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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://mapdldocs.pyansys.com/examples/gallery_examples/00-mapdl-examples/pressure_vessel.html | 1,656,179,826,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036077.8/warc/CC-MAIN-20220625160220-20220625190220-00080.warc.gz | 421,366,510 | 8,899 | Pressure Vessel#
This example demonstrates how to create a basic pressure vessel and apply a pressure to it.
Also shown here: - Various ways of accessing stress results from MAPDL. - Comparison between PRNSOL, VGET (efficient wrapping), and the legacy reader. - Notes regarding FULL vs. POWER graphics when using PRNSOL.
```import numpy as np
from ansys.mapdl.core import launch_mapdl
# start mapdl, enter the preprocessor, and set the units
mapdl = launch_mapdl()
mapdl.clear()
mapdl.prep7()
# US Customary system using inches (in, lbf*s2/in, s, °F).
mapdl.units("BIN")
```
Out:
```U.S. CUSTOMARY INCH UNITS SPECIFIED FOR INTERNAL
LENGTH = INCHES (IN)
MASS = LBF-S**2/IN
TIME = SECONDS (SEC)
TEMPERATURE = FAHRENHEIT
TOFFSET = 460.0
FORCE = LBF
HEAT = IN-LBF
PRESSURE = PSI (LBF/IN**2)
ENERGY = IN-LBF
POWER = IN-LBF/SEC
INPUT UNITS ARE ALSO SET TO BIN
```
Set the materials and element type
```mapdl.et(1, "SOLID285")
mapdl.mp("EX", 1, 10e6)
mapdl.mp("PRXY", 1, 0.3)
mapdl.mp("DENS", 1, 0.1)
print(mapdl.mplist())
```
Out:
```MATERIAL NUMBER 1
TEMP EX
0.1000000E+08
TEMP DENS
0.1000000
TEMP PRXY
0.3000000
```
Create the Geometry
```# area generation
height = 10
inner_width = 2.5
outer_width = 3.5
mapdl.rectng(inner_width, outer_width, 0, height)
mapdl.cyl4(0, height, inner_width, 0, outer_width, 90)
# combine areas
mapdl.aplot(color="grey", background="w", show_area_numbering=True)
# Generate a cylindrical volume by rotating an area pattern about an axis
mapdl.vrotat(a_comb, pax1=6, arc=90)
mapdl.vplot(color="grey", background="w", show_area_numbering=True, cpos="zy")
```
Create the mesh
```mapdl.smrtsize(1)
mapdl.esize(0.25, 0)
mapdl.mshape(1, "3D")
mapdl.mshkey(0)
mapdl.vmesh("ALL")
mapdl.eplot(color="grey", background="w")
```
Solve
```# boundary condition selection
mapdl.geometry.area_select([3, 5, 7])
mapdl.da("ALL", "SYMM")
mapdl.allsel()
# apply pressure
mapdl.geometry.area_select([1, 6])
mapdl.sfa("ALL", 1, "PRES", 1000)
mapdl.allsel()
# solver
mapdl.run("/SOL")
mapdl.antype(0)
mapdl.outres("ALL", "ALL")
mapdl.run("/STATUS,SOLU")
sol_output = mapdl.solve()
mapdl.finish()
```
Out:
```FINISH SOLUTION PROCESSING
***** ROUTINE COMPLETED ***** CP = 101.185
```
Post-Processing#
Enter the MAPDL post-postprocessing routine (/POST1) and obtain the von-mises stress for the single static solution. Here, we use MAPDL directly to obtain the results using a wrapper around the VGET method to efficiently obtain results without outputting to disk.
```# enter the postprocessing routine
mapdl.post1()
mapdl.set(1, 1)
# results directly from MAPDL's VGET command
# VGET, __VAR__, NODE, , S, EQV
nnum = mapdl.mesh.nnum
von_mises_mapdl = mapdl.post_processing.nodal_eqv_stress()
# we could print out the solution for each node with:
print(f"\nNode Stress (psi)")
for node_num, stress_value in zip(nnum[:5], von_mises_mapdl[:5]):
print(f"{node_num:<5d} {stress_value:.3f}")
print("...")
# or simply get the maximum stress value and corresponding node
idx = np.argmax(von_mises_mapdl)
node_num = nnum[idx]
stress_value = von_mises_mapdl[idx]
print(f"\nMaximum Stress")
print(f"Node Stress (psi)")
print(f"{node_num:<5d} {stress_value:.3f}")
```
Out:
```Node Stress (psi)
1 3301.403
2 1906.958
3 1387.545
4 1373.553
5 867.288
...
Maximum Stress
Node Stress (psi)
1004 3483.042
```
Plot the results
```mapdl.post_processing.plot_nodal_eqv_stress(cpos="zy")
```
We could, alternatively, get the exact same results by directly accessing the result file using the legacy file reader ansys-mapdl-reader.
```# access the result
result = mapdl.result
# Get the von mises stess and show that this is equivalent to the
# stress obtained from MAPDL.
nnum, stress = result.principal_nodal_stress(0)
von_mises = stress[:, -1] # von-Mises stress is the right most column
min_von_mises, max_von_mises = np.min(von_mises), np.max(von_mises)
print("All close:", np.allclose(von_mises, von_mises_mapdl))
```
Out:
```All close: True
```
That these results are equivalent to results from PRNSOL.
Note
Enabling POWER GRAPHICS with `mapdl.graphics('POWER')` will change the averaging scheme.
```mapdl.header("OFF", "OFF", "OFF", "OFF", "OFF", "OFF")
table = mapdl.prnsol("S", "PRIN").splitlines()[1:]
prnsol_eqv = np.genfromtxt(table)[:, -1] # eqv is the last column
# show these are equivalent (RTOL due to rounding within PRNSOL)
print("All close:", np.allclose(von_mises, prnsol_eqv, rtol=1e-4))
print(f"LEGACY Reader and MAPDL VGET Min: {min_von_mises}")
print(f"PRNSOL MAPDL Min: {prnsol_eqv.min()}")
print()
print(f"LEGACY Reader and MAPDL VGET Min: {max_von_mises}")
print(f"PRNSOL MAPDL Min: {prnsol_eqv.max()}")
```
Out:
```All close: True
LEGACY Reader and MAPDL VGET Min: 691.8562018090861
PRNSOL MAPDL Min: 691.8562017
LEGACY Reader and MAPDL VGET Min: 3483.0418952187233
PRNSOL MAPDL Min: 3483.041895
```
stop mapdl
```mapdl.exit()
```
Total running time of the script: ( 0 minutes 6.020 seconds)
Gallery generated by Sphinx-Gallery | 1,608 | 5,174 | {"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-2022-27 | latest | en | 0.463722 |
https://www.coursehero.com/file/200538/124quiz5solutions/ | 1,519,214,108,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813608.70/warc/CC-MAIN-20180221103712-20180221123712-00438.warc.gz | 797,620,408 | 48,002 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
124quiz5solutions
# 124quiz5solutions - Math 124 Quiz 5 Oct 20 2006 SOLUTIONS 1...
This preview shows pages 1–2. Sign up to view the full content.
Math 124 Quiz 5 Oct. 20, 2006 SOLUTIONS 1) Let g ( x ) = a sin( x + 1) + 3, where a is a non-zero constant. (a) Find the local linearization L ( x ) for values of x near - 1. (Note your answer will be in terms of a ). (b) Use the local linearization to approximate g ( - . 95). Solutions: (a) First, g 0 ( x ) = a cos( x + 1). Then L ( x ) = g ( - 1) + g 0 ( - 1)( x - ( - 1)) = 3 + a ( x + 1) (b) g ( - 0 . 95) L ( - 0 . 95) = 3 + a ( - 0 . 95 + 1) = 3 + 0 . 05 a 2) The graph of a function f ( x ) is given below. The line y = x + 1 is the tangent line to f ( x ) at the point (3 , 4). (a) What is a in the tangent line approximation formula in this situation? (b) Find f ( a ) and f 0 ( a ). (c) Estimate f (2 . 9). (d) Is your estimate for (c) an underestimate or overestimate? Give an explanation. Solutions: (a) 3 (b) f ( a ) = 4 and f 0 ( a ) = 1 , i.e. the slope of the tangent line. (c) Note that here L ( x ) = x + 1, the y -value of the tangent line. Hence, f (2 . 9) L (2 . 9) = 2 . 9 + 1 = 3 . 9 (d) The above estimate is an
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 2
124quiz5solutions - Math 124 Quiz 5 Oct 20 2006 SOLUTIONS 1...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 578 | 1,665 | {"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-2018-09 | latest | en | 0.724908 |
https://www.solvedlib.com/chapter-2-section-23-additional-question-02-a,382787 | 1,695,971,270,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510498.88/warc/CC-MAIN-20230929054611-20230929084611-00844.warc.gz | 1,069,892,816 | 16,197 | # Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 100 gal of fresh water. Then water containinglb of
###### Question:
Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 100 gal of fresh water. Then water containinglb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 10 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min. Round your answer to two decimal places lbs
#### Similar Solved Questions
##### Lab 11 Analysis Qucstion 3 Homtcork unansucrcoComparing the-pGLO plates on LB/Amp (left} and LB only (right), whichof the statements bclow best explain why thesetwo plates Jre used a5 controls is INCORRECT?~RGLo~PGLoLB/AMPSticctan ansae and *bminkeyboaid navigation; use the up} down antowkeys [0 sclecte Fme
Lab 11 Analysis Qucstion 3 Homtcork unansucrco Comparing the-pGLO plates on LB/Amp (left} and LB only (right), whichof the statements bclow best explain why thesetwo plates Jre used a5 controls is INCORRECT? ~RGLo ~PGLo LB/AMP Sticctan ansae and *bmin keyboaid navigation; use the up} down antowkeys ...
##### Problem 4Let (X, p) be metric space: Recall that uniformly continuous function f X 7 R extends uniquely to a (uniformly_ continuous g C 7 R, where C is the completion of X and X € C, SO g(x) f(z) for each € X Explain why unique, continuous extension exists also for X = Q and C = R and any continuous f : Q - R.
Problem 4 Let (X, p) be metric space: Recall that uniformly continuous function f X 7 R extends uniquely to a (uniformly_ continuous g C 7 R, where C is the completion of X and X € C, SO g(x) f(z) for each € X Explain why unique, continuous extension exists also for X = Q and C = R and a...
##### 5. A1,00 Msolution of an unknown monoprotic acld Is Utrated with Naoh; When 85.0%4,9f Ehls Weak acid has reacted the PH of the resulting solutlon I5 5.33. Determlne the Kjof this Weakiacla (2 Marks)6. Calculate Ehe velumne 0j 0.25 M sodlum dlhvdrogen phosphate that must be added Eo }50 mL of,0150 M spdium hvdrogen phosphate to prapare 3 Duffer Isolutlonlaripk xoollci 5 murki)
5. A1,00 Msolution of an unknown monoprotic acld Is Utrated with Naoh; When 85.0%4,9f Ehls Weak acid has reacted the PH of the resulting solutlon I5 5.33. Determlne the Kjof this Weakiacla (2 Marks) 6. Calculate Ehe velumne 0j 0.25 M sodlum dlhvdrogen phosphate that must be added Eo }50 mL of,0150 ...
##### Part A What is the speed of an electron with a de Broglie wavelength of 0.17nm...
Part A What is the speed of an electron with a de Broglie wavelength of 0.17nm ? Express your answer using two significant figures. Part B What is the speed of a proton with a de Broglie wavelength of 0.17nm ? Express your answer using two significant figures....
##### ~5 10Let T : Fs _ } F4 be a linear trausformation with [Tlsand supposc that RREF([Tls)=What is the range of T?Range(T)b) Is T' onto? Briefly explain.What is the nullspace of T?N(T)d) Is T one-to-one? Briedy explain.
~5 10 Let T : Fs _ } F4 be a linear trausformation with [Tls and supposc that RREF([Tls)= What is the range of T? Range(T) b) Is T' onto? Briefly explain. What is the nullspace of T? N(T) d) Is T one-to-one? Briedy explain....
##### Given the following differential equation: dx Ax + eAt _ (2sin(3t) + bcos(t)) , x € Rn dtandy (t) e-Atc (t) _ What is dy? dtdy dt~2Ae-Atc(t) + 2sin(3t) + 5cos(t)dy dt2Ae-Atz(t) + 2sin(3t) + 5cos(t)dy ~2sin(3t) 5cos(t)dy dt2sin(3t) 5cos(t)
Given the following differential equation: dx Ax + eAt _ (2sin(3t) + bcos(t)) , x € Rn dt and y (t) e-Atc (t) _ What is dy? dt dy dt ~2Ae-Atc(t) + 2sin(3t) + 5cos(t) dy dt 2Ae-Atz(t) + 2sin(3t) + 5cos(t) dy ~2sin(3t) 5cos(t) dy dt 2sin(3t) 5cos(t)... | 1,285 | 3,913 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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} | 3.328125 | 3 | CC-MAIN-2023-40 | latest | en | 0.873605 |
https://db0nus869y26v.cloudfront.net/en/Vertical_deflection | 1,713,837,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00504.warc.gz | 174,173,426 | 12,928 | The vertical deflection (VD) or deflection of the vertical (DoV), also known as deflection of the plumb line and astro-geodetic deflection, is a measure of how far the gravity direction at a given point of interest is rotated by local mass anomalies such as nearby mountains. They are widely used in geodesy, for surveying networks and for geophysical purposes.
The vertical deflection are the angular components between the true zenithnadir curve (plumb line) tangent line and the normal vector to the surface of the reference ellipsoid (chosen to approximate the Earth's sea-level surface). VDs are caused by mountains and by underground geological irregularities and can amount to angles of 10 in flat areas or 20–50″ in mountainous terrain).[citation needed]
The deflection of the vertical has a north–south component ξ (xi) and an east–west component η (eta). The value of ξ is the difference between the astronomic latitude and the geodetic latitude (taking north latitudes to be positive and south latitudes to be negative); the latter is usually calculated by geodetic network coordinates. The value of η is the product of cosine of latitude and the difference between the astronomic longitude and the longitude (taking east longitudes to be positive and west longitudes to be negative). When a new mapping datum replaces the old, with new geodetic latitudes and longitudes on a new ellipsoid, the calculated vertical deflections will also change.
Determination
The deflections reflect the undulation of the geoid and gravity anomalies, for they depend on the gravity field and its inhomogeneities.
Vertical deflections are usually determined astronomically. The true zenith is observed astronomically with respect to the stars, and the ellipsoidal zenith (theoretical vertical) by geodetic network computation, which always takes place on a reference ellipsoid. Additionally, the very local variations of the vertical deflection can be computed from gravimetric survey data and by means of digital terrain models (DTM), using a theory originally developed by Vening-Meinesz.
VDs are used in astrogeodetic levelling: as a vertical deflection describes the difference between the geoidal and ellipsoidal normal direction, it represents the horizontal spatial gradient of the geoid undulations of the geoid (i.e., the separation between geoid and reference ellipsoid).
In practice, the deflections are observed at special points with spacings of 20 or 50 kilometers. The densification is done by a combination of DTM models and areal gravimetry. Precise vertical deflection observations have accuracies of ±0.2″ (on high mountains ±0.5″), calculated values of about 1–2″.
The maximal vertical deflection of Central Europe seems to be a point near the Großglockner (3,798 m), the highest peak of the Austrian Alps. The approx. values are ξ = +50″ and η = −30″. In the Himalaya region, very asymmetric peaks may have vertical deflections up to 100″ (0.03°). In the rather flat area between Vienna and Hungary the values are less than 15", but scatter by ±10″ for irregular rock densities in the subsurface.
More recently, a combination of digital camera and tiltmeter have also been used, see zenith camera.[1]
Application
Vertical deflections are principally used in four matters:
1. For precise calculation of survey networks. The geodetic theodolites and levelling instruments are oriented with respect to the true vertical, but its deflection exceeds the geodetic measuring accuracy by a factor of 5 to 50. Therefore, the data would have to be corrected exactly with respect to the global ellipsoid. Without these reductions, the surveys may be distorted by some centimeters or even decimeters per km.
2. For the geoid determination (mean sea level) and for exact transformation of elevations. The global geoidal undulations amount to 50–100 m, and their regional values to 10–50 m. They are adequate to the integrals of VD components ξ,η and therefore can be calculated with cm accuracy over distances of many kilometers.
3. For GPS surveys. The satellites measurements refer to a pure geometrical system (usually the WGS84 ellipsoid), whereas the terrestrial heights refer to the geoid. We need accurate geoid data to combine the different types of measurements.
4. For geophysics. Because VD data are affected by the physical structure of the Earth's crust and mantle, geodesists are engaged in models to improve our knowledge of the Earth's interior. Additionally and similar to applied geophysics, the VD data can support the future exploration of raw materials, oil, gas or ores.
Historical implications
Vertical deflections were used to measure Earth's density in the Schiehallion experiment.
Vertical deflection is the reason why modern prime meridian passes more than 100 m to the east of the historical astronomic prime meridian in Greenwich.[2]
The meridian arc measurement made by Nicolas-Louis de Lacaille north of Cape Town in 1752 (de Lacaille's arc measurement) was affected by vertical deflection.[3] The resulting discrepancy with Northern Hemisphere measurements was not explained until a visit to the area by George Everest in 1820; Maclear's arc measurement resurvey ultimately confirmed Everest's conjecture.[4]
Errors in the meridian arc of Delambre and Méchain determination, which affected the original definition of the metre,[5] were long known to be mainly caused by an uncertain determination of Barcelona's latitude later explained by vertical deflection.[6][7][8] When these errors where acknowledged in 1866,[9] it became urgent to proceed to a new measurement of the French arc between Dunkirk and Perpignan. The operations concerning the revision of the French arc linked to Spanish triangulation were completed only in 1896. Meanwhile the French geodesists had accomplished in 1879 the junction of Algeria to Spain, with the help of the geodesists of the Madrid Institute headed by the late Carlos Ibañez Ibáñez de Ibero (1825-1891), who had been president of the International Geodetic Association (now called International Association of Geodesy), first president of the International Committee for Weights and Measures and one of the 81 initial members of the International Statistical Institute.[10] Until Hayford ellipsoid was calculated in 1910, vertical deflections were considered as random errors.[11] Plumb line deviations were identified by Jean Le Rond d'Alembert as an important source of error in geodetic surveys as early as 1756, a few years later, in 1828, Carl Friedrich Gauss proposed the concept of geoid.[12][13]
References
1. ^ Hirt, C.; Bürki, B.; Somieski, A.; Seeber, G. N. (2010). "Modern Determination of Vertical Deflections Using Digital Zenith Cameras" (PDF). Journal of Surveying Engineering. 136: 1–12. doi:10.1061/(ASCE)SU.1943-5428.0000009. hdl:20.500.11937/34194.
2. ^ Malys, Stephen; Seago, John H.; Palvis, Nikolaos K.; Seidelmann, P. Kenneth; Kaplan, George H. (1 August 2015). "Why the Greenwich meridian moved". Journal of Geodesy. 89 (12): 1263. Bibcode:2015JGeod..89.1263M. doi:10.1007/s00190-015-0844-y.
3. ^ "Arc of the Meridian". Astronomical Society of South Africa. Retrieved 27 August 2020.
4. ^ Warner, Brian (1 April 2002). "Lacaille 250 years on". Astronomy and Geophysics. 43 (2): 2.25–2.26. doi:10.1046/j.1468-4004.2002.43225.x.
5. ^ Alder, K. (2002). The Measure of All Things: The Seven-year Odyssey and Hidden Error that Transformed the World. Free Press. ISBN 978-0-7432-1675-3. Retrieved 2020-08-02.
6. ^ Jean-Étienne Duby, Rapport sur les travaux de la Société de Physique et d’Histoire naturelle de Genève de juillet 1860 à juin 1861 par M. le Pasteur Duby. Lu à la séance du 13 juin 1861, in Mémoires de la Société de physique et d’histoire naturelle de Genève, 16 (1861-1862), 196-197.
7. ^ Vaníček, Petr; Foroughi, Ismael (2019). "How gravity field shortened our metre". Journal of Geodesy. 93 (9): 1821–1827. Bibcode:2019JGeod..93.1821V. doi:10.1007/s00190-019-01257-7. ISSN 0949-7714. S2CID 146099564.
8. ^ Levallois, Jean-Jacques (1991). "La méridienne de Dunkerque à Barcelone et la déterminiation du mètre (1972-1799)". E-Periodica (in French). doi:10.5169/seals-234595. Retrieved 2022-12-23.
9. ^ Hirsch, Adolphe (1865). "Sur les progrès des travaux géodésiques en Europe". E-Periodica (in French). doi:10.5169/seals-88030. Retrieved 2022-12-23.
10. ^ Clarke, Alexander Ross; Helmert, Friedrich Robert (1911). "Earth, Figure of the" . Encyclopædia Britannica. Vol. 8 (11th ed.). pp. 801–813. see page 811
11. ^ Géodésie in Encyclopedia Universalis. Encyclopedia Universalis. 1996. pp. Vol 10, p. 302. ISBN 978-2-85229-290-1. OCLC 36747385.
12. ^ d'Alembert, Jean Le Rond (1756). "Article Figure de la Terre, (Astron. Géog. Physiq. & Méch.), vol. VI (1756), p. 749b–761b". enccre.academie-sciences.fr. Retrieved 2022-12-23.
13. ^ US Department of Commerce, National Oceanic and Atmospheric Administration. "What is the geoid?". geodesy.noaa.gov. Retrieved 2022-12-23.
• The NGS website gives vertical deflection anywhere in the United States here and here. | 2,331 | 9,086 | {"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-2024-18 | latest | en | 0.894792 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/74/2/a/a/ | 1,716,938,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00650.warc.gz | 735,328,822 | 59,134 | # Properties
Label 74.2.a.a Level $74$ Weight $2$ Character orbit 74.a Self dual yes Analytic conductor $0.591$ Analytic rank $0$ Dimension $2$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [74,2,Mod(1,74)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(74, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("74.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$74 = 2 \cdot 37$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 74.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$0.590892974957$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{13})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{2} - x - 3$$ x^2 - x - 3 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: yes Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = \frac{1}{2}(1 + \sqrt{13})$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - q^{2} + (\beta + 1) q^{3} + q^{4} - \beta q^{5} + ( - \beta - 1) q^{6} + ( - 2 \beta + 2) q^{7} - q^{8} + (3 \beta + 1) q^{9} +O(q^{10})$$ q - q^2 + (b + 1) * q^3 + q^4 - b * q^5 + (-b - 1) * q^6 + (-2*b + 2) * q^7 - q^8 + (3*b + 1) * q^9 $$q - q^{2} + (\beta + 1) q^{3} + q^{4} - \beta q^{5} + ( - \beta - 1) q^{6} + ( - 2 \beta + 2) q^{7} - q^{8} + (3 \beta + 1) q^{9} + \beta q^{10} - \beta q^{11} + (\beta + 1) q^{12} + (\beta - 1) q^{13} + (2 \beta - 2) q^{14} + ( - 2 \beta - 3) q^{15} + q^{16} - 6 q^{17} + ( - 3 \beta - 1) q^{18} + 2 q^{19} - \beta q^{20} + ( - 2 \beta - 4) q^{21} + \beta q^{22} + (3 \beta - 3) q^{23} + ( - \beta - 1) q^{24} + (\beta - 2) q^{25} + ( - \beta + 1) q^{26} + (4 \beta + 7) q^{27} + ( - 2 \beta + 2) q^{28} + ( - 3 \beta + 3) q^{29} + (2 \beta + 3) q^{30} + ( - \beta + 2) q^{31} - q^{32} + ( - 2 \beta - 3) q^{33} + 6 q^{34} + 6 q^{35} + (3 \beta + 1) q^{36} + q^{37} - 2 q^{38} + (\beta + 2) q^{39} + \beta q^{40} + (3 \beta + 3) q^{41} + (2 \beta + 4) q^{42} + (2 \beta - 4) q^{43} - \beta q^{44} + ( - 4 \beta - 9) q^{45} + ( - 3 \beta + 3) q^{46} + 2 \beta q^{47} + (\beta + 1) q^{48} + ( - 4 \beta + 9) q^{49} + ( - \beta + 2) q^{50} + ( - 6 \beta - 6) q^{51} + (\beta - 1) q^{52} - 6 q^{53} + ( - 4 \beta - 7) q^{54} + (\beta + 3) q^{55} + (2 \beta - 2) q^{56} + (2 \beta + 2) q^{57} + (3 \beta - 3) q^{58} + (2 \beta + 6) q^{59} + ( - 2 \beta - 3) q^{60} + (5 \beta - 4) q^{61} + (\beta - 2) q^{62} + ( - 2 \beta - 16) q^{63} + q^{64} - 3 q^{65} + (2 \beta + 3) q^{66} + ( - 5 \beta + 8) q^{67} - 6 q^{68} + (3 \beta + 6) q^{69} - 6 q^{70} + 6 q^{71} + ( - 3 \beta - 1) q^{72} + ( - \beta - 10) q^{73} - q^{74} + q^{75} + 2 q^{76} + 6 q^{77} + ( - \beta - 2) q^{78} + (7 \beta - 7) q^{79} - \beta q^{80} + (6 \beta + 16) q^{81} + ( - 3 \beta - 3) q^{82} + ( - 4 \beta + 12) q^{83} + ( - 2 \beta - 4) q^{84} + 6 \beta q^{85} + ( - 2 \beta + 4) q^{86} + ( - 3 \beta - 6) q^{87} + \beta q^{88} - 4 \beta q^{89} + (4 \beta + 9) q^{90} + (2 \beta - 8) q^{91} + (3 \beta - 3) q^{92} - q^{93} - 2 \beta q^{94} - 2 \beta q^{95} + ( - \beta - 1) q^{96} + ( - 8 \beta + 2) q^{97} + (4 \beta - 9) q^{98} + ( - 4 \beta - 9) q^{99} +O(q^{100})$$ q - q^2 + (b + 1) * q^3 + q^4 - b * q^5 + (-b - 1) * q^6 + (-2*b + 2) * q^7 - q^8 + (3*b + 1) * q^9 + b * q^10 - b * q^11 + (b + 1) * q^12 + (b - 1) * q^13 + (2*b - 2) * q^14 + (-2*b - 3) * q^15 + q^16 - 6 * q^17 + (-3*b - 1) * q^18 + 2 * q^19 - b * q^20 + (-2*b - 4) * q^21 + b * q^22 + (3*b - 3) * q^23 + (-b - 1) * q^24 + (b - 2) * q^25 + (-b + 1) * q^26 + (4*b + 7) * q^27 + (-2*b + 2) * q^28 + (-3*b + 3) * q^29 + (2*b + 3) * q^30 + (-b + 2) * q^31 - q^32 + (-2*b - 3) * q^33 + 6 * q^34 + 6 * q^35 + (3*b + 1) * q^36 + q^37 - 2 * q^38 + (b + 2) * q^39 + b * q^40 + (3*b + 3) * q^41 + (2*b + 4) * q^42 + (2*b - 4) * q^43 - b * q^44 + (-4*b - 9) * q^45 + (-3*b + 3) * q^46 + 2*b * q^47 + (b + 1) * q^48 + (-4*b + 9) * q^49 + (-b + 2) * q^50 + (-6*b - 6) * q^51 + (b - 1) * q^52 - 6 * q^53 + (-4*b - 7) * q^54 + (b + 3) * q^55 + (2*b - 2) * q^56 + (2*b + 2) * q^57 + (3*b - 3) * q^58 + (2*b + 6) * q^59 + (-2*b - 3) * q^60 + (5*b - 4) * q^61 + (b - 2) * q^62 + (-2*b - 16) * q^63 + q^64 - 3 * q^65 + (2*b + 3) * q^66 + (-5*b + 8) * q^67 - 6 * q^68 + (3*b + 6) * q^69 - 6 * q^70 + 6 * q^71 + (-3*b - 1) * q^72 + (-b - 10) * q^73 - q^74 + q^75 + 2 * q^76 + 6 * q^77 + (-b - 2) * q^78 + (7*b - 7) * q^79 - b * q^80 + (6*b + 16) * q^81 + (-3*b - 3) * q^82 + (-4*b + 12) * q^83 + (-2*b - 4) * q^84 + 6*b * q^85 + (-2*b + 4) * q^86 + (-3*b - 6) * q^87 + b * q^88 - 4*b * q^89 + (4*b + 9) * q^90 + (2*b - 8) * q^91 + (3*b - 3) * q^92 - q^93 - 2*b * q^94 - 2*b * q^95 + (-b - 1) * q^96 + (-8*b + 2) * q^97 + (4*b - 9) * q^98 + (-4*b - 9) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - 2 q^{2} + 3 q^{3} + 2 q^{4} - q^{5} - 3 q^{6} + 2 q^{7} - 2 q^{8} + 5 q^{9}+O(q^{10})$$ 2 * q - 2 * q^2 + 3 * q^3 + 2 * q^4 - q^5 - 3 * q^6 + 2 * q^7 - 2 * q^8 + 5 * q^9 $$2 q - 2 q^{2} + 3 q^{3} + 2 q^{4} - q^{5} - 3 q^{6} + 2 q^{7} - 2 q^{8} + 5 q^{9} + q^{10} - q^{11} + 3 q^{12} - q^{13} - 2 q^{14} - 8 q^{15} + 2 q^{16} - 12 q^{17} - 5 q^{18} + 4 q^{19} - q^{20} - 10 q^{21} + q^{22} - 3 q^{23} - 3 q^{24} - 3 q^{25} + q^{26} + 18 q^{27} + 2 q^{28} + 3 q^{29} + 8 q^{30} + 3 q^{31} - 2 q^{32} - 8 q^{33} + 12 q^{34} + 12 q^{35} + 5 q^{36} + 2 q^{37} - 4 q^{38} + 5 q^{39} + q^{40} + 9 q^{41} + 10 q^{42} - 6 q^{43} - q^{44} - 22 q^{45} + 3 q^{46} + 2 q^{47} + 3 q^{48} + 14 q^{49} + 3 q^{50} - 18 q^{51} - q^{52} - 12 q^{53} - 18 q^{54} + 7 q^{55} - 2 q^{56} + 6 q^{57} - 3 q^{58} + 14 q^{59} - 8 q^{60} - 3 q^{61} - 3 q^{62} - 34 q^{63} + 2 q^{64} - 6 q^{65} + 8 q^{66} + 11 q^{67} - 12 q^{68} + 15 q^{69} - 12 q^{70} + 12 q^{71} - 5 q^{72} - 21 q^{73} - 2 q^{74} + 2 q^{75} + 4 q^{76} + 12 q^{77} - 5 q^{78} - 7 q^{79} - q^{80} + 38 q^{81} - 9 q^{82} + 20 q^{83} - 10 q^{84} + 6 q^{85} + 6 q^{86} - 15 q^{87} + q^{88} - 4 q^{89} + 22 q^{90} - 14 q^{91} - 3 q^{92} - 2 q^{93} - 2 q^{94} - 2 q^{95} - 3 q^{96} - 4 q^{97} - 14 q^{98} - 22 q^{99}+O(q^{100})$$ 2 * q - 2 * q^2 + 3 * q^3 + 2 * q^4 - q^5 - 3 * q^6 + 2 * q^7 - 2 * q^8 + 5 * q^9 + q^10 - q^11 + 3 * q^12 - q^13 - 2 * q^14 - 8 * q^15 + 2 * q^16 - 12 * q^17 - 5 * q^18 + 4 * q^19 - q^20 - 10 * q^21 + q^22 - 3 * q^23 - 3 * q^24 - 3 * q^25 + q^26 + 18 * q^27 + 2 * q^28 + 3 * q^29 + 8 * q^30 + 3 * q^31 - 2 * q^32 - 8 * q^33 + 12 * q^34 + 12 * q^35 + 5 * q^36 + 2 * q^37 - 4 * q^38 + 5 * q^39 + q^40 + 9 * q^41 + 10 * q^42 - 6 * q^43 - q^44 - 22 * q^45 + 3 * q^46 + 2 * q^47 + 3 * q^48 + 14 * q^49 + 3 * q^50 - 18 * q^51 - q^52 - 12 * q^53 - 18 * q^54 + 7 * q^55 - 2 * q^56 + 6 * q^57 - 3 * q^58 + 14 * q^59 - 8 * q^60 - 3 * q^61 - 3 * q^62 - 34 * q^63 + 2 * q^64 - 6 * q^65 + 8 * q^66 + 11 * q^67 - 12 * q^68 + 15 * q^69 - 12 * q^70 + 12 * q^71 - 5 * q^72 - 21 * q^73 - 2 * q^74 + 2 * q^75 + 4 * q^76 + 12 * q^77 - 5 * q^78 - 7 * q^79 - q^80 + 38 * q^81 - 9 * q^82 + 20 * q^83 - 10 * q^84 + 6 * q^85 + 6 * q^86 - 15 * q^87 + q^88 - 4 * q^89 + 22 * q^90 - 14 * q^91 - 3 * q^92 - 2 * q^93 - 2 * q^94 - 2 * q^95 - 3 * q^96 - 4 * q^97 - 14 * q^98 - 22 * q^99
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−1.30278 2.30278
−1.00000 −0.302776 1.00000 1.30278 0.302776 4.60555 −1.00000 −2.90833 −1.30278
1.2 −1.00000 3.30278 1.00000 −2.30278 −3.30278 −2.60555 −1.00000 7.90833 2.30278
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$+1$$
$$37$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 74.2.a.a 2
3.b odd 2 1 666.2.a.j 2
4.b odd 2 1 592.2.a.f 2
5.b even 2 1 1850.2.a.u 2
5.c odd 4 2 1850.2.b.i 4
7.b odd 2 1 3626.2.a.a 2
8.b even 2 1 2368.2.a.s 2
8.d odd 2 1 2368.2.a.ba 2
11.b odd 2 1 8954.2.a.p 2
12.b even 2 1 5328.2.a.bf 2
37.b even 2 1 2738.2.a.l 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
74.2.a.a 2 1.a even 1 1 trivial
592.2.a.f 2 4.b odd 2 1
666.2.a.j 2 3.b odd 2 1
1850.2.a.u 2 5.b even 2 1
1850.2.b.i 4 5.c odd 4 2
2368.2.a.s 2 8.b even 2 1
2368.2.a.ba 2 8.d odd 2 1
2738.2.a.l 2 37.b even 2 1
3626.2.a.a 2 7.b odd 2 1
5328.2.a.bf 2 12.b even 2 1
8954.2.a.p 2 11.b odd 2 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{3}^{2} - 3T_{3} - 1$$ acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(74))$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$(T + 1)^{2}$$
$3$ $$T^{2} - 3T - 1$$
$5$ $$T^{2} + T - 3$$
$7$ $$T^{2} - 2T - 12$$
$11$ $$T^{2} + T - 3$$
$13$ $$T^{2} + T - 3$$
$17$ $$(T + 6)^{2}$$
$19$ $$(T - 2)^{2}$$
$23$ $$T^{2} + 3T - 27$$
$29$ $$T^{2} - 3T - 27$$
$31$ $$T^{2} - 3T - 1$$
$37$ $$(T - 1)^{2}$$
$41$ $$T^{2} - 9T - 9$$
$43$ $$T^{2} + 6T - 4$$
$47$ $$T^{2} - 2T - 12$$
$53$ $$(T + 6)^{2}$$
$59$ $$T^{2} - 14T + 36$$
$61$ $$T^{2} + 3T - 79$$
$67$ $$T^{2} - 11T - 51$$
$71$ $$(T - 6)^{2}$$
$73$ $$T^{2} + 21T + 107$$
$79$ $$T^{2} + 7T - 147$$
$83$ $$T^{2} - 20T + 48$$
$89$ $$T^{2} + 4T - 48$$
$97$ $$T^{2} + 4T - 204$$ | 5,434 | 10,185 | {"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} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.518335 |
https://cs.stackexchange.com/questions/20102/query-about-fp-divide-latency-and-initiation-interval | 1,718,930,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00680.warc.gz | 161,708,621 | 38,570 | # Query about FP Divide latency and Initiation Interval
Latency is defined as the number of intervening cycles between an instruction that produces a result and an instruction that uses the result. The initiation or repeat interval is the number of cycles that must elapse between issuing two operations of a given type.
In the Appendix C, pages 52-54 of Computer Architecture: A Quantitative approach 5th edition, it was mentioned that
1. the FP adder is of 4 stages and pipelined, so latency =3(No. of pipeline stages -1) and initiation interval =1 (as pipelined)
2. the FP multiply is of 7 stages and pipelined, so latency =6(No. of pipeline stages -1) and initiation interval =1 (as pipelined)
So for 3. The FP divide of 24 stages and unpipelined, so latency =23(No. of pipeline stages -1) and initiation interval =24 (as unpipelined). But it was mentioned that the latency was 24 and initiation interval was 25.
Why is this?
Can anyone explain me? As I missing something | 238 | 982 | {"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-2024-26 | latest | en | 0.958658 |
https://www.physicsforums.com/threads/differing-first-principle-models-for-maxwell-boltzmann-statistics.973287/page-2 | 1,560,720,069,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998298.91/warc/CC-MAIN-20190616202813-20190616224813-00135.warc.gz | 843,931,943 | 21,132 | # I Differing first-principle models for Maxwell-Boltzmann statistics?
#### Dale
Mentor
I think even real electrons in a beer-electron-pong set-up would result in the dice model.
If that were true then we would be able to observe substantial departures from equipartition
#### greswd
If that were true then we would be able to observe substantial departures from equipartition
If we fire 9 electrons, one at a time, at a set-up of 6 holes where the electron has a 1/6th chance of entering any one hole, we can see from there that it is no different from beer-pong.
#### Dale
Mentor
If we fire 9 electrons, one at a time, at a set-up of 6 holes where the electron has a 1/6th chance of entering any one hole, we can see from there that it is no different from beer-pong.
If that were true then there would be a violation of equipartition. I don't have any evidence to suggest that is correct, do you?
#### greswd
If that were true then there would be a violation of equipartition. I don't have any evidence to suggest that is correct, do you?
But you can imagine a fair set-up, with a 1/6th chance each, like a fair die.
And the electrons are fired one at a time. It is virtually identical to beer pong.
#### Dale
Mentor
It is virtually identical to beer pong.
With the very big difference of distinguishability, right? You are still envisioning that the electrons are indistinguishable, but ping pong balls are not. That has testable consequences.
#### greswd
With the very big difference of distinguishability, right? You are still envisioning that the electrons are indistinguishable, but ping pong balls are not. That has testable consequences.
But can you imagine how it would differ?
We set it up with a fair 1/6th chance each, or very closely to perfect fairness.
But somehow the electrons behave differently and don't behave according to those odds.
That would be very strange. Electrons can indeed behave very strangely, or non-classically, but not in this manner.
#### DrClaude
Mentor
If I understand correctly, in the dice / beer pong analogy, you want to label the units of energy, distinguishing which is where?
If that is the case, you need to understand that keep track of the energy is only accounting (see the Feynman lectures). The state of the economy does not depend on which dollar is in which bank account.
#### greswd
If I understand correctly, in the dice / beer pong analogy, you want to label the units of energy, distinguishing which is where?
If that is the case, you need to understand that keep track of the energy is only accounting (see the Feynman lectures). The state of the economy does not depend on which dollar is in which bank account.
I mentioned removing all forms of labels in #15, and also mentioned throwing all 9 balls at the same time, albeit "ghostly" balls, in #10.
#### DrClaude
Mentor
I mentioned removing all forms of labels in #15, and also mentioned throwing all 9 balls at the same time, albeit "ghostly" balls, in #10.
Then I must admit I don't understand the discussion.
#### Dale
Mentor
Electrons can indeed behave very strangely, or non-classically, but not in this manner.
I think they do behave non-classically in exactly this manner. At least I am not aware of any evidence of a violation of the equipartition theorem. Are you? You seem very convinced by this, but the consequences would be easily observable.
#### greswd
I think they do behave non-classically in exactly this manner. At least I am not aware of any evidence of a violation of the equipartition theorem. Are you? You seem very convinced by this, but the consequences would be easily observable.
But electrons, no matter how non-classical, can't change the 1/6th odds of the fair setup. And its one at a time so two electrons can't interfere with one another.
So it could be that Maxwell-Boltzmann statistics are right but with the wrong explanation.
Or because they don't refer to electrons, but quanta of energy, which somehow behave differently.
#### Dale
Mentor
And its one at a time so two electrons can't interfere with one another.
If you detect after each electron where it went then that makes it distinguishable. If you fire them in one at a time so there is no interference but don't observe their locations until the end then I think they are indistinguishable and would have the corresponding statistics, not the beer pong statistics.
I don't think it works the way you seem to think, and I am not aware of any evidence suggesting it works the way you suggest which I think would be big-news kind of evidence. And you don't seem to be aware of any such evidence either.
It is certainly also possible that I am misunderstanding or misapplying the equipartition theorem, but as far as I know distinguishability has some pretty dramatic physical consequences.
#### greswd
If you detect after each electron where it went then that makes it distinguishable. If you fire them in one at a time so there is no interference but don't observe their locations until the end then I think they are indistinguishable and would have the corresponding statistics, not the beer pong statistics.
But if its like the double-slit experiment, with the beer cups being like the slits, the electrons just pass right through the slits, and there is nothing to observe at the end.
Kinda like having cups with the bottoms cut off, the ping-pong balls just fall right through, and at the end of the experiment, all we're left with is 6 empty cups.
#### Dale
Mentor
But if its like the double-slit experiment, with the beer cups being like the slits, the electrons just pass right through the slits, and there is nothing to observe at the end.
I didn't say anything about the double slit experiment. It is not the only experiment that invalidates the notion of counterfactual definiteness. There is no counterfactual definiteness in QM, regardless of if you are describing electrons in wells or photons through slits.
#### greswd
I didn't say anything about the double slit experiment. It is not the only experiment that invalidates the notion of counterfactual definiteness. There is no counterfactual definiteness in QM, regardless of if you are describing electrons in wells or photons through slits.
If you detect after each electron where it went then that makes it distinguishable. If you fire them in one at a time so there is no interference but don't observe their locations until the end then I think they are indistinguishable and would have the corresponding statistics, not the beer pong statistics.
noted, not double-slit. what kind of experimental set-up do you think would yield the corresponding statistics?
because I'm wondering what it means to not observe during the firing and observing at the end of it all.
#### Dale
Mentor
noted, not double-slit. what kind of experimental set-up do you think would yield the corresponding statistics?
because I'm wondering what it means to not observe during the firing and observing at the end of it all.
I would think of some sort of potential well created by a circuit where you could turn the individual wells on or off as needed. They would need to be deep enough potential wells that 9 electrons plus or minus would not alter the probabilities. Then you can turn off the wells one at a time and count how many electrons leave each well.
#### greswd
I would think of some sort of potential well created by a circuit where you could turn the individual wells on or off as needed. They would need to be deep enough potential wells that 9 electrons plus or minus would not alter the probabilities. Then you can turn off the wells one at a time and count how many electrons leave each well.
That means that somehow the electrons are defying the setup that every well has a 1/6th chance. Interesting, someone should totally conduct this experiment.
#### Dale
Mentor
That means that somehow the electrons are defying the setup that every well has a 1/6th chance.
Actually, no it doesn't mean that. If you go to all of the 2002 distinct configurations you still find that each well is equally likely to contain any given amount of energy. The 1/6 chance is not violated. All that is changed is the distinction if a well with one unit of energy got the first unit or the last unit. In either case it still had the same 1/6 chance to get it but those two scenarios are actually the same scenario.
"Differing first-principle models for Maxwell-Boltzmann statistics?"
### Physics Forums Values
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving | 1,948 | 8,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-26 | latest | en | 0.963554 |
https://answerbun.com/physics/good-quantum-numbers-in-l-s-coupling/ | 1,669,561,349,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00557.warc.gz | 131,535,979 | 16,011 | # Good Quantum numbers in $L$-$S$ coupling
Physics Asked by 1MegaMan1 on December 6, 2020
I’m having a hard time understanding a few things about L-S and j-j coupling in 2 (or more) electron atoms. What I picked up from our lectures is the following:
The Hamiltonian can be “broken up” into 3 parts, a Central field part, a residual electrostatic part and a Spin-Orbit part:
$$hat{H} = hat{H}_{CF}+Deltahat{H}_{RE} + Deltahat{H}_{SO}$$
where we can treat $$Deltahat{H}_{RE}$$ and $$Deltahat{H}_{SO}$$ as small perturbations. Which of these terms has to be tackled first is decided by their relative size, so that $$Deltahat{H}_{RE} > Deltahat{H}_{SO}$$ leads to what we call the LS-coupling scheme, where we treat $$Deltahat{H}_{RE}$$ first, and $$Deltahat{H}_{RE} < Deltahat{H}_{SO}$$ gives us the jj-coupling scheme, where $$Deltahat{H}_{SO}$$ is treated first.
In the LS-coupling scheme we find that in order to diagonalise $$Deltahat{H}_{RE}$$ we have to pick a basis of eigenstates labeled by the quantum numbers $$|L,M_L,S,M_Srangle$$ as these diagonalise the perturbation $$Deltahat{H}_{RE}$$. This then leads to energy shift, which depend on $$L$$ and $$M$$, $$Delta E_{RE} = Delta E_{RE}left(L,Sright)$$, and the introduction of term symbols.
Now we treat the smaller perturbation $$Deltahat{H}_{SO}$$. According to our lectures the following is true:
$$Deltahat{H}_{SO} = sum_i beta_i underline{hat{l}}_i cdot underline{hat{s}}_i$$
and also that:
$$langle underline{hat{l}}_i cdot underline{hat{s}}_i rangle propto langle underline{hat{L}}cdot underline{hat{S}}rangle = frac{1}{2} langle hat{J}^2 – hat{L}^2 – hat{S}^2 rangle$$
This seems to imply to me that, actually, the basis states of $$|L,S,J,M_J rangle$$ would diagonalise both the Residual Electrostatic perturbation and also the Spin-Orbit interaction at the same time.
This must obviously be wrong as if it was possible to diagonalise both perturbations at the same time using the same set of eigenfunctions, there would be no need to distinguish between the two cases of L-S coupling and j-j coupling.
I’d greatly appreciate it, if someone could explain to me where I went to wrong, or was able to point me to some resources explaining this in more detail.
Spin-orbit coupling is not coupling the total $$L$$ and $$S ,$$ but rather it is coupling the (lowercase) $$l$$ and $$s ;$$ i.e. it couples the spin angular momentum and the orbital angular momentum of each electron individually, not the total $$L$$ and $$S .$$
Answered by The Notorious on December 6, 2020
## Related Questions
### Accelerating observers in special relativity
2 Asked on November 19, 2020
### Entropic force in polymers
1 Asked on November 19, 2020
### Time-independent Klein-Gordon PDE
3 Asked on November 19, 2020 by jepsilon
### Conditions for water to go down from high places
1 Asked on November 18, 2020 by user12479
1 Asked on November 18, 2020 by a-quarky-name
### What is the amount of mass ($rm kg$) that the Sun loses daily in the irradiation form?
1 Asked on November 18, 2020 by joo-bosco
### How can I calculate the $D$ field when all the free charge is on a conductor embedded in a dielectric?
0 Asked on November 18, 2020 by moca-aoba
### How can we detect antihydrogen?
2 Asked on November 17, 2020
### Quantum model of the atom
3 Asked on November 17, 2020
### Difference between two derivations of length contraction?
3 Asked on November 17, 2020 by heatherfield
### Inequality between Holevo capacity and coherent information of quantum channels
0 Asked on November 16, 2020 by zouba
### About the $sigma_{xy}$ in the integer quantum Hall effect (or quantum anomalous Hall effect)
0 Asked on November 16, 2020 by fbs147
### Does the Higgs field explain inertia?
1 Asked on November 16, 2020 by john-eastmond
### Counting number of states for fermions
0 Asked on November 15, 2020
### Angular momentum singlet states for arbitrary representations
1 Asked on November 15, 2020 by toms
### Obtaining curved space Dirac equation from action (tetrad formalism)
2 Asked on November 15, 2020 by pam
### Particle in a potential well in mathematica
0 Asked on November 15, 2020 by user276420
### Turning a bra made up of a tensor product of two bras into a ket (and vice-versa)
1 Asked on November 15, 2020 by dja | 1,240 | 4,336 | {"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": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-49 | latest | en | 0.854517 |
https://wikidiff.com/category/terms/octadic | 1,627,505,683,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00123.warc.gz | 626,259,873 | 8,101 | #### Eightfold vs Octadic - What's the difference?
is that eightfold is eight times as much; multiplied by eight while octadic is of or pertaining to an octad; eightfold.
is eight times as much. | 47 | 196 | {"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-2021-31 | latest | en | 0.973069 |
http://www.codeblogbt.com/archives/546300 | 1,553,499,885,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203842.71/warc/CC-MAIN-20190325072024-20190325094024-00359.warc.gz | 255,004,507 | 10,721 | Article From:https://www.cnblogs.com/lllxq/p/9967752.html
### Problem G: Approximate Research
Time limit: 1 SecMemory limit: 128 MB
Submission: 30Solution: 25
#### Title Description
Scientists’explorations on the planet Samuel have been enriched with energy reserves, making it possible for the long-term operation of the large computer “Samuel II” on the space station. As a result of last year’s hard work and good results, the League was allowed to use “Samuel II”Mathematical research.
Xiaolian recently studied the problem of approximation. He counted the number of approximations of each positive number N and expressed it in F (N). For example, the approximate number of 12 is 1, 2, 3, 4, 6, 12. So f (12) =6. The following table gives some values of f(N):
Now the League hopes to use “Samuel II” to count the sum of F (1) to f (N) and M.
#### input
Only one line and one integer N (0< N< 1000000)
#### output
There is only one line of output, which is the sum of integers M, i. e. f (1) to f (N).
```3
```
#### sample output
`5Train of thought:1.The rule of tabulation: f[x]=(num_of {prime_fac[i]}+1)2.The contribution of approximate number I to the answer is N/i ans = sum (N/i)ACCode:`
```#include <iostream>
#include<cstdio>
#include<cmath>
typedef long long ll;
using namespace std;
bool isprime[1000010];
ll cnt=0,prime[1000010];
ll num[1000010];
void get_prime(){
for(int i=0;i<=1000000;i++) isprime[i]=1;
isprime[0]=isprime[1]=0;
for(ll i=2;i<=1000000;i++){
if(isprime[i]) {
prime[++cnt]=i;num[i]=2;
for(ll j=i+i;j<=1000000;j+=i) {
isprime[j]=0;
ll tmp=0,cop=j;
while(cop%i==0) cop/=i,tmp++;
num[j]*=(tmp+1);
}
}
}
}
int main()
{
for(ll i=0;i<=1000000;i++) num[i]=1;
get_prime();
ll n;scanf("%lld",&n);
ll ans=0;
for(ll i=1;i<=n;i++){
ans+=num[i];
}
printf("%lld\n",ans);
return 0;
}```
### Question D: shuffle
Time limit: 1 SecMemory limit: 128 MB
Submission: 18Solution: 14
#### Title Description
In recognition of Xiaolian’s contribution to the exploration of Samuel, Xiaolian was invited to participate in the close manned exploration of Samuel. Because the planet Samuel is so far away that scientists have to spend quite a long time in the spaceship, Xiaolian proposes to use poker cards to travel long distances.The boredom of time. After playing a few games, people thought that simply playing cards was too easy for a highly intelligent businessman like them. A new method of playing cards has been proposed. A shuffle of playing cards is defined as dividing a stack of N (even N) cards into upper and lower stacks on average.Take the first sheet of the following stack as the first sheet of the new stack, then take the first sheet of the upper stack as the second sheet of the new stack, and then take the second sheet of the next stack as the third sheet of the new stack… So alternate until all cards are finished. For a stack of six cards, 12,3456. The process of shuffling is as follows:
As can be seen from the figure, after a shuffle, the sequence 123,4456 changed to 415,263. Of course, a shuffle of the resulting sequence will turn to 246 135. The game is like this, if given a stack of poker cards of length N, and the cardsFace size increases continuously from 1 to N (regardless of color). For such a stack of poker cards, M shuffles are performed. The first scientist to tell the size of the L card in the shuffled poker sequence won. Can you help Lenovo win the game?
#### input
There are three integers separated by spaces representing N, M and L (where 0 < N < 10 ^ 10, 0 < M < 10 ^ 10, and N is even).
#### output
One-line output specifies the face size of the playing card.
```6 2 3
```
#### sample output
`6Thought: Find the law:N=12The change relation is1->2->4->8->3->6->12->11->9->5->10->7->1（i->jRepresents that position I moves to position J after a shuffle.It can be seen that position x moves to x*2% (N+1) after a shuffle.So after M shuffles position x will move to x*pow(2,M)%(N+1)Location x moves M times to position L to find position xThat is to say, the solution of equation x*pow(2,M)%(N+1)=L xACCode:`
```#include <iostream>
#include<cstdio>
typedef long long ll;
using namespace std;
ll qpow(ll a,ll b,ll mod){
ll ret=1;
while(b){
if(b&1) ret=ret*a%mod;
a=a*a%mod;
b>>=1;
}
return ret;
}
void exgcd(ll a,ll b,ll& d,ll& x,ll& y){
if(!b){ d=a; x=1; y=0;}
else{ exgcd(b,a%b,d,y,x); y-=x*(a/b); }
}
int main()
{
ll n,m,l;
scanf("%lld%lld%lld",&n,&m,&l);
if(n==1) {printf("1\n"); return 0;}
ll a=qpow(2,m,n+1);
ll b=(n+1);
//printf("a=%lld b=%lld\n",a,b);
ll d,x,y;
exgcd(a,b,d,x,y);
x=x*(l/d),y=y*(l/d);
//printf("x=%lld y=%lld\n",x,y);
ll k=b/d;
//printf("%lld\n",k);
if(x<=0) x=x+(x/k)*k;
if(x>n) x=x-((x-n)/k)*k;
while(x<=0) x+=k;
while(x>n) x-=k;
printf("%lld\n",x);
return 0;
}```
### Question M: gold coins
Time limit: 1 SecMemory limit: 128 MB
Submission: 118Solution: 44
#### Title Description
There are n people sitting on the round table. Each person has a certain number of gold coins. The total number of gold coins can be divided by n. Everyone can give some gold coins to his neighbours, eventually making the number of gold coins equal for everyone. Your task is to find the minimum number of gold coins that have been transferred.
#### input
The first action integer n (n> = 3), the following N lines each line a positive integer, according to the counter-clockwise order to give each person has the number of gold coins.
3<=N<=100000,Total gold coins & lt; = 10 ^ 9
#### output
Minimum number of gold coins to be transferred
```4
1
2
5
4
```
```4
```
#### Tips
Let four persons be numbered 1,2,3,4. The third person gives the second person two gold coins (into 1,4,3,4), the second person and the fourth person gives the first person one gold coin respectively.
`Train of thought:Let Xi denote I - & gt; I + 1 gold coin (i + 1 - & gt; I gold coin is - xi)bea1+xn-x1=ave;a2+x1-x2=ave;...an+xn-1-xn=ave;So:x1=a1-ave+xn;x2=a1+a2-2*ave+xn;x3=a1+a2+a3-3*ave+xn;...xn-1=a1+a2+..+an-1-(n-1)*ave+xn;So min {| x1 |+ | x2 |+ | x3 | +... + | xn |}= min {| ave-a1-xn |+ | 2 * ave-a1-a2-xn |+ | (n-1) * ave a1-a2-. - an-1-xn |+ | 0-0Xn|}ACCode:`
```#include <iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
typedef long long ll;
using namespace std;
ll a[100005],c[100005];
int main()
{
ll n;scanf("%lld",&n);
ll sum=0;
for(ll i=1;i<=n;i++){
scanf("%lld",&a[i]);
sum+=a[i];
}
ll ave=sum/n;
for(ll i=1;i<=n-1;i++){
c[i]=c[i-1]+(ave-a[i]);
}
c[n]=0;
sort(c+1,c+1+n);
ll x=c[(n+1)/2];
ll ans=0;
for(int i=1;i<=n;i++){
ans+=abs(c[i]-x);
}
printf("%lld\n",ans);
return 0;
}
``` | 2,054 | 6,592 | {"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} | 3.40625 | 3 | CC-MAIN-2019-13 | latest | en | 0.675663 |
https://www.physicsforums.com/threads/work-done-by-an-ideal-gas.370434/ | 1,548,108,069,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583814455.32/warc/CC-MAIN-20190121213506-20190121235506-00176.warc.gz | 907,115,905 | 11,812 | Work done by an Ideal Gas
1. Jan 18, 2010
Last edited: Jan 18, 2010
2. Jan 18, 2010
I'm trying to calculate the work done when 50.0 g of tin dissolves in excess acid at 1.00 atm and 298.15 K
$$Sn(s)+2H^+(aq)-->Sn^{2+}(aq)+H_2(g)$$
Here is what I've done:
$$\Delta V=\frac{\Delta n RT}{P}$$
$$\Delta n=n_{H_2(g)(after)}-n_{H_2(g)(before)}=2 mol H_2$$
It seems that the moles of hydrogen gas will apply pressure to the air, so the work will be
$$w=-P\Delta V$$
$$\frac{-P(2 mol H_2)RT}{P}$$
$$=-(2 mol H_2)RT$$
This makes sense in terms of dimensions, but I don't have the answer in my book, so I can't tell if I did this right. | 242 | 638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-04 | latest | en | 0.866035 |
https://medical-dictionary.thefreedictionary.com/categorical+data | 1,624,608,635,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00554.warc.gz | 350,837,974 | 11,869 | # categorical data
Also found in: Dictionary, Thesaurus, Legal, Encyclopedia.
Related to categorical data: Numerical data
## categorical data
A term used in the context of a clinical trial for data evaluated by sorting values—e.g., severe, moderate and mild—into various categories.
References in periodicals archive ?
Clustering Categorical Data. Let X = {[X.sub.1], [X.sub.2], ..., [X.sub.N]} be a categorical dataset with N data points [X.sub.i].
He, "Density-based clustering algorithm for numerical and categorical data with mixed distance measure methods," Control Theory and Applications, vol.
Categorical data, also called nominal data, are an either/or type of data: males or females, fell or did not fall, patients had a total knee arthroplasty or a total hip arthroplasty (TKA/THA) or they did not.
The process of converting numerical to categorical data is called binning or discretization [11].
In a complementary line of research, the authors have developed a model for the analysis of categorical data with an endogenous weighting system (see Herrero & Villar, 2013) permitting one to address this problem from a slightly different viewpoint (see Herrero, Mendez & Villar, 2014).
Although it is assumed in categorical data analysis that each observed variable has an underlying scale that is both continuous and normally distributed, especially in scales with five or more categories (Byrne, 2006), problems begin to emerge as the observed item distributions diverge widely from a normal distribution.
Six aspects of the urinalysis were measured as categorical data. Information provided by the researchers stated the aspects of the urinalysis that were measured but not why they were ordinal measurements.
To test the hypotheses, an analysis of categorical data was used.
The obtained data are usually "categorical data" or "ordinal categorical data," which are collected based on a scale of "strongly agree," "agree," "have no opinion," "disagree," and "strongly disagree." Because most data in traditional statistical methods are interval data, researchers often assign these ordinal categorical data a score first, convert them into interval data, and then conduct further statistical analyses, such as factor analysis, principal analysis, and discriminate analysis.
However, methods for imputing categorical data are still experimental in some software releases.
The topics include a survey of partitional and hierarchical clustering algorithms, non-negative matrix factorizations for clustering, clustering categorical data, time-series data clustering, network clustering, uncertain data clustering algorithms, semi-supervised clustering, and educational and software resources for data clustering.
Site: Follow: Share:
Open / Close | 549 | 2,757 | {"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-2021-25 | latest | en | 0.924536 |
https://web2.0calc.com/questions/algebra-please-help_6 | 1,601,118,784,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00637.warc.gz | 678,577,363 | 6,377 | +0
# Algebra - Please Help!!!
0
32
2
+22
I pick two whole numbers x and y between 1 and 10 inclusive (not necessarily distinct). My friend picks two numbers x-4 and 2y-1. If the product of my friend's numbers is one greater than the product of my numbers, then what is the product of my numbers?
Sep 6, 2020
### 2+0 Answers
#1
0
The product of your numbers is 8.
Sep 6, 2020
#2
+22
0
Work? How?
CapriSun Sep 6, 2020 | 143 | 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} | 3.015625 | 3 | CC-MAIN-2020-40 | longest | en | 0.907827 |
https://www.sawaal.com/quantitative-aptitude-arithmetic-ability-questions-and-answers/if-160y2-x1x-160then-160what-160is-160the-160value-160of-1601y1-160-1602y1y2-1-160_21584 | 1,656,793,288,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00292.warc.gz | 1,025,681,225 | 14,577 | 0
Q:
A) (1+x)(2-x)2x-1 B) (1-x)(2+x)x-1 C) (1+x)(2-x)1-2x D) (1+x)(1-2x)2-x
Explanation:
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
What is the average of Export (in tonnes) during the five financial years ?
A) 1279.5 B) 1552.4 C) 1025.9 D) 1279.6
Explanation:
0 26
Q:
There was 25% off on bags. A lady bought a bag and got additional 20% discount for paying in cash. She paid ₹480. What was the price tag (in ₹) on the bag?
A) 950 B) 825 C) 800 D) 750
Explanation:
1 162
Q:
What is the area of a rhombus(in cm ) whose side is 20 cm and one of the diagonals is 24 cm ?
A) 396 B) 392 C) 350 D) 384
Explanation:
0 97
Q:
The simplified value of is A circle is inscribed in a triangle ABC.It touches sides AB, BC and AC at points P, Q and R respectively. If BP = 9 cm, CQ = 10 cm and AR = 11 cm, then the perimeter (in cm)of the triangle ABC is:
A) 72.5 B) 57.5 C) 60 D) 75
Explanation:
1 106
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
What is the average of total Import and Export (in tonnes) during the five financial year ?
A) 2508.8 B) 2279.5 C) 2552.4 D) 2325.9
Explanation:
0 171
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
What is the ratio of total Imports to total Exports during 2013-2014, 2015-2016 and 2017-2018?
A) 4175 : 4011 B) 4011 : 4175 C) 3619 : 3604 D) 3604 : 3073
Answer & Explanation Answer: C) 3619 : 3604
Explanation:
0 118
Q:
In which financial year the total of the Exports and Imports is the lowest ?
A) 2013-2014 B) 2017-2018 C) 2015-2016 D) 2014-2015
Explanation:
0 156
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
A) 2016-2017 B) 2014-2015 C) 2015-2016 D) 2017-2018 | 779 | 2,254 | {"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} | 3.90625 | 4 | CC-MAIN-2022-27 | latest | en | 0.753029 |
https://www.justanswer.com/homework/cdc3m-final-product-2-specific-gravity.html | 1,558,304,610,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255182.37/warc/CC-MAIN-20190519221616-20190520003616-00385.warc.gz | 829,334,992 | 51,759 | Homework
Related Homework Questions
RR Jha- you answered a question for me last night but I am
RR Jha- hi- you answered a question for me last night but I am not sure if this is the correct answer for exactly what I was looking for...can I ask again? it was the specific gravity problem … read more
R.R. Jha
Tutor
Bachelor's Degree
5,870 satisfied customers
I am doing a report for a school assignment and I had to
Good morning, I am doing a report for a school assignment and I had to pick out 2 computers and determine which CPU is faster and whether or not the clock rate itself makes the CPU faster. I chose the… read more
RJ
Teacher
Brd field social studies
7 satisfied customers
does bulk density of a particle impact its mass?i am trying
does bulk density of a particle impact its mass? i am trying to work out if a particle that is smaller, but has higher bulk density, can have higher kinetic energy? comparing: Garnet 30/60#, bulk dens… read more
Ray Atkinson
Bachelor's Degree
1,687 satisfied customers
Dear RRJha, Thanks for your help. Please answer question below and show the working out calculations. Kind regard This question carries 30% of the marks for this assignment. Your answer to each of par… read more
R.R. Jha
Tutor
Bachelor's Degree
5,870 satisfied customers
You are an astronaut who has been sent to a large rocky asteroid
You are an astronaut who has been sent to a large rocky asteroid which is essentially spherical in shape. You are standing on the surface of the asteroid and you need to make an estimate of its radius… read more
R.R. Jha
Tutor
Bachelor's Degree
5,870 satisfied customers
I am trying to find the equation needed to calculate the necessary
I am trying to find the equation needed to calculate the necessary pressure to cause a mist to come out of a hose with holes in it. I know it has something to do with density of the liquid and the hei… read more
David Coleman
Solicitor
Bachelor's Degree
97 satisfied customers
Physics Help (explanations must be shown) __________________________ A
Physics Help: (explanations must be shown) __________________________ A helicopter is rising at 4.7 m/s when a bag is dropped from it. (Assume that the positive direction is upward.) (a) After 2.0 s, … read more
David
Post-Doctoral Degree
2,112 satisfied customers
Greetings I need some help with the following question below.
Greetings I need some help with the following question below. Can You help me. I am a MBA student and this is a critical grade for me. How would I calculate center of gravity, location break-even anal… read more
Dr Jim
Doctoral Degree
Tevin Trader starts a merchandising business on December 1 and enters into three inventory purchases: Tevin Trader starts a merchandising business on December 1 and enters into three inventory purchas… read more
Manal Elkhoshkhany
Tutor
4,556 satisfied customers
Cost Accounting 1. A plant that uses process costing has
Cost Accounting 1. A plant that uses process costing has 8,000 units in beginning work in process, 15,000 more started, and 5,000 units in the ending work in process. Using this info, answer the follo… read more
Falak Naz
Master's Degree
324 satisfied customers
Gravimetric Analysis for Sulfate Data Sample A Mass of
Gravimetric Analysis for Sulfate: Data: Sample A: Mass of sample 0.3078g, Mass of crucible 20.3821g, Mass of crucible & BaSO4 20.7979g, Mass of BaSO4 0.4158g. Sample B: Mass of sample 0.3300g., Mass o… read more
David the Scientist
Post-Doctoral Degree
A stunt pilot who has been diving her airplane vertically pulls out
A 53.0 kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. a) If the plane''s speed at the lowest point of the circ… read more
Scott
Master's Degree
16,043 satisfied customers
My son has a rocket kit for his science project for
My son has a rocket kit for his science project for the science fair. we launched the rocket after putting the kit together, but I''m not sure what my Question, Prediction, Variables/controls, Results… read more
Dean Mason
design engineer
currently working on Masters (M.Div
94 satisfied customers
A piece of wood that floats on water has a mass of 0.0175 ...
A piece of wood that floats on water has a mass of 0.0175 kg. A lead sinker is tied to the wood, and the apparent mass with the wood in air and the lead sinker submerged in water is 0.0765 kg. The app… read more
mit2008
Master's Degree
54 satisfied customers
Chemistry lab exercise 1. A 0.750 g piece of nickel ...
Chemistry lab exercise: 1. A 0.750 g piece of nickel was added to a solution of HCL. The Ni completely reacted and 2.577 g of product, a nickel chloride were recovered. PLEASE SHOW WORK. a) How many g… read more
NAL
1,470 satisfied customers
You removed 100cm2 each from the brisket, flank, and rump ...
You removed 100cm2 each from the brisket, flank, and rump for a 300 cm2 composite sample. After sample collection, you returned to the laboratory, added 45ml of diluent to the cores, and stomached to … read more
NAL
1,470 satisfied customers
Hi i have to move any object i want using a falling ...
Hi i have to move any object i want using a falling 1kg weight as the power source (Gravity) but i dont not see this working all the designs i have tried all failed. the weight must be on the object o… read more
David DeKeyser
Home improvement specialist
68 satisfied customers
Any object can become airborne if the wind velocity is
Any object can become airborne if the wind velocity is sufficient. The following is a formula that provides a reasonable estimate of liftoff in mph: V=(195 x WEIGHT / AREA))^1/2 (WEIGHT (lbs) AREA (sq… read more
Alex
Master's Degree
378 satisfied customers
Disclaimer: Information in questions, answers, and other posts on this site ("Posts") comes from individual users, not JustAnswer; JustAnswer is not responsible for Posts. Posts are for general information, are not intended to substitute for informed professional advice (medical, legal, veterinary, financial, etc.), or to establish a professional-client relationship. The site and services are provided "as is" with no warranty or representations by JustAnswer regarding the qualifications of Experts. To see what credentials have been verified by a third-party service, please click on the "Verified" symbol in some Experts' profiles. JustAnswer is not intended or designed for EMERGENCY questions which should be directed immediately by telephone or in-person to qualified professionals.
Ask-a-doc Web sites: If you've got a quick question, you can try to get an answer from sites that say they have various specialists on hand to give quick answers... Justanswer.com.
...leave nothing to chance.
Traffic on JustAnswer rose 14 percent...and had nearly 400,000 page views in 30 days...inquiries related to stress, high blood pressure, drinking and heart pain jumped 33 percent.
Tory Johnson, GMA Workplace Contributor, discusses work-from-home jobs, such as JustAnswer in which verified Experts answer people’s questions.
I will tell you that...the things you have to go through to be an Expert are quite rigorous.
## What Customers are Saying:
Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help.
Mary C.Freshfield, Liverpool, UK
This expert is wonderful. They truly know what they are talking about, and they actually care about you. They really helped put my nerves at ease. Thank you so much!!!!
AlexLos Angeles, CA
Thank you for all your help. It is nice to know that this service is here for people like myself, who need answers fast and are not sure who to consult.
GPHesperia, CA
I couldn't be more satisfied! This is the site I will always come to when I need a second opinion.
JustinKernersville, NC
Just let me say that this encounter has been entirely professional and most helpful. I liked that I could ask additional questions and get answered in a very short turn around.
EstherWoodstock, NY
Thank you so much for taking your time and knowledge to support my concerns. Not only did you answer my questions, you even took it a step further with replying with more pertinent information I needed to know.
RobinElkton, Maryland
He answered my question promptly and gave me accurate, detailed information. If all of your experts are half as good, you have a great thing going here.
DianeDallas, TX
< Previous | Next >
## Meet the Experts:
LogicPro
Engineer
5,894 satisfied customers
Expert in Java C++ C C# VB Javascript Design SQL HTML
Manal Elkhoshkhany
Tutor
4,556 satisfied customers
More than 5000 online tutoring sessions.
Linda_us
Finance, Accounts & Homework Tutor
3,138 satisfied customers
Post Graduate Diploma in Management (MBA)
Chris M.
M.S.W. Social Work
2,699 satisfied customers
Master's Degree, strong math and writing skills, experience in one-on-one tutoring (college English)
F. Naz
Chartered Accountant
2,194 satisfied customers
Experience with chartered accountancy
Bizhelp
CPA
1,887 satisfied customers
Bachelors Degree and CPA with Accounting work experience
Seanna
Tutor
1,781 satisfied customers
3,000+ satisfied customers, all topics, A+ work
< Previous | Next >
Disclaimer: Information in questions, answers, and other posts on this site ("Posts") comes from individual users, not JustAnswer; JustAnswer is not responsible for Posts. Posts are for general information, are not intended to substitute for informed professional advice (medical, legal, veterinary, financial, etc.), or to establish a professional-client relationship. The site and services are provided "as is" with no warranty or representations by JustAnswer regarding the qualifications of Experts. To see what credentials have been verified by a third-party service, please click on the "Verified" symbol in some Experts' profiles. JustAnswer is not intended or designed for EMERGENCY questions which should be directed immediately by telephone or in-person to qualified professionals.
Show MoreShow Less | 2,347 | 10,070 | {"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-2019-22 | latest | en | 0.953128 |
https://www.teach-nology.com/teachers/lesson_plans/holidays/fall/fallsale.html | 1,713,961,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00086.warc.gz | 926,873,721 | 4,374 | ### Lesson Plan Title : Fall Clothing Sale
Overview and Purpose: In this activity, students practice their math skills while shopping for a new fall wardrobe. They have to work with a group to create a new fall wardrobe within a budget using store circulars.
Objective: The student will be able to use store circulars to shop for a new fall wardrobe while staying within a budget.
Resources:
Several store circulars that have fall clothing
Chart paper
Scissors
Glue
Activities:
Divide your students into groups of three. Pass out one piece of chart paper per group and instruct them to create a new fall wardrobe for under \$100. Give each group a list of items they must purchase. Items can include jackets, pants, sweaters, and boots. Have each group cut out the items (with the prices) that they are going to buy and paste them on their piece of chart paper.
After each group is finished, have them present their wardrobe to the class. They should have a list of the prices and be able to show if they were able to stay under a \$100.
Wrap Up:
This task can be made more challenging for advanced students by telling them all the items are on sale for 20% off their listed price. | 253 | 1,193 | {"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-2024-18 | latest | en | 0.959974 |
https://theintactone.com/2019/03/05/qtm-u4-topic-5-application-of-poisson-and-exponential-distribution-in-estimating-arrival-rate-and-service-rate/ | 1,675,339,105,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00202.warc.gz | 594,483,031 | 97,733 | # Application of Poisson and Exponential distribution in estimating Arrival Rate and Service Rate
Both the Poisson and Exponential distributions play a prominent role in queuing theory. The Poisson distribution counts the number of discrete events in a fixed time period; it is closely connected to the exponential distribution, which (among other applications) measures the time between arrivals of the events. The Poisson distribution is a discrete distribution; the random variable can only take nonnegative integer values. The exponential distribution can take any (nonnegative) real value.
Considering a problem of determining the probability of n arrivals being observed during a time interval of length t, where the following assumptions are made.
1. Probability that an arrival is observed during a small time interval (say of length v) is proportional to the length of interval. Let the proportionality constant be l, so that the probability is lv.
2. Probability of two or more arrivals in such a small interval is zero.
3. Number of arrivals in any time interval is independent of the number in nonoverlapping time interval.
These assumptions may be combined to yield what probability distributions are likely to be, under Poisson distribution with exactly n customers in the system.
Suppose function P is defined as follows:
P (n customers during period t) = the probability that n arrivals will be observed in a time interval of length t
This is the Poisson probability distribution for the discrete random variable n, the number of arrivals, where the length of time interval, t is assumed to be given. This situation in queuing theory is called Poisson arrivals. Since the arrivals alone are considered (not departures), it is called a pure birth process.
The time between successive arrivals is called inter-arrival time. In the case where the number of arrivals in a given time interval has Poisson distribution, inter-arrival times can be shown to have the exponential distribution. If the inter-arrival times are independent random variables, they must follow an exponential distribution with density f(t) where,
### Poisson and Exponential Distribution Practice Problems
The practice problems of poisson and exponential distributions are given below
Example:Â In a factory, the machines break down and require service according to a Poisson distribution at the average of four per day. What is the probability that exactly six machines break down in two days?
In the output screen, for n = 6 the probability, Pn is given as 0.12214.
Example:Â On an average, 6 customers arrive in a coffee shop per hour. Determine the probability that exactly 3 customers will reach in a 30 minute period, assuming that the arrivals follow Poisson distribution.
Similarly, when the time taken to serve different customers are independent, the probability that no more than t periods would be required to serve a customer is given by exponential distribution as follows:
p(not more than t time period) = 1 – e– mt where m = average service rate
Example:Â A manager of a fast food restaurant observes that, an average of 9 customers is served by a waiter in a one-hour time period. Assuming that the service time has an exponential distribution, what is the probability that
1. A customer shall be free within 12 minutes.
2. A customer shall be serviced in more than 25 minutes.
Solution:
(a) Given, ÎĽ = 9 customers / hour
t = 15 minutes = 0.25 hour
Therefore, p (less than 15 minutes) = l – e– mtÂ
= 1– e– 9 × 0.25
= 0.8946
(b) Given, ÎĽ = 9 customers / hour
t = 25 minutes = 0.4166 hour
Therefore, P (more than 25 minutes) = l – e– mtÂ
= 1– e– 9 × 0.4166
= 0.0235
To analyze queuing situations, the questions of interest that are typically concerned with measures of queuing system performance include,
• What will be the waiting time for a customer before service is complete?
• What will be the average length of the queue?
• What will be the probability that the queue length exceeds a certain length?
• How can a system be designed at minimum total cost?
• How many servers should be employed?
• Should priorities of the customers be considered?
• Is there sufficient waiting area for the customers?
## One thought on “Application of Poisson and Exponential distribution in estimating Arrival Rate and Service Rate”
error: Content is protected !! | 965 | 4,407 | {"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} | 3.90625 | 4 | CC-MAIN-2023-06 | longest | en | 0.896847 |
http://www.mem50212.com/MDME/MEMmods/MEM23061A/Hounsfield/Hounsfield.html | 1,696,330,729,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00645.warc.gz | 71,793,515 | 4,645 | MDME: MANUFACTURING, DESIGN, MECHANICAL ENGINEERING
# HOUNSFIELD TEST
The Hounsfield Tester can do a variety of tests on a small test-piece. It is mostly used for tensile testing. It doesn't measure elasticity very well.
## Finding UTS using the Hounsfield Tensometer
In this lab we are measuring the UTS - Ultimate Tensile Strength. This is the highest stress a material can take before it breaks. We will also analyse the test to determine the YS - Yield Strength.
Notes: Hounsfield Tensometer Guide: Hounsfield Contents.pdf
#### Axial Stress Theory
Axial stress acts along an axis, which is really just a shorthand way of saying Tensile or Compressive Stress.
Axial Stress: s = F /A s = Axial Stress (MPa) A = Cross-sectional Area (mm2) F = Force (N) e = Axial strain (no units) L = Gauge length - originally (mm) x = extension (mm) E = Modulus of Elasticity (MPa) Axial Strain: e = x / L Modulus of Elasticity: E = ss / e s
By recording the force as we increase the load on the specimen, the highest force is used to determine UTS.
#### Hounsfield Laboratory Procedure
Before starting the lab, you must calculate the maximum load that can be applied to the specimen, assuming a sensible UTS of the material. Note that in practice, the UTS is nearly always higher than the published values. This is because the published data is usually a 'guaranteed' value, or the lowest value you would rarely, if ever, encounter. See Materials.htm for our list of material data.
1. Take measurements of the specimen. Diameter, Length etc. Ascertain the material as best you can.
3. Apply load to about 75% of Yield to ensure the specimen it fully seated in the jaws.
4. Release the load and set the mercury indicator.
5. Add paper to the chart roller and begin the test (recording the load as you go) until the specimen breaks. Blank graph paper is here: High quality (1MB), Low quality (0.3MB).
6. Remove paper, measure the % elongation of the specimen.
7. Attempt to record all sources of error relevant to the measurements taken (Force, Extension)
## Laboratory Report
Determine the following:
1. Ultimate Tensile Stress (UTS), MPa
2. Yield Stress (YS), MPa
3. % elongation
4. Discuss the issues regarding the horizontal axis of the graph plot (elongation) and why it is not possible to get a reliable Strain or Modulus of Elasticity (E).
5. Explain two possible ways to (approximately) calibrate the elongation axis.
The report should follow the Laboratory Report guide. Include an approximate error analysis.
Results
The following images show plots of Houndsfield tests. Choose the appropriate test for your class (or as instructed).
Thurs night Class. Nov 2015
Mon Night Class. Sep 8 2014.
Mon Night Class. Sep 9 2013.
Thurs Day Class. Feb 21 2013. Larger Image here.
Monday Night Class. Aug 6 2012. Larger Image here.
Note: Specimen reached limit of machine and did not break. Please make comments about what this test can tell us, but complete your lab assignment using this specimen here.
Thursday Day Class July 26 2012. Larger image here
Thurs Class June 14 2012
Mon night class: Sep 05, 2011 (larger image here)
Monday PM classOctober 26, 2010Original Diam 5.05mmOriginal Length ID = 27Original Length OD = 36.6Final Length ID = 30.89Final Length OD = 40.03Final Diam 4.5mm
Final Length = 34.7mm
Break Diameter (necking) = 3.8mm
The vertical axis is calibrated as 5kN on the mercury scale = 8cm on the graph paper.
Test 1: Monday night class: Oct 27
Parallel red curves have been superimposed to estimate the contraction after breakage, from which the x axis (extension) can be calibrated.
Test 2: Tuesday night class: Oct 28
Test 3: Wed day class: Oct 29
#### Questions:
Assignment:
##### Relevant pages in MDME
• Mechanical Properties practice test: 10101cp | 973 | 3,805 | {"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-2023-40 | latest | en | 0.880872 |
https://web2.0calc.com/questions/solve-the-system-using-gauss-or-gauss-jordan-elimination | 1,548,278,006,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584350539.86/warc/CC-MAIN-20190123193004-20190123215004-00482.warc.gz | 694,151,741 | 6,362 | +0
# Solve the system using Gauss or Gauss - Jordan elimination
0
439
1
+88
2x + y = - 4
-2y + 4z = 0
3x - 2z = - 11
x=? y=? z=?
Mar 24, 2015
#1
+95361
+18
I didn't know how to do this so I gave myself a quick lesson from here
Watch th video and work through the video example
then put it together with your knowledge of how elimination method to solve simultaneous equations works.
$$\begin{pmatrix} 2 &1&0&|-4 \\ 0 &-2&4&|+0 \\ 3 &0&-2&|-11 \\ \end{pmatrix}$$
First you need to get rid of the 3 in the bottom left corner.
1) multiply R1 by3 and R3 by 2 WRITE down the new matrix
2) R1-R3 replaces R3 WRITE down the new matrix (the 3 has been replaced with a 0)
NOW there is a 3 in R3, C2 You have to get rid of it
3) mult R2 by 3 and mult R3 by 2 WRITE down the new matrix
4) R2+R3 replaces R3 WRITE down the new matrix (the 6 has been replaced with a 0)
NOW the diagonal must only contain ones
5) Multiply R1 by 1/6, Multiply R2 by -1/6 and multiply R3 by 20
And you should be able to take it from there
x=-3, y=2, and z=1
Mar 24, 2015
#1
+95361
+18
I didn't know how to do this so I gave myself a quick lesson from here
Watch th video and work through the video example
then put it together with your knowledge of how elimination method to solve simultaneous equations works.
$$\begin{pmatrix} 2 &1&0&|-4 \\ 0 &-2&4&|+0 \\ 3 &0&-2&|-11 \\ \end{pmatrix}$$
First you need to get rid of the 3 in the bottom left corner.
1) multiply R1 by3 and R3 by 2 WRITE down the new matrix
2) R1-R3 replaces R3 WRITE down the new matrix (the 3 has been replaced with a 0)
NOW there is a 3 in R3, C2 You have to get rid of it
3) mult R2 by 3 and mult R3 by 2 WRITE down the new matrix
4) R2+R3 replaces R3 WRITE down the new matrix (the 6 has been replaced with a 0)
NOW the diagonal must only contain ones
5) Multiply R1 by 1/6, Multiply R2 by -1/6 and multiply R3 by 20
And you should be able to take it from there | 683 | 2,080 | {"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} | 4.53125 | 5 | CC-MAIN-2019-04 | latest | en | 0.852747 |
https://tradingkrfkq.netlify.app/tutela31947ro/how-to-times-and-divide-indices-guh.html | 1,679,676,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00107.warc.gz | 666,328,852 | 10,612 | ## How to times and divide indices
Multiplication is one of the four elementary mathematical operations of arithmetic, with the One can only meaningfully add or subtract quantities of the same type but can multiply or divide quantities of different types. with successive integer values substituted for the index of multiplication, starting from the lower bound
How do I multiply and divide indices? First you need to understand what an index (or power) is. It's just a compact way of writing a number multiplied by itself a To divide exponents (or powers) with the same base, subtract the exponents. Division is the opposite of multiplication, so it makes sense that because you add We know that: In general: This formula tells us that when dividing powers with the same base, the index in the denominator is subtracted from the index in the Rewrite products or quotients of powers, including negative powers.
## 13 May 2013 In this lesson we look at how to Divide Algebra Expressions which contain exponents. We also introduce the Dividing Exponents “Subtraction”
Exponents are shorthand for repeated multiplication of the same thing by itself. For instance, the shorthand for multiplying three copies of the number 5 is shown 16 Oct 2019 How to Divide Exponents. Dividing expressions that have exponents is easier than it looks. As long as you're working with the same base, Multiplication is one of the four elementary mathematical operations of arithmetic, with the One can only meaningfully add or subtract quantities of the same type but can multiply or divide quantities of different types. with successive integer values substituted for the index of multiplication, starting from the lower bound To multiply or divide numbers with exponents, if the base numbers are different, you must simplify each number with an exponent first and then perform the
### When multiplying and dividing and working out the power of a power, you must follow the three rules of indices: Multiplying. When multiplying indices, you add
A knowledge of powers, or indices as they are often called, is essential for an that if we are multiplying expressions such as these then we add the indices Here we are dividing by a fraction, and to divide by a fraction we need to invert and. For example,. (3x2y)3 = 33(x2)3y3 = 27x6y3. Exercises. 1. Simplify the following expressions, leaving only positive indices in the answer. (a) 42 × 4−3. Proceed from left to right for multiplication and division. Solve addition and subtraction last after parentheses, exponents, roots and multiplying/dividing. Again, 22 Mar 2015 How do you simplify 2√3? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division 20 Jun 2012 Rule 2: You can combine bases when multiplying or dividing terms, provided the exponents are the same. Simply multiply or divide the bases.
### 20 Jun 2012 Rule 2: You can combine bases when multiplying or dividing terms, provided the exponents are the same. Simply multiply or divide the bases.
22 Mar 2015 How do you simplify 2√3? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division 20 Jun 2012 Rule 2: You can combine bases when multiplying or dividing terms, provided the exponents are the same. Simply multiply or divide the bases. 13 May 2013 In this lesson we look at how to Divide Algebra Expressions which contain exponents. We also introduce the Dividing Exponents “Subtraction” Click on any of the questions below to watch explanation videos. Simplify the following and leave the answer showing only positive exponents:
## It tells you how many time to multiply the base by itself. For example, 8 2 means to multiply 8 by itself twice to get 16, and 10 3 means 10 • 10 • 10 = 1,000. When you have negative exponents, the negative exponent rule dictates that, instead of multiplying the base the indicated number of times, you divide the base into 1 that number of
Understanding base numbers and powers (indices). Explains why you can add the powers together when multiplying powers of the same base number, and similarly why you can subtract the powers when Dividing exponents with different bases. When the bases are different and the exponents of a and b are the same, we can divide a and b first: a n / b n = (a / b) n. Example: 6 3 / 2 3 = (6/2) 3 = 3 3 = 3⋅3⋅3 = 27 . When the bases and the exponents are different we have to calculate each exponent and then divide: a n / b m. Example: 6 2 / 3 Dividing is the inverse (opposite) of Multiplying. A negative exponent means how many times to divide by the number. Example: 8 -1 = 1 ÷ 8 = 1/8 = 0.125 This worksheet covers the rules regarding multiplying and dividing indices with the same base, including when the variable has a coefficient. Many of the questions involve negative powers.
21 Jan 2020 An overview of indices, and how to multiply, divide, and raise them to an index. Roots and radicals are introduced. Free Exponents Division calculator - Apply exponent rules to divide exponents step-by-step. Simplify algebraic equations by multiplying and dividing variables with exponents . This BrainPOP algebra movie gives you the power! Tim and Moby multiply and divide numbers with exponents. Exponents are shorthand for repeated multiplication of the same thing by itself. For instance, the shorthand for multiplying three copies of the number 5 is shown 16 Oct 2019 How to Divide Exponents. Dividing expressions that have exponents is easier than it looks. As long as you're working with the same base, | 1,258 | 5,615 | {"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} | 4.8125 | 5 | CC-MAIN-2023-14 | latest | en | 0.901547 |
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2004.SU/Stein/assign1/assign1main.html | 1,542,182,038,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741660.40/warc/CC-MAIN-20181114062005-20181114084005-00324.warc.gz | 180,800,097 | 1,667 | Assignment 1
Exploring graphs
By
Mandy Stein
Problem 6.
Graph
What do you expect for the graph of
or
First, let's look at the graphs separately:
Looking at the graphs we observe a few trends. The first thing observed is that when the exponent is 2 or 4 the graphs are closed and when the exponent is 1, 3, or 5 the graph is continuous. This leads to the assumption that if the exponent is even the graph will be closed and if the exponent is odd the graph will be continuous. The second thing observed is that when the exponent equals 2 the graph is a circle and as the exponent increases the graph becomes more square-like.
Using the observations from the graphs, we can assume that the graph of will be closed and square-like and that the graph of will be continuous and square-like.
Let's look at the graphs and see if our assumptions are correct:
The graphs look like what we expected.
Here is a graph of all six equations. This allows you to see the graphs becoming more square-like as the exponent increases. | 226 | 1,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-47 | latest | en | 0.962786 |
http://math.stackexchange.com/questions/216580/can-we-define-topology-recursively | 1,469,386,091,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00225-ip-10-185-27-174.ec2.internal.warc.gz | 160,437,293 | 19,108 | # Can we define topology recursively?
Let $X$ be a topological space with a fixed topology $\mathscr{T}$. We know that the following are equivalent for all $U \subseteq X$.
1. $U \in \mathscr{T}$.
2. For all $x \in U$ there is $U_{x} \in \mathscr{T}$ such that $x \in U_{x} \subseteq U$.
My question is, is it okay to define topology as follows?
Definition. Let $X$ be a set. A subset of power set $\mathscr{T} \subseteq \mathcal{P}(X)$ is called a topology on $X$ if for every $U \in \mathscr{T}$ the following is true: for all $x \in U$, there exists $U_{x} \in \mathscr{T}$ such that $x \in U_{x} \subseteq U$.
If this is okay, I don't know what made most of people introduce topology with three axioms. Is it because this definition is self-referencing? Before I sat in my introductiory topolgy course, I always thought that this was going to be introduced in the beginning of the class rather than more axiomatic definition.
-
What is $U$ in your definition? – Chris Eagle Oct 18 '12 at 22:57
Your definition does not require the union of open sets to be open. Isn't this something desirable? – Andrés E. Caicedo Oct 18 '12 at 22:57
The role of axioms in mathematics is to give a rigorous framework to describe properties of objects. Some long time ago, people figured what are the properties we expect open sets to have, and ended up with a very concise way of saying that. Not everything needs to be recursive, and not everything can be recursive. – Asaf Karagila Oct 18 '12 at 23:00
@Chris Thanks, I meant to say for all $U \in\mathscr{T}$. Let me fix it right away. – user123454321 Oct 18 '12 at 23:02
@AndresCaicedo - the definition had a typo. Sorry for the confusion. – user123454321 Oct 18 '12 at 23:04
## 2 Answers
No. For one, any covering of $X$ would be a topology. Alternatively, the set of singletons (not the discrete topology) would be a topology on every space by your logic.
You need to be able to take unions and stay inside the topology to do anything remotely interesting.
-
That makes sense. So recursive definition has more sets in $\mathscr{T}$. What a stupid question. I totally forgot the fact that we are already given more information that $\mathscr{T}$ is a topology in the very first logical equivalence. Thanks! – user123454321 Oct 18 '12 at 23:15
This definition only guarantee you that all $U \subset \mathscr{T}$ are open, this is because in the definition of topology is important:
1. Both the empty set and $X$ are elements of $\mathscr{T}$
2. Any union of elements of $\mathscr T$ is an element of $\mathscr{T}$
3. Any intersection of finitely many elements of $\mathscr{T}$ is an element of $\mathscr{T}$
This is no recursively. And the first part tell you that if you have any set, maybe a closed set, you can find a open set of any point of that set that belong to the fixed topology.
-
I don't think this answer is relevant to what I asked. See anon's answer. – user123454321 Oct 18 '12 at 23:19
"And the first part tell you that if you have any set, maybe a closed set, you can find a open set of any point of that set that belong to the fixed topology." - What do you mean? – user123454321 Oct 19 '12 at 0:17
If U is open then $U \in \mathscr{T}$, but if U is closed for all $x \in U$, you can find $U_{x} \in \mathscr{T}$ which is open by definition. – José Ramírez Oct 19 '12 at 0:26
Note that I did not use the definition of the word "closed" as one should not use it as we are discussing the definition of the word "open." – user123454321 Oct 19 '12 at 3:17
I used the set closed as a example and it's evident the definition. – José Ramírez Oct 19 '12 at 3:56 | 1,012 | 3,625 | {"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} | 2.953125 | 3 | CC-MAIN-2016-30 | latest | en | 0.909007 |
https://codegolf.stackexchange.com/questions/42234/coding-a-recursive-function-for-highest-possible-input | 1,718,267,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00538.warc.gz | 158,581,587 | 43,083 | Coding a recursive function for highest possible input
Challenge
You are given the following function:- which is the same as:-
with the base cases q(r, b, L) = 1 whenever r ≤ L, q(r, 0, L) = 0, if r > L and q(r, 0, L) = 1, if r ≤ L.
Your task is to code a program that takes r as input, and outputs the value of q(r, r - L, L) for all L taking integer values from 1 to (r-1), where r is any nonnegative integer.
Example 1
Input
Enter the value of r: 2
Output
q(2,1,1) = 0.3333333333
Example 2
Input
Enter the value of r: 3
Output
q(3,2,1) = 0.1
q(3,1,2) = 0.5
Winning criterion
The code that can correctly output q(r, r-L, L) for all L taking integer values from 1 to (r-1), for the highest value of r, in less than 300 seconds. In case of a tie, the code with lesser runtime will be considered. As this is a runtime-based challenge, I shall test all submissions on my machine.
• Your sum/product notations both use the letter k as their variable. I think this is technically valid, but it might be easier to read if you used two different letters. Commented Dec 7, 2014 at 19:03
• Which of the two base cases takes priority when r < L and b = 0? And what value of b will be used in the winning criterion? Commented Dec 7, 2014 at 23:21
• @PeterTaylor let me guess: are you working on a closed-form? Commented Dec 8, 2014 at 7:42
• q(r, r-1, 1) == 2(r!)^2/(2r)! (oeis.org/A001700) and r(r, 1, r-1) == (r-1)/(r+1) (trivial). Haven't checked the others yet. Commented Dec 8, 2014 at 17:23
• In the interests of clarity, q(r,r-1,1) is the reciprocal of A001700. Commented Dec 10, 2014 at 10:12
Java
I've applied some algebraic transformation and dynamic programming.
public class CodeGolf42234 {
public static void main(String[] args) {
int r = Integer.parseInt(args[0]);
for (int L = 1; L < r; L++) System.out.println(qDouble5(r, r-L, L));
}
private static double qDouble5(int _r, int _b, int L) {
double[][] q = new double[_r+1][_b+1];
for (int r = 0; r <= L; r++) {
for (int b = 0; b < q[r].length; b++) q[r][b] = 1;
}
for (int r = L + 1; r < q.length; r++) {
for (int b = 1 + (r - L - 1)/L; b < q[r].length; b++) {
double sum = 0, m = 1;
for (int k = 0; k <= L; k++) {
sum += m * q[r-k][b-1];
m = m * (r - k) / (r + b - 1 - k);
}
q[r][b] = sum * b / (r + b);
}
}
return q[_r][_b];
}
}
• Thank you for the submission. On my system (Windows, Netbeans IDE), the highest value of r that can be evaluated using this code, in under 300 seconds (297 seconds) is 731. Good job, Peter! :) Commented Dec 10, 2014 at 10:36
Java
Here's a recursive solution. Just like you wanted. You can always uncomment the output and display it if you want.
import java.util.ArrayList;
import java.util.Scanner;
{
static ArrayList<Double> answers = new ArrayList<Double>();
public static void main(String[]args)
{
Scanner scanner = new Scanner(System.in);
int r = scanner.nextInt();
for(int b=r-1; b!=0; b--)
for(int L=r-1; L!=0; L--)
// int i=-1;
// for(int b=r-1; b!=0; b--)
// for(int L=r-1; L!=0; L--)
// System.out.println("q("+r+", "+b+", "+L+") = "+answers.get(++i));
}
public static double formulate(double r, double b, double L)
{
if(b==0)
if(r>L)
return 0.0;
else
return 1;
if(r<=L)
return 1;
double partone = (b/(r+b));
double result = 0.0;
for(int k=1; k<=L; k++)
{
result+=summation(r, b, k) * (b/(r+b-k)) * formulate(r-k, b-1, L);
}
return (partone * formulate(r, b-1, L)) + result;
}
public static double summation(double r, double b, int k)
{
double rb = r+b;
double mul=r/rb;
for(int i=1; i<k; i++)
{
mul*=(r-i)/(rb-i);
}
return mul;
}
} | 1,203 | 3,588 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.809844 |
https://www.edureka.co/community/30102/logits-softmax-softmax_cross_entropy_with_logits-python?show=30104 | 1,566,163,699,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314130.7/warc/CC-MAIN-20190818205919-20190818231919-00025.warc.gz | 804,283,780 | 20,588 | I was going through the tensorflow API docs here. In the tensorflow documentation, they used a keyword called logits. What is it? In a lot of methods in the API docs it is written like
`tf.nn.softmax(logits, name=None)`
If what is written is those logits are only Tensors, why keeping a different name like logits?
Another thing is that there are two methods I could not differentiate. They were
```tf.nn.softmax(logits, name=None)
tf.nn.softmax_cross_entropy_with_logits(logits, labels, name=None)```
What are the differences between them? The docs are not clear to me. I know what tf.nn.softmaxdoes. But not the other. An example will be really helpful.
Nov 12, 2018 in Python 479 views
Suppose you have two tensors, where y_hat contains computed scores for each class (for example, from y = W*x +b) and y_true contains one-hot encoded true labels.
```y_hat = ... # Predicted label, e.g. y = tf.matmul(X, W) + b
y_true = ... # True label, one-hot encoded```
If you interpret the scores in y_hat as unnormalized log probabilities, then they are logits.
Additionally, the total cross-entropy loss computed in this manner:
```y_hat_softmax = tf.nn.softmax(y_hat)
total_loss = tf.reduce_mean(-tf.reduce_sum(y_true * tf.log(y_hat_softmax), [1]))```
is essentially equivalent to the total cross-entropy loss computed with the function softmax_cross_entropy_with_logits():
`total_loss = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(y_hat, y_true))`
• 3,500 points
+1 vote
What do you mean by python scripting? What is a script and a module in python?
A module is a file containing a ...READ MORE
+1 vote
What is the difference between range and xrange functions in Python 2.X?
xrange only stores the range params and ...READ MORE
What is a Tuple in Python and how to use it?
it is a python sequency which stores ...READ MORE
In Python, what is difference between Array and List?
Lists and arrays are used in Python ...READ MORE
+1 vote
how can i count the items in a list?
Syntax : list. count(value) Code: colors = ['red', 'green', ...READ MORE
how do i use the enumerate function inside a list?
can you give an example using a ...READ MORE
Lowercase in Python
You can simply the built-in function in ...READ MORE | 554 | 2,261 | {"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} | 2.78125 | 3 | CC-MAIN-2019-35 | latest | en | 0.845714 |
https://henrigougaud.info/essay-become/rectangles-common-core-geometry-homework-answers-18043.php | 1,638,976,937,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00074.warc.gz | 374,265,020 | 8,863 | # Rectangles common core geometry homework answers
12:02
759
Saturday, April 24, 2021
4%
281 Voices
Bookmark this to easily find it later. Then send your curated collection to your children, or put together your own custom lesson plan. My Education. Log in with different email For more assistance contact customer service. Common Core. Third Grade Math.
## Common Core Grade 5 Math (Worksheets, Homework, Solutions, Examples, Lesson Plans)
Identify and describe shapes. Analyze, compare, create, and compose shapes. Reason with shapes and their attributes. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Recognize that equal shares of identical wholes need not have the same shape. Grade 3 Reason with shapes and their attributes.
### Common Core Grade 4 Math (Homework, Lesson Plans, & Worksheets)
Understand that shapes in different categories e. Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories. Here is a collection of our common core aligned worksheets for core standard 3. A brief description of the worksheets is on each of the worksheet widgets.
Kindergarten Identify and describe shapes. Analyze, compare, create, and compose shapes. For example, "Can you join these two triangles with full sides touching to make a rectangle? Describe the whole as two of, or four of the shares.
Category: essay become
Topic сomments
(11)
Jonathan T.
25.04.2021
And it also it is acceptable.
Rodrigo J.
26.04.2021
your effort of love shall not be for Granted
Damien S.
26.04.2021
highly recommended 👍,🏻,👍,🏻,👍,🏻,👍,🏻,👍,🏻,👍,🏻,👍,🏻,
Dean N.
27.04.2021
Good course! I have learned a lot!
Oscar P. L.
28.04.2021
When I have the other task from my professor, I will use this service for sure.
Jaye R.
29.04.2021
She made a lot of work and very fast.
Billy K.
30.04.2021
You could see their photo, experience & reviews you could go into the service with your eyes wide open.
Timothy C. J.
30.04.2021
Brilliant !!
Oswaldo C.
30.04.2021
Your writers really can pull offf any level of difficulty and they seem to be very competent.
Rodolfo A.
30.04.2021
Great refresher course!
Joshua Q.
01.05.2021
But my grades are excellent, and its just because of Write My Essay 247.
Comment on the essay:
Research Paper from category
Related Essays Trending Now
henrigougaud.info | 656 | 2,506 | {"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} | 3.125 | 3 | CC-MAIN-2021-49 | latest | en | 0.887718 |
https://faculty.kfupm.edu.sa/MATH/ffairag/math101_101/hw/index.htm | 1,675,641,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500294.64/warc/CC-MAIN-20230205224620-20230206014620-00421.warc.gz | 271,486,964 | 5,737 | Math 101 (101)
Suggested Homework and Recitation Problems
Section Homework Recitation CAS* 2.1 3, 5 2.2 2, 6, 9, 14, 14, 29, 30 4, 16, 28, 32 - 2.3 2, 4, 9, 20, 23, 26, 37, 44, 48, 49, 55 10, 15, 29, 38, 51, 56 - 2.4 3, 4, 18, 20 1, 2, 16, 21 - 2.5 4, 10, 14, 16, 20, 26, 34, 39, 42, 43(a,c), 48 3, 12, 19, 27, 43(b), 50 30 2.6 4, 9, 18, 24, 26, 33, 36, 42, 47, 50 3, 7, 23, 441, 49 - 2.7 3, 19(a,b), 15, 19, 23(a), 29, 34, 38 11, 12, 17, 20, 31 - 2.8 4, 25, 36, 41, 45, 49, 52, 54 3, 12, 43, 48, 53 30 3.1 10, 24, 32, 35, 46, 51, 58, 60, 62(b), 70, 73 23, 30, 50,(a,b), 68, 75 48 3.2 10, 24, 28, 34, 44(b,c), 48(b), 55, 58 20, 30, 47, 50© 40 3.3 4, 16, 18, 22, 30, 34, 41, 48, 51 19, 31, 42, 45 - 3.4 19, 36, 39, 46, 50, 53, 61, 71, 75 65, 74, 76 - 3.5 10, 19, 26, 35, 46, 53, 67(a,b) 34, 47, 65, 68 - 3.6 4, 11, 16, 22, 25, 30, 33, 38, 46, 50, 52 16, 32, 42, 53 - 3.7 1, 7 4, 5 - 3.9 4, 10, 12, 13, 15, 29, 35 5, 9, 41 - 3.10 4, 9, 11(b), 16, 20, 25, 34 2, 10, 24, 35 5 3.11 3(a), 4(b), 10, 13, 19, 20, 23(a,e), 30, 40, 42 1(b), 6(b), 17, 21, 37, 45 - 4.1 4, 8, 10, 22, 33, 39, 42, 50, 58, 68(b) 14, 28, 44, 74 - 4.2 4, 6, 12, 14, 18, 24 2, 5, 16, 20, 30 - 4.3 2, 6, 8, 14, 16, 20, 25, 37, 46, 49 35, 40, 47, 50 56 4.4 2, 4, 12, 22, 28, 31, 35, 45, 47, 60, 64 13, 30, 44, 52, 53 - 4.5 6, 10, 26, 34, 37, 50, 58, 65 18, 36, 67, 70 - 4.7 6, 11, 14, 19, 25, 27, 33, 35, 39, 50 12, 24, 46, 52 - 4.8 2, 6, 8, 12 1, 7, 11 - 4.9 12, 16, 32, 33, 42, 44, 50, 61 5, 17, 36, 49, 62 -
* CAS problems require the use of a technology tool (e.g., graphing calculators or computers). You are encouraged to do these problems in order to enhance your understanding of the concepts involved.
Tips on how to enhance your problem-solving abilities:
1. Do all the homework assignments on time.
2. You are urged to practice (but not memorize) more problems than the above lists.
3. You should always try to solve a problem on your own before reading the solution or asking for help.
4. If you find it difficult to handle a certain type of problems, you should try more problems of that type.
5. You should try the recitation problems before coming to class.
6. You are encouraged to solve some of the review problems at the end of each chapter.
7. The practice you get doing homework and reviewing the class lectures and recitations will make exam problems easier to tackle.
8. Try to make good use of the office hours of your instructor. | 1,290 | 2,487 | {"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} | 3.671875 | 4 | CC-MAIN-2023-06 | latest | en | 0.805646 |
http://forums.wolfram.com/mathgroup/archive/2011/Feb/msg00022.html | 1,722,807,899,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00469.warc.gz | 12,896,554 | 8,615 | MousePosition with PlotLegends
• To: mathgroup at smc.vnet.net
• Subject: [mg116088] MousePosition with PlotLegends
• From: "E. Martin-Serrano" <eMartinSerrano at telefonica.net>
• Date: Tue, 1 Feb 2011 06:55:02 -0500 (EST)
```Hi,
The following is an example of the problem I am facing in using
MousePosition[] over graphics containing legends, with Mathematica 7.01.
(W-XP) SP3.
In[1]:= Needs["PlotLegends`"]
(* Graphics primitives making up a legend *)
In[2]:=
g=Graphics[Legend[{{Graphics[{Blue,Disk[{0,0},1]}],Sin[x]>0},{Graphics[{Red,
Disk[{0,0},1]}],Sin[x]<0}}] ];
In[3]:= (* This gives the full PlotRange definition for the resulting
graphics *)
In[4]:= PlotRange/.FullOptions[g]
Out[4]= {{-1.,-0.15},{-1.05,-0.4}}
In[5]:= (*This gives the simplify PlotRange definition for the resulting
graphics *)
In[6]:=
Cases[AbsoluteOptions[g],HoldPattern[PlotRange->pr_]:>pr[[2]],1,1][[1]]
Out[6]= {-1.05,-0.4}
In[7]:= (* This renders the picture with the MousePosition["Graphics"]
feature included *)
In[8]:=
{Graphics[Legend[{{Graphics[{Blue,Disk[{0,0},1]}],Sin[x]>0},{Graphics[{Red,D
isk[{0,0},1]}],Sin[x]<0}}] ], Dynamic@MousePosition["Graphics"]}
Out[8]:= (* the picture in Out[8] has been deleted, you can get it by
executing In[8] *)
(* But the coordinates given by MousePosition["Graphics"] when moving the
mouse over the graphics does not coincide with the coordinates corresponding
to the range limits given by the PlotRange setting got above*)
No attempt with the MousePosition different options has helped me to solve
the issue, and I never had this kind of problem until I used MousePosition
and PlotLegends togheter.
Any help will be welcome
E. Martin-Serrano
___________________________________________________
This e-mail and the documents attached are confidential and intended solely
for the addressee; it may also be privileged. If you receive this e-mail in
error, please notify the sender immediately and destroy it. As its integrity
cannot be secured on the Internet, E. Mart=EDn-Serrano liability cannot be
triggered for the message content. Although the sender endeavors to maintain
a computer virus-free network, the sender does not warrant that this
transmission is virus-free and will not be liable for any damages resulting
from any virus transmitted.
Este mensaje y los ficheros adjuntos pueden contener informaci=F3n
anteriormente pueden estar protegidos por secreto profesional y en
cualquier caso el mensaje en su totalidad est=E1 amparado y protegido por la
legislaci=F3n vigente que preserva el secreto de las comunicaciones, y por la
legislaci=F3n de protecci=F3n de datos de car=E1cter personal. Si usted recibe
este correo electr=F3nico por error, gracias por informar inmediatamente al
remitente y destruir el mensaje. Al no estar asegurada la integridad de este
mensaje sobre la red, E. Mart=EDn-Serrano no se hace responsable por su
contenido. Su contenido no constituye ning=FAn compromiso para el remitente,
salvo ratificaci=F3n escrita por ambas partes. Aunque se esfuerza al m=E1ximo
por mantener su red libre de virus, el emisor no puede garantizar nada al
respecto y no ser=E1 responsable de cualesquiera da=F1os que puedan resultar de
una transmisi=F3n de virus.
```
• Prev by Date: statistical comparison of parameters from two applications of NonlinearModelFit
• Next by Date: ascii quotes in Program cell
• Previous by thread: Re: statistical comparison of parameters from two applications of NonlinearModelFit
• Next by thread: Re: MousePosition with PlotLegends | 962 | 3,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.744042 |
https://www.slideserve.com/anoush/straight-line-drawings-of-planar-graphs-part-i | 1,527,451,663,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00120.warc.gz | 859,583,381 | 15,986 | Straight line drawings of planar graphs – part I
1 / 17
# Straight line drawings of planar graphs – part I - PowerPoint PPT Presentation
Straight line drawings of planar graphs – part I. Roeland Luitwieler. Outline. Introduction The shift method The canonical ordering The shift algorithm Next time: the realizer method. Introduction. This presentation is about: Straight line grid drawings of planar graphs Minimized area
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about 'Straight line drawings of planar graphs – part I' - anoush
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
### Straight line drawings of planar graphs – part I
Roeland Luitwieler
Outline
• Introduction
• The shift method
• The canonical ordering
• The shift algorithm
• Next time: the realizer method
Introduction
• Straight line grid drawings of planar graphs
• Minimized area
• De Fraysseix, Pach & Pollack, 1988:
• The shift method: 2n–4 x n–2 grid
• Schnyder, 1990:
• The realizer method: n–2 x n–2 grid
• Both are quite different, but can be implemented as linear time algorithms
Introduction
• Assumptions (can be taken care of in linear time)
• n ≥ 3 (trivial otherwise)
• Graph is maximal planar (=triangulated)
• Graph has a topological embedding
• Terminology
• Given an ordering on the vertices of graph G:
• Gk is the induced subgraph of vertices v1, …, vk
• Co(G) is the outer cycle of a 2-connected graph G
• Given a cycle C with u and v on it, but (u,v) not on it:
• The edge (u,v) is a chord of C
The canonical ordering
• An ordering of vertices v1, …, vn is canonical if for each index k, 3 ≤ k ≤ n:
• Gk is 2-connected and internally triangulated
• (v1, v2) is on Co(Gk)
• If k+1≤n, vk+1 is in the exterior face of Gk and its neighbours in Gk appear on Co(Gk) consecutively
The canonical ordering
• Lemma: every triangulated plane graph has a canonical ordering
• Trivial for n=3, so assume n≥4
• Proof by reverse induction
• Base case (k=n): conditions hold since Gn = G
• Co(Gn) consists of v1, v2 and vn
• We now assume that vn, vn-1, ..., vk+1 for k+1≥4 have been appropriately chosen and that the conditions hold for k
• To prove: the conditions hold for k–1
The canonical ordering
• If we can find a vertex (= vk) on Co(Gk) which is not an end of a chord, the conditions hold
• Follows from Gk–1 = Gk – vk
• Such a vertex always exists, because a chord always skips at least one vertex on the cycle
The canonical ordering
• A canonical ordering of a triangulated plane graph can be found in linear time
• Use labels for vertices
• Label -1 means not yet visited
• Label 0 means visited once
• Label i means visited more than once and there are i intervals of visited neighbours in the circular order around the vertex
• Algorithm:
• Let v1 and v2 be two consecutive vertices on Co(G) in counter clockwise order
• Label all other vertices -1
• Process v1
• Process v2
• for k = 3 to n do
• Chose a vertex v with label 1
• vk = v
• Process vk
The canonical ordering
• (Sub-algorithm) Process v:
• For all neighbours w of v not yet processed do
• If label -1 then label(w) = 0
• If label 0 then (w has one processed neighbour u)
• If w next to u in circular order of neighbours of v then label(w) = 1
• Otherwise label(w) = 2
• Otherwise (label i)
• If vertices next to w in circular order of neighbours of v both have been processed then label(w) = i–1
• If neither has been processed yet then label(w) = i+1
• Correctness
• All processed vertices form Gk and the chosen vertex is vk+1
• Verify the conditions hold now
• Successful termination guaranteed by previous proof
The shift algorithm
• General idea:
• Insert vertices in canonical order in drawing
• Invariants when inserting vk:
• x(v1) = 0, y(v1) = 0, x(v2) = 2k–6, y(v2) = 0
• x(w1) < … < x(wt) where Co(Gk–1) = w1, …, wt with w1 = v1, wt = v2
• Each edge on Co(Gk–1) except (v1, v2) is drawn by a straight line having slope +1 or –1
The shift algorithm
• General idea:
• Some vertices need to be shifted when inserting vk
The shift algorithm
• To obtain a linear time algorithm, x-offsets are used in stead of x-coordinates
• We use a binary tree to represent shift dependencies
• Each node shifts whenever its parent shifts
• Store x-offset and y-coordinate in each node
• Using the tree, the x-offsets can be accumulated in linear time
The shift algorithm
• Compute x-offsets and y-coordinates:
• Compute values and
create the tree for G3
• For k = 4 to n do
• Let w1, …, wt = Co(Gk–1) with w1 = v1, wt = v2
• Let wp, wp+1, …, wq be the neighbours of vk on Co(Gk–1)
• Increase Δx(wp+1) and Δx(wq) by one
• Compute Δx(vk) and y(vk)
• Update the tree accordingly
The shift algorithm
• Updating
the tree
The shift algorithm
• Computing Δx(vk) and y(vk)
• Δx(wp, wq) = Δx(wp+1) + ... + Δx(wq)
• Δx(vk) = ½ {Δx(wp, wq) + y(wq) – y(wp)}
• y(vk) = ½ {Δx(wp, wq) + y(wq) + y(wp)}
• Δx(wq) = Δx(wp, wq) – Δx(vk)
• If p+1 <> q then Δx(wp+1) = Δx(wp+1) – Δx(vk)
Conclusions
• The shift method uses
• The canonical ordering
• A 2n–4 x n–2 grid, obviously
• Linear time
• Next time: the realizer method
• Questions?
References
• H. de Fraysseix, J. Pach and R. Pollack, How to draw a planar graph on a grid, Combinatorica 10 (1), 1990, pp. 41–51.
• D. Harel and M. Sardas, An Algorithm for Straight-Line Drawing of Planar Graphs, Algorithmica 20, 1998, pp. 119–135.
• T. Nishizeki and Md. S. Rahman, Planar Graph Drawing, World Scientific, Singapore, 2004, pp. 45–88.
• W. Schnyder, Embedding planar graphs on the grid, in: Proceedings of the First ACM-SIAM Symposium on Discrete Algorithms, 1990, pp. 138–148. | 1,797 | 6,086 | {"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-2018-22 | longest | en | 0.786414 |
https://ccssmathanswers.com/into-math-grade-4-module-10-lesson-2-answer-key/ | 1,713,905,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00150.warc.gz | 136,425,557 | 58,635 | # Into Math Grade 4 Module 10 Lesson 2 Answer Key Identify Factors
We included HMH Into Math Grade 4 Answer Key PDF Module 10 Lesson 2 Identify Factors to make students experts in learning maths.
## HMH Into Math Grade 4 Module 10 Lesson 2 Answer Key Identify Factors
I Can use division to determine whether a number is a factor of another number.
1. Mrs. Aisha owns a bakery. She wants to put all of these blueberry muffins in a display case with the same number of muffins on each shelf. Which display case should Mrs. Aisha use?
Display case Mrs. Aisha should use = 4 shelves because in 3 shelves they cannot fit correctly.
Explanation:
Number of blueberry muffins she has = 16.
Multiples of 16:
1 × 16 = 16.
2 × 8 = 16.
4 × 4 = 16.
Number of muffins on each shelf = 4 selves = 4 × 4.
Turn and Talk Why can putting 16 objects in 3 and 4 equal rows help you know whether 3 or 4 is a factor of 16?
Putting 16 objects in 3 and 4 equal rows can help to know whether 3 or 4 is a factor of 16 because it can be divisible by that number.
Explanation:
Multiples of 16:
1 × 16 = 16.
4 × 4 = 16.
Build Understanding
Question 1.
Mrs. Aisha makes a new kind of granola bar for her customers to try. She wants to display 40 bars in equal rows on a tray. Can Mrs. Aisha place all the bars in 5 equal rows?
Use a visual model to represent the problem.
A. What shape does your visual model show? ______
B. What does this shape tell you about whether or not all the granola bars can be placed in equal rows?
_______________________________
_______________________________
C. Can Aisha put all the granola bars in 5 rows? How does your visual model show this?
_______________________________
_______________________________
D. Is 5 a factor of 40? How do you know?
_______________________________
_______________________________
Yes, Mrs. Aisha can place all the bars in 5 equal rows.
Explanation:
Number of bars in equal rows on a tray she wants to display = 40.
Mrs. Aisha place all the bars in 5 equal rows.
Multiples of 40:
1 × 40 = 40.
2 × 20 = 40.
4 × 10 = 40.
5 × 8 = 40.
Turn and Talk Why can you write a division equation and a multiplication equation to model this problem?
We can write a division equation and a multiplication equation to model this problem, because both are same to find how many bars comes in five rows as we know the total bars count.
Explanation:
We can write a division equation and a multiplication equation to model this problem, because both are same to find how many bars comes in five rows as we know the total bars count.
2. Mrs. Aisha has 35 loaves of French bread. She wants to put the same number of loaves in 3 baskets. Can she put all the loaves in the baskets with none left over?
A. Why can you use division to solve this problem?
_______________________________
_______________________________
B. Complete the division. Show your work.
C. Can all the loaves be put in 3 baskets with the same number in each basket? How do you know?
_______________________________
_______________________________
D. Is 3 a factor of 35? How do you know?
_______________________________
_______________________________
_______________________________
11 baskets Loaves can be kept and 2 will be remaining.
Explanation:
Number of loaves of French bread Mrs. Aisha has = 35.
Number of loaves in baskets she wants to put the same = 3.
Factors of 3: 3,6,9,12,15,18,21,24,27,30,33,36.
35 is not a factor of 3.
Hence, in 11 baskets they can be kept and 2 will remain.
Connect to Vocabulary
A factor of a number divides that number evenly. The quotient is a whole number and the remainder is 0.
Turn and Talk How does using division to find factors of a number relate to using an array or area model to find factors of a number?
We can use division to find factors of a number relate to using an array or area model to find factors of a number by division method.
Explanation:
Arrays are useful models for multiplication which can be used in a variety of ways, ranging from highly structured lessons to games and open investigations. An array is formed by arranging a set of objects into rows and columns. Each column must contain the same number of objects as the other columns, and each row must have the same number as the other rows.
Step It Out
3. Is 6 a factor of 84?
Use divisibility rules to answer and explain.
Connect to Vocabulary
A number is divisible by a second number if the quotient is a whole number and the remainder is 0. So, if a number is divisible by , then is a factor of that number.
A. Is 84 divisible by 2? _____
B. Is 84 divisible by 3? _____
C. Is 84 divisible by 6? ____
D. Is 6 a factor of 84? _____
So, if a number is divisible by 2 and 3, then 6 is a factor of that number.
Yes, 6 is a factor of 84.
Explanation:
A. 84 divisible by 2:
=> 84 is even, Hence 84 is divisible by 2.
b. 84 divisible by 3.
Sum of digits = 8 + 4 = 12.
=> 3 × 4 = 12.
Hence, 84 is divisible by 3 because sum of digits is a multiple of 3.
C. 84 divisible by 6.
Sum of digits = 8 + 4 = 12.
=> 2 × 6 = 12.
3 × 4 = 12.
84 is divisible by 6 because it is divisible by 2 and 3.
D. 6 a factor of 84.
Multiples of 6:
6,12,18,24,30,36,482,48,54,60,66,72,78,84.
Check Understanding Math Board
Question 1.
Mrs. Aisha wants to put 51 bagels in bins, with the same number in each bin. Should she use 3 or 4 bins? Use factors to justify your answer.
_______________________________
_______________________________
_______________________________
_______________________________
she should use 3 bins to put 51 bagels in bins, with the same number in each bin.
Explanation:
Number of bagels in bins Mrs. Aisha wants to put = 51.
Multiples of 51:
1 × 51 = 51.
3 × 17 = 51.
3 is a factor of 51 not 4.
Question 2.
Use Tools There are 26 members in the marching band. Can all the members march in 4 equal rows?
• Use square tiles to represent the problem. Then draw an array to show your work.
• Can 26 band members march in 4 equal rows? Use factors or divisibility to explain.
4 is not the factor of 26. They cannot march in 4 equal rows.
Explanation:
There are 26 members in the marching band.
All the members march in 4 equal rows.
=> Multiples of 26:
1 × 26 = 26.
2 × 13 = 26.
Use a visual model to answer and explain.
Question 3.
Is 6 a factor of 32?
No, 32 is not a factor of 6.
Explanation:
Multiples of 32:
1 × 32 = 32.
2 × 16 = 32.
4 × 8 = 32.
Question 4.
Is 5 a factor of 25?
Yes, 5 a factor of 25.
Explanation:
Multiples of 25:
1 × 25 = 25.
5 × 5 = 25.
Question 5.
Use Repeated Reasoning Explain whether or not the following statement is true: Since 89 is not divisible by 2, then 89 is not divisible by any multiple of 2.
Yes, the following statement is true because 89 cannot be divided by any factor of 2 as 89 is not an even number.
Explanation:
89 is not divisible by 2, then 89 is not divisible by any multiple of 2.
Question 6.
Use Structure There are 72 students in the after-school program. They will play one of the sports listed in the table. The program leader wants all the students to be on a team and all the teams to have the same number of players. Which sport should they play? Why?
_______________________________
_______________________________
Volley ball sport they should play because number of players on each team 9 in 12 teams.
Explanation:
Number of students in the after-school program = 72.
Multiples of 72:
1 × 72 = 72.
2 × 36 = 72.
3 × 24 = 72.
4 × 18 = 72.
6 × 12 = 72.
8 × 9 = 72.
Use division to answer and explain.
Question 7.
Is 7 a factor of 91?
Explanation:
91 ÷ 7 = 13.
Question 8.
Is 8 a factor of 94?
Explanation:
94 ÷ 8 = 11$$\frac{6}{8}$$
Use divisibility rules to answer and explain.
Question 9.
Is 5 a factor of 80?
________________
________________
Yes, 5 is a factor of 80.
Explanation:
Multiples of 80:
1 × 80 = 80.
2 × 40 = 80.
4 × 20 = 80.
5 × 16 = 80.
8 × 10 = 80.
Question 10.
Is 9 a factor of 88?
________________
________________
No, 9 is not a factor of 88.
Explanation:
Factors of 88:
1 × 88 = 88.
2 × 44 = 88.
4 × 22 = 88.
8 × 11 = 88.
I’m in a Learning Mindset!
How effective was using division to determine whether or not a number is a factor of another number? | 2,253 | 8,182 | {"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} | 5 | 5 | CC-MAIN-2024-18 | latest | en | 0.845216 |
https://www.physicsforums.com/threads/few-questions.159913/ | 1,653,314,762,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00707.warc.gz | 1,082,725,322 | 16,187 | # Few Questions
Gold Member
A string, 50cm long, has a stone of 100g tied to its end, and it is swung in a horizontal circle of radius 30cm. Calculate the tension in the string.
Because the string is longer than the radius of the circle, the path of the string must form a cone shape. Using the measurements, and forming a triangle, i found that the angle the string is forming is ~37 degrees to the vertical.
With this question, am i able to say that the vertical component of the tension force is the weight force from the stone? So ~0.98N? And then use that, along with the angle, to find the tension in the rope?
When you throw an orange up vertically from a moving ute, will it land back at the same spot from which it was thrown? explain.
For this question, i said if air resistance is neglected, and the use it travelling with uniform velocity, it will land on the same spot, since the orange will initially have the same horizontal velocity as the ute, and unless another force acts, they will both continute to have the same velocity.
A basketball player shoots for goal. The ball goes through the ring without touching it. If the ball is thrown from 50 degrees above the horizontal, and the basket is 2.5m in front of the player, and 1.3m higher than the point from which the ball was thrown, at what speed was the ball released?
I started by finding the time it takes for the ball to travel 2.5m in the horizontal plane.
$$\begin{array}{c} t = \frac{{s_h }}{{v_h }} \\ = \frac{{2.5}}{{v\cos 50}} \\ \end{array}$$
At that point in time, the vertical displacement should be 1.3m.
$$\begin{array}{c} s = ut + \frac{{at^2 }}{2} \\ 1.3 = \frac{{2.5v\sin 50}}{{v\cos 50}} - 4.9\left( {\frac{{2.5}}{{v\cos 50}}} \right)^2 \\ v = \pm 6.64 \\ \end{array}$$
So the initial velocity of the ball was 6.64 m/s at 50 degrees to the horizontal.
_____________________________________________
These questions didnt come with answers, and im not really sure if ive done them correctly. If you could just look over them, and tell me if ive done anything wrong, that would be great.
Dan.
A string, 50cm long, has a stone of 100g tied to its end, and it is swung in a horizontal circle of radius 30cm. Calculate the tension in the string.
Because the string is longer than the radius of the circle, the path of the string must form a cone shape. Using the measurements, and forming a triangle, i found that the angle the string is forming is ~37 degrees to the vertical.
With this question, am i able to say that the vertical component of the tension force is the weight force from the stone? So ~0.98N? And then use that, along with the angle, to find the tension in the rope?
Yep, this seems correct to me. The vertical component of the tension force is the weight of the stone, the horizontal component is then the centripetal force of the circular motion.
When you throw an orange up vertically from a moving ute, will it land back at the same spot from which it was thrown? explain.
For this question, i said if air resistance is neglected, and the use it travelling with uniform velocity, it will land on the same spot, since the orange will initially have the same horizontal velocity as the ute, and unless another force acts, they will both continute to have the same velocity.
If the frame of reference is connected with the ute (the ute is travelling with uniform velocity and air resistance is neglected), it'll land on the same spot. However, if it is connected with the road, it won't land on the same spot - due to the initial horizontal velocity.
danago;1267728[b said:
A basketball player shoots for goal. The ball goes through the ring without touching it. If the ball is thrown from 50 degrees above the horizontal, and the basket is 2.5m in front of the player, and 1.3m higher than the point from which the ball was thrown, at what speed was the ball released?[/b]
I started by finding the time it takes for the ball to travel 2.5m in the horizontal plane.
$$\begin{array}{c} t = \frac{{s_h }}{{v_h }} \\ = \frac{{2.5}}{{v\cos 50}} \\ \end{array}$$
At that point in time, the vertical displacement should be 1.3m.
$$\begin{array}{c} s = ut + \frac{{at^2 }}{2} \\ 1.3 = \frac{{2.5v\sin 50}}{{v\cos 50}} - 4.9\left( {\frac{{2.5}}{{v\cos 50}}} \right)^2 \\ v = \pm 6.64 \\ \end{array}$$
So the initial velocity of the ball was 6.64 m/s at 50 degrees to the horizontal.
Just quickly flicking through your calcs on this one your ball does travel 2.5m in the horizontal but travels further than 1.3m in the vertical. This is because the ball has to come down through the top of the hoop.
Look at this as a ballistic trajectory question. The ball will travel an arc where the peak of the arc (max vertical motion) is less than 2.5m from the start point, the maximum range (max horizontal motion) is greater than 2.5m.
I haven't got my dynamics books to hand, what you need to do is work out the equation for a projectile at any point in the arc. possibly somebody who does have a dynamics book, or a large brain than mine can give you a pointer.
Gold Member
Yea i realise it travels further than 1.3m, but its displacement will still be 1.3m in the vertical plane wont it? The ring is 1.3m from the point of release, so wont the vertical displacement need to be 1.3m as it goes through?
Thanks for the help by the way everyone. | 1,383 | 5,361 | {"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} | 4.34375 | 4 | CC-MAIN-2022-21 | latest | en | 0.929578 |
https://ru.coursera.org/learn/quantum-computing-lfmu/reviews | 1,638,977,302,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00088.warc.gz | 550,990,419 | 123,670 | Вернуться к Quantum Computing. Less Formulas - More Understanding
# Отзывы учащихся о курсе Quantum Computing. Less Formulas - More Understanding от партнера Санкт-Петербургский государственный университет
4.8
звезд
Оценки: 108
Рецензии: 38
## О курсе
This is yet one more introductory course on quantum computing. Here I concentrate more on how the mathematical model of quantum computing grows out from physics and experiment, while omitting most of the formulas (when possible) and rigorous proofs. On the first week I try to explain in simple language (I hope) where the computational power of a quantum computer comes from, and why it is so hard to implement it. To understand the materials of this week you don't need math above the school level. Second and third weeks are about the mathematical model of quantum computing, and how it is justified experimentally. Some more math is required here. I introduce the notion of a linear vector space, discuss some simple differential equations and use complex numbers. The forth week is dedicated to the mathematical language of quantum mechanics. You might need this if you want to dig deeper into subject, however I touch only the tip of the iceberg here. On the week 5 I finally introduce some simple quantum algorithms for cryptography and teleportation....
## Лучшие рецензии
BT
25 сент. 2021 г.
Loved the course and recommend for all, especially for those who are just starting their journey in quantum computing or those who just wonder how weird the quantum world really is.
GR
10 дек. 2020 г.
I like everything about this course except the title, there are many formulas. But really well explained, I enjoyed it a lot and I think I learned something. I recommend it.
Фильтр по:
## 1–25 из 40 отзывов о курсе Quantum Computing. Less Formulas - More Understanding
автор: Vishal B
15 окт. 2020 г.
An awesome introductory course!
Having looked the the course "Introduction to Quantum Computing" by Professor Сысоев Сергей Сергеевич and finding it overwhelming at start >.< this is a welcome and a more friendly introduction to the same aforesaid course. It could be, as someone above pointed out, called as the Introduction to the Introduction to Quantum Computing course with well layed out and clear explanation to a lot of concepts.
I am extremely glad to note that professor Сысоев Сергей Сергеевич has hugely improved on his on camera presence and is much more fun and free in this new course. I totally liked his camera presence and the way he explains is stellar as always in both the courses, but this one shows he has worked on in perfecting his presence on video. I am looking forward to gladly look into more courses by him as I really like the way he explains and put forth concepts, clarifying everything and explaining with clear justifications. Thank you so much professor. I will be now advancing into your next course as I believe most of the perquisites and the things that overwhelmed me have been cleared and I hope to complete it soon :D
автор: Jayesh P
30 сент. 2020 г.
Very good course with lot of knowledge packed in brief video sessions. It may require multiple reading/watching but at end it makes lot of sense. The contents of this course are good for both entry level person as well as someone who knows basics, As a computer software engineer I found it one of the best courses available online on Quantum Computing. Thank you for creating this course !!
автор: QUAGLIA R
22 сент. 2020 г.
This course keeps its promises, few math and more focus to explain the key concepts that enable quantum computing.
The name could be "The Introduction to The Introduction to Quantum Computing", the other course of the same professor.
If you completed the Introduction first, like in my case, you'll probably find this course easy and fast to pass. Anyway it worth it to do it because here are given additional "tools", like the Bloch Sphere, and other gates and algorithms descriptions not included in the Introduction.
I particularly appreciated the professor's effort in being more communicative - if you also do the Introduction you will notice differences in tone of voice and emphasis, which can also be seen in this course as a video of the Introduction has also been included here.
One thingh that remains to improve are the sub-titles, that are automatically generated by some imperfetc tool producing errors, sometimes funny (e.g.: Quantum Computing that becomes Quantum Confusion) but sometimes missleading (e.g.: near the end of the video about wavefuntions). I would suggest either to change voice recognition tool or to review and correct the result before publishing the course.
автор: Thomas A T
11 янв. 2021 г.
This is a good first course for a difficult subject. Keep in mind "Less Formulas" in the title is a relative term. As the Professor of the course said, if you have a problem with any of the math you should look for other explanations of the math at issue on YouTube so that you actually understand the current concept before moving on. It is very helpful to review the same video again and if still not clear to listen to other videos to see if they explain it in a way that resonates with you. It is a good course, although in difficult subject so some effort is required. I am planning to take other of the QC courses from this professor.
автор: Oskar
16 дек. 2020 г.
I really appreciate the way course was delivered. Great, mind opening examples supported with precise language and comments.
автор: Yevonnael A
26 нояб. 2020 г.
A very nice introduction to the subject! Now I will take a harder course. Thank you!
автор: Fred v d B
5 мар. 2021 г.
The course would be easier to follow if it started with an overview that shows how the various elements of the course are related and fit together into a complete picture. I had to listen to the course twice and for the first 3 weeks three times to build up this picture.
автор: Oosman B
3 дек. 2021 г.
This is a baby-level course. But don't say we got nothing out of it! In fact, many a time Sergey gave us impressive information and insights, drawing on his mathematical expertise and an understanding-focused approach to the mathematical language (although he is a mathematician!) I really like his approach—he likes to list fact-by-fact instead of convoluted arguments. Please note, just because you have completed this course, doesn't mean you are ready to understand Quantum Mechanics! More abstract Mathematics will be required but doing this course first is useful in gaining ideas on how to instantly apply what you will learn in Mathematics for Quantum Mechanics courses. PS.: His graphics are beautiful, and he's overall a charming guy. LOL!
автор: Rodrigo G
15 июля 2021 г.
The course is indeed a help to understand some formulae and applications found in introductory books. Doing the exercises as suggested helped me to detect misconceptions I had on the field with 3-qubits systems. Thank you very much, Professor Sysoev!
A private note: I did detect some typos in the English transcription. Some of them were quite critical, when the transcriptor confused negative with possitive sentences. I do not remember the exact phrase but it was something like: "The vectors are (not) perpendicular", with the "not" added by the trascriptor. I strongly suggest to dully double check all transcription, as it may propagate to other languages.
автор: Kris L
5 февр. 2021 г.
This was my first Coursera course and it was a hard subject I really wanted to learn more about. I wasn't disappointed. It was truly excellent. Professor Sysoev really helped advance my understanding of quantum mechanics and quantum computing, without overwhelming amounts of advanced math. There certainly were times, especially week four, where I was stretched to my personal mathematic abilities, but even then I was able to follow the concepts and continue. Professor Sysoev is a great teacher, with the right mix of clarity, example and humor to make this challenging subject approachable. I highly recommend this course!
автор: Amanda A
30 мар. 2021 г.
I found this course to be a really accessible way to get into the exciting principles behind quantum computing. My AP Calc skills are very rusty, but I have to say that the Professor makes the dense math in lecture 4 clear and relatively easy to wrap your head around. The lectures are broken up into comfortable lengths, and I found his presentation of the material clear and concise. Overall, this has been a really nice way to dip my toes into this fascinating field - and tempts me to take some more math in order to have a go at the more advanced version of this course.
автор: Denis C
1 дек. 2021 г.
I wish there are more courses like this one that gives a basic understanding of complex things like Quantum Computing. The course is well designed and thought through. Before the course, quantum physics sounds like a spell to me, and now I can understand how all this works and even apply some simple computation formulas and algorithms. Next step, I'm going to try applying my new knowledge to programming a quantum computer (luckily we already have some these days).
The course was really inspiring. Thank you so much for teaching!
автор: E S
2 нояб. 2021 г.
Very helpful course to understand quantum computign from the basic, but week 4 (quantum teleportation) needs more explanation for the calculation. The part of the course is difficult to a true beginner like me.
NOTE: This course is "Less" formulas, not "No" formula. You need to understand basic-level complex number and liner algebra (and Euler's formula, if possible). If you are not confident about these, I recommend you to be familier to these topics.
автор: Kyle C
13 апр. 2021 г.
Fantastic course for someone who has minimal training in math but wants to better understand quantum computing. Professor Sysoev made the subject approachable and even fun. This course inspired me to try to learn the math I need to take his next course, Introduction to Quantum Computing. Thank you!
автор: simon s
26 февр. 2021 г.
Bravo!!! I am a PhD student in this area and watched the videos and are awesome, really well explained. I already know a little about the subject and most of the explanations are some of the best I have seen. Really great work.
26 сент. 2021 г.
Loved the course and recommend for all, especially for those who are just starting their journey in quantum computing or those who just wonder how weird the quantum world really is.
автор: Guillem C R
11 дек. 2020 г.
I like everything about this course except the title, there are many formulas. But really well explained, I enjoyed it a lot and I think I learned something. I recommend it.
автор: Diana M
19 апр. 2021 г.
Great course! Awesome opportunity to learn from the top Russian university, professor Sysoev presents this difficult topic in an engaging style.
автор: Hélder M
4 сент. 2021 г.
Great course for understanding the basics of quantum computing. I also like the book suggestions given in the course.
автор: Al M
29 окт. 2021 г.
Great, approachable course as an introduction to quantum systems and universal gate quantum computing.
автор: Darío A A S
17 мая 2021 г.
A very useful course, it was more than I think I could learn in a very brief course. 100% satisfied
автор: Poonyarak C
14 сент. 2021 г.
Great course, very informative, easy to understand and enjoyable. Thank you
автор: Paulo H d S
12 апр. 2021 г.
A very nice introduction to the quantum computing. Really well explained.
автор: Ralf B
24 июня 2021 г.
An excellent course! The contents are explained very clearly and well.
автор: Edgar A J P
20 мар. 2021 г.
Buen curso, vale la pena seguirlo, el instructor enseña bien. | 2,597 | 11,814 | {"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-2021-49 | latest | en | 0.783025 |
blog.edutoolbox.de | 1,513,284,124,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550986.47/warc/CC-MAIN-20171214202730-20171214222730-00198.warc.gz | 36,266,807 | 19,137 | # Find position of bit in byte with one bit set
Gegeben ist ein Byte mit einem gesetzten Bit: i.e. 1,2,4,8,16,32,64,128
Die Position des gesetzten Bits soll gefunden werden:
i.e. Byte = 1 enstpricht Position 1
i.e. Byte = 64 entspricht Position 7
… ended up with deBruijn (without knowing then …)
``````
package main
/*
int pos(int a){
int p = (a & 3) | ((a >> 1) & 6) | ((a >> 2) & 5) | ((a >> 3) & 4) | ((a >> 4) & 15) | ((a >> 5) & 2) | ((a >> 6) & 1);
return p;
}
*/
import "C"
import "fmt"
func main() {
a := 1
for i := 0; i < 8; i++ {
e := (a & 1) | (a & 2) | (a >> 1 & 4) | (a >> 4 & 8) // 1, 2, 4, 8
e |= ((a >> 1) & 2) | (a>>2)&1 // 3
e |= ((a >> 2) & 4) | (a>>4)&1 // 5
e |= ((a >> 3) & 4) | (a>>4)&2 // 6
e |= ((a >> 4) & 4) | ((a >> 5) & 2) | (a>>6)&1 // 7
fmt.Println(e)
a <<= 1
}
b := 1
for i := 0; i < 8; i++ {
a := b
e := a & 3 // (a & 1) | (a & 2) // 1, 2
a >>= 1 // b>>1
e |= a & 6 // (a & 4) | (a & 2) // 4, 3
a >>= 1 // b>>2
e |= a & 5 // (a & 1) | (a & 4) // 3, 5
// a >>= 1 // b>>3
e |= (a >> 1 & 4) // 6
a >>= 2 // b>>4
//e |= (a & 1) // 5
//e |= (a & 2) // 6
//e |= (a & 4) // 7
//e |= (a & 8) // 8
e |= a & 15 // 5, 6, 7, 8
e |= (a>>1)&2 | (a>>2)&1 // 7
fmt.Println(e)
b <<= 1
}
c := 1
for i := 0; i < 8; i++ {
a := c
e := (a & 3) | ((a >> 1) & 6) | ((a >> 2) & 5) | ((a >> 3) & 4) | ((a >> 4) & 15) | ((a >> 5) & 2) | ((a >> 6) & 1)
fmt.Println(e, int(C.pos(C.int(a))))
c <<= 1
}
// faster ! really this is as fast as
// 1 bitwise shift operation,
// 1 bitwise 'AND' operation
// .. followed by one array uplook
// (found out this is deBruijn style)
// 111 101 001 100 010 110 011 000 '1101000111' (839*(1<<i))>>6)&7
//
var deBruijnMap = [8]byte{4, 5, 2, 6, 3, 1, 8, 7} // ->{5, 2, 4, 0, 1, 3, 7, 6}
const mult = 839
for i := 0; i < 8; i++ {
a := 1 << uint(i)
fmt.Println(deBruijnMap[((mult*a)>>6)&7])
}
}
``````
# Go – running fixed number of concurrent go routines
This snippet is about the control of running a fixed number of concurrent worker functions concurrently.
Lets start with a basic version, that does no error checking at all and that will simply panic in case one of the concurrent processes crashes.
Snippet 1
``````
package main
import "fmt"
func main(){
parallel(func(x int) error {
fmt.Println("Parameter:", x)
return nil
}, []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 3)
}
func parallel(fn func(int) error, sq []int, p int) error {
var q = make(chan bool, p)
defer func() { close(q) }()
for _, s := range sq {
q <- true
go func(s int) {
fn(s)
<-q
}(s)
}
for len(q) > 0 {
}
return nil
}
``````
The function parallel takes the function to run in parallel (fn), a slice of parameters to start the concurrent workers (sq) with and the number of workers, you want to run in parallel. First a channel q of the capacity p (the number of concurrent processes) is instantiated. The defer function will care for proper closing. The for-loop takes care for starting the go routines with the appropriate values. By filling up the channel q the loop execution is blocked if the capacity is exhausted and p processes are running. Each return of a goroutine receives once from q and thus allows for a new loop until all parameter values are issued to the worker functions. The first for-loop exits and the second for-loop cares that all coroutines to finish before parallel returns.
Make sure your function fn in calling parallel(fn, sq, p) meets the function type (=signature) and the value table is adjusted properly then you are done with.
Note, that Snippet 1 panics if any of the concurrent processes crashes. Furthermore the second for loop will run in an endless loop, if one of the go routines didn’t return because ‚<-q' won't happen than. Such a code works well if the function called concurrently is guaranteed to non-panic or block (or if the project is an a state, where such panicking helps to improve logic). Snippet 2 cares for recovery and returns general error information and information on the parameter value that resulted in the crash.
Snippet 2
``````
package main
import "fmt"
import "errors"
func main(){
err, errlist := parallel(func(x int) error {
fmt.Println(x, 100/x)
return nil
}, []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0}, 3)
if err != nil {
fmt.Printf("Error! -> %v\n", err)
for _, err := range errlist {
fmt.Println(" ", err)
}
}
}
func parallel(fn func(int) error, sq []int, p int) (err error, errlist []error) {
var q = make(chan bool, p)
var e = make(chan error, len(sq))
defer func() {
close(q)
close(e)
}()
for _, s := range sq {
q <- true
go func(s int) {
defer func() {
if P := recover(); P != nil {
e <- errors.New(fmt.Sprintf("func(%v) crashed: %v", s, P))
}
<-q
}()
fn(s)
}(s)
}
for len(q) > 0 {
}
for len(e) > 0 {
err = errors.New(fmt.Sprintf("%d concurrent process crashed", len(e)))
errlist = append(errlist, <-e)
}
return err, errlist
}
------------------------------
Output from running this code.
3 33
4 25
5 20
6 16
1 100
7 14
2 50
9 11
10 10
8 12
Error! -> 1 concurrent process crashed
func(0) crashed: runtime error: integer divide by zero
``````
This gives better control on crashes and allows the main process to proceed even in case one of the concurrent functions crashes. Anyway it does not care properly for a concurrent function not returning.
Furthermore errors are printed out after all processes finished, which might be inconvenient on long running processes.
In the next Snippet 3 errors are channeled out to a concurrent running receiver loop.
``````
package main
import "fmt"
import "errors"
func main(){
// -> routine 4 concurrent error / result output
errChan := make(chan error, 11)
//ergChan := make(chan int, 11) // enable concurrent result output
alldone := make(chan bool, 1)
go func() {
for done := false; !done; {
select {
case e := <-errChan:
fmt.Println("Error -> ", e)
// case ee := <-ergChan: // enable concurrent result output
// fmt.Println("Result -> ", ee)
case done = <-alldone:
break
}
}
}()
// <-
parallel(func(x int) error {
r := 100 / x
fmt.Println(x, r)
// ergChan <- r
return nil
}, []int{1, 2, 3, 0, 4, 5, 0, 6, 0, 7, 8, 9, 10}, 3, errChan)
// let error output goroutine exit
close(alldone)
// ....
}
// func parallel with errors returned through an error channel
func parallel(fn func(int) error, sq []int, p int, eC chan error) {
var q = make(chan bool, p)
defer func() {
close(q)
}()
for _, s := range sq {
q <- true
go func(s int) {
defer func() {
if P := recover(); P != nil {
eC <- errors.New(fmt.Sprintf("func(%v) crashed: %v", s, P))
}
<-q
}()
eC <- fn(s)
}(s)
}
for len(q) > 0 {
}
}
``````
I am aware that there is a number possibilities to make sure the number of concurrent running processes remains stable including the use of patterns to loop with a partial number of wait groups or monitoring a running counter and the like. None the less i think such use of channels is a particular elegant way to manage this problem.
# GO: passing Arrays & Slices to functions
When passing Arrays as parameters to functions some specialities need to be cared of. Go’s documentation stated, that arrays will be copied.
You would assume, the following function would, according to the steps, work like this: copy the array (when passing), sort the copy and return the median without changing the original array ‚l‘.
``````
package main
import (
"fmt"
"sort"
)
func arrayMedian(list []int) (median int) {
lenAR := len(list)
if lenAR == 0 {
return 0
}
sort.Ints(list)
if lenAR%2 == 0 {
return (list[lenAR/2-1] + list[lenAR/2]) / 2
}
return list[lenAR/2]
}
func main(){
l := []int{0,3,4,1,0,9,6,3,9,8}
fmt.Println(l, arrayMedian(l), l)
}
``````
Surprise – if looking at the output, you’ll find the original array ‚l‘ is sorted now.
Why? I guess, passing l by ‚arrayMedian(l)‘ didn’t pass the original array (or generates a copy) but passes a slice over ‚l‘ (rsp: a copy of the slice over ‚l‘), what actually means, that pointers to the ‚l’s list members are passed. The manipulation (here: sorting) is performed on the original array.
What you need to do is this instead:
``````
func arrayMedian(list []int) (median int) {
lenAR := len(list)
if lenAR == 0 {
return 0
}
ar := make([]int, lenAR)
copy(ar, list[:])
sort.Ints(ar)
if lenAR%2 == 0 {
return (ar[lenAR/2-1] + ar[lenAR/2]) / 2
}
return ar[lenAR/2]
}
``````
Though you produce a copy manually, sort the copy and return the result by leaving the original array ‚l‘ unchanged.
# How to include binary data into HTML5
… without getting cross-site-scripting problems on local files / server less testing.
For some 3D data representation i needed some MB binary 3D data to be included in the HTML side for further use in threejs. The file should be useful for standalone opening after email-sending – loading data from a fileserver wasn’t a an option though. On the other hand writing 8bit data points into asciitext wasn’t an option either, because such file fast becomes to big to be handled properly: each datapoint 8bit needs to be represented by 1-3bytes plus separator or by 3 bytes fixed length and both would pump up the file size. Another option would be a base64 encoding of the raw binary data which also might be compressed (zipped) before encoding resulting in an about 1.3 larger b64 asciitext, that can be unzipped and parsed to regain the data. This solution is smart but involves „unzipping to get the asciitext“ meaning that there has to be imported/bundled a fast and reliable javascript zip-library.
This is what i finally came up with:
1. write the 8bit binary data into a png file
2. base64 encode the resulting run-length compressed png file
3. use `<img src="data:image/png;base64,....">` to include that into the html5 file
4. create a hidden canvas of appropriate dimension and use context.drawImage and context.getImageData to extract the binary data which is now stored in an Uint8ClampedArray
5. calculate the needed Float32Array(s) for threejs from that Uint8ClampedArray
Note on step 1: you’ve the choice to use all 24bit of the pixel matrix in png (four 8bit binary data -> 1 pixel in png) and have some data length information elsewhere in your javascript to manage padding problems that might come up eventually. The lazy way is to encode only 1 Byte per pixel (grayscale png) and rely on the run length compression to deal with the resulting duplications. In the latter case, make sure you extract each forth value from the canvas context.getImageData method result.
This python code will provide you with the base64 encoded png data:
``````
from PIL import Image
import base64
fle = file("testdata.bin") # binary data: 12960000 bytes
im.save("testdata.png")
b64 = '<img id="dimg" src="data:image/png;base64,%s">'%b64
file('testdata64.html','wb').write(b64).close()
``````
# a faster bloomfilter for go/golang
a bloom filter is a structure that is designed to register bit pattern in a way, you can securely say the bit pattern you present to the bloom filter is unique and unknown to the filter if the filter returns that. Unfortunately you cannot invert this phrase: if the bloom filter response marks the bit pattern to be included („known“) to the filter, there is a certain possibility that this is factual wrong.
To get an idea about bloomfilters and how they work have a look here:
https://en.wikipedia.org/wiki/Bloom_filter
and here: https://de.wikipedia.org/wiki/Bloom_filter
and do not miss the great links provided at the end of the articles.
There is a range of bloom filters implemented in go/golang (see https://github.com/search?utf8=✓&q=bloom+filter+language%3AGo&type=Repositories&ref=advsearch&l=Go).
This said i had my own ideas about what is fast in bloom filters and the result is this:
https://github.com/AndreasBriese/bbloom
# go / golang (super) fast PRNG: BREEZE
i became unhappy with the usual go PRNGs including salsa and the like. Despite knowing about being foolish to take such a task i did create a deterministic chaos based random generater (CBRNG) that produces random bits (tested with the NIST package) and is faster than go’s standard random generator, salsa20 and of course way faster than go’s crypto/random package.
you may find this and a lot of information about the how and why and however else and speed …
at my github site: https://github.com/AndreasBriese/breeze
and there is a pdf summary too: https://github.com/AndreasBriese/breeze/blob/master/breeze.pdf
!! Do not oversee the fact, it is not so far cryptoanalysed by any third party !!
# Hash mich – Nachtrag zu PRNGs als Hashgeneratoren
In Ergänzung zu den Hashfunktionen hier
Auch Pseudo-Random-Number-Generatoren PRNG können grundsätzlich dazu eingesetzt werden, Hashwerte zu erzeugen.
PRNGs erzeugen ausgehend von einem Seed eine feste Abfolge von zufällig verteilten Zahlen, die man dann zu einem Hashwert beliebiger Länge kombinieren kann. Ein Problem können sehr kurze Strings sein, die gehasht werden sollen, da, wie gesagt, erst eine ausreichend großer Seed nötig ist, um den PRNG zu initialisieren.
SipHash https://131002.net/siphash/ von Jean-Philippe Aumasson und Daniel J. Bernstein (der u. a. Salsa20 PRNG für eine kryptologische Stream-Verschlüsselung entwickelt hat, ist ein solcher PRNG, der für kurze Eingaben optimiert wurde. SipHash ist nach Angaben der Verfasser etwa genauso schnell wie MurmurHash3.
Update:
SipHash hat mich überzeugt. Es ist schnell und anscheinend ziemlich Kollisionsresistent. Ich konnte BruteForce jedenfalls keine Kollisionen für strings mit Länge 8 und 16 finden.
# Hash mich – eine kleine Sammlung zu Hash-Algorithmen
Hashfunktionen begegnet man überall, wo es darum geht, Zeichenketten (z.B. Nutzernamen oder Passwörter) oder beliebige Daten (z.B. Download – Archive) voneinander zu unterscheiden. Oft werden Hashs im Hintergrund gerechnet, ohne dass der User das sieht. Anders beim Download von Programmen über das Internet: Hier wird auf der Downloadseite oft ein Hashwert des Downloads (syn. Hash- oder Programm-„Signatur“) angegeben, damit man nach dem Herunterladen (und vor dem Starten des Programms) noch einmal prüfen kann, ob das Programm vollständig und unverändert geladen wurde. Hashfunktionen sind vor allem im Verborgenen unverzichtbar – sie übernehmen das Encoding/Decoding von Zeichenfolgen zu Hashtables, die der schnellen Zuweisung von Befehlen zu Codes im Kern der Programmiersprachen und beim Kompilieren dienen (Wer mehr dazu wissen will, findet hier einen Videomitschnitt vom 29. CCC-Kongress, der das Dilemma (Hash-Table Flooding) „schlechter“ Hashfunktionen verdeutlicht.
Das Berechnen des Hashwerts kann eine rechenintensive Aufgabe sein. Ein Hash-Algorithmus liest Daten schrittweise ein und verrechnet den Stream zu einem Hashwert, der je nach Güte und Art des Hashprogramms in einen „seltenen“ oder sogar „einzigartigen“ Zahlenwert besteht (oft im Hexadezimalformat ausgegeben).
Hashfunktionen allgemein werden nach der Länge des Ausgabewertes (Angabe in Bit) unterschieden, kryptologische Hashfunktionen liefern Hashs mit einer fixen Länge (z.B. SHA256->256Bits ; SHA512->512Bits) , andere Funktionen liefern Hashwerte bis zu einem Maximalwert.
Kryptologisch sichere Hashfunktionen sind unter anderem dahingehend optimiert, dass es keine – nicht einmal theoretische – Kollisionen gibt (Kollisionen sind gleiche Hashwerte für ungleiche Eingangsdaten). Außerdem sind die Algorithmen so gestaltet, dass zwei Eingaben, die sich nur sehr geringfügig unterscheiden, bereits stark unterschiedliche Hashergebnisse (Signaturen) ergeben. Dabei ist es wichtig, dass der Rechenweg so konzipiert ist, dass sie sogenannte „Einwegfunktionen“ darstellen und vom Hash nicht auf die Eingabedaten zurück gerechnet werden kann. Will man also von einer kryptologischen Signatur auf den Eingabewert zurückschließen, so ginge dies nur, wenn man eine Tabelle mit allen möglichen Eingabewerten und zugehörigen Hash-Ausgaben hat. Liegen solche Tabellen (z.B. „Rainbow-Tables“ für Strings in typischer Passwortlänge) einmal vor, so ist diese Hashfunktion natürlich nicht mehr sicher und sollte im sicherheitsrelevanten Zusammenhängen – zum Beispiel fürs Speichern von Passwörtern – nicht weiter benutzt werden: MD5, SHA-1; Weiteres siehe: https://de.wikipedia.org/wiki/Kryptologische_Hashfunktion ).
Erwähnenswert: Zu den ganz neuen kryptologischen Hashfunktionen gehören BLAKE2s (32 Bit Maschinen) und BLAKE2b (64 Bit Maschinen) https://blake2.net, die Jean-Philippe Aumasson et al. unter Verwendung von Daniel Bernsteins ChaCha Code entwickelt haben. Blake2 Hashfunktionen sind so schnell wie MD5 oder SHA-1, im Unterschied zu vorgenannten sind sie jedoch (derzeit) kryptologisch sicher.
Nicht kryptologische Hashfunktionen
Im Gegenzug sind aufgrund zusätzlicher Rechenschritte kryptologische Hashfunktionen vergleichsweise langsamer als nicht kryptologische Hashfunktionen (Faktor 30x-300x). Für die Verarbeitung nicht sicherheitsrelevanter Daten, bei denen kryptologische Eigenschaften des Hashs keine Rolle spielen, werden darum einfachere (= schnellere) Hashfunktionen eingesetzt. Eventuell ist auch Kollisionsfreiheit weniger wichtig als Geschwindigkeit oder das Risiko von Kollisionen wird verringert, indem mehrere (zwei +) Hashs mit verschiedenen schnellen, aber nicht kollisionssicheren Hashfunktionen berechnet und kombiniert werden. Trotzdem speilt Kollisionssicherheit eine grosse Rolle und die neuesten Hash-Algorithmen punkten hier deutlich.
0815 Nutzen: Hashen jenseits von Passwort/Nutzernamen
Nehmen wir an, eine Reihe Bilder, die sich eventuell nur um den Farbwert einzelner Pixel unterscheiden, sollen daraufhin geprüft werden, ob sie exakt einer Referenz (Original) entsprechen. Ein „naiver“ Ansatz wäre, jedes der Bilder Pixel für Pixel mit der Referenz zu vergleichen. Geht es aber nur darum festzustellen, ob sich die Bilder von der Referenz unterscheiden, so ist die Verwendung eines Hashes die effektivere Lösung.
``````
Erstelle einen Hash der Referenz
Für jedes Bild
erstelle einen Hasch
vergleiche den Bild-Hash mit dem Referenz-Hash
gib zurück: gleich / ungleich
``````
Habe ich einen schnellen Hash-Algorithmus so kann ich die Vergleichsdaten auch in Segmente unterteilen und dann die Segment-Hashs untersuchen und vergleichsweise resourcen-sparsam und schnell Unterschiede lokalisieren.
``````
Erstelle einen Hash der Referenz
Für n-Segmente der Referenz
erstelle je einen Referenz-Segment-Hash
Für jedes Bild
erstelle einen Hasch
vergleiche den Bild-Hash mit dem Referenz-Hash
wenn ungleich
für n-Segmente des Bildes
erstelle Bild-Segment-Hash
vergleiche Bild-Segment-Hash mit Referenz-Segment-Hash
gib zurück: ungleiche Segmente
``````
Statt ganze Bilder, Pixel für Pixel mit einer Referenz zu vergleichen, ist dann nur noch ein Vergleich der Segmente, die ungleiche Hashwerte hatten, nötig, um unterschiedliche Pixel exakt zu lokalisieren.
Welche Hashfunktion ist die richtige für meinen Zweck
Viele Hashfunktionen sind optimiert für die schnelle Verarbeitung großer Datenmengen. Das geschieht z.B. durch die Verarbeitung von Byte-Gruppen in einem Durchgang. Dass die Hashrechnung selbst etwas aufwendiger ist, fällt dabei dann weniger ins Gewicht, als bei Funktionen, die einen Stream/String Byte für Byte verrechnen. In der Gegenüberstellung zu Googles xxHash http://code.google.com/p/xxhash/ finden sich Vertreter dieses (neueren) Hashtyps auf den oberen sechs Plätzen der Rankingtabelle. Siehe auch hier: http://brainspec.com/blog/2012/10/29/fast-hashing-with-cityhash/ Zur Entwicklung von Hashfunktionen neueren Typs, die bei großen Datenmengen auch die Möglichkeiten aktueller Prozessoren zur parallelen Berechnungen ausnutzen: http://web.stanford.edu/class/ee380/Abstracts/121017-slides.pdf
Für eine Aufgabe, wie die oben geschilderten Vergleiche von Bilddateien, können diese „modernen“ hochperformanten Hashfunktionen gut geeignet sein. Wenn man allerdings nur kurze Strings hashen muss (z.B. IP-Adressen oder Log-Einträge) und ohnehin nur 32Bit oder 64Bit Ergebnisse braucht, waren die Hash-Funktionen 2. und 3. Generation bei meinen eigenen Benchmarks allesamt langsamer als SDBM, FNV oder DJB2, da die Byte-Gruppen-Verarbeitung einen recht großen Overhead (Initialisierung, Finalisieren) mitbringt.
Schnelle, nicht kryptologische Hashfunktionen
Im Folgenden eine Liste mit Links zu verschiedenen Webseiten, die sich mit Hash-Berechnungen beschäftigen:
FNV-Hash Family http://www.isthe.com/chongo/tech/comp/fnv/index.html based on prime calculations.
Bob Jenkins Sites http://burtleburtle.net/bob/
• Hash-/Block-Cipher http://burtleburtle.net/bob/hash/index.html
• lookUP3 http://burtleburtle.net/bob/c/lookup3.c
• SpookyhashV2 (2012) http://burtleburtle.net/bob/hash/spooky.html
und Grundlagen: http://burtleburtle.net/bob/hash/integer.html Integer Hashes mit verschiedenen Beispielen.
• Pearson Hash mit random Hashtable (klassisches BASIC) http://burtleburtle.net/bob/hash/pearson.html
• Arash Partow Zusammen-&Gegenüberstellung klassischer Hashfunktionen http://partow.net/programming/hashfunctions/index.html
Die Seite diskutiert verschiedenste Hashfunktionen und erlaubt den Download in verschiedenen Programmiersprachen z.B. CPP oder Python:
``````
// citation: An algorithm proposed by Donald E. Knuth in The Art Of Computer Programming Volume 3,
// under the topic of sorting and search chapter 6.4.
def DEKHash(key):
hash = len(key);
for i in range(len(key)):
hash = ((hash << 5) ^ (hash >> 27)) ^ ord(key[i])
return hash
``````
DEKHash in Go:
``````
func (bl Bloom) DEKHash(b *[]byte) (hash uint64) {
hash := uint64(len(*b))
for _, c := range *b {
hash = ((hash << 5) ^ (hash >> 27)) ^ uint64(c)
}
return hash
}
``````
http://www.cse.yorku.ca/~oz/hash.html – DJB2 und SDBM-Hash Funktionen in C
http://code.google.com/p/xxhash/ xxHash und ein Benchmarking verschiedener Hash-Funktionen: xxHash, cityHash, lookUp3, SpookyHash2, Murmur3, FNV …
Eine ausführliche Diskussion verschiedener Hashfunktionen mit vielen Links findet sich auch hier:http://blog.reverberate.org/2012/01/state-of-hash-functions-2012.html
# Fast prime-generators with python
Euler project „Project Euler“ http://projecteuler.net inspired some coders to think about prime generation in python. Even if the primes are already known, the pure python implementations are interesting to get information about fast computation models for big number arrays.
Das „Project Euler“ http://projecteuler.net hat eine Reihe Programmierer inspiriert, sich mit der Frage der Erzeugung von Primzahlen auseinanderzusetzen. Das ist in erster Linie ein akademisches Unterfangen, weil die entsprechenden Primzahlen (<10**10 u.Ä.) natürlich bereits bekannt sind. Trotzdem eine spannende Aufgabe, fand ich, und habe eine Routine programmiert, von der ich hoffte, sie könne im Vergleich mithalten, indem sie durch Deduction die Anzahl der Kalkulationen drückt. Hat nur bedingt geklappt.
Hier ist der Sieger-Code von Robert William Hanks:
``````
def rwh_primes(n):
# http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a list of primes < n """
sieve = [True] * n
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
return [2] + [i for i in xrange(3,n,2) if sieve[i]]
``````
``````
def rwh_primes1(n):
# http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a list of primes < n """
sieve = [True] * (n/2)
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i/2]: sieve[i*i/2::i] = [False] * ((n-i*i-1)/(2*i)+1)
return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]]
``````
``````
def rwh_primes2(n):
# http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Input n>=6, Returns a list of primes, 2 <= p < n """
correction = (n%6>1)
n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
sieve = [True] * (n/3)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)/3) ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1)
sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1)
return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]
``````
However – meine Lösung ist auf einem MBPro15/i7 als schnellste hinter RWHs Lösungen 1 und 2 dennoch ziemlich gut weggekommen:
``````
def primeSieveSeq(MAX_Int):
if MAX_Int > 5*10**8:
import ctypes
int16Array = ctypes.c_ushort * (MAX_Int >> 1)
sieve = int16Array()
#print 'uses ctypes "unsigned short int Array"'
else:
sieve = (MAX_Int >> 1) * [False]
#print 'uses python list() of long long int'
if MAX_Int < 10**8:
sieve[4::3] = [True]*((MAX_Int - 8)/6+1)
sieve[12::5] = [True]*((MAX_Int - 24)/10+1)
r = [2, 3, 5]
n = 0
for i in xrange(int(MAX_Int**0.5)/30+1):
n += 3
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
n += 2
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
n += 1
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
n += 2
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
n += 1
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
n += 2
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
n += 3
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
n += 1
if not sieve[n]:
n2 = (n << 1) + 1
r.append(n2)
n2q = (n2**2) >> 1
sieve[n2q::n2] = [True]*(((MAX_Int >> 1) - n2q - 1) / n2 + 1)
if MAX_Int < 10**8:
return [2, 3, 5]+[(p << 1) + 1 for p in [n for n in xrange(3, MAX_Int >> 1) if not sieve[n]]]
n = n >> 1
try:
for i in xrange((MAX_Int-2*n)/30 + 1):
n += 3
if not sieve[n]:
r.append((n << 1) + 1)
n += 2
if not sieve[n]:
r.append((n << 1) + 1)
n += 1
if not sieve[n]:
r.append((n << 1) + 1)
n += 2
if not sieve[n]:
r.append((n << 1) + 1)
n += 1
if not sieve[n]:
r.append((n << 1) + 1)
n += 2
if not sieve[n]:
r.append((n << 1) + 1)
n += 3
if not sieve[n]:
r.append((n << 1) + 1)
n += 1
if not sieve[n]:
r.append((n << 1) + 1)
except:
pass
return r
``````
Auf jeden Fall bleibt festzuhalten, dass in python der Fill mit Array Access über Indexranges array[start:stop:step] = [True]*(((MAX_Int >> 1) – n2q – 1) / n2 + 1) jeder Schleifenkontrolle in Bezug auf processing-speed deutlich überlegen ist.
# Speed-comparison of two different designs of a prime-number-generator
Eventually i needed some prime numbers to play around with Bloom filtering.
The Python – script (see below) will produce primes below 2**x base number by stepping backward using the step parameter.
python primes.py <number of file> <2**x exponent x> <step> <number of primes to produce>
if the <number of file> parameter is a character output to a file will be omitted and the primes will print to stdout instead.
<number of primes to produce> defaults to 10
I tested the performance of a single threaded and multi-threaded approach on some PC of various age. The outcome was interesting:
On one hand the script performance benefits very much from improvements of CPU architecture. Nonetheless clock speed and number of real cores seem to be of highest importance.
``````
#! /usr/bin/python
# procuces scripts < 2**x by testing in a by
import os
import sys
import subprocess
import pprint
import time
cnt = totcnt = 0
erg = []
# check 4 gmpy
try:
# raise
from gmpy import mpz
print "Message: Using gmpy/GMP mpz.is_prime instead of openSSL prime. (up to 10times faster)"
except:
print """Message:
If you try to produce many or really large primes,
consider installing GMP ( http://gmplib.org )
and gmpy/gmpy2 ( http:http://www.aleax.it/gmpy.html )"""
def __init__(self, num):
self.num = num
def run(self):
global erg, cnt
prm = check(self.num)
if prm:
erg.append(prm)
cnt += 1
def check(num):
r = None
try:
if mpz(num).is_prime() == 1:
r = num
except:
print '#', num, '#'
else:
e = subprocess.Popen(['/usr/bin/openSSL', 'prime', num], stdout=subprocess.PIPE)
if e.communicate()[0][-9:-7] == "is":
r = num
return r
def longnumget(start, step):
start = str(start+step)
step = 2*step
lenstep = 2*len(str(step))
def numSeperator(numAsString, lenstep):
formatter = '%%0%ii' % lenstep
frst, secnd = numAsString[:-lenstep], start[-lenstep:]
diff = int(secnd)-step
secnd = formatter % diff
r = frst + secnd
if diff < 0:
r = numSeperator(numAsString, lenstep+1)
return r
while True:
start = numSeperator(start, lenstep)
while start[-1] == '5':
start = numSeperator(start, lenstep)
yield start
def shortnumget(start, step):
start = start+step
step = 2*step
while True:
start -= step
while str(start)[-1] == '5':
start -= step
yield str(start)
if __name__ == '__main__':
if len(sys.argv) == 5:
p, seqNum, base2exp, step, count = map(str, sys.argv)
if count:
count = int(count) or 10
step = abs(int(step))
if step % 2 == 0:
print("Step must be odd!")
sys.exit()
if int(base2exp) < 4000:
numget = shortnumget
else:
numget = longnumget
print "custom string-cut partitional substraction"
startnum = mpz('0b1'+int(base2exp)*'0', 0)
else:
startnum = eval('2**'+base2exp)
nums = numget(startnum, step)
sTime = time.time()
while cnt <= count:
# uncomment the next 2 lines to test of the single threaded approach
# check(nums.next())
# totcnt += 1
#
# uncomment next 5 lines to test the threaded approach
t = checkr(nums.next())
t.start()
totcnt += 1
print 'Time elapsed:', time.time() - sTime, 'seconds'
print totcnt, 'numbers checked for prime'
print len(erg), 'primes found (', round(100.0*len(erg)/totcnt, 1), '%)'
try:
seqNum = int(seqNum)
with file('primes_%s_%sK_%s.txt' % (seqNum, count/1000, base2exp), 'wb') as outFle:
for i in xrange(len(erg)):
print >> outFle, '2**%s-%s' % (base2exp, startnum - int(erg[i])), erg[i], hex(int(erg[i]))
except Exception as e:
pp = pprint.PrettyPrinter()
pp.pprint(erg)
`````` | 9,377 | 30,418 | {"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} | 3.171875 | 3 | CC-MAIN-2017-51 | longest | en | 0.273978 |
https://estebantorreshighschool.com/equation-help/oscillation-frequency-equation.html | 1,695,744,254,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510214.81/warc/CC-MAIN-20230926143354-20230926173354-00256.warc.gz | 271,219,616 | 10,889 | What is the formula for oscillation?
The Equation of Motion The period of this sytem (time for one oscillation) is T=2πω=2π√Lg.
What is meant by frequency of oscillation?
The frequency of oscillation is the number of full oscillations in one time unit, say in a second. A pendulum that takes 0.5 seconds to make one full oscillation has a frequency of 1 oscillation per 0.5 second, or 2 oscillations per second.
How do you calculate frequency?
How to calculate frequencyDetermine the action. Decide what action you want to use to determine the frequency. Select the length of time. Select the length of time over which you will measure the frequency. Divide the numbers. To calculate frequency, divide the number of times the event occurs by the length of time.
How do you find the frequency of oscillation on a graph?
Determining wave frequency from a graphDetermining wave frequency from a graph.f • Frequency = #of cycles/time • Measured in Hertz (Hz)• 1 cycle = 1 full wave to repeat itself.31 2 4 5 6 7 8 9 10 11 12 Time in seconds 3 cycles.from 0 to 12 seconds 31 2 4 5 6 7 8 9 10 11 12 Time in seconds 0.f=3 /12 s = ¼ Hz 31 2 4 5 6 7 8 9 10 11 12 Time in seconds.
WHAT IS A in SHM?
Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.
What are the types of oscillation?
OscillationsSimple Harmonic Motion.Damped Simple Harmonic Motion.Forced Simple Harmonic Motion.Force Law for Simple Harmonic Motion.Velocity and Acceleration in Simple Harmonic Motion.Some Systems executing Simple Harmonic Motion.Energy in Simple Harmonic Motion.Periodic and Oscillatory Motion.
You might be interested: Chemical equation for respiration
What are the two types of frequency?
The wave period and the wave frequency are the two phenomena of the oscillations. The wave period is the difference between the wave and the wave frequency is the number of waves per units time.
What is frequency example?
A frequency is the number of times a data value occurs. For example, if ten students score 80 in statistics, then the score of 80 has a frequency of 10. Frequency is often represented by the letter f.
What are the types of frequency?
Types of Frequency DistributionGrouped frequency distribution.Ungrouped frequency distribution.Cumulative frequency distribution.Relative frequency distribution.Relative cumulative frequency distribution.
What is the fundamental frequency equation?
The fundamental frequency (n = 1) is ν = v/2l. The higher frequencies, called harmonics or overtones, are multiples of the fundamental. It is customary to refer to the fundamental as the first harmonic; n = 2 gives the second harmonic or first overtone, and so on.
How do you calculate relative frequency and frequency?
Remember, you count frequencies. To find the relative frequency, divide the frequency by the total number of data values. To find the cumulative relative frequency, add all of the previous relative frequencies to the relative frequency for the current row.
What is the frequency of a graph?
frequency: The frequency of a trigonometric function is the number of cycles it completes in a given interval. This interval is generally 2π radians (or 360º) for the sine and cosine curves. A sinusoidal curve is the graph of the sine function in trigonometry.
Releated
Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […] | 948 | 4,093 | {"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} | 3.8125 | 4 | CC-MAIN-2023-40 | latest | en | 0.87573 |
https://discuss.interviewbit.com/t/it-says-didnt-complete-in-the-allocated-time-limit-i-have-used-only-one-while-loop/58262 | 1,620,916,482,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989814.35/warc/CC-MAIN-20210513142421-20210513172421-00625.warc.gz | 238,147,782 | 3,568 | # It says 'didn't complete in the allocated time limit'. I have used only one while loop
#1
ListNode* Solution::solve(ListNode* A) {
ListNode *B;
ListNode *C;
ListNode *D;
ListNode *E;
ListNode *next;
B = NULL;
C = A;
``````while(C != NULL){
if(C->val == 1) {
next = C->next;
C->next = B;
B = C;
C = next;
}
else{
next = C->next;
E = D;
D = C;
D -> next = B;
E -> next = D;
D = E;
}
}``````
#2
Your condition is C!=null, and none of your increments MOVE c
#3
First is You are not moving your c fwd and is it necessary to change the whole node i.e you may simply swap the value:)
#4
In else, you are not moving C. So try moving C in else case. | 207 | 653 | {"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-2021-21 | latest | en | 0.843474 |
https://bytes.com/topic/c/answers/217235-define-side-effects | 1,575,911,024,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00148.warc.gz | 307,011,649 | 8,952 | 440,333 Members | 1,843 Online
Need help? Post your question and get tips & solutions from a community of 440,333 IT Pros & Developers. It's quick & easy.
# define side effects
P: n/a This is one of the posts that i got. ------------------------------ A "side effect" of an operation is something that *happens*, not something that *is produced*. Examples: In the expression 2+2, the value 4 *is produced*. Nothing *happens*. Thus, 4 is the value of the expression, and it has no side effects. In the expression g=2.0, the value 2.0 is produced. What *happens* is that 2.0 is assigned to g. Thus, 2.0 is the value of the expression, and its side effect is to assign 2.0 to g. In the expression (int)g, the value 2 is produced. Nothing happens. Thus, 2 is the value of the expression (int)g, and it has no side effects. In the expression (a=1,++a), the value 2 is produced. What happens is that first 1 is assigned to a, and then a is incremented; those are the side effects of the expression. ----------------------------- This seems to be a good definition in most of the cases. I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 -> 2 is that happens. I am really confused. Can some one give me a more clear definition of side effects ? If i apply the same definition to i=j++ where j=3 then i=3 and j=4 are the side effects and value 3,4 are produced. Am i correct ? Can someone throw more light on how to define side effects without ambiguity ? - Nik Nov 14 '05 #1
5 Replies
P: n/a ni*****@gamebox.net (Niklaus) wrote: This is one of the posts that i got. ------------------------------ A "side effect" of an operation is something that *happens*, not something that *is produced*. Examples: In the expression 2+2, the value 4 *is produced*. Nothing *happens*. Thus, 4 is the value of the expression, and it has no side effects. In the expression g=2.0, the value 2.0 is produced. What *happens* is that 2.0 is assigned to g. Thus, 2.0 is the value of the expression, and its side effect is to assign 2.0 to g. In the expression (int)g, the value 2 is produced. Nothing happens. Thus, 2 is the value of the expression (int)g, and it has no side effects. In the expression (a=1,++a), the value 2 is produced. What happens is that first 1 is assigned to a, and then a is incremented; those are the side effects of the expression. ----------------------------- This seems to be a good definition in most of the cases. I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 -> 2 is that happens. No; the 2.0 is not changed. Nothing happens "behind the scenes"; the value 2.0 does get truncated, but only for the direct reason of calculating the value of the expression. Assigning this truncated value to k _is_ a side effect. I am really confused. Can some one give me a more clear definition of side effects ? Well, according to the Standard, # [#2] Accessing a volatile object, modifying an object, # modifying a file, or calling a function that does any of # those operations are all side effects, which are changes # in the state of the execution environment. Whether that is more clear, well... it's unambiguous, anyway. If i apply the same definition to i=j++ where j=3 then i=3 and j=4 are the side effects and value 3,4 are produced. Am i correct ? No; 4 is never produced. j is increased to 4, but that value is never passed on to any other sub-expression; its previous value, 3, is. Richard Nov 14 '05 #2
P: n/a rl*@hoekstra-uitgeverij.nl (Richard Bos) wrote in message news:<40***************@news.individual.net>... ni*****@gamebox.net (Niklaus) wrote: This is one of the posts that i got. ------------------------------ A "side effect" of an operation is something that *happens*, not something that *is produced*. Examples: In the expression 2+2, the value 4 *is produced*. Nothing *happens*. Thus, 4 is the value of the expression, and it has no side effects. In the expression g=2.0, the value 2.0 is produced. What *happens* is that 2.0 is assigned to g. Thus, 2.0 is the value of the expression, and its side effect is to assign 2.0 to g. In the expression (int)g, the value 2 is produced. Nothing happens. Thus, 2 is the value of the expression (int)g, and it has no side effects. In the expression (a=1,++a), the value 2 is produced. What happens is that first 1 is assigned to a, and then a is incremented; those are the side effects of the expression. ----------------------------- This seems to be a good definition in most of the cases. I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 -> 2 is that happens. No; the 2.0 is not changed. Nothing happens "behind the scenes"; the value 2.0 does get truncated, but only for the direct reason of calculating the value of the expression. Assigning this truncated value to k _is_ a side effect. I am really confused. Can some one give me a more clear definition of side effects ? Well, according to the Standard, # [#2] Accessing a volatile object, modifying an object, # modifying a file, or calling a function that does any of # those operations are all side effects, which are changes # in the state of the execution environment. Whether that is more clear, well... it's unambiguous, anyway. If i apply the same definition to i=j++ where j=3 then i=3 and j=4 are the side effects and value 3,4 are produced. Am i correct ? No; 4 is never produced. j is increased to 4, but that value is never passed on to any other sub-expression; its previous value, 3, is. Richard regarding side - effects what exactly does this sequence 'point mean' Nov 14 '05 #3
P: n/a Gautam scribbled the following: rl*@hoekstra-uitgeverij.nl (Richard Bos) wrote in message news:<40***************@news.individual.net>... ni*****@gamebox.net (Niklaus) wrote: > This is one of the posts that i got. > ------------------------------ > A "side effect" of an operation is something that > *happens*, not something that *is produced*. Examples: > In the expression 2+2, the value 4 *is produced*. Nothing > *happens*. > Thus, 4 is the value of the expression, and it has no side effects. > In the expression g=2.0, the value 2.0 is produced. What *happens* > is that 2.0 is assigned to g. Thus, 2.0 is the value of the > expression, > and its side effect is to assign 2.0 to g. > In the expression (int)g, the value 2 is produced. Nothing happens. > Thus, 2 is the value of the expression (int)g, and it has no side > effects. > In the expression (a=1,++a), the value 2 is produced. What happens > is that first 1 is assigned to a, and then a is incremented; those are > the side effects of the expression. > ----------------------------- > > This seems to be a good definition in most of the cases. > > I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 > -> 2 is that happens. No; the 2.0 is not changed. Nothing happens "behind the scenes"; the value 2.0 does get truncated, but only for the direct reason of calculating the value of the expression. Assigning this truncated value to k _is_ a side effect. > I am really confused. Can some one give me a more clear definition > of side effects ? Well, according to the Standard, # [#2] Accessing a volatile object, modifying an object, # modifying a file, or calling a function that does any of # those operations are all side effects, which are changes # in the state of the execution environment. Whether that is more clear, well... it's unambiguous, anyway. > If i apply the same definition to i=j++ where j=3 then i=3 and j=4 > are the side effects and value 3,4 are produced. Am i correct ? No; 4 is never produced. j is increased to 4, but that value is never passed on to any other sub-expression; its previous value, 3, is. regarding side - effects what exactly does this sequence 'point mean' It's a point during the evaluation of an expression, when all side effects are guaranteed to have taken place. Sequence points include: - The terminating ; in a statement - The && and || operators - The ?: operator - The , operator Also, AFAIK when a function is called, its entry point forms a sequence point for the expressions in its arguments. -- /-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\ \-- http://www.helsinki.fi/~palaste --------------------- rules! --------/ "You could take his life and..." - Mirja Tolsa Nov 14 '05 #4
P: n/a Simply The same as in medicine. A drug cures one problem, but causes another. In code you fix one bug, but cause another. Nov 14 '05 #5
P: n/a On Sat, 8 May 2004, Neil Kurzman wrote only: Simply The same as in medicine. A drug cures one problem, but causes another. In code you fix one bug, but cause another. An interesting quotation, but why do you say so? If this is supposed to be a new topic of discussion (incidentally, one more suited to comp.programming than comp.lang.c, which is dedicated specifically to fixing bugs in *C* code :) then you might have considered starting a new thread rather than piggybacking on an existing one. We've seen people here before who apparently thought they should "conserve threads" by posting irrelevant replies to old threads; that's really not necessary. And if you *didn't* mean your reply to be irrelevant to the old thread, then you ought to have quoted some context so people could tell to what you were responding. Google "usenet faq" for more information. -Arthur Nov 14 '05 #6
### This discussion thread is closed
Replies have been disabled for this discussion. | 2,464 | 9,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-51 | latest | en | 0.908471 |
https://bulleintime.com/flintstone/associative-property-of-addition-example-problems.php | 1,726,627,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00425.warc.gz | 130,716,590 | 8,245 | Associative Property Lesson Plan lavijm0 Learning Through Examples 1. Prove that (4 + 5) + 9 = 4 + (5 + 9) by using associative property of addition In this problem the associative property of addition
## Properties of Addition Practice Quiz
Associative Property of Addition You Certainly Need to. The associative property in Addition ♥ Addition how this property works with a more visual example: of Addition and Multiplication in a Problem;, Addition properties worksheets include practice problems based on commutative, associative, for kids to set the own example for each property..
Examples Associative Property. I model how to show the work for each of the problems so that, for example, (about the commutative property/ about how addition ... the commutative property, the associative that are associative are addition and multiplication. Addition is associative because, for example, the problem
We provide FREE Solved Math problems with easy to comprehend examples and video tutorials to help understand complex mathematical concepts. In addition to Learning Through Examples 1. Prove that (4 + 5) + 9 = 4 + (5 + 9) by using associative property of addition In this problem the associative property of addition
Select a problem set using the buttons above, Identify the property being illustrated by each statement as either associative or commutative. Select a problem set using the buttons above, Identify the property being illustrated by each statement as either associative or commutative.
Glossary of Properties . you cannot distribute division over addition. Let's try an example: 14 / (5 + 2 The addition property of equality says that if a Discover (and save!) your own Pins on Pinterest. Property Of Addition Example Associative Property, of addition help students solve addition problems.
This is known as the commutative property of multiplication and is quite similar to the associative property of addition. For example, to Solve Problems in Property Worksheets for Practice. the correct equation in this associative property of addition student with three addends in each addition problem.
A look at the Associative, Distributive and Commutative Properties --examples, with practice problems Property Example with Addition; This prealgebra lesson defines and explains the associative property of multiplication . Advertisement. Text block . Pre The Associative Property of Addition.
A look at the Associative, Distributive and Commutative Properties --examples, with practice problems Property Example with Addition; What word comes to mind when you hear “associative?” The Associative Property of Addition. Example 1: (14 + 6) Word Problems;
Find and save ideas about Associative property example on Pinterest. Practice math problems like Relate Subtraction to Addition (Think Addition Strategy) The associative, commutative, and distributive properties of algebra are Example. Problem . 3.5 in two different ways using the associative property of addition.
Addition properties worksheets include practice problems based on commutative, associative, for kids to set the own example for each property. ... Associative Property: In this lesson students will be exploring the associative property of addition. Write this addition problem on the poster and
If you have an addition problem such as The associative property states that in addition and The Associative Property: Definition and Examples Related Solve multiplication problems using macaroni! Associative Property Of Addition Example Associative Property, Properties Of Addition,
### Associative Property of Addition Fun & Educational Stuff
Associative Property of Addition Fun & Educational Stuff. Solve multiplication problems using macaroni! Associative Property Of Addition Example Associative Property, Properties Of Addition,, Discover (and save!) your own Pins on Pinterest. Property Of Addition Example Associative Property, of addition help students solve addition problems..
### Properties of Addition and Subtraction TutorVista
Associative Property of Addition Fun & Educational Stuff. associative property of addition math distributive property definition with examples practice problems . associative property of multiplication Learning Through Examples 1. Prove that (4 + 5) + 9 = 4 + (5 + 9) by using associative property of addition In this problem the associative property of addition.
• Associative Property of Addition Definition & Worksheets
• Associative Property of Addition Definition & Worksheets
• Properties of Addition and Subtraction TutorVista
• Just keep in mind that you can use the associative property with addition and property works in the following examples: problem 5 – (4 – 0) = (5 Solve multiplication problems using macaroni! Associative Property Of Addition Example Associative Property, Properties Of Addition,
The associative property in Addition ♥ Addition how this property works with a more visual example: of Addition and Multiplication in a Problem; Solve multiplication problems using macaroni! Associative Property Of Addition Example Associative Property, Properties Of Addition,
If you have an addition problem such as The associative property states that in addition and The Associative Property: Definition and Examples Related Let's take a look at the associative property for addition. the commutative and associative properties can easily be confused. Examples: (5 x 4) x 25 = 500
Glossary of Properties . you cannot distribute division over addition. Let's try an example: 14 / (5 + 2 The addition property of equality says that if a Learning Through Examples 1. Prove that (4 + 5) + 9 = 4 + (5 + 9) by using associative property of addition In this problem the associative property of addition
Discover (and save!) your own Pins on Pinterest. Property Of Addition Example Associative Property, of addition help students solve addition problems. Glossary of Properties . you cannot distribute division over addition. Let's try an example: 14 / (5 + 2 The addition property of equality says that if a
Select a problem set using the buttons above, Identify the property being illustrated by each statement as either associative or commutative. Commutative Property Of Addition Example Commutative Property Of Addition, Properties Of Addition, Addition Example Associative Property, ADDITION PROBLEMS
The properties of addition are listed below, such as commutative property, associative property, Example Problem for Addition and Subtraction Properties: Commutative Property Of Addition Example Commutative Property Of Addition, Properties Of Addition, Addition Example Associative Property, ADDITION PROBLEMS
Associative Property. Both addition and multiplication to share their own examples of the associative property. Commutative and Associative Problems.pdf. Learn properties of integers with the help of examples and explanations. In Associative Property of Addition, Properties of Integers; Word problems;
The associative property of addition simply says that the way in which you What is the Associative Property of For example, in a problem like 3 x 4 The associative property of addition simply says that the way in which you What is the Associative Property of For example, in a problem like 3 x 4
Find and save ideas about Associative property example on Pinterest. Practice math problems like Relate Subtraction to Addition (Think Addition Strategy) What word comes to mind when you hear “associative?” The Associative Property of Addition. Example 1: (14 + 6) Word Problems;
Any time they refer in a problem to using the Distributive Property, the Distributive Property refers to both addition and the Associative Property is the The associative, commutative, and distributive properties of algebra are Example. Problem . 3.5 in two different ways using the associative property of addition.
## Basic Algebra/Introduction to Basic Algebra Ideas/Solving
Properties of Addition and Subtraction TutorVista. Glossary of Properties . you cannot distribute division over addition. Let's try an example: 14 / (5 + 2 The addition property of equality says that if a, What word comes to mind when you hear “associative?” The Associative Property of Addition. Example 1: (14 + 6) Word Problems;.
### Associative Property of Addition You Certainly Need to
Associative Property of Addition Fun & Educational Stuff. The best source for free properties of addition and properties of Example (Hover to the which digit is missing from an associative property problem., Select a problem set using the buttons above, Identify the property being illustrated by each statement as either associative or commutative..
Commutative Property Of Addition Example Commutative Property Of Addition, Properties Of Addition, Addition Example Associative Property, ADDITION PROBLEMS Any time they refer in a problem to using the Distributive Property, the Distributive Property refers to both addition and the Associative Property is the
associative property of addition math distributive property definition with examples practice problems . associative property of multiplication Select a problem set using the buttons above, Identify the property being illustrated by each statement as either associative or commutative.
Discover (and save!) your own Pins on Pinterest. Property Of Addition Example Associative Property, of addition help students solve addition problems. This is known as the commutative property of multiplication and is quite similar to the associative property of addition. For example, to Solve Problems in
Glossary of Properties . you cannot distribute division over addition. Let's try an example: 14 / (5 + 2 The addition property of equality says that if a This prealgebra lesson defines and explains the associative property of multiplication . Advertisement. Text block . Pre The Associative Property of Addition.
Basic Algebra/Introduction to Basic Algebra Ideas/Solving Equations Using Properties of Mathematics. Associative Property of Addition Example Problems The associative property of addition simply says that the way in which you What is the Associative Property of For example, in a problem like 3 x 4
Let's take a look at the associative property for addition. the commutative and associative properties can easily be confused. Examples: (5 x 4) x 25 = 500 Solve multiplication problems using macaroni! Associative Property Of Addition Example Associative Property, Properties Of Addition,
... Associative Property: In this lesson students will be exploring the associative property of addition. Write this addition problem on the poster and The associative property in Addition ♥ Addition how this property works with a more visual example: of Addition and Multiplication in a Problem;
Both ways of approaching the problem gives the same The above example illustrates the associative property of addition Examples to illustrate which property. Let's take a look at the associative property for addition. the commutative and associative properties can easily be confused. Examples: (5 x 4) x 25 = 500
Solve multiplication problems using macaroni! Associative Property Of Addition Example Associative Property, Properties Of Addition, Associative Property. Both addition and multiplication to share their own examples of the associative property. Commutative and Associative Problems.pdf.
Discover (and save!) your own Pins on Pinterest. Property Of Addition Example Associative Property, of addition help students solve addition problems. Glossary of Properties . you cannot distribute division over addition. Let's try an example: 14 / (5 + 2 The addition property of equality says that if a
### Associative Property of Addition Math Pinterest
Associative Property Lesson Plan lavijm0. Let's take a look at the associative property for addition. the commutative and associative properties can easily be confused. Examples: (5 x 4) x 25 = 500, Discover (and save!) your own Pins on Pinterest. Property Of Addition Example Associative Property, of addition help students solve addition problems..
### Properties of Addition and Subtraction TutorVista
Associative Property Lesson Plan lavijm0. The associative, commutative, and distributive properties of algebra are Example. Problem . 3.5 in two different ways using the associative property of addition. Find and save ideas about Associative property example on Pinterest. Practice math problems like Relate Subtraction to Addition (Think Addition Strategy).
• Associative Property of Addition Fun & Educational Stuff
• Associative Property of Addition You Certainly Need to
• Glossary of Properties . you cannot distribute division over addition. Let's try an example: 14 / (5 + 2 The addition property of equality says that if a A look at the Associative, Distributive and Commutative Properties --examples, with practice problems Property Example with Addition;
Practice changing the grouping of factors in multiplication problems and see how it affects the product. Associative law of addition. Associative property of Property Worksheets for Practice. the correct equation in this associative property of addition student with three addends in each addition problem.
Both ways of approaching the problem gives the same The above example illustrates the associative property of addition Examples to illustrate which property. Let's take a look at the associative property for addition. the commutative and associative properties can easily be confused. Examples: (5 x 4) x 25 = 500
The best source for free properties of addition and properties of Example (Hover to the which digit is missing from an associative property problem. The associative property of addition simply says that the way in which you What is the Associative Property of For example, in a problem like 3 x 4
The following equation demonstrates the use of the Associative Property. Example Problems The Associative Property of Addition is one The Seven Property Worksheets for Practice. the correct equation in this associative property of addition student with three addends in each addition problem.
associative property of addition math distributive property definition with examples practice problems . associative property of multiplication ... the commutative property, the associative that are associative are addition and multiplication. Addition is associative because, for example, the problem
Select a problem set using the buttons above, Identify the property being illustrated by each statement as either associative or commutative. Just keep in mind that you can use the associative property with addition and property works in the following examples: problem 5 – (4 – 0) = (5
The best source for free properties of addition and properties of Example (Hover to the which digit is missing from an associative property problem. Any time they refer in a problem to using the Distributive Property, the Distributive Property refers to both addition and the Associative Property is the
Addition properties worksheets include practice problems based on commutative, associative, for kids to set the own example for each property. Solve multiplication problems using macaroni! Associative Property Of Addition Example Associative Property, Properties Of Addition,
This prealgebra lesson defines and explains the associative property of multiplication . Advertisement. Text block . Pre The Associative Property of Addition. ... Associative Property: In this lesson students will be exploring the associative property of addition. Write this addition problem on the poster and
Examples Associative Property. I model how to show the work for each of the problems so that, for example, (about the commutative property/ about how addition Any time they refer in a problem to using the Distributive Property, the Distributive Property refers to both addition and the Associative Property is the | 3,096 | 15,967 | {"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} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.909951 |
https://www.airmilescalculator.com/distance/lft-to-sfo/ | 1,611,835,239,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704843561.95/warc/CC-MAIN-20210128102756-20210128132756-00215.warc.gz | 651,170,433 | 59,149 | # Distance between Lafayette, LA (LFT) and San Francisco, CA (SFO)
Flight distance from Lafayette to San Francisco (Lafayette Regional Airport – San Francisco International Airport) is 1810 miles / 2914 kilometers / 1573 nautical miles. Estimated flight time is 3 hours 55 minutes.
Driving distance from Lafayette (LFT) to San Francisco (SFO) is 2113 miles / 3401 kilometers and travel time by car is about 35 hours 29 minutes.
## Map of flight path and driving directions from Lafayette to San Francisco.
Shortest flight path between Lafayette Regional Airport (LFT) and San Francisco International Airport (SFO).
## How far is San Francisco from Lafayette?
There are several ways to calculate distances between Lafayette and San Francisco. Here are two common methods:
Vincenty's formula (applied above)
• 1810.410 miles
• 2913.572 kilometers
• 1573.203 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1807.173 miles
• 2908.362 kilometers
• 1570.390 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Lafayette Regional Airport
City: Lafayette, LA
Country: United States
IATA Code: LFT
ICAO Code: KLFT
Coordinates: 30°12′19″N, 91°59′15″W
B San Francisco International Airport
City: San Francisco, CA
Country: United States
IATA Code: SFO
ICAO Code: KSFO
Coordinates: 37°37′8″N, 122°22′30″W
## Time difference and current local times
The time difference between Lafayette and San Francisco is 2 hours. San Francisco is 2 hours behind Lafayette.
CST
PST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 201 kg (443 pounds).
## Frequent Flyer Miles Calculator
Lafayette (LFT) → San Francisco (SFO).
Distance:
1810
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1810
Round trip? | 508 | 2,052 | {"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-2021-04 | latest | en | 0.796983 |
https://socratic.org/questions/how-do-you-rationalize-the-denominator-sqrt6-sqrt13-3 | 1,701,637,734,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00853.warc.gz | 594,441,368 | 5,692 | # How do you rationalize the denominator sqrt6/(sqrt13+3)?
##### 1 Answer
Jun 2, 2015
Multiply both numerator (top) and denominator (bottom) by the conjugate $\left(\sqrt{13} - 3\right)$ ...
$\frac{\sqrt{6}}{\sqrt{13} + 3}$
$= \frac{\sqrt{6}}{\sqrt{13} + 3} \cdot \frac{\sqrt{13} - 3}{\sqrt{13} - 3}$
$= \frac{\sqrt{6} \cdot \left(\sqrt{13} - 3\right)}{\left(\sqrt{13} + 3\right) \left(\sqrt{13} - 3\right)}$
$= \frac{\sqrt{6} \cdot \left(\sqrt{13} - 3\right)}{13 - 9}$
$= \frac{\sqrt{6} \cdot \left(\sqrt{13} - 3\right)}{4}$ | 226 | 532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 4.21875 | 4 | CC-MAIN-2023-50 | latest | en | 0.189948 |
http://mathhelpforum.com/discrete-math/144726-solving-recurrence-systems-print.html | 1,529,674,505,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864482.90/warc/CC-MAIN-20180622123642-20180622143642-00358.warc.gz | 197,799,701 | 4,506 | # Solving recurrence systems
• May 14th 2010, 09:59 AM
RedKMan
Solving recurrence systems
Solve the recurrence system below and dtermine a formula for the time complexity function T(n).
T(0) = 4.
T(n) = 5 + T(n - 1) (n > 0)
My effort at solving is below...
T(3) = 5 + T(2).
T(2) = 5 + T(1).
T(1) = 5 + T(0).
T(3) + T(2) + T(1) = 3 * 5 + T(2) + T(1) + T(0)
T(3) = 3 * 5 + T(0)
T(3) = 15 + 4.
Now try to obtain a formula for T(n) in terms of n.
T(n) = 5 + T(n - 1).
T(n - 1) = 5 + T(n - 2).
T(1) = 5 + T(0).
T(0) = 4.
T(n) + T(n - 1) + T(1) + T(0) = 3 * 5 + T(n - 1) + T(n - 2) + T(0) + 4.
**THIS IS WHERE MY BRAIN HITS MELTING POINT AND THE COOLANT IS RELEASED.
I'm pretty sure the formula should be T(n) = 5 * n + 4. So,
T(0) = 5 * 0 + 4 = 4
T(3) = 5 * 3 + 4 = 19.
I need to prove it though :(
• May 14th 2010, 10:14 AM
Plato
$\displaystyle \begin{array}{rcl} {T_4 } & = & {5 + T_3 } \\ {} & = & {5 + \left[ {5 + T_2 } \right]} \\ {} & = & {5 + \left[ {5 + \left[ {5 + T_1 } \right]} \right]} \\ {} & = & {\left( 3 \right)5 + 4} \\ \end{array}$
Now is not clear what the pattern is?
• May 14th 2010, 10:40 AM
Soroban
Hello, RedKMan!
You're making it much too hard . . .
Quote:
Solve the recurrence system and determine a formula for the function $\displaystyle T(n).$.
$\displaystyle T(0) = 4,\quad T(n) \:=\: T(n - 1) + 5, \quad n > 0$
Crank out the first few terms . . .
. . $\displaystyle \begin{array}{c|c} n & T(n) \\ \hline 0 & 4 \\ 1 & 9 \\ 2 & 14 \\ 3 & 19 \\ 4 & 24 \\ \vdots & \vdots \end{array}$
The sequence starts at 4 and "goes up by 5."
It is clear that the function is: .$\displaystyle T(n) \;=\;4 + 5n$
• May 14th 2010, 01:16 PM
RedKMan
How does this look below:-
Now try to obtain a formula for T(n) in terms of n.
T(n) = 5 + T(n - 1).
T(n - 1) = 5 + T(n - 2).
T(2) = 5 + T(1)
T(1) = 5 + T(0).
T(0) = 4.
Sum each side and equate result...
T(n) + T(n - 1) + T(2) + T(1) + T(0) = T(n - 1) + T(n - 2) + T(1) + T(0) + 4 + (n - 1) * 5 + 4.
Subtract terms that appear on both sides...
T(n) = (n - 1) * 5 + 4 = 5n - 5 + 4 = 5n + 4 = 4 + 5n.
I'm a bit dubious about my algebra above, I feel like I'm close, I can taste it. Is it close / correct?
• May 14th 2010, 05:35 PM
Renji Rodrigo
You can do this
$\displaystyle t(k+1)=t(k)+5$
so
$\displaystyle t(k+1)-t(k) =5$
apply the summation $\displaystyle \sum^{n-1}_{k=0}$
on the both sides, the first is a telescoping summation
$\displaystyle \sum^{n-1}_{k=0} t(k+1)-t(k) = t(n)-t(0)=t(n)-4 =\sum^{n-1}_{k=0} 5=5n$
so
$\displaystyle t(n)=5n +4$
• May 14th 2010, 08:28 PM
undefined
Quote:
Originally Posted by RedKMan
I'm pretty sure the formula should be T(n) = 5 * n + 4. So,
T(0) = 5 * 0 + 4 = 4
T(3) = 5 * 3 + 4 = 19.
I need to prove it though :(
Well I would hope that after reading these posts and thinking about them, you will no longer be "pretty sure" but rather 100% sure that the solution is T(n) = 5n + 4.
As for proof, if you're comfortable with inductive proofs, this one is very easy. Using mathematical induction is the typical way to prove these kinds of things.
• May 15th 2010, 01:05 AM
RedKMan
Problem is I need to use algebra to arrive at the formula rather than guessing and then using induction to prove it. Thats why I'm hoping someone can check my,
T(n) = 5 + T(n - 1).
T(n - 1) = 5 + T(n - 2).
T(2) = 5 + T(1)
T(1) = 5 + T(0).
T(0) = 4.
Sum each side and equate result...
T(n) + T(n - 1) + T(2) + T(1) + T(0) = T(n - 1) + T(n - 2) + T(1) + T(0) + 4 + (n - 1) * 5 + 4.
Subtract terms that appear on both sides...
T(n) = (n - 1) * 5 + 4 = 5n - 5 + 4 = 5n + 4 = 4 + 5n.
Thus T(n) = 4 + 5n
• May 15th 2010, 01:21 AM
undefined
Quote:
Originally Posted by RedKMan
Problem is I need to use algebra to arrive at the formula rather than guessing and then using induction to prove it.
There is no guessing involved. You can see that the expression is T(n) = 5n + 4 if you just look hard enough.
It's like this sequence:
a(0) = 1
a(n) = 2*a(n-1), n>0
I can just see that this is a(n) = 2^n and there is absolutely no guessing involved.
Although to be fair, guessing is a valid approach for many of these types of problems.
Quote:
Originally Posted by RedKMan
Thats why I'm hoping someone can check my,
T(n) = 5 + T(n - 1).
T(n - 1) = 5 + T(n - 2).
T(2) = 5 + T(1)
T(1) = 5 + T(0).
T(0) = 4.
Sum each side and equate result...
T(n) + T(n - 1) + T(2) + T(1) + T(0) = T(n - 1) + T(n - 2) + T(1) + T(0) + 4 + (n - 1) * 5 + 4.
...
Where did this last equation come from??? When I sum each side I get
T(n) + T(n - 1) + T(2) + T(1) + T(0) = 5 + T(n - 1) + 5 + T(n - 2) + 5 + T(1) + 5 + T(0) + 4
• May 15th 2010, 01:42 AM
RedKMan
Your correct, when you sum each side you do get,
T(n) + T(n - 1) + T(2) + T(1) + T(0) = 5 + T(n - 1) + 5 + T(n - 2) + 5 + T(1) + 5 + T(0) + 4
Removing all the terms that appear on both sides, I get,
T(n) + T(2) = 5 + 5 + T(n - 2) + 5 + 5 + 4
Problem is I think I need to get rid of that T(2) term on the right some how??
• May 15th 2010, 01:48 AM
undefined
Quote:
Originally Posted by RedKMan
Your correct, when you sum each side you do get,
T(n) + T(n - 1) + T(2) + T(1) + T(0) = 5 + T(n - 1) + 5 + T(n - 2) + 5 + T(1) + 5 + T(0) + 4
Removing all the terms that appear on both sides, I get,
T(n) + T(2) = 5 + 5 + T(n - 2) + 5 + 5 + 4
Problem is I think I need to get rid of that T(2) term on the right some how??
I could be wrong, but I don't think your approach will succeed.
If you really dislike induction, Renji Rodrigo's post looks like a great option for you.
• May 15th 2010, 06:05 AM
RedKMan
I've gone with induction and it all works out.
With the time complexity function being T(n) = 5n + 4.
I've worked out the Big Oh value to be:-
f(n) = 5n. Giving O(n) of linear type.
Does that look right?
Thanks for all the help everyone. | 2,379 | 5,803 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-26 | latest | en | 0.662458 |
https://sp.okwave.jp/qa/q5890941.html | 1,656,311,781,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00256.warc.gz | 599,641,415 | 26,214 | • ベストアンサー
• すぐに回答を！
# 卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の
noname#111372
• 英語
• 回答数1
• 閲覧数66
• ありがとう数3
## 質問者が選んだベストアンサー
• ベストアンサー
• 回答No.1
### 関連するQ&A
• 卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の
卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の文章です。 Consider a building that has two floors, with an escalator to carry people from one floor to the other. Ignoring queuing delays, the response time for a passenger is the mean time taken by the escalator to ascend or descend one floor. The throughput (bandwidth) is the mean number of passengers that can be loaded or per second. Suppose that an average of five people step on the escalator in one second, and that the escalator takes an average of 10 seconds to go up one floor. The response time for a passenger, therefore, is 10 seconds, and the throughput of the escalator is 5 passengers/second. Thus, the degree of parallelism, which is the mean number of passengers carried simultaneously, is 5*10=50. To see this, mark a passenger with a daub of red paint as she steps on the escalator. In the ten seconds that she takes to reach the top, we expect that fifty more passengers boarded the escalator. Thus, when she steps off, the escalator carries an average of fifty passengers, which is its degree of parallelism.
• 翻訳をお願いしたいです。コンピューター関係の書物の文章です。
翻訳をお願いしたいです。コンピューター関係の書物の文章です。 Consider a building that has two floors, with an escalator to carry people from one floor to the other. Ignoring queuing delays, the response time for a passenger is the mean time taken by the escalator to ascend or descend one floor. The throughput (bandwidth) is the mean number of passengers that can be loaded or per second. Suppose that an average of five people step on the escalator in one second, and that the escalator takes an average of 10 seconds to go up one floor. The response time for a passenger, therefore, is 10 seconds, and the throughput of the escalator is 5 passengers/second. Thus, the degree of parallelism, which is the mean number of passengers carried simultaneously, is 5*10=50. To see this, mark a passenger with a daub of red paint as she steps on the escalator. In the ten seconds that she takes to reach the top, we expect that fifty more passengers boarded the escalator. Thus, when she steps off, the escalator carries an average of fifty passengers, which is its degree of parallelism.
• 卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の
卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の文章です。 If, in a system, on an average, 20 tasks complete in 10 seconds, and each task takes 3 seconds, what is the degree of parallelism? Solution : The throughput is 20/10 task/second, and the response time is 3 seconds. Thus, the degree of parallelism = 20/10*3=6. In other words, an average of six tasks execute in parallel in the system.
• 翻訳をお願いしたいです。コンピューター関係の書物の文章です。
翻訳をお願いしたいです。コンピューター関係の書物の文章です。 A system designer must typically optimize one or more performance metrics given a set of resource constrains. A performance metric measures some aspect of a system's performance, such as throughput, response time, cost development time, or mean time between failures(we will define these metrics more formally in Section 6.2). A resource constraint is a limitation on a resource, such as time, bandwidth, or computing power, that the design must obey.
• 卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の
卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の文章です。 Time can constrain a design in many ways. For example, a user may require a task to complete before a given time, or may want to limit the time taken for a packet to travel from a source to a destination. At a different level, there may be a time constraint on how long it can take to design and build a system (time-to-market). Or, we may want to maximize the mean time between failures. We now study some standard ways to measure the use of time in a system.
• 卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の
卒論に使用するため、翻訳をお願いしたいです。コンピューター関係の書物の文章です。 We can describe most computer system resources as a combination of five common resources: time, space, computation, money, and labor. We now study these resources in more detail, with examples of how they arise in real-world problems, and a description of associated performance metrics. Our definitions of these resources are purposely vague, since the exact definition varies with the problem.
• 翻訳をお願いしたいです。コンピューター関係の書物の文章です。
翻訳をお願いしたいです。コンピューター関係の書物の文章です。 If we could quantify and control every aspect of a system, then system design would be a relatively simple matter. Unfortunately there are several practical reasons why system design is both an art and a science. First, although we can quantitatively measure some aspects of system performance, such as throughput or response time, we cannot measure others, such as simplicity, scalability, modularity, and elegance. Yet a designer must make a series of trade- offs among these intangible quantities, appealing as much to good sense and personal choice as performance measurements. Second, rapid technological change can make constraint assumptions obsolete. A designer must not only meet the current set of design constraints, but also anticipate how future changes in technology might affect the design. The future is hard to predict, and a designer must appeal to instinct and intuition to make a design "future-proof." Third, market conditions may dictate that design requirements change when part of the design is already complete. Finally, international standards, which themselves change over time, may impose irksome and arbitrary constraints. These factors imply that, in real life, a designer is usually confronted with a complex, underspecified, multifactor optimization problem. In the face of these uncertainties, prescribing the one true path to system design is impossible.
• 翻訳をお願いしたいです。コンピューター関係の書物の文章です。
翻訳をお願いしたいです。コンピューター関係の書物の文章です。 System design is important not only in computer systems, but also in other areas, such as automobile design. For example, a car designer might try to maximize the reliability of a car(measured in the mean time between equipment failures) that costs less than \$10,000 to build. In this example, the mean time between failures measures performance, and the resource constraint is money. In real life, of course, designs must try to simultaneously optimize many, possibly conflicting metrics (such as reliability, performance, and recyclability) while satisfying many constraints (such as the price of the car and the time allowed for the design).
• 翻訳をお願いしたいです。コンピューター関係の書物の文章です。
翻訳をお願いしたいです。コンピューター関係の書物の文章です。 Nevertheless, it is still, possible to identify some principles of good design that have withstood the test of time and are applicable in a variety of situations. In Section6.2, we will study some common resources, so that the reader can get some intuition in identifying them in real systems. We will then build up, in Section6.3, a set of tool to help us trade freely available (unconstrained) resources for scarce (constrained) ones. Properly applied, these tools allow us to match the design to the constraints at hand. Finally, in Section6.4, we will outline a methodology for performance analysis and tuning. This methodology helps pinpoint problems in a design and build a more efficient and robust system.
• 翻訳をお願いしたいです。コンピューター関係の書物の文章です。翻訳サイト
翻訳をお願いしたいです。コンピューター関係の書物の文章です。翻訳サイトのコピペはご遠慮ください。 In any system, some resources are less constrained than others. We call the most constrained resource in a system(or the binding constraint) its bottleneck. System performance improves if and only if we devote additional resources to a bottlenecked resource. Conversely, decreasing the amount of an unconstrained resource does not adversely affect performance. When we relieve one bottleneck, however, it is possible for another resource to become a bottleneck. Thus, we must remove the bottlenecks one by one until all the resources are equally constrained. We call such a system a balanced system. A balanced system is optimal, in that we fully utilize every component. However, in practice, we rarely achieve balanced systems. Rapid changes in technology, market constraints, and customer expectations mean that a system's components are almost constantly in flux, with the bottleneck moving from place to place in the system. | 2,092 | 8,265 | {"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} | 2.90625 | 3 | CC-MAIN-2022-27 | latest | en | 0.712793 |
https://www.mrexcel.com/board/threads/sumifs-tab-references-and-indirect.1172035/ | 1,627,364,228,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00487.warc.gz | 837,959,301 | 17,461 | # SUMIFS, tab references and INDIRECT?
#### hochstas
##### New Member
Hello,
I use a number of versions of the formula below on a spreadsheet on a sharepoint site that several people are working in each day. Rows are continuously deleted and data is continuously being pasted in.
=SUMIFS('Open Calls Report'!\$P\$2:\$P\$9978,'Open Calls Report'!\$N\$2:\$N\$9978,B27,'Open Calls Report'!\$H\$2:\$H\$9978,\$C\$24)
It works perfectly for my needs except for 2 problems:
1. As rows are deleted and added, I continually lose range. I've taken to setting the ranges from 2 to 10000 so I don't have to fix the formulas as often. There's normally between 4-700 rows in the spreadsheet. I just did this this morning and already down to 9978. I heard that INDIRECT may solve that problem, but I've tried to modify my formula and I don't think I'm understanding the syntax correctly. This is the primary problem and I'd be thrilled if I could just keep the cells fixed between 2 and 1000.
2. Secondary item that could save some time if fixable. The report that we run to paste additional rows into 'open calls' tab always has the values in column P as "numbers formatted as text". So each morning I have to open the spreadsheet in desktop version, highlight all of column P, click the little exclamation point box and choose "convert to number". Is there anything that can be added to the formula to add the number regardless of whether they're formatted as numbers or text?
Thanks in advance for the help. I'm an excel novice and it took me forever to research enough to figure out how to create the original formula.
### Excel Facts
Select a hidden cell
Somehide hide payroll data in column G? Press F5. Type G1. Enter. Look in formula bar while you arrow down through G.
#### mrshl9898
##### Well-known Member
You don't really need to fix the range on a SUMIFS in my experience,
=SUMIFS('Open Calls Report'!\$P:\$P,'Open Calls Report'!\$N:\$N,B27,'Open Calls Report'!\$H:\$H,\$C\$24)
Not sure about the second issue.
#### hochstas
##### New Member
You don't really need to fix the range on a SUMIFS in my experience,
=SUMIFS('Open Calls Report'!\$P:\$P,'Open Calls Report'!\$N:\$N,B27,'Open Calls Report'!\$H:\$H,\$C\$24)
Not sure about the second issue.
Thanks! That solved my first issue!
Replies
3
Views
117
Replies
1
Views
473
Replies
1
Views
263
Replies
5
Views
378
Replies
4
Views
159
1,136,445
Messages
5,675,899
Members
419,591
Latest member
mersanko
### We've detected that you are using an adblocker.
We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com.
### Which adblocker are you using?
1)Click on the icon in the browser’s toolbar.
2)Click on the icon in the browser’s toolbar.
2)Click on the "Pause on this site" option.
Go back
1)Click on the icon in the browser’s toolbar.
2)Click on the toggle to disable it for "mrexcel.com".
Go back
### Disable uBlock Origin
Follow these easy steps to disable uBlock Origin
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
### Disable uBlock
Follow these easy steps to disable uBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back | 885 | 3,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.934964 |
http://www.lmfdb.org/NumberField/9.1.2366667072.1 | 1,603,592,293,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885126.36/warc/CC-MAIN-20201025012538-20201025042538-00118.warc.gz | 152,294,298 | 9,528 | # Properties
Label 9.1.2366667072.1 Degree $9$ Signature $[1, 4]$ Discriminant $2366667072$ Root discriminant $11.00$ Ramified primes $2, 3, 7, 11$ Class number $1$ Class group trivial Galois group $S_3\wr S_3$ (as 9T31)
# Related objects
Show commands for: SageMath / Pari/GP / Magma
## Normalizeddefining polynomial
sage: x = polygen(QQ); K.<a> = NumberField(x^9 - x^8 - 3*x^6 + x^5 + 3*x^4 + 3*x^3 - 4*x^2 + 1)
gp: K = bnfinit(x^9 - x^8 - 3*x^6 + x^5 + 3*x^4 + 3*x^3 - 4*x^2 + 1, 1)
magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![1, 0, -4, 3, 3, 1, -3, 0, -1, 1]);
$$x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$$
sage: K.defining_polynomial()
gp: K.pol
magma: DefiningPolynomial(K);
## Invariants
Degree: $9$ sage: K.degree() gp: poldegree(K.pol) magma: Degree(K); Signature: $[1, 4]$ sage: K.signature() gp: K.sign magma: Signature(K); Discriminant: $$2366667072$$$$\medspace = 2^{6}\cdot 3^{4}\cdot 7^{3}\cdot 11^{3}$$ sage: K.disc() gp: K.disc magma: Discriminant(Integers(K)); Root discriminant: $11.00$ sage: (K.disc().abs())^(1./K.degree()) gp: abs(K.disc)^(1/poldegree(K.pol)) magma: Abs(Discriminant(Integers(K)))^(1/Degree(K)); Ramified primes: $2, 3, 7, 11$ sage: K.disc().support() gp: factor(abs(K.disc))[,1]~ magma: PrimeDivisors(Discriminant(Integers(K))); $|\Aut(K/\Q)|$: $1$ This field is not Galois over $\Q$. This is not a CM field.
## Integral basis (with respect to field generator $$a$$)
$1$, $a$, $a^{2}$, $a^{3}$, $a^{4}$, $a^{5}$, $a^{6}$, $\frac{1}{3} a^{7} - \frac{1}{3} a^{5} - \frac{1}{3} a^{4} + \frac{1}{3} a^{3} - \frac{1}{3} a^{2} + \frac{1}{3} a + \frac{1}{3}$, $\frac{1}{3} a^{8} - \frac{1}{3} a^{6} - \frac{1}{3} a^{5} + \frac{1}{3} a^{4} - \frac{1}{3} a^{3} + \frac{1}{3} a^{2} + \frac{1}{3} a$
sage: K.integral_basis()
gp: K.zk
magma: IntegralBasis(K);
## Class group and class number
Trivial group, which has order $1$
sage: K.class_group().invariants()
gp: K.clgp
magma: ClassGroup(K);
## Unit group
sage: UK = K.unit_group()
magma: UK, f := UnitGroup(K);
Rank: $4$ sage: UK.rank() gp: K.fu magma: UnitRank(K); Torsion generator: $$-1$$ (order $2$) sage: UK.torsion_generator() gp: K.tu[2] magma: K!f(TU.1) where TU,f is TorsionUnitGroup(K); Fundamental units: $$a$$, $$a^{8} - \frac{2}{3} a^{7} - \frac{10}{3} a^{5} - \frac{1}{3} a^{4} + \frac{7}{3} a^{3} + \frac{14}{3} a^{2} - \frac{5}{3} a - \frac{2}{3}$$, $$a^{8} - 3 a^{5} - 2 a^{4} + a^{3} + 4 a^{2}$$, $$\frac{4}{3} a^{8} - \frac{1}{3} a^{7} - \frac{1}{3} a^{6} - 4 a^{5} - \frac{7}{3} a^{4} + \frac{10}{3} a^{3} + \frac{17}{3} a^{2} - \frac{1}{3}$$ sage: UK.fundamental_units() gp: K.fu magma: [K!f(g): g in Generators(UK)]; Regulator: $$16.5992891922$$ sage: K.regulator() gp: K.reg magma: Regulator(K);
## Class number formula
$\displaystyle\lim_{s\to 1} (s-1)\zeta_K(s) \approx\frac{2^{1}\cdot(2\pi)^{4}\cdot 16.5992891922 \cdot 1}{2\sqrt{2366667072}}\approx 0.531790262147$
## Galois group
$S_3\wr S_3$ (as 9T31):
sage: K.galois_group(type='pari')
gp: polgalois(K.pol)
magma: GaloisGroup(K);
A solvable group of order 1296 The 22 conjugacy class representatives for $S_3\wr S_3$ Character table for $S_3\wr S_3$ is not computed
## Intermediate fields
Fields in the database are given up to isomorphism. Isomorphic intermediate fields are shown with their multiplicities.
## Sibling fields
Degree 12 sibling: data not computed Degree 18 siblings: data not computed Degree 24 siblings: data not computed Degree 27 siblings: data not computed Degree 36 siblings: data not computed
## Frobenius cycle types
$p$ $2$ $3$ $5$ $7$ $11$ $13$ $17$ $19$ $23$ $29$ $31$ $37$ $41$ $43$ $47$ $53$ $59$ Cycle type R R ${\href{/LocalNumberField/5.6.0.1}{6} }{,}\,{\href{/LocalNumberField/5.3.0.1}{3} }$ R R ${\href{/LocalNumberField/13.9.0.1}{9} }$ ${\href{/LocalNumberField/17.6.0.1}{6} }{,}\,{\href{/LocalNumberField/17.2.0.1}{2} }{,}\,{\href{/LocalNumberField/17.1.0.1}{1} }$ ${\href{/LocalNumberField/19.9.0.1}{9} }$ ${\href{/LocalNumberField/23.6.0.1}{6} }{,}\,{\href{/LocalNumberField/23.2.0.1}{2} }{,}\,{\href{/LocalNumberField/23.1.0.1}{1} }$ ${\href{/LocalNumberField/29.6.0.1}{6} }{,}\,{\href{/LocalNumberField/29.3.0.1}{3} }$ ${\href{/LocalNumberField/31.4.0.1}{4} }{,}\,{\href{/LocalNumberField/31.2.0.1}{2} }^{2}{,}\,{\href{/LocalNumberField/31.1.0.1}{1} }$ ${\href{/LocalNumberField/37.9.0.1}{9} }$ ${\href{/LocalNumberField/41.4.0.1}{4} }{,}\,{\href{/LocalNumberField/41.3.0.1}{3} }{,}\,{\href{/LocalNumberField/41.2.0.1}{2} }$ ${\href{/LocalNumberField/43.4.0.1}{4} }{,}\,{\href{/LocalNumberField/43.2.0.1}{2} }^{2}{,}\,{\href{/LocalNumberField/43.1.0.1}{1} }$ ${\href{/LocalNumberField/47.6.0.1}{6} }{,}\,{\href{/LocalNumberField/47.3.0.1}{3} }$ ${\href{/LocalNumberField/53.4.0.1}{4} }{,}\,{\href{/LocalNumberField/53.3.0.1}{3} }{,}\,{\href{/LocalNumberField/53.2.0.1}{2} }$ ${\href{/LocalNumberField/59.6.0.1}{6} }{,}\,{\href{/LocalNumberField/59.3.0.1}{3} }$
In the table, R denotes a ramified prime. Cycle lengths which are repeated in a cycle type are indicated by exponents.
sage: p = 7; # to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
sage: [(e, pr.norm().valuation(p)) for pr,e in K.factor(p)]
gp: p = 7; \\ to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
gp: idealfactors = idealprimedec(K, p); \\ get the data
gp: vector(length(idealfactors), j, [idealfactors[j][3], idealfactors[j][4]])
magma: p := 7; // to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
magma: idealfactors := Factorization(p*Integers(K)); // get the data
magma: [<primefactor[2], Valuation(Norm(primefactor[1]), p)> : primefactor in idealfactors];
## Local algebras for ramified primes
$p$LabelPolynomial $e$ $f$ $c$ Galois group Slope content
$2$2.3.0.1$x^{3} - x + 1$$1$$3$$0$$C_3$$[\ ]^{3} 2.6.6.2x^{6} - x^{4} - 5$$2$$3$$6$$A_4\times C_2$$[2, 2]^{6}$
$3$3.2.1.1$x^{2} - 3$$2$$1$$1$$C_2$$[\ ]_{2} 3.3.0.1x^{3} - x + 1$$1$$3$$0$$C_3$$[\ ]^{3}$
3.4.3.2$x^{4} - 3$$4$$1$$3$$D_{4}$$[\ ]_{4}^{2} 77.3.0.1x^{3} - x + 2$$1$$3$$0$$C_3$$[\ ]^{3}$
7.6.3.1$x^{6} - 14 x^{4} + 49 x^{2} - 1372$$2$$3$$3$$C_6$$[\ ]_{2}^{3} 11$$\Q_{11}$$x + 3$$1$$1$$0$Trivial$[\ ]$
11.2.0.1$x^{2} - x + 7$$1$$2$$0$$C_2$$[\ ]^{2} 11.6.3.1x^{6} - 22 x^{4} + 121 x^{2} - 11979$$2$$3$$3$$C_6$$[\ ]_{2}^{3}$
## Artin representations
Label Dimension Conductor Defining polynomial of Artin field $G$ Ind $\chi(c)$
* 1.1.1t1.a.a$1$ $1$ $x$ $C_1$ $1$ $1$
1.231.2t1.a.a$1$ $3 \cdot 7 \cdot 11$ $x^{2} - x + 58$ $C_2$ (as 2T1) $1$ $-1$
1.3.2t1.a.a$1$ $3$ $x^{2} - x + 1$ $C_2$ (as 2T1) $1$ $-1$
1.77.2t1.a.a$1$ $7 \cdot 11$ $x^{2} - x - 19$ $C_2$ (as 2T1) $1$ $1$
2.231.6t3.f.a$2$ $3 \cdot 7 \cdot 11$ $x^{6} - 9 x^{3} + 1$ $D_{6}$ (as 6T3) $1$ $0$
* 2.231.3t2.a.a$2$ $3 \cdot 7 \cdot 11$ $x^{3} - x^{2} + 3$ $S_3$ (as 3T2) $1$ $0$
3.133056.4t5.a.a$3$ $2^{6} \cdot 3^{3} \cdot 7 \cdot 11$ $x^{4} - 2 x^{3} + 6 x^{2} + 4 x - 2$ $S_4$ (as 4T5) $1$ $1$
3.10245312.6t11.a.a$3$ $2^{6} \cdot 3^{3} \cdot 7^{2} \cdot 11^{2}$ $x^{6} + 2 x^{4} + 15 x^{2} + 3$ $S_4\times C_2$ (as 6T11) $1$ $1$
3.3415104.6t8.d.a$3$ $2^{6} \cdot 3^{2} \cdot 7^{2} \cdot 11^{2}$ $x^{4} - 2 x^{3} + 6 x^{2} + 4 x - 2$ $S_4$ (as 4T5) $1$ $-1$
3.44352.6t11.a.a$3$ $2^{6} \cdot 3^{2} \cdot 7 \cdot 11$ $x^{6} + 2 x^{4} + 15 x^{2} + 3$ $S_4\times C_2$ (as 6T11) $1$ $-1$
* 6.10245312.9t31.a.a$6$ $2^{6} \cdot 3^{3} \cdot 7^{2} \cdot 11^{2}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
6.60744454848.18t300.a.a$6$ $2^{6} \cdot 3^{3} \cdot 7^{4} \cdot 11^{4}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
6.546700093632.18t319.a.a$6$ $2^{6} \cdot 3^{5} \cdot 7^{4} \cdot 11^{4}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
6.92207808.18t311.a.a$6$ $2^{6} \cdot 3^{5} \cdot 7^{2} \cdot 11^{2}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
8.622...576.24t2893.b.a$8$ $2^{12} \cdot 3^{6} \cdot 7^{6} \cdot 11^{6}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
8.17703899136.12t213.a.a$8$ $2^{12} \cdot 3^{6} \cdot 7^{2} \cdot 11^{2}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
12.828...256.36t2217.a.a$12$ $2^{18} \cdot 3^{9} \cdot 7^{7} \cdot 11^{7}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $2$
12.139...264.36t2214.a.a$12$ $2^{18} \cdot 3^{9} \cdot 7^{5} \cdot 11^{5}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $-2$
12.206...976.36t2210.a.a$12$ $2^{24} \cdot 3^{10} \cdot 7^{6} \cdot 11^{6}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
12.828...256.36t2216.a.a$12$ $2^{18} \cdot 3^{9} \cdot 7^{7} \cdot 11^{7}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $-2$
12.139...264.18t315.a.a$12$ $2^{18} \cdot 3^{9} \cdot 7^{5} \cdot 11^{5}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $2$
16.110...336.24t2912.a.a$16$ $2^{24} \cdot 3^{12} \cdot 7^{8} \cdot 11^{8}$ $x^{9} - x^{8} - 3 x^{6} + x^{5} + 3 x^{4} + 3 x^{3} - 4 x^{2} + 1$ $S_3\wr S_3$ (as 9T31) $1$ $0$
Data is given for all irreducible representations of the Galois group for the Galois closure of this field. Those marked with * are summands in the permutation representation coming from this field. Representations which appear with multiplicity greater than one are indicated by exponents on the *. | 4,868 | 9,794 | {"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": 2, "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-2020-45 | latest | en | 0.23686 |
https://nrich.maths.org/public/leg.php?code=20&cl=4&cldcmpid=2422 | 1,568,606,416,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572484.20/warc/CC-MAIN-20190916035549-20190916061549-00054.warc.gz | 592,477,197 | 6,173 | # Search by Topic
#### Resources tagged with Calculating with fractions similar to And So on - and on -and On:
Filter by: Content type:
Age range:
Challenge level:
### There are 18 results
Broad Topics > Fractions, Decimals, Percentages, Ratio and Proportion > Calculating with fractions
### And So on - and on -and On
##### Age 16 to 18 Challenge Level:
Can you find the value of this function involving algebraic fractions for x=2000?
### There's a Limit
##### Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Archimedes and Numerical Roots
##### Age 14 to 16 Challenge Level:
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
### More Twisting and Turning
##### Age 11 to 16 Challenge Level:
It would be nice to have a strategy for disentangling any tangled ropes...
### Not Continued Fractions
##### Age 14 to 18 Challenge Level:
Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?
### All Tangled Up
##### Age 14 to 18 Challenge Level:
Can you tangle yourself up and reach any fraction?
### Comparing Continued Fractions
##### Age 16 to 18 Challenge Level:
Which of these continued fractions is bigger and why?
### Fair Shares?
##### Age 14 to 16 Challenge Level:
A mother wants to share a sum of money by giving each of her children in turn a lump sum plus a fraction of the remainder. How can she do this in order to share the money out equally?
### Lower Bound
##### Age 14 to 16 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Fracmax
##### Age 14 to 16 Challenge Level:
Find the maximum value of 1/p + 1/q + 1/r where this sum is less than 1 and p, q, and r are positive integers.
### Harmonic Triangle
##### Age 14 to 16 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### The Harmonic Triangle and Pascal's Triangle
##### Age 16 to 18
The harmonic triangle is built from fractions with unit numerators using a rule very similar to Pascal's triangle.
##### Age 11 to 16 Challenge Level:
The items in the shopping basket add and multiply to give the same amount. What could their prices be?
### Countdown Fractions
##### Age 11 to 16 Challenge Level:
Here is a chance to play a fractions version of the classic Countdown Game.
### Investigating the Dilution Series
##### Age 14 to 16 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Reductant Ratios
##### Age 16 to 18 Challenge Level:
What does the empirical formula of this mixture of iron oxides tell you about its consituents?
### Ratios and Dilutions
##### Age 14 to 16 Challenge Level:
Scientists often require solutions which are diluted to a particular concentration. In this problem, you can explore the mathematics of simple dilutions
### A Swiss Sum
##### Age 16 to 18 Challenge Level:
Can you use the given image to say something about the sum of an infinite series? | 785 | 3,243 | {"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} | 4.15625 | 4 | CC-MAIN-2019-39 | latest | en | 0.856972 |
https://discuss.leetcode.com/topic/15061/this-algorithms-fails-for-a-test-case-but-seems-correct-please-help | 1,508,279,429,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00656.warc.gz | 672,609,289 | 14,714 | # This algorithms fails for a test case but seems correct..please help
• ``````public class Solution {
public int minSubArrayLen(int s, int[] nums) {
Arrays.sort(nums);
int sum=0;
int count = 0;
int counter = 0;
int maxsum =0;
if(nums.length==0)
return 0;
for(int i=0;i<nums.length;i++)
{maxsum=maxsum+nums[i];
if(nums[i]==s)
{counter = i;sum =sum+ nums[counter];}
}
System.out.println(maxsum);
if(maxsum<s)
return 0;
if(counter ==0)
{counter = nums.length -1;sum =sum+ nums[counter];}
for(int i = counter-1;i>=0;i--)
{
if(sum>=s)
return ++count;
else if(sum<s)
{sum = sum+nums[i];
count++;
}
}
return count;
}
``````
}
• It failed for the following test case:
{5334,6299,4199,9663,8945,3566,9509,3124,6026,6250,7475,5420,9201,9501,38,5897,4411,6638,9845,161,9563,8854,3731,5564,5331,4294,3275,1972,1521,2377,3701,6462,6778,187,9778,758,550,7510,6225,8691,3666,4622,9722,8011,7247,575,5431,4777,4032,8682,5888,8047,3562,9462,6501,7855,505,4675,6973,493,1374,3227,1244,7364,2298,3244,8627,5102,6375,8653,1820,3857,7195,7830,4461,7821,5037,2918,4279,2791,1500,9858,6915,5156,970,1471,5296,1688,578,7266,4182,1430,4985,5730,7941,3880,607,8776,1348,2974,1094,6733,5177,4975,5421,8190,8255,9112,8651,2797,335,8677,3754,893,1818,8479,5875,1695,8295,7993,7037,8546,7906,4102,7279,1407,2462,4425,2148,2925,3903,5447,5893,3534,3663,8307,8679,8474,1202,3474,2961,1149,7451,4279,7875,5692,6186,8109,7763,7798,2250,2969,7974,9781,7741,4914,5446,1861,8914,2544,5683,8952,6745,4870,1848,7887,6448,7873,128,3281,794,1965,7036,8094,1211,9450,6981,4244,2418,8610,8681,2402,2904,7712,3252,5029,3004,5526,6965,8866,2764,600,631,9075,2631,3411,2737,2328,652,494,6556,9391,4517,8934,8892,4561,9331,1386,4636,9627,5435,9272,110,413,9706,5470,5008,1706,7045,9648,7505,6968,7509,3120,7869,6776,6434,7994,5441,288,492,1617,3274,7019,5575,6664,6056,7069,1996,9581,3103,9266,2554,7471,4251,4320,4749,649,2617,3018,4332,415,2243,1924,69,5902,3602,2925,6542,345,4657,9034,8977,6799,8397,1187,3678,4921,6518,851,6941,6920,259,4503,2637,7438,3893,5042,8552,6661,5043,9555,9095,4123,142,1446,8047,6234,1199,8848,5656,1910,3430,2843,8043,9156,7838,2332,9634,2410,2958,3431,4270,1420,4227,7712,6648,1607,1575,3741,1493,7770,3018,5398,6215,8601,6244,7551,2587,2254,3607,1147,5184,9173,8680,8610,1597,1763,7914,3441,7006,1318,7044,7267,8206,9684,4814,9748,4497,2239}
s=697439
expected:132
output:80
• I have the same problem of this testing case
• I have figured it out. I misunderstood the requirement. It is the subarray not the number of the elements.
• public class Solution {
public int minSubArrayLen(int s, int[] nums)
{
int n = nums.length;
if(n == 0)
return 0;
int len = n + 1;
int start = 0;
int sum = 0;
for(int i = 0;i <= n - 1;i++)
{
sum += nums[i];
while(sum >= s)
{
len = (len < i - start + 1) ? len:i - start + 1;
sum -= nums[start];
start++;
}
}
return (len == n + 1) ? 0:len;
}
}
• I have the same error as yours. I don't understand. What is the reason?
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 1,412 | 3,043 | {"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} | 2.65625 | 3 | CC-MAIN-2017-43 | longest | en | 0.600918 |
https://stats.stackexchange.com/questions/411570/prove-that-the-pitman-estimator-is-itself-complete-sufficient | 1,566,104,087,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313617.6/warc/CC-MAIN-20190818042813-20190818064813-00354.warc.gz | 644,876,193 | 29,710 | # Prove that the Pitman estimator is itself complete sufficient
It's a homework question, but I just have no idea about it ...
Let $X_{1} ,... , X_{n}$ be random variables according to a distribution having joint density $f(x_{1}-\theta ,...,x_{n}-\theta )$,where $\theta \in R$ is a location parameter. Assume that there exists a complete sufficient statistics $S(X_{1} ,... , X_{n} )$ for $\theta$. Prove that the Pitman estimator
$\delta^{*}(X_{1},...,X_{n})=\frac{\int _{R}\theta f(X_{1}-\theta ,...,X_{n}-\theta) d\theta }{\int _{R} f(X_{1}-\theta ,...,X_{n}-\theta) d\theta }$
is itself complete sufficient.
• Add the self-study tag. – Michael Chernick Jun 5 at 3:20
• Please use MathJax for typesetting math here. – StubbornAtom Jun 6 at 13:10 | 237 | 755 | {"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": 6, "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-2019-35 | longest | en | 0.734895 |
https://www.eng-tips.com/viewthread.cfm?qid=401365 | 1,653,691,054,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00212.warc.gz | 833,312,263 | 13,521 | ×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS
Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.
#### Posting Guidelines
Promoting, selling, recruiting, coursework and thesis posting is forbidden.
# use A / Control in the frequency domain combined with A / vibration
## use A / Control in the frequency domain combined with A / vibration
(OP)
I have a simple 1 DOF mass spring damper system. The mass is forced by a swept sine force (5-2000 Hz). I used A/Vibration and I found the charateristic behaviour of this kind of system with an acceleration peack in the resonance frequency. Now I want to add a closed loop control aimed to modulate the input force in order to maintain the mass acceleration to a required value (e.g. 1g). So I need to use A/Control in the frequency domain using as input of the Adams plant a vibration input channel and as output a vibration output channel. Someone can help me, is there any tutorial or guide useful to solve this problem?
Thanks, Vincenzo.
### RE: use A / Control in the frequency domain combined with A / vibration
Are you sure you want the controller to work in the frequency domain? That sounds like a typical feedback loop.
Anyway 4 examples are given in the controls folder, I'd have thought running them would be helpful. I use neither product so that's all i know.
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
### RE: use A / Control in the frequency domain combined with A / vibration
(OP)
Yes I'm sure i need to work in the frequency domain, I need to sweep all the frequencies from 5 to 2000 Hz.
I have already seen all the examples but all are in the time domain.
I hope that someone can help me.
Cheers,
Vincenzo
#### Red Flag This Post
Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.
#### Red Flag Submitted
Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.
#### Resources
Low-Volume Rapid Injection Molding With 3D Printed Molds
Learn methods and guidelines for using stereolithography (SLA) 3D printed molds in the injection molding process to lower costs and lead time. Discover how this hybrid manufacturing process enables on-demand mold fabrication to quickly produce small batches of thermoplastic parts. Download Now
Examine how the principles of DfAM upend many of the long-standing rules around manufacturability - allowing engineers and designers to place a part’s function at the center of their design considerations. Download Now
Taking Control of Engineering Documents
This ebook covers tips for creating and managing workflows, security best practices and protection of intellectual property, Cloud vs. on-premise software solutions, CAD file management, compliance, and more. Download Now
Close Box
# Join Eng-Tips® Today!
Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.
Here's Why Members Love Eng-Tips Forums:
• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...
Register now while it's still free! | 793 | 3,624 | {"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-2022-21 | longest | en | 0.894767 |
https://estudyassistant.com/mathematics/question17133756 | 1,627,987,203,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154457.66/warc/CC-MAIN-20210803092648-20210803122648-00014.warc.gz | 246,094,034 | 17,737 | , 01.08.2020 14:01 tryintopassenioryear
# Another plan to secure the roller coaster involves placing two concrete struts on either side of the center of the leg of the roller coaster to add reinforcement against southerly winds in the region. Again, using the center of the half-circle as the origin, the struts are modeled by the equations and. A vertical reinforcement beam will extend from one strut to the other when the two cables are 2 feet apart. Recall that a reinforcement beam will extend from one strut to the other when the two struts are 2 feet apart. Algebraically determine the x -value of where the beam should be placed. (15 points) Explain where to place the beam. (10 points)
Answers: 2
### Another question on Mathematics
Mathematics, 21.06.2019 16:20
7.(03.01 lc)which set represents the range of the function shown? {(-1, 5), (2,8), (5, 3), 13, -4)} (5 points){-1, 2, 5, 13){(5, -1), (8, 2), (3,5), (-4, 13)){-4, 3, 5, 8}{-4, -1, 2, 3, 5, 5, 8, 13}
Answers: 3
Mathematics, 21.06.2019 17:00
Adifferent website gives the cost for the first train as £56.88 and the second train as £20.11 with a £9.50 charge for cycle storage for the whole trip. how much would the journey cost you?
Answers: 1
Mathematics, 21.06.2019 19:00
What is the correlation coefficient between the variables? round to three decimal places. enter your answer in the box
Answers: 2
Mathematics, 21.06.2019 19:00
The temperature in a laboratory must be between 70 and 74. fahrenheit, inclusive. if tis the temperature in the laboratory, which of the following inequalities represents this situation?
Answers: 3
You know the right answer?
Another plan to secure the roller coaster involves placing two concrete struts on either side of the...
Questions on the website: 13584521 | 508 | 1,771 | {"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} | 3.453125 | 3 | CC-MAIN-2021-31 | latest | en | 0.876131 |
https://betterlesson.com/lesson/resource/2842102/adding-10-booklet-pdf | 1,498,592,998,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00503.warc.gz | 752,589,928 | 20,681 | ## Adding 10 Booklet.pdf - Section 3: Independent Practice
Adding 10 Booklet.pdf
Adding 10 Booklet.pdf
# Add 10 Books
Unit 15: Base 10 Bonanza
Lesson 2 of 5
## Big Idea: The CCSS emphasizes that students should be able to mentally add and subtract 10, without having to count. Help students develop a conceptual understanding of repeatedly adding 10s in this lesson-this promotes mental math!
Print Lesson
7 teachers like this lesson
Standards:
Subject(s):
55 minutes
### Amanda Cole
##### Similar Lessons
###### 10 More and 10 Less
1st Grade Math » Numbers and Place Value
Big Idea: This lesson is More or Less a great lesson! Engage students in finding 10 more and 10 less than a number using these fun activities.
Favorites(21)
Resources(12)
Lakeland, FL
Environment: Urban
###### Assessment of Unit Concepts: Addition and Subtraction Story Problems
1st Grade Math » The Number 10 and the Addition and Subtraction Concept
Big Idea: Students will complete a few tasks that will allow you to assess students' abilities to solve addition and subtraction problems and identify the strategy that each student is using.
Favorites(1)
Resources(33)
Waitsfield, VT
Environment: Suburban
###### 10 More, 10 Less
1st Grade Math » Place Value
Big Idea: My class has been learning about place value and the meaning of tens and ones. This lesson will develop knowledge of adding or subtracting ten from a 2-digit number.
Favorites(6)
Resources(19)
Oklahoma City, OK
Environment: Urban
sign up or
Something went wrong. See details for more info
Nothing to upload details close | 370 | 1,575 | {"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} | 2.90625 | 3 | CC-MAIN-2017-26 | longest | en | 0.85873 |
https://jp.mathworks.com/matlabcentral/profile/authors/14783252 | 1,716,736,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00862.warc.gz | 285,008,921 | 23,099 | # nune pratyusha
Last seen: 約1年 前 2020 年からアクティブ
Followers: 0 Following: 0
バッジを表示
#### Feeds
Empty sym: 0-by-1
clc clear all close all syms x1 x2 x3 x4 t tspan=[0 100]; [solx1 solx2 solx3 solx4] = solve([-x1+x2==0,-x1+x2.*x3+(0.0001+...
1年以上 前 | 1 件の回答 | 0
### 1
I am getting Error using deval
function yd = hyperchaos(t,y) %%%% selecting some cases for publication global aa bb dd aa =3.55; dd = 0; mm=1; ...
1年以上 前 | 1 件の回答 | 1
### 1
I am getting error like invalid expression
function Le=lyapunovk(n,rhs_ext_fcn,tstart,stept,tend,ystart,R); n1=n; n2=n1*(n1+1); % Number of steps nit = round((tend-tst...
1年以上 前 | 1 件の回答 | 0
### 1
Lyapunov exponent for fractional order differential equation
you have to download fde12.m file and put all programs in one folder then run. It will work
2年弱 前 | 0
How to plot two parameter bifurcation diagram
How to plot two parameter (2D) bifurcation plot in matlab. Please give me the suggestion
2年弱 前 | 1 件の回答 | 0
### 1
Without using drawnow command i want plot from below code
function run_LE_FO_p1(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,p_min,p_max,n); figure(); hold on; ne=3; ext_fcn=@LE_RF_p1...
2年弱 前 | 1 件の回答 | 0
### 1
I am not getting correct plot
for plotting purpose i am using draw now() command because without draw now() plot is not coming. But plot is not looks good an...
2年弱 前 | 1 件の回答 | 1
### 1
Not enough input arguments.
i am getting error like: >> run_Lp xstart = 0 0 1 Not enough input arguments. Error in l_ext (line 11) f(2)=-...
2年弱 前 | 1 件の回答 | 0
### 1
Index exceeds the number of array elements. Index must not exceed 1.
function run_LE_FO_a(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,a_min,a_max,n); figure(); hold on; ne=4; ext_fcn=@LE_RF_a; ...
### 1
i am not getting plot
function run_LE_FO_p1(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,p_min,p_max,n); hold on; ne=3; ext_fcn=@LE_RF_p1; t_start=...
### 1
Operator '*' is not supported for operands of type 'function_handle'.
clc; clear all; format compact; format long e; set(0,'defaultaxesfontsize',16.5,'defaultaxeslinewidth',0.8,... 'defaul...
### 1
how to correlation plots from image
but i am getting different hystogram plots using below code A = imread('Figure11(b).png'); Ref = imread('Figure(b).png')...
how to correlation plots from image
I = imread('Figure_1.png'); J = medfilt2(I); R = corr2(I,J) it showing error like Error using medfilt2 Expected input...
### 2
i am not getting chaotic plot because of warning
function dx = compactchua(t,x) dx=zeros(4,1); dx(1) = -(1/(8200*47*1e-9))*x(2); dx(2) = (((x(3)-x(2))/2000) -(x(4).^3*4.7e3+1...
2年以上 前 | 1 件の回答 | 0
### 1
i am not getting chaotic plot because of warning
function dx = compactchua(t,x) dx=zeros(4,1); dx(1) = -(1/(8200*47*1e-9))*x(2); dx(2) = (((x(3)-x(2))/2000) -(x(4).^3*4.7e3...
2年以上 前 | 1 件の回答 | 0
### 1
i am getting not enough input arguments error for the below code
function dx = m(t,x) dx=zeros(4,1); dx(1) = -(1/(8200*47*1e-9))*x(2); dx(2) = (((x(3)-x(2))/2000) -(0.667e-3 + 0.029e-3*x(1...
2年以上 前 | 1 件の回答 | 0
### 1
in nonlinear dynamics how to find lyapunov exponent?
how to find lyapunov exponets for three dimensional differential equations
2年以上 前 | 1 件の回答 | 0
### 1
how to plot below bar graphs
i don't know it is in research paper
2年以上 前 | 0
how to plot below bar graphs
how to plot below mentioned bar graphs
2年以上 前 | 1 件の回答 | 0
### 1
how to convert picoscope data to matlab
when i am saving picoscope data with .mat extension its opening in microsoft access table. How to import picoscope data to matla...
3年弱 前 | 1 件の回答 | 0
### 1
how to plot graph with cycles
3年弱 前 | 1 件の回答 | 0
### 1
how to generate below sawtooth pulse
and what is pulse number and how it generates in matlab
### 2
plot from discrete points
Freq point F=10 [0.1,0.00001] F=20 [0. 4,0.0003] F=35 [0.106, 0.0001] F=40 [-0.0009,-0.012] F=70 [0...
### 1
error using integral function
clc clear all close all a=7.2*10^-6; Gm=2.5*10^-2; b=4.7; sp=2.75*10^-5; yon=6*10^-2; son=4.5*10^-1; B=10^-4; A=10^-10...
### 2
error using ode23
function dydt = parasi12(t1,y1) a=7.2*10^-6; Gm=2.5*10^-2; Sp=2.75*10^-5; yon=6*10^-2; b=4.7; son=4.5*10^-1; B=10^-4; vm...
### 1
bifurcation analysis of differential dynamical system
my system is like k(t)=g(x,l)l(t) dx/dt=f(x,l) for this type of system how to plot bifurcation diagram
### 1
how to plot bifurcation diagram in matlab
a=7.2*10^-6; Gm=2.5*10^-2 Sp=2.75*10^-5; yon=6*10^-2; b=4.7; son=4.5*10^-1; B=10^-4; vm=0:0.5:2.5; gp=sinh(vm); A=10^-1...
### 1
sawtooth wave modulation i want mean voltage will be change with respect to frequency
t=0:1e-5:0.1; f=10 v=sawtooth(2*pi*f*t); v=0.3*(v+1); k=mean(v) my output is coming like k=0.299 i want to change mean vol...
### 1
Error when solving differential equations
function dydt = vdpi1(t,y) syms n a=3.5*10^-6; Gm=0.02 Sp=2.75*10^-5; yon=0.01; b=3.1; son=0.10; B=90; % vm=0:0.5:2.5; ...
### 1
how to solve non linear differential equations
dx(t)/dt=y(t) dy(t)/dt=-(1/r+g+a+b|x(t)|)y(t)/c-x(t)/(lc) t=-0.05:0.01:0.05 r = 1430; a = -0.0683; b = 0.0676; ... | 1,955 | 5,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.417475 |
http://www.codingforums.com/showthread.php?t=282113&goto=nextoldest | 1,386,933,181,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164931675/warc/CC-MAIN-20131204134851-00056-ip-10-33-133-15.ec2.internal.warc.gz | 293,077,414 | 19,007 | Flash Website Builder- Trendy Site Builder is a Flash Site Building tool that helps users build stunning websites. Check Out Custom Custom Logo Design by LogoBee. Website Design and Free Logo Templates available.
CodingForums.com JavaScript Function: Generate a random integer within specified range AND digit limit
User Name Remember Me? Password
Before you post, read our: Rules & Posting Guidelines
Enjoy an ad free experience by logging in. Not a member yet? Register.
11-14-2012, 07:05 AM PM User | #1 genstonewall New to the CF scene Join Date: Nov 2012 Posts: 1 Thanks: 0 Thanked 0 Times in 0 Posts JavaScript Function: Generate a random integer within specified range AND digit limit I need a function that generates a completely random integer (very important) within a user specified number range (between -9999 to 9999) and a user specified digit limit (between 1 and 4 digits). Example 1: If the user wants a number between -9999 and 9999 that's 4 digits, the following numbers would be eligible choices -9999 to -1000 and 1000 to 9999. Example 2: If the user wants a number between 25 and 200 that's 2 OR 3 digits, the following numbers would be eligible choices 25 to 200. I wrote a function that works but I am not sure if it's the best solution? There's duplicate code and I don't think it's completely random? Thanks. Code: ```// Generates a random integer // Number range // Min (-9999-9999) // Max (-9999-9999) // Digit limit // Min (1-4) // Max (1-4) function generateRandomInteger(minNumber, maxNumber, minDigits, maxDigits) { // Generate a random integer in the number range var num = Math.floor(Math.random() * (maxNumber - minNumber)) + minNumber; // Find number of digits var n = num.toString(); n = n.length; // If number is negative subtract 1 from length because of "-" sign if (num < 0) { n--; } // End: find number of digits while ((n > maxDigits) || (n < minDigits)) { // Generate a random integer in the number range num = Math.floor(Math.random() * (maxNumber - minNumber)) + minNumber; // Find number of digits var n = num.toString(); n = n.length; // If number is negative subtract 1 from length because of "-" sign if (num < 0) { n--; } // End: find number of digits } return num; }```
11-14-2012, 08:54 AM PM User | #2 Philip M Supreme Master coder! Join Date: Jun 2002 Location: London, England Posts: 17,490 Thanks: 200 Thanked 2,471 Times in 2,449 Posts As you say, the code can be simplified, but otherwise seems to work OK. The number generated is random (actually pseudo-random, but that is the best you can do!) Code: `````` You need to check that the user data is consistent, e.g. not 10,99,3,4 or an infinite loop is set up. "Never attribute to malice that which can be adequately explained by stupidity." - Napoleon Bonaparte __________________ All the code given in this post has been tested and is intended to address the question asked. Unless stated otherwise it is not just a demonstration. Last edited by Philip M; 11-14-2012 at 09:31 AM..
11-14-2012, 09:25 AM PM User | #3 Labrar New Coder Join Date: Jun 2008 Posts: 61 Thanks: 0 Thanked 12 Times in 12 Posts What about Code: ```function randomizer( min, max, digits ) { if(isNaN(digits)){digits=4;} if( min > max ) { return( -1 ); } if( min == max ) { return( min ); } var rands=( min + parseInt( Math.random() * ( max-min+1 ) ) ); rands=rands.toString(); var returner=''; for(var i =0; i
11-14-2012, 09:32 AM PM User | #4 Philip M Supreme Master coder! Join Date: Jun 2002 Location: London, England Posts: 17,490 Thanks: 200 Thanked 2,471 Times in 2,449 Posts Labrar - please post Javascript code within code tags. Not php tags! That is likely to confuse beginners. __________________ All the code given in this post has been tested and is intended to address the question asked. Unless stated otherwise it is not just a demonstration.
11-14-2012, 10:04 AM PM User | #5 Labrar New Coder Join Date: Jun 2008 Posts: 61 Thanks: 0 Thanked 12 Times in 12 Posts Sorry. How? is it [JS], [JavaScript] ??? Edit: got it Is there any button to do this? Last edited by Labrar; 11-14-2012 at 10:07 AM..
11-14-2012, 10:12 AM PM User | #6 minder Banned Join Date: Oct 2012 Posts: 81 Thanks: 0 Thanked 4 Times in 4 Posts Edit: wrong solution Last edited by minder; 11-14-2012 at 10:34 AM..
11-14-2012, 10:21 AM PM User | #7 devnull69 Senior Coder Join Date: Dec 2010 Posts: 2,355 Thanks: 11 Thanked 558 Times in 551 Posts Not if you read and understand the OP's post He wants to exclude those random numbers that are not of a certain length Example: Random number with 3 to 4 digits between -10000 and 10000 would exclude those numbers that are only two digits (like -12 or 98)
11-14-2012, 10:34 AM PM User | #8 minder Banned Join Date: Oct 2012 Posts: 81 Thanks: 0 Thanked 4 Times in 4 Posts yep you're right - deleted my post
11-14-2012, 10:46 AM PM User | #9 felgall Master Coder Join Date: Sep 2005 Location: Sydney, Australia Posts: 6,099 Thanks: 0 Thanked 579 Times in 569 Posts There's no need for a loop. The following code does it with only one call to Math.random as the number returned always maps to within the required range with the same distribution regardless of what numbers are entered. Code: ```function generateRandomInteger(minNumber, maxNumber, minDigits, maxDigits) { var minNum, maxNum, minDigNum, maxDigNum, range, num; minNum = Math.min(minNumber, maxNumber); maxNum = Math.max(minNumber, maxNumber); minDigNum = Math.pow(10,minDigits-1); if (minNum<0) minDigNum *= -1; else minDigNum = Math.max(maxNum, minDigNum); maxDigNum = Math.pow(10,maxDigits-1); if (maxNum<0) { maxDigNum *= -1; if (maxDigNum minDigNum) num += maxDigNum; return num; }``` __________________ Stephen Learn Modern JavaScript - http://javascriptexample.net/ Helping others to solve their computer problem at http://www.felgall.com/ Beginners need to advise whether they want to learn "Latin" JavaScript for Netscape 3 or "Italian" JavaScript for modern browsers. Last edited by felgall; 11-14-2012 at 10:49 AM..
11-14-2012, 11:04 AM PM User | #10 Philip M Supreme Master coder! Join Date: Jun 2002 Location: London, England Posts: 17,490 Thanks: 200 Thanked 2,471 Times in 2,449 Posts Yet another good example of a different way to skin a cat. __________________ All the code given in this post has been tested and is intended to address the question asked. Unless stated otherwise it is not just a demonstration.
Bookmarks
Thread Tools Rate This Thread Rate This Thread: 5 : Excellent 4 : Good 3 : Average 2 : Bad 1 : Terrible
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home :: Client side development JavaScript programming DOM and JSON scripting Ajax and Design JavaScript frameworks Post a JavaScript HTML & CSS XML Flash & ActionScript Adobe Flex Graphics and Multimedia discussions General web building Site reviews Building for mobile devices :: Server side development Apache configuration Perl/ CGI PHP Post a PHP snippet MySQL Other Databases Ruby & Ruby On Rails ASP ASP.NET Java and JSP Other server side languages/ issues ColdFusion Python :: Computing & Sciences Computer Programming Computer/PC discussions Geek News and Humour Web Projects and Services Marketplace Web Projects Small projects (quick fixes and changes) Medium projects (new script, new features, etc) Large Projects (new web application, complex features etc) Unknown sized projects (request quote) Vacant job positions Looking for work/ for hire Project collaboration/ partnership Paid work offers and requests (Now CLOSED) Career, job, and business ideas or advice Domains, Sites, and Designs for sale Domains for sale Websites for sale Design templates and graphics for sale :: Other forums Member Offers Forum feedback and announcements
All times are GMT +1. The time now is 12:13 PM. | 2,133 | 8,328 | {"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} | 3.125 | 3 | CC-MAIN-2013-48 | latest | en | 0.650488 |
https://numbersworksheet.com/multiplying-3-digit-numbers-worksheet-pdf/ | 1,723,439,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00016.warc.gz | 330,703,150 | 13,834 | # Multiplying 3 Digit Numbers Worksheet Pdf
This multiplication worksheet is focused on educating individuals how you can mentally increase complete figures. Pupils may use custom grids to put particularly a single concern. The worksheets also deal withfractions and decimals, and exponents. There are also multiplication worksheets using a distributed residence. These worksheets certainly are a have to-have for the math concepts class. They may be found in school to learn to mentally flourish entire numbers and line them up. Multiplying 3 Digit Numbers Worksheet Pdf.
## Multiplication of entire phone numbers
If you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets will help you expert this standard idea. It is possible to decide to use one particular digit multipliers or two-digit and three-digit multipliers. Abilities of 10 can also be an incredible alternative. These worksheets will assist you to process long practice and multiplication studying the figures. They are also a terrific way to support your child understand the necessity of understanding the various kinds of entire amounts.
## Multiplication of fractions
Having multiplication of fractions on a worksheet can help professors plan and make lessons efficiently. Using fractions worksheets enables educators to quickly assess students’ idea of fractions. Students can be pushed to finish the worksheet in just a particular efforts and then mark their answers to see in which they need more instructions. Individuals can usually benefit from expression conditions that relate maths to real-lifestyle scenarios. Some fractions worksheets incorporate types of comparing and contrasting numbers.
## Multiplication of decimals
When you multiply two decimal amounts, be sure to class them up and down. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. By way of example, 01 x (11.2) by 2 could be comparable to 01 x 2.33 by 11.2 unless this product has decimal locations of lower than two. Then, the merchandise is round towards the nearest complete number.
## Multiplication of exponents
A math concepts worksheet for Multiplication of exponents will help you practice multiplying and dividing phone numbers with exponents. This worksheet will also offer things that will require pupils to multiply two different exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. Aside from, you can also get into particular recommendations on the worksheet by itself. When you’re finished, you can just click “Produce” along with the worksheet will probably be downloaded.
## Department of exponents
The standard tip for division of exponents when multiplying numbers is to subtract the exponent from the denominator from your exponent in the numerator. You can simply divide the numbers using the same rule if the bases of the two numbers are not the same. By way of example, \$23 divided by 4 will equivalent 27. This method is not always accurate, however. This system can cause misunderstandings when multiplying amounts which are too large or not big enough.
## Linear capabilities
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So the total rent would be \$470. A linear purpose of this particular type has got the type f(by), where by ‘x’ is the amount of time the car was hired. Furthermore, they have the form f(by) = ax b, where by ‘b’ and ‘a’ are true figures. | 703 | 3,616 | {"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} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.908117 |
http://mathhelpforum.com/advanced-algebra/180963-prove-finite-cyclic-group-isomorphic.html | 1,480,915,450,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541525.50/warc/CC-MAIN-20161202170901-00228-ip-10-31-129-80.ec2.internal.warc.gz | 178,590,865 | 10,306 | # Thread: Prove that the finite cyclic group is isomorphic
1. ## Prove that the finite cyclic group is isomorphic
Please can you help me with the following question:
Prove that every finite cyclic group is isomorphic to Z/(n) for nEZ
(where Z is an integer)
I really don't know where to start on this question but I know that if the group is bijective then it is isomorphic. Do I need to use this as a starting point?
Any help would be really greatful
2. Originally Posted by mathspeep
Please can you help me with the following question:
Prove that every finite cyclic group is isomorphic to Z/(n) for nEZ
(where Z is an integer)
I really don't know where to start on this question but I know that if the group is bijective then it is isomorphic. Do I need to use this as a starting point?
Any help would be really greatful
hi mathspeep!
welcome to the forum.
there are problems with your notations. you must first get those right. moreover you write "nEZ (where Z is an integer)". this does not make much sense. read the original problem. understand what is being asked... take your time and when you start to see that the problem really makes sense then re-attempt it. if then you can't do it then re-post it along with your attempt. Me and others will be happy to help.
3. if G is cyclic, it has a single generator. let's call it x.
can you think of a mapping that takes x^k to an element of Z/nZ that might be a homomorphism?
(use the properties of exponents. if G is isomorphic to Z/nZ, what should the order of x be?). | 372 | 1,538 | {"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} | 3.171875 | 3 | CC-MAIN-2016-50 | longest | en | 0.969462 |
https://demtutoring.com/answered/womensstudies/q5837 | 1,675,663,949,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00653.warc.gz | 221,110,268 | 3,740 | Flash Speed Questions
The solution time is much shorter than you think.
+ name: fraction word problems solve each problem. answer as a mixed mumber (if possible). 1) olivia needed 4 1/6 feet of thread to finish a pillow she was making. if she has 4 times as much thread as she needs, what is the length of the thread she has? 2. Get the answer
Category: womensstudies | Author: Mona Eva
Related Questions
+ 1101 06 pm guadal the mean value of land and buildings per acre fro
chemistry5 Minutes ago
+ 8.6 8.7 8.8 8.9 9.0 kyle truncated the decimal expansion of x to fi
geography7 Minutes ago
+ clever portal ? e multiplying and dividing rationa x mentum. com/co
ecology12 Minutes ago
art
Abraham Uilleam 55 Minutes ago
+ jesse decided to rate restaurants on a scale of 0 to 50. his ratings for 15 restaurants are shown in the table. restaurants 5 25 13 49 26 34 5 3 8
geography
Giiwedin Frigyes 1 Hours ago
+ marcus states that the polynomial expression 3r3 - 4x²y y3 + 2 is in standard form. ariel states that it should be y3 – 4x+y + 3x3 +2. explain whic
statistics
Sagi Boris 1 Hours ago
+ put the following decimals in order, starting with the smallest at the top. smallest 5.07 5.76 50.67 55.706 50.765 | 365 | 1,219 | {"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-2023-06 | latest | en | 0.890874 |
http://www.instructables.com/answers/Transformers-how-do-they-work/ | 1,513,155,902,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522343.41/warc/CC-MAIN-20171213084839-20171213104839-00010.warc.gz | 374,964,337 | 10,392 | # Transformers, how do they work?
Im trying to make my cheap laptop cooling stand use stand alone power. ie, not using the USB power. i obviously cant wire up that little fan motor to standard 120v. SO, i have this transformer here, i pulled it out of an electric razor. It makes sense that i could use this in my fan project. it has six, now de-soldered, pins, three on each side. how would i wire up the neg and pos of the power supply and the fan to this part?
PS. due to my webcam, it make the images blurry. the serial is:
4222 018 47541
SA-0101
If that helps any
Jack A Lopez5 years ago
A transformer converts an AC signal into another AC signal at a different voltage level. For example a transformer like the one you've got there might take 120 VAC on its primary side, and give 6 VAC on its secondary side, so that the ratio of the magnitude of the voltage on the primary side, to that on the secondary side is 20:1.
I claim that this ratio of input AC voltage, to output AC voltage, is a property of the transformer, and it is also the ratio of the number of turns on the primary, to the number of turns on the secondary. If you actually want to know how and why this works, the Wikipedia article on transformers might be a good place to start:
http://en.wikipedia.org/wiki/Transformer
Regarding this plan to drive a fan with your transformer: you did not mention the kind of fan you are using, but if you pulled it out of an old computer, it probably wants DC, not higher than about 12 VDC.
Because the fan wants DC, you will need some kind of rectifier circuit, to change the low voltage AC from your transformer, to DC. In fact the board on the electric razor, from which you removed the transformer, probably has such a rectifier circuit, consisting of maybe 2 or 4 diodes, plus a big electrolytic capacitor. This section of the Wikipedia article on rectifiers,
http://en.wikipedia.org/wiki/Rectifier#Full-wave_rectification
shows two of the usual circuits used for converting AC from a transformer into DC. The rectifier circuit your razor used was probably either the one using 4 diodes in a "bridge" shape, or the one using 2 diodes connected to a center-tapped transformer, if your transformer has a center-tapped secondary.
rickharris5 years ago
steveastrouk5 years ago
It would be better to just find an old router power supply brick and use that. The codes on the transformer don't help. | 569 | 2,421 | {"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 | 3 | CC-MAIN-2017-51 | latest | en | 0.891041 |
http://stackoverflow.com/questions/2459295/invertible-stft-and-istft-in-python/2470279 | 1,406,059,465,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997862711.66/warc/CC-MAIN-20140722025742-00030-ip-10-33-131-23.ec2.internal.warc.gz | 200,515,287 | 19,501 | # Invertible STFT and ISTFT in Python
Is there any general-purpose form of short-time Fourier transform with corresponding inverse transform built into SciPy or NumPy or whatever?
There's the pyplot `specgram` function in matplotlib, which calls `ax.specgram()`, which calls `mlab.specgram()`, which calls `_spectral_helper()`:
``````#The checks for if y is x are so that we can use the same function to
#implement the core of psd(), csd(), and spectrogram() without doing
#extra calculations. We return the unaveraged Pxy, freqs, and t.
``````
but
This is a helper function that implements the commonality between the 204 #psd, csd, and spectrogram. It is NOT meant to be used outside of mlab
I'm not sure if this can be used to do an STFT and ISTFT, though. Is there anything else, or should I translate something like these MATLAB functions?
I know how to write my own ad-hoc implementation; I'm just looking for something full-featured, which can handle different windowing functions (but has a sane default), is fully invertible with COLA windows (`istft(stft(x))==x`), tested by multiple people, no off-by-one errors, handles the ends and zero padding well, fast RFFT implementation for real input, etc.
-
I'm looking for exactly the same thing, similar to Matlab's "spectrogram" function. – khpeek May 10 at 21:08
– endolith May 10 at 21:54
Here is my Python code, simplified for this answer:
``````import scipy, pylab
def stft(x, fs, framesz, hop):
framesamp = int(framesz*fs)
hopsamp = int(hop*fs)
w = scipy.hamming(framesamp)
X = scipy.array([scipy.fft(w*x[i:i+framesamp])
for i in range(0, len(x)-framesamp, hopsamp)])
return X
def istft(X, fs, T, hop):
x = scipy.zeros(T*fs)
framesamp = X.shape[1]
hopsamp = int(hop*fs)
for n,i in enumerate(range(0, len(x)-framesamp, hopsamp)):
x[i:i+framesamp] += scipy.real(scipy.ifft(X[n]))
return x
``````
Notes:
1. The list comprehension is a little trick I like to use to simulate block processing of signals in numpy/scipy. It's like `blkproc` in Matlab. Instead of a `for` loop, I apply a command (e.g., `fft`) to each frame of the signal inside a list comprehension, and then `scipy.array` casts it to a 2D-array. I use this to make spectrograms, chromagrams, MFCC-grams, and much more.
2. For this example, I use a naive overlap-and-add method in `istft`. Due to windowing and frame overlap, the magnitude of the reconstructed signal is not the same as that of the input signal. You will need to adjust that yourself, somehow.
3. There are probably more principled ways of computing the ISTFT. This example is mainly meant to be educational.
A test:
``````if __name__ == '__main__':
f0 = 440 # Compute the STFT of a 440 Hz sinusoid
fs = 8000 # sampled at 8 kHz
T = 5 # lasting 5 seconds
framesz = 0.050 # with a frame size of 50 milliseconds
hop = 0.020 # and hop size of 20 milliseconds.
# Create test signal and STFT.
t = scipy.linspace(0, T, T*fs, endpoint=False)
x = scipy.sin(2*scipy.pi*f0*t)
X = stft(x, fs, framesz, hop)
# Plot the magnitude spectrogram.
pylab.figure()
pylab.imshow(scipy.absolute(X.T), origin='lower', aspect='auto',
interpolation='nearest')
pylab.xlabel('Time')
pylab.ylabel('Frequency')
pylab.show()
# Compute the ISTFT.
xhat = istft(X, fs, T, hop)
# Plot the input and output signals over 0.1 seconds.
T1 = int(0.1*fs)
pylab.figure()
pylab.plot(t[:T1], x[:T1], t[:T1], xhat[:T1])
pylab.xlabel('Time (seconds)')
pylab.figure()
pylab.plot(t[-T1:], x[-T1:], t[-T1:], xhat[-T1:])
pylab.xlabel('Time (seconds)')
``````
-
Is there an unsimplified version online you can link to? – endolith Jul 31 '11 at 21:02
Not off the top of my head. But is there anything wrong with the above code? You can modify it, if necessary. – Steve Tjoa Aug 2 '11 at 4:41
No, but you said "simplified for this answer", so I assumed this was an abridged version of something else you wrote – endolith Aug 2 '11 at 4:44
Sorry for the confusion. Yes, simplified from my application-specific needs. Example features: if the input is a stereo signal, make it mono first; plot the spectrogram over a given frequency and time range; plot the log-spectrogram; round `framesamp` up to the nearest power of two; embed `stft` inside a `Spectrogram` class; etc. Your needs may differ. But the core code above still gets the job done. – Steve Tjoa Aug 2 '11 at 5:09
Thanks for this code. Just a remark : what happens in `stft` if x is not a multiple of the `hop` length ? Shouldn't the last frame be zero-padded ? – Basj Nov 19 '13 at 14:08
Here is the STFT code that I use. STFT + ISTFT here gives perfect reconstruction (even for the first frames). I slightly modified the code given here by Steve Tjoa : here the magnitude of the reconstructed signal is the same as that of the input signal.
``````import scipy, numpy as np
def stft(x, fftsize=1024, overlap=4):
hop = fftsize / overlap
w = scipy.hanning(fftsize+1)[:-1] # better reconstruction with this trick +1)[:-1]
return np.array([np.fft.rfft(w*x[i:i+fftsize]) for i in range(0, len(x)-fftsize, hop)])
def istft(X, overlap=4):
fftsize=(X.shape[1]-1)*2
hop = fftsize / overlap
w = scipy.hanning(fftsize+1)[:-1]
x = scipy.zeros(X.shape[0]*hop)
wsum = scipy.zeros(X.shape[0]*hop)
for n,i in enumerate(range(0, len(x)-fftsize, hop)):
x[i:i+fftsize] += scipy.real(np.fft.irfft(X[n])) * w # overlap-add
wsum[i:i+fftsize] += w ** 2.
pos = wsum != 0
x[pos] /= wsum[pos]
return x
``````
-
Found another STFT, but no corresponding inverse function:
``````def stft(x, w, L=None):
...
return X_stft
``````
• w is a window function as an array
• L is the overlap, in samples
-
I also found this on GitHub, but it seems to operate on pipelines instead of normal arrays:
http://github.com/ronw/frontend/blob/master/basic.py#LID281
``````def STFT(nfft, nwin=None, nhop=None, winfun=np.hanning):
...
return dataprocessor.Pipeline(Framer(nwin, nhop), Window(winfun),
RFFT(nfft))
def ISTFT(nfft, nwin=None, nhop=None, winfun=np.hanning):
...
return dataprocessor.Pipeline(IRFFT(nfft), Window(winfun),
``````
-
Neither of the above answers worked well OOTB for me. So I modified Steve Tjoa's.
``````import scipy, pylab
import numpy as np
def stft(x, fs, framesz, hop):
"""
x - signal
fs - sample rate
framesz - frame size
hop - hop size (frame size = overlap + hop size)
"""
framesamp = int(framesz*fs)
hopsamp = int(hop*fs)
w = scipy.hamming(framesamp)
X = scipy.array([scipy.fft(w*x[i:i+framesamp])
for i in range(0, len(x)-framesamp, hopsamp)])
return X
def istft(X, fs, T, hop):
""" T - signal length """
length = T*fs
x = scipy.zeros(T*fs)
framesamp = X.shape[1]
hopsamp = int(hop*fs)
for n,i in enumerate(range(0, len(x)-framesamp, hopsamp)):
x[i:i+framesamp] += scipy.real(scipy.ifft(X[n]))
# calculate the inverse envelope to scale results at the ends.
env = scipy.zeros(T*fs)
w = scipy.hamming(framesamp)
for i in range(0, len(x)-framesamp, hopsamp):
env[i:i+framesamp] += w
env[-(length%hopsamp):] += w[-(length%hopsamp):]
env = np.maximum(env, .01)
return x/env # right side is still a little messed up...
``````
- | 2,073 | 7,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-23 | latest | en | 0.796262 |
https://justaaa.com/statistics-and-probability/180028-assume-that-adults-have-iq-scores-that-are | 1,713,044,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00316.warc.gz | 296,899,771 | 8,995 | Question
Assume that adults have IQ scores that are normally distributed with a mean of 97.8 and...
Assume that adults have IQ scores that are normally distributed with a mean of
97.8
and a standard deviation
21.2
Find the first quartile
Upper Q 1
which is the IQ score separating the bottom 25% from the top 75%. (Hint: Draw a graph.)
The first quartile is
(Type an integer or decimal rounded to one decimal place as needed.)
Given that, mean (μ) = 97.8 and
standard deviation = 21.2
X ~ N(97.8, 21.2)
We want to find, the value of x such that, P(X < x) = 0.25
First we find, the z-score such that, P(Z < z) = 0.25
Using standard normal z-table we get, z-score corresponding probability of 0.25 is z = -0.67
Therefore, the first quartile is 83.6 | 228 | 768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-18 | latest | en | 0.909265 |
https://metanumbers.com/24255 | 1,628,089,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00105.warc.gz | 398,429,838 | 11,012 | ## 24255
24,255 (twenty-four thousand two hundred fifty-five) is an odd five-digits composite number following 24254 and preceding 24256. In scientific notation, it is written as 2.4255 × 104. The sum of its digits is 18. It has a total of 6 prime factors and 36 positive divisors. There are 10,080 positive integers (up to 24255) that are relatively prime to 24255.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 18
• Digital Root 9
## Name
Short name 24 thousand 255 twenty-four thousand two hundred fifty-five
## Notation
Scientific notation 2.4255 × 104 24.255 × 103
## Prime Factorization of 24255
Prime Factorization 32 × 5 × 72 × 11
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 1155 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 24,255 is 32 × 5 × 72 × 11. Since it has a total of 6 prime factors, 24,255 is a composite number.
## Divisors of 24255
1, 3, 5, 7, 9, 11, 15, 21, 33, 35, 45, 49, 55, 63, 77, 99, 105, 147, 165, 231, 245, 315, 385, 441, 495, 539, 693, 735, 1155, 1617, 2205, 2695, 3465, 4851, 8085, 24255
36 divisors
Even divisors 0 36 18 18
Total Divisors Sum of Divisors Aliquot Sum τ(n) 36 Total number of the positive divisors of n σ(n) 53352 Sum of all the positive divisors of n s(n) 29097 Sum of the proper positive divisors of n A(n) 1482 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 155.74 Returns the nth root of the product of n divisors H(n) 16.3664 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 24,255 can be divided by 36 positive divisors (out of which 0 are even, and 36 are odd). The sum of these divisors (counting 24,255) is 53,352, the average is 1,482.
## Other Arithmetic Functions (n = 24255)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 10080 Total number of positive integers not greater than n that are coprime to n λ(n) 420 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2689 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 10,080 positive integers (less than 24,255) that are coprime with 24,255. And there are approximately 2,689 prime numbers less than or equal to 24,255.
## Divisibility of 24255
m n mod m 2 3 4 5 6 7 8 9 1 0 3 0 3 0 7 0
The number 24,255 is divisible by 3, 5, 7 and 9.
• Arithmetic
• Abundant
• Polite
## Base conversion (24255)
Base System Value
2 Binary 101111010111111
3 Ternary 1020021100
4 Quaternary 11322333
5 Quinary 1234010
6 Senary 304143
8 Octal 57277
10 Decimal 24255
12 Duodecimal 12053
20 Vigesimal 30cf
36 Base36 ipr
## Basic calculations (n = 24255)
### Multiplication
n×i
n×2 48510 72765 97020 121275
### Division
ni
n⁄2 12127.5 8085 6063.75 4851
### Exponentiation
ni
n2 588305025 14269338381375 346102802440250625 8394723473188278909375
### Nth Root
i√n
2√n 155.74 28.9468 12.4796 7.53287
## 24255 as geometric shapes
### Circle
Diameter 48510 152399 1.84821e+09
### Sphere
Volume 5.97713e+13 7.39286e+09 152399
### Square
Length = n
Perimeter 97020 5.88305e+08 34301.7
### Cube
Length = n
Surface area 3.52983e+09 1.42693e+13 42010.9
### Equilateral Triangle
Length = n
Perimeter 72765 2.54744e+08 21005.4
### Triangular Pyramid
Length = n
Surface area 1.01897e+09 1.68166e+12 19804.1
## Cryptographic Hash Functions
md5 87be96790ea277c50f980ce4df0b4412 166ee73e931fecc04ccc54f663910b847366cda8 9af03e9f6437853761cf05e74545bcdcde9953344a544eb3d72a7d8fcf8e3a7e 47d68435c337c15de09f9aac11f0a70fc7b9942f2805ad362f8bf3229dc6ef0c364bf6b73a751b308c0980910f74e479aa1307b486f5f1b330fc71ab4c56ccea 99272c4a1f6b662522d0efb5121abe540d30f63d | 1,538 | 4,204 | {"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} | 3.625 | 4 | CC-MAIN-2021-31 | latest | en | 0.777689 |
http://resolutionofriemannhypothesis.blogspot.com/2018/06/dynamic-perspective-on-addition-and.html | 1,548,249,632,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584332824.92/warc/CC-MAIN-20190123130602-20190123152602-00151.warc.gz | 184,488,465 | 15,069 | ## Sunday, June 10, 2018
### Dynamic Perspective on Addition and Multiplication
Things are not always what they seem!
I mentioned this yesterday with reference to the standard analytic treatment of number 1 + 1 = 2.
Indeed based on our common-sense intuitions of reality this statement seems so obvious that it is frequently used as a metaphor for self-evident truth.
In fact however it reveals the reduced nature of accepted mathematical understanding, which through millennia now of unchallenged usage has become so deeply ingrained in our thinking that we have become completely blind as to its limitations.
The conventional understanding of this simple number relationship is based on mere analytic appreciation (where the “whole” number 2, represents the quantitative sum of its part unit members).
So the unit numbers here are viewed as absolutely independent of each other (in a homogeneous manner). In other words, as there is no way to distinguish as between the two units, they thereby lack qualitative distinction.
However if the individual units were indeed absolutely independent in such a manner, there would be no way of combining them to obtain the new collective identity of 2!
Therefore the very ability to relate the two unit numbers presupposes a distinct qualitative identity of relational interdependence.
In the conventional treatment of addition, this key fact is completely overlooked.
In other words the qualitative aspect of number (i.e. relational interdependence) is simply reduced in an absolute quantitative manner.
This then enables the convenient fiction to be maintained that numbers can be successfully combined with each other (while preserving their absolute quantitative identity).
Thus from this perspective 1 + 1 = 2 (understood in absolute quantitative terms).
However when one addresses the gross inconsistency entailed here, then the very nature of number is thereby radically changed.
For when we accept that all numbers necessarily possess a qualitative aspect of relational interdependence, then the corresponding quantitative aspect (of distinct identity) can only be properly conceived in a relatively independent fashion.
Thus all numbers necessarily possess both an analytic aspect (of quantitative independence) and a holistic aspect (of qualitative interdependence) respectively, which are dynamically related to each other in complementary fashion.
So let us now return to our simple identity, 1 + 1 = 2.
Again, in terms of appropriate dynamic appreciation, each unit is relatively independent in a quantitative manner, while equally possessing a qualitative identity of shared relational interdependence.
Whereas the quantitative aspect is directly related to the cardinal, the qualitative aspect is directly related to the corresponding ordinal nature of number respectively.
So again from the quantitative perspective, the two units are given a separate individual identity (in actual terms); however from the qualitative perspective, they are given a shared collective identity (in a potential manner).
So with respect to number recognition, each individual unit must be quantitatively distinguished (in an actual manner).
However equally, the two units must be given a qualitative identity of shared relational interdependence (in potential terms). From this perspective ordinal notions of 1st and 2nd with respect to the two units are viewed as fully interchangeable with each other.
At a deeper psychological level, whereas the analytic (quantitative) aspect of number relates directly to rational (conscious) recognition, the holistic (qualitative) aspect relates by contrast directly to intuitive (unconscious) appreciation.
So in conventional mathematical practice, the reduction of the holistic (qualitative) aspect of number recognition, in analytic (quantitative) terms, parallels precisely the corresponding psychological reduction of intuitive (unconscious) appreciation in a rational (conscious) manner.
Properly therefore, a comprehensive understanding of number entails both conscious and unconscious aspects of appreciation where both need to be explicitly recognised.
However once more, in conventional mathematical terms, the holistic (unconscious) aspect is formally reduced in an analytic (conscious) manner.
Therefore the true nature of number is inherently dynamic with quantitative notions of relative independence continually interacting with qualitative notions of relative interdependence respectively.
And in this new dynamic context, the two relative notions of number independence and interdependence are in turn intimately linked with the mathematical operations of addition and multiplication respectively.
1 + 1 = 2.
More fully we can express this relationship as,
11 + 11 = 21.
And using the Peano postulates all the natural numbers can ultimately be derived from successively adding 1 to a previous natural number (starting with 1).
So to combine numbers in a quantitative manner (through the operation of addition) we must explicitly define them with respect to the default dimension of 1 (which in this context remains implicit in understanding). In this way all natural numbers are viewed as points on the same real number line.
Thus in this context, 2 (i.e. 21) represents the quantitative sum of the two base units.
However when we now, in relative terms, combine the two units through multiplication, a different qualitative notion of number emerges, which can be expressed as follows,
11 * 11 = 12.
So in dynamic complementary fashion the focus is explicitly on the dimensional units (with the corresponding base units now remaining implicit).
Thus in this dynamically related context 2 (i.e. 12) represents the qualitative product of the two dimensional units).
What this means in effect is that these two units are now understood in holistic ordinal terms (as relatively interdependent) whereby 1st and 2nd positions are freely interchangeable with each other.
So just as we recognise that what are designated right and left turns at a crossroads can switch with each other depending on the direction from which the crossroads is approached, we likewise recognise that the two dimensional units can likewise switch ordinal positions (depending on context).
However as always in dynamic interactive terms, reference frames can switch.
Thus there is a valid sense in which addition can likewise be identified in qualitative terms with the interdependence of its unit members.
So 11 + 11 = 21. So here we are explicitly recognising in a holistic manner the qualitative interdependence of the two base units (with quantitative recognition now implicit).
And finally there is equally a valid sense in complementary fashion, through which multiplication can be likewise identified with the quantitative independence of its dimensional units (i.e. where each dimensional unit is recognised as separate).
Thus 11 * 11 = 12.
Therefore what we have demonstrated here is the following:
1) The number 2 - and by extension all natural numbers - can be given both analytic (quantitative) and holistic (qualitative) interpretations with respect to both base and dimensional usage. Whereas the analytic aspect corresponds directly with rational, the holistic aspect by contrast corresponds directly with intuitive type recognition.
2) The relationship between both is of a dynamic interactive nature with quantitative and qualitative aspects understood in complementary fashion.
So when the analytic (quantitative) aspect of number is made explicit, the corresponding holistic (qualitative) aspects remains, in this context implicit.
However, when the holistic (qualitative) aspect of number is now in turn made explicit, the corresponding analytic (quantitative) aspect is now implicit.
3) What is extremely important to bear in mind is that in this new dynamic framework, addition and multiplication become the very means by which the switching from quantitative to qualitative (and in turn qualitative to quantitative) recognition takes place.
So addition and multiplication operate as dynamic complementary partners in the very recognition of number.
Thus if we return to the number 2 for a moment, there is of course a valid sense in which we can recognise its two units as quantitatively independent of each other; however there is an equally valid sense in which we can recognise that these two units are likewise interdependent (and thereby interchangeable with each other) in an ordinal manner. So in this context 2 necessarily contains unique 1st and 2nd units.
And both of these interpretations are necessarily related in two-way fashion with each other.
Thus we cannot explicitly recognise the quantitative independence of the two units (without implicit recognition of their corresponding qualitative interdependence).
And we cannot explicit recognise the qualitative interdependence of the two units as, interchangeably, 1st and 2nd respectively (without implicit recognition of their quantitative independence).
Thus properly understood, the number 2 - and by extension every other number - is necessarily of a dynamic interactive nature, comprising analytic (quantitative) and holistic (qualitative) aspects in a complementary manner. | 1,701 | 9,288 | {"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} | 3.46875 | 3 | CC-MAIN-2019-04 | latest | en | 0.953565 |
https://brilliant.org/discussions/thread/fermats-last-theorem-x/ | 1,529,548,452,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864019.29/warc/CC-MAIN-20180621020632-20180621040632-00137.warc.gz | 570,237,169 | 13,359 | # Fermat's Last Theorem
After reading Arthur C. Clarke and Frederik Pohl's The Last Theorem,I thought about the problem for a while and I found that the difference between two numbers raised to the power n ( $$c^{n} = (a+b)^{n} - a^{n}$$ ) can be expressed as $$\displaystyle \sum_{i=0}^n b \times a^{n-1} \times ( \frac{a+b}{a} )^{i}$$.So if it can be proved that $$\displaystyle \sum_{i=0}^n b \times a^{n-1} \times ( \frac{a+b}{a} )^{i}$$ cannot be equal to a number raised to the $$n$$th power then this problem could be solved.
Note by Tan Li Xuan
4 years, 4 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$ | 503 | 1,440 | {"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": 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} | 3.734375 | 4 | CC-MAIN-2018-26 | latest | en | 0.784262 |
https://mathesis-online.com/scalar-product | 1,726,886,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00510.warc.gz | 340,961,932 | 45,667 | Select Page
The scalar or dot product of two vectors in real space is a real number that takes into account the direction, sense and magnitude of both vectors.
## 1.The natural scalar product in the Euclidean plane
### 1.1.From the distance between two points to the scalar product
In the Euclidean plane $\mathbb R^2$, the so-called “natural (Euclidean) distance” between two points $M=(a,b)$ and $N=(c,d)$ is defined as the number $\sqrt{(a-c)^2+(b-d)^2}$. This is because this length is interpreted as the length of the hypotenuse of a right triangle, which can be calculated using the Pythagorean theorem (see The Euclidean Plane). By analogy, in Euclidean (three-dimensional) space $\mathbb R^3$ we also define the natural distance between two points $A=(a,b,c)$ and $B=(x,y,z)$ as the square root of the sum of the squares of the differences of the coordinates. It is thus the real number $d(A,B)=\sqrt{(a-x)^2+(b-y)^2+(c-z)^2}$. Thanks to this definition one can define for example a sphere of centre $A=(a,b,c)$ and of radius $R\geq 0$ as the set of points $M=(x, y,z) of R^3$ such that $d(A,M)=R$, which translates into the equation $(a-x)^2+(b-y)^2+(c-z)^2=R^2$.
If we return to the plane $\mathbb R^2$, we find that the expression under the sign $\sqrt{}$ is of the form $x^2+y^2$ for the real numbers $x=a-c$ and $y=b-d$. In other words, instead of considering the distance as associated with the two points $M$ and $N$, we can consider it as associated with the vector $\vec{MN}=(x,y)\in\mathbb R^2$: this is what we call the norm of the vector $\vec{MN}$, which is noted $||vec{MN}||$. This norm, which measures th magnitude of a vector, is associated with a natural operation between two vectors $\vec u=(x_1,y_1)\in\mathbb R^2$ and $\vec v=(x_2,y_2)\in\mathbb R^2$, which is called the scalar or dot product of $\vec u$ and $\vec v$. This scalar product is defined as the real number $\vec u.\vec v=x_1x_2+y_1y_2$, and we see that the norm of a vector $\vec u=(x,y)$, i.e. the distance $\sqrt{x^2+y^2}$ betweeen its two extremities, is the square root of its scalar product with itself, i.e. $\sqrt{\vec u.\vec v}$.
In the following figure, the coordinates $(c-a,d-b)$ of the vector $\vec{MN}$ are readable from the coordinates of the points $M$ ($a,b)$ and $N$ ($c,d)$. The norm of the vector is then the distance between $M$ and $N$, i.e. $\sqrt{(c-a)^2+(d-b)^2}$.
### 1.2.A real product with a geometric meaning
Why is this operation called a product? Because it has properties similar to the product of two numbers. For example, if $a,b,c\in\mathbb R$ are three real numbers, we have the well-known property of distributivity of multiplication over addition, i.e. the equality $c\times (a+b)=c\times a+c\times b$. In the same way, for three vectors $\vec u=(a,b)$, $\vec v=(x,y)$ and $w=(z,t)$ of $\mathbb R^2$, we have $\vec w.(\vec u+\vec v)=z\times (a+x)+t\times (b+y)=(za+tb)+(zx+ty)=(\vec w.\vec u)+(\vec w.\vec v).$ We see that this distributivity of the dot product on the addition of vectors comes in fact from the distributivity of $\times$ on $+$ in the set $\mathbb R$. In the same way, if $a$ is a real number, one can easily show that $\vec u.(a\vec v)=(a\vec u).\vec v$, which is rather similar to the commutativity and associativity of the product of two numbers. Let us also note that the scalar product is symmetrical, i.e. that $\vec u.\vec v=\vec v.\vec u$.
In the plane, the scalar product has an indirect geometric interpretation. If we consider the (non-oriented) angle $(\vec u,\vec v)$ formed by the vectors $\vec u$ and $\vec v$, we can give a trigonometric expression of the dot product of $\vec u$ and $\vec v$ in the form $\vec u. \vec v=||\vec u||\times ||\vec v||\times \cos (\vec u,\vec v)$. From there, if we consider that two vectors determine a parallelogram, then the scalar product $\vec u. \vec v$ is the signed area (i.e. taking into account the sign, i.e. a positive or negative number) of the parallelogram determined, not by the vectors $\vec u$ and $\vec v$, but by $\vec u$ and the vector $\vec w$ obtained by applying a rotation of angle $\pi/2$ (a quarter turn to the left) to the vector $\vec v$. The area is signed because the cosine of the angle between $\vec u$ and $\vec v$ is positive or negative, depending on whether the (non-oriented) angle is acute (between $0$ and $\pi/2$) or obtuse (between $\pi/2$ and $\pi$).
On the following figure, one can visualize the scalar product $\vec u.\vec v=||\vec u||\times ||\vec v||\times \cos\alpha$ of the vectors $\vec u$ and $\vec v$ as the signed area of the parallelogram $ABCD$ constructed from the vectors $\vec u$ and $\vec w$, where $\vec w$ is obtained from $\vec v$ by an clockwise rotation of angle $\pi/2$, and where $\alpha$ is the angle between $\vec u$ and $\vec v$. The sign of the dotr product is positive here, because the angle $\alpha$ is acute.
### 1.3.A measure of orthogonality
Thus, the scalar product of two vectors is a real number that combines both the direction and the sense of two vectors through the angle they form. It can therefore be used as a criterion for determining the orthogonality (the vector analogue of perpendicularity) of two vectors, and thus of two lines for example. In this sense, it is a “measure” of the orthogonality of two vectors, and can even be used to define when two vectors are orthogonal.
Two vectors $\vec u$ and $\vec v$ are, in fact, said to be orthogonal, by definition, when the (non-oriented) angle they form is a right angle. Using the trigonometric expression of the scalar product $\vec u.\vec v=||\vec u||\times ||vec v||\times \cos(\vec u,\vec v)$, we see that this scalar product is zero exactly when one of the vectors $\vec u$ or $\vec v$ is of zero norm, or when $\cos(\vec u,\vec v)=0$. If therefore none of the vectors $\vec u$ and $\vec v$ are zero, their scalar product is zero if, and only if, they are orthogonal. This makes it possible, for example, to calculate directly the equation of a line orthogonal to a given line when a directing vector of the line is known.
In the following figure, which repeats the first, we show how the scalar product can be used to find a Cartesian equation of the line $D$ orthogonal to the vector $\vec{MN}$ at $M$. A point $P$ with coordinates $(x,y)$ is on this line $D$ if and only if the vectors $\vec{MN}$ and $\vec{MP}$ are orthogonal, in other words if and only if $\vec{MN}.\vec{MP}=0$, or $(c-a)(x-a)+(d-b)(y-b)=0$. By developing, we obtain a Cartesian equation of $D$ of the form $(c-a)x+(d-b)y-a(c-a)-b(d-b)=0$, whose coefficients are written from the coordinates of $M$ and $N$.
## 2.Scalar products in higher dimensions
### 2.1.The natural scalar product in dimension 3
In the $3$-dimensional real space $\mathbb R^3$, the geometry of angles is rather more complicated, but we can still define a natural dot product. Indeed, the analytical definition of the scalar product in the plane (i.e. the one using coordinates) is inspired by the distance between two points, and uses coordinates. It is possible to “extend” it to vectors in dimension $3$.
We have already given an expression for the distance between two points in three-dimensional space, from which we can derive the norm of a vector $\vec v=(x,y,z)\in\mathbb R^3$: this is the positive real number $||\vec v||=\sqrt{x^2+y^2+z^2}$. As in the case of the plane, the scalar product of two vectors $\vec u=(a,b,c)$ and $\vec v=(x,y,z)$ of $\mathbb R^3$ is then defined as $\vec u.\vec v=ax+by+cz.$ We can easily verify by calculation that it has the same properties as the scalar product in the plane.
In the following figure, the $x$ axis (abscissa) is shown in red and the $y$ axis (ordinate) in green. We have represented two points of the Euclidean space $\mathbb R^3$, the point $M$ with coordinates $(1,-4,3)$ and the point $N$ with coordinates $(2,0,-1)$. The Euclidean norm of the vector $\vec{MN}$ (represented by the purple arrow) is the distance between $M$ and $N$, i.e. $\sqrt{(1-2)^2+(-4-0)^2+(3-(-1))^2}=\sqrt{1+16+16}=\sqrt{33}$.
### 2.2.The scalar product in higher dimensions
More generally, if $n$ is a non-zero natural number, let us call “$n$-dimensional real space” the set $\mathbb R^n$ of $n$-tuples of real numbers, i.e. of finite sequences $(x_1,x_2,\ldots,x_n)$ of real numbers. For $n=2$, we obtain the Euclidean plane $\mathbb R^2$ and for $n=3$, we obtain the Euclidean space $\mathbb R^3$. We can then define, in an analogous way, the scalar product of two vectors $\vec u(x_1,x_2,\ldots,x_n)$ and $\vec v=(y_1,y_2,\ldots,y_n)$ of $\mathbb R^n$, as the real number $\vec u.\vec v=x_1y_1+x_2y_2+ldots+x_ny_n.$ We also have the Euclidean norm $||\vec u||=\sqrt{x_1^2+x_2^2+\ldots+x_n^2}$ which measures the “magnitude” of a vector $\vec u$ and the natural Euclidean distance between two points. In these spaces, the dot product is also used to define a notion of orthogonality, and there is always a version of the Pythagorean theorem.
For $n=4$, we could consider the points or vectors $(t,x,y,z)\in \mathbb R^4$ as the elements of a “mathematical space-time” (where the first coordinate $t$ would represent an ideal time and the three others $x,y,z$ a place in space). We can therefore conceive an element of $\mathbb R^4$ as an “event”. The Euclidean distance between two points is then a distance between two events, which takes into account both space and time (this being said, another form of distance is used in relativity theory). Associated with this scalar product and distance is a Euclidean geometry in dimension $4$, where one can find for example the analogue of the cube (called hypercube or tesseract) and the analogue of the sphere.
### 2.3.Other scalar products and analogues
Scalar products are examples of functions that are called symmetric bilinear forms. The scalar product presented here is said to be “natural” because it is directly defined from the natural description of the real spaces $\mathbb R^n$, but there exist on each such space an infinite number of other scalar products, which are all other ways of measuring orthogonality and distances, and thus of doing Euclidean geometry.
There are also other types of bilinear forms that are not dot products and allow us to develop similar geometric notions. We have mentioned the Minkowski space-time, which can be thought of as the real space $\mathbb R^4$ on which we have an analogue of the scalar product defined for two vectors $\vec u=(t,x,y,z)$ and $\vec v=(t’,x’,y’,z’)$ by $\vec u.\vec v=tt’-xx’-yy’-zz’$ (compare with the formula for the scalar product). Such a “structure” allows to give a geometrical representation of the physical properties of space-time in special relativity theory (see Wikipedia, Minkowski space).
Finally, there is also a natural complex scalar product, which can be defined in an elementary way on complex spaces $\mathbb C^n$, and which uses complex conjugation. If $z=(z_1,\ldots,z_n),w=(w_1,\ldots,w_n)\in \mathbb C^n$ are two vectors, we define it as $z.w=z_1\overline{w_1}+\ldots +z_n\overline{w_n}$, where $\overline w$ denotes the conjugate of the complex number $w$. Its properties are analogous, but more complex (!), to those of the natural real scalar product, which it extends in a way. Note that it is also possible to define scalar products, real or complex, in infinite dimensional spaces (such as certain function spaces), where one can export notions of Euclidean geometry to do analysis for example. But such a presentation is beyond the scope of this article. | 3,176 | 11,549 | {"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} | 4.59375 | 5 | CC-MAIN-2024-38 | latest | en | 0.884234 |
http://imath.co.za/Gr10Math/ans10M10facdif2sq.html | 1,579,895,148,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250625097.75/warc/CC-MAIN-20200124191133-20200124220133-00066.warc.gz | 80,313,769 | 2,212 | #### The factors of the difference of two squares - answers.
1.
(a + b)(a - b)
2.
(c + d)(c - d)
3.
(m + n)(m - n)
4.
(p - q)(p + q)
5.
(x + y)(x - y)
6.
(a + 2)(a - 2)
7.
(b - 4)(b + 4)
8.
(c - 6)(c + 6)
9.
(d - 10)(d + 10)
10.
(h - 15)(h + 15)
11.
(m - 12)(m + 12)
12.
(n - 17)(n + 17)
1
1
1
1
1
1
13.
(p - ---- )(p + ---- )
14.
(q + ---- )(q - ---- )
15.
(x - --- )(x + --- )
2
2
5
5
6
6
4
4
7
7
2b
2b
16.
(y - --- )(y + --- )
17.
(z - --- )(z + --- )
18.
(a - ----- )(a + ----- )
5
5
6
6
9
9
19.
(x - 0,1 )(x + 0,1 )
20.
(y - 0,3 )(y + 0,3 )
21.
( z - 0,16 )( z + 0,16 )
22.
(a + 0,4)(a - 0,4)
23.
(3b - 4)(3b + 4)
24.
(5c - 7)(5c + 7)
25.
(4m - 7n)(4m + 7n)
26.
(5n - 9p)(5n + 9p)
27.
(11pq - 15xy)(11pq + 15xy)
28.
(a 2 - b 2 )(a 2 + b 2 )
= (a - b)(a + b)(a 2 + b 2 )
29.
(p 3 - q 3 )(p 3 + q 3 )
= (p - q)(p 2 + pq + q 2 )(p + q)(p 2 - pq + q 2 )
30.
(x 4 - y 4 )(x 4 + y 4 )
= (x 2 - y 2 )(x 2 + y 2 )(x 4 + y 4 )
= (x - y)(x + y)(x 2 + y 2 )(x 4 + y 4 )
To the top Exercise Exercises - Grade 10 Exercises - Grade 11 Exercises - Grade 12 Homepage | 577 | 1,058 | {"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-2020-05 | longest | en | 0.333432 |
https://www.coursehero.com/file/6108979/Make-up-2/ | 1,519,236,419,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813691.14/warc/CC-MAIN-20180221163021-20180221183021-00346.warc.gz | 875,828,737 | 26,220 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
Make up 2
# Make up 2 - Rehman(aar638 – Makeup Exam 2 – sachse...
This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Rehman (aar638) – Makeup Exam 2 – sachse – (56620) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the slope in the x-direction at the point P (0 , 2 , f (0 , 2)) on the graph of f when f ( x, y ) = 2(2 x + y ) e − xy . 1. slope =- 10 2. slope =- 4 correct 3. slope =- 8 4. slope =- 12 5. slope =- 6 Explanation: The graph of f is a surface in 3-space and the slope in the x-direction at the point P (0 , 2 , f (0 , 2)) on that surface is the value of the partial derivative f x at (0 , 2). Now f x = 4 e − xy- 2(2 xy + y 2 ) e − xy . Consequently, at P (0 , 2 , f (0 , 2)) slope =- 4 . 002 10.0 points Determine the second partial f xy of f when f ( x, y ) = 6 x 2 y + y 2 6 x . 1. f xy = 12 x- y 2. f xy = 12 x y 2 + y 3 x 2 3. f xy =- 12 x y 2- y 3 x 2 correct 4. f xy = 12 x y 2- y 3 x 2 5. f xy = 12 x + y Explanation: Differentiating with respect to x , we obtain f x = 12 x y- y 2 6 x 2 , and so after differentiation with respect to y we see that f xy =- 12 x y 2- y 3 x 2 . 003 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay A 3 xy 2 9 + x 2 dA over the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 4 ,- 4 ≤ y ≤ 4 bracerightBig , integrating first with respect to y . 1. I = 32 ln parenleftBig 25 9 parenrightBig 2. I = 32 ln parenleftBig 25 18 parenrightBig 3. I = 64 ln parenleftBig 25 9 parenrightBig correct 4. I = 64 ln parenleftBig 25 18 parenrightBig 5. I = 32 ln parenleftBig 9 25 parenrightBig 6. I = 64 ln parenleftBig 9 25 parenrightBig Explanation: The double integral over the rectangle A can be represented as the iterated integral I = integraldisplay 4 parenleftbiggintegraldisplay 4 − 4 3 xy 2 9 + x 2 dy parenrightbigg dx , Rehman (aar638) – Makeup Exam 2 – sachse – (56620) 2 integrating first with respect to y . Now after integration with respect to y with x fixed, we see that integraldisplay 4 − 4 3 xy 2 9 + x 2 dy = bracketleftBig xy 3 9 + x 2 bracketrightBig 4 − 4 = 128 x 9 + x 2 . But integraldisplay 4 128 x 9 + x 2 dx = bracketleftBig 64 ln(9 + x 2 ) bracketrightBig 4 . Consequently, I = 64 ln parenleftBig 25 9 parenrightBig . 004 10.0 points Find the value of the double integral I = integraldisplay integraldisplay A (8 x- y ) dxdy when A is the region braceleftBig ( x, y ) : y ≤ x ≤ √ y, ≤ y ≤ 1 bracerightBig . 1. I = 7 10 2. I = 1 3. I = 9 10 4. I = 4 5 5. I = 3 5 correct Explanation: The integral can be written as the repeated integral I = integraldisplay 1 bracketleftBigg integraldisplay √ y y (8 x- y ) dx bracketrightBigg dy....
View Full Document
{[ snackBarMessage ]}
### Page1 / 8
Make up 2 - Rehman(aar638 – Makeup Exam 2 – sachse...
This preview shows document pages 1 - 3. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 1,121 | 3,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-09 | latest | en | 0.651289 |
http://squidarth.com/rc/math/2018/06/24/fourier.html | 1,579,629,084,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604849.31/warc/CC-MAIN-20200121162615-20200121191615-00056.warc.gz | 159,321,398 | 8,063 | # The Discrete Cosine Transform in Action
Sid Shanker -
Last week, Lucy Zhang and I worked on compressing images with the Discrete Cosine Transform (DCT).
I was pretty blown away by how well this ended up working, and in this post, I’m going to explain the intuition behind why. If you’re curious about the work that we did, check out our code.
But first, what is compression?
## A note on compression
In addition to being a major plot element on HBO’s Silicon Valley, compression is also the process of reducing the amount of space needed to represent data. In practice, that “data” could be text, images, video, audio, or really anything else. There are two main types, lossless, and lossy. In lossless compression, 100% original data can be recovered in the decompression process, while with lossy compression, “uninteresting” pieces of the original data are thrown out, while the important structure is retained. What consitutes “uninteresting” pieces of data is up to the compression algorithm used.
Compression with DCTs is a form of lossy compression–here are our results:
Before:
After:
As you can see, the compressed image is a little fuzzier than the original, but the core features of the image are still recognizable.
Compressed even further:
Because we are using lossy compression, the amount of “loss” we are willing to tolerate is configurable. Here, we compress the image even further. There is a very distracting blur on the image now, and a lot of detail is missing. However, the core structures of the image, like the baboon’s nose and eyes, can still be discerned.
This is pretty incredible. Without our algorithm “knowing” anything about baboons, it was able to “figure out” what the core structures of the image are.
# What is a Fourier transform?
The DCT is a a type of Fourier-related transformation. A Fourier transform is the process of decomposing some kind of digital signal into the sum of some trignometric functions.
So for instance, if you have audio signal like the following:
By Fourier1789 [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)], from Wikimedia Commons
applying a Fourier transform would yield a bunch of different sine and cosine waves of different frequencies that sum together to be the equivalent of the original audio wave.
It’s called a transform because it transforms the data from one form, the amplitude over time, into a list of coefficients, where each coefficient corresponds to a different trignometric function and governs “how much” of that function to use in order to get the original wave.
By Phonical [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)], from Wikimedia Commons
It’s worth noting that these are equivalent ways of encoding the sound wave. This is a fundamental property of Fourier transforms–no data is lost converting between these two “formats” of encoding the data.
# What does this have to do with images?
Similarly to audio, image data can also be modeled by waves. Let’s start by understanding how to deal with images as numerical data.
For grayscale images, we can model them as a matrix of values, where the element at position (i,j) in the matrix corresponds to the pixel at position (i,j) in the image, and the value of that matrix element is the pixel’s intensity. 0 corresponds to black, and 255 corresponds to white.
It may not be obvious that any image data is “wave-like”, so let’s try graphing some image data!
From the baboon image, let’s take a row of pixels from halfway down the image (it’s 256 pixels long, so this picture is only the first few):
Graphing the values of an entire row of pixels, we get a picture that looks like this:
This is pretty cool! Remembering that 255 represents the color white, we get two nice peaks, which represent the nose of the baboon, which has some dark coloring in the middle.
It’s clear from this picture that the pixels follow a wave-like pattern, and it’s not hard to imagine that this could be represented as the sum of some trignometric functions.
## What does this have to do with compression?
There are some clues in the graph above as to how figuring out how the decomposition of that picture into different sine and cosine waves might be useful in compressing the image.
Specifically, it’s clear that some of the structure in the graph is very important to preserve. Notably, it’s very important to preserving the overall structure of the image to preserve the peaks in the middle of the graph, as we’d want any compressed image to still have a nose.
However, some of the higher frequency components, like the smaller changes in amplitude leading up to those peaks, are less important, and could be removed without us losing the notion of a “nose”. Once we have decomposed the image into a bunch of trig functions, it becomes easy to remove higher frequency functions that don’t contribute as much to the core structures of the image. As I mentioned in a previous section, it is a property of Fourier transforms that they are reversable–given the trig functions and the coefficients, the original data can be recovered. So after removing the higher frequency components, a new, compressed image can be generated.
# How do we perform the “decomposition”?
That concludes the main intuition behind why breaking down an image in a sum of trignometric function can help in image compression.
The next question, then, is how exactly one goes about decomposing the image data into component trig functions.
The Discrete Cosine Transform
The mechanism that we’ll be using for decomposing the image data into trignometric functions is the Discrete Cosine Transform. In this post, I won’t be going deep into how the math works, and will be a little hand-wavy, so if you’re interested in going further, the wikipedia page is a great starting point.
I’ll start by considering how this would for a single row of pixels.
The DCT is a linear transformation that transforms a vector of length n containing “amplitudes”, and returns a different vector of length n containing the coefficients for n different cosine functions. Therefore, it is encoded by an n x n matrix, in which each row corresponds with a cosine function of a different frequency.
### Why use n cosine waves?
An important feature that is required in order for our image compression to work properly is that the transformation that we use to decompose the data into cosine functions must be invertible. If used fewer or greater than n cosine functions, then we would not be able to represent the transformation as a square matrix, and it would therefore not be invertibile, which is the key to being able to get our data back in terms of amplitudes after converting it to cosine coefficients.
## Generating the cosine functions
Alright, so we know now that the DCT for a length n array of pixels will be represented by an n x n matrix, where each row in the matrix corresponds to a different cosine wave. The next question then is how do you represent a cosine wave as a row in a matrix?
If X is the transformation matrix, the matrix values are given by:
Here, i, indicates the row we are on, and j the column within that row. N is the total size of the matrix, so both i and j range from 0 to N.
If we think of each row as corresponding to a different cosine function, we can see that higher values of i correspond to cosine waves of higher frequency. To prove this to ourselves, we can graph some rows for some values of i.
## Performing the decomposition
The next step, after computing the DCT matrix, is actually performing the decomposition and figuring out the right coefficients for each of the component waves. The short answer here is that this is a simple matter of matrix multiplication–we take our input vector of pixel intensities $v_1$, and multiply it by X, and get our coefficient vector $v_2$, where each value in $v_2$ corresponds to a coefficient.
But what is really happening here? When we perform matrix multiplication we are for each row in X taking the dot product of $X_i$ and $v_1$. The dot product can be interpreted as a measure of similarity between two vectors- -how much of $v_1$ is in $X_i$? If the pixel data is coincident with the values in one particular wave, the value of the dot products, and if it is orthogonal, it will be 0. Therefore, by computing this dot product, we can figure out what coefficient to use for that particular wave.
## Generalizing this to two dimensions
Great, so now we can get our decomposition for a single row of pixels. Alas, single rows of pixels tend to not be super interesting, how do we get this to work in two dimensions?
The naive solution is to just apply this transform for each row of pixels. However, the issue with this is that there is signal and structure in the columns that we miss if we don’t take that into account. This can be remedied by performing the DCT twice–once along the rows, and then again along the columns.
## Performing the compression
Once we’ve computed the DCT transform in two dimensions, what we’re left with is an n x n matrix–with each value in the matrix corresponding to the coefficient of a cosine function, rather than the amplitude values of the original matrix.
Let’s say the original resolution of the image is 256 x 256 pixels. This means that the DCT transform matrix will have dimensions of 256 x 256. The way the compression works is that you take the K most signalful cosine waves, and save the coefficients for those to disk. Say, for instance, we choose to save the coefficients for 100 functions. What we end up saving to disk is a 100 x 100 matrix, instead of the full uncompressed 256 x 256 matrix.
A 5x5 DCT matrix compressed to 3x3
When we want to read the file back off the disk, pad the matrix with 0’s to get a 256 x 256 matrix, and then put it through the inverse DCT to get the compressed image.
What one chooses to compress to completely depends on the application and use case for the compression. The tradeoff for different sizes of K is that with higher values of K, the quality of the image will be higher, while for lower values, the size on disk will be lower.
# Why the DCT?
The DCT is only one of many Fourier-related transforms, each of which decomposes a signal into the sum of different combinations of sine and cosine wave. It’s worth noting that image compression can be done with other types of transforms, for instance, the Discrete Fourier Transform (DFT).
The main difference between the DCT and DFT is that the DFT does not just use cosine waves–the waves that it uses are a combination of sine and cosine waves. I’m not going to get into the details as to why here, but in practice, the DCT typically allows for more signal to be retained with fewer functions.
# Final Thoughts
I hope that you are as blown away by the power of the DCT as me. If you’re curious about digging more deeply into the math behind the DCT, and proving to yourself the statements that I waved my hands over, I’d highly recommend the wikipedia page as a good starting point.
My biggest takeaway from this exercise is how incredible it is that a fairly simple math operation can be used to used to discern major features from an image. And even more impressive is that the DCT can be used to store the image in a more semantically meaningful way that in an array of pixels–it can actually be stored in a way that encodes information about broader structures of the image. I’m very curious now as to whether this kind of encoding is used to prepare data for machine learning algorithms.
I’d like to also thank Lucy again for introducing me to this topic and for working on this project with me.
Sid Shanker <sid.p.shanker at gmail.com> | 2,535 | 11,743 | {"found_math": true, "script_math_tex": 7, "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-2020-05 | latest | en | 0.940454 |
https://brilliant.org/discussions/thread/possible-violations-of-the-second-law-of/ | 1,627,333,929,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.92/warc/CC-MAIN-20210726183622-20210726213622-00532.warc.gz | 153,962,969 | 18,256 | # Possible Violations of the Second Law of Thermodynamics (and Perpetual Motion)
One statement of the second law of thermodynamics is that the entropy of a closed system is always increasing with natural thermodynamic processes. Originally, this was a purely empirical observation, so some physicists set out to disprove the law in hopes of finding a more general law. None truly succeeded in doing so. It is still, to a great extent, an empirical law, but we understand the math behind it better as a consequence of statistical thermodynamics.
For the past 20 years, researchers have tried to find exceptions to the second law. Surprisingly, some have (more or less) succeeded. It has been about a year since one of the most recent examples: here. It has been about 13 years since one of the most widely circulated examples: here. A possible fundamental violation of the second law happened here.
Now I ask you the following questions:
$\text{1) Out of the 3 examples above, which (if any) of them}$$\text{ actually violated the classical second law, and to what extent?}$
$\text{2) Can you think of any potential violations of the 2nd law?}$
$\text{3) Due to these violations, is it actually possible to}$ $\text{achieve a perpetual motion machine?}$
Answer whichever of these you want to.
Note by Caleb Townsend
6 years, 5 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
As the Second Law is in essence a statistical one, there is a non-zero probability that it can be "violated", i.e., that the entropy of an isolated system can spontaneously increase. As your examples show, this can be found to occur over short periods of time for microscopic systems. The Fluctuation Theorem quantifies the probability that this increase can occur. The last example you cite is particularly impressive; apparently absolute zero isn't absolute after all. These examples also show how exciting the future is for nanotechnology; if a perpetual motion machine is possible, this is the field in which it will be found.
- 6 years, 5 months ago
The second law of thermodynamics is a circular law. Yes heat generally flows from hot to cold. It only makes sense. No need for the second law. Moreover common senses: Heat will generally flow from where the thermal energy density is highest to lowest. Nothing brilliant here.
As for the second law's accomplice, i.e. entropy: the entropy of a system of molecules increases. Molecules will tend to disperse. Nothing scared in that. Its no different than billiard balls on a table - hit the balls and they disperse. So what. Yet the idea that entropy increases is taken to have a demigod status, while the dispersion of molecules is not. Come on people think.
The second law was formulated to explain work lost by cyclic heat engines in the 19th century, i.e the Carnot engine. Wherein a beautiful eloquent math was equated to PdV, that being work that is commonly lost, or if you prefer we often write: TdS=E+PdV.
And since this lost work is continually experimentally found to be equal to: PdV, well we all believe in what is taught verbatim. Come on it only equals PdV because in the 19th century we equated it to PdV
We all have been taught that the second law is demigod law. It is not. It is a circular law, with very limited validity. everything that we claim is due to the second law can be explained in other simpler terms.
May I suggest that you see: Peer reviewed: Note the referees biggest concern was that this paper takes down 150 yrs of indoctrination based upon such a simple argument. 1) http://physicsessays.org/browse-journal-2/product/1173-24-kent-w-mayhew-second-law-and-lost-work.html
2) https://www.createspace.com/5277845
3) www.new-thermodynamics.com
Concerning 1) where the referee was more concerned with the fact that such a simple argument can take down the second law. Has he never heard of Occam's razor.
Anyhow the cat is hopefully out of the bag and may we dispose of this second law BS once and for all.
sincerely Kent Mayhew
- 6 years, 4 months ago
I agree, I think the second law is equivalent to the statement "hot is hotter than cold by definition," although it helped shape early thermodynamics. Furthermore I think entropy is a bad approximation of the truth, since macroscopic mechanics (and the ideal gas law) don't really apply to gases as easily. But now we have supercomputers that can accurately simulate heat transfers throughout a gas and entropy seems silly in retrospect.
My main problem with the second law, though, is how professors cling to it as a constant axiom of physics, like Newton's laws of motion. I don't think it will remain on its throne for much longer. In the words of a contemporary physicist, news of inconstancy may lie ahead.
Now I have a question for you Kent: did you join Brilliant for the purpose of this discussion? If so, I recommend that you look around the website, mostly the community section. There are thousands of community-designed problems in different fields of mathematics and science (and most recently, finance). There are also many community discussions such as this one. I guarantee there must be at the very least $100$ problems that pique your curiosity, so perhaps check the topics that you are interested in, and give a problem a go.
- 6 years, 4 months ago
I joined brilliant mainly for the purpose of spreading the word against the second law, or at least opening the eyes of others to the possibilities that the second law is a bogus circular based law and to the fact that (okay my belief) entropy is nothing more than a mathematical contrivance.
Certainly my interests extend far beyond thermodynamics, so I am open to other discussions. I realize that there are hundreds of other problems.
I do not mean to be rude but at this time in my life I am concentrating upon showing others that both entropy and the second law are the two biggest blunders ever in science. Of course getting others to understand why is another thing. And the biggest issue is how does one get the powers to be ie those who hold strong positions in science to expunge their theoretical stances.
The only way as I see it is to start a bunch of grass roots thought groups such as your
I will try this with you. If I ask you: 1) Does an expanding system displace the Earth's atmosphere? 2) Does the atmosphere have mass?
What would you answer, I assume yes.
Then is not the displacement of this mass (our atmosphere) upwardly against gravity? If you say yes then it requires work. The amount of word for an ideal gas can be shown to be :PdV
If you answer yes and accept what I say then you are ahead of the experts in thermodynamics.
Because to acknowledge that an expanding system must displace our atmosphere's mass hence requires work is something an expert in thermodynamics does not want to acknowledge. They prefer to write TdS=dE+PdV and then use a whole bunch of partial derivatives to confuse us into thinking they are right when they are not
PdV is actually the basis of lot work or if you prefer work lost by expanding systems. Please understand that a reason that entropy and the second law were conceived in the 19th century by the likes of, Clausius, Boltzman, Lord Kelvin and Maxwell ( to name a few) was to explain why energy was lost by cyclic heat engines - namely the Carnot engine. Moreover they formulated the fundamental of statistical mechanics/physics and equated things to PdV or if you prefer as I said the set: TdS=dE+PdV
No wonder thermodynamics has become such an illogical mess
Another issue now becomes; like a heard of lemmings we all (including me) fell into the void of believing what we were taught in university without ever questioning it's sanity. why because we spent so much time trying to figure out how to navigate thru all these shuffling partial derivatives that we lost sight of any logic
It turns out that entropy is a mathematical contrivance with limited real meaning - now wonder everyone has a different interpretation as to its meaning.
And that the second law is limited in its truest context - so limited that it should not apply to anything , yet we have been taught that it applies to everything
Furthermore entropy and the second law should not apply to the universe, like the big bang.
Entropy has nothing to do with the arrow of time
No wonder paradoxes arise when we apply the second law to black holes and to think Hawkins got a noble prize for his blunder in these matters
the list goes on and on- why because it is the biggest scientific blunder ever beheld by the human race - it actually makes us look like a ship of fools to any (assuming it exists) intelligent life in this universe - I am assuming that intelligent life exists somewhere but not necessarily here
If you care to have further discussion I am open.
If you feel it does not warrant discussion in Brilliant then I accept that to and will move on
Sincerely Kent
- 6 years, 4 months ago
I am sorry to simply re-post but i want to try something with you. (anyhow since it is now out in my peer reviewed paper, there can be no harm done: my paper is entitled "Second Law and Lost Work" in March 2015 edition of journal "Physics Essays"
I found Brilliant by googling the second law: The article above states "One statement of the second law of thermodynamics is that the entropy of a closed system is always increasing with natural thermodynamic processes."
The statement is wrong in that it should say: the second law of thermodynamics is that the entropy of an isolated closed system is always increasing with natural thermodynamic processes.
You see "isolated" is fundamental to all of this. An isolated closed system does not exchange energy (or interact) with its surroundings. Now consider an expanding system here on earth. since this expanding system here on Earth must displace our atmosphere, then it must be doing work here on Earth - that being the work required to upwardly lift our atmosphere's gases. Since the expanding system here on Earth is doing work onto its surroundings (the atm) then it cannot be an isolated system.
In other words there are very few useful systems here on Earth that the second law even applies to. By useful I mean systems that we can use to power machinery and/or move men. so if the second law does not apply to useful systems (cyclic engines, expanding systems etc etc) here on Earth, then why do we constantly use it to explain so much.
So as much as you may consider my addition of isolated to the statement on this page, it is a fundamental addition that renders the second law of thermodynamics into a useless law.
Moreover the original statement included the word entropy. a word that has little meaning or should I say means something different to everyone. I assume that by the context of the word, they are implying entropy to mean "the randomness of molecules in incessant motion" (the mid 20th century meaning) or perhaps Frank Lamberts definition " the dispersal of a system's molecule's energy"
In so far as I love those definitions can someone please tell me what randomness has to do with energy?
Perhaps entropy should just be the heat capacity of an inhomogeneous closed system i.e. the old: TS=systems energy
sounds great but consider a N molecule ideal gas; by kinetic theory its energy is : 3NkT/2 so does that mean that I can write TS=3NkT/2?
sounds fine by what about enthalpy TS=E+PV which then is transformed by the ideal gas law PV=NkT So now can I write: TS = NkT
If so then why does it not match what I wrote earlier for the total energy of the gas
The point I am making is that entropy cannot be two different things. It cannot
1) be something which when multiplied by temperature gives a systems enthalpy and at the same time 2) be something which when multiplied by temperature gives a system's energy
Sure it can be one or the other but it cannot be both. and then I invite you to tell me what does enthalpy or the energy of a gas have to do with randomness anyhow.
cheers Kent
- 6 years, 4 months ago | 3,097 | 13,684 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.936484 |
https://dh-data.org/posts/2022-01-30-velocity-of-money/ | 1,718,807,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00686.warc.gz | 176,238,290 | 19,000 | # velocity of money
MV = PY illustrates the problem but is tautological
David Harper https://www.bionicturtle.com/
2022-01-03
The quantity equation is given by M * V = P * Y. My motivation here is: although recent CPI (inflation) numbers are higher (> 5.0%), they aren’t nearly as high as you might expect given the dramatic jump in the quantity of money (I’m just looking at M1 here). A common explanation I’ve read is the same that Cathie Wood recently gave (in tweet reply to Jack Dorsey now-infamous hyperinflation claim) here at her twitter account. She said “velocity is still falling.”
FRED key (private), echo=FALSE
``````library(fredr)
library(tidyverse)
library(patchwork)
library(ggthemes)
obs_start = as.Date("2016-07-01")
obs_end = as.Date("2021-04-01") # not used
Y_realGDP = "GDPC1"; P_deflator = "GDPDEF"
M_quantity = "M1NS"; V_velocity = "M1V"
fredr_set_key(davidh_key)
params <- list(
series_id = c(Y_realGDP, P_deflator, M_quantity, V_velocity),
frequency = "q"
)
t1 <- pmap_dfr(
.l = params,
.f = ~ fredr(series_id = .x, frequency = .y)
)
t2 <- t1 %>% pivot_wider(names_from = series_id) %>% select(!starts_with("realtime"))
t3 <- t2 %>% filter(date >= obs_start)
a1 = 8000; b1 = 100
sz_line = 1.5
clr_Y = "darkorange2"
clr_P = "firebrick1"
clr_M = "darkgreen"
clr_V = "darkorchid3"
p1 <- ggplot(t3, aes(x = date)) +
geom_line(aes(y = GDPC1), size = sz_line, color = clr_Y) +
geom_line(aes(y = GDPDEF * b1 + a1), size = sz_line, color = clr_P) +
scale_y_continuous(
name = "Real GDP (Y)",
sec.axis = sec_axis(~ (. - a1)/b1, name = "Price Level (P)")
) +
theme_few() +
theme(
axis.title.x = element_blank(),
axis.title.y.left = element_text(color = clr_Y),
axis.text.y.left = element_text(color = clr_Y),
axis.title.y.right = element_text(color = clr_P),
axis.text.y.right = element_text(color = clr_P)
)
a2 = 0; b2 = 1/1000
p2 <- ggplot(t3, aes(x = date)) +
geom_line(aes(y = M1V), size = sz_line, color = clr_V) +
geom_line(aes(y = a2 + M1NS * b2), size = sz_line, color = clr_M) +
scale_y_continuous(
name = "Velocity (V)",
sec.axis = sec_axis(~ (. - a2)/b2, name = "Quantity of Money (M)")
) +
theme_few() +
theme(
axis.title.x = element_blank(),
axis.title.y.left = element_text(color = clr_V),
axis.text.y.left = element_text(color = clr_V),
axis.title.y.right = element_text(color = clr_M),
axis.text.y.right = element_text(color = clr_M)
)
p1 + p2
`````` | 746 | 2,396 | {"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-2024-26 | latest | en | 0.664902 |
biochemistryquestions.wordpress.com | 1,561,634,746,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001089.83/warc/CC-MAIN-20190627095649-20190627121649-00130.warc.gz | 374,341,750 | 22,357 | # Total Oxidation of a 17 carbon fatty acid, including oxidation of the resulting Propionyl CoA
Beta-oxidation of fatty acids with an odd number of carbons.
Energetic balance of the total (and I mean total) oxidation of a fatty acid with an odd number of carbons.
Oxidation of a fatty acid with 17 atoms of carbon.
Activation of the fatty acid to Acyl CoA = -2 ATP
Number of rounds in the Beta oxidation
(17/2) -1.5 = 8.5-1.5 = 7
7 rounds x 5 ATP/ round = 35 ATP
Number of units of Acetyl CoA produced = 7 Acetyl CoA
7 Acetyl CoA x 12 ATP/Acetyl CoA = 84 ATP
Propionyl CoA up to Succinyl CoA = -1ATP
Succinyl CoA up to Malate = 3 ATP
Malate up to Pyruvate (1 NADPH.H+)
Pyruvate up to Acetyl Co A = 3 ATP
Acetyl CoA oxidation in the Krebs Cycle = 12 ATP
Total of ATP (considering the total oxidation of Propionyl CoA converted to Malate and then from Malate to Pyruvate and then from Pyruvate to Acetyl CoA = -2 + 35 +84 -1 + 3 +3 +12 = 134 ATP
In summary (following the equivalence of 1 NADH.H+ yielding 3 ATP in the Respiratory Chain and 1 FADH2 yielding 2 ATP):
-Calculate the number of rounds of the fatty acid in the Beta-oxidation:
Number of rounds = (Number of carbons/2) -1.5
-The number of Acetyl CoA is the same as the number of rounds
-Subtract 2 ATP that were used in the initial activation of the fatty acid.
-Multiply the number of rounds x 5 ATP/round.
-Multiply the number of Acetyl CoA x 12 ATP/Acetyl CoA.
-Add 17 ATP produced in the total oxidation of Propionyl CoA to CO2
To practice this kind of exercise, I suggest that you do the calculations using now the criteria that considers that each NADH.H+ oxidized in the Respiratory Chain yields 2.5 ATP and each FADH2 yields 1.5 ATP
# Energetic Balance of the Total (and I mean Total) Oxidation of a Fatty Acid with an Odd number of Carbons.
In previous posts we have discussed how the fatty acids with an odd number of carbons chain, release 1 unit of Acetyl CoA and 1 unit of Propionyl CoA, instead of the two Acetyl CoA units released when a fatty acid with an even number of carbons is beta-oxidized.
Let’s use some examples, (we represent here just the carbons in the chains):
Example 1: a fatty acid with 6 carbons (Hexanoic acid)
C-C-C-C-C-C-C
During the Beta-oxidation, three units of Acetyl CoA are released (two carbons each):
C-C/C-C/C-C
Example 2: A fatty acid with 7 carbons (Heptanoic acid):
C-C-C-C-C-C-C
During the Beta-oxidation two units of Acetyl CoA and one unit of Propionyl CoA are released (two units of two carbons and one unit of three carbons):
C-C-C/C-C/C-C
As discussed previously in another post, the Acetyl CoA are oxidized in the Krebs Cycle, but the Propionyl CoA is used in the formation of Succinyl CoA, in a process that consumes 1 ATP (-1 ATP).
The Succinyl CoA can continue in the Krebs Cycle and form Oxalacetate. Oxalacetate will react with Acetyl CoA (Citrate Synthase reaction) to form Citrate, following the reactions in the Krebs Cycle. If it happens, we can consider that the atoms of carbons of the Propionyl CoA have followed an anaplerotic pathway (to be used in the Kreb’s Cycle without being consumed).
BUT these carbons could also be completely oxidized if they follow this sequence of reactions:
Succinyl CoA + GDP + (P) – -> Succinate + CoA + GTP (it is equivalent to 1 ATP)
Succinate + FAD – – – – >Fumarate + FADH2 (It generates 2 ATP in the Respiratory Chain)
Fumarate + H2O—–> Malate
But the Malate can now diffuse from the matrix through the mitochondrial membranes and be decarboxylated (under the action of the cytoplasmatic malic enzyme) to Pyruvate (and production of 1 CO2).
Pyruvate can return to the interior of the mitochondria, where another decarboxylation occurs, this time under the action of the Pyruvate dehydrogenase complex, (with production of another CO2) and the formation of Acetyl CoA, whose Acetyl group will be oxidized in the Cycle producing other two molecules of CO2.
We can see that through this sequence of reactions it is possible the total oxidation of the three original carbons of the Propionate to 3 molecules of CO2! (To avoid confusions, observe that, yes, there are 4 decarboxylations, but one of the CO2 does not come originally from the Propionyl CoA, but from the carboxylation process in the conversion of Propionyl CoA to Succinyl CoA)
Which would be the energetic balance of the total oxidation of an odd chain fatty acid considering this sequence of reactions?
Let’s see:
Propionyl CoA to Succinyl Co A = -1 ATP
In the mitochondria, (following the reactions of the Kreb’s Cycle up to Malate):
Succinyl CoA + GDP + (P) –> Succinate +CoA + GTP (it is equivalent to 1 ATP)
Succinate + FAD ——– – – – > Fumarate + FADH2 (It generates 2 ATP in the Respiratory Chain)
Fumarate + H2O————–>Malate
In the cytoplasm:
Malate + NADP+ – – – >Pyruvate + NADPH.H+ + CO2 (we will not consider this reduced cofactor in the balance since NADPH.H+ is not a source of energy, but a source of reduction equivalents for synthetic reactions)
In the mitochondria again:
Pyruvate + CoA + NAD+ —-> Acetyl CoA + CO2 + NADH.H+ (Observe that this NADH.H+ is generated inside the mitochondria, so it yields 3 ATP)
The Acetyl CoA produced in the previous reaction, when oxidized in the Krebs Cycle: 12 ATP
Therefore, considering this metabolic way,
-1 +1 +2 +3 + 12 = 17 ATP as a result of the total oxidation of the Propionyl CoA generated by the beta-oxidation of a fatty acid of odd number of carbons.
Therefore, for calculating the energetic balance we should add 17 ATPs from the oxidation of the Propionyl CoA, to the ATPs generated in the Beta-oxidation, and the ATPs generated as a result of the oxidation in the Krebs Cycle of the Acetyl CoA units formed during the Beta-oxidation of the odd chain fatty acid. ( We should recall also that 2 ATPs are consumed in the initial activation of the fatty acid)
In our next post we will analyze the oxidation of the heptadecanoic acid (17 carbons) as an example of the application of these calculations.
# Oxidation of a fatty acid with 17 atoms of carbon
(This post analise the energetic balance considering that the Propionyl CoA follows an anaplerotic fate)
Apply the equations described in the previous post:
N= Number of Carbons
(N/2) -1.5 = Number of rounds in Beta-oxidation
(N/2) -1.5 = Number of acetyl CoA produced in Beta-oxidation
So, in terms of production and consumption of ATP of ATPs, the oxidation of a 17-carbons fatty acid will show the following energetic balance:
Activation of a fatty acid to Acyl CoA = -2 ATP
Number of rounds in Beta-Oxidation:
(17/2) – 1.5 = 8.5 -1.5 = 7
7 rounds x 5 ATP/round = 35 ATP
Number of produced Acetyl CoA: 7 Acetyl CoA
7 Acetyl CoA x 12 ATP/Acetyl CoA = 84 ATP
Additionally, the Beta-oxidation has produced 1 Propionyl CoA. The conversion of Propionyl CoA to Succinyl CoA, as described in a former post, will consume 1 ATP (Consider -1 ATP).
As described in a previous post:
We can consider the conversion of Propionyl CoA to Succinyl CoA as an anaplerotic pathway, in which case, the molecule of Succinyl CoA continue in the Krebs Cycle generating Oxalacetate in the following sequence of reactions:
Succinyl CoA + GDP + (P) —> Succinate + CoA + GTP (equivalent to 1 ATP)
Succinate + FAD ———-> Fumarate + FADH2 (Generates 2 ATPs in the Respiratory Chain)
Fumarate + H2O————– > Malate
Malate + NAD+ —————-> Oxalacetate + NADH.H+ (Generates 3 ATPs in the Respiratory Chain) “
Total = (-1+1+2+3) = 5 ATPs
Total of ATPs produced (considering the anaplerotic fate of the Propionyl CoA turned into Succinyl CoA) = -2+35+84-1+6 = 122 ATP
BUT…
We may also consider a total oxidation that would include the carbon atoms of the Propionyl CoA!
In our next post, we will consider what happens if, instead of the Succinyl CoA following an anaplerotic pathway, it follows a path that allow the carbon atoms of Propionyl CoA end up being oxidized to CO2. It would allow a real total oxidation of all the original carbons of the heptadecanoic acid or any other fatty acid with an odd number of carbons.
# Oxidation of fatty acids with an odd number of carbons
Reviewing the subject, I found a lack of detailed information in the literature available on the Internet, and even in texts of Biochemistry that are frequently used in the Schools of Medicine and Biochemistry courses in other academic careers. This lack of information is the result of the fact that the bulk of fatty acids in our bodies, (and in the diet we consume) are generally fatty acids with an even number of carbons.
Referring to the oxidation of odd-chain fatty acids, texts and articles usually are limited to report that in the last round of the Beta-oxidation of fatty acids of this type, one Propionyl CoA and one Acetyl CoA are produced, and then these texts generally describe the conversion of Propionyl CoA to Succinyl CoA, but without specifically mentioning the ATP balance in these reactions.
Therefore, and answering your questions, I have included in this post the energetic considerations to take into account when analyzing the ATP production in the oxidation of fatty acids of an odd number of carbons:
N= number of carbons
(N/2) –1.5 = Number of rounds in the Beta-oxidation
(N/2) –1.5 = Number of acetyl CoA produced in the Beta-oxidation
Additionally, 1 Propionyl CoA (3-carbon Acyl CoA) is produced in the last round.
Based on a yield of 3 ATP per NADH.H+ and 2 ATP per FADH2 that are oxidized in the respiratory chain:
-Multiply the number of turns in Beta-oxidation x 5 ATP / turn
Multiply the number of Acetyl CoA x 12 ATP / Acetyl CoA (since each Acetyl CoA yields 12 ATP when oxidized in the Krebs Cycle)
Subtract now two ATP (-2 ATP) consumed in the initial activation of the fatty acid (see the related post for explanation)
But also, in this process, as was written before, 1 Propionyl CoA is released, because in the last round of the Beta-oxidation, instead of obtaining two Acetyl CoA (as with the even chain fatty acids), the odd fatty acids now yields 1 Acetyl CoA and 1 Propionyl CoA.
What happens with this Propionyl CoA?
The Propionyl CoA must undergo a carboxylation in a sequence of reactions requiring Biotin and Vitamin B12. These reactions produce Succinyl CoA.
Note also that this sequence of reaction requires the consumption of 1 ATP (then consider it as -1 ATP )
What happens to the carbon atoms of the Succinyl CoA?
Let’s discuss an anaplerotic fate for the Succinyl CoA:
We can consider the conversion of propionyl CoA to Succinyl CoA as an anaplerotic pathway, in which case, the molecule of Succinyl CoA continue in the Krebs Cycle generating Oxalacetate in the following sequence of reactions:
Succinyl CoA + GDP + (P) —> Succinate + CoA + GTP (equivalent to 1 ATP)
Succinate + FAD ———-> Fumarate + FADH2 (Generates 2 ATP in the Respiratory Chain)
Fumarate + H2O————– > Malate
Malate + NAD+ —————-> Oxalacetate + NADH.H+ (Generates 3 ATP in the Respiratory Chain)
Remember that by definition the products of the anaplerotic reactions are incorporated into the Krebs Cycle, increasing its activity, but without being oxidized. In this case, because of the sequence of reactions experienced by the Succinyl CoA up to Oxalacetate, we can consider that the incorporation of the Succinyl CoA originated from the Propionyl CoA, has generated 6 additional ATPs.
Therefore, considering the “anaplerotic” fate of the Propionyl CoA:
– Add (-1 +1 +2 +3) = 5 ATP to the previous calculations.
In our next post, we will use an example. We will apply this information in the calculation of the energetic balance of the Beta-oxidation of the decaheptanoic acid (17 Carbons), assuming that the propionyl CoA has followed the anaplerotic pathway up to oxalacetate, as described in this post.
# How to calculate QUICKLY the energetic balance of the total oxidation of a fatty acid
The basis for the calculation and how to do it have been explained in detail in other post.
Anyway, since time is an issue in exams, some readers are interested in knowing a method of doing the calculation in a real fast way.
For obtaining a quick answer when calculating the energetic balance of the total oxidation of a fatty acid (and also for verifying your answers if you use some other method of calculation) you can apply these formulas:
a) If your course use for the reduced cofactors these energetic yields
Use this formula:
[(n/2) -1] x (14) +8 = Number of ATP formed
Where n= Number of carbons
Example:
Balance of the total oxidation of palmitic acid (16 carbons):
(16/2)-1] x (14) +8 = 106 ATP
b) If the equivalence used for the energetic yields in your course are:
Then apply this formula:
[(n/2) -1] x 17 +10 = Number of ATP formed
For the total oxidation of palmitic acid, following the former energetic yielding criteria:
[(16/2)-1] x 17 + 10 = 129 ATP
Thanks for the questions and comments!
# How to calculate the energetic balance of the total oxidation of a fatty acid
The total oxidation of a fatty acid comprehends different processes:
1. – The activation of the fatty acid
2. – Beta-Oxidation
3. – Krebs Cycle.
For being metabolized, a fatty acid should experiment activation:
Fatty acid + CoA + ATP —-à Acyl CoA + AMP + 2(P)
The activation of the fatty acid requires 1 molecule of ATP, but since two energy rich bonds are hydrolyzed (the ATP is hydrolyzed to AMP and 2 (P) ) for energetic balance purposes it is considered that 2 ATP have been consumed in this activation process)
The formed acyl CoA will experiment different oxidation reactions. These reactions occur in the Beta-carbon. That is why the process is called Beta-oxidation
Recalling:
CH3 -……………………………….. -CH2 – CH2 – CH2 – CH2 -COOH
(Omega carbon) -Delta-Gamma-Beta-Alpha-Carboxyl
In Beta-oxidation (a mitochondrial process) the acyl CoA is totally oxidized to Acetyl groups in form of Acetyl CoA units.
Beta-Oxidation
How many Acetyl CoA units will be formed?
Since the Acetyl group of the acetyl coA is formed by two carbons, we should divide the number of carbons in the acyl group between two.
Miristic acid (14 carbons): 14 carbons /2 = 7 Acetyl CoA
Palmitic acid (16 carbons): 16 carbons/2 = 8 Acetyl CoA
In order to be completed degraded to Acetyl CoA, the fatty acid in form of acyl CoA should experiment several “rounds” in the Beta-oxidation process. En each round, an Acetyl CoA is released and, a NADH.H+ and a FADH2 are produced.
We know already how many acetyls CoA are formed from each fatty acid: n/2, where n is the number of carbons.
Now, how many rounds are necessary for converting all the fatty acid to acetyl CoA?
Since in the last round we already obtain two Acetyl CoA, the number of necessary rounds is (n/2) -1
Observe:
Miristic acid (14 carbons)
1st round:
Produce one acyl CoA of 12 carbons and one Acetyl CoA + NAD H.H+ + FADH2
2nd round:
Produce one acyl CoA of 10 carbons and one Acetyl CoA + NAD H.H+ + FADH2
3rd round:
Produce one acyl CoA of 8 carbons and one Acetyl CoA + NAD H.H+ + FADH2
4th round:
Produce one acyl CoA of 6 carbons and one Acetyl CoA + NAD H.H+ + FADH2
5th round:
Produce one acyl CoA of 4 carbons and one Acetyl CoA + NAD H.H+ + FADH2
6th round:
Produce one acyl CoA of 2 carbons and one Acetyl CoA + NAD H.H+ + FADH2
But the acyl CoA of 2 carbons is already an Acetyl CoA, so it is the last round!
So, after 4 rounds, all the Miristic acid has been converted to Acetyl CoA.
Then, what we have obtained as a result of the beta-oxidation of Miristic acid?
7 acetyl CoA
The acetyl CoA units are oxidized up to CO2 and H2O in the Krebs Cycle.
In terms of ATP, the yielding depends on the kind of yielding that is used for reduced cofactors:
a) If during your course it is considered that each NADH.H+ yields 2.5 ATP and each FADH2 yields 1.5 ATP, then
7 acetyl CoA x 10 ATP/AcetylCoA in the Krebs Cycle: 70 ATP
Minus 2 ATP used in the activation (Miristic acid to miristyl CoA)
70+11+9-2 = 92 ATP
b) If during your course it is considered that each NADH yields 3 ATP and each FADH2 yields 2 ATP, then
7 acetyl CoA x 12 ATP/AcetylCoA in the Krebs Cycle: 84 ATP
Minus 2 ATP used in the activation (Miristic acid to Miristyl coA)
84+18+12-2 = 112 ATP
How to calculate the energetic balance of any fatty acid?
Step 1.- Number of Carbons/2 = Number of Acetyl CoA formed.
Step 2.- Number of rounds in the Beta-oxidation necessary for converting the whole fatty acid to Acetyl Co A units: Number of Acetyl CoA minus 1 [(n/2)-1]
Step 3 (Option a) If you consider that each NADH yields 2.5 ATP and each FADH2 yields 1.5 ATP then multiply the number of rounds times 4 and multiply the number of Acetyl CoA times 1o
OR
Step 3 (Option b).- If you consider that each NADH yields 3 ATP and each FADH2 yields 2 ATP then multiply the number of rounds times 5 and multiply the number of Acetyl CoA times 12.
Step 4.- Take two ATP that were used for the activation of the Fatty Acid
Example:
Fatty acid with 12 Carbons
– Option (a):
Step 1: 12/2 = 6 Acetyl CoA
Step 2: 6-1 = 5 rounds
Step 3 (Option a): (5 x 4) + (6 x10)
Step 4 = -2
Total: 78 ATP
– Option (b):
Step 1: 12/2 = 6 Acetyl CoA
Step 2: 6-1 = 5 rounds
Step 3 (Option b): (5 x 5) + (6 x12)
Step 4 = -2
Total: 95 ATP
I would like now that the reader calculates the energetic balance of the total oxidation of a fatty acid with 18 carbons. Assume that the oxidation of each NADH.H+ yields 3 ATP and that each FADH2 yields 2 ATP.
# How to calculate the energetic balance of the total oxidation of a fatty acid?
The total oxidation of a fatty acid comprehends different processes:
1.- The activation of the fatty acid
2.- Beta-Oxidation
3.- Kreb’s Cycle.
For being metabolized, a fatty acid should experiment an activation:
Fatty acid + CoA + ATP —-à Acyl CoA + AMP + 2(P)
The activation of the fatty acid requires 1 molecule of ATP, but since two energy rich bonds are hydrolized (the ATP is hydrolyzed to AMP and 2 (P) ) for energetic balance purposes it is considered that 2 ATP have been consumed in this activation process)
The formed acyl CoA will experiment different oxidation reactions. These reaction occurs in the Beta-carbon. That is why the process is called Beta-oxidation
Recalling:
CH3 -……………………………….. -CH2 – CH2 – CH2 – CH2 -COOH
(Omega carbon) -Delta-Gamma-Beta-Alpha-Carboxyl
In Beta-oxidation (a mitochondrial process) the acyl CoA is totally oxidazed to Acetyl groups in form of Acetyl CoA units.
How many Acetyl CoA units will be formed?
Since the Acetyl group of the acetyl coA is formed by two carbons, we should divide the number of carbons in the acyl group between two.
Miristic acid (14 carbons): 14 carbons /2 = 7 Acetil CoA
Palmitic acid (16 carbons): 16 carbons/2 = 8 Acetyl CoA
In order to be completed degraded to Acetyl CoA, the fatty acid in form of acyl CoA should experiment several “rounds” in the Beta-oxidation process. En each round, an Acetyl CoA is released and, a NADH.H+ and a FADH2 are produced.
We know already how many acetyl Coa are formed from each fatty acid: n/2, where n is the number of carbons.
Now, how many rounds are necessary for converting all the fatty acid to acetyl CoA?
Since in the last round we already obtain two Acetyl CoA, the number of necessary rounds is (n/2) -1
Observe:
Miristic acid (14 carbons)
1st round:
Produce one acyl CoA of 12 carbons and one Acetyl CoA + NAD H.H+ + FADH2
2nd round:
Produce one acyl CoA of 10 carbons and one Acetyl CoA + NAD H.H+ + FADH2
3rd round:
Produce one acyl CoA of 8 carbons and one Acetyl CoA + NAD H.H+ + FADH2
4th round:
Produce one acyl CoA of 6 carbons and one Acetyl CoA + NAD H.H+ + FADH2
5th round:
Produce one acyl CoA of 4 carbons and one Acetyl CoA + NAD H.H+ + FADH2
6th round:
Produce one acyl CoA of 2 carbons and one Acetyl CoA + NAD H.H+ + FADH2
But the acyl CoA of 2 carbons is already an Acetyl CoA, so it is the last round!
So, after 4 rounds, all the Miristic acid has been converted to Acetyl CoA.
Then, what we have obtained as a result of the beta-oxidation of Miristic acid?
7 acetyl CoA
The acetyl CoA units are oxidized up to CO2 and H2O in the Krebs Cycle.
In terms of ATP, the yielding depends on the kind of yielding that is used for reduced cofactors:
a) If during your course it is considered that each NADH.H+ yields 2.5 ATP and each FADH2 yields 1.5 ATP, then
7 acetyl CoA x 10 ATP/AcetylCoA in the Krebs Cycle: 70 ATP
Minus 2 ATP used in the activation (Miristic acid to miristyl CoA)
70+11+9-2 = 88 ATP
b) If during your course it is considered that each NADH yields 3 ATP and each FADH2 yields 2 ATP, then
7 acetyl CoA x 12 ATP/AcetylCoA in the Krebs Cycle: 84 ATP
Minus 2 ATP used in the activation (Miristic acid to Miristyl coA)
84+18+12-2 = 112 ATP
How to calculate the energetic balance of any fatty acid?
Step 1.- Number of Carbons/2 = Number of Acetyl CoA formed.
Step 2.- Number of rounds in the Beta-oxidation necessary for converting the whole fatty acid to Acetyl Co A units: Number of Acetyl CoA minus 1 [(n/2)-1]
Step 3 (Option a) If you consider that each NADH yields 2.5 ATP and each FADH2 yields 1.5 ATP then multiply the number of rounds times 4 and multiply the number of Acetyl CoA times 1o
OR
Step 3 (Option b).- If you consider that each NADH yields 3 ATP and each FADH2 yields 2 ATP then multiply the number of rounds times 5 and multiply the number of Acetyl CoA times 12.
Step 4.- Take two ATP that were used for the activation of the Fatty Acid
Example:
Fatty acid with 12 Carbons
– Option (a):
Step 1: 12/2 = 6 Acetyl CoA
Step 2: 6-1 = 5 rounds
Step 3 (Option a): (5 x 4) + (6 x10)
Step 4 = -2
Total: 78 ATP
– Option (b):
Step 1: 12/2 = 6 Acetyl CoA
Step 2: 6-1 = 5 rounds
Step 3 (Option b): (5 x 5) + (6 x12)
Step 4 = -2
Total: 95 ATP
I would like now that the reader calculates the energetic balance of the total oxidation of a fatty acid with 18 carbons. Assume that the oxidation of each NADH.H+ yields 3 ATP and that each FADH2 yields 2 ATP. | 6,409 | 22,225 | {"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-2019-26 | latest | en | 0.803759 |
https://www.physicsforums.com/threads/spinning-puck-rotational-motion.552347/ | 1,652,766,856,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00094.warc.gz | 1,122,862,574 | 14,176 | # Spinning puck rotational motion
## Homework Statement
A puck, mass m, on the end of a (thin, light) string rotates in a circle of radius r_0
at a speed v_0
on a
frictionless table. The radius of the circle is slowly reduced from its initial value by pulling the string
through a hole in the table.
A. Hence write down an expression for the speed of the mass when the radius is reduced to some radius
r
B. Write down an expression for the tension in the string and show it goes like 1/r^3
v=rw
## The Attempt at a Solution
i split the above problem into two systems. rotational and kinematic. the kinetic energy is
0.5mrw^2-0.5^mr_0w^2. For the rotational system i am unsure how this is linked to tension ?. Is my starting point correct ?.
Spinnor
Gold Member
What is the angular momentum of the system before the string is shortened?
What is the tension?
After the string is shortened does the angular momentum change?
If not, then you can figure the velocity as a function of the strings length and with that figure out the tension.
Good luck! | 259 | 1,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-21 | longest | en | 0.886314 |
http://www.jiskha.com/display.cgi?id=1365449463 | 1,495,880,233,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608936.94/warc/CC-MAIN-20170527094439-20170527114439-00369.warc.gz | 676,781,683 | 3,629 | # chemistry
posted by on .
Which weak acid would be best to use when preparing a buffer solution with a pH of 8.50?
1) an acid with Ka=3.3x10^-9
2) an acid with Ka=5.0x10^-4
3) an acid with KA=6.3x10^-6
4) an acid with Ka=5.0x10^-7
5) an acid with Ka=4.0x10^-5
6) an acid with Ka=2.0x10^-10 | 119 | 292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-22 | latest | en | 0.881902 |
https://vlacs.org/courses/elementary-math-grade-k/ | 1,725,991,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00475.warc.gz | 571,095,231 | 24,349 | Elementary
1.0 Credit
36 weeks
Open
The Elementary math courses inspire students to become critical thinkers and problem solvers. The learners use math as a tool to make sense of and understand the world around them. The courses include media that uses sight and sound to engage students. For example, rhymes, chants, songs, and videos help teach and practice foundational math skills. Students are provided with many practice opportunities that involve both on-screen and off-screen activities.
Please view the Elementary Parents Guide for Grades K-2 with guidance on helping your student transition to online learning and thrive at VLACS.
### Major Topics and Concepts
Segment One
Counting and writing numbers 1–5
Rearranging groups with 2–5 objects
Identifying and writing 0
Using equal, greater than, and less than to compare groups to 5
Comparing groups to 5
Comparing numbers to 5
Counting and writing numbers 6–10
Rearranging groups with 6–10 objects
Using equal, greater than, and less than to compare groups to 10
Counting to compare groups to 10
Counting forward and backward within 10
Identifying 2-D shapes
Finding, sorting, and comparing 2-D shapes
Making composite shapes
Identifying 3-D shapes
Finding, sorting, and comparing 3-D shapes
Sorting objects into categories
Counting and comparing categories by counting
Segment Two
Identifying and using the plus sign
Representing subtraction as taking apart
Identifying and using the minus sign
Solving subtraction equations within 10
Decomposing and representing numbers 5–10
Developing subtraction fluency within 10
Solving real-world subtraction word problems
Counting, writing, and rearranging numbers 11–20
Representing numbers 11–20
Counting to 100 by ones and tens
Counting from any number to 100
Comparing objects by using length, height, volume, and weight
Using nonstandard measurement
### Course Materials
A pupil may enter kindergarten if his/her chronological age will be five before October 1st of the year of entering school. For a complete description of the policy, please see Policy JEB Age of Entrance.
To achieve success, students are expected to submit work in each course weekly. Students can learn at their own pace; however, “any pace” still means that students must make progress in the course every week. To measure learning, students complete self-checks, practice lessons, multiple choice questions, projects, discussion-based assessments, and discussions. Students and families are expected to maintain regular contact with teachers because, when teachers, students, and parents work together, students are successful.
Required Materials – Please view the list of materials before registering. | 582 | 2,693 | {"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} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.905579 |
https://community.ibm.com/community/user/datascience/discussion/level-of-mathematical-skill-for-ai | 1,670,128,691,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00705.warc.gz | 214,656,786 | 251,874 | # Data and AI Learning Group
View Only
## Level of Mathematical skill for AI
• #### 1. Level of Mathematical skill for AI
Posted Mon April 20, 2020 12:10 PM
To what level does one needs to learn the mathematical skills like linear algebra / probability / statistics / calculus for AI/ML...Any resources which one can share on the above topics ...Thanks in Advance ...
------------------------------
Naresh Luthra
------------------------------
• #### 2. RE: Level of Mathematical skill for AI
Posted Tue April 21, 2020 07:18 AM
Hi Naresh!
The basics of calculus, algebra, linear algebra are going to be important. Then, statistics and probability theory should be studied deeper than that.
There are some great courses in coursera on these topics like:
• Introduction to Probability and Data.
• Data Science Math Skills.
• Inferential Statistics.
I hope this helps.
Best
------------------------------
------------------------------
• #### 3. RE: Level of Mathematical skill for AI
Posted Tue April 21, 2020 12:54 PM
That is a really good list to start off @Kalyan Prasad. Thanks.
------------------------------
Leandro Santos
------------------------------
• #### 4. RE: Level of Mathematical skill for AI
Posted Tue April 21, 2020 01:50 PM
@Leandro Santos : I am glad that it helps you. Good luck!
------------------------------
------------------------------
• #### 5. RE: Level of Mathematical skill for AI
Posted Wed April 22, 2020 07:12 AM
Hi Naresh,
Truth is almost every model in AI or Machine Learning is a product of Linear Algebra, informed by Statistics and Probability . If you have a good understanding of these three subjects (Which frankly are relatively easy to grasp unlike Calculus), you'd be fine, no matter how complicated the algorithm may be. Note that deep learning is simply repeated, complex Linear Algebra, some times referred to as Matrix Algebra. In deep learning each node or neuron is simply divided into two parts, the first part computes the complex Linear Algebra, while the second part applies an activation function to the first part such as maybe Sigmoid, Tanh, Relu or Leaky-Relu activation functions.
Now about Calculus, if you already know high school calculus, just do a refresher. If you cannot remember or did not take calculus in school, you do not NEED to know calculus to do AI. If you know Linear Algebra, Statistics and Probability you can grasp any basics or Advanced knowledge of AI or Machine Learning without knowing beyond little about Calculus.
Where knowledge of Calculus is needed in AI is during Back-Propagation, which is when the model computes the Loss, then moves back from prediction to inputs one step at a time, calculating the derivatives or partial derivative outputs of each neuron in the deep neural network. This is the most complicated calculation in the entire field of deep learning or AI in general. So once you understand the general idea of back propagation and how derivatives simply mean slopes and how based on these derivatives the weights and biases of the entire neural networks are updated iteratively until the model accurately and confidently learns all the right parameters. Then you can simply gloss over or avoid going to learn Calculus, which in my opinion is more difficult to grasp than understanding the basics of Linear Algebra, Statistics and Probability.
I'm saying without knowing anything substantial about Calculus itself you can advance far into AI by simply understanding the entire forward propagation, loss computation and back-propagation steps to updating new weights and biases. Plus a good understanding of Linear Algebra, Statistics and Probability.
If you want to find out more take Andrew NG machine learning course or google ML crash course, then take the deep learning course from deeplearning.ai
Cheers.
------------------------------
Lawrence Krukrubo
------------------------------
• #### 6. RE: Level of Mathematical skill for AI
Posted Wed April 22, 2020 11:09 AM
What a great thread! I am later on in my career and am currently trying to shift to this field. Where I need to up my math was on my mind as well.
Thanks OP for asking the question and those that provided such great answers.
------------------------------
Lennea Kernaz
------------------------------ | 896 | 4,316 | {"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-2022-49 | latest | en | 0.824534 |
https://goprep.co/q19-pqrs-is-a-cyclic-quadrilateral-such-that-pr-is-a-i-1njony | 1,606,466,856,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141191511.46/warc/CC-MAIN-20201127073750-20201127103750-00635.warc.gz | 309,296,091 | 44,170 | Q. 195.0( 1 Vote )
# PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR=67° and ∠SPR=72°, then ∠QRS=A. 41°B. 23°C. 67°D. 18°
Given that,
QPR =67°
SPR = 72°
SPQ = QPR + SPR
= 67o + 72o
= 139°
SPQ + QRS = 180° (Opposite angles of cyclic quadrilateral)
139o + QRS = 180°
QRS = 41°
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
RD Sharma | Extra Qs. of Cyclic Quadrilaterals31 mins
Important Questions on Circles29 mins
NCERT | Imp. Qs. on Circles43 mins
Quiz | Lets Roll on a Circle45 mins
Proof of Important Theorems of Circles45 mins | 230 | 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} | 3.390625 | 3 | CC-MAIN-2020-50 | longest | en | 0.794124 |
http://mathhelpforum.com/calculus/197361-grow-rate-big-omega-polynomial-print.html | 1,529,510,793,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00315.warc.gz | 198,993,668 | 3,033 | Grow Rate and Big Omega of Polynomial
• Apr 15th 2012, 10:21 PM
jfk
Growth Rate and Big Omega of Polynomial
Hello everybody,
I'm dealing with this problem already for a couple of days and I got really stuck, I just need to understand in plain English or Spanish (lol) how to deal with this problem.
Here is the problem:
Let $\displaystyle f(n)$ be a polynomial of order $\displaystyle k$, that is $\displaystyle f(n)=b_{k}n^{k} + b_{k-1}n^{k-1} +...+ b_{0}$.
Prove that $\displaystyle f(n) = \Omega(n^{k})$.
Note: $\displaystyle b_{k} > 0$, but $\displaystyle b_{i}$ may be negative for $\displaystyle 0\le{i}<k$.
Hint: use the highest index for which $\displaystyle b_{s}< 0$.
[How can I prove that? and what is $\displaystyle b_{s}$???]
• Apr 16th 2012, 04:21 AM
emakarov
Re: Grow Rate and Big Omega of Polynomial
Quote:
Originally Posted by jfk
what is $\displaystyle b_{s}$???]
The hint is to use the highest index $\displaystyle s$ for which $\displaystyle b_{s}< 0$.
My idea, rather, is to break $\displaystyle b_kn^k$ into k + 1 equal parts and to prove that each part is greater than $\displaystyle |b_i|n^i$ for all 0 <= i < k (starting from some n). Then even if all non-leading coefficients are negative,
$\displaystyle b_{k}n^{k}+\sum_{i=0}^{k-1}b_in^i \ge b_{k}n^{k}-\sum_{i=0}^{k-1}|b_i|n^i \ge b_{k}n^{k}-k\cdot b_{k}n^{k}/(k+1) =(b_{k}/(k+1))n^{k}$
So it is sufficient to find N such that for all 0 <= i < k and all n > N we have $\displaystyle |b_i|n^i\le b_kn^k/(k+1)$. | 514 | 1,497 | {"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} | 3.953125 | 4 | CC-MAIN-2018-26 | latest | en | 0.814 |
https://rdrr.io/rforge/parfm/man/parfm.html | 1,519,559,844,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816370.72/warc/CC-MAIN-20180225110552-20180225130552-00202.warc.gz | 780,168,106 | 17,024 | # parfm: Parametric Frailty Models In parfm: Parametric Frailty Models
## Description
Fits parametric Frailty Models by maximizing the marginal likelihood.
## Usage
1 2 3 4 5 6 7 8 parfm(formula, cluster = NULL, strata = NULL, data, inip = NULL, iniFpar = NULL, dist = c("weibull", "inweibull", "frechet", "exponential", "gompertz", "loglogistic", "lognormal", "logskewnormal"), frailty = c("none", "gamma", "ingau", "possta", "lognormal", "loglogistic"), method = "nlminb", maxit = 500, Fparscale = 1, showtime = FALSE, correct = 0)
## Arguments
formula A formula object, with the response on the left of a ~ operator, and the terms on the right. The response must be a survival object as returned by the Surv() function. The status indicator 'event' in the Surv object must be 0=alive, 1=dead. cluster The name of a cluster variable in data. strata The name of a strata variable in data. data A data.frame in which to interpret the variables named in the formula. inip The vector of initial values. First components are for the baseline hazard parameters according to the order given in 'details'; Other components are for the regression parameters according to the order given in 'formula'. iniFpar The initial value of the frailty parameter. dist The baseline hazard distribution. One of weibull, inweibull, frechet, exponential, gompertz, lognormal, loglogistic, or logskewnormal. frailty The Frailty distribution. One of: none, gamma, ingau, possta or lognormal. method The optimisation method from the function optimx(). maxit Maximum number of iterations (see optimx()). Fparscale the scaling value for the frailty parameter in optimx(). Optimisation is performed on Fpar/Fparscale. showtime Show the execution time? correct A correction factor that does not change the marginal log-likelihood except for an additive constant given by #clusters * correct * log(10). It may be useful in order to get finite log-likelihood values in case of many events per cluster with Positive Stable frailties. Note that the value of the log-likelihood in the output is the re-adjusted value.
## Details
Baseline hazards
The Weibull hazard (dist="weibull") is
h(t; rho, lambda) = rho * lambda * t^(rho - 1)
with rho, lambda > 0.
The inverse Weibull (or Fréchet) hazard (dist="inweibull" or dist="frechet") is
h(t; rho, lambda) = rho * lambda * t^(-rho - 1) / (exp(lambda t^-rho) - 1)
with rho, lambda > 0.
The exponential hazard (dist="exponential") is
h(t; lambda) = lambda
with lambda > 0.
The Gompertz hazard (dist="gompertz") is
h(t; gamma, lambda) = lambda * exp(gamma * t)
with gamma, lambda > 0.
The lognormal hazard (dist="lognormal") is
h(t; mu, sigma) = phi ( (log(t) - mu) / sigma ) / {sigma * t * [1 - Phi( (log(t)- mu) / sigma ) )]}
with mu in R, sigma > 0, and where phi and Phi are the density and distribution functions of a standard Normal random variable.
The loglogistic hazard (dist="loglogistic") is
h(t; alpha, kappa) = [exp(alpha) * kappa * t^(kappa - 1)] / [1 + exp(alpha) * t^(kappa)]
with alpha in R, kappa > 0.
The log-skew-normal hazard (dist="logskewnormal") is obtained as the ratio between the density and the cumulative distribution function of a log-skew normal random variable (Azzalini, 1985), which has density
f(t; ξ, ω, α) = \frac{2}{ω t} φ≤ft(\frac{\log(t) - ξ}{ω}\right) Φ≤ft(α\frac{\log(t)-ξ}{ω}\right)
with xi in R, omega > 0, alpha in R, and where phi and Phi are the density and distribution functions of a standard Normal random variable. Of note, if alpha=0 then the log-skew-normal boils down to the log-normal distribution, with mu=xi and sigma=omega.
Frailty distributions
The gamma frailty distribution (frailty="gamma") is
f(u) = [theta^(-1 / theta) * u^(1 / theta - 1) * exp(-u / theta)] / [Gamma(1 / theta)]
with theta > 0 and where Gamma(.) is the gamma function.
The inverse Gaussian frailty distribution (frailty="ingau") is
f(u) = 1 / sqrt(2 * pi * theta) * u^(-3 / 2) * exp[- (u - 1)^2 / (2 * theta * u)]
with theta > 0.
The positive stable frailty distribution (frailty="poosta") is
f(u) = - 1 / (pi * u) * sum_(k=1, 2, ..., infiny) [Gamma(k * (1 - nu) + 1) / k! * (-u^(nu - 1))^k * sin((1 - nu) * k * pi)]
with 0 < nu < 1.
The lognormal frailty distribution (frailty="lognormal") is
f(u) = 1 / sqrt(2 * pi * theta) * u^(-1) * exp[- (log u)^2 / (2 * theta)]
with theta > 0. As the Laplace tranform of the lognormal frailties does not exist in closed form, the saddlepoint approximation is used (Goutis and Casella, 1999).
## Value
An object of class parfm.
## Author(s)
Federico Rotolo [aut, cre], Marco Munda [aut], Andrea Callegaro [ctb]
## References
Munda M, Rotolo F, Legrand C (2012). parfm: Parametric Frailty Models in R. Journal of Statistical Software, 51(11), 1-20. DOI 10.18637/jss.v051.i11
Duchateau L, Janssen P (2008). The frailty model. Springer.
Wienke A (2010). Frailty Models in Survival Analysis. Chapman & Hall/CRC biostatistics series. Taylor and Francis.
Goutis C, Casella G (1999). Explaining the Saddlepoint Approximation. The American Statistician 53(3), 216-224. DOI 10.1080/00031305.1999.10474463
Azzalini A (1985). A class of distributions which includes the normal ones. Scandinavian Journal of Statistics, 12(2):171-178. URL http://www.jstor.org/stable/4615982
select.parfm, ci.parfm, predict.parfm
## Examples
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 #------Kidney dataset------ data(kidney) # type 'help(kidney)' for a description of the data set kidney$sex <- kidney$sex - 1 parfm(Surv(time,status) ~ sex + age, cluster = "id", data = kidney, dist = "exponential", frailty = "gamma") parfm(Surv(time,status) ~ sex + age, cluster = "id", data = kidney, dist = "exponential", frailty = "lognormal") parfm(Surv(time,status) ~ sex + age, cluster = "id", data = kidney, dist = "weibull", frailty = "ingau") parfm(Surv(time,status) ~ sex + age, cluster = "id", data = kidney, dist="gompertz", frailty="possta", method="CG") parfm(Surv(time,status) ~ sex + age, cluster = "id", data = kidney, dist="gompertz", frailty="logskewnormal") #-------------------------- #------Asthma dataset------ data(asthma) head(asthma) # type 'help(asthma)' for a description of the data set asthma$time <- asthma$End - asthma$Begin parfm(Surv(time, Status) ~ Drug, cluster = "Patid", data = asthma, dist = "weibull", frailty = "gamma") parfm(Surv(time, Status) ~ Drug, cluster = "Patid", data = asthma, dist = "weibull", frailty = "lognormal") parfm(Surv(Begin, End, Status) ~ Drug, cluster = "Patid", data = asthma[asthma$Fevent == 0, ], dist = "weibull", frailty = "lognormal", method = "Nelder-Mead") #--------------------------
parfm documentation built on May 31, 2017, 2:48 a.m. | 2,077 | 6,779 | {"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} | 2.609375 | 3 | CC-MAIN-2018-09 | longest | en | 0.692365 |
https://swizec.com/blog/week-19-relativity-theory-and-time-perception/ | 1,720,920,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00024.warc.gz | 501,135,097 | 127,578 | swizec.com
#### Senior Mindset Book
Get promoted, earn a bigger salary, work for top companies
# Week 19: Relativity theory and time perception
[This post is part of an ongoing challenge to understand 52 papers in 52 weeks. You can read previous entries, here, or subscribe to be notified of new posts by email]
"When a man sits with a pretty girl for an hour, it seems like a minute. But let him sit on a hot stove for a minute and it's longer than any hour." ~ Albert Einstein
Time sometimes flies by, sometimes drags along shuffling its feet. But how does that work? How come context affects the way we perceive time?
Neurobiological evidence has been found that people use different areas of their brain to time different lengths of time, but the traditional view of psychological time is still that there is a timer, a counter, and simple arithmetic. In Relativity Theory and Time Perception: Single or Multiple Clocks? Buhusi and Meck use rats to see whether their brain uses independent clocks or not.
## The experiment
They used a tri-lever set up to train rats to recognise three different time intervals - 10s, 30s and 90s - using something called a tri-peak procedure.
The paper treats this as common understanding in psychology research, so it's never explained in great detail. From what I can tell, a light in rats' cages acted as a signal. When it turned on, the experiment started.
When it either turned off or flashed again after X seconds, I'm not sure, it counted as a signal or a gap. If the rat pulled the correct lever, it was given a food pellet.
Now I'm not sure whether the light was still used for anything other than starting the trial after the rats had learned how to time correctly, but the variable we're looking at is by how much they miss the target time.
The fun part is inserting gaps into the timing. First a small gap, then a larger and larger gap. When this happens rats can do one of three things with their internal clock:
• run, timer ignores the gap
• stop, timer is paused for the duration of the gap, then continues
• reset, timer starts from scratch after the gap
I'm not sure how researches communicated the gap to their rats, but if the same effect happens for all three intervals, rats are using a simple clock+accumulator setup. If the actions happen independently for each timer, then they are using several parallel clocks.
Put simply, if you miss the first lever by 10s and still hit the third lever perfectly, then you have independent clocks. If you miss the first lever by the same amount as the last lever, then you aren't.
## Effect of gaps on response rates
In baseline tests response rates were just around the target times, which means rats successfully learned how to time the three intervals.
But when researchers inserted a fixed gap at the 15 second mark, the three timing functions behaved differently. The effect of a gap seems to increase with the relative duration of a gap to the timed interval.
A 10s gap delayed the 10s timer by 30s (reset rule), by 20s the 30s clock (intermediate between stop and reset), and by 10s the 90s clock (stop rule). This suggests the clocks are operated independently.
To see if this affects only responses, or actual timers, researchers tried progressively expanding the gap. A 1s gap had no effect on timing. A 3s gap reset the 10s clock, but did not affect the other two. A 10s gap reset the 10s and 30s clocks, and a 30s gap reset all clocks.
One way to model this effect is to think of gaps as spending resources in the brain. This is essentially memory decay rate, which is proportional to the relative gap to criterion duration - -salience*g/Tk, where g is the gap and Tk is each clock's criterion.
If the criterion is short, g/Tk is large and memory decays quickly (the reset rule), if it's long then the ratio is smaller and memory decays slower (run or stop rule). Using this we can simulate what happens in experiments - clocks resetting in a hierarchical manner as the gap increases.
## Psychological models of interval timing
Several models have been proposed to explain data from past interval-timing experiments, but they fail to predict the multiple parallel clocks behaviour.
A switch model, which assumes that a switch stops accumulation of time in working memory, would predict that all three clocks should use the stop rule when a gap occurs. But present data clearly indicates that rats flexibly change behaviour of each clock independently.
Another model says that instructional-ambiguity would reset clocks based on how similar the gap is to the timed signal. If it's similar, the clock stops, if not, it resets. But in this experiment the signal and gap were identical and rats still used different rules for different clocks.
The passive memory-decay model proposed by Cabeza de Vaca is a step in the right direction, but predicts clock behaviour should depend solely on gap duration. This experiment shows it depends on criterion duration as well.
The resource allocation model suggested by the authors, assumes that during a gap resources are diverted from timing, which means working memory can't maintain its current time. The gap's effect is primarily a function of the contrast between the gap and timed interval. Memory decays with a ratio of g/Tk.
Better still, this model can be used to model any type of distractor, not just timing intervals.
## Conclusion
The authors found that multiple independent clocks are likely always running in our brain, timing separate events depending on the relative context they occur in. Their findings also have direct support from neurobiological studies showing that physically separate neurons handle the different clocks.
I'm not sure what all this entails, but it's definitely interesting. And I guess practical for multitasking.
Published on April 14th, 2014 in 52papers52weeks, Learning, Personal, Papers
#### Continue reading about Week 19: Relativity theory and time perception
Semantically similar articles hand-picked by GPT-4
### Senior Mindset Book
Get promoted, earn a bigger salary, work for top companies
Have a burning question that you think I can answer? Hit me up on twitter and I'll do my best.
Who am I and who do I help? I'm Swizec Teller and I turn coders into engineers with "Raw and honest from the heart!" writing. No bullshit. Real insights into the career and skills of a modern software engineer.
Want to become a true senior engineer? Take ownership, have autonomy, and be a force multiplier on your team. The Senior Engineer Mindset ebook can help 👉 swizec.com/senior-mindset. These are the shifts in mindset that unlocked my career.
Curious about Serverless and the modern backend? Check out Serverless Handbook, for frontend engineers 👉 ServerlessHandbook.dev
Want to Stop copy pasting D3 examples and create data visualizations of your own? Learn how to build scalable dataviz React components your whole team can understand with React for Data Visualization
Want to get my best emails on JavaScript, React, Serverless, Fullstack Web, or Indie Hacking? Check out swizec.com/collections
Did someone amazing share this letter with you? Wonderful! You can sign up for my weekly letters for software engineers on their path to greatness, here: swizec.com/blog
Want to brush up on your modern JavaScript syntax? Check out my interactive cheatsheet: es6cheatsheet.com
By the way, just in case no one has told you it yet today: I love and appreciate you for who you are ❤️
Created by Swizec with ❤️ | 1,628 | 7,542 | {"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-2024-30 | latest | en | 0.955601 |
https://www.speedsolving.com/forum/threads/2x2-eg-erik-gunnar-method.17876/ | 1,505,821,196,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685129.23/warc/CC-MAIN-20170919112242-20170919132242-00517.warc.gz | 857,512,230 | 15,778 | # 2x2 EG (Erik-Gunnar) Method
Discussion in 'General Speedcubing Discussion' started by hansho13, Dec 22, 2009.
Welcome to the Speedsolving.com. You are currently viewing our boards as a guest which gives you limited access to join discussions and access our other features. By joining our free community of over 30,000 people, you will have access to post topics, communicate privately with other members (PM), respond to polls, upload content and access many other special features. Registration is fast, simple and absolutely free so please, join our community today!
If you have any problems with the registration process or your account login, please contact us and we'll help you get started. We look forward to seeing you on the forums!
Already a member? Login to stop seeing this message.
1. ### hansho13Member
10
0
Dec 22, 2009
hansho13
does anybody know a good site for the EG (Erik-Gunnar) Method for the 2x2 rubik's cube?
2. ### EscherBabby
3,364
54
Jul 23, 2008
WCA:
2008KINN01
RowanKinneavy
3. ### CitricAcidMember
OK, thank you for your helpful criticism.
Removing post.
4. ### hansho13Member
10
0
Dec 22, 2009
hansho13
2x2 program?
also i read somewhere there was a program or something called ron's something i cant quite remember the name. does anybody know what im talking about?
5. ### EscherBabby
3,364
54
Jul 23, 2008
WCA:
2008KINN01
RowanKinneavy
Here you go:
http://www.speedcubing.com/CubeSolver/MiniCubeSolver.html
You can also use Cube Explorer, just grey out the non corner cubies.
I'm pretty sure you can use this too
6. ### hansho13Member
10
0
Dec 22, 2009
hansho13
do you know a program that simulates pyraminx?
Edit: OK never mind about the pyraminx. So just to make sure i know what the EG method is, you solve the white side(or any other side) randomly and have one algorithm to fully solve the white side and LL(last layer). Correct?
Last edited: Dec 24, 2009
7. ### VishalMember
90
0
Jun 12, 2010
Florida us
Pyraminx lag generator
I would live to know where you can generate pyraminx algs
May 17, 2008
9. ### TMOYMember
Jun 29, 2008
WCA:
2008COUR01
He must be tryihg to adapt EG to the pyra
10. ### VishalMember
90
0
Jun 12, 2010
Florida us
I am trying to adapt eg to the pyraminx if any one
Wants to help that would be great and the pyraminx solver is pretty much crap it doesn't give you the algs is at a bad angle and the color scheme is messed up
11. ### riffzMember
Oct 3, 2008
WCA:
2009HOLT01
riffz
Like solve one first layer edge and know algs for when the remaining two are switched? I can't say for sure but I'm pretty the algs would not be too nice. Polish V is probably a much more worthwhile endeavor, although I guess it couldn't hurt to know these algs.
10
0
Jan 17, 2009
WCA:
2008VORJ01
EG Algs
I'm looking for a collection of EG 2 algs. I know why they are hard to find, but does anyone have a full set anywhere?
342
0
May 29, 2010
UK
WCA:
2010REES03
15. ### HexiMember
63
0
Dec 2, 2010
Prague
TheHexiii
EG Method
Hey, I wanna learn the EG method... but i think, i don't understand it... I deduced from the tutorials, that i have to do one side (not 1 layer, only side), and then "x2" and do one of that algorithms: EG 1 , but when I do that, I can't find situation, that is on my cube. So the question is: What am I doin' wrong? Thx for answers.
676
0
Sep 9, 2010
Erzz197
17. ### HexiMember
63
0
Dec 2, 2010
Prague
TheHexiii
What's the difference between EG 1 and EG 2?
18. ### RyanReese09Premium Member
May 16, 2010
Whiteford, MD, USA
WCA:
2010REES01
RyanReese09
EG1 I believe is when you have 2 adjacent corner swap in the side you complete, and then you put htat face on D and solve the entire cube from there
EG2 is when you have diagonal corner swap in the side, and you put that faceo n D and solve hte cube from there
CLL is when you have a completeed layer and you solve teh last layer in 1 alg
I might have switched eg1/2 up, but they are correct definitions
edit-I was right
http://www.speedcubing101.com/eg-1.html
19. ### HexiMember
63
0
Dec 2, 2010
Prague
TheHexiii
ok, but still i can't find my situation:
Yellow | GreenBlue | Red
uhm.. what's wrong with the table?
Last edited: Dec 3, 2010
Sep 8, 2008
Emmaus, PA
WCA:
2010KOTC01 | 1,293 | 4,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-39 | latest | en | 0.890907 |
https://catalog.foothill.edu/course-outlines/MATH-2A/ | 1,725,781,008,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00822.warc.gz | 133,067,804 | 6,001 | # MATH 2A: DIFFERENTIAL EQUATIONS
## Foothill College Course Outline of Record
Foothill College Course Outline of Record
Effective Term: Summer 2024
Units: 5
Hours: 5 lecture per week (60 total per quarter)
Prerequisite: MATH 1C.
Advisory: Demonstrated proficiency in English by placement via multiple measures OR through an equivalent placement process OR completion of ESLL 125 & ESLL 249; not open to students with credit in MATH 12A.
Degree & Credit Status: Degree-Applicable Credit Course
Foothill GE: Non-GE
Transferable: CSU/UC
Repeatability: Not Repeatable
## Student Learning Outcomes
• Students will model continuous processes using differential equations and use the model to answer related questions.
• Students will develop conceptual understanding of mathematical modeling of continuous processes and their rates of change. They will learn to demonstrate and communicate this understanding in a variety of ways, such as: reasoning with definitions and theorems, connecting concepts, and connecting multiple representations, as appropriate.
• Students will demonstrate the ability to solve differential equations and verify their solutions analytically, numerically, graphically, and qualitatively.
## Description
Differential equations and selected topics of mathematical analysis.
## Course Objectives
The student will be able to:
1. Classify differential equations by order, linearity, separability, exactness, coefficient functions, homogeneity, type of any nonhomogeneities, and other qualities.
2. Identify appropriate analytic, numerical, and graphical techniques for solving or approximating solutions to differential equations of the particular classes specified in the expanded description of course content.
3. Solve differential equations with appropriate analytic techniques.
4. Approximate solutions to differential equations with appropriate numeric techniques.
5. Investigate solutions to differential equations with appropriate graphical techniques.
6. Verify solutions to differential equations analytically, numerically, graphically, and qualitatively.
7. Write differential equations and initial value problems to model phenomena in the physical, life, and social sciences.
8. Interpret solutions to differential equations and initial value problems in context.
9. Discuss differential equations and their solutions in accurate mathematical language and notation.
10. Investigate solutions to differential equations using at least one numerical or graphing utility.
## Course Content
1. Classes of differential equations
1. First order
1. Separable
2. Linear
3. Exact
2. Second order
1. Linear
2. Constant coefficient
3. Polynomial coefficient
3. Higher-order linear
4. Autonomous
5. Homogeneous
6. Nonhomogeneous
1. Polynomial
2. Exponential
3. Sinusoid
4. Other continuous functions
5. Discontinuous functions
6. Impulses
2. Initial value problems
1. Existence and uniqueness theorem
1. Applications
3. Classes of solutions
1. General solution
2. n-parameter family of solutions
3. Particular solution
4. Unique solution
5. Equilibrium solution
6. Linearly independent solutions with Wronskian
4. Systems of linear differential equations
5. Techniques for solving differential equations
1. Separation of variables
2. Integrating factor
3. Total differential
4. Characteristic equations
1. Distinct real roots
2. Repeated real roots
3. Complex roots
4. Fundamental solutions
5. Superposition principle
6. Undetermined coefficients
7. Variation of parameters
8. Annihilator method
9. Reduction of order
10. Laplace transforms
11. Power series
12. Method of Frobenius
13. Matrix methods
6. Approximation of solution
1. Numerical approximation
1. Euler's method
2. Improved Euler's method
2. Graphical approximation
1. Direction field
2. Phase line
3. Phase plane
7. Applications selected from the following topics:
1. Population models
1. Predator-prey models
2. Thresholds and carrying capacities
2. Growth and decay
3. Mixing problems
4. Spring-mass systems
1. Undamped
2. Damped
5. Electrical circuits
1. Inductor-capacitor
2. Resistor-inductor-capacitor
6. Newton's laws
1. Falling bodies
2. Pendulums
3. Cooling
7. Orthogonal trajectories
8. Torricelli's law
9. Financial applications
1. Compound interest
2. Time value of money
3. Annuities
10. Communication models
2. Mass marketing
11. Public health models
1. Epidemics
2. Health care utilization
Not applicable.
## Special Facilities and/or Equipment
1. Access to graphing technology, such as a graphing calculator or graphing software.
2. When taught online/hybrid: students need internet access and access to course management system and specific software related to the course.
## Method(s) of Evaluation
Methods of Evaluation may include but are not limited to the following:
Homework
Class participation
Term paper(s)
Presentation(s)
Computer lab assignment(s)
Term project
Quizzes
Unit exam(s)
Proctored comprehensive final examination
## Method(s) of Instruction
Methods of Instruction may include but are not limited to the following:
Lecture
Discussion
Cooperative learning exercises
## Representative Text(s) and Other Materials
Nagle R., E. Saff, and D. Snyder. Fundamentals of Differential Equations, 9th ed.. 2018.
Boyce W., R. DiPrima, and D. Meade. Elementary Differential Equations, 11th ed.. 2017.
These texts are the most recent editions available; we will adopt the next edition of each as it becomes available.
## Types and/or Examples of Required Reading, Writing, and Outside of Class Assignments
1. Homework problems covering subject matter from text and related material ranging from 15-30 problems per week. Students will need to employ critical thinking in order to complete assignments
2. Five hours per week of lecture covering subject matter from text and related material. Reading and study of the textbook, related materials, and notes
3. Student projects covering subject matter from textbook and related materials. Projects will require students to discuss mathematical problems, write solutions in accurate mathematical language and notation, and interpret mathematical solutions. Projects may require the use of a computer algebra system such as Mathematica or MATLAB
4. Worksheets: Problems and activities covering the subject matter. Such problems and activities will require students to think critically. Such worksheets may be completed inside and/or outside of class
Mathematics | 1,384 | 6,436 | {"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 | 3 | CC-MAIN-2024-38 | latest | en | 0.878195 |
https://doubtnut.com/question-answer/find-the-distance-between-the-points-cos-theta-sin-theta-sin-theta-cos-theta-8488202 | 1,618,298,137,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072175.30/warc/CC-MAIN-20210413062409-20210413092409-00170.warc.gz | 319,889,299 | 49,304 | Class 12
MATHS
Cartesian Systems Of Rectangular Coordinates - For Boards
Find the distance between the points : (cos theta, sin theta), (sin theta, cos theta)
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 19-2-2021
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
178.3 K+
8.9 K+
Related Videos
3137386
5.3 K+
107.0 K+
1:28
1413700
8.5 K+
171.7 K+
2:25
2342231
25.1 K+
502.7 K+
2:19
56705135
600+
13.8 K+
1413701
7.4 K+
149.0 K+
2:09
5931307
101.7 K+
118.2 K+
4:30
1736014
304.1 K+
401.6 K+
2:10
1457942
46.4 K+
168.5 K+
6:16
92138771
56.9 K+
235.9 K+
2:43
54830246
4.5 K+
89.2 K+
1:32
10873
15.1 K+
301.4 K+
2:34
61733860
10.0 K+
199.7 K+
5:09
61733865
9.8 K+
196.8 K+
5:23
36165580
19.4 K+
388.9 K+
0:59
68155
150.5 K+
336.7 K+
4:56
Very Important Questions
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.9 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.7 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.4 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.4 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.2 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.2 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.1 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.1 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.1 K+ Views
Class 12th
CARTESIAN SYSTEMS OF RECTANGULAR COORDINATES - FOR BOARDS
1.1 K+ Views | 686 | 1,790 | {"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-2021-17 | latest | en | 0.508018 |
https://www.excelguru.ca/forums/showthread.php?1408-How-to-pull-out-cards-with-balances | 1,632,069,626,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00581.warc.gz | 785,298,679 | 16,212 | # Thread: How to pull out cards with balances?
1. ## How to pull out cards with balances?
I figured out how to count how many accounts I have open with balances. What I want to do next is start working on a snowball calculator. Before I can do that though I need to pull out to a separate tab the following
Issuing Bank, Card Name. Balance Limit Utilization, Minimum Payment and Amount to goal.
All this information is on the Credit Cards Data tab
My ultimate goal right now is to have the following snowball methods
1) Snowball by Utilization Goal
2) Snowball by balance smallest to largest
3) Snowball by balance largest to smallest
4) Snowball by interest rate lowest to highest
5) Snowball by interest rate highest to lowest
I figured I can break that out by a sort possibly by a radio button to choose a pre-defined sort on the Payment calculation sheet then put the results back to the credit card data sheet. I would use the values on the bank balances tab minus the Fixed Monthly payment tab and what is left will go to the credit card payments
I hope this makes sense as I am tired and my brain is a bit foggy right now.
2. Sorry Emanuel, I'm not quite following this. My brain is also a bit foggy at the moment too though.
I see the Snowball column, but the Payment Calculation Sheet is blank.
Can you be a bit more detailed with the logic you're trying to impart here?
3. Ok I had sleep and can type coherently now.
I *think* the best way to do what I want is to have a vba macro or script "pull"from the credit cards data page the following
Card Name, Util %, % to pay to goal, and minimum payment only for cards with balances
Then after sorting the accounts on this tab based on the sort method I choose have it subtract the total of minimum payments from the total liquid value value.
Then taking the amount left after the minimum payment apply that to the card with balances based on the sort order until each goal is reached and money is left or the payment money for the month runs out.
4. Hi Emanuel,
I just cobbled this together to start the extraction of the info. Tell me if this is along the right direction...
After that, I'd need a bit of clarification on the next step... what sort method are you after?
Code:
```Sub Extract()
Dim ary(3) As Variant
Dim cl As Range
Dim wsSource As Worksheet
Dim wsTarget As Worksheet
Set wsSource = Worksheets("Credit Cards Data")
Set wsTarget = Worksheets("Payment Calculation Sheet")
With wsSource
For Each cl In .Range("A2:A" & .Range("A" & .Rows.Count).End(xlUp).Row)
'Check if card balance is > 0
With cl
If .Offset(0, 4).Value > 0 Then
ary(0) = .Offset(0, 2)
ary(1) = .Offset(0, 6)
ary(2) = .Offset(0, 7)
ary(3) = .Offset(0, 9)
With wsTarget
.Range("A" & .Rows.Count).End(xlUp).Offset(1, 0).Resize(1, 4) = ary()
End With
End If
End With
Next cl
End With
End Sub```
5. Originally Posted by Ken Puls
Hi Emanuel,
I just cobbled this together to start the extraction of the info. Tell me if this is along the right direction...
After that, I'd need a bit of clarification on the next step... what sort method are you after?
Code:
```Sub Extract()
Dim ary(3) As Variant
Dim cl As Range
Dim wsSource As Worksheet
Dim wsTarget As Worksheet
Set wsSource = Worksheets("Credit Cards Data")
Set wsTarget = Worksheets("Payment Calculation Sheet")
With wsSource
For Each cl In .Range("A2:A" & .Range("A" & .Rows.Count).End(xlUp).Row)
'Check if card balance is > 0
With cl
If .Offset(0, 4).Value > 0 Then
ary(0) = .Offset(0, 2)
ary(1) = .Offset(0, 6)
ary(2) = .Offset(0, 7)
ary(3) = .Offset(0, 9)
With wsTarget
.Range("A" & .Rows.Count).End(xlUp).Offset(1, 0).Resize(1, 4) = ary()
End With
End If
End With
Next cl
End With
End Sub```
Ken I will try this tonight and i will answer your question tonight
6. Ok when I run it it pulls data but now what I'm looking for entirely just some minor issues.
1) I need the issuing bank along with the card name for it to make sense to me not just the card name
2) The values are not showing in the right format. They are just numeric and not percentage or currency
After manually putting in a header and formatting the cells so they look like
CARD Percent used Goal to pay to Minimum Payment Platinum 31.60% 29.00% \$28 Costco 1.26% 5.00% \$35.00 Zync 0.00% 0.00% \$33.08 Cash Back 0.15% 0.00% \$8.00 Freedom 34.15% 29.00% Slate 31.47% 28.00% Discover IT 3.61% 0.00% Gas Card 1.44% 0.00% \$9.39 CashRewards 92.95% 90.00% \$298.00 Visa 28.50% 28.00% \$61.00
I want to be able to sort the card by any of the following
Balances both High to Low and Low to High
Interest rate both lowest to highest and highest to lowest.
Utilization percentage both lowest to highest and highest to lowest.
I then want to make a money waterfall and take the money I have left after the minimum payment and apply it to each card until it hits the goal then take anything left and apply it to the next card and so one until either all the cards are at goal or the money is at zero
7. Hi Emanuel,
I've modified the macro to add the formatting, and sort by the Minimum payment first:
Code:
```Sub Extract()
Dim ary(4) As Variant
Dim cl As Range
Dim wsSource As Worksheet
Dim wsTarget As Worksheet
Dim lCol As Long
Set wsSource = Worksheets("Credit Cards Data")
Set wsTarget = Worksheets("Payment Calculation Sheet")
With wsTarget
'Clear out the old data
.Cells.ClearContents
.Range("A1:E1") = Array("Issuing Bank", "Card", "Percent Used", "Goal to pay to", "Minimum Payment")
'Format columns
.Columns("C:D").NumberFormat = "0.00%"
.Columns("E:E").Style = "Currency"
.Columns("A:E").EntireColumn.AutoFit
End With
With wsSource
'Fill table
For Each cl In .Range("A2:A" & .Range("A" & .Rows.Count).End(xlUp).Row)
'Check if card balance is > 0
With cl
If .Offset(0, 4).Value > 0 Then
ary(0) = .Offset(0, 0)
ary(1) = .Offset(0, 2)
ary(2) = .Offset(0, 6)
ary(3) = .Offset(0, 7)
ary(4) = .Offset(0, 9)
With wsTarget
.Range("A" & .Rows.Count).End(xlUp).Offset(1, 0).Resize(1, 5) = ary()
End With
End If
End With
Next cl
End With
With wsTarget
.Sort.SortFields.Clear
Key:=Range("E2:E" & .Range("E" & .Rows.Count).End(xlUp).Row), _
SortOn:=xlSortOnValues, _
Order:=xlDescending, _
DataOption:=xlSortNormal
With .Sort
.SetRange Range("A1:E" & wsTarget.Range("E" & wsTarget.Rows.Count).End(xlUp).Row)
.Orientation = xlTopToBottom
.SortMethod = xlPinYin
.Apply
End With
End With
End Sub```
From here, I'm not quite sure on how to proceed. Since we can only sort by one thing to start with, it would seem to make sense to me that any additional sorts would be manually triggered. As far as the money to apply, where is that, and how do you see it being applied to the schedule, exactly? Are you thinking of a formula that shows ever decreasing amounts available or... Will that money be provided at runtime, is it alreayd somewhere in your file?
8. That code works perfectly.
The money to apply is the total liquid value on the Bank Balances tab in cell F1
I was thinking of having a drop down box with the different sort options and whatever one is chosen is the sort that's applied.
After giving this some thought I think the amount that can be applied to the bills should be the liquid amount after subtracting the fixed monthly bills which is already it's own tab.
The total left in fixed bills is found in cell M4 on the Fixed Monthly Payments Tab the only problems with that tab are
1) It's adding future payments example the car and insurance payments are not due this month but it's adding them in
2) I have no idea how to make it reset that the payment is due when the month changes again without manually removing the value in Amt Paid This Month
See the Red Text on Payment Calculation sheet for my thoughts about the "waterfall" | 2,105 | 7,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.95412 |
https://finmas.in/elliot-wave/2-2-extension | 1,701,768,849,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00819.warc.gz | 302,621,269 | 6,541 | • Mon - Fri 9.00 AM - 06.00 PM - Sat 10.00 AM-12.00 PM
• +91 99400 77330
# 2.2 Extension
• home
• 2.2 Extension
## Extension
Most impulses contain what Elliott called an extension. Extensions are elongated impulses with exaggerated subdivisions. The vast majority of impulse waves do contain an extension in one and only one of their three actionary subwaves. At times, the subdivisions of an extended wave are nearly the same amplitude and duration as the other four waves of the larger impulse, giving a total count of nine waves of similar size rather than the normal count of “five” for the sequence. In a nine-wave sequence, it is occasionally difficult to say which wave extended. However, it is usually irrelevant anyway, since under the Elliott system, a count of nine and a count of five have the same technical significance. The diagrams in Figure 1-5, illustrating extensions, will clarify this point.
Figure 5
The fact that extensions typically occur in only one actionary subwave provides a useful guide to the expected lengths of upcoming waves. For instance, if the first and third waves are of about equal length, the fifth wave will likely be a protracted surge. (In waves below Primary degree, a developing fifth wave extension will be confirmed by new high volume, as described in Lesson 13 under “Volume.”) Conversely, if wave three extends, the fifth should be simply constructed and resemble wave one.
In the stock market, the most commonly extended wave is wave 3. This fact is of particular importance to real time wave interpretation when considered in conjunction with two of the rules of impulse waves: that wave 3 is never the shortest actionary wave, and that wave 4 may not overlap wave 1. To clarify, let us assume two situations involving an improper middle wave, as illustrated in Figures 1-6 and 1-7.
Figure 1-6 Figure 1-7 Figure 1-8
In Figure 1-6, wave 4 overlaps the top of wave 1. In Figure 1-7, wave 3 is shorter than wave 1 and shorter than wave 5. According to the rules, neither is an acceptable labeling. Once the apparent wave 3 is proved unacceptable, it must be relabeled in some way that is acceptable. In fact, it is almost always to be labeled as shown in Figure 1-8, implying an extended wave (3) in the making. Do not hesitate to get into the habit of labeling the early stages of a third wave extension. The exercise will prove highly rewarding, as you will understand from the discussion under Wave Personality in Lesson 14. Figure 1-8 is perhaps the single most useful guide to real time impulse wave counting in this course.
Extensions may also occur within extensions. In the stock market, the third wave of an extended third wave is typically an extension as well, producing a profile such as shown in Figure 1-9. Figure 1-10 illustrates a fifth wave extension of a fifth wave extension. Extended fifths are fairly uncommon except in bull markets in commodities covered in Lesson 28.
. | 667 | 2,954 | {"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-2023-50 | latest | en | 0.921961 |
https://calculus7.org/tag/rate-of-approximation/ | 1,603,737,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891624.95/warc/CC-MAIN-20201026175019-20201026205019-00322.warc.gz | 249,750,327 | 28,440 | Wild power pie
Many people are aware of ${\pi}$ being a number between 3 and 4, and some also know that ${e}$ is between 2 and 3. Although the difference ${\pi-e}$ is less than 1/2, it’s enough to place the two constants in separate buckets on the number line, separated by an integer.
When dealing with powers of ${e}$, using ${e>2}$ is frequently wasteful, so it helps to know that ${e^2>7}$. Similarly, ${\pi^2<10}$ is way more precise than ${\pi<4}$. To summarize: ${e^2}$ is between 7 and 8, while ${\pi^2}$ is between 9 and 10.
Do any two powers of ${\pi}$ and ${e}$ have the same integer part? That is, does the equation ${\lfloor \pi^n \rfloor = \lfloor e^m \rfloor}$ have a solution in positive integers ${m,n}$?
Probably not. Chances are that the only pairs ${(m,n)}$ for which ${|\pi^n - e^m|<10}$ are ${m,n\in \{1,2\}}$, the smallest difference attained by ${m=n=1}$.
Indeed, having ${|\pi^n - e^m|<1}$ implies that ${|n\log \pi - m|, or put differently, ${\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n \,\pi^n}}$. This would be an extraordinary rational approximation… for example, with ${n=100}$ it would mean that ${\log \pi = 1.14\ldots}$ with the following ${50}$ digits all being ${0}$. This isn’t happening.
Looking at the continued fraction expansion of ${\log \pi}$ shows the denominators of modest size ${[1; 6, 1, 10, 24, \dots]}$, indicating the lack of extraordinarily nice rational approximations. Of course, can use them to get good approximations, ${\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n^2}}$, which leads to ${\pi^n\approx e^m}$ with small relative error. For example, dropping ${24}$ and subsequent terms yields the convergent ${87/76}$, and one can check that ${\pi^{76} = 6.0728... \cdot 10^{37}}$ while ${e^{87} = 6.0760...\cdot 10^{37}}$.
Trying a few not-too-obscure constants with the help of mpmath library, the best coincidence of integer parts that I found is the following: the 13th power of the golden ratio ${\varphi = (\sqrt{5}+1)/2}$ and the 34th power of Apèry’s constant ${\zeta(3) = 1^{-3}+2^{-3}+3^{-3}+4^{-4}+\dots}$ both have integer part 521.
Down with sines!
Suppose you have a reasonable continuous function ${f}$ on some interval, say ${f(x)=e^x}$ on ${[-1,1]}$, and you want to approximate it by a trigonometric polynomial. A straightforward approach is to write
$\displaystyle f(x) \approx \frac12 A_0+\sum_{n=1}^N (A_n\cos n \pi x +B_n \sin n \pi x)$
where ${A_n}$ and ${B_n}$ are the Fourier coefficients:
$\displaystyle A_n= \int_{-1}^1 e^x \cos \pi n x \,dx = (-1)^n \frac{e-e^{-1}}{1+ \pi^2 n^2 }$
$\displaystyle B_n= \int_{-1}^1 e^x \sin \pi n x \,dx = (-1)^{n-1} \frac{\pi n (e-e^{-1})}{1+ \pi^2 n^2 }$
(Integration can be done with the standard Calculus torture device). With ${N=4}$, we get
$\displaystyle e^{x} \approx 1.175 -0.216\cos\pi x +0.679 \sin \pi x +0.058\cos 2\pi x -0.365 \sin 2\pi x -0.026\cos 3\pi x +0.247 \sin 3\pi x +0.015\cos 4\pi x -0.186 \sin 4\pi x$
which, frankly, is not a very good deal for the price.
Still using the standard Fourier expansion formulas, one can improve approximation by shifting the function to ${[0,2]}$ and expanding it into the cosine Fourier series.
$\displaystyle f(x-1) \approx \frac12 A_0+\sum_{n=1}^N A_n\cos \frac{n \pi x}{2}$
where
$\displaystyle A_n= \int_{0}^2 e^{x-1} \cos \frac{\pi n x}{2} \,dx = \begin{cases} \dfrac{e-e^{-1}}{1+ \pi^2 k^2 } \quad & n=2k \\ {} \\ (-1)^k \dfrac{4(e+e^{-1})}{4+ \pi^2 (2k-1)^2 } \quad & n = 2k-1 \end{cases}$
Then replace ${x}$ with ${x+1}$ to shift the interval back. With ${N=4}$, the partial sum is
$\displaystyle e^x \approx 1.175 -0.890\cos \frac{\pi(x+1)}{2} +0.216\cos \pi(x+1)-0.133\cos \frac{3\pi(x+1)}{2} +0.058\cos 2\pi (x+1)$
which gives a much better approximation with fewer coefficients to calculate.
To see what is going on, one has to look beyond the interval on which ${f}$ is defined. The first series actually approximates the periodic extension of ${f}$, which is discontinuous because the endpoint values are not equal:
Cosines, being even, approximate the symmetric periodic extension of ${f}$, which is continuous whenever ${f}$ is.
Discontinuities hurt the quality of Fourier approximation more than the lack of smoothness does.
Just for laughs I included the pure sine approximation, also with ${N=4}$.
Improving the Wallis product
The Wallis product for ${\pi}$, as seen on Wikipedia, is
${\displaystyle 2\prod_{k=1}^\infty \frac{4k^2}{4k^2-1} = \pi \qquad \qquad (1)}$
Historical significance of this formula nonwithstanding, one has to admit that this is not a good way to approximate ${\pi}$. For example, the product up to ${k=10}$ is
${\displaystyle 2\,\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdot 8 \cdot 10 \cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16\cdot 16\cdot 18 \cdot 18\cdot 20\cdot 20}{1\cdot 3\cdot 3\cdot 5 \cdot 5\cdot 7\cdot 7\cdot 9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15\cdot 17\cdot 17\cdot 19\cdot 19\cdot 21} =\frac{137438953472}{44801898141} }$
And all we get for this effort is the lousy approximation ${\pi\approx \mathbf{3.0677}}$.
But it turns out that (1) can be dramatically improved with a little tweak. First, let us rewrite partial products in (1) in terms of double factorials. This can be done in two ways: either
${\displaystyle 2\prod_{k=1}^n \frac{4k^2}{4k^2-1} = (4n+2) \left(\frac{(2n)!!}{(2n+1)!!}\right)^2 \qquad \qquad (2)}$
or
${\displaystyle 2\prod_{k=1}^n \frac{4k^2}{4k^2-1} = \frac{2}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \qquad \qquad (3)}$
Seeing how badly (2) underestimates ${\pi}$, it is natural to bump it up: replace ${4n+2}$ with ${4n+3}$:
${\displaystyle \pi \approx b_n= (4n+3) \left(\frac{(2n)!!}{(2n+1)!!}\right)^2 \qquad \qquad (4)}$
Now with ${n=10}$ we get ${\mathbf{3.1407}}$ instead of ${\mathbf{3.0677}}$. The error is down by two orders of magnitude, and all we had to do was to replace the factor of ${4n+2=42}$ with ${4n+3=43}$. In particular, the size of numerator and denominator hardly changed:
${\displaystyle b_{10}=43\, \frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdot 8 \cdot 10 \cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16\cdot 16\cdot 18 \cdot 18\cdot 20\cdot 20}{3\cdot 3\cdot 5 \cdot 5\cdot 7\cdot 7\cdot 9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15\cdot 17\cdot 17\cdot 19\cdot 19\cdot 21\cdot 21} }$
Approximation (4) differs from (2) by additional term ${\left(\frac{(2n)!!}{(2n+1)!!}\right)^2}$, which decreases to zero. Therefore, it is not obvious whether the sequence ${b_n}$ is increasing. To prove that it is, observe that the ratio ${b_{n+1}/b_n}$ is
${\displaystyle \frac{4n+7}{4n+3}\left(\frac{2n+2}{2n+3}\right)^2}$
which is greater than 1 because
${\displaystyle (4n+7)(2n+2)^2 - (4n+3)(2n+3)^2 = 1 >0 }$
Sweet cancellation here. Incidentally, it shows that if we used ${4n+3+\epsilon}$ instead of ${4n+3}$, the sequence would overshoot ${\pi}$ and no longer be increasing.
The formula (3) can be similarly improved. The fraction ${2/(2n+1)}$ is secretly ${4/(4n+2)}$, which should be replaced with ${4/(4n+1)}$. The resulting approximation for ${\pi}$
${\displaystyle c_n = \frac{4}{4n+1} \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \qquad \qquad (5)}$
is about as good as ${b_n}$, but it approaches ${\pi}$ from above. For example, ${c_{10}\approx \mathbf{3.1425}}$.
The proof that ${c_n}$ is decreasing is familiar: the ratio ${c_{n+1}/c_n}$ is
${\displaystyle \frac{4n+1}{4n+5}\left(\frac{2n+2}{2n+1}\right)^2}$
which is less than 1 because
${\displaystyle (4n+1)(2n+2)^2 - (4n+5)(2n+1)^2 = -1 <0 }$
Sweet cancellation once again.
Thus, ${b_n<\pi for all ${n}$. The midpoint of this containing interval provides an even better approximation: for example, ${(b_{10}+c_{10})/2 \approx \mathbf{3.1416}}$. The plot below displays the quality of approximation as logarithm of the absolute error:
• yellow dots show the error of Wallis partial products (2)-(3)
• blue is the error of ${b_n}$
• red is for ${c_n}$
• black is for ${(b_n+c_n)/2}$
And all we had to do was to replace ${4n+2}$ with ${4n+3}$ or ${4n+1}$ in the right places.
Dirichlet vs Fejér
Convolution of a continuous function ${f}$ on the circle ${\mathbb T=\mathbb R/\mathbb (2\pi \mathbb Z)}$ with the Fejér kernel ${\displaystyle F_N(x)=\frac{1-\cos (N+1)x}{(N+1)(1-\cos x)}}$ is guaranteed to produce trigonometric polynomials that converge to ${f}$ uniformly as ${N\rightarrow\infty}$. For the Dirichlet kernel ${\displaystyle D_N(x)=\frac{\sin((N+1/2)x)}{\sin(x/2)}}$ this is not the case: the sequence may fail to converge to ${f}$ even pointwise. The underlying reason is that ${\int_{\mathbb T} |D_N|\rightarrow \infty }$, while the Fejér kernel, being positive, has constant ${L^1}$ norm. Does this mean that Fejér’s kernel is to be preferred for approximation purposes?
Let’s compare the performance of both kernels on the function ${f(x)=2\pi^2 x^2-x^4}$, which is reasonably nice: ${f\in C^2(\mathbb T)}$. Convolution with ${D_2}$ yields ${\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)D_2(x-t)\,dt = \frac{7\pi^4}{15} -48 \cos x +3 \cos 2x }$. The trigonometric polynomial is in blue, the original function in red:
I’d say this is a very good approximation.
Now try the Fejér kernel, also with ${N=2}$. The polynomial is ${\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)K_2(x-t)\,dt = \frac{7\pi^4}{15} - 32 \cos x + \cos 2x }$
This is not good at all.
And even with ${N=20}$ terms the Fejér approximation is not as good as Dirichlet with merely ${N=2}$.
The performance of ${F_{50}}$ is comparable to that of ${D_2}$. Of course, a ${50}$-term approximation is not what one normally wants to use. And it still has visible deviation near the origin, where the function ${f}$ is ${C^\infty}$ smooth:
In contrast, the Dirichlet kernel with ${N=4}$ gives a low-degree polynomial
${\displaystyle \frac{7\pi^4}{15} -48 \cos x +3 \cos 2x -\frac{16}{27}\cos 3x+\frac{3}{16}\cos 4x}$ that approximates ${f}$ to within the resolution of the plot:
What we have here is the trigonometric version of Biased and unbiased mollification. Convolution with ${D_N}$ amounts to truncation of the Fourier series at index ${N}$. Therefore, it reproduces the trigonometric polynomials of low degrees precisely. But ${F_N}$ performs soft thresholding: it multiplies the ${n}$th Fourier coefficient of ${f}$ by ${(1-|n|/(N+1))^+}$. In particular, it transforms ${\cos x}$ into ${(N/(N+1))\cos x}$, introducing the error of order ${1/N}$ — a pretty big one. Since this error is built into the kernel, it limits the rate of convergence no matter how smooth the function ${f}$ is. Such is the price that must be paid for positivity.
This reminds me of a parenthetical remark by G. B. Folland in Real Analysis (2nd ed., page 264):
if one wants to approximate a function ${f\in C(\mathbb T)}$ uniformly by trigonometric polynomials, one should not count on partial sums ${S_mf}$ to do the job; the Cesàro means work much better in general.
Right, for ugly “generic” elements of ${C(\mathbb T)}$ the Fejér kernel is a safer option. But for decently behaved functions the Dirichlet kernel wins by a landslide. The function above was ${C^2}$-smooth; as a final example I take ${f(x)=x^2}$ which is merely Lipschitz on ${\mathbb T}$. The original function is in red, ${f*D_4}$ is in blue, and ${f*F_4}$ is in green.
Added: the Jackson kernel ${J_{2N}}$ is the square of ${F_{N}}$, normalized. I use ${2N}$ as the index because squaring doubles the degree. Here is how it approximates ${f(x)=2\pi^2 x^2-x^4}$:
The Jackson kernel performs somewhat better than ${F_N}$, because the coefficient of ${\cos x}$ is off by ${O(1/N^2)}$. Still not nearly as good as the non-positive Dirichlet kernel.
Biased and unbiased mollification
When we want to smoothen (mollify) a given function ${f:{\mathbb R}\rightarrow{\mathbb R}}$, the standard recipe suggests: take the ${C^{\infty}}$-smooth bump function
$\displaystyle \varphi(t) = \begin{cases} c\, \exp\{1/(1-t^2)\}\quad & |t|<1; \\ 0 \quad & |t|\ge 1 \end{cases}$
where ${c}$ is chosen so that ${\int_{{\mathbb R}} \varphi=1}$ (for the record, ${c\approx 2.2523}$).
Make the bump narrow and tall: ${\varphi_{\delta}(t)=\delta^{-1}\varphi(t/\delta)}$. Then define ${f_\delta = f*\varphi_\delta}$, that is
$\displaystyle f_\delta(x) = \int_{\mathbb R} f(x-t) \varphi_\delta(t)\,dt = \int_{\mathbb R} f(t) \varphi_\delta(x-t)\,dt$
The second form of the integral makes it clear that ${f_\delta}$ is infinitely differentiable. And it is easy to prove that for any continuous ${f}$ the approximation ${f_\delta}$ converges to ${ f}$ uniformly on compact subsets of ${{\mathbb R}}$.
The choice of the particular mollifier given above is quite natural: we want a ${C^\infty}$ function with compact support (to avoid any issues with fast-growing functions ${f}$), so it cannot be analytic. And functions like ${\exp(-1/t)}$ are standard examples of infinitely smooth non-analytic functions. Being nonnegative is obviously a plus. What else to ask for?
Well, one may ask for a good rate of convergence. If ${f}$ is an ugly function like ${f(x)=\sqrt{|x|}}$, then we probably should not expect fast convergence. But is could be something like ${f(x)=|x|^7}$; a function that is already six times differentiable. Will the rate of convergence be commensurate with the head start ${f\in C^6}$ that we are given?
No, it will not. The limiting factor is not the lack of seventh derivative at ${x=0}$; it is the presence of (nonzero) second derivative at ${x\ne 0}$. To study this effect in isolation, consider the function ${f(x)=x^2}$, which has nothing beyond the second derivative. Here it is together with ${f_{0.1}}$: the red and blue graphs are nearly indistinguishable.
But upon closer inspection, ${f_{0.1}}$ misses the target by almost ${2\cdot 10^{-3}}$. And not only around the origin: the difference ${f_{0.1}-f}$ is constant.
With ${\delta=0.01}$ the approximation is better.
But upon closer inspection, ${f_{0.01}}$ misses the target by almost ${2\cdot 10^{-5}}$.
And so it continues, with the error of order ${\delta^2}$. And here is where it comes from:
$\displaystyle f_\delta(0) = \int_{\mathbb R} t^2\varphi_\delta(t)\,dt = \delta^{-1} \int_{\mathbb R} t^2\varphi(t/\delta)\,dt = \delta^{2} \int_{\mathbb R} s^2\varphi(s)\,ds$
The root of the problem is the nonzero second moment ${\int_{\mathbb R} s^2\varphi(s)\,ds \approx 0.158}$. But of course, this moment cannot be zero if ${\varphi}$ does not change sign. All familiar mollifiers, from Gaussian and Poisson kernels to compactly supported bumps such as ${\varphi}$, have this limitation. Since they do not reproduce quadratic polynomials exactly, they cannot approximate anything with nonzero second derivative better than to the order ${\delta^2}$.
Let’s find a mollifier without such limitations; that is, with zero moments of all orders. One way to do it is to use the Fourier transform. Since ${\int_{\mathbb R} t^n \varphi(t)\,dt }$ is a multiple of ${\widehat{\varphi}^{(n)}(0)}$, it suffices to find a nice function ${\psi}$ such that ${\psi(0)=1}$ and ${\psi^{(n)}(0)=0}$ for ${n\ge 1}$; the mollifier will be the inverse Fourier transform of ${\psi}$.
As an example, I took something similar to the original ${\varphi}$, but with a flat top:
$\displaystyle \psi(\xi) = \begin{cases} 1 \quad & |\xi|\le 0.1; \\ \exp\{1-1/(1-(|\xi|-0.01)^2)\} \quad & 0.1<|\xi|<1.1\\ 0\quad & |\xi|\ge 1.1 \end{cases}$
The inverse Fourier transform of ${\psi}$ is a mollifier that reproduces all polynomials exactly: ${p*\varphi = p}$ for any polynomial. Here it is:
Since I did not make ${\psi}$ very carefully (its second derivative is discontinuous at ${\pm 0.01}$), the mollifier ${\varphi}$ has a moderately heavy tail. With a more careful construction it would decay faster than any power of ${t}$. However, it cannot be compactly supported. Indeed, if ${\varphi}$ were compactly supported, then ${\widehat{\varphi}}$ would be real-analytic; that is, represented by its power series centered at ${\xi=0}$. But that power series is
$\displaystyle 1+0+0+0+0+0+0+0+0+0+\dots$
The idea of using negative weights in the process of averaging ${f}$ looks counterintuitive, but it’s a fact of life. Like the appearance of negative coefficients in the 9-point Newton-Cotes quadrature formula… but that’s another story.
Credit: I got the idea of this post from the following remark by fedja on MathOverflow:
The usual spherical cap mollifiers reproduce constants and linear functions faithfully but have a bias on quadratic polynomials. That’s why you cannot go beyond ${C^2}$ and ${\delta^2}$ with them. | 5,434 | 16,691 | {"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": 213, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-45 | latest | en | 0.914396 |
http://www.twistypuzzles.com/forum/viewtopic.php?f=15&t=24405&start=0 | 1,406,204,330,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997888866.9/warc/CC-MAIN-20140722025808-00184-ip-10-33-131-23.ec2.internal.warc.gz | 333,090,993 | 11,224 | Online since 2002. Over 3300 puzzles, 2600 worldwide members, and 270,000 messages.
TwistyPuzzles.com Forum
It is currently Thu Jul 24, 2014 7:18 am
All times are UTC - 5 hours
Page 1 of 1 [ 14 posts ]
Print view Previous topic | Next topic
Author Message
Post subject: Jason's RhombicultimatePosted: Mon Oct 01, 2012 1:59 am
Joined: Sat Apr 21, 2007 11:21 pm
Location: Marin, CA
This is a shape modification of my Pentultimate (mechanism v2.5), using caps to create a rhombic triacontahedron.
This modification was first suggested to my knowledge by Robert Webb in 2003 here.
One interesting thing about this modification is that it doesn't change shape as many other pentultimate shape modifications do. It's always a rhombic triacontahedron after every turn.
Every piece orientation matters, as opposed to the Pentultimate where only the triangles matter, or the Icosamate, where only the pentagons matter.
At the end of the video I'm attempting to show how well the puzzle can turn by using only one finger with fast turns. It gets a little loud...
Thanks for looking!
EDIT: I have learned that TomZ made this puzzle first, in 2010.
_________________
Jason Smith posted here as 'io' through 2012.
Visit Jason Smith's PuzzleForge on Shapeways!
Jason Smith's Puzzles - YouTube Channel.
Last edited by JasonSmith on Mon Oct 01, 2012 10:41 am, edited 1 time in total.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 2:19 am
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
What a beauty!
One of the things that gets me about solving a Pentultimate is that physically a half-turn (36 degrees) puts the dodecahedron into a shape that almost feels right. It seems the rhombic triacontahedron has this property and the way you've stickered your second photo that much more confusing.
Edit: corrected my reference to a rhombic dodecahedron. The puzzle is a rhombic triacontahedron.
_________________
Prior to using my real name I posted under the account named bmenrigh.
Last edited by Brandon Enright on Mon Oct 01, 2012 10:31 am, edited 1 time in total.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 2:50 am
Joined: Wed Dec 14, 2011 12:25 pm
Location: Finland
Do opposite faces have same colors? I think it's just been twisted half a turn from solved state.
_________________
My pen-and-paper puzzles
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 2:57 am
Joined: Sat Mar 22, 2003 9:11 am
Location: Marin, CA
Fun info about this stickering: Each pentagon has each of the five colors, which could be arranged in 4*3*2=24 possible arrangements, and there are 12 faces, so all the even permutations are used. Each triangle has three colors, and there are 5*4*3/3 = 20 possible colorings of those, and there are 20 triangles, so each possible one is included exactly once.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 9:47 am
Joined: Wed Jan 07, 2009 6:46 pm
Location: San Francisco, CA
How beautiful! And it turns like a dream, too!
Bram wrote:
Each pentagon has each of the five colors
I may be missing something here, but what pentagons? If you mean the vertices with 5-fold symmetry, then how could you say that each one has each of the five colors? There are more than 5 colors... Can you clarify?
_________________
Eitan = "EIGHT-ahn"
Check out my video: Twisty Puzzles a la Vi
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 10:24 am
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Coaster1235 wrote:
Do opposite faces have same colors? I think it's just been twisted half a turn from solved state.
Yeah it was half-twisted. What I was trying to say is that doing this on a dodecahedral pentultimate already "feels right" and his choice of stickering makes it look right too.
_________________
Prior to using my real name I posted under the account named bmenrigh.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 10:28 am
Joined: Mon Aug 02, 2004 7:03 am
Location: Koblenz, Germany
Seems like Tom van der Zanden was faster:
http://twistypuzzles.com/cgi-bin/puzzle.cgi?pkey=2634
Anyway: Variants of the Pentultimate are always amazing.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 10:35 am
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Bram wrote:
Fun info about this stickering: Each pentagon has each of the five colors, which could be arranged in 4*3*2=24 possible arrangements, and there are 12 faces, so all the even permutations are used. Each triangle has three colors, and there are 5*4*3/3 = 20 possible colorings of those, and there are 20 triangles, so each possible one is included exactly once.
I'm with you if this puzzle were stickered with only 5 colors but it looks like Jason has used 10 or 15.
_________________
Prior to using my real name I posted under the account named bmenrigh.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 10:44 am
Joined: Sat Apr 21, 2007 11:21 pm
Location: Marin, CA
Robert Webb did suggest a 5-color sticker pattern, but I didn't try it here. I just used 15 colors, with matching opposite faces to get the interesting half turn look.
Andreas, thanks for the info about TomZ's version. I didn't know I was second to get this one made. Apologies for not mentioning that, Tom.
_________________
Jason Smith posted here as 'io' through 2012.
Visit Jason Smith's PuzzleForge on Shapeways!
Jason Smith's Puzzles - YouTube Channel.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 11:03 am
Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
I'm happy that you made it, but sad because I can't afford to buy it, much less the whole family.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 11:28 am
Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
io wrote:
Every piece orientation matters, as opposed to the Pentultimate where only the triangles matter, or the Icosamate, where only the pentagons matter.
Unless I'm mistaken, the only platonic solid for which this is true is, of all things, the octahedron. In any other platonic shape, the Pentultimate has either non-orientable pieces or duplicate pieces.
(I'm guessing the center triangles on the octahedral version wouldn't lie flat after a 120-degree rotation... I could be wrong though.)
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 11:34 am
Joined: Wed Dec 14, 2011 12:25 pm
Location: Finland
Jared wrote:
(I'm guessing the center triangles on the octahedral version wouldn't lie flat after a 120-degree rotation... I could be wrong though.)
I'm pretty sure they do remain flat.
_________________
My pen-and-paper puzzles
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 12:40 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Jared wrote:
io wrote:
Every piece orientation matters, as opposed to the Pentultimate where only the triangles matter, or the Icosamate, where only the pentagons matter.
Unless I'm mistaken, the only platonic solid for which this is true is, of all things, the octahedron. In any other platonic shape, the Pentultimate has either non-orientable pieces or duplicate pieces.
(I'm guessing the center triangles on the octahedral version wouldn't lie flat after a 120-degree rotation... I could be wrong though.)
The rhombic triacontahedron is not a platonic solid (although it's closely related).
I'm not sure there is really a strict definition for how a shape-mod of the Pentultimate should look in the other platonic solids. For a dodecahedron the obvious choice (and arguably the one "correct" choice) has each cutting plane parallel to a face and bisecting the puzzle. For an Icosahedron you could also argue that the one correct choice simply transforms the dodecahedron into its dual.
For a tetrahedron, cube, and octahedron I think the choice is much less clear. If you Fisher / Axis / Ghost / whatever a cubic version of the Pentultimate you could probably make every piece unique and show orientation.
_________________
Prior to using my real name I posted under the account named bmenrigh.
Top
Post subject: Re: Jason's RhombicultimatePosted: Mon Oct 01, 2012 12:52 pm
Joined: Wed Nov 24, 2010 11:12 am
Location: Hong Kong/Beijing
Now we have Pentultimate in a Dodecahedron shape, a ball shape &......Rhombic shape!
Top
Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
Page 1 of 1 [ 14 posts ]
All times are UTC - 5 hours
#### Who is online
Users browsing this forum: Google [Bot] and 11 guests
You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum
Search for:
Jump to: Select a forum ------------------ Announcements General Puzzle Topics New Puzzles Puzzle Building and Modding Puzzle Collecting Solving Puzzles Marketplace Non-Twisty Puzzles Site Comments, Suggestions & Questions Content Moderators Off Topic | 2,459 | 9,242 | {"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-2014-23 | latest | en | 0.905913 |
https://metanumbers.com/465037 | 1,632,438,730,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057479.26/warc/CC-MAIN-20210923225758-20210924015758-00200.warc.gz | 439,744,452 | 7,390 | # 465037 (number)
465,037 (four hundred sixty-five thousand thirty-seven) is an odd six-digits composite number following 465036 and preceding 465038. In scientific notation, it is written as 4.65037 × 105. The sum of its digits is 25. It has a total of 2 prime factors and 4 positive divisors. There are 444,796 positive integers (up to 465037) that are relatively prime to 465037.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 25
• Digital Root 7
## Name
Short name 465 thousand 37 four hundred sixty-five thousand thirty-seven
## Notation
Scientific notation 4.65037 × 105 465.037 × 103
## Prime Factorization of 465037
Prime Factorization 23 × 20219
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 465037 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 465,037 is 23 × 20219. Since it has a total of 2 prime factors, 465,037 is a composite number.
## Divisors of 465037
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 485280 Sum of all the positive divisors of n s(n) 20243 Sum of the proper positive divisors of n A(n) 121320 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 681.936 Returns the nth root of the product of n divisors H(n) 3.83314 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 465,037 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 465,037) is 485,280, the average is 121,320.
## Other Arithmetic Functions (n = 465037)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 444796 Total number of positive integers not greater than n that are coprime to n λ(n) 20218 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 38764 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 444,796 positive integers (less than 465,037) that are coprime with 465,037. And there are approximately 38,764 prime numbers less than or equal to 465,037.
## Divisibility of 465037
m n mod m 2 3 4 5 6 7 8 9 1 1 1 2 1 6 5 7
465,037 is not divisible by any number less than or equal to 9.
## Classification of 465037
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (465037)
Base System Value
2 Binary 1110001100010001101
3 Ternary 212121220121
4 Quaternary 1301202031
5 Quinary 104340122
6 Senary 13544541
8 Octal 1614215
10 Decimal 465037
12 Duodecimal 1a5151
20 Vigesimal 2i2bh
36 Base36 9ytp
## Basic calculations (n = 465037)
### Multiplication
n×y
n×2 930074 1395111 1860148 2325185
### Division
n÷y
n÷2 232518 155012 116259 93007.4
### Exponentiation
ny
n2 216259411369 100568627884805653 46768133005666366454161 21748912268556070056743668957
### Nth Root
y√n
2√n 681.936 77.4752 26.1139 13.5987
## 465037 as geometric shapes
### Circle
Diameter 930074 2.92191e+06 6.79399e+11
### Sphere
Volume 4.21261e+17 2.7176e+12 2.92191e+06
### Square
Length = n
Perimeter 1.86015e+06 2.16259e+11 657662
### Cube
Length = n
Surface area 1.29756e+12 1.00569e+17 805468
### Equilateral Triangle
Length = n
Perimeter 1.39511e+06 9.36431e+10 402734
### Triangular Pyramid
Length = n
Surface area 3.74572e+11 1.18521e+16 379701
## Cryptographic Hash Functions
md5 2222cdac54698d657fd145421bddc848 5de3fb450bf4204ca0f2225b5743f8cc8fc0fbbc 678757906facf2b9f8a7c94fe58a211a992dddcaf5cae5e3052e1b1fdb0bd9bd 4b018f0cc28498458f086389f1b175b81b9ce9db4d600577da5c1aebf26dc22a74948e9f232eba2d7dbe23e25dd2ea23f34757879b5ce1b40cd1c9658b58344c b465cee86ffe15b7e8ea17ba77a49ec9b7dce521 | 1,482 | 4,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-39 | latest | en | 0.807782 |
https://forum.arduino.cc/t/bmp180-barometer-calibration/884543 | 1,680,099,774,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00039.warc.gz | 309,511,651 | 5,500 | # BMP180 Barometer Calibration?
While working on improving a portable weather station project, I noticed discrepancies with barometer measurements from the BMP180 sensor upon cross-referencing my data and locally reported weather data. Barometric pressure in my area generally is in the 1014 β 1018 Millibar range, readings from my sensor indicated an average of 951 millibar. After investigating if I had a bad sensor and going over numerous ways to calculate something to calibrate the BMP180 I came up with a solution using standard deviation that could hopefully shed light on this topic (if itβs been done or considered before? I donβt quite know). More accurate measurements were noticed after converting the default pascal(Pa) to millibar(Mbar = Pa / 100) then to inHg(Inch of mercury = Mbar / 33.864). Below is the calculation with (basePressure = 29.92inHg) as the "standard" pressure:
mBar = pressure / 100;
sensorPressure = mBar / 33.86;
sumCalc = basePressure + sensorPressure;
meanCalc = sumCalc / 2;
diff1 = meanCalc - sensorPressure;
diff2 = meanCalc - sensorPressure;
diff1Sq = diff1 * diff1;
diff2Sq = diff2 * diff2;
sumDiff = diff1Sq + diff2Sq;
variance = sumDiff;
realPressure = sensorPressure + variance;
This calculates the variance, then adds it to the measurement from readPressure as the value for realPressure. So far the measurements have been very close to matching with locally reported weather data. Feel free to shed more wisdom. Thanks!
Are you aware that air pressure is usually reported after correcting to sea level?
The correction depends on a number of factors (mostly altitude), and is described here.
1 Like
Ok, that makes sense. I wasn't aware of some other vital aspects of it. I'm fairly new to weather projects and this helped alot. Thank you
This topic was automatically closed 120 days after the last reply. New replies are no longer allowed. | 439 | 1,904 | {"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-2023-14 | latest | en | 0.926686 |
https://christopherberry.ca/how-to-predict-how-many-visits-a-website-will-receive-on-a-given-day/ | 1,723,197,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640762343.50/warc/CC-MAIN-20240809075530-20240809105530-00861.warc.gz | 138,431,829 | 15,452 | Predictive analytics is somewhat mysterious. So, let’s shed some light on it.
(Note that I’m simplifying this quite a bit to be accessible.)
The first step in predictive analytics is to understand what you’re predicting. We’ll call this the Y variable.
In this instance, ‘how many visits from Boston can I expect on a given day’. My Y will be ‘Visits’.
Have some discipline. I see way too many analysts change the Y variable before their investigation is through.
The second step is to identify all the variables that might be associated with a variation in Y. These might include factors like paid media, search, new visits, returning visits – and date. Then there are paid campaigns, posting new content, social campaigns, traditional media spend, promotions, and so on. Day of the week is another key variable, along with statutory holidays, and extending out to other factors like weather and creativity.
The third step is to extract, transform, and load the data you CAN actually access. You can spend months fighting to build an absolute complete model, or, you CAN start putting together a story with the facts that are available. I chose action over inertia. You should too.
That date field is usually pretty bad to extract, transform, and load. There are functions both in excel and SPSS that handle dates with some difficulty. Devils abound in the details around ‘the date where in the world’. If your installation is set to Eastern Time, and most of your traffic comes from Australia, you’ll be one day lagged. You ought to adjust the figures using the appropriate offset.
The figure below is what I could extract from Google Analytics in about an hour. (Collinearity abounds!)
The fourth step is to run the math against your model.
I use SPSS to run a regression. If you don’t have SPSS, you can try using open source programs like Octave or R. The reason for using software is because it’s annoying to do by hand. I didn’t enjoy a copy of SPSS at my first research position, so I had to code out linear regression in Excel. I learned a lot, but it is not expedient!
The figure below is the output from the software.
The way to read the table is Y = Constant + B1(X1) + B2(X2).
So, Visits = 4.888 – 1.872 (istheweekend).
If it’s the weekend, I can predict Visits = 4.888 – 1.872 (1). Which equals 3 visits.
If it’s not the weekend, I can predict Visits = 4.888 – 1.872(0). Which equals 4.888.
Not bad for Boston traffic! And I understand the impact of a single variable on visits.
My dataset is incredibly spikey. So, what’s causing some of that spikyness? I went through all the dates that I posted new content – reran the math, and got the table below.
The model above is the best. It explains 12.7% of the variance in the set.
The equation is: Visits = 4.496 -1.76(istheweekend) + 2.482(newpost).
I can tell – according to this version of reality – that if I want the maximum bump from Boston, posting during the weekday is best. And I can tell the proportional impact of each variable.
Sometimes this answer is good enough. There are more advanced methods – like curvilinear regression, machine learning, and neural networks. There are ways to introduce more variables into the equation. But typically – this method is sufficient to get a first idea about the relationships among variables and their relative importance, rooted in fact, as opposed to gut bias.
The fifth step is to make decisions based on scenarios.
If you take this equation and plot it out, you can engage in a few what-if’s. Would writing more weekend friendly material result in a lower Beta? Would increasing the frequency of new posts drastically improve the performance of the website? If so, by how much? The size of the newpost beta, as compared to the total number of Boston visits per day hints at that relative strength.
That’s the power of predictive analytics.
Category : | 877 | 3,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-33 | latest | en | 0.939716 |
https://www.thestudentroom.co.uk/showthread.php?t=2285043&page=6 | 1,524,155,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936981.24/warc/CC-MAIN-20180419150012-20180419170012-00634.warc.gz | 896,277,520 | 44,394 | x Turn on thread page Beta
You are Here: Home >< Maths
# AQA Core 3 - Thursday 6th June 2013 (AM) - Official Thread watch
1. For the last question the working is:
u=x^1/2
Integral = (1/x^1/2) X 1/(1+x^1/2)
du/dx= 1/2x^1/2
2du/dx=1/x^1/2
Integral =(1/1+u) X 2du/dx
limits are now 2 and 1
Integral = 2/1+u
= 2ln(1+u)
2. (Original post by edina101morris)
hang on, i think y=(2x-3)
then think f-1(x) was (e^x+3)/2
not sure cant quite remeber
I knew there was a 3 somewhere I couldn't remember, I think you are right might have put that as well.
3. (Original post by stuart_aitken)
Ummmmm. I didn't draw it. My graph software did!
There's the logic/
y = pi - cos^-1(x)
It's define for -1<x<1, because cos# cannot be outside this range. Therefore the limits for x are -1 to +1. So for the important values:
x= -1
y = pi - cos^-1(-1) = pi - pi = 0
(-1,0)>>>end point
x = 0
y = pi - cos^-1(0) = pi - pi/2) = pi/2
(0,pi/2)>>>y-intercep
x = 1,
y = pi-cos^1(1) = pi - 0 = pi
(1,pi)>>>end point
As can be seen from these results, the graph I drew is surely correct
Sorry you are right
4. (Original post by sophonax1)
For the last question the working is:
u=x^1/2
Integral = (1/x^1/2) X 1/(1+x^1/2)
du/dx= 1/2x^1/2
2du/dx=1/x^1/2
Integral =(1/1+u) X 2du/dx
limits are now 2 and 1
Integral = 2/1+u
= 2ln(1+u)
Yay! But tbh, I'm not worried about integrals as I just checked em on my calculator.
5. [QUOTE=Samtaylor47;43004668]
(Original post by Dinomouse)
I got 160pi/3 for the vase question and 5.05 and 1.23 for the 7 mark question with sec squared in.
For the volume question, I got the answer as 73/3 pi and checked in my calculator and it was the same?
This question was one i was fairly sure with haha. I don't know how you got the same answer with 73/3 pi, maybe a typing into calculator mistake?
6. (Original post by ryanc_95)
for the last one i got 2ln2+2. A few people have also said they got this, is this correct or is it 2ln(3/2)?
I got this!!!
Posted from TSR Mobile
7. What do you think method marks will be like on the last question?
I managed to work out that 2du = 1/X^0.5 dx but then when it came to integrating it did it all wrong :/
8. Okay so aqa have shown us how much they like graphs this year.
9. I just rearranged the whole equation to find x^2 and then integrated it?
10. (Original post by Dirtybit)
Okay so aqa have shown us how much they like graphs this year.
they went a little overboard...
11. For inequalities did u guys get x less than or equal to one and then x is greater than or equal to one ?
And for the question on the intercepts about ln bla bla the 2nd last part where it says find fg(x) in a certain form, anyone remember the answer?
12. For the last question I did everything right but put du/dx instead of dx/du I then followed all this through and got a very complicated answer. How many (out of 7) marks would I get?
13. Think it will be about 70 for 90ums and 64 for 80ums. What does everyone else think? I need an A* and I think I've dropped about 10 marks.
14. One of the many questions that confused me was the secx 5 one I only got two answers and that seems wrong for a 7 mark question... can any one clarify please
15. (Original post by esen212)
One of the many questions that confused me was the secx 5 one I only got two answers and that seems wrong for a 7 mark question... can any one clarify please
I got that too. You had to discard one of the values and the remaining value yielded 2 x solutions.
16. (Original post by Dirtybit)
For inequalities did u guys get x less than or equal to one and then x is greater than or equal to one ?
And for the question on the intercepts about ln bla bla the 2nd last part where it says find fg(x) in a certain form, anyone remember the answer?
I got x<=1 and x>=3. For the fg(x) part, I don't remember too well but ln8/2??
17. Guys what was the first integration question on the last page?
18. (Original post by esen212)
One of the many questions that confused me was the secx 5 one I only got two answers and that seems wrong for a 7 mark question... can any one clarify please
I got the same, cant have cos-1 of -3.
19. (Original post by esen212)
One of the many questions that confused me was the secx 5 one I only got two answers and that seems wrong for a 7 mark question... can any one clarify please
I got 2, there were to values for cosx but one had no solutions, I think?? Answer was 5.05 and 1.??
20. Came out thinking it was just about OK. Then discussed with friends and now I'm worried. Definitely got the final answers to the vase question and last integration by substitution question wrong. And there were many question where we had to draw graphs. Not sure if they were right either...
Posted from TSR Mobile
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: January 26, 2014
Today on TSR
### Negatives of studying at Oxbridge
What are the downsides?
### Grade 9 in GCSE English - AMA
Poll
Useful resources
Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR
## Groups associated with this forum:
View associated groups
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 1,608 | 5,532 | {"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-2018-17 | latest | en | 0.952089 |
http://stackoverflow.com/questions/13480026/interplating-3d-matrix-by-cubic-spline | 1,467,225,860,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397795.31/warc/CC-MAIN-20160624154957-00046-ip-10-164-35-72.ec2.internal.warc.gz | 310,491,931 | 15,888 | interplating 3d matrix by cubic spline
Map provides a 185 x 250 x 3 matrix Map representing an image and “image coordinates” xi and yi. interpolation coordinates 5 < x < 180 and 5 < y < 245, both evenly spaced with spacing 0.1. Interpolate each of the 3 blocks of size 185 x 250 of the matrix Map first in one direction, then the other, to produce a 1751 x 2401 x 3 matrix InterpolatedMap. Set any values larger than 1 to 1, and any less than 0 to 0. Use the image command to compare the original image Map to the image in InterpolatedMap.
I have generated code MyCubicInterpolation which works in following way, xi = linspace(-1,1,150); zi = sin((pi.*xi)/2);
x = linspace(-1,1,200);
y = MyCubicSpline(xi,zi,x);
Now I dont understand my which one will be my 1st direction and which one my second direction, should I select entire block Map(:,:,1) out of three. In short I can not understand next steps.
- | 253 | 910 | {"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-2016-26 | latest | en | 0.841772 |
http://www.excel-exercise.com/ | 1,513,318,848,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567042.50/warc/CC-MAIN-20171215060102-20171215080102-00070.warc.gz | 351,488,362 | 16,436 | Compare 2 columns
How to compare 2 columns in Excel? It’s very simple 😄😎👍 I will show you in this post how to do this with the functions VLOOKUP for the research ISNA for the test IF to customize the final result 😉 But first, let’s start with a live demo 💛❤💙💜💚 Live demo In this worksheet (yes, it’s …
Using conditional formatting to highlight dates
Date functions in Excel make it is possible to perform date calculations, like addition or subtraction, resulting in automated or semi-automated worksheets. The NOW function, which calculates values based on the current date and time, is a great example of this. Taking this functionality a step further, when you mix date functions with conditional formatting, …
Calculation without equal
I always wonder why, in most of the offices, there is a calculator close to a computer. It’s totally ridiculous especially if you have Excel on your computer. But a secretary told me that it’s not convenient for her to start with the calculation with the symbol = In fact, there is a …
Create an Offset Vlookup
In the case you have a table that contains the same ID for many – or empty cells – it is strongly recommended to reorder your data. To perform this modification, we will use 3 functions , INDEX , MATCH and OFFSET. Problem to be solved We have a table (column A:D ) that lists the sales …
Extract without duplicates with formula
It is very easy to extract without duplicates data from a table through the menu Data>Remove duplicates. It works great for one or more columns but the thing about this tool is that you can not control data that will be delete. I recently worked on a customer address file by keeping only the most …
Create an online survey with Excel
Since it is possible to use Excel from a web browser, new special features are available with Excel Online. This is particularly the case for creating, posting and collecting the results of a questionnaire. And not only is it very easy to implement a surver but it is free. Click on this link to see …
How to round in k\$ without formula?
In many companies, it is difficult to present in a worksheet, big numbers. So, to avoid that, it is better to present your figure rounded in kilo dollar (k\$) or million (m\$). But how to round in k\$? In this article you will find 3 techniques to display a result in k\$ (or m\$ for the millions]. Each …
Major update
Recently Microsoft has announced a new public update to Excel 2013. This new release includes major stability and usability improvements in Self Service BI scenarios – especially when Power Query, Power Pivot and Power View are involved. For this one we eliminated over 120 bugs, crashes and hangs. On the other hand, we introduced two new …
How to Lookup Values to left column?
The VLOOKUP function searches for data in a table by returning data always to the right of the column containing your ID. However, it is sometimes necessary to us to return the value that is located into the left columns of your ID. As the VLOOKUP function is not able to do, we can solve the problem … | 677 | 3,076 | {"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-2017-51 | latest | en | 0.892434 |
http://isabelle.in.tum.de/repos/isabelle/diff/c26eeb000470/src/ZF/Constructible/Relative.thy | 1,571,348,689,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986676227.57/warc/CC-MAIN-20191017200101-20191017223601-00259.warc.gz | 103,307,781 | 2,912 | src/ZF/Constructible/Relative.thy
changeset 13363 c26eeb000470 parent 13353 1800e7134d2e child 13382 b37764a46b16
``` 1.1 --- a/src/ZF/Constructible/Relative.thy Tue Jul 16 16:28:49 2002 +0200
1.2 +++ b/src/ZF/Constructible/Relative.thy Tue Jul 16 16:29:36 2002 +0200
1.3 @@ -28,6 +28,15 @@
1.4 successor :: "[i=>o,i,i] => o"
1.5 "successor(M,a,z) == is_cons(M,a,a,z)"
1.6
1.7 + number1 :: "[i=>o,i] => o"
1.8 + "number1(M,a) == (\<exists>x[M]. empty(M,x) & successor(M,x,a))"
1.9 +
1.10 + number2 :: "[i=>o,i] => o"
1.11 + "number2(M,a) == (\<exists>x[M]. number1(M,x) & successor(M,x,a))"
1.12 +
1.13 + number3 :: "[i=>o,i] => o"
1.14 + "number3(M,a) == (\<exists>x[M]. number2(M,x) & successor(M,x,a))"
1.15 +
1.16 powerset :: "[i=>o,i,i] => o"
1.17 "powerset(M,A,z) == \<forall>x[M]. x \<in> z <-> subset(M,x,A)"
1.18
1.19 @@ -161,15 +170,6 @@
1.20 --{*omega is a limit ordinal none of whose elements are limit*}
1.21 "omega(M,a) == limit_ordinal(M,a) & (\<forall>x[M]. x\<in>a --> ~ limit_ordinal(M,x))"
1.22
1.23 - number1 :: "[i=>o,i] => o"
1.24 - "number1(M,a) == (\<exists>x[M]. empty(M,x) & successor(M,x,a))"
1.25 -
1.26 - number2 :: "[i=>o,i] => o"
1.27 - "number2(M,a) == (\<exists>x[M]. number1(M,x) & successor(M,x,a))"
1.28 -
1.29 - number3 :: "[i=>o,i] => o"
1.30 - "number3(M,a) == (\<exists>x[M]. number2(M,x) & successor(M,x,a))"
1.31 -
1.32 is_quasinat :: "[i=>o,i] => o"
1.33 "is_quasinat(M,z) == empty(M,z) | (\<exists>m[M]. successor(M,m,z))"
1.34
1.35 @@ -177,7 +177,7 @@
1.36 "is_nat_case(M, a, is_b, k, z) ==
1.37 (empty(M,k) --> z=a) &
1.38 (\<forall>m[M]. successor(M,m,k) --> is_b(m,z)) &
1.39 - (is_quasinat(M,k) | z=0)"
1.40 + (is_quasinat(M,k) | empty(M,z))"
1.41
1.42 relativize1 :: "[i=>o, [i,i]=>o, i=>i] => o"
1.43 "relativize1(M,is_f,f) == \<forall>x[M]. \<forall>y[M]. is_f(x,y) <-> y = f(x)"
1.44 @@ -669,8 +669,10 @@ | 813 | 1,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.303603 |
https://www.northernarchitecture.us/earthbag-building/foundation-and-stem-wall.html | 1,579,795,851,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00518.warc.gz | 1,002,592,576 | 7,970 | ## Foundation and Stem Wall
We began our bag work right on the ground of this excavation;these bags became the foundation of our structure. The first row of bags can be filled with gravel to inhibit capillary action from the ground up into the earthen walls. The continuing bag work up to grade consisted of the earthen fill we had prepared previously (Fig. 12.6).
Use the compass arm to delineate the shape of the structure. The angle bracket set on the horizontal arm denotes where the inside circumference of the finished, tamped bags should be. Use this arm when initially placing each bag. Set the filled, untamped bag about one inch (2.5 cm) outside this angle bracket. This will allow for expansion of the bag once it is tamped. The horizontal compass arm must be used for each bag placed around the circle (Fig. 12.7).
12.7: Use the compass arm to delineate the contour of each bag to conform to the circle.
When this first row of earth-filled bags have been placed and tamped, swing the compass arm around the perimeter to see how well you have done. You may need to further tamp a few bags to meet the angle bracket on the horizontal arm. Or possibly the bags came in too far and need to be pounded outward a little so the angle arm can pass without binding on the bags. Adjust the bags, not the arm. This first row will tell you how the following rows need to be modified to adjust to the proper diameter of the compass arm (Fig. 12.8).
If you read the section in Chapter 3 concerning the building compass, you will have bound or taped a level onto the horizontal arm of the compass. As you rotate the compass arm around to check the
proper placement of each bag in the circle, also check the level of each bag in respect to the other bags. The idea is to keep each row as level as possible. Measure the height of the bags after they have been tamped. The bags we used at this stage of the building process tamped down to a thickness of five inches (12.5 cm). Once the average thickness is determined, it's simply a matter of raising the horizontal arm of the compass the corresponding amount for the next row. Tamp the bags until they are level with this setting.
Several options may be considered for creating the stem wall. Two rows of concrete, stabilized earth-bags, or gravel-filled tires, are all effective stem wall options. If using exterior rigid foam insulation, install way too big stem seated on gravel sill
Hz0-
brane tucked under-stern wall bag
rz Tu V
brane tucked under-stern wall bag gravel bag
12.10: Honey House foundation detail.
interior floor
12.9: Continue the bag work until the bags are up to or just below grade. This is where the stem wall of the structure will begin.
the foam high enough to protect the stem wall. Another option at this point, instead of insulation or in addition to it, is to install a moisture barrier around the perimeter of the bag work up to the top of the stem wall. (All of this is covered in excruciating detail in Chapter 4).
For the below-ground bag work that we are describing, the dome walls are designed to be vertically plumb (like a yurt, or what we refer to as kiva-style) to provide a little additional interior height. Think of it as a knee wall in an attic. Choose a stem wall height that is appropriate for your climate -usually the wetter and colder the environment, the higher the stem wall. Your individual circumstances are paramount in making these choices (Fig. 12.9).
12.11: Way too big bags on gravel sill at grade.
For the Honey House dome, we began the below-grade bag work using a standard 50-lb. bag that tamped to about 15 inches (37.5 cm) wide. For the stem wall we switched to the larger bags we call way-too-big. We maintained the same interior diameter of the bags below, allowing the way-too-big bags to extend beyond the outside perimeter of the lower bags. We accomplished this by backfilling and tamping the outside space below grade (after installing a moisture barrier) with gravel, up to the level where the stem wall bags begin (Fig. 12.10 & 12.11).
12.12: We made a top, or plan view, drawing to delineate how we wanted the windows and door oriented by sighting the compass arm down the center of the box forms.
12.13: Use the compass to align box forms.
12.12: We made a top, or plan view, drawing to delineate how we wanted the windows and door oriented by sighting the compass arm down the center of the box forms. | 988 | 4,443 | {"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-2020-05 | latest | en | 0.934432 |
https://math.stackexchange.com/questions/2571371/fundamental-group-of-circle-union-a-line | 1,696,206,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510942.97/warc/CC-MAIN-20231002001302-20231002031302-00751.warc.gz | 417,260,166 | 35,710 | # Fundamental group of circle union a line
Let $X$ be the union of the unit circle centered at 0 and the line segment between the points $(1,0)$ and $(2,0)$. What is the universal covering of $X$? Compute the fundamental group of $X$.
My Attempt:
The fundamental group seems easy enough. Since the line segment between the two points has the point $(1,0)$ as its deformation retract, the fundamental group of $X$ will simply be the fundamental group of the circle which is $\mathbb{Z}$.
As for the universal cover, can we take the usual covering of the circle which is $\mathbb{R}$? Is there a way to formally show this?
• The universal cover will look like $A \cup B$ where $A$ is the $y$-axis and $B$ is the set of $\{(t,n) : t \in \Bbb R, n \in \Bbb N\}$. Dec 18, 2017 at 5:13
• @NicolasHemelsoet can you explain? Dec 18, 2017 at 5:19
• Oh sorry I was thinking to a different space (I was thinking to the union of a circle with a tangent line). But are you sure about the fundamental group ? If seems to me that $X$ retracts on a wedge of two circles. Dec 18, 2017 at 5:22
• @NicolasHemelsoet How do you figure? I thought any line was null-homotopic, so it can be homotoped to a point. So all we are left with is the circle. Is this bad logic? Dec 18, 2017 at 5:26
• You are right, but when you do this there is still a little portion of circle which will become a full circle. For example move the line so that $X$ becomes $S^1$ with the vertical segment from $(0,-1)$ to $(0,1)$. Now, you can retracts this segment and you see that you have indeed two circles intersecting in one point. This gives $\pi_1(X)$ is the free group on two generators. Dec 18, 2017 at 5:29
## 1 Answer
Let $\widetilde{X}$ denote the universal covering of $X$. Then we can write $\widetilde{X}$ as a subset of $\mathbb{R}^2$: $$\widetilde{X} = \{(x,0)\in\Bbb{R}^2~:~x\in\Bbb{R}\} \cup \{(n,t)\in\Bbb{R}^2~:~n\in\mathbb{Z}, \; 0\leq t \leq 1\}.$$
A picture of $X$ and $\widetilde{X}$ is below. One should be able to construct the covering map from these descriptions.
• Thank you Adam. This makes sense. How about the fundamental group. Am I right in saying it is $\mathbb{Z}$? Dec 18, 2017 at 20:04
• Yes, the fundamental group is isomorphic to $\Bbb{Z}$. Dec 18, 2017 at 20:08
• @adam lowrance wouldn’t The real line work as well? Since this spiky space can be retracted to it
– Zee
Dec 19, 2017 at 4:20
• The real line is not the universal cover of $X$ because a covering map must be a local homeomorphism. Informally, the point $(1,0)$ in $X$ locally has a "Y" shape and so the same must be true of its preimage in any covering space. On the other hand, one can prove that $X$ deformation retracts onto the circle $S^1$, and so $\pi_1(X)\cong \mathbb{Z}$. A major part of the proof that $\pi_1(X)\cong\mathbb{Z}$ involves the universal cover of $S^1$, which is the real line $\mathbb{R}$. Dec 19, 2017 at 4:33
• +1. Was trying to figure out how to say what @AdamLowrance posted but it's already a great answer. Dec 20, 2017 at 3:55 | 954 | 3,024 | {"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} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.903305 |
https://de.mathworks.com/matlabcentral/profile/authors/1932847-hananeh?s_tid=cody_local_to_profile | 1,585,677,184,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00407.warc.gz | 435,795,447 | 18,310 | Community Profile
# Hananeh
### university of illinois at chicago
12 total contributions since 2014
View details...
Contributions in
View by
Solved
Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
etwa 5 Jahre ago
Solved
Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
etwa 5 Jahre ago
Solved
Rounding
Round 10.67 and make 'y' equal to that number.
etwa 5 Jahre ago
Solved
Remainder
Make 'y' equal to the remainder of 27 divided by 5. When x=27 and t=5
mehr als 5 Jahre ago
Solved
Evaluating a polynomial
Given the following polynomial and the value for x, determine y. y = 3x^5 – x^3 + 8x – 3 Example x = 1 y = 3 - 1 +...
mehr als 5 Jahre ago
Problem
Evaluating a polynomial
Given the following polynomial and the value for x, determine y. y = 3x^5 – x^3 + 8x – 3 Example x = 1 y = 3 - 1 +...
mehr als 5 Jahre ago | 1
Solved
length of a vector
Find twice the length of a given vector.
mehr als 5 Jahre ago
Problem
length of a vector
Find twice the length of a given vector.
mehr als 5 Jahre ago | 0
Problem
size of a vector
Find twice the size of a given vector.
mehr als 5 Jahre ago | 0
Problem
size of a vector
Find twice the size of a given vector.
mehr als 5 Jahre ago | 0
Solved
Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
mehr als 5 Jahre ago
Solved
Max of a Vector
Write a function to return the max of a vector
fast 6 Jahre ago | 532 | 1,711 | {"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-2020-16 | latest | en | 0.427059 |
http://forum.math.toronto.edu/index.php?action=profile;area=showposts;u=2150 | 1,611,697,797,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803737.78/warc/CC-MAIN-20210126202017-20210126232017-00191.warc.gz | 39,612,464 | 5,250 | Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
Messages - yuhan cheng
Pages: [1]
1
Quiz-4 / tut0402 quiz4
« on: October 18, 2019, 02:01:58 PM »
22.$$y^{\prime \prime}+2 y^{\prime}+2 y=0, y(\pi / 4)=2, y^{\prime}(\pi / 4)=-2$$
$$\begin{array}{c}{y^{\prime \prime}+2 y^{\prime}+2 y=0} \\ {r^{2}+2 r+2=0} \\ {r=-1 \pm i} \\ {y=c_{1} e^{-t} \cos t+c_{2} e^{-t} \sin t}\end{array}$$
We know that general solution is $y=c_{1} e^{-t} \cos t+c_{2} e^{-t} \sin t$
Initial condition $y(\pi / 4)=2$
$$\begin{array}{l}{c_{1} e^{-\pi / 4} \cdot \frac{\sqrt{2}}{2}+c_{2} e^{-\pi / 4} \cdot \frac{\sqrt{2}}{2}=2} \\ {c_{1}+c_{2}=2 \sqrt{2} e^{\pi / 4}}\end{array}$$
$$\begin{array}{l}{y^{\prime}=-c_{1} e^{-t} \cos t-c_{1} e^{-t} \sin t-c_{2} e^{-t} \sin t+c_{2} e^{-t} \cos t} \\ {y^{\prime}(\pi / 4)=-2} \\ {-c_{1} e^{-\pi / 4} \frac{\sqrt{2}}{2}-c_{1} e^{-\pi / 4} \frac{\sqrt{2}}{2}-c_{2} e^{-\pi / 4} \frac{\sqrt{2}}{2}+c_{2} e^{-\pi / 4} \frac{\sqrt{2}}{2}=-2} \\ {-\sqrt{2} c_{1}=-2 e^{\pi / 4}} \\ {c_{1}=\sqrt{2} e^{\pi / 4}} \\ {c_{2}=2 \sqrt{2} e^{\pi / 4}-\sqrt{2} e^{\pi / 4}=\sqrt{2} e^{\pi / 4}}\end{array}$$
Solution: $y=\sqrt{2} e^{-t+\pi / 4} \cos t+\sqrt{2} e^{-t+\pi / 4} \sin t$
2
Quiz-3 / TUT 5102 QUIZ3
« on: October 11, 2019, 02:00:45 PM »
$y''-2y'-2y=0$
$y''-2y'-2y=0$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation
$r^2-2r-2=0$
We use the quadratic formula which is
$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Hence,
$\left\{ \begin{array}{c} r_1=1+\sqrt3\\ r_2=1-\sqrt3\\ \end{array} \right.$
Since the general solution has the form of
$y=c_1e^{r_1t}+c_2e^{r_2t}$
Therefore, the general solution of the given differential equation is
$y=c_1e^{(1+\sqrt3)t}+c_2e^{(1-\sqrt3)t}$
3
Quiz-3 / TUT 5102 QUIZ3
« on: October 11, 2019, 02:00:00 PM »
12. $y^{\prime \prime}+3 y^{\prime}=0, y(0)=-2, y^{\prime}(0)=3$
Find roots of characteristic equation:
$$\begin{array}{l}{r^{2}+3 r=0} \\ {r(r+3)=0} \\ {r_{1}=0} \\ {r_{2}=-3}\end{array}$$
General solution of equation is:
$$y=c_{1}+c_{2} e^{-3 t}$$
Initial terms:
$$\begin{array}{l}{y(0)=-2} \\ {-2=c_{1}+c_{2} e^{0}} \\ {-2=c_{1}+c_{2}}\end{array}$$
$$\begin{array}{l}{y^{\prime}(0)=3} \\ {-3 c_{2} e^{-3 t}=3} \\ {-3 c_{2} e^{0}=3} \\ {-3 c_{2}=3}\end{array}$$
$$\begin{array}{l}{c_{2}=-1} \\ {c_{1}+c_{2}=-2} \\ {c_{1}=-2+1=-1}\end{array}$$
Solution of problem: $$y=-1-e^{-3 t}$$
4
Quiz-2 / TUT0402 quiz2
« on: October 04, 2019, 02:04:45 PM »
$$1+(\frac{x}{y}-\sin(y))y^{\prime}=0.$$
\noindent Let
$$M(x,y)=1\quad \text{and}\quad N(x,y) =(\frac{x}{y}-\sin(y))$$
Then
$$\frac{\partial}{\partial y}M(x,y)=0\quad\text{and}\quad \frac{\partial}{\partial x}N(x,y)=\frac{1}{y}$$
We can see that this equation is not exact, however, note that
$$\frac{d\mu}{d y}=(\frac{N_x-M_y}{M})\mu=\frac{\mu}{y}\quad\rightarrow\quad \mu=y$$
Multiplying our original euqation by$\mu(y)$, we have
$$y+(x-y\sin(y))y^{\prime}=0$$
We can see that this equaion is exact, since
$$\frac{\partial}{\partial y}(y)=1=\frac{\partial}{\partial x}(x-y\sin(y))$$
Thus, there exists a function $\psi(x,y)$ such that
\begin{align}
\psi_x (x,y)&=y\\
\psi_y (x,y)&=x-y\sin(y)
\end{align}
Integating (1) with respect to x, we get
$$\psi(x,y)=xy+h(y)$$
for some arbitary function $h$ of $y$. Next, differentiating with respect to $y$, and equating with (2), we get
$$\psi_y(x,y)=x+h^{\prime}(y)$$
Therefore,
$$h^{\prime}(y)=-y\sin(y)\quad\rightarrow\quad h(y)=s\int y\sin(y)dy=y\cos(y)-\sin(y)$$
and we have
$$\psi(x,y)=xy+y\cos(y)-\sin(y)$$
Thus, the solutions of the differential equation are given implicity by
$$xy+y\cos(y)-\sin(y)=C$$
5
Quiz-2 / TUT0402 quiz2
« on: October 04, 2019, 02:00:32 PM »
tut 0402 quiz2
6
Quiz-1 / TUT0402 QUIZ1
« on: September 27, 2019, 02:00:00 PM »
tut 0402 quiz 1
Pages: [1] | 1,787 | 3,888 | {"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": 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} | 3.9375 | 4 | CC-MAIN-2021-04 | latest | en | 0.534526 |
http://mathhelpforum.com/calculus/11280-calc-gradient-question.html | 1,481,140,208,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542244.4/warc/CC-MAIN-20161202170902-00507-ip-10-31-129-80.ec2.internal.warc.gz | 178,067,688 | 10,334 | Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
2. Originally Posted by bobby77
Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
$
\nabla f=[2\,x\,y^3\,z^6,\ 3\,x^2\,y^2\,z^6,\ 6\,x^2\,y^3\,z^5]
$
so at the point in question:
$
\nabla f=[-2,\ 3,\ -6]
$
,
so the unit vector in the direction that $f$ is increasing most rapidly is:
$\frac{\nabla f}{|\nabla f|}=[-2/7,\ 3/7,\ -6/7]$,
and the rate of increase in this direction is $|\nabla f|=7$
RonL
3. Originally Posted by bobby77
Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
That is the gradient at the point.
$\nabla f = 2xy^3z^6\bold{i}+3x^2y^2z^6\bold{j}+6x^2y^3z^5\bol d{k}$
Substitute, in the values of $x,y,z$ to get the optimal vector it increases by.
$2(1)(-1)^3(1)^6\bold{i}+3(1)^2(-1)^2(1)^6\bold{j}+6(1)^2(-1)^3(1)^5\bold{k}=-2\bold{i}+3\bold{j}-6\bold{k}$
The norm of this vector is $\sqrt{2^2+3^2+6^2}=7$.
Thus, the unit vector is,
$\bold{u}=-\frac{2}{7}\bold{i}+\frac{3}{7}\bold{j}-\frac{6}{7}\bold{k}$
This is the direction of maximal increase. | 550 | 1,360 | {"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": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-50 | longest | en | 0.755593 |
https://community.deeplearning.ai/t/wk2-mobilenet-architecture-video-residual-skip-connection/223954 | 1,712,938,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00496.warc.gz | 155,766,273 | 6,948 | # Wk2, MobileNet architecture video - Residual/skip connection
In the video at around 3:30 the skip connection is depicted as skipping the input and output “blocks”. Is this correct!? I thought the residual connection should skip the “bottleneck” block.
I think you are just misinterpreting the graphical presentation there. I suggest you watch it again with the following idea in mind: the “bottleneck block” is not a single layer. It is a sequence of layers. The diagram they show has the “skip” layer diverging right before the “expansion” step that marks the start of the bottleneck block and it rejoins right after the “shrinking” step that is the last step of the bottleneck block. So it is precisely skipping around the bottleneck block. That is exactly what they are depicting there.
The other point is that this is just one such instance. The full MobilNet consists of multiple of those concatenated and some other layers as well.
That’s exactly how it is shown in the first part of the video (prior to 3:30) which makes perfect sense. However, it appears that starting at 3:30 the skip connection picks up before the input block and plugs back in right after the output.
My point is still that you are misinterpreting what you are seeing. What is the “input” block in your definition? I claim that is what I am calling the “expansion” block that is the first step of the bottleneck layer. The skip connection takes the same input that the expansion block is getting and passes it around and rejoins after the “shrink” step at the end of the bottleneck block.
Or maybe the simpler way to state my point is that it looks like the problem is that we are disagreeing about the definition of the “bottleneck block”. My claim is that it is exactly everything that is being skipped by the residual connection in that diagram. That’s the point.
Here is a screenshot of the confusing diagram in the video.
Thanks, @paulinpaloalto! I understand what you are saying. Perhaps, the diagram in the video loosely shows the arrow originating prior to the n x n x 3 input “block” and rejoining right after the n x n x 3 output block. The correct depiction of the skip connection in this case would be if the arrow originated right after the input block (but before the expansion block) and rejoined right after the projection, but before the output block. Right?
Sorry, I still think the diagram is correct as written, but it’s been a while since I listened to this lecture. I will need to watch it again and perhaps actually look at the MobilNet code in this weeks assignments to make sure I understand what is going on here.
Ok, I think this boils down to how to interpret what Prof Ng means by that diagram. I looked at `model.summary()` of the MobilNet model in the Transfer Learning assignment. He’s not really showing a normal ConvNet diagram here as he did in C4 W1 where each block shows the output of that block. The entire diagram shows one complete “Bottleneck” section. The input to that whole section is n x n x 3 and that same input is also the input to the “skip” connection. Then in the bottleneck layer, you do the following operations:
1. The expansion, which is a 1 x 1 convolution with 18 output filters. So each filter is 1 x 1 x 3. That produces the second box on the diagram.
2. To get from the second to the third box in that diagram, you do a “depthwise” convolution. The input and output of that operation are both n x n x 18.
3. Then you do another 1 x 1 convolution to reduce the channels back to 3. So there are 3 filters there, each of dimension 1 x 1 x 18. That produces the output which is n x n x 3.
4. The “skip input” shaped n x n x 3 is added to the output of step 3) to form the output of the entire operation.
So that is what the above diagram actually means. With that interpretation, I claim that my way of reading it matches the way the picture is drawn. You are simply interpreting the meaning of the boxes in the drawing differently, but we are both saying the same thing in terms of the actual operations that happen.
Ok, thanks @paulinpaloalto! | 929 | 4,098 | {"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} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.951126 |
https://programmer.ink/think/pytorch-quick-start.html | 1,632,269,410,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00703.warc.gz | 521,724,247 | 4,666 | # pyTorch quick start
Posted by slashpine on Thu, 13 Feb 2020 08:02:29 +0100
Preface:
The epidemic is too serious to start school. It may only start in March or even April. This winter vacation is really long. Use this time to learn something at home. Self discipline can let me make full use of my time; self discipline can let me supplement my shortcomings; self discipline can let me study the theoretical basis in depth; self discipline can let me further improve the code ability.
This year, I hope that I can be more self disciplined and step by step close to qualified Algorithm Engineers.
In recent days, I'm very interested in the book "hands on deep learning pyTorch edition", but I'm not familiar with pyTorch. So let's quickly learn. As the introduction to pyTorch, I attach the official link of pyTorch https://pyTorch.org/tutorials/starter/blitz/sensor_tutorial.html#sphx-glr-begin-blitz-sensor-tutorial-py
## 1. pyTorch
pyTorch is similar to numpy, while Tensors in pyTorch are similar to darray in numpy.
## 2. Basic operation
(1) Declares an uninitialized matrix, but does not contain a known value before use.
```x = torch.empty(5, 3)
print(x)
```
```tensor([[2.3576e-36, 0.0000e+00, 0.0000e+00],
[0.0000e+00, 0.0000e+00, 0.0000e+00],
[0.0000e+00, 0.0000e+00, 2.8026e-45],
[0.0000e+00, 1.1210e-44, 0.0000e+00],
[1.4013e-45, 0.0000e+00, 0.0000e+00]])
```
(2) Constructing a random initialization matrix
```x = torch.rand(5, 3)
print(x)
```
```tensor([[0.2031, 0.8195, 0.2181],
[0.4732, 0.4602, 0.8097],
[0.6037, 0.4483, 0.2570],
[0.1677, 0.1944, 0.7259],
[0.3056, 0.8363, 0.1282]])
```
(3) Construct a matrix whose elements are all zero and whose type is long
```x = torch.zeros(5, 3, dtype=torch.long)
print(x)
```
```tensor([[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0]])
```
(4) Construct tensors directly from data
```x = torch.tensor([5.5, 3])
print(x)
```
```tensor([5.5000, 3.0000])
```
(5) Create tensors based on existing tensors (unless the user provides a new value, these methods will reuse the properties of existing tensors, such as dtype)
```x = x.new_ones(5, 3, dtype=torch.double) # new_* methods take in sizes
print(x)
x = torch.randn_like(x, dtype=torch.float) # override dtype! Same as Xsize above
print(x)
```
```tensor([[1., 1., 1.],
[1., 1., 1.],
[1., 1., 1.],
[1., 1., 1.],
[1., 1., 1.]], dtype=torch.float64)
tensor([[-0.0929, 0.9791, 0.1481],
[-1.7100, 0.4316, -0.4209],
[-0.6479, 1.5173, -0.3646],
[-1.3288, 2.6447, -0.6539],
[-0.0300, 1.8167, 0.8633]])
```
```print(x.size())
torch.Size([5, 3])
```
(6) Vector addition, subtraction, multiplication and division
```y = torch.rand(5, 3)
#Law 1
print(x + y)
#Law two
#Law three
result = torch.empty(5, 3)
print(result)
#Law four
print(y)
```
The result is the same after operation
```tensor([[ 0.4184, 0.6502, 0.3735],
[ 2.0532, 0.1959, 0.6630],
[ 1.6664, 0.4967, 0.7961],
[ 0.8452, 1.1509, 0.6831],
[ 1.0740, 0.8284, -0.2371]])
tensor([[ 0.4184, 0.6502, 0.3735],
[ 2.0532, 0.1959, 0.6630],
[ 1.6664, 0.4967, 0.7961],
[ 0.8452, 1.1509, 0.6831],
[ 1.0740, 0.8284, -0.2371]])
tensor([[ 0.4184, 0.6502, 0.3735],
[ 2.0532, 0.1959, 0.6630],
[ 1.6664, 0.4967, 0.7961],
[ 0.8452, 1.1509, 0.6831],
[ 1.0740, 0.8284, -0.2371]])
tensor([[ 0.4184, 0.6502, 0.3735],
[ 2.0532, 0.1959, 0.6630],
[ 1.6664, 0.4967, 0.7961],
[ 0.8452, 1.1509, 0.6831],
[ 1.0740, 0.8284, -0.2371]])
```
(7) Index that can be similar to NumPy
```print(x[:, 1])
```
```tensor([ 0.5996, -0.0382, 0.4017, 0.5467, 0.2213])
```
(8) Resize
To adjust the tensor size, you can use torch.view:
```x = torch.randn(4, 4)
y = x.view(16)
z = x.view(-1, 8) # -1 means to infer from other dimensions
print(x.size(), y.size(), z.size())
```
```torch.Size([4, 4]) torch.Size([16]) torch.Size([2, 8])
```
(9) If there is only one element's tensor, use. item() to get the value
```x = torch.randn(1)
print(x)
print(x.item())
```
```tensor([-0.4073])
-0.40732377767562866
```
(10) The transformation of Torch tensor and NumPy array
```# torch to numpy
a = torch.ones(5)
print(a)
b = a.numpy()
print(b)
tensor([1., 1., 1., 1., 1.])
[1. 1. 1. 1. 1.]
```
```# numpy to torch
import numpy as np
a = np.ones(5)
b = torch.from_numpy(a) | 1,744 | 4,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-39 | latest | en | 0.822258 |
https://edurev.in/t/120042/NCERT-Solutions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-Exercise-52 | 1,713,336,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00415.warc.gz | 186,831,516 | 66,978 | NCERT Solutions: Arithmetic Progressions (Exercise 5.2)
# NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions (Exercise 5.2)
Q1: Fill in the blanks in the following table, given that a is the first term, d is the common difference and an the nth term of the A.P.
Sol:
(i) Given:
First term, a = 7
Common difference, d = 3
Number of terms, n = 8,
We have to find the nth term, an = ?
As we know, for an A.P.,
an = a+(n−1)d
Putting the values:
⇒ 7+(8 −1) 3
⇒ 7+(7) 3
⇒ 7+21 = 28
Hence, an = 28
(ii) Given:
First term, a = -18
Common difference, d = ?
Number of terms, n = 10
Nth term, an = 0
As we know, for an A.P.,
an = a+(n−1)d
Putting the values,
⇒ 0 = − 18 +(10−1)d
⇒ 18 = 9d
⇒ d = 18/9 = 2
Hence, common difference, = 2
(iii) Given:
First term, a = ?
Common difference, d = -3
Number of terms, n = 18
Nth term, an = -5
As we know, for an A.P.,
an = a+(n−1)d
Putting the values,
⇒ −5 = a+(18−1) (−3)
⇒ −5 = a+(17) (−3)
⇒ −5 = a−51
⇒ a = 51−5 = 46
Hence, a = 46
(iv) Given:
First term, a = -18.9
Common difference, d = 2.5
Number of terms, n = ?
Nth term, an = 3.6
As we know, for an A.P.,
⇒ an = a +(n −1)d
Putting the values,
⇒ 3.6 = − 18.9+(n −1)2.5
⇒ 3.6 + 18.9 = (n−1)2.5
⇒ 22.5 = (n−1)2.5
⇒ (n – 1) = 22.5/2.5
⇒ n – 1 = 9
⇒ n = 10
Hence, n = 10
(v) Given:
First term, a = 3.5
Common difference, d = 0
Number of terms, n = 105
Nth term, an = ?
As we know, for an A.P.,
an = a+(n −1)d
Putting the values,
⇒ an = 3.5+(105−1) 0
⇒ an = 3.5+104×0
⇒ an = 3.5
Hence, an = 3.5
Q2: Choose the correct choice in the following and justify:
(i) 30th term of the A.P: 10,7, 4, …, is
(a) 97
(b) 77
(c) −77
(d) −87
Sol:
Given here:
A.P. = 10, 7, 4, …
Therefore, we can find,
First term, a = 10
Common difference, d = a2 − a= 7−10 = −3
As we know, for an A.P.,
an = a +(n−1)d
Putting the values;
⇒ a30 = 10+(30−1)(−3)
⇒ a30 = 10+(29)(−3)
⇒ a30 = 10−87 = −77
Hence, the correct answer is option c.
(ii) 11th term of the A.P. -3, -1/2, ,2 …. is
(a) 28
(b) 22
(c) – 38
(d)
Given here:
A.P. = -3, -1/2, ,2 …
Therefore, we can find,
First term a = – 3
Common difference, d = a2 − a1 = (-1/2) -(-3)
⇒ (-1/2) + 3 = 5/2
As we know, for an A.P.,
an = a+(n−1)d
Putting the values;
⇒ a11 = 3+(11-1)(5/2)
⇒ a11 = 3+(10)(5/2)
⇒ a11 = -3+25
⇒ a11 = 22
Hence, the answer is option B.
Q3: In the following APs find the missing term in the boxes.
Sol:
(i) For the given A.P., 2, 2, 26
The first and third term are;
a = 2
a3 = 26
As we know, for an A.P.,
an = a+(n −1)d
Therefore, putting the values here,
⇒ a3 = 2+(3-1)d
⇒ 26 = 2+2d
⇒ 24 = 2d
⇒ d = 12
⇒ a2 = 2+(2-1)12
⇒ a= 14
Therefore, 14 is the missing term.
(ii) For the given A.P., , 13, ,3
a2 = 13 and
a4 = 3
As we know, for an A.P.,
an = a+(n−1) d
Therefore, putting the values here,
⇒ a2 = a +(2-1)d
⇒ 13 = a+d ………………. (i)
⇒ a4 = a+(4-1)d
⇒ 3 = a+3d ………….. (ii)
On subtracting equation (i) from (ii), we get,
⇒ – 10 = 2d
⇒ d = – 5
From equation (i), putting the value of d,we get
⇒ 13 = a+(-5)
⇒ a = 18
⇒ a3 = 18+(3-1)(-5)
⇒ 18+2(-5) = 18-10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For the given A.P.,
a = 5 and
a4 = 19/2
As we know, for an A.P.,
an = a+(n−1)d
Therefore, putting the values here,
⇒ a4 = a+(4-1)d
⇒ 19/2 = 5+3d
⇒ (19/2) – 5 = 3d
⇒ 3d = 9/2
⇒ d = 3/2
⇒ a2 = a+(2-1)d
⇒ a2 = 5+3/2
⇒ a2 = 13/2
⇒ a3 = a+(3-1)d
⇒ a3 = 5+2×3/2
⇒ a3 = 8
Therefore, the missing terms are 13/2 and 8 respectively.
(iv) For the given A.P.,
a = −4 and
a6 = 6
As we know, for an A.P.,
an = a +(n−1) d
Therefore, putting the values here,
⇒ a6 = a+(6−1)d
⇒ 6 = − 4+5d
⇒ 10 = 5d
⇒ d = 2
⇒ a2 = a+d = − 4+2 = −2
⇒ a3 = a+2d = − 4+2(2) = 0
⇒ a4 = a+3d = − 4+ 3(2) = 2
⇒ a5 = a+4d = − 4+4(2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v) For the given A.P.,
a2 = 38
a6 = −22
As we know, for an A.P.,
an = a+(n −1)d
Therefore, putting the values here,
⇒ a2 = a+(2−1)d
⇒ 38 = a+d ……………………. (i)
⇒ a6 = a+(6−1)d
⇒ −22 = a+5d …………………. (ii)
On subtracting equation (i) from (ii), we get
⇒ − 22 − 38 = 4d
⇒ −60 = 4d
⇒ d = −15
⇒ a = a2 − d = 38 − (−15) = 53
⇒ a3 = a + 2d = 53 + 2 (−15) = 23
⇒ a4 = a + 3d = 53 + 3 (−15) = 8
⇒ a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.
Q4: Which term of the A.P. 3, 8, 13, 18, … is 78?
Sol:
Given the A.P. series as3, 8, 13, 18, …
First term, a = 3
Common difference, d = a2 − a1 = 8 − 3 = 5
Let the nth term of given A.P. be 78. Now as we know,
an = a+(n−1)d
Therefore,
⇒ 78 = 3+(n −1)5
⇒ 75 = (n−1)5
⇒ (n−1) = 15
⇒ n = 16
Hence, 16th term of this A.P. is 78.
Q5: Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
Given, 7, 13, 19, …, 205 is the A.P
Therefore,
The first term, a = 7
Common difference, d = a2 − a1 = 13 − 7 = 6
Let there are n terms in this A.P.
an = 205
As we know, for an A.P.,
⇒ an = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
⇒ 198 = (n − 1) 6
⇒ 33 = (n − 1)
⇒ n = 34
Therefore, this given series has 34 terms in it.
First term, a = 18
Common difference, d = a2-a=
⇒ d = (31-36)/2 = -5/2
Let there are n terms in this A.P.
an = 205
As we know, for an A.P.,
an = a+(n−1)d
⇒ -47 = 18+(n-1)(-5/2)
⇒ -47-18 = (n-1)(-5/2)
⇒ -65 = (n-1)(-5/2)
⇒ (n-1) = -130/-5
⇒ (n-1) = 26
⇒ n = 27
Therefore, this given A.P. has 27 terms in it.
Q6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
For the given series, A.P. 11, 8, 5, 2..
First term, a = 11
Common difference, d = a2−a1 = 8−11 = −3
Let −150 be the nth term of this A.P.
As we know, for an A.P.,
an = a+(n−1)d
⇒ -150 = 11+(n -1)(-3)
⇒ -150 = 11-3n +3
⇒ -164 = -3n
⇒ n = 164/3
Clearly, n is not an integer but a fraction.
Therefore, – 150 is not a term of this A.P.
Q7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Sol:
Given that,
11th term, a11 = 38
and 16th term, a16 = 73
We know that,
an = a+(n−1)d
⇒ a11 = a+(11−1)d
⇒ 38 = a+10d ………………………………. (i)
In the same way,
⇒ a16 = a +(16−1)d
⇒ 73 = a+15d ………………………………………… (ii)
On subtracting equation (i) from (ii), we get
⇒ 35 = 5d
⇒ d = 7
From equation (i), we can write,
⇒ 38 = a+10×(7)
⇒ 38 − 70 = a
⇒ a = −32
a31 = a +(31−1) d
⇒ − 32 + 30 (7)
⇒ − 32 + 210
⇒ 178
Hence, 31st term is 178.
Q8: An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Sol:
Given that,
3rd term, a3 = 12
50th term, a50 = 106
We know that,
an = a+(n−1)d
a3 = a+(3−1)d
12 = a+2d ……………………………. (i)
In the same way,
a50 = a+(50−1)d
106 = a+49d …………………………. (ii)
On subtracting equation (i) from (ii), we get
94 = 47d
d = 2 = common difference
From equation (i), we can write now,
12 = a+2(2)
a = 12−4 = 8
a29 = a+(29−1) d
a29 = 8+(28)2
a29 = 8+56 = 64
Therefore, 29th term is 64.
Q9: If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero?
Sol:
Given that,
3rd term, a3 = 4
and 9th term, a9 = −8
We know that,
an = a+(n−1)d
Therefore,
a3 = a+(3−1)d
4 = a+2d ……………………………………… (i)
a9 = a+(9−1)d
−8 = a+8d ………………………………………………… (ii)
On subtracting equation (i) from (ii), we will get here,
−12 = 6d
d = −2
From equation (i), we can write,
4 = a+2(−2)
4 = a−4
a = 8
Let nth term of this A.P. be zero.
a= a+(n−1)d
0 = 8+(n−1)(−2)
0 = 8−2n+2
2n = 10
n = 5
Hence, 5th term of this A.P. is 0.
Q10: If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Sol:
We know that, for an A.P series:
an = a+(n−1)d
a17 = a+(17−1)d
a17 = a +16d
In the same way,
a10 = a+9d
As it is given in the question,
a17 − a10 = 7
Therefore,
(a +16d)−(a+9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.
Q11: Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th term?
Sol:
Given A.P. is 3, 15, 27, 39, …
first term, a = 3
common difference, d = a2 − a1 = 15 − 3 = 12
We know that,
an = a+(n−1)d
Therefore,
a54 = a+(54−1)d
⇒ 3+(53)(12)
⇒ 3+636 = 639
a54 = 639
We have to find the term of this A.P. which is 132 more than a54, i.e.771.
Let nth term be 771.
an = a+(n−1)d
771 = 3+(n −1)12
768 = (n−1)12
(n −1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.
Or another method is;
Let nth term be 132 more than 54th term.
n = 54 + 132/2
= 54 + 11 = 65th term
Q12: Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Sol:
Let, the first term of two APs be a1 and a2 respectively
And the common difference of these APs be d.
For the first A.P.,we know,
an = a+(n−1)d
Therefore,
a100 = a1+(100−1)d
= a1 + 99d
a1000 = a1+(1000−1)d
a1000 = a1+999d
For second A.P., we know,
an = a+(n−1)d
Therefore,
a100 = a2+(100−1)d
= a2+99d
a1000 = a2+(1000−1)d
= a2+999d
Given that, difference between 100th term of the two APs = 100
Therefore, (a1+99d) − (a2+99d) = 100
a1−a2 = 100……………………………………………………………….. (i)
Difference between 1000th terms of the two APs
(a1+999d) − (a2+999d) = a1−a2
From equation (i),
This difference, a1−a= 100
Hence, the difference between 1000th terms of the two A.P. will be 100.
Q13: How many three-digit numbers are divisible by 7?
Sol:
First three-digit number that is divisible by 7 are;
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
Therefore, 105, 112, 119, …
All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
As we know, the largest possible three-digit number is 999.
When we divide 999 by 7, the remainder will be 5.
Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
first term, a = 105
common difference, d = 7
an = 994
n = ?
As we know,
an = a+(n−1)d
994 = 105+(n−1)7
889 = (n−1)7
(n−1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
Q14: How many multiples of 4 lie between 10 and 250?
Sol:
The first multiple of 4 that is greater than 10 is 12.
Next multiple will be 16.
Therefore, the series formed as;
12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows, now;
12, 16, 20, 24, …, 248
Let 248 be the nth term of this A.P.
first term, a = 12
common difference, d = 4
an = 248
As we know,
an = a+(n−1)d
248 = 12+(n-1)×4
236/4 = n-1
59 = n-1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.
Q15: For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Sol:
Given two APs as; 63, 65, 67,… and 3, 10, 17,….
Taking first AP,
63, 65, 67, …
First term, a = 63
Common difference, d = a2−a1 = 65−63 = 2
We know, nth term of this A.P. = an = a+(n−1)d
an= 63+(n−1)2 = 63+2n−2
an = 61+2n ………………………………………. (i)
Taking second AP,
3, 10, 17, …
First term, a = 3
Common difference, d = a2 − a1 = 10 − 3 = 7
We know that,
nth term of this A.P. = 3+(n−1)7
an = 3+7n−7
an = 7n−4 ……………………………………………………….. (ii)
Given, nth term of these A.P.s are equal to each other.
Equating both these equations, we get,
61+2n = 7n−4
61+4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other.
Q16: Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Sol:
Given,
Third term, a3 = 16
As we know,
a +(3−1)d = 16
a+2d = 16 ………………………………………. (i)
It is given that, 7th term exceeds the 5th term by 12.
a7 − a5 = 12
[a+(7−1)d]−[a +(5−1)d]= 12
(a+6d)−(a+4d) = 12
2d = 12
d = 6
From equation (i), we get,
a+2(6) = 16
a+12 = 16
a = 4
Therefore, A.P. will be 4, 10, 16, 22, …
Q17: Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Sol:
Given A.P. is3, 8, 13, …, 253
Common difference, d= 5.
Therefore, we can write the given AP in reverse order as;
253, 248, 243, …, 13, 8, 5
Now for the new AP,
first term, a = 253
and common difference, d = 248 − 253 = −5
n = 20
Therefore, using nth term formula, we get,
a20 = a+(20−1)d
a20 = 253+(19)(−5)
a20 = 253−95
a = 158
Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253 is 158.
Q18: The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Sol:
We know that, the nth term of the AP is;
an = a+(n−1)d
a4 = a+(4−1)d
a4 = a+3d
In the same way, we can write,
a8 = a+7d
a6 = a+5d
a10 = a+9d
Given that,
a4+a8 = 24
a+3d+a+7d = 24
2a+10d = 24
a+5d = 12 …………………………………………………… (i)
a6+a10 = 44
a +5d+a+9d = 44
2a+14d = 44
a+7d = 22 …………………………………………………… (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get,
a+5d = 12
a+5(5) = 12
a+25 = 12
a = −13
a2 = a+d = − 13+5 = −8
a3 = a2+d = − 8+5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.
Q19: Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Sol:
It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.
Therefore, after 1995, the salaries of each year are:
5000, 5200, 5400, …
Here, first term, a = 5000
and common difference, d = 200
Let after nth year, his salary be Rs 7000.
Therefore, by the nth term formula of AP,
an = a+(n−1) d
7000 = 5000+(n−1)200
200(n−1)= 2000
(n−1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.
Q20: Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Sol:
Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.
Hence,
First term, a = 5
and common difference, d = 1.75
Also given,
a= 20.75
Find, n = ?
As we know, by the nth term formula,
an = a+(n−1)d
Therefore,
20.75 = 5+(n -1)×1.75
15.75 = (n -1)×1.75
(n -1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n -1 = 9
n = 10
Hence, n is 10.
The document NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions (Exercise 5.2) is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
122 videos|474 docs|105 tests
### Up next
Test | 15 ques
Doc | 6 pages
Test | 10 ques
## Mathematics (Maths) Class 10
122 videos|474 docs|105 tests
### Up next
Test | 15 ques
Doc | 6 pages
Test | 10 ques
Explore Courses for Class 10 exam
### Top Courses for Class 10
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; | 6,503 | 14,863 | {"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} | 4.90625 | 5 | CC-MAIN-2024-18 | latest | en | 0.68694 |
https://electronics.stackexchange.com/questions/tagged/fixed-point?sort=votes&pageSize=50 | 1,571,617,076,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00077.warc.gz | 475,542,902 | 30,154 | The Stack Overflow podcast is back! Listen to an interview with our new CEO.
# Questions tagged [fixed-point]
A number with a fixed number of digits before or after an implied decimal point. It contrasts with the floating point number representation. Often used in place of floating point numbers if hardware floating point unit is not present.
12 questions
Filter by
Sorted by
Tagged with
3k views
### Why does signed (2's complement) binary multiplication have different procedure than unsigned?
The 2's complement binary multiplication does not have same procedure as unsigned if the both operands do not have the same sign. What is the logic behind that? Does special consideration apply to ...
2k views
### How to break multi digit number into separate digits in VHDL?
I found the method for c language.But I do not know how to perform this in VHDL. Let a fixed point number (12 downto -19) like 3456.478396 I need break this number entirely into separate numbers 3456....
2k views
### Use floating values in VHDL code
Assume I want to have a continous sine signal as input to my VHDL code. The values will be of type float since it will take on non-integer values for example: 10.5 mA. How do I manage these numbers ...
36 views
### Interpretation of I2S data
I am working with a TI PCM1807 ADC (datasheet: https://www.ti.com/lit/ds/symlink/pcm1807.pdf) and try do some signal processing on a FPGA. The PCM1807 output is I2S. I was wondering how to interpret ...
437 views
### Range of unsigned fixed point division in VHDL
I was thinking about the range a result signal should have to acommodate an unsigned fixed point division. Suppose we have: SIGNAL a : UFIXED (3 DOWNTO -3); SIGNAL b : UFIXED (4 DOWNTO -2); am I ...
576 views
### RFFT on 8192 samples in Q15 with CMSIS
I need to perform an FFT on a block of 8192 samples on an STM32F446 microcontroller. For that I wanted to use the CMSIS DSP library as it's available easily and optimised for the STM32F4. My 8192 ...
898 views
### Is it possible to show fixed point numbers as base 10 in modelsim wave?
If a person is creating a system using fixed point numbers a decimal point is implied. In this case, if one is going to use the wave window to see the result, it will be beneficial to see the actual ...
1k views
### How to represent Negative real numbers in Fixed Point representation
I find this module for the addition of two Fixed Point Numbers. Manual for using this Module: https://opencores.org/project,verilog_fixed_point_math_library,manual How to add -1.5 (or any negative ...
91 views
### Two single-precision versus one double-precision unit
Suppose you are designing a floating point unit, and it is desired that it be capable of both single precision and double precision operation, in the former case not by simply expanding the single ...
599 views
### Can fixed point division be implemented using a divider that outputs quotient and remainder?
From what I have seen, division is a highly expensive operation in terms of time or area (tradeoff). It is usually implemented as an operation of continuous subtraction of a number from another number ... | 724 | 3,156 | {"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-2019-43 | latest | en | 0.876058 |
https://errong.win/maximum-subarray-vi/ | 1,726,141,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00831.warc.gz | 209,095,320 | 5,764 | # Question
Given an array of integers. find the maximum XOR subarray value in given array.
What's the XOR: https://en.wikipedia.org/wiki/Exclusive_or
# Notice
Expected time complexity O(n).
Let's say F(L,R) is XOR of subarray from L to R.
Here we use the property that F(L,R)=F(1,R) XOR F(1,L-1). How?
Let's say our subarray with maximum XOR ended at position i. Now, we need to maximise F(L,i) ie. F(1,i) XOR F(1,L-1) where L<=i. Suppose, we inserted F(1,L-1) in our trie for all L<=i, then it's just problem1.
``````class Solution {
public:
/*
* @param : the array
* @return: the max xor sum of the subarray in a given array
*/
int maxXorSubarray(vector<int> &nums) {
// write code here
int ans = INT_MIN;
int pre = 0;
Trie trie;
for (auto n : nums) {
pre = pre ^ n;
trie.insert(pre);
ans = max(ans, trie.query(pre));
}
return ans;
}
private:
class Trie {
public:
Trie() {
root = new TrieNode(0);
INTBITS = sizeof(int) * 8;
}
void insert(int x) {
TrieNode* iter = root;
for (int i = INTBITS - 1; i >= 0; i--) {
bool v = x & (1 << i);
if (iter->arr[v] == 0)
iter->arr[v] = new TrieNode(0);
iter = iter->arr[v];
}
iter->val = x;
}
int query(int x) {
TrieNode* iter = root;
for (int i = INTBITS - 1; i >= 0; i--) {
bool v = x & (1 << i);
if (iter->arr[1-v] != 0)
iter = iter->arr[1-v];
else if (iter->arr[v] != 0)
iter = iter->arr[v];
}
return x ^ iter->val;
}
private:
class TrieNode {
public:
int val;
TrieNode *arr[2];
TrieNode(int v)
: val(v) {
arr[0] = arr[1] = 0;
}
};
TrieNode* root;
int INTBITS;
};
};
`````` | 522 | 1,520 | {"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-2024-38 | latest | en | 0.387484 |
https://math.answers.com/basic-math/What_are_the_multiples_of_5_that_are_prime | 1,702,087,722,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00529.warc.gz | 422,896,211 | 45,677 | 0
# What are the multiples of 5 that are prime?
Updated: 11/29/2022
Wiki User
12y ago
There are no multiples of 5 that are prime.
Wiki User
12y ago
Vecnussy McQuire
Lvl 2
1y ago
2
Anonymous
Lvl 1
3y ago
5 15 25 35 45 55 65
Earn +20 pts
Q: What are the multiples of 5 that are prime?
Submit
Still have questions?
Related questions
### What is a prime multiples for 75?
there are no prime multiples for 75 the prime factors or 75 are 5, 5, and 3
### Is the number 5 prime or composite?
It is prime because its multiples are 1 and itself that is a prime number
### What are the first two multiples of 5?
There are only two multiples of five. The multiples are 1 and 5. 5 is a prime number. A prime number is when a number is or can only be multiplied by one or itself. (In this case itself is five.)
### What prime factors are common to all multiples of 10?
2 and 5 are the prime factors that are common to all multiples of 10.
### What are the multiples of 3 5 and7?
Since all 3 are prime numbers LCM = 3 * 5 * 7 = 105 All multiples of 105 are multiples of 3, 5 and 7
### What are the multiples of 72 that are prime?
There are no multiples of 72 that are prime.
2 and 5
### What do the prime factors 3 5 and 7 go into?
Multiples of 105.
2 and 5
### What prime numbers are less than 200 and are multiples of 5?
There is only one prime number that fits this condition... the number 5 !
### What are 5 prime multiples and square numbers of 50?
25 is a square number that is a factor of 50 and a multiple of the prime number 5.
### Are all multiples of a prime number are prime numbers?
No, multiples of prime numbers are composite. | 469 | 1,662 | {"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-2023-50 | latest | en | 0.933002 |
https://www.tutorialspoint.com/questions/category/scikit-learn | 1,669,495,203,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00393.warc.gz | 1,112,737,148 | 11,306 | How to use DBSCAN clustering algorithm in Python Scikit-learn?
Updated on 04-Oct-2022 08:48:56
DBSCAN stands for Density-based spatial clustering of applications with noise. This algorithm is based on the intuitive notion of “clusters” & “noise” that clusters are dense regions of the lower density in the data space, separated by lower density regions of data points. Scikit-learn have sklearn.cluster.DBSCAN module to perform DBSCAN clustering. There are two important parameters namely min_samples and eps used by this algorithm to define dense. Higher value of parameter min_samples or lower value of the parameter eps will give an indication about the higher density of data points which is necessary to form a cluster. Steps We can ... Read More
How to use Affinity Propagation clustering algorithm in Python Scikit-learn?
Updated on 04-Oct-2022 08:46:43
Affinity propagation clustering algorithm is based on the concept of ‘message passing’ between different pairs of samples until convergence. It does not require the number of clusters to be specified before running the algorithm. One of the biggest disadvantages of this algorithm is its time complexity which is of the order $0(N^2T)$ Scikit-learn have sklearn.cluster.AffinityPropagation module to perform Affinity Propagation clustering in Python. Steps We can follow the below given steps to perform Affinity Propagation clustering algorithm in Python Scikit-learn − Step 1 − Import necessary libraries. Step 2 − Set the figure size. Step 3 − Define binary classification dataset having ... Read More
How to use K-Means clustering algorithm in Python Scikit-learn?
Updated on 04-Oct-2022 08:43:08
K-Means clustering algorithm computes the centroids and iterates until it finds optimal centroid. It requires the number of clusters to be specified that’s why it assumes that they are already known. The main logic of this algorithm is to cluster the data separating samples in n number of groups of equal variances by minimizing the criteria known as the inertia. The number of clusters identified by algorithm is represented by ‘K. Scikit-learn have sklearn.cluster.KMeans module to perform K-Means clustering algorithm in Python. Example For the example below, we will create a test binary classification dataset by using the make_classification() function. ... Read More
How to implement linear classification with Python Scikit-learn?
Updated on 04-Oct-2022 08:40:49
Linear classification is one of the simplest machine learning problems. To implement linear classification, we will be using sklearn’s SGD (Stochastic Gradient Descent) classifier to predict the Iris flower species. Steps You can follow the below given steps to implement linear classification with Python Scikit-learn − Step 1 − First import the necessary packages scikit-learn, NumPy, and matplotlib Step 2 − Load the dataset and build a training and testing dataset out of it. Step 3 − Plot the training instances using matplotlib. Although this step is optional, it is good practice to plot the instances for more clarity. Step 4 − Create ... Read More
How to transform Scikit-learn IRIS dataset to 2-feature dataset in Python?
Updated on 04-Oct-2022 08:38:18
Iris, a multivariate flower dataset, is one of the most useful Pyhton scikit-learn datasets. It has 3 classes of 50 instances each and contains the measurements of the sepal and petal parts of three Iris species namely Iris setosa, Iris virginica, and Iris versicolor. Along with that Iris dataset contains 50 instances from each of these three species and consists of four features namely sepal_length (cm), sepal_width (cm), petal_length (cm), petal_width (cm). We can use Principal Component Analysis (PCA) to transform IRIS dataset into a new feature space with 2 features. Steps We can follow the below given steps to ... Read More
How to transform Sklearn DIGITS dataset to 2 and 3-feature dataset in Python?
Updated on 04-Oct-2022 08:35:06
Sklearn DIGITS dataset has 64 features as each image of the digit is of size 8 by 8 pixels. We can use Principal Component Analysis (PCA) to transform Scikit-learn DIGITS dataset into new feature space with 2 features. Transforming 64 features dataset to 2-feature dataset will be a big reduction in the size of data and we’ll lose some useful information. It will also impact the classification accuracy of ML model. Steps to Transform DIGITS Dataset to 2-feature Dataset We can follow the below given steps to transform DIGITS dataset to 2-feature dataset using PCA − First, import the ... Read More
How to perform dimensionality reduction using Python Scikit-learn?
Updated on 04-Oct-2022 08:32:09
Dimensionality reduction, an unsupervised machine learning method is used to reduce the number of feature variables for each data sample selecting set of principal features. Principal Component Analysis (PCA) is one of the popular algorithms for dimensionality reduction available in Sklearn. In this tutorial, we perform dimensionality reduction using principal component analysis and incremental principal component analysis using Python Scikit-learn (Sklearn). Using Principal Component Analysis (PCA) PCA is a statistical method that linearly project the data into new feature space by analyzing the features of original dataset. The main concept behind PCA is to select the “principal” characteristics of the ... Read More
How to implement Random Projection using Python Scikit-learn?
Updated on 04-Oct-2022 08:29:24
Random projection is a dimensionality reduction and data visualization method to simplify the complexity of highly dimensional data. It is basically applied to the data where other dimensionality reduction techniques such as Principal Component Analysis (PCA) can not do the justice to data. Python Scikit-learn provides a module named sklearn.random_projection that implements a computationally efficient way to reduce the data dimensionality. It implements the following two types of an unstructured random matrix − Gaussian Random Matrix Sparse Random Matrix Implementing Gaussian Random Projection For implementing Gaussian random matrix, random_projection module uses GaussianRandomProjection() function which reduces the dimensionality by ... Read More
How to build Naive Bayes classifiers using Python Scikit-learn?
Updated on 04-Oct-2022 08:25:42
Naïve Bayes classification, based on the Bayes theorem of probability, is the process of predicting the category from unknown data sets. Scikit-learn has three Naïve Bayes models namely, Gaussian Naïve Bayes Bernoulli Naïve Bayes Multinomial Naïve Bayes In this tutorial, we will learn Gaussian Naïve Bayes and Bernoulli Naïve Bayes classifiers using Python Scikit-learn (Sklearn). Gaussian Naïve Bayes Classifier Gaussian naïve bayes classifier is based on a continuous distribution characterized by mean and variance. With the help of an example, let’s see how we can use the Scikit-Learn Python ML library to build a Gaussian Naïve Bayes classifier. ... Read More
How to create a random forest classifier using Python Scikit-learn?
Updated on 04-Oct-2022 08:22:46
Random forest is a supervised machine learning algorithm that is used for classification, regression, and other tasks by creating decision trees on data samples. After creating the decision trees, a random forest classifier collects the prediction from each of them and selects the best solution by means of voting. One of the best advantages of a random forest classifier is that it reduces overfitting by averaging the result. That is the reason we get better results as compared to a single decision tree. Steps to Create Random Forest Classifier We can follow the below steps to create a random forest ... Read More | 1,623 | 7,728 | {"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} | 2.609375 | 3 | CC-MAIN-2022-49 | latest | en | 0.834282 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.