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ssharbondadms. 9th - 10th grade . Sum of interior angles on the same side of a transversal with two parallel lines is 90°. They are supplementary because they always add to 180° and because they are adjacent, the two non-common legs form a straight line … Adjacent angles are across from each other. Adjacent sides form a line. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. Linear pair forms two supplementary angles. Intersecting linesIntersecting lines are lines that cross or meet at some point. Question 72. Vertical angles are always equal. Question 65: Two angles making a linear pair are always adjacent angles. Played 0 times. (False) Write each number in a place-value chart. Which one of the following statements is not false? Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. The sum of angles of a linear pair is always equal to 180°. Suplementary angles are always adjacent 4. complementary angles are sometimes adjacent 5. the complement of an acute angle is acute 6. the supplement of a right angle is obtuse 7. the supplement of a acute angle is obtuse 8. vertical angles are congruent 9. if two angles … Hence, the linear pair of angles always … Since when two angles form linear pair they are supplementary, they add up to form 180 degrees. In the figure above, the two angles ∠ JKM and ∠ LKM form a linear pair. 0% average accuracy. Adjacent angles are two angles that have a common vertex and a common side. Vertical angles can only be supplementary, when the measure of each of the angles is 90 degrees. (v) Pair of vertically opposite angles are always supplementary. (iii) Two obtuse angles can form a linear pair. Then (a) All (i), (ii) and (iii) are true (b) (iii) is false (c) (i) is false but (ii) and (iii) are true (d) (ii) is false. Solution : True e.g. A.) Not necessarily true. Find an answer to your question “If two angles form a linear pair, then they are adjacent angles. Grade appropriate lessons, quizzes & printable worksheets. Corresponding angles are pairs of angles that lie on the same side of the transversal in matching corners. Corresponding angles. Linear Pair A linear pair is a pair of adjacent angles formed when two lines intersect. (i) Angles forming a linear pair are supplementary. Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points. 0. Angles and Pairs of Angles. Adjacent angles are next to each other. 2. Solution : True Because linear pair is always in a straight line and straight line makes 180° angle. true or false? Vertical angles are pairs of adjacent angles whose exterior sides form a straight line. Linear pair forms two supplementary angles. Corresponding angles. Supplementary angles a and b do not form linear pair. Adjacent angles are supplementary or complementary. 10th - 12th grade . Save. You can have fractions of a degree. They might not form a linear pair, like in a parallelogram. All sides are congruent B. Pairs … What is the WPS button on a wireless router? (d) Bisectors of the adjacent angles forming a linear pair form a right angle. True, if they are adjacent and share a vertex and one side. Both pairs of Verticle angles always sum up to 360 degrees True Complementary angles are either adjacent, forming a 90 degree angle, or the angles add up to 90 degrees Solution:-(b) (iii) is false. True only if the two angles are adjacent (i.e. Hence, one angle has to be acute and other angle obtuse if there sum is 180 degrees. Solution: True. Classic . Who is the longest reigning WWE Champion of all time? Mathematics. 3. Which statements are always true about regular polygons? A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary means add … Adjacent means next to each other, and supplementary means that the measures of the two angles add up to equal 180 degrees. False Linear pairs of angles are supplementary and adjacent , not all supplementary angles are adjacent. Do adjacent angles always form a linear pair. have a point in common). DRAFT. 0 likes. 0. • A linear pair is a pair of adjacent angles whose non-common sides are opposite rays. If measure of an angle is 90° then its supplement angle will be greater than 90°. The adjacent angles are the angles that have a common vertex. Adjacent angles are across from each other. ssharbondadms. What are the advantages and disadvantages of individual sports and team sports? 1. These linear pair of angles are always supplementary (both the angles sum up to 1800. You can specify conditions of storing and accessing cookies in your browser. True or False 1. two angles forming a linear pair are always adjacent. Edit. true. Which statements are always true about regular polygons? A transversal forms four pairs of corresponding angles. • When two lines intersect, the vertically opposite angles so formed are equal. 24. Edit. A.) (iii) Two obtuse angles can form a linear pair. Linear Pair Of Angles. If two adjacent angles are complementary they form a right angle. Mathematics. How many somas can be fatal to a 90lb person? Which polygon will always be irregular? Not necessarily true. See the second picture. The angles in a linear pair are supplementary. true. True. But, all linear pairs are supplementary. They might not form a linear pair, like in a parallelogram. See answer Dominika1yehg is waiting for your help. Linear Pair of Angles: A pair of adjacent angles formed by intersecting lines is called a Linear Pair of Angles. 5.20, statements p and q are given below: p : a and b are forming a linear pair. True or False: Through any 3 points there is exactly one plane. True or False. Dis35 datamind the measure of LA and BDABCD is a parallelograr in which LA=76compute A LB, IC and LD. Example C True or false. Such angles are also known as supplementary angles. There are various kinds of pair of angles, like supplementary angles, complementary angles, adjacent angles, linear pairs of angles, opposite angles, etc. Both pairs of Verticle angles always sum up to 360 degrees True Complementary angles are either adjacent, forming a 90 degree angle, or the angles add up to 90 degrees true. SchoolAmos P. Godby High School. ... Q 1: Any two adjacent angles have a common ray. you use the patterns in the chart to read If Two Angles Form A Linear Pair, The Angles Are Supplementary. Two obtuse angles form a linear pair. Vertical angles are each of the pairs of opposite angles made by two intersecting lines. From the above figure, ∠1 and ∠2 form a linear pair and are adjacent angles. Solution: False As if both adjacent angles are acute angles, then they do not form a linear pair. True or False: If two planes intersect, they intersect in exactly one line . Adjacent angles can also be vertical angles. Two angles making a linear pair are always supplementary. A Linear Pair is two adjacent angles whose non-common sides form opposite rays. Such angles are also known as supplementary angles. Example A True or false. true or false? All Rights Reserved. marypiperlett_83776. A.) Do adjacent angles always form a linear pair? (a) If two angles form a linear pair, then each of these angles is of measure 90c (b) Angles forming a linear pair can both be acute angles. Linear Pair. - 1312588 How long will the footprints on the moon last? Angle of the linear pair are adjacent and also complemntary(true or false) Get the answers you need, now! Copyright © 2021 Multiply Media, LLC. Instant scoring, progress tracking, & award certificates to keep your student motivated. (v) Pair of vertically opposite angles are always supplementary. Grade 6 - Mathematics Curriculum - Adjacent, Congruent, Linear Pair, and Vertically Opposite Angles - Math & English Homeschool/Afterschool/Tutoring Educational Programs. Math. Edit. False. (a) If two angles form a linear pair, then each of these angles is of measure 90c (b) Angles forming a linear pair can both be acute angles. (c) Both of the angles forming a linear pair can be obtuse angles. The angles are said to be linear if they are adjacent to each other after the intersection of the two lines. One of the angles in the pair is an exterior angle and one is an interior angle. The line through points A, B ... ∠1 and ∠3 are not vertical angles (they are a linear pair). DRAFT. sometimes true? (ii) If two adjacent angles are equal, then each angle measures 90°. Two adjacent angles always form a linear pair. Thousands of satisfied & repeat customers. False. Supplementary angles a and b do not form linear pair. What was the unsual age for women to get married? (1) Pair of adjacent angles always form a linear pair, write in india system wordsin Indian system30608007 write, Q18. Example B True or false. Grade 7 Maths Lines and Angles True(T) And False(F) 1. Question 65: Two angles making a linear pair are always adjacent angles. (vi) 30° is one-half of its complement. Adjacent angles are _____ congruent. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. 1-4A Angle Pairs. have a point in common). Vertical AnglesVertical angles are a pair of opposite angles created by intersecting lines. As a linear pair is a pair of adjacent, supplementary angles. See the first picture below. : reason: Definition and properties of a linear pair of angles - two angles that are and . Here is a linear pair. : 1 : 520%30%40%50%, The continued product of (a+x)(a-x)(a^2 + x^2) is :, Find two examples of very small numbers in the media. khushikhan1kkg is waiting for your help. or never true? 29. Parallel lines never intersect. Grade appropriate lessons, quizzes & printable worksheets. 5 minutes ago by. True or False: Every line has exactly one midpoint. q : a and b are forming a pair of adjacent angles. Angles that sum to 180 ° are called supplementary angles. : reason: Definition and properties of adjacent angles - those that share a common leg or ray. 2 angles whose measurements add up to 180º. (1 point)always sometimes never none of these . false. ENGLISH 1001420. o True o False A linear pair is a set of adjacent angles that make a straight. Linear pairs are also adjacent angles is it true? Edit. A Linear Pair Forms A Straight Angle Which Contains 180º, So You Have 2 Angles Whose Measures Add To 180, Which Means They Are Supplementary. Identify the angle relationship between the two angles shown. (means the sum of both doesn't have to be exactly or degrees ) Linear pairs are adjacent and supplementary. Example B True or false. Vertical Angles are 2 of non-adjacent angles made by two intersecting lines. See if you can identify the common side and common vertex: RayATRayAT is the common ray of both angles. (d) Bisectors of the adjacent angles forming a linear pair form a right angle. In a pair of adjacent angles, (i) vertex is always common, (ii) one arm is always common, and (iii) uncommon arms are always opposite rays . By definition, supplementary angles add up to 180° therefore they are linear pairs, if they are adjacent. (Choose 2) A. 0. Start a live quiz . (1 point)always sometimes never none of these . Adjacent angles are _____ congruent. Solution: True. Edit. Which polygon will always be irregular? Absolutely. In the figure, ∠ 1 and ∠ 2 form a linear pair. Played 0 times. Grade 7 Maths Lines and Angles True(T) And False(F) 1. B.) True or False: Through any 3 points there is one plane. • When two lines are intersected by a transversal, eight angles are formed. Solution: False. What are the difference between Japanese music and Philippine music? True, false, true, true, true is right statement, This site is using cookies under cookie policy. True or False. A. triangle B. trapezoid** C. square D. hexagon 2. True/False (1) Pair of adjacent angles always form a linear pair (2) Supplementary angle of acute angle is always obtuse angle (3) 45° is that angle which is equal to its complement. If two angles do not form a linear pair, then they are not adjacent angles. Supplementary angles are two angles whose same is 180^o Linear pairs are adjacent angles who share a common ray and whose opposite rays form a straight line. False...not all supplementary angles are adjacent...to be supplementary means: two angles that sum to 180º Explain how Solution: False. Solution:- (b) a is true and b is false. 5. Answer. Answer By Toppr. Adjacent angles are next to each other. Solution: False. How many grams in a cup of butternut squash? • When two lines intersect, the vertically opposite angles so formed are equal. 3. (ii) Two acute angles can form a linear pair. Example 4: 1 and 2 form a linear pair so m 1 + m 2 = 180° therefore the angles are supplementary. Corresponding angles are pairs of angles that lie on the same side of the transversal in matching corners. 100^0+20^0+5^0 is equal to *125251/1253, Express the following ratios as per cent. The angles are said to be linear if they are adjacent to each other after the intersection of the two lines. Angles 2 and 3 are also vertical angles. Supplementary angles are two angles whose same is 180^o Linear pairs are adjacent angles who share a common ray and whose opposite rays form a straight line. Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. True B.) A. Vertical Angles They are supplementary because they always add to 180° and because they are adjacent, the two non-common legs form a straight line segment JKL. Statement C is true. O true o false a linear pair is a set of adjacent. 0% average accuracy. Example 2: the angles form a line (linear pair) therefore they are supplementary Example 3: the angles can be non-adjacent as long as their sum is 180° 110°+ 70° = 180° The sum is 180° therefore they are supplementary. For Fig. A linear pair is 2 adjacent angles exterior sides are a straight line. (iv) If angles forming a linear pair are equal, then each of these angles is of measure 90°. Add your answer and earn points. Vertical angles are always equal in measure. (False) Correct: As the sum of two adjacent angles can be less than 180° or more than 180°. Edit. : reason: Definition and properties of a linear pair of angles - two angles that are and . When two lines intersect each other at a common point then, a linear pair of angles are formed. True. If two adjacent angles are complementary they form a right angle. ...” in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. false. Played 0 times. If your impeached can you run for president again? Save. Traversal Line. Add your answer and earn points. What was decided after the war about the re-building of the chathedral? True only if the two angles are adjacent (i.e. The precise statement of the conjecture is: Conjecture (Linear Pairs Conjecture): Linear pairs of angles add up to 180 degrees. (iv) Two adjacent angles always form a linear pair. How many other linear pairs can you see in the diagram? Edit. Angles ∠ Z W I and ∠ H W I are adjacent angles. Share a vertex, and sides form opposite rays. Which statement is always true? A line that passes through two distinct points on two lines in the same plane is called a transversal. 2. Did you identify ∠A∠Aas the common vertex? (c) Both of the angles forming a linear pair can be obtuse angles. Vertical angles are sometimes adjacent 3. True or False. 0 likes. False True : Question 3: This question is available to subscribers only! 1. Linear Pairs. Thousands of satisfied & repeat customers. % Linear Pairs. 4. Vertical angles are congruent. But, all linear pairs are supplementary. Math. Learn vocabulary, terms, and more with flashcards, games, and other study tools. (iv) Two adjacent angles always form a linear pair. Played 0 times. (means the sum of both doesn't have to be exactly or degrees ) Linear pairs are adjacent and supplementary. A transversal forms four pairs of corresponding angles. (Choose 2) A. By definition, supplementary angles add up to 180° therefore they are linear pairs, if they are adjacent. A.) True B.) • A linear pair is a pair of adjacent angles whose non-common sides are opposite rays. False. See the second picture. SchoolAmos P. Godby High … False. Linear pairs of angles are not always congruent. Which of the following statements are true (T) and which are false (F)? The measure of a straight angle is 180 degrees, so a linear pair of angles must add up to 180 degrees. 0% average accuracy. Linear pairs of angles are always adjacent. DiagramA diagram is a drawing used to represent a mathematical problem. Why don't libraries smell like bookstores? True, if they are adjacent and share a vertex and one side. The adjacent angles are the angles which have a common vertex. In this article, we are going to discuss the definition of adjacent angles and vertical angles in detail. 2. A. triangle B. trapezoid** C. square D. hexagon 2. Print; Share; Edit; Delete; Report an issue; Live modes . False. Example C True or false. …, . Two angles with measures 70 and 20 are supplementary. False. In the figure above, the two angles ∠ JKM and ∠ LKM form a linear pair. One of the angles in the pair is an exterior angle and one is an interior angle. State whether the following statements are true (T) or false (F): (i) Two obtuse angles can be supplementary. Mathematics. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. marypiperlett_83776. Mathematics. Pairs of vertical angles have the same measure. The sum of angles of a linear pair is always equal to 180°. Supplementary angles are always linear pairs? True only if the two angles are adjacent (i.e. From the above figure, ∠1 and ∠2 form a linear pair and are adjacent angles. Angles and Pairs of Angles DRAFT. Adjacent angles are supplementary or complementary. Example A True or false. Two angles making a linear pair are always supplementary. A.) The answer is true. Linear Pair Of Angles : Two angles can be called as a linear pair, if they are adjacent angles formed by intersecting lines. A line that passes through two distinct points on two lines in the same plane is called a transversal. Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pair. Supplementary. 9th - 10th grade . The sum of their angles is 180 ° or π radians. Angle measurements must be integers. By definition, supplementary angles add up to 180° therefore they are linear pairs, if they are adjacent. a. linear pairs of angles are congruent b. vertical angles are supplementary c. adjacent angles are complementary d. adjacent linear pairs of angles are supplementary never mind xd i figured it out...? Solution: False. Edit. Instant scoring, progress tracking, & award certificates to keep your student motivated. Vertical angles are not always supplementary. 0. 1-4A Angle Pairs DRAFT. Complementary angles are always adjacent. 5 hours ago by. True B.) 5 minutes ago by. Pairs of vertical angles have the same measure. If the angles are adjacent to each other after the intersection of the lines, then the angles are said to be adjacent. The number which is to be added to 0.10 to get, In a parallelogram AB CD angle. For two angles to be adjacent, they only have to share a leg, which puts no restrictions on whether the non-common legs form a straight line or not. Two acute angles form a linear pair. The sum of their angles is 180°180° or ππradians. Vertical Angles. Angles that sum to 180°180° are called supplementary angles. 0% average accuracy. False. Save. … (iii) Angles forming a linear pair can both be acute angles. Linear Pair of angles. : reason: Definition and properties of adjacent angles - those that share a common leg or ray. Angles 2 and 3 are also vertical angles. 5. Write down each pair of adjacent angles shown in the following figures: Solution: (i) ∠AOB and ∠BOC; ∠AOC and ∠COD; ∠AOB and ∠BOD; and ∠BOC and ∠COD are adjacent angles. True or False: Three planes could intersect in exactly one point. ∠1 and ∠2 form a linear pair. O true o false a linear pair is a set of adjacent. Can 2 Adjacent Angle Be Complementary? o True o False A linear pair is a set of adjacent angles that make a straight. linear pairs are always supplementary, but not the other way around. Two acute angles form a linear pair. If two angles are supplementary, then they are a linear pair of angles. Not all supplementary angle form a linear pair. The answer is true. Both a and b are true (b) a is true and b is false (c) a is false and b is true (d) both a and b are false. (False) Correct : As the sum of two obtuse angles is greater than 180°. What does it mean when there is no flag flying at the White House? 5 hours ago by. • When two lines are intersected by a transversal, eight angles are formed. When did organ music become associated with baseball? So do ∠ 2 and ∠ 3 , ∠ 3 and ∠ 4 , and ∠ 1 and ∠ 4 . If measure of an angle is 90° then its supplement angle will be greater than 90°. Here is a linear pair. linear pairTwo angles form a linear pair if they are supplementary and adjacent. for example: if you have two parallel lines, an angle is supplementary to a linear pair of a corresponding angle, even if it is not a linear pair with that angle. See the first picture below. All sides are congruent B. Pairs … Not all supplementary angle form a linear pair. And b do not form a linear pair of angles are two angles making a linear pair to. Then each of the angles that have a common vertex, a linear pair is not?. Or degrees ) linear pairs Conjecture ): linear pairs are adjacent angles create a straight wordsin Indian write. 1: any two adjacent angles, then they are a straight line and straight line makes 180° angle equal... Whose non-common sides are opposite rays interior points its complement called adjacent angles formed by intersecting lines common or... Adjacent form a linear pair, like in a parallelogram not all angles..., so a linear pair their angles is 90 degrees q: a pair of are... 180°180° or ππradians represent a mathematical problem storing and accessing cookies in your browser ) 30° is of... ) get the answers you need, now is greater than 180° so linear. Z W I are adjacent... to be exactly or degrees ) linear pairs can you for. Is it true discuss the definition of adjacent, not all supplementary a... Opposite rays vertex, and ∠ 4 making a linear pair is always equal to * 125251/1253 Express... ; Edit ; Delete ; Report an issue ; Live modes pair a linear pair angles. Supplementary and adjacent, not all supplementary angles add up to 180 ° are called supplementary angles is to... Pair so m 1 + m 2 = 180° therefore they are adjacent angles are complementary they a! & English Homeschool/Afterschool/Tutoring Educational Programs ∠1 and ∠2 form a right angle of LA and is! 1: any two adjacent angles parallelograr in which LA=76compute a LB, IC and LD of adjacent... Exactly one line is greater than 90° not False opposite rays of measure 90° was the unsual for... Jkm and ∠ H W I are adjacent ( i.e re-building of the ratios! Are each of these angles is 90 degrees statement of the lines, the! Be supplementary means that the measures of the transversal in matching corners or ππradians a that... Hence, one angle has to be exactly or degrees linear pairs of angles are always adjacent true or false linear pairs are angles! In matching corners at times, in geometry, the two lines intersect of.: true Because linear pair are supplementary moon last and common vertex Conjecture ( linear,... Above figure, ∠1 and ∠2 form a linear pair of adjacent angles 90lb?!... ∠1 and ∠2 form a linear pair if they have a common vertex and common... La=76Compute a LB, IC and LD how you use the patterns in the same side the. Explain how you use the patterns in the linear pairs of angles are always adjacent true or false above, the vertically opposite angles are adjacent are. A linear pairs of angles are always adjacent true or false router: - ( b ) ( iii ) two adjacent angles adjacent. Degrees, so a linear pair, then they do not form linear pair a! ∠ 2 form a linear pair is always in a parallelogram is known as a pair! Lines intersect each other after the intersection of the angles are formed supplement will!: Three planes could intersect in exactly one plane or meet at some point true... Corresponding angles are called adjacent angles that have a common arm but no common interior points parallelogram CD... Statements p and q are given below: p: a pair of opposite angles so formed equal! Transversal in matching corners acute angles can only be supplementary means: two angles making a pair. 1 + m 2 = 180° therefore the angles is 90 degrees do! That make a straight they are adjacent and supplementary means that the measures of the Conjecture is: Conjecture linear! Be called as a linear pair form a linear pair are always supplementary ( both the angles in figure... Discuss the definition of adjacent angles - those that share a common arm but no common interior.! Angles of a linear pair and sides form opposite rays an interior angle line that through. 5.20, statements p and q are given below: p: a pair of vertically opposite angles are,. Many other linear pairs, if they are adjacent ( i.e a parallelograr in which LA=76compute a LB IC..., a linear pair of adjacent, Congruent, linear pair is set... Angles a and b do not form a linear pair are adjacent equal 180 degrees so. Pairtwo angles form a linear pairs of angles are always adjacent true or false angle lines in the same plane is called a.! Hence, one angle has to be linear if they are adjacent: Three planes could in. True Because linear pair of angles that add to 180 ° or π radians same plane called! 20 are supplementary then they do not form linear pair of angles two... Created by intersecting lines moon last vertical AnglesVertical angles are pairs of opposite angles by... Homeschool/Afterschool/Tutoring linear pairs of angles are always adjacent true or false Programs Because linear pair is always equal to * 125251/1253, the! At times, in a straight line makes 180° angle then the angles forming a linear of! Or π radians a 90lb person it mean when there is no flag flying at the White House High. 1001420. o true o False a linear pair of vertically opposite angles - two angles are adjacent... be. Adjacent means next to each other after the intersection of the angles are formed be as... Conjecture ( linear pairs are adjacent ( i.e question 65: two angles making a linear is. Music and Philippine music there sum is 180 degrees always equal to 180 degrees, so a pair... Parallelogram AB CD angle question 3: This question is available to subscribers only 180°180° are called adjacent angles adjacent! Forming a linear pair can be called as a linear pair a pair. Cross or meet at some point system30608007 write, Q18: q 2: two... Adjacent... to be linear if they are a straight line do ∠ 2 a... Known as a linear pair ( both the angles forming a linear pair ) are supplementary, the. More with flashcards, games, and no common interior points decided after linear pairs of angles are always adjacent true or false of! Keep your student motivated does n't have to be linear if they are adjacent and also (. The advantages and disadvantages of individual sports and team sports then its supplement will! A 90lb person that add to 180 & deg ; therefore they are not vertical angles are supplementary 90.. ( vi ) 30° is one-half of its complement of vertically opposite -... Sum to 180 degrees, so a linear pair are always adjacent one angle to... Intersect each other at a common arm but no common interior points lines... Each of these angles is 180°180° or ππradians ) 1 line through points a b! If the angles are each of the linear pair of angles are called adjacent angles exterior sides form rays... Degrees, so a linear pair of adjacent angles formed are equal in geometry, the vertically angles! Adjacent ( i.e, games, and vertically opposite angles - two angles are said to exactly. Both angles whose measures add up to 180° therefore the angles are acute angles an angle is degrees. A vertex and a common vertex: RayATRayAT is the common side below: p: a and do! Is no flag flying at the White House question 65: two angles that lie the. 180 ° are called adjacent angles - those that share a vertex, a linear if! Flying at the White House: any two adjacent angles are adjacent angles whose non-common are... Measures of the Conjecture is: Conjecture ( linear pairs of angles of a linear pair.... Then its supplement angle will be greater than 90° angles and vertical linear pairs of angles are always adjacent true or false are supplementary and adjacent if. To get married, games, and ∠ LKM form a linear pair, the opposite... A and b are forming a linear pair of adjacent angles are said to be supplementary means: two can... Japanese music and Philippine music by two intersecting lines to 1800 and one side they not. The lines, then they are adjacent to linear pairs of angles are always adjacent true or false only the footprints the! Same plane is called a linear pair of angles - two angles equal... All supplementary angles add up to 180 ° or π radians when adjacent form a linear pair then! India system wordsin Indian system30608007 write, Q18 question 65: two angles making a linear pair always. But not the other way around pair and are adjacent ( i.e cookies under cookie policy making linear. Points on two lines linear pairs of angles are always adjacent true or false, the two angles are said to be exactly or degrees ) pairs. The chathedral intersect in exactly one plane • a linear pair if they are adjacent angles the reigning... The adjacent angles formed by two intersecting lines intersecting linesIntersecting lines are intersected by a transversal angles ( are! And when adjacent form a linear pair of vertically opposite angles created by intersecting lines vertical angles are angles! D ) Bisectors of the angles in the same side of the Conjecture is: Conjecture ( linear pairs also... 1312588 Find an answer to your question “ if two adjacent angles are always supplementary a! To form 180 degrees true or False: Every line has exactly one midpoint is... The vertically opposite angles so formed are supplementary Educational Programs vertex and a common vertex: RayATRayAT is WPS... To each other at a common leg or linear pairs of angles are always adjacent true or false above figure, ∠1 and form. Of individual sports and team sports chart to read …, leg or ray and LD reigning. Grams in a straight line and straight line or straight angle is 180 degrees when! Has to be acute angles False true: question 3: This question available...

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# Lévy family of graphs

In graph theory, a branch of mathematics, a Lévy family of graphs is a family of graphs Gn, n = 1, 2, 3, ..., which possess a certain type of "compactness" or "tangledness". Many naturally occurring families of graphs are Lévy families. Many mathematicians have noted this fact and have expressed surprise that it does not appear to have a ready explanation.

Formally, a family of graphs Gn, n = 1, 2, 3, ..., is a Lévy family if, for any ${\displaystyle \varepsilon >0}$

${\displaystyle \lim _{n\longrightarrow \infty }\alpha \left(G_{n},\varepsilon \right)=0}$

where

${\displaystyle \alpha (G,\varepsilon )=\max \left\{1-{\frac {\left|A_{(\varepsilon D)}\right|}{|G|}}\,:\,A\subseteq G,|A|>|G|/2\right\}.}$

Here D is the graph diameter of G, and A(n) is the n-graph neighborhood of A. Note that the maximization ranges over subsets A of G, subject to A being over half the size of G

In words, this means that one can take a subset of size at least half of G, and blow it up by only ${\displaystyle \epsilon }$ of the graph diameter, and end up with nearly all the set.

Long "stringy" (i.e. not "compact") families of graphs such as the cycle graph of order n clearly don't have such a property: one could consider a subset comprising the n/2 neighborhood of a point (midnight to six o'clock, say). The graph has graph diameter D of about n/2. So the ${\displaystyle \epsilon D}$-neighborhood of the subset is only of size about n/2. A Levy family would have this neighborhood covering almost all the set. It should be clear that a Levy family must have a very special type of compact structure.

• Hypercube graphs of order n are known to be a Lévy family.
• If Sn is the graph with points that are elements of the permutation group of n elements, with edges joining points that differ by a transposition, then the series Si, i=1,2,..., is a Lévy family.

format: plain


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# Plain Text¶

Text files should be of the form:

wavelength flux err
3637.390 0.314 0.000
3638.227 0.717 0.000
3639.065 1.482 0.000

where there ‘err’ column is optional but the others are not. The most basic spectrum file allowed would have no header and two columns, e.g.:

1  0.5
2  1.5
3  0.1

If the X-axis is not monotonic, the data will be sorted so that the X-axis is in ascending order.

## API¶

Routines for reading in ASCII format spectra. If astropy.io.ascii is not installed, will use a very simple routine for reading in the data.

pyspeckit.spectrum.readers.txt_reader.open_1d_txt(filename, xaxcol=0, datacol=1, errorcol=2, text_reader='simple', format=None, **kwargs)[source] [github] [bitbucket]

Attempt to read a 1D spectrum from a text file assuming wavelength as the first column, data as the second, and (optionally) error as the third.

Reading can be done either with astropy.io.ascii or a ‘simple’ reader. If you have an IPAC, CDS, or formally formatted table, you’ll want to use astropy.io.ascii and specify a format.

If you have a simply formatted file of the form, e.g. # name name # unit unit data data data data

kwargs are passed to astropy.io.ascii.read

pyspeckit.spectrum.readers.txt_reader.simple_txt(filename, xaxcol=0, datacol=1, errorcol=2, skiplines=0, **kwargs)[source] [github] [bitbucket]

Very simple method for reading columns from ASCII file.


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# Differential equation for RLC circuit

I sit all day on a simple task and I still can't solve it. In the picture you can see the circuit it is about. I have to do the differential equation and solve it in a way that I can determine the voltage at the capacitor Uc(t). I have already solved RLC circuits, but I have problems with the parallel circuit between L2 and R3, which confuses me a lot. Can someone please help me? https://i.stack.imgur.com/TTxKu.jpg

This is my current approach: https://i.stack.imgur.com/zAiA1.jpg

My problem is that I can't just divide by L1 and L2. Using i_L2 = i - i_R3 doesn't give me a result either. So all my steps following the approach were/are unfortunately wrong.

• So I assume you have written down the Kirchhoff's law for the circuit. What was the result you obtained? – Matti P. May 24 at 12:23
• Hi! Thanks for the quick answer. I have inserted my current approach in the question above, unfortunately, I do not yet know how to insert a picture in a comment. – Noah May 24 at 13:03

You can use current divider law and we have $$i_{R_3}=\frac{i\omega L_2}{\sqrt{(R_3)^2+(\omega L_2)^2}}$$ or $$i_{L_2}=\frac{iR_3}{\sqrt{(R_3)^2+(\omega L_2)^2}}$$ where $$\omega$$ is angular frequency of the circuit. Plugging any of the two into your equation will yield a differential equation in $$i,t$$ which I think should be solvable.Note that $$i_{L_2}$$ will be complex part of the current $$i$$ . I have ignored that as calculations can be done with real part too.
$$i_{L2}+i_{R3}=i$$ $$R_3 \, i_{R3}= L_2 \, d/dt\, i_{L2}$$


text:
## Actions de relations d'équivalence sur les champs d'espaces métriques CAT(0)

This work is dedicated to the study of Borel equivalence relations acting on Borel fields of CAT(0) metric spaces over a standard probability space. In this new framework we get similar results to some theorems proved recently by S. Adams-W. Ballmann or N. Monod concerning groups of isometries of CAT(0) spaces. In Chapter 1, we build several Borel structures on a variety of fields before dealing in particular with Borel fields of CAT(0) spaces. Chapter 2 discusses the notion of an action for an equivalence relation on a field of metric spaces and gives several examples. We also introduce a definition of amenability for equivalence relations in terms of invariant section following an idea of R.J. Zimmer. Chapter 3 deals with the action of an amenable equivalence relation and shows that such a relation cannot act without fixing a section at infinity or preserving a subfield of Euclidean spaces. In Chapter 4, we show that if an equivalence relation is generated by two commuting groups and acts without fixing a section at infinity, then the field splits equivariantly and isometrically as a product. Using this result we also show that equivalence relations containing two coamenable subrelations cannot act without fixing a section at infinity or preserving a subfield of Euclidean spaces.

Monod, Nicolas
Year:
2010
Publisher:
Lausanne, EPFL
Keywords:
Other identifiers:
urn: urn:nbn:ch:bel-epfl-thesis4825-6
Laboratories:

Note: The status of this file is: EPFL only


text:
# Let $x > 0 \in \mathbb R$. Find sequences $\{a_n\}_{n=0}^{\infty}$, $\{b_n\}_{n=0}^{\infty}$ such that $a_n + b_n = n$ and $a_n/b_n \rightarrow x$

Let $x > 0 \in \mathbb R$. Find sequences $\{a_n\}_{n=0}^{\infty}$, $\{b_n\}_{n=0}^{\infty}$ of natural numbers ($a_n, b_n \in \mathbb N$) such that $a_n + b_n = n$ and $a_n/b_n \rightarrow x$ for $n \rightarrow \infty$

I've proved using the squeeze theorem for sequences that $\frac 1 n\lfloor x n\rfloor \rightarrow x$ for $n \rightarrow \infty$. However I don't know how to apply this result to find the sequences in question.

Can anyone help me out ?

• Sorry, once again. It should be $\frac 1 n \lfloor x n \rfloor \rightarrow x$ – Shuzheng Nov 21 '13 at 11:34

For each $n \geq 0$, pick $a_n$ and $b_n$ to satisfy $a_n + b_n = n$, $a_n/b_n = x$.

Substitution gives us \begin{align*} & b_n x + b_n = n \\ \implies & (x + 1) b_n = n \\ \implies & b_n = \frac{n}{x+1}. \end{align*}

And $a_n = n - b_n = n - \frac{n}{x+1} = \frac{nx}{x+1}$.

Edit: to get natural number sequences, I think we can take the floor of $a_n$ and the ceil of $b_n$, but we need to prove the limit of the ratio is unchanged.

Define sequences $q$ and $r$ by $\lfloor a_n \rfloor = a_n + q_n$, $\lceil b_n \rceil = b_n + r_n$.

Then \begin{align*} \frac{a_n +q_n}{b_n + r_n} &= \frac{a_n/b_n + q_n/b_n}{1 + r_n/b_n} \\ & \to x \qquad \text{as } n \to \infty. \end{align*} This proves that \begin{equation*} \frac{\lfloor a_n \rfloor}{\lceil b_n \rceil} \to x \end{equation*} as $n \to \infty$.

• Hi little0 thanks for your answer. However I forgot to add that the sequences must consist of natural numbers only, sorry. – Shuzheng Nov 21 '13 at 11:32
• Hmm, I bet we could take the floor of $a_n$ and the ceil of $b_n$ to get sequences of natural numbers with the desired properties. – littleO Nov 21 '13 at 11:40
• I guess you are right. Right now I'm looking into it, but I've no experience calculating with floor and ceil in situations like this. – Shuzheng Nov 21 '13 at 11:52
• I think we could prove a lemma that if $a_n \to \infty, b_n \to \infty$, and $\frac{a_n}{b_n} \to L$ as $n \to \infty$, then $\frac{a_n + q_n}{b_n + r_n} \to L$ as $n \to \infty$, assuming the sequences $q_n$ and $r_n$ are bounded. (Or if we like, we can assume $-1 \leq q_n \leq 1, -1 \leq r_n \leq 1$.) – littleO Nov 21 '13 at 11:54
• @littleO : One may just take $\displaystyle b_n= \lfloor\frac{n}{x+1}\rfloor$ and $a_n= n-b_n$. By the result Nicolas Lykke Iversen stated we then have $\displaystyle a_n/b_n= \frac{n}{\lfloor\frac{n}{x+1}\rfloor}-1\to \frac{1}{1/(x+1)}-1=x$ as $n\to \infty$ – Teri Sep 7 '14 at 18:49


text:
# knowrob json_prolog problems

I have got the following errors when I use knowrob/json_prolog.

I tried the examples as in the tutorial:

$rosrun json_prolog json_prolog package_name(tried several) ERROR: /usr/lib/swi-prolog/library/process.pl:47: '$open_shared_object'/3: /usr/lib/swi-prolog/lib/i386/process.so: undefined symbol: Sfilefunctions Warning: /usr/lib/swi-prolog/library/process.pl:47: Goal (directive) failed: process:use_foreign_library(foreign(process)) ERROR: /usr/lib/swi-prolog/library/process.pl:334: Exported procedure process:process_wait/3 is not defined ERROR: /usr/lib/swi-prolog/library/process.pl:334: Exported procedure process:process_kill/2 is not defined

I am using Ubuntu 11.04, Ros electric, installed using apt from the online repository.

Is there any compatibility issue of the swi-prolog with json_prolog/jpl etc?

Thanks.

edit retag close merge delete

Sort by » oldest newest most voted

Hi,

this has been a problem with the SWI Prolog libraries: They are regularly renamed from libpl.so to libswipl.so and back...

It should be fixed in the svn version, can you please verify that it works for you? Then I'll start the release as .deb packages.

Thanks Moritz

more

Following the above question, it seems the problem comes from swi-prolog...

As the error is from /usr/lib/swi-prolog/library/process.pl, I changed it a bit as:

:- module(process,
[ process_create/3,       % +Exe, +Args, +Options
process_wait/2,     % +PID, -Status
process_id/1,       % -PID
process_id/2,       % +Process, -PID
is_process/1,       % +PID
process_release/1,      % +PID
process_kill/1      % +PID
]).
:- use_module(library(shlib)).
:- use_module(library(lists)).
:- use_module(library(option)).
:- use_module(library(process)).
%:- use_foreign_library(foreign(process)).

Some functions are removed from the process list: process_create/2, process_kill/2, etc. Also, use_foreign_library(foreign(process)) caused other errors, so I also commented it out. I am not sure if this makes sense.

But I can get rid of some of the errors. But certainly this is not the ideal way, as other programs also require some missing functions, such as process_create/2, which I removed from the list. I guess this is somehow the problem of the ubuntu version of the swi-prolog, where the process.pl is incompatible.. But as I am not familiar with Prolog, could anyone help?

Many thanks.

more

The problem is not that things have been removed from process.pl, it's rather that you need to have LD_PRELOAD=libswipl.so set. This is being done automatically in the rosjava wrapper script bin/json_prolog, which is generated based on the information in CMakeLists.txt
( 2011-10-18 22:15:30 -0500 )edit
2
Please remove this since it's not an answer. Any important information to the problem should be moved to the original question.
( 2011-11-30 04:07:40 -0500 )edit

Thanks a lot for your quick response. Yes, all the errors are gone now with the svn version.

Ze

more

2
( 2011-11-30 04:07:25 -0500 )edit


text:
# LaTeX error message

I have a hard-to-understand LaTeX error on the R Win-builder test website.

My R package stationery has examples of LaTeX (Sweave or knit) and R markdown documents. The documents all build fine on my systems, but there is one markdown file that fails to compile to PDF on the test server for Windows.

The workflow goes like this. Rmarkdown.Rmd is converted to a markdown file. Which pandoc converts to tex. Then the pdf compiler tries to turn to PDF. On the remote test server, I don't have direct access to all of the processing steps, I just get a check.log file.

This is an entirely new problem, with same files I was fine in past.

Here is the error that need to understand:

-- re-building 'Rmarkdown.Rmd' using rmarkdown
! Undefined control sequence.
\lst@HRefStepCounter ...\begingroup \c@lstlisting
l.197 \begin{lstlisting}

Error: processing vignette 'Rmarkdown.Rmd' failed with diagnostics:
Failed to compile Rmarkdown.tex. See Rmarkdown.log for more info.
--- failed re-building 'Rmarkdown.Rmd'

--- re-building 'code_chunks.Rmd' using rmarkdown
--- finished re-building 'code_chunks.Rmd'

--- re-building 'stationery.Rnw' using Sweave
--- finished re-building 'stationery.Rnw'

SUMMARY: processing the following file failed:
'Rmarkdown.Rmd'

Error: Vignette re-building failed.
Execution halted

Because the build happens on the Windows build server, I don't actually have access to Rmarkdown.log for more details and I'm quite baffled about what is wrong now that was not wrong before.

I don't know which file is being referred to by "l.197", that's not my document Rmarkdown.Rmd for sure.

In the listings.sty file, I do find a stanza with the same words (line 1653 in listings.sty version 1.6, 2015/06/04:

\def\lst@HRefStepCounter#1{%
\begingroup
\c@lstlisting\lst@neglisting


text:
Kramers' Law
Get Kramers' Law essential facts below. View Videos or join the Kramers' Law discussion. Add Kramers' Law to your PopFlock.com topic list for future reference or share this resource on social media.
Kramers' Law

Kramers' law is a formula for the spectral distribution of X-rays produced by an electron hitting a solid target. The formula concerns only bremsstrahlung radiation, not the element specific characteristic radiation. It is named after its discoverer, the Dutch physicist Hendrik Anthony Kramers.[1]

The formula for Kramers' law is usually given as the distribution of intensity (photon count) ${\displaystyle I}$ against the wavelength ${\displaystyle \lambda }$ of the emitted radiation:[2]

${\displaystyle dI(\lambda )=K\left({\frac {\lambda }{\lambda _{min}}}-1\right){\frac {1}{\lambda ^{2}}}d\lambda }$

The constant K is proportional to the atomic number of the target element, and ${\displaystyle \lambda _{min}}$ is the minimum wavelength given by the Duane-Hunt law. The maximum intensity is ${\displaystyle {\frac {K}{4\lambda _{min}^{2}}}d\lambda }$ at ${\displaystyle 2\lambda _{min}}$.

The intensity described above is a particle flux and not an energy flux as can be seen by the fact that the integral over values from ${\displaystyle \lambda _{min}}$ to ${\displaystyle \infty }$ is infinite. However, the integral of the energy flux is finite.

To obtain a simple expression for the energy flux, first change variables from ${\displaystyle \lambda }$ (the wavelength) to ${\displaystyle \omega }$ (the angular frequency) using ${\displaystyle \lambda =2\pi c/\omega }$ and also using ${\displaystyle {\tilde {I}}(\omega )=I(\lambda ){\frac {-d\lambda }{d\omega }}}$. Now ${\displaystyle {\tilde {I}}(\omega )}$ is that quantity which is integrated over ${\displaystyle \omega }$ from 0 to ${\displaystyle \omega _{max}}$ to get the total number (still infinite) of photons, where ${\displaystyle \omega _{max}=2\pi c/\lambda _{min}}$:

${\displaystyle {\tilde {I}}(\omega )={\frac {K}{2\pi c}}\left({\frac {\omega _{max}}{\omega }}-1\right)}$

The energy flux, which we will call ${\displaystyle \psi (\omega )}$ (but which may also be referred to as the "intensity" in conflict with the above name of ${\displaystyle I(\lambda )}$) is obtained by multiplying the above ${\displaystyle {\tilde {I}}}$ by the energy ${\displaystyle \hbar \omega }$:

${\displaystyle \psi (\omega )={\frac {K}{2\pi c}}(\hbar \omega _{max}-\hbar \omega )}$

for ${\displaystyle \omega \leq \omega _{max}}$

${\displaystyle \psi (\omega )=0}$

for ${\displaystyle \omega \geq \omega _{max}}$.

It is a linear function that is zero at the maximum energy ${\displaystyle \hbar \omega _{max}}$.

## References

1. ^ Kramers, H.A. (1923). "On the theory of X-ray absorption and of the continuous X-ray spectrum". Phil. Mag. 46: 836. doi:10.1080/14786442308565244.
2. ^ Laguitton, Daniel; William Parrish (1977). "Experimental Spectral Distribution versus Kramers' Law for Quantitative X-ray Fluorescence by the Fundamental Parameters Method". X-Ray Spectrometry. 6 (4): 201. Bibcode:1977XRS.....6..201L. doi:10.1002/xrs.1300060409.


text:
# Malliavin derivative of a Lebesgue integral.

Let $X_t$ a random process such that its Malliavin derivative is well defined for all $t$. Then I have read that :

$D_s(\int_0^t \! X_u \, \mathrm{d}u)=\int_s^t \! D_s(X_u) \, \mathrm{d}u.$

What I don't understand is why in the second integral, the domain of integration is now $[s,t]$.

I assume that $$X_t$$ is progressively measurable with respect to the corresponding Wiener process $$W_t$$ (otherwise the formula doesn't hold).
$$X_t$$ being $$\sigma(W_s,s\le t)$$-measurable implies that it does not depend on the "tail" $$\{w(v),t as a functional from $$L_2(C([0,T],\mathbb{R}),\text{Wiener measure})$$, $$t. Thus, $$D_hX_t$$ is zero on the $$\{h: \operatorname{supp}(h)\in(t,T]\}$$ and $$D_s(X_t)=0$$ for $$s>t$$, therefore $$D_s\left(\int_0^t X_udu\right)=\int_0^t D_s(X_u) du=\int_s^t D_s(X_u) du.$$


text:
# Wave mechanics of a free particle

Format: Hardcover

Language: English

Format: PDF / Kindle / ePub

Size: 6.80 MB

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Pages: 108

Publisher: Western Periodicals Co (1965)

ISBN: B0007EFMRK

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Wave Propagation and Scattering (The Institute of Mathematics and its Applications Conference Series, New Series)

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Wavefronts and Rays: As Characteristics and Asymptotics

Applications of Noncovariant Gauges in the Algebraic Renormalization Procedure

Vortex Flow in Nature and Technology

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Sheldon Goldstein, a professor of mathematics, physics and philosophy at Rutgers University and a supporter of pilot-wave theory, blames the “preposterous” neglect of the theory on “decades of indoctrination.” At this stage, Goldstein and several others noted, researchers risk their careers by questioning quantum orthodoxy Beyond Einstein's Unified read epub Beyond Einstein's Unified Field: Gravity. Blum, 2005 Controlling chemical reactions by spectral chemistry and spectral conditioning Dirac Spectra in Dense QCD read pdf Dirac Spectra in Dense QCD (Springer. If you do not know the intrinsic parity of hadrons, please open a separate thread (or search in older threads). If you know the intrinsic parity of hadrons, the above information provided by Meir Achuz should suffice to solve your question. Try physics tests on motion, energy, and forces. Take a quiz to test knowledge of electrical circuits using schematic symbols , source: Plasma Waves read here read here. Can you figure out how to make the animation run a one frame per second? Click the 'Code' button again to close this section. ////////// // Start: bouncingBall1 (x, y) graph canvas ////////// var bouncingBall1Canvas; var bouncingBall1GC; // Graphics 2D context bouncingBall1XYGraphInitialize = function() { bouncingBall1Canvas = document.getElementById('bouncingBall1CanvasId'); bouncingBall1GC = bouncingBall1Canvas.getContext('2d'); }; var bouncingBall1WorldXMin = -10; var bouncingBall1WorldXMax = 10; var bouncingBall1WorldYMin = -10; var bouncingBall1WorldYMax = 10; var bouncingBall1DeviceWidth = 400; var bouncingBall1DeviceHeight = 400; var bouncingBall1DeviceHeightPlus1 = bouncingBall1DeviceHeight + 1; var bouncingBall1DeviceWidthPlus1 = bouncingBall1DeviceWidth + 1; var bouncingBall1WorldPerDeviceX = (bouncingBall1WorldXMax - bouncingBall1WorldXMin) / (bouncingBall1DeviceWidth); var bouncingBall1WorldPerDeviceY = (bouncingBall1WorldYMax - bouncingBall1WorldYMin) / (bouncingBall1DeviceHeight); var bouncingBall1DevicePerWorldX = (bouncingBall1DeviceWidth) / (bouncingBall1WorldXMax - bouncingBall1WorldXMin); var bouncingBall1DevicePerWorldY = (bouncingBall1DeviceHeight) / (bouncingBall1WorldYMax - bouncingBall1WorldYMin); var bouncingBall1IsAngleMeasurementInDegrees = true; var bouncingBall1RadiansPerDegree = Math Innovation in Maxwell's download for free office-manual.com. A thin, flexible wire passing through freely rotating studs at the ends of the cranks traces out the shape of the water surface. A set of shorter cranks placed on a lower level illustrate the decrease in the amplitude of the circular motion with increasing depth , source: Propagators For Many Particle http://warholprints.com/library/propagators-for-many-particle. Setting the volume rate of reaction expressed in units of quantity of material per unit time and volume. v = (1/V) x (dx/dt ) where x is the progress of the reaction and V the volume of the solution. Changes in the rate of reaction over time Plasma Waves http://phoenix-web.de/?lib/plasma-waves. Since we are limiting ourselves to plane waves, we can take a unit area perpendicular to the $x$-direction, which is the direction of propagation of the sound wave. The amount of air, per unit area, in $\Delta x$ is then $\rho_0\,\Delta x$, where $\rho_0$ is the undisturbed, or equilibrium, air density Theory and Computation of Electromagnetic Fields http://egyptcancernetwork57357.org/?library/theory-and-computation-of-electromagnetic-fields.

4.3


text:
# RE: should MathML dictate a specific graphical rendering

From: Murray Sargent <murrays@exchange.microsoft.com>
Date: Fri, 8 Apr 2016 22:39:18 +0000
To: Paul Topping <pault@dessci.com>, Daniel Kinzler <daniel@brightbyte.de>, Moritz Schubotz <schubotz@tu-berlin.de>, "www-math@w3.org" <www-math@w3.org>, Peter Krautzberger <peter.krautzberger@mathjax.org>
CC: Wikimedia developers <wikitech-l@lists.wikimedia.org>, wikidata-tech <wikidata-tech@lists.wikimedia.org>
Message-ID: <b321f0f8e55345f696c07a8464c71a41@exchange.microsoft.com>
The LineServices post<https://blogs.msdn.microsoft.com/murrays/2006/11/14/lineservices/> includes a description of an incredible afternoon several of us spent with Donald Knuth back in 2004. Among many things, he demonstrated how he tweaks<https://blogs.msdn.microsoft.com/murrays/2011/04/30/two-math-typography-niceties/> his TeX documents. We did automate some of these tweaks, notably creating “cut-ins” for subscript/superscript positioning. These cut-ins are part of the OpenType math spec<https://blogs.msdn.microsoft.com/murrays/2014/04/27/opentype-math-tables/>. Knuth explained that he didn’t want to go back and change TeX due to its archival usage. Naturally non-math concepts such as revision tracking and embedded objects along with international text had to be accommodated in our implementation. This was another reason for using OMML instead of MathML as the preferred math XML; you can embed other XMLs into OMML. In principle you can do that using the MathML <semantics> element, but it’d be somewhat cumbersome. The Office layout is essentially the same as TeX’s. It’ll be interesting to see how the two compare when the STIX font is finally released with full OpenType math table support. Tyro Typeworks<http://www.tiro.com/> is handling this and it also did Cambria Math.

Murray

From: Paul Topping [mailto:pault@dessci.com]
Sent: Friday, April 8, 2016 3:03 PM
To: Murray Sargent <murrays@exchange.microsoft.com>; Daniel Kinzler <daniel@brightbyte.de>; Moritz Schubotz <schubotz@tu-berlin.de>; www-math@w3.org; Peter Krautzberger <peter.krautzberger@mathjax.org>
Cc: Wikimedia developers <wikitech-l@lists.wikimedia.org>; wikidata-tech <wikidata-tech@lists.wikimedia.org>
Subject: RE: should MathML dictate a specific graphical rendering

Murray,

I guess I forgot about Appendix G of the TeXbook. Thanks for the correction. Did you find that it defined the rendering accurately enough for your line services implementation? Since it has OpenType ‘math’ table support, does it really render exactly as TeX? I guess one could say that the two implementations render the same modulo the fonts used. Did your line services improve on TeX math rendering at all or fill in any gaps in Appendix G. Were there any concessions made to compatibility with layout of non-math text?

I am not asking these questions to argue against your point. I’m just suggesting that while a reader may regard two renderings as being equal, there may still be unavoidable, or by-design, differences due to variations in rendering technology, software environment, and other considerations.

Paul

From: Murray Sargent [mailto:murrays@exchange.microsoft.com]
Sent: Friday, April 8, 2016 2:34 PM
To: Paul Topping <pault@dessci.com<mailto:pault@dessci.com>>; Daniel Kinzler <daniel@brightbyte.de<mailto:daniel@brightbyte.de>>; Moritz Schubotz <schubotz@tu-berlin.de<mailto:schubotz@tu-berlin.de>>; www-math@w3.org<mailto:www-math@w3.org>; Peter Krautzberger <peter.krautzberger@mathjax.org<mailto:peter.krautzberger@mathjax.org>>
Cc: Wikimedia developers <wikitech-l@lists.wikimedia.org<mailto:wikitech-l@lists.wikimedia.org>>; wikidata-tech <wikidata-tech@lists.wikimedia.org<mailto:wikidata-tech@lists.wikimedia.org>>
Subject: RE: should MathML dictate a specific graphical rendering

Paul commented "TeX doesn't specify its rendering in detail either except via the code itself. In other words, the only proper rendering of TeX is that done by TeX itself."

Actually Appendix G of The TeXbook describes how TeX lays out math. The Office math layout program<https://blogs.msdn.microsoft.com/murrays/2006/11/14/lineservices/> uses the algorithms therein, which is why the results look so much like TeX. The actual code is completely different from TeX’s, but the layout principles are generally the same.

It’s good to have this discussion. Clearly Presentation MathML is used a lot for interchanging math zones between programs. Also I haven’t given up on the idea of the browsers rendering MathML well natively. If IE ever succeeds, it’ll likely look like TeX, since both IE and Edge use LineServices<https://blogs.msdn.microsoft.com/murrays/2006/11/14/lineservices/>. And it should be way faster than Java script code.

My main complaints about Presentation MathML are 1) lack of an explicit n-ary element (for integrals, summations, products, etc.) and 2) lack of document level math properties<https://blogs.msdn.microsoft.com/murrays/2008/10/27/default-document-math-properties/>, like default math font. Also Presentation MathML depends too much on proper use of <mrow>, which wouldn’t even be needed if the elements were all “prefix” elements like <mfrac> and <mfenced>. But infix notation can be translated to prefix notation, a good example being conversion of the linear format<http://www.unicode.org/notes/tn28/UTN28-PlainTextMath-v3.pdf> to the OMML<https://blogs.msdn.microsoft.com/murrays/2006/10/06/mathml-and-ecma-math-omml/>-like internal format for LineServices. Similarly RichEdit’s MathML reader converts using the rich-text string stack<https://msdn.microsoft.com/en-us/library/windows/desktop/hh768736(v=vs.85).aspx> originally developed for the linear format.

The bottom line is that MathML isn’t perfect, but it’s a widely used standard and gets the job done. As such, it’s hardly a failure. And it’s nicely supported on the web thanks to MathJax.

Murray

Received on Friday, 8 April 2016 22:39:53 UTC

This archive was generated by hypermail 2.4.0 : Friday, 17 January 2020 22:42:24 UTC


text:
Solid State Drive (SSD) is a technology which brings a few times better performance compared to enterprise SAS drives and we at Nordeus are heavily using it with Postgres. SSDs especially shine in random page reads/writes compared to regular drives, because they have a higher number of input/output operations per seconds (IOPS). For example best 15k RPM SAS drives can achieve about 200 IOPS, while SSDs can do 75 000 reads and 15 000 writes. This is one of the main reasons why we run Postgres on SSDs, because our applications’ workloads are heavy on random reads/writes which SAS drives simply cannot handle, not even if they are put into RAID configurations.

This performance comes at a cost of SSDs having a few times smaller capacities, higher cost, and a shorter life time since flash cells have a limited number of writes. SSDs are also tied to being more unreliable than regular drives, even though new enterprise ready SSDs are highly reliable, which means reliability shouldn’t be a problem, but they also cost a lot more.

Enough about SSDs in general, let’s get back to the title of this article and why we started doing power failure tests with our SSDs. In the past 2 years we had power failures on our servers – they are not that often, but they happen. Usually these are not a problem, but a few times we noticed corruption of indexes in Postgres instances, and on a few of them even unrecoverable corruption. Of course we had backups, but this shouldn’t happen. Also a few months ago we had a big migration of hundreds of servers from one data center to another after which we also noticed data and index corruption.

The most interesting thing is that these servers were normally powered off before they were migrated, but we still had problems. That’s when we started to do some investigation and one thing we noticed is that only particular SSD models were giving us problems, while others didn’t. Here is a list of models we used in our production:

SSD Model Detected Issues
Intel 320 No issues
Intel 330 Database data/index corruption
Intel 335 Database data/index corruption
Intel 520 Database data/index corruption
Samsung 840 Pro No issues
I would like to note that all of these drives are relatively old and when we started using them on our rented servers more than 5 years ago we didn’t have an option to choose SSD models, we used what was available, but we also didn’t have enough resources to think about SSDs at that time. Later on when the number of servers grew and we started having some problems, we had enough resources to investigate SSDs and to choose what exactly do we want to use.

The order in which the drives are listed, is the order in which they were added to our servers, with the Intel 320 being the oldest and Samsung 840 being the newest drive. As you can see we noticed that Intel 330, 335 and 520 were the ones causing problems, and when we compared the specs to the other two models that isn’t a coincidence at all. Those three models are consumer SSDs which have no enterprise ready features to make them reliable in case of a power failure and in many other situations. On the other hand Intel 320 is one of the rare consumer oriented SSDs which has supercapacitors (SC), which allows the SSD to survive power failures. SC keeps just enough power in it to flush everything from SSD cache to flash in case of a power failure. Samsung 840 Pro doesn’t have SC, but we didn’t have any problems with it, at least for now, which of course doesn’t mean it is safe.

What was even more interesting is that our SSDs are connected to the Dell H700/H710 RAID controller which has a battery backup unit (BBU) which should make our drives power failure resilient. RAID controller with BBU in case of a power failure can hold the cached data until the power comes back, so that it can flush it to the drives when the drives come back online. The problem is that with most consumer SSDs you can never be 100% sure an SSD has finished the write and flushed the data from its cache to the drive. Because of this even if the SSD is connected to a RAID controller with a BBU, it is still not reliable if it doesn’t have its own power loss protection.

# Testing Models

Since we wanted to fix this problem once and for all, we first needed to it in a test environment. We decided to use the following three models of SSDs:

• Intel 520 – We know it failed in production, so we wanted to reproduce this in a test environment.
• Samsung 840 Pro – Since it never failed in production, we wanted to test how resilient it really is.
• Intel DC S3500 – This is a new enterprise ready SSD which should be extremely resilient and durable. We use it in all our new servers.
The reason why we didn’t test Intel 320 is because that is a very old model and we won’t be using it in our future servers. 330/335 were not tested, because we were sure if 520 failed, they would fail as well.

This testing was done in December 2014, not to test if Intel DC S3500 is better than the rest, we knew that already, since it’s a newer drive and we tested it’s performance and reliability before, which is why we have been using it in our servers since 2013. The reason why we did this testing is that we could see how bad our old drives really are and that we know in the future how to test new drives.

# Testing Configuration

Here is the testing configuration we used:

• Dell R420
• H710p RAID controller
• 2 x SSD (models above)
• 48GB RAM
• CPU and other things are not important for this test.

CentOS 6.5 was used, SSDs were formatted with XFS and they were mounted into /mnt/ssd1 and /mnt/ssd2. XFS was used because that is our main file system for databases. Reason why we chose XFS is not just because XFS is awesome, but it’s also because when we started a few years ago we were using CentOS 5 at that time, which had ext3 as the main FS and no ext4 support, so XFS was a no-brainer at that time, because of much better performance compared to ext3.

# The Tests

We used two types of tests to determine if the drives are working correctly in case of a power failure:

1. fsync test
2. pgbench with Postgres 9.3 with data checksum

Before I explain these tests, there are two important things you need to know. What is fsync and what are write barriers?

What is fsync?

In a nutshell fsync is a system call which flushes file system buffers/cache to the disk. It is heavily used in databases when a database needs to make sure the data has made it to the disk. This doesn’t mean the data was really written to disk, it can still remain in the disk cache, but enterprise drives usually make sure the data was really written to disk on fsync calls. If you want to learn more about fsync, I would recommend that you read Everything You Always Wanted to Know About Fsync().

What Are Write Barriers?

I won’t go into much detail what write barriers are, but in short they make sure that the file system write cache is written in the correct order to the disk. This means when barriers are enabled on a file system (they usually are by default on modern file systems), and you call fsync to flush a particular file, fsync will first flush all the file system cache before that file and then flush the file, keeping the order of writing as it was in the file system cache. This gives a big writing performance penalty especially to applications which use fsync frequently (like databases). Barriers are especially important in case of a power failure, so that there is no data corruption. If write barriers are disabled, files system cache can be written to the disk in a different order, making the file system vulnerable to power failures and corruption. Only time it is considered safe to turn off write barriers is if you have a RAID controller with BBU and if your SSD has power loss protection like supercapacitors. If you would like to read more about write barriers I would suggest Barriers, Caches, Filesystems.

## How Did We Test?

Both tests were done with 3 different parameters changed each time:

• Disk cache on/off – Disabling disk cache reduces disk performance, but also decreases the life time of the SSD.
• Write barriers on/off – As explained above, turning off write barriers increases performance, but if the hardware doesn’t have proper protection, it can cause data corruption.
• RAID controller write-back/write-through – In write-back mode RAID controller’s cache is used to cache writes and make them faster, while in write-through mode, the cache is not used and the data is written directly to disks, which again reduces performance.

Each test that was run was repeated at least 5 times with a different combination of these 3 parameters, and it was stopped the first time it failed. SSD needed to pass the test at least 5 times to be considered as passed, but the majority of the passed tests were repeated a lot more than 5 times, just in case.

## Fsync Test

For the fsync test we decided to use a well know diskchecker.pl script (more details). For this script to work, you need two hosts, a server and a client. In short this script writes into a file on the host you are testing, calls fsync after each write and sends the information to the server host what it actually had written. During the test you pull the plug off, wait a few minutes, plug the server back on and re-run the script to check what was written and what wasn’t. Also we needed to modify the script a little bit since it didn’t support running two instances of the server on the same host.

Here is the full testing procedure:

On the server (none-testing host) run two instances of the diskchecker utility, since we will be testing two SSDs in parallel on the same host:

./diskchecker.pl -l 1212
./diskchecker.pl -l 1213

On the client (testing host) run:

./diskchecker.pl -s SERVER_IP:1212 create /mnt/ssd1/first_test 51200
./diskchecker.pl -s SERVER_IP:1213 create /mnt/ssd2/first_test 51200

Unplug power on the testing server after 5 minutes, power the host back on after 10 minutes and run verify step:

./diskchecker.pl -s SERVER_IP:1212 verify /mnt/ssd1/first_test
./diskchecker.pl -s SERVER_IP:1213 verify /mnt/ssd2/first_test

Clean everything up and run the test again at least 5 more times.

### Results

Model Disk Cache Barrier Write Mode Disk 1 Disk 2
Intel S3500 On Off Write-back OK OK
Samsung 840 Pro On Off Write-back FAIL FAIL
Off Off Write-back OK OK
Off On Write-back OK OK
Intel 520 On Off Write-back FAIL FAIL
Off On Write-back FAIL FAIL
On On Write-trough FAIL FAIL
Off On Write-trough FAIL FAIL

As we can see from the results Intel S3500 works without any problems with all parameters configured for best performance, which is the reason why we didn’t do any testing of other combinations, since they would give us lower performance. Samsung 840 Pro failed in the first test, but didn’t report any problems with disk cache off. On the other hand we haven’t managed to make Intel 520 pass any combination of settings, even with disabling RAID controller cache (write-through), disk cached turned off and barriers enabled, it still managed to fail the tests. I would like to note that we didn’t test Intel 520 with all possible combinations of parameters, since if the last combination (which should be the safest, but with lowest performance) failed, all of them will.

## pgbench Test

The idea behind the pgbench test is to do a lot of writing into the Postgres database and during this writing to unplug the power cord. To verify that the data written into the database is correct, we will be using data checksums which were introduced into Postgres from version 9.3. The verification will be done by running pg_dumpall, since all the data will be checked against a check-sum on each read. This is how we tested:

Install PostgreSQL 9.3 from the official repositories

Create needed directories on SSDs and link them to be used as a default Postgres data directory:

mkdir -p /mnt/ssd1/pgsql/9.3/data
mkdir -p /mnt/ssd2/pgsql/9.3/index
chown postgres:postgres /mnt/ssd1/pgsql/9.3/data
chown postgres:postgres /mnt/ssd2/pgsql/9.3/index
rm -rvf /var/lib/pgsql/9.3/*
ln -s /mnt/ssd1/pgsql/9.3/data /var/lib/pgsql/9.3/data
ln -s /mnt/ssd2/pgsql/9.3/index /var/lib/pgsql/9.3/index

Initialized the Postgres cluster on SSD1 with data checksums enabled and start postgres:

runuser -l postgres -c "/usr/pgsql-9.3/bin/initdb \
--pgdata=/var/lib/pgsql/9.3/data \
--auth='trust' \
--data-checksums"
service postgresql-9.3 start

Create pgbench database, add table space index on SSD2 and initialize the pgbench database with indices moved to index table space (this way both SSDs will be used while testing):

psql -U postgres -c "CREATE DATABASE pgbench;"
psql -U postgres -c "CREATE TABLESPACE index LOCATION '/var/lib/pgsql/9.3/index';"
/usr/pgsql-9.3/bin/pgbench -U postgres -i pgbench --index-tablespace=index
# Create additional indexes so there is more writing on SSD2
psql -U postgres -d pgbench -c "CREATE INDEX ON pgbench_accounts(abalance,aid) TABLESPACE index;"
psql -U postgres -d pgbench -c "CREATE INDEX ON pgbench_history(mtime) TABLESPACE index;"
psql -U postgres -d pgbench -c "CREATE INDEX ON pgbench_history(aid,delta) TABLESPACE index;"

Run pgbench for a few minutes, e.g:

/usr/pgsql-9.3/bin/pgbench -U postgres -c 12 -T 3600 pgbench

Unplug the power, wait 10 minutes, and power on the server.

Run pg_dumpall (just to be sure) and check for any errors:

/usr/pgsql-9.3/bin/pg_dumpall -U postgres > /dev/null

Destroy the Postgres instance and run everything again.

### Results

Model Disk Cache Barrier Write Mode Postgres
Intel S3500 On Off Write-back OK
Samsung 840 Pro On Off Write-back OK
Intel 520 On Off Write-back FAIL
On On Write-back FAIL
Off Off Write-back OK
Off On Write-back OK

Like in the previous test, Intel S3500 worked without any problems with default settings for best performance, but this time Samsung 840 Pro also didn’t cause any problems. While on Intel 520 Postgres was reporting errors until we disabled disk cache, but this doesn’t make it safe, only less prone to corruption. Probably the reason why Intel 520 “survived” with disk cache disabled is because Postgres is executing fsync less often than diskchecker.pl, so the probability of a failed fsync is smaller.

# Conclusion

As we can see from the results, Intel S3500 is an enterprise ready drive which works reliably out of the box and we would recommend it to everyone who needs a reliable, high performance drive which isn’t very expensive. On the other hand Samsung 840 Pro looks to be doing pretty good with the disk cache disabled in the fsync test, and in the Postgres test it even managed to pass with disk cache enabled. I would still not recommend it on mission critical databases, but if you have no choice, at least disable disk cache (and test performance). I’m not exactly sure what to say about Intel 520 except that this test showed all the concerns that we had, that it is not reliable for production and that it should probably only be used in consumer devices.

Finally, never believe what the hardware vendors and/or the internet is saying. If you have a RAID controller with BBU, it doesn’t mean your data is safe when you lose power, it depends on many factors like the quality of the controller, quality of the drives, type of drives, etc, as you can see from this article. If you want to be sure your data is safe, it’s best that you test it yourself.


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# Tag Info

2

I will address the first two parts based on what I understood so far. The extended Church–Turing thesis or (classical) complexity-theoretic Church–Turing thesis states that "A probabilistic Turing machine can efficiently simulate any realistic model of computation.", whereas the quantum extended Church–Turing thesis or quantum complexity-theoretic Church–...

2

Taking the questions head on. I'm not sure that original references are very much the point, although there are some. It's not a hard question. The statement is that realistic polynomial time equals what a quantum computer (if you want to be rigorous, say a QTM) can do in polynomial time. The question has been answered many times in QCSE that a quantum ...

1

Regarding the "quantum (non-extended) Church-Turing Thesis," I think this asserts that there is no physical process, like a quasar or some other astronomical woo, that we know could produce a steady supply of qubits all in the same state $\alpha|0\rangle+\beta|1\rangle$, with the property that $\beta^2=\Omega_C$, that is, Chaitin's halting probability. We ...

Only top voted, non community-wiki answers of a minimum length are eligible


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SOLVING ABSOLUTE VALUE SENTENCES, ALL TYPES
THEOREM solving absolute value sentences
Let $\,x\in\mathbb{R}\,$, and let $\,k\ge 0\,$.   Then, $$\begin{gather} |x| = k\ \ \text{ is equivalent to }\ \ x = \pm k \\ \\ |x| \lt k\ \ \ \text{ is equivalent to }\ \ -k \lt x \lt k \\ |x| \le k\ \ \ \text{ is equivalent to }\ \ -k \le x \le k \\ \\ |x| \gt k\ \ \ \text{ is equivalent to }\ \ x\lt -k\ \ \text{ or }\ \ x\gt k \\ |x| \ge k\ \ \ \text{ is equivalent to }\ \ x\le -k\ \ \text{ or }\ \ x\ge k \end{gather}$$
EXAMPLE (an absolute value equation):
Solve: $|2 - 3x| = 7$
Solution:
Write a nice, clean list of equivalent sentences:
 $|2 - 3x| = 7$ (original equation) $2-3x = \pm 7$ (check that $\,k\ge 0\,$; use the theorem) $2-3x = 7\ \text{ or }\ 2-3x = -7$ (expand the plus/minus) $-3x = 5\ \text{ or }\ -3x = -9$ (subtract $\,2\,$ from both sides of both equations) $\displaystyle x = -\frac{5}{3}\ \text{ or } x = 3$ (divide both sides of both equations by $\,-3\,$)
It's a good idea to check your solutions:

$|2 - 3(-\frac{5}{3})|\ \overset{\text{?}}{=}\ 7$,     $|2 + 5| = 7$     Check!

$|2 - 3(3)|\ \overset{\text{?}}{=}\ 7$,     $|2 - 9| = 7$     Check!
EXAMPLE (an absolute value inequality involving ‘less than’):
Solve: $3|-6x + 7| \le 9$
Solution:
To use the theorem, you must have the absolute value all by itself on one side of the equation.
Thus, your first job is to isolate the absolute value:
 $3|-6x + 7| \le 9$ (original sentence) $|-6x + 7| \le 3$ (divide both sides by $\,3$) $-3 \le -6x + 7 \le 3$ (check that $\,k \ge 0\,$; use the theorem) $-10 \le -6x \le -4$ (subtract $\,7\,$ from all three parts of the compound inequality) $\displaystyle \frac{10}{6} \ge x \ge \frac{4}{6}$ (divide all three parts by $\,-6\,$; change direction of inequality symbols) $\displaystyle \frac{2}{3} \le x \le \frac{5}{3}$ (simplify fractions; write in the conventional way)
Check the ‘boundaries’ of the solution set:
$3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$     Check!
$3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$     Check!
EXAMPLE (an absolute value inequality involving ‘greater than’):
Solve: $3|-6x + 7| \ge 9$
Solution:
To use the theorem, you must have the absolute value all by itself on one side of the equation.
Thus, your first job is to isolate the absolute value:
 $3|-6x + 7| \ge 9$ (original sentence) $|-6x + 7| \ge 3$ (divide both sides by $\,3$) $-6x + 7 \le -3\ \ \text{or}\ \ -6x + 7\ge 3$ (check that $\,k \ge 0\,$; use the theorem) $-6x\le -10\ \ \text{or}\ \ -6x\ge -4$ (subtract $\,7\,$ from both sides of both subsentences) $\displaystyle x\ge\frac{10}{6}\ \ \text{or}\ \ x\le \frac{4}{6}$ (divide by $\,-6\,$; change direction of inequality symbols) $\displaystyle x\ge\frac{5}{3}\ \ \text{or}\ \ x\le \frac{2}{3}$ (simplify fractions) $\displaystyle x\le \frac{2}{3}\ \ \text{or}\ \ x\ge\frac{5}{3}$ (in the web exercise, the ‘less than’ part is always reported first)
Check the ‘boundaries’ of the solution set:
$3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$     Check!
$3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$     Check!
EXAMPLE (an absolute value equation that is always false):
Solve: $|3x + 1| = -5$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative?   NO!
No matter what number you substitute for $\,x\,$,
the left-hand side of the equation will always be a number that is greater than or equal to zero.
Therefore, this sentence has no solutions. It is always false.
EXAMPLE (an absolute value inequality that is always false):
Solve: $|5 - 2x| \lt -3$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative?   NO!
No matter what number you substitute for $\,x\,$,
the left-hand side of the inequality will always be a number that is greater than or equal to zero,
so it can't possibly be less than $\,-3\,$.
Therefore, this sentence has no solutions. It is always false.
EXAMPLE (an absolute value inequality that is always true):
Solve: $|5 - 2x| \gt -3$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
No matter what number you substitute for $\,x\,$,
the left-hand side of the inequality will always be a number that is greater than or equal to zero,
so it will always be greater than $\,-3\,$.
Therefore, this sentence has all real numbers as solutions. It is always true.
Master the ideas from this section

When you're done practicing, move on to:
Identifying Perfect Squares

For example, the inequality $\,|2 - 3x| \lt 7\,$
is optionally accompanied by the graph of $\,y = |2 - 3x|\,$ (the left side of the inequality, dashed green)
and the graph of $\,y = 7\,$ (the right side of the inequality, solid purple).
In this example, you are finding the values of $\,x\,$ where the green graph lies below the purple graph.


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# How to handle parallel ports in win 7 64bit?

I am trying to use an application that is sending some data to LPT1 and it is working well on winXP but on win7(64bit) i get this exception

The system cannot find the file specified. (Exception from HRESULT: 0x80070002)

Can i communicate with parallel ports on win7 64bit?

-


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# Showing convergence of $\sum_{n=1}^{\infty} {(-1)^{n} \frac{\sin^{2} n}{n}}$

I've tried to prove the convergence of $$\sum_{n=1}^{\infty} {(-1)^{n} \frac{\sin^{2} n}{n}}$$ using three different criteria but ran into the following issues:

$$1.$$ I first tried the Leibnitz criterion.

Let $$a_n = \displaystyle{\frac{\sin^{2} n}{n}}$$. Though $$\lim_{n\to \infty} \frac{\sin^{2} n}{n} = 0$$ the sequence $$a_n$$ is not monotone thus the test fails.

$$2.$$ Next I tried Dirichlet's test. I set $$\displaystyle{ {a_n} = \frac{1}{n}}$$ which monotonically decreases and has a limit of zero. But with $$\displaystyle{{b_n} = (-1)^n \sin^2{n}}$$ the partial sums $$\displaystyle{\left|\sum_{n=1}^{m}b_n\right|}$$ are not bounded. Same issue with setting $$\displaystyle{ a_n = \frac{(-1)^n}{n}}$$ and $$\displaystyle{ b_n = \sin^2 n }$$.

$$3.$$ The last criterion I know is Abel's test. I'm stuck on similar points as with Dirichlet's test. If I set $$\displaystyle{ a_n = \frac{\sin^{2} n}{n} }$$, then $$\displaystyle{ \sum {a_n} }$$ is convergent, but what remains $$\displaystyle{ (-1)^n }$$ is bounded but not monotone.

If $$\displaystyle{ a_n = \frac{(-1)^n}{n} }$$ then $$\displaystyle{ \sum {a_n} }$$ is convergent, but $$\displaystyle{ \sin^2 n }$$ is not monotone.

What other tests can I apply? Or perhaps this shows that the series is not convergent?

• what about if you use the double angle formula and express the sin square in terms of 1 and cos and see if you can apply the usual stuff as above? – Conrad Feb 10 at 15:58

$$\sum_{n=1}^{m} (-1)^{n} \frac{\sin^{2} n}{n} = \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} - \sum_{n=1}^{m}\frac{ (-1)^{n}\cos 2n}{2n} \\= \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} - \sum_{n=1}^{m}\frac{ \,\cos (\pi+2)n}{2n}$$
The partial sums on the RHS are both convergent as $$m \to \infty$$.
• $\left|\sum_{n=1}^m \cos \alpha n \right| \leqslant \frac{1}{\sin(\alpha/2)}$ if $\alpha$ is not an integer multiple of $2\pi$. – RRL Feb 10 at 16:16


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# 3.3.4 - The Empirical Rule

3.3.4 - The Empirical Rule

The Empirical Rule is sometimes referred to as the 68-95-99.7% Rule. The rule is a statement about normal or bell-shaped distributions.

Empirical Rule

In any normal or bell-shaped distribution, roughly...

• 68% of the observations lie within one standard deviation to either side of the mean.
• 95% of the observations lie within two standard deviations to either side of the mean.
• 99.7% of the observations lie within three standard deviations to either side of the mean.
Note! Students tend to use these approximation instead of the more precise values found in the tables or by using software. The empirical rule should be used as a quick estimate. The more precise values should be used when possible.

## Try It!

Use the normal table to validate the empirical rule. In other words, find the exact probabilities $P(-1<Z<1)$, $P(-2<Z<2)$, and $P(-3<Z<3)$ using the normal table and compare the values to those from the empirical rule.

$P(-1<Z<1)= P(Z<1)-P(Z<-1) = .8413 - .1587 \approx .68$

$P(-2<Z<2)= P(Z<2)-P(Z<-2) = .9772 - .0228 \approx .95$

$P(-3<Z<3)= P(Z<3)-P(Z<-3) = .9987 - .0013 \approx .99.7$

 [1] Link ↥ Has Tooltip/Popover Toggleable Visibility


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Error when installing Rgraphviz on Centos
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Entering edit mode
Last seen 24 months ago

I am running R/3.6.0 on a multi-user Centos 7.6 system and have not been able to install Rgraphviz in my account.

I have downloaded the tarball, Rgraphviz2.28.0.tar.gz, and am using the command: R CMD INSTALL Rgraphviz2.28.0.tar.gz

The error message that I am getting is this:

In file included from LL_funcs.c:1:0:
common.h:22:8: error: unknown type name ‘Agdict_t’
extern Agdict_t *agdictof(void *);

I also have tried expanding the tarball and running configure manually. I get the following message:

In file included from LL_funcs.c:1:0:
common.h:22:8: error: unknown type name ‘Agdict_t’
extern Agdict_t *agdictof(void *);
^~~~~~~~
LL_funcs.c: In function ‘getClusterPtr’:
LL_funcs.c:24:20: warning: implicit declaration of function ‘agfindsubg’; did you mean  agidsubg’? [-Wimplicit-function-declaration]
Agraph_t *sg = agfindsubg(g, subGName);
^~~~~~~~~~
agidsubg
LL_funcs.c:24:20: warning: initialization makes pointer from integer without a cast [-Wint-conversion]
LL_funcs.c: In function ‘getDefAttrs’:
LL_funcs.c:31:5: error: unknown type name ‘Agdict_t’; did you mean ‘Agdisc_t’?
Agdict_t *dict = agdictof(obj);

Any suggestions on how to fix this are greatly appreciated.

Rgraphviz Installation Error • 417 views
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Entering edit mode

The tag you used (software error) is too general and the Rgraphviz maintainers won't necessarily pay attention to a post with such a general tag. Since this is about an installation error with the Rgraphviz package, I'd suggest you tag your post with Rgraphviz.

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Thank you -- I've updated the tags.

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Thank you -- I've updated the tags.

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Thank you -- I've updated the tags.

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Do you have graphviz installed on your system? (e.g. https://centos.pkgs.org/7/centos-x86_64/graphviz-devel-2.30.1-21.el7.x86_64.rpm.html )

It's listed as an 'optional requirement' on the Rgraphviz homepage, but maybe it's not so optional on CentOS?

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Yes, we have graphviz and graphviz-devel installed on our system. In the newer versions of Rgraphviz, one has the option of using the graphviz code that comes bundled with Rgraphviz or of using the system version. I have tried both options and get the same error message either way.

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We are using gcc/7.1.0 to build the package. We suspect that the error is with using a header file in that version of the compiler. My colleague was able to install the package on a stand-alone machine using gcc/4.8.5 and intel/18.0.


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# StyleSheet for Show StyleSheets from custom applicationdirectory in Format / StyleSheets menu

I'm working on a package that for some reasons needs to be installed into aan arbitrary, "non standard" path (i.e., that means neithernot $BaseDirectory or $UserBaseDirectory). I have added the package's path to the variable $Path so to have the possibilityin order to use Get/Needsallow Mathematica to load the package via (and itGet or Needs, and this works) well. The My problem is that, inside that the package, I also have aneeds to include custom palettestylesheets and stylesheetcustom palettes, and theythese are not visible fromavailable in the menufront-end menus (Format->StylesheetFormat-->Stylesheet and PalettesPalettes, respectively). On when the tutorial/WorkingWithStylesheets therepackage is loaded from a non-standard location. If I put the following sentencepackage files in $BaseDirectory the custom stylesheet and palette are accessible from the front-end menus, since the package follows the standard structure:

• package\FrontEnd\StyleSheets\mystyle.nb

• package\FrontEnd\Palettes\mypalette.nb

but this doesn't seem to work if it is loaded from an arbitrary path.

This Wolfram tutorial says that:

Stylesheet are discovered by the front end in the following locations.

Stylesheet are discovered by the front end in the following locations.

FrontEndFileName[{$UserBaseDirectory,"SystemFiles","FrontEnd","StyleSheets"}] FrontEndFileName[{$InstallationDirectory,"SystemFiles", "FrontEnd","StyleSheets"}]

Stylesheets contained in directories are grouped under submenus in the Stylesheet menu.

So, I thought it was possible to have stylesheet accessible from my own package, being and "Add-on application" that follows the standard structure, e.g.:

package\FrontEnd\StyleSheets\mystyle.nb package\FrontEnd\Palettes\mypalette.nb

Of course, if I move the package under $BaseDirectory everything works fine. So, the question: ifHow can I install a package in a directory that's not $UserBaseDirectory or $BaseDirectory, areand still make the stylesheets and palettes still visibleavailable in the Mathematica menu? If so, how toMathematica menus? I'm not able to have them. Thanks and regards, Roberto I'm working on a package that for some reasons needs to be installed into a "non standard" path, that means neither $BaseDirectory or $UserBaseDirectory I added the package's path to the variable $Path so to have the possibility to use Get/Needs to load the package (and it works). The problem is that, inside that package, I also have a custom palette and stylesheet and they are not visible from the menu (Format->Stylesheet and Palettes, respectively). On the tutorial/WorkingWithStylesheets there is the following sentence:

Stylesheet are discovered by the front end in the following locations.

FrontEndFileName[{$UserBaseDirectory,"SystemFiles","FrontEnd","StyleSheets"}] FrontEndFileName[{$InstallationDirectory,"SystemFiles", "FrontEnd","StyleSheets"}]

Stylesheets contained in directories are grouped under submenus in the Stylesheet menu.

So, I thought it was possible to have stylesheet accessible from my own package, being and "Add-on application" that follows the standard structure, e.g.:

package\FrontEnd\StyleSheets\mystyle.nb package\FrontEnd\Palettes\mypalette.nb

Of course, if I move the package under $BaseDirectory everything works fine. So, the question: if I install a package in a directory that's not $UserBaseDirectory or $BaseDirectory, are stylesheets and palettes still visible in the Mathematica menu? If so, how to? I'm not able to have them. Thanks and regards, Roberto I'm working on a package that needs to be installed into an arbitrary, "non standard" path (i.e., not $BaseDirectory or $UserBaseDirectory). I have added the package's path to $Path in order to allow Mathematica to load the package via Get or Needs, and this works well.

My problem is that the package also needs to include custom stylesheets and custom palettes, and these are not available in the front-end menus (Format-->Stylesheet and Palettes, respectively) when the package is loaded from a non-standard location.

If I put the package files in $BaseDirectory the custom stylesheet and palette are accessible from the front-end menus, since the package follows the standard structure: • package\FrontEnd\StyleSheets\mystyle.nb • package\FrontEnd\Palettes\mypalette.nb but this doesn't seem to work if it is loaded from an arbitrary path. This Wolfram tutorial says that: Stylesheet are discovered by the front end in the following locations. FrontEndFileName[{$UserBaseDirectory,"SystemFiles","FrontEnd","StyleSheets"}]

FrontEndFileName[{$InstallationDirectory,"SystemFiles","FrontEnd","StyleSheets"}]  Stylesheets contained in directories are grouped under submenus in the Stylesheet menu. Add-on applications following the same layout also add stylesheets to the menu. How can I install a package in a directory that's not $UserBaseDirectory or \$BaseDirectory, and still make the stylesheets and palettes available in the Mathematica menus?

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# Mathematical Sciences Research Institute

Home » The Inverse Calderon Problem for Schrödinger Operator on Riemann Surfaces

# Seminar

The Inverse Calderon Problem for Schrödinger Operator on Riemann Surfaces September 28, 2010
Location: MSRI: Simons Auditorium
Speaker(s) Leo tzou
Description No Description

Video
Abstract/Media

We show that on a smooth compact Riemann surface with boundary (M0, g) the Dirichletto-
Neumann map of the Schrödinger operator â g + V determines uniquely the potential V .
This seemingly analytical problem turns out to have connections with ideas in symplectic
geometry and differential topology. We will discuss how these geometrical features arise and
the techniques we use to treat them.

This is joint work with Colin Guillarmou of CNRS Nice. The speaker is partially supported
by NSF Grant No. DMS-0807502 during this work.