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http://www.chegg.com/homework-help/questions-and-answers/spring-spring-constant-k-mass-m-comprise-simple-harmonic-oscillator-oscillator-attached-st-q951771 | 1,475,090,730,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661763.18/warc/CC-MAIN-20160924173741-00067-ip-10-143-35-109.ec2.internal.warc.gz | 393,045,980 | 13,701 | A spring (spring constant K),and mass m,comprise a simple harmonic oscillator. The oscillator is attached to string of linear mass density ยต under tension t such that when the mass oscillates a wave travels along the string. the string is fixed at both ends.
(a) Find the wavelength,? ,of the wave.
(b)Find the distance L between the ends of the string so that the n=4 harmonic standing wave is established in the string.
(c) The mass m is replaced with a different mass M with everything else remaining the same.Find an exoression for M such that the n=3 harmonic standing wave is established on the string. Use the lenght you found in (b). | 144 | 642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-40 | latest | en | 0.899143 |
https://www.topperlearning.com/forums/home-work-help-19/if-x3-ax2-bx-6-has-x-2-as-a-factor-and-leaves-a-re-mathematics-polynomials-48582/reply | 1,506,009,043,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687833.62/warc/CC-MAIN-20170921153438-20170921173438-00129.warc.gz | 873,429,098 | 41,008 | Question
Sun June 26, 2011 By:
# if (x3 + ax2 + bx + 6) has (x-2) as a factor and leaves a remainder 3 when divided by (x-3), find the value of a and b ? how to solve this question ??
Sun June 26, 2011
Let p(x) = x3 + ax2 + bx +6
(x-2) is a factor of the polynomial x3 + ax2 + b x +6
p(2) = 0
p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0
7 +2 a +b = 0
b = - 7 -2aÂ…(i)
x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3
11+3a +b =1 Þ3a+b =-10Þb=-10-3a….(ii)
Equating the value of b from (ii) and (i) , we have
(- 7 -2a) = (-10 – 3a)
a = -3
Substituting a = -3 in (i), we get
b = - 7 -2(-3) = -7 + 6 = -1
Thus the values of a and b are -3 and -1 respectively.
Related Questions
Sat September 09, 2017
# Please tell the answer as fast as possible !
Fri August 18, 2017
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## #1 2011-11-26 01:08:42
juantheron
Power Member
Offline
### vector
The vector
bisects the angle between the vectors
and
.
determine a unit vector along
.
## #2 2013-04-14 05:58:45
Nehushtan
Power Member
Offline
### Re: vector
Let the angle between
and
be
. Taking the dot product gives
Then the angle between
and
is
and taking the dot product gives
Thus you can take any x, y satisfying the last equation and z such that
– for example,
,
,
.
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May 4, 2016
# Posts by Youssef
Total # Posts: 7
physics
. A long string carries a wave; a 6-m segment of the string contains four complete wavelengths and has a mass of 180 g. The string vibrates sinusoidal with a frequency of 50 Hz and a peak-to-valley displacement of 15 cm. (The “peak-to- valley” distance is the ...
March 24, 2015
Physics
Rutherford, Geiger, and Marsden (his students) fired a beam of alpha particles at a gold foil. The alpha particles had an energy of about 3.6 million electron volts, a charge of +2e, and a mass of about four AMU (one Atomic Mass Unit is 1.66 x 10-27 kg. Gold has atomic number ...
February 10, 2013
Physics
Three charges are located at the corners of a rectangle as follows: Charge A: lower left corner, -3.0 μC Charge B: upper left corner, -6.1 μC Charge C: upper right corner, +2.7 μC The distance between A and B is 0.16 m and between B ...
February 10, 2013
Physics
As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and the cloud to be the other plate. The altitude of the cloud is 600m and its area is 6.5 km2. How much charge can the cloud hold before the dielectric strength of air is ...
February 10, 2013
Physics
A positive and a negative charge are separated a distance r. What can we say about the electric potential energy of this two-particle system, according to our normal definitions of potential energy and work? Select the best answer. a. It is positive b. It is negative c. It is ...
February 10, 2013
math
i want a mathematic equation to apply on the number 579494 and get as a result the number 573 ,they told me that its simple but i can't figure it out(if somebody find many ways please give it all ).thanks
May 20, 2011
math
how to get 537 out of 579494 ?
May 19, 2011
1. Pages:
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https://git.musl-libc.org/cgit/musl/plain/src/math/log1p.c?id=5aac5e2189f322a54a49958d928f30e1c9505561 | 1,679,376,884,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00386.warc.gz | 306,256,513 | 3,201 | /* origin: FreeBSD /usr/src/lib/msun/src/s_log1p.c */ /* * ==================================================== * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. * * Developed at SunPro, a Sun Microsystems, Inc. business. * Permission to use, copy, modify, and distribute this * software is freely granted, provided that this notice * is preserved. * ==================================================== */ /* double log1p(double x) * * Method : * 1. Argument Reduction: find k and f such that * 1+x = 2^k * (1+f), * where sqrt(2)/2 < 1+f < sqrt(2) . * * Note. If k=0, then f=x is exact. However, if k!=0, then f * may not be representable exactly. In that case, a correction * term is need. Let u=1+x rounded. Let c = (1+x)-u, then * log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u), * and add back the correction term c/u. * (Note: when x > 2**53, one can simply return log(x)) * * 2. Approximation of log1p(f). * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) * = 2s + 2/3 s**3 + 2/5 s**5 + ....., * = 2s + s*R * We use a special Reme algorithm on [0,0.1716] to generate * a polynomial of degree 14 to approximate R The maximum error * of this polynomial approximation is bounded by 2**-58.45. In * other words, * 2 4 6 8 10 12 14 * R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s * (the values of Lp1 to Lp7 are listed in the program) * and * | 2 14 | -58.45 * | Lp1*s +...+Lp7*s - R(z) | <= 2 * | | * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. * In order to guarantee error in log below 1ulp, we compute log * by * log1p(f) = f - (hfsq - s*(hfsq+R)). * * 3. Finally, log1p(x) = k*ln2 + log1p(f). * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) * Here ln2 is split into two floating point number: * ln2_hi + ln2_lo, * where n*ln2_hi is always exact for |n| < 2000. * * Special cases: * log1p(x) is NaN with signal if x < -1 (including -INF) ; * log1p(+INF) is +INF; log1p(-1) is -INF with signal; * log1p(NaN) is that NaN with no signal. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. * * Note: Assuming log() return accurate answer, the following * algorithm can be used to compute log1p(x) to within a few ULP: * * u = 1+x; * if(u==1.0) return x ; else * return log(u)*(x/(u-1.0)); * * See HP-15C Advanced Functions Handbook, p.193. */ #include "libm.h" static const double ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */ ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */ two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */ Lp1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */ Lp2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */ Lp3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */ Lp4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */ Lp5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */ Lp6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */ Lp7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */ double log1p(double x) { double hfsq,f,c,s,z,R,u; int32_t k,hx,hu,ax; GET_HIGH_WORD(hx, x); ax = hx & 0x7fffffff; k = 1; if (hx < 0x3FDA827A) { /* 1+x < sqrt(2)+ */ if (ax >= 0x3ff00000) { /* x <= -1.0 */ if (x == -1.0) return -two54/0.0; /* log1p(-1)=+inf */ return (x-x)/(x-x); /* log1p(x<-1)=NaN */ } if (ax < 0x3e200000) { /* |x| < 2**-29 */ /* raise inexact */ if (two54 + x > 0.0 && ax < 0x3c900000) /* |x| < 2**-54 */ return x; return x - x*x*0.5; } if (hx > 0 || hx <= (int32_t)0xbfd2bec4) { /* sqrt(2)/2- <= 1+x < sqrt(2)+ */ k = 0; f = x; hu = 1; } } if (hx >= 0x7ff00000) return x+x; if (k != 0) { if (hx < 0x43400000) { STRICT_ASSIGN(double, u, 1.0 + x); GET_HIGH_WORD(hu, u); k = (hu>>20) - 1023; c = k > 0 ? 1.0-(u-x) : x-(u-1.0); /* correction term */ c /= u; } else { u = x; GET_HIGH_WORD(hu,u); k = (hu>>20) - 1023; c = 0; } hu &= 0x000fffff; /* * The approximation to sqrt(2) used in thresholds is not * critical. However, the ones used above must give less * strict bounds than the one here so that the k==0 case is * never reached from here, since here we have committed to * using the correction term but don't use it if k==0. */ if (hu < 0x6a09e) { /* u ~< sqrt(2) */ SET_HIGH_WORD(u, hu|0x3ff00000); /* normalize u */ } else { k += 1; SET_HIGH_WORD(u, hu|0x3fe00000); /* normalize u/2 */ hu = (0x00100000-hu)>>2; } f = u - 1.0; } hfsq = 0.5*f*f; if (hu == 0) { /* |f| < 2**-20 */ if (f == 0.0) { if(k == 0) return 0.0; c += k*ln2_lo; return k*ln2_hi + c; } R = hfsq*(1.0 - 0.66666666666666666*f); if (k == 0) return f - R; return k*ln2_hi - ((R-(k*ln2_lo+c))-f); } s = f/(2.0+f); z = s*s; R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7)))))); if (k == 0) return f - (hfsq-s*(hfsq+R)); return k*ln2_hi - ((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f); } | 1,981 | 5,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-14 | latest | en | 0.797327 |
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A234809 a(n) = |{0 < k < n: p = k + phi(n-k) and 2*(n-p) + 1 are both prime}|, where phi(.) is Euler's totient function. 1
0, 0, 1, 2, 1, 3, 1, 4, 1, 1, 1, 5, 3, 7, 3, 1, 1, 7, 5, 9, 4, 2, 1, 9, 5, 2, 4, 3, 1, 10, 5, 14, 2, 2, 2, 1, 6, 14, 5, 4, 1, 15, 5, 16, 5, 5, 3, 17, 8, 4, 5, 6, 3, 17, 7, 5, 2, 6, 6, 17, 11, 25, 3, 5, 3, 1, 11, 25, 4, 4, 4, 22, 10, 26, 6, 7, 8, 3, 9, 26, 7, 9, 6, 25, 8, 3, 7, 9, 10, 25, 15, 6, 2, 9, 9, 2, 13, 29, 3, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 COMMENTS Conjecture: a(n) > 0 for all n > 2. Clearly, this implies Lemoine's conjecture which states that any odd number 2*n + 1 > 5 can be written as 2*p + q with p and q both prime. See also A234808 for a similar conjecture. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE a(5) = 1 since 1 + phi(4) = 3 and 2*(5-3) + 1 = 5 are both prime. a(16) = 1 since 7 + phi(9) = 13 and 2*(16-13) + 1 = 7 are both prime. a(41) = 1 since 7 +phi(34) = 23 and 2*(41-23) + 1 = 37 are both prime. a(156) = 1 since 131 + phi(25) = 151 and 2*(156-151) + 1 = 11 are both prime. MATHEMATICA f[n_, k_]:=k+EulerPhi[n-k] p[n_, k_]:=PrimeQ[f[n, k]]&&PrimeQ[2*(n-f[n, k])+1] a[n_]:=a[n]=Sum[If[p[n, k], 1, 0], {k, 1, n-1}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000010, A000040, A046927, A234470, A234475, A234514, A234567, A234615, A234694, A234808 Sequence in context: A213594 A325252 A243334 * A135591 A114897 A210954 Adjacent sequences: A234806 A234807 A234808 * A234810 A234811 A234812 KEYWORD nonn AUTHOR Zhi-Wei Sun, Dec 30 2013 STATUS approved
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Last modified December 9 09:21 EST 2019. Contains 329877 sequences. (Running on oeis4.) | 1,005 | 2,377 | {"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-2019-51 | latest | en | 0.838022 |
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Percents and fractionsorksheets the best image toorksheet percentages pdf converting percent. Math worksheets percentages to fractions worksheet pdf converting tes changing percent fraction. Percents to fractions worksheet math worksheets download them and try solve percentages decimals ks2 converting. 5th grade math worksheetss decimals and percents download percentages to worksheet pdf percentage decimal.
The loan deferment process involves contacting the lender, submitting a deferment application, and undergoing the application process. The actual process can vary by lender. Other factors taken into account include the borrower's credit history, type of loan, and number of payments being deferred. Approval can take less than 24 hours to several weeks.
Borrowers should create a folder to store loan document records, along with a record of phone and email correspondence. Always keep track of phone conversations by writing down a summary of the call, date, time, and name of the bank representative spoken with. When important documents are mailed, invest in the extra protection of tracking receipts. Certified letters should be sent with a return receipt request in case it is necessary to provide evidence the documents were received.
Besides a standard loan for which a loan agreement is drawn up, there is another popular type of loan, the demand loan. That is a short term loan, with a period of repayment for up to 180 days. The date for the repayment of the loan is not fixed, and the interest rate for it is a floating one. The demand loan offers advantages for both borrowers and lenders. The lender can demand the repayment of the loan at any time, and on the other hand, the borrower does not need to adhere to a repayment in installments, as the repayment should be made for the entire amount. Furthermore, demand loans are easier to qualify for.
The content of the loan agreement includes the precise details of the agreement that has transpired between lender and debtor. Not only that, the agreement should also consider the standard government laws that is already in effect or established. Since the laws were created to protect all people, it is beneficial to both parties concerned. | 442 | 2,341 | {"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-2018-47 | longest | en | 0.945463 |
http://www.puzzles-world.com/2017/03/a-guy-walks-into-store-and-steals-100.html | 1,500,727,687,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424060.49/warc/CC-MAIN-20170722122816-20170722142816-00670.warc.gz | 553,786,455 | 19,299 | # A guy walks into a store and steals a \$100 bill
### Can you Solve this ?
A guy walks into a store and
steals a \$100 bill from the register
without the owner's knowledge.
He then buys \$70 worth of goods
using the \$100 bill and the owner
give \$30 in change.
How much money did the owner lose ? | 77 | 302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-30 | longest | en | 0.937703 |
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# 6.
## 045: Automata, Computability, and
Complexity (GITCS)
Class 16
Nancy Lynch
## Today: More NP-Completeness
Topics:
3SAT is NP-complete
Clique and VertexCover are NP-complete
More examples, overview
Hamiltonian path and Hamiltonian circuit
Traveling Salesman problem
More examples, revisited
Sipser Sections 7.4-7.5
Garey and Johnson
Next:
Sipser Section 10.2
3SAT is NP-Complete
NP-Completeness
Definition: Language B is NP-complete if both
of the following hold:
(a) B NP, and
(b) For any language A NP, A p B.
NP
## Definition: Language B is NP-hard if, for any
language A NP, A p B.
3SAT is NP-Complete
SAT = { < > | is a satisfiable Boolean formula }
Boolean formula: Constructed from literals using
operations, e.g.:
= x ( ( y z ) (y z ) ) ( x z )
## A Boolean formula is satisfiable iff there is an assignment
of 0s and 1s to the variables that makes the entire formula
evaluate to 1 (true).
Theorem: SAT is NP-complete.
3SAT: Satisfiable Boolean formulas of a restricted kind--conjunctive normal form (CNF) with exactly 3 literals per
clause.
Theorem: 3SAT is NP-complete.
Proof:
3SAT NP: Obvious.
3SAT is NP-hard:
3SAT is NP-hard
Clause: Disjunction of literals, e.g., ( x1 x2 x3 )
CNF: Conjunction of such clauses
Example:
( x1 x2 ) ( x1 x2 ) ( x1 x2 x3 ) ( x3 )
3-CNF:
{ < > | is a CNF formula in which each clause has
exactly 3 literals }
CNF-SAT: { < > | is a satisfiable CNF formula }
3-SAT: { < > | is a satisfiable 3-CNF formula }
= SAT 3-CNF
Theorem: 3SAT is NP-hard.
Proof: Show CNF-SAT is NP-hard, and CNF-SAT p
3SAT.
CNF-SAT is NP-hard
Theorem: CNF-SAT is NP-hard.
Proof:
We wont show SAT p CNF-SAT.
Instead, modify the proof that SAT is NP-hard, so that it
shows A p CNF-SAT, for an arbitrary A in NP, instead
of just A p SAT as before.
Weve almost done this: formula w is almost in CNF.
Its a conjunction w = cell start accept move.
And each of these is itself in CNF, except move .
move is:
a conjunction over all (i,j)
of disjunctions over all tiles
of conjunctions of 6 conditions on the 6 cells:
xi,j,a1 xi,j+1,a2 xi,j+2,a3 xi+1,j,b1 xi+1,j+1,b2 xi+1,j+2,b3
CNF-SAT is NP-hard
Show A p CNF-SAT.
w is a conjunction w = cell start accept move, where
each is in CNF, except move .
move is:
a conjunction ( ) over all (i,j)
of disjunctions ( ) over all tiles
of conjunctions ( ) of 6 conditions on the 6 cells:
xi,j,a1 xi,j+1,a2 xi,j+2,a3 xi+1,j,b1 xi+1,j+1,b2 xi+1,j+2,b3
We want just of .
Can use distributive laws to replace ( of ) with ( of ),
which would yield overall of , as needed.
In general, transforming ( of ) to ( of ), could cause
formula size to grow too much (exponentially).
However, in this situation, the clauses for each (i,j) have
total size that depends only on the TM M, and not on w.
So the size of the transformed formula is still poly in |w|.
CNF-SAT is NP-hard
Theorem: CNF-SAT is NP-hard.
Proof:
Modify the proof that SAT is NP-hard.
w = cell start accept move.
Can be put into CNF, while keeping the size of
the transformed formula poly in |w|.
Shows that A p CNF-SAT.
Since A is any language in NP, CNF-SAT is NPhard.
3SAT is NP-hard
Proved: Theorem: CNF-SAT is NP-hard.
Now: Theorem: 3SAT is NP-hard.
Proof:
## Use reduction, show CNF-SAT p 3SAT.
Construct f, polynomial-time computable, such that w
CNF-SAT if and only if f(w) 3SAT.
If w isnt a CNF formula, then f(w) isnt either.
If w is a CNF formula, then f(w) is another CNF formula,
this one with 3 literals per clause, satisfiable iff w is
satisfiable.
f works by converting each clause to a conjunction of
clauses, each with 3 literals (add repeats to get 3).
Show by example: (a b c d e) gets converted to
(a r1) ( r1 b r2) ( r2 c r3) ( r3 d r4)
( r4 e)
f is polynomial-time computable.
3SAT is NP-hard
Proof:
Show CNF-SAT p 3SAT.
Construct f such that w CNF-SAT iff f(w) 3SAT; converts each
clause to a conjunction of clauses.
f converts w = (a b c d e) to f(w) =
(a r1) (r1 b r2) (r2 c r3) (r3 d r4) (r4 e)
Claim w is satisfiable iff f(w) is satisfiable.
:
Given a satisfying assignment for w, add values for r1, r2, , to
satisfy f(w).
Start from a clause containing a literal with value 1---there must be
one---make the new literals in that clause 0 and propagate
consequences left and right.
Example: Above, if c = 1, a = b = d = e = 0 satisfy w, use:
(a r1) (r1 b r2) (r2 c r3) (r3 d r4) (r4 e)
0 1
0 0 1
0 1 0
1 0 0
1 0
3SAT is NP-hard
Proof:
Show CNF-SAT p 3SAT.
Construct f such that w CNF-SAT iff f(w) 3SAT;
converts each clause to a conjunction of clauses.
f converts w = (a b c d e) to f(w) =
(a r1) (r1 b r2) (r2 c r3) (r3 d r4)
(r4 e)
Claim w is satisfiable iff f(w) is satisfiable.
:
Given satisfying assignment for f(w), restrict to satisfy w.
Each ri can make only one clause true.
Theres one fewer ri than clauses; so some clause must
be made true by an original literal, i.e., some original
literal must be true, satisfying w.
3SAT is NP-hard
Theorem: CNF-SAT is NP-hard.
Theorem: 3SAT is NP-hard.
Proof:
Constructed polynomial-time-computable f such
that w CNF-SAT iff f(w) 3SAT.
Thus, CNF-SAT p 3SAT.
Since CNF-SAT is NP-hard, so is 3SAT.
NP-Complete
## CLIQUE and VERTEX-COVER
CLIQUE = { < G, k > | G is a graph with a k-clique }
k-clique: k vertices with edges between all pairs in
the clique.
Theorem: CLIQUE is NP-complete.
Proof:
To show CLIQUE is NP-hard, show 3SAT p CLIQUE.
Need poly-time-computable f, such that w 3SAT iff f(w)
CLIQUE.
f must map a formula w in 3-CNF to <G, k> such that w is
satisfiable iff G has a k-clique.
Show by example:
(x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
CLIQUE is NP-hard
Proof:
## Show 3SAT p CLIQUE; construct f such that w 3SAT
iff f(w) CLIQUE.
f maps a formula w in 3-CNF to <G, k> such that w is
satisfiable iff G has a k-clique.
(x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
Graph G: Nodes for all (clause, literal) pairs, edges
between all non-contradictory nodes in different
clauses.
C1
k: Number of clauses
x1
x2
x3
x3
C3
x1
x2
x2
x1
x3
C2
CLIQUE is NP-hard
Graph G: Nodes for all (clause, literal) pairs, edges
between all non-contradictory nodes in different clauses.
k: Number of clauses
(x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
Claim (general): w satisfiable iff G has a k-clique.
:
Assume the formula is
satisfiable.
Satisfying assignment
gives one literal in
each clause, all with
assignments.
Yields a k-clique.
C1
x1
x2
x3
x3
x1
x2
x2
C3
x1
x3
C2
CLIQUE is NP-hard
Example:
(x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
Satisfiable, with satisfying assignment x1 = 1, x2 = x3= 0
Yields 3-clique:
:
Assume the formula is
x1
satisfiable.
Satisfying assignment
gives one literal in
x3
each clause, all with
x2
x1
assignments.
C3
Yields a k-clique.
C1
x2
x3
x1
x2
x3
C2
CLIQUE is NP-hard
Graph G: Nodes for all (clause, literal) pairs, edges
between all non-contradictory nodes in different clauses.
k: Number of clauses
(x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
Claim (general): w satisfiable iff G has a k-clique.
:
Assume a k-clique.
Yields one node per
clause, none
Yields a consistent
assignment satisfying
all clauses of w.
C1
x1
x2
x3
x3
x1
x2
x2
C3
x1
x3
C2
CLIQUE is NP-hard
Graph G: Nodes for all (clause, literal) pairs, edges
between all non-contradictory nodes in different clauses.
k: Number of clauses
Claim (general): w satisfiable iff G has a k-clique.
So, 3SAT p CLIQUE.
Since 3SAT is NP-hard,
so is CLIQUE.
So CLIQUE is NP-complete.
C1
x1
x2
x3
x3
x1
x2
x2
C3
x1
x3
C2
VERTEX-COVER is NP-complete
VERTEX-COVER =
{ < G, k > | G is a graph with a vertex cover of size k }
Vertex cover of G = (V, E): A subset C of V such
that, for every edge (u,v) in E, either u or v C.
Theorem: VERTEX-COVER is NP-complete.
Proof:
Show VERTEX-COVER is NP-hard.
That is, if A NP, then A p VERTEX-COVER.
We know A p CLIQUE, since CLIQUE is NP-hard.
Recall CLIQUE p VERTEX-COVER.
By transitivity of p, A p VERTEX-COVER, as needed.
VERTEX-COVER is NP-complete
Theorem: VERTEX-COVER is NP-complete.
More succinct proof:
## VC NP; show VC is NP-hard.
CLIQUE is NP-hard.
CLIQUE p VC.
So VC is NP-hard.
## In general, can show language B is NP-complete by:
Showing B NP, and
Showing A p B for some known NP-hard problem A.
More Examples
## More NP-Complete Problems
[Garey, Johnson] show hundreds of problems are
NP-complete.
All but 3SAT use the polynomial-time reduction
method.
3SAT
Examples:
CLIQUE
VERTEXCOVER
HAMILTONIAN
PATH/CIRCUIT
TRAVELING
SALESMAN
SUBSETSUM
Etc.
SET
PARTITION
MULTIPROCESSOR
SCHEDULING
3SAT
CLIQUE
VERTEXCOVER
As we just
showed.
HAMILTONIAN
PATH/CIRCUIT
TRAVELING
SALESMAN
## Will do this now.
SUBSETSUM
Etc.
SET
PARTITION
MULTIPROCESSOR
SCHEDULING
Recitation?
A B means A p B.
Hardness propagates to the right in p, downward along
tree branches.
3SAT p HAMILTONIAN
PATH/CIRCUIT
## 3SAT p HAMILTONIAN PATH/CIRCUIT
Two versions of the problem, for directed and undirected
graphs.
Consider directed version; undirected shown by reduction
from directed version.
DHAMPATH = { <G, s, t> | G is a directed graph, s and t
are two distinct vertices, and there is a path from s to t in G
that passes through each vertex of G exactly once }
DHAMPATH NP: Guess path and verify.
3SAT p DHAMPATH:
3CNF
Digraph, s,t
f
3SAT
f
DHAMPATH
## 3SAT p HAMILTONIAN PATH/CIRCUIT
DHAMPATH = { <G, s, t> | G is a directed graph, s and t
are two distinct vertices, and there is a path from s to t in G
that passes through each vertex of G exactly once }
3SAT p DHAMPATH:
Map a 3CNF formula to <G, s, t> so that is satisfiable if and only
if G has a Hamiltonian path from s to t.
In fact, there will be a direct correspondence between a satisfying
assignment for and a Hamiltonian path in G.
3CNF
Digraph, s,t
f
3SAT
f
DHAMPATH
3SAT p DHAMPATH
Map a 3CNF formula to <G, s, t> so that is satisfiable if
and only if G has a Hamiltonian path from s to t.
Correspondence between satisfying assignment for and
Hamiltonian path in G.
Notation:
## Write = (a1 b1 c1) (a2 b2 c2) (ak bk ck)
k clauses C1, C2, , Ck
Variables: x1, x2,, xl
Each aj, bj, and cj is either some xi or some xi.
## Digraph is constructed from pieces (gadgets), one for each
variable xi and one for each clause Cj.
Row contains 3k+1 nodes,
not counting endpoints.
Notation:
3SAT p DHAMPATH
## = (a1 b1 c1) (a2 b2 c2) (ak bk ck)
k clauses C1, C2, , Ck
Variables: x1, x2,, xl
Each aj, bj, and cj is either some xi or some xi.
or
Notation:
3SAT p DHAMPATH
## = (a1 b1 c1) (a2 b2 c2) (ak bk ck)
k clauses C1, C2, , Ck
Variables: x1, x2,, xl
Each aj, bj, and cj is either some xi or some xi.
Just a single node.
Cj
## Putting the pieces together:
Put variables gadgets in order x1, x2, , xl, top to bottom,
identifying bottom node of each gadget with top node of the next.
Make s and t the overall top and bottom node, respectively
3SAT p DHAMPATH
s
## Putting the pieces
together:
C1
in order x1, x2, , xl,
identifying bottom
node of each with top
node of the next.
Make s and t the
overall top and bottom
node.
C2
Ck
## We still must connect
t
3SAT p DHAMPATH
Divide the 3k+1 nodes in the cross-bar of xis gadget into k
pairs, one per clause, separated by k+1 separator nodes:
C1
C2
## If xi appears in Cj, add edges between the
Cj node and the nodes for Cj in the
crossbar, going from left to right.
Ck
Cj
left-to-right.
Cj
3SAT p DHAMPATH
C1
C2
Ck
Cj
## If xi appears in Cj, add edges L to R.
Allows detour to Cj while traversing crossbar L to R.
Cj
## If xi appears in Cj, add edges R to L.
Allows detour to Cj while traversing crossbar R to L.
## If both xi and xi appear, add both sets of
edges.
This completes the construction of G, s, t.
Cj
Cj
Example
= (x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
s
x1
x1
C1
x2
x3
x2
C2
x3
C3
t
Example
= (x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
s
C1
x1
x1
x2
x2
C2
x3
x3
C3
t
Example
= (x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
s
C1
x1
x2
C2
x1
x2
x3
C3
t
x3
## The entire graph G
= (x1 x2 x3) (x1 x2 x3) (x1 x2 x3)
s
x1
x1
C1
x2
x3
x1
x2
x2
C2
x3
x1
x2
x3
C3
t
x3
3SAT p DHAMPATH
Claim: is satisfiable iff the graph G has a Hamiltonian
path from s to t.
Proof:
Assume is satisfiable; fix a particular satisfying assignment.
L to R through gadgets for xis that are set true.
R to L through gadgets for xis that are set false.
This visits all nodes of G except the Cj nodes.
For these, we must take detours.
Cj
For any particular clause Cj:
At least one of its literals must be set true; pick one.
If its of the form xi, then do:
Cj pair in xi row
## Works since xi = true means we traverse this crossbar L to R.
3SAT p DHAMPATH
Claim: is satisfiable iff the graph G has a Hamiltonian
path from s to t.
Proof:
Assume is satisfiable; fix a particular satisfying assignment.
L to R through gadgets for xis that are set true.
R to L through gadgets for xis that are set false.
This visits all nodes of G except the Cj nodes.
For these, we must take detours.
Cj
For any particular clause Cj:
At least one of its literals must be set true; pick one.
If its of the form xi, then do:
Cj pair in xi row
## Works since xi = false means we traverse this crossbar R to L.
3SAT p DHAMPATH
Claim: is satisfiable iff the graph G has a Hamiltonian
path from s to t.
Proof:
Assume G has a Hamiltonian path from s to t, get a satisfying
assignment for .
If the path is normal (goes in order through the gadgets, top to
bottom, going one way or the other through each crossbar, and
detouring to pick up the Cj nodes), then define the assignment by:
Set each xi true if path goes L to R through xis gadget, false if it
goes R to L.
Why is this a satisfying assignment
for ?
Consider any clause Cj.
The path goes through its node in
one of two ways:
Cj
Cj
Cj pair in xi row
Cj pair in xi row
3SAT p DHAMPATH
Claim: is satisfiable iff the graph G has a Hamiltonian
path from s to t.
Proof:
Assume G has a Hamiltonian path from s to t, get a satisfying
assignment for .
If the path is normal, then define the assignment by:
Set each xi true if path goes L to R through xis gadget, false if it
goes R to L.
Cj
Cj
To see that this satisfies , consider
or
any clause C .
j
## The path goes through Cjs node by:
If the first, then:
xi is true, since path goes L-R.
By the way the detour edges are
set, Cj contains literal xi.
So Cj is satisfied by xi.
Cj pair in xi row
Cj pair in xi row
3SAT p DHAMPATH
Claim: is satisfiable iff the graph G has a Hamiltonian
path from s to t.
Proof:
Assume G has a Hamiltonian path from s to t, get a satisfying
assignment for .
If the path is normal, then define the assignment by:
Set each xi true if path goes L to R through xis gadget, false if it
goes R to L.
Cj
Cj
To see that this satisfies , consider
or
any clause C .
j
## The path goes through Cjs node by:
If the second, then:
xi is false, since path goes R-L.
By the way the detour edges are
set, Cj contains literal xi.
So Cj is satisfied by xi.
Cj pair in xi row
Cj pair in xi row
3SAT p DHAMPATH
Claim: is satisfiable iff the graph G has a Hamiltonian
path from s to t.
Proof:
Assume G has a Hamiltonian path from s to t.
If the path is normal, then it yields a satisfying assignment.
It remains to show that the path is normal (goes in order through
the gadgets, top to bottom, going one way or the other through
each crossbar, and detouring to pick up the Cj nodes),
The only problem (hand-waving) is if a detour doesnt work right,
but jumps from one gadget to another, e.g.:
But then the Ham. path could never reach a2:
Cj
Can reach a2 only from a1, a3,
and (possibly) Cj.
xi
a1 a2 a3
elsewhere.
And reaching a2 from a3 leaves
xi
nowhere to go from a2, stuck.
Summary: DHAMPATH
We have proved 3SAT p
DHAMPATH.
So DHAMPATH is NP-complete.
Can prove similar result for
DHAMCIRCUIT = { <G> | G is a
directed graph, and there is a circuit in
G that passes through each vertex of
G exactly once }
Theorem: 3SAT p DHAMCIRCUIT.
Proof:
## Same construction, but wrap around,
identifying s and t nodes.
Now a satisfying assignment for
corresponds to a Hamiltonian circuit.
s
Identify these two s nodes.
## UHAMPATH and UHAMCIRCUIT
Same questions about paths/circuits in undirected graphs.
UHAMPATH = { <G, s, t> | G is an undirected graph, s and
t are two distinct vertices, and there is a path from s to t in
G that passes through each vertex of G exactly once }
UHAMCIRCUIT = { <G> | G is an undirected graph, and
there is a circuit in G that passes through each vertex of G
exactly once }
Theorem: Both are NP-complete.
Obviously in NP.
To show NP-hardness, reduce the digraph versions of the
problems to the undirected versions---no need to consider
Boolean formulas again.
DHAMPATH p UHAMPATH
DHAMCIRCUIT p UHAMCIRCUIT
DHAMPATH p UHAMPATH
UHAMPATH = { <G, s, t> | G is an undirected graph, s and
t are two distinct vertices, and there is a path from s to t in
G that passes through each vertex of G exactly once }
Map <G, s, t> (directed) to <G, s, t > (undirected) so that
<G, s, t> DHAMPATH iff <G, s, t > UHAMPATH.
Example:
s
3
s
u
t
w
v
u1
v1
w1
u2
v2
w2
u3
v3
w3
t1
DHAMPATH p UHAMPATH
s3
s
u
t
In general:
w
v
u1
v1
w1
u2
v2
w2
u3
v3
w3
t1
Replace each vertex x other than s, t with vertices x1, x2, x3, connected
in a line.
Replace s with just s3, t with just t1.
For each directed edge from x to y in G, except incoming edges of s
and outgoing edges of t, include undirected edge between x3 and y1.
Dont include anything for incoming edges of s or outgoing edges of t--not needed since they cant be part of a Ham. path in G from s to t.
DHAMPATH p UHAMPATH
s3
s
u
t
In general:
w
v
u1
v1
w1
u2
v2
w2
u3
v3
w3
t1
## Replace each vertex x other than s, t with x1---x2---x3.
Replace s with s3, t with t1.
For each directed edge from x to y in G, except incoming edges of s
and outgoing edges of t, include x3---y1.
## G = the resulting undirected graph; s = s3; t = t1
Claim G has directed Hamiltonian path from s to t iff G has
an undirected Hamiltonian path from s to t.
Idea: Indices 1,2,3 enforce consistent direction of traversal.
Proof LTTR (in book).
Summary: UHAMPATH
We have proved DHAMPATH p UHAMPATH.
So UHAMPATH is NP-complete.
Can prove similar result for
UHAMCIRCUIT = { <G> | G is an undirected
graph, and there is a circuit in G that passes
through each vertex of G exactly once }
Theorem: DHAMCIRCUIT p UHAMCIRCUIT.
Proof:
Similar construction.
## Traveling Salesman Problem (TSP)
Variant of UHAMCIRCUIT.
n cities = vertices, in a complete (undirected) graph.
Each edge (u,v) has a cost, c(u,v), a nonnegative integer.
Salesman should visit all cities, each just once, at low cost.
Express as a language:
TSP = { <G, c, k> | G = (V,E) is a complete graph, c: E N,
k N, and G has a cycle visiting each node exactly once,
with total cost k }
Theorem: TSP is NP-complete.
Proof:
TSP NP: Guess tour and verify.
TSP is NP-hard: Show UHAMCIRCUIT p TSP.
Map <G> (undirected graph) to <G, c, k> so that G has a Ham.
circuit iff G with cost function c has a tour of total cost at most k.
UHAMCIRCUIT p TSP
TSP = { <G, c, k> | G = (V,E) is a complete graph, c: E
N, k N, and G has a cycle visiting each node exactly
once, with total cost k }
Map <G> (undirected graph) to <G, c, k> so that G has a
Ham. circuit iff G with cost function c has a tour of total
cost k.
Define mapping so that a Ham. circuit corresponds closely
with a tour of cost k.
G = (V, E), where V = V, all vertices of G, E = all edges
(complete graph).
c(u,v) = 1 if (u, v) E, 0 if (u,v) E.
k = 0.
v
v
Example:
u
0
0
0
0
UHAMCIRCUIT p TSP
TSP = { <G, c, k> | G = (V,E) is a complete graph,
c: E N, k N, and G has a cycle visiting each
node exactly once, with total cost k }
Map <G> (undirected graph) to <G, c, k>:
G = (V, E), where V = V, all vertices of G, E = all
edges (complete graph).
c(u,v) = 1 if (u, v) E, 0 if (u,v) E.
k = 0.
## Claim: G has a Ham. circuit iff G with cost
function c has a tour of total cost k.
Proof:
## If G has a Ham. circuit, all its edges have cost 0 in G
with c, so we have a circuit of cost 0 in G.
Tour of cost 0 in G must consist of edges of cost 0,
which are edges in G.
## More Examples, Revisited
SUBSET-SUM
SUBSET-SUM = {<S,t> | S is a multiset of
N, t N, and t is expressible as the sum of
some of the elements of S }
Example: S = { 2, 2, 4, 5, 5, 7 }, t = 13
<S, t > SUBSET-SUM, because 7 + 4 + 2 = 13
## Theorem: SUBSET-SUM is NP-complete.
Proof:
Show 3SAT p SUBSET-SUM.
Tricky, detailed, see book.
PARTITION
PARTITION = { <S> | S is a multiset of N and S
can be split into multisets S1 and S2 having equal
sums }
Example: S = { 2, 2, 4, 5, 5, 7 }
S PARTITION, since the sum is odd
Example: T = { 2, 2, 5, 6, 9, 12 }
T PARTITION, since 2 + 2 + 5 + 9 = 6 + 12.
Theorem: PARTITION is NP-complete.
Proof:
Show SUBSET-SUM p PARTITION.
Simplein recitation?
MULTIPROCESSOR SCHEDULING
MPS = { <S, m, D > |
S is a multiset of N (represents durations for
m N (number of processors), and
and S can be written as S1 S2 Sm such
that, for every i, sum(Si) D }
## Theorem: MPS is NP-complete.
Proof:
Show PARTITION p MPS.
Simplein recitation?
Next time
Probabilistic Turing Machines and
Probabilistic Time Complexity Classes | 7,009 | 21,129 | {"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} | 3.875 | 4 | CC-MAIN-2020-05 | latest | en | 0.8364 |
https://forum.alphasoftware.com/archive/index.php?t-23014.html&s=149e3d2c7a9f5ca82be5da92aa0a5def | 1,540,264,750,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00288.warc.gz | 677,993,888 | 2,488 | PDA
View Full Version : Problem:Decimal Time to Standard Time
ABC123
GregOnline
09-10-2001, 12:21 PM
I have been writing a gateway for a payroll program, and the data is exported from the timeclock in decimal (9.50 rather than 9:30)
Is there any way to convert these times with a calculated field?
I broke out the decimal portion with the INT() command, but then when I total all the individual lines for an employee it doesn't recognize that the integer portion should increase after 59 minutes, so I end up with something like this: 77:70.
Stephen Williams
09-10-2001, 12:31 PM
Mess with this...
TOSECONDS(END)-TOSECONDS(START))/3600
Stephen Williams
09-10-2001, 12:35 PM
And going the other way...
Jhrs = int(Flh/Calc->Size)
Jmins = mod(Flh/Calc->Size,1)*60
Time = alltrim(str(Calc->Jhrs))+":"+alltrim(str(Calc->Jmins))
Stephen Williams
09-10-2001, 12:36 PM
OOps - going too fast here, FLH is a decimal hour quantity...
Stephen Williams
09-10-2001, 12:41 PM
Oh dear - I am adjusting for crew sizes and all sorts of stuff and nonsense
From my interactive editor
Decimal_time = 9.50
Time = alltrim(str(int(Decimal_time)))+":"+alltrim(str(mod(Decimal_time,1)*60))
?time
= "9:30"
csda1
09-10-2001, 03:22 PM
Stephen,
No, NO, NOOOOO! {:?)
As a duly deputized member of the function police, I sentence you to 50 re-reading of the XBasic manual! However, since you do contribute to the board, I'll ask the judge to reduce your sentence to 25 re-readings of the manual!!!!
To convert decimal minutes you would simply use
TOTIME(decimaltime*3600,FormatCode,DecimalPlaces)
which in most cases would be something like
timetext=TOTIME(9.50*3600,2,0)
To convert the other way,
TOSECONDS(timetext)/3600
Regards,
Ira J. Perlow
Computer Systems Design & Associates
csda@mediaone.net | 519 | 1,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-43 | latest | en | 0.763875 |
https://iaras.org/home/caijmcm/fault-tolerance-of-hypercubes-and-folded-hypercubes | 1,723,239,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00563.warc.gz | 243,605,966 | 7,972 | ## Fault Tolerance of Hypercubes and Folded Hypercubes
AUTHOR(S): Litao Guo TITLE Fault Tolerance of Hypercubes and Folded Hypercubes PDF KEYWORDS Interconnection networks; Fault tolerance; r-component edge connectivity ABSTRACT Connectivity is a vital metric to explore fault tolerance and reliability of network structure based on a graph model. There are many kinds of connectivity to measure the fault tolerance and reliability of networks, such as classic connectivity, super connectivity, extraconnectivity. In this paper we focus on the number of components of graph which is called component connectivity. Let G = (V, E) be a connected graph. A r-component cut of G is a set of vertices whose deletion results in a graph with at least r components. r-component connectivity cκr (G) of G is the size of the smallest r-component cut. The r-component edge connectivity cλr (G) can be defined similarly. In this paper, we determine the r-component edge connectivity of hypercubes and folded hypercubes:(1) cλ2(Qn) = λ(Qn) = n for n ≥ 2. (2) cλ3(Qn) = 2n − 1 for n ≥ 2. (3) cλ4(Qn) = 3n − 2 for n ≥ 2. (4) cλ2(F Qn) = n + 1 for n ≥ 3. (5) cλ3(F Qn) = 2n + 1 for n ≥ 3. (6) cλ4(F Qn) = 3n + 1 for n ≥ 3. Cite this paper Litao Guo. (2017) Fault Tolerance of Hypercubes and Folded Hypercubes. International Journal of Mathematical and Computational Methods, 2, 72-75 | 389 | 1,369 | {"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-2024-33 | latest | en | 0.829253 |
https://physics.stackexchange.com/questions/416534/why-the-volume-of-a-region-is-not-a-diffeomorphism-invariant-lqg | 1,561,597,478,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000609.90/warc/CC-MAIN-20190626234958-20190627020958-00506.warc.gz | 539,755,997 | 36,578 | # Why the volume of a region is not a diffeomorphism invariant? (LQG)
In loop quantum gravity, the volume operator for a given region is not a diffeomorphism invariant. But classically we know that volume is a scalar quantity under a diffeomorphism even if we take the full manifold or any region.
• Volume is not invariant under a diffeomorphism, only under an isometry. – Javier Jul 13 '18 at 21:34
• But the volume element in n-manifold is n-form which is a scalar quantity under any (passive or active diffeomorphism) – o.nemoul Jul 14 '18 at 10:51
• @ravjotsk n-forms are scalar quantities, but the components of the n-form who transform as a tensor. The volume element in a 4d- metric space (M,g) is: sqrt( det(g) )*d^4x which is a scalar quantity. – o.nemoul Jul 14 '18 at 19:44
• @o.nemoul My bad. I’ll remove my comment since it’s not useful. However I thought that the term scalars was used for tensors of rank zero and an n-form is an alternating tensor of covariant rank-n. Although the tensor itself as an object remains invariant under coordinate transformations, the fact that its components change means its not a scalar (which should have the same value no matter what coordinates you use) – ravjotsk Jul 14 '18 at 20:12
Why would you think that volume is not diffeo-invariant?
If the region moves along with the diffemorphisms (aka passive diffeomorphisms), the volume is invariant. In LQG, the volume is also invariant (if you disagree, please explain why).
If the region doesn't move with diffeomorphisms (aka active diffeomorphisms), the volume changes. That is because the region here is only a region in the coordinate space, which doesn't have a well-defined notion of volume. The same is also true in LQG.
LQG is really not different from General Relavitity when it comes to diffeomorphism invariance.
• Volume is not invariant under Lorentz transformations (through the contractrion factor). If the diffeomorphism invariance is meant only for the 3d spacial submanifold(then volume would be invariant) of full diffeomorphism invariant for the 4d pseudo riemannanin manifold (then 3 volume.isnt invariantn): I have.no.idea. – lalala Jul 15 '18 at 16:14
• @lalala Lorentz transformations are active diffeomorphisms. Volume is not invariant under those neither in GR or in LQG, as I said in my answer. – Solenodon Paradoxus Jul 15 '18 at 16:18
• Ah now I see what you meant. I thought the passive/active wording.unusual so.I didnt really understood your point – lalala Jul 15 '18 at 16:46
• @lalala but in LQG volume is also not invariant under the spatial diffeomorphism. – o.nemoul Jul 15 '18 at 17:51
• @SolenodonParadoxus if you mean that the fixed region is in the coordinates chart and not in the real manifold, i will agree with you. But why do we consider the region inside the coordinates system? – o.nemoul Jul 15 '18 at 17:54 | 764 | 2,866 | {"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-2019-26 | longest | en | 0.911857 |
https://topqa.wiki/how-many-days-until-may-6th/ | 1,708,590,524,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00723.warc.gz | 610,811,262 | 14,425 | # How many days until may 6th
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Below is a list of the best how to remember port and starboard voted by users and compiled by us, invite you to learn together | 2,630 | 9,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-10 | latest | en | 0.910613 |
http://www.gap-system.org/Gap3/Manual3/C065S117.htm | 1,371,669,278,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709037764/warc/CC-MAIN-20130516125717-00084-ip-10-60-113-184.ec2.internal.warc.gz | 463,191,989 | 1,231 | 65.117 UpperBound
`UpperBound( n, d, q )`
`UpperBound` returns the best known upper bound A(<n>,<d>) for the size of a code of length n, minimum distance d over a field of size q. The function `UpperBound` first checks for trivial cases (like <d>=1 or <n>=<d>) and if the value is in the built-in table. Then it calculates the minimum value of the upper bound using the methods of Singleton (see UpperBoundSingleton), Hamming (see UpperBoundHamming), Johnson (see UpperBoundJohnson), Plotkin (see UpperBoundPlotkin) and Elias (see UpperBoundElias). If the code is binary, A(<n>, 2*l-1) = A(<n>+1, 2*l), so the `UpperBound` takes the minimum of the values obtained from all methods for the parameters (<n>, 2*l-1) and (<n>+1, 2*l).
``` gap> UpperBound( 10, 3, 2 );
85
gap> UpperBound( 25, 9, 8 );
1211778792827540 ```
GAP 3.4.4
April 1997 | 263 | 844 | {"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-2013-20 | latest | en | 0.594781 |
https://tskinnersbec.edublogs.org/2011/11/ | 1,709,457,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00481.warc.gz | 569,052,454 | 20,875 | # Hon Chemistry 11-30-11 Intro to Radioactivity
HON CHEMISTRY: Welcome to nuclear chemistry! No, really, I think you’ll be surprised how much nuclear chemistry is already a part of your everyday life. And now you know where E=mc2 came from!! Here’s the lecture from Wednesday; we’ll finish the properties and begin talking about nuclear equations tomorrow. By the way, you need to read about the life of Marie Cure. Fascinating woman of science with an incredible story!
# PHYSICS 11-30-11 Power & the Horsepower Lab
PHYSICS: Great job on the horsepower lab. Lotta noise – I’m thinking some of you took out some of your physics phrustrations on the stairs! Post your own horsepower numbers here. Did you notice that not all of the stairs are the same height? BTW: 1 Hp = 746 W and 1 kg = 2.2 lb. Post your lab results here. Who has the most power? And why?
Here’s the lecture from today on power. Great start on the problems!
# Physics 11-29-11 Chapter 5 Review Worksheet Problems
PHYSICS – Hey guys, it was good to get back to exercising the old gray matter today! Here are some of the problems from the first chapter 5 review worksheet: 3, 4, 5, 2, and 10. Watch out for 2 and 10! You did a great job on them with me in class, but make double sure you can work them also by yourself. That’s homework tonight, re-work by yourself the ones you didn’t get before. And then do it again til you make sure you have it! 🙂
I’ve never seen Old Faithful in person, have you? So we calculated the velocity of the water coming from the bottom – reckon how much power that would be? Power? Let’s talk about that tomorrow!
# Chemistry 11-29-11 Dalton’s Atomic Theory vs. Modern Atomic Theory
CHEMISTRY: Wow! Pennies turned to gold? Tell me again – how you could you prove that it was or wasn’t? Hey – what a lot of theory today! So the smallest thing that can be that thing is an atom. How do we know that? Well, we have to start at the beginning, and that’s Dalton’s Atomic Theory. Great job today comparing it with the Modern Atomic Theory. Keep those examples in mind when you’re studying for the test – especially when it comes to explaining those laws.
Did you hear about the make-up lab change? It will be this Thursday at 7:15 A.M. – last one of the year!
By the way – when you listen to the vodcast, that awful clicking noise you hear just might have to do with my necklace sliding back and forth across the mic. Sorry about that! 🙂
# Physics Roller Coasters & Worksheet
PHYSICS – Hey guys, great job on the roller coasters today! Who’s your vote for King of the Roller Coasters and why?
Did you get a copy of the correct worksheet. You want the one that says Chapter 5 Review Worksheet. Click here for a copy if you need one. Check the newly revised syllabus for the exact numbers you are supposed to do. See you tomorrow? Hmmmmmm……
# Happy Thanksgiving & Welcome Back!!
I hope you had a wonderful Thanksgiving! Do you know what I’m very thankful for? You! Thanks for the hard work you’ve put into chemistry and physics. Keep it up and don’t give up.
Are you ready to begin again? Get ready for two intense and phenomenal weeks! Stay strong and trust in the truth that will never fail, that will endure forever!
“Enter into His gates with thanksgiving and into His courts with praise. Be thankful unto Him and bless His name, for the Lord is good, His mercy is everlasting, and His truth endures to all generations.”
# Chemistry 11-18-11 Chapter 7 Test Review
CHEMISTRY: It’s been a crazy week, huh? Good luck in studying for the test! Here’s the review we did today. Also, did you find the Chapter 7 Stuff to Know sheet on Edline? Click here on the name, if you need one. I know it seems like a lot, but you can do it! (Remember that you don’t have to know diatomic molecules.)
First priority – make sure that you have memorized EVERYTHING. Then, go to sciencegeek.net and make sure you can write and name chemical formulas. Practice, practice, practice!!! Then start practicing the different kinds of problems – do at least three of each one of them. And also try those on sciencegeek.net. That’s always some good practice. Good luck – I’ll be praying for you! You can do it!!
# Physics 11-18-11 Law of Conservation of Energy
PHYSICS: So why is the first hill of a roller coaster always the highest? Here’s the lecture from Monday on the conservation of energy. Great job on the concepts and problems today – let’s put it to practice next week, and even this weekend on that take-home lab. Hey, if you were to design a roller coaster, what would it look like?!? 🙂
# Hon Chemistry 11-18-11 Average Atomic Mass
HON CHEMISTRY: Awwww….aren’t they cute! So what do you think the average atomic mass of puppies is? Here’s the lecture from Friday on calculating the average atomic mass of isotopes. Isotopes…. not puppies. 🙂
Have a great weekend! And to the cast and crew of Father Knows Best, break a leg!
flickr photo by Xanboozled
# Physics 11-17-11 Potential Energy & the Work Energy Theorem
PHYSICS: Speaking of balancing acts, this has been one more week, hasn’t it – and it’s only Thursday! Thanks to all of you who gave blood or tried to give blood today! Here’s the lecture from Thursday on potential energy. And here’s something to think about for tomorrow – why is the first hill of a roller coaster always the highest? | 1,302 | 5,363 | {"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-2024-10 | latest | en | 0.930996 |
http://scienceblogs.com/principles/2010/01/26/playing-with-graphs-people-in/ | 1,467,035,365,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396027.60/warc/CC-MAIN-20160624154956-00105-ip-10-164-35-72.ec2.internal.warc.gz | 274,731,080 | 21,220 | # Playing With Graphs: People in Albany Don’t Own Kindles
A few days back, Matthew Beckler added the Kindle edition to his sales rank tracker for How to Teach Physics to Your Dog. Given my well-known love for playing with graphs of data, it was inevitable that I would plot both of these in a variety of ways.
So, what do we learn from this? Well, we learn that people in the Albany. NY area don’t own Kindles:
OK, maybe that’s not obvious to everybody…
When you look at that graph, the blue line is the Amazon sales rank of the physical book edition, while the red line is the Amazon sales rank of the Kindle edition. The two track each other pretty well for a while, but diverge dramatically after about 48 hours, with the physical book sales rank shooting into the triple digits, while the Kindle sales rank stayed around 3,500. So, what happened at 48 hours?
The only significant development that I’m aware of that took place around then (which was about 1pm ET Sunday) is that both the Albany Times Union and the Schenectady Gazette ran articles about the book (the Times Union had a piece about the book written for them by a freelance writer, the Gazette went with the AP review, plus a notice of this weekend’s signing). Nothing else happened around that time that I know of.
So, this tells us that notices in the local papers were enough to drive up the sales rank of the physical book, but not the Kindle edition. So, people in the Albany area don’t own Kindles. Or, to be more precise, people in the Albany area who read print newspapers (neither the Gazette nor the Times Union put the book on their web sites) don’t own Kindles.
Amazing what you can learn from looking at graphs.
The other obvious thing that you can do with these data is to look at what relationship, if any, exists between the book sales rank and the Kindle sales rank. The easiest way to get at this is to plot one on the vertical axis and the other on the horizontal axis:
I’ve divided the data into two sets for this graph. The purple points at the first 48 hours of the data set, and the green are the last 48 hours. For the first 48 hours, they track each other pretty well– a straight line drawn through the purple points would come close to most of them, and has a slope close to 1 (1.27, to be precise). It’s not perfect, but it’s plenty good enough for social science.
The green points are way off that line, for the most part, but there’s a big clump of them over in the upper left, that would fit reasonably well to a line with a slope of a bit less than 3 (making a rough cut of that group gives a slope of 2.93). Those points are the stretch from Sunday afternoon through Monday night, when the physical book rank was at its highest point.
There’s also a sort of a tail connecting the Sunday-Monday group, as the book rank drifts back up to more or less where it was before the dramatic spike. At the time of this writing, the book rank is back up to 2500, which is about where it’s been since the AP review ran.
So that’s this week’s thrilling installment of Playing With Graphs…
1. #1 Jérôme ^
January 26, 2010
Sales rank is probably not the good data to plot here; sales volume would be more fitting (you might recover it by assuming that it follows a Pareto law).
January 26, 2010
It’s not perfect, but it’s plenty good enough for social science.
I LOL’ed.
3. #3 Danny
January 26, 2010
What an irresponsible analysis.
You’re supposed to be in a position of scientific authority here, yet you post a couple charts then unapologetically, without qualification, jump to some obscure conclusion because it makes for a sensationalist headline? This is a terrible example to be setting for your readers.
You want to conclude from these two charts that “people in the Albany area who read print newspapers don’t own Kindles.”?
There are plenty of other explanations. Let me try a few:
1. Something else you don’t know about (god forbid) caused a few more people to buy print versions.
2. A larger chunk of the population (which includes disproportionately fewer Kindle owners) buys books on Sundays.
3. People who own Kindles are less likely than non-Kindle owners to be convinced by a stupid article they read in the local newspaper to buy the book.
4. People who read the newspaper and _and_ own a Kindle still prefer to buy some books in print.
5. Random variation.
You’re not being a scientist. You’re being a sensationalist. Then we wonder why science is so misrepresented in the media…
4. #4 idlemind
January 26, 2010
Well I thought it was funny…
5. #5 tde
January 26, 2010
6. Scientists should never, ever be funny.
6. #6 Danny
January 26, 2010
Ok, then you’re not a very good blogger for failing to recognize that sarcasm doesn’t work on the internet. Clearly the first commenter thought this was a serious analysis.
7. #7 Skribb
January 26, 2010
If I were a social scientist, I might examine the comments of this post and come to the unequivocally true conclusion that readers of science themed blogs are incapable of recognizing hyperbole and sarcasm, despite when the blogger is laying it on awful thick.
@1 and especially @3: Oh come on! He was drawing absolute conclusions from a scatter plot. It is blatantly obvious that his conclusions are meant to be comical and an attempt to poke at the soft sciences.
8. #8 Sven DIMIlo
January 27, 2010
For variables like these, without any obvious cause-effect relationship, you should actually be calculating the slope for a reduced-major-axis regression, rather than simple Least Squares. The RMA slope is always higher.
But the slopes are irrelevant here anyway; it’s the correlation coefficients we want.
Yes, I’m having fun.
9. #9 BdN
January 27, 2010
Ah! I’ll repeat what idlemind wrote : I thought it was funny… Though maybe a little unfair to social scientists. They would at least run a covariance matrix…
Ok, then you’re not a very good blogger for failing to recognize that sarcasm doesn’t work on the internet.
Or maybe you’re not a good enough blog reader ? Or not a regular reader of this particular blog ? Ever heard of inside jokes ?
And you left out one of the more plausible explanations. “Or, to be more precise, people in the Albany area who read print newspapers […] don’t own Kindles.” : there may be a link between usage of technology for reading newspapers and usage of technology for reading in general. Or maybe those who buy for immediate reading have different habits or buy at different hours of the day or maybe it has to do with part of the population of the country waking up at another time or people seeing the print edition of the newspaper liked better the black and white picture of the book rather than the color one,etc.
Or maybe I’m reading it backwards since sale rank is inverse number of copies sold : the more copies, the higher rank. 1000 copies = 3000th rank 3000 copies = 1000th rank. According to this, Mr. Orzel is misreading the “data”, if we can call it so, since the spike in rank at 48 hours means the exact opposite : it “fell” to the 3 digits area. And the Kindle version “rose”.
So everybody in Albany who reads print papers DO OWN a Kindle.
And the second graph is also interesting : it clearly shows that the book (both versions) now sell more than in the beginning. And that, when more books are bought, well, more books are bought. That’s a good lesson.
10. #10 BdN
January 27, 2010
Or maybe I am the one misreading the graph…
*runs hiding under his bed*
11. #11 Mark
January 27, 2010
Ok, then you’re not a very good blogger for failing to recognize that sarcasm doesn’t work on the internet.
Sarcasm doesn’t work on the internet? Oh really. Fantastic. Thanks so much for telling me.
Seems to be working just fine. Great post, btw.
12. #12 shaodonglin
March 1, 2010
To the forecasting of earthquakes in Chile and Argentina newspapers, e-mail.
Argentina Newspapers: Buenos Aires province, and the helicopter crash in Santiago, Chile earthquake-related. Crash area meteorological disasters will occur. Sao Paulo, Rio de Janeiro, the Amazon River in central Peru has also crash will occur. Details http://www.shaodl.com/sspl7.htm
This is before the earthquake to the United Nations in Haiti, Dominica, Jamaica, newspapers, e-mail. http://www.dominicantoday.com/dr/world/2009/10/11/33512/UN-recovers-11-victims-from-Haiti-plane-crash
English Channel, California, Ecuador, Peru,south and north, New Zealand, Italy south — north, Mexico, Spain, Turkey southern ,Morocco, Caucasus, northern Iran, in recent earthquake. | 1,992 | 8,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-26 | longest | en | 0.963064 |
http://www.waldomaths.com/Newton1L.jsp | 1,512,950,418,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948511435.4/warc/CC-MAIN-20171210235516-20171211015516-00733.warc.gz | 472,863,885 | 4,150 | # The Newton-Raphson Method
This applet investigates the Newton-Raphson iterative method for approximating roots of equations, using differentiation
###### Author and programmer: Ron Barrow
UK Year 13,KS5, GCE A Level Core Mathematics
## How to Use this Applet
The function whose graph is drawn is a cubic of the form:
y = Ax3 + Bx2 + Cx + D, and you can vary this function by dragging the A, B, C and D sliders at the bottom.
The Newton-Raphson Method involves the iterative formula:
xn + 1 = xn - f(xn)/f '(xn)
which is used to try to solve the equation f(x) = 0.
The starting value for the iteration is represented by the light blue circle drawn on the x-axis. This circle can be dragged left and right with the mouse, to see how the iteration depends on the starting value.
You can also vary the number of iterations performed. The slider at the top right changes the number of iterations rapidly, but you can use the "Iter+" and "Iter-" buttons at the very top to change the number of iterations one at a time.
The program enables you to investigate how this iteration converges to a solution to the equation (if it does!), which solution is found (if any), and how quickly or slowly the convergence happens.
Play around - there are some fascinating patterns to observe! | 295 | 1,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-51 | latest | en | 0.889138 |
https://breakingdownfinance.com/finance-topics/performance-measurement/active-share/ | 1,726,134,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651440.11/warc/CC-MAIN-20240912074814-20240912104814-00054.warc.gz | 127,570,334 | 49,364 | # Active share
Active share is a measure used by fund managers that are benchmark-aware, meaning that they try to be active and ensure that they deviate sufficiently from the benchmark. Active share and active risk are two measures used to determine the acceptable level of deviation from the benchmark.
On this page, we discuss the active share formula, illustrate the approach using an example and finally provide an Excel spreadsheet that implements the approach. The spreadsheet can be downloaded at the bottom of the page.
## Active share formula
Active share (AS) measures the degree to which the number and sizing of the positions in a manager’s portfolio are different from those of a benchmark. It is given by the following equation:
where n is the total number of securities in the benchmark or the portfolio,
## Active share example
Next, let’s consider an example of how to calculate AS. The following table presents a case where the AS is calculated for a portfolio consisting of 6 securities.
To do the necessary calculations, we need to have the weights of the securities in the benchmark and the weights of the securities in the portfolios. We can then calculate the absolute value of the deviations, sum everything and divide by 2. This yields an AS of 20%. The spreadsheet is available for download at the bottom of the page.
## Summary
We discussed active share (AS), an important measure used to determine whether or not a mutual fund manager is actively managing a portfolio. The higher the AS, the more the portfolio deviates from the benchmark and thus the more active the fund manager. It is important to keep in mind that that AS should always be considered together with the tracking error or active risk. That’s because it is possible to construct portfolios with a high AS and a lot tracking error (by using securities that are not in the benchmark). | 366 | 1,889 | {"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-2024-38 | latest | en | 0.904924 |
https://www.physicsforums.com/threads/twice-of-supersymmetric-transformation-translation.798229/ | 1,508,524,524,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824293.62/warc/CC-MAIN-20171020173404-20171020193404-00223.warc.gz | 988,809,098 | 16,999 | Twice of supersymmetric transformation = translation
Tags:
1. Feb 16, 2015
wphysics
Hello, I have one conceptual question. I have been working on Supersymmetry.
Now, I understand that twice of supersymmetric transformation is equivalent to translation mathematically(naively).
However, I don't quite understand why this should be the case conceptually. Supersymmetric transformation is that boson to fermion and fermion to boson, meaning that it changes spin by 1/2. How can twice of it be equivalent to just translation?
I just want a conceptual answer, so it does not need to be super consistent.
2. Feb 16, 2015
ChrisVer
Twice a supersymmetric transformation = supersymmetric transformation. I don't know why you say that you have just translation. Otherwise you wouldn't be able to form a group.
What is equivalent to a translation is the commutator of two supersymmetric transformations.
3. Feb 16, 2015
wphysics
In Super Poincare algebra, we know that the anti commutator of two weyl spinors is proportional to translation. This is the equivalence that I am talking.
Thanks
4. Feb 16, 2015
ChrisVer
So you are asking why $\{ Q_a , \bar{Q}_\dot{b} \} = 2 \sigma^\mu_{a\dot{b}} P_\mu$ ?
5. Feb 16, 2015
wphysics
Yes! I am wondering why it is the case conceptually.
6. Feb 16, 2015
ChrisVer
But looking at it, the anticommutator will give you something like $(1/2 , 0) \times (0 ,1/2)$ (the times symbol looks weird here but I hope you understand what I wanted to point out), so it seems natural for me to give you a normal Lorentz repr (and not an irrep). So the $\sigma^\mu$ and the momentum.
7. Feb 16, 2015
wphysics
Yes. That explanation is the most common one. Let me put it in this way. For spin 0 particle, supersymmetric transformation transforms it to spin 1/2 particle. If we do again, it should be back to spin 0 and effectively, its effect is translation.
I cannot draw this picture in my head physically.
8. Feb 16, 2015
ChrisVer
9. Feb 16, 2015
ChrisVer
That's the reason why you can look at supersymmetry as a generalization of spacetime symmetry. And when you write a SUSY element in exponential form, you don't only add the spinor generators Q,Q* but also the momentum. I think it's intrinsic to it for closure reasons. Otherwise you can try to have a look at the Coleman-Mandula theorem.
10. Feb 17, 2015
haushofer
You can regard supertransformations Q as translations in the fermionic direction, and P as translations in spacetime. Both are as such translations in superspace.
11. Feb 25, 2015
Roy_1981
@wphysics
The best way to think about your query, in my opinion is look at the converse of the situation i.e. a susy transformation is DEFINED as a "square root" of a translation. This is exactly analogous to the way we define/picture spinors as "square root" of a vector following Dirac (who as the folklore has it arrived at the Dirac spinor equation by taking a "square root" of Klein-Gordon operator).
There is no simple way to motivate why a square of susy transformation is translation, but as some other members have pointed out, perhaps we need to augment our spacetime with Grassmann directions in addition to usual spacetime i.e. picture a larger spacetime (superspace) . And then the anticommutator of two susy transformation can be seen as a round trip in the Grassman direction, and this round trip does not bring you at the starting point, instead you get a shift along the usual spacetime direction i.e. a translation. This is bit like moving around in a helix every 360 degree doesn't return you to your original position but you arrive a point veritcally shifted over the starting point along the axis.
12. Feb 25, 2015
ChrisVer
Well it is still not enough, because you don't make two fermions... it's a QQ* combination (left/right-handed fermions)...
13. Feb 25, 2015
Roy_1981
You are semi-correct :). You compose two supersymmetry and then undo it by acting in reverse order (for bosons this AB-BA, for fermionic operations, its AB+BA). This set of two operations and then the reveres constitutes the departure from "parallelogram law" with the departure/defect being a translation operation. Two fermions of opposite handedness indeed can be turned into a vector using Pauli matrices ($\sigma^{\mu}$). | 1,080 | 4,291 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-43 | longest | en | 0.926483 |
https://frugalwiz.com/the-basics-of-domino/ | 1,695,483,263,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00832.warc.gz | 316,964,647 | 18,843 | # The Basics of Domino
The game of domino is a family of tile-based games that is played with a set of rectangular tiles, each with two square ends marked with the number of spots on each side. Traditionally, players used two pairs of dominoes to form sets, and then used a score to determine which pair was superior. This article will introduce the basics of the game of domino, and explain the rules of the game. In addition to the rules, you will also learn the variations and the pieces of the game.
## Rules
The main objective of the Rules of Domino is to form ‘cells’ of dominoes, each with an area of half a domino tile. Every ‘cell’ a player creates earns one point. The graphic illustration below shows various examples of cell formation and tactics, using Game Option 1.
## Variations
The word “domino” derives from a Venetian Carnival costume. A black robe and white mask were worn by a player to participate in the festivities. No language has an equivalent word for “polyomino”. There are several popular variations of the game. Some of the most famous are the Domino Whist, Texas 42, and the Mexican Train. Other popular variations include Double Fives and Threes.
## Pieces
A family of tile-based games, pieces of domino are rectangular pieces with square ends. Each spot on each domino represents a number of points. A player takes turns attempting to score as many points as possible by arranging the pieces of the same color in a row. Attempting to achieve a perfect score by stacking all the pieces of the same color will win you the game. If you’ve been playing dominoes for some time, you’ll recognize the rules.
## Score
Dominoes, also known as bones or cards, are dice with squares on either side. They are usually twice as long as they are wide. There is a central line, and each side is divided into two squares, called ends. The value of each side is based on how many spots are on that side. In the most popular variant, there are six pips on each side, and on the other, there are no pips at all. The sum of the pips on either side is called its weight.
## Free-agency period
The NBA free-agency period follows a predictable pattern. Teams identify players they want and move on to the next best available player if they can’t get that player from their current team. In 2018, the domino effect may begin with Kyle Lowry. If he signs with a new team, he could start a domino effect among free agents. But that’s not the only scenario for this free-agency period.
## Origin
The origin of the domino is shrouded in mystery. Although some historians believe that the game was first played as far back as 1120 AD in China, others believe it may have been first played in Asia or Egypt. The earliest known set of dominoes was found in the tomb of Tutankhamen, king of the 18th dynasty, around 1355 BC. While the origins of dominoes are unknown, the game has remained a popular pastime and can be enjoyed by all ages.
## Gameplay
The basic rule in dominoes is to place a tile onto the playing surface so that two adjacent ends touch. Doubles may be played only once, or they can be placed at both ends. When the player puts a tile in a double’s hand, it is considered to have “stitched up” the end of the chain. If both ends of the chain are obstructed, the player with the lowest hand wins. | 758 | 3,328 | {"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} | 3.578125 | 4 | CC-MAIN-2023-40 | longest | en | 0.969762 |
https://www.unitsconverters.com/en/Micro/Ms2-To-Microm/S2/Utu-7446-3424 | 1,627,999,114,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00463.warc.gz | 1,034,342,371 | 38,534 | Formula Used
1 Micron per Square Millisecond = 1 Meter per Square Second
1 Meter per Square Second = 1000000 Micrometer per Square Second
1 Micron per Square Millisecond = 1000000 Micrometer per Square Second
## μ/ms² to µm/s² Conversion
The abbreviation for μ/ms² and µm/s² is micron per square millisecond and micrometer per square second respectively. 1 μ/ms² is 1000000 times bigger than a µm/s². To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including μ/ms² to µm/s² conversion.
## Micron per Square Millisecond to µm/s²
Check our Micron per Square Millisecond to µm/s² converter and click on formula to get the conversion factor. When you are converting acceleration from Micron per Square Millisecond to µm/s², you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## μ/ms² to Micrometer per Square Second
The formula used to convert μ/ms² to Micrometer per Square Second is 1 Micron per Square Millisecond = 1000000 Micrometer per Square Second. Measurement is one of the most fundamental concepts. Note that we have Fahrenheit as the biggest unit for length while Per Degree Celsius is the smallest one.
## Convert μ/ms² to µm/s²
How to convert μ/ms² to µm/s²? Now you can do μ/ms² to µm/s² conversion with the help of this tool. In the length measurement, first choose μ/ms² from the left dropdown and µm/s² from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from µm/s² to μ/ms²? You can check our µm/s² to μ/ms² converter.
How to convert μ/ms² to µm/s²?
The formula to convert μ/ms² to µm/s² is 1 Micron per Square Millisecond = 1000000 Micrometer per Square Second. μ/ms² is 1000000 times Bigger than µm/s². Enter the value of μ/ms² and hit Convert to get value in µm/s². Check our μ/ms² to µm/s² converter. Need a reverse calculation from µm/s² to μ/ms²? You can check our µm/s² to μ/ms² Converter.
How many m/s² is 1 μ/ms²?
1 μ/ms² is equal to 1 m/s². 1 μ/ms² is 1 times Smaller than 1 m/s².
How many km/s² is 1 μ/ms²?
1 μ/ms² is equal to 0.001 km/s². 1 μ/ms² is 1000 times Smaller than 1 km/s².
How many µm/s² is 1 μ/ms²?
1 μ/ms² is equal to 1000000 µm/s². 1 μ/ms² is 1000000 times Bigger than 1 µm/s².
How many mi/s² is 1 μ/ms²?
1 μ/ms² is equal to 0.000621371192237334 mi/s². 1 μ/ms² is 1609.344 times Smaller than 1 mi/s².
## μ/ms² to µm/s² Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like acceleration finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like μ/ms² to µm/s² through multiplicative conversion factors. When you are converting acceleration, you need a Micron per Square Millisecond to Micrometer per Square Second converter that is elaborate and still easy to use. Converting μ/ms² to Micrometer per Square Second is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Micron per Square Millisecond to µm/s², this tool is the answer that gives you the exact conversion of units. You can also get the formula used in μ/ms² to µm/s² conversion along with a table representing the entire conversion.
Let Others Know | 971 | 3,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-31 | latest | en | 0.766418 |
http://www.haskell.org/hoogle/?hoogle=map+-base+-opengl&start=21 | 1,369,241,564,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702019913/warc/CC-MAIN-20130516110019-00008-ip-10-60-113-184.ec2.internal.warc.gz | 493,161,901 | 5,948 | # map -base -opengl
An efficient implementation of ordered maps from keys to values (dictionaries). This module re-exports the value lazy Lazy API, plus several value strict functions from Strict. These modules are intended to be imported qualified, to avoid name clashes with Prelude functions, e.g. > import qualified Data.Map as Map The implementation of Map is based on size balanced binary trees (or trees of bounded balance) as described by: * Stephen Adams, "Efficient sets: a balancing act", Journal of Functional Programming 3(4):553-562, October 1993, http://www.swiss.ai.mit.edu/~adams/BB/. * J. Nievergelt and E.M. Reingold, "Binary search trees of bounded balance", SIAM journal of computing 2(1), March 1973. Note that the implementation is left-biased -- the elements of a first argument are always preferred to the second, for example in union or insert. Operation comments contain the operation time complexity in the Big-O notation (http://en.wikipedia.org/wiki/Big_O_notation).
A Map from keys k to values a.
O(n) map f xs is the ByteString obtained by applying f to each element of xs
O(n) map f t is the Text obtained by applying f to each element of t. Subject to fusion. Performs replacement on invalid scalar values.
O(n*min(n,W)). map f s is the set obtained by applying f to each element of s. It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y
O(n) map f xs is the ByteString obtained by applying f to each element of xs.
O(n) map f xs is the ByteString obtained by applying f to each element of xs. This function is subject to array fusion.
O(n). Map a function over all values in the map. > map (++ "x") (fromList [(5,"a"), (3,"b")]) == fromList [(3, "bx"), (5, "ax")]
O(n). Map a function over all values in the map. > map (++ "x") (fromList [(5,"a"), (3,"b")]) == fromList [(3, "bx"), (5, "ax")]
O(n*log n). map f s is the set obtained by applying f to each element of s. It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y
O(n). The function mapAccum threads an accumulating argument through the map in ascending order of keys. > let f a b = (a ++ b, b ++ "X") > mapAccum f "Everything: " (fromList [(5,"a"), (3,"b")]) == ("Everything: ba", fromList [(3, "bX"), (5, "aX")])
O(n). The function mapAccum threads an accumulating argument through the map in ascending order of keys. > let f a b = (a ++ b, b ++ "X") > mapAccum f "Everything: " (fromList [(5,"a"), (3,"b")]) == ("Everything: ba", fromList [(3, "bX"), (5, "aX")])
O(n) Like a combination of map and foldl'. Applies a function to each element of a Text, passing an accumulating parameter from left to right, and returns a final Text. Performs replacement on invalid scalar values.
The mapAccumL function behaves like a combination of map and foldl; it applies a function to each element of a ByteString, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new ByteString.
The mapAccumL function behaves like a combination of map and foldl; it applies a function to each element of a ByteString, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new list.
The mapAccumL function behaves like a combination of map and foldl; it applies a function to each element of a ByteString, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new ByteString.
The mapAccumL function behaves like a combination of map and foldl; it applies a function to each element of a ByteString, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new list.
The mapAccumR function behaves like a combination of map and a strict foldr; it applies a function to each element of a Text, passing an accumulating parameter from right to left, and returning a final value of this accumulator together with the new Text. Performs replacement on invalid scalar values.
The mapAccumR function behaves like a combination of map and foldr; it applies a function to each element of a ByteString, passing an accumulating parameter from right to left, and returning a final value of this accumulator together with the new ByteString.
The mapAccumR function behaves like a combination of map and foldr; it applies a function to each element of a ByteString, passing an accumulating parameter from right to left, and returning a final value of this accumulator together with the new ByteString.
O(n). The function mapAccumR threads an accumulating argument through the map in descending order of keys.
O(n). The function mapAccumR threads an accumulating argument through the map in descending order of keys.
O(n). The function mapAccumWithKey threads an accumulating argument through the map in ascending order of keys. > let f a k b = (a ++ " " ++ (show k) ++ "-" ++ b, b ++ "X") > mapAccumWithKey f "Everything:" (fromList [(5,"a"), (3,"b")]) == ("Everything: 3-b 5-a", fromList [(3, "bX"), (5, "aX")])
O(n). The function mapAccumWithKey threads an accumulating argument through the map in ascending order of keys. > let f a k b = (a ++ " " ++ (show k) ++ "-" ++ b, b ++ "X") > mapAccumWithKey f "Everything:" (fromList [(5,"a"), (3,"b")]) == ("Everything: 3-b 5-a", fromList [(3, "bX"), (5, "aX")])
Constructs a new array derived from the original array by applying a function to each of the elements.
Apply a function to transform the result of a continuation-passing computation. * (mapCont f m) = f . runCont
Apply a function to transform the result of a continuation-passing computation. * (mapContT f m) = f . runContT
O(n). Map values and separate the Left and Right results. > let f a = if a < "c" then Left a else Right a > mapEither f (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (fromList [(3,"b"), (5,"a")], fromList [(1,"x"), (7,"z")]) > > mapEither (\ a -> Right a) (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (empty, fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")])
O(n). Map values and separate the Left and Right results. > let f a = if a < "c" then Left a else Right a > mapEither f (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (fromList [(3,"b"), (5,"a")], fromList [(1,"x"), (7,"z")]) > > mapEither (\ a -> Right a) (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (empty, fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")])
O(n). Map keys/values and separate the Left and Right results. > let f k a = if k < 5 then Left (k * 2) else Right (a ++ a) > mapEitherWithKey f (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (fromList [(1,2), (3,6)], fromList [(5,"aa"), (7,"zz")]) > > mapEitherWithKey (\_ a -> Right a) (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (empty, fromList [(1,"x"), (3,"b"), (5,"a"), (7,"z")])
O(n). Map keys/values and separate the Left and Right results. > let f k a = if k < 5 then Left (k * 2) else Right (a ++ a) > mapEitherWithKey f (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (fromList [(1,2), (3,6)], fromList [(5,"aa"), (7,"zz")]) > > mapEitherWithKey (\_ a -> Right a) (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")]) > == (empty, fromList [(1,"x"), (3,"b"), (5,"a"), (7,"z")])
Map the unwrapped computation using the given function. * (mapErrorT f m) = f (runErrorT >
Lift a unary operation to the new monad.
Constructs a new array derived from the original array by applying a function to each of the indices.
O(n*min(n,W)). mapKeys f s is the map obtained by applying f to each key of s. The size of the result may be smaller if f maps two or more distinct keys to the same new key. In this case the value at the greatest of the original keys is retained. > mapKeys (+ 1) (fromList [(5,"a"), (3,"b")]) == fromList [(4, "b"), (6, "a")] > mapKeys (\ _ -> 1) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 1 "c" > mapKeys (\ _ -> 3) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 3 "c"
O(n*log n). mapKeys f s is the map obtained by applying f to each key of s. The size of the result may be smaller if f maps two or more distinct keys to the same new key. In this case the value at the greatest of the original keys is retained. > mapKeys (+ 1) (fromList [(5,"a"), (3,"b")]) == fromList [(4, "b"), (6, "a")] > mapKeys (\ _ -> 1) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 1 "c" > mapKeys (\ _ -> 3) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 3 "c"
O(n*min(n,W)). mapKeysMonotonic f s == mapKeys f s, but works only when f is strictly monotonic. That is, for any values x and y, if x < y then f x < f y. The precondition is not checked. Semi-formally, we have: > and [x < y ==> f x < f y | x <- ls, y <- ls] > ==> mapKeysMonotonic f s == mapKeys f s > This means that f maps distinct original keys to distinct resulting keys. This function has slightly better performance than mapKeys. > mapKeysMonotonic (\ k -> k * 2) (fromList [(5,"a"), (3,"b")]) == fromList [(6, "b"), (10, "a")]
O(n). mapKeysMonotonic f s == mapKeys f s, but works only when f is strictly monotonic. That is, for any values x and y, if x < y then f x < f y. The precondition is not checked. Semi-formally, we have: > and [x < y ==> f x < f y | x <- ls, y <- ls] > ==> mapKeysMonotonic f s == mapKeys f s > This means that f maps distinct original keys to distinct resulting keys. This function has better performance than mapKeys. > mapKeysMonotonic (\ k -> k * 2) (fromList [(5,"a"), (3,"b")]) == fromList [(6, "b"), (10, "a")] > valid (mapKeysMonotonic (\ k -> k * 2) (fromList [(5,"a"), (3,"b")])) == True > valid (mapKeysMonotonic (\ _ -> 1) (fromList [(5,"a"), (3,"b")])) == False | 2,852 | 9,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2013-20 | longest | en | 0.859886 |
https://academickids.com/encyclopedia/index.php/C%2A-algebra | 1,717,093,585,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00830.warc.gz | 62,932,458 | 9,838 | C*-algebra
C*-algebras are an important area of research in functional analysis. A C*-algebra can be defined concretely as a complex algebra A of linear operators on a complex Hilbert space with two additional properties:
• A is closed under the operation taking adjoints of operators.
It is generally believed that C*-algebras were first considered primarily for their use in quantum mechanics to model algebras of physical observables. This line of research began in an extremely rudimentary form with Werner Heisenberg's matrix mechanics and in a more mathematically developed form with Pascual Jordan around 1933. Subsequently John von Neumann attempted to establish a general framework for these algebras which culminated in a series of papers on rings of operators. These papers considered a special class of C*-algebras which are now known as von Neumann algebras.
Around 1943, the work of Gel'fand, Mark Naimark and Irving Segal yielded an abstract characterisation of C*-algebras making no reference to operators.
C*-algebras are now an important tool in the theory of unitary representations of locally compact groups, and are also used in algebraic formulations of quantum mechanics.
Contents
Abstract characterization
We begin with the abstract characterization of C*-algebras given in the 1943 paper by Gel'fand and Naimark.
A C*-algebra A is a Banach algebra over the field of complex numbers, together with a map * : AA called involution. The image of an element x of A under involution is written x*. Involution has the following properties:
• For all x, y in A:
[itex] (x + y)^* = x^* + y^* \quad [itex]
[itex] (x y)^* = y^* x^*. \quad [itex]
• For every λ in C and every x in A:
[itex] (\lambda x)^* = \overline{\lambda} x^*. [itex]
• For all x in A
[itex] (x^*)^* = x. \quad [itex]
• The C* condition holds for all x in A:
[itex] \|x x^* \| = \|x\|^2. [itex]
Any C*-algebra is automatically a B*-algebra, since the C* condition implies that
[itex] \|x \| = \|x^*\| [itex]
for all x in A. However, not every B*-algebra is a C*-algebra.
A bounded linear map π : AB between B*-algebras A and B is called a *-homomorphism if
• For x and y in A
[itex] \pi(x y) = \pi(x) \pi(y). \quad [itex]
• For x in A
[itex] \pi(x^*) = \pi(x)^*. \quad [itex]
In the case of C*-algebras, the boundedness condition is superfluous. In fact, any *-homomorphism between C*-algebras is contractive. If π is bijective, then its inverse is also a *-homomorphism and π is called a *-isomorphism and A and B are said to be *-isomorphic.
Examples
Finite-dimensional C*-algebras
The algebra Mn(C) of n-by-n matrices over C becomes a C*-algebra if we consider matrices as operators on the Euclidean space Cn and use the operator norm ||.|| on matrices. The involution is given by the conjugate transpose. More generally, one can consider finite direct sums of matrix algebras.
Theorem. A finite-dimensional C*-algebra A is canonically isomorphic to a finite direct sum
[itex] A = \bigoplus_{e \in \min A } A e[itex]
where min A is the set of minimal nonzero self-adjoint central projections of A. Each C*-algebra Ae is isomorphic (in a noncanonical way) to the full matrix algebra Mdim(e)(C). The finite family indexed on min A given by {dim(e)}e is called the dimension vector of A. This vector uniquely determines the isomorphism class of a finite-dimensional C*-algebra.
C*-algebras of operators
The prototypical example of a C*-algebra is the algebra L(H) of continuous linear operators defined on a complex Hilbert space H; here x* denotes the adjoint operator of the operator x : HH. In fact, every C*-algebra A is *-isomorphic to a norm-closed adjoint closed subalgebra of L(H) for a suitable Hilbert space H; this is the content of the Gelfand-Naimark theorem.
Commutative C*-algebras
Let X be a locally compact Hausdorff space. The space C0(X) of complex-valued continuous functions on X that vanish at infinity (defined in the article on local compactness) form a commutative C*-algebra C0(X) under pointwise multiplication and addition. The involution is pointwise conjugation. C0(X) has a multiplicative unit element iff X is compact. As does any C*-algebra, C0(X) has an approximate identity. In the case of C0(X) this is immediate: consider the directed set of compact subsets of X, and for each compact K let fK be a function of compact support which is identically 1 on K. Such functions exist by the Tietze-Urysohn theorem which applies to locally compact Hausdorff spaces. {fK}K is an approximate identity.
The Gelfand representation states that every commutative C*-algebra is *-isomorphic to the algebra C0(X), where X is the space of characters equipped with the weak* topology. Furthermore if C0(X) is isomorphic to C0(Y) as C*-algebras, it follows that X and Y are homeomorphic. This characterization is one of the motivations for the noncommutative topology and noncommutative geometry programs.
The C*-algebra of compact operators
Let H be a separable infinite-dimensional Hilbert space. K(H) is the algebra of compact operators on H. It is a norm closed subalgebra of L(H). K(H) is also closed under involution; hence it is a C*-algebra. Though K(H) does not have an identity element; an approximate identity for K(H) can be easily displayed. To be specific, H is isomorphic to the space of square summable sequences l2, so we may assume that H = l2. For each natural number n let Hn be the subspace of sequences of l2 which vanish for indices kn and let en be the orthogonal projection onto Hn. The sequence {en}n is an approximate identity for K(H).
The quotient of L(H) by K(H) is the Calkin algebra.
C*-enveloping algebra
Given a B*-algebra A with an approximate identity, there is (up to C*-isomorphism) unique C*-algebra E(A) and *-morphism π from A into E(A) which is universal, that is every other B*-morphism π': AB factors uniquely through π. E(A) is called the C*-enveloping algebra of the B*-algebra A.
Of particular importance is the C*-algebra of a locally compact group G. This is defined as the enveloping C*-algebra enveloping algebra of the group algebra of G. The C*-algebra of G provides context for general harmonic analysis of G in the case G is non-abelian. In particular, the dual of locally compact group is defined to the primitive ideal space of the group C*-algebra. See spectrum of a C*-algebra.
von Neumann algebras
von Neumann algebras, known as W* algebras before the 1960s, are a special kind of C*-algebra. They are required to be closed in a topology which is weaker than the norm topology. Their study is a specialized area of functional analysis in itself.
C*-algebras and quantum field theory
In quantum field theory, one typically describes a physical system with a C*-algebra A with unit element; the self-adjoint elements of A (elements x with x* = x) are thought of as the observables, the measurable quantities, of the system. A state of the system is defined as a positive functional on A (a C-linear map φ : AC with φ(u u*) > 0 for all uA) such that φ(1) = 1. The expected value of the observable x, if the system is in state φ, is then φ(x).
Properties of C*-algebras
C*-algebras have a large number of properties which are technically convenient. These properties can be established by use the continuous functional calculus or by reduction to commutative C*-algebras. In the latter case, we can use the fact that the structure of these is completely determined by the Gelfand isomorphism.
• Any *-morphism between C*-algebras has norm ≤ 1.
• The algebraic quotient of a C*-algebra by a closed proper two-sided ideal is a C*-algebra in a unique way.
• The set of elements of a C*-algebra A of the form x*x forms a closed convex cone. This cone is identical to the elements of the form x x*. Elements of this cone are called non-negative (or sometimes positive, even though this terminology conflicts with its use for elements of R.)
• The set of self-adjoint elements of a C*-algebra A naturally has the structure of an partially ordered vector space; the ordering is usually denoted ≥. In this ordering, a self-adjoint element x of A satisfies x ≥ 0 iff x is non-negative. Two self-adjoint elements x and y of A satisfy xy if x - y ≥ 0.
• Any C*-algebra has an approximate identity. In fact, there is a directed family {eλ}λ ∈ I of self-adjoint elements of A such that
[itex] x e_\lambda \rightarrow x [itex]
[itex] 0 \leq e_\lambda \leq e_\mu \leq 1\quad \mbox{ whenever } \lambda \leq \mu. [itex]
In case A is separable, A has a sequential approximate identity.
References
• A. Connes, Noncommutative geometry, Academic Press, 1994. This book is widely regarded as a source of new research material, providing much supporting intuition. ISBN 0-121-85860-X
• J. Dixmier, Les C*-algèbres et leurs représentations, Gauthier-Villars, 1969. This is a somewhat dated reference, but is still considered as a high-quality technical exposition. It is available in English from North Holland press.
• G. Emch, Algebraic Methods in Statistical Mechanics and Quantum Field Theory, Wiley-Interscience, 1972. Mathematically rigorous reference which provides extensive physics background.
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy | 2,415 | 9,309 | {"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-2024-22 | latest | en | 0.920779 |
https://electronics.stackexchange.com/questions/127993/latching-solenoid-drive-voltage-problem | 1,558,377,522,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256100.64/warc/CC-MAIN-20190520182057-20190520204057-00280.warc.gz | 460,657,876 | 34,550 | # Latching Solenoid Drive Voltage Problem
I have a circuit which drives a latching selonoid valve(valve changes state for every 50ms pulse)
Below there is an easy go schematic of the circuit. Basically a 3.7V LiPo Battery supplies for 3.3V mcu regulator and a boost converter of 9V's (LM2623).
To switch the valve mcu enables the boost converter and waits for 10ms for stabile 9V. And then set drive pulses to valve driver.
When I connect the board to MSP430 Launchpad(debugger, vcc=3.5V) and in debug state; if I run the code, valve perfectly switches between state. In the switching time, voltage on coil only decreases to ~7V.
When I disconnect the debugger and run with only battery, code tries to swith the valve but, it not switches. And voltage on coil decreases till ~5v or so.
How can this extra voltage decrease can be compansated. There is already 100uF connected to valve driver supply.
simulate this circuit – Schematic created using CircuitLab
You can calculate the required capacitor from:
C= $\frac{I \cdot T}{\Delta V}$
Where I is the coil current, T is the pulse time, and $\Delta V$ is the allowable voltage droop.
For example, if the current is 1A, the allowable droop is 1V and time is 50ms, then a 50mF capacitor (50,000uF) is required. Naturally, a larger capacitor will take more than 10ms to charge if the battery cannot supply enough current to hold the supply up for 50ms. | 358 | 1,410 | {"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} | 2.828125 | 3 | CC-MAIN-2019-22 | longest | en | 0.871453 |
http://peroglyfer.se/2018/01/page/2/ | 1,550,495,917,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247486480.6/warc/CC-MAIN-20190218114622-20190218140622-00200.warc.gz | 201,473,232 | 13,324 | # Machine Learning Insight – Every Parameter Is A Network Including The Bias
The idea is simple.
I guess the brain does not store a lot of hard coded numbers so why should we. Assigning 1’s to biases seems to be a wasteful thing.
The philosophy of machine learning could perhaps be to take advantage of everything. So from this I’m going to test what predictive biases or state converged biases could do. So y,bias = model(x,bias).
# Product Idea – Turning The Screen 180 Degrees For A Drawing Tablet Setup
The idea is simple.
Why not take advantage of laptops with 180 degree turnable screens. Then you can use it as a tablet overlay display. Quite cool idea.
This makes a cheap drawing tablet feel professional. Works fine with Linux Budgie. Just rotate the screen in the Nvidia settings or similar.
So if you have an old laptop you dont use anymore. This could bring some joy.
# Math Idea – Mandelbrot Fractal With Recursive c
I was wondering if I could use machine learning with fractals.
So inspired by this I reused the c0 in z(i+1) = z(i)**2 + c0. Changing the c0 for every iteration with some rule seems to have ?split the set.
I was thinking maybe there exist a network model function for c0 for which you can adapt the set or image more.
After some guessing I came up with the “Linux Penguin Light saber Fractal”
Enjoy
# Machine Learning – Is Image Raytracing Using Machine Learning Possible?
Why does image of a glass ball on a surface look like it does?
To every complex question there is the network answer. It makes sense to a network.
So I will test if I can calculate a simple raytracing scenario using machine learning. The network looks as follows. The atoms are my weights with its parameters.
The idea is to put the output image as an internal layer just before the output layer neuron. The last output neuron is then a truth value.
For the truth I will have to test a little. But to simplify. All light from the emission input image should correspond to the value in the output neuron. Here I will test the sum( input pixel value energy ) = sum ( output image pixel value energy ). Here the internal image layer is bigger than the input image so the energy will have to be distributed. Another truth is that I got index of refraction for my object. So some of my parameter values are already given for my image layer in air and object as glass.
Further if I put a ”circular” layer around the raytracing network maybe I can use that as a similar truth calculation.
A truth calculation is just that you know the output for a given input. So all black input should give an all black or zero output.
From the last image of the spherical surrounding space layer. I guess that if the energy is to be distributed over an infinite amount of neurons. Then the energy on all neurons would get to zero amounts. Some equal split.
But the sum is to be equal to the input energy so here you get another truth maybe.
If there exist input energy above zero then the emission layer depending on the starting position. Some distance from the center point. Will have a distribution of that energy on the outer capture layer neurons. Like a ?normal distribution maybe.
Hmm if you place the object in the center point it could cause a problem. Equal distribution. So I wonder if you can take a second outer layer and generate some difference. Like two eyes are separated from each other. Here you got two separated outer capture layers at some random distance apart.
# Machine Learning Idea – The Light Sigmoid Or Dual Linearity Function
I was looking into my network theory. When it occurred to me that the sigmoid activation function bare resemblance to the light trajectory in water.
So for the universe system to calculate what happens inside the material it ?uses a dual linear function as an output changing function.
So I will test the light activation function with one or two parameters for the angles in the machine learning network model.
# Machine Learning Idea – Calculating With Complex Roots
I was thinking. If you calculate the model with a implicit error function. Then its possible to get complex numbers. However complex numbers provide additional ’svängrum’ (rotational room) for the model.
So I made the implicit error function give two solutions where it could give a complex number. It turned out. The solution gave a double root. That is. Two equal complex numbers. Then all I had to do was to take the absolute length of the number.
# Machine Learning Idea – Implicit Error Function
I’m currently doing some calculation on weather data. Just a time series.
I thought I get inspired by an equation.
dataIn – dataOut = dataModel
After some guessing I get the error equation. Here dataIn, dataout, … are >= 0. They are scaled from 0 to 1. I make some odd looking equation. sqrt(dataIn) – sqrt(dataOut) = sqrt(dataModel). Even though it does not follow above equation.
dataIn + dataOut – 2*((dataIn*dataOut)**.5) = dataModel(dataIn)
Here dataIn + dataOut – 2*((dataIn*dataOut)**0.5) is my target value for my model function for a given dataIn. This because I could get ‘nan’ otherwise where the model update can get negative for dataModel values.
So iterating through all my dataIn(i), dataOut(i) values I get the parameters for my model.
Then to predict a value for a given value x as dataIn. I input that x in my model(dataIn = x). Get an equation:
x + dataOut – 2*((x*dataOut)**.5) = model(x)
From this equation I solve the possible dataOut values as one of the predicted value.
# Newspaper And Magazine Innovation – Include Python Coding Courses In Machine Learning?
As a way to make money on online magazines and newspapers I wonder if innovation can help. What about coding. In particular machine learning.
In Machine learning its not that hard to get results. The code is short. The first time it might not be so great but with imagination you get better. When you do find the ?right system and model its instant satisfaction.
So for course assignments you have in data, training data and test data at hand. Its a rewarding process of finding the model parameters. Perfect for learning how to code.
So why is this important. Learning to code is only as important as your ability to imagine the possibilities with machine learning. Its not about robots. Its about engineering and physics and our future ability to tackle climate change.
# Idea – Python Plots As Good As A Christmas Card !?
I was wondering. For data plots in python. Why should they look so 90’s. I don’t know. But if blender was around at that time it could look this this render.
So why not put blender to use in 2D/3D data plotting in python.
from matplotlib.blender import blenderplot I whish | 1,452 | 6,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-09 | latest | en | 0.911863 |
http://www.ck12.org/book/CK-12-Middle-School-Math-Concepts-Grade-7/section/6.14/ | 1,469,899,699,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258936356.77/warc/CC-MAIN-20160723072856-00172-ip-10-185-27-174.ec2.internal.warc.gz | 364,770,091 | 46,176 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.14: Percent of Increase
Difficulty Level: At Grade Created by: CK-12
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Practice Percent of Increase
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On March 8, 2010, Cento Coffee Shop began charging $2.50 for its drip coffee. Prior to that date, the shop charged$2.25 for its drip coffee. What was the percent increase in the drip coffee’s cost on March 8, 2010?
In this concept, you will learn how to find the percent of increase.
### Finding Percent of Increase
Sometimes, you will have a price that has been increased by a specific amount. Other times you can observe that a price has increased over time. Past pricing is often thought of in this way. When you compare a past price and an increased current price, you can figure out the percent by which a price has increased. You call this percent the percent of increase.
The percent of increase from one amount to another is the ratio of the amount of increase to the original amount.
To find the percent of increase, follow these steps.
First, find the amount of increase by subtracting the original price from the new price.
Next, write a fraction in which the numerator is the amount of increase and the denominator is the original amount.
Percent of increase=Amount of IncreaseOriginal Amount\begin{align*}\text{Percent of increase} = \frac{\text{Amount of Increase}} {\text{Original Amount}}\end{align*}
Then, write the fraction as a percent.
Now let’s apply these steps.
Find the percent of increase from 20 to 35.
First, subtract 20 from 35.
3520=15\begin{align*}35 - 20 = 15\end{align*}
Next, write the fraction: Percent of increase=Amount of IncreaseOriginal Amount=1520\begin{align*}\text{Percent of increase} = \frac{\text{Amount of Increase}} {\text{Original Amount}} = \frac{15}{20}\end{align*}
Then:
One Way
152020x20x20x====x1001,5001,5002075\begin{align*}\begin{array}{rcl} \frac{15}{20}&=&\frac{x}{100} \\ 20x &=& 1,500 \\ \frac{\cancel{20}x}{\cancel{20}} &=& \frac{1,500}{20} \\ x &=& 75 \end{array}\end{align*}Another Way
1520=15÷520÷5=344)3.00¯¯¯¯¯¯¯¯¯¯¯¯ 0.75Divide to 2 decimal places.0.75=75%\begin{align*}&\frac{15}{20}=\frac{15 \div 5}{20 \div 5}=\frac{3}{4} \\ & \overset{ \ \ 0.75}{4 \overline{ ) {3.00 \;}}} \quad \leftarrow \text{Divide to} \ 2 \ \text{decimal places.} \\ &0.75 = 75 \%\end{align*}
The answer is the percent of increase from 20 to 35 is 75%.
Notice that you could solve for the percent in two different ways. One was to use a proportion and the other was to simply divide. Either way, you will get the same answer.
### Examples
#### Example 1
Earlier, you were given a problem about Cento Coffee House.
On March 8, 2010, the price of a drip coffee went from $2.25 to$2.50. What was the percent increase in the drip coffee’s cost?
First, figure out the amount of the increase by subtracting.
2.502.25=0.25\begin{align*}2.50 - 2.25 = 0.25\end{align*}
Next, divide this number by the original amount.
0.25÷2.25=0.11\begin{align*}0.25 \div 2.25 = 0.11\end{align*}
Finally, convert the decimal to a percent.
0.11=11%\begin{align*}0.11 = 11\%\end{align*}
The answer is the price of the drip coffee increased by 11%.
#### Example 2
Let’s look at another example.
Find the percent of increase from 24 to 72.
First, subtract 24 from 72.
7224=48\begin{align*}72 - 24 = 48\end{align*}
Next, write the fraction: Percent of increase=Amount of IncreaseOriginal Amount=4824\begin{align*}\text{Percent of increase} = \frac{\text{Amount of Increase}} {\text{Original Amount}} = \frac{48}{24}\end{align*}
Then:
One Way
482424x24x24x====x1004,8004,80024200\begin{align*}\begin{array}{rcl} \frac{48}{24}&=&\frac{x}{100} \\ 24x &=& 4,800 \\ \frac{\cancel{24}x}{\cancel{24}} &=& \frac{4,800}{24} \\ x &=& 200 \end{array}\end{align*}
Another Way
48242==48÷2424÷24=21200%\begin{align*}\begin{array}{rcl} \frac{48}{24}&=&\frac{48 \div 24}{24 \div 24}=\frac{2}{1} \\ 2 &=& 200 \% \end{array}\end{align*}
The answer is the percent increase from 24 to 72 is 200%.
Yes, sometimes the percent of increase can be greater than 100%!
#### Example 3
Find the percent of increase from 45 to 50. You may round to the nearest whole percent when needed.
First, subtract 45 from 50.
5045=5\begin{align*}50 - 45 = 5\end{align*}
Next, write the fraction: Percent of increase=Amount of IncreaseOriginal Amount=545\begin{align*}\text{Percent of increase} = \frac{\text{Amount of Increase}} {\text{Original Amount}} = \frac{5}{45}\end{align*}
Then solve the proportion:
54545xx===x10050011.11\begin{align*}\begin{array}{rcl} \frac{5}{45} &=& \frac{x}{100} \\ 45x &=& 500 \\ x &=& 11.11 \end{array}\end{align*}
The answer is rounded to the nearest whole percent, the percent increase from 45 to 50 is 11%.
#### Example 4
Find the percent of increase from $1.00 to$1.75. You may round to the nearest whole percent when needed.
First, subtract 1.00 from 1.75.
1.751.00=0.75\begin{align*}1.75 - 1.00 = 0.75\end{align*}
Next, write the fraction: Percent of increase=Amount of IncreaseOriginal Amount=0.751.00\begin{align*}\text{Percent of increase} = \frac{\text{Amount of Increase}} {\text{Original Amount}} = \frac{0.75}{1.00} \end{align*}
Then, divide:
0.751.00=0.75=75%\begin{align*}\frac{0.75}{1.00} = 0.75 = 75 \%\end{align*}
The answer is the percent increase from $1.00 to$1.75 is 75%.
#### Example 5
Find the percent of increase from 34 to 60. You may round to the nearest whole percent when needed.
First, subtract 34 from 60.
6034=26\begin{align*}60 - 34 = 26\end{align*}
Next, write the fraction: Percent of increase=Amount of IncreaseOriginal Amount=2634\begin{align*}\text{Percent of increase} = \frac{\text{Amount of Increase}} {\text{Original Amount}} = \frac{26}{34}\end{align*}
Then, solve the proportion:
263434xx===x1002,60076.47\begin{align*}\begin{array}{rcl} \frac{26}{34} &=& \frac{x}{100} \\ 34x &=& 2,600 \\ x &=& 76.47 \end{array} \end{align*}The answer is rounded to the nearest whole percent, the percent increase from 34 to 60 is 76%.
### Review
Find the percent of increase given the original amount. You may round to the nearest whole percent when necessary.
1. From 25 to 40
2. From 15 to 30
3. From 18 to 50
4. From 22 to 80
5. From 16 to 18
6. From 3 to 10
7. From 85 to 100
8. From 75 to 90
9. From 26 to 36
10. From 100 to 125
11. From 100 to 150
12. From 125 to 175
13. From 175 to 200
14. From 200 to 225
15. From 225 to 275
### My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Percent of Increase
The percent of increase is the percent that a value has increased by.
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# What are 1.76 Meters to Feet? – Meters to Feet Table
Welcome to 1.76 meters per inch. Inputs in meters are often written with the unit symbol m, while outputs in the usual US unit inch are abbreviated as o ″. You will find everything to do with 1.76m in ″, including the converter and model.
You’ve come to the right place if you’re looking for something that means 1.76 meters in inches. You can replace the length in meters in the calculator below with 1.76. Our application then performs the calculations automatically.
## 1.76 Meters to Feet
1.76 meters to feet shows you how many feet are equal to 1.76 meters as well as in other units such as miles, inches, yards, centimeters, and kilometers
## How to Convert Meters to Feet?
1 meter is equal to 3.280839895 feet:
• 1m = 100cm/(2.54cm/in)/(12in/ft) = 3.280839895ft
The distance d in feet (ft) is equal to the distance d in meters (cm) times 3.280839895, that conversion formula:
• d(ft) = d(m) × 3.280839895
## How many Feet in a Meter?
One Meter is equal to 3.28084 Feet:
• 1m = 1m × 3.280839895 = 3.28084ft
## How many Meters in a Foot?
One Foot is equal to 0.3048 Meters:
• 1ft = 1ft × 0.3048 = 0.3048m
## How to Convert 30 Meters to Feet?
• d(ft) = 30(m) × 3.280839895 = 98.4252ft
## Formula
Ft = meters × 3.28084. According to the ‘meters to feet’ conversion formula, if you want to convert 1.76 (one point seven six) Meters to Feet, you must multiply 1.76 by 3.28084.
Complete solution:1.76 meters × 3.28084=5.77′If you want to convert 1.76 M to both Feet and Inches parts, then you first have to calculate the whole number part for Feet by rounding 1.76 × 3.28084 fractions down.
• And then convert the remainder of the division to Inches by multiplying by 12 (according to the Feet to Inches conversion formula)
Complete solution: ( 1.76 meters × 3.28084 )=5′get the Inches Part((1.76 × 3.28084) – 5′) * 12=(5.774 – 5′) * 12=0.774 * 12=9.29″so the full record will look like 5′9.29″
## How many Feet are 1.76 Meters?
1.76 meters equals 5.77 feet because 1 meter is equal to roughly 3.28 feet. To convert 1.76 meters to feet, multiply by 3.28.
Other Conversions Feet: 5.77428 Meters: 1.76 Miles 0.00109 inches: 69.29138 Yards: 1.92475 Kilometers: 0.00176 Centimeters: 176.00000
## Meters to Feet Conversion Formula
• [X] ft = 3.2808398950131 × [Y] m
• where [X] is the result in ft and [Y] is the amount of m we want to convert
## 1.76 Meters to Feet Conversion breakdown and explanation
1.76 m to ft conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result).
• Every display form has its advantages, and in different situations, a particular condition is more convenient than another.
• For example, using scientific notation when working with big numbers is recommended due to more effortless reading and comprehension. Usage of fractions is recommended when more precision is needed.
• If we want to calculate how many Feet are 1.76 M, we have to multiply 1.76 by 1250 and divide the product by 381. So for 1.76 we have: (1.76 × 1250) ÷ 381 = 2200 ÷ 381 = 5.7742782152231 Feet
Hence 1.76 m = 5.7742782152231 ft
## Meters to Feet Table
Meters Feet 1.60 m 5.249 ft 1.61 m 5.282 ft 1.62 m 5.315 ft 1.63 m 5.348 ft 1.64 m 5.381 ft 1.65 m 5.413 ft 1.66 m 5.446 ft 1.67 m 5.479 ft 1.68 m 5.512 ft 1.69 m 5.545 ft 1.70 m 5.577 ft 1.71 m 5.610 ft 1.72 m 5.643 ft 1.73 m 5.676 ft 1.74 m 5.709 ft 1.75 m 5.741 ft 1.76 m 5.774 ft 1.77 m 5.807 ft 1.78 m 5.840 ft 1.79 m 5.873 ft 1.80 m 5.906 ft 1.81 m 5.938 ft 1.82 m 5.971 ft 1.83 m 6.004 ft 1.84 m 6.037 ft 1.85 m 6.070 ft 1.86 m 6.102 ft 1.87 m 6.135 ft 1.88 m 6.168 ft 1.89 m 6.201 ft 1.90 m 6.234 ft
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January 30, 2024 | 1,601 | 4,868 | {"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} | 3.84375 | 4 | CC-MAIN-2024-10 | latest | en | 0.879197 |
http://www.docstoc.com/docs/28462927/SYSTEMS-OF-LINEAR-EQUATIONS-%C3%A2%E2%82%AC%E2%80%9C-WORD-PROBLEMS | 1,398,024,179,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00580-ip-10-147-4-33.ec2.internal.warc.gz | 396,805,771 | 15,562 | # SYSTEMS OF LINEAR EQUATIONS – WORD PROBLEMS
Document Sample
``` SYSTEMS OF LINEAR EQUATIONS – WORD PROBLEMS
1. It takes Cathy 1.5 hours to paddle her canoe 6 miles upstream. Then she turns her canoe around and paddles
6 miles downstream in 1 hour. What is the rate of the current? What is Cathy’s paddling rate in still water?
2. With a tailwind, a jet flew 2000 miles in 4 hours. The jet’s return trip against the same wind required 5
hours. Find the jet’s speed and the wind speed.
3. With a tailwind, a helicopter flies 300 miles in 1.5 hours. When the helicopter flies back against the same
wind, the trip takes 3 hours. What is the helicopter’s speed? What is the wind’s speed?
4. Allyson paddles her canoe 9 miles upstream in 4.5 hours. The return trip downstream takes her 1.5 hours.
What is the rate at which Allyson paddles in still water? What is the rate of the current?
5. With a tailwind, a plane makes a 3000-mile trip in 5 hours. On the return trip, the plane flies against the
same wind and covers the 3000 miles in 6 hours. What is the speed of the wind?
****************************************************************************************
6. A chemist mixed a 15% glucose solution with a 35% glucose solution. This mixture produced 35 liters of a
19% solution. How many liters of each solution did the chemist use in the mixture?
7. A 4% salt solution is mixed with a 15% salt solution. How many milliliters of each solution are needed to
obtain 600 milliliters of a 10% salt solution?
8. A jar contains quarters and dimes. There are 15 more quarters than dimes. The total value of the coins is
\$23. How many of each coin are there?
9. Donnell wants to make a 2-pound mixture of cashews and pecans that costs \$2.60 per pound. Cashews cost
\$2.50 per pound and pecans costs \$3.00 per pound. How many pounds of each should he use?
10. At the Snack Shack, dried cherries cost \$3.50 per pound. Dried apricots cost \$1.50 per pound. The store’s
owner wants to make 10 pounds of a cherry-apricot mixture that costs \$2.70 per pound. How many pounds of
cherries and apricots should the owner use to make the mixture?
****************************************************************************************
11. The sum of the digits of a two-digit number is 14. When the digits are reversed, the new number is 36
more than the original number. What is the original number?
12. The sum of the digits of a two-digit number is 10. If 18 is added to the number, the digits will be reversed.
Find the number.
13. The sum of the digits of a two-digit number is 14. The first digit is 4 less than twice the second digit. What
is the number?
14. Alex is 6 years older than Frank. The sum of their ages is 50. Find Alex’s age and Frank’s age.
15. Leticia is 21 years older than Katie. In 2 years, Leticia will be twice as old as Katie. Find Leticia’s current
age and Katie’s current age.
NOTES: SYSTEM OF LINEAR EQUATIONS
I. SOLVING RATE PROBLEMS 2. With a tailwind, an airplane makes a 900-mile trip in 2.25 hours.
On the return trip, the plane flies against the wind and makes the trip
1. Ben paddles his kayak along a 9-mile course on a river. Going in 3 hours. What is the plane’s speed? What is the wind speed?
upstream, it takes him 6 hours to complete the course. Going
downstream, it takes him 2 hours to complete the same course. What Let ___ = __________________________________________
is the rate of the kayak? What is the rate of the current?
Let ___ = __________________________________________
Let ___ = __________________________________________
Let ___ = __________________________________________
Equations:
Solve:
Equations:
Solve:
II. SOLVING MIXTURE PROBLEMS 2. A coin bank contains 250 dimes and quarters worth a total of
\$39.25. Write and solve a system of linear equations to find how
1. A pharmacist wants to mix an ointment that is 9% ointment zinc many dimes and quarters there are in the coin bank.
oxide with an ointment with an ointment that is 15% zinc ointment to
make 30 grams of an ointment that is 10% zinc oxide. How many Let ___ = __________________________________________
grams of each ointment should the pharmacist mix together?
Let ___ = __________________________________________
Let ___ = __________________________________________
Let ___ = _________________________________________
Equations:
Solve:
Equations:
Solve:
III. SOLVING NUMBER-DIGIT PROBLEMS 2. The sum of the digits of a two-digit number is 10. When the digits
are reversed, the new number is 54 more than the original number.
1. The sum of the digits of a two-digit number is 17. When the digits What is the original number?
are reversed, the new number is 9 more than the original number.
What is the original number? Check your answer. Let ___ = __________________________________________
Let ___ = __________________________________________ Let ___ = __________________________________________
Let ___ = __________________________________________ Original number: _________________ New number: _____________
Original number: _________________ New number: _____________
Equations:
Equations:
Solve:
Solve: | 1,262 | 5,417 | {"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-2014-15 | longest | en | 0.916906 |
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Ordinary Mad Minute Multiplication Worksheets. This series of multiplication worksheets is similar to the Rocket Math worksheets or Mad Minute Multiplication drills used by many schools. Each of these 80 or 100 problem worksheets are designed to be completed in roughly two minutes and they can provide a challenge at home that makes the one minute multiplication worksheets at school feel like a breeze. Mad Math Minute Multiplication Tables 9 9 x 7 63 9 x 4 36 9 x 1 9 9 x 8 72 9 x 9 81 9 x 5 45 9 x 2 18 9 x 3 27 9 x 6 54 9 x 1 9 9 x 4 36 9 x 8 72 9 x 7 63 9 x 9 81 9 x 5 45 9 x 4 36 9 x 6 54 9 x 1 9 9 x 3 27 9 x 8 72 9 x 2. These multiplication worksheets may be configured for 2 3 or 4 digit multiplicands being multiplied by multiples of ten that you choose from a table.
3rd Grade Multiplication Worksheets Best Coloring Pages For Kids 3rd Grade Math Worksheets Subtraction Worksheets Math For Kids From pinterest.com
Multiplication worksheets 12 times tables multiplication worksheets 5 times table and multiplication fact sheet are three main things we want to present to you based on the post title. These worksheets are slightly longer versions of the one minute multiplication drills in the previous section. You may vary the numbers of problems on the worksheet from 15 to 27. When done on a regular basis these timed worksheets will help students improve speed and accuracy with basic adding subtracting multiplication and division facts. Teachers Parents and Students can print these worksheets and make copies. These one minute timed multiplication worksheets all come with a corresponding printable answer page.
### Decimal Word Problems Multiplication Math Worksheets Kids Multiplying.
These multiplication worksheets are appropriate for Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade and 5th Grade. You begin with easy facts in the earier worksheets and once the student can complete each level with only one or two mistakes they advance to higher level where more multiplication facts enter the landscape. Teachers Parents and Students can print these worksheets and make copies. Teachers should also check out Multiplication Lesson Plans. Nov 15 2017 - A printable multiplication worksheet for practicing mad minute drills of 1 to 5 minutes. Displaying 8 worksheets for Mad Minute Addition And Subtraction Mixed.
Source: pinterest.com
Displaying 8 worksheets for Mad Minute Addition And Subtraction Mixed. You begin with easy facts in the earier worksheets and once the student can complete each level with only one or two mistakes they advance to higher level where more multiplication facts enter the landscape. Doubles to Double 20 Halves to Half of 40. Early Multiplication and Division Skills. Multiplication Basic Multiplication Multi-Digit Order of Operations.
Source: pinterest.com
18102015 While we talk about Mad Minute Multiplication Worksheets 6-9 below we will see various variation of photos to complete your ideas. When done on a regular basis these timed worksheets will help students improve speed and accuracy with basic adding subtracting multiplication and division facts. Students are given a short period of time usually three minutes or so to complete as many problems as they can. Most quizzes are EXACTLY one. Choose the number of problems and number ranges.
Source: pinterest.com
The worksheets are designed for students to complete multiplication drills of 1 to 5 minutes in length. Free multiplication and division worksheets interactive activities games and other resources for 5 to 11 year olds. 04052020 Multiplication Worksheets Mad Minute May 4 2020. Teachers Parents and Students can print these worksheets and make copies. First print the multiplication table in slide No.
Source: pinterest.com
Students are given a short period of time usually three minutes or so to complete as many problems as they can. 18102015 While we talk about Mad Minute Multiplication Worksheets 6-9 below we will see various variation of photos to complete your ideas. These worksheets are slightly longer versions of the one minute multiplication drills in the previous section. Mad Math Minute Multiplication Tables 9 9 x 7 63 9 x 4 36 9 x 1 9 9 x 8 72 9 x 9 81 9 x 5 45 9 x 2 18 9 x 3 27 9 x 6 54 9 x 1 9 9 x 4 36 9 x 8 72 9 x 7 63 9 x 9 81 9 x 5 45 9 x 4 36 9 x 6 54 9 x 1 9 9 x 3 27 9 x 8 72 9 x 2. Provided that you tend not to overload your young children with worksheets nearly all of them get pleasure from the challenge of beating their very best time.
Source: pinterest.com
You may vary the numbers of problems on the worksheet from 15 to 27. Doubles to Double 20 Halves to Half of 40. Worksheets Working with free multiplication worksheets is an excellent approach to add some range for your homeschooling. A multiplication math drill is a worksheet with all of the single digit problems for multiplication on one page. You may vary the numbers of problems on the worksheet from 15 to 27.
Source: pinterest.com
These worksheets are slightly longer versions of the one minute multiplication drills in the previous section. These one minute timed multiplication worksheets all come with a corresponding printable answer page. Students are given a short period of time usually three minutes or so to complete as many problems as they can. Teachers Parents and Students can print these worksheets and make copies. You may vary the numbers of problems on the worksheet from 15 to 27.
Source: pinterest.com
06082020 Gallery of Printable Multiplication Mad Minute Pin On Charts Throughout Printable Multiplication Mad Minute 7Th Grade Math Minutes Worksheet Printable Worksheets And In Printable Multiplication Mad Minute Minute Math Multiplication Worksheets Mad Minute Inside Printable Multiplication Mad Minute 428 Addition Worksheets For You To Print Right Now In Printable Multiplication Mad Minute. These multiplication worksheets may be configured for 2 3 or 4 digit multiplicands being multiplied by multiples of ten that you choose from a table. Doubles to double 6 Doubles to Double 12 Halves to Half of 24. Most quizzes are EXACTLY one. Provided that you tend not to overload your young children with worksheets nearly all of them get pleasure from the challenge of beating their very best time.
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This multiplication math drill worksheet is appropriate for Kindergarten 1st Grade 2nd Grade and 3rd Grade. The worksheets are designed for students to complete multiplication drills of 1 to 5 minutes in length. Decimal Word Problems Multiplication Math Worksheets Kids Multiplying. Worksheets Working with free multiplication worksheets is an excellent approach to add some range for your homeschooling. Displaying 8 worksheets for Mad Minute Addition And Subtraction Mixed.
Source: pinterest.com
First print the multiplication table in slide No. 18102015 While we talk about Mad Minute Multiplication Worksheets 6-9 below we will see various variation of photos to complete your ideas. 04052020 Multiplication Worksheets Mad Minute May 4 2020. Use it to help students learn their multiplication facts. Teachers Parents and Students can print these worksheets and make copies.
Source: pinterest.com
Multiplication Basic Multiplication Multi-Digit Order of Operations. These 1 minute timed multiplication worksheets are downloadable and printable. You begin with easy facts in the earier worksheets and once the student can complete each level with only one or two mistakes they advance to higher level where more multiplication facts enter the landscape. Students are given a short period of time usually three minutes or so to complete as many problems as they can. You can choose the number of problems number ranges and header styles for each worksheet that is dynamically generated to your specifications.
Source: pinterest.com
Multiplication Basic Multiplication Multi-Digit Order of Operations. Worksheets Working with free multiplication worksheets is an excellent approach to add some range for your homeschooling. You begin with easy facts in the earier worksheets and once the student can complete each level with only one or two mistakes they advance to higher level where more multiplication facts enter the landscape. Choose the number of problems and number ranges. You may vary the numbers of problems on the worksheet from 15 to 27.
Source: pinterest.com
Coloring Pages Book Multiplication Worksheets Free Printable Report Grade Puzzles Converting Fractions Decimals Worksheet Logical Thinking Answers Science Variable Multiplying. These worksheets are slightly longer versions of the one minute multiplication drills in the previous section. Worksheets are Mad minutes Mad minute math multiplication work Minute. Mad Math Minute Multiplication Tables 9 9 x 7 63 9 x 4 36 9 x 1 9 9 x 8 72 9 x 9 81 9 x 5 45 9 x 2 18 9 x 3 27 9 x 6 54 9 x 1 9 9 x 4 36 9 x 8 72 9 x 7 63 9 x 9 81 9 x 5 45 9 x 4 36 9 x 6 54 9 x 1 9 9 x 3 27 9 x 8 72 9 x 2. These multiplication worksheets may be configured for 2 3 or 4 digit multiplicands being multiplied by multiples of ten that you choose from a table.
Source: pinterest.com
Addition Basic Facts Mad Minute Drills. Most quizzes are EXACTLY one. Choose the number of problems and number ranges. Nov 15 2017 - A printable multiplication worksheet for practicing mad minute drills of 1 to 5 minutes. Doubles to double 6 Doubles to Double 12 Halves to Half of 24.
Source: pinterest.com
26052019 The free worksheets below offer students plenty of opportunities to hone their multiplication skills. These multiplication worksheets are appropriate for Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade and 5th Grade. Doubles to Double 20 Halves to Half of 40. The worksheets are designed for students to complete multiplication drills of 1 to 5 minutes in length. Free multiplication and division worksheets interactive activities games and other resources for 5 to 11 year olds.
Source: pinterest.com
Most quizzes are EXACTLY one. Teachers Parents and Students can print these worksheets and make copies. These 1 minute timed multiplication worksheets are downloadable and printable. A multiplication math drill is a worksheet with all of the single digit problems for multiplication on one page. These multiplication worksheets are appropriate for Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade and 5th Grade.
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Mar 22 . 10 min read | 2,759 | 12,582 | {"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-2021-21 | longest | en | 0.849009 |
hamradiooutsidethebox.ca | 1,680,439,884,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00172.warc.gz | 338,673,968 | 25,477 | # The Weird Antenna Wire End Effect
I have been giving some thought to antenna wires recently. I was particularly interested in the reason for them being shorter than we would expect if we simply calculate their length by dividing signal propagation speed by the frequency of the signal.
The length (L) of a half wave wire can be calculated using the simple formula: c/2f=L where c is the propagation speed of an electromagnetic wave in free space. The rounded value of c is 984 million feet per second, so a half wave wire at 3.5MHz will be 984/(2×3.5)=140 feet.
If we cut and trim a half wave wire for the 80m band we find the resonant length is shorter than the value calculated above. This is because a half wave wire is electrically longer than its physical length due to something called the “End Effect”. This is the capacitive coupling of an unterminated wire end to ground.
An 80m half-wave is also a full wave on 40m (A in figure 1). On 40m it is comprised of an inner half wave wire (B in figure 1) and an outer half wave wire (C+D in figure 1). The inner half wave wire is terminated by the outer half wave wire and is therefore unaffected by the End Effect.
The outer half wave wire is unterminated so it is shortened by about 3% due to the antenna End Effect. Wire insulation, velocity factor etc result in a total shortening of about 5%.
The shortened length is calculated by the formula 468/f. This means that the inner half wave wire (which should be defined by the formula 492/f) is too short by an amount 468/492=95%.
So a half wave wire that is tuned for resonance on 80m will NOT be resonant on 40m despite a precise harmonic relationship between the two bands. The bottom of the 40m band (in North America) is 7.0MHz which is exactly twice the frequency of the bottom of the 80m band – 3.5MHz.
There is also a precise harmonic relationship between the 80m band, the 40m band and the 20m band. The bottom end of the 20m band – 14.0MHz – is exactly twice the frequency of the bottom of the 40m band and exactly four times the frequency of the bottom of the 80m band. But a half wave wire cut for resonance on 80m will similarly NOT be resonant on 20m.
If we refer to figure 2 we can see why. The portion of the wire “B” is now a full wave on 20m and it is terminated by the half wave sections “C” and “D”. So B is too short by the same 5%.
Now suppose we take a half wave wire and turn it vertically. We now have a half wave vertical antenna. Does the End Effect still apply? Does it apply if the “other half” of the dipole is virtual (i.e. a ground-mounted quarter wave vertical)?
We still have an unterminated wire end at the top of the antenna (it could be a metal whip). We still use the same 5% shortening factor in calculating its length. Let’s take it one step further. The End Effect is caused by a capacitive coupling between an unterminated wire end and the ground. The capacitive shortening effect is only about 3%, but could we increase the capacitive coupling to make the shortening even greater?
The answer is a definite yes. We can create a larger capacitive coupling to ground using something called a “capacitance hat“. A capacitance hat, also known as a capacity hat, or top hat can be used to significantly shorten a vertical radiator.
A capacitance hat is made with radial spokes or a circular conductor (or similar arrangements). The diameter of the hat is roughly equal to the amount by which the radiating element can be shortened.
This is illustrated in figure 3. A half wave wire that is erected as an inverted-V has a quarter-wave vertical section (B) and a quarter wave horizontal section (A). If the quarter wave horizontal section is arranged so that it lies centered on top of the quarter wave vertical section, we now have a quarter wave vertical antenna (D) with a capacitance hat (C) that will theoretically be resonant at the same frequency as the half-wave inverted-V antenna.
In practical terms, a single wire capacitance hat may not be very efficient, so several such wires forming a circular arrangement around the plane of the vertical element may be required.
If we stand on our heads and take another look, a quarter wave vertical antenna with a quarter wave capacitance hat is an upside down version of a quarter wave vertical antenna with a quarter wave radial field.
The main difference is the vertical with the “cap hat” is resonant at twice the frequency of its upside down cousin. Could we build a dual-band antenna and change bands by turning it upside down? Impractical perhaps but it kinda makes sense in an upside down sorta way.
These are my thoughts on the topic and, since this is not a scientific treatise, feel free to debate or disagree. | 1,084 | 4,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-14 | longest | en | 0.952269 |
https://nus.kattis.com/courses/CS2040C/CS2040C_S1_AY2122/assignments/hmgvis/problems/fire3 | 1,679,888,910,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00534.warc.gz | 487,765,759 | 7,824 | Hide
Problem FFire!
Joe works in a maze. Unfortunately, portions of the maze have caught on fire, and the owner of the maze neglected to create a fire escape plan. Help Joe escape the maze.
Given Joe’s location in the maze and which squares of the maze are on fire, you must determine whether Joe can exit the maze before the fire reaches him, and how fast he can do it.
Joe and the fire each move one square per minute, vertically or horizontally (not diagonally). The fire spreads all four directions from each square that is on fire. Joe may exit the maze from any square that borders the edge of the maze. Neither Joe nor the fire may enter a square that is occupied by a wall.
Input
The first line of input contains the two integers $R$ and $C$, separated by spaces, with $1 \le R, C \le 1000$. The following $R$ lines of input each contain one row of the maze. Each of these lines contains exactly $C$ characters, and each of these characters is one of:
• #, a wall
• ., a passable square
• J, Joe’s initial position in the maze, which is a passable square
• F, a square that is on fire
There will be exactly one J in the input.
Output
Output a single line containing “IMPOSSIBLE” if Joe cannot exit the maze before the fire reaches him, or an integer giving the earliest time Joe can safely exit the maze, in minutes.
Sample Input 1 Sample Output 1
4 4
####
#JF#
#..#
#..#
3
Sample Input 2 Sample Output 2
3 3
###
#J.
#.F
IMPOSSIBLE | 364 | 1,456 | {"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} | 2.734375 | 3 | CC-MAIN-2023-14 | latest | en | 0.889368 |
https://ja.scribd.com/document/258982638/Chapter-7-chap-seven | 1,560,633,926,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997335.70/warc/CC-MAIN-20190615202724-20190615224724-00324.warc.gz | 480,057,355 | 58,873 | You are on page 1of 11
# Financial Management: Principles and Applications, 11e (Titman)
## Chapter 7 An Introduction to Risk and Return-History of Financial Market Returns
7.1 Realized and Expected Rates of Return and Risk
1) You purchased the stock of Sargent Motors at a price of \$75.75 one year ago today. If you sell
the stock today for \$89.00, what is your holding period return?
A) 35.00%
B) 12.50%
C) 17.50%
D) 25.00%
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
2) You have invested in a project that has the following payoff schedule:
Probability of
Payoff
Occurrence
\$40
.15
\$50
.20
\$60
.30
\$70
.30
\$80
.05
What is the expected value of the investment's payoff? (Round to the nearest \$1.)
A) \$60
B) \$65
C) \$58
D) \$70
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
3) If there is a 20% chance we will get a 16% return, a 30% chance of getting a 14% return, a
40% chance of getting a 12% return, and a 10% chance of getting an 8% return, what is the
expected rate of return?
A) 12%
B) 13%
C) 14%
D) 15%
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
1
## Principles: Principle 2: There Is a Risk-Return Tradeoff
4) You are considering investing in a project with the following possible outcomes:
Probability of Investment
States
Occurrence
Returns
State 1: Economic boom 15%
16%
State 2: Economic growth 45%
12%
State 3: Economic decline 25%
5%
State 4: Depression
15%
-5%
Calculate the expected rate of return for this investment.
A) 9.8%
B) 7.0%
C) 8.3%
D) 6.3%
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
5) Spartan Sofas, Inc. is selling for \$50.00 per share today. In one year, Spartan will be selling
for \$48.00 per share, and the dividend for the year will be \$3.00. What is the cash return on
Spartan stock?
A) 0%
B) 2%
C) 6%
D) 10%
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
6) What is the standard deviation of an investment that has the following expected scenario? 18%
probability of a recession, 2.0% return; 65% probability of a moderate economy, 9.5% return;
17% probability of a strong economy, 14.2% return.
A) 3.68%
B) 1.23%
C) 8.47%
D) 6.66%
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: standard deviation
Principles: Principle 2: There Is a Risk-Return Tradeoff
2
7) You are considering investing in a firm that has the following possible outcomes:
Economic boom: probability of 25%; return of 25%
Economic growth: probability of 60%; return of 15%
Economic decline: probability of 15%; return of -5%
What is the expected rate of return on the investment?
A) 15.0%
B) 11.7%
C) 14.5%
D) 25.0%
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
8) Which of the following best measures the risk of holding an asset in isolation (i.e., stand-alone
risk)?
A) The mean co-variance
B) The standard deviation
C) The coefficient of optimization
D) The standard asset pricing model
E) The correlation
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: standard deviation
Principles: Principle 2: There Is a Risk-Return Tradeoff
9) The holding period return is always positive.
Diff: 1
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
10) Because returns are more certain for the least risky investments, the required return on these
investments should be higher than the required returns on more risky investments.
Diff: 1
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
3
11) Even though an investor expects a positive rate of return, it is possible that the actual return
will be negative.
Diff: 1
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
12) The expected rate of return is the weighted average of the possible returns for an investment.
Diff: 1
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
13) The expected rate of return is the sum of each possible return times it likelihood of
occurrence.
Diff: 1
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
14) The higher the standard deviation, the less risk the investment has.
Diff: 1
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: standard deviation
Principles: Principle 2: There Is a Risk-Return Tradeoff
15) Using the following information for McDonovan, Inc.'s stock, calculate their expected return
and standard deviation.
State
Probability
Return
Boom
20%
40%
Normal
60%
15%
Recession
20%
(20%)
Answer: Ki = (Ki)(Pi) = (.20)(40%) + (.60)(15%) + (.20)(-20%)
= 8% + 9% - 4% = 13%
2
i = ((Ki K) Pi).5
i = ((40%-13%)2(.2) + (15%-13%)2 (.6) + (-20%-13%)2 (.2)).5 = 19.13%
Diff: 2
Topic: 7.1 Realized and Expected Rates of Return and Risk
Keywords: standard deviation
Principles: Principle 2: There Is a Risk-Return Tradeoff
4
## 7.2 A Brief History of Financial Market Returns
1) The risk-return tradeoff tells us that expected returns should be higher on investments that
have higher risk.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: risk
Principles: Principle 2: There Is a Risk-Return Tradeoff
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: risk
Principles: Principle 2: There Is a Risk-Return Tradeoff
3) Less risky investments have lower standard deviations than do more risky investments.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: risk, return
Principles: Principle 2: There Is a Risk-Return Tradeoff
4) Investments in emerging markets have higher volatility than do U.S. Stocks.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: standard deviation
Principles: Principle 2: There Is a Risk-Return Tradeoff
5) Expected return and realized return are the same thing.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: holding period return
Principles: Principle 2: There Is a Risk-Return Tradeoff
6) Historically, in the United States stocks have had higher returns and greater volatility than
have government bonds.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: risk, return
Principles: Principle 2: There Is a Risk-Return Tradeoff
5
## 7) Treasury Bills have less default risk than do Government Bonds.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: risk, return
Principles: Principle 2: There Is a Risk-Return Tradeoff
8) Investors are always rewarded for taking higher risk with higher realized returns.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: risk, return
Principles: Principle 2: There Is a Risk-Return Tradeoff
9) Investors make different investment choices partially because individuals do not all have the
same tolerance for risk.
Diff: 1
Topic: 7.2 A Brief History of Financial Market Returns
Keywords: investor tolerance
Principles: Principle 2: There Is a Risk-Return Tradeoff
7.3 Geometric vs. Arithmetic Average Rates of Return
1) Marcus Berger invested \$9842.33 in Hawkeyehats, Inc. four years ago. He sold the stock
today for \$11,396.22. What is his geometric average return?
A) There is insufficient information to derive an answer.
B) 2.98%
C) 3.73%
D) 3.95%
Diff: 1
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: holding period return
Principles: Principle 1: Money Has a Time Value
2) Marcus Berger invested \$9842.33 in Hawkeyehats, Inc. four years ago. He sold the stock
today for \$11,396.22. What is his arithmetic average return?
A) There is insufficient information to derive an answer.
B) 2.98%
C) 3.73%
D) 3.95%
Diff: 1
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: arithmetic average return
Principles: Principle 1: Money Has a Time Value
6
## Use the following to answer the following question(s).
Roddy Richards invested \$12014.88 in Wolverine Meat Distributors (W.M.D.) five years ago.
The investment had yearly arithmetic returns of -9.7%, -8.1%, 15%, 7.2%, and 15.4%.
3) What is the arithmetic average return of Roddy Richard's investment?
A) 2.42%
B) 3.96%
C) 5.18%
D) 15.1%
Diff: 2
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: arithmetic average return
Principles: Principle 1: Money Has a Time Value
4) What is the geometric average return of Roddy's Richard's investment?
A) 3.38%
B) 4.63%
C) 6.96%
D) 8.78%
Diff: 2
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: compound interest
Principles: Principle 1: Money Has a Time Value
5) How much money did Roddy Richards receive when he sold his shares of W.M.D.?
A) \$12,014.88
B) \$12,398.42
C) \$13,663.47
D) \$14,184.73
Diff: 2
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: compound interest
Principles: Principle 1: Money Has a Time Value
7
## Use the following information to answer the following question(s).
Susan Bright will get returns of 18%, -20.3%, -14%, 17.6%, and 8.3% in the next five years on
her investment in CoffeeTown, Inc. stock, which she purchases for \$73,419.66 today.
6) What is the arithmetic average return on her stock if she sells it five years from today?
A) 1.92%
B) 3.98%
C) 6.47%
D) 7.11%
Diff: 2
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: arithmetic average return
Principles: Principle 1: Money Has a Time Value
7) What is the geometric average return on her stock if she sells it five years from today?
A) -2.33%
B) .59%
C) 3.67%
D) 4.88%
Diff: 2
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: geometric average return
Principles: Principle 1: Money Has a Time Value
8) How much will Susan's stock be worth if she sells it five years from today?
A) \$71,423.85
B) \$73,419.66
C) \$75,628.75
D) \$80,333.40
Diff: 2
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: holding period return
Principles: Principle 1: Money Has a Time Value
9) Arithmetic average rate of return takes compounding into effect.
Diff: 1
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: compound interest
Principles: Principle 1: Money Has a Time Value
8
10) An investor who wishes to hold a stock for five years will be most interested in geometric
average rather than in the arithmetic average return.
Diff: 1
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: compound interest
Principles: Principle 1: Money Has a Time Value
11) If an investor holds a stock for six years, the value at the end of six years will be the initial
cost times (1 + geometric average return)to the sixth power.
Diff: 1
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: compound interest
Principles: Principle 1: Money Has a Time Value
12) If an investor holds a stock for three years, the value at the end of three years will always be
the initial cost of the stock times (1 + arithmetic average return) to the third power.
Diff: 1
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: compound interest
Principles: Principle 1: Money Has a Time Value
13) Why do the arithmetic average return and the geometric return differ?
Answer: The arithmetic average return does not take what the value of the investment was at the
start of each period. Hence, even though a company may have the same arithmetic return for two
consecutive years, the dollar amount of those returns will be different in later years than in the
first year. For instance, if the investor started with \$1,000, and earned 20% the first year, lost
20% the second year, and earned 15% the third year, the average arithmetic return would be 5%,
and the 20% gain the first year would be \$200, but the 20% loss the second year would be \$240.
The investment would be worth \$1104 after three years, giving an average geometric return of
3.35%, different from the average arithmetic return.
Diff: 2
Topic: 7.3 Geometric vs. Arithmetic Average Rates of Return
Keywords: compound interest
Principles: Principle 1: Money Has a Time Value
9
## 7.4 What Determines Stock Prices?
1) Each of the following would tend to weaken the Efficient Market Hypothesis EXCEPT:
A) There is publicly available information that Boeing Aircraft has procured a contract to build
25 planes for the U.S. Government and the price of Boeing quickly goes up.
B) ACG, Inc. performed well for the past six months, but they just lost a major distribution
contract, but the price of ACG stock continues to go up.
C) Louisville Slugger, Inc., gets a contract to supply bats for Little League play, a contract it
never had before, and stock price remains stable.
D) Disney corporation, a growth company, opens a new theme park, which investors expect will
do tremendously well, and the stock price stays stable, while Urban Electric Company, which has
a set infrastructure, and generates 95% of its earnings from assets it owns, outperforms Disney.
Diff: 1
Topic: 7.4 What Determines Stock Prices?
Keywords: efficient markets
Principles: Principle 4: Market Prices Reflect Information
2) Stock prices go up when there is positive information about a company, and go down when
there is negative information about the company.
Diff: 1
Topic: 7.4 What Determines Stock Prices?
Keywords: efficient markets
Principles: Principle 4: Market Prices Reflect Information
3) An investor with access to all publicly available information will be able to make higher than
expected profit if the market has semi-strong efficiency.
Diff: 1
Topic: 7.4 What Determines Stock Prices?
Keywords: efficient markets
Principles: Principle 4: Market Prices Reflect Information
4) If a market has weak form efficiency, an investor can make higher than expected profits by
studying the past price patterns of a stock.
Diff: 1
Topic: 7.4 What Determines Stock Prices?
Keywords: efficient markets
Principles: Principle 4: Market Prices Reflect Information
10
5) If an individual with inside information can make higher than expected profits, the market is
no more than semi-strong form efficient.
Diff: 1
Topic: 7.4 What Determines Stock Prices?
Keywords: efficient markets
Principles: Principle 4: Market Prices Reflect Information
6) Under the efficient market hypothesis, would securities be properly priced.
Answer: If markets were perfectly efficient, then investors would price a stock based on the
company's expected future cash flows, so at any time the security would be properly priced. If
good news becomes available, that would tend to increase the expected cash flows to a company,
the stock price will go up, meaning that the new price is then the proper price for the stock.
Diff: 2
Topic: 7.4 What Determines Stock Prices?
Keywords: efficient markets
Principles: Principle 4: Market Prices Reflect Information
7) Are markets moving toward being more efficient or toward being less efficient?
Answer: Empirical evidence shows that since about the year 2000 pricing anomalies have
diminished considerably. Hedge funds have been trying to exploit pricing inefficiencies, and by
doing so, eliminate the inefficiencies. Hence, the market appears to be becoming more efficient
over time.
Diff: 2
Topic: 7.4 What Determines Stock Prices?
Keywords: efficient markets
Principles: Principle 4: Market Prices Reflect Information
11 | 4,347 | 16,090 | {"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-2019-26 | latest | en | 0.825697 |
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 609 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-04 | latest | en | 0.86807 |
https://radiantlondon.com/how-the-lottery-works/ | 1,726,642,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00379.warc.gz | 441,231,386 | 9,740 | # How the Lottery Works
A lottery is a form of gambling where you purchase a ticket for a chance to win a prize. The prizes are usually cash or goods. In the United States, lotteries raise billions of dollars each year. Some people play the lottery just for fun, while others believe that it is their only chance of a better life. Whatever the reason, it is important to understand how the lottery works before you spend your hard-earned money. Here are some tips that will help you make an informed decision about whether or not to play.
Many people think that the odds of winning are the same for every drawing. However, that is not true. In reality, the odds of winning are always different depending on how many tickets are sold for a particular drawing. Using this information, you can predict the odds of winning a particular drawing. This can be done by looking at a history of past results. You can also use probability theory to calculate the odds of winning. The odds of winning are a mathematical concept that is based on the Law of Large Numbers. This means that the more tickets are sold, the higher the chances of someone winning.
Lotteries have been around for centuries. The Old Testament contains a verse instructing Moses to take a census of Israel and divide the land by lot. Lotteries were also used by Roman emperors as a way to give away property and slaves. In modern times, the lottery is a popular way to raise money for a variety of purposes, including education and public works projects. In the United States, lottery games are legal in most states and are regulated by state law.
The first state-sponsored lotteries were held in the 15th century in the Low Countries, where they were a common method of raising funds for local needs. Town records from Ghent, Utrecht, and Bruges refer to public lotteries as early as 1445.
In the United States, lotteries began to be widely popular in the 1800s. They became especially popular in the northeastern United States, where they helped fund several American colleges. In addition, the lottery was a painless way for colonists to pay taxes to Britain.
While it is possible to win a huge sum of money in a lottery, there are some risks involved. Generally, you should only gamble with money that you can afford to lose. This will prevent you from going into debt. It is also important to treat the lottery as entertainment and not a career choice. You should budget for your lottery entertainment like you would a trip to the movies.
The lottery is a great way to raise money for public school districts. Each county receives a share of the Lottery’s funding based on its Average Daily Attendance or full-time enrollment for community college and higher education institutions. You can view the latest Lottery contributions to a particular county by clicking on a map or entering a county name in the search box. | 593 | 2,892 | {"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-2024-38 | latest | en | 0.981794 |
https://proofwiki.org/wiki/Real_Number_between_Zero_and_One_is_Greater_than_Power/Natural_Number/Proof_1 | 1,680,163,656,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00534.warc.gz | 544,859,549 | 11,526 | # Real Number between Zero and One is Greater than Power/Natural Number/Proof 1
## Theorem
Let $x \in \R$.
Let $0 < x < 1$.
Let $n$ be a natural number.
Then:
$0 < x^n \le x$
## Proof
For all $n \in \N$, let $\map P n$ be the proposition:
$0 < x < 1 \implies 0 < x^n \le x$
### Basis for the Induction
$\map P 1$ is true, since $0 < x < 1 \implies 0 < x^1 \le x$ by definition of exponent of $1$.
This is our basis for the induction.
### Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
$0 < x < 1 \implies 0 < x^k \le x$
Then we need to show:
$0 < x < 1 \implies 0 < x^{k + 1} \le x$
### Induction Step
This is our induction step:
$\ds 0 < x < 1$ $\leadsto$ $\ds 0 < x^k \le x$ Induction Hypothesis $\ds$ $\leadsto$ $\ds 0 < x^{k + 1} \le x \cdot x$ Real Number Ordering is Compatible with Multiplication $\ds$ $\leadsto$ $\ds 0 < x^{k + 1} \le x$ Multiple of Positive Real Number with Number Less Than One is Less Than Real Number
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
$\forall n \in \N: 0 < x < 1 \implies 0 < x^n \le x$
Hence the result.
$\blacksquare$ | 446 | 1,293 | {"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} | 4.65625 | 5 | CC-MAIN-2023-14 | latest | en | 0.683152 |
https://slideplayer.com/slide/3869608/ | 1,575,750,290,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540501887.27/warc/CC-MAIN-20191207183439-20191207211439-00273.warc.gz | 542,555,862 | 26,884 | # Clustering Beyond K-means
## Presentation on theme: "Clustering Beyond K-means"— Presentation transcript:
Clustering Beyond K-means
David Kauchak CS 451 – Fall 2013
Administrative Final project Final exam next week
Presentations on Friday 3 minute max 1-2 PowerPoint slides. me by 9am on Friday What problem you tackled and results Paper and final code submitted on Sunday Final exam next week
Assign/cluster each example to closest center Recalculate centers as the mean of the points in a cluster
Problems with K-means Determining K is challenging Spherical assumption about the data (distance to cluster center) Hard clustering isn’t always right Greedy approach
Problems with K-means What would K-means give us here?
Assumes spherical clusters
k-means assumes spherical clusters!
K-means: another view
K-means: another view
K-means: assign points to nearest center
Iteratively learning a collection of spherical clusters
EM clustering: mixtures of Gaussians
Assume data came from a mixture of Gaussians (elliptical data), assign data to cluster with a certain probability EM k-means
EM clustering Two main differences between K-means and EM clustering:
Very similar at a high-level to K-means Iterate between assigning points and recalculating cluster centers Two main differences between K-means and EM clustering: We assume elliptical clusters (instead of spherical) It is a “soft” clustering algorithm
Soft clustering p(red) = 0.8 p(blue) = 0.2 p(red) = 0.9 p(blue) = 0.1
EM clustering soft assigned points to each cluster
Start with some initial cluster centers Iterate: soft assigned points to each cluster recalculate the cluster centers Calculate: p(θc| x) the probability of each point belonging to each cluster Calculate new cluster parameters, θc maximum likelihood cluster centers given the current soft clustering
Figure from Chris Bishop
Step 1: soft cluster points
Which points belong to which clusters (soft)? Figure from Chris Bishop
Step 1: soft cluster points
Notice it’s a soft (probabilistic) assignment Figure from Chris Bishop
Step 2: recalculate centers
What do the new centers look like? Figure from Chris Bishop
Step 2: recalculate centers
Cluster centers get a weighted contribution from points Figure from Chris Bishop
keep iterating… Figure from Chris Bishop
Model: mixture of Gaussians
How do you define a Gaussian (i.e. ellipse)? In 1-D? In M-D?
Gaussian in 1D parameterized by the mean and the standard deviation/variance
Gaussian in multiple dimensions
Covariance determines the shape of these contours We learn the means of each cluster (i.e. the center) and the covariance matrix (i.e. how spread out it is in any given direction)
Step 1: soft cluster points
soft assigned points to each cluster Calculate: p(θc|x) the probability of each point belonging to each cluster How do we calculate these probabilities?
Step 1: soft cluster points
soft assigned points to each cluster Calculate: p(θc|x) the probability of each point belonging to each cluster Just plug into the Gaussian equation for each cluster! (and normalize to make a probability)
Step 2: recalculate centers
calculate new cluster parameters, θc maximum likelihood cluster centers given the current soft clustering How do calculate the cluster centers?
Fitting a Gaussian What is the “best”-fit Gaussian for this data?
10, 10, 10, 9, 9, 8, 11, 7, 6, … Recall this is the 1-D Gaussian equation:
Fitting a Gaussian What is the “best”-fit Gaussian for this data?
10, 10, 10, 9, 9, 8, 11, 7, 6, … The MLE is just the mean and variance of the data! Recall this is the 1-D Gaussian equation:
Step 2: recalculate centers
maximum likelihood cluster centers given the current soft clustering How do we deal with “soft” data points?
Step 2: recalculate centers
maximum likelihood cluster centers given the current soft clustering Use fractional counts!
E and M steps: creating a better model
EM stands for Expectation Maximization Expectation: Given the current model, figure out the expected probabilities of the data points to each cluster p(θc|x) What is the probability of each point belonging to each cluster? Maximization: Given the probabilistic assignment of all the points, estimate a new model, θc Just like NB maximum likelihood estimation, except we use fractional counts instead of whole counts
Similar to k-means p(θc|x) Iterate:
Assign/cluster each point to closest center Recalculate centers as the mean of the points in a cluster Expectation: Given the current model, figure out the expected probabilities of the points to each cluster p(θc|x) Maximization: Given the probabilistic assignment of all the points, estimate a new model, θc
E and M steps Expectation: Given the current model, figure out the expected probabilities of the data points to each cluster Maximization: Given the probabilistic assignment of all the points, estimate a new model, θc Iterate: each iterations increases the likelihood of the data and guaranteed to converge (though to a local optimum)!
EM EM is a general purpose approach for training a model when you don’t have labels Not just for clustering! K-means is just for clustering One of the most general purpose unsupervised approaches can be hard to get right!
EM is a general framework
Create an initial model, θ’ Arbitrarily, randomly, or with a small set of training examples Use the model θ’ to obtain another model θ such that Σi log Pθ(datai) > Σi log Pθ’(datai) Let θ’ = θ and repeat the above step until reaching a local maximum Guaranteed to find a better model after each iteration i.e. better models data (increased log likelihood) Where else have you seen EM?
EM shows up all over the place
Training HMMs (Baum-Welch algorithm) Learning probabilities for Bayesian networks EM-clustering Learning word alignments for language translation Learning Twitter friend network Genetics Finance Anytime you have a model and unlabeled data!
Other clustering algorithms
K-means and EM-clustering are by far the most popular for clustering However, they can’t handle all clustering tasks What types of clustering problems can’t they handle?
Non-gaussian data What is the problem?
Similar to classification: global decision (linear model) vs. local decision (K-NN) Spectral clustering
Spectral clustering examples
Ng et al On Spectral clustering: analysis and algorithm
Spectral clustering examples
Ng et al On Spectral clustering: analysis and algorithm
Spectral clustering examples
Ng et al On Spectral clustering: analysis and algorithm | 1,446 | 6,525 | {"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-2019-51 | longest | en | 0.846245 |
https://allinonehomeschool.com/using-similar-figures/ | 1,701,179,779,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00520.warc.gz | 115,812,544 | 23,486 | # Using Similar Figures
In earlier lessons, you learned that similar figures are figures that have the same shape — they have congruent corresponding angles and proportional corresponding sides. You learned to set up a proportion to find the missing side lengths. When two polygons are similar, the symbol ~ is used. When you know that polygons are similar, this allows you to find unknown lengths and angle measures. In this lesson, you will learn that the ratio of the lengths of corresponding sides in similar figures is sometimes called their scale factor.
## Investigate
When you compare similar shapes to find their scale factor, it is often easier to put the larger side first in the ratio. Scale factors can be expressed as a ratio or as a percent. Take a look at the following examples.
### Example 1
Look at the triangles. We can see that are corresponding sides. Therefore, the scale factor is:
Remember that when you compare similar triangles to find their scale factor, you sometimes put the larger side first in the ratio, so using the same triangles, you could say that the scale factor is:
The scale factors for the triangles above could also be written as 80% or 125%. This simply means that one triangle is 80% as large as the other, or that one triangle is 125% larger than the other.
The best strategy for getting the order correct is to write when you are not sure how the question is to be answered.
### Example 2
These quadrilaterals are similar. Find their scale factor.
Remember to apply the definition of scale factor and use division.
Find a pair of corresponding side lengths. All lengths are given in DEFG, but only has its length given in KLMN. Use the pair .
Calculate the scale factor:
Solution: The scale factor is 65%
## Investigate
#### If similar figures have scale factor r, then the perimeters of those figures are in ratio r.
Let’s see exactly what this means.
### Example 1
Look at the rectangles to the left.
Apply the definition of the perimeter rule for scale factor for the rectangles.
First, find the scale factor, using one of the pairs of corresponding sides.
. The scale factor is 4.
Next, find the perimeter of the small rectangle.
2(l) + 2(w) = 2(1) + 2(3) = 2 + 6 = 8
Next, multiply the perimeter by the scale factor.
8 * 4 = 32
The perimeter of the large rectangle is 32.
You can check this strategy by finding the length of each side and getting the perimeter of each rectangle. It works every time!
### Example 2
Let’s try this with a triangle.
A right triangle has sides of 3 cm, 4 cm, and 5 cm. The perimeter of this triangle would be 3 + 4 + 5 = 12. Can you find the perimeter of a triangle with a scale factor of 150%?
Apply the rule for finding perimeter of similar figures. If the perimeter of the original triangle is 12 and the scale factor is 150%, just multiply 12 by 1.5.
12 * 1.5 = 18. The perimeter of the new triangle would be 18. If you were asked to find the sides of the new triangle:
Start with the original triangle side lengths and multiply each one by 150%.
3 * 1.5 = 4.5
4 * 1.5 = 6
5 * 1.5 = 7.5
The side lengths of the new triangle would be 4.5 cm, 6 cm, and 7.5 cm. If you add these sides together, the perimeter would be 18 cm.
## Investigate
### Example 1
Look at the rectangles to the left.
Use the corresponding sides to determine the scale factor:
. The scale factor is 4.
Next, find the area of the small rectangle.
Next, square the scale factor and multiply the area of the small rectangle by .
3 * 16 = 48
The area of the large rectangle is .
You can check this strategy by finding the length of each side of the rectangle and getting the area – It works every time!
### Example 2
A triangle has area of 20 sq. ft. A similar triangle is constructed using a 10% scale factor. What is the area of the new triangle?
1. Apply the area rule.
2. The triangles have scale factor r = 10% or 0.1
3. Square the scale factor: 0.1 * 0.1 = 0.01
4. Multiply the scale factor by the area of the original triangle.
5. 20 * 0.01 = 0.2 sq. ft.
6. The new triangle has an area of 0.2 square feet.
### Example 3
A rectangle has an area of 30 feet. A similar rectangle has an area of 270 feet. What is the scale factor?
1. First, divide 270 by 30 = 9.
2. Then, find the square root of that number. .
3. Therefore, the scale factor of these two rectangles is 3.
Let’s quickly review these strategies and rules:
1. To get the scale factor, simply write the ratio of the two figures as New over Original. (unless otherwise specified).
2. To get the perimeter ratio, multiply the perimeter by the scale factor.
3. To get the area ratio, multiply the area by the scale factor squared (for 2 dimensional figures).
4. To find the scale factor from the area, work backwards – find the area ratio and UNsquare it (find its square root).
## Similar Figures Practice
Quadrilateral ABCD ~ JKLM. Complete each of the following:
1. The scale factor (%) of JKLM to ABCD is _____
2. BC = ______
3. JM = ______
4. JK = ______
5. The ratio of the perimeter of JKLM to the perimeter of ABCD = ______
6. The ratio of the area of JKLM to the area of ABCD = ________
(source) | 1,297 | 5,192 | {"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-2023-50 | latest | en | 0.931104 |
http://logic-1.butterfill.com/units/unit_02.html | 1,723,105,836,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00364.warc.gz | 12,271,767 | 5,319 | Press the right key for the next slide (or swipe left)
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## Quick Intro to awFOL
John is square
Square( a )
John is to the left of Ayesha
LeftOf( a , b )
John is square or Ayesha is trinagulra
Square( a ) Trinagulra( b )
name (refers to an object)
predicate (refers to a property)
connective (joins sentences)
sentence (can be true or false)
atomic sentence (no connectives)
non-atomic sentence (contains connectives)
Our approach to studying logic will involve a formal language called awFOL. FOL' stands for first order language, and I call this particular first-order language awFOL` because, like nearly all first-order languages used in textbooks, it’s awful. (Where are the binary quantifiers? Why are brackets used with two completely different meanings? ...)
The language of the textbook is called ‘FOL’. ‘awFOL’ is basically the same as FOL except that you can replace symbols with words which makes typing it easier. Also 'FOL' is a really stupid name because there are lots of first-order languages. It's a bit like I ask you what language you speak and instead of saying 'Farsi' or 'English' or 'Cantonese' you say 'Language, I speak Language'. But this is trivial, it doesn't really matter what you call things. Let's move on.
As I was saying, for the purposes of logic we are going to use a formal language. In order to get a sense for this language, let's compare it to English. Take a look at this sentence, John is square.
This is a sentence.
For now a sentence is just something capable of being true or false. (In a longer course we would define what it is to be a sentence more carefully.)
In English there are names ...
... these are terms that function to refer to objects.
There are also predicates, like 'Square'.
Predicates are things that refer to properties. In this case the property is that of being square.
Take a look at this sentence.
Some properties relate several things; for example, being 'to the left of' involves two things rather than one. The expressions for these relational properties are also called predictaes.
By the way, this is also a sentence containing multiple names, 'John' and 'Ayesha'.
Now have a look at this sentence, 'John is square or Ayehsa is triangular' ... or, as I perfer to say, 'trinagulra'. (Did you spot the mistake? Well done.)
Consider the word 'or' in this sentence. It isn't a name or a predicate. It doesn't refer to an object, nor to a property.
Instead its function is to join two sentences, making a new one. We'll call things like this 'connectives'. A connective is anything that you can combine with zero or more sentences and to make a new sentence.
Here's another piece of terminology: a sentence with one or more connectives is 'non-atomic'
And, as you'd expect, a sentence with no connectives is 'atomic';
Now let's see how these sentences look in our formal language, awFOL.
Here's how the equivalent of 'John is square' looks in awFOL.
The whole thing is a sentence of awFOL.
The letter 'a' is a name; just like the English name 'John', the function of 'a' is to refer to an object (in this case, John)
And 'Square( )' is the predicate.
What about 'John is to the left of Ayesha', how can we say something like this in awFOL?
Here's the equivalent of 'John is to the left of Ayesha in awFOL'
Again, the single letters a and b are names.
And 'LeftOf( )' is the predicate. Note that, as in English, the order of the names matters. It affects who we are saying is to the left of who.
Lastly, what is the equivalent of the third sentence in awFOL?
Much as you would expect.
This is a non-atomic sentence (because it contains a connective).
Note that where the English 'or' appears, we use a special symbol. This symbol doesn't do exactly what the English 'or' does, as we'll see later.
Alles klar? Molto bene.
You might be thinking that this English sentence looks, well, ...
... a lot like this awFOL sentence. What's the point of learning a formal language? How will it help us to understand logic?
(It's a bit tricky to answer this question as I haven't yet said what logic is.)
A formal langauge enables us to avoid ambiguity, e.g.:
We need a formal language because ambiguity is awkward to deal with theoretically
\begin{quote}
This is a hospital where doctors are trained.
\end{quote}
A formal langauge also enables us to some avoid appearance--reality problems:
Appearance and reality. We need a formal language because we want a guarantee that a sentence which seems to express a proposition really does express a proposition.
\begin{quote}
Many more people have been to Paris than I have.
\end{quote}
Finally, consider these sentences.
Ayesha doesn’t know diddly squat about logic
Ayesha does know diddly squat about logic
The only difference is an extra negation in the first sentence. Normally you might think that adding a negation changes the meaning, and does so systematically. But this is not true of natural languages like English. We can construct our formal language so that it is true, thereby making our lives simpler insofar as we are interested in reflecting on inferential relations among sentences.
1.1--1.5
*1.6
1.8--1.10 | 1,270 | 5,343 | {"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} | 3.328125 | 3 | CC-MAIN-2024-33 | latest | en | 0.94586 |
https://www.javatpoint.com/computer-graphics-3d-transformations | 1,679,374,271,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00373.warc.gz | 894,243,751 | 11,705 | # Three Dimensional Transformations
The geometric transformations play a vital role in generating images of three Dimensional objects with the help of these transformations. The location of objects relative to others can be easily expressed. Sometimes viewpoint changes rapidly, or sometimes objects move in relation to each other. For this number of transformation can be carried out repeatedly.
## Translation
It is the movement of an object from one position to another position. Translation is done using translation vectors. There are three vectors in 3D instead of two. These vectors are in x, y, and z directions. Translation in the x-direction is represented using Tx. The translation is y-direction is represented using Ty. The translation in the z- direction is represented using Tz.
If P is a point having co-ordinates in three directions (x, y, z) is translated, then after translation its coordinates will be (x1 y1 z1) after translation. Tx Ty Tz are translation vectors in x, y, and z directions respectively.
x1=x+ Tx
y1=y+Ty
z1=z+ Tz
Three-dimensional transformations are performed by transforming each vertex of the object. If an object has five corners, then the translation will be accomplished by translating all five points to new locations. Following figure 1 shows the translation of point figure 2 shows the translation of the cube.
## Matrix representation of point translation
Point shown in fig is (x, y, z). It become (x1,y1,z1) after translation. Tx Ty Tz are translation vector.
Example: A point has coordinates in the x, y, z direction i.e., (5, 6, 7). The translation is done in the x-direction by 3 coordinate and y direction. Three coordinates and in the z- direction by two coordinates. Shift the object. Find coordinates of the new position.
Solution: Co-ordinate of the point are (5, 6, 7)
Translation vector in x direction = 3
Translation vector in y direction = 3
Translation vector in z direction = 2
Translation matrix is
Multiply co-ordinates of point with translation matrix
= [5+0+0+30+6+0+30+0+7+20+0+0+1] = [8991]
x becomes x1=8
y becomes y1=9
z becomes z1=9
Next Topic3D Scaling | 502 | 2,140 | {"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} | 4.71875 | 5 | CC-MAIN-2023-14 | latest | en | 0.900783 |
https://www.physicsforums.com/threads/motion-problem.193434/ | 1,531,857,196,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589892.87/warc/CC-MAIN-20180717183929-20180717203929-00047.warc.gz | 944,614,822 | 16,627 | # Homework Help: Motion problem
1. Oct 23, 2007
### motoxtremechick
A pedestrian is running at his maximum speed of 6.0 m/s to catch a bus stopped by a traffic light. When he is 25 meters from the bus, the light changes and the bus accelerates uniformly at 1.0 m/s2. Find either {A} how far he has to run to catch the bus or {B} his frustration distance (closest approach).
I have used d=Vf^2-Vi^2/2a and gotten 18 meters but im not sure if im correct..
2. Oct 23, 2007
### pooface
I dont think this is correct. You can see that the pedestrian has to run more than 25 meters to catch up to the bus so 18 cannot be the answer.
I suggest you seperately write down your data for the pedestrian and the bus.
3. Oct 23, 2007
### motoxtremechick
ok so
BUS
V=1.0m/s^2
PED
a=6.0m/s
d=25m
could i possibly use d=Vave(t) to find time and use 25 meters as distance or is that wrong?..
4. Oct 23, 2007
### pooface
Bus has an acceleration of 1m/s^2, not velocity. It also has an initial velocity of 0 since it was at rest.
Pedestrian has a constant velocity of 6m/s not acceleration. The acceleration is 0 because V is constant. And if V is constant then Vo and Vf are the same.
you do not know the displacement, you only know that when the bus begins to acceleration the pedestrian is 25 metres behind. So it is a catch up game.
5. Oct 23, 2007
### motoxtremechick
right right my bad on the A and V mess ups sorry.
and then would i use the d=Vf^2-Vi^2/2a to find a of the pedestrian?
putting 25m into d?
6. Oct 23, 2007
### pooface
Well you can see if you use that formula you will get d=0/2a. Not what you want. Secondly 'a' of the pedestrian is 0 because he is running at a constant velocity.
you dont have displacement, so you cant use 25m.
displacement of the pedestrian is
d=d+25
and displacement of the bus is d=d.
This is because the pedestrian is 25 meters behind the bus. If the bus displaced 'd' metres then the pedestrian must displace 'd +25m' at their meeting point if there is one.
Think about equating certain equations to find the time they will be at the SAME distance(d).
7. Oct 23, 2007
### motoxtremechick
o alright.
thank you
im sorry im just way lost and i guess not thinking straight
the only equations i have are the kinematics equations and i cant think of which one to apply that uses what i have/dont have
8. Oct 23, 2007
### pooface
Analyze each equation, try to equate the d's. So effectively you will have to move the 25 over. Don't be afraid! I am here to help.
PED
d=d+25
Vo=6
Vf=6
a=0
t=?
BUS
d=d
Vo=0
Vf=?
a=1
t=?
More than enough information here
9. Oct 23, 2007
### motoxtremechick
Thank you for bearing with me [=!!
so just to get to the 25 meter mark from 0 the pedestrian would have to run 150 seconds....
wait
so i did d+25=Vf^2-Vi^2/2a?
so d+25=18m
d=-7 so he is 7m away from the bus? so 25 plus the extra 7 the bus travles so he is 32m away from the bus?
10. Oct 23, 2007
### pooface
The pedestrian is running at 6 metres/second. To get to 25 metres, it will take him about 4.something seconds right? But during those 4.something seconds the bus is travelling forwards.
Try a different equation. You will need two equations for 'd'. One for the pedestrian and one for the bus.
Since 'd' is the same, you can equation the two equations and solve for something else.
example:
Bus: d=vot+0.5at^2
Ped: d+25=vot+0.5at^2
We dont have d, but we know they have to be the same.
vot+0.5at^2 = vot+0.5at^2 - 25
effectively we have eliminated d, and can solve for t.
Go back to the starting of 'example:' and see if you can simplify those equations with the data known.
11. Oct 23, 2007
### motoxtremechick
ooo rigght..geez
hmm well i am very horrible at this
i dont know how to get t out of the equation effectively to solve for it..
would i just divide both sides by t^2 then...no see i am just digging a deeper hole with this problem
12. Oct 23, 2007
### pooface
look back at the known data from post 8.
13. Oct 23, 2007
### motoxtremechick
well
0m/s(t)+0.5(1.0m/s^2)(t^2)=6m/s(t)+0.5(0m/s^2)(t^2)-25
then by taking out the ones that would equal 0 out i reduced it to..
.5m/s^2(t^2)=6m/s(t)-25
hmm
14. Oct 23, 2007
### pooface
cool looks like you have a quadratic!
What is t(the time the boy would catch up to the bus)?
15. Oct 23, 2007
### motoxtremechick
i got t=-.064 and t=.3054
?
16. Oct 23, 2007
### pooface
Equation is:
0.5t^2-6t+25 = 0
How did you get those values? Did you use the quadratic formula?
17. Oct 23, 2007
### motoxtremechick
wait i think i switched my values
so i got
3.27 and -15.27
18. Oct 23, 2007
### pooface
If I plug these values into the equation for t, I dont get zero.
19. Oct 23, 2007
### motoxtremechick
so
a=.5
b=-6
c=25
you cant have a - square root right
wouldnt those values give you a negative under the root sign?
20. Oct 23, 2007
### pooface
Yes they would! So what happened here?
21. Oct 23, 2007
### motoxtremechick
it couldnt have happened?
22. Oct 23, 2007
### pooface
Exactly. This reveals that the pedestrian never caught up to the bus. Which leads to part B of the problem, what is the closest he gets? I guess that was the subliminal hint that he wouldn't make it.
23. Oct 23, 2007
### motoxtremechick
haha oooo man ok wow
so now to figure out the closest..?? it wouldnt be the 25 meters would it?
24. Oct 23, 2007
### pooface
I can tell you the bus driver makes a narrow escape.
hehehe. | 1,710 | 5,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-30 | latest | en | 0.945262 |
https://www.chemzipper.com/2023/08/calculate-average-atomic-mass-of.html | 1,721,756,715,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00849.warc.gz | 591,200,378 | 18,765 | Welcome to Chem Zipper.com......: Calculate the average atomic mass of silicon if relative abundance is 92.23 ,isotope Si^28, 4.77 % isotope = Si^29 and 3% isotope = Si^39 .
## Tuesday, August 29, 2023
### Calculate the average atomic mass of silicon if relative abundance is 92.23 ,isotope Si^28, 4.77 % isotope = Si^29 and 3% isotope = Si^39 .
Answer Key: The average atomic mass of silicon is 28.1 amu | 126 | 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} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.613819 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/AG01/%233.12.trs.Thm17:POLO_FILTER:NO.html.lzma | 1,718,514,514,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00744.warc.gz | 81,972,469 | 2,185 | Term Rewriting System R:
[y, n, x]
app(nil, y) -> y
reverse(nil) -> nil
shuffle(nil) -> nil
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
APP(add(n, x), y) -> APP(x, y)
Furthermore, R contains three SCCs.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Argument Filtering and Ordering`
` →DP Problem 2`
` ↳AFS`
` →DP Problem 3`
` ↳AFS`
Dependency Pair:
APP(add(n, x), y) -> APP(x, y)
Rules:
app(nil, y) -> y
reverse(nil) -> nil
shuffle(nil) -> nil
The following dependency pair can be strictly oriented:
APP(add(n, x), y) -> APP(x, y)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(APP(x1, x2)) = x1 + x2 POL(add(x1, x2)) = 1 + x1 + x2
resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> APP(x1, x2)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 4`
` ↳Dependency Graph`
` →DP Problem 2`
` ↳AFS`
` →DP Problem 3`
` ↳AFS`
Dependency Pair:
Rules:
app(nil, y) -> y
reverse(nil) -> nil
shuffle(nil) -> nil
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳Argument Filtering and Ordering`
` →DP Problem 3`
` ↳AFS`
Dependency Pair:
Rules:
app(nil, y) -> y
reverse(nil) -> nil
shuffle(nil) -> nil
The following dependency pair can be strictly oriented:
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(REVERSE(x1)) = x1 POL(add(x1, x2)) = 1 + x1 + x2
resulting in one new DP problem.
Used Argument Filtering System:
REVERSE(x1) -> REVERSE(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳AFS`
` →DP Problem 5`
` ↳Dependency Graph`
` →DP Problem 3`
` ↳AFS`
Dependency Pair:
Rules:
app(nil, y) -> y
reverse(nil) -> nil
shuffle(nil) -> nil
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳AFS`
` →DP Problem 3`
` ↳Argument Filtering and Ordering`
Dependency Pair:
Rules:
app(nil, y) -> y
reverse(nil) -> nil
shuffle(nil) -> nil
The following dependency pair can be strictly oriented:
The following usable rules using the Ce-refinement can be oriented:
reverse(nil) -> nil
app(nil, y) -> y
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(reverse(x1)) = x1 POL(SHUFFLE(x1)) = 1 + x1 POL(nil) = 0 POL(app(x1, x2)) = x1 + x2 POL(add(x1, x2)) = 1 + x1 + x2
resulting in one new DP problem.
Used Argument Filtering System:
SHUFFLE(x1) -> SHUFFLE(x1)
reverse(x1) -> reverse(x1)
app(x1, x2) -> app(x1, x2)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳AFS`
` →DP Problem 3`
` ↳AFS`
` →DP Problem 6`
` ↳Dependency Graph`
Dependency Pair:
Rules:
app(nil, y) -> y
reverse(nil) -> nil | 1,010 | 3,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.521621 |
http://clay6.com/qa/125958/a-particle-of-mass-0-3-kg-is-subjected-to-a-force-f-kx-with-k-15-n-m-what-w | 1,601,175,176,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400250241.72/warc/CC-MAIN-20200927023329-20200927053329-00015.warc.gz | 28,562,278 | 6,104 | # A particle of mass $0.3 \;kg$ is subjected to a force $F = -kx$ with $k = 15\; N/m$. What will be its initial acceleration, if it is released from a point 20 cm away from the origin ?
( A ) $10\;m/s^2$
( B ) $5\;m/s^2$
( C ) $15\;m/s^2$
( D ) $3\;m/s^2$ | 107 | 255 | {"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} | 2.6875 | 3 | CC-MAIN-2020-40 | latest | en | 0.854438 |
https://forum.lazarus.freepascal.org/index.php/topic,65991.0.html?PHPSESSID=91mi14rvcdoenoo1fret5s8e66 | 1,716,694,406,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00087.warc.gz | 218,998,704 | 11,744 | ### Bookstore
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### Recent
#### abk.964
• Newbie
• Posts: 4
« on: January 25, 2024, 11:04:23 am »
I would like to add IFERROR formula. Al least with some limits.
Something like this:
Code: Pascal [Select][+][-]
1. procedure fpsIFERROR(var Result: TsExpressionResult; const Args: TsExprParameterArray);
2. // IFERROR( value; value_if_error )
3. // If "value" is an error value (#N/A, #VALUE!, #REF!, #DIV/0!, #NUM!, #NAME?
4. // or #NULL), this function will return Args[1]. Otherwise, it will return Args[0].
5. var
6. cell: PCell;
7. begin
8. if (Args[0].ResultType = rtCell) then
9. begin
10. cell := ArgToCell(Args[0]);
11. if (cell <> nil) and (cell^.ContentType = cctError) and (cell^.ErrorValue <= errArgError)
12. then Result := Args[1]
13. else result := Args[0];
14. end else
15. if (Args[0].ResultType = rtError) then
16. Result := Args[1]
17. else
18. Result := Args[0];
19. end;
20.
Code: Pascal [Select][+][-]
1. AddFunction(cat, 'IFERROR', 'S', '?S', INT_EXCEL_SHEET_FUNC_IFERROR, @fpsIFERROR);
2.
Const:
Code: Pascal [Select][+][-]
1. INT_EXCEL_SHEET_FUNC_IFERROR = 370;
2.
Not sure, that it's right.
Any corrections are welcome.
PS. I use xlsx files.
#### wp
• Hero Member
• Posts: 12013
« Reply #1 on: January 26, 2024, 04:34:00 pm »
Yes it seems to work for xlsx and ods. However, it does not work for xls; the file created cannot be read by Excel without error, maybe because the BIFF format does not support it? It is possible to re-enter the formula in Excel (I have v2016), but when saving it tells that it cannot write the formula and must modify it. Looking at the written xls file with the BIFFExplorer it can be seen that the formula identifier code has been written as value 255, and a "defined name" record has been created with the value "_xlfn.IFERROR". fpspreadsheet does not support defined names, so far. But since Excel does not seem to be able to assign any code to it, it does not matter. It is my impression that 255 is the identifier for "formula unknown". My question is: where did you get the value INT_EXCEL_SHEET_FUNC_IFERROR=370 from?
xlsx and ods files, however, can be written correctly. Since they are more important than xls, I applied your code to the svn version.
#### abk.964
• Newbie
• Posts: 4
« Reply #2 on: January 30, 2024, 08:21:05 am »
I'd got 370 as some unused number, so asking for correction. Let`s set 255.
Is result and parameters type right in AddFunction? ('S' and '?S')
Can range be 1 parameter of function IFERROR?
#### wp
• Hero Member
• Posts: 12013
« Reply #3 on: January 30, 2024, 10:07:30 am »
Is result and parameters type right in AddFunction? ('S' and '?S')
The input parameters certainly are correct since the first parameter is a cell and the second one a message text. Not sure about the result parameter, I tend to set it to '?' because it returns the cell when there is no error.
Can range be 1 parameter of function IFERROR?
You mean whether it is possible to use a range in the first parameter? I think, in Excel you can, but not in fpspreadsheet which does not support array formulas.
• Newbie
• Posts: 4 | 933 | 3,223 | {"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 | 0.709752 |
https://dsp.stackexchange.com/questions/87999/what-would-be-an-example-of-something-digital-which-isnt-electronic/88016 | 1,726,779,546,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00442.warc.gz | 195,637,153 | 56,559 | # What would be an example of something digital which isn't electronic?
The terms digital and electronic are often used interchangeably but I know that it's not correct because something can be digital but not electronic.
Something can be digital in the sense that it's discrete both in time and magnitude.
A picture of a ruler appearing each second is discrete in time because this timing is countable and it's also discrete in value/amplitude/magnitude because it is splitted into countable (in this case, two or more) parts.
A picture like that is likely to be electronic or displayed on an electronic device.
What would be an example of something digital which isn't electronic?
The ruler itself (given all times it appeared in the minds of people)?
• Abacus. Book of tables. Commented May 28, 2023 at 10:44
• Paper punch cards? Commented May 28, 2023 at 19:30
• "Electronic" refers to the medium where the data is stored, not the nature (format) of the data. You can store digital data onto any non-digital or non-electronic medium, e.g. writing down the bit stream of jpeg on paper, pressing a CD/DVD, or printing out a QR code, or punch a program onto paper cards. The data is still digital but the medium is clearly not related to electrons. Commented May 29, 2023 at 5:12
• relay logic is also discrete and not electronic
– Ben
Commented May 29, 2023 at 23:06
• Baseball could be considered digital, the score by inning has a discrete time basis (t-axis or x-axis) , and the values (y-axis) are always integers. I saw an answer with a scoreboard flipper that was written before my comment. Commented Jun 1, 2023 at 17:39
Boolean algebra? That's not electronic, that's pure math and logic. It can just be and commonly is implemented with electronic circuits, but you can just as well make a logical AND gate with mechanical elements, whether they are cogwheels, or valves with flowing water or air pressure.
In Boolean algebra, there are variables that are either true or false. Statements can be constructed to be true or false, such as is it day or night, does it rain, or is it cloudy.
However, as you already might have seen, just like in electronics, there is always a gray area between day and night, where given a bunch of people they will not all agree on the exact moment when day turns into night or vice versa.
Which means, even electronic circuits are not strictly digital in as even digital logic is implemented by using analog components and circuits.
• I never learned enough math to get to the algebra part. Is there an example from daily life? From the sapce we may see without light pollution? Perhaps even the universe itself is "digital" in the sense we can count its age and it has countable parts?
– somo
Commented May 27, 2023 at 19:11
• Yes, an abacus is digital calculator. You can count things with it. For some reason I just used two binary states as an example, but all integers are digital. Your fingers are digits too, so digital. Commented May 27, 2023 at 19:25
• I assume that fingers are "digital" because they are discrete in time (exists as long as the person exists, a countable time) and discrete in value (we can count a number, say, five, of fingers).
– somo
Commented May 27, 2023 at 20:37
• Boolean algebra is actually a topic in philosophy. In Introduction to Logic or similarly titled course. Different symbols, but they got truth tables and the same relationship rules including DeMorgan's Theorem. Commented May 27, 2023 at 23:13
• @somo Most of all, fingers are digital, because the latin word "digitus" (which is the root of the word digital) means finger. Commented May 28, 2023 at 19:54
Counting with my fingers.
Also when I was a kid in the '60s, all of the adding machines and cash registers were digital and mechanical.
• If I understand correctly you say that act of counting is "digital" and not necessarily by the number of fingers but by any number. A shopkeeper counting (countable) merchandise (at least once) is a "digital situation"?
– somo
Commented May 27, 2023 at 20:24
• If they're counting, yes. If they're measuring length or weight, no. Commented May 27, 2023 at 23:16
• Etymologically correct! Commented May 28, 2023 at 3:27
• me: spends month cranking out dissertation, 24 updoots in 1.7 years -- rbj: "cOuNtinG wIth mY fiNGeRs", 25 in 6 days -- also your second-highest Commented Jun 2, 2023 at 5:48
• I don't get it either. Commented Jun 2, 2023 at 5:53
A scoreboard flipper (or whatever it's called) immediately came to my mind:
• See also my comment on the original post question. Commented Jun 1, 2023 at 17:40
The difference engine and a number of other pre-electronic mechanical calculators are probably the closest to what we consider nowadays digital. And they are, well, pre-electronic.
A mechanical calculator, or calculating machine, is a mechanical device used to perform the basic operations of arithmetic automatically, or (historically) a simulation such as an analog computer or a slide rule. Most mechanical calculators were comparable in size to small desktop computers and have been rendered obsolete by the advent of the electronic calculator and the digital computer.
The abacus is the oldest example that many will recognize, but Pascal's calculator is pretty similar in user experience to a modern day basic calculator, while being entirely mechanical, working by gears and springs.
The word digital literally comes from the Latin "digitus" meaning a finger or toe, which eventually entered English with both literal and figurative uses. By the 14th century, the word "digit" was used in English with the meaning of a natural and/or whole number up to ten through analogy (people would commonly count things and do minor arithmetic with their fingers). Quite literally, the original digital processing was done on people's fingers in prehistory, and then later in antiquity by written and mechanical devices that modeled fingers. When people counted with fingers, they regarded each of them as either up or down, which also means that digital data is inherently expressible in (or convertible to) a binary format!
The metaphor, then, is that "digital" data is data that can be counted on fingers, or, by extension, expressed exactly in numerals or another format that can be modeled losslessly by a sufficient number of fingers, each of them regarded as held either up or down (and not anywhere in between). That means that "digital" data is more or less synonymous with discrete data.
Examples of non-electronic digital data would thus include:
• Any data displayed by the holding up or down of various fingers
• Any data expressed exactly in numbers, such as 3, 55.2, or -0.99
• Data expressed in any form that has a lossless conversion back and forth to one of the above, such as:
• Data expressed on an abacus (an ancient digital mechanical calculator)
• Tally marks
• True/false or multiple-choice answers on a test
• Printouts of pre-calculated values, such as times tables, log tables, or trig tables
• Printed bus and train schedules (e.g. the R2 bus leaves Maple Park Square at 8:03 AM M-F heading northbound toward University Center East)
• Sheet music
• All forms of manual signboards that express or encode exact numbers, such as the scoreboard flipper mentioned by Steve
• Textual information that can be transcribed exactly into a word processor or text editor
Some of the categories above, of course, may have multiple encodings. For example, textual data may be encoded as ASCII, EBCDIC, or Unicode. The critical requirement is that at least one encoding exists which converts both back and forth without loss.
By contrast, the following data would not normally be considered digital:
• Positions on a ruler
• Positions on a slide rule
• Positions on a mercury thermometer, barometer, speedometer, or other physical analog gauge
• Most forms of graphic art
The above data and their associated devices are considered analog. The mechanical digital calculators mentioned by robert bristow-johnson would contrast with the analog slide rule.
Regarding the music angle, this corresponds nicely with digital versus analog music in the electronic world. A digital representation of music (e.g. an mp3 or a MIDI file) is nothing more than a numerical representation of how to play back the music, in other words, it is an electronic form of sheet music, providing discrete instructions to the computer or player on how to play it back.
Regarding written words and art, the difference is whether it is possible to transcribe exactly. The exact idiosyncrasies of the handwriting are irrelevant to the work. A handwritten novel, for example, can be converted to a text file without loss of any information considered relevant, since the exact form of the handwriting is considered outside of the scope of the novel's content. Calligraphy, however, involves appreciation of the relative positions of the loops, dots, and other forms made with the pen on paper (or other writing medium) and thus a computer transcript does not completely encode it.
Steve Mould on YouTube built a water computer using siphons that does simple binary addition: Water computer (YouTube)
There's a game called Turing Tumble (Wikipedia) that is a mechanical computer using marbles and can supposedly do anything a modern computer can do given a large enough playfield
Finally, there's an educational toy(?) called Spintronics (Product site) (not to be confused with the field in physics) that allows the construction of mechanical analog circuits. However, Steve Mould showed it could be used to create flip-flops and XOR gates, indicating it could be used to create things like a binary adder: Spintronics (YouTube). There very well could be someone that has done something more elaborate, I just don't have time to try and find it right now
A telegraph. It is considered as the first digital device, and was so named in 1837. Digital because a finger, or digit, was used to communicate.
• Telegraph used electric current to transfer the signals, so I wouldn't count it as not electronic. Commented May 29, 2023 at 9:50
• @Ruslan not by the generally accepted conventions for the terms. Electronic devices are devices which don't require converting electric energy into other forms to function, but are based on the behaviour of the electrons themselves, starting in 1904 with the vacuum tube; the electric telegraph was based on electromechanical relays (digital devices converting electric current to the mechanical motion of a switch) not electronics. Commented May 30, 2023 at 11:21
Perhaps the nearest mechanical equivalent to electronic logic would Fluidics. This is basically hydraulic logic, with no moving parts beyond the fluid itself. The Coanda effect is used to make bistable logic.
Although no logic systems of any appreciable size are built with fludics, a few fluidic components often find their way into the business end of many systems that handle fluids.
Fluidics have the potential to operate at very high temperatures, and in very high radiation environments, so have often been touted as a control method for use within reactors and other difficult environments.
digital in the sense that it's discrete both in time and magnitude
A gravity swing pendulum clock divides up time in discrete intervals. Through a set of gears the pendulum clock is able to rotate two sticks around the center of a circle. This results in the magnitude, namely, the time of day, as perceived by us with the help of numbers on the circle's perimeter.
A pneumatic player piano energized by a foot pedal is another digital device. Notes and combinations of notes are represented by holes in a paper scroll at regular or non- regular time intervals. https://en.m.wikipedia.org/wiki/Player_piano
There is an electromechanical computer having no electronics. It was built in 1941. https://en.m.wikipedia.org/wiki/Z3_(computer)#:~:text=The%20Z3%20was%20a%20German,of%20about%205%E2%80%9310%20Hz.
• I think that most authorities would call the pendulum clock an analog device because it uses analog means (pointers moving around a dial) to display the time. Even though its hands actually move in discrete steps, that's more of a defect than a feature. If you could find or build a flip-down clock that was powered by a spring or by weights, and was regulated by a pendulum, then that truly would be a digital clock. Commented May 28, 2023 at 12:59
• Agree that most of the pendulum clock is an analog device. Some have a cog that by design creates the discrete time units I was referring to. Here's a 10 sec video. videos.pond5.com/… Commented May 29, 2023 at 6:35
• Not "some." The "cog" in that video is the escapement wheel, invented in 1657. A wheel like that has been at the heart of practically every pendulum regulated or spring regulated clock that's been manufactured since that date. Commented May 29, 2023 at 13:46
The alphabet. Any of them.
Formalized to the point of being "digital" many centuries before electronics ever existed.
Digital hydraulics. In contrast to classical hydraulic valves, that are proportional ("analog"), digital hydraulics uses a number of on/off valves in parallel to control the flow of hydraulic fluid. It is said to improve efficiency.
• Or, for that matter, digital pneumatics. Some pipe organs open and close wind chests with a pedal that "seems" analog, but actually opens and closes a few switches that send a binary value to a wind chest that uses bellows and levers to convert a binary-weighted value into a shutter angle. Commented May 29, 2023 at 19:38
Similar to mechanical calculators mentioned in another answer, the idea has been taken to the extreme to make entire programmable mechanical computers. See my own question on Retrocomputing.SE. To quote the accepted answer there:
While the term mechanical computer may sound not much, the 170 was an unusual beast and maybe the peak of mechanical computing. In 1955 it for sure outclassed all electronic computers of similar or even lager size.
It features one or more 12 digit (decimal) wide ALU and up to 5 result registers per ALU all connected over a mechanical parallel bus.
Programming of the 170 was done by fitting pins into a plug-board.
A program could have up to 53 steps made up one (or sometimes more) of the 90 instructions.
Digital? Check. Non-electronic? Also check. Absolutely fascinating machines.
To return to the original question, the simplest thing that is not electronic, and digital, would be the die we roll in games, or the coil we flip. Statistically, if the object has n faces, we would expect a good die to give us a probability of 1/n to achieve one of these faces. The next thing is the representation that we have on one of these faces, this might be a number, or some other unique object. So, if we take a coin, and we decorate it with a 0 and a 1, then we get a binary number stream from repeatedly flipping the coin. If we take the die, with 6 faces, and we decorate it with 0,1,2,3,4,5, we would generate a number in this sixfold basis. I am sure there are many other examples, but this is all you really need to get started.
Strictly speaking, the Bombes of Bletchly Park were electromechanical rather than being electronic, and were about as complex digital technology got before electronics.
The bombe was an electro-mechanical device used by British cryptologists to help decipher German Enigma-machine-encrypted secret messages during World War II.
Each machine was about 7 feet (2.1 m) wide, 6 feet 6 inches (1.98 m) tall, 2 feet (0.61 m) deep and weighed about a ton. On the front of each bombe were 108 places where drums could be mounted. The drums were in three groups of 12 triplets. Each triplet, arranged vertically, corresponded to the three rotors of an Enigma scrambler. The bombe drums' input and output contacts went to cable connectors, allowing the bombe to be wired up according to the menu. The 'fast' drum rotated at a speed of 50.4 rpm in the first models and 120 rpm in later ones, when the time to set up and run through all 17,576 possible positions for one rotor order was about 20 minutes. wikipedia
The USA also made bombes, but those variants employed thermionic valves, the first electronic device. The computers made in Bletchly Park also were valve-based. | 3,714 | 16,360 | {"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-2024-38 | latest | en | 0.955998 |
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# Algebra: Logarithms
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1. The value of e ln 1+ ln 2+ ln 3 is ?
2. If logN(25) - logN(81) = 2, then N is?
3. Solve for x: log5( x) + log5(2x +13) = log5(24)
4. Find the coefficient of the fourth term in the expansion of (2x +3y)^11
##### Solution Summary
Logarithm problems are solved. Algebraic functions are analyzed for logarithms.
##### Solution Preview
1.)
exp(ln(1) +ln(2) + ln(3))
= exp(ln(1*2*3)) (because ln(a) + ln(b) = ln(a*b))
= exp(ln(6))
= 6 (because e^(log_base_e (x)) = x) --Answer
2.)
log_base_N(25) - log_base_N(81) = 2
=> log_base_N(25/81) = 2 (because log(m) - log(n) = log(m/n))
=>N^2 = 25/81 (because ...
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4.0
2,600 個評分
645 條評論
## 課程概述
This intermediate-level course introduces the mathematical foundations to derive Principal Component Analysis (PCA), a fundamental dimensionality reduction technique. We'll cover some basic statistics of data sets, such as mean values and variances, we'll compute distances and angles between vectors using inner products and derive orthogonal projections of data onto lower-dimensional subspaces. Using all these tools, we'll then derive PCA as a method that minimizes the average squared reconstruction error between data points and their reconstruction. At the end of this course, you'll be familiar with important mathematical concepts and you can implement PCA all by yourself. If you’re struggling, you'll find a set of jupyter notebooks that will allow you to explore properties of the techniques and walk you through what you need to do to get on track. If you are already an expert, this course may refresh some of your knowledge. The lectures, examples and exercises require: 1. Some ability of abstract thinking 2. Good background in linear algebra (e.g., matrix and vector algebra, linear independence, basis) 3. Basic background in multivariate calculus (e.g., partial derivatives, basic optimization) 4. Basic knowledge in python programming and numpy Disclaimer: This course is substantially more abstract and requires more programming than the other two courses of the specialization. However, this type of abstract thinking, algebraic manipulation and programming is necessary if you want to understand and develop machine learning algorithms....
## 熱門審閱
JS
2018年7月16日
This is one hell of an inspiring course that demystified the difficult concepts and math behind PCA. Excellent instructors in imparting the these knowledge with easy-to-understand illustrations.
NS
2020年6月18日
Relatively tougher than previous two courses in the specialization. I'd suggest giving more time and being patient in pursuit of completing this course and understanding the concepts involved.
## 176 - Mathematics for Machine Learning: PCA 的 200 個評論（共 641 個）
2020年5月6日
This course was tough but awesome. Lots of things i learnt from this course. Great course indeed and worth doing.
2021年5月17日
Undoubtedly one of the best courses I have taken on mathematics for Machine Learning with world-class teachers.
2021年2月15日
one of the best course to learn whats happening in machine learning and how it make sense through mathematics.
2020年7月30日
The PCA part Was a bit tricky barely handle the concepts.
thank you imperial team for such interactive course
2019年8月21日
One of the most challenging course in my life - almost impossible without python and mathematics background.
2020年8月25日
Need more Effort to grasp the materials explained_-" you need to be patience,the lecturer is really on top
2020年7月29日
Excellent course ... Quite challenging, a little difficult but I have learned a lot ... Thank you ...
2019年9月6日
Amazing course and provides basic introduction for the PCA. Need for programming help in this course.
2020年2月24日
Great course. I appreciate the rigor and clear mathematical explanations provided by Dr. Deisenroth.
2019年2月25日
exellent course! nice python wokring enviroment and very good explanation at each topic. thank you!
2020年1月18日
Excellent and to-the-point explanations, useful assignments to make the concepts etched in memory
2020年6月20日
This course helped me in getting a deeper knowledge on Principal Component Analysis. Thank You.
2018年10月16日
concise and to the point. Might want to introduce a bit the technique of Lagrangin multiplier
2021年5月2日
This was an amazing course, I really enjoyed it and learn a lot!
Thank you so much, greetings
2021年3月27日
I'm struggle with assigments of week 4 about implementing PCA. But, yeaah finally i got this
2020年12月3日
This course cleared so many concepts and enabled me to further master the subject on my own.
2020年3月17日
I learnt a lot from this course and now I think I am much more familiar with this algorithm.
2020年4月22日
extremely informative and really help me understand the basic math in Machine learning
2020年4月17日
Course was challenging, so does the math. It was a very excellent learning experience!
2019年11月14日
This course is also so helpful, and the lecturer is so predominant on what he taught.
2019年10月20日
Truly hardcore course if your are a noob in reduced order modelling. Very challenging
2020年8月8日
Algebra, Calculus and PCA
These are all excellent, if you have mathematics knowledge
2019年11月5日
Excellent course and extremely difficult one to grasp at one go. Regards Arijit Bose
2018年5月25日
Very hard to follow, but you need to do it to understand machine learning very well.
2019年7月27日
I have thoroughly enjoyed every course of this specialization. Thank you very much. | 1,137 | 4,895 | {"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-2021-25 | latest | en | 0.860236 |
http://www.physicsforums.com/showthread.php?p=3692367 | 1,386,894,541,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164770786/warc/CC-MAIN-20131204134610-00064-ip-10-33-133-15.ec2.internal.warc.gz | 604,446,084 | 14,728 | # Axiomatic derivation of [X,P]?
by Gerenuk
Tags: axiomatic, derivation
P: 1,057 How do you derive what quantum mechanical momentum is, from some axioms about reality? Therefore how do you justify one of the following more or less equivalent statements: [X,P]=ih =exp(ikx) psi(x,k)=exp(ikx) I've seen argumentations why quantum mechanics is set up with the mathematical framework it has (Hilbert space etc.), but no an explanation for momentum.
P: 1,480
Quote by Gerenuk How do you derive what quantum mechanical momentum is, from some axioms about reality? Therefore how do you justify one of the following more or less equivalent statements: [X,P]=ih =exp(ikx) psi(x,k)=exp(ikx) I've seen argumentations why quantum mechanics is set up with the mathematical framework it has (Hilbert space etc.), but no an explanation for momentum.
It comes from translations that we can perform to move from x="here" to x="there".
Less flippantly, one can start with the Galilean group of transformations (for the nonrelativistic case), and note that the probabilities we measure in experiments are invariant under these transformations.
Ballentine's textbook does a reasonably good job of developing (nonrelativistic) QM from this perspective.
P: 1,057
Quote by strangerep It comes from translations that we can perform to move from x="here" to x="there". [...]
Sounds good! I'll surely get this book.
Could you give me a hint, why translation operators and momentum in reality as we know it are related?
Translations probably yield the above form, but where does reality turn into translations? And can I derive classical mechanics from translations as well?
Now I even wonder... what is the essence of momentum anyway? Is it more than just postulating that it is conserved?
P: 1,742
## Axiomatic derivation of [X,P]?
The operator of momentum P is *defined* as a generator of space translations. Then the translation by a finite distance $a$ is represented by the exponential function of the generator $e^{\frac{i}{\hbar}Pa}$. The action of this transformation on the position operator X should result in a shift
$$e^{-\frac{i}{\hbar}Pa} X e^{\frac{i}{\hbar}Pa} = X-a$$
From this postulate you can get the desired commutator
$$[X,P] = i \hbar$$
Eugene.
P: 835 This may be interesting: http://www.hep.upenn.edu/~rreece/doc...esentation.pdf The equivalence you mentioned is arrived at midway.
P: 1,057 OK, from these ideas I understand why translation operations give the quantum mechanics form of momentum. So how are translation generators now consistent with the momentum we learned at school?
P: 835 I think that requires some knowledge about Lie Groups. However, I found this short explanation. I haven't looked through it properly, but I looks like it answers your question. http://www.dfcd.net/articles/momentumop.pdf
HW Helper Sci Advisor P: 11,722 [x,p]=ihbar 1 is the cornerstone of quantum mechanics. It's a postulate, unless one starts off with the Galilei group. So axiomatic derivation would mean to go along Ballentine's arguments and refine them using functional and harmonic analysis.
P: 1,480
Quote by Gerenuk OK, from these ideas I understand why translation operations give the quantum mechanics form of momentum. So how are translation generators now consistent with the momentum we learned at school?
Ballentine also covers this. In sect 3.3 he covers the generators of the Galilei group, one of which is the P mentioned above, but at that point its meaning is merely geometrical, and not yet identified with the "momentum" concept from classical dynamics.
Then in sect 3.4 he proceeds to identify these generators the dynamical variables from classical mechanics, using the postulate that the dynamics of a free particle are invariant under the full Galilei group.
(Sorry, but the entire argument is too long to reproduce here. Maybe Amazon or Google Books will let you read some of those sections of Ballentine.)
P: 1,057
Quote by strangerep [...] Then in sect 3.4 he proceeds to identify these generators the dynamical variables from classical mechanics, using the postulate that the dynamics of a free particle are invariant under the full Galilei group. (Sorry, but the entire argument is too long to reproduce here. Maybe Amazon or Google Books will let you read some of those sections of Ballentine.)
It's OK. I'll get the book, so no need to reproduce what is well written somewhere else ;)
So what is actually the definition of the classical momentum? I don't know what it means (yet), but you say momentum is a complete and direct consequence of the Galilei group, which itself is more or less simple translations?
For the relativistic case I do the same argument with the Poincare group?
Does Ballentine give all the neccessary maths I need to understand full how that emerges?
A more philosophical question:
Does all this treatment mean, that everything has to be a particle and not some sort of field?
Isn't it that when you have invariants, it can mean that your coordinate system is overdetermined? As an example take x, y, z coordinates for a sphere which are overdetermined and contraints, in contrast to euler angles. Does such an idea of other coordinate exists where the Galilei invariance is a natural consequence of the mathematical representation?
P: 1,480
Quote by Gerenuk So what is actually the definition of the classical momentum? I don't know what it means (yet), but you say momentum is a complete and direct consequence of the Galilei group, which itself is more or less simple translations?
For the free nonrelativistic case, we just use p = mv = m dq/dt. To get a
velocity operator from Galilean generators, we just use
$$V = i[H,Q]$$
(since d/dt of an operator corresponds to commutation with the Hamiltonian H).
But this is a bit of an oversimplification.
For less trivial interacting systems, the distinction between "position" and
"momentum" gets a bit blurred -- if you've done any Hamiltonian dynamics
perhaps you've heard of canonical transformations which mix position and
momentum, but preserve Hamilton's dynamical equations? But I was only talking
For the relativistic case I do the same argument with the Poincare group?
In the relativistic case, things are trickier since there's no position
operator in the basic algebra. One can be built up, but not for all
combinations of mass and spin. E.g., a position operator for the photon
remains problematic to this day.
But the basic idea is the same: determine the unitary irreducible
representations of the Poincare group. That's the "Wignerian" approach.
Does Ballentine give all the necessary maths I need to understand full how that emerges?
It's not a maths textbook, but anyone with reasonable proficiency in calculus
should be able to cope. He covers the basics of Lie group ideas when
introducing the Galilei group, but some prior exposure to group theory is
never wasted. Since I don't know what your math background is, I can't be more
specific.
Anyway, you can always ask here on PF if you get stuck. :-)
A more philosophical question: Does all this treatment mean, that everything has to be a particle and not some sort of field?
No. We're really finding representations of Lie algebras as operators on a
Hilbert space. In some case these representations turn out to be
finite-dimensional, others infinite-dimensional. Some of the latter are field
theories. Fields can have momentum too... :-)
Isn't it that when you have invariants, it can mean that your coordinate system is overdetermined? As an example take x, y, z coordinates for a sphere which are overdetermined and constraints, in contrast to euler angles. Does such an idea of other coordinate exists where the Galilei invariance is a natural consequence of the mathematical representation?
In the group theoretic development of quantum theory we proceed from the other
direction. One finds a maximal set of commuting operators within the algebra
of dynamical variables, which includes the invariants (called Casimirs) and
one other. Analysis of the spectrum of these operators determines the
necessary dimension of the Hilbert space.
P: 1,057
Quote by strangerep It's not a maths textbook, but anyone with reasonable proficiency in calculus should be able to cope. He covers the basics of Lie group ideas when introducing the Galilei group, but some prior exposure to group theory is never wasted. Since I don't know what your math background is, I can't be more specific. Anyway, you can always ask here on PF if you get stuck. :-)
Oh, it gets more and more interesting and I suppose it's time for me to study books! I've ordered Ballentine, yet I believe it won't outline all ideas you have presented here?!
I have a very good understanding of undergrad physics and math and a few more topics I looked up for interest. I don't know Lie groups and I had only one simple course on group theory.
Could you recommend books I should read to fully understand the answers you have given? It should be a "physicists" book, by which I mean I care about most steps of derivations, but not about mathematical details (convergence, abstract definitions, etc.)
If you give several suggestions, I will pick the most clear for me from the library.
Maybe one more question:
Are dimensions of space in any way predicted by these methods?
P: 1,480
Quote by Gerenuk Oh, it gets more and more interesting and I suppose it's time for me to study books! I've ordered Ballentine, yet I believe it won't outline all ideas you have presented here?! I have a very good understanding of undergrad physics and math and a few more topics I looked up for interest. I don't know Lie groups and I had only one simple course on group theory. Could you recommend books I should read to fully understand the answers you have given? It should be a "physicists" book, by which I mean I care about most steps of derivations, but not about mathematical details (convergence, abstract definitions, etc.) If you give several suggestions, I will pick the most clear for me from the library.
Start with Ballentine's "QM - A Modern Development", and see whether his treatment of Lie groups via the example of the Galilei group is sufficient for you to get through the rest of the book.
If not, or maybe later, you might take a look at Greiner & Muller, "QM - Symmetries".
Maybe one more question: Are dimensions of space in any way predicted by these methods?
If you mean the 3+1 dimensionality of space-time, then no (afaik).
P: 22 The first thing Ballentine says about a particle with no internal degrees of freedom (pag.80) is that the operators Q,P form an irreducible set. I think it is a crucial point, but he doesn't say why we are assuming this. What exactly means that Q and P are irreducible? We know only that Q is a position operator and P is the generator of spatial translations. We don't know anything about compatible observables.
P: 1,057
Quote by strangerep Start with Ballentine's "QM - A Modern Development", and see whether his treatment of Lie groups via the example of the Galilei group is sufficient for you to get through the rest of the book. If not, or maybe later, you might take a look at Greiner & Muller, "QM - Symmetries". If you mean the 3+1 dimensionality of space-time, then no (afaik).
A revivial of the thread... Meanwhile I've had a look at Ballentine and I'm quite pleased with that type of derivation :) However, I'd find it better to see it more general and rigorous derivation (but still understandable for a physicist :) ). Perhaps even one for the Poincare group. Maybe I can grab hold of the other book.
Even better if the derivation was in terms of the density matrix. Could I avoid treating the wavefunction and it's phase invariance this way??
P: 1,480
Quote by Gerenuk A revivial of the thread... Meanwhile I've had a look at Ballentine and I'm quite pleased with that type of derivation :) However, I'd find it better to see it more general and rigorous derivation (but still understandable for a physicist :) ). Perhaps even one for the Poincare group. Maybe I can grab hold of the other book.
I mentioned Greiner & Muller only because it contains some introductory material on Lie groups in a physics-relevant context. I don't think you'll find the same kind of derivation of ##[X,P]## there, nor much about Poincare.
Even better if the derivation was in terms of the density matrix. Could I avoid treating the wavefunction and it's phase invariance this way??
No. The central element in the quantum Galilei algebra (which later gets identified with mass) is a consequence of the requirement that probabilities be invariant. For some groups it can be disposed of by redefining some generators, but not in the Galilei case.
BTW, if you're now interested in delving a bit deeper into this question, you might consider exactly what's going on a bit later in Ballentine when he deals with the external field case. It involves a modification of the Hamiltonian, and some other adjustments.
This leads to a more general perspective that what we're doing is simply finding a quantization of a particular dynamics (meaning that we take classical dynamical variables with their Poisson algebra, and attempt to represent them as operators on a Hilbert space). This yields additional insight into the free case: instead of starting with static Euclidean 3-space, plus time, and then invoking free Newtonian dynamics to identify the geometric quantities with dynamical ones, we could just as well start with the dynamical equations of a free particle and try to find the largest group of transformations that preserve these equations. One finds the Galilei group as a subgroup. Of course, this is to be expected since our intuitive picture of 3D space is constructed by extrapolation of the motion of free particles... :-)
Separately, (and I don't know how much you follow other threads here), but I was recently made aware of how the central-extended Galilei algebra (with ##[X,P] \propto 1##) can be regarded as a consequence of relativistic Poincare invariance -- when one takes the nonrelativistic limit of low velocity. See Kaiser: http://arxiv.org/abs/0910.0352 , sect 4.2, esp p95 et seq. I'm starting to prefer Kaiser's explanation over the standard one.
P: 1,480
Quote by naffin The first thing Ballentine says about a particle with no internal degrees of freedom (pag.80) is that the operators Q,P form an irreducible set. I think it is a crucial point, but he doesn't say why we are assuming this. What exactly means that Q and P are irreducible?
Did you read Ballentine's appendices A & B ?
Complaints about mis-sequencing on this point have been made before: proper understanding of appendix B relies on having studied at least some of ch 4 -- which comes later than p80.
We know only that Q is a position operator and P is the generator of spatial translations. We don't know anything about compatible observables.
At that point, we know at least that Q and P are linearly independent generators. To go from this to irreducibility, one must understand that irreducibility means that no subspace of the space of states is left invariant by these operators. But detailed understanding of this point needs material on wave functions in the subsequent ch4. Herein lies, perhaps, a weakness in this sequence of presentation -- but a more advanced approach based on general quantization of Hamiltonian dynamics would be too difficult at that stage for the intended readership, imho.
P: 1,057
Quote by strangerep No. The central element in the quantum Galilei algebra (which later gets identified with mass) is a consequence of the requirement that probabilities be invariant. For some groups it can be disposed of by redefining some generators, but not in the Galilei case.
Sure, but density matrix and wave functions should be equivalent representations (the former being slightly more general). The density matrix has phase invariance included just by its mathematical form. So can you express the whole derivation with density matrices?
Btw, he proves [X,Px]=[Y,Py]=const only? So, the classical case with zero as a constant is included?
Quote by strangerep BTW, if you're now interested in delving a bit deeper into this question, you might consider exactly what's going on a bit later in Ballentine when he deals with the external field case. It involves a modification of the Hamiltonian, and some other adjustments.
I've seen that and it seems possible interactions have to be of mathematical "electric-field or magnetic field type". No other are possible?
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View Full Version : Elasticity tensor calculation
Paul_S
2007-08-07, 05:38
Hello everybody,
I am trying to write UMAT subroutine for Abaqus and have a great problem now to understand the principle of Elasticity tensor definition (ET). Let as consider material description of ET that is defined as
ET(i,j,k,l) = 4 * d[ d[psi]/d[Cij] ] / d[Ckl]
where: d[A]/d[b] mean derivation of A with respect of b
psi means energy density function
Cij mean (i,j) components of Right C-G tensor
What I really don't understand is how to do derivation (e.g. analyticaly) to respect the SYMMETRY OF C! Let me consider the following HYPOTHETICAL example:
psi = psi (C11,C12,C21,C33) = 2*C12 + 3*C12*C21 + 5*C22
(note that psi is symmetric with respect to C12 and C21)
HOW TO COMPUTE THE ET components? I did e.g. the following:
ET(1,2,1,2) = 4* d [ d[psi]/d[C12] ] / d[C12] = 4* d [3*C21] / d[C12] = 0
ET(1,2,2,1) = 4* d [ d[psi]/d[C12] ] / d[C21] = 4* d [3*C21] / d[C21] = 4*3 = 12
HOWEVER this result does not corresponds to my expectation of symmetry of ET tensor :-(
I think that the following relation should be held: ET(1,2,1,2) = ET(1,2,2,1)
May be I could consider symmetry by setting C21=C12 and re-express the form of psi to the form of psi=psi(C11,C12,C22) and calculate only componets of ET regarding to C12. However I am not sure if this approach is correct (gives me results that does not correspond to my refference results - factor of 2 or 4 higher).
Could anybody, please, give me an advice how to correctly do the general analytical calculation of ET based on components of C?
Paul
Jorgen
2007-08-08, 19:09
Well, isn't your hypothetical example fishy in the sense that the strain energy function is not symmetrical in C12 and C21? Do you claim that your function psi is physical??
- Jorgen
Paul_S
2007-08-10, 03:30
Thanks a lot for your response, Jorgen,
I have to apologize!!! - I made a typing mistake .... the correct form of (hypothetical) strain energy function should be:
psi = psi (C11,C12,C21,C33) = 2*C11 + 3*C12*C21 + 5*C22
... that is (now really) symmetrical in C12 and C21.
I am sorry to make you confused and hope that this is the only mistake in my above contribution.
However the problem is still the same:
ET(1,2,1,2) = 0
ET(1,2,2,1) = 12
that is not symmetrical!
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LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
M.Sc. DEGREE EXAMINATION – MATHEMATICS
FOURTH SEMESTER – APRIL 2012
# MT 4811 – OPERATIONS RESEARCH
Date : 18-04-2012 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
Answer ALL the questions
All questions carry equal marks
I a) Explain sensitivity analysis. Is it really useful to a company?
(or)
1. b) Explain branch and bound method. (5)
1. c) Find the optimum integer solution to the following LPP.
(15)
(or)
1. Solve the problem
Discuss the effect of changing the requirement vector from on the optimum solution.
II a) What is goal programming? How is it useful for a manufacturing company?
(or)
1. b) Mention the differences between LP and GP approach. (5)
1. c) A firm produces two products A and B. Each product must be processed through two departments. Department I has 30 hours of production capacity per day, and department II has 60 hours. Each unit of Product A requires 2 hours in department I and 6 hours in department II. Each unit of product B requires 3 hours in department I and 4 hours in department II. Management has rank ordered the following goals it would like to achieve in determining the daily product mix.
P1 : Minimize the underachievement of joint total production of 10 units.
P2 : Minimize the underachievement of producing 7 units of product B.
P3 : Minimize the underachievement of producing 8 units of product A. Formulate
this problem as a GP problem and illustrate with graph. (15)
(or)
1. d) A factory can manufacture two products A and B. The profit on a unit of A is `.80 and of B is `.40. The maximum demand of A is 6 units per day and of B is 8 units. The manufacturer has set up a goal of achieving a profit of `.640 per day. Formulate the problem as goal programming and solve it. (15)
III a) Explain the following terms in inventory: setup cost, holding cost, lead time, optimal cost and stock out cost.
(or)
1. b) Explain the term Price Break. Is it advisable to accept it always? (5)
1. c) Group the items given below into an ABC classification.
Item No. Units Unit cost in Rs. 1 7000 5 2 2400 3 3 1500 10 4 600 22 5 3800 1.50 6 40000 0.50 7 60000 0.20 8 3000 3.50 9 300 8.00 10 29000 0.40 11 11500 7.10 12 4100 6.20
Explain by graphical representation.
(or)
1. d) (i) A company operating 50 weeks in a year is concerned about its stock of
copper cables. One meter costs `.240 and there is a demand for 8000 meters per
week. The setup cost is `.2,700 and the holding cost is 25 % of one meter cost.
Assuming no shortages are allowed, find the optimal inventory policy. Also find
the number of orders and total inventory cost.
(ii) Plastic drums are produced at the rate of 50 items per day. The demand
occurs at the rate of 25 items per day. If the setup cost is `.1000 per setup and
holding cost is `.1.00 per unit of item per day find the economic lot size for one
run, assuming that the shortages are not allowed. Also find the time of cycle
and the minimum total cost of one production run. (8+7)
1. a) Explain optimistic time and pessimistic time in network model.
(or)
1. b) Explain Kendall’s notation for representing queuing models. (5)
1. c) Use Branch and Bound technique to solve the following:
(15)
(or)
1. d) With usual notation show that the probability distribution of queue length is
given by where .
V a) Write Kuhn-Tucker conditions for a quadratic programming problem.
(or)
1. b) State Wolfe’s algorithm. (5)
1. c) Using Kuhn-Tucker conditions
(15)
(or)
1. d) State the special features of dynamic programming technique. Find the shortest
route for traveling from city 1 to 10 using dynamic programming technique.
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# LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034.
M.A. DEGREE EXAMINATION – ECONOMICS
SECOND SEMESTER – APRIL 2003
## ST 2900 / s 872 – OPERATIONS RESEARCH
30.04.2003
1.00 – 4.00 Max : 100 Marks
### PART – A (5´ 4=20 marks)
Answer any FIVE questions.
1. Define Operations Research and highlight any four significant features.
2. How will you define basic solution to a system of ‘m’ simultaneous linear equations in ‘n’ unknowns (m < n)?
3. State the transportation problem mathematically as a linear programming problem.
4. Define : (i) Game (ii) Fair game (iii) Strictly determinable game (iv) Saddle point.
5. Using dominance method solve the game:
Player B
Player A
1. Write a note on (i) CPM (ii)
2. Define : (i) Setup cost (ii) Holding cost (iii) Shortage cost (iv) Lead time.
### Answer any FOUR questions.
1. Find all the basic feasible solutions of the equations:.
2. Use graphical method to solve the following:
Maximize
Subject to the constraints :
1. Write the simplex algorithm used in finding an optimum solution to a linear programming problem.
2. Explain the ABC Inventory system.
1. Four professors are each capable of teaching any one of four different courses. Class preparation time in hours for different topics varies from professor to professor and is given in the table below. Each professor is assigned only one course. Determine an assignment schedule so as to minimize the total course preparation time for all courses:
Professor Linear Programmes Queueing Theory Dynamic Programme Regression Analysis A 2 10 9 7 B 15 4 14 8 C 13 14 16 11 D 4 15 13 9
1. A small maintenance project consists of the following ten jobs whose precedence relations are identified by their node number:
Job (i, j) (1, 2) (2, 3) (2, 4) (3, 5) (3, 6) Duration (days) 2 3 5 4 1 Job (i, j) (4, 6) (4, 7) (5, 8) (6, 8) (7,8) Duration (days) 6 2 8 7 4
• Draw an arrow diagram.
• Find the critical path.
1. Explain multiple item static model with storage limitation.
### PART – C (2´20=40 marks)
Answer any TWO questions.
1. Use Big M method to
Minimize
Subject to the constraints :
1. Consider the following transportation table showing production and transportation costs along with the supply and demand positions of factories/distribution centres:
M1 M2 M3 M4 Supply F1 4 6 8 13 500 F2 13 11 10 8 700 F3 14 4 10 13 300 F4 9 11 13 3 500 Demand 250 350 1050 200
Find an initial basic feasible solution by using VAM.
• Find an optimal solution for the above problem.
1. Solve the game graphically :
Player B
Player A
1. Explain single item static model with one price break with the necessary diagram in detail.
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## Loyola College B.Sc. Statistics April 2006 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – STATISTICS
AC 23
SIXTH SEMESTER – APRIL 2006
# ST 6601 – OPERATIONS RESEARCH
(Also equivalent to STA 601)
Date & Time : 21-04-2006/9.00-12.00 Dept. No. Max. : 100 Marks
SECTION A (10 x 2 =20)
Answer ALL the questions. Each carries two Marks
1. Define saddle point of a pay-off matrix
2. How many basic cells will be there in a starting solution where there are 4 destinations and 3 origins in a transportation problem ?
3. Distinguish between “pure” and “mixed” strategies
4. What is a minimal spanning tree ?
5. How do you identify the critical activities of a project ?
6. Give the formula for free float and total float corresponding to an activity defined by arcs i and j
7. How will you conclude the presence of more than one optima for an LPP ?
8. What is an unbalanced transportation problem ?
9. Give the fundamental difference between PERT and CPM
10. In what way Floyd’s algorithm is superior to Dijkstra’s algorithm ?
SECTION B (5 x 8 =40)
Answer any FIVE of the following. Each carries EIGHT marks
1. Diamond is an aspiring young student of Loyola College. He realizes that “all work and no play makes him a dull boy”. As a result, Diamond wants to apportion his available time of about 10 hours a day between work and play. He estimates that play is twice as much fun as work. He also wants to study at least as much as he plays. However, Diamond realizes that if he is going to get all his homework assignments done, he can not play more than 4 hours a day. How should Diamond allocate his time to maximize his pleasure from both work and play ?
1. Comment on the solution of the following LPP
Maximize
Subject to
1. Specify the range for value of the game in the following case assuming that the payoff is for player A
B1 B2 B3 B4
A1 1 9 6 0
A2 2 3 8 4
A3 -5 -2 10 -3
A4 7 4 -2 -5
1. Use the Hungarian method to solve the following Assignment Problem
\$3 \$9 \$2 \$3 \$7
\$6 \$1 \$5 \$6 \$6
\$9 \$4 \$7 \$10 \$3
\$2 \$5 \$4 \$2 \$1
\$9 \$6 \$2 \$4 \$5
1. Write the following problem in standard form :
Maximize
Subject to
unrestricted
and also form the initial simplex table (No need to solve the problem)
1. Obtain the minimal spanning tree corresponding to the following network
3
1
6
4
9
5
10
7 5 8
3
1. Draw the time schedule for the following network and identify red flagged activities
25
5
B F
1. Express a transportation problem as Linear programming problem.
SECTION C (2x 20=40)
Answer any TWO of the following. Each carries TWENTY marks.
1. Solve the following problem using big-M method
Maximize
subject to
1. Solve the following network problem using Floyd’s algorithm
5
3
6 4
10
15
1. Solve the following 2 x 4 game graphically
B1 B2 B3 B4
A1 2 2 3 -1
A2 4 3 2 6
1. Find the critical path corresponding to the following project network and draw the time schedule
` A 5 J 7 K 3
C 8
B 10 F 10 L 8
D 9 E 4
I 1 H 5 M 4
G 3
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## Loyola College B.Sc. Statistics April 2007 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – STATISTICS
AC 23
SIXTH SEMESTER – APRIL 2007
# ST 6601 – OPERATIONS RESEARCH
Date & Time: 18/04/2007 / 9:00 – 12:00 Dept. No. Max. : 100 Marks
SECTION A
Answer all questions. (10×2=20)
1. Define: Basic Feasible Solution.
2. Mention any two limitations in solving a LPP by graphical method.
3. Express the following LPP in its standard form:
Maximize Z = 3x1 + 2x2
subject to x1 + x2 ≤ 2; 5x1 + 4x2 ≥4; x1,x2 ≥ 0;
1. What is the difference between Big-M and Two Phase method?
2. What is an unbalanced Transportation Problem?
3. Comment on the following statement:
“Assignment problem is a particular case of Transportation problem”.
1. Define: Activity and Node with reference to network analysis.
2. Draw the network, given the following precedence relationships:
Event : 1 2,3 4 5 6 7
Preceded by: — 1 2,3 3 4,5 5,6
1. Verify whether saddle point exists for the following game:
Player B
Player A B1 B2 B3
A1 5 8 1
A2 10 16 13
A3 25 22 20
1. Give the Laplace criteria for making decisions under uncertainty.
SECTION B
Answer any FIVE questions. (5×8=40)
1. Show graphically that the maximum or minimum values of the objective function for the following problem are same:
Maximize (Minimize) Z = 5x1 + 3x2
Subject to: x1 + x2 ≤ 6;
2x1 + 3x2 ≥ 3;
x1 ≥ 3;
x2 ≥ 3;
x1,x2 ≥ 0;
1. Use Big M- method to maximize Z = 2x1 + 3x2 subject to the constraints:
x1 + 2x2 ≤ 4; x1 + x2 =3; x1,x2 ≥ 0
1. Explain the MODI algorithm to obtain an optimum solution for a transportation problem.
1. Solve the following assignment problem.
1 2 3 4 5 A 3 8 2 10 3 B 8 7 2 9 7 C 6 4 2 7 5 D 8 4 2 3 5 E 9 10 6 9 10
1. Explain the Floyd’s method of finding the shortest route between any two given nodes.
2. Draw the network based on the following information and determine the critical path.
Activity Duration (in days) Activity Duration (in days)
(1,2) 3 (3,4) 3
(1,3) 1 (3,7) 10
(1,4) 15 (4,5) 10
(1,6) 7 (4,7) 22
(2,3) 8 (5,6) 5
(2,5) 10 (5,7) 12
(6,7) 7
1. A farmer wants to decide which of the three crops he should plant on his 100 acre farm. The profit from each is dependent on the rainfall during the growing season. The farmer has categorized the amount of rainfall as high, medium and low. His estimated profit for each crop is shown in the table below:
Rainfall Crop A Crop B Crop C
High 8000 3500 5000
Medium 4500 4500 5000
Low 2000 5000 4000
If the farmer wishes to plant only one crop, decide which should be his ‘best
Crop’ using Maximin, Savage and Hurwicz (α = 0.5) rule
1. Use simplex method to find the inverse of the following matrix
A =
SECTION C
Answer any TWO questions. (2×20=40)
1. a.) Explain the mathematical formulation of a Linear Programming Problem.
b.) Use two phase method to verify that there does not exist a feasible solution
to the following LPP:
Maximize z = 2x1 + 3x2 + 5x3 subject to
3x1 + 10x2 +5x3 ≤ 15;
33x1 – 10x2 + 9x3 ≤ 33;
x1 + 2x2 + x3 ≥ 4;
x1 , x2 , x3 ≥ 0.
1. Obtain an optimum solution to the following transportation problem where the entries denote the unit cost of transporting commodities from source to destination:
From↓To→ I II III IV Supply A 2 3 11 7 6 B 1 0 6 1 1 C 5 8 15 9 10 Demand 7 5 3 2
Find initial solution by least cost method and Vogle’s method and use the best
among them as the starting solution.
1. a.) Solve the following traveling salesman problem so as to minimize the total
distance traveled:
To
From A B C D E
A — 20 4 10 —
B 20 — 5 — 10
C 4 5 — 6 6
D 10 — 6 — 20
E — 10 6 20 —
• Explain the calculations in PERT to identify the critical path of a project.
1. ) Consider the following data that gives the distance between pairs of cities
1, 2…,6:
Route: (1,2) (1,3) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5)
Distance: 5 1 1 5 2 2 1 3 Route: (3,6) (4,6) (5,6)
Distance: 4 4 3
Use Dijkstra’s algorithm to find the shortest route between the cities
i.) 1 and 4 ii.) 1 and 6.
b.) Solve graphically the 2 x 4 game whose pay-off matrix is given below
Player B
Player A 1 3 11 7
8 5 2 5
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## Loyola College B.Sc. Statistics April 2008 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – STATISTICS
# NO 29
SIXTH SEMESTER – APRIL 2008
# ST 6601 – OPERATIONS RESEARCH
Date : 21/04/2008 Dept. No. Max. : 100 Marks
Time : 9:00 – 12:00
SECTION – A
Answer ALL questions: (10 x 2 = 20)
1. Define basic feasible solution and optimal solution in a linear programming problem.
2. Rewrite into standard form:
1. What is the role of artificial variables in linear programming?
2. Write the dual of :
1. Explain a transshipment problem
2. What is an unbalanced transportation problem?
3. Distinguish between CPM and PERT in network analysis.
4. Define activity and event in network analysis.
5. Define decision under uncertainity
6. Give an example of a 3 x 3 game without a saddle point.
SECTION – B
Answer any FIVE questions. (5 x 8 = 40)
1. The standard weight of special type of brick is 5kg and it contains two basic ingredients B1 and B2. B1 costs Rs. 5/kg and B2 costs Rs. 8/kg. The strength considerations dictate that the brick contains not more than 4 kgs of B1 and a minimum of 2 kgs of B2. Find graphically the minimum cost of the brick satisfying the above conditions.
2. Show that the following linear programming problem has an unbounded solution.
1. Explain how you will solve a travelling salesman problem.
2. Draw the network diagram given the following activities and find the critical path.
Job : A B C D E F G H I J K
Time (days) : 13 8 10 9 11 10 8 6 7 14 18
Immediate – A B C B E D,F E H G,I J
Predecessor :
1. Explain the maximal flow problem.
1. Solve the following problem for the minimum time:
I II III IV V A 16 13 17 19 20 B 14 12 13 16 17 C 14 11 12 17 18 D 5 5 8 8 11 E 3 3 8 8 10
1. A decision problem is expressed in the following pay-off table (Rs.’ 000s). What is the maximax and maximin payoff action?
Strategy
State I II III A – 1880 1840 1800 B – 1620 1600 1640 C – 1400 1420 1720
1. How will you solve a game problem using linear programming?
SECTION – C
Answer any TWO questions. (2 x 20 = 40)
1. a) Discuss in brief duality in linear programming (8)
2. b) Solve using dual simplex method:
1. a) Explain Vogel’s method of finding an initial solution to a transportation problem. (8)
1. b) Use least cost method to find initial basic feasible solution and then find the minimum transportation cost for the following problem.
Destinations
D1 D2 D3 D4 Supply 01 1 2 1 4 30 Origins 02 3 3 2 1 50 03 4 2 5 9 20 Demand 20 40 30 10 100
1. a) Explain i) total float ii) free float and
iii)independent float in network analysis (8)
1. b) The time estimates in weeks for the activities of a PERT network are given below:
Activity: 1-2 1-3 1-4 2-5 3-5 4-6 5-6 Opt.time: 1 1 2 1 2 2 3 Most likely time 1 4 2 1 5 5 6 Pessimistic time: 7 7 8 1 14 8 5
1. Draw the network diagram and find the critical path
2. Find the expected length and variance of the critical path.
• What is the probability that the project is completed in 13 weeks? (12)
1. a) Explain Laplace and Hurintz criteria in decision theory
1. b) Solve the following game graphically.
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## Loyola College B.Sc. Statistics April 2009 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – STATISTICS
YB 29
SIXTH SEMESTER – April 2009
# ST 6601 – OPERATIONS RESEARCH
Date & Time: 21/04/2009 / 9:00 – 12:00 Dept. No. Max. : 100 Marks
PART A
Answer ALL questions: (10 x 2 = 20)
1. Define Operations Research.
2. How will you identify an infeasible solution in linear programming?
3. Distinguish between Big M and Two phase methods.
4. Write the dual of
Max z = x1 + x2
subject to
x1, x2 ≥ 0
1. What is a traveling salesman problem?
2. Define a transhipment problem.
3. Define a critical path.
4. What are the time estimates used in PERT?
5. What is meant by decision under uncertainty?
6. Define a two-person zero sum game
PART B
Answer any FIVE questions: (5 x 8 = 40)
1. A person requires 10, 12 and 12 units of chemicals A, B and C respectively for his garden. The liquid product contains 5, 2 and 1 units of A, B and C respectively per jar. A dry product contains 1, 2 and 4 units of A, B and C per carton. If the liquid product sells for Rs 3 per jar and dry product sells for Rs 2 per carton, how many of each should he purchase in order to minimize the cost using graphical method?
2. How will you identify unbounded and alternate solutions in a linear programming problem?
3. Explain the dual simplex method of solving a linear programming problem.
4. Four professors are each capable of teaching any one of the following courses. The table below shows the class preparation time in hours by the professors on the topics. Find the assignment of the courses to the professors to minimize the preparation time.
Professor Linear Programming Queuing Theory Dynamic Programming Inventory Control A 2 10 9 7 B 15 4 14 8 C 13 14 16 11 D 4 15 13 9
1. Distinguish between PERT and CPM.
1. For the following network determine the maximum flow and optimum flow in each link:
1. Find the optimum strategy and value of the game for which there is no saddle point given the payoff matrix of A:
1. Explain minimax and savage criteria in decision making with suitable examples.
PART – C
Answer any TWO questions: (2 x 20 = 40)
1. (a) Explain primal-dual relationship in linear programming.
(b) Use penalty method to
Max z = 2x1 + x2 + 3x3
subject to
x1 + x2 + 2x3 ≤ 5
2x1 + 3x2 + 4x3 = 12
x1 , x2 , x3 ≥ 0
1. (a) Solve the following transportation problem with cost coefficients, demand and supply as
given below
(b) Solve using simplex method
Max Z = 10x1 + x2 + 2x3
subject to
x1 + x2 – 2x3 £10
4x1 + x2 + x3 £ 20
x1, x2, x3 ≥ 0
1. (a) State the rules for drawing the network diagram.
(b) The following table lists the jobs in a network along with the time estimates.
Job 1 – 2 1 – 6 2 – 3 2 – 4 3 – 5 4 – 5 6 – 7 5 – 8 7 – 8 Optimistic time (days) 3 2 6 2 5 3 3 1 4 Most likely time (days) 6 5 12 5 11 6 9 4 19 Pessimistic time (days) 15 14 30 8 17 15 27 7 28
(i) Draw the network diagram
(ii) Calculate the length and variance of the critical path
(iii)Find the probability of completing the project before 31 days.
1. (a) A business man has three alternatives each of which can be followed by any one of the four possible events. The conditional payoff in rupees for each action – event combination are as given below:
Alternative Payoff A B C D X 8 0 10 6 Y -4 12 18 -2 Z 14 6 0 8
Determine which alternative should the business man choose if he adopts (i) Hurwicz Criterion, the degree of optimism being 0.7 (ii) Laplace Criterion
(b) Solve following game graphically:
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## Loyola College B.Sc. Statistics April 2011 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B. Ss. DEGREE EXAMINATION – STATISTICS
SIXTH SEMESTER – April 2011
ST6604 / ST6601 – OPERATIONS RESEARCH
Time: 3 hrs Max.:100 Marks
PART A
Answer ALL questions: (10 x 2 = 20)
1. Define Basic feasible solution.
2. What do you mean by Slack variable?
3. When do we go for artificial variables?
4. What are the two rules for selecting the entering and leaving variables?
5. Write the Standard form for the given Linear programming problem.
1. What do you understand by transportation problem?
2. What is a traveling salesman problem?
3. Define a Two- person zero sum game.
4. Distinguish between PERT and CPM.
5. What do you mean by Minimax criterion?
PART B
Answer any FIVE questions: (5 x 8 = 40)
1. Let us assume that you have inherited Rs. 1,00, 000 from your father – in – law that can be inversted in a combination of only two stock portfolios, with the maximum investment allowed in either portfolio set at Rs. 75, 000. The first portfolio has an average of 10%, whereas the second has 20%. In terms of risk factors associated with these portfolios, the first has a risk rating of 4 (on a scale of 0 to 10) and the second has a risk rating of 9. since you wish to maximize yours returns, you will not accept an average rate of return below 12% or a risk factor above 6. Hence, you face the important question. How much should you invest in each portfolio?
Formulate this as a Linear Programming Problem.
1. Explain the Two Phase method of solving an LPP.
1. Find the initial basic feasible solution for the following transportation problem using Vogel’s Approximation method.
From To 9 12 9 6 9 10 5 7 3 7 7 5 5 6 6 5 9 11 3 11 2 6 8 11 2 2 10 9 4 4 6 2 4 2
1. Solve the following linear programming problem by graphical method:
Maximize z = x1 + 2x2
Subject to
x1 + 2x2 £ 15
2x1 + x2 £ 20
x1 + x2 ³ 1
x1, x2 ³ 0
1. Consider a firm having two factories. The firm is to ship its products from the factories to three retail stores. The number of units available at factories X and Y are 200 and 300 respectively, while those demanded at retail stores A, B and C are 100, 150 and 250 respectively. Rather than shipping directly from factories to retail stores, it is added to investigate the possibility of transhipment. The transportation cost (in rupees) per unit is given below:
Factory Retail Store X Y A B C Factory X 0 8 7 8 9 Y 6 0 5 4 3 Retail Store A 7 2 0 5 1 B 1 5 1 0 4 C 8 9 7 8 0
Find the optimal shipping schedule.
1. Solve the following game graphically:
Player B
Player A
1. Determine the total, free and independent floats and identify the critical path for the given project network.
1. Differentiate between Simplex method and Dual Simplex method.
PART C
Answer any TWO questions: (2 x 20 = 40)
1. a) Give the Dual for the following Primal.
Maximize z = 5x1 + 2x2 Subject to 6x1 + x2 ≥ 6 4x1 + 3x2 ≥ 12 x1 + 2x2 ≥ 4 x1, x2 ≥ 0
19 b) Solve the given problem using Big M method.
Minimize z = x1 + 2x2 + 3x3 – x4
Subject to
x1 + 2x2 + 3x3 = 15
2x1 + x2 + 5x3 = 10
x1 + 2x2 + x3 + x= 10
x1, x2, x3, x4 ≥ 0 (5 + 15)
1. a) Reduce a m x n game to a linear programming problem.:
b) Solve the following assignment problem to find the minimum total expected cost.
Men I II III IV Job A 42 35 28 21 B 30 25 20 15 C 30 25 20 15 D 24 20 16 12
21. a) State the rules for drawing the network diagram.
21. b) A project consists of eight activities with the following relevant information:
Activity Immediate Predecessor Estimated duration (days) Optimistic Most likely Pessimistic A – 1 1 7 B – 1 4 7 C – 2 2 8 D A 1 1 1 E B 2 5 14 F C 2 5 8 G D, E 3 6 15 H F, G 1 2 3
(i) Draw the PERT network and find out the expected project completion time and variance
(ii) What duration will have 95% confidence for project completion?
\$\$\$\$\$\$\$\$
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## Loyola College B.Sc. Statistics April 2012 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B. Ss. DEGREE EXAMINATION – STATISTICS
SIXTH SEMESTER – April 2012
ST6604 / ST6601 – OPERATIONS RESEARCH
Date: 18-04-2012 Dept. No. Max.:100 Marks
Time: 1.00 – 4.00
PART – A
Answer ALL questions: (10 x 2 = 20)
1. What are the different phases of Operations Research?
2. What is a degenerate solution?
3. Give the dual for the following primal:
1. When can dual simplex method be applied to a LPP?
2. Define Two- person zero sum game.
3. Mention the two phases in the two-phase method.
4. When is there a necessity for artificial variables?
5. What are the methods used to solve a mixed strategy game?
6. What is CPM?
7. What is Minimax criterion?
PART – B
Answer any FIVE questions: (5 x 8 = 40)
1. Explain simplex method for solving an LPP.
2. The manager of an oil refinery must decide on the optimum mix of two possible blending processes of which the input and output production runs are as follows:
Process Input Output Crude A Crude B Gasoline X Gasoline Y 1 6 4 6 9 2 5 6 5 5
The maximum amounts available of crudes A & B are 250 units and 200 units respectively. Market demand shows that at least 150 units of gasoline X and 130 units of gasoline Y must be produced. The profits per production run from process 1 and process 2 are Rs. 4 and Rs. 5 respectively. Formulate the problem for maximizing the profit.
1. What are the steps involved in solving an unbalanced transportation problem to get the optimum solution?
1. A departmental head has four subordinates and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimate of the time each man would take to perform each task, is given in the matrix below:
Tasks Men E F G H A 18 26 17 11 B 13 28 14 26 C 38 19 18 15 D 19 26 24 10
How should the tasks be allocated, one to a man, so as to minimize the total man-hours?
1. Solve the following graphically:
Player B
Player A
1. Differentiate between Simplex method and Dual Simplex method.
2. Explain Laplace and Savage criterion in detail.
3. Explain PERT algorithm in detail.
PART – C
Answer any TWO Questions: ( 2 x 20 = 40 marks)
1. a) Explain Primal – Dual Relationship. (5)
1. b) Use Duel Simplex method to solve the following LPP. (15)
Minimize z = 6x1 + 7x2 + 3x3 + 5x4
Subject to
5x1 + 6x2 – 3x3 + 4x4 ≥ 12
x2 – 5x3 – 6x≥ 10
2x1 + 5x2 + x3 + x4 ≥ 8
x1, x2, x3, x4 ≥ 0
1. a) Show that for any zero-sum two-person game where there is no saddle point and for
which A’s payoff matrix is
(15)
the optimal strategies (x1, x2) and (y1, y2) for A and B respectively are determined by
What is the value of the game?
1. 20. b) What is the principle of dominance in game theory? (5)
1. Find the initial basic feasible solution for the given transportation problem using:
(i) North West Carner rule (ii) Vogels method (iii) Least Cost method.
From To Available A B C I 50 30 220 1 II 90 45 170 3 III 250 200 50 4 Requirement 4 2 2
And obtain the optimal solution.
1. a) What are the rules for constructing a network diagram? (5)
22. b) A project consists of eight activities with the following relevant information:
Activity Immediate Predecessor Estimated duration (days) Optimistic Most likely Pessimistic A – 1 1 7 B – 1 4 7 C – 2 2 8 D A 1 1 1 E B 2 5 14 F C 2 5 8 G D, E 3 6 15 H F, G 1 2 3
(i) Draw the PERT network and find out the expected project completion time.
(ii) What duration will have 95% confidence for project completion?
(iii) If the average duration for activity F increases to 14 days, what will be its effect on the expected project completion time, which will have 95% confidence?
(15)
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## Loyola College B.Sc. Mathematics April 2007 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
CV 16
B.Sc. DEGREE EXAMINATION –MATHEMATICS
FIFTH SEMESTER – APRIL 2007
MT 5504 – OPERATIONS RESEARCH
Date & Time: 03/05/2007 / 1:00 – 4:00 Dept. No. Max. : 100 Marks
SECTION –A
## Answer All: 2 x 10 = 20
1. Define Operations Research.
2. What are the three methods to find Initial Basic Feasible solution in Transportation problem ?
3. Solve the Transportation problem by Least Cost Method.
A1 A2 A3 Supply B1 4 6 2 5 B2 3 1 5 15 B3 4 5 3 15 Demand 10 10 10
1. Solve the game:
2 1 4 1 4 3 2 2 6
1. Define Unbalanced situation in Transportation problem.
2. Define Feasible Solution.
3. Solve the Assignment problem
3 7 5 4 7 2 5 4 6
1. What is Dummy activity in Network problem ?
2. Define Optimistic Time Estimate.
3. Define Economic Order Quantity.
SECTION –B
## Answer any five: 5x 8 = 40
1. Find the Initial Basic Feasible solution in Transportation problem using i) North West Corner Rule ii) Least Cost Method.
A1 A2 A3 A4 Supply B1 4 2 1 3 20 B2 8 4 2 4 20 B3 1 2 3 4 30 B4 5 2 4 6 20 Demand 10 30 10 40
1. Using Graphical method solve the Linear Programming Problem
Max z = 2x1+4x2 subject to the constraints
2x1+4x≤ 5 , 2x1+4x2 ≤ 4, x1, x2 ≥ 0.
1. Solve the Assignment problem
M1 M2 M3 M4 M5 J1 9 22 58 11 19 J2 43 78 72 50 63 J3 41 28 91 37 45 J4 74 42 27 49 39 J5 36 11 57 22 25
1. Solve using Matrix Oddment method
-1 2 1 1 -2 2 3 4 -3
1. Define critical path and draw the Network diagram for
Activity: A B C D E F G H I J K
Immediate predecessor: – – – A B B C D E H,I F,G
1. Solve using Dominance property
1 7 3 4 5 6 4 5 7 2 0 3
1. Solve the Transportation problem
A1 A2 A3 A4 Supply B1 1 2 1 4 30 B2 3 3 2 1 50 B3 4 2 5 9 20 Demand 20 40 30 10
1. The probability distribution of monthly sales of certain item is as follows:
Number of items: 0 1 2 3 4 5 6
P(d) : 0.02 0.05 0.30 0.27 0.20 0.10 0.06
The cost of carrying inventory is Rs 10 per unit per month . Find the
shortage cost for one item for one unit of time.
(P.T.O)
SECTION –C
## Answer any two: 2x 20 = 40
1. Solve the following Linear Programming Problem using Simplex method
Max z = 3x1+2x2 subject to the constraints
x1+2x≤ 6 , 2x1+x2 ≤ 8, -x1+x2 ≤ 1, x2 ≤ 2, x1, x2 ≥ 0. (20)
20 a) Solve the Transportation problem to maximize the profit
A1 A2 A3 A4 Supply B1 40 25 22 33 100 B2 44 35 30 30 30 B3 38 38 38 30 70 Demand 40 20 60 30
1. b) Solve the following traveling sales man problem
A B C D E A – 3 6 2 3 B 3 – 5 2 3 C 6 5 – 6 4 D 2 2 6 – 6 E 3 3 4 6 –
(10+10)
21 a) Solve the game graphically
1 0 4 -1 -1 1 2 5
1. b) The annual demand for an item is 3200 units, the unit cost is Rs 6 and inventory
Carrying charges 25% per annum. If the cost of one procurement is Rs 150. Find
1. i) Economic Order Quantity ii) Time between two consecutive orders
iii) Number of orders per year iv) The optimal total cost (10+10)
22 a) Draw the Network diagram ,the Critical path ,the project duration and the total float
for the following activities
Activity: 1-2 2-3 3-4 3-7 4-5 4-7 5-6 6-7
Duration: 3 4 4 4 2 2 3 2
1. b) What is the probability that the project will be completed in 27 days? Draw the
network diagram also .
Activity: 1-2 1-3 1-4 2-5 2-6 3-6 4-7 5-7 6-7
T0 : 3 2 6 2 5 3 3 1 2
Tm : 6 5 12 5 11 6 9 4 5
Tp : 15 14 30 8 17 15 27 7 8
(10+10)
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## Loyola College B.Sc. Mathematics Nov 2008 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
# AB 21
FIFTH SEMESTER – November 2008
# MT 5504 – OPERATIONS RESEARCH
Date : 07-11-08 Dept. No. Max. : 100 Marks
Time : 9:00 – 12:00
SECTION A
Answer all the questions. (10 X 2 = 20)
1. Define feasible solution of a general Linear Programming Problem.
2. Give the symmetric form of a LPP.
3. State the necessary and sufficient condition for a Transportation problem to have a feasible solution.
4. How can you convert a maximization assignment problem to a minimization problem?
5. What is a two person zero sum game.
6. Define EOQ.
7. Define total float of an activity.
8. What is payoff matrix?
9. What is a project? List the 3 main phases of a project.
10. What are the different types of inventory?
SECTION B
Answer any five questions.(5 X 8 = 40)
1. Solve the following L.P.P by the Graphical method:
Max Z = 3x1 + 2x2 subject to
-2x1 + x2 ≤ 1,
x1≤ 2,
x1 + x2 ≤ 3 and x1, x2 ≥ 0.
1. Find the initial basic feasible solution for the following transportation problem by least cost method.
To Supply
From 1 2 1 4 30
3 3 2 1 50
4 2 5 9 20
Demand 20 40 30 10
1. Solve the following 2 X 2 game:
B
A 5 1
4 0
1. Construct the network for the project whose activities are given below and compute the total, free and independent float of each activity and hence determine the critical path and the project duration:
Activity: 0-1 1-2 1-3 2-4 2-5 3-4 3-6 4-7 5-7 6-7
Duration:3 8 12 6 3 3 8 5 3 8
1. For an item, the production is instantaneous. The storage cost of one item is Re. one per month and the set up cost is 25 per run. If the demand is 200 units per month, find the optimum quantity to be produced per set-up and hence determine the total cost of storage and set-up per month.
1. A commodity is to be supplied at a constant rate of 200 units per day. Supplies for any amounts can be had at any required time, but each ordering costs Rs. 50. Cost of holding the commodity in inventory is Rs. 2.00 per unit per day while the delay in the supply of the items induces a penalty of Rs.10 per unit per delay of one day. Formulate the average cost function of this situation and find the optimal policy (q,t) where t is the reorder cycle period and q is the inventory level after reorder. What should be the best policy if the penalty cost becomes infinite?
1. The processing time in hours for the jobs when allocated to the different machines is indicated below. Assign the machines for the jobs so that the total processing time is minimum.
Machines Jobs M1 M2 M3 M4 M5 J1 9 22 58 11 19 J2 43 78 72 50 63 J3 41 28 91 37 45 J4 74 42 27 49 39 J5 36 11 57 22 25
1. Solve the following:
Maximize
Z=15x1 + 6x2+ 9x3 + 2x4
Subject to 2x1 + x2+ 5x3+6x4 ≤ 20
3x1+x2+3x3+25x4 ≤ 24
7x1 + x4 ≤70
x1, x2, x3, x4 ≥ 0.
SECTION C
Answer any two questions: (2 X 20 = 40)
19a) Use Penalty method to solve Z=2x1 + x2+ x3
Subject to 4x1 + 6x2+ 3x3 ≤ 8
3x1– 6x2– 4x3 ≤ 1
2x1 + 3x2 – 5x3 ≥ 4
x1, x2, x3 ≥ 0.
19b) A person wants to decide the constituents of a diet which will fulfill his daily requirements of essential nutrition at the minimum cost. The choice is to be made from four different types of foods. The yields per unit of these foods are given in the following table:
Food type Yield/unit Cost / unit (Rs) Proteins Fats Carbohydrates 1 3 2 6 45 2 4 2 4 40 3 8 7 7 85 4 6 5 4 65 Minimum requirement 800 200 700
Formulate the linear programming model for the [problem.
20a) Solve the following Transportation problem to minimize the total cost of transportation:
Destination origin D1 D2 D3 D4 supply O1 14 56 48 27 70 O2 82 35 21 81 47 O3 99 31 71 63 93 Demand 70 35 45 60 210
20b) Solve the following Travelling salesman problem:
A B C D From A – 46 16 40 B 41 – 50 40 C 82 32 – 60 D 40 40 36 –
21a) Three time estimates (in months) of all activities of a project are given below:
Time in months Activity a m b 1-2 0.8 1.0 1.2 2-3 3.7 5.6 9.9 2-4 6.2 6.6 15.4 3-4 2.1 2.7 6.1 4-5 0.8 3.4 3.6 5-6 0.9 1.0 1.1
Find the expected duration and standard deviation of each activity,
Construct the project network,
Determine the critical path, expected project length and expected variance of the project length.
21b) For the pay-off matrix given below, decide optimum strategies for A and B
B
• 2
A 1 200 80
2 110 170
22a) Explain the purchase inventory model with n-price breaks.
22b) Find the optimal order quantity for a product for which the price break is as follows:
Quantity Unit cost
0 ≤ Q1 < 50 Rs. 10
50 ≤Q2<100 Rs. 9
100 ≤Q3 Rs. 8
The monthly demand for the product is 200 units, the cost of the storage is 25% of the unit cost and ordering cost is Rs. 20 per order.
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## Loyola College B.Sc. Mathematics April 2009 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
ZA 31
FIFTH SEMESTER – April 2009
# MT 5507 / 5504 – OPERATIONS RESEARCH
Date & Time: 17/04/2009 / 9:00 – 12:00 Dept. No. Max. : 100 Marks
SECTION A
Answer ALL Questions (10 x 2 = 20)
1. Write down any two uses of Operations Research.
2. Define Slack variables.
3. Which is the necessary and sufficient condition for the transportation problem to have
a feasible solution?
1. Describe a traveling salesman problem.
2. What is meant by mixed strategy?
3. What is the value of the game?
4. Define Total float
5. Explain: Minimal Spanning tree problem?
6. Give any two reasons for maintaining inventory
7. Define Recorder level.
SECTION B
Answer ANY FIVE Questions. (5 x 8 = 40)
1. Solve the following Linear Programming Problem by graphical method
Maximize Z = 5x1 + 8x2
Subject to:
15x1 + 10x≤ 180
10x1 + 20x≤ 200
15x1 + 20x≤ 210
and x1, x≥ 0
1. Use simplex method to solve the following LPP
Maximize Z = 4x1 + 10x2
Subject to:
2x1 + x≤ 50
2x1 + 5x≤ 100
2x1 + 3x≤ 90
and x1, x≥ 0
1. Find the initial basic feasible solution for the following transportation problem
D1 D2 D3 D4 Supply S1 21 16 25 13 11 S2 17 18 14 23 13 S3 32 27 18 41 19 Demand 6 10 12 15
1. Write the algorithm for solving Assignment problem.
1. Solve the following game using dominance property.
I II III Row mini. I 1 7 2 1 II 6 2 7 2 III 6 1 6 1 Column max. 6 7 7
1. Draw the network for the project whose activities with their predecessor relationships
are given below:
A, C, D can start simultaneously; E >B, C; F, G >D; H, I > E ,F ; J >I, G ;
K > H; B > A.
1. The annual demand of a product is 10,000 units, each unit cost Rs.100 if orders
placed in quantities below 200 units but for orders of 200 or above, the price is Rs.95,
the annual inventory holding cost is 10% of the value of the item, and the ordering
cost is Rs.5 per order. Find the economic lot size.
1. The demand for an item in a company is 18,000 units per year and the company can
produce the item at a rate of 3000 per month. The cost of one set up is Rs. 500 and
the holding cost of one unit per month is 15 paise. The shortage cost of one unit is
Rs. 20 per month. Determine the optimum manufacturing quantity and the number
of shortage. Also determine the manufacturing time and time between setups.
SECTION C
Answer ANY TWO Questions. (2 x 20 = 40)
1. Solve the following Linear Programming Problem by Dual Simplex method
Minimize Z = x1 + x2
Subject to:
2x1 + x≥ 2
-x1 – x2 ≥ 1
and x1, x≥ 0
1. (a) Solve the following transportation problem using Least Cost Method to find the
Initial Basic Feasible solution.
A1 A2 A3 A4 A5 Supply B1 4 1 2 6 9 100 B2 1 4 7 3 8 120 B3 7 2 4 7 7 120 Demand 40 20 70 90 90
.
(b)Solve the following traveling sales man problem
M1 M2 M3 M4 M5 J1 9 22 58 11 19 J2 43 78 72 50 63 J3 41 28 91 37 45 J4 74 42 27 49 39 J5 36 11 57 22 25
(10 + 10)
1. (a) Solve the following game graphically
B1 B2 B3 B4 A1 1 0 4 -1 A2 -1 1 2 5
(b) In a game of matching points with 2 players suppose A wins one unit value when
there are 2 heads, wins nothing when there are 2 tails, and loses ½ unit value when
there are 1 head and 1 tail. Determine the pay-off matrix, the best strategy for each
player and the value of the game. (10+10)
22 (a) Draw the network, determine the critical path , project duration and the total float for the
following activities .
Activity 1-2 2-3 3-4 3-7 4-5 4-7 5-6 6-7 Duration 3 4 4 4 2 2 3 2
(b) ABC manufacturing company purchases 9,000 parts of a machine for its annual
requirement, ordering one month’s usage at a time. Each part costs Rs.20.
The ordering cost per order is Rs.15, and the carrying charges are 15% of the
average inventory per year.
You have been asked to suggest a more economical purchasing policy for the
company. What advice would you offer and how much would it save the company
per year? (10+10)
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## Loyola College B.Sc. Mathematics April 2012 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
FIFTH SEMESTER – APRIL 2012
# MT 5507/MT 5504 – OPERATIONS RESEARCH
Date : 30-04-2012 Dept. No. Max. : 100 Marks
Time : 9:00 – 12:00
PART – A
Answer ALL questions: (10×2=20 marks)
1. A firm plans to purchase at least 200 quintals of scrap containing high quality metal X and low quality metal Y. It decides that the scrap to be purchased must contain at least 100 quintals of X and not more than 35 quintals of Y. The firm can purchase the scrap from two suppliers (A and B) in unlimited quantities. The percentage of X and Y metals in terms of weight in the scrap supplied by A and B is given below:
Metals Supplier A Supplier B X 25% 75% Y 10% 20%
The price of A’s scrap is Rs.200 per quintal and that of B is Rs. 400 per quintal. The firm wants to determine the quantities that it should buy from the two suppliers so that total cost is minimized. Construct a LP model.
1. Give at least four different contrasting properties of primal and dual of general LP problems
2. What is meant by unbalanced transportation problem?
3. Name four different methods to solve an assignment problem.
4. Consider the game with the following payoff table:
Player B Player A B1 B2 A1 2 6 A2 -2 λ
Show that the game is strictly determinable, whatever λ may be.
1. When no saddle point is found in a payoff matrix of a game, how do we find the value of the game?
2. Give an example of weighted graph which is not a tree. Find two different minimal spanning trees of the weighted graph you constructed.
3. Why is in PERT network each activity time assume a Beta-distribution?
4. What are the basic information required for an efficient control of inventory?
5. What is Economic Order Quantity (EOQ) concept?
PART – B
Answer any FIVE questions: (5×8=40 marks)
1. Using Simplex Method, solve the following Problem and give your comments on the solution: Maximize Z= 6x1+4x2, Subject to: x1+x2£ 5, x2≥ 8, x1, x2≥ 0.
2. Prove that dual of the dual is the primal.
Tasks Clerks A B C D 1 4 7 5 6 2 – 8 7 4 3 3 – 5 3 4 6 6 4 2
1. Consider a problem of assigning four clerks to four tasks. The time (hours) required to complete the task is given below:
Clerk 2 cannot be assigned to task A and clerk 3 cannot be assigned to task B. Find all the optimum assignment schedules.
1. A company has three production facilities S1, S2 and S3 with production capacity of 7, 9 and 18 units (in 100s) per week of a product, respectively. These units are to be shipped to four warehouses D1, D2, D3 and D4 with requirement of 5, 6, 7 and 14 units (in 100s) per week, respectively. The transportation costs (in rupees) pre unit between factories to warehouses are given in the table below:
D1 D2 D3 D4 Capacity S1 19 30 50 10 7 S2 70 30 40 60 9 S3 40 8 70 20 18 Demand 5 8 7 14 34
Use North-West Corner Method to find an initial basic feasible solution to the transportation problem.
1. A soft drink company calculated the market share of two products against its major competitor having three products and found out the impact of additional advertisement in any one of its products against the other
Company B Company A B1 B2 B3 A1 6 7 15 A2 20 12 10
What is the best strategy for the company as well as the competitor? What is the payoff obtained by the company and the competitor in the long run? Use graphical method to obtain the solution.
1. Solve the following game after reducing it to a 2×2 game
Player B Player A B1 B2 B3 A1 1 7 2 A2 6 2 7 A3 5 1 6
1. An assembly is to be made from two parts X and Y. Both parts must be turned on a lathe and Y must be polished whereas X need not be polished. The sequence of activities together with their predecessors is given below.
Activity Description Predecessor Activity A Open work order – B Get material for X A C Get material for Y A D Turn X on lathe B E Turn Y on lathe B, C F Polish Y E G Assemble X and Y D, F H Pack G
Draw a network diagram of activities for the project.
1. The production department for a company requires 3600 kg of raw material for manufacturing a particular item per year. It has been estimated that the cost of placing an order is Rs. 36 and the cost of carrying inventory is 25% of the investment in the inventories. The price is Rs. 10 per kg. Calculate the optimal lot size, optimal order cycle time, minimum yearly variable inventory cost and minimum yearly total inventory cost.
PART – C
Answer any TWO questions: (2×20=40 marks)
1. (a) Use graphical method and solve the LPP: Maximize Z= 15x1+10x2, Subject to: 4x1+6x2£ 360,
3x1£ 180, x2≥ 8, 5x2£ 200, x1, x2≥ 0. (7 marks)
(b) Use simplex method to solve the LPP: Maximize Z= 3x1+5x2+4x3, Subject to: 2x1+3x2£ 8,
2x2+5x3£ 10, 3x1+2x2+4x3£ 15, x1, x2, x3≥ 0. (13 marks)
1. (a) ABC limited has three production shops supplying a product to five warehouses. The cost of production varies from shop to shop and cost of transportation form one shop to a warehouse also varies. Each shop has a specific production capacity and each warehouse has certain amount of requirement. The costs of transportation are given below:
Warehouse Shop I II III IV V Supply A 6 4 4 7 5 100 B 5 6 7 4 8 125 C 3 4 6 3 4 175 Demand 60 80 85 105 70 400
The cost of manufacturing the product at different production shops is given by:
Shop Variable Cost CCost Fixed Cost A 14 7000 B 16 4000 C 15 5000
Find the optimal quantity to be supplied from each shop to different warehouses at minimum total cost. (10 marks)
(b) Two firms A and B make smart phones and tablets. Firm A can make either 150 smart phones in a
week or an equal number of tablets, and make a profit of Rs. 400 per smart phone and Rs. 300 per
tablet. Firm B can, on the other hand, make either 300 smart phones, or 150 smart phones and 150
tablets, or 300 tablets per week. It also has the same profit margin on the two products as A. Each
week there is a market of 150 smart phones and 300 tablets and the manufacturers would share
market in the proportion in which they manufacture a particular product. Write the payoff matrix
of A per week. Obtain graphically A’s and B’s optimal strategies and value of the game.
(10 marks)
1. A small project is composed of 7 activities whose time estimates are listed in the table below. Activities are identified by their beginning(i) and ending(j) node numbers.
Activity Estimated Duration (weeks) (i–j) Optimistic Most Likely Pessimistic 1-2 1 1 7 1-3 1 4 7 1-4 2 2 8 2-5 1 1 1 3-5 2 5 14 4-6 2 5 8 5-6 3 6 15
• Draw the network of the activities in the project (5 marks)
• Find the expected duration and variance for each activity. (5 marks)
• What are the expected project length and critical path? (5 marks)
• Calculate the standard deviation of the project length. (5 marks)
1. (a) The annual demand of a product is 10,000 units. Each unit costs Rs.100 if orders are placed in
quantities below 200 units but for orders of 200 or above the price is Rs. 95. The annual inventory
holding costs is 10% of the value of the item and the ordering cost is Rs. 5 per order. Find the
economic lot size. (8 marks)
(b) A dealer supplies you the following information with regard to a product dealt in by her: Annual
demand=10000 units; Ordering Cost=Rs 10/order; Price=Rs.20/unit; Inventory carrying cost=20%
of the value of the inventory per year. The dealer is considering the possibility of allowing some
backorder to occur. She has estimated that the annual cost of backordering will be 25% of the
value of inventory.
• What should be the optimum number of units of the product she should buy in one lot?
• What quantity of the product should be allowed to be backordered, if any?
• What would be the maximum quantity of inventory at any time of the year?
• Would you recommend her to allow backordering? If so, what would be the annual cost saving by adopting the policy of backordering? (12 marks)
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## Loyola College B.Sc. Mathematics Nov 2012 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
FIFTH SEMESTER – NOVEMBER 2012
# MT 5507/MT 5504 – OPERATIONS RESEARCH
Date : 06/11/2012 Dept. No. Max. : 100 Marks
Time : 9:00 – 12:00
PART – A
Answer any ALL questions: (10 x 2 = 20 Marks)
1. Define the following: (i) Basic solution (ii) Basic feasible solution
2. Express the following linear programming problem into standard form:
Maximize
Subject to
3
1. What is a transportation problem?
2. Give the mathematical formulation of an assignment problem.
3. Define a pure strategy in game theory.
4. Define a saddle point.
5. Define a spanning tree in a network.
6. Define a critical path in a network.
7. What is the Economic order quantity?
8. Differentiate the deterministic and the probabilistic demand inventory models.
PART – B
Answer any FIVE questions: (5 x 8 = 40 Marks)
1. Use the graphical method to solve the following linear programming problem.
Minimize
Subject to
1. Solve the following LPP by dual simplex method.
Maximize
Subject to
1. Determine an initial basic feasible solution to the following transportation problem by Vogel Approximation Method.
Available
Requirement 6 10 12 15
1. Solve the following assignment problem.
Jobs
I II III
Men
1. Solve the following game graphically
1. A project consists of a series of tables labeled A, B, …, H, I with the following relationships (W < X, Y means X&Y cannot start until W is completed; X, Y < W means W cannot start until both X&Y are completed). With this notations construct the network diagram having the following constraints: A < D, E; B, D < F; C < G; B < H; F, G < I
1. Determine the critical path of the following network.
1. A particular item has a demand of quantity 9000 units/year. The cost of the one procurement is Rs.100 and the holding cost per unit is Rs.2.40 per year. The replacement is instantaneous and no shortages are allowed. Determine
• the economic lot size
• the number of orders per year
• the time between orders
PART – C
Answer any TWO questions: (2 x 20 = 40 Marks)
1. a) Find the optimal solution for the following transportation problem using MODI method.
D1 D2 D3 D4 Supply
S1 19 30 50 10 7 S2 70 30 40 60 9 S3 40 8 70 20 18 Demand 5 8 7 14 34
1. Use the penalty (Big-M) method to solve the following LP problem. (10)
Minimize
Subject to
1. a) Define the Total float, free float and Independent float. (6)
1. b) The following indicates the details of the activities of a project.
The durations are in days. (14)
Activities TO TM TP 1 – 2 4 5 6 1 – 3 8 9 11 1 – 4 6 8 12 2 – 4 2 4 6 2 – 5 3 4 6 3 – 4 2 3 4 4 – 5 3 5 8
• Draw the network
• Find the critical path
• Find the mean and standard deviation of the project completion time
1. a) Reduce the following game to game and hence find the optimum strategies and the value
of the game. (12)
Player B
I II III IV I 3 2 4 0 II 3 4 2 4 III 4 2 4 0 IV 0 4 0 8
Player A
1. b) Solve the following unbalanced assignment problem of minimizing total time for doing all the jobs. (8)
Jobs Operators 1 2 3 4 5 1 6 2 5 2 6 2 2 5 8 7 7 3 7 8 6 9 8 4 6 2 3 4 5 5 9 3 8 9 7 6 4 7 4 6 8
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## Loyola College B.Sc. Computer Science April 2011 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – COMPUTER SCIENCE
FIFTH SEMESTER – APRIL 2011
# CS 5402 – OPERATIONS RESEARCH
Date : 12-04-2011 Dept. No. Max. : 100 Marks
Time : 9:00 – 12:00
PART-A
Answer ALL questions 10*2=20
1. Write the steps involved in L.P model formulation.
2. What is pivot element?
3. What is the difference between regular simplex method and the dual simplex method?
4. Define an assignment problem.
5. List out the methods of solving Transportation problem
1. What is a sequencing problem?
2. Define optimistic & pessimistic time.
1. What is dummy activity?
2. What is Holding Cost?
3. Write the formula to calculate EOQ for the system with constant demand rate.
PART-B
Answer ALL questions. 5*8=40
11 a) A company makes two kinds of leather belts. Belt A is a high quality belt, and belt
B is of lower quality. The respective profits are Rs.4 and Rs. 3 per belt. Each belt
of type A requires twice as much time as a belt of type B, and if all belts were of
type B, the company could make 1000 per day. The supply of leathers is sufficient
for only 800 belts per day (Both A and B combined). Belt A requires a fancy buckle
and only 400 per day are available. There are only 700 buckles a day available for
belt B. Formulate the mathematical model to determine the optimal product mix .
(OR)
11 b) Solve the following L.P.P graphically.
Max Z = 10 x1 +15x2
Subject to 2 x1 +x2 ≤ 26
2 x1 +4x2 ≤ 56
– x1 +x2 ≤ 5
x1 ,x2 ≥0
12 a) Construct the dual to the primal problem
Max Z = x1 +2x2 +3 x3
Subject to 3 x1 +x+ x3 ≤ 12
x1 +2x2 +4 x3 ≤ 20
2 x1 +5x2 x3 ≤ 18
x1 ,x2, x3 ≥0
(OR)
12 b) Find an initial allocation by Vogel’s approximation method for the following transportation problem
D E F G Availability A 11 13 17 14 250 B 16 18 14 10 300 C 21 24 13 10 400 Demand 200 225 275 250
13a) Five men are available to do five different jobs. From past records, the time (in hours) that each man take to do each job is known and given in the table
Jobs Men I II III IV V A 2 9 2 7 1 B 6 8 7 6 1 C 4 6 5 3 1 D 4 2 7 3 1 E 5 3 9 5 1
(OR)
13 b) Find the sequence that minimizes the total elapsed time required to complete the following tasks on two machines
Task : A B C D E F G H I
Machine I : 2 5 4 9 6 8 7 5 4
Machine II : 6 8 7 4 3 9 3 8 11
14a) A project consists of a series of activities called A,B,..,I with the following
relationship<X,Y means X and Y cannot start until W is completed with this
notation construct a network diagram having the following constraints.
A<D, A<E; B<F; D<F; C<G; C<H;F<I;G<I;
Job: A B C D E F G H I
Activity:8 10 8 10 16 17 18 14 9
(days)
(OR)
14 b) (i) Write down the difference between PERT & CPM. (4)
(ii) Define the following terms:
a)dummy activity b) Total float (4)
15a) Manufacture has to supply 600 units of his product/year. Shortages are not allowed
and storage cost amounts to Rs.0.60/unit/year. The set up cost/run is Rs.80.Find the
optimum run size and the minimum average yearly cost.
(OR)
15b) A commodity is to be supplied at the rate of 200 units per day. Ordering cost is Rs.50 and the holding cost is Rs.2 per day. The delay in supply induces penalty of Rs10 per unit per delay of one day. Find the optimal policy and reorder cycle period.
PART-C
Answer ANY TWO questions 2*20=40
16 a) Use Simplex method to solve the following L.P.P
Max Z = 5 x1 +3x2
Subject to x1 +x2 ≤ 2
5 x1 +2x2 ≤ 10
3x1 +8x2 ≤ 12
x1 ,x2 ≥0
1. b) Determine an initial basic feasible solution to the following transportation problem
by using (a) North west corner rule (b) Least cost method
Destination Source D1 D2 D3 D4 Supply S1 21 16 15 3 11 S2 17 18 14 23 13 S3 32 27 18 41 19 Demand 6 10 12 15
17 a) A marketing manager has 5 salesmen and 5 sales districts. Considering the capabilities of the salesman and the nature of districts, the marketing manager estimates that sales per month (in hundred rupees) for each salesman in each district would be as follows:
Salesman Sales District A B C D E 1 32 38 40 28 40 2 40 24 28 21 36 3 41 27 33 30 37 4 22 38 41 36 36 5 29 33 40 35 39
What is the maximum sale that may be expected if an optimum assignment is made?
17 b)Find the sequence that minimizes the total time required in performing the following
job on three machines in order ABC .A processing time (in hours) are given in the
following table.
Jobs :1 2 3 4 5
Machine A :8 10 6 7 11
Machine B :5 6 2 3 4
Machine C :4 9 8 6 5
18a) A stockiest has to supply 12,000 units of a product per year to his customer. The
demand is fixed and known and the shortage cost is assumed is to be infinite. The
inventory holding cost is Re.0.20 per unit per month and the ordering cost per order
is Rs.350. Determine the following
(i) The optimum lot size q0
(ii) Optimum scheduling period t0
(iii) Minimum total variable yearly cost
18b) A company uses annually 24,000 units of raw material which costs Rs1.25/unit
placing each order cost Rs.22.50 and the carrying cost is 5.4%/year of the average
inventory. Find the total cost including the cost of material.
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## Loyola College B.Sc. Computer Science April 2012 Operations Research Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – COMPUTER SCIENCE
FIFTH SEMESTER – APRIL 2012
# CS 5402/5503 – OPERATIONS RESEARCH
Date : 27-04-2012 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
PART-A
Answer ALL questions: 10 X 2=20
1. Write a General form of LPP.
2. What is optimal solution?
3. Why we should prefer dual problem to solve LPP?
4. Define traveling salesman problem.
5. List out the methods of solving Transportation problem.
6. Define Activity & Node
7. What is a sequencing problem?
8. What is setup cost?
9. What is reordering level?
10. When replacement is made?
PART-B
Answer All questions: 5 X 8=40
11 a) ) A Company produces refrigerators in unit I and heaters in unit II. The two
products are produced and sold in a weekly basis. Weekly production cannot
exceed 25 in unit I and 36 in unit II. Formulate this problem as an LP model
(OR)
11 b Solve the following l.p.p graphically.
Max Z = 10 x1 +15x2
Subject to 2 x1 +x2 ≤ 26
2 x1 +4x2 ≤ 56
– x1 +x2 ≤ 5
x1 ,x2 ≥0
12 a) Construct the dual to the primal problem
Minimum Z = x1 +x2 +x3
Subject to x1 –3x+4 x3 = 5
x1 -2x2 ≤ 3
2 x2 – x3 ≥ 4
x1 ,x2 ≥0 x3 unrestricted in sign.
(OR)
12 b) Obtain the initial solution of the following transportation problem by the north-west corner rule and matrix minima given that (i) the requirements are 40, 90 and 100 units and (ii) the supply are 90, 70 and 70.
Source Destination S1 S2 S3 D1 15 28 27 D2 24 24 25 D3 22 25 20
13 a) A department has five employees with five jobs to be performed . From past records, the time (in hours) that each man take to do each job is known and given in the table
Employee Jobs I II III IV V A 10 5 13 15 16 B 3 9 18 13 6 C 10 7 2 2 2 D 7 11 9 7 12 E 7 9 10 4 12
How should the jobs be allotted on per employee, so as to minimize the total number of hours
(OR)
13 b) Find the sequence that minimizes the total elapsed time required to complete the following tasks on two machines
Task : A B C D E F G H I
Machine I : 2 5 4 9 6 8 7 5 4
Machine II : 6 8 7 4 3 9 3 8 11
14 a) A project consists of a series of activities called A,B,..,I with the following relationship<X,Y means X and Y cannot start until W is completed with this notation construct a network diagram having the following constraints.
A<D,E; B,D<F; C<G; B<H; F,G<I;
Time: A B C D E F G H I
Activity:23 8 20 16 24 18 19 4 10
.
(OR)
14 b ) (i) Write down the difference between PERT & CPM. (4)
(ii) Define the following terms:
a)dummy activity b) total float .
15a) Manufacture has to supply 600 units of his product/year. Shortages are not allowed and storage cost amounts to Rs.0.60/unit/year.The set up cost/run is Rs.80.Determine(i) optimum run size (ii) the minimum average yearly cost.
(OR)
15 b) The annual demand for an item is 3200 units. The unit cost is Rs. 6/- and
inventory carrying charges 25% per annum. If the cost of one procurement is
Rs. 150/- determine (i) Economic order quality (ii) time between two
consecutive orders (iii) number of order per year (iv) the optimal total cost
PART-C
Answer any TWO 2 X 20=40
16 a) Use simplex method to solve the following L.P.P
Maximize Z = x1 +2x2+x3
Subject to 2 x1 +x2 – x3 ≤ 2
-2 x1 +x2 -5x3 ≥-6
4 x1 +x2 +x3 ≤ 6
x1 ,x2 ,x3≥0
1. b) Determine an initial basic feasible solution to the following transportation problem
by using (a) North west corner rule (b) Least cost method(c)Vogel’s
approximation.
Destination Source D1 D2 D3 D4 Supply S1 21 16 15 3 11 S2 17 18 14 23 13 S3 32 27 18 41 19 Demand 6 10 12 15
17 a) Find the sequence that minimizes the total time required in performing the following job on three machines in order ABC .A processing time (in hours) are given in the following table.
Jobs :1 2 3 4 5
Machine A :8 10 6 7 11
Machine B :5 6 2 3 4
Machine C :4 9 8 6 5
1. b) The project has the following time schedules.
Activity 1-2 1-6 2-3 2-4 3-5 4-5 6-7 5-8 7-8 t0 3 2 6 2 5 3 3 1 4 tm 6 5 12 5 11 6 9 4 19 tp 15 14 30 8 17 15 27 7 28
1. Draw the Project Network
2. Find the critical path.
.
18 a) A stockiest has to supply 12,000 units of a product per year to his customer. The
demand is fixed and known and the shortage cost is assumed is to be infinite. The
inventory holding cost is Re.0.20 per unit per month and the ordering cost per order
is Rs.350. Determine the following
(i) The optimum lot size q0
(ii) Optimum scheduling period t0
(iii) Minimum total variable yearly cost.
1. b) ) Given the following data:
Job 1 2 3 4 5 6 Machine A 12 10 9 14 7 9 Machine B 7 6 6 5 4 4 Machine C 6 5 6 4 2 4
Order of Processing : ACB
Determine the optimal sequence & the total elapsed time associated with it.
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© Copyright Entrance India - Engineering and Medical Entrance Exams in India | Website Maintained by Firewall Firm - IT Monteur | 20,388 | 72,217 | {"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} | 3.59375 | 4 | CC-MAIN-2023-14 | latest | en | 0.868746 |
http://slideplayer.com/slide/277335/ | 1,529,828,127,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866888.20/warc/CC-MAIN-20180624063553-20180624083553-00435.warc.gz | 289,808,131 | 22,365 | # Forces act in pairs.
## Presentation on theme: "Forces act in pairs."— Presentation transcript:
Forces act in pairs
Force engine = thrust friction = drag
support = gravity
Horizontal forces acting on a moving car thrust( force of the engine) drag( air resistance)
Forces acting on an aeroplane lift, drag, gravity, thrust L=W and T=D when the plane is flying at a constant velocity
First Law of Motion: Balanced Forces balanced forces = no change in velocity
A stationary object will never move as long as the forces acting on it are balanced.
Resultant force gives direction of acceleration
The forces acting on an object moving at a constant velocity(steady speed) are also balanced forces. If the object starts to accelerate or decelerate then the forces will no longer be equal or balanced.
What is happening to the helicopter in each picture?
Thrust equals weight. Helicopter hovers at a constant altitude.
Weight greater than thrust. Helicopter stays on the helipad or drops in altitude. Thrust is greater than weight. Helicopter takes off and increases in altitude. | 230 | 1,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-26 | longest | en | 0.904817 |
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What are digital elevation models, and why are they used so often in spatial analyses?
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2
How are digital elevation data created?
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3
Write the definition of slope and aspect, and the mathematical formulas used to derive them from digital elevation data
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4
11.4 - 11.9 page 511
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5
Plot a graph of slope in degrees (on x axis) against slope in percent (y axis). Which is usually larger, slope as degrees, or slope expressed as percent?
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What is an elevation contour?
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7
11.2 - 11.13
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8
What is the formula to calculate a contour height from two measured elevation points?
...
9
11.15 - 11.19
...
10
What are plan curvature and profile curvature, and how do they differ?
...
11
Define the following: solar zenith angle, solar azimuth angle, and solar incidence angle
...
12
draw a diagram illustrating the solar incidence angle, and identify what site/terrain factors affect the solar incidence angle
...
13
what are viewsheds, when are they used and how are they calculated?
...
14
What is a shaded relief map, also known as a hillshade surface? How are the values for each cell of the hillshade surface calculated?
... | 337 | 1,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-34 | latest | en | 0.898982 |
https://stats.stackexchange.com/questions/179900/optimizing-a-support-vector-machine-with-quadratic-programming | 1,660,067,286,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00039.warc.gz | 514,613,795 | 66,892 | # Optimizing a Support Vector Machine with Quadratic Programming
I'm trying to understand the process for training a linear support vector machine. I realize that properties of SMVs allow them to be optimized much quicker than by using a quadratic programming solver, but for learning purposes I'd like to see how this works.
## Training Data
set.seed(2015)
df <- data.frame(X1=c(rnorm(5), rnorm(5)+5), X2=c(rnorm(5), rnorm(5)+3), Y=c(rep(1,5), rep(-1, 5)))
df
X1 X2 Y
1 -1.5454484 0.50127 1
2 -0.5283932 -0.80316 1
3 -1.0867588 0.63644 1
4 -0.0001115 1.14290 1
5 0.3889538 0.06119 1
6 5.5326313 3.68034 -1
7 3.1624283 2.71982 -1
8 5.6505985 3.18633 -1
9 4.3757546 1.78240 -1
10 5.8915550 1.66511 -1
library(ggplot2)
ggplot(df, aes(x=X1, y=X2, color=as.factor(Y)))+geom_point()
## Finding the Maximum Margin Hyperplane
According to this Wikipedia article on SVMs, to find the maximum margin hyperplane I need to solve
$$\arg\min_{(\mathbf{w},b)}\frac{1}{2}\|\mathbf{w}\|^2$$ subject to (for any i = 1, ..., n) $$y_i(\mathbf{w}\cdot\mathbf{x_i} - b) \ge 1.$$
How do I 'plug' my sample data into a QP solver in R (for instance quadprog) to determine $\mathbf{w}$?
• You have to solve the dual problem
– user83346
Nov 3, 2015 at 18:08
• @fcop can you elaborate? What is the dual in this case? How do I solve using R? etc.
– Ben
Nov 3, 2015 at 18:24
HINT:
Quadprog solves the following:
\begin{align*} \min_x d^T x + 1/2 x^T D x\\ \text{such that }A^T x \geq x_0 \end{align*}
Consider $$x = \begin{pmatrix} w\\ b \end{pmatrix} \text{and } D=\begin{pmatrix} I & 0\\ 0 & 0 \end{pmatrix}$$
where $I$ is the identity matrix.
If $w$ is $p \times 1$ and $y$ is $n \times 1$:
\begin{align*} x &: (2p+1) \times 1 \\ D &: (2p+1) \times (2p+1) \end{align*}
On similar lines: $$x_0 = \begin{pmatrix} 1\\ 1 \end{pmatrix}_{n \times 1}$$
Formulate $A$ using the hints above to represent your inequality constraint.
• I'm lost. what is $d^T$?
– Ben
Nov 4, 2015 at 4:17
• What is the coefficient of $w$ in your objective function? Not $||w||^2_2$ but $w$? Nov 4, 2015 at 4:19
• Appreciate the help. I thought I figured this out but when I set D = the matrix you suggest quadprog returns the error "matrix D in quadratic function is not positive definite!"
– Ben
Nov 4, 2015 at 4:53
• HACK: Perturb $D$ by adding a small value say $1e-6$ on the diagonal Nov 4, 2015 at 4:59
Following rightskewed's hints...
library(quadprog)
# min(−dvec^T b + 1/2 b^T Dmat b) with the constraints Amat^T b >= bvec)
Dmat <- matrix(rep(0, 3*3), nrow=3, ncol=3)
diag(Dmat) <- 1
Dmat[nrow(Dmat), ncol(Dmat)] <- .0000001
dvec <- rep(0, 3)
Amat <- as.matrix(df[, c("X1", "X2")])
Amat <- cbind(Amat, b=rep(-1, 10))
Amat <- Amat * df\$Y
bvec <- rep(1, 10)
solve.QP(Dmat,dvec,t(Amat),bvec=bvec)
plotMargin <- function(w = 1*c(-1, 1), b = 1){
x1 = seq(-20, 20, by = .01)
x2 = (-w[1]*x1 + b)/w[2]
l1 = (-w[1]*x1 + b + 1)/w[2]
l2 = (-w[1]*x1 + b - 1)/w[2]
dt <- data.table(X1=x1, X2=x2, L1=l1, L2=l2)
ggplot(dt)+geom_line(aes(x=X1, y=X2))+geom_line(aes(x=X1, y=L1), color="blue")+geom_line(aes(x=X1, y=L2), color="green")+
geom_hline(yintercept=0, color="red")+geom_vline(xintercept=0, color="red")+xlim(-5, 5)+ylim(-5, 5)+
labs(title=paste0("w=(", w[1], ",", w[2], "), b=", b))
}
plotMargin(w=c(-0.5065, -0.2525), b=-1.2886)+geom_point(data=df, aes(x=X1, y=X2, color=as.factor(Y))) | 1,374 | 3,417 | {"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} | 3.484375 | 3 | CC-MAIN-2022-33 | longest | en | 0.696376 |
https://ch.mathworks.com/matlabcentral/cody/problems/54-maximum-running-product-for-a-string-of-numbers/solutions/2032236 | 1,576,128,764,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00428.warc.gz | 319,975,470 | 17,918 | Cody
# Problem 54. Maximum running product for a string of numbers
Solution 2032236
Submitted on 22 Nov 2019 at 16:37 by Asif Newaz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
s = '123454321'; i_correct = 3; assert(isequal(running_product(s),i_correct))
a = 9 b = 120 b = 120 480 b = 120 480 720 b = 120 480 720 480 b = 120 480 720 480 120 i = 3
2 Pass
s = '5820974944592307816406286208998628034825342117067'; i_correct = 28; assert(isequal(running_product(s),i_correct))
a = 49 b = 0 b = 0 0 b = 0 0 0 b = 0 0 0 0 b = 0 0 0 0 9072 b = 0 0 0 0 9072 4032 b = 0 0 0 0 9072 4032 2880 b = 0 0 0 0 9072 4032 2880 6480 b = 0 0 0 0 9072 4032 2880 6480 1440 b = 0 0 0 0 9072 4032 2880 6480 1440 1080 b = 0 0 0 0 9072 4032 2880 6480 1440 1080 0 b = 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 b = 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 b = 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 b = 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Column 16 1344 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 17 1344 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 18 1344 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 19 1344 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 20 1344 0 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 21 1344 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 22 1344 0 0 0 0 0 1152 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 23 1344 0 0 0 0 0 1152 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 24 1344 0 0 0 0 0 1152 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 25 1344 0 0 0 0 0 1152 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 26 1344 0 0 0 0 0 1152 0 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 27 1344 0 0 0 0 0 1152 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 28 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 29 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 Column 31 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 Columns 31 through 32 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 Columns 31 through 33 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 Columns 31 through 34 0 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 Columns 31 through 35 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 Columns 31 through 36 0 0 0 0 0 960 b = Columns 1 through 15 0 0 0 0 9072 4032 2880 6480 1440 1080 0 0 0 0 0 Columns 16 through 30 1344 0 0 0 0 0 1152 0 0 0 0 0 31104 7776 6912 Columns 31 through 37 0 0 0 0 0 960 960 b = Columns 1 through 15...
3 Pass
s = '141592653589793238462643383279502884197169399399999'; i_correct = 47; assert(isequal(running_product(s),i_correct))
a = 51 b = 180 b = 180 360 b = 180 360 540 b = 180 360 540 2700 b = 180 360 540 2700 1620 b = 180 360 540 2700 1620 900 b = 180 360 540 2700 1620 900 3600 b = 180 360 540 2700 1620 900 3600 5400 b = 180 360 540 2700 1620 900 3600 5400 7560 b = 180 360 540 2700 1620 900 3600 5400 7560 22680 b = 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 b = 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 b = 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 b = 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 b = 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Column 16 1152 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 17 1152 1152 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 18 1152 1152 2304 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 19 1152 1152 2304 1152 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 20 1152 1152 2304 1152 864 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 21 1152 1152 2304 1152 864 432 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 22 1152 1152 2304 1152 864 432 1728 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 23 1152 1152 2304 1152 864 432 1728 864 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 24 1152 1152 2304 1152 864 432 1728 864 432 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 25 1152 1152 2304 1152 864 432 1728 864 432 1008 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 26 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 27 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 28 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 29 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 Column 31 0 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 Columns 31 through 32 0 0 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 Columns 31 through 33 0 0 512 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 Columns 31 through 34 0 0 512 2304 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 Columns 31 through 35 0 0 512 2304 2016 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 Columns 31 through 36 0 0 512 2304 2016 252 b = Columns 1 through 15 180 360 540 2700 1620 900 3600 5400 7560 22680 13608 3402 1134 1296 576 Columns 16 through 30 1152 1152 2304 1152 864 432 1728 864 432 1008 3024 1890 0 0 0 Columns 31 through 37 0 0 512 2304 2016 252 378 b =...
4 Pass
s = '7831652712019091456485669234603486104543266482133936072602'; i_correct = 21; assert(isequal(running_product(s),i_correct))
a = 58 b = 1008 b = 1008 720 b = 1008 720 180 b = 1008 720 180 420 b = 1008 720 180 420 420 b = 1008 720 180 420 420 140 b = 1008 720 180 420 420 140 0 b = 1008 720 180 420 420 140 0 0 b = 1008 720 180 420 420 140 0 0 0 b = 1008 720 180 420 420 140 0 0 0 0 b = 1008 720 180 420 420 140 0 0 0 0 0 b = 1008 720 180 420 420 140 0 0 0 0 0 0 b = 1008 720 180 420 420 140 0 0 0 0 0 0 0 b = 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 b = 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Column 16 480 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 17 480 3840 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 18 480 3840 4800 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 19 480 3840 4800 5760 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 20 480 3840 4800 5760 5760 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 21 480 3840 4800 5760 5760 12960 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 22 480 3840 4800 5760 5760 12960 3240 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 23 480 3840 4800 5760 5760 12960 3240 1944 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 24 480 3840 4800 5760 5760 12960 3240 1944 1296 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 25 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 26 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 27 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 28 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 29 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 Column 31 576 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 Columns 31 through 32 576 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 Columns 31 through 33 576 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 Columns 31 through 34 576 0 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 Columns 31 through 35 576 0 0 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 Columns 31 through 36 576 0 0 0 0 0 b = Columns 1 through 15 1008 720 180 420 420 140 0 0 0 0 0 0 0 0 1080 Columns 16 through 30 480 3840 4800 5760 5760 12960 3240 1944 1296 1296 0 0 0 0 0 Columns 31 through 37 576 0 0 0 ...
5 Pass
s = '70066063155881748815209209628292540917153643678925903600113305305488'; i_correct = 44; assert(isequal(running_product(s),i_correct))
a = 68 b = 0 b = 0 0 b = 0 0 0 b = 0 0 0 0 b = 0 0 0 0 0 b = 0 0 0 0 0 0 b = 0 0 0 0 0 0 450 b = 0 0 0 0 0 0 450 600 b = 0 0 0 0 0 0 450 600 1600 b = 0 0 0 0 0 0 450 600 1600 1600 b = 0 0 0 0 0 0 450 600 1600 1600 2240 b = 0 0 0 0 0 0 450 600 1600 1600 2240 1792 b = 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 b = 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 b = 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Column 16 1280 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 17 1280 640 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 18 1280 640 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 19 1280 640 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 20 1280 640 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 21 1280 640 0 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 22 1280 640 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 23 1280 640 0 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 24 1280 640 0 0 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 25 1280 640 0 0 0 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 26 1280 640 0 0 0 0 0 0 0 0 1728 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 27 1280 640 0 0 0 0 0 0 0 0 1728 1728 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 28 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 29 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 Column 31 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 Columns 31 through 32 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 Columns 31 through 33 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 Columns 31 through 34 0 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 Columns 31 through 35 0 0 0 0 0 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 Columns 31 through 36 0 0 0 0 0 315 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 1792 1792 1792 Columns 16 through 30 1280 640 0 0 0 0 0 0 0 0 1728 1728 576 1440 720 Columns 31 through 37 0 0 0 0 0 315 105 b = Columns 1 through 15 0 0 0 0 0 0 450 600 1600 1600 2240 1792 17...
6 Pass
s = '11111'; i_correct = 1; assert(isequal(running_product(s),i_correct))
a = 5 b = 1 i = 1 | 8,683 | 16,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-51 | latest | en | 0.541339 |
https://www.shaalaa.com/question-bank-solutions/fundamental-theorem-arithmetic-consider-number-12-n-where-n-natural-number-check-whether-there-any-value-n-n-which-12-n-ends-digital-zero_5575 | 1,542,797,987,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00134.warc.gz | 982,045,148 | 14,495 | Account
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# Solution - Consider the number 12^n where n is a natural number. Check whether there is any value of n ∈ N for which 12^n ends with the digital zero. - CBSE Class 10 - Mathematics
#### Question
Consider the number 12n where n is a natural number. Check whether there is any value of n ∈ N for which 12n ends with the digital zero.
#### Solution
We know if any number ends with the digit zero it is always divisible by 5.
If 12n ends with the digit zero, it must divisible by 5.
This is possible only if prime factorisation of 12n contains the prime number 5.
Now, 12 = 2 x 2 x 3 = 22 x 3
⇒ 12n = (22 x 3)n = 22n x 3n
i.e., prime factorisation of 12n does not contain the prime number 5.
⇒ There is no value of n ∈ N for which 12n ends with the digit zero.
Is there an error in this question or solution?
#### Reference Material
Solution for question: Consider the number 12^n where n is a natural number. Check whether there is any value of n ∈ N for which 12^n ends with the digital zero. concept: Fundamental Theorem of Arithmetic. For the course CBSE
S | 310 | 1,114 | {"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} | 3.96875 | 4 | CC-MAIN-2018-47 | latest | en | 0.799316 |
https://devhubby.com/thread/how-to-check-for-nan-in-numpy | 1,726,004,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00403.warc.gz | 184,512,239 | 25,824 | # How to check for nan in Numpy?
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by roberto , in category: Python , 2 years ago
How to check for nan in Numpy?
, 2 years ago
@roberto To check if any value in a Numpy array is nan, you can use the isnan() method. This method returns a new array with the same shape as the original array, but with boolean values indicating whether each element is nan or not. For example:
``` 1 2 3 4 5 6 7 8 9 10 ``` ```import numpy as np # Create an array with some nan values arr = np.array([1, 2, np.nan, 4, 5]) # Check for nan values using isnan() result = np.isnan(arr) # Output: [False, False, True, False, False] print(result) ```
Member
by kendrick , 2 years ago
@roberto you can use the any() method to check if any element in the array is nan. This method returns a single boolean value indicating whether any element in the array is nan. For example:
```1 2 3 4 5 6 7 8 9 ``` ```import numpy as np # Create an array with some nan values arr = np.array([1, 2, np.nan, 4, 5]) # Check if any value is nan using any() result = np.any(np.isnan(arr)) print(result) ```
This code will print True, indicating that the arr array contains at least one nan value. | 336 | 1,171 | {"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-2024-38 | latest | en | 0.625523 |
https://web2.0calc.com/questions/2e-3t-400-2 | 1,723,380,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00618.warc.gz | 476,972,436 | 5,456 | +0
# 2e 3t = 400 |:2
-1
2720
1
+12529
2e3t = 400 |:2
e3t = 200
lne2t =ln200
2tlne=ln200 lne=1
2t =ln200 |:2
t=ln200/2
t=2,649158683...
Apr 7, 2017 | 99 | 165 | {"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-2024-33 | latest | en | 0.153714 |
https://www.coursehero.com/file/6518873/Lecture10-Subplots-Roundofferror/ | 1,524,627,034,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00223.warc.gz | 757,441,614 | 197,127 | {[ promptMessage ]}
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Lecture+10--Subplots+_+Roundoff+error
# Lecture+10--Subplots+_+Roundoff+error - Passing Functions...
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Passing Functions to M-File Example: to solve the system of ODEs * Sol: create a function rigid containing the equations function dy = rigid(t,y) dy = zeros(3,1); % a column vector dy(1) = y(2)*y(3); dy(2) = -y(1)*y(3); dy(3) = -0.51*y(1)*y(2); * Use error tolerances using the odeset command solve on a time interval [0 12] with an initial condition vector y= [0 1 1] at time 0.
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Passing Functions to M-File >>options = odeset('RelTol',1e-4,'AbsTol',[1e-4 1e-4 1e-5]); >>[T,Y] = ode45( @rigid ,[0 12],[0 1 1],options); * Plotting the columns of the returned array Y vs T shows the solution >>plot(T,Y(:,1),'-',T,Y(:,2),'-.',T,Y(:,3),'.') 0 2 4 6 8 10 12 -1.5 -1 -0.5 0 0.5 1 1.5
Bungee Jumper: Euler’s Method function [x, y] = Euler( f , tspan, y0, n) % solve y' = f(x,y) with initial condition y(a) = y0 % using n steps of Euler's method; step size h = (b-a)/n % tspan = [a, b] a = tspan(1); b = tspan(2); h = (b - a) / n; x = (a+h : h : b); y(1) = y0 + h*feval(f, a, y0); for i = 2 : n y(i) = y(i-1) + h *feval( f, x(i-1), y(i-1)); end x = [ a x ]; y = [y0 y ]; function f = bungee_f( t,v) % Solve dv/dt = f for bungee jumper velocity g = 9.81; m = 68.1; cd = 0.25; f = g - cd* v^2/m; Use “ feval ” for function evaluation 2 1 1 ( , ) ( ) ( ) ( , )( ) d i i i i i i c dv g v f v t dt m v t v t f v t t t + + = - = + -
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MATLAB M-File: Bungee Jumper
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{[ snackBarMessage ]} | 697 | 1,878 | {"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-2018-17 | latest | en | 0.575081 |
https://kidsworksheetfun.com/parallel-lines-and-transversals-worksheet-answer-key-with-work/ | 1,721,856,818,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00020.warc.gz | 302,372,027 | 26,909 | # Parallel Lines And Transversals Worksheet Answer Key With Work
Parallel lines and transversals unit vocabulary assignment and puzzles this is a introductory vocabulary assignment for a unit on parallel lines and transversals. In the diagram on the next page line t is a transversal of lines q and r.
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Parallel lines and transversals worksheet answer key with work. Then make a conjecture about their angle measures. Answer sheet for parallel cut by a transversal displaying top 8 worksheets found for this concept. Word problems on sets and venn diagrams.
Identify the pairs of angles in the diagram. Some of the worksheets for this concept are chapter 3 parallel lines and transversals answer pdf 3 parallel lines and transversals geometry answer key parallel lines and transversals pdf work section 3 1 parallel lines and transversals parallel lines and transversals answer key lesson. Your students will love working with these task cards.
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Use these 40 task cards with your students to help them practice solving word problems that include parallel lines cut by a transversal. 2 in the diagram below identify the relationship between each pair of angles. Time and work word problems.
Identify the pairs of angles in the diagram. Notice that line t forms a total of eight angles with lines q and r.
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https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Fluid_Mechanics_(Bar-Meir)/02%3A_Review_of_Thermodynamics/2.1%3A_Introductory_Remarks | 1,709,524,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00448.warc.gz | 238,606,417 | 28,285 | # 2.1: Introductory Remarks
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
In this chapter, a review of several definitions of common thermodynamics terms is presented. This introduction is provided to bring the student back to current place with the material. | 427 | 1,238 | {"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} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.319049 |
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# If ((x+4)^2)^1/2=3, which of the following could be the
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If ((x+4)^2)^1/2=3, which of the following could be the [#permalink]
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10 Dec 2012, 00:58
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If $$\sqrt{(x+4)^2} = 3$$, which of the following could be the value of x - 4?
A. -11
B. -7
C. -4
D. -3
E. 5
I am confused the way this was solved. Please provide your details solutions, post which I will post my specific question.
[Reveal] Spoiler: OA
Last edited by Bunuel on 10 Dec 2012, 01:07, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If ((x+4)^2)^1/2=3, which of the following could be the [#permalink]
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10 Dec 2012, 01:11
If $$\sqrt{(x+4)^2} = 3$$, which of the following could be the value of x - 4?
A. -11
B. -7
C. -4
D. -3
E. 5
$$\sqrt{(x+4)^2} = 3$$ --> $$(x+4)^2=9$$ --> $$x+4=3$$ or $$x+4=-3$$ --> $$x=-1$$ or $$x=-7$$ --> $$x-4=-5$$ or $$x-4=-11$$.
Hope it's clear.
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Re: If ((x+4)^2)^1/2=3, which of the following could be the [#permalink]
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10 Dec 2012, 01:33
Bunuel wrote:
If $$\sqrt{(x+4)^2} = 3$$, which of the following could be the value of x - 4?
A. -11
B. -7
C. -4
D. -3
E. 5
$$\sqrt{(x+4)^2} = 3$$ --> $$(x+4)^2=9$$ --> $$x+4=3$$ or $$x+4=-3$$ --> $$x=-1$$ or $$x=-7$$ --> $$x-4=-5$$ or $$x-4=-11$$.
Hope it's clear.
Bunuel, at 1st place why are you squaring both sides when we know that square root have a value of 1/2 which eventually cancels out the square.
\sqrt{(x+4)^2=3}
so this leaves me with x+4=3
and x=-1.
What is wrong with my approach. I know that I don't have an option of -5 in the answer but still my solution is correct.
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Re: If ((x+4)^2)^1/2=3, which of the following could be the [#permalink]
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10 Dec 2012, 01:43
davidfrank wrote:
Bunuel wrote:
If $$\sqrt{(x+4)^2} = 3$$, which of the following could be the value of x - 4?
A. -11
B. -7
C. -4
D. -3
E. 5
$$\sqrt{(x+4)^2} = 3$$ --> $$(x+4)^2=9$$ --> $$x+4=3$$ or $$x+4=-3$$ --> $$x=-1$$ or $$x=-7$$ --> $$x-4=-5$$ or $$x-4=-11$$.
Hope it's clear.
Bunuel, at 1st place why are you squaring both sides when we know that square root have a value of 1/2 which eventually cancels out the square.
\sqrt{(x+4)^2=3}
so this leaves me with x+4=3
and x=-1.
What is wrong with my approach. I know that I don't have an option of -5 in the answer but still my solution is correct.
That's not correct.
$$\sqrt{x^2} =|x|$$ (check here: if-x-3-4-and-y-2-5-what-is-the-value-of-110071.html#p880130), thus $$\sqrt{(x+4)^2} =|x+4|$$ not x+4.
From $$|x+4|=3$$ --> $$x=-1$$ or $$x=-7$$.
Hope it's clear.
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Re: If ((x+4)^2)^1/2=3, which of the following could be the [#permalink]
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10 Dec 2012, 03:44
Ans: Squaring both the sides we get (x+4)^2=9, therefore (x+4)=3,-3. So (x-4) could be -5 or -11. The answer is (A).
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Re: If ((x+4)^2)^1/2=3, which of the following could be the [#permalink]
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Re: If ((x+4)^2)^1/2=3, which of the following could be the [#permalink] 15 Oct 2014, 15:34
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At the MSRI in Berkeley, there is a marble sculpture by Helaman Ferguson showing Klein’s quartic surface.
This is a Riemann surface of genus 3 with 168 automorphisms. Our Euclidean brains have a hard time seeing all these. Let’s start with an automorphism of order 7, and a tiling of the plane by π/7 triangles:
Fourteen of them fit around a common vertex (at the center of our hyperbolic universe), and the black geodesic indicates how to identify edges of the green-yellow 14-gon (repeat the pattern by 2π/7 rotations). Euler will tell you that the identification space has genus 3. A little miracle is that these π/3 triangles fit nicely into a tiling by π/3 heptagons. This becomes evident like so:
The geodesic we used to indicate the 14-gon identification pattern becomes a geodesic in the heptagon tiling that passes through edge midpoints of eight consecutive heptagons, and all such geodesics will be closed on the identification space. This allows to define this surface also as an identification space of 24 heptagons (using the same geodesics). As this description is intrinsic to the heptagon tiling, it is invariant under all symmetries of that tiling, which include rotations of order 2 and order 3, in addition to the order 7 rotation.
Why is this surface called a quartic? Replacing the hyperbolic π/7 triangles with Euclidean (1,2,4)π/7 triangles in three different ways and keeping the identifications, we obtain three different translation structures on the Klein quartic, which define a basis of holomorphic 1-forms. Playing with their divisors show that these 1-forms satisfy the equation x³y+y³z+z³x, showing that the canonical curve of Klein’s surface is a quartic curve in the complex projective plane. | 421 | 1,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-30 | latest | en | 0.930472 |
https://it.scribd.com/doc/307588778/mult-intelligence-quiz | 1,596,933,762,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738366.27/warc/CC-MAIN-20200808224308-20200809014308-00207.warc.gz | 361,008,615 | 80,624 | Sei sulla pagina 1di 2
# Name ______________________________
## Getting To Know You Survey
Bodily-Kinesthetic
Word
Music
Art
## Body People Self
Intrapersonal
Visual-Spatial
Nature Math
Interpersonal
Musical-Rhythmic
0-5
Verbal-Linguistic
## Which of the following are true about you?
Naturalist
Fold the paper on the dark vertical line so that the eight columns
on the right are folded back. Then read each statement below.
Rate each statement from 0 to 5 according to how well the
description fits you (0 = Not at All to 5 = Very True) Next unfold
the paper and transfer each number over to the outlined block
on the same row. Finally, add the numbers in each column to find
the total score for each multiple intelligence area. The highest
possible score in one area is 15. How many ways are you smart?
Mathematical-Logical
Directions:
## I enjoy singing and I sing well.
I love crossword puzzles and other word games.
I like spending time by myself.
Charts, maps, and graphic organizers help me learn.
I learn best when I can talk over a new idea.
I enjoy art, photography, or doing craft projects.
I often listen to music in my free time.
I get along well with different types of people.
I often think about my goals and dreams for the future.
I enjoy studying about the earth and nature.
I enjoy caring for pets and other animals.
I love projects that involve acting or moving.
Written assignments are usually easy for me.
I can learn new math ideas easily.
I play a musical instrument (or would like to).
I am good at physical activities like sports or dancing.
I like to play games involving numbers and logic.
My best way to learn is by doing hands-on activities.
I love painting, drawing, or designing on the computer.
I often help others without being asked.
I enjoy being outside in all types of weather.
I love the challenge of solving a difficult math problem.
Having quiet time to think over ideas is important to me.
I read for pleasure every day.
Totals
2011 and 2015 ~ Laura Candler ~ Teaching Resources ~ www.lauracandler.com
Name ______________________________
Intrapersonal
Interpersonal
Bodily-Kinesthetic
Totals
Visual-Spatial
0-5
Musical-Rhythmic
## Which of the following are true about you?
Verbal-Linguistic
Fold the paper on the dark vertical line so that the eight columns
on the right are folded back. Then read each statement below.
Rate each statement from 0 to 5 according to how well the
description fits you (0 = Not at All to 5 = Very True) Next unfold
the paper and transfer each number over to the outlined block
on the same row. Finally, add the numbers in each column to find
the total score for each multiple intelligence area. The highest
possible score in one area is 15. How many ways are you smart?
Naturalist
Directions:
Mathematical-Logical
4
3
4
4
5
3
3
4
2
4
5
0
4
4
3
4
2
2
5
4
5
8
14
13
13
Nature Math
Word
Music
Art
11 | 712 | 2,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-34 | latest | en | 0.943339 |
https://ell.stackexchange.com/questions/290090/plural-or-single-at-velocities-or-at-velocity/290112 | 1,718,210,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00656.warc.gz | 204,201,711 | 36,588 | # Plural or single—at velocities or at velocity [duplicate]
Hypothetical Case
There are 2 types of cars that move at constant velocity on the road. The first type move at velocity 90 (km/hr), and the second type move at velocity 95 (km/hr).
How it is proper in the formal writing:
The cars move at velocities 90 (km/hr), and 95 (km/hr).
or
The cars move at velocity 90 (km/hr), and 95 (km/hr).
I don't know because they are 2 velocities–plural, but the second option seems more English style.
Edit
replace 'same' by 'constant' as
• The speed of a car depends on the driver - do you mean their maximum speeds? Jun 28, 2021 at 14:41
• Unless this is physics, we say: move at speeds of x and y, Jun 28, 2021 at 15:05
• If these were two velocities, then it would be better to say velocities than velocity in this context. However, these are speeds not velocities. The word velocity means something slightly different. Jun 29, 2021 at 5:34
• You mean speed, not velocity. Jun 29, 2021 at 9:47
That's a needless problem. When you say X km/h, that's a velocity. Just say
The cars move at 90 km/hr and 95 km/hr.
However, if you want to include the word velocity, a more idiomatic expression would be
The cars move at velocities of 90 and 95 km/hr.
Note that the units don't need to be repeated when the values are right next to each other.
• 1) Have to write velocity (vector) to emphisize velocity, because otherwise a student may choose speed (scalar) 2) Why 2 option is wrong? (The cars move at velocity 90 (km/hr), and 95 (km/hr))
– Ben
Jun 28, 2021 at 23:05
• If you use option 2, you should use plural "velocities". Omitting "of" is possible, but it's easier to read and understand with "of". Putting units in parentheses isn't correct, and repeating the units when they don't change between figures isn't necessary and decreases readability. As to point 1), yes, there's a difference between the scalar and vector quantities, but unless the element of direction is important in the statement, it might as well be omitted. Since a road is more or less one dimensional, if the cars are going the same direction, you might as well use "speed" or nothing. Jun 28, 2021 at 23:44
• On the other hand, on a linear road, if the cars are going in opposite directions, the "velocities" should be stated as 95 km/hr and -90 km/hr, or vice versa. Jun 28, 2021 at 23:45
• @Ben, if you are going to be pedantic, I have to say that 90 km/hr is not a velocity -- velocity is a vector. 90 km/hr is simply a speed. Jun 29, 2021 at 0:20
• If the velocities (rather than the speeds) are so important, why not tell the students the velocities? For example, "The cars move at velocities of 90 km/hr due east and 95 km/hr due north." Jun 29, 2021 at 11:48
What is clearest is to write
The first and second cars move at speeds of 90 and 95 km/hr respectively
It corresponds exactly to the mathematical notation
||v_1||= 90 km/hr and ||v_2|| = 95 km/hr
The so-called velocities that you are talking about are irrelevant because what is being discussed are speeds. There are no directional components involved. | 830 | 3,104 | {"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-2024-26 | latest | en | 0.946752 |
http://mathhelpforum.com/discrete-math/207031-reflexive-anti-symmetric-3.html | 1,527,019,422,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864872.17/warc/CC-MAIN-20180522190147-20180522210147-00310.warc.gz | 188,350,565 | 17,955 | 1. ## Vacuously True and Anti-symmetry
As reference for others who like me may have been unclear about the term "vacuously true" used in the thread:
(If A then B) is vacuously true if hypothesis A is false.
For example: Let E be the empty set and S any set. Then E is a subset of S:
(If x ϵ E then x ϵ S) is vacuously true because x ϵ E is false. (Reference for others who may also have been un)
Note the last 2 lines in the truth table from post #26 are vacuously true:
Originally Posted by Hartlw
Def: aRb is True if it is a member of R. aRb is False if it is not a member of R.
Def: R is anti-symmetric iff, for all (a,b)* belonging to R, the logical implication A→B is true, where A = (aRb and bRa) and B = (a=b).
A B A→B
T T T aRb and bRa and a=b
T F F
F T T aRb and a=b
F F T
R is anti-symmetric iff it is reflexive.
Found a clear, precise, non-trivial definition of anti-symmetry in Kelly (General Topology), pg 9: “R is anti-symmetric iff it is never the case that both xRy and yRx. “
* parentheses added to clarify. Pointed out by Deveno.
2. ## Re: Vacuously True and Anti-symmetry
Originally Posted by Hartlw
Found a clear, precise, non-trivial definition of anti-symmetry in Kelly (General Topology), pg 9: “R is anti-symmetric iff it is never the case that both xRy and yRx. “
Umm...Is there an echo in here? Yes, Hartlw. You got it.
-Dan
PS Of course you still have the possibility that xRx.
3. ## Re: Vacuously True and Anti-symmetry
Originally Posted by topsquark
Umm...Is there an echo in here? Yes, Hartlw. You got it.
Actually, I didn’t “get it.” This is “getting it:”
By definition of antisymmetric from first post, and standard definition (non math-logical) of “If Then” (→), R is antisymmetric if aRb and bRa and a=b.
The notion that one has to regard Truth Function concepts and Truth Tables from mathematical logic in ordinary and mathematical writing would make writing and reading virtually impossible. I caught myself doing it and thought, this is ridiculous. “If-Then “ is a fundamental, primitive, lanquage concept which occurrs repeatedly in any math text with a clear understanding of what it means. (If A then B) is true when A implies B and false when A is true and B is false. It has the force of y=f(x). Getting the same result for “antisymmetric” from a math logic point of view is interesting and forestalls an incomplete or vaque “math logic Proof” of a different conclusion.
Finally, showing that a so-called standard definiton is trivial, at the least, is worthy of note, and doesn’t deserve to get buried.
4. ## Re: Vacuously True and Anti-symmetry
Originally Posted by Hartlw
Finally, showing that a so-called standard definiton is trivial, at the least, is worthy of note, and doesn’t deserve to get buried.
I had promised myself to stay out of this. But when I saw your posting "Found a clear, precise, non-trivial definition of anti-symmetry in Kelly (General Topology), pg 9: “R is anti-symmetric iff it is never the case that both xRy and yRx. “
I simply did not think that could be correct.
John L Kelley belongs to one of the most important lines of 20th century topology. He was even the chair of mathematics at Berkeley.
So I cannot believe he would have written that, that is totally out of the mainstream, some have said it is even wrong. On that same page Kelley goes on to write that $\displaystyle \mathcal{R}\cap\mathcal{R}^{-1}=\emptyset$ if $\displaystyle \mathcal{R}$ is anti-symmetric.
But most standard set theory texts (see Charles Pinter) have this theorem:
If $\displaystyle \mathcal{R}$ is anti-symmetric then $\displaystyle \mathcal{R}\cap\mathcal{R}^{-1}\subseteq\Delta_A=\{(x,x):x\in A\}$.
So I think that you are very unwise to rely on Kelley's definition.
5. ## Re: Reflexive and Anti-symmetric
There goes my nice post #33 into oblivion. Sic Transit Gloria.
"On that same page Kelley goes on to write that .... if R is anti-symmetric." That's preciseley why it makes sense. And like a true master, his definition is consumateley understandable. Who says it's wrong.
"But most standard set theory texts (see Charles Pinter) have this theorem:" Based on what (intelligible) definition of anti-symmetry?
EDIT: Just occurred to me that 0 is a subset of what you have in brackets (don't know how to access it from here, and I can't copy the whole formula), ie, Kelly's def is consistent with Pinter's theorem, ie, Pinters Theorem does not prove Kelly wrong.
6. ## Re: Reflexive and Anti-symmetric
Originally Posted by Hartlw
"But most standard set theory texts (see Charles Pinter) have this theorem:" Based on what (intelligible) definition of anti-symmetry?
Here is an exact definition of anti-symmetry from Pinter.
$\displaystyle \mathcal{G}$ is a relation in $\displaystyle A$, then $\displaystyle \mathcal{G}$ is anti-symmetric if $\displaystyle (a,b)\in \mathcal{G}~\&~(b,a)\in \mathcal{G}$ then $\displaystyle a=b$.
I dare you to find a definition of anti-symmetry equivalent of Kelley's anywhere else.
7. ## Re: Reflexive and Anti-symmetric
Originally Posted by Plato
Here is an exact definition of anti-symmetry from Pinter.
$\displaystyle \mathcal{G}$ is a relation in $\displaystyle A$, then $\displaystyle \mathcal{G}$ is anti-symmetric if $\displaystyle (a,b)\in \mathcal{G}~\&~(b,a)\in \mathcal{G}$ then $\displaystyle a=b$.
I dare you to find a definition of anti-symmetry equivalent of Kelley's anywhere else.
Your (Pinter's) definition is the same one we have been using throughout this thread, which has conclusiveley been proved trivial, ie, R is reflexive iff R is anti-symmetric, ie, WRONG. See posts 26 & 33.
By the way, Pinters Set Theory (1971) is out of print and only a few very expensive copies are left. Standard?
8. ## Re: Reflexive and Anti-symmetric
Originally Posted by Hartlw
Your (Pinter's) definition is the same one we have been using throughout this thread, which has conclusiveley been proved trivial, ie, R is reflexive iff R is anti-symmetric, ie, WRONG. See posts 26 & 33.
By the way, Pinters Set Theory (1971) is out of print and only a few very expensive copies are left. Standard?
The Pinter is out of print. But it is widely referenced in later texts. He did his PhD in Paris with some of the most important figures in set theory. He did not get it wrong.
As for Kelley, on page 9 of General Topology he writes "$\displaystyle \mathcal{R}$ is anti-symmetric iff it is never the case that both $\displaystyle x\mathcal{R}y\text{ and }y\mathcal{R}y$" Had he stopped there, the the Moore tradition of precise language use would have made that a subset of Pinter's definition( the and implies two). However, he did not stop there. The very next sentence is "In other words, $\displaystyle \mathcal{R}$ is anti-symmetric iff $\displaystyle \mathcal{R}\cap\mathcal{R}^{-1}$ is void".
Once again here is a standard example.
Let $\displaystyle A=\{1,2,3\}$ and define $\displaystyle \mathcal{R}=\{(1,2),(2,3),(3,3}\}$. Now using the standard definition $\displaystyle \mathcal{R}$ is anti-symetric, BUT $\displaystyle \mathcal{R}\cap\mathcal{R}^{-1}=\{(3,3)\}$. So Kelley's is not standard or even a subset of the standard.
9. ## Re: Reflexive and Anti-symmetric
Originally Posted by Hartlw
Your (Pinter's) definition is the same one we have been using throughout this thread, which has conclusiveley been proved trivial, ie, R is reflexive iff R is anti-symmetric, ie, WRONG. See posts 26 & 33.
By the way, Pinters Set Theory (1971) is out of print and only a few very expensive copies are left. Standard?
from: http://www.cse.unl.edu/~choueiry/S06...outNoNotes.pdf
$\displaystyle \text{A relation }R\ \text{on a set }A\ \text{is called } \emph{antisymmetric}\ \text{if}$
$\displaystyle \forall a,b \left[((a,b) \in R \wedge (b,a) \in R) \implies a=b \right]$
$\displaystyle \text{for all }a,b \in A$.
i urge you to read some of his comments. this definition is STANDARD in mathematics, it is the same one as Pinter gives, and can be found on Wikipedia as well.
this DOES NOT IMPLY that R is reflexive. i will say this again, for emphasis:
an anti-symmetric relation may, or may not, contain diagonal elements (such as (a,a)). the pairs: {(a,a), (a,a)} (the colors are just to highlight the fact that we have made two choices, here: first a, then a again) are the ONLY "symmetric pairs" {(b,a),(a,b)} that occur in R.
sometimes the term "strictly anti-symmetric" (or less commonly "asymmetric") is used for anti-symmetric and irreflexive relations. this is what Kelly is talking about (and what is happening when we replace "≤" with "<" or talk about PROPER subsets or divisors, instead of just subsets or divisors).
what YOU said:
Originally Posted by Hartlw
R is antisymmetric if aRb and bRa and a=b
is NOT THE SAME as:
R is antisymmetric if aRb and bRa IMPLIES a=b.
the truth-tables for "and" and "implies" are very different.
10. ## Re: Vacuously True and Anti-symmetry
Originally Posted by Hartlw
The notion that one has to regard Truth Function concepts and Truth Tables from mathematical logic in ordinary and mathematical writing would make writing and reading virtually impossible. I caught myself doing it and thought, this is ridiculous. “If-Then “ is a fundamental, primitive, lanquage concept which occurrs repeatedly in any math text with a clear understanding of what it means. (If A then B) is true when A implies B and false when A is true and B is false.
I assume that the last phrase also means that "if A, then B" is true when it is not the case that A is true and B is false, i.e., when A is false or B is true. But this is exactly the truth table definition of "if ... then ...". So, it's not hard after all. And without a clear understanding when "If A, then B" is true and when it is false it is impossible to do any mathematics. For example, it is impossible even to reason about the claim "For any integer n, if n is divisible by 4, then n is even."
11. ## Re: Reflexive and Anti-symmetric
Plato,
What Halmos, Quine, and others mean to say is:
R is antisymmetric if aRb and not bRa, or aRa.* Ex: { (3,1), (1,2), (1,1)}
What Kelly, Tarski, Kleene and others say is:
R is antisymmetric if aRb and not bRa. Ex: {(3,1), (1,2)}
Since they are definitions, which one is “right” depends on context in which they are used. There could be a problem if a proof invokes antisymmetry without specifying which definition.
* This is not the same as the definition R is antisymmetric if [aRb and bRa → a=b].
[aRb and bRa → a=b] is true if (aRb and bRa) is true or vacuously true if (aRb and bRa) is false. In either case a=b and reflexive = antisymmetric.
12. ## Re: Vacuously True and Anti-symmetry
Originally Posted by emakarov
And without a clear understanding when "If A, then B" is true and when it is false it is impossible to do any mathematics. For example, it is impossible even to reason about the claim "For any integer n, if n is divisible by 4, then n is even."
It's easy. (If A then B) is true when A true implies B true and false when A true implies B false. That is the standard format of any theorem in mathematics. If you assume A is true and can show B is true, the theorem is true. If you assume A is true and can show B is false, the theorem is False.
"For any integer n, if n is divisible by 4, then n is even," Depends on n, ie, (If Then) is meaningless in this case without further qualification.
13. ## Re: Reflexive and Anti-symmetric
Originally Posted by Hartlw
Plato,
What Halmos, Quine, and others mean to say is:
R is antisymmetric if aRb and not bRa, or aRa.* Ex: { (3,1), (1,2), (1,1)}
What Kelly, Tarski, Kleene and others say is:
R is antisymmetric if aRb and not bRa. Ex: {(3,1), (1,2)}
Since they are definitions, which one is “right” depends on context in which they are used. There could be a problem if a proof invokes antisymmetry without specifying which definition.
you were doing fine up to here
Originally Posted by Hartlw
* This is not the same as the definition R is antisymmetric if [aRb and bRa → a=b].
[aRb and bRa → a=b] is true if (aRb and bRa) is true or vacuously true if (aRb and bRa) is false. In either case a=b and reflexive = antisymmetric.
no, [aRb and bRa → a=b] is true IF (aRb and bRa) is false (whether or NOT a = b)
OR
if (aRb and bRa) is true AND a=b is true.
the case where (aRb and bRa) is true is really the "uninteresting" one, but it is ONLY in this case we may conclude a = b.
there are THREE ways in which (aRb and bRa) may fail to be true:
1. aRb is false, and bRa is false. in this case R tells us NOTHING about the relationship between a and b.
2. aRb is false, and bRa is true. (see next case)
3. aRb is true, and bRa is false.
in both of these cases (2&3), we cannot have a = b, for if we did, we would have: aRa is true, and aRa is false, a contradiction (by substituting a ( = b) for b in the two formulae aRb and bRa).
so in 3 out of the 4 cases for the possible truth-values of (aRb & bRa), a might not equal b. in 2 of them, a and b MUST be unequal. in one case (1) we can't tell, because it might be that:
a)aRa is false, so when a = b aRb and bRa are BOTH false
b)aRa is true, but aRb is false for some b different from a, and bRa is false for the same pair {a,b}.
c)aRa is false, aRb is false for some b distinct from a, and bRa is also false.
you also don't seem to really comprehend "vacuous truth". this is not the same as having a FALSE premise, it means that there are NO instances where the premise is true.
let me illustrate the difference: suppose our set is S = {1,2}.
here is relation R1 = {(1,2),(2,2)}
if we take a = 1, b = 2, the premise aR1b & bR1a is false (1(R1)2 is true, but 2(R1)1 is false). this is case (2) above, and we know from the (Halmos) definition of anti-symmetry that a cannot equal b. we do indeed, however, have an instance that makes aR1b and bR1a true, the pair (2,2).
here is R2 = { } (no elements at all).
here we have "vacuous truth", there aren't any elements (a,b) to test! the premise "doesn't apply". we can say with conviction R2 is "anti-symmetric" but that's not saying a whole lot. it's symmetric, too (it is, however, NOT reflexive, because neither (1,1) nor (2,2) are elements of R2).
14. ## Re: Reflexive and Anti-symmetric
Originally Posted by Hartlw
Def: aRb is True if it is a member of R. aRb is False if it is not a member of R.
Def: R is anti-symmetric iff, for all (a,b) belonging to R, the logical implication A→B is true, where A = (aRb and bRa) and B = (a=b).
A B A→B
T T T aRb and bRa and a=b
T F F
F T T aRb and a=b
F F T
R is anti-symmetric iff it is reflexive.
I see my mistake. I forgot to fill in the fourth line , so the truth table should be:
A B A→B
T T T aRb and bRa and a=b
T F F
F T T aRb and a=b
F F T aRb only and a unequal b.
So the definition "R is antisymmetric if [aRb and bRa imply b=a]" is only true if the implication is the Truth Table Function of Mathematical Lanquage. If that isn't specified the definition is not specified because in standard lanquage, math or otherwise, (If A then B) may or may not be true if A is false. It depends on A and B. So "if A then B" is only unequivocally true if A is true.
15. ## Re: Reflexive and Anti-symmetric
well, here's the thing: in logic, we can't assign a truth value of "i dunno". there's true, and there's false, and that's all that is on the menu.
yes, the statement A implies B is usually only MEANINGFUL to us (in ordinary language) if indeed A is true. if A ISN'T true, we can't really be certain of our "derivation" (or deduction) of B.
an example i have seen before is this one:
your professor says, "if you get an A on the final, i will give you an A for the course".
well, if you get an A on the final, and the prof gives you a B, it's clear he was lying.
and if you get an A on the final, and he gives you an A as a grade, he was being truthful.
but what if you get a B on the final? can you say with any certainty what your grade will be, even if you know the professor never lies?
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MCQ▾
### Question
A scored 5 points during a cricket game. B scored 8 times as many points as A. Which of these shows the total number of points that B scored?
### Choices
Choice (4) Response
a.
The difference between 8 and 5
b.
The sum of 5 and 8
c.
The difference between 22 and 15
d.
The product of 5 and 8
## Question number: 17
» Mathematics » Multiplication
MCQ▾
### Question
Karan wants to buy 5 pens. Each pens costs Rs. 10. If she uses a coupon for ’ Rs. 10 off’, how much will Karan owe for the 5 pens?
### Choices
Choice (4) Response
a.
Rs. 100
b.
Rs. 40
c.
Rs. 109
d.
Rs. 55
## Question number: 18
» Mathematics » Multiplication
MCQ▾
### Question
Mona bought two sandwiches and a small coke. The sandwich cost Rs. 20 each and the coke costs Rs. 15. How much did her meal cost?
### Choices
Choice (4) Response
a.
Rs. 55
b.
Rs. 85
c.
Rs. 65
d.
Rs. 70
## Question number: 19
» Mathematics » Multiplication
MCQ▾
### Question
Which of the following is equal to 100 50?
### Choices
Choice (4) Response
a.
(100 5)
b.
(20 30)
c.
(20 (70
d.
(50 20)
## Question number: 20
» Mathematics » Multiplication
MCQ▾
### Question
Rinki bought these erasers, each eraser cost Rs. 4. How much did the eraser cost all together?
### Choices
Choice (4) Response
a.
Rs. 52
b.
Rs. 48
c.
Rs. 50
d.
Rs. 56
f Page | 542 | 1,802 | {"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-2018-22 | latest | en | 0.818906 |
https://dlmf.nist.gov/13.23#E7 | 1,632,506,263,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057564.48/warc/CC-MAIN-20210924171348-20210924201348-00494.warc.gz | 253,993,959 | 14,533 | # §13.23 Integrals
## §13.23(i) Laplace and Mellin Transforms
For the notation see §§15.1, 15.2(i), and 10.25(ii).
13.23.1 $\int_{0}^{\infty}e^{-zt}t^{\nu-1}M_{\kappa,\mu}\left(t\right)\mathrm{d}t=\frac% {\Gamma\left(\mu+\nu+\tfrac{1}{2}\right)}{\left(z+\frac{1}{2}\right)^{\mu+\nu+% \frac{1}{2}}}\*{{}_{2}F_{1}}\left({\tfrac{1}{2}+\mu-\kappa,\tfrac{1}{2}+\mu+% \nu\atop 1+2\mu};\frac{1}{z+\frac{1}{2}}\right),$ $\Re\mu+\nu+\tfrac{1}{2}>0$, $\Re z>\tfrac{1}{2}$.
13.23.2 $\int_{0}^{\infty}e^{-zt}t^{\mu-\frac{1}{2}}M_{\kappa,\mu}\left(t\right)\mathrm% {d}t=\Gamma\left(2\mu+1\right)\left(z+\tfrac{1}{2}\right)^{-\kappa-\mu-\frac{1% }{2}}\*\left(z-\tfrac{1}{2}\right)^{\kappa-\mu-\frac{1}{2}},$ $\Re\mu>-\tfrac{1}{2}$, $\Re z>\tfrac{1}{2}$,
13.23.3 $\frac{1}{\Gamma\left(1+2\mu\right)}\int_{0}^{\infty}e^{-\frac{1}{2}t}t^{\nu-1}% M_{\kappa,\mu}\left(t\right)\mathrm{d}t=\frac{\Gamma\left(\mu+\nu+\frac{1}{2}% \right)\Gamma\left(\kappa-\nu\right)}{\Gamma\left(\frac{1}{2}+\mu+\kappa\right% )\Gamma\left(\frac{1}{2}+\mu-\nu\right)},$ $-\tfrac{1}{2}-\Re\mu<\Re\nu<\Re\kappa$.
13.23.4 $\int_{0}^{\infty}e^{-zt}t^{\nu-1}W_{\kappa,\mu}\left(t\right)\mathrm{d}t=% \Gamma\left(\tfrac{1}{2}+\mu+\nu\right)\Gamma\left(\tfrac{1}{2}-\mu+\nu\right)% \*{{}_{2}{\mathbf{F}}_{1}}\left({\tfrac{1}{2}-\mu+\nu,\tfrac{1}{2}+\mu+\nu% \atop\nu-\kappa+1};\tfrac{1}{2}-z\right),$ $\Re\left(\nu+\tfrac{1}{2}\right)>|\Re\mu|$, $\Re z>-\tfrac{1}{2}$,
13.23.5 $\int_{0}^{\infty}e^{\frac{1}{2}t}t^{\nu-1}W_{\kappa,\mu}\left(t\right)\mathrm{% d}t=\frac{\Gamma\left(\frac{1}{2}+\mu+\nu\right)\Gamma\left(\frac{1}{2}-\mu+% \nu\right)\Gamma\left(-\kappa-\nu\right)}{\Gamma\left(\frac{1}{2}+\mu-\kappa% \right)\Gamma\left(\frac{1}{2}-\mu-\kappa\right)},$ $|\Re\mu|-\tfrac{1}{2}<\Re\nu<-\Re\kappa$.
13.23.6 $\frac{1}{\Gamma\left(1+2\mu\right)2\pi\mathrm{i}}\int_{-\infty}^{(0+)}e^{zt+% \frac{1}{2}t^{-1}}t^{\kappa}M_{\kappa,\mu}\left(t^{-1}\right)\mathrm{d}t=\frac% {z^{-\kappa-\frac{1}{2}}}{\Gamma\left(\frac{1}{2}+\mu-\kappa\right)}I_{2\mu}% \left(2\sqrt{z}\right),$ $\Re z>0$.
13.23.7 $\frac{1}{2\pi\mathrm{i}}\int_{-\infty}^{(0+)}e^{zt+\frac{1}{2}t^{-1}}t^{\kappa% }W_{\kappa,\mu}\left(t^{-1}\right)\mathrm{d}t=\frac{2z^{-\kappa-\frac{1}{2}}}{% \Gamma\left(\frac{1}{2}+\mu-\kappa\right)\Gamma\left(\frac{1}{2}-\mu-\kappa% \right)}K_{2\mu}\left(2\sqrt{z}\right),$ $\Re z>0$.
For additional Laplace and Mellin transforms see Erdélyi et al. (1954a, §§4.22, 5.20, 6.9, 7.5), Marichev (1983, pp. 283–287), Oberhettinger and Badii (1973, §1.17), Oberhettinger (1974, §§1.13, 2.8), and Prudnikov et al. (1992a, §§3.34, 3.35). Inverse Laplace transforms are given in Oberhettinger and Badii (1973, §2.16) and Prudnikov et al. (1992b, §§3.33, 3.34).
## §13.23(ii) Fourier Transforms
13.23.8 $\frac{1}{\Gamma\left(1+2\mu\right)}\int_{0}^{\infty}\cos\left(2xt\right)e^{-% \frac{1}{2}t^{2}}t^{-2\mu-1}M_{\kappa,\mu}\left(t^{2}\right)\mathrm{d}t=\frac{% \sqrt{\pi}e^{-\frac{1}{2}x^{2}}x^{\mu+\kappa-1}}{2\Gamma\left(\frac{1}{2}+\mu+% \kappa\right)}W_{\frac{1}{2}\kappa-\frac{3}{2}\mu,\frac{1}{2}\kappa+\frac{1}{2% }\mu}\left(x^{2}\right),$ $\Re\left(\kappa+\mu\right)>-\tfrac{1}{2}$.
For additional Fourier transforms see Erdélyi et al. (1954a, §§1.14, 2.14, 3.3) and Oberhettinger (1990, §§1.22, 2.22).
## §13.23(iii) Hankel Transforms
For the notation see §10.2(ii).
13.23.9 $\int_{0}^{\infty}e^{-\frac{1}{2}t}t^{\mu-\frac{1}{2}(\nu+1)}M_{\kappa,\mu}% \left(t\right)J_{\nu}\left(2\sqrt{xt}\right)\mathrm{d}t=\frac{\Gamma\left(1+2% \mu\right)}{\Gamma\left(\frac{1}{2}-\mu+\kappa+\nu\right)}\*e^{-\frac{1}{2}x}x% ^{\frac{1}{2}(\kappa-\mu-\frac{3}{2})}\*M_{\frac{1}{2}(\kappa+3\mu-\nu+\frac{1% }{2}),\frac{1}{2}(\kappa-\mu+\nu-\frac{1}{2})}\left(x\right),$ $x>0$, $-\tfrac{1}{2}<\Re\mu<\Re\left(\kappa+\tfrac{1}{2}\nu\right)+\tfrac{3}{4}$,
13.23.10 $\frac{1}{\Gamma\left(1+2\mu\right)}\int_{0}^{\infty}e^{-\frac{1}{2}t}t^{\frac{% 1}{2}(\nu-1)-\mu}M_{\kappa,\mu}\left(t\right)J_{\nu}\left(2\sqrt{xt}\right)% \mathrm{d}t=\frac{e^{-\frac{1}{2}x}x^{\frac{1}{2}(\kappa+\mu-\frac{3}{2})}}{% \Gamma\left(\frac{1}{2}+\mu+\kappa\right)}\*W_{\frac{1}{2}(\kappa-3\mu+\nu+% \frac{1}{2}),\frac{1}{2}(\kappa+\mu-\nu-\frac{1}{2})}\left(x\right),$ $x>0$, $-1<\Re\nu<2\Re\left(\mu+\kappa\right)+\tfrac{1}{2}$.
13.23.11 $\int_{0}^{\infty}e^{\frac{1}{2}t}t^{\frac{1}{2}(\nu-1)-\mu}W_{\kappa,\mu}\left% (t\right)J_{\nu}\left(2\sqrt{xt}\right)\mathrm{d}t=\frac{\Gamma\left(\nu-2\mu+% 1\right)}{\Gamma\left(\frac{1}{2}+\mu-\kappa\right)}\*e^{\frac{1}{2}x}x^{\frac% {1}{2}(\mu-\kappa-\frac{3}{2})}\*W_{\frac{1}{2}(\kappa+3\mu-\nu-\frac{1}{2}),% \frac{1}{2}(\kappa-\mu+\nu+\frac{1}{2})}\left(x\right),$ $x>0$, $\max(2\Re\mu-1,-1)<\Re\nu<2\Re\mu-\kappa+\tfrac{3}{2}$,
13.23.12 $\int_{0}^{\infty}e^{-\frac{1}{2}t}t^{\frac{1}{2}(\nu-1)-\mu}W_{\kappa,\mu}% \left(t\right)J_{\nu}\left(2\sqrt{xt}\right)\mathrm{d}t=\frac{\Gamma\left(\nu-% 2\mu+1\right)}{\Gamma\left(\frac{3}{2}-\mu-\kappa+\nu\right)}\*e^{-\frac{1}{2}% x}x^{\frac{1}{2}(\mu+\kappa-\frac{3}{2})}\*M_{\frac{1}{2}(\kappa-3\mu+\nu+% \frac{1}{2}),\frac{1}{2}(\nu-\mu-\kappa+\frac{1}{2})}\left(x\right),$ $x>0$, $\max(2\Re\mu-1,-1)<\Re\nu$.
For additional Hankel transforms and also other Bessel transforms see Erdélyi et al. (1954b, §8.18) and Oberhettinger (1972, §1.16 and 3.4.42–46, 4.4.45–47, 5.94–97).
## §13.23(iv) Integral Transforms in terms of Whittaker Functions
Let $f(x)$ be absolutely integrable on the interval $[r,R]$ for all positive $r, $f(x)=O\left(x^{\rho_{0}}\right)$ as $x\to 0+$, and $f(x)=O\left(e^{-\rho_{1}x}\right)$ as $x\to+\infty$, where $\rho_{1}>\frac{1}{2}$. Then for $\mu$ in the half-plane $\Re\mu\geq\mu_{1}>\max\left(-\rho_{0},\Re\kappa-\frac{1}{2}\right)$
13.23.13 $\displaystyle g(\mu)$ $\displaystyle=\frac{1}{\Gamma\left(1+2\mu\right)}\int_{0}^{\infty}f(x)x^{-% \frac{3}{2}}M_{\kappa,\mu}\left(x\right)\mathrm{d}x,$ 13.23.14 $\displaystyle f(x)$ $\displaystyle=\frac{1}{\pi\mathrm{i}\sqrt{x}}\int_{\mu_{1}-\mathrm{i}\infty}^{% \mu_{1}+\mathrm{i}\infty}\mu g(\mu)\Gamma\left(\tfrac{1}{2}+\mu-\kappa\right)W% _{\kappa,\mu}\left(x\right)\mathrm{d}\mu.$
For additional integral transforms see Magnus et al. (1966, p. 189), Prudnikov et al. (1992b, §§4.3.39–4.3.42), and Wimp (1964).
## §13.23(v) Other Integrals
Additional integrals involving confluent hypergeometric functions can be found in Apelblat (1983, pp. 388–392), Erdélyi et al. (1954b), Gradshteyn and Ryzhik (2000, §7.6), and Prudnikov et al. (1990, §§1.13, 1.14, 2.19, 4.2.2). See also (13.16.2), (13.16.6), (13.16.7). Generalized orthogonality integrals (33.14.13) and (33.14.15) can be expressed in terms of Whittaker functions via the definitions in that section. | 3,222 | 6,679 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 158, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-39 | latest | en | 0.282151 |
https://energychampbtv.com/qa/quick-answer-can-bank-gives-100-percent-home-loan.html | 1,618,240,048,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00399.warc.gz | 301,885,445 | 8,711 | # Quick Answer: Can Bank Gives 100 Percent Home Loan?
## How do I buy a house with no money?
A no down payment mortgage allows first-time home buyers and repeat home buyers to purchase property with no money required at closing except standard closing costs.
Other options, including the FHA loan, the HomeReady™ mortgage and the Conventional 97 loan offer low down payment options with a little as 3% down..
## How much loan I can get if my salary is 60000?
If you take a personal loan for a maximum of 5 years, then your loan amount will be ₹ 36,000*12*5 = ₹ 21,60,000. However, the multiplier is 20, then the loan amount will be ₹ 60,000*20 = ₹ 12,00,000. Therefore, the amount you will get on ₹ 60,000 salary is ₹ 12,00,000.
## What is the maximum limit of home loan?
Your Home Loan Eligibility will be calculated after deductions of the EMIs that you are paying. Generally, the banks provide maximum upto 85% of loan against the value of property. Therefore, if you want a home loan for buying a property of Rs. 50 lakhs, the maximum amount you can get is 85% of that ie 42.50 lakhs.
## Do banks give 100 percent bonds?
“Banks are willing to approve 100% bonds if they can see that you have a clean credit history and can comfortably afford the monthly repayment instalments.
## Is it possible to buy a house with no deposit?
Yes, you can, but you will need a guarantor. Most people who get no deposit loans are first home buyers who will live in the homes they purchase. Most lenders prefer these types of buyers as they usually pay their loans on time. … Some banks also offer 105% loans as long as you have a parent who can act as guarantor.
## What is the EMI for 20 lakhs home loan?
Housing Loan Interest CalculatorEMI for various home loan amounts15 years20 years₹ 16 Lakh₹ 14,159₹ 12,166₹ 20 Lakh₹ 17,698₹ 15,207₹ 25 Lakh₹ 22,123₹ 19,009₹ 30 Lakh₹ 26,547₹ 22,8111 more row
## How much loan I can get if my salary is 25000?
The take-home salary will determine the EMI amount you can afford and thus the total loan amount you can borrow. For instance, if your take-home salary is Rs. 25,000, you can avail as much as Rs. 18.64 lakh as a loan to purchase a home worth Rs.
## How much home loan can I get if my salary is 12000?
Salary of 12000, Am I eligible for Loan? Check here onlineProductSalaryMax. loan amountPersonal Loan1200065000Home Loan12000600000Car Loan12000230000Credit Card12000
## Can we get 100% home loan?
Thus 100% loan for purchase of home is not possible but as explained earlier in the article, upto 90% loan is possible based on specific guidelines. Don’t worry if you are short of cash for the down-payment of your dream home. A personal loan is just a click away with Money View loans app.
## How much loan can I get on 35000 salary?
If you are taking a home loan for 35,000 salary, you can get a maximum loan amount of Rs. 20,16,481 at say an 8.5% interest rate for a tenure of 20 years. In this situation, the home loan EMI amount you would pay is not more than Rs. 17,500.
## Which bank gives home loans easily?
Best Bank for Lowest Home Loan Interest Rate. Citibank banks are the best choice for home loans with their lowest interest on home loan starting from 6.75%. Axis bank, ICICI Bank, and Kotak Bank are the best banks for home loan as they have quick loan disbursal with low-interest rates.
## How do you qualify for a 100% mortgage?
100% Financing: The USDA Home Loan That’s why this loan type is also known as the rural development loan. To qualify, you have to have enough income to support your house payment, but not too much income. You have to be within limits set by USDA. You also must buy a home that is within USDA’s geographical boundaries.
## How much loan can I get if my salary is 1 lakh?
Home LoanBanksMinimum Income CriteriaMaximum Loan AmountLIC HFLFor Salaried: ₹15,000 per month For Self-employed: ₹2,00,000 per annumMinimum ₹1 lakh to 85% of total cost of the propertyState Bank of India (SBI)For Salaried: ₹1,20,000 (p.a.) For Self-employed: ₹2,00,000 (p.a.)Up to ₹2 crores3 more rows
## How much loan can I get if my salary is 15000?
Salary of 15000, Am I eligible for Loan? Check here onlineProductSalaryMin. loan amountPersonal Loan1500040000Home Loan15000300000Car Loan15000100000Credit Card15000
## How can I get maximum home loan from Bank?
Any one or a combination of these methods can help you improve your overall home loan eligibility significantly.Clear your existing loans. … Improve your CIBIL score. … Take joint loans. … Go for a step-up home loan. … Opt for a longer tenure. … Additional source of income. | 1,199 | 4,618 | {"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-2021-17 | latest | en | 0.952846 |
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# Simple Harmonic Motion Calculator
A kind of periodic motion in which the restoring force acting is directly proportional to the displacement and acts in the opposite direction to that of displacement is called as simple harmonic motion. The important factors associated with this oscillatory motion are the amplitude and frequency of the motion. Use the simple harmonic motion calculator to calculate the angular frequency, velocity, and acceleration known the input values frequency, displacement and time.
Hz
m
sec
m/s
mm/s2
A kind of periodic motion in which the restoring force acting is directly proportional to the displacement and acts in the opposite direction to that of displacement is called as simple harmonic motion. The important factors associated with this oscillatory motion are the amplitude and frequency of the motion. Use the simple harmonic motion calculator to calculate the angular frequency, velocity, and acceleration known the input values frequency, displacement and time.
#### Formula:
ω = 2πf v = Aω cos (ωt) a= - d ω2 sin (ωt) Where, ω = Angular Frequency f = Frequency v = Velocity A = Ampltude t = Time d = Displacement a = Acceleration
#### Example
Solve for acceleration, velocity and angular frequency using simple harmonic equations. Given :
Frequency = 10 Hz
Displacement = 8 mm
Time = 5 seconds
Solution:
ω = 2x3.14 x 10 | 288 | 1,377 | {"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} | 3.828125 | 4 | CC-MAIN-2023-40 | latest | en | 0.879161 |
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• Print publication year: 2016
• Online publication date: July 2016
# 4 - The counterexamples
## Summary
In this chapter, I will present two prima facie counterexamples to Jeffrey's PMCEU of the kind inspired by Newcomb's paradox, briefly describe a number of others of the same general kind and try to clarify their structure. Newcomb's paradox itself will be discussed in Chapter 8.
The prima facie counterexamples are best understood as establishing a prima facie conflict between PMCEU and another intuitively plausible and well-respected principle of choice which, though not very broadly applicable, is almost certainly true. The second principle is a revised version of Savage's principle of dominance that takes into account the possibility that the states are not act-independent. Henceforth, when referring to the principle of dominance (PDOM, for short), I shall be referring to the revised version, stated below.
PDOM will here be given a propositional formulation. That is, the acts, states and outcomes referred to in my statement of the principle will be propositions, à la Jeffrey, to the effect that the agent performs such and such an act, that such and such state of the world obtains and that such and such an outcome accrues to the agent, respectively. To state PDOM, I must first give the following definition: Where A and B are any available acts and {F 1, … Fr } is any set of mutually exclusive and collectively exhaustive decision-relevant propositions, A dominates B in the agent's preferences relative to {F 1, … Fr } if, and only if, D(A & F i ) > D(B & F i ), for i = 1, … r. PDOM states: If (i) some act, A, dominates every other act in your preferences relative to some partition and (ii) you believe that which act you perform does not causally affect which element of that partition is true, then do A.
Let the states and outcomes considered by the agent be as labeled in the section on Jeffrey in the previous chapter. Then a consequence of PDOM, a weaker version, relies on the following definition: For any acts A and B, A dominates B in the agent's preferences if, and only if, for any two (maximally specific) decision-relevant propositions of the forms A & S i & O j and B & S i & O j , D(A & S i , & O j ) > D(B & S i & O j ). | 525 | 2,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-34 | latest | en | 0.920174 |
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# 1.
Moist air exists at 40°C dry-bulb temperature, 20°C thermodynamic wet-bulb temperature, and 101.325 kPa pressure. Determine the humidity ratio, enthalpy, dew-point temperature, relative humidity, and specific volume.
Solution: Locate state point on the psychrometric chart at the intersection of 40°C dry-bulb temperature and 20°C thermodynamic wet-bulb temperature lines. Read humidity ratio W = 6.5 g (water)/kg (dry air). The enthalpy can be found by using two triangles to draw a line parallel to the nearest enthalpy line [60 kJ/kg (dry air)] through the state point to the nearest edge scale. Read h = 56.7 kJ/kg (dry air). Dew-point temperature can be read at the intersection of W = 6.5 g (water)/kg (dry air) with the saturation curve. Thus, td = 7°C. Relative humidity can be estimated directly. Thus, = 14%.
Specific volume can be found by linear interpolation between the volume lines for 0.88 and 0.90 m3/kg (dry air). Thus, v = 0.896 m3/kg (dry air). 2. Moist air, saturated at 2°C, enters a heating coil at a rate of 10 m3/s. Air leaves the coil at 40°C. Find the required rate of heat addition.
Solution:
The above figure schematically shows the solution. State 1 is located on the saturation curve at 2°C. Thus, h1 = 13.0 kJ/kg (dry air), W1 = 4.3 g (water)/kg (dry air), and v1 = 0.784 m3/kg (dry air). State 2 is located at the intersection of t = 40°C and W2 = W1 = 4.3 g (water)/kg (dry air). Thus, h2 = 51.6 kJ/kg (dry air). The mass flow of dry air is:
mda = 10 / 0.784 = 12.76 kg/s (dry air)
The rate of heat addition = 1q2 = 12.76 (51.6 – 13.0) = 492 kW
h2 = 29.3 – 29. Moist air at 30°C dry-bulb temperature and 50% rh enters a cooling coil at 5 m3/s and is processed to a final saturation condition at 10°C. W1 = 13.11] = 197 kW . From Table 2.5 kJ/kg (dry air) and W2 = 7. Solution: The above figure shows the schematic solution.0133 – 0.877 m3/kg (dry air).3 kJ/kg (dry air).00766) 42.70 [(64. Find the kW of refrigeration required.70 kg/s (dry air) The kW of refrigeration = 1q2 = 5.3 g (water)/kg (dry air).66 g (water)/kg (dry air).11 kJ/kg (water).877 = 5. and v1 = 0.5) – (0. State 2 is located on the saturation curve at 10°C. The mass flow of dry air is: mda = 5 / 0. State 1 is located at the intersection of t = 30°C and = 50%. Thus. h1 = 64. hw2 = 42.3. Thus.
858 m3/kg (dry air). and the values t3 = 19. mda1 = 2 / 0. .6°C found.284 kg/s (dry air) By comparing the proportion of the lines.25 / 0. State 3 is located. Using a ruler.25 m3/s of recirculated air at 25°C dry-bulb temperature and 50% rh.789 = 2. and v2 = 0. States 1 and 2 are located on the ASHRAE chart. Consequently. Solution: The above figure shows the schematic solution.4.742 times the length of entire line 1–2. Find the dry-bulb temperature and thermodynamic wet-bulb temperature of the resulting mixture. revealing that v1 = 0.858 = 7.789 m3/kg (dry air).5°C and t3 * = 14. the length of line segment 1–3 is 0.535 kg/s (dry air) mda 2 = 6. A stream of 2 m3/s of outdoor air at 4°C dry-bulb temperature and 2°C thermodynamic wet-bulb temperature is adiabatically mixed with 6. Therefore.
the enthalpy of the steam hg = 2691 kJ/kg (water). This second line is the condition line. Therefore. State 2 is established at the intersection of the condition line with the horizontal line extended from the saturation curve at 13°C (td2 = 13°C). From a steam properties table. Find the final dry-bulb temperature of the moist air and the rate of steam flow. Thus. Draw a second line parallel to the reference line and through the initial state point of the moist air. The required steam flow is.691 kJ/g (water). according to the psychrometric equation. m w = mda = (W2 – W1 ) = 2 × 1000 (0.5. Solution: The above figure shows the schematic solution.0093 – 0. The rate of dry airflow is 2 kg/s (dry air). the condition line on the psychrometric chart connecting States 1 and 2 must have a direction: h/ W = 2. Values of W2 and W1 can be read from the chart.0018) = 15. t2 = 21°C. First.0 kg/s (steam) . Moist air at 20°C dry-bulb and 8°C thermodynamic wet-bulb temperature is to be processed to a final dew-point temperature of 13°C by adiabatic injection of saturated steam at 110°C.691 kJ/g (water) The condition line can be drawn with the h/ W protractor. establish the reference line on the protractor by connecting the origin with the value h/ W = 2. | 1,368 | 4,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-47 | latest | en | 0.853953 |
https://crypto.stackexchange.com/questions/89446/why-do-problems-for-post-quantum-algorithms-have-to-be-np-hard/89451 | 1,716,522,690,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00601.warc.gz | 159,662,005 | 43,365 | # Why do Problems for Post-Quantum algorithms have to be NP-Hard?
The mathematical problems used for Post-Quantum Cryptography problems I came across, are NP-complete, e.g.
• Solving quadratic equations over finite fields
• short lattice vectors and close lattice vectors
• bounded distance decoding over finite fields
At least the general version of these is NP-complete
I am asking myself why these mathematical problems need to be NP-complete (in the general version) and how this is even useful, when instances are used that are not NP-complete
I am unaware of cryptography that is hard solely assuming that $$P\neq NP$$, so I believe you are misunderstanding something. I know the story best when it comes to lattices, so I'll discuss why lattices are solely "adjacent" to an $$NP$$-hard problem.
The story is rather simple, but also technical. Let $$\mathsf{LWE}[n, \sigma, q]$$ be the average-case LWE problem in dimension $$n$$, standard deviation $$\sigma$$, and moduli $$q$$. Regev's quantum worst-case to average-case reduction for LWE states that:
$$\mathsf{SIVP}_{\tilde{O}(nq/\sigma)} \leq \mathsf{LWE}[n, \sigma, q]$$
Where $$\mathsf{SIVP}_\gamma$$ is the short independent vectors problem (think of it like a generalization of the shortest vector problem to finding $$n > 1$$ short linearly independent vectors, where $$n$$ is the dimension of the lattice). Note that the $$\gamma$$ here is an approximation factor that is allowed in the problem.
What is the precise complexity of $$\mathsf{SIVP}_\gamma$$? This highly depends on the parameter $$\gamma$$, but the following should suffice for this post. It is known that $$\mathsf{SIVP}_{\tilde{\Omega}(\sqrt{n})}$$ is in $$\mathsf{AM}\cap co\mathsf{AM}$$. This implies that it is not $$\mathsf{NP}$$-hard unless the polynomial hierarchy collapses to some finite (it looks like 2nd?) level, which complexity theorists view as being unlikely.
This essentially means that while $$\mathsf{SIVP}_\gamma$$ is known to be NP-hard for some $$\gamma$$, the $$\gamma$$ used in lattice cryptography is such that we view it as extremely unlikely that $$\mathsf{SIVP}_\gamma$$ will be NP-hard. Still, for basic protocols one can generally take $$\gamma$$ to be some small polynomial, so $$\mathsf{SIVP}_\gamma$$ is "close" to an NP-hard problem, especially as the smallest approximation factor that we have polynomial-time algorithms for is something sub-exponential iirc.
More generally, NP hardness is the wrong thing to look at in cryptography. What people actually want is some notion of average-case hardness. In lattice cryptography one can formally connect that with worst-case hardness, but not every area in cryptography has such reductions. In areas that don't, the particular worst-case hardness of the problem is not very important --- a problem can have some instances that are very hard while still being bad for cryptography, as it may be hard to generate a hard instance. What is more important is specifying some plausible average-case hard distribution and examining the particular hardness of this.
• A minor quibble on an otherwise great answer: in the SIVP approx factor, the $O$ should be $\tilde{O}$ (or add a tiny $\epsilon$ to the exponent), and it should be mentioned that this reduction is quantum. Apr 17, 2021 at 19:29
1. They don't need to be: isogeny-based cryptography has no connection to any NP-complete problems, as far as I am aware.
2. Generally you want the underlying mathematical problem to be hard, and you can't get "harder" than NP, since (to be very imprecise) the secret key of a public-key cryptosystem acts like a "witness" for any hard problem you would want to solve. My guess for "why" would be that an NP-complete problem is a good place to start looking for hard problems, so historically that's where they came from.
3. For whether it's useful, I'm not sure there is a definitive answer. Certainly it makes me more confident in the schemes. As an example of why, all the best (generic) attacks against lattice crypto are based on just solving SVP. Unless P=NP I can be reasonably confident that these attacks can't get that much better, and it restricts the avenues of possible alternative attacks to, e.g., an LWE-specific attack that doesn't extend to solve SVP generally. Such an algorithm just feels weird, so that makes me more inclined to believe the scheme is generally secure.
4. This isn't exactly what you asked, but you might also wonder: why not use a scheme based on an NP-complete problem, rather than instances that are not NP-complete? The reason is that NP-complete problems are worst-case hard, not average-case hard. When you use a scheme, you want to be sure that your key, as an instance of some computational problem, is hard; it's no good if you have a guarantee that someone else's key is hard to break. Hence, we need to restrict to subclasses of NP-complete problems. (Lattices go from SVP to LWE; codes go from general decoding to decoding a specific family like Goppa codes; multivariate gets restricted to polynomials formed by some particular process like oil-and-vinegar). The way I think of it is that NP-complete problems must be versatile enough that any other problem can reduce to them, and that versatility creates a lot of easy instances of the problem. I've heard second-hand that there may be deep complexity theory reasons we shouldn't expect an average-case hard NP-complete problem to exist, but that's way beyond my understanding and.
### Problems for post-quantum algorithms do not need to be NP-hard.
The goal of post-quantum cryptography is for the cryptographic scheme to be secure against quantum computers.
For example, a cryptographic scheme in which the hardest part of the decryption algorithm is the factoring of integers is not post-quantum secure because Shor's algorithm for factoring numbers allows you to do the decryption with a cost that is a polynomial of the key size no matter how large the key size gets.
Likewise, many schemes based on elliptic curves can be decrypted by Shor's algorithm for finding discrete logarithms with again a polynomial cost.
However, there's many problems that are not known to be NP-hard for which there is still no efficient quantum algorithm!
As long as decrypting the cryptographic scheme requires some computation for which there is not yet any quantum algorithm that can do the computation efficiently, it can be considered a "post-quantum" scheme, regardless of whether or not there's an NP-hard problem involved. One example would be Supersingular isogeny key exchange which was invented in 2011 by the professor who taught my cryptography class (David Jao) and two of his colleagues. | 1,521 | 6,712 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.875567 |
https://boyslife.org/hobbies-projects/funstuff/13770/fun-games-to-play-in-the-car/comment-page-12/ | 1,603,206,685,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872746.20/warc/CC-MAIN-20201020134010-20201020164010-00611.warc.gz | 245,822,620 | 16,639 | Stuck in the backseat on a long drive? Here are some classic (and free!) games to help travel time go by faster.
Already a seasoned traveler? Take our Summer Vacation Hot Spot quiz
For long trips, the License Plate Game might be the thing. In this version, call out letters from the license plates you see, then make as many ridiculous phrases as possible. For example, AST = Aardvarks Singling Tenderly or After-School Trampolining. You get the idea.
## ALPHABET GAME
The Alphabet Game is good along roads with a lot of signs. Find a word with an “A,” call it out, then look for a word with “B” and so on. First one to the end of the alphabet wins. You might want to skip trying to find “Q” and “Z.” Or not.
## LINES AND DOTS
Use graph paper for a couple of travel games. With Lines and Dots, draw lines on the paper. Take turns drawing one side of a square at a time, trying to make boxes. When you complete a box, put your initials in it. Whoever makes the most boxes in 15 minutes or so wins.
## INFINITE TIC-TAC-TOE
Another game with graph paper is Infinite Tic-Tac-Toe. In this game, the first person to get five in a row wins, and there are no borders except for the edge of the page.
Share these summer vacation jokes for a break between games
## I SPY
In I Spy, pick out something you see nearby (that other players can also see), but keep what it is to yourself. Say, “I spy with my little eye …” and then provide one description of the object. For instance, with the windshield wipers you might say, “I spy something long” or “something black.” After each hint, players take a turn guessing what the object is.
## ROCK-PAPER-SCISSORS AND 20 QUESTIONS
Rock-Paper-Scissors and 20 Questions hardly need explaining. There are variations of these games, such as Foot-Cockroach-Nuclear Bomb, and 20 Questions about movies you’ve seen or books you’ve read, instead of some random object.
#### 10 Comments on 6 Fun Games to Play in the Car
1. Cool games
2. GILthechamp123 // January 2, 2018 at 8:53 pm // Reply
FYI don’t play bonk it…. very annoying!
3. these will be great for my road trip from Miami to Detroit.
4. HesitantKikoNiko // July 19, 2017 at 2:23 pm // Reply
These games are very cool!
5. HesitantKikoNiko // July 19, 2017 at 2:21 pm // Reply
Wow! These games kill time in the car big time!!!
6. you guys are to funny for me
7. Thanks! Since i am going on a train trip tommorow (it’s pretty much the same) and it has no wi-fi, I needed some games to play with my younger sister and mum.
8. I have a rock-paper-scissors version! Water Lightning Rubber. Rubber beats Lightning, Lightning beats water, water beats rubber
9. Alphabet Shopping game: Each person as they start “shopping” says, “I went to the store and bought…” They can buy anything in the world but it has to be in alphabetical order. And it continues with the next person relating what the previous person(s) have bought. As such: person 1: I went to the store and bought an aardvark. Person 2: I went to the store and bought an aardvark, and a banana. Person 3: I went to the store and bought an aardvark, a boat, and a chisel. Etc….. | 816 | 3,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-45 | latest | en | 0.921592 |
http://www.gurufocus.com/term/e10/SONC/E10/Sonic%2BCorporation | 1,485,073,324,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281419.3/warc/CC-MAIN-20170116095121-00207-ip-10-171-10-70.ec2.internal.warc.gz | 483,046,598 | 29,638 | Switch to:
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Sonic Corp (NAS:SONC)
E10
\$0.85 (As of Nov. 2016)
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. E10 is the average of the inflation adjusted earnings of a company over the past 10 years.
Sonic Corp's adjusted earnings per share data for the three months ended in Nov. 2016 was \$0.280. Add all the adjusted EPS for the past 10 years together and divide 10 will get our e10, which is \$0.85 for the trailing ten years ended in Nov. 2016.
As of today, Sonic Corp's current stock price is \$25.40. Sonic Corp's E10 for the quarter that ended in Nov. 2016 was \$0.85. Sonic Corp's Shiller P/E Ratio of today is 29.88.
During the past 13 years, the highest Shiller P/E Ratio of Sonic Corp was 44.40. The lowest was 8.82. And the median was 21.05.
Definition
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. When we calculate the today's Shiller P/E ratio of a stock, we use todays price divided by E10.
What is E10? How do we calculate E10?
E10 is the average of the inflation adjusted earnings of a company over the past 10 years. Lets use an example to explain.
If we want to calculate the E10 of Wal-Mart (WMT) for Dec. 31, 2010, we need to have the inflation data and the earnings from 2001 through 2010.
We adjusted the earnings of 2001 earnings data with the total inflation from 2001 through 2010 to the equivalent earnings in 2010. If the total inflation from 2001 to 2010 is 40%, and Wal-Mart earned \$1 a share in 2001, then the 2001s equivalent earnings in 2010 is \$1.4 a share. If Wal-Mart earns \$1 again in 2002, and the total inflation from 2002 through 2010 is 35%, then the equivalent 2002 earnings in 2010 is \$1.35. So on and so forth, you get the equivalent earnings of past 10 years. Then you add them together and divided the sum by 10 to get E10.
For example, Sonic Corp's adjusted earnings per share data for the three months ended in Nov. 2016 was:
Adj_EPS = Earnigns per Share / CPI of Nov. 2016 (Change) * Current CPI (Nov. 2016) = 0.28 / 241.353 * 241.353 = 0.280
Current CPI (Nov. 2016) = 241.353.
Sonic Corp Quarterly Data
201408 201411 201502 201505 201508 201511 201602 201605 201608 201611 per share eps 0.34 0.18 0.14 0.38 0.5 0.24 0.22 0.31 0.52 0.28 CPI 237.852 236.151 234.722 237.805 238.316 237.336 237.111 240.229 240.849 241.353 Adj_EPS 0.345 0.184 0.144 0.386 0.506 0.244 0.224 0.311 0.521 0.28 201202 201205 201208 201211 201302 201305 201308 201311 201402 201405 per share eps 0.03 0.24 0.24 0.11 0.06 0.26 0.21 0.14 0.07 0.3 CPI 227.663 229.815 230.379 230.221 232.166 232.945 233.877 233.069 234.781 237.9 Adj_EPS 0.032 0.252 0.251 0.115 0.062 0.269 0.217 0.145 0.072 0.304 200908 200911 201002 201005 201008 201011 201102 201105 201108 201111 per share eps 0.27 0.1 -0.01 0.18 0.07 0.12 0.07 -0.08 0.2 0.09 CPI 215.834 216.33 216.741 218.178 218.312 218.803 221.309 225.964 226.545 226.23 Adj_EPS 0.302 0.112 -0.011 0.199 0.077 0.132 0.076 -0.085 0.213 0.096 200702 200705 200708 200711 200802 200805 200808 200811 200902 200905 per share eps 0.09 0.31 0.33 0.22 0.15 0.28 0.33 0.12 0.14 0.27 CPI 203.499 207.949 207.917 210.177 211.693 216.632 219.086 212.425 212.193 213.856 Adj_EPS 0.107 0.36 0.383 0.253 0.171 0.312 0.364 0.136 0.159 0.305
Add all the adjusted EPS together and divide 10 will get our e10.
Explanation
If a company grows much fast than inflation, E10 may underestimate the company's earnings power. Shiller P/E Ratio can seem to be too high even the actual P/E is low.
For the Shiller P/E, the earnings of the past 10 years are inflation-adjusted and averaged. The result is used for P/E calculation. Since it looks at the average over the last 10 years, the Shiller P/E is also called PE10.
The Shiller P/E was first used by professor Robert Shiller to measure the valuation of the overall market. The same calculation is applied here to individual companies.
Sonic Corp's Shiller P/E Ratio of today is calculated as
Shiller P/E Ratio = Share Price / E10 = 25.40 / 0.85 = 29.88
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
During the past 13 years, the highest Shiller P/E Ratio of Sonic Corp was 44.40. The lowest was 8.82. And the median was 21.05.
Be Aware
Shiller P/E Ratio works better for cyclical companies. It gives you a better idea on the company's real earnings power.
Related Terms
Shiller P/E Ratio
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Sonic Corp Annual Data
Aug07 Aug08 Aug09 Aug10 Aug11 Aug12 Aug13 Aug14 Aug15 Aug16 e10 0.60 0.71 0.74 0.74 0.75 0.75 0.76 0.79 0.81 0.85
Sonic Corp Quarterly Data
Aug14 Nov14 Feb15 May15 Aug15 Nov15 Feb16 May16 Aug16 Nov16 e10 0.79 0.78 0.77 0.79 0.81 0.81 0.81 0.83 0.85 0.85
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,773 | 5,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-04 | longest | en | 0.93894 |
http://isomorphismes.tumblr.com/post/12382581136/alphabet-space-order-set-intelligibility-reading-mathema | 1,398,272,282,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223203235.2/warc/CC-MAIN-20140423032003-00203-ip-10-147-4-33.ec2.internal.warc.gz | 172,655,086 | 39,513 | ## Talking in Tuples
#### Mathematics is the *Most* Different Language.
In Can we make mathematics intelligible?, R P Boas jokes:
There is a test for identifying some of the future professional mathematicians at an early age. These are the students who instantly comprehend a sentence beginning “Let X be an ordered quintuple (a, T, πσ, 𝔅), where …”
I’ll try to explain what mathematicians mean when they write this way.
Letters
Think about a set containing the letters {A, B, C, D, E, F, G}. As written it’s equivalent to the set {F, C, E, G, A, B}, so the set doesn’t communicate the order information we know “should” go along with these letters. To express that, we should talk about the pair ( {A,B,C,D,E,F,G}, 𝓞) where 𝓞 is the ordering A < B < C < D < E < F < G.
Would it have been clearer if I’d pasted the definition of the ordering into the interior of the pair, instead of using 𝓞 as a shorthand? I’m not sure. Part of the way you have to learn to read mathematics papers is mentally substituting shorthands for definitions wherever they appear.
Since no one reading this has to look up the enumerative definition of the alphabet, let me just use the shorthand 𝓪 for the set containing each of the letters and 𝓞ʹ for the well-known ordering of the letters A < B < C … < X < Y < Z (remember, I already used 𝓞 so now I have to add a prime to differentiate this new, larger ordering). Now I can just write (𝓪, 𝓞ʹ) for the ordered alphabet.
The Next Letter
So what if I wanted to talk about “the letter after Q” ? Using the current pair (𝓪, 𝓞ʹ) this concept is undefined. In order to include “after” as a concept in the space I am developing, I need to expand the pair to a triple.
Now, how should I add in the concept of “after”? I could parsimoniously add only the +1 operation. But I may want to talk about “the fourth letter after Q” as well. Should that be four iterations of +1 (i.e., 𝔰∘𝔰∘𝔰∘𝔰 where 𝔰 is the successor function)? It will be annoying enough to write out a definition of 𝔰 that clearly states “C = B+1, S = R+1, ” and so on. Deary me. I wouldn’t want to have to enumerate that for +3, +13, and so on. I don’t have an infinite amount of either ink or patience.
I’ll leave it to function composition and just define the output (image) of the +1 operator 𝔰 as “The letter to the right of the input, under the ordering 𝓞ʹ.” That doesn’t sound formal enough to be correct, but I’ll stop there.
An Ordered Triple
Now “we have” a triple (𝓪, 𝓞ʹ, 𝔰) containing a set 𝓪, an order 𝓞ʹ, and a function 𝔰. This is still the alphabet we’re trying to talk about here, right? In fact I’m starting to doubt if even a triple is enough, because the triple doesn’t contain either < or ∘, and those are symbols I’ve used so they’d better belong to the universe.
So I’ll say the alphabet is defined as a quintuple (𝓪, 𝓞ʹ, 𝔰, <,) containing a set, an order, a unary function 𝔰, and two binary functions > and . Phew. Please tell me I’m done!
You know what, I just thought of something else. What about the letter before? Argh! It’s so simple (the alphabet) and yet so difficult (defining an appropriate k-tuple). Alrighty, I learned a shortcut in school for this: I’ll define an inverse operation 𝔰¹. And everybody knows what I mean before I say it so I’ll just stop here.
Now, consider the alphabet. The alphabet is defined as a quintuple (𝓪 , 𝓞ʹ , 𝔰 , < , , 𝔰 ⁻¹) . Or maybe I should say it’s a triple? (𝓪 , (𝓞ʹ, <) , (𝔰, ∘, 𝔰⁻¹) ). One has options.
I still haven’t captured everything we know about the alphabet. I couldn’t do Excel spread sheets with Z < AA < AB < AC < … < AZ < BA < BB < BC < … BZ < …. Nor could I take account of cryptographic rules where Z loops around: Z+1=A, and A < B < C < … X < Y < Z < A (←a non-wellfounded set). I didn’t include the Alphabet Song or the pronunciation of the letters (have you been saying “zed” or “zee”?), nor did I include vowel/consonant classification, rhyming info, or an IPA-style breakdown of the phonemes each letter can make (and in English there are many phonemes per letter). But I did include a bunch of known information about the alphabet into the logical universe .
So here’s the point of this example. Even to express a simple concept that everyone knows — the alphabet — as well as what are normally implicit mappings and relationships — you have to explicitly include those facts in a tuple to be logically complete.
Mathematics is a totally different language than English. It’s more different from English than is Mandarin, Pormpuraaw, Tagalog, Aymara, Farsi, or Pirahã. That means you can think different thoughts once you learn mathematics. You can fathom what was unfathomable. Conceive what was inconceivable. See what was invisible. It also means that learning to “speak” this way sounds very strange.
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10. isomorphismes posted this | 1,487 | 5,326 | {"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-2014-15 | latest | en | 0.898558 |
http://www.ck12.org/algebra/Graphs-in-the-Coordinate-Plane/?by=community | 1,493,579,647,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125841.92/warc/CC-MAIN-20170423031205-00380-ip-10-145-167-34.ec2.internal.warc.gz | 471,601,730 | 15,322 | # Graphs in the Coordinate Plane
## Graph functions using value tables or slope/intercept
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Graphs in the Coordinate Plane
Learn to graph points and functions in the coordinate plane.
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## The Coordinate Plane
Graph ordered pairs on a coordinate grid with four quadrants. Use your knowledge of ordered pairs to locate items on a map.
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## Relationships and Graphs
by Math Writing Team //at grade
Learn to graph points and functions in the coordinate plane.
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## Solution Sets and Graphs of Linear Equations
Use tables of values and rules to graph functions. Determine solutions to a linear function.
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## Graphs in the Coordinate Plane
Use tables of values and rules to graph functions.
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## The Coordinate Plane
by Victor Brice Hunt //at grade
Graph ordered pairs on a coordinate grid with four quadrants. Use your knowledge of ordered pairs to locate items on a map.
MEMORY METER
This indicates how strong in your memory this concept is
0 | 369 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-17 | latest | en | 0.861543 |
https://www.unitsconverters.com/en/Gr/Daa-To-Gr/Km2/Utu-7531-7533 | 1,696,386,453,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00299.warc.gz | 1,155,558,543 | 35,214 | Formula Used
1 Grains Per Square Centimeter = 1000 Grain Per Decare
1 Grains Per Square Centimeter = 1000000 Grain Per Square Kilometer
1 Grain Per Decare = 10000000 Grain Per Square Kilometer
gr/daa to gr/km² Conversion
The abbreviation for gr/daa and gr/km² is grain per decare and grain per square kilometer respectively. 1 gr/daa is 10000000 times bigger than a gr/km². To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including gr/daa to gr/km² conversion.
Grain Per Decare to gr/km²
Check our Grain Per Decare to gr/km² converter and click on formula to get the conversion factor. When you are converting area density from Grain Per Decare to gr/km², you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
gr/daa to Grain Per Square Kilometer
The formula used to convert gr/daa to Grain Per Square Kilometer is 1 Grain Per Decare = 10000000 Grain Per Square Kilometer. Measurement is one of the most fundamental concepts. Note that we have Grains Per Square Centimeter as the biggest unit for length while Grain Per Square Mile (US Survey) is the smallest one.
Convert gr/daa to gr/km²
How to convert gr/daa to gr/km²? Now you can do gr/daa to gr/km² conversion with the help of this tool. In the length measurement, first choose gr/daa from the left dropdown and gr/km² from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from gr/km² to gr/daa? You can check our gr/km² to gr/daa converter.
gr/daa to gr/km² Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area density finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like gr/daa to gr/km² through multiplicative conversion factors. When you are converting area density, you need a Grain Per Decare to Grain Per Square Kilometer converter that is elaborate and still easy to use. Converting gr/daa to Grain Per Square Kilometer is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Grain Per Decare to gr/km², this tool is the answer that gives you the exact conversion of units. You can also get the formula used in gr/daa to gr/km² conversion along with a table representing the entire conversion.
Let Others Know | 655 | 2,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-40 | latest | en | 0.836755 |
https://www.jiskha.com/display.cgi?id=1224714712 | 1,516,482,932,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889733.57/warc/CC-MAIN-20180120201828-20180120221828-00088.warc.gz | 923,639,570 | 3,993 | # MATH
posted by .
I AM A NUMBER. ROUNDING TO THE NEAREST THOUSAND MAKES ME 1,000. ALL MY DIGITS ARE EVEN, THEIR SUM IS 12, THEY ARE IN ORDER FROM GREATEST TO LEAST, AND NO DIGIT IS REPEATED. WHAT NUMBER AM I?
• MATH -
Between 500 and 1499 so three or 4 digits.
Try 3
864 no
642 ah hah
• MATH -
642
• MATH -
i dont know
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More Similar Questions | 692 | 2,683 | {"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-2018-05 | latest | en | 0.855488 |
https://daitips.com/what-is-the-rule-of-7/ | 1,656,397,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00187.warc.gz | 250,674,389 | 15,246 | # What Is The Rule Of 7?
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## What Is The Rule Of 7?
The rule of seven simply says that the prospective buyer should hear or see the marketing message at least seven times before they buy it from you. There may be many reasons why number seven is used. … Traditionally, number seven have been given precedence over other numbers by many cultures.
## What is the rule of 7 in writing?
a. Use no more than 7 words per line and no more than 7 lines per visual. b. If you need more words, make sub-points below the main point.
## Where does the rule of 7 come from?
The Marketing Rule of 7
The Rule of 7 states that a prospect needs to “hear” the advertiser’s message at least 7 times before they’ll take action to buy that product or service. The Marketing Rule of 7 is a marketing maxim developed by the movie industry in the 1930s.
## Does Rule of 7 still apply?
In the 1970’s, the average consumer in the U.S. saw around 500 ads per day. That number has since increased to more than 5,000 ads per day in 2017. … Not only does the ‘Marketing Rule of 7′ no longer apply, there are few traditional marketing methods that are working in our modern, ad saturated world.
## What is the rule of 7 in PMI?
The Rule of Seven states that seven data points trending in one direction (up or down) or seven data points on one side of the mean indicate that the process isn’t random. This means that you should check the measurement to determine whether something is influencing the process.
## How does the rule of 7 work?
With an estimated annual return of 7%, you’d divide 72 by 7 to see that your investment will double every 10.29 years. In this equation, “T” is the time for the investment to double, “ln” is the natural log function, and “r” is the compounded interest rate.
## What is the 5 7 rule?
His response was “5 of 7”. You should aim to have 50% or more of your users coming back to your app 5 out of 7 days. In other words, if you are trying to build an app that is about a daily habit, you should get a sense of how many users are close to using it as a daily habit.
## What is the theory of seven?
Social media ghostwriter for leading entrepreneurs. My Theory of Seven says that anytime you have to communicate with a large group of people, you should do so as though everyone is seven years old. This doesn’t mean talking down to people; it means being so interesting, clear and simple that you hold their attention.
## What are the 7 touchpoints?
Seven Touches: A Basic Marketing Principle in Action
• A physical connection, such as meeting at a networking event.
• Seeing an ad, either physical or digital.
• Seeing your logo, maybe as a sponsor or on a brochure.
• Seeing your social media posts in a news stream.
• Receiving your e-newsletter or other email marketing piece.
## What is the rule of 7 in Bridge?
The Rule of 7 says to hold up twice (7-5). The Rule of Thinking says to win the first heart and don’t hold up. From the lead of the deuce (4th best), declarer knows the hearts are splitting 4-4.
## How many times does someone need to hear something before they remember it?
Studies have shown that people need to see a message at least seven times before it sinks in. It supports the notion that people learn, and therefore remember, by repetition.
## How many times does someone have to see something before they buy it?
The Rule of Seven is an old marketing adage. It says that a prospect needs to see or hear your marketing message at least seven times before they take action and buy from you. Now the number seven isn’t cast in stone.
## How many times does a person have to see something before they buy it?
The Marketing Rule of 7 states that a prospect needs to “hear” the advertiser’s message at least 7 times before they’ll take action to buy that product or service. It’s a marketing maxim developed by the movie industry in the 1930s.
## What is the rule of seven quizlet?
Rule Seven. If either premise is negative, the conclusion must be negative.
## When analyzing the control chart what is the rule of 7?
Rule of seven is a rule of thumb or heuristic. On a control chart, when seven consecutive data points fall on the same side of the mean, either above or below, the process is said to be out of control and in need of adjustment. All the seven points may be within the control limits.
## What is Pareto chart in PMP?
A Pareto chart is a specific type of histogram that ranks causes or issues by their overall influence. … A Pareto chart is named after Pareto’s Law that states that a relatively small number of causes will typically produce a large majority of the problems or defects.
## What is the rule of divisibility by 7?
Divisibility rules for numbers 1–30
Divisor Divisibility condition Examples
7 Subtracting 2 times the last digit from the rest gives a multiple of 7. (Works because 21 is divisible by 7.) 483: 48 − (3 × 2) = 42 = 7 × 6.
Subtracting 9 times the last digit from the rest gives a multiple of 7. 483: 48 − (3 × 9) = 21 = 7 × 3.
## What is the rule of 72 how is it calculated?
The Rule of 72 is a calculation that estimates the number of years it takes to double your money at a specified rate of return. If, for example, your account earns 4 percent, divide 72 by 4 to get the number of years it will take for your money to double.
## How long will it take for \$7000 to double at the rate of 8 %?
The rule says that to find the number of years required to double your money at a given interest rate, you just divide the interest rate into 72. For example, if you want to know how long it will take to double your money at eight percent interest, divide 8 into 72 and get 9 years.
## What does the 7×7 rule mean?
The 7×7 rule is simple: For every slide, use no more than seven lines of text — or seven bullet points — and no more than seven words per line. Slide titles aren’t included in the count.
## What is the 5 by 5 rule in Powerpoint?
Follow the 5/5/5 rule
To keep your audience from feeling overwhelmed, you should keep the text on each slide short and to the point. Some experts suggest using the 5/5/5 rule: no more than five words per line of text, five lines of text per slide, or five text-heavy slides in a row.
## What is the 6 by 6 rule for Powerpoint presentations?
Stick to the basics when it comes to transitions between slides. A good way to keep yourself in line is by remembering the 666 rule. Presentation University recommends slides shave no more than six words per bullet, six bullets per image and six word slides in a row.
## What is the marketing rule of 7?
The principle
The marketing rule of 7’s states that a potential customer must see a message at least 7 times before they’ll be provoked to take an action.
## How many touches does it take to turn a lead?
Qualified leads are more likely to convert to buyers
The six to eight touches it takes to qualify a lead are crucial components of the lead nurturing process, allowing marketing the opportunity to educate and inform prospects as they move through each stage in the buying journey.
## Why are during the sale touch points important?
It helps look closely at each phase of the buyer journey and the points where customers interacted with your brand. Touchpoint mapping is essential since it allows brands to understand customer experience at every step and how it can be improved.
## What does 8 ever 9 never mean in bridge?
“Eight ever, nine never,” is the old saying, meaning that with eight cards missing the queen you should finesse against it, but with nine you should play for the drop.
## What is the rule of 9 in Bridge?
The Rule of 9 may help one decide whether to pass for penalty or bid. To use the rule, add the level of the contract, the number of the trump, and the number of trump honors held including the ten. If this sum is nine or more, pass the takeout double for penalty.
## What is the rule of 8 in bridge?
The Rule of 8 is a means of deciding whether to bid over an opponent’s 1NT opener. The key to this system is distribution; overcaller should hold a 6-card or longer suit or two 5-card suits (rarely make a bid with 5-4 shape).
## How many times does something need to be repeated for a child to learn it?
“A baby needs 1,000 repetitions to learn a word; by the time he’s a toddler, he might need 50 repetitions; and when he’s in kindergarten, he may need only a few repetitions to master it because the brain connections have been laid out.” Parents help this process by singing the same song, reading the same story, …
## How effective is repetition?
Repetition is a persuasive technique often used by politicians, journalists, and advertisers – but why is it so effective? According to several psychological studies, repeating simple words and phrases can convince us that they are true, even if they aren’t.
## Does repeating something make you believe it?
Repetition makes statements easier to process relative to new, unrepeated statements, leading people to believe that the repeated conclusion is more truthful. The illusory truth effect has also been linked to hindsight bias, in which the recollection of confidence is skewed after the truth has been received.
## How many impressions should you have before a sale?
How many touches are required on average before the sale? In the case of an impulse purchase, only one contact is needed, but this usually happens inside a favorite store where there is already a level of confidence and trust, or the price is so low that it does not matter to the buyer.
## What is the rule of three in marketing?
The “rule of three” is based on the principle that things that come in threes are inherently funnier, more satisfying, or more effective than any other number. When used in words, either by speech or text, the reader or audience is more likely to consume the information if it is written in threes.
## How many times does someone see an ad before they click?
Modern research believes that the average consumer needs to view an ad at least 7-8 times before it’ll really sink in.
## What are the rules of validity?
VALIDITY REQUIREMENT FOR THE CATEGORICAL ARGUMENT
• The argument must have exactly three terms.
• Every term must be used exactly twice.
• A term may be used only once in any premise.
• The middle term of a syllogism must be used in an unqualified or universal sense.
See more articles in category: Uncategorized | 2,401 | 10,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-27 | latest | en | 0.956621 |
https://intro.nyuadim.com/2023/09/19/ocean-waves/ | 1,725,802,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00480.warc.gz | 303,633,329 | 17,304 | # Ocean Waves
Concept: For this project, my inspiration came from an image I found on an Instagram account. Initially, I attempted to create a vertical wave effect and incorporate snow, but it didn’t achieve the desired visual result. As a result, I decided to shift my focus towards designing the sand and the waves, which evoked memories of my childhood spent at the beach. While I initially tried to create bird-like shapes, I ultimately chose to keep the design more abstract, allowing for interpretation by the user.
https://www.instagram.com/ag.lr.88/
Code Snippet: I take pride in the code that generates the sand grains because I achieved pleasing results by fine-tuning the parameters. I’m also pleased with how I implemented the wave motion, giving it a slower and smoother appearance, in addition to using the concept of Perlin Noise in general.
```function Wave() {
this.yPos = 0.001;
this.display = function () {
fill(118, 170, 206);
// Drawing a polygon out of the wave points
beginShape();
strokeWeight(13);
stroke(255);
let xPos = 0;
// Iterate over horizontal pixels
for (let x = 0; x <= width; x += 10) {
// Calculate a y value according to noise and map it
let y = map(noise(xPos, this.yPos), 0, 0.8, 200, 380);
// Set the vertex
vertex(x, y);
// Increment x dimension for noise
xPos += 0.05;
}
// Increment y dimension for noise
this.yPos += 0.002;
vertex(width, height);
vertex(0, height);
endShape(CLOSE);
};
}
// Create sand grains using a loop
for (let x = 0; x < width; x += grainSize) {
for (let y = 0; y < height; y += grainSize) {
let grainColor = color(random(235, 245), random(220, 230), random(180, 190)); // Light sand grain color
texture.fill(grainColor);
texture.noStroke();
let grainSizeX = random(grainSize * 1, grainSize * 3); // Adjust the X grain size
let grainSizeY = random(grainSize * 1, grainSize * 2); // Adjust the Y grain size
texture.ellipse(x + random(-grainSize / 4, grainSize / 4), y + random(-grainSize / 4, grainSize / 4), grainSizeX, grainSizeY);
}
}
return texture;
}```
Reflection:This project marked my first time using code snippets from online examples in P5.js. I enjoyed working with these examples and modifying them to match my creative vision. However, some of the functions and P5.js features used in these examples were unfamiliar and a bit complex for my current level of expertise. I spent a considerable amount of time exploring my own ideas from scratch, sometimes overlooking the new properties introduced in the examples. This experience highlighted the importance of experimenting with P5.js properties independently before incorporating more advanced techniques that may surpass my current skill level.
References:
https://p5js.org/examples/
https://www.schemecolor.com/beach-front.php | 681 | 2,773 | {"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-2024-38 | latest | en | 0.665114 |
https://nrich.maths.org/public/leg.php?code=31&cl=2&cldcmpid=6605 | 1,526,871,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863923.6/warc/CC-MAIN-20180521023747-20180521043747-00567.warc.gz | 609,961,834 | 9,541 | # Search by Topic
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### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Pair Sums
##### Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### Cunning Card Trick
##### Stage: 3 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Wild Jack
##### Stage: 2 Challenge Level:
A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Criss Cross Quiz
##### Stage: 2 Challenge Level:
A game for 2 players. Practises subtraction or other maths operations knowledge.
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Reverse Trick
##### Stage: 2 Challenge Level:
Tell your friends that you have a strange calculator that turns numbers backwards. What secret number do you have to enter to make 141 414 turn around?
### Read, Write and Give Values
##### Stage: 2 Challenge Level:
Generate large numbers then give the values of each digit.
### Tug Harder!
##### Stage: 2 Challenge Level:
In this game, you can add, subtract, multiply or divide the numbers on the dice. Which will you do so that you get to the end of the number line first?
### Number Juggle
##### Stage: 2 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### Magic Squares 4x4
##### Stage: 2 Challenge Level:
Fill in the numbers to make the sum of each row, column and diagonal equal to 34. For an extra challenge try the huge American Flag magic square.
### The Deca Tree
##### Stage: 2 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### 1, 2, 3 Magic Square
##### Stage: 2 Challenge Level:
Arrange three 1s, three 2s and three 3s in this square so that every row, column and diagonal adds to the same total.
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### Rod Measures
##### Stage: 2 Challenge Level:
Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### The Twelve Pointed Star Game
##### Stage: 2 Challenge Level:
Have a go at this game which involves throwing two dice and adding their totals. Where should you place your counters to be more likely to win?
### Which Symbol?
##### Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Being Resourceful - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration. | 2,028 | 8,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.853267 |
https://www.physicsforums.com/threads/the-notion-of-independance-of-equations.163411/ | 1,701,947,296,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00387.warc.gz | 1,044,180,310 | 15,185 | # The notion of independance of equations
• quasar987
Essentially what you are saying is that if the rank of the matrix A is maximal, then the equations are independant.
#### quasar987
Homework Helper
Gold Member
I got here in my classical mechanics textbook a set of k equations
$$f_{\alpha}(x_1,...,x_N)=0, \ \ \ \ \ \ \alpha=1,...,k$$
and it is said that these k equations are independant when the rank of the matrix
$$A_{\alpha i}=\left(\frac{\partial f_{\alpha}}{\partial x_i}\right)$$
is maximal, i.e. equals k.
Could someone explain why this definition makes sense. I.e. why does it meet the intuitive notion of independance, and exactly what this notion of independance is when we're talking about equations. Some references would be nice to!
Thank you all.
Let us assume that there exists a continuous set of solutions about a solution point $\vec{x}_{0}[/tex] Then, we would have for some perturbation vector [itex]d\vec{x}$ that
$$f_{\alpha}(\vec{x}_{0}+d\vec{x})=0$$
Now, rewriting the left-hand side we get in the limit of a tiny perturbation:
$$f_{\alpha}(\vec{x}_{0})+A_{\alpha{i}}dx_{i}=0\to{A}_{\alpha{i}}dx_{i}=0$$
Thus, if we are to ensure that there does NOT exist some non-zero perturbation vector in the neighbourhood of a solution $\vec{x}_{0}$, we must require that A is invertible.
This is in tune with standard ideas of linear independence.
I'm not sure what you mean by the "intuitive notion" of independence but the standard definition (from linear algebra) is that the only way we can have $a_1f_1(x_1,...,x_N)+ a_2f_2(x_1,...,x_N)+ \cdot\cdot\cdot+ a_kf_k(x_1,...x_N)= 0$ for all values of $x_1,...,x_N$ is to have $a_1= a_2= \cdot\cdot\cdot= a_k= 0$. That is the same as saying that the only solution to the system of equations
$a_1f_1(x_1,...,x_N)+ a_2f_2(x_1,...,x_N)+ \cdot\cdot\cdot+ a_kf_k(x_1,...x_N)= 0$
$a_1f_1_{x_1}(x_1,...,x_N)+ a_2f_2_{x_1}(x_1,...,x_N)+ \cdot\cdot\cdot+ a_kf_k_{x_1}(x_1,...x_N)= 0$
...
$a_1f_1_{x_N}(x_1,...,x_N)+ a_2f_2_{x_N}(x_1,...,x_N)+ \cdot\cdot\cdot+ a_kf_k_{x_N}(x_1,...x_N)= 0$
for any specific values of the xs has a unique solution. That is true if and only if the coefficient matrix, which is just the matrix you cite, has rank k.
I believe that is pretty much what arildno is saying from a slightly different point of view. | 744 | 2,310 | {"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} | 3.75 | 4 | CC-MAIN-2023-50 | latest | en | 0.806938 |
https://socratic.org/questions/how-do-you-find-the-sum-of-5x-2y-and-2xy-2-x-2y | 1,725,776,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00011.warc.gz | 536,198,160 | 5,804 | How do you find the sum of 5x^2y and 2xy^2 + x^2y?
$5 {x}^{2} y + {x}^{2} y + 2 x {y}^{2} = 6 {x}^{2} y + 2 x {y}^{2}$
$= 2 x y \cdot \left(3 x + y\right)$ | 92 | 156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-38 | latest | en | 0.381845 |
http://stackoverflow.com/questions/5232986/evaluate-all-possible-interpretations-in-ocaml | 1,419,478,413,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447544739.15/warc/CC-MAIN-20141224185904-00011-ip-10-231-17-201.ec2.internal.warc.gz | 93,566,747 | 18,194 | # Evaluate all possible interpretations in OCaml
I need to evaluate whether two formulas are equivalent or not. Here, I use a simple definition of formula, which is a prefix formula.
For example, `And(Atom("b"), True)` means `b and true`, while `And(Atom("b"), Or(Atom("c"), Not(Atom("c"))))` means `(b and (c or not c))`
My idea is simple, get all atoms, apply every combination (for my cases, I will have 4 combination, which are true-true, true-false, false-true, and false-false). The thing is, I don't know how to create these combinations.
For now, I have known how to get all involving atoms, so in case of there are 5 atoms, I should create 32 combinations. How to do it in OCaml?
-
Ok, so what you need is a function `combinations n` that will produce all the booleans combinations of length `n`; let's represent them as lists of lists of booleans (i.e. a single assignment of variables will be a list of booleans). Then this function would do the job:
``````let rec combinations = function
| 0 -> [[]]
| n ->
let rest = combinations (n - 1) in
let comb_f = List.map (fun l -> false::l) rest in
let comb_t = List.map (fun l -> true::l) rest in
comb_t @ comb_f
``````
There is only one empty combination of length `0` and for `n > 0` we produce combinations of `n-1` and prefix them with `false` and with `true` to produce all possible combinations of length `n`.
You could write a function to print such combinations, let's say:
``````let rec combinations_to_string = function
| [] -> ""
| x::xs ->
let rec bools_to_str = function
| [] -> ""
| b::bs -> Printf.sprintf "%s%s" (if b then "T" else "F") (bools_to_str bs)
in
Printf.sprintf "[%s]%s" (bools_to_str x) (combinations_to_string xs)
``````
and then test it all with:
``````let _ =
let n = int_of_string Sys.argv.(1) in
let combs = combinations n in
Printf.eprintf "combinations(%d) = %s\n" n (combinations_to_string combs)
``````
to get:
``````> ./combinations 3
combinations(3) = [TTT][TTF][TFT][TFF][FTT][FTF][FFT][FFF]
``````
-
this is incredible! could you explain to me the meaning of @ there? is it the same as List.append? and one more, what's the meaning of fun l -> false::l ? – zfm Mar 8 '11 at 14:21
@zfm: sure, `@` is indeed a shortcut for `List.append` (see caml.inria.fr/pub/docs/manual-ocaml/libref/Pervasives.html). `fun l -> false::l` is an anonymous function. You could write `let ff l = false::l in List.map ff rest` but instead you can "inline" the function as I did above. Hope that helps. – akoprowski Mar 8 '11 at 15:12
If you cannot hold all the combinations in memory, check out a post I made awhile ago (with some ocaml) for creating an index of the combination in lexicographical order, stackoverflow.com/questions/127704/… – nlucaroni Mar 8 '11 at 15:23
If you think of a list of booleans as a list of bits of fixed length, there is a very simple solution: Count!
If you want to have all combinations of 4 booleans, count from 0 to 15 (2^4 - 1) -- then interpret each bit as one of the booleans. For simplicity I'll use a for-loop, but you can also do it with a recursion:
``````let size = 4 in
(* '1 lsl size' computes 2^size *)
for i = 0 to (1 lsl size) - 1 do
(* from: is the least significant bit '1'? *)
let b0 = 1 = ((i / 1) mod 2) in
let b1 = 1 = ((i / 2) mod 2) in
let b2 = 1 = ((i / 4) mod 2) in
(* to: is the most significant bit '1'? *)
let b3 = 1 = ((i / 8) mod 2) in
(* do your thing *)
compute b0 b1 b2 b3
done
``````
Of course you can make the body of the loop more general so that it e.g. creates a list/array of booleans depending on the size given above etc.; The point is that you can solve this problem by enumerating all values you are searching for. If this is the case, compute all integers up to your problem size. Write a function that generates a value of your original problem from an integer. Put it all together.
This method has the advantage that you do not need to first create all combinations, before starting your computation. For large problems this might well save you. For rather small size=16 you will already need 65535 * sizeof(type) memory -- and this is growing exponentially with the size! The above solution will require only a constant amount of memory of sizeof(type).
And for science's sake: Your problem is NP-complete, so if you want the exact solution, it will take exponential time.
-
I agree that this problem is NP-complete, however since the size will not very big (less than 10 variables) for any formulas given, I don't really care about the space consuming here. – zfm Mar 9 '11 at 20:47 | 1,292 | 4,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2014-52 | latest | en | 0.830044 |
https://mathemerize.com/graph-of-a-parabola/ | 1,723,220,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00899.warc.gz | 311,801,820 | 30,883 | # Graph of a Parabola – Types of Parabolas
A parabola is locus of a point which moves in a plane, such that its distance from a fixed point called focus is equal to its perpendicular distance from a fixed straight line called directrix. Graph of a Parabola and their types are shown below.
## Basic Concepts of a Parabola
(a) Focal distance : The distance of a point on parabola from the focus is called focal distance of the point.
(b) Focal chord : A chord of parabola, which passes through focus is called focal chord.
(c) Double ordinate : A chord of the parabola perpendicular to the axis of the symmetry is called double ordinate.
(d) Latus rectum : A double ordinate passing through focus or a focal chord perpendicular to axis of parabola is called latus rectum.
## Types and Graph of a Parabola :
Four standard forms of parabola are $$y^2$$ = 4ax ; $$y^2$$ = -4ax ; $$x^2$$ = 4ay ; $$x^2$$ = -4ay. All of them with their graphs are given below :
(i) Parabola $$y^2 = 4ax$$ :
Vertex O is (0, 0)
Focus S is (a, 0)
Axis is y = 0
Directrix ZZ’ is x = -a
Focal length = x + a
Length of Latus rectum = 4a
(ii) Parabola $$y^2 = -4ax$$ :
Vertex O is (0,0)
Focus S is (-a,0)
Axis is y = 0
Directrix ZZ’ is x = a
Focal length = x – a
Length of Latus rectum = 4a
(iii) Parabola $$x^2 = 4ay$$ :
Vertex O is (0,0)
Focus S is (0,a)
Axis is x = 0
Directrix ZZ’ is y = -a
Focal length = y + a
Length of Latus rectum = 4a
(iv) Parabola $$x^2 = -4ay$$ :
Vertex O is (0,0)
Focus S is (0,-a)
Axis is x = 0
Directrix ZZ’ is y = a
Focal length = y – a
Length of Latus rectum = 4a
Hope you learnt basic concepts and graph of a parabola, learn more concepts of parabola and practice more questions to get ahead in the competition. Good luck! | 553 | 1,763 | {"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} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.903189 |
https://www.newscientist.com/article/mg21228421-100-enigma-number-1676/ | 1,721,892,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00534.warc.gz | 785,254,301 | 59,078 | # Enigma Number 1676
7 December 2011
## Pick ‘n’ mix
There is always the same number of sweets – fewer than 50 – in a box of Tweets. Some are red and the rest are green. The proportion of reds is always the same.
Two lads, Dip and Flip, had a box each last week and a box each this week. Dip picks sweets at random from his box without looking in. Last week, when he had just eaten his last red one, he noticed he had just one green left. This week, when he had eaten his last red sweet, there were two greens left. Last week’s situation is four times as likely as this week’s.
Until he runs out of one colour, Flip picks his sweets by tossing a coin: heads he eats a red one and tails he eats a green one. Last week he had just eaten his last red one and he counted the number of green ones left. This week, when he had eaten his last red sweet there was one more green sweet left than in the previous week. In fact, having last week’s number left is four times as likely as having this week’s.
Tell me how many red sweets and how many green ones make up a box of Tweets.
WIN £15 will be awarded to the sender of the first correct answer opened on Wednesday 1 February. The Editor’s decision is final. Please send entries to Enigma 1676, New Scientist, Lacon House, 84 Theobald’s Road, London WC1X 8NS, or to enigma@newscientist.com (please include your postal address).
Answer to 1670 How mean? Frack’s average was 25¾; Dessie’s, 24.75
The winner Jan Giezen of Maassluis, the Netherlands | 378 | 1,498 | {"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 | 0.98415 |
https://math.answers.com/Q/What_is_13_out_of_25_written_as_a_percent | 1,652,940,931,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00719.warc.gz | 455,751,561 | 38,277 | 0
# What is 13 out of 25 written as a percent?
Wiki User
2010-04-07 03:07:04
Start out by converting 13 out of 25 as a fraction: 13/25
Then convert to a decimal by dividing 13 by 25: .52
Then convert to percent by moving the decimal point 2 places to the right and changing the decimal point to a % sign: 52%
Wiki User
2010-04-07 03:07:04
Study guides
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818 Reviews | 132 | 404 | {"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} | 3.375 | 3 | CC-MAIN-2022-21 | latest | en | 0.870704 |
https://www.explainxkcd.com/wiki/index.php?title=Talk:882:_Significant&diff=prev&oldid=80798 | 1,606,858,009,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00557.warc.gz | 609,609,209 | 9,351 | # Difference between revisions of "Talk:882: Significant"
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Those lazy scientists are playing minecraft instead of curing cancer! Lynch 'em! Davidy22[talk] 00:35, 11 January 2013 (UTC)
But I heard that Minecraft cures cancer... Scientists! Investigate! <off: cheers from active group, boos from the control group> 178.99.81.144 19:31, 30 April 2013 (UTC)
You know this experiment isn't conducted properly when you know you're in the control group. Troy (talk) 05:24, 4 March 2014 (UTC)
So you have to somehow convince them they are playing Minecraft, when in fact they are not. That's easy, select people who have never played the game. But what if KNOWING the game is Minecraft is what cures cancer? Oh boy... Cflare (talk) 13:57, 15 August 2014 (UTC)
Um, I take it that whoever explained this comic can't tell the difference between < and >, as the fact that the confidence was changed wasn't mentioned in the article... 76.246.37.141 23:19, 20 September 2013 (UTC)
Yes, I also figured out this today, green is lower than 0.05, on other colors there is just a confidence that it's NOT lower than 0.05. The newspaper did add this remaining 19 panels to 95%. The article is marked as incomplete, it needs a major rewrite.--Dgbrt (talk) 19:12, 3 October 2013 (UTC)
This explanation seems to misinterpret α. α is the chance of rejecting a true null hypothesis, a false positive. The 5% here is α. The correct interpretation of it is that if the null hypothesis is true, there is a 5% chance that we will mistakenly reject it. P in "P<0.05" is the chance that, if the null hypothesis is true, a result as extreme as, or more extreme than, the result we get from this experiment. α is not the chance that, given our current data, the null hypothsis is true. We wish to know what that is, but we do not know.108.162.215.72 08:52, 16 May 2014 (UTC)
In layman's terms, the comic appears to misrepresent what "95% confidence" (p <0.05) means. The statistic "p < 0.05" means that when we find a correlation based on data, that correlation will be a false positive fewer than 5 percent of the time. In other words, when we observe the correlation in the data, that correlation actually exists in the real world at least 19 out of 20 times. It does not mean that 1 out of every 20 tests will produce a false positive. This comic displays a pretty significant failure in understanding of Bayesian mathematics. The 5% chance isn't a 5% chance that any test will produce a (false) positive; it's a 5% chance that a statistical positive is a false positive. 108.162.219.196 (talk) (please sign your comments with ~~~~)
Is the "e" in "News" supposed to look like an epsilon (and the "w" a rotated epsilon)? 108.162.250.222 15:00, 15 December 2014 (UTC) | 739 | 2,770 | {"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-2020-50 | latest | en | 0.977437 |
http://www.starlight-publishing.com/Matter/PartI/I4/I42Magnetic.html | 1,582,434,618,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145746.24/warc/CC-MAIN-20200223032129-20200223062129-00067.warc.gz | 233,039,428 | 3,393 | THE NATURE OF MATTER
PART I THE BACKGROUND OF MATTER
EVIDENCE OF A BACKGROUND OF MATTER
CHAPTER 4
Next section: Section 2 Cosmic Microwave Background Radiation
MAGNETIC FLUX
Section 1
Human beings can already observe the effects of currents in the universal ocean in our everyday life. As is shown is detail in Part III Chapter 7 Magnetic Flux., the flow of the neutral massless background oceanic particles manifests in our daily lives as the flow of magnetic flux.
Magnetic flux is proposed to be the flow of background pure matter particles.
The ebb and flow of the electromagnetic interaction parallels the alternating motion of the flow of charged matter and the reaction of the background ocean in response to the flow of the charged matter. The motion of charged matter and the flow of background oceanic particles in response to the motion of charged matter is proposed to be the electromagnetic interaction.
The proposition is made that the momentum energy of charged matter in motion pushes against the background causing momentum energy to be transferred to the background oceanic particles, causing the background oceanic particles to have momentum. Then, since the energy actually hosted by the charged particle that was caused the flow of the background particles, that same energy is regained as the momentum energy of the charged particle.
Magnetic Flux is discussed in detail in Part III, Chapter 7, Magnetic Flux.
MAGNETIC FLUX FROM CURRENT IN A WIRE
When electrical current flows in a wire, it generates a magnetic field around the wire. The direction of force in the magnetic field follows the left hand rule due to the negative charge of the electrons in the current. The left hand rule applies for negative current with the thumb of the left hand is pointed in the direction that the negative current is flowing, and the left hand fingers point in the north direction of the magnetic flux caused by the current.
An electron flowing through a symbolic wire.
In the above 'head on view' illustration of an electron flowing in a wire, the electron is traveling toward the reader and heading out of the paper toward the reader. In this 'head on view', the electron spins clockwise and exhibits left hand spin. (The left hand thumb is pointed toward the reader.)
Electrons have the quantum mechanical property of intrinsic angular momentum or spin, and a magnetic dipole. The magnetic dipole has the same axis as the spin axis.
The electromotive force that causes the electrons to flow in the wire also orients the electrons so that they travel in the direction of north on their magnetic dipoles. There is a correlation between the left hand angular momentum of the electrons and the left hand flow of magnetic flux caused by their motion.
It is proposed here that the left hand energy rotation about the electron, or intrinsic angular momentum, causes the flow of the background particles, which manifest as magnetic flux.
The quantum angular momentum and the magnetic flux around the wire have the same direction. Suppose the energy rotation around the electron caused background ocean particles to have momentum. Both the energy rotation and momentum of background particles would then have left hand spin. Suppose the energy rotation around the electron causes momentum energy to be imparted onto the massless neutral oceanic particles causing them to move around the wire. The motion of the oceanic particles would then be the same thing as magnetic flux.
The lines of force of the electron are indicated by the horizontal and vertical lines. The heavy circle is the wire boundary. The direction of the magnetic field around the wire is indicated by the diamonds with the north and south magnetic poles labeled. The energy rotation of the electron (angular moment) causes the background field particles to flow in a clockwise motion indicated by the two bent arrows. The small arrow beside each oceanic particle indicates the north end of the magnetic dipole of the background oceanic particle.
CHAPTER 4
EVIDENCE OF A BACKGROUND OF MATTER
The sections in this chapter are:
Title Page
of the Nature of Matter | 798 | 4,153 | {"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-2020-10 | latest | en | 0.894806 |
https://practicetestgeeks.com/parcc-grade-3-math-practice-test-questions/ | 1,713,069,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00746.warc.gz | 424,419,222 | 71,790 | # PARCC Grade 3 Math Practice Test Questions
0%
#### 1. Steve had to pay a library fee. The amount he paid is shown below. How much did he pay?
Correct! Wrong!
1. C: There are 2 dollar bills, which represent 2 dollars. There are also 2 quarters, 2 nickels, 1 dime, and 4 pennies. Two quarters are worth \$0.50 since each is worth \$0.25 (2 x0.25=0.50) , 2 nickels are worth \$0.10 since each is worth \$0.05 (2 x0.05=0.10) , 1 dime is worth \$0.10, and 4 pennies are worth \$0.04 since each is worth \$0.01. The sum of the coins can be found by writing: \$0.50+\$0.10+\$0.10+\$0.04, which equals \$0.74. The sum of the two dollar bills and the coins can be written as: \$2.00+\$0.74. Thus, he paid \$2.74.
#### 2. What number sentence is illustrated by the diagram below?
Correct! Wrong!
2. C: The diagram shows 32 counters divided into 4 groups, with 8 counters in each group. Therefore, the total number of counters, 32, is divided by 4, giving a quotient of 8, which is written as: 32÷4=8.
#### 3. Which shape has a number of vertices that is equal to two times the number of vertices found on a triangle?
Correct! Wrong!
3. C: A vertex is a point where two or more edges meet. A triangle has 3 vertices. Two times that would be 6 vertices. The figure shown for Choice C is a triangular prism, which is the only figure that has 6 vertices. A triangular pyramid (Choice A) has 4 vertices, a cube (Choice B) has 8 vertices, and a square pyramid (Choice D) has 5 vertices.
#### 4. What is the perimeter of the figure shown below?
Correct! Wrong!
4. B: The perimeter is the distance around the figure. So, if you add up all of the numbers you get 22 in.
#### 5. What number does Point P represent?
Correct! Wrong!
5. B: Each increment represents one-half. This can be determined by counting that there are 3 marks, or 4 spaces, that lie between the difference of two wholes, as in between 10 and 12. Thus, one increment past 10, where Point P is located, represents 10 1/2. | 571 | 1,988 | {"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} | 4.5625 | 5 | CC-MAIN-2024-18 | latest | en | 0.950486 |
http://brainden.com/forum/topic/17796-re-rollo/page/11/?tab=comments | 1,603,142,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00406.warc.gz | 21,457,258 | 29,297 | BrainDen.com - Brain Teasers
# Re-ROLLO
## Recommended Posts
EXTRA
POINT
LINES
• Replies 872
• Created
#### Popular Posts
USUAL
Ooh. It's been a while. ROLLO - Logic: At this point, I don't need logic.
foyer - 1 proves the Y since other 4 letters are known word is GLYPH gl was proven since brass = 0
_ _ _ _ _
QUOTE - 1
EXTRA - 1
POINT - 0
LINES - 0
Edited by Logophobic
BUTTE
EXIST
SNARE
_ _ _ _ _
QUOTE - 1
EXTRA - 1
POINT - 0
LINES - 0
BUTTE - 0
EXIST - 1
SNARE - 0
AXMEN
_ _ _ _ _
QUOTE - 1
EXTRA - 1
POINT - 0
LINES - 0
BUTTE - 0
EXIST - 1
SNARE - 0
AXMEN - 0
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EXIST = 1
But AXMEN, POINT = 0 rules out X, I, T so E is first or S is 4th
POISE
if 0 then S is not 4th so E is 1st
If 1 then S is 4th as POINT, SNARE = 0rule out other 4 letters.
Additionally A is 5th from EXTRA = 1 and E would be proven to not be 1st from above, while AXMEN, BUTTE, SNARE = 0 rule out XTR
EPOCH
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Aww, I was going to wait for results
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QUITE
butte-0 disproves the -u-te, quote-1 means either Q or --O-- are correct. If quite-1 then Q is proven, if it is - then --O-- is proven
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E P O C H
QUOTE - 1
EXTRA - 1
POINT - 0
LINES - 0
BUTTE - 0
EXIST - 1
SNARE - 0
AXMEN - 0
POISE - 0 maurice +5
EPOCH - 5 araver +25?
Woot woot.
_ _ _ _ _
MOUSE
FLOCK
THIRD
DUSTY
CRYER
_ _ _ _ _
MOUSE - 1
FLOCK - 0
THIRD - 0
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MOUSY - if 0 then E is last letter, as only change from MOUSE - 1
##### Share on other sites
_ _ _ _ E
MOUSE - 1
FLOCK - 0
THIRD - 0
DUSTY - 0
CRYER - 0
MOUSY - 0 mmjay +5
PIQUE
##### Share on other sites
_ _ _ _ E
MOUSE - 1
FLOCK - 0
THIRD - 0
DUSTY - 0
CRYER - 0
MOUSY - 0 mmjay +5
PIQUE - 1
SCARE
##### Share on other sites
_ _ _ _ E
MOUSE - 1
FLOCK - 0
THIRD - 0
DUSTY - 0
CRYER - 0
MOUSY - 0 mmjay +5
PIQUE - 1
SCARE - 1
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× | 877 | 2,308 | {"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-2020-45 | latest | en | 0.787516 |
https://community.ultimaker.com/topic/10937-cura-150201-problem-witj-minimal-layer-time/ | 1,713,192,984,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00009.warc.gz | 156,794,308 | 45,433 | # Cura 15.02.01 - Problem witj Minimal Layer Time
## Recommended Posts
Posted (edited) · Cura 15.02.01 - Problem witj Minimal Layer Time
I think it's the oposite. For layer time x = or less then lift head or pause. So if the layer time you set it's 5 tehn all the layers with that or less time used will have a pause or lift of the head. It doesn't set the minimum layer time but it affects layers with that time or less. So you must set 4 seconds if that's what it takes to print it. Anyway for small items it's really good to print more than one even 3 being that small to allow the material to set. Specially if it's pla
Edited by Guest
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Posted · Cura 15.02.01 - Problem witj Minimal Layer Time
Hi,
(copied to here, since I posted in the wrong category)
Today I wanted to print a tripod / photo screw.
I started the first print and saw that the thread is not very clean, looks like two or more layers melt together.
I watched the print process and could see that the printer takes only 4 seconds for one layer for the shaft and thread layers.
Now I increased the 'Minimal Layer Time' to 15 seconds, as you can see here:
Then I did a 2nd print but still the layer time for shaft and thread is only about 4 seconds, not 15 what I set under 'Minimal Layer Time'.
I checked several times, and saved the file with 15 sec. 'Minimal Layer Time' using a different filename, but UM2 does not increase the layer time :-(
What I am missing?
Thanks...
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Posted · Cura 15.02.01 - Problem witj Minimal Layer Time
I think it's the oposite. For layer time x = or less then lift head or pause. So if the layer time you set it's 5 tehn all the layers with that or less time used will have a pause or lift of the head. It doesn't set the minimum layer time but it affects layers with that time or less. So you must set 4 seconds if that's what it takes to print it. Anyway for small items it's really good to print more than one even 3 being that small to allow the material to set. Specially if it's pla
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Posted · Cura 15.02.01 - Problem witj Minimal Layer Time
My post banished
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Posted · Cura 15.02.01 - Problem witj Minimal Layer Time
Cura never reduces the printing speed below the "minimum speed" setting, no matter of the value for the minimum layer time.
"Cool head lift" affects the surface quality too much (speaking from my own experience).
Probably the best solution is - as neotko suggested - print more then one at a time and use a nozzle temperature from the lower end of the scale.
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Posted · Cura 15.02.01 - Problem witj Minimal Layer Time
What tinkergnome said. Click the "..." button next to "enable cooling fan" and you will see more settings.
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Posted · Cura 15.02.01 - Problem witj Minimal Layer Time
Also, please, do not rely on that screw to hold anything valuable. It WILL break if you put a bit of load on it.
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Posted · Cura 15.02.01 - Problem witj Minimal Layer Time
Hi IRobertI,
sorry I don't know what you mean. Which screw please?
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• Cura 5.7 is here and it brings a handy new workflow improvement when using Thingiverse and Cura together, as well as additional capabilities for Method series printers, and a powerful way of sharing print settings using new printer-agnostic project files! Read on to find out about all of these improvements and more.
• 7 replies
• S-Line Firmware 8.3.0 was released Nov. 20th on the "Latest" firmware branch.
(Sorry, was out of office when this released)
This update is for...
All UltiMaker S series
New features
Temperature status. During print preparation, the temperatures of the print cores and build plate will be shown on the display. This gives a better indication of the progress and remaining wait time. Save log files in paused state. It is now possible to save the printer's log files to USB if the currently active print job is paused. Previously, the Dump logs to USB option was only enabled if the printer was in idle state. Confirm print removal via Digital Factory. If the printer is connected to the Digital Factory, it is now possible to confirm the removal of a previous print job via the Digital Factory interface. This is useful in situations where the build plate is clear, but the operator forgot to select Confirm removal on the printer’s display. Visit this page for more information about this feature.
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• Create New... | 1,109 | 4,623 | {"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-2024-18 | latest | en | 0.931639 |
http://mathoverflow.net/questions/21665/the-geometry-behind-the-icm-2010-logo/21999 | 1,467,230,083,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397797.77/warc/CC-MAIN-20160624154957-00178-ip-10-164-35-72.ec2.internal.warc.gz | 203,330,856 | 17,984 | # The geometry behind the ICM 2010 Logo
The logo for this year's ICM show the inequality $|\tau(n)| \leq n^{11/2} d(n)$ where $\sum \tau(n)q^n = q \prod_{n \neq 1} (1 - q^n)^{24}$ is the tau function. Wikipedia says this bound was conjectured by Ramanujan (appropriate for a conference in Hyderabad) and proven by Deligne in '74 in the process of proving as a corollary of the Weil Conjectures (which I also don't get). The background of the ICM logo looks like Ford circles (or sun rays). What is the hyperbolic geometry behind the Tau Conjecture and its proof?
EDIT: It would also be nice to see the proof of this bound, but the Weil conjectures and l-adic cohomology are a topic in themselves.
-
There is no hyperbolic geometry behind the conjecture... So please explain what do you like to have: the precise link between Weil conjectures and the growth of the coefficients of cusp forms? – Wadim Zudilin Apr 17 '10 at 14:42
I am not too familiar with modular forms. However, I know the generating function of the tau function is the Dedekind eta function, which is a modular form of weight 1/2. This is the type of geometric relation I was expecting. Cusp forms are defined by the vanishing of the 0-th Fourier coefficient and I guess the tau function is an example of this. The Weil conjectures do not involve hyperbolic geometry and I don't know why they come into play here. – john mangual Apr 17 '10 at 15:38
I would only correct that the generating function of Ramanujan's tau is the 24th power of Dedekind's function, so it's a cusp form of weight 12. A simple argument due to Hecke (see, e.g., Serre's Course in Arithmetic) shows that the $n$-th Fourier coefficient of a cusp form $f(z)$ of weight $2k$ is $O(n^k)$. If $f$ admits an additional arithmetic structure, namely if its $L$-series admits an Euler-product expansion, then one gets $O(n^{k-1/2}d(n))$. – Wadim Zudilin Apr 17 '10 at 23:29
The logo has a piece of the complex upper half plane divided into fundamental domains for the action of $SL_2(\mathbb{Z})$ by Möbius transformations (which are hyperbolic isometries - see the appropriate entry in McMullen's gallery or Wikipedia). There are no Ford circles in sight, but you may have been confused by the semicircular regions on the bottom. Those regions are unions of fundamental domains, and are cut out by geodesics in the $SL_2(\mathbb{Z})$ orbit of $1/2 \leftrightarrow \infty$ (which forms part of the boundary of a few fundamental domains). Ford circles come from the $SL_2(\mathbb{Z})$ orbit of the horocycle $\operatorname{Im}(\tau) = 1$, and horocycles have constant nonzero geodesic curvature (imagine driving with your steering wheel turned a bit to the right, but never returning to where you started).
The generating function for $\tau$ is the 24th power of Dedekind's $\eta$ function (often written as the discriminant form $\Delta$), and it is a function on the upper half plane that is invariant under the weight 12 action of $SL_2(\mathbb{Z})$. Up to normalization, it is the unique lowest weight nonzero level 1 cusp form, and this makes it automatically a Hecke eigenform. Both Mordell's proof of the earlier Ramanujan conjectures and Deligne's proof of the growth conjecture use this fact in an essential way. I should note that Deligne proved the Ramanujan conjecture as a corollary of the Weil conjectures, not "in the process".
The connection between hyperbolic geometry and the Ramanujan conjecture is not particularly strong, as far as I know (but I would be happy to be shown the errors of my ways).
-
One way to answer this question is as follows: Ramanujan's conjecture is a special case of a much more general conjecture that any cuspdial automorphic representation of $GL_n$ over a number field is tempered. This is a technical but fundamental notion, which in the special case of the automorphic representation of $GL_2$ attached to the $\Delta$ function, reduces to Ramanujan's original conjecture. In fact, many people working in the theory of automorphic representations refer to this very general conjecture simply as the Ramanujan conjecture.
When applied to other cuspforms on $GL_2$ (namely Maass forms) it includes Selberg's conjecture that on congruence quotients of the upper half-plane, the spectrum of the hyperbolic Laplacian is bounded below by $1/4$.
The appearance of hyperbolic geometry can be understood in the following way: the quotient of $SL_n(\mathbb R)$ by $SO(n)$ is a non-compact symmetric space, which in the particular case of $SL_2$ is the hyperbolic plane. So, while the particular appearance of hyperbolic geometry may be a bit of a red herring, the appearence of highly symmetric geometry is a reflection of the group representation theory that is underlying the picture.
As of the current moment, no purely representation-theoretic approach to the (general form of) Ramanujan's conjecture is known. (Or rather, a proof strategy involving what is called symmetric power functoriality is known, but the requisite results on symmmetric power functoriality seem very much out of reach at the moment). The only cases that are proved at the moment are cases when one can relate the group theoretic picture of automorphic forms to algebraic geometry (first over $\mathbb C$, then over a number field, and then ultimately over finite fields, so that the Weil conjecture apply). This is how Deligne's proof proceeds. This connection between the geometry of symmetric spaces and arithmetic and geometry over finite fields is one of the profound points of investigation of modern number theory, but despite many positive results related to it, it remains essentially mysterious, even to experts.
- | 1,398 | 5,686 | {"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} | 3.015625 | 3 | CC-MAIN-2016-26 | latest | en | 0.916505 |
https://community.qlik.com/t5/QlikView-Layout-Visualizations/QlikSense-Pie-Chart-without-Dimension/td-p/1460104 | 1,545,003,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827998.66/warc/CC-MAIN-20181216213120-20181216235120-00090.warc.gz | 566,383,407 | 41,712 | # QlikView Layout & Visualizations
Discussion Board for collaboration on QlikView Layout & Visualizations.
New Contributor II
## QlikSense - Pie Chart without Dimension
HI Team,
I Need to create a Pie chart without dimension.
Exp = Sum(A)+sum(B)+sum(C)+Sum(D)
Valued Contributor
## Re: QlikSense - Pie Chart without Dimension
Dimension
=ValueList('A','B','C','D')
Measure
=If(ValueList('A','B','C','D')='A',Sum(A),
If(ValueList('A','B','C','D')='B',Sum(B),
If(ValueList('A','B','C','D')='C',Sum(C),
If(ValueList('A','B','C','D')='D',Sum(D)))))
I hope this helps,
Cheers,
Luis | 178 | 593 | {"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-2018-51 | latest | en | 0.509587 |
https://apps.apple.com/us/app/numbers-learning-recognition/id1136681674?ls=1 | 1,566,728,297,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323246.35/warc/CC-MAIN-20190825084751-20190825110751-00469.warc.gz | 366,859,187 | 49,546 | ## Description
Developed school education systems use many techniques and practices to enforce the numbers recognition and counting sequence. These practices include small interactive exercises in the form of games. These exercises sub-consciously teaches basic numbers and counting in a fun to learn methodology.
We’ve combined the 6 proven techniques (Number recognition activities) bundled in one app so the parents and teachers can help their young ones in early stages. The app is rich in features and contains interactive exercises that helps to involve in the learning process while playing.
The names and a brief introduction of the activities are as follows.
▪ Number Flashcards (1-100)
▪ Touch the Number
▪ Catch the Ladybugs
▪ Touch the Crabs
▪ Count the Objects
▪ Drag & Match Numbers
1) Flashcards
Flashcard uses spaced repetition technique and promotes the mental process of active recall. The flashcards are proven valuable. This technique performs well if Flashcards are displayed in a sequence. We have provided some very useful settings to change the colours and fonts to make them suitable for one’s needs.
2) Touch the Number
Touch the Number is a simple activity that initially displays 3 numbers and requires to touch the correct one. Upon 5 consecutive correct attempts, the level increases and asks to choose from more and bigger numbers.
The app’s AI (Artificial Intelligence) is designed carefully after repeatedly practicing and observing behaviour from different age groups.
3) Catch the Ladybugs
The idea behind this activity is to reinforcing the current COUNT of the touched ladybugs. This activity starts with 3 ladybugs which are running around. The user is supposed to touch (catch) them. Over successful catches, the number of ladybugs increases to a moderate level. The ladybugs also goes off screen, so the anticipation and curiosity helps to reinforce the count.
4) Touch the Crabs
This activity is absolutely similar to the ladybugs except it presents animated crabs. Both objects (ladybugs and crabs) are considered to grab the attention. Both activities reinforce counting.
5 ) Count the Objects
This activity is a small quiz. It presents a number of different objects and gives four options to choose their total count. When the object is touched, it transforms into its current count. This small quiz actively teaches the difference of numbers and counting.
6) Drag & Match Numbers
This activity provides numbers in pairs to drag and match. To make it a bit easier or challenging, the pairs can be assigned similar or different colours. The numbers can also be displayed in a row or distributed randomly to find on the screen. It is always a fun activity for youngsters to drag and match the objects.
The app is fairly customisable to suite the need for the individual preference. Several Fonts suitable for early education are bundled to use as per the requirement.
It is a must have educational app which is provided Free with limited functionality for evaluation. It contains advertisement and some functionalities can be used temporarily by watching Ads. However we strongly recommend that you practice the ads-free complete app. The app can be unlocked with a one time In-App Purchase.
The app is ideal to be practiced in the classroom and is equally beneficial to be practiced in the home. It is designed to consume the least possible amount of disk space, thus can be kept in the device for as long as required.
We develop, proof-read the content and test our apps by involving several experienced teachers. We also perform usage tests with different age groups for the optimum quality assurance.
There is a great effort put in to produce this app. We thank you for your interest to download and evaluate it. We look forward for your input to make the app even better. Please consider supporting us by rating the app and writing a small review. It really helps us to produce new and better apps.
## What’s New
Version 2.0
Imroved user interface
Improved Layouts
App Updated to support iPhone X, XS, XSMAX, iPad Pro
App Updted to support latest iOS
Several small Improvements and bug fixes
## Ratings and Reviews
4.8 out of 5
54 Ratings
54 Ratings
Tokunagachan ,
### Activities are good and well thought out
I like that there are a variety of activities so that my child has choices and is less likely to get bored. Games have enough depth and difficulty levels to make payment worth it (to remove Ads).
Solid educational app.
Lucidreamer1971 ,
### Fantastic
Kindergarten teacher here. I was looking for something like this simple app. It has everything you need to practice with your children in the classroom. I recommend to other teachers also. It is equally good for the parents who wants to introduce to basic numbers to their toddlers.
Nice work.
Jayleon! ,
### Good early childhood education app!!
This app is really amazing, my little cousin just learn numbers and counting so fast after using this app. The colourful pictures and attractive teaching modes just keep children's attention on it. I really think this app is one of great app for early childhood education~nice one!!!
## Information
Seller
Naveed Abbas
Size
69.8 MB
Category
Education
Compatibility
Requires iOS 9.0 or later. Compatible with iPhone, iPad, and iPod touch.
Languages
English, French, German, Italian, Spanish
Age Rating
Rated 4+
Copyright
© Copyright 2019 Holiday Educationist. All rights reserved.
Price
Free
In-App Purchases
1. Remove Ads \$3.99 | 1,130 | 5,522 | {"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} | 3.375 | 3 | CC-MAIN-2019-35 | latest | en | 0.927879 |
https://puzzling.stackexchange.com/questions/51270/what-does-11 | 1,652,793,533,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00385.warc.gz | 548,117,537 | 71,790 | What does 1+1=?
Most of this is is flavour, with a section of puns setting the scene at the start, and then the story itself in the middle. Scroll down for the actual puzzle. You may be interested to know that all the facts about the brain are true.
Everything in this story is true, except for the bits that aren't. That's probably not very helpful...
Someplace, somewhere, at exactly sometime before noon, there is a building which specialised in doing just one thing: being a hospital, university and medical research facility.
This medical building employed people. It employed chemists in their element, physicists of little matter and biologists studying blood cells, but in vain.
It made sure it didn't employ poets. The building had decided that anyone who writes inverse must be a backward person.
It did however employ mathematicians, the small fraction of the world who knew there was a fine line between a denominator and a numerator.
One of these mathematicians was employed as a maths teacher for the university students. He was such a good teacher, he actually could control the class and teach them things. He wasn't like the cross-eyed teacher down the corridor who couldn't control his pupils. He wasn't like the teacher in the room opposite who was rather average because he was so mean. And he definitely wasn't like the teacher upstairs who had problems which showed the first sine of madness. No, this teacher was a good teacher. He was also a doctor, and he was very good at that too. He never lost his patients.
Today he was teaching his students about mathematical fallacy. Quite illogical if you ask me.
The teacher turned and wrote What is 1+1? on the blackboard. He turned to the class and asked what was the answer. Someone shouted out "2!".
"Alright then, let me prove it equals 1."
He turned back to the blackboard and wrote the following:
Now he looked at the class and was about to point out the mistake, where dividing by (a-b) is dividing by zero. He was though interrupted as a panting doctor burst into the room.
"Dr, Patient 3145 is awake and walking round the hospital."
The class stirred. Everyone had heard of Patient 3145. No-one knew who he was, just that he had been trapped in an icy lake and suffered extreme brain damage. The doctors, including the teacher, had tried some experimental treatment on him and it had worked. But with some side effects.
During his coma, his brain had shown signs of hyperactivity. His treatment had greatly increased the rate of neurogenesis, usually very small in adults, and as a result, instead of losing around 7000 neurons per day, the average amount for an adult over 20, he was gaining a couple of thousand. This in turn was opening up new neural pathways, allowing quicker thought and reactions, more control of the body and the ability to multitask. Not only this, but his axons were firing quicker across synapses, neurotransmitters moving faster and information being transferred quicker. He was now estimated to be the most intelligent human in history.
On the few times he had woken, he hadn't done that much, but he had shown signs of heightened intelligence.
In the end however this treatment would be a failure. The reproduction of brain cells was causing immense pressure on his brain, which would eventually shut itself down.
But now it seems he had woken and had gained enough strength to walk. And as if on cue, he walked into the classroom.
Everyone fell silent watching him. He walked up to the board, looked at it for a second, then picked up a chalk. For a few seconds no-one could see what he was drawing, but then he stepped away and walked out of the room to reveal a beautifully intricate pattern:
(Click for a larger image of the main pattern, image has been updated to give a hint and to fix an error)
The students nor the teacher could work out its message. Maybe you can?
What does 1+1=?
Hint:
This will be helpful for spotting something in the puzzle indicating the next step (Now Solved)
Hint for last step:
Something like Excel can create something useful to do with what ChrisHappy found in his answer (3x16) which can be filled with information from the triangles. Then you have the final answer.
Bigger Hint:
Try filling a 3x16 grid in with the triangles
• Note: Not all puns are mine, I found some online Apr 24, 2017 at 5:57
• Ow. Ow. Ow..... Apr 24, 2017 at 6:54
• Very nice visuals! Intricate pattern indeed... Apr 24, 2017 at 7:46
• @BmyGuest thanks! Tbh I'm surprised no-ones got past the first step yet! Apr 24, 2017 at 15:56
1+1 =
10
The circle
I noticed that, there are 8 smaller circles on the lines. Which clues for binary. White being 1 and black being zero. The double bar seems to be a start.
01101011 01101111 01101111 01101100 01110011 01101111 01101111 01100111 01100110 01101001 01101110 01100101 01110011 01101111 01101000 01100011
OP confirms, there might be a little error where S should have been R. (OP: Now corrected image) Which gives - CHOSEN IF GO OR LOOK, which was meant to be LOOK OR GO IF CHOSEN
Taking first letter of each word gives - LOGIC
This means, we need LOGIC to solve the next step.
Next step with the help of Chris's answer here -
Utilizing LOGIC Gates and treating the white squares as 1 and black as 0, and applying NOR logic from the symbol in the picture we get -
1 0 -> 0
1 1 -> 0
0 0 -> 1
0 0 -> 1
0 1 -> 0
1 1 -> 0
0 0 -> 1
0 0 -> 1
0 1 -> 0
0 0 -> 1
0 0 -> 1
0 0 -> 1
0 0 -> 1
0 1 -> 0
1 0 -> 0
1 1 -> 0
1 1 -> 0
1 1 -> 0
0 0 -> 1
0 0 -> 1
0 1 -> 0
0 1 -> 0
1 0 -> 0
0 0 -> 1
0 1 -> 0
1 0 -> 0
0 0 -> 1
0 0 -> 1
1 0 -> 0
0 0 -> 1
0 0 -> 1
0 1 -> 0
00110011 01111000 00110001 00110110 -> 3x16
So, the second part leads to 3x16 = 48
Which leads to the number of triangles left which is 48
Final part
The bigger hint reveals that we need to fill the triangles in a grid of 16*3 with Black and White blocks. But here's another trick. We need to interpret filled triangles as Black while empty ones as White as they represent NOT gates.
Doing this we get -
Which is $1 + 1 = 10$
OP Clarifications:
This is a great answer by Tech, but I'll clarify a couple of things.
The two circles to the side of the main one show you where to start and which way to read. One circle has I on the outer ring, taken works in through the rings up to V. This is indicating that you start on the outer ring of the circle and move inwards. The other circle points to 12 o'clock and indicates clockwise. This is telling you to start at 12 o'clock on the main circle and read round clockwise.
You get the NOR gate from the symbols on the outer ring with the sets of binary. That is the symbol for the NOR gate, so the outer ring gives NOR LOGIC.
Applying that to the next wrong gives you the size of a grid. Next filling it with the triangles, but inverting the colours because a triangle represents the NOT gate, you get the final answer.
I came up with the answer after seeing a poster saying
'There are 10 types of people in this world, those who understand binary and those who don't'
And thought it the perfect starting point for a puzzle. Hope you enjoyed :)
(If you are still unsure on something or want me to make something clearer, ask me in the comments)
• That circle with V likely means you have 4 circles you need to decypher with V in the center. The other means you start at top each time? So, figure out what those squares mean, then outwards triangles and finally inwards triangles. Assuming all are binary, it likely generates more text. There are 2x8 on the inner, 4x 8 on the next and 8x8 on the outermost. So, 14 more letters to find. Apr 24, 2017 at 8:34
• @ZizyArcher Thanks. Already working on collecting the text out of it. Apr 24, 2017 at 8:36
• (continue...) though, I am ending up getting gibberish. So, either I am doing something wrong or may be there is an error somewhere. Apr 24, 2017 at 8:44
Partial
Through hint and a techy idiot ;)
The top left circle:
Displays the code number of values in Roman Numerals: I, II, III, IV, V (Zoom in)
Bottom left circle:
Displays the order which to decode the code: Clockwise.
Through @Techidiot's answer, we got the first solution: LOGIC
However, the second solution is 3x16...I think. Why?
The two digits hinted that it would not be binary and the hint hinted that it had something to do with logic. Used this JSfidle to play around with them.
Note: There's are five patterns. This caused me to look for the fifth one and notice the outer lines form a pattern: l___l_l_ l_l_____ => 01110101 01011111 => u_. Pretty sure that I messed that up.
• Great! Exactly what I needed someone to find. Next step: try seeing how many triangles remain ;) Apr 24, 2017 at 17:36
• Yeah the thing you noticed doesn't mean anything. And there may be 5 parts to the puzzle (indicated by the numerals), but actually the fifth isn't in the circle. Once you work out what '3x16' is referring to, then you can get the final solution. (Hint: Use Excel or something similar for final part) Apr 24, 2017 at 17:59
Partial thoughts
Since no one else has mentioned it yet, the issue with the algebra is that since b=1 (step 7) and a=b (step 1), then
(a-b) = 0 so the process to get from step 4 to step 5 involves dividing by zero, which is an illegal operation.
No idea if this has any bearing on the final solution, but it's worth noting.
• This is actually mentioned in the puzzle itself: "Now he looked at the class and was about to point out the mistake, where dividing by (a-b) is dividing by zero." Apr 24, 2017 at 13:49
• Whoops. Don't know how I missed that on my first reading. Oh, well. I'll leave this here in case anyone else missed that mention, too. Apr 24, 2017 at 13:52
• Nope sorry not relevant Apr 24, 2017 at 14:53 | 2,630 | 9,978 | {"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} | 2.9375 | 3 | CC-MAIN-2022-21 | longest | en | 0.99266 |
https://www.learnnext.com/CBSE/Class-12/Maths/Relations-and-Functions/Inverse-of-a-Bijective-Function/L-1803760.htm | 1,576,202,745,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2019-51/segments/1575540548537.21/warc/CC-MAIN-20191213020114-20191213044114-00292.warc.gz | 776,484,098 | 50,300 | ]]>
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# Inverse of a Bijective Function
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#### Inverse of a Bijective Function - Lesson Summary
Let f: A → B be a function. If, for an arbitrary x ∈ A we have f(x) = y ∈ B, then the function, g: B → A, given by g(y) = x, where y ∈ B and x ∈ A, is called the inverse function of f.
Ex:
Define f: A → B such that
f(2) = -2, f(½) = -2, f(½) = -½, f(-1) = 1, f(-1/9) = 1/9
Here f one-one and onto.
g(-2) = 2, g(-½) = 2, g(-½) = ½, g(1) = -1, g(1/9) = -1/9
g is the inverse of f.
A function, f: A → B, is said to be invertible, if there exists a function, g : B → A, such that g o f = IA and f o g = IB.
The function, g, is called the inverse of f, and is denoted by f -1.
Ex:
Let 2 ∈ A.Then gof(2) = g{f(2)} = g(-2) = 2
Let -2 ∈ B.Then fog(-2) = f{g(-2)} = f(2) = -2.
Example:
Let P = {y ϵ N: y = 3x - 2 for some x ϵN}. Show that a function, f : N → P, defined by f (x) = 3x - 2, is invertible, and find f-1.
Solution:
Let us consider an arbitrary element, y ϵ P.
⇒ y=3x-2
⇒ x = (y+2)/3
Let us define g : P → N by g(y) = (y+2)/3
Let x ∈ N.
Then g o f (x) = g (f (x)) = g (3x - 2)
= (3x-2+2)/3 = x
This shows g o f = IN …(i)
Let y ∈ P.
Then fog (y) = f (g (y)) = f((y+2)/3)
= 3((y+2)/3) - 2 = y
This shows fog = IP …(ii)
Hence, f is invertible and g is the inverse of f.
Theorem:
Let f : X → Y and g : Y → Z be two invertible (i.e. bijective) functions. Then g o f is also invertible with (g o f)-1 = f -1o g-1.
Proof:
Given, f and g are invertible functions.
To prove that g o f is invertible, with (g o f)-1 = f -1o g-1.
It is sufficient to prove that:
i. (f -1 o g-1) o (g o f) = IX, and
ii. (g o f)o( f -1o g-1) = IZ.
Now, ( f -1 o g-1) o (g o f) = {( f -1 o g-1) o g} o f {'.' l o (m o n) = (l o m) o n}
= { f -1 o (g-1 o g)} o f
= ( f -1oIY)of {'.' g-1og = IY}
= f-1of { '.' f-1oI = f-1}
= IX
Hence, (f -1o g-1)o(g o f) =IX …… ( i)
Similarly, (g o f)o( f -1o g-1)
={(g o f) o f--1} o g-1
={g 0 (f o f--1} o g-1
=(g 0 Ix) o g-1
=g o g-1 = Iz
(g o f)o( f -1o g-1) =IZ ……. (ii)
From equations (i) and (ii),
(g o f)-1 = f -1 o g-1
Hence, the composition of two invertible functions is also invertible. | 1,085 | 2,526 | {"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} | 4.5 | 4 | CC-MAIN-2019-51 | longest | en | 0.827981 |
https://www.lmfdb.org/EllipticCurve/2.2.12.1/1024.1/m/2 | 1,582,661,601,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00153.warc.gz | 797,623,664 | 58,877 | # Properties
Base field $$\Q(\sqrt{3})$$ Label 2.2.12.1-1024.1-m2 Conductor $$(32)$$ Conductor norm $$1024$$ CM yes ($$-4$$) base-change no Q-curve yes Torsion order $$4$$ Rank not available
# Related objects
Show commands for: Magma / SageMath / Pari/GP
## Base field $$\Q(\sqrt{3})$$
Generator $$a$$, with minimal polynomial $$x^{2} - 3$$; class number $$1$$.
magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![-3, 0, 1]);
sage: x = polygen(QQ); K.<a> = NumberField(x^2 - 3)
gp: K = nfinit(a^2 - 3);
## Weierstrass equation
$$y^2 = x^{3} + \left(2 a - 4\right) x$$
magma: E := ChangeRing(EllipticCurve([0, 0, 0, 2*a - 4, 0]),K);
sage: E = EllipticCurve(K, [0, 0, 0, 2*a - 4, 0])
gp: E = ellinit([0, 0, 0, 2*a - 4, 0],K)
This is a global minimal model.
sage: E.is_global_minimal_model()
## Invariants
$$\mathfrak{N}$$ = $$(32)$$ = $$\left(a + 1\right)^{10}$$ magma: Conductor(E); sage: E.conductor() $$N(\mathfrak{N})$$ = $$1024$$ = $$2^{10}$$ magma: Norm(Conductor(E)); sage: E.conductor().norm() $$\mathfrak{D}$$ = $$(512)$$ = $$\left(a + 1\right)^{18}$$ magma: Discriminant(E); sage: E.discriminant() gp: E.disc $$N(\mathfrak{D})$$ = $$262144$$ = $$2^{18}$$ magma: Norm(Discriminant(E)); sage: E.discriminant().norm() gp: norm(E.disc) $$j$$ = $$1728$$ magma: jInvariant(E); sage: E.j_invariant() gp: E.j $$\text{End} (E)$$ = $$\Z[\sqrt{-1}]$$ ( Complex Multiplication ) magma: HasComplexMultiplication(E); sage: E.has_cm(), E.cm_discriminant() $$\text{ST} (E)$$ = $N(\mathrm{U}(1))$
## Mordell-Weil group
Rank not available.
magma: Rank(E);
sage: E.rank()
Regulator: not available
magma: gens := [P:P in Generators(E)|Order(P) eq 0]; gens;
sage: gens = E.gens(); gens
magma: Regulator(gens);
sage: E.regulator_of_points(gens)
## Torsion subgroup
Structure: $$\Z/2\Z\times\Z/2\Z$$ magma: T,piT := TorsionSubgroup(E); Invariants(T); sage: T = E.torsion_subgroup(); T.invariants() gp: T = elltors(E); T[2] $\left(-a + 1 : 0 : 1\right)$,$\left(0 : 0 : 1\right)$ magma: [piT(P) : P in Generators(T)]; sage: T.gens() gp: T[3]
## Local data at primes of bad reduction
magma: LocalInformation(E);
sage: E.local_data()
prime Norm Tamagawa number Kodaira symbol Reduction type Root number ord($$\mathfrak{N}$$) ord($$\mathfrak{D}$$) ord$$(j)_{-}$$
$$\left(a + 1\right)$$ $$2$$ $$4$$ $$I_{4}^*$$ Additive $$1$$ $$10$$ $$18$$ $$0$$
## Galois Representations
The mod $$p$$ Galois Representation has maximal image for all primes $$p$$ except those listed.
prime Image of Galois Representation
$$2$$ 2Cs
For all other primes $$p$$, the image is the normalizer of a split Cartan subgroup if $$\left(\frac{ -1 }{p}\right)=+1$$ or the normalizer of a nonsplit Cartan subgroup if $$\left(\frac{ -1 }{p}\right)=-1$$.
## Isogenies and isogeny class
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2.
Its isogeny class 1024.1-m consists of curves linked by isogenies of degrees dividing 4.
## Base change
This curve is not the base-change of an elliptic curve defined over $$\Q$$. It is a $$\Q$$-curve. | 1,136 | 3,077 | {"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} | 3.28125 | 3 | CC-MAIN-2020-10 | latest | en | 0.296869 |
http://jgomezdans.github.io/eoldas_release/example1.html | 1,558,613,486,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257243.19/warc/CC-MAIN-20190523103802-20190523125802-00088.warc.gz | 109,712,644 | 21,927 | # A simple observation operator example, running in eoldas¶
## Purpose of this section¶
This section of the user guide will take a simple example of combining two cost functions in a variational DA sense: an Identity (or Prior) operator and a smoothness (derivative) constraint. The examples used here are of filtering noisy time series of NDVI data.
As well as learning about these concepts if you are unfamiliar with them, this section will also take you through some examples of running the eoldas to solve this sort of problem. Towards the end of the section, we delve into writing some python code to make use of eoldas, and also mention other ways of interacting with it.
We learn that at the heart of eoldas is one or more configuration files that allow us to solve problems without the need for any code writing (though you can if you want to!).
### Introducing the observation operator concept¶
In the previous sections, we have assumed that the state of the land surface can be observed directly, and that these observations are only limited by noise in their acquisition. In general, the state of the surface and the observations will be different entities. For example, the state of the surface may include parameters such as leaf area index, chlorophyll concentration or soil moisture, whereas the observations may be of directional surface reflectance, top of atmosphere radiance or microwave temperatures. The link between these magnitudes is the observation operator, which maps the land surface parameters into quantities that are measured by a sensor. Many types of observation operator are available: from the statistical to the physics based. The simplest case, however, is the identity operator. In fact, we have already used it, as we have assumed that the observations are just direct measurements of the state vector.
### A simple data assimilation example using an identity observation operator¶
The simplest observation operator is the identity observation operator, where the observations are identical to the state vector components. We can see the use of this operator as a way of optimally smoothing univariate timeseries, for example NDVI. Additionally, the use of a DA system allows to interpolate where data points are not available. The following demonstrates how the EOLDAS prototype can be used for this task, and also allows us to explore the use of the weak assimilation paradigm conveniently.
The main way to run EOLDAS is via one or more configuration files. This is partly to make sure that a record exists of a particular experimental setup and partly to allow flexible running of the system without the need for further coiding by the user (unless he/she wants to add new classes or methods).
Here is the start of the configuration file that we are going to use:
# An EOLDAS confioguration file for
# a simple Identity operator and a regulariser
[parameter]
location = ['time']
limits = [[1,365,1]]
names = gamma_time,NDVI
solve = 0,1
datatypes = x
[parameter.result]
filename = output/Identity/NDVI_Identity.params
format = 'PARAMETERS'
help_filename='Set the output state filename'
[parameter.x]
datatype = x
names = $parameter.names default = 200,0 help_default="Set the default values of the states" apply_grid = True sd = [1.]*len($parameter.names)
bounds = [[0.000001,1000000],[-1,1]]
This sets up the section [parameter], which described the state variables within EOLDAS.
The section must contain fields describing the locational information of the data to be used as well as the names of the state variables we will consider. In this case, we have two state variables gamma_time and NDVI that we wish to estimate over the (time) range 1 to 365 (inclusive) in steps of 1 (day).
The subsection parameter.solve gives a list of codes indicating whether we wish to solve for each parameter or not.
A code of 0 tells the EOLDAS not to solve, i.e. just to use the data that are read in or any other default values.
A code of 1 tells EOLDAS to solve for that state variable at all locations (times here).
A code of 2 tells EOLDAS to solve for the state variable, but to maintain a single value over all locations.
The section parameter.result gives information on any output file name and format for the state.
Finally in this section, the field parameter.x sets up data and conditions for the state vector x. Remember that it is this state vector x that we will solve for in the EOLDAS. Here, we specify that it is of datatype x (i.e. the x state vector), that it has the same names as we set up in parameter.names, that default values to be assigned are 25 and 0 for the two variables respectively. If any data are read in, these override the default values, but in this case, we simply start with the defaults.
The text help_default allows the field parameter.x.default to be set from the command line with the option --parameter.x.default=a,b.
The flag apply_grid, which is the default, tells the EOLDAS to produce the state vector on a grid over the bounds defined in parameter.limits.
Finally, we define default uncertainty information for the state vector (this does not directly affect the running of the EOLDAS (parameter.x.sd), and define the bounds for each state vector (use None if no bound is to be used). Here, we set the lower bound as 0 and the upper bound as 1 for both parameters.
The next section sets up general conditions:
[general]
doplot=True
help_do_plot='do plotting'
In this case, we set a flag to do plotting when the results are written out. Plotting will use the filenames in any state variable section e.g. parameter.result.filename to generate a series of plots with filenames the same as this data filename. We will see some examples later.
The next section sets up the operators that we want to define here.
[operator]
modelt.name=DModel_Operator
modelt.datatypes = x
obs.name=Operator
obs.datatypes = x,y
Here, we define two operators, DModel_Operator and Operator. These names refer directly to python classes for the operators in EOLDAS. The base class is ‘Operator’ which implements the Identity operator. All other classes are derived from this. The differential operator works only on the x state vector, which is equlivalent to defining $$\mathbf{y_{obs}}=0$$. The operator ‘Operator’ access both x and y data if it to act as a Prior constraint, so we set up x and y datatypes.
Next we set the details of these operators. First, the differential operator:
[operator.modelt.x]
names = $parameter.names sd = [1.0]*len($operator.modelt.x.names)
datatype = x
[operator.modelt.rt_model]
model_order=1
help_model_order='The differential model order'
wraparound=periodic,365
where we specify which state vector elements this operator has access to (all of those in parameter.names here) and set up the default uncertainty and datatype.
We then set the parameters specific to the ‘model’ $$H(\mathbf{x})$$, in this case the order of the differential model (2 here) and the edge conditions (periodic, with a period of 200 (days)).
Finally, we set up the operator operator.obs, specifically, parameters for its x and y state vectors.
[operator.obs.x]
names = $parameter.names[1:] sd = [1.0]*len($operator.obs.x.names)
help_sd='Set the observation sd'
datatype = x
[operator.obs.y]
names = $parameter.names[1:] sd = [0.15]*len($operator.obs.x.names)
help_sd="set the sd for the observations"
datatype = y
state = data/Identity/random_ndvi1.dat
help_state = "Set the y state vector"
[operator.obs.y.result]
filename = output/Identity/NDVI_fwd.params
specifying default uncertainty information, data types and any required output files. Here, we wish to write out the results in operator.obs.y, so we specify a filename for this. Again, we see the use of a ‘help’ variable, which here allows operator.obs.y.result.filename to be set from the command line. This interfacing to the command line means that a single configuration file can generally serve for multiple experiments and the user does not need to keep generating new ones.
The main program can be accessed in various ways. One way is to write some front end code that calls the eoldas python code.
An example is solve_eoldas_identity.py that includes three sections:
First, some code to generate a synthetic dataset.
import pdb
import numpy as np
def create_data ( n_per=4, noise=0.15, obs_off=0.33, \
window_size=0, order=4):
"""
Create synthetic "NDVI-like" data for a fictitious time series. We return
the original data, noisy data (using IID Gaussian noise), the QA flag as well
as the time axis.
Missing observations are simulated by drawing a random number between 0 and 1
and checking against obs_off.
Parameters
----------
n_per : integer
Observation periodicity. By default, assumes every 8 days
noise : float
The noise standard deviation. By default, 0.15
obs_off : float
The threshold to decide on missing observations in the time series.
window_size : integer, odd
window size for savitzky_golay filtering. A large window size will lead
to larger data gaps by correlating the noise. Set to zero by default
which applies no smoothing.
order : integer
order of the savitzky_golay filter. By default 4.
"""
from savitzky_golay import savitzky_golay
import numpy as np
doys = np.arange ( 1, 365+1, n_per)
ndvi_clean = np.clip(np.sin((doys-1)/72.), 0,1)
ndvi = np.clip(np.sin(doys/72.), 0,1)
# add Gaussian noise of sd noise
ndvi = np.random.normal(ndvi,noise,ndvi.shape[0])
# set the qa flags for each sample to 1 (good data)
qa_flag = np.ones_like ( ndvi).astype( np.int32 )
passer = np.random.rand( ndvi.shape[0])
if window_size >0:
# force odd
window_size = 2*(window_size/2)+1
passer = savitzky_golay(passer, window_size=window_size, order=order)
# assign a proportion of the qa to 0 from an ordering of the smoothed
# random numbers
qa_flag[np.argsort(passer)[:passer.size * obs_off]] = 0
return ( doys, ndvi_clean, ndvi, qa_flag )
Here, we generate some NDVI data which has a trajectory of a sine wave for the first half of the year and is flat at zero for the second half. The sampling is controlled by n_per and obs_off. The parameter obs_off randomly removes a proportion of the data. If window_size and order are set then a savitzky_golay filter is used to induce correlation in the timing of the samples that are removed from thie dataset (qa=0). This mimics what we practically have in Optical Earth Observation with temporal correlation in cloud cover.
The next section calculates the ‘ideal’ value of $$\gamma$$ by calculating the root mean squared deviation of the original dataset. We use this ideal gamma here for to demonstrate the physical meaning of the gamma value, though in practice this would be unknown. This section also writes the dataset to a temporary file in ‘BRDF’ format.
All of this so far standard python coding, though such datasets could be generated in many other ways. The final section interfaces to the top level of the eoldas code, which is what is of immediate concern in this tutorial.
import sys,tempfile
this = sys.argv[0]
import eoldas
# SD of noise
noise=0.15
# nominal sampling period
n_per=7
# proportion of missing samples
obs_off=0.33
# order of differential model (integer)
model_order=1
# sgwindow is larger to create larger data gaps
sgwindow=10
# set up data for this experiment
file, ideal_gamma,doys,ndvi_clean,ndvi,qa_flag = \
prepare(noise=noise,n_per=n_per,\
obs_off=obs_off,model_order=model_order,\
sgwindow=sgwindow)
# set gamma to less than the the theoretical value
gamma = ideal_gamma*0.33
# set op file names
xfile = 'output/Identity/NDVI_Identity.params'
yfile = 'output/Identity/NDVI_fwd.params'
# initialise optuions for DA overriding any in config files
cmd = 'eoldas ' + \
' --conf=config_files/eoldas_config.conf --conf=config_files/Identity.conf ' + \
' --logfile=mylogs/Identity.log ' + \
' --calc_posterior_unc ' + \
' --parameter.solve=0,1 ' + \
' --parameter.result.filename=%s '%xfile +\
' --parameter.x.default=%f,0.0 '%(gamma) + \
' --operator.obs.y.result.filename=%s'%yfile +\
' --operator.obs.y.state=%s'%file+\
' --operator.modelt.rt_model.model_order=%d '%model_order
# initialise eoldas
self = eoldas.eoldas(cmd)
# solve DA
self.solve(write=True)
The first part of the code extends the system path for where it searches for libraries. This is done relative to where solve_eoldas_identity.py is (in the bin directory of the distribution). After that, we set up values for the parameters for generating the synthetic dataset.
The interface to the eoldas here is mainly to make a string with a number of flags. The most important flag is --conf=config_files/Identity.conf which specifies the configuration file for this experiment. In addition, --conf=eoldas_config.conf is given, which specifies a system default configuration file, eoldas_config.conf.
So, the simplest ‘top level’ interface to eoldas from python code involves:
gamma = ideal_gamma*0.33
# set op file names
xfile = 'output/Identity/NDVI_Identity.params'
yfile = 'output/Identity/NDVI_fwd.params'
# initialise optuions for DA overriding any in config files
cmd = 'eoldas ' + \
' --conf=config_files/eoldas_config.conf --conf=config_files/Identity.conf ' + \
' --logfile=mylogs/Identity.log ' + \
' --calc_posterior_unc ' + \
' --parameter.solve=0,1 ' + \
' --parameter.result.filename=%s '%xfile +\
' --parameter.x.default=%f,0.0 '%(gamma) + \
' --operator.obs.y.result.filename=%s'%yfile +\
' --operator.obs.y.state=%s'%file+\
' --operator.modelt.rt_model.model_order=%d '%model_order
and
' --operator.modelt.rt_model.model_order=%d '%model_order
# initialise eoldas
self = eoldas.eoldas(cmd)
# solve DA
self.solve(write=True)
where we set up the text string, initiate the eoldas object (eoldas.eoldas) and then call the eoldas.solve() method.
This command string is clearly of some importance. The flags --logfile=mylogs/Identity.log and --calc_posterior_unc correspond to items in the [general] section of the configuration file. Here, eoldas_config.conf contains the lines:
[general]
help_datadir ="Specify where the data and or conf files are"
here = os.getcwdu()
grid = True
is_spectral = True
calc_posterior_unc=False
help_calc_posterior_unc ="Switch to calculate the posterior uncertainty"
write_results=True
help_write_results="Flag to make eoldas write its results to files"
init_test=False
help_init_test="Flag to make eoldas run a test of the cost functions before proceeding with DA"
doplot=False
plotmod=100000000
plotmovie=False
[general.optimisation]
# These are the default values
iprint=1
gtol=1e-3
help_gtol = 'set the relative tolerance for termination'
maxIter=1e4
maxFunEvals=2e4
plot=0
# see http://openopt.org/NLP#Box-bound_constrained
solverfn=scipy_lbfgsb
randomise=False
help_randomise='Randomise the starting point'
To understand how e.g the flag --calc_posterior_unc can be used we can look at:
calc_posterior_unc=False
help_calc_posterior_unc ="Switch to calculate the posterior uncertainty"
where we set the default value of the item (False) and also have a ‘help’ statement which allows this value to be overridden. You shouldn’t normally need to change things in the system configuration file eoldas_config.conf. This flag controls whether we calculate the posterior iuncertainty or not. The default is False because it can be quite computationally expensive and is often best done in a post processing step.
The flag --operator.obs.y.state=filename refers specifically to the section operator.obs.y.state of the configuration, which is something we have set up in Identity.conf with a ‘help’ field, so we can override the default value set in the configuration file.
[operator.obs.y]
names = $parameter.names[1:] sd = [0.15]*len($operator.obs.x.names)
help_sd="set the sd for the observations"
datatype = y
If we run solve_eoldas_identity.py, it sends logging information to mylogs/Identity.log and reports (to the stderr) the progress of the optimisation, ‘f’ being the total of all of the $$J$$ terms for this configuration. It should converge to a solution within some tens of iterations and result in a final value of $$J$$ of around 1800. We set the name of the logfile in solve_eoldas_identity.py:
if window_size >0:
There is a lot of detail in the log file about exactly what value terms are set to and the progress of the eoldas. It also contains information on the individual $$J$$ terms:
2012-06-14 15:15:57,455 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 293.213607
2012-06-14 15:15:57,456 - eoldas.solver.eoldas.solver-modelt-x - INFO - J = 34.354830
2012-06-14 15:15:57,457 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 196.657803
2012-06-14 15:15:57,458 - eoldas.solver.eoldas.solver-modelt-x - INFO - J = 87.225910
2012-06-14 15:15:57,459 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 112.026943
2012-06-14 15:15:57,461 - eoldas.solver.eoldas.solver-modelt-x - INFO - J = 113.479088
2012-06-14 15:15:57,462 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 78.729997
2012-06-14 15:15:57,463 - eoldas.solver.eoldas.solver-modelt-x - INFO - J = 93.761052
2012-06-14 15:15:57,464 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 43.172989
2012-06-14 15:15:57,466 - eoldas.solver.eoldas.solver-modelt-x - INFO - J = 99.078418
2012-06-14 15:15:57,467 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 40.564246
2012-06-14 15:15:57,468 - eoldas.solver.eoldas.solver-modelt-x - INFO - J = 74.685441
2012-06-14 15:15:57,469 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 39.829009
2012-06-14 15:15:57,470 - eoldas.solver.eoldas.solver-modelt-x - INFO - J = 63.881006
2012-06-14 15:15:57,471 - eoldas.solver.eoldas.solver-obs-x - INFO - J = 27.079490
A log file is important to the running of eoldas, as the processing can take quite some time for some problems.
The state vector results will be written to output/Identity/NDVI_Identity.params because we first specified this in Identity.conf:
[parameter.result]
filename = output/Identity/NDVI_Identity.params
format = 'PARAMETERS'
(actually, since we also set the comamnd ' --operator.obs.y.result.filename=%s'%yfile in the code we have written, this will override what is in the parameter file).
The output file format is a general ‘PARAMETERS’ format that should look something like this:
The first line of the file is a header statement that describes the location fields (‘time’ here) and state variables (gamma_time and NDVI here) and subsequent lines give the values for those fields. This format can be used for input data, but it cannot at present be used to define a full input covariance matrix, only the standard deviation for each observation as in this example.
A graph of the result is in output/Identity/NDVI_Identity.params.plot.x.png:
This result is quite interesting for understanding how our DA system works: The plot shows the mean of the estimate of the NDVI state vector in the lower panel (the upper panel shows the value of gamma_time which we did not solve for) as a red line as we solve for the state every day (of 365 days). The 95% confidence interval is shown shaded in grey. We will show this below in a more refined plot of the results.
We also specified the ‘y’ data to be written out (in Identity.conf):
help_state = "Set the y state vector"
[operator.obs.y.result]
filename = output/Identity/NDVI_fwd.params
so this goes to a file output/Identity/NDVI_fwd.params unless the flag --operator.obs.y.result.filename=somethingElse.dat is used. The format of this file is the same as above, but it shows the retrieved NDVI data since this reports $$H(x)$$.
#PARAMETERS time mask NDVI sd-NDVI
1.000000 1.000000 0.211816 0.000000
8.000000 1.000000 0.275085 0.000000
15.000000 1.000000 0.336462 0.000000
22.000000 1.000000 0.401630 0.000000
29.000000 1.000000 0.455163 0.000000
36.000000 1.000000 0.542784 0.000000
43.000000 1.000000 0.610468 0.000000
50.000000 1.000000 0.669554 0.000000
92.000000 1.000000 0.834135 0.000000
The original dataset is output form convenience in the same format as the y state, being output/Identity/NDVI_fwd.params_orig here:
#PARAMETERS time mask NDVI sd-NDVI
1.000000 1.000000 0.315700 0.150000
8.000000 1.000000 0.286300 0.150000
15.000000 1.000000 0.313996 0.150000
22.000000 1.000000 0.470529 0.150000
29.000000 1.000000 0.253425 0.150000
36.000000 1.000000 0.660812 0.150000
43.000000 1.000000 0.661460 0.150000
50.000000 1.000000 0.856904 0.150000
92.000000 1.000000 0.894455 0.150000
Various other graphics are output for a ‘y’ state:
A plot at each location, in output/Identity/NDVI_fwd.params.plot.y.png (not of much relevance in this example):
A plot of the observations and modelled values as a function of location (the observations are the green dots) in output/Identity/NDVI_fwd.params.plot.y2.png:
and a plot of the x state vector associated with this operator (NDVI here).
## Example plotting data from the output files¶
The above plots are automatically generated by eoldas provided general.doplot is True but these are intended as quicklooks, and users are likely to want to form their own plots.
An example of this is implemented in example1plot.py:
#!/usr/bin/env python
import numpy as np
import pylab as plt
#
# Some plotting of the synthetic and retrieved data
# generate the clean data
doys = np.arange ( 1, 365+1, 1)
ndvi_clean = np.clip(np.sin((doys-1)/72.), 0,1)
sdata = np.array([np.array(i.split()).astype(float) for i in state[1:]])
sNdvi = sdata[:,2]
sdNdvi = sdata[:,4]
# noisy sample data
ndata = np.array([np.array(i.split()).astype(float) for i in noisy[1:]])
nNdvi = ndata[:,2]
sdnNdvi = ndata[:,3]
ndoys = ndata[:,0]
# retrieved
plt.fill_between(doys,y1=sNdvi-1.96*sdNdvi,y2=sNdvi+1.96*sdNdvi,facecolor='0.8')
p1 = plt.plot(doys,sNdvi,'r')
# original
p2 = plt.plot(doys,ndvi_clean,'g')
# samples used
p3 = plt.errorbar(ndoys,nNdvi,yerr=sdnNdvi*1.96,fmt='bo')
plt.legend([p2,p3,p1],['Original state','Sampled noisy state','Retrieved state'])
#plt.show()
plt.savefig('images/example1plot.png')
We can see that smoothers of this sort have some difficulty maintaining sudden changes, although this is quite challenging here given the level of noise amd the rather large data gaps.
More importantly, we have used the smoother to interpolate over the missing observations and to reduce uncertainty.
If we inspect the file output/Identity/NDVI_Identity.params <output/Identity/NDVI_Identity.params>:
#PARAMETERS time gamma_time NDVI sd-gamma_time sd-NDVI
1.000000 42.912561 0.211816 0.000000 0.063006
2.000000 42.912561 0.220854 0.000000 0.064164
3.000000 42.912561 0.229893 0.000000 0.065042
4.000000 42.912561 0.238931 0.000000 0.065651
5.000000 42.912561 0.247970 0.000000 0.066000
6.000000 42.912561 0.257008 0.000000 0.066092
7.000000 42.912561 0.266046 0.000000 0.065928
8.000000 42.912561 0.275085 0.000000 0.065507
9.000000 42.912561 0.283853 0.000000 0.066408
alongside the original data output/Identity/NDVI_fwd.params_orig <output/Identity/NDVI_Identity.params>:
#PARAMETERS time mask NDVI sd-NDVI
1.000000 1.000000 0.315700 0.150000
8.000000 1.000000 0.286300 0.150000
15.000000 1.000000 0.313996 0.150000
22.000000 1.000000 0.470529 0.150000
29.000000 1.000000 0.253425 0.150000
36.000000 1.000000 0.660812 0.150000
43.000000 1.000000 0.661460 0.150000
50.000000 1.000000 0.856904 0.150000
92.000000 1.000000 0.894455 0.150000
We note that the original state here (green line) lies entirely within the 95% CI. The error reduction has been of the order of 2.2 (compare sd-NDVI after the DA with that prior to it). We can see that the uncertainty at the datapoints has been reduced from 0.15 (that of the input data) to typically around 0.065. This grows slightly (to around 0.10) when the data gaps are large.
The impact of ‘filtering’ in this way (optimising a fit of the model to the observations) is to smooth the data and reduce uncertainty in the output. The reduction in uncertainty is related to the amount of smoothing that we apply. The second effect is to interpolate between observations, with uncertainty growing where we have no observations.
You might try changing the value of gamma used and seeing the effect on the results.
We could do this by modifying a few lines of solve_eoldas_identity.py to produce solve_eoldas_identity1.py:
if __name__ == "__main__":
import sys,tempfile
this = sys.argv[0]
import eoldas
# SD of noise
noise=0.15
# nominal sampling period
n_per=7
# proportion of missing samples
obs_off=0.33
# order of differential model (integer)
model_order=1
# sgwindow is larger to create larger data gaps
sgwindow=10
# set up data for this experiment
file, ideal_gamma,doys,ndvi_clean,ndvi,qa_flag = \
prepare(noise=noise,n_per=n_per,\
obs_off=obs_off,model_order=model_order,\
sgwindow=sgwindow)
# set gamma to less than the the theoretical value
gamma = ideal_gamma*0.45
# set op file names
xfile = 'output/Identity/NDVI_Identity1.params'
yfile = 'output/Identity/NDVI_fwd1.params'
# initialise optuions for DA overriding any in config files
which writes out to output/Identity/NDVI_Identity1.params and output/Identity/NDVI_fwd1.params and has a gamma value that is 0.45/0.33 of that previously used.
Now, plotting this using example1plot1.py:
This is possibly a better result, but in fact what we see is further limitation of the model that we have chosen here: we enforce wraparound (i.e. the NDVI at day 1 is expected to be the same as at day 365) and we enforce smoothness (so as we increase the gamma, the smoothness, we over-smooth at the sudden change that occurs half way through the year).
We could remove the wraparound condition, but in practice, it is better simply to weaken this constraint. We have done this in solve_eoldas_identity2.py:
if __name__ == "__main__":
import sys,tempfile
this = sys.argv[0]
import eoldas
# SD of noise
noise=0.15
# nominal sampling period
n_per=7
# proportion of missing samples
obs_off=0.33
# order of differential model (integer)
model_order=1
# sgwindow is larger to create larger data gaps
sgwindow=10
# set up data for this experiment
file, ideal_gamma,doys,ndvi_clean,ndvi,qa_flag = \
prepare(noise=noise,n_per=n_per,\
obs_off=obs_off,model_order=model_order,\
sgwindow=sgwindow)
# set gamma to less than the the theoretical value
gamma = ideal_gamma*0.33
# set op file names
xfile = 'output/Identity/NDVI_Identity2.params'
yfile = 'output/Identity/NDVI_fwd2.params'
# initialise optuions for DA overriding any in config files
cmd = 'eoldas ' + \
' --conf=config_files/eoldas_config.conf --conf=config_files/Identity.conf ' + \
' --logfile=mylogs/Identity.log ' + \
' --calc_posterior_unc ' + \
' --parameter.solve=0,1 ' + \
' --parameter.result.filename=%s '%xfile +\
' --parameter.x.default=%f,0.0 '%(gamma) + \
which write out to output/Identity/NDVI_Identity2.params and output/Identity/NDVI_fwd2.params and has the same (higher) gamma value used above.
Now, plotting this using example1plot2.py:
If you re-run these scripts several times, so that you see different configurations for the temporal sampling, you will notice that the interpolation sometimes behaves well over the entire dataset (for some given gamma) and sometimes doesn’t. For example:
## Interfacing a little more deeply with the eoldas code¶
Whilst considering writing wrapper codes around eoldas functionality and outputs it is instructive to explore some of the data structure available.
We can re-use the example solve_eoldas_identity.py developed above and access some of the data structure as shown in solve_eoldas_identity_a.py.
#!/usr/bin/env python
import pdb
import numpy as np
if __name__ == "__main__":
import sys,tempfile
this = sys.argv[0]
import eoldas
from solve_eoldas_identity import *
# import the setup methods from solve_eoldas_identity
import pylab as plt
# SD of noise
noise=0.15
# nominal sampling period
n_per=7
# proportion of missing samples
obs_off=0.33
# order of differential model (integer)
model_order=1
# sgwindow is larger to create larger data gaps
sgwindow=10
# set up data for this experiment
file, ideal_gamma,doys,ndvi_clean,ndvi,qa_flag = \
prepare(noise=noise,n_per=n_per,\
obs_off=obs_off,model_order=model_order,\
sgwindow=sgwindow)
# set gamma to thge theoretical value
gamma = ideal_gamma
# set op file names
xfile = 'output/Identity/NDVI_Identity_a.params'
yfile = 'output/Identity/NDVI_fwd_a.params'
# initialise options for DA overriding any in config files
# make sure we use some different output file names to othe scripts
cmd = 'eoldas ' + \
' --conf=config_files/eoldas_config.conf --conf=config_files/Identity.conf ' + \
' --logfile=mylogs/Identity.log ' + \
' --calc_posterior_unc ' + \
' --parameter.solve=0,1 ' + \
' --parameter.result.filename=%s '%xfile +\
' --parameter.x.default=%f,0.0 '%(gamma) + \
' --operator.obs.y.result.filename=%s'%yfile +\
' --operator.obs.y.state=%s'%file+\
' --operator.modelt.rt_model.model_order=%d '%model_order
# initialise eoldas
self = eoldas.eoldas(cmd)
# solve DA
self.solve(write=True)
# now pull some data out of the eoldas
# the 'root' of the DA data structure is in self.solver.root
root = self.solver.root
# The state vector data are stored in root.x
# with ancillary information in root.x_meta
# so the state vector is e.g. in root.x.state
# and the names are in root.x_meta.state
state_names = root.x_meta.state
state = root.x.state
# The sd representation of the posterior is in root.x.sd
# This is all set up in eoldas_Solver.py
# All storage is of type ParamStorage, an extended
# dictionary structure. You can explore it interactively
# with e.g. root.dict().keys() or self.keys() since
# self here is a straight dictionary
sd = root.x.sd
# The full inverse Hessian is in self.solver.Ihessian
Ihessian = self.solver.Ihessian
# A mask to reduce this to only the state variables
# being targeted (solve == 1) is through a call to:
# This is of shape (365,365) here.
# so now lets produce an image of it
# to visualise the structure
fig = plt.figure()
cax = ax.imshow(NDVI_Ihessian,interpolation='nearest')
ax.set_title('Posterior Uncertainty matrix for NDVI')
# Add colorbar, make sure to specify
# tick locations to match desired ticklabels
cbar = fig.colorbar(cax) #), ticks=[-1, 0, 1])
#cbar.ax.set_yticklabels(['< -1', '0', '> 1'])# vertically oriented colorbar
# see http://matplotlib.sourceforge.net/plot_directive/mpl_examples/pylab_examples/colorbar_tick_labelling_demo.py
# save it
plt.savefig('output/IHessianNDVI_expt1.png')
The comments in the code should be self explanatory and anyone interested in delving much further into the eoldas codes should see the full class documentation. Here, we can see at least how to access the posterior estimate of the state and its uncertainty. We write out the uncertainty to an image using matplotlib (pylab):
Now, this is a very interesting figure for understanding how these multiple constraints are interacting. The observation uncertainty is just described by standard deviation, so lies along the leading diagonal of the a priori uncertainty. Further, it only exists where there are data points. The impact of applying the (regularisation) model constraint is to reduce the uncertainty at the observation points as we would expect. We can see a ‘sausage’ pattern from above in this figure quite clearly. The ‘pinch points’ are when the sample points are dense. Where there are large data gaps (from our simulated cloud impacts here) the uncertainty is higher, but it is ‘spread out’ from the leading diagonal by applying temporal covariance. The impact of the filtering is large where the observation impact is low (or non existant). We will see these same effects in many DA experiments.
## Running EOLDAS from the command line¶
An alternative to writing your own python code for the front end is to use eoldas.py, which can be direcly run from the command line.
Help on command line options is available by typing::
eoldas_run.py --help
As an aid to setting up the correct python variables etc, a front end script, eoldas can also be accessed (in the bin directory).
eoldas_run.py --conf=confs/Identity.conf
N.B. Make sure the eoldas_run.py script is in your path. If you install the python packages for a single user, in UNIX it will usually be under ~/.local/bin/. You may want to add that path to your users’ path.
### Application to MODIS reflectance data¶
We can now apply the concepts demonstrated above to real EO data from the MODIS sensors.
This is actually quite trivial to achieve as it involves the same basic configuration file as previously. This time, we will access it from eoldas_run.py.
## The MODIS data file format¶
An example data file then is data/modis_botswana.dat which is MODIS reflectance data for a site in Botswana. The format of this file is ‘BRDF’, which looks like:
BRDF 67 7 645 858.5 469 555 1240 1640 2130 0.003 0.004 0.004 0.015 0.013 0.01 0.006
181 1 62.830002 -83.040001 37.910000 23.219999 0.032900 0.068600 0.018000 0.028400 0.098100 0.097600 0.081900
182 1 33.980000 99.540001 44.970001 39.290001 0.050500 0.091800 0.030000 0.045300 0.123200 0.137100 0.124800
184 1 51.660000 99.300003 46.889999 42.349998 0.056200 0.106000 0.029500 0.053800 0.151600 0.154600 0.141400
185 1 32.880001 -81.900002 40.540001 31.510000 0.039200 0.069600 0.022900 0.034400 0.089800 0.102500 0.092800
186 1 63.049999 100.470001 48.910000 45.169998 0.058200 0.138600 0.038900 0.055200 0.184600 0.183800 0.143100
187 1 6.430000 -79.459999 42.090000 35.259998 0.043700 0.074600 0.026400 0.038700 0.103100 0.118400 0.103600
189 1 22.389999 97.720001 43.790001 38.750000 0.044500 0.088400 0.027100 0.040200 0.129500 0.135800 0.127600
190 1 57.400002 -82.720001 38.029999 26.480000 0.050700 0.087700 0.024600 0.045200 0.119900 0.118000 0.098100
191 1 44.040001 99.419998 45.619999 41.990002 0.062800 0.107800 0.035600 0.057100 0.141600 0.154800 0.137600
192 1 42.689999 -82.820000 39.290001 30.719999 0.041300 0.078800 0.022700 0.036600 0.106400 0.109000 0.096100
193 1 58.139999 99.800003 47.560001 45.000000 0.063500 0.124200 0.034900 0.060800 0.169700 0.170900 0.143400
194 1 20.129999 -82.099998 40.730000 34.689999 0.038600 0.073200 0.024000 0.035100 0.106900 0.111700 0.105200
196 1 8.940000 96.330002 42.330002 38.389999 0.052200 0.090000 0.030300 0.047200 0.118400 0.131700 0.123100
197 1 62.480000 -83.120003 36.610001 25.559999 0.053800 0.092700 0.025000 0.049200 0.118100 0.120200 0.093200
198 1 34.730000 99.230003 44.060001 41.840000 0.054500 0.103800 0.030300 0.051400 0.144800 0.147200 0.133300
199 1 50.549999 -82.849998 37.750000 30.030001 0.044100 0.082200 0.025500 0.041300 0.104800 0.114500 0.093900
200 1 52.070000 100.010002 45.910000 45.029999 0.074300 0.132000 0.037400 0.069800 0.174200 0.169700 0.143300
201 1 32.130001 -82.889999 39.080002 34.240002 0.041500 0.077000 0.026200 0.038900 0.071200 0.108400 0.096300
202 1 63.400002 100.080002 47.880001 48.009998 0.082600 0.155400 0.043600 0.073800 0.191000 0.187000 0.137200
203 1 5.270000 -76.970001 40.590000 38.189999 0.053000 0.092500 0.026800 0.047500 0.122900 0.121400 0.109300
205 1 23.219999 97.779999 42.220001 41.840000 0.067000 0.113900 0.034600 0.060000 0.143100 0.148500 0.131800
206 1 56.959999 -83.139999 35.910000 29.340000 0.065800 0.102000 0.033800 0.062000 0.117800 0.113800 0.087400
The first line is a header, as with the previous ‘PARAMETERS’ format we saw. The first word on the header must be BRDF (a # can be included for compatitbilty with some other formats).
The second item on the header line is the number of samples in the file (92 here). The third item is the number of wavebands represented (7 here) then this is followed by wavelengths, wavebands, or waveband identifiers.
If a single float number is given, it is assumed to represent a narrow waveband centred around that wavelength. Wavelength is assumed to be in nm. If it is two float numbers connected by a dash (e.g. 45-550) then it is taken to be a tophat function representation of a waveband, where the two number represent the minimum and maximum wavelength respectively. If some other code is found (e.g. B2), it is assumed to be a tag associated with a particular waveband. In this case, any spectral libraries loaded in the configuration files are searched to see if they contain a definition of this waveband. If nothing is found and it cannot be interpreted as a number, it is assigned an arbitrary wavelength (index starting from 0) and should not be interpreted as a wavelength identifier.
The 7 columns following the wavebands specify the standard deviation in the reflectance data. All subsequent lines contain reflectance, as a function of ‘locational’, ‘control’ and spectral information. The first column here defines location (time), columns 2 to 6 are ‘control’ variables: things the observation waries as a function of that are not spectral or locational. In this case, they are (all angles are in degrees):
# a data mask (1 for good), # view zenith angle, # view azimuth angle, # solar zenith angle, # solar azimuth angle.
The final 7 columns give the reflectance in the 7 wavebands for each location defined.
The data span days of year 181 to 272 inclusive. There is a lot of day to day variation in the data, but a clear underlying trend in the Near infrared. We will set ourselves the task of trying to use eoldas to filter the dataset so that the trend becomes more apparent. We will also try to extrapolate from the sample days to the whole year (which is clearly quite a challenge, but instructive for understanding uncertainty).
We will apply the same data assimilation components as previously, i.e. an Identity operator for the NIR reflectance (in other words, we take the reflectance as a prior estimate of what we wish to estimate) and a regularisation filter implemented as a constraint on first order derivatives at lag one day. We will suppose that we know the uncertainty in this model (the temporal constraint) to be expressed as a standard deviation of .002 (1/500), i.e. the root mean squared deviation from one day to the next is expected to be around .002. An examination of the dataset will show that the NIR reflectance roughly increases from around 0.1 to 0.2 over about 100 days, so we could use a gamma value of 1000. However, we don’t want to impose that constraint too strongly, so we will instead use a gamma of 500. We are not trying to fit a trendline here, just to smooth the dataset so that we can detect the underlying behaviour. The result should however not be very sensitive to this guess and can always be explored using approaches such as cross validation.
## The configuration file¶
We have a configuration file set up in config_files/Identity2.conf. This provides a full description of the problem we wish to solve, which is to reduce noise in the NIR observation series.
# An EOLDAS configuration file for
# a simple Identity operator and a regulariser
[parameter]
location = ['time']
limits = [[1,365,1]]
help_limits="define the limits for the required locational information"
names = "gamma_time 858.5".split()
solve = 0,1
datatypes = x
[parameter.result]
filename = output/Identity/MODIS_botswana.params
format = 'PARAMETERS'
help_filename='Set the output state filename'
[parameter.x]
datatype = x
names = $parameter.names default = 500,0.1 state= data/modis_botswana.dat help_default="Set the default values of the states" apply_grid = True sd = [1.]*len($parameter.names)
bounds = [[0.000001,1000000],[0,1]]
[general]
doplot=True
help_do_plot='do plotting'
calc_posterior_unc=False
[operator]
modelt.name=DModel_Operator
modelt.datatypes = x
obs.name=Operator
obs.datatypes = x,y
[operator.modelt.x]
names = $parameter.names sd = [1.0]*len($operator.modelt.x.names)
datatype = x
[operator.modelt.rt_model]
model_order=1
help_model_order='The differential model order'
wraparound=periodic,365
#wraparound=None
[operator.obs.x]
names = $parameter.names[1:] sd = [1.0]*len($operator.obs.x.names)
help_sd='Set the observation sd'
datatype = x
[operator.obs.y]
names = $parameter.names[1:] sd = [0.015]*len($operator.obs.x.names)
#help_sd="set the sd for the observations"
datatype = y
state = data/modis_botswana.dat
help_state = "Set the y state vector"
[operator.obs.y.result]
filename = output/Identity/Botswana_fwd.params
format = 'PARAMETERS'
help_filename = 'Set the fwd modelling result filename'
In this, we can see the required location field defined in the [parameters] section (i.e. parameters.location=[[1,365,1]]). The control variables are associated with the observation operator, here given as [mask,vza,vaa,sza,saa].
To run eoldas with this configuration then all we need do is type:
eoldas_run.py --conf=config_files/eoldas_config.conf \
--conf=config_files/Identity2.conf --calc_posterior_unc
This writes the files:
output/Identity/MODIS_botswana.params: state estimation (smoothed reflectance)
181.000000 1.000000 500.000000 0.095014 0.000000 0.005513
182.000000 1.000000 500.000000 0.095247 0.000000 0.005255
183.000000 1.000000 500.000000 0.095541 0.000000 0.005063
184.000000 1.000000 500.000000 0.095835 0.000000 0.004840
185.000000 1.000000 500.000000 0.095949 0.000000 0.004669
186.000000 1.000000 500.000000 0.096530 0.000000 0.004541
187.000000 1.000000 500.000000 0.096364 0.000000 0.004451
188.000000 1.000000 500.000000 0.096585 0.000000 0.004396
189.000000 1.000000 500.000000 0.096806 0.000000 0.004293
190.000000 1.000000 500.000000 0.097176 0.000000 0.004216
191.000000 1.000000 500.000000 0.097714 0.000000 0.004161
output/Identity/Botswana_fwd.params: forward modelling of y
#PARAMETERS time mask,vza,vaa,sza,saa 858.5 sd-858.5
181.000000 0.000000 0.095014 0.000000
182.000000 0.000000 0.095247 0.000000
184.000000 0.000000 0.095835 0.000000
185.000000 0.000000 0.095949 0.000000
186.000000 0.000000 0.096530 0.000000
187.000000 0.000000 0.096364 0.000000
189.000000 0.000000 0.096806 0.000000
190.000000 0.000000 0.097176 0.000000
191.000000 0.000000 0.097714 0.000000
as well as the orignal data in output/Identity/Botswana_fwd.params_orig
#PARAMETERS time mask,vza,vaa,sza,saa 858.5 sd-858.5
181.000000 0.000000 0.068600 0.015000
182.000000 0.000000 0.091800 0.015000
184.000000 0.000000 0.106000 0.015000
185.000000 0.000000 0.069600 0.015000
186.000000 0.000000 0.138600 0.015000
187.000000 0.000000 0.074600 0.015000
189.000000 0.000000 0.088400 0.015000
190.000000 0.000000 0.087700 0.015000
191.000000 0.000000 0.107800 0.015000
with appropriate graphics for the state:
and the y data:
The resultant state data are quite instructive: where we have observations, the uncertainty is reduced from 0.015 to around 0.004 (the actual degree of noise reduction depends on the value of gamma used). Where there are no data, the uncertainty grows to around 0.017. It is slightly reduced at the year start/end because of the wrparound condition used here.
In this case, we have used a first order differential constraint with a periodic boundary condition. These are quite important in this case: we only have observations in a limited time window, so using a periodic boundary condition is one way to place some form of constraint at what happens when we have no data. The first order differential model will in essence perform a linear interpolation where there are no data, which is probably appropriate for this case. You can try changing the model order to see what happens. Run e.g.:
eoldas_run.py --conf=config_files/eoldas_config.conf \
--conf=config_files/Identity2.conf --calc_posterior_unc \
--operator.modelt.rt_model.model_order=2 \
--parameter.x.default=5000,0.1 \
--operator.obs.y.result.filename=output/Identity/Botswana_fwd.params2 \
--parameter.result.filename=output/Identity/MODIS_botswana.params2
and have a look at output/Identity/MODIS_botswana.params2.plot.x.png:
The assumed behaviour is quite different to the first order differential constraint outside of the observations. With a high value of gamma, the result is essentially a straight line where there are observations. The influence of the wraparound condition is also clear here.
If we used a lower gamma, we would see some features of using a second order model:
eoldas_run.py --conf=config_files/eoldas_config.conf \
--conf=config_files/Identity2.conf --calc_posterior_unc \
--operator.modelt.rt_model.model_order=2 \
--parameter.x.default=200,0.1 \
--operator.obs.y.result.filename=output/Identity/Botswana_fwd.params3 \
--parameter.result.filename=output/Identity/MODIS_botswana.params3
Where there are observations, the second order difference constraint would have high frequency oscillations. This can be a positive feature or an annoyance: it all depends on how you expect the function to behave. It is not an arbitrary choice: the user has imposed a particular expectation of behaviour here through the model. There clearly needs to be some evidence for choosing one form of model over another, although that is not always straightforward.
One other interesting feature of this result is that the mean estimate is bounded (0.0,1.0) which is reasonable for reflectance data (theoretically, BRF can go above 1, but this is rarely observed). This means that the mean value (the red line) is forced to stay above 0 (days 100 to 150) even though the apparent trend from the observations might otherwise make it go below zero. This condition is imposed in the configuration by the line:
bounds = [[0.000001,1000000],[0,1]]
One further thing to note about this result is that when the uncertainty is so large (as is the case here when we are extrapolating) that the confidence interval spans more than the entire data range (0,1) the mean and standard deviation described in the Gaussian statistics are a little meaningless or at least should be interpreted with caution. It might in this case be better to describe the data (in the extrapolated region) as simply being somewhere between the physical limits.
A final comment is that actually, a large amount of the departure of the signal from our smooth trajectory, the high frequency variation that we have treated as noise here, can most likely be described by considering the physics os scattering by the surface (there are, on the whole, BRDF effects). We will return to this later.
As well as raising a few issues with regard to model selection, we have demonstrated further use of the eoldas command line for quickly changing some terms in an experiment. However, it is all very well in showing that eoldas can ingest satellite data and apply constraints to provide an estimate of state variables, but the state variables here (reflectance) do not directly help us monitor the properties of the land surface or link to process models.
To do that using EO data, we generally need more complex obeseravtion operators. These are dealt with in the next section. | 13,322 | 47,657 | {"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} | 2.921875 | 3 | CC-MAIN-2019-22 | latest | en | 0.914041 |
https://achtungbeauty.com/qa/what-is-12am-in-24hr-time.html | 1,603,542,222,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882581.13/warc/CC-MAIN-20201024110118-20201024140118-00028.warc.gz | 193,290,411 | 8,137 | # What Is 12am In 24hr Time?
## Is 12am in the morning?
12am is the exact moment the 12th hour of the morning finishes (am), and similarly for pm.
Therefor 12am is midday and pm would start straight after.
The confusion has started since the invention of the digital clock..
## Is it 12 am tomorrow or today?
Strictly speaking, there is no 12 AM just as there is no 12 PM. The correct usages are “midnight” and “noon” respectively.
## Why is 12 pm noon?
The abbreviation am stands for ante-meridiem (before the Sun has crossed the meridian line) and pm stands for post-meridiem (after the Sun has crossed the meridian line). At 12 noon, the Sun is at its highest point in the sky and directly over the meridian. It is therefore neither ‘ante’ (am) nor ‘post’ (pm).
## Is it 12am 0000 or 2400?
12 A.M. is often referred to as both 0000 and 2400 hours. However, Clocks that display Military Time always display it as 0000.
## How can I remember 12 AM and PM?
If midnight were 12 PM, then the minute before 1 AM would be 12:59 PM, which wouldn’t make sense. The words come from Latin, but you can remember PM with an English mnemonic as “post-midday” — anything after noon is post-midday, including 12:00:01 and 12:59 (one minute before 1 PM).
## Why is 12am at night?
Midnight is exactly 180 degrees apart from noon because the earth is round in the network of geographical coordinates the meridian of Greenwich indicated 12:00 noon Any point on the surface of planet earth at 180 degrees of the meridian of Greenwich or “0” is marking midnight this means that it is neither am nor pm.
## Does 12pm come after 11am?
As standard 12:00 AM is taken as the start of the day. Time of the day is divided in two halves (24hour= 12 hours before midday + 12 hours after midday) so the 11 AM is 11th hour before midday as clock strikes 12 again it changes to PM (after midday) so it’s 12 PM after 11 AM and 12 AM after 11 PM.
## What time is 12am?
Another convention sometimes used is that, since 12 noon is by definition neither ante meridiem (before noon) nor post meridiem (after noon), then 12am refers to midnight at the start of the specified day (00:00) and 12pm to midnight at the end of that day (24:00).
## How many hours is 12am to 12pm?
12 hoursHow many hours from 12am to 12pm? There are 12 hours from 12am to 12pm.
## Is it Am Night or Day?
Using numbers from 1 to 12, followed by am or pm, the 12-hour clock system identifies all 24 hours of the day. For example, 5 am is early in the morning, and 5 pm is late in the afternoon; 1 am is one hour after midnight, while 11 pm is one hour before midnight.
## Is 12 am midnight the next day?
12:00am is midnight. 12:00pm is noon. Midnight is usually defined as 12:00:00 a.m., even though logically it’s neither “ante-meridiem” nor “post-meridiem”. In fact a new day is usually defined as beginning at 12 a.m. exactly, even though no time has elapsed in the new day at that time. | 809 | 2,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-45 | latest | en | 0.942547 |
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