bracelets (necklaces for part b and bracelets for part c) of length $p$ with $p$ prime can be made with beads of two possible colors such that there are at least three beads of the first color used. – JMoravitz Feb 5 '20 at 19:47
• Almost exactly the same question was posted just a week ago but then deleted when I answered it. a) Does this happen to be from an ongoing contest? If so, are you aware of the contest problem policy? b) If it isn't, and I post an answer, do you promise not to delete the question? :-) – joriki Feb 5 '20 at 19:47
• @joriki it is not a contest problem. I haven't seen the other post / don't see why it was deleted. This is a problem from a university class :) – Student_514 Feb 5 '20 at 19:51
The polygons are in correspondence with the subsets of the vertices with at least $$3$$ elements. Thus, as JMoravitz noted in a comment, this can be viewed as counting binary necklaces and bracelets of length $$p$$ with at least three beads of the first colour, where the first colour signifies a vertex that’s included and the second colour signifies a vertex that isn’t included in the polygon.
For part a), there is no symmetry to take into account, so the result is just $$2^p-\binom p2-\binom p1-\binom p0$$. (Your result is correct, you just didn't use the fact that $$\sum_k\binom pk=2^k$$, the total number of subsets of the set of $$k$$ vertices.)
For part b), with rotational symmetry, you failed to properly account for the fact that the polygon that includes all vertices has only one rotational equivalent, not $$p$$. Your final result is correct, but only because you replaced $$\frac{\binom pp}p=\frac1p$$ by $$1$$. If you count correctly, you have $$2^p-\binom p2-\binom p1-\binom p0-\binom pp$$ polygons that form classes of $$p$$ rotational equivalents each, and $$\binom pp=1$$ polygon that’s in a class of its own, so the total count is | {
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$$\frac{2^p-\binom p2-\binom p1-\binom p0-\binom pp}p+1=\frac{2^p-2}p-\frac{p+1}2\;.$$
For part c), with rotational and reflectional symmetry, it becomes easier to perform the count using Burnside’s lemma. The general result is given in the Wikipedia article linked to above. In your case, since $$p$$ is an odd prime, there are
$$\begin{eqnarray} B_2(p) &=& \frac12N_2(p)+\frac12\cdot2^{\frac{p+1}2} \\ &=& \frac12\cdot\frac1p\left(1\cdot2^p+(p-1)\cdot2^1\right)+2^{\frac{p-1}2} \\ &=& \frac{2^{p-1}-1}p+1+2^{\frac{p-1}2} \end{eqnarray}$$
binary bracelets. We need to subtract the $$1$$ bracelet with $$0$$ elements of the second colour, the $$1$$ bracelet with $$1$$ element of the second colour and the $$\frac{p-1}2$$ bracelets with $$2$$ elements of the second colour, so the number of polygons up to rotations and reflections is
$$\frac{2^{p-1}-1}p+2^{\frac{p-1}2}-\frac{p+1}2\;.$$ | {
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## Section5.9Numerical Integration
###### Motivating Questions
• How do we accurately evaluate a definite integral such as $\int_0^1 e^{-x^2} \, dx$ when we cannot use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative? Are there ways to generate accurate estimates without using extremely large values of $n$ in Riemann sums?
• What is the Trapezoid Rule, and how is it related to left, right, and middle Riemann sums?
• How are the errors in the Trapezoid Rule and Midpoint Rule related, and how can they be used to develop an even more accurate rule?
When we first explored finding the net signed area bounded by a curve, we developed the concept of a Riemann sum4.2 as a helpful estimation tool and a key step in the definition of the definite integral. Recall that the left, right, and middle Riemann sums of a function $f$ on an interval $[a,b]$ are given by
\begin{align} L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x,\label{E-Left}\tag{5.15}\\ R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x,\label{E-Right}\tag{5.16}\\ M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x\text{,}\label{E-Mid}\tag{5.17} \end{align}
where $x_0 = a\text{,}$ $x_i = a + i\Delta x\text{,}$ $x_n = b\text{,}$ and $\Delta x = \frac{b-a}{n}\text{.}$ For the middle sum, we defined $\overline{x}_{i} = (x_{i-1} + x_i)/2\text{.}$ | {
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A Riemann sum is a sum of (possibly signed) areas of rectangles. The value of $n$ determines the number of rectangles, and our choice of left endpoints, right endpoints, or midpoints determines the heights of the rectangles. We can see the similarities and differences among these three options in Figure5.77, where we consider the function $f(x) = \frac{1}{20}(x-4)^3 + 7$ on the interval $[1,8]\text{,}$ and use 5 rectangles for each of the Riemann sums.
Part (d) of Example4.36 explored how to determine if $L_n$ and $R_n$ are underestimates or overestimates of $\int_a^b f(x) \, dx$ when $f(x)$ is increasing. Similar statements can be made when $f(x)$ is decreasing. These are summarized next.
###### Error types when approximating area using left and right Riemann sums
• If $f(x)$ is increasing on $[a,b] \text{,}$ then $L_n$ underestimates $\int_a^b f(x) \, dx$ and $R_n$ overestimates $\int_a^b f(x) \, dx\text{.}$
• If $f(x)$ is decreasing on $[a,b] \text{,}$ then $L_n$ overestimates $\int_a^b f(x) \, dx$ and $R_n$ underestimates $\int_a^b f(x) \, dx\text{.}$
While it is a good exercise to compute a few Riemann sums by hand, just to ensure that we understand how they work and how varying the function, the number of subintervals, and the choice of endpoints or midpoints affects the result, using computing technology is a quick way to determine $L_n\text{,}$ $R_n\text{,}$ and $M_n\text{.}$ Any computer algebra system will offer this capability.
In this section we explore several different alternatives for estimating definite integrals. Our main goal is to develop formulas to estimate definite integrals accurately without using a large numbers of rectangles.
###### Example5.78
As we begin to investigate ways to approximate definite integrals, it will be insightful to compare results to integrals whose exact values we know. To that end, the following sequence of questions centers on $\int_0^3 x^3 \, dx\text{.}$ | {
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1. Use the applet at http://gvsu.edu/s/a9 with the function $f(x) = x^3$ on the window of $x$ values from $0$ to $3$ to compute $L_3\text{,}$ the left Riemann sum with three subintervals.
$L_3=9$
Solution
In addition to changing $f(x)$ to $f(x)=x^3$ and the number of subintervals to $3 \text{,}$ make sure to change the Sample Point Placement so that "Relative" is checked and the slide bar is all the way to the left. You'll see that now it is sampling $f(x)$ values from the left side of each sub-interval.
\begin{equation*} L_3=9 \end{equation*}
2. Likewise, use the applet to compute $R_3$ and $M_3\text{,}$ the right and middle Riemann sums with three subintervals, respectively.
$M_3=19.125 \ , R_3=36$
Solution
First change $f(x)$ and the number of subintervals to $f(x)=x^3$ and $n=3\text{.}$ For $R_3\text{,}$ make sure to change the Sample Point Placement so that "Relative" is checked and the slide bar is all the way to the right. You'll see that now it is sampling $f(x)$ values from the right side of each subinterval. For $M_3\text{,}$ make sure to change the Sample Point Placement so that "Relative" is checked and the slide bar is in the middle and the "midpoint" label shows. You'll see that now it is sampling $f(x)$ values from the middle of each interval.
\begin{equation*} M_3=19.125 \ , R_3=36 \end{equation*}
3. Use the Fundamental Theorem of Calculus to compute the exact value of $I = \int_0^3 x^3 \, dx\text{.}$
Solution
\begin{align*} I&=\int_0^3 x^3 \, dx \\ &= \left.\frac{1}{4}x^4 \right|_{0}^{3} \\ &=\frac{3^4}{4}-\frac{0^4}{4}\\ &=20.25 \end{align*}
4. We define the error in an approximation of a definite integral to be the difference between the integral's exact value and the approximation's value. What is the error that results from using $L_3\text{?}$ From $R_3\text{?}$ From $M_3\text{?}$
\begin{equation*} E_{L,3}= 11.25 \end{equation*}
\begin{equation*} E_{R,3}=-15.75 \end{equation*}
\begin{equation*} E_{M,3}=1.125 \end{equation*}
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Denote the error from each approximation $E_{L,3}, E_{R,3}, \text{ and } E_{M,3}\text{.}$ Then
\begin{equation*} E_{L,3}=I-L_3=20.25-9 =11.25 \end{equation*}
\begin{equation*} E_{R,3}=I-R_3=20.25-36=-15.75 \end{equation*}
\begin{equation*} E_{M,3}=I-M_3=20.25-19.125=1.125 \end{equation*}
5. In what follows in this section, we will learn a new approach to estimating the value of a definite integral known as the Trapezoid Rule. The basic idea is to use trapezoids, rather than rectangles, to estimate the area under a curve. What is the formula for the area of a trapezoid with bases of length $b_1$ and $b_2$ and height $h\text{?}$
Solution
The area of a trapezoid with bases of length $b_1$ and $b_2$ and height $h$ is
\begin{equation*} A=\frac{h(b_1+b_2)}{2}\text{.} \end{equation*}
6. Working by hand, estimate the area under $f(x) = x^3$ on $[0,3]$ using three subintervals and three corresponding trapezoids. What is the error in this approximation? How does it compare to the errors you calculated in (d)?
\begin{equation*} T_3=\frac{45}{2}=22.5 \end{equation*}
\begin{equation*} E_{T,3}=-2.25 \end{equation*}
Solution
The left-most trapezoid has base lengths 0 and 1 and height 1, so the area of the first trapezoid is $\frac{1}{2} (0+1)\cdot 1=\frac{1}{2}\text{.}$ The middle trapezoid has base lengths 1 and 8 and height 1, so the area of the second trapezoid is $\frac{1}{2}(1+8) \cdot 1=\frac{9}{2} \text{.}$ The right-most trapezoid has base lengths 8 and 27 and height 1, so the area of the third trapezoid is $\frac{1}{2}(8+27)\cdot 1=\frac{35}{2}\text{.}$ Therefore, the approximate area under the graph of $f(x)=x^3$ from $0$ to $3$ using the trapezoid rule with 3 subintervals is $T_3= \frac{1}{2} + \frac{9}{2} + \frac{35}{2}=\frac{45}{2}=22.5\text{.}$ | {
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The error is $E_{T,3}=20.25-22.5=-2.25 \text{.}$ Using trapezoids creates a smaller error compared to $L_3$ and $R_3 \text{.}$ The magnitude of the error from approximating using $M_3$ is half the magnitude of the error from approximating using $T_3 \text{,}$ but they have opposite signs.
### SubsectionThe Trapezoid Rule
So far, we have used the simplest possible quadrilaterals (that is, rectangles) to estimate areas. It is natural, however, to wonder if other familiar shapes might serve us even better.
An alternative to $L_n\text{,}$ $R_n\text{,}$ and $M_n$ is called the Trapezoid Rule. Rather than using a rectangle to estimate the (signed) area bounded by $y = f(x)$ on a small interval, we use a trapezoid. For example, in Figure5.80, we estimate the area under the curve using three subintervals and the trapezoids that result from connecting the corresponding points on the curve with straight lines.
The biggest difference between the Trapezoid Rule and a Riemann sum is that on each subinterval, the Trapezoid Rule uses two function values, rather than one, to estimate the (signed) area bounded by the curve. For instance, to compute $D_1\text{,}$ the area of the trapezoid on $[x_0, x_1]\text{,}$ we observe that the left base has length $f(x_0)\text{,}$ while the right base has length $f(x_1)\text{.}$ The height of the trapezoid is $x_1 - x_0 = \Delta x = \frac{b-a}{3}\text{.}$ The area of a trapezoid is the average of the bases times the height, so we have
\begin{equation*} D_1 = \frac{1}{2}(f(x_0) + f(x_1)) \cdot \Delta x\text{.} \end{equation*}
Using similar computations for $D_2$ and $D_3\text{,}$ we find that $T_3\text{,}$ the trapezoidal approximation to $\int_a^b f(x) \, dx$ is given by
\begin{align*} T_3 &= D_1 + D_2 + D_3\\ &= \frac{1}{2}(f(x_0) + f(x_1)) \cdot \Delta x + \frac{1}{2}(f(x_1) + f(x_2)) \cdot \Delta x + \frac{1}{2}(f(x_2) + f(x_3)) \cdot \Delta x\text{.} \end{align*} | {
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Because both left and right endpoints are being used, we recognize within the trapezoidal approximation the use of both left and right Riemann sums. Rearranging the expression for $T_3$ by removing factors of $\frac{1}{2}$ and $\Delta x \text{,}$ grouping the left endpoint and right endpoint evaluations of $f\text{,}$ we see that
$$T_3 = \frac{1}{2} \left[ f(x_0) + f(x_1) + f(x_2) \right] \Delta x + \frac{1}{2} \left[ f(x_1) + f(x_2) + f(x_3) \right] \Delta x\text{.}\label{E-Trap3}\tag{5.18}$$
We now observe that two familiar sums have arisen. The left Riemann sum $L_3$ is $L_3 = f(x_0) \Delta x + f(x_1) \Delta x + f(x_2) \Delta x\text{,}$ and the right Riemann sum is $R_3 = f(x_1) \Delta x + f(x_2) \Delta x + f(x_3) \Delta x\text{.}$ Substituting $L_3$ and $R_3$ for the corresponding expressions in Equation(5.18), it follows that $T_3 = \frac{1}{2} \left[ L_3 + R_3 \right]\text{.}$ We have thus seen a very important result: using trapezoids to estimate the (signed) area bounded by a curve is the same as averaging the estimates generated by using left and right endpoints.
###### The Trapezoid Rule
The trapezoidal approximation, $T_n\text{,}$ of the definite integral $\int_a^b f(x) \, dx$ using $n$ subintervals is given by the rule
\begin{align*} T_n =\mathstrut \amp \left[\frac{1}{2}(f(x_0) + f(x_1)) + \frac{1}{2}(f(x_1) + f(x_2)) + \cdots + \frac{1}{2}(f(x_{n-1}) + f(x_n)) \right] \Delta x.\\ =\mathstrut \amp \sum_{i=0}^{n-1} \frac{1}{2}(f(x_i) + f(x_{i+1})) \Delta x\text{.} \end{align*}
Moreover, $T_n = \frac{1}{2} \left[ L_n + R_n \right]\text{.}$
###### Example5.81
In this example, we explore the relationships among the errors generated by left, right, midpoint, and trapezoid approximations to the definite integral $\int_1^2 \frac{1}{x^2} \, dx$
1. Use the First FTC to evaluate $\int_1^2 \frac{1}{x^2} \, dx$ exactly. | {
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1. Use the First FTC to evaluate $\int_1^2 \frac{1}{x^2} \, dx$ exactly.
2. Use appropriate computing technology to compute the following approximations for $\int_1^2 \frac{1}{x^2} \, dx\text{:}$ $T_4\text{,}$ $M_4\text{,}$ $T_8\text{,}$ and $M_8\text{.}$
3. Recall that the error of an approximation is the difference between the exact value of the definite integral and the resulting approximation. For instance, if we let $E_{T,4}$ represent the error that results from using the trapezoid rule with 4 subintervals to estimate the integral, we have
\begin{equation*} E_{T,4} = \int_1^2 \frac{1}{x^2} \, dx - T_4\text{.} \end{equation*}
Similarly, we compute the error of the midpoint rule approximation with 8 subintervals by the formula
\begin{equation*} E_{M,8} = \int_1^2 \frac{1}{x^2} \, dx - M_8\text{.} \end{equation*}
Based on your work in (a) and (b) above, compute $E_{T,4}\text{,}$ $E_{T,8}\text{,}$ $E_{M,4}\text{,}$ $E_{M,8}\text{.}$
4. Which rule consistently over-estimates the exact value of the definite integral? Which rule consistently under-estimates the definite integral?
5. What behavior(s) of the function $f(x) = \frac{1}{x^2}$ lead to your observations in (d)?
Hint
1. $\frac{1}{x^2} = x^{-2}\text{.}$
2. Use a computational device.
3. Use a computational device.
4. Which estimate is larger than the true value of the definite integral?
5. Note that how the curve bends makes a big difference in whether the trapezoid rule over- or under-estimates the value of the definite integral.
1. $\int_1^2 \dfrac{1}{x^2} dx = \dfrac{1}{2}\text{.}$
2. The table below gives values of the trapezoid rule and corresponding errors for different $n$-values.
$n$ $T_n$ $E_{T,n}$ $4$ $0.50899$ $-0.50899$ $8$ $0.50227$ $-0.50227$ $16$ $0.50057$ $-0.50057$
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$n$ $M_n$ $E_{M,n}$ $4$ $0.49555$ $0.00445$ $8$ $0.49887$ $0.00113$ $16$ $0.49972$ $0.00028$
4. The trapezoid rule overestimates; the midpoint rule underestimates.
5. $f(x) = \dfrac{1}{x^2}$ is concave up on $[1, 2]\text{.}$
Solution
1. $\int_1^2 \dfrac{1}{x^2} dx = \left. -x^{-1} \right|_1^2 = -\dfrac{1}{2} + 1 = \dfrac{1}{2}\text{.}$
2. The table below gives values of the trapezoid rule and corresponding errors for different $n$-values.
$n$ $T_n$ $E_{T,n}$ $4$ $0.50899$ $-0.50899$ $8$ $0.50227$ $-0.50227$ $16$ $0.50057$ $-0.50057$
3. The table below gives values of the midpoint rule and corresponding errors for different $n$-values.
$n$ $M_n$ $E_{M,n}$ $4$ $0.49555$ $0.00445$ $8$ $0.49887$ $0.00113$ $16$ $0.49972$ $0.00028$
4. From the errors in comparison to the known exact value, we see that the trapezoid rule overestimates this definite integral and the midpoint rule underestimates this definite integral.
5. The graph of the function given by $f(x) = \dfrac{1}{x^2}$ is concave up on the interval $[1, 2]\text{.}$ Because of this fact, we can see graphically that the line forming the top of each trapezoid lies fully above the curve, and thus the trapezoid rule overestimates the true value of the definite integral. Later in this section we'll see graphically why this concavity makes the midpoint rule an underestimate.
### SubsectionComparing the Midpoint and Trapezoid Rules | {
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### SubsectionComparing the Midpoint and Trapezoid Rules
We know from the definition of the definite integral that if we let $n$ be large enough, we can make any of the approximations $L_n\text{,}$ $R_n\text{,}$ and $M_n$ as close as we'd like (in theory) to the exact value of $\int_a^b f(x) \, dx\text{.}$ Thus, it may be natural to wonder why we ever use any rule other than $L_n$ or $R_n$ (with a sufficiently large $n$ value) to estimate a definite integral. One of the primary reasons is that as $n \to \infty\text{,}$ $\Delta x = \frac{b-a}{n} \to 0\text{,}$ and thus in a Riemann sum calculation with a large $n$ value, we end up multiplying by a number that is very close to zero. Doing so often generates roundoff error, because representing numbers close to zero accurately is a persistent challenge for computers.
Hence, we explore ways to estimate definite integrals to high levels of precision, but without using extremely large values of $n\text{.}$ Paying close attention to patterns in errors, such as those observed in Example5.81, is one way to begin to see some alternate approaches.
To begin, we compare the errors in the Midpoint and Trapezoid rules. First, consider a function that is concave up on a given interval, and picture approximating the area bounded on that interval by both the Midpoint and Trapezoid rules using a single subinterval. | {
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As seen in Figure5.82, it is evident that whenever the function is concave up on an interval, the Trapezoid Rule with one subinterval, $T_1\text{,}$ will overestimate the exact value of the definite integral on that interval. From a careful analysis of the line that bounds the top of the rectangle for the Midpoint Rule (shown in magenta), we see that if we rotate this line segment until it is tangent to the curve at the midpoint of the interval (as shown at right in Figure5.82), the resulting trapezoid has the same area as $M_1\text{,}$ and this value is less than the exact value of the definite integral. Thus, when the function is concave up on the interval, $M_1$ underestimates the integral's true value.
The preceding discussion explores how to determine if $M_n$ and $T_n$ are underestimates or overestimates of $\int_a^b f(x) \, dx$ if $f(x)$ is concave up. Similar statements can be made when $f(x)$ is concave down. These are summarized next.
###### Error types when approximating using Midpoint and Trapezoid Rules
• If $f(x)$ is concave up on $[a,b] \text{,}$ then $M_n$ underestimates $\int_a^b f(x) \, dx$ and $T_n$ overestimates $\int_a^b f(x) \, dx\text{.}$
• If $f(x)$ is concave down on $[a,b] \text{,}$ then $M_n$ overestimates $\int_a^b f(x) \, dx$ and $T_n$ underestimates $\int_a^b f(x) \, dx\text{.}$
Next, we compare the size of the errors between $M_n$ and $T_n\text{.}$ Again, we focus on $M_1$ and $T_1$ on an interval where the concavity of $f$ is consistent. In Figure5.83, where the error of the Trapezoid Rule is shaded in red, while the error of the Midpoint Rule is shaded lighter red, it is visually apparent that the error in the Trapezoid Rule is more significant. To see how much more significant, let's consider two examples and some particular computations.
If we let $f(x) = 1-x^2$ and consider $\int_0^1 f(x) \,dx\text{,}$ we know by the First FTC that the exact value of the integral is | {
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\begin{equation*} \int_0^1 (1-x^2) \, dx = x - \frac{x^3}{3} \bigg\vert_0^1 = \frac{2}{3}\text{.} \end{equation*}
Using appropriate technology to compute $M_4\text{,}$ $M_8\text{,}$ $T_4\text{,}$ and $T_8\text{,}$ as well as the corresponding errors $E_{M,4}\text{,}$ $E_{M,8}\text{,}$ $E_{T,4}\text{,}$ and $E_{T,8}\text{,}$ as we did in Example5.81, we find the results summarized in Table5.84. We also include the approximations and their errors for the example $\int_1^2 \frac{1}{x^2} \, dx$ from Example5.81.
For a given function $f$ and interval $[a,b]\text{,}$ $E_{T,4} = \int_a^b f(x) \,dx - T_4$ calculates the difference between the exact value of the definite integral and the approximation generated by the Trapezoid Rule with $n = 4\text{.}$ If we look at not only $E_{T,4}\text{,}$ but also the other errors generated by using $T_n$ and $M_n$ with $n = 4$ and $n = 8$ in the two examples noted in Table5.84, we see an evident pattern. Not only is the sign of the error (which measures whether the rule generates an over- or under-estimate) tied to the rule used and the function's concavity, but the magnitude of the errors generated by $T_n$ and $M_n$ seems closely connected. In particular, the errors generated by the Midpoint Rule seem to be about half the size of those generated by the Trapezoid Rule.
That is, we can observe in both examples that $E_{M,4} \approx -\frac{1}{2} E_{T,4}$ and $E_{M,8} \approx -\frac{1}{2}E_{T,8}\text{.}$ This property of the Midpoint and Trapezoid Rules turns out to hold in general: for a function of consistent concavity, the error in the Midpoint Rule has the opposite sign and approximately half the magnitude of the error of the Trapezoid Rule. Written symbolically,
\begin{equation*} E_{M,n} \approx -\frac{1}{2} E_{T,n}\text{.} \end{equation*}
This important relationship suggests a way to combine the Midpoint and Trapezoid Rules to create an even more accurate approximation to a definite integral.
### SubsectionSimpson's Rule | {
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### SubsectionSimpson's Rule
When we first developed the Trapezoid Rule, we observed that it is an average of the Left and Right Riemann sums:
\begin{equation*} T_n = \frac{1}{2}(L_n + R_n)\text{.} \end{equation*}
If a function is always increasing or always decreasing on the interval $[a,b]\text{,}$ one of $L_n$ and $R_n$ will over-estimate the true value of $\int_a^b f(x) \, dx\text{,}$ while the other will under-estimate the integral. Thus, the errors found in $L_n$ and $R_n$ will have opposite signs; so averaging $L_n$ and $R_n$ eliminates a considerable amount of the error present in the respective approximations. In a similar way, it makes sense to think about averaging $M_n$ and $T_n$ in order to generate a still more accurate approximation.
We've just observed that $M_n$ is typically about twice as accurate as $T_n\text{.}$ This leads to an approximation method known as Simpson's Rule 14Thomas Simpson was an 18th century mathematician; his idea was to extend the Trapezoid rule, but rather than using straight lines to build trapezoids, to use quadratic functions to build regions whose area was bounded by parabolas (whose areas he could find exactly). Simpson's Rule is often developed from the more sophisticated perspective of using interpolation by quadratic functions. which is a weighted average of the Midpoint and Trapezoid approximations.
###### Simpson's Rule
The Simpson's Rule approximation $S_{2n}$ of the area $\int_a^b f(x) \ dx$ is the weighted average
$$S_{2n} = \frac{2M_n + T_n}{3}\text{.}\label{E-Simpson}\tag{5.19}$$
where $M_n$ and $T_n$ are the Midpoint and Trapezoid rule approximations using $n$ subintervals. | {
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where $M_n$ and $T_n$ are the Midpoint and Trapezoid rule approximations using $n$ subintervals.
Note that we use $S_{2n}$ rather that $S_n$ since the $n$ points the Midpoint Rule uses are different from the $n$ points the Trapezoid Rule uses, and thus Simpson's Rule is using $2n$ points at which to evaluate the function. We build upon the results in Table5.84 to see the approximations generated by Simpson's Rule. In particular, in Table5.85, we include all of the results in Table5.84, but include additional results for $S_8 = \frac{2M_4 + T_4}{3}$ and $S_{16} = \frac{2M_8 + T_8}{3}\text{.}$
The results seen in Table5.85 are striking. If we consider the $S_{16}$ approximation of $\int_1^2 \frac{1}{x^2} \, dx\text{,}$ the error is only $E_{S,16} = 0.0000019434\text{.}$ By contrast, $L_8 = 0.5491458502\text{,}$ so the error of that estimate is $E_{L,8} = -0.0491458502\text{.}$ Moreover, we observe that generating the approximations for Simpson's Rule is almost no additional work: once we have $L_n\text{,}$ $R_n\text{,}$ and $M_n$ for a given value of $n\text{,}$ it is a simple exercise to generate $T_n\text{,}$ and from there to calculate $S_{2n}\text{.}$ Finally, note that the error in the Simpson's Rule approximations of $\int_0^1 (1-x^2) \, dx$ is zero!15Similar to how the Midpoint and Trapezoid approximations are exact for linear functions, Simpson's Rule approximations are exact for quadratic and cubic functions. See additional discussion on this issue later in the section and in the exercises.
These rules are not only useful for approximating definite integrals such as $\int_0^1 e^{-x^2} \, dx\text{,}$ for which we cannot find an elementary antiderivative of $e^{-x^2}\text{,}$ but also for approximating definite integrals when we are given a function through a table of data.
###### Example5.86 | {
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###### Example5.86
A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. Assume that $v$ is continuous, always decreasing, and always decreasing at a decreasing rate, as is suggested by the data.
seconds, $t$ Velocity in ft/sec, $v(t)$ $0$ $100$ $0.3$ $99$ $0.6$ $96$ $0.9$ $90$ $1.2$ $80$ $1.5$ $50$ $1.8$ $0$
1. Plot the given data on the set of axes provided in Figure5.88 with time on the horizontal axis and the velocity on the vertical axis.
2. What definite integral will give you the exact distance the car traveled on $[0,1.8]\text{?}$
3. Estimate the total distance traveled on $[0,1.8]$ by computing $L_3\text{,}$ $R_3\text{,}$ and $T_3\text{.}$ Which of these under-estimates the true distance traveled?
4. Estimate the total distance traveled on $[0,1.8]$ by computing $M_3\text{.}$ Is this an over- or under-estimate? Why?
5. Using your results from (c) and (d), improve your estimate further by using Simpson's Rule.
6. What is your best estimate of the average velocity of the car on $[0,1.8]\text{?}$ Why? What are the units on this quantity?
Hint
1. Plot the data.
2. What are the units of $v(t) \cdot \Delta t\text{?}$
3. Recall the standard rules for sums that produce $L_3\text{,}$ $R_3\text{,}$ $T_3\text{.}$
4. Think about concavity to decide if $M_3$ is an over- or under-estimate.
5. Recall how $S_3$ is a weighted average of $T_3$ and $M_3\text{.}$
6. Simpson's Rule gives the best estimate for a function of consistent concavity.
1. Plot the data.
2. $\int_0^{1.8} v(t) dt\text{.}$
3. \begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft }\text{.} \end{align*}
$R_3$ and $T_3$ are underestimates.
4. $M_3 = 143.4 \text{ ft }$ ; overestimate.
5. $S_6 = 140.8 \text{ ft } \text{.}$ | {
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4. $M_3 = 143.4 \text{ ft }$ ; overestimate.
5. $S_6 = 140.8 \text{ ft } \text{.}$
6. Simpson's rule gives the best approximation of the distance traveled, $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}$
Solution
1. Plot the data.
2. Since the velocity is always positive, the definite integral that will give the exact distance traveled by the car on the interval $[0, 1.8]$ is
\begin{equation*} \int_0^{1.8} v(t) dt\text{.} \end{equation*}
3. The estimates of $\int_0^{1.8} v(t) dt$ are
\begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft }\text{.} \end{align*}
$R_3$ is an underestimate of the distance traveled since $v(t)$ is decreasing. $T_3$ is an underestimate of the distance traveled since $v(t)$ is concave down.
4. Another estimate of the distance traveled is
\begin{equation*} M_3 = 143.4 \text{ ft }\text{.} \end{equation*}
This is an overestimate since $v(t)$ is concave down.
5. For Simpson's Rule, we see that
\begin{equation*} S_6 = \frac{2}{3}M_3 + \frac{1}{3}T_3 = 140.8 \text{ ft }\text{.} \end{equation*}
6. Simpson's rule gives the best approximation of the distance traveled since it is a weighted average of the midpoint and trapezoid rules and uses more information about the velocity than the other methods. The units on each of the estimates, including Simpson's Rule, are "feet", since ft/sec $\cdot$ sec = ft. Thus, the best approximation we have generated is that $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}$
### SubsectionComparing $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ $M_n\text{,}$ and $S_{2n}\text{.}$
As we conclude our discussion of numerical approximation of definite integrals, it is important to summarize general trends in how the various rules over- or under-estimate the true value of a definite integral, and by how much. To revisit some past observations and see some new ones, we consider the following example.
###### Example5.89 | {
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###### Example5.89
Consider the functions $f(x) = 2-x^2\text{,}$ $g(x) = 2-x^3\text{,}$ and $h(x) = 2-x^4\text{,}$ all on the interval $[0,1]\text{.}$ For each of the questions that require a numerical answer in what follows, write your answer exactly in fraction form.
1. On the three sets of axes provided in Figure5.90, sketch a graph of each function on the interval $[0,1]\text{,}$ and compute $L_1$ and $R_1$ for each. What do you observe?
2. Compute $M_1$ for each function to approximate $\int_0^1 f(x) \,dx\text{,}$ $\int_0^1 g(x) \,dx\text{,}$ and $\int_0^1 h(x) \,dx\text{,}$ respectively.
3. Compute $T_1$ for each of the three functions, and hence compute $S_2$ for each of the three functions.
4. Evaluate each of the integrals $\int_0^1 f(x) \,dx\text{,}$ $\int_0^1 g(x) \,dx\text{,}$ and $\int_0^1 h(x) \,dx$ exactly using the First FTC.
5. For each of the three functions $f\text{,}$ $g\text{,}$ and $h\text{,}$ compare the results of $L_1\text{,}$ $R_1\text{,}$ $M_1\text{,}$ $T_1\text{,}$ and $S_2$ to the true value of the corresponding definite integral. What patterns do you observe?
Hint
1. For each estimate, just one function evaluation is needed.
2. Use the midpoint rule with $n=1\text{.}$
3. Remember that both the trapezoid and Simpson's rule can be executed using (weighted) averages of known values.
4. Find antiderivatives to evaluate the integrals exactly.
5. Think about trends in over- and under-estimates.
1. For $L_1$ and $T_1\text{:}$
The values of $L_1$ and $R_1$ are the same for all three.
2. For the $M_1\text{,}$
3. For $T_1$ and $S_2\text{,}$ | {
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2. For the $M_1\text{,}$
3. For $T_1$ and $S_2\text{,}$
4. \begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5} \end{align*}
5. Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson's rule is exact for both $f$ and $g\text{,}$ while a slight overestimate of $\int_0^1 h(x) dx\text{.}$
Solution
1. For the left and right endpoint rules, we see that
Thus, we observe that despite the fact the functions are all different, the values of $L_1$ and $R_1$ are the same for all three.
2. For the midpoint rule, we find that
3. For the trapezoid rule and Simpson's rule,
4. The exact values of the three definite integrals are
\begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5}\\ \amp \approx 1.6667 \amp \amp = 1.75 \amp \amp = 1.8 \end{align*}
5. We observe that each of the left endpoint rule results are overestimates, each of the right endpoint rules are underestimates, each of the midpoint rules are overestimates, and each of the trapezoid rules are underestimates. These results hold because each of the three functions are both decreasing and concave down. For Simpson's rule, we see that the result is exact for both $f$ and $g\text{,}$ while Simpson's rule is a slight overestimate of $\int_0^1 h(x) dx\text{.}$
The results seen in Example5.89 generalize nicely. For instance, if $f$ is decreasing on $[a,b]\text{,}$ $L_n$ will overestimate the exact value of $\int_a^b f(x) \,dx\text{,}$ and if $f$ is concave down on $[a,b]\text{,}$ $M_n$ will overestimate the exact value of the integral. An excellent exercise is to write a collection of scenarios of possible function behavior, and then categorize whether each of $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ and $M_n$ is an over- or under-estimate. | {
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