url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://nrich.maths.org/7176/solution | 1,632,659,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00109.warc.gz | 453,539,088 | 4,510 | N000ughty Thoughts
How many noughts are at the end of these giant numbers?
DOTS Division
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
Mod 3
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
Indivisible
Age 14 to 16 ShortChallenge Level
Lining up in rows of 3:
Lining up in rows of each number:
What comes next in each pattern?
Next number is a multiple of 3, 4, 5, 6
Lowest common multiples of 3, 4, 5, 6: 60, 120, 180, etc.
Fewer than 100 students $\therefore$ there are 59 students.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection. | 249 | 817 | {"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} | 3.390625 | 3 | CC-MAIN-2021-39 | latest | en | 0.876167 |
https://www.weegy.com/Home.aspx?ConversationId=0031685E | 1,537,362,167,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156224.9/warc/CC-MAIN-20180919122227-20180919142227-00511.warc.gz | 917,797,756 | 9,058 | -2x>1/9
Question
Updated 3/7/2014 9:26:02 PM
Flagged by yeswey [3/7/2014 9:13:12 PM], Unflagged by yeswey [3/7/2014 9:13:28 PM]
Original conversation
User: -2x>1/9
Weegy: What kind of math problem is this please?
kikaycore|Points 2460|
User: solve
Weegy: solve (slv, s?lv) v. solved, solv?ing, solves v.tr. 1. To find a solution to. 2. To work out a correct solution to (a problem).
sipichapie|Points 2224|
User: solve -2x>1/9
Weegy: What kind of math problem is this please?
kikaycore|Points 2460|
User: -2x=1/9
Weegy: What kind of math problem is this please?
kikaycore|Points 2460|
User: algebra
Weegy: Do you have an algebra question?
sharpies|Points 5754|
User: yes
Weegy: I am not sure what you are referring to, is "yes" a reply to a previous answer you received?
farawayspices|Points 10|
User: yes
Weegy: I am not sure what you are referring to, is "yes" a reply to a previous answer you received?
farawayspices|Points 10|
User: algebra problem
Weegy: Do you have an algebra question?
sharpies|Points 5754|
User: yes
Weegy: I am not sure what you are referring to, is "yes" a reply to a previous answer you received?
farawayspices|Points 10|
User: yes
Weegy: I am not sure what you are referring to, is "yes" a reply to a previous answer you received?
farawayspices|Points 10|
User: solve -2x>1/9
Weegy: What kind of math problem is this please?
kikaycore|Points 2460|
User: 8x-(2x+5)=43
Weegy: 8x-(2x+5)=43 = 6x-48=0 x=8
User: 5/2x+1/4x=11/4x
Question
Updated 3/7/2014 9:26:02 PM
Flagged by yeswey [3/7/2014 9:13:12 PM], Unflagged by yeswey [3/7/2014 9:13:28 PM]
Rating
3
-2x>1/9
2x < -1/9
x < -1/18
Confirmed by andrewpallarca [3/8/2014 2:37:08 PM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,217 | 3,413 | {"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} | 3.078125 | 3 | CC-MAIN-2018-39 | latest | en | 0.886262 |
http://www.chegg.com/homework-help/questions-and-answers/clock-placed-satellite-orbits-theearth-period-120-min-radius805535-106-m-time-interval-clo-q1256372 | 1,394,270,510,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999654282/warc/CC-MAIN-20140305060734-00034-ip-10-183-142-35.ec2.internal.warc.gz | 278,569,206 | 7,187 | # Clock and Orbit Relativity
0 pts ended
A clock is placed in a satellite that orbits the
earth with a period of 120 min at the radius
8.05535 × 106 m.
By what time interval will this clock dif-
fer from an identical clock on the earth af-
ter 1 y? Assume that special relativity ap-
plies and neglect general relativity. There are
365.25 days in a year. A useful approximation
is sqrt(1-x)~1-.5x for x<<<1. | 113 | 408 | {"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-2014-10 | latest | en | 0.845971 |
https://www.redhotpawn.com/forum/posers-and-puzzles/water-pinwheel.103562 | 1,561,637,191,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001138.93/warc/CC-MAIN-20190627115818-20190627141818-00083.warc.gz | 867,714,399 | 6,755 | # Water pinwheel
PBE6
Posers and Puzzles 12 Nov '08 21:32
1. PBE6
Bananarama
12 Nov '08 21:32
I remember reading about this problem in noted theoretical physicist Richard Feynman's book "Surely You're Joking, Mr. Feynman!", but I don't think he included the answer! Instead of going to the library to check, I thought I'd ask everyone here. 😀
Consider a hollow tube with two smooth right-angled bends in it, shaped roughly like a squared off "S" like this:
_
_|
This tube is connected to a hose which attaches right in the middle of the tube, and the tube if free to rotate about its centre. The tube is submerged in a tank of water, and the hose is turned on. The water shooting out of both ends of the tube will push back on the tube making it rotate opposite to the direction of the water (i.e. if you are looking down at the tube and water is shooting out to the top-right and bottom-left, the tube will rotate counterclockwise).
The question Mr. Feynman tried to answer (with spectacularly funny results, but no answer) was: what happens if you run it in reverse, and start sucking water into the tube instead?
2. sonhouse
Fast and Curious
13 Nov '08 05:03
Originally posted by PBE6
I remember reading about this problem in noted theoretical physicist Richard Feynman's book "Surely You're Joking, Mr. Feynman!", but I don't think he included the answer! Instead of going to the library to check, I thought I'd ask everyone here. 😀
Consider a hollow tube with two smooth right-angled bends in it, shaped roughly like a squared off "S" like t ...[text shortened]... s: what happens if you run it in reverse, and start sucking water into the tube instead?
You empty the tank?
3. AThousandYoung
13 Nov '08 05:12
Originally posted by PBE6
I remember reading about this problem in noted theoretical physicist Richard Feynman's book "Surely You're Joking, Mr. Feynman!", but I don't think he included the answer! Instead of going to the library to check, I thought I'd ask everyone here. 😀
Consider a hollow tube with two smooth right-angled bends in it, shaped roughly like a squared off "S" like t ...[text shortened]... s: what happens if you run it in reverse, and start sucking water into the tube instead?
It should spin the other way I would think. Why not?
4. PBE6
Bananarama
13 Nov '08 18:22
Originally posted by AThousandYoung
It should spin the other way I would think. Why not?
Apparently, all three possible answers (same direction, opposite direction, doesn't move) have all been discussed and "demonstrated" to some degree.
http://en.wikipedia.org/wiki/Feynman_sprinkler | 668 | 2,600 | {"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} | 3.546875 | 4 | CC-MAIN-2019-26 | latest | en | 0.958533 |
https://publiclab.org/notes/bhickman/04-21-2014/wheetrometer-2-0 | 1,721,556,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00341.warc.gz | 415,031,601 | 25,048 | # WheeTrometer 2.0
by bhickman | 21 Apr 04:11
WheeTrometer 2.0
## What we want to do:
We want to provide an affordable, open source spectrophotometer for measuring absorption of visible light by solutions. The goal is to have 1 nm wavelength resolution and 12 bit intensity resolution.
## Background:
A description of our first spectrophotometer can be found here. The wavelength resolution of this instrument was not as good as we hoped, which we believe was due to no focusing element (mirror or lenses) after the diffraction grating. In this report, we have also altered the arrangement of the grating, collimating lens, and slit after a learning a little more about optics.
## Attempt and results:
The spectrophotometer is housed in a project box. It uses a wooded base as the optical platform, a wooden cuvette holder, a slit made of two razor blades, a collimating lens, diffraction grating (1000 lines/mm), a focusing mirror, and a TSL1406R 768 X 1 element photodiode array (pda). Data from the pda is read by a Tiva C series LaunchPad and sent to a computer. A general schematic of the spectrometer is shown below.
In the figure above f1 is the focal length of the collimating lens (27 mm) and f2 is the focal length of the focusing mirror (50 mm). By placing the slit a distance of f1 away from the collimating lens, the image of the slit is focused at infinity (i.e. collimates the light). Also by placing the PDA a distance of f2 away from the focusing mirror, the diverging light from the diffraction grating is focused at a position on the PDA dependent upon the angel it is diffracted.
A couple key equations for the alignment of the spectrometer are:
d sin(θ)=mλ (1)
h= ftan(α) (2)
In Equation 1, d is the width between the slits on the diffraction grating, θ is the angle of diffraction, m is the order of diffraction, and λ is the wavelength being diffracted. This equation allows calculation of the angle and distance of the focusing mirror from the diffraction grating so that only the wavelengths of interest are falling on the PDA.
Equation 2 is a general concave mirror equation. In Equation 2 h is the height of an image above the focal point, f is the focal length of the mirror and α is the angle of deviation from normal from the mirror (Shown below).
With these two equations and a little trig (maybe a lot…) the grating, focusing mirror, and PDA can be aligned so that light of the desired wavelengths are focused on the PDA.
A few picture of the spectrometer are shown below.
As shown below, a hole was drilled in the top of the housing to allow insertion of a cuvette. A second hole (in the back side) allows light to enter the spectrometer. In the previous design, the light source was inside the housing. This new configuration allows for calibration using a fluorescent light, then operation using a second light source, such as a halogen lamp. Wires connecting the PDA to the LaunchPad exit via a third hole.
The cuvette holder, slit, collimating lens, and grating were mounted on a block of wood. The focusing mirror and PDA were mounted directly to the bottom piece of the project box.
The cuvette holder (shown below) is a block of wood with two perpendicular holes, one to accept the cuvette and the other to allow light through.
The slit was made from two razor blades mounted on a block with a hole drilled through it. The razor blades were screwed to the block, aligned, and then epoxied to keep them in place. To allow adjustment of the slit position, the block is held in place by clamping it to the wooden platform using a piece of plastic with screws on each side. The collimating lens and grating are both mounted to a third block of wood. A hole slightly smaller than a collimating lens was drilled in this block and the lens fitted so that its edge was even with the edge of the block. The grating was simply glued to the other end of the block.
The focusing mirror and PDA were mounted using similar methods. The mirror was first mounted on a piece of metal using epoxy glue as shown below.
The metal plate was mounted to a piece of plastic with four screws and the plastic was mounted on an L bracket with three screws. Each of these screws had a spring spacer between the L bracket and the plastic so that the angel of the mirror could be adjusted. The other arm of the L bracket was bolted to the bottom of the project box.
The PDA was mounted similarly. First the PDA was mounted to a piece of plastic with two bolts, using spacers to keep a gap for the electronics. The plastic was then mounted to an L bracket which was then bolted to the bottom of the project box.
With the current design I have been able to get ~5 to 10 nm resolution. The screen shot below illustrates the performance of the spectrometer, tested using a fluorescent light. As you can see the two peaks at approx. 542 nm and 546 nm (mercury and terbium according to Wikipedia) are not resolved.
## Questions and Next Steps:
• Use cuvette holder made by Dr. Summers (Actually fits a cuvette snugly) • Improve slit design. One of the major contributors to decreased resolution • Improve alignment of optics to further maximize resolution • Use a 2048 element PDA rather than the current 768 element PDA. This PDA is cheaper and may improve wavelength resolution • Write up the math for figuring out the alignment of the optics • Clean up the code and post on GitHub
## Acknowledgements:
Dr. Jack Summers of Western Carolina University
This is a neat project-- you might try one of our photo-printed slits on acetate to replace the razor blades. if you need custom sizes, we use the services of Camera Graphics. If your order is too small I can stick it in a batch to make it more reasonable.
Thanks for the tip. I'm going to order some and try them out.
Super cool project, thanks for sharing your work and results, very encouraging. Couple of questions, why the razor blades for slits, is the depth aspect important or would a laser cut slit in a 2D sheet material work well? Where can you buy the 2048 pixel sensor?
Thanks,
J
Is this a question? Click here to post it to the Questions page. | 1,395 | 6,152 | {"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} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.929368 |
https://technodocbox.com/C_and_CPP/122814769-Logic-design-dr-mahmoud-abo_elfetouh.html | 1,623,757,994,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00414.warc.gz | 501,604,287 | 26,238 | # LOGIC DESIGN. Dr. Mahmoud Abo_elfetouh
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1 LOGIC DESIGN Dr. Mahmoud Abo_elfetouh
2 Course objectives This course provides you with a basic understanding of what digital devices are, how they operate, and how they can be designed to perform useful functions. The course is intended to give you an understanding of Binary systems, Boolean algebra, digital design techniques, logic gates, logic minimization, standard combinational circuits, sequential circuits, flip-flops, synthesis of synchronous sequential circuits, and arithmetic circuits.
3 Contents Week No. Topic Lecture Practical Total 1 Number Systems and Codes Number Systems and Codes Boolean Algebra and Logic Simplification Minimization Techniques- Karnaugh Map Minimization Techniques- Karnaugh Map Logic Gates Arithmetic Circuits-Adders Mid- Term Exam
4 Contents Week No. Topic Lecture Practical Total 9 Arithmetic Circuits -Subtracter Combinational Circuits Combinational Circuits Flip-Flops Flip-Flops Counters - Registers Memory Devices Final- Term Exam
5 Textbook Logic and Computer Design Fundamentals, 4th Edition by M. Morris Mano and Charles R. Kime, Prentice Hall, 2008
6 Chapter 1 Number Systems
7 1. Number Systems Location in course textbook Chapt. 1
8 Common Number Systems System Base Symbols Used by humans? Used in computers? Decimal 10 0, 1, 9 Yes No Binary 2 0, 1 No Yes Octal 8 0, 1, 7 No No Hexadecimal 16 0, 1, 9, A, B, F No No
9 Quantities/Counting (1 of 3) Decimal Binary Octal Hexadecimal p. 33
10 Quantities/Counting (2 of 3) Decimal Binary Octal Hexadecimal A B C D E F
11 Quantities/Counting (3 of 3) Decimal Binary Octal Hexadecimal Etc.
12 Conversion Among Bases The possibilities: Decimal Octal Binary Hexadecimal pp
13 Quick Example = = 31 8 = Base
14 Decimal to Decimal (just for fun) Decimal Octal Binary Hexadecimal Next slide
15 Weight => 5 x 10 0 = 5 2 x 10 1 = 20 1 x 10 2 = Base
16 Binary to Decimal Decimal Octal Binary Hexadecimal
17 Binary to Decimal Technique Multiply each bit by 2 n, where n is the weight of the bit The weight is the position of the bit, starting from 0 on the right Add the results
18 Example Bit => 1 x 2 0 = 1 1 x 2 1 = 2 0 x 2 2 = 0 1 x 2 3 = 8 0 x 2 4 = 0 1 x 2 5 =
19 Octal to Decimal Decimal Octal Binary Hexadecimal
20 Octal to Decimal Technique Multiply each bit by 8 n, where n is the weight of the bit The weight is the position of the bit, starting from 0 on the right Add the results
21 Example => 4 x 8 0 = 4 2 x 8 1 = 16 7 x 8 2 =
23 Hexadecimal to Decimal Technique Multiply each bit by 16 n, where n is the weight of the bit The weight is the position of the bit, starting from 0 on the right Add the results
24 Example ABC 16 => C x 16 0 = 12 x 1 = 12 B x 16 1 = 11 x 16 = 176 A x 16 2 = 10 x 256 =
25 Decimal to Binary Decimal Octal Binary Hexadecimal
26 Decimal to Binary Technique Divide by two, keep track of the remainder First remainder is bit 0 (LSB, least-significant bit) Second remainder is bit 1 Etc.
27 Example =? =
28 Decimal to Octal Decimal Octal Binary Hexadecimal
29 Decimal to Octal Technique Divide by 8 Keep track of the remainder
30 Example =? =
32 Decimal to Hexadecimal Technique Divide by 16 Keep track of the remainder
33 Example =? = D = 4D2 16
34 Octal to Binary Decimal Octal Binary Hexadecimal
35 Octal to Binary Technique Convert each octal digit to a 3-bit equivalent binary representation
36 Example =? =
38 Hexadecimal to Binary Technique Convert each hexadecimal digit to a 4-bit equivalent binary representation
39 Example 10AF 16 =? A F AF 16 =
40 Binary to Octal Decimal Octal Binary Hexadecimal
41 Binary to Octal Technique Group bits in threes, starting on right Convert to octal digits
42 Example =? =
44 Binary to Hexadecimal Technique Group bits in fours, starting on right Convert to hexadecimal digits
45 Example =? B B = 2BB 16
47 Octal to Hexadecimal Technique Use binary as an intermediary
48 Example =? E = 23E 16
50 Hexadecimal to Octal Technique Use binary as an intermediary
51 Example 1F0C 16 =? 8 1 F 0 C F0C 16 =
54 Common Powers (1 of 2) Base 10 Power Preface Symbol pico p 10-9 nano n 10-6 micro 10-3 milli m 10 3 kilo k 10 6 mega M 10 9 giga G tera T Value
55 Common Powers (2 of 2) Base 2 Power Preface Symbol 2 10 kilo k 2 20 mega M 2 30 Giga G Value What is the value of k, M, and G? In computing, particularly w.r.t. memory, the base-2 interpretation generally applies
56 Example In the lab 1. Double click on My Computer 2. Right click on C: 3. Click on Properties / 2 30 =
57 Exercise Free Space Determine the free space on all drives on a machine in the lab Drive Bytes Free space GB A: C: D: E: etc.
58 Review multiplying powers For common bases, add powers a b a c = a b+c = 2 16 = 65,536 or = = 64k
59 Fractions Decimal to decimal (just for fun) 3.14 => 4 x 10-2 = x 10-1 = x 10 0 = pp
60 Fractions Binary to decimal => 1 x 2-4 = x 2-3 = x 2-2 = x 2-1 = x 2 0 = x 2 1 = pp
61 Fractions Decimal to binary x x x x x x etc. p. 50
62 Fractions Octal to decimal => 2 x 8-2 = x 8-1 = x 8 0 = x 8 1 = pp
63 Fractions Decimal to octal x x x x x x etc. p. 50
64 Fractions Hexadecimal to decimal 2B.84 => 4 x 16-2 = x 16-1 = 0.5 B x 16 0 = x 16 1 = pp
65 Fractions Decimal to Hexadecima x x x x x x etc. p. 50
67 Exercise Convert Answer Decimal Binary Octal Hexadecimal D.CC D C C.82
68 Binary Addition (1 of 2) Two 1-bit values A B A + B two pp
69 Binary Addition (2 of 2) Two n-bit values Add individual bits Propagate carries E.g.,
70 Multiplication (2 of 3) Binary, two 1-bit values A B A B
71 Binary Subtraction (1 of 2) Two 1-bit values Borrow 1 A B A - B pp
72 Binary Subtraction (2 of 2) Two n-bit values Subtract individual bits Propagate borrows E.g.,
73 Binary Subtraction (2 of 2) Two n-bit values Subtract individual bits Propagate borrows E.g.,
74 Subtraction with Complements Complements are used for simplifying the subtraction operations. There are two types of complements for each base-r system: the r's complement and the (r l)'s complement. 2's complement and 1's complement for binary numbers, and the 10's complement and 9's com plement for decimal numbers.
75 9's complement The 9's complement of a decimal number is obtained by subtracting each digit from 9. The 9's complement of is: = The 9's complement of is: =
76 Binary numbers, the 1's complement The 1's complement of a binary number is formed by changing 1's to 0's and 0's to 1's. Examples: The 1's complement of is The 1's complement of is
77 10's complement The 10's complement can be formed by leaving all least significant 0's un changed, subtracting the first nonzero least significant digit from 10, and subtracting all higher significant digits from 9. The 10's complement of is The 10's complement of is
78 2's complement The 2's complement can be formed by leaving all least significant 0's and the first 1 unchanged, and replacing 1's with 0's and 0's with 1's in all other higher significant digits. The 2's complement of is The 2's complement of is
79 Subtraction with Complements The subtraction of two n-digit unsigned numbers M N in base r can be done as follows: Add M to the r's complement of N. If M N, the sum will produce an end carry, r n, which is discarded; what is left is the result M - N. If M < N, the sum does not produce an end carry. To obtain the answer in a familiar form, take the r's complement of the sum and place a negative sign in front.
80 Examples to illustrate the procedure Given the two binary numbers; X = and Y = , perform the subtraction: (a) X Y (b) Y X using 2's complements.
81 X Y= X = 's complement of Y = Sum = Discard end carry = Answer: X Y =
82 Y X= Y = 's complement of X = Sum = There is no end carry. Answer: Y - X = -(2's complement of ) =
83 Thank you Next topic
84 Binary Codes Binary codes are codes which are represented in binary system with modification from the original ones. Binary codes are classified as: Weighted Binary Systems Non Weighted Codes
85 Weighted Binary Systems Weighted binary codes are those which obey the positional weighting principles, Each position of the number represents a specific weight. The codes 8421, 2421, 5421, and 5211 are weighted binary codes.
86 Weighted Binary Systems
87 8421 Code/BCD Code The BCD (Binary Coded Decimal) is a straight assignment of the binary equivalent. It is possible to assign weights to the binary bits according to their positions. The weights in the BCD code are 8,4,2,1. Example: The bit assignment 1001, can be seen by its weights to represent the decimal 9 because: 1x8+0x4+0x2+1x1 = 9 Ex. number 12 is represented in BCD as [ ]
88 2421 Code 2421 Code This is a weighted code, its weights are 2, 4, 2 and 1. A decimal number is represented in 4-bit form and the total four bits weight is = 9. Hence the 2421 code represents the decimal numbers from 0 to 9.
89 5211 Code 5211 Code This is a weighted code, its weights are 5, 2, 1 and 1. A decimal number is represented in 4-bit form and the total four bits weight is = 9. Hence the 5211 code represents the decimal numbers from 0 to 9.
90 Reflective Code Reflective Code A code is said to be reflective when code for 9 is complement for the code for 0, and so is for 8 and 1 codes, 7 and 2, 6 and 3, 5 and 4. Codes 2421, 5211, and excess-3 are reflective, whereas the 8421 code is not.
91 Sequential Codes Sequential Codes A code is said to be sequential when two subsequent codes, seen as numbers in binary representation, differ by one. This greatly aids mathematical manipulation of data. The 8421 and Excess-3 codes are sequential, whereas the 2421 and 5211 codes are not.
92 Excess-3 Code Excess-3 Code Excess-3 is a non weighted code used to express decimal numbers. The code derives its name from the fact that each binary code is the corresponding 8421 code plus 0011(3). Example: 1000 of 8421 = 1011 in Excess-3
93 Error Detecting and Correction Codes For reliable transmission and storage of digital data, error detection and correction is required.
94 Error Detecting Codes When data is transmitted from one point to another there are chances that data may get corrupted. To detect these data errors, we use special codes, which are error detection codes.
95 Parity check In parity codes, every binary message is checked if they have even number of ones or even number of zeros. Based on this information an additional bit is appended to the original data. At the receiver side, once again parity is calculated and matched with the received parity, and if they match, data is ok, otherwise data is corrupt. There are two types of parity: Even parity and Odd Parity
96 Parity There are two types of parity: Even parity: Checks if there is an even number of ones; if so, parity bit is zero. When the number of ones is odd then parity bit is set to 1. Message xyz Even parity code xyz p
97 Parity Odd Parity: Checks if there is an odd number of ones; if so, parity bit is zero. When number of ones is even then parity bit is set to 1. Message xyz Odd parity code xyz p
98 Alphanumeric Codes The binary codes that can be used to represent all the letters of the alphabet, numbers and mathematical symbols, punctuation marks, are known as alphanumeric codes or character codes. These codes enable us to interface the input-output devices like the keyboard, printers, video displays with the computer.
99 ASCII Code ASCII Code ASCII stands for American Standard Code for Information Interchange. It has become a world standard alphanumeric code for microcomputers and computers. It is a 7-bit code representing 2 7 = 128 different characters. These characters represent 26 upper case letters (A to Z), 26 lowercase letters (a to z), 10 numbers (0 to 9), 33 special characters and symbols and 33 control characters.
100 ASCII Code The 7-bit code is divided into two portions, The leftmost 3 bits portion is called zone bits and the 4-bit portion on the right is called numeric bits. Character A B 3 7-bit ASCII
101 ASCII Code An 8-bit version of ASCII code is known as ASCII-8. The 8-bit version can represent a maximum of 256 characters.
102 EBCDIC Code EBCDIC Code EBCDIC stands for Extended Binary Coded Decimal Interchange. It is mainly used with large computer systems like mainframes. EBCDIC is an 8-bit code and thus accommodates up to 256 characters. An EBCDIC code is divided into two portions: 4 zone bits (on the left) and 4 numeric bits (on the right).
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ECPE 170 Jeff Shafer University of the Pacific Exam 1 Review 2 Exam 1 Basics Topics Chapter 2 Data representabons Chapter 3 Digital logic Part of Chapter 4 Basic organizabon and memory systems Nothing | 9,674 | 39,043 | {"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-2021-25 | latest | en | 0.741416 |
http://math.mit.edu/~djk/18_01/chapter28/section03.html | 1,542,228,266,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114213308-00311.warc.gz | 199,295,540 | 1,314 | ## 28.3 Arc Length in Polar Coordinates
The circumference or length around a circle of radius r is 2r, or r per radian.
The length for an angle d is therefore rd.
Length in the r direction is just dr; since the r and directions are always orthogonal, we have:
This function can be integrated over a curve to give its length.
Spiral
Four leaves clover
Cardioid | 89 | 366 | {"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-2018-47 | latest | en | 0.907011 |
https://brainmass.com/statistics/sample-size-determination/minimum-sample-size-big-corporation-14905 | 1,713,398,579,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00546.warc.gz | 130,249,846 | 7,038 | Purchase Solution
# Minimum sample size BIG Corporation
Not what you're looking for?
A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of eight ounces of coffee per cup. If it dispenses more that that on average, the corporation may lose money, and if it dispenses less, the customers may complain.
BIG Corporation would like to estimate the mean amount of coffee, M , dispensed per cup by this machine. BIG will choose a random sample of cup amounts dispensed by this machine and use this sample to estimate M . Assuming that the standard deviation of cup amounts dispensed by this machine is 0.35 ounces, what is the minimum sample size needed in order for BIG to be confident that its estimate is within 0.08 ounces of M ?
Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).
Please see the attached file for full problem description.
##### Solution Summary
The solution calculates the sample size required for estimating the mean amount of coffee.
##### Solution Preview
Mean=M = 8 ounces
Standard deviation =s= 0.35 ounces
(sample mean-M )= 0.08 ounces (to ensure estimate is within 0.08 ounces of M )
sx=standard error of ...
##### Terms and Definitions for Statistics
This quiz covers basic terms and definitions of statistics.
##### Measures of Central Tendency
This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.
##### Measures of Central Tendency
Tests knowledge of the three main measures of central tendency, including some simple calculation questions. | 360 | 1,834 | {"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-2024-18 | latest | en | 0.920496 |
http://msdn.microsoft.com/tr-tr/library/windowsazure/hh689810.aspx | 1,394,842,281,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678695499/warc/CC-MAIN-20140313024455-00098-ip-10-183-142-35.ec2.internal.warc.gz | 89,642,689 | 15,131 | SATıŞ: 1-800-867-1389
EN
Bu konu henüz değerlendirilmedi - Bu konuyu değerlendir
Expressions
Updated: November 21, 2013
Expressions Map Operators
The following table lists the Expression Map Operations available within a Transform:
Map Operation Description Parameters Output
Arithmetic Expression
Evaluates a mathematical expression using inputs and constants. Arithmetic Expressions consists of the following operators:
• Subtraction
• Multiplication
• Division
• Modulo
• Absolute Value
• Maximum
• Minimum
• Round
• Square Root
Can have 0 to 100 optional input parameters:
Input A numeric value. Arithmetic Expression A mathematical expression defined using the inputs and constants.
Note
The maximum length for an expression is 1024 characters.
A numeric value that is the result of the computation.
Logical Expression
Evaluates a condition and outputs the Boolean value of the evaluation. Logical Expression consists of the following operators:
• Relational Operators:
>
<
>=
<=
==
!=
• Logical Negation (!)
• Conditional AND (&&)
Conditional OR (||)
Can have 0 to 100 optional input parameters:
Input Can be a numeric value, string value or Boolean value. Logical Expression An expression defined using the inputs and constants that evaluate to a Boolean value.
Note
The maximum length for an expression is 1024 characters.
True is returned if the Logical Expression returns true. Otherwise, False is returned.
If-Then-Else Expression
Evaluates a statement that results in one of two possible outputs.
Can have 0 to 100 optional input parameters:
Input Can be a numeric value, string value or Boolean value. Condition An expression defined using the inputs and constants. Then Value If statement or expression is true, this value is used. Else Value If statement or expression is false, this value is used.
Note
The maximum length for an expression is 1024 characters.
The result is based on a true or false evaluation of the conditional expression.
If true, the Then Value is used. If false, the Else Value is used.
Conditional Assignment
Returns a value from one of two input parameters. If the first input value is True, then a node is created in the output document with the second input value. If the first input value is False, then the corresponding node is not created in the output document.
Requires exactly two input parameters:
Condition An expression that results in a Boolean value. Can be one of the following: Link from tree node Link from a Map Operation Assign Value The value assigned to the destination node if the condition is true.
Note
This Map Operation can only be connected to a destination tree node.
If the Condition value is "true", then a node is created with the Assign Value input value.
The following table lists additional functions that can be used with any Map Operation:
Function Expression Description
Exists
Exists(Source_Node_Name)
Requires a single input that is the element name in the source document. If the element exists, True is returned. Otherwise, False is returned.
IsDate
IsDate(Input1)
Requires a single input of the type string. The DateTime.TryParse() method is used to parse the input into a DateTime object. If the input is parsed successfully, True is returned. Otherwise, False is returned.
IsEmpty
IsEmpty(Input1)
Requires a single input of the type string. If the string is null or empty, True is returned. Otherwise, False is returned. If the input is not a string object, True is returned.
IsNil
IsNil(Source_Node_Name)
Requires a single argument that is the element name in the source document. If the element exists and xsi:nil is set to True, then True is returned. Otherwise, False is returned.
IsNumber
IsNumber(Input1)
Requires a single input of the type string. The Double.TryParse() method is used to parse the input into a double. If the input is parsed successfully, True is returned. Otherwise, False is returned.
Note
The comma “,” is supported as the thousands separator and the period “.” is supported as a decimal point.
Important
All Map Operation and functions can be used within other Map Operation and functions except Exists and IsNil. Exists and IsNil point to a single node in the source document.
Error and Data Handling
BizTalk Services provides the ability to configure how an error is handled and how an empty or null node is handled. The error handling behavior of the following Expression Map Operations is configurable:
• Logical Expression
• Arithmetic Expression
• If-Then-Else Expression
Steps:
1. Open a BizTalk Service project or the BizTalk Service Artifacts project in Visual Studio.
2. Double-click a Transform (.trfm) to open the Transform Designer.
3. In the Transform toolbar, click Settings.
Error Handling tab
In the Error Handling tab, the following Expression Map Operations have two Behavior options:
• Logical Expression:
• Fail map: The entire Transform is aborted. Since Transforms are executed within a pipeline, an error occurs within the pipeline and the error is then sent back to the client that sent the message.
• Output default value false: If the Map Operation fails, False is returned as the output.
• Arithmetic Expression:
• Fail map: The entire Transform is aborted. Since Transforms are executed within a pipeline, an error occurs within the pipeline and the error is then sent to the client that sent the message.
• Output default value NaN: If the Map Operation fails, NaN (Not a Number) is returned as the output.
• Output default value 0: If the Map Operation fails, zero (0) is returned as the output.
• If-Then-Else Expression:
• Fail map: The entire Transform is aborted. Since Transforms are executed within a pipeline, an error occurs within the pipeline and the error is then sent back to the client that sent the message.
• Output Null/Zero/False based on type of output: If the Map Operation fails, Null/Zero/False is returned as the output based on type of output.
Null/Empty Data Handling tab
In the Null/Empty Data Handling tab, there are three options:
• Consider empty nodes in cumulative operations: By default, this is not checked. When not checked, no empty nodes are included in the iteration. When checked, all nodes, including empty nodes, are included in the iteration.
EXAMPLE: There is a document with 10 <record> nodes. Three of these <record> nodes are empty. When Consider empty nodes in iterations is not checked, the Map Operation returns a value of seven. When Consider empty nodes in iterations is checked, the Map Operation returns a value of 10.
• Consider empty nodes in iterations: By default, this is not checked. When not checked, no empty nodes are included in the iteration. When checked, all nodes, including empty nodes, are included in the iteration.
EXAMPLE: there is a document with 10 <record> nodes. Three of these <record> nodes are empty. When Consider empty nodes in iterations is not checked, the Map Operation iterates seven times for the non-empty nodes. As a result, seven <record> nodes are generated in the Target. When Consider empty nodes in iterations is checked, the Map Operation iterates 10 times for all nodes, including the empty nodes. As a result, 10 <record> nodes are generated in the Target.
• Target Node Generation: If empty nodes are configured to be considered, you choose to generate an empty node in the output or to not generate an empty node in the output. Specifically:
• Do not generate empty nodes: Default option.
• Generate empty nodes
EXAMPLE: There is a document with 10 <record> nodes. Three of these <record> nodes are empty. Consider empty nodes in cumulative operations or Consider empty nodes in iterations are not checked, the Map Operation iterates seven times for the non-empty nodes. As a result, seven <record> nodes are generated in the Target. When Consider empty nodes in iterations is checked, the Map Operation iterates 10 times for all nodes, including the empty nodes. As a result, 10 <record> nodes are generated in the Target.
In This Section
Examples of these Map Operations: | 1,685 | 8,112 | {"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-2014-10 | latest | en | 0.674169 |
https://rustgym.com/leetcode/278 | 1,660,425,625,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00494.warc.gz | 448,788,757 | 3,290 | You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
Suppose you have `n` versions `[1, 2, ..., n]` and you want to find out the first bad one, which causes all the following ones to be bad.
You are given an API `bool isBadVersion(version)` which returns whether `version` is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
Example 1:
```Input: n = 5, bad = 4
Output: 4
Explanation:
Then 4 is the first bad version.
```
Example 2:
```Input: n = 1, bad = 1
Output: 1
```
Constraints:
• `1 <= bad <= n <= 231 - 1`
``````struct Solution {
}
impl Solution {
fn new(bad: i32) -> Self {
}
#[allow(non_snake_case)]
fn isBadVersion(&self, version: i32) -> bool {
}
}
impl Solution {
fn first_bad_version(&self, n: i32) -> i32 {
let mut lower = 1;
let mut upper = n;
while lower < upper {
let mid = lower + (upper - lower) / 2;
lower = mid + 1;
} else {
upper = mid;
}
}
lower
}
}
#[test]
fn test() { | 332 | 1,182 | {"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} | 3.03125 | 3 | CC-MAIN-2022-33 | latest | en | 0.835739 |
https://elizabethsid.org/and-pdf/1242-dynamics-questions-and-answers-pdf-809-618.php | 1,653,553,711,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00345.warc.gz | 281,265,873 | 8,268 | Friday, May 14, 2021 10:58:34 AM
# Dynamics Questions And Answers Pdf
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Published: 14.05.2021
Engineering Mechanics - Statics and Dynamics. Units and dimensions, Properties of fluids. Dynamics Exam1 and Problem Solutions 1. This is why you remain in the best website to look the unbelievable book to have. Even though small the free section features an impressive range of fiction and non-fiction.
## Dynamics Extra Study Questions Short Answer
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## Engineering Mechanics Dynamics Problems And Solutions Pdf
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Стратмор покачал головой: - Это внешний файл. Она ждала чего угодно, но только не. - Внешний файл.
Хейл наклонил голову набок, явно заинтригованный такой скрытностью. И, как бы желая обратить все в игру, сделал еще один шаг. Но он не был готов к тому, что произошло в следующее мгновение.
Вскоре она едва заметно кивнула и широко улыбнулась. - Дэвид, ты превзошел самого .
Пусть остается. - Стратмор кивнул в сторону лаборатории систем безопасности. - Чатрукьян уже, надеюсь, ушел.
Я думаю, он собирался оставаться поблизости и вовремя все это остановить. Глядя на экран, Фонтейн увидел, как полностью исчезла первая из пяти защитных стен. - Бастион рухнул! - крикнул техник, сидевший в задней части комнаты.
Слушаю, Джабба. Металлический голос Джаббы заполнил комнату: - Мидж, я в главном банке данных. У нас тут творятся довольно странные вещи.
Загорелое лицо консьержа расплылось еще шире. - Si, si, senor. Мануэль - это. Чего желаете. - Сеньор Ролдан из агентства сопровождения Белена сказал мне, что вы… Взмахом руки консьерж заставил Беккера остановиться и нервно оглядел фойе.
Тело же его было бледно-желтого цвета - кроме крохотного красноватого кровоподтека прямо над сердцем. Скорее всего от искусственного дыхания и массажа сердца, - подумал Беккер. - Жаль, что бедняге это не помогло. Он принялся рассматривать руки покойного.
Дэвид положил трубку. Она долго лежала без сна, ожидая его звонка.
Слова Стратмора внезапно были прерваны постукиванием по стеклянной стене Третьего узла. Они обернулись. Сотрудник отдела обеспечения системной безопасности Фил Чатрукьян, приникнув лицом к стеклу, отчаянно барабанил по нему, стараясь разглядеть, есть ли кто-нибудь внутри. Он что-то говорил, но сквозь звуконепроницаемую перегородку слов не было слышно. У него был такой вид, словно он только что увидел привидение.
Звук показался очень далеким, едва различимым в шуме генераторов. Она никогда раньше не слышала выстрелов, разве что по телевизору, но не сомневалась в том, что это был за звук. Сьюзан словно пронзило током. В панике она сразу же представила себе самое худшее. Ей вспомнились мечты коммандера: черный ход в Цифровую крепость и величайший переворот в разведке, который он должен был вызвать.
Слова приятеля его очень удивили. Дело в том, что АНБ не только существовало, но и считалось одной из самых влиятельных правительственных организаций в США и во всем мире. Уже больше полувека оно занималось тем, что собирало электронные разведданные по всему миру и защищало американскую секретную информацию. | 1,489 | 5,682 | {"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-2022-21 | latest | en | 0.890273 |
https://www.teacherspayteachers.com/Product/Multiplication-and-Division-Ultimate-BUNDLE-3842780 | 1,540,235,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515375.86/warc/CC-MAIN-20181022180558-20181022202058-00543.warc.gz | 1,072,238,226 | 27,292 | # Multiplication and Division Ultimate BUNDLE
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18 Products in this Bundle
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Bundle Description
This Ultimate Bundle features everything you need to teach any level of multiplication and division to all learners - fantastic for beginners, special education, and intervention students. Students learn to multiply multi-digit numbers and solve long division problems using visual strategies!
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Choose worksheets with organizers or graph paper, challenge students with gradual release by selecting just the right level of support for your individual students.
So many levels to choose from!
The organizers and worksheets offer NEW, unique strategies for teaching multiplication and long division using colors and shapes to visually teach math! These are concrete methods that help students visualize the process. This method uses graphic organizers allowing students to visualize the step-by-step process.
Once students are familiar with the multiplication and division processes, move them to graph paper for the next level of independence!
All division worksheets and organizers in this pack have 1-digit divisors.
-------------------------------------------------------------------------------------------------------------------------------
Benefits of this resource:
→ Alignment of Numbers and Columns
→ Neatness and Number Size - improves graphomotor skills
→ Effective for visual learners and visual discrimination
→ Organized Writing = Organized Thinking
Remember - proper alignment is absolutely critical to success with multiplication and division!
-------------------------------------------------------------------------------------------------------------------------------
Lots of practice and great intervention - fabulous for new learners, special education, at-risk students, Response to Intervention, RTI, and strugglers! Scaffold for students who need additional help!
This BUNDLE of differentiated multiplication and division organizers, worksheets, and graph paper worksheets will save you time as you prepare your students with much needed repetition, while reinforcing positive math skills!
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The graph paper long division kits are a bit different from typical worksheets. You will choose for your students either the basic worksheet (typical long division problems on graph paper) OR the guided practice worksheets - providing the quotient for students to use to show all of the work to solve the problem.
These multiplication and long division Organizers and Graph Paper worksheets are great for visual learners.
**Please view the PREVIEW to see the organizers and worksheet styles and skills covered.**
-------------------------------------------------------------------------------------------------------------------------------
You'll LOVE this MEGA BUNDLE!
“These are amazing! This may be my favorite thing I have ever purchased on TPT!!! It is great for my struggling intervention students and also helpful for my dyslexic students.” –Branden P.
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-------------------------------------------------------------------------------------------------------------------------------
These multiplication and long division graphic organizers and graph paper worksheets are teaching aids like you've never seen before! They’re designed to make math a bit easier for students to understand and learn.
Note: These resources are not editable by computer, but made to be written on by hand.
Please note: If you purchase this pack, you will NOT need to purchase any of the separate packs!
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 805 | 4,858 | {"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-2018-43 | latest | en | 0.816816 |
https://es.mathworks.com/matlabcentral/answers/1592054-signal-amplitude-using-fft-conflicting-results-and-misunderstandings | 1,686,224,339,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00633.warc.gz | 258,936,153 | 30,003 | # signal amplitude using FFT - conflicting results and misunderstandings
6 views (last 30 days)
LO on 21 Nov 2021
Edited: dpb on 24 Nov 2021
I am trying to measure the amplitude of different frequencies composing a signal using some code found here on this post.
Using this code
Fs=20000;
nfft = length(seg1); % calculate lenght of signal, seg1 is my signal
res = fft(seg1,nfft)/ nfft; % calculate fft over the whole segment normalizing the fft
f = Fs/2*linspace(0,1,nfft/2+1); % choosing correct frequency axes
res = res(1:nfft/2+1);
figure(2), plot(f,abs(res));
xlabel('Frequency in Hz');ylabel('Amplitude (mV)');
xlim([200 600])
ytix = yticks*1000; % convert signal in mV (the original signal units are in V)
yticklabels(ytix);
I get this
It seems fine, now I cannot verify if the signal amp is really around 45 mV but I could test it.
Just for proofing, I checked the help of the FFT function in MATLAB and I found this code to calculate signal power
figure(3)
Fs = 20000;
L = length(seg1); % length of signal
n = 2^nextpow2(L); % determine adjustable window size based on powers of 2
dim=2;
Y = fft(seg1,n,dim); % calculate fft of signal over the whole segment
P2 = abs(Y/L); % calculate signal power
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:(L/2))/L;
plot(f,P1)
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
xlim([200 600])
what I get is this
I think the result is the same and it is just a matter of converting the units in the right way. Besides the 10^3 conversion, I think there is a 2x factor that needs to be removed somewhere...however I am not sure which would be the more correct way to calculate it.
While in the first code power is simply calculated as the absolute value of the fft transformed signal, In the second case it seems that, besides that (see the line commented with "calculate signal power"), the signal power is also multiplied per 2. As a result I get a double of the first power measurement in the second plot (note: the units in this second plot are not converted in millivolts but if one does it, the value seems to me double). And in fact I checked it, and it is the case.
Just wondering what would be the correct way to calculate signal amplitude. My signal is a sinewave voltage signal at 400 Hz. The amplitude should be in the range of few tens of mV (so the first calculation could be realistic... but the second too).
Thank you for your feedback !
dpb on 21 Nov 2021
Edited: dpb on 24 Nov 2021
The latter accounts that the FFT and PSD are two-sided (hence the "2" in the variable P2 for the PSD). The total energy is thus spread over [-Fmax +Fmax] or, in your case of a single frequency in two peaks, one at +/-400 Hz.
Hence, the total energy in the signal is twice the energy in each of those two peaks, ergo, this would indicate your input sine wave was of amplitude 90 mV rather than 45 mV.
It should be noted that the factor of 2 is applied only on the PSD elements (2:end-1) of P1 -- this is owing to the fact that the DC and Fmax components are unique in the two-sided PSD/FFT and so contain the total energy for those frequencies already. This can be confirmed empirically by comparing the DC component of a signal with a nonzero mean to the mean of the input time signal.
See the example at doc FFT from whence the latter code snippet was derived (recognizable by the variable naming) that illustrates producing the exact amplitudes of the input signal for a dual-component sine wave with two frequencies and amplitudes of 1.0 and 0.7.
Also NB: that the peak will only match the input exactly when the frequency resolution of the FFT is such that the bin center frequency of one of the FFT/PSD bins matches EXACTLY the input frequency of the signal; if this is not the case, then one will have energy content in surrounding bins as well and the total energy will be the integral of the peak, not just the peak value.
I've written and posted examples of these issues at length on Answers altho I didn't think to keep a library of links to those at hand...
##### 2 CommentsShow 1 older commentHide 1 older comment
dpb on 24 Nov 2021
The given normalization will produce the peak amplitude of the input time series, yes.
This is only one possible way to normalize.
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Translated by | 1,140 | 4,502 | {"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} | 3.1875 | 3 | CC-MAIN-2023-23 | longest | en | 0.868653 |
https://physics.stackexchange.com/questions/151383/solving-special-function-equations-using-lie-symmetries | 1,726,613,287,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00658.warc.gz | 405,464,533 | 41,262 | # Solving Special Function Equations Using Lie Symmetries
The lie group + representation theory approach to special functions & how they solve the ode's arising in physics is absolutely amazing. I've given an example of it's power below on Bessel's equation.
Kaufman's article describes algebraic methods for dealing with Hermite, Legendre & Associated Legendre. Can we take the other special functions mentioned in this paper, obtainable as linear combinations of the conformal symmetries of the Laplacian (expressed as lie algebra elements), and obtain their solution analogously to how Bessel is solved below? I believe it's something like a geometric interpretation of Weisner's method.
Bessel's equation seems to be saying: find a function in the plane such that when we shift it right, then shift it back left again, all locally (i.e. differentially) in polar coordinates, we get the same function:
(c.f. Killingbeck, Mathematical Technique's and Applications, sec. 8.21).
The idea is to take Bessel's equation, factor it, add an extra variable to make the factors parameter independent so that they become elements of a lie algebra, identify the meaning of those factors, in this case notice the Lie algebra factors are translations in Polar coordinates, and realize it's just a differential expression of a symmetry.
Bessel's equation arises from $LRv = v$ when you express $L$ & $R$ in polar coordinates. It makes sense to express them in polar coordinates since Bessel arises from separating the Laplacian assuming cylindrical symmetry, and the $LRv = v$ assumption (not $LRv = w$) is motivated by the symmetry of the Laplacian.
Using this idea we can, for some reason, actually solve Bessel's equation with a picture!:
We just want to shift $\mathcal{J}_n(r,\phi)$ in the x-direction using the operator $e^{a\tfrac{\partial}{\partial x}}$ expressed in polar coordinates: $e^{\tfrac{a}{2}(\mathcal{L}-\mathcal{R})}$ and realize it will be equal to $\mathcal{J}_n(r',\phi')$:
So the last line comes from dragging all this to the origin and putting it along the x-axis, here we see the geometric meaning of Bessel functions!
Hypergeometric is supposed to be related to $SL(2,R)$ symmetries, Bessel to translational planar symmetries, the Gamma function related to linear symmetries $y = ax + b$, etc... Is there an easy unified geometric exposition on how to deal with these babies?
It'd be great to understand the other equations, their formulation and solution, with a geometric interpretation like this one.
References:
1. Killingbeck, Mathematical Technique's and Applications, sec. 8.21
2. Vilenkin, Representation of Lie Groups and Special Functions Vol. 1
3. Vilenkin, Special Functions and Theory of Group Representations
4. Miller, Lie Theory and Special Functions
5. Kaufman, Special Functions of Mathematical Physics from the Viewpoint of Lie Algebra | 660 | 2,882 | {"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} | 3.671875 | 4 | CC-MAIN-2024-38 | latest | en | 0.887218 |
https://blog.lboro.ac.uk/cmc/2021/03/16/mixed-ability-maths-groups-influence-pupils-mindsets-teachers-mindsets-and-teachers-beliefs-and-practices/ | 1,722,758,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00681.warc.gz | 110,170,746 | 20,981 | Centre for Mathematical Cognition Blog
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# Mixed-ability Maths Groups Influence Pupils’ Mindsets, Teachers’ Mindsets and Teachers’ Beliefs and Practices
16 March 2021
7 mins
Written by Tom Francome. Tom is a PhD student and a Senior Enterprise Fellow at the CMC at Loughborough University. If you are interested in this blog post and would like to get in touch, please email him directly at T.J.Francome@lboro.ac.uk, or comment below to start a conversation.
The Coronavirus outbreak showcased the adaptability and resilience of teachers. As students across the country returned to classrooms for face-to-face teaching, one consequence of social-distancing measures is that many mathematics classes are now mixed-attainment groups – a practice uncommon in UK secondary schools. The prospect of mixed-attainment mathematics classes might daunt some teachers. However, this can be viewed as an opportunity, rather than a problem.
Setting and streaming are not effective strategies for raising attainment. Everyone in Scandinavia teaches mixed-attainment mathematics, but there are fewer examples of good practice in the UK, so little evidence of benefits to encourage sceptical stakeholders. This “vicious circle” of factors deters UK secondary schools from teaching maths in mixed-attainment groups and maintains the status quo of ‘ability grouping’, which depends upon the misguided beliefs that abilities are ‘fixed’ and can be assessed accurately. Usual setting practices mean up to 50% of pupils are put in the “wrong” set and are rarely moved. This can have dire consequences:
“If three pupils with the same scores on entrance to school were placed in different sets, one in a top set, one in a middle set and one in a low set, the performance of the pupil in a top set would be significantly higher and that of the pupil in the bottom set significantly lower.” (Ireson, Hallam, Hack, Clark, & Plewis, 2002, p. 311)
It is desirable for pupils to believe that mathematical ability increases as a result of effort and effective teaching (a growth mindset). The alternative is that you have a ‘fixed’ ability which is preserved by avoiding challenges and potential failures. Of course, growth mindset interventions do not tend to work beyond the short-term; perhaps suggesting that long-term structural changes may be needed to promote and maintain growth mindsets. Unfortunately, research suggests setting practices can create fixed mindsets. Effectively telling some pupils:
“You’re good at mathematics… so you don’t have to try.”
or
“You’re not good at mathematics… so there’s no point in trying.”
Following on from this research, we conducted a small-scale study looking at the effects of setting versus mixed-attainment groups. We compared beliefs and practices in mathematics in two schools: School M taught in mixed-ability groups and School S taught in setted groups. We surveyed 286 year 7 pupils (age 11/12) and 12 teachers via questionnaires, lesson observations, and interviews.
Teachers of mixed groups believed more strongly that effort could increase ability, compared to teachers who taught setted groups. Pupils in both schools reported growth-mindsets but the beliefs were stronger for mixed-attainment groups who had stronger views that intelligence is improvable, were more strongly motivated by ‘learning goals’, and had stronger beliefs that effort led to improvement. Pupils in both schools wanted challenging work, wanted to learn through mistakes and wanted to discuss their work with others. Data suggested that pupils in mixed-attainment groups were more likely to be given or seek out challenging work, and to believe these tasks would help them learn.
Teachers from both schools expressed similar beliefs about the way they worked with pupils, but this was not supported by the pupil feedback. Mixed-attainment lessons tended to involve pupils discussing ideas collaboratively in pairs/small groups, using mistakes and misconceptions as learning opportunities, and using substantial tasks (i.e., bigger and more challenging tasks, which were accessible at different levels, and may generate more mistakes). Lessons with setted classes tended to involve pupils working mostly on their own, following methods shown by the teacher, and closely following textbooks/worksheets.
“I like discussing my answers with other classmates because I like to see if we came up with similar strategies” (Pupil M29)
“I work hard and I sometimes make mistakes but [the teacher] helps me learn from them.” (Pupil M117)
“My maths lessons are fun and interesting. My maths lessons are helping me get better at maths” (Pupil M12)
“In my maths lessons we always have a worksheet to do but before we start our teacher gives us some examples on the board.” (Pupil S100)
“In my maths class I sit alone and get on with my work” (Pupil S127).
“I’m always doing work at my level” (Pupil S135)
Our observations corroborated the pupil reports and showed that pupils in the mixed-attainment groups spent a far greater proportion of time working collaboratively.
Our observations corroborated the pupil reports and showed that pupils in the mixed-attainment groups spent a far greater proportion of time working collaboratively.
Table 1 shows that pupils in both schools spent similar proportions of time in whole-class teaching and consulting with peers (such as checking they had the same answer). The vast majority of the remaining time was spent on individual work in School S and on collaborative work in School M.
A small study like this cannot be generalised but it raises some important questions. Can mixed-attainment groups encourage teachers to believe in pupils’ ability to improve mathematically? Can mixed-attainment groups change teaching practices in ways that support the learning of all pupils? When teachers work with mixed-attainment groups, they have to take account of pupils’ prior experiences in their planning. Teaching “at a particular level” is unlikely to succeed and so offering substantial tasks can allow each pupil to feel challenged mathematically. Different ideas arise and need discussion beyond ‘the answer’ so collaboration can be genuine. Pupils are more likely to make mistakes working on substantial tasks than following small steps and this allows greater opportunity for pupils to learn from their mistakes. Such a variety of pupil perspectives allows for greater opportunity to make connections between the different mathematical aspects of their work.
In contrast, setted classrooms can give teachers false impressions of the pupils being “at the same level”. This may lead to more procedural teaching where pupils are more likely to reproduce steps without error. This often means pupils do not feel challenged, and do not gain the benefits which come with learning from mistakes. There can also be less variety in the mathematics taking place within a lesson and so less opportunity for making connections between different topics. A consequence of this is that the actual experiences pupils have of learning mathematics, and their beliefs about what mathematics is, may be at odds with the beliefs expressed by teachers within the school.
This study also offered some evidence that grouping practices could influence pupils’ mindsets, teachers’ mindsets and teachers’ beliefs and practices when teaching mathematics. We found that both the experience of learning mathematics and pupil outcomes were improved if pupils believed they could improve, were less reluctant to engage with challenging problems, and persevered following setbacks (growth mindsets). ‘Mixed’ pupils had stronger growth-mindsets, ‘mixed’ teachers held more ‘connectionist’ beliefs and also had stronger growth-mindsets. As one mixed-attainment teacher said, “I think the most important lesson for anyone to learn in maths is the harder you work at it, the better you’ll do” (Teacher M4).
If current circumstances necessitate mixed-attainment groups, then we hope that this leads to more collaboration between teachers, generates more good-practice examples of mixed-attainment teaching and makes the transition to mixed-attainment groups less daunting for other schools. What started out as a pandemic-induced necessity, may just be the catalyst we need for improving pupils’ experiences of learning mathematics.
For further details, see: Francome, T., & Hewitt, D. (2018). “My math lessons are all about learning from your mistakes”: how mixed-attainment mathematics grouping affects the way students experience mathematics. Educational Review. https://doi.org/10.1080/00131911.2018.1513908 Available for free: here
###### Centre for Mathematical Cognition
We write mostly about mathematics education, numerical cognition and general academic life. Our centre’s research is wide-ranging, so there is something for everyone: teachers, researchers and general interest. This blog is managed by Dr Bethany Woollacott, a research associate at the CMC, who edits and typesets all posts. Please email b.woollacott@lboro.ac.uk if you have any feedback or if you would like information about being a guest contributor. We hope you enjoy our blog!
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http://www.troublefreepool.com/threads/60698-How-accurate-is-the-CYA-test | 1,513,401,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948583808.69/warc/CC-MAIN-20171216045655-20171216071655-00532.warc.gz | 485,544,796 | 11,930 | # Thread: How accurate is the CYA test?
1. ## How accurate is the CYA test?
I've been doing this for over a year and I still have trouble deciding on my CYA level. The black spot on the bottom becomes invisible, then I think I have added too much, then I take out a little and can't decide if I really see a shadow or not. Do other people have this problem? Maybe I am a slow learner, or just old and dumb.
If 10 experienced pool owners on the forum each tested the same water for CYA, do you think they would all get the same reading? If not, do you think the readings would all be with in 5 of each other? How far apart would you guess the highest and lowest reading would be if we could do this experiment?
2. ## Re: How accurate is the CYA test?
It is +/- 15ppm. Are you pouring the solution back and fork to see if you keep getting the same reading? Or take an average.
3. ## Re: How accurate is the CYA test?
I would guess we'd all be within 5 of each other.
If you want to perfect your testing technique, order the CYA standard solution the next time you order refills. Mix it up as if it were your pool water and see if you get 50. Experiment with the lighting, how far you hold it, anything. Whatever it takes so that you hit 50 +/-every time. Then when you test your water, do it exactly the same.
4. ## Re: How accurate is the CYA test?
Thanks for the info. I'll try the CYA standard solution.
5. ## Re: How accurate is the CYA test?
The +/-15 ppm is the range Taylor gives and is like a 95th percentile for any of the measured CYA levels. It most likely applies to a high CYA level where 15 ppm isn't a great distance in the tube so its much more likely to have that size of error. For lower CYA levels in the 50 ppm or so range, I'd expect the error to be less than 10 ppm and for experienced people they'll likely be within +/- 5 of their average among them. Once you get used to it, it's a pretty consistent test, assuming you've got reasonably consistent lighting conditions such as standing outside with the sun to your back with your body shading the tube.
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• | 538 | 2,234 | {"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-2017-51 | longest | en | 0.966938 |
https://puzzlefry.com/puzzles/what-did-the-wise-man-say-to-them/ | 1,726,538,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00035.warc.gz | 430,954,746 | 33,735 | # What did the wise man say to them?
1,500.0K Views
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can.
What did the wise man say to them?
pnikam Expert Asked on 24th August 2015 in
Switch the camels
anikam Expert Answered on 24th August 2015.
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Matemaattis-luonnontieteellinen Matematiikan ja tilastotieteen laitos Jere Meril¨ ainen On Hedging and Pricing of Options Matematiikka Pro gradu -tutkielma Lokakuu 2019 73 s. Hedging, Pricing, Option, Mathematical finance, Arbitrage, Martingale, Financial market model Kumpulan tiedekirjasto In this thesis we cover some fundamental topics in mathematical finance and construct market models for the option pricing. An option on an asset is a contract giving the owner the right, but not the obligation, to trade the underlying asset for a fixed price at a future date. Our main goal is to find a price for an option that will not allow the existence of an arbitrage, that is, a way to make a riskless profit. We will see that the hedging has an essential role in this pricing. Both the hedging and the pricing are very import tasks for an investor trading at constantly growing derivative markets. We begin our mission by assuming that the time parameter is a discrete variable. The ad- vantage of this approach is that we are able to jump into financial concepts with just a small quantity of prerequisites. The proper understanding of these concepts in discrete time is crucial before moving to continuous-time market models, that is, models in which the time parameter is a continuous variable. This may seem like a minor transition, but it has a significant impact on the complexity of the mathematical theory. In discrete time, we review how the existence of an equivalent martingale measure characte- rizes market models. If such measure exists, then market model does not contain arbitrages and the price of an option is determined by this measure via the conditional expectation. Furthermore, if the measure also unique, then all the European options (ones that can be exercised only at a pre- determined time) are hedgeable in the model, that is, we can replicate the payoffs of those options with strategies constructed from other assets without adding or withdrawing capital after initial in- vestments. In this case the market model is called complete. We also study how the hedging can be done in incomplete market models, particularly how to build risk-minimizing strategies. After that, we derive some useful tools to the problems of finding optimal exercise and hedging strategies for American options (ones that can be exercised at any moment before a fixed time) and introduce the Cox-Ross-Rubinstein binomial model to use it as a testbed for the methods we have developed so far. In continuous time, we begin by constructing stochastic integrals respect to the Brownian motion, which is a stochastic component in our models. We then study important properties of stochastic integrals extensively. These help us comprehend dynamics of asset prices and portfolio values. In the end, we apply the tools we have developed to deal with the Black-Scholes model. Particularly, we use the Itˆ o’s lemma and the Girsanov’s theorem to derive the Black-Scholes partial differential equation and further we exploit the Feynman-Kac formula to get the celebrated Black-Scholes formula. Tiedekunta/Osasto — Fakultet/Sektion — Faculty Laitos — Institution — Department Tekij¨ a — F¨ orfattare — Author Ty¨ on nimi — Arbetets titel — Title Oppiaine — L¨ aro¨ amne — Subject Ty¨ on laji — Arbetets art — Level Aika — Datum — Month and year Sivum¨ ar¨ a — Sidoantal — Number of pages Tiivistelm¨ a — Referat — Abstract Avainsanat — Nyckelord — Keywords ailytyspaikka — F¨ orvaringsst¨ alle — Where deposited Muita tietoja — ¨ Ovriga uppgifter — Additional information HELSINGIN YLIOPISTO — HELSINGFORS UNIVERSITET — UNIVERSITY OF HELSINKI
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Jere Merilainen
Matematiikka
Hedging, Pricing, Option, Mathematical finance, Arbitrage, Martingale, Financial market model
Kumpulan tiedekirjasto
In this thesis we cover some fundamental topics in mathematical finance and construct market
models for the option pricing. An option on an asset is a contract giving the owner the right, but
not the obligation, to trade the underlying asset for a fixed price at a future date. Our main goal
is to find a price for an option that will not allow the existence of an arbitrage, that is, a way
to make a riskless profit. We will see that the hedging has an essential role in this pricing. Both
the hedging and the pricing are very import tasks for an investor trading at constantly growing
derivative markets.
We begin our mission by assuming that the time parameter is a discrete variable. The ad-
vantage of this approach is that we are able to jump into financial concepts with just a small
quantity of prerequisites. The proper understanding of these concepts in discrete time is crucial
before moving to continuous-time market models, that is, models in which the time parameter is a
continuous variable. This may seem like a minor transition, but it has a significant impact on the
complexity of the mathematical theory.
In discrete time, we review how the existence of an equivalent martingale measure characte-
rizes market models. If such measure exists, then market model does not contain arbitrages and
the price of an option is determined by this measure via the conditional expectation. Furthermore,
if the measure also unique, then all the European options (ones that can be exercised only at a pre-
determined time) are hedgeable in the model, that is, we can replicate the payoffs of those options
with strategies constructed from other assets without adding or withdrawing capital after initial in-
vestments. In this case the market model is called complete. We also study how the hedging can be
done in incomplete market models, particularly how to build risk-minimizing strategies. After that,
we derive some useful tools to the problems of finding optimal exercise and hedging strategies for
American options (ones that can be exercised at any moment before a fixed time) and introduce the
Cox-Ross-Rubinstein binomial model to use it as a testbed for the methods we have developed so far.
In continuous time, we begin by constructing stochastic integrals respect to the Brownian
motion, which is a stochastic component in our models. We then study important properties of
stochastic integrals extensively. These help us comprehend dynamics of asset prices and portfolio
values. In the end, we apply the tools we have developed to deal with the Black-Scholes model.
Particularly, we use the Ito’s lemma and the Girsanov’s theorem to derive the Black-Scholes
partial differential equation and further we exploit the Feynman-Kac formula to get the celebrated
Black-Scholes formula.
Tekija — Forfattare — Author
Oppiaine — Laroamne — Subject
Tyon laji — Arbetets art — Level Aika — Datum — Month and year Sivumaara — Sidoantal — Number of pages
Tiivistelma — Referat — Abstract
Avainsanat — Nyckelord — Keywords
HELSINGIN YLIOPISTO — HELSINGFORS UNIVERSITET — UNIVERSITY OF HELSINKI
University of Helsinki
Master’s thesis
Author: Jere Merilainen
Supervisor: Dario Gasbarra
October 16, 2019
Contents
1 Introduction 1 1.1 Discrete time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Continuous time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 On bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Discrete-time financial market 6 2.1 Some probability preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.2 Market dynamics and information . . . . . . . . . . . . . . . . . . . . . . . 8 2.3 Self-financing strategies and discounting . . . . . . . . . . . . . . . . . . . 10 2.4 Arbitrage and martingale measures . . . . . . . . . . . . . . . . . . . . . . 11 2.5 Local arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.6 Geometric interpretation of arbitrage . . . . . . . . . . . . . . . . . . . . . 17 2.7 European options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.8 Pricing and hedging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.9 Market completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.10 Hedging in incomplete market . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.11 Change of numeraire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.12 American options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.13 Binomial model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.13.1 Market model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 2.13.2 Option valuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 2.13.3 Hedging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3 Continuous-time financial market 47 3.1 Brownian motion and semimartingales . . . . . . . . . . . . . . . . . . . . 47 3.2 Stochastic integrals and Ito calculus . . . . . . . . . . . . . . . . . . . . . . 50 3.3 Girsanov’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.4 Feynman-Kac formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.5 Pricing and free lunch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.6 Black-Scholes model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.6.1 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.6.2 Black-Scholes partial differential equation . . . . . . . . . . . . . . . 66 3.6.3 Hedging and pricing . . . . . . . . . . . . . . . . . . . . . . . . . . 67 3.6.4 Beyond Black-Scholes . . . . . . . . . . . . . . . . . . . . . . . . . . 70
Chapter 1
Introduction
Consider the stocks of Finnish retailing conglomerate Kesko Corporation which are traded in the Helsinki Stock Exchange for the price of 58 euros per share at the time of writing. You are offered a contract (or an option) in which you have an opportunity to buy one stock of Kesko for the price of 60 euros at a future date. How much would you pay for this contract? Such an innocent question takes us to a long journey to the magnificent theory of the mathematical finance.
Let us try to unravel the problem. Our profit from the contract depends on the stock price so it would be beneficial to model the movements of the stock price somehow. As rational traders we do not want to pay too high a price for the contract and hence allowing seller to make a riskless profit at our expense. On the other hand, no-one is willing to sell us the contract with too low price. But what determines aforementioned ”too high” and ”too low” prices? The answer is often given by hedging.
Assume then, that we can replicate (or hedge) our profit from the contract with a portfolio constructed solely from other assets without adding or withdrawing capital after an initial investment. Let us assume further that we know the prices of those other assets. In this case the initial value of such replicating portfolio must coincide with the price of the contract. Otherwise there would be a possibility to make a riskless profit. A strategy that exploits this possibility is called an arbitrage.
Let us justify the claim we made. If the initial value of the replicating portfolio is higher, then we can short sell the portfolio and use the received money to buy the contract. If the initial value is lower, then we short sell the contract and buy the replicating portfolio. In both cases, we can make a positive profit without initial capital or the risk of losing money. This simplified principle describes the relationship between hedging and pricing and the idea will be carried throughout the thesis.
In this thesis we want to give pricing and hedging problems a rigorous (yet clear) mathematical treatment. We begin by restricting ourselves to financial market models with a discrete time variable. This allows us to dive into financial topics quickly without going into the rather demanding theory of stochastic integrals as in continuous-time mod- els. It is important to understand financial notions in discrete time so that one’s ”forest’s scenery is not obscured by the sight of the trees” when one moves on to mathematically more challenging continuous-time models.
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1.1 Discrete time
We begin the chapter concerning discrete time by recalling the concepts of Lp-spaces and the conditional expectation. Then we directly move on to financial topics by mathema- tizing discrete-time financial market models. These models include assumptions of the underlying probability space, stochastic processes which represent asset prices, and the information structure. In the following sections, we study the properties that financial models may have, particularly arbitrage-freeness and completeness.
An important notion, concerning aforementioned properties, is self-financing strategy, that is, strategy which is build without adding or withdrawing capital after an initial investment. A model is called arbitrage-free if there does not exist self-financing strategies that can yield positive profit without initial capital or risk of loosing money. Further the model is called complete if every (European) claim can be replicated with a self-financing strategy.
Both arbitrage-freeness and completeness are characterized by the existence of an equivalent martingale measure, that is, a probability measure which has property that discounted asset price processes are martingales under this measure. This means that price processes, which are scaled by another asset price (called numeraire), are constant on average. We call the equivalent martingale measure risk-neutral, since it does not reward for risk-taking, that is, every asset price have the same expected value (equalling numeraire’s expected value) under this measure despite the riskiness of the asset.
We will see that the existence of an equivalent martingale measure precludes the exis- tence of an arbitrage. Furthermore, the existence of such measure is entirely determined by the arbitrage-freeness of a market model. This is called the first fundamental the- orem of asset pricing. If the existing measure is also unique, then the market model is complete and vice versa. This claim is called the second fundamental theorem of as- set pricing. These elegant theorems connect a rather abstract mathematical notion to practical financial concepts.
In the later section, we will introduce different types of options, particularly European call and put options. Aforementioned options give the buyer a right, but not an obligation, to trade the underlying asset for a fixed price at predetermined time. Prices of these mentioned options link to each other via so-called put-call parity, hence it suffices to find either one of the prices in order to know them both. In the one-period market model, the price of an option can be solved conveniently with a geometric analysis.
The equivalent martingale measure proves its usefulness once again when it comes to the pricing of options. We prove that the conditional expectation of the discounted payoff of an option, with respect to mentioned measure, yields such a price that arbitrage- freeness of a market model is preserved. If the market model is complete or the option is hedgeable, then the price is unique. If the option can not be replicated, then the prices belong to an open interval. Notice that there is always numeraire attached to an equivalent martingale measure. But we will see that it can be changed, by defining the Radon-Nikodym derivative a right way, so that the returned measure is another equivalent martingale measure.
Often market models are not complete, therefore we build ways hedge options, when the full replication is not possible. These techniques, particularly risk-minimizing meth-
2
ods, depend highly on the features of a underlying L2-space. We will also familiarize our- selves with the super-hedging, that is, a way to surpass future claims with a self-financing strategy.
The American options generalize the European ones in a such way that they can be exercised at any moment before (and including) the fixed expiration time. This feature excites the problem of finding an optimal exercise strategy along with more complicated pricing and hedging problems. To tackle these problems, we introduce tools called the Snell envelope, that is, a supermartingale which dominates the payoff of an American option and the Doob’s decomposition, that is, a way to decompose a stochastic process in to a martingale and predictable part.
The last topic we will cover in discrete time is called the Cox-Ross-Rubinstein binomial market model. In this simple model, we have two asset, one risky asset (say a stock) and one riskless asset (say a bank account). The bank account yields constant return, while the stock price has exactly two possible states in the next point of time: either it has gone up or down. Assuming reasonable conditions, we can find a unique equivalent martingale measure in this model, hence it is both arbitrage-free and complete. We will use this knowledge to derive the Cox-Ross-Rubinstein pricing formula for a European call option, which is the discrete-time analogue of the well-known Black-Scholes formula.
1.2 Continuous time
In the continuous time, we begin by introducing the Brownian motion and some of its well-known properties, for example, its quadratic variation up to time t is simply t. The Brownian motion will be a stochastic component in our continuous-time stock price pro- cesses. In the same section, we generalize concept of martingale to local martingale and even further to semimartingale. These notions appear, when we define stochastic integrals, that is, integrals with stochastic process as integrator.
Stochastic integrals and their properties are important, since they help us comprehend dynamics of asset prices and portfolio values. Particularly advantageous is to know, when a process defined by a stochastic integral is a local martingale or a (true) martingale and how to calculate the quadratic variation of such process. We will define and study stochastic integrals extensively. After that, we are ready to define Ito processes and express essential results from the Ito’s calculus called the Ito’s formula and the stochastic integration by parts.
An Ito process consists of drift (or the direction of movement) and diffusion (or the stochastic fluctuation of movement) terms, where the latter is defined by a stochastic integral. The Ito process is a (local) martingale if and only if it has a null drift. Ito’s formula presents dynamics for a certain kind of function, that is given Ito processes as arguments. It can be seen as the stochastic calculus version of the chain rule, where the extra term comes from the non-zero quadratic variation.
In the following sections, we introduce two major theorems. The first one offers a way to change a probability measure so that we can eliminate the drift from the Ito process without affection for the diffusion term by changing the Brownian motion. This is called the Girsanov’s theorem and it can help us find a risk-neutral measure that turns stock
3
processes into (local) martingales and give the explicit stock price dynamics under this new measure.
The other theorem that we will motivate is called the Feynman-Kac formula. It constructs solution to deterministic partial differential equation with certain boundary condition by mixing stochastics into it. The link between deterministic problem and a stochastic process, in this theorem, results fundamentally from the Ito’s formula. These type of boundary value problems are common in the mathematical finance, hence the Feynman-Kac formula, that originates from a physics problem, is useful to us.
In the final section, we introduce the Black-Scholes model. In this model we have two asset, one risky asset (say a stock) and one riskless asset (say a bank account). The bank account is assumed to compound constant interest continuously, while the stock price follows the geometric Brownian motion. We start by using Girsanov’s theorem to get rid of the drift from the discounted stock price process. This stands for risk-neutral approach.
Then we use these risk-neutral stock price dynamics along with Ito’s formula to derive the Black-Scholes partial differential equation for the value of the self-financing Marko- vian strategy, that is, a strategy which depends solely on the current value of the stock price process and the time parameter. We did this because, then we know that if a Euro- pean option is replicable with a self-financing Markovian strategy, then the value of the replicating portfolio must satisfy the Black-Scholes partial differential equation and the solution to this yields the price for the option.
Fortunately, the Black-Scholes model is both arbitrage-free and complete (assumimg some extra restrictions to these definitions). This means that we can find a unique arbitrage-free price for a European option by solving the Black-Scholes partial differ- ential equation with a boundary conditions given by the replication condition, that is, the terminal value of the replication portfolio must coincide with the payoff of the option. The solution to this boundary value problem is given by the Feynman-Kac formula. Us- ing this knowledge, we derive the celebrated Black-Scholes formula, that is, formula for the arbitrage-free price of the European call option. In the end, we discuss some of the shortages of the Black-Scholes model and suggest further topics.
1.3 On bibliography
The most important sources of this thesis, in discrete time, are ”Stochastic finance: An Introduction in Discrete Time” (2016) by Follmer and Schied [10], ”Mathematics of Fi- nancial Markets” (1999) by Elliot and Kopp [9], and ”Financial Mathematics: Theory and Problems for Multi-period Models” (2012) by Pascucci and Runggaldier [16]. The corresponding sources, in continuous time, are ”Arbitrage Theory in Continuous Time” (2004) by Bjork [2], ”Stochastic Processes” (2011) by Bass [1], and ”PDE and Martingale Methods in Option Pricing” (2011) by Pascucci [15].
The following reviews are personal opinions of the writer of this thesis. It was sur- prising to discover that there exists so many books on the same topics but completely different level of rigour and generality. The best first book in (multi-period) discrete-time mathematical finance is Pascucci and Runggaldier [16] followed by Elliot and Kopp [9]. The first-mentioned is low on details and leaves gaps that have to be filled up by other
4
sources afterwards. The latter does not have this problem but it is a bit harder to grasp. The book by Follmer and Schied [10] contains almost 600 pages and it can be con-
sidered as the cornucopia of the discrete-time mathematical finance. It might not be the most accessible book at first reading, since it contains a rather precise treatment of math- ematical finance. But after familiarising oneself, for example, with [16] or [9], Follmer and Schied turns into the tremendous source of information.
”The Mathematics of Arbitrage” (2006) by Delbaen and Schachermayer [7] and ”Fun- damentals and Advanced Techniques in Derivatives Hedging” by Bouchard [3] are great for mathematicians but rather abstract by nature. I would highly recommend reading something more intuitive (to understand the larger picture of finance) before entering these detail-oriented books. The latter deals with the most general case of discrete-time financial markets.
Both the discrete-time and continuous-time mathematical finance demand a strong background on the probability theory. In this thesis, the most important probability prerequisites for mathematical finance are given but obviously we are not able to give the exhaustive knowledge of probability theory. Good sources to strengthen one’s scholarship are ”Probability: Theory and Examples” (2010) by Durrett [8] and ”Probability with Martingales” (1991) by Williams [19].
In continuous time, it was harder to find a suitable source. Overall the books were much more disorganized than discrete-time books. The book by Bjork [2] is great on intuition but does not quench the thirst for mathematical details. I found the book by Pascucci [15] the most enjoyable to read, therefore the continuous-time part leans heavily on this source. The continuous-time mathematical finance requires a sound background on the stochastic calculus. I would recommend ”Introduction to Stochastic Calculus for Finance: A New Didactic Approach” (2007) by Dieter Sondermann for the first book in stochastic calculus, even though this is not listed in the bibliography at the end of the thesis.
Although the lectures and notes in the University of Helsinki are not mentioned as a direct reference in the bibliography, those have had a big influence on the thesis. Partic- ularly influential were lectures notes on financial economics by Harri Nyrhinen, lectures on mathematical finance and discussions after and during the lectures by Dario Gasbarra, and lecture notes on mathematical finance by Tommi Sottinen.
This thesis contains constant balancing between rigour and clarity. Our main goal is to yield applicable results to the problems of pricing and hedging of options. Therefore some irrelevant details are omitted but references to these details are given. Nevertheless the self-contained theory is included to support the results.
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Discrete-time financial market
In this chapter, we assume that the time parameter is a discrete variable with distinct values 0, 1, 2, . . . , T , where T ∈ N.
2.1 Some probability preliminaries
Before going to financial applications, we recall some notions from probability theory that we will need later. Other preliminaries will be introduced as we progress through thesis. We restrict ourselves to real-valued random variables and stochastic processes unless otherwise stated.
Definition 2.1.1. Let (,F ,P) be a probability space. Then for every p ∈ [1,∞] we denote
L(p)(,F ,P) := {X : → R | X is F -measurable, Xp <∞}, (2.1.2)
where the p-norm · p of random variable X is defined by
Xp := E(|X|p) 1 p =
(∫
Xp := inf{M ≥ 0 | P(|X| > M) = 0} (2.1.4)
when p =∞. Further X is called essentially bounded if X∞ <∞. ♦
Technically · p is only a seminorm in L(p)(,F ,P). If we want a (true) norm, we need to be little more explicit. Let us denote
X0 = {X : → R | X is F -measurable, X = 0 P-almost surely} (2.1.5)
and define equivalence class [X] := X+X0 for X ∈ L(p)(,F ,P). Thus for every Y ∈ [X] it holds that Y = X P-almost surely. Now we are ready for conclusive definition.
6
Lp(,F ,P) := {[X] := X + X0 | X ∈ L(p)(,F ,P)} (2.1.7)
and define [X]p := Xp and E([X]) := E(X) for [X] ∈ Lp(,F ,P). ♦
We will often use notation Lp(P) or even Lp for Lp(,F ,P) if some of ,F , and P are clear from the context. Recall that p-norm · p is a true norm in Lp and that Lp
has some useful features. For example, for every p ∈ [1,∞] we have that Lp equipped with p-norm is complete normed space (or Banach space), that is, every Cauchy-sequence converges. Particularly we will use L2 space which is complete inner product space (or Hilbert space) when it is equipped with inner product
[X], [Y ] := E(XY ) =
X(ω)Y (ω)dP(ω) for [X], [Y ] ∈ L2(,F ,P). (2.1.8)
We notice that inner product induces norm to L2, since [X]2 = √ [X], [X]. Addition-
ally, we will denote the space (of equivalence classes) of random variables by
L0(,F ,P) := {[X] := X + X0 | X : → R, X is F -measurable}. (2.1.9)
From now on, we will leave out square brackets [ · ], when we refer to elements of Lp or L0
spaces. So instead of talking about equivalence classes, we talk about underlying random variables for simplicity. Also for multidimensional random variable X = (X1, . . . , Xd) we use X ∈ Lp to indicate that X i ∈ Lp for every i = 1, . . . , d. In the next definition we denote by 1A : → {0, 1} the indicator random variable, such that 1A(ω) = 1 if ω ∈ A and 1A(ω) = 0 otherwise.
Definition 2.1.10. Let (,F ,P) be a probability space, and G ⊆ F a sub-sigma-algebra of F . A Conditional expectation of random variable X ∈ L1(,F ,P) given G, denoted by E(X|G), is (G-measurable) random variable in L1(,G,P) which satisfies
E(1AE(X|G)) = E(1AX) for each A ∈ G (2.1.11)
The conditional expectation has lots of useful properties. Next lemma covers the ones we will need later. For the proof of existence of conditional expectation and its properties, see for example Chapter 9 of [19].
Lemma 2.1.12. (Properties of conditional expectation) Let us assume that X, Y ∈ L1(,F ,P) and a, b ∈ R, then (using the notation of Definition 2.1.10) we have (P-almost surely)
(i) linearity: E(aX + bY |G) = aE(X|G) + bE(Y |G),
(ii) independency property: if X is independent of G, then E(X|G) = E(X),
(iii) ”taking out what is known”: if X is G-measurable, then E(XY |G) = XE(Y |G), particularly E(X|G) = X,
(iv) tower property: if H is a sub-sigma-algebra of G, then E(E(X|G)|H) = E(X|H), particularly E(E(X|G)) = E(X).
We will later introduce other probability measures. Notation EP indicates that the
expectation is to be taken under the measure P. A plain symbol E means that expectation is to be taken under the initial probability measure P.
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2.2 Market dynamics and information
Consider a probability space (,F ,P). We assume that sample space has a finite number of elements and that F is the power set of with P({ω}) > 0 for any ω ∈ . The prices of assets (for example stock prices) change in time and can be considered as stochastic processes. Elements ω ∈ are called scenarios since they correspond to different scenarios of the possible changes of the asset prices. The initial probability measure P is called objective or physical measure. Let us assume that t0 < t1 < · · · < tN are real-valued and denote by
T = {t0, t1, . . . , tN}, the set of the trading times, where N ∈ N determines the number of trading periods. Without loss of generality, we assume that T = {0, 1, . . . , T}, where t0 = 0 can be interpreted as today’s date and tN = T ∈ N as the expiration date of a derivative (more on this later).
We suppose that the market consists of a non-risky asset (bank account) S0 and d ∈ N number of risky assets (stocks) S1, . . . , Sd. Each Si = {Sit | t ∈ T} is a positively real- valued stochastic process, where random variable Sit > 0 denotes the price of an asset at time t ∈ T for every i = 0, 1, . . . , d. We assume that the non-risky asset has the following deterministic dynamics
S0 0 = 1 and S0
t = rtS 0 t−1 for t ∈ T \ {0}, (2.2.1)
where Xt = Xt−Xt−1 and rt > −1 denotes risk-free interest rate in the period (t−1, t]. In a similar fashion, the risky assets have the following stochastic dynamics
Si0 > 0 and Sit = Ri tS
i t−1 for t ∈ T \ {0}, (2.2.2)
where Ri t is a random variable that represents the rate of return of the asset i in the
period (t− 1, t]. The dynamics of the risky asset can be written in a demonstrative form Sit = (1 + Ri
t)S i t−1. We could also calculate the rate of return Ri
t = (Sit − Sit−1)/Sit−1, provided that we knew the prices of the asset i at times t− 1 and t.
We make a natural assumption that the investor’s decisions at moment t ∈ T can be based on information available up to the moment t and not the future information. Furthermore, we assume that the information is non-degreasing in time so that investors learn without forgetting. The information structure available to the investors is given by an increasing and finite sequence F0 ⊂ F1 ⊂ F2 ⊂ . . . ⊂ FT = F of sub-sigma-algebras of F . In the information structure we assume F0 to be trivial, that is, F0 = {∅,}. We call an increasing family of sigma-algebras a filtration F = {Ft | t ∈ T} on (,F ,P) and a probability space equipped with the filtration F a stochastic basis (,F ,F,P).
A stochastic process X = {Xt | t ∈ T} is called adapted (to the filtration F), if Xt
is Ft-measurable for every t ∈ T. This means that, if X is adapted, then the value Xt(ω) is known to us at time t. We assume every price process Si is adapted and thus every the rate of return process Ri is adapted. Let us denote Rt = (R1
t , . . . , R d t ) and
St = (S0 t , S
1 t , . . . , S
d t ). Since in the market, based on (2.2.2), the sequence Rt will be the
only source of randomness, we assume that Ft is generated by Rt, so that Ft = FRt = σ{Rk | k ≤ t} for t ∈ T \ {0}. By the bijective correspondence between the processes Ri
and Si, we have Ft = FRt = FSt . Finally we have defined a market model (,F ,P,T,F, S).
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Remark 2.2.3. At the beginning of the section, we assumed that we are operating in the finite probability space, that is, sample space has finitely many elements. Consequently, every Lp(,F ,P) with p ∈ [1,∞] contains the same random variables. Furthermore, in the market model defined above, the price of an asset i at time t, explicitly Sit : → R, belongs to Lp(,Ft,P) for every p ∈ [1,∞]. Although this discrete time market model is not the most general setting, the results in the following sections usually hold in the more general discrete time models. We will use the notion of Lp spaces occasionally anyway to remind us of the generality of results. The advantage of our market model is that, it provides intuition and clarity for the mathematics of financial concepts. 4
Example 2.2.4. Let us define two-period market model with two assets, stock S and bank account B. Now T = {0, 1, 2} and we define
= {(ω1, ω2) ∈ R2 |ω1, ω2 ∈ {0, 1}} = {(0, 0), (1, 0), (0, 1), (1, 1)},
where we assume that P({ω}) > 0 for all ω ∈ . Let us suppose that the bank account yields with constant interest rate r = 1/10, so that B has dynamics
Bt = (1 + r)t, t = 0, 1, 2
and assume that the stock price has the following dynamics
S0 = 150, S1(ω1) = [1 +R(ω1)]S0 and S2(ω1, ω2) = [1 +R(ω2)]S1(ω1),
with the rate of return given by
R(ωt) =
2
5 , ωt = 0
for t = 1, 2. Figure 2.1 visualizes the evolution of the stock price process. Each node in the tree corresponds to a stock price at given time and ”the state of the world”. Filtration F = {Ft | t = 0, 1, 2} is given by
F0 = {∅,} F1 =
} ∪ F0
} ∪ F0
F2 = 2,
where F2 is the power set of , that is, it contains all the subsets of so in this case 16 different subsets. Intuitively F0 does not contain any information about the market, F1
contains information whether the stock price has gone up or down in the first time period and, F2 contains all the information, that can be received from the market. 4
9
150
120
96
S2
Figure 2.1: The possible paths of the stock price process form a tree-shaped structure.
2.3 Self-financing strategies and discounting
Consider Rd+1-valued stochastic process θ = {(θ0 t , θ
1 t , . . . , θ
d t ) | t ∈ T}, where θit represents
the amount of asset Si hold during the period (t − 1, t]. Aforementioned vector-valued process θ is called a strategy (or a portfolio) and the value of the portfolio θ at time t (precisely just after the asset values have changed to St) is given by
V θ t = θt · St :=
d∑ i=0
where we used notation St = (S0 t , S
1 t , . . . , S
d t ) for price vector at time t and notation θt =
(θ0 t , θ
1 t , . . . , θ
d t ) for strategy in time interval (t− 1, t]. The value V θ
0 is the investor’s initial investment. The investor chooses his/hers time t portfolio θt as soon as the stock prices at time t−1 are known. Then he/she holds this portfolio the time interval (t−1, t]. Therefore we require that strategy is a predictable stochastic process, that is, X = {Xt | t ∈ T} is predictable, if Xt+1 is Ft-measurable for every t ∈ T \ {T}. Remark 2.3.2. Note that we allow negative values for θit. We interpret negative amount of asset as short-selling or borrowing of assets. 4
Definition 2.3.3. A strategy θ is self-financing if it satisfies
θt+1 · St = θt · St, (2.3.4)
for every t ∈ T \ {T}. ♦
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From (2.3.1) we know that θt · St is the value of the portfolio at time t. So in the self-financing strategy, the new portfolio θt+1 (for time interval (t, t + 1]) is created with investor’s capital from the portfolio at time t without adding new capital outside of the portfolio and without withdrawing capital from the portfolio.
Remark 2.3.5. Since there is no information available before time t = 0, we set θ0 = θ1. 4 Let us assume that θ is a self-financing strategy. Hence
V θ t = θt · St − θt−1 · St−1
= θt · St − θt · St−1
= θt ·St. (2.3.6)
We define the gain process associated with θ by setting
Gθ 0 = 0 and Gθ
t = t∑
u=1
θu ·Su for t ∈ T \ {0}. (2.3.7)
Now we see that θ is self-financing if and only if
V θ t = V θ
0 +Gθ t (2.3.8)
for all t ∈ T. For a rational investor money available today is worth more than the identical sum in
the future because money has potential to grow in value over a given period of time. If we want to compare two sums from the different moments of time, we have to take into account this time value of money. Let us introduce discounted price of the asset i and discounted value of the portfolio θ which are defined by
Sit := Sit S0 t
and V θ t := θt · St (2.3.9)
for t ∈ T, respectively. Furthermore, if θ is self-financing, then (2.3.6) and (2.3.8) yield
V θ t = V θ
0 + t∑
u=1
t (2.3.10)
for all t ∈ T. In equations (2.3.9) the asset S0 is called a numeraire. Numeraire is an asset in terms of whose price the relative prices of all other asset are expressed. In other words, other prices are normalized by the price of the numeraire asset. We could have taken any of the assets as a numeraire as long as the price of chosen asset is strictly positive. Choosing the right numeraire can make computations much easier.
2.4 Arbitrage and martingale measures
An arbitrage is a strategy that can yield positive profit without initial capital or risk of losing money. Loosely speaking the arbitrage creates money out of thin air. With rational
11
investors in the market, there should not exist arbitrages. One can argue that if arbitrage existed, then many of investors would exploit it and thus the prices of assets would correct themselves, by the law of supply and demand, in a such way that the arbitrage would vanish. This reasoning restricts our attention to markets with no arbitrages.
Definition 2.4.1. Self-financing and predictable strategy θ is an arbitrage if the following conditions hold
V θ 0 = 0, P(V θ
T ≥ 0) = 1 and P(V θ T > 0) > 0. (2.4.2)
We say that a market model is arbitrage-free if there does not exist an arbitrage. ♦
Next we will start to lay the foundation of an important concept called equivalent martingale measure. We will see that the existence of such measure is closely connected with the existence of an arbitrage.
Definition 2.4.3. A stochastic process M = {Mt | t ∈ T} is called a martingale (with respect to (F,P)) if Mt ∈ L1(,Ft,P) for all t ∈ T and
E(Mt|Ft−1) = Mt−1 for every t ∈ T \ {0}. (2.4.4)
A supermartingale is defined similarly, except that the last condition is replaced by E(Mt|Ft−1) ≤ Mt−1, and a submartingale is defined by replacing it by E(Mt|Ft−1) ≥ Mt−1. ♦
Note that for a martingale M it holds that E(Mt|Ft−1) = 0 for all t ∈ T \ {0}. Thus, by the law of total expectation, E(Mt) = 0 and hence E(Mt) = E(Mt−1) for all t ∈ T \ {0}, so that a martingale is ”constant on average”. Similarly, a supermatingale decreases, and a submartingale increases on average. If we assume that gambler’s wealth process is a martingale, then we can interpret that gambler is playing a ”fair” game. A supermartingale would model ”unfavourable” and a submartingale ”favourable” game.
Furthermore, for every t ∈ T \ {T}, we have
Mt = E(Mt+1|Ft) = E(E(Mt+2|Ft+1)|Ft) = E(Mt+2|Ft),
by the tower property of conditional expectation. We could continue the procedure above (until we reach time T in the expectation) and hence we have the following useful result: Mu = E(Mt|Fu) for every u ≤ t, when u, t ∈ T. We also need the following lemma of martingales later:
Lemma 2.4.5. Let c ∈ R be a constant and let also M1 and M2 be martingales (with respect to (F,P)), then c+M1 := {c+M1
t | t ∈ T} and M1 +M2 := {M1 t +M2
t | t ∈ T} are martingales (with respect to (F,P)).
Proof. The claim follows immediately from the linearity of conditional expectation.
Definition 2.4.6. Let M = {Mt | t ∈ T} be a (super)martingale and φ = {φt | t ∈ T} be a predictable stochastic process. Process X = φ • M given by
X0 = 0 and Xt = t∑
u=1
is the (super)martingale transformation of M by φ. ♦
12
The martingale transform φ • M is the discrete analogue of the stochastic integral∫ φ dM . We can think φu as stake on game at time u. Since φ is predictable, the value of
φu can be decided based on the information up to (and including) time u− 1. The return of the game in time interval (u−1, u] is φu(Mu−Mu−1) (= φuMu) and the total return up to time t is
(φ • M)t = φ1M1 + φ2M2 + . . .+ φtMt = t∑
u=1
φuMu.
Proposition 2.4.8. (”We can’t beat the system with a finite capital”) Let process φ be predictable and bounded, so that for some K ∈ [0,∞), |φu(ω)| ≤ K for every u ∈ T and ω ∈ .
(i) If M is a martingale, then φ • M is a martingale.
(ii) If M is a supermartingale and φ is non-negative, then φ • M is a supermartingale.
Proof. Write X = φ • M . Since φ is bounded and Ft−1-measurable,
E(Xt −Xt−1|Ft−1) = φtE(Mt −Mt−1|Ft−1) = 0 (and ≤ 0 for a supermartingale).
Consider discounted gain process Gθ t of self-financing strategy θ. We can write the
process as
Gθ t =
t∑ u=1
d∑ i=0
(θi • Si)t. (2.4.9)
Thus if S is martingale under some probability measure Q, then the discounted gain process of a self-financing strategy is finite sum of martingales transforms and hence a martingale itself. Recall that Gθ
0 = 0, therefore EQ(Gθ t ) = EQ(Gθ
0) = 0, since Gθ is a
martingale. Recall also the equation (2.3.10) that is V θ t = V θ
0 +Gθ t , thus EQ(V θ
T ) = EQ(V θ 0 ).
This precludes the existence of an arbitrage with respect to probability measure Q: if V θ
0 = 0 and Q(V θ T ≥ 0) = 1, but EQ(V θ
T ) = 0, it follows that Q(V θ T = 0) = 1. This
remains true for P, if these two probability measures (P and Q) agree on which events have probability zero. If such measure Q can be found, then no self-financing and predictable strategy θ can be an arbitrage, thus the market model is arbitrage-free.
Definition 2.4.10. Probability measures P and P, on the same measurable space (,F), are called equivalent if and only if
P(A) = 0 ⇐⇒ P(A) = 0 for every A ∈ F .
We use notation P ∼ P for equivalent measures. ♦
Remark 2.4.11. If sample space has finite (or countable) amount of elements then P ∼ P if and only if P({ω}) = 0 ⇐⇒ P({ω}) = 0 for every ω ∈ . Recall that we exclude
from the consideration ω such that P({ω}) = 0. So P ∼ P if and only if P({ω}) > 0 for every ω ∈ . 4
13
Definition 2.4.12. An equivalent martingale measure with numeraire S0 is a probability measure Q on (,F) such that St ∈ L1(,Ft,Q) for all t ∈ T,
Q ∼ P and St−1 = EQ(St|Ft−1) for all t ∈ T \ {0}, (2.4.13)
that is, Q is such that discounted price process S is (F,Q)-martingale. We denote the set of all equivalent martingale measures (with essentially bounded
density) by Mb. We will deal with the notion of density later and often do not mention about the density, when referring to these measures. ♦
An equivalent martingale measure is often called risk neutral measure. The latter name comes from the fact that such measure does not reward for risk-taking. The expected return is same (the return of the non-risky asset) for every asset under this measure despite the riskiness of an asset. This does not mean that we believe that we live in a risk neutral world but we will still carry out the computations as if we lived. The advantage of such an approach comes when we valuate options since every investor agrees on the valuation regardless of their attitude towards risk and the price does not allow arbitrages.
Example 2.4.14. Let us find out if market model defined in Example 2.2.4 is arbitrage- free by searching the equivalent martingale measure Q. Recall that
= {(0, 0), (1, 0), (0, 1), (1, 1)}
and P({ω}) > 0 for all ω ∈ so we denote conveniently
qij = Q({(i, j)}) > 0 for i, j = 0, 1.
Since Q must be probability measure, we have Q() = 1, that is
q00 + q10 + q01 + q11 = 1.
Recall that F1 = {∅, A0, A1,}, where{ A0 = {S1 = 120} = {ω ∈ |S1(ω) = 120} = {(0, 0), (0, 1)} A1 = {S1 = 210} = {(1, 0), (1, 1)}.
Hence EQ(S2|F1) is a random variable such that
EQ(S2|F1)(ω) =
{ EQ(S2|A0), ω ∈ A0
EQ(S2|A1), ω ∈ A1,
where on the right hand side the expectations are calculated with respect to measure Q( · |Ai), defined by
Q({ω}|Ai) = Q({ω} ∩ Ai)
Q(Ai) , ∀ω ∈
14
for i = 0, 1. Now if there exists equivalent martingale measure (with numeraire B), then EQ(S1) = EQ(S1|F0) = S0
EQ(S2) = EQ(S2|F0) = S0
EQ(S2|F1)(ω) = S1(ω), for ω ∈ .
We can take out the deterministic discounting factor and use the stock prices shown in the tree in Example 2.2.4 to calculate expectations so the equations above yield four equations depending on qij:
10 11 · [210(q10 + q11) + 120(q01 + q00)] = 150
100 121 · [294q11 + 168(q01 + q10) + 96q00] = 150
100 121 · [168 q01
q01+q00 + 96 q00
10 · 210.
We can choose any 3 out of these 4 equations along with aforementioned ∑1
i,j=0 qij = 1 and solve for qij. We find out that there exists unique symmetric equivalent martingale measure Q (with numeraire B) satisfying
q00 = q10 = q01 = q11 = 1
4 .
Thus the market model in Example 2.2.4 is arbitrage-free. 4
It can be proven that the existence of an equivalent martingale measure does not only imply the absence of arbitrage but also the reverse implication is true. Next important theorem states this fact with technical details considering Radon-Nikodymin derivative (or density) which is introduced in Section 2.11.
Theorem 2.4.15. (First fundamental theorem of asset pricing) A discrete-time market model is arbitrage-free if and only if there exists at least one equivalent martingale measure Q with essentially bounded density, that is, dQ/dP ∈ L∞(P).
Proof. ”⇐”: In the earlier discussion we already concluded that the existence of equivalent martingale measure ensures that our market model is arbitrage-free. ”⇒”: In this direction the proof is more technical and we will skip it. Nevertheless, we will mention that in general probability space the proof of the theorem requires well-known result from functional analysis called the Hahn-Banach separation theorem. The general version of the proof can be found, for example, from the first chapter of [3]. In a finite probability space, one can use the special case of the Hahn-Banach separation theorem called the separating hyperplane theorem. See, for example, the proof of Theorem 3.2.2 in [9].
With respect to risk neutral measure, the processes of the discounted prices of assets are martingales. The next proposition shows that this holds also for any self-financing and predictable strategy.
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Proposition 2.4.16. Let Q be an equivalent martingale measure and θ a predictable and self-financing strategy with value V θ. Then we have
V θ t−1 = EQ(V θ
t |Ft−1) and particularly V θ 0 = EQ(V θ
t ) (2.4.17)
for every t ∈ T \ {0}.
Proof. For a self-financing strategy θ the equation (2.3.6) states that V θ t = V θ
t−1 +θt ·St. Taking the conditional expectation, we get
EQ(V θ t |Ft−1) = V θ
t−1 + θt · EQ(St|Ft−1) =0
= V θ t−1
Corollary 2.4.18. In an arbitrage-free market, if two predictable self-financing strategies θ and φ have the same final value V θ
T = V φ T almost surely, then they are also such that
V θ t = V φ
t almost surely for every t ∈ T.
Proof. Since the market is arbitrage-free, Theorem 2.4.15 states that there exists an equiv- alent martingale measure Q. Thus by Proposition 2.4.16
V θ t = EQ(V θ
T |Ft) = EQ(V φ T |Ft) = V φ
t
2.5 Local arbitrage
In earlier sections, we considered only times 0 and T as we dealt with the definition of arbitrage. In this section we will observe the intuitive matter that arbitrage-free markets are arbitrage-free for every time period [t, t+1] before T . First we formulate arbitrage-free condition explicitly.
Definition 2.5.1. A no arbitrage condition denoted by (NA) holds if for every self- financing and predictable strategies θ for which V θ
0 = 0 the following holds:
V θ T ≥ 0 P-a.s. ⇒ V θ
T = 0 P-a.s.
In a similar fashion we define no arbitrage condition in one period.
Definition 2.5.2. A local no arbitrage condition denoted by (NA)t for t ∈ T \ {T} is defined by:
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Theorem 2.5.3. (NA) holds if and only if (NA)t holds for every t ∈ T \ {T}
Proof. ”⇒”: Let us make counter-assumption that there exists such ξ ∈ L0(Ft,P) that (NA)t does not hold for some t ∈ T \{T}. Now strategy φt = ξ and φs = 0 for s ∈ T \{t} violates the (NA) condition. ”⇐”: Let us assume that (NA) does not hold, that is there exists self-financing predictable strategy φ such that V φ
0 = 0, V φ T ≥ 0 (P-a.s.) and V φ
T > 0 with positive probability. We define
u = inf{t ∈ T \ {T} : P(V φ t+1 ≥ 0) = 1, P(V φ
t+1 > 0) > 0}. Recall that the value of self-financing portfolio increases only via increments in asset prices, that is
V φ u+1 = V φ
u+1 − V φ u = φu+1 ·Su+1.
Now let us first consider the case, when V φ u ≤ 0 (P-a.s.) which follows that φu+1·Su+1 ≥ 0
(P-a.s.) and strictly positive with positive probability. We can choose ξ = φu+1 and have
immediately that ξ ·St+1 ≥ 0 (P-a.s.) and strictly positive with positive probability. So the (NA)t condition is not satisfied in this case at time t = u.
Then let us consider the remaining case, when V φ u > 0 with positive probability. From
the way we defined u, we get that also V φ u < 0 with positive probability. In this case we
choose ξ = φu+11{V φu <0}, which yields
ξ ·Su+1 = (φu+1 ·Su+1)1{V φu <0} = (V φ u+1 − V φ
u )1{V φu <0} ≥ 0 P-a.s.
and strictly positive with positive probability, since P(V φ u < 0) > 0. Again we have been
able to create an arbitrage at time t = u. Thus the proof is complete.
2.6 Geometric interpretation of arbitrage
In this section we restrict ourselves to one-period and study arbitrage-free prices from the geometric perspective. The following notation is used throughout the section. We assume that S = (S1, . . . , Sd) is the vector of final prices of risky assets with initial prices π = (π1, . . . πd). For non-risky numeraire asset the final price is denoted by S0 and initial price π0. Other notations remain the same.
Let us introduce Borel probability measure µ on Rd such that
µ(B) = P(S ∈ B) for all Borel sets B ⊂ Rd
Definition 2.6.1. Let A ⊂ Rd be the smallest closed set such that µ(Ac) = 0. We call set A the support of µ and denote it by supp(µ). ♦
For every Borel probability measure on Rd unique support exists (see Proposition 1.45 in [10]). We say that set C in an underlying vector space is convex if
x, y ∈ C and λ ∈ [0, 1] ⇒ (λx+ (1− λ)y) ∈ C.
The smallest convex set containing A ⊂ Rd is called the convex hull of A. Next definition will give us useful equivalent characterization of the convex hull.
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Definition 2.6.2. The convex hull of nonempty set A ⊂ Rd is denoted by conv(A) and defined by
conv(A) =
In the geometric characterization of arbitrage-free prices, we are only interested in prices that belong ”strictly” to conv(supp(µ)) and thus guarantee the existence of an equivalent martingale measure. So we need the next definition.
Definition 2.6.3. The relative interior of a convex set C ⊂ Rd denoted by ri(C) is the set of all points x ∈ C such that for all y ∈ C there exists ε > 0 with (x−ε(y−x)) ∈ C ♦
The next theorem is the main result of this section. We are only interested in applica- tions of the final result and thus will not motivate this result. Reader is refered to Section 1.4 of [10] for details.
Theorem 2.6.4. Let µ be the distribution of the vector of discounted final prices S of risky assets. Then one-period market model is arbitrage-free if and only if the vector of initial prices π belongs to ri(conv(supp(µ))), that is the relative interior of the convex hull of the support of µ.
Example 2.6.5. Suppose random variable X is binomially distributed, that is
P(X = k) =
with parameter p = 1 2
and n = 4. Let B be the non-risky numeraire asset with constant final and initial price B = πB = 1 (for example a bank account with zero interest rate) and risky asset (for example stock) S = 1
2 X with initial price πS. We introduce European call
option (see Section 2.7 for details) C = (S−K)+, with strike price K = 1 and initial price πC . Let us set = {ω0, ω1, ω2, ω3, ω4} and define {ωk} = {ω ∈ |X(ω) = k} = {X = k} thus
P({ωk}) = P(X = k) for k ∈ {0, 1, 2, 3, 4}.
In Table 2.1 we have presented realizations for the asset prices with respect to different scenarios. We define
µ({(πS, πC)}) = P((S, C) = (πS, πC))
and notice that
supp(µ) = {(0, 0), (1/2, 0), (1, 0), (3/2, 1/2), (2, 1)},
since these are the only pairs with positive probability. The support of µ is drawn in Figure 2.2 as thick points and the convex hull of the support is the triangle generated by those points including the boundary presented by dashed lines. Now by Theorem 2.6.4 we know that the pairs of arbitrage-free prices (πS, πC) ∈ R2 belong to coloured area seen
18
k = 0 k = 1 k = 2 k = 3 k = 4
P({ωk}) 1/16 1/4 3/8 1/4 1/16
X(ωk) 0 1 2 3 4
B(ωk) 1 1 1 1 1
S(ωk) 0 1/2 1 3/2 2
C(ωk) 0 0 0 1/2 1
Table 2.1: Realizations of final asset prices.
in Figure 2.2 excluding the boundary points shown as dashed lines. Explicitly the area is determined by 3 lines: the upper boundary πC = 1
2 πS and lower boundaries πC = πS − 1
and πC = 0 so the area is
{(πS, πC) ∈ R2 |πC > 0, πC > πS − 1, πC < 1
2 πS}
2 πS}
For example, if we fix the stock price πS = 3/2, then we see that the option price πC
belongs to the open interval (1 2 , 3
4 ), when the market is arbitrage-free.
Figure 2.2 is also useful for finding out the super- or sub-hedging strategy (see Def- inition 2.8.1) for the option. The hypotenuse of the triangle generated by dashed lines gives guideline how to determine the cheapest superhedge. Let us define portfolio con- sisting of η amount of money in bank account B and ψ amount of stock S. The value of superhedging portfolio must satisfy
ηB + ψS(ω) ≥ C(ω) for all ω ∈ . (2.6.6)
Now choosing the portfolio suggested by the hypotenuse, we get η = 0 and ψ = 1/2. This is a superhedging strategy since ψS(ω) is clearly larger than (or equal to) C(ω) for ω ∈ {ω0, ω1, ω2} and also for ω ∈ {ω3, ω4} it meets the demands. From the figure we see that there is no cheaper superhedging strategy available. 4
2.7 European options
Definition 2.7.1. A European derivative with underlying assets S = (S1, . . . , Sd) is a random variableX ∈ L0(,F ,P) which is measurable with respect to FST = σ{St | t ≤ T}. We denote XT := X for the payoff (or the claim) of the derivative at time T . ♦
In this thesis we are particularly interested in the one type of derivatives, options. An option on an asset is a contract giving the owner the right, but not the obligation, to trade the asset for a fixed price at a future date. We call last date on which option can
19
20
be exercised (according to its terms) the expiration date T . The option is European, if it can only be exercised at the fixed expiration date T . Let us denote Si for a price process of a underlying risky asset of an option in this section.
A European call option gives the buyer (of the option) a right, but not an obligation, to buy the asset whose real price is SiT at time T at the strike price K. If the real price of the asset (at time T ) is higher than the strike price, then the buyer of the option can buy the asset at price K and immediately sell it at price SiT . Thus the buyer gets the difference SiT −K as a payoff. If the real price of the asset is lower than the strike price, then the option expires worthless. Therefore we can write the payoff of a European call option as
CT = (SiT −K)+,
where we used notation (·)+ = max{ · , 0}. A European put option gives the buyer (of the option) a right, but not an obligation,
to sell the asset whose real price is SiT at time T at the strike price K. If the real price of the asset (at time T ) is lower than the strike price, then the buyer of the option can buy the asset at price SiT and immediately sell it at price K. Thus the buyer gets the difference K −SiT as a payoff. If the real price of the asset is higher than the strike price, then the option expires worthless. Therefore we can write the payoff of a European put option as
PT = (K − SiT )+.
Consider the relationship between the payoffs of the call and the put option. We see that the following equality holds
CT − PT = (SiT −K)+ − (K − SiT )+ = SiT −K,
regardless of which one of the prices (real or strike) is higher. Let us further denote cCt and cPt for the price of the call option and the put option at time t ∈ T, respectively. If we assume that market is arbitrage-free, then there exists equivalent martingale measure Q (with numeraire S0) so that
cCt = S0 tEQ(CT |Ft) and cPt = S0
tEQ(PT |Ft)
are the arbitrage-free time t prices of those options (see the next section for details). Hence evidently we have the following result:
Proposition 2.7.2. (Put-Call parity) In an arbitrage-free market model
cCt − cPt = Sit − S0 t
S0 T
K, (2.7.3)
for every t ∈ T.
Previous proposition enables us to concentrate our attention on call options alone, since we can always (in an arbitrage-free market) calculate the price of a put option by means of the price of a call option.
21
In general, derivative X is called path-independent if the payoff of X only depends on the time T values of assets. That is, if the payoff of X can be represented in the form XT = f(ST ) for some function f . For example, a European call option falls into this category.
On the contrary, path-depend derivative X has a payoff that can depend also on the earlier values of assets. For example, the payoff of an Asian call option with strike price K can depend on the average price asset i. Thus it is path-depend with the payoff
XT =
( 1
T
. (2.7.4)
Loosely speaking, European and American (type which is introduced later) call and put options are sometimes called vanilla options. More complex, often path-dependent, op- tions are called exotic options. There exist a vast number of different types of options for all kinds of purposes.
2.8 Pricing and hedging
Definition 2.8.1. We say that derivative X is replicable (or hedgeable), if there exists predictable self-financing strategy θ that takes at expiration date T the same value of a derivative X, that is
V θ T = XT (2.8.2)
almost surely. We call such θ a replicating strategy for X. In the cases of V θ T ≥ XT and
V θ T ≤ XT we call θ a super - and sub-replicating strategy, respectively. ♦
Example 2.8.3. Let f : R → R be a convex and smooth function and XT = f(SiT ) derivative with underlying asset Si
XT = f(SiT ) = f(0) +
) dx
(SiT − k)+f ′′(k)dk
We notice that X can be replicated by depositing f(0) amount to a bank account (or borrowing from the bank in case of negative value), purchasing f ′(0) amount of asset (or short-selling in case of negative value) and trading European call options with different
22
strike prices. In practice, though, there are only finitely many strike prices available for options in the market so the integral must be approximated with a sum.
Moreover we can relax the smoothness condition, but then we have to consider left and right differential coefficients and different measure for the integral. 4
Recall that we denote the set of equivalent martingale measures by
} .
Set Mb is convex, meaning that if Q1,Q2 ∈Mb and λ ∈ [0, 1], then also
λQ1 + (1− λ)Q2 ∈Mb.
We define the set of arbitrage-free initial prices of discounted derivative X by
Π(X) = { EQ(XT ) | Q ∈Mb and EQ(XT ) <∞
} and the lower and upper boundaries of Π(X) by
πXinf = inf Q∈Mb
EQ(XT )
respectively. From the convexity of Mb and the linearity of expectation, we deduce that Π(X) is an interval and if Mb is a singleton (a set with exactly one element), then Π(X must be also singleton (only one measure uniquely defines expectation). We will later see further that if Mb = {Q}, then all derivatives are replicable (and that this also holds on the reversed direction). Let us compile previous observations with some extra conditions:
Proposition 2.8.4. Let X be discounted derivative, then either
(i) X is replicable and the set of arbitrage-free initial prices is singleton
Π(X) = {πX}, and πX ≥ 0.
(ii) X is not replicable and the set of arbitrage-free initial prices is open interval
Π(X) = (πXinf , π X sup), and πXinf < πXsup.
Proof. See the proof of Theorem 5.33 in [10] for openness of Π(X) and some technical details.
Example 2.8.5. Consider the market model described in Example 2.6.5. Let us try to replicate option C with a portfolio consisting of η amount of money in bank account B and ψ amount of stock S. The final value of such portfolio must coincide with the final value C. So in order to replicate
ηB + ψS(ω) = C(ω) (2.8.6)
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has to hold for all ω ∈ . We can use the table of asset prices from aforementioned example, thus equation (2.8.6) yields
η = 0, ω = ω0
η + ψ = 0, ω = ω2
η + 2ψ = 1, ω = ω4.
From the first two equation, we get that η = ψ = 0 which contradicts the last equa- tion. Hence C is not replicable in this model (and clearly C is not either). Notice that discounting in this model does not change values since B = πB = 1.
Let us fix the initial price of stock πS = 3/2 and calculate the set of arbitrage-free initial prices for the discounted option. We use the conditions of equivalent martingale measure Q (with qi := Q({ωi}) ∈ (0, 1), for i = 0, 1, 2, 3, 4) to find bounds for πC :{
EQ(S) = πS ⇐⇒ 1 2 q1 + q2 + 3
2 q3 + 2q4 = 3
4 q3 which together with (†) yields the upper bound:
πC = 1
Since Q is a probability measure, we must have ∑4
i=0 qi = 1 and hence−q2 > q1+q3+q4−1, so (∗) gives
2q4 = 3
2 − 1
2 q1 −
That is q4 > 1 2 − 1
2 q3, so again combined with (†) we get the lower bound:
πC = 1
We have observed that non-replicable discounted option C has arbitrage-free initial prices that belong to open interval Π(C) = (1
2 , 3
4 ). This coincides with Proposition 2.8.4 and
the result we got in Example 2.6.5 by geometrically studying the arbitrage-free prices. Although we did not explicitly write down the different measures, we implicitly used two different measures to determine bounds for the interval, when we gave estimates for parameters qi to get inequations above. 4
Recall that Corollary 2.4.18 guarantees that, if the final portfolio values coincide in the arbitrage-free market, then the values have to coincide at any earlier moment of time. This statement is sometimes called the principle of no arbitrage. This means that it suffices to find the initial value of a replicating strategy for a derivative in order to price the derivative. Let us denote the family of predictable and self-financing strategies by Θ and the families of super- and sub-replicating strategies for the derivative X by
Θ+ X = {θ ∈ Θ | V θ
T ≥ XT} and Θ−X = {θ ∈ Θ | V θ T ≤ XT}.
24
Lemma 2.8.7. For every equivalent martingale measure Q, we have
sup θ∈Θ−X
θ∈Θ+ X
V θ t ,
Proof. Let θ ∈ Θ+ X so that XT ≤ V θ
T for every θ. Thus, by Proposition 2.4.16, we have
EQ(XT |Ft) ≤ EQ(V θ T |Ft) = V θ
t ,
and in a similar fashion we can prove the reversed inequality for θ ∈ Θ−X .
For a fixed equivalent martingale measure Q with numeraire S0, let us define a process by setting
cXt (Q) := S0 tEQ(XT |Ft). (2.8.8)
We say that process cX(Q) = (cXt (Q))t∈T is the Q-risk neutral price of X. Notice that it indeed depends on the chosen equivalent martingale measure Q. In the following result we see that, if X is replicable, then every Q agree with the same price and thus the price is independent of Q and unique. This is the basis of arbitrage-free pricing.
Theorem 2.8.9. If X is a replicable derivative in an arbitrage-free market, then for every replicating strategy θ and for every equivalent martingale measure Q, we have
EQ(XT |Ft) = V θ t , t ∈ T. (2.8.10)
We call the process (cXt )t∈T defined by cXt := V θ t the arbitrage-free price of X.
Proof. Let θ and φ be replicating strategies for X. Since the market is arbitrage-free and time T values of θ, φ, and X coincide, according to Corollary 2.4.18 also all earlier values match one another. Furthermore, because θ ∈ Θ+
X ∩Θ−X , Lemma 2.8.7 yields
EQ(XT |Ft) = V θ t
for every martingale measure Q.
Example 2.8.11. Consider model defined in Example 2.2.4. We add a new asset, Eu- ropean call option C2 = (S2 − K)+ with strike price K = 140, to the model. We know stock values S2 at time t = 2 so we can straightaway calculate the payoff of the option:
C2(ω) =
28, ω ∈ {(1, 0), (0, 1)} 0, ω = (0, 0)
Let us find replicating strategy θ = {θt | t = 1, 2} = {(ηt, ψt) | t = 1, 2} of C2 using bank account B and stock S for. The value of strategy θ has to equal the payoff of the option at time t = 2, that is
V θ 2 (ω1, ω2) = η2(ω1)B2 + ψ2(ω1)S2(ω1, ω2) = C2(ω1, ω2) ∀ω = (ω1, ω2) ∈ , (2.8.12)
25
where we denoted η2(ω1) := η2(ω1, ω2) and ψ2(ω1) := ψ2(ω1, ω2) since θ is predictable and thus η2 and ψ2 do not depend on ω2. Now equation (2.8.12) gives us four condition for η2
and ψ2, with different ω1, ω2 ∈ {0, 1}. Let us first fix ω1 = 1, that yields{ η2(1) · 1.21 + ψ2(1) · 294 = 154, when ω2 = 1
η2(1) · 1.21 + ψ2(1) · 168 = 28, when ω2 = 0,
from which we can solve η2(1) ≈ −115.7 and ψ2(1) = 1. In a similar fashion we fix ω1 = 0 and get corresponding two equations from which we solve η2(0) ≈ −30.9 and ψ2(0) ≈ 0.4.
Recall that replicating strategy must be self-financing, that is
η1B1 + ψ1S1(ω1) = η2(ω1)B1 + ψ2(ω1)S1(ω1), for ω1 = 0, 1,
where η1 and ψ1 are deterministic. From the self-financing condition, we can solve η1 ≈ −73.3 and ψ1 ≈ 0.8.
So we can replicate C2 by first, at time t = 0, investing η1 ≈ −73.3 in the bank account (borrowing money) and buying ψ1 ≈ 0.8 amount of stock. Then, at time t = 1, we choose corresponding amounts
η2(ω1) ≈
{ 0.4, ω1 = 0
1, ω1 = 1,
depending on whether the stock price has gone up (ω1 = 1) or down (ω1 = 0). Now we can use theorem 2.8.9 and replicating strategy to determine the arbitrage-free price of C2. According to aforementioned theorem, the value of the replicating strategy at times t = 0, 1 is the arbitrage-free price of the option at those times. At time t = 0, the price is
cC2 0 = V θ
0 = η1B0 + ψ1S0 ≈ 43.4.
This is also the price of the hedging for the option. At time t = 1, we calculate the price similarly
cC2 1 (ω1) = V θ
1 = η1B1 + ψ1S1(ω1) ≈
{ 82.7, ω1 = 1
12.7, ω1 = 0,
which naturally also depends on whether the stock price has gone up or down. 4
Often derivative X is not replicable so there is no replicating strategy for X and hence Theorem 2.8.9 cannot be used. Fortunately, formula (2.8.8) provides way to price X so that the market remains arbitrage-free, although the price is not unique since it depends on Q. The next corollary predicates that the aforementioned formula does not yield unique price for non-replicable X.
Corollary 2.8.13. In an arbitrage-free market, derivative X is replicable if and only if EQ(XT ) has the same value for every equivalent martingale measure Q.
Proof. ”⇒”: Follows directly from Theorem 2.8.9. ”⇐”: Follows directly from Proposition 2.8.4.
26
As we already discussed earlier, we have:
Proposition 2.8.14. For any equivalent martingale measure Q, the market model with riskless asset S0 and risky assets S1, . . . , Sd and the price process cX(Q) (defined by (2.8.8)) is arbitrage-free.
Proof. Since cX(Q) is a Q-martingale (by its definition) then the augmented market with cX(Q) has the same equivalent martingale measure Q. Hence by the first fundamental theorem of asset pricing (Theorem 2.4.15) the augmented market is arbitrage-free.
2.9 Market completeness
Recall from the previous section, that a replicable (European) derivative has unique arbitrage-free price process determined by the equivalent martingale measure. In some market models, it is possible to replicate every (European) derivative and thus finding right prices is particularly convenient.
Definition 2.9.1. We say that a market model is complete if every European derivative is replicable in it. On the contrary, if there exist a European derivative that can not be replicated, we say that the market model is incomplete. ♦
Let us for a moment assume that we have an arbitrage-free and complete market model. From the first fundamental theorem, we have that there exist at least one equivalent martingale measure. Let us assume that Q and Q are such probability measures in our model. By the completeness of model, derivative X = 1A with A ∈ F is replicable. Thus by Corollary 2.8.13, we have
EQ(X) = EQ(X) ⇐⇒ EQ(1A) = EQ(1A) ⇐⇒ Q(A) = Q(A) (2.9.2)
for every A ∈ F . So the completeness (together with arbitrage-freeness) implies unique- ness of equivalent martingale measure. The same corollary (2.8.13) also implies the con- verse. That is, if there exists only one equivalent martingale measure in model, then the expectation is always unique for each derivative and hence every (European) derivative is replicable.
Theorem 2.9.3. (Second fundamental theorem of asset pricing) An arbitrage-free discrete- time market model is complete if and only if there exist a unique equivalent martingale measure.
Proof. Discussion above motivates the result with a small amount of details, since it relies on the mentioned corollary. For more details, see for example, the proof of Theorem 4.1.2 in [9].
The completeness of a market model requires a certain structure from the underlying probability space. We will next study this required structure.
Definition 2.9.4. Consider probability space (,F ,P). Set A ⊂ in F is called an atom if P(A) > 0 and for any subset B ⊂ A in F with P(B) < P(A) it holds that P(B) = 0. ♦
27
A complete market model must have such underlying probability space, that can be separated into finitely many atoms. Also the the number of atoms can be (d+ 1)T at most, where d ∈ N is the number of risky asset. We will not prove this requirement. See, for example, the ending of the proof of Theorem 5.37 in [10].
Another property of an arbitrage-free and complete market tells that every martingale (under Q) can be represented in terms of discounted asset price processes. This prop- erty can be extended to continuous-time models by replacing martingale transforms with stochastic integrals.
Proposition 2.9.5. (Martingale representation property) An arbitrage-free market model with equivalent martingale measure Q is complete if and only if each (F,Q)-martingale M = {Mt | t ∈ T} can be represented in the form
Mt = M0 + t∑
(φi • Si)t, (2.9.6)
where φ is some predictable process.
Proof. See, for example, the proof of Theorem 5.38 in [10] or Proposition 4.2.1 in [9].
2.10 Hedging in incomplete market
This section depends highly on the properties of the underlying L2 Hilbert space which was introduced in the preliminary section. We review different ways to hedge deriva- tives, when the full replication with a self-financing strategy is not possible. We no longer restrict ourselves to self-financing portfolios. Let us consider trading strategy ξ = {(ξ0
t , ξ 1 t , . . . , ξ
d t ) | t ∈ T}, where ξ0 is adapted and ξi is predictable process for all
i = 1, 2, . . . , d. We assume that value of such portfolio is given by
V ξ 0 = ξ0
t S 0 t + ξ∗t · St ∀t ∈ T \ {0},
where S0 denotes price process of riskless asset and S denotes the price process of risky assets contrary to earlier sections and ξ∗ tags portfolio of risky assets. For simplicity and instead of discounting we set S0
t = 1 for all t ∈ T. We will carry out this convention throughout the section. The gain process of strategy ξ is now given by
Gξ 0 = 0 and Gξ
t = t∑
u=1
ξ∗u ·Su ∀t ∈ T \ {0},
where Su = Su−Su−1. We want to keep up with cumulated cost of the hedging strategy, so we introduce the cost process of strategy ξ denoted by Cξ and defined by
Cξ t = V ξ
t −G ξ t , ∀t ∈ T.
Remark 2.10.1. Earlier we considered only self-financing strategies for which the cost process would have been constant (the initial value of the portfolio). 4
28
We assume that X is a European derivative for which holds X ∈ L2(,FT ,P). Like- wise we demand that St ∈ L2(,Ft,P) for all t ∈ T. Furthermore, we say that strategy ξ is L2-admissible strategy for X if
V ξ T = X P-a.s. and V ξ
t ∈ L2(,Ft,P) ∀t ∈ T.
We want to find L2-admissible strategy ξ such that for all t ∈ T \ {T}
E[(C ξ t+1)2|Ft] ≤ E[(Cξ
t+1)2|Ft] P-a.s. (2.10.2)
for each L2-admissible strategy ξ. Such strategy ξ is called locally risk-minimizing . As before, we use to represent change of value of the process in question at given time,
that is, C ξ t+1 = C ξ
t+1−C ξ t . Using the expression (2.10.12), we can equivalently solve our
minimizing problem by finding minimum for
E [( V ξ t+1 − (V ξ
t + ξ∗t+1 ·St+1) )2|Ft
] . (2.10.3)
This minimizing is related to linear least squares method for estimating the unknown parameters (in this particular case these would be V ξ
t and ξ∗t+1) in a linear regression
model, provided that we knew value V ξ t+1. Next definition is a substitute for self-financing
condition that was used particularly in complete markets.
Definition 2.10.4. Strategy ξ is called mean self-financing if it is L2-admissible and
E(Cξ t+1|Ft) = 0, P-a.s.
for all t ∈ T \ {T}. ♦
We notice that mean self-financing strategy has cost process with martingale property. Recall that martingale constant on average, so on average the cost of hedging will be the initial value of the portfolio. We will need a few definition before stating the main result. Let us assume that X, Y ∈ L2(,F ,P) =: L2(P), then conditional covariance of X and Y with respect to sigma-algebra G ⊂ F is denoted and defined by
Cov(X, Y |G) := E(XY |G)− E(X|G)E(Y |G) (2.10.5)
Similarly conditional variance of X ∈ L2(P) is denoted and defined by
Var(X|G) := E(X2|G)− (E(X|G))2. (2.10.6)
In multidimensional case, where X = (X1, . . . , Xd) and Y,X1, . . . , Xd ∈ L2(P), we have
Cov(X, Y |G) := [Cov(X1, Y |G), . . . ,Cov(Xd, Y |G)]. (2.10.7)
Also for multidimensional X we have that Cov(X|G) is (d × d)-matrix with elements Cov(Xi, Xj|G) for i, j = 1, . . . , d.
29
Definition 2.10.8. Two F-adapted processes X = {Xt | t ∈ T} and Y = {Yt | t ∈ T} are called strongly orthogonal to each other under P if
E(Xt+1Yt+1|Ft) = 0, P-a.s. (2.10.9)
for every t ∈ T \ {T}. ♦
Strongly orthogonal condition resembles orthogonality in L2 with respect to inner product X, Y = E(XY ) with X, Y ∈ L2. Hence the term strongly orthogonal. We are finally ready to state the main result for locally risk-minimizing strategies.
Theorem 2.10.10. Let us assume that ξ is L2-admissible strategy. Then ξ is locally
risk-minimizing if and only if ξ is mean self-financing and its cost process C ξ is strongly orthogonal to the process of risky asset prices S.
Proof. We start by writing
E[(C ξ t+1)2|Ft] = Var(C ξ
t+1|Ft) + E(C ξ t+1|Ft)2 (2.10.11)
and try to find conditions for ξ and V ξ so that both terms on right-hand side minimize. By the definition of cost process, we have a useful expression
C ξ t+1 = C ξ
t+1 − C ξ t = V ξ
t+1 − V ξ t − ξ∗t+1 ·St+1. (2.10.12)
Now we can write the second term of the right-hand side of equation (2.10.11) as
E(C ξ t+1|Ft)2 = [E(V ξ
t+1|Ft)− V ξ t − ξt+1 · E(St+1|Ft)]2. (2.10.13)
Assuming that t and V ξ t+1 are fixed, this is minimized when
V ξ t = E(V ξ
t+1|Ft)− ξ∗t+1 · E(St+1|Ft). (2.10.14)
If this is satisfied, we notice that
E(C ξ t+1|Ft) = 0, (2.10.15)
that is, ξ is mean self-financing. Next let us consider the first term of the right-hand side of equation (2.10.11). Again we use expression (2.10.12) and then use basic properties of conditional variance and conditional covariance.
Var(C ξ t+1|Ft) = Var(V ξ
t+1 − V ξ t − ξ∗t+1 ·St+1|Ft)
= Var(V ξ t+1|Ft) + Var(ξ∗t+1 ·St+1|Ft)
− 2Cov(V ξ t+1, ξ
∗ t+1 ·St+1|Ft)
= Var(V ξ t+1|Ft) + (ξ∗t+1)TCov(St+1|Ft)ξ∗t+1
− 2Cov(V ξ t+1, St+1|Ft)ξ∗t+1 (2.10.16)
30
We notice that right-hand side of (2.10.16) is of form f(x) = c + xTAx − 2bTx where
]T .
Then we have ∂ ∂x f(x) = 2Ax−2b, when A is symmetric matrix. Recall from linear algebra
that f is minimized (we ignore explicit conditions) when ∂ ∂x f(x) = 0, that is Ax− b = 0.
Hence we must have almost surely
Cov(St+1|Ft)ξ∗t+1 − Cov(V ξ t+1, St+1|Ft) = 0. (2.10.17)
By equation (2.10.12) we can write
V ξ t+1 = C ξ
t+1 + V ξ t + ξ∗t+1 ·St+1 (2.10.18)
Thus the second term on the right-hand side of equation (2.10.17) can be written as
Cov(V ξ t+1, St+1|Ft) = Cov(C ξ
t+1, St+1|Ft) + Cov(St+1|Ft)ξ∗t+1, (2.10.19)
using the basic properties of conditional covariance. Hence the equation (2.10.17) is equivalently
Cov(C ξ t+1, St+1|Ft) = 0. (2.10.20)
Recall that have the mean self-financing condition (equation (2.10.15)) for optimal ξ so (2.10.20) yields
E(C ξ t+1St+1|Ft) = 0, (2.10.21)
that is, the cost process of strategy ξ is strongly orthogonal to S.
The proof above gives a recursive way of solving the minimizing problem (2.10.3). Particularly equations (2.10.14) and (2.10.17) are useful. Let us restric ourselves to case,
where there is only one risky asset in the market and put V ξ T = X for derivative X. Then
mentioned useful equations yield
V ξ t = E(V ξ
t+1|Ft)− ξ1 t+1E(St+1|Ft) with ξ1
t+1 = Cov(V ξ
, (2.10.22)
that is, a recursive way to find locally risk-minimizing strategy. Here we put ξ0 t :=
V ξ t − ξ1
t S 1 t to complete the strategy. However we have to assume further that there exist
constant c ∈ R such that (E(St|Ft−1))2 ≤ cVar(St|Ft−1) almost surely in order to meet L2-admissibility. See Proposition 10.10 in [10] for details.
31
Another minimizing problem arises, when we want to find self-financing strategy qξ, that minimizes quadratic hedging error
E[(X − V qξ T )2] (2.10.23)
for derivative X. Note that minimizing the quadratic hedging error corresponds to mini-
mizing L2-norm X − V qξ T 2, which is the ”distance” between X and V
qξ T . Recall that the
value of self-financing strategy qξ can be written in terms of gain process as
V qξ t = V
Ξ := {ξ | ξ is predictable, Gξ t ∈ L2(P) ∀t ∈ T}.
Strategy qξ ∈ Ξ is called quadratic risk-minimizing , if it minimizes quadratic hedging error for (European) derivative X, that is,
E[(X − (V qξ
0 +Gξ T ))2] (2.10.25)
for all ξ ∈ Ξ. Solution to this minimizing problem is closely closely connected with local risk-minimizing problem. Let us again restrict to case with only one risky asset and one riskless asset in the market. Assume further that
Var(St|Ft−1)
E(St|Ft−1)2
is deterministic for all t ∈ T \ {0}. Then (we state without proving) the quadratic risk-
minimizing strategy qξ is given by
V qξ
t := ξ1 t +
[ V ξ t−1 − (V
qξ 0 +G
qξ t−1) ] ,
(2.10.26)
where ξ refers to locally risk-minimizing strategy given in (2.10.22). Another imperfect way to hedge derivatives in incomplete market would be super-
hedging which we already encountered with in Section 2.8. In super-hedging, we try to find the cheapest (by initial price) self-financing strategy that surpasses the payoff of the derivative. The upside of this approach is, that we are always (at least in theory) able to stay on the ”safe side”, that is, there is no risk of losing money more than was planned beforehand. The downside of this approach is that even the cheapest super-hedge may still require a large initial investment so it is expensive compared to other methods. More about super-hedging can be read from Chapter 7 of [10].
Assuming that investor does not want to pay the initial amount of capital required to build a superhedge and he/she is willing to carry some risk, there is some alternatives to consider. We will introduce two of these alternatives. In quantile hedging one
32
tries to find self-financing strategy ξ that maximizes the probability of successful super- hedge P(V ξ
T ≥ X) over the strategies that cost (initially) less than super-hedging strategy. Yet another imperfect hedging strategy is called shortfall risk-minimizing . In this method, one tries to find strategy ξ that minimizes E[(X − V ξ
T )+]. We will not determine how to build these alternative strategies. More about these approaches can be found from Chapter 8 of [10].
2.11 Change of numeraire
The existence of a concept introduced in the next important lemma is well-known result called Radon-Nikodym theorem. We will omit the proof.
Lemma 2.11.1. Let Q ∼ P be two probability measures on (,F). Then there exists random variable Z ≥ 0 in L1(,F ,P) such that
EQ(X) = EP(ZX) ∀X ∈ L1(,F ,Q)
We denote
Z := dQ dP
and call Z the Radon-Nikodym derivative of measure Q with respect to measure P.
Remark 2.11.2. (i) Expectation under P of Radon-Nikodym derivative Z is one, since
EP(Z) =
Z(ω)dP(ω) =
Z(ω) = dQ dP
∀ω ∈ .
4
Lemma 2.11.3. (Bayes’ formula) Let Q ∼ P be two probability measures on (,F) and assume that Z is Radon-Nikodym derivative of Q with respect to P. If G is a sub-σ-algebra of F , then
EQ(X|G) = EP(XZ|G)
EP(Z|G) , ∀X ∈ L1(,F ,Q). (2.11.4)
Proof. We have prove to that EP(XZ|G) = EQ(X|G)EP(Z|G). We prove the aforemen- tioned by the definition of conditional expectation 2.1.10. Let G ∈ G, by the properties of conditional expectation and Radon-Nikodym derivative, we get∫
G
X1GdQ =
∫ G
XZdP.
33
In this section let us write Y = {Yt | t ∈ T} for the value of an arbitrary self- financing and predictable strategy or one of the assets S1, . . . , Sd. From the Proposition 2.4.16 we know that if Q is an equivalent martingale measure with numeraire S0, then Y = {Yt/S0
t | t ∈ T} is Q-martingale. The next proposition will show how a new measure QY with numeraire Y should be defined so that it becomes equivalent martingale measure in the market model.
Proposition 2.11.5. In an arbitrage-free market model, let Q be an equivalent martingale measure with numeraire S0. Let also Y = {Yt | t ∈ T} be a positive process such that Y is a Q-martingale. Then the measure QY defined by
dQY
dQ =
( S0
0
Y0
|Ft ) , t ∈ T (2.11.7)
for every X ∈ L1(,FT ,Q). Therefore QY is an equivalent martingale measure with numeraire Y .
Proof. Let us denote
) Yt S0 t
, for t ∈ T.
Since Y discounted with S0 is Q-martingale, we have EQ(ZT |Ft) = Zt. Now by Bayes’ formula (Lemma 2.11.3) we get
YtEQY
( X
so QY is equivalent martingale measure with Y as numeraire.
In earlier section, we have restricted ourselves to deterministic interest rates. In reality this is not the case but interest rates are stochastic. That makes pricing the derivatives more challenging, since we have to consider the joint distribution discounting factor and the payoff of the derivative, when we are computing the expectation under equivalent martingale measure. An important concept called forward measure can sometimes ease the computation. The forward measure uses suitable bond as a numeraire so that the need for joint distribution disappears.
As we already mentioned, numeraire Y does not have to be a single asset. It can be entire portfolio of asset. This is particularly useful, when one tries to price a swap
34
option (or swaption), that is, an option giving its owner the right but not the obligation to exchange cash flows by contract with another party. In this situation a convenient choice of numeraire could be a portfolio consisting of zero-coupon bonds maturing at times different times. We will not review these concepts in discrete-time market models.
Example 2.11.8. Consider market model defined in Example 2.2.4, where we have two asset S and B. Recall that in another Example 2.2.4, we found equivalent martingale measure Q with numeraire B. Measure Q is unique and symmetric with
Q({ω}) = 1
4 , for each ω ∈ = {(0, 0), (1, 0), (0, 1), (1, 1)}.
One might ask: does there exist equivalent martingale measure with S as numeraire? The answer is yes, since St(ω) > 0 for all t = 0, 1, 2 and ω ∈ , furth | 23,805 | 83,788 | {"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} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.857446 |
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Cot
http://functions.wolfram.com/01.09.27.0077.01
Input Form
Cot[z] == (Sqrt[-z^2]/z) (1/Sqrt[1 - Sech[I z]^2]) /; Im[z] != 0
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["Cot", "[", "z", "]"]], "\[Equal]", RowBox[List[FractionBox[RowBox[List[SqrtBox[RowBox[List["-", SuperscriptBox["z", "2"]]]], " "]], RowBox[List["z", " "]]], FractionBox["1", SqrtBox[RowBox[List["1", "-", SuperscriptBox[RowBox[List["Sech", "[", RowBox[List["\[ImaginaryI]", " ", "z"]], "]"]], "2"]]]]]]]]], "/;", RowBox[List[RowBox[List["Im", "[", "z", "]"]], "\[NotEqual]", "0"]]]]]]
MathML Form
cot ( z ) - z 2 z 1 1 - sech 2 ( z ) /; Im ( z ) 0 Condition z -1 z 2 1 2 z -1 1 1 -1 z 2 1 2 -1 z 0 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["Cot", "[", "z_", "]"]], "]"]], "\[RuleDelayed]", RowBox[List[FractionBox[SqrtBox[RowBox[List["-", SuperscriptBox["z", "2"]]]], RowBox[List["z", " ", SqrtBox[RowBox[List["1", "-", SuperscriptBox[RowBox[List["Sech", "[", RowBox[List["\[ImaginaryI]", " ", "z"]], "]"]], "2"]]]]]]], "/;", RowBox[List[RowBox[List["Im", "[", "z", "]"]], "\[NotEqual]", "0"]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 616 | 1,669 | {"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-2024-22 | latest | en | 0.230772 |
https://www.examtestprep.com/IGCSE/Mathematics-0580/Paper-1/Solved-Specimen-Questions/Part-27.html | 1,606,190,606,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141171077.4/warc/CC-MAIN-20201124025131-20201124055131-00148.warc.gz | 679,401,939 | 16,832 | # IGCSE Mathematics Paper-1: Specimen Questions with Answers 78 - 81 of 324
## Question 78
Edit
### Write in Short
Solve the equation
### Explanation
Here, equation is
## Question 79
Edit
### Write in Short
AB is a chord of a circle, O is center. Angle OAB = . Calculate angle AOB.
### Explanation
Here,
As we can see in the figure AB is chord and O is center, so OA and OB will be radius of the circle.
So, these radiuses will make same angle with points of chord.
So, Angle OAB = , so angle OBA =
Now, in any triangle addition of all angles will give
So, in
.
## Question 80
Edit
### Describe in Detail
Essay▾
The population of India was 1.252 billion in 2013. It is expected that this population will increase by 30 % by the year 2015. Calculate the expected population in 2015.
### Explanation
Here, Population is. (1 billion = 1000 million)
Expected population will increase by 30%.
So, increment in population will: .
So, expected population in 2015 will be
## Question 81
Edit
### Write in Short
As shown in the diagram DE is parallel to BC. ADB and AEC are straight lines.
(a) Fill in the blanks with any of the words given: Congruent equilateral isosceles similar
Triangle ADE and triangle ABC are________.
(b) Angle What is the size of angle DBC?
### Explanation
Here,
(a) Here, Triangle ADE is similar to triangle ABC because there all three angles are same in both as below:
Angle A is common in both so it is same, angle ADB and ABC are corresponding angles and AED and ACB are also corresponding angles, so these will be also same.
While both have no equal sides, so it can’t be called as congruent triangles.
• Two triangles are called as similartriangles:
• If all angles are same in both OR
• If all pairs of sides are in same proportion OR
• If two pairs of sides are in same proportion and one angle is also same in both.
• Two triangles are called as congruent triangles when both have all equal sides and equal angles.
• A triangle with two equal sides and the angles opposite the equal sides are also equal is called as Isosceles.
(b) Here, angle , angle DBC =?
Now, BD is straight line, so angle
So, angle DBC = | 543 | 2,178 | {"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} | 4.59375 | 5 | CC-MAIN-2020-50 | longest | en | 0.912295 |
https://de.mathworks.com/matlabcentral/answers/270643-how-to-create-a-smoothed-histogram-and-compute-it-s-derivative-magnitude-in-matlab | 1,638,668,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00196.warc.gz | 274,287,722 | 23,539 | # How to create a smoothed histogram and compute it's derivative magnitude in matlab
141 views (last 30 days)
Tolentee on 29 Feb 2016
Edited: Mike Garrity on 2 Mar 2016
How can I created a smoothed histogram in matlab and I will also like to know how I can compute it's derivative magnitude.
(Assume I want to do that for a vector e.g. y = randn(1,500))
Mike Garrity on 2 Mar 2016
Edited: Mike Garrity on 2 Mar 2016
If you have the statistics toolbox, you might want to consider ksdensity.
x = [randn(300,1); 5+randn(300,1)];
histogram(x,'Normalization','probability')
[f,xi] = ksdensity(x);
hold on
plot(xi,f)
Richa Gupta on 2 Mar 2016
If you have the Curve Fitting Toolbox, you can use the 'smooth' function for creating a smoothed histogram. For computing the derivative magnitude, you can use the 'diff' function. A sample code is below:
y = randn(1,500);
x = smooth(y);
z = diff(y); | 257 | 890 | {"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-2021-49 | latest | en | 0.878866 |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-4-practice-exercises-page-279/115 | 1,575,565,282,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540481281.1/warc/CC-MAIN-20191205164243-20191205192243-00521.warc.gz | 736,609,181 | 12,611 | ## University Calculus: Early Transcendentals (3rd Edition)
$\dfrac{1}{2}e^{t}+e^{-t}+c$
Calculate the anti-derivative. Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $\int e^{-x} dx=-e^{-x}+c$ where $c$ is a constant of proportionality. Then $\int (\dfrac{1}{2}e^t-e^{-t}) dt=\int \dfrac{1}{2}e^t dt-\int e^{-t} dt$ As we know $\int e^{-x} dx=-e^{-x}+c$ or, $= \dfrac{1}{2}e^{t}+e^{-t}+c$ | 181 | 401 | {"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} | 3.9375 | 4 | CC-MAIN-2019-51 | latest | en | 0.49397 |
https://forextester.com/forum/viewtopic.php?f=9&p=60417&sid=5ce9845ab0ed166747e93a58b65922a4 | 1,638,086,053,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00000.warc.gz | 326,964,050 | 6,820 | ## how to set precision for double in C++
Examples step by step how to make your own automated strategy or user indicator
Message
Author
Ktanger
Posts: 1
Joined: Mon Apr 18, 2016 2:15 pm
### how to set precision for double in C++
Hi
Please check below MQL4 programming first:
if (NormalizeDouble(PriceGapStart, 3) >= 1.500)
{
...
}
the question is:
I want to find solution in C++ to set precision for double data like MQL4 above "NormalizeDouble(PriceGapStart, 3)" so that i can compare the result, any one can help me?
violajsilver
Posts: 5
Joined: Mon May 09, 2016 11:42 pm
### Re: how to set precision for double in C++
You can set the precision directly on std::cout and used the std::fixed format specifier.
Code: Select all
`double d = 3.14159265358979;cout.precision(17);cout << "Pi: " << fixed << d << endl;`
You can #include <limits> to get the maximum precision of a float or double.
Code: Select all
`#include <limits>typedef std::numeric_limits< double > dbl;double d = 3.14159265358979;cout.precision(dbl::max_digits10);cout << "Pi: " << fixed << d << endl;`
mark1205
Posts: 1
Joined: Wed Aug 31, 2016 4:05 am
### Re: how to set precision for double in C++
std::cout << std::setprecision (std::numeric_limits<double>::digits10 + 1)
<< 3.14159265358979
<< std::endl;
Jamesstewart01
Posts: 1
Joined: Thu Jan 25, 2018 1:21 am
Contact:
### Re: how to set precision for double in C++
thanks for the post
zoraya
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Contact:
### Re: how to set precision for double in C++
Thanks for the information!
4waytechnologies
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Location: California
Contact:
### Re: how to set precision for double in C++
Thanks for the information!
GragMilligan
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### Re: how to set precision for double in C++
std::cout << std::setprecision (std::numeric_limits<double>::digits10 + 1)
<< 3.14159265358979
<< std::endl;
Its works really. Accepting you really wanted to research articles forming and other making works out, we recommend that you visit some great site, actually look at this. I'm an understudy, yet I oftentimes ask him for help since it saves time and gives me results. | 673 | 2,288 | {"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-2021-49 | latest | en | 0.632701 |
https://mazurekgravity.in/question/for-the-crossed-four-bar-mechanism-shown-in-fig-1-the-dimensions-of-various-links-are-cd-65-mm-c-a-60-mm-db-80-mm-and-ab55-mm-find-the-angular-velocity-and-angular-accelerations-of-the-li/ | 1,680,174,435,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949181.44/warc/CC-MAIN-20230330101355-20230330131355-00470.warc.gz | 452,384,504 | 22,504 | # For the crossed four bar mechanism shown in Fig.1, The dimensions of various links are: CD = 65 mm; C A = 60 mm; DB = 80 mm; and AB=55 mm. Find the angular velocity and angular accelerations of the links A B and DB, if the crank CA rotates at 100 rpm. in the anticlockwise direction. USING RELATIVE VELOCITY METHOD
Question-AnswerCategory: Theory Of MachinesFor the crossed four bar mechanism shown in Fig.1, The dimensions of various links are: CD = 65 mm; C A = 60 mm; DB = 80 mm; and AB=55 mm. Find the angular velocity and angular accelerations of the links A B and DB, if the crank CA rotates at 100 rpm. in the anticlockwise direction. USING RELATIVE VELOCITY METHOD
RS Khurmi asked 11 months ago
For the crossed four bar mechanism shown in Fig.1, The dimensions of various links are: CD = 65 mm; C A = 60 mm; DB = 80 mm; and AB=55 mm. Find the angular velocity and angular accelerations of the links A B and DB, if the crank CA rotates at 100 rpm. in the anticlockwise direction. USING RELATIVE VELOCITY METHOD
Mazurek Gravity answered 11 months ago
USING RELATIVE VELOCITY METHOD
Therefore
Angular velocity of AB = 10π rad/sec (CCW)
Angular velocity of DB = 5.625π rad/sec (CCW)
Angular acceleration of AB = 1000(π)² rad/sec²
Angular acceleration of DB = (7000/55)π² rad/sec² | 360 | 1,290 | {"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-2023-14 | latest | en | 0.810576 |
http://slideshowes.com/doc/239118/the-dewey-decimal-classification-system | 1,495,496,844,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607242.32/warc/CC-MAIN-20170522230356-20170523010356-00492.warc.gz | 319,927,622 | 12,336 | ```The Dewey Decimal
Classification System
How the Library is
Organized
Melvil Dewey 1851-1931
Melvil Dewey devised a
system of organizing
books which is used in
many libraries. He
divided books up into 10
main subjects, and then
assigned these subjects
a number. That way all
the books on a topic
would be together on the
shelf.
If books weren’t
shelved this way,
libraries would be in
total chaos!
What is a Call Number?
Every book in the library is given a unique call
number that is an address for locating the book
on the shelf.
The call number is located on the spine of a
book.
The call number is made up of two parts: the
Dewey Decimal classification number and the
first three letters of the author’s last name.
The Key to Library Call Numbers
CALL NUMBER
TYPE OF BOOK
HOW IT IS SHELVED
E
ALPHABETICALLY BY
AUTHOR NAME
FIC
EVERYBODY =
PICTURE BOOKS
FICTION = CHAPTER
000-999
NONFICTION
DEWEY DECIMAL NUMBER,
THEN AUTHOR NAME
92 OR 921 OR
B
BIOGRAPHY
DEWEY DECIMAL NUMBER,
THEN SUBJECT’S NAME
REF
REFERENCE
DEWEY DECIMAL NUMBER,
THEN AUTHOR’S NAME
ALPHABETICALLY BY
AUTHOR NAME
The Difference Between
Fiction and Nonfiction
Fiction
Books that are stories
They are not true, but
might be based on true
events.
Nonfiction
real things, people,
events and places.
They are true, except
for fairy and folk tales
in 398 and literature in
the 800s.
Fiction and Nonfiction
Call Numbers
Fiction call numbers
begin with E or FIC or F,
and then the first three
letters of the author’s
last name. It’s still a call
number even though
there are no numbers in
it!
Nonfiction call numbers
begin with a Dewey
number, and then the
first three letters of the
author’s last name.
599
E
FIC
SEU
PAU
GIB
Just Remember…….
FICTION CALL
NUMBERS HAVE
ONLY LETTERS!
NONFICTION
CALL NUMBERS
HAVE NUMBERS
AND LETTERS!
+
The Dewey System Has 10 Main Classes
for Organizing Nonfiction Books.
Dewey #
10 Main Classes
000-099 General Works
Kinds of Books
Encyclopedias, almanacs, record
books
100-199 Philosophy and Psychology
Paranormal phenomena, ethics,
how we think
200-299 Religion
300-399 Social Science
Bible, mythology, religions
400-499 Language
English and foreign languages, sign
language, dictionaries
500-599 Natural Science
Math, chemistry, biology, weather,
plants, animals
600-699 Applied Science
Inventions, transportation, cooking,
pets
700-799 Fine Arts and Recreation
Crafts, painting, music, games,
sports
800-899 Literature
900-999 History and Geography
Poetry, plays
Government, holidays, fairy tales,
education, community
famous people
Each of the 10 Main Classes
can be further divided:
…to Specific
500
Natural Science
590
Animals
599
Mammals
599.8
Primates
599.88
Apes
Dewey and Decimals
Think of a Dewey number as if it were dollars
and cents! For example:
595.23
Comes Before
595.30
When in doubt, add a 0 (zero) to even out the
digits after the decimal.
Locating a Book
on the Shelf
Remember this rule for how books
are usually shelved!
You start at the left on the top
shelf, and move to the right
until the shelf ends. Then you
go to the next shelf beneath
that, and do the same-left to
right, top to bottom.
What Else Might I Find
In My Library???
Magazines
Beginning Chapter Books
Short Fiction
Series
Music
Books on Tape
Maps and Globes
Computers
Try making a map of the library to show the
following sections:
Everybody, Fiction, Nonfiction, Biographies and
Reference
Mark your favorite section on the map.
Include the circulation desk where you check out
and return books.
Write down the name of the Library Technician who
Presentation created by:
Julie Favero and Diana Dorney,
Librarians,
Lake Oswego School District
Garnetta Wilker,
District Librarian,
Oregon Trail School District
September 2004
``` | 1,093 | 3,758 | {"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} | 3.15625 | 3 | CC-MAIN-2017-22 | longest | en | 0.875359 |
http://clay6.com/qa/40177/a-variable-name-in-certain-computer-language-must-be-either-a-alphabet-or-a | 1,513,374,668,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579567.73/warc/CC-MAIN-20171215211734-20171215233734-00443.warc.gz | 59,507,870 | 27,166 | # A variable name in certain computer language must be either a alphabet or a alphabet followed by a decimal digit.Total number of different variable names that can exist in that language is equal to
$\begin{array}{1 1}(A)\;280\\(B)\;290\\(C)\;286\\(D)\;296\end{array}$
Total variables if only alphabet is used =26
Total variables if alphabets and digits both are used =$26\times 10$
$\Rightarrow 260$
Total variables =$26(1+10)$
$\Rightarrow 26(11)$
$\Rightarrow 286$
Hence (C) is the correct answer. | 141 | 503 | {"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} | 2.890625 | 3 | CC-MAIN-2017-51 | longest | en | 0.648636 |
http://webapi.bu.edu/freezing-point-of-naphthalene-graph.php | 1,713,885,485,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00332.warc.gz | 27,474,133 | 4,941 | Freezing point of naphthalene graph. What is the melting point and freezing point of naphthalene? 2022-10-09
Freezing point of naphthalene graph Rating: 7,9/10 1320 reviews
The freezing point of a substance is the temperature at which it changes from a liquid to a solid. The freezing point of naphthalene, also known as mothballs, can be represented on a graph as a horizontal line at a specific temperature.
To construct a graph of the freezing point of naphthalene, one would need to perform a series of experiments to determine the temperature at which naphthalene changes from a liquid to a solid. This can be done by placing a sample of naphthalene in a container with a thermometer and gradually decreasing the temperature until the naphthalene begins to solidify. The temperature at which this occurs can then be recorded and plotted on the graph as a horizontal line.
One interesting aspect of the freezing point of naphthalene is that it is not a constant value, but rather it is dependent on the pressure at which the naphthalene is being held. At high pressures, the freezing point of naphthalene will be higher than at low pressures. This relationship can also be represented on a graph, with the freezing point plotted as a function of pressure.
In addition to the freezing point, other physical properties of naphthalene such as its melting point, boiling point, and vapor pressure can also be plotted on a graph. These properties can provide valuable information about the behavior of naphthalene under different conditions, and can be used to predict how it will behave in a variety of applications.
Overall, the freezing point of naphthalene can be represented on a graph as a horizontal line at a specific temperature, and this line can be plotted as a function of pressure to show the relationship between the freezing point and pressure. Understanding the physical properties of naphthalene can be useful for a variety of applications, from pest control to industrial uses.
Determining the Freezing Point of Pure webapi.bu.edu
Purpose To determine the freezing point of a known substance, naphthalene II. Observe the change in temperature from 90o to 70o Celsius, recording the temperature at regular intervals, preferably 15 seconds. Mixing salt and ice decreases the freezing temperature of water, lowering the freezing point of the new mixture. What happen to the temperature of naphthalene during melting? Over a time period of 12 minutes and 30 seconds, we recorded the temperature at regular 15 second intervals, and, with this data, constructed a chart showing the general curve. Sublimation is what the process of the dried ice with water being poured on it.
Next
Freezing Point of Naphthalene Lab Answers
To measure the evaporation rate it is important to know that a half pound of dry ice will be placed outside and timed. We also learned how to determine the amount of heat each substance absorbed and compare which one absorbed the most heat. Once the temperature has fallen to 70o, melt the naphthalene which is now frozen to remove the thermometer. Follow the graphing guidelines in the instructions. In this experiment it is prepared by the reaction of p-aminophenol with acetic anhydride, as illustrated below.
Next
Answered: Freezing point of mixture (from graph…
The hypotheses is that the dry ice in the outside will evaporation faster. As heat is taken away from a liquid, the temperature of that liquid decreases as the substance begins to freeze. The way we did this was first finding the temperature at which ice crystals form for just regular BHT. In experiment 3, the melting point ranges of the crude and purified benzoic acid were both determined by using Melt Temp Apparatus. The temperatures from Point B to C remain constant because the heat energy absorbed by the particles was used to overcome the forces between the particles so that the solid can turn into liquid.
Next
Answered: reezing point of pure naphthalene T1 =…
In this experiment, we record out data on a computer program that allows us to draw a graph to be able to measure its trend. Data Time Elapsed Temperature of Naphthalene Time Temperature Initial 0:00 100oC 7:00 78. Purpose To determine the freezing point of a known substance, naphthalene II. As the temperature rises, the naphthalene will eventually start to melt. Ignite the Bunsen burner and using direct heat melt the naphthalene powder until it completely turns to a liquid. Upon inspection of the graph and our data chart, we found the experimental freezing point of naphthalene to be around 79oC.
Next
Freezing Point of Naphthalene
The melting point of a substance depends on pressure and is usually specified at a standard pressure such as 1 atmosphere or 100 kPa. At the melting point the solid and liquid phase exist in equilibrium. Transcribed Image Text: Table 1: Time and Temperature for melting of Naphthalene and the mixture Unknown letter Mass of unknown compound 0. Freezing point is the temperature at which a liquid becomes a solid at normal atmospheric pressure. This lab experiment tries to answer the question: what happens to the temperature of naphthalene as it melts? We found that the temperature at which crystallization of BHT was 68. Conclusions In this lab, we heated the known substance naphthalene in a test tube to approximately 100oC and observed its temperature while it cooled to approximately 70oC.
Next
What is the Hypothesis of the melting and freezing point of naphthalene?
Introduction: Acetaminophen is commonly used as an analgesic to reduce pain and fever. This lab experiment tries to answer the question: what is the freezing point of an unknown liquid naphthalene? The freezing point temperature will be found by cooling down the substance in a bath filled ice and salt mixture past its freezing point. Luckily thanks to our TA we were given temperature probes so we could pinpoint exact temperatures. For this experiment, we used freezing point depression to understand and be able to determine the molar mass of an unknown sample. Use a clamp to hold the thermometer in place. We were tasked with finding the molar mass of an unknown compound using the colligative property of Freezing Point Depression.
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What is the melting point and freezing point of naphthalene?
Why temperature is constant during melting and freezing of naphthalene take place? Thinking fast on your feet, and being the the smart chemistry student you are, you and your friends go buy oil, water and dry ice. Assemble the Bunsen burner, attaching one end of the hose to the burner and the other to a Use a clamp to hold the thermometer in place. Purity will then be determined by melting point. When every senior is trying to come up with the perfect prank to play on the school. Considering the time that it takes for the balloons to blow up and the capacity they can hold, you were able to calculate the time in which it took to… Comparing the Coolant Effects of Dry Ice and Ice What we learned in this lab is how to compare the abilities of dry ice and solid water to act as heat absorbers. You will graph the temperature changes before drawing any conclusions.
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Artikelen
Another fundamental principle that is discussed in this lab is the principle where we relate the freezing point depression of a pure solvent to the molality of the solution. So in this experiment the main topics are chemical compounds, temperature, and evaporation. We also learned to determine which substance was the most cost effective. As with the melting point, increased pressure usually raises the freezing point. We then heated and melted the two substances… Freezing Point Depression Lab The goal of this experiment is to find the molar mass of an unknown substance by measuring the freezing point depression of a known amount in an aqueous solution. The melting point of an organic solid can be determined by introducing a tiny amount into a small capillary tube, attaching this to the stem of a thermometer centred in a heating bath, heating the bath slowly, and observing the temperatures at which melting begins and is complete.
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Freezing point depression is a colligative property of solutions. The temperature at which the solvent starts to freeze will be specified as the freezing point of the solution. The melting point of pure naphthalene is 80. Observe the change in temperature from 90o to 70o Celsius, recording the temperature at regular intervals, preferably 15 seconds. Colligative properties of a solution depend on the amount of solute and solvent molecules and not the specific properties of the molecules. Properly dispose of the naphthalene liquid as instructed by the teacher. Place the thermometer in the crystals so that it is surrounded by the naphthalene powder but not touching the sides or bottom of the test tube.
Next | 1,877 | 8,879 | {"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-2024-18 | latest | en | 0.935439 |
https://www.physicsforums.com/threads/problem-with-part-of-c-program.274621/ | 1,527,445,212,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869732.36/warc/CC-MAIN-20180527170428-20180527190428-00048.warc.gz | 805,186,847 | 16,255 | # Homework Help: Problem with part of c program
1. Nov 24, 2008
### sulev8
I have a problem with c program calculation part . i have a mathematic funxion which equals to y. And i need to calculate the value. While x has a conditon thats value will increase 15 times. Like x1=a ; x2=a+h x3=a+h+(c*h) ; x4=a+h+(c*h)+((c*c)*h) ; x5=a+h+(c*h)+((c*c)*h)+((c*c*c)*h) and so on to X15. x1 which = A ,c,h and Ymin value is inserted by the program user. I Have come to conclusion , that i can write a formula for x value increase , which is Xi=Xi-1 +(c^i-2 * H) and it goes from x2 to x15 .
My problem is that how i can write this to c language , what line i should write that the program would calculate x1 by just adding to the y= formula and from x2 to x15 it would use the condition that Xi=Xi-1 +(c^i-2 * H).
And please dont give me just advice with this , i would just like to know how should i mark this in C language.
2. Nov 24, 2008
### Staff: Mentor
Moved to Homework Help forums.
sulev8, we do not give out answers to homework questions. Show us your attempt at the program first, and then we can offer tutorial help.
3. Nov 24, 2008
### sulev8
i dont want u to solve the whole program , this is just the part of a program , i am asking how i should write these lines into my program .Its the calculation part of the program , and i need some help with it. As i wrote i gave out the solution , now i need help how to put it down , just these few lines.
As i see it was another useless attempt to seek help from here , its like a basic answer in this forum.
// calculation part
void calculation (float xva[15], float yva[15])
{
int i;
i=0;
while (i<15)
{
xva=xva[i-1]+(pow(c,i-2)*h);
if (xva!=0)
{
/* if x aint 0 , calculate value*/
yva=(xva*xva)+(xva/2)-sqrt(1/(2*xva));
}
i++;
}
}
this is what i have made , but i dont get if the program takes into count that x1 stays same as A and that it ( xva=xva[i-1]+(pow(c,i-2)*h);) should start from x2 to x15
4. Nov 24, 2008
### sulev8
why cant u just comment my work ,and help me by giving me how it should be , and tell whats missing
5. Nov 24, 2008
### mgb_phys
What are c and h - where are the defined, initialised ?
You can't call this with i<2, you start at i=0;
Last edited: Nov 24, 2008
6. Nov 24, 2008
### wildman
Well, one thing can cause you trouble:
sqrt(1/(2*xva));
You check for zero, but since xva is a float, it still can be a very small number (not zero) and blow up on you. Keep in mind that while a human would know that .000000000001 is basically zero, to a computer it is not.
you also don't want to initialize i to zero. Otherwise, the i-2 factor will end up at -2. You want to initialize it to 2.
Finally, use a for statement instead of a while: Like for(i=2;i < 15;i++) | 832 | 2,763 | {"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} | 3.21875 | 3 | CC-MAIN-2018-22 | longest | en | 0.913604 |
https://www.prettymotors.com/how-much-dirt-can-a-dump-truck-haul/ | 1,723,010,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00137.warc.gz | 736,835,939 | 24,078 | # How Much Dirt Can a Dump Truck Haul?
How much dirt can a dump truck haul depends on several factors, including terrain shape and weather conditions. Generally, eight to twelve trips should be sufficient to fill the capacity of the dump truck. One way to estimate the amount of dirt you can haul with a dump truck is to imagine how much space one load takes up inside the vehicle. Then, multiply that number by two. That’s how much dirt a standard dump truck can haul per truckload.
Dump trucks are large, open-box vehicles that can carry anything from a few cubic yards to several cubic yards. They are commonly equipped with hydraulic arms that lift the front of the bed. When the bed is lifted, the dirt is dumped out of the truck. Dump trucks can carry anywhere from 10 to 14 cubic yards of dirt. Their capacity will vary, however, depending on their usage and the amount of dirt they are carrying.
## What is the Capacity of a Dump Truck?
A dump truck’s capacity varies depending on its model, size, and type. Larger dump trucks typically carry between 10 and 14 cubic yards of fine material. You can use this figure to calculate how much material your truck can handle, and to figure out how much it weighs per ton of resource it is capable of transporting. Listed below are some tips on how to calculate a dump truck’s capacity.
First, it’s important to know how much material each cubic yard can hold. Most of these materials are sold in cubic yards. The cubic yard capacity of a dump truck is indicated in the owner’s manual. A cubic yard of compact dirt weighs two to three thousand pounds, while heavier materials can weigh up to three thousand pounds. To find out the cubic yard capacity of a dump truck, determine how much material the truck can hold.
Once you know how much material a dump truck can carry, you can start planning your materials. Most dump trucks are configured to take advantage of the weight limits for specific highways, bridges, and buildings. This means that the load limit for your truck may be lower than the maximum allowable weight for the material you plan to haul. You can also check the weight limit for your truck on an official document or on the Internet.
## How Many Yards is a 1 Ton Dump Truck?
What does a one ton dump truck have to carry? Generally, a dump truck’s capacity ranges from ten to sixteen cubic yards. For example, a one ton dump truck can haul about six cubic yards of gravel. A similar amount of gravel, but less compacted, will weigh around three tons. For this reason, full-size dump trucks typically have a capacity of between ten and sixteen cubic yards.
READ ALSO: Why Does My Truck Battery Keep Dying?
The capacity of a full-size dump truck varies. Most standard full-size dump trucks hold between ten and 16 cubic yards. Smaller trucks can hold as little as one and a half cubic yards. The amount of material a dump truck can carry is directly influenced by the size of the truck’s bed. Knowing this can help you determine how many trips a truck will need to haul the material you need.
Another important consideration when purchasing a dump truck is its weight capacity. A standard commercial dump truck can hold between ten and eighteen cubic yards of dirt. Larger dump trucks can hold as much as 30,000 pounds of material in one trip, while smaller trucks may be limited to carrying less than one ton of material. Whether you need a large or a small truck will depend on the job site.
## How Big is a 16 Yard Dump Truck?
How much can a 16-yard dump truck haul? The actual weight of a full-sized dump truck varies widely. The ADT Volvo Construction Equipment A60H can carry 60 tons, while other manufacturers produce dump trucks ranging from 25 to 45 tons. Doosan, for example, manufactures a 31-ton dump truck. John Deere manufactures four models in the 26-to-46-ton range. Caterpillar has seven dump body models.
The weight capacity of a dump truck is determined by the weight and size of its bed. A full-sized truck can carry 12 to 16 cubic yards of materials. Smaller versions of dump trucks carry up to thirteen or fourteen cubic yards. Crushed-stone gravel can weigh up to 3000 pounds per cubic yard. Depending on the type of material being hauled, a standard dump truck may be just right for your needs.
Another option is a 16-18 ft. Load King tri-axle dump truck. Its length is typically 16′-18′ and is capable of moving sand, gravel, and other materials. Dump trucks in this size range come with motorized mesh tarps. They are also available in a large number of different sizes. Ultimately, you will need to decide what kind of truck is best for your project.
## How Do You Calculate Dump Truck Loads?
When renting a dump truck, you should know how much it can carry. Dump truck capacities are measured in cubic yards. Each cubic yard weighs a certain amount depending on the material it is made of and the amount it can carry. A single cubic yard of compact dirt weighs between two and three hundred pounds, while one cubic yard of sand can weigh up to three thousand pounds. Depending on the density and the type of material, a dump truck may have a capacity of fifteen or sixteen tons.
READ ALSO: How Much Cement Truck Cost?
To calculate the weight of a dump truck, you must know the number of cubic yards per cubic yard. For example, a 20-ton truck can carry up to ten to fourteen cubic yards of dirt, which is about 10 to 14 tons per truckload. A fifteen-ton dump truck, on the other hand, can hold about eight to twelve cubic yards. The exact number depends on the amount of material you’re dumping, but this is a good guideline to follow.
## How Many Tons Can a Dump Truck Carry?
How many tons a dump truck can carry depends on its size, weight capacity, and bed size. A larger dump truck can carry up to 14 tons of material, while smaller ones only carry between six and 12 cubic yards. These capacities can be useful for estimating the amount of material to be hauled and the weight per unit of resource. Here’s how to calculate how much your dump truck can carry.
To estimate how much a dump truck can carry, check its gross vehicle weight rating, which can be found on its registration. Then, multiply that number by the area it can carry. A dump truck with 6.5 cubic yards of material will weigh about 13 tons when fully loaded. Similarly, a dump truck with five tires can carry up to 12 tons. A dump truck with 10 tires can carry 15 tons of material.
A typical dump truck has a capacity of approximately fourteen tons of material. The weight of a dump truck’s bed is determined by the volume of the material it can carry. A small dump truck can hold about seven to eight cubic yards of material. The cubic yard capacity of a dump truck depends on the size and construction materials of the truck’s body. A commercial dump truck’s bed typically holds between 10 and 14 cubic yards, or about one to three tons of material.
## How Much Can a 6 Axle Dump Truck Haul?
The length and width of a typical short-bed pickup bed are important factors in determining how much dirt a truck can haul. A typical yard of dirt measures approximately 108 square feet. A yard of topsoil measures approximately 55 square feet. And, a yard of gardening dirt measures approximately 41 square feet. If you’re wondering how much dirt a 6-axle dump truck can haul, the answer is about 40 cubic yards.
READ ALSO: Can You Make an SUV into a Truck?
Before you start to search for a dump truck, it’s important to measure the truck’s bed size. Then, multiply that number by 27 to get the cubic yard capacity. This figure will be useful when you need to know the weight capacity of your vehicle. The maximum amount of dirt a 6 axle dump truck can haul will vary based on the model. A standard tri-axle dump truck can carry about twelve cubic yards of dirt. A quad-axle truck is capable of carrying about 14 cubic yards of dirt.
The gross vehicle weight rating or GVWR of a dump truck is the maximum load capacity it can carry. This figure is based on the weight of the material and is often the same as the truck’s gross vehicle weight rating (GVWR). The weight of the material itself may be significantly greater. Dump trucks are capable of hauling dirt and other materials up to 13 tons. However, soil can weigh as much as a thousand pounds per cubic yard.
## How Many Cubic Yards are in a 15 Ton Dump Truck?
When measuring the cubic yards in a dump truck, it’s important to consider what you’re transporting. The size of the truck bed and weight rating are the two most important factors in cubic yard capacity. The truck bed should be measured from one side of the inside wall to the other, and then converted to feet. If your truck can only handle a certain amount of material, you may need a larger one.
To determine how many cubic yards are in a 15-ton dump truck, divide the length and width of your project by 12. Then, multiply those values by 12 to get the total volume. In this case, one cubic yard equals 46,656 cubic inches. You can also calculate the volume of your project by multiplying the width and depth of the area by the cubic yards of the material.
A dump truck’s capacity depends on the manufacturer of its body and the buyer’s needs. If you’re hauling gravel, crushed stone, or mulch, you’re probably only half or two-thirds full. Lighter materials can be loaded to the maximum capacity, but you’ll want to retain some of the material by using a canvas or screen cover. To find out the cubic yard capacity of your truck, measure the cargo box and the bed to determine its cubic yard capacity. | 2,077 | 9,617 | {"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} | 3.046875 | 3 | CC-MAIN-2024-33 | latest | en | 0.955603 |
https://math.stackexchange.com/questions/1865411/explanations-for-a-diagonal-process-construction-of-a-sequence | 1,713,208,058,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817014.15/warc/CC-MAIN-20240415174104-20240415204104-00424.warc.gz | 350,323,614 | 35,491 | # Explanations for a "diagonal process" construction of a sequence
I am reading Peter Lax's "Functional analysis".
Let $y_n$ be a bounded sequence of vectors in a Banach reflexive space, $X, Y$ their closed linear span. Take a countable set $m_j$ of applications belonging to the dual $Y^*$ of $Y$ ($Y^*$ is separable).
The author then says that we can apply the "classical diagonal process" to obtain a subsequence $z_n$ of vectors such that $\lim_n m_j(z_n)$ exists for every $j$.
I do not understand how he builds this subsequence.
• How is $X$ defined ? Jul 20, 2016 at 15:16
• X is a reflexive Banach space Jul 20, 2016 at 15:29
The key ingredient is the observation that if a sequence is convergent, then any subsequence is also convergent.
1. First, you construct recursively a sequence of subsequences of $y_n$ as follows.
2. To begin, put $z_{n,0}:=y_n$.
3. Given $(z_{n,j})_n$, you can find (by compactness) a subsequence $(z_{n,j+1})_n$ such that $\lim_{n}m_j(z_{n,j+1})$ exists.
4. Note that then all $\lim_n m_{j'}(z_{n,j+1})$ exist, for $j'\leq j$ (because in this case $(z_{n,j+1})_n$ is a subsequence of $(z_{n,j'+1})_n$).
5. Finally, put $z_n:=z_{n,n}$ (i.e. $z_n$ is the diagonal of $z_{n,j}$, hence the name). To see that $\lim_n m_j(z_n)$ exists, note that except for finitely many elements, $z_n$ is a subsequence of $z_{n,j}$.
You don't need $X$ to be reflexive, or even a Banach space for this argument. All that really matters is that all the functions $m_j$ map the sequence $y_n$ into compact metric spaces (so that we can use compactness in step 3.) -- the topology on $X$ (or its linear structure) is completely irrelevant beyond this, guaranteed by the fact that bounded functionals by definition map bounded sequences to bounded subsets of the real line.
• Thank you very much Tomasz, now it is clear. It has been really instructive. Jul 20, 2016 at 18:35
• The only thing it would like to understand better is passage 4. Could you be more explicit please? Jul 20, 2016 at 19:14
• @GiovanniSiclari: There you go. Jul 21, 2016 at 9:25
• Thank you. This morning I have finally understood it. Jul 21, 2016 at 9:27 | 657 | 2,149 | {"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} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.914181 |
http://www.campusplace.co.in/2014/01/clocks-introduction.html | 1,550,285,459,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479838.37/warc/CC-MAIN-20190216024809-20190216050809-00173.warc.gz | 318,782,440 | 44,773 | # 16. Aptitude Questions -- CLOCKS - Introduction and Important Theory
Introduction:
A clock is a devise which shows time in consists of two hands the hour hand and the minute hand. The hour hand (short hand) indicates time in hours and the minute had or long hand indicates time in minutes. For simplicity we will not consider the third hand – the second hand which is also presenting many clocks and watches.
Usually clocks are circular in shape but we can see many other different shapes also. Irrespective of the shape of the dial of the clocks, the minute hand and hour hand always rotate in circular direction.
The circumference of a dial clock is divided into 12 equal parts called hour spaces.
Each hour space is further subdivided into 5 equal parts called minute spaces.
Thus the complete circumference of a dial of a clock is divided into 12 X 5 = 60 minute spaces.
In an hour the hour hand crosses 5 minute spaces. While the minute hand crosses 60 minute spaces therefore is obvious that in 60 minutes, the minute hand gains(60-5=55) minute spaces or minutes over the hour hand. This is very important and useful conclusion and therefore should be remembered while solving the problems on clocks.
Some important facts:
1 . In one hour the minute hand makes 3600
In 60 minutes the minute hand makes 3600
So in one minute the minute hand makes
360 / 60
that is 60
Therefore each minutes space = 3600 i.e., minute spaces are 60 apart.
2. In one hour the hour hand makes
360 / 12
i.e., 300
In 60 minutes the hour hand makes 300
So in one minute the hour hand makes
30 / 60
that is 1/20
Therefore each hour space = 300 i.e., hour spaces are 30apart.
3. The minute hand moves through 60 in each minute whereas the hour hand moves through 1/2 0 in each minute.
Thus in one minute the minute hand gains 5 ½ 0 than the hour hand.
4. When the 2 hands are coincident, they are at 00 are 0 minutes apart.
i) In every hour the hour hand and minute hand coincide only once.
ii) In twelve hours the 2 hands coincide 11 times. Because there is common a position 12’O clock between 11 and 1’ O clock.
iii) In 24 hours the 2 hands coincide 22 times
5. When the 2 hand are in opposite direction they are 180/6 i.e., 30 minutes apart.
i) It happens only once in an hour.
ii) In every hour the hour hand and minute hand are in opposite direction only once.
iii) In 12 hours the 2 hands are in opposite direction 11 times. Because there is common a position 6’ O clock between 5 and 11’ O clock.
iv) In 24 hours the 2 hands opposite direction 22 times.
6. When the 2 hands at right angle (900) or15 minutes spaces apart.
i) It happens 2 times in every hour.
ii) In 12 hours the hands are at right angle only 22 times. It is so because there are 2 common positions in every 12 hours i.e.,3’O clock & 9’O Clock.
iii) In 24 hours the 2 hands at right angle 44 times.
7. The minute and hour hands of a clock coincide for every 65
5 / 11
minutes.
Too Fast:
When the clock indicates time more then the correct time it is said to be running too fast by the difference between the correct time and the time indicated by the clock. For example, the clock indicates 9:30 am when the correct time is 9:15 am it is set to be 15 minutes fast
Too slow:
When the clock indicates time less than the correct time it is said to be running too slow by the difference between the correct time and the time indicated by the clock. For example, the clock indicates 9:30 am when the correct time is 9:45 am it is set to be 15 minutes slow
Note : The minute hand moves 12 times as fast as the hour hand.
Note : The clock has two hands—the hour hand and the minute . The hour hand (or short hand) indicates time in hours and the minute hand (or long hand) indicates time in minutes. In an hour, the hour hand covers 5 minute spaces while the minute hand covers 60 minute spaces. Thus, in one hour or 60 minutes, the minute hand gains 55 minute spaces over the hour hand.
This is very important and useful conclusion .It must be always remembered while solving problems on clocks.
For examples and solved problems on clocks , go through
Solved Problems on Clocks
For more problems and On-line tests on Quantitative aptitude, visit http://www.9exams.com | 1,103 | 4,388 | {"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-2019-09 | longest | en | 0.865811 |
https://byjusexamprep.com/average-questions-with-short-tricks-i | 1,721,270,714,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00833.warc.gz | 127,806,244 | 62,421 | # Average Questions with Short Tricks!
By Avinash Kumar|Updated : June 24th, 2021
Time management is one of the most important tasks in all competitive examinations. We all struggle to attempt as many questions as possible in the given time to have a niche over other competitors. So, keeping in mind the needs of aspirants, we have as come up with “Important Questions with Short Tricks” to solve Average questions. To make the chapter easy for you all, we are providing you all some Short Tricks to solve the questions in the Quant Section.
What is Average?
The result obtained by adding several quantities together and then dividing this total by the number of quantities is called Average.
The main term of average is an equal distribution of value among all which may distribute persons or things. We obtain the average of a number using a formula that is the sum of observations divided by the Number of observations.
Here are average based some facts, formulas and shortcut tricks examples. Given below are some more example of practising.
Formula:
• Average: = (Sum of observations / Number of observations).
Find the Average Speed
• If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr then the average speed during the whole journey is given by 2xy/(x+y)
• If a person covers A km at x km/hr and B km at y km/hr and C km at z km/hr, then the average speed in covering the whole distance is- 3xyz/(xy + yz + zx)
• When a person leaves the group and another person joins the group in place of that person then-
• If the average age is increased,
Age of new person = Age of separated person + (Increase in average × total number of persons)
• If the average age is decreased,
Age of new person
= Age of separated person - (Decrease in average × total number of persons)
When a person joins the group-In case of an increase in average
• Age of new member = Previous average + (Increase in average × Number of members including new member)
In case of a decrease in average
• Age of new member = Previous average - (Decrease in average × Number of members including new member)
In the Arithmetic Progression, there are two cases when the number of terms is odd and second one is when the number of terms is even.
• So when the number of terms is odd the average will be the middle term.
• when the number of terms is even then the average will be the average of two middle terms.
Question 1. The average of 8 numbers is 20. The average of first two numbers is 15.5 and the average of next three number is If the 6th number is 4 and 7 less by the 7th and the 8th numbers respectively then what will be the 8th number?
(1) 25
(2) 22
(3) 35
(4) 30
Solution:
Short Trick:
Sum of all = 160
Sum of first 5 = 95 coming from
sum of last 3 = 65
x + x + 4 + x + 7 = 65
18 + 22 + 25
Question 2: The average temperature from Monday to Wednesday is 370C while the average temperature from Tuesday to Thursday is 340C. The temperature of Thursday is times to that of Monday. Find the temperature of Thursday?
(1) 360C
(2) 330C
(3) 370C
(4) 340C
Solution:
Short Trick:
Mon + Tues + Wed = 111
Tues + Wed + Thurs + 102 (diff. = 9)
Thursday = Thursday =
Question 3: Of the three numbers whose average is 60, the first is one-fourth of the sum of the others. The first number is
(1) 30
(2) 36
(3) 42
(4) 45
Solution:
x + y + z = 180 …………(i)
then 4x + y + z
Putting the value of y + z in
Equation …………….(i) x + 4x = 180
5x = 180 x = 36
Sum =
Question 4: The average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of the failed candidates is 15, what is the number of candidates who passed the examination?
(1) 100
(2) 120
(3) 140
(4) 160
Solution:
Let the number of passed candidates be x
Then total marks = 120 x 35 = 39x + (120 – x) x 15
Or, 4200 = 39x + 1800 – 15x
Or, 24x = 2400
x = 100
Number of passed candidates = 100
Short Trick:
Pass Fail
39 15
\ /
35
/ \
20 4
5 : 1
Hence, total number of passed candidates
Question 5: The average weight of 8 persons is increased by 2.5 kg when one of them who weight 56 kg is replaced by a new man. The weight of the new man is
(1) 73 kg
(2) 72 kg
(3) 76 kg
(4) 80 kg
Solution:
Aw2 – AW1 = 2.5
Total weight1 – Total weight = 20.0 kgs
This difference is because of the new man.
Hence the weight of the new man = 56 + 20 = 76 kgs.
Short Trick:
56 + 8 x 2.5
= 76 kgs
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Unformatted text preview: 25-2Chapter 25(b)(c)(d)Keep constant, where tis the time the shutter is open. soThe closest shutter speed setting is 25.5. Set Up:The image is real since it is projected onto the film. Solve:Reflect:The object distance sis much greater than so the image is just beyond the focal point of the lens.25.6. Set Up:The image is real since it is projected onto the film. is the distance from the lens to the film plane.Solve: (a)Since the image is real the lens must be convex. (b)For For gives The range of distances is 50 mm to 56 mm.25.7. Set Up:The photocells must be at the image location.Solve: (a)(b)The image is inverted and 3.33 cm tall. so the image is real.(c)Reflect:Since the image is projected onto the photocell, it must be real. Any real image formed by a single lens isinverted.25.8. Set Up:The image is real since it is projected onto the film and is the distance from the lens to the film....
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## This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.
Ask a homework question - tutors are online | 363 | 1,373 | {"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} | 2.671875 | 3 | CC-MAIN-2017-22 | longest | en | 0.910569 |
http://www.physicsforums.com/showthread.php?t=312391 | 1,386,876,289,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164692455/warc/CC-MAIN-20131204134452-00091-ip-10-33-133-15.ec2.internal.warc.gz | 482,957,818 | 8,981 | # Vector Calculus, sketching regions in R3
by fredrick08
Tags: calculus, regions, sketching, vector
P: 376 solved
P: 376 or if possible, be able to tell me, how to draw regions in maple? so i can at least picture it, and try to understand how to draw it.
P: 376 ok from that, i think i can see that x is from 0 to root(1-y^2), y is from 0 to 1 and z is from 0 to 2y.. so they are bounds of my region.
P: 376
## Vector Calculus, sketching regions in R3
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P: 38,428
Quote by fredrick08 1. The problem statement, all variables and given/known data sketch the following region in R3 V={$$\widehat{}r$$:y$$\geq$$0,0$$\leq$$x2+y2$$\leq$$1,0$$\leq$$z$$\leq$$2y} You have $y\ge 0$, $0\le x^2+ y^2\le 3. The attempt at a solution ok I havn't done this a lot, and dont quite understand, i think that when y=0, 0$$\leq$$x2$$\leq$$1... and 0$$\leq$$z$$\leq$$2y... but i dont understand how to sketch these, in previous examples, we had z=f(x,y)... but this one is different... could someone please explain how to sketch this... coz im not sure.. x2+y2 is a dome shape that is greater then 0 but less the 1... but i just cant picture/sketch it.. apparently it is meant to look like a wedge shape, can anyone help me plz. Thanks. Quote by fredrick08 please anyone? Don't expect a response within a few hours. People aren't just sitting at their computers waiting for problems. I like to sleep at night! You have [itex]y\ge 0$, $0\le x^2+ y^2\le 1$, $z\le 2y$
first draw an xy-plane. $y\ge 0$ means your graph is in the upper half plane. the 0 in $0\le x^2+ y^2\le 1$ doesn't really tell us anything because $x^2+ y^2$ can't be negative anyway. But $\x^2+ y^2= 1$ is a circle with center at (0,0) and radius 1. Since y must be positive, draw the upper semi-circle. Now draw a yz-plane beside your xy-graph. z= 2y is a line from (0,0) to (2, 1). If you now imagine that z-axis coming directly up from your first graph, with the y-axes aligned, you should see that plane forms a "top" on the parabolic cylinder. Yes, it is basically a "wedge" shape with the x= 0 plane forming one side, the plane top going down to cut the parabola. It is NOT a bounded region because there is no "bottom" unless you left out z= 0, say.
P: 376 ok sorry, i have an anxiety problem.. thankyou very much, i will try and draw it and post it up.
P: 376 ok, skip on the drawing part, lol i cant draw at all, so to confirm the shape is a .... quarter of a cone, with base radius 1, and height 2? very well explained, by the way = )
P: 376 or maybe not, because z=2y cuts through the origin and ends at (2,1,0) but a cone would have and apex at (0,0,0) instead of a cone, does it look like a pyramid with curved surface... oh i don't Im confused now
P: 376 ok, been thinking, i think i got it now its an upside down quarter cone, with z edges of the line z=y... therefore the apex would be at (0,0,0)
P: 376 no wait, coz if was a cone that would imply that z=2x aswelll.... omg i dont know, help....
Related Discussions Calculus & Beyond Homework 6 Calculus 3 Calculus & Beyond Homework 3 Introductory Physics Homework 1 | 968 | 3,118 | {"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} | 3.6875 | 4 | CC-MAIN-2013-48 | longest | en | 0.939184 |
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a. calculate and interpret the net present value (NPV) and the internal rate of return (IRR) of an investment;
b. contrast the NPV rule to the IRR rule, and identify problems associated with the IRR rule;
A company should choose those capital investment processes that maximize shareholder wealth.
The net present value (NPV) of an investment is the present value of its cash inflows minus the present value of its cash outflows. The internal rate of return (IRR) is the discount rate that makes net present value equal to 0.
According to the NPV rule, a company should accept projects where the NPV is positive and reject those in which the NPV is negative. A positive NPV suggests that cash inflows outweigh cash outflows on a present value basis. That is, the positive cash flows are sufficient to repay the initial investment along with the capital costs (opportunity cost) associated with the project. If the company must choose between two, mutually-exclusive projects, the one with the higher NPV should be chosen.
• According to the IRR Rule, a company should accept projects where the IRR is greater than the discount rate used (WACC) and reject those in which the IRR is less than the discount rate. An IRR greater than the WACC suggests that the project will more than repay the capital costs (opportunity costs) incurred.
There are three problems associated with IRR as a decision rule.
• Reinvestment
The IRR is intended to provide a single number that represents the rate of return generated by a capital investment. As such, it is an easy number to interpret and understand. However, calculation of the IRR assumes that all project cash flows can be reinvested to earn a rate of return exactly equal to the IRR itself. In other words, a project with an IRR of 6% assumes that all cash flows can be reinvested to earn exactly 6%. If the cash flows are invested at a rate lower than 6%, the realized return will be less than the IRR. If the cash flows are invested at a rate higher than 6%, the realized return will be greater than the IRR.
• Scale
In most cases, NPV and IRR rules provide the same recommendation as to whether to accept or reject a given capital investment project. However, when choosing between two mutually-exclusive projects (ranking), NPV and IRR rules may provide conflicting recommendations. In such cases, the NPV rule's recommendation should take precedence.
One of the situations in which IRR is likely to contradict NPV is when there are two mutually-exclusive projects of greatly differing scale: one that requires a relatively small investment and returns relatively small cash flows, and another that requires a much larger investment and returns much larger cash flows.
• Timing
The other situation in which IRR is likely to contradict NPV is when there are two mutually-exclusive projects whose cash flows are timed very differently: one that receives its largest cash flows early in the project and another that receives its largest cash flows late in the project.
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### ?LASQ3 could cause NaNs appearance in ?GESVD and ?GESDD
Posted: Fri Jun 29, 2012 1:30 am
Hi,
We recently noticed that some of matrices could cause ?GESVD to produce NaNs on output. One example is a matrix with all elements being equal to 1. After bidiagonalization it naturally has a lot of denormalized numbers in it. But not every sequence of such numbers fails, and appearance of a failing sequence affected by matrix sizes, compiler and it’s options used to compile BLAS, and so on. However the rootcause is clear.
NaNs appear in ?LASQ3 on call tree ?GESVD->?BDSQR->?LASQ1->?LASQ2->?LASQ3, because operation on line 271 of ?LASQ3 which is T = HALF*( ( Z( NN-7 )-Z( NN-3 ) )+Z( NN-5 ) ) could cause T to be equal to ZERO for very close denormalized values in Z after taking the half. Followed operations with use of the T which is equal to ZERO causes NaN appearance: S = something/T and next step in computing S contains S/T.
It looks like the preceding condition Z( NN-5 ).GT.Z( NN-3 )*TOL2 is weak, and doesn’t take into account possibility of T to become ZERO.
Easy fix could be just replacing:
IF( Z( NN-5 ).GT.Z( NN-3 )*TOL2 ) THEN
T = HALF*( ( Z( NN-7 )-Z( NN-3 ) )+Z( NN-5 ) )
With:
T = HALF*( ( Z( NN-7 )-Z( NN-3 ) )+Z( NN-5 ) )
IF( Z( NN-5 ).GT.Z( NN-3 )*TOL2 .AND. T.NE.ZERO ) THEN
Thanks,
Alexander
### Re: ?LASQ3 could cause NaNs appearance in ?GESVD and ?GESDD
Posted: Fri Jun 29, 2012 7:22 am
Alexander,
Thanks for finding this. I will put your solution in the svn and it will be in the next release.
Regards,
Rodney | 484 | 1,561 | {"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-2018-26 | latest | en | 0.884578 |
queerneuromathgirl.blog | 1,686,060,468,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00464.warc.gz | 530,695,790 | 35,017 | ## First Quarter Research Progress and Ideas
To be honest, I spent most of my first quarter of graduate school on classes, seminars, and getting adjusted to the new environment. However, I did start attending research meetings in a group I am interested in, and I have some ideas for a potential project. I am very excited about beginning this project, and I hope that this coming quarter, I will be able to make more progress. Luckily, there is a postdoc in the group who is also excited about it, and he has been very thorough in providing me with papers to read and feedback on my work. I will begin to describe my progress briefly.
The group I have been working in studies a wide range of systems such as predator-prey dynamics, multi-drug interaction, the relationship between sleep and metabolic rate, and cardiovascular networks. Since there are so many diverse projects happening in our group, our group meetings are split by topic. The sub-group I joined focuses on networks. So far, they have been mostly focusing on cardiovascular networks. They develop models that describe these networks, such as the scaling laws that describe changes in the radius and length of vessels across levels of the network. Then, they test these models against data extracted from 3D images.
Since my primary interest in biology is in neuroscience, I approached the group to find out if there were any projects in neuroscience. The PI told me that although there are currently no projects in neuroscience in this group, there are mathematical similarities between neuronal networks and cardiovascular networks, and he saw a future in extending the image analysis of cardiovascular networks to neurons.
We can think of a network of neurons like the cardiovascular system, a resource distribution network that is subject to biological and physical constraints. Deriving a power law relationship between radius and length of successive levels of a vascular network relies on minimizing the power lost due to dissipation while maintaining the assumptions that the network is of a fixed size, a fixed volume, and space filling. This calculation is carried out using the method of Lagrange multipliers, and assuming that the flow rate is constant. The power loss due to dissipation in the cardiovascular network is $P = \dot{Q_0}^2 Z_{net}$, where $\dot{Q_0}$ is the volume flow rate of blood and $Z_{net}$ is the resistance to blood flow in the network. For a neuronal network, we will use an analogous equation, $P = I_0^2 R_{net}$, where $I_0$ is the current, and $R_{net}$ is the resistance to current flow in the network. We will carry out the Lagrange multiplier calculations in a similar fashion to the calculations for cardiovascular networks.
For cardiovascular networks, we use the Poiseuille formula for resistance, which is the hydrodynamic resistance to blood flow in the network. According to this formula, the impedance at a level k in the network is given by $Z_k = \frac{8 \mu l_k}{\pi r_k^4}$. We can reduce $\frac{8 \mu}{\pi}$ to a single constant C, so this is equivalent to $Cl_k r_k^{-4}$. Thus, the resistance is proportional to the product of some powers of the length and the radius. If we want to consider a general formula for the resistance, we can consider a formula with powers p and q of of length and radius respectively. That is, our resistance formula at level k is $R_k = \Tilde{C} l_k^p r_k^q$.
We define the objective function as follows:
$P = I_0^2 R_{net} + \lambda V + \lambda_M M + \sum_{k=0}^{N} \lambda_k n^k l_k^3$
This objective function arises from the fact we want to minimize power loss, the first term, while imposing the three constraints that correspond to the last three terms: size, volume, and space filling. Each constraint corresponds to a Lagrange multiplier. The last constraint comes from the fact that a resource distribution network must feed every cell in the body. This, each branch at the end of the network feeds a group of cells called the service volume, $v_N$, where N is the terminal level, and the number of vessels at that level is $N_N$, so the total volume of living tissue is $V_{tot} = N_N v_N$. If we assume that this argument holds over all network levels, we have $N_N v_N = N_{N-1} v_{N-1} = … = N_0 v_0$. We assume that the service volumes vary in proportion to $l_k^3$, so the total volume is proportional to $N_kl_k^3$. Our objective function has N terms related to space filling, since the space filling constraint must be satisfied at each level k. We assume that the branching ratio is constant, so the number of vessels at level k is $n^k$. We can define the volume as $\sum_{k=0}^N N_k \pi r_k^2 l_k$.
Note that we are defining the constraints the same we we did for vascular networks, but it is unclear whether these assumptions are accurate for neuronal networks. However, for the sake at arriving at a preliminary theoretical result for the scaling of neuronal networks, we will keep constraints.
The total resistance at each level is the resistance for a single vessel divided by the total number of vessels, that is, $R_{k, tot} = \frac{\Tilde{C} l_k^p r_k^q}{n^k}$. The net resistance of the network is the sum of the resistances at each level, so $R_{net} = \sum_{k = 0}^N \frac{\Tilde{C} l_k^p r_k^q}{n^k}$. If we define new Lagrange multipliers, $\lambda’ = \pi \lambda$, we can rewrite the objective function as follows:
$P = I_0^2 \sum_{k = 0}^N \frac{\Tilde{C} l_k^p r_k^q}{n^k} + \lambda’ \sum_{k=0}^N n^k r_k^2 l_k + \lambda’_M M + \sum_{k=0}^{N} \lambda’_k n^k l_k^3$
To normalize further, we can divide by the constant $I_0^2\Tilde{C}$, since the current is constant, and absorbing this constant into new definitions of the Lagrange multipliers, we get:
$P = \sum_{k = 0}^N \frac{l_k^p r_k^q}{n^k} + \Tilde{\lambda} \sum_{k=0}^N n^k r_k^2 l_k + \Tilde{\lambda}_M M + \sum_{k=0}^{N} \Tilde{\lambda}_k n^k l_k^3$
To find the radius scaling ratio, we will minimize P with respect to $r^k$, at an arbitrary level k, and set the result to 0. Thus, we can find a formula for a Lagrange multiplier and derive the scaling law.
So we have:
$\frac{dP}{dr_k} = \frac{l_k^p qr_k^{q-1}}{n^k} + 2 \Tilde{\lambda} n^k r_k l_k = 0$
Solving for the Lagrange multiplier, we have:
$\Tilde{\lambda} = -\frac{qr_k^{q-1}l_k^p}{2n^{2k} r_k l_k} = \frac{\frac{-q}{2}}{n^{2k}l_k^{1-p}r_k^{2-q}}$
Since this is a constant, the denominator must be constant across levels. So
$\frac{n^{2(k+1)}l_{k+1}^{1-p}r_{k+1}^{2-q}}{n^{2k}l_{k}^{1-p}r_{k}^{2-q}} = 1$
It is useful to consider the case where the resistance is related to the length linearly, that is, for p =1. Thus, we obtain the scaling ratio:
$\frac{n^{2(k+1)}r_{k+1}^{2-q}} {n^{2k}r_{k}^{2-q}} = 1 \rightarrow \frac{r_{k+1}}{r_k} = n^{\frac{-2}{2-q}}$
To find the length scaling ratio, we will minimize P with respect to $l^k$, at an arbitrary level k, and set the result to 0. Thus, we can find a formula for a Lagrange multiplier, using the formula above, and derive the scaling law.
So we have:
$\frac{dP}{dl_k} = \frac{pl_k^{p-1}r_k^{q}}{n^k} + \Tilde{\lambda} n^k r_k^2 + 3\Tilde{\lambda_k} n^k l_k^2 = 0$
Solving for the Lagrange multiplier, we have:
$\Tilde{\lambda_k} = \frac{-\frac{pl_k^{p-1}r_k^{q}}{n^k} – \Tilde{\lambda} n^k r_k^2}{3n^k l_k^2}$
Substituting $\Tilde{\lambda}$, as calculated before:
$\Tilde{\lambda_k} = \frac{-\frac{pl_k^{p-1}r_k^{q}}{n^k} + \frac{q r_k^2}{2n^{k}l_k^{1-p}r_k^{2-q}} }{3n^k l_k^2} = \frac{(\frac{q}{2} – p)pr_k^q l_k^{p-1}}{3n^{2k} l_k^2} = \frac{q-2p}{6} \frac{1}{n^{2k}l_k^{3-p}r_k^{-q}}$
Since this is a constant, the denominator must be constant across levels. So
$\frac{n^{2(k+1)}l_{k+1}^{3-p}r_{k+1}^{-q}}{n^{2k}l_{k}^{3-p}r_{k}^{-q}} = 1$
In the case where p=1, we have
$\frac{n^{2(k+1)}l_{k+1}^{2}r_{k+1}^{-q}}{n^{2k}l_{k}^{2}r_{k}^{-q}} = 1\rightarrow (\frac{l_{k+1}}{l_k})^2 = n^{-2} (\frac{r_{k+1}}{r_k})^q$
Substituting the scaling law for radius, we have:
$(\frac{l_{k+1}}{l_k})^2 = n^{-2} (n^{\frac{-2}{2-q}})^q \rightarrow \frac{l_{k+1}}{l_k} = n^{-1 – \frac{q}{2-q}} \rightarrow \frac{l_{k+1}}{l_k} = n^{\frac{-2}{2-q}}$
We can test these calculations for our vascular networks calculation, where q = -4. Our scaling laws for radius and length are $\frac{r_{k+1}}{r_k} = \frac{l_{k+1}}{l_k} = n^{-1/3}$, as expected.
We will now attempt to repeat these calculations using a resistance formula specific to neuronal networks.
We think of the resistance to blood flow as the resistance due to the viscosity of the fluid. For neuronal networks, we can think of axons and dendrites as wires through which current is flowing. The resistance as the resistance to current flow through the “wire” due to intrinsic properties of the wire. The resistance is given by $R_k = \frac{\rho l_k }{A}$, where A is the cross-sectional area of the wire, and $l_k$ is the length of the segment at that level. $\rho$ is the intrinsic resistivity of the axon or dendrite, and we are assuming that $\rho$ is constant, meaning that the material is uniform. If we assume that the axons or dendrites are cylindrical, we can define the cross-sectional area as $\pi r_k^2$ for level k, so the resistance for level k is given by $R_k = \frac{\rho l_k }{\pi r_k^2}$.
Assuming that the branching ratio is constant, the number of branches at each level is $n^k$, and the total resistance at each level is $R_{k,tot} = \frac{\rho l_k }{\pi r_k^2 n^k}$. The net resistance is the sum across all levels, that is $R_{net} = \sum_{k=0}^N\frac{\rho l_k }{\pi r_k^2 n^k}$.
Our objective function for this case can be derived in a similar manner as in the general case, setting $\Tilde{C} = \frac{\rho}{\pi}$, setting p = 1, and q = -2, based on the constants and powers for our specific resistance equation. Thus, we have the objective function
$P = \sum_{k = 0}^N \frac{l_k}{r_k^2 n^k} + \Tilde{\lambda} \sum_{k=0}^N n_k r_k^2 l_k + \Tilde{\lambda}_M M + \sum_{k=0}^{N} \Tilde{\lambda}_k n^k l_k^3$
To find the radius scaling ratio, we will minimize P with respect to $r^k$, at an arbitrary level k, and set the result to 0. Thus, we can find a formula for a Lagrange multiplier and derive the scaling law.
So we have:
$\frac{dP}{dr_k} = \frac{-2l_k}{n^k r_k^3} + 2 \Tilde{\lambda} n^k r_k l_k = 0$
Solving for the Lagrange multiplier, we have:
$\Tilde{\lambda} = \frac{1}{n^{2k}r_k^{4}}$
Since this is a constant, the denominator must be constant across levels. So
$\frac{n^{2(k+1)}r_{k+1}^{4}}{n^{2k}r_{k}^{4}} = 1$
Thus, we can solve for the scaling ratio:
$\frac{r_{k+1}}{r_k} = (n^{-2})^{1/4} = n^{-1/2}$
To find the length scaling ratio, we will minimize P with respect to $l^k$, at an arbitrary level k, and set the result to 0. Thus, we can find a formula for a Lagrange multiplier, using the formula above, and derive the scaling law.
So we have:
$\frac{dP}{dl_k} = \frac{1}{n^k r_k^2} + \Tilde{\lambda} n^k r_k^2 + 3\Tilde{\lambda_k} n^k l_k^2 = 0$
Solving for the Lagrange multiplier, we have:
$\Tilde{\lambda_k} = \frac{-\frac{1}{n^k r_k^2} – \Tilde{\lambda} n^k r_k^2}{3n^k l_k^2}$
Substituting $\Tilde{\lambda}$, as calculated before:
$\Tilde{\lambda_k} = \frac{-\frac{1}{n^k r_k^2} – \frac{1}{n^{k}r_k^{2}} }{3n^k l_k^2} = – \frac{2}{3n^{2k}l_k^2 r_k^2}$
Since this is a constant, the denominator must be constant across levels. So
$\frac{n^{2(k+1)}l_{k+1}^{2}r_{k+1}^{2}}{n^{2k}l_{k}^{2}r_{k}^{2}} = 1$
Thus, substituting in the scaling ratio for radius, we can solve for the scaling ratio for length:
$(\frac{l_{k+1}}{l_k})^2 = n^{-2} (\frac{r_{k+1}}{r_k})^{-2} = n^{-2} (n^{-1/2})^{-2} = n^{-1} \rightarrow \frac{l_{k+1}}{l_k} = n^{-1/2}$
Note that these scaling laws are consistent for the theoretical predictions from our general formulas, for q = -2.
Some of the assumptions we have made for the purpose of these calculations are as follows:
• The current flow is constant across all levels of the network
• The axons and dendrites are cylindrical
• The material of the axons and dendrites is uniform and can be linked to a constant of specific resistivity
• The network has a fixed size
• The network is contained within a fixed volume
• The network is space filling
• The branching ratio is constant
Particularly in the case of the volume and space-filling constraints, and the constant branching ratio, it is unclear if a neuronal network has the same properties that we assume hold for vascular networks. In addition, it is unclear whether it is reasonable to assume that the current flow is constant. Thus, it might be worth reexamining these constraints and assumptions to add more biologically realistic and relevant ones.
Moreover, instead of focusing on this optimization problem of minimizing power loss, it might be more fruitful to examine a different optimization problem, such as minimizing the time for a signal to travel from one end to another end of the network.
These scaling laws give us some preliminary ideas to work with. We can try using image analysis techniques to measure length and radii of segments of axons and dendrites across levels in images and see whether information extracted from the data supports our theoretical conclusions.
References
Savage, Van M., Deeds, Eric J., Fontana, Walter. (2008). Sizing up Allometric Scaling Theory. PLOS Computational Biology.
Johnston, Daniel, Wu, Samuel Miao-Sin . (2001). Foundations of Cellular Neurophysiology. MIT Press.
## Network Dynamics, Biophysics, and Mental Illness
[latexpage] This past fall was my first quarter of graduate school, and one of our core courses was Deterministic Models in Biology. For our final project, we chose a quantitative biology paper on a topic of our interest and presented on it to the class. The paper I chose was a review paper, Psychiatric Illnesses as Disorders of Network Dynamics by Daniel Durstewitz, Quentin J.M. Huys, and Georgia Koppe. My undergraduate research focused on the dynamics of neurons at the molecular level, and this paper helped me connect it to specific characteristics of mental illnesses.
This paper proposes that since observable cognitive and emotional states rely on the underlying dynamics of neuronal networks, we should use Dynamical Systems Theory (DST) to characterize, diagnose, and develop therapeutic strategies for mental illness.
The central idea of DST is that there is a set of differential equations that evolve in time. A set of dynamical equations could look as follows:
$\frac{dx_1}{dt} = \dot{x_1} = f_1(x_1, … , x_M, t; \boldsymbol{\theta} )$
$\frac{dx_2}{dt} = \dot{x_2} = f_1(x_1, … , x_M, t; \boldsymbol{\theta})$
$\vdots$
$\frac{dx_M}{dt} = \dot{x_M} = f_M(x_1, … , x_M, t; \boldsymbol{\theta})$
The variables $x_1, x_2, … x_M$ represent the dynamical variables such as voltage or neural firing rate. These equations describe how each of these variables change over time. $\boldsymbol{\theta}$ represents parameters, fixed values that are properties of the system that do not change over time.
We define a fixed point as the point at which the derivatives of all of the variables are equal to 0. Fixed points are stable if activity converges towards them, and unstable if activity diverges from them. Stable fixed points are called attractors. We can define the basin of attraction as the set of points from which activity converges towards the attractor.
The figure below shows an example of a phase plane, a representation of a space spanned by the two variables of a system. Note that it is possible to use dimensionality reduction methods to obtain visual representations for higher dimensional systems. The arrows show the activity of the system. The blue and orange curves represent nullclines, and along each of these curves, the derivative of one of the variables is 0. The green line represents the barrier between the two basins of attractions. It is possible to cross over this barrier as a result of either external influences or random fluctuations.
I will discuss some basic neuroscience before going into the dynamics of mental illnesses. There are many ion currents that pass through a neuron membrane such as sodium, potassium, and calcium. The dynamics of these ions are driven by electrochemical gradients. Spiking activity occurs when there is a rapid influx of sodium ions, producing the spike followed by an efflux of potassium ions, returning the membrane potential to the threshold potential.
We can think of a neuron membrane as a capacitor, where positive and negative charges are accumulated on either side. The current is the rate of charge flowing per time, $I = \frac{dq}{dt}$, and the charge of a capacitor is defined as q = CV. The current through the membrane is this $I_m = C_m \frac{dV_m}{dt}$. We can think of this system as the circuit shown below:
Because of charge conservation, the sum of the currents across the capacitor and each of the resistors must be 0. In mathematical terms, this is $C_m \frac{dV_m}{dt} = -\sum_i I_i$.
If we approximate each of these currents as ohmic, they will satisfy Ohm’s law, V = IR, meaning that the current is proportional to the difference between the membrane voltage and the threshold voltage by a factor of 1/R, or in other words, the conductance.
If the conductance were constant over time, these would be linear. However, the conductance depends on the proportion of ion channels that are open and the proportion of channels that are closed, called the gating variables. For example, a sodium current can be described as
$I_{Na} = g_{max}m^3h(V_m – E_{Na})$
In this system, m and h are the gating variables, and they vary from 0 to 1, and $g_{max}$ is the maximal conductance.
We can think of the dynamical equations for the gating variables as the result of a mass equation. Consider the reaction
$Closed \rightleftharpoons Open$
Suppose $\alpha$ is the rate of opening of a channel, or the forward reaction above, and $\beta$ is the rate of closing, the reverse reaction above, and both of these rates depend on the voltage. If m represents the proportion of channels that are open, the derivative over time is equal to the forward rate times the concentration of reactants minus the reverse rate times the concentration of products. In other words:
$\frac{dm}{dt} = \alpha(V_m)(1-m) – \beta (V_m)m$
Another form of this dynamical equation commonly seen in the literature is:
$\frac{dm}{dt} = \frac{m_{\infty}(V_m) – m}{\tau_{Na}(V_m)}$
$\tau_{Na}$ is the voltage-dependent time constant, and $m_{\infty}$ is the steady-state proportion of open channels as a function of voltage.
The dynamical equation for voltage for the simple NaKL model is as follows:
$\frac{dV}{dt} = g_{Na}m^3h(E_{Na}-V) + g_K n^4 (E_K -V) + g_L (E_L – V) + I_{inj}C^{-1}$
Neuronal networks are the result of multiple neurons connected to one another through synapses. Pre-synaptic neurons deliverer chemicals, called neurotransmitters, to post-synaptic neurons. Some neurotransmitters are excitatory, such as NMDA (N-Methyl-D-aspartic acid), meaning they increase the likelihood of spiking activity, and others are inhibitory, such as GABA (gamma-aminobutyric acid), meaning that they decrease the likelihood of spiking activity. To describe the networks of neuronal networks, each individual neuron has a voltage equation as illustrated above, with additional terms relating to its synaptic currents. These currents depend on the synaptic conductance, the difference between the membrane voltage and the threshold voltage, the strengths of the synaptic connections, and the fraction of open channels for each receptor. The dynamical equation for the fraction of open channels usually depends on properties of the presynaptic neuron.
So far, the variables we have been considering have been the voltage and the gating variables. In order to discuss the dynamics of mental illness, we must think about another important variable: firing rate. This simply describes the rate of voltage spikes over time. Below is an example of a phase plane, where the vertical axis is the average firing rate of inhibitory neurons, and the horizontal axis is the average firing rate of excitatory neurons.
In this system, the fixed points can be thought of as memories or goal-states, and we can use this system to consider the effects of the underlying dynamics on working memory or decision making. Increasing the depth of the basin of attraction can have the effect of increasing the stability of the state, while flattening the basin of attraction reduces the stability of the state.
This paper highlights the key role of dopamine in altering these attractor dynamics. Stimulating the D1 dopamine receptors has the effect of increasing firing activity of both excitatory (NMDA) and inhibitory (GABA) neurons. This alters the parameters of the system, in particular, the strengths of synaptic connections, over time. As a result, the basins of attraction are deepened, and the state is more stable and robust to external perturbations or noise fluctuations.
Stimulation of the D2 dopamine receptors has the opposite effect, flattening the basins of attractions. These flat attractor landscapes could lead to disorganized or spontaneous thoughts that can be experienced as hallucinations that are characteristic of schizophrenia. This can also explain the high distractibility in attention-deficit hyperactive disorder (ADHD). On the other hand, Obsessive Compulsive Disorder (OCD), a disorder characterized by rumination, invasive and recurrent obsessions and compulsions, can be linked to deep basins of attractions that are robust to potential distractors. Major Depressive Disorder characterized by a coexistence of rumination and a negative mood with lack of concentration and distractibility, and one can think of it as an imbalance between multiple attractor states.
The main point this review paper aims to illustrate is that in order to characterize and develop treatments for mental illnesses, one must consider the underlying network dynamics. The suggested role of dopamine in altering the depth of basins of attractions suggest that we might try to target the dynamics of schizophrenia patients, for example, through dopaminergic drugs.
I found the process of reading this review paper and the sources it cited extremely helpful for me in improving my understanding of neurons, neuronal networks, biophysics, and nonlinear dynamics, and linking my previous understanding of neurons to cognitive processes, something that I had not fully understood before. Because the review paper goes over the general information, I read many of the papers it cited to find the basis behind some of its claims. However, I still do not clearly understand the mechanism behind the changes in the attractor dynamics. I would like to learn more about how the parameters are changed, and how these changes, in turn, alter the attractor landscapes.
At this point, I believe that the connection between these dynamics and mental illnesses as presented in this paper seems rather speculative. However, I think that as more data is collected and analyzed, and further models are developed to understand the dynamics of neuronal networks, we can glean more insight towards understanding and developing treatments for mental illnesses.
References:
Durstewitz, D., Huys, Quentin J. M., Koppe, Georgia. (2018). Psychiatric Illnesses as Disorders of Network Dynamics. doi: https://arxiv.org/pdf/1809.06303.pdf
Durstewitz, D. (2009). Implications of synaptic biophysics for recurrent network dynamics and active memory. Neural Networks, 22(8), 1189-1200.
Durstewitz, D., Seamans, J. K. (2008). The dual-state theory of prefrontal cortex dopamine function with relevance to catechol-o-methyltransferase genotypes and schizophrenia. Biological Psychiatry, 64(9), 739-749.
Durstewitz, D. (2006). A few important points about dopamine’s role in neural network dynamics. Pharmacopsychiatry, 39(S 1), 72-75.
Izhikevich, E. M. (2007). Dynamical Systems in Neuroscience: MIT Press.
Johnston, Daniel, Wu, Samuel Miao-Sin . (2001) Foundations of Cellular Neurophysiology. MIT Press.
Rolls, E. T., Loh, M., Deco, G. (2008). An attractor hypothesis of obsessive-compulsive disorder. European Journal of Neuroscience, 28(4), 782-793. doi: 10.1111/j.1460-9568.2008.06379.x
Strogatz, S. H. (2018). Nonlinear dynamics and chaos: with applications to physics, biology, chemistry, and engineering: CRC Press.
[latexpage]My most important experience in undergrad was working in a group in theoretical physics studying neurons, both on the level of individual neurons and beginning to build simple models for neuronal networks. My group studied a range of nonlinear dynamical systems, and my research focused on dynamics at the molecular level.
When I first began working in the group, my primary prior experience had been undergraduate coursework in chemistry. I had taken only lower-level undergrad courses in math, physics, and to a lesser extent, biology, and my only programming experience was one week of an online course in Python. It definitely didn’t feel like enough at first, and it was an extremely steep learning curve. After my two years of working there, I picked up a lot of skills in programming, learned some basic neuroscience and physics concepts, was able to put the material from my coursework in mathematics, numerical analysis, and programming into practice, and most importantly, learned how to teach myself new material on the fly.
The data I had access to for my research was current and voltage data from current clamp experiments. This means that during the experiment, a current was injected into a cell, and the resulting potential was measured at discrete time intervals of 0.02 milliseconds. Although we only had data from one of the variables, since the dynamical equation of voltage depends on the dynamics of the gating variables and a set of parameters such as the maximal conductances of the ion channels, we can extract this information from the voltage time series. We do this by minimizing a cost function, which has terms for both measurement error and model error. We fix the measurement error and begin with an initial model error, obtaining an initial guess for the minimum, and then me slowly enforce the model constraints until we arrive at a global minimum. We use this state to estimate the most likely values of parameters and time series for the variables.
The first project I worked on was estimating parameter values for induced human neurons. Our experimental collaborators in neuroscience were able to create these cells by converting human skin stem cells to cells with neuronal properties. They were able to obtain current and voltage data through current-clamp experiments. The goal of the project was to estimate parameters for both healthy cells and cells from Alzheimer’s patients. In comparing the results, if we are able to find separation in the parameter space, we might even use this to classify unknown cells based on their current and voltage activity. Moreover, we can learn more about the dynamics and modify our model for induced human neurons as needed.
To test the validity of our estimates, we use the parameter estimates at the end of our time window and use the model to integrate forward the voltage equation, obtaining a time series prediction for voltage. If these predictions match the data closely, we can place more confidence in our estimates.
Using the simple NaKL model, where we were only considering sodium and potassium currents, we got the following results for predictions:
As we can see, although the model predicts the spiking regions well, the subthreshold regions are less accurate. As a result, I tried adding a hyperpolarization-activated inward current to the voltage equation, which added two more variables to the system. The results of the predictions using the estimated parameters were as follows:
Another project I started working on was modeling the network of neurons in HVC, the premotor nucleus of a songbird called a zebra finch. Songbirds are good models for human language learning because male songbirds spend their youth listening to a tutor, producing syllables and listening to themselves, and eventually establishing a pattern of song syllables unique to themselves.
Within HVC, there are three types of neurons. The $HVC_{Ra}$ neurons lead to the premotor pathway for the song, the $HVC_{X}$ neurons are essential for learning and memory, and the $HVC_I$ neurons have inhibitory connections with the other two types of neurons.
We built a simple model of the connections with the following assumptions, determined from the results of in vivo experiments:
1. $HVC_I$ neurons have only inhibitory connections with the others
2. $HVC_{RA}$ and $HVC_X$ neurons have only excitatory connections with $HVC_I$ neurons
3. $HVC_{RA}$ have a sequence of excitatory connections with each other that store the bird’s own song
4. There are no direct connections between $HVC_{RA}$ and $HVC_X$ neurons
5. There can be multiple inhibitory connections on a single $HVC_X$ neuron
6. The auditory input, which is converted to a current, directly influences all of these neurons to some extent
Below is an illustration of the simplest form of our model, with only three neurons of each type:
When I was working in the group, we did not yet have experimental data. However, we attempted to create simulated data with pre-determined parameters and use our methods to estimate them. We planned to use the results of these twin experiments to design experiments for our collaborators.
We used song recordings from the lab and extracted pressure wave data from the mp3 files, and then used a transfer function to convert this to a current. Then, we used this current and parameters values we determined, integrating the model’s dynamical equations and obtaining time series data for voltage and the gating variables. In this model, there are nine neurons, and each of these has its own voltage equation and corresponding gating variable equations.
I was only able to complete the twin experiments for this simple model before coming to grad school, but during my time in the group, I developed a script in C that would automatically write the model equations and organize the relevant information into the files we need for data assimilation.
My code makes use of the connection matrix, where the column on the left refers to the presynaptic neuron and the column on the right refers to the postsynaptic neuron, and the synaptic connections strengths are either 0, signaling no connection, or 1, signaling a connection. The code asks the user to manually list the connections using coordinates.
The code can easily be modified for more complex models, such as varying the size of the connection matrix, or varying the strengths of the synaptic connections. When I first wrote the files for data assimilation for this model with a network that has three neurons of each type, it took a couple weeks to complete manually, with some trial and error. My hope that this code will make it more efficient to run twin experiments for larger and more complex models.
I am happy with the research experiences I have had in undergrad, and I feel that it has prepared me to approach independent research here in graduate school. However, our models are very simple and not very biologically realistic. Since my program has a greater emphasis on not only physics, but biological training, I will be able to understand the properties and behavior of neurons at a deeper level, and develop models that are not simply mathematically elegant, but capture the essence of the biology as accurately as possible.
References
Armstrong, E., Abarbanel, H. D. (2016). Model of the songbird nucleus HVC as a network of central pattern generators. Journal of neurophysiology, 116(5), 2405-2419.
Daou, A., Ross, M., Johnson, F., Hyson, R., Bertram, R. (2013). Electrophysiological characterization and computational models of HVC neurons in the zebra finch. Journal of neurophysiology, 110, 1227-1245.
Long, M. A., Jin, D. Z., Fee, M. S. (2010). Support for a synaptic chain model of neuronal sequence generation. Nature, 468(7322), 394.
Mooney, R., Prather, J. F. (2005). The HVC microcircuit: the synaptic basis for interactions between song motor and vocal plasticity pathways. Journal of Neuroscience, 25(8), 1952-1964.
Williams, H. (2004). Birdsong and singing behavior. Annals of the New York Academy of Sciences, 1016(1), 1-30. | 8,118 | 32,559 | {"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": 2, "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-2023-23 | latest | en | 0.963734 |
https://academicassignmentexperts.com/economics-homework-help-813-2/ | 1,722,827,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00461.warc.gz | 62,219,549 | 17,409 | # Economics homework help
Week 2 Assignment: Need to be done MS Word
Question 1:
Given: A statistics student is currently performing in accordance with the data below.
Midterm 10% 100
Discussion 30% 0
Assignment 20% 88
Project 20% 90
Quizzes 20% 70
a. What is the mean for the grade?
b. What is the weighed mean for the grade?
c. Why are the answers to a and b different?
d. What would you suggest to this student to raise his/her grade? Why?
Question 2
P(A) = 0.55, P(B) = 0.35, and P(A ∩ B) = 0.15
What does P(A) mean?
What does P(A ∩ B) mean?
Do A and B overlap? In other words, is it possible for A and B to happen at the same time?
What is the probability of Event A happening, given that event B already happened? (Show equation and work)
Question 3 – Each question has only one answer. There should be a lot of work to see – include problem set up and work starting at #5.
Three vitamin and four sugar tablets identical in appearance are in a box. Nothing else is in the box. One tablet is taken at random and given to Person A. Another tablet is then selected and given to Person B.
What is the probability that
Person A was given a sugar tablet or a vitamin tablet?
Person A was given a rock?
Person A was given a vitamin tablet?
Person A was given a sugar tablet GIVEN THAT Person B ALREADY was given a vitamin tablet. (capital letters are hints)
Person A was given a vitamin tablet and Person B was given a sugar tablet?
Neither was given a vitamin tablet?
Both were given vitamin tablets?
Exactly one person was given a vitamin tablet?
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If you are convinced that our writer has not followed your requirements, feel free to ask for a refund. | 479 | 1,977 | {"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-2024-33 | latest | en | 0.951633 |
https://socratic.org/questions/how-do-you-solve-e-x-4 | 1,529,299,961,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00169.warc.gz | 718,485,242 | 85,574 | # How do you solve e^x=4 ?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
3
Burglar Share
Jun 25, 2016
The solution is $x = 1.38$
#### Explanation:
The operation inverse of the exponential is the logarithm.
Then we apply the natural logarithm ($\ln$) on both sides of the equation:
${e}^{x} = 4$
$\ln \left({e}^{x}\right) = \ln \left(4\right)$
$x = \ln \left(4\right) \setminus \approx 1.38$.
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• 34 minutes ago | 245 | 770 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-26 | longest | en | 0.657045 |
https://www.numwords.com/words-to-number/en/3030 | 1,606,518,650,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194634.29/warc/CC-MAIN-20201127221446-20201128011446-00374.warc.gz | 797,820,142 | 2,559 | NumWords.com
How to write Three thousand thirty in numbers in English?
We can write Three thousand thirty equal to 3030 in numbers in English
< Three thousand twenty-nine :||: Three thousand thirty-one >
Six thousand sixty = 6060 = 3030 × 2
Nine thousand ninety = 9090 = 3030 × 3
Twelve thousand one hundred twenty = 12120 = 3030 × 4
Fifteen thousand one hundred fifty = 15150 = 3030 × 5
Eighteen thousand one hundred eighty = 18180 = 3030 × 6
Twenty-one thousand two hundred ten = 21210 = 3030 × 7
Twenty-four thousand two hundred forty = 24240 = 3030 × 8
Twenty-seven thousand two hundred seventy = 27270 = 3030 × 9
Thirty thousand three hundred = 30300 = 3030 × 10
Thirty-three thousand three hundred thirty = 33330 = 3030 × 11
Thirty-six thousand three hundred sixty = 36360 = 3030 × 12
Thirty-nine thousand three hundred ninety = 39390 = 3030 × 13
Forty-two thousand four hundred twenty = 42420 = 3030 × 14
Forty-five thousand four hundred fifty = 45450 = 3030 × 15
Forty-eight thousand four hundred eighty = 48480 = 3030 × 16
Fifty-one thousand five hundred ten = 51510 = 3030 × 17
Fifty-four thousand five hundred forty = 54540 = 3030 × 18
Fifty-seven thousand five hundred seventy = 57570 = 3030 × 19
Sixty thousand six hundred = 60600 = 3030 × 20
Sitemap | 387 | 1,268 | {"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} | 3.1875 | 3 | CC-MAIN-2020-50 | latest | en | 0.761368 |
https://www.physicsforums.com/threads/transform-formula-into-another-form.902598/ | 1,508,440,195,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823360.1/warc/CC-MAIN-20171019175016-20171019195016-00419.warc.gz | 975,575,571 | 17,996 | # Transform formula into another form
1. Feb 4, 2017
### fonseh
1. The problem statement, all variables and given/known data
I need help when changing the formula of the yellow circled part to red-circled part ...
2. Relevant equations
3. The attempt at a solution
I tried , but i didnt get what the author got .... Here's my working ... Which part of my working is wrong ?
File size:
48.4 KB
Views:
32
File size:
28.2 KB
Views:
29
2. Feb 4, 2017
### Staff: Mentor
Moved to Calculus and Beyond section.
@fonseh, many helpers won't respond if the problem statement and work are shown only in attached images.
3. Feb 4, 2017
### haruspex
I cannot make sense of the text either. It looks quite wrong. Certainly a du has been omitted from the right hand side, and I cannot see how that 1-u term can appear. Perhaps the N exponent is in the wrong place. Maybe some approximation has been made, but that should be stated.
Your own working has y on the right instead of dy (=y0du).
4. Feb 4, 2017
### haruspex
The closest I can get is $\frac{y_0du}{S_0}[\left(1-\frac {1}{1-u^N}\right)$ $(1-(\frac{y_c}{y_0})^Mu^{-M})^{-1}]$.
If we use the binomial approximation for the last (...)-1 term, we get:
$\frac{y_0du}{S_0}[\left(1-\frac {1}{1-u^N}\right)$ $(1+(\frac{y_c}{y_0})^Mu^{-M})]$.
Multiplying the brackets out produces something quite close to the text. The remaining differences look like typos.
5. Feb 6, 2017
### fonseh
Can you show your working pls ? I tried it many times , but didnt get the red part , i gt (1-(1/u)^N ) instead
6. Feb 6, 2017
### fonseh
This is actually hydraulic formula , should i move it to another thread ? Or this is just purely mathematics derivation ? I'm not sure
7. Feb 7, 2017
### Staff: Mentor
It's fine here. The gist of the question is mathematical in nature. | 535 | 1,814 | {"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} | 3.234375 | 3 | CC-MAIN-2017-43 | longest | en | 0.87849 |
http://practicalmethod.com/2010/10/hand-foot-connection-online-video-trailer/?replytocom=4132 | 1,591,502,037,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00157.warc.gz | 98,275,253 | 13,818 | # Hand-Foot Connection Online Video Trailer
by on 2010/10/28
Author: Chen Zhonghua Length: 60 minutes Language: English Chapters: 1 Year: 2010 Location: Edmonton
In 60 minutes, Master Chen Zhonghua clearly demonstrated the hand-foot relationships in every move of the entire Chen Style Taijiquan Practical Method Yilu routine. Demonstrating the principles of stick energy to push, and rope energy to pull.
Hand-Foot Connection Online Video
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Mark Hanley May 27, 2020 at 4:47 am
Thanks Kelvin
Mark Hanley May 26, 2020 at 3:33 pm
This video was excellent
Starting from 7:39 Master Chen is talking about hand-out in the positive circle, He indicates that the shoulder and dantien squeeze into the line to help moving hand -out. These are both large areas. I cannot move squeeze the back shoulder without moving my body, I can squeeze starting from the back pectoral I can achieve the squeeze without moving is this ok? Master Chen also said the the dantien area is the other part of the squeeze would this be more from the front half of the body?
Master Chen suggests both variations of these moves in positive are throughout the form. Am I correct to say the squeeze is in every move that has a line between the back foot and opposite front hand?
Kelvin Ho May 26, 2020 at 11:21 pm
What Master Chen described in the video was the guiding concept. You may have a top triangle and a bottom triangle sharing two points and one side. The shared side is the line connecting the two points. The two points in this example will be the front hand and the rear foot. You can squeeze the top triangle down while you squeeze the bottom triangle up. Since the rear foot can’t go anywhere, the energy is squeezed to the front hand. What you can do today is what you can do at this point. As long as it follows the guiding concept, it is ok and you will get better at it over time.
> Am I correct to say the squeeze is in every move that has a line between the back foot and opposite front hand?
Yes, and it’s more than that. It’s the same for any three points you want to make for the triangle. Please see:
http://practicalmethod.com/2014/07/triangle_double_lock_single_lock/
Vicent den hengst May 18, 2020 at 7:24 pm
Master Chen relates the 0 – 1 computer principle to the two main actions with regard to the hand foot connection around 10minutes from start. A highly recommendable purchase!!!
vincent den hengst May 15, 2020 at 9:53 am
Absolutey fundamental explanations of the hand – foot connection for yilu. Good base for understanding of having power through my overal structure and the unbroken connections between my hand and feet (which are making way more sense to me now).
This video forms a complete and clear instructional set of how to connect hands to feet on in every yilu 81 move and to make sure the connection can never be broken. It makes a great combination with the four video Energy Alignment series.
Samvanity May 26, 2016 at 11:44 pm
Thanks Master Chen. This is a very important video, especially important for beginners. Getting the Waist-Kua movements correct is vital. The movements are nearly invisible to the naked eyes. If you don’t get the Waist-Kua movements right you’ll be tossing all the way from beginning to finish in your Yilu practice. And it’s nearly impossible to get right without watching this video.
I’ve purchased many videos and this one is ranked top in terms of importance.
charlie wishon July 15, 2014 at 6:57 am
Awesome lesson. Further discourse on understanding tai ji principles. A great compliment to the energy alignment series, and yilu detailed instruction . A must have ! Thankyou Master Chen Zhonghua
chenstylejohn October 23, 2013 at 4:51 pm
Wonderful, thank you!
As usual, Master Chen’s use of tools and down-to-Earth style of explanation makes these difficult subjects much easier to grasp.
Highly recommended!
lrkays June 12, 2013 at 4:27 pm
Wow, what a gem. Almost passed this by thinking it would be to basic, this video is a classic and belongs in everyone’s library. Thank you Master Chen for sharing your knowledge.
gigi June 5, 2012 at 10:45 pm
That stretch is felt in the fingers doing yilu?
And is it possilile to strech only the forearm and feel the stretch
in the fingers thinking (wrong ) that everything is streched?
bruce.schaub March 23, 2012 at 7:48 am
Fantastic video!!! Thank you Master Chen. This clearly shows how to utilize the lines established between hand and foot to generate Taiji power….and how important our attention to them really are. The step by step guidance through the Yilu of where all those connections should be generated seems so absolutely essential. Not to mention all the other important pieces of information on switching using kua, anchor points and anchor lines, the track, etc. Really incredible how much great taiji instruction is packed into this 1 hour lesson!
bh7pea March 22, 2012 at 7:17 pm
Do you have Chinese subtitles for this video? I want to buy this video but cannot understand the language.
Chen Zhonghua March 22, 2012 at 8:26 pm
Sorry we don’t have Chinese subtitiles for this videos. We are planning to make videos in Chinese language in the summer of 2012.
Michael Winkler December 29, 2010 at 5:19 pm
Again, this is a very valuable video! When already learned the coreograohy of the yilu, with this video one can go through the whole form again, following Master Chen’s clear instructions according to the connection of hand and foot.
Following these instructions I have the feeling to gain lots of more content into my yilu. I’m really looking forward to see Master Chen the next time for hands on instructions, and meanwhile I just feel perfect support through these videos (I did also appreciate “Yilu-Detailled Instructions”, “Energy Alignment”, the “Tossing Online Video” and much more of all the online videos).
One can get a very good and close idea what and how to train with this video.
Thanks to Master Chen “Josef” Zhonghua very much for all this support!
chenstylejohn November 8, 2010 at 6:19 am
Really very helpful to see his use of the kua & how he switches power to one side or the other! Clearly demonstrated, useful analogies, helpful discussion…
Overall a fantastic video (not recommended for the beginner). Thank you Master Chen!
Xavier Santiago November 2, 2010 at 2:52 pm
Another valuable mini-lesson by Shifu Chen Zhonghua. In my personal training, one of the concepts that has given me the most difficulty in putting to practice is how to connect the hand to the foot. I previously tried it in many ways but found that I was actually causing the opposite effect and would end up using upper body power in my practice. This video’s simple format has allowed me to see clearly where I was wrong and how to fix my circles in order to have true lower body power with the hand/foot connection. I highly recommend this video lesson.
greadore October 30, 2010 at 9:56 pm
Very good video. I have learned Yang style and another Chen style in the past and will try and learn the Practical Method Yilu from the resource material you have available. This really helped explain the concept of double weighting which I was never clear about. The concept of pivot points was enlightening as well. The Practical Method really makes sense to me and I am so glad I have found your site and thank you for being so open to sharing your information.
Chen Zhonghua October 31, 2010 at 1:06 am
Hong’s own writing on Double Heavy will give you a clear idea.
http://practicalmethod.com/1986/10/problems-in-learning-taiji-by-hong-junsheng/
James Chan October 30, 2010 at 6:50 pm
Great demonstration of gears, rotation, leverage, displacement, separation and connection. I see them in Master Chen’s movement.
Allan Belsheim October 29, 2010 at 9:39 am
This video is such an excellent source for explaining what other instructor’s can’t explain. Master Chen has again revealed his great depth of knowledge and shares it with us. A must have video! | 1,894 | 8,100 | {"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} | 2.734375 | 3 | CC-MAIN-2020-24 | longest | en | 0.941513 |
https://www.physicsforums.com/threads/question-relating-newtons-first-law-of-motion.644551/ | 1,519,333,984,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814290.11/warc/CC-MAIN-20180222200259-20180222220259-00477.warc.gz | 857,125,844 | 16,503 | # Question relating Newton's First Law of motion!
1. Oct 16, 2012
### ehabmozart
Hi there! I kept on reading my textbook and i found one simple confusion in Newton's First law of motion. I want to consider a situation where a space shuttle is moving in the outerspace (assume there is no gravity) under the influence of the engines. What will happen when the engine stops functioning? It is given that the space shuttle will move with a constant velocity. Now, my opinion would be that there was an initial forward force given by the engine, no it is taken off. Is is like pulling the space shuttle. So it ultimately must end up by a stop! This is confusing me. I would really be thankful to whoever helps me in clarifying my confusion. Thanks in advance. I need elaborate answers if u please :)
2. Oct 16, 2012
### mikelepore
The intuition that once you stop the force the object will stop is an illusion produced
by the fact that we grow up in an environment where there is friction everywhere. When we stop pushing an object, it may only move a short distance before friction slows it down to a stop. But friction is also a force. If no force at all, including friction, acts on an object, it will move with a constant velocity forever. The ancient people, even the most educated, believed the idea that seems like common sense but is incorrect -- until Galileo figured out the real answer in the 1600s.
3. Oct 16, 2012
### stoner420
space is a fun place to do physics because we are allowed to make up situations that we normally couldn't do on earth.
So lets say you are going in your shuttle off earth into space and at t=20 your engine turns off, and you get past earth's gravity, you are in free floating space. you could then just keep on going that same speed for ever and ever until you bumped into something, or near something real big. (I think we have satelights pointed at us sending information that use no fuel to keep on going. Because there is no need)
Thinking it would slow down is common sense nowadaysz because if your rolling down the street on a scooter eventually your going to slow down. There are so many different frictions to think about when on a scooter that we take for granted. The fact that there is air that is trying to get around us slows us down, plus friction between the wheel and the ground, and then the wheel bearing.
4. Oct 16, 2012
### davenn
yup it sure is we can only sorta create the conditions of Newtons first law of every action has an equal and opposite reaction
At NASA they use smooth floors and have air cusioned units that will glide over the surface with a minimal amount of friction its a very good approximation :)
very difficult to get away from the effects of Earth's gravity. You much be a VERY long way away. Remember that even in orbit satellites, shuttle and its astronauts are in a constant state of freefall due to gravity
Satellites do have small rocket thrusters ( usually just gas exhausts) these are there to periodically boost the satellite's orbit. Most satellites (in lower earth orbit regions) are not that far above the Earth's atmosphere and the small amount of resistance from that and loss of orbital velocity means they do slowly loose orbital height and need the occassional boost to get them back in position.
The life of the satellite comes to an end when there is no more propellant on the satellite to do the boosts in orbit and then they are on that inevidable spiral back into the atmosphere and burn up
Dave
5. Oct 17, 2012
### HomogenousCow
When you stop accelerating an object, it will continue to move at the velocity it had before.
For example, say I had a rock and attached a string to it, then went and spun the rock with the string (Think David and Goliath),after a while the string snapped and ceased to pull on the rock, the rock will fly out tangent to the circular trajectory it used to be on. | 863 | 3,915 | {"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} | 3.28125 | 3 | CC-MAIN-2018-09 | longest | en | 0.959964 |
https://www.teacherspayteachers.com/Product/Multiplication-3-Digits-x-3-Digit-Set-36-Math-Ladders-2769743 | 1,537,479,454,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156613.38/warc/CC-MAIN-20180920195131-20180920215531-00076.warc.gz | 877,642,847 | 19,074 | # Multiplication - 3 Digits x 3 Digit - Set 3.6 {Math Ladders}
Subject
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Product Rating
4.0
4 Ratings
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PDF (Acrobat) Document File
2 MB|10 Sets
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Product Description
Get your students practicing math skills with a fun, self checking, center activity. Math Ladders can be used a partner work, but are designed for independent practice on specific skills. Each packet includes a single skill on a variety of different levels to make differentiation easier for you, the teacher. Just keep the table of contents handy and assign students to work (or move up or down in the skill level) on the one they need. Problems are written horizontally so students must line up the digits correctly, adding another layer of accountability for the student.
To play, students solve the problem on the START card. They find the answer on the left side of one of the other cards and place it above their original problem. Then, they solve the problem on their new card and continue until they reach the top FINISH card.
To Set up as a Center:
-Print game boards for each person at the center
-Print enough sets for each person at the center, cut apart & keep in a snack-sized baggie
-Print answer sheets for each student
-Print one answer key to keep at the center (if you want them to check their answers)
Alternatively, you can print a class set of the game boards & the cards and have students cut & paste their assignment.
Download a FREEBIE Addition Math Ladder to get a full center and an idea of how the activity works. The skills practiced in this set are listed below and part of the preview file.
Skills in this Set:
Set 3.6 A – Factors are Multiples of 100
Set 3.6 B – Factors are Multiples of 100
Set 3.6 C – Second Factor is a Multiple of 100
Set 3.6 D - All Digits Are Low, No Zeros
Set 3.6 E – No Zeros
Set 3.6 F – No Zeros
Set 3.6 G – Each Problem Has a Zero in One Factor
Set 3.6 H - Mixed
Set 3.6 I - Mixed
Set 3.6 J - Mixed
CCSS.MATH.CONTENT.5.NBT.B.5
Fluently multiply multi-digit whole numbers using the standard algorithm.
Don't forget to grab extra Math Ladder boards (for free!) from my store. I'll be adding more of these (always free!) as the holidays come around, so don't forget to check back.
--------------------------------------------------------------------------------------
Thanks for visiting and checking out my products!
Total Pages
10 Sets
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Teaching Duration
30 minutes
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\$3.00 | 592 | 2,462 | {"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-2018-39 | latest | en | 0.8868 |
http://www.studentoffortune.com/question/1989049/ACC-206-Chapter-8 | 1,369,105,576,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699675907/warc/CC-MAIN-20130516102115-00060-ip-10-60-113-184.ec2.internal.warc.gz | 742,442,091 | 7,683 | This question's due date has already passed. You may post a tutorial, but there's no guarantee that the original asker will purchase the tutorial. But other people might!
Question
\$20.00ACC 206 Chapter 8
• Closed, but you can still post tutorials
• Due on Dec. 18, 2012
• Asked on Dec 17, 2012 at 11:28:54PM
Rating :No Rating
Tutorials Posted: 0
Q:
Chapter Eight Problems
Please complete the following 5 exercises below in either Excel or a word document (but must be single document). You must show your work where appropriate (leaving the calculations within Excel cells is acceptable). Save the document, and submit it in the appropriate week using the Assignment Submission button.
Chapter 8 Exercise 1:
1. Basic present value calculations
Calculate the present value of the following cash flows, rounding to the nearest dollar:
1. A single cash inflow of \$12,000 in five years, discounted at a 12% rate of return.
2. An annual receipt of \$16,000 over the next 12 years, discounted at a 14% rate of return.
3. A single receipt of \$15,000 at the end of Year 1 followed by a single receipt of \$10,000 at the end of Year 3. The company has a 10% rate of return.
4. An annual receipt of \$8,000 for three years followed by a single receipt of \$10,000 at the end of Year 4. The company has a 16% rate of return.
Chapter 8 Exercise 4:
4. Cash flow calculations and net present value
On January 2, 19X1, Bruce Greene invested \$10,000 in the stock market and purchased 500 shares of Heartland Development, Inc. Heartland paid cash dividends of \$2.60 per share in 19X1 and 19X2; the dividend was raised to \$3.10 per share in 19X3. On December 31, 19X3, Greene sold his holdings and generated proceeds of \$13,000. Greene uses the net-present- value method and desires a 16% return on investments.
1. Prepare a chronological list of the investment's cash flows. Note: Greene is entitled to the 19X3 dividend.
2. Compute the investment's net present value, rounding calculations to the nearest dollar.
3. Given the results of part (b), should Greene have acquired the Heartland stock? Briefly explain.
Chapter 8 exercise 5:
5. Straightforward net present value and internal rate of return
The City of Bedford is studying a 600-acre site on Route 356 for a new landfill. The startup cost has been calculated as follows:
Purchase cost: \$450 per acre
Site preparation: \$175,000
The site can be used for 20 years before it reaches capacity. Bedford, which shares a facility in Bath Township with other municipalities, estimates that the new location will save \$40,000 in annual operating costs.
1. Should the landfill be acquired if Bedford desires an 8% return on its investment? Use the net-present-value method to determine your answer.
2. Compute the internal rate of return on this project.
Chapter 8 Problem 1:
1. Straightforward net-present-value and payback computations
STL Entertainment is considering the acquisition of a sight-seeing boat for summer tours along the Mississippi River. The following information is available:
Cost of boat \$500,000 Service life 10 summer seasons Disposal value at the end of 10 seasons \$100,000 Capacity per trip 300 passengers Fixed operating costs per season (including straight-line depreciation) \$160,000 Variable operating costs per trip \$1,000 Ticket price \$5 per passenger
All operating costs, except depreciation, require cash outlays. On the basis of similar operations in other parts of the country, management anticipates that each trip will be sold out and that 120,000 passengers will be carried each season. Ignore income taxes.
Instructions:
By using the net-present-value method, determine whether STL Entertainment should acquire the boat. Assume a 14% desired return on all investments,- round calculations to the nearest dollar.
Chapter 8 Problem 4:
4. Equipment replacement decision
Columbia Enterprises is studying the replacement of some equipment that originally cost \$74,000. The equipment is expected to provide six more years of service if \$8,700 of major repairs are performed in two years. Annual cash operating costs total \$27,200. Columbia can sell the equipment now for \$36,000; the estimated residual value in six years is \$5,000.
New equipment is available that will reduce annual cash operating costs to \$21,000. The equipment costs \$103,000, has a service life of six years, and has an estimated residual value of \$13,000. Company sales will total \$430,000 per year with either the existing or the new equipment. Columbia has a minimum desired return of 12% and depreciates all equipment by the straight-line method.
Instructions:
1. By using the net-present-value method, determine whether Columbia should keep its present equipment or acquire the new equipment. Round all calculations to the nearest dollar, and ignore income taxes.
2. Columbia's management feels that the time value of money should be considered in all long-term decisions. Briefly discuss the rationale that underlies management's belief. | 1,129 | 5,039 | {"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} | 3.609375 | 4 | CC-MAIN-2013-20 | latest | en | 0.918224 |
http://www.stata.com/statalist/archive/2006-08/msg00102.html | 1,440,898,505,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064865.49/warc/CC-MAIN-20150827025424-00216-ip-10-171-96-226.ec2.internal.warc.gz | 707,715,209 | 3,928 | # st: boxcox and test
From David Airey To statalist@hsphsun2.harvard.edu Subject st: boxcox and test Date Fri, 4 Aug 2006 15:09:09 -0500
If I do a simple linear model with two factors and an interaction, I can test the terms for the interaction using "test".
When I do the same thing after a boxcox model, I cannot do the same test using "test". Anyone understand why?
. xi: regress la i.gf*i.gd
i.gf _Igf_0-2 (naturally coded; _Igf_0 omitted)
i.gd _Igd_0-2 (naturally coded; _Igd_0 omitted)
i.gf*i.gd _IgfXgd_#_# (coded as above)
Source | SS df MS Number of obs = 599
-------------+------------------------------ F( 8, 590) = 5.10
Model | 56.1889027 8 7.02361283 Prob > F = 0.0000
Residual | 811.983485 590 1.37624319 R-squared = 0.0647
Total | 868.172387 598 1.45179329 Root MSE = 1.1731
------------------------------------------------------------------------ ------
la | Coef. Std. Err. t P>|t| [95% Conf. Interval]
------------- +----------------------------------------------------------------
_Igf_1 | -.2585714 .2684938 -0.96 0.336 -. 7858914 .2687485
_Igf_2 | .1110119 .2789682 0.40 0.691 -. 4368797 .6589035
_Igd_1 | .4551786 .2575938 1.77 0.078 -. 0507339 .9610911
_Igd_2 | 1.160559 .3301336 3.52 0.000 . 5121789 1.808939
_IgfXgd_1_1 | .1260119 .3122241 0.40 0.687 -.4871941 . 7392178
_IgfXgd_1_2 | -.7744614 .3856281 -2.01 0.045 -1.531832 -. 0170906
_IgfXgd_2_1 | -.5876191 .3386546 -1.74 0.083 -1.252734 . 0774962
_IgfXgd_2_2 | -1.562642 .4148084 -3.77 0.000 -2.377323 -. 7479615
_cons | 1.178571 .2217015 5.32 0.000 . 7431513 1.613992
------------------------------------------------------------------------ ------
. test _IgfXgd_1_1 _IgfXgd_1_2 _IgfXgd_2_1 _IgfXgd_2_2
( 1) _IgfXgd_1_1 = 0
( 2) _IgfXgd_1_2 = 0
( 3) _IgfXgd_2_1 = 0
( 4) _IgfXgd_2_2 = 0
F( 4, 590) = 4.77
Prob > F = 0.0009
. xi: boxcox la i.gf*i.gd, model(lhsonly) lrtest nolog
i.gf _Igf_0-2 (naturally coded; _Igf_0 omitted)
i.gd _Igd_0-2 (naturally coded; _Igd_0 omitted)
i.gf*i.gd _IgfXgd_#_# (coded as above)
Fitting comparison model
Fitting full model
Fitting comparison models for LR tests
Iteration 0: log likelihood = -941.52802
Iteration 1: log likelihood = -729.28071
Iteration 2: log likelihood = -712.61307
Iteration 3: log likelihood = -712.61277
Iteration 4: log likelihood = -712.61277
Iteration 0: log likelihood = -941.13797
Iteration 1: log likelihood = -728.93451
Iteration 2: log likelihood = -712.11616
Iteration 3: log likelihood = -712.10848
Iteration 4: log likelihood = -712.10848
Iteration 0: log likelihood = -942.63844
Iteration 1: log likelihood = -729.92924
Iteration 2: log likelihood = -712.52812
Iteration 3: log likelihood = -712.51975
Iteration 4: log likelihood = -712.51975
Iteration 0: log likelihood = -947.26615
Iteration 1: log likelihood = -734.06093
Iteration 2: log likelihood = -716.36727
Iteration 3: log likelihood = -716.36482
Iteration 4: log likelihood = -716.36482
Iteration 0: log likelihood = -941.14027
Iteration 1: log likelihood = -728.82513
Iteration 2: log likelihood = -712.0049
Iteration 3: log likelihood = -712.00324
Iteration 4: log likelihood = -712.00324
Iteration 0: log likelihood = -943.09805
Iteration 1: log likelihood = -730.50236
Iteration 2: log likelihood = -713.29541
Iteration 3: log likelihood = -713.29536
Iteration 0: log likelihood = -942.58206
Iteration 1: log likelihood = -730.14825
Iteration 2: log likelihood = -713.22914
Iteration 3: log likelihood = -713.21722
Iteration 4: log likelihood = -713.21722
Iteration 0: log likelihood = -948.17624
Iteration 1: log likelihood = -736.09204
Iteration 2: log likelihood = -719.30185
Iteration 3: log likelihood = -719.30096
Iteration 4: log likelihood = -719.30096
Number of obs = 599
LR chi2(8) = 32.15
Log likelihood = -711.88092 Prob > chi2 = 0.000
------------------------------------------------------------------------ ------
la | Coef. Std. Err. z P>|z| [95% Conf. Interval]
------------- +----------------------------------------------------------------
/theta | .0766468 .0412206 1.86 0.063 -. 004144 .1574377
------------------------------------------------------------------------ ------
Estimates of scale-variant parameters
-------------------------------------------------------------
| Coef. chi2(df) P>chi2(df) df of chi2
-------------+-----------------------------------------------
Notrans |
_Igf_1 | -.2285903 1.464 0.226 1
_Igf_2 | .1323904 0.455 0.500 1
_Igd_1 | .2050526 1.278 0.258 1
_Igd_2 | .6985838 8.968 0.003 1
_IgfXgd_1_1 | .1086209 0.245 0.621 1
_IgfXgd_1_2 | -.456846 2.829 0.093 1
_IgfXgd_2_1 | -.389851 2.673 0.102 1
_IgfXgd_2_2 | -1.130939 14.840 0.000 1
_cons | -.0550481
-------------+-----------------------------------------------
/sigma | .8250483
-------------------------------------------------------------
---------------------------------------------------------
Test Restricted LR statistic P-Value
H0: log likelihood chi2 Prob > chi2
---------------------------------------------------------
theta = -1 -1057.5776 691.39 0.000
theta = 0 -713.61738 3.47 0.062
theta = 1 -941.05759 458.35 0.000
---------------------------------------------------------
. test _IgfXgd_1_1 _IgfXgd_1_2 _IgfXgd_2_1 _IgfXgd_2_2
( 1) [Notrans]_IgfXgd_1_1 = 0
( 2) [Notrans]_IgfXgd_1_2 = 0
( 3) [Notrans]_IgfXgd_2_1 = 0
( 4) [Notrans]_IgfXgd_2_2 = 0
Constraint 1 dropped
Constraint 2 dropped
Constraint 3 dropped
Constraint 4 dropped
F( 0, 3) = .
Prob > F = .
*
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* http://www.ats.ucla.edu/stat/stata/ | 1,975 | 5,584 | {"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} | 2.734375 | 3 | CC-MAIN-2015-35 | longest | en | 0.3113 |
https://woodworking.stackexchange.com/questions/908/what-general-considerations-do-i-need-to-take-into-account-for-wood-movement/910?s=2%7C1.2481 | 1,719,143,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00845.warc.gz | 543,831,272 | 43,505 | # What general considerations do I need to take into account for wood movement?
I know that wood movement is an issue, but it's a bit vague for me where I should be concerned with it.
This page Suggests that there is almost no movement in the longitudinal direction, but that there is significant movement in radial and tangential directions respective to the grain.
Are there particular orientations which I should avoid, e.g. radial in one piece in a plane with tangential of another?
With what orientations is it safe to use screws without hardware that would allow movement?
• Suppose it is worth mentioning that it depends on the grain direction of your wood. Longitudinal and tangential movements differ depending on that. Then there is temperature, humidity, assembly location and permanent location....my god so much to consider. Look forward to an answer
– Matt
Commented Apr 13, 2015 at 0:56
• Longitudinal should be "with" the grain, radial is inward from the curve, and tangential is tangent to the curve ... I'm not sure what you mean by "it depends on the grain direction of the wood?" Do I need to clarify the question? Commented Apr 13, 2015 at 1:10
• I was more trying to suggest that there could be too many conditions to account for. Not all boards react the same way to their environment. I was basing that comment on what I was reading about grain direction on wood. Your question is fine.
– Matt
Commented Apr 13, 2015 at 2:04
That page you link to is correct, wood movement is almost completely across the grain. Longitudinal or long-grain movement is so slight it can nearly always be ignored.
Just to note, we must remember movement is both expansion and contraction, which typically happens seasonally:
Obviously there are specific instances where movement cannot be accurately predicted. Where the grain shifts direction, as much as 90° within the same board (e.g. wood taken from various crotches within the tree) and particularly where there is wildly changing grain (e.g. in a burr or burl).
Are there particular orientations which I should avoid, e.g. radial in one piece in a plane with tangential of another?
This is the standard cross-grain situation you want to try to avoid where possible. So the answer is basically yes but it's complex because you can't avoid it in many cases.
Where you have no choice but to make a cross-grain join, you must build in some means to allow for movement, for example by fixing loosely as with various tabletop fasteners, fixing at one end of the joint only or allowing something to float within a groove or rebate (US: rabbet), as in the floating panel in frame-and-panel construction.
Unless you take these sort of steps to avoid an issue you can take it that something will happen, not that it might.
With what orientations is it safe to use screws without hardware that would allow movement?
That depends for a start on how you define hardware. The standard 'buttons' used to attach tabletops and allow movement are wood, and are made by the woodworker, but are technically hardware.
Rather than think about the orientations it is safe to use screws visualise the expansion and contraction of the wood and how you allow this to happen after fixing. So essentially screws could be safe in any joint they can be used in, as long as they're utilised appropriately.
A good example is the top to any table, which typically will be cross-grain to some portion of the underframe or apron assembly. One standard way is to fix firmly in the centre and allow the front and back edges to move. Where you want an edge to remain fixed relative to the piece (e.g. a table intended to go flush against the wall) you fix firmly there and allow movement to occur out from that edge:
Screws are used here both for the points where firm fixing is desired as well as where movement is being allowed for, by screwing directly into the tabletop through the apron and through some form of movable fixing/fastener respectively.
• So these all have to do with large surfaces, but there are plenty of instances where wood is joined cross-grain on smaller surfaces. The simplest example would be a lap joint on a small frame, at which point I have movement up and down, as well as outward/inward from the center of the frame, do these not put stresses on the joints, or is it that they are all expanding/contracting in roughly the same shape and so the stress on the joint is minimized? Commented Apr 13, 2015 at 13:40
• @Daniel B, movement is proportional to the size of the piece of wood. Small pieces = small movements, which is why we can get away with things like corner lap joints on things like picture frames (species and cut of the wood, as well as the type of glue used are also factors). But even at that scale joints can be pushed apart, or pull away from a glue line due to shrinkage. I have a lap tray (you can visualise the approximate size) where all 4 corners are now broken in some way. The gaps were large during the winter but are closing up as the moisture level in the air goes up as it gets warmer. Commented Apr 13, 2015 at 14:42
• So what's really important is the size of the joint. If I tried to screw every corner of the table top down, it'd snap screws or split wood because there's a potential for(for a 4 foot width) as much(or more) as an inch of movement, but if I make an open frame of the same size, the only movement is the radial expansion of the boards, which wouldn't be more than 1/64th, a negligible amount. Is that about right? Commented Apr 13, 2015 at 15:29
• @Daniel B, yes that's about right. I'm not sure if you'd ever see a full 25mm or 1" of expansion on a tabletop of that width though! Commented Apr 14, 2015 at 7:15
• @Daniel B, I think the takeaway from this is to err on the side of caution. Particularly since it often doesn't require much effort to make allowance for movement so it's really lazy not to :-) Commented Apr 14, 2015 at 14:22
Lets use a worse case scenario. A table top 30" wide made from flat-sawn wood connected to a cross brace (so the grains of the two pieces run perpendicular). If you look at how table tops are connected, they are usually just attached in the middle, allowing the top to expand to each side. In this case, the table could expand or contract up to 3/8 inch in total width.
In a best case, you have a 4 sided box. Even though the sides will expand and contract, they should all change the same amount at the same rate, so the joinery can be tight.
Wood moves more in areas that are cold and dry in the winter then hot and humid in the summer, than in areas that are relatively consistent year around. Items stored in a climate controlled house won't see nearly as much movement as others.
You can minimize the effects of wood movement by choosing quartersawn and rift-sawn boards.
Graphus gave a great answer covering many of the design considerations and pretty much covered the strategies for dealing with movement.
Shrinkage and expansion also vary by material (wood, MDF, OSB, plywood...), species, climate, and application, as well as other factors. If the humidity level is relatively consistent, you won't need to allow for as much movement as if you're building a piece of outdoor furniture which will be subjected to the elements, for example.
If you aren't sure how much movement to account for in your project, you can use the Shrinkulator to get an estimate based on the wood species you're using. | 1,669 | 7,472 | {"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-2024-26 | latest | en | 0.965943 |
https://codescracker.com/python/program/python-program-convert-kilometres-to-miles.htm | 1,657,093,629,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104668059.88/warc/CC-MAIN-20220706060502-20220706090502-00198.warc.gz | 211,182,538 | 10,274 | Tutorials
Examples
Tests
# Python Program to Convert Kilometer to Miles
In this article, we've created some programs in Python, to convert Kilometer value entered by user at run-time to its equivalent value in Miles. Here are the list of programs:
• Kilometer to Miles without Function
• Using Function
• Using Class
Before creating these programs, let's see the formula used for conversion.
### Kilometer to Miles Formula
The Kilometer to miles formula is:
M = K * (0.621371)
Here M indicates to the value in miles and K indicates to the value in Kilometer.
## Kilometer to Miles without Function
To convert Kilometer to miles in Python, you have to ask from user to enter the value of distance in kilometre, then convert that distance into miles using above formula, as shown in the program given below:
print("Enter Kilometer Value: ")
km = float(input())
miles = km * (0.621371)
print("\nEquivalent Value in Miles = ", miles)
Here is the initial output produced by this Python program:
Type value in Kilometer say 10 and press ENTER key to find and print its equivalent value in miles as shown in the snapshot given below:
#### Modified Version of Previous Program
This program uses end= to skip printing of an automatic newline using print(). The str() method is used to convert any type of value to a string type. Since + operator is used to concatenate same type of values. And {:.2f} with format() is used to print the value of variable provided as argument of format(), upto only two decimal places.
print("Enter Kilometer Value: ", end="")
km = float(input())
miles = km * (0.621371)
print("\n" + str(km) + " Kilometer = " + "{:.2f}".format(miles) + " Miles")
Here is its sample run with same user, that is 10 as of previous program's sample run:
## Kilometer to Miles using Function
This program is created using a user-defined function named KiloToMile(). The function takes a value (Kilometer) as its argument and returns its equivalent value in miles.
def KiloToMile(k):
return k * 0.621371
print("Enter Kilometer Value: ", end="")
km = float(input())
m = KiloToMile(km)
print("\n" + str(km) + " Kilometer = " + "{:.2f}".format(m) + " Miles")
## Kilometer to Miles using Class
This program uses class and object, and object-oriented feature of Python to do the same job as of previous program.
class CodesCracker:
def KiloToMile(self, k):
return k * 0.621371
print("Enter Kilometer Value: ", end="")
km = float(input())
ob = CodesCracker()
m = ob.KiloToMile(km)
print("\n" + str(km) + " Kilometer = " + "{:.2f}".format(m) + " Miles")
In above program, an object named ob is created of class CodesCracker to access its member function named KiloToMile() using dot (.) operator.
Python Online Test
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# Common Errors in Statistics and How to Avoid Them - Good P.I
Good P.I Common Errors in Statistics and How to Avoid Them - Wiley publishing , 2003. - 235 p.
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q = q (F, f) = Ex t~F Q yy 0, h F yt 0)).
In addition, I call the quantity
A A 1 n
qapp = q yy,F) = Ex 0 - F Q yy 0, h F(t 0)) = - X Q yyi, h F (ti))
n i=1
the apparent error of hf. The difference
r(f, F) = q(F, F)- q(F, F)
is the excess error of hf. The expected excess error is
r = EP~FR (F, F),
where the expectation is taken over F, which is obtained from x1,. . . , xn generated by F. In Section 4, I will clarify the distinction between excess error and expected excess error. I will consider estimates of the expected excess error, although what we would rather have are estimates of the excess error.
I will consider three estimates (the bootstrap, the jackknife, and crossvalidation) of the expected excess error. The bootstrap procedure for estimating r = EF~FR(F, F) replaces F with F. Thus
Foot = ef ,~f r (f *, F),
where F* is the empirical distribution function of a random sample x*,..., x* from F. Since F is known, the expectation can in principle be calculated. The calculations are usually too complicated to perform analytically, however, so we resort to Monte Carlo methods.
1. Generate x*, . . . , x*, a random sample from F. Let F* be the empirical distribution of x*, ..., x*.
2. Construct h F*, the realized prediction rule based on x*, ..., x*.
3. Form
APPENDIX B EXCESS ERROR ESTIMATION IN FORWARD LOGISTIC REGRESSION 175
R* = q (*, F)- q (*, F*)
1 n 1 n
= -XQ(j;,ni*{ti))-nXOh*, h^h*)) (2-1)
4. Repeat 1-3 a large number R times to get R*,R*. The bootstrap estimate of expected excess error is
See Efron (1982) for more details.
The jackknife estimate of expected excess error is
rjack = (n - !)((.) - R),
where F() is the empirical distribution function of (xi,. . ., x;-1, xi+1, . . ., xn), and
Efron (1982) showed that the jackknife estimate can be reexpressed as
Let the training sample omit patients one by one. For each omission, apply the prediction rule to the remaining sample and count the number (0 or 1) of errors that the realized prediction rule makes when it predicts the omitted patient. In total, we apply the prediction rule n times and predict the outcome of n patients. The proportion of errors made in these n predictions is the cross-validation estimate of the error rate and is the first term on the right-hand side. [Stone (1974) is a key reference on cross-validation and has a good historical account. Also see Geisser
The cross-validation estimate of expected excess error is
1 n 1 n
rcross = - Y Q ((, hr (-'(ti )) X Q hi , hi (ti )).
(1975).]
176 APPENDIX B EXCESS ERROR ESTIMATION IN FORWARD LOGISTIC REGRESSION
3. CHRONIC HEPATITIS: AN EXAMPLE
We now discuss a real prediction rule. From 1975 to 1980, Peter Gregory (personal communication, 1980) of Stanford Hospital observed n = 155 chronic hepatitis patients, of which 33 died from the disease. On each patient were recorded p = 19 covariates summarizing medical history, physical examinations, X rays, liver function tests, and biopsies. (Missing values were replaced by sample averages before further analysis of the data.) An effective prediction rule, based on these 19 covariates, was desired to identify future patients at high risk. Such patients require more aggressive treatment.
Gregory used a prediction rule based on forward logistic regression. We assume x1 = (t1, yi),. . . , xn = (tn, yn) are independent and identically distributed such that conditional on t, yi is Bernoulli with probability of
success 6(t), where logit 6(t i) = A, + tb, and where A is a column vector of p elements. If (/J0, A) is an estimate of (/0, A), then 6 (t0), such that logit 6 (t0) = A, + t0 A, is an estimate of 6(t0). We predict death if the estimated probability 6(t0) of death were greater than -.:
hF(to) = 1 if 6(to)> 2, i.e., Ao +1oA > 0
= 0 otherwise. (3.1)
Gregory’s rule for estimating (/0, A) consists of three steps.
1. Perform an initial screening of the variables by testing H0: bj = 0 in the simple logistic model, logit 0(t0) = b + t0jbj, for j = 1, ..., p separately at level a = 0.05. Retain only those variables j for which the test is significant. Applied to Gregory’s data, the initial screening retained 13 variables, 17, 12, 14, 11, 13, 19, 6, 5, 18, 10, 1, 4, 2, in increasing order of p-values.
2. To the variables that were retained in the initial screening, apply forward logistic regression that adds variables one at a time in the following way. Assume variables ji, j2,..., jP are already added to the model. For each remaining j, test H0: bj = 0 in the linear logistic model that contains variables j1, j2,..., jp1, j together with the intercept. Rao’s (1973, pp. 417-420) efficient score test requires calculating the maximum likelihood estimate only under H0. If the most significant variable is significant at a = 0.05, we add that variable to the model as variable jP +1 and start again. If none of the remaining variables is significant at a = 0.05, we stop. From the aforementioned 13 variables, forward logistic regression applied to Gregory’s data chose four variables (17, 11, 14, 2) that are, respectively, albumin, spiders, bilirubin, and sex.
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AP Calculus AB/BC
Schedule and Assignments
Applications of Derivatives - Part 1. AP Calculus AB/BC. Schedule and Assignments. Chapter 3 – Applications of Derivatives (Part 1) Date.
Chapter 3 – Applications ofDerivatives (Part 1)
1. The AP exams will ask you to find derivatives using the various techniques and rules including: The Power Rulefor integer, rational (fractional) exponents, expressions with radicals. Derivatives of sum, differences, products, and quotients. The Chain Rulefor composite functions.
2. AP Calculus Unit #4-Applications of the Derivative Part I Smac Key.pdf Author: sean.mcconnell Created Date: 12/5/2018 12:55:12 PM.
Date Topics ThursdayOctober 29 Unit 2 Derivatives TestHomework: None FridayOctober 30 3.a Indeterminate Forms and L’Hôpital’s RuleFRQ/MC # 8 DueHomework: 3.b Video and Survey MondayNovember 2 3.b Maximum and Minimum ValuesHomework: 3.c Video and Survey TuesdayNovember 3 3.c Increasing and Decreasing FunctionsHomework: 3.d Video and Survey WednesdayNovember 4 3.d Concavity and the Second Derivative TestHomework: Section 3.5 # 1, 3, 11, 13, 41, 43 ThursdayNovember 5 Mixed Review (Matching Activity)Foldable QuizHomework: Worksheet Prove It! FridayNovember 6 Mixed Review (Stations)FRQ/MC # 9 DueHomework: Finish Station Activity MondayNovember 9 3.e Curve SketchingHomework: Section 3.6 # 1, 2 TuesdayNovember 10 Mixed ReviewHomework: Functions, Graphs and Derivatives Paragraph, Study for Quiz (Review Notes, Complete Textbook Problems) WednesdayNovember 11 Quiz on Sections 3.1 through 3.6Homework: 3.f Video and Survey
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### Section 4-14 : Business Applications
In the final section of this chapter let’s take a look at some applications of derivatives in the business world. For the most part these are really applications that we’ve already looked at, but they are now going to be approached with an eye towards the business world.
Let’s start things out with a couple of optimization problems. We’ve already looked at more than a few of these in previous sections so there really isn’t anything all that new here except for the fact that they are coming out of the business world.
Example 1 An apartment complex has 250 apartments to rent. If they rent (x) apartments then their monthly profit, in dollars, is given by, [Pleft( x right) = - 8{x^2} + 3200x - 80,000]
How many apartments should they rent in order to maximize their profit?
Show Solution
All that we’re really being asked to do here is to maximize the profit subject to the constraint that (x) must be in the range (0 le x le 250).
First, we’ll need the derivative and the critical point(s) that fall in the range (0 le x le 250).
[P'left( x right) = - 16x + 3200hspace{0.5in} Rightarrow hspace{0.25in} 3200 - 16x = 0hspace{0.25in}, Rightarrow hspace{0.25in} x = frac{{3200}}{{16}} = 200]
Since the profit function is continuous and we have an interval with finite bounds we can find the maximum value by simply plugging in the only critical point that we have (which nicely enough in the range of acceptable answers) and the end points of the range.
[Pleft( 0 right) = - 80,000hspace{0.5in}Pleft( {200} right) = 240,000hspace{0.5in}Pleft( {250} right) = 220,000]
So, it looks like they will generate the most profit if they only rent out 200 of the apartments instead of all 250 of them.
Note that with these problems you shouldn’t just assume that renting all the apartments will generate the most profit. Do not forget that there are all sorts of maintenance costs and that the more tenants renting apartments the more the maintenance costs will be. With this analysis we can see that, for this complex at least, something probably needs to be done to get the maximum profit more towards full capacity. This kind of analysis can help them determine just what they need to do to move towards that goal whether it be raising rent or finding a way to reduce maintenance costs.
Note as well that because most apartment complexes have at least a few units empty after a tenant moves out and the like that it’s possible that they would actually like the maximum profit to fall slightly under full capacity to take this into account. Again, another reason to not just assume that maximum profit will always be at the upper limit of the range.
Let’s take a quick look at another problem along these lines.
Example 2 A production facility is capable of producing 60,000 widgets in a day and the total daily cost of producing (x) widgets in a day is given by, [Cleft( x right) = 250,000 + 0.08x + frac{{200,000,000}}{x}]
How many widgets per day should they produce in order to minimize production costs?
Problem Statement. Show Solution
Here we need to minimize the cost subject to the constraint that (x) must be in the range (0 le x le 60,000). Note that in this case the cost function is not continuous at the left endpoint and so we won’t be able to just plug critical points and endpoints into the cost function to find the minimum value.
Let’s get the first couple of derivatives of the cost function.
[C'left( x right) = 0.08 - frac{{200,000,000}}{{{x^2}}}hspace{0.5in}C'left( x right) = frac{{400,000,000}}{{{x^3}}}]
The critical points of the cost function are,
[begin{align*}0.08 - frac{{200,000,000}}{{{x^2}}} & = 0 0.08{x^2} & = 200,000,000 {x^2} & = 2,500,000,000hspace{0.25in} Rightarrow hspace{0.25in}x = pm sqrt {2,500,000,000} = pm ,50,000end{align*}]
Now, clearly the negative value doesn’t make any sense in this setting and so we have a single critical point in the range of possible solutions : 50,000.
Now, as long as (x > 0) the second derivative is positive and so, in the range of possible solutions the function is always concave up and so producing 50,000 widgets will yield the absolute minimum production cost.
Recall from the Optimization section we discussed how we can use the second derivative to identity the absolute extrema even though all we really get from it is relative extrema.
Now, we shouldn’t walk out of the previous two examples with the idea that the only applications to business are just applications we’ve already looked at but with a business “twist” to them.
There are some very real applications to calculus that are in the business world and at some level that is the point of this section. Note that to really learn these applications and all of their intricacies you’ll need to take a business course or two or three. In this section we’re just going to scratch the surface and get a feel for some of the actual applications of calculus from the business world and some of the main “buzz” words in the applications.
Let’s start off by looking at the following example.
Example 3 The production costs per week for producing (x) widgets is given by, [Cleft( x right) = 500 + 350x - 0.09{x^2},0 le x le 1000]
Answer each of the following questions.
### 4 Applications Of The Derivative Part 1ap Calculus Calculator
1. What is the cost to produce the 301st widget?
2. What is the rate of change of the cost at (x = 300)?
Show All SolutionsHide All Solutionsa What is the cost to produce the 301st widget? Show Solution
We can’t just compute (Cleft( {301} right)) as that is the cost of producing 301 widgets while we are looking for the actual cost of producing the 301st widget. In other words, what we’re looking for here is,
[Cleft( {301} right) - Cleft( {300} right) = 97,695.91 - 97,400.00 = 295.91]
So, the cost of producing the 301st widget is \$295.91.
b What is the rate of change of the cost at (x = 300)? Show Solution
In this part all we need to do is get the derivative and then compute (C'left( {300} right)).
[C'left( x right) = 350 - 0.18xhspace{0.5in} Rightarrow hspace{0.5in}C'left( {300} right) = 296.00]
Okay, so just what did we learn in this example? The cost to produce an additional item is called the marginal cost and as we’ve seen in the above example the marginal cost is approximated by the rate of change of the cost function, (Cleft( x right)). So, we define the marginal cost function to be the derivative of the cost function or, (C'left( x right)). Let’s work a quick example of this.
Example 4 The production costs per day for some widget is given by, [Cleft( x right) = 2500 - 10x - 0.01{x^2} + 0.0002{x^3}]
What is the marginal cost when (x = 200), (x = 300) and (x = 400)?
Show Solution
So, we need the derivative and then we’ll need to compute some values of the derivative.
[begin{align*}C'left( x right) & = - 10 - 0.02x + 0.0006{x^2} C'left( {200} right) & = 10hspace{0.5in}C'left( {300} right) = 38hspace{0.5in}C'left( {400} right) = 78end{align*}]
So, in order to produce the 201st widget it will cost approximately \$10. To produce the 301st widget will cost around \$38. Finally, to product the 401st widget it will cost approximately \$78.
Note that it is important to note that (C'left( n right)) is the approximate cost of producing the ({left( {n + 1} right)^{{mbox{st}}}}) item and NOT the nth item as it may seem to imply!
Let’s now turn our attention to the average cost function. If (Cleft( x right)) is the cost function for some item then the average cost function is,
[overline{C}left( x right) = frac{{Cleft( x right)}}{x}]
Here is the sketch of the average cost function from Example 4 above.
We can see from this that the average cost function has an absolute minimum. We can also see that this absolute minimum will occur at a critical point when (overline C'left( x right) = 0) since it clearly will have a horizontal tangent there.
Now, we could get the average cost function, differentiate that and then find the critical point. However, this average cost function is fairly typical for average cost functions so let’s instead differentiate the general formula above using the quotient rule and see what we have.
[overline{C}'left( x right) = frac{{x,C'left( x right) - Cleft( x right)}}{{{x^2}}}]
Now, as we noted above the absolute minimum will occur when (overline C'left( x right) = 0) and this will in turn occur when,
[x,C'left( x right) - Cleft( x right) = 0hspace{0.5in} Rightarrow hspace{0.5in}C'left( x right) = frac{{Cleft( x right)}}{x} = overline Cleft( x right)]
So, we can see that it looks like for a typical average cost function we will get the minimum average cost when the marginal cost is equal to the average cost.
We should note however that not all average cost functions will look like this and so you shouldn’t assume that this will always be the case.
Let’s now move onto the revenue and profit functions. First, let’s suppose that the price that some item can be sold at if there is a demand for (x) units is given by (pleft( x right)). This function is typically called either the demand function or the price function.
The revenue function is then how much money is made by selling (x) items and is,
[Rleft( x right) = x,pleft( x right)]
### 4 Applications Of The Derivative Part 1ap Calculus Solver
The profit function is then,
[Pleft( x right) = Rleft( x right) - Cleft( x right) = x,pleft( x right) - Cleft( x right)]
Be careful to not confuse the demand function, (pleft( x right)) - lower case (p), and the profit function, (Pleft( x right)) - upper case (P). Bad notation maybe, but there it is.
Finally, the marginal revenue function is (R'left( x right)) and the marginal profit function is (P'left( x right)) and these represent the revenue and profit respectively if one more unit is sold.
Let’s take a quick look at an example of using these.
### 4 Applications Of The Derivative Part 1ap Calculus 14th Edition
Example 5 The weekly cost to produce (x) widgets is given by [Cleft( x right) = 75,000 + 100x - 0.03{x^2} + 0.000004{x^3}hspace{0.5in}0 le x le 10000]
and the demand function for the widgets is given by,
[pleft( x right) = 200 - 0.005xhspace{0.5in}0 le x le 10000]
Determine the marginal cost, marginal revenue and marginal profit when 2500 widgets are sold and when 7500 widgets are sold. Assume that the company sells exactly what they produce.
Show Solution
Okay, the first thing we need to do is get all the various functions that we’ll need. Here are the revenue and profit functions.
[begin{align*}Rleft( x right) & = xleft( {200 - 0.005x} right) = 200x - 0.005{x^2} Pleft( x right) & = 200x - 0.005{x^2} - left( {75,000 + 100x - 0.03{x^2} + 0.000004{x^3}} right) & = - 75,000 + 100x + 0.025{x^2} - 0.000004{x^3}end{align*}]
Now, all the marginal functions are,
[begin{align*}C'left( x right) & = 100 - 0.06x + 0.000012{x^2} R'left( x right) & = 200 - 0.01x P'left( x right) & = 100 + 0.05x - 0.000012{x^2}end{align*}]
The marginal functions when 2500 widgets are sold are,
[C'left( {2500} right) = 25hspace{0.5in}R'left( {2500} right) = 175hspace{0.5in}P'left( {2500} right) = 150]
The marginal functions when 7500 are sold are,
[C'left( {7500} right) = 325hspace{0.5in}R'left( {7500} right) = 125hspace{0.5in}P'left( {7500} right) = - 200]
So, upon producing and selling the 2501st widget it will cost the company approximately \$25 to produce the widget and they will see an added \$175 in revenue and \$150 in profit.
On the other hand, when they produce and sell the 7501st widget it will cost an additional \$325 and they will receive an extra \$125 in revenue, but lose \$200 in profit.
### 4 Applications Of The Derivative Part 1ap Calculus 2nd Edition
We’ll close this section out with a brief discussion on maximizing the profit. If we assume that the maximum profit will occur at a critical point such that (P'left( x right) = 0) we can then say the following,
[P'left( x right) = R'left( x right) - C'left( x right) = 0hspace{0.5in} Rightarrow hspace{0.5in}R'left( x right) = C'left( x right)]
We then will know that this will be a maximum we also were to know that the profit was always concave down or,
[P'left( x right) = R'left( x right) - C'left( x right) < 0hspace{0.5in} Rightarrow hspace{0.5in}R'left( x right) < C'left( x right)]
So, if we know that (R'left( x right) < C'left( x right)) then we will maximize the profit if (R'left( x right) = C'left( x right)) or if the marginal cost equals the marginal revenue.
In this section we took a brief look at some of the ideas in the business world that involve calculus. Again, it needs to be stressed however that there is a lot more going on here and to really see how these applications are done you should really take some business courses. The point of this section was to just give a few ideas on how calculus is used in a field other than the sciences. | 4,117 | 15,012 | {"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} | 3.484375 | 3 | CC-MAIN-2022-05 | latest | en | 0.840939 |
https://sound.stackexchange.com/questions/52850/expected-behavior-of-two-dissimilar-dynamic-mics-connected-to-one-input-with-a | 1,721,915,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00061.warc.gz | 461,617,741 | 40,974 | # Expected behavior of two dissimilar dynamic mics connected to one input with a "Y" cable
I understand that it is not uncommon to use a "Y" cable to plug two similar dynamic mics into a single input in certain situations (e.g. individually miked rack toms). What happens if two dynamic microphones with dissimilar impedances are connected in this way? Does the impedance mismatch affect the relative sensitivities of the two mics?
• "I understand that it is not uncommon to use a "Y" cable to plug two similar dynamic mics into a single input in certain situations (e.g. individually miked rack toms)." I'm not sure where you got that understanding from, but I've never even heard of anyone doing this in about 30 years of recording and engineering. Commented Jul 1 at 19:17
I don't think using a y cable to combine 2 mics is that common, I'm in agreement with Todd Willcox's comment that I've never actually seen that in many years recording and engineering.
But, if the y cable is correctly wired to combine the 2 mics, so their ground, +ve and -ve (ie the signal, and the flipped signal) are all connected accordingly then it shouldn't pose much of a problem. Mic inputs are designed to have a fairly high input impedance (relatively).
The ideal state (for signal transfer) is a low impedance source (the mic) feeding a high impedance input (the mic pre), that way the most voltage is dropped in the circuitry you want the signal to be passed to. This is not the case when you want to use electrical energy as some kind of power, where you want the impedances to be matched, but this is not relevant with transferring a signal to an amplification stage.
You need to know what output impedance your mic has, but most recording mics these days have a low output impedance, 50Ω - 1000Ω. Going into a mixer with a similar or roughly higher input impedance, say 1kΩ - 10kΩ, will be fine for one mic. Combining 2 mics, well, it would depend heavily on the circuitry of the mic in question but at most the impedances would sum, even in that case they would likely transfer fine to a mic pre stage. They may even combine to represent a lower impedance than either mic has individually, which would slightly improve transfer.
Generally, bad sound from impedance mismatch starts to become noticeable when the impedances are out by a few order of magnitudes from ideal, so I wouldn't worry too much, just try it and see if it sounds odd.
The range a mic pre is expected to deal with will probably not be challenged with two normal low impedance mics. If you are using piezo etc, well thats a whole different area where more knowledge of the circuitry would be needed to predict what would happen..
By using a Y cable, you are placing the microphones in parallel, in terms of an electric circuit. This means that if two like microphones were connected, the amount of current flowing through both microphones would be halved (equal amount of current through each microphone). The overall circuit resistance would also be halved. Now resistance and impedance are not entirely the same, depending on the active and/passive components used in thr microphones. You certainly don’t want to try this for microphones that use phantom power, without ensuring you are providing enough power. Dissimilar microphones would complicate this. Assuming you are using the same passive microphones, the sensitivity of the microphones with less current throughput will affect signal gain. Over driving the circuit may cause signal quality loss (distortion) and damage components. Reduce the gain to zero or power the mixer off before connecting them in to the Y cable. Do the same before unplugging them. Some mixing desks will cope quite well because they are designed to deal with a much broader impedance range, and microphone arrays. The overall impedance will likely not change linearly when using different types of microphones. As the components will react differently and alter the current flow accordingly. Even if the mixing desk can handle the different impedances, it doesn’t mean your microphones will perform as well in parallel.
• Welcome to Sound Design Stack Exchange. Please read our tour and How to Answer pages to understand what we require in an answer post. Currently, your post does not meet requirements so is likely to be downvoted and may be deleted. You can edit to improve by describing the expected behaviour as the question asks. Commented Jul 1 at 18:59 | 947 | 4,467 | {"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} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.950092 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=141&t=59974&p=229599 | 1,606,815,516,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00030.warc.gz | 384,712,734 | 11,333 | ## cell potential
$\Delta G^{\circ} = -nFE_{cell}^{\circ}$
josmit_1D
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am
### cell potential
is there a maximum cell potential that can be reached?
Jasmine Fendi 1D
Posts: 108
Joined: Sat Aug 24, 2019 12:15 am
### Re: cell potential
Yes! The max potential of a cell is the maximum potential difference between two half reactions in a galvanic cell. I think you can calculate this by manipulating different equations.
805303639
Posts: 50
Joined: Thu Jul 11, 2019 12:16 am
Been upvoted: 1 time
### Re: cell potential
The maximum cell potential occurs at the instant when the anode and cathode are first connected and the flow of electrons begins. Beyond that point, the concentrations in the half-reactions change. The production of product decreases the cell potential per increasing Q in the Nernst equation: E= E^0 -RT/nF ln Q.
gabbymaraziti
Posts: 111
Joined: Wed Sep 18, 2019 12:19 am
### Re: cell potential
Cell potential is greatest when the cell reaction first begins to occur. Following this, the concentrations change and the potential is no longer at its maximum value. | 309 | 1,131 | {"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": 0, "codecogs_latex": 1, "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-2020-50 | latest | en | 0.890016 |
https://beta.geogebra.org/m/KZHUpfwQ | 1,679,648,400,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00407.warc.gz | 156,771,582 | 9,851 | # How to inscribe a circle in a circular sector?
Author:
kokosek
Topic:
Circle
First we have to have a sector with three points - A, B and C. The circle must be in the middle of the sector so we make a bisector from angle CAB. It intersects the circle in point D. Then we make a tangent to the original circle through point D. We extend the segment AC to be a half line so that it intersects the tangent in point E. We then make another bisector from angle AED. It intersects the other bisector in point F which is the center for out inscribed circle. All we have to do is make a circle from point F with radius of segment FD. | 149 | 627 | {"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} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.888572 |
https://lw2.issarice.com/posts/oF6D8SieWbMPob9r6/no-ai-is-just-as-energy-efficient-as-your-brain | 1,722,837,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00039.warc.gz | 299,199,555 | 5,532 | # No - AI is just as energy-efficient as your brain.
post by Maxwell Clarke (maxwell-clarke) · 2023-05-24T02:30:24.526Z · LW · GW · 7 comments
This is simply because AIs run on electricity.
Our brains use sunlight via photosynthesis via dietary energy intake. From sunlight to dietary energy this is about 0.25-0.5% energy efficient. Let's say it's 0.35% efficient. (This is then our complete sunlight-to-capability efficiency for a human)
AI systems use sunlight via solar energy via electrical consumption. From sunlight to GPU this is about 10-18% efficient let's say it's 13% efficient. (Data: typical solar panel efficiency = 15-20% efficient. Typical electricity distribution efficiency = 90% efficient). 13% is then AI sunlight-to-electricity efficiency, but not yet sunlight-to-capability efficiency.
Let's calculate the final part. We need to assume an amount of compute for a human-equivalent AI system. I will assume that (note, at inference time) we can run a human-equivalent AI system on 10 NVIDIA 4070 GPUs at full power, which each consume ~200 Watts, so 2 kW in total. In contrast the typical human consumes 100 watts. 100W divided by 2 kW gives us 5% as AI electricity-to-capability efficiency, which we then multiply to get the total AI efficiency. 5% electricity-to-capability efficiency times 13% sunlight-to-electricity efficiency = 0.65% sunlight-to-capability efficiency.
So humans are 0.35% efficient, and AIs are 0.65% efficient.
By these assumptions, AI is somewhat more efficient in terms of real energy input - here, energy from the sun. This is the number that the economy is going to care about - why use land for crops, when you can use it for solar panels?
Please pick apart my numbers and assumptions in the comments. Thanks.
## 7 comments
Comments sorted by top scores.
comment by Joey Marcellino · 2023-05-24T14:07:40.034Z · LW(p) · GW(p)
The conclusion seems rather to be "human metabolism is less efficient than solar panels," which, while perhaps true, has limited bearing on the question of whether or not the brain is thermodynamically efficient as a computer when compared to current or future AI. The latter is the question that recent discussion has been focused on, and to which the "No - " in the title makes it seem like you're responding.
Moreover, while a quick Google search turns up 100W as the average resting power output of a person, another search suggests the brain is only responsible for about 20% of energy consumption per time. Adding this to your analysis gives .13% "efficiency" in the sense that you're using it, so the brain still outperforms AI even on this admittedly rather odd sunlight-to-capability metric.
Replies from: maxwell-clarke
comment by Maxwell Clarke (maxwell-clarke) · 2023-05-24T18:58:32.590Z · LW(p) · GW(p)
Well, yes, the point of my post is just to point out that the number that actually matters is the end-to-end energy efficiency — and it is completely comparable to humans.
The per-flop efficiency is obviously worse. But, that's irrelevant if AI is already cheaper for a given task in real terms.
I admit the title is a little clickbaity but i am responding to a real argument (that humans are still "superior" to AI because the brain is more thermodynamically efficient per-flop)
comment by Anon User (anon-user) · 2023-05-27T19:17:03.966Z · LW(p) · GW(p)
I made a similar point (but without specific numbers - great to have them!) in a comment https://www.lesswrong.com/posts/Lwy7XKsDEEkjskZ77/?commentId=nQYirfRzhpgdfF775 [LW · GW] on a post that posited human brain energy efficiency over AIs as a core anti-doom argument, and I also think that the energy efficiency comparisons are not particularly relevant either way:
Humanity is generating and consuming enormous amount of power - why is the power budget even relevant? And even if it was, energy for running brains ultimately comes from Sun - if you include the agriculture energy chain, and "grade" the energy efficiency of brains by the amount of solar energy it ultimately takes to power a brain, AI definitely has a potential to be more efficient. And even if a single human brain is fairly efficient, the human civilization is clearly not. With AI, you can quickly scale up the amount of compute you use, but scaling beyond a single brain is very inefficient.
comment by Gunnar_Zarncke · 2023-05-24T15:41:51.387Z · LW(p) · GW(p)
If you go this way, you have to include the energy cost of growing a human on the one hand and building and deploying the solar cells and the chip factories on the other.
Replies from: maxwell-clarke
comment by Maxwell Clarke (maxwell-clarke) · 2023-05-24T19:02:06.308Z · LW(p) · GW(p)
Yes, that's fair. I was ignoring scale but you're right that it's a better comparison if it is between a marginal new human and a marginal new AI.
comment by mako yass (MakoYass) · 2023-05-24T02:48:29.343Z · LW(p) · GW(p)
Can't you make human food production a lot more efficient with biotech? Algae, for instance? Spirulina maybe? Tastes bitter, grows fast, highly nutritious. (Are plants or algae as efficient at generating sugars from sunlight as new forms of life evolved to directly use electricity from a solar panel would be?)
Even if that wasn't practical for humans, if such an organism would be very easily imaginable I think that still gives us some weird biopunk menu options for the medium-term future of intelligence?
Replies from: maxwell-clarke
comment by Maxwell Clarke (maxwell-clarke) · 2023-05-24T02:55:39.350Z · LW(p) · GW(p)
I saw some numbers for algae being 1-2% efficient but it was for biomass rather than dietary energy. Even if you put the brain in the same organism, you wouldn't expect as good efficiency as that. The difference is that creating biomass (which is mostly long chains of glucose) is the first step, and then the brain must use the glucose, which is a second lossy step.
But I mean there is definitely far-future biopunk options eg. I'd guess it's easy to create some kind of solar panel organism which grows silicon crystals instead of using chlorophyll. | 1,488 | 6,098 | {"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-2024-33 | latest | en | 0.917227 |
https://stats.stackexchange.com/questions/22736/chances-of-meeting-people-whos-initials-contain-a-j | 1,576,100,003,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00084.warc.gz | 552,900,918 | 31,046 | # Chances of meeting people who's initials contain a J
I am not a statistics expert but was interested in the validity of the reasoning below.
Somebody argued that many people had the letter 'J' in their initials. Can someone comment on the validity of the following approach and how it could be improved, notably what probability distribution would a list of firs tnames sorted by popularity follows?
According to wikipedia the most common first names for males in the US according to the 1990 census are, in that order:
'James','John','Robert','Michael','William','David','Richard','Charles','Joseph', and 'Thomas'
Let's use an approximate probability distribution (any better way?) that gives the probabilities to these names according to their order i (starting at 0) following:
P(name_i) = 0.15 / (i + 2)
This is a probability distribution for at least 1000 names (sums to 1), and starts like this:
0.0749, 0.0499, 0.0374, 0.0299, 0.0249, ...
meaning that someone has 7.49% chances of being called James. The sum of the probabilities of having a name starting with J from these list is 14% (P(James)+P(John)+P(Joseph)), lets call it Pj. Of course Pj is in fact higher because more names than these 8 in the list of 1000 may start with J.
Now let's assume a child is given three first names, the probability that at least of them starts with J given pj is 36%!! (from (1-(1-pj)^3)). So every time you meet a new person, you have one third chances to see him having a J on his business card. Let's call this probability pIj, probability of at least one initial starting with a J.
However, when you say overwhelming majority, let's assume that you mean 8 out of 10. The probability that out of 10 people you meet in a row 8 have a name starting with a J is, using binomial distribution, only 0.5% I'm afraid, by comb(10,8)pIj^8(1-pIj)^2 .
However, if you settle with 5 our of 10, your chances rise to 16%.
FYI the original question and post can be found here.
• It seems to me that you systematically replace information that could be obtained from data by assumptions that have little or no objective support, including the assumed probability distribution, that you meet people at random, that names are assigned independently, that you can ignore the J's in the next 992 names, etc. This kind of exercise can be useful to get a sense of what a solution might look like and what assumptions are important to verify, but it's unrealistic to suppose that the actual chances you obtain can be trusted. – whuber Feb 13 '12 at 22:51
• You're right, and I guess that the more experienced you are the more assumptions get realistic. The tradeoff is between getting a sense of what the solution might look like and doing the research, which is not justified when the question is of anecdotal interest only. Something akin to estimates and order-of-magnitude calculations in engineering. – Vladtn Feb 14 '12 at 14:35
• Actual frequencies of male given names in the 1990 US census can be found at names.mongabay.com/male_names.htm – Pere Dec 11 '16 at 22:24 | 749 | 3,065 | {"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} | 4.03125 | 4 | CC-MAIN-2019-51 | latest | en | 0.951222 |
https://jeeplasmachine.com/qa/quick-answer-what-does-z-mean-on-a-tire.html | 1,611,457,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00020.warc.gz | 413,450,649 | 9,638 | # Quick Answer: What Does Z Mean On A Tire?
## Does Zr mean run flat?
The following letter(s) describe the construction: R is for radial; ZR stands for a Z-rated radial (see service description); and ZRF means Z-rated run-flat (rare).
The last two-digit number is the diameter of the applicable wheel, in inches..
## What does 35 mean on tires?
The 35 represents the aspect ratio of the tire or the sidewall’s height as a percentage of the tread width. This means the tire’s sidewall is 35 percent of the tread’s width, which calculates to 82.25 mm for the tire shown.
## What does XL mean on a tire?
Extra LoadReinforced or XL (Extra Load) tires are specially reinforced tires. They can carry heavier loads than tires of the same size. Reinforced tires are designated on the sidewall by the letters “RF”; extra load tires with the letters “XL.” Reinforced and XL tires require higher inflation pressures than standard tires.
## What does ZR stand for?
AcronymDefinitionZRZirconiumZRZone Rating (baseball statistic; used to measure fielding ability)ZRZettai Ryouiki (anime)ZRFreezing Rain (Weather Symbol)7 more rows
## Are 325 tires the same as 35?
325/60R20 is 35.4 inches tall and a 13 in. section width versus a 35 12.50R20 which is 35 inches tall and a 13 inch section width. So you gain about half an inch in tire height.
## Should I get H or V rated tires?
To support running at higher speeds, V-rated tires will have a stiffer sidewall and slightly firmer ride than H-rated tires. For normal driving conditions the H-rated tire will provide a more comfortable ride and the V-rated should give slightly better handling.
## Can I use 235 tires instead of 225?
Obviously, you will need a certain kind of tire depending on how you use your vehicle. Due to this, a 235 tire might not be better than a 225 tire, or vice versa, but depending on your situation, one tire might serve you better than the other.
## What is the difference between R and Zr in tires?
ZR basically means the tire is adapted for speeds above 149mph. ZR tires includes the speed ratings V(149mph), W(168mph) and Y(186mph). The R stands for radial. Prime members enjoy FREE Delivery and exclusive access to music, movies, TV shows, original audio series, and Kindle books.
## What do the 3 numbers mean on tire size?
B: TIRE WIDTH The three-digit number following the letter is the tire’s width (from side to side, looking at the tire head on) in millimeters. … In other words, it’s sidewall height divided by tire width. In this example, the aspect ratio is 65, meaning the sidewall is 65 percent as high as the tire is wide.
## What does 97w XL mean on a tire?
Some vehicles are equipped with “XL” tires. It doesn’t mean that they’re extra large, but it does mean that they are extra-load tires.
## What are the best 35 inch tires?
Best 35-inch Tires Reviews 2020Best 35-inch Tires Reviews 2020. … Toyo Tire Open Country M/T – Best Mud Terrain Tires. … Federal Couragia M/T Mud-Terrain Tire. … Mastercraft Courser MXT Mud Terrain Radial Tire. … Ironman All Country M/T all_season Radial Tire-LT35/12.50R20.
## Does tire speed rating matter?
The speed rating tells you the speed the tire can safely maintain over time. A higher speed rating usually means you will have better control and handling at higher speeds – and that the tire can take the extra heat. As a general rule, tires with higher speed ratings also handle better at slower speeds.
## What does the 265 mean in tire size?
265. This number indicates that your tire has a width of 265 millimeters. 70. This number means that your tire has an aspect ratio of 70%. In other words, your tire’s sidewall height (from the edge of the rim to the tire’s tread) is 70% of the width.
## Do XL tires ride rough?
The XL tires have a stiffer sidewall and will ride a bit more ‘rough’. Probably will respond quicker too as they have stiffer sidewalls.
## Are extra load Tyres better?
The advantage of tyres with greater load-bearing capacity as a possible alternative to standard tyres is that they offer a higher buffer up to the maximum load. … If you like driving off-road or you regularly drive in mountainous areas with a packed car, then XL tyres are always better.
## Is a 315 tire a 35?
A 315/75R16 tire has nearly the same diameter as a 35×12. 50 tire and is commonly referred to as a “metric 35.”
## Should I get H or T rated tires?
H rated tires have stronger construction than T rated tires. The concern may not be speed although a Subaru can exceed 118 MPH hence those tires however it may be load rating of the tires. … Not only is the “T” rating good for 118 mph, it’s good for CONTINUOUS use at 118 mph.
## What is the best tire speed rating?
Speed RatingL75 mph120 km/hT118 mph190 km/hU124 mph200 km/hH130 mph210 km/hV149 mph240 km/h6 more rows
## What is the ZR rating on a tire?
ZR shown on the sidewall of your tire is a reference to the tire’s speed rating.. Historically, the speed rating ZR on a tire meant that the construction of the tire could reach of 150 miles or 240 kilometers per hour. Today, a tire labeled ZR, has more capabilities than before.
## Can I use 225 tires instead of 215?
Unless you’re exploring the limits of the car’s performance, most drivers probably wouldn’t notice a difference between a 225 and a 215, all else being equal. If the outer circumference is a little different, the speedo may be a tiny bit off. | 1,316 | 5,426 | {"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} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.924266 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese683.htm | 1,696,475,570,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00513.warc.gz | 650,768,342 | 6,304 | 3.683 $$\int \frac{1}{(d+e x)^{5/2} \sqrt{a+c x^2}} \, dx$$
Optimal. Leaf size=382 $\frac{2 \sqrt{-a} \sqrt{c} \sqrt{\frac{c x^2}{a}+1} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right ),-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \sqrt{a+c x^2} \sqrt{d+e x} \left (a e^2+c d^2\right )}-\frac{8 \sqrt{-a} c^{3/2} d \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \sqrt{a+c x^2} \left (a e^2+c d^2\right )^2 \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}}-\frac{8 c d e \sqrt{a+c x^2}}{3 \sqrt{d+e x} \left (a e^2+c d^2\right )^2}-\frac{2 e \sqrt{a+c x^2}}{3 (d+e x)^{3/2} \left (a e^2+c d^2\right )}$
[Out]
(-2*e*Sqrt[a + c*x^2])/(3*(c*d^2 + a*e^2)*(d + e*x)^(3/2)) - (8*c*d*e*Sqrt[a + c*x^2])/(3*(c*d^2 + a*e^2)^2*Sq
rt[d + e*x]) - (8*Sqrt[-a]*c^(3/2)*d*Sqrt[d + e*x]*Sqrt[1 + (c*x^2)/a]*EllipticE[ArcSin[Sqrt[1 - (Sqrt[c]*x)/S
qrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*e)])/(3*(c*d^2 + a*e^2)^2*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c
]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^2]) + (2*Sqrt[-a]*Sqrt[c]*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*S
qrt[1 + (c*x^2)/a]*EllipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*
e)])/(3*(c*d^2 + a*e^2)*Sqrt[d + e*x]*Sqrt[a + c*x^2])
________________________________________________________________________________________
Rubi [A] time = 0.265522, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {745, 835, 844, 719, 424, 419} $-\frac{8 \sqrt{-a} c^{3/2} d \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \sqrt{a+c x^2} \left (a e^2+c d^2\right )^2 \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}}-\frac{8 c d e \sqrt{a+c x^2}}{3 \sqrt{d+e x} \left (a e^2+c d^2\right )^2}-\frac{2 e \sqrt{a+c x^2}}{3 (d+e x)^{3/2} \left (a e^2+c d^2\right )}+\frac{2 \sqrt{-a} \sqrt{c} \sqrt{\frac{c x^2}{a}+1} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} F\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \sqrt{a+c x^2} \sqrt{d+e x} \left (a e^2+c d^2\right )}$
Antiderivative was successfully verified.
[In]
Int[1/((d + e*x)^(5/2)*Sqrt[a + c*x^2]),x]
[Out]
(-2*e*Sqrt[a + c*x^2])/(3*(c*d^2 + a*e^2)*(d + e*x)^(3/2)) - (8*c*d*e*Sqrt[a + c*x^2])/(3*(c*d^2 + a*e^2)^2*Sq
rt[d + e*x]) - (8*Sqrt[-a]*c^(3/2)*d*Sqrt[d + e*x]*Sqrt[1 + (c*x^2)/a]*EllipticE[ArcSin[Sqrt[1 - (Sqrt[c]*x)/S
qrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*e)])/(3*(c*d^2 + a*e^2)^2*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c
]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^2]) + (2*Sqrt[-a]*Sqrt[c]*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*S
qrt[1 + (c*x^2)/a]*EllipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*
e)])/(3*(c*d^2 + a*e^2)*Sqrt[d + e*x]*Sqrt[a + c*x^2])
Rule 745
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[c/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])
Rule 835
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])
Rule 844
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && !IGtQ[m, 0]
Rule 719
Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*a*Rt[-(c/a), 2]*(d + e*x)^m*Sqrt[
1 + (c*x^2)/a])/(c*Sqrt[a + c*x^2]*((c*(d + e*x))/(c*d - a*e*Rt[-(c/a), 2]))^m), Subst[Int[(1 + (2*a*e*Rt[-(c/
a), 2]*x^2)/(c*d - a*e*Rt[-(c/a), 2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(1 - Rt[-(c/a), 2]*x)/2]], x] /; FreeQ[{a,
c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m^2, 1/4]
Rule 424
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]
Rule 419
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] && !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])
Rubi steps
\begin{align*} \int \frac{1}{(d+e x)^{5/2} \sqrt{a+c x^2}} \, dx &=-\frac{2 e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right ) (d+e x)^{3/2}}-\frac{(2 c) \int \frac{-\frac{3 d}{2}+\frac{e x}{2}}{(d+e x)^{3/2} \sqrt{a+c x^2}} \, dx}{3 \left (c d^2+a e^2\right )}\\ &=-\frac{2 e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right ) (d+e x)^{3/2}}-\frac{8 c d e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right )^2 \sqrt{d+e x}}+\frac{(4 c) \int \frac{\frac{1}{4} \left (3 c d^2-a e^2\right )+c d e x}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{3 \left (c d^2+a e^2\right )^2}\\ &=-\frac{2 e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right ) (d+e x)^{3/2}}-\frac{8 c d e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right )^2 \sqrt{d+e x}}+\frac{\left (4 c^2 d\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+c x^2}} \, dx}{3 \left (c d^2+a e^2\right )^2}-\frac{c \int \frac{1}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{3 \left (c d^2+a e^2\right )}\\ &=-\frac{2 e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right ) (d+e x)^{3/2}}-\frac{8 c d e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right )^2 \sqrt{d+e x}}+\frac{\left (8 a c^{3/2} d \sqrt{d+e x} \sqrt{1+\frac{c x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{2 a \sqrt{c} e x^2}{\sqrt{-a} \left (c d-\frac{a \sqrt{c} e}{\sqrt{-a}}\right )}}}{\sqrt{1-x^2}} \, dx,x,\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )}{3 \sqrt{-a} \left (c d^2+a e^2\right )^2 \sqrt{\frac{c (d+e x)}{c d-\frac{a \sqrt{c} e}{\sqrt{-a}}}} \sqrt{a+c x^2}}-\frac{\left (2 a \sqrt{c} \sqrt{\frac{c (d+e x)}{c d-\frac{a \sqrt{c} e}{\sqrt{-a}}}} \sqrt{1+\frac{c x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+\frac{2 a \sqrt{c} e x^2}{\sqrt{-a} \left (c d-\frac{a \sqrt{c} e}{\sqrt{-a}}\right )}}} \, dx,x,\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )}{3 \sqrt{-a} \left (c d^2+a e^2\right ) \sqrt{d+e x} \sqrt{a+c x^2}}\\ &=-\frac{2 e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right ) (d+e x)^{3/2}}-\frac{8 c d e \sqrt{a+c x^2}}{3 \left (c d^2+a e^2\right )^2 \sqrt{d+e x}}-\frac{8 \sqrt{-a} c^{3/2} d \sqrt{d+e x} \sqrt{1+\frac{c x^2}{a}} E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \left (c d^2+a e^2\right )^2 \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}} \sqrt{a+c x^2}}+\frac{2 \sqrt{-a} \sqrt{c} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}} \sqrt{1+\frac{c x^2}{a}} F\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \left (c d^2+a e^2\right ) \sqrt{d+e x} \sqrt{a+c x^2}}\\ \end{align*}
Mathematica [C] time = 1.66069, size = 494, normalized size = 1.29 $\frac{2 \left (-e^2 \left (a+c x^2\right ) \left (a e^2+c d (5 d+4 e x)\right )+\frac{c (d+e x) \left (i (d+e x)^{3/2} \left (4 i \sqrt{a} \sqrt{c} d e-a e^2+3 c d^2\right ) \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right ),\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )+4 d e^2 \left (a+c x^2\right ) \sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}+4 \sqrt{c} d (d+e x)^{3/2} \left (\sqrt{a} e-i \sqrt{c} d\right ) \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right )|\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )\right )}{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}\right )}{3 e \sqrt{a+c x^2} (d+e x)^{3/2} \left (a e^2+c d^2\right )^2}$
Antiderivative was successfully verified.
[In]
Integrate[1/((d + e*x)^(5/2)*Sqrt[a + c*x^2]),x]
[Out]
(2*(-(e^2*(a + c*x^2)*(a*e^2 + c*d*(5*d + 4*e*x))) + (c*(d + e*x)*(4*d*e^2*Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]*(a
+ c*x^2) + 4*Sqrt[c]*d*((-I)*Sqrt[c]*d + Sqrt[a]*e)*Sqrt[(e*((I*Sqrt[a])/Sqrt[c] + x))/(d + e*x)]*Sqrt[-(((I*
Sqrt[a]*e)/Sqrt[c] - e*x)/(d + e*x))]*(d + e*x)^(3/2)*EllipticE[I*ArcSinh[Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]/Sqr
t[d + e*x]], (Sqrt[c]*d - I*Sqrt[a]*e)/(Sqrt[c]*d + I*Sqrt[a]*e)] + I*(3*c*d^2 + (4*I)*Sqrt[a]*Sqrt[c]*d*e - a
*e^2)*Sqrt[(e*((I*Sqrt[a])/Sqrt[c] + x))/(d + e*x)]*Sqrt[-(((I*Sqrt[a]*e)/Sqrt[c] - e*x)/(d + e*x))]*(d + e*x)
^(3/2)*EllipticF[I*ArcSinh[Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]/Sqrt[d + e*x]], (Sqrt[c]*d - I*Sqrt[a]*e)/(Sqrt[c]
*d + I*Sqrt[a]*e)]))/Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]))/(3*e*(c*d^2 + a*e^2)^2*(d + e*x)^(3/2)*Sqrt[a + c*x^2]
)
________________________________________________________________________________________
Maple [B] time = 0.284, size = 1904, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(e*x+d)^(5/2)/(c*x^2+a)^(1/2),x)
[Out]
2/3*(3*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*x
*a*c*d*e^3*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-
a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)+EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-
c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*x*a*e^4*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*
c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)*(-a*c)^(1/2)+3*EllipticF((-(e*x+d)*c/
((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*x*c^2*d^3*e*(-(e*x+d)*c/((-a*c
)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e
-c*d))^(1/2)+EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1
/2))*x*c*d^2*e^2*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((
c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)*(-a*c)^(1/2)-4*EllipticE((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/
2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*x*a*c*d*e^3*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-
c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)-4*EllipticE(
(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*x*c^2*d^3*e*(-(e*x
+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-
a*c)^(1/2)*e-c*d))^(1/2)+3*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d)
)^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-
((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*a*c*d^2*e^2+(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(
-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)*EllipticF((-(e*x+
d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*(-a*c)^(1/2)*a*d*e^3+3*(-
(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e
/((-a*c)^(1/2)*e-c*d))^(1/2)*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^
(1/2)*e+c*d))^(1/2))*c^2*d^4+(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*
d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),
(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*(-a*c)^(1/2)*c*d^3*e-4*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(
1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)*Elli
pticE((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*a*c*d^2*e^2-
4*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2
))*e/((-a*c)^(1/2)*e-c*d))^(1/2)*EllipticE((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a
*c)^(1/2)*e+c*d))^(1/2))*c^2*d^4-4*x^3*c^2*d*e^3-x^2*a*c*e^4-5*x^2*c^2*d^2*e^2-4*x*a*c*d*e^3-a^2*e^4-5*a*c*d^2
*e^2)/(c*x^2+a)^(1/2)/(a*e^2+c*d^2)^2/(e*x+d)^(3/2)/e
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{2} + a}{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(5/2)/(c*x^2+a)^(1/2),x, algorithm="maxima")
[Out]
integrate(1/(sqrt(c*x^2 + a)*(e*x + d)^(5/2)), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + a} \sqrt{e x + d}}{c e^{3} x^{5} + 3 \, c d e^{2} x^{4} + 3 \, a d^{2} e x + a d^{3} +{\left (3 \, c d^{2} e + a e^{3}\right )} x^{3} +{\left (c d^{3} + 3 \, a d e^{2}\right )} x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(5/2)/(c*x^2+a)^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(c*x^2 + a)*sqrt(e*x + d)/(c*e^3*x^5 + 3*c*d*e^2*x^4 + 3*a*d^2*e*x + a*d^3 + (3*c*d^2*e + a*e^3)*
x^3 + (c*d^3 + 3*a*d*e^2)*x^2), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + c x^{2}} \left (d + e x\right )^{\frac{5}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)**(5/2)/(c*x**2+a)**(1/2),x)
[Out]
Integral(1/(sqrt(a + c*x**2)*(d + e*x)**(5/2)), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{2} + a}{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(5/2)/(c*x^2+a)^(1/2),x, algorithm="giac")
[Out]
integrate(1/(sqrt(c*x^2 + a)*(e*x + d)^(5/2)), x) | 8,651 | 15,776 | {"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} | 3.34375 | 3 | CC-MAIN-2023-40 | latest | en | 0.14556 |
http://jamesrahn.com/pages/pre-calculus/ti-84.htm | 1,550,612,707,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247492825.22/warc/CC-MAIN-20190219203410-20190219225410-00491.warc.gz | 141,044,923 | 4,307 | ## Section 19 for the TI-84C
Some of these operations you should be familiar with on the TI-83 are listed below.
As a student continues their study of mathematics in an Advanced Placement Calculus class the graphing calculator will continue to help the student build a deeper understanding for the function numerically and graphically. Once the student has developed these two foundational viewpoints the student will have less difficulty understanding the analytically study of the same mathematical concepts.
It doesn't matter which graphing calculator you use, all of the following skills are important on any graphing calculator.
1. Working with the calculator in the correct mode
Function Mode: (Func)
In most cases the student will want to make sure their calculator is in the Function mode. Although there will be a few times when parametric, polar, and sequence mode will also be used.
Sequential Mode vs. Simultaneous Mode:
It usually does not matter which of these two modes a student selects. Each has their advantages--you should study how two functions are graphed under each of these modes and make your choice as which one you would use in a particular situation.
Floating Decimal Point:
Students should round off all their final answers to THREE decimal places, but this means that throughout a problem a student should work with more than three decimal places (possibly five). Then when the final answer is stated the answer should be rounded to THREE decimal places.
Dot vs. Connected Mode: Students should realize that they are working with a discrete calculator which locates a finite number of points associated with a function and connects these finite number of points to create a smooth graph (connected mode). As a student traces along the graph they can read the values of x the calculator used to generate the representative graph. If a student places a calculator in dot mode they will see the set of points the calculator generated to create a picture of the graph.
2. Window dimensions:
The graphing calculator can graph a function in many different windows. You can select minimum and maximum x- and y-values. Most graphing calculators allow you to view a graph in a standard window {-10, 10, 1, -10, 10, 1}, but this is not necessary the best window to view all functions. For example the function y=sin x would be better viewed in a window {-8.63, 8.63, 1, -4, 4, 1}. Students should consider the behavior of a function and its range in building a window.
The TI-83 graphing calculator does have one zoom feature which makes building a window easy. It is 0: ZoomFit. You must first select a domain and then the calculator will determine what function values are needed to view the entire graph. Suppose you had the equation and wanted to view it in the domain . First enter the equation in a y-slot and then set Xmin=0, Xmax=10, and Xscl=1. Press Zoom, and select choice 0: ZoomFit.
From the graph you can see that the function takes on many negative function values: . But it is not real clear what the function is doing about x=0. To view the behavior around x=0 change Xmax=2 and re-build a window using ZoomFit.
This view shows the behavior the function between .
3. Table set and Table
To study a function numerically a student will often build a table. To build a table automatically the student must tell the calculator the minimum value of x in the table and the increments between the x values.
Once these values are selected, and the student selects TABLE, the calculator can build an automatic table.
If a student wants to build a table with x values which are not equally spaced they may elect to build a table by ask mode. When this mode is selected, and the student selects TABLE the student will input the x values and the calculator will determine the corresponding y-values for the selected x-values.
4. Y= Key
When a student enters an equation in one of the function slots, the calculator will graph the equation in the selected window. Very often a student forgets that an equation has been entered in one of the slots when working on a graph in a statistical mode. These equations can be turned off so they don't graph at an inappropriate time.
5. Stat Plots
From the Y= screen it is possible to tell if a statistical plot is turned on or off. Always check the top of the screen to see if any of the STAT Plots are turned on before graphing an equation.
6. Returning to the Home Screen
Students should know that 2nd Quit will always return them to the HOME screen for entering numerical calculations and commands.
7. Entering numerical expressions
In most cases the graphing calculator usually follows the order of operations learned by most students as part of a pre-algebra course. But there are a few exceptions. Students should practice entering some of the following expressions to see whether they get the correct values. (Answer have been rounded off to three decimal places of accuracy.)
Expression Answer -2 ~1.754 -32 ~.374 ~-0.180
Check Your Understanding: Build an appropriate window to view each of the following graphs. Be prepared to explain why you have selected each window. In some questions a dot mode should be selected, while in other cases the connected mode should be selected. The graph should show the behavior of the function for the given domain.
1. Graph the function for the real numbers -20 < x < 20. After graphing the function in the given domain, find the value of y(100).
2. Graph the rational function for the real numbers -0.5 < x < 4. After graphing the function in the given domain, find the value of y(4)-y(2).
3. Graph the function for the real numbers 1 < x < 10. After graphing the function in the given domain, find the value of
4. First graph the function for the real numbers 1 < x < 5. Then set up a table to find the values of for the values x = 2.4, 2.5, 2.6, 2.7, 2.8, 2.9 and 3.
Rev. 03/21/16 | 1,340 | 5,985 | {"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-2019-09 | latest | en | 0.938789 |
https://physics.stackexchange.com/questions/199366/what-does-general-relativity-predict-for-spacetime-without-energy | 1,653,252,749,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00772.warc.gz | 513,098,152 | 69,351 | # What does general relativity predict for spacetime without energy?
I am kind of new to GR but I have been familiar with the concepts for a long time, I am getting used to the mathematics just now. My question is, what would GR predict if we would have an empty universe? No energy whatsoever, only a spacetime continuum?
What I mean by energy is any type of energy, including mass, so a 0 curvature universe.
• That's Minkowski space! Either that or any other Ricci-flat spacetime, such as spacetimes with gravitational waves. Aug 8, 2015 at 20:17
• So you want to know what $G_{\mu\nu}=0$ (or equivalently $T_{\mu\nu}=0$) represents? Aug 8, 2015 at 20:19
• Yes, 0 curvature Aug 8, 2015 at 22:00
• @Darawan 0 Riemann curvature? The unique solution to $\mathrm{Riem}[g]=0$ is the Minkowski metric. 0 Ricci curvature? See Wiki. (Note that because the metric is nondegenerate, the scalar curvature can only vanish iff the Ricci curvature vanishes.) Aug 9, 2015 at 3:46
• @Slereah OP is not asking for a vanishing energy momentum tensor, but rather for a spacetime without any kind of energy at all. It is possible for a vaccum solution to have nonzero energy, for instance the ADM energy of Schwarzschild spacetime is just the mass of the body. Aug 9, 2015 at 3:52
Solutions to $G_{\mu\nu}=0$ are called vacuum solutions in GR, it follows mathematically that this happens if and only if the Ricci tensor vanishes, i.e. the solutions are exactly the Ricci flat Lorentzian manifolds. In most known explicit examples only some region is Ricci flat (e.g. around Schwarzschild or Kerr black holes), but some global vacuum solutions without singularities are known. Their existence contradicts the strong Mach principle, which implies that universe without matter (singularities are interpreted as degenerated matter) must be the flat Minkowski spacetime. One example is the Ozsváth–Schücking vacuum, describing a sinusoidal gravitational wave, others are given by a family of Kasner vacua, describing quaint expanding universes without matter. In Kasner universes the expansion can never be isotropic, in fact if the volume overall is expanding with time at least one spatial direction will be contracting.
Keep in mind however that "only a spacetime continuum" (zero stress-energy and hence Einstein tensor) does not imply "no energy whatsoever", because gravitational field itself can do work, and therefore carries energy. For instance, Ozsváth–Schücking waves transport energy just as electromagnetic waves would. "No energy whatsoever" means that not only $G_{\mu\nu}=0$ but even $\partial_{\alpha}g_{\mu\nu}=0$, i.e. the spacetime is locally flat Minkowski. Even those can be peculiar though, for instance the locally flat Deutsch–Politzer space contains closed timelike curves ("time machines").
• Well doesn't this give a clue to what dark energy and dark matter is then? Thank you for you answer, I think I've got it. Aug 9, 2015 at 10:42
• $g_{\mu\nu}=0$ is not Minkowski space. Aug 9, 2015 at 11:56
• @0celo7 Sorry, forgot the $\nabla$. Aug 10, 2015 at 19:19
• Then what's $\nabla$? If it's the covariant derivative as usual, then $\nabla g_{\mu\nu} = 0$ in any spacetime whatsoever in GTR because the connection is Levi-Civita. Maybe it would work better in terms of vanishing Weyl curvature as the additional condition, $C_{mu\nu\sigma\rho} = 0$, because the Weyl curvature is part of the Ricci curvature not determined by the stress-energy of sources. Aug 10, 2015 at 19:42
• @Conifold: the clear way of writing this would be $\partial_{a}g_{bc} =0$ Aug 10, 2015 at 23:23
OP is looking for the vacuum solutions to the Einstein field equations. Including only the cosmological constant $\Lambda$, the EFE become the Lambdavacuum field equations, $$R_{\mu\nu}=\left(\frac12R-\Lambda\right)g_{\mu\nu}\tag{1}$$ with $R_{\mu\nu}$ the Ricci curvature tensor and $R$ the Ricci scalar. Solutions to this depend on the sign/value of $\Lambda$.
For $\Lambda\neq0$, the spacetime must be treated as curved, as (1) does not admit flat spacetime solutions (cf. Padmanabhan (2003), pdf). Solutions to (1) result in either de Sitter space (for $\Lambda>0$) or anti-de Sitter space (for $\Lambda<0$). See also this NED article (also by Padmanabhan) and this Physics.SE post.
For $\Lambda=0$, the vacuum field equations can be solved with flat Minkowski space, Schwarzschild space or Kerr space (provided we are looking at the space outside some sphere of non-zero radius). Solutions of this case would be singularity free, Ricci flat but not necessarily Riemann flat. See also this Physics.SE post.
The Einstein Field equations are $G_{\mu\nu}=8\pi GT_{\mu\nu}$.
An empty universe would be one where $T_{\mu\nu}=0$ The Einstein field equations would than read
$G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=0$.
The 00 term of this (for an FLRW metric) is
$\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{\kappa}{a^2}$.
You say that you want the density of the universe to be 0, so $\rho=0$. Setting the curvature to be flat, positive, or negative ($\kappa = 0/+1/-1$, respectively), gives you three differential equations to solve for the scale factor as a function of time ($a(t)=...$). | 1,426 | 5,205 | {"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} | 2.96875 | 3 | CC-MAIN-2022-21 | latest | en | 0.892295 |
https://allpaper.org/2020/05/12/isds351-julia-juice-bar-interest-rate-and-amount-of-loan-paper/ | 1,603,646,392,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00324.warc.gz | 206,622,356 | 11,421 | Chat with us, powered by LiveChat ISDS351 Julia Juice Bar Interest Rate and Amount of Loan Paper | All Paper
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Assignment 3M.S. Excel 2013 or 2016 Julia’s Juice Bar Learning Objectives Goal SeekWhat-if sensitivity analysisCharting Use my instructional video Lesson 4ABC to complete the following: A worksheet Lastnamefirstname_A4.xls has been started for you and you need to complete it based on the following assumptions for a company called Julia’s Juice Bar. Julia’s juice bar is trying to trim down their annual maintenance expenses to 10% of the total expenses. Use Goal Seek to calculate the new amount.Julia would like to have a pie chart that would show the summary of annual percentage of expenses. Create a 3D pie chart with the given data.Julia is considering expanding her business and is planning on taking a loan for \$425,000.00 and is willing to pay no more than \$9000 per month as monthly payment. Further, she wishes to repay the loan in 4 years and is therefore willing to lower her budget if need be. You are to help her come up with the right amount to borrow. What is your recommendation. Indicate this in the space provided in the spreadsheet. Create a pdf of the entire workbook and upload it in the link provided by the due date posted. Take the following options to create the pdf version of your file. Go to backstage view, export, create pdf, options, in publish what section in the dialog box, select entire work book, then select ok and publish. Good Luck!
instructions_for_assignment_3.docx
lastnamefirstname_a3.xlsx
Unformatted Attachment Preview
1
Prof. M. Krishnamurthi
ISDS 351.
Assignment 3 M.S. Excel 2013 or 2016
Julia’s Juice Bar
Learning Objectives
1. Goal Seek
2. What-if sensitivity analysis
3. Charting
Use my instructional video Lesson 4ABC to complete the following:
A worksheet Lastnamefirstname_A4.xls has been started for you and you need to
complete it based on the following assumptions for a company called Julia’s Juice
Bar.
1. Julia’s juice bar is trying to trim down their annual maintenance expenses
to 10% of the total expenses. Use Goal Seek to calculate the new amount.
2. Julia would like to have a pie chart that would show the summary of annual
percentage of expenses. Create a 3D pie chart with the given data.
3. Julia is considering expanding her business and is planning on taking a loan
for \$425,000.00 and is willing to pay no more than \$9000 per month as
monthly payment.
4. Further, she wishes to repay the loan in 4 years and is therefore willing to
lower her budget if need be. You are to help her come up with the right
amount to borrow. What is your recommendation. Indicate this in the
Create a pdf of the entire workbook and upload it in the link provided by the
due date posted.
Go to backstage view, export, create pdf, options, in publish what section in the
dialog box, select entire work book, then select ok and publish.
Good Luck!
Julia’s Juice Bar
Projected Decrease in Expenses
Annual Total
Employee Salaries
Equipment
Produce
Marketing
Maintenance
Totals
\$
\$
\$
\$
\$
2,678,512
2,904,123
1,345,689
3,456,123
3,412,902
Goal: Decrease Maintenance to 10%
Goal Amount
Percent of
Total
Costa Mesa Store Loan Options
Amount of Loan
Period (years)
Interest rate (per year)
Payment (per month)
\$425,000
4.00
4%
Option #1 Reduce Loan Amount
Amount of Loan
Period (years)
Interest rate (per year)
Payment (per month)
Option #2 Increase Number of Years
Amount of Loan
Period (years)
Interest rate (per year)
Payment (per month) | 863 | 3,563 | {"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-2020-45 | latest | en | 0.921142 |
http://mathhelpforum.com/calculus/13088-line-integrals-print.html | 1,527,011,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864837.40/warc/CC-MAIN-20180522170703-20180522190703-00508.warc.gz | 185,548,804 | 3,593 | # Line integrals
• Mar 28th 2007, 05:23 PM
pakman
Line integrals
F, i, j, r, q, p are vectors
F(x,y) = (2xy)i - (x^2)j
div F(x,y) = 2y
curl F(x,y) = -4x which is not equal to zero, allowing no potential.
Now this is the confusing part. Work = 25/3
Here is how they got it
Row (or atleast I think that's what it is) is parameterized by r(t) = p + t(q-p)
x(t) = 1+t
y(t) = 3+t
p=<1,3>
q=<2,4>
0 <= t <= 1
INT(F dot dr) = INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
u = 1+t
du = dt
then on from there to get 25/3
My question is... what is all of that stuff for Work? I don't understand where the r(t) = p + t(q-p) comes from... same goes for all the values of x(t), y(t) etc...
Someone help please! I had to postpone my finals due to military training and now I've forgotten so much.
• Mar 28th 2007, 06:38 PM
ThePerfectHacker
Quote:
Originally Posted by pakman
F, i, j, r, q, p are vectors
F(x,y) = (2xy)i - (x^2)j
div F(x,y) = 2y
curl F(x,y) = -4x which is not equal to zero, allowing no potential.
Now this is the confusing part. Work = 25/3
.
Since it is not path independent, you need to specify what path are you intergrating?
• Mar 29th 2007, 05:24 AM
topsquark
Quote:
Originally Posted by ThePerfectHacker
Since it is not path independent, you need to specify what path are you intergrating?
I don't really understand the parametrization part, but isn't he specifying a path there? (It's just a guess, it's too confusing for me to follow.)
-Dan
• Mar 29th 2007, 06:55 AM
qpmathelp
Quote:
Originally Posted by pakman
F, i, j, r, q, p are vectors
F(x,y) = (2xy)i - (x^2)j
div F(x,y) = 2y
curl F(x,y) = -4x which is not equal to zero, allowing no potential.
Now this is the confusing part. Work = 25/3
Here is how they got it
Row (or atleast I think that's what it is) is parameterized by r(t) = p + t(q-p)
x(t) = 1+t
y(t) = 3+t
p=<1,3>
q=<2,4>
0 <= t <= 1
INT(F dot dr) = INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
u = 1+t
du = dt
then on from there to get 25/3
My question is... what is all of that stuff for Work? I don't understand where the r(t) = p + t(q-p) comes from... same goes for all the values of x(t), y(t) etc...
Someone help please! I had to postpone my finals due to military training and now I've forgotten so much.
INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
just expand and integrate term by term you do not need a substitution
if you are asking about r(t) = p + t(q-p), i think it is the equation of a line passing through points with position vector p and q where t is a parameter
you might have been asked to evaluate the work done in moving from p=<1,3> to q=<2,4>
substituting into r(t) = p + t(q-p) and simplifying,
you get r(t) =<1+t , 3+t>
if r(t) = <x(t),y(t)>
<x(t),y(t)>=<1+t , 3+t> will give you x(t) = 1+t y(t) = 3+t
work done is line integral of F.dr
replace the x and y in F(x,y) = (2xy)i - (x^2)j with x(t) = 1+t y(t) = 3+t
and take the dot product with r(t) =<1+t , 3+t>
to get INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
when t =0 and 1, you get p and q from r(t) =<1+t , 3+t> hence the limits
• Mar 29th 2007, 07:03 AM
ecMathGeek
Quote:
Originally Posted by qpmathelp | 1,126 | 3,131 | {"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} | 3.53125 | 4 | CC-MAIN-2018-22 | latest | en | 0.887838 |
https://www.mathworks.com/matlabcentral/cody/problems/2022-find-a-pythagorean-triple/solutions/710076 | 1,611,072,583,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00720.warc.gz | 878,346,018 | 18,723 | Cody
# Problem 2022. Find a Pythagorean triple
Solution 710076
Submitted on 3 Aug 2015 by Monica Roberts
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
%% a = 1; b = 2; c = 3; d = 4; flag_correct = false; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
2 Pass
%% a = 2; b = 3; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
3 Pass
%% a = 3; b = 4; c = 5; d = 6; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
4 Pass
%% a = 3; b = 4; c = 4.5; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
5 Pass
%% a = 3; b = 3.5; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
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Start Hunting! | 350 | 1,043 | {"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} | 3.1875 | 3 | CC-MAIN-2021-04 | latest | en | 0.584429 |
https://se.mathworks.com/matlabcentral/answers/710143-show-intersection-of-two-spheres?s_tid=prof_contriblnk | 1,623,586,127,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00548.warc.gz | 454,650,476 | 22,824 | # Show intersection of two spheres
17 views (last 30 days)
ercan duzgun on 6 Jan 2021
Edited: Adam Danz on 6 Jan 2021
How can I show the intersections of two spheres on MATLAB plot?
My code is this:
close all;clear all;clc;
pA1=[-2.9,-0.9,0]';
pA2=[-1.2,3.0,0]';
pB1=[1.3,-2.3,0]';
pB2=[-1.2,-3.7,0]';
pC1=[2.5,4.1,0]';
pC2=[3.2,1.0,0]';
patch([pA1(1),pA2(1),pC1(1),pC2(1),pB1(1),pB2(1)],[pA1(2),pA2(2),pC1(2),pC2(2),pB1(2),pB2(2)],0)
grid on; axis equal;hold on;
text(pA1(1),pA1(2),'A1');text(pA2(1),pA2(2),'A2')
text(pB1(1),pB1(2),'B1');text(pB2(1),pB2(2),'B2')
text(pC1(1),pC1(2),'C1');text(pC2(1),pC2(2),'C2')
L1=5.0;L2=4.5;L3=5.0;L4=5.5;L5=5.5;L6=5.7;
%kure ciziyor
[x y z] = sphere(128);x=L1*x;y=L1*y;z=L1*z;
h = surfl(x+pA1(1), y+pA1(2), z+pA1(3));
set(h, 'FaceAlpha', 0.1)
[x y z] = sphere(128);x=L2*x;y=L2*y;z=L2*z;
h = surfl(x+pA2(1), y+pA2(2), z+pA2(3));
set(h, 'FaceAlpha', 0.5)
patch([pA1(1),pA2(1),pC1(1),pC2(1),pB1(1),pB2(1)],[pA1(2),pA2(2),pC1(2),pC2(2),pB1(2),pB2(2)],0)
Adam Danz on 6 Jan 2021
> How can I show the intersections of two spheres on MATLAB plot?
This has been addressed in the forum several times:
R2019a
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https://www.sarthaks.com/2543425/the-ratio-of-radius-of-two-circles-is-1-4-what-will-be-the-ratio-of-the-areas-of-the-circle | 1,680,242,892,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00263.warc.gz | 1,078,609,028 | 15,012 | # The ratio of radius of two circles is 1 ∶ 4, what will be the ratio of the areas of the circle?
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The ratio of radius of two circles is 1 ∶ 4, what will be the ratio of the areas of the circle?
1. 1 ∶ 4
2. 1 ∶ 16
3. 16 ∶ 1
4. 4 ∶ 1
by (30.0k points)
selected
Correct Answer - Option 2 : 1 ∶ 16
Given:
Ratio of the radius of the circles = 1 ∶ 4
Formula used:
Area of the circle = πr2
Calculation:
Let the radius of the circles be x and 4x units
Area of circle1 = πx2
Area of circle2 = π(4x)2
Ratio of areas = πx2 ∶ π(4x)2
⇒ Ratio of area = x2 ∶ 16x2
∴ Ratio of the area of the two circles is 1 ∶ 16 | 257 | 641 | {"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} | 3.984375 | 4 | CC-MAIN-2023-14 | latest | en | 0.81696 |
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Topic: LOGIC & MATHEMATICS
Replies: 96 Last Post: Jun 6, 2013 5:19 AM
Messages: [ Previous | Next ]
Charlie-Boo Posts: 1,635 Registered: 2/27/06
Re: LOGIC & MATHEMATICS
Posted: May 28, 2013 2:41 PM
On May 26, 3:43 pm, Zuhair <zaljo...@gmail.com> wrote:
> On May 26, 5:03 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>
>
>
>
>
>
>
>
>
> > On 26/05/2013 7:54 AM, Peter Percival wrote:
>
> > > Nam Nguyen wrote:
>
> > >> Such application isn't purely logical. Finiteness might be a purely
> > >> logical concept but recursion isn't: it requires a _non-logical_
> > >> concept (that of the natural numbers).
>
> > > What do you mean when you write "Finiteness might be a purely logical
> > > concept"?
>
> > FOL requires the concept of finite strings of symbols, the concept of
> > there being only finitely many steps in making logically inferences;
> > and as such the word "finite" is a priori we have to accept we
> > understand its meaning and usage: we can't define that "finite" priori
> > further.
>
> FOL requires the concept of finite strings of symbols that are
> RECURSIVELY formed.
> The same is applicable here. The question of recursion being involved
> in those logical theories doesn't render them non logical!
>
> The theory that I displayed here is as logical as first order logical
> is.
>
> Zuhair
>
>
>
>
>
>
>
> > --
> > ----------------------------------------------------
> > There is no remainder in the mathematics of infinity.
>
> > NYOGEN SENZAKI
> > ----------------------------------------------------
Recursion is required only because the set of strings (and various
subsets of them) is infinite.
C-B
Date Subject Author
5/26/13 Zaljohar@gmail.com
5/26/13 namducnguyen
5/26/13 Zaljohar@gmail.com
5/26/13 namducnguyen
5/26/13 Peter Percival
5/26/13 namducnguyen
5/26/13 Peter Percival
5/26/13 namducnguyen
5/26/13 Zaljohar@gmail.com
5/28/13 Charlie-Boo
5/28/13 Charlie-Boo
5/26/13 Zaljohar@gmail.com
5/27/13 zuhair
5/27/13 fom
5/27/13 Zaljohar@gmail.com
5/27/13 fom
5/28/13 namducnguyen
5/28/13 Zaljohar@gmail.com
5/28/13 namducnguyen
5/29/13 Peter Percival
5/30/13 namducnguyen
5/30/13 Peter Percival
5/30/13 Peter Percival
5/30/13 namducnguyen
5/31/13 Peter Percival
5/30/13 Bill Taylor
5/30/13 Peter Percival
5/30/13 Zaljohar@gmail.com
5/30/13 Zaljohar@gmail.com
5/30/13 namducnguyen
5/31/13 Peter Percival
5/31/13 Zaljohar@gmail.com
5/31/13 LudovicoVan
5/31/13 fom
5/28/13 Peter Percival
5/28/13 namducnguyen
5/27/13 Charlie-Boo
5/27/13 fom
5/28/13 Charlie-Boo
5/28/13 fom
6/4/13 Charlie-Boo
6/4/13 fom
6/5/13 Zaljohar@gmail.com
5/28/13 Zaljohar@gmail.com
5/28/13 LudovicoVan
5/28/13 ross.finlayson@gmail.com
5/28/13 LudovicoVan
5/28/13 LudovicoVan
5/28/13 fom
5/29/13 LudovicoVan
5/29/13 fom
5/30/13 LudovicoVan
5/29/13 fom
5/30/13 LudovicoVan
5/30/13 fom
5/31/13 LudovicoVan
5/31/13 Zaljohar@gmail.com
5/31/13 LudovicoVan
5/31/13 ross.finlayson@gmail.com
6/1/13 LudovicoVan
6/1/13 namducnguyen
6/1/13 ross.finlayson@gmail.com
6/2/13 LudovicoVan
6/2/13 ross.finlayson@gmail.com
6/3/13 Shmuel (Seymour J.) Metz
6/3/13 ross.finlayson@gmail.com
6/4/13 LudovicoVan
6/4/13 namducnguyen
6/4/13 Peter Percival
6/5/13 Shmuel (Seymour J.) Metz
6/5/13 fom
6/6/13 Peter Percival
5/31/13 fom
6/1/13 LudovicoVan
6/1/13 fom
6/2/13 ross.finlayson@gmail.com
6/2/13 fom
6/2/13 Herman Rubin
6/2/13 fom
6/2/13 LudovicoVan
6/3/13 Herman Rubin
6/3/13 Peter Percival
6/4/13 Herman Rubin
6/4/13 Peter Percival
6/4/13 Peter Percival
6/1/13 fom
6/1/13 LudovicoVan
6/1/13 namducnguyen
6/5/13 Peter Percival
6/1/13 fom
6/2/13 LudovicoVan
6/2/13 fom
5/28/13 Zaljohar@gmail.com
5/28/13 Charlie-Boo
5/27/13 Zaljohar@gmail.com
5/28/13 Charlie-Boo
5/30/13 Zaljohar@gmail.com | 1,503 | 3,936 | {"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-2018-13 | longest | en | 0.878735 |
https://math.stackexchange.com/questions/3099913/inequality-with-absolute-values-xy-1xy-x-1x-y-1y?noredirect=1 | 1,568,956,293,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573832.23/warc/CC-MAIN-20190920050858-20190920072858-00098.warc.gz | 555,736,729 | 27,616 | # Inequality with absolute values |x+y|/(1+|x+y|) <= |x|/(1+|x|) +|y|/(1+|y|) [duplicate]
$$\frac{|x+y|}{1+|x+y|}\leq \frac{|x|}{1+|x|} +\frac{|y|}{1+|y|}$$
How can i solve this inequality? I have solved it in a long way but i guess there should be an easier way
## marked as duplicate by Arthur, Martin R, Community♦Feb 4 at 14:52
The easy way is to get rid of the absolute values by considering the cases $$x\geq 0$$ and $$x<0$$ for $$|x|$$, $$y\geq 0$$ and $$y<0$$ for $$|y|$$, and $$x+y\geq 0$$ and $$x+y<0$$ for $$|x+y|$$.
Hence, you get $$8$$ cases. | 227 | 559 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-39 | latest | en | 0.760229 |
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A117586 Coefficients of q in series expansion of Zagier's identity. 0
0, -1, -2, -1, -1, 2, 0, 4, 1, 2, 1, 2, -4, 1, -1, -5, -2, -1, -3, -1, -2, -2, 5, 0, -1, 1, 8, 0, 3, 2, 2, 2, 3, 0, 4, -7, 0, 0, 2, -3, -8, -2, -1, -3, -2, -4, 0, -3, -3, -2, -1, 7, -1, 0, 1, -1, 0, 12, 2, 2, 0, 4, 3, 4, 0, 2, 4, 3, 0, 5, -12, 2, 0, 1, -1, 1, -3, -11, -1, -2, -6, 2, -4, -3, -3, -4, -2, 1, -5, -3, -3, -2, 11, 2, -2, -3, 2, 0, 0, 3, 12, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS Robin Chapman, Franklin's argument proves an identity of Zagier, Electron. J. Combin. 7 #R54 (2000). Eric Weisstein's World of Mathematics, Zagier's Identity FORMULA Negative of sequence is convolution of A010815 with A046746. - Michael Somos, Jan 07 2015 a(n) = A067661(n) - A067659(n) [Chapman]. - George Beck, May 06 2017 EXAMPLE G.f. = - x - 2*x^2 - x^3 - x^4 + 2*x^5 + 4*x^7 + x^8 + 2*x^9 + x^10 + ... MATHEMATICA Flatten[{0, CoefficientList[Series[-Sum[x^(n - 1)*(QPochhammer[x^(n + 1), x]^2/QPochhammer[x^(n), x]), {n, 1, 101}], {x, 0, 100}], x]}] (* Mats Granvik, Jan 05 2015 *) a[ n_] := SeriesCoefficient[ Sum[ QPochhammer[ x] - QPochhammer[ x, x, k], {k, 0, n}], {x, 0, n}]; (* Michael Somos, Jan 07 2015 *) a[ n_] := SeriesCoefficient[ -Sum[ QPochhammer[ x^k, x] x^k / (1 - x^k)^2, {k, n}], {x, 0, n}]; (* Michael Somos, Jan 07 2015 *) CROSSREFS Cf. A046746. Sequence in context: A123223 A088226 A244658 * A307988 A268917 A176811 Adjacent sequences: A117583 A117584 A117585 * A117587 A117588 A117589 KEYWORD sign AUTHOR Eric W. Weisstein, Mar 29 2006 STATUS approved
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Last modified March 30 07:43 EDT 2020. Contains 333119 sequences. (Running on oeis4.) | 897 | 2,100 | {"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} | 3.390625 | 3 | CC-MAIN-2020-16 | latest | en | 0.464639 |
https://study.com/academy/lesson/writing-solving-addition-word-problems-with-one-variable.html | 1,576,398,441,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307797.77/warc/CC-MAIN-20191215070636-20191215094636-00099.warc.gz | 541,075,078 | 43,140 | # Writing & Solving Addition Word Problems with One Variable
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• 0:01 Addition Word Problem
• 0:38 Writing Your Algebraic…
• 2:13 Solving Your Problem
• 3:17 Example
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After watching this video lesson, you will be able to write and solve addition word problems. Learn how to read your word problem, write out the algebraic equation, and then solve it for the unknown variable.
## Addition Word Problem
In this video lesson, you will learn how to take a word problem that you are given, understand it, write out the algebraic equation, and then solve it to find your answer. In this lesson, we focus specifically on addition word problems. Word problems are math problems given in words. Usually, they best describe a real-world scenario that you can picture in your head, and they will ask you to find an unknown number. For example, you might see the following: 'Suzie and Jenny together have \$50 to go out for a nice dinner together. If Suzie put in \$23, how much did Jenny put in?'
## Writing Your Algebraic Equation
In order to solve this problem, you first need to turn this problem into an algebraic equation. An algebraic equation is a mathematical statement using numbers, variables, and symbols that include an equals sign. Yes, you need to write an algebraic equation for the given information. To do this, you need to fully understand the problem. Try to visualize what is going on in the problem. Once you fully understand what is going on and what is being asked, then you can go ahead and write out your algebraic equation, putting in as much of the given information as possible.
Reading through the problem, you see that the problem tells you that the combined amount from both Suzie and Jenny is \$50. What does it mean to be combined together? It means that there are things that have been added together. In this case, you're adding money, one amount from Suzie and another amount from Jenny.
What else does the problem say? It says that Suzie's part is \$23. It asks about Jenny's part. So, Jenny's part is the unknown number that you need to find, or the variable. Now you have all the information you need to write your algebraic equation. You use an x for Jenny's portion. You know you are supposed to add Suzie's and Jenny's portions together and that their total equals \$50, so you write 23 + x = 50. Looking at what you just wrote, you see that it expresses the same thing the problem does. You are adding Suzie's and Jenny's portions together, and the total should equal \$50.
## Solving Your Problem
Now that you have your algebraic equation, you can go ahead and solve it. You want to isolate the variable, the x in this case, so that it is by itself. To do this, you look to see what numbers are currently attached to the variable. Then you perform the inverse operation to both sides of the equation to detach the numbers from the variable. This inverse operation must be performed on both sides of the equation to get a correct answer. Because your problem is an addition problem, the inverse operation here is subtraction.
So, you subtract whatever is being added to the variable from both sides of the equation. Then you evaluate the problem to get your answer. You see that right now you have a 23 being added to the variable. This means that you need to subtract 23 from both sides. 23 + x - 23 = 50 - 23. Evaluating this, you get x = 27. Your answer is 27. Our variable represents Jenny's part, so you know that Jenny's contribution to the nice dinner fund is \$27.
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Support | 1,152 | 5,198 | {"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} | 4.125 | 4 | CC-MAIN-2019-51 | longest | en | 0.910399 |
http://www.enworld.org/forum/showthread.php?51163-Math-question-Finding-the-Size-of-a-shadow | 1,524,632,372,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947693.49/warc/CC-MAIN-20180425041916-20180425061916-00059.warc.gz | 390,698,243 | 31,051 | [Math question] Finding the Size of a shadow
1. [Math question] Finding the Size of a shadow
I was wondering if anyone can help me. I need to figure out a way to find the size a shadow that a person or creature will cast in relation to a light source. At first I thought it would be similair to the equation for light intensity. I= K/d (squared) but I know that it is wrong.
2. This is a trigonometry problem. At noon (light directly above), you cast no shadow (more or less). At dawn or dusk (light on the horizon), you cast an infinitely long shadow (on the floor, pretending it's a perfectly flat plane). At halfway between dawn and noon or noon and dusk, you cast a shadow equal to your height.
Calculate the tangent of the light-source's angle with respect to the shadow-surface, and multiply that by the subject's height.
Calculate the tangent of the light-source's angle with respect to the shadow-surface, and multiply that by the subject's height.
I'm may be wrong, but wouldn't it be the inverse tangent of the light source's angle?
4. 1. What do you mean by 'size'? Length? Area?
2. What sort of surface is the shadow cast on? Flat? Angular? What angle is it in relation to the caster?
3. What sort of light source? Diffuse? Point? Where is it?
You need to give us the information you're starting with, and the result you need, or else we're going to have to fully spec out a raytracer for you.
5. What do I mean by size, lenght, area. Well that is always going to change. I was reading the spell Shadow Strike from Relics and rituals and it says that you cast the spell and turn the subjects shadow into a conduit for dealing damage. I could find no rules governing the size of the subject shadow. So wanting to take full advantage of the spell, and of course not wanting to get the smake down each time I use it, I need to find out how to figure out the lenght of shadow a person or creature casts.
6. The Size of a Shadow?
Monster Manual says "Medium".
-Hyp.
7. Technically, light goes on forever, in a straight line. So, without any further obstruction, a lateral light-source illuminating an object would cast a shadow reaching to the horizon, until the curvature of the planet interferes with its (projected) path.
That's a totally useless answer from a gaming perspective, though, so just pick a number
8. Originally posted by Dareoon Dalandrove
What do I mean by size, lenght, area. Well that is always going to change. I was reading the spell Shadow Strike from Relics and rituals and it says that you cast the spell and turn the subjects shadow into a conduit for dealing damage. I could find no rules governing the size of the subject shadow. So wanting to take full advantage of the spell, and of course not wanting to get the smake down each time I use it, I need to find out how to figure out the lenght of shadow a person or creature casts.
For Simplicity sake, I wold use the following:
Late Morning/Early Afternoon Shadow occoupies 5 ft square
Early Morning/Late Afternoon - Shadow occupies 10 ft square
Now if you want to get complicated:
Shadow length = heighth of object*SIN(Angle of Sun off of Horizon)
9. Completely disregarding longitude, latitude and time of day, as well as all other geometry, as a DM I would go with a simple factor based length of shadow.
Length = Height x (0.25 per hour before/past noon)
For example, 6 foot character stands around at 10 AM.
I would only include 6 AM to 6 PM, so the maximum factor would be 1.5xheight. Also if you are farther north of the equator you could add to the equation a bit:
Length = Height x (0.25 + 0.25 per hour before/past noon)
So at noon that same 6 foot guy now has a shadow of 1.5 feet at noon, and 9 feet at 6 AM/PM.
And in the extreme north the shadows would be even longer so have fun telling your DM that your shadow extends to the horizon at dawn and dusk. Your DM might say "it most certainly does, however it is so diffuse and weak that you can only use the first X feet of length." Or at least that's what I might say.
If you have another light source, say indoors, and you want to find your shadow on a wall, I would use some simple geometry to get a rough idea. The shadow will change size dependant on distance of the light source from the character and distance of the light source from the wall or floor. Have fun doing the math.
10. The tip of the shadow will form a right triangle with the light source.
Taking that fact into account, the object's highest point will also form a right triangle with the tip of it's shadow.
It's all relative to the angle of the light object from the object's heighst point. | 1,084 | 4,645 | {"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} | 3.515625 | 4 | CC-MAIN-2018-17 | latest | en | 0.941749 |
https://www.jiskha.com/questions/93012/22-mL-of-0-37-mol-per-litre-acetic-acid-is-titrated-by-way-of-a-standardized-0-29 | 1,556,139,392,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578656640.56/warc/CC-MAIN-20190424194348-20190424220348-00471.warc.gz | 737,077,169 | 5,126 | # chemistry
22 mL of 0.37 mol per litre acetic acid is titrated by way of a standardized 0.29 mol/L KOH solution. Calculate the pH of the solution after roughly 18 mL of the solution of KOH is added. The Ka of acetic acid proves to be 1.8 x 10^-5
1. 👍 0
2. 👎 0
3. 👁 52
1. The equation is
CH3COOH + NaOH ==> CH3COONa + HOH
mol CH3COOH initially = M x L = ??
mol NaOH initially = M x L
Look to see how much has reacted, how much sodium acetate (the salt) has been formed, and how much CH3COOH remains unreacted. You will find that there is salt and an excess of CH3COOH and this constitutes a buffer solution (a weak acid and its conjugate base). Then use the Henderson-Hasselbalch equation and calculate pH. Post your work if you get stuck.
1. 👍 0
2. 👎 0
posted by DrBob222
2. dear DR Bod
I had the same problem and worked it out as follows:
the equation: CH3COOH + NaOH ==> CH3COONa + H20
MOL CH3COOH = M x L
MOL NaON = M x L
now, to find how much sodium acetate formed, do i have to add the moles of NaOH and CH3COOH together? and how do i find the amount of CH3COOH that has not reacted?
1. 👍 0
2. 👎 0
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# How many degrees are there in a 12 sided polygon?
Updated: 12/20/2022
Wiki User
14y ago
To calculate this, you must find out how many triangles you can draw into the shape. Do this by taking the total number of sides (s) and taking away 2. This leaves you with how many triangles you can draw in the shape (t) from one vertex to another. (Not including outside lines!). Then, multiply the number of triangles in the shape by 180. This leaves you with how many degrees are inside that shape. (a)
t = s-2
a = t X 180
This means there are 1800 degrees in a 12 sided polygon.
Wiki User
14y ago
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Related questions
### How many degrees are in a 12 sided polygon?
1800 degrees (that is the sum of the internal angles of the 12-sided polygon).
1,440 Degrees
### What is each interior angle of a 12 sided polygon?
Each interior angle in 12-sided polygon is 150 degrees.
### What is the size of each interior angle of a 12 sided polygon?
If it is a regular 12 sided polygon then each interior angle is 150 degrees.
### How many degrees each side in a 12 sided polygon?
Degrees are used to measure angles, not the lengths of sides.
### What is the sum of the measure of the interior angles of a 12-sided polygon?
The sum of the interior angles of a 12 sided polygon add up to 1800 degrees.
### What is the measure of an exterior angle in a 12 sided polygon?
Providing that it is a regular 12 sided polygon then each exterior angle measures 30 degrees
### What is the sum of the measure of the interior angles of a 12 Sided polygon?
The interior angles of a 12 sided polygon add up to: (12-2)*180 = 1800 degrees
### In the exterior side how many degrees are there in a regular 12 sided polygon?
Each exterior angle of a regular 12-gon measures 30 degrees.
### What is the name the 12 sided polygon?
The name of a 12 sided polygon is Dodecagon.
### What is the interior angle of a 12- sided shape?
The total interior angle of a 12 sided polygon is = (12 - 2) 180 degrees = 1800 degrees. The internal angle at each vertex of an regular dodecagon is equal to = = 150 degrees. While the exterior angle at a single vertex is = = 30 degrees. The total exterior angle of a 12 sided polygon is 360 degrees.
### How many sides does a polygon have if one of the interior angles is 30 degrees?
No such regular polygon exist so it has to be some kind of irregular polygon but if each exterior angle was 30 degrees then it would be a regular 12 sided polygon | 635 | 2,564 | {"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} | 4.21875 | 4 | CC-MAIN-2024-38 | latest | en | 0.925152 |
http://xn--e1ajelebdlch4i0a.xn--p1ai/how-to-trade/trade-fred-binary-option.php | 1,606,924,801,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00109.warc.gz | 170,344,664 | 5,782 | Now that we have a basic idea on how binary option trades work, let’s take a look at a simple example.
Let’s say, you decide to trade EUR/USD with the assumption that price will rise.
The pair’s current price is 1.3000, and you believe that after one hour, EUR/USD will be higher than that level.
You then look at your trading platform and see that the broker’s payout is 79% on a one hour option contract with a target strike of 1.3000.
After much deliberation, you finally decide to buy a “call” (or “up”) option and risk a \$100.00 premium.
You could say it’s similar to going “long” on EUR/USD on the spot forex market.
Ending Scenarios After Entering a CALL OptionGain/Loss
Expiry price is above the strike price
(in-the-money)
\$100.00 x 79% = \$79
\$100.00 + \$79.00 = \$179.00
You gain \$179.00 on your account.
Expiry price is equal to or below the strike price
(out-of-the-money)
As you can see from the calculations above, the risk you take is limited to the premium paid on the option.
You cannot lose more than your stake.
Unlike in spot forex trading, where your losses can get bigger the further the trade goes against you (which is why using stops are crucial), the risk in binary options trading is absolutely limited.
Payouts in Binary Options
Now that we’ve looked at the mechanics of a simple binary trade, we think it’s high time for you to learn how payouts are calculated.
More often than not, the payout will be determined by the size of your capital at risk per trade, whether you’re in- or out-of-the-money when the trade is closed, the type of option trade, and your broker’s commission rate.
In the example given above, you bet \$100 that EUR/USD will close above 1.3000 after an hour with your broker offering a 79% payout rate.
Let’s say that your analysis was spot on and your trade ends up being in-the-money. You would then get a payout of \$179.
\$100 (your initial investment) + \$79 (79% of your initial capital) = \$179
Easy peasy, right?
50\$ to 7 000\$ in 10 minutes - Binary Options Latest Strategy 2019
Don’t get too excited just yet! You should know that there’s no one-size-fits-all formula for calculating payouts.
There are a few other factors that affect them.
Factors in Payout Calculations
Each broker has its own payout rate.
For starters, Forex Ninja’s intel shows that most brokers offer somewhere between 70% and 75% for the most basic option plays while there are those who offer as low at 65%.
Various factors come into play when determining the percentage payout.
The underlying asset traded and the time to expiration are a couple of big components to the equation.
Normally, a market that is relatively less volatile and an expiration time that is longer usually means a lower percentage payout.
Next, the broker’s “commission” is also factored into the payout rate.
After all, brokers are providing a service for you, the trader, to play out your ideas in the market so they should be compensated for it.
The commission rate does vary widely among brokers, but since there are so many binary options brokers out there (and more coming along), the rates should become increasingly competitive over time.
When a Binary Option Trade is Closed
As mentioned before, binary options are typically “all-or-nothing” trading instruments in that the payout or loss is only given at contract expiration, but there are a few brokers that allow you to close a binary option trade ahead of expiration.
This usually depends on the type of option, and usually it’s only available within a certain timeframe (e.g., available 5 minutes after an option trade opens, up until 5 minutes before an option expiration).
The trade-off for this flexible feature is that brokers who do allow early trade closure tend to have lower payout rates.
When trading with a binary option broker that allows early closure of an option trade, the value of the option tends to move along with the value of the underlying asset.
For example, with a “put” (or “down”) option play, the value of the option contract increases as the market moves below the target (strike) price.
This means that, depending on how far it has moved passed the strike, the closing value of the option may be more than the risk premium paid (but never greater than the agreed maximum payout).
Conversely, if the underlying market moved higher, further out-of-the-money, the value of the option contract decreases and the option buyer would be returned much less than the premium paid if he/she closed early.
Of course, in both cases, the broker commission is factored into the payout of an option trade when closed early.
So before you decide to jump head first into trading binary options, make sure you do your research and find out what your broker’s payout rates and conditions are! | 1,060 | 4,817 | {"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 | latest | en | 0.923063 |
https://teswesm.com/mchoice/from-a-point-p-on-a-level-ground-the-angle-of-elevation-of-the-top-of-a-tower-is-30-if-the-tower-is/13784 | 1,628,197,652,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00492.warc.gz | 547,920,877 | 8,784 | # Height & Distance Mcqs
Q. From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is :_________?
a. 149 m
b. 156 m
c. 173 m
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https://www.jobne.in/reasoning-ability-quiz-for-upcoming-exams-4/ | 1,701,727,894,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00880.warc.gz | 931,302,504 | 31,903 | Reasoning Ability Quiz for Upcoming Exams
Directions (1-5): Study the following information carefully and answer the given questions.
Eight boxes P, Q, R, S, T, U, V and W are placed one above the another, but not necessarily in the same order. Two boxes are placed between U and T. U is placed above T. One box is placed between U and V. Three boxes are placed between P and W. P is placed immediately below U. Two boxes are placed between R and W. Q is placed above S.
Q1. Which of the following Box is placed at top?
(a) Q
(b) V
(c) U
(d) W
(e) None of these
Q2. How many boxes are placed between Box T and Box W?
(a) None
(b) Two
(c) One
(d) Four
(e) Three
Q3. Which of the following Box is placed at bottom?
(a) W
(b) R
(c) T
(d) S
(e) None of these
Directions (4-8): Study the following information and answer the following questions:
In a certain code language-
‘man play for improve’ is written as ‘fm lb pd ub’,
‘boy for help other’ is written as ‘ub ty qb gb’,
‘improve other by work’ is written as ‘fd pd nu ty ’,
‘boy see work done’ is written as ‘qb nu vf mb’,
Q4. What is the code for ‘see’ in the given code language?
(a) qb
(b) vf
(c) nu
(d) mb
(e) Cannot determined
Q5. Which among the following word is written as ‘gb’ in the given code language?
(a) work
(b) help
(c) man
(d) boy
(e) None of these
Q6. Which among the following word is written as ‘fd pd’ in the given code language?
(a) play improve
(b) man done
(c) by work
(d) improve by
(e) None of these
Q7.What is the possible code for ‘other time’ in the given code language?
(a) wc fd
(b) mb vf
(c) ty wc
(d) rb mb
(e) ty ub
Q8. Which among the following word is written as ‘fm’ in the given code language?
(a) man
(b) boy
(c) done
(d) play
(e) Either (a) or (d)
Q9. How many pairs of letters are there in the word “VALUATION” each of which have as many letters between them (both forward and backward directions) in the word as they have between them in the English alphabetical series?
(a) None
(b) Two
(c) Three
(d) More than three
(e) None of these
Q10. Pointing towards photograph a person said “This girl is daughter in law of my father”. That person has no brother. How is person related to that girl?
(a) Father
(b) Father in law
(c) Grandfather
(d) Son
(e) None of these
Directions (11-13): Study the information carefully and answer the questions given below.
Six persons M, N, O, P, Q, and S are different weight. P is heavier than N. M is lighter than S. Q is heavier than N. M is not lighter than P and Q. S is not heaviest, and Q is not second lightest. The weight of heaviest is 70kg.
Q11. Who among the following person is third heaviest?
(a) M
(b) S
(c) P
(d) Q
(e) Can’t be determined
Q12. Who among the following is the lightest person?
(a) M
(b) N
(c) Q
(d) P
(e) None of these
Q13. If weight of M is 67kg what may be the weight of S?
(a) 60kg
(b) 63kg
(c) 61kg
(d) 69kg
(e) None of these
Q14. If in the number 39475132, 2 is multiply to each of the digit which is less than 4 and 3 is subtracted from each of the digit which is more than 4 and equal to 4 then how many digits are repeating in the number thus formed?
(a) None
(b) Four
(c) One
(d) Three
(e) Two
Q15. If all the letters in the word CONSTITUTION are arranged in alphabetical order from left to right, then how many alphabets remains same in their position?
(a) Three
(b) One
(c) None
(d) Two
(e) four
.
.
.
.
.
.
Solutions | 1,005 | 3,408 | {"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} | 3.25 | 3 | CC-MAIN-2023-50 | longest | en | 0.935711 |
https://cpep.org/mathematics/2835106-marking-city-is-5-inches-away-from-jamming-city-on-the-map-what-is-the.html | 1,695,861,565,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00757.warc.gz | 205,226,869 | 7,402 | 15 September, 09:33
# Marking City is 5 inches away from Jamming City on the map. What is theactual distance between the two cities?2in : 35miles
+5
1. 15 September, 09:51
0
87.5 miles
Step-by-step explanation:
If 2 inches is 35 miles then 4 inches will be 70 inches
to get the last inch all you have to do is divide the 35 in have to get 17.5
70 + 17.5 = 87.5 | 123 | 366 | {"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} | 3.734375 | 4 | CC-MAIN-2023-40 | latest | en | 0.889099 |
http://www.barnesandnoble.com/w/bears-odd-bears-even-harriet-ziefert/1003248855?ean=9780140385397&itm=1&usri=andrea+baruffi | 1,462,010,406,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111809.10/warc/CC-MAIN-20160428161511-00214-ip-10-239-7-51.ec2.internal.warc.gz | 368,811,042 | 22,361 | Bears Odd, Bears Even
Editorial Reviews
Children's Literature - Ru Story-Huffman
This beginning to read book illustrates the math concepts of even and odd. Using short sentences, rhymes, and repetition, Ziefert has presented a good book which is both educational and fun. In addition to teaching about math, Bears ODD, Bears EVEN will encourage children to read on their own, as it utilizes words they may already know, plus new challenges. Each two-page spread features large text, plenty of white space around the words, and illustrations that provide some visual clues to the concept being introduced. As they read this text, children are asked questions and given the opportunity to reinforce the learning process. In addition to learning to distinguish between odd and even, children are presented with some basic numeral concepts. The final page features a section on more math fun in which children are encouraged to explore their world using numbers. Ziefert, the creator of Viking's 'Hello Reading' series, has scored another hit.
School Library Journal
PreS-Gr 2--This math book falls short of the mark. The text, which describes brown bears and polar bears engaged in various activities, has a distracting, uneven rhythm due to its constant shift between rhyme and prose. Although some of the rules for odd and even are pointed out (e.g., "Two, four, six, eight, ten--even numbers"), many questions go unanswered, such as "Count the bikes./Count the wheels./Which number is odd?/Which is even?" The illustrations serve the text in simple demonstrations of even and odd (e.g., four polar bears play on a giant number "4"; five brown bears paint on an oversized numeral "5"). However, the pictures do little to illuminate readers' understanding of the tougher concepts such as "Odd plus even--always odd!" These are pictured merely as equations the bears write on a chalkboard (e.g., 5+4=9). Stick with Stuart Murphy and G. Brian Karas's Give Me Half! (HarperCollins, 1996), Stuart Murphy and Lois Ehlert's A Pair of Socks (HarperCollins, 1996), and Pat Hutchins's The Doorbell Rang (Greenwillow, 1986) and wait for a better book that focuses on odd and even.--Gale W. Sherman, Pocatello Public Library, ID
Product Details
ISBN-13:
9780140385397
Publisher:
Penguin Publishing Group
Publication date:
08/28/1997
Series:
Pages:
32
Product dimensions:
6.00(w) x 9.00(h) x 0.10(d)
Lexile:
Age Range:
5 - 8 Years
Customer Reviews
Average Review:
Write a Review
and post it to your social network | 582 | 2,507 | {"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-2016-18 | latest | en | 0.941683 |
https://catonmat.net/ | 1,601,307,118,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00101.warc.gz | 310,196,947 | 31,219 | # Twentieth site in online tools network: onlineINTEGERtools.com
At Browserling we're building a network of online tools websites. Each site focuses on one and only one tool category. Each tool does one and only one thing.
Today we're releasing the twentieth site – Online Integer Tools.
Online Integer Tools offers a collection of simple, free, and easy to use utilities for crunching integers. There are no intrusive ads, popups, or other garbage, just integer utilities that work in your browser. Load integers and instantly get the result!
www.onlineINTEGERtools.com
Here's a list of all integer tools so far:
Here are the upcoming integer tools:
• Draw von Neumann Universe
• Let Zalgo Destroy Integers
• Create Integer n-Tuples
• Convert Integers to Unary Base
• Convert Unary Base to Integers
• Convert Integers to Binary Base
• Convert Binary Base to Integers
• Convert Integers to Octal Base
• Convert Octal Base to Integers
• Convert Integers to Hex Base
• Convert Hex Base to Integers
• Convert Integers to Any Base
• Convert Integers to Base64
• Convert Base64 to Integers
• Convert Integers to HTML Entities
• Convert HTML Entities to Integers
• Create Random Integers
• Create Palindromic Integers
• Check if an Integer is Symmetric
• Generate an Integer Matrix
• Rewrite an Integer as a Sum
• Rewrite an Integer as a Product
• Rotate Integers
• Shift Integer Digits
• Subtract Integers
• Multiply Integers
• AND Integers
• OR Integers
• XOR Integers
• NOT Integers
• NAND Integers
• NOR Integers
• XNOR Integers
• Find the Running Total
• Find the Running Product
• Convert an Integer to Ordinal
• Convert an Oridinal to Integer
• Calculate Integer Digit Sum
• Filter Integers
• Merge Integers
• Truncate Integers
• Right-align Integers
• Center Integers
• Take Absolute Value of Integers
• Make All Integers Negative
• Convert Fractions to Integers
• Analyze an Integer
• Find and Replace Integers
• Generate Integers from Regex
• Create an Integer Array
The first nineteen websites in the network are:
The next few sites are onlineHASHtools, onlineGIFtools, onlineBITMAPtools, onlinePDFtools, onlineBROWSERtools, onlineCRYPTOtools, onlineAUDIOtools, onlineCSStools, onlineJStools, and a dozen more.
See you next time!
# Announcing TuxURLs.com
I and my team at Browserling just released TuxUrls.com. It's a neat Linux news aggregator that collects news titles from the top 20 Linux blogs and websites and has a super fast search. Check it out!
Other websites in our URLs network are:
• MathUrls – a mathematics news aggregator
• FinUrls – a finance news aggregator
• SciUrls – a science news aggregator
• TechUrls – a technology news aggregator
• DevUrls – a developer news aggregator
Check them out as well and see you next time!
# The most beautiful fractal I have ever created
As usual, I was spending my morning creating random fractals and suddenly this artwork came into existence. I think it's the most beautiful fractal I have ever created.
This artwork is a random form of the canopy fractal. If you follow this link, you can adjust the fractal's parameters and create your own fractal artwork. Let's see what you can make!
# Read man pages with vim
## Surround
With Surround, you can quickly delete, change, and add matching pairs of surrounding symbols around text. Let's say you have a string "hello world". By pressing ds", you'll delete the quotes and the string will become hello world. By pressing cs"' you'll change double quotes to single quotes and the string will become 'hello world'. You can also press cs"<div> and that will change quotes to <div></div> HTML tags and the string will become <div>hello world</div>. If you have a plain string hello world, then to wrap it in quotes, visually select it with v and then press S and enter the wrapping symbol. If you have a single word hello, then to wrap it in single quotes, press ysiw'.
## Swap
Swap lets you swap words or comma separated function arguments quickly. For example, if you have a function print(a, b, c), then you can press g> and it will become print(b, a, c). Press g< and it will become print(a, b, c) again. You can also enter visual swap mode by pressing gs. Then use hjkl keys to swap arguments around.
## Syntax-css3
Syntax highlighting plugin for CSS3.
## Syntax-dockerfile
Syntax highlighting plugin for Dockerfiles.
## Syntax-html5
Syntax highlighting plugin for HTML5.
## Syntax-i3config
Syntax highlighting plugin for i3 window manager configuration files.
## Syntax-javascript
Syntax highlighting plugin for JavaScript.
## Syntax-json
Syntax highlighting plugin for JSON.
## Syntax-nginx
Syntax highlighting plugin for Nginx configuration files.
## Syntax-php
Syntax highlighting plugin for PHP.
## Syntax-tmux
Syntax highlighting plugin for tmux configuration files.
## System-copy
Terminal vim doesn't have the * or + registers that are used for copying and pasting to/from system clipboard. This plugin adds the cp shortcut that copies the selected text or a motion to system clipboard by calling xsel external utility. It also has the cv shortcut for pasting from system clipboard.
## Tabular
Tabular does the same as EasyAlign (above). It aligns text in columns. It has an advantage over EasyAlign that you can immediately pass it a regular expression for the alignment. The position where the regular expression matches is where the alignment happens. For example, :Tabular /regex will create neat columns of regex.
## Tagalong
This plugin makes it easy to rename pairs of opening/closing HTML tags. You only have to rename one of them and this plugin will automatically rename the other. For example, if you have a <div>...</div> and you rename the opening <div> to <section>, then the closing tag will automatically be renamed to </section> and you'll get <section>...</section> in the output.
## Targets
One of the steps in reaching vim mastery is learning to use text objects. Text objects let you operate on entire text constructs rather than individual characters. Vim ships with many text objects already – you can access words, sentences, paragraphs, and code blocks. This plugin adds a dozen more text objects. For example, you can delete comma separated items with di, or you can change the next and previous comma separated items with cin, and cil,, you can change sum elements with ci+, and much more.
## Terminus
Terminus improves your vim experience in the terminal. Often, when you paste a multi-line text, vim adds indents to each new line and your text runs to the right. It fixes this problem by enabling bracketed paste. With bracketed paste, the entire pasted text is treated as a single blob and vim doesn't indent each line. It also adds caret cursor for the insert mode, handles terminal focus events, and it improves the mouse support by enabling the sgr mouse mode.
## Textobj-user
It's very difficult to write your own text objects from scratch because you have to replicate precise vim behavior when a text object is invoked and know all the gotchas and pitfalls. This plugin provides a quick and easy interface that lets you implement your own text objects. There's an entire ecosystem of "textobj" plugins that use this plugin to add various useful text objects.
## Textobj-entire
This plugin uses Textobj-user (above) to create a text objects ie and ae that operate on the entire file. For example, to select the entire file, you can do vae, to delete the entire file, you can do vaed or dae.
## Textmanip
This plugin lets you visually move the selected text. For example, you can visually select a word, and then use the ctrl+h and ctrl+l shortcuts to move it to the left or right. Similarly, you can move entire lines down and down with the ctrl+j and ctrl+k shortcuts.
## Thesaurus-query
With this plugin, you can access Libreoffice thesaurus. Once you configure it, you can use the :Thesaurus word command to find all words related to the word word and you can also use :ThesaurusW to find all related words of the word under the cursor.
## Tmux-complete
When you're in the flow, a lot is happening at once and you have many tmux windows and panes open. One with git log, another with tests, another one with a man page, etc. Often, you need to get info from a tmux pane into vim. Usually, you have to use your mouse to copy it in but that is very ineffective. With this plugin you can just press ctrl+x ctrl+u in insert mode and complete words from any tmux windows/panes. It will offer a list of all words from all tmux windows/panes.
## Traces
When substituting text with the :s/old/new command, you can't see the changes until you execute it. This plugin previews the old match as you type it as well as the substitution part new and you get a visual feedback that shows if you're doing it right.
Ultisnips is a snippet engine. When you press the tab key, it looks at the last typed token and expands it to a snippet. To make it work, you need to create a language.snippets file and write your snippets there. For example, you can create php.snippets and put an if snippet there if ($1) {$0 }. Now when you're programming PHP and type if and press tab, it will expand to if (|) { } and your cursor will be where the | character is. If you press tab again, it will jump to 0 token between the curly braces if () { | }. ## Undoquit Often, you're just too quick and close a window that you didn't want to close. With this plugin, you can now hit ctrl+w+u and undo a closed window. ## Unicode This plugin lets you quickly search and insert Unicode characters. For example, if you want to insert a Unicode rabbit symbol, you can type rabbit and then press ctrl+x ctrl+z. The typed text rabbit will get substituted with a rabbit emoji. ## Unimpaired Many commands come in pairs. For example, :bn and :bp to go to the next/previous buffer, :cn and :cp that go to the next/previous quickfix list item, :ln and :lp that go to the next/previous location list item, etc. This plugin adds quick shortcuts for these commands. You can now ]b and [b to go to the next/previous buffer, ]q and [q to go to the next/previous quickfix entry, ]l and [l to o to the next/previous location list entry, etc. ## Vimade When you're working with multiple split windows, after a short distraction, it's easy to forget which split you're working in. This plugin fades all splits so that the currently focused split is clearly visible. ## Visual-repeat The Repeat plugin (above) repeats normal mode commands when you press .. This plugin extends Repeat to work in the visual mode. When you select a visual region and press . it will run the normal mode commands only on the selected area. ## Visual-split Often, you need to keep a comment, a function definition, or a code fragment above the fold so that it's always visible. The usual approach is to split the window with ctrl+w+s and then resize it smaller with 10ctrl+w+-. This plugin merges these two actions. You can now visually select the area and press ctrl+w+gss. The plugin will split the window and resize it to exactly fit the selected lines. ## Writeable-search Writeable-search is similar to CtrlSF (above). It lets you quickly find results and immediately edit them in the results window. When you run the :WritableSearch pattern command, it will grep all files in the current directory for the pattern and open a new tab with the results. You can then edit the results and when you do a :w in the same results window, it will update files with changes. If you already have a list of things to fix in the quickfix window, then you can transfer them to this plugin for editing via the :WritableSearchFromQuickfix command. See you next time! # My Linux and Vim Notes These are my Linux and Vim notes. You may find something useful in this list. • flameshot - neat screenshot app • peek - screen recorder that I haven't tried yet • ppa-purge - delete installed ppa packages • light - app to change screen brightness • lenovo-throttling-fix - fix throttling on thinkpads • mypaint - neat infinite canvas painting app • pinta - paint app that crashes • ranger - file manager for the shell • fzf - find items quickly in lists • nnn - another file manager for the shell • xrandr - manage monitors • arandr - gui for xrandr • pdfmod - pdf editor that i haven't tried yet • vimium - chrome plugin for vim keyboard shortcuts • vimperator - firefox plugin for vim keyboard shortcuts • subuser - run apps in docker, haven't tried yet • ag - silver searcher, faster grep but today ripgrep is better • icdiff - prettier diff • xsel - use system clipboard form the shell • xclip - xsel alternative • syncthing - synchronize files between computers • unison - also synchronize files between computers • sway - i3 window manager for wayland • tmux-xpanes - split terminal into rectangles and run commands in them • github.com/jarun - clever guy who wrote nnn • luke smith - interesting videos about linux • byobu - predefined tmux and screen configs • rat poison - another tiling window manager • gmic - image manipulation from the shell • things every hacker once knew - interesting blog post by esr • syntastic - vim lint plugin • vimawesome.com - vim plugins ranked by usage • vim after 15 years - someone's experience about using vim for 15 years • tmux-resurrect - restore tmux windows as they were on restart • scratchpad in i3 - hide a window and then bring it back • i3bar - default status bar for i3 • i3status - system status string generator for i3bar • i3blocks - alternative system status string generator for i3bar • polybar - another system status string generator for status bars • dmenu - default i3 app launcher • rofi - alternative app launcher • i3-layout-manage - save and load i3 layouts • xprop - print x windows info • font-awesome cheat sheet - find icons fast • going mouseless - youtube video about using linux without a mouse • xss-lock - screen lock utility for x • xset - change x settings such as keyboard speed • x power tools - a book about x windows tools • xlsfonts - list x windows bitmap fonts • xlfd - naming scheme for bitmap fonts • xfs - modern fonts for x or something like that • stephane chezelas - interesting stackoverflow answers • yesmeck - interesting tmux.conf on github • showkey - detect key presses and print keyboard keys • xev - same but only works in x windows • tmux2 - a book about mouse free development • dunst - send notifications to the screen from console and scripts • wmctrl - advanced control of x windows via command line • sxhkd - window manager independent keyboard shortcut manager • tao of tmux - a book that teaches tmux • xcb - alternative library to xlib for writing x windows apps • xcb tutorial - a short tutorial on xcb • entr - watch a file for changes and run commands • toe - print terminal types • infocmp - print terminal info • text terminal howto - linux howto about text terminals • man console_codes - man page of linux console codes and escape sequences • ESC[XmTEXTESC[0m - print TEXT in color X in terminal • dcvim - plugin to use vim keys in double commander • ack - improved grep • ctags - search code for keywords, short for exuberant ctags • cscope - more detailed code search for keywords and function names • universal ctags - new and maintained version of exuberant ctags • gnu global source tagging system - ctags/cscope alternative • exa - ls alternative, not sure why • ripgrep - fastest grep • fzf.vim - vim plugin for fzf • bat - cat replacement with syntax highlighting • surfingkeys - vimium alternative for chrome • brookhong.github.io - blog of author of surfingkeys • alacritty - gpu accelerated terminal • kitty - another gpu accelerated terminal • evil - emacs plugin that makes it work like vim • org mode - note taking plugin for emacs • spacemacs - emacs clone with vim features • ctrl+j in bash - accept ctrl+r history search entry • ctrl+g in bash - discard ctrl+r history search entry • ctrl+w in bash - delete the current word • ctrl+y in bash - put the deleted word back • going all in with neovim - youtube video about neovim • mosh tech talk - google tech talk about mosh • x window system design principles - youtube video about x window history • real story behind wayland and x - youtube video about wayland and x • tacit programming - synonym for point-free programming, not sure what that is • stack-oriented programming - programming with stacks, something advanced • architecture of open source apps - book about popular open source apps • thinkpad x210 - old laptop remake by 51nb enthusiast group from china • lstopo - print device tree • debootstrap.sh - a script to create debian root structure in an arbitrary directory • shell wtf's - oilshell blog posts about shell gotchas • flatpak - new app package format for linux • debian popularity contest - public stats about most downloaded debian packages • nwallace dotfiles - interesting dotfiles of github user nwallace • cool uses for fzf - google this to find cool uses of fzf • tcplife - bpf to trace tcp connections • bcc - collection of bpf tools • vim.reversed.top - a list of tools that have vim interface • reversed.top - interesting blog about unix things • wine tricks - wine helper program to install libraries • i3-vim-focus - integrate vim splits with i3 • vimcolors.com - collection of vim color schemes with previews • vim-bufferline - vim plugin that shows open buffers at the bottom • defaults.vim - file with default vim settings that comes with vim • :term in neovim - command to open a terminal inside neovim • vimcasts - website with short vim video tutorials • vifm.vim - use vifm file manager in vim • bufexplorer.vim - vim plugin to quickly switch between buffers • ctrlp.vim - vim plugin that lets you open files by fuzzy matching on ctrl+p • command-t.vim - vim plugin similar to ctrlp but does the same with ctrl+t • vim-fugitive - vim plugin to use git from vim • ack.vim - vim plugin to use ack from vim • unimpaired.vim - adds pairs of commands to vim such as [b and ]b, [s and ]s, etc • fzf.vim - use fzf in vim to open files and find buffers • sensible.vim - modern default settings for vim • supertab.vim - autocomplete plugin for vim that uses tab as the completion key • zoomwin.vim - vim plugin to maximize the current window • quick fix window - special vim window that can be used to list errors and jump to them • airline.vim - pretty status bar for vim • amix vim configs - interesting vim config of github user amix • solarized color scheme - popular nice looking color scheme for vim and terminal • tagbar.vim - vim plugin that shows functions on the right side of the current file • vim-css-color - vim plugin that shows css colors in vim • goyo.vim - vim plugin that centers text and removes line numbers, status bar, etc • coding in goyo mode - something to try • ale.vim - vim lint plugin, alternative to syntastic • statico dotfiles - very complex and interesting dotfiles of github user statico • :helpgrep - vim command to grep its help files • vim-textobj-user - vim plugin to define your own text objects • :ls - vim command to list currently open buffers • :b substr - vim command to jump to buffer that matches "substr" • ctrl+] - press this in vim's help to jump to definition of the word under cursor • ctrl+t - press this in vim's help to jump back from previous ctrl+] press • ctags -R . - build tags file for the current directory and all subdirectories • i_ctrl+x-ctrl+k - in vim's insert mode, show dictionary suggestions • i_ctrl+x-ctrl+t - in vim's insert mode, show thesaurus suggestions • :only - vim command to close all windows and maximize the current one • :tabonly - vim command to close all tabs and maximize the current one • i_ctrl+o - in vim's insert mode, switch to command mode for 1 command • ctrl+^ - switch between last two open buffers in vim • vim 8.2 adds popup windows and text properties • :match and :highlight - vim commands to color patterns • :set undofile - if undo file is set in vim, you can undo after closing a file • i_ctrl+R - in vim's insert mode, insert a register • /r/neovim - neovim subreddit • sneak.vim - vim plugin to quickly find two character patterns • vim-easy-align - vim plugin to align text in columns • lion.vim - another vim plugin to align text in columns • ultisnips.vim - code snippet plugin for vim • vim to emacs in 14 days - blog post about moving to emacs from vim in 14 days • denite.nvim - a modern vim/neovim plugin to create you own interfaces • neosnippet.vim - another snippets vim plugin • deoplete.vim - another completion vim plugin • you don't grok vim - someone's answer on stack overflow that explains how vim works • iw - vim motion that selects a word • aw - vim motion that also selects a word, including trailing whitespace • is - vim motion to select a sentence • as - also vim motion to select a sentence, including trailing whitespace • '[ - start of the last change in vim • '] - end of the last change in vim • clap.vim - vim plugin interactive finder that uses :popup • LSP - language server protocol used for auto hinting and linting • tree-sitter - incrementally parse source code, not sure • jutin m keyes - neovim lead maintainer • sink.io - justin's blog • github.com/justinmk/notes - justin's tech notes, especially vim notes • c_ctrl+f - open a list of command history in vim command mode • unix as ide - blog post series about using unix tools for programming • :syntax keyword WordError teh - mark the word "teh" red in vim • brew.sh/analytics/install - top brew installs • neovim diff.txt - neovim documentation page about vim and neovim differences • ]) - in vim jump to next ), doesn't work for me • ]( - in vim jump to previous (, doesn't work for me • ]/ - in vim jump to end of comment, doesn't work • [/ - in vim jump to beginning of comment, doesn't work, not sure • learn vim script the hard way - good book about vim script • nixers.net - active unix forum • nixers.net podcast - unix podcast • fzf in a floating window - cool idea • xi text editor - a new text editor by raph levien • design of lua - pdf document about lua • vim hall of wtf - blog post about complex code in vim • neoterm.vim - vim/neovim plugin that improves built-in terminal usage • skim - fzf alternative written in rust • gq-motion - in vim format code that is matched by motion • :ptag tag - open a tag in preview window • ctrl+w+z - close the preview window • ctrl+w+} - run :ptag on the word under the cursor • :pclose - close the preview window • highlightyank.vim - vim plugin to highlight yanked text region • meld - neat visual diff tool • beyond compare - another visual diff tool • fd - find command alternative • vis.vim - vim plugin to run a command on visually selected area • tmate - tmux fork that lets you share terminal • vimux - create tmux splits and run tmux commands from vim • trunk based development - no feature branches, just master branch • coderay - syntax highlighting app • risc-v - open source instruction set for cpus • look "word" - unix command that prints lines that begin with a word • icegiant - new cpu cooler • ctrl+e - in vim in ctrl+x mode, discard selection • ctrl+y - in vim in ctrl+x mode, accept selection • ]m - in vim jump to end of a method, doesn't work for me, not sure • [m - in vim jump to beginning of a method, not working • rpcinfo -p host - show rpc status of host • tldp - linux documentation project • jq - json filtering utility for the shell • fslint - app to find duplicate files • sjl bitbucket dot files - interesting dotfiles of user steve losh • zzapper.co.uk/vimtips.html - 16 years of vim commands • junegunn dotfiles - interesting dotfiles of fzf author junegunn • nnoremap a b, nnoremap b a - swap keys a and b in vim • i_ctrl+k - enter a digraph in vim • git reflog - show reflog info, useful for redoing commits • git rebase --exec 'cmd' - exec a command after each commit while rebasing • n_+ - in vim pressing + will go to next non-blank on next-line • g; - in vim go to previous change • g, - in vim go to next change • n_ctrl+e - in vim scroll up the screen by one line • n_ctrl+y - in vim scroll down the screen by one line • i_ctrl_x+ctrl+e - in vim's insert mode, scroll up the screen by one line • i_ctrl_x+ctrl+y - in vim's insert mode, scroll down the screen by one line • zENTER - in normal mode in vim, scroll current line to top of the screen • zt - same as zENTER • z. - in vim's normal mode, center the current line on the screen • zz - same as z. • z- - in vim's normal mode, put the curernt line at the bottom of the screen • zb - same as z- • gd - in vim, go to definition of the word under cursor in current file • power of g command - vim wiki summary of g command • ". - vim register that contains last inserted text, doesn't work for me • i_ctrl+a - repeat last inserted text in vim in insert mode, doesn't work for me • i_ctrl+@ - repeat last inserted text and leave insert mode, not working • i_ctrl+x_ctrl+n - appears to do the same as i_ctrl+n, not sure how this is different • z= - in vim, show spelling suggestions for the word under the cursor • vim buffer faq - faq about working with vim buffers • bind -m "mode" - create a readline key binding that only works in "mode" • docker multi-stage builds - build a docker container from another container • invoke-rc.d - same as service command • git commit --amend - edit commit message of the previous commit • offline imap - use imap mailboxed without the internet • mbsync - offline imap alternative • notmuch - search your mailboxes • sc-im - command line spreadsheet • gnu units - command line utility to convert between various units • z - remember visited directories and quickly jump to them, best with fzf • autojump - z alternative • fasd - z alternative • v - quickly open last edited files in vim, best with fzf • bfs - find alternative to search directories level by level • awesome-shell - curated list of awesome shell stuff • tabular.vim - vim plugin to align text in columns • bufkill.vim - vim plugin to kill the buffer but don't close window • closetag.vim - vim plugin to autoclose html tags • broot - another file manager for the terminal • tint2 - taskbar for x windows • youcompleteme.vim - lightweight vimcompletesme.vim plugin alternative • pathpicker fpp - interactively select files in the terminal and edit them • shellcheck - bash script linter • the art of command line - concise command line tutorial on github • asciinema - record and share command line sessions as movies • zoom tmux pane - you can zoom a tmux pane in and out with prefix-z • interactive filter: a new standard tool - summary of interactive filters on lobste.rs • aria2 - lightweight download app that also supports bittorrent • linux problems on the desktop - article about why linux on desktop is garbage • prettyping - wrapper around ping to make its output pretty • dwdiff - tool that prints word-level diff • vimdiff - diff files using vim • colordiff - wrapper around diff to add color to it • xbindkeys - app to bind keys to do anything you want in x • progress - utility to monitor cp, mv progress • httpie - a curl alternative written in python • curlie - a curl wrapper with httpie syntax • oh my zsh - scripts to make zsh pretty • mycli - much nicer mysql command line tool • pgcli - same as mycli but for postgres • litecli - same as mycli but for sqlite • ts task spooler - queue and run tasks from command line • w3mimgdisplay - show images in the terminal • ueberzug - also show images in the terminal • heidisql - free open source gui for mariadb, mysql, and postgresql • shortcutfoo.com - learn keyboard shortcuts of various apps by repeating them • grc - colorize terminal output using regular expressions • supercat - grc alternative • fselect - find files using sql-like queries • newsboat - terminal rss reader • glances - top/htop alternative • up pipe plumber - instantly preview results when you pipe programs • autossh - automatically restart ssh sessions and ssh tunnels when they die • sshuttle - vpn over ssh • lib.rs - search rust libraries by category • procs - ps alternative • ctrlsf.vim - finds patterns in vim and show them on the left for quick refactoring • chars - command line program to print unicode character info • ascii - command line program to print ascii table • vim-qf - vim plugin to make it easier to work with quick fix window • readrust.net - rust news and articles • gutentags.vim - vim plugin to regenerate ctags files • codi.vim - vim plugin to interactively write and run code snippets • jedi.vim - vim plugin to use jedi autocompletion in python • keepass + fzf - cool idea • qutebrowser - web browser that uses vim modes and vim keyboard shortcuts • fzf instead of dmenu - cool idea • bash search history via fzf - cool idea • notational velocity - popular note taking app • notational-fzf-vim - vim plugin that implements notational velocity with fzf • bmore - binary more • bvi - binary vi • favorite terminal apps - forum thread on nixers • diff-so-fancy - another pretty diff program • what's your vim setup like - lobsters thread • peekaboo.vim - preview vim registers on the right side before inserting them • awesome cli apps - github repo with a list of cool cli apps • tldr.sh - man pages through examples • learnbyexample - clever github user with many tutorial repos • insect - scientific calculator • seven habits of effective text editing with vim - talk by bram moolenaar • rga ripgrep-all - ripgrep in pdfs too • sxiv - very simple image viewer for x • mdcat - cat markdown files • dust - du alternative • diskus - another du alternative • websocat - netcat for websockets • machakann vimrc - advancd vimrc configuration by github user macakann • history and effective use of vim - detailed post about vim • surf - minimalistic web browser • thundering herd - computer science term when all processes wake up at once • fzf-mru - vim plugin that adds mru list of edited files accessible via fzf • targets.vim - vim plugin that adds more text objects, similar to surround.vim • context.vim - vim plugin that shows where you're at in nested code • tmux-complete.vim - complete text in vim from all tmux panes • silicon.vim - generate code screenshots from vim • bluz71 blog - interesting vim articles • greg hurrell vimcasts - on youtube and wincent.com • ranger + fzf + ripgrep - cool idea • fzf wiki - lots of fzf ideas • fzf + bat - while searching, preview files via bat in fzf preview window • fzf + bfs - combine these to have a better better file selection list order • go in nerdtree - open file but leave cursor in nerdtree • markus kuhn - author of fixed x fonts • 6x13 - default fixed x font with alias "fixed" • bdf - bitmap x font format • pcf - newer bitmap x font format • switch.vim - vim plugin to switch between predefined often used patterns • vim-multiple-cursors - vim plugin to emulate multiple cursors • terryma dotfiles - many interesting dotfiles of github user terryma • vim-markdown - markdown syntax highlighting plugin for vim • cfilter.vim - vim plugin to filter and narrow down quick fix entries • zenburn color scheme - popular color scheme for vim • vim-move - vim plugin to quickly move lines of text around • gotbletu youtube channel - linux app reviews • traces.vim - live preview changes as you type :s/foo/bar • neomake.vim - to be explored, a vim plugin that runs commands asynchronously? • vim-galore - amazing vim tutorial • github.com/topics/vim - most popular repositories with tagged vim • github.com/topics/neovim - most popular repositories with tagged neovim • terminus font - neat fixed bitmap font • proggy font - another neat fixed bitmap font • hack font - neat ttf font • chris siebenmann blog - advanced unix blog • nick janetakis - interesting youtube videos about vim • vim-obsession - vim plugin to manage vim sessions • fzf + wordnet - cool idea • nord color scheme - neat color scheme for vim and terminal • vim-conflicted - vim plugin that makes resolving git merge conflicts easier • vimconf jp videos - vim conference in japan on youtube • quickrun.vim - vim plugin to quickly run contents of file through a program • run or raise - run an app if it wasn't yet running, or focus it if it's running • neg-serg on github - user who made a cool i3 mod with run or raise and more • jumpapp - app that implements run or raise • marathon - another app that implements run or raise • shortcut.vim - preview available vim commands as you type them • codeface github repo - user's chrissimpkins collection of programming fonts • comp.fonts usenet faq - great explanation of font terminology • vim-qfedit - vim plugin to edit quickfix window entries • apprentice color scheme - nice dark color scheme • vimways - vim blog by romainl and robertmeta • xargs -I '{}' command '{}' - replace '{}' in command arguments with xargs input • bufselect.vim - bufexplorer alternative • philrunninger vim files - on github • shift+i in nerd tree - show hidden files • bitmap-fonts github repository by user tecate - fonts collection with screenshots • vim.wasm - vim implementation in wasm, works in a browser • vim-wordmotion - vim plugin that makes 'w' key work with camelcase words • notion window manager - static tiling window manager, successor of ion wm • echodoc.vim - vim plugin to display function prototypes in command line • neosolarized - color scheme for neovim • vim-submode - vim plugin to create your own modes • xnest - run x in x • xephyr - modern way to run x in x • paperwm - scrollable tiling window manager, interesting • 10gui.com - similar idea to paperwm • gilesorr wm summary - a list of all window managers and their descriptions • firefox public data report - firefox user hardware stats • termite - modal terminal emulator github.com/thestinger/termite • i3spin - alt+tab for i3 • i3ipc-python - python ipc bindings for i3, control i3 via python • set shortmess-=s - in vim, show match count even if there are 0 matches • i3-easyfocus - draws a letter on each window for quickly focusing • nq - queue tasks from command line, ts alternative • git-annex - add large files to git without tracking them • git-lfs - git-annex alternative • git-vfs - work with large git repos without creating a local repo copy • fdupes - find duplicate files • remacs - emacs rewritten in rust • commandlinefu top commands - useful commands to know • servo - a multi-threaded web browser engine written in rust • gtalug youtube channel - many linux videos • num utils - command line utilities for working with numbers • multitail - tail multiple files at once by splitting terminal in multiple windows • httping - ping via http • lnav - read multiple log files at once and correlate timestamps • :normal x - in vim, run normal mode command x • linux.conf.au videos from 2010 to 2020 on youtube • moreutils - additional unix utilities • chronic in moreutils - run command quietly unless it fails • combine in moreutils - combine lines of files with and, or, not, xor operators • errno in moreutils - print error codes and names • ifdata in moreutils - print network interface info • ifne in moreutils - run a program if stdin is not empty • isutf8 in moreutils - check if a file has a valid utf8 encoding • mispipe in moreutils - pipe two commands and return exit code of first command • parallel in moreutils - run commands in parallel, often confused with gnu parallel • pee in moreutils - pipe stdin to multiple commands • sponge in moreutils - write stdin to a file • ts in moreutils - timestamp input • vidir in moreutils - rename and delete files and directories in vi • vipe in moreutils - insert vi in a pipe • zrun in moreutils - uncompress compressed files in command arguments • freebsd on a desktop - blog series at vermaden.wordpress.com • linux app of the year - yearly question on linuxquestions • muttator - vim keyboard shortcuts for thunderbird • neomutt - mutt fork • zero, one, infinity - allow either 0 of something, or 1, or infinity, but not 2, 3, etc • necromancer dos navigator - file manager, mc alternative • pure bash bible - how to do everything in bash • shellharden - show bash script errors in red • how to do things safely in bash - shellharden tutorial on how to write safe bash • qdir - quad file manager with 4 splits, works in wine • tabbed - put apps that support xembed feature in tabs • pspg - view data in mysql and postgres databases • what are your favorite terminal programs - hacker news thread • iftop - top for network interfaces • atop - top/htop alternative • nethogs - top for process network activity • q - command line tool that parses tabular data with sql queries • gron - convert json to a format that can be grepped • wuzz - ncurses interface for http debugging • usql - utility to connect to any sql or nosql database • drawille - python library for drawing in the terminal using braille charset • dbeaver - gui for working with mysql and postgre, heidisql alternative • bpython - curses repl for python with suggestions • prompt-toolkit - python library for building interactive cli apps • noti - send a notification when a process completes • peco - fzf alternative • stdbuf - coreutils utility to modify buffering of commands • findmnt - find a mounted filesystem • rlwrap - add readline support to interactive cli programs that don't use it • colrm 1 5 - remove first five characters from input • namei - resolve symlinks and print full pathname of a file • tc netem - emulate network delay and packet loss • inconsolation.wordpress.com - blog about cli apps • must-watch-css - a list of css talks to watch on github • useful registry keys - wine doc about windows registry keys • eurobsdconf youtube videos - to watch talks on zfs, rust • vim-sandwich - vim plugin similar to surround.vim • inferno - os that is descendant of plan9 • 9front - fork of plan9 • acme - default text editor on plan9 • ben eater crc videos on youtube - crc explained • brodie robertson - youtube videos about linux • wireguard - modern and simple vpn • delimitmate.vim - auto-pairs.vim alternative, automatically close ", (, ', etc • undotree.vim - vim plugin to visualize vim's undo tree • learnxinyminutes.com - learn things fast via examples • fcat - fast cat implementation • hello rust - a live coding youtube show about rust • peep - like less but creates a tiny preview window • fauxclip.vim - vim plugin to copy/paste things to/from system clipboard • tre - tree command alternative • qfenter.vim - vim plugin to open a quickfix entry in any open split • autocwd.vim - vim plugin that lets you change currently working dir of buffers • vim-lucius - a nice dark and light color scheme for vim • vim-forest-night - a nice dark color scheme for vim • tmux-fzf - tmux plugin that lets you fuzzy find sessions, windows and panes • vim-dirvish - select files and browse directories in vim, nerdtree alternative • rsi.vim - vim plugin that adds readline key bindings for vim • vim-sayonara - better bufkill.vim plugin • wildfire.vim - vim plugin to incrementally select closest text objects • vim-indent-object - plugin that adds indent text object, useful when writing python • flygrep.vim - vim plugin to asynchronously grep text • fz - add fzf matching to z • grabc - x utility to grab a pixel color • qalc - awesome cli calculator • vifm - another command line file manager • fff - another command line file manager • lf - another command line file manager • derek banas youtube lua tutorial - learn lua in 1 hr • z.lua - z alternative written in lua • lexima.vim - auto-pairs alternative • coc-pairs.vim - auto-pairs alternative • pear-tree.vim - auto-pairs alternative • howivim.com - vim stories • the patient vimmer - advanced vim tutorial by romainl • learnvim.txt - vimdoc tutorial by barry arthur • of-vim-and-vigor.blogspot.com - barry arthur's vim blog • vimgolf.com - vim golfing • aerc - terminal email client, mutt/neomutt alternative • last status line for vim - blog post about how to customize vim's status line • cb.vu/unixtoolbox.xhtml - hundreds of unix commands explained • vim-table-mode - vim plugin to draw ascii tables in vim • zoom.vim - another vim plugin for zooming one window by dhruvasagar • caw.vim - another commenting vim plugin • editorconfig - config file for text editors to maintain consistent coding style • nrrwrgn.vim - vim plugin that allows focusing on a narrow region of text • vim-speeddating - vim plugin to inc/dec dates and times with ctrl+a/ctrl+z • tracyone dotfiles - github user tracyone has advanced dotfiles • gina.vim - vim plugin to use git from vim, fugitive alternative • gpick - color picker for x • cppman - c++ documentation in the command line • vim-cppman - vim plugin for cppman • vim-yoink - vim plugin to maintain yank history • vital.vim - collection of vim subroutines by japan vim user group • diffchar.vim - vim plugin that visualizes diff changes char by char • a-list-of-vims-lists - summary of all lists that vim internally maintains • vim-foldsearch - show only lines that match a pattern • area-41 - templates for creating vim plugins • :Cfilter - vim command to filter quickfix window entries • ferret - vim plugin to quickly search files and display results in quickfix window • quickfix-reflector - qfedit alternative, vim plugin to edit quickfix results • foo.vim - vim plugin with examples of how to write vim functions • quickfixsigns.vim - add marks next to line numbers for quickfix results • yankring.vim - vim plugin that maintains yank history • :h yankring-tutorial - yankring tutorial • slime.vim - send code to repl from vim • easyclip + neoyank - plugin combo for clipboard/yank management • vinegar.vim - vim plugin that improves built-in netrw plugin usage • slimv.vim - slime.vim alternative with swank protocol support • spacevim - vim distribution with latest features and plugins • mintree.vim - minimalistic reimplementation of nerdtree vim plugin • delta - another pretty diff tool • :colder - after filtering quickfix list, use this command to go back to unfiltered • defx.nvim - vim/neovim file explorer, alternative to nerdtree • vim-sort-motion - vim plugin that sorts data in a motion • vim-markbar - peekaboo for vim marks • :he quickref - concise summary of vim commands • vim-altr - vim plugin to alternate between related files • vim-tabpagecd - vim plugin to set the cwd for a tab • vim-smartinput - vim plugin auto-pairs and delimitmate alternative • whileimautomaton.net - kana natsuno vim blog • vim-floaterm - vim plugin that creates a floating terminal window • i3-workspace-groups - create per project workspace groups • chrisbra vim_faq - github user chrisbra up to date vim faq • vis - vi-like editor based on plan9 structural regular expressions • dvtm - tmux alternative • abduco - add sessions to dvtm • repo.or.cz - public git repo, interesting to sort repos by date • how to boost your vim productivity - blog post about vim and tmux productivity • kakoune - text editor that is similar to vim • vimlondon videos - vim videos to watch on vimeo • gi - vim command to enter insert mode in the last insert position • i3ass - i3 assistance tools, written using i3 ipc • budlabs - i3ass author's youtube channel • i3run from i3 assistance tools - run or raise an app in i3 • michael stapelberg - author of i3, @zekjur on twitter • garbage podcast - unix podcast • colorizer.vim - vim plugin to colorize css colors and ansi escape codes • ranger + autojump - cool idea • read man pages in vim - cool idea • qtile - tiling window manager written and configured in python • vimpager - vim plugin to make vim act like a pager • vim-man - vim plugin to read man pages • dispatch.vim - vim plugin to run make and tests asynchronously • tmux-navigator - vim plugin to navigate between vim splits and tmux panes • udiskie - auto mount usb • checkinstall - utility to track make install files • arch linux aur - sort by votes to find interesting apps • bracketed paste - terminal escape sequences for raw pasting • htop-vim - htop with vim keybindings • set scrollbind - make vim scroll two windows synchronously • debian stretch - codename for debian 9 • debian buster - codename for debian 10 • debian bullseye - codename for debian 11 • debian bookworm - codename for debian 12 • debian sid - unstable rolling debian branch, don't know what that means • autocutsel - sync cutbuffer and clipboard and optionally also primary • vim.org scripts page - sort by downloads to find interesting plugins • vim.fandom.com - vim tips wiki • alan-kay.fandom.com - alan kay wiki • 2bwm - floating keyboard-driven window manager • desed - gui for debugging sed scripts • sshfs - mount filesystems over ssh • grex - generate regexes from data • ubuntu bionic beaver - code name for ubuntu 18.04 • ubuntu focal fossa - code name for ubuntu 20.04 • gnu stow - utility to quickly symlink configs from dotfiles • ieee 754 - floating point standard • whatfiles - strace like utility for tracking file open/read/write/delete • xkeysnail - flexible utility to remap keyboard keys • tdrop - drop down any app • s4cmd - faster s3cmd implementation • s5cmd - faster s4cmd implementation See you next time! # Browserling is now a top 20k website in the world According to Alexa, Browserling is now a top 20k website in the world. 10k positions up from 30k a few months ago. It's yet another small step for a ling and a giant leap for ling-kind. alexa.com/siteinfo/browserling.com My goal is to make Browserling a top 10k website and beyond. Thanks for following along my adventure and see you next time! I'm just getting started here. # My Book is in Linux & BSD Humble Book Bundle My book Perl One-Liners together with many other great books is in Linux & BSD Book Bundle put together by my publisher No Starch Press. You can get it for just1.
10,000 bundles sold in the first few days! Make it 10,001 and get it at Linux & BSD Book Bundle.
See you next time!
# Nineteenth site in online tools network: onlineUNICODEtools.com
At Browserling we're building a network of online tools websites. Each site focuses on one and only one tool category. Each tool does one and only one thing.
Today we're releasing the nineteenth site – Online Unicode Tools.
Online Unicode Tools is a collection of simple, free and easy to use utilities for working with Unicode data. There are no ads, popups, or other garbage. Just Unicode utilities that work in your browser. Load Unicode and instantly get the result!
www.onlineUNICODEtools.com
Here's a list of all Unicode tools so far:
Here are the upcoming Unicode tools:
• Find the Names of Unicode Symbols
• URL-encode Unicode
• URL-decode Unicode
• Convert Binary to Unicode
• Convert Octal to Unicode
• Convert Decimal to Unicode
• Convert Hex to Unicode
• Convert Unicode to Any Base
• Convert Any Base to Unicode
• Convert Unicode to ASCII
• Convert ASCII to Unicode
• Convert Unicode to Latin1
• Convert Latin1 to Unicode
• Convert Unicode to Bytes
• Convert Bytes to Unicode
• Sort Unicode
• Validate Unicode
• Convert Unicode to Punycode
• Convert Punycode to Unicode
• Decode Base64 to Unicode
• Encode Unicode to Data URI
• Decode Data URI to Unicode
• Convert HTML to Unicode
• Convert UTF8 to Unicode
• Convert UTF16 to Unicode
• Convert UTF32 to Unicode
• Convert Unicode to Uppercase
• Convert Unicode to Lowercase
• Convert Unicode to Randomcase
• Convert Unicode to Lowercase
• JSON Stringify Unicode
• JSON Parse Unicode
• Analyze Unicode
• Let Zalgo Destroy Unicode
The first eighteen websites in the network are:
The next few sites are onlineHASHtools, onlineGIFtools, onlineBITMAPtools, onlinePDFtools, onlineBROWSERtools, onlineCRYPTOtools, onlineAUDIOtools, onlineCSStools, onlineJStools, and a dozen more.
See you next time!
Thanks for reading my articles. If you enjoyed them and would like to receive my new articles automatically, you can subscribe via rss feed or email. | 11,095 | 47,887 | {"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": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-40 | longest | en | 0.554585 |
https://whomadewhat.org/how-many-feet-are-in-a-10th-of-a-mile/ | 1,695,958,798,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510481.79/warc/CC-MAIN-20230929022639-20230929052639-00759.warc.gz | 669,610,718 | 12,598 | 528 feet
Moreover, How many feet is exactly a mile?
In respect to this, How many feet is 2 miles?
How do you convert miles into feet?
Furthermore, How do you convert miles into yards and feet?
## What is the scale factor to convert miles to feet?
How to convert miles to feet? There are 5280 feet in 1 mile. To convert miles to feet, multiply the mile value 5280. For example, to find out how many feet there are in 2 miles, multiply 5280 by 2, that makes 10560 feet in 2 miles.
## How do you change miles to feet?
To convert miles to feet, multiply the mile value 5280. For example, to find out how many feet there are in 2 miles, multiply 5280 by 2, that makes 10560 feet in 2 miles.
## How many distance make a mile?
The mile, sometimes the international mile or statute mile to distinguish it from other miles, is a British imperial unit and US customary unit of distance; both are based on the older English unit of length equal to 5,280 English feet, or 1,760 yards.The mile, sometimes the international mile or statute mile to distinguish it from other miles, is a British imperial unit and US customary unit of distance; both are based on the older English unitEnglish unitEnglish units are the units of measurement used in England up to 1826 (when they were replaced by Imperial units), which evolved as a combination of the Anglo-Saxon and Roman systems of units.https://en.wikipedia.org › wiki › English_unitsEnglish units – Wikipedia of length equal to 5,280 English feet, or 1,760 yards.
## How do you convert miles to yard?
To convert a mile measurement to a yard measurement, multiply the length by the conversion ratio. The length in yards is equal to the miles multiplied by 1,760.
## How many feet is a mole?
There are 5280 feet in 1 mile.
## How do you convert 10 miles to feet?
To convert 10 mi to ft multiply the length in miles by 5280.0. The 10 mi in ft formula is [ft] = 10 * 5280.0. Thus, for 10 miles in foot we get 52800.0 ft.
## What is the conversion factor for feet?
Multiply by to obtain
———————– —————- ———–
feet 30.48* centimeters
feet 0.0003048* kilometers
feet2 0.3048 meters
feet (U.S. survey foot) 0.304800609601** meters
## What makes up 1 mile?
Mile, any of various units of distance, such as the statute mile of 5,280 feet (1.609 km). It originated from the Roman mille passus, or “thousand paces,” which measured 5,000 Roman feet.
## How many miles are in 1 mile?
mile
——————
1 mi. or mi in …
SI units
imperial/US units
U.S. survey mile
## Which is more 26400 feet or 5 miles?
26400 feet = 5 miles Formula: multiply the value in feet by the conversion factor ‘0.00018939393939394’. So, 26400 feet = 26400 × 0.00018939393939394 = 5 miles.
## How many hours are in 1 mile?
It is a slow walking pace. One hour walking at 1 mph moves you 1 mile. Miles per hour is often used for car speeds. One minute at 60 mph will move you 1 mile.
## How many feet are in exactly 2 miles?
mi ft
—- ——
2.00 10,560
2.01 10,613
2.02 10,666
2.03 10,718
Join our community, and help us by sharing this post ! | 844 | 3,056 | {"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-2023-40 | latest | en | 0.924599 |
https://adreasnow.com/academic/Cheat%20Sheets%20and%20Play/SDL/Katya%20Pas/02/ | 1,725,967,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00105.warc.gz | 62,007,060 | 17,925 | # Principles of Quantum Chemistry¶
## First Principle - Dual Nature of Electrons¶
Any matter can act as a wave, based on the equation:
$\lambda=\frac{h}{m\cdot v}=\frac{h}{p}$
• For a car with a massive mass, this is quite tiny - $$2\e{-28}\:nm$$
• For an electron it is not - $$0.3\:nm$$
## Second Principle - Uncertainty Principle¶
There is a natural floor to the precision we can get for the properties of a quantum particle, governed by the equations:
$\begin{gather} \Delta x\cdot \Delta p \geq\frac{h}{2\pi}\\ \Delta E \cdot \Delta t \geq\frac{h}{2\pi} \end{gather}$
In quantum chemistry, we describe the wave as a wavefunction, with $$i$$ bringing in the imaginary component and $$a_0$$ giving the real component:
$\Psi(x,t)=a_0\cdot e^{2\pi i \big(\frac{x}{\lambda}-v\cdot t\big)}$
To get any useful values out of this, we need to square the wavefunction (multiply by it’s complex conjugate, as $$\sqrt{-1}^2=-1$$):
$\begin{gather} |\Psi|^2=\text{real component}\\ |\Psi|^2=\psi^*\cdot\psi=\frac{1}{\sqrt{2\pi}}e^{\big(-\frac{x^2}{2}\big)} \end{gather}$
This gives us a probability density for the wavefunction
## Third Principle - Derivation of the Schrödinger Equation¶
When we substitute in the kinetic energy $$T$$ and the potential energy $$V$$ to the Schrödinger equation, with our wavefunction, we get the time dependent Schrödinger equation:
$\hat{H}\Psi(x,y,z,t)=i\hbar\frac{\partial}{\partial t}\Psi(x,y,z,t)$
## Fourth Principle - Time Dependence¶
$$\hat{H}$$ is an operator (indicated by the hat). An operator is a function that transforms one function into another by a certain set of rules.
An operator acting on an eigenfunction (such as $$\Psi$$) returns an eigenvalue (scalar) multiplied by the original eigenfunction
\begin{align} <operator><eigenfunction>&=<eigenvlue><eigenfunction>\\ \hat{H}\Psi&=E\Psi \end{align}
There are operators for every quantum property, such as dipole moment and polarisability, though not all experimental properties can be attributed to an operator and thus cannot be accurately measured with the Schrödinger equation.
When we remove the time dependence of the Schrödinger equation, we can get the time independent equation:
$\hat{H}\Psi(x,y,z)=E\Psi(x,y,z)$
## Fifth Principle - Average Values¶
The properties that are calculate from the Schrödinger equation will not be logical/useful unless we average them:
$\langle E\rangle=\frac{\int_{-\infin}^{\infin}\psi^*\hat{H}\psi dr}{\int_{-\infin}^{\infin}\psi^*\psi dr}$
This separates the eigenvalue from the returned eigenfunction. | 728 | 2,563 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.729264 |
http://betterlesson.com/lesson/resource/3275533/unit-6-warm-up-5-pptx | 1,488,130,226,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172018.95/warc/CC-MAIN-20170219104612-00518-ip-10-171-10-108.ec2.internal.warc.gz | 24,944,429 | 20,984 | ## unit 6 warm up 5.pptx - Section 1: Curriculum Reinforcer
unit 6 warm up 5.pptx
unit 6 warm up 5.pptx
# The Shape Data Makes: Dot Plots
Unit 6: Statistics
Lesson 7 of 12
## Big Idea: What are the statistics on seeing spots? Displaying data on dot plots.
Print Lesson
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Subject(s):
70 minutes
### Michelle Braggs
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Environment: Urban | 345 | 1,380 | {"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} | 2.9375 | 3 | CC-MAIN-2017-09 | latest | en | 0.842977 |
https://opentextbc.ca/physicstestbook2/chapter/temperature/ | 1,558,602,165,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257197.14/warc/CC-MAIN-20190523083722-20190523105722-00449.warc.gz | 594,506,239 | 34,134 | Temperature, Kinetic Theory, and the Gas Laws
Temperature
OpenStaxCollege
Learning Objectives
• Define temperature.
• Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales.
• Define thermal equilibrium.
• State the zeroth law of thermodynamics.
The concept of temperature has evolved from the common concepts of hot and cold. Human perception of what feels hot or cold is a relative one. For example, if you place one hand in hot water and the other in cold water, and then place both hands in tepid water, the tepid water will feel cool to the hand that was in hot water, and warm to the one that was in cold water. The scientific definition of temperature is less ambiguous than your senses of hot and cold. Temperature is operationally defined to be what we measure with a thermometer. (Many physical quantities are defined solely in terms of how they are measured. We shall see later how temperature is related to the kinetic energies of atoms and molecules, a more physical explanation.) Two accurate thermometers, one placed in hot water and the other in cold water, will show the hot water to have a higher temperature. If they are then placed in the tepid water, both will give identical readings (within measurement uncertainties). In this section, we discuss temperature, its measurement by thermometers, and its relationship to thermal equilibrium. Again, temperature is the quantity measured by a thermometer.
Misconception Alert: Human Perception vs. Reality
On a cold winter morning, the wood on a porch feels warmer than the metal of your bike. The wood and bicycle are in thermal equilibrium with the outside air, and are thus the same temperature. They feel different because of the difference in the way that they conduct heat away from your skin. The metal conducts heat away from your body faster than the wood does (see more about conductivity in Conduction). This is just one example demonstrating that the human sense of hot and cold is not determined by temperature alone.
Another factor that affects our perception of temperature is humidity. Most people feel much hotter on hot, humid days than on hot, dry days. This is because on humid days, sweat does not evaporate from the skin as efficiently as it does on dry days. It is the evaporation of sweat (or water from a sprinkler or pool) that cools us off.
Any physical property that depends on temperature, and whose response to temperature is reproducible, can be used as the basis of a thermometer. Because many physical properties depend on temperature, the variety of thermometers is remarkable. For example, volume increases with temperature for most substances. This property is the basis for the common alcohol thermometer, the old mercury thermometer, and the bimetallic strip ([link]). Other properties used to measure temperature include electrical resistance and color, as shown in [link], and the emission of infrared radiation, as shown in [link].
The curvature of a bimetallic strip depends on temperature. (a) The strip is straight at the starting temperature, where its two components have the same length. (b) At a higher temperature, this strip bends to the right, because the metal on the left has expanded more than the metal on the right.
Each of the six squares on this plastic (liquid crystal) thermometer contains a film of a different heat-sensitive liquid crystal material. Below , all six squares are black. When the plastic thermometer is exposed to temperature that increases to , the first liquid crystal square changes color. When the temperature increases above the second liquid crystal square also changes color, and so forth. (credit: Arkrishna, Wikimedia Commons)
Fireman Jason Ormand uses a pyrometer to check the temperature of an aircraft carrier’s ventilation system. Infrared radiation (whose emission varies with temperature) from the vent is measured and a temperature readout is quickly produced. Infrared measurements are also frequently used as a measure of body temperature. These modern thermometers, placed in the ear canal, are more accurate than alcohol thermometers placed under the tongue or in the armpit. (credit: Lamel J. Hinton/U.S. Navy)
Temperature Scales
Thermometers are used to measure temperature according to well-defined scales of measurement, which use pre-defined reference points to help compare quantities. The three most common temperature scales are the Fahrenheit, Celsius, and Kelvin scales. A temperature scale can be created by identifying two easily reproducible temperatures. The freezing and boiling temperatures of water at standard atmospheric pressure are commonly used.
The Celsius scale (which replaced the slightly different centigrade scale) has the freezing point of water at and the boiling point at . Its unit is the degree Celsius. On the Fahrenheit scale (still the most frequently used in the United States), the freezing point of water is at and the boiling point is at . The unit of temperature on this scale is the degree Fahrenheit. Note that a temperature difference of one degree Celsius is greater than a temperature difference of one degree Fahrenheit. Only 100 Celsius degrees span the same range as 180 Fahrenheit degrees, thus one degree on the Celsius scale is 1.8 times larger than one degree on the Fahrenheit scale
The Kelvin scale is the temperature scale that is commonly used in science. It is an absolute temperature scale defined to have 0 K at the lowest possible temperature, called absolute zero. The official temperature unit on this scale is the kelvin, which is abbreviated K, and is not accompanied by a degree sign. The freezing and boiling points of water are 273.15 K and 373.15 K, respectively. Thus, the magnitude of temperature differences is the same in units of kelvins and degrees Celsius. Unlike other temperature scales, the Kelvin scale is an absolute scale. It is used extensively in scientific work because a number of physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature. The kelvin is the SI unit used in scientific work.
Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales, rounded to the nearest degree. The relative sizes of the scales are also shown.
The relationships between the three common temperature scales is shown in [link]. Temperatures on these scales can be converted using the equations in [link].
Temperature Conversions
To convert from . . . Use this equation . . . Also written as . . .
Celsius to Fahrenheit
Fahrenheit to Celsius
Celsius to Kelvin
Kelvin to Celsius
Fahrenheit to Kelvin
Kelvin to Fahrenheit
Notice that the conversions between Fahrenheit and Kelvin look quite complicated. In fact, they are simple combinations of the conversions between Fahrenheit and Celsius, and the conversions between Celsius and Kelvin.
Converting between Temperature Scales: Room Temperature
“Room temperature” is generally defined to be . (a) What is room temperature in ? (b) What is it in K?
Strategy
To answer these questions, all we need to do is choose the correct conversion equations and plug in the known values.
Solution for (a)
1. Choose the right equation. To convert from to , use the equation
2. Plug the known value into the equation and solve:
Solution for (b)
1. Choose the right equation. To convert from to K, use the equation
2. Plug the known value into the equation and solve:
Converting between Temperature Scales: the Reaumur Scale
The Reaumur scale is a temperature scale that was used widely in Europe in the 18th and 19th centuries. On the Reaumur temperature scale, the freezing point of water is and the boiling temperature is . If “room temperature” is on the Celsius scale, what is it on the Reaumur scale?
Strategy
To answer this question, we must compare the Reaumur scale to the Celsius scale. The difference between the freezing point and boiling point of water on the Reaumur scale is . On the Celsius scale it is . Therefore . Both scales start at for freezing, so we can derive a simple formula to convert between temperatures on the two scales.
Solution
1. Derive a formula to convert from one scale to the other:
2. Plug the known value into the equation and solve:
Temperature Ranges in the Universe
[link] shows the wide range of temperatures found in the universe. Human beings have been known to survive with body temperatures within a small range, from to to ). The average normal body temperature is usually given as (), and variations in this temperature can indicate a medical condition: a fever, an infection, a tumor, or circulatory problems (see [link]).
This image of radiation from a person’s body (an infrared thermograph) shows the location of temperature abnormalities in the upper body. Dark blue corresponds to cold areas and red to white corresponds to hot areas. An elevated temperature might be an indication of malignant tissue (a cancerous tumor in the breast, for example), while a depressed temperature might be due to a decline in blood flow from a clot. In this case, the abnormalities are caused by a condition called hyperhidrosis. (credit: Porcelina81, Wikimedia Commons)
The lowest temperatures ever recorded have been measured during laboratory experiments: at the Massachusetts Institute of Technology (USA), and at Helsinki University of Technology (Finland). In comparison, the coldest recorded place on Earth’s surface is Vostok, Antarctica at 183 K , and the coldest place (outside the lab) known in the universe is the Boomerang Nebula, with a temperature of 1 K.
Each increment on this logarithmic scale indicates an increase by a factor of ten, and thus illustrates the tremendous range of temperatures in nature. Note that zero on a logarithmic scale would occur off the bottom of the page at infinity.
Making Connections: Absolute Zero
What is absolute zero? Absolute zero is the temperature at which all molecular motion has ceased. The concept of absolute zero arises from the behavior of gases. [link] shows how the pressure of gases at a constant volume decreases as temperature decreases. Various scientists have noted that the pressures of gases extrapolate to zero at the same temperature, . This extrapolation implies that there is a lowest temperature. This temperature is called absolute zero. Today we know that most gases first liquefy and then freeze, and it is not actually possible to reach absolute zero. The numerical value of absolute zero temperature is or 0 K.
Graph of pressure versus temperature for various gases kept at a constant volume. Note that all of the graphs extrapolate to zero pressure at the same temperature.
Thermal Equilibrium and the Zeroth Law of Thermodynamics
Thermometers actually take their own temperature, not the temperature of the object they are measuring. This raises the question of how we can be certain that a thermometer measures the temperature of the object with which it is in contact. It is based on the fact that any two systems placed in thermal contact (meaning heat transfer can occur between them) will reach the same temperature. That is, heat will flow from the hotter object to the cooler one until they have exactly the same temperature. The objects are then in thermal equilibrium, and no further changes will occur. The systems interact and change because their temperatures differ, and the changes stop once their temperatures are the same. Thus, if enough time is allowed for this transfer of heat to run its course, the temperature a thermometer registers does represent the system with which it is in thermal equilibrium. Thermal equilibrium is established when two bodies are in contact with each other and can freely exchange energy.
Furthermore, experimentation has shown that if two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system C, then A is also in thermal equilibrium with C. This conclusion may seem obvious, because all three have the same temperature, but it is basic to thermodynamics. It is called the zeroth law of thermodynamics.
The Zeroth Law of Thermodynamics
If two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C.
This law was postulated in the 1930s, after the first and second laws of thermodynamics had been developed and named. It is called the zeroth law because it comes logically before the first and second laws (discussed in Thermodynamics). An example of this law in action is seen in babies in incubators: babies in incubators normally have very few clothes on, so to an observer they look as if they may not be warm enough. However, the temperature of the air, the cot, and the baby is the same, because they are in thermal equilibrium, which is accomplished by maintaining air temperature to keep the baby comfortable.
Does the temperature of a body depend on its size?
No, the system can be divided into smaller parts each of which is at the same temperature. We say that the temperature is an intensive quantity. Intensive quantities are independent of size.
Section Summary
• Temperature is the quantity measured by a thermometer.
• Temperature is related to the average kinetic energy of atoms and molecules in a system.
• Absolute zero is the temperature at which there is no molecular motion.
• There are three main temperature scales: Celsius, Fahrenheit, and Kelvin.
• Temperatures on one scale can be converted to temperatures on another scale using the following equations:
• Systems are in thermal equilibrium when they have the same temperature.
• Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy.
• The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C.
Conceptual Questions
What does it mean to say that two systems are in thermal equilibrium?
Give an example of a physical property that varies with temperature and describe how it is used to measure temperature.
When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why.
If you add boiling water to a cup at room temperature, what would you expect the final equilibrium temperature of the unit to be? You will need to include the surroundings as part of the system. Consider the zeroth law of thermodynamics.
Problems & Exercises
What is the Fahrenheit temperature of a person with a fever?
Frost damage to most plants occurs at temperatures of or lower. What is this temperature on the Kelvin scale?
To conserve energy, room temperatures are kept at in the winter and in the summer. What are these temperatures on the Celsius scale?
and
A tungsten light bulb filament may operate at 2900 K. What is its Fahrenheit temperature? What is this on the Celsius scale?
The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale?
One of the hottest temperatures ever recorded on the surface of Earth was in Death Valley, CA. What is this temperature in Celsius degrees? What is this temperature in Kelvin?
(a) Suppose a cold front blows into your locale and drops the temperature by 40.0 Fahrenheit degrees. How many degrees Celsius does the temperature decrease when there is a decrease in temperature? (b) Show that any change in temperature in Fahrenheit degrees is nine-fifths the change in Celsius degrees.
(a)
(b)
(a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value?
Glossary
temperature
the quantity measured by a thermometer
Celsius scale
temperature scale in which the freezing point of water is and the boiling point of water is
degree Celsius
unit on the Celsius temperature scale
Fahrenheit scale
temperature scale in which the freezing point of water is and the boiling point of water is
degree Fahrenheit
unit on the Fahrenheit temperature scale
Kelvin scale
temperature scale in which 0 K is the lowest possible temperature, representing absolute zero
absolute zero
the lowest possible temperature; the temperature at which all molecular motion ceases
thermal equilibrium
the condition in which heat no longer flows between two objects that are in contact; the two objects have the same temperature
zeroth law of thermodynamics
law that states that if two objects are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object | 3,346 | 16,815 | {"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} | 3.359375 | 3 | CC-MAIN-2019-22 | latest | en | 0.932472 |
http://www.studyalways.com/forum/5145-body-on-application-constant-force-moves.aspx | 1,524,297,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945082.84/warc/CC-MAIN-20180421071203-20180421091203-00123.warc.gz | 529,600,303 | 9,705 | # A body on application of constant force moves with uniform
Options :
1. Speed
2. Velocity
3. Acceleration
4. Momentum
Explanation:
Speed is the distance travelled per unit time. In everyday use and in kinematics, the speed of an object is the magnitude of its velocity (the rate of change of its position); it is thus a scalar quantity. Newton's First Law of Motion states that 'Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it.' Based on that principle, a body on application of force will move with uniform speed, since speed is a scalar quantity. A scalar quantity or parameter has no directional component, only magnitude.
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A leader without a clear vision and plans only abuses his power because visions, dreams and plans are the fulcrum along which the loads of success will spine by your own efforts. And where power is abused, there is manipulation instead of inspiration.
-Israelmore Ayivor | 318 | 1,460 | {"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-2018-17 | latest | en | 0.922481 |
http://freetofindtruth.blogspot.com/2014/02/alfred-hitchcock-numerology.html | 1,585,670,737,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00115.warc.gz | 65,872,248 | 23,624 | ## Thursday, February 13, 2014
### 33 44 | Alfred Hitchcock Numerology, Life and Death
Alfred Hitchcock was born on August 13, 1899. The numerology of this day is 39.
• 8/13/1899 = 8+1+3+1+8+9+9 = 39
Alfred Hitchcock began making movies in Europe in 1922, but his career wouldn't take off until he moved to the United States. In March of 1939, Hitchcock would land in the movie making city of Hollywood and enhance his filmmaking career after signing a seven year contract with a major motion picture studio. At the time, he and Walt Disney are said to have had the greatest vision for what the movies could becomes for the masses. As I write this, I'm wondering if Alfred Hitchcock was aware of this numerology, and it factored into this movie title, The 39 Steps.
Hitchcock would become famous for his murder mystery films, as well as other great films. Interestingly enough, the number 44 seems to have much to do with murder and assassination, and the name Hitchcock, has a numerology of 44.
• Alfred = 1+3+6+9+5+4 = 28
• Hitchcock = 8+9+2+3+8+3+6+3+2 = 44
• Alfred Hitchcock = 72
• Alfred = 1+12+6+18+5+4 = 46
• Hitchcock = 8+9+20+3+8+3+15+3+11 = 80
• Alfred Hitchcock = 126
Is it possible to think, that the numerology of his name, possessed him to make the films that he went on to make? Is it possible his name was chosen for this very reason; that is to say that he would be the father of murder mysteries? Better yet, Alfred Hitchcock made 44 films during the sound era, and eight prior, unless you count his lost films, then ten.
Making the Alfred Hitchcock numerology all the more curious, is the numerology of the day he passed on. While it is true that he died at a point when his life was in decline, is it possible that someone helped accelerate the process? Hitchcock died on April 29, 1980; a date with a numerology of 33.
• 4/29/1980 = 4+2+9+1+9+8+0 = 33
• 4/29/1980 = 4+29+19+80 = 132
• 4/29/80 = 4+29+80 = 113
• Mainstream = 113
• Green Screen = 113
• Dishonest = 113
• April 29 is the 119th day of the year
Even if you just add 4 for April to 29, the day, you also get 33.
• April 29 = 4/29 = 4+29 = 33
Here is the N.Y. Times obituary for Alfred Hitchcock. Notice he is referred to as a master in the title, and recall that the number 33, is the master teacher number. | 714 | 2,304 | {"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-2020-16 | latest | en | 0.958117 |
https://essaysresearch.org/on-a-workday-the-average-decibel-level-of-a-busy-street-is-70-db-with-100-cars-passing-a-given-point-every-minute/ | 1,701,634,707,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00219.warc.gz | 283,496,203 | 9,672 | # On a workday the average decibel level of a busy street is 70 dB, with 100 cars passing a given point every minute.
On a workday the average decibel level of a busy street is 70 dB, with 100 cars passing a given point every minute.
On a workday the average decibel level of a busy street is 70 dB, with 100 cars passing a given point every minute. If the number of cars is reduced to 15 every minute on a weekend, what is the decibel level of the street?
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On a workday the average decibel level of a busy street is 70 dB, with 100 cars passing a given point every minute. was first posted on August 19, 2020 at 2:29 pm. | 233 | 885 | {"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} | 2.84375 | 3 | CC-MAIN-2023-50 | latest | en | 0.899868 |
https://book.huihoo.com/data-structures-and-algorithms-with-object-oriented-design-patterns-in-csharp/html/page65.html | 1,579,889,430,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00054.warc.gz | 350,573,016 | 2,384 | Data Structures and Algorithms with Object-Oriented Design Patterns in C#
In this section we determine the asymptotic behavior of logarithms. Interestingly, despite the fact that diverges as n gets large, for all integers . Hence, . Furthermore, as the following theorem will show, raised to any integer power is still O(n).
Theorem For every integer , .
extbfProof This result follows immediately from Theorem and the observation that for all integers ,
This observation can be proved by induction as follows:
Base Case Consider the limit
for the case k=1. Using L'Hôpital's rule we see that
Inductive Hypothesis Assume that Equation holds for . Consider the case k=m+1. Using L'Hôpital's rule we see that
Therefore, by induction on m, Equation holds for all integers .
For example, using this property of logarithms together with the rule for determining the asymptotic behavior of the product of two functions (Theorem ), we can determine that since , then . | 215 | 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} | 3.140625 | 3 | CC-MAIN-2020-05 | latest | en | 0.87757 |
https://www.ellipsix.net/blog/tagged/teaching.html | 1,713,885,746,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00351.warc.gz | 670,945,789 | 4,098 | 1. 2011
May
06
## Teaching physics
Best XKCD ever:
I can’t count the number of times I’ve tried to explain to people that all the analogies we use are just analogies, and that if you think about them too hard they don’t make sense, and that’s okay because physics is all about the math anyway.
2. 2010
Nov
19
## Teaching vectors
I’ve noticed that (many of) my intro-level physics students, who have typically just learned about vectors within the past year, have a hard time working with them in equations. For example, given an equation like $$\sum\vec{F} = m\vec{a}$$, I’ll typically think of the force vectors in a geometrical representation, as arrows. Vector addition is done by starting each vector from the endpoint of the previous one in the sum, and scalar multiplication just means scaling the length of the arrow. But almost invariably, whenever I try to walk a student through this representation, they don’t get it. It seems like they’re really not prepared to accept the idea of doing math on things that they can’t plug into their calculators.
What I’ve found to work is this: I tell the students that whenever they see a vector equation, write one copy of it for each dimension. Newton’s second law, for example, becomes a set of three: $$\sum F_x = m a_x$$, $$\sum F_y = m a_y$$, and $$\sum F_z = m a_z$$. That way, all they have to deal with are simple numerical equations. Sure, they miss out on understanding the true nature … | 348 | 1,453 | {"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} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.945936 |
https://quizlet.com/explanations/questions/find-the-following-specified-function-value-ysqrt-3-x7-for-x-14-ebbce6a7-05fcf456-4624-46a9-9989-84a977753995 | 1,701,936,526,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00804.warc.gz | 553,958,646 | 71,566 | Try the fastest way to create flashcards
Question
# Find the following specified function value.$y=\sqrt{-3 x+7}$ for x = -14
Solution
Verified
Step 1
1 of 2
Substitute $x$ with $-14$ in the function and simplify.
$y(-14)=\sqrt{-3\cdot (-14)+7}=\sqrt{42+7}=\sqrt{49}=7$
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5,067 solutions | 236 | 768 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 25, "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-2023-50 | latest | en | 0.537407 |
https://ko.webqc.org/balance.php?reaction=C2H5OH%28l%29+%2B+O2%28g%29+%3D+CO2%28g%29+%2B+H2O%28g%29 | 1,590,782,090,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406365.40/warc/CC-MAIN-20200529183529-20200529213529-00414.warc.gz | 409,879,272 | 13,859 | #### 온라인 화학반응식 계산기
균형 방정식:
C2H5OH(l) + 3 O2(g) = 2 CO2(g) + 3 H2O(g)
반응 방식: 이중 치환
Reaction stoichiometry 한계 반응물
화합물계수몰 질량무게
C2H5OH(l)146.06844
O2(g)331.9988
CO2(g)244.0095
H2O(g)318.01528
단위: 몰 질량 - g/mol, 질량 - g.
무료로 제공되는 이 화학 소프트웨어를 친구들에게 알리세요!
이 균형 방정식에 직접 링크:
화학 방정식을 균형에 대한 지침 :
• Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
• Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
• To enter an electron into a chemical equation use {-} or e
• To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
• Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
• Compound states [like (s) (aq) or (g)] are not required.
• If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
• Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.
• Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents.
완전한 비례값이 계산된 화학 방정식의 예 : 화학 방정식 시약 (전체 방정식을 권장)의 예 : 저희 화학반응식 계산기에 만족하셨다면 만족도 평가를 남겨주세요
오늘 계산된 화학반응식
온라인 화학 도구 메뉴로 돌아가기 | 565 | 1,560 | {"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-2020-24 | latest | en | 0.406067 |
https://oeis.org/A207825 | 1,624,601,662,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00346.warc.gz | 397,364,955 | 3,932 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A207825 Primes p such that (p+k)/(k+1) is a prime number for k = 1,...,6. 3
5516281, 18164161, 51755761, 175472641, 322030801, 668745001, 859992841, 1163859481, 1566071641, 3662011081, 4285181881, 4609064881, 4630772161, 5329427041, 5601138481, 5934191761, 7123263841, 7321991041, 7391371681 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The subsequence of p in A208455 such that (p+6)/7 is a prime. Also, the subsequence of terms of A071369 which end in the digit "1". All terms are of the form 2520m+1. The m-values are listed in A208549. LINKS FORMULA A207825(n) = A208549(n)*2520 + 1. PROG (PARI) {my(p=1); until(, isprime(p+=2520) | next; for(j=2, 7, isprime(p\j+1)|next(2)); print1(p", "))} CROSSREFS Sequence in context: A206176 A205430 A208455 * A204802 A227931 A254195 Adjacent sequences: A207822 A207823 A207824 * A207826 A207827 A207828 KEYWORD nonn AUTHOR M. F. Hasler, Feb 28 2012 STATUS approved
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Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
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Last modified June 25 02:03 EDT 2021. Contains 345445 sequences. (Running on oeis4.) | 494 | 1,461 | {"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-2021-25 | latest | en | 0.662504 |
https://www.thestudentroom.co.uk/showthread.php?t=186596 | 1,542,350,884,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742978.60/warc/CC-MAIN-20181116045735-20181116071735-00529.warc.gz | 1,026,923,644 | 40,394 | You are Here: Home >< Maths
# c3 help! watch
1. k, i'm pretty sure this is easy, but can't seem 2 work it out!
Q7). The points P and Q lie on the curve with equation y=e(.5x)
The x-coordinate of P and Q are ln4 and ln16 respectively.
a). Find and equation for the line PQ
c). Calculate the length, to 3 sig. figures of the line segment PQ
2. a) Find the y coordinates of P and Q. With this you can find the gradient of the line PQ by m= (y_q - y_p)/(x_q-x_p).
Then put the coordinates of P or Q into
y-y_1 = m(x-x_1)
c) Use Pythagoras.
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Top tips from students who have already aced their exams | 351 | 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} | 2.984375 | 3 | CC-MAIN-2018-47 | latest | en | 0.87061 |
https://www.quickpaycheckadvance.com/cash-advance-union-city-tn-cash-advance-near-me-bad-credit.html | 1,548,114,669,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00407.warc.gz | 918,608,134 | 5,831 | Shop for the credit offer with the lowest cost. Compare the APR and the finance charge, which includes loan fees, interest and other credit costs. You are looking for the lowest APR. Military personnel have special protections against super-high fees or rates, and all consumers in some states and the District of Columbia have some protections dealing with limits on rates. Even with these protections, payday loans can be expensive, particularly if you roll-over the loan and are responsible for paying additional fees. Other credit offers may come with lower rates and costs.
In the traditional retail model, borrowers visit a payday lending store and secure a small cash loan, with payment due in full at the borrower's next paycheck. The borrower writes a postdated check to the lender in the full amount of the loan plus fees. On the maturity date, the borrower is expected to return to the store to repay the loan in person. If the borrower does not repay the loan in person, the lender may redeem the check. If the account is short on funds to cover the check, the borrower may now face a bounced check fee from their bank in addition to the costs of the loan, and the loan may incur additional fees or an increased interest rate (or both) as a result of the failure to pay.
###### Jump up ^ \$15 on \$100 over 14 days is ratio of 15/100 = 0.15, so this is a 14-day rate. Over a year (365.25 days) this 14-day rate can aggregate to either 391% (assuming you carry the \$100 loan for a year, and pay \$15 every 14 days: 0.15 x (365.25/14) = 3.91, which converts to a percentage increase (interest rate) of: 3.91 x 100 = 391%) or 3733% (assuming you take out a new loan every 14 days that will cover your principal and "charge", and every new loan is taken at same 15% "charge" of the amount borrowed: (1 + 0.15)365.25/14 − 1 = 37.33, which converts to a percentage increase (interest rate) of: 37.33 x 100 = 3733%).
Tennessee: The State of Tennessee requires a minimum principal reduction. In order to comply with the minimum state-required principal reduction, Speedy Cash requires that minimum payments include a principal reduction of 2% or \$2.50 for Customers who get paid bi-weekly/twice-a-month, or 4% or \$5 for Customers who get paid monthly, whichever is greater.
Check City realizes that payday may not always come on the day that you need it to. With a fast, convenient payday loan from Check City you won’t have to worry. A payday loan is surprisingly easy to qualify for and affordable.* When used effectively, a payday loan is a superior alternative to late, over draft, over the limit, and reconnection fees as well as the long term affects that such fees can have over time due to changes in your credit score. A payday loan also gives you access to money you need to take advantage of great sales or limited time offers that you would otherwise miss out on if you waited till payday.
The content on this page provides general consumer information. It is not legal advice or regulatory guidance. The CFPB updates this information periodically. This information may include links or references to third-party resources or content. We do not endorse the third-party or guarantee the accuracy of this third-party information. There may be other resources that also serve your needs.
The content on this page provides general consumer information. It is not legal advice or regulatory guidance. The CFPB updates this information periodically. This information may include links or references to third-party resources or content. We do not endorse the third-party or guarantee the accuracy of this third-party information. There may be other resources that also serve your needs.
They are far superior to their online counterparts. This is an expensive loan; of course, but the customer service is excellent and the reps are extremely professional, yet pleasant and personable. Review the website and you'll agree there aren't hidden fees. The reps are "very up front" and knowledgeable. Totally satisfied with my experience so far. Just saying.....
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We want to hear from you and encourage a lively discussion among our users. Please help us keep our site clean and safe by following our posting guidelines, and avoid disclosing personal or sensitive information such as bank account or phone numbers. Any comments posted under NerdWallet's official account are not reviewed or endorsed by representatives of financial institutions affiliated with the reviewed products, unless explicitly stated otherwise.
Payday loans should be used for short-term financial needs only and not as a long-term financial solution. Any advance of money obtained through a payday loan is not intended to meet long-term financial needs. A payday loan should only be used to meet immediate short-term cash needs. Refinancing a payday loan rather than paying the debt in full when due will require the payment of additional charges. Customers with credit difficulties should seek credit counseling before entering into any payday loan transaction.
Prior to 2009 regulation of consumer credit was primarily conducted by the states and territories. Some states such as New South Wales and Queensland legislated effective annual interest rate caps of 48%.[53] In 2008 the Australian states and territories referred powers of consumer credit to the Commonwealth. In 2009 the National Consumer Credit Protection Act 2009 (Cth) was introduced, which initially treated payday lenders no differently from all other lenders. In 2013 Parliament tightened regulation on the payday lending further introducing the Consumer Credit and Corporations Legislation Amendment (Enhancements) Act 2012 (Cth) which imposed an effective APR cap of 48% for all consumer credit contracts (inclusive of all fees and charges). Payday lenders who provided a loan falling within the definition of a small amount credit contract (SACC), defined as a contract provided by a non authorised-deposit taking institution for less than \$2,000 for a term between 16 days and 1 year,[54] are permitted to charge a 20% establishment fee in addition to monthly (or part thereof) fee of 4% (effective 48% p.a.).[55] Payday lenders who provide a loan falling within the definition of a medium amount credit contract (MACC), defined as a credit contract provided by a non-deposit taking institution for between \$2,000–\$5,000 may charge a \$400 establishment fee in addition to the statutory interest rate cap of 48%. Payday lenders are still required to comply with Responsible lending obligations applying to all creditors. Unlike other jurisdictions Australian payday lenders providing SACC or MACC products are not required to display their fees as an effective annual interest rate percentage.[citation needed]
Payday loans are also much more expensive than other methods of borrowing money. In most cases, the annual percentage rate (APR) on a payday loan averages about 400%, but the APR is often as high as 5,000%. APRs for credit cards can range from about 9% to 30%; personal loans generally have lower APRs than credit cards. If possible, it is better to use a credit card or tap into your savings in the event of an emergency.
*Approval depends upon meeting legal, regulatory and underwriting requirements. If approved, online loans are funded the next business day. All times and dates are based on Eastern Standard Time (EST). Check `n Go and third party lenders may, at their discretion, verify application information by using national databases that may provide information from one or more national credit bureaus, and Check `n Go or third party lenders may take that into consideration in the approval process.
Affiliate Disclosure: There are links on this site that can be defined as affiliate links. This means that I may receive a small commission (at no cost to you) if you purchase something when clicking on the links that take you through to a different website. By clicking on the links, you are in no way obligated to buy.
Please Note:The material on this site is provided for informational purposes only and is not financial advice. Always consult a professional before making any financial decisions. | 1,756 | 8,560 | {"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-2019-04 | latest | en | 0.953192 |
https://convertoctopus.com/2802-milliliters-to-quarts | 1,708,687,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00844.warc.gz | 189,200,746 | 7,348 | ## Conversion formula
The conversion factor from milliliters to quarts is 0.0010566882049662, which means that 1 milliliter is equal to 0.0010566882049662 quarts:
1 ml = 0.0010566882049662 qt
To convert 2802 milliliters into quarts we have to multiply 2802 by the conversion factor in order to get the volume amount from milliliters to quarts. We can also form a simple proportion to calculate the result:
1 ml → 0.0010566882049662 qt
2802 ml → V(qt)
Solve the above proportion to obtain the volume V in quarts:
V(qt) = 2802 ml × 0.0010566882049662 qt
V(qt) = 2.9608403503154 qt
The final result is:
2802 ml → 2.9608403503154 qt
We conclude that 2802 milliliters is equivalent to 2.9608403503154 quarts:
2802 milliliters = 2.9608403503154 quarts
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 quart is equal to 0.33774195217702 × 2802 milliliters.
Another way is saying that 2802 milliliters is equal to 1 ÷ 0.33774195217702 quarts.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two thousand eight hundred two milliliters is approximately two point nine six one quarts:
2802 ml ≅ 2.961 qt
An alternative is also that one quart is approximately zero point three three eight times two thousand eight hundred two milliliters.
## Conversion table
### milliliters to quarts chart
For quick reference purposes, below is the conversion table you can use to convert from milliliters to quarts
milliliters (ml) quarts (qt)
2803 milliliters 2.962 quarts
2804 milliliters 2.963 quarts
2805 milliliters 2.964 quarts
2806 milliliters 2.965 quarts
2807 milliliters 2.966 quarts
2808 milliliters 2.967 quarts
2809 milliliters 2.968 quarts
2810 milliliters 2.969 quarts
2811 milliliters 2.97 quarts
2812 milliliters 2.971 quarts | 558 | 1,886 | {"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 | 4 | CC-MAIN-2024-10 | latest | en | 0.764698 |
https://www.jiskha.com/display.cgi?id=1353345494 | 1,516,248,668,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887065.16/warc/CC-MAIN-20180118032119-20180118052119-00018.warc.gz | 941,292,651 | 3,813 | # Physics
posted by .
The surface of the Sun is approximately 5700 K and the temperature of the Earth's surface is about 290 K. What total entropy change occurs when 1000 J of heat of energy is transferred from the Sun to the Earth?
• Physics -
Flows to the Earth
ΔS(E)=1000/290 =3.448 J/K
Flows out of the Sun
ΔS(S)=-1000/5700 =- 0.1754J/K
ΔS= ΔS(E)+ ΔS(S)= 3.448-0.1754=3.2726 J/K
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# A building with 600 floors has two elevators. The floors
Author Message
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Manager
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A building with 600 floors has two elevators. The floors [#permalink] 12 Dec 2006, 00:02
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A building with 600 floors has two elevators. The floors are evenly spaced, such that an elevator must travel 40 feet to reach an adjacent floor. Elvator A travels at a rate of 10 feet per second, whereas elevator B travels at a rate of 20 feet per second. If elevator A begins going down from the 450th floor at the same time that elevator B begins traveling up from the 5th floor, at what floor will the two elevators pass each other, assuming neither makes any stops?
Director
Joined: 05 Feb 2006
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Kudos [?]: 52 [0], given: 0
A travels 1 floor in 4 seconds. = 1/4
B travels 1 floor in 2 seconds. = 1/2
Common rate=1/4+1/2=3/4
There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)
So A will travel:
1/4*592=d
d=148 floors
B will travel:
1/2*592=d
d=296 floors
They will meet at:
A: 450-148=302
B: 5+296=301
They will pass each other at approximately 302nd floor....
Senior Manager
Joined: 20 Feb 2006
Posts: 373
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SimaQ wrote:
There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)
I think the part highlighted is wrong because if A was on the 10th floor and B was on the 5th floor then the amount of distance between them is 5*40
If you draw it:
10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th
Now you can see there are five lots of 40ft inbetween the 10th and 5th floor so if we want to calculate the distance in feet it should be (10-5)*40.
Furthermore, if you used the original rationale then the no. of floors worth of distance between the 10th and 9th floor would be (10-9+1)*40 = 80 ---- when the actual distance is 40
Senior Manager
Joined: 01 Sep 2006
Posts: 302
Location: Phoenix, AZ, USA
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Both traveling opp direction
relative speed = (10/40)+(20/40) floor per sec= 3/4 floor per sec
Distance to be traveled 450-5 = 445 floor
3/4 in 1 sec
445 in 445/(3/4)= 593 sec
Floor by B= 593*1/2=296
Now its at 296+5= 301 floor
B 593/4 = 148.25 450-148.25 = 301.5
Director
Joined: 05 Feb 2006
Posts: 901
Followers: 1
Kudos [?]: 52 [0], given: 0
SimaQ wrote:
There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)
I think the part highlighted is wrong because if A was on the 10th floor and B was on the 5th floor then the amount of distance between them is 5*40
If you draw it:
10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th
Now you can see there are five lots of 40ft inbetween the 10th and 5th floor so if we want to calculate the distance in feet it should be (10-5)*40.
Furthermore, if you used the original rationale then the no. of floors worth of distance between the 10th and 9th floor would be (10-9+1)*40 = 80 ---- when the actual distance is 40
I converted everything into floors in order not to mess with feet.... That way it is easier to calculate....
You would have been right if i used the rate of A and B in feet rather than floors.....
Senior Manager
Joined: 08 Jun 2006
Posts: 342
Location: Washington DC
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Distance between 450th and 5th floor = (445 -1) * 40ft = 444 * 40
Both the lifts together will cover this distance in 444 * 40 / (20 + 10) sec = 148 * 4 sec
B traveled = 148 * 4 * 20 ft
Number of floors B traveled = 148 * 4 * 20 /40 = 296
They will meet = 5 + 296 + 1 = 302
(Consider the 1st floor that is not at 40ft, the distance between 1st and 2nd flr is 40ft. If a lift travels 40ft, it will travel from 1st flr to 2nd flr)
Senior Manager
Joined: 20 Feb 2006
Posts: 373
Followers: 1
Kudos [?]: 2 [0], given: 0
SimaQ wrote:
SimaQ wrote:
There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)
I think the part highlighted is wrong because if A was on the 10th floor and B was on the 5th floor then the amount of distance between them is 5*40
If you draw it:
Even if you're talking about floors then there are
10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th
Now you can see there are five lots of 40ft inbetween the 10th and 5th floor so if we want to calculate the distance in feet it should be (10-5)*40.
Furthermore, if you used the original rationale then the no. of floors worth of distance between the 10th and 9th floor would be (10-9+1)*40 = 80 ---- when the actual distance is 40
I converted everything into floors in order not to mess with feet.... That way it is easier to calculate....
You would have been right if i used the rate of A and B in feet rather than floors.....
I'm still not sure - between 10th & 5th there are 4 floors between them (9,8,7,6) or 5 floors worth of space (200ft). Using your calc it would give (10-5)+1 = 6 floors?
10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th
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## ForceIndex
by bug man, 5316 days ago
The Force index is an index developed by Dr Alexander Elder. It measures the strength of bulls and bears in the market. The formula uses both close prices and volume data. It is calculated by subtracting the yesterday close price from today (current bar) close price, then multiplying the result by today volume. The values of this index could be positive or negative depending on whether today close price is higher or lower than yesterday close price.
The raw values of the Force index should be plotted as a histogram. But because the raw data is erratic, the index is better smoothed by an exponential moving average. The length of the moving average determines whether you want to measure bulls and bears strength in the short, intermediate or long term. Generally a two-day EMA is used for the short term while a 13-day exponential moving average is used for the intermediate term.
The Force index tells us that the bulls are in control if the signal is above the zero line, the bears are in control if the signal is below the zero line. Near the zero level, neither the bulls nor the bears are in control.
A divergence between the EMA and the stock prices could also be used as signal entries, these divergences could indicate that the security trend is about to change its direction.
What is this?
Type: Trading Indicator Object ID: 132 Country: All Market: All Style: Technical Analysis
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Trading financial instruments, including foreign exchange on margin, carries a high level of risk and is not suitable for all investors. The high degree of leverage can work against you as well as for you. Before deciding to invest in financial instruments or foreign exchange you should carefully consider your investment objectives, level of experience, and risk appetite. The possibility exists that you could sustain a loss of some or all of your initial investment and therefore you should not invest money that you cannot afford to lose. You should be aware of all the risks associated with trading and seek advice from an independent financial advisor if you have any doubts. | 628 | 3,092 | {"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-2024-10 | latest | en | 0.87593 |
https://www.aqua-calc.com/one-to-one/density/grain-per-us-quart/kilogram-per-cubic-centimeter/1 | 1,582,704,164,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146187.93/warc/CC-MAIN-20200226054316-20200226084316-00051.warc.gz | 629,623,035 | 8,719 | # 1 grain per US quart [gr/US qt] in kilograms per cubic centimeter
## gr/US qt to kg/cm³ unit converter of density
1 grain per US quart [gr/qt] = 6.8472244 × 10-8 kilogram per cubic centimeter [kg/cm³]
### grains per US quart to kilograms per cubic centimeter density conversion cards
• 1
through
25
grains per US quart
• 1 gr/qt to kg/cm³ = 6.8472244 × 10-8 kg/cm³
• 2 gr/qt to kg/cm³ = 1 × 10-7 kg/cm³
• 3 gr/qt to kg/cm³ = 2 × 10-7 kg/cm³
• 4 gr/qt to kg/cm³ = 3 × 10-7 kg/cm³
• 5 gr/qt to kg/cm³ = 3 × 10-7 kg/cm³
• 6 gr/qt to kg/cm³ = 4 × 10-7 kg/cm³
• 7 gr/qt to kg/cm³ = 5 × 10-7 kg/cm³
• 8 gr/qt to kg/cm³ = 5 × 10-7 kg/cm³
• 9 gr/qt to kg/cm³ = 6 × 10-7 kg/cm³
• 10 gr/qt to kg/cm³ = 7 × 10-7 kg/cm³
• 11 gr/qt to kg/cm³ = 8 × 10-7 kg/cm³
• 12 gr/qt to kg/cm³ = 8 × 10-7 kg/cm³
• 13 gr/qt to kg/cm³ = 9 × 10-7 kg/cm³
• 14 gr/qt to kg/cm³ = 1 × 10-6 kg/cm³
• 15 gr/qt to kg/cm³ = 1 × 10-6 kg/cm³
• 16 gr/qt to kg/cm³ = 1.1 × 10-6 kg/cm³
• 17 gr/qt to kg/cm³ = 1.2 × 10-6 kg/cm³
• 18 gr/qt to kg/cm³ = 1.2 × 10-6 kg/cm³
• 19 gr/qt to kg/cm³ = 1.3 × 10-6 kg/cm³
• 20 gr/qt to kg/cm³ = 1.4 × 10-6 kg/cm³
• 21 gr/qt to kg/cm³ = 1.4 × 10-6 kg/cm³
• 22 gr/qt to kg/cm³ = 1.5 × 10-6 kg/cm³
• 23 gr/qt to kg/cm³ = 1.6 × 10-6 kg/cm³
• 24 gr/qt to kg/cm³ = 1.6 × 10-6 kg/cm³
• 25 gr/qt to kg/cm³ = 1.7 × 10-6 kg/cm³
• 26
through
50
grains per US quart
• 26 gr/qt to kg/cm³ = 1.8 × 10-6 kg/cm³
• 27 gr/qt to kg/cm³ = 1.8 × 10-6 kg/cm³
• 28 gr/qt to kg/cm³ = 1.9 × 10-6 kg/cm³
• 29 gr/qt to kg/cm³ = 2 × 10-6 kg/cm³
• 30 gr/qt to kg/cm³ = 2.1 × 10-6 kg/cm³
• 31 gr/qt to kg/cm³ = 2.1 × 10-6 kg/cm³
• 32 gr/qt to kg/cm³ = 2.2 × 10-6 kg/cm³
• 33 gr/qt to kg/cm³ = 2.3 × 10-6 kg/cm³
• 34 gr/qt to kg/cm³ = 2.3 × 10-6 kg/cm³
• 35 gr/qt to kg/cm³ = 2.4 × 10-6 kg/cm³
• 36 gr/qt to kg/cm³ = 2.5 × 10-6 kg/cm³
• 37 gr/qt to kg/cm³ = 2.5 × 10-6 kg/cm³
• 38 gr/qt to kg/cm³ = 2.6 × 10-6 kg/cm³
• 39 gr/qt to kg/cm³ = 2.7 × 10-6 kg/cm³
• 40 gr/qt to kg/cm³ = 2.7 × 10-6 kg/cm³
• 41 gr/qt to kg/cm³ = 2.8 × 10-6 kg/cm³
• 42 gr/qt to kg/cm³ = 2.9 × 10-6 kg/cm³
• 43 gr/qt to kg/cm³ = 2.9 × 10-6 kg/cm³
• 44 gr/qt to kg/cm³ = 3 × 10-6 kg/cm³
• 45 gr/qt to kg/cm³ = 3.1 × 10-6 kg/cm³
• 46 gr/qt to kg/cm³ = 3.1 × 10-6 kg/cm³
• 47 gr/qt to kg/cm³ = 3.2 × 10-6 kg/cm³
• 48 gr/qt to kg/cm³ = 3.3 × 10-6 kg/cm³
• 49 gr/qt to kg/cm³ = 3.4 × 10-6 kg/cm³
• 50 gr/qt to kg/cm³ = 3.4 × 10-6 kg/cm³
• 51
through
75
grains per US quart
• 51 gr/qt to kg/cm³ = 3.5 × 10-6 kg/cm³
• 52 gr/qt to kg/cm³ = 3.6 × 10-6 kg/cm³
• 53 gr/qt to kg/cm³ = 3.6 × 10-6 kg/cm³
• 54 gr/qt to kg/cm³ = 3.7 × 10-6 kg/cm³
• 55 gr/qt to kg/cm³ = 3.8 × 10-6 kg/cm³
• 56 gr/qt to kg/cm³ = 3.8 × 10-6 kg/cm³
• 57 gr/qt to kg/cm³ = 3.9 × 10-6 kg/cm³
• 58 gr/qt to kg/cm³ = 4 × 10-6 kg/cm³
• 59 gr/qt to kg/cm³ = 4 × 10-6 kg/cm³
• 60 gr/qt to kg/cm³ = 4.1 × 10-6 kg/cm³
• 61 gr/qt to kg/cm³ = 4.2 × 10-6 kg/cm³
• 62 gr/qt to kg/cm³ = 4.2 × 10-6 kg/cm³
• 63 gr/qt to kg/cm³ = 4.3 × 10-6 kg/cm³
• 64 gr/qt to kg/cm³ = 4.4 × 10-6 kg/cm³
• 65 gr/qt to kg/cm³ = 4.5 × 10-6 kg/cm³
• 66 gr/qt to kg/cm³ = 4.5 × 10-6 kg/cm³
• 67 gr/qt to kg/cm³ = 4.6 × 10-6 kg/cm³
• 68 gr/qt to kg/cm³ = 4.7 × 10-6 kg/cm³
• 69 gr/qt to kg/cm³ = 4.7 × 10-6 kg/cm³
• 70 gr/qt to kg/cm³ = 4.8 × 10-6 kg/cm³
• 71 gr/qt to kg/cm³ = 4.9 × 10-6 kg/cm³
• 72 gr/qt to kg/cm³ = 4.9 × 10-6 kg/cm³
• 73 gr/qt to kg/cm³ = 5 × 10-6 kg/cm³
• 74 gr/qt to kg/cm³ = 5.1 × 10-6 kg/cm³
• 75 gr/qt to kg/cm³ = 5.1 × 10-6 kg/cm³
• 76
through
100
grains per US quart
• 76 gr/qt to kg/cm³ = 5.2 × 10-6 kg/cm³
• 77 gr/qt to kg/cm³ = 5.3 × 10-6 kg/cm³
• 78 gr/qt to kg/cm³ = 5.3 × 10-6 kg/cm³
• 79 gr/qt to kg/cm³ = 5.4 × 10-6 kg/cm³
• 80 gr/qt to kg/cm³ = 5.5 × 10-6 kg/cm³
• 81 gr/qt to kg/cm³ = 5.5 × 10-6 kg/cm³
• 82 gr/qt to kg/cm³ = 5.6 × 10-6 kg/cm³
• 83 gr/qt to kg/cm³ = 5.7 × 10-6 kg/cm³
• 84 gr/qt to kg/cm³ = 5.8 × 10-6 kg/cm³
• 85 gr/qt to kg/cm³ = 5.8 × 10-6 kg/cm³
• 86 gr/qt to kg/cm³ = 5.9 × 10-6 kg/cm³
• 87 gr/qt to kg/cm³ = 6 × 10-6 kg/cm³
• 88 gr/qt to kg/cm³ = 6 × 10-6 kg/cm³
• 89 gr/qt to kg/cm³ = 6.1 × 10-6 kg/cm³
• 90 gr/qt to kg/cm³ = 6.2 × 10-6 kg/cm³
• 91 gr/qt to kg/cm³ = 6.2 × 10-6 kg/cm³
• 92 gr/qt to kg/cm³ = 6.3 × 10-6 kg/cm³
• 93 gr/qt to kg/cm³ = 6.4 × 10-6 kg/cm³
• 94 gr/qt to kg/cm³ = 6.4 × 10-6 kg/cm³
• 95 gr/qt to kg/cm³ = 6.5 × 10-6 kg/cm³
• 96 gr/qt to kg/cm³ = 6.6 × 10-6 kg/cm³
• 97 gr/qt to kg/cm³ = 6.6 × 10-6 kg/cm³
• 98 gr/qt to kg/cm³ = 6.7 × 10-6 kg/cm³
• 99 gr/qt to kg/cm³ = 6.8 × 10-6 kg/cm³
• 100 gr/qt to kg/cm³ = 6.8 × 10-6 kg/cm³
• gr/qt stands for gr/US qt
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st/in to sl/cm conversion table, st/in to sl/cm unit converter or convert between all units of linear density measurement.
#### Calculators
Body Mass Index calculator using person's weight and height | 2,927 | 6,117 | {"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} | 2.84375 | 3 | CC-MAIN-2020-10 | latest | en | 0.119047 |
https://ansanswers.com/mathematics/question12832602 | 1,627,999,147,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00718.warc.gz | 113,298,654 | 16,537 | , 22.06.2019 20:20 epicriderexe
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https://community.qlik.com/thread/215963 | 1,516,168,419,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886815.20/warc/CC-MAIN-20180117043259-20180117063259-00315.warc.gz | 660,430,552 | 23,725 | 6 Replies Latest reply: May 8, 2016 12:28 PM by Steve Br.
# Dividing values
Hello!
I have a data file about production and two of the columns are time (minutes) and cost. It also contains info about plants, workers, etc. I would like to include a KPI that shows the cost per minute and changes when I select different plants on my visualization. So I need to divide cost by time but I have no idea how to include this as a measure. The values are non-integer. Can anybody help me out here please?
Thanks!
PS: I am only using Qlik Sense for two days now, so please keep in mind that I'm a total noob.
• ###### Re: Dividing values
One of the first thing to learn in Qlik is how to handle date and time values.
Data Types in QlikView
Get the Dates Right
Why don’t my dates work?
If you are looking at time always in terms of minutes, you could interprete / transform your time field so that it shows a numeric value in minutes. If it's currently a text value like '05:30', maybe like
Interval#(TIMEFIELD, 'mm:ss')*24*60 as TimeInMinutes,
...
FROM ...;
Then you can use a KPI expression like
=Sum(COSTFIELD) / Sum(TimeInMinutes)
to get the average cost per minute
• ###### Re: Dividing values
The time is shown in minutes so i don't need this LOAD expression, right? Just for future reference, where exactly should It be entered. And is there an easier way adjust settings like this, e.g. something like the format cells feature in Excel?
Now back to my question. When I enter =Sum(Cost) / Sum(Time) as an expression in the KPI filed Add measure, I get an error saying: Error in expression: ')' expected. Don't know what is wrong.
btw. the data looks like this:
Plant | Worker | Time | Cost
Pl.1 | W1 | 150 | 77,8
Pl.1 | W2 | 176 | 83,2
Pl.1 | W3 | 147 | 82,2
Pl.2 | W4 | 119 | 90,6
...............
Pl.35 | W453 | 155 | 67,8
Thanks again and sorry for the lame questions.
• ###### Re: Dividing values
Not sure what the issue is.
Attached a sample QVF with your sample table.
If this is not helping you, can you post your sample app?
• ###### Re: Dividing values
Awesome, it worked now! Thank you very much for the sample file!
Could you please also answer my question about this LOAD expression and if there is an easier way to change the time values in Qlik Sense. I am pretty sure that I'll need it in the future. Thanks!
• ###### Re: Dividing values
You need to understand how QlikView stores date and time values. See my previously posted links.
You would interpret source data that is not automatically read in as date / time using interpretation functions like Date#() , Timestamp#() or Time#() in the LOAD script.
You can set default formats in the LOAD script that helps QV to interpret your source data correctly.
• ###### Re: Dividing values
Okay, thank you so much! | 734 | 2,888 | {"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-2018-05 | longest | en | 0.837265 |
https://answerofmath.com/solved-likelihood-ratio-test-on-a-single-model/ | 1,696,261,633,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00186.warc.gz | 111,691,073 | 19,353 | # Solved – Likelihood ratio test on a single model
If i were To perform a likelihood ratio test where I compare two models A and B I would basically try to find out which of these models are the better one of these models fits the data best. But if i were to perform a LR test on only one of these models, what am i actually trying to find out? and by that i mean, what is the Null hypothesis? is it something in the line of H0: "the model fits data well"?
Contents
SmallChess is right when they say you can't do an LRT with a single model.
However, you can compare your model to an intercept only model, which would be equivalent to testing if all the coefficients except the intercept are 0.
(As an aside, you could specify that all coefficients, including the intercept, are 0, but this is not usually done).
Under this perspective, the null hypothesis remains unchanged; you are still determining if the change in log-likelihood by regressing your outcome onto your covariates is more different than would be expected if the covariates were not actually informative (note that the likelihood can ALWAYS be made smaller, even when the covariates are truly not informative). More concretely,
$$H_0: beta_i = 0 text{ for all covariates } x_i$$
$$H_A: exists beta_jneq 0 text{ for some covariate } x_j$$.
Here is an example. I'll create a linear regression where the coefficients are all 0, except the intercept.
``set.seed(0) library(tidyverse) d = data_frame( x1 = rnorm(100), x2 = rnorm(100) ) X = model.matrix(~ ., data = d) d\$y = X%*% c(1,0,0) + rnorm(100) model = glm(y~., data = d) model_null = glm(y~1, data = d) anova(model_null,model, test = 'LRT') Analysis of Deviance Table Model 1: y ~ 1 Model 2: y ~ x1 + x2 Resid. Df Resid. Dev Df Deviance Pr(>Chi) 1 99 88.533 2 97 86.211 2 2.322 0.2708 ``
We fail to reject the null because the change in likelihood (in this case deviance because I used the glm command) is not large enough.
Note that when I change one of the coefficients to a non-zero number
``d\$y = X%*% c(1,0,1) + rnorm(100) model = glm(y~., data = d) model_null = glm(y~1, data = d) anova(model_null,model, test = 'LRT') Analysis of Deviance Table Model 1: y ~ 1 Model 2: y ~ x1 + x2 Resid. Df Resid. Dev Df Deviance Pr(>Chi) 1 99 203.36 2 97 118.14 2 85.219 6.406e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 ``
We reject the null because the resulting change in likelihood (again, here it is deviance) is larger than chance would allow at the 0.05 level.
Rate this post | 777 | 2,662 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-40 | latest | en | 0.932551 |
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