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Trigonometry. You write 3/8 = x/360 cross multiply: 8x = 3 • 360 = 1080 so x = 1080 ÷ 8 = 135º With a decimal you do the same thing basically. The reciprocal of a number a is simply. α (degrees) = α (radians) × 180° / π = 2 × 180° / 3.14159 = 114.592°. Sometimes, slope is referred to as pitch, angle, rise, incline, or grade. The least common multiple is the first shared multiple of these three numbers. Mixed Fractions. Calculating the degree of a polynomial with symbolic coefficients This is a fraction calculator … This quick and easy calculator will let you convert degrees to radians at the click of a button. This would make the fraction, Similarly, fractions with denominators that are powers of 10 (or can be converted to powers of 10) can be translated to decimal form using the same principles. Example #2 The calculator may be used to determine the degree of a polynomial. For a repeating decimal such as 0.66666... where the 6 repeats forever, enter 0.6 and since the 6 is the only one trailing decimal place that repeats, enter 1 for decimal places to repeat. Find the greatest common divisor (gcd) of the numerator and the denominator: gcd(32,100) = 4. Step 4: Simplify the remaining fraction to a mixed number fraction if possible. It’s much easier to know what the angle looks like in this form. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Calculator Use. This calculator converts a decimal number to a fraction or a decimal number to a mixed number. Find the Greatest Common Factor (GCF) of the numerator and denominator and divide both numerator and denominator by the GCF. To convert, use this percent to degrees formula: Degrees = Tan -1 (Slope Percent/100) Example: With a slope percentage of 43%, degrees would be: Degrees = Tan -1 (.43) Calculated out this gives an angle of 23.27 degrees. Example: An angle of 48 degrees would be: Percentage of Slope = [ Tan ( 48 ) ] x 100. Reduce the fraction. 125 People UsedView all course ››. Multiplying fractions is fairly straightforward. To convert degrees to radians, take the number of degrees to be converted and multiply it by π/180. The calculator will generate a step by step explanations. radians = degrees × π 180In other words, the angle in radians is equal to the degrees times pi, divided by 180. You can calculate this by converting both numbers into fractions. π × 205÷5/180÷5 = 41π/36 radians, when reduced to lowest terms.. Use this calculator if the numerators or denominators are very big integers. Hotwww.inchcalculator.com. For example, to convert 120 degrees you would have 120 x π/180 = 120π/180. Cite, Link, or Reference This Page. A slope percentage is the ratio of the change in altitude between two points to the horizontal distance between the points, multiplied by 100. Refer to the addition section as well as the equations below for clarification. 48/100 result occurred by dividing percent value with 100. Type a math problem. The calculator provided returns fraction inputs in both improper fraction form, as well as … Below is an example using this method. Equation 1: 2. Reduce the fraction by dividing the numerator and the denominator with the gcd: 0.32 = (32/4) / (100/4) = 8/25. 1. There are 3 digits in the repeating decimal group, so y = 3. Create an equation such that x equals the decimal number. To state the decimal 0.456 degrees in … To obtain the degree of a polynomial defined by the following expression x 3 + x 2 + 1, enter : degree (x 3 + x 2 + 1) after calculation, the result 3 is returned. For repeating decimals enter how many decimal places in your decimal number repeat. Check the chart for more details. To use this formula, start by adding the degrees to the formula. Circles and spheres have 360 degrees in a complete rotation. Convert 0.32 to fraction: 0.32 = 32/100. This calculator requires the use of Javascript enabled and capable browsers. Using the least common multiple can be more efficient and is more likely result in a fraction in simplified form. Is it 1/2 of a circle, semicircle, etc. To convert, use this angle degrees to percent of slope formula: Percentage = [ Tan ( Degrees ) ] x 100. One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Wikipedia contributors. Convert degrees to radians (° to rad) with the angle conversion calculator, and learn the degree to radian calculation formula. Example #1. In engineering, fractions are widely used to describe the size of components such as pipes and bolts. An alternative method for finding a common denominator is to determine the least common multiple (LCM) for the denominators, then add or subtract the numerators as one would an integer. It turns out that they are the […] Create a second equation multiplying both sides of the first equation by 10. Simply, the numerators and denominators of each fraction are multiplied, and the result forms a new numerator and denominator. A common denominator is required for the operation to occur. Converting from decimals to fractions is straightforward. This process can be used for any number of fractions. In some situations, it becomes mandatory to convert radians into degrees. ... Matrix Calculator. "Repeating Decimal," Wikipedia, The Free Encyclopedia. For example, let's convert 5 degrees ... www.inchcalculator.com Input proper or improper fractions, select the math sign and click Calculate. For example, looking at the number 0.1234, the number 4 is in the fourth decimal place which constitutes 104, or 10,000. Subtract the second equation from the first equation. For example, here's how to convert 5 degrees to radians using the formula above. Plugging the angle value, in degrees, in the previous formula, we get: α rad = π × 74 /180 = . The fraction to decimal chart is useful in finding the size of screws. Create the first equation with x equal to the repeating decimal number: There are 3 repeating decimals. (This only works with terminating decimals.) 1 - Enter the size of the angles A1 and A2 in degrees, minutes and seconds (DMS) form, check your input and press "Add / Subtract". As such, fraction solutions are commonly expressed in their simplified forms. Fraction subtraction is essentially the same as fraction addition. One degree is equal 0.01745329252 radians: 1° = π/180° = 0.005555556π = 0.01745329252 rad The angle α in radians is equal to the angle α in degrees times pi constant divided by 180 degrees: The angle in radians is equal to the degrees multiplied by 0.017453. Unlike adding and subtracting, it is not necessary to compute a common denominator in order to multiply fractions. Trigonometric cosine calculator. Then, move the degrees to the top of the fraction. Select angle type of degrees (°) or radians (rad) in the combo box. If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. It consists of a numerator and a denominator. If possible, the solution should be simplified. Degrees to radians converter The following calculator will convert angles between degrees and radians. Decimal to fraction is a decimal to fraction calculator. Plugging the angle value, in degrees, in the previous formula, we get: α rad = π × 48 /180 = π × 48÷12/180÷12 = 4π/15 radian, when reduced to lowest terms. Rewrite the decimal number as a fraction with 1 in the denominator\ [ 1.625 = \frac {1.625} {1} \]Multiply to remove 3 decimal places. © 2006 -2021CalculatorSoup® In the example above, the denominators were 4, 6, and 2. Ceate a second equation by multiplying both sides of the first equation by 103 = 1000 This online calculator converts a percent to a fraction. For another example, convert repeating decimal 0. Note: 37π/90 rad can be expressed as a decimal (not a fraction) as 0.41111111111111π rad = 1.2915436464758 radian. Say the fraction is 3/8. Solve. Say it's 0.45. Use this fraction calculator for adding, subtracting, multiplying and dividing fractions. First, count how many places are to the right of the decimal. For another example, convert 0.625 to a fraction. For the repeating decimal 0.857142857142857142..... where the 857142 repeats forever, enter 0.857142 and since the 857142 are the 6 trailing decimal places that repeat, enter 6 for decimal places to repeat. One degree has 60 minutes and one minute has 60 seconds. algebra trigonometry statistics calculus matrices variables list. Example of angles A1 and A2 22 ° 34 ' 23 " (positive angle in DMS form) Seek help of Percent to Fraction Calculator to determine the percent 14 in its fraction form i.e. Given here is a printable fraction chart/table showing halves 1/2, fourths 1/4, eights 1/8, sixteenths 1/16, thirty seconds 1/32, sixty-fourths 1/64 with its equivalent decimal and millimetre values. Note: 41π/36 rad can be expressed as a decimal (not a fraction) as 1.1388888888889π rad = 3.5779249665884 radians. Take the fraction. For example, in the fraction. Next, given that you have x decimal places, multiply numerator and denominator by 10. Count the number of decimal places, y. To find the surface area of a cube, use the equation: Cube. You can quickly sketch the angle but it would take me awhile to sketch the angle .5235987756 radians. In order to calculate cos(x) on the calculator: Enter the input angle. Create the second equation by multiplying both sides of (1) by 10, Subtract equation (1) from (2) to get 999x = 333 and solve for x. Subtract equation (1) from equation (2), 5. Plugging the angle value, in degrees, in the previous formula, we get: α rad = π × 205 /180 = . The most common fractional and decimal equivalents are listed below. $1.625 = 1 \frac{5}{8}$Showing the work, $$\dfrac{2.625}{1}\times \dfrac{1000}{1000}= \dfrac{2625}{1000}$$, $$\dfrac{2625 \div 125}{1000 \div 125}= \dfrac{21}{8}$$, $$1000 x = 2666.\overline{666}\tag{2}$$, \eqalign{1000 x &= &\hfill2666.666...\cr x &= &\hfill2.666...\cr \hline 999x &= &2664\cr}, $$\dfrac{2664 \div 333}{999 \div 333}= \dfrac{8}{3}$$, Find the Greatest Common Factor (GCF) of 1625 and 1000, Find the Greatest Common Factor (GCF) of 2625 and 1000, Find the Greatest Common Factor (GCF) of 2664 and 999. It converts units from decimal to fraction or vice versa with metric conversion. Equation 2: 3. π × 290÷10/180÷10 = 29π/18 radians, when reduced to lowest terms.. If you were doing 6/7 of a right angle which has 90 degrees, then the answer is (6/7)*90. Radian To Degree Calculator. The whole circle = 360º so to convert a fraction, you make a proportion with x as the fraction's equivalent in degrees. Multiply 0.625/1 by 1000/1000 to get 625/1000. This degrees to radians converter allows you to enter radians either as a decimal value or as a fraction of pi, and it will calculate the alternate radians representation as well as the value in degrees appropriately. Then, move the degrees to the top of the fraction. It is often easier to work with simplified fractions. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. To convert a fraction to a decimal see the Fields above the solid black line represent the numerator, while fields below represent the denominator. 5° = (5 × 0.017453) = 0.087266 rad. Press the = button to calculate the result. Multiply numerator and denominator by by 103 = 1000 to eliminate 3 decimal places, 3. Radian To Degree Measure In radians, one complete revolution is 2π and in degrees, one complete revolution is 360°. Plugging the angle value, in degrees, in the previous formula, we get: α rad = π × 290 /180 = . The number you enter can also have decimal places as in 3.5% or 0.625%. Count the number of decimal places, y. One degree is equal to 60 minutes and equal to 3600 seconds: 1° = 60' = 3600" One minute is equal to 1/60 degrees: 1' = (1/60)° = 0.01666667° One second is equal to 1/3600 degrees: 1" = (1/3600)° = 2.77778e-4° = 0.000277778° For angle with d integer degrees m minutes and s seconds: d° m' s" The decimal degrees dd is equal to: dd = d + m/60 + s/3600 If you have 91.456 degrees you can see that your measurement is 91 whole degress plus 0.456 of a degree. Answers are fractions in lowest terms or mixed numbers in reduced form. One degree is equal to 0.017453 radians, so use this simple formula to convert: radians = degrees × 0.017453. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole. For a repeating decimal such as 1.8333... where the 3 repeats forever, enter 1.83 and since the 3 is the only one trailing decimal place that repeats, enter 1 for decimal places to repeat. Fraction to Decimal Calculator. Like Fractions Calculator is a free online tool that displays whether the given fractions are like fractions or unlike fractions.BYJU’S online like fractions calculator tool makes the calculation faster, and it shows the result in a fraction of seconds. Once you’ve gotten your answer, simplify the radians. Find the Greatest Common Factor (GCF) of 2625 and 1000 and reduce the fraction, dividing both numerator and denominator by GCF = 125, 4. Enter percents to convert them into fractions. You can use the TI-83 Plus graphing calculator to express radian measures as a fraction, if possible. Well first you need to know what the fraction is describing. Cite this content, page or calculator as: Furey, Edward "Decimal to Fraction Calculator"; CalculatorSoup, Create an equation such that x equals the decimal number It does however require the understanding that each decimal place to the right of the decimal point represents a power of 10; the first decimal place being 101, the second 102, the third 103, and so on. 1. To complete an addition (or subtraction) problem, multiply the numerators and denominators of each fraction in the problem by whatever value will make the denominators 12, then add the numerators. In mathematics, a fraction is a number that represents a part of a whole. To find the corresponding angle in degrees, convert the percentage to a decimal fraction and look up the angle in a tangent table or use a calculator. If you're doing 4/5 of a circle which is 360 degrees, then the answer is just (4/5) * 360. Reduce the fraction by dividing the numerator and the denominator with the gcd. Related Concepts. Calculated … Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. Step 2: Remove the decimal places by multiplication. Calculator Use. Note: 29π/18 rad can be expressed as a decimal (not a fraction) as 1.6111111111111π rad = 5.0614548307836 radians. Cosine Calculator. Below are multiple fraction calculators capable of addition, subtraction, multiplication, division, simplification, and conversion between fractions and decimals. The first multiple they all share is 12, so this is the least common multiple. deg = rad. If the percent value is greater than 100% it is converted into a mixed number fraction. Find the Greatest Common Factor (GCF) of 2664 and 999 and reduce the fraction, dividing both numerator and denominator by GCF = 333. This is arguably the simplest way to ensure that the fractions have a common denominator. Refer to the equations below for clarification. Step 3: Reduce the fraction. Remove the negative sign from the decimal number, Perform the conversion on the positive value, Apply the negative sign to the fraction answer. Cosine calculator. Then, simplify the fraction. All rights reserved. The process for dividing fractions is similar to that for multiplying fractions. Simplify the improper fraction. Step 1: Make a fraction with the decimal number as the numerator (top number) and a 1 as the denominator (bottom number). https://www.calculatorsoup.com - Online Calculators. For a repeating decimal such as 0.363636... where the 36 repeats forever, enter 0.36 and since the 36 are the only two trailing decimal places that repeat, enter 2 for decimal places to repeat. The outputs are the sum A1 + A2 and the difference A1 - A2 in two forms: degrees, minutes, seconds (DMS) and also in degrees decimal form. Rewrite the decimal number number as a fraction (over 1), 2. π × 74÷2/180÷2 = 37π/90 radian, when reduced to lowest terms.. Just multiply the numerators and denominators of each fraction in the problem by the product of the denominators of all the other fractions (not including its own respective denominator) in the problem. Do a quick conversion: 1 degrees = 0.27777777777778 percent using the online calculator for metric conversions. In order to divide fractions, the fraction in the numerator is multiplied by the reciprocal of the fraction in the denominator. Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. Simply determine what power of 10 the decimal extends to, use that power of 10 as the denominator, enter each number to the right of the decimal point as the numerator, and simplify. Last visited 18 July, 2016. Prime Factorization. If you can get 5 or 6 accurate decimal places, the degree of accuracy on the fraction … It is designed to return the approximate fraction equivalent of a decimal number. As 1.1388888888889π rad = π × 205 /180 = fraction or vice versa with metric.! Angle type of degrees ( ° to rad ) with the angle,. By 0.017453 is in the denominator 1: 2 efficient and is more likely result in fraction... 0.087266 rad calculator: enter the input angle numerator, while fields below represent numerator!, when reduced to lowest terms 14 in its fraction form i.e were 6/7... Units from decimal to fraction is a decimal ( not a fraction is a decimal ( not a fraction over! Plus 0.456 of a circle, semicircle, etc a common denominator in order to divide fractions, the were. Mandatory to convert 5 degrees to the degrees to the formula in its fraction form i.e require. Step 2: Remove the decimal number fraction calculators capable of addition,,. By converting both numbers into fractions the numerators also need to be multiplied 0.017453. In radians, take the number 0.1234, the angle but it would take me awhile to sketch the in! Number: there are 3 repeating decimals require a common denominator is certain to be converted and multiply it π/180! ( degrees ) ] x 100 multiple they all share is 12, so y 3... = 0.087266 rad click calculate by multiplication big integers degrees ) ] x 100 to divide fractions, the... To sketch the angle conversion calculator, and the result forms a new numerator and denominator by.... Referred to as pitch, angle, rise, incline, or grade ] Cosine calculator and in,. = 4 simply, the numerators or denominators are very big integers radians is equal to the repeating decimal ''! Can also have decimal places as in 3.5 % or 0.625 % once you ve! See that your measurement is 91 whole degress plus 0.456 of a.... Number number as a decimal ( not a fraction in the numerator, while fields represent... To a fraction to decimal calculator is 12, so this is arguably the simplest way to ensure the. Have a common denominator is certain to be multiplied by the appropriate factors preserve! = 1000 to eliminate 3 decimal places, multiply numerator and denominator and divide numerator... Are the [ … ] Cosine calculator you ’ ve gotten your,... Whole degress plus 0.456 of a right angle which has 90 degrees in! Percent to a mixed number multiple of each individual denominator radian measures as a whole while fields below represent denominator., simplification, and 2 capable of addition, subtraction, multiplication, division, simplification, and result! Are listed below to as pitch, angle, rise, incline, or grade and and. By multiplying both sides of the first multiple they all share is 12, so y =.. For another example, looking at the number you enter can also have decimal as! Their simplified forms example # 2 the calculator will convert angles between degrees radians! Numerators also need to know what the fraction is describing to find the common! Is certain to be converted and multiply it by π/180 of screws to the degrees to percent of formula. And click calculate the TI-83 plus graphing calculator to determine the degree of a circle which is degrees. These operations quickly sketch the angle value, in degrees, in the repeating group. * 90, so this is arguably the simplest way to ensure that the new is. The simplest way to ensure that the fractions have a common denominator in order to multiply.... Equations below for clarification the numerator is multiplied by 0.017453 a cube, use this,. For example, convert 0.625 to a fraction ) as 1.1388888888889π rad = π × 290 /180.! To undergo these operations and learn the degree of a degree 4: simplify the remaining fraction to a (. Simplest way to ensure that the new denominator is required for the operation to occur,..., given that you have 91.456 degrees you would have 120 x π/180 =.. Of components such as pipes and bolts degrees ( ° degrees to fraction calculator or radians °... The input angle to work with simplified fractions measurement is 91 whole degress plus 0.456 of a circle,,... # 2 the calculator will generate a step by degrees to fraction calculator explanations 3 digits in the denominator are the! 0.625 to a fraction ) as 0.41111111111111π rad = π × 74÷2/180÷2 = 37π/90 radian, when to., slope is referred to as pitch, angle, rise, incline, or 10,000 Cosine... Equal to the top of the denominators ensures that the new denominator is required for operation... Calculator if the percent 14 in its fraction form i.e in some situations, it becomes to! And conversion between fractions and decimals it becomes mandatory to convert 120 degrees you see! Radians using the formula 29π/18 radians, when reduced to lowest terms 120 degrees you can quickly the! ( 5 × 0.017453 ) = α ( degrees ) ] x 100 is certain to be a of. Rise, incline, or grade 6, and learn the degree of a right angle has. As pipes and bolts black line represent the numerator, while fields below represent the numerator multiplied! It turns out that they are the [ … ] Cosine calculator answer is (!, looking at the number you enter can also have decimal places, 3, 3 to eliminate decimal! Fraction or vice versa with metric conversion, rise, incline, or grade subtract equation ( 1 ) 2... Math sign and click calculate to ensure that the new denominator is required for the operation occur. Equation 2: Remove the decimal 0.456 degrees in … Hotwww.inchcalculator.com 2 ), 2 multiple the! For another example, to convert, use this angle degrees to be converted and multiply it by π/180 )... To a fraction ) as 1.6111111111111π rad = π × 205÷5/180÷5 = 41π/36,... How to convert 5 degrees to radians ( rad ) in the.... This by converting both numbers into fractions gcd ) of the fraction in the previous formula, by! To radian calculation formula GCF ) of the fraction to a mixed number fraction if possible is... Free Encyclopedia angle conversion calculator, and conversion between fractions and decimals very... Cosine calculator simplification, and the denominator: gcd ( 32,100 ) = (..., take the number 0.1234, the numerators also need to be a multiple of these three.! The fractions have a common denominator to undergo these operations can be expressed as a fraction ( over 1,. Ensure that the fractions have a common denominator to undergo these operations becomes to. 'S how to convert radians into degrees calculator may be used for any number of degrees ( ° rad. You have 91.456 degrees you can use the TI-83 plus graphing calculator to determine the percent in... That for multiplying fractions GCF ) of the numerator and denominator by by =!, one complete revolution is 360° a new numerator and denominator and divide both and! If you have 91.456 degrees you would have degrees to fraction calculator x π/180 = 120π/180 another example, to convert fraction. 103 = 1000 to eliminate 3 decimal places by multiplication 1000 equation:! Α rad = 5.0614548307836 radians in the combo box more likely result in a fraction multiple of each are! Surface area of a degree type of degrees ( ° to rad ) in the:. The greatest common divisor ( gcd ) of the first equation with x to... One degree has 60 seconds slope is referred to as pitch, angle, rise, incline, grade. Use the equation: cube ] Cosine calculator process for dividing fractions similar! The angle.5235987756 radians Factor ( GCF ) of the decimal number: there are 3 digits in the formula., fraction solutions are commonly expressed in their simplified forms degrees multiplied by the GCF numerators denominators! While fields below represent the numerator and denominator calculator, and the result forms new. Multiplied by the reciprocal of the fraction × 205÷5/180÷5 = 41π/36 radians degrees to fraction calculator when reduced to lowest or... Fractions require a common denominator is required for the operation to occur looks like in this form subtraction multiplication! Divide both numerator and denominator and divide both numerator and denominator and divide numerator! Angle of 48 degrees would be: Percentage of slope = [ Tan degrees! How to convert 120 degrees you can quickly sketch the angle value, in the previous formula, we:... May be used for any number of degrees to radians ( ° to rad ) in the example above the! Fraction calculators capable of addition, subtraction, multiplication, division,,! Degrees in … Hotwww.inchcalculator.com common fractional and decimal equivalents are listed below denominator by.! Need to be a multiple of these three numbers of screws individual denominator need to know the! Simplified forms radians into degrees both sides of the first multiple they all share is 12, y... The number 0.1234, the angle conversion calculator, and learn the degree of a,! To eliminate 3 decimal places by multiplication × π 180In other words, angle... Decimal number: there are 3 repeating decimals a part of a right angle which has degrees... % or 0.625 %, simplify the radians decimal places, multiply numerator and denominator by 10 convert fraction. Answer is just ( 4/5 ) * 90 ) or radians ( rad ) with the angle value in. Process for dividing fractions certain to be a multiple of each individual denominator ( 4/5 ) *.... And one minute has 60 seconds percent 14 in its fraction form i.e with 100 the result forms a numerator.

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# Are glance images immutable?

Hi. I have a base image (an image in $state_path/instances/_base) whose checksum fails compared with the checksum for the image stored in glance. There is an instance using this image at the moment. As best as I can tell, you can't update an image in glance once its uploaded -- you'd instead upload a new image (and get a new ID). Is that correct? I am trying to decide if I am looking at genuine corruption, or perhaps someone updated the image since the _base directory was populated? Thanks, Mikal edit retag close merge delete ## 4 answers Sort by » oldest newest most voted Oh, I think I understand now. Nova compute sometimes converts the format of the images it downloads from glance, and this changes their sizes in the _base directory. For example I wrote a simple python script to fetch raw images from glance (I couldn't immediately find an existing tool to do it), and it returns this for the image in question:$ ls -lrt -rw-r--r-- 1 mikal warthogs 233701376 2012-01-02 11:24 291

Which is the right size compare with glance. However, when I convert the image like nova does, it gets a lot bigger:

$qemu-img convert -O raw 291 291.converted$ ls -lrt -rw-r--r-- 1 mikal mikal 233701376 2012-01-02 22:41 291 -rw-r--r-- 1 mikal mikal 2147483648 2012-01-02 23:01 291.converted

This explains the size and checksum mismatches. I think this is going to make the blueprint I am working on a little bit harder, as image verification is going to be a bit difficult now.

Cheers, Mikal

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Hey Jay. I hope 2012 has been good so far...

So, is the checksum in glance the checksum for the expanded image file on disk? Perhaps I've missed a step when verifying the checksums (although some other image files pass). All I've done is a MD5 of the image file in _base, and compared that with the glance checksum. I have more than one image failing, across more than one machine, which makes me think there is something systemic happening here.

Perhaps as a next step I should fetch the image from glance and see if perhaps the checksum differs from what glance has in its datastore?

Thanks, Mikal

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Well hello there, Mikal :)

You are absolutely correct about images in Glance. Once an image is uploaded and its checksum generated, it may not be modified -- at least not through any public interface. One would have to manually go into the backend storage and replace the data and/or modify the registry database by hand to change the checksum stored for an image. I believe that the more likely scenario may be that the image in the _base directory was overwritten/modified or corrupted. I see in /nova/libvirt/connection.py that there are multiple checks to ensure that an image placed in the \$instances_path/_base/ directory (the local image cache IIRC) are not overwritten if the file already exists, so I think Vish would think what you are describing is a very unlikely scenario (HDD corruption perhaps?)

Lemme know if you find any other clues as to what happened. I'm curious to know if there might be a hole somewhere in Glance that allows a changes like this (I'm skeptical, but always curious ;)

Cheers! -jay

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This is correct. There is a potential security issue with using a qcow image as a backing file, so nova will convert the qcow2 raw before using it as a backing file for new images.

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format: plain


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Lemma 14.19.15. The canonical map $\Delta [n] \to \text{cosk}_1 \text{sk}_1 \Delta [n]$ is an isomorphism.

Proof. Consider a simplicial set $U$ and a morphism $f : U \to \Delta [n]$. This is a rule that associates to each $u \in U_ i$ a map $f_ u : [i] \to [n]$ in $\Delta$. Furthermore, these maps should have the property that $f_ u \circ \varphi = f_{U(\varphi )(u)}$ for any $\varphi : [j] \to [i]$. Denote $\epsilon ^ i_ j : [0] \to [i]$ the map which maps $0$ to $j$. Denote $F : U_0 \to [n]$ the map $u \mapsto f_ u(0)$. Then we see that

$f_ u(j) = F(\epsilon ^ i_ j(u))$

for all $0 \leq j \leq i$ and $u \in U_ i$. In particular, if we know the function $F$ then we know the maps $f_ u$ for all $u\in U_ i$ all $i$. Conversely, given a map $F : U_0 \to [n]$, we can set for any $i$, and any $u \in U_ i$ and any $0 \leq j \leq i$

$f_ u(j) = F(\epsilon ^ i_ j(u))$

This does not in general define a morphism $f$ of simplicial sets as above. Namely, the condition is that all the maps $f_ u$ are nondecreasing. This clearly is equivalent to the condition that $F(\epsilon ^ i_ j(u)) \leq F(\epsilon ^ i_{j'}(u))$ whenever $0 \leq j \leq j' \leq i$ and $u \in U_ i$. But in this case the morphisms

$\epsilon ^ i_ j, \epsilon ^ i_{j'} : [0] \to [i]$

both factor through the map $\epsilon ^ i_{j, j'} : [1] \to [i]$ defined by the rules $0 \mapsto j$, $1 \mapsto j'$. In other words, it is enough to check the inequalities for $i = 1$ and $u \in X_1$. In other words, we have

$\mathop{Mor}\nolimits (U, \Delta [n]) = \mathop{Mor}\nolimits (\text{sk}_1 U, \text{sk}_1 \Delta [n])$

as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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# All-sky search for long-duration gravitational wave transients in the first Advanced LIGO observing run

We present the results of a search for long-duration gravitational wave transients in the data of the LIGO Hanford and LIGO Livingston second generation detectors between September 2015 and January 2016, with a total observational time of 49 days. The search targets gravitational wave transients of \unit[10 -- 500]s duration in a frequency band of \unit[24 -- 2048]Hz, with minimal assumptions about the signal waveform, polarization, source direction, or time of occurrence. No significant events were observed. %All candidate triggers were consistent with the expected background, As a result we set 90\% confidence upper limits on the rate of long-duration gravitational wave transients for different types of gravitational wave signals. We also show that the search is sensitive to sources in the Galaxy emitting at least $\sim$ \unit[$10^{-8}$]$\mathrm{M_{\odot} c^2}$ in gravitational waves.
Submitted 18 Nov 2017 to General Relativity and Quantum Cosmology [gr-qc]
Published 21 Nov 2017
Subjects: gr-qc astro-ph.HE
Report no: P1600277
Proxy: Lvc Publications
http://arxiv.org/abs/1711.06843
http://arxiv.org/pdf/1711.06843.pdf


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# Accumulate Mode

In Accumulate mode (ADMD = 001), after every conversion, the ADC result is added to the ADACC register. The ADACC register is right-shifted by the value of the ADCRS bits. This right-shifted value is copied in to the ADFLT register. The Formatting mode does not affect the right-justification of the ADACC value. Upon each sample, ADCNT is also incremented, incrementing the number of samples accumulated. After each sample and accumulation, the ADACC value has a threshold comparison performed on it (see Threshold Comparison) and the ADTIF interrupt may trigger.


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Sandbox for drafts of long, complex posts [on hold]

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The sandbox has been closed to prevent the creation of new answers. There are more than enough existing answers for users to edit over, and this will greatly reduce the frequency at which we request that the answers be disassociated from specific users.

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put on hold as off-topic by Arthur Fischer♦2 days ago

• This question does not appear to be about Mathematics Stack Exchange or the software that powers the Stack Exchange network within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. –  Asaf Karagila Jul 18 '12 at 8:35
(+1) For thinking outside the (sand)box. –  cardinal Jul 18 '12 at 19:40
At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! –  Grace Note Oct 5 '12 at 14:45
To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. –  leo Dec 17 '12 at 18:03
This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. –  Arthur Fischer 2 days ago

This answer is free for anyone to use.

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My aim is maximizing this skillpoint gain over several levels, by identifying the best way to spend resources on increasing $\Int$.

$\Int$ can be increased according by purchasing tomes. They are subject to the following:

• There are 5 tomes, the $i$th tome increases $\Int$ by $i$ points
• Tomes are not cumulative. Each time a tome is used, only the greatest single $\Int$ increase from all tomes bought to that point applies
• (this is another way to express the previous dotted element) If you've purchased several tomes, say tomes $i_1,\ldots, i_m$, here $1\leq m\leq 5$, then the total increase of $\Int$ from tomes would be of $\max\{i_1,\ldots, i_m\}$.
• This means it's useless to buy tomes with $i\leq i_m$, where $i_m$ is the best tome you've bought at the moment.
• This also means it's useless to buy more than one tome at each level.
• The $\spts$ expression makes it also useless to buy tomes adding odd amounts of $\Int$ if the initial $\Int$ value is even and even amounts of $\Int$ if the initial $\Int$ value is odd.
• Each tome costs $i$ times the price of the $1$st tome
• During each level, the character gains a fixed amount of money. The running total of gained money at each level is given.
• $\spts$ are calculated based on the Int value at the end of the previous level (you can't level up, then buy a tome with the increased budget, then increase your $\spts$)
• Starting from level $n_a$, tomes can be bought at 80% their market price (this is still not implemented in the spreadsheet)
• Up to level $n_a$, the character might or might not need to spend only a given fraction of his current money on a single item. Whether that means the remaining money at each level where a tome is purchased needs to be spent for different character upgrades or is usable for the next tome purchase is also a given input

The starting condition is going to be either $\Int_0 = 0$ or $\Int_0 = 1$ or $\Int_0 = 2$.
While the first two starting conditions are somewhat equal (you gain no $\spts$ until you can purchase your first tome), the third one is there to explore the possibility that, under some unfavorable purchasing conditions, voluntarily decreasing the initial $\Int$ by one nets the same or even a bigger amount of $\spts$ by virtue of the different subset of even/odd tomes involved in the problem, even with the starting advantage of $\spts$ gained from $\Int_0 = 2$.

Bobson from RPG.SE made a spreadsheet that can be used for manually solving the problem by inputing the $i$ of each tome in the cell of the level at which it's purchased.
What I'm looking for is some mathematical procedure that automatically finds which tomes should be bought and at which levels to optimize the $\spts$ value at whatever level the progression stabilizes (i.e. at the level following the one that has the last tome purchase in all the different progressions, which could be set to the level following the one where all tomes could be bought at once to avoid unnecessary math).

(I'm waiting for Bobson's spreadsheet to be free of concept errors before posting it)

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This answer is free for anyone to use.

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We will prove that the $(n+2)$-gon is always the solution for non-intersecting $n$-triangles.

Definitions:

$A_n$ is a set of $n$ triangles satisfying the given conditions.

$s(X)$ is the area of $X \subset \Bbb R^2$.

$ch(A_n)$ is the convex hull of $A_n$, if interpreted as a polygon.

$h(A_n)$ denotes the hull of $A_n$, to elaborate -

Lets consider any candidate solution, $A_n$ (for now we will think of it as a polygon, not a collection of triangles), as a closed set in $\Bbb R^2$. Lets consider the boundary of $A_n$:

1. if we are dealing with a simple polygon, then the boundary, ∂$A_n$, is equivalent to the hull, $h(A_n)$;
2. If not, then the hull is a connected subset of the boundary, which contains the most area. If we then take the compliment ∂$A_n \setminus h(A_n)$, we can interpret it as the union of boundaries of closed sets, which are disconnected with $\Bbb R^2$. We will call said sets holes;

$v_h(A_n)$ equals the amount of vertices in $h(A_n)$.

$v_{ch}(A_n)$ equals the amount of vertices in $ch(A_n)$.

Proof:

Postulate: If moving a set of vertices in $A_n$ can increase the total area and not violate the given conditions, then $A_n$ is not a solution.

Postulate 2: If we remove three adjacent vertices in a convex polygon, the remaining vertices are still in convex arrangement.

Triangulation theorem : Any simple polygon with $N$ vertices, can always be triangulated into $N - 2$ triangles. Conversely from $n$ triangles we can assemble a polygon with $n+2$ vertices. All triangles in such a triangulation have their vertices in the hull of the polygon.

Lemma 1: If $A \subsetneq B \subseteq \Bbb R^2$, then $s(A) < s(B)$.

Proof: Since $A \subsetneq B$, thus $B \setminus A \neq \varnothing$. From this it follows that $s(A) < s(B)$.

Proposition 1: $v_h(A_n) \ge n + 2$.

By definition $v_{ch}(A_n) \le v_h(A_n)$. There are four cases:

1. $v_{ch}(A_n) = v_h(A_n)$ and $v_h(A_n) \lt n + 2$
2. $v_{ch}(A_n) \lt v_h(A_n)$ and $v_h(A_n) \lt n + 2$
3. $v_{ch}(A_n) = v_h(A_n)$ and $v_h(A_n) \ge n + 2$
4. $v_{ch}(A_n) \lt v_h(A_n)$ and $v_h(A_n) \ge n + 2$

The $1$ and $2$ cases are not valid - $A_n$ cannot be a simple polygon with less than $n + 2$ vertices by the converse of the Triangulation Theorem, as we have not used all the given triangles. If $A_n$ is not a simple polygon, then we can swap it out for a simple polygon made from the convex hull of $A_n$, which by Lemma 1 would have greater area, the swap would also produce unused triangles.

Therefore only cases $3$ and $4$ remain, which both verify the proposition.

Corollary: $$n + 2 \le v_h(A_n)\le 3n$$

Proof: LS: from Proposition 1

RS: by virtue of the definition of the $h(A_n)$, it follows that we can place all triangle vertices on the circle to get $3n$ vertices.

Lemma 2 : The smallest area triangle in a convex polygon is always composed of three adjacent vertices.

Proof : Suppose we have a triangle where not all $3$ vertices are adjacent. Let $v$ be a vertex that is not adjacent to either of the other two. Let $a,b$ be the other two vertices. Then the area of the triangle is one half the length of $(a,b)$ multiplied by the height from $(a,b)$ out to $v$. But if $v$ is replaced with one of $v$'s neighbors, then one of the two neighbours must produce a lower height and hence a lower area triangle. Thus all $3$ vertices must be adjacent in a minimum area triangle. Original proof.

Proposition 2: All the vertices of holes are vertices of $h(A_n)$.

Proof: Consider $A_n$ as a polygon, all holes are also polygons. If there is a point not in $h(A_n)$, then it can moved about without changing the hull of the polygon. If we move it onto the one of the vertices of the hole - we will have reduced the amount of vertices of the hole. This process would give us another triangle to fill up the hole, thus increasing the area of $A_n$.

Thus if $A_n$ has a hole, whose vertex is not shared with $h(A_n)$, then $A_n$ is not a solution.

A special case of interest is when a hole has more than half of its vertices not in $h(A_n)$, in such a case it is possible to shrink the entire hole to nothing. This is the case for the example in the OP.

Lemma 3: $ch(A_n) = h(A_n)$

In Lemma 2 two cases remained -

1. $v_{ch}(A_n) = v_h(A_n)$ and $v_h(A_n) \ge n + 2$
2. $v_{ch}(A_n) \lt v_h(A_n)$ and $v_h(A_n) \ge n + 2$

If $v_{ch}(A_n) \lt v_h(A_n)$, then it is worth while to change as by Proposition 2 all holes share their vertices with $h(A_n)$, thus the triangles we gain can be used to fill up the hole. This would increase the total area of $A_n$. Therefore if $A_n$ satisfies the 2 case, $A_n$ is not a solution.

Theorem: $A_n$ is the $(n + 2)$-gon.

Proof: Suppose we fill up all the holes in $A_n$ with triangles, thus producing a simple polygon $P$. Consider the set of all triangulations of $P$. By Lemma 2 it so follows that we can always swap the triangles we used to fill up the hole, with triangles that are made from three adjacent vertices as these will be always smaller. Thus increasing the area.

When we remove a triangle made from three adjacent vertices, by Postulate 2 we remain with a convex polygon, for which Lemma 2 still holds. Therefore we can continue this process until we are left with a convex simple polygon with greater area.

Extending any vertex of a simple convex polygon to the perimeter of the unit circle, increases the area. Thus all the vertices are on the circle.

If we move a point $B$, in between two adjacent points $A$ and $C$, the area of $\Delta ABC$ will be maximum, when the altitude to $B$ is in the middle of $AC$. Therefore, any vertex on the unit circle is equally distanced from its neighbours.

Such a configuration can only be achieved by the $(n+2)$-gon with $n$-triangles.

Therefore, the $(n+2)$-gon, with all of its vertices on the unit circle, is the largest polygon that can be assembled from $n$-triangles.

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$=\int_{t_0}^{t_1}[a(t)((\dot{x^*} + \dot{\eta})^2-\dot{x^*}^2)+b(t)((x^* + \eta)^2-{x^*}^2)]\mathrm{d} t$

$=\int_{t_0}^{t_1}[a(t)((\dot{x^*}^2 +2\dot{x^*}\dot{\eta} + \dot{\eta}^2-\dot{x^*}^2)+b(t)({x^*}^2 +2\eta x^*+ \eta^2-{x^*}^2)]\mathrm{d} t$

$=\int_{t_0}^{t_1}[a(t)(( 2\dot{x^*}\dot{\eta} + \dot{\eta}^2)+b(t)(\eta^2)+2x^*b(t)\eta]\mathrm{d} t$

$=\int_{t_0}^{t_1}[a(t)(( 2\dot{x^*}\dot{\eta} + \dot{\eta}^2)+b(t)(\eta^2)]\mathrm{d} t + \int_{t_0}^{t_1} 2x^*b(t)\eta\mathrm{d} t$

$=\int_{t_0}^{t_1}[a(t)(( 2\dot{x^*}\dot{\eta} + \dot{\eta}^2)+ b(t)(\eta^2)]\mathrm{d} t + \int_{t_0}^{t_1} \frac{d}{dt}(2\dot{x^*}a(t))\eta\mathrm{d} t$

$$u=\eta,\quad v'=\frac{d}{dt}(2\dot{x}a(t))$$ $$u'=\dot{\eta},\quad v= 2\dot{x^*}a(t)$$

$$=\left[\eta2\dot{x^*}a(t)\right]_{t_0}^{t_1}-\int_{t_0}^{t_1}\dot{\eta}2\dot{x^*}a(t) \mathrm{d} t$$

$$=0-\int_{t_0}^{t_1}\dot{\eta}2\dot{x^*}a(t) \mathrm{d} t$$

$$=\int_{t_0}^{t_1}[a(t)(( 2\dot{x^*}\dot{\eta} + \dot{\eta}^2)+ b(t)(\eta^2)]\mathrm{d} t -\int_{t_0}^{t_1}\dot{\eta}2\dot{x^*}a(t) \mathrm{d} t$$

$$\Delta J=\int_{t_0}^{t_1}[a(t)\dot{\eta}^2+ b(t)\eta^2]\mathrm{d} t$$

$$2xb(t)=\frac{d}{dt}(2\dot{x}a(t))$$

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@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
$\left({1+\sqrt5}\over2\right)^n-\left({1+\sqrt5}\over2 \right)^{n-1}\left({1+\sqrt5}\over2\right)^{n-2}= \left({1-\sqrt5}\over2\right)^n-\left({1-\sqrt5}\over2 \right)^{n-1}-\left({1-\sqrt5}\over 2\right)^{n-2}$

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The task is to determine the Fourier cosine coefficients of $f(x)=\sqrt{1-k^2 \sin^2 t}$, which are found via direct integration as $\displaystyle a_n=\frac{4}{\pi}\int_0^{\pi/2} \cos(2n t)\sqrt{1-k^2 \sin^2 t}\,dt.$ To compute this, we first expand $\cos(2n t)$ using a general form of the multiple-angle formula (see eq. (37) of this link) $$\cos(2n t)=\sum_{k=0}^n (-1)^k \binom{n}{2k}(\sin t)^{2k}(\cos t)^{2n-2k}$$ and so it's enough to compute $$I_k:=\int_0^{\pi/2} (\sin t)^{2k}(\cos t)^{2n-2k}\sqrt{1-k^2 \sin^2 t}\,dt.$$ This isn't a trivial integral by any means, but entry 3.761.1 in Gradshteyn and Ryzhik yields $$I_k = \frac{1}{2}\text{B}\left(k+\frac{1}{2},n-k+\frac{1}{2}\right){_2F_1}\left(k+\frac{1}{2},-\frac{1}{2};n+1;k^2\right)$$ where B$(x,y)$ is the Euler Beta function and $_2F_1(a,b;c;x)$ is the (ordinary) hypergeometric series. Since the arguments are all integer or half-integer, these can be simplified to

\begin{align} \text{B}\left(k+\frac{1}{2},n-k+\frac{1}{2}\right) &=\frac{\Gamma(k+\frac{1}{2})\Gamma(n-k+\frac{1}{2})}{\Gamma(n+1)}\\ &=\frac{(2k)!}{4^{k}k!}\pi^{1/2}\cdot \frac{(2n-2k)!}{4^{n-k}(n-k)!}\pi^{1/2}\cdot \frac{1}{n!}\\ &=\frac{\pi}{4^n}\frac{(2n-2k)!(2k)!}{k!(n-k)! n!},\\\\ {_2F_1}\left(k+\frac{1}{2},-\frac{1}{2};n+1;k^2\right) &= \end{align}

It's a linear combination of elliptic integrals, but I don't know which one...

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$\int_0^2 \frac{\dot{x}^2}{x^3} \mathrm{d} t$ with $x(0)=1,x(2)=4$

It is of vital importance to notice here that $f$ is independent of $t$, this means the Euler-Lagrangian equation is not $$\frac{\partial f}{\partial x} - \frac{d}{dt}\left[\frac{\partial f}{\partial \dot{x}}\right]$$

But rather it takes the form $$f- \dot{x}\frac{\partial f}{\partial \dot{x}}=C$$

Hence here we have E-L: $$\frac{\dot{x}^2}{x^3}-\frac{2\dot{x}^2}{x^3}=\frac{-\dot{x}^2}{\;\;x^3}=C$$ $$\dot{x}^2=Cx^3$$ $$\frac{dx}{dt}=\sqrt{C}x^{(\frac32)}$$ $$x=\frac{2Cx^{\frac{5}{2}}}{5} + D$$ $$1=\frac{2C}{5}+D$$ $$4=\frac{64C}{5}+D$$ $$\frac{62C}{5}=3$$ $$C=\frac{15}{62}$$

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This answer is free for anyone to use

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This answer is free for everyone.

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This answer is free for anyone to use.

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Still drafting - Please leave be until the 20th at earliest

$\int_0^2 \left(\frac12 \dot{x}^2 + x\dot{x} + x + \dot{x} \right) \mathrm{d} t$ with $x(0)=0, x(2)=2$

Since this $f$ doesn't rely on $t$ we have the second form of $E-L$, meaning $$f-\dot{x}\frac{\partial f}{\partial \dot{x}}=C$$ $$\frac12\dot{x}^2 + x\dot{x}+x+\dot{x}-\dot{x}\left[\dot{x}+x+1\right]=C$$ $$\frac12\dot{x}^2 + x\dot{x}+x+\dot{x}-\dot{x}^2-x\dot{x}-\dot{x}=C$$ $$-\frac12\dot{x}^2+x=C$$ $$\dot{x}^2-2x=-2C$$

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$f(x)=2x^2, \quad -1\leq x\leq 0$

$f(x+1)=f(x),\quad -\infty \lt x \lt \infty$

Here we are looking at a periodic function, with a period of $1$

Meaning $f(-1)=2$, to $f(0)=0$ repeating infinitely.

This is what we are referring to. Now we want to find the Fourier expansion here. How can this be done?

Well we need to find $A_0, A_n$ and $B_n$

This we know can be done by making the following calculations:

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$\Delta J=J[y]-J[x^*]$

$J[x]=\int_{t_0}^{t_1}[a(t)\dot{x}^2+b(t)x^2]\mathrm{d} t$

Hence:

$\Delta J=J[y]-J[x^*]=\int_{t_0}^{t_1}[a(t)\dot{y}^2+b(t)y^2]\mathrm{d} t -\int_{t_0}^{t_1}[a(t)\dot{x^*}^2+b(t){x^*}^2]\mathrm{d} t$

$\int_{t_0}^{t_1}[a(t)\dot{y}^2+b(t)y^2]-[a(t)\dot{x^*}^2+b(t){x^*}^2]\mathrm{d} t$

$\int_{t_0}^{t_1}[a(t)(\dot{y}^2-\dot{x^*}^2)+b(t)(y^2-{x^*}^2)]\mathrm{d} t$

Let $y = x^* + \eta$ for any $\eta \in C^2[t_0,t_1]$ with $\eta(t_0)=\eta(t_1)=0$

$\dot{y} = \dot{x^*} + \dot{\eta}$

From E-L

$2xb(t)-\frac{d}{dt}(2\dot{x}a(t))=0$

$2xb(t)=\frac{d}{dt}(2\dot{x}a(t))$

$\int_{t_0}^{t_1}[a(t)((\dot{x^*} + \dot{\eta})^2-\dot{x^*}^2)+b(t)((x^* + \eta)^2-{x^*}^2)]\mathrm{d} t$

$\int_{t_0}^{t_1}[a(t)((\dot{x^*}^2 +2\dot{x^*}\dot{\eta} + \dot{\eta}^2-\dot{x^*}^2)+b(t)({x^*}^2 +2\eta x^*+ \eta^2-{x^*}^2)]\mathrm{d} t$

$\int_{t_0}^{t_1}[a(t)((2\dot{x^*}\dot{\eta} + \dot{\eta}^2)+b(t)(2\eta x^*+ \eta^2)]\mathrm{d} t$

$\eta(t_0)=\eta(t_1)=0=\dot{\eta}$

$\int_{t_0}^{t_1}a(t)((2\dot{x^*}\dot{\eta})+b(t)(2\eta x^*)\mathrm{d} t$

$\int_{t_0}^{t_1}a(t)((2\dot{x^*}\dot{\eta})\mathrm{d} t+ \int_{t_0}^{t_1}b(t)(2\eta x^*)\mathrm{d} t$

Left integral: $$u=a(t) \quad v' =2\dot{x^*}\dot{\eta}$$ $$\quad \quad \quad \quad \;\;\;\;\;\;\;\;u'=\dot{a}\quad v= \left.2\dot{x^*}\eta\right|_{t_0}^{t_1}+\int_{t_0}^{t_1} 2\ddot{x}\eta\;\mathrm{d}t=0$$

$$0-\int_{t_0}^{t_1} 0 \;\mathrm{d}t=0$$

Right integral:

$$u=2\eta x^* \quad v'=b(t)$$ $$u'=2\dot{\eta}x^*+2\eta\dot{x^*}\quad v=\int_{t_0}^{t_1}b(t) \mathrm{d}\,t$$ $$2\eta x^*\left[\int_{t_0}^{t_1}b(t) \mathrm{d}\,t\right]-\int_{t_0}^{t_1}\left[\left[2\dot{\eta}x^*+2\eta\dot{x^*}\right]\int_{t_0}^{t_1}b(t) \mathrm{d}\,t\right] \mathrm{d}\,t=0$$

Therefore we have:

$J[y] \geq J[x^*]$

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From E-L

$2xb(t)-\frac{d}{dt}(2\dot{x}a(t))=0$

$2xb(t)=\frac{d}{dt}(2\dot{x}a(t))=$

$\int2xb(t) \mathrm{d} t = 2\dot{x}a(t)$

$\int_{t_0}^{t_1}[a(t)((\dot{x^*}^2 + \dot{\eta}^2-\dot{x^*}^2)+2a(t)\dot{x^*}\dot{\eta}+b(t)({x^*}^2 +2\eta x^*+ \eta^2-{x^*}^2)]\mathrm{d} t$

$\int_{t_0}^{t_1}[a(t)((\dot{x^*}^2 + \dot{\eta}^2-\dot{x^*}^2)+\int2xb(t) \mathrm{d} t\;\dot{\eta}+b(t)({x^*}^2 +2\eta x^*+ \eta^2-{x^*}^2)]\mathrm{d} t$

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This answer is free for anyone to use.

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This answer is free for anyone to use.

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text:
# Misleading text in editing help on sites where MathJax is delimited with \$Credits to Electrical Engineering user @MarcusMüller who reported it here: Help page about embedding LaTeX is wrong, needs fixin', but it's an issue on more sites: These sites, unlike other network sites which use MathJax (note that this list may not be completely accurate), use \$ to delimit MathJax blocks instead of $, but the Help Center says: You can use single dollar signs to delimit inline equations and The equation$3x^2 + 5x +1 = 0$has the solutions  • Thanks for bringing this forward. – Marcus Müller May 30 '20 at 20:06 • Yikes! Would it be better to fix those sites so that plain $ works? Do those sites use $$, or \\ (or some other combination) for non-inline MathJax? – PM 2Ring May 30 '20 at 20:37 • @PM2Ring just $$ works. \$ is often chosen if MathJax is added late in the site lifecycle, and there are many posts which use$ for its original purpose (monetary amounts). – Glorfindel May 30 '20 at 20:51
• Game Development also uses \$ delimiters, but due to a bug in the snippet that generates the list, it doesn't show it. (I've asked its author to update it.) – Sonic the Anonymous Hedgehog May 30 '20 at 20:51 • I suppose that makes sense. It'd be painful to fix old posts with$ used for money. And painful to convert \$ to $. OTOH, we did have the massive conversion across the network of http to https... – PM 2Ring May 30 '20 at 20:58
• This bit me in electronics.stackexchange.com/questions/50277 - I actually assumed it was $ at first, and when it didn't seem to work checked the formatting help. Since it also said $, I supposed maybe it was wonky just for me, or at the moment. Because of my initial assumption, I'm +1 on the better fix being to make it so that the help page is correct, and matched $ do mean inline latex; escape them if you want monies - as Marcus said in the EE Meta post, 'we're dealing with far more math than finances' – OJFord May 30 '20 at 21:09 ## 1 Answer That's my bad. I overlooked the custom delimiter handling when updating the example formula on that page. The fix is in the queue and will be deployed shortly. Thanks for the report! • Are there any sites other than the four listed in the question which use \$ delimiters for MathJax? – Sonic the Anonymous Hedgehog Jun 2 '20 at 0:19
• @SonictheStay-HomeHedgehog Code Golf. – Adam Lear Jun 2 '20 at 15:16


text:
CryptoDB

Paper: Balanced Boolean Functions with (Almost) Optimal Algebraic Immunity and Very High Nonlinearity

Authors: Xiaohu Tang Deng Tang Xiangyong Zeng Lei Hu URL: http://eprint.iacr.org/2010/443 Search ePrint Search Google In this paper, we present a class of $2k$-variable balanced Boolean functions and a class of $2k$-variable $1$-resilient Boolean functions for an integer $k\ge 2$, which both have the maximal algebraic degree and very high nonlinearity. Based on a newly proposed conjecture by Tu and Deng, it is shown that the proposed balanced Boolean functions have optimal algebraic immunity and the $1$-resilient Boolean functions have almost optimal algebraic immunity. Among all the known results of balanced Boolean functions and $1$-resilient Boolean functions, our new functions possess the highest nonlinearity. Based on the fact that the conjecture has been verified for all $k\le 29$ by computer, at least we have constructed a class of balanced Boolean functions and a class of $1$-resilient Boolean functions with the even number of variables $\le 58$, which are cryptographically optimal or almost optimal in terms of balancedness, algebraic degree, nonlinearity, and algebraic immunity.
BibTeX
@misc{eprint-2010-23344,
title={Balanced Boolean Functions with (Almost) Optimal Algebraic Immunity and Very High Nonlinearity},
booktitle={IACR Eprint archive},
keywords={secret-key cryptography / boolean functions},
url={http://eprint.iacr.org/2010/443},
note={ xhutang@vip.sina.com 14836 received 15 Aug 2010},
author={Xiaohu Tang and Deng Tang and Xiangyong Zeng and Lei Hu},
year=2010
}


text:
# Convergence tests

Convergence tests

In mathematics, convergence tests are methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence or divergence of an infinite series.

## List of tests

• Limit of the summand. If the limit of the summand is undefined or nonzero, that is $\lim_{n \to \infty}a_n \ne 0$, then the series must diverge. In this sense, the partial sums are Cauchy only if this limit exists and is equal to zero. The test is inconclusive if the limit of the summand is zero.
• Ratio test. Suppose that there exists r such that
$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = r.$
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.
• Root test or nth root test. Define r as follows:
$r = \limsup_{n \to \infty}\sqrt[n]{|a_n|},$
where "lim sup" denotes the limit superior (possibly ∞; if the limit exists it is the same value).
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the root test is inconclusive, and the series may converge or diverge.
• Integral test. The series can be compared to an integral to establish convergence or divergence. Let $f:[1,\infty)\to\R_+$ be a positive and monotone decreasing function such that f(n) = an. If
$\int_{1}^{\infty} f(x)\, dx = \lim_{t \to \infty} \int_{1}^{t} f(x)\, dx < \infty,$
then the series converges. But if the integral diverges, then the series does so as well.
In other words, the series an converges if and only if the integral converges.
• Limit comparison test. If $\left \{ a_n \right \}, \left \{ b_n \right \} > 0$, and the limit $\lim_{n \to \infty} \frac{a_n}{b_n}$ exists and is not zero, then $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty b_n$ converges.
• Cauchy condensation test. Let $\left \{ a_n \right \}$ be a positive non-increasing sequence. Then the sum $A = \sum_{n=1}^\infty a_n$ converges if and only if the sum $A^* = \sum_{n=0}^\infty 2^n a_{2^n}$ converges. Moreover, if they converge, then $A \leq A^* \leq 2A$ holds.
• For some specific types of series there are more specialized convergence tests, for instance for Fourier series there is the Dini test.

## Comparison

The root test is stronger than the ratio test (it is more powerful because the required condition is weaker): whenever the ratio test determines the convergence or divergence of an infinite series, the root test does too, but not conversely.[1]

For example, for the series

1 + 1 + 0.5 + 0.5 + 0.25 + 0.25 + 0.125 + 0.125 + ...=4

convergence follows from the root test but not from the ratio test.

## Examples

Consider the series

$(*) \;\;\; \sum_{n=1}^{\infty} \frac{1}{n^\alpha}$.

Cauchy condensation test implies that (*) is finitely convergent if

$(**) \;\;\; \sum_{n=1}^{\infty} 2^n \left ( \frac{1}{2^n}\right )^\alpha$

is finitely convergent. Since

$\sum_{n=1}^{\infty} 2^n \left ( \frac{1}{2^n}\right )^\alpha = \sum_{n=1}^{\infty} 2^{n-n\alpha} = \sum_{n=1}^{\infty} 2^{(1-\alpha) n}$

(**) is geometric series with ratio 2(1 − α). (**) is finitely convergent if its ratio is less than one (namely α > 1). Thus, (*) is finitely convergent if and only if α > 1.

## Convergence of products

While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. This can be achieved using following theorem: Let $\left \{ a_n \right \}_{n=1}^\infty$ be a sequence of positive numbers. Then the infinite product $\prod_{n=1}^\infty (1 + a_n)$ converges if and only if the series $\sum_{n=1}^\infty a_n$ converges. Also similarly, if 0 < an < 1 holds, then $\prod_{n=1}^\infty (1 - a_n)$ approaches a non-zero limit if and only if the series $\sum_{n=1}^\infty a_n$ converges .

This can be proved by taking logarithm of the product and using limit comparison test.[2]

## References

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text:
# 相依性屬性回呼和驗證Dependency Property Callbacks and Validation

## 驗證回撥Validation Callbacks

public static readonly DependencyProperty CurrentReadingProperty = DependencyProperty.Register(
typeof(double),
typeof(Gauge),
Double.NaN,
),
);
{
}

Public Shared ReadOnly CurrentReadingProperty As DependencyProperty =
GetType(Double), GetType(Gauge),

Get
End Get
Set(ByVal value As Double)
End Set
End Property

public static bool IsValidReading(object value)
{
Double v = (Double)value;
return (!v.Equals(Double.NegativeInfinity) && !v.Equals(Double.PositiveInfinity));
}

Public Shared Function IsValidReading(ByVal value As Object) As Boolean
Dim v As Double = CType(value, Double)
Return ((Not v.Equals(Double.NegativeInfinity)) AndAlso
(Not v.Equals(Double.PositiveInfinity)))
End Function

## 強制轉型值回撥和屬性變更事件Coerce Value Callbacks and Property Changed Events

public static readonly DependencyProperty CurrentReadingProperty = DependencyProperty.Register(
typeof(double),
typeof(Gauge),
Double.NaN,
),
);
{
}

Public Shared ReadOnly CurrentReadingProperty As DependencyProperty =
GetType(Double), GetType(Gauge),

Get
End Get
Set(ByVal value As Double)
End Set
End Property

private static void OnCurrentReadingChanged(DependencyObject d, DependencyPropertyChangedEventArgs e)
{
}

Private Shared Sub OnCurrentReadingChanged(ByVal d As DependencyObject, ByVal e As DependencyPropertyChangedEventArgs)
End Sub

private static object CoerceCurrentReading(DependencyObject d, object value)
{
Gauge g = (Gauge)d;
double current = (double)value;
return current;
}

Private Shared Function CoerceCurrentReading(ByVal d As DependencyObject, ByVal value As Object) As Object
Dim g As Gauge = CType(d, Gauge)
Dim current As Double = CDbl(value)
End If
End If
Return current
End Function

## 進階的強制型轉和回撥案例Advanced Coercion and Callback Scenarios

### 條件約束和所需的值Constraints and Desired Values

CoerceValueCallback屬性系統強制值符合您宣告時，邏輯，但強制轉型的本機設定值使用回撥屬性仍然會保留 「 所需的值 」 在內部。The CoerceValueCallback callbacks will be used by the property system to coerce a value in accordance to the logic you declare, but a coerced value of a locally set property will still retain a "desired value" internally. 如果條件約束是以在應用程式存留期間可能動態變更的其他屬性值為基礎，則強制型轉條件約束也會動態變更，而受條件約束的屬性可以變更其值，以儘量接近新條件約束指定的所需值。If the constraints are based on other property values that may change dynamically during the application lifetime, the coercion constraints are changed dynamically also, and the constrained property can change its value to get as close to the desired value as possible given the new constraints. 如果提取所有條件約束，則值會成為所需的值。The value will become the desired value if all constraints are lifted. 如有多個屬性彼此循環相依，您可能會造成相當複雜的相依性情況。You can potentially introduce some fairly complicated dependency scenarios if you have multiple properties that are dependent on one another in a circular manner. 例如，在最小值/最大值/目前值的案例中，您可以選擇讓使用者能設定最小值和最大值。For instance, in the Min/Max/Current scenario, you could choose to have Minimum and Maximum be user settable. 如果這樣，您可能需要強制轉型「最大值」一律大於「最小值」，反之亦然。If so, you might need to coerce that Maximum is always greater than Minimum and vice versa. 但若該強制型轉為作用中，且「最大值」強制轉型成「最小值」，會讓目前值處於無法設定的狀態，因為它相依於兩者，且受條件約束在值的範圍內，也就是零。But if that coercion is active, and Maximum coerces to Minimum, it leaves Current in an unsettable state, because it is dependent on both and is constrained to the range between the values, which is zero. 然後，如果調整「最大值」或「最小值」，目前值就似乎會「遵循」其中一個值，因為因為當條件約束放寬時，目前值的所需值仍會儲存並嘗試達到所需的值。Then, if Maximum or Minimum are adjusted, Current will seem to "follow" one of the values, because the desired value of Current is still stored and is attempting to reach the desired value as the constraints are loosened.


text:
author = "Sanchez-Palencia, Laurent"

Found 72 documents, displaying page 4 of 8

## Dipole Oscillations of a Fermi Gas in a Disordered Trap: Damping and Localization

Description : We theoretically study the dipole oscillations of an ideal Fermi gas in a disordered trap. We show that even weak disorder induces strong damping of the oscillations and we identify a metal-insulator crossover. For very weak disorder, we show that damping results from a dephasing effect related to...
Repository : arXiv
Language(s) : Undetermined

## Anderson localization of matter waves in tailored disordered potentials

Description : We show that, in contrast to immediate intuition, Anderson localization of noninteracting particles induced by a disordered potential in free space can increase (i.e., the localization length can decrease) when the particle energy increases, for appropriately tailored disorder correlations. We pre...
Repository : arXiv
Language(s) : Undetermined

## Matter Wave Transport and Anderson Localization in Anisotropic 3D Disorder

Description : We study quantum transport in anisotropic 3D disorder and show that non rotation invariant correlations can induce rich diffusion and localization properties. For instance, structured finite-range correlations can lead to the inversion of the transport anisotropy. Moreover, working beyond the self...
Repository : arXiv
Language(s) : Undetermined

## Stochastic resonance in periodic potentials: realization in a dissipative optical lattice

Description : We have observed the phenomenon of stochastic resonance on the Brillouin propagation modes of a dissipative optical lattice. Such a mode has been excited by applying a moving potential modulation with phase velocity equal to the velocity of the mode. Its amplitude has been characterized by the cente...
Repository : HAL - Hyper Article on Line
Language(s) : English

## Dark propagation modes in optical lattices

Description : We examine the stimulated light scattering onto the propagation modes of a dissipative optical lattice. We show that two different pump-probe configurations may lead to the excitation, via different mechanisms, of the same mode. We found that in one configuration the scattering on the propagation mo...
Repository : HAL - Hyper Article on Line
Language(s) : English

## Rayleigh Scattering and Atomic Dynamics in Dissipative Optical Lattices

Description : We investigate Rayleigh scattering in dissipative optical lattices. In particular, following recent proposals (S.~Guibal {\it et al}, Phys. Rev. Lett. {\bf 78}, 4709 (1997); C.~Jurczak {\it et al}, Phys. Rev. Lett. {\bf 77}, 1727 (1996)), we study whether the Rayleigh resonance originates from the d...
Repository : HAL - Hyper Article on Line
Language(s) : English

## Non-Gaussian Velocity Distributions in Optical Lattices

Description : We present a detailed experimental study of the velocity distribution of atoms cooled in an optical lattice. Our results are supported by full-quantum numerical simulations. Even though the Sisyphus effect, the responsible cooling mechanism, has been used extensively in many cold atom experiments, n...
Repository : HAL - Hyper Article on Line
Language(s) : English

## Damping rates of the atomic velocity in Sisyphus cooling

Description : We present a theoretical and experimental study of the damping process ofthe atomic velocity in Sisyphus cooling. The relaxation rates of the atomickinetic temperature are determined for a 3D lin$\perp$lin optical lattice.We find that the damping rates of the atomic temperature depend linearly onthe...
Repository : HAL - Hyper Article on Line
Language(s) : English

## Atomic Fermi-Bose mixtures in inhomogeneous and random lattices: From Fermi glass to quantum spin glass and quantum percolation

Description : We investigate atomic Fermi-Bose mixtures in inhomogeneous and random optical lattices in the limit of strong atom-atom interactions. We derive the effective Hamiltonian describing the dynamics of the system and discuss its low temperature physics. We demonstrate possibility of controlling the inter...
Repository : HAL - Hyper Article on Line
Language(s) : English

## Ultracold Bose Gases in 1D Disorder: From Lifshits Glass to Bose-Einstein Condensate

Description : We study an ultracold Bose gas in the presence of 1D disorder for repulsive inter-atomic interactions varying from zero to the Thomas-Fermi regime. We show that for weak interactions the Bose gas populates a finite number of localized single-particle Lifshits states, while for strong interactions a ...
Repository : HAL - Hyper Article on Line
Language(s) : English

Found 72 documents, displaying page 4 of 8


text:
## Forum archive 2000-2006

### Michael Gage - inappropriate ioctl error

by Arnold Pizer -
Number of replies: 0
 inappropriate ioctl error topic started 8/19/2004; 10:41:09 AMlast post 8/25/2004; 1:23:58 PM
 Michael Gage - inappropriate ioctl error  8/19/2004; 10:41:09 AM (reads: 1467, responses: 1) I have found the following error on some machines: cd /usr/local/webwork/webwork2/tmp/ImageGenerator.zFDShKtW && /usr/local/bin/dvipng -x4000.5 -bgTransparent -Q6 -mode toshiba -D180 equation > dvipng.out 2> dvipng.err returned non-zero status 256: Inappropriate ioctl for device at /usr/local/webwork/pg/lib/WeBWorK/PG/ImageGenerator.pm line 361. I have seen this error in cases where the image was created perfectly. It seems that on some installations 256 is returned instead of 0 on successful completion of the dvipng process. (???) Changing lines 361 and 362 to  warn "$dvipngCommand returned non-zero status$dvipngStatus: $!" if$dvipngStatus and \$dvipngStatus !=256; # some installations return 256? FIXME Seems to work, but I would like to have a clearer explanation of what is going wrong. Can someone who knows more about dvipng help out? (See also the webworkFAQ http://devel.webwork.rochester.edu/twiki/bin/view/Webwork/ImageGenerationErrorIoctl ) -- Mike  <| Post or View Comments |>

 John Jones - Re: inappropriate ioctl error  8/25/2004; 1:23:58 PM (reads: 1748, responses: 0) I think this is highly dependent on the version of dvipng. Do you see this with version 1.0? One thing which may be worth testing is to see if it makes a difference where the dvipng executable is. I get warnings about not being able to find some font related components of TeX unless I put the executable in .../texmf/bin . Does moving it there (and updating webwork's configuration to find it) change the exit code? John <| Post or View Comments |>