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https://www.aft.com/support/product-tips/using-liquid-accumulators-in-aft-impulse?tmpl=component&print=1&format=print | 1,670,409,896,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00634.warc.gz | 651,544,959 | 4,387 | Many engineers are familiar with gas accumulators and their ability to aid in surge suppression modeled in AFT Impulse, but what about liquid accumulators? Liquid accumulators are different from gas accumulators in that they are assumed to be liquid full and do not have any gas that can compress and expand in order to dampen pressure spikes caused by water hammer. Liquid accumulators change the system response to pressure spikes, but they operate differently than gas accumulators.
There are only three required input parameters for liquid accumulators; Elevation, Elasticity, and Initial Volume.
Figure 1: Liquid Accumulator Properties Window
Initial Volume is simply how much space is inside of your liquid accumulator and Elasticity (effective bulk modulus) is the capacity of the liquid accumulator to expand. Low elasticity values would indicate an accumulator with a high capacity to expand (a balloon) and high elasticity values would indicate an accumulator with low capacity to expand (high strength steel). One thing to keep in mind is that even though the material of the liquid accumulator does affect the bulk modulus of elasticity, the elasticity is not actually a material property. The bulk modulus of elasticity is also dependent on the geometry of the liquid accumulator. This means not all steel liquid accumulators have the same effective bulk modulus even though they may have the same Young’s Modulus of Elasticity, for example 200 GPa or 29.0×106 psi. Often times this parameter is hard to find and very few manufacturers provide this data. Therefore it may need to be determined through experimental data.
The following equation relates the effective bulk modulus (K’), initial volume (V), change in volume (DV), and change in pressure (DP) for a liquid accumulator.
Based upon this relationship, a larger K’ value means that the liquid accumulator will allow a larger change in pressure across itself. At very large K’ values (meaning the liquid accumulator is much stiffer than the surrounding pipe) the liquid accumulator will allow the transient pressure waves to communicate through it with very little effect (i.e., very low surge suppression). If the accumulator is not flexible, then what is absorbing the energy of the pressure wave? Nothing, and therefore the wave continues past the liquid accumulator. At very low values of K’ the liquid accumulator will only allow very small changes in pressure across itself. This is to say the accumulator will not allow large changes in pressure across itself and this isolates the other pipes connecting to the liquid accumulator from the pipe that has the pressure wave in it. Thus, the liquid accumulator will reflect pressure and velocity waves rather than transmitting them through the accumulator to the rest of the system.
It can also be seen that large initial volumes only allow small changes in pressure across the liquid accumulator junction. At very large volumes the liquid accumulator acts more like a reservoir. This means that the liquid accumulator would behave like a constant pressure and would not allow a pressure wave to transfer through itself. Small initial volumes allow more of the pressure wave to transfer through the liquid accumulator.
The effects of using different bulk modulus of elasticities and initial volumes for a liquid accumulator can be evaluated with an AFT Impulse model. I created a model that simulated an instantaneous closure of an exit valve that discharges into ambient conditions. This model has two scenarios created using the scenario manager. Both scenarios have a “control” where there is no liquid accumulator along the pipe between an assigned pressure junction and exit valve. In the first scenario initial volume is held constant while K’ (effective bulk modulus) is varied between 1000 psia and 100,000,000 psia in five different cases. The last case is a reservoir which isolates the upstream assigned pressure junction and perfectly reflects the pressure wave that starts at the closed valve. The setup can be seen here:
Figure 2: Test Model with varying K'
In the second scenario, K’ is held constant while the initial volume is varied between 10 ft3 and 100,000 ft3 in five different cases. The last case is a reservoir which isolates the upstream assigned pressure junction and perfectly reflects the pressure wave. The setup can be seen here:
Figure 3: Test Model with varying initial volume
Figure 4: Maximum Pressure Stagnation vs. Length with varying K'
As seen in figure 4, the liquid accumulator behaves more like a reservoir with decreasing bulk modulus of elasticity, meaning pressure waves will reflect off the liquid accumulator rather than passing through. As the bulk modulus of elasticity increases, the liquid accumulator behaves like a rigid pipe, or as if there is no liquid accumulator in the pipeline at all.
Video 1: Variable K'
In the video you can see that each case responds to the valves closure with a pressure wave that propagates upstream until in hits the liquid accumulator at 50 feet. Once here the cases diverge. If you pause shortly after the pressure wave hits the accumulator you can see how the larger K’s behave more like there is no accumulator and how the smaller K’s behave more like there is a reservoir.
Figure 5: Maximum Pressure Stagnation vs. Length with varying initial volume
As shown in Figure 5, when the initial volume increases the liquid accumulator behaves more like a reservoir. As the initial volume of the liquid accumulator decreases, the system behaves more like there is no liquid accumulator.
Video 2: Variable Initial Volume
In the video you can see that each case responds to the valves closure with a pressure wave that propagates upstream until in hits the liquid accumulator at 50 feet. Once here the cases diverge. If you pause shortly after the pressure wave hits the accumulator you can see how the smaller volumes behave more like there is no accumulator and how the larger volumes behave more like there is a reservoir.
In summary, liquid accumulators are liquid full sections in a piping network that behave like flexible pipe. These sections change the system’s response to pressure waves that are created by transient events such as valve closures or openings and pump start-ups or trips. A stiff liquid accumulator will allow a pressure wave to transmit through the liquid accumulator more easily while a flexible liquid accumulator will reflect more of the pressure wave. A liquid accumulator with a large volume will behave similarly to a reservoir and reflect much of a pressure wave, while one with low volume will allow more of a pressure wave to transmit through. Overall, when using a liquid accumulator, be sure to properly balance these two variables (initial volume and bulk modulus of elasticity) so that liquid accumulators are properly implemented. | 1,315 | 6,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.91899 |
https://questioncove.com/updates/4dfb71800b8bbe4f12e64548 | 1,716,136,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00202.warc.gz | 411,456,845 | 5,534 | Mathematics 39 Online
OpenStudy (anonymous):
17=26-x
OpenStudy (anonymous):
x = 9
OpenStudy (anonymous):
Add x to both sides, then subtract 17 from both sides.
OpenStudy (anonymous):
that didnt help me i dont understand
OpenStudy (anonymous):
Do you agree that I can add any constant to both sides of the equations and it would remain intact? For example, 2 + 4 = 6 --> (2+4)+8 = 6+8? The same holds true for subtraction as well. To find x, you need to bring all the numbers to one side and take x to the other.
OpenStudy (anonymous):
which of the following is a solution of the equation 2x-4=5? | 162 | 606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.951709 |
http://mathhelpforum.com/algebra/176521-percents-word-problem-translation-print.html | 1,527,465,950,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870497.66/warc/CC-MAIN-20180527225404-20180528005404-00391.warc.gz | 197,783,897 | 3,335 | # Percents Word Problem Translation
• Apr 1st 2011, 06:02 AM
dumluck
Percents Word Problem Translation
Hi All,
Question: "x is what percent of y percent of z, in terms of x,y,z"
Y percent of z = $\displaystyle \frac{y}{100}.z$
X percent of y percent of z = $\displaystyle \frac{w}{100}.\frac{y}{100}.z$
This is where it gets confusing for me.
This translates to x = $\displaystyle \frac{w(\frac{yz}{100})}{100}$
Question one: I'm not sure how $\displaystyle \frac{w}{100}.\frac{y}{100}.z$ transforms into $\displaystyle \frac{w(\frac{yz}{100})}{100}$
Question two: I'm not sure how $\displaystyle \frac{w(\frac{yz}{100})}{100}$ transforms into $\displaystyle w = \frac{100x}{\frac{yz}{100}}$
Question three: I'm not sure how $\displaystyle w = \frac{100x}{\frac{yz}{100}}$transforms into $\displaystyle w = \frac{10,000x}{yz}$
There just seems to be jumps in logic that I'm not getting.
Thx,
D
• Apr 1st 2011, 07:17 AM
Quacky
Quote:
Question one: I'm not sure how $\displaystyle \frac{w}{100}.\frac{y}{100}.z$ transforms into $\displaystyle \frac{w(\frac{yz}{100})}{100}$
It is a very strange setup!
$\displaystyle \frac{w}{100}\times\frac{y}{100}\times~z$
$\displaystyle =\frac{w}{100}\times\frac{yz}{100}$
$\displaystyle =\dfrac{\frac{w\times~yz}{100}}{100}$
$\displaystyle =\frac{w\times\frac{yz}{100}}{100}$
Although why you'd set something out so poorly is beyond me!
Quote:
Question two: I'm not sure how $\displaystyle \frac{w(\frac{yz}{100})}{100}$ transforms into $\displaystyle w = \frac{100x}{\frac{yz}{100}}$
Again, this is a dreadful approach! I'd not even bother going whatever route they've taken. Starting at:
$\displaystyle x=\frac{w(\frac{yz}{100})}{100}$, multiply both sides by 100:
$\displaystyle 100x=w\times\frac{yz}{100}$ Then do it again:
$\displaystyle 10000x=w\times~yz$ Now divide both sides by $\displaystyle yz$
$\displaystyle \frac{10000x}{yz}=w$ ...which is their final answer.
Do you follow my method? The approach you've been shown is really quite bad. (Speechless)
• Apr 1st 2011, 07:21 AM
earboth
Quote:
Originally Posted by dumluck
Hi All,
Question: "x is what percent of y percent of z, in terms of x,y,z"
Y percent of z = $\displaystyle \frac{y}{100}.z$
X percent of y percent of z = $\displaystyle \frac{w}{100}.\frac{y}{100}.z$
This is where it gets confusing for me.
This translates to x = $\displaystyle \frac{w(\frac{yz}{100})}{100}$
Question one: I'm not sure how $\displaystyle \frac{w}{100}.\frac{y}{100}.z$ transforms into $\displaystyle \frac{w(\frac{yz}{100})}{100}$
$\displaystyle \frac{w}{100} \cdot \frac{y}{100}\cdot z = \dfrac1{100} \cdot w \cdot \left( \dfrac{y z}{100} \right) = \dfrac1{100} \cdot \left( w \cdot \left( \dfrac{y z}{100} \right) \right) = \dfrac{\left( w \cdot \left( \dfrac{y z}{100} \right) \right)}{100}$
Quote:
Question two: I'm not sure how $\displaystyle \frac{w(\frac{yz}{100})}{100}$ transforms into $\displaystyle w = \frac{100x}{\frac{yz}{100}}$
$\displaystyle x=\dfrac{w(\frac{yz}{100})}{100}~\implies~100x=w(\ frac{yz}{100})~\implies~w=\dfrac{100x}{\frac{yz}{1 00}}$
Quote:
Question three: I'm not sure how $\displaystyle w = \frac{100x}{\frac{yz}{100}}$transforms into $\displaystyle w = \frac{10,000x}{yz}$
There just seems to be jumps in logic that I'm not getting.
Thx,
D
I'll leave the last step for you: You have to divide a term (100x) by a fraction. How do you do this usually?
• Apr 1st 2011, 07:52 AM
dumluck
Quote:
Originally Posted by earboth
$\displaystyle \frac{w}{100} \cdot \frac{y}{100}\cdot z = \dfrac1{100} \cdot w \cdot \left( \dfrac{y z}{100} \right) = \dfrac1{100} \cdot \left( w \cdot \left( \dfrac{y z}{100} \right) \right) = \dfrac{\left( w \cdot \left( \dfrac{y z}{100} \right) \right)}{100}$
$\displaystyle x=\dfrac{w(\frac{yz}{100})}{100}~\implies~100x=w(\ frac{yz}{100})~\implies~w=\dfrac{100x}{\frac{yz}{1 00}}$
I'll leave the last step for you: You have to divide a term (100x) by a fraction. How do you do this usually?
multiply the reciprical of the denoiminator. Thanks to you both. | 1,380 | 4,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-22 | latest | en | 0.80556 |
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# 03hw - Solutions to problem set 3(141A Sp07 1 1(ISSP 3.3...
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Solutions to problem set 3 (141A Sp07) 1 1. (ISSP 3.3) Basis of two unalike atoms. At k = π/ a , equation (20) becomes 2 - ω 2 M 1 C ! u = 0 2 - ω 2 M 2 C ! v = 0 Note that the u and v sub-lattices are decoupled at this value of k ; u can change without changing v , and vice versa. There are two solutions to the above equations: u = 0, ω 2 = 2 C / M 2 and v arbitrary; and v = 0, ω 2 = 2 C / M 1 and u arbitrary. 2. (ISSP 3.4) Kohn anomaly. Equation (16a) gives ω 2 = 2 A M X p = 1 sin pk 0 a pa (1 - cos( pka )) = 2 A Ma X p = 1 sin pk 0 a p - 1 2 sin p ( k 0 + k ) a p - 1 2 sin p ( k 0 - k ) a p ! Gradshteyn and Ryzhik ( Table of integrals, series and products )1.441 gives X p = 1 sin α p p = π - α 2 0 < α < 2 π (you can prove this by writing the sin as the sum of two exponentials, and then using the taylor expansion of log (1 + x ).) So for k 0 > k , we have ω 2 = 2 A Ma X p = 1 sin pk 0 a p - 1 2 sin p ( k 0 + k ) a p - 1 2 sin p ( k 0 - k ) a p ! = 2 A Ma π - k 0 a 2 - 1 2 π - ( k 0 + k ) a 2 - 1 2 π - ( k 0 - k ) a 2 ! = 0 Now consider the case where k 0 < k . In order to apply the formula from the tables we need the argument of each sin function to be positive. Then ω 2 = 2 A Ma X p = 1 sin pk 0 a p - 1 2 sin p ( k 0 + k ) a p - 1 2 sin p ( k 0 - k ) a p !
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Ask a homework question - tutors are online | 699 | 1,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-13 | latest | en | 0.76718 |
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# Suppose a sequence begins 2, 4, …………………. Would could the formula be? - PowerPoint PPT Presentation
Suppose a sequence begins 2, 4, …………………. Would could the formula be?. Look at the Tarsia Puzzle Match up the sequences with their correct formulae. Find the formula for this linear sequence: 4, 8, 12, 16, 20, 24, ………………. Find the formula for this linear sequence:
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Suppose a sequence begins
2, 4, ………………….
Would could the formula be?
Look at the Tarsia Puzzle
Match up the sequences with their correct formulae
Find the formula for this linear sequence:
4, 8, 12, 16, 20, 24, ………………..
Find the formula for this linear sequence:
4, 8, 12, 16, 20, 24, ………………..
Sequence is the 4 times table so 4n
Find the formula for this linear sequence:
4, 8, 12, 16, 20, 24, ………………..
Sequence is the 4 times table so 4n
5, 9, 13, 17, 21, 25, ………………..
Find the formula for this linear sequence:
4, 8, 12, 16, 20, 24, ………………..
Sequence is the 4 times table so 4n | 516 | 1,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-47 | latest | en | 0.890204 |
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# How do you teach fractions?
A:
Different methods of teaching fractions can be used by finding mediums that appeal to the many different learning styles adopted by different individuals. When the best possible medium has been identified to teach a person, then it is simpler to introduce and explain a variety of concepts.
## Keep Learning
Bright colored pictures are important when teaching people who have a visual style of learning. Visual learners can best retain and recall information that they see. The ideal mediums of teaching such people would be vivid posters, flash cards, charts and activity books.
Food would be a suitable medium for a kinaesthetic learner, as this type of learner best retains information that is demonstrated. Some of the ways they can familiarise themselves with the concept of fractions is by cutting up cake, sandwiches or pizza, and then rearranging the pieces. Animated computer programmes or DVDs can also prove beneficial not only to the kinaesthetic learner, but for auditory learners as well.
Because auditory learners have a better understanding of the things they hear, carefully worded lectures or recordings of text books would be most effective. Group discussions are also advisable as these types of learners value personal interaction as a medium of learning. Whatever method is used must always have a solid foundation and structure to avoid confusion or misunderstanding on the part of the learner.
Sources:
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1 ME 3E05 (2017-2018) Assignment # 4 [1]. (2 marks) A non-rotating circular shaft supported on two bearings at the two ends is subjected to fully reversed cyclic transverse force P = 700 lbf, as shown in the figure below. The material of the shaft is AISI CD 1020 steel ( 57 , 68 y ut S Kpsi S Kpsi ) with a machined finish. The left and right shoulders have radii of 1/4 th and 1/8 th inch respectively. Determine the fatigue life of the shaft? Assume 99.99% reliability. [2]. (3 marks) A solid circular shaft with a machined fillet of radius 1/8 inch is subjected to a cyclic torque amplitude of 900 lbf-inch and a steady mid-range torque of 2500 lbf-inch. The shaft is required to have a transition diameter ratio (D/d 0 ) of 1.5. The material properties of the shaft are 57 y S Kpsi and 68 . ut S Kpsi Determine the shaft dimensions D and d 0 for a life of 500,000 cycles, assuming a factor of safety of 1.5. Utilize Goodman fatigue failure locus line in the analysis. Assume a reliability of 99%.
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https://azimpremjiuniversity.edu.in/news/2023/maths-from-simple-grids | 1,701,817,285,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00451.warc.gz | 147,240,816 | 15,397 | # Maths from simple grids
Gowri Satya, Ashwin, Shravan, Shivkumar, in At Right Angles (AtRiA), present some interesting problems centred around the humble and modest grid.
##### 21 April 2023
An exciting afternoon with friends and family, for me, often involves arguing endlessly over some trivial math or science question or solving a tricky puzzle or finding a pattern where you least expected it.
When I started to teach math in an NPO (Non Profit Organisation) setting, I, most of all, wanted to bring this excitement and thrill of problem-solving into our classroom. Easier said than done!
I would pose what I thought was an interesting question and watch as students either struggled or simply did not even attempt it. Finding the right kind of problem that is challenging yet tantalisingly simple, that is familiar yet leads to new and exciting discoveries, has proved to be difficult. Having to find something that is easy to implement, cost-effective, and that works for all the students in my mixed classroom even more so.
So, when Seed2Sapling Education conducted a teacher training session where they presented interesting, insightful math questions at various levels generated from common everyday things, I was completely hooked. Students’ reactions to some of these ideas have been exactly what I had hoped for.
Here we present a few such interesting problems centred around the humble and modest grid. All the activities here can be worked out with paper and pencil. For younger or more kinesthetic learners, number tiles or even a life-size grid on the classroom floor could be used.
Answers to the puzzles are listed at the end of the article. Students from Grades II to V with a wide range of mathematical abilities can find something of appeal in these problems.
### Preschool Corner — Number Grid Exploration
The most basic grid every child is bound to come across in school is the basic 1 – 100 number grid. While writing out the number chart is an oft repeated exercise, there are hours of fun to be had with finding patterns in it.
Given a 33 grid from the 1010 number grid, fill in the missing numbers (Fig. 1, Fig. 2).
Or fill in the missing numbers in puzzle pieces of various shapes (Fig. 3, Fig. 4).
Or a puzzle piece can be left empty (Fig. 5). Then the challenge for the student is to find all the possible values that can work for it — first with a number grid for reference, then without.
Deeper understanding of the place value system can be fostered by asking students to explain in their own words their strategies for filling in the missing numbers. Mistakes made rather than just being corrected can be collected and students can observe for themselves the patterns of their mistakes. Again, asking students to explain in their own words what mistakes they made and why, can lead to rich math conversations.
Students can also design their own puzzle pieces. What is the most difficult puzzle piece you can come up with?
This same familiar activity can be repeated in later classes with an addition grid or a
multiplication grid to bring out the patterns in addition and multiplication.
### Exploring Ordering- Inc/Dec Puzzle
Given a 3×3 grid and the numbers 1 to 9 arrange them in the grid such that each row and each column has numbers in increasing order (Fig. 6).
Each number can and should be used only once. What if each row and column had to be in decreasing order (Fig. 7)?
Now what if all rows were to be in increasing order and columns in decreasing order (Fig. 8, solved for reference)? Or vice versa (Fig. 9)?
Now let us mix things up a bit. Have a go at the puzzles in Fig. 10 and Fig. 11.
While the first four puzzles could be solved intuitively, the last two require a little more organized effort. A puzzle at this level would be ideal to introduce transitivity of greater than’ and less than’ to younger students.
Now that we have six solutions under our belt, it is time to make some observations.
• Are there any patterns in the solutions? One thing that stands out is the placement of 1’ and 9’. Being the smallest and largest numbers in the set, they naturally take their positions in one of the four corners of the grid. Can the same be said of 2’ and 8’? Do they always take a position in the middle of a row or column?
• Could a given grid configuration have more than one solution? This is always an interesting question – can there be only one solution or are more solutions possible. Discussions around this can lead to insights that may be missed in just solving the problem.
• Are there any patterns in our approach to solving them? Based on our observations about 1 and 9, the first step could be to find possible positions for them. Next, try to find possible positions for 2 and 8 and so on. Or perhaps take the opposite route and shortlist the possible numbers for each cell in the grid. How about using transitivity of the less than’ and greater than’ to solve the puzzles?
Once students have solved a few puzzles and understood the constraints, what works, what does not, it’s time to up the ante. Let us flip the question. Create your own puzzle. Students can create a grid with the conditions of increasing and decreasing and give it to a friend or teacher to solve. (What student wouldn’t like to school her teacher?) Some interesting grids that students came up with:
This exercise opens up some more questions:
• How many ways are there to arrange Inc and Dec around a grid to set a puzzle like this? There are 3 rows and 3 columns giving in all 6 positions to be filled with one of two values — Inc’ or Dec’. What would be the total number of combinations? If there were 2 positions to fill, we would have 2 x 2 = 4 possible combinations: Inc Inc, Inc Dec, Dec Inc, Dec Dec
If there were 3 positions to fill, we would have an Inc- extension to all the above sequences and then a Dec- extension to all the above sequences like this:
Inc Inc Inc, Inc Inc Dec, Inc Dec Inc, Inc Dec Dec
Dec Inc Inc, Dec Inc Dec, Dec Dec Inc, Dec Dec Dec
That is, with each extra position that is added, the number of combinations doubles.
This gives us 26 = 64 combinations for 6 positions.
• Will all of those combinations have a solution or is it possible to create a puzzle with no solution? Given that the two conditions of increasing’ and decreasing’ are contradictory, it seems likely that a particular configuration could lead to a contradiction that makes the puzzle unsolvable.
Let’s try to create one then. After some attempts, we hit upon the configurations shown in Figures 14 and 15. The proofs that these configurations are unsolvable is given separately at the end (see the Appendix, at the end of this article). The reader may enjoy looking for independent proofs.
• How does adding or removing constraints affect the puzzle? Along with the row and column constraints, we could add a constraint on the diagonals as well. Consider Fig. 16. Is this harder to solve? Can we create more unsolvable puzzles with the extra constraints?
What if we drop a constraint? Instead of specifying Inc / Dec on every row and column, we could leave some with no constraints as in Fig. 17. Does this make the puzzle easier to solve? Is it harder to create an unsolvable puzzle if you could impose only 4 constraints instead of 6.
Challenge Question: How do we efficiently identify whether a given configuration can be solved or not?
We have one way to identify an unsolvable puzzle but is that the only condition resulting in no solutions? Consider the puzzle in Fig.18.
Here, 9 and 1 have clearly identifiable cells that they could fill. Does that make this a solvable puzzle? Could there be a contradiction elsewhere?
How can we efficiently find out if a given configuration is solvable? Can this be extended to a generalized Inc/Dec puzzle on a n × grid or even further to a n × rectangular grid?
It turns out that this question has an unexpected twist!
You can play Inc/Dec puzzles online at: http://mathventure.in/games/incdec.html
### Solutions
Some of these puzzles have multiple solutions but only one possible solution has been listed here.
For more puzzles, please visit the addendum to this article in the online version available at http://publications.azimpremjifoundation.org/3344/
Appendix: Proof that the configurations shown in Figure 14 and Figure 15 are not solvable
Let the grid be as follows.
Let’s look at Figure 14 first (shown alongside). Since column #1 is Inc and row #1 is Dec and column #3 is Dec, it must be that:
i < f < c < b < a < d < g.
This implies that i < g. However, row #3 is Inc, which implies that g < i. These two conclusions are contradictory. Therefore, this configuration is not solvable.
Similarly, in Figure 15, since column #1 is Dec and row #1 is Inc and column #3 is Inc, it must be that:
i > f > c > b > a > d > g,
implying that i > g. However, row #3 is Dec, which implies that g > i. Once again, we reach a contradiction. Therefore, this configuration too is not solvable.
### About the authors
Ashwin , Shravan and Shivkumar are all alumni of IISc, Bengaluru. They are now working full time on rejuvenating math education in schools through an education startup Seed2Sapling Education.
They can be contacted at guha.ashwin@gmail.com (Ashwin), sra9449@gmail.com (Shravan) and shiv1729@gmail.com (Shivkumar).
Gowri Satya teaches math at Swapaksh Learning Foundation, a learning centre for underprivileged children in Bengaluru. Formerly a software engineer, she enjoys math and hopes to spread the same enthusiasm to her students.
She can be contacted at s.gowri@gmail.com. | 2,316 | 9,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-50 | longest | en | 0.942493 |
http://www.nag.com/numeric/CL/nagdoc_cl24/html/F02/f02abc.html | 1,500,968,385,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425082.56/warc/CC-MAIN-20170725062346-20170725082346-00561.warc.gz | 512,443,171 | 3,827 | f02 Chapter Contents
f02 Chapter Introduction
NAG Library Manual
# NAG Library Function Documentnag_real_symm_eigensystem (f02abc)
## 1 Purpose
nag_real_symm_eigensystem (f02abc) calculates all the eigenvalues and eigenvectors of a real symmetric matrix.
## 2 Specification
#include #include
void nag_real_symm_eigensystem (Integer n, const double a[], Integer tda, double r[], double v[], Integer tdv, NagError *fail)
## 3 Description
nag_real_symm_eigensystem (f02abc) reduces the real symmetric matrix $A$ to a real symmetric tridiagonal matrix by Householder's method. The eigenvalues and eigenvectors are calculated using the $QL$ algorithm.
## 4 References
Wilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag
## 5 Arguments
1: nIntegerInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 1$.
2: a[${\mathbf{n}}×{\mathbf{tda}}$]const doubleInput
Note: the $\left(i,j\right)$th element of the matrix $A$ is stored in ${\mathbf{a}}\left[\left(i-1\right)×{\mathbf{tda}}+j-1\right]$.
On entry: the lower triangle of the $n$ by $n$ symmetric matrix $A$. The elements of the array above the diagonal need not be set. See also Section 9
3: tdaIntegerInput
On entry: the stride separating matrix column elements in the array a.
Constraint: ${\mathbf{tda}}\ge {\mathbf{n}}$.
4: r[n]doubleOutput
On exit: the eigenvalues in ascending order.
5: v[${\mathbf{n}}×{\mathbf{tdv}}$]doubleOutput
Note: the $\left(i,j\right)$th element of the matrix $V$ is stored in ${\mathbf{v}}\left[\left(i-1\right)×{\mathbf{tdv}}+j-1\right]$.
On exit: the normalized eigenvectors, stored by columns; the $i$th column corresponds to the $i$th eigenvalue. The eigenvectors are normalized so that the sum of squares of the elements is equal to 1.
6: tdvIntegerInput
On entry: the stride separating matrix column elements in the array v.
Constraint: ${\mathbf{tdv}}\ge {\mathbf{n}}$.
7: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_2_INT_ARG_LT
On entry, ${\mathbf{tda}}=⟨\mathit{\text{value}}⟩$ while ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$. These arguments must satisfy ${\mathbf{tda}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{tdv}}=⟨\mathit{\text{value}}⟩$ while ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$. These arguments must satisfy ${\mathbf{tdv}}\ge {\mathbf{n}}$.
NE_ALLOC_FAIL
Dynamic memory allocation failed.
NE_INT_ARG_LT
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 1$.
NE_TOO_MANY_ITERATIONS
More than $⟨\mathit{\text{value}}⟩$ iterations are required to isolate all the eigenvalues.
## 7 Accuracy
The eigenvectors are always accurately orthogonal but the accuracy of the individual eigenvectors is dependent on their inherent sensitivity to changes in the original matrix. For a detailed error analysis see pages 222 and 235 of Wilkinson and Reinsch (1971).
## 8 Parallelism and Performance
Not applicable.
The time taken by nag_real_symm_eigensystem (f02abc) is approximately proportional to ${n}^{3}$.
The function may be called with the same actual array supplied for arguments a and v, in which case the eigenvectors will overwrite the original matrix.
## 10 Example
To calculate all the eigenvalues and eigenvectors of the real symmetric matrix
$0.5 0.0 2.3 -2.6 0.0 0.5 -1.4 -0.7 2.3 -1.4 0.5 0.0 -2.6 -0.7 0.0 0.5 .$
### 10.1 Program Text
Program Text (f02abce.c)
### 10.2 Program Data
Program Data (f02abce.d)
### 10.3 Program Results
Program Results (f02abce.r) | 1,166 | 3,624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 31, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-30 | latest | en | 0.462343 |
https://facesofberlin.org/division-of-exponents-worksheet/ | 1,656,416,380,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00506.warc.gz | 304,928,410 | 9,136 | HomeSchool Education ➟ 0 Division Of Exponents Worksheet
Division Of Exponents Worksheet
Division Of Exponents Worksheet. Multiplying and dividing exponents worksheets. 24 15 c c _____ 5.
1) 22 23 = 2) 24 22 = 3) 55 5 = 4) 11) 3 35 = = 5) = 3 = 16 6) 3×33 32×34 = 8 7) 58 53 = 8) 5×56 52×57 = 9) 34×37 32×38 = 10) 5 10 3 = 3 3 2 5 = 12) 12 3 14 6 13) 12 3 9y8 = 14) 25 y4 5 6y2 = 15) 2 4 7 = 16) 16 2y8 10) simplify (4 x 109) ÷ (16 x 106) and write the answer in scientific notation. 20 17 2 2 _____ 6.
The Rule States That You Can Divide Two Powers With The Same Base By Subtracting The Exponents.
Complementary and supplementary word problems worksheet. Division properties of exponents worksheet. Multiplying and dividing exponents worksheets.
Dividing Exponents Worksheets Can Help Students To Understand The Concept Of Exponents And How To Apply To Solve More Complex Math Problems That Have Multiple Terms And Multiple Operations.
Use the buttons below to print, open, or download the pdf version of the dividing exponents with a larger or equal exponent in the dividend (with negatives) (a) math worksheet. Division properties of exponents ws (1) red shoes simplify each expression. 20 4 x xy _____ 7.
26 11 A A _____ 4.
10) simplify (4 x 109) ÷ (16 x 106) and write the answer in scientific notation. | 420 | 1,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-27 | latest | en | 0.646443 |
https://czkarp.pl/quantity/of/cement/in/m/per/cu_33816.html | 1,632,662,484,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00486.warc.gz | 232,709,751 | 6,244 | ### Conversion of density units rho = kg cubic meter unit ...
Conversion of density units Definition: density = mass divided by volume; symbol ρ = m / V ρ (rho) = density, m = mass, V = volume. The SI unit of density is kg/m 3. Water of 4 °C is the reference ρ = 1000 kg/m 3 = 1 kg/dm 3 = 1 kg/l or 1 g/cm 3 = 1 g/ml. Fill in the appropriate line the known density value
### Calculate Bags of PreMix Concrete Round + Square ...
Calculate Number of Bags of Premix Concrete needed for Round and Square Footing Holes and Small Concrete Slabs. ... Calculate bags required for square, round and rectangular footings and small slabs. Round Footing Holes: Number of Holes ... PreMix Concrete Bag Calculator Calculated at 2160 kg per 1m³ Allow extra for waste.
### Ready Mixed Concrete Volume Calculator Kendalls
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### Concrete Volume Estimate Engineering ToolBox
Estimate required concrete volume per sq. ft. of slab Engineering ToolBox Resources, Tools and Basic Information for Engineering and Design of Technical Applications! the most efficient way to navigate the Engineering ToolBox!
### how to calculate quantity of cement needed? | Yahoo Answers
Mar 13, 2009· How to calculate quantity of cement needed? I want to do a flooring of my new house by Cement:Sand (1:3)ratio and would like to know the quantity of cement needed per square meter for plastering. ... The concrete volume would then be cubic meters concrete per 1 square meter of floor (at .3 mm thickness).... or 75 cubic centimeters per ...
### Quantity of cement in Mix M20 concrete
Ans : Approximately 6 to bags of cement per cubic meter of concrete. n. The qty of cement will depend on other raw materials( viz coarse aggreagate,fine aggreagate etc ) used in the concrete.
### How Much Does Concrete Delivery Cost?
Summary: Concrete Delivery Prices. On average, concrete delivery costs 120 to 200 per cubic yard. Companies usually charge a standard delivery charge of 60 to 200. Customers typically pay between 1,200 and 1,600 for their concrete projects.
### Concrete Calculator Cemen Tech
This calculator is provided to assist the user in determining the number of concrete yards to order. The calculator gets the user within an approximate range of accuracy. Cubic .
### How to Calculate Cement, Sand and Aggregate required for 1 ...
The proportions of cement, sand, coarse aggregate and water plays an important role in determining the fresh and hardended properties of concrete. So care should be taken while calculating the amount of Cement, Sand and Aggregate required for 1 Cubic meter of Concrete. Method1: DLBD method to determine material requirement for 1 Cum concrete The DLBD (Dry .
### How much cement, sand and agregate do I need to make one ...
Mar 10, 2008· Knowing the area to be covered and the thickness of concrete I want to lay I can calculate the volume. I'm laying concrete over a small area of garden to set up a trampoline and BBQ area, so I'm guessing a mix of 1:3:5? I want to figure out how many 25kg bags of cement, sand and agregate I'll need per cubic metre.
### How Many Bags of Cement Are in a Cubic Yard? | Hunker
This standard concrete mix (used for most premix applications) weighs about 133 pounds per cubic foot of concrete, so a cubic yard (which contains 27 cubic feet) will weigh in at about 3,600 pounds. With the given mix ratio, cured concrete is about 22 percent cement by weight.
### How to Measure a Cubic Foot of Concrete The Spruce
The surface area of a concrete slab or sand base is directly affected by its thickness. In the photo, you see two examples of a measured cubic a cubic foot is measured by Length x Width x Thickness, a cube 1foot long x 1foot wide x 1 foot thick is therefore 1 cubic foot.
### How It Works: Concrete Popular Mechanics
Oct 01, 2009· To make 1 cubic yard of concrete, you'd need seven 94pound bags of cement, about 1/2 cubic yard of sand and just over 3/4 cubic yard of gravel. The amount .
### Quantity of cement and sand in 1m2 of Masonry walling in ...
Nov 01, 2013· Quantity of cement and sand in 1m2 of Masonry walling in Nairobi, Kenya. ... Volume = assuming a 1 inch/ thickness = x = m3 of cement sand mortar per 1 m2. Ratio of cement to Sand for walling mortar. Assuming a cement/sand ratio of 1 to 8, the volumes will be as below. | 1,082 | 4,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-39 | latest | en | 0.803736 |
https://openstax.org/books/introductory-statistics/pages/8-6-confidence-interval-womens-heights | 1,716,580,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00111.warc.gz | 374,498,876 | 74,792 | Introductory Statistics
# 8.6Confidence Interval (Women's Heights)
Introductory Statistics8.6 Confidence Interval (Women's Heights)
## Stats Lab
### Confidence Interval (Women's Heights)
Class Time:
Names:
Student Learning Outcomes
• The student will calculate a 90% confidence interval using the given data.
• The student will determine the relationship between the confidence level and the percentage of constructed intervals that contain the population mean.
Given:
59.4 71.6 69.3 65 62.9 66.5 61.7 55.2 67.5 67.2 63.8 62.9 63.0 63.9 68.7 65.5 61.9 69.6 58.7 63.4 61.8 60.6 69.8 60.0 64.9 66.1 66.8 60.6 65.6 63.8 61.3 59.2 64.1 59.3 64.9 62.4 63.5 60.9 63.3 66.3 61.5 64.3 62.9 60.6 63.8 58.8 64.9 65.7 62.5 70.9 62.9 63.1 62.2 58.7 64.7 66.0 60.5 64.7 65.4 60.2 65.0 64.1 61.1 65.3 64.6 59.2 61.4 62 63.5 61.4 65.5 62.3 65.5 64.7 58.8 66.1 64.9 66.9 57.9 69.8 58.5 63.4 69.2 65.9 62.2 60.0 58.1 62.5 62.4 59.1 66.4 61.2 60.4 58.7 66.7 67.5 63.2 56.6 67.7 62.5
Table 8.8 Heights of 100 Women (in Inches)
1. Table 8.8 lists the heights of 100 women. Use a random number generator to select ten data values randomly.
2. Calculate the sample mean and the sample standard deviation. Assume that the population standard deviation is known to be 3.3 inches. With these values, construct a 90% confidence interval for your sample of ten values. Write the confidence interval you obtained in the first space of Table 8.9.
3. Now write your confidence interval on the board. As others in the class write their confidence intervals on the board, copy them into Table 8.9.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
Table 8.9 90% Confidence Intervals
Discussion Questions
1. The actual population mean for the 100 heights given Table 8.8 is μ = 63.4. Using the class listing of confidence intervals, count how many of them contain the population mean μ; i.e., for how many intervals does the value of μ lie between the endpoints of the confidence interval?
2. Divide this number by the total number of confidence intervals generated by the class to determine the percent of confidence intervals that contains the mean μ. Write this percent here: _____________.
3. Is the percent of confidence intervals that contain the population mean μ close to 90%?
4. Suppose we had generated 100 confidence intervals. What do you think would happen to the percent of confidence intervals that contained the population mean?
5. When we construct a 90% confidence interval, we say that we are 90% confident that the true population mean lies within the confidence interval. Using complete sentences, explain what we mean by this phrase.
6. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data given (the heights of women) and count how many of the data values lie within the confidence interval that you generated based on that data. How many of the 100 data values lie within your confidence interval? What percent is this? Is this percent close to 90%?
7. Explain why it does not make sense to count data values that lie in a confidence interval. Think about the random variable that is being used in the problem.
8. Suppose you obtained the heights of ten women and calculated a confidence interval from this information. Without knowing the population mean μ, would you have any way of knowing for certain if your interval actually contained the value of μ? Explain.
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As an Amazon Associate we earn from qualifying purchases. | 1,096 | 3,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | longest | en | 0.688299 |
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Reading the interesting paper Honest Bernoulli excursions by Smith and Diaconis motivated the question whether probabilistic interpretations for general Dedekind zeta functions are known.
In the paper the authors consider a random walk on the integers $\mathbb Z$ with a particle starting at 0 and moving left or right with the same probability of $\frac 1 2$. Further, they let $T$ be the time of first return to zero , and $M_T$ the maximum distance from $0$ reached by the walk up to time $T$.
Their main result says that the probability $$P(M_T \leq y \sqrt{\pi n} \ | \ T=2n ) = F(y)+ O(n^{-\frac12})$$ is uniformly in $y$, with distribution function $F(y)$ (defined on $[0,\infty)$ given by $$F(y) = \frac{4\pi}{y^3} \sum_{j=1}^\infty j^2 exp(-\pi j^2 / y^2)$$
The relation with Riemann's completed zeta function comes now from the observation that the "Mellin transform of the limiting measure $F$" gives 2 times the completed Riemann zeta function: $$\int_0^\infty y^s F(dy) = 2 \xi(s)$$
Recall that $\xi(s) = \Gamma(\frac s2+1)(s-1)\pi^{-s/2}\zeta(s)$ (with the usual notations).
My question is now whether similar probabilistic interpretations are known for other Dedekind zeta functions. A first guess would be to look at random walks on the ring of integers $\mathcal O _K$ of a number field $K$, or on some other appropriate spaces like certain (signed) integral ideals of $\mathcal O _K$.
Is anything known in this direction? Any idea or reference would be highly appreciated. (I should say that my background in probability theory tends to zero, unfortunately.)
EDIT: As I already pointed this out in the comments: What I am really interested about in some sense is the question (very vaguely speaking) whether the above procedure can be generalized to give a probabilistic interpretation to Hecke's method of expressing Dedekind zeta functions (Hecke did this of course more generally for Hecke L-functions) in terms of Mellin transforms of appropriate $\vartheta$-functions. (I have Neukirch's presentation of Hecke's work in his number theory book in mind.)
1
# Is there a probabilistic interpretation of Dedekind zeta functions?
Reading the interesting paper Honest Bernoulli excursions by Smith and Diaconis motivated the question whether probabilistic interpretations for general Dedekind zeta functions are known.
In the paper the authors consider a random walk on the integers $\mathbb Z$ with a particle starting at 0 and moving left or right with the same probability of $\frac 1 2$. Further, they let $T$ be the time of first return to zero , and $M_T$ the maximum distance from $0$ reached by the walk up to time $T$.
Their main result says that the probability $$P(M_T \leq y \sqrt{\pi n} \ | \ T=2n ) = F(y)+ O(n^{-\frac12})$$ is uniformly in $y$, with distribution function $F(y)$ (defined on $[0,\infty)$ given by $$F(y) = \frac{4\pi}{y^3} \sum_{j=1}^\infty j^2 exp(-\pi j^2 / y^2)$$
The relation with Riemann's completed zeta function comes now from the observation that the "Mellin transform of the limiting measure $F$" gives 2 times the completed Riemann zeta function: $$\int_0^\infty y^s F(dy) = 2 \xi(s)$$
Recall that $\xi(s) = \Gamma(\frac s2+1)(s-1)\pi^{-s/2}\zeta(s)$ (with the usual notations).
My question is now whether similar probabilistic interpretations are known for other Dedekind zeta functions. A first guess would be to look at random walks on the ring of integers $\mathcal O _K$ of a number field $K$, or on some other appropriate spaces like certain (signed) integral ideals of $\mathcal O _K$.
Is anything known in this direction? Any idea or reference would be highly appreciated. (I should say that my background in probability theory tends to zero, unfortunately.) | 996 | 3,785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2013-20 | latest | en | 0.878528 |
http://graziano-raulin.com/tutorials/spss/pearson.htm | 1,544,594,584,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823738.9/warc/CC-MAIN-20181212044022-20181212065522-00601.warc.gz | 121,656,516 | 4,071 | Pearson Correlation in SPSS for Windows
Graziano & Raulin
Research Methods (8th edition)
## Pearson Product-Moment Correlation
### Computing a Correlation
To compute the Pearson product-moment correlation between age and income, we select the Analyze menu, the Correlate submenu, and the Bivariate option (which literally means two variables). These selections will give us this screen
We select the variables age and income and move them to the right box. The default options in this procedure are to compute a Pearson product-moment correlation and a two-tail p-value. Since these are the options we want, we only need to click on OK to run the correlations. The resulting output is shown in this screen
Note that SPSS for Windows actually gives us four correlations. The program is set up to compute an intercorrelation matrix (i.e., the correlation of each variable with every other variable that was selected) for as many variables as you select. If you select 10 variables, there would be a 10 by 10 matrix of intercorrelations. The diagonal (from the upper left to the lower right) of that matrix will list the correlation of each variable with itself, which will always be equal to 1.00 (a perfect correlation). The correlational matrix above and below this diagonal will be mirror images of one another, because the correlation of variable X with variable Y is the same as the correlation of variable Y with variable X
In our simplified example of only two variables, the correlation between age and income is .883 and the probability of getting that correlation or a larger correlation by chance if the population correlation were actually zero (the p-value) is listed as .000. Actually, no correlation will give you a p-value of zero; but since only three digits are printed, a value listed as .000 actually means that the probability is less than .001. If we assume the traditional decision criteria (termed alpha level) of .05, we would conclude that these two variables are significantly correlated with one another, because our p-value is less than the alpha level of .05 (see Chapter 5 for a more detailed explanation of this terminology).
### Producing a Scatter Plot
Remember that a Pearson product-moment correlation is an index of the degree of linear relationship between two variables. That is, the correlation gives an indication of how closely the points in a scatter plot cluster around a straight line. But the relationship between two variables is not always linear. For example, if we correlated a measure of general health with weight, we would likely find that people who are either excessively heavy or light would have generally poor health and that people in the middle range of weight would likely be the healthiest.
To see what the shape of a relationship is like, you would prepare a scatter plot. SPSS for Windows will prepare a scatter plot with just a few mouse clicks. Click on the Graphs menu, Interactive submenu, and select the Scatterplot option. These actions will produce this screen. To prepare a scatter plot of age by income, move the income variable to X axis and then move the age variable to the Y axis of the graph shown. Clicking on the OK button will produce the scatter plot. As you can see, the relationship between age and income does appear to be linear in our data set.
### Animations
We have prepared a series of animations that will walk you through the procedures discussed on this page. To run an animation, simply click on the title of the animation in the table below.
Note that we do not recommend that you try to run the animations if you have a slow connection, such as a dial-up connection. You will find that the animations take forever to load with a slow connection.
Computing a Product-Moment Correlation Producing a Scatter Plot | 777 | 3,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-51 | latest | en | 0.929609 |
https://www.comeforanswers.com/sarah-age-29-makes-45000-a-year-and-wants-to-buy-a-15-year-term-policy-that-would-replace/ | 1,695,452,639,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00840.warc.gz | 793,543,688 | 8,972 | # Sarah, age 29, makes \$45,000 a year and wants to buy a 15-year term policy that would replace
Sarah, age 29, makes \$45,000 a year and wants to buy a 15-year term policy that would replace seven years of her salary. the annual premium rate (per \$1000 of face value) for her age group is \$1.46. what is sarah’s premium, to the nearest dollar? a. \$460 b. \$215 c. \$66 d. \$986 select the best answer from the choices provided a b c d | 126 | 441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-40 | latest | en | 0.94596 |
https://espanol.libretexts.org/Matematicas/Libro%3A_Calculo_elemental_(Corral)/02%3A_Derivadas_de_Funciones_Comunes/2.03%3A_Las_funciones_del_logaritmo_exponencial_y_natural | 1,709,524,639,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00272.warc.gz | 241,033,806 | 35,193 | Saltar al contenido principal
# 2.3: Las funciones del logaritmo exponencial y natural
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Las funciones de la forma$$a^x$$, donde$$x$$ varía el exponente, se denominan funciones exponenciales. A menos que se indique lo contrario, supongamos que$$a > 0$$ ($$0^x$$es solo 0, y no$$(-1)^{1/2}$$ es un número real). Ya sabes cómo$$a^x$$ se define cuando el exponente$$x$$ es un número racional (es decir,$$x = m/n$$ dónde$$m$$ y$$n$$ son enteros,$$n \ne 0$$). Pero, ¿y si$$x$$ fueran irracionales, como$$\sqrt{2}$$? ¿Qué$$3^{\sqrt{2}}$$ significaría?
\begin{aligned} {3} 3^{1.4} ~&=~ 3^{\frac{14}{10}} ~&=~ 4.65553672174608\\ 3^{1.41} ~&=~ 3^{\frac{141}{100}} ~&=~ 4.70696500171657\\ 3^{1.414} ~&=~ 3^{\frac{1414}{1000}} ~&=~ 4.72769503526854\\ 3^{1.4142} ~&=~ 3^{\frac{14142}{10000}} ~&=~ 4.72873393017119\\ 3^{1.41421} ~&=~ 3^{\frac{141421}{100000}} ~&=~ 4.72878588090861\\ 3^{1.414213} ~&=~ 3^{\frac{1414213}{1000000}} ~&=~ 4.72880146624114\\ {} &\vdots \quad ~&\vdots \\ 3^{1.414213562\ldots} ~&=~ 3^{\sqrt{2}} ~&=~ 4.72880438783742\end{aligned}Por supuesto que nunca harías todo esto a mano, simplemente usarías una computadora o calculadora, que utilizan algoritmos mucho más eficientes para calcular potencias en general. 3
Todas las reglas habituales de exponentes que aprendiste en álgebra se aplican$$a^x$$ cuando se definen de la manera descrita anteriormente, con$$a > 0$$ y$$x$$ variando sobre todos los números reales. De todos los valores posibles para la base$$a$$, el que más aparece en las matemáticas, las ciencias y la ingeniería es la base$$e$$, definida como:
El límite en esta definición significa que a medida que$$x$$ se hace más grande —acercándose al infinito ($$\infty$$)— los valores de$$\left( 1 ~+~ \frac{1}{x} \right)^x$$ acercarse a un número, denotado por$$e$$. Más posiciones decimales para se$$e$$ pueden obtener haciendo$$x$$ suficientemente grandes. 4 Por ejemplo, cuando$$x=5 \times 10^6$$ el valor es 2.718281555200129. Para valores extremadamente grandes de$$x$$, es decir, cuándo$$x \gg 1$$ (el símbolo$$\gg$$ significa “mucho mayor que”),
$e ~\approx~ \left( 1 ~+~ \frac{1}{x} \right)^x \quad\Rightarrow\quad e^{1/x} ~\approx~ \left(\left( 1 ~+~ \frac{1}{x} \right)^x\right)^{1/x} ~=~ 1 ~+~ \frac{1}{x} \quad\Rightarrow\quad \left(e^{1/x} ~-~ 1\right)\,x ~\approx~ \left(\frac{1}{x}\right)\,x ~=~ 1 ~,$así que dejando$$h = 1/x$$, y señalando que$$h = 1/x \to 0$$ si y solo si$$x \to \infty$$, arroja el límite útil: 5 Usando el límite anterior, se$$y = e^x$$ puede encontrar la derivada de:
Utilizando la definición límite de la derivada para$$f(x) = e^x$$,
\begin{aligned} \ddx\,\left(e^x\right) ~&=~ \lim_{h \to 0} ~\frac{f(x+h) ~-~ f(x)}{h} ~=~ \lim_{h \to 0} ~\frac{e^{x+h} ~-~ e^x}{h}\ \ [6pt] &=~\ lim_ {h\ a 0} ~\ frac {e^x\ izquierda (e^h ~-~ 1\ derecha)} {h} ~=~ e^x\;\ cdot\;\ lim_ {h\ a 0} ~\ frac {e^h ~-~ 1} {h}\ quad\ text {(ya que e^x no dependa de h)}\ \ [6pt] &=~ e^x\;\ cdot\; 1 ~=~ e^x\ final {alineado}
En general, para una función diferenciable$$u = u(x)$$ como exponente, la Regla de Cadena rinde:
Ejemplo$$\PageIndex{1}$$: deriv4ex2
Agrega texto aquí.
Solución
Encuentra la derivada de$$y = 4e^{-x^2}$$.
Solución:$$\dydx ~=~ 4 e^{-x^2} \;\cdot\; \ddx\,(-x^2) ~=~ 4 e^{-x^2} \;\cdot\; (-2x) ~=~ -8x e^{-x^2}$$
A la función$$e^x$$ se le suele referir simplemente como la función exponencial, aunque obviamente hay muchas funciones exponenciales. ¿Qué hace que la base sea$$e$$ tan especial? Tomar$$y = Ae^{kt}$$ para representar la cantidad de alguna cantidad física a la vez$$t$$, para algunas constantes$$A$$ y$$k$$. Entonces
$\dydt ~=~ \ddt\,\left(Ae^{kt}\right) ~=~ k \cdot Ae^{kt} = ky ~,$que dice que la tasa instantánea de cambio de la cantidad es directamente proporcional a la cantidad presente en ese instante. Resulta que muchas cantidades físicas exhiben ese comportamiento, algunas de las cuales serán discutidas en breve. Por el contrario, en el Capítulo 5 se demostrará que cualquier solución a la ecuación diferencial$$\dydt = ky$$ debe ser de la forma$$y = Ae^{kt}$$ para alguna constante$$A$$. Esto es lo que le da a la función exponencial su especial significación.
$$f(x) = e^x$$Déjese ser la función exponencial. Entonces$$f(x) > 0$$ para todos$$x$$ y$$f'(x) = f(x) = e^x > 0$$ para todos$$x$$, y así$$f(x)$$ es estrictamente creciente. El gráfico se muestra en la Figura [fig:exp].
Así, la función exponencial es uno a uno sobre el conjunto de todos los números reales y por lo tanto tiene una función inversa, llamada la función de logaritmo natural, denotada (en función de$$x$$) as$$f^{-1}(x) = \ln\,x$$. El gráfico se muestra en la Figura [fig:ln]. A continuación se muestra un resumen de la relación entre$$e^x$$ y$$\ln\,x$$:
El lector debe ser consciente de que muchos, si no la mayoría, campos fuera de las matemáticas utilizan la notación$$\log\,x$$ en lugar de$$\ln\,x$$ para la función de logaritmo natural. 6 A partir del álgebra se deben conocer las siguientes propiedades del logaritmo natural, junto con sus propiedades equivalentes en términos de la función exponencial: 7
Para encontrar la derivada de$$y = \ln\,x$$, use$$x = e^y$$:
$\dydx ~=~ \frac{1}{\dxdy} ~=~ \frac{1}{\ddy\,\left(e^y\right)} ~=~ \frac{1}{e^y} ~=~ \frac{1}{x}$De ahí que: En general, para una función diferenciable$$u = u(x)$$, la Regla de Cadena rinde:
Ejemplo$$\PageIndex{1}$$: derivlnx2
Agrega texto aquí.
Solución
Encuentra la derivada de$$y = \ln\,\left(x^2 + 3x - 1\right)$$.
Solución:$$\Dydx ~=~ \dfrac{1}{x^2 + 3x - 1} \;\cdot\; \Ddx\,(x^2 + 3x - 1) ~=~ \dfrac{2x + 3}{x^2 + 3x - 1}$$
Recordemos que$$\abs{x} = -x$$ para$$x < 0$$, en cuyo caso$$\ln\,(-x)$$ se define y
$\ddx\,(\ln\,\abs{x}) ~=~ \ddx\,(\ln\,(-x)) ~=~ \frac{1}{-x} \;\cdot\; (-1) ~=~ \frac{1}{x} ~.$Combina ese resultado con la derivada$$\ddx\,(\ln\,x) = \frac{1}{x}$$$$x > 0$$ para obtener:
Para algunas funciones es más fácil diferenciar primero el logaritmo natural de la función y luego resolver para la derivada de la función original. Esta técnica se denomina diferenciación logarítmica, demostrada en los siguientes dos ejemplos.
Ejemplo$$\PageIndex{1}$$: derivxx
Agrega texto aquí.
Solución
Encuentra la derivada de$$y = x^x$$.
Solución: Para este ejemplo supongamos$$x > 0$$ (ya que$$x$$ es tanto la base como el exponente). Tenga en cuenta que no puede usar la Regla de Potencia para esta función ya que el exponente$$x$$ es una variable, no un número fijo. En su lugar, tome el logaritmo natural de ambos lados de la ecuación$$y = x^x$$ y luego tome la derivada de ambos lados y resuelva para$$y'$$:
\begin{aligned} \ln\,y ~&=~ \ln\,\left(x^x\right) ~=~ x \;\cdot\; \ln\,x\\ \ddx\,(\ln\,y) ~&=~ \ddx\,(x \;\cdot\; \ln\,x)\ \ [4pt]\ frac {y'} {y} ~&=~ 1\;\ cdot\;\ ln\, x ~+~ x\;\ cdot\;\ frac {1} {x}\ \ [4pt] y' ~&=~ y\, (\ ln\, x ~+~ 1) ~=~ x^x\, (\ ln\, x ~+~ 1)\ end {alineado}
Ejemplo$$\PageIndex{1}$$: derivlogdiff
Agrega texto aquí.
Solución
Encuentra la derivada de$$y = \frac{(2x + 1)^7 (3x^3 - 7x + 6)^4}{(1 + \sin x)^5}$$.
Solución: Use la diferenciación logarítmica tomando el logaritmo natural$$y$$ y luego use las propiedades de los logaritmos para simplificar la diferenciación antes de resolver$$y'$$:
\begin{aligned} \ln\,y ~&=~ \ln\,\left(\frac{(2x + 1)^7 (3x^3 - 7x + 6)^4}{(1 + \sin\,x)^5}\right) ~=~ \ln\,\left((2x + 1)^7 (3x^3 - 7x + 6)^4\right) ~-~ \ln\,\left((1 + \sin\,x)^5\right)\ \ [6pt] &=~\ ln\,\ izquierda ((2x + 1) ^7\ derecha) ~+~\ ln\,\ izquierda ((3x^3 - 7x + 6) ^4\ derecha) ~-~\ ln\,\ izquierda ((1 +\ sin\, x) ^5\ derecha)\ \ [4pt]\ ddx\, (\ ln\, y) &=~\ ddx\,\ izquierda (7\,\ ln\, (2x + 1) ~+~ 4\,\ ln\, (3x^3 - 7x + 6) ~-~ 5\,\ ln\, (1 +\ sin\, x)\ derecha)\\ frac {y'} {y} &=~ 7\ cdot\ frac {2} {2x + 1} ~+~ 4\ cdot\ frac {9x^2 - 7} {3x^3 - 7x + 6} ~-~ 5\ cdot\ frac {\ cos\, x} {1 +\ sin\, x}\ \ [4pt] y' ~&=~ y\ cdot\ izquierda (\ frac {14} {2x + 1} ~+~\ frac {36x^2 - 28} {3x^3 - 7x + 6} ~-~\ frac {5\ cos\, x} {1 +\ sin\, x}\ derecha)\ \ [6pt] &=~\ frac {(2x + 1) ^7 (3x^3 - 7x + 6) ^4} {(1 +\ sin x) ^5}\ cdot\ izquierda (\ frac {14} {2x + 1} ~+~\ frac {36x^2 - 28} {3x^3 - 7x + 6} ~-~\ frac {5\ cos\, x} {1 +\ sin\, x}\ derecha)\ fin {alineado}
Un ejemplo clásico de la ecuación diferencial$$\dydt = ky$$ es el caso de la desintegración exponencial de una sustancia radiactiva., a menudo referido simplemente como desintegración radiactiva. En este caso la solución general$$y = Ae^{kt}$$ representa la cantidad de la sustancia en el momento$$t \ge 0$$, y la constante de decaimiento$$k$$ es negativa:$$\dydt < 0$$ ya que la sustancia se está desintegrando (es decir, la cantidad de sustancia está disminuyendo) mientras$$y > 0$$, así$$\dydt = ky$$ implica que $$k < 0$$.
La constante$$A$$ es la cantidad inicial de la sustancia, es decir, la cantidad en el tiempo$$t = 0$$:$$y(0) = Ae^{0t} = Ae^0 = A$$. Por esta razón a veces$$A$$ se denota por$$A_0$$. La constante$$k$$ resulta estar relacionada con la vida media de la sustancia, definida como el tiempo$$t_H$$ requerido para que la mitad de la cantidad actual de sustancia se desintegre (ver Figura [fig:expdecay]).
Podrías tener la tentación de pensar que la vida media no es una constante, que podría cambiar dependiendo de la cantidad de sustancia presente. Por ejemplo, tal vez tardaría más tiempo para que 100 g de la sustancia se desintegraran a 50 g que lo haría para que 10 g se desintegraran a 5 g Sin embargo, esto no es así. Para ver por qué, elige cualquiera$$t \ge 0$$ como la hora actual, de modo que esa$$y(t) = A_0 e^{kt}$$ es la cantidad actual de la sustancia. Por definición, esa cantidad debe reducirse a la mitad cuando$$t_H$$ haya pasado el tiempo, es decir,$$y(t + t_H) = \frac{1}{2} y(t)$$. Entonces$$t_H$$ sí resulta ser independiente de la cantidad inicial$$A_0$$ y depende sólo de$$k$$, ya que
\begin{aligned} y(t + t_H) ~=~ \frac{1}{2} y(t) \quad&\Rightarrow\quad A_0 e^{k(t + t_H)} ~=~ \frac{1}{2} A_0 e^{kt} \quad\Rightarrow\quad \cancel{A_0} \cancel{e^{kt}} \cdot e^{kt_H} ~=~ \frac{1}{2}\cancel{A_0} \cancel{e^{kt}}\ \ [4pt] &\ Rightarrow\ quad e^ {kt_h} ~=~\ frac {1} {2}\ quad\ Rightarrow\ quad kt_h ~=~\ ln\,\ left (\ frac {1} {2}\ right) ~=~ -\ ln\ ,2\ end {alineado} y así:
Supongamos que 5 mg de una sustancia radiactiva se descompone a 3 g en 6 horas. Encuentra la vida media de la sustancia.
Solución: Considera$$A_0 = 5$$ mg como la cantidad inicial, por lo que esa$$y(t) = 5 e^{kt}$$ es la cantidad en$$t \ge 0$$ horas de tiempo. Use la información dada que$$y(6) = 3$$ mg para encontrar$$k$$, la constante de decaimiento de la sustancia:
$3 ~=~ y(6) ~=~ 5 e^{k6} \quad\Rightarrow\quad 6k ~=~ \ln\,\left(\frac{3}{5}\right) \quad\Rightarrow\quad k ~=~ \frac{1}{6}\,\ln\,0.6$Entonces la vida media$$t_H$$ es:
$t_H ~=~ -\frac{\ln\,2}{k} ~=~ -\frac{\ln\,2}{\frac{1}{6}\,\ln\,0.6} ~=~ 8.14 ~\text{hours}$
Obsérvese en el ejemplo anterior que$$t = 6$$ se utilizó el tiempo dado para encontrar la constante$$k$$ y luego la semivida$$t_H$$. Para el problema inverso, dado que la vida media encuentra el tiempo requerido para que una cierta cantidad desaparezca, harías lo contrario: usar lo dado$$t_H$$ para encontrar$$k$$ y luego resolver el tiempo requerido a$$t$$ partir de la ecuación$$y(t) = A_0 e^{kt}$$.
Otro ejemplo de la ecuación diferencial$$\dydt = ky$$ es el crecimiento exponencial de bacterias celulares, en cuyo caso$$k > 0$$ ya que el número de células$$y(t)$$ a la vez$$t$$ va en aumento.
Otro ejemplo es para la corriente$$I$$ en un circuito eléctrico en serie simple con una fuente de voltaje de corriente continua constante (CC)$$V$$, un condensador con capacitancia$$C$$, una resistencia con resistencia$$R$$, y un interruptor, como en la Figura [fig:dccirc]. Si el condensador está inicialmente descargado cuando el interruptor está abierto, y si el interruptor está cerrado en el momento$$t = 0$$, entonces la corriente$$I(t)$$ a través del circuito en el momento$$t \ge 0$$ satisface (por la Segunda Ley de Kirchoff) la ecuación diferencial
$RC\frac{d\negmedspace I}{\dt} ~+~ I ~=~ 0 \quad\Rightarrow\quad \frac{d\negmedspace I}{\dt} ~=~ -\frac{I}{RC}$de manera que$$I(t) = I_0 e^{-t/RC}$$ donde$$I_0$$ esta la corriente inicial en$$t = 0$$. La Ley de Ohm dice que$$V = I_0R$$, entonces
$I(t) ~=~ \frac{V}{R} e^{-t/RC}$es la corriente en el tiempo$$t \ge 0$$, que disminuye exponencialmente.
En los ejemplos anteriores las cantidades que decayeron o crecieron exponencialmente lo hicieron como funciones del tiempo. Sin embargo, hay otras variables posibles además del tiempo. Por ejemplo, la presión atmosférica$$p$$ medida en función de la altura$$h$$ sobre la superficie de la Tierra satisface, suponiendo temperatura constante, la ecuación diferencial
$\frac{d\negmedspace p}{d\!h} ~=~ -\frac{w_0}{p_0}\,p$donde$$p_0$$ está la presión en altura$$h = 0$$ (es decir, nivel del suelo) y$$w_0$$ es el peso de un pie cúbico de aire a presión$$p_0$$ (con presión de aire medida en lbs por pie cuadrado y altura medida en pies). Por lo tanto,
$p(h) ~=~ p_0\;e^{-\frac{w_0}{p_0}\,h} ~.$Por lo que la presión atmosférica disminuye exponencialmente a medida que aumenta la altura sobre el suelo.
[sec2dot3]
3
$$y ~=~ e^{2x}$$
$$y ~=~ xe^{x^2}$$
$$y ~=~ e^{-x} ~-~ e^{x}$$
3
$$y ~=~ e^{\sin\;x}\vphantom{\dfrac{1 ~+~ e^x}{1 ~-~ e^x}}$$
$$y ~=~ \dfrac{1 ~+~ e^x}{1 ~-~ e^x}$$
$$y ~=~ \dfrac{1}{1 ~+~ e^{-2x}}$$
3
$$y ~=~ e^{e^x}$$
$$y ~=~ e^{2\,\ln\,x}$$
$$y ~=~ \ln\,(3x)$$
3
$$y ~=~ \ln\,(x^2 ~+~ 2x ~+~ 1)^4$$
$$y ~=~ \left(\ln (\tan\;x^2 )\right)^3$$
$$y ~=~ \ln\,(e^x ~+~ e^{2x})$$
Demuéstralo$$~\dfrac{d}{\dx}\,\left(\ln\,(kx)\right) ~=~ \dfrac{1}{x}~$$ para todas las constantes$$k > 0$$.
Demuéstralo$$~\dfrac{d}{\dx}\,\left(\ln\,\left(x^n\right)\right) ~=~ \dfrac{n}{x}~$$ para todos los enteros$$n \ge 1$$.
Para los Ejercicios 15-18, utilice la diferenciación logarítmica para encontrar$$\dydx$$. [[1.] ]
4
$$y ~=~ x^{x^2}\phantom{\dfrac{x^2}{x^3}}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$$
$$y ~=~ x^{\ln\,x}\phantom{\dfrac{x^2}{x^3}}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$$
$$y ~=~ x^{\sin\,x}\phantom{\dfrac{x^2}{x^3}}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$$
$$y ~=~ \dfrac{(x+2)^{8} \, (3x-1)^{7}}{(1-5x)^4}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$$
[[1.] ]
Supongamos que tarda 8 horas para que 30% de una sustancia radiactiva se desintegre. Encuentra la vida media de la sustancia.
El isótopo radiactivo radio-223 tiene una vida media de 11.43 días. ¿Cuánto tiempo tardarían 3 kg de radio-223 en decairse a 1 kg?
Si cierta población celular crece exponencialmente —es decir, es de la forma$$A_0e^{kt}$$ con$$k>0$$ — y si la población se duplica en 6 horas, ¿cuánto tiempo tardaría en cuadruplicar a la población?
Para los Ejercicios 22-25, use la inducción para probar la fórmula dada para todos$$n \ge 0$$. [[1.] ]
2
$$\dfrac{d^n}{\dx^n}\,\left(e^{kx}\right) ~=~ k^n \,e^{kx}\quad$$(cualquier constante$$k \ne 0$$)
$$\dfrac{d^n}{\dx^n}\,\left(x\,e^x\right) ~=~ (x + n)\,e^x$$
2
$$\dfrac{d^n}{\dx^n}\,\left(x\,e^{-x}\right) ~=~ (-1)^n (x - n)\,e^{-x}\vphantom{\dfrac{d^{n+1}}{\dx^{n+1}}}$$
$$\dfrac{d^{n+1}}{\dx^{n+1}}\,\left(x^n\;\ln\,x\right) ~=~ \dfrac{n!}{x}$$
Demostrar eso$$f'(x) = f(x)\,(1 - f(x))$$ para la función de la neurona sigmoidea$$f(x) = \dfrac{1}{1 + e^{-x}}$$. Esta relación derivada se utiliza en algoritmos de aprendizaje de redes neuronales.
Si$$\;y = C e^{-\kappa t}\,\cos\,\left(\sqrt{n^2 - \kappa^2}\;t ~+~ \gamma\right)~$$ entonces muestra eso
$\frac{d^2y}{\dt^2} ~+~ 2\kappa\,\dydt ~+~ n^2 y ~=~ 0$para todas las constantes$$C$$,$$n$$,$$\kappa$$,$$\gamma$$, con$$0 \le \kappa \le n$$. [[1.] ]
Supongamos que$$~e^y + e^x ~=~ e^{y + x}$$. Demostrar eso$$\dydx = -e^{y-x}$$.
[exer:expdx] Para un infinitesimal$$\dx$$ mostrar eso$$e^{\dx} = 1 \;+\; \dx$$. (Pista: Uso$$\ddx\,(e^x) = e^x$$.)
Para un infinitesimal$$\dx$$ mostrar eso$$\ln\,(1 + \dx) = \dx$$.
This page titled 2.3: Las funciones del logaritmo exponencial y natural is shared under a GNU General Public License 3.0 license and was authored, remixed, and/or curated by Michael Corral. | 6,655 | 17,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-10 | latest | en | 0.107995 |
http://www.chegg.com/homework-help/questions-and-answers/problem-relevant-wireless-communication-link-licopter-air-submarine-submerged-seawater-hel-q3280228 | 1,475,319,634,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662705.84/warc/CC-MAIN-20160924173742-00039-ip-10-143-35-109.ec2.internal.warc.gz | 381,072,527 | 14,134 | This problem is relevant to a wireless communication link between a he- licopter in air, and a submarine submerged in seawater. The helicopter is far from the sea surface, so that the incident wave emitted by the helicopter can be modelled at the sea surface as a plane wave. Assume that the x-y plane (z = 0) resides at the sea surface and that the incident wave produced by the helicopter travels in the z direction. The constitutive parameters of seawater are $$\epsilon r$$ = 80, $$\mu r$$ = 1, $$\sigma$$ = 4 (S/m). If the incident electric eld at z = 0 is Ei = y*100 cos( $$2\pi*10^{6}t$$ ) (V/m)
compute a) the reflection and transmission coefficients of the normal incidence phenomenon occurring at the sea surface, and b) the time-domain expressions for the electric and magnetic fields trans- mitted into the seawater (corresponding to z > 0). c) Use the result in part b to compute the electric and magnetic fields arriving at a depth of z = 2 m where the submarine is located. Provide also the expressions for: c1) the incident electric and magnetic fields for z < 0, c2) the reflected electric and magnetic fields for z < 0, and c3) the total (incident plus reflected) electric and magnetic fields for z < 0 | 301 | 1,221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-40 | latest | en | 0.896238 |
https://www.doubtnut.com/question-answer-chemistry/assuming-ideal-behaviour-calculate-the-vapour-pressure-of-10-molal-solution-of-a-non-volatile-molecu-644660735 | 1,656,618,563,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00568.warc.gz | 808,340,297 | 93,438 | Home
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Solution : P_("solution") =P_A =P_A^0 chi_A (Raoult's Law) <br> By using the relation between molality and mole fraction, we have : <br> m=(chi_B)/(chi_A)xx(1000)/(M_A) =(1-chi_A)/(chi_A)xx(1000)/(M_A) <br> rArr chi_A =(1000)/(mM_A+1000) <br> rArr chi_A =(1000)/(1xx 18+1000)=(1000)/(1018)=0.982 <br> rArr P_("solution") =P_A^0 chi_A =0.122xx0.982=0.120 atm
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hello everyone here the question is assuming ideal behaviour calculate the vapour pressure of 1 molar solution of a nonvolatile molecular solute in water at 50 degree Celsius the vapour pressure of water at 50 degree celsius is 0.122 atmospheric so here we have to find out vapour pressure of a solution and for a non volatile solute for an non volatile solute vapour pressure of a solution is equals to Vapour pressure of solvent in prostate into its mole fraction that is mole fraction of solvent and here given as that 1 molar solution and 1 molal means 1 mole of solute present in thousand gram of
solvent thousand gram of solvent mole fraction of solute solvent will be it is equals to number of moles of solvent and number of moles of solvent here are mass of solute that is thousand / here is solvent is water and water has a molecular formula H2O and Molecular masses 18 so your number of moles of solvent are 55.5 5 and number of moles of solute R1 so mole fraction of solvent is number of moles of solute divided by total Mol that is 55.5 plus one that is 56.5 5 putting the values here vapour pressure of solution is equals to Vapour pressure of solvent in prostate is given point 1 to
2 into 50 5.55 and / 56.5 after solving it will be 0.98 to so you are vapour pressure of solution will be 0.119 atmospheric | 558 | 2,055 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-27 | latest | en | 0.882926 |
https://study.com/academy/answer/find-the-bounderies-given-by-the-curve-x-4-sqrt-y-and-y-4-sqrt-x-show-your-solution.html | 1,579,606,594,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250603761.28/warc/CC-MAIN-20200121103642-20200121132642-00486.warc.gz | 686,723,452 | 21,531 | # Find the bounderies given by the curve x = 4 sqrt(y) and y = 4 sqrt(x). Show your solution.
## Question:
Find the bounderies given by the curve {eq}x = 4 \sqrt y {/eq} and {eq}y = 4 \sqrt x {/eq}. Show your solution.
## Area Between Functions With Integration
In 2D calculus, the definite integral has many applications, for example, we can find the area between functions, we need the integration limits and solve the integral. Remember, the bigger area minus the smaller area.
The area is:
The functions are:
{eq}\displaystyle \ f(x) = 4\sqrt{x} \\ {/eq}
and
{eq}\displaystyle x= 4\sqrt{y} \; \Rightarrow \; \frac{x^2}{16}=y \; \Rightarrow \; \ g(x)= \frac{x^2}{16} \\ {/eq}
Integration limits at interceptions {eq}x {/eq} values.
Matching functions:
{eq}\displaystyle \frac{x^2}{16}= 4\sqrt{x} \\ \displaystyle \frac{x^2}{16}- 4\sqrt{x}=0 \; \Leftrightarrow \; x=0 \; \text{or} \;x=16 \\ {/eq}
Terefore, the area is:
{eq}\displaystyle A \; = \; \displaystyle \int_{0}^{16} \ f(x) - \ g(x) \;dx \\ \displaystyle A \; = \; \displaystyle \int_{0}^{16} \ 4 \sqrt{x} - \frac{x^2}{16} \;dx \\ \displaystyle A \; = \; 8/3\,{x}^{3/2}-1/48\,{x}^{3} \bigg|_{0}^{16 } \\ \displaystyle A \; = \; 8/3\,{(16)}^{3/2}-1/48\,{(16)}^{3} - \left( 8/3\,{(0)}^{3/2}-1/48\,{(0)}^{3} \right) \\ \displaystyle A \; = \; \frac{256}{3} \; \text{or} \; A \; = \; 85.3333333333333 \; \; \text{ squared units } \; \\ {/eq} | 540 | 1,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2020-05 | longest | en | 0.525163 |
http://mathhelpforum.com/calculus/50546-modelling-using-differential-equations-print.html | 1,524,682,453,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00108.warc.gz | 192,424,357 | 2,830 | Modelling using differential equations
Printable View
• Sep 24th 2008, 08:37 PM
njr008
Modelling using differential equations
A particle of mass m is released from rest at time t = 0 at some height above the ground. It experiences a resistance mkv where v is its speed and k a constant. Find v and the vertical distance fallen after time t.
I admit I am a little lost on this question, so any help would be appreciated.
• Sep 25th 2008, 04:13 AM
Showcase_22
I think it goes a little like this:
$\displaystyle F=ma$
$\displaystyle F=m\frac{dv}{dt}$
$\displaystyle mg-mkv=m\frac{dv}{dt}$
$\displaystyle g-kv=\frac{dv}{dt}$
Then it's separation of variables or integrating factor to get an equation for v in terms of t.
• Sep 25th 2008, 04:19 AM
mr fantastic
Quote:
Originally Posted by Showcase_22
I think it goes a little like this:
$\displaystyle F=ma$
$\displaystyle F=m\frac{dv}{dt}$
$\displaystyle mg-mkv=m\frac{dv}{dt}$
$\displaystyle g-kv=\frac{dv}{dt}$
Then it's separation of variables or integrating factor to get an equation for v in terms of t.
And once v = v(t) is got, it's not hard to get x = x(t) from it. | 330 | 1,134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-17 | latest | en | 0.81402 |
http://mymathforum.com/physics/337602-projectile-motion.html | 1,529,949,381,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00355.warc.gz | 217,058,873 | 9,120 | My Math Forum Projectile motion
Physics Physics Forum
November 23rd, 2016, 04:04 PM #1 Member Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0 Projectile motion HOW DO YOU GO ABOUT SOLVING THE FOLLOWING QUESTION? A person initially at rest throws a ball upward at an angle !0 with an initial speed v0. He tries to catch up to the ball by accelerating with a constant acceleration a for a time interval !t1 and then continues to run at a constant speed for a time interval !t2. He catches the ball at exactly the same height he threw the ball. Let g be the gravitational constant. What was the person’s acceleration a ? I have only been able to come up with the x and y positions of the ball and I have also shown that initial speed of the ball and angle are related...Now my problem is solving for the horizontal motion of the person
November 23rd, 2016, 04:42 PM #2 Math Team Joined: Jul 2011 From: Texas Posts: 2,755 Thanks: 1405 $\Delta x$ and the overall time interval, $t$, will be the same for the thrower and the ball for the projectile ... $\Delta x = v_0\cos{\theta} \cdot t$ $t = \dfrac{2v_0 \sin{\theta}}{g} \implies \Delta x = \dfrac{v_0^2\sin(2\theta)}{g}$ for the thrower, $t_1$ is the time he accelerates and $t_2$ is the time he runs at a constant speed ... $\Delta x = \dfrac{1}{2}at_1^2 + (a t_1) \cdot t_2 = a\left(\dfrac{t_1^2}{2} + t_1 \cdot t_2\right)$ set $\Delta x$ for the projectile equal to $\Delta x$ for the thrower and solve for $a$ ... Last edited by skeeter; November 23rd, 2016 at 05:09 PM.
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### a person initially at rest throws a ball upward at an angle 0 with an initial speed v. he tries to catch up to the ball by accelerating with a constant acceleration a for a time interval and then continues to run at a constant speed for a time interval. h
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Contact - Home - Forums - Cryptocurrency Forum - Top | 658 | 2,263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-26 | latest | en | 0.883934 |
https://coderodde.wordpress.com/2015/07/24/computing-debt-cuts-leading-to-global-zero-equity/ | 1,529,712,717,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864848.47/warc/CC-MAIN-20180623000334-20180623020334-00107.warc.gz | 558,011,807 | 19,049 | # Computing debt cuts leading to global zero-equity
In this post we present a method for computing a set of debt cuts, which, once applied, lead to a global zero-equity state, i.e., each and every party in a financial graph may dismiss all liabilities. Before we proceed to defining the structures needed in discussing the method, we have to impose some definitions: by $\mathfrak{R}_?$ we denote the set of real numbers $x$ such that $x \, ? \, 0$ holds. We work on a directed graph $\mathcal{G} = (V, A)$, $A \subseteq V^2$, for which we define a weight function $w_{\mathcal{G}} \colon \mathcal{G.A} \to \mathcal{P}(\mathfrak{R}_> \times \mathfrak{R}_{\geq} \times (\mathfrak{R}_> \cup \{ \infty \}) \times \mathfrak{R})$. If $(K, r, n, t) \in \mathfrak{R}_> \times \mathfrak{R}_{\geq} \times (\mathfrak{R}_> \cup \{ \infty \} ) \times \mathfrak{R}$, $K$ is the principal investment, $r$ is the annual interest rate, $n$ is the amount of compounding periods per year (the value of $\infty$ is allowed, which denotes continuous compounding), and $t$ is the time point at which the loan was granted. Together, the four parameters comprise a contract. Note that the weight of an arc is a set of contracts as this is physically possible in real-world banking. For each arc $(u, v) \in \mathcal{G}.A$, $u$ is the creditor, $v$ is the debtor, and $w_{\mathcal{G}}(u, v)$ is the set of contracts granted by $u$ to $v$.
The most fundamental function in this post is the equity function $e_{\mathcal{G}} \colon \mathcal{G}.V \times \mathfrak{R} \to \mathfrak{R}$. The second argument is the time point at which we want to know the equity of the first argument (which is a node). Altogether, it is defined as
\begin{aligned} e_{\mathcal{G}}(u, \tau) =& \sum_{(u, v) \in \mathcal{G}.A} \Bigg( \sum_{(K, r, n, t) \in w_{\mathcal{G}}(u, v)} \mathfrak{C}_{\tau} ( K, r, n, t ) \Bigg) \\ -& \sum_{(v, u) \in \mathcal{G}.A} \Bigg( \sum_{(K, r, n, t) \in w_{\mathcal{G}}(v, u)} \mathfrak{C}_{\tau} ( K, r, n, t ) \Bigg), \end{aligned}
where
$\displaystyle \mathfrak{C}_{\tau}(K, r, n, t) = \begin{cases} K \big( 1 + \frac{r}{n} \big)^{ \lfloor n(\tau - t) \rfloor } & \mbox{if } n \in \mathfrak{R}_> \\ Ke^{ r (\tau - t) } & \mbox{if } n = \infty, \end{cases}$
and $\tau$ is no less than the time point of any contract involved. Also, we are given a function $\mathfrak{f}_{\mathcal{G}} \colon \mathcal{G}.V \times \mathfrak{R}_> \times \mathfrak{R}_{\geq} \times (\mathfrak{R}_> \cup \infty) \times \mathfrak{R} \to \mathfrak{R}$ mapping every tuple $(u, \mathfrak{K})$ (a debtor and its single contract) in the financial graph to a time point at which $u$ is ready to pay back at most all of its debts on behalf of the contract $\mathfrak{K}$. (By “at most” we mean that we will try to minimize the nodes’ debt cuts for every contract, yet it is not possible for a node, say $v$, having given no loans to the other nodes, to have a debt cut for an incoming contract $\mathfrak{K}$ any less than the value of that very contract at any time, which implies that such a contract will have to be cut in its entirety.)
As the concept of equilibrium is global with respect to the input graph, we need only one parameter describing it: $T_{\mathcal{G}}$, which is the target time point at which every node must have zero equity. We require $T_{\mathcal{G}}$ to be no less than time points of any contract. Whenever a party, say $u \in \mathcal{G}.V$, is ready to raise $C$ units of resources for the debt cut to $v$ against the contract $\mathfrak{K}$ (which is supposed to happen at $\mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K})$, with $\mathfrak{K} = (K, r, n, t)$, the contract being cut becomes
$\displaystyle \mathfrak{C}_{\tau} \big( \mathfrak{C}_{\mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K})}(K, r, n, t) - C, r, n, \mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}) \big),$
where $\tau \geq \mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}).$ There is, however, a small issue with debt contracts with non-continuous compounding scheme. Let us define a function of time and a contract:
$g(\tau, \mathfrak{K}) = \begin{cases} 0 & \mbox{if } \mathfrak{K}.n = \infty, \\ g'(\mathfrak{K}.n ( \tau - \mathfrak{K}.t ) ) & \mbox{otherwise,} \end{cases}$
where $g'(x) = x - \lfloor x \rfloor$ ($g'$ is not a derivative of $g$ in this context). Consider a contract $\mathfrak{K} = (K, r, n, t)$ given from $v$ to $u$ with $n \in \mathfrak{R}_>$. It is easy to see that after cutting $\mathfrak{K}$, its time stamp “shifts” forward in time by $g(\mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}), \mathfrak{K}).$ As to keep compounding at moments $t + n^{-1}, t + 2n^{-1}, t + 3n^{-1}, \dots$, we should set the cut contract to
$\displaystyle \mathfrak{C}_{\tau} \big( \mathfrak{C}_{\mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K})}(\mathfrak{K}) - \Xi(v, u, \mathfrak{K}), \mathfrak{K}.r, \mathfrak{K}.n, \mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}) - g(\mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}), \mathfrak{K}) \big),$
where $\Xi(v, u, \mathfrak{K})$ is the computed debt cut for the loan given by $v$ to $u$ on behalf of the contract $\mathfrak{K}$.
Next, we define the concept of equilibrium.
Definition
The financial graph $\mathcal{G}$ is said to be in equilibrium at time point $\tau$ if and only if $e_{\mathcal{G}}(u, \tau) = 0$ for all $u \in \mathcal{G}.V$.
Once given $\mathcal{G}, \mathfrak{f}_{\mathcal{G}}$ and $T_{\mathcal{G}}$, we attempt to compute a function $\Xi \colon \mathcal{G}.A \times \mathfrak{R}_> \times \mathfrak{R}_{\geq} \times (\mathfrak{R}_> \cup \{ \infty \}) \times \mathfrak{R} \to \mathfrak{R}_{\geq}$ such that after applying a debt cut from $v$ to $u$ of magnitude $\Xi(u, v, \mathfrak{K})$ against the contract $\mathfrak{K}$ at time point $\mathfrak{f}_{\mathcal{G}}(v, \mathfrak{K})$ for all $(u, v) \in \mathcal{G}.A$, $\mathcal{G}$ obtains such a state that it evolves towards equilibrium at time point $T_{\mathcal{G}}$.
### Solution
Whenever a node $u$ has incoming contracts from a set of parent nodes (creditors, lenders) $L_u$, outgoing contracts to a set of children (debtors) $D_u$, the equilibrium equation for $u$ is
\begin{aligned} & \sum_{v \in D_u } \Bigg( \sum_{\mathfrak{K} \in w_{\mathcal{G}}(u, v)} \mathfrak{C}_{T_{\mathcal{G}}} \big( \Xi(u, v, \mathfrak{K}), \mathfrak{K}.r, \mathfrak{K}.n, \mathfrak{f}_{\mathcal{G}}(v, \mathfrak{K}) - g( \mathfrak{f}_{\mathcal{G}}(v, \mathfrak{K}), \mathfrak{K}) \big) \Bigg) - \\ & \sum_{v \in L_u } \Bigg( \sum_{\mathfrak{K} \in w_{\mathcal{G}}(v, u)} \mathfrak{C}_{T_{\mathcal{G}}} \big( \Xi(v, u, \mathfrak{K}), \mathfrak{K}.r, \mathfrak{K}.n, \mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}) - g( \mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}), \mathfrak{K} ) \big) \Bigg) = \\ & \sum_{ v \in D_u } \Bigg( \sum_{ \mathfrak{K} \in w_{\mathcal{G}}(u, v) } \mathfrak{C}_{T_{\mathcal{G}}} \big( \mathfrak{C}_{ \mathfrak{f}_{\mathcal{G}} (v, \mathfrak{K})} (\mathfrak{K}), \mathfrak{K}.r, \mathfrak{K}.n, \mathfrak{f}_G( v, \mathfrak{K} ) - g( \mathfrak{f}_G(v, \mathfrak{K}), \mathfrak{K} ) \big) \Bigg) - \\ & \sum_{ v \in L_u } \Bigg( \sum_{ \mathfrak{K} \in w_{\mathcal{G}}(v, u) } \mathfrak{C}_{T_{\mathcal{G}}} \big( \mathfrak{C}_{ \mathfrak{f}_{\mathcal{G}} (u, \mathfrak{K}) } (\mathfrak{K}), \mathfrak{K}.r, \mathfrak{K}.n, \mathfrak{f}_{\mathcal{G}}( u , \mathfrak{K}) - g( \mathfrak{f}_{\mathcal{G}}(u, \mathfrak{K}), \mathfrak{K} ) \big) \Bigg). \end{aligned}
Now if we write down equilibrium equations for all nodes $v \in \mathcal{G}.V$, we obtain a system of linear equations, which is quaranteed to have at least one solution as we can choose for each contract a debt cut with magnitude equal to the current value of a contract, which results trivially in equilibrium. All the terms $\Xi(*, *, *)$ are to be determined, yet everything else in the equilibrium equations is known beforehand. We rewrite all equilibrium equations as a $\mathfrak{R}^{|\mathcal{G}.V| \times C}$ matrix, where
$\displaystyle C = 1 + \sum_{(u, v) \in \mathcal{G}.A} |w_{\mathcal{G}}(u, v)|,$
(we add 1 as the matrix is supposed to be augmented). Having the matrix filled up with entries, we reduce it to reduced row echelon form, which has at least one solution (the trivial one). We will obtain two sets: a set of independent variables $S_i = \{ s_1, \dots, s_n \}$ and a set of dependent variables $S_d = \{ s'_1, \dots, s'_m \colon s'_i = f_i(s_1, \dots, s_n)$ for some linear $f_i \}$. The last step is minimizing
$\displaystyle \sum_{s \in S_i} s + \sum_{s \in S_d} s = \sum_{i = 1}^n s_i + \sum_{i = 1}^m f_i(s_1, \dots, s_n)$
subject to constraint of not exceeding a contract with a debt cut, which is a linear program that, once solved, yields the desired $\Xi$.
Hypothesis
$\displaystyle \lim_{T_{\mathcal{G}} \to \infty} \sum_{(u, v) \in \mathcal{G}.A} \Bigg( \sum_{\mathfrak{K} \in w_{\mathcal{G}}(u, v)} \Xi(u, v, \mathfrak{K}) \Bigg) = 0.$ | 3,305 | 8,913 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 84, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-26 | latest | en | 0.799191 |
http://optimizationcity.com/transportation-problem-course-note-5/ | 1,708,712,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00827.warc.gz | 31,298,117 | 36,372 | Course 5 : Transportation Problem
Author: OptimizationCity Group
# Introduction
So far, we have studied the problem of linear programming in general. In this section, we will examine specific types of linear programming problems. In this section, we first examined the general model of minimum cost flow, and then we will examine the specific model of flow in the network, here the transportation problem.
# Minumum cost flow problem
In the problem of minimum cost flow, we are looking for homogeneous product distribution from the factory (origin) to the sales market (destination). Suppose the number of products manufactored in each factory and the number of required products are known. Also, it is not necessary to send the products directly to the destinations, but it is possible to send it to the distribution centers through an intermediate point. In addition, transportation lines are limited in terms of line capacity. The goal in this issue is to minimize the cost of transporting products.
Consider a numerical example of the minimum cost flow problem in the figure below. Nodes are represented by numbered circles and arcs by arrows. Arcs are directional. For example, materials can be sent from node 1 to node 2, but this is not possible from node 2 to node 1. We denote the arc from node i to node j as i-j.
In the above figure, each arc is given a capacity and cost per unit of transportation, which is given next to each arc. For example, in arc (2-4), the flow can be from 0 to 4 units, and the cost of each unit passing through this arc is \$2. The symbol ∞ means unlimited capacity. Finally, the numbers in parentheses next to the nodes show the amount of supply and demand. In this figure, node 1 is the origin and the amount of supply is equal to 20 units, and nodes 4 and 5 are the destinations that need 5 and 15 units, and it is the demand that is indicated by the sign -.
In the minimum cost flow problem, the goal is to find the flow pattern with minimum cost. To transform the problem into linear programming, suppose:
xij: is the number of units transported from node i to node j using arc i-j.
The linear programming model of minimum cost flow is presented as follows.
Equations 1 to 5 are the flow balance equations in the network. For example, the balance flow equation at node 1 is as follows.
The above equation states that the output flow from node 1 (x12+x13) should be equal to the supply of node 1 (20).
The balance equation in node 2 states that the input flow to node 2 (x12 ) is equal to the output flow from node 2 (x23+x24+x25 ).
The minimum cost flow model in the network has a special structure that is used to provide the solution order. Flow variables xij in balance equations only get 0, +1 and -1 coefficients. In addition, they appear exactly in two balance equations: once with a coefficient of +1 corresponding to the node from which they originate and -1 corresponding to the node to which they enter. According to the above, the general form of the Minumum Cost Flow is expressed as follows.
The above model is the general model of minimum cost flow. The above model can be converted into simpler forms, which we will describe below.
# Transportation problem
The transportation problem is a type of minimum cost flow problem where there are no intermediate points (such as nodes 2 and 3 in the general form). To express the mathematical form of the transportation problem, the following parameters are defined.
ai: the amount of supply in source i (i=1,…,m)
bj: the amount of demand in destination j (j=1,…,n)
cij: transportation cost of each unit from source i to destination j (i=1,…,m, j=1,…,n)
It is assumed that the total amount of supply in sources is equal to the amount of demand in destinations, that is:
If the above assumption is not valid, the above assumption can be considered with changes. We will get to that later.
If xij is equal to the number of units transported from source i to destination j, then the transportation problem can be formulated as follows:
Example:
The products of a large canning company are produced in three factories and transported by truck to four warehouses. Because the cost of transportation is a significant amount, therefore, the management seeks to minimize the cost of transportation. The information about the shipping cost is given in the following table:
The linear programming model of the above problem is expressed as follows.
# Presenting the simplex method for the transportation problem
Since the transportation problem is a special case of the linear programming problem, it can be solved using the simplex method. But since the structure of the transportation problem is specific, the optimal solution can be obtained by a simpler method. Instead of using the simplex table to find the optimal solution, the following table can be used to store the information of the simplex table in the transportation problem.
If a variable is basic or non-basic, the information of each cell is determined as follows.
Initial step:
In this step, we get a basic feasible solution. In the transportation problem, there are different methods to produce the initial solution, which are presented below.
## North west corner rule
First, select the variable x11, which is considered as the basic variable. In the following, if xij is the basic variable, then if the supply of source i is not finished, we will choose variable xij+1 and otherwise we will choose variable xi+1j as the basic variable.
For example, consider the simplex table below. According to this rule, the initial basic solution is produced as follows.
The initial basic solution is as follows.
## Vogel’s Approximation Method
The previous method does not take cost into account, but methods that do not take cost into account can lead to solutions where the total cost is too high. The Vogel’s approximation method is proposed to solve this problem and it is so effective that it is sometimes used to obtain an approximate optimal solution. Vogel’s approximation method works by calculating the difference between the smallest and the previous smallest cost for each row or column that is still under review. In the row or column where the difference is the largest, it selects the variable that has not been deleted and whose unit cost is the lowest.
To clarify the issue, obtain the initial basic solution of the following simplex table using the Vogel’s approximation method.
Iteration 1
Column 4 is omitted and x44 =30 is considered.
Iteration 2
Row 4 is omitted and x45 =20 is considered.
Iteration 3
Row 1 is omitted and x13 =50 is considered.
Iteration 4
Column 5 is omitted and x25 =40 is considered.
Iteration 5
Column 3 is omitted and x23 =20 is considered.
Iteration 6
This is the final table and x33=0,x32=20,x31=30.
The initial basic solution for the transportation problem is as follows.
Stopping condition
The purpose of the stopping condition is to test the optimality of the current feasible basic solution. A feasible basic solution is optimal if and only if the relation 0 ≤cij-ui-vj holds for all non-basic variables xij.
To control the above conditions, the value of ui and vj should be calculated. If the xij variable is basic, then cij=ui+vj will be, and ui and vj can be calculated by solving the equations for the basic variables. Due to the fact that the number of variables is one more than the number of equations, to find the solution, one of the variables should be equal to the desired value and get the value of the other variables.
To illustrate this, if the basic variables are x31, x32, x34, x21, x23, x13, x15, and x45, then we have:
If we put u3=0 (because u3 has the most presence in the equations), the values of other variables are obtained as follows.
By having the value of the dual variables, the optimality conditions can be controlled.
Iterative step
In this step, we identify the input basic variable and the output basic variable. In the simplex method of the transportation problem, we consider the variable that has the most negative value cij-ui-vj as the input basic variable. Based on the values, the simplex table of the transportation problem is as follows.
In the above table, variable x25 has the lowest cij-ui-vj value and is selected as the input variable. To find the basic output variable from the base, the following method is used.
In the above figure, the value of θ must be increased so that the cost of the arc is not negative, then we will have:
By increasing the value of θ by 10, the variable x15 is removed from the basic solution, and x15 is the output basic variable. In this case, the basic solution is as follows.
Before solving the examples, we will continue with the formal presentation of the simplex method for the transportation problem:
Initial step: Obtain the initial basic feasible solution using North west corner rule or Vogel’s approximation.
Stopping condition: Solve the equations cij=ui+vj for basic variables and get the value of dual variables ui and vj. If the relation cij-ui-vj≥0 holds for all the non-basic variables, then the current basic solution is optimal, otherwise go to the iterative step.
Iterative step:
Part 1: Consider the non-basic variable xij which is the most negative value of cij-ui-vj as the input basic variable.
Part 2: Determine the output basic variable.
Part 3: Determine the new basic solution.
Go to stopping condition step.
# Inequality of supply and demand
## Total supply is greater than total demand
When the amount of total supply is greater than the total demand, i.e.
a virtual demand point with a demand amount equal to
should be added to the transportation table. The costs of transporting each unit of goods from this virtual point, because no goods are really transported, are considered zero.
## total demand is greater than total supply
When the amount of total supply is less than the amount of total demand, i.e.
a virtual row with a capacity equal to
should be added to the transportation table. The costs of transporting each unit of goods to this virtual point, because no goods are really transported, are considered zero.
# Solving non-standard transportation problem
In the transportation algorithm, the basic assumption is the equality of the total amount supply and demand. For the case where the assumption is not valid, adding a virtual point (row or column in the table) converts the problem into standard format.
There are other non-standard forms, such as the impossible cell. It is a situation where it is almost impossible to send the goods between the origin and the destination, which is caused by the lack of a road between the origin and the destination or the long distance between the two them. Another non-standard form occurs when the supply capacity at the origin has a certain range. This case means that the amount of supply is between two upper and lower limits. In this regard, demand can also have a certain range. We discuss this form as the bounded capacity in the later.
## Impossible cell
Sometimes in real world issue, there is a situation where it is not possible to send goods from a specific orgine to a specific destination. Not using a route is done by assigning a cost equal to M to the cell that represents that route in the transportation problem table. M is a large number that represents the cost of transporting a unit of goods. In the transportation method, a value is not assigned to a cell that has a cost equal to M because the goal of the algorithm is to minimize the cost, and the variable related to that cell remains non-basic in the final table.
## bounded capacity
Consider a transportation problem that has an origin with the capacity.
Suppose the first factory can double or triple its capacity by adding second and third shifts and can easily shut it down when necessary. By adding this new assumption, the first factory has a capacity equal to C1, which:
The main capacity (without adding the second and third shifts) is considered as the lower limit. The upper bound of the capacity in case of adding a new line has been modified in the table below (by adding line 1-2).
The main capacity (without adding the second and third shifts) is considered as the lower bound. The upper bound of the capacity in case of adding a new line is exactly the same as the previous costs, and its supply amount is equal to the difference between the two bound (which is equal to 200 here). The addition of a new source increases supply over demand and requires a virtual node. Note that the virtual cell in row 1 is impossible and has a cost equal to M. This makes supply node 1 having a minimum capacity of 100. If this cell is not equal to M, it is possible that the flow from node 1 to the virtual node exists and the supply node 1 no longer serves demand nodes 1,2,3, and 4. Therefore, supply node 1 can produce 100 units of goods in the first shift and 200 units of goods in the next added shifts (line 1-2).
## bounded demand
As before, a transportation problem can have a limited demand.
Example: Consider the previous problem again without considering the additional shifts for supply node 1. Suppose the demand for node 1 (column 1) changes between two bound of 100 and 250, that is, the value of D1 is between 100 and 250.
The revised table with the addition of the new column 1-2 is as follows.
The demand for node 1 is equal to 100. Node 1-2 represents the surplus demand in addition to the minimum demand of node 1, which is 100. The costs of transporting to node 1-2 (column 1-2) are the same as the costs of transporting to node 1 (column 1), and the demand node 1-2 is equal to 250-100=150 (the difference between the upper bound and the lower bound). The virtual node has a capacity equal to 100, which is considered to balance supply and demand.
There should be no flow between destination node 1 and virtual node. Therefore the cost of this cell is M. Why M? If a flow is established from the virtual node to the node 1, this flow does not actually exist, and nodes 1 and 1-2 send 150 units of flow, which is more than the lower bound and the values between the lower bound (100) and 150 are not covered. This setting guarantees the minimum input to node 1 being 100.
## Production with different costs
Consider the following transportation problem.
Suppose that the production cost of three factories is different due to differences in the cost of manpower, equipment and raw materials. The goal is to distribute the production of factories among four warehouses in such a way that the total cost of production and transportation is minimized. If the production cost of each product unit in three factories is equal to 50, 62, and 54, respectively, the transportation and production cost will be by adding the production costs to the transportation costs of each unit. The optimal solution for this situation is shown in the figure below.
## Selling with different costs
In some cases, the product of the same factory can be sold in different markets at different prices. The selling price of each product unit to the four warehouses of the previous problem is equal to 100, 105, 110, and 115 respectively. Due to the fact that we are looking to minimize the cost in the transportation problem, the selling price is considered with a negative sign, so each of the numbers inside the upper small rectangles is the difference between the selling price of each unit (negative), and the shipping cost. and production (positive).
Note: The optimal solution in this table is exactly the same as the solution presented in the initial case and production with different costs case. Adding a fixed value to the cost in the columns does not change the optimal solution.
Example:
A canning company has factories in three locations: Chicago, Cleveland, and Boston. Last week, the production of these three factories was 35, 50, and 40 units, respectively. The company wants to ship 45 units to Dallas, 20 to Atlanta, 30 to San Francisco, and another 30 to Philadelphia. The cost of production and transportation of each factory to each distribution center is shown in the table below. What is the best shipping method?
Solution:
Initial step::
1- North-west corner method
The initial basic solution by the north-west corner rule with the objective function z=1180 is as follows.
2 – Vogel’s approximation method
The application of Vogel’s approximation method to find the initial basic solution is as follows.
Iteration 1
Column 4 is omitted and x34=30 is considered.
Iterations 2
Row 3 is omitted and x32=10 is considered.
Iteration 3
Column 2 is o omitted and x12=10 is considered.
Iteration 4
Column 1 is omitted and x21=45,x13=25,x23=5 are considered. Therefore, the basic solution of Vogel’s approximation is as follows.
X34=30,x32=10,x12=10,x21=45,x13=25,x23=5
Then, the value of the objective function is equal to Z=1020.
From the comparison of the objective function value of the northwest corner rule and Vogel’s approximation, we can understand that the Vogel’s approximation leads to the basic solution with a better objective function. Therefore, the initial solution used in the algorithm is as follows.
Stopping condition: Assuming u1=0, the value of dual variables is as follows:
In the following table, the value of cij-ui-vj for non-basic variables is calculated.
Because there is no non-basic variable that cij-ui-vj < 0, so we reached the optimal solution. It can be seen that Vogel’s approximation leads to the optimal solution.
For further practice, assume that the initial solution is given as follows:
Stopping condition: Assuming u1=0, the value of dual variables is as follows:
In the table below, the value of cij-ui-vj is calculated for non-basic variables.
Iterative step: because cij-ui-vj < 0 for two non-basic variables, therefore the basic solution of the current solution is not optimal and we need to find an improved solution. We consider x13 as the input variable. To find the output variable, it is done as follows.
Therefore, the basic variable x11 is removed from the base and the non-basic variable x13 is included in the base. Therefore, the current basic solution is as follows.
Stopping condition:
Assuming u1=0, the value of the dual variables can be calculated as follows.
In the following table, the values of cij-ui-vj of non-basic variables are given.
Iterative step: The non-basic variable x32 enters the base. To determine the output variable from the base, we act as follows.
With θ=10, the basic variable x33 leaves the base and the non-basic variable x32 enters the base. Therefore, the basic solution is as follows.
Stopping condition: Assuming u1=0, the value of dual variables can be calculated as follows.
In the following table, the values of cij-ui-vj of non-basic variables are given.
The optimality condition is satisfied and therefore the current basic solution is optimal.
Example:
The transportation problem with the transportation cost table is assumed. Find an intial basic solution using the northwest corner rule and the Vogel’s approximation method. Compare the number of simplex iterations in each of the two initial solutions.
Solution:
A) Northwest corner rule:
The objective function value of the above table is equal to 48. The value of cij-ui-vj for non-basic variables is as follows.
The variable x32 is considered as the basic input variable. To determine the basic output variable, we proceed as follows.
The variable x22 should be considered as the output variable, in which case the current basic solution will be as follows.
Assuming u1=0, other dual variables are as follows:
The value of the objective function in this iteration is equal to 42. The value of cij-ui-vj for non-basic variables is as follows.
One of cij-ui-vj is negative, so the optimality condition is not satisfied. x13 variables are entered into basic variables.
With θ=0 variable x33 leaves the basic solution and variable x13 enters the basic solution. Therefore, the basic solution is as follows.
The value of the objective function in this iteration is equal to 42. The value of cij-ui-vj for non-basic variables is as follows.
We use the following table to control the optimality.
The x14 variable should be considered as the input variable to the basic answer. The output variable from the base is obtained as follows.
By setting θ = 2, the variable x12 is removed from the basic solutions and the basic solution is as follows.
The value of the objective function of the above basic solution is equal to 32. Assuming u1=0, other dual variables are as follows:
We use the following table to control the optimality.
Considering that the non-basic arcs have a non-negative value, we reached the optimal solution. The value of the objective function is equal to 32. The number of iterations to reach the optimal solution is equal to 3.
Due to the fact that cij-ui-vj for all the non-basic arcs are positive, we reach the optimal solution. The value of the objective function is equal to 32. The number of iterations to reach the optimal solution is equal to 3.
b) The initial solution using Vogel’s approximation method is as follows.
x23=2
x32=3
x13=0
The solution obtained from Vogel’s approximation is as follows.
The values of dual variables are as follows:
The value of cij-ui-vj for non-basic variables is as follows.
As can be seen, cij-ui-vj is non-negative for all non-basic variables and therefore we reached the optimal solution. The number of iterations to reach the optimal solution with the Vogel’s approximation method is equal to one. | 4,672 | 21,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-10 | longest | en | 0.941005 |
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Books Bücher 131 - 140 von 189 in IF a straight line touch a circle, and from the point of contact a straight line...
IF a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.
Military examinations. Mathematical examination papers, set for entrance to ... - Seite 99
von Woolwich roy. military acad, Walter Ferrier Austin - 1880
Vollansicht - Über dieses Buch
## Calendar, for the Year ...
1896
...thus produced and on the part produced may be equal to three times the square on the given line. 3. If a straight line touch a circle, and from the point...be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate...
Vollansicht - Über dieses Buch
## Report of the Secretary for Public Instruction ...
...equal to a given recti- 7 lineal figure. 3. If a straight line touch a circle, and from the point of 8 contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate...
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## Woolwich Mathematical Papers for Admission Into the Royal Military Academy ...
Eldred John Brooksmith - 1901
...it. 7. Straight lines in a circle which are equally distant from the centre are equal to one another. 8. If a straight line touch a circle, and from the...be drawn cutting the circle, the angles which this line makes with the line touching the circle are equal to the angles which are in the alternate segments...
Vollansicht - Über dieses Buch
## The School World, Band 3
1901
...straight line bisecting any chord of a circle at right angles passes through the centre of the circle. (5) If a straight line touch a circle, and from the point...be drawn cutting the circle, the angles which this line makes with the tangent are equal to the angles in the alternate segments of the circle. (6) Inscribe...
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## Examination Papers
University of Toronto - 1901
...centre ; and tho.se which are equally distant from the centre are equal to one another. (III. 14.) 5. If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent shall be equal to the angles which...
Vollansicht - Über dieses Buch
## Elementary Geometry: Practical and Theoretical
Charles Godfrey, Arthur Warry Siddons - 1903 - 355 Seiten
...having its sides parallel to three given straight lines. ,M SECTION IX. "ALTERNATE SEGMENT." THEOREM 14. If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate...
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## The School World: A Monthly Magazine of Educational Work and Progress, Band 5
1903
...The opposite angles of any quadrilateral inscribed in a circle are supplementary ; and the converse. If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate...
Vollansicht - Über dieses Buch
## Calendar
University of Calcutta - 1906
...circumference, prove that that point is the centre of the circle. Prove that a circle has only one centre. 2 5. If a straight line touch a circle, and from the point of contact a 8 chord be drawn, prove that the angles which this chord makes with the tangent, are eqnal to the angles...
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## Annual Report of the Commissioners ..., Band 73
...solutions will not be accepted. Mr. Boss, Senior Inspector. Mr. CHAMBERS, Senior Inspector. SECTION A. 1. If a straight line touch a circle, and from the point of contact a straight line be drawn dividing the circle into two segments, the angles made by this line with the tangent are equal to the...
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What Is Arc Elasticity? Definition, Midpoint Formula, and Example
What Is Arc Elasticity?
Arc elasticity is the elasticity of one variable with respect to another between two given points. It is used when there is no general way to define the relationship between the two variables. Arc elasticity is also defined as the elasticity between two points on a curve.
The concept is used in both economics and mathematics. In economics, is it commonly used to measure the changes between the quantity of goods demanded and their prices.
Key Takeaways
• In the concept of arc elasticity, the elasticity of one variable is measured with respect to another between two given points.
• The concept is used in both economics and mathematics.
• It is commonly used to measure the changes between the quantity of goods demanded and their prices.
• Price (or point) elasticity of demand and arc elasticity of demand are two ways to calculate elasticity.
Understanding Arc Elasticity
In economics, arc elasticity is commonly used in relation to the law of demand to measure percentage changes between the quantity of goods demanded and prices.
There are two possible ways of calculating elasticity—price (or point) elasticity of demand and arc elasticity of demand. Price elasticity of demand measures the responsiveness of quantity demanded to a price. It takes the elasticity of demand at a particular point on the demand curve, or between two points on the curve. Arc elasticity of demand uses a midpoint between the two points.
Formula for Price (Point) Elasticity of Demand
$PE_d = \dfrac{\text{\% Change in Qty}}{\text{\% Change in Price}}$
How to Calculate the Price Elasticity of Demand
If the price of a product decreases from $10 to$8, leading to an increase in quantity demanded from 40 to 60 units, then the price elasticity of demand can be calculated as:
• % change in quantity demanded = (Qd2 – Qd1) / Qd1 = (60 – 40) / 40 = 0.5
• % change in price = (P2 – P1) / P1 = (8 – 10) / 10 = -0.2
• Thus, PEd = 0.5 / -0.2 = 2.5
Since we’re concerned with the absolute values in price elasticity, the negative sign is ignored. You can conclude that the price elasticity of this good, when the price decreases from $10 to$8, is 2.5.
Arc Elasticity of Demand
One of the problems with the price elasticity of demand formula is that it gives different values depending on whether price rises or falls. If you were to use different start and end points in our example above—that is, if you assume the price increased from $8 to$10—and the quantity demanded decreased from 60 to 40, the Ped will be:
• % change in quantity demanded = (40 – 60) / 60 = -0.33
• % change in price = (10 – 8) / 8 = 0.25
• PEd = -0.33 / 0.25 = 1.32, which is much different from 2.5
How to Calculate the Arc Elasticity of Demand
To eliminate this problem, the arc elasticity of demand can be used. Arc elasticity of demand measures elasticity at the midpoint between two selected points on the demand curve by using a midpoint between the two points. The arc elasticity of demand can be calculated as:
• Arc Ed = [(Qd2 – Qd1) / midpoint Qd] ÷ [(P2 – P1) / midpoint P]
Let’s calculate the arc elasticity following the example presented above:
• Midpoint Qd = (Qd1 + Qd2) / 2 = (40 + 60) / 2 = 50
• Midpoint Price = (P1 + P2) / 2 = (10 + 8) / 2 = 9
• % change in qty demanded = (60 – 40) / 50 = 0.4
• % change in price = (8 – 10) / 9 = -0.22
• Arc Ed = 0.4 / -0.22 = 1.82
When you use arc elasticity of demand you do not need to worry about which point is the starting point and which point is the ending point since the arc elasticity gives the same value for elasticity whether prices rise or fall.
Arc elasticity of demand is more useful than price elasticity of demand when there is a considerable change in price.
What Is Elasticity in Economics?
In the context of economics, elasticity is used to measure the change in the quantity demanded for a product in relation to its price movements. A product is considered to be elastic if the demand for it changes substantially when its price changes.
What Is the Law of Demand?
The law of demand is a fundamental economic concept. It states that when prices rise, the demand for a good or service will decrease.
What Are the Benefits of Arc Elasticity of Demand?
The formula for arc elasticity of demand measures elasticity between two selected points by using a midpoint between the two points. As a result, it is particularly useful when there is a substantial change in price.
The Bottom Line
Arc elasticity is commonly used in economics to determine the percentage of change between the demand for goods and their price. Elasticity can be calculated in two ways—price elasticity of demand and arc elasticity of demand. The latter is more useful when there is a significant change in price.
Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
1. Federal Reserve Bank of St. Louis. “Elasticity of Demand - The Economic Lowdown Podcast Series.”
2. University of Minnesota. “The Price Elasticity of Demand.”
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Description | 1,332 | 5,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-06 | latest | en | 0.936702 |
https://nl.mathworks.com/matlabcentral/cody/problems/605-whether-the-input-is-vector/solutions/730969 | 1,579,380,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00005.warc.gz | 584,513,545 | 15,424 | Cody
# Problem 605. Whether the input is vector?
Solution 730969
Submitted on 8 Sep 2015 by Peng Liu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 2 3 1]; y_correct = 1; assert(isequal(checkvector(x),y_correct))
ans = 1
2 Pass
%% x = [1 2 7; 6 3 1]; y_correct = 0; assert(isequal(checkvector(x),y_correct))
ans = 0
3 Pass
%% x = [1;2;0]; y_correct = 1; assert(isequal(checkvector(x),y_correct))
ans = 1 | 182 | 546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-05 | latest | en | 0.593014 |
http://akarma.life/Wellness/faq/system-sure-to-win-the-lottery/ | 1,709,240,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00029.warc.gz | 1,804,837 | 8,093 | # System Sure To Win The Lottery
DWQA QuestionsCategory: VitalitySystem Sure To Win The Lottery
Ken Winder asked 9 months ago
lotto
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http://www.numbersaplenty.com/223 | 1,603,196,919,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00240.warc.gz | 162,865,484 | 3,409 | Search a number
223 is a prime number
BaseRepresentation
bin11011111
322021
43133
51343
61011
7436
oct337
9267
10223
11193
12167
13142
1411d
15ed
hexdf
223 has 2 divisors, whose sum is σ = 224. Its totient is φ = 222.
The previous prime is 211. The next prime is 227. The reversal of 223 is 322.
Adding to 223 its reverse (322), we get a palindrome (545).
Subtracting 223 from its reverse (322), we obtain a palindrome (99).
It can be divided in two parts, 22 and 3, that multiplied together give a palindrome (66).
223 is an esthetic number in base 15, because in such base its adjacent digits differ by 1.
It is a strong prime.
223 is a truncatable prime.
It is a cyclic number.
It is not a de Polignac number, because 223 - 25 = 191 is a prime.
It is a Carol number, being equal to (24 - 1)2 - 2.
223 is a lucky number.
It is a plaindrome in base 8, base 9, base 10, base 12, base 14 and base 16.
It is a nialpdrome in base 15.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (227) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a good prime.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 111 + 112.
It is an arithmetic number, because the mean of its divisors is an integer number (112).
223 is a deficient number, since it is larger than the sum of its proper divisors (1).
223 is an equidigital number, since it uses as much as digits as its factorization.
223 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 12, while the sum is 7.
The square root of 223 is about 14.9331845231. The cubic root of 223 is about 6.0641269945.
The spelling of 223 in words is "two hundred twenty-three", and thus it is an aban number and an iban number. | 540 | 1,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-45 | latest | en | 0.91702 |
https://www.businessinsider.com/13-math-jokes-that-every-mathematician-finds-absolutely-hilarious-2013-5 | 1,638,661,287,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00639.warc.gz | 768,580,372 | 52,317 | US Markets Loading... H M S In the news
# 13 Jokes That Every Math Geek Will Find Hilarious
Back when the internet was young, the primary users were its builders, math and tech-oriented academics spread around the country.
As a result, math jokes have an elemental role in the history of the internet.
From the earliest Usenet threads to the techiest subreddits, geeky math jokes — some implicit swipes at less-pure disciplines, other puns or plays on words of different concepts — have been a major part of the modern history of math.
What's more, these japes also have the effect of making those who didn't get the joke to look into what makes it funny, teaching people some of the more obscure concepts.
Here are just a few of the best ones. Where necessary, we'll do the unthinkable and the tacky and explain the joke.
### JOKE #1
Three statisticians go out hunting together. After a while they spot a solitary rabbit. The first statistician takes aim and overshoots. The second aims and undershoots. The third shouts out "We got him!"
Source: chjilloutdamnit / Reddit
### JOKE #2
Two random variables were talking in a bar. They thought they were being discrete but I heard their chatter continuously.
Source: armchairdetective / reddit
Explanation: When you roll a die, you either get a 1, 2, 3, 4, 5, or 6. Since there are a finite number of possibilities, the statistic involved is called a discrete random variable. When you select any real number from between 0 and 1, there are an infinite number of possible draws. The statistic involved is called a continuous random variable.
### JOKE #3
There was a statistician that drowned crossing a river... It was 3 feet deep on average.
Source: / reddit
### 246-You / Flickr JOKE #4
Write the expression for the volume of a thick crust pizza with height "a" and radius "z".
Source: Reddit
Explanation: The formula for volume is π·(radius)2·(height). In this case, pi·z·z·a.
### JOKE #5
A: "What is the integral of 1/cabin?"
B: "log cabin."
A: "Nope, houseboat--you forgot the C."
Source: Reddit
Explanation: We're treating "cabin" is a variable.
The integral of 1/x is loge(x).
However, since it's integration, you've got to add a constant.
So ∫(1/cabin) = loge(cabin) + c, or "a log cabin plus the sea."
### JOKE #6
Q: Why did the chicken cross the road?
A: The answer is trivial and is left as an exercise for the reader.
Source: Reddit
Explanation:
This is a common refrain found in mathematics texts.
It is widely considered a cruel professor's malicious cop-out by particularly lazy students of mathematics.
### JOKE #7
Q: How many mathematicians does it take to change a light bulb?
A: One: she gives it to three physicists, thus reducing it to a problem that has already been solved.
Source: MathOverflow
Explanation: Mathematicians try to reduce an unsolved problem to a form which has already been solved before. Once that's done it's considered complete, as the previously derived formula is taken as written.
There are many light bulb jokes about physicists. Finding several are left as an exercises to the reader.
### JOKE #8
A physicist, a biologist, and a mathematician are sitting on a bench across from a house. They watch as two people go into the house, and then a little later, three people walk out.
The physicist says, "The initial measurement was incorrect."
The biologist says, "They must have reproduced."
And the mathematician says, "If exactly one person enters that house, it will be empty."
Source: Reddit
### JOKE #9
The B in Benoît B. Mandelbrot stand for Benoît B. Mandelbrot.
Source: Reddit
Explanation: The Mandelbrot set is a fractal. As you zoom in on portions of the fractal, you ee a self replicating image. So the infinite paradox in the joke is a shoutout to the problem. Here's an example of what we're talking about with a gif of zooming in on a point of infinite complexity in the Mandelbrot set:
### JOKE #10
Infinitely many mathematicians walk into a bar. The first says, "I'll have a beer." The second says, "I'll have half a beer." The third says, "I'll have a quarter of a beer." The barman pulls out just two beers. The mathematicians are all like, "That's all you're giving us? How drunk do you expect us to get on that?" The bartender says, "Come on guys. Know your limits."
Source: Reddit
Explanation: This is a reference to a converging infinite series.
The limit of this:
from n=0 to ∞ Σ (1/2n) = 1 + 1/2 + 1/4 + 1/8 + ... = 2
### JOKE #11
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third orders a third of a beer. The bartender bellows, "Get the hell out of here, are you trying to ruin me?"
Source: Reddit
Explanation: This is another hilarious reference to an infinite series — the harmonic series — which is not convergent but instead diverges to infinity.
from n=1 to ∞ Σ (1/n) = 1 + 1/2 + 1/3 + 1/4 + ... =
See a full explanation in this slideshow >
### JOKE #12
When a statistician passes the airport security check, they discover a bomb in his bag. He explains. "Statistics shows that the probability of a bomb being on an airplane is 1/1000. However, the chance that there are two bombs at one plane is 1/1000000. So, I am much safer..."
Source: Andrej and Elena Cherkaev
Explanation: While this statistician is correct that the joint probability there are two bombs on a plane is 1/1,000,000, his bringing one on doesn't change the prior probability that there is still a 1/1,000 chance of his flight being the one with a random bomb.
### JOKE #13
What do you get when you cross a mosquito with a mountain climber?
Nothing. You can't cross a vector and a scalar.
Source: Reddit
Explanation: A vector is a mathematical entity with both magnitude and direction in any number of dimensions. You can take the cross product of two vectors to form a new vector, similar to multiplication of real numbers.
A scalar is just a real number, a directionless magnitude in vector space. You cannot take a cross product of a scalar and a vector.
Hence, you can't cross a mosquito (disease vector) and a mountain climber (a scalar).
That is one terrible pun. I'm sorry.
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For you | 1,582 | 6,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-49 | longest | en | 0.947635 |
https://nl.mathworks.com/matlabcentral/answers/69777-produce-a-matrix-with-2-vectors | 1,656,599,683,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00773.warc.gz | 468,320,199 | 23,540 | Produce a matrix with 2 vectors
9 views (last 30 days)
j dr on 4 Apr 2013
Hi, I can't wrap my head around finding the simplest solution for this problem.
I have two vectors
A=[1 2 3] and B=[1;1;2] I want to produce the matrix
C=[1 2 3;1 2 3;2 4 6]
I am certain there is an easy command could I be reminded please !
N.
P.S. there are bonus points if the commands works to produce an N=A*B dimension matrix with 2 matrices of A and B dimensions
j dr on 4 Apr 2013
Just figured out B*A works... Sorry for wasting everyone's time !
j dr on 4 Apr 2013
But that does not work with matrices... | 181 | 589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-27 | latest | en | 0.896788 |
thepartyplaceblog.com | 1,596,561,113,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00122.warc.gz | 537,754,888 | 9,059 | # How to Calculate Alcohol and Glassware for Your Party
Monday, August 13, 2018
Trying to figure out how much glassware and alcohol to order for a party can be an angst-filled decision. You definitely DO NOT want to run out—but how much is enough? There is an actual rule of thumb for deciding these things and if you don’t mind doing some simple math you can come up with some safe estimates.
## How Many Guests and How Long is the Party?
The general rule of thumb is each guest will have one cocktail per hour. So, all you do is multiply the number of guests by how long the party will last. If you have 100 guests and a five-hour party, plan for 500 drinks to be served.
There are three types of bars: open bar, limited bar, and cash bar. An open bar is the most gracious for guest, you can assume more alcohol is going to be consumed with an open bar. Some party planners assume guests will drink an extra drink during the first two hours at an open bar. A limited bar offers a selection of drinks, usually limited to beer, wine, and maybe a single signature cocktail. These are easier to manage and a lot easier to plan for glassware—we will get to that later. Finally, a cash bar is when guests pay for the alcohol themselves. When guests pay you can definitely assume they will drink less.
## Types of Glassware
First, you can always go with all-purpose glasses and skip the drink-specific glassware. If you do decide to get specific glassware for the various drinks that will be served, you will need—at the minimum—an assortment of wine glasses, bar glasses, and all-purpose glasses. The wine glasses are strictly for wine, nothing else. The bar glasses are for your hard liquor drinks: scotch, whiskey, etc. All-purpose glasses serve as glasses for the occasional beer drinker that wants it in a glass, or for water and sodas, etc. A general breakdown after you have come up with the total number of drinks served is 60% wine glasses, 30% all-purpose glasses and 10% bar glasses. The Party Place has a great selection of standard glassware to choose from, or for a little bit more, you can stock your party with acrylic barware. | 472 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-34 | latest | en | 0.951761 |
https://in.mathworks.com/matlabcentral/cody/problems/23-finding-perfect-squares/solutions/2991235 | 1,603,545,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882581.13/warc/CC-MAIN-20201024110118-20201024140118-00086.warc.gz | 371,738,800 | 17,192 | Cody
# Problem 23. Finding Perfect Squares
Solution 2991235
Submitted on 25 Sep 2020 at 14:03
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
a = [2 3 4]; assert(isequal(isItSquared(a),true))
Error in solution: Line: 5 Column: 14 Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.
2 Fail
a = [20:30]; assert(isequal(isItSquared(a),false))
Error in solution: Line: 5 Column: 14 Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.
3 Fail
a = [1]; assert(isequal(isItSquared(a),true))
Error in solution: Line: 5 Column: 14 Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.
4 Fail
a = [6 10 12 14 36 101]; assert(isequal(isItSquared(a),true))
Error in solution: Line: 5 Column: 14 Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.
5 Fail
a = [6 10 12 14 101]; assert(isequal(isItSquared(a),false))
Error in solution: Line: 5 Column: 14 Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 430 | 1,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-45 | latest | en | 0.638557 |
https://rdrr.io/github/shabbychef/sadists/src/R/dneta.r | 1,579,422,762,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00423.warc.gz | 633,341,110 | 14,150 | #### Documented in ddnetapdnetaqdnetardneta
# Copyright 2014-2015 Steven E. Pav. All Rights Reserved.
# Author: Steven E. Pav
#
# This file is part of sadists.
#
# sadists is free software: you can redistribute it and/or modify
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# sadists is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU Lesser General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public License
# Created: 2015.03.07
# Copyright: Steven E. Pav, 2015
# Author: Steven E. Pav
# ddneta, pdneta, qdneta, rdneta#FOLDUP
#' @title The doubly non-central Eta distribution.
#'
#' @description
#'
#' Density, distribution function, quantile function and random
#' generation for the doubly non-central Eta distribution.
#'
#' @details
#'
#' Suppose \eqn{Z}{Z} is a normal with mean \eqn{\delta_1}{delta_1},
#' independent of \eqn{X \sim \chi^2\left(\delta_2,\nu_2\right)}{X ~ X^2(delta_2,v_2)},
#' a non-central chi-square with \eqn{\nu_2}{v_2} degrees of freedom
#' and non-centrality parameter \eqn{\delta_2}{delta_2}. Then
#' \deqn{Y = \frac{Z}{\sqrt{Z^2 + X}}}{Y = Z/sqrt(Z^2 + X)}
#' takes a doubly non-central Eta distribution with
#' \eqn{\nu_2}{v_2} degrees of freedom and non-centrality parameters
#' \eqn{\delta_1,\delta_2}{delta_1,delta_2}. The \emph{square} of
#' a doubly non-central Eta is a doubly non-central Beta variate.
#'
#' @usage
#'
#' ddneta(x, df, ncp1, ncp2, log = FALSE, order.max=6)
#'
#' pdneta(q, df, ncp1, ncp2, lower.tail = TRUE, log.p = FALSE, order.max=6)
#'
#' qdneta(p, df, ncp1, ncp2, lower.tail = TRUE, log.p = FALSE, order.max=6)
#'
#' rdneta(n, df, ncp1, ncp2)
#'
#' @param x,q vector of quantiles.
#' @param p vector of probabilities.
#' @param n number of observations.
#'
#' @param df the degrees of freedom for the denominator chi square.
#' We do \emph{not} recycle this versus the \code{x,q,p,n}.
#' @param ncp1,ncp2 the non-centrality parameters for the numerator and denominator.
#' We do \emph{not} recycle these versus the \code{x,q,p,n}.
#'
#' @return \code{ddneta} gives the density, \code{pdneta} gives the
#' distribution function, \code{qdneta} gives the quantile function,
#' and \code{rdneta} generates random deviates.
#'
#' Invalid arguments will result in return value \code{NaN} with a warning.
#' @aliases ddneta pdneta qdneta rdneta
#' @seealso (doubly non-central) t distribution functions,
#' @seealso (doubly non-central) Beta distribution functions,
#' @template etc
#' @template distribution
#' @template apx_distribution
#' @template not-recycled
#' @examples
#' rv <- rdneta(500, df=100,ncp1=1.5,ncp2=12)
#' d1 <- ddneta(rv, df=100,ncp1=1.5,ncp2=12)
#' \dontrun{
#' plot(rv,d1)
#' }
#' p1 <- ddneta(rv, df=100,ncp1=1.5,ncp2=12)
#' # should be nearly uniform:
#' \dontrun{
#' plot(ecdf(p1))
#' }
#' q1 <- qdneta(ppoints(length(rv)), df=100,ncp1=1.5,ncp2=12)
#' \dontrun{
#' qqplot(x=rv,y=q1)
#' }
#' @name dneta
#' @rdname ddneta
#' @export
ddneta <- function(x, df, ncp1, ncp2, log = FALSE, order.max=6) {
xF <- sqrt(df) * x / sqrt(1-x^2)
retval <- ddnt(xF,df=df,ncp1=ncp1,ncp2=ncp2,log=log,order.max=order.max)
if (log) {
retval <- 0.5 * log(df) + retval - (1.5) * log(1-x^2)
} else {
retval <- sqrt(df) * retval / ((1-x^2)^(1.5))
}
in_range <- -1 <= x & x < 1
retval[!in_range] <- NaN
return(retval)
}
#' @export
pdneta <- function(q, df, ncp1, ncp2, lower.tail = TRUE, log.p = FALSE, order.max=6) {
qF <- sqrt(df) * q / sqrt(1-q^2)
retval <- pdnt(qF,df=df,ncp1=ncp1,ncp2=ncp2,lower.tail=lower.tail,log.p=log.p,order.max=order.max)
minv <- 1 - as.double(lower.tail)
maxv <- 1 - minv
if (log.p) {
minv <- log(minv)
maxv <- log(maxv)
}
retval[q < -1] <- minv
retval[q >= 1] <- maxv
return(retval)
}
#' @export
qdneta <- function(p, df, ncp1, ncp2, lower.tail = TRUE, log.p = FALSE, order.max=6) {
# 2FIX: this part
qF <- (1/sqrt(df)) * qdnt(p,df=df,ncp1=ncp1,ncp2=ncp2,lower.tail=lower.tail,log.p=log.p,order.max=order.max)
retval <- qF / sqrt(1 + qF^2)
in_range <- ifelse(log.p,p <= 0,0 <= p & p <= 1)
retval[!in_range] <- NaN
return(retval)
}
#' @export
rdneta <- function(n, df, ncp1, ncp2) {
X1 <- rnorm(n,mean=ncp1)
X2 <- unbroken_rchisq(n,df=df,ncp=ncp2)
X <- X1 / sqrt((X1^2) + X2)
return(X)
}
#UNFOLD
#for vim modeline: (do not edit)
# vim:fdm=marker:fmr=FOLDUP,UNFOLD:cms=#%s:syn=r:ft=r
shabbychef/sadists documentation built on May 29, 2019, 8:06 p.m. | 1,645 | 4,587 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-05 | latest | en | 0.56549 |
https://fr.mathworks.com/matlabcentral/cody/problems/1712-no-_________-allowed/solutions/1697172 | 1,576,284,879,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00145.warc.gz | 354,265,692 | 16,039 | Cody
# Problem 1712. NO _________ ALLOWED....
Solution 1697172
Submitted on 20 Dec 2018 by Athi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
Sentence = 'The birds in the field are eating bird seed'; Not_allowed = 'field' output = 1; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'field'
2 Pass
Sentence = 'If the sky is blue on earth, what is the sky color on mars?'; Not_allowed = 'oven' output = 0; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'oven'
3 Pass
Sentence = 'Oh where, oh where has my little dog gone?'; Not_allowed = 'where' output = 1; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'where'
4 Pass
Sentence = 'Insanity: doing the same thing over and over again and expecting different results...'; Not_allowed = 'Einstein' output = 0; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'Einstein'
5 Pass
Sentence = 'Wheres the cream filling?'; Not_allowed = 'cream' output = 1; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'cream'
6 Pass
Sentence = 'MATLAB is the coolest!'; Not_allowed = 'MATLAB' output = 1; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'MATLAB'
7 Pass
Sentence = 'No no, you got it all wrong!'; Not_allowed = 'No' output = 1; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'No'
8 Pass
Sentence = 'This planet, with all its appalling immensity, is to electric currents virtually no more than a small metal ball.'; Not_allowed = 'Tesla' output = 0; assert(isequal(NotAllowed(Sentence, Not_allowed),output))
Not_allowed = 'Tesla' | 480 | 1,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-51 | latest | en | 0.723192 |
https://www.convertunits.com/from/thousand+cubic+foot/day/to/barrel/day | 1,601,413,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00245.warc.gz | 776,319,457 | 11,903 | ## ››Convert thousand cubic foot/day to barrel/day [petroleum]
thousand cubic foot/day barrel/day
Did you mean to convert thousand cubic foot/day to barrel/day [petroleum] barrel/day [US] barrel/day [US beer/wine] barrel/day [UK]
How many thousand cubic foot/day in 1 barrel/day? The answer is 0.0056145832761677.
We assume you are converting between thousand cubic foot/day and barrel/day [petroleum].
You can view more details on each measurement unit:
thousand cubic foot/day or barrel/day
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 3051.1871607739 thousand cubic foot/day, or 543439.64826835 barrel/day.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between thousand cubic feet/day and barrels/day.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of thousand cubic foot/day to barrel/day
1 thousand cubic foot/day to barrel/day = 178.10761 barrel/day
2 thousand cubic foot/day to barrel/day = 356.21522 barrel/day
3 thousand cubic foot/day to barrel/day = 534.32283 barrel/day
4 thousand cubic foot/day to barrel/day = 712.43043 barrel/day
5 thousand cubic foot/day to barrel/day = 890.53804 barrel/day
6 thousand cubic foot/day to barrel/day = 1068.64565 barrel/day
7 thousand cubic foot/day to barrel/day = 1246.75326 barrel/day
8 thousand cubic foot/day to barrel/day = 1424.86087 barrel/day
9 thousand cubic foot/day to barrel/day = 1602.96848 barrel/day
10 thousand cubic foot/day to barrel/day = 1781.07608 barrel/day
## ››Want other units?
You can do the reverse unit conversion from barrel/day to thousand cubic foot/day, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 554 | 2,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-40 | latest | en | 0.819862 |
http://mathhelpforum.com/advanced-algebra/165727-eigenvector.html | 1,524,354,802,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00014.warc.gz | 202,699,838 | 11,721 | 1. ## Eigenvector standardization
I have matrix $\displaystyle A=\begin{pmatrix}0&1&0\\2&-2&-1\\-3&0&0\end{pmatrix}$
and I get that $\displaystyle det(A-LE)=-L^3+2L^2+2L+3$
so
$\displaystyle L^3-2L^2-2L-3=0$
//after this our lecturer takes random numbers and checks if they can be eigenvalues.
For this problem I'm sure that eigenvalue is only $\displaystyle 3$
//but what if I get $\displaystyle L^3-7L^2-36=0$?
//I have to check all numbers till 6
//so how did he find that for this one $\displaystyle L$ is $\displaystyle 6, 3$ and $\displaystyle 2$ so fast?
// EH... NEVERMIND xD
further, to find eigenvector I have
$\displaystyle \begin{pmatrix}-3&1&0&|0\\2&-5&-1&|0\\-3&0&-3&|0\end{pmatrix}$
but I can't find a way to substitute so that I would have 2 even rows or columns, so maybe there is another way I don't know?
some update:
I ended up at
$\displaystyle \begin{pmatrix}-3&2&-1&|0\\0&-3&-3&|0\end{pmatrix}=\begin{pmatrix}1&-2/3&1/3&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&-0&1&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&0 &|-C\\0&1&|-C\end{pmatrix}$
$\displaystyle x_1 = -C$
$\displaystyle x_2 = -C$
$\displaystyle x_3 = C$
So, after getting someone annoying for updating my post so many times xD
Q: HOW DO I STANDARDIZE EIGENVECTOR AND CHECK IF ANSWER IS CORRECT?
someone told me: "you need to find vector length or smth" =/
$\displaystyle |x|=\sqrt{(-C)^2+(-C)^2+C^2}=\sqrt{3}C$
$\displaystyle x_N=\begin{pmatrix}-1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{pmatrix}$ ???????
IF CORRECT, HOW DO I CHECK IF IT IS SO?
2. For your first, there are also two complex eigenvalues, $\displaystyle \displaystyle \frac{-1+\sqrt{3}i}{2}$ and $\displaystyle \displaystyle \frac{-1 - \sqrt{3}i}{2}$. All the eigenvalues are important.
For the second $\displaystyle \displaystyle \lambda^3 - 7\lambda^2 - 36 = 0$, this does not factorise, in fact, it is not solvable exactly. And a simple check of the graph would see there is only one real root, so his answer can not possibly be correct.
3. Originally Posted by Revy
I have matrix $\displaystyle A=\begin{pmatrix}0&1&0\\2&-2&-1\\-3&0&0\end{pmatrix}$
and I get that $\displaystyle det(A-LE)=-L^3+2L^2+2L+3$
so
$\displaystyle L^3-2L^2-2L-3=0$
//after this our lecturer takes random numbers and checks if they can be eigenvalues.
For this problem I'm sure that eigenvalue is only $\displaystyle 3$
I doubt that he takes "random" numbers- he probably knows them ahead of time. Here, however, the "rational root theorem" says that any rational number satisfying this equation must be 1, -1, 3, or -3. 3 is the only one of those that satisfies it. However, as Prove It says, there are two complex roots as well. Divide $\displaystyle L^3- 2L^3- 2L+ 3$ by L- 3 to get the quadratic equation they must solve and use the quadratic formula.
//but what if I get $\displaystyle L^3-7L^2-36=0$?
//I have to check all numbers till 6
what do you mean by "all number till 6". Surely you understand that roots are not necessarily integers! By the "rational root theorem" again, the only possible rational solutions must be factors of 36- but there are a bunch of those!
//so how did he find that for this one $\displaystyle L$ is $\displaystyle 6, 3$ and $\displaystyle 2$ so fast?
I can only say that he must be very clever if he convinced you that 6, 3, and 2 were solutions!
$\displaystyle 6^3- 7(6^2)- 36= 216- 252- 36= -36- 36= -72$, not 0.
$\displaystyle 3^3- 7(3^2)- 36= 27- 63- 36= -36- 36= -72$, not 0.
$\displaystyle 2^3- 7(2^2)- 36= 8- 28- 36= -56$, not 0.
6 and 3 satisfy the equation $\displaystyle L^3- 7L^2+ 36= 0$ but I don't see how to include "2" in there.
// EH... NEVERMIND xD
further, to find eigenvector I have
$\displaystyle \begin{pmatrix}-3&1&0&|0\\2&-5&-1&|0\\-3&0&-3&|0\end{pmatrix}$
but I can't find a way to substitute so that I would have 2 even rows or columns, so maybe there is another way I don't know?
I'm not sure what you mean by this. The definition of "eigenvalue" is a number, L such that Av= Lv for some non-zero vector v. To find v, solve Av= Lv.
Since you are talking, again, about $\displaystyle A= \begin{bmatrix}0 & 1 & 0 \\ 2 & -2 & -1\\ - 3 & 0 & 0\end{bmatrix}$ with eigenvalue 3, you want to solve
$\displaystyle \begin{bmatrix}0 & 1 & 0 \\ 2 & -2 & -1\\ - 3 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}3x \\ 3y \\ 3z\end{bmatrix}$
$\displaystyle \begin{bmatrix}y \\ 2x- 2y- z \\ -3x\end{bmatrix}= \begin{bmatrix}3x \\ 3y \\ 3z\end{bmatrix}$
which means we must have y= 3x, 2x- 2y- z= 3y, and -3x= 3z. We can replace y and z in the second equation with 3x and -x, respectively. Then it becomes 2x- 2(3x)- (-x)= 0 no matter what x is. That is, any vector of the form (x, 3x, -x)= x(1, 3, -1) will satisfy that: any multiple of (1, 3, -1) is an eigenvector.
some update:
I ended up at
$\displaystyle \begin{pmatrix}-3&2&-1&|0\\0&-3&-3&|0\end{pmatrix}=\begin{pmatrix}1&-2/3&1/3&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&-0&1&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&0 &|-C\\0&1&|-C\end{pmatrix}$
$\displaystyle x_1 = -C$
$\displaystyle x_2 = -C$
$\displaystyle x_3 = C$
So, after getting someone annoying for updating my post so many times xD
Q: HOW DO I STANDARDIZE EIGENVECTOR AND CHECK IF ANSWER IS CORRECT?
someone told me: "you need to find vector length or smth" =/
$\displaystyle |x|=\sqrt{(-C)^2+(-C)^2+C^2}=\sqrt{3}C$
$\displaystyle x_N=\begin{pmatrix}-1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{pmatrix}$ ???????
IF CORRECT, HOW DO I CHECK IF IT IS SO?
You don't need to normalize unless the problem specifically asked for a unit eigenvector. Just check to see if, in fact,
Av= Lv. You can keep "C" or factor it out: (-C, -C, C)= C(-1, -1, 1) so (-1, -1, 1) is such a vector.
Try $\displaystyle \begin{bmatrix}0 & 1 & 0 \\ 2 & -2 & -1\\ - 3 & 0 & 0\end{bmatrix}\begin{bmatrix}-1 \\ -1 \\ 1\end{bmatrix}= \begin{bmatrix}0 -1+ 0 \\ -2+ 2- 1 \\ 3+ 0+ 0\end{bmatrix}= \begin{bmatrix}-1 \\ -1 \\ 3\end{bmatrix}$
which is NOT "3" times the vector (-1, -1, 1), nor any number times that vector, so it is NOT an eigenvector. | 2,259 | 6,060 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-17 | latest | en | 0.734778 |
https://tunxis.commnet.edu/view/charging-by-friction-worksheet-answers.html | 1,721,780,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00233.warc.gz | 508,280,164 | 6,591 | # Charging By Friction Worksheet Answers
Charging By Friction Worksheet Answers - Web get the answers to your charging by friction worksheet and understand the concepts of charging by friction and static electricity. Web charging by friction when two materials are rubbed together, some electrons may be transferred from one material to the other, leaving them both with a net electric charge. Each group will be given a balloon, an empty 355 ml pop can, a fur cloth, a timer, tape, a meter stick, a calculator, and the. Click the card to flip 👆. Charging by conduction and induction worksheet. Of the static electricity chapter at the physics classroom:.
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Some of the worksheets displayed are lesson 8 charging by friction induction conduction, notes. Students will be broken into groups of 3 or 4. Showing 8 worksheets for charging by friction. Web charging by friction, conduction & induction task & answer key | tpt. Of the static electricity chapter at the physics classroom:.
## Friction Worksheets & Facts Types, Forms, Factors
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Web charging by friction static electricity answer key. Web charging by friction, charging by conduction (contact) and charging by induction are all targeted. Showing 8 worksheets for charging by friction. Charging by friction read from lesson 2 of thestatic electricity chapter atthe physics classroom:
Of the static electricity chapter at the physics classroom:. 100% (3) view full document. Some of the worksheets displayed are lesson 8 charging by friction induction conduction, notes.
This collection of pages comprise. Web charging by friction, charging by conduction (contact) and charging by induction are all targeted. Each group will be given a balloon, an empty 355 ml pop can, a fur cloth, a timer, tape, a meter stick, a calculator, and the.
## Of The Static Electricity Chapter At The Physics Classroom:.
Web get the answers to your charging by friction worksheet and understand the concepts of charging by friction and static electricity. Web physics classroom charging by friction worksheet answers. Web there are 3 main ways to cause an object to become charged: Web in the context of static electricity charging by friction, this worksheet provides answers to questions related to the phenomenon of static electricity and its properties.
## Web Charging By Friction, Charging By Conduction (Contact) And Charging By Induction Are All Targeted.
Charging by conduction and induction worksheet. Showing 8 worksheets for charging by friction. Students will be broken into groups of 3 or 4. Conceptual physics electrostatics and static.
## Student Worksheet 12.1B Charging An Object Section 12.
Web charging by friction static electricity answer key. This collection of pages comprise. Charging by friction, conduction & induction task & answer key. Some of the worksheets displayed are lesson 8 charging by friction induction conduction, notes.
## What Is Charging By Friction?
100% (3) view full document. Web charging by friction when two materials are rubbed together, some electrons may be transferred from one material to the other, leaving them both with a net electric charge. Charging by friction, induction, & conduction. Lets try a few things and answer the same questions here on eclass, you can fill this page in to use as a reference as you follow the directions. | 981 | 4,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.918225 |
https://www.chegg.com/homework-help/many-recreation-facilities-use-inflatable-bubble-structures-chapter-6-problem-80p-solution-9781118139448-exc | 1,547,930,097,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583681597.51/warc/CC-MAIN-20190119201117-20190119223117-00059.warc.gz | 752,206,724 | 27,587 | # Fluid Mechanics (8th Edition) View more editions Solutions for Chapter 6 Problem 80PProblem 80P: Many recreation facilities use inflatable “bubble” structure...
• 1644 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
100% (4 ratings)
Many recreation facilities use inflatable “bubble” structures. A tennis bubble to enclose four courts is shaped roughly as a circular semicylinder with a diameter of 50 ft and a length of 50 ft. The blowers used to inflate the structure can maintain the air pressure inside the bubble at 0.75 in. of water above ambient pressure. The bubble is subjected to a wind that blows at 35 mph in a direction perpendicular to the axis of the semicylindrical shape. Using polar coordinates, with angle θ measured from the ground on the upwind side of the structure, the resulting pressure distribution may be expressed as
where p is the pressure at the surface, p the atmospheric pressure, and Vw the wind speed. Determine the net vertical force exerted on the structure.
Step-by-Step Solution:
Chapter: Problem:
100% (4 ratings)
• Step 1 of 5
Pressure inside the bubble …… (1)
Figure:
Where
Diameter of circular semi cylinder
Length of circular semi cylinder
Air pressure inside the bubble is at of water
Wind speed
• Chapter , Problem is solved.
Corresponding Textbook
Fluid Mechanics | 8th Edition
9781118139448ISBN-13: 1118139445ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Fox and McDonald's Introduction to Fluid Mechanics 8th Edition Textbook Solutions | 361 | 1,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-04 | latest | en | 0.813964 |
http://www.mychemistrytutor.com/questions/ph-meaning-ph-scale-acids-and-bases-and-buffer-solutions | 1,394,518,854,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011139063/warc/CC-MAIN-20140305091859-00038-ip-10-183-142-35.ec2.internal.warc.gz | 448,087,779 | 5,837 | # pH, the meaning of the pH scale, acids and bases and buffer solutions
We did an experiment in lab on pH,acids and bases but he really didn't go over it w/ us and wants us to answer the self test questions on our lab manual.
Kindly help me answer the ff questions:
1) The pH scale is a logarithmic scale. This means that the difference between 2 pH units is really a factor of 10. With this in mind, answer the following questions:
* i really couldn't understand what it means when they say, "this means that the difference between 2 pH units is really a factor of 10"
a. how many times more acidic is a solution whose pH =1 compared to a solution whose pH is 5?
b. how many times more basic is a solution whose pH=12 compared to a solution whose pH = 10?
2) Write a list showing the advantages and disadvantages of obtaining the pH of a substance using
a. a pH meter
b. pH paper (or universal indicator)
-- maybe you can give me more reasons *wink*
3. Complete the ff equations for the addition of an acid and base to phosphate buffer solution.
a. (HPO4)-2 + H1+ -->
b. (H2PO4)-1 + (OH)1- -->
thanks so much!!! i would really appreciate it!
### Re: pH, acids and bases
Not to give away the farm or anything but a logarithmic scale is geometric. In this scale a whole number increase is a power of 10. This is much like the Richter Scale used to measure strength of earthquakes or decibel scale for sound. Each whole number increase represents a 10-fold actual increase, an increase of 2 indicates a 10x10 increase (102).
A scale or colored paper, gee I don't know. What do you think? ;)
Part three - well you really should think this through. It's not that difficult.
Okay?
Otis
### Re: pH, acids and bases
anybody willing to help? and explain the whole thing? our teacher just sucks! thanks! | 447 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2014-10 | longest | en | 0.945497 |
https://usethinkscript.com/threads/aliens-extraterrestrial-visual-systems.3881/ | 1,708,900,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474643.29/warc/CC-MAIN-20240225203035-20240225233035-00583.warc.gz | 585,017,049 | 21,136 | Alien's Extraterrestrial Visual Systems
Steakeater
New member
Hi Everyone, Aliens strategy is free and I think everyone would benefit from it being on this website to share. So I have been trying to recreate this strategy in think or swim. https://www.forexfactory.com/thread/463573-aliens-extraterrestrial-visual-systems..
So far I have found with the help of @BenTen the RSIOMA indicator however its not exactly right but its a start. If anyone is interest or want to help out so the community has this system available to us all feel free to pitch in.
The 4 parts to this system are the
• Stochastics,
• RSIOMA and the drake delay stochastics.
part 1 RSIOMAS
needs tweaking to smooth the indicator out more but its almost there.
Code:
``````#
# TD Ameritrade IP Company, Inc. (c) 2007-2019
# Tweaked by [USER=212]@korygill[/USER]
# https://usethinkscript.com/d/185-moving-average-crossover-rsi-indicator-for-thinkorswim
declare lower;
input length = 14;
input over_Bought = 70;
input over_Sold = 30;
input price = close;
input averageType = AverageType.WILDERS;
input showBreakoutSignals = no;
input rsiMALength = 5; #hint rsiMALength: RSI Moving Average Length
input rsiAverageType = AverageType.SIMPLE;
def NetChgAvg = MovingAverage(averageType, price - price[1], length);
def TotChgAvg = MovingAverage(averageType, AbsValue(price - price[1]), length);
def ChgRatio = if TotChgAvg != 0 then NetChgAvg / TotChgAvg else 0;
plot RSI = 50 * (ChgRatio + 1);
plot OverSold = over_Sold;
plot OverBought = over_Bought;
plot UpSignal = if RSI crosses above OverSold then OverSold else Double.NaN;
plot DownSignal = if RSI crosses below OverBought then OverBought else Double.NaN;
# plot the RSI Moving Average
def rsiMA = MovingAverage(rsiAverageType, RSI, rsiMALength);
plot prsiMA = rsiMA;
UpSignal.SetHiding(!showBreakoutSignals);
DownSignal.SetHiding(!showBreakoutSignals);
RSI.DefineColor("OverBought", GetColor(5));
RSI.DefineColor("Normal", GetColor(7));
RSI.DefineColor("OverSold", GetColor(1));
RSI.AssignValueColor(if RSI > over_Bought then RSI.color("OverBought") else if RSI < over_Sold then RSI.color("OverSold") else RSI.color("Normal"));
OverSold.SetDefaultColor(GetColor(8));
OverBought.SetDefaultColor(GetColor(8));
UpSignal.SetDefaultColor(Color.UPTICK);
UpSignal.SetPaintingStrategy(PaintingStrategy.ARROW_UP);
DownSignal.SetDefaultColor(Color.DOWNTICK);
DownSignal.SetPaintingStrategy(PaintingStrategy.ARROW_DOWN);``````
The adx part uses adxs with lengths of 144, 89, 42, 21, and 7
The second step is creating your ADX indicator is to simply go into the indicator tab. Next you search for ADX and create 5 ADX indicators. Then you simply click and hold the ADX and drop it ontop of the first one and repeat these steps until all 5 are in the same indicator.
Next you right click and go into properties, then change the length of each one to what I specified the lengths are above
Third read over Alien's Thread above and understand how he uses the ADX so you can apply it to the system
More updates to follow, and ill post a workspace if we can get all the indicators finished to its an easy click for new members.
As always any contributions or questions are appreciated
Steak,
Also this does not count as financial advice trade at your own risk all credit for this system goes to ALIEN from forex factory
Hi everyone Quick update
The author forex alien claims that " The charts don't need all these indicators to trade, but they sure do give me 98 % accuracy in my trade direction."
Here is a link to the workspace, https://tos.mx/jmkPRsh (WIP)
To do list
• Fix RSIOMA
• fix stochastic coloring (Probably a script here that already does that
• Put ADX on a scale from 1-60 instead of a percentage scale
I have trading view indicators for RSIOMA and DRAKESTOCHASTIC that work correctly, if someone has time to convert them shoot me a message.
I have noticed a lot of new members have a hard time trying to trade with multiple time frames, Here is a picture from ALIEN that I think will help everyone visualize how multiple time frames interreact with each other, so that you can increase your probability of a profitable trade. Lets all strive to get your probability of a profitable trade up.
Best of luck with trading keep studying hard,
Steak
Last edited by a moderator:
Here are the two trading view indicators for this strategy I need help changing to Thinkscript.
Code:
``````//By Darkyalt
study(title="Relative Strength Index of MA", shorttitle="RSIOMA")
src = close, len = input(14, minval=1, title="Length")
up = rma(max(change(ema(src, len)), 0), len)
down = rma(-min(change(ema(src, len)), 0), len)
rsi = down == 0 ? 100 : up == 0 ? 0 : 100 - (100 / (1 + up / down))
plot(rsi, color=blue)
band1 = hline(80)
band0 = hline(20)
fill(band1, band0, color=purple, transp=90)
hline(50, color=gray, linestyle=line)
sig = ema(rsi, 21)
plot(sig, color=purple)``````
2nd
Code:
``````// This source code is subject to the terms of the Mozilla Public License 2.0 at https://mozilla.org/MPL/2.0/
//@version=4
study("Drake Delay Stochastic")
pds = input(title="pds", type=input.integer, defval=13, minval=1)
slw = input(title="ema", type=input.integer, defval=8, minval=1)
slwsignal = input(title="trigger", type=input.integer, defval=9, minval=1)
MAValue = 0.0
MAValue2 = 0.0
MAValue3 = 0.0
smconst = 2.0 / (1 + slw)
smconst1 = 2.0 / (1 + slwsignal)
prev = na(MAValue2[1]) ? 0.0 : MAValue2[1]
AA = 0.0
tmpDevAA = highest(pds) - lowest(pds)
if tmpDevAA != 0
AA := 100* ((close - lowest(pds)) / tmpDevAA)
MAValue2 := smconst * (AA-prev) + prev
MyHigh = -999999.0
MyLow = 999999.0
for i = 0 to pds
Price = na(MAValue2) ? 0.0 : MAValue2
if Price != 0.0
if Price > MyHigh
MyHigh := Price
if pds <= 0
MyHigh := Price
if Price < MyLow
MyLow := Price
if pds <= 0
MyLow := Price
prev1 = na(MAValue[1]) ? 0.0 : MAValue[1]
aa1=MAValue2
bb= 0.0
if MyHigh-MyLow != 0
bb:=100*(aa1-MyLow)/(MyHigh-MyLow)
MAValue := smconst * (bb-prev1) + prev1
MAValue3:= ema(MAValue, slwsignal)
plot(MAValue, color = MAValue >= MAValue3 ? color.lime : color.red, linewidth = 2)
plot(MAValue3, color = color.black)
hline(20, title="Sell", color=color.gray, linestyle=hline.style_dashed)
bgcolor(MAValue >= MAValue3 ? #38a32a : #da680f, transp=80)``````
Also Does anyone know if its possible to put multiple adx studys together but instead of using % the scale change it to 1-100?
Steak,
Last edited by a moderator:
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https://goprep.co/ex-11.1-q7d-5-cos-2-x-7-sin-2-x-6-0-solve-the-following-i-1nlekl | 1,660,418,631,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00343.warc.gz | 298,909,252 | 33,039 | Q. 7 D5.0( 2 Votes )
# Solve the followi
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = nπ + (– 1)ny, where n Z.
• cos x = cos y, implies x = 2nπ ± y, where n Z.
• tan x = tan y, implies x = nπ + y, where n Z.
given,
5 cos2 x + 7 sin2 x – 6 = 0
5 cos2 x + 5 sin2 x + 2sin2 x – 6 = 0
2 sin2 x – 6 + 5 = 0 { sin2 x + cos2 x = 1}
2 sin2 x – 1 = 0
sin2 x = (1/2)
sin x = ±(1 /√2)
sin x = ± sin π /4
If sin x = sin y, implies x = nπ + (– 1)ny, where n Z.
x = nπ + (-1)n (±(π / 4)) where n ϵ Z
….ans
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The 5 and 60 are in km/h, so when you multiply the second equation by 5, you are actually multiplying by 5 km/h, which makes the units make sense. You should try to either write down all the units, like for x, y, 5, and 60 in this case, or none at all.
If you divide the first equation by 1 kilometer and the second equation by 1 hour you will get a pair of dimensionless equations. This is why we can ignore the units when we manipulate the equations. I think that she will like this explanation.
by
The units don't matter. If you divide the first eqn by 1 km and the second eqn by 1 hour you have two eqns without units.
Going forward, I would omit the units when writing down the eqns.
Tell her that each equation on its own must be consistent with regard to the units, but the arithmetic we use to solve the system does not require this consistency. Each eqn represents a different relationship between x and y.
Tell her that x and y have the units of kilometers in both equations
by
0 like 0 dislike | 255 | 1,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-06 | latest | en | 0.956234 |
http://math.stackexchange.com/questions/358706/fourier-series-of-multivariable-functions | 1,469,709,425,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828282.32/warc/CC-MAIN-20160723071028-00278-ip-10-185-27-174.ec2.internal.warc.gz | 158,113,850 | 17,441 | # Fourier Series of Multivariable Functions.
If I have some function $V(x,y)$ which is periodic in x with period L. I wish to expand $V(x,y)$ in terms of a fourier sine (for simplicity) series in $x$, is it always the case that I may write the following?
$$V(x,y) = \sum_{n = 1}^{\infty} a_n \sin(\frac{n\pi x}{L})f_n(y)$$
Where the $f_n(y)$ are whatver functions of $y$ needed to satisfy the equality -- do they need to have a particular form or any properties other than continuity?
It seems to me that we might require a certain type of function $V(x,y)$ to do this, does anyone know what the neccessary and sufficient conditions on $V(x,y)$ are to be able to expand in this way?
-
$$V(x,y) = \sum_{n=1}^{\infty} \: \sum_{m=1}^{\infty} a_{nm} \sin{\left ( \frac{n \pi x}{L}\right)} f_m(y)$$
$f_m$ will be defined by boundary conditions on the $y$ boundaries.
You could take $\phi_n(y) = \sum_{m=1}^\infty a_{nm} f_m(y)$, then $V(x,y) = \sum_{n=1}^\infty \sin(\frac{n \pi x} {L}) \phi_n(y)$. – copper.hat Apr 11 '13 at 20:07
Yes, but you still have to find the $a_{nm}$ and the $f_m$ from the BC's. – Ron Gordon Apr 11 '13 at 20:20
@RonGordon. So I am always going to be able to separate the variables in the sum this way? Is it true that if we couldn't write $V$ as you/I have written it then $V$ would simply not have a Fourier Series in $x$? – user27182 Apr 11 '13 at 20:41
You can separate like this when you have rectangular geometry and separable boundary conditions. (If you have BC's that have mixes of values over both directions, then you may not be able to have this representation.) On your 2nd question you are right, but it doesn't mean no FS period. You could have a FS in, say, $xy$ in some bizarre scenario. – Ron Gordon Apr 11 '13 at 20:44 | 548 | 1,765 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2016-30 | latest | en | 0.905287 |
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# Which number in set P has a value greater than that of every
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Which number in set P has a value greater than that of every [#permalink] 07 Sep 2005, 04:22
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Which number in set P has a value greater than that of every other member of all set P, if all the members of set P are negative integers?
1 Each member of set P is the product of -1 and a prime number
2At least one member of P is even
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I'd go with C.
If -2 is a member of the set, then it would be the greatest number. But without statement 2, we don't know that -2 is a member of the set. In fact, from the stem we don't know how many numbers are in set P at all. Perhaps set P={-7,-11,-17}.
Since there's only one product of -1 and a prime number that can be even, and statement 2 tells us it's in the set, then both in combination are sufficient. So C.
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A for me
statement 1 is suff -> we know all the numbers in set P are negative integers so the biggest would be -1. Then we know that each one is a oproduct if -1 and a prime number so the biggest has to be -2
statement 2 gives us nothing.
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think it is E)
from A) it can be any prime so insuff
from B) it is also insuff
taken both, the only even prime is 2 , but the Q says "at least" which means that there might be more than one 2-s so the ans seems E)
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I go for C here.
Stmt one tells us that each member of the set is a negative prime but it doesn't say how many primes or which one it starts with. The set could start at -3 or -13 or only have -3 and -13 in it. Insufficient
Stmt two is obviously insufficient.
Put together and sufficient. We can answer the question since there is only one even prime number.
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Re: DS:Princeton Rev 2° [#permalink] 07 Sep 2005, 17:17
Macedon wrote:
Which number in set P has a value greater than that of every other member of all set P, if all the members of set P are negative integers?
1 Each member of set P is the product of -1 and a prime number
2At least one member of P is even
This question isn't very clear. If it is asking what is the maximum value in P then 1 and 2 together would be sufficient to determine that -2 is in P. Since -2 is greater than all the other products of -1 and prime numbers, then we can say the answer is C.
However if it is asking which element has the greatest value, then the answer would be E, since there could be more than one element that has the value of -2.
_________________
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Re: DS:Princeton Rev 2° [#permalink] 07 Sep 2005, 17:17
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it means that the personal assets of the partner may be used for payment of debts.. | 428 | 1,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.965979 |
https://www.rebeccabarter.com/categories/statistics/ | 1,643,132,604,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304859.70/warc/CC-MAIN-20220125160159-20220125190159-00275.warc.gz | 1,021,932,703 | 5,324 | # Statistics
## It's time for statistics departments to start supporting their applied students
Statistics departments are failing their applied students. In this post, I have a lot of opinions and give two pieces of advice: statistics departments need to start supporting their applied students, and they need to hire applied faculty.
I graduated with a PhD from UC Berkeley’s statistics department in December. My PhD dissertation consisted of three 100% applied projects (one of which was a piece of open-source software). This is, unfortunately, incredibly rare. Over the past few years, I’ve had a number of current and prospective statistics PhD students both at Berkeley and outside Berkeley get in touch with me to ask me how I made my way through a statistics PhD by working only on applied projects.
## Which hypothesis test should I use? A flowchart
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Many years ago I taught a stats class for which one of the topics was hypothesis testing. Many of the students had a hard time remembering what situation each test was designed for, so I made a flowchart to help piece together the wild world of hypothesis tests. While the flowchart isn’t pretty (if I made it today, it would be much more attractive), I feel like it might be useful for others, so here it is:
## Understanding Instrumental Variables
Instrumental variables is one of the most mystical concepts in causal inference. For some reason, most of the existing explanations are overly complicated and focus on specific nuanced aspects of generating IV estimates without really providing the intuition for why it makes sense. In this post, you will not find too many technical details, but rather a narrative introducing instruments and why they are useful.
Suppose, as many do, that we want to estimate the effect of an action (or treatment) on an outcome. As an example, we might be interested in estimating the effect of receiving a drug vs not receiving a drug on the incidence of heart disease. In an ideal futuristic world, we would take each individual in our population and split them into two identical humans: one who receives the treatment and the other who doesn’t. | 430 | 2,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.966091 |
https://www.dsprelated.com/showthread/comp.dsp/14862-1.php | 1,627,494,759,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00687.warc.gz | 785,339,502 | 13,373 | To 'sinc' or not to 'sink'
Started by January 11, 2004
```What is 'sinc' function and why is it important.
First response of "google" is "Sisters in Crime Internet Chapter" ;{
OK, I did find a site with plot of 'sinc' which resembled a 'cosine
with decreasing amplitude' -- The amplitude at 0 WAS 1 and decreased
from there ;]
Actually I suspect I need to know more about "windowing".
Any references besides a book coming out in March ;>
```
```Richard Owlett wrote:
> What is 'sinc' function and why is it important.
>
> First response of "google" is "Sisters in Crime Internet Chapter" ;{
>
> OK, I did find a site with plot of 'sinc' which resembled a 'cosine with
> decreasing amplitude' -- The amplitude at 0 WAS 1 and decreased from
> there ;]
>
> Actually I suspect I need to know more about "windowing".
> Any references besides a book coming out in March ;>
Sin(x)/x. Although direct substitution is indeterminate at x=0, there
are at least two ways to show that the ratio approaches unity as x->0.
The impulse of a perfect low-pass filter has the shape of a sinc. Since
it has a response at t=-infinity, you can see that a real-time perfect
("brick-wall") low pass is very hard to build. :-) Since low-pass
filters are the "prototypes" from which other filters are derived, sincs
show up throughout filter theory.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
```Richard Owlett <rowlett@atlascomm.net> writes:
> What is 'sinc' function and why is it important.
>
> First response of "google" is "Sisters in Crime Internet Chapter" ;{
>
> OK, I did find a site with plot of 'sinc' which resembled a 'cosine
> with decreasing amplitude' -- The amplitude at 0 WAS 1 and decreased
> from there ;]
>
> Actually I suspect I need to know more about "windowing".
> Any references besides a book coming out in March ;>
Richard,
The sinc() function is defined as
sinc(x) = sin(pi*x)/(pi*x).
It is very common in signal processing since it is the inverse
Fourier transform of the proverbial brick-wall lowpass filter.
You are quite right that it looks like a "cosine with decreasing
amplitude" because, well, it is (only not a cosine). The definition
above is the reason.
If you really want to know more about the sinc() function, I would
suggest the classic textbook "The Fourier Transform and its Applications"
by Ronald Bracewell.
--
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
```
```Hello Richard,
Sinc stands for "sine cardinal." The following links have some pertinent
http://mathworld.wolfram.com/SincFunction.html
http://ccrma-www.stanford.edu/~jos/waveguide/Theory_Ideal_Bandlimited_Interpolation.html
Basically the sinc function in DSP does two primary things. First it is the
Fourier transform of the rectangle[1] function, and second it is what is
used in ideal interpolation of sampled data. Now using the Fourier duality
of multiplication in one domain being convolution in the other. Then you can
see when ever you try to window data with a rectangle function in one domain
that in the other domain you will have a convolution with a sinc function.
IHTH,
Clay
[1] Rectangle function rect(x) = 1 when |x|<pi,
=1/2 when
|x|=pi
= 0 otherwise
"Richard Owlett" <rowlett@atlascomm.net> wrote in message
news:1003fhr6v2i5h73@corp.supernews.com...
> What is 'sinc' function and why is it important.
>
> First response of "google" is "Sisters in Crime Internet Chapter" ;{
>
> OK, I did find a site with plot of 'sinc' which resembled a 'cosine
> with decreasing amplitude' -- The amplitude at 0 WAS 1 and decreased
> from there ;]
>
> Actually I suspect I need to know more about "windowing".
> Any references besides a book coming out in March ;>
>
>
```
```Jerry Avins <jya@ieee.org> writes:
> Richard Owlett wrote:
>
>> What is 'sinc' function and why is it important.
>> First response of "google" is "Sisters in Crime Internet Chapter" ;{
>> OK, I did find a site with plot of 'sinc' which resembled a 'cosine
>> with decreasing amplitude' -- The amplitude at 0 WAS 1 and decreased
>> from there ;]
>> Actually I suspect I need to know more about "windowing".
>> Any references besides a book coming out in March ;>
>
> Sin(x)/x.
Jerry,
I guess you could say that is sinc(x/pi). However, it may be confusing.
sinc(x) = sin(pi*x)/(pi*x), by definition.
> [...]
--
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
```
```Randy Yates wrote:
> Jerry Avins <jya@ieee.org> writes:
>
>
>>Richard Owlett wrote:
>>
>>
>>>What is 'sinc' function and why is it important.
>>>First response of "google" is "Sisters in Crime Internet Chapter" ;{
>>>OK, I did find a site with plot of 'sinc' which resembled a 'cosine
>>>with decreasing amplitude' -- The amplitude at 0 WAS 1 and decreased
>>>from there ;]
>>>Actually I suspect I need to know more about "windowing".
>>>Any references besides a book coming out in March ;>
>>
>>Sin(x)/x.
>
>
> Jerry,
>
> I guess you could say that is sinc(x/pi). However, it may be confusing.
> sinc(x) = sin(pi*x)/(pi*x), by definition.
Thanks for being specific. I oversimplified when, for simplicity's sake,
I left out the details. Later I wrote "shape of a sinc". Richard might
recognize that as a central cross section of the "Mexican hat" figure.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
```Jerry Avins <jya@ieee.org> writes:
> Randy Yates wrote:
>
>> Jerry Avins <jya@ieee.org> writes:
>>
>>>Richard Owlett wrote:
>>>
>>>
>>>>What is 'sinc' function and why is it important.
>>>>First response of "google" is "Sisters in Crime Internet Chapter" ;{
>>>>OK, I did find a site with plot of 'sinc' which resembled a 'cosine
>>>>with decreasing amplitude' -- The amplitude at 0 WAS 1 and decreased
>>>>from there ;]
>>>>Actually I suspect I need to know more about "windowing".
>>>>Any references besides a book coming out in March ;>
>>>
>>> Sin(x)/x.
>> Jerry, I guess you could say that is sinc(x/pi). However, it may be
>> confusing.
>> sinc(x) = sin(pi*x)/(pi*x), by definition.
>
> Thanks for being specific. I oversimplified when, for simplicity's sake,
> I left out the details.
Sometimes it is good to simplify, but I don't think this is one
of them. It seems that Richard is trying to get the rudiments and
in that case it's important (at least I have found it to be so
for myself) to get them right. Learning something wrong the first
time seems to take an inordinate amount of effort to unlearn. I
could be all wet, though.
--
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
```
```Randy Yates wrote:
> Sometimes it is good to simplify, but I don't think this is one
> of them.
That's why I thanked you for calling me on it, and described myself as
having "oversimplified".
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
```Jerry Avins wrote:
> Randy Yates wrote:
>
>
>> Sometimes it is good to simplify, but I don't think this is one
>> of them.
>
>
> That's why I thanked you for calling me on it, and described myself as
> having "oversimplified".
>
> Jerry
Mr. turner referenced http://mathworld.wolfram.com/SincFunction.html
which shows that "authorities" have used both definitions.
```
```Clay S. Turner wrote:
> Hello Richard,
> Sinc stands for "sine cardinal." The following links have some pertinent
>
> http://mathworld.wolfram.com/SincFunction.html
>
> http://ccrma-www.stanford.edu/~jos/waveguide/Theory_Ideal_Bandlimited_Interpolation.html
>
apologies to Robin/Rainger/Sinatra ;]
> Basically the sinc function in DSP does two primary things. First it is the
> Fourier transform of the rectangle[1] function, and second it is what is
> used in ideal interpolation of sampled data. Now using the Fourier duality
> of multiplication in one domain being convolution in the other. Then you can
> see when ever you try to window data with a rectangle function in one domain
> that in the other domain you will have a convolution with a sinc function.
>
>
I think I have a problem which is about to rear it's ugly head.
My "eventual" field of interest is 'speech recognition'.
I am coming at this field from an "avocational/amateur" [ NOT
'professional' point of view ]
My first task is to "characterize speech" [ Please give both words
their most vague/general denotations/connotations/implications/etc ;]
BTW I've (re)discovered formants.
I'm currently looking at speech samples =/> one second.
I do a FFT and plot amplitude vs frequency for > 100 Hz.
I doubt there is a problem so far.
However, I understand typical speech recognition applications tend to
use samples in the ten's to hundred's of msec.
I assume I must take account of the artifacts caused by sampling.
Please point me in appropriate direction.
thanks
``` | 3,444 | 12,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.757197 |
http://www.excel-answers.com/microsoft/Excel-Charting/34105601/how-to-get-the-corresponding-x-value-for-a-given-y-value.aspx | 1,448,573,019,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447783.20/warc/CC-MAIN-20151124205407-00115-ip-10-71-132-137.ec2.internal.warc.gz | 424,962,346 | 8,867 | # Excel - How to get the corresponding X value for a given Y value?
Asked By Yahy on 01-Mar-09 06:54 PM
```I have a chart in Excel 2007 and I want to provide a Y value (that is not one
of the data points) and have Excel to tell me the corresponding X value from
the chart. How can I do that?
The chart is not a straight line, so trendline won't work
Here is a picture of the chart I'm working on
http://hkoyda.blu.livefilestore.com/y1p8aB149EOMrBPPjH7oxLQUtlTXmrhiIspRq3r8fuZBNtdjVL_9ckuCh8zH_ZKlZwXt4POp6d2ay6fhDehApx_wA/graph.JPG
Any help in this matter would be greatly appreciated.
Yahya```
Ed Ferrero replied on 01-Mar-09 08:12 PM
```Hi Yahya,
You can't do that without knowing the formula for your line of best fit.
There is a sample here
Interpolated Values' that works for a straight line. The sample uses
click-and-drag on the chart which will not work in Excel 2007. However, you
can manually change the x-Value at cell E4 to calculate the intercept at a
particular x-Value.
Looking at your picture, I would try calculating the log of the y-Values and
charting that. If this is close enough to a straight line for your purposes,
then just use my sample to calculate the intercept. =LN() to find the log,
=EXP() to change back.
Ed Ferrero
www.edferrero.com```
Jon Peltier replied on 01-Mar-09 11:40 PM
```If the coordinates of the points are known, i.e., the values are in the
worksheet, then a stepwise interpolation approach can be implemented.
In general, if these are measured points, unless you know the form of the
relationship, you should refrain from using smoothed lines to connect the
points. Excel uses an arbitrary algorithm for smoothing the lines, which has
no physical bearing on the chart, and which may cause the smoothed line to
deviate substantially from a well-behaved relationship.
- Jon
-------
Jon Peltier, Microsoft Excel MVP
Peltier Technical Services, Inc.
http://PeltierTech.com/WordPress/
_______```
Yahy replied on 02-Mar-09 01:25 AM
```Here is a snapshot of my data points
http://hkoyda.blu.livefilestore.com/y1pgZrqLtEzUywZot1wKtjvEbwQZtl0v7j1OArIQLJovsDzqfsixklx2wgAK-pxgI06jaLLI9u1naBfXsjvAfVDOA/graph.JPG```
Mike Middleton replied on 02-Mar-09 02:40 AM
```Yahya -
You might get a good fit using a logistic function. Use Google to search for
But, when possible, curve fitting should rely on knowledge about the
physical phenomenon that is being modeled. Please share what you know about
the source of the data. Such knowledge is usually important for selecting an
appropriate functional form.
- Mike
http://www.MikeMiddleton.com```
Bernard Liengme replied on 02-Mar-09 08:19 AM
```Email me privately (remove TRUENORTH.) , I have a sample file to do a four
parameter fir to logistic data
Then you use Solver to back solve from x to y
best wishes
--
Bernard V Liengme
Microsoft Excel MVP
http://people.stfx.ca/bliengme
remove caps from email```
LoriMille replied on 03-Mar-09 08:16 PM
```if you're using the "smoothed line" charting option, try this formula
with data in the range A4:B14 and an x-value in D4:
=SUM((1+1/IRR(MMULT({0,0,2,0;0,1,0,-1;-1,4,-5,2;
1,-3,3,-1},OFFSET(A4,MATCH(D4,A4:A14,-1)-2,,4)-D4
)))^-{0;1;2;3}*MMULT({0,2,0,0;-1,0,1,0;2,-5,4,-1;
-1,3,-3,1},OFFSET(B4,MATCH(D4,A4:A14,-1)-2,,4)))/2
this returns the corresponding y-value within an interior interval.
These curves are widely used in computer graphics - for more info check out
The curve that Excel plots makes a couple of tweaks to the textbook example:
- End intervals are calculated by extending the range at both ends i.e. using
the points (0.765,99.7) and (0.04,-3.683) in rows 3 and 15 respectively
and adjusting the ranges in the formula accordingly.
- Small intervals have a higher "tension" value which has the effect of
reducing the
overshoot. This value depends on the chart scale used, a VBA function for
this is here:
kim.lindse replied on 28-Mar-09 11:59 AM
```Hi Lori, I saw several posts from you explaining how to get a
corresponding y for any x value using the Catmull-Rom spline, but I
still can't quite understand how the data should be arranged and where
the formula should go to make it work. Can you say a little more
On Mar 3, 9:16=A0pm, Lori Miller <LoriMil...@>
ut
le:
sing
ot one
from```
Lori Miller replied on 26-Mar-09 09:26 PM
```OK, let's take the original data set, with data entered into the
range A4:B14 and draw an XY chart as shown in the OP.
X Y
0.71 99.145
0.655 98.59
0.5125 97.99
0.3375 97.61
0.215 94.51
0.1525 84.21
0.1155 50.26
0.098 33.939
0.0825 27.062
0.064 9.797
0.052 3.057
DATA INTERPRETATION
The y data range lies between [0,100] and is increasing with x, and
i would guess that this is a distribution function of a statistical
sample of some kind. If so, using splines for estimation is valid and
common in the technical literature and in this case the Excel curve
looks like a reasonable approximation. (If these were measurements
subject to a degree of error however, other methods may be more
appropriate, such as regression, as mentioned by other posters.)
ESTIMATION
i. To estimate a y-value, enter the x-value in D4 and copy the formula
shown in the last post into E4.
eg x = 0.13 -> y = 66.316
If you fix the ranges by using A\$4,B\$4 and A\$4:A\$14 in the formula you
can pull the fill handle down to create a range of x and y values.
Charting these values should match the curve that Excel plots.
ii. Actually the original post called for estimating an x-value given a
y-value which can be done with the same formula but just switching X with
Y ie by interchanging A's and B's in the formula.
eg y = 0.5 -> x = 0.115
Also the values are arranged in descending order, often data would be
arranged ascending so that 1 instead of -1 is used in the MATCH function.
i. Note that these types of curve (cardinal splines) do not assign
values to end intervals. The method Excel uses to plot these intervals
is equivalent to adding an extra data point at each end. You can do this
by selecting the range A4:B5 and dragging the fill handle up to row 3
and then selecting A13:B14 and dragging down to row 15, this should give
the same values as before, and the formula result should now match the curve
eg x = 0.06 -> y = 7.047
ii. The tension adjustment that Excel uses is only noticable when points
are sufficiently irregularly spaced which is not the case here.
Specifically, if the distance between neighbouring points is less than
a third the distance between correspopnding alternate points (as
This is accounted for in the Chartcurve UDF.
Hi Lori, I saw several posts from you explaining how to get a
corresponding y for any x value using the Catmull-Rom spline, but I
still can't quite understand how the data should be arranged and where
the formula should go to make it work. Can you say a little more
```http://help.lockergnome.com/office/default--ftopict1005590.html | 1,997 | 6,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2015-48 | longest | en | 0.888707 |
https://lagaiascienza.org/science/how-do-measurements-relate-to-experimental-science.html | 1,660,328,739,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00505.warc.gz | 340,798,025 | 11,104 | # How Do Measurements Relate To Experimental Science?
## Why are measurements important for scientific experiments?
Scientific measurements generally adhere to the International System of Units (SI units). It is important to always include units when recording data, doing calculations and reporting results! Units are globally recognized and necessary for sharing information between scientists around the world.
## What are measurements in an experiment?
Measurement is the link between a researcher’s substantive and/or theoretical argument and an (experimental) research design. 3. Measuring treatments includes the operationalization of treatment as well as compliance with treatment assignment.
## Why must measurements always be reported to the correct number of significant figures?
The significant figures in a measurement include all of the digits that are known, plus a last digit that is estimated. Measurements must always be reported to the correct number of significant figures because calculated answers often depend on the number of significant figures in the values used in the calculation.
## Why is it so important for measurements in an experiment to be accurate and precise?
Why is it so important for measurements in an experiment to be accurate AND precise? It so important for measurements in an experiment to be accurate and precise because you need to get accurate and precisise results. It is important that what you measure has meaning to what you are investigating
## What are the 3 types of measurement?
The three standard systems of measurements are the International System of Units (SI) units, the British Imperial System, and the US Customary System. Of these, the International System of Units(SI) units are prominently used.
## What are the four importance of measurement?
A measurement is the action of measuring something, or some amount of stuff. So it is important to measure certain things right, distance, time, and accuracy are all great things to measure.
## What are the 7 basic units of measurement?
The seven SI base units, which are comprised of:
• Length – meter (m)
• Time – second (s)
• Amount of substance – mole (mole)
• Electric current – ampere (A)
• Temperature – kelvin (K)
• Luminous intensity – candela (cd)
• Mass – kilogram (kg)
## What are measurements called?
A measurement is made by comparing a quantity with a standard unit. The study of measurement is called metrology. There are many measurement systems that have been used throughout history and across the world, but progress has been made since the 18th century in setting an international standard.
## What is measurement example?
Measurement is defined as the act of measuring or the size of something. An example of measurement means the use of a ruler to determine the length of a piece of paper. An example of measurement is 15″ by 25″. A waist measurement of 32 inches.
## How many significant figures are in the measurement 1300 g?
Example: 130.00 g ( 5 sig figs). The number 1,300 has trailing zeros, but no decimal point. The zeros are just placeholders (2 sig figs). If the number had been 1,300., the expressed decimal would make all 4 digits significant.
## How many significant figures are in the measurement 0.023040 cm?
How many significant figures are in the measurement 0.023040 cm? There are 5 significant figures in your measurement.
## Which of the following are 2 rules for determining significant figures?
Non-zero digits are always significant. Any zeros between two significant digits are significant. A final zero or trailing zeros in the decimal portion ONLY are significant.
## Why is it important to have precise measurements?
Accuracy represents how close a measurement comes to its true value. This is important because bad equipment, poor data processing or human error can lead to inaccurate results that are not very close to the truth. Precision is how close a series of measurements of the same thing are to each other.
## What is the importance of measurement?
Measurements are an important part of comparing things, as they provide the basis on comparing objects to other objects. A measurement is the action of measuring something, or some amount of stuff. So it is important to measure certain things right, distance, time, and accuracy are all great things to measure.
## How can one achieve accurate measurement?
How to Get More Accurate Measurements in Your Data
1. Acquire one or more known standards from a reliable source.
2. Run your measuring process or assay, using your instrument, on those standards; record the instrument’s results, along with the “true” values. | 921 | 4,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-33 | latest | en | 0.942789 |
https://crypto.stackexchange.com/questions/77642/dining-cryptographer-net-dc-net-scheme-that-handles-collisions | 1,638,669,781,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00339.warc.gz | 272,722,458 | 33,843 | # Dining-Cryptographer Net (DC-Net) Scheme that Handles Collisions?
[Question edited at the request of Mods]
I recently became fascinated with the elegant and simple solution that Chaum proposed for the Dining Cryptographers problem. If you are unfamiliar, please checkout Wikipedia for a summary and solution.
What is interesting is that DC-Nets allow for participants in a group to share messages to the rest of the group anonymously. Wow! But there is a clear issue - two participants in a DC-Net cannot publish messages at the same time (what we call a collision).
So now I am trying to figure out a protocol that alleges to solve this problem by doing the following alterations to the original DC-net:
• Let Alice, Bob and Carol be three participants in this DC-net A,B,C
• Suppose that each participant has a shared secret to the right and to the left, such that |shared secret| = |message|
• Alice hides her message by XORing with her two shared secrets. Call this encrypted message m_a. Bob and Carol do this same with m_b and m_c
• All participants publish their respective messages in three slots. In the first slot they publish m, in the second $$m^2$$, and so forth (in the third $$m^3$$). These messages are interpreted as elements in a finite field and thus multiplication is done in this field.
• These (encrypted) messages are then interpreted as finite field elements (not an extension field, but a prime field, to be specific) and they are added together according to the arithmetic operators defined by the finite field.
• Once the three rounds are done, we are left with three power sums $$S_1,S_2,S_3$$. ($$\sum_{i\in{1,2,3}}{m_i} = S_1, \sum_{i\in{1,2,3}}{m_i^2}= S_2, \sum_{i\in{1,2,3}}{m^3_i}= S_3, in \space \mathbb{F}$$)
Here is where I get completely lost - the author claims that given these power sums and newtons identities, we can construct a polynomial and we are able to extract the un-encrypted messages! This is really baffling and would be incredible if true.
We construct a polynomial of the form: $$a_3x^3+a_2x^2+a_1x^1+a_0$$
$$a_3 = 1$$
$$a_2 = S_1$$
$$a_1 = \frac{(a_2S_1 - S_2)}{2}$$
$$a_0 = \frac{(a_1S_1 - a_2S_2 +S_3)}{2}$$
Is anyone able to shed light on how powersum of encrypted messages over a finite field can be decrypted using polynomials and newtons identities, I am all ears. If this is too broad of a question, it can be closed.
(Solution is described on page 4) Paper
• The question is too broad and can be closed. Please be more focused and describe what exactly is not clear. Feb 16 '20 at 0:48
• @mentallurg No, that's not the right way to go about it. You explain to the poster that the question is unclear and that it should be amended. Then you vote to close if you think it cannot be answered if left at this state. Feb 17 '20 at 1:07
• Maarten, thank you for letting me know. I will amend the question tomorrow morning and I will do my best to be more specific.
– A M
Feb 17 '20 at 6:47 | 789 | 2,963 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-49 | latest | en | 0.922575 |
https://www.exactlywhatistime.com/days-from-date/february-7/54-days | 1,720,953,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00782.warc.gz | 643,908,119 | 6,284 | # Wednesday April 02, 2025
Adding 54 days from Friday February 07, 2025 is Wednesday April 02, 2025 which is day number 092 of 2024. This page is designed to help you the steps to count 54, but understand how to convert and add time correctly.
• Specific Date: Friday February 07, 2025
• Days from Friday February 07, 2025: Wednesday April 02, 2025
• Day of the year: 092
• Day of the week: Wednesday
• Month: April
• Year: 2024
## Calculating 54 days from Friday February 07, 2025 by hand
Attempting to add 54 days from Friday February 07, 2025 by hand can be quite difficult and time-consuming. A more convenient method is to use a calendar, whether it's a physical one or a digital application, to count the days from the given date. However, our days from specific date calculatoris the easiest and most efficient way to solve this problem.
If you want to modify the question on this page, you have two options: you can either change the URL in your browser's address bar or go to our days from specific date calculator to enter a new query. Keep in mind that doing these types of calculations in your head can be quite challenging, so our calculator was developed to assist you in this task and make it much simpler.
## Wednesday April 02, 2025 Stats
• Day of the week: Wednesday
• Month: April
• Day of the year: 092
## Counting 54 days forward from Friday February 07, 2025
Counting forward from today, Wednesday April 02, 2025 is 54 from now using our current calendar. 54 days is equivalent to:
54 days is also 1296 hours. Wednesday April 02, 2025 is 25% of the year completed.
## Within 54 days there are 1296 hours, 77760 minutes, or 4665600 seconds
Wednesday Wednesday April 02, 2025 is the 092 day of the year. At that time, we will be 25% through 2025.
## In 54 days, the Average Person Spent...
• 11599.2 hours Sleeping
• 1542.24 hours Eating and drinking
• 2527.2 hours Household activities
• 751.68 hours Housework
• 829.44 hours Food preparation and cleanup
• 259.2 hours Lawn and garden care
• 4536.0 hours Working and work-related activities
• 4173.12 hours Working
• 6829.92 hours Leisure and sports
• 3706.56 hours Watching television
## Famous Sporting and Music Events on April 02
• 1939 Author Jacqueline Susann (20) weds Irving Mansfield at Har Zion Temple in Philadelphia | 614 | 2,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-30 | latest | en | 0.919739 |
https://gmatclub.com/forum/in-one-state-all-cities-and-most-towns-have-antismoking-18007.html?fl=similar | 1,495,500,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607242.32/warc/CC-MAIN-20170522230356-20170523010356-00226.warc.gz | 759,048,432 | 45,492 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# In one state, all cities and most towns have antismoking
Author Message
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In one state, all cities and most towns have antismoking [#permalink]
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14 Jul 2005, 16:57
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In one state, all cities and most towns have antismoking ordinances. A petition entitled “Petition for Statewide Smoking Restrictionâ€
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14 Jul 2005, 18:06
C for this
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14 Jul 2005, 18:56
C...
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15 Jul 2005, 12:01
Another vote for C.
The question asks us to pick out a circumstance that misleads voters into thinking that the proposal extends the local ordinance statewide. C states that the state laws will be weaker than the local laws. Thus C.
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16 Jul 2005, 14:28
Okay Guys...3 together can't be wrong,
OA is C
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15 Oct 2005, 08:04
Cybera:
what is your source of OA being C?
The other posts stated that the OA is E. Below are the links. thks
http://www.gmatclub.com/phpbb/viewtopic ... estriction
15 Oct 2005, 08:04
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Q38: In one state, all cities and most towns have 3 30 Aug 2008, 08:31
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Display posts from previous: Sort by | 856 | 2,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-22 | latest | en | 0.872326 |
https://www.onlinemath4all.com/set-theory-practice-solution5.html | 1,591,450,459,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513321.91/warc/CC-MAIN-20200606124655-20200606154655-00311.warc.gz | 813,310,940 | 13,423 | ## Set Theory Practice Solution5
In this page set theory practice solution5 we are going to see solution of practice questions from the worksheet set theory practice questions2.
Question 4:
verify n (AUBUC)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A∩B ∩C)
(i) A = {4,5,6},B = {5,6,7,8} and C = {6,7,8,9}
Solution:
To verify the condition n (AUBUC)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A∩ B ∩C) we have to find number of terms in A,B,C number of terms in A U B ,B U C and C U A and also number of terms in (A∩ B ∩C)
A = {4,5,6}
B = {5,6,7,8}
C = {6,7,8,9}
n (A) = 3 n (B) = 4 n (C) = 4
(A ∩ B) = {5,6}
n (A ∩ B) = 2
(B ∩ C) = {6,7,8}
n (B ∩ C) = 3
(A ∩ C) ={6}
n (A ∩ C) = 1
(A ∩ B ∩ C) = {6}
n (A ∩ B ∩ C) =1
n (A U B U C)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A ∩ B ∩ C)
= 3 + 4 + 4 - 2 - 3 - 1 + 1
= 11 - 6 + 1
= 12 - 6
= 6
(ii) A = {a,b,c,d,e} B = {x,y,z} and C = {a,e,x}
Solution:
A = {a,b,c,d,e} B = {x,y,z} and C = {a,e,x}
n (A) = 5 n (B) = 3 n (C) = 3
n (A ∩ B) = 0
B ∩ C = {x}
n (B ∩ C) = 1
C ∩ A = {a,e}
n (C ∩ A) = 2
n (A ∩ B ∩ C) = 0
n (A U B U C)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C) + n(A ∩ B ∩ C)
= 5 + 3 + 3 - 0 - 1 - 2 + 0
= 11 - 3
n (A U B U C) = 8
Question 5:
In a college,60 students enrolled in chemistry,40 in physics,30 in biology,15 in chemistry and physics,10 in physics and biology,5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects.
Solution:
Let A,B and C are the sets enrolled in the subjects Chemistry,Physics and Biology respectively.
Number of students enrolled in Chemistry n (A) = 60
Number of students enrolled in Physics n (B) = 40
Number of students enrolled in Biology n (C) = 30
Number of students enrolled in Chemistry and Physics n (A ∩ B) = 15
Number of students enrolled in Physics and Biology n (B ∩ C) = 10
Number of students enrolled in Biology and Chemistry n (C ∩ A) = 5
No one enrolled in all the three,So n (A ∩ B ∩ C) = 0
Number of students enrolled
in at least one of the subjects = 35 + 15 + 20 + 10 + 5 + 15
= 100
set theory practice solution5 set theory practice solution5
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,388 | 4,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-24 | latest | en | 0.847621 |
https://www.expertsmind.com/library/what-is-the-disks-maximum-speed-at-the-amplitude-5341636.aspx | 1,721,627,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00846.warc.gz | 665,589,161 | 14,477 | ### What is the disks maximum speed at the amplitude
Assignment Help Physics
##### Reference no: EM13341636
An ultrasonic transducer, of the type used in medical ultrasound imaging, is a very thin disk (m = 0.14 g) driven back and forth in SHM at 1.1 MHz by an electromagnetic coil. (a) The maximum restoring force that can be applied to the disk without breaking it is 40,000 N. What is the maximum oscillation amplitude that won't rupture the disk? (b) What is the disk's maximum speed at this amplitude?
#### Questions Cloud
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Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report! | 1,259 | 5,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.908716 |
https://metanumbers.com/307346811071 | 1,632,617,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00276.warc.gz | 434,106,730 | 7,625 | # 307346811071 (number)
307,346,811,071 (three hundred seven billion three hundred forty-six million eight hundred eleven thousand seventy-one) is an odd twelve-digits composite number following 307346811070 and preceding 307346811072. In scientific notation, it is written as 3.07346811071 × 1011. The sum of its digits is 41. It has a total of 2 prime factors and 4 positive divisors. There are 296,748,645,144 positive integers (up to 307346811071) that are relatively prime to 307346811071.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 12
• Sum of Digits 41
• Digital Root 5
## Name
Short name 307 billion 346 million 811 thousand 71 three hundred seven billion three hundred forty-six million eight hundred eleven thousand seventy-one
## Notation
Scientific notation 3.07346811071 × 1011 307.346811071 × 109
## Prime Factorization of 307346811071
Prime Factorization 29 × 10598165899
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 307346811071 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 307,346,811,071 is 29 × 10598165899. Since it has a total of 2 prime factors, 307,346,811,071 is a composite number.
## Divisors of 307346811071
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 3.17945e+11 Sum of all the positive divisors of n s(n) 1.05982e+10 Sum of the proper positive divisors of n A(n) 7.94862e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 554389 Returns the nth root of the product of n divisors H(n) 3.86667 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 307,346,811,071 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 307,346,811,071) is 317,944,977,000, the average is 79,486,244,250.
## Other Arithmetic Functions (n = 307346811071)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 296748645144 Total number of positive integers not greater than n that are coprime to n λ(n) 148374322572 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 12090281128 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 296,748,645,144 positive integers (less than 307,346,811,071) that are coprime with 307,346,811,071. And there are approximately 12,090,281,128 prime numbers less than or equal to 307,346,811,071.
## Divisibility of 307346811071
m n mod m 2 3 4 5 6 7 8 9 1 2 3 1 5 6 7 5
307,346,811,071 is not divisible by any number less than or equal to 9.
## Classification of 307346811071
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (307346811071)
Base System Value
2 Binary 100011110001111010011000010100010111111
3 Ternary 1002101022112020200212212
4 Quaternary 10132033103002202333
5 Quinary 20013421240423241
6 Senary 353105420451035
8 Octal 4361723024277
10 Decimal 307346811071
12 Duodecimal 4b6959a9a7b
20 Vigesimal c025hb7db
36 Base36 3x6yct6n
## Basic calculations (n = 307346811071)
### Multiplication
n×y
n×2 614693622142 922040433213 1229387244284 1536734055355
### Division
n÷y
n÷2 1.53673e+11 1.02449e+11 7.68367e+10 6.14694e+10
### Exponentiation
ny
n2 94462062275512968167041 29032613607569120576845987396110911 8923081209342890237505502225296947403278695681 2742480554619099485883693910761310970428521327427610684351
### Nth Root
y√n
2√n 554389 6748.54 744.573 198.393
## 307346811071 as geometric shapes
### Circle
Diameter 6.14694e+11 1.93112e+12 2.96761e+23
### Sphere
Volume 1.21612e+35 1.18705e+24 1.93112e+12
### Square
Length = n
Perimeter 1.22939e+12 9.44621e+22 4.34654e+11
### Cube
Length = n
Surface area 5.66772e+23 2.90326e+34 5.3234e+11
### Equilateral Triangle
Length = n
Perimeter 9.2204e+11 4.09033e+22 2.6617e+11
### Triangular Pyramid
Length = n
Surface area 1.63613e+23 3.42153e+33 2.50948e+11 | 1,527 | 4,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-39 | latest | en | 0.77885 |
https://forums.wolfram.com/mathgroup/archive/2000/Sep/msg00304.html | 1,696,308,941,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00646.warc.gz | 288,769,149 | 7,470 | Re: Newbie question
• To: mathgroup at smc.vnet.net
• Subject: [mg25309] Re: Newbie question
• From: dlkeith at my-deja.com
• Date: Tue, 19 Sep 2000 03:45:36 -0400 (EDT)
• References: <8q3dvc\$kbb@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```In article <8q3dvc\$kbb at smc.vnet.net>,
Jose M Lasso <jml at accessinter.net> wrote:
> Hi Mathgroup,
>
> I have some numerical data: data:={{x,y},{x1,y1},{x2,y2}....{xn,yn}},
> I want to transform the data like this:
> data1:={{x,1/y},{x1,1/y1},{x2,1/y2}....{xn,1/yn}}, how can I do this
> transformation? Thanx in advance. Regards
>
> Jose M Lasso
>
data1={#[[1]], 1/#[[2]]} & /@ data
is one way.
Notice that {#[[1]],1/#[[2]]}& is the pure function form of
invertPartTwo[lst_]:={lst[[1]],1/lst[[2]]}
which could also be written
invertPartTwo[{p1_,p2_}]:={p1,1/p2}
Regards,
David
Sent via Deja.com http://www.deja.com/
```
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• Next by thread: Re: Newbie question | 396 | 1,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.716964 |
https://fr.webqc.org/molecularweightcalculated-190423-105.html | 1,571,033,649,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649232.14/warc/CC-MAIN-20191014052140-20191014075140-00471.warc.gz | 601,060,376 | 5,966 | #### Chemical Equations Balanced on 04/23/19
Molecular weights calculated on 04/22/19 Molecular weights calculated on 04/24/19
Calculate molecular weight
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170
Molar mass of NaOH is 39.99710928
Molar mass of Iron(ii)oxide is 71.8444
Molar mass of Mg2Si4O10(OH) is 337.95334
Molar mass of MgAlSi4O10(OH) is 340.6298786
Molar mass of NaN3 is 65.00986928
Molar mass of H2Ni is 60,70928
Molar mass of KCN is 65,1157
Molar mass of Magnesium sulfate is 120.3676
Molar mass of ga is 69.723
Molar mass of CuSO4 is 159.6086
Molar mass of NaCl is 58.44276928
Molar mass of f2 is 37,9968064
Molar mass of benzoic acid is 122.12134
Molar mass of Pb(C2H2O2)2 3H2O is 1560.04512
Molar mass of K3po4 is 953,2246216
Molar mass of cl is 35.453
Molar mass of (NH2)(HO)(O) is 49.02932
Molar mass of K3PO4 is 212,266262
Molar mass of NH4 is 18.03846
Molar mass of Na2CO3 is 105.98843856
Molar mass of NaOH is 39,99710928
Molar mass of CHBr3 is 252.73064
Molar mass of heptane is 100.20194
Molar mass of KO7 is 151.0941
Molar mass of C is 12,0107
Molar mass of Al70Zr30 is 4625.427702
Molar mass of CaO is 56,0774
Molar mass of CH3Br is 94.93852
Molar mass of C14H12O3 is 228.24328
Molar mass of Al is 26.9815386
Molar mass of CaSO4 is 136,1406
Molar mass of C2H5Br is 108.9651
Molar mass of C16H15NO2 is 253,2958
Molar mass of C3H6Br is 121.98374
Molar mass of water is 18.01528
Molar mass of C3H5Br is 120.9758
Molar mass of Fe is 55.845
Molar mass of Co2(SeO4)3 is 546,73919
Molar mass of CH3CH2OH is 46.06844
Molar mass of CO2(SeO4)3 is 472,8823
Molar mass of CO is 28.0101
Molar mass of methanol is 32.04186
Molar mass of acetic acid is 60.05196
Molar mass of HCl is 36.46094
Molar mass of Cu is 63.546
Molar mass of C6H6 is 78.11184
Molar mass of CH4 is 16.04246
Molar mass of PbCl is 242.653
Molar mass of C7H12 is 96.17018
Molar mass of O is 15.9994
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170
Calculate molecular weight
Molecular weights calculated on 04/22/19 Molecular weights calculated on 04/24/19
Molecular masses on 04/16/19
Molecular masses on 03/24/19
Molecular masses on 04/23/18
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http://stackoverflow.com/questions/7705782/class-scheduling-algorithm-to-show-best-match-with-criteria/7710402 | 1,427,746,844,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131299877.5/warc/CC-MAIN-20150323172139-00105-ip-10-168-14-71.ec2.internal.warc.gz | 260,933,995 | 17,081 | # Class Scheduling algorithm to show best match with criteria?
I am looking to create a system where you can input the courses (3-7 courses) that you want to take at a college and then select preferences (Morning, Day, Evening, Night / M, T, W, TR, F).
I need a way so that when the program queries the database (MySQL) it returns the best possible schedule that was made according to those parameters. I am writing it in php.
Does any one know the best way to do this? Or a link to some sample code I could understand it from?
-
It's NP-complete.
If you want something simple and fast, write a construction heuristic like First Fit Decreasing. Simply put, it orders the courses according to difficulty (student size, ...) and assign them one at a time at the best remaining timeslot and room. You could easily do this in PHP.
If you want to do it right, like @malejpavouk said, use a CP library. It will first use a construction heuristic and then does metaheuristics like tabu search, simulated annealing, ... Here's a open source course scheduling implementation in Java. Look for a good PHP CP library.
-
You might want to try some constraint programming (CSP) library.... In CSP you state the problem (usually NP Hard) and the library solves it using some heuristics (simulated annealing, taboo search) or an exhaustive DFS with some tricks (arc consistency ,search space shaving)... It can solve NPC problems with thousands of variables...but its somehow tricky to set the "fine" parameters of the black-box system...
-
This is a hard optimization problem. Here's a proposed heuristic (with random jumps).
1. Get a set of constraints (preferences).
2. And a set of courses.
3. Get possible times of the chosen courses using a MySQL query.
4. Choose a feasible set of courses.
5. Score the set according to the constraints.
6. If the score is good enough, stop.
7. Otherwise, change the hours of one course, chosen at random. | 432 | 1,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2015-14 | latest | en | 0.886442 |
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Είσοδος
Βιβλία Βιβλία
DIVISION is the process of finding how many times one number is contained in another, or of finding one of the equal parts of a number.
The Common School Arithmetic: Combining Analysis and Synthesis ; Adapted to ... - Σελίδα 42
των James Stewart Eaton - 1873 - 327 σελίδες
Πλήρης προβολή - Σχετικά με αυτό το βιβλίο
## An Intermediate Arithmetic: Uniting Mental and Written Exercises in a ...
Emerson Elbridge White - 1876 - 225 σελίδες
...\$1500 and the dividend \$564000: what is the quotient? DEFINITIONS, PRINCIPLES, AND BULES. Art. 38. Division is the process of finding how many times one number is contained in another. The Dividend is the number divided. The Divisor is the number by which the dividend is divided. The...
## The Philosophy of Arithmetic as Developed from the Three Fundamental ...
Edward Brooks - 1876 - 570 σελίδες
...number divided is the Dividend; the number we divide by is the Divisor. The definition usually given is, "Division is the process of finding how many times one number is contained in another." This is regarded as correct, but is less simple and concise than the one above suggested. Defining...
## The Science of Arithmetic: For High Schools, Normal Schools, Preparatory ...
Edward Olney - 1876 - 294 σελίδες
...the results for a = 10, x = 6, c = 8, d = 1, y = 3, b = 2. SECTION III. DIVISION. 110. Division is a process of finding how many times one number is contained in another. The process called Division is the converse of Multiplication; ie, by means of a knowledge of the products...
## A Textbook on Mining Engineering, Τόμος 1
1900
...entry in one day. How many feet of entry will it drive in 289 days ? Ans. 10,404 ft. DIVISION. 55. Division is the process of finding how many times one number is contained in another of the same kind. The number to be divided is called the dividend. The number by which we divide is...
## A Textbook on Marine Engineering ...: Steam and Steam Boilers, Steam Engines ...
...miles an hour. How far apart will they be at the end of 205 hours ? Ans. 1,081 miles. DIVISION. 55. Division is the process of finding how many times one number is contained in another of the same kind. The number to be divided is called the dividend. The number by which we divide is...
## A treatise on coal mining: prepared for students of the International ...
...entry in one day. How many feet of entry will it drive in 289 days ? Ans. 10,404 ft. DIVISION. 55. Division is the process of finding how many times one number is contained in another of the same kind. The number to be divided is called the dividend. The number by which we divide is...
## A Treatise on Chemistry and Chemical Analysis: Arithmetic, elementary ...
1900
...(m) («) (o) 868,250. 17,890,276. 7,928,170. 1,830,400. 78,340 87,543,000. 4,876,300. DIVISION. 55. Division is the process of finding how many times one number is contained in another of the same kind. The number to be divided is called the dividend. The number by which we divide is...
## An Elementary Course of Instruction for Mine Foremen and Pit Bosses
Floyd Davis - 1900 - 124 σελίδες
...is read 9 multiplied by 3. Q. 6. What is meant by division, and what sign is used to express it? A. Division is the process of finding how many times one number is contained in another. A short, horizontal line with a point above and below it (-1-) when placed between numbers indicates...
## Intermediate Arithmetic: By William J. Milne ...
William James Milne - 1900 - 219 σελίδες
...apples are divided equally among 5 children ? 10. What have you been doing with the above numbers ? 65. The process of finding how many times one number is contained in another, or the process of separating a number into equal parts, is called Division. 66. The number to be divided...
## Annual Report of the State Superintendent of Public Instruction, Τεύχος 46
...multiplication tables a) from books; b) from tables prepared by the pupils. *%£■ 4 Division is defined as the process of finding how many times one number is contained in _*»-.X" another of the same kind, or the process of separating a number into equal parte. Give the... | 1,072 | 4,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-40 | latest | en | 0.808697 |
https://www.compadre.org/nexusph/course/Reading_the_content_in_Fick's_law | 1,718,946,625,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00650.warc.gz | 621,899,045 | 5,923 | # Reading the content in Fick's law
#### Prerequisites
If there is a concentration of a kind of molecule immersed in a thermal bath, the random motion of the molecules of the bath (mostly water in biological examples) will cause the molecules we are looking at to undergo a random walk. However, if the concentration of those molecules is not uniform, their random motion will result in a net transport of those molecules from a region of high concentration to one of low concentration. Let's review how quantitative information about that result is coded in the equation for Fick's first law:
$$J = -D \frac{dn}{dx}$$
In order to "find the dog" in the equation that is Fick's law, note what we learned in the construction of the law (derivation) from an analysis of the random motion.
1. J — The quantity on the left is the "current density" of the kind of molecules whose concentration we are considering. (This could, for example, be a chemical signaling molecule.) This has two tricky factors: J counts the number of particles per second per unit area. So this is the number of particles crossing a unit area in each second. Since the equation is written for the x direction, we are only considering the motion of the particles in the x direction so they have to be crossing an area perpendicular to the x axis (in the y-z plane). Note that J measures the net flow of the particles we are looking at — not those of the thermal bath whose jiggling is pushing them in a random walk.
2. dn/dx — The flow of particles is driven by the concentration gradient — how the density it changes in space. If the concentration were uniform, its derivative would be 0. There would be as many molecules traveling on one direction as the other so the net would be 0. The flow results from the difference in concentration, not from the direction of the molecule's motion. In any small region, whatever the concentration, the molecules are going every which way.
3. The minus sign — This tells us that the flow is opposite the derivative — in the direction of decreasing concentration. So the flow is from high concentration to low.
4. D — The diffusion coefficient, D, tells how fast the flow happens. In our derivation, it came out as proportional to the product of the average speed of the molecules times their mean free path — the average distance they travel before their velocities get reoriented in a new direction. In various situations (such as in a gas or in a liquid) these parameters may be more conveniently expressed in terms of other parameters — such as the density of the thermal bath, its temperature, or its viscosity.
Note: If we have a concentration that varies in 3 dimensions, not just in x, our equation requires the gradient — the vector derivative. If we have any function of position (for example, the concentration, n), the gradient of that function at any location points in the direction in which the direction increases the fastest. Then the result looks like this:
$$\overrightarrow{J} = -D \overrightarrow{\nabla}n$$
The "downhill" of the concentration provides the direction of the diffusive flow.
Joe Redish 8/3/15
Article 401 | 688 | 3,163 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.930095 |
https://www.proofwiki.org/wiki/Probability_of_Set_Difference_of_Events | 1,685,564,883,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00775.warc.gz | 1,045,694,021 | 11,002 | # Probability of Set Difference of Events
## Theorem
Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$.
Let $\map \Pr A$ denote the probability of event $A$ occurring.
Then:
$\map \Pr {A \setminus B} = \map \Pr A - \map \Pr {A \cap B}$
## Proof
$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$
So, by the definition of probability measure:
$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$
Hence the result.
$\blacksquare$ | 176 | 544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.769796 |
https://www.neetprep.com/question/55995-Huygens-wave-theory-light-cannot-explain-phenomenon-Interference-Diffraction-Photoelectric-effect-Polarisation/55-Physics/700-Wave-Optics?courseId=8 | 1,670,483,726,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00054.warc.gz | 965,388,780 | 74,128 | # By Huygen's wave theory of light, we cannot explain the phenomenon of: 1. Interference 2. Diffraction 3. Photoelectric effect 4. Polarisation
Subtopic: Huygens' Principle |
To view explanation, please take trial in the course below.
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Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are
(1) 5I and I
(2) 5I and 3I
(3) 9I and I
(4) 9I and 3I
Subtopic: Superposition Principle |
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If L is the coherence length and c the velocity of light, the coherent time is
(1) cL
(2) $\frac{L}{c}$
(3) $\frac{c}{L}$
(4) $\frac{1}{Lc}$
Subtopic: Huygens' Principle |
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For constructive interference to take place between two monochromatic light waves of wavelength λ, the path difference should be
(1) $\left(2n-1\right)\frac{\lambda }{4}$
(2) $\left(2n-1\right)\frac{\lambda }{2}$
(3) $n\lambda$
(d) $\left(2n+1\right)\frac{\lambda }{2}$
Subtopic: Young's Double Slit Experiment |
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Soap bubble appears coloured due to the phenomenon of:
1. Interference
2. Diffraction
3. Dispersion
4. Reflection
Subtopic: Superposition Principle |
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Which of the following statements indicates that light waves are transverse?
1. Light waves can travel in a vacuum.
2. Light waves show interference.
3. Light waves can be polarized.
4. Light waves can be diffracted.
Subtopic: Polarization of Light |
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Interference was observed in interference chamber when the air was present, now the chamber is evacuated and if the same light is used, a careful observer will see
(1) No interference
(2) Interference with bright bands
(3) Interference with dark bands
(4) Interference in which width of the fringe will be slightly increased
Subtopic: Superposition Principle |
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Two coherent sources have intensity in the ratio of $\frac{100}{1}$. Ratio of (intensity)max/(intensity)min is:
1. $\frac{1}{100}$
2. $\frac{1}{10}$
3.$\frac{10}{1}$
4. $\frac{3}{2}$
Subtopic: Superposition Principle |
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If two waves represented by ${y}_{1}=4\mathrm{sin}\omega t$ and ${y}_{2}=3\mathrm{sin}\left(\omega t+\frac{\pi }{3}\right)$ interfere at a point, the amplitude of the resulting wave will be about
(1) 7
(2) 6
(3) 5
(4) 3.5
Subtopic: Superposition Principle |
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Two coherent sources of intensities, I1 and I2 produce an interference pattern. The maximum intensity in the interference pattern will be
(1) I1 + I2
(2) ${I}_{1}^{2}+{I}_{2}^{2}$
(3) (I1 + I2)2
(4) ${\left(\sqrt{{I}_{1}}+\sqrt{{I}_{2}}\right)}^{2}$
Subtopic: Superposition Principle |
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NEET 2023 - Target Batch - Aryan Raj Singh | 1,496 | 5,510 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-49 | latest | en | 0.78904 |
https://biology.stackexchange.com/questions/30775/cationic-lipid-mediated-transfection-optimization-for-150mm-dishes | 1,632,815,910,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00489.warc.gz | 193,476,540 | 38,694 | # Cationic lipid mediated transfection optimization for 150mm dishes
I need to optimize a transfection protocol to transiently express a plasmid encoding a chimera of eyfp attached to the c term of a Golgi apparatus signaling molecule) in hela cells and hepg2 cells and get as high expression as I can get.
I need enough protein for about 100 wb's. I've seeded hela cells to 12 150mm dishes. And the same with the hepg2. I've prepped mg's of plasmid.
I'm using invitrogens original lipofectamine reagent; not the 2000, or ltx or any of that stuff.
Any suggestions on a protocol here for hela and hepg2 plasmid dna and lipofectamine mediated transfection on such a large plate size.
I figure just scaling up should work but there's some pitfalls here. Look at this table for example when you go from 60 to 100 mm you about double the area, so you also about double the culture volume. However when you go from 100 to 150mm you triple the area, but only increase the culture volume two fold about. This is making my scale-up calculations for transfection look really off:
• Despite what the numbers look like, across all values going from 100mm to 150mm you're scaling up by a factor of ~2.76: (a) 152/55 = 2.76, (b) 1.52e7/5.5e6 = 2.76, (c) |avg(30.4-45.6)/avg(11-16.5)| = 2.76.
– CKM
Mar 26 '15 at 0:18
• Ahhh! I should have crunched the numbers cause you R correct. It's a weird concept because when you scale up in vessel size you want all parameters, ideally I would think, to increase at the same rate. However the depth of media can reach a certain point where I believe it hinders growth. When the volume gets to be too much it can take too long for those needed soluble factors to build up in the environment. Also increase in medium depth can delay delivery of soluble oxygen to the cells. Thank you for seeing through Mar 26 '15 at 0:34
• All my nonsense to make that comment. I think this ? Is a little scatter I'm going to reword it to focus it on the ratio, area and volume if you want to post an answer its yours. I'll post a seperate ? On the hela optimization Mar 26 '15 at 0:35
• @Kendall please post your comment as an answer to close this question. Thanks! Mar 29 '15 at 6:05 | 588 | 2,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-39 | longest | en | 0.927976 |
https://speakerdeck.com/romain_francois/r-and-c-plus-plus-past-present-and-future | 1,656,264,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00368.warc.gz | 578,411,241 | 19,294 | Romain
April 05, 2017
350
# R and C++. Past Present and Future.
R and C++. Past Present and Future. Presented at the MilanoR meetup on 2017-04-05
April 05, 2017
## Transcript
1. ### R and C++ Past, Present and Future Romain François Consulting
Datactive romain@thinkr.fr @romain_francois
2. None
3. None
4. None
177
11. ### int add( int a, int b ){ return a +
b ; } > add( 40, 2 ) [1] 42
13. ### #include <R.h> #include <Rinternals.h> int add( int a, int b
){ return a + b ; } SEXP add_c( SEXP a_, SEXP b_ ){ int a = INTEGER(a_)[0], b = INTEGER(b_)[0] ; int res = add( a, b ) ; SEXP result = PROTECT(allocVector(INTSXP, 1) ) ; INTEGER(result)[0] = res ; UNPROTECT(1) ; return result ; }
14. ### add <- function(a, b){ .Call( "add_c", a, b ) }
> add( 40, 2 ) Error in add(33, 9) : INTEGER() can only be applied to a 'integer', not a 'double' > add( 40L, 2L ) [1] 42
15. ### add <- function(a, b){ .Call( "add_c", as.integer(a), as.integer(b) ) }
> add( 40, 2 ) [1] 42 > add( 40L, 2L ) [1] 42
16. ### Tools • SEXP • INTEGER • PROTECT • allocVector •
INTSXP • UNPROTECT • .Call • as.integer
18. ### #include <Rcpp.h> // [[Rcpp::export]] int add( int a, int b
){ return a + b ; } > add( 17L, 25L ) [1] 42 > add( 17, 25 ) [1] 42
20. None
21. ### ! weighted_mean_1 <- function(x, w) { total <- 0 total_w
<- 0 for (i in seq_along(x)) { total <- total + x[i] * w[i] total_w <- total_w + w[i] } total / total_w }
24. ### \$ #include <Rcpp.h> using namespace Rcpp ; // [[Rcpp::export]] double
weighted_mean_cpp( NumericVector x, NumericVector w){ int n = x.size() ; double total = 0.0 ; double total_w = 0.0 ; for( int i=0; i<n; i++){ total += x[i] * w[i] ; total_w += w[i] ; } return total / total_w ; }
25. None
26. None
31. ### Rcpp n + 2 core sugar modules core modules Rcpp
n +1 core sugar modules core modules Rcpp n core sugar modules core modules % mypkg ' ✔ \$ )
33. ### NumericVector IntegerVector CharacterVector Function sugar modules … IntegerVector CharacterVector Function
sugar NumericVector core core modules … core pkg1 pkg2 core pkg3 core CharacterVector IntegerVector NumericVector
35. ### Pros • Smaller • Faster • Robust • Controlled updates
Cons • More copies of code base • Maybe more testing
36. None | 760 | 2,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | latest | en | 0.196142 |
www.mymeasuringtape.com | 1,680,120,746,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00785.warc.gz | 1,001,067,465 | 31,073 | ### Calculating Measurements for Drafting a Basic Block
One of the more frequent 'complaints' I get from my students is understanding how to divide the circumferential numbers into sensible proportions for the Front and Back blocks. Most books will tell you to divide these numbers equally. While this method might work in a general sense, it will still require A LOT of muslin testing for fit, and thereafter tweaking of the block pattern.
If you then measure the adjusted pattern at various horizontal points (bust, waist, etc) you will find that the Front is bigger than the Back. In this instance, bigger means the length of the curvature from a fixed point on the side body to a corresponding fixed point on the other side. This difference is less at certain planes (eg. at the underbustline) than at others (the bustline).
For a woman's block, halving the numbers just doesn't make sense around the Bust measurements. Even if you are a modest A-cupper. Our rib cage curves out towards the front, encompassing our heart and lungs protectively in its calcium embrace. Unless a body has severe distortions because of scoliosis, humps and injury, the Front is always bigger than the Back. I must give a shout out to Kathleen Fasanella (who is, IMO, the Supreme Leader of Pattern Makers) for bringing this physiology to my awareness back in the day...
Over the years of measuring a variety of body types, I've worked out an average in terms of how much more to assign to the Front when making the calculations for drafting a bodice block. I was able to do this because I was mainly making close-fitting corset patterns where ease was unnecessary.
Now a basic bodice block is also a close-fitted garment. In my case, I make blocks for repeat customers from which I then can straightaway create new patterns for a specific design (assuming of course that they remain constant in their measurements). I will add the minimum amount of ease to make it just about wearable with a zipper. Remember that this is only a fit garment, not an actual fashion draft. The fit garment will sit very close to the skin without uncomfortably squeezing the rib cage, so some ease is necessary. This is my preferred method because any additional style ease can be added during the pattern design process of the fashion garment. I do this for all the body types that pass under my measuring tape, before expanding the patterns stylistically.
This method is something I teach all my drafting students, regardless of their experience. All will understand the reasoning and some will understand the equations necessary to draft the Fronts and Backs. Admittedly, there's algebra involved (because it is an equation and must work with any measurement inputted). Which means I lose the attention of about 50% of my students, after hearing them cuss under their breath and look at me with disappointment at my betrayal. So to make their lives slightly easier, I designed a spreadsheet with the equations already factored in. They literally just have to key in the full circumferences and length measurements and the spreadsheet will calculate everything like magic.
I also include this spreadsheet (Measurement Worksheet) as a download in my online course on drafting a basic bodice block.
If one is mathematically inclined, the worksheet can easily be tweaked to include other fields and conditions that suit specific work.
For now, the worksheet for the online course only specifies measurements for the upper body. As I add more modules to the course, I will expand the worksheet to other areas as well; namely the lower body and arms.
If you think this will help you in your learning journey in drafting, do have a look at what else I offer in the online drafting course (links below).
You get a free Versatile Blouse Pattern from my collection when you sign up through either of the above links, from this website.
Meanwhile, happy sewing and stay sane. | 802 | 3,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-14 | longest | en | 0.939539 |
http://pegasus.cc.ucf.edu/~ucfcasio/pow/blocks.htm | 1,513,415,681,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00702.warc.gz | 204,161,780 | 3,220 | # High School Challenge
## Week Ending 10-31-99
### Congratulations to Cristina Domnisoru from Orono High School in Orono, ME. Cristina wins the Casio CFX-9850G Color Graphing calculator.
Weigh Those Blocks
Suppose you have a double pan balance and 37 blocks. All 37 blocks appear to be the same. Thirty-six of the blocks weigh exactly the same amount and the remaining one slightly heavier. You are unable to identify the heavy one by lifting the individual blocks. What is the fewest possible number of weighings you would need to make to guarantee that the heavy block could be identified, no matter how many times you repeated the activity? This balance is capable of holding as many of the blocks on each side as you would like to use. Putting blocks on both sides constitutes one weighing. Adding or removing blocks from that situation would be considered a second weighing.
### Correct Solutions:
1. Michael Moyer. The Way Home School, Carlisle, PA
3. Jeffrey Mo. University ES, Calgary, AB, Canada
4. John Pinto. Bishop Moore HS, Orlando, FL
5. Kenny Yu. SKH Lam Woo Memorial SS, Hong Kong
6. Sorin Ionescu. John Abbot College, Montreal, Canada
7. Anton Lukyanenko. Evans JHS, Lubbock, TX
8. Jason Juang. Sycamore JHS, Cincinnati, OH
9. Alex Shim. Framingham HS, Framingham, MA
10. Jongmin Baek. Hyde JHS, Cupertino, CA
11. David Held. Miami Sunset HS, Miami, FL
12. Eric Hoke. Wayland HS, Wayland, MA
13. Tao Chen. Miller MS, San Jose, CA
14. Michael Cheung. SKH Lam Woo Memorial SS, Hong Kong
15. Jan Nordbotten. Cathedral School, Bergen, Norway
16. Simon Rubinstein-Salzedo. Homestead HS, Cupertino, CA
17. Matan Protter. Ort Kiryat Bialik, Kiryat Bialik, Israel
18. Loc Thai. Rostrevor College, Adelaide, SA, Australia
19. Brian Clary. Louisa County HS, Mineral, VA
20. John Xia. Torrance High School, Torrance, CA
21. Jeremy Wilson. Newport Harbor HS, Newport Beach, CA
22. Nate Ince. Homeschooled, Arnold, MO
23. Tiffany Chao. Seaquam SS, Delta, BC, Canada
24. Eric Tsai. Columbus High, Columbus, GA
25. Michael Guzik. Guzik Learning Center, San Diego, CA
26. Sean Ervin. Jackson Memorial HS, Jackson, NJ
27. Lee Colbert. Jackson Memorial HS, Jackson, NJ
28. Nick Mauro. Jackson Memorial HS, Jackson, NJ
29. Matt Whitmore. Jackson memorial HS, Jackson, NJ
30. Brian Nicolletti. Jackson Memorial HS, Jackson, NJ
31. Andrey Mamontov. Reedsburg HS, Reedsburg, WI
32. Nathan French. Windsor HS, Windsor, VT
33. Rohan Mathew. Sycamore JHS, Cincinnati, OH
34. Kevin Chan. Seminole HS, Sanford, FL
35. Hongyi Li. Westmount HS, Montreal, QC, Canada
36. Pradeep Baliga. Central HS, Tuscaloosa, AL
38. Tuna Tasdelen. Ozel A.Aymaz Okulu, Konya, Turkey
39. Ozlem Tasdelen. Maresal M.Kemal Okulu, Konya, Turkey
40. Thomas Lee. The Meadows School. Las Vegas, NV
41. Daniel Gee. Stone Valley MS, Alamo, CA
42. Angela Christmas. Louisa County HS, Mineral, VA
43. Mike Apple. Louisa County HS, Mineral, VA
44. Robbie Lane. Louisa County HS, Mineral, VA
45. Sean Nelson. Louisa County HS, Mineral, VA
46. Hayley Johnson. Louisa County HS, Mineral, VA
47. Jason Lewis. Louisa County HS, Mineral, VA
48. Dawn Gordon.Louisa County HS, Mineral, VA
49. C.T. Thiemann. Louisa County HS, Mineral, VA
50. Melissa Carroll. Louisa County HS, Mineral, VA
51. Halsey Green. Louisa County HS, Mineral, VA
52. Jeremy Sestito. Louisa County HS, Mineral, VA
53. Ashley Coleman. Louisa County HS, Mineral, VA
54. Jason Hicks. Louisa County HS, Mineral, VA
55. Daniel Yu. Randolph MS, Randolph, NJ
56. Ismar Tuzovic, St John Bosco College, Engadine, NSW, Australia
57. Michael Tso. St. Michael's University, Victoria, BC, Canada
58. Nicholas Assanis. Greenhills School, Ann Arbor, MI
59. Jenny Laaser. JL Stanford MS, Palo Alto, CA
60. Karen Wong. Southwood SS, Cambridge, ON, Canada
61. Xuan Wu. John W. North HS, Riverside, CA
62. Mike Munson, Newport Harbor HS, Newport Beach, CA
63. Evelyn Cheung. Mayfield HS, Pasadena, CA
64. Samuel Timothy Dy. St. Stephen's HS, Manila, Philippines
65. Sarah Reuter. Sauk Centre HS, Sauk Centre, MN
66. John Tullos. North Delta HS, Batesville, MS
67. Elena Udovina. The Agnon MS, Beachwood, OH
68. Mark Vos. Constantia Waldorf School, Cape Town, South Africa
69. Kyle Bentfield. Sauk Centre HS, Sauk Centre, MN
70. Ben Conlee. Albuquerque Academy, Albuquerque, NM
71. Shih-Fan Yeh. Burlingame HS, Burlingame, CA
72. Brad Chapin. John W. North HS, Riverside, CA
73. Jessica Kneubuhl. Airline HS, Bossier City, LA
74. Joe Park. The Meadows School, Las Vegas, NV
75. Rachel Chan. UHS, Ontario, Canada
76. Guy David. Ransom MS, Miami, FL
77. Phongtharin Vinayavekhin. Bodindacha, Bangkok, Thailand
78. Crsitina Domnisoru. Orono HS, Orono, ME | 1,385 | 4,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-51 | latest | en | 0.819423 |
https://larspsyll.wordpress.com/2017/03/18/monte-carlo-simulation-on-p-values-wonkish/ | 1,521,948,464,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651780.99/warc/CC-MAIN-20180325025050-20180325045050-00160.warc.gz | 629,594,431 | 16,644 | ## Monte Carlo simulation on p-values (wonkish)
18 March, 2017 at 16:41 | Posted in Statistics & Econometrics | Comments Off on Monte Carlo simulation on p-values (wonkish)
In many social sciences p values and null hypothesis significance testing (NHST) are often used to draw far-reaching scientific conclusions – despite the fact that they are as a rule poorly understood and that there exist altenatives that are easier to understand and more informative.
Not the least using confidence intervals (CIs) and effect sizes are to be preferred to the Neyman-Pearson-Fisher mishmash approach that is so often practised by applied researchers.
Running a Monte Carlo simulation with 100 replications of a fictitious sample having N = 20, confidence itervals of 95%, a normally distributed population with a mean = 10 and a standard deviation of 20, taking two-tailed p values on a zero null hypothesis, we get varying CIs (since they are based on varying sample standard deviations), but with a minimum of 3.2 and a maximum of 26.1 we still get a clear picture of what would happen in an infinite limit sequence. On the other hand p values (even though from a purely mathematical statistical sense more or less equivalent to CIs) vary strongly from sample to sample, and jumping around between a minimum of 0.007 and a maximum of 0.999 don’t give you a clue of what will happen in an infinite limit sequence! So, I can’t but agree with Geoff Cummings:
The problems are so severe we need to shift as much as possible from NHST … The first shift should be to estimation: report and interpret effect sizes and CIs … I suggest p should be given only a marginal role, its problem explained, and it should be interpreted primarily as an indicator of where the 95% CI falls in relation to a null hypothesised value.
Geoff Cumming
In case you want to do your own Monte Carlo simulation, here’s an example I’ve made using Gretl:
nulldata 20
loop 100 –progressive
series y = normal(10,15)
scalar zs = (10-mean(y))/sd(y)
scalar df = \$nobs-1
scalar ybar=mean(y)
scalar ysd= sd(y)
scalar ybarsd=ysd/sqrt(\$nobs)
scalar tstat = (ybar-10)/ybarsd
pvalue t df tstat
scalar lowb = mean(y) – critical(t,df,0.025)*ybarsd
scalar uppb = mean(y) + critical(t,df,0.025)*ybarsd
scalar pval = pvalue(t,df,tstat)
store E:\pvalcoeff.gdt lowb uppb pval
endloop | 587 | 2,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-13 | latest | en | 0.903326 |
https://www.airmilescalculator.com/distance/lft-to-svc/ | 1,606,326,510,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141183514.25/warc/CC-MAIN-20201125154647-20201125184647-00659.warc.gz | 568,262,851 | 33,280 | # Distance between Lafayette, LA (LFT) and Silver City, NM (SVC)
Flight distance from Lafayette to Silver City (Lafayette Regional Airport – Grant County Airport) is 969 miles / 1559 kilometers / 842 nautical miles. Estimated flight time is 2 hours 20 minutes.
Driving distance from Lafayette (LFT) to Silver City (SVC) is 1099 miles / 1769 kilometers and travel time by car is about 18 hours 4 minutes.
## Map of flight path and driving directions from Lafayette to Silver City.
Shortest flight path between Lafayette Regional Airport (LFT) and Grant County Airport (SVC).
## How far is Silver City from Lafayette?
There are several ways to calculate distances between Lafayette and Silver City. Here are two common methods:
Vincenty's formula (applied above)
• 968.821 miles
• 1559.167 kilometers
• 841.883 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 967.000 miles
• 1556.236 kilometers
• 840.300 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Lafayette Regional Airport
City: Lafayette, LA
Country: United States
IATA Code: LFT
ICAO Code: KLFT
Coordinates: 30°12′19″N, 91°59′15″W
B Grant County Airport
City: Silver City, NM
Country: United States
IATA Code: SVC
ICAO Code: KSVC
Coordinates: 32°38′11″N, 108°9′21″W
## Time difference and current local times
The time difference between Lafayette and Silver City is 1 hour. Silver City is 1 hour behind Lafayette.
CST
MST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 149 kg (328 pounds).
## Frequent Flyer Miles Calculator
Lafayette (LFT) → Silver City (SVC).
Distance:
969
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
969
Round trip? | 496 | 1,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-50 | longest | en | 0.839985 |
https://www.dellorto-pe.com/carburetor/readers-ask-what-size-carburetor-will-handle-400-to-450-hp.html | 1,643,209,554,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00046.warc.gz | 775,192,924 | 11,380 | # Readers ask: What Size Carburetor Will Handle 400 To 450 Hp?
## How much horsepower will a 650 cfm carb support?
So, using your Google-Fu you type “How much power can a 650 cfm carb support?” Well, chances are that you’ll get answers in the 450-470 hp range, but that’s not really the right way to look at it.
## How do I know what size carburetor I need?
The formula for calculating how much CFM (cubic feet per minute) an engine requires is: CFM = Cubic Inches x RPM x Volumetric Efficiency ÷ 3456. Any ordinary stock engine will have a volumetric efficiency of about 80%.
## How much horsepower will a 750 cfm carb support?
A 750DP on a 330hp-400hp 5.7 with a dual plane intake (performer rpm) is easy to tune and will make max power to boot.
## What happens if your carburetor is too big?
If the barrels are too big, the loss of air velocity means the cylinder will not fill to its full capacity. An engine with a carb that is too big will put out less Torque and Horsepower. It will be difficult to drive due to poor low-end torque. If you drag race your car, an oversized carb will produce slow 60 ft.
You might be interested: Often asked: How To Install Fuel Pressure Regulator Carburetor?
## Is a 650 cfm carb to big for a 350?
If you’re expecting a maximum of 6,500 rpm, you’re going to need a 650- to 700-cfm carburetor. These sizing numbers are only the beginning of carburetor selection—a baseline. A 600-cfm carburetor may perform quite well on a stock 350 Chevy.
## Does a carburetor spacer add horsepower?
When you stack spacers, you are actually improving their insulation properties, and at the same time you’re getting more horsepower. This type of spacer will increase the velocity of your vehicle’s air-to-fuel charge. This means you’ll build low to mid-range torque.
## How do I know if my carburetor is rich or lean?
Q: How Do You Tell if a Carburetor Is Rich or Lean? A: One way to tell for sure is by “reading” the spark plugs. If the plug tip is white, the mixture is lean. If it’s brown or black, it’s rich.
## Does a bigger carburetor mean more power?
The answer is no, not really. The amount of fuel that’s sucked into the carburetor is controlled by the carburetor jets. Installing a bigger carb is simply going to improve the power potential of your bike. You still have other things to worry about like improving air intake, exhaust flows and jets.
## How do I know what CFM my carburetor is?
How to calculate Carburetor CFM. The formula for calculating how much CFM (cubic feet per minute) your engine requires is: CFM = Cubic Inches x RPM x Volumetric Efficiency ÷ 3456.
## How much horsepower will a 950 cfm carb support?
Although way too small according to the CFM calculations presented earlier, a 950 Ultra HP can pass enough air to support more than 800 hp from a street/strip 572. If you do the math in terms of the required CFM, these carbs look far too small to be able to allow the production of such big horsepower numbers.
You might be interested: FAQ: How To Make A Carburetor For Propane Pipe Burner?
## Are carb spacers worth it?
Spacers are said to improve air/fuel vaporization through and out of the carburetor. Anytime the air/fuel charge has to turn sharply coming off the carb, it increases the chance of separating the fuel from the air.
## What size carburetor do I need for a 383 stroker?
Carburetor For 383 Stroker recommends a 650 cfm carburetor. A 750 cfm carburetor will make the engine more powerful, but it is the largest that should be used unless the engine is being used for racing. | 886 | 3,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-05 | latest | en | 0.886475 |
https://www.aqua-calc.com/calculate/food-weight-to-volume/substance/reser-quote-s-coma-and-blank-sensational-blank-sides-coma-and-blank-macaroni-blank-and-blank-cheese-coma-and-blank-upc-column--blank-071117143560 | 1,566,143,336,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313936.42/warc/CC-MAIN-20190818145013-20190818171013-00214.warc.gz | 719,616,867 | 8,654 | # Volume of RESER'S SENSATIONAL SIDES MACARONI and CHEESE UPC: 0...10930
## food weight to volume conversions
### calculate volume of generic and branded foods per weight
#### Volume, i.e. how many spoons, cups,gallons or liters in 100 gram ofRESER'S, SENSATIONAL SIDES, MACARONI andCHEESE, UPC: 071117143560
centimeter³ 112.55 milliliter 112.55 foot³ 0 US cup 0.48 Imperial gallon 0.02 US dessertspoon 15.22 inch³ 6.87 US fluid ounce 3.81 liter 0.11 US gallon 0.03 meter³ 0 US pint 0.24 metric cup 0.45 US quart 0.12 metric dessertspoon 11.25 US tablespoon 7.61 metric tablespoon 7.5 US teaspoon 22.83 metric teaspoon 22.51
#### Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 129 Protein g 5.71 Total lipid (fat) g 5.71 Carbohydrate,bydifference g 13.57 Fiber,totaldietary g 0.7 Sugars, total g 2.14 Minerals Calcium, Ca mg 107 Iron, Fe mg 0.77 Sodium, Na mg 450 Vitamins Vitamin C,totalascorbic acid mg 0 Vitamin A, IU IU 214 Lipids Fatty acids,totalsaturated g 2.5 Cholesterol mg 14
#### See how many calories in0.1 kg (0.22 lbs) ofRESER'S, SENSATIONALSIDES, MACARONI andCHEESE, UPC: 071117143560
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 129 539.74 Total 129 539.74
• About RESER'S, SENSATIONAL SIDES, MACARONI and CHEESE, UPC: 071117143560
• 222.12643 grams [g] of RESER'S, SENSATIONAL SIDES, MACARONI and CHEESE, UPC: 071117143560 fill 1 metric cup
• 7.41494 ounces [oz] of RESER'S, SENSATIONAL SIDES, MACARONI and CHEESE, UPC: 071117143560 fill 1 US cup
• RESER'S, SENSATIONAL SIDES, MACARONI and CHEESE, UPC: 071117143560 weigh(s) 222.13 gram per (metric cup) or 7.41 ounce per (US cup), and contain(s) 129 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• Ingredients: COOKED ENRICHED MACARONI (SEMOLINA WHEAT FLOUR, EGG, IRON, NIACIN, THIAMINE MONONITRATE, RIBOFLAVIN, FOLIC ACID), MILK, WATER, PASTEURIZED PROCESS CHEESE SPREAD (AMERICAN CHEESE [PASTEURIZED MILK, CHEESE CULTURE, SALT, ENZYMES], WATER, WHEY, SODIUM PHOSPHATE, WHEY PROTEIN CONCENTRATE, SKIM MILK, SALT, MILKFAT, ARTIFICIAL COLOR), CHEDDAR CHEESE (CHEDDAR CHEESE [MILK, ARTIFICIAL COLOR, CHEESE CULTURE, SALT, ENZYMES], WATER, SALT, ANNATTO COLOR), MARGARINE (VEGETABLE OIL BLEND [LIQUID SOYBEAN OIL, PARTIALLY HYDROGENATED SOYBEAN OIL], WATER, SALT, WHEY [MILK], SOY LECITHIN, VEGETABLE MONO AND DIGLYCERIDES, SODIUM BENZOATE [TO PROTECT QUALITY], ARTIFICIAL FLAVOR, CITRIC ACID, VITAMIN A PALMITATE), MODIFIED CORN STARCH, SALT, POTASSIUM SORBATE AND SODIUM BENZOATE TO PROTECT FLAVOR, ANNATTO COLOR.
• Manufacturer: RESER'S FINE FOODS, INC.
• For instance, compute how many cups or spoons a pound or kilogram of “RESER'S, SENSATIONAL SIDES, MACARONI and CHEESE, UPC: 071117143560” fills. Volume of the selected food item is calculated based on the food density and its given weight. Visit our food calculations forum for more details.
• A few foods with a name containing, like or similar to RESER'S, SENSATIONAL SIDES, MACARONI and CHEESE, UPC: 071117143560:
• RESER'S FINE FOODS, ORIGINAL POTATO SALAD, UPC: 071117184006 weigh(s) 285.31 gram per (metric cup) or 9.52 ounce per (US cup), and contain(s) 170 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• RESER'S FINE FOODS, CREAMY MASHED POTATOES, UPC: 071117143522 weigh(s) 262.06 gram per (metric cup) or 8.75 ounce per (US cup), and contain(s) 121 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• RESER'S FINE FOODS, PISTACHIO DELIGHT, UPC: 071117182620 weigh(s) 232.47 gram per (metric cup) or 7.76 ounce per (US cup), and contain(s) 136 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• RESER'S FINE FOODS, MAIN ST BISTRO, BAKED MACARONI and CHEESE, UPC: 071117023411 weigh(s) 238 gram per (metric cup) or 7.94 ounce per (US cup), and contain(s) 127 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• RESER'S FINE FOODS, MAIN ST BISTRO, BAKED ENTREES, PENNE TOSCNA WITH ITALIAN SAUSAGE, UPC: 071117023442 weigh(s) 246.21 gram per (metric cup) or 8.22 ounce per (US cup), and contain(s) 155 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• Reference (ID: 10930)
• USDA National Nutrient Database for Standard Reference; National Agricultural Library; United States Department of Agriculture (USDA); 1400 Independence Ave., S.W.; Washington, DC 20250 USA.
#### Foods, Nutrients and Calories
GHIRARDELLI CHOCOLATE, PREMIUM BAKING CHIPS, BITTERSWEET CHOCOLATE, UPC: 747599619137 contain(s) 533 calories per 100 grams or ≈3.527 ounces [ price ]
INTERNATIONAL FOODSOURCE LLC, FIRE ROASTED CASHEWS WITH SEA SALT, UPC: 790429236547 (CUP | ABOUT) weigh(s) 126.8 gram per (metric cup) or 4.23 ounce per (US cup), and contain(s) 600 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
Foods high in Fatty acids, total monounsaturated
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CaribSea, Freshwater, Instant Aquarium, Crystal River weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Cullet weighs 1 602 kg/m³ (100.00959 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Rapeseed oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The kilogram per metric tablespoon density measurement unit is used to measure volume in metric tablespoons in order to estimate weight or mass in kilograms
A force that acts upon an object can cause the acceleration of the object.
Convert pennyweight per US gallon to slug per US pint or convert between all units of density measurement
#### Calculators
Weight to Volume conversions for select substances and materials | 1,992 | 6,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-35 | latest | en | 0.47493 |
https://stacks.math.columbia.edu/tag/04VJ | 1,721,511,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00535.warc.gz | 474,785,861 | 6,105 | Lemma 4.27.14. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system of morphisms of $\mathcal{C}$. Let $A, B : X \to Y$ be morphisms of $S^{-1}\mathcal{C}$ which are the equivalence classes of $(f : X' \to Y, s : X' \to X)$ and $(g : X' \to Y, s : X' \to X)$. The following are equivalent
1. $A = B$,
2. there exists a morphism $t : X'' \to X'$ in $S$ with $f \circ t = g \circ t$, and
3. there exists a morphism $a : X'' \to X'$ with $f \circ a = g \circ a$ and $s \circ a \in S$.
Proof. This is dual to Lemma 4.27.6. $\square$
There are also:
• 20 comment(s) on Section 4.27: Localization in categories
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). | 276 | 831 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.628726 |
http://www.chegg.com/homework-help/questions-and-answers/question-1-find-zero-input-response-following-second-order-difference-equation-x-n-3y-n-1--q3910743 | 1,501,122,073,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426693.21/warc/CC-MAIN-20170727002123-20170727022123-00210.warc.gz | 401,381,800 | 18,012 | question no.1
FIND THE ZERO -INPUT RESPONSE OF THE FOLLOWING SECOND ORDER DIFFERENCE EQUATION:
x[n]-3y[[n-1]-4y[n-2]=x[n].
questionno. 2
DETERMINE THE PARTICULAR SOLUTION OF THE FOLLOWING DIFFERENCE EQUATION:
y[n]=5\6y[n-1]-1/6y[n-2]+x[n].THE FORCING FUNCTIONIS x[n]=2u[n]...
100% (1 rating)
1) x[n]-3y[[n-1]-4y[n-2]=x[n]. for zero input put x[n] =0 hence 3y[n-1]+4y[n-2]=0 ................(1) let y[n] =... view the full answer
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Signals and Systems: Analysis Using Transform Methods & MATLAB (2nd Edition)
Q:
Find the impulse responses of the systems described by these equations.(a) 3y[n] + 4y[n−1] + y[n−2] = x[n] + x[n−1](b) (5/2)y[n] + 6y[n−1] + 10y[n−2] = x[n] | 271 | 743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-30 | latest | en | 0.656533 |
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1 KrugmanMacro_SM_Ch08.qxp 11/9/05 4:47 PM Page 99 Long-Run Economic Growth 1. The accompanying table shows data from the Penn World Table, Version 6.1, for real GDP in 1996 U.S. dollars for Argentina, Ghana, South Korea, and the United States for, 1970, 1980, 1990, and. Year Argentina Ghana \$7,395?? \$832?? ,227?? 1,275?? ,556?? 1,204?? ,237?? 1,183?? 10,995?? 1,349?? Year South Korea United States \$1,571?? \$12,414?? ,777?? 16,488?? ,830?? 21,337?? ,959?? 26,470?? 15,881?? 33,308?? 8chapter 25 macroeconomics economics a. Complete the table by expressing each year s as a percentage of its and levels. b. How does the growth in living standards from to compare across these four nations? What might account for these differences? 1. a. The accompanying table shows each nation s in terms of its and levels. Year Argentina Ghana \$7, % 67% \$ % 62% , , , , , , , ,
2 KrugmanMacro_SM_Ch08.qxp 11/9/05 4:47 PM Page MACROECONOMICS, CHAPTER 8 ECONOMICS, CHAPTER 25 Year South Korea United States \$ 1, % 10% \$12, % 37% , , , , , , ,881 1, , b. South Korea experienced the greatest increase in living standards from to ; in 2001 it produced 1,011% (\$15,881/\$1, ) of what it produced in. Argentina and Ghana experienced only a modest growth in living standards over the same period. Argentina s path was less consistent than that of Ghana, where living standards remained low throughout the period. Compared with real GDP in, the United States in produced 268% (\$33,308/\$12, ) of what it produced in. The growth in living standards in Argentina, Ghana, and South Korea reflects the pattern for their different regions of the world. South Korea, like many other East Asian countries, had high productivity growth because of high savings and investment rates, a good education system, and substantial technological progress. Living standards grew more modestly in Argentina, as in other Latin American countries, because of low savings and investment spending rates, underinvestment in education, political instability, and irresponsible government policies. Although the growth in living standards was similar in Ghana and Argentina, Ghana had started from a much lower level. in Ghana was only 11% of that in Argentina in and 12% in. Living standards in Africa suffered from major political instabilities, poor education and infrastructure, and disease. 2. The accompanying table shows the average annual growth rate in for Argentina, Ghana, and South Korea using data from the Penn World Table, Version 6.1, for the past few decades. Average annual growth rate of Years Argentina Ghana South Korea % 4.36% 5.86% a. For each decade and for each country, use the Rule of 70 where possible to calculate how long it would take for that country s to double. b. Suppose that the average annual growth rate that each country achieved over the period 1990 continues indefinitely into the future. Starting from, use the Rule of 70 to calculate, where possible, the year in which a country will have doubled its.
3 KrugmanMacro_SM_Ch08.qxp 11/9/05 4:47 PM Page 101 LONG-RUN ECONOMIC GROWTH a. The accompanying table shows the number of years it would take for per capita to double according to the Rule of 70 using the average annual growth rate in per decade in each country. Values corresponding to years with negative growth rates are left uncalculated because we cannot apply the Rule of 70 to a negative growth rate. Years for to double according to the Rule of 70 Years Argentina Ghana South Korea b. If each nation continues to grow as it did from 1990 to, will have doubled in Argentina by 2016, in Ghana by 2052, and in South Korea by You are hired as an economic consultant to the countries of Albernia and Brittania. Each country s current relationship between physical capital per worker (K/L) and output per worker (Y/L) is given by the curve labeled Productivity 1 in the accompanying diagram. Albernia is at point A and Brittania is at point B. per worker (Y/L) B B Productivity 1 (Y/L) A A (K/L) A (K/L) B Physical capital per worker a. In the relationship depicted by the curve Productivity 1, what factors are held fixed? Do these countries experience diminishing returns to physical capital per worker? b. Assuming that the amount of human capital per worker and the technology are held fixed in each country, can you recommend a policy to generate a doubling of in each country? c. How would your policy recommendation change if the amount of human capital per worker and the technology were not fixed? Draw a curve on the diagram that represents this policy for Albernia. 3. a. The curve reflecting the relationship between physical capital per worker (K/L) and output per worker (Y/L) is drawn holding human capital per worker and technology fixed. Both Albernia and Brittania experience diminishing returns to physical capital since in both countries equal successive increases in physical capital per worker holding human capital per worker and technology constant will result in smaller and smaller increases in per worker. b. Albernia should increase its physical capital per worker to (K/L) B. Brittania will have to add a huge amount of physical capital per worker.
4 102_KrugmanMacro_SM_Ch08.qxp 11/10/05 1:07 PM Page MACROECONOMICS, CHAPTER 8 ECONOMICS, CHAPTER 25 c. If it were possible to increase the amount of human capital per worker or improve the technology, or both, then Productivity 1 could shift to Productivity 2 and Albernia could double per worker without a change in the physical capital per worker. On the accompanying diagram, Albernia would move from point A to point C. per worker Productivity 2 (Y/L) B C B Productivity 1 (Y/L) A A (K/L) A (K/L) B Physical capital per worker 4. Why would you expect in California and Pennsylvania to exhibit convergence but not in California and Baja California, a state of Mexico that borders the United States? What changes would allow California and Baja California to converge? 4. According to the conditional convergence hypothesis, other things equal, countries with relatively low tend to have higher rates of growth than countries with relatively high. We can apply this hypothesis to regions as well. It is more likely that the factors that affect growth will be equal in California and Pennsylvania: both states have similar educational systems, infrastructure, rule of law, and so on. But that is not true of California and Baja California: in comparing them, the factors that affect growth are not likely to be equal. California and Baja California have very different educational systems, different infrastructures, and there are differences in how the rule of law is applied. So it is less likely that they will converge. For California and Baja California to converge in, they would have to become more similar in the factors that affect growth. 5. The economy of Profunctia has estimated its aggregate production function, when holding human capital per worker and technology constant, as Y L = 100 K L
5 KrugmanMacro_SM_Ch08.qxp 11/9/05 4:47 PM Page 103 LONG-RUN ECONOMIC GROWTH 103 Y is, L is the number of workers, and K is the quantity of physical capital. Given that Profunctia has 1,000 workers, calculate per worker and the quantity of physical capital per worker for the differing amounts of physical capital shown in the accompanying table. K L K/L Y/L \$0 1,000?? 10 1,000?? 20 1,000?? 30 1,000?? 40 1,000?? 50 1,000?? 60 1,000?? 70 1,000?? 80 1,000?? 90 1,000?? 100 1,000?? a. Plot the aggregate production function for Profunctia. b. Does the aggregate production function exhibit diminishing returns to physical capital? Explain your answer. 5. a. The accompanying table and diagram show the aggregate production function for Profunctia. K L K/L Y/L \$0 1,000 \$0.00 \$ , , , , , , , , , , per worker, Y/L \$ \$ Physical capital per worker, K/L
6 KrugmanMacro_SM_Ch08.qxp 11/11/05 5:31 PM Page MACROECONOMICS, CHAPTER 8 ECONOMICS, CHAPTER 25 b. The aggregate production function does exhibit diminishing returns to physical capital. For example, the table shows that as K increases from \$30 to \$40, Y/L increases by \$2.68, but as K increases from \$70 to \$80, Y/L increases only by \$ The Bureau of Labor Statistics regularly releases the Productivity and Costs report for the previous month. Go to and find the latest report. (On the Bureau of Labor Statistics home page, click on Productivity under Latest Numbers and then choose the latest Productivity and Costs report.) What were the percent changes in business and nonfarm business productivity for the previous quarter? How does the percent change in that quarter s productivity compare to previous data? 6. Answers will vary with the latest data. For the third quarter of 2005, business and nonfarm business productivity grew by 4.8% and 4.1%, respectively. These were higher than the productivity growth figures for the second quarter of 2005, which were 0.81% and 2.1%, respectively. 7. What roles do physical capital, human capital, technology, and natural resources play in influencing long-run economic growth of aggregate output? 7. Physical capital, human capital, technology, and natural resources play important roles in influencing long-run growth in. Increases in both physical capital and human capital help a given labor force to produce more over time. Although economic studies have suggested that increases in human capital may explain increases in productivity better than increases in physical capital per worker, technological progress is probably the most important driver of productivity growth. While natural resources played a prominent role historically in determining productivity, they play a less important role in increasing productivity than do increases in human or physical capital in most countries today. 8. Through its policies and institutions, how has the United States influenced U.S. longrun economic growth? Why might persistently large borrowing by the U.S. government ultimately limit long-run economic growth in the future? 8. Institutions and policies in the United States have greatly aided U.S. economic growth. The country has been politically stable, and its laws and institutions protect private property. The economy has attracted significant savings, both domestic and foreign, that have allowed investment spending to spur the growth of the capital stock and fund research and development. The government has directly supported economic growth through its support of public education as well as research and development. However, the government s persistently large borrowing may reduce private investment spending (a phenomenon known as crowding out ), consequently slowing economic growth. 9. Over the next 100 years, in Groland is expected to grow at an average annual rate of 2.0%. In Sloland, however, growth is expected to be somewhat slower, at an average annual growth rate of 1.5%. If both countries have a today of \$20,000, how will their differ in 100 years? (Hint: A country that has a today of \$x and grows at y% per year will achieve a of \$x ( y) z in z years.)
7 KrugmanMacro_SM_Ch08.qxp 11/9/05 4:47 PM Page 105 LONG-RUN ECONOMIC GROWTH If in Groland grows at an average annual rate of 2.0%, in 100 years will be \$144,893 [\$20,000 (1 0.02) 100 ]. At an average annual rate of growth of 1.5%, in Sloland in 100 years will be \$88,641 [\$20,000 ( ) 100 ]. Although both nations start with the same real GDP today, the differential growth rates will result in living standards in Sloland that are 61.2% (\$88,641/\$144, ) of those in Groland. 10. The accompanying table shows data from the Penn World Table, Version 6.1, for real GDP U.S. in France, Japan, the United Kingdom, and the United States in 1950 and. Complete the table. Have these countries converged economically? 1950 Percentage Percentage of U.S. of U.S. France \$5,561? \$22,254? Japan 2,445? 24,495? United Kingdom 7,498? 22,849? United States 10,601? 33,308? 10. The accompanying table shows U.S. in France, Japan, and the United Kingdom as a percentage of in the United States U.S. U.S. France \$5, % \$22, % Japan 2, , United Kingdom 7, , United States 10, , in France and Japan, the two nations with the lowest in 1950, closed some of the gap in living standards with the United States. Japan s grew from only 23.1% of that in the United States to 73.5%, and France s rose from 52.5% to 66.8%. But living standards in the United Kingdom relative to those in the United States actually declined; fell from 70.7% of that in the United States to 68.6%. France and Japan have converged, but the United Kingdom has not.
8 KrugmanMacro_SM_Ch08.qxp 11/9/05 4:47 PM Page MACROECONOMICS, CHAPTER 8 ECONOMICS, CHAPTER The accompanying table shows data from the Penn World Table, Version 6.1, for real GDP U.S. for Argentina, Ghana, South Korea, and the United States in and. Complete the table. Have these countries converged economically? Percentage Percentage of U.S. of U.S. Argentina \$7,395? \$10,995? Ghana 832? 1,349? South Korea 1,571? 15,881? United States 12,414? 33,308? 11. The accompanying table shows U.S. in Argentina, Ghana, and South Korea as a percentage of in the United States. U.S. U.S. Argentina \$7, % \$10, % Ghana , South Korea 1, , United States 12, , There is little evidence of convergence for either Argentina or Ghana. Living standards in both nations declined relative to those in the United States. In Argentina fell from 59.6% of that of the United States to 33.0%; Ghana s fell from 6.7% to 4.0%. But South Korea s showed signs of convergence with those in the United States; rose from 12.7% of that in the United States to 47.7%.
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ACS news SalarieS of chemists fall Unemployment reaches new heights in 2009 as recession hits profession hard The economic recession has taken its toll on chemists. Despite holding up fairly well in previous | 9,504 | 41,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-26 | longest | en | 0.805099 |
http://stackoverflow.com/questions/1861612/sieve-of-eratosthenes-algorithm?answertab=votes | 1,395,048,989,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678705051/warc/CC-MAIN-20140313024505-00017-ip-10-183-142-35.ec2.internal.warc.gz | 139,282,663 | 17,804 | Sieve of Eratosthenes Algorithm
I did some searching and was not able to find any information regarding this implementation versus every other one I have seen.
``````function sieve(\$top)
{
for(\$i = 11; \$i<\$top; \$i+=2)
{
if(\$i % 3 == 0 || \$i % 5 == 0
|| \$i % 7 == 0)
{
continue;
}
echo "\$i <br />";
}
}
``````
Yeah I know it just prints it out, but that's not the important part. What is the major pitfall whether it be time or other?
EDIT: Are there any other issues beyond scalability? Also again thanks for the comments about moving forward with prime finding.
-
Probably just a typo, but you have `for(\$1 ...)` instead of `for(\$i ...)` in line 3. – Daniel Vandersluis Dec 7 '09 at 17:45
The first non prime number this code will output is 221. 221 is 13*17. – Greg Dec 7 '09 at 17:49
Damnit. I mean 169 (13*13). – Greg Dec 7 '09 at 17:51
@Greg - I think you mean 121 (11*11)... – cjk Dec 7 '09 at 17:54
I appreciate the comments, I am going to make the quick edit to \$1 to \$i. Also does anyone have any new information beyond the scalability issue? If nothing else arises by the end of the day I will mark ck's as the excepted answer. Thanks again – Woot4Moo Dec 7 '09 at 18:01
The major pitfall of this is it doesn't scale. Once the numbers are large enough anything will be returned. You list of modulus excluders needs to grow with the search.
-
From what I have seen that is also the major pitfall for this algorithm in general. I hear 10 ^ 9 is the real breaking point for this algorithm in general. – Woot4Moo Dec 7 '09 at 17:48
your version fails at about 10^2 :) – Peter Recore Dec 7 '09 at 18:03
@Peter: actually, as 7 is the largest known prime he is sieving against, anything above 49 is not guaranteed. @Woot4Moo: is this something you are doing as practice, as you are quite a long way off reinventing the wheel if you need a full solution. I wrote myself a prime-finder jsut to test CPU operational speed and multithreading, and rather than use the SOE, I just did number crunching checking for any lack of remainder when dividing by 3 += 2... – cjk Dec 7 '09 at 21:06
@ck it was just something to get used to the syntax of php. No real application in what I do on a daily basis. – Woot4Moo Dec 8 '09 at 4:20
It's limited to prime numbers up to 11. To extend it any further you need to add `|| \$u % 11 == 0 || \$i % 13 == 0 ...` etc
-
My thought is since anything divisible by 2,3,5, or 7 is not prime I would not need to do % 11 or % 13. Since with integer multiplication I do not have to worry. Are there any specific situations where I could multiply 2 numbers that would not make that statement evaluate to true? – Woot4Moo Dec 7 '09 at 17:46
I'm not sure what you mean, but I think a counter example is 121, which is 11*11 and is not divisible by 2, 3, 5, or 7. You need to check against each prime less than the square root of a number in order to determine whether it's prime. – StrixVaria Dec 7 '09 at 17:51
@Woot4Moo - you are right, anything that is divisible by 2, 3, 5 or 7 is not prime, but the reverse is not true. Greg's comment to your question has the example of 221 which is not divisible by any of these, but is also not prime. – cjk Dec 7 '09 at 17:53
thank you, that is one of those things that I was hoping for some confirmation on. – Woot4Moo Dec 7 '09 at 17:58
You can refer to Sieve of Eratosthenes on Wikipedia; and this link for a PHP implementation.
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https://www.coursehero.com/file/6303552/Week6/ | 1,495,883,249,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608936.94/warc/CC-MAIN-20170527094439-20170527114439-00253.warc.gz | 1,086,312,675 | 72,951 | Week6 - coordinates are zero In particular W 1 contains the...
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Solutions for Assignment 6 Applied Linear Algebra MATH 232 (Fall 2008) Section 2.8 Section 2.9
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Assignment 6 Applied Linear Algebra Math 232(Fall 2008) Additional question: Let W 1 = x y z : x = 0 and y = 0 and W 2 = x y z : x = 0 or y = 0 . Is W 1 a subspace of R 3 ? Same question for W 2 . Justify your answers. A1. W 1 is a subspace. Notice that W 1 is just the set of vectors whose first two
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Unformatted text preview: coordinates are zero. In particular, W 1 contains the zero vector and is closed under scalar multiplication and addition. Notice that if the vectors 1 and 1 are both in W 2 , but their sum, 1 1 , is not. Hence W 2 is not a subspace. 1...
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Ask a homework question - tutors are online | 365 | 1,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-22 | longest | en | 0.913384 |
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# Understanding 2/5 as a Decimal: Simplifying Fraction to Decimal Conversion
Published:
## Introduction
Have you ever encountered the fraction 2/5 and wondered how to express it as a decimal? Converting fractions to decimals is a fundamental mathematical skill that comes in handy in various real-world scenarios. In this article, we’ll delve into the process of converting the fraction 2/5 as a decimal equivalent. Whether you’re a student grappling with math problems or someone who simply wants to understand fractions better, this article will guide you through the steps of this conversion with clarity and simplicity.
1. The Basics of Fractions and Decimals
• Understanding Numerators and Denominators
• Introduction to Decimal Representation
2. Converting Fractions to Decimals: A Step-by-Step Guide
• Long Division Method
• Converting Common Fractions
3. Breaking Down 2/5: Fraction Analysis
• Numerator and Denominator Exploration
• Insights into the Fraction 2/5
4. Converting 2/5 to a Decimal
• Step 1: Long Division Setup
• Step 2: Long Division Calculation
• Step 3: Interpreting the Result
5. Why Decimal Conversions Matter
• Practical Applications in Daily Life
• Mathematical Significance and Problem Solving
6. Comparing Decimals and Fractions
• Understanding Equivalencies
• Choosing the Appropriate Representation
7. Visualizing 2/5 as a Decimal
• Using Visual Models for Better Grasp
• Representing 2/5 on the Number Line
8. Practice Makes Perfect
• Interactive Exercises for Decimal Conversion
9. The Connection Between Fractions, Decimals, and Percentages
• Expressing 2/5 as a Percentage
10. Real-World Examples
• Applying 2/5 as a Decimal in Practical Scenarios
• Problem-Solving and Critical Thinking
11. Common Misconceptions and Pitfalls
• Addressing Common Errors in Conversion
• Clarifying Misunderstandings
12. Tips for Quick Mental Conversions
• Handy Techniques for Rapid Calculations
• Building Confidence in Math Skills
13. The Beauty of Mathematical Abstraction
• Appreciating the Logic Behind Conversions
• How Math Shapes Our Understanding
14. Exploring Further: Fractions and Beyond
• Advanced Topics in Fraction Mathematics
• Navigating More Complex Conversions
15. Conclusion
• Embracing the Power of Decimal Conversion
• Confidence in Handling Fractions and Decimals
## Converting 2/5 to a Decimal: Step by Step
Let’s dive into converting the fraction 2/5 into a decimal using the long division method.
Step 1: Long Division Setup
1. Write down 2 as the dividend and 5 as the divisor.
2. Place a decimal point above the dividend.
Step 2: Long Division Calculation
1. Divide 2 by 5. The result is 0 with a remainder of 2.
2. Bring down a zero after the remainder to continue the division.
Step 3: Interpreting the Result
1. Divide 20 (the remainder 2 followed by a zero) by 5. The result is 4.
2. The decimal representation of 2/5 is 0.4.
## Why Decimal Conversions Matter
Decimal conversions are essential in various real-life situations. They allow us to express fractions in a way that’s easier to work with in calculations. Whether you’re dealing with measurements, recipes, or financial calculations, decimals provide a more precise representation.
Consider baking a cake: if a recipe calls for 2/5 cups of flour, knowing that it’s equivalent to 0.4 cups as a decimal makes measurements simpler and ensures accurate results.
## Practice and Application
To reinforce your understanding, practice converting different fractions to decimals. Try to convert 3/5, 4/7, and 1/3 into their decimal equivalents. This will help solidify your skills and build confidence in performing conversions quickly and accurately.
## Conclusion
Converting 2/5 to a decimal might seem like a small mathematical task, but it’s a skill that holds practical significance. Understanding how to convert fractions to decimals opens doors to smoother calculations and a better grasp of numerical relationships. Embrace the power of decimal conversions and enhance your math skills in everyday scenarios.
## FAQs
1. Is 0.4 the only decimal representation of 2/5? No, 2/5 can also be expressed as 0.40 or 0.400, with the zero padding indicating additional decimal places.
2. Can fractions with larger denominators be converted similarly? Yes, fractions with larger denominators can be converted using the same long division method, but the calculations might be more intricate.
3. Are decimals used outside of mathematics? Absolutely, decimals are widely used in various fields including science, finance, and statistics for accurate data representation.
4. How do decimals relate to percentages? Decimals and percentages are closely related. For instance, 0.4 as a decimal is equivalent to 40% as a percentage. | 1,064 | 4,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-33 | latest | en | 0.743888 |
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Explore Related Concepts
Advantages of Heating Effect of Electric Current
Advantages of heating effect of electric current is a tool which describes the useful part of the heating effect of the electric current. First of all before we understand the heating effect of electric current, it is important to understand the meaning of electric current. In simple words, electric current is basically described as the flow of electrons. So if the electrons flow than it means that the electric current is flowing. More the number of free electrons more the electric current will flow. This is all about the basics of electric current.
Now in our day to day life we have observed that the electrical appliances after continuous operation become hot. Hence, it is important to understand the reason for the heating of electrical appliances hot after continuous operation.
The reason behind the same can be explained on the basis of a very important law and that law is known as joule law. Joule law describes that the heat produced is directly proportional to square of current, resistance and is also directly proportional to the time. If H is the heat produced, I is the electric current, R is the resistance and t is the time taken, then the joule law can be expressed as follows
H is directly proportional to I ^2 * R * t.
Based on this law now it is clear that the heating effect of electrical appliances is all due to the joules law and it basically depends upon the current, resistance and time taken. From the above all detailed concepts, we are now in position to describe the advantages of heating effect of electric current. The advantages are based on the equipment design. The following equipments are designed based on the heating effect of electric current and these are the biggest advantages of heating effect of electric current.
1) Electric iron
2) Heater
a) First type is Space heaters
b) Second type is convection heater
c) Third type is forced convection heaters.
d) Fourth type is electric under floor heating (domestic)
e) Fifth type is Fan/forced heaters
f) Sixth type is immersion heater
g) Seventh type is heat pumps and
h) Last type is Electrode heater.
3) Electric tower rails
4) Electric Air curtains
5) Electric water heaters and
6) Electric fires
The above advantages of heating effect of electric current shows that there are lots of advantages of heating effect of electric current. But without the thorough knowledge of this subject we shall not deal with electric current, otherwise this may be dangerous even we can also lost our valuable life. So the basic building block that is the Joule’s law shall be studied in a detailed way and before going for the working model it is important to go for prototype. Prototype is basically a conceptual model which shall be made before actually working on the working model.
This is because that there is no danger involved in making prototype. And the other advantage is that we can take a prior feel before actually going to the working model and if any correction is required we can easily make it in working model on the basis of proto type. So this is all about advantages of heating effect of electric current. | 640 | 3,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-22 | latest | en | 0.955963 |
https://ebooksumo.com/search/analysis-and-simulation-of-noise-in-nonlinear-electronic-circuits-and-systems/ | 1,659,919,788,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00025.warc.gz | 243,494,615 | 10,879 | # Analysis and Simulation of Noise in Nonlinear Electronic Circuits and Systems
In electronic circuit and system design, the word noise is used to refer to any undesired excitation on the system.
Author: Alper Demir
Publisher: Springer Science & Business Media
ISBN: 9781461560630
Category: Technology & Engineering
Page: 275
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In electronic circuit and system design, the word noise is used to refer to any undesired excitation on the system. In other contexts, noise is also used to refer to signals or excitations which exhibit chaotic or random behavior. The source of noise can be either internal or external to the system. For instance, the thermal and shot noise generated within integrated circuit devices are in ternal noise sources, and the noise picked up from the environment through electromagnetic interference is an external one. Electromagnetic interference can also occur between different components of the same system. In integrated circuits (Ies), signals in one part of the system can propagate to the other parts of the same system through electromagnetic coupling, power supply lines and the Ie substrate. For instance, in a mixed-signal Ie, the switching activity in the digital parts of the circuit can adversely affect the performance of the analog section of the circuit by traveling through the power supply lines and the substrate. Prediction of the effect of these noise sources on the performance of an electronic system is called noise analysis or noise simulation. A methodology for the noise analysis or simulation of an electronic system usually has the following four components: 2 NOISE IN NONLINEAR ELECTRONIC CIRCUITS • Mathematical representations or models for the noise sources. • Mathematical model or representation for the system that is under the in fluence of the noise sources.
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# Modeling Simulation and Optimization of Integrated Circuits
References [1] W. Anzill, F.X. Kärtner, P. Russer: Simulation of the phase noise of oscillators in the frequency domain, ... A. Sangiovanni-Vincentelli: Analysis and Simulation of noise in nonlinear electronic circuits and systems, ...
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The third Conference on Mathematical Models and Numerical Simulation in Electronic Industry brought together researchers in mathematics, electrical engineering and scientists working in industry. The contributions to this volume try to bridge the gap between basic and applied mathematics, research in electrical engineering and the needs of industry.
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# Scientific Computing in Electrical Engineering
Analysis and simulation of noise in nonlinear electronic circuits and systems. Kluwer Academic Publishers, 1998. ... CAD-based electric-circuit modeling in industry I. mathematical structure and index of network equations. Surv. Math.
Author: Wilhelmus H. Schilders
Publisher: Springer Science & Business Media
ISBN: 9783642558726
Category: Mathematics
Page: 420
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# RF and Microwave Oscillator Design
[ 92 ] Odyniec , M. , “ Nonlinear Oscillators Theory and Design , ” IEEE MTT - S , Workshop on Recent Advances in ... [ 100 ] Demir , A. , “ Analysis and Simulations of Noise in Nonlinear Electronic Circuits and Systems , ” Ph.D. thesis ...
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Publisher: Artech House
ISBN: 1580537685
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This groundbreaking book is the first to present the state of the art in microwave oscillator design with an emphasis on new nonlinear methods. A compilation of pioneering work from experts in the field, it also provides rigorous theory and historical background. Invaluable for professionals at all levels of design expertise, this volume helps you to bridge the gap between design practice and new powerful design methods, learn all aspects of modern oscillator design and review practical designs and experimental results of fixed-frequency, high-Q, low-noise oscillators.
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# Scientific Computing in Electrical Engineering
PAMM, 4(1), 15–18, (2004) [DS98] Demir, A., Sangiovanni-Vincentelli, A.: Analysis and simulation of noise in nonlinear electronic circuits and systems. Kluwer Academic Publishers (1998) [DW03] Denk, G., Winkler, R.: Modeling and ...
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This book is a collection of papers presented at the last Scientific Computing in Electrical Engineering (SCEE) Conference, held in Sicily, in 2004. The series of SCEE conferences aims at addressing mathematical problems which have a relevancy to industry. The areas covered at SCEE-2004 were: Electromagnetism, Circuit Simulation, Coupled Problems and General mathematical and computational methods.
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# Numerical Methods in Electromagnetics
Analysis and Simulation of Noise in Nonlinear Electronic Circuits and Systems (Kluwer, Boston). DUTTON, R.W., TROYANOVSKY, B., YU, Z., ARNBORG, T., ROTELLA, F., MA, G., SATO-IWANAGA, J. (1997). Device simulation for RF applications.
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Page: 928
View: 212
This special volume provides a broad overview and insight in the way numerical methods are being used to solve the wide variety of problems in the electronics industry. Furthermore its aim is to give researchers from other fields of application the opportunity to benefit from the results wich have been obtained in the electronics industry. * Complete survey of numerical methods used in the electronic industry * Each chapter is selfcontained * Presents state-of-the-art applications and methods * Internationally recognised authors
Categories: Technology & Engineering
# High Performance CMOS Range Imaging
In: Custom Integrated Circuits, 1999. Proceedings of the IEEE 1999. 1999, pp. 385–393. Alper Demir and Alberto Sangiovanni-Vincentelli. Analysis and Simulation of Noise in Nonlinear Electronic Circuits and Systems. Second printing.
Author: Andreas Süss
Publisher: CRC Press
ISBN: 9781315643885
Category: Computers
Page: 262
View: 751
This work is dedicated to CMOS based imaging with the emphasis on the noise modeling, characterization and optimization in order to contribute to the design of high performance imagers in general and range imagers in particular. CMOS is known to be superior to CCD due to its flexibility in terms of integration capabilities, but typically has to be
Categories: Computers
# High Performance Scientific And Engineering Computing
Demir A. , Sangiovanni - Vincentelli A. L. ( 1998 ) Analysis and Simulation of Noise in Nonlinear Electronic Circuits and Systems . Kluwer Academic Publishers 2. Denk G. , Hillermeier C. , Schäffler St. ( 2000 ) Ein Online - Verfahren ...
Author: Michael Breuer
Publisher: Springer Science & Business Media
ISBN: 3540429468
Category: Computers
Page: 408
View: 99
Categories: Computers
# IEEE Transactions on Circuits and Systems
[ 41 ] A. Demir and A. Sangiovanni - Vincentelli , Analysis and Simulation of Noise in Nonlinear Electronic Circuits and Systems . Norwell , MA : Kluwer , 1998 . [ 42 ] A. Demir , “ Phase noise in oscillators as differential - algebraic ...
Author:
Publisher:
ISBN: UIUC:30112081627207
Category: Electric circuits
Page:
View: 476
Categories: Electric circuits
# A Short History of Circuits and Systems
Hand analysis of such large nonlinear circuits was no longer feasible so attention turned to developing efficient ... Circuit simulation was subdivided into four key tasks in Rohrer's MNAbased CANCER software at UC Berkeley in the early ...
Author: Franco Maloberti
Publisher: Stylus Publishing, LLC
ISBN: 9788793609860
Category: Technology & Engineering
Page: 344
View: 196
After an overview of major scientific discoveries of the 18th and 19th centuries, which created electrical science as we know and understand it and led to its useful applications in energy conversion, transmission, manufacturing industry and communications, this Circuits and Systems History book fills a gap in published literature by providing a record of the many outstanding scientists, mathematicians and engineers who laid the foundations of Circuit Theory and Filter Design from the mid-20th Century. Additionally, the book records the history of the IEEE Circuits and Systems Society from its origins as the small Circuit Theory Group of the Institute of Radio Engineers (IRE), which merged with the American Institute of Electrical Engineers (AIEE) to form IEEE in 1963, to the large and broad-coverage worldwide IEEE Society which it is today.Many authors from many countries contributed to the creation of this book, working to a very tight time-schedule. The result is a substantial contribution to their enthusiasm and expertise which it is hoped that readers will find both interesting and useful. It is sure that in such a book omissions will be found and in the space and time available, much valuable material had to be left out. It is hoped that this book will stimulate an interest in the marvellous heritage and contributions that have come from the many outstanding people who worked in the Circuits and Systems area.
Categories: Technology & Engineering | 2,047 | 9,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-33 | latest | en | 0.893547 |
http://crazy-russian-youngs.info/bozi/nominal-rate-of-return-calculator-hymy.php | 1,560,681,928,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00205.warc.gz | 39,341,594 | 6,243 | Select Page
# Nominal rate of return calculator
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## Effective Interest Rate Formula
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1. Nominal Annual Interest Rate Formulas:
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The real rate of return formula is the sum of one plus the nominal rate divided by the sum of one plus the inflation rate which then is subtracted by one. For example, if you have a nominal rate of return of 6% on a investment in a period when inflation is averaging 2%, your real rate of return is %.
1. Real Rate of Return Formula
You need to provide the and debt will help you and Inflation Rate. For quick calculation, an individual bank and take a loan, real rate of return by interest rate of the loan nominal rate - inflation rate. Calculates annual nominal yield. For example, detailed data on a mutual fund Bing Yahoo to achieve your goals and. The wooden boat school. Press 0then CFj. Customize Calculations - unlimited. Let us now do the. | 685 | 3,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-26 | latest | en | 0.927302 |
https://rdrr.io/cran/exams/man/matrix_to_schoice.html | 1,529,708,774,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864822.44/warc/CC-MAIN-20180622220911-20180623000911-00305.warc.gz | 688,274,514 | 15,835 | # matrix_to_schoice: Generate Single- and Multiple-Choice Question Lists for... In exams: Automatic Generation of Exams in R
## Description
Functions for generating single- and multiple-choice question lists for a matrix solution. (Optimized for integer matrices.)
## Usage
```1 2 3 4 5 6 7``` ```matrix_to_schoice(x, y = NULL, lower = FALSE, name = "a", delta = 0.5, digits = 0) matrix_to_mchoice(x, y = NULL, lower = FALSE, name = "a", comparisons = c("==", "<", ">", "<=", ">=")) det_to_schoice(x, y = NULL, range = NULL, delta = 0.5, digits = 0) ```
## Arguments
`x` matrix (correct result). `y` numeric vector (optional) with (potentially) wrong solutions/comparisons. `lower` logical. Should only elements from the lower triangle be assessed? `name` character. Base name for matrix elements. `delta` numeric. Minimal distance between solutions. `digits` integer. Digits that should be displayed. `comparisons` character. Vector of logical comparisons that should be employed. `range` numeric vector of length 2 (optional) with range of random wrong solutions.
## Details
The function `matrix_to_schoice` (or equivalently `matrix2schoice`) can be used for generating a single-choice question list for a correct result matrix `x`. One element is picked randomly from the matrix and chosen to be the correct solution. Other values from the observed absolute range are used as wrong solutions by default (if `y` does not provide an alternative list of potential solutions).
The function `matrix_to_mchoice` (or equivalently `matrix2mchoice`) can be used for generating a multiple-choice question list for a correct result matrix `x`. Each item from the question list is a logical comparison of one matrix element with a comparion value. By default the comparisons are picked randomly from the observed absolute range (unless `y` specifies a different list of comparisons).
The function `det_to_schoice` (or equivalently `det2schoice`) can be used for generating a single-choice question list for the determinant of a 2x2 matrix. It has been optimized for matrices with single-digit integer elements. It may not yield very balanced random solutions for other scenarios.
## Value
`matrix_to_schoice`/`matrix2schoice` returns a list with the following components:
`index` numeric vector with matrix index of the correct solution chosen. `name` character with LaTeX code for the correct matrix element chosen. `solutions` a logical vector of length 5 indicating the `correct` solution, `questions` a character vector of length 5 with question list.
`matrix_to_mchoice`/`matrix2mchoice` returns a list with the following components:
`solutions` a logical vector of length 5 indicating the `correct` solution, `questions` a character vector of length 5 with question list. `explanations` a character vector of length 5 with explanations why the solutions are correct or wrong.
`det_to_schoice`/`det2schoice` returns a list with the following components:
`solutions` a logical vector of length 5 indicating the `correct` solution, `questions` a character vector of length 5 with question list.
`num_to_schoice`
## Examples
```1 2 3 4 5 6 7 8``` ```A <- matrix(c(-9, 0, 5, -2), ncol = 2) matrix_to_schoice(A) matrix_to_mchoice(A) det_to_schoice(A) B <- matrix(1:9, ncol = 3) matrix_to_schoice(B) matrix_to_mchoice(B) ```
### Example output
```\$index
row col
1 1
\$name
[1] "\$a_{11}\$" "\$a_{11}= -9\$"
\$solutions
[1] FALSE FALSE TRUE FALSE FALSE
\$questions
[1] "\$ 5\$" "\$-1\$" "\$-9\$" "\$ 0\$" "\$-2\$"
\$questions
[1] "\$a_{12} > 5\$" "\$a_{22} > -8\$" "\$a_{21} < 0\$" "\$a_{11} > -9\$"
[5] "\$a_{12} > 5\$"
\$solutions
[1] FALSE TRUE FALSE FALSE FALSE
\$explanations
[1] "\$a_{12} = 5 \\not> 5\$" "\$a_{22} = -2\$"
[3] "\$a_{21} = 0 \\not< 0\$" "\$a_{11} = -9 \\not> -9\$"
[5] "\$a_{12} = 5 \\not> 5\$"
\$solutions
[1] FALSE FALSE FALSE TRUE FALSE
\$questions
[1] "\$56\$" "\$-64\$" "\$-9\$" "\$18\$" "\$60\$"
\$index
row col
2 3
\$name
[1] "\$a_{23}\$" "\$a_{23}= 8\$"
\$solutions
[1] FALSE TRUE FALSE FALSE FALSE
\$questions
[1] "\$ 4\$" "\$ 8\$" "\$ 7\$" "\$ 9\$" "\$-3\$"
\$questions
[1] "\$a_{33} \\ge -7\$" "\$a_{31} \\le 3\$" "\$a_{13} < 6\$" "\$a_{22} \\ge -3\$"
[5] "\$a_{11} > 0\$"
\$solutions
[1] TRUE TRUE FALSE TRUE TRUE
\$explanations
[1] "\$a_{33} = 9\$" "\$a_{31} = 3\$" "\$a_{13} = 7 \\not< 6\$"
[4] "\$a_{22} = 5\$" "\$a_{11} = 1\$"
```
exams documentation built on May 12, 2018, 1:05 a.m. | 1,392 | 4,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-26 | latest | en | 0.780567 |
https://math.answers.com/Q/What_are_the_three_basic_parts_of_a_circle | 1,712,958,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00812.warc.gz | 345,915,882 | 47,060 | 0
# What are the three basic parts of a circle?
Updated: 12/17/2022
Wiki User
7y ago
Its circumference
Its diameter
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7y ago
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Q: What are the three basic parts of a circle?
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### What divids a circle into three parts?
A circle divides a plane into three parts.
### What are three basic parts of volcano?
the lava, the mantle, the outside. so there you are you have the three basic parts to a volcano.
### What 3 parts does a circle divide a plane into?
The three parts are:the interior of the circle, the points on the circle (points on its circumference) the outside of the circle.
### What does a circle divide into three parts?
A circle divides a plane into three parts.
### How do you divide a circle into eight parts using three straight lines?
You can cut a round cake into eight parts with three cuts, but you can't cut a circle into eight parts with three straight lines.
### How do you divide a circle with 9cm diameter in three equal parts?
Construct a circle with a 4.5 radius. The circle's circumference is 360 degrees. So mark out 3 by 120 degrees on the circumference and join them to the centre of the circle which will divide the circle into three equal parts.
A circle can.
### Where are the three basic parts of an atom labeled?
The three basic parts of an atom is the nucleus,protons and neutrons
### Does a composite cell have three basic parts?
A composite cell has three basic parts consisting of the nucleus, cell membrane, and cytoplasm. A cell is the structural and functional unit of an organism.
### What are the 3 basic parts of a triangle?
Three vertices, three line segments, three angles.
### What is three basic parts of a computer?
Processor ports and monitor
### The three basic parts of a mushroom are?
stem, cap and gill | 420 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-18 | latest | en | 0.925354 |
http://web.lemoyne.edu/giunta/chm151L/expquiz1q.htm | 1,508,252,790,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00499.warc.gz | 354,358,557 | 4,007 | ### Exponential notation
Write using correct exponential notation:
1) 43 = x10
2) 0.00000269 = x10
3) 432,506 = x10
4) 0.00900 = x10
5) 93,000,000 = x10
6) 2,400,000,000 = x10
7) 0.0000000158 = x10
8) 56,000 = x10
9) 186,000 = x10
10) 432x10-4 = x10
11) 0.00378x10-3 = x10
12) 400x102 = x10 | 148 | 292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-43 | latest | en | 0.450352 |
http://math.about.com/od/reference/a/gr5.htm | 1,371,609,167,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707440258/warc/CC-MAIN-20130516123040-00041-ip-10-60-113-184.ec2.internal.warc.gz | 155,838,509 | 7,980 | • Share
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## 5th Grade Math Course of Study
The following list provides you with the basic concepts that should be attained by the end of the school year. Mastery of the concepts at the previous grade is assumed.
Number
• Read print numbers to 100 000 and locate, compare, order, represent, estimate and identify numbers to 100 000 using regular and expanded forms
• Full understanding of place value to the right and left of 0 - 4 places
• Count by 3's, 4's, 6's, 7's, 8's, 9's, and 10's, 11's and 12's to 144
• Multiplication facts are committed to memory up to the 12 X tables(Understand the division facts also)
• Understand decimals to the thousandths 0.013 and be able to add and subtract decimals.
• Demonstrate a solid understanding of fractions and their related decimals to 100dths.
• Multiply and divide decimals
• Communicate mathematical thinking in problem solving - selecting the appropriate strategies
• Select the appropriate problem solving methods in word problems for the above operations
Measurement
• Complete understanding of inches, feet, yards, miles, millimeters, centimeters, meters, kilometers and apply these terms to problem solving activities
• Measure accuratey, and make appropriate estimations of which units of measure apply.
• Construct or illustrate items using a variety of units of measure
• Estimate and round accurately
• Read and write dates using a variety of methods (Jan. 10, 2002, 02/10/02 etc.)
• Money amounts to \$1000.00 in making change and in problem solving
• Investigate and solve measurement problems with circumference, perimeter, volume, capacity and area and explain the rules and apply the formulas
Geometry
• Identify, sort, classify, construct, measure, and apply a variety of geometric shapes and figures and problems
• Full understanding of geometric properties and relationships
• Classify triangles by side properties and types (obtuse, isosceles) etc.
• Identify the 2-D nets that the solids are represented by and construct the nets
• Measure and construct a variety of triangle and angles with the protractor
• Explore and discover tiling patterns that cover a plane and tesselations
• Understand the coordinate system on both maps and grids
Algebra/Patterning
• Identify, create, analyze and extend patterns and describe the rules with more than one variable
• Determine the values in equations when there are missing terms in the four operations and provide the rules
• Determine the amount in missing values when given an equation that involve more than 1 operation
• Demonstrate equivalence in equations with the 4 operations
Probability
• Design surveys, collect the data and record it appropriate, be able to discuss the findings
• Construct a variety of graphs and label them appropriately and state the difference between selecting one graph over another
• Discuss the real world need for data and the collection of data
• Read, analyze and interpret data in a variety of graphs etc.
• Us tree diagrams to organize data, make decision on data gathered and sorted and record the results
• Conduct probability experiments and apply logical reasoning to the outcomes
• Predict probability based on background information
Pre-K Kdg. Gr. 1 Gr. 2 Gr. 3 Gr. 4 Gr. 5 Gr. 6 Gr. 7 Gr. 8 Gr. 9 Gr. 10 Gr.11 Gr. 12
Deb Russell
Mathematics Guide
Explore Mathematics
By Category | 751 | 3,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2013-20 | latest | en | 0.881912 |
https://www.idsemergencymanagement.com/2020/01/18/what-is-input-output-theory/ | 1,653,665,309,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00207.warc.gz | 935,528,344 | 14,782 | # What is input output theory?
Jan 18, 2020
## What is input output theory?
Input–output theory, developed by Collett and Gardiner in 1984, is a way to treat the interaction of a system with a thermal bath, in which the bath is modeled as a quantum field [128, 199]. Input–output theory was originally derived by considering the damping of a mode of an optical cavity.
## What is input process and output process?
Inputs are the conditions that exist prior to group activity, whereas processes are the interactions among group members. Outputs are the results of group activity that are valued by the team or the organization.
## What is input and output of a system?
An input is data that a computer receives. An output is data that a computer sends. Computers only work with digital information. Any input that a computer receives must be digitised. Often data has to be converted back to an analogue format when it’s output, for example the sound from a computer’s speakers.
## What are the main features of input-output analysis?
As such, it has three main elements; Firstly, the input-output analysis concentrates on an economy which is in equilibrium. Secondly, it does not concern itself with the demand analysis. It deals exclusively with technical problems of production. Lastly, it is based on empirical investigation.
## Which is the input in input-output approach?
Input-output tables are the foundation of input-output analysis, depicting rows and columns of data that quantify the supply chain for all of the sectors of an economy. Three types of impacts are modeled in input-output analysis. They are direct impact, indirect impact, and induced impact.
## What is the process of input?
Input: It is captures the data from user, or it is the process of accepting data or information, by using input the computer can do any process. Process: It is the process to convert the input into output. Storage: It stores the data or information or instructions, for future use.
## What is input and output examples?
Difference between Input and Output devices:
INPUT DEVICE OUTPUT DEVICE
The design of input devices are more complex. The design of output devices are less complex.
Ex: Keyboard, Image Scanner, Microphone, Pointing device, Graphics tablet, Joystick. Ex: Monitor, Printers, Plotters, Projector, Speakers.
## What is the purpose of input-process-output?
A computer program or any other sort of process using the input-process-output model receives inputs from a user or other source, does some computations on the inputs, and returns the results of the computations. The system divides the work into three categories: A requirement from the environment (input)
## What is an input-output diagram?
An Input-Output Diagram is a simple high-level representation of a system that shows: The major inputs to a system and their suppliers. The major outputs from a system and their customers. The major components of the system necessary for it to achieve its purpose through transforming inputs to outputs.
## Why is the Input Process Output Model important?
The input-process-output model has historically been the dominant approach to understanding and explaining team performance and continues to exert a strong influence on group research today. The framework is based on classic systems theory, which states that the general structure of a system is as important in determining how effectively it
## What are the inputs and outputs of a system?
Inputs and Outputs Output is the information produced by a system or process from a specific input. Within the context of systems theory, the inputs are what are put into a system and the outputs are the results obtained after running an entire process or just a small part of a process.
## How does a system receive input from the environment?
Systems receive input from the environment either as information or in the form of resources. The systems then process the input internally, which is called throughput, and release outputs into the environment in an attempt to restore equilibrium to the environment.
## Which is the end product of processing to get an output?
Throughput – The input will be processed to get an output. A simple logical explanation. This process which the system employs to get a desired output can be termed throughput Output – It is the end product of the processing by the system | 855 | 4,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-21 | longest | en | 0.921567 |
https://expertinstudy.com/t/bOCg52LXOHx | 1,675,032,329,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00360.warc.gz | 263,880,675 | 4,336 | Ottavia Jan 1, 2021
# Ravi has 36 toys Sham has 22 toys. How many toysare there
The time for answering the question is over
TOTAL TOYS ARE 58.
Ravi has = 36 toys
Sham has = 22 toys
Total toys = 36 + 22
= 58
Ans : 58 toys are there in total.
441
Christian
Jan 1, 2021 | 97 | 276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-06 | latest | en | 0.959391 |
https://yorkporc.wordpress.com/2014/02/09/klein-group-tunny-square-lie-algebra-decibannage/ | 1,524,735,842,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00356.warc.gz | 917,176,964 | 13,203 | ## Klein Group, Tunny square, Lie algebra, decibannage
When we looked at the Tunny sqaure in the General Report (circa 1945), we quickly figured that it was related to the 4 –operator Klein Group. (It sticks out!…)
I see now how I need to improve the labels, so we accommodate the adjoint space, too. That is our picture (for the4-char group version of the algebra) SHOULD HAVE BEEN
Now the theory of repeats etc has a strong basis. Its really the commutator algebra in which the action of the normalizer compresses the original plaintext (down to) to the decibannage. Each commutator, being independent further compresses the length of the plaintext superpositions down, until there is nothing left to do which means the lengths in phase space are now associated with the radius of the normalizer line of the 7-point algebra – which have DEPENDENT terms (by definition). This means there is no further independence (or mutual information measure) to be obtained from the plaintext.
Fascinating to how these were little matrices, in a unique basis, all along.
We can assume that the still “secret” report from the Testery would show that folks were not ONLY doing mental arithmatic with 2 characters (having memorized their xor); but also had tables of larger and larger matrices, as above. Which leads to today….where its easy to do some algebras with even huge matrices, in fields Much larger than 2**5.
I suspect that element 1, in the picture I drew above is still wrong. It REALLY ought to be labeled S, not S squared. S squared is the group identity, is is a matrix of all \. But, near enough for me. We get the core idea!
Even the definition of S+ holds out, since xor is the same for subtraction or addition:
(this is bullshit, note. two matrices have constant product, the weight difference)
Now the theory of 2, 4, 8, 16 counts makes some theoretical sense, being an act of calculating at different levels of granularity/accuracy. One uses the 2-width matrix, the 16 width matrix… etc, which compares 2 and upto 16 of the source bits in a buffer , pairwise.
Putting this thought train together with that of the previous train (looking at amplitudes and intensity of light, detected by modern photodiodes rather than the 1945 era 5205 machine), let’s not forget our article that makes it all rather practical:
https://yorkporc.wordpress.com/2011/07/16/from-5205-to-balanced-homodyne-detection/
In some ways, the 2-,4-,8- counts are a little like a MRI machine, that slices cross-sectionally, thinking in terms of quantum “tomography”
We can show that product of two matrices do give a constant weight profile (the difference between each component). The profile is the diagonal of the matrix not used (ignoring the identity matrix). If one multiples with the identity matrix, the profile is that of the non-identity argument. Hmm! | 658 | 2,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-17 | latest | en | 0.9508 |
http://arnienumbers.blogspot.com/2010/11/recurrence-relation-for-abundancy-index.html | 1,532,358,637,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596542.97/warc/CC-MAIN-20180723145409-20180723165409-00264.warc.gz | 23,010,083 | 13,182 | ## 5.11.10
### A Recurrence Relation for the Abundancy Index
C. A Recurrence Relation for the Abundancy Index
Fix a prime p, and let k be a positive integer.
Write Sigma(p^k) = 1 + p + p^2 + ... + p^(k - 1) + p^k as:
Sigma(p^k) = (1 + p + p^2 + ... + p^(k - 1)) + p^k
Thus: Sigma(p^k) = Sigma(p^(k - 1)) + p^k
Consequently, F(p, k) = Sigma(p^k) satisfies the recurrence relation:
(!) F(k) = F(k - 1) + g(k) (note: F(k) = F(p, k) -- we omit p because it is fixed)
where g(k) is the exponential function g(k) = p^k (where p is held constant).
The recurrence relation for F gives rise to an equivalent recurrence relation for f:
We have: Sigma(p^k) = Sigma(p^(k - 1)) + p^k
Divide through by (p^k): I(p^k) = (1/p)I(p^(k - 1)) + 1
Multiply through by (p): p*I(p^k) = I(p^(k - 1)) + p
Thus, the resulting recurrence relation for f(p, k) = I(p^k) is given by:
(@) p*f(k) = f(k - 1) + p (note: f(k) = f(p, k) -- we omit p because
it is fixed)
Since I(1) = 1, for consistency in our recurrence equations, we can set:
f(0) = 1
With this initial seed, our recursion goes thus:
p*f(1) = f(0) + p = 1 + p
f(1) = (1 + p)/p = (1/p) + 1
p*f(2) = f(1) + p = (1 + p)/p + p = (1 + p + p^2)/p
f(2) = (1 + p + p^2)/p^2 = (1/p)^2 + (1/p) + 1
p*f(3) = f(2) + p = (1 + p + p^2)/p^2 + p
= (1 + p + p^2 + p^3)/p^2
f(3) = (1 + p + p^2 + p^3)/p^3
= (1/p)^3 + (1/p)^2 + (1/p) + 1
...
...
...
Thus, roughly we have:
f(p, k) = Big-OH((1/p)^k)
when p is fixed.
That is, the abundancy index is a (rational) NON-polynomial function of k for constant p. [Please see comment #2.]
Indeed, we have the closed form:
I(p^k) = (p^(k + 1) - 1)/((p^k)(p - 1))
where of course you will be able to recognize the familiar formula:
Sigma(p^k) = (p^(k + 1) - 1)/(p - 1)
which is derived via finite geometric series. | 729 | 1,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-30 | latest | en | 0.761023 |
https://mathematica.stackexchange.com/questions/231493/how-do-i-enforce-this-boundary-condition | 1,716,646,690,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058829.24/warc/CC-MAIN-20240525131056-20240525161056-00843.warc.gz | 325,870,950 | 39,859 | # How do I enforce this boundary condition?
I am solving this differential equation:
(1 - 2 M/r) D[(1 - (2 M)/r) D[q[r], r], r] - (1 - 2 M/r) ((l (l + 1))/r^2 - (6 M)/r^3) q[r]
with the boundary conditions that q[0]==0 and $$q\rightarrow\dfrac{1}{r^l}$$ at infinity. How do I enforce the second condition?
• Do you need a symbolic solution or numeric solution? If the latter, what's the value of those paremeters? Oct 9, 2020 at 2:25
• @xzczd I need a numerical solution. You can set M=1 and l=2 Oct 9, 2020 at 4:41
• If $l=2$, isn't the b.c. at infinity equivalent to $q(\infty)=0$? Oct 9, 2020 at 5:08
• @xzczd I am following this paper articles.adsabs.harvard.edu/pdf/1978ApJ...224..643C (page 4 of 25) They say to impose that b.c. Oct 9, 2020 at 5:39
• The output of AsymptoticDSolveValue[{(1 - (2 M)/r) D[(1 - (2*M)/r)*D[q[r], r], r] - (1 - (2 M)/r) ((l (l + 1))/r^2 - (6 M)/r^3) q[r] == 0(*,q[0]\[Equal]0*)}, q@r, {r, Infinity, 1}] seems to suggest such solution doesn't exist… Perhaps there's some deeper math here? Or the paper is wrong? Oct 9, 2020 at 6:00
Mathematica's result doesn't really make sense with the paper you reference. We can solve the ode without boundary conditions.
ode = (1 - (2*M)/r)*D[(1 - (2*M)/r)*D[q[r], r], r] -
(1 - (2*M)/r)*((l*(l + 1))/r^2 - (6*M)/r^3)*q[r] == 0
\$Assumptions = r >= 0 && M > 0
sol = DSolve[ode, q[r], r] // Flatten // FunctionExpand
We get hypergeometric functions, but they simplify if we define $$l$$.
For $$l=2$$
sol /. l -> 2 // Simplify
$$\left\{q(r)\to c_2 \left( \begin{array}{cc} \{ & \begin{array}{cc} \frac{r^3}{8 M^3} & r<2 M \\ 0 & r>2 M \\ \text{Indeterminate} & \text{True} \\ \end{array} \\ \end{array} \right)-\frac{c_1 r^3}{8 M^3}\right\}$$
We can make $$q=0$$ for $$r<2M$$ by setting $$c_1=c_2$$, but for $$r>2M$$ we get a solution proportional to $$r^3$$ which will not satisfy your boundary at infinity.
• DSolveValue[ode /. l -> 2, q[r], r] gives a different answer, r^3 C[1] + (C[2] (12 M^4 + 8 M^3 r + 6 M^2 r^2 + 6 M r^3 - 3 r^4 Log[r] + 3 r^4 Log[-2 M + r]))/(96 M^5 r). The discrepancy is due to a bug in Mathematica that I reported to Wolfram Inc over three years ago. See 138440. Simply stated, Mathematica has given you only one independent solution instead of two due to this bug. I agree, though, that this problem has no solution for the boundary conditions specified. Oct 10, 2020 at 1:14 | 898 | 2,390 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.770142 |
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# Simplified method of measuring currency risk
Moderator
When banks analyze currency risk, we prefer to use advanced simulation tools to get accurate results. Our selected method is to have the analysis software generate 2000 different scenarios for future currency development and monitor the distributions worst case scenarios. We normally look at the 5% scenarios where a company has its largest loss, when assessing the risk.
There are simpler ways of doing this kind of analysis, and get a fairly accurate result. The interesting question is: How far can a currency rate move within 90% likelihood (a 90% confidence level)?
Described in a more operational context, a case like this could be:
A company with EUR as it functional currency is expecting to receive a 10 million USD payment in one year. How much can the value of this USD amount change within 90% likelihood?
If we use the currency market to find the forward rate for EURUSD in one year, we can calculate the expected value of the USD payment. If the forward rate is at 1,2200, the expected value of the USD payment is EUR 8.196.721. This amount is possible to lock in using currency forward contracts.
The company is evaluating what could happen with the value of this USD payment if the currency rate is not locked in with currency forwards. What is the expected risk?
Another element of information needed to assess the risk, is the market volatility for EURUSD. This is the currency markets expected standard deviation for the year to come. The volatility is a measure on how much the market rate can move during a specified time period. This volatility is observable in the currency options market. It might be found through different sources of financial information (Thomson Reuters, Bloomberg, or your bank). If you can’t find the accurate market volatility, and you are look at currency rates between developed countries, you might use 10%. The actual EURUSD 1 year volatility is at 7,5% (November 2017).
You are now ready to calculate the simplified risk on your expected USD payment. The formula to use is this:
The number 1,65 in formula is a scaling factor.
If we input the information we have gathered, the formula might look like this:
We then end up with the risk amount of EUR 1.014.344.
The market has provided us with the information that there is a 90% likelihood that the value of this USD payment will not change with more than +/- EUR 1.014.344 through one year. There is a 5% likelihood of an increase in value with more than +EUR 1.014.344. This 5% is not a concern, since the company receives much higher values than expected.
There is a 5% likelihood of a decrease in with value with more than -EUR 1.014.344. This is the company’s worst case risk.
Hence there is a 95% likelihood that the company will not lose more than EUR 1.014.344 on this USD payment in one year. You now have facts about the risk and can now asses if you want or need to hedge all or parts of the expected payment.
Contributor
Hi Ståle,
Thanks for sharing!
I am wondering whether this simplified method can be applied when measuring VaR as well? I.e. balance sheet exposures which are basis for translation risk? In that case would it just be to replace "expected payment" with "expected exposure", applying the same parameters and then ending up with a proxy on expected decrease in balance sheet exposure?
Also, if the timeperiod is less than one year, e.g. 1-3 months, would it still be reasonable to apply 10 % as expected volatility for developed currencies?
Appreciate any sort of feedback.
Best regards, Tom-André W. Hansen (PRA Group Europe AS)
Moderator
Hi Tom-André
Good to hear from you.
The methodologies in Cash Flow at Risk and Value at Risk are basically the same.
The cash Flow at Risk modell is used to measure risk connected with cash flows in the future, while Value at Risk is used for balance risk.
Time horison in these two analysis might be different. A cash flow at Risk analysis often covers cash flows through one year or more, while a Value at Risk modell on easily sellable assets might look at value changes through a shorter time (maybe a week).
The replacement you propose in the formula will work fine. The time horizon you select will be based on how easy it is to sell the assets.
We have been going through some of the market volatilities today (current market conditions).
Most of the currency volatilites in developed economies are lower than 10%.
If you look at current USDNOK volatilites, the one year voltility is at 9,2%, while the one month volatility is 8,2%.
EURNOK volatilites are even lower.
You overshoot your risk estimates with a 10% volatility.
If you need accurate markets volatilites, feel free to send me an email.
Best regards, Ståle
Ståle Johansen, Risk Advisory, DNB Markets
Contributor
Hi
Thank you for article.
Excuse ignorance but how do you get the scaling factor of 1.65
Thanking You
R
Moderator
Hi Raelliss
The scaling factor comes from the normal distribution. The assumption that the currency rate changes are normally distributed will define a 90% confidence level. You might find the scaling from finding a Standard Normal Distribution Table, and look for the one sided 45% likelyhood, which is 1.65 standard deviation from the expected value (the forward rate).
BR
Stale | 1,213 | 5,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-05 | latest | en | 0.91402 |
https://rdrr.io/cran/icd/man/icd_get_billable.html | 1,508,265,525,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822480.15/warc/CC-MAIN-20171017181947-20171017201947-00458.warc.gz | 801,023,405 | 14,775 | icd_get_billable: Get billable ICD codes In icd: Tools for Working with ICD-9 and ICD-10 Codes, and Finding Comorbidities
Description
Get billable ICD codes, implicitly, this refers to an ICD implementation which is specialized for a country, typically for billing, e.g. ICD-9-CM in the USA.
Usage
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19``` ```icd_get_billable(...) ## Default S3 method: icd_get_billable(x, short_code = icd_guess_short(x), ...) ## S3 method for class 'icd9cm' icd_get_billable(x, short_code = icd_guess_short(x), invert = FALSE, icd9cm_edition = icd9cm_latest_edition(), ...) ## S3 method for class 'icd9' icd_get_billable(...) ## S3 method for class 'icd10cm' icd_get_billable(x, short_code = icd_guess_short(x), invert = FALSE, icd10cm_edition = "2016", ...) ## S3 method for class 'icd10' icd_get_billable(x, short_code = icd_guess_short(x), invert = FALSE, icd10cm_edition = "2016", ...) ```
Arguments
`x` input vector of ICD codes `short_code` single logical value which determines whether the ICD-9 code provided is in short (`TRUE`) or decimal (`FALSE`) form. Where reasonable, this is guessed from the input data. `invert` Single logical value. Returns the inverse of the result. E.g. if seeking valid ICD-9 codes, the invalid ones are returned. `icd9cm_edition` e.g. "32", not ICD-9 vs ICD-10
Methods (by class)
• `default`: Get billable ICD codes, guessing whether ICD-9 or ICD-10, and code short vs decimal type.
• `icd9cm`: Get billable ICD-9-CM codes
• `icd9`: Get billable ICD-9 codes, which is currently implemented assuming ICD-9-CM
• `icd10cm`: Get billable, i.e. leaf nodes from ICD-10-CM
• `icd10`: Get billable, i.e. leaf nodes from ICD-10-CM
icd documentation built on May 30, 2017, 1:48 a.m. | 563 | 1,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-43 | longest | en | 0.673367 |
https://www.airmilescalculator.com/distance/lhr-to-ldy/ | 1,628,011,818,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00606.warc.gz | 593,702,607 | 31,511 | # Distance between London (LHR) and Derry (LDY)
Flight distance from London to Derry (London Heathrow Airport – City of Derry Airport) is 371 miles / 598 kilometers / 323 nautical miles. Estimated flight time is 1 hour 12 minutes.
Driving distance from London (LHR) to Derry (LDY) is 492 miles / 791 kilometers and travel time by car is about 9 hours 41 minutes.
## Map of flight path and driving directions from London to Derry.
Shortest flight path between London Heathrow Airport (LHR) and City of Derry Airport (LDY).
## How far is Derry from London?
There are several ways to calculate distances between London and Derry. Here are two common methods:
Vincenty's formula (applied above)
• 371.463 miles
• 597.812 kilometers
• 322.793 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 370.646 miles
• 596.496 kilometers
• 322.082 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A London Heathrow Airport
City: London
Country: United Kingdom
IATA Code: LHR
ICAO Code: EGLL
Coordinates: 51°28′14″N, 0°27′42″W
B City of Derry Airport
City: Derry
Country: United Kingdom
IATA Code: LDY
ICAO Code: EGAE
Coordinates: 55°2′34″N, 7°9′39″W
## Time difference and current local times
There is no time difference between London and Derry.
BST
BST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 80 kg (176 pounds).
## Frequent Flyer Miles Calculator
London (LHR) → Derry (LDY).
Distance:
371
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
371
Round trip? | 476 | 1,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-31 | latest | en | 0.8395 |
https://lips.cs.princeton.edu/the-gumbel-max-trick-for-discrete-distributions/ | 1,713,210,222,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817014.15/warc/CC-MAIN-20240415174104-20240415204104-00463.warc.gz | 338,970,978 | 4,017 | # The Gumbel-Max Trick for Discrete Distributions
Ryan Adams · April 6, 2013
It often comes up in neural networks, generalized linear models, topic models and many other probabilistic models that one wishes to parameterize a discrete distribution in terms of an unconstrained vector of numbers, i.e., a vector that is not confined to the simplex, might be negative, etc. A very common way to address this is to use the "softmax" transformation:
\begin{align*}
\pi_k &= \frac{\exp\{x_k\}}{\sum_{k'=1}^K\exp\{x_{k'}\}}
\end{align*}
where the $x_k$ are unconstrained in $\mathbb{R}$, but the $\pi_k$ live on the simplex, i.e., $\pi_k \geq 0$ and $\sum_{k}\pi_k=1$. The $x_k$ parameterize a discrete distribution (not uniquely) and we can generate data by performing the softmax transformation and then doing the usual thing to draw from a discrete distribution. Interestingly, it turns out that there is an alternative way to arrive at such discrete samples, that doesn't actually require constructing the discrete distribution.
It turns out that the following trick is equivalent to the softmax-discrete procedure: add Gumbel noise to each $x_k$ and then take the argmax. That is, add independent noise to each one and then do a max. This doesn't change the asymptotic complexity of the algorithm, but opens the door to some interesting implementation possibilities. How does this work? The Gumbel distribution with unit scale and location parameter $\mu$ has the following PDF:
\begin{align*}
f(z\,;\,\mu) &= \exp\{-(z-\mu) - \exp\{-(z-\mu)\}\}.
\end{align*}
The CDF of the Gumbel is
\begin{align*}
F(z\,;\,\mu) &= \exp\{-\exp\{-(z-\mu)\}\}
\end{align*}
Now, imagine that the $k$th of our Gumbels, with location $x_k$, resulted in an outcome $z_k$. The probability that all of the other $z_{k'\neq k}$ are less than this is
\begin{align*}
\Pr(z_k \text{ is largest}\,|\, z_k, \{x_{k'}\}^K_{k'=1}) &= \prod_{k'\neq k}\exp\{-\exp\{-(z_k-x_{k'})\}\}
\end{align*}
We know the marginal distribution over $z_k$ and we need to integrate it out to find the overall probability:
\begin{align*}
\Pr(\text{$k$ is largest}\,|\,\{x_{k'}\}) =
\int \exp\{-(z_k-x_k)-\exp\{-(z_k-x_k)\}\}\\
\times \prod_{k'\neq k}\exp\{-\exp\{-(z_k-x_{k'})\}\} \,\mathrm{d}z_k\quad
\end{align*}
With a little bit of algebra, we get:
\begin{align*}
\Pr(\text{$k$ is largest}\,|\,\{x_{k'}\}) =&
\int \exp\{-z_k + x_k \\
\Pr(\text{$k$ is largest}\,|\,\{x_{k'}\}) = \frac{\exp\{x_k\}}{\sum_{k'=1}^K\exp\{x_{k'}\}} | 785 | 2,478 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | latest | en | 0.856565 |
http://gdevtest.geeksforgeeks.org/minimum-replacements-to-make-adjacent-characters-unequal-in-a-ternary-string/ | 1,555,835,283,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530505.30/warc/CC-MAIN-20190421080255-20190421102255-00299.warc.gz | 70,154,191 | 24,687 | # Minimum replacements to make adjacent characters unequal in a ternary string
Given a string of ‘0’, ‘1’ and ‘2’. The task is to find the minimum number of replacements such that the adjacent characters are not equal.
Examples:
Input: s = “201220211”
Output: 2
Resultant string after changes is 201210210
Input: s = “0120102”
Output: 0
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: The following problem can be solved using greedy method. We can greedily compare every adjacent pair. If the adjacent pairs which is character at ith and i-1th are same, then replace ith th charcater with a character which is not equal to the character at i-1th and i+1th index. In case of the last adjacent pair, just replace it with the character which is not equal to the character at i-1th index.
Below is the implementation of the above approach:
## C++
`// C++ program to count the minimal ` `// replacements such that adjacent characters ` `// are unequal ` `#include ` `using` `namespace` `std; ` ` ` `// Function to count the number of ` `// minimal replacements ` `int` `countMinimalReplacements(string s) ` `{ ` ` ``// Find the length of the string ` ` ``int` `n = s.length(); ` ` ` ` ``int` `cnt = 0; ` ` ` ` ``// Iterate in the string ` ` ``for` `(``int` `i = 1; i < n; i++) { ` ` ` ` ``// Check if adjacent is similar ` ` ``if` `(s[i] == s[i - 1]) { ` ` ``cnt += 1; ` ` ` ` ``// If not the last pair ` ` ``if` `(i != (n - 1)) { ` ` ` ` ``// Check for character which is ` ` ``// not same in i+1 and i-1 ` ` ``for` `(``auto` `it : ``"012"``) { ` ` ``if` `(it != s[i + 1] && ` ` ``it != s[i - 1]) { ` ` ``s[i] = it; ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``else` `// Last pair ` ` ``{ ` ` ``// Check for character which is ` ` ``// not same in i-1 index ` ` ``for` `(``auto` `it : ``"012"``) { ` ` ``if` `(it != s[i - 1]) { ` ` ``s[i] = it; ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``return` `cnt; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ``string s = ``"201220211"``; ` ` ``cout << countMinimalReplacements(s); ` ` ``return` `0; ` `} `
## Java
`// Java program to count the minimal ` `// replacements such that adjacent ` `// characters are unequal ` `class` `GFG ` `{ ` ` ` ` ``static` `final` `int` `MAX = ``26``; ` ` ` ` ``// Function to count the number of ` ` ``// minimal replacements ` ` ``static` `int` `countMinimalReplacements(``char``[] s) ` ` ``{ ` ` ``// Find the length of the String ` ` ``int` `n = s.length; ` ` ` ` ``int` `cnt = ``0``; ` ` ` ` ``// Iterate in the String ` ` ``for` `(``int` `i = ``1``; i < n; i++) ` ` ``{ ` ` ` ` ``// Check if adjacent is similar ` ` ``if` `(s[i] == s[i - ``1``]) ` ` ``{ ` ` ``cnt += ``1``; ` ` ` ` ``// If not the last pair ` ` ``if` `(i != (n - ``1``)) ` ` ``{ ` ` ` ` ``// Check for character which is ` ` ``// not same in i+1 and i-1 ` ` ``for` `(``char` `it : ``"012"``.toCharArray()) ` ` ``{ ` ` ``if` `(it != s[i + ``1``] ` ` ``&& it != s[i - ``1``]) ` ` ``{ ` ` ``s[i] = it; ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ``else` `// Last pair ` ` ``{ ` ` ``// Check for character which is ` ` ``// not same in i-1 index ` ` ``for` `(``char` `it : ``"012"``.toCharArray()) ` ` ``{ ` ` ``if` `(it != s[i - ``1``]) ` ` ``{ ` ` ``s[i] = it; ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ``return` `cnt; ` ` ``} ` ` ` ` ``// Driver Code ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``String s = ``"201220211"``; ` ` ``System.out.println(countMinimalReplacements(s.toCharArray())); ` ` ``} ` `} ` ` ` `// This code is contributed by 29AjayKumar `
## Python3
# Python 3 program to count the minimal
# characters are unequal
# Function to count the number of
# minimal replacements
def countMinimalReplacements(s):
# Find the length of the string
n = len(s)
cnt = 0
# Iterate in the string
for i in range(1, n):
# Check if adjacent is similar
if (s[i] == s[i – 1]):
cnt += 1;
# If not the last pair
if (i != (n – 1)):
# Check for character which is
# not same in i+1 and i-1
s = list(s)
for j in “012”:
if (j != s[i + 1] and
j != s[i – 1]):
s[i] = j
break
s = ”.join(s)
# Last pair
else:
# Check for character which is
# not same in i-1 index
s = list(s)
for k in “012”:
if (k != s[i – 1]):
s[i] = k
break
s = ”.join(s)
return cnt
# Driver Code
if __name__ == ‘__main__’:
s = “201220211”
print(countMinimalReplacements(s))
# This code is contributed by
# Surendra_Gangwar
## C#
`// C# program to count the minimal ` `// replacements such that adjacent ` `// characters are unequal ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ``static` `readonly` `int` `MAX = 26; ` ` ` ` ``// Function to count the number of ` ` ``// minimal replacements ` ` ``static` `int` `countMinimalReplacements(``char``[] s) ` ` ``{ ` ` ``// Find the length of the String ` ` ``int` `n = s.Length; ` ` ` ` ``int` `cnt = 0; ` ` ` ` ``// Iterate in the String ` ` ``for` `(``int` `i = 1; i < n; i++) ` ` ``{ ` ` ` ` ``// Check if adjacent is similar ` ` ``if` `(s[i] == s[i - 1]) ` ` ``{ ` ` ``cnt += 1; ` ` ` ` ``// If not the last pair ` ` ``if` `(i != (n - 1)) ` ` ``{ ` ` ` ` ``// Check for character which is ` ` ``// not same in i+1 and i-1 ` ` ``foreach` `(``char` `it ``in` `"012"``.ToCharArray()) ` ` ``{ ` ` ``if` `(it != s[i + 1] && ` ` ``it != s[i - 1]) ` ` ``{ ` ` ``s[i] = it; ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ``else` `// Last pair ` ` ``{ ` ` ``// Check for character which is ` ` ``// not same in i-1 index ` ` ``foreach` `(``char` `it ``in` `"012"``.ToCharArray()) ` ` ``{ ` ` ``if` `(it != s[i - 1]) ` ` ``{ ` ` ``s[i] = it; ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ``return` `cnt; ` ` ``} ` ` ` ` ``// Driver Code ` ` ``public` `static` `void` `Main(String[] args) ` ` ``{ ` ` ``String s = ``"201220211"``; ` ` ``Console.WriteLine(countMinimalReplacements(s.ToCharArray())); ` ` ``} ` `} ` ` ` `// This code is contributed by Rajput-Ji `
## PHP
` `
Output:
```2
```
Time Complexity : O(n)
My Personal Notes arrow_drop_up | 2,587 | 8,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-18 | latest | en | 0.718504 |
https://www.csdn.net/tags/MtzaUg4sNzk2LWJsb2cO0O0O.html | 1,656,288,541,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00610.warc.gz | 789,924,787 | 30,778 | • 2019-08-31 16:43:38
1.对于方阵A，如果为非奇异方阵，则存在逆矩阵inv(A)
2.对于奇异矩阵或者非方阵，并不存在逆矩阵，但可以使用pinv(A)求其伪逆
若A为非奇异矩阵，请不要使用pinv求逆，虽然计算结果相同，即
inv( A ) = pinv( A )
但pinv的计算复杂度较高。
更多相关内容
• I am using Octave 3.8.1, a Matlab-like program. I'd like to generalize 1/x to the case where x may be a scalar or a matrix. Replacing 1/x with inv(x) or pinv(x) works for most x, except:octave:1> ....
I am using Octave 3.8.1, a Matlab-like program. I'd like to generalize 1/x to the case where x may be a scalar or a matrix. Replacing 1/x with inv(x) or pinv(x) works for most x, except:
octave:1> 1/inf
ans = 0
octave:2> pinv([inf])
ans = NaN
octave:3> inv([inf])
warning: inverse: matrix singular to machine precision, rcond = 0
ans = Inf
Should I convert NaN to 0 afterwards to get this to work? Or have I missed something? Thanks!
解决方案
The Moore–Penrose pseudo inverse, which is the basis for Matab and octave's pinv, is implemented via completely different algorithm than the inv function. More specifically, singular value decomposition is used, which require's finite-valued matrices (they also can't be sparse). You didn't say if your matrices are square or not. The real use of pinv is for solving non-square systems (over- or underdetermined).
However, you shouldn't be using pinv or inv for your application, no matter the dimension of your matrices. Instead you should use mldivide (octave, Matlab), i.e., the backslash operator, \. This is much more efficient and numerically robust.
A1 = 3;
A2 = [1 2 1;2 4 6;1 1 3];
A1inv = A1\1
A2inv = A2\eye(size(A2))
The mldivide function handles rectangular matrices too, but you will get different answers for underdetermined systems compared to pinv because the two use different methods to choose the solution.
A3 = [1 2 1;2 4 6]; % Underdetermined
A4 = [1 2;2 4;1 1]; % Overdetermined
A3inv = A3\eye(min(size(A3))) % Compare to pinv(A3), different answer
A4inv = A4\eye(max(size(A4))) % Compare to pinv(A4), same answer
If you run the code above, you'll see that you get a slightly different result for A3inv as compared to what is returned by pinv(A3). However, both are valid solutions.
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• 一、pinv()原型 函数形式：pinv(J, G, P)； 核心算法： 解释：G是正定矩阵，P是矩阵J的伪逆矩阵。当G为恒定常数时，以下等式成立： 函数原型： template<typename DerivedA, typename DerivedB, typename ...
## 一、pinv()原型
函数形式：pinv(J, G, P)；
核心算法：
解释：G是正定矩阵，P是矩阵J的伪逆矩阵。当G为恒定常数时，以下等式成立：
函数原型：
template<typename DerivedA, typename DerivedB, typename DerivedC>
void pinv(const MatrixBase<DerivedA>& J, const MatrixBase<DerivedB>&G, MatrixBase<DerivedC>& P)
{
MatrixXd J_temp(2, 2);
J_temp = J.transpose()*G*J;
P = (A_temp.inverse())*J.transpose()*G;
}
二、验证上述等式
直接上代码：
template<typename DerivedA, typename DerivedB, typename DerivedC>
void pinv(const MatrixBase<DerivedA>& A, const MatrixBase<DerivedB>&G, MatrixBase<DerivedC>& B)
{
//这里就是在算这个投影矩阵p，最后的结果用B输出来，这里的A就是那个Γ，这里的G就是那个G
MatrixXd A_temp(2, 2);
A_temp = A.transpose()*G*A;
B = (A_temp.inverse())*A.transpose()*G;
}
template<typename DerivedA, typename DerivedB>
void pinv2(const MatrixBase<DerivedA>& A, MatrixBase<DerivedB>& B) {
MatrixXd A_temp(2, 2);
A_temp = A.transpose()*A;
B = (A_temp.inverse())*A.transpose();
}
int main()
{
Vector2d ratio;
Matrix3d m_ratio;
MatrixXd projection(2, 3);
MatrixXd a(3, 2);
a<< 3, 0,
88, 0,
0, 1;
pinv2(a, projection);// 求a的伪逆矩阵projection
cout << "projection = " << endl << projection << endl << endl;
/*** 验证当G为任意常数时，pinv()等价于pinv2() **************************************/
MatrixXd projection2(2, 3);
Matrix2d G;
G << 5, 0，
0，5,;
pinv(a, G, projection2);
cout << "projection2 = " << endl << projection2 << endl;
}
打印结果如下：
当我们改变G是值的时候，打印结果依然不变，可以证明：
当G为恒定常数时，
## 三、上述pinv()算法是否与Matlab中的pinv()函数一样？
思路：我们以求b矩阵的伪逆矩阵为例，看看与在Matlab中调用pinv()所得结果是否一致。
b << 1, 2, 3,
4, 5, 6,
0, 0, 1;
（1）首先利用我们前面介绍的算法，看看结果如何：
int main()
{
MatrixXd projection3(3, 3);
MatrixXd projection4(3, 3);
MatrixXd b(3, 3);
Matrix3d G;
G<< 50, 0，0,
0，50, 0,
0, 0, 50;
b << 1, 2, 3,
4, 5, 6,
0, 0, 1;
pinv2(b, projection3);
pinv(b, G, projection4);
cout << "projection3 = " << endl << projection3 << endl << endl;
cout << "projection4 = " << endl << projection4 << endl;
cout << "b.inverse() = " << endl << b.inverse() << endl;
}
我们顺便把b的逆矩阵也求出来了，结果为：
可以发现伪逆矩阵和逆矩阵结果是一样的。
（2）再看看在Matlab中的计算结果：
（3）结论
【1】可以发现Matlab计算出来的pinv(b)与前面的算法计算的结果一样。
所以，可以得出，我们写的pinv()函数与Matlab调用的pinv()是一样的。
【2】我们还可以发现，伪逆矩阵和逆矩阵也相同。有这样的一个结论：
当A矩阵是方阵，且可逆时，A的伪逆矩阵=A的逆矩阵。
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• matlab中pinv代码机器学习线性回归梯度下降 #Machine Learning Course Coursera作者：Andrew NG 它包含具有单变量和多变量的线性回归和梯度下降的矢量化实现。 它还包括通过正则方程法的解决方案。 #Regression...
• ## pinv(matlab 伪逆)
千次阅读 2021-01-27 07:42:02
老师给了一个程序，看到pinv的时候有点糊涂，“伪”逆矩阵到底什么意思啊。pinv(B)求的是矩阵B的Moore-Penrose逆，是B的一种广义逆，也就是你说的伪逆，该广义逆满足四个条件：A*B*A = A B*A*B = B A*B 是海森矩阵 B...
老师给了一个程序，看到pinv的时候有点糊涂，“伪”逆矩阵到底什么意思啊。
pinv(B)求的是矩阵B的Moore-Penrose逆，是B的一种广义逆，也就是你说的伪逆，该广义逆满足四个条件：A*B*A = A B*A*B = B A*B 是海森矩阵 B*A是海森矩阵。这个在.
打开Pinv函数，发现里面分情况讨论：若N>M；则共轭转置后再求解，否则使。
就是“伪”逆阵。求逆阵要求方阵嘛，这个可以对非方阵求逆。也就是说pinv(A)*A = I转置的原因就是要保证矩阵的行数不小于列数，这样使得转置是稳定且唯一的。我举.
opencv或者其它函数库中有类似matlab的pinv函数吗
有的
从数值计算的角度讲，大多数矩阵运算的复杂度都是O(n^3)(矩阵乘法的复杂度就是O(n^3))求伪逆的复杂度不超过求特征值和特征向量的复杂度，所以仍然是O(n^3)
谁能指导左除和pinv的在解方程组方面主要区别在那？自己好久了一直没弄明。
这个大体说来是采用的算法不同，左除采用的是高斯消元法，而pinv采用的是求伪逆的方法(里面具体怎么求的我不知道)进行的，区别应该就在这里吧
pinv是求矩阵的伪逆的函数 伪逆是对于非方阵和奇异矩阵来说的
想通过svd函数分解出的U、S、V自己编程实现求取奇异矩阵的广义逆矩阵，。
假定拟计算一般矩阵A的Moore-Penrose广义逆A+,1)对A做SVD:A = U S V， 其中 U, V为酉方阵， S为一般对角阵；2)将S非零元取逆， 零元不变， 然后专置得到一个一.
您好我有一个64*62的矩阵 我用matlab 里的pinv函数求完广义逆过后值变得非.
假定拟计算一般矩阵a的moore-penrose广义逆a+,1)对a做svd:a = u s v， 其中 u, v为酉方阵， s为一般对角阵；2)将s非零元取逆， 零元不变， 然后专置得到一个一般对.
inv代表逆，pinv是伪逆当一个矩阵不是满秩的时候，如果要求逆的话，只能用伪逆pinv来求了
使用pinv进行矩阵求逆是有效方便的，但是一直不明白pinv和inv函数的区别，此外matlab有时会显示警告信息，指出计算不需要求逆，使用符号“/”或“\”会避免求逆，.
在高等数学中没有关于这样的矩阵逆的求法，但在MATLAB中有关于这样矩阵的逆的求法，不过是叫求伪逆矩阵，是用函数命令pinv()进行计算.如A=[1;2];B=pinv(A)运行-.
奇异矩阵怎么用SVD求逆？matlab上用svd()分解成三个[U S V]，然后怎么.
既然是奇异矩阵就没有逆，只能求Moore-Penrose逆如果已经得到了[U, S, V] = svd(A)A的Moore-Penrose逆就是V*pinv(S)*U'当然要有特殊需求才会这样做，不然直接用pinv(.
矩阵论中的东西
逆矩阵： 当矩阵所形成的方程，称为矩阵方程，如ax=b. 其中：a为线性议程组的系数矩阵x为线性方程组的未知矩阵.而b为线性方程组的右端项矩阵(也称常数矩阵) 定义.
A是一个3行5列的矩阵，X和B是一个向量，A和B已知，用matlab编写程序求.
可以先通过X0 = A\B求得一个特解，然后调用XX = null(A)求得“零空间”的正交基(也就是基础解系)，最后可以得到原方程组的通解。
3 2 1 3 1 5 3 2 1 用这个(A|E)这个来求！！
这个矩阵不可逆，可以用matlab求其伪逆。 具体的伪逆形式不止一种，要根据你的实际情况来决定用哪种伪逆形式。 用matlab中pinv函数求出A的伪逆为X 0.1218 -0.0256.
超定方程 对于方程组Ax=b,A为n*m矩阵，如果A列满秩，且n>m。则方程组没有精确. 还可以用广义逆来求，即x=pinv(A)，所得的解不一定满足Ax=b,x只是知最小二乘意义.
SMI也可以~~~主要是 需要 零陷方向图~~
%%%LCMV在多个来波方向约束下波束形成%%%clc;clear all;close all;ima=sqrt(-1)；. %求信号相关矩阵R=pinv(Rx)； %相关矩阵求逆a1theta=exp(ima*2*pi*d_lamda*sin(.
可多加个条件|A|=|B|，就是只通过ri+krj这种初等行或列变换得到，应该可以不.
a,b相似，则存在可逆矩阵p，使得b=p^(-1)ap 则b*=(p^(-1)ap)*=p*a*(p^(-1))*=p*a*(p*)^(-1) 因此b*与a*相似
w=pinv(Y3'*Y3)*Y3'*train_y; M=(dbn.rbm{1,1}.W)'*(dbn.rbm{1,2}.W)'*(dbn.rbm{1.
1，首先定义了一个dog类，相当于javabean，用来存放和获取狗的名字和体重， 2， 从控制台分别输入4条狗的名字和体重 3，定义四条狗的对象，把2输入的4条狗的名字.
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• matlab中pinv代码#rl ##要点 在Simple Walker上使用S〜AC 使用Wouter的论文在Matlab中实现LWR（第III-C节） 在摆锤上尝试LWR〜AC 在Simple Walker上尝试LWR〜AC ===== ##使用模型要做的新点 将每个控制步骤的计划...
• matlab中pinv代码FasPI 概述 FastPI（Fast PseudoInverse）是一种用于计算现实世界优化问题使用的稀疏特征矩阵的伪逆的新颖方法。 基于观察到许多现实世界的特征矩阵稀疏且高度偏斜的事实，FastPI重新排序并...
• 对于方阵A，如果为非奇异方阵，则存在逆矩阵inv(A)对于奇异矩阵或者非方阵，并不存在逆矩阵，但可以使用pinv(A)求其伪逆inv：inv(A)*B实际上可以写成A\BB*inv(A)实际上可以写成B/A这样比求逆之后带入精度要高A\B=...
• 当输入的矩阵行数等于列数时 采用inv求矩阵的逆，即 Y=inv(X) 其中X为要求逆的矩阵，Y 为 X 矩阵的逆。 计算一个 3×3 矩阵的逆矩阵。 X = [1 0 2; -1 5 0;...pinv 将 A 小于容差的奇异值视为零。
• 2、matlab中广义伪逆的求法pinv(M)为求矩阵M的广义逆。3、MATLAB中e的意思matlab中的e则有不同的意思，如果e的前面没有系数，那么命令行的e则是未定义的不明字符。下面的例子也足够说明。如果前面有系数，且后面有...
• ## Matlab中的pinv和inv
千次阅读 2013-06-25 14:18:38
1.对于方阵A，如果为非奇异方阵，则存在逆矩阵inv(A) 2.对于奇异矩阵或者非方阵，并不存在逆矩阵，但可以使用pinv(A)求其伪逆
• 任意m*n矩阵的奇异值分解，c++语言实现
• matlab中pinv代码快速耐用PCA的IRCUR 这是Matlab的快速非凸鲁棒主成分分析（RPCA）算法的仓库，它被称为迭代鲁棒CUR（IRCUR）[1]。 为了正确显示数学符号，可能必须安装MathJax插件。 例如， 。 稳健的主成分分析 ...
• matlab中pinv代码使用说明 该代码优化了单个线圈的尺寸和位置，以提高多线圈匀场设置的匀场能力。 要求： 具有以下工具箱的Matlab Optimization_Toolbox statistics_toolbox distrib_computing_toolbox 火车数据 ...
• 对于奇异矩阵或者非方阵，并不存在逆矩阵，但可以使用pinv(A)求其伪逆 inv： inv(A)B 实际上可以写成A\B Binv(A) 实际上可以写成B/A 这样比求逆之后带入精度要高 A\B=pinv(A)B&nbsp; A/B=Apinv(B) pinv： X=...
• 伪逆矩阵函数pinv伪逆矩阵的MATLAB定义.ppt(2)正交(QR)分解函数 将矩阵A分解为一个正交矩阵与另一个矩阵的乘积称为矩阵A的正交分解。 格式一：[Q, R]=qr(A) 功能：产生与A同维的上三角矩阵R和一个实正交矩阵或复归...
• matlab中pinv代码<使用L1范数惩罚进行颜色再现的光谱反射率重建的代码和数据> 计算色彩科学工具可以在中找到。 作者是斯蒂芬·韦斯特兰教授。 PRMLT是Chen mo博士为一书编写的matlab代码。 ===================...
• A\B=pinv(A)*B A/B=A*pinv(B) 可见，'\'用的是高斯消元法。 同时还查到： Square Matrices If A is symmetric and has real, positive diagonal elements, MATLABattempts a Cholesky factorization. If the ...
• I had a matrix D which is m*n and i am calculating the pseudo inverse using the formula inv(D'*D)*D' but it is not generating the same result as pinv(D).I need the term inv(D'*D) which i require for i...
• 1引言Matlab是一款高性能的数值计算和可视化软件，集成数值分析、矩阵计算、信号运算、信号处理和图形显示于一体，构成了一个方便的、界面友好的用户环境。目前，基于Matlab的语音识别开发平台虽然在可读性、可移植...
• matlab中pinv代码HR_BP_BLE_Android_Project 心率和血压监测系统 抽象的 在当今的情况下，医学与计算机科学与电子领域已尽可能地融合在一起，但是真正的挑战是将技术集成到设备，以使它们尽可能容易地被广大民众...
• matlab中pinv代码空间音频听力测试环境（SALTE）-音频渲染器 SALTE是一款听力测试软件，旨在帮助将来进行空间音频系统的研究和开发。 该工具由专用的音频渲染引擎组成，用于进行空间音频收听实验。 空间音频听力...
• matlab中pinv代码fea-1D-bar 用MATLAB编写的代码来解决FEA 1D线性条形元素问题 与其他贡献者一起创建为迷你项目： 大师普拉萨斯P 戴维·史密斯·孙达辛格 戈皮·卡西克（Gopi Karthick R） 脚本文件： Linear_bar_...
• 在了解逆和伪逆之前，我们先来了解一下是线性代数里面奇异矩阵和非奇异矩阵是什么。 首先，矩阵是方阵，即矩阵的行数和列数相等的矩阵。若行数和列数不相等，那就谈不上奇异矩阵和非奇异矩阵。 ...
• matlab中pinv代码企业资源计划 ERP方法的实现-精确重新分配伪逆方法 这是基于Johannes Stephan和Walter Fichter的论文的代码实现： 标题：“线性致动系统的快速精确重新分配伪逆方法” DOI：10.1109 / TCST.2017....
• matlab中pinv代码具有多个变量的线性回归 在这个项目，我们将使用多个变量实施线性回归以预测房屋价格。 该项目是的可选练习。 任务描述如下：假设您正在出售房屋，并且想知道一个好的市场价格。 实现此目的的一...
• (Note that MATLAB® uses 1 based indexing while Python uses 0 based indexing, [ 1 2 3; 4 5 6 ] array([[1.,2.,3.], [4.,5.,6.]]) mat([[1.,2.,3.], [4.,5.,6.]]) or mat("1 2 3; 4 5 6") 2x3 ...
• MATLAB中：左右除法、逆inv、广义逆pinv的区别
...
matlab 订阅 | 5,196 | 9,754 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-27 | latest | en | 0.746952 |
https://www.indagro.net/water-supply/what-is-recovery-heaters-for-light-duty-hot-water-supply-service.html | 1,656,575,907,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00128.warc.gz | 867,181,939 | 11,978 | # What Is Recovery Heaters For Light-duty Hot Water Supply Service.?
## What is a recovery heater?
Here’s the official definition: A water heater’s “recovery time” (also called “recovery rate“) is the amount of hot water (in gallons) a tank water heater can provide in just one hour after being completely drained. Recovery rate basically gives you an idea of how fast a water heater can heat water.
## How does a quick recovery water heater work?
“Quick Recovery” water heaters usually refer to electric water heaters that are equipped with dual elements. This allows the upper element to heat a much smaller volume of water (about 1/4 of the tank’s capacity) before the lower element takes over.
## What is a good recovery rate for a hot water heater?
water heaters will have recovery rates in the range of 40 gph, but there are heavy-duty models that go as high as 50 to 60 gph. There are also a few companies, such as Bock Water Heaters, with residential products that have powerful burners which enable very high recovery rates (142 to 159 gph).
You might be interested: Often asked: What Problem Was Caused By An Uncontrolled Water Supply In Sumer?
## What is a high recovery water heater?
A standard water heater has a recovery rate of approximately forty gallons an hour. A high-recovery unit recovers 55 gallons of water an hour. From these two recovery rates, we can deduce that high recovery water heaters are great for homes with many occupants or families.
## How long does it take a hot water heater to recover?
A gas water heater will take about an hour to recover, while an electric water heater will take about two hours. If you have a high recovery water heater, which normally has a larger tank, it will heat the water in much less time, on average about 20 minutes.
## How is water heater recovery rate calculated?
The recovery rate is the amount of hot water the water heater is capable of producing in a given period of time. Take this & divide by 8.33 (pounds per gallon of water) & then divide by the temperature rise required.
## Why does it take so long for me to get hot water?
Why does it take so long for hot water to come out of your shower and other faucets? There are several reasons; the distance from the water heater, the diameter of the piping, and the flow rate of the water. The further the hot water has to flow, the longer it takes to heat up the shower faucet.
## What is the difference between 4500 watt and 5500 watt water heater?
The significant difference between a 4500 and a 5500 Watt water heater is how much time each water heater takes to bring the water to its optimum heating level. Generally, a 4500 Watt heater heats about 18 to 25 gallons every hour. 5500 Watt water heaters can heat about 25 to 35 gallons of water every hour.
You might be interested: Question: How Long Should Water Supply Line Be?
## How do I increase my water heater?
You can remove sediment by draining your tank and flushing it out with cold water. We recommend draining and flushing your water heater at least once a year. Removing sediment buildup will help your heater run much more efficiently.
## How long can you shower with a 50 gallon water heater?
A 50 gallon hot water heater with a temperature set around 120-140 degrees Fahrenheit will allow you to shower for around 17 minutes before you will run out of hot water. A average shower time is just under 8 minutes and uses about 16-17 gallons of water depending on your shower head and water flow.
## How many BTU’s do I need for my hot water heater?
Input ranges from about 32,000 on a 30-gallon unit to 88,000 on a 100-gallon tank. A common input is 34,000 BTUs on a 40-gallon tank and 36,000 BTUs on a 50-gallon tank. The higher the BTU input and efficiency, the faster the recovery.
## How long does it take for a 50 gallon water heater to heat up?
A 50-gallon hot water heater with 5,500-watt elements set to 120 degrees takes about 1 hour and 20 minutes to heat water coming in to the unit at 60 degrees. Conversely, when the water entering this same tank is 40 degrees, it takes 1 hours, 47 minutes to heat it up.
## Which type of water heater typically has the highest recovery rate?
Electric residential water heaters are generally considered to have a 100% recovery efficiency. This is because immersion style elements place all the heat into the water and there is no flue.
You might be interested: FAQ: Where Does Your Community Get Its Water Supply?
## What brand of water heater is the most reliable?
15 Best & Most Reliable Water Heater Brands in the World
• A.O. Smith.
• Rheem. Rheem was founded in 1925 by two brothers, Donald and Richard Rheem – and with financial backing from a third brother, William.
• Kenmore.
• Bradford White.
• American Standard.
• American Water Heaters.
• Bosch.
• EcoSmart.
## Is a short or tall water heater better?
Budget-minded consumers should note: a short water heater is much more energy-efficient than a tall heater —up to 40% more efficient. Additionally, short water heaters provide a better gallons per minute rate than tall water heaters. | 1,152 | 5,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-27 | latest | en | 0.948684 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/3332/1/w/a/2059/1/ | 1,679,629,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00748.warc.gz | 910,227,956 | 71,809 | # Properties
Label 3332.1.w.a.2059.1 Level $3332$ Weight $1$ Character 3332.2059 Analytic conductor $1.663$ Analytic rank $0$ Dimension $4$ Projective image $D_{8}$ CM discriminant -4 Inner twists $4$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [3332,1,Mod(491,3332)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(3332, base_ring=CyclotomicField(8))
chi = DirichletCharacter(H, H._module([4, 0, 3]))
N = Newforms(chi, 1, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("3332.491");
S:= CuspForms(chi, 1);
N := Newforms(S);
Level: $$N$$ $$=$$ $$3332 = 2^{2} \cdot 7^{2} \cdot 17$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 3332.w (of order $$8$$, degree $$4$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$1.66288462209$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{8})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{4} + 1$$ x^4 + 1 Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image: $$D_{8}$$ Projective field: Galois closure of 8.2.3089659810545728.1
## Embedding invariants
Embedding label 2059.1 Root $$0.707107 + 0.707107i$$ of defining polynomial Character $$\chi$$ $$=$$ 3332.2059 Dual form 3332.1.w.a.1471.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q+(-0.707107 - 0.707107i) q^{2} +1.00000i q^{4} +(-0.292893 - 0.707107i) q^{5} +(0.707107 - 0.707107i) q^{8} +(0.707107 - 0.707107i) q^{9} +O(q^{10})$$ $$q+(-0.707107 - 0.707107i) q^{2} +1.00000i q^{4} +(-0.292893 - 0.707107i) q^{5} +(0.707107 - 0.707107i) q^{8} +(0.707107 - 0.707107i) q^{9} +(-0.292893 + 0.707107i) q^{10} -2.00000i q^{13} -1.00000 q^{16} +(0.707107 - 0.707107i) q^{17} -1.00000 q^{18} +(0.707107 - 0.292893i) q^{20} +(0.292893 - 0.292893i) q^{25} +(-1.41421 + 1.41421i) q^{26} +(0.707107 + 1.70711i) q^{29} +(0.707107 + 0.707107i) q^{32} -1.00000 q^{34} +(0.707107 + 0.707107i) q^{36} +(-1.70711 + 0.707107i) q^{37} +(-0.707107 - 0.292893i) q^{40} +(-0.707107 + 1.70711i) q^{41} +(-0.707107 - 0.292893i) q^{45} -0.414214 q^{50} +2.00000 q^{52} +(-1.00000 - 1.00000i) q^{53} +(0.707107 - 1.70711i) q^{58} +(0.707107 - 1.70711i) q^{61} -1.00000i q^{64} +(-1.41421 + 0.585786i) q^{65} +(0.707107 + 0.707107i) q^{68} -1.00000i q^{72} +(-0.292893 - 0.707107i) q^{73} +(1.70711 + 0.707107i) q^{74} +(0.292893 + 0.707107i) q^{80} -1.00000i q^{81} +(1.70711 - 0.707107i) q^{82} +(-0.707107 - 0.292893i) q^{85} +(0.292893 + 0.707107i) q^{90} +(-0.292893 - 0.707107i) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q - 4 q^{5}+O(q^{10})$$ 4 * q - 4 * q^5 $$4 q - 4 q^{5} - 4 q^{10} - 4 q^{16} - 4 q^{18} + 4 q^{25} - 4 q^{34} - 4 q^{37} + 4 q^{50} + 8 q^{52} - 4 q^{53} - 4 q^{73} + 4 q^{74} + 4 q^{80} + 4 q^{82} + 4 q^{90} - 4 q^{97}+O(q^{100})$$ 4 * q - 4 * q^5 - 4 * q^10 - 4 * q^16 - 4 * q^18 + 4 * q^25 - 4 * q^34 - 4 * q^37 + 4 * q^50 + 8 * q^52 - 4 * q^53 - 4 * q^73 + 4 * q^74 + 4 * q^80 + 4 * q^82 + 4 * q^90 - 4 * q^97
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/3332\mathbb{Z}\right)^\times$$.
$$n$$ $$785$$ $$885$$ $$1667$$ $$\chi(n)$$ $$e\left(\frac{7}{8}\right)$$ $$1$$ $$-1$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ −0.707107 0.707107i −0.707107 0.707107i
$$3$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$4$$ 1.00000i 1.00000i
$$5$$ −0.292893 0.707107i −0.292893 0.707107i 0.707107 0.707107i $$-0.250000\pi$$
−1.00000 $$\pi$$
$$6$$ 0 0
$$7$$ 0 0
$$8$$ 0.707107 0.707107i 0.707107 0.707107i
$$9$$ 0.707107 0.707107i 0.707107 0.707107i
$$10$$ −0.292893 + 0.707107i −0.292893 + 0.707107i
$$11$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$12$$ 0 0
$$13$$ 2.00000i 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$14$$ 0 0
$$15$$ 0 0
$$16$$ −1.00000 −1.00000
$$17$$ 0.707107 0.707107i 0.707107 0.707107i
$$18$$ −1.00000 −1.00000
$$19$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$20$$ 0.707107 0.292893i 0.707107 0.292893i
$$21$$ 0 0
$$22$$ 0 0
$$23$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$24$$ 0 0
$$25$$ 0.292893 0.292893i 0.292893 0.292893i
$$26$$ −1.41421 + 1.41421i −1.41421 + 1.41421i
$$27$$ 0 0
$$28$$ 0 0
$$29$$ 0.707107 + 1.70711i 0.707107 + 1.70711i 0.707107 + 0.707107i $$0.250000\pi$$
1.00000i $$0.5\pi$$
$$30$$ 0 0
$$31$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$32$$ 0.707107 + 0.707107i 0.707107 + 0.707107i
$$33$$ 0 0
$$34$$ −1.00000 −1.00000
$$35$$ 0 0
$$36$$ 0.707107 + 0.707107i 0.707107 + 0.707107i
$$37$$ −1.70711 + 0.707107i −1.70711 + 0.707107i −0.707107 + 0.707107i $$0.750000\pi$$
−1.00000 $$\pi$$
$$38$$ 0 0
$$39$$ 0 0
$$40$$ −0.707107 0.292893i −0.707107 0.292893i
$$41$$ −0.707107 + 1.70711i −0.707107 + 1.70711i 1.00000i $$0.5\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$42$$ 0 0
$$43$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$44$$ 0 0
$$45$$ −0.707107 0.292893i −0.707107 0.292893i
$$46$$ 0 0
$$47$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$48$$ 0 0
$$49$$ 0 0
$$50$$ −0.414214 −0.414214
$$51$$ 0 0
$$52$$ 2.00000 2.00000
$$53$$ −1.00000 1.00000i −1.00000 1.00000i 1.00000i $$-0.5\pi$$
−1.00000 $$\pi$$
$$54$$ 0 0
$$55$$ 0 0
$$56$$ 0 0
$$57$$ 0 0
$$58$$ 0.707107 1.70711i 0.707107 1.70711i
$$59$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$60$$ 0 0
$$61$$ 0.707107 1.70711i 0.707107 1.70711i 1.00000i $$-0.5\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$62$$ 0 0
$$63$$ 0 0
$$64$$ 1.00000i 1.00000i
$$65$$ −1.41421 + 0.585786i −1.41421 + 0.585786i
$$66$$ 0 0
$$67$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$68$$ 0.707107 + 0.707107i 0.707107 + 0.707107i
$$69$$ 0 0
$$70$$ 0 0
$$71$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$72$$ 1.00000i 1.00000i
$$73$$ −0.292893 0.707107i −0.292893 0.707107i 0.707107 0.707107i $$-0.250000\pi$$
−1.00000 $$\pi$$
$$74$$ 1.70711 + 0.707107i 1.70711 + 0.707107i
$$75$$ 0 0
$$76$$ 0 0
$$77$$ 0 0
$$78$$ 0 0
$$79$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$80$$ 0.292893 + 0.707107i 0.292893 + 0.707107i
$$81$$ 1.00000i 1.00000i
$$82$$ 1.70711 0.707107i 1.70711 0.707107i
$$83$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$84$$ 0 0
$$85$$ −0.707107 0.292893i −0.707107 0.292893i
$$86$$ 0 0
$$87$$ 0 0
$$88$$ 0 0
$$89$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$90$$ 0.292893 + 0.707107i 0.292893 + 0.707107i
$$91$$ 0 0
$$92$$ 0 0
$$93$$ 0 0
$$94$$ 0 0
$$95$$ 0 0
$$96$$ 0 0
$$97$$ −0.292893 0.707107i −0.292893 0.707107i 0.707107 0.707107i $$-0.250000\pi$$
−1.00000 $$\pi$$
$$98$$ 0 0
$$99$$ 0 0
$$100$$ 0.292893 + 0.292893i 0.292893 + 0.292893i
$$101$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$102$$ 0 0
$$103$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$104$$ −1.41421 1.41421i −1.41421 1.41421i
$$105$$ 0 0
$$106$$ 1.41421i 1.41421i
$$107$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$108$$ 0 0
$$109$$ 0.292893 0.707107i 0.292893 0.707107i −0.707107 0.707107i $$-0.750000\pi$$
1.00000 $$0$$
$$110$$ 0 0
$$111$$ 0 0
$$112$$ 0 0
$$113$$ 0.707107 + 0.292893i 0.707107 + 0.292893i 0.707107 0.707107i $$-0.250000\pi$$
1.00000i $$0.5\pi$$
$$114$$ 0 0
$$115$$ 0 0
$$116$$ −1.70711 + 0.707107i −1.70711 + 0.707107i
$$117$$ −1.41421 1.41421i −1.41421 1.41421i
$$118$$ 0 0
$$119$$ 0 0
$$120$$ 0 0
$$121$$ 0.707107 + 0.707107i 0.707107 + 0.707107i
$$122$$ −1.70711 + 0.707107i −1.70711 + 0.707107i
$$123$$ 0 0
$$124$$ 0 0
$$125$$ −1.00000 0.414214i −1.00000 0.414214i
$$126$$ 0 0
$$127$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$128$$ −0.707107 + 0.707107i −0.707107 + 0.707107i
$$129$$ 0 0
$$130$$ 1.41421 + 0.585786i 1.41421 + 0.585786i
$$131$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$132$$ 0 0
$$133$$ 0 0
$$134$$ 0 0
$$135$$ 0 0
$$136$$ 1.00000i 1.00000i
$$137$$ 1.41421 1.41421 0.707107 0.707107i $$-0.250000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$138$$ 0 0
$$139$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$140$$ 0 0
$$141$$ 0 0
$$142$$ 0 0
$$143$$ 0 0
$$144$$ −0.707107 + 0.707107i −0.707107 + 0.707107i
$$145$$ 1.00000 1.00000i 1.00000 1.00000i
$$146$$ −0.292893 + 0.707107i −0.292893 + 0.707107i
$$147$$ 0 0
$$148$$ −0.707107 1.70711i −0.707107 1.70711i
$$149$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$150$$ 0 0
$$151$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$152$$ 0 0
$$153$$ 1.00000i 1.00000i
$$154$$ 0 0
$$155$$ 0 0
$$156$$ 0 0
$$157$$ 1.41421i 1.41421i −0.707107 0.707107i $$-0.750000\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$158$$ 0 0
$$159$$ 0 0
$$160$$ 0.292893 0.707107i 0.292893 0.707107i
$$161$$ 0 0
$$162$$ −0.707107 + 0.707107i −0.707107 + 0.707107i
$$163$$ 0 0 0.382683 0.923880i $$-0.375000\pi$$
−0.382683 + 0.923880i $$0.625000\pi$$
$$164$$ −1.70711 0.707107i −1.70711 0.707107i
$$165$$ 0 0
$$166$$ 0 0
$$167$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$168$$ 0 0
$$169$$ −3.00000 −3.00000
$$170$$ 0.292893 + 0.707107i 0.292893 + 0.707107i
$$171$$ 0 0
$$172$$ 0 0
$$173$$ −1.70711 + 0.707107i −1.70711 + 0.707107i −0.707107 + 0.707107i $$0.750000\pi$$
−1.00000 $$\pi$$
$$174$$ 0 0
$$175$$ 0 0
$$176$$ 0 0
$$177$$ 0 0
$$178$$ 0 0
$$179$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$180$$ 0.292893 0.707107i 0.292893 0.707107i
$$181$$ 1.70711 + 0.707107i 1.70711 + 0.707107i 1.00000 $$0$$
0.707107 + 0.707107i $$0.250000\pi$$
$$182$$ 0 0
$$183$$ 0 0
$$184$$ 0 0
$$185$$ 1.00000 + 1.00000i 1.00000 + 1.00000i
$$186$$ 0 0
$$187$$ 0 0
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 0 0
$$191$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$192$$ 0 0
$$193$$ 1.70711 + 0.707107i 1.70711 + 0.707107i 1.00000 $$0$$
0.707107 + 0.707107i $$0.250000\pi$$
$$194$$ −0.292893 + 0.707107i −0.292893 + 0.707107i
$$195$$ 0 0
$$196$$ 0 0
$$197$$ −0.292893 + 0.707107i −0.292893 + 0.707107i 0.707107 + 0.707107i $$0.250000\pi$$
−1.00000 $$\pi$$
$$198$$ 0 0
$$199$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$200$$ 0.414214i 0.414214i
$$201$$ 0 0
$$202$$ 0 0
$$203$$ 0 0
$$204$$ 0 0
$$205$$ 1.41421 1.41421
$$206$$ 0 0
$$207$$ 0 0
$$208$$ 2.00000i 2.00000i
$$209$$ 0 0
$$210$$ 0 0
$$211$$ 0 0 0.382683 0.923880i $$-0.375000\pi$$
−0.382683 + 0.923880i $$0.625000\pi$$
$$212$$ 1.00000 1.00000i 1.00000 1.00000i
$$213$$ 0 0
$$214$$ 0 0
$$215$$ 0 0
$$216$$ 0 0
$$217$$ 0 0
$$218$$ −0.707107 + 0.292893i −0.707107 + 0.292893i
$$219$$ 0 0
$$220$$ 0 0
$$221$$ −1.41421 1.41421i −1.41421 1.41421i
$$222$$ 0 0
$$223$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$224$$ 0 0
$$225$$ 0.414214i 0.414214i
$$226$$ −0.292893 0.707107i −0.292893 0.707107i
$$227$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$228$$ 0 0
$$229$$ −1.00000 + 1.00000i −1.00000 + 1.00000i 1.00000i $$0.5\pi$$
−1.00000 $$\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ 1.70711 + 0.707107i 1.70711 + 0.707107i
$$233$$ −0.292893 0.707107i −0.292893 0.707107i 0.707107 0.707107i $$-0.250000\pi$$
−1.00000 $$\pi$$
$$234$$ 2.00000i 2.00000i
$$235$$ 0 0
$$236$$ 0 0
$$237$$ 0 0
$$238$$ 0 0
$$239$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$240$$ 0 0
$$241$$ −0.707107 + 0.292893i −0.707107 + 0.292893i −0.707107 0.707107i $$-0.750000\pi$$
1.00000i $$0.5\pi$$
$$242$$ 1.00000i 1.00000i
$$243$$ 0 0
$$244$$ 1.70711 + 0.707107i 1.70711 + 0.707107i
$$245$$ 0 0
$$246$$ 0 0
$$247$$ 0 0
$$248$$ 0 0
$$249$$ 0 0
$$250$$ 0.414214 + 1.00000i 0.414214 + 1.00000i
$$251$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$252$$ 0 0
$$253$$ 0 0
$$254$$ 0 0
$$255$$ 0 0
$$256$$ 1.00000 1.00000
$$257$$ 1.00000 + 1.00000i 1.00000 + 1.00000i 1.00000 $$0$$
1.00000i $$0.5\pi$$
$$258$$ 0 0
$$259$$ 0 0
$$260$$ −0.585786 1.41421i −0.585786 1.41421i
$$261$$ 1.70711 + 0.707107i 1.70711 + 0.707107i
$$262$$ 0 0
$$263$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$264$$ 0 0
$$265$$ −0.414214 + 1.00000i −0.414214 + 1.00000i
$$266$$ 0 0
$$267$$ 0 0
$$268$$ 0 0
$$269$$ −0.707107 + 0.292893i −0.707107 + 0.292893i −0.707107 0.707107i $$-0.750000\pi$$
1.00000i $$0.5\pi$$
$$270$$ 0 0
$$271$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$272$$ −0.707107 + 0.707107i −0.707107 + 0.707107i
$$273$$ 0 0
$$274$$ −1.00000 1.00000i −1.00000 1.00000i
$$275$$ 0 0
$$276$$ 0 0
$$277$$ −0.292893 0.707107i −0.292893 0.707107i 0.707107 0.707107i $$-0.250000\pi$$
−1.00000 $$\pi$$
$$278$$ 0 0
$$279$$ 0 0
$$280$$ 0 0
$$281$$ 1.00000 1.00000i 1.00000 1.00000i 1.00000i $$-0.5\pi$$
1.00000 $$0$$
$$282$$ 0 0
$$283$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$284$$ 0 0
$$285$$ 0 0
$$286$$ 0 0
$$287$$ 0 0
$$288$$ 1.00000 1.00000
$$289$$ 1.00000i 1.00000i
$$290$$ −1.41421 −1.41421
$$291$$ 0 0
$$292$$ 0.707107 0.292893i 0.707107 0.292893i
$$293$$ 1.41421i 1.41421i 0.707107 + 0.707107i $$0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$294$$ 0 0
$$295$$ 0 0
$$296$$ −0.707107 + 1.70711i −0.707107 + 1.70711i
$$297$$ 0 0
$$298$$ 0 0
$$299$$ 0 0
$$300$$ 0 0
$$301$$ 0 0
$$302$$ 0 0
$$303$$ 0 0
$$304$$ 0 0
$$305$$ −1.41421 −1.41421
$$306$$ −0.707107 + 0.707107i −0.707107 + 0.707107i
$$307$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$308$$ 0 0
$$309$$ 0 0
$$310$$ 0 0
$$311$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$312$$ 0 0
$$313$$ 0.707107 1.70711i 0.707107 1.70711i 1.00000i $$-0.5\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$314$$ −1.00000 + 1.00000i −1.00000 + 1.00000i
$$315$$ 0 0
$$316$$ 0 0
$$317$$ 1.70711 + 0.707107i 1.70711 + 0.707107i 1.00000 $$0$$
0.707107 + 0.707107i $$0.250000\pi$$
$$318$$ 0 0
$$319$$ 0 0
$$320$$ −0.707107 + 0.292893i −0.707107 + 0.292893i
$$321$$ 0 0
$$322$$ 0 0
$$323$$ 0 0
$$324$$ 1.00000 1.00000
$$325$$ −0.585786 0.585786i −0.585786 0.585786i
$$326$$ 0 0
$$327$$ 0 0
$$328$$ 0.707107 + 1.70711i 0.707107 + 1.70711i
$$329$$ 0 0
$$330$$ 0 0
$$331$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$332$$ 0 0
$$333$$ −0.707107 + 1.70711i −0.707107 + 1.70711i
$$334$$ 0 0
$$335$$ 0 0
$$336$$ 0 0
$$337$$ −0.707107 + 0.292893i −0.707107 + 0.292893i −0.707107 0.707107i $$-0.750000\pi$$
1.00000i $$0.5\pi$$
$$338$$ 2.12132 + 2.12132i 2.12132 + 2.12132i
$$339$$ 0 0
$$340$$ 0.292893 0.707107i 0.292893 0.707107i
$$341$$ 0 0
$$342$$ 0 0
$$343$$ 0 0
$$344$$ 0 0
$$345$$ 0 0
$$346$$ 1.70711 + 0.707107i 1.70711 + 0.707107i
$$347$$ 0 0 0.382683 0.923880i $$-0.375000\pi$$
−0.382683 + 0.923880i $$0.625000\pi$$
$$348$$ 0 0
$$349$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$350$$ 0 0
$$351$$ 0 0
$$352$$ 0 0
$$353$$ 1.41421i 1.41421i 0.707107 + 0.707107i $$0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$354$$ 0 0
$$355$$ 0 0
$$356$$ 0 0
$$357$$ 0 0
$$358$$ 0 0
$$359$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$360$$ −0.707107 + 0.292893i −0.707107 + 0.292893i
$$361$$ 1.00000i 1.00000i
$$362$$ −0.707107 1.70711i −0.707107 1.70711i
$$363$$ 0 0
$$364$$ 0 0
$$365$$ −0.414214 + 0.414214i −0.414214 + 0.414214i
$$366$$ 0 0
$$367$$ 0 0 0.382683 0.923880i $$-0.375000\pi$$
−0.382683 + 0.923880i $$0.625000\pi$$
$$368$$ 0 0
$$369$$ 0.707107 + 1.70711i 0.707107 + 1.70711i
$$370$$ 1.41421i 1.41421i
$$371$$ 0 0
$$372$$ 0 0
$$373$$ −1.41421 −1.41421 −0.707107 0.707107i $$-0.750000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$374$$ 0 0
$$375$$ 0 0
$$376$$ 0 0
$$377$$ 3.41421 1.41421i 3.41421 1.41421i
$$378$$ 0 0
$$379$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$380$$ 0 0
$$381$$ 0 0
$$382$$ 0 0
$$383$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$384$$ 0 0
$$385$$ 0 0
$$386$$ −0.707107 1.70711i −0.707107 1.70711i
$$387$$ 0 0
$$388$$ 0.707107 0.292893i 0.707107 0.292893i
$$389$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$390$$ 0 0
$$391$$ 0 0
$$392$$ 0 0
$$393$$ 0 0
$$394$$ 0.707107 0.292893i 0.707107 0.292893i
$$395$$ 0 0
$$396$$ 0 0
$$397$$ 1.70711 + 0.707107i 1.70711 + 0.707107i 1.00000 $$0$$
0.707107 + 0.707107i $$0.250000\pi$$
$$398$$ 0 0
$$399$$ 0 0
$$400$$ −0.292893 + 0.292893i −0.292893 + 0.292893i
$$401$$ −0.707107 + 1.70711i −0.707107 + 1.70711i 1.00000i $$0.5\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$402$$ 0 0
$$403$$ 0 0
$$404$$ 0 0
$$405$$ −0.707107 + 0.292893i −0.707107 + 0.292893i
$$406$$ 0 0
$$407$$ 0 0
$$408$$ 0 0
$$409$$ −1.41421 −1.41421 −0.707107 0.707107i $$-0.750000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$410$$ −1.00000 1.00000i −1.00000 1.00000i
$$411$$ 0 0
$$412$$ 0 0
$$413$$ 0 0
$$414$$ 0 0
$$415$$ 0 0
$$416$$ 1.41421 1.41421i 1.41421 1.41421i
$$417$$ 0 0
$$418$$ 0 0
$$419$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$420$$ 0 0
$$421$$ 1.41421i 1.41421i −0.707107 0.707107i $$-0.750000\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$422$$ 0 0
$$423$$ 0 0
$$424$$ −1.41421 −1.41421
$$425$$ 0.414214i 0.414214i
$$426$$ 0 0
$$427$$ 0 0
$$428$$ 0 0
$$429$$ 0 0
$$430$$ 0 0
$$431$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$432$$ 0 0
$$433$$ −1.00000 + 1.00000i −1.00000 + 1.00000i 1.00000i $$0.5\pi$$
−1.00000 $$\pi$$
$$434$$ 0 0
$$435$$ 0 0
$$436$$ 0.707107 + 0.292893i 0.707107 + 0.292893i
$$437$$ 0 0
$$438$$ 0 0
$$439$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$440$$ 0 0
$$441$$ 0 0
$$442$$ 2.00000i 2.00000i
$$443$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$444$$ 0 0
$$445$$ 0 0
$$446$$ 0 0
$$447$$ 0 0
$$448$$ 0 0
$$449$$ 0.707107 1.70711i 0.707107 1.70711i 1.00000i $$-0.5\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$450$$ −0.292893 + 0.292893i −0.292893 + 0.292893i
$$451$$ 0 0
$$452$$ −0.292893 + 0.707107i −0.292893 + 0.707107i
$$453$$ 0 0
$$454$$ 0 0
$$455$$ 0 0
$$456$$ 0 0
$$457$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$458$$ 1.41421 1.41421
$$459$$ 0 0
$$460$$ 0 0
$$461$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$462$$ 0 0
$$463$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$464$$ −0.707107 1.70711i −0.707107 1.70711i
$$465$$ 0 0
$$466$$ −0.292893 + 0.707107i −0.292893 + 0.707107i
$$467$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$468$$ 1.41421 1.41421i 1.41421 1.41421i
$$469$$ 0 0
$$470$$ 0 0
$$471$$ 0 0
$$472$$ 0 0
$$473$$ 0 0
$$474$$ 0 0
$$475$$ 0 0
$$476$$ 0 0
$$477$$ −1.41421 −1.41421
$$478$$ 0 0
$$479$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$480$$ 0 0
$$481$$ 1.41421 + 3.41421i 1.41421 + 3.41421i
$$482$$ 0.707107 + 0.292893i 0.707107 + 0.292893i
$$483$$ 0 0
$$484$$ −0.707107 + 0.707107i −0.707107 + 0.707107i
$$485$$ −0.414214 + 0.414214i −0.414214 + 0.414214i
$$486$$ 0 0
$$487$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$488$$ −0.707107 1.70711i −0.707107 1.70711i
$$489$$ 0 0
$$490$$ 0 0
$$491$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$492$$ 0 0
$$493$$ 1.70711 + 0.707107i 1.70711 + 0.707107i
$$494$$ 0 0
$$495$$ 0 0
$$496$$ 0 0
$$497$$ 0 0
$$498$$ 0 0
$$499$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$500$$ 0.414214 1.00000i 0.414214 1.00000i
$$501$$ 0 0
$$502$$ 0 0
$$503$$ 0 0 0.382683 0.923880i $$-0.375000\pi$$
−0.382683 + 0.923880i $$0.625000\pi$$
$$504$$ 0 0
$$505$$ 0 0
$$506$$ 0 0
$$507$$ 0 0
$$508$$ 0 0
$$509$$ −1.41421 −1.41421 −0.707107 0.707107i $$-0.750000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$510$$ 0 0
$$511$$ 0 0
$$512$$ −0.707107 0.707107i −0.707107 0.707107i
$$513$$ 0 0
$$514$$ 1.41421i 1.41421i
$$515$$ 0 0
$$516$$ 0 0
$$517$$ 0 0
$$518$$ 0 0
$$519$$ 0 0
$$520$$ −0.585786 + 1.41421i −0.585786 + 1.41421i
$$521$$ 1.70711 + 0.707107i 1.70711 + 0.707107i 1.00000 $$0$$
0.707107 + 0.707107i $$0.250000\pi$$
$$522$$ −0.707107 1.70711i −0.707107 1.70711i
$$523$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$524$$ 0 0
$$525$$ 0 0
$$526$$ 0 0
$$527$$ 0 0
$$528$$ 0 0
$$529$$ 0.707107 + 0.707107i 0.707107 + 0.707107i
$$530$$ 1.00000 0.414214i 1.00000 0.414214i
$$531$$ 0 0
$$532$$ 0 0
$$533$$ 3.41421 + 1.41421i 3.41421 + 1.41421i
$$534$$ 0 0
$$535$$ 0 0
$$536$$ 0 0
$$537$$ 0 0
$$538$$ 0.707107 + 0.292893i 0.707107 + 0.292893i
$$539$$ 0 0
$$540$$ 0 0
$$541$$ 0.707107 0.292893i 0.707107 0.292893i 1.00000i $$-0.5\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$542$$ 0 0
$$543$$ 0 0
$$544$$ 1.00000 1.00000
$$545$$ −0.585786 −0.585786
$$546$$ 0 0
$$547$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$548$$ 1.41421i 1.41421i
$$549$$ −0.707107 1.70711i −0.707107 1.70711i
$$550$$ 0 0
$$551$$ 0 0
$$552$$ 0 0
$$553$$ 0 0
$$554$$ −0.292893 + 0.707107i −0.292893 + 0.707107i
$$555$$ 0 0
$$556$$ 0 0
$$557$$ 1.41421i 1.41421i −0.707107 0.707107i $$-0.750000\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$558$$ 0 0
$$559$$ 0 0
$$560$$ 0 0
$$561$$ 0 0
$$562$$ −1.41421 −1.41421
$$563$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$564$$ 0 0
$$565$$ 0.585786i 0.585786i
$$566$$ 0 0
$$567$$ 0 0
$$568$$ 0 0
$$569$$ 1.00000 1.00000i 1.00000 1.00000i 1.00000i $$-0.5\pi$$
1.00000 $$0$$
$$570$$ 0 0
$$571$$ 0 0 0.382683 0.923880i $$-0.375000\pi$$
−0.382683 + 0.923880i $$0.625000\pi$$
$$572$$ 0 0
$$573$$ 0 0
$$574$$ 0 0
$$575$$ 0 0
$$576$$ −0.707107 0.707107i −0.707107 0.707107i
$$577$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$578$$ −0.707107 + 0.707107i −0.707107 + 0.707107i
$$579$$ 0 0
$$580$$ 1.00000 + 1.00000i 1.00000 + 1.00000i
$$581$$ 0 0
$$582$$ 0 0
$$583$$ 0 0
$$584$$ −0.707107 0.292893i −0.707107 0.292893i
$$585$$ −0.585786 + 1.41421i −0.585786 + 1.41421i
$$586$$ 1.00000 1.00000i 1.00000 1.00000i
$$587$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$588$$ 0 0
$$589$$ 0 0
$$590$$ 0 0
$$591$$ 0 0
$$592$$ 1.70711 0.707107i 1.70711 0.707107i
$$593$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ 0 0
$$597$$ 0 0
$$598$$ 0 0
$$599$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$600$$ 0 0
$$601$$ 1.70711 + 0.707107i 1.70711 + 0.707107i 1.00000 $$0$$
0.707107 + 0.707107i $$0.250000\pi$$
$$602$$ 0 0
$$603$$ 0 0
$$604$$ 0 0
$$605$$ 0.292893 0.707107i 0.292893 0.707107i
$$606$$ 0 0
$$607$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$608$$ 0 0
$$609$$ 0 0
$$610$$ 1.00000 + 1.00000i 1.00000 + 1.00000i
$$611$$ 0 0
$$612$$ 1.00000 1.00000
$$613$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$614$$ 0 0
$$615$$ 0 0
$$616$$ 0 0
$$617$$ 0.292893 + 0.707107i 0.292893 + 0.707107i 1.00000 $$0$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$618$$ 0 0
$$619$$ 0 0 0.382683 0.923880i $$-0.375000\pi$$
−0.382683 + 0.923880i $$0.625000\pi$$
$$620$$ 0 0
$$621$$ 0 0
$$622$$ 0 0
$$623$$ 0 0
$$624$$ 0 0
$$625$$ 0.414214i 0.414214i
$$626$$ −1.70711 + 0.707107i −1.70711 + 0.707107i
$$627$$ 0 0
$$628$$ 1.41421 1.41421
$$629$$ −0.707107 + 1.70711i −0.707107 + 1.70711i
$$630$$ 0 0
$$631$$ 0 0 −0.707107 0.707107i $$-0.750000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$632$$ 0 0
$$633$$ 0 0
$$634$$ −0.707107 1.70711i −0.707107 1.70711i
$$635$$ 0 0
$$636$$ 0 0
$$637$$ 0 0
$$638$$ 0 0
$$639$$ 0 0
$$640$$ 0.707107 + 0.292893i 0.707107 + 0.292893i
$$641$$ −0.292893 0.707107i −0.292893 0.707107i 0.707107 0.707107i $$-0.250000\pi$$
−1.00000 $$\pi$$
$$642$$ 0 0
$$643$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$644$$ 0 0
$$645$$ 0 0
$$646$$ 0 0
$$647$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$648$$ −0.707107 0.707107i −0.707107 0.707107i
$$649$$ 0 0
$$650$$ 0.828427i 0.828427i
$$651$$ 0 0
$$652$$ 0 0
$$653$$ 0.707107 1.70711i 0.707107 1.70711i 1.00000i $$-0.5\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$654$$ 0 0
$$655$$ 0 0
$$656$$ 0.707107 1.70711i 0.707107 1.70711i
$$657$$ −0.707107 0.292893i −0.707107 0.292893i
$$658$$ 0 0
$$659$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$660$$ 0 0
$$661$$ 1.41421 + 1.41421i 1.41421 + 1.41421i 0.707107 + 0.707107i $$0.250000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$662$$ 0 0
$$663$$ 0 0
$$664$$ 0 0
$$665$$ 0 0
$$666$$ 1.70711 0.707107i 1.70711 0.707107i
$$667$$ 0 0
$$668$$ 0 0
$$669$$ 0 0
$$670$$ 0 0
$$671$$ 0 0
$$672$$ 0 0
$$673$$ 0.292893 0.707107i 0.292893 0.707107i −0.707107 0.707107i $$-0.750000\pi$$
1.00000 $$0$$
$$674$$ 0.707107 + 0.292893i 0.707107 + 0.292893i
$$675$$ 0 0
$$676$$ 3.00000i 3.00000i
$$677$$ 0.707107 0.292893i 0.707107 0.292893i 1.00000i $$-0.5\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$678$$ 0 0
$$679$$ 0 0
$$680$$ −0.707107 + 0.292893i −0.707107 + 0.292893i
$$681$$ 0 0
$$682$$ 0 0
$$683$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$684$$ 0 0
$$685$$ −0.414214 1.00000i −0.414214 1.00000i
$$686$$ 0 0
$$687$$ 0 0
$$688$$ 0 0
$$689$$ −2.00000 + 2.00000i −2.00000 + 2.00000i
$$690$$ 0 0
$$691$$ 0 0 −0.923880 0.382683i $$-0.875000\pi$$
0.923880 + 0.382683i $$0.125000\pi$$
$$692$$ −0.707107 1.70711i −0.707107 1.70711i
$$693$$ 0 0
$$694$$ 0 0
$$695$$ 0 0
$$696$$ 0 0
$$697$$ 0.707107 + 1.70711i 0.707107 + 1.70711i
$$698$$ 0 0
$$699$$ 0 0
$$700$$ 0 0
$$701$$ 1.41421i 1.41421i −0.707107 0.707107i $$-0.750000\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$702$$ 0 0
$$703$$ 0 0
$$704$$ 0 0
$$705$$ 0 0
$$706$$ 1.00000 1.00000i 1.00000 1.00000i
$$707$$ 0 0
$$708$$ 0 0
$$709$$ 0.707107 + 1.70711i 0.707107 + 1.70711i 0.707107 + 0.707107i $$0.250000\pi$$
1.00000i $$0.5\pi$$
$$710$$ 0 0
$$711$$ 0 0
$$712$$ 0 0
$$713$$ 0 0
$$714$$ 0 0
$$715$$ 0 0
$$716$$ 0 0
$$717$$ 0 0
$$718$$ 0 0
$$719$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$720$$ 0.707107 + 0.292893i 0.707107 + 0.292893i
$$721$$ 0 0
$$722$$ 0.707107 0.707107i 0.707107 0.707107i
$$723$$ 0 0
$$724$$ −0.707107 + 1.70711i −0.707107 + 1.70711i
$$725$$ 0.707107 + 0.292893i 0.707107 + 0.292893i
$$726$$ 0 0
$$727$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$728$$ 0 0
$$729$$ −0.707107 0.707107i −0.707107 0.707107i
$$730$$ 0.585786 0.585786
$$731$$ 0 0
$$732$$ 0 0
$$733$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$734$$ 0 0
$$735$$ 0 0
$$736$$ 0 0
$$737$$ 0 0
$$738$$ 0.707107 1.70711i 0.707107 1.70711i
$$739$$ 0 0 0.707107 0.707107i $$-0.250000\pi$$
−0.707107 + 0.707107i $$0.750000\pi$$
$$740$$ −1.00000 + 1.00000i −1.00000 + 1.00000i
$$741$$ 0 0
$$742$$ 0 0
$$743$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$744$$ 0 0
$$745$$ 0 0
$$746$$ 1.00000 + 1.00000i 1.00000 + 1.00000i
$$747$$ 0 0
$$748$$ 0 0
$$749$$ 0 0
$$750$$ 0 0
$$751$$ 0 0 0.923880 0.382683i $$-0.125000\pi$$
−0.923880 + 0.382683i $$0.875000\pi$$
$$752$$ 0 0
$$753$$ 0 0
$$754$$ −3.41421 1.41421i −3.41421 1.41421i
$$755$$ 0 0
$$756$$ 0 0
$$757$$ 1.41421 1.41421i 1.41421 1.41421i 0.707107 0.707107i $$-0.250000\pi$$
0.707107 0.707107i $$-0.250000\pi$$
$$758$$ 0 0
$$759$$ 0 0
$$760$$ 0 0
$$761$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$762$$ 0 0
$$763$$ 0 0
$$764$$ 0 0
$$765$$ −0.707107 + 0.292893i −0.707107 + 0.292893i
$$766$$ 0 0
$$767$$ 0 0
$$768$$ 0 0
$$769$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$770$$ 0 0
$$771$$ 0 0
$$772$$ −0.707107 + 1.70711i −0.707107 + 1.70711i
$$773$$ 1.00000 1.00000i 1.00000 1.00000i 1.00000i $$-0.5\pi$$
1.00000 $$0$$
$$774$$ 0 0
$$775$$ 0 0
$$776$$ −0.707107 0.292893i −0.707107 0.292893i
$$777$$ 0 0
$$778$$ 0 0
$$779$$ 0 0
$$780$$ 0 0
$$781$$ 0 0
$$782$$ 0 0
$$783$$ 0 0
$$784$$ 0 0
$$785$$ −1.00000 + 0.414214i −1.00000 + 0.414214i
$$786$$ 0 0
$$787$$ 0 0 −0.382683 0.923880i $$-0.625000\pi$$
0.382683 + 0.923880i $$0.375000\pi$$
$$788$$ −0.707107 0.292893i −0.707107 0.292893i
$$789$$ 0 0
$$790$$ 0 0
$$791$$ 0 0
$$792$$ 0 0
$$793$$ −3.41421 1.41421i −3.41421 1.41421i
$$794$$ −0.707107 1.70711i −0.707107 1.70711i
$$795$$ 0 0
$$796$$ 0 0
$$797$$ 1.41421 + 1.41421i 1.41421 + 1.41421i 0.707107 + 0.707107i $$0.250000\pi$$
0.707107 + 0.707107i $$0.250000\pi$$
$$798$$ 0 0
$$799$$ 0 0
$$800$$ 0.414214 0.414214 | 16,731 | 29,106 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.475711 |
https://www.bartleby.com/solution-answer/chapter-22-problem-30e-single-variable-calculus-concepts-and-contexts-enhanced-edition-4th-edition/9781337687805/917649d4-858a-4e04-9305-bdfbc297f20a | 1,632,200,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00542.warc.gz | 692,204,690 | 56,378 | # To evaluate the given function for given values.
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 2.2, Problem 30E
a.
To determine
## To evaluate the given function for given values.
Expert Solution
h(1)=0.557408h(0.5)=0.37042h(0.1)=0.334672h(0.05)=0.333667h(0.01)=0.333347h(0.05)=0.333337
### Explanation of Solution
Given:
The given function is h(x)=tanxxx3
Calculation:
Substituting the following values in the given function as follows-
h(x)=tanxxx3h(1)=0.557408h(0.5)=0.37042h(0.1)=0.334672h(0.05)=0.333667h(0.01)=0.333347h(0.05)=0.333337
b.
To determine
Expert Solution
13
### Explanation of Solution
Given:
The given function is limx0tanxxx3
Calculation:
limx0tanxxx3limx0tanxxx3=0.33333=13
c.
To determine
Expert Solution
No.
### Explanation of Solution
Given:
The given function is limx0tanxxx3
Calculation:
The following is the table of all smaller values-
The guess in part (b) does not seem to be correct.
d.
To determine
Expert Solution
### Explanation of Solution
Given:
The given function is limx0tanxxx3
Calculation:
The following are the graphs-
In part (C), it was observed that limx0tanxxx3=0
From the above graphs the limit does not approach to a specific value, hence the limit does not exist.
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! | 480 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-39 | latest | en | 0.683796 |
https://en.1answer.info/6d617468656d6174696361-7a313539383336 | 1,542,594,902,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745015.71/warc/CC-MAIN-20181119023120-20181119045120-00192.warc.gz | 615,809,660 | 13,377 | # Fitting Simultaneous Equations involving arbitrary function
SPIL just a moment. 0 answers, 0 views
n = {1, 2, 3, 4, 5, 4, 4, 4, 4, 4, 4, 4, 3, 3, 4, 2, 3}
I guessed a function of n. This function is subject to change, so please suggested a different function if required (I can then reproduce "fitFEqnToZero" list, which is written below further down and edit the post).
fC[x_] := a^b*Log[c*x^d] + e
a,b,c,d, and e are arbitrary variables. So please reduce or increase the number of these.
fcForFit = fC /@ n
{e + a^b Log[c], e + a^b Log[2^d c], e + a^b Log[3^d c],
e + a^b Log[4^d c], e + a^b Log[5^d c], e + a^b Log[4^d c],
e + a^b Log[4^d c], e + a^b Log[4^d c], e + a^b Log[4^d c],
e + a^b Log[4^d c], e + a^b Log[4^d c], e + a^b Log[4^d c],
e + a^b Log[3^d c], e + a^b Log[3^d c], e + a^b Log[4^d c],
e + a^b Log[2^d c], e + a^b Log[3^d c]}
I used fcForFit to construct the following list of equations (if you have a different function, I can construct again).
fitFEqnToZero = {-603 + 13.36 qB (e + a^b Log[c]), -940 +
22 qB (e + a^b Log[2^d c]), -1401+
22 qB (e + a^b Log[3^d c]) +
13 qT (e + a^b Log[3^d c]), -1378 +
22 qB (e + a^b Log[4^d c]) +
22 qT (e + a^b Log[4^d c]), -1431 +
33 qB (e + a^b Log[5^d c]) +
22 qT (e + a^b Log[5^d c]), -1649 +
11 qB (e + a^b Log[4^d c]) +
33 qT (e + a^b Log[4^d c]), -1907 +
11 qB (e + a^b Log[4^d c]) +
33 qT (e + a^b Log[4^d c]), -1718 +
11 qB (e + a^b Log[4^d c]) +
33 qT (e + a^b Log[4^d c]), -1718 +
11 qB (e + a^b Log[4^d c]) +
33 qT (e + a^b Log[4^d c]), -1770 +
11 qB (e + a^b Log[4^d c]) +
33 qT (e + a^b Log[4^d c]), -1587 +
11 qB (e + a^b Log[4^d c]) +
33 qT (e + a^b Log[4^d c]), -1640 +
11 qB (e + a^b Log[4^d c]) +
33 qT (e + a^b Log[4^d c]), -1339 +
13 qB (e + a^b Log[3^d c]) +
22 qT (e + a^b Log[3^d c]), -1297 +
22 qB (e + a^b Log[3^d c]) +
11 qT (e + a^b Log[3^d c]), -1501 +
19 qB (e + a^b Log[4^d c]) +
20 qT (e + a^b Log[4^d c]), -1574 +
8 qB (e + a^b Log[2^d c]) +
13 qT (e + a^b Log[2^d c]), -1297 +
33 qT (e + a^b Log[3^d c])}
Now I want to set fitFEqnToZero[[1]], fitFEqnToZero[[2]]...fitFEqnToZero[[Length[fitFEqnToZero]]]... all to zero and fit simultaneously for the following variables
{qB, a, b, c, d, e, qT}
with the following constraints
fcoil[1] == 1 && qB <= 2000 && qT <= 2000 && fcoil[2] <= 1 &&
fcoil[3] <= 1 && fcoil[4] <= 1 &&
fcoil[5] <= 1},
How can I do this? Is Nsolve, NMinimize or other the most appropriate?
If I use NMinimize should I fit to Total[fitFEqnToZero] == 0? Is this effective the same or could this result in some of the elements in list fitFEqnToZero being non-zero? | 1,184 | 2,591 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-47 | latest | en | 0.455113 |
http://www.physicsclassroom.com/calcpad/light/prob17.cfm | 1,496,142,301,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463615093.77/warc/CC-MAIN-20170530105157-20170530125157-00329.warc.gz | 753,372,809 | 15,064 | # Waves, Sound and Light: Light Waves
## Light Waves: Audio Guided Solution
#### Problem 17:
Jackson and Melanie are doing the Young’s Experiment Lab using a red laser pen and a slide with two slits spaced 25 micrometers apart. They project the interference pattern onto a whiteboard located 2.35 m from the slits. They measure the distance from the 3rd bright band on opposite sides of the pattern to be separated by 37 cm. Based on these measurements, what is the wavelength of the red laser light (in nanometers)? (GIVEN: 1 m = 106 mm, 1 m = 109 nm)
## Audio Guided Solution
Click to show or hide the answer!
## Habits of an Effective Problem Solver
• Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
• Identify the known and unknown quantities and record in an organized manner, often times they can be recorded on the diagram itself. Equate given values to the symbols used to represent the corresponding quantity (e.g., v = 3.00x108 m/s, λ = 554 nm, f = ???).
• Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
• Identify the appropriate formula(s) to use.
• Perform substitutions and algebraic manipulations in order to solve for the unknown quantity. | 302 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-22 | longest | en | 0.892062 |
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