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https://physics.stackexchange.com/questions/422703/what-is-the-interpretation-of-langle-ab-rangle | 1,571,105,512,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00093.warc.gz | 609,304,465 | 28,904 | # What is the interpretation of $|\langle AB\rangle|$? [closed]
what is the interpretation of $|\langle AB \rangle|$ ? since $|\langle AB \rangle|^2 = \langle AB \rangle [\langle AB \rangle]^\dagger$ , could $|\langle AB \rangle| = \sqrt(|\langle AB \rangle|^2)$ ?
PS. : feel free to correct me
## closed as unclear what you're asking by Emilio Pisanty, probably_someone, Norbert Schuch, John Rennie, AccidentalFourierTransformAug 14 '18 at 19:59
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• Your question is rather unclear - please use clearer notation, using MathJax wherever necessary. – Emilio Pisanty Aug 14 '18 at 15:01
• Should this be migrated to math? – Aaron Stevens Aug 14 '18 at 17:37
• What is $\langle AB \rangle$? Is it the expectation value of the product of two observables? Is it a misnotated inner product of two states (which should be written $\langle A | B \rangle$)? More generally, are $A$ and $B$ states, operators, observables, or something else entirely? – probably_someone Aug 14 '18 at 17:41
• yes it is the expectation value of the operator AB , calculated on some system state $|\psi \rangle$ . when we generally type $\langle AB \rangle$ , it generally means the expectation value of the product of two operators A and B – gautam bhuyan Aug 15 '18 at 7:45
$| \langle AB \rangle|=|\langle \Psi|AB|\Psi \rangle|$ for some operators $A,B$ acting on state $|\Psi \rangle$. Using for example a position representation of $A,B$ we can use a wave function i.e. $\Psi= \langle q|\Psi \rangle$ where $|q \rangle$ is a position eigenket (in some dimension, nothing is specified so it is hard to imagine the dimensionality of the problem 1D, 3D etc.). Then we get:
$| \langle A_{pos}B_{pos} \rangle|=|\int\Psi^* A_{pos} B_{pos}\Psi d^3x|$ where $A_{pos}$is the position representation of the operator $A$. If $A$ already is in position representation, i.e. $\hat{x}=x,\hat{p}=-i\hbar \partial_x$ for 1D then you could say $A_{pos}=A$ and ignore the first part.
• But $|\langle AB \rangle |$ is not same as the expectation of the operator AB calculated on some system state $| \psi \rangle$ , it should mean something else right ? – gautam bhuyan Aug 15 '18 at 7:35
• thats's why i am asking how do i interpret this , as modulus of some complex number or square root of the $l_2$ norm , i.e , the square root of $||AB |\Psi \rangle||^2$ ? – gautam bhuyan Aug 15 '18 at 11:15
• how can $|\langle AB \rangle |$ be equal to $\langle AB \rangle$ ? – gautam bhuyan Aug 15 '18 at 11:24
• That is only true if $AB$ is Hermitian, so that the expectation value is real! – Dani Aug 15 '18 at 11:31 | 832 | 2,897 | {"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} | 3.3125 | 3 | CC-MAIN-2019-43 | latest | en | 0.787462 |
http://mathoverflow.net/revisions/4024/list | 1,369,124,675,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699798457/warc/CC-MAIN-20130516102318-00035-ip-10-60-113-184.ec2.internal.warc.gz | 166,235,024 | 5,423 | 2 added 446 characters in body
This is true for quadratic polynomials with constant term +1. Any such polynomial is the determinant of a matrix in SL2 ℤ (e.g. using the companion matrix as indicated by Ben Webster). It's well known that any such matrix is conjugate to a multiple product of [[1,1],[0,1]] and [[1,0],[1,1]] (upper and lower triangular unipotent matrices). I'm not sure the original reference for this fact, but a reference is Proposition 2.1 of this paper.
I believe that your criterion implies that the maximal root of the polynomial is a Perron number. If so, then Lind has shown that every Perron number occurs as the spectral radius of a non-negative integral Perron-Frobenius matrix (and therefore the spectral radius of a recurrent digraph). This only implies that the polynomial divides the characteristic polynomial of the matrix - there might be other factors.
Added comment: The general quadratic case might be possible to work out using Markov partitions of the induced map of a torus.
I forgot about the cyclotomic case, which can occur if the matrix is not Perron-Frobenius. If the polynomial is irreducible, I think the condition implies that the maximal norm roots are complex Perron numbers (or cyclotomic). These crop up in work of Kenyon on self-similar tilings (MR1392326 (97j:52025) ).
1
This is true for quadratic polynomials with constant term +1. Any such polynomial is the determinant of a matrix in SL2 ℤ (e.g. using the companion matrix as indicated by Ben Webster). It's well known that any such matrix is conjugate to a multiple product of [[1,1],[0,1]] and [[1,0],[1,1]] (upper and lower triangular unipotent matrices). I'm not sure the original reference for this fact, but a reference is Proposition 2.1 of this paper.
I believe that your criterion implies that the maximal root of the polynomial is a Perron number. If so, then Lind has shown that every Perron number occurs as the spectral radius of a non-negative integral Perron-Frobenius matrix (and therefore the spectral radius of a recurrent digraph). This only implies that the polynomial divides the characteristic polynomial of the matrix - there might be other factors. | 496 | 2,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2013-20 | latest | en | 0.908641 |
https://www.mathworks.com/matlabcentral/cody/problems/1974-length-of-a-short-side/solutions/1621682 | 1,600,431,140,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187390.18/warc/CC-MAIN-20200918092913-20200918122913-00102.warc.gz | 1,269,892,383 | 18,168 | Cody
# Problem 1974. Length of a short side
Solution 1621682
Submitted on 4 Sep 2018 by ilker golcuk
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
b = 1; c = 2; a_correct = sqrt(3); tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance);
2 Pass
b = 4; c = 5; a_correct = 3; tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance);
3 Pass
b = 12; c = 13; a_correct = 5; tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance);
4 Pass
b = 8; c = 10; a_correct = 6; tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance); | 248 | 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} | 2.890625 | 3 | CC-MAIN-2020-40 | latest | en | 0.657917 |
https://ask.cvxr.com/t/cvx-invalid-quadratic-form-s-not-a-square/3160 | 1,680,434,802,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00153.warc.gz | 148,106,897 | 4,593 | # CVX Invalid quadratic form(s): not a square
Hi, can you help me for the following question?
When I use CVX to solve the following problem, I encounter the following question:
``````R_min=1;
M=1;
K=1;
Nt=4;
eta=1;
psi_s=0.001;
psi_e_k=0.001;
P_In_m=0.01;
P_th=1;
sigma_s=10^(-5);
sigma_e=10^(-5);
sigma_p_s=10^(-7);
N=1000;
power=1;
h=wgn(Nt,1, power,'linear','complex');
g=wgn(Nt,1, power,'linear','complex');
xi_e_k=0.1*mean((abs(g)).^2);
q=wgn(Nt,1, power,'linear','complex');
xi_p_m=0.1*mean((abs(q)).^2);
beta=1.0001;
cvx_begin SDP
variable W(4,4) complex semidefinite;
variable Si(4,4) complex semidefinite;
variable rho(1) nonnegative;
variable omega(1) nonnegative;
variable a(1) nonnegative;
variable lambda(1) nonnegative;
minimize(trace(W+Si));
subject to
[omega*eye(Nt)-(W-(beta-1)*Si) -(W-(beta-1)*Si)*g;-g'*(W-(beta-1)*Si) (beta-1)*sigma_e-g'*(W-(beta-1)*Si)*g-omega*xi_e_k]==semidefinite(Nt+1);
[a*eye(Nt)+(W+Si) (W+Si)*g; g'*(W+Si) g'*(W+Si)*g+sigma_e-psi_e_k*(eta^(-1))-a*xi_e_k]==semidefinite(Nt+1);
[lambda*eye(Nt)-(W+Si) -(W+Si)*q; -q'*(W+Si) P_In_m-q'*(W+Si)*q-lambda*xi_p_m]==semidefinite(Nt+1);
rho*trace(W+(1-(2^R_min)*beta)*Si)+(1-(2^R_min)*beta)*(rho*sigma_s+sigma_p_s)>=0
(1-rho)*eta*(trace((W+Si)*h*h')+sigma_s)>=psi_s;
trace(W+Si)<=Pth;
0<rho<1;
cvx_end
``````
The constraints:
``````(1-rho)*eta*(trace((W+Si)*h*h')+sigma_s)>=psi_s;
trace(W+Si)<=Pth;
``````
Error using . (line 262)*
Disciplined convex programming error:
** Invalid quadratic form(s): not a square.**
can you help me?
Thanks
Thanks, I have solved the problem with your help.
Thanks very much | 635 | 1,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.547612 |
http://homeenergypros.lbl.gov/forum/topics/using-average-monthly?groupUrl=buildingsimulations&xg_source=activity&groupId=6069565%3AGroup%3A3935&id=6069565%3ATopic%3A4061&page=2 | 1,503,202,053,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105970.61/warc/CC-MAIN-20170820034343-20170820054343-00563.warc.gz | 182,902,599 | 23,458 | Using Average Monthly Temperature to Model Home Energy Use
Discuss how monthly electric and gas billing data can be combined with average monthly temperature data to model a building's energy use, quantify energy conservation opportunities and verify performance improvements.
The following steps are used determine to determine the characteristic curve for a particular building:
Step One - Normalize electricity and/or gas usage in terms of Watts/Ft2 (or Watts/m2).
Electricity, W/Ft2 = ([Monthly kWh] x 1000 Watts/kW) / ([Days/Bill] x 24 Hr/Day x [Bldg. Area])
Natural Gas,
W/Ft2 = ([Monthly Therms] x 100,000 BTU/Therm)
(3.413 BTU/Watt x [Days/Bill] x 24 Hr/Day x [Bldg. Area])
Step Two - Determine average monthly temperature for each billing period.
Step Three - Plot each month's normalized energy use verses that month's average monthly temperature.
After performing this anlysis on many buildings, we have discovered consistent trends. A typical all-electric home with electric resistance heat will use about 3.5 Watts/Ft2 at 20F and approximately 0.75 Watt/Ft2 around 60F, rising to about 1 Watt/Ft2 at 70F. An all-electric home with an air-source heat pump will be similar except the home will use about 2.5 Watts/Ft2 at 20F instead of 3.5 for the electric resistance home.
The slope of the line between 20F and 50F corresponds to the heat loss coefficient of the building due to conduction, infiltration and ventilation. Reducing the slope by a prescribed amount will predict the energy savings that will occur by replacing windows, adding insulation, reducing infiltration or reducing ventilation air.
The intersection of this line with the X axis, typically 70F, indicates the average temperature maintained inside the home. (Although, as Michael Blasnik correctly states in his comments below, the intersection of the UA line can be significantly influenced by solar gains or non-electric loads within the home.) Shifting the line to the left will predict the savings that occur by setting the thermostat back to a lower temperature.
I prefer to use watts per square foot (or watts per square meter) instead of using British Thermal Units (BTU's), kilowatt hours (kWh's) or Joules to normalize a building’s monthly or annual energy use for the following reasons:
1. A watt is one of the few metric measurements that is commonly used and understood by most Americans. (75 watt light bulb, 1,500 watt heater, etc.) Because lighting and heating loads are often stated in terms of watts, it is relatively easy to quantify their contribution to a building’s normalized overall energy when it is also represented in terms of average watts per unit area.
2. Even though a watt is a measurement of power, not energy, I find that energy represented in terms of average watts to be more intuitive than BTU’s, kWh or Joules.
3. Switching between metric and non-metric values is relatively easy. The rule of the thumb conversion between square meters and square feet is a simple factor of ten (1 meter by 1 meter = 10.75 square feet). One watt per square foot is 10.76 watt per square meter.
4. Direct solar gain at the surface of the earth at noon on a clear day near the equator is roughly 1,000 watts per square meter, or approximately 100 watts per square foot. These rule-of-thumb values make it easier to relate solar heat gains to other heat gains within a home.
The graph below shows the metered AC electrical energy generated by a 6.5 kW solar PV array that has been normalized in terms of watts per unit area of a 2,000 sq. ft. all-electric house. Although the energy generated each month by the solar array is not perfectly correlated with average monthly temperature, it shows just how far this house on the east side of the Cascades in the Pacific Northwest has to go to reach net zero energy.
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It seems that we are getting closer to agreement on issue #2. In your initial post you said "The intersection of this line with the X axis, typically 70F, indicates the average temperature maintained inside the home." I disagreed with that claim. In your latest post you say "the x-intercept, when corrected for solar gain, is directly related to interior space temperature". This is a very different claim. I never said that the X intercept isn't related to or affected by the interior temperature. I just said that it wasn't equal to the interior temperature. Now that you have changed your claim, we are pretty much in agreement on this point.
Average monthly temperatures will not represent usage patterns as well as degree day methods. It seems like your main argument against degree days is that they are harder to use. I don't find it to be a problem -- it can be readily automated in whatever analysis tool you use. If you don't want to bother that's up to you -- the temperature plots and fits are useful, just not as good.
Your claims about assessing savings based on one or two months of data are not bolstered by showing one building,. I didn't say that 1 or 2 months of data isn't potentially useful for assessing savings, just that it can be misleading and has considerably larger uncertainty than using more data -- this year they went away for Christmas break and last year they hosted extended family....wow look at all the savings from changing a light bulb...;) I think it boils down to what sorts of claims can and should be made about what billing data can and can't tell us about an individual home. I think you overstate the level of detail and certainty while I exercise a little more caution.
Here is a plot that shows the electricity generated by a 6.5 kW solar PV array compared to the energy consumed by an all-electric 2,000 square foot home that is heated by an air source heat pump. The energy generated by the solar PV array is based on actual monthly kWh production meter reads and has been normalized in terms of the home's square footage.
Typical electricity billing information is shown in yellow below.The information highlighted in yellow above is shown in the bar graph below. The bar chart below does not account for different billing days in each month or for weather differences that occur each month.
The above chart can be normalized for weather variations and different number of billing days in each month by plotting the average kW for each month versus each month's average temperature.
Average kW is equal to: kW = [kWh/Month] / ([Days/Month] x 24 Hr/Day).
The above chart can now be analyzed in terms of heating, cooling and baseload energy consumption.
The above chart can be used to calculate the savings of installing a ductless heat pump or other heat related or air conditioning based conservation measure, such as a ductless heat pump.
The "Generic Billing Analysis" spreadsheet that I uploaded came up as a .zip file instead of an Excel spreadsheet. The spreadsheet below is another attempt to upload the Excel spreadsheet that can be used to conduct a billing analysis of gas and electric bills.
Attachments:
Discussion Forum
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Trying to find %age energy saving from variety of upgrades in 10K sf multifamily house
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How about simulation in the multifamily context?
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Thursday | 1,816 | 8,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-34 | latest | en | 0.858203 |
https://cris.openu.ac.il/ar/publications/on-the-integral-dicycle-packings-and-covers-and-the-linear-orderi | 1,723,035,604,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00088.warc.gz | 145,420,428 | 10,916 | # On the integral dicycle packings and covers and the linear ordering polytope
Zeev Nutov, Michal Penn
نتاج البحث: نشر في مجلةمقالةمراجعة النظراء
## ملخص
The linear ordering polytope PLOn is defined as the convex hull of the incidence vectors of the acyclic tournaments on n nodes. It is known that for every facet of PLOn, there corresponds a digraph inducing it. Let D be a digraph that induces a facet-defining inequality for PLOn, that is nonequivalent to a trivial inequality or to a 3-dicycle inequality. We show that for such a digraph the following holds: the value τ of a minimum integral dicycle cover is greater than the value τ* of a minimum dicycle cover. We show that τ* can be found by minimizing a linear function over a polytope which is defined by a polynomial number of constraints. Let v denote the value of a maximum integral dicycle packing. We prove that if D is a certain digraph with a two-node cut satisfying τ = v in each part, then τ = v in D as well. Dridi's description of PLO5 enables a simple derivation of the fact that τ = v for any digraph on 5 nodes. Combining these results with the theorem of Lucchesi and Younger for planar digraphs as well as Wagner's decomposition, we obtain that τ = v in K3.3-free digraphs. This last result was proved recently by Barahona et al. (1990) using polyhedral techniques while our proof is based mainly on combinatorial tools.
اللغة الأصلية الإنجليزيّة 293-309 17 Discrete Applied Mathematics 60 1-3 https://doi.org/10.1016/0166-218X(94)00060-Q نُشِر - 23 يونيو 1995 نعم
### ملاحظة ببليوغرافية
Funding Information:
Let Dn be the complete digraph on n nodes. Let P~.o denote the convex hull of all the incidence vectors of arc sets of linear orderings of the nodes of D. (i.e. these are exactly the incidence vectors of the acyclic tournaments of Dn). P~.o is called the linear ordering polytope. The linear ordering problem consists of maximizing a linear function over P~.o. The investigation of the linear ordering problem is motivated by *Corresponding author. tpart of this work was done as part of the author's M.Sc. thesis, done under the supervision of Dov Monderer and Michal Penn, in the Department of Applied Mathematics. Technion, Haifa, Israel. "Research of this author was partially supported by the Mendence France Fellowship Trust.
## بصمة
أدرس بدقة موضوعات البحث “On the integral dicycle packings and covers and the linear ordering polytope'. فهما يشكلان معًا بصمة فريدة. | 635 | 2,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.872782 |
http://mathhelpforum.com/algebra/28495-algebra-story-problem-need-help-asap-print.html | 1,508,789,294,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826283.88/warc/CC-MAIN-20171023183146-20171023203146-00633.warc.gz | 225,158,842 | 4,192 | # Algebra story problem. Need Help ASAP
• Feb 17th 2008, 04:27 PM
armath
Algebra story problem. Need Help ASAP
ok...you have some $1 bills,$5 bills, and $10 bills. There are 225 bills total and they are worth$596. If you have seven times as many $1 as you do$10, how many of each bill do you have?????
i have tried this problem over 20 times!!!!!!(Thinking)
Sincerly
armath
• Feb 17th 2008, 04:39 PM
mr fantastic
Quote:
Originally Posted by armath
ok...you have some $1 bills,$5 bills, and $10 bills. There are 225 bills total and they are worth$596. If you have seven times as many $1 as you do$10, how many of each bill do you have?????
i have tried this problem over 20 times!!!!!!(Thinking)
Sincerly
armath
Let x = number of $1 bills, y = number of$5 bills and z = number of $10 bills. Then: x + y + z = 225 .... (1) x + 5y + 10z = 596 .... (2) x = 7z .... (3) I suggest substituting (3) into (1) and (2) to eliminate x. Now solve (1) and (2) simultaneously for z and y. Now sub the value of x into (3) to get x. • Feb 17th 2008, 05:11 PM angel.white Quote: Originally Posted by armath ok...you have some$1 bills, $5 bills, and$10 bills. There are 225 bills total and they are worth $596. If you have seven times as many$1 as you do \$10, how many of each bill do you have?????
i have tried this problem over 20 times!!!!!!(Thinking)
Sincerly
armath
You have 2 equations here, let "A" stand for the number of 1 dollar bills you have, "B" stand for the number of 5 dollar bills you have, and C stand for the number of 10 dollar bills you have.
Then you know that the total number of bills is 225, so A + B + C = 225
And you know that each A is worth 1 dollar, each B is worth 5 dollars, and each C is worth 10 dollars, and that you have 596 dollars total. So A + 5B + 10C = 596
So set the equations next to eachother:
$\begin{array}{ccccc}
A&+B&+C&=&225\\
A&+5B&+10C&=&596\\
\end{array}$
Now, we have three variables, but only 2 equations. But we know that there are 7 times as many ones as tens. This means that the number of ones is equal to 7 times the number of tens. (Think if there was 1 ten, then there would need to be 7 ones) so A = 7C. This means we can substitute out A:
$\begin{array}{ccccc}
7C&+B&+C&=&225\\
(7C)&+5B&+10C&=&596\\
\end{array}$
Simplify
$\begin{array}{ccccc}
&B&+8C&=&225\\
&5B&+17C&=&596\\
\end{array}$
Now you can solve this any way you like, I'll do it with substitution. Since B + 8C = 225, it we can see that B = 225-8C So lets substitute that into our other equation:
$5(225-8C) +17C = 596$
$1125-40C +17C = 596$
$-23C = -529$
$C = 23$
Now we can plug that back into our other equations:
$\begin{array}{ccccc|cccccc}
A&+B&+C&=&225& \Longrightarrow& A&+B&+23&=&225\\
A&+5B&+10C&=&596&&A&+5B&+230&=&596\\
\end{array}$
And simplify
$\begin{array}{cccc}
A&+B&=&202\\
A&+5B&=&-366\\
\end{array}$
Again, you can solve this any way you like, I'll use matrices this time:
$\left(\begin{array}{cccc}
A&+B&=&202\\
A&+5B&=&366\\
\end{array}\right)$
Subtract line 1 from line 2:
$\left(\begin{array}{cccc}
A&+B&=&202\\
&4B&=&164\\
\end{array}\right)$
Divide line 2 by 4
$\left(\begin{array}{cccc}
A&+B&=&202\\
&B&=&41\\
\end{array}\right)$
And substitute The value of B into the first equation:
$\left(\begin{array}{cccc}
A&+41&=&202\\
&B&=&41\\
\end{array}\right)$
Subtract 41 from line 1
$\left(\begin{array}{cccc}
A&&=&161\\
&B&=&41\\
\end{array}\right)$
So we have A=161, B=41, and C=23
We can verify our answer by plugging them back into our equations:
$\begin{array}{ccccc|cccccc}
A&+B&+C&=&225&\Longrightarrow&161&+41&+23&=&225\ \
A&+5B&+10C&=&596&&161&+5(41)&+10(23)&=&596\\
\end{array}$
You can use your calculator to verify that these answers are correct (they are, I checked them, but you should get into that practice).
And so we only need to verify that they also meet the last criteria that A=7C, so we plug the values in: 161=7(23) and we see that this is correct as well. So we know that our answer is correct.
Some parting advice: Make sure you set them up right, ie seven times as many ones as tens might make you want to say 7A=C, but plug in a value for A, and you will see that C is actually 7 times bigger than A, so it must be the other way around that A=7C. Second, when manipulating equations, use one to manipulate the other, you can never use an equation to manipulate itself, because you will end up with A=A or 0=0 or some other such useless equation. So if you solve for a variable in one equation, use that on the other equation. | 1,504 | 4,516 | {"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": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-43 | longest | en | 0.927058 |
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posted by .
Calculate the tidal force experienced by Io. How does it compare to the tidal force experienced by the Moon due to the Earth? What would the Earth-Moon distance (i.e., distance between their centres) need to be in order for the Moon to experience similar tidal forces to those experienced by Io due to Jupiter?
• astronomy -
According to
http://en.wikipedia.org/wiki/Tidal_force ,
the force (or differential acceleration between opposite sides of a spherical body) is
2 G r M/R^3
where R is the distance to the body exerting the gravitational force, M is its mass, G is the universal constant of gravity, and r is the radius of the body for which the tidal force is being calculated.
The ratio of tidal forces of Io to those of our moon is
Force(Io)/Force(moon) = (r/r')(M/M')*(R'/R)^3
where
M = Jupiter mass
M' = Earth mass
R = Io-Jupiter distance
R' = Earth-Moon distance
Use that formula to answer both questions. There are a lot of numbers to look up.
For your second question, assume the tidal force ratio is 1 and solve for the R'/R value needed to make that happen.
• astronomy -
Thank you very much!
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https://electronics.stackexchange.com/tags/power-amplifier/hot | 1,575,823,334,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540511946.30/warc/CC-MAIN-20191208150734-20191208174734-00105.warc.gz | 354,011,454 | 27,698 | # Tag Info
8
The classes were originally arranged to describe the conduction angle for a single quadrant of a power amplifier. In class-A, active conduction occurs throughout all $360^\circ$ of the period. In class-B, active conduction occurs for $180^\circ$, or one-half of the period. In class-C, active conduction occurs for $\lt 180^\circ$ (but more than $0^\... 7 The diodes are part of a RF switch that switches between RX/TX modes of the transceiver (See also my answer to another question). This can be accomplished by changing the DC potential at the RX LINE side of the diodes. You can see a RF filter of the TX LINE side consisting of R150 C174, C176 and L40. I'm sure there is a similar filter on the RX line side (... 6 Simply redraw the circuit and you will see from where this base current is coming from. simulate this circuit – Schematic created using CircuitLab As you can see the$Q_2$base current is coming from$\textrm{Vee}$voltage source. And flows this path: +Vee--->R1---> Q2 base-emitter junction--->-Vee And remember that in the electronic ... 5 That the amplifier needs a DC supply, not AC. The + G - is a strong give away. An AC power input would likely have a pair of ~ symbols. Also an AC input would need at least a bridge rectifier and a big capacitor. I spot neither near the power supply connector. 5 Yes, LM386 is designed to supply far more current than most op-amps. However, LM386 should not be viewed as an op-amp. It is designed to drive low-resistance loads like a speaker, while op-amps have far more applications. LM386 does have differential inputs (inverting and non-inverting). And it does have a single output, like an op amp. But its internal ... 4 The collector of$Q_{19}$cannot be more than a diode drop above the bottom rail, because its base is tied to the collector.$D_5$then prevents the collector of$Q_{12}$from being more than two diode drops above the bottom rail. Since the$Q_6$(I was wrong to write$Q_8$before, in comments) base is tied there too, this prevents the base of$...
4
They are pin diode switches, when TX is enabled, current flows via R150,L40 forward biasing the diodes and effectively making them transparent to the RF power. Remove the bias (Or better reverse bias the diodes) and they present as small capacitors to the RF effectively isolating the PA from the aerial on RX. I expect there is a similar arrangement ...
4
It all depends on how much distortion you can tolerate, and any other side effects that may occur when saturating the amplifier (harmonics, etc.). You might be fine running several dB over the P1dB point. Or maybe you need to stay several dB below. You might just need to experiment. If your FMCW radar can tolerate some distortion, then you might be able to ...
4
The two circuits are both based on emitter followers. As you may recall, the emitter follower has a gain of about 1 and an offset of one base-emitter voltage, Vbe. The followers are arranged as a complementary pair. In both cases the purpose of the diodes D1 and D2 is to insert a bias that parallels the Vbe of each transistor. (Vbe is about the same as a ...
3
This should explain your questions: - EDITED SECTION I was clearly wrong about C5 - upon further inspection it has nothing to do with being part of a tank circuit so apologies for that. And L1 is therefore not a tank inductor but just a collector load of about 240 ohms at midband broadcast FM frequencies. Here's another attempt to fix my earlier stupidity:...
3
They both must have opposite direction and must differ from each other and the 0 must be on load. What makes you think that the output currents of the NPN and PNP transistor have a different direction? You should be careful with the direction of currents and also separate the direction of a DC current (bias current) and the direction of a change in current (...
3
Have you plotted the stability circles? Do a load pull and plot these, they contain useful information. Low frequency instability is often down to one of a few things: Source impedance goes high Z at low frequency, unterminating the input: Either provide a low frequency termination or an input pad of a few dB or so to reduce the mismatch at the input. A ...
3
That board uses an INA-02186 device, which is not a bipolar transistor but a complete MMIC gain block encapsulated in a typical transistor package. This "performance figure" comes from a random ebay seller (to whom, by the way, I'm not related in any way): which, as you can see, is nothing but a snapshop from the INA-02186 datasheet: The INA-02186 is pre-...
3
Class A = both transistors are ON all the time. Class AB = both transistors are ON at idle, then up to a certain output current. When output current is higher than a certain limit, one of the transistors turns off. Class B = either one transistor or the other is ON, but not both. The transistor that is ON is determined by output current polarity. Class C ...
3
When the base is driven at the positive peak, the emitter-follower's emitter current is at a maximum because it must supply both the current sink current as well as the current through $R_1$. But this is also at the same point where the $V_\text{CE}$ is at a minimum. So there is a power minima here. When the base is driven at the negative peak, the ...
3
Found a datasheet for the NTE1337. It is a replacement for the STK086. The datasheet for the NTE1337 includes this drawing: It shows the package size, hole sizes and locations, and the pin spacing (as well as pin dimensions.) The measurements seem to match those that @KingDuken found, but are easier to read. You will need all that it has to offer since ...
3
1.6W mean's the peak voltage is about 12V. How is it possible with Vs = 5V? Look at the data sheet application circuit diagram below. I've added a red box and a blue box: - The red box is around L1 (an inductor) and when you have a collector of a transistor being pulled up to Vs, the peak-to-peak voltage that can be attained is nearly twice Vs. So, if Vs ...
3
The Sziklai topology places the output BJTs within a local NFB loop. It's enough that the quiescent current is approximately 20 times less sensitive to the output BJT temperature changes. Partly because of this, you do NOT need to include the power output BJTs on a shared and monitored heat sink. It's the driver BJTs that need to be monitored for ...
3
110 watts into an 8 ohm load implies an RMS current of 3.71 amps. That current, through a single 0.22 ohm resistor implies a power of 3.025 watts. Considering that this power is shared by 4 resistors, the average power per resistor is 0.76 watts. But then one has to also consider the biasing of the output stage and this will cause a certain amount of DC ...
3
Always read the datasheet. The NE5534 has an input bias current of 500-2000nA. (500nA)(22k$\Omega$) = 11mV. So you're getting just about exactly the expected voltage across R1. If you put a resistor between your output and the inverting input of the op-amp, then it'll have the same bias current, $\pm$ the input offset current (which is 20nA up to a ...
2
In this common source stage, the inductor acts as a current source and capacitor act as a voltage source Nonsense ! The inductor is a low impedance for DC, meaning it allows a DC biasing current to flow while keeping the drain of the MOSFET at a DC level equal to VDD. The capacitor does the opposite, it blocks DC current and only allows AC signal to go to ...
2
I think you need to use ideal baluns for that.
2
When you try to understand how commercial amps work from reputable Japanese designs, and see the purpose of each component, you may learn how to improve quality. Of course there are many details in thermal matching and hFE binning needed as well. Here TR110 acts as the Vbe multiplier which must be thermally attached to the output stages. Tuning is ...
2
Why all books told that Av differential of an differential input stage in power amp is very big? I find "all books" a dangerous statement. It really depends on the input stage if it has a large voltage gain (Av). Also you're talking about Power amplifiers for audio, you should have mentioned that because audio is low frequency. There are also Power ...
2
You have mentioned the expression Ad = Rc/re. Please note that this expression is (1) a very rough approximation for a common emitter amplifier with heavy negative feedback (due to re) and (2) does not apply for a diff. amplifier. Neither for a simple one (two or three transistors only) nor for the one shown by you. This is because the resistance in the ...
2
If my memory serves me well the first stage gain is equal to: $$A_{V1} = gm*r_\pi = \frac{r_\pi Q6}{re2}$$ $$re2 = \frac{26mV}{I_{C2}}=\frac{26mV}{290\mu A} = 90 \Omega$$ $$r_\pi Q6 = (\beta+1)*re6 = 150 * 4.8\Omega = 720\Omega$$ Hence first stage gain is: $$A_{V1} = 8V/V$$ Q6 stage voltage gain is large but will drop due to $R_L$ loading effect....
2
There is a lot of overlap in what manufacturers describe as gain blocks and power amplifiers in the +10dBm to +20dBm output region. It would be an unusual power amplifier with less than +10dBm output, and gain blocks don't often have >+20dBm output power. As well as being lower power, gain blocks tend to emphasise good port matching and good isolation, as ...
2
I'll just write out the first few answers I was working on, before seeing G36's response (which you should definitely read.) The leftmost two form a high pass filter with about $4\:\textrm{Hz}$ as the corner frequency. The other two form a low pass filter with more than $100\:\textrm{kHz}$ as the corner frequency. Together, I'd have to play with how the ...
2
1 - 18kΩ resistor together with 2.2µF capacitor for a high pass filter. The Fc frequency is equal $F_C\approx\frac{0.16}{RC} = 4\textrm{Hz}$ Additional 1kΩ resistor form a low pass filter together with 1.2nF $F_C\approx\frac{0.16}{RC} = 134\textrm{kHz}$. And this filter stops any RF signal, so the RF signal is not being amplified by the amp. 2 - ...
2
To keep the quiescent current low, the left hand transistors run at a low current, usually <1mA. To provide usable output current, usually >20mA, to drive the load and feedback resistors, the right hand transistors provide current gain. That's not a 'power amplifier', but you do see that sort of transistor arrangement on the outputs of power amplifiers.
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# asif iqbal
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#### Problem 2129. Sum of odd numbers in a matrix
Created by: Koteswar Rao Jerripothula
#### Problem 2104. construct matrix with identical rows
Created by: sea knowledge
#### Problem 5. Triangle Numbers
Created by: Cody Team
Tags math, triangle, nice
#### Problem 25. Remove any row in which a NaN appears
Created by: Cody Team
#### Problem 10. Determine whether a vector is monotonically increasing
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#### Problem 23. Finding Perfect Squares
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#### Problem 2031. Half-Swap
Created by: Khizar
#### Problem 30. Sort a list of complex numbers based on far they are from the origin.
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#### Problem 838. Check if number exists in vector
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#### Problem 180. Omit columns averages from a matrix
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Tags columns, matrix
#### Problem 17. Find all elements less than 0 or greater than 10 and replace them with NaN
Created by: Cody Team
#### Problem 6. Select every other element of a vector
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#### Problem 16. Return the largest number that is adjacent to a zero
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Tags vectors
#### Problem 26. Determine if input is odd
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#### Problem 1. Times 2 - START HERE
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Tags intro, math, easy
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Tags matrices
#### Problem 2. Make the vector [1 2 3 4 5 6 7 8 9 10]
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Tags basic, basics, colon
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#### Problem 19. Swap the first and last columns
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#### Problem 8. Add two numbers
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#### Problem 3. Find the sum of all the numbers of the input vector
Created by: Cody Team
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Created by: Roy Fahn
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http://math.stackexchange.com/questions/tagged/circle+packing-problem | 1,408,511,380,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500800168.29/warc/CC-MAIN-20140820021320-00213-ip-10-180-136-8.ec2.internal.warc.gz | 131,584,534 | 14,302 | # Tagged Questions
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### How many circles of radius r fit in a single bigger circle of radius R?
Is there any formula to calculate how many circles of radius r fit in a single bigger circle of radius R? I'd apreciate if it didn't involve advanced math, like calculus (unless there is no other way, ...
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247 views
### Applonius Circle/ Ford Circle / Infinite GP / Circle Packing
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### Circle Packing: Unsolved Problem in Geometry?
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### What is the relative behaviour when a center circle surrounded by 6 circles is (recursively) replaced by 6 circles
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### Smallest Circle that encircles $4$ circles
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### Circle Packing Algorithm
I have question related to circle-packing. The problem is to find the circle of minimum radius enclosing four non-overlapping circles of arbitrary radius. I have to write a program in C for this ...
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### Packing squares into a circle
I need determine the maximum number of squares of the given size that can be packed into a circle of the given radius. Squares can be rotated. I'm not sure how complex this problem is and i can find ... | 791 | 3,410 | {"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} | 3.578125 | 4 | CC-MAIN-2014-35 | latest | en | 0.890248 |
http://math.stackexchange.com/questions/208126/probability-of-picking-kth-item-at-odd-turn | 1,469,301,907,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823387.9/warc/CC-MAIN-20160723071023-00007-ip-10-185-27-174.ec2.internal.warc.gz | 162,532,505 | 21,599 | # Probability of picking kth item at odd turn
There are n items in a line. Items are to be picked one by one and removed. So after each turn one item is removed and the next item is picked from remaining items.
Constraint is that only first item or last item from the remaining items can be picked with probability 0.5.
The last item remaining will be picked by probability 1.
So what is the probability that kth item is picked at odd number of turn means probability of picking kth item in 1st, 3rd ,5th turn and so on.
For clarification, if n=2, then in first turn either item can be picked with probability 0.5 leaving one item which will be picked in next turn.
if n>2, second item can't be picked in first turn, it can be picked in second turn or more turns.
-
Can there be recurrence for this? – user1515905 Oct 6 '12 at 9:17
I guess the probability is $1/2$. – Berci Oct 6 '12 at 9:28
@Berci, for an item at the end of the line, the probability of being picked on an odd turn is strictly greater than the probability of being picked on the first turn, which is already 1/2. – Gerry Myerson Oct 6 '12 at 9:59
@GerryMyerson Last item can also be picked at first turn but not the inner items. – user1515905 Oct 6 '12 at 10:03
@user, yes --- what's your point? – Gerry Myerson Oct 6 '12 at 10:04
Let $P(n,k)$ denote the desired probability for odd number of steps. Then we have $$P(n,k,even) = 1-P(n,k)$$ $$P(n,k) = P(n,n+1-k)$$ $$P(n,k) =0\ \text{ if }\ k\le 0\ \text{ or }\ k>n$$ $$P(1,1) = 1$$
\begin{align*} P(n,k ) &=\frac12\cdot (1- P(n-1,k))+\frac12\cdot (1-P(n-1,k-1)) \\ &=1-\frac12\cdot\left(P(n-1,k)+P(n-1,k-1)\right) \end{align*}
So, it is a recurrence, similar to Pascal's triangle.
For example, it gives $P(2,1) = 1/2$, $P(3,1)=1-1/2\cdot(P(2,1)+P(2,0))=3/4$ $P(3,2)=1-1/2\cdot(P(2,2)+P(2,1)) = 1/2$, $P(4,1)=1-1/2\cdot3/4 = 5/8$, $P(4,2)=1-1/2\cdot(1/2+3/4)=3/8$...
So, something like (multiplied the $n$th row by $\displaystyle\frac1{2^{n-1}}$): $$\begin{bmatrix} 1 \\ 1&1\\ 3&2&3\\ 5&3&3&5 \\ 11&8&10&8&11\\ \vdots&&&&& \ddots \end{bmatrix}$$
Update: Meanwhile I calculated the generator function (better starting with $n=k=0$, but doesn't really matter): $$F(x,y):=\sum_{n,k} P(n,k)\cdot x^ny^k$$ By the recurrence law, we get $$F(x,y)=\sum_{n,k}x^ny^k-\frac12xy\cdot F(x,y)-\frac12x\cdot F(x,y)$$ And hence, using $\sum x^ny^k =\frac1{1-x}\cdot\frac1{1-y}$, we arrive to $$F(x,y)= \frac{1}{(1-x)(1-y)(1+(xy+x)/2)}$$ Then use $\frac1{1-z}$ and the binomial thm for $(1+y)^k$. Unless I miscalculated, it yields $$P(u,v) = \sum_{k\le u,\,l\le v}\frac1{(-2)^k}\cdot\begin{pmatrix}k\\l\end{pmatrix}$$
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Whoa! This is actually the same problem as math.stackexchange.com/questions/206551/… – Gerry Myerson Oct 6 '12 at 12:21
Thanks for such a descriptive solution. – user1515905 Oct 6 '12 at 22:19
I'll do a special case. Consider an item at an end of the line. It's picked at the first turn with probability 1/2; at the third turn with probability 1/8; fifth turn, 1/32, and so on. As $n\to\infty$, the probability such an item is chosen at an odd turn approaches $$1/2+1/8+1/32+\cdots=2/3$$
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That's fit for first item among infinite items. But for kth item among n items, the answer will depend on k and n. – user1515905 Oct 6 '12 at 10:08
No kidding. But the journey of a thousand miles begins with a single step. – Gerry Myerson Oct 6 '12 at 10:24
Define
$P(S_i)=Probability\ of\ success\ in\ i'th\ turn\ (We\ pick\ K'th\ element\ in\ this\ turn)$
$P(F_i)=Probability\ of\ fail\ in\ i'th\ turn\ (We\ don't\ pick\ K'th\ element\ in\ this\ turn)$
so for all $i<n$ we have:
$P(S_i) = P(F_{i-1}) * {{1}\over{2}}$
but for $i = n$ (last turn), we have only one item and we pick it with probabilty $1$ so we have:
$P(S_n) = P(F_{n-1}) * 1$
$P(F_n) = 0$
in similar manner we define
$P(F_i) = {1\over2} * P(F_{i-1})$
so we have
$P(S_1) = {1\over2} \\P(F_1) = {1\over2} \\P(S_2) = {1\over2}*P(F_1) = {1\over4}\\ P(F_2) = {1\over2}*P(F_1) = {1\over4}\\...\\ P(S_i) = {1\over{2^i}}\\P(F_i)={1\over{2^i}}\\... \\ P(S_{n-1})={1\over{2^{n-1}}}\\P(F_{n-1})={1\over{2^{n-1}}}\\ P(S_n) = {1\over{2^{n-1}}}\\P(F_n)=0$
now probability of picking k'th element in the odd turn is equal to some over all $P(S_i)$ where $i$ is odd
Finally we have:
if $n$ is even we have
$\Sigma_{i=1,3,...,n-1}{P(S_i)} = \Sigma_{i=1,3,...,n-1}{1\over{2^i}}$
if $n$ is odd we have
$[\Sigma_{i=1,3,...,n-2}{P(S_i)}] + P(S_n) = [\Sigma_{i=1,3,...,n-2}{1\over{2^i}}] + {1\over{2^{n-1}}}$
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For a given case, if the numbers are not too big, you can do this by hand. Take, for example, $n=17$ items, with $k=6$; we want the probability that the 6th item is picked on an odd turn. Each turn removes either the first remaining item ($F$) or the last remaining item ($L$), with equal probability. So we just enumerate the different ways of arriving at item 6 after an odd number of turns:
$5F$ and $1L$ in any order, followed by an $F$: $p={6\choose1}/2^7$
$5F$ and $3L$ in any order, followed by an $F$: $p={8\choose3}/2^9$
$5F$ and $5L$ in any order, followed by an $F$: $p={10\choose5}/2^{11}$
$5F$ and $7L$ in any order, followed by an $F$: $p={12\choose7}/2^{13}$
$5F$ and $9L$ in any order, followed by an $F$: $p={14\choose9}/2^{15}$
$1F$ and $11L$ in any order, followed by an $L$: $p={12\choose1}/2^{13}$
$3F$ and $11L$ in any order, followed by an $L$: $p={14\choose3}/2^{15}$
$5F$ and $11L$ in any order, and then only item 6 remains: $p={16\choose5}/2^{16}$
But I can't see any way to get a nice formula for the sum of these probabilities.
- | 2,108 | 5,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2016-30 | latest | en | 0.859857 |
https://math.stackexchange.com/questions/445185/to-which-extent-distribution-of-riemann-non-trivial-zeros-follow-a-gauss-process | 1,563,500,605,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00033.warc.gz | 468,117,662 | 36,348 | To which extent distribution of Riemann non-trivial zeros follow a gauss process?
I am trying to clearer and preciser understand
to which extent the distribution of the non-trivial zeros of the Riemann $\zeta$-function follow a Gauss process?
Yet, what I figured out from readnigs, is that such a process acts only (conjectured to be) at the level of the neighborhood of the zeros. Not sure how neighborhood can be interpreted, or is the radius even larger?
• I mean the distance or radius (or some metric) from a non-trivial zero on the $1/2$ axis. There is no philosophy behind enxtent. – al-Hwarizmi Jul 16 '13 at 17:25
Assume the Riemann Hypothesis. Let $\rho = \tfrac 12 + i\gamma$ denote zeros and $N(T)$ be the number of zeros with $\gamma \in [T,2T]$. Let $\psi = \psi(T)$ be a function tending to infinity with $t$, arbitrarily slowly. One can prove that, $$\frac{1}{N(T)}\# \bigg \{ T \leq \gamma \leq 2T: \frac{N \big (\gamma + \frac{\pi \psi}{\log T} \big )- N \big (\gamma - \frac{\pi \psi}{\log T} \big ) - \psi}{\sqrt{\psi}} \in [\alpha,\beta] \bigg \} \rightarrow \int_{\alpha}^{\beta} e^{-u^2/2} \frac{du}{\sqrt{2\pi}}$$ Maybe I got the constant in the variance slightly wrong. The statement is no longer true if $\psi$ does not tend to infinity. | 388 | 1,267 | {"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} | 3.09375 | 3 | CC-MAIN-2019-30 | latest | en | 0.828769 |
https://www.rationalresponders.com/forum/the_rational_response_squad_radio_show/freethinking_anonymous/11322 | 1,581,900,887,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00184.warc.gz | 914,335,498 | 11,646 | # Probability, Razors, and Definitions (a mini-essay I've started).
CrimsonEdge
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Probability, Razors, and Definitions (a mini-essay I've started).
Note that this essay is not supposed to be scientifically sound, nor is it supposed to be accurate to a millionth of a degree. The point of this essay is to demonstrate probability, how occam's razor should be used, and to explain that a fallacy of definition is one of the major problems that we face against claims of Intelligent Design. I would appreciate critiques, comments, criticism, etc. This essay, again, is not supposed to be accurate. Instead it is to demonstrate a very basic principle and idea of the Big Bang and how Occam's Razor applies to it. Please excuse any formatting and spelling errors, this is a very bare first draft.
Quote:
First, lets start with some basic definitions. I will be using the Meriam-Webster online dictionary as a reference for any and all definitions being used so there is absolutely no confusion.
Probability - 1: the quality or state of being probable 2: something (as an event or circumstance) that is probable 3 a (1): the ratio of the number of outcomes in an exhaustive set of equally likely outcomes that produce a given event to the total number of possible outcomes (2): the chance that a given event will occur b: a branch of mathematics concerned with the study of probabilities
Occam's Razor - : a scientific and philosophic rule that entities should not be multiplied unnecessarily which is interpreted as requiring that the simplest of competing theories be preferred to the more complex or that explanations of unknown phenomena be sought first in terms of known quantities
Faith - 2 a (1): belief and trust in and loyalty to God (2): belief in the traditional doctrines of a religion b (1): firm belief in something for which there is no proof (2): complete trust3: something that is believed especially with strong conviction; especially : a system of religious beliefs <the Protestant faith> {The first definition on the website, not pasted here, does not apply}
Chance - 1 a: something that happens unpredictably without discernible human intention or observable cause b: the assumed impersonal purposeless determiner of unaccountable happenings : luck <an outcome decided by chance> c: the fortuitous or incalculable element in existence
Luck - 2: favoring chance; also : success <had great luck growing orchids>
Probability and chance are not the same. Chance is representative of luck. The probability of something happening is not the same as chance. For example, the probability of a coin landing on heads is, roughly, .50... or 1 in 2. This does not mean that if we flip the coin 100 times that it will land on heads 50 times. There is a chance that it will, but it's not probable. Probabilities are not accurate. This is something that needs to be understood. They are not representative of the future, nor are they accurate for predicting the outcome of things. The probability of the coin landing on heads is 50%, however, there is a chance that it will not. If you do not understand the difference, read the definition and apply them to different sentences. "The probability of me rolling a 6 on this 10 sided dice is 1/10, however, chances are it will not."
Luck is also unrelated to probability as luck favors an outcome. For example, saying that a hole in one is lucky would be accurate. Saying that it is a probable outcome is not. Consider the following statements after seeing an amazing shot in golf, such as this: http://www.youtube.com/watch?v=DaSQ20nH6VU ...
"What are the chances of that?"
"What a lucky shot!"
"The probability of those events happening is very small! I am amazed that the events folded out as they did!"
As you can see, the three statements, while meaning the same thing, all ask, or state, seperate things. One asks what the probability is while still playing on the luck factor, one tosses out probability and focuses on the event, the other states that the probability of the sequence of events was very small, however, they happened anyway. While the three words can be used interchangabley in some instances, they can not in others. The places they can not be used are in both a math and science classroom.
For further information on probability, please go here: http://physics.mtsu.edu/~phys2020/Lectures/L6-L11/L8/Prob/prob.html
So, what is the point of all of this? To show what scientists mean by probability. They do not mean chance, nor do they mean luck. This is a fallacy of definition, which is something many have problems with. It's understanding to commit this fallacy when words mean, roughly, the same thing as eachother or when one word has many definitions (especially archaic definitions or slang... think 'cool'. The problem, though, is that this fallacy is commited frequently and is the core of many of the misunderstandings of various positions one has regarding things like biology, weather men, and especially the tonality of instruments such as the guitar. Any musician knows what I'm talking about as many 'textural' words are used to describe sound. Things like chunky, warm, muddy, etc. Another common word people commit fallacies with is the word theory, but that is not the subject of this mini-essay.
Another error that many commit is in regards to the 'Conservation of mass', the 'Conservation of energy' and the 'Big Bang theory'. First and foremost, lets start with a very simple equation that everyone has seen and heard. E=MC(2). E=the energy equivallent to the mass. M=mass. C=The speed of light in (in a vacuum). In short, Mass and energy are two ways of representing the same thing. For more information on the Mass-energy equivalency, read here: http://en.wikipedia.org/wiki/Mass-energy_equivalence
So, let us start with a basic understanding of what the Big Bang Theory actualy states. To sum it up simply, it states that there was great density and temperature in one spot. Density is another word for mass (in physics... which is what is being talked about) and temperature is relative to energy. As you can see, 'nothing', according to this model, never existed. There has always been mass and energy, in one form or another. This idea is very hard to grasp, however, everybody uses this idea for the universe. Regardless of your theory of how things came to be, there is always an being/force that has been around forever. And yes, forever is another word for infinite.
This is where Occam's Razor steps in and an understanding of how it should be used, as well as how it works, is required for almost all scientific and philosophical debate and conversation. Occam's Razor can be applied many ways. Why did I change the channel just now? Well, there are an infinite number of possibilities, however, which is more probable? Well, let us examine some of the reasons why I might have.
1. I was bored of what was on T.V. so I changed it to something that had the potential to be entertaining.
2. I detest a certain show that was on.
3. A Bugbear told me to do it or he would kill me.
While, yes, there are an infinite number of possibilities, I chose these three to represent a wide spectrum. Two are more probable than the other. Which can you guess is the oddball? If you said #3 then you are correct. Using probability coupled with personal experience, you've come to the conclusion that a Bugbear did not tell me what to do. This has nothing to do with knowing what a Bugbear is, what it looks like, etc. This has nothing to do with me being able to prove that a Bugbear exists, let alone was standing next to me in my house threatining me over something trivial.
How can this be applied to current popular models of how the universe (as we know it currently) came into existance? Well, lets examine three different positions.
1. The universe has always existed, simply in a different state than its current one before the Big Bang.
2. The universe was created by God in six days (whichever translation you want to use for the word day is fine) as a way to test what it created.
3. The universe was danced into existance by Nataraja (shiva in his/her dancing form) as part of its duties as a diety.
Which one of these sounds the most absurd to you? Well, let us examine them from the bottom up.
Number 3 seems the most absurd to almost everyone in the world. A dancing god with multiple forms and duties? What kind of a god is that? Anyway, let us Occam's Razor this bad boy and go back one further step than the universe. When we go back that step, we are left with some questions. Why does this god have duties? Why must it dance as a duty? Why does it create and destroy life when it dances? Who is requiring this duty? Has this being always existed? If so, then what created it?
Number 2 is the most common conclusion we see today, but does it create more questions? Well first, the idea of an ever-lasting universe is thrown out the window, so the first, and most obvious question would be if God has always existed. Has he? If so, then did something create him (etc.)?If not, why is he exempt from the creation rule? Why did he create something only to test what he created (I don't care about the issues of omni-max here as they are irrelevant to what is being demonstrated)? Why six days? Does God exist inside of the universe he created? If so, was his creation a requirement for his existance? Has time always existed because God has always existed or did he create it? If he created it, how did he exist before it?
Number 1 is the most shunned upon by most people, atleast in the U.S. So, what questions does it create? What was the state of the universe before the Big Bang? Did time time exist before the Big Bang? Did our current laws of physics exist before the Big Bang? If so, what was/were the event/s that triggered it?
There are more for all three, but these are the more important ones. With the basic understandings of what probability is, coupled with Occam's Razor and a small amount of critical thinking, we can all logically infer that #1 sprouts the least number of questions and leaves many of the other questions (such as what created this, what created that, why, etc.) and takes many of the human factors out of it... such as who and why. It simply leaves the where, how, and when.
This is very important to note when dealing with things that are supernatural or non-human. Who and why are no longer important factors to the equation, further making the probability of it happening greater. There is no luck or chance involved, simply probability. Further, the probability of something happening, while already knowing the outcome, is 100%. This is something to understand about equating the probability that life appeared somewhere... which is my next point.
When scientists calculate the probability of something happening, especially life appearing on other planets, they do not go in to it with the foreknowledge that life exists elsewhere. They take the current understanding of the processes required for life and apply it to the planet they are studying. For example, when scientists equate the probability of finding life on Europa, they know not if there is life on the planet, nor if life on other planets exist. However, they take into account the factors that go into it, such as water, and create an equation that gives them the probability of life on the planet. The same can be applied to every planet, including ours, however, we do not the exact state of the planet when life formed, nor do we know exactly when this was. We have very, very, good estimates that are probably right, however, we do not know for sure. Without this exact data, we are left with a period of time where many things could have happened, all including many different unknowns.
This, however, does not mean that it is improbable for life to form on Earth, neither does it mean that it is improbable for life to form into its current state on Earth. In fact, it has happened so it is, infact, probable that it did. Regardless of the improbability of an event occuring, if the event occured then it means that it was probable. This is where many commit the fallacy.
In regards to faith. If you have faith, you do not have evidence to support you. No scientist has faith that the theories they test are correct. They are completely indifferent (or should be) to the outcome of the test. Faith takes no place in science and has no part of science at all. Remember, the definition for faith includes believing without evidence or believing although evidence supports otherwise. This is fine. If you have faith in something then more power to you, however, because you have faith does not make it wrong. Further, because you have faith in something (or that something is wrong) does not mean what you believe should be taught anywhere by anyone at any time. Faith is just that. Faith.
While the probability that your faith is right is very very small, there is a small chance that it is... and you might be lucky, but chances are you won't be. Playing Pascal's Wager won't help you either, but that is neither here nor there. | 2,880 | 13,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.931948 |
https://headinside.blogspot.com/2012/11/bayes-theorem.html | 1,686,355,814,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00375.warc.gz | 330,285,275 | 270,197 | 0
## Bayes' Theorem
Published on Sunday, November 11, 2012 in , , , ,
Ever wonder what happens to those amazing breakthroughs you hear about on the news, but never hear about again? Somehow, when they're finally released, the amazing qualities of, say, that new wonder drug, never seem to reduce the suffering the way most people hoped.
Look through the reports on the test results of those breakthroughs, and you'll frequently find one line that says p < 0.05. In other words, the tests indicate that the results reported on in the report had only a 5% chance of happening randomly.
If I flip a coin 20 times, and heads shows up 15 or more times (in other words, greater than 14 times), we can work out that there is roughly a 2.07% chance of that happening at random. Reporting on this, we'd note that p < 0.05, and use this to justify examining whether the coin is really fair.
That works great for events dealing with pure randomness, such as coins, but how do you update the probabilities for non-random factors? In other words, how do you take new knowledge into account as you go? This is where Bayes' theorem comes in. It's named after Thomas Bayes, who developed it in the mid-1700s, but the basic idea has been around for some time.
You should be familiar, of course, with the basic formula for determining the probability of a targeted outcome:
The following video describes the process of Bayes' theorem without going into any more mathematics than the above formula, using the example of an e-mail spam filter:
To get into the mathematical theorem itself, it's important to understand a few things. First, Bayes' theorem pays close attention to the differences between the event (an e-mail actually being spam or not, in the above video) and the test for that event (whether a given e-mail passes the spam test or not). It doesn't assume that the test is 100% reliable for the event.
BetterExplained.com's post An Intuitive (and Short) Explanation of Bayes’ Theorem takes you from this premise and a similar example, all the way up to the formula for Bayes' theorem. It's interesting to note that it's effectively the same as the classic probability formula above, but modified to account for new knowledge.
The following video uses another example, and is also simple to follow, but delves into the math as well as the process. Understanding the process first, and then seeing how the math falls into place helps make it clear:
The tree structure used in this video helps dramatize one clear point. Bayes' theorem allows you to see a particular result, and make an educated guess as to what chain of events led to that result.
The p < 0.05 approach simply says “We're at least 95% certain that these results didn't happen randomly.” The Bayes' theorem approach, on the other hand, says “Given these results, here are the possible causes in order of their likelihood.”
If I shuffle a standard 52-card deck, probability tells us that the odds of the top card being an Ace of Spades is 1/52. If I turn up the top card and show you that it's actually the 4 of Clubs, our knowledge not only chance the odds of the top card being the Ace of Spades to 0/52, but gives us enough certain data we can switch to employing logic. Having seen the 4 of Clubs on top and knowing that all the cards in the deck are different, I can logically conclude that the 26th card in the deck is NOT the 4 of Clubs.
We can switch from probability to logic in this manner because we've gone from randomness to certainty. What if I don't introduce certainty, however? What if I look at the top card without showing it to you, and only state that it's an Ace?
This is the strength of Bayes' theorem. It bridges the ground between probability with logic, by allowing you to update probabilities based on your current state of knowledge, not just randomness. That's really the most important point about Bayes' theorem.
There's much more to Bayes' theorem than I could convey in a short blog post. If you're interested in a more in-depth look, I suggest the YouTube video series Bayes' Theorem for Everyone. I think you'll find it surprisingly fascinating. | 928 | 4,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-23 | latest | en | 0.961866 |
https://unix.stackexchange.com/questions/386641/sum-multiple-columns | 1,569,268,874,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514578201.99/warc/CC-MAIN-20190923193125-20190923215125-00038.warc.gz | 711,677,604 | 30,445 | # Sum multiple columns
I need to do multiple sums; my input file is:
``````DATE|NATION|CITY|FILES|REVENUE|FREQUENCY|INVESTMENT
20170807|USA|VIRGINIA|TIMES|1919150|1779|282075
20170808|USA|NYC|TIMES|361625|1592|0
``````
1. Sum \$5 in every uniq of \$1 (date)
2. sum \$5 in every uniq where \$4=="TIMES"
3. sum \$5 in every uniq where \$4=="ROADS"
4. sum \$5 in every uniq where \$4=="ROADS" and \$3=="NYC"
5. arrange based on column \$1
my expected output
``````DATE|REV|TIMES|ROADS|ROADS&NYC
20170807|2012027|1919150|92877|92877
20170808|494626|361625|133001|0
``````
I only know how to sum based on 1 column
``````awk -F"|" '{FS=OFS="|"}{col[\$1]+=\$5} END {for (i in col) print i, col[i]}'
``````
Complex gawk solution to reach the goal:
``````awk 'BEGIN{ FS=OFS="|" }NR==1{ next }{ sum[\$1]+=\$5 }
\$4=="TIMES"{ d[\$1]["t"]+=\$5 }
\$4=="ROADS"{ d[\$1]["r"]+=\$5 }\$3=="NYC" && \$4=="ROADS"{ d[\$1]["r&n"]+=\$5 }
for(i in d) print i, sum[i], int(d[i]["t"]), int(d[i]["r"]), int(d[i]["r&n"]) }' file
``````
The output:
``````DATE|REV|TIMES|ROADS|ROADS&NYC
20170807|2012027|1919150|92877|92877
20170808|494626|361625|133001|0
``````
• it works, thanks alot. I wish I can learn faster in this forum. Thanks alot – dars Aug 17 '17 at 12:45
Here is `awk` solution:
``````awk -F'|' 'NR>1{I[\$1]+=\$5}
(\$4=="TIMES"){T[\$1]+=\$5}
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on constructing hyper operations for bases > eta JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 04/07/2015, 01:47 AM (This post was last modified: 04/07/2015, 02:04 AM by JmsNxn.) Hello everyone. So as I put in my last thread here, I have a solution to hyper operations for bases between 1 and $\eta$. The natural question is if we can extend our results. As fivexthethird pointed out we can apply these ideas to the super root function. Then as Tommy pointed out this can then again result in a provable recursion. To be clean and precise, I will explain the methodology here. If $^s (srt(s,x)) = x$ and if $\frac{d^{s-1}}{dw^{s-1}}|_{w=0} \sum_{n=0}^\infty srt(n+1,x) \frac{w^n}{n!}$ converges and exists then quite exactly since $srt(n+1,x)$ is the inverse of $x^{x^{x^{...n+1\,times...}}}$ it is calcuable for $x>\eta$ and then $srt(s,x)=\frac{d^{s-1}}{dw^{s-1}}|_{w=0} \sum_{n=0}^\infty srt(n+1,x) \frac{w^n}{n!}$ Inverting this function $ssrt$ gives us a solution to tetration, proving the recursion with a little breath of thought. The problem is proving convergence of the differintegral. If we can do this we can iterate the procedure and solve for pentation then hexation and then septation, etc... This will give us a solution to hyper operations with a base on the positive real line. Not only that, an incredibly economical solution. I've begun work on this and I think the procedure for proving convergence is the only problem, and then I think I got the knack of the result. I'm performing a similar induction schema and I think the result only needs a little bit of analysis magic that I've yet to find. But I'm sure it's out there. I'm posting this thread to see if anyone has any ideas. I'm very excited about this result and it tittilates me to see it proved. I am wanting to collaborate on a paper with whomever wants to work on these ideas. I think some of the steps may be more complicated than first expected. I'm sort of starting with the idea of convergence, as I am sure that is the hardest hill to climb. Thanks again, and hope you guys can help. « Next Oldest | Next Newest »
Messages In This Thread on constructing hyper operations for bases > eta - by JmsNxn - 04/07/2015, 01:47 AM RE: on constructing hyper operations for bases > eta - by marraco - 04/08/2015, 09:18 PM
Possibly Related Threads... Thread Author Replies Views Last Post On my old fractional calculus approach to hyper-operations JmsNxn 14 3,804 07/07/2021, 07:35 AM Last Post: JmsNxn hyper 0 dantheman163 2 5,504 03/09/2021, 10:28 PM Last Post: MphLee On to C^\infty--and attempts at C^\infty hyper-operations JmsNxn 11 4,299 03/02/2021, 09:55 PM Last Post: JmsNxn Constructing real tetration solutions Daniel 4 6,194 12/24/2019, 12:10 AM Last Post: sheldonison Thoughts on hyper-operations of rational but non-integer orders? VSO 2 3,901 09/09/2019, 10:38 PM Last Post: tommy1729 Could there be an "arctic geometry" by raising the rank of all operations? Syzithryx 2 4,318 07/24/2019, 05:59 PM Last Post: Syzithryx Hyper-volume by integration Xorter 0 3,068 04/08/2017, 01:52 PM Last Post: Xorter Hyper operators in computability theory JmsNxn 5 9,844 02/15/2017, 10:07 PM Last Post: MphLee Recursive formula generating bounded hyper-operators JmsNxn 0 3,343 01/17/2017, 05:10 AM Last Post: JmsNxn holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 28,981 08/22/2016, 12:19 AM Last Post: JmsNxn
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Here is a solution of two rings and a closed loop with 15 arcs, all matching their paths. See how the rings have symmetrical path patterns and contain one "double" tile each. Double tiles have an arc that connects the same level — 1-1, 2-2, 3-3, 4-4 and 5-5. All the other tiles step up or down between different levels. See also how on the larger loop the arcs turn in and out, so that converse edges meet — 1 to 5, 2 to 4 — while 3 meets 3 either way. That's a valuable hint for solving, saving the 3's to the last. | 154 | 545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.933051 |
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Several other blogs have been talking about Bret Victor’s Kill Math website, including its Scrubbing Calculator. I’d like to talk about how the Scrubbing Calculator is both very similar to and very different from an approach to solving word problems we call “Guess-Check-Generalize”. Here’s a graphic from a sample problem solved Scrubbingly.
The challenge is to find the height of each bar, given the information about other heights. When I first taught Algebra 1, my approach to this would be to get students to “translate” the problem into algebra, trying to get them to write an equation that would be true for the right height. And the results were a mixed bag, for a lot of reasons that might be good for a different post. I think there’s something inherently challenging about trying to write a fully symbolic statement immediately from a problem situation.
The concept of guess-check-generalize starts by changing the nature of the problem. The question to start with changes:
from What is the correct bar height? …
to Is 100 the correct bar height?
Here, 100 could have been any number at all, it’s a total guess. (Some teachers using this method ask students to write down their first guess before even presenting the problem, since students may be afraid to guess incorrectly.)
Now we see if the guess is right. Up until now, I agree completely with the philosophy of the Scrubbing Calculator: make a guess at the bar height, then see if it’s right. This is where things get interesting, because there’s more than one way to check the guess. The most conventional way is to add up the heights on the right side, and a student might do this:
60 + 100 + 20 + 100 + 20 + 100 + 20 + 100 + 20 + 100 + 20 + 100 + 20 + 100 + 20 + 100 + 20 + 100 + 140 = nope
It doesn’t actually matter what that equals, as long as it doesn’t equal 768. Guess-check-generalize is about determining a process you can use to check any guess; then, the process you’ve described becomes an equation to solve. And the process can evolve from one guess to another, as students realize they’ve used the same number 8 times or that this thing is twice that thing.
So 100 was wrong; take a second guess. It doesn’t have to be a better guess, because you’re not trying to nail the numeric answer, you’re trying to nail the process of checking a guess. Let’s guess 36. Checking this guess a student might notice they could combine some terms from before:
$60 + (9 \cdot 36) + (8 \cdot 20) + 140 = 684, \text{nope}$
No more guessing. The third guess is $h$, a variable. (Students may need more guesses, especially at first; eventually some only need one or zero guesses.) Take all the places the guess was found and replace them with the variable, noting that the correct guess yields 768:
$60 + (9 \cdot h) + (8 \cdot 20) + 140 = 768$
Solving that equation and bringing the answer back into context are still issues, but I always found the largest difficulty with the dreaded “word problem” is an inability to take the situation and make a mathematical statement about it. When almost every real mathematical situation an adult encounters is a “word problem”, this is a major issue that needs to be addressed.
Here’s why I think guess-check-generalize is a good way of dealing with word problems.
• The method is general in nature. The method presented here works equally well for linear and nonlinear situations, for problems with a variable on each side of the equation, for rates, coin values, painting houses, counting beans, whatever. This is a general-purpose tool that is useful over many years, including some surprising topics like generating the equations of lines and circles. (More on this some other time.)
• This is what people do with problems. When a problem is new or overly complicated, picking a few cases and following them through leads to an understanding of what happens in general. Traditional word problem methods expect students to have the generalization at the ready, and it just doesn’t work that way in reality. The concept of generalizing from repeated example is a fundamental one that all students should learn, not just those heading into STEM careers.
• Students have a simple place to start from. By asking students to guess at the answer, the difficulty level of word problems can be reduced by 2 or 3 grade levels immediately. Students with language difficulty can learn what is happening by calculating with numbers, connecting the new language to the calculations they know, then advancing to symbols when appropriate.
• There are no black boxes. Students construct equations and can understand where they come from. Multiple equations with the same answer can be found from different techniques used on the same problem, leading to good discussions about the basic moves of algebra and how different equations and formulas are related.
• Connections between arithmetic and algebra are reinforced. Bret Victor says this: “We are accustomed to assuming that variables must be symbols. But this isn’t true — a variable is simply a number that varies.” I’d like this to change. Too many students only see variables as symbols for manipulation, and not as numbers that vary. Students make mistakes with variables they would never make with numbers. When this happens, it is because they don’t see that the symbol represents a number. Since arithmetic is at the heart of guess-check-generalize, students are asked to solidify their number skill and sense. Students begin to guess “nice” numbers, like a multiple of 3 when they see that dividing by 3 will be part of the process.
It is on this last point that I disagree deeply with the philosophy of the Scrubbing Calculator; students don’t really do any of the calculating. In the end, a student might see that the answer produced by Scrubbing works, but if there is more than one answer, there’s no way for a student to discern this. If the problem changes slightly from its original form (say, to a 1024-high screen), the Scrubbing solution method is to start from scratch, which doesn’t help students generalize toward functions and formulas (in this case, a relationship between the screen height and the bar height).
What if the correct answer to the equation is $\sqrt 2$ or even $\frac 2 3$? I don’t see how the Scrubbing Calculator could get these answers. I agree that too many students don’t see the real meaning of a variable, but this is no reason to ditch symbolic algebra, this is a reason to make the connections between arithmetic and algebra as strong as possible, as often as possible.
The Scrubbing Calculator’s method is an opportunity for students to make deep connections between arithmetic and algebra, between real problems and symbolic algebra. I’m disappointed that its intended purpose is to remove symbolic algebra altogether, because it could be pretty cool. What do you think?
For homework, solve this problem using guess-check-generalize or come up with a better one. No scrubbing, please!
Nancy takes a long car trip from Boston. In one direction she drives at an average speed of 60 miles per hour, and in the other direction she drives at an average speed of 50 miles per hour. She’s in the car a total of 38 hours for the round trip. How far from Boston was her destination? (Bonus: what city did she drive to?)
## Induction through Failure
I was inspired by a recent post from Ben Blum-Smith about induction to talk about an approach to induction I learned from Al Cuoco. I don’t know its origins, but it really does a nice job of dealing with the issue Ben brings up: you assume you’re right then prove it. But if you’re already right, why prove it?
Our curriculum (CME Project Precalculus Investigation 4A, for those playing at home) ties induction to the process of finding closed-form rules to match recursive definitions, such as this one:
$f(n) = f(n-1) + 5, f(0) = 3$
What is $f(23)$? All the formula says is that $f(23) = f(22) + 5$, so that’s no good… except that it is, since now you do the same thing to $f(22)$, eventually working your way back to the “base case”. This actually works out pretty nicely:
$f(23) = f(22) + 5$
$= [f(21) + 5] + 5$
$= f(20) + 5 + 5 + 5$
$\vdots$
$= f(0) + 5 + 5 + \cdots + 5$
But how many fives? Why is chasing it all the way back to $f(0)$ a good idea? What would change with $f(117)$? with $f(n)$? This works to find closed-form rules for recursions surprisingly often.
Another way to find $f(23)$ is to count off outputs: $f(1), f(2), \dots$ until you reach 23. More likely, you’ll figure out a pattern to the outputs before you get there… 3, 8, 13, 18…
So now you’ve got these two rules:
$f(n) = f(n-1) + 5, f(0) = 3 \qquad \text{and} \qquad g(n) = 5n + 3$
If I want to prove these two rules will always agree (whenever f is defined, anyway) it’s time for induction. Before that, we really want to be sure the rules agree. This is where technology comes in: using Excel or the TI-Nspire or several other tools, these functions can be entered then compared. (For the Nspire, see page 3 of this document for an example.) Now use the technology to compare $f(23)$ and $g(23)$ … $f(500)$ and $g(500)$ … $f(100000)$ and $g(100000).$
No matter what piece of technology you use, at some point it is going to stop saying that these two functions agree. (On Nspire hardware, this happens somewhere around 100, but depends on the device’s memory.) Say for example that the technology agrees that $f(105) = g(105)$ but $f(106)$ is undefined. How could you show that $f(106) = g(106)$ anyway?
By doing this, the “induction step” process is one that actually happens with a numeric value. So let’s work this one out… evaluating $f(106)$ fails, but we know from its definition that
$f(106) = f(105) + 5$
Oh and we also know that $f(105) = g(105)$, because the functions agreed up to that point. So:
$f(106) = g(105) + 5$
$= (5 \cdot 105 + 3) + 5$
$= 5 \cdot 106 + 3$
$= g(106)$
The induction step happens naturally, based on the failure of the technology, and based on a numeric example. (Now, try to show that $f(107) = g(107)$…)
The steps taken above also evolve quickly into a general argument for $f(n)$ simply by using $n$ instead of 106, and noting that 105 should be $(n-1)$.
$f(n) = f(n-1) + 5$
$= g(n-1) + 5$
$= (5(n-1) + 3) + 5$
$= 5n + 3$
$= g(n)$
And, boom goes the dynamite.
I feel this method does a nice job of dealing with the “proving what you know to be true” issue that surrounds induction. This method can also help with some of the issues around base cases, because it encourages the checking of several cases before trying to complete an inductive proof. Lastly, the extension from the numeric calculation of $f(106)$ to the algebraic calculation of $f(n)$ lines up well with one of Common Core’s mathematical practices, “Look for and express regularity in repeated reasoning“. After all, that’s what induction is all about…
## Deriving the Quadratic Formula (the easy way)
In Bowen’s post from April 25, he showed a great method for factoring non-monic quadratics. Using that same method, you can derive the quadratic formula pretty cleanly, without lots of fractions, rationalizations, and the like.
The goal is to find a solution to the generic quadratic equation:
$ax^2 + bx + c = 0$
We can’t factor using sums and products, so we resort to completing the square. The equation would be much easier to work with if it were monic. We could divide through by $a$, but then there’s lots of fractions to keep track of. Instead, let’s multiply both sides by $a$ and make it a quadratic in $ax$:
$a^2x^2 + abx + ac = 0$
Hmmm. Again, things would be so much easier to complete the square if that middle term were even. We can multiply both sides by 2 to do that, but then the first term isn’t a perfect square. Fine, let’s multiply both sides by 4 then:
$4a^2x^2 + 4abx + 4ac = 0$
Now, just rewrite that a little bit:
$(2ax)^2 + 2b(2ax) + 4ac = 0$
and now we can put our fingers over $2ax$ and see this as a simpler monic:
$F^2 + 2bF + 4ac = 0$
And now, complete the square. First, let’s put that constant on the other side:
$F^2 + 2bF = -4ac$
To get a complete square on the left side, we need a $b^2$, so add it to each side:
$F^2 + 2bF + b^2 = b^2 - 4ac$
The left side is now a perfect square… and the right side looks familiar… Let’s factor the left side:
$(F + b)^2 = b^2 - 4ac$
To solve for $F$, take the square root of each side:
$F + b = \pm\sqrt{b^2 - 4ac}$
and subtract $b$ from each side:
$F = - b \pm \sqrt{b^2 - 4ac}$
What was $F$ again? Lift that finger…oh, yeah, $2ax$:
$2ax = - b \pm \sqrt{b^2 - 4ac}$
So finish this off by dividing each side by $2a$:
$x = \dfrac{- b \pm \sqrt{b^2 - 4ac}}{2a}$
And there you have it… the quadratic formula, with no fractions until the final step. Enjoy!
## Counting with Polynomials
This is to expand on a recent comment I made to “Factoring non-monic quadratics.”
We hear a lot these days about “modeling with mathematics.” One aspect of this is the use of formal calculations as a modeling device. An example, pretty common in many high school programs, is the use of the binomial theorem to compute combinations. So, you can use $(t+h)^{10}$ to compute the number of ways that you can get, say, 4 heads and 6 tails when you toss a fair coins 10 times. Another example is when you use powers of a matrix to get Fibonacci numbers. The calculations not only give you answers, they let you derive properties of the phenomena they model.
One of my favorite uses of formal calculations comes from a project that my daughter gave to her sixth grade class some year ago: compute the most likely sum (and the distribution of sums) when several dice are thrown. For two dice, the problem is fairly standard, and most kids make a 2 by 2 table of all possibilities. For three dice, her students invented all kinds of clever representations, some very algebraic — not in notation, but in spirit. This triggered an idea that has become a recurring theme in our high school program. It goes like this:
If you use expansion boxes to multiply polynomials, you see that the expansion of
$(x+x^2+x^3+x^4+x^5+x^6)^2$
contains exactly the same numbers as the 2-dimensional table that records the possible outcomes when you roll two dice. In other words, the number of ways of rolling a 5 when you throw two dice is the coefficient of $x^5$ in
$(x+x^2+x^3+x^4+x^5+x^6)^2$
It’s the number of ways you can make 5 as a sum of two integers, each between 1 and 6. The same reasoning shows that the coefficient of $x^k$ in
$(x+x^2+x^3+x^4+x^5+x^6)^m$
gives the number of ways you can roll a $k$ when $m$ dice are thrown.
From here, you can use the structure of the expression $(x+x^2+x^3+x^4+x^5+x^6)^m$
to get results about the distribution of sums. For example:
• There are $6^m$ possible sums, because this is the sum of the coefficients, and the sum of the coefficients comes from putting $x=1$
• There are as many even sums as odd sums (replace $x$ by $-1$).
• The distribution of sums on three dice whose faces are labeled $\{0, 2, 3, 4, 5, 5\}, \{0, 1, 1, 2, 2, 2\}, \text{and} \{1, 2, 3, 6, 6, 6\}$ can be read off from the coefficients of the product of three different polynomials.
• Various factorizations of $(x+x^2+x^3+x^4+x^5+x^6)^m$ give you different information about the distributions. One interesting thing to try is to rearrange the factors of $(x+x^2+x^3+x^4+x^5+x^6)^2$ to try and produce two dice, different from the standard ones, whose distribution is the same as the standard distribution. (This may or may not be possible, we’re not telling.)
All this is a preview, accessible to high school students, of the incredibly useful field of algebraic combinatorics and generating functions.
Al
## Factoring Non-Monic Quadratics
Regardless of the course, non-monic factoring was always a thorny issue in my teaching. My students never seemed to get “good” at it, even though they seemed alright at monic factoring (“monic” just means the first coefficient is 1, like $x^2 + 14x + 48$). This topic made me really question why I was teaching it, for several reasons:
• The very next thing was the quadratic formula, and for most of the things non-monic factoring could be useful for, I felt the quadratic formula would be just as good.
• I couldn’t find many places later in my curriculum where non-monic factoring was being used, so it seemed like a topic taught for a single purpose.
• The methods I saw for factoring non-monic quadratics had little or nothing to do with the methods for factoring monic quadratics.
• The method I learned and first taught amounted to trial-and-error.
And maybe you know this method too: to factor $6x^2 + 31x + 35$, you write down all the factors of 6, separately write down all the factors of 35, and start making pairs. Eventually you either find the pair that works, or you run out of pairs:
$(x+1)(6x + 35) = 6x^2 + 41x + 35$, nope
$(x+5)(6x + 7) = 6x^2 + 37x + 35$, nope
$(x+7)(6x + 5) = 6x^2 + 47x + 35$, nope, keep trying…
And I was polite in picking 6 and 35 here, two numbers with only two prime factors each! I feel this method is a mathematical nightmare. Keep testing, keep checking. And don’t give the ones that aren’t factorable, since the only way to know it’s unfactorable is to test all the possibilities, and that’s just mean.
A year or two later, I learned and taught the “key number method”: multiply the coefficients of “a” and “c” ($6 \cdot 35 = 210$). Then you break up the middle term ($31x$) into two pieces whose coefficients multiply to 210:
$6x^2 + 31x + 35 = 6x^2 + 10x + 21x + 35$
Then “group” in pairs and a miracle occurs:
$(6x^2 + 10x) + (21x + 35) = 2x(3x + 5) + 7(3x + 5)$
$= (3x + 5)(2x + 7)$
This worked a lot better for my students, by which I mean they got correct answers faster and with greater accuracy. But the core of this method is the “miracle” that splitting the $31x$ in this exact, specific way will do great things. It works because it works. (There are better explanations, but my students just memorized what to do.)
One advantage of the key number method is it can be applied to monics, too, visualizing the “sum and product” concept:
$x^2 + 14x + 48 = x^2 + 6x + 8x + 48$
$= x(x+6) + 8(x+6)$
$= (x+6)(x+8)$
But this generally comes after the fact: I didn’t teach students to factor monics in this way.
While working on CME Project, I learned (through Al Cuoco and Jeremy Kahan, a field test teacher) about a “scaling” method that uses monic factoring as the core of non-monic factoring. It feels a lot more natural, cements monic factoring, and fits tightly with Mathematical Practice #7, “Look for and make use of structure.” It starts with specific non-monics like this one:
$25x^2 + 70x + 48$
Try factoring that for a second using either of the methods presented above. It’s messy! But, would you believe this is actually a monic quadratic? It’s just got a different variable: $5x$.
$25x^2 + 70x + 48 = (5x)^2 + 14(5x) + 48$
Now cover your finger over each $5x$: it reads $finger^2 + 14 \cdot finger + 48$. Doesn’t matter what’s under the finger: it factors!
$F^2 + 14F + 48 = (F+6)(F+8)$
And you’re done when you lift your finger, remembering that $F = 5x$. In teaching, I used capital letters for these substitutions, to remind students that there was more work to be done later.
$(5x+6)(5x+8)$
How fast was that? And understandable, too! The core concept of a replacement of variable (the book calls this “chunking”) plays forward deeply into later topics and courses: when I say I used capital letters for substitutions, I generally was doing that with Precalculus or Calculus students, but the concept can be seen much, much earlier. By using it frequently, it becomes a tool students actively look to use when they see something complicated.
But I fudged the example: it’s got $25x^2$. How about that original one, $6x^2 + 31x + 35$? It doesn’t have a perfect square term, but … wishful thinking … we can make one by multiplying through by 6, then paying it back later.
$6(6x^2 + 31x + 35) = (6x)^2 + 31(6x) + 210$
$\mathbf{= F^2 + 31F + 210}$
$\mathbf{= (F+10)(F+21)}$
Note that this method includes the step that was part of the “key number method”: the 210 is produced by multiplying the coefficients of “a” and “c”, but this time there is a more mathematical reason for doing so. And the payoff is the same, since we then need two numbers that add to 31 and multiply to 210 — but we use the monic factoring method to perform that step. This cements monic factoring skills, as it becomes part of the process in the later topic. And now the miracle, as we have common factors in the two right-side terms:
$6(6x^2 + 31x + 35) = (6x+10)(6x+21)$
$= 2(3x+5) \cdot 3(2x + 7)$
$= 6(3x+5)(2x+7)$
And now you zap the 6 from each side and you’re done:
$6x^2 + 31x + 35 = (3x+5)(2x+7)$
It was shocking to me that this method works at all, and especially shocking that it always works: any factorable non-monic quadratic can be dealt with using this method. And variable substitution is a natural method used in other places: completing the square is a variable replacement using $\left(x - \frac b 2\right)$ as the variable … $x^4 - 1$ is a difference of squares … circles and ellipses all relate to the unit circle $x^2 + y^2 = 1$… trigonometric equations are just regular equations when you cover your hand over the “$\sin x$” part … a z-score is a linear substitution … and others.
The biggest benefit of presenting substitution methods as early as possible is that students learn a general-purpose tool they can apply repeatedly across grades and topics. I also think it makes quadratic factoring easier and faster to teach. What do you think?
Next: how this method can be used to develop the quadratic formula…
(If you know how to better display equations easily in places like this, let me know. The LaTeX equations look pretty bad in the vertical alignment category, and I had to force a white background on each equation. As long as it’s readable, I guess, but somehow I think it could be better. Thanks to Mark Betnel for the pointer to the LaTeX commands available.) | 6,039 | 22,364 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 120, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.925465 |
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# hw3_0 - EE 378 Handout #3 Statistical Signal Processing...
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Unformatted text preview: EE 378 Handout #3 Statistical Signal Processing Wednesday, April 25, 2007 Homework Set #3 Due: Wednesday, May 2, 2007. Announcement: You can hand in the HW either after class or deposit it, before 5pm, in the Homework in box in the 378 drawer of the class file cabinet on the second floor of the Packard Building. 1. For a stationary zero-mean process, X n with power spectral density S xx ( e jω ) and k th- order correlation matrix R k , show that λ max λ min ≤ S max S min where S max and S min are the largest and smallest values of S xx ( e jω ) and λ max and λ min are the largest and smallest eigenvalues of the k th-order correlation matrix, R k 2. The power spectral density S xx ( e jω ) of every real-valued WSS process is real, even, and nonnegative. In this problem you will show that, conversely, if S xx ( e jω ) is a real, even, nonnegative function with R π- π S xx ( e jω ) dw < ∞ , then S xx ( e jω ) is the psd for some WSS random process. Let us consider the case that 1 π Z π- π S xx ( e jω ) dw = 1 . Define the random process X ( n ) = cos(Ω n + Θ) , where Ω is a random variable on (- π,π ] with a pdf, 1 π S xx ( e jω ) and Θ ∼ U[0 , 2 π ] are independent.independent....
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## This note was uploaded on 03/03/2011 for the course EE 378 taught by Professor Weissman,i during the Spring '07 term at Stanford.
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hw3_0 - EE 378 Handout #3 Statistical Signal Processing...
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Ask a homework question - tutors are online | 521 | 1,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.849569 |
https://en.wikipedia.org/wiki/Base_12 | 1,498,302,464,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320257.16/warc/CC-MAIN-20170624101204-20170624121204-00631.warc.gz | 749,437,519 | 38,452 | # Duodecimal
(Redirected from Base 12)
The duodecimal system (also known as base 12 or dozenal) is a positional notation numeral system using twelve as its base. In this system, the number ten may be written by a rotated "2" (2) and the number eleven by a rotated "3" (3). This notation was introduced by Sir Isaac Pitman.[1] These digit forms are available as Unicode characters on computerized systems since June 2015[2] as ↊ (Code point 218A) and ↋ (Code point 218B), respectively.[3] Other notations use "A", "T", or "X" for ten and "B" or "E" for eleven. The number twelve (that is, the number written as "12" in the base ten numerical system) is instead written as "10" in duodecimal (meaning "1 dozen and 0 units", instead of "1 ten and 0 units"), whereas the digit string "12" means "1 dozen and 2 units" (i.e. the same number that in decimal is written as "14"). Similarly, in duodecimal "100" means "1 gross", "1000" means "1 great gross", and "0.1" means "1 twelfth" (instead of their decimal meanings "1 hundred", "1 thousand", and "1 tenth").
The number twelve, a superior highly composite number, is the smallest number with four non-trivial factors (2, 3, 4, 6), and the smallest to include as factors all four numbers (1 to 4) within the subitizing range. As a result of this increased factorability of the radix and its divisibility by a wide range of the most elemental numbers (whereas ten has only two non-trivial factors: 2 and 5, and not 3, 4, or 6), duodecimal representations fit more easily than decimal ones into many common patterns, as evidenced by the higher regularity observable in the duodecimal multiplication table. As a result, duodecimal has been described as the optimal number system.[4] Of its factors, 2 and 3 are prime, which means the reciprocals of all 3-smooth numbers (such as 2, 3, 4, 6, 8, 9...) have a terminating representation in duodecimal. In particular, the five most elementary fractions ( 12, 13, 23, 14 and 34) all have a short terminating representation in duodecimal (0.6, 0.4, 0.8, 0.3 and 0.9, respectively), and twelve is the smallest radix with this feature (because it is the least common multiple of 3 and 4). This all makes it a more convenient number system for computing fractions than most other number systems in common use, such as the decimal, vigesimal, binary, octal and hexadecimal systems. Although the trigesimal and sexagesimal systems (where the reciprocals of all 5-smooth numbers terminate) do even better in this respect, this is at the cost of unwieldy multiplication tables and a much larger number of symbols to memorize.
## Origin
In this section, numerals are based on decimal places. For example, 10 means ten, 12 means twelve.
Languages using duodecimal number systems are uncommon. Languages in the Nigerian Middle Belt such as Janji, Gbiri-Niragu (Gure-Kahugu), Piti, and the Nimbia dialect of Gwandara;[5] the Chepang language of Nepal[6] and the Mahl language of Minicoy Island in India are known to use duodecimal numerals.
Germanic languages have special words for 11 and 12, such as eleven and twelve in English. However, they are considered to come from Proto-Germanic *ainlif and *twalif (respectively one left and two left), both of which were decimal.[7]
Historically, units of time in many civilizations are duodecimal. There are twelve signs of the zodiac, twelve months in a year, and the Babylonians had twelve hours in a day (although at some point this was changed to 24). Traditional Chinese calendars, clocks, and compasses are based on the twelve Earthly Branches. There are 12 inches in an imperial foot, 12 troy ounces in a troy pound, 12 old British pence in a shilling, 24 (12×2) hours in a day, and many other items counted by the dozen, gross (144, square of 12) or great gross (1728, cube of 12). The Romans used a fraction system based on 12, including the uncia which became both the English words ounce and inch. Pre-decimalisation, Ireland and the United Kingdom used a mixed duodecimal-vigesimal currency system (12 pence = 1 shilling, 20 shillings or 240 pence to the pound sterling or Irish pound), and Charlemagne established a monetary system that also had a mixed base of twelve and twenty, the remnants of which persist in many places.
Table of units from a base of 12
Relative
value
French unit
of length
English unit
of length
English unit
of weight
Roman unit
of weight
English unit of mass
120 pied foot pound libra
12−1 pouce inch ounce uncia slinch
12−2 ligne line 2 scruples 2 scrupulum slug
12−3 point point seed siliqua
The importance of 12 has been attributed to the number of lunar cycles in a year, and also to the fact that humans have 12 finger bones (phalanges) on one hand (three on each of four fingers).[8] It is possible to count to 12 with the thumb acting as a pointer, touching each finger bone in turn. A traditional finger counting system still in use in many regions of Asia works in this way, and could help to explain the occurrence of numeral systems based on 12 and 60 besides those based on 10, 20 and 5. In this system, the one (usually right) hand counts repeatedly to 12, displaying the number of iterations on the other (usually left), until five dozens, i. e. the 60, are full.[9][10]
## Notations and pronunciations
### Symbols
In a duodecimal place system twelve is written as 10, but there are numerous proposals for how to write ten and eleven.[11] The simplified notations use only basic and easy to access letters such as T and E (for ten and eleven), X and Z, t and e, d and k, other use A and B or a and b like in the hexadecimal system. Some employ Greek letters such as δ (standing for Greek δέκα 'ten') and ε (for Greek ένδεκα 'eleven'), or τ and ε.[11] Frank Emerson Andrews, an early American advocate for duodecimal, suggested and used in his book New Numbers an X and a script E (ℰ, U+2130).[12]
The Dozenal Society of Great Britain proposes a rotated digit two 2 (↊, U+218A) for ten and a reversed or rotated digit three 3 (↋, U+218B) for eleven.[11] This notation was introduced by Sir Isaac Pitman.[11][13]
The Dozenal Society of America uses and instead, the symbols devised by William Addison Dwiggins.[11][14]
Other proposals are more creative or aesthetic, for example, Edna Kramer in her 1951 book The Main Stream of Mathematics [sic] used a six-pointed asterisk (sextile) ⚹ for ten and a hash (or octothorpe) # for eleven.[11] The symbols were chosen because they are available in typewriters and already present in telephone dials.[11] This notation was used in publications of the Dozenal Society of America in the period 1974–2008.[15][16]
### Pronunciation
The Dozenal Society of America suggests the pronunciation of ten and eleven as "dek" and "el", each order has its own name and the prefix e- is added for fractions.[14][17] The overall system is:[14]
Duodecimal Name Decimal Duodecimal fraction Name
1 one 1
10 do 12 0;1 edo
100 gro 144 0;01 egro
1,000 mo 1,728 0;001 emo
10,000 do-mo 20,736 0;000,1 edo-mo
100,000 gro-mo 248,832 0;000,01 egro-mo
1,000,000 bi-mo 2,985,984 0;000,001 ebi-mo
1,000,000,000 tri-mo 5,159,780,352 0;000,000,001 etri-mo
Multiple digits in this are pronounced differently. 12 is "one do two", 30 is "three do", 100 is "one gro", ↋↊9 is "el gro dek do nine", ↋8,65↊,300 is "el do eight bi-mo, six gro five do dek mo, three gro", and so on.[17]
The case for the duodecimal system was put forth at length in F. Emerson Andrews' 1935 book New Numbers: How Acceptance of a Duodecimal Base Would Simplify Mathematics. Emerson noted that, due to the prevalence of factors of twelve in many traditional units of weight and measure, many of the computational advantages claimed for the metric system could be realized either by the adoption of ten-based weights and measure or by the adoption of the duodecimal number system.
A duodecimal clockface as in the logo of the Dozenal Society of America, here used to denote musical keys
The Dozenal Society of America and the Dozenal Society of Great Britain promote widespread adoption of the base-twelve system. They use the word "dozenal" instead of "duodecimal" because the latter comes from Latin roots that express twelve in base-ten terminology.
The renowned mathematician and mental calculator Alexander Craig Aitken was an outspoken advocate of the advantages and superiority of duodecimal over decimal:
The duodecimal tables are easy to master, easier than the decimal ones; and in elementary teaching they would be so much more interesting, since young children would find more fascinating things to do with twelve rods or blocks than with ten. Anyone having these tables at command will do these calculations more than one-and-a-half times as fast in the duodecimal scale as in the decimal. This is my experience; I am certain that even more so it would be the experience of others.
— A. C. Aitken, "Twelves and Tens", in The Listener, January 25, 1962
But the final quantitative advantage, in my own experience, is this: in varied and extensive calculations of an ordinary and not unduly complicated kind, carried out over many years, I come to the conclusion that the efficiency of the decimal system might be rated at about 65 or less, if we assign 100 to the duodecimal.
— A. C. Aitken, The Case Against Decimalisation (Edinburgh / London: Oliver & Boyd, 1962)
In Jorge Luis Borges' short story Tlön, Uqbar, Orbis Tertius Herbert Ashe, a melancholy English engineer, working for the Southern Argentine Railway company, is converting a duodecimal number system to a hexadecimal system. He leaves behind on his death in 1937 a manuscript Orbis Tertius that posthumously identifies him as one of the anonymous authors of the encyclopaedia of Tlön.
In Leo Frankowski's Conrad Stargard novels, Conrad introduces a duodecimal system of arithmetic at the suggestion of a merchant, who is accustomed to buying and selling goods in dozens and grosses, rather than tens or hundreds. He then invents an entire system of weights and measures in base twelve, including a clock with twelve hours in a day, rather than twenty-four hours.[citation needed]
In Lee Carroll's Kryon: Alchemy of the Human Spirit, a chapter is dedicated to the advantages of the duodecimal system. The duodecimal system is supposedly suggested by Kryon (a fictional entity believed in by New Age circles) for all-round use, aiming at better and more natural representation of nature of the Universe through mathematics. An individual article "Mathematica" by James D. Watt (included in the above publication) exposes a few of the unusual symmetry connections between the duodecimal system and the golden ratio, as well as provides numerous number symmetry-based arguments for the universal nature of the base-12 number system.[18]
In "Little Twelvetoes", American television series Schoolhouse Rock! portrayed an alien child using base-twelve arithmetic, using "dek", "el" and "doh" as names for ten, eleven and twelve, and Andrews' script-X and script-E for the digit symbols.[19]
### In computing
In March 2013, a proposal was submitted to include the digit forms for ten and eleven propagated by the Dozenal Societies of Great Britain and America in the Unicode Standard.[20] Of these, the British forms were accepted for encoding as characters at code points U+218A turned digit two (↊) and U+218B turned digit three (↋) They have been included in the Unicode 8.0 release in June 2015.[2][21]
Few fonts support these new characters, but Abibas, EB Garamond, Everson Mono, Squarish Sans CT, and Symbola do.
Also, the turned digits two and three are available in LaTeX as \textturntwo and \textturnthree.[22]
### Duodecimal metric systems
Systems of measurement proposed by dozenalists include:
• Tom Pendlebury's TGM system[23]
• Takashi Suga's Universal Unit System[24]
## Comparison to other numeral systems
A duodecimal multiplication table
The number 12 has six factors, which are 1, 2, 3, 4, 6, and 12, of which 2 and 3 are prime. The decimal system has only four factors, which are 1, 2, 5, and 10, of which 2 and 5 are prime. Vigesimal (base 20) adds two factors to those of ten, namely 4 and 20, but no additional prime factor. Although twenty has 6 factors, 2 of them prime, similarly to twelve, it is also a much larger base, and so the digit set and the multiplication table are much larger. Binary has only two factors, 1 and 2, the latter being prime. Hexadecimal (base 16) has five factors, adding 4, 8 and 16 to those of 2, but no additional prime. Trigesimal (base 30) is the smallest system that has three different prime factors (all of the three smallest primes: 2, 3 and 5) and it has eight factors in total (1, 2, 3, 5, 6, 10, 15, and 30). Sexagesimal—which the ancient Sumerians and Babylonians among others actually used—adds the four convenient factors 4, 12, 20, and 60 to this but no new prime factors. The smallest system that has four different prime factors is base 210 and the pattern follows the primorials. In all base systems, there are similarities to the representation of multiples of numbers which are one less than the base.
## Conversion tables to and from decimal
To convert numbers between bases, one can use the general conversion algorithm (see the relevant section under positional notation). Alternatively, one can use digit-conversion tables. The ones provided below can be used to convert any duodecimal number between 0.01 and ƐƐƐ,ƐƐƐ.ƐƐ to decimal, or any decimal number between 0.01 and 999,999.99 to duodecimal. To use them, the given number must first be decomposed into a sum of numbers with only one significant digit each. For example:
123,456.78 = 100,000 + 20,000 + 3,000 + 400 + 50 + 6 + 0.7 + 0.08
This decomposition works the same no matter what base the number is expressed in. Just isolate each non-zero digit, padding them with as many zeros as necessary to preserve their respective place values. If the digits in the given number include zeroes (for example, 102,304.05), these are, of course, left out in the digit decomposition (102,304.05 = 100,000 + 2,000 + 300 + 4 + 0.05). Then the digit conversion tables can be used to obtain the equivalent value in the target base for each digit. If the given number is in duodecimal and the target base is decimal, we get:
(duodecimal) 100,000 + 20,000 + 3,000 + 400 + 50 + 6 + 0.7 + 0.08 = (decimal) 248,832 + 41,472 + 5,184 + 576 + 60 + 6 + 0.583333333333... + 0.055555555555...
Now, because the summands are already converted to base ten, the usual decimal arithmetic is used to perform the addition and recompose the number, arriving at the conversion result:
Duodecimal -----> Decimal
100,000 = 248,832
20,000 = 41,472
3,000 = 5,184
400 = 576
50 = 60
+ 6 = + 6
0.7 = 0.583333333333...
0.08 = 0.055555555555...
--------------------------------------------
123,456.78 = 296,130.638888888888...
That is, (duodecimal) 123,456.78 equals (decimal) 296,130.638 ≈ 296,130.64
If the given number is in decimal and the target base is duodecimal, the method is basically same. Using the digit conversion tables:
(decimal) 100,000 + 20,000 + 3,000 + 400 + 50 + 6 + 0.7 + 0.08 = (duodecimal) 49,ᘔ54 + Ɛ,6ᘔ8 + 1,8ᘔ0 + 294 + 42 + 6 + 0.849724972497249724972497... + 0.0Ɛ62ᘔ68781Ɛ05915343ᘔ0Ɛ62...
However, in order to do this sum and recompose the number, now the addition tables for the duodecimal system have to be used, instead of the addition tables for decimal most people are already familiar with, because the summands are now in base twelve and so the arithmetic with them has to be in duodecimal as well. In decimal, 6 + 6 equals 12, but in duodecimal it equals 10; so, if using decimal arithmetic with duodecimal numbers one would arrive at an incorrect result. Doing the arithmetic properly in duodecimal, one gets the result:
Decimal -----> Duodecimal
100,000 = 49,ᘔ54
20,000 = Ɛ,6ᘔ8
3,000 = 1,8ᘔ0
400 = 294
50 = 42
+ 6 = + 6
0.7 = 0.849724972497249724972497...
0.08 = 0.0Ɛ62ᘔ68781Ɛ05915343ᘔ0Ɛ62...
--------------------------------------------------------
123,456.78 = 5Ɛ,540.943ᘔ0Ɛ62ᘔ68781Ɛ05915343ᘔ...
That is, (decimal) 123,456.78 equals (duodecimal) 5Ɛ,540.943ᘔ0Ɛ62ᘔ68781Ɛ059153... ≈ 5Ɛ,540.94
### Duodecimal to decimal digit conversion
Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec.
100,000 248,832 10,000 20,736 1,000 1,728 100 144 10 12 1 1 0.1 0.083 0.01 0.00694
200,000 497,664 20,000 41,472 2,000 3,456 200 288 20 24 2 2 0.2 0.16 0.02 0.0138
300,000 746,496 30,000 62,208 3,000 5,184 300 432 30 36 3 3 0.3 0.25 0.03 0.02083
400,000 995,328 40,000 82,944 4,000 6,912 400 576 40 48 4 4 0.4 0.3 0.04 0.027
500,000 1,244,160 50,000 103,680 5,000 8,640 500 720 50 60 5 5 0.5 0.416 0.05 0.03472
600,000 1,492,992 60,000 124,416 6,000 10,368 600 864 60 72 6 6 0.6 0.5 0.06 0.0416
700,000 1,741,824 70,000 145,152 7,000 12,096 700 1008 70 84 7 7 0.7 0.583 0.07 0.04861
800,000 1,990,656 80,000 165,888 8,000 13,824 800 1152 80 96 8 8 0.8 0.6 0.08 0.05
900,000 2,239,488 90,000 186,624 9,000 15,552 900 1,296 90 108 9 9 0.9 0.75 0.09 0.0625
ᘔ00,000 2,488,320 ᘔ0,000 207,360 ᘔ,000 17,280 ᘔ00 1,440 ᘔ0 120 10 0.ᘔ 0.83 0.0ᘔ 0.0694
Ɛ00,000 2,737,152 Ɛ0,000 228,096 Ɛ,000 19,008 Ɛ00 1,584 Ɛ0 132 Ɛ 11 0.Ɛ 0.916 0.0Ɛ 0.07638
### Decimal to duodecimal digit conversion
Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod.
100,000 49,ᘔ54 10,000 5,954 1,000 6Ɛ4 100 84 10 1 1 0.1 0.12497 0.01 0.015343ᘔ0Ɛ62ᘔ68781Ɛ059
200,000 97,8ᘔ8 20,000 Ɛ,6ᘔ8 2,000 1,1ᘔ8 200 148 20 18 2 2 0.2 0.2497 0.02 0.02ᘔ68781Ɛ05915343ᘔ0Ɛ6
300,000 125,740 30,000 15,440 3,000 1,8ᘔ0 300 210 30 26 3 3 0.3 0.37249 0.03 0.043ᘔ0Ɛ62ᘔ68781Ɛ059153
400,000 173,594 40,000 1Ɛ,194 4,000 2,394 400 294 40 34 4 4 0.4 0.4972 0.04 0.05915343ᘔ0Ɛ62ᘔ68781Ɛ
500,000 201,428 50,000 24,Ɛ28 5,000 2,ᘔ88 500 358 50 42 5 5 0.5 0.6 0.05 0.07249
600,000 24Ɛ,280 60,000 2ᘔ,880 6,000 3,580 600 420 60 50 6 6 0.6 0.7249 0.06 0.08781Ɛ05915343ᘔ0Ɛ62ᘔ6
700,000 299,114 70,000 34,614 7,000 4,074 700 4ᘔ4 70 5ᘔ 7 7 0.7 0.84972 0.07 0.0ᘔ0Ɛ62ᘔ68781Ɛ05915343
800,000 326,Ɛ68 80,000 3ᘔ,368 8,000 4,768 800 568 80 68 8 8 0.8 0.9724 0.08 0.0Ɛ62ᘔ68781Ɛ05915343ᘔ
900,000 374,ᘔ00 90,000 44,100 9,000 5,260 900 630 90 76 9 9 0.9 0.ᘔ9724 0.09 0.10Ɛ62ᘔ68781Ɛ05915343ᘔ
### Conversion of powers
Exponent b=2 b=3 b=4 b=5 b=6 b=7
Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod.
b6 64 54 729 509 4,096 2454 15,625 9,061 46,656 23,000 117,649 58,101
b5 32 28 243 183 1,024 714 3,125 1,985 7,776 4,600 16,807 9,887
b4 16 14 81 69 256 194 625 441 1,296 900 2,401 1,481
b3 8 8 27 23 64 54 125 ᘔ5 216 160 343 247
b2 4 4 9 9 16 14 25 21 36 30 49 41
b1 2 2 3 3 4 4 5 5 6 6 7 7
b−1 0.5 0.6 0.3 0.4 0.25 0.3 0.2 0.2497 0.16 0.2 0.142857 0.186ᘔ35
b−2 0.25 0.3 0.1 0.14 0.0625 0.09 0.04 0.05915343ᘔ0
Ɛ62ᘔ68781Ɛ
0.027 0.04 0.0204081632653
06122448979591
836734693877551
0.02Ɛ322547ᘔ05ᘔ
644ᘔ9380Ɛ908996
741Ɛ615771283Ɛ
Exponent b=8 b=9 b=10 b=11 b=12
Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod.
b6 262,144 107,854 531,441 217,669 1,000,000 402,854 1,771,561 715,261 2,985,984 1,000,000
b5 32,768 16,Ɛ68 59,049 2ᘔ,209 100,000 49,ᘔ54 161,051 79,24Ɛ 248,832 100,000
b4 4,096 2,454 6,561 3,969 10,000 5,954 14,641 8,581 20,736 10,000
b3 512 368 729 509 1,000 6Ɛ4 1,331 92Ɛ 1,728 1,000
b2 64 54 81 69 100 84 121 ᘔ1 144 100
b1 8 8 9 9 10 11 Ɛ 12 10
b−1 0.125 0.16 0.1 0.14 0.1 0.12497 0.09 0.1 0.083 0.1
b−2 0.015625 0.023 0.012345679 0.0194 0.01 0.015343ᘔ0Ɛ6
2ᘔ68781Ɛ059
0.00826446280
99173553719
0.0123456789Ɛ 0.00694 0.01
## Fractions and irrational numbers
### Fractions
Duodecimal fractions may be simple:
• 1/2 = 0.6
• 1/3 = 0.4
• 1/4 = 0.3
• 1/6 = 0.2
• 1/8 = 0.16
• 1/9 = 0.14
• 1/10 = 0.1 (note that this is a twelfth, 1/ is a tenth)
or complicated:
• 1/5 = 0.249724972497... recurring (rounded to 0.24ᘔ)
• 1/7 = 0.186ᘔ35186ᘔ35... recurring (rounded to 0.187)
• 1/ = 0.1249724972497... recurring (rounded to 0.125)
• 1/Ɛ = 0.111111111111... recurring (rounded to 0.111)
• 1/11 = 0.0Ɛ0Ɛ0Ɛ0Ɛ0Ɛ0Ɛ... recurring (rounded to 0.0Ɛ1)
• 1/12 = 0.0ᘔ35186ᘔ35186... recurring (rounded to 0.0ᘔ3)
Examples in duodecimal Decimal equivalent
1 × (5/8) = 0.76 1 × (5/8) = 0.625
100 × (5/8) = 76 144 × (5/8) = 90
576/9 = 76 810/9 = 90
400/9 = 54 576/9 = 64
1ᘔ.6 + 7.6 = 26 22.5 + 7.5 = 30
As explained in recurring decimals, whenever an irreducible fraction is written in radix point notation in any base, the fraction can be expressed exactly (terminates) if and only if all the prime factors of its denominator are also prime factors of the base. Thus, in base-ten (= 2×5) system, fractions whose denominators are made up solely of multiples of 2 and 5 terminate: 1/8 = 1/(2×2×2), 1/20 = 1/(2×2×5) and 1/500 = 1/(2×2×5×5×5) can be expressed exactly as 0.125, 0.05 and 0.002 respectively. 1/3 and 1/7, however, recur (0.333... and 0.142857142857...). In the duodecimal (= 2×2×3) system, 1/8 is exact; 1/20 and 1/500 recur because they include 5 as a factor; 1/3 is exact; and 1/7 recurs, just as it does in decimal.
The number of denominators which give terminating fractions within a given number of digits, say n, in a base b is the number of factors (divisors) of bn, the nth power of the base b (although this includes the divisor 1, which does not produce fractions when used as the denominator). The number of factors of bn is given using its prime factorization.
For decimal, 10n = 2n * 5n. The number of divisors is found by adding one to each exponent of each prime and multiplying the resulting quantities together. Factors of 10n = (n+1)(n+1) = (n+1)2.
For example, the number 8 is a factor of 103 (1000), so 1/8 and other fractions with a denominator of 8 can not require more than 3 fractional decimal digits to terminate. 5/8 = 0.625ten
For duodecimal, 12n = 22n * 3n. This has (2n+1)(n+1) divisors. The sample denominator of 8 is a factor of a gross (122 = 144), so eighths can not need more than two duodecimal fractional places to terminate. 5/8 = 0.76twelve
Because both ten and twelve have two unique prime factors, the number of divisors of bn for b = 10 or 12 grows quadratically with the exponent n (in other words, of the order of n2).
### Recurring digits
The Dozenal Society of America argues that factors of 3 are more commonly encountered in real-life division problems than factors of 5.[25] Thus, in practical applications, the nuisance of repeating decimals is encountered less often when duodecimal notation is used. Advocates of duodecimal systems argue that this is particularly true of financial calculations, in which the twelve months of the year often enter into calculations.
However, when recurring fractions do occur in duodecimal notation, they are less likely to have a very short period than in decimal notation, because 12 (twelve) is between two prime numbers, 11 (eleven) and 13 (thirteen), whereas ten is adjacent to the composite number 9. Nonetheless, having a shorter or longer period doesn't help the main inconvenience that one does not get a finite representation for such fractions in the given base (so rounding, which introduces inexactitude, is necessary to handle them in calculations), and overall one is more likely to have to deal with infinite recurring digits when fractions are expressed in decimal than in duodecimal, because one out of every three consecutive numbers contains the prime factor 3 in its factorization, whereas only one out of every five contains the prime factor 5. All other prime factors, except 2, are not shared by either ten or twelve, so they do not influence the relative likeliness of encountering recurring digits (any irreducible fraction that contains any of these other factors in its denominator will recur in either base). Also, the prime factor 2 appears twice in the factorization of twelve, whereas only once in the factorization of ten; which means that most fractions whose denominators are powers of two will have a shorter, more convenient terminating representation in duodecimal than in decimal representation (e.g. 1/(22) = 0.25 ten = 0.3 twelve; 1/(23) = 0.125 ten = 0.16 twelve; 1/(24) = 0.0625 ten = 0.09 twelve; 1/(25) = 0.03125 ten = 0.046 twelve; etc.).
Values in bold indicate that value is exact.
Fraction Prime factors Positional representation Positional representation Prime factors Fraction of the denominator of the denominator Decimal base Prime factors of the base: 2, 5 Prime factors of one below the base: 3 Prime factors of one above the base: 11 All other primes: 7 Duodecimal base Prime factors of the base: 2, 3 Prime factors of one below the base: Ɛ Prime factors of one above the base: 11 All other primes: 7 1/2 2 0.5 0.6 2 1/2 1/3 3 0.3 0.4 3 1/3 1/4 2 0.25 0.3 2 1/4 1/5 5 0.2 0.2497 5 1/5 1/6 2, 3 0.16 0.2 2, 3 1/6 1/7 7 0.142857 0.186ᘔ35 7 1/7 1/8 2 0.125 0.16 2 1/8 1/9 3 0.1 0.14 3 1/9 1/10 2, 5 0.1 0.12497 2, 5 1/ᘔ 1/11 11 0.09 0.1 Ɛ 1/Ɛ 1/12 2, 3 0.083 0.1 2, 3 1/10 1/13 13 0.076923 0.0Ɛ 11 1/11 1/14 2, 7 0.0714285 0.0ᘔ35186 2, 7 1/12 1/15 3, 5 0.06 0.09724 3, 5 1/13 1/16 2 0.0625 0.09 2 1/14 1/17 17 0.0588235294117647 0.08579214Ɛ36429ᘔ7 15 1/15 1/18 2, 3 0.05 0.08 2, 3 1/16 1/19 19 0.052631578947368421 0.076Ɛ45 17 1/17 1/20 2, 5 0.05 0.07249 2, 5 1/18 1/21 3, 7 0.047619 0.06ᘔ3518 3, 7 1/19 1/22 2, 11 0.045 0.06 2, Ɛ 1/1ᘔ 1/23 23 0.0434782608695652173913 0.06316948421 1Ɛ 1/1Ɛ 1/24 2, 3 0.0416 0.06 2, 3 1/20 1/25 5 0.04 0.05915343ᘔ0Ɛ62ᘔ68781Ɛ 5 1/21 1/26 2, 13 0.0384615 0.056 2, 11 1/22 1/27 3 0.037 0.054 3 1/23 1/28 2, 7 0.03571428 0.05186ᘔ3 2, 7 1/24 1/29 29 0.0344827586206896551724137931 0.04Ɛ7 25 1/25 1/30 2, 3, 5 0.03 0.04972 2, 3, 5 1/26 1/31 31 0.032258064516129 0.0478ᘔᘔ093598166Ɛ74311Ɛ28623ᘔ55 27 1/27 1/32 2 0.03125 0.046 2 1/28 1/33 3, 11 0.03 0.04 3, Ɛ 1/29 1/34 2, 17 0.02941176470588235 0.0429ᘔ708579214Ɛ36 2, 15 1/2ᘔ 1/35 5, 7 0.0285714 0.0414559Ɛ3931 5, 7 1/2Ɛ 1/36 2, 3 0.027 0.04 2, 3 1/30
The duodecimal period length of 1/n are
0, 0, 0, 0, 4, 0, 6, 0, 0, 4, 1, 0, 2, 6, 4, 0, 16, 0, 6, 4, 6, 1, 11, 0, 20, 2, 0, 6, 4, 4, 30, 0, 1, 16, 12, 0, 9, 6, 2, 4, 40, 6, 42, 1, 4, 11, 23, 0, 42, 20, 16, 2, 52, 0, 4, 6, 6, 4, 29, 4, 15, 30, 6, 0, 4, 1, 66, 16, 11, 12, 35, 0, ... (sequence A246004 in the OEIS)
The duodecimal period length of 1/(nth prime) are
0, 0, 4, 6, 1, 2, 16, 6, 11, 4, 30, 9, 40, 42, 23, 52, 29, 15, 66, 35, 36, 26, 41, 8, 16, 100, 102, 53, 54, 112, 126, 65, 136, 138, 148, 150, 3, 162, 83, 172, 89, 90, 95, 24, 196, 66, 14, 222, 113, 114, 8, 119, 120, 125, 256, 131, 268, 54, 138, 280, ... (sequence A246489 in the OEIS)
Smallest prime with duodecimal period n are
11, 13, 157, 5, 22621, 7, 659, 89, 37, 19141, 23, 20593, 477517, 211, 61, 17, 2693651, 1657, 29043636306420266077, 85403261, 8177824843189, 57154490053, 47, 193, 303551, 79, 306829, 673, 59, 31, 373, 153953, 886381, 2551, 71, 73, ... (sequence A252170 in the OEIS)
### Irrational numbers
As for irrational numbers, none of them have a finite representation in any of the rational-based positional number systems (such as the decimal and duodecimal ones); this is because a rational-based positional number system is essentially nothing but a way of expressing quantities as a sum of fractions whose denominators are powers of the base, and by definition no finite sum of rational numbers can ever result in an irrational number. For example, 123.456 = 1 × 102 + 2 × 101 + 3 × 100 + 4 × 1/101 + 5 × 1/102 + 6 × 1/103 (this is also the reason why fractions that contain prime factors in their denominator not in common with those of the base do not have a terminating representation in that base). Moreover, the infinite series of digits of an irrational number does not exhibit a strictly repeating pattern; instead, the different digits often succeed in a seemingly random fashion. The following chart compares the first few digits of the decimal and duodecimal representation of several of the most important algebraic and transcendental irrational numbers. Some of these numbers may be perceived as having fortuitous patterns, making them easier to memorize, when represented in one base or the other.
Algebraic irrational number In decimal In duodecimal
2 (the length of the diagonal of a unit square) 1.41421356237309... (≈ 1.4142) 1.4Ɛ79170ᘔ07Ɛ857... (≈ 1.5)
3 (the length of the diagonal of a unit cube, or twice the height of an equilateral triangle of unit side) 1.73205080756887... (≈ 1.732) 1.894Ɛ97ƐƐ968704... (≈ 1.895)
5 (the length of the diagonal of a 1×2 rectangle) 2.2360679774997... (≈ 2.236) 2.29ƐƐ132540589... (≈ 2.2ᘔ)
φ (phi, the golden ratio = ${\displaystyle \scriptstyle {\frac {1+{\sqrt {5}}}{2}}}$) 1.6180339887498... (≈ 1.618) 1.74ƐƐ6772802ᘔ4... (≈ 1.75)
Transcendental irrational number In decimal In duodecimal
π (pi, the ratio of circumference to diameter) 3.1415926535897932384626433
8327950288419716939937510...
(≈ 3.1416)
3.184809493Ɛ918664573ᘔ6211Ɛ
Ɛ151551ᘔ05729290ᘔ7809ᘔ492...
(≈ 3.1848)
e (the base of the natural logarithm) 2.718281828459045... (≈ 2.718) 2.8752360698219Ɛ8... (≈ 2.875)
The first few digits of the decimal and duodecimal representation of another important number, the Euler–Mascheroni constant (the status of which as a rational or irrational number is not yet known), are:
Number In decimal In duodecimal
γ (the limiting difference between the harmonic series and the natural logarithm) 0.57721566490153... (≈ 0.577) 0.6Ɛ15188ᘔ6760Ɛ3... (≈ 0.7)
## References
1. ^ Pitman, Isaac (ed.): A triple (twelve gross) Gems of Wisdom. London 1860
2. ^ a b "Unicode 8.0.0". Unicode Consortium. Retrieved 2016-05-30.
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4. ^ George Dvorsky (2013-01-18). "Why We Should Switch To A Base-12 Counting System". Retrieved 2013-12-21.
5. ^ Matsushita, Shuji (1998). Decimal vs. Duodecimal: An interaction between two systems of numeration. 2nd Meeting of the AFLANG, October 1998, Tokyo. Archived from the original on 2008-10-05. Retrieved 2011-05-29
6. ^ Mazaudon, Martine (2002). "Les principes de construction du nombre dans les langues tibéto-birmanes". In François, Jacques. La Pluralité (PDF). Leuven: Peeters. pp. 91–119. ISBN 90-429-1295-2
7. ^ von Mengden, Ferdinand (2006). "The peculiarities of the Old English numeral system". In Nikolaus Ritt, Herbert Schendl, Christiane Dalton-Puffer, Dieter Kastovsky. Medieval English and its Heritage: Structure Meaning and Mechanisms of Change. Studies in English Medieval Language and Literature. 16. Frankfurt: Peter Lang Pub. pp. 125–45.
von Mengden, Ferdinand (2010). Cardinal Numerals: Old English from a Cross-Linguistic Perspective. Topics in English Linguistics. 67. Berlin; New York: De Gruyter Mouton. pp. 159–161.
8. ^ Nishikawa, Yoshiaki (2002). "ヒマラヤの満月と十二進法 (The Full Moon in the Himalayas and the Duodecimal System)". Archived from the original on March 29, 2008. Retrieved 2008-03-24
9. ^ Ifrah, Georges (2000). The Universal History of Numbers: From prehistory to the invention of the computer. John Wiley and Sons. ISBN 0-471-39340-1. Translated from the French by David Bellos, E.F. Harding, Sophie Wood and Ian Monk.
10. ^ Macey, Samuel L. (1989). The Dynamics of Progress: Time, Method, and Measure. Atlanta, Georgia: University of Georgia Press. p. 92. ISBN 978-0-8203-3796-8
11. De Vlieger, Michael (2010). "Symbology Overview" (PDF). The Duodecimal Bulletin. 4X [59] (2).
12. ^ Andrews, Frank Emerson (1935). New Numbers: How Acceptance of a Duodecimal (12) Base Would Simplify Mathematics. p. 52.
13. ^ Pitman, Isaac (1947). "A Reckoning Reform [reprint from 1857]" (PDF). The Duodecimal Bulletin. 3 (2).
14. ^ a b c "Mo for Megro" (PDF). The Duodecimal Bulletin. 1 (1). 1945.
15. ^ "Annual Meeting of 1973 and Meeting of the Board" (PDF). The Duodecimal Bulletin. 25 [29] (1). 1974.
16. ^ De Vlieger, Michael (2008). "Going Classic" (PDF). The Duodecimal Bulletin. 49 [57] (2).
17. ^ a b Zirkel, Gene (2010). "How Do You Pronounce Dozenals?" (PDF). The Duodecimal Bulletin. 4E [59] (2).
18. ^ Carroll, Lee (1995). Kryon—Alchemy of the Human Spirit. The Kryon Writings, Inc. ISBN 0-9636304-8-2.
19. ^ "Little Twelvetoes"
20. ^ Karl Pentzlin (2013-03-30). "Proposal to encode Duodecimal Digit Forms in the UCS" (PDF). ISO/IEC JTC1/SC2/WG2, Document N4399. Retrieved 2016-05-30.
21. ^ "The Unicode Standard, Version 8.0: Number Forms" (PDF). Unicode Consortium. Retrieved 2016-05-30.
22. ^ Scott Pakin (2009). "The Comprehensive LATEX Symbol List" (PDF). Retrieved 2016-05-30.
23. ^ Pendlebury, Tom (May 2011). "TGM. A coherent dozenal metrology based on Time, Gravity and Mass" (PDF). The Dozenal Society of Great Britain.
24. ^ Suga, Takashi (2002). "Proposal for the Universal Unit System".
25. ^ http://www.dozenal.org/articles/DSA-DozenalFAQs.pdf | 11,575 | 32,897 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.5 | 4 | CC-MAIN-2017-26 | latest | en | 0.929227 |
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Author(s): No creator set | 1,606 | 7,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2013-20 | latest | en | 0.923839 |
https://wrfranklin.org/pmwiki/pmwiki.php/Research/Siting | 1,670,663,479,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710421.14/warc/CC-MAIN-20221210074242-20221210104242-00723.warc.gz | 668,547,681 | 10,375 | This page summarizes the siting research of Prof W. Randolph Franklin, Rensselaer Polytechnic Institute, Troy NY 12180, (web: http://franklin.org/ ) and colleagues, including Prof Marcus Andrade at the Universidade Federal de Viçosa (MG), Brasil. Much of this research was performed in 2005–2009 as a DARPA/DSO/Geo* project. Since then, work has continued as a small part of various NSF grants, such as IIS-1117277.
# Algorithms
Our research has produced the following efficient algorithms (with implementations) related to siting observers on a raster (e.g., DEM) terrain.
1. Visibility index computation. This determines how good a given observer is, by computing what fraction of a circle of specified radius, centered on the observer, that is visible by that observer.
2. Viewshed computation. This computes the exact region of the terrain that a given observer can see.
3. Multi-observer siting. For given K, find quasi-optimal sites (or locations) for K observers, to maximize the area of the union of their viewsheds.
4. ... with intervisibility. Constrain the multi-observer siting so that the graph of observer – observer visibility is connected. (The vertices of this graph are the observers. There is an edge between two observers if they are in each others' viewsheds.)
5. Path planning to avoid the observers, aka the smugglers and border guards problem. Given some terrain, and some observers that have been jointly sited as in item 3 above, and given a source and a goal pixel, determine an optimal path between the source and goal. The cost function has three components, in order from most to least important.
1. Try very hard to avoid being seen.
2. Minimize uphill travel. (Since downhill travel is free, the cost is not symmetric and so not a metric function.)
3. Minimize horizontal distance traveled.
# Properties
General properties of all the algorithms are as follows.
1. The terrain data structure is a grid of elevation posts or pixels.
2. The observer is allowed to be a specified distance above the local terrain, as would be the case for a lighthouse on a reef.
3. The targets, whose visibility is desired, are also allowed to be a specified distance above their local terrain, as for lookouts on ships.
4. The target must be within a given radius of interest of the observer to be visible.
5. The terrain can be hi-res — 3600x3600 pixels is easy to process.
6. The radius of interest can be large, easily hundreds of pixel widths. (These last two advantages are not always enjoyed by competing algorithms).
7. Since these problems usually take exponential time to optimize perfectly, and since the data has a large error bar in z and a low resolution in (x,y), we use Monte Carlo algorithms and heuristics rather than theoretically optimal algorithms.
8. We implement everything and test it on multiple datasets.
9. Our code is freely available for non-profit research and development.
# Applications of multiobserver siting
This multiple observers case is particularly interesting and complex, and has many applications. A cell phone provider wishes to install multiple towers so that at least one tower is visible (in a radio sense) from every place a customer's cellphone might be. Here, the identities of the observers of highest visibility index are of more interest than their exact visibility indices, or than the visibility indices of all observers. One novel future application of siting radio transmitters will occur when the moon is settled. The moon has no ionosphere to reflect signals, and no stable satellite orbits. The choices for long-range communication would seem to include either a lot of fiber optic cable or many relay towers. That solution is the multiple observer visibility problem.
As another example, a military planner needs to put observers so that there is nowhere to hide that is not visible from at least one. This leads to a corollary application, where the other side's planner may want to analyze the first side's observers to find places to hide. In this case, the problem is to optimize the targets' locations, instead of the observers'.
Again, a planner for a scenic area may consider each place where a tourist might be to be an observer, and then want to locate ugly infrastructure, such as work yards, at relatively hidden sites. We may wish site a forest clearcut to be invisible to observers driving on a highway sited to give a good view. Finally, an architect may be trying to site a new house while following the planning board's instruction that, You can have a view, but you can't be the view.
Speed of execution on large datasets is of more importance than may be apparent. Many prototype implementations, altho demonstrated on small datasets, do not scale up well. Some preliminary published algorithms may even be exponential if performing a naive search. Therefore, we strive for the best time possible.
In addition, large datasets may contain cases, which did not occur in the small test sets, that require tedious special programming by the designer. In a perfect software development process, all such cases would have been theoretically analyzed and treated. However, in the real world, testing on the largest available datasets increases some confidence in the program's correctness.
Next, a large enough quantitative increase in execution speed leads to a qualitative increase in what we can do. Only if visibility can be computed efficiently, can it be used in a subroutine that is called many times, perhaps as as part of a search, to optimize the number of observers. This becomes more important when a more realistic function is being optimized, such as the total cost. E.g., for radio towers, there may be a tradeoff between a few tall and expensive towers, and many short and cheap ones. Alternatively, certain tower locations may be more expensive because of the need to build a road. We may even wish to add redundancy so that every possible target is visible from at least two observers. In all these cases, where a massive search of the solution space is required, success depends on each query being as fast as possible.
Finally, altho the size of available data is growing quickly, it is not necessarily true that available computing power is keeping pace. There is a military need to offload computations to small portable devices, such as a Personal Digital Assistant (PDA). A PDA's computation power is limited by its battery, since, approximately, for a given silicon technology, each elemental computation consumes a fixed amount of energy. Batteries are not getting better very quickly; increasing the processor's cycle speed just runs down the battery faster.
There is also a compounding effect between efficient time and efficient space. Smaller data structures fit into cache better, and so page less, which reduces time. The point of all this is that efficient software is at least as important now as ever.
Our computations are performed on a full resolution cell. That is, on a 1201x1201 cell, the visibility indices of all 1.4M points are estimated. For each tentative observer, as accurate a viewshed as possible is computed. This point is important since some other systems appear to compute low-resolution approximations to viewsheds.
Our system scales up well. Elevations cells of resolution over 2000x2000 are no problem.
# Viewshed examples
The sample cell used here is the USGS level-1 Lake Champlain West DEM. It has elevations ranging from the highest point in New York State, down to a large lake only 200 ft above sea level. Here is the cell (the image is scaled down by half to fit the display). North is to the right. The shading was done with Povray.
Lake Champlain West DEM
This test uses a radius of interest of 100 postings, and an observer and target height of 30 meters.
The time on a dual Xeon to site the observers was about 2 minutes. Actually, I do most computations on my laptop.
These images are a montage of some of the viewsheds of tentative observers. Observe the level of detail. Compare this to the viewsheds as determined by other systems.
Viewshed montage
Since the montage images may be scaled down, here is one viewshed at double full resolution, to show the detail.
Viewshed at full resolution
# Successively Adding Observers, with Intervisibility
The following is a montage of the increasing visible area in the cell as more and more observers are sited. To save space, we generally show a new image only after 3 new observers.
We added a new constraint: each new observer must be visible to at least one observer that has already been added. Therefore all the observers can communicate with line-of-sight radio.
The images also have black dots for the observer positions, perimeter circles at a distance of the ROI from each included observer, and heavy black lines joining each pair of visible observers.
Following that is one cumulative viewshed at its original resolution.
Finally, here is a video showing the observers being added. The black areas are not visible from any observer. As more observers are added, the remaining hidden patches tend to be small and well separated.
# Efficiency
The system is very fast (a few minutes for a level 1 DEM). Indeed generating the images took several times longer than siting the observers.
The system takes advantage of multiple processors. On a dual processor Xeon system with hyperthreading, 4-way parallelism is used for most of the computations (altho a hyperthreaded processor runs only about 30% faster than a non-hyperthreaded one.)
Efficiency is more important than some people realize. A sufficient quantitative improvement in speed leads to a qualitative change in what can be done. More experiments become possible. The efficient system can become a base on which to build new, higher level, tools.
# Application to terrain compression
Evaluating our lossy terrain compression algorithms is one of our major siting applications. Assume that original terrain, {${\cal T}_0$} is lossily compressed then reconstructed to give terrain {${\cal T}_1$}.
The obvious compression criterion is to measure the RMS elevation error in {${\cal T}_1$} compared to {${\cal T}_0$}. However that is naive.
We used siting, by siting observers, on both {${\cal T}_0$} and {${\cal T}_1$}. However, comparing the two sets of locations is wrong. The problem is that two quite different sets of locations might be equally good; they might have the same joint viewshed.
It would also be wrong to evaluate, on {${\cal T}_1$}, sitings and paths computed on {${\cal T}_1$}. A malicious compression program might output a completely wrong terrain, so that anything computed on it is meaningless. But, how would we know that such a result is meaningless? We did the following to evaluate the terrain compression.
1. On {${\cal T}_0$}, solve the multiple observer siting problem, with solution set {${\cal S}_0$} and cost {$c_0$}.
2. On {${\cal T}_1$}, solve the multiple observer siting problem, with solution set {${\cal S}_1$} and cost {$c_1$}.
3. Transfer {${\cal S}_1$} to {${\cal T}_0$} and compute its cost, {$c_{1\rightarrow0}$}.
4. Compare {$c_{1\rightarrow0}$} to {$c_0$}; if the former is not much larger than the latter, then {${\cal T}_1$} is a good approximation to {${\cal T}_0$}.
5. Paths can also be used, instead of observer sites, to evaluate the terrain, as the following figure shows.
Smugglers’ Path Planning on 16x Compressed “Scooped” Terrain Representation
# Surprising results
1. Even for single observer siting, there can be little correlation between observer elevation and visibility index.
2. Viewsheds can have remarkably large error bars. In one experiment, the visibility of one half of all the pixels in a cell varied depending on how terrain elevation was interpolated between two adjacent posts. That is a necessary operation because lines of sight usually run between posts.
# Open research questions
1. Precision of the results. Study this in more detail. The big input is the limited information about the terrain. That is one motivation for my long-term project on Mathematics of Terrain. Perhaps statistical techniques like resampling might be of some use.
2. Application of limited precision to making the algorithm faster. Since viewsheds and visibility indexes are imprecise, there may be faster algorithms whose outputs are just as good. (The obsolete engineering analogy is to hand-calculate using only the necessary number of significant digits.)
3. Dynamic input. Smugglers and border guards is a two ply game. A possible third ply by the guards would be to add observers to cut off long smugglers' paths rather than merely to maximize joint viewshed. In turbulent regions of the world, a possible fourth ply, by the smugglers, would be to attempt to delete some guards.
# Publications and talks
The online version of this page contains links to most of the papers, and to any posters and talk slides.
1. bibtexsummary:[/wrf.bib,magalhaes-ijcisim-2011]
2. bibtexsummary:[/wrf.bib,magalhaes-his-2010]
4. bibtexsummary:[/wrf.bib,tracy-acmgis-2008]
5. bibtexsummary:[/wrf.bib,wrf-fwcg-2008]
6. bibtexsummary:[/wrf.bib,tracy-fwcg-2008]
7. bibtexsummary:[/wrf.bib,dt-wrf-spie-2007]
8. bibtexsummary:[/wrf.bib,acmgis07]
9. bibtexsummary:[/wrf.bib,unigis-2006]
This summarizes RPI's role in the GeoStar project as of 2006.
10. bibtexsummary:[/wrf.bib,wrf-sdh2006]
11. bibtexsummary:[/wrf.bib,wrf-cv-siting-isprs]
12. bibtexsummary:[/wrf.bib,wrf-siting-apr2004]
13. bibtexsummary:[/wrf.bib,wrf-site]
14. bibtexsummary:[/wrf.bib,fr-hinbv-94]
15. bibtexsummary:[/wrf.bib,wrf-savannah]
# Software
A testbed consisting of a sequence of interacting programs has been produced. To learn about it, read here:
• sections 1 and 2 of the SDH2002 paper, linked above.
If those are not satisfactory, then read here:
The source code, makefiles and sample data are here: http://wrfranklin.org/wiki/Main/siting/site.tgz. My development environment is linux and c++.
The source code, plus executables and sample computed viewsheds and images are in this 126MB tarball: http://wrfranklin.org/wiki/Main/siting/site-big.tgz. | 3,204 | 14,179 | {"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} | 2.859375 | 3 | CC-MAIN-2022-49 | latest | en | 0.918106 |
https://bendwavy.org/klitzing/incmats/3tisdip-blend.htm | 1,624,449,358,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00321.warc.gz | 130,361,261 | 1,751 | Acronym ... Name three tisdip blend,"coord-planes squares star" atop cube Circumradius sqrt(5/6) = 0.912871 Confer "thah-squares star" || cube "coord-axes edge star" || cube
This polychoron classifies as wild, because each square of the base cube shares its edges with a lacing square and a triangle of an according tutrip each, and that tripple of incident faces would all be corealmic.
Incidence matrix according to Dynkin symbol
```"coord-planes squares star" || cube → sqrt(3)/2 = 0.866025
12 * ♦ 2 2 0 | 1 4 1 0 | 2 2 0
* 8 | 0 3 3 | 0 6 3 3 | 3 3 1
-----+----------+-----------+------
2 0 | 12 * * | 1 2 0 0 | 2 1 0
1 1 | * 24 * | 0 2 1 0 | 1 2 0
0 2 | * * 12 | 0 2 1 2 | 2 2 1
-----+----------+-----------+------
4 0 | 4 0 0 | 3 * * * | 2 0 0
2 2 | 1 2 1 | * 24 * * | 1 1 0
1 2 | 0 2 1 | * * 12 * | 0 2 0
0 4 | 0 0 4 | * * * 6 | 1 0 1
-----+----------+-----------+------
4 4 | 4 4 4 | 1 4 0 1 | 6 * * cube
4 4 | 2 8 4 | 0 4 4 0 | * 6 * tutrip
0 8 | 0 0 12 | 0 0 0 6 | * * 1 cube
``` | 518 | 1,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.64914 |
https://onlinetexttoolz.com/hexa-to-hex | 1,696,365,080,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00386.warc.gz | 477,704,585 | 13,056 | # Online HEXA To HEX Converter
HEXA to HEX is a useful tool for those who work with computers, as it allows for the conversion of numbers from one system to another. It is often used to convert data from one computer system to another, as computers use different systems to store information. HEXA to HEX is a process that involves the conversion of a number from one system to another, such as from HEXAdecimal to HEXAdecimal (HEX to HEX). It is a very useful tool for those who are working with computer systems, as it can help them to quickly and accurately convert data from one system to another.
Binary to HEXAdecimal conversion is a process of converting binary numbers into their equivalent HEXAdecimal numbers. HEXAdecimal numbers are used in computers and electronic devices to represent information in a more concise and readable format. Binary numbers are the representation of data in the form of a series of zeros and ones, while HEXAdecimal numbers are a base 16 number system that is easier to read and more efficient for computers to process.
The process of converting binary to HEXAdecimal is fairly straightforward. The first step is to break the binary number up into groups of four digits, also known as a "nibble". For example, if the binary number is 11011010, it can be broken up into 1101 and 1010. Each of these four-digit numbers is then converted to its corresponding HEXAdecimal value. The value of 1101 is D, and the value of 1010 is A. Therefore, the HEXAdecimal value of 11011010 is DA.
When converting from HEXAdecimal to binary, the process is reversed. The HEXAdecimal number is broken up into two-digit groups and each group is converted to its corresponding binary value. For example, the HEXAdecimal number 1F can be broken up into 1 and F, with 1 corresponding to 0001 and F corresponding to 1111. Therefore, the binary value of 1F is 00011111.
There are a variety of online tools available to assist with binary to HEXAdecimal conversion. These tools can easily convert one number system to the other and generally provide a visual representation of the conversion process. Additionally, many programming languages have built-in functions for binary to HEXAdecimal conversion which can be used for more complex conversions.
In summary, binary to HEXAdecimal conversion is a simple process that can be done manually or with the help of online tools or programming language functions. It is a useful process for representing information in a more efficient and readable format.
HEXAdecimal arithmetic is a type of numerical system used in computing and other applications where a base 16 number system is ideal. The HEXAdecimal system is based on the traditional binary system where each digit in a number represents a power of two. However, instead of the binary system having two digits, 0 and 1, the HEXAdecimal system has six digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. Each HEXAdecimal digit can represent a range of numbers from 0 to 15, which is an advantage over the binary system because it requires fewer digits to represent the same range of numbers.
In order to convert from one form of a number to the other, the two must first be expressed in their respective bases. This is done using the multiplication method. For example, if we wish to convert from HEXAdecimal to binary, we must first multiply each HEXAdecimal digit by the appropriate power of sixteen. This will give us the equivalent binary number. To convert from binary to HEXAdecimal, the same method is used but with the appropriate power of two.
Once the two numbers are expressed in the desired base, the conversion is relatively simple. For example, if we wish to convert the HEXAdecimal number FFFF to binary, we must first multiply each HEXAdecimal digit by its power of sixteen. We then add up the resulting binary numbers to get the equivalent binary number. In this case, FFFF would be equivalent to 1111111111111111 in binary.
In order to convert from binary to HEXAdecimal, the same method is used but with the appropriate power of two. For example, if we wish to convert the binary number 1111111111111111 to HEXAdecimal, we must first multiply each binary digit by its power of two. We then add up the resulting HEXAdecimal numbers to get the equivalent HEXAdecimal number. In this case, 1111111111111111 would be equivalent to FFFF in HEXAdecimal.
Online HEXA to HEX conversion is available on many websites. All you need to do is enter the two numbers in their respective bases and the website will automatically convert them for you. This makes it easy to quickly convert from one base to another without having to manually calculate the power of
HEXAdecimal representation of data is an important aspect of computer programming. HEXAdecimal is a base 16-number system, meaning it uses 16 distinct symbols to represent different values. The symbols 0-9 represent the values 0-9, while A-F represents the values 10-15. HEXAdecimal is often used to represent data because it can represent a large amount of data in a small amount of space. HEXAdecimal is also easier to read and write than regular decimal, which is why it is commonly used in computer programming.
Online HEXA to HEX is a tool that can be used to convert data between HEXAdecimal and decimal formats. This tool can be used for both encoding and decoding data, making it useful for a variety of applications. By encoding data in HEXAdecimal format, it is possible to reduce the amount of data that needs to be transferred over a network, which can be beneficial for applications that require large amounts of data to be transferred quickly.
Using Online HEXA to HEX, users can quickly and easily convert decimal values to their HEXAdecimal equivalents. The tool also allows users to convert HEXAdecimal values back into decimal values. This means that users can quickly and easily transfer data between two different formats, saving time and energy. Furthermore, the tool can be used to encode and decode data, allowing users to make sure that their data is secure and unreadable by unauthorized persons.
Online HEXA to HEX is a useful tool for anyone working with data or who needs to transfer data between two different formats. With this tool, users can quickly and easily convert data between decimal and HEXAdecimal formats, allowing them to store and transfer data more efficiently. Furthermore, the tool can be used to encode and decode data, ensuring that data remains secure and unreadable by unauthorized persons.
HEXAdecimal color codes are the industry standard for web development. They are used to specify the color of elements on a webpage and have become the standard in part because of their simplicity and wide availability. HEXAdecimal codes are composed of six characters, including the letters A-F and the numbers 0-9. These codes are the representation of a color in the HEXAdecimal system. When used in web design, they allow a designer to precisely specify the color of a particular element on the page.
HEXAdecimal codes are often written in shorthand notation. For example, the code #FF0000 is written as #F00 to save time. This shorthand form makes these codes easier to read and write and can be easily remembered. HEXAdecimal codes are also widely used in other areas such as printing and graphic design.
The HEXAdecimal system is also used to represent other kinds of data, such as character encoding and binary numbers. HEXAdecimal is a base 16 number system, which means that each character in the code is a multiple of 16. By using this system, it is possible to represent much larger numbers than would be possible using the decimal system.
HEXAdecimal codes can be difficult to read and interpret and can be prone to errors. To help make the codes easier to use, there are a number of online tools that convert HEXAdecimal codes to other formats. These tools allow users to quickly and easily convert HEXAdecimal codes into RGB or HSL values. For example, a HEXAdecimal code such as #FF0000 can be converted into RGB values of (255, 0, 0).
These online tools also allow users to quickly and easily convert HEXAdecimal codes into other formats, such as HTML or CSS. This makes it easier to use the HEXAdecimal codes in web development. HEXAdecimal codes can also be used to create color palettes. By entering a few codes, users can quickly and easily create a set of colors for use on a website or in a design project.
Overall, HEXAdecimal color codes provide an efficient and accurate way to represent color on the web. Online tools make them even easier to use, allowing users to quickly and easily convert HEXAdecimal codes into other formats and create color palettes. HEXAdecimal codes are an important part of a web.
HEXAdecimal file formats are a type of format used to store data in a computer. They are made up of 16 different symbols, 0 through 9 and A through F, which represent the 16 different values of the HEXAdecimal system. HEXAdecimal files are used to store and transfer data, such as images, music, videos, and programs. They are often used in computing because they are easier to read and understand than binary files.
HEXAdecimal files are often used for computer programming. When a programmer wants to write a program, they will break it down into instructions and store them in a HEXAdecimal file. The instructions are written in HEXAdecimal, so the computer can understand them. This makes programming easier and faster, as the programmer only needs to write the instructions in HEXAdecimal and the computer can understand them.
HEXAdecimal files are also used to store data. For example, a Photoshop file can be stored in a HEXAdecimal file. This makes it easier to transfer the file to another computer. HEXAdecimal files are also used to store music, videos, and other types of data.
HEXAdecimal files can be converted to other file formats. This is done using a HEXAdecimal to HEXAdecimal converter. The converter takes a HEXAdecimal file and converts it into a different file format, such as a JPEG or a PDF. This makes it easier to share HEXAdecimal files with others, as they can be converted to a more common file format.
HEXAdecimal files can also be converted to other HEXAdecimal files. This is done by using an online HEXA to HEX converter. This converter takes a HEXAdecimal file and converts it into another HEXAdecimal file. This is useful when transferring data between different computers, as the HEXAdecimal files can be converted to a format that is compatible with the other computer.
HEXAdecimal files are an important part of computing, as they are used to store and transfer data. They are also used in computer programming, as they are easier to read and understand than binary files. HEXAdecimal files can be converted to other file formats, such as JPEGs or PDFs, and they can also be converted to other HEXAdecimal files. Online HEXA to HEX
HEXAdecimal encoding and decoding algorithms are an essential set of tools when it comes to data transmission and storage. HEXAdecimal, or ‘HEX’ for short, is a numerical representation of data that uses sixteen unique symbols to represent binary code. It is used in a variety of applications, from web page coding to programming languages. One of the most common uses of HEXAdecimal is in online HEXA to HEX conversion.
Prior to the development of online HEXA to HEX conversion tools, encoding and decoding data using HEXAdecimal was a laborious process. It involved manually converting each character into a unique HEXAdecimal code, then writing the code down and repeating the process for each character. This was a time-consuming task and, in many cases, errors were made which resulted in incorrect conversions.
Online HEXA to HEX conversion tools have revolutionized the way HEXAdecimal encoding and decoding is done. These tools allow users to quickly and easily convert any string of characters or numbers into its corresponding HEXAdecimal code. This process is often referred to as ‘HEX-encoding’ and is used for a variety of purposes, including data storage, network communications, and more.
HEXA to HEX conversion tools are available in a variety of forms, from online tools to software applications and even dedicated hardware devices. Online tools are usually the most convenient and cost-effective option, as they don’t require any additional software or hardware and can be used without any prior knowledge of HEXAdecimal encoding. These tools are often used by web developers and other professionals who need to quickly and accurately convert data into a format that can be easily transmitted and read by other devices.
HEXA to HEX conversion tools are an essential part of any data storage or transmission system, as they allow data to be accurately converted from one form to another. With the help of these tools, data can be securely stored, transmitted, and read quickly and accurately, without the risk of errors or data corruption.
HEXAdecimal representation is a popular form of numeric representation that is used in many applications involving computers. It is also used in many other fields such as engineering and mathematics. HEXAdecimal representation is a base-16 system that uses sixteen distinct symbols, 0-9 and A-F, to represent numbers. This system is much more compact than a decimal system, which uses ten symbols and is much easier to use in computers due to its straightforwardness.
HEXAdecimal representation is used in a variety of different applications. In the field of computer science, HEXAdecimal representation is used to simplify the process of data input and output. HEXAdecimal is the primary way of inputting and outputting data in computers. HEXAdecimal is also used in many software applications such as HTML, CSS, and JavaScript as the primary form of numeric representation. HEXAdecimal is also used in programming languages such as C, C++, and Java.
In the field of engineering, HEXAdecimal is used to represent numbers in a more efficient and compact manner. HEXAdecimal is used in the design of electrical circuits, as well as in the design of mechanical systems. HEXAdecimal representation of numbers is also used in the design of chemical formulas, as well as in the calculation of distances between points in three-dimensional space.
HEXAdecimal representation is also used in mathematics. HEXAdecimal representation of numbers is used in the calculation of complex mathematical equations. HEXAdecimal is also used in the calculation of mathematical constants, such as the value of pi, and in the calculation of the exponential function.
HEXAdecimal compression techniques are becoming increasingly popular for increasing storage space and reducing file size in a variety of applications, including computer graphics, data storage, and audio and video streaming. HEXAdecimal compression is a process of encoding data into a HEXAdecimal format, which is a representation of binary data using only sixteen symbols, 0-9 and A-F. The process of HEXAdecimal compression takes the data or file and compresses it by eliminating redundancies and unused bytes, resulting in a much smaller file size.
HEXAdecimal compression is a great way to store large amounts of data or files in a much smaller space, as it can reduce file size by up to 60% or more. It also saves time when transmitting data, as the smaller file size means less time spent downloading or uploading the data. Furthermore, HEXAdecimal compression is also useful for encrypting and decrypting data, as it can make it much more difficult for unauthorized users to access the data.
Online HEXAdecimal compression is becoming increasingly popular as it offers a convenient way to compress or decompress files without the need to download or install additional software. Online HEXAdecimal compression tools are generally easy to use and can compress and decompress files quickly and efficiently. Additionally, some online HEXAdecimal compression tools offer additional features, such as the ability to password-protect files or encrypt data for added security.
Overall, HEXAdecimal compression techniques are becoming increasingly popular for a variety of applications. HEXAdecimal compression can reduce the file size by up to 60% or more, making it a great way to store large amounts of data in a much smaller space. Additionally, online HEXAdecimal compression tools offer a convenient way to quickly and easily compress or decompress files without the need to download or install additional software.
## HEXAdecimal in Networking and Security
HEXAdecimal is a number system used in networking and security that consists of 16 characters, 0-9 and A-F. It is used for many different purposes in these fields, such as representing data in a more compact form, making it easier to read and understand, and providing a secure way to store data. In networking, HEXAdecimal is used to represent IP addresses, MAC addresses, and subnet masks. In security, HEXAdecimal is used for encryption and hashing, as well as for representing data in a more secure form.
HEXAdecimal is also widely used in online applications. It is used to represent data stored in databases and in web pages and is often used to create unique strings of data for authentication purposes. HEXAdecimal is also used to represent binary data, as it is a more efficient way to represent binary data than ASCII or other encoding methods. HEXAdecimal is also used to represent color values in web pages and can be used to create unique IDs for objects.
One of the most common uses of HEXAdecimal in online applications is converting from HEXAdecimal to other number systems. This is often done with a HEXAdecimal to decimal converter, or a HEXAdecimal to binary converter.
HEXAdecimal to decimal converters are used to convert HEXAdecimal numbers into decimal numbers, and HEXAdecimal to binary converters are used to convert HEXAdecimal numbers into binary numbers. HEXAdecimal to HEXAdecimal converters are also available, which can be used to convert from one HEXAdecimal number to another.
HEXAdecimal is an extremely useful system for networking and security and is widely used in online applications. It is easy to use and understand and provides a secure way to store data. HEXAdecimal to HEXAdecimal converters can be used to quickly and easily convert from one HEXAdecimal number to another, making it an invaluable tool for online applications.
The use of HEXAdecimal in automotive systems has been one of the most important developments in the modern automotive industry. HEXAdecimal is an encoding scheme used to represent numbers and characters in binary form, which makes it easier to store and manipulate data. It is widely used in the automotive industry to represent engine control parameters, such as fuel mix, spark timing, and ignition timing.
HEXAdecimal is also used to represent common automotive components, such as fuel injectors, spark plugs, and air filters. It is also used to represent the physical size of components, such as wheel sizes, engine sizes, and transmission types.
HEXAdecimal is a vital component of automotive systems because it allows for a more efficient and reliable way to store and manipulate data. It is also more secure than other encoding schemes because it is more difficult for Digital Intruders to decode. Additionally, it is easier for automotive engineers to implement and understand, because it is based on a relatively simple set of rules.
In order to convert data from HEXAdecimal to other formats, such as binary, decimal, or ASCII, special software is required. This software is known as an online HEX converter, and it can be used to quickly and easily convert data from one format to another. HEXAdecimal is also used in the automotive industry to represent common commands, such as start/stop, reset, and various types of diagnostic data. Online HEX converters are also used to convert between different automotive systems, such as OBD-II and CAN bus.
Overall, the use of HEXAdecimal in automotive systems has enabled engineers to quickly and easily store and manipulate data. It has also allowed for a more secure way to store and manipulate data, which has enabled the automotive industry to advance at a rapid pace. By using online HEX converters, engineers can quickly and easily convert between different automotive systems, allowing for more efficient and reliable data management.
#### Conclusion
In conclusion, HEXA to HEX is a useful tool for converting numbers from HEXAdecimal to HEXAdecimal. It is easy to use and can be used for a variety of tasks. HEXA to HEX is a great way to quickly convert numbers from base 16 to base 16 without having to perform any calculations. It is a reliable and efficient tool and can be used for a variety of purposes.
### What is HEXA?
HEXA is a base-16 number system, also known as HEXAdecimal, used to represent numbers in computing systems. It is similar to the binary system used in computing but uses 16 symbols instead of 2. HEXA is represented by the symbols 0-9 and A-F.
### How do I convert HEXA to base
To convert a HEXA number to Base 10, you need to multiply each digit of the HEXA number with the corresponding power of 16 and sum the results. For example, if the HEXA number is "1A" then the base 10 equivalent is 26 since 1 x 16^1 + 10 x 16^0 = 26.
### How do I convert HEXA to HEX?
HEXA to HEX conversion can be done by simply writing down the HEXA number and removing any leading 0s (zeros). For example, if the HEXA number is "0F0A" then the HEX equivalent is "F0A".
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HEXA to HSLA
Convert your HEXA color format to HSLA format.
0 | 4,586 | 22,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.922381 |
https://nl.mathworks.com/matlabcentral/cody/problems/460-replace-may-with-april/solutions/82494 | 1,590,552,446,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392057.6/warc/CC-MAIN-20200527013445-20200527043445-00502.warc.gz | 467,656,977 | 15,631 | Cody
# Problem 460. Replace May with April
Solution 82494
Submitted on 26 Apr 2012 by Binbin Qi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 'The flowers may bloom in May'; y_correct = 'The flowers may bloom in April'; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
%% x = 'May I come to visit you in April?'; y_correct = 'April I come to visit you in April?'; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
%% x = 'May I come to visit you in April?'; y_correct = 'April I come to visit you in April?'; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
%% x = 'April is the cruelest month. Maybe not, though.'; y_correct = 'April is the cruelest month. Aprilbe not, though.'; assert(isequal(your_fcn_name(x),y_correct))
5 Pass
%% x = 'I read with dismay about your defeat last May.'; y_correct = 'I read with dismay about your defeat last April.'; assert(isequal(your_fcn_name(x),y_correct))
6 Pass
%% x = 'Moynihan for Mayor! Hurray for My Man Moynihan!'; y_correct = 'Moynihan for Aprilor! Hurray for My Man Moynihan!'; assert(isequal(your_fcn_name(x),y_correct)) | 360 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-24 | latest | en | 0.825377 |
https://freakonometrics.hypotheses.org/8657 | 1,709,330,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475701.61/warc/CC-MAIN-20240301193300-20240301223300-00233.warc.gz | 272,471,037 | 45,997 | # Non-observable vs. observable heterogeneity factor
This morning, in the ACT2040 class (on non-life insurance), we’ve discussed the difference between observable and non-observable heterogeneity in ratemaking (from an economic perspective). To illustrate that point (we will spend more time, later on, discussing observable and non-observable risk factors), we looked at the following simple example. Let $X$ denote the height of a person. Consider the following dataset
> Davis=read.table(
+ "http://socserv.socsci.mcmaster.ca/jfox/Books/Applied-Regression-2E/datasets/Davis.txt")
There is a small typo in the dataset, so let us make manual changes here
> Davis[12,c(2,3)]=Davis[12,c(3,2)]
Here, the variable of interest is the height of a given person,
> X=Davis$height If we look at the histogram, we have > hist(X,col="light green", border="white",proba=TRUE,xlab="",main="") Can we assume that we have a Gaussian distribution ? $X\sim\mathcal{N}(\theta,\sigma^2)$Maybe not… Here, if we fit a Gaussian distribution, plot it, and add a kernel based estimator, we get > (param <- fitdistr(X,"normal")$estimate)
> f1 <- function(x) dnorm(x,param[1],param[2])
> x=seq(100,210,by=.2)
> lines(x,f1(x),lty=2,col="red")
> lines(density(X))
If you look at that black line, you might think of a mixture, i.e. something like
$X\sim p_1\cdot\mathcal{N}(\theta_1,\sigma_1^2)+p_2\cdot\mathcal{N}(\theta_2,\sigma_2^2)$
(using standard mixture notations). Mixture are obtained when we have a non-observable heterogeneity factor: with probability $p_1$, we have a random variable $\mathcal{N}(\mu_1,\sigma_1^2)$ (call it type [1]), and with probability $p_2$, a random variable $\mathcal{N}(\mu_2,\sigma_2^2)$ (call it type [2]). So far, nothing new. And we can fit such a mixture distribution, using e.g.
> library(mixtools)
> mix <- normalmixEM(X)
number of iterations= 335
> (param12 <- c(mix$lambda[1],mix$mu,mix$sigma)) [1] 0.4002202 178.4997298 165.2703616 6.3561363 5.9460023 If we plot that mixture of two Gaussian distributions, we get > f2 <- function(x){ param12[1]*dnorm(x,param12[2],param12[4]) + (1-param12[1])*dnorm(x,param12[3],param12[5]) } > lines(x,f2(x),lwd=2, col="red") lines(density(X)) Not bad. Actually, we can try to maximize the likelihood with our own codes, > logdf <- function(x,parameter){ + p <- parameter[1] + m1 <- parameter[2] + s1 <- parameter[4] + m2 <- parameter[3] + s2 <- parameter[5] + return(log(p*dnorm(x,m1,s1)+(1-p)*dnorm(x,m2,s2))) + } > logL <- function(parameter) -sum(logdf(X,parameter)) > Amat <- matrix(c(1,-1,0,0,0,0, + 0,0,0,0,1,0,0,0,0,0,0,0,0,1), 4, 5) > bvec <- c(0,-1,0,0) > constrOptim(c(.5,160,180,10,10), logL, NULL, ui = Amat, ci = bvec)$par
[1] 0.5996263 165.2690084 178.4991624 5.9447675 6.3564746
Here, we include some constraints, to insurance that the probability belongs to the unit interval, and that the variance parameters remain positive. Note that we have something close to the previous output.
Let us try something a little bit more complex now. What if we assume that the underlying distributions have the same variance, namely
$X\sim p_1\cdot\mathcal{N}(\theta_1,\sigma^2)+p_2\cdot\mathcal{N}(\theta_2,\sigma^2)$
In that case, we have to use the previous code, and make small changes,
> logdf <- function(x,parameter){
+ p <- parameter[1]
+ m1 <- parameter[2]
+ s1 <- parameter[4]
+ m2 <- parameter[3]
+ s2 <- parameter[4]
+ return(log(p*dnorm(x,m1,s1)+(1-p)*dnorm(x,m2,s2)))
+ }
> logL <- function(parameter) -sum(logdf(X,parameter))
> Amat <- matrix(c(1,-1,0,0,0,0,0,0,0,0,0,1), 3, 4)
> bvec <- c(0,-1,0)
> (param12c= constrOptim(c(.5,160,180,10), logL, NULL, ui = Amat, ci = bvec)$par) [1] 0.6319105 165.6142824 179.0623954 6.1072614 This is what we can do if we cannot observe the heterogeneity factor. But wait… we actually have some information in the dataset. For instance, we have the sex of the person. Now, if we look at histograms of height per sex, and kernel based density estimator of the height, per sex, we have So, it looks like the height for male, and the height for female are different. Maybe we can use that variable, that was actually observed, to explain the heterogeneity in our sample. Formally, here, the idea is to consider a mixture, with an observable heterogeneity factor: the sex, $X\sim p_H\cdot\mathcal{N}(\theta_H,\sigma_H^2)+p_F\cdot\mathcal{N}(\theta_F,\sigma_F^2)$ We now have interpretation of what we used to call class [1] and [2] previously: male and female. And here, estimating parameters is quite simple, > (pM <- mean(sex=="M")) [1] 0.44 > (paramF <- fitdistr(X[sex=="F"],"normal")$estimate)
mean sd
164.714286 5.633808
> (paramM <- fitdistr(X[sex=="M"],"normal")\$estimate)
mean sd
178.011364 6.404001
And if we plot the density, we have
> f4 <- function(x) pM*dnorm(x,paramM[1],paramM[2])+(1-pM)*dnorm(x,paramF[1],paramF[2])
> lines(x,f4(x),lwd=3,col="blue")
What if, once again, we assume identical variance? Namely, the model becomes
$X\sim p_H\cdot\mathcal{N}(\theta_H,\sigma^2)+p_F\cdot\mathcal{N}(\theta_F,\sigma^2)$Then a natural idea to derive an estimator for the variance, based on previous computations, is to use
$\sigma^2=\frac{1}{n-2}\left(\sum_{i:H} [X_i-\overline{X}_H]^2+\sum_{i:F} [X_i-\overline{X}_F]^2\right)$
The code is here
> s=sqrt((sum((height[sex=="M"]-paramM[1])^2)+sum((height[sex=="F"]-paramF[1])^2))/(nrow(Davis)-2))
> s
[1] 6.015068
and again, it is possible to plot the associated density,
> f5 <- function(x) pM*dnorm(x,paramM[1],s)+(1-pM)*dnorm(x,paramF[1],s)
> lines(x,f5(x),lwd=3,col="blue")
Now, if we think a little about what we’ve just done, it is simply a linear regression on a factor, the sex of the person,
$X=\mu_H\cdot\boldsymbol{1}(H)+\mu_F\cdot\boldsymbol{1}(F)+\varepsilon$
where $\varepsilon\sim\mathcal{N}(0,\sigma^2)$. And indeed, if we run the code to estimate this linear model,
> summary(lm(height~sex,data=Davis))
Call:
lm(formula = height ~ sex, data = Davis)
Residuals:
Min 1Q Median 3Q Max
-16.7143 -3.7143 -0.0114 4.2857 18.9886
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 164.7143 0.5684 289.80 <2e-16 ***
sexM 13.2971 0.8569 15.52 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 6.015 on 198 degrees of freedom
Multiple R-squared: 0.5488, Adjusted R-squared: 0.5465
F-statistic: 240.8 on 1 and 198 DF, p-value: < 2.2e-16
we get the same estimators for the means and the variance as the ones obtained previously. So, as mentioned this morning in class, if you have a non-observable heterogeneity factor, we can use a mixture model to fit a distribution, but if you can get a proxy of that factor, that is observable, then you can run a regression. But most of the time, that observable variable is just a proxy of a non-observable one…
Cite this blog post
Arthur Charpentier (2013, September 11). Non-observable vs. observable heterogeneity factor. Freakonometrics. Retrieved March 1, 2024, from https://doi.org/10.58079/our8
## 5 thoughts on “Non-observable vs. observable heterogeneity factor”
1. DavidP says:
Working through your code from scratch (twice!), I find my mixture distribution is radically different from yours (and each of mine differs one from the other!).
For example, the first time I ran the code, lambda was about 0.15; the second time, about 0.6.
1. Mixture estimation is a difficult topic… and indeed, numerical computations are rather unstable.
2. Joel Mayer says:
I’m barely literate in Mathematica so I’m just barely able to follow your line of reasoning in R. Still, I have a question (suggestion). Wouldn’t it be better to handle biological distributions with the classical binomial? Seems to me when I see columns I see ‘sources of error’ and I have a better chance of assigning a gene to each ‘source of error’ than when I’m working with smooth approximators like the normal curve.
3. Al says:
Great explanation, very helpful. Thanks for posting
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http://www.ehow.com/video_4753972_what-difference-between-current-volts.html | 1,484,756,881,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00576-ip-10-171-10-70.ec2.internal.warc.gz | 430,034,207 | 15,891 | # What Is the Difference Between Current & Volts?
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Current represents the flow of electricity and depends on resistance, while voltage is the electrical pressure within an electrical circuit. Discover more about the role of resistance in electrical currents with information from a math and science teacher in this free video on electricity.
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## Video Transcript
Hi, I'm Steve Jones and I'm going to tell you the difference between current and voltage. It might surprise you that on the board I have a diagram of a hosepipe. We will see that in a minute. Voltage is a kind of electrical pressure. It's how hard the electricity is pushed around the circuit, and it's measured in Volts. Current, however, is the flow, and this depends on resistance. And this is an important term in all of electricity, so resistance is the resistance to the flow of current. Now, if we have a hosepipe, for example, now we can have a pressure applied to the hose (water pressure) but the amount of flow the current or the flow depends on how big the opening is. So if I increase the resistance then the flow is going to go down. So this is the important idea that you can get from a hosepipe that you can imagine in your mind, and then transfer this to the same idea in a piece of electrical wire. If the electrical wire has a high resistance, you need a big pressure to push electric current through it. So this is why there is a difference between these two terms. Voltage is about how hard you are pushing. Current is about how much actually goes through. So that is the basic difference between current and voltage in electrical circuits.
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• by
• Mar 02, 2014
• GRE Blog, GRE Tutor
There certainly is a lot to know when preparing for the GRE. However, it is imperative that you are studying highly tested facts versus ambiguous material that rarely shows up on the test. It’s easy to think that you simply need to focus on high school math concepts such as geometry, algebra, proportions, fractions, percents, decimals, and the order of operations (PEMDAS ), but it simply isn’t true. That list is not exhaustive and it is extremely difficult to revisit four years of high school math plus a few university courses in the limited time you have to prepare for the GRE. So, we made things easier! Despite the fact that everyone’s exam is distinct, there are commonly tested concepts that will help rack up valuable Test Day points. These are must know facts that are commonly tested on the GRE.
Here are some must-know math facts that will help you navigate the GRE.
• A factor of a number is any positive integer that can be multiplies by an integer to equal the number. For example, the factors of 12 are 1,2,3,4,6,12.
• A multiple of a number is the product of that number and any other whole number. For example, some multiples of 12 are 12, 24, 36, 48 …
• Remember that there are few factors and many multiples.
• A prime number is a positive integer greater than 1 that is divisible by 1 and itself.
For example: 2, 3, 5, 7, 11 ….
• The number one is NOT prime.
• The number two is the only even prime number.
• The number two is the smallest prime number.
• Zero is not prime.
• Negative numbers are not prime.
• Every positive, nonprime number greater than 1 can be expresses as the product of a series of prime numbers.
Common GRE percent conversions
You are permitted to use a calculator on the GRE. However, knowing common percents, decimal and fractions conversions can save a lot of time. It is very helpful on Test Day to be able to convert fractions to decimals and percents and vice versa. Here is a chart with commonly used percents:
5% .05 1/20
10% .1 1/10
12.5% .125 1/8
16 2/3% .167 1/6
20% .2 1/5
25% .25 ¼
30% .3 3/10
33 1/3% .33 1/3
50% .5 1/2
GRE Ratios
The GRE can describe ratios in various forms. For example, the ratio of X to Y can be written as X/Y or X:Y. If you are given a ratio and an actual number of items that corresponds to one of the components of the ratio, you can determine the number of items represented by each of the other components of the ratio. Consider the problem below:
Column A Column B
The ratio of boys to girls in a classroom
is 2:5 and there are 35 girls in the class.
The # of boys in the class 14
First translate boys and girls into a fraction and plug in the numbers. So boys/girls = 2/5. Then plug in the variable over the total amount of girls and set the two fractions equal to each other. So: 2/5 = B/35. Then cross multiply to get (5B) = 70, so B = 14. These columns are equal so the correct answer is (C).
Join us next time for GRE triangle basics.
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https://www.stata.com/statalist/archive/2012-02/msg00859.html | 1,590,994,461,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00176.warc.gz | 915,962,945 | 4,735 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: weighted time dependent Cox model
From Steve Samuels To statalist@hsphsun2.harvard.edu Subject Re: st: weighted time dependent Cox model Date Sat, 18 Feb 2012 22:54:51 -0500
```In Stata, -enter- refers to the earliest entry time for a subject, and -exit- to the last observed time. (This is in the Manual entry for -stset-). Use this -stset- statement:
****************************************************
stset exit [pweight=weight], fail(event) id(id)
****************************************************'
Steve
sjsamuels@gmail.com
On Feb 18, 2012, at 6:07 PM, Ehsan Karim wrote:
Dear Stata list,
I am trying to reproduce the weighted time dependent Cox model
(Andersen–Gill format with IPTW) results in Stata that are originally
obtained from R using same dataset, but so far getting the estimates
different. Could anyone indicate what I could be done to fix this?
Any suggestions/references will be highly appreciated.
Thanks,
Ehsan
##########################################
# R results: coef -0.288 se 0.174
##########################################
> msmc = coxph(Surv(enter, exit, event) ~ tx + cluster(id), robust = TRUE, data = dataset, weights = weight)
> summary(msmc)
n= 14372, number of events= 131
coef exp(coef) se(coef) robust se z Pr(>|z|)
tx -0.2882 0.7496 0.1745 0.1964 -1.467 0.142
exp(coef) exp(-coef) lower .95 upper .95
tx 0.7496 1.334 0.5101 1.102
Concordance= 0.534 (se = 0.021 )
Rsquare= 0 (max possible= 0.136 )
Likelihood ratio test= 2.8 on 1 df, p=0.09455
Wald test = 2.15 on 1 df, p=0.1423
Score (logrank) test = 2.75 on 1 df, p=0.09739, Robust = 2.11 p=0.1459
##########################################
# Stata results: coef -.605 se .643
##########################################
. use http://stat.ubc.ca/~e.karim/dataset, clear
(6 vars, 14372 obs)
. stset exit [pweight=weight], fail(event) exit(exit) id(id) enter(enter)
id: id
failure event: event != 0 & event < .
obs. time interval: (exit[_n-1], exit]
enter on or after: time enter
exit on or before: time exit
weight: [pweight=weight]
------------------------------------------------------------------------------
14372 total obs.
12872 obs. begin on or after exit
------------------------------------------------------------------------------
1500 obs. remaining, representing
1500 subjects
17 failures in single failure-per-subject data
1500 total analysis time at risk, at risk from t = 0
earliest observed entry t = 0
last observed exit t = 1
. stcox tx, nohr robust nolog
failure _d: event
analysis time _t: exit
enter on or after: time enter
exit on or before: time exit
id: id
weight: [pweight=weight]
(sum of wgt is 1.5062e+03)
Cox regression -- Breslow method for ties
No. of subjects = 1506.227846 Number of obs = 1500
No. of failures = 17.16531456
Time at risk = 1506.227846
Wald chi2(1) = 0.89
Log pseudolikelihood = -124.48415 Prob > chi2 = 0.3464
(Std. Err. adjusted for 1500 clusters in id)
------------------------------------------------------------------------------
| Robust
_t | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
tx | -.6053404 .6428669 -0.94 0.346 -1.865336 .6546555
------------------------------------------------------------------------------
##########################################
# dataset (partial)
##########################################
. list
+-------------------------------------------------------------------+
| id tx event enter exit weight _st _d _t _t0 |
|-------------------------------------------------------------------|
1. | 1 0 0 0 1 1.356058 1 0 1 0 |
2. | 1 0 0 1 2 1.356058 0 . . . |
3. | 1 0 0 2 3 1.356058 0 . . . |
4. | 1 1 0 3 4 1.356058 0 . . . |
5. | 1 0 0 4 5 1.356058 0 . . . |
6. | 1 0 0 5 6 1.356058 0 . . . |
7. | 1 0 0 6 7 1.356058 0 . . . |
8. | 1 0 0 7 8 1.356058 0 . . . |
9. | 1 0 0 8 9 1.356058 0 . . . |
10. | 1 1 0 9 10 1.356058 0 . . . |
|-------------------------------------------------------------------|
11. | 2 0 0 0 1 1.57168 1 0 1 0 |
12. | 2 0 0 1 2 1.57168 0 . . . |
13. | 2 1 0 2 3 1.57168 0 . . . |
14. | 2 0 0 3 4 1.57168 0 . . . |
15. | 2 1 0 4 5 1.57168 0 . . . |
16. | 2 1 0 5 6 1.57168 0 . . . |
17. | 2 1 0 6 7 1.57168 0 . . . |
18. | 2 0 0 7 8 1.57168 0 . . . |
19. | 2 0 0 8 9 1.57168 0 . . . |
20. | 2 0 0 9 10 1.57168 0 . . . |
|-------------------------------------------------------------------|
21. | 3 0 0 0 1 .983771 1 0 1 0 |
22. | 3 0 0 1 2 .983771 0 . . . |
23. | 3 0 0 2 3 .983771 0 . . . |
24. | 3 0 0 3 4 .983771 0 . . . |
25. | 3 1 0 4 5 .983771 0 . . . |
26. | 3 0 0 5 6 .983771 0 . . . |
27. | 3 0 0 6 7 .983771 0 . . . |
28. | 3 0 0 7 8 .983771 0 . . . |
29. | 3 1 0 8 9 .983771 0 . . . |
30. | 3 0 0 9 10 .983771 0 . . . |
--more--
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* http://www.ats.ucla.edu/stat/stata/
``` | 2,192 | 6,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-24 | latest | en | 0.655444 |
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### Author Topic: top and bottom spacings on frames (Read 2114 times)
#### cbinstrasburg
• New Bee
• Posts: 22
##### top and bottom spacings on frames
« on: February 14, 2012, 10:55:11 PM »
I am new to bee keeping and have been introduced to this by a dear friend. I have ordered several boxes of bees to be delivered in April. Being mostly retired I have lots of time and tools on my hands I am considering making my bee equipment. I have studied several drawings and understand them. The one thing I lack understanding on is the frames. I understand the width and side spacings but the top and bottom spacings are not clear. In calculating from the box and frame drawings there is space on top and on bottom...is this normal. Shouldn't the frames be flush with the top or bottom. Can someone or more someones :) direct me on what the norm or what should be done. I realize what ever way it is done has to be done on all to keep the bee space in the correct size.
TIA
Carl
#### The Bix
• House Bee
• Posts: 427
• Gender:
##### Re: top and bottom spacings on frames
« Reply #1 on: February 15, 2012, 12:11:32 AM »
Carl,
I just measured one of my medium 10-frame Langstroth boxes and a corresponding frame. The box and the frame came from Western Bee Supply, who I believe is the supplier for all of Dadant's woodenware. Anyway, in this particular example, there is 1/8" of space from the top of the top bar to the top of the box and there is also 1/8" of space from the bottom of the bottom bar to the bottom of the box. Obviously, this leaves 1/4" between layers. I know that, "A sample of one, doth not a conclusion make", but I'm willing to bet that this is pretty standard.
-John
« Last Edit: February 15, 2012, 12:26:43 AM by The Bix »
#### Michael Bush
• Universal Bee
• Posts: 16121
• Gender:
##### Re: top and bottom spacings on frames
« Reply #2 on: February 15, 2012, 12:47:37 AM »
There should be 1/4" at the top (a 5/8" frame rest rabbet) and 1/8" at the bottom.
My website: bushfarms.com/bees.htm en espanol: bushfarms.com/es_bees.htm auf deutsche: bushfarms.com/de_bees.htm
My book: ThePracticalBeekeeper.com
-------------------
"Everything works if you let it."--James "Big Boy" Medlin
#### cbinstrasburg
• New Bee
• Posts: 22
##### Re: top and bottom spacings on frames
« Reply #3 on: February 15, 2012, 01:21:01 AM »
ok...so the thickness of the top bar would have to be 3/8" to have 1/4" on top with a 5/8" rabbit. Is this common because the drawings I have show the top bars to be 1/2" to 7/16". If this is not right where can I find a drawing that shows top bars 3/8" thick.
Carl
#### Michael Bush
• Universal Bee
• Posts: 16121
• Gender:
##### Re: top and bottom spacings on frames
« Reply #4 on: February 15, 2012, 04:38:10 AM »
The ears on most top bars are tapered and where they rest they are usually 3/8".
My website: bushfarms.com/bees.htm en espanol: bushfarms.com/es_bees.htm auf deutsche: bushfarms.com/de_bees.htm
My book: ThePracticalBeekeeper.com
-------------------
"Everything works if you let it."--James "Big Boy" Medlin
#### cbinstrasburg
• New Bee
• Posts: 22
##### Re: top and bottom spacings on frames
« Reply #5 on: February 15, 2012, 12:48:29 PM »
Michael and Bix thanks for the information.
Carl
#### cbinstrasburg
• New Bee
• Posts: 22
##### Re: top and bottom spacings on frames
« Reply #6 on: February 17, 2012, 10:45:39 AM »
Another challenging question :) In the aspect of keeping 3/8" bee space. All measurements of Langstroth hives come from the plan from this forum and most of the others I found.
long side of box
overall width of frame @ side bars is 17 3/4"...box ID dim. is 18 3/8" difference of 5/8"...5/16" each side.
narrow side of box
10 frames x 1 3/8" is 13 3/4"...box ID dim. is 14 3/4" difference of 1"...1/2" each side.
top frames
side bars 1 3/8"...top bar 1 1/16" difference of 5/16"
None of these differences are 3/8"
#1 Why the difference from one side to the other on the frames? :?
#2 Do the bees move through the top bars?
Don't understand the 3/8" bee space concept here.
TIA
Carl
« Last Edit: February 17, 2012, 11:12:01 PM by cbinstrasburg »
#### Michael Bush
• Universal Bee
• Posts: 16121
• Gender:
##### Re: top and bottom spacings on frames
« Reply #7 on: February 17, 2012, 11:53:20 PM »
A bee space is between 1/4" and 3/8". The mean of those is 5/16".
3/8" is Maximum bee space.
1/4" is Minimum bee space.
5/16" is in the middle between maximum and minimum beespace.
My website: bushfarms.com/bees.htm en espanol: bushfarms.com/es_bees.htm auf deutsche: bushfarms.com/de_bees.htm
My book: ThePracticalBeekeeper.com
-------------------
"Everything works if you let it."--James "Big Boy" Medlin
#### cbinstrasburg
• New Bee
• Posts: 22
##### Re: top and bottom spacings on frames
« Reply #8 on: February 18, 2012, 12:07:57 AM »
Thanks Michael...but why is there a one inch difference in the box ID and the total width of ten frames on the Langstroth on the drawings...that is my question.
#### Michael Bush
• Universal Bee
• Posts: 16121
• Gender:
##### Re: top and bottom spacings on frames
« Reply #9 on: February 18, 2012, 02:35:17 AM »
You have to have enough space to move a frame over to get them out. There is always extra space on the outside edges. If there wasn't you'd have a lot of difficulty removing the first frame. Frames should be pushed together tightly in the center.
My website: bushfarms.com/bees.htm en espanol: bushfarms.com/es_bees.htm auf deutsche: bushfarms.com/de_bees.htm
My book: ThePracticalBeekeeper.com
-------------------
"Everything works if you let it."--James "Big Boy" Medlin
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[–] 21 points22 points (2 children)
None of the stuff you took, except for complex analysis (and linear algebra, because is basic) will be of much use and neither will Hilbert spaces. You need to study differential geometry and (complex) algebraic geometry. You probably going to need some abstract algebra and maybe some topology too. So, basically, the first year or so of graduate school in math.
[–]Functional Analysis 1 point2 points (1 child)
We get introductory courses in complex algebraic geometry and differential geometry as early as 2nd year undergrad here.
[–]Combinatorics 0 points1 point (0 children)
If you don't mind me asking, what text did you use and what were your school's prerequisites for taking differential geometry? How far did you get in that course (past Gauss-Bonnet thm?)
I just finished working through the first 4 chapters of Do Carmo (skipping some stuff) as a junior math major and well, I felt pretty underprepared. While both my linear algebra and vector calc were pretty weak going in, I'm wondering how a uni manages to get people doing differential geometry their second year.
[–] 9 points10 points (4 children)
I won't pretend to know a whole lot about the subject but from what little I do know you'd need to learn:
1) Smooth Manifolds - No idea what book. Everyone seems to use Guilleman-Pollack, but I hated it. You might like it.
2) Complex Analytic Manifolds - No idea what book.
3) Complex Algebraic Geometry (this is because CY-manifolds seem to be analogues of K3 surfaces, which I believe are all algebraic. This might be horribly wrong. In any case I hear a lot of the algebraic geometry people at my uni. talking about these bastards) - The only book out there that I know of is Griffiths-Harris. It's HARD, and everyone seems to use it.
[–]Differential Geometry 5 points6 points (0 children)
For (1): The book Smooth manifolds by John Lee. For (2) and (3) Huybrechts is an alternative to Griffiths-Harris (though it has some pretty bad errors).
Moroianu has a very nice set of notes online called "Lectures on Kahler Geometry" (Calabi-Yau manifolds are special types of Kahler manifolds). This introduces all the differential geometry and complex geometry needed and has a chapter on Calabi-Yau manifolds. It might be a little terse for someone who hasn't seen this stuff before but it should be useful and also serve as a nice guide. It also has a lot of nice exercises.
[–] 0 points1 point (2 children)
I have heard of Griffiths-Harris, but never really looked into it. After reading the table of contents, it seems like you really weren't kidding about the HARD part. I've used books that others complain about before and thought they weren't as bad as their reputation. I have a feeling I might agree with you on Griffiths-Harris.
I thought it was interesting that they do roughly 700 pages of algebraic geometry without introducing commutative algebra. At which point, they move from the basic intro to commutative algebra to coherent sheaves in under 30 pages. Though, they use sheaves before that, so maybe by that point rings would be a complete triviality.
[–] 1 point2 points (1 child)
[–] 0 points1 point (0 children)
Those are my favorite books to read.
[–] 5 points6 points (2 children)
Griffiths-Harris Principles of Algebraic Geometry is a pretty standard reference for people starting to learn complex algebraic geometry. I'd say Chapter 0 of that textbook is about the bare minimum needed to really understand the definition of a CY manifold, and then there are many directions you can go in once you understand the definition. So look at Chapter 0 and see how accessible it is for you right now, and try to work backwards from it if it is above your level right now.
You will probably find that you still have quite awhile to go. 1-2 courses in each of manifold topology, algebra, algebraic topology, and differential geometry will get you there.
A more physics-oriented reference would be Geometry, Topology, and Physics by Nakahara. You will be able to 'do physics' much faster this way, but you won't know anything about how to do things rigorously. In any case, it will give you a decent bird's-eye view of what you need to learn in the next 2 years or so before you can start to study CY manifolds.
[–][S] 0 points1 point (0 children)
thanks! checking it out now. i think i can grasp this...
[–] 0 points1 point (0 children)
Nakahara looks awesome. Glad you mentioned it.
[–]Mathematical Physics 2 points3 points (2 children)
You should really get to grips with manifolds first.
There's a string theory textbook by Becker, Becker and Schwarz which has a reasonably basic and very unrigorous introduction to stringy geometry. It's very terse but if you can do all the exercises then you might find it useful. (A little bird told me you can find a pdf online.)
But as someone doing research in string theory, my advice is not to bother with CY manifolds at your stage. They play a very small role (they permit compactifications which break most but not all of the supersymmetry) and require a disproportionate amount of effort for the payoff you'd get.
If you want to get your teeth into some nice maths of string theory, try to understand the identification of string worldsheets with Riemann surfaces (complex curves) and how that arises. Have you checked out Zwiebach's book?
[–][S] 0 points1 point (1 child)
can i jump right into zweibach's text without differential geometry, abstract algebra, etc. ?
[–]Mathematical Physics 0 points1 point (0 children)
Zwiebach is aimed at advanced Physics undergrads so yes, but also for that reason it gives only a basic understanding of bosonic string theory, not even superstrings, but it's the best jumping-off-point. Also read David Tong's notes which are a little more advanced. Polchinski is probably the gold standard, but I don't think you'd get much from it without a working knowledge of QFT (I didn't).
[–] 1 point2 points (0 children)
Dennis Westra has a short (50-page) introduction to Calabi-Yau manifolds here: http://www.mat.univie.ac.at/~westra/calabiyau3.pdf | 1,483 | 6,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-26 | latest | en | 0.962173 |
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Roulette is thrilling game where the spin of wheel can yield up to thirty-five times your original profit. The game of roulette. Is probably the ultimate casino symbol. No one thinks of casino without visualizing the easy spin of the roulette wheel and. Resting of the small ball. Essentially roulette is game of high thrills but may not pay as much as people expect it to.In the game of roulette unlike several casino card games it is easy to calculate odds. Whether it is American.
Roulette or European roulette the odds are calculable through simple formula. However the catch is to transform those odds into actual wins.Calculating Roulette OddsUnlike multiple card-deck games roulette has thirty-eight numerical slots. This makes it easier to predict the odds of any particular number winning. Of course odds are only mathematical possibility and not strategy. In conventional American roulette the housed edge is about 5.26%. This edge is applicable for every bet except five number bets.The house edge defines the odds that favor the house or are against the player. This means that if you place \$1 bet the chances of hitting the number you bet on is one.
In thirty-eight. If you do hit the number the casino only pays out \$35. However the point about understanding the odds is to realize that for every thirty-eight spins you will lose \$2. This \$2 is the casino charge. If you work on the math you will calculate that \$2 is about 5.26% of \$38. Calculating odds is as simple as that. Mathematically the formula is: Expectancy probability or winning winning amount probability of losing losing amount. The expectancy is likely to come to negative value. It is negative since the formula calculates the expectancy. From the player’s perspective. This negative value denotes the fact that you.
Are negatively likely to win or the house has the expectancy edge.The exception to these odds calculation is the five numbers bet. This bet is worse than the other is when it comes to winning odds since it has much higher house advantage. The house edge on five-number bet is calculated to be about 7.89%.European RouletteThe.
European version of roulette has slightly. Better odds since it lacks the extra space in the double zero slot. This way the European wheel has only 37 slots. Although the game pays out the same value the 37 slots make the game more suitable for players from the probability view.European roulette is far more lucrative than American roulette especially since in American roulette there is no way to alter the odds of any game. No strategy is able to guarantee better odds. In fact experts advise players to seek out easier variations of the game where the odds might be better. This is probably.
The only way you can feel reassured that you have better chance of winning.Roulette BetsThe best kinds of. Bets in roulette are those that are have similar payouts as odds. In most versions or bets in roulette there is wide gap in the payout offered and the probably odds. The best bet is one where this gap is less wide. An example of likely bets is even bets on low high even or odd numbers. Even placing even bets on red and black would give you payout of to 1.As far as roulette systems go they are reasonably useless. All roulette systems that claim to be based on mathematical data are in fact just using odds to show attractive strategies. Odds calculation in roulette is simple and largely theoretical. No odds are guarantee since they only reflect chance. The roulette wheel is unaware of its own odds.
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# help plzzzzz
+1
312
4
How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among 4 people if each person can receive any number of bracelets, including 0 ? Each bracelet must be given to someone.
Dec 28, 2018
#1
+4569
+1
See here: https://web2.0calc.com/questions/hlelp-plzzz
Dec 28, 2018
#2
+23593
0
Dec 28, 2018
#3
+23593
0
R Y G B
1 1 1 1 Person 1 gets all 4 R Y G B
1 1 1 2 Person 1 gets R Y G and person 2 gets B (& Peeps 3 &4 get 0: No Bracelet)
.
.
.
.
.
4 4 4 4 where the numbers represent the person who gets the R Y G B bracelet....
So there is 4 * 4 * 4 * 4 =256 combos
( if a person is not reperesented by a number, it is zero: No Bracelet)
Dec 28, 2018
edited by Guest Dec 28, 2018
edited by Guest Dec 28, 2018
#4
+4569
+2
This is also distinguishability with duplicates, so \(4^4=\boxed{256}.\)
.
Dec 28, 2018 | 364 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-24 | latest | en | 0.666448 |
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Category
Enter the grades and the weighted semester grades in the tool and it will calculate the semester grades.
📕 First quarter
%
%
📗 Second quarter
%
%
🖋️ Final exam
%
%
The final semester grade calculator calculates the overall semester grade, which helps you simplify the grading process. The grade semester calculator offers convenience and efficiency to both students and instructors. It gives 100% accurate calculations in less than a minute.
## How to Calculate Semester Grades?
You can calculate semester grades with the assessment of the student's overall performance in the semester. The grade semester average calculator evaluates the percentage of marks for the first quarter, second quarter and final exam.
The percentage of each semester has a specific weightage in the final examination.
You can calculate semester grades as follows:
• Determine Weightings
• Calculate Assessment Scores
• Sum the Weighted Scores
## Formula for Weighted Semester Grade:
Our calculator uses the following formula to calculate weighted semester grades with accuracy:
Each of the grading systems can be different from other grading system. The semester grade calculator high school evaluates marks on the basis of the weight of the grade in each semester.
## Example:
Let's suppose there is a first quarter worth 30% (0.30) with a score of 85 and a second quarter worth 20% (0.20) with an average score of 92. The final exam worth is 50% (0.50) with a score of 78. Then how to calculate your semester grade?
Sol:
Weighted Semester Grade = (0.30 * 85) + (0.20 * 92) + (0.50 * 78)**
Weighted Semester Grade = 25.5 + 18.4 + 39**
Weighted Semester Grade = 82.9 %
So, the weighted semester grade is 82.9 %.
The semester grading pie chart is represented as
The semester grade calculator is a valuable online tool for students and educators to calculate final semester grades precisely. The main reason beyond that there can be different grading systems and weightage of marks.
The grading systems vary somewhat depending on the institution. The semester grading chart used in schools and universities is as follows:
Letter Grade Grade Points Percentage A+ 4.0 97-100 A 4.0 93-96 A- 3.7 90-92 B+ 3.3 87-89 B 3.0 83-86 B- 2.7 80-82 C+ 2.3 77-79 C 2.0 73-76 C- 1.7 70-72 D+ 1.3 67-69 D 1.0 63-66 D- 0.7 60-62 F 0.0 Below 60
There can be a difference between the weightage of each quarter. The 2 quarter exam calculator allows the user to adjust the weightage of each semester.
## How Does Semester Grade Calculator Work?
The semester exam calculator takes into account each quarter marks the final exam marks separately.
Let's see how!
Input:
• Enter the grades of the first quarter, second quarter and final exam
• Enter the relative weightage of grades
Output:
• Pie Chart of Weightage
## FAQs:
### What is the Main Difference Between Cumulative GPA and Semester GPA?
The cumulative GPA is a calculation of all the courses in school or a college. The semester GPA is the average course grade for the current semester.
### How to Calculate the Final Grade in a Points-Based System?
• Determine the point values
• Record all the points earned
• Add points and possible points
• Divide the total points and possible points
Calculate semester grades on the basis of the points and then convert them into the percentage with the semester test calculator.
### What Is a 3.0 GPA?
A 3.0 unweighted GPA means a student has earned a B average across the course studied during a specific class.
### What Is the Highest GPA In the US?
A 4.0 GPA is equivalent to acquiring an A in all of your classes. This is the highest possible score. | 861 | 3,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-38 | latest | en | 0.886442 |
https://drewherringracing.com/qa/how-do-you-convert-mm-to-cm.html | 1,606,779,623,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00291.warc.gz | 265,334,717 | 6,891 | # How Do You Convert Mm To Cm?
## What is 650mm in CM?
65 cm650 mm is 65 cm..
## How do you convert mm to cm examples?
To convert millimeters to centimeters, multiply the millimeter value by 0.1 or divide by 10. For example, to convert 5 mm to cm, multiply 5 by 0.1, that makes 0.5 cm is 5 mm.
## What is 230mm in CM?
230 mm to cm conversion. A millimeter, or millimetre, is a unit of length equal to one thousandth of a meter. A centimeter, or centimetre, is a unit of length equal to one hundredth of a meter….Convert 230 Millimeters to Centimeters.mmcm230.0023230.0123.001230.0223.002230.0323.00396 more rows
## Is 1m 100cm?
Convert meter to cm, centimeters to meter (1m = 100cm)
## What is 1000 kg called?
metric tonThe metric ton used in most other countries is 1,000 kg, equivalent to 2,204.6 pounds avoirdupois.
## What is 10 cm called?
decimeterA decimeter is a unit of length in the metric system. The term “Deci” means one-tenth, and therefore decimetre means one-tenth of a meter. Since a meter is made up of 100 cm, one-tenth of 100 cm is 10 cm. Thus one decimeter measures 10 cm.
## What is the size of 1 cm?
0.3937 inchOne centimeter equals 0.3937 inch.
## Which is larger 14mm or 1 cm?
Lenses are measured in milimeter not in centimeters. And 1 cm is wider than 14mm as 1cm have 10mm’s. 14mm is longer, because, 10mm = 1cm. So, 14mm = 1.4cm, which is longer than 1cm.
## Is 50mm a 5cm?
Millimeters to centimeters conversion tableMillimeters (mm)Centimeters (cm)30 mm3 cm40 mm4 cm50 mm5 cm60 mm6 cm17 more rows
## What size is 200mm in CM?
Convert 200 Millimeters to Centimetersmmcm200.0020200.0120.001200.0220.002200.0320.00396 more rows
## Which is bigger M or CM?
In addition to the difference in the basic units, the metric system is based on 10s, and different measures for length include kilometer, meter, decimeter, centimeter, and millimeter. … This means that a meter is 100 times larger than a centimeter, and a kilogram is 1,000 times heavier than a gram.
## How many cm means 1 meter?
100 centimetresYou can do this by multiplying 2.9 by 100 as there are 100 centimetres in a metre. To convert a number back from centimetres back into metres divide by 100. | 673 | 2,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | longest | en | 0.84546 |
https://gotmyhomework.com/microeconomics/ | 1,624,245,117,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488262046.80/warc/CC-MAIN-20210621025359-20210621055359-00497.warc.gz | 265,109,380 | 10,438 | # Microeconomics
Assignment 2: Market Forms
For this assignment you will do a significant portion of work in MS Excel and import it into an MS Word document for submission. You will use the data below to address Price and Output decisions faced by firms that are not in pure competition. Some numbers may be rounded.
Table 1 Output Average Fixed cost Average Variable Cost Average Total Cost Marginal Cost Price Total Revenue Marginal Revenue 0 \$ 345.00 1 \$ 180.00 \$ 135.00 \$ 315.00 \$ 300.00 2 \$ 90.00 \$ 127.50 \$ 217.50 \$ 249.00 3 \$ 60.00 \$ 120.00 \$ 180.00 \$ 213.00 4 \$ 45.00 \$ 112.50 \$ 157.50 \$ 189.00 5 \$ 36.00 \$ 111.00 \$ 147.00 \$ 165.00 6 \$ 30.00 \$ 112.50 \$ 142.50 \$ 144.00 7 \$ 25.71 \$ 115.70 \$ 141.41 \$ 126.00 8 \$ 22.50 \$ 121.90 \$ 144.40 \$ 111.00 9 \$ 20.00 \$ 130.00 \$ 150.00 \$ 99.00 10 \$ 18.00 \$ 139.50 \$ 157.50 \$ 87.00
1. Complete Table-1. Summarize your calculations.
2. Prepare a chart showing:
• Average Fixed Costs
• Average Variable Costs
• Average Total Costs
• Marginal Revenue
• Marginal Costs
3. Using the data in the table and on your graph, explain the profit maximizing, or loss minimizing level of output.
4. Define a normal profit and an economic profit. Are normal profits being earned in this example? Are economic profits present for this firm in this example? Explain your answers.
5. Given the data in the table and the graph, what type of market structurecould this be in the short run? Explain your answers.
6. If the data in Table-1 represents the long run, what type of firm must this data represent? Explain your answers.
Save your MS Word document using the filename LastnameFirstInitial_M4A2 and submit it to the M4: Assignment 2 Dropbox by Wednesday, December 10, 2014.
Quotations, paraphrases, and ideas you get from books, articles, or other sources of information should be cited using APA style.
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### Black Scholes vs Binomial Model
I'm trying to confirm my understanding of the 2 models. It is my understanding that the black-scholes is a special case of a binomial model with infinite steps. Does this mean that if I were to start ...
333 views
### Intuition behind American Option pricing
The price of an American option is given by $$V_n = \max\left(G_n,\frac{pV_{n +1}H^d + qV_{n + 1}H^u}{1 + r}\right)$$ where p, q are the risk neutral probabilities. I have two questions: How can ...
1k views
### Does the Binomial Pricing Model require a no-arbitrage assumption?
In a binomial option model, if we take the uptick as 6%, downtick as 5% (assume equally probable), and RFR of 6% (continuous compounding), then we have a violation of $0 < d < 1 + r < u$. ...
132 views
### Looback Put Option - finding the number of paths that reach each level
In a 4 period binomial model, I have a lookback put option that pays $\left [M_{4}-4 \right ]^{+}$, where $M_{4}$ is the maximum price reached during the sequence of 4 trials. Lets say the starting ...
2k views
152 views
179 views
### Error on Paul Wilmott Section 5.2?
I gave this a long and hard thought because Paul Wilmott is a respected quant and I don't want to criticize his book, but am I correct in concluding that this section contains lots of errors? These ...
405 views
### Real Options: Calculating the “option to switch use” using binomial lattices
I'm currently looking into calculating the "option to switch use" to determine the benefit of the ability to switch between two technologies at any point in time (american option). This is also called ...
278 views
### Proof of optimal exercise time theorem for American derivative security in N-period binomial asset-pricing model
At least two textbooks (Shreve's Stochastic Calculus for Finance - I, theorem 4.4.5 or Campolieti & Makarov's Financial Mathematics, proposition 7.8) prove the optimal exercise theorem that says ...
3k views
### Pricing American Put Options via Binomial Tree in Matlab
I currently am completing a Computational Finance Assignment, and am trying to figure out how to alter this Matlab code which prices a European put or call option, in order to price an American Put ...
205 views
### Call option pricing using CCR model - derivation problem
I'm viewing the following derivation of a Call Option price using the CRR model. There is one piece of the derivation which I cannot understand. \begin{align} C_0 &= e^{-rT} \sum_{i=0}^{N} (S_{0}\...
325 views
54 views
1k views
### Calculating expected shortfall
I'm trying to calculate the expected shortfall for the below scenario. I don't understand why the 1.04% probability of 0 bonds defaulting is used as a weight when calculating ES, since the binomial ...
505 views
### Divergence between binomial pricing and monte carlo simulation for vanilla european call?
I notice a divergence in my own code, but it's evident even in public code: http://www.thalesians.com/finance/index.php/Knowledge_Base/Finance/Option_Pricing_in_Python_and_Simple_English Pricing a ...
80 views
### Option pricing models relation between theoretical and actual price
I have trying to figure out the relationship between theoretical option price and actual market price spotted from market which is determined by supply and demand. I yet cannot understand how to ... | 834 | 3,482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-21 | latest | en | 0.911218 |
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## 6 posts in this topic
Posted · Report post
Not a pack of lies ,
15 cards - one suit Ace to King plus two Jokers. All shuffled and placed on the table in a row. It does not matter if placed left to right or right to left as no card occupies it's original position either way, nor starts or finishes with two jokers. No court card (face card) or joker are at either end, nor in adjacent positions. The jokers have six cards between them. The card on the far left is one higher than on the far right, and is adjacent to a court card. Seven is two places left of two. Three is two places left of the rightmost Joker. The Ace is two places to the right of the King, and two left of the leftmost joker. Ten is two places to the left of five and three to the right of six. The Queen is two left of the Jack and four to the right of the Ace. The four is further left than the seven
Which order is each card if viewed L-R
I put the numbers in wriiten form and ran the whole thing to one paragraph to be awkward. You might want to rearrange it into numbers/names eg A,2,3,4,5, Jo, Ja etc and use seperate lines for each sentance.
I know I hav'nt named all the cards, that would be too EZ.
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Posted · Report post
9 K 4 A 7 JK 2 Q 6 J 3 10 JK 5 8
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Posted · Report post
It does not matter if placed left to right or right to left as no card occupies it's original position either way, nor starts or finishes with two jokers.
Can you make this two separate sentences?
"no card ... starts or ends with two jokers" doesn't make sense.
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Posted · Report post
Can you make this two separate sentences?
"no card ... starts or ends with two jokers" doesn't make sense.
10 min edit has gone!
But to clarify There are no jokers in position 1,2 or 14,15
I had spent a lot of time edditing but over cooked it!
Barne012 did it so it an't be too bad
Average time is 45 mins on this.
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Posted · Report post
Only 9 and 8 were left out.And I found the combination to be
9 K 4 A 7 Jo 2 Q 6 J 3 10 Jo 5 8
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Posted · Report post
I drew and labeled a 15x15 grid.
Then it took
9 K 4 A 7 Joker 2 Q 6 J 3 10 Joker 5 8
Nice puzzle.
0
## Create an account
Register a new account | 628 | 2,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | latest | en | 0.940689 |
http://www.javaprogrammingforums.com/%20algorithms-recursion/31440-compiler-printingthethread.html | 1,526,936,960,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00408.warc.gz | 389,992,262 | 3,221 | # compiler
• September 11th, 2013, 12:27 PM
jamarrufo
compiler
Using Arrays with Sorting and Searching Algorithms
(based on Module 4 material)
1)
This program has six required outputs and involves searching and sorting an array of integers.
Write a Java application that initializes an array with the following numbers, in this order:
23, 17, 5, 90, 12, 44, 38, 84, 77, 3, 66, 55, 1, 19, 37, 88, 8, 97, 25, 50, 75, 61, and 49
Then display the unsorted values. This is required output #1 of 6 for this program.
Using a sequential search of the unsorted array, determine and report the 1
-relative (i.e. 1, 2, 3, 4, etc.)Positions of the following numbers in the array (or -1 if not found), and the number of searches required to locate the numbers: 25,30, 50, 75, and 92.
This is required output #2 of 6.
Then display the total number of searches for all five numbers.
This is required output #3 of 6.
Sort the numbers using any algorithm of your choice and then display the sorted array.
This is required output #4 of 6.
Using a binary search of the sorted array, determine and report the 1-relative positions of the following numbers in the array (or -1 if not found), and the number of searches required to locate the numbers: 25, 30, 50, 75, and 92.
This is required output #5 of 6.Finally; display the total number of searches for all five numbers.
This is required output #6 of 6.
(There are six required sets of output as numbered in the above paragraphs.
Can you tell me if this code is right to the about problem it compile right but I believe I have an problem tank you
import java.util.Arrays;
public class arrayCount
{
/*
outputs the 1-relative index of the query in the array if found, -1 otherwise
*/
public int sortCount;
public static int search(int[] arr, int term){
for(int i = 0; i < arr.length; ++i) { // iterate through array
if (arr[i] == term) { // if term found
return ++i; } } // return 1-relative index of found term
return -1; } // if loop doesn't find anything, return -1
public static void main(String[] args) {
int arrayNum[] = {23, 17, 5, 90, 12, 44, 38, 84, 77, 3, 66, 55, 1, 19, 37, 88, 8, 97, 25, 50, 75, 61, 49};
int numLocation;
int numSearch = 0;
String str = "Unsorted values: ";
// step 1
for(int i = 0; i < arrayNum.length; ++i)
str = str + " " + arrayNum[i];
System.out.println(str);
// step 2
int searches[] = {25,30,50, 75, 92}; // initialize numbers to search
for (int i = 0; i < searches.length; ++i){ // iterate through searches array
numLocation = search(arrayNum, searches[i]);
System.out.println("Search for " + searches[i] + ": " // print search term and
+ numLocation);
if (numLocation >= 0){ numSearch = numSearch + numLocation;
System.out.println("Number of searches " + numLocation);}
else {numSearch = numSearch +arrayNum.length;
System.out.println("Number of searches " + arrayNum.length);}
}
System.out.println("Total Number of Searches " + numSearch); // result of search
Arrays.sort(arrayNum);
String two = "Sorted values: ";
// step 4
for(int i = 0; i < arrayNum.length; ++i)
two = two + " " + arrayNum[i];
System.out.println(two);
int searchesTwo[] = {25,30,50,75,92};
int numsearchTwo = 0;
for (int i = 0; i < searchesTwo.length; ++i){
numLocation = Arrays.binarySearch(arrayNum, searchesTwo[i]);
numLocation = numLocation + 1;
if (numLocation >= 0){ numsearchTwo = numsearchTwo + numLocation ;
System.out.println("Search for " + searchesTwo[i] + ": "
+ numLocation);
System.out.println("Number of searches " + numLocation) ;}
else {numsearchTwo = numsearchTwo +arrayNum.length;
System.out.println("Search for " + searchesTwo[i] + ": -1" );
System.out.println("Number of searches " + arrayNum.length);}
}
System.out.println("Total Number of Searches " + numsearchTwo);
}
}
• September 11th, 2013, 12:35 PM
KevinWorkman
Re: compiler
You forgot the highlight tags, which makes your code pretty hard to read.
What makes you think you have a problem?
• September 11th, 2013, 09:16 PM
jamarrufo
Re: compiler
I have found the problem, I program only 4 and no 6 tank you
Jesus M | 1,170 | 4,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-22 | latest | en | 0.816166 |
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We thought you may like to see a picture of the Greek mathematician Pythagoras.
Level 7-8 Shapes - Pythagoras
You've looked a lot at shapes in KS3 Maths. This particular quiz on shapes looks at the special number relationship linking the lengths of the sides of a right-angled triangle. This relationship is named after the ancient Greek mathematician, Pythagoras.
Pythagoras' Theorem says that, in right-angled triangles: 'The square on the hypotenuse (that's the longest side which will be opposite the right angle) is equal to the sum of the squares on the other two sides'. Because this rule is true for every right-angled triangle, you can use it to test whether a triangle whose angles you don't know (so long as you do know the lengths of the sides) has a right angle. That will come in handy in this quiz!
You may find it helpful to draw some diagrams while doing this quiz. So, get your pencil and paper out and see how well you can do. Take your time and read each question carefully before you choose your answers. Good luck!
1.
In a right-angled triangle the sides forming the 90o angle are 6 cm and 8 cm. What is the length of the third side?
10 cm
12 cm
14 cm
16 cm
36 + 64 = 100; ?100 = 10. The longest side in a right-angled triangle is called the hypotenuse
2.
The cross-section of a porch roof is an isosceles triangle with height 0.8 m and base 3.0 m. How long is each sloping roof section?
3.1 m
2.3 m
1.9 m
1.7 m
I hope you remembered the rules for isosceles triangles
3.
The foot of a ladder of length 3.75 m is 1.4 m away from a vertical wall. How high up the wall will the ladder reach (to the nearest 1 cm)?
3.84 m
3.48 m
3.24 m
3.18 m
This assumes that the ground is perfectly level!
4.
In a triangle, base 25 cm and height 12 cm, the perpendicular from the base to the opposite vertex divides the base in the ratio 3:2. How long are the other two sides (to 1dp)?
15.6 cm and 19.2 cm
16.3 cm and 18.9 cm
15.6 cm and 16.3 cm
18.9 cm and 19.2 cm
Split the triangle into two right-angled triangles with bases 15 cm and 10 cm
5.
The following sets of numbers represent the lengths of the sides of a triangle. Which is not a right-angled triangle?
15, 20, 25
18, 24, 30
16, 30, 36
15, 36, 39
Trial and error is a very useful tool in maths
6.
If the longest side of a right-angled triangle is c cm and the other sides are a cm and b cm, Pythagoras' Theorem states that .......
a2 x b2 = c2
(a + b)2 = c2
a2 + b2 = c2
a2 - b2 = c2
Remember, Pythagoras' Theorem only applies to right-angled triangles
7.
A right-angled isosceles triangle has hypotenuse of length 15 cm. What are the angles at either end of the hypotenuse?
15o and 75o
30o and 60o
45o and 45o
20o and 70o
The side length is irrelevant. Any right-angled isosceles triangle has two angles of 45o
8.
A rectangle has length 2.4 m and width 0.7 m. How long is its diagonal?
2.0 m
2.5 m
3.0 m
3.5 m
The diagonal is the hypotenuse of a right-angled triangle
9.
The hypotenuse of a right-angled triangle is 26 cm. One of the shorter sides is 10 cm. How long is the third side?
30 cm
24 cm
16 cm
8 cm
262 - 102 = 242
10.
A triangle has sides of length 4 cm, 5 cm and 6 cm. Which of these statements is not true?
The three angles add up to 180o
All three angles are acute
The triangle is scalene
The largest angle is 90o
(4 x 4) + (5 x 5) = 41 but 6 x 6 = 36 so the triangle cannot contain a right angle
You can find more about this topic by visiting BBC Bitesize - Pythagoras
Author: Frank Evans
© Copyright 2016-2024 - Education Quizzes
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##### How to solve this equation?
label Mathematics
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5
4x^2 - 8x + 4 = 8
Oct 14th, 2017
4(x^2 - 2x +1) = 8
x^2 - 2x + 1 = 2
(x-1)^2 = 2
x - 1 = sqrt(2) or x - 1 = sqrt(2)
x = 1+sqrt(2) or x = 1-sqrt(2)
Oct 11th, 2014
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Oct 14th, 2017
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Oct 14th, 2017
Oct 17th, 2017
check_circle | 170 | 375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-43 | latest | en | 0.780397 |
http://www.pedagonet.com/puzzles/jamming1.html | 1,508,384,881,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823220.45/warc/CC-MAIN-20171019031425-20171019051425-00372.warc.gz | 530,630,250 | 3,660 | Videos Test Yourself Wordcipher MathFacts Math Diagnostics Math Tricks Daily PhysEd Class Photos Worksheets Math Genius Musiclopedia More Resources Teacher Timesavers Study Guides
Making Jam Thanks Jelly Preserve Answer: x = small containers y = medium containers z = large container Therefore: 11x + 8y + z = 8,4 x + 2y + z = 2x + 4y 8x + 2y = 2x + 4y. Remove z in the first two equations and calculate x and y Solution is : x = 0,2 kg y = 0,6 kg z = 1,4 kg | 134 | 459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-43 | latest | en | 0.574161 |
http://aven.amritalearning.com/index.php?sub=101&brch=304&sim=1579&cnt=3799 | 1,542,228,306,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114213308-00078.warc.gz | 32,369,537 | 7,465 | Factors of Algebraic Expressions
## Factors of natural numbers
You will remember what you learnt about factors in Class VI. Let us take a natural number, say 30, and write it as a product of other natural numbers, say
30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter.
## Factors of algebraic expressions
We have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e.,
5xy = 5× x× y
Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. We say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y.
Next consider the expression 3x (x + 2). It can be written as a product of factors. 3, x and (x + 2)
3x(x + 2) =3× x×(x + 2)
The factors 3, x and (x +2) are irreducible factors of 3x (x + 2).
Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form
as 10x (x + 2) (y + 3) = 2 × 5 × x × (x + 2) × ( y + 3).
## What is Factorisation?
When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, 5x2 y , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. Their factors can be just read off from them, as we already know.
On the other hand consider expressions like 2x + 4, 3x + 3y, x2 + 5x, x2 + 5x + 6. It is not obvious what their factors are. We need to develop systematic methods to factorise these expressions, i.e., to find their factors. This is what we shall do now.
## Method of common factors
• We begin with a simple example: Factorise 2x + 4.
We shall write each term as a product of irreducible factors;
2x = 2 × x
4 = 2 × 2
Hence 2x + 4 = (2 × x) + (2 × 2)
Notice that factor 2 is common to both the terms.
Observe, by distributive law
2 × (x + 2) = (2 × x) + (2 × 2)
Therefore, we can write
2x + 4 = 2 × (x + 2) = 2 (x + 2)
Thus, the expression 2x + 4 is the same as 2 (x + 2). Now we can read off its factors:
they are 2 and (x + 2). These factors are irreducible.
Next, factorise 5xy + 10x.
The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x
Observe that the two terms have 5 and x as common factors. Now,
5xy + 10x = (5 × x × y) + (5 × x × 2)
= (5x × y) + (5x × 2)
We combine the two terms using the distributive law,
(5x× y) + (5x× 2) = 5x × ( y + 2)
Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.)
Example : Factorise 12a2b + 15ab2
Solution: We have 12a2b = 2 × 2 × 3 × a × a × b
15ab2 = 3 × 5 × a × b × b
The two terms have 3, a and b as common factors.
Therefore, 12a2b + 15ab2 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
= 3ab × (4a + 5b)
= 3ab (4a + 5b) (required factor form)
Example 2: Factorise 10x2 – 18x3 + 14x4
Solution: 10x2 = 2 × 5 × x × x
18x3 = 2 × 3 × 3 × x × x × x
14x4 = 2 × 7 × x × x × x × x
The common factors of the three terms are 2, x and x.
Therefore, 10x2 – 18x3 + 14x4 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x)
+ (2 × x × x × 7 × x × x)
= 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)]
= 2x2 × (5 – 9x + 7x2) = 2x2 (7x2 − 9x + 5)
## Factorisation by regrouping terms
Look at the expression 2xy + 2y + 3x + 3. You will notice that the first two terms have common factors 2 and y and the last two terms have a common factor 3. But there is no single factor common to all the terms. How shall we proceed?
Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)
Similarly, 3x + 3 = (3 × x) + (3 × 1)
= 3 × (x + 1) = 3 ( x + 1)
Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)
Observe, now we have a common factor (x + 1) in both the terms on the right hand
side. Combining the two terms,
2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)
The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its
factors are (x + 1) and (2y + 3). Note, these factors are irreducible.
## What is regrouping?
Suppose, the above expression was given as 2xy + 3 + 2y + 3x; then it will not be easy to see the factorisation. Rearranging the expression, as 2xy + 2y + 3x + 3, allows us to form groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping.
Regrouping may be possible in more than one ways. Suppose, we regroup the expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try:
2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3
= x × (2y + 3) + 1 × (2y + 3)
= (2y + 3) (x + 1)
The factors are the same (as they have to be), although they appear in different order
Example : Factorise 6xy – 4y + 6 – 9x.
Solution:
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor 2y;
6xy – 4y = 2y (3x – 2) (a)
What about the last two terms? Observe them. If you change their order to
– 9x + 6, the factor ( 3x – 2) will come out;
–9x + 6 = –3 (3x) + 3 (2)
= – 3 (3x – 2) (b)
Step 3 Putting (a) and (b) together,
6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).
## Factorisation using identities
We know that (a + b)2 = a2 + 2ab + b2 (I)
(a – b)2 = a2 – 2ab + b2 (II)
(a + b) (a – b) = a2 – b2 (III)
The following solved examples illustrate how to use these identities for factorisation. What we do is to observe the given expression. If it has a form that fits the right hand side of one of the identities, then the expression corresponding to the left hand side of the identity gives the desired factorisation.
Example : Factorise x2 + 8x + 16
Solution: Observe the expression; it has three terms. Therefore, it does not fit
Identity III. Also, it’s first and third terms are perfect squares with a positive sign before
the middle term. So, it is of the form a2 + 2ab + b2 where a = x and b = 4
such that a2 + 2ab + b2 = x2 + 2 (x) (4) + 42
= x2 + 8x + 16
Since a2 + 2ab + b2 = (a + b)2,
by comparison x2 + 8x + 16 = ( x + 4)2 (the required factorisation)
Example : Factorise a2 – 2ab + b2 – c2
Solution: The first three terms of the given expression form (a – b)2. The fourth term is a square. So the expression can be reduced to a difference of two squares.
Thus, a2 – 2ab + b2 – c2 = (a – b)2– c (Applying Identity II)
= [(a – b) – c) ((a – b) + c)] (Applying Identity III)
= (a – b – c) (a – b + c) (required factorisation)
Notice, how we applied two identities one after the other to obtain the required factorisation
## Factors of the form ( x + a) ( x + b)
Let us now discuss how we can factorise expressions in one variable, like x2 + 5x + 6, y2 – 7y + 12, z2 – 4z – 12, 3m2 + 9m + 6, etc. Observe that these expressions are not of the type (a + b) 2 or (a – b) 2, i.e., they are not perfect squares. For example, in
x2 + 5x + 6, the term 6 is not a perfect square. These expressions obviously also do not fit the type (a2 – b2) either.
They, however, seem to be of the type x2 + (a + b) x + a b.We may therefore, try to use Identity IV studied in the last chapter to factorise these expressions:
(x + a) (x + b) = x2 + (a + b) x + ab (IV)
For that we have to look at the coefficients of x and the constant term. Let us see how it is done in the following example.
Example: Find the factors of y2 –7y +12.
Solution: We note 12 = 3 × 4 and 3 + 4 = 7. Therefore,
y2 – 7y+ 12 = y2 – 3y – 4y + 12
= y (y –3) – 4 (y –3) = (y –3) (y – 4)
Note:, this time we did not compare the expression with that in Identity (IV) to identif a and b. After sufficient practice you may not need to compare the given expressions for their factorisation with the expressions in the identities; instead you can proceed directly as we did above
Example : Find the factors of 3m2 + 9m + 6.
Solution: We notice that 3 is a common factor of all the terms.
Therefore, 3m2 + 9m + 6 = 3(m2 + 3m + 2)
Now, m 2 + 3m + 2 = m2 + m + 2m + 2 (as 2 = 1 × 2)
= m(m + 1)+ 2( m + 1)
= (m + 1) (m + 2)
Therefore, 3m2 + 9m + 6 = 3(m + 1) (m + 2)
Cite this Simulator: | 3,367 | 9,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2018-47 | longest | en | 0.94432 |
https://bedtimemath.org/fun-math-saving-energy/ | 1,680,263,850,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00568.warc.gz | 151,917,821 | 18,418 | # Watch Those Lights!
Today is “National Cut Your Energy Costs” Day, the day when we think about using less electricity in our homes, gasoline in our cars, and other forms of energy. Do you know what things in your house use the most energy? Here at Bedtime Math we watched our electricity meter while running one appliance at a time, and found out that the clothes dryer uses about 1/5 of all our energy — and that we use more over the years because we have more kids now, which means more laundry, which means more loads in the machine every week. And as everybody gets bigger and dirtier, there’s more and more laundry. What will help someday is to save costs by finding energy that doesn’t cost much to begin with. One solar panel 2-3 feet long can make up to 300 watts and run a few light bulbs; better yet, a megawatt of energy from a wind farm can power 400 homes. If we can lasso the sun and wind even more cheaply, maybe someday we can dry our clothes as much as we want.
Wee ones: Try to find 3 light bulbs in your home that are on right now.
Little kids: If it’s sunny from 1:00 in the afternoon till sunset at 8:00 pm, how many hours can you leave the lights off? Bonus: If 1 giant windmill, or “turbine,” on a wind farm can power 10 houses, how many houses can 10 turbines run? Count up by 10s!
Big kids: If a solar panel can run 4 light bulbs and your home has 32 bulbs in the various rooms, how many panels do you need to stick on your roof to run them all? Bonus: If you see in the dark using night vision goggles instead of the 6 60-watt light bulbs in your kitchen, how many watts are you saving?
Wee ones: Count them as you find them: 1, 2, 3.
Little kids: 7 hours. Bonus: 100 houses: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.
Big kids: 8 panels. Bonus: 360 watts.
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Vehicles and Transportation | 510 | 1,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.932728 |
https://www.kodytools.com/units/inertia/from/lbkm2/to/grkm2 | 1,716,770,056,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00445.warc.gz | 736,451,779 | 18,497 | # Pound Square Kilometer to Grain Square Kilometer Converter
1 Pound Square Kilometer = 7000 Grain Square Kilometers
## One Pound Square Kilometer is Equal to How Many Grain Square Kilometers?
The answer is one Pound Square Kilometer is equal to 7000 Grain Square Kilometers and that means we can also write it as 1 Pound Square Kilometer = 7000 Grain Square Kilometers. Feel free to use our online unit conversion calculator to convert the unit from Pound Square Kilometer to Grain Square Kilometer. Just simply enter value 1 in Pound Square Kilometer and see the result in Grain Square Kilometer.
Manually converting Pound Square Kilometer to Grain Square Kilometer can be time-consuming,especially when you don’t have enough knowledge about Moment of Inertia units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Pound Square Kilometer to Grain Square Kilometer converter tool to get the job done as soon as possible.
We have so many online tools available to convert Pound Square Kilometer to Grain Square Kilometer, but not every online tool gives an accurate result and that is why we have created this online Pound Square Kilometer to Grain Square Kilometer converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Pound Square Kilometer to Grain Square Kilometer (lb·km2 to gr·km2)
By using our Pound Square Kilometer to Grain Square Kilometer conversion tool, you know that one Pound Square Kilometer is equivalent to 7000 Grain Square Kilometer. Hence, to convert Pound Square Kilometer to Grain Square Kilometer, we just need to multiply the number by 7000. We are going to use very simple Pound Square Kilometer to Grain Square Kilometer conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Pound Square Kilometer} = 1 \times 7000 = \text{7000 Grain Square Kilometers}$$
## What Unit of Measure is Pound Square Kilometer?
Pound square kilometer is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one kilometer distance from the rotation axis and having a mass of one pound.
## What is the Symbol of Pound Square Kilometer?
The symbol of Pound Square Kilometer is lb·km2. This means you can also write one Pound Square Kilometer as 1 lb·km2.
## What Unit of Measure is Grain Square Kilometer?
Grain square kilometer is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one kilometer distance from the rotation axis and having a mass of one grain.
## What is the Symbol of Grain Square Kilometer?
The symbol of Grain Square Kilometer is gr·km2. This means you can also write one Grain Square Kilometer as 1 gr·km2.
## How to Use Pound Square Kilometer to Grain Square Kilometer Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Pound Square Kilometer and in the first input field, enter a value.
• From the second dropdown, select Grain Square Kilometer.
• Instantly, the tool will convert the value from Pound Square Kilometer to Grain Square Kilometer and display the result in the second input field.
## Example of Pound Square Kilometer to Grain Square Kilometer Converter Tool
Pound Square Kilometer
1
Grain Square Kilometer
7000
# Pound Square Kilometer to Grain Square Kilometer Conversion Table
Pound Square Kilometer [lb·km2]Grain Square Kilometer [gr·km2]Description
1 Pound Square Kilometer7000 Grain Square Kilometer1 Pound Square Kilometer = 7000 Grain Square Kilometer
2 Pound Square Kilometer14000 Grain Square Kilometer2 Pound Square Kilometer = 14000 Grain Square Kilometer
3 Pound Square Kilometer21000 Grain Square Kilometer3 Pound Square Kilometer = 21000 Grain Square Kilometer
4 Pound Square Kilometer28000 Grain Square Kilometer4 Pound Square Kilometer = 28000 Grain Square Kilometer
5 Pound Square Kilometer35000 Grain Square Kilometer5 Pound Square Kilometer = 35000 Grain Square Kilometer
6 Pound Square Kilometer42000 Grain Square Kilometer6 Pound Square Kilometer = 42000 Grain Square Kilometer
7 Pound Square Kilometer49000 Grain Square Kilometer7 Pound Square Kilometer = 49000 Grain Square Kilometer
8 Pound Square Kilometer56000 Grain Square Kilometer8 Pound Square Kilometer = 56000 Grain Square Kilometer
9 Pound Square Kilometer63000 Grain Square Kilometer9 Pound Square Kilometer = 63000 Grain Square Kilometer
10 Pound Square Kilometer70000 Grain Square Kilometer10 Pound Square Kilometer = 70000 Grain Square Kilometer
100 Pound Square Kilometer700000 Grain Square Kilometer100 Pound Square Kilometer = 700000 Grain Square Kilometer
1000 Pound Square Kilometer7000000 Grain Square Kilometer1000 Pound Square Kilometer = 7000000 Grain Square Kilometer
# Pound Square Kilometer to Other Units Conversion Table
ConversionDescription
1 Pound Square Kilometer = 453592370 Gram Square Meter1 Pound Square Kilometer in Gram Square Meter is equal to 453592370
1 Pound Square Kilometer = 45359237000 Gram Square Decimeter1 Pound Square Kilometer in Gram Square Decimeter is equal to 45359237000
1 Pound Square Kilometer = 4535923700000 Gram Square Centimeter1 Pound Square Kilometer in Gram Square Centimeter is equal to 4535923700000
1 Pound Square Kilometer = 453592370000000 Gram Square Millimeter1 Pound Square Kilometer in Gram Square Millimeter is equal to 453592370000000
1 Pound Square Kilometer = 453592370000000000000 Gram Square Micrometer1 Pound Square Kilometer in Gram Square Micrometer is equal to 453592370000000000000
1 Pound Square Kilometer = 453.59 Gram Square Kilometer1 Pound Square Kilometer in Gram Square Kilometer is equal to 453.59
1 Pound Square Kilometer = 4.5359237e+26 Gram Square Nanometer1 Pound Square Kilometer in Gram Square Nanometer is equal to 4.5359237e+26
1 Pound Square Kilometer = 542491959.6 Gram Square Yard1 Pound Square Kilometer in Gram Square Yard is equal to 542491959.6
1 Pound Square Kilometer = 703069579639.16 Gram Square Inch1 Pound Square Kilometer in Gram Square Inch is equal to 703069579639.16
1 Pound Square Kilometer = 4882427636.38 Gram Square Foot1 Pound Square Kilometer in Gram Square Foot is equal to 4882427636.38
1 Pound Square Kilometer = 175.13 Gram Square Mile1 Pound Square Kilometer in Gram Square Mile is equal to 175.13
1 Pound Square Kilometer = 453592.37 Kilogram Square Meter1 Pound Square Kilometer in Kilogram Square Meter is equal to 453592.37
1 Pound Square Kilometer = 45359237 Kilogram Square Decimeter1 Pound Square Kilometer in Kilogram Square Decimeter is equal to 45359237
1 Pound Square Kilometer = 4535923700 Kilogram Square Centimeter1 Pound Square Kilometer in Kilogram Square Centimeter is equal to 4535923700
1 Pound Square Kilometer = 453592370000 Kilogram Square Millimeter1 Pound Square Kilometer in Kilogram Square Millimeter is equal to 453592370000
1 Pound Square Kilometer = 453592370000000000 Kilogram Square Micrometer1 Pound Square Kilometer in Kilogram Square Micrometer is equal to 453592370000000000
1 Pound Square Kilometer = 0.45359237 Kilogram Square Kilometer1 Pound Square Kilometer in Kilogram Square Kilometer is equal to 0.45359237
1 Pound Square Kilometer = 4.5359237e+23 Kilogram Square Nanometer1 Pound Square Kilometer in Kilogram Square Nanometer is equal to 4.5359237e+23
1 Pound Square Kilometer = 542491.96 Kilogram Square Yard1 Pound Square Kilometer in Kilogram Square Yard is equal to 542491.96
1 Pound Square Kilometer = 703069579.64 Kilogram Square Inch1 Pound Square Kilometer in Kilogram Square Inch is equal to 703069579.64
1 Pound Square Kilometer = 4882427.64 Kilogram Square Foot1 Pound Square Kilometer in Kilogram Square Foot is equal to 4882427.64
1 Pound Square Kilometer = 0.17513299315538 Kilogram Square Mile1 Pound Square Kilometer in Kilogram Square Mile is equal to 0.17513299315538
1 Pound Square Kilometer = 453592370000 Milligram Square Meter1 Pound Square Kilometer in Milligram Square Meter is equal to 453592370000
1 Pound Square Kilometer = 45359237000000 Milligram Square Decimeter1 Pound Square Kilometer in Milligram Square Decimeter is equal to 45359237000000
1 Pound Square Kilometer = 4535923700000000 Milligram Square Centimeter1 Pound Square Kilometer in Milligram Square Centimeter is equal to 4535923700000000
1 Pound Square Kilometer = 453592370000000000 Milligram Square Millimeter1 Pound Square Kilometer in Milligram Square Millimeter is equal to 453592370000000000
1 Pound Square Kilometer = 4.5359237e+23 Milligram Square Micrometer1 Pound Square Kilometer in Milligram Square Micrometer is equal to 4.5359237e+23
1 Pound Square Kilometer = 453592.37 Milligram Square Kilometer1 Pound Square Kilometer in Milligram Square Kilometer is equal to 453592.37
1 Pound Square Kilometer = 4.5359237e+29 Milligram Square Nanometer1 Pound Square Kilometer in Milligram Square Nanometer is equal to 4.5359237e+29
1 Pound Square Kilometer = 542491959598.12 Milligram Square Yard1 Pound Square Kilometer in Milligram Square Yard is equal to 542491959598.12
1 Pound Square Kilometer = 703069579639160 Milligram Square Inch1 Pound Square Kilometer in Milligram Square Inch is equal to 703069579639160
1 Pound Square Kilometer = 4882427636383.1 Milligram Square Foot1 Pound Square Kilometer in Milligram Square Foot is equal to 4882427636383.1
1 Pound Square Kilometer = 175132.99 Milligram Square Mile1 Pound Square Kilometer in Milligram Square Mile is equal to 175132.99
1 Pound Square Kilometer = 453592370000000 Microgram Square Meter1 Pound Square Kilometer in Microgram Square Meter is equal to 453592370000000
1 Pound Square Kilometer = 45359237000000000 Microgram Square Decimeter1 Pound Square Kilometer in Microgram Square Decimeter is equal to 45359237000000000
1 Pound Square Kilometer = 4535923700000000000 Microgram Square Centimeter1 Pound Square Kilometer in Microgram Square Centimeter is equal to 4535923700000000000
1 Pound Square Kilometer = 453592370000000000000 Microgram Square Millimeter1 Pound Square Kilometer in Microgram Square Millimeter is equal to 453592370000000000000
1 Pound Square Kilometer = 4.5359237e+26 Microgram Square Micrometer1 Pound Square Kilometer in Microgram Square Micrometer is equal to 4.5359237e+26
1 Pound Square Kilometer = 453592370 Microgram Square Kilometer1 Pound Square Kilometer in Microgram Square Kilometer is equal to 453592370
1 Pound Square Kilometer = 4.5359237e+32 Microgram Square Nanometer1 Pound Square Kilometer in Microgram Square Nanometer is equal to 4.5359237e+32
1 Pound Square Kilometer = 542491959598120 Microgram Square Yard1 Pound Square Kilometer in Microgram Square Yard is equal to 542491959598120
1 Pound Square Kilometer = 703069579639160000 Microgram Square Inch1 Pound Square Kilometer in Microgram Square Inch is equal to 703069579639160000
1 Pound Square Kilometer = 4882427636383100 Microgram Square Foot1 Pound Square Kilometer in Microgram Square Foot is equal to 4882427636383100
1 Pound Square Kilometer = 175132993.16 Microgram Square Mile1 Pound Square Kilometer in Microgram Square Mile is equal to 175132993.16
1 Pound Square Kilometer = 453.59 Ton Square Meter1 Pound Square Kilometer in Ton Square Meter is equal to 453.59
1 Pound Square Kilometer = 45359.24 Ton Square Decimeter1 Pound Square Kilometer in Ton Square Decimeter is equal to 45359.24
1 Pound Square Kilometer = 4535923.7 Ton Square Centimeter1 Pound Square Kilometer in Ton Square Centimeter is equal to 4535923.7
1 Pound Square Kilometer = 453592370 Ton Square Millimeter1 Pound Square Kilometer in Ton Square Millimeter is equal to 453592370
1 Pound Square Kilometer = 453592370000000 Ton Square Micrometer1 Pound Square Kilometer in Ton Square Micrometer is equal to 453592370000000
1 Pound Square Kilometer = 0.00045359237 Ton Square Kilometer1 Pound Square Kilometer in Ton Square Kilometer is equal to 0.00045359237
1 Pound Square Kilometer = 453592370000000000000 Ton Square Nanometer1 Pound Square Kilometer in Ton Square Nanometer is equal to 453592370000000000000
1 Pound Square Kilometer = 542.49 Ton Square Yard1 Pound Square Kilometer in Ton Square Yard is equal to 542.49
1 Pound Square Kilometer = 703069.58 Ton Square Inch1 Pound Square Kilometer in Ton Square Inch is equal to 703069.58
1 Pound Square Kilometer = 4882.43 Ton Square Foot1 Pound Square Kilometer in Ton Square Foot is equal to 4882.43
1 Pound Square Kilometer = 0.00017513299315538 Ton Square Mile1 Pound Square Kilometer in Ton Square Mile is equal to 0.00017513299315538
1 Pound Square Kilometer = 2267961850 Carat Square Meter1 Pound Square Kilometer in Carat Square Meter is equal to 2267961850
1 Pound Square Kilometer = 226796185000 Carat Square Decimeter1 Pound Square Kilometer in Carat Square Decimeter is equal to 226796185000
1 Pound Square Kilometer = 22679618500000 Carat Square Centimeter1 Pound Square Kilometer in Carat Square Centimeter is equal to 22679618500000
1 Pound Square Kilometer = 2267961850000000 Carat Square Millimeter1 Pound Square Kilometer in Carat Square Millimeter is equal to 2267961850000000
1 Pound Square Kilometer = 2.26796185e+21 Carat Square Micrometer1 Pound Square Kilometer in Carat Square Micrometer is equal to 2.26796185e+21
1 Pound Square Kilometer = 2267.96 Carat Square Kilometer1 Pound Square Kilometer in Carat Square Kilometer is equal to 2267.96
1 Pound Square Kilometer = 2.26796185e+27 Carat Square Nanometer1 Pound Square Kilometer in Carat Square Nanometer is equal to 2.26796185e+27
1 Pound Square Kilometer = 2712459797.99 Carat Square Yard1 Pound Square Kilometer in Carat Square Yard is equal to 2712459797.99
1 Pound Square Kilometer = 3515347898195.8 Carat Square Inch1 Pound Square Kilometer in Carat Square Inch is equal to 3515347898195.8
1 Pound Square Kilometer = 24412138181.92 Carat Square Foot1 Pound Square Kilometer in Carat Square Foot is equal to 24412138181.92
1 Pound Square Kilometer = 875.66 Carat Square Mile1 Pound Square Kilometer in Carat Square Mile is equal to 875.66
1 Pound Square Kilometer = 16000000 Ounce Square Meter1 Pound Square Kilometer in Ounce Square Meter is equal to 16000000
1 Pound Square Kilometer = 1600000000 Ounce Square Decimeter1 Pound Square Kilometer in Ounce Square Decimeter is equal to 1600000000
1 Pound Square Kilometer = 160000000000 Ounce Square Centimeter1 Pound Square Kilometer in Ounce Square Centimeter is equal to 160000000000
1 Pound Square Kilometer = 16000000000000 Ounce Square Millimeter1 Pound Square Kilometer in Ounce Square Millimeter is equal to 16000000000000
1 Pound Square Kilometer = 16000000000000000000 Ounce Square Micrometer1 Pound Square Kilometer in Ounce Square Micrometer is equal to 16000000000000000000
1 Pound Square Kilometer = 16 Ounce Square Kilometer1 Pound Square Kilometer in Ounce Square Kilometer is equal to 16
1 Pound Square Kilometer = 1.6e+25 Ounce Square Nanometer1 Pound Square Kilometer in Ounce Square Nanometer is equal to 1.6e+25
1 Pound Square Kilometer = 19135840.74 Ounce Square Yard1 Pound Square Kilometer in Ounce Square Yard is equal to 19135840.74
1 Pound Square Kilometer = 24800049600.1 Ounce Square Inch1 Pound Square Kilometer in Ounce Square Inch is equal to 24800049600.1
1 Pound Square Kilometer = 172222566.67 Ounce Square Foot1 Pound Square Kilometer in Ounce Square Foot is equal to 172222566.67
1 Pound Square Kilometer = 6.18 Ounce Square Mile1 Pound Square Kilometer in Ounce Square Mile is equal to 6.18
1 Pound Square Kilometer = 1000 Kilopound Square Meter1 Pound Square Kilometer in Kilopound Square Meter is equal to 1000
1 Pound Square Kilometer = 100000 Kilopound Square Decimeter1 Pound Square Kilometer in Kilopound Square Decimeter is equal to 100000
1 Pound Square Kilometer = 10000000 Kilopound Square Centimeter1 Pound Square Kilometer in Kilopound Square Centimeter is equal to 10000000
1 Pound Square Kilometer = 1000000000 Kilopound Square Millimeter1 Pound Square Kilometer in Kilopound Square Millimeter is equal to 1000000000
1 Pound Square Kilometer = 1000000000000000 Kilopound Square Micrometer1 Pound Square Kilometer in Kilopound Square Micrometer is equal to 1000000000000000
1 Pound Square Kilometer = 0.001 Kilopound Square Kilometer1 Pound Square Kilometer in Kilopound Square Kilometer is equal to 0.001
1 Pound Square Kilometer = 1e+21 Kilopound Square Nanometer1 Pound Square Kilometer in Kilopound Square Nanometer is equal to 1e+21
1 Pound Square Kilometer = 1195.99 Kilopound Square Yard1 Pound Square Kilometer in Kilopound Square Yard is equal to 1195.99
1 Pound Square Kilometer = 1550003.1 Kilopound Square Inch1 Pound Square Kilometer in Kilopound Square Inch is equal to 1550003.1
1 Pound Square Kilometer = 10763.91 Kilopound Square Foot1 Pound Square Kilometer in Kilopound Square Foot is equal to 10763.91
1 Pound Square Kilometer = 0.00038610215854245 Kilopound Square Mile1 Pound Square Kilometer in Kilopound Square Mile is equal to 0.00038610215854245
1 Pound Square Kilometer = 1000000 Pound Square Meter1 Pound Square Kilometer in Pound Square Meter is equal to 1000000
1 Pound Square Kilometer = 100000000 Pound Square Decimeter1 Pound Square Kilometer in Pound Square Decimeter is equal to 100000000
1 Pound Square Kilometer = 10000000000 Pound Square Centimeter1 Pound Square Kilometer in Pound Square Centimeter is equal to 10000000000
1 Pound Square Kilometer = 1000000000000 Pound Square Millimeter1 Pound Square Kilometer in Pound Square Millimeter is equal to 1000000000000
1 Pound Square Kilometer = 1000000000000000000 Pound Square Micrometer1 Pound Square Kilometer in Pound Square Micrometer is equal to 1000000000000000000
1 Pound Square Kilometer = 1e+24 Pound Square Nanometer1 Pound Square Kilometer in Pound Square Nanometer is equal to 1e+24
1 Pound Square Kilometer = 1195990.05 Pound Square Yard1 Pound Square Kilometer in Pound Square Yard is equal to 1195990.05
1 Pound Square Kilometer = 1550003100.01 Pound Square Inch1 Pound Square Kilometer in Pound Square Inch is equal to 1550003100.01
1 Pound Square Kilometer = 10763910.42 Pound Square Foot1 Pound Square Kilometer in Pound Square Foot is equal to 10763910.42
1 Pound Square Kilometer = 0.38610215854245 Pound Square Mile1 Pound Square Kilometer in Pound Square Mile is equal to 0.38610215854245
1 Pound Square Kilometer = 32200000 Poundal Square Meter1 Pound Square Kilometer in Poundal Square Meter is equal to 32200000
1 Pound Square Kilometer = 3220000000.05 Poundal Square Decimeter1 Pound Square Kilometer in Poundal Square Decimeter is equal to 3220000000.05
1 Pound Square Kilometer = 322000000004.97 Poundal Square Centimeter1 Pound Square Kilometer in Poundal Square Centimeter is equal to 322000000004.97
1 Pound Square Kilometer = 32200000000497 Poundal Square Millimeter1 Pound Square Kilometer in Poundal Square Millimeter is equal to 32200000000497
1 Pound Square Kilometer = 32200000000497000000 Poundal Square Micrometer1 Pound Square Kilometer in Poundal Square Micrometer is equal to 32200000000497000000
1 Pound Square Kilometer = 32.2 Poundal Square Kilometer1 Pound Square Kilometer in Poundal Square Kilometer is equal to 32.2
1 Pound Square Kilometer = 3.2200000000497e+25 Poundal Square Nanometer1 Pound Square Kilometer in Poundal Square Nanometer is equal to 3.2200000000497e+25
1 Pound Square Kilometer = 38510879.49 Poundal Square Yard1 Pound Square Kilometer in Poundal Square Yard is equal to 38510879.49
1 Pound Square Kilometer = 49910099820.97 Poundal Square Inch1 Pound Square Kilometer in Poundal Square Inch is equal to 49910099820.97
1 Pound Square Kilometer = 346597915.42 Poundal Square Foot1 Pound Square Kilometer in Poundal Square Foot is equal to 346597915.42
1 Pound Square Kilometer = 12.43 Poundal Square Mile1 Pound Square Kilometer in Poundal Square Mile is equal to 12.43
1 Pound Square Kilometer = 7000000000 Grain Square Meter1 Pound Square Kilometer in Grain Square Meter is equal to 7000000000
1 Pound Square Kilometer = 700000000000 Grain Square Decimeter1 Pound Square Kilometer in Grain Square Decimeter is equal to 700000000000
1 Pound Square Kilometer = 70000000000000 Grain Square Centimeter1 Pound Square Kilometer in Grain Square Centimeter is equal to 70000000000000
1 Pound Square Kilometer = 7000000000000000 Grain Square Millimeter1 Pound Square Kilometer in Grain Square Millimeter is equal to 7000000000000000
1 Pound Square Kilometer = 7e+21 Grain Square Micrometer1 Pound Square Kilometer in Grain Square Micrometer is equal to 7e+21
1 Pound Square Kilometer = 7000 Grain Square Kilometer1 Pound Square Kilometer in Grain Square Kilometer is equal to 7000
1 Pound Square Kilometer = 7e+27 Grain Square Nanometer1 Pound Square Kilometer in Grain Square Nanometer is equal to 7e+27
1 Pound Square Kilometer = 8371930324.11 Grain Square Yard1 Pound Square Kilometer in Grain Square Yard is equal to 8371930324.11
1 Pound Square Kilometer = 10850021700043 Grain Square Inch1 Pound Square Kilometer in Grain Square Inch is equal to 10850021700043
1 Pound Square Kilometer = 75347372916.97 Grain Square Foot1 Pound Square Kilometer in Grain Square Foot is equal to 75347372916.97
1 Pound Square Kilometer = 2702.72 Grain Square Mile1 Pound Square Kilometer in Grain Square Mile is equal to 2702.72
1 Pound Square Kilometer = 31080.96 Slug Square Meter1 Pound Square Kilometer in Slug Square Meter is equal to 31080.96
1 Pound Square Kilometer = 3108095.64 Slug Square Decimeter1 Pound Square Kilometer in Slug Square Decimeter is equal to 3108095.64
1 Pound Square Kilometer = 310809564.27 Slug Square Centimeter1 Pound Square Kilometer in Slug Square Centimeter is equal to 310809564.27
1 Pound Square Kilometer = 31080956427 Slug Square Millimeter1 Pound Square Kilometer in Slug Square Millimeter is equal to 31080956427
1 Pound Square Kilometer = 31080956427000000 Slug Square Micrometer1 Pound Square Kilometer in Slug Square Micrometer is equal to 31080956427000000
1 Pound Square Kilometer = 0.031080956427 Slug Square Kilometer1 Pound Square Kilometer in Slug Square Kilometer is equal to 0.031080956427
1 Pound Square Kilometer = 3.1080956427e+22 Slug Square Nanometer1 Pound Square Kilometer in Slug Square Nanometer is equal to 3.1080956427e+22
1 Pound Square Kilometer = 37172.51 Slug Square Yard1 Pound Square Kilometer in Slug Square Yard is equal to 37172.51
1 Pound Square Kilometer = 48175578.81 Slug Square Inch1 Pound Square Kilometer in Slug Square Inch is equal to 48175578.81
1 Pound Square Kilometer = 334552.63 Slug Square Foot1 Pound Square Kilometer in Slug Square Foot is equal to 334552.63
1 Pound Square Kilometer = 0.012000424366029 Slug Square Mile1 Pound Square Kilometer in Slug Square Mile is equal to 0.012000424366029
1 Pound Square Kilometer = 453592370000000 Gamma Square Meter1 Pound Square Kilometer in Gamma Square Meter is equal to 453592370000000
1 Pound Square Kilometer = 45359237000000000 Gamma Square Decimeter1 Pound Square Kilometer in Gamma Square Decimeter is equal to 45359237000000000
1 Pound Square Kilometer = 4535923700000000000 Gamma Square Centimeter1 Pound Square Kilometer in Gamma Square Centimeter is equal to 4535923700000000000
1 Pound Square Kilometer = 453592370000000000000 Gamma Square Millimeter1 Pound Square Kilometer in Gamma Square Millimeter is equal to 453592370000000000000
1 Pound Square Kilometer = 4.5359237e+26 Gamma Square Micrometer1 Pound Square Kilometer in Gamma Square Micrometer is equal to 4.5359237e+26
1 Pound Square Kilometer = 453592370 Gamma Square Kilometer1 Pound Square Kilometer in Gamma Square Kilometer is equal to 453592370
1 Pound Square Kilometer = 4.5359237e+32 Gamma Square Nanometer1 Pound Square Kilometer in Gamma Square Nanometer is equal to 4.5359237e+32
1 Pound Square Kilometer = 542491959598120 Gamma Square Yard1 Pound Square Kilometer in Gamma Square Yard is equal to 542491959598120
1 Pound Square Kilometer = 703069579639160000 Gamma Square Inch1 Pound Square Kilometer in Gamma Square Inch is equal to 703069579639160000
1 Pound Square Kilometer = 4882427636383100 Gamma Square Foot1 Pound Square Kilometer in Gamma Square Foot is equal to 4882427636383100
1 Pound Square Kilometer = 175132993.16 Gamma Square Mile1 Pound Square Kilometer in Gamma Square Mile is equal to 175132993.16 | 6,296 | 24,318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-22 | latest | en | 0.846647 |
https://www.physicsforums.com/threads/what-constitutes-something-being-coordinate-free.667932/ | 1,725,963,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00421.warc.gz | 898,506,103 | 21,060 | # What Constitutes something being coordinate free
• saminator910
In summary: If you have a coordinate system, you can never be coordinate free. Coordinate free means that you don't specify a coordinate system.
saminator910
What Constitutes something being "coordinate free"
People say that exterior calculus ie. differentiating and integrating differential forms, can be done without a metric, in without specifying a certain coordinate system. I don't really get what qualifies something to be 'coordinate free', I mean in the differential forms I do, one still references components ie. x1,x2, etc., yet I never specified a metric, so is this classified as 'coordinate free'. Also, how does one do differential geometry without a coordinate system, in my mind once you don't specify a coordinate system or a metric, and things become vague, it sort of turns into differential topology, is there a 'middle ground' I am missing, keep in mind I have never taken a coarse in differential geometry. Also, in differential geometry, it has always been pertinent to give specific parametrization in order to find tangent vectors, metrics, etc.
If you use xi, that (implicitly) means that you have chosen a coordinate system, so it is not coordinate free.
To give a very simple example, consider a linear transformation (e.g. rotation) T on a vector x. You could write this in coordinate-free form as
x' = T x
This does not depend on which basis you choose for the space that x lives in - it just means: apply this transformation.
When you calculate the result on a vector, you usually pick a coordinate system by choosing a set of basis vectors (x, y, z-axis) and write the action of T as a matrix M.
You then calculate
$$\mathbf{x'}_i = \sum_{j = 1}^n M_{ij} x_j$$
This is not coordinate-free, because both the components of x and x' as well as the entries of M depend on the coordinate system.
The advantage of the coordinate-free form is that it looks the same in any choice of basis. If you and I both chose a different coordinate system and wrote down M, we would get two different bunches of numbers but it would not be immediately clear that we're talking about the same "physical" operation.
Another example: Let X be the set of all quadratic polynomials from R to R and define T by T(f)= df/dx+ f. That is a "coordinate free" definition. If I had chosen $\{1, x, x^2\}$ as basis (essentially choosing a "coordinate system" by choosing a basis), say that T(1)= 1, T(x)= x+ 1, T(x2)= x2+ 2x, then say T is "extended by linearity", T(af+ bg)= aT(f)+ bT(g), that is not "coordinaate free" because I have used a coordinate system (a basis) to define it. Or course, those are exactly the same definition.
It would be interesting to sort out the distinction between "having a coordinate system" and "having a metric". The two definitions are not identical, but what common situations allow us to proceed from having one to having the other?
Aren't they two completely different things?
Sure, for a lot of "common" metric spaces the metric is defined in terms of coordinates. But for Rn, for example, that's mostly because people usually learn Pythagoras before inner products, so
$$\sum_i (y_i - x_i)^2$$
is a little more intuitive than
$$\langle \vec y - \vec x, \vec y - \vec x\rangle$$
CompuChip said:
Aren't they two completely different things?
Of course they are, that's why sorting out their relation is complicated. The idea of a metric on a set of things is standardized, but I'm not sure whether there is an standard definition for a set of things to "have a coordinate system".
The pythagorean idea won't necessarily work for a coordinate system where the same thing can have two different coordinates (as is the case in the polar coordinate system).
Okay so, your standard differential form, written with coordinates say for example a standard 1 form, $\alpha = \sum^{n}_{i=1}f_{i}du_{i}$, you are still referencing a it's local coordinates, but you don't necessarily need to know what the metric is ie. euclidean space vs. it being embedded in some other manifold, so the coordinates don't necessarily need any intrinsic value. then if you know the $u_{i}$ coordinates in terms of euclidean or other coordinates, you pullback/pushforward the form. Does this count as 'coordinate free' since you don't specify a basis: the coordinates in the form aren't in terms of anything.
Stephen Tashi said:
It would be interesting to sort out the distinction between "having a coordinate system" and "having a metric". The two definitions are not identical, but what common situations allow us to proceed from having one to having the other?
Once we have a coordinate system, so that every point, x, can be identified with $(x_1, x_2, ..., x_n)$, there is a "natural" metric, $d(x,y)= \sqrt{(x_1-y_1)^2+ (x_2- y_2)^2+ ...+ (x_n- y_n)^2}$.
Given a metric space, even finite dimensional $R^n$, we would need a choice of "origin", and n-1 "directions" for the coordinate axes (after choosing n-1 coordinate directions, the last is fixed) in order to have a coordinate system. So a "coordinate system" is much more restrictive, and stronger, than a "metric space".
HallsofIvy said:
Once we have a coordinate system, so that every point, x, can be identified with $(x_1, x_2, ..., x_n)$, there is a "natural" metric, $d(x,y)= \sqrt{(x_1-y_1)^2+ (x_2- y_2)^2+ ...+ (x_n- y_n)^2}$.
But does the definition of a "coordinate system" ( if there is such a standard definition) include the idea that each element of "the space" can be identified with a unique finite vector of coordinates? And must the vector consist of a vector of real numbers?
okay, so now can someone give an example of using a differential form without a metric?
## What does it mean for something to be coordinate free?
Being coordinate free means that a mathematical or scientific concept or equation does not rely on a specific set of coordinates or reference frame to be defined or calculated.
## Why is it important for something to be coordinate free?
Having a coordinate free concept or equation allows for more general and universal applicability, as it is not limited to a specific coordinate system. This makes it easier to apply to different scenarios and simplifies calculations.
## How can I tell if something is coordinate free?
One way to determine if something is coordinate free is to see if it can be expressed using only mathematical objects or operations that are invariant under coordinate transformations. Another way is to see if it remains valid and applicable in different coordinate systems.
## What are some examples of coordinate free concepts or equations?
Some common examples of coordinate free concepts include vector spaces, tensors, and differential forms. Specific equations that are coordinate free include Maxwell's equations in electromagnetism and Einstein's field equations in general relativity.
## What are the benefits of using coordinate free concepts?
Using coordinate free concepts allows for a more elegant and concise formulation of mathematical and scientific ideas. It also makes it easier to generalize and extend these concepts to new situations without needing to change the underlying equations. Additionally, it can help identify and eliminate errors caused by coordinate system dependencies.
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991 | 1,757 | 7,530 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-38 | latest | en | 0.89651 |
https://forums.tesla.com/en_AU/forum/forums/model-x-dual-motor-allwheel-drive-does-mean-car-will-have-2-identical-motors-so-same-mo | 1,490,449,430,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188926.39/warc/CC-MAIN-20170322212948-00641-ip-10-233-31-227.ec2.internal.warc.gz | 783,328,222 | 12,200 | Forums
Model X with Dual Motor All-Wheel Drive. Does this mean that this car will have 2 identical motors, so same motor front as rear?
The main question is will these 2 electric motors be identical (and therefore, will they have the same capacity) or will they not be identical (and therefore, if they are different, will they not have the same capacity)? What will it be? Does anyone know anything about this?
cprenzl | 31 January, 2013
Go watch the Model X reveal. Should answer that better but, short answer: No
Benz | 31 January, 2013
@ cprenzl
I have just been watching the video of the reveal of the Tesla Model X. I have been listening to what Elon Musk has told about the Dual Motor AWD. The only interesting thing about the Dual Motor AWD that he mentioned was that "you can dynamically shift power from front to rear, depending upon what has got the most traction". But he did not say anything about the type of those two motors, and if they were identical or not. So, after having watched this video of the release of the Tesla Model X, my question has not been answered. But I still am interested why your answer is: "No"?
llehctim | 31 January, 2013
"The second motor enables more than all-weather, all-road capabilities: it increases torque by 50%."
I don't know much about the science of torque versus wattage of motors, but this suggests to me that the front motor is half as strong.
Alternatively the rear motor could be smaller for the AWD than the 2WD and still give rise to that statement.
Under violent acceleration (or steep uphill) I expect most of the weight to be on the rear wheels. Hence if Tesla is optimizing for 0-60 times then I would have the expectation of the weaker front motor.
My wild guesses/speculation as to what Tesla would use to maximize reuse from Model S of motors and inverters... assuming X is ~10% heavier than S
Model S motor/ | 0-60 | Model X | 0-60
inverter | | |
---------------------------------------
S 60 5.9 60 6.5
S 85 5.6 85 opt1 6.2 (likely)
P 85 4.4 85 opt2 4.9 (unlikely)
S85 mot/inv + weak front 85 AWD 5.6*1.1*2/3=4.1... with a weaker inverter maybe makes this one ~4.4
P85 mot/inv + weak front P85AWD 4.4*1.1*2/3=3.3 !
The 2/3 is based on the 50% extra torque statement.
Tesla was also conservative about S 0-60 times from what we see users/reviewers reporting.
Note with the 4WD options I have no idea how the inverters would work (ianaee) and these may be limited by the batteries instead of inverters/motors.
Timo | 31 January, 2013
It is also possible that battery just can't deliver enough power that two motors can get twice as much torque, but still have twice as much torque when not used at full power.
Or alternatively, if the torque is measured at the wheels, then reduction gear is lower: instead of "fixed gear with 9.73:1 reduction ratio" you have "fixed gear with 7.3:1 reduction ratio". That would allow higher top speed because torque drop and the absolute limit of RPM comes in play at higher speeds.
600Nm*9.73 = 5838Nm, 2*600Nm*7.3 = 8760N, 8760Nm/5838Nm = 1.50.
5838Nm /2 = 2919Nm/wheel, 8760Nm /4 = 2190Nm/wheel (still happily burns rubber with TC off).
Truth is probably somewhere between those two.
Benz | 1 February, 2013
Can we forward this question to the people from Tesla Motors? Maybe they can shine some more light on this matter?
RDoc | 4 February, 2013
Looking at the Model X platform photo on the Tesla site, the one with no body, it looks to me like the front motor is half the rear. The rear motor actually has 2 motors feeding the power train. It looks to me like the front motor is just one of those.
Benz | 4 February, 2013
I have seen that photo particular photo. But I do not have the knowledge to say something like that.
Brian H | 4 February, 2013
No, that "2nd motor" in the rear is actually the PEM, the controller and DC/AC inverter for the motor(s). Only one is needed to handle the front and rear.
Benz | 5 February, 2013
@ Brian H
So, do you know the answer to my question (are both motors identical or not)?
Brian H | 5 February, 2013
No, deep proprietary corporate secret. Occam's Design Razor suggests the differences would be in gearing and the controller, though.
Brian H | 5 February, 2013
Occam's Design Razor: Do not multiply devices and design details beyond necessity.
scottf200 | 6 February, 2013
This video indicate they would be used for different purposes as well.
Benz | 6 February, 2013
I have watched the video. And I have understood that with a second motor you add more mass to the EV. But this does NOT result in a lower range!!!! That's very interesting. And the two motors will not always be working equally hard. They will be working individually depending on the several different driving situations/conditions. That's also very interesting to know. I think that I am slowly getting more and more positively attracted to this dual motor AWD option. It sounds great, but still I would like to test drive a Tesla Model X AWD. Not before September, I am afraid.
Brian H | 6 February, 2013
The motors are light, about 130 lb for the single MS motor. The two in the AWD could be each lighter than that. IAC, not significant in the total car weight ... a rounding error, basically.
Timo | 6 February, 2013
My cautious estimate for entire drivetrain minus battery is 150kg, so with 150kg added to already over 2100 kg car would be adding about 7% weight. That would result in about 2-3% less range, which is small enough that efficiency difference caused by load difference between two and one motors could negate that entirely.
Brian H | 6 February, 2013
Adding one motor does not duplicate the drivetrain. I doubt even 150 lbs. would be added.
Vawlkus | 6 February, 2013
Umm..... actually Brian, it WOULD duplicate the drivetrain, or else the second motor wouldn't be able to do very much :P
Don't forget, the front wheels have to pivot, so the drivetrain must be modified to allow for that steering. If anything, I think the front drivetrain would weigh a little more than the rear one does, just for that alone.
Timo | 6 February, 2013
Also heavy duty cabling (around 1000A cables are not light) and liquid cooling would need to be added in addition to more complex axle system. Only thing that doesn't add itself there is the PEM assuming single PEM can handle two motors. If not then that too would need to be added.
Brian H | 6 February, 2013
Both motors would be able to be lighter, and the photo referred to by RDoc above shows no PEM up front. The current motor reputedly weighs 130 lb. So where is this 150 kg. (330 lbs) coming from? That's at least 200 lb. too high.
Benz | 6 February, 2013
@ Timo
"About 2-3% less range" is good.
But there sure is another thing: Air Drag. The Tesla Model S has an air drag figure of 0.24 (that is extremely low, for this kind of vehicle). We all know that the Tesla Model X is less earodynamic than the Tesla Model S. And this is also going to result into a lesser range. But will that figure also be around 3% or is 10% more likely. I really do not have any clue on this issue. But I sure would like to know more about it. Anyone?
Timo | 7 February, 2013
Brian H; the motor alone is only small portion of the whole drivetrain assembly. I would assume transmission casing alone weights 50lbs, structures required to fix that motor in place (so that torque doesn't rip it loose), axles, cables, liquid cooling fluids and tubes would easily add up to 100 lbs if not more. Add in 130 lbs from the motor and you have that around 300 lbs. | 1,929 | 7,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-13 | latest | en | 0.953464 |
https://www.excelhow.net/tag/excel-and | 1,685,505,062,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00236.warc.gz | 806,079,497 | 52,018 | ## How to Statistic Pass 4 Out of 5 Rounds in a Competition by Excel Functions
Suppose you are counting the numbers of participants who are the winners in a competition. And the passing line is provided as pass 4 out of 5 rounds of games. You need to know if participant pass the game or not firstly. If you are familiar with the functions in Excel and can use them expertly, you can work out this issue easily. By the way, if you are confused of choosing and applying which function among all Excel functions, this article will help you to solve your problem.
In this article, through explaining the example below, we will introduce you to apply IF and COUNTIF functions together to return win or lose for each participant. You can get the basic knowledge and usage of these two functions in this article.
## EXAMPLE
In this case, we want to know if participants win or lose the game. “Win” is returned if participant passed four rounds of games with a high score; otherwise “Failed” is returned.
In this article, to approach our goal, we applied IF function with COUNTIF function inside.
## SOLUTION
In this instance, the final output in “Win of Lose” column is “Win” or “False”, the result depends on if participants passed any four rounds of games with high a high score. To output either A or B based on one condition, we can apply IF function here. This function is frequently used in the situations like ‘to return A or B by a given precondition”.
To count number of rounds with a high score, we can select COUNTIF function as it can count with one condition properly.
Besides, COUNTIF has some related functions like COUNT and COUNTIFS. COUNT function returns number of cells which contain a numeric value; COUNTIF is COUNT+IF, it returns number of cells with one condition or criterion; if there are multiple criteria, we can apply COUNTIFS.
## FORMULA with IF & COUNTIF FUNCTION
In H2, we input the formula =IF(COUNTIF(B2:F2,”>7″)>=4,”Win”,”Lose”). After typing, press Enter, verify that “Win” is displayed in H2. As Ada has a high score in round-1, round-2, round-4 and round-5, she passed the game with 4-rounds high scores, so “Win” is returned for her.
Drag down handle to fill other cells with the same formula in “Win or Lose” column.
Michelle lose the game as she only passes the round-3 and round-5 with a high score.
## FUNCTION INTRODUCTION
In this instance, returning “Win” or “False” depends on the result of whether participants passed four rounds of games with a high score, the final output “Win” of “False” comes from IF function. To determine if participants passed the game with 4 or more high scores, we applied COUNTIF function here.
a. COUNTIF returns the number of cells which contain a number and the number meets one condition or criterion. The condition or criterion is determined by our input in “criteria” argument.
Syntax:
=COUNTIF(range, criteria)
Example.
b. IF function returns “true value” or “false value” based on the result of provided logical test. It is one of the most popular function in Excel.
Syntax:
=IF(logical_test,[value_if_true],[value_if_false])
Example.
## FORMULA EXPLANATION
=IF(COUNTIF(B2:F2,">7")>=4,"Win","Lose")
// logical test is “COUNTIF(B2:F2,”>7″)>=4”
// value of true is “Win”
// value of false is “Lose”
c. COUNTIF(B2:F2,”>7″) is used for counting the number of cells from range B2:F2 with the condition number in cell should be “greater than 7”. In this case, this expression is used for counting the rounds number with a high score 8, 9 or 10 for Ada. Expand values in B2:F2, we get below expression:
COUNTIF({10,9,7,8,10},”>7″) // returns 4
d. COUNTIF returned value “4” comes to the logical expression “COUNTIF(B2:F2,”>7″)>=4” inside IF function. In this logical test “4>=4” is true, so the value of true “Win” is returned by this formula finally.
=IF(4>=4,”Win”,”Lose”) // returns “Win”
## EXPAND
If we add two new conditions that round-1 and round-2 must pass with high score, current formula cannot satisfy our demand. Now, the passing line for “Win” is updated to must meet below three conditions:
1. The initial condition “COUNTIF(B2:F2,”>7″)>=4”.
2. Round-1 has a high score “B2>7”.
3. Round-2 has a high score “C2>7”.
Here, we can apply another function AND to concentrate the three conditions in one expression. AND function can return a “TRUE” if all conditions are met.
Update formula to =IF(AND(COUNTIFS(B2:F2,”>7″)>=4,B2>7,C2>7),”Win”,”Lose”).
### Related Functions
• Excel COUNTIF function
The Excel COUNTIF function will count the number of cells in a range that meet a given criteria. This function can be used to count the different kinds of cells with number, date, text values, blank, non-blanks, or containing specific characters.etc.= COUNTIF (range, criteria)…
• Excel COUNTIFS function
The Excel COUNTIFS function returns the count of cells in a range that meet one or more criteria. The syntax of the COUNTIFS function is as below:= COUNTIFS(criteria_range1, criteria1, [criteria_range2, criteria2]…)…
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.The syntax of the IF function is as below:= IF (condition, [true_value], [false_value])….
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…)…
## How to Calculate Weekly Average by Formula in Excel
In our daily life we often create tables for recording the money we spend for lunch or dinner or something else per date. This can help us know the total cost or average cost per date or other unit clearly via some statistic related functions in excel, and then we can cut back on unnecessary expenses and save our money.
Normally, one method for statistic in excel is creating pivot table. we can create a pivot table to auto calculate the total cost or average cost base on the values recorded in original table, and we can do filter by date or month or quarter to show different average cost for different period. Details see below screenshot.
Obviously, Week is not listed in default grouping date list.
If you want to know the method for counting average data by week (without using pivot table), you can read below article, it can help you to solve your question.
## Calculate Weekly Average by Formula in Excel
In Pre-Condition:
Prepare a simple table for demonstration. See screenshot below. You can see that we spend 10 or 5 or other values for lunch on different dates, some dates are included in one week. We want to count the weekly average for spend.
Step 1: Create another table with three columns Week No., Year and Average. This table is used for recording the week number, year and the weekly average for launch spend.
Step 2: In E2, enter the formula =WEEKNUM(C2,2).
For function WEEKNUM, we set the second parameter as 2, that means week begins on Monday.
Step 3: Click Enter to get returned value. We get 2 in this case. That means the date belongs to the second week in year 2019.
Step 4: In F2, enter the formula =Year(C2). We can get Year 2019.
Step 5: In G2, enter the formula =IF(AND(E2=E1,F2=F1),””,AVERAGEIFS(\$B\$2:\$B\$10,\$E\$2:\$E\$10,E2,\$F\$2:\$F\$10,F2)). Then we get the returned value 10.
Comment: In this formula, we use AVERAGEIFS function to calculate the average by specific criteria, and use IF function to classify dates by week and then get according average. If (E2=E1,F2=F1) is true, then we get blank value, otherwise we will get the weekly average.
Step 6: Drag the fill handle to fill the following cells in the table. We get weekly average values for all weeks.
Comment: If you want to ignore Year, you can just enter the formula =AVERAGEIFS(\$B\$2:\$B\$10,\$E\$2:\$E\$10,E2). Then we can get below table. You can depend on your requirement to determine whether add or remove criteria.
### Related Functions
• Excel WEEKNUM function
The Excel WEEKNUM function returns the week number of a specific date, and the returned value is ranging from 1 to 53.The syntax of the WEEKNUM function is as below:=WEEKNUM (serial_number,[return_type])…
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.The syntax of the IF function is as below:= IF (condition, [true_value], [false_value])….
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…)…
## How to Check If value Is between Two Numbers in Excel
This post will guide you how to check if a cell value is between two numbers in Excel. How do I build an IF statement that test if a cell value is between two given values in formula in Excel.
## Check Value If It Is between Two Numbers
Assuming that you have list of data in range A1:C4, which contain numbers. And you need to test if a numeric value in Cell C2 is between two numbers in Range: A2:B2. How to do it. You can use IF function and the AND function to build an IF statement to check values. Like this:
=IF(AND(C2>A2,C2<B2),"True","False")
Type this formula into a blank cell which you want to place the result, and press Enter key on your keyboard. Then drag the AutoFill Handle down to cell C4 to apply this formula.
Note: this formula is only valid when the Cell value in A2 is smaller than the cell value in B2.
You can also use another formula based on the AND function, the MIN function and the MAX function to achieve the same result. Like this:
=AND(C2>MIN(A2,B2),C2<MAX(A2,B2))
### Related Functions
• Excel MIN function
The Excel MIN function returns the smallest numeric value from the numbers that you provided. Or returns the smallest value in the array.The MIN function is a build-in function in Microsoft Excel and it is categorized as a Statistical Function.The syntax of the MIN function is as below:= MIN(num1,[num2,…numn])….
• Excel MAX function
The Excel MAX function returns the largest numeric value from the numbers that you provided. Or returns the largest value in the array.= MAX(num1,[num2,…numn])…
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.The syntax of the IF function is as below:= IF (condition, [true_value], [false_value])….
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…)…
## Ignoring Blank or Zero Cells with Conditional formatting
This post will guide you how to make conditional formatting ignore blank cells or zero cell in Excel. How do I force blank cells or zero cells to be ignored in conditional formatting in Excel. How to ignore blank cells or blank cells when applying conditional formatting in Excel.
## Ignoring Blank with Conditional Formatting
When you create conditional formatting rules for a list of selected data, and you want to ignore all Blank cells, How to achieve it. Just do the following steps:
#1 select the source data that you want to apply the conditional Formatting.
#2 go to HOME tab, click Conditional Formatting command under Styles group. And select New Rule from the popup menu list. And the New Formatting Rule dialog will open.
#3 select Use a formula to determine which cells to format in the Select a Rule Type section.
#4 type the following formula into the Format values where this formula is true text box. Click Ok button.
=ISBLANK(B1)=TRUE
Note: the B1 is the first cell of the selected range of cells.
## Ignoring zero with Conditional Formatting
If you want to ignore zero values with conditional Formatting, you can do the following steps:
#1 select the range of cells C2:C5
#2 go to HOME tab, click Conditional Formatting command under Styles group, and select New Rule, and the New Formatting Rule dialog will open.
#3 select Use a formula to determine which cells to format in the Select a Rule Type section.
#4 type the following formula into the Format values where this formula is true text box.
=AND(C2<>0,C2<=SMALL(IF(C\$2:C\$5<>0,\$C\$2:\$C\$5),2))
#5 click Format button, and switch to Fill tab in Format Cells dialog, select one color as the background color. Click Ok button.
#6 click Ok button. You will see that all zero values are ignored by conditional formatting.
### Related Functions
• Excel ISBLANK function
The Excel ISBLANK function returns TRUE if the value is blank or null.The syntax of the ISBLANK function is as below:= ISBLANK (value)…
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.The syntax of the IF function is as below:= IF (condition, [true_value], [false_value])….
• Excel SMALL function
• The Excel SMALL function returns the smallest numeric value from the numbers that you provided. Or returns the smallest value in the array.The syntax of the SMALL function is as below:=SMALL(array,nth) …
• Excel AND function
• The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…)…
## Returning Value if the Dates Fall between Two Dates
This post will guide you how to calculate value between two dates if the dates falls those two dates in Excel. How do return a value if the dates fall between two given dates in excel. How to use a formula to determine whether a date falls between two dates in Excel. How to check if a date falls between a range of two dates in Excel.
## Determine Whether a Date Falls between Two Dates
Assuming that you have a list of data that contain dates in one column or range, and you want to check those dates if fall between two given dates in range C1:D1. How to achieve it. You can create a formula based on the IF function, and the AND function. Like this:
=IF(AND(A1>\$C\$1,A1<\$D\$1),TRUE,FALSE)
Type this formula into Cell B1, and press Enter key in your keyboard, and then drag the AutoFill Handle from Cell B1 to B5.
If the data falls between the given two dates, and this formula return TRUE, otherwise, return FALSE.
### Related Functions
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.The syntax of the IF function is as below:= IF (condition, [true_value], [false_value])….
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
## Check If Multiple Cells are Equal In Excel
This post will guide you how to check if the values of multiple cells are equal with a formula in Excel. How do I verify that multiple cellsare the same in Excel. How to compare three or more cells in Excel to see if they are the same with a formula.
## Check If Multiple Cells are Equal
Assuming that you have a list of data in range A1:C1, and you want to compare if these cells are equal, if so, then return True, otherwise, return False. How to achieve it.
You need to create an Excel array formula based on the AND function and the EXACT function. Just like this:
=AND(EXACT(A1:C1,A1))
Type this formula into a blank cell, and press Ctrl +Shift +Enter shortcuts in your keyboard.
There is another formula based on the COUNTIF function to achieve the same result. Like this:
=COUNTIF(A1:C1,A1)=3
### Related Functions
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel COUNTIF function
The Excel COUNTIF function will count the number of cells in a range that meet a given criteria. This function can be used to count the different kinds of cells with number, date, text values, blank, non-blanks, or containing specific characters.etc.= COUNTIF (range, criteria)…
• Excel EXACT function
The Excel EXACT function compares if two text strings are the same and returns TRUE if they are the same, Or, it will return FALSE.The syntax of the EXACT function is as below:= EXACT (text1,text2)…
## Find Duplicate Values in Two Columns
This tutorial will show you how to find duplicate values in two given columns in Excel. Or How to compare two columns and find duplicate entries in Microsoft Excel worksheet. And you will learn two methods to compare columns so that the duplicate values can be highlighted or listed in a range of Cells.
Assuming that you need to compare two columns (Column A and Column B) to get the duplicate values in your worksheet, you can find duplicate values in two columns with Excel Formula, or Excel VBA Macro code.
## Method 1: Find duplicate values in two columns with Excel Formula
To compare two given columns and find duplicate values in Excel, you can use a combination of the IF function, the ISERROR function, and the MATCH function to create an Excel formula. So you can use the following formula:
=IF(ISERROR(MATCH(A1,\$B\$1:\$B\$4,0))," ",A1)
Now you need to type this formula in Cell C1, press Enter key, drag AutoFill Handle down to Cell C2:C4, you will see all of the duplicated values are displayed in Column C.
## Method 2: Find duplicate values in two columns with VBA Macro code
If you are familiar with the programming language and you can use a Visual Basic Macro to compare the value in two columns and then find duplicate values, just refer to the following steps:
1# click on “Visual Basic” command under DEVELOPER Tab.
2# then the “Visual Basic Editor” window will appear.
3# click “Insert” ->”Module” to create a new module
4# paste the below VBA code into the code window. Then clicking “Save” button.
Sub FindUplicatesinTwoColumns()
Set Range1 = Application.Selection
Set Range1 = Application.InputBox("Select the first range in one column:", "FindUplicatesinTwoColumns", Range1.Address, Type:=8)
Set Range2 = Application.InputBox("Select the second range in another column:", "FindUplicatesinTwoColumns", Type:=8)
For Each R1 In Range1
xValue = R1.Value
For Each R2 In Range2
If xValue = R2.Value Then
If R3 Is Nothing Then
Set R3 = R1
Else
Set R3 = Application.Union(R3, R1)
End If
End If
Next
Next
R3.Interior.ColorIndex=3
End Sub
5# back to the current worksheet, then run the above excel macro.
6# Select the first range A1:A4 in Column A, click OK button, then select the second range B1:B4 in Column B, click OK button.
7# let’s see the result:
## Method 3: Find duplicate values in two columns with Conditional Formatting feature
You can use conditional formatting with on a formula based on the COUNTIF function and the AND function to find the duplicate values in two specified columns and then highlighted them. Just do it following:
1# Select the entire Column A via click on the Column Header and then the column A will be highlighted
2# on the HOME tab, click the Conditional Formatting command under Styles group. Then select New Rules… from the drop-down menu list.
3# select Use a formula to determine which cells to format as Rule Type in the New Formatting Rule window
4# Type the following formula in the Format values where this formula is true: box
=COUNTIf(\$B:\$B, \$A1)
5# click the Format… button, then the Format Cells window will appear.
6# in the “Format Cells” window, switch to the Fill tab, choose the background color, and then click OK button.
7#you will be back to the New Formatting Rule windows and you can check a preview of the formatting you have selected. Then click OK button.
Then the conditional formatting rule will be applied to all values in two columns and highlighted the duplicate values.
### Related Functions
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.The syntax of the IF function is as below:= IF (condition, [true_value], [false_value])….
• Excel ISERROR function
The Excel ISERROR function used to check for any error type that excel generates and it returns TRUE for any error type, and the ISERR function also can be checked for error values except #N/A error, it returns TRUE while the error is #N/A. The syntax of the ISERROR function is as below:= ISERROR (value)….
• Excel MATCH function
The Excel MATCH function search a value in an array and returns the position of that item.The syntax of the MATCH function is as below:= MATCH (lookup_value, lookup_array, [match_type])….
• Excel COUNTIF function
The Excel COUNTIF function will count the number of cells in a range that meet a given criteria. This function can be used to count the different kinds of cells with number, date, text values, blank, non-blanks, or containing specific characters.etc.= COUNTIF (range, criteria)…
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…)…
### Related Posts
• Highlight Rows
You will learn that how to change the color of the entire rows if the value of cells in a specified column meets your conditions, such as, if the value of cells is equal to or greater than a certain number or text values, then excel should be highlight entire rows or change a row color as you need.…
• Find Duplicate Rows
If you want to check the entire row that duplicated or not, if True, then returns “duplicates” value, otherwise, returns “no duplicates”. You can create a formula based on the IF function and the SUMPRODUCT function..…
• Highlight Duplicate Rows
this post will talk that how to highlight entire rows that are duplicates in excel 2016, 2013 or lower version. Or how to change the background color of duplicate rows..…
• Highlight duplicate values
this post will teach you how to highlight duplicate values in the range of cells in excel. Normally, you may be need to identify duplicate values with a range of cells in Excel. And there is one of the fasted way that is using conditional formatting feature in Microsoft Excel……
## Check if Cell Contains Certain Values but do not Contain Others Values
In the previous post, we only talked that how to check a cell if contains one of several values from a range in excel. And this post explains that how to check a cell if it contains certain values or contains one of several values but do not contain other certain values in another range or a list in excel.
## Check if Cell Contains Certain Values but do not Contain Others Values
Supposing that you have a list of text strings in the range B1:B3 and you need to check each Cell that if it contains one of several values in a range D2:D4 but do not contains any of values in range E2:E3. If TRUE, then returns TRUE value, otherwise, returns FALSE value.
To check a cell to see if it contains certain string but do not contain others. You can use a combination of the AND function, the COUNT function and the SEARCH function to create a new array formula as follows:
=AND(COUNT(SEARCH(\$D\$2:\$D\$4,B1))>0, COUNT(SEARCH(\$E\$2:\$E\$3,B1))=0 )
Let’s see how this formula works:
= SEARCH(\$D\$2:\$D\$4,B1)
The SEACH function returns position of the first character of find_text in a text string. And this formula will search each value in the range D2:D4 inside within_text in Cell B1, then returns position of each text string in Cell B1, so it will return an array result like this:
{1;7;12}
= SEARCH(\$E\$2:\$E\$3,B1)
This SEARCH formula will search each value in range E2:E3 inside within_text in Cell B1, and then returns position of the first position of each find_text in Cell B1. When no match is found, the SEARCH function will return the #VALUE error. So it will return an array result like this:
{#VALUE!;#VALUE!;1}
=COUNT(SEARCH(\$D\$2:\$D\$4,B1))>0
The COUNT function will count the number of cells that contain numbers in an array returned by the SEARCH function. If it is greater than 0, then it indicated that at least one values is found in Cell B1.
=COUNT(SEARCH(\$E\$2:\$E\$4,B1))=0
If the number of cells that contain numbers is equal to 0, then it means that none of values can be found in Cell B1.
=AND(COUNT(SEARCH(\$D\$2:\$D\$4,B1))>0, COUNT(SEARCH(\$E\$2:\$E\$4,B1))=0 )
If both two COUNT function return TRUE value, the AND function returns TRUE. And if either COUNT function is FALSE, this formula returns FALSE.
### Related Functions
• Excel SEARCH function
The Excel SEARCH function returns the number of the starting location of a substring in a text string.The syntax of the SEARCH function is as below:= SEARCH (find_text, within_text,[start_num])…
• Excel COUNT function
The Excel COUNT function counts the number of cells that contain numbers, and counts numbers within the list of arguments. It returns a numeric value that indicate the number of cells that contain numbers in a range…
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…)…
## Excel IF Function With Numbers
Normally, If you want to write an IF formula for numbers in combining with the logical operators in excel, such as: greater than, less than, greater than or equal to, less than or equal to, equal to, not equal to.etc.
The generic IF formula with logical operator for number values is as below:
## Excel IF formula for range of numbers
If you want to check if a cell values is between two values or checking for the range of numbers or multiple values in cells, at this time, we need to use AND or OR logical function in combination with the logical operator and IF function.
For example, you need to check if the value in cell B1 is between values in Cell A1 and A3, then we can write down the following excel IF formula:
We can enter the above formula into the formula bar in the cell C1 and then press the Enter key. And rag the Fill Handle to the range C1:C3.it will apply for the other cells for this formula.
## Excel IF formula for negative numbers
If you want to create an IF formula to display some text string in cells if the value in cell falls between the range negative numbers -10 to positive numbers 10.
Actually, whatever negative numbers or positive numbers, we still can use AND or OR logical function to create a logical test in IF formula. we can write down the following IF formula:
Also, we can use another excel function ABS() to achieve the same results. Let’s see the below IF formula in combination with ABS function.
### Related Formulas
• Excel IF formula with operator : greater than,less than
Now we can use the above logical test with greater than operator. Let’s see the below generic if formula with greater than operator:=IF(A1>10,”excelhow.net”,”google”) …
• Excel IF function with text values
If you want to write an IF formula for text values in combining with the below two logical operators in excel, such as: “equal to” or “not equal to”.
• Excel IF function with Dates
If you have a list of dates and then want to compare to these dates with a specified date to check if those dates is greater than or less than that specified date. …
• Excel IF Function With Numbers
If you want to check if a cell values is between two values or checking for the range of numbers or multiple values in cells, at this time, we need to use AND or OR logical function in combination with the logical operator and IF function…
### Related Functions
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE….
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
• Excel ABS function
The Excel ABS function returns the absolute value of a number
## Excel IF formula with NOT logical function
This post will guide you how to use excel if formula with NOT logical function in Microsoft excel.
## NOT Logical Function
The syntax of NOT logical function is as follow:
=NOT(logical_value)
The NOT logical function will return the reversed logical value. If the logical_value is TRUE, then FALSE is returned. If the logical_value is FALSE, then TRUE is returned. Let’s see the below two examples:
=NOT(FALSE)
=NOT(2+1=3)
Both of the above excel formulas will return TRUE.
## Excel IF Formula Combining With NOT
If you want to check several test conditions in an excel formula,then take different actions. You can use NOT function in combination with the AND or OR logical function in excel IF function.
Let’s see the following excel if formula:
=IF(NOT(OR(B1="red",B1="black")),TRUE,FALSE)
In the above formula, the excel formula will exclude both red and blank colors. It will check cells in column B if the values is NOT red or blank. TRUE or FALSE will be returned.
Actually, we also can use the Not equal to (<>) logical operator to achieve the same requirement , For example, we can write down the following excel IF formula to reflect the above logic test.
=IF(AND(B1<>red,B1<>black)),TRUE,FALSE)
## NOT Function combining with ISBLANK function
As you know, the function ISBLANK(B1) will return TRUE if the cell B1 is blank. If we use NOT function to enclose ISBLANK(B1) function, it will reverse the result, if the cell is blank, then returns FALSE.
Let’s create a nested IF statement with the NOT and ISBLANK functions as follow:
=IF(NOT(ISBLANK(B1)),B1*5,””)
For above IF statement, IF the cell B1 is not blank, then multiply the number in B1 by 5.we can enter the above formula into cell C1, and then drag the Fill Handle to the range C1:C3.it will apply for the other cells for this formula.
### Related Functions
• Excel OR function
The Excel OR function returns TRUE if any of the conditions are TRUE in logic test. Otherwise, it returns FALSE.
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE….
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
• Excel ISBLANK function
The Excel ISBLANK function returns TRUE if the value is blank or null.The syntax of the ISBLANK function is as below:= ISBLANK (value)…
• Excel NOT function
The Excel NOT function returns the opposite of a given logical or Boolean value. For example, if you supplied with the value TRUE, the NOT function will return FALSE; If you supplied with the value FALSE, and the NOT function will TRUE. The syntax of the NOT function is as below:=NOT(logical)…
## Excel IF formula with operator : greater than,less than
This post will guide you how to use if function with “greater than”, “greater than or equal to”, ”less than” and “less than or equal to” logical operators in excel.
## Greater than, Greater than or equal to, Less than, less than or equal to
The “greater than”(>)operator will compare the size of two different values in cells and then return TRUE if the first value in cells compared is larger than the second values in cells. FALSE if they are not.
The “greater than or equal to” (>=) operator will return TRUE if the first value in cells is larger than the second or if the two values are equal.
The “Less than” operator returns TRUE if the first value in cell is smaller than the second value in cells.
The “Less than or equal to” operator returns TRUE if the first value in cell is smaller than the second of the two values are equal.
Those excel comparison operators mostly used to compare numbers, date and time values.
Also, we can use those operators to compare text string, and it will compare the value of the first letters in the text string. And if the first letters are same, it will compare the second letters and so on.
Let’s see the below excel if formula with greater than operator:
Formula example Description =A1>10 Returns True if a number value in cell A1 is greater than 10, FALSE if they are not. =A1>=(B1*2) Returns True if a number value in cell A1 is greater than or equal to the value of B1*2, FALSE if they are not. =A1
## The Generic IF formula with Greater than operator
Now we can use the above logical test with greater than operator. Let’s see the below generic if formula with greater than operator:
If you want to take the different action when a value in cells is greater than a certain value (it can be number, text, date or another function), then you can use the IF function with greater than operator to make a logical test and return one value if the logical test condition is true, or return another value if the logical test condition is false.
For the above IF formula, it just tests the value in cell A1 to check if the value is greater than 10. If TRUE, the IF formula will return “excelhow.net”. If FALSE, the IF function will return “google”.
You might want to return “excelhow.net” if the test condition is TRUE. Or return nothing if the test returns FALSE. You can use an empty string (“”) as False value in the IF function, see below if formula:
=IF(A1>10,”excelhow.net”,””)
## The Generic IF formula with Greater than operator
### Related Functions
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE….
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
• Excel DATEVALUE Function
The Excel DATEVALUE function returns the serial number of date. And it can be used to convert a date represented as text format into a serial number that recognizes as a date format.The syntax of the DATEVALUE function is as below:=DATEVALUE(date_text)…
## Excel IF formula with AND & OR logical functions
If you want to test the result of cells based on several sets of multiple test conditions, you can use the IF function with the AND and OR functions at a time.
Assuming that you have the following test logic to check the exam scores:
Logic test1: 60<=B1<=70
Logic test2: 60<=C1<=70
If any of the above conditions are met, then mark the result as “good”, otherwise, mark the result as “bad”. We need to use two AND functions to reflect the above two logical test and enclose them in the OR function as one logical test in the Excel IF formula. Let’s write down the below IF formula with AND & OR logical functions:
Let’s see another IF formula with AND and OR functions as follows:
=IF(AND(B1="tom",OR(C1>=60,D1>=60)),"pass","")
When B1 is equal to “tom”, and either C1 is greater than or equal to “60” or D1 is greater than or equal to 60, the formula returns “pass”. Otherwise, the formula will return an empty string.
If we just want to check the cells where the value in column B is “tom” AND the score in column C is greater than or equal to 60. If you want to return TRUE, you can use the below logical test statement with the AND function:
AND(B1="tom",C1>=60)
Next, we can extend the above statement, we want to check the cells if the values in column B is “tom” AND the scores in column C or column D is greater than or equal to 60. Then we can nest the OR function in the above AND function like below:
AND(B1="tom",OR(C1>=60,D1>=60))
### Related Formulas
• Excel IF formula with equal to logical operator
The “Equal to” logical operator can be used to compare the below data types, such as: text string, numbers, dates, Booleans. This section will guide you how to use “equal to” logical operator in excel IF formula with text string value and dates value…
• Excel IF formula with OR logical function
If you want to check if one of several conditions is met in your excel workbook, if the test is TRUE, then you can take certain action. You can use the IF function combining with OR function to construct a condition statement…
• Excel IF formula with AND logical function
You can use the IF function combining with AND function to construct a condition statement. If the test is FALSE, then take another action…
### Related Functions
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel OR function
The Excel OR function used to test multiple conditions and returns TRUE if any of the conditions are TRUE. Otherwise, it will return FALSE. The syntax of the OR function is as below:=OR(logical1, [logical2], …)…
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE….
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
## Excel IF formula with OR logical function
This post will guide you how to use IF function combining with OR logical function in Excel.
If you want to check if one of several conditions is met in your excel workbook, if the test is TRUE, then you can take certain action. You can use the IF function combining with OR function to construct a condition statement. If the test is FALSE, then take another action.
## Excel formula using IF & OR function
The syntax of OR function in excel is as follow:
=OR(condition1,[condition2],...)
The Excel OR function returns TRUE if any of the conditions are TRUE in logic test. Otherwise, it returns FALSE.
Now we want to check the results of two exam scores where the first score in cell A1 is greater than or equal to 40 OR the second score in cell B1 is greater than or equal to 50. If any of conditions are TRUE, then returns the text of “good”, and if FALSE returns “bad”.
Based on the above logic, we can write down the below IF formula:
We can enter the above formula into cell D1, and then drag the Fill Handle to the range D1:D3.it will apply for the other cells for this formula.
So the difference from the IF&AND formula discussed in the previous post is that the IF&OR formula will return TRUE if any of conditions are true.
In the above formula, we just want to check both A1 and B1 cells, the Excel will take an action when the result of test condition is TRUE. In this case, we will add “good” text in the D1 cell. If the test condition is FALSE, assuming that you do not want to display anything in the cell D1, then we can write down the following excel if formula:
=IF(OR(A1=”40”,B1=”50”),”good”,””)
Normally, you can use as many AND/OR logical functions as your logic requires, but there is limitation for argument numbers, we can follow the below limitation for excel formula:
1. No more than 255 arguments in Microsoft excel 2016,2013,2010 and 2007
2. No more than 8192 characters in Microsoft excel 2016,2013,2010 and 2007
3. No more than 30 arguments and the length of formula should not exceed 1024 characters in Microsoft excel 2003 or lower version.
### Related Functions
• Excel OR function
The Excel OR function returns TRUE if any of the conditions are TRUE in logic test. Otherwise, it returns FALSE.
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE….
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
## Excel IF formula with AND logical function
This post will guide you how to use IF function combining with AND logical function in Excel.
If you want to check if a value in cell match two or more conditions at the same time, if the test is TRUE, then you can take certain action. You can use the IF function combining with AND function to construct a condition statement. If the test is FALSE, then take another action.
## Excel formula using IF & AND function
The syntax of AND function in excel is as follow:
=AND(condition1,[condition2],...)
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.
Now we want to check the results of two exam scores where the first score in cell A1 is greater than or equal to 60 AND the second score in cell B1 is greater than or equal to 80. If both conditions are TRUE, then returns the text of “good”, and if FALSE returns “bad”. Based on the above logic, we can write down the below IF formula:
We can enter the above formula into cell D1, and then drag the Fill Handle to the range D1:D3.it will apply for the other cells for this formula.
The logic test will check the two different cells in the above IF formula, if we can run two or more tests on the same cell. Of course yes, for example, if you want to check cell A1 if the value in cell A1 is greater than 60 but less than or equal to 70. If the test is TRUE, then take one action.
The logical test we can use for above logic as below:
=AND(A1>60, A1<=70)
IF formula:
=IF(AND(A1>60,A1<=70),”good”,””)
In this formula, if the test is FALSE, we just add an empty string (“”), IF the test is TRUE, the test string of “good” will appear.
Note: The AND function will check all test conditions in excel, even if one of the already tested conditions has a value of FALSE. This behavior is different with other programming languages. If any previous test conditions return FALSE, the subsequent conditions will not be tested.
For example: If the value of Cell B1 is 0, then the below IF&AND formula will return error message as “Divide by Zero Error”.
=IF(AND(B1<>0,(1/B1)>0.2),”A”,”D”))
As mentioned above, the AND function will check all conditions. To avoid this issue, we can use a nested IF function and check if the value of Cell B1 is not equal to 0 in the first IF statement.
The nested if formula is as below:
=IF(B1<>0,IF((1/B1>0.2),”A”,”D”))
### Related Formulas
• Excel IF formula with equal to logical operator
The “Equal to” logical operator can be used to compare the below data types, such as: text string, numbers, dates, Booleans. This section will guide you how to use “equal to” logical operator in excel IF formula with text string value and dates value…
• Excel IF formula with OR logical function
If you want to check if one of several conditions is met in your excel workbook, if the test is TRUE, then you can take certain action. You can use the IF function combining with OR function to construct a condition statement…
• Excel IF formula with AND & OR logical functions
If you want to test the result of cells based on several sets of multiple test conditions, you can use the IF function with the AND and OR functions at a time…
### Related Functions
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE….
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
## Excel nested if statements with ranges
Many people usually asked that how to write an excel nested if statements based on multiple ranges of cells to return different values in a cell? How to nested if statement using date ranges? How to use nested if statement between different values in excel 2013 or 2016?
This post will guide you how to understand excel nested if statements through some classic examples.
## Nested IF statements based on multiple ranges
Assuming that you want to reflect the below request through nested if statements:
a) If 1<B1<0, Then return 0.1
b) If 0.99<B1<5, then return 0.15
c) If 4.99<B1<15, then return 0.2
d) If 14.99<B1<30, then return 0.5
So if B1 cell was “14.5”, then the formula should be returned “0.2” in the cell.
From above logic request, we can get that it need 4 if statements in the excel formula, and there are multiple ranges so that we can combine with logical function AND in the nested if statements. The below is the nested if statements that I have tested.
=IF(AND(B1>0,B1<1),0.1,IF(AND(B1>0.99, B1<5),0.15, IF(AND(B1>4.99,B1<15),0.2, IF(AND(B1>14.99,B1<30),0.5,””))))
If you don’t want to use AND function in the above nested if statement, can try the below formula:
=IF(B1<1,0.1, If(B1<5,0.15,IF(B1<15,0.2,IF(B1<30,0.5,””))))
## Nested IF Statement Using Date Ranges
I want to write a nested IF statement to calculated the right Quarter based one the criteria in the below table.
According to the request, we need to compare date ranges, such as: if B1<E2<C1, then return A1.
So we can consider to use AND logical function in the nested if statement. The formula is as follows:
=IF(AND(E2>B1,E2<C1),A1,IF(AND(E2>B2,E2<C2),A2,IF(AND(E2>B3,E2<C3),A3,IF(AND(E2>B4,E2<C3),A4))))
Or we can use INDEX function and combine with MATCH Function to get the right quarter.
=INDEX(\$A\$1:\$A\$4,MATCH(E2,\$B\$1:\$B\$4,1))
## Nested IF Statements For A Range Of Cells
If you have the following requirement and need to write a nested IF statement in excel:
If any of the cells A1 to C1 contain “excelhow”, then return “excelhow” in the cell E1.
If any of the cells A1 to C1 contain “google”, then return “excelhow” in the cell E1.
If any of the cells A1 to C1 contain “ibm”, then return “ibm” in the cell E1.
If any of the cells A1 to C1 contain “Cloud”, then return “ibm” in the cell E1.
How to check if cell ranges A1:C1 contain another string, using “COUNTIF” function is a good choice.
Countif function: Counts the number of cells within a range that meet the given criteria
Let’s try to test the below nested if statement:
## Nested IF Statement between different values
Assuming you have the following different range values, if the B1 has the value 65, then expected to return “under average”in cell C1, if the Cell B1 has the value 75, then return “average” in cell C1. And if the Cell B1 has the value 85, then return “above average” in the cell C1.
0-70 under average 71-80 average 81-100 above average
How do I format the nested if statement in Cell C1 to display the right value? Just try to use the below nested if function or using INDEX function.
=IF(B1< 0,"",IF(B1<= 70, "under average",IF(B1<=80, "Average", IF(B1<=100,"Above Average",""))))
### Related Functions
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.The syntax of the IF function is as below:= IF (condition, [true_value], [false_value])….
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel COUNTIF function
The Excel COUNTIF function will count the number of cells in a range that meet a given criteria.This function can be used to count the different kinds of cells with number, date, text values, blank, non-blanks, or containing specific characters.etc.The syntax of the COUNTIF function is as below:= COUNTIF (range, criteria) …
• Excel INDEX function
The Excel INDEX function returns a value from a table based on the index (row number and column number)The INDEX function is a build-in function in Microsoft Excel and it is categorized as a Lookup and Reference Function.The syntax of the INDEX function is as below:= INDEX (array, row_num,[column_num])…
• Excel MATCH function
The Excel MATCH function search a value in an array and returns the position of that item.The MATCH function is a build-in function in Microsoft Excel and it is categorized as a Lookup and Reference Function.The syntax of the MATCH function is as below:= MATCH (lookup_value, lookup_array, [match_type])….
## Excel Nexted IF Functions (Statements) Tutorial (15 IF Formulas examples)
This tutorial will guide you how to use nested Excel IF function (include multiple If statements in excel formula) with syntax and provide about 15 nested IF formula examples with the detailed explanation in Microsoft excel.
## Description
The Excel IF function perform a logical test to return one value if the condition statement is TRUE and return another value if the condition statement is FALSE. The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.
The excel if function just only test one condition and if you want to deal with more than one condition and return different actions depending on the result of the tests, then you need to include several IF statements (functions) in one excel IF formula, these multiple IF statements are also called Excel Nested IF formula(Nested IFs). It’s also similar with IF-THEN-ELSE statement.
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions.
The most recent versions of Excel (Excel 2016, Excel 2013, Excel 2010 and Excel 2007) allow 64 IF functions (statements) in one formula, and it was only possible up to 7 nested IF functions (statements) in Excel 2003 and lower version.
## Syntax
The syntax of Nested IF function is as below:
=IF(Condition_1,Value_if_True_1,IF(Condition_2,Value_if_True_2,Value_if_False_2))
Where the Nested IF function argument is:
Condition_1 – The condition that you want to test in the first IF statement.
Value_if_True_1 – The value that is returned if first IF statement is True. If the condition_1 return False, then move into the next IF function.
Condition_2 – The condition that you want to test in the second IF statement.
Value_if_True_2 – The value that is returned if second IF statement is True.
Value_if_False_2 – The value is returned if second IF statement is False.
This is equivalent to the following IF THEN ELSE statement:
IF Condition_1 THEN
Value_if_True_1
ELSEIF Condition_2
Value_if_True_2
ELSE
Value_if_False_2
END IF
## Examples of Nested IF function (Statement) in Excel
The below examples will show you how to use Excel Nested IF function with the detailed explanation of their syntax and logic.
Example 1# The most basic Nested IF function with one level of nesting
If you want to write a nested if function to test the following calculation logic for assigning value in the cell A1.
IF A1 =="excelhow" THEN
return "excel"
ELSEIF A1 == "excelhow.net" THEN
return "MS excel"
ELSE
return "MS"
END IF
we can write a nested IF function based on the above logic as follows:
=IF(A1="excelhow", "excel", IF(A1="excelhow.net","MS excel","MS"))
In above Nested IF formula, the nested if function is shown with bold style and it is inside the outer IF function. we can see that if A1 is not equal to the “excelhow“, then the second nested IF function will be test. and if second IF condition statement return FALSE, then the entire IF function will return “MS” value.
Example 2# The Nested IF function with two levels of nesting
Assuming that you want to test more than one condition statement in the above nested if function, add one condition to test if the value of the cell A1 reference is equal to “www.excelhow.net” , If TRUE, then return “Microsot excel“.
The calculation Logic is as below:
IF A1 =="excelhow" THEN
return "excel"
ELSEIF A1 == "excelhow.com" THEN
return "MS excel"
ELSEIF A1 == "www.excelhow.net" THEN
return "Microsoft excel"
ELSE
return "MS"
END IF
we can add one more IF statement inside the second IF function in the above excel nested if formula in example1. let’s see the below nested if function with tow level nesting:
=IF(A1="excelhow", "excel", IF(A1="excelhow.com","MS excel",IF(A1="www.excelhow.net","Microsoft excel","MS")))
In the above nested excel IF formula, the first nested if function is marked with red color, and the second nested excel if function is marked with blue color.
If the both first and second conditions are False and the third IF condition will be check, IF A1 is equal to “www.excelhow.net” , then return “Microsoft excel“, or the entire nested IF formula will return “MS“.
Example 3# Describes the each IF function contained in the nested IF function
We will use one typical example of excel nested if function to describe each IF function included in the nested if function.
Assuming that you need to assign a grade based on a score with the following test conditions:
Let’s write a nested if function based on the above logic as follows:
=IF(A1>=80, "excellent", IF(A1>=60, "good", IF(A1>0, "bad", "no valid score")))
For the above excel if formula, lets describe it for each IF function statement.
1# IF Cell A1 is greater than or equal to 80, then the formula will return “excellent” or move to the second If function.
2# If Cell A1 is greater than or equal to 60, then the formula will return “good” or move to the third IF function
3# IF Cell A1 is greater than 0, then the formula will return “bad”, or the IF function will return “no valid score”.
Example 4# Describes each If function in the excel Nested IF Statement (another simple example of if function)
Let’s describe the below Nested IF Function example:
=IF(A1<=6,60, IF(A1<=8,80,IF(A1<=10,100,200)))
a) If cell A1 is equal to 6 or less than 6, then return value 60 in cell C1. Let’s see below screenshot.
b) If Cell A1 is greater than 6 and less or equal to 8, then retrun value 80 in Cell C1.
c) If cell A1 is greater than 8 and less than or equal to 10, then return value 100 in cell C1.
d) If cell A1 is greater than 10 , then the Nested if function will return the last value “200”in cell C1.
Example 5# Excel Nested IF function with arithmetic operator (+, -, * , /)
Assuming that you want to write a Nested If function to reflect the following logic tasks:
a) IF Cell A1 is less than 10, then multiply by 10.
b) IF Cell A1 is greater than or equal to 10 but less than 20, then add 20
c) IF Cell A1 is greater than or equal to 20 but less than 30, then minus 20
d) IF Cell A1 is greater than or equal to 30 but less than 50, then divided by 20
The nested IF formula is as follows:
=IF(A1<10,A1*10,IF(A1<20,A1+20,IF(A1<30,A1-20,IF(A1<50,A1/20))))
a1) if Cell A1 is less 10 (A1=5), then the first If condition matched and will take multiply action, A1 * 10=5*10=50, so it will return 50 in the cell C1
b1) if Cell 10<=A1<20 (A1=15), then the second if condition matched and will take add action, A1+20=15+20=35, so it will return 35 in the cell C1.
c1) if Cell 20<=A1<30(A1=25), then the third if condition matched and will take minus action, A1-20=25-20=5, so it will return 5 in the cell C1.
d1) if Cell 30<=A1<50 (A1=35), then the forth if condition matched and will take divide action, A1/20=35/20=1.75, so it will return 1.75 in the cell C1.
Example 6# Excel Nested IF function with logical function –AND
Assuming that you need a nested if function to reflect the following logic:
a) IF A1+B1 is less than 10, then return 10
b) IF A1+B1 is greater than 10 but less than or equal to 20, then return 20
c) IF A1+B1 is greater than 20 but less than or equal to 30, then return 30.
d) IF A1+B1 is greater than 30, then return 200.
Let’s write the following nested IF formula in the cell C1:
=IF(A1+B1<10,10, IF(A1+B1<=20,20, IF(A1+B1<=30,30,200)))
The above formula just use basic nested IF function syntax, we also can use logic function to re-write it, the nested if formula with AND function is as follows:
=IF((A1+B1)<10,10,IF(AND((A1+B1)>10,(A1+B1)<=20),20, IF(AND((A1+B1)>20,(A1+B1)<=30),30,200)))
The above nested IF formula combined with two AND function.
In the second IF Statement, AND((A1+B1)>10,(A1+B1)<=20) will check if 10<A1+B1<=20, If TRUE, then the formula will return 20.
In the third IF Statement, AND((A1+B1)>20,(A1+B1)<=30) will check if 20<A1+B1<=30, If TRUE, then the formula will return 30.
Example 7# Excel Nested IF function with logical function –OR
Assuming that you need a nested if function to reflect the following logic:
a) IF Cell A1=5, return A1/B1
b) IF Cell A1=10, return A2/B2
c) IF Cell A1=15 or A1=20, return A3/B3
In Cell C1, we can write the below nested if formula based on the above conditions.
=IF(A1=5,A1/B1,IF(A1=10,A2/B2,IF(OR(A1=15,A1=20),A3/B3)))
One OR function be used in the above excel nested if function, it will check if A1=15 or A1=20, if TRUE, then return A3/B3.
Example 8# Excel nested if function with text and logical function AND
Wrote a nested if function with text to reflect the following logic:
a) If Cell A1=”E” and Cell B1=5, then return “Excel”
b) If Cell A1=”P” and Cell B1=3, then return “PPT”
c) If Cell A1=”W” and Cell B1=4, then return “Word”
d) Else return “Access”
In Cell C1, try to enter into the following excel nested If formula with AND function:
=IF(AND(A1="E",B1=5),"Excel", IF(AND(A1="P",B1=3),"PPT", IF(AND(A1="W",B1=4),"Word","Access")))
Example 9# Excel nested if function with ISBLANK function and logical function AND
a) If you want to wrote a nested if function with ISBLANK function and logical function AND to reflect the following logic:
b) If both Cell A1 and Cell B1 are empty, then return “”
c) If only Cell A1 is empty, then return B1-today()
d) If both two cells A1 and B1 are not empty, then return “excel” string.
In Cell C1, use the following excel nested If formula with ISBLANK and AND function:
=IF(AND(ISBLANK(A1),ISBLANK(B1)),"",IF(ISBLANK(A1),B1-TODAY(),"excel"))
Example 10# Using nested IF functions to check grade level based on student’s score(multiple IF statements)
The logic is as below:
We will write a nested If function that reflect the above logic, and will check if the score is below 50, If TRUE, it is considered as “Fail”. If FALSE, move into the next IF statement to test if the score is between 51 and 60 and it is considered as “Grade C”. If False, we will move into another IF statement to check if the score is between 61 and 80, IF True and it is considered as “Grade B”. If FASLSE, just check the rest conditions.
We can use a nested if formula as follows:
Example 11# Nested IF function for checking two Empty Cells
Let’s see the below image a product table of a company (need to create an excel table firstly):
a) If we need to check both “Price” cell and “Quantity” cell are empty, If True, then return empty. If the only “price” cell is empty only, IF True, return empty.
b) If the only “Quantity” cell is empty, IF True, return empty.
c) If both “price” and “Quantity” are not empty, then return multiply Price * Quantity as subtotal value.
So To check both “Price” and Quantity cells, we can use table header name as condition variable to test each Price cells or Quantity cells, so we can write the nested if formula as follows:
=IF([Price]="","",IF([Quantity]="","",[Price]*[Quantity]))
Just using the above excel if formula in the subtotal cells, the formula will check the first IF statement if Price Cell is empty, IF TRUE, then will return empty (“”) in the subtotal cell. IF FALSE, then move to the next IF statement and so on. Last, IF neither cell is empty, then will return the value of multiply [Price]*[Quantity] in subtotal cell.
Of course, we can use another nested if function to achieve the above calculation logic (easy to understand).
=IF(ISBLANK(B2),"",IF(ISBLANK(C2),"", B2*C2))
OR
=IF(B2="","",IF(C2="","", B2*C2))
## Nested IF Functions Order
There is one important thing that need us keep in mind when write Excel Nested IF Function, it is the order of nested IF function. It can nested up to 64 If statements, and how to order multiple IF condition statements, it is key point. Or the wrong result will be returned. The point is that excel nested if function will test the first if condition in the order, once any condition is met, and the subsequent if conditions will not be checked.
So let’s remember the below rules while writing excel nested if function:
• The most important condition First or Harder Test First
Let’s see the below example what it means:
Example 12#
There are two test conditions in the following excel nested if function:
When using this formula in the cell B3, If the amount in cell B1 is 95, then “excellent” would be returned. because it is greater than 90. And the second IF condition will not be evaluated.
However, if the order of nested if statment are reversed as follows:
The above formula would check for the condition B1>=60 first, if the amount in cell B1 is 95, then the value “good” would be returned in cell B3. Because the Cell B1 match the first test condition, and it will not check the second if condition and will return the incorrect result.
## Nested IF Function Alternatives
To make your excel formulas more efficiency and fast, you can try to use the following alternatives to excel nested if function.
1)
Excel nested if function can be easily replaced with the VLOOUP, Lookup, INDEX/MATCH or CHOOSE functions.
Example 13# Use VLookup function instead of nested IF function
Nested IF function:
Vlookup function:
=VLOOKUP(39,A1:B3,2,FALSE)
Example 14# Use CHOOSE function instead of nested if function
Nested If unction:
Using CHOOSE function as follows:
2) Use IFS instead of nested if function
3) Use the CONCATENATE function or the concatenate operator (&).
Example 15#
Nested IF function:
Using CONCATENATE function as follows:
Question1: Is there any tool to help write Excel formulas and nested Ifs?
This is an Excel formula with nested IF statements:
=IF((B2="East"),4,IF((B2="West"),3,IF((B2="North"),2,IF((B2="South"),1,""))))
To essentially accomplish this:
If cell B2 = "East"
return "4"
ElseIf cell B2 = "West"
return "3"
ElseIf cell B2 = "North"
return "2"
ElseIf cell B2 = "South"
return "1"
Else
return ""
Can Excel formulas be written in such a “more readable” manner and converted to the official syntax? Is there any tool to help write Excel formulas?
Answer: Excel Formula Formatter add-in by Rob van Gelder, mentioned at Daily Dose of Excel.
Excel’s formula bar ignores line feeds and white space, so you can Alt+Enter and spacebar to format the formulas however you like. I’ve tried it and I quickly stopped doing it. Too much spacebar-ing, especially if you need to edit.
Question2: I am working on an excel file, and i am trying to use a nested if formula to achieve what i would like.
i have two columns:A B. And the condition is this: if the value in a2=a3, then check if the minus of b2 and b3 is certain value, and if it is, put a yes, else put a no. this will iterate till the end of the excel file.
so far here is what i have. not sure how to use the excel formulas. any help is much appreciated.
if(a2=a3,b2-b3=5 or b2-b3=-5 or b2-b3=20 or b2-b3=-20, "yes", "no")
Answer: you should be able to use the OR function within your nested if formula to test for “B2-B3=5,B2-B3=-5,B2-B3=20,B2-B3=-20” as follows:
=IF(A2=A3,IF(OR(B2-B3=5,B2-B3=-5,B2-B3=20,B2-B3=-20),"yes","no"),"no")
### Related Functions
• Excel IF function
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE….
• Excel And function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE…
• Excel Concat Function
The excel CONCAT function combines 2 or more strings or ranges together. The syntax of the CONCAT function is as below: =CONCAT (text1,[text2],…) …
• Excel Vlookup Function
The Excel VLOOKUP function lookup a value in the first column of the table and return the value in the same row based on index_num position.The syntax of the VLOOKUP function is as below:= VLOOKUP (lookup_value, table_array, column_index_num,[range_lookup])…
• Excel Choose Function
The Excel CHOOSE function returns a value from a list of values. The CHOOSE function is a build-in function in Microsoft Excel and it is categorized as a Lookup and Reference Function.The syntax of the CHOOSE function is as below:=CHOOSE (index_num, value1,[value2],…)…
• Excel ISBLANK function
The Excel ISBLANK function returns TRUE if the value is blank or null.The syntax of the ISBLANK function is as below:= ISBLANK (value)…
## Excel IF Function
This post will guide you how to use Excel IF function with syntax and examples in Microsoft excel.
## Description
The Excel IF function perform a logical test to return one value if the condition is TRUE and return another value if the condition is FALSE.
The IF function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.
The IF function is available in Excel 2016, Excel 2013, Excel 2010, Excel 2007, Excel 2003, Excel XP, Excel 2000, Excel 2011 for Mac.
## Syntax
The syntax of the IF function is as below:
= IF (condition, [true_value], [false_value])
Where the IF function arguments are:
Condition -This is a required argument. A user-defined condition that is to be tested.
True_value – This is an optional argument. The value that is returned if condition evaluates to TRUE.
False_value – This is an optional argument. The value that is returned if condition evaluates to FALSE.
The below examples will show you how to use Excel IF Function to return one value if the condition is TRUE or FALSE.
Note: the above excel formula will test a condition “b1>90”, if the condition is true, then “good” will return or “bad” will return.
## Nested IF statements
The excel if function just only test one condition and if you want to deal with more than one condition and return different actions depending on the result of the tests, then you need to include several IF statements (functions) in one excel IF formula, these multiple IF statements are also called Excel Nested IF formula(Nested IFs).
For example:
=IF(A1>=80, "excellent", IF(A1>=60, "good", IF(A1>0, "bad", "no valid score")))
More reading: Excel Nexted IF Functions (Statements) Tutorial
## Excel Logical operators
When you are writing an If statement, you may need to use any of the following excel logical operators:
Operator Meaning Example Description = Equal To A1=B1 Returns True if a value in cell A1 is equal to the values in cell B1; FALSE if they are not. > Greater Than A1>B1 Returns True if a value in cell A1 is greater than the values in cell B1; FALSE if they are not. >= Greater Than or Equal to A1>=B1 Returns True if a value in cell A1 is greater than or equal to the values in cell B1; FALSE if they are not. < Less Than A1 Not Equal to A1<>B1 Returns True if a value in cell A1 is not equal to the values in cell B1; FALSE if they are not.
Let’s see some examples for logical operators in Excel formula:
A1 B1 8 5
=(A1>B1)
Output: TRUE
=(A1<B1)
Output: FALSE
=(A1>=B1)
Output: TRUE
=(A1<>8)
Output: FALSE
=(A1<>B1)
Output: TRUE
=(IF A1>3, “TRUE”, “FALSE”)
Output: TRUE
Question1: I want to write an IF Formula based on the below test criteria in the Microsoft excel.
If the number in Cell B1 is greater than 30, then I want it to return A
If the number in cell B1 is between 20 and 30, then I want it to return B
If the number in Cell B1 is below 20, then I want it to return C
The below formula is what I currently write:
=IF(B1>30,"A",IF(B1<30,B1>20,"B","C"))
When I entered the above formula in the Cell C1, and it will return an Error.
Answer: You can use IF function in combination with AND function to reflect the above logic. So you can do it using the following formula:
=IF(B1>30,"A",IF(AND(B1>=20,B1<=30),"B","C"))
Or you can use another IF formula without AND function as follows:
=IF(B1>30,"A",IF(B1<20,"C","B"))
Question 2: I want to write a excel Formula for sales rep, If the sales are less than 10K, then the member will get no commission. If the sales are between 10 and 50K, then the member will get a 3% commission. If the sales are more than 50k, then the member will get 5% commission. Could you help me?
Answer: Based on the above description, we can use the following excel IF formula:
=IF(B1<10,0,IF(B1<50,B1*3%,B1*5%))
Question 3: I want to select students for the scholarship in school, and based on the student’s scores and attendance value, if the student scores more than 85 and has the attendance of more than 85%, then he/she will get the scholarship. Please help me how to write this formula based on my test criteria.
Answer: you can use the IF function with the AND function to check whether both of two conditions are met or not. If the test are met, then return “Yes”, otherwise returns “No”.
So you can use the following IF formula in Excel:
=IF(AND(B1>85, C1>85%),"Yes","No")
Question 4: I am trying to write an excel formula based on the following logic:
I want to enter a formula in Cell C1 that will Add B1, B2 and B3 or multiply B1 by 1, multiply B2 by 2, multiply B2 by 3, which action will be taken based on what you put in Cell D1.
Appreciated for any help. Thanks
Answer: Based on the above logic description, you need to use SUM function to count the sum of B1, B2 and B3. We can use the following IF formula:
Question 5: I am trying to write a formula in excel to check the employee number field and returns the relevant band of employee. I already have one IF formula, but I always return the first value of “A”. Could you help?
=IF(B1>=10,"A",IF(B1>=20,"B",IF(B1>=50,"C")))
Answer: when you write an IF nested formula, you need to start with the largest number first or start the smallest number first with less than or equal to operator.
1# start with the largest number first, we can write down the below formula:
=IF(B1>=50,"C",IF(B1>=20,"B",IF(B1>=10,"A")))
2# start with the smallest number first, you need to change the >= to <= , just see the below formula:
=IF(B1<=10,"A",IF(B1<=20,"B",IF(B1<=50,"C")))
Question 6: I want to write a Formula in excel to return “bad” if the cell B1 is either <100 or >500, otherwise, it should be returned the value of cell B1.
Answer: For the above logic test, we should use the OR function in the IF condition, so we can write down the below IF formula:
=IF(OR(B1>500,B1<100),"good",B1)
Question 7: I want to write a formula in excel using IF function to check if the cell B1 >10 and B1<20. If TRUE, returns “good”, If FALSE, returns “bad”
Answer: You can use IF function in combination with AND function to check the value of Cell B1, if cell B1 is greater than 10 and cell B1 is less than 20, then returns “good”, otherwise, it will return “bad”.
Question 8: I need to write a nested IF statement in excel 2013 to check the following logic:
If Cell B1 is less than 5, then multiply it by 5.
If Cell B1 is greater than or equal to 5 but less than 10, then multiply it by 10.
If cell B1 is greater than 10, then multiply it by 15.
Answer: This is a generic nested IF formula in excel, we can write the below nested IF statement to reflect the logic.
=IF(B1<5, B1*5, IF(B1<10, B1*10, B1*15))
Question 9: I want to write a formula to match the following logic in MS Excel 2013:
If A1*B1 <=10, then returns A
If A1*B1 >10 but A1*B1 <=20, then returns B
If A1*B1 >20 but A1*B1 <=30, then returns C
If A1*B1>30, then returns D
Answer: You can use IF function to build a Nested IF statement in combination with AND function to achieve it. Let’s write down the following IF formula:
=IF(A1*B1<=10,"A", IF(AND(A1*B1>10,A1*B1<=20),"B", IF(AND(A1*B1>20,a1*B1<=30,"C","D"))))
Question 10: I want to write an IF function to check if the value in cell B1 is blank or Text string or Numeric value, if the cell B1 is empty, then return “blank”, if the cell B1 is a text string, then return “Text”, if the value in cell B1 is a numeric value, then return “number”. Any help for this formula, thanks.
Answer: Based on the description, you need to use ISBLANK function to check if the value in cell b1 is blank or not. And need to use ISTEXT function to check if the value in cell B1 is a text string or not and need to use ISNUMBER function to check if the value in cell B1 is a number or not. So you can write a nested IF statements in combination with ISBLANK, ISTEXT and ISNUMBER functions in Excel as follows:
=IF(ISBLANK(B1),"blank", IF(ISTEXT(B1),"Text", IF(ISNUMBER(B1),"number")))
Question 11: I want to write a formula in excel to calculate the bonus for employees in company, if the employee salary is greater than or equal to \$2000, then the bonus will be 10% of the salary , otherwise, the bonus will be 5% of the employee salary.
Answer: we need to check if the salary in Cell B1 (it’s the salary of the first employee) is greater than or equal to 2000. It the condition test is TRUE, then returns B1*10%, otherwise returns b1*5%. So we can write down the following IF formula in excel:
=IF(B1>=2000, B1*10%, B1*5%)
Question 12: In Microsoft Excel 2013, I want to create an IF function to check if any employee who have at least 10 years of experience and whose salary is greater than \$5000, If TRUE, then the bonus will be 20% of salary.
Answer: you can use IF function in combination with AND function to check if the value in cell B1 is greater than or equal to 10 and the value in Cell C1 is greater than 5000. If both of two conditions are TRUE, then returns C1*20%, otherwise returns “No Bonus”.
Let’s write down the following IF statement as follows:
=IF(AND(B1>=10,C1>5000),C1*20%, "No Bonus")
Question 13: I want to create an IF formula in Excel 2013 to check the following text logic:
If B1<10, then multiply B1 by 1%, but the returned value is not less than 50.
If B1>10, then multiply B1 by 2%, but the returned value is not greater than 100.
Answer: To reflect the first test condition, you need to use MAX function to match the condition. To reflect the second test condition, you need to use MIN function to match that the returned value should not be greater than 100. So we can use the following excel IF formula:
=IF(B1<10, MAX(50,B1*1%), IF(B1>10, MIN(100,B1*5%)))
Question 14: I want to use IF function to check if B1 is greater than 10 and B2 is greater than 20 and B3 is less than 30, if TRUE, then returns “good”, otherwise, it should be returned “bad”. So How to create the IF formula based on the above test criteria to check three conditions at the same time.
Answer: you need to use AND function within the IF function in excel to create an IF formula as follows:
Question 15: I have an IF formula that might cause a division by zero error, I don’t know how to avoid this kind of errors in Excel formula.
Answer: You can use ISERROR function to catch this kind of errors then use the IF function to check the returned values from ISERROR function to avoid an error. So we can write an IF function in combination with ISERROR function in excel. For example, we can use the following IF formula:
=IF(ISERROR(B1/C1),0, B1/C1)
The ISERROR function will return TRUE when trying to divide B1 by 0.
Question 16: I need to create a formula in Excel 2013 to reflect the following logic:
If B1=Excel, then return E
If B1=word, then return W
If B1=access, then return A
Answer: You can use IF function to create a nested IF statements as follows:
=IF(B1="Excel","E",IF(B1="word","W",IF(B1="access","A")))
Question 17: I have a worksheet that containing cells that are formatted as date format. I want to write an IF formula to check the first value in Cell (month part). So I am trying to create the following IF formula:
=IF(LEFT(B1,1)=8, "August","Null")
The returned results are always “Null”.
Answer: As the dates are not recognized as string, so you cannot use the LEFT function to exact the first value in the dates. At this moment, you need to use Month function to convert the date to its month number in Excel. So we can write down the below IF formula in combination with Month function:
=IF(MONTH(B1)=8, "August","Null")
Question18: I am trying to create an IF formula to check if the time value in cells are greater than 10.00h for the following date format: 08/11/2018 09:43.50. Appreciative of any help.
Answer: In excel, you can use HOUR, MINUTE, SECOND functions to compare the date or time value. so if you want to compare the value in Cell B1 if it is greater than 10h, then you can use HOUR function within IF function, just like this: HOUR(B1)>10, so we can write down the below formula:
=IF(HOUR(B1)>10, "greater","less")
Question 19: I want to create an IF function and need to combine with another RAND function. The formula just works find for the first IF_VALUE_TRUE statement, but the formula works not good for IF_VALUE_FALSE statement. Here is the IF formula I have got:
=IF(ISBLANK(B1),"","=rand()")
In the above IF function, it will return empty string when the value in cell B1 is blank. Otherwise, it should add a random function, the problem is that the cell doesn’t run the RAND function. So how can I fix it? Thanks
Answer: when you write an IF formula, if you want to enclose others function, you need not to add quotes around the RAND () function, so just remove the quotes and equal sign in your IF function. Just like the below IF formula:
=IF(ISBLANK(B1),"",RAND())
Question 20: I am trying to write an IF formula in Excel 2013 to reflect the following logic:
If B1 is greater than or equal to 50 and C1 is 0
OR
If B1 is greater than or equal to 30 and C1 is greater than or equal to 1
OR
If B1 is greater than or equal to 20 and C1 is 2
Then take the following action: D2/E2
Otherwise, return FALSE
I have wrote the following IF formula, but it doesn’t work at all.
=IF(AND(B1>=50,C1>=0),OR(AND(B1>=30,C1>=1)),OR(AND(B1>=20,C1>=2)),D2/E2,"FALSE")
Answer: you need to nest your different AND function within an OR function in the IF formula. So you can try the below IF formula:
=IF(OR(AND(B1>=50,C1>=0),AND(B1>=30,C1>=1),AND(B1>=20,C1>=2)), D2/E2,"FALSE")
Question 21: I want to create an IF function in combination with MID function in excel 2013. it need to check if the value in one specified cell is TRUE, then return the first six characters from another cell. Otherwise return empty value.
Answer: you just need to add MID function within IF function in excel, and do not add any quotes around MID function. So you can use the following IF formula to achieve your request:
=IF(B1=TRUE, MID(C1,1,5), "")
Question 22: I have a excel worksheet as below:
A B
——
20 O
30 V
10 T
50 T
I want to create an IF formula to reflect the following logic:
IF cell A1 is less than or equal to 30 and Cell B1 is equal to “O” or “V”, if TRUE, then returns 300, otherwise returns 400.
IF Cell A1 is greater than 30 and Cell B1 is equal to “O” or “V”, then returns 500, otherwise returns 600.
I have wrote the below tow IF formulas for the above two conditions as follows:
=IF(AND(A1<=30,OR(B1="O",B1="V")),300,400)
=IF(AND(A1>30,OR(B1="O",B1="V")),500,600)
I am able to check the above two IF formula and the returned results is OK… But I am not able to combine the above two IF formula into a single IF formula. So anybody can help? Many thanks
Answer: Based on the above logic, you can use the below IF formula to combine with above two IF formulas:
=IF(A1<=30,IF(OR(B1="O",B1="V"),300,400),IF(OR(B1="O",B1="V"),500,600))
Question 23: I am trying to write an IF function to prevent zero and negative values in cells. What I would like is that if the value in cell B1 is less than or equal to 0, then it should be returned “Null” otherwise, it should return the calculation of Cells value, like as:B2*(C2-D2)*E2.
The below IF formula is what I have:
=IF(B1<=0,"Null","B2*(C2-D2)*E2")
When I run the formula above, it only returns my calculation string and do not take the actual calculation.
Answer: In excel, the double quotes make any values in between be recognized as Text string. So if you want to take calculation for your IF formula, just remove the double quotes. Let’s see the modified IF formula as follows:
=IF(B1<=0,"Null",B2*(C2-D2)*E2)
Question 24: I want to create an new IF function to check if the value in cells is Saturday or Sunday, If TRUE, returns “yes”, otherwise, returns “No”. And I am using the following IF formula, but I get an error, so what’s wrong for this formula?
=IF((OR(\$B1="Saturday","\$B1="Sunday"),"yes","no"))
Answer: you need to use the WEEKDAY function within IF function to handle the dates if they are in date format, like as: 11/8/2018.
So you can use the below IF function to achieve your logic:
=IF(WEEKDAY(\$B\$1,2)>5,"yes","no")
You can also use TEXT function within the IF function to achieve the same results, just like the below IF formula:
=IF(LEFT(TEXT(\$B\$1,"ddd"))="S","yes","no")
Question 25: I want to write an IF function in excel to check if the first character in one Cell is equal to 5, then the returned value should be the five rightmost characters of that cell, otherwise, the returned value should be the four rightmost characters. I wrote one IF formula as follows, but it doesn’t work.
=IF(LEFT(B1,1)=5, RIGHT(B1,5),RIGHT(B1,4))
In the above IF function, it always return the rightmost four characters even though the first character in Cell B1 starts with “S”. Please help me to fix it.
Answer: you should know the returned value of the LEFT function firstly in excel. As the LEFT function will return a Text value, so you also need to provide a string for comparison, so adding quotes to enclose it. Just like the below IF function:
=IF(LEFT(B1,1)="5", RIGHT(B1,5),RIGHT(B1,4))
There is another way to achieve the same results, you can use NUMBERVALUE function to convert the result of LEFT function to a numeric value, like the below IF function:
=IF(NUMBERVALUE(LEFT(B1,1))=5,RIGHT(B1,5),RIGHT(B1,4))
Question 26: I have 2 columns contain date and time or just only contain time. And I want to check if the times of column A is greater than the times of column B. the key issue may be that column A has the date and time. I wrote the following IF formula to run it in Cell C1, but it returned the inaccurate results.
=IF(A1>B1,"yes","no")
Any help would be appreciated… Many thanks!
Answer: the date part of the value in column A is the integer, while the time is the decimal. You can use the following IF formula:
=IF(A1-INT(A1)>B2),"yes","no")
Question 27: The below are the results that I expected, and I want cell B1 to B4 can detect the string from A1 to A4 automatically and return the same string value plus the severity level when the test match. For example, If A1 is equal to “critical”, then it should be returned “critical severity 1” in the cell B1. Etc.
And I am using the following IF formula, but it does not work at all. Please help to fix it. Thanks
=IF(A1="Critical","Critical Severity 1",""),IF(A1="High","High Severity 2",""),IF(A1="Medium","Medium Severity 3",""),IF(A1="Low","Low Severity 4","")
A B critical critical severity 1 high high severity 2 medium Medium severity 3 low low severity 4
Answer: you can try to run the following IF formula in excel:
=IF(A1="critical","critical Severity 1",IF(A1="high","high Severity 2",IF(A1="medium","medium Severity 3",IF(A1="low","low Severity 4",""))))
Question 28: I am working on an excel file and want to create a new IF formula to reflect the following logic:
If the value in Cell A1 is equal to the value in Cell A2, then check if the minus of B1 and B2 is equal to a special value, and if the condition is TRUE, returns “yes”, otherwise, returns “no”. Here is the IF formula I have:
=if(A1=A2,B1-B2=5 or B1-B2=-5 or B1-B2=20 or B1-B2=-20, "yes", "no")
Any help is appreciated…Thanks
Answer: you need to use OR function with IF function to create a nested IF statement to achieve your request. So you can try the following Excel IF formula in your excel file.
=IF(A1=A2,IF(OR(B1-B2=5,B1-B2=-5,B1-B2=20,B1-B2=-20),"yes","no"),"no")
Question 29: I want to create an IF formula to check the range of cells in excel. I have scores of different subject for a student, and want to check if any one of scores is less than 60, If TRUE, then return “BAD”. Can this logic be done with the IF statements in excel 2013?
Answer: you can use COUNTIF function within IF function to create a generic IF formula as follows:
Question 30: I am trying to create a new IF statement so that when the formula is looking at Row A and Row B, the returned values should be shown in the Row C. the following logic need to be checked:
IF the value in the Row A is equal to “NA” And the value in the Row B is equal to “NA”, then return “NA” value in Row C.
IF the value in the Row A is equal to “NA”, and the value in the Row B is equal to “denied”, then return “denied” value in Row C.
IF the value in the Row A is equal to “allowed” and the value in the Row B is equal to “NA”, then return “allowed” value in Row C.
Here is my formula:
=IF(AND(A1 = "NA", B1 = "NA"),"NA",IF(OR(A1="denied",B1 ="denied"),"denied", "NA"))
I don’t know how to include “allowed” in the IF formula above, anyone can help this, many thanks.
Answer: This is a typical nested if statement, you can use OR function within IF function in excel. We can write down this nested IF formula as follows:
=IF(OR(A1="denied", B1="denied"), "denied", IF(OR(A1="allowed", B1="allowed"), "allowed", "NA"))
### Related Functions
• Excel AND function
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.The syntax of the AND function is as below:= AND (condition1,[condition2],…) …
• Excel ISBLANK function
The Excel ISBLANK function returns TRUE if the value is blank or null.The syntax of the ISBLANK function is as below:= ISBLANK (value)…
• Excel COUNTIF function
The Excel COUNTIF function will count the number of cells in a range that meet a given criteria.= COUNTIF (range, criteria) …
• Excel ISERROR function
The Excel ISERROR function returns TRUE if the value is any error value except #N/A. The ISERROR function is a build-in function in Microsoft Excel and it is categorized as an Information Function. The syntax of the ISERROR function is as below: = ISERROR (value) …
• Excel LEFT function
The Excel LEFT function returns a substring (a specified number of the characters) from a text string, starting from the leftmost character.The syntax of the LEFT function is as below:= LEFT(text,[num_chars])…
• Excel MID function
The Excel MID function returns a substring from a text string at the position that you specify. The MID function is a build-in function in Microsoft Excel and it is categorized as a Text Function. The syntax of the MID function is as below: = MID (text, start_num, num_chars) …
• Excel TEXT function
The Excel TEXT function converts a numeric value into text string with a specified format. The syntax of the TEXT function is as below:= TEXT (value, Format code)…
• Excel RIGHT function
The Excel RIGHT function returns a substring (a specified number of the characters) from a text string, starting from the rightmost character…
### More Excel IF Formula Examples
• Excel nested if function
The nested IF function is formed by multiple if statements within one Excel if function. This excel nested if statement makes it possible for a single formula to take multiple actions…
• Excel IF formula with Equal to logical operators
The “Equal to” logical operator can be used to compare the below data types, such as: text string, numbers, dates, Booleans. This section will guide you how to use equal to logical operator in excel IF formula with text string value and dates value…
• Excel IF formula with greater than logical operators
How to use if function with greater than, greater than or equal to, less than and less than or equal to logical operators in excel. Let’s see the below generic if formula with greater than operator: =IF(A1>10,”excelhow.net”,”google”) …
• Excel IF formula with AND logical function
You can use the IF function combining with AND function to construct a condition statement. If the test is FALSE, then take another action. The syntax of AND function in excel is as follow: =AND(condition1,[condition2],…)…
• Excel IF function with OR logical function
If you want to check if one of several conditions is met in your excel workbook, if the test is TRUE, then you can take certain action. You can use the IF function combining with OR function to construct a condition statement…
• Excel IF function with “NOT” logical function
If you want to check several test conditions in an excel formula, then take different actions. You can use NOT function in combination with the AND or OR logical function in excel IF function…
• Excel IF formula with AND & OR logical functions
If you want to test the result of cells based on several sets of multiple test conditions, you can use the IF function with the AND and OR functions at a time…
• Excel IF Function With Numbers
If you want to check if a cell values is between two values or checking for the range of numbers or multiple values in cells, at this time, we need to use AND or OR logical function in combination with the logical operator and IF function…
• Excel IF function with text values
If you want to write an IF formula for text values in combining with the below two logical operators in excel, such as: “equal to” or “not equal to”…
• Excel IF function with Dates
If you have a list of dates and then want to compare to these dates with a specified date to check if those dates is greater than or less than that specified date. …
• Excel IF function check if the cell is blank or not-blank
If you want to check the value in one cell if it is blank or empty cell, then you can use IF function in combination with ISBLANK function or logical operator (equal to) in excel…
• Excel nested if statements with ranges
Many people usually asked that how to write an excel nested if statements based on multiple ranges of cells to return different values in a cell? How to nested if statement using date ranges? How to use nested if statement between different values in excel 2013 or 2016…
• Count Cells That Contain Specific Text
This post will discuss that how to count the number of cells that contain specific text or certain text in a specified cells of range in Excel. How to get the total number of cells that contain certain text.……
• Sum Values by Group
Assuming that you have a table that contain the product name and its sales result. And you have group by those values based on the product name, then you want to sum values based one each product name. You can create a new Excel formula based on the IF function and SUMIF function.…
• Get nth Largest Value with Duplicates
If you want to get the largest unique value in a data set with duplicates, you can create a new excel array formula based on the MAX function and the IF function..…
• Find the Largest Value Based on Multiple Criteria
Assuming that you have a list of data that you want to find the largest value based on the product “excel” and the sales region “east”. You can create a new excel formula based on the SUMPRODUCT function and the LARGE function..……
• Create a Five Star Rating System
How do I change the five points system to five star rating in excel. How to use the conditional formatting function to create a five star rating system in excel…..
• Rank values in a column based a specific value in another column
Assuming that you have a list of data contains two columns and the first column is product list and another is Sales number. You want to rank the sales number of a specified product name. You can try to write a complex formula based on the IF function and the COUNTIFS function to achieve the result..…
• Comparing Columns Using Conditional Formatting Icon Sets
how to compare the adjacent cells in the different columns using Conditional Formatting Icon Sets in Excel. How to compare Columns or rows using Conditional Formatting Icon Sets to show increase or decrease status in your current worksheet..…
• Show Only Positive Values
You can create a formula based on the IF function, and the SUM function to sum all values in the range A1:C5 and just show only positive values…
• Count Unique Values Using Pivot Table
You can insert a 3rd or helper column with a formula to check if the value is unique in the selected range of cells, and the create pivot table based on the 1st and 3rd column to count unique values..…
• Compare Dates
Assuming that you have a list of data that contain date values in Excel, you can use the IF function to create a formula to achieve it. If the date is greater that the given date value, then return True. Otherwise, it returns False….
• Excel Vlookup Return True or False
you can use the VLOOKUP function to look for a value in a column in a table and then returns TRUE from a given column in that table if it finds something. If it doesn’t, it returns FALSE …
• VLOOKUP Return Multiple Values Horizontally
You can create a complex array formula based on the INDEX function, the SMALL function, the IF function, the ROW function and the COLUMN function to vlookup a value and then return multiple corresponding values horizontally in Excel.…
• Copy and Paste Only Non-blank Cells
If you want only copy non-blank cells in a range in Excel, you need to select the non-blank cells firstly, then press Ctrl +C keys to copy the selected cells. So how to only select all non-blank cells in the selected range in your worksheet..…
• Changing Negative Number to Zero in Excel
If you want to change all negative numbers to zero value from a cell in Excel, you can use a formula based on the MAX function or IF function.…
• Count Dates in Given Year/Month/Day in Excel
You can create a formula based on the SUMPRODUCT function and the YEAR function to count dates by a give year….
• Generate All Possible Combinations of Two Lists
You can use a formula based on the IF function, the ROW function, the COUNTA function, The INDEX function and the MOD function to get a list of all possible combinations from those two list….
• Find Missing Numbers in a Sequence in Excel
You can use an excel array formula based on the SMALL function, the IF function, the ISNA function, the MATCH function, and the ROW function to find missing numbers in a sequence…
## Excel And Function
This post will guide you how to use Excel AND function with syntax and examples in Microsoft excel.
### Description
The Excel AND function returns TRUE if all of arguments are TRUE, and it returns FALSE if any of arguments are FALSE.
The AND function is a build-in function in Microsoft Excel and it is categorized as a Logical Function.
The AND function is available in Excel 2016, Excel 2013, Excel 2010, Excel 2007, Excel 2003, Excel XP, Excel 2000, Excel 2011 for Mac.
### Syntax
The syntax of the AND function is as below:
= AND (condition1,[condition2],…)
Where the AND function arguments are:
Condition1 -This is a required argument. The first arguments to test if it is true.
Condition2 – Optional
### Example
The below examples will show you how to use Excel AND Function to test whether all of conditions are TRUE.
#1 To Test if B1 is greater than 10 AND less than 20, if so, returns TRUE value, just using the following excel formula: =AND(B1>10, B1<20)
### More Excel AND Function Examples
• Conditional Formatting date with red Amber or Green
How to highlight the date with red if the cell dates is past now(). How to highlight the date with amber if the cell date is past now but within the next 6 months from now(). How to highlight the date with green if the cell date is more than 6 months from now…
• Check If Multiple Cells are Equal In Excel
Assuming that you have a list of data in range A1:C1, and you want to compare if these cells are equal, if so, then return True, otherwise, return False. You need to create an Excel array formula based on the AND function and the EXACT function… | 24,313 | 98,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.913631 |
https://www.instasolv.com/question/12-x-13-x-73-9-x-co-is-if-x-14-57-and-y-lutte-36-a-0-b-1-c-2-d-3-7rxmjz | 1,611,253,047,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00023.warc.gz | 823,064,344 | 10,745 | 12(-x)+13(-x)+ / 73(9-x)+...+ co is...
Question
# 12(-x)+13(-x)+ / 73(9-x)+...+ co is If x= 14-57 and y = lutte 36 (A) 0 (B) 1 (C) 2 (D) 3
JEE/Engineering Exams
Maths
Solution
112
4.0 (1 ratings) | 100 | 197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.342112 |
https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH104/December_2013/Question_02_(c) | 1,713,769,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818081.81/warc/CC-MAIN-20240422051258-20240422081258-00292.warc.gz | 531,405,749 | 11,744 | # Science:Math Exam Resources/Courses/MATH104/December 2013/Question 02 (c)
MATH104 December 2013
Other MATH104 Exams
### Question 02 (c)
Consider the function
${\displaystyle f(x)={\frac {x^{2}}{x^{2}-4}}.}$
Its first and second derivatives are given by
${\displaystyle f'(x)=-{\frac {8x}{(x^{2}-4)^{2}}},\quad f''(x)={\frac {8(3x^{2}+4)}{(x^{2}-4)^{3}}}}$
(c) On which intervals is ${\displaystyle f(x)}$ increasing? On which intervals is ${\displaystyle f(x)}$ decreasing?
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
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MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Critical points and intervals of increase and decrease, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag
Math Learning Centre A space to study math together. Free math graduate and undergraduate TA support. Mon - Fri: 12 pm - 5 pm in LSK 301&302 and 5 pm - 7 pm online. Private tutor We can help to | 541 | 2,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | latest | en | 0.92852 |
https://www.jiskha.com/display.cgi?id=1286327230 | 1,503,435,194,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112682.87/warc/CC-MAIN-20170822201124-20170822221124-00717.warc.gz | 889,123,268 | 3,922 | # MATH!! NEED HELP! If I don't pass this, Ill fail!!
posted by .
-20 = 7 + x/2
need to show work!
:(
I don't get it!
Help!?
• MATH!! NEED HELP! If I don't pass this, Ill fail!! -
-20 = 7 + x/2
-20 - 7 = x/2
-27 = x/2
Multiply both sides of the equation by 2.
-54 = x
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### 32.1: AC Sources TABLE OFCONTENTS X ## Chapter 1: Units, Dimensions, and Measurements 301.1: The Scope of Physics301.2: Orders of Magnitude301.3: Units and Standards of Measurement301.4: Base Quantities and Derived Quantities301.5: Conversion of Units301.6: Accuracy and Precision301.7: Random and Systematic Errors301.8: Rules for Significant Figures301.9: Significant Figures in Calculations301.10: Dimensional Analysis301.11: Solving Problems in Physics ## Chapter 2: Vectors and Scalars 302.1: Introduction to Scalars302.2: Introduction to Vectors302.3: Vector Components in the Cartesian Coordinate System302.4: Polar and Cylindrical Coordinates302.5: Vector Algebra: Graphical Method302.6: Vector Algebra: Method of Components302.7: Scalar Product (Dot Product)302.8: Vector Product (Cross Product) ## Chapter 3: Motion Along a Straight Line 303.1: Position and Displacement303.2: Average Velocity303.3: Instantaneous Velocity - I303.4: Instantaneous Velocity - II303.5: Average Acceleration303.6: Instantaneous Acceleration303.7: Kinematic Equations - I303.8: Kinematic Equations - II303.9: Kinematic Equations - III303.10: Kinematic Equations: Problem Solving303.11: Free-falling Bodies: Introduction303.12: Free-falling Bodies: Example303.13: Velocity and Position by Graphical Method303.14: Velocity and Position by Integral Method ## Chapter 4: Motion in Two or Three Dimensions 304.1: Position and Displacement Vectors304.2: Average and Instantaneous Velocity Vectors304.3: Acceleration Vectors304.4: Direction of Acceleration Vectors304.5: Projectile Motion304.6: Projectile Motion: Equations304.7: Projectile Motion: Example304.8: Uniform Circular Motion304.9: Non-uniform Circular Motion304.10: Relative Velocity in One Dimension304.11: Relative Velocity in Two Dimensions ## Chapter 5: Newton's Laws of Motion 305.1: Force305.2: Types of Forces305.3: Newton's First Law: Introduction305.4: Newton's First Law: Application305.5: Internal and External Forces305.6: Newton's Second Law305.7: Mass and Weight305.8: Weightlessness305.9: Newton's Third Law: Introduction305.10: Newton's Third Law: Examples305.11: Drawing Free-body Diagrams: Rules305.12: Free Body Diagrams: Examples305.13: Inertial Frames of Reference305.14: Non-inertial Frames of Reference ## Chapter 6: Application of Newton's Laws of Motion 306.1: First Law: Particles in One-dimensional Equilibrium306.2: First Law: Particles in Two-dimensional Equilibrium306.3: Second Law: Motion under Same Force306.4: Second Law: Motion under Same Acceleration306.5: Frictional Force306.6: Static and Kinetic Frictional Force306.7: Dynamics of Circular Motion306.8: Dynamics Of Circular Motion: Applications ## Chapter 7: Work and Kinetic Energy 307.1: Work307.2: Positive, Negative, and Zero Work307.3: Energy307.4: Types of Potential Energy307.5: Types of Kinetic Energy307.6: Kinetic Energy - I307.7: Kinetic Energy - II307.8: Work-energy Theorem307.9: Power307.10: Power Expended by a Constant Force ## Chapter 8: Potential Energy and Energy Conservation 308.1: Gravitational Potential Energy308.2: Elastic Potential Energy308.3: Conservative Forces308.4: Non-conservative Forces308.5: Conservation of Energy308.6: Conservation of Energy: Application308.7: Force and Potential Energy in One Dimension308.8: Force and Potential Energy in Three Dimensions308.9: Energy Diagrams - I308.10: Energy Diagrams - II ## Chapter 9: Linear Momentum, Impulse and Collisions 309.1: Linear Momentum309.2: Force and Momentum309.3: Impulse309.4: Impulse-Momentum Theorem309.5: Conservation of Momentum: Introduction309.6: Conservation of Momentum: Problem Solving309.7: Types Of Collisions - I309.8: Types of Collisions - II309.9: Elastic Collisions: Introduction309.10: Elastic Collisions: Case Study309.11: Collisions in Multiple Dimensions: Introduction309.12: Collisions in Multiple Dimensions: Problem Solving309.13: Center of Mass: Introduction309.14: Significance of Center of Mass309.15: Rocket Propulsion in Empty Space - I309.16: Rocket Propulsion In Empty Space - II309.17: Rocket Propulsion in Gravitational Field - I309.18: Rocket Propulsion in Gravitational Field - II ## Chapter 10: Rotation and Rigid Bodies 3010.1: Angular Velocity and Displacement3010.2: Angular Velocity and Acceleration3010.3: Rotation with Constant Angular Acceleration - I3010.4: Rotation with Constant Angular Acceleration - II3010.5: Relating Angular And Linear Quantities - I3010.6: Relating Angular And Linear Quantities - II3010.7: Moment of Inertia3010.8: Moment of Inertia and Rotational Kinetic Energy3010.9: Moment of Inertia: Calculations3010.10: Parallel-axis Theorem3010.11: Moment of Inertia of Compound Objects ## Chapter 11: Dynamics of Rotational Motions 3011.1: Torque3011.2: Net Torque Calculations3011.3: Work and Power for Rotational Motion3011.4: Work-Energy Theorem for Rotational Motion3011.5: Angular Momentum: Single Particle3011.6: Angular Momentum: Rigid Body3011.7: Conservation of Angular Momentum3011.8: Conservation of Angular Momentum: Application3011.9: Gyroscope3011.10: Gyroscope: Precession ## Chapter 12: Equilibrium and Elasticity 3012.1: Static Equilibrium - I3012.2: Static Equilibrium - II3012.3: Center of Gravity3012.4: Finding the Center of Gravity3012.5: Rigid Body Equilibrium Problems - I3012.6: Rigid Body Equilibrium Problems - II3012.7: Stress3012.8: Strain and Elastic Modulus3012.9: Elasticity3012.10: Plasticity ## Chapter 13: Fluid Mechanics 3013.1: Density3013.2: Pressure of Fluids3013.3: Pascal's Law3013.4: Application of Pascal's Law3013.5: Buoyancy3013.6: Archimedes' Principle3013.7: Density and Archimedes' Principle3013.8: Laminar and Turbulent Flow3013.9: Equation of Continuity3013.10: Bernoulli's Equation3013.11: Bernoulli's Principle3013.12: Viscosity3013.13: Poiseuille's Law and Reynolds Number ## Chapter 14: Gravitation 3014.1: Newton's Law of Gravitation3014.2: Gravity between Spherical Bodies3014.3: Acceleration due to Gravity on Earth3014.4: Acceleration due to Gravity on Other Planets3014.5: Apparent Weight and the Earth's Rotation3014.6: Variation in Acceleration due to Gravity near the Earth's Surface3014.7: Potential Energy due to Gravitation3014.8: Escape Velocity3014.9: Circular Orbits and Critical Velocity for Satellites3014.10: Energy of a Satellite in a Circular Orbit3014.11: Kepler's First Law of Planetary Motion3014.12: Kepler's Second Law of Planetary Motion3014.13: Kepler's Third Law of Planetary Motion3014.14: Tidal Forces3014.15: Schwarzschild Radius and Event Horizon3014.16: Detection of Black Holes3014.17: Principle of Equivalence3014.18: Space-Time Curvature and the General Theory of Relativity ## Chapter 15: Oscillations 3015.1: Simple Harmonic Motion3015.2: Characteristics of Simple Harmonic Motion3015.3: Oscillations about an Equilibrium Position3015.4: Energy in Simple Harmonic Motion3015.5: Simple Harmonic Motion and Uniform Circular Motion3015.6: Simple Pendulum3015.7: Torsional Pendulum3015.8: Damped Oscillations3015.9: Types of Damping3015.10: Forced Oscillations3015.11: Concept of Resonance and its Characteristics ## Chapter 16: Waves 3016.1: Travelling Waves3016.2: Wave Parameters3016.3: Equations of Wave Motion3016.4: Velocity and Acceleration of a Wave3016.5: Kinetic and Potential Energy of a Wave3016.6: Energy and Power of a Wave3016.7: Interference and Superposition of Waves3016.8: Reflection of Waves3016.9: Propagation of Waves3016.10: Standing Waves3016.11: Modes of Standing Waves - I ## Chapter 17: Sound 3017.1: Sound Waves3017.2: Perception of Sound Waves3017.3: Speed of Sound in Solids and Liquids3017.4: Speed of Sound in Gases3017.5: Sound Intensity3017.6: Sound Intensity Level3017.7: Sound Waves: Interference3017.8: Sound Waves: Resonance3017.9: Doppler Effect - I3017.10: Doppler Effect - II3017.11: Shock Waves ## Chapter 18: Temperature and Heat 3018.1: Temperature and Thermal Equilibrium3018.2: Zeroth Law of Thermodynamics3018.3: Thermometers and Temperature Scales3018.4: Gas Thermometers and the Kelvin Scale3018.5: Thermal Expansion3018.6: Thermal Stress3018.7: Heat Flow and Specific Heat3018.8: Phase Changes3018.9: Mechanisms of Heat Transfer I3018.10: Mechanisms of Heat Transfer II ## Chapter 19: The Kinetic Theory of Gases 3019.1: Ideal Gas Equation3019.2: Van der Waals Equation3019.3: pV-Diagrams3019.4: Kinetic Theory of an Ideal Gas3019.5: Molecular Kinetic Energy3019.6: Distribution of Molecular Speeds3019.7: Phase Diagram ## Chapter 20: The First Law of Thermodynamics 3020.1: Thermodynamic Systems3020.2: Work Done During Volume Change3020.3: Path Between Thermodynamics States3020.4: Internal Energy3020.5: First Law of Thermodynamics3020.6: Cyclic Processes And Isolated Systems3020.7: Isothermal Processes3020.8: Isochoric and Isobaric Processes3020.9: Heat Capacities of an Ideal Gas I3020.10: Heat Capacities of an Ideal Gas II3020.11: Heat Capacities of an Ideal Gas III3020.12: Adiabatic Processes for an Ideal Gas3020.13: Pressure and Volume in an Adiabatic Process3020.14: Work Done in an Adiabatic Process ## Chapter 21: The Second Law of Thermodynamics 3021.1: Reversible and Irreversible Processes3021.2: Heat Engines3021.3: Refrigerators and Heat Pumps3021.4: Statements of the Second Law of Thermodynamics3021.5: The Carnot Cycle3021.6: Efficiency of The Carnot Cycle3021.7: The Carnot Cycle and the Second Law of Thermodynamics3021.8: Entropy3021.9: Entropy Change in Reversible Processes3021.10: Entropy and the Second Law of Thermodynamics ## Chapter 22: Electric Charges and Fields 3022.1: Electric Charges3022.2: Sources and Properties of Electric Charge3022.3: Conductors and Insulators3022.4: Charging Conductors By Induction3022.5: Coulomb's Law3022.6: Coulomb's Law and The Principle of Superposition3022.7: Electric Field3022.8: Electric Field of Two Equal and Opposite Charges3022.9: Continuous Charge Distributions3022.10: Electric Field Lines3022.11: Properties of Electric Field Lines3022.12: Electric Dipoles and Dipole Moment3022.13: Induced Electric Dipoles ## Chapter 23: Gauss's Law 3023.1: Electric Flux3023.2: Gauss's Law3023.3: Gauss's Law: Spherical Symmetry3023.4: Gauss's Law: Cylindrical Symmetry3023.5: Gauss's Law: Planar Symmetry3023.6: Electric Field Inside a Conductor3023.7: Charge on a Conductor3023.8: Electric Field at the Surface of a Conductor ## Chapter 24: Electric Potential 3024.1: Electric Potential Energy3024.2: Electric Potential Energy in a Uniform Electric Field3024.3: Electric Potential Energy of Two Point Charges3024.4: Electric Potential and Potential Difference3024.5: Finding Electric Potential From Electric Field3024.6: Equipotential Surfaces and Field Lines3024.7: Equipotential Surfaces and Conductors3024.8: Determining Electric Field From Electric Potential ## Chapter 25: Capacitance 3025.1: Capacitors and Capacitance3025.2: Spherical and Cylindrical Capacitor3025.3: Capacitors in Series and Parallel3025.4: Energy Stored in a Capacitor3025.5: Capacitor With A Dielectric3025.6: Dielectric Polarization in a Capacitor ## Chapter 26: Current and Resistance 3026.1: Electrical Current3026.2: Drift Velocity3026.3: Current Density3026.4: Resistivity3026.5: Resistance3026.6: Ohm's Law3026.7: Electrical Power ## Chapter 27: Direct-Current Circuits 3027.1: Electromotive Force3027.2: Resistors In Series3027.3: Resistors In Parallel3027.4: Kirchhoff's Rules3027.5: Galvanometer3027.6: Ammeter3027.7: RC Circuits: Charging A Capacitor3027.8: RC Circuits: Discharging A Capacitor ## Chapter 28: Magnetic Forces and Fields 3028.1: Magnetism3028.2: Magnetic Fields3028.3: Magnetic Field Lines3028.4: Magnetic Flux3028.5: Motion Of A Charged Particle In A Magnetic Field3028.6: Magnetic Force On A Current-Carrying Conductor3028.7: Force On A Current Loop In A Magnetic Field3028.8: Torque On A Current Loop In A Magnetic Field ## Chapter 29: Sources of Magnetic Fields 3029.1: Magnetic Field due to Moving Charges3029.2: Biot-Savart Law3029.3: Magnetic Field Due To A Thin Straight Wire3029.4: Magnetic Force Between Two Parallel Currents3029.5: Magnetic Field Of A Current Loop3029.6: Ampere's Law3029.7: Ampere's Law: Problem-Solving3029.8: Solenoids3029.9: Magnetic Field of a Solenoid3029.10: Toroids3029.11: Diamagnetism3029.12: Paramagnetism3029.13: Ferromagnetism ## Chapter 30: Electromagnetic Induction 3030.1: Induction3030.2: Faraday's Law3030.3: Lenz's Law3030.4: Motional Emf3030.5: Induced Electric Fields3030.6: Displacement Current3030.7: Significance of Displacement Current3030.8: Electromagnetic Fields3030.9: Maxwell's Equation Of Electromagnetism3030.10: Symmetry in Maxwell's Equations3030.11: Electric Generator: Alternator3030.12: Back EMF ## Chapter 31: Inductance 3031.1: Mutual Inductance3031.2: Self-Inductance3031.3: Inductors3031.4: Energy In A Magnetic Field3031.5: RL Circuits3031.6: Current Growth And Decay In RL Circuits3031.7: LC Circuits3031.8: Oscillations In An LC Circuit3031.9: RLC Series Circuits ## Chapter 32: Alternating-Current Circuits 3032.1: AC Sources3032.2: Resistor in an AC Circuit3032.3: Capacitor in an AC Circuit3032.4: Inductor in an AC Circuit3032.5: RLC Series Circuits: Introduction3032.6: RLC Series Circuits: Impedance3032.7: Power in an AC Circuit3032.8: Resonance in an AC Circuit ## Chapter 33: Electromagnetic Waves 3033.1: Electromagnetic Waves3033.2: The Electromagnetic Spectrum3033.3: Plane Electromagnetic Waves I3033.4: Plane Electromagnetic Waves II3033.5: Propagation Speed of Electromagnetic Waves3033.6: Electromagnetic Waves in Matter3033.7: Energy Carried By Electromagnetic Waves3033.8: Intensity Of Electromagnetic Waves3033.9: Momentum And Radiation Pressure Full Table of Contents
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AC Sources
### 32.1: AC Sources
Direct current is a flow of electric charge in only one direction and has a steady state of constant voltage in the circuit. Rectifiers, batteries, commutator-equipped generators, and fuel cells are some examples of devices that generate direct current. Nowadays, most applications use a time-varying voltage source. Alternating current is a flow of electric charge that periodically reverses direction. An alternating current is produced by an alternating emf that is generated in a power plant. If the alternating current source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Some electrical devices, such as fans, bulbs, air conditioners, and motors, run on alternating current. Different alternating current voltages and frequencies are used in homes, offices, and other places around the world. The potential difference between two electrical outlets in a typical home alternates sinusoidally with a frequency of 60 Hz or 50 Hz and an amplitude of 156 V or 311 V depending on whether you live in the United States or Europe, respectively. The electrical outlets in the US and Europe have a potential difference of 120 V or 220 V, respectively. However, these voltages are the typical voltages in the outlets, not the peak values. When a resistor is connected to an alternating current voltage source, the voltage and current vary in time across the resistor. The voltage fluctuates sinusoidally with time at a fixed frequency on either the battery terminals or the resistor. | 4,229 | 15,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.681862 |
https://pakebooks.com/07-a-a-adeyanju-a-t-ademola-b-s-ogundarepdf | 1,675,486,827,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00294.warc.gz | 468,612,909 | 15,879 | ## A differential equation is said to be an integro-differential equation (IDE) if it contains the integrals of the unknown function. When the current state of such an integro-differential equation now depends on the previous states, it is known to be a time-delay integro-differential equation. Indeed, it is a well-known fact that stability and boundedness properties of solutions of second order (also higher order) ordinary differential equations and integro-differential equations with or without delay have many applications in many fields of science and technology such as biology, medicine, engineering, informa- tion system, control theory and financial mathematics. Therefore, the study of their qualitative properties has attracted the attention of many researchers, see ( [1] - [41]) and references contained in them. Readers are referred to [3] for an exposi- tory treatment of Volterra integral and differential equations.
A differential equation is said to be an integro-differential equation (IDE) if it
contains the integrals of the unknown function. When the current state of such an
integro-differential equation now depends on the previous states, it is known to be
a time-delay integro-differential equation.
Indeed, it is a well-known fact that stability and boundedness properties of
solutions of second order (also higher order) ordinary differential equations and
integro-differential equations with or without delay have many applications in many
fields of science and technology such as biology, medicine, engineering, informa-
tion system, control theory and financial mathematics. Therefore, the study of
their qualitative properties has attracted the attention of many researchers, see ( [1]
- [41]) and references contained in them. Readers are referred to [3] for an exposi-
tory treatment of Volterra integral and differential equations. | 374 | 1,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.933735 |
https://www.tek.com/fr/blog/measuring-power-using-your-keithley-dmms-ratio-function | 1,721,077,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00096.warc.gz | 891,152,468 | 13,333 | Contactez-nous
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# Measuring power using your Keithley DMM’s Ratio function
By Andrew Kirby
One interesting application of the Ratio function, measuring power using a digital multimeter such as the DMM6500, is achieved with the help of TSP scripting and a low ohm current sensing resistor. The script leverages the fact that the Ratio function stores a voltage measurement for both sense and input voltages in a single reading, displaying Input voltage over sense voltage.
The Ratio function compares the voltage across the input terminals to that on the sense terminals and outputs their quotient, input voltage divided by sense voltage. Because this measurement encodes two separate voltage readings, some interesting things can be done if TSP scripting is employed.
For example, in the application below, the Ratio function measures power by placing an external low ohm shunt resistor between the sense terminals. This means you can measure current using the sense terminal and voltage on the input. With these two values you can calculate power consumption by any segment of a given circuit
Before that can happen however, you must first decode the ratio measurement to extract the voltage and current readings. The voltage on the sense terminals is stored in the “Extra Value” field in a “Full” style buffer, which is used to store the readings (in this case, called readingBuffer). From there, you need to multiply the ratio reading with the corresponding sense voltage reading to get the voltage on the input terminals.
Now that both voltages are stored, you can divide the sense voltage measurement by the shunt resistor value to obtain current. Finally, because both current and voltage at a given point in time have been obtained, you can calculate the power being consumed by simply multiplying the two together. The power reading can be output to an active writable buffer (powerBuffer) to be displayed on the screen. This may seem a little complex, but in reality, it only takes a couple dozen lines of code, as shown below.
This script will work with any Keithley DMM with TSP support. To learn about the full range Tektronix and Keithley digital multimeters including the new DMM6500, go to https://www.tek.com/digital-multimeter. | 549 | 2,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.78869 |
http://stackoverflow.com/questions/5750937/how-to-convert-22-7-to-ieee-754-floating-point | 1,427,882,468,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131304412.34/warc/CC-MAIN-20150323172144-00007-ip-10-168-14-71.ec2.internal.warc.gz | 264,078,235 | 16,468 | # How to convert 2+(2/7) to IEEE 754 floating point
Can someone explain to me the steps to convert a number in decimal format (such as 2+(2/7)) into IEEE 754 Floating Point representation? Thanks!
-
Are you asking about an arithmetic expression evaluator in assembly code, or what? – 500 - Internal Server Error Apr 21 '11 at 23:46
arithmetic expression – Casey Flynn Apr 21 '11 at 23:50
You mean other than just `2.0+(2.0/7.0)`? Do you want the binary representation of those numbers in IEEE754 and a description of how the addition and divide works, or something else? – Chris Dodd Apr 22 '11 at 16:56
First, `2 + 2/7` isn't in what most people would call "decimal format". "Decimal format" would more commonly be used to indicate a number like:
``````2.285714285714285714285714285714285714285714...
``````
Even the `...` is a little bit fast and loose. More commonly, the number would be truncated or rounded to some number of decimal digits:
``````2.2857142857142857
``````
Of course, at this point, it is no longer exactly equal to `2 + 2/7`, but is "close enough" for most uses.
We do something similar to convert a number to a IEEE-754 format; instead of base 10, we begin by writing the number in base 2:
``````10.010010010010010010010010010010010010010010010010010010010010...
``````
Next we "normalize" the number, by writing it in the form `2^e * 1.xxx...` for some exponent `e` (specifically, the digit position of the leading bit of our number):
``````2^1 * 1.0010010010010010010010010010010010010010010010010010010010010...
``````
At this point, we have to choose a specific IEEE-754 format, because we need to know how many digits to keep around. Let's choose "single-precision", which has a 24-bit significand. We round the repeating binary number to 24 bits:
``````2^1 * 1.00100100100100100100100 10010010010010010010010010010010010010...
24 leading bits bits to be rounded away
``````
Because the trailing bits to be rounded off are larger than `1000...`, the number rounds up to:
``````2^1 * 1.00100100100100100100101
``````
Now, how does this value actually get encoded in IEEE-754 format? The single-precision format has a leading signbit (zero, because the number is positive), followed by eight bits that contain the value `127 + e` in binary, followed by the fractional part of the significand:
``````0 10000000 00100100100100100100101
s exponent fraction of significand
``````
In hexadecimal, this gives `0x40124925`.
- | 699 | 2,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2015-14 | latest | en | 0.864058 |
https://web2.0calc.com/questions/a-bit-confused-on-what-to-do | 1,642,623,842,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00267.warc.gz | 615,836,287 | 5,699 | +0
# A bit confused on what to do
+1
172
1
+50
What is the largest number of regions formed by 6 planes in space?
This was part b of the problem, I already solved part a and I am confused on why it isn't the same answer as part a.
Part a's question:
By drawing 6 lines in the plane, what is the largest number of regions we can create?
I got an answer of 22 using a recursive formula: f(n) = f(n-1) + n.
Can anyone give me any hints on how to do part b?
all help is appreciated - Zekken4717
p.s this problem isn't due at any time so I have time to do it but quicker is better :)
p.s 22 is the right answer for part a.
Aug 8, 2021
edited by Zekken Aug 8, 2021
edited by Zekken Aug 8, 2021
#1
+1
When a plane intersects a different plane, we get two more regions, so
f(n) = f(n - 1) + 2(n - 1).
Then
f(1) = 2
f(2) = 4
f(3) = 8
f(4) = 14
f(5) = 22
f(6) = 32
Aug 8, 2021 | 309 | 891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-05 | latest | en | 0.945529 |
https://qliktech-public-v7.hosted.jivesoftware.com/thread/311176 | 1,544,537,410,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823621.10/warc/CC-MAIN-20181211125831-20181211151331-00495.warc.gz | 727,117,776 | 21,553 | 1 Reply Latest reply: Aug 16, 2018 11:08 AM by Bala Bhaskar
# How to get revenue bracket for the matrix?
Hi All,
Please look at the attachment and I will make buckets like below
I am trying like below expression, but not working
If(RBYM_Revenue>0 and RBYM_Revenue>=50000,'0-50000',
If(RBYM_Revenue>500000 and RBYM_Revenue>=1000000,'0-100000'))
• add a filter box to select a revenue bracket for the matrix (0-50,000, 50,000-100,000, 100,000- 500,000, 5000,000, 1m, 1m-5m,5m -10m, 10m-20m, 20m+)
Thanks,
Manoj
• ###### Re: How to get revenue bracket for the matrix?
May be this:
(0-50,000, 50,000-100,000, 100,000- 500,000, 5000,000, 1m, 1m-5m,5m -10m, 10m-20m, 20m+)
If(RBYM_Revenue > 0 and RBYM_Revenue <= 50,000,'0-50,000',
If(RBYM_Revenue > 50,000 and RBYM_Revenue <= 100,000,'50,000-100,000',
If(RBYM_Revenue > 100,000 and RBYM_Revenue <= 500,000,'100,000-500,000',
If(RBYM_Revenue <= 5000,000,'5000,000',
If(RBYM_Revenue <= 1,0000,000,'1M',
If(RBYM_Revenue > 1,0000,000 and RBYM_Revenue <= 5,0000,000,'1M-5M',
If(RBYM_Revenue > 5,0000,000 and RBYM_Revenue <= 10,0000,000,'5M-10M',
If(RBYM_Revenue > 10,0000,000 and RBYM_Revenue <= 20,0000,000,'10M-20M','20M+')))))))) | 505 | 1,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-51 | latest | en | 0.733489 |
https://electronics.stackexchange.com/questions/416539/what-does-resistor-as-a-reference-means | 1,656,522,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00327.warc.gz | 274,928,325 | 67,073 | # What does "resistor as a reference" means?
I have a book with some projects examples that use Arduino, one of the projects use piezo to detect vibrations. The book says the following:
When plugged into 5V, the sensor (piezo) can detect vibrations that can be read by the Arduino's analog inputs. You'll need to plug in high value resistor (like 1-megaohm) as reference to ground for this to work well.
The book also says in other page:
Lower resistor values will make the piezo less sensitive to vibrations.
What does this mean? What does "resistor as a reference" means? And why would the piezo be less sensitive with less value resistors?
I don't have any experience in electronics, I started learning one week ago, and resistors are one of the most confusing things for me.
Project's circuit scheme:
• piezo parts are high impedance and dont generate much current but with small currents going thru high R values like 1 Meg, you get V=IR ( sorry megaohm) Jan 12, 2019 at 7:25
• If you add the schematic diagram into your question we may be able to help you better. Jan 12, 2019 at 10:06
• @Transistor Done Jan 12, 2019 at 11:05
• Where are you getting these weird white-on-brown schematics, anyway? They look like they're printed on wood or something. Jan 12, 2019 at 13:08
• @DaveTweed from the book I mentioned in the question, the book has some projects and in each project there is scheme for the circuit used in the project, I captured it. Jan 12, 2019 at 13:11
In very simple terms, the piezo transducer can give out highish voltages when subject to vibrations. The piezo transducer has extremely high internal resistance - you can try to measure it on the highest range of your multimeter ohms range - so the current it can give out is very low. As a result we call this a "high-impedance source". (Impedance can be thought of as resistance to current flow.)
Your microcontroller analog input has an input range of 0 to 5 V DC and to get best use of this our signal into it should be in that range of voltages. A 0 to 0.5 V signal, for example, would only use 1/10 of the range of the ADC (analog to digital converter) and for your microcontroller I think that would give you 102 possible values out of 1024.
To maximise the voltage from your weak transducer you need to feed the small current into a high value resistor. With the 1 MΩ value in your circuit we can see from Ohm's Law that you will reach 5 V when the current from the transducer is $$\ I = \frac {V}{R} = \frac {5}{1M} = 5 \ \mu \text A \$$.
The design of the circuit isn't great. The transducer may give out negative voltages or voltages higher than 5 V and they are relying on the microcontroller's internal protection diodes to prevent damage to the IC. A commercial design would be likely to include external protection.
You'll need to plug in high value resistor (like 1-megaohm) as reference to ground for this to work well.
I think we've covered this. The resistor converts the current from the transducer to a voltage for the microcontroller.
Lower resistor values will make the piezo less sensitive to vibrations.
We've covered this. A lower value resistor will result in a lower voltage for a given current from the transducer.
What does "resistor as a reference" means?
It is a poor choice of words. It is really a "load" for the piezo to convert current to voltage.
And why would the piezo be less sensitive with less value resistors?
Explained above.
Resistors are one of the most confusing things for me.
Resistors can be used to limit current in a circuit, decrease the voltage or to convert current into voltage. Keep up your studies and you will start to see where they are used and why as you read more circuits. | 904 | 3,730 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.945339 |
https://aptitude.gateoverflow.in/8439/cat-2021-set-2-quantitative-aptitude-question-14 | 1,670,579,339,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00630.warc.gz | 129,215,740 | 20,470 | 225 views
Let $\text{D}$ and $\text{E}$ be points on sides $\text{AB}$ and $\text{AC},$ respectively, of a triangle $\text{ABC},$ such that $\text{AD}$ : $\text{BD} = 2 : 1$ and $\text{AE}$ : $\text{CE} = 2 : 3.$ If the area of the triangle $\text{ADE}$ is $8 \; \text{sq cm},$ then the area of the triangle $\text{ABC, in sq cm},$ is
Let’s first draw the diagram.
We know that, the area of $\triangle \text{ADE} = \frac{\text{AD}}{\text{AB}} \times \frac{\text{AE}}{\text{AC}} \times \text{Area of}\; \triangle \text{ABC}$
$\Rightarrow 8=\frac{2}{3} \times \frac{2}{5} \times \text{Area of}\; \triangle \text{ABC}$
$\Rightarrow \boxed{\text{Area of}\; \triangle \text{ABC} = 30\; \text{cm}^{2}.}$
Correct Answer $:30$
10.3k points
1 vote
1
1 vote | 280 | 756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-49 | latest | en | 0.431403 |
http://de.metamath.org/mpeuni/21wlkd.html | 1,721,724,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00351.warc.gz | 8,556,828 | 6,602 | Mathbox for Alexander van der Vekens < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > 21wlkd Structured version Visualization version GIF version
Theorem 21wlkd 41143
Description: Construction of a walk from two given edges in a graph. (Contributed by Alexander van der Vekens, 5-Feb-2018.) (Revised by AV, 23-Jan-2021.) (Proof shortened by AV, 14-Feb-2021.) (Revised by AV, 24-Mar-2021.)
Hypotheses
Ref Expression
21wlkd.p 𝑃 = ⟨“𝐴𝐵𝐶”⟩
21wlkd.f 𝐹 = ⟨“𝐽𝐾”⟩
21wlkd.s (𝜑 → (𝐴𝑉𝐵𝑉𝐶𝑉))
21wlkd.n (𝜑 → (𝐴𝐵𝐵𝐶))
21wlkd.e (𝜑 → ({𝐴, 𝐵} ⊆ (𝐼𝐽) ∧ {𝐵, 𝐶} ⊆ (𝐼𝐾)))
21wlkd.v 𝑉 = (Vtx‘𝐺)
21wlkd.i 𝐼 = (iEdg‘𝐺)
Assertion
Ref Expression
21wlkd (𝜑𝐹(1Walks‘𝐺)𝑃)
Proof of Theorem 21wlkd
Dummy variable 𝑘 is distinct from all other variables.
StepHypRef Expression
1 21wlkd.p . . . 4 𝑃 = ⟨“𝐴𝐵𝐶”⟩
2 s3cli 13476 . . . 4 ⟨“𝐴𝐵𝐶”⟩ ∈ Word V
31, 2eqeltri 2684 . . 3 𝑃 ∈ Word V
43a1i 11 . 2 (𝜑𝑃 ∈ Word V)
5 21wlkd.f . . . 4 𝐹 = ⟨“𝐽𝐾”⟩
6 s2cli 13475 . . . 4 ⟨“𝐽𝐾”⟩ ∈ Word V
75, 6eqeltri 2684 . . 3 𝐹 ∈ Word V
87a1i 11 . 2 (𝜑𝐹 ∈ Word V)
91, 521wlkdlem1 41132 . . 3 (#‘𝑃) = ((#‘𝐹) + 1)
109a1i 11 . 2 (𝜑 → (#‘𝑃) = ((#‘𝐹) + 1))
11 21wlkd.s . . 3 (𝜑 → (𝐴𝑉𝐵𝑉𝐶𝑉))
12 21wlkd.n . . 3 (𝜑 → (𝐴𝐵𝐵𝐶))
13 21wlkd.e . . 3 (𝜑 → ({𝐴, 𝐵} ⊆ (𝐼𝐽) ∧ {𝐵, 𝐶} ⊆ (𝐼𝐾)))
141, 5, 11, 12, 1321wlkdlem10 41142 . 2 (𝜑 → ∀𝑘 ∈ (0..^(#‘𝐹)){(𝑃𝑘), (𝑃‘(𝑘 + 1))} ⊆ (𝐼‘(𝐹𝑘)))
151, 5, 11, 1221wlkdlem5 41136 . 2 (𝜑 → ∀𝑘 ∈ (0..^(#‘𝐹))(𝑃𝑘) ≠ (𝑃‘(𝑘 + 1)))
16 21wlkd.v . . . . 5 𝑉 = (Vtx‘𝐺)
17161vgrex 25679 . . . 4 (𝐴𝑉𝐺 ∈ V)
18173ad2ant1 1075 . . 3 ((𝐴𝑉𝐵𝑉𝐶𝑉) → 𝐺 ∈ V)
1911, 18syl 17 . 2 (𝜑𝐺 ∈ V)
20 21wlkd.i . 2 𝐼 = (iEdg‘𝐺)
211, 5, 1121wlkdlem4 41135 . 2 (𝜑 → ∀𝑘 ∈ (0...(#‘𝐹))(𝑃𝑘) ∈ 𝑉)
224, 8, 10, 14, 15, 19, 16, 20, 211wlkd 40895 1 (𝜑𝐹(1Walks‘𝐺)𝑃)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ≠ wne 2780 Vcvv 3173 ⊆ wss 3540 {cpr 4127 class class class wbr 4583 ‘cfv 5804 (class class class)co 6549 1c1 9816 + caddc 9818 #chash 12979 Word cword 13146 ⟨“cs2 13437 ⟨“cs3 13438 Vtxcvtx 25673 iEdgciedg 25674 1Walksc1wlks 40796 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-ifp 1007 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-er 7629 df-map 7746 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-card 8648 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-3 10957 df-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-fzo 12335 df-hash 12980 df-word 13154 df-concat 13156 df-s1 13157 df-s2 13444 df-s3 13445 df-1wlks 40800 This theorem is referenced by: 21wlkond 41144 2trld 41145 umgr2adedgwlk 41152
Copyright terms: Public domain W3C validator | 2,627 | 4,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.226684 |
http://www.numbersaplenty.com/546875 | 1,581,877,585,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141396.22/warc/CC-MAIN-20200216182139-20200216212139-00162.warc.gz | 231,693,938 | 3,660 | Search a number
546875 = 577
BaseRepresentation
bin10000101100000111011
31000210011122
42011200323
5120000000
615415455
74435250
oct2054073
91023148
10546875
1134396a
1222458b
13161bc4
14103427
15ac085
hex8583b
546875 has 16 divisors (see below), whose sum is σ = 781248. Its totient is φ = 375000.
The previous prime is 546869. The next prime is 546881. The reversal of 546875 is 578645.
Multipling 546875 by its sum of digits (35), we get a square (19140625 = 43752).
546875 divided by its sum of digits (35) gives a 6-th power (15625 = 56).
It is an interprime number because it is at equal distance from previous prime (546869) and next prime (546881).
It is an ABA number since it can be written as A⋅BA, here for A=7, B=5.
It is not a de Polignac number, because 546875 - 24 = 546859 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (35).
It is a d-powerful number, because it can be written as 52 + 49 + 84 + 67 + 72 + 54 .
It is a Duffinian number.
Its product of digits (33600) is a multiple of the sum of its prime divisors (12).
It is a plaindrome in base 12.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 78122 + ... + 78128.
It is an arithmetic number, because the mean of its divisors is an integer number (48828).
546875 is a Friedman number, since it can be written as 7*5^(6-8+5+4), using all its digits and the basic arithmetic operations.
2546875 is an apocalyptic number.
546875 is a deficient number, since it is larger than the sum of its proper divisors (234373).
546875 is an frugal number, since it uses more digits than its factorization.
546875 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 42 (or 12 counting only the distinct ones).
The product of its digits is 33600, while the sum is 35.
The square root of 546875 is about 739.5099728875. The cubic root of 546875 is about 81.7766577547.
The spelling of 546875 in words is "five hundred forty-six thousand, eight hundred seventy-five". | 628 | 2,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.900979 |
https://nl.mathworks.com/matlabcentral/cody/solutions/1834483 | 1,575,594,234,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482954.0/warc/CC-MAIN-20191206000309-20191206024309-00189.warc.gz | 489,226,275 | 16,179 | Cody
# Problem 327. TRON challenge
Solution 1834483
Submitted on 2 Jun 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
urlwrite('https://sites.google.com/a/alfnie.com/alfnie/software/SetSolutionScore.p?attredirects=0&d=1','SetSolutionScore.p'); SetSolutionScore(100); doDraw=false; %set to true for display params=struct('n',50,'str','NESW','D',[-1,50,1,-50],'wins',[0,0,0]); rand('state',0); for nboards=1:100, % initialize board params.board=zeros(params.n); params.board([1,end],:)=-3; params.board(:,[1,end])=-3; [i,j]=find(params.board|1); [nill,pos1]=max((i==2|j==2)+~params.board(:)+rand(params.n*params.n,1)); params.board(pos1)=1; [nill,pos2]=max((i==2|j==2)+~params.board(:)+rand(params.n*params.n,1)); params.board(pos2)=2; while 1 % blue's next movement params.state=rand('state'); assignin('caller','params',params); d1=tron(params.board); params=evalin('caller','params'); rand('state',params.state); % red's next movement i=find(params.board==2); [nill,j]=max(~params.board(i+params.D)+.5*(params.board(i-params.D)==-2)); d2=params.str(j); % evaluate movements d1=findstr(params.str,d1); d2=findstr(params.str,d2); assert(~isempty(d1)); params.board(pos1)=-1; params.board(pos2)=-2; pos1=pos1+params.D(d1); pos2=pos2+params.D(d2); if pos1==pos2||(params.board(pos1)&¶ms.board(pos2)), %tie params.wins(2)=params.wins(2)+1; if doDraw, disp('tie'); end break; elseif params.board(pos1), %loose if doDraw, disp('loose'); end params.wins(3)=params.wins(3)+1; break; elseif params.board(pos2), %win if doDraw, disp('win'); end params.wins(1)=params.wins(1)+1; break; end params.board(pos1)=1; params.board(pos2)=2; if doDraw image(4+params.board); axis equal off; colormap([1,1,1;.5,0,0;0,0,.5;0,0,0;0,0,1;1,0,0]); set(gcf,'color','k'); drawnow; end end if doDraw, pause; end end disp(sprintf('%d wins; %d ties; %d looses',params.wins)); % Score = 100+#Looses-#Wins SetSolutionScore(100+params.wins*[-1;0;1]); assert(params.wins(1)>=90,sprintf('%d wins; %d ties; %d looses',params.wins));
36 wins; 27 ties; 37 looses
36 wins; 27 ties; 37 looses | 728 | 2,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.191639 |
https://jsxgraph.uni-bayreuth.de/wiki/index.php?title=Five_Circle_Theorem&oldid=2912 | 1,632,607,237,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057775.50/warc/CC-MAIN-20210925202717-20210925232717-00366.warc.gz | 361,009,350 | 6,644 | # Five Circle Theorem
### The underlying JavaScript code
var brd = JXG.JSXGraph.initBoard('jxgbox',{boundingbox:[-5,5,5,-5]});
var p = [];
var l = [];
var i = [];
var c = [];
var j = [];
p[0] = brd.createElement('point',[-2.5,-3],{name:''});
p[1] = brd.createElement('point',[-0,4],{name:''});
p[2] = brd.createElement('point',[2.5,-3],{name:''});
p[3] = brd.createElement('point',[-4,0],{name:''});
p[4] = brd.createElement('point',[4,0],{name:''});
for (k=0;k<5;k++) {
l[k] = brd.createElement('segment',[p[k],p[(k+1)%5]],{});
}
for (k=0;k<5;k++) {
i[k] = brd.createElement('point',[brd.intersection(l[k],l[(k+2)%5],0)],{name:''});
}
for (k=0;k<5;k++) {
c[k] = brd.createElement('circumcircle',[p[k],i[k],i[(k+2)%5]],{});
c[k][1].setProperty({strokeColor:'gray',strokeWidth:1});
c[k][0].setProperty({visible:false});
l[k].setProperty({strokeColor:'gray',strokeWidth:1});
}
for (k=0;k<5;k++) {
j[k] = brd.createElement('point',[brd.intersection(c[k][1],c[(k+2)%5][1],0)],{style:4,name:''});
}
cc = brd.createElement('circumcircle',[j[0],j[2],j[3]],{strokeColor:'red'});
cc[1].setProperty({strokeColor:'red',strokeWidth:2}); | 417 | 1,131 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 2.765625 | 3 | CC-MAIN-2021-39 | longest | en | 0.272461 |
https://bdnyc.org/blog/tag/photometry/ | 1,726,574,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00027.warc.gz | 112,241,485 | 16,200 | # Ground Based Photometry
Here I want to calculate some photometric points from spectra for comparison with published values for a bunch of known brown dwarfs.
In order to get the true magnitude $$m$$ for an object, I first need to calculate the instrumental magnitude $$m_\text{inst}$$ and then correct for a number of effects. That is, I calculate the apparent magnitude from a particular place on the Earth and then add corrections to determine what it would be if we measured from space.
After all our corrections are made, the magnitude is given by:
$$!m=m_\text{inst}-ZP_m-k_m\cdot X$$
### Instrumental Magnitude
The first term on the right in the equation above is the magnitude measured by the instrument on the ground, given by:
$$!m_\text{inst}=-2.5\log\left(\int f_\lambda (\lambda)\left(\frac{\lambda}{hc}\right)S_m(\lambda)d\lambda\right)$$
Where $$f_\lambda (\lambda)$$ is the energy flux density of the source in units of [erg s-1 cm-2 A-1] and $$S_m(\lambda)$$ is the scalar filter throughput for the band of interest.
Since I will be comparing my calculated magnitudes to photometry taken with photon counting devices, the factor of $$\frac{\lambda}{hc}$$ converts $$f_\lambda (\lambda)$$ to a photon flux density in units [photons s-1 cm-2 A-1].
### Zero Point Correction
The second term in our magnitude equation is a first order correction to compare $$m_\text{inst}$$ to some standard we define as zero. I will use a flux calibrated spectrum of the A0 star Vega to calculate the zero point magnitude for the band:
$$!ZP_m=-2.5\log\left(\int f_{\lambda\text{ Vega}}(\lambda)\left(\frac{\lambda}{hc}\right)S_m(\lambda)d\lambda\right)$$
Just as we obtained our instrumental magnitude above.
### Extinction Correction
The third term is to correct for the extinction of the source flux due to atmospheric absorption. We can get closer to the true apparent magnitude (above the atmosphere) by adding an extinction term:
$$!k_m\cdot\sec (z)=k_m\cdot X$$
Where $$k_m$$ is the extinction coefficient for the band of interest and $$\sec(z)=X$$ is the airmass.
The airmass is the optical path length of the atmosphere, which attenuates the source flux depending on its angle from the zenith $$z$$. Approximating the truly spherical atmosphere as plane-parallel, the airmass goes from $$X=1$$ at $$z=0^\circ$$ to $$X=2$$ at $$z=60^\circ$$. At zenith angles greater than that, the plane-parallel approximation falls apart and the airmass term gets complicated.
Where the airmass is the amount of atmosphere in the line of sight, the extinction coefficient is the amount by which the incident light is attenuated as it travels through the airmass. The extinction coefficient is related to the optical depth $$\tau$$ of the atmosphere as:
$$!m-m_0=-2.5\log\left(\frac{I}{I_0}\right)=-2.5\log (e^{-\tau X})=1.086\cdot\tau\cdot X=k_m\cdot X$$
Where $$m$$ and $$m_0$$ are the magnitudes below and above the atmosphere respectively.
### Example: J21512543-2441000
As an example, I’d like to calculate the 2MASS J-band magnitude of the brown dwarf at 21h51m25.43s -24d41m00s given a low resolution NIR energy flux density from the SpeX Prism instrument on the 3m NASA Infrared Telescope Facility.
Interpolating the filter throughput to the object spectrum and then integrating as in the equation above, I get $$J_\text{inst}=10.046$$ as my instrumental magnitude in the J-band.
Performing the same procedure on the flux calibrated spectrum of Vega, I get $$ZP_J=-5.721$$ for my J-band zero point magnitude.
Checking the FITS file header, I will use $$X=1.444625$$ for the airmass. The mean extinction coefficient for the MKO system J-band is given as $$k_J=0.0153$$ in Tokunaga & Vacca (2007), making the atmospheric correction term $$k_J\cdot X=0.0221$$.
The corrected magnitude is then:
$$!J=J_\text{inst}-ZP_J-k_J\cdot X=10.046-(-5.721)-0.0221=15.745$$
Which is only 0.007 magnitudes off from the value of $$J=15.752$$ from the 2MASS catalog.
### Remaining Problems
As shown in the example above, this works… but not for every object.
2MASS apparent J magnitudes vs. my calculated apparent j magnitudes for 67 brown dwarfs. The solid black line is for perfect agreement and the dashed line is a best fit of the data.
2MASS apparent H magnitudes vs. my calculated apparent h magnitudes for 67 brown dwarfs. The solid black line is for perfect agreement and the dashed line is a best fit of the data.
I whittled down my sample of 875 to only those objects with flux units and airmass values taken at Mauna Kea so that I could use the same extinction coefficient and make sure they are all in the same units of [erg s-1 cm-2 A-1].
Then I pulled the 2MASS catalog J and H magnitudes with uncertainties for these remaining objects and plotted them against my calculated values with uncertainties.
To the left are the plots of the 67 objects that fit the selection criteria in J-band (above) and H-band (below).
Though it’s not the biggest sample, the deviation of the best fit line from unity suggests I’m off by a factor of 0.9 from the 2MASS catalog value across the board.
But more worrisome is the fact that most of the calculated magnitudes are not within the errors of the 2MASS magnitudes. This deviation ranges from very good agreement of a few thousandths of a magnitude up to the worst offenders of about 0.8 mags.
# Filter Effective Wavelength(s)
The effective wavelength of a filter for narrow band photometry can easily be approximated by a constant and just looked up when needed. For broad band photometry, however, the width of the filter and the amount of flux in the band being measured actually come into play.
The effective wavelength of a filter is given by:
$$!\lambda_\text{eff}=\frac{\int \lambda \text{ }f_\lambda (\lambda)\text{ }S(\lambda)\text{ } d\lambda}{\int f_\lambda (\lambda)\text{ } S(\lambda)\text{ } d\lambda}$$
Where $$S(\lambda)$$ is the scalar filter throughput and $$f_\lambda$$ is the flux density in units of [erg s-1 cm-2 A-1] or [photons s-1 cm-2 A-1] depending upon whether you are using an energy measuring or a photon counting detector, respectively.
Here are the results for 67 brown dwarfs with complete spectrum coverage of the 2MASS J-band:
Effective wavelength values for the 2MASS J-band filter. Blue and green circles indicate lambda calculated using photon flux densities (PFD) and energy flux densities (EFD) respectively. Filled circles are for 67 confirmed brown dwarfs. Open circles are for Vega.
The red line on the plot shows the specified value given by 2MASS. For fainter objects like brown dwarfs (filled circles), the calculated effective wavelength of the J-band filter can shift redward by as much as 150 angstroms. Vega (open circles) shifts it blueward by about 30 angstroms.
The difference is small but measurable and demonstrates the dependence of the filter width, source spectrum, and detector type on the effective wavelength $$\lambda_\text{eff}$$ while doing broad band photometry.
# Brown Dwarf Synthetic Photometry
The goal here was to get the synthetic colors in the SDSS, 2MASS and WISE filters of ~2000 model objects generated by the PHOENIX stellar and planetary atmosphere software.
Since it would be silly (and incredibly slow… and much more boring) to just calculate and store every single color for all 12 filter profiles, I wrote a module to calculate colors a la carte.
### The Filters
I got the J, H, and K band relative spectral response (RSR) curves in the 2MASS documentation, the u, g, r, i and z bands from the SDSS documentation, and the W1, W2, W3, and W4 bands from the WISE documentation.
I dumped all my .txt filter files into one directory and wrote a function to grab them all, pull out the wavelength and transmission values, and output the filter name in position [0], x-values in [1], and y-values in [2]:
def get_filters(filter_directory): import glob, os files = glob.glob(filter_directory+'*.txt') if len(files) == 0: print 'No filters in', filter_directory else: filter_names = [os.path.splitext(os.path.basename(i))[0] for i in files] RSR = [open(i) for i in files] filt_data = [filter(None,[map(float,i.split()) for i in j if not i.startswith('#')]) for j in RSR] for i in RSR: i.close() RSR_x = [[x[0] for x in i] for i in filt_data] RSR_y = [[y[1] for y in i] for i in filt_data] filters = {} for i,j,k in zip(filter_names,RSR_x,RSR_y): filters[i] = j, k, center(i) return filters
### Calculating Apparent Magnitudes
We can’t have colors without magnitudes so here’s a function to grab the Teff and log g specified spectra, and calculate the apparent magnitudes in a particular band:
def mags(band, teff='', logg='', bin=1): from scipy.io.idl import readsav from collections import Counter from scipy import trapz, log10, interp s = readsav(path+'modelspeclowresdustywise.save') Fr, Wr = [i for i in s.modelspec['fsyn']], [i for i in s['wsyn']] Tr, Gr = [int(i) for i in s.modelspec['teff']], [round(i,1) for i in s.modelspec['logg']] # The band to compute RSR_x, RSR_y, lambda_eff = get_filters(path)[band] # Option to specify an effective temperature value if teff: t = [i for i, x in enumerate(s.modelspec['teff']) if x == teff] if len(t) == 0: print "No such effective temperature! Please choose from 1400K to 4500K in 50K increments or leave blank to select all." else: t = range(len(s.modelspec['teff'])) # Option to specify a surfave gravity value if logg: g = [i for i, x in enumerate(s.modelspec['logg']) if x == logg] if len(g) == 0: print "No such surface gravity! Please choose from 3.0 to 6.0 in 0.1 increments or leave blank to select all." else: g = range(len(s.modelspec['logg'])) # Pulls out objects that fit criteria above obj = list((Counter(t) & Counter(g)).elements()) F = [Fr[i][::bin] for i in obj] T = [Tr[i] for i in obj] G = [Gr[i] for i in obj] W = Wr[::bin] # Interpolate to find new filter y-values I = interp(W,RSR_x,RSR_y,left=0,right=0) # Convolve the interpolated flux with each filter (FxR = RxF) FxR = [convolution(i,I) for i in F] # Integral of RSR curve over all lambda R0 = trapz(I,x=W) # Integrate to find the spectral flux density per unit wavelength [ergs][s-1][cm-2] then divide by R0 to get [erg][s-1][cm-2][cm-1] F_lambda = [trapz(y,x=W)/R0 for y in FxR] # Calculate apparent magnitude of each spectrum in each filter band Mags = [round(-2.5*log10(m/F_lambda_0(band)),3) for m in F_lambda] result = sorted(zip(Mags, T, G, F, I, FxR), key=itemgetter(1,2)) result.insert(0,W) return result
### Calculating Colors
Now we can calculate the colors. Next, I wrote a function to accept any two bands with options to specify a surface gravity and/or effective temperature as well as a bin size to cut down on computation. Here’s the code:
def colors(first, second, teff='', logg='', bin=1): (Mags_a, T, G) = [[i[j] for i in get_mags(first, teff=teff, logg=logg, bin=bin)[1:]] for j in range(3)] Mags_b = [i[0] for i in get_mags(second, teff=teff, logg=logg, bin=bin)[1:]] colors = [round(a-b,3) for a,b in zip(Mags_a,Mags_b)] print_mags(first, colors, T, G, second=second) return [colors, T, G]
The PHOENIX code gives the flux as Fλ in cgs units [erg][s-1][cm-2][cm-1] but as long as both spectra are in the same units the colors will be the same.
### Makin’ It Handsome
Then I wrote a short function to print out the magnitudes or colors in the Terminal:
def print_mags(first, Mags, T, G, second=''): LAYOUT = "{!s:10} {!s:10} {!s:25}" if second: print LAYOUT.format("Teff", "log g", first+'-'+second) else: print LAYOUT.format("Teff", "log g", first) for i,j,k in sorted(zip(T, G, Mags)): print LAYOUT.format(i, j, k)
### The Output
Then if I just want the J-K color for objects with log g = 4.0 over the entire range of effective temperatures, I launch ipython and just do:
In [1]: import syn_phot as s In [2]: s.colors('J','K', logg=4) Teff -------- log g -------- J-K 1400.0 ------ 4.0 ---------- 4.386 1450.0 ------ 4.0 ---------- 4.154 ... 4450.0 ------ 4.0 ---------- 0.756 4500.0 ------ 4.0 ---------- 0.733
Similarly, I can specify just the target effective temperature and get the whole range of surface gravities. Or I can specify an effective temperature AND a specific gravity to get the color of just that one object with:
In [3]: s.colors('i','W2', teff=3050, logg=5) Teff -------- log g -------- J-K 3050.0 ------ 5.0 ---------- 3.442
I can also reduce the number of data points in each flux array if my sample is very large. I just have to specify the number of data points to skip with the “bin” optional parameter. For example:
In [4]: s.colors('W1','W2', teff=1850, bin=3)
This will calculate the W1-W2 color for all the objects with Teff = 1850K and all gravities, but only take every third flux value.
I also wrote functions to generate color-color, color-parameter and color-magnitude plots but those will be in a different post.
### Plots!
Here are a few color-parameter animated plots I made using my code. Here’s how I made them. Click to animate!
And here are a few colorful-colorful color-color plots I made:
### Plots with observational data
Just to be sure I’m on base, here’s a color-color plot of J-H vs. H-Ks for objects with a log surface gravity of 5 dex (blue dots) plotted over some data for the Chamaeleon I Molecular Cloud (semi-transparent) from Carpenter et al. (2002).
The color scale is for main sequence stars and the black dots are probable members of the group. Cooler dwarfs move up and to the right.
And here’s a plot of J-Ks vs. z-Ks as well as J-Ks vs. z-J. Again, the blue dots are from my synthetic photometry code at log(g)=5 and the semi-transparent points with errors are from Dahn et al. (2002). | 3,846 | 13,920 | {"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": 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} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.794775 |
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60 GHz Wireless – A Reality Check
The wireless revolution has been fascinating to watch. Radio (and micro) waves are transforming the way we live our lives. However, I’m increasingly seeing indications the hype may be getting ahead of itself and we’re beginning to have inflated expectations (c/o the hype cycle) about wireless broadband. In this post, I’d like to revisit some of my prior posts on the subject in light of something that has recently come to my attention: 60 GHz wireless.
Wavelength Matters
As I outlined in Physics of Wireless Broadband, the most important property that determines the propagation characteristics of radio (and micro) waves is its wavelength. Technical news and marketing materials about wireless broadband refer to frequency, but there is a simple translation to wavelength (in cm) given by:
\begin{aligned} \lambda(cm) = \frac{30}{f(GHz)}. \end{aligned}
Ages ago when I generated those SAR images, cell phones operated at 900 MHz (0.9 GHz) corresponding to a wavelength of about 33 cm. More recent 3G and 4G wireless devices operate at higher carrier frequencies up to 2.5 GHz corresponding to a shorter wavelength of 12 cm. Earlier this month, the FCC announced plans to release bandwidth at 5 GHz (6 cm).
This frequency creep is partially due to the issues related to the ultimate wireless bandwidth speed limit I outlined, but is also driven by a slight misconception that can be found on Wikipedia:
A key characteristic of bandwidth is that a band of a given width can carry the same amount of information, regardless of where that band is located in the frequency spectrum.
Although this is true from a pure information theoretic perspective, when it comes to wireless broadband, the transmission of information is not determined by Shannon alone. One must also consider Maxwell and there are far fewer people in the world that understand the latter than the former.
The propagation characteristics of 2G radio waves at 900 MHz (33 cm) are already quite different than 3G/4G microwaves at 2.5 GHz (12 cm) not to mention the newly announced 5 GHz (6 cm). That is why I was more than a little surprised to learn that organizations are seriously promoting 60 GHz WiFi. Plugging 60 GHz into our formula gives a wavelength of just 5 mm. This is important for three reasons: 1) Directionality and 2) Penetration, and 3) Diffraction.
Directionality
As I mentioned in Physics of Wireless Broadband, in order for an antenna to broadcast fairly uniformly in all directions, the antenna length should not be much more than half the carrier wavelength. At 60 GHz, this means the antenna should not be much larger than 2.5 mm. This is not feasible due to the small amount of energy transmitted/received by such a tiny antenna.
Consequently, the antenna would end up being very directional, i.e. it will have preferred directions for transmission/reception, and you’ll need to aim your wireless device toward the router. With the possible exception of being in an empty anechoic chamber, the idea that you’ll be able to carry around a wireless device operating at 60 GHz and maintain a good connection is wishful thinking to say the least.
Penetration
If directionality weren’t an issue, the transmission characteristics of 60 GHz microwaves alone should dampen any hopes for gigabit wireless at this frequency. Although the physics of transmission is complicated, as a general rule of thumb, the depth at which electromagnetic waves penetrate material is related to wavelength. Early 2G (33 cm) and more recent 3G/4G (12 cm) do a decent job of penetrating walls and doors, etc.
At 60 GHz (5 mm), the signal would be severely challenged to penetrate a paperback novel much less chairs, tables, or cubical walls. As a result, to receive full signal strength, 60 GHz wireless requires direct unobstructed line of sight between the device and router.
Photon Torpedoes vs. Molasses
The more interesting aspects of wireless signal propagation are diffraction and reflection, both of which can be understood via Huygen’s beautiful principle and both of which depend on wavelength. Wireless signals do a reasonably good job of oozing around obstacles if the wavelength is long compared to the size of the obstacle, i.e. at low frequencies. Wireless signal propagation is much better at lower frequencies because the signal can penetrate walls and doors and for those obstacles that cannot be penetrated, you still might receive a signal because the signal can ooze around corners.
As the frequency of the signal increases, the wave stops behaving like molasses oozing around and through obstacles, and begins acting more like photon torpedoes bouncing around the room like particles and shadowing begins to occur. At 60 GHz, shadowing would be severe and communication would depend on direct line of sight or indirect line of sight via reflections. However, it is important to keep in mind that each time the signal bounces off an obstacle, the strength is significantly weakened.
What Does it all Mean?
The idea that we can increase wireless broadband speeds simply by increasing the available bandwidth indefinitely is flawed because you must also consider the propagation characteristics of the carrier frequency. There is only a finite amount of spectrum available that has reasonable directionality, penetration, and diffraction characteristics. This unavoidable inherent physical limitation will lead us eventually to the ultimate wireless broadband speed limit. There is no amount of engineering that can defeat Heisenberg.
There are ways to obtain high bandwidth wireless signals, but you must sacrifice directionality. The extreme would be direct line of sight laser beam communications. Two routers can certainly communicate at gigabit speeds and beyond if they are connected by laser beams. Of course, there can be no obstacles between the routers or the signal will be lost. I can almost imagine a future-esque Star Wars-like communication system where individual mobile devices are, in fact, tracked with laser beams, but I don’t see that ever becoming a practical reality.
We still have some time before we reach this ultimate wireless broadband limit, but to not begin preparing for it now is irresponsible. The only future-proof technology is fiber optics. Communities should avoid the temptation to fore go fiber plans in favor of wireless because those who do so will soon bump into this wireless broadband limit and need to roll out fiber anyway.
Written by Eric
January 21, 2013 at 9:15 am
WSJ: “Culprit in Wi-Fi Failures: Chicken Wire”
On Thursday, the Wall Street Journal published an article
Culprit in Wi-Fi Failures: Chicken Wire
that is consistent with the theme I’ve been talking about lately. This is a perfect example of what I outlined in
The Ultimate Wireless Broadband Speed Limit
where I wrote:
Now consider two people trying to communicate via radio waves, but they are separated by a wall of metal. No dice. The radio waves cannot penetrate. Now, puncture a small hole in the wall. To be concrete, lets say we are communicating at a frequency of 3 GHz with a corresponding wavelength of 10 cm. If the hole is 1 cm in diameter, it is difficult for the radio signal to “ooze” through that tiny hole. Remember, the “size” of anything as far as a radio wave is concerned is only as a ratio of its wavelength. In this case, the hole is $.1\lambda$.
The ability of a wave to ooze through the hole depends on the size of the hole relative to its wavelength. Roughly speaking, when the size of the hole is greater than $.5\lambda$ it has a much easier time oozing through it.
This example is actually pretty close to the situation described in the Wall Street Journal article. Wi-Fi operates at a frequency of roughly 2.5 GHz with a wavelength of roughly 13 cm. I’m guessing the holes in chicken wire are probably close to 3 cm in diameter or roughly $.25\lambda$. Since the hole is smaller than $.5\lambda$, the Wi-Fi signal cannot easily penetrate the chicken wire.
The photo in the Wall Street Journal illustrates another point I was trying to make. Although the 2.5 GHz Wi-Fi signal cannot penetrate the chicken wire, the MUCH higher frequency visible light does easily penetrate it, i.e. you can see through the chicken wire. The physics here is similar to why AM radio signals in underground parking lots are much worse than FM radio signals. The radio can see the FM radio signals, but the lower frequency AM radio signals get blocked.
Written by Eric
January 3, 2010 at 12:26 pm
The Ultimate Wireless Broadband Speed Limit
This is a follow-up to my previous posts
as well as some of the comments those posts generated.
First of all, I was pleased that loganb (Comment on 12/29, 9:27 PM EST) brought Shannon into the picture because information theory is important. As Rod pointed out, “information is physical” so the limits of wireless broadband communications will involve a knowledge of both Maxwell and Shannon. In fact, if any grad student out there was interested in both information theory and wireless communications, it would be a fun project to try to determine this ultimate wireless broadband speed limit.
The Problem
If two computers were to communicate via laser beam and they had direct line of sight access, the speeds at which those two computers could communicate over the airwaves should approach those possible via fiber optics.
Of course that is not what one means by wireless broadband. We do not think of a PC communicating to a tower via laser beam with direct line of sight. So this is not the “ultimate speed limit” I’m talking about.
Instead, we want to think of a region of space containing multiple wireless broadband devices. What is the maximum “density of information” available in the air within this region. For concreteness, we could consider
$\text{1 unit} = 10 m\times 10 m\times 10 m$
and
$\text{1 test region} = \text{10 units}\times\text{10 units}\times\text{10 units}$
What is the ultimate physical maximum number of bits that can be communicated via wireless broadband devices within one test region?
My Guestimate
My “guestimate” is that in a test region the ultimate wireless broadband speed limit, i.e. the maximum number of bits that can be communicated via wireless broadband devices is 1,000 Mbps. This is the total number of bits available to everyone within the test region. To get the speed available to any one wireless broadband user, we simply take this speed limit and divide by the number of users simultaneously downloading (or uploading) content.
For example, if my guestimate is correct and there were 100 people simultaneously using their wireless broadband devices, then with a perfectly design wireless network (*cough*) each person would have access to speeds of 10 Mbps. This is still pretty high and most in the US would drool for these speeds (although people in Korea, Japan, and parts of Europe would yawn).
I don’t ever expect routine wireless broadband access in excess of 10 Mbps in crowded tech savvy cities and even this speed assumes perfect network design. It will take a while for us to reach this ultimate speed limit as determined by fundamental physics, but we will eventually. With bandwidth demands doubling every couple of years, it is easy to imagine running into this speed limit within 5 years.
A Note on Radio Wave Oozing
Here is a comment from Zathras:
“The behavior of a wave depends on its frequency. At low frequencies, radio waves are kind of like molasses. They can ooze around corners and through buildings.
As frequencies increase, the waves start acting more like laser beams.”
This is absolute nonsense. For starters, go into a parking garage and compare your AM radio reception to your FM. The AM won’t come in, despite being lower frequency than the FM. There is no “molasses effect” here. Yes, I realize that these are lower frequencies than the cell phone ones, but since the above quote is stated in such absolute terms, it is still an effective counterexample.
The reasons for the interference are more complex. It has to do with the interference with vibrational frequencies that the molecules in the barriers have. The multi-GHz range is full of resonant frequencies for molecules. It also has to do with the fact that higher frequencies attenuate more in conducting metals than lower frequencies do. It certainly has nothing to do with any laser beam/molasses nonsense.
The actual physics of radio waves is more complex than can be communicated in a few paragraphs. However, the “oozing” analogy does take you some distance in understanding what is going on. Unlike molasses, radio waves can ooze through some materials such as glass, dry wall, wood, etc. One material that radio waves cannot penetrate is metal. In fact, the better the material is at conducting electricity, the worse radio waves are at penetrating it. For example, the earth is fairly good at conducting electricity so it is difficult to send radio waves through the earth.
Now consider two people trying to communicate via radio waves, but they are separated by a wall of metal. No dice. The radio waves cannot penetrate. Now, puncture a small hole in the wall. To be concrete, lets say we are communicating at a frequency of 3 GHz with a corresponding wavelength of 10 cm. If the hole is 1 cm in diameter, it is difficult for the radio signal to “ooze” through that tiny hole. Remember, the “size” of anything as far as a radio wave is concerned is only as a ratio of its wavelength. In this case, the hole is $.1 \lambda$.
The ability of a wave to ooze through the hole depends on the size of the hole relative to its wavelength. Roughly speaking, when the size of the hole is greater than $.5\lambda$ it has a much easier time oozing through it.
Keep in mind that most buildings have some kind of metal exterior and especially underground parking lots are surrounded by rebar, etc.
Now, an AM radio wave is roughly 1,000 kHz (or .001 GHz) with a corresponding wavelength of 300 meters. For an AM radio wave to ooze through a hole, that hole would have to be more than 150 meters wide. This is why AM does not penetrate very far into a tunnel.
On the other hand, an FM radio wave is roughly 100 MHz (or .1 GHz) with a corresponding wavelength of 3 meters. Although still fairly big, FM radio waves can still ooze through garage doors in underground parking lots, through building windows, into tunnels, etc.
Wireless broadband at 2.5 GHz or 12 cm can “ooze” though windows etc, but as I said, they ooze too fast and have a harder time turning corners. Hence, at these higher frequencies, we begin to see shadowing etc.
Written by Eric
December 31, 2009 at 12:34 pm
With a title like the one I’ve chosen, I’m sure only the die hards will get even this far, so this post will be a somewhat technical follow-up to my informal post yesterday that was picked up by Felix Salmon and also noted by Paul Krugman.
In graduate school, I had fun building large-scale simulations where I modeled the propagation of radio waves transmitted from a cell phone through a 1 cubic millimeter resolution model of a human head
SAR distributions for the sagittal slice
SAR distributions for the frontal slice
SAR distributions for the coronal slice
These simulations are generally extremely accurate. The physics is well understood and can be modeled with a high degree of confidence. One of my favorite stories is when my office mate was modeling an aircraft and comparing results to measurements, he noticed the angle seemed off by .5 degrees. He called up the lab and, sure enough, the measurement was off by .5 degrees from what was specified. The simulation was more accurate than the measurement.
As mentioned yesterday, the behavior of a radio wave depends on its frequency and hence its wavelength. The two are related by
$\lambda = \frac{c}{f},$
where $\lambda$ is the wavelength, $f$ is the frequency, and $c$ is the speed of light. To simplify things, we can write that formula as
$\lambda (cm) = \frac{30}{f(GHz)}.$
For example, if the frequency is 1 GHz, i.e. $f(GHz) = 1$, then the wavelength is 30 cm, i.e. $\lambda(cm) = 30.$ When we double the frequency to 2 GHz, the wavelength reduces by half to 15 cm.
The wavelength is an important number to keep in mind because radio waves interact more strongly with objects whose size is roughly on the order of the wavelength of the radio wave. When I was in grad school, the frequency we were looking at was 900 MHz (.9 GHz) with a corresponding wavelength of roughly 33 cm. In a comment on Felix’s blog, Mark states:
It’s amazing that Phorgy can make so many technical errors and still make you worry that he’s right. For example:
“That is why the…7-800 MHz range is so valuable for cell applications.”
The 700 MHz band has never been used for cell phone communications in the US. It was auctioned off, but the spectrum is unused at present.
There is a big difference between what is valuable and what is available. Things are valuable sometimes precisely because they are not available. My statement was about the frequencies at which cell applications would be better off. I could have and maybe should have made the range a little broader, say 700-900 MHz, but that was not the point I was trying to make. The point is that radio waves propagate nicely, i.e. they ooze well, in the 700-800 MHz range. Higher than that and we begin to see directionality creep in, i.e. the radio waves begin to have preferred directions and the coverage becomes less uniform. For a very nice interactive demonstration of this, have a look at this:
Radiation Pattern of a Linear Antenna
The important number in that demonstration is the “Dipole Length (Wavelength)”. This is the length of the antenna relative to the wavelength. So in that demo, setting the slider t0 .5 means the antenna is half the length of the wave. For numbers up to 1.0, the pattern is fairly uniform, but once you get above 1.0, you start to see nulls where there is no radiation. This is one source of directionality in wireless signals.
From Felix’s article, we have:
For one thing, Phorgy’s limit of 1,000mps in total for a few city blocks is I think far higher than anything AT&T is currently able to provide. With what Baruch calls “compression, prioritisation, all that level 4-7 stuff you can do at the packet level” (don’t ask me), you can serve a lot of people with that kind of bandwidth.
The number I gave (1,000 mbps) was a “guestimate”, but it wasn’t a wild guestimate. When/if the formal studies are done, I am confident the number will not be too far off from this. This number includes “compression, prioritisation” and even polarization and modulation. I’m not talking about spectrum here, I am talking about the total availability of bits to everyone within a given vicinity. The number available to any one person will be simply this number divided by the number of people simultaneously downloading stuff within this vicinity. This is similar to the early days of cable modems. You could tell your neighbor was downloading a pirated movie because your connection drops to a crawl.
To be sure, my note was meant to convey an important message and sometimes a degree of license is warranted. No one should believe that any phone company has built out enough towers to reach the ultimate wireless speed limit, but how many people knew there was a wireless speed limit?
A couple years ago, I was telling people that in a few years, we would begin to see the limits of wireless broadband. I think we are beginning to see it, but we still have a way to go before we truly hit that ultimate speed limit. But we will.
There is no number of towers or “Wi-Fi” hot spots that will overcome this physical limitation. For one thing, you cannot put Wi-Fi hotspots too close together or they start to interfere with one another. You can be clever and switch to a neighboring channel, but that again only delays the inevitable.
Like another comment by loganb on Felix’s blog points out:
The real limit isn’t Heisenberg’s, it’s (Claude) Shannon’s, and those limits only apply to limits on the capacity of a given base station.
This is a great point, but I would say the two go hand in hand. Shannon and Heisenberg together determine the ultimate limit of how much information can by communicated via wireless broadband within a given vicinity.
One day, in the not too distant future, instead of going into a cafe and hopping online via a Wi-Fi hotspot, that same cafe will have a fiber optic plug next to the salt shaker.
Written by Eric
December 30, 2009 at 10:28 am
Hello Heisenberg: “New York City not ready for the iPhone”
Interesting story from The Consumerist:
AT&T Customer Service: “New York City Is Not Ready For The iPhone”
Recall an earlier article of mine (from July 2007):
I chose a technically incorrect term “Wi-Fi” because that is what most people were talking about back then, but the subject was more generally about “wireless broadband”.
It is not that NYC isn’t ready for the iPhone. It is that NYC was the first to bump up against the inherent physical limitation of wireless broadband. There is no number of towers that will be able to accommodate hundreds or thousands of people within a small vicinity all expecting reasonable wireless bandwidth. There is a little thing called the Heisenberg uncertainty principle that no amount of marketing or engineering will be able to get around.
Edit: Here is a copy of a comment below in response to the questions:
I’ll try to write a separate article, but this is about the physics of waves.
Unlike finance, the physics of electromagnetic waves is well understood. Computer programs can be written to model radio waves to many digits of accuracy.
The behavior of a wave depends on its frequency. At low frequencies, radio waves are kind of like molasses. They can ooze around corners and through buildings. That is why the (relatively low frequency) 7-800 MHz range is so valuable for cell applications.
In recent years, the frequencies of cell phones and even more recently, smart phones, has increased from a little over 1 GHz to over 2.5 GHz (and beyond).
(Note: Your microwave oven operates at the same frequency as most smart phones now.)
As frequencies increase, the waves start acting more like laser beams. They no longer ooze around corners. You start to get “shadows” or dead spots with no signal. It becomes more difficult for the signals to penetrate walls etc. These problems get worse the higher you go in frequency.
An extreme case is an actual laser. Here, it becomes more difficult to distinguish the wave dynamics from particle dynamics. Like in Star Wars, the laser beams can bounce around like particles.
So we have two extremes: low frequency molasses waves and high frequency laser beams. As bandwidth demands increase, we begin moving the dial away from molasses (where we have good wireless signals) to laser beams (where we have dark spots, shadows, with no signal, etc).
There are many clever modulation tricks that delay the inevitable, but the basic rule is that you cannot defeat Heisenberg. This is an imprecise (but I hope effective) analogy that relates to the fact that at lower frequency (and longer wavelength, i.e. larger “effective size” of the wave), you have more certainty as to “where the photon is” (because it is coming from a relatively smaller antenna) you have more uncertainty about where it goes, i.e. it goes everywhere like a good wireless signal should. At higher frequency (and shorter wavelength), the antenna is relatively larger (compared to the wave) so we know less precisely where the photons are, hence we have more certainty as to where they are going, i.e. in a straight line instead of around a corner, which is undesirable for a wireless signal.
This physical fact does not deter marketing people. You can easily set up a demonstration on a van driving down the highway at 65 mph with an antenna mounted on top downloading web content at 100-1000 mbps. Don’t fall for this trick! They are essentially shining a laser beam at the van and tracking it down the road. Ask them to do the same demo with 1000 vans stuck in LA traffic. Forget about it.
I’d guestimate that a practical limit for the available wireless bandwidth in the air in a vicinity of say a couple square NYC blocks would be 1000 mbps. This is the TOTAL BANDWIDTH AVAILABLE FOR EVERYONE WITHIN A FEW NYC BLOCKS. So now divide 1000 mpbs by the number of people downloading stuff wirelessly taking into consideration highrise buildings, etc. That is probably a decent estimate of what the long term limits of wireless broadband would be.
So you can see, for the early adopters, wireless bandwidth is great! “Geez! This wireless is faster than my ethernet!” But once you start adding some real traffic, say 100 or 1000s of people all downloading stuff wireless within a small vicinity and you can imagine that we will easily bump up against the basic physical limitations as communicated by my good friend James Clerk Maxwell.
Written by Eric
December 27, 2009 at 4:42 pm
The new corporate trend: “onshoring”
Ha! What did I tell you? It is wayyy past my bedtime and I should be sleeping (especially since I’m sick and feel like I’ve been run over by a truck), but just stumbled on an interesting article from one of my favorite economic development blogs: Design Nine. He points to an article in the journalgazette.net.
Recall that in the midst of all the market doom and gloom, I recently said:
I’m actually ironically optimistic about the outlook for suburban and rural economic development. A weaker dollar will make outsourcing less attractive. That will bring manufacturing jobs back home. I can imagine a boon in suburban and rural development. Just imagine if communities developed decent broadband via fiber-to-the-home/business. Suddenly, there will be attractive jobs and living standards in affordable places.
I admit that the quote is a bit misleading because Andrew Cohill has influenced the way I think about things, but still timely I think. Here is an excerpt from the article:
Small-town America: The new Bangalore?
[snip]
Onshoring, in fact, is becoming trendy.
Some U.S. companies recently have pulled back from India to set up shop in rural areas where access to high-speed broadband connections isn’t the problem it was just a few years ago and where lower real-estate prices and wages are attractive.
Note that the key to onshoring is an investment in telecommunications infrastructure. In particular, fiber-to-the-home. It is quite sad to see so many municipalities rest their hopes on wireless broadband. That will only end in tears as the reality of wireless broadband becomes apparent. Any community that is not investing now in fiber will lose out on an important opportunity that is now beginning to present itself: onshoring. As Andrew will tell you more eloquently than I could, an intelligent investment in a communities future MUST involve a combination of both fiber and wireless and I would put wireless as a distant second. I can explain in gory detail why wireless will fail if you like (I did my PhD in the subject), but for now need to hit the sack.
Go onshoring! Go USA!
Written by Eric
December 4, 2007 at 12:50 am
WP: Japan’s Warp-Speed Ride to Internet Future
An important factor for long term economic growth is technological innovation. An important factor for technological innovation is increasingly access to high speed internet connections. The US has been falling behind Japan, South Korea, and parts of Europe in terms of access to fiber optic connections to the home. I alluded to this here and here and probably elsewhere.
In the next couple of years, I suspect you’ll see the number of articles like the one pointed out by Barry Ritholtz over on The Big Picture
Japan’s Warp-Speed Ride to Internet Future
TOKYO — Americans invented the Internet, but the Japanese are running away with it.
Broadband service here is eight to 30 times as fast as in the United States — and considerably cheaper. Japan has the world’s fastest Internet connections, delivering more data at a lower cost than anywhere else, recent studies show.
Written by Eric
September 1, 2007 at 5:13 pm | 6,050 | 28,403 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.927591 |
https://www.intmath.com/blog/letters/the-intmath-newsletter-oct-2007-787 | 1,675,066,364,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00106.warc.gz | 841,304,851 | 33,080 | Search IntMath
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# The IntMath Newsletter - Oct 2007
By Murray Bourne, 14 Oct 2007
=========================
1. This month's math tip
2. What motivates you?
3. Latest Poll
4. From the math blog
=========================
## 1. THIS MONTH'S MATH TIP - MISTAKES
One of my math teachers used to say to us:
"It doesn't matter how many mistakes you make in my class - as long as you learn something from each one."
Unfortunately, many students think that mathematics is some rigid system where there is only one correct answer and only one way to find that answer, and that mathematicians never make mistakes. But this is not always so.
Mathematicians have been making mistakes for centuries. Sometimes the mistakes didn't matter much, but sometimes they affected a lot of people. Let's recall two of the biggest math mistakes.
a. The Earth is the Center of the Solar System Mistake
This mistake was probably more to do with religious dogma than mathematician's errors. However, assuming the Earth is at the center, it was up to the mathematicians to describe the resulting weird motions of the planets. Many mathematicians blindly accepted the incorrect idea (and remember most of them had no choice - if you disagreed you would be burned at the stake.) It became a lot easier when they started to get it right in the 16th century - that the sun is at the center and the planets revolve around it in elliptical orbits.
b. The September Calendar Mistake
There was a serious problem with calendars during the Middle Ages. The seasons were 'shifting' - that is, summer was starting on the wrong date each year. The problem occurred because the calendar they were using since Julius Caesar's day was 365.25 days long. It had leap years every 4th year. But we now know that the year is actually close to 365.2425 days long - slightly shorter than 365.25. We allow for the difference by having a 'leap century' every 400 years (the last one was in 2000).
The Catholics fixed the problem and adopted a new 'Gregorian' calendar in October 1582. But the Protestant British refused to follow suit until they realised they had to do something with the 11-day shift in the calendar. So in the year 1752, they chopped out 11 days between September 2 and 14 so it looked like this:
September 1752:
Su Mo Tu We Th Fr Sa
30 31 01 02 14 15 16
17 18 19 20 21 22 23
24 25 26 27 28 29 30
The early mathematicians made mistakes, learned from them, fixed them and moved on.
Next time you make a mistake in mathematics, consider it a good sign:
(i) You have at least tried the problem.
(ii) You have realised there is a mistake.
(iii) You have thought about why it was wrong and have done some more learning to fix the error. You may even see another way to do the problem, and this may make more sense to you than the textbook answer.
(iv) By going through the process of finding the correct answer, you are more likely to remember how to do it next time.
Mistakes in math are not the end of the world - they are the beginning of real learning.
## 2. WHAT MOTIVATES YOU?
What's your passion? Think of passion as something you do because it excites you, you really want to do it, you believe in it and nobody has to tell you to do it. And now think about all the things you learn by following that passion.
Maybe you are into surfing and you have become an expert on surf techniques, or maybe board construction - and you have certainly learned to swim well! Perhaps you love computer games and have learned many strategies and skills because you enjoy moving on to the next game level.
Conclusion: Passion is the energy needed for learning.
Now, think about your feelings towards study. Where do those feelings come on a "passion scale"? Is it a 10? Perhaps 5? Or maybe a 2?
If we let ourselves become passionate about the things we are studying - yes, math included - then we will enjoy it more and will perform much better. Resist the peer pressures to hate study and let yourself enjoy it. Avoid people who constantly say "this sucks - why are we doing this?" You are more likely to be successful at something if you mix with others who are good at it and enjoy it.
You have the power to determine your own passion. And that includes your passion for math!
## 3. LATEST POLL
The latest poll on Interactive Mathematics asks if the math you are currently studying is useful to you or not. You can answer on any page in Interactive Mathematics
The results so far have been surprising to me (you can see the results after you have voted).
## 4. RECENTLY ON THE MATH BLOG
1) MUSIC IS MATH - BOARDS OF CANADA
This week's Friday math movie is a music video from Boards of Canada.
2) NOT KNOT (PARTS 1 AND 2)
Not Knot is a mind-expanding set of computer animations that demonstrate hyperbolic space.
3) GOOGLE TRENDS - INSIGHTS INTO WHAT'S HOT (math trends)
Google Trends gives us an interesting insight into what's hot on the Internet. It is a valuable tool for educators to spot changing trends in different fields.
4) DANICA MCKELLAR - MATH WONDERGIRL
Daniel McKellar is best known for her TV roles, but did you know she has published a mathematics paper and written a math advice book?
Be the first to comment below.
### Comment Preview
HTML: You can use simple tags like <b>, <a href="...">, etc.
To enter math, you can can either:
1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)
NOTE: You can mix both types of math entry in your comment.
From Math Blogs | 1,403 | 5,958 | {"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": 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-2023-06 | longest | en | 0.97177 |
https://www.palaceofchance.com/casino-games/sic-bo/ | 1,484,756,109,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00044-ip-10-171-10-70.ec2.internal.warc.gz | 967,286,051 | 12,975 | # Sic Bo
Find a Game
• ##### Popular Games
Originated hundreds of years ago in China, Sic Bo (Dice Pair in English) has swept the Western hemisphere to offer a blast from the past with this ancient dice game and here at Palace of Chance, we’re in love with this fantastic casino novelty. Three dices are thrown and the player must predict the number that will land on each dice. The rules for this game are pretty simple, if you guess correctly you’ll win and if you are wrong then the opposite will happen.
### How to Play Sic Bo
The game is played by rolling the dice; it’s the bet and outcome that dictates your winnings. This exciting casino game allows multiple bets, which vastly increases the potential prizes, however, there are a number of ways that you can work the dice in your favor. To get started, make your bet, click the ‘Roll’ button to start shaking the three dice in the container and let the game begin.
### Player’s Turn
You begin the game by choosing which section to put your chips. The bet that you place is based on what you think that the outcome is for the different combinations of the numbers on the three dice. Pressing the ‘Roll Button’ will make the three dices to shake in their container. Once they are rolled the results are shown on the right hand side of the screen.
### You Win When
#### Sic Bo Number Bet
Depending on how many times a number appears on the dice, if you bet on that particular number then you will win. Choose between 1 and 6 and place your chips where you feel the dice are going to land. If your number shows up once then you will get a payout of 1 to 1, 2 times and you will get 2x payout and a lucky 3 you will get a payout of 3x.
#### Small or Big Sic Bo Bet
You can check on the status bar as to whether it is small or big. If the total of the three dice is 4 to 10 then you win the small bet. If the total sum of the three dice is 11 to 18 then you win the big bet. If all the dice have the same number then you won’t win anything. The payout ratio in this option is 1 to 1.
#### Sic Bo Pair Bet
This option allows the player to bet on any possible 15 dice outcome. This can result in huge winnings as payouts can be up to five times your bet if the dice land on both of the numbers that you predicted.
#### Sic Bo Total Bet
In this option the player predicts the sum total of all what all the three dice will land on. Totals can be between 4 and 17.
There is also the Sic Bo Double Pair which allows you to predict a particular pair that will come up, Sic Bo Triple which as the name suggests allows you to predict a specific triple of numbers to show up on the dice or the Sic Bo Any Triple Bet which allows the player to place a bet on any numbers show up in unison on all of the three dices. All of these great features are what makes Sic Bo one of the newest, most exciting casino game releases at Palace of Chance.
### Related Games
Sic Bo Game Screen
Sic Bo Options Screen
Sic Bo Bet Options | 683 | 2,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.929561 |
http://math.stackexchange.com/users/24852/hqt?tab=activity&sort=all&page=4 | 1,429,348,308,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246634257.45/warc/CC-MAIN-20150417045714-00289-ip-10-235-10-82.ec2.internal.warc.gz | 174,117,317 | 12,662 | hqt
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Jun8 comment Lagrange's method to find min/max question can you give me an example, please. My teacher says the thing opposite :( that we can sure that no max for that function (in above example) Jun8 asked Lagrange's method to find min/max question Jun7 accepted Partial Derivation: $\lim_{(x,y)\to(0,0)}\frac{x^2+\sin^2y}{2x^2+y}$ Jun7 asked Partial Derivation: $\lim_{(x,y)\to(0,0)}\frac{x^2+\sin^2y}{2x^2+y}$ Jun4 accepted Sum of two divergence series is always divergence series? Jun3 comment Sum of two divergence series is always divergence series? @Marvis Oh. sorry so much :( I always think two examples is same :( Jun3 comment Sum of two divergence series is always divergence series? Take opposite of series and plus together is too special in my opinion. Jun3 comment Sum of two divergence series is always divergence series? @DavidMitra yes. I have thought your example before, but it's too special. Jun3 asked Sum of two divergence series is always divergence series? May30 accepted Maple: how to solving composite function May28 comment Maple: how to solving composite function No. I don't think it's just for fun. it's nice. but I don't know your solution compare to first one (above post) will be same performance or not :) May28 comment Maple: how to solving composite function Can you explain a bit at line 3 and 4,please. I know how they work, but I really don't understand why it works right. I'm thinking about your nice solution so much. Thanks :) May28 asked Maple: how to solving composite function May22 accepted Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ dx}$ May22 accepted Compute lim from Graph May22 accepted Min/Max of $f(x,y) = e^{xy}$ where $x^3+y^3=16$ May20 comment Min/Max of $f(x,y) = e^{xy}$ where $x^3+y^3=16$ Oh. It likes x=y=2 case when I use Lagrange Multiplier. But at min case, I don't have any idea how to solve by Lagrange Multiplier May20 awarded Editor May20 revised Min/Max of $f(x,y) = e^{xy}$ where $x^3+y^3=16$ added 25 characters in body May20 comment Min/Max of $f(x,y) = e^{xy}$ where $x^3+y^3=16$ @N.I ah, I understand. Because I just view fomular from other post , I don't really know different and \$ :D | 609 | 2,238 | {"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": 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} | 3.40625 | 3 | CC-MAIN-2015-18 | longest | en | 0.906968 |
http://www.slideserve.com/braden/chapter-4-axial-load | 1,493,447,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123276.44/warc/CC-MAIN-20170423031203-00539-ip-10-145-167-34.ec2.internal.warc.gz | 688,086,262 | 17,706 | # Chapter 4 Axial Load - PowerPoint PPT Presentation
1 / 12
Chapter 4 Axial Load. Saint - Venant's Principle. Saint- Venant's Principle claims that localized effects caused by any load acting on a body will dissipate or smooth out within regions that are at a sufficient distance from the point of load application
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- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
### Saint -Venant's Principle
• Saint-Venant's Principle claims that localized effects caused by any load acting on a body will dissipate or smooth out within regions that are at a sufficient distance from the point of load application
• A general rule that applies to many cases of loading and member geometry, this distance can be considered to be at least equal to the largest dimension of the loaded cross section
• The resulting stress distribution in these regions will be the same as that caused by any other statically equivalent load applied to the body within the same localized area
### Elastic Deformation of an Axially Loaded Bar
• Bar is assumed to have a cross-sectional area that varies along its length, external end loads (P1 and P2), and an external load that varies along its length (such as the weight of a vertically oriented bar)
• Using the method of sections, a differential element (infinitesimally thin slice) of length dx and cross-sectional area A(x) is isolated from the bar at an arbitrary position x
• Resultant axial force P(x) varies along its length due to the external loading
• The relative displacement δ of one end of the bar with respect to the other end
### Special Cases of Axially Loaded Bars
• Constant load and cross-sectional area and homogeneous material (E is constant),
• Abrupt changes in axial force, cross-sectional area, and/or modulus of elasticity from one region of the bar to the next
• Can apply above equation to each segment of the bar where these quantities are all constant
### Sign Convention
• Force and displacement are considered positive if they cause tension and elongation
• Force and displacement are considered negative if they cause compression and contraction
### Relative Displacement and the Sign Convention
• δA/D relative displacement (displacement of A with respect to D)
• Problems, page 131
### Principle of Superposition
• Principle of superposition - the resultant stress or displacement at a point can be determined by first finding the stress or displacement caused by each component load acting separately on the member and then, algebraically adding the contributions caused by each of the components
• The loading must be linearly related to the stress or displacement that is to be determined (i.e. σ = P/A or δ = PL/AE must hold)
• The loading must not significantly change the original geometry or configuration of the member (deformations are typically so small that changes in geometry or configuration can be considered to be insignificant, with one exception being the deformations anticipated in buckling analysis)
### Statically Indeterminate Axially Loaded Member
• See "Equilibrium of a Rigid Body" topic in Statics (see "Redundant supports" slide)
• Equilibrium equation is insufficient to find all the reactions, FB + FA - P = 0
• Compatibility or kinematic condition
• Specifies conditions for displacement
• δA/B = 0
• Using
### Force Method of Analysis for Axially Loaded Members
• Flexibility or Force Method of Analysis
• Choose any one of the two supports as "redundant"
• Temporarily remove its effect from the member
• Using principle of superposition, the actual member is equivalent to the one subjected to the external load (with redundant force removed) plus the member subjected only to the redundant force
• Equilibrium equation can then be used to determine FA
• Problems, page 145
### Thermal Stress
• If the temperature increases, generally a material expands, whereas if the temperature decreases, a material will contract
• The deformation equals δT = αΔT L
• δT = change in length of the member due to a temperature change
• α = linear coefficient of thermal expansion (CTE)
• ΔT = change in temperature of the member
• L = length of the member
• If the change in temperature or the CTE vary along the length then
• Problems, page 155
### Stress Concentrations
• Stress concentrations occur at locations where the cross-sectional area abruptly changes (i.e. holes and transition regions)
• Stress concentration factors, K
• K = σmax / σavg the ratio of the maximum stress to the average stress acting at the smallest cross section
• K is determined empirically or
through application of the
theory of elasticity
• K is commonly plotted as a
function of geometric features
### Stress Concentration Factors
Change in cross section Hole
Problems, page 169 | 1,129 | 5,308 | {"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} | 2.953125 | 3 | CC-MAIN-2017-17 | longest | en | 0.914353 |
https://doisinkidney.com/code/depth-comonads/DepthComonads.Dec.html | 1,679,961,806,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00091.warc.gz | 255,272,522 | 2,991 | ```{-# OPTIONS --cubical --safe #-}
Reflects : Type a → Bool → Type a
Reflects A true = A
Reflects A false = ¬ A
record Dec {a} (A : Type a) : Type a where
constructor _because_
field
does : Bool
why : Reflects A does
open Dec public
pattern yes p = true because p
pattern no ¬p = false because ¬p
map-reflects : (A → B) → (¬ A → ¬ B) → ∀ {d} → Reflects A d → Reflects B d
map-reflects {A = A} {B = B} to fro {d = d} = bool {P = λ d → Reflects A d → Reflects B d} fro to d
map-dec : (A → B) → (¬ A → ¬ B) → Dec A → Dec B
map-dec to fro dec .does = dec .does
map-dec to fro dec .why = map-reflects to fro (dec .why)
T? : (b : Bool) → Dec (T b)
T? b .does = b
T? false .why ()
T? true .why = _
Dec→Stable : Dec A → Stable A
Dec→Stable (no ¬x) ¬¬x = ⊥-elim (¬¬x ¬x)
Dec→Stable (yes x) ¬¬x = x
``` | 319 | 810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-14 | latest | en | 0.542389 |
https://www.flightpedia.org/convert/55-milligals-to-inches-per-minute-per-second.html | 1,722,927,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00006.warc.gz | 583,339,191 | 4,330 | Convert 55 Milligals to Inches per Minute per Second
55 Milligals (mGal)
1 mGal = 0.023622 ipm/s
=
1.29921 Inches per Minute per Second (ipm/s)
1 ipm/s = 42.3333 mGal
Data Acceleration converter
Q: How many Milligals in a Inch per Minute per Second?
The answer is 42.3333 Inch per Minute per Second
Q: How do you convert 55 Milligal (mGal) to Inch per Minute per Second (ipm/s)?
55 Milligal is equal to 1.29921 Inch per Minute per Second. Formula to convert 55 mGal to ipm/s is 55 * 0.02362205 | 162 | 499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.529599 |
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# Pre-Algebra : Solving Equations and Writing Equations from Word Problems (7 Problems)
Solve the equations:
1- 3.14 = Z/-3.7
2- -4.7t = 12.22
3- m+2.7 = -9.3
4- -38=z minus 20.5
Write an algebraic equation and solve.
1- Years of stress from his BLA students have reduced the number of hair on Mr. Sit's head to 7560 strands. if he lost 963 strands during his time at BLA, how much hair did Mr. Sit have before he started at BLA?
2- Leah, Melissa, and Natalie are going to a dance party next month. However, they do not have any rhythm or beat. Because they know that he is a dancing machine, they beg mr. Sit for some dance lessons. After eleven lessons with dance master Smit, they are jamming dancers. If Mr Smit charged each of them \$825 for the lessons, how much was each lesson?
3- Mr Smit went to stop and shop to pick up some snacks. He bought three cans of spam for \$3 each, a gallon of prune juice, two cans of anchioves for \$2.75 each, and a box of jello for \$.85 if he spent a total of \$18.20 how much ws the prune juice?
#### Solution Preview
Please see the attached file for the complete solution.
Thanks for using BrainMass.
1. 3.14 = Z/-3.7
2. -4.7t = 12.22
divide both sides by -4.7,
then, t = 12.22/ (-4.7)
= -2.6
3. m+2.7 = -9.3
substract 2.7 from both sides,
then m = -9.3-2.7
= -12
4. -38=z minus 20.5 | 426 | 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} | 4.5 | 4 | CC-MAIN-2017-22 | longest | en | 0.949439 |
https://www.encyclopedia.com/science-and-technology/mathematics/mathematics/division | 1,726,467,060,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00870.warc.gz | 704,024,334 | 21,286 | # Division
views updated Jun 11 2018
# Division
KEY TERMS
Resources
Division is the mathematical operation that is the inverse of multiplication. If one multiplies 47 by 92 and, then, divides by 92, the result is the original 47. In general, (ab)/b = a. Likewise, if one divides first and, then, multiplies, the two operations nullify each other: (a/b)b = a. This latter relationship can be taken as the definition of division: a/b is a number which, when multiplied by b, yields a.
In the real world using ordinary arithmetic, division is used in two basic ways. The first is to partition a quantity of something into parts of a known size, in which case the quotient represents the number of parts, for example, finding how many three-egg omelets can be made from a dozen eggs. The second is to share a quantity among a known number of shares, as in finding how many eggs will be available per omelet for each of five people. In the latter case, the quotient represents the size of each share. For omelets, if made individually one could use two eggs each for the five people and still have two eggs left over, or you could put all the eggs in one bowl, so each person would get 2.4 (12/5) eggs.
The three components of a division situation can represent three distinct categories of things. While it would not make sense to add dollars to earnings-per-share, one can divide dollars by earnings-per-share and have a meaningful result (in this case, shares). This is true, too, in the familiar rate-distance-time relationship R = D/T. Here the categories are even more distinct. Distance is measured with a tape; time by a clock; and rate is the result of these two quantities.
Another example would be in preparing a quarterly report for share holders; a company treasurer would divide the total earnings for the quarter by the number of shares in order to compute the earnings-per-share. On the other hand, if the company wanted to raise \$6, 000, 000 in new capital by issuing new shares, and if shares were currently selling for \$181/8, the treasurer would use division to figure out how many new shares would be needed, i.e., about 330, 000 shares.
Division is symbolized in two ways, with the symbol ÷ and with a bar, horizontal or slanted. In a/b or a ÷ b, a is called the dividend; b, the divisor; and the entire expression, the quotient.
Division is not commutative; 6/4 is not the same as 4/6. It is not associative; (8 ÷ 4) ÷ 2 is not the same as 8 ÷ (4 ÷ 2). For this reason, care must be used when writing expressions involving division, or interpreting them. An expression such as is meaningless. It can be given meaning by making one bar noticably longer than the other to indicate that 3/4 is to be divided by 7. The horizontal bar also acts as a grouping symbol. In the expressions the division indicated by the horizontal bar is the last operation to be performed.
In computing a quotient, one uses an algorithm, which finds an unknown multiplier digit by digit or term by term.
In the algorithm on the left, one starts with the digit 7 (actually 0.7) because it is the biggest digit one can use so that 4× 7 is 30 or less. That is followed by 5 (actually 0.05) because it is the biggest digit whose product with the divisor equals what remains of the dividend, or less. Thus, one has found (0.7 + 0.05) which, multiplied by 4 equals 3. In the algorithm on the right, one does the same thing, but with polynomials. One finds the polynomial of largest degree whose product with the divisor is equal to the dividend or less. In the case of polynomials, less is measured by the degree of the polynomial remainder rather than its numerical value. Had the dividend been x2 - 4, the quotient would still have been x - 1, with a remainder of -3, because any other quotient would have left a remainder whose degree was greater than or equal to that of the divisor.
These last two examples point out another way in which division is a less versatile operation than multiplication. If one is working with integers, one can always multiply two of them and have an integer for a result. That is not so with division. Although 3 and 4 are integers, their quotient is not. Likewise, the product of two polynomials is always a polynomial, but the quotient is not. Occasionally it is, as in the example above, but had one tried to divide x2 - 4 by x + 1, the best one could have done would have been to find a quotient and remainder, in this case a quotient of x - 1 and a remainder of -3. Many sets that are closed with respect to multiplication (i.e., multiplication can always be completed without going outside the set) are not closed with respect to division.
One number that can never, ever be used as a divisor is zero. The definition of division says that (a/b)b = a, but the multiplicative property of zero says that (a/b) × 0 = 0. Thus, when one tries to divide a number such as 5 by zero, one is seeking a number whose product with 0 is 5. No such number exists. Even if the dividend were zero as well, division by zero would not work. In that case, one would have (0/0)0 = 0, and 0/0 could be any number whatsoever.
Unfortunately, division by zero is a trap one can fall into without realizing it. If one divides both sides of the equation x2 - 1 = 0 by x - 1, the resulting equation, x + 1 = 0, has one root, namely -1. The original equation had two roots, however, -1 and 1. Dividing by x - 1 caused one of the roots to disappear, specifically the root that made x - 1 equal to zero.
The division algorithm shown above on the left converts the quotient of two numbers into a decimal, and if the division does not come out even, it does so only approximately. If one uses it to divide 2 by 3, for instance, the quotient is 0.33333. . . with the 3s repeating indefinitely. No matter where one stops, the quotient is a little too small. To arrive at an exact quotient, one must use fractions. Then the answer, a/b, looks exactly like the problem, a/b, but since one uses the bar to represent both division and the separator in the ratio form of a rational number, that is the way it is.
The algorithm for dividing rational numbers and leaving the quotient in ratio form is actually much
## KEY TERMS
Dividend The number that is divided or partitioned in a quotient.
Divisor The number of parts into which a dividend is partitioned.
Quotient The result of dividing one number by another.
Remainder The part of the dividend left over when an exact quotient cannot be computed.
simpler. To divide a number by a number in ratio form, one simply multiplies by its reciprocal. That is, (a/b) ÷ (c/d) = (a/b)(d/c).
## Resources
### BOOKS
Bittinger, Marvin L, and Davic Ellenbogen. Intermediate Algebra: Concepts and Applications. 6th ed. Reading, MA: Addison-Wesley Publishing, 2002.
Grahm, Alan. Teach Yourself Basic Mathematics. Chicago, IL: McGraw-Hill Contemporary, 2001.
Lorenz, Falko. Algebra. New York: Springer, 2006. Setek, William M. Fundamentals of Mathematics. Upper Saddle River, NJ: Pearson Prentice Hall, 2005.
Weisstein, Eric W. The CRC Concise Encyclopedia of Mathematics. Boca Raton, FL: Chapman & Hall/CRC, 2003.
J. Paul Moulton
# Division
views updated Jun 11 2018
# Division
Division is the mathematical operation that is the inverse of multiplication . If one multiplies 47 by 92 then divides by 92, the result is the original 47. In general, (ab)/b = a. Likewise, if one divides first then multiplies, the two operations nullify each other: (a/b)b = a. This latter relationship can be taken as the definition of division: a/b is a number which, when multiplied by b, yields a.
In the real world using ordinary arithmetic , division is used in two basic ways. The first is to partition a quantity of something into parts of a known size, in which case the quotient represents the number of parts, for example, finding how many three-egg omelets can be made from a dozen eggs. The second is to share a quantity among a known number of shares, as in finding how many eggs will be available per omelet for each of five people. In the latter case the quotient represents the size of each share. For omelets, if made individually you could use two eggs each for the five people and still have two left over, or you could put all the eggs in one bowl, so each person would get 22⁄3 eggs.
The three components of a division situation can represent three distinct categories of things. While it would not make sense to add dollars to earnings-per-share, one can divide dollars by earnings-per-share and have a meaningful result (in this case, shares). This is true, too, in the familiar rate-distance-time relationship R = D/T. Here the categories are even more distinct. Distance is measured with a tape; time by a clock; and rate is the result of these two quantities.
Another example would be in preparing a quarterly report for share holders; a company treasurer would divide the total earnings for the quarter by the number of shares in order to compute the earnings-per-share. On the other hand, if the company wanted to raise \$6,000,000 in new capital by issuing new shares, and if shares were currently selling for \$181⁄8, the treasurer would use division to figure out how many new shares would be needed, i. e., about 330,000 shares.
Division is symbolized in two ways, with the symbol ÷ and with a bar, horizontal or slanted. In a/b or a ÷ b, a is called the dividend; b, the divisor; and the entire expression, the quotient.
Division is not commutative; 6/4 is not the same as 4/6. It is not associative; (8 ÷ 4) ÷ 2 is not the same as 8 ÷ (4 ÷ 2). For this reason care must be used when writing expressions involving division, or interpreting them. An expression such as
is meaningless. It can be given meaning by making one bar noticably longer than the other
to indicate that 3/4 is to be divided by 7. The horizontal bar also acts as a grouping symbol. In the expressions
The division indicated by the horizontal bar is the last operation to be performed.
In computing a quotient one uses an algorithm , which finds an unknown multiplier digit by digit or term by term.
In the algorithm on the left, one starts with the digit 7 (actually 0.7) because it is the biggest digit one can use so that 4 × 7 is 30 or less. That is followed by 5 (actually 0.05) because it is the biggest digit whose product with the divisor equals what remains of the dividend, or less. Thus one has found (.7 + 0.05) which, multiplied by 4 equals 3. In the algorithm on the right, one does the same thing, but with polynomials . One finds the polynomial of biggest degree whose product with the divisor is equal to the dividend or less. In the case of polynomials, "less" is measured by the degree of the polynomial remainder rather than its numerical value. Had the dividend been x2 - 4, the quotient would still have been x - 1, with a remainder of -3, because any other quotient would have left a remainder whose degree was greater than or equal to that of the divisor.
These last two examples point out another way in which division is a less versatile operation than multiplication. If one is working with integers , one can always multiply two of them an have an integer for a result. That is not so with division. Although 3 and 4 are integers, their quotient is not. Likewise, the product of two polynomials is always a polynomial, but the quotient is not. Occasionally it is, as in the example above, but had one tried to divide x2 - 4 by x + 1, the best one could have done would have been to find a quotient and remainder, in this case a quotient of x - 1 and a remainder of -3. Many sets that are closed with respect to multiplication (i.e., multiplication can always be completed without going outside the set) are not closed with respect to division.
One number that can never, ever be used as a divisor is zero . The definition of division says that (a/b)b = a, but the multiplicative property of zero says that (a/b) × 0 = 0. Thus, when one tries to divide a number such as 5 by zero, one is seeking a number whose product with 0 is 5. No such number exists. Even if the dividend were zero as well, division by zero would not work. In that case one would have (0/0)0 = 0, and 0/0 could be any number whatsoever.
Unfortunately division by zero is a trap one can fall into without realizing it. If one divides both sides of the equation x2 - 1 = 0 by x - 1, the resulting equation, x + 1 = 0, has one root, namely -1. The original equation had two roots, however, -1 and 1. Dividing by x - 1 caused one of the roots to disappear, specifically the root that made x - 1 equal to zero.
The division algorithm shown above on the left converts the quotient of two numbers into a decimal, and if the division does not come out even, it does so only approximately. If one uses it to divide 2 by 3, for instance, the quotient is 0.33333... with the 3s repeating indefinitely. No matter where one stops, the quotient is a little too small. To arrive at an exact quotient, one must use fractions. Then the "answer," a/b, looks exactly like the "problem," a/b, but since we use the bar to represent both division and the separator in the ratio form of a rational number , that is the way it is.
The algorithm for dividing rational numbers and leaving the quotient in ratio form is actually much simpler. To divide a number by a number in ratio form, one simply multiplies by its reciprocal . That is, (a/b) ÷ (c/d) = (a/b)(d/c).
## Resources
### books
Bittinger, Marvin L., and Davic Ellenbogen. Intermediate Algebra: Concepts and Applications. 6th ed. Reading, MA: Addison-Wesley Publishing, 2001.
Eves, Howard Whitley. Foundations and Fundamental Concepts of Mathematics. NewYork: Dover, 1997.
Grahm, Alan. Teach Yourself Basic Mathematics. Chicago: McGraw-Hill Contemporary, 2001.
Weisstein, Eric W. The CRC Concise Encyclopedia of Mathematics. New York: CRC Press, 1998.
J. Paul Moulton
## KEY TERMS
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Dividend
—The number that is divided or partitioned in a quotient.
Divisor
—The number of parts into which a dividend is partitioned.
Quotient
—The result of dividing one number by another.
Remainder
—The part of the dividend left over when an exact quotient cannot be computed.
# division
views updated May 21 2018
di·vi·sion / diˈvizhən/ • n. 1. the action of separating something into parts, or the process of being separated: the division of the land into small fields | a gene that helps regulate cell division. ∎ the distribution of something separated into parts: the division of his estates between the two branches of his family. ∎ an instance of members of a legislative body separating into two groups to vote for or against a bill: the new clause was agreed without a division. ∎ the action of splitting the roots of a perennial plant into parts to be replanted separately, as a means of propagation: the plant can also be easily increased by division in autumn. ∎ Logic the action of dividing a wider class into two or more subclasses.2. disagreement between two or more groups, typically producing tension or hostility: a growing sense of division between north and south | a country with ethnic and cultural divisions. 3. the process or skill of dividing one number by another. See also long division, short division. ∎ Math. the process of dividing a matrix, vector, or other quantity by another under specific rules to obtain a quotient.4. each of the parts into which something is divided: the main divisions of the book. ∎ a major unit or section of an organization, typically one handling a particular kind of work: a retail division. ∎ a group of army brigades or regiments: an infantry division. ∎ a number of teams or competitors grouped together in a sport for competitive purposes according to such characteristics as ability, size, or geographic location: the team will finish in fifth place in Division One. ∎ a part of a county, country, or city defined for administrative or political purposes: a licensing division of a district. ∎ Bot. a principal taxonomic category that ranks above class and below kingdom, equivalent to the phylum in zoology. ∎ Zool. any subsidiary category between major levels of classification.5. a partition that divides two groups or things: the villagers lived in a communal building and there were no solid divisions between neighbors.PHRASES: division of labor the assignment of different parts of a manufacturing process or task to different people in order to improve efficiency.
# division
views updated Jun 27 2018
division A category used traditionally in the classification of plants that consists of one or several similar classes. An example is the Spermatophyta (seed-bearing plants). In modern classification systems the phylum has replaced the division.
# division
views updated May 17 2018
division (di-vizh-ŏn) n. the separation of an organ or tissue into parts by surgery. | 4,034 | 17,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.951767 |
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Who discovered the secret? (Posted on 2016-12-27)
Four colleagues were suspicious to have discovered a secret of the highest level of classified information by outwitting the military computers. At the hearing, they have been asked who discovered the secret. But the transcript of the hearing only shows four statements, one made by each of the suspects. We do not know (and may even doubt) whether the minute taker has recorded word-for-word what they said during the hearing. But suppose, the minute taker has truly captured (maybe unintentionally) the right logical essence in all their speaking. Thereafter, according to the documents, no one in the commission had any idea what to do with the statements of the hearing.
Would you have been able to find out who discovered the secret, based on these four recorded statements?
Atkins: It's not true that if I'm lying, Stone is also lying or Yates has discovered the secret.
Evert: It's a lie that if I'm not telling the truth, Atkins is telling the truth or Stone has not discovered the secret.
Yates: It is not the case that if I'm lying, Atkins and Evert have discovered the secret.
Stone: If I'm lying, then either Atkins is telling the truth or Evert is lying or Yates has not discovered the secret.
No Solution Yet Submitted by ollie No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
solution | Comment 2 of 4 |
In order for A's statement to be true, the "If I'm lying..." conditional would have to be false, but to falsify that, A would have to be lying. So A's statement can't be true and must be false. So the "If I'm lying..." conditional is true, and indeed the antecedent is true so the consequent is also: either S is also lying or Y has discovered the secret.
By the same reasoning with E's statement, A is telling the truth or S has not discovered the secret. Well, A is not telling the truth, so S has not discovered the secret.
Same reasoning with Y's statement: A and E have discovered the secret.
If S is telling the truth, the rest of his statement doesn't really tell you anything. If he's lying than A is lying (we knew that already) and E is lying and Y has discovered the secret.
Bottom line:
I can tell that A and E have discovered the secret, and also Y has discovered the secret. S is the only one not to have discovered the secret.
Posted by Charlie on 2016-12-27 09:45:00
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# How can I Identify the P,Q,R,S and T waves in all 12 leads of ECG
7 views (last 30 days)
Naveen F on 17 Oct 2020
Commented: Naveen F on 17 Oct 2020
Hi all,
I am currenly working on diagnosing heart diseases using 12 lead ecgs.
for this project i use preprocessed data which has 500hz sampling rate.
Here i attached a sample file from the database i used.
as a first step seperated the 12 leads ecg
Data = load('HR00001.mat'); % 12x5000 double
ecg = Data.val(1,1:5000); % 1x5000 double
my next step is to identify the P,Q,R,S and T waves. As a beginner i found it quite challenging.
I wrote a code which only able to find R waves in the lead I but its totally useless when it comes to other leads.
Can some one help me a step by guidance for detect P,Q,R,S and T in different leads.
Data = load('HR00001.mat'); % 12x5000 double
ecg = Data.val(1,1:5000); % 1x5000 double
peaks_1 = ecg>500;
f_s = 500;
time = (1:numel(ecg))/f_s;
[peaks_2,pos_peaks] = findpeaks(ecg,'MINPEAKDISTANCE',10,'MINPEAKHEIGHT',500);
plot(time,ecg,'b',pos_peaks/500,(peaks_2),'ro')
axis ([0 10 -300 800])
Please help me in this issue . your help will will highly appreciated.
thanks and best regards.
#### 6 Comments
Show 3 older comments
Naveen F on 17 Oct 2020
That so true sir, I tried lots of proposed codes in the matlab forum but the problem that i am going to address is quite complex one. I really appreciate your guidance.
I access the source ( Braunwald's Heart Disease: A Textbook of Cardiovascular Medicine ) from google scholar since the website you have mentioned is currenlty not working or it blocks the IPs from my region . I hope this will help me to solve my issue.
Thanks and best regards.
Star Strider on 17 Oct 2020
Naveen —
My pleasure!
I have a hardcopy of the latest edition in my physical library. If you are doing anything related to heart disease, I suggest that you get a copy or refer to one in your university library It is an invaluable resource.
Naveen F on 17 Oct 2020
Thank you very much for the suggestion Star Strider sir, I will deffinitely send an email to our uni librarian regarding this resource.
Thanks and best regards.
Sign in to comment.
### Answers (1)
Cris LaPierre on 17 Oct 2020
Edited: Cris LaPierre on 17 Oct 2020
If you would like an interactive way to try to detect peaks, I suggest using the Find Local Extrema task in the live editor. It allows you to use a graphical interface to set the peak detection parameters. Once you get something you like, you can convrt the task to code, which allows you to build a script to run on multiple data sets.
It might be important to set your expectations here. This is not a silver bullet. You will still need to gain an understanding of what it is doing to use it appropriately, especially if you are trying to make medical diagnoses from your analysis.
#### 1 Comment
Naveen F on 17 Oct 2020
Thank you very much pitching your answer for my question. I know nothing comes on a silver palate. I'll put my fullest effort on this and find a way to solve my problem using this method. I hope this method will help me to solve my issue.
keep up the good work.
Note : Still i can ask any follow up questions regarding this problem right even after i accept this answer ? (just ask because i am new to this forum )
Thanks and best regards.
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The normal distribution is
f(x) = (1σ ) e-(x-µ) ²2σ ²
Notation Probability density functionThe red curve is the standard normal distribution Cumulative distribution function µ µ µ σ² 0 0
Curve Equation
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PDF
Half Normal Distribution Equations
• The parameter μ in this formula is the mean or expectation of the distribution (and also its median and mode).
• The parameter σ is its standard deviation; its variance is therefore σ' 2. A random variable with a Gaussian distribution is said to be normally distributed and is called a normal deviate. | 140 | 614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-21 | latest | en | 0.858736 |
https://math.stackexchange.com/questions/1614300/confusion-on-taking-the-second-partial-derivative | 1,566,350,544,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00017.warc.gz | 544,602,070 | 30,730 | Confusion on taking the second partial derivative:
From this previous question I gained understanding of why the chain rule for a function $u(x,y)=f(p)$ is expressed as $$\frac{\partial u}{\partial x}=\frac{\mathrm{d}f(p)}{\mathrm{d}p}\times \frac{\partial p}{\partial x}$$ where $p=p(x,y)$
Now suppose that $u(x,y)=f(ax+by)$ therefore $$\color{blue}{\frac{\partial u}{\partial x}=a\frac{\mathrm{d}f(p)}{\mathrm{d}p}\tag{A}}$$ where $p=ax + by$
I need to compute $$\frac{\partial^2 u}{\partial x^2}$$
So my attempt is to recognize that $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)$$ Therefore from equation $\color{blue}{\mathrm{(A)}}$ it should follow that $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}\left(a\frac{\mathrm{d}f(p)}{\mathrm{d}p}\right)$$ but according to my book source this is wrong and the correct answer is $$\frac{\partial^2 u}{\partial x^2}=\color{#A0F}{a^2}\color{#F80}{\frac{\mathrm{d^2}f(p)}{\mathrm{d}p^2}}$$
Could someone please kindly explain how this equation was obtained?
Specifically, I have no idea where the $\color{#A0F}{a^2}$ and the $\color{#F80}{\dfrac{\mathrm{d^2}f(p)}{\mathrm{d}p^2}}$ came from.
$\color{red}{\text{For some more context here is an image of the book source:}}$
Neither you nor the book are wrong. The book just takes the calculation a step further: $$\frac{\partial}{\partial x}\left(a\frac{df}{dp}\right) =a\frac{\partial}{\partial x}\left(\frac{df}{dp}\right).$$ Treating $df/dp$ as a function $g(p) = \frac{df}{dp}(p)$, we apply the same reasoning you initially applied to $f$ to obtain $$\frac{\partial}{\partial x}\left(\frac{df}{dp}\right) = \frac{\partial}{\partial x}g(p) = \frac{dg}{dp}\frac{\partial p}{\partial x} = a\frac{d^2f}{dp^2}.$$ Putting this back into the first equation, $$\frac{\partial}{\partial x}\left(a\frac{df}{dp}\right) = a^2\frac{d^2f}{dp^2}.$$ | 645 | 1,917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-35 | latest | en | 0.763944 |
https://physics.stackexchange.com/questions/339890/newtonian-mechanics-how-does-something-like-a-car-wheel-roll/378623 | 1,725,851,904,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00239.warc.gz | 435,418,800 | 44,816 | # Newtonian Mechanics - How does something like a car wheel roll?
Ok so please bear with me here and what is almost certainly a really stupid question, partly because I don't quite know how to ask it.
When you have a wheel such as one attached to a car, a torque is applied to it from the engine. It's my understanding that the wheel would slide over the road and the car would not move if it were not for static friction which is of course the concept of rolling without slipping.
The problems I am having is when I consider what the magnitude of the force exerted on the road from the wheel would be, or specifically the equal and opposite force exerted back onto the wheel from the road via static friction. The reason for this is because I want to say that it is just the torque exerted on the wheel by the engine divided by the radius of the wheel but that seems to lead to a ridiculous conclusion. If I follow that logic, then the equal and opposite force exerted on the wheel by the road, multiplied by the radius of the wheel to get the counter-torque, is equal to the torque exerted on the wheel by the engine but in the opposite direction. That of course would mean that there is no net torque and the wheels would never move (unless the force were greater than that available from static friction but then it would just slip) which means that the car would never be able to move! So simply put, what am I missing that is almost certainly staring me in the face (again)? Clearly things roll and cars move so I know i'm going horribly and embarrassingly wrong somewhere.
Also, I often hear people say that static friction is the force responsible for propelling a car forward. How is that the case when it acts in the direction opposite to the direction that the wheels are trying to roll? That's sort of the same question but I think it may help highlight my misunderstandings of rotational motion in this area.
I never knew wheels could be so complicated! At least for me anyway.
The static friction will be pointing forwards not backwards. It comes into existence to counteract sliding when the torque pulls the wheel around - the wheel tries to push the ground backwards and static friction is the ground's equal and opposite response.
That static friction is then the only force on the wheel. Newton's 2nd law says that the wheel will then accelerate forward.
Update
Since this means that there is translational acceleration $a$, there must be angular acceleration $\alpha$ of the wheel as well. The well spins faster and faster while the car goes faster and faster. This is because of the so-called geometric bond between wheel and ground:
$$a=R\alpha$$
You can't have $a$ without $\alpha$ in this case.
This means that the torques do no balance out, which seemed to be the assumption that caused your whole confusion. The applied torque from the engine will not equal the torque from friction. It will be a bit bigger in order to also cause some angular acceleration. You must use $$\sum \tau =I\alpha$$ And not $$\sum\tau=0\;.$$
This equation along with the geometric bond and Newton's 2nd law should be enough equations for you to solve for the static friction.
• Right so the static friction will point forwards on the bottom of the wheel, but that part of the wheel is trying to move backwards due to the torque from the engine. That's part of the problem I am trying to understand. Commented Jun 17, 2017 at 16:39
• @An_African_Ape If you force the wheel to turn but the bottom point is stuck, then the centre will "fall" forward. Moving the wheel Centre means moving the car. This is essentially how the forward motion happens. Commented Jun 17, 2017 at 16:53
• Your phrase "if you force the wheel to turn" implies a net torque and that is the main problem I am having. How can there be a net torque if the force exerted on the bottom of the wheel by static friction (equal and opposite to the force exerted onto the road by static friction) is equal to the torque from the engine on the wheel divided by its radius? It seems as though the static friction force from the road to the bottom of the wheel multiplied by the wheels radius is equal to the torque from the engine but in the opposite direction so they cancel out. Clearly that is not the case but how? Commented Jun 17, 2017 at 17:29
• @An_African_Ape Aha, I see your point now. You think that the applied torque will be cancelled out by the torque from static friction. This is not the case. When your car speeds up, it accelerates translationally and the wheel rotation accelerates as well (angular acceleration). The sum of torques is not equal to zero but to $\sum\tau=I\alpha$ in the same way that the sum of forces is not zero but equal to $\sum F=ma$. The reason we know for sure that the angular acceleration must be present is because of the connection $a=\alpha R$. Commented Jun 17, 2017 at 19:30
• @An_African_Ape I have added an update to the answer explaining this. Commented Jun 17, 2017 at 19:42
Torque due to frictional force is equal to applied torque on wheel only when the vehicle is moving at a constant velocity. That means the traction force completely balances all the resistance forces and hence there is no linear or angular acceleration.
Net force = Total Traction force-Total Resistance Force
Acceleration = 0 means Total Traction Force = Total Resistance Force
In normal case (neglecting resistance forces): $$F_tR=T-I\alpha$$ where $F_t$ is the traction force, $R$ is the radius of the wheel, $T$ is the applied torque on the wheel, $I$ is the moment of inertia of the wheel, and $\alpha$ is the angular acceleration of the wheel.
• You can format mathematical expression much more cleanly using mathjax. If you click edit you can see a simple example applied to your answer. Commented Jan 8, 2018 at 10:34
With problems like this, a diagram always helps:
In this front wheel drive car, the green arrow represents the direction in which the engine is turning the wheel.
The red arrow is the reaction force of the road onto the wheel - and that is a net external force on the car-plus-wheels. If there are no other forces of friction, this red force is what accelerates the car.
If you are interested in analyzing the torque more carefully, you have to account for the fact that if the car is accelerating, the inertial reaction of the center of mass (which is higher than the surface of the road) will tend to shift some of the weight to the rear wheels of the car; this ensures that there is once again no net torque, so the car doesn't flip over (if there is a net torque the angular momentum of the car will change; in fact if the car is accelerating there will of course be a change in the angular momentum of the wheels, so there is some small net torque needed).
I hope I didn't end up confusing you more...
It appears as though you're taking the origin to be the center of the wheel in your analysis. In this case, you have a torque both from the axle (applied very close to but not at the center, so nonzero) and from static friction at point of contact between the tire and road.
Instead, take as the origin the point of contact between the tire and the road. Then, the torque from static friction is zero, since $\vec r = \vec 0$, and the only torque on the tire is that supplied by the axle, which gives a nonzero net torque in the direction one would expect.
If you were to analyze it in your original frame, it's important to realize that the friction opposes the motion of the tire (consider the case of a crate sitting in a truck bed), not the torque, therefore $\tau_{axle} \neq \tau_{friction}$. | 1,703 | 7,651 | {"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} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.979728 |
https://studysoup.com/tsg/1061782/vector-mechanics-for-engineers-statics-10-edition-chapter-7-problem-7-2 | 1,632,254,814,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00397.warc.gz | 588,094,391 | 12,949 | ×
Get Full Access to Vector Mechanics For Engineers: Statics - 10 Edition - Chapter 7 - Problem 7.2
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Solved: Determine the internal forces (axial force, shearing force, and bending moment)
ISBN: 9780077402280 491
Solution for problem 7.2 Chapter 7
Vector Mechanics for Engineers: Statics | 10th Edition
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Problem 7.2
Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of 6.78.
Step-by-Step Solution:
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ISBN: 9780077402280
The full step-by-step solution to problem: 7.2 from chapter: 7 was answered by , our top Engineering and Tech solution expert on 03/19/18, 04:25PM. This textbook survival guide was created for the textbook: Vector Mechanics for Engineers: Statics, edition: 10. The answer to “Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of 6.78.” is broken down into a number of easy to follow steps, and 23 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 9 chapters, and 1417 solutions. Since the solution to 7.2 from 7 chapter was answered, more than 238 students have viewed the full step-by-step answer. Vector Mechanics for Engineers: Statics was written by and is associated to the ISBN: 9780077402280.
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Unlock Textbook Solution | 602 | 2,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.860029 |
https://topic.alibabacloud.com/a/baidu-interview-questions_8_8_31950173.html | 1,725,857,818,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00850.warc.gz | 545,737,423 | 17,249 | # Baidu interview questions
Source: Internet
Author: User
Baidu interview questions are for some reference only.
1. Complete the Function
Size_t Foo (unsigned int * A1, size_t al1, unsigned int * A2, size_t Al)
Both A1 and A2 are unsigned arrays, al1 and Al are the length of arrays, and the length of arrays is an even number.
An unsigned array consists of a number range. For example:
A1 is 0, 1, 3, 6, 10, 20
A2 is 0, 1, 20, 50, 4, 5
A1 indicates the following range: [] [] []
A2 indicates the following range [] [] []
The overlapping part of A1 and A2 is [] [], and its length is 2.
The Foo function must return the length of the overlap interval. 2 In the preceding example.
Requirements:
Write out the original code of function Foo. In addition, the efficiency is as high as possible, and the Code complexity analysis is given.
Restrictions:
The length of al1 and Al 2 cannot exceed 1 million. In addition, the intervals of the same Array may overlap.
For example, A1 may be 9,100, or 80.
The storage space is as small as possible.
Solution: sort the intervals of two arrays first. Sort the first number of intervals as the key, and then construct a binary sorting tree for an array, search the intervals in another array in the binary sorting tree. Time complexity (nlogn ).
Someone proposed the complexity of log (+ n), that is, to sort the two arrays. The sorting rules are as follows. Then, the two arrays are merged and the result is the result.
Two people are in a queue. We think that the sequence from low to high is correct, but some people do not follow the order. If we say that the people in front are higher than those in the back (the two are the same height and think it is appropriate), we think that these two people are "troublemakers", for example, there is a sequence:
176,178,180,170,171
<176,170>, <176,171>, <178,170>, <178,171>, <180,170>, <180,171>,
So, now we provide an integer sequence. Please find out the number of these troublemakers (only the number of troublemakers is given, without a specific pair)
Requirements:
Input:
Is a file (in), each row of the file is a sequence. The sequences are all numbers separated by commas.
Output:
It is a file (out) with a number for each behavior, indicating the logarithm of the troublemakers.
Describe in detail about your own solutions and some key points of your implementation. The implementation code is provided, and the time complexity is analyzed.
Restrictions:
The maximum number of numbers in each line is 100000, and the maximum number is 6 digits. There is no memory usage limit for the process sequence.
Basic Idea: divide the entire sequence into multiple incremental sequences, and then start from the second sequence, compare each number with the previous sequence, the time complexity is
2. The following two questions are selected. Please answer one question based on your situation (4th answers for the Web direction and 3rd answers for other positions ).
3
Consider an online friend system. The system maintains a friend list for each user. The list can contain a maximum of 500 friends. Friends must be other users in the system. Friend relationships are unidirectional. User B is a friend of user a, but user A is not necessarily a friend of user B.
The user is represented as an ID. The text format of the friend list data is as follows:
2 567,890
31, 66
14 567
78 10000
...
Each row has two columns. The first column is the user ID and the second column is the friend ID. Different IDs are separated by commas (,), and IDs are sorted in ascending order. Columns are separated by "T.
Requirements:
Design an appropriate index data structure to complete the following queries:
If user a and user B are given, check whether there is such a relationship between user a and user B: B is a two-dimensional friend of user a (friend's friend ).
In the preceding example, 10000 is a two-dimensional friend of 1, because 10000 is a friend of 1 and is a friend of 78.
Describe in detail about your own solutions and some key points of your implementation. The pseudo-code is provided to implement the indexing and query processes, and the space and time complexity are described.
Restrictions:
The number of users is no more than 10 million, with an average of 50 friends.
4
Link mode: user (userid, username), article (ArticleID, userid, title, content), vote (ArticleID, score), and user is a user relationship, article indicates the relationship between the articles published by the user, vote indicates the relationship between the votes received by the articles, title indicates the article title, and score indicates the number of votes received.
(1) query all user names that have not published any articles in SQL;
(2) query the titles of all articles with more than 100 votes in SQL, and sort them in reverse order;
(3) use SQL to query the user names whose number of articles in the departure table is greater than 5 and whose average number of articles is greater than 100, sorted in descending order by average number of votes;
(4) design the primary keys, foreign keys, and indexes of these tables, and point out the indexes used for the preceding three queries.
(5) When the number of users exceeds 10 million and the number of articles exceeds 0.1 billion, how should we consider storage and performance improvement and optimization?
Related Keywords:
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 1,415 | 6,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.937692 |
https://hchapman.org/talks/altknots/ | 1,713,321,425,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00734.warc.gz | 272,750,985 | 7,216 | # Alternating knots are rare
Harrison Chapman
hchapman.github.io/talks/altknots
OIST Geometry and Topology of Manifolds Group
18th February 2019
A knot or link type is alternating if it admits a diagram whose crossings alternate between over- and underpasses.
## Thistlethwaite '98
Let $$\altlinkclass$$ be the set of all alternating link types,
with counts $$\altlinkcount$$ indexed by crossing number
If $$L_n$$ is the number of all link types, the theorem says; $\lim_{n\to\infty}{L_n^{1/n}} > \lim_{n\to\infty}{\altlinkcount^{1/n}}$
## Question (Classical)
Does the same result hold for alternating knot types?
## Alternating knots should be rare
Data of Hoste, Thistlethwaite, and Weeks '98 suggests
alternating knot types are rare
$$n$$ # Knots # Alternating % Alternating
633100%
777100%
8211886%
$$\vdots$$
15253 29385 26351%
161 388 705379 79938%
## Weak & Insufficient Reasoning
Every $$n$$-crossing link diagram has $$2^n$$ crossing assignments, of which precisely two are alternating.
Must pay very close attention to equivalence classes of diagrams
## Simplification from Combinatorics
Objects with fewer symmetries are easier to enumerate
Sundberg and Thistlethwaite '98
## Tait Flyping Conjecture
Ingredient 1: The Tait Flyping Conjecture holds for alternating links;
## Menasco and Thistlethwaite '91
Any two alternating link or tangle diagrams are related by a sequence of flypes if and only if they represent the same link or tangle type.
Zinn-Justin and Zuber '10
## Tait Flyping Conjecture & Knots
Tait Flyping Conjecture still holds for alternating knots
## Menasco and Thistlethwaite '91
Any two alternating knot diagrams are related by a sequence of flypes if and only if they represent the same knot type.
Ingredient 2: Alternating link and tangle types can be enumerated;
## Sundberg and Thistlethwaite '98
The numbers $$\tanglecount$$ of alternating tangle types have a well-understood generating function (g.f.), $\tanglegenfun{z} = \sum_{\color{tangle}\substack{\text{alternating}\\\text{link types } {{\color{black}L}}}}{z^{\mathop{cr}(L)}} = \sum_{n}^{\infty} {\tanglecount z^n}.$
## Sundberg and Thistlethwaite '98
The counts $$\tanglecount$$ grow asymptotically as,
$\tanglecount \sim \frac{3 \,c_1}{4 \sqrt \pi} n^{-5/2} \altlinkgrowth^{n-3/2},$
for known constants $$c_1$$ and $$\altlinkgrowth$$.
Zinn-Justin and Zuber '02 reprove this using
techniques from random matrix theory
## The generating function $$\tanglegenfun{z}$$
The g.f. of alternating tangle types $$\tanglegenfun{z}$$ satisfies: $\tanglegenfun{z} = \algtanglegenfun{z}{\bpotanglegenfun{\tanglegenfun{z}}}$ where;
• $$z$$ marks crossings,
• $$\zeta$$ marks slots for tangles (or templates)
• $$\bpotanglegenfun{\zeta}$$ is the g.f. of basic polyhedral templates
• $$\algtanglegenfun{z}{\zeta}$$ is the g.f. of flype-equivalence classes of algebraic templates,
## Basic polyhedral templates
Sundberg and Thistlethwaite '98 Basic polyhedral templates, counted by $$\bpotanglegenfun{\zeta}$$
Bijection with rooted $$c$$-nets studied by Tutte '63: $\bpotanglegenfun{\zeta} = {\scriptsize \frac{1}{2(\zeta+2)^3}\left( (1-4\zeta)^{3/2} + (2\zeta^2 - 10\zeta - 1) \right) - \frac{2}{1+\zeta} - \zeta + 2}$ (formula due to Sundberg and Thistlethwaite)
## Algebraic templates
Sundberg and Thistlethwaite '98 Flype equivalence classes of algebraic templates, counted by $$\algtanglegenfun{z}{\zeta}$$
## Key observation
Flyping can only occur within an algebraic template
## Algebraic templates
Sundberg and Thistlethwaite '98
Admit recursive decomposition into horizontal templates: $\algtanglegenfun{z}{\zeta} = \hortanglegenfun{z}{\zeta} + \hortanglegenfun{z}{\zeta} + z - \zeta,$ where $$\hortanglegamma$$ satisfies $\hortanglegamma = z\algtanglegenfun{z}{\zeta} + \frac{\hortanglegamma^2}{1-\hortanglegamma} + \zeta$
## Counts of tangles and links
Counts of alternating tangle types $$\tanglecount$$ and
alternating link types $$\altlinkcount$$ are related;
## Counts
$\frac{\tanglecountdec}{8(2n-3)} \le \altlinkcount \le \frac{\tanglecountdec}2$
## Exponential growth rates
$\lim_{n\to\infty}{\altlinkcount^{1/n}} = \lim_{n\to\infty}{\tanglecount^{1/n}} = \altlinkgrowth$
## Enumeration of alternating knots
Little is known about the counts $$\altknotcount$$ of alternating knot types;
## Schaeffer and Zinn-Justin '04
For some constants $$\tau$$ and $$\gamma$$, $\altknotcount \sim C \tau^n n^{\gamma-3}$
## Currently
The g.f. $$\sum{\altknotcount}z^n$$ is not well understood and the radius of convergence is not known
## Thistlethwaite '98
The set of tangles $$\superaltlinkclass$$ generated by introducing the above (non-alternating) tangle into the set of algebraic templates still satisfies the flyping conjecture
The g.f. of this superclass of tangle types is obtained by replacing $\algtanglegenfun{z}{\zeta} \quad \text{with} \quad \beta(z,\zeta) = \algtanglegenfun{z}{\zeta} + z^{13}.$ This g.f. has strictly smaller radius of convergence than that of alternating tangle types $$\tanglegenfun{z}$$, so:
## Thistlethwaite '98
Alternating tangle (and link) types are exponentially rare.
## Observation
The above tangle can be found in knot types, and is contained in a superclass $$\superaltknotclass$$ of knot diagrams which still satisfy the flyping conjecture
Cannot quantify the ocurrences with g.f's
(not enough understanding of knots)
## We have:
Alternating knot types of $$\altknotclass$$ satisfy the flyping conjecture, as do a superclass $$\superaltknotclass$$.
## We lack:
Enumeration of alternating knot types, or knots in $$\superaltknotclass$$.
## Pattern theorem
Pattern theorems quantify the density of structure:
## C. '18
Let $$P$$ be a reduced alternating tangle admitting no interior flypes that may be found in alternating knot diagrams.
There exist constants $$\occurconst > 0$$, and $$N \in \NN$$ such that for all $$n \ge N$$, all but exponentially few alternating knot types contain $$\ge \occurconst n$$ copies of $$P$$.
## Pattern theorem implies rarity
$$R =$$ $$\overline R =$$
1. There is $$\occurconst$$ so that the tangle $$R$$ occurs $$\occurconst n$$ times in almost all alternating knot types.
2. In $$\superaltknotclass$$, the tangles $$R$$ and $$\overline R$$ are equiprobable.
3. Knot types which contain $$\overline R$$ are non-alternating.
4. To each alternating knot type in $$\altknotclass$$, there are at least $$2^{\occurconst n}-1$$ more non-alternating knot types in $$\superaltknotclass$$.
We conclude:
## C. '18
Alternating knot types are exponentially rare among all knot types.
## Requirements for a pattern theorem
Pattern theorem requires a way of adding
a pattern to alternating knot types that:
1. Only produces alternating knot types
2. Has a linear number ($$\propto n$$) of attachment sites, and
3. Can be un-done, yielding the original knot type
## The idea
$\text{# lone crossings} + \sum_{\substack{\text{all horiz.}\\\text{tangles }\gamma}}{\text{# sites in }\gamma} = n$
## Millett and Jablan '09
The probability that a minimal prime knot diagram contains a trefoil knotted segment goes to one as the crossing number goes to infinity
## C. '18
The probability that an alternating knot type has a minimal prime diagram containing a trefoil knotted segment goes to one as the crossing number goes to infinity
## Proof
The following pattern introduces a trefoil segment:
## Classical
The crossing number of knots is additive with respect to connected sum.
Of 1,388,705 prime knot types of 16 crossings,
Almost all prime knot types are hyperbolic.
## Malyutin '16
Either of these conjectures contradicts the other.
It's not unreasonable that Adams's conjecture is false and that current data is only representative of small crossing number
## Conjecture
Can we disprove Adams's conjecture, possibly using pattern-theoretic methods?
## Take-aways & conclusion
1. Alternating links have the flype conjecture and a complete enumeration
2. Alternating knots share the result on flypes but lack precise enumeration
3. Absent enumerations, pattern theorems can quantify presence of structure:
• Rarity of alternating knots
• Certainty of subknotting
4. Pattern theorems can likely be proven for other classes in topology, geometry, etc...
# Thank you!
C. 2018. On the structure and scarcity of alternating knots. Submitted. arXiv: 1804.09780 | 2,361 | 8,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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 | 3 | CC-MAIN-2024-18 | latest | en | 0.736753 |
http://math.stackexchange.com/questions/82838/4-color-theorem-on-surfaces/83385 | 1,469,358,731,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823996.40/warc/CC-MAIN-20160723071023-00220-ip-10-185-27-174.ec2.internal.warc.gz | 156,011,659 | 22,931 | # 4-Color Theorem on Surfaces
Is it true, that the minimum number of colors to color a map is still 4, even if the surface you draw the map upon is more complex than a sphere?
What if we have “Planar” graphs on Möbius strips or on a torus ?
-
In the first edition of Stan Wagon's "Mathematica in Action", there is some code for generating a picture of a 7-colored torus. – J. M. Nov 17 '11 at 0:13
There is an entire book on this, Topological Graph Theory by Jonathan L. Gross and Thomas W. Tucker – Will Jagy Jan 3 '12 at 22:18
No. The correct replacement for $4$ for all compact surfaces without boundary except the Klein bottle is given by the Ringel-Youngs theorem (formerly the Heawood conjecture): it takes $$\left\lfloor \frac{7 + \sqrt{49 - 24 \chi}}{2} \right\rfloor$$
colors except for the Klein bottle where it takes $6$, where $\chi$ is the Euler characteristic of the surface. For the torus, $\chi = 0$, so it takes $7$ colors. I don't know the answer off the top of my head for a Möbius strip.
-
The presence of boundaries doesn't affect the answer; for the purposes of this question, a Mobius strip is the same as a projective plane. – Gerry Myerson Nov 16 '11 at 23:13
Just for the record: for the torus, the answer is 7; it's a theorem, not a conjecture. – lhf Nov 17 '11 at 0:27
Yes, the Heawood "conjecture", like Bertrand's "postulate", is actually a theorem (as Qiaochu knows). – Gerry Myerson Nov 17 '11 at 0:57
Is there any intuition for why the Klein bottle should be exceptional? – Jason DeVito Nov 17 '11 at 3:04
A good place to read about questions related to coloring graphs on surfaces (as well issues about the genus of a graph) is the book by Mohar and Thomassen, Graphs on Surfaces, John Hopkins U. Press, 2001. Chapter 4 has a discussion of Ringel and Youngs solution of the Heawood conjecture and Chapter 8 deals with coloring problems. – Joseph Malkevitch Nov 17 '11 at 4:04
Though others have already given more ambitious answers, it may yet be worthwhile to explain why "pure thought" shows that the answer to the OP's question in the first paragraph must be no.
The key observation is this: every finite graph can be embedded in some subsurface of $\mathbb{R}^3$. To do this, start with a 2:1 planar immersion of the graph, i.e., a map which is locally an embedding into the plane at globally has, at worst, two edges crossing at any given point. (It is an easy exercise to prove that such things exist for any finite graph.) Now imagine the graph is the blueprint for a highway: it will be a bad idea for the highway to have self-intersections, so what do we do? Well, as in the real world, whenever the highway would cross itself, we can build an overpass, lifting one branch up into the third dimension so that it does not cross the other. In topological terms, we add a handle to the plane in which the graph is embedded. Thus we succeed in embedding the graph in a surface, which is in fact the $g$-handled torus (where $g$ is the number of planar crossings) minus a single point (because we are starting with the plane rather than the $2$-sphere).
[This argument actually proves something quantitative: we define the crossing number of a finite graph to be the minimal number of crossings in any $2:1$ planar immersion, and we define the embedding genus of a finite graph to be the minimum genus of a compact surface into which $G$ embeds (thus planar graphs are precisely those of genus zero). Then the above argument shows that the embedding genus of $G$ is less than or equal to the crossing number of $G$.]
For any $n \in \mathbb{Z}^+$, the complete graph $K_n$ has chromatic number $n$. Thus the map coloring numbers of a compact orientable surface of genus $g$ must tend to infinity with $g$. (The precise formula is given in Qiaochu's answer.) In fact, in the same 1968 paper in which they proved Heawood's Conjecture, Ringel and Youngs gave a precise formula for the embedding genus of $K_n$: for $n \geq 3$ it is $\frac{(n-3)(n-4)}{12}$. (Clearly it is $0$ for $1 \leq n \leq 4$.)
1) I say "embedding genus" where most graph theorists say "genus". As an arithmetic geometer who encounters graphs while studying degeneration of algebraic curves, I reserve the term "genus" of a graph for what is classically called the cyclomatic number, i.e., the first Betti number of the corresponding cell complex.
2) I know less about this subject than the above answer might indicate. For instance, I found the computation of the embedding genus of $K_n$ via an internet search while writing this answer. In particular I have not read the paper of Ringel and Youngs. Anyone who wishes to add further information on this subject will be educating me first of all...
-
Apparently, that code for 7-coloring a torus from Wagon's book that I was speaking of in the comments is a bit old; here's an update for newer versions of Mathematica:
spiral[theta0_, theta1_, phi0_, phi1_, color_] :=
ParametricPlot3D[{
(4 + Cos[1/143 (144 phi + 12 theta)]) Cos[1/143 (12 phi + 144 theta)],
(4 + Cos[1/143 (144 phi + 12 theta)]) Sin[1/143 (12 phi + 144 theta)],
Sin[1/143 (144 phi + 12 theta)]},
{theta, theta0, theta1}, {phi, phi0, phi1},
Mesh -> False, PlotStyle -> color]
Show[
spiral[Pi/12, (21*Pi)/12, Pi/3, 2*Pi, White],
spiral[-((3*Pi)/12), -(Pi/12), Pi/2, (13*Pi)/6, Green],
spiral[-(Pi/12), Pi/12, Pi/3, 2*Pi, Brown],
spiral[-(Pi/6), 0, Pi/6, Pi/3, Yellow],
spiral[Pi/6, (23*Pi)/12, Pi/6, Pi/3, Blue],
spiral[0, Pi/6, Pi/6, Pi/3, Purple],
spiral[Pi/12, (11*Pi)/6, 0, Pi/6, Red],
Axes -> None, Boxed -> False, Lighting -> "Neutral",
PlotRange -> All, ViewPoint -> {2, 0.2, 1.2}]
-
yery nice example. thanks – draks ... Nov 18 '11 at 7:02
Thanks for updating my code. I have left this torus example out of the newest edition of my book.
Slightly related: Mathematica has a database (GraphData) of 6000+ graphs. I have just gone through ALL the planar ones in that database and determined the chromatic number (typically 3 or 4). In particular I did all the polyhedra skeletons and their duals, motivated by a (Trubridge) light fixture I purchased. The main algorithm is Kempe's method, described in detail in my book (Mathematica in Action). Using ILP (integer-linear programming) is also possible on modest sized graphs to determine chromatic numbers, or edge chromatic numbers, etc. I don't know if there is anything especially interesting about the polyhedral case: among the 1100 chromatic numbers I found for such "polyhedral graphs", about 100 were 4 and 1100 were 3. For more info on any of this, contact me.
For named polyhedron graphs only the following were 4-chromatic: "IcosahedralGraph", "PentakisDodecahedralGraph", "SmallTriakisOctahedralGraph", "SnubDodecahedralGraph", "TetrahedralGraph", "TriakisIcosahedralGraph", "TriakisTetrahedralGraph".
As for relevance to the original question, I can say that my Kempe implementation is pretty fast for planar graphs, having succeeded on the graph of all 3300 counties in the US. As for the famous counterexamples, I conjecture that one can always get around them with Kempe by simply permuting the vertices and trying again if one reaches an impasse.
Stan Wagon (wagon a t macalester d o t e d u )
- | 1,987 | 7,214 | {"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} | 2.953125 | 3 | CC-MAIN-2016-30 | latest | en | 0.8919 |
https://brilliant.org/problems/an-interesting-sort/ | 1,490,611,534,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189471.55/warc/CC-MAIN-20170322212949-00349-ip-10-233-31-227.ec2.internal.warc.gz | 792,990,544 | 12,750 | # An Interesting Sort
Consider the following algorithm for sorting a list:
1. Shuffle the elements into random order
2. Iterate through the list to check if it is sorted.
3. If it isn't sorted, go back to step 1.
Assume that shuffling and checking the list are both linear operations in terms of the number of elements in the list.
Let O(f(n)) be the average case performance of this algorithm, where n is the number of elements in the list. What is f(4)/f(1) ?
For example: if the average case performance is O($$n^2$$), then $$f(n) = n^2$$, and $$f(4)/f(1) = 16/1 = 16$$.
× | 161 | 581 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-13 | longest | en | 0.862538 |
https://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=23553&nr_roman=(X)(X)MMMDLIII&lang=en | 1,606,836,642,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00653.warc.gz | 793,662,551 | 9,890 | # Convert number: 23,553 in Roman numerals, how to write?
## Latest conversions of Arabic numbers to Roman numerals
501,454 = (D)MCDLIV Dec 01 15:30 UTC (GMT) 23,553 = (X)(X)MMMDLIII Dec 01 15:30 UTC (GMT) 1,141,974 = (M)(C)(X)(L)MCMLXXIV Dec 01 15:30 UTC (GMT) 200,779 = (C)(C)DCCLXXIX Dec 01 15:30 UTC (GMT) 134,826 = (C)(X)(X)(X)M(V)DCCCXXVI Dec 01 15:30 UTC (GMT) 2,158,737 = (M)(M)(C)(L)(V)MMMDCCXXXVII Dec 01 15:30 UTC (GMT) 195,547 = (C)(X)(C)(V)DXLVII Dec 01 15:30 UTC (GMT) 9 = IX Dec 01 15:30 UTC (GMT) 841,235 = (D)(C)(C)(C)(X)(L)MCCXXXV Dec 01 15:30 UTC (GMT) 794,998 = (D)(C)(C)(X)(C)M(V)CMXCVIII Dec 01 15:30 UTC (GMT) 97 = XCVII Dec 01 15:30 UTC (GMT) 100,414 = (C)CDXIV Dec 01 15:30 UTC (GMT) 1,003,121 = (M)MMMCXXI Dec 01 15:30 UTC (GMT) converted numbers, see more...
## The set of basic symbols of the Roman system of writing numerals
• ### (*) M = 1,000,000 or |M| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.
(*) These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: "(" and ")", because:
• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and
• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line "|" and the Roman numeral "I" (1).
(*) An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate "1,000 times". See below...
Logic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.
(*) At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.
Thus, initially, the largest number that could be written using Roman numerals was:
• MMMCMXCIX = 3,999. | 726 | 2,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.898491 |
http://oeis.org/A281581 | 1,547,873,460,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662124.0/warc/CC-MAIN-20190119034320-20190119060320-00060.warc.gz | 160,376,573 | 4,138 | This site is supported by donations to The OEIS Foundation.
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A281581 a(n) = (15*2^(2*n+2) + 15*2^(n+2) + 5*2^(n+3)*3^(n+1) - 24*5^(n+1))/120. 2
1, 4, 21, 127, 807, 5179, 33111, 210067, 1321887, 8255899, 51225351, 316067107, 1941032367, 11873549419, 72394874391, 440204293747, 2670669533247, 16172309991739, 97779619272231, 590423692897987, 3561340764760527, 21462312506478859 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Seiichi Manyama, Table of n, a(n) for n = 0..1285 Index entries for linear recurrences with constant coefficients, signature (17,-104,268,-240). FORMULA G.f.: ( 1-13*x+57*x^2-82*x^3 ) / ( (6*x-1)*(4*x-1)*(2*x-1)*(5*x-1) ). - R. J. Mathar, Mar 19 2017 a(n) = 6^n +2^(n-1)-5^n+4^n/2 . - R. J. Mathar, Mar 19 2017 MATHEMATICA Table[(15*2^(2*n+2) + 15*2^(n+2) + 5*2^(n+3)*3^(n+1) - 24*5^(n+1)) / 120, {n, 0, 21] (* Indranil Ghosh, Mar 05 2017 *) PROG (PARI) a(n) = (15*2^(2*n+2) + 15*2^(n+2) + 5*2^(n+3)*3^(n+1) - 24*5^(n+1)) / 120; for (n=0, 21, print1(a(n), ", ")); \\ Indranil Ghosh, Mar 05 2017 (Python) def A281581(n): return (15*2**(2*n+2) + 15*2**(n+2) + 5*2**(n+3)*3**(n+1) - 24*5**(n+1)) / 120 # Indranil Ghosh, Mar 05 2017 (Ruby) def A281581(n) (0..n).map{|i| (15 * 2 ** (2 * i + 2) + 15 * 2 ** (i + 2) + 5 * 2 ** (i + 3) * 3 ** (i + 1) - 24 * 5 ** (i + 1)) / 120} end CROSSREFS Row n=5 of A283272. Sequence in context: A211249 A185047 A032326 * A007345 A255673 A099250 Adjacent sequences: A281578 A281579 A281580 * A281582 A281583 A281584 KEYWORD nonn,easy AUTHOR Seiichi Manyama, Mar 05 2017 STATUS approved
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Last modified January 18 23:05 EST 2019. Contains 319282 sequences. (Running on oeis4.) | 889 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-04 | latest | en | 0.500546 |
http://math.stackexchange.com/questions/119481/trigonometric-identity-possible-error | 1,467,107,105,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396872.10/warc/CC-MAIN-20160624154956-00081-ip-10-164-35-72.ec2.internal.warc.gz | 196,473,242 | 18,644 | # Trigonometric identity, possible error
I need to prove the following trigonometric identity: $$\frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2}) =\sin^2(\alpha)$$
I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it be a textbook error?
-
Have you tried using $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$? – Alex Becker Mar 12 '12 at 23:14
I should only use the properties of the trigonometric functions(even, odd, periodic). Sum and difference identities are not allowed. – Adam Mar 12 '12 at 23:18
I don't think what you have is correct. $\alpha = 0$ gives the lhs $\neq$ rhs – user17762 Mar 12 '12 at 23:20
If you can't use sum/difference identities, I don't think you'll be able to do anything with that term with the $5\pi/6$. – Gerry Myerson Jul 2 '12 at 5:24
Since all the trig values are squared, it seems as though the exercise is simply playing with shifts by odd or even multiples of $\pi/2$.
Loosely,
• Shifting by "$\frac{\pi}{2} \cdot \text{odd}$" switches "sin" and "cos" (and possibly affects the sign)
• Shifting by "$\frac{\pi}{2} \cdot \text{even}$" ($=$ "$\pi \cdot \text{any}$") preserves "sin" and "cos" (and possibly affects sign)
• Negating the argument preserves "sin" and "cos" (and possibly affects sign)
Since squaring eliminates sign considerations, we have, simply:
\begin{align} \mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{odd} \right) &= \mathrm{cotrig}^2\theta \\ \mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{even} \right) &= \mathrm{trig}^2\left( \pm \; \theta \pm \pi \cdot \text{any} \right) = \mathrm{trig}^2\theta \end{align}
where each "$\pm$" is independent, "any" means (of course) "any integer", and "trig" can in fact be any of the six trig functions.
This makes pretty quick work of the simplification process ... \begin{align} \frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\left(\alpha-4\pi\right)} - \cot^2\left(\alpha-11\pi\right) \; \sin^2\left(-\alpha-\frac{13\pi}{2}\right) &\stackrel{?}{=} \sin^2\alpha \\[1em] \frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha &\stackrel{?}{=} \sin^2\alpha \end{align} ... right up to the point at which the process shudders to a halt.
Given the nature of all the other terms (and @Adam's comment that sum and difference identities are not allowed), I suspect that "$\frac{5\pi}{6}$" is a typo of "$\frac{5\pi}{2}$", which would get us a little further ...
$$\frac{\cos^2\alpha}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha = 1 - \cot^2\alpha \;\cos^2\alpha \stackrel{?}{=} \sin^2\alpha$$
... but we hit another snag. Could it be that "$\sin^2\left(-\alpha-\frac{13\pi}{2}\right)$" is a typo of "$\cos^2(...)$"? If so, then that factor should've simplified to "$\sin^2\alpha$", and we'd have
$$1 - \cot^2\alpha \;\sin^2\alpha = 1 - \cos^2\alpha = \sin^2\alpha$$
as desired.
(It's also possible that, instead of a sin-cos typo, "$\cot$" is a typo for "$\tan$", but it seems like that would be an easier one for the OP to notice.)
-
Some important translations:
$$\tag 1\sin(x\pm 2 \pi) = \sin x$$ $$\tag {1'}\cos(x\pm 2 \pi) = \cos x$$ $$\tag 2\cot(x\pm \pi)= \cot x$$ $$\tag {2'}\tan(x\pm \pi)= \tan x$$
$$\tag 3 \sin \left(\frac \pi 2 -x \right)=\cos x$$ $$\tag 4 \cos \left(\frac \pi 2 -x \right)=\sin x$$ $$\tag 5 \sin(\pi-x)=\sin x$$ and $$\tag 6\cos(\pi-x)=-\cos x$$
and $$\tag 7 \sin(-x)=-\sin x$$ $$\tag 8 \cos (-x) = \cos x$$
Although taking $\alpha =0$ reveals that the equality doesn't hold, assume that there is no typo, then, we could move on as follows.
You have that
$$\frac{\sin^2 \left(\frac{5\pi}{6} - \alpha \right)}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2 \left(-\alpha - \frac{13\pi}{2}\right) =\sin^2(\alpha)$$
Using the above, we can write
\eqalign{ & {\sin ^2}\left( {\frac{{5\pi }}{6} - \alpha } \right) = {\left[ { - \sin \left( {\alpha - \frac{{5\pi }}{6}} \right)} \right]^2} = {\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) \cr & {\cot ^2}(\alpha - 11\pi ) = {\cot ^2}\left( {\alpha - 10\pi } \right) = \cdots = {\cot ^2}\alpha \cr & {\sin ^2}\left( { - \alpha - \frac{{13\pi }}{2}} \right) = {\left[ { - \sin \left( {\alpha + \frac{{13\pi }}{2}} \right)} \right]^2} = \sin {\left( {\alpha + \frac{{13\pi }}{2}} \right)^2} \cr & {\cos ^2}(\alpha - 4\pi ) = {\cos ^2}(\alpha - 2\pi ) = {\cos ^2}\alpha \cr}
so that we have
$$\frac{{{{\sin }^2}\left( {\alpha - \frac{{5\pi }}{6}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\alpha$$
Now
$${\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{{3\pi }}{6} - \frac{{2\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{\pi }{3} - \frac{\pi }{2}} \right) = {\left( { - 1} \right)^2}{\sin ^2}\left( {\frac{\pi }{2} - \left( {\alpha - \frac{\pi }{3}} \right)} \right) = {\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right)$$
and
\eqalign{ & {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\left( {\alpha + \frac{{12\pi }}{2} + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + 6\pi + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + \frac{\pi }{2}} \right) \cr & = {\sin ^2}\left( {\frac{\pi }{2} - \left( { - \alpha } \right)} \right) = {\cos ^2}\left( { - \alpha } \right) = {\cos ^2}\alpha \cr}
so that you have
$$\frac{{{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\cos ^2}\alpha = {\sin ^2}\alpha$$
Now, solving for $${{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}$$
gives
$${\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right) = {\cos ^2}\alpha {\sin ^2}\alpha + {\cos ^6}\alpha \frac{1}{{{{\sin }^2}\alpha }}$$
There is some typo in your excercise, since letting $\alpha =0$ gives $1/4$ on the LHS and is not defined for the RHS. When you discover what the typo is, move on with the listed translations.
- | 2,276 | 5,987 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2016-26 | latest | en | 0.766497 |
http://what-when-how.com/Tutorial/topic-1114galv/Database-Design-and-Relational-Theory-165.html | 1,670,397,324,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711150.61/warc/CC-MAIN-20221207053157-20221207083157-00622.warc.gz | 55,371,958 | 5,282 | Databases Reference
In-Depth Information
If (a) supplier s1 supplies part p1 to project j2 and (b) supplier s2 supplies part p1 to project j1 ( s1 s2 , j1
j2 ), then (c) no part, not even p1 , can be supplied by supplier s1 to project j1 .
If (a) supplier s1 supplies part p2 to project j1 and (b) supplier s2 supplies part p1 to project j1 ( s1 s2 , p1
p2 ), then (c) no project, not even j1 , can be supplied by supplier s1 with part p1 .
In fact, these three business rules can all be combined into one, as follows. Let's agree, just for the moment,
to say each tuple of relvar SPJ′ represents a shipment (by some supplier of some part to some project). Then there
cannot exist three distinct shipments x , y , and z such that x and y involve the same supplier, y and z involve the same
part, and z and x involve the same project.
There's still another point to be made in connection with the SPJ′ example. Refer again to the analysis that
led to the first of the foregoing three business rules. That analysis showed that tuple t3 can't appear together with
tuples t1 and t2 . It follows, therefore, that SPJ′ suffers from an insertion anomaly, despite the fact that it's in RFNF
(and the fact that no tuple forcing JD holds, therefore). By contrast, it doesn't suffer from a deletion
anomaly─assuming, that is, that the only constraints to which it's subject are the stated FD and JD (and logical
consequences thereof). So one difference between 5NF and RFNF is this: Even though both are redundancy free,
5NF guarantees “no insertion anomalies” while RFNF doesn't─assuming, again, that FDs and JDs are the only
constraints under consideration.
Of course, it's tempting to conclude from the SPJ′ example that relvars that are in RFNF and not 5NF are
likely to be rare in practice. Nevertheless, there's a clear logical difference between the two normal forms, and thus,
from the point of view of reducing redundancy at least, it's really RFNF and not 5NF that ought to be the target to
be aimed for. (As a bonus, I note that RFNF is a little easier to test for, too, than 5NF is.)
Note: As a matter of fact, the SPJ′ example reinforces the foregoing observation in another way also. Since
the relvar is subject to the JD {{SNO,PNO},{PNO,JNO},{JNO,SNO}}, it can be nonloss decomposed into its
projections on {SNO,PNO}, {PNO,JNO}, and {JNO,SNO}, respectively. Those projections are each “all key,” and
in fact in 5NF. However, that decomposition “loses” the FD {SNO,PNO} → {JNO}! As we saw in Chapter 6,
losing dependencies is generally not recommended. Hence relvar SPJ′ illustrates the point that not only is 5NF
sometimes too strong, but sometimes it might be positively contraindicated.
DOMAIN-KEY NORMAL FORM
Domain-key normal form (DK/NF) differs from all of the normal forms discussed in this topic so far in that it's not
defined in terms of FDs, MVDs, and JDs, as such, at all. 15 DK/NF is really a kind of “ideal” normal form: It's
desirable because, by definition, a relvar in DK/NF is guaranteed to be free of certain update anomalies; sadly,
however, it's not always achievable, nor has the question “Exactly when can it be achieved?” been answered. Be
that as it may, let's investigate.
DK/NF is defined in terms of domain constraints and key constraints . Key constraints are already familiar,
of course (they were defined in Chapter 5). As for domain constraints, I remind you that domain is essentially just
another word for type (see the answer to Exercise 2.4 in Appendix D). It follows that a domain constraint ought
logically to be the same thing as a type constraint; in other words, it ought simply to be a specification of the set of
values that constitute the type in question (see SQL and Relational Theory for further explanation of this concept).
However, the term is being used in the present context in a slightly special sense. To be specific, a domain
constraint , as that term is used here, is a constraint to the effect that values of a given attribute are taken from some
15 Well … it's defined in terms of key constraints, as we'll see, and key constraints in turn are a special case of FDs, so this remark is perhaps a
little economical with the truth.
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## Iteration
Here we have our two intervals on our single, purple line ‘permanently’ end-to-end linked by conscious construction. They are “daisy-chained”. And we at once find that we have a new perspectivity, centered on M, making the interval PS equivalent to the black interval and, thereby, identical to the red and blue intervals, too!
We see that we can join N to S, and get a new equivalent black interval, and from P form a new perspectivity on it, to get a new, green interval. We see we can repeat this process, indefinitely, and develop a set of intervals in one line, all identical with a single interval (PS), and so get our projective ruler.
There are exactly two places on the purple line, called “invariants”, beyond which this, “back-and-forth, up-and-down”, process of interval replication cannot go. In fact, they “end” the entire ruler. Can you find them?
You can explore all this ruler's manifestations
here. | 264 | 1,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-43 | latest | en | 0.861547 |
https://it.mathworks.com/matlabcentral/answers/1599789-how-to-display-the-3d-graph-for-this-problem?s_tid=srchtitle | 1,642,619,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00509.warc.gz | 371,713,723 | 30,866 | # how to display the 3d graph for this problem?
5 views (last 30 days)
i was able to plot the graph in 2d but in 3d i am unable to plot the graph around the red line
(use this code and execute and see)
clc
clear all
syms x
f = tan(pi*x/4)
fL = [0 1]
yr = 0
iL = [0 1]
finverse(f)
Volume = pi*int((finverse(f))^2,iL(1),iL(2));
disp(['Volume is: ', num2str(double(Volume))])
fx = inline(vectorize(f));
xvals = linspace(fL(1),fL(2),201);
xvalsr = fliplr(xvals);
xivals = linspace(iL(1),iL(2),201);
xivalsr = fliplr(xivals);
xlim = [fL(1)-0.5, fL(2)+0.5];
ylim = fx(xlim);
figure(1)
subplot(2,1,1)
hold on
plot(xvals, fx(xvals),'-b','LineWidth',2)
plot([yr yr],[fL(1) fL(2)],'-r','LineWidth',2)
fill([xvals, xvalsr],[fx(xvals), ones(size(xvalsr))*yr],[0.8 0.8 0.8],'FaceAlpha',0.8)
legend('Function Plot','Axis of Rotation','Filled Region')
title('Function y=f(x) and Region')
xlabel('x-axis')
ylabel('y-axis')
subplot(2,1,2)
hold on
plot(xivals,fx(xivals),'-b','LineWidth',2)
plot(-xivals,fx(xivals),'-m','LineWidth',2)
plot([iL(1) iL(2)],[yr yr],'-r','LineWidth',2)
fill([xivals, xivalsr],[fx(xivals), ones(size(xivalsr))*yr], [0.8 0.8 0.8])
fill([-xivals, -xivalsr],[ones(size(xivals))*yr, fx(xivalsr)], [1 0.8 0.8])
title('Rotated Region in xy-Plane')
xlabel('x-axis')
ylabel('y-axis')
figure(2)
[X,Y,Z] = cylinder(xivals-yr,100);
hold on
Z = iL(1)+ Z.*(iL(2)-iL(1));
surf(Z,Y,X+yr,'EdgeColor','none','FaceColor','flat','FaceAlpha',0.6)
plot([yr yr],[iL(1) iL(2)],'-r','LineWidth',2)
title('Volume generated by revolving a curve about the y-axis')
xlabel('x-axis')
ylabel('y-axis')
zlabel('Z-axis')
view(-22,32)
the code needs correction after the figure(2) line only or maybe before too i may be wrong
Star Strider on 30 Nov 2021
Try it with these changes —
hs = surf(X,Y,Z,'EdgeColor','none','FaceColor','flat','FaceAlpha',0.6);
hs.ZData = hs.ZData-0.5;
hs.YData = hs.YData+0.5;
rotate(hs, [1 0 0], 90)
I already made them in the code. This just nudges the cone to where it should be, and then rotates it.
syms x
f(x) = tan(pi*x/4)
f(x) =
fL = [0 1]
fL = 1×2
0 1
yr = 0
yr = 0
iL = [0 1]
iL = 1×2
0 1
finverse(f)
ans(x) =
Volume = pi*int((finverse(f))^2,iL(1),iL(2));
disp(['Volume is: ', num2str(double(Volume))])
Volume is: 1.2492
fx(x) = f(x); % Inline Is Not Good Programming Practise
xvals = linspace(fL(1),fL(2),201);
xvalsr = fliplr(xvals);
xivals = linspace(iL(1),iL(2),201);
xivalsr = fliplr(xivals);
xlim = [fL(1)-0.5, fL(2)+0.5];
ylim = fx(xlim);
figure(1)
subplot(2,1,1)
hold on
plot(xvals, fx(xvals),'-b','LineWidth',2)
plot([yr yr],[fL(1) fL(2)],'-r','LineWidth',2)
fill([xvals, xvalsr],[fx(xvals), ones(size(xvalsr))*yr],[0.8 0.8 0.8],'FaceAlpha',0.8)
legend('Function Plot','Axis of Rotation','Filled Region')
title('Function y=f(x) and Region')
xlabel('x-axis')
ylabel('y-axis')
subplot(2,1,2)
hold on
plot(xivals,fx(xivals),'-b','LineWidth',2)
plot(-xivals,fx(xivals),'-m','LineWidth',2)
plot([iL(1) iL(2)],[yr yr],'-r','LineWidth',2)
fill([xivals, xivalsr],[fx(xivals), ones(size(xivalsr))*yr], [0.8 0.8 0.8])
fill([-xivals, -xivalsr],[ones(size(xivals))*yr, fx(xivalsr)], [1 0.8 0.8])
title('Rotated Region in xy-Plane')
xlabel('x-axis')
ylabel('y-axis')
figure(2)
[X,Y,Z] = cylinder(xivals-yr,100);
hold on
Z = iL(1)+ Z.*(iL(2)-iL(1));
hs = surf(X,Y,Z,'EdgeColor','none','FaceColor','flat','FaceAlpha',0.6);
hs.ZData = hs.ZData-0.5;
hs.YData = hs.YData+0.5;
rotate(hs, [1 0 0], 90)
plot([yr yr],[iL(1) iL(2)],'-r','LineWidth',2)
title('Volume generated by revolving a curve about the y-axis')
xlabel('x-axis')
ylabel('y-axis')
zlabel('Z-axis')
view(-22,32)
% view(80,30)
grid on
.
Star Strider on 1 Dec 2021
As always, my pleasure!
No worries, and my time was definitely not wasted in responding to the posted problem. I simply had a great deal of trouble understanding what the desired result is (and I am still not certain that I have).
.
can you just rotate the function y=tan(pi*x/4) about y-axis or x=0 and plot in 3D.?
imshow(I)
see that graph and pls help me :). | 1,505 | 3,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.665771 |
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# Complete linear algebra: theory and implementation in code
Complete linear algebra: theory and implementation in code -
Learn concepts in linear algebra and matrix analysis, and implement them in MATLAB and Python.
Preview this course
What you'll learn
• Understand theoretical concepts in linear algebra, including proofs
• Implement linear algebra concepts in scientific programming languages (MATLAB, Python)
• Apply linear algebra concepts to real datasets
• Ace your linear algebra exam!
• Apply linear algebra on computers with confidence
• Gain additional insights into solving problems in linear algebra, including homeworks and applications
• Be confident in learning advanced linear algebra topics
• Understand some of the important maths underlying machine learning
• The math underlying most of AI (artificial intelligence)
Requirements
• Basic understanding of high-school algebra (e.g., solve for x in 2x=5)
• Interest in learning about matrices and vectors!
• (optional) Computer with MATLAB, Octave, or Python (or Jupyter)
Description
You need to learn linear algebra!
Linear algebra is perhaps the most important branch of mathematics for computational sciences, including machine learning, AI, data science, statistics, simulations, computer graphics, multivariate analyses, matrix decompositions, signal processing, and so on.
You need to know applied linear algebra, not just abstract linear algebra!
The way linear algebra is presented in 30-year-old textbooks is different from how professionals use linear algebra in computers to solve real-world applications in machine learning, data science, statistics, and signal processing. For example, the "determinant" of a matrix is important for linear algebra theory, but should you actually use the determinant in practical applications? The answer may surprise you, and it's in this course!
If you are interested in learning the mathematical concepts linear algebra and matrix analysis, but also want to apply those concepts to data analyses on computers (e.g., statistics or signal processing), then this course is for you! You'll see all the maths concepts implemented in MATLAB and in Python.
Unique aspects of this course
Clear and comprehensible explanations of concepts and theories in linear algebra.
Several distinct explanations of the same ideas, which is a proven technique for learning.
Visualization using graphs, numbers, and spaces that strengthens the geometric intuition of linear algebra.
Implementations in MATLAB and Python. Com'on, in the real world, you never solve math problems by hand! You need to know how to implement math in software!
Beginning to intermediate topics, including vectors, matrix multiplications, least-squares projections, eigendecomposition, and singular-value decomposition.
Strong focus on modern applications-oriented aspects of linear algebra and matrix analysis.
Intuitive visual explanations of diagonalization, eigenvalues and eigenvectors, and singular value decomposition.
Improve your coding skills! You do need to have a little bit of coding experience for this course (I do not teach elementary Python or MATLAB), but you will definitely improve your scientific and data analysis programming skills in this course. Everything is explained in MATLAB and in Python (mostly using numpy and matplotlib; also sympy and scipy and some other relevant toolboxes).
Benefits of learning linear algebra
Understand statistics including least-squares, regression, and multivariate analyses.
Improve mathematical simulations in engineering, computational biology, finance, and physics.
Understand data compression and dimension-reduction (PCA, SVD, eigendecomposition).
Understand the math underlying machine learning and linear classification algorithms.
Deeper knowledge of signal processing methods, particularly filtering and multivariate subspace methods.
Explore the link between linear algebra, matrices, and geometry.
Gain more experience implementing math and understanding machine-learning concepts in Python and MATLAB.
Linear algebra is a prerequisite of machine learning and artificial intelligence (A.I.).
Why I am qualified to teach this course:
I have been using linear algebra extensively in my research and teaching (in MATLAB and Python) for many years. I have written several textbooks about data analysis, programming, and statistics, that rely extensively on concepts in linear algebra.
So what are you waiting for??
Watch the course introductory video and free sample videos to learn more about the contents of this course and about my teaching style. If you are unsure if this course is right for you and want to learn more, feel free to contact with me questions before you sign up.
I hope to see you soon in the course!
Mike
Who this course is for:
• Anyone interested in learning about matrices and vectors
• Students who want supplemental instruction/practice for a linear algebra course
• Engineers who want to refresh their knowledge of matrices and decompositions | 939 | 5,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-06 | latest | en | 0.88165 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=142&t=44411&p=153615 | 1,582,825,211,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146744.74/warc/CC-MAIN-20200227160355-20200227190355-00426.warc.gz | 427,502,654 | 10,966 | ## Nernst equation
$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$
Ashley McClearnen 1B
Posts: 59
Joined: Fri Sep 28, 2018 12:26 am
### Nernst equation
Can the equations E=E0-(RT/nF)lnQ and E=E0-(0.0592/n)lnQ be used interchangeably?
Chase Yonamine 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am
### Re: Nernst equation
I believe you meant E=E0-.05916/n logQ. We get 0.05916 from the gas constant (R) x room temperature in kelvin (K) x 2.303logx (which is the conversion from natural log to log base 10) all divided by Faraday's constant. So this equation is the same as E=E0-RT/nf lnQ only when temperature is at room temperature, and you will be using log instead of ln. You can find ph easier with the E=E0-.05916/n logQ equation.
Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”
### Who is online
Users browsing this forum: Joowon Seo 3A and 2 guests | 312 | 963 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-10 | latest | en | 0.795596 |
sportscentertltc.com | 1,685,342,734,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644683.18/warc/CC-MAIN-20230529042138-20230529072138-00714.warc.gz | 607,224,930 | 11,255 | # Methods for raking in the cash from slot machines
The simple answer to this issue is that you have to risk a lot to win a lot. But if you want to make a lot of money with the greatest possible chances, you’ll need to maximise the volatility of your approach.
It would seem that the best way to win large is to lose as much money as possible as rapidly as possible. Not sure how to make sense of it. We need to elaborate.
From a statistical perspective, you may expect to lose money on any bet you make in the long term. To put it another way, the RTP works like this. In order to lessen your overall loss, you should cut down on the amount of money you bet. If you want to minimise the amount of money you lose gambling, you should try to deplete your bankroll with as few bets as possible. If you do this and the RTP remains the same, statistical likelihood will reward you with a higher payout if you win สล็อตค่ายนอกเว็บตรง
Following is a comprehensive breakdown of the Smart Gambler strategy:
• Find a slot machine that lets you gamble your profits.
• Select a play time limit that suits your needs. You may roughly estimate the total number of rounds by dividing this time by the average duration of a round.
• Verify your budget and divide it by the total number of rounds you established in the previous stage. If you have a budget of \$100, then \$11 is the result of dividing that amount by 900. The minimum wager for each round is \$0.10.
• Choose the amount of money that you’d be happy to walk away with. Make a note of this number or commit it to memory.
Put in the calculated initial bet of \$0.10 and get the game going.
If you are confident in your winning hand, you may hit the gamble button. You will double your bet until you have amassed a sufficient fortune. Make off with the cash and have a party on me.
You’ve been knocked out of the running.
In this updated version of the game, doubling down is no longer an option
The number of consecutive bets in certain games is limited. If you are disappointed with your win and want to keep gambling, you will have to find another way to do it. The many options are as follows:
Bet the whole of your profit on a single number in roulette. Decide on a kind of wager that, should you win, will net you the sum you have established as your goal.
Invest the whole profit on a risky bet on the following turn. In the event of victory, refer back to item #6. (repeat till you reach your goal)
Have you ever visited a site that promised a secret slot strategy that would result in a substantial windfall if you tried it? Nothing even substantially comparable can be found in this article. The long game usually favours Slots. | 591 | 2,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.947057 |
https://multi-converter.com/revolution-per-second-to-revolution-per-minute | 1,679,526,061,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.97/warc/CC-MAIN-20230322211955-20230323001955-00571.warc.gz | 429,876,113 | 6,809 | # Revolution per Second to Revolution per Minute
Convert RPS to RPM
Change to Revolution per Minute to Revolution per Second
Share:
## How to convert Revolution per Second to Revolution per Minute
1 [Revolution per Second] = 60 [Revolution per Minute]
[Revolution per Minute] = [Revolution per Second] * 60
To convert Revolution per Second to Revolution per Minute multiply Revolution per Second * 60.
## Example
96 Revolution per Second to Revolution per Minute
96 [RPS] * 60 = 5760 [RPM]
## Conversion table
Revolution per Second Revolution per Minute
0.01 RPS0.6 RPM
0.1 RPS6 RPM
1 RPS60 RPM
2 RPS120 RPM
3 RPS180 RPM
4 RPS240 RPM
5 RPS300 RPM
10 RPS600 RPM
15 RPS900 RPM
50 RPS3000 RPM
100 RPS6000 RPM
500 RPS30000 RPM
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https://www.bartleby.com/questions-and-answers/an-archer-pulls-her-bowstring-back0.390m-by-exerting-a-force-that-increases-uniformly-from-zero-to21/09dbcde0-7063-4fc3-b228-a03b1ed9f033 | 1,580,143,096,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700988.64/warc/CC-MAIN-20200127143516-20200127173516-00132.warc.gz | 773,562,072 | 23,369 | # An archer pulls her bowstring back 0.390 m by exerting a force that increases uniformly from zero to 217 N.(a) What is the equivalent spring constant of the bow?(b) How much work does the archer do in pulling the bow?
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An archer pulls her bowstring back 0.390 m by exerting a force that increases uniformly from zero to 217 N.
(a) What is the equivalent spring constant of the bow?
(b) How much work does the archer do in pulling the bow?
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Step 1
(a) Spring constant is,
Step 2
(b) Work done is equal to the elast...
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Tagged in | 204 | 832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-05 | latest | en | 0.914972 |
https://www.geeksforgeeks.org/python-numpy-np-multinomial-method/ | 1,643,275,702,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00635.warc.gz | 821,061,489 | 21,613 | # Python | Numpy np.multinomial() method
• Last Updated : 13 Oct, 2019
With the help of `np.multinomial()` method, we can get the array of multinomial distribution by using `np.multinomial()` method.
Syntax : `np.multinomial(n, nval, size)`
Return : Return the array of multinomial distribution.
Example #1 :
In this example we can see that by using `np.multinomial()` method, we are able to get the multinomial distribution array using this method.
`# import numpy``import` `numpy as np`` ` `# using np.multinomial() method``gfg ``=` `np.random.multinomial(``8``, [``0.1``, ``0.22``, ``0.333``, ``0.4444``], ``2``)`` ` `print``(gfg)`
Output :
[[1 4 2 1]
[0 0 3 5]]
Example #2 :
`# import numpy``import` `numpy as np`` ` `# using np.multinomial() method``gfg ``=` `np.random.multinomial(``12``, [``0.1``, ``0.12``, ``0.123``, ``0.1234``, ``0.12345``], ``3``)`` ` `print``(gfg)`
Output :
[[2 0 1 1 8]
[0 1 1 1 9]
[1 1 1 0 9]]
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My Personal Notes arrow_drop_up | 397 | 1,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-05 | latest | en | 0.607459 |
https://rdrr.io/bioc/peco/man/rotation.html | 1,669,475,857,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00511.warc.gz | 541,960,473 | 8,283 | # rotation: Rotate circular variable shift_var to minimize distance... In peco: A Supervised Approach for **P**r**e**dicting **c**ell Cycle Pr**o**gression using scRNA-seq data
## Description
Because the origin of the cell cycle phases is arbitrary, we transform the angles prior to computing the distance (rotation and shifting) to minimize the distance between two vectors. After this, one can apply circ_dist to compute the distance between the output value and ref_var.
## Usage
`1` ```rotation(ref_var, shift_var) ```
## Arguments
`ref_var` A vector of reference angles. `shift_var` A vector of angles to be compared to ref_var.
## Value
The transformed values of shift_var after rotation and shifting.
Matthew Stephens
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13``` ```# a vector of angles theta_ref <- seq(0,2*pi, length.out=100) # shift the origin of theta_ref to pi theta_compare <- shift_origin(theta_ref, origin = pi) # rotate theta_compare in a such a way that the distance # between theta_ref and thet_compare is minimized theta_compare_rotated <- rotation(ref_var=theta_ref, shift_var=theta_compare) par(mofrow=c(1,2)) plot(x=theta_ref, y = theta_compare) plot(x=theta_ref, y = theta_compare_rotated) ```
peco documentation built on Nov. 8, 2020, 8:16 p.m. | 335 | 1,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.653417 |
http://openstudy.com/updates/512c12e5e4b02acc415d9847 | 1,444,247,735,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737882743.52/warc/CC-MAIN-20151001221802-00033-ip-10-137-6-227.ec2.internal.warc.gz | 243,398,138 | 10,477 | pooja195 2 years ago For the given angle measure, find the measure of a supplementary angle and the measure of a complementary angle, if possible. 4. 30° 5. 170°
1. pooja195
@jim_thompson5910
2. jim_thompson5910
4 supplementary: 180 - 30 = ??? complementary: 90 - 30 = ??? do the same for # 5
3. pooja195
150 and 60 5.10 and 80 right??
4. jim_thompson5910
oh i didn't see 170 for #5 there's no possible complementary angle to 170 degrees since 170 is larger than 90
5. jim_thompson5910
but the first three answers are correct
6. jim_thompson5910
the last answer would simply be "not possible"
7. jim_thompson5910
or "NA" or something like that
8. pooja195
so 10 is right
9. jim_thompson5910
yes 10 is the supplement of 170
10. pooja195
oh i see thnx | 243 | 769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2015-40 | longest | en | 0.844619 |
https://math.stackexchange.com/questions/4492339/curvature-on-an-associated-vector-bundle | 1,708,545,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00469.warc.gz | 401,264,985 | 35,754 | # Curvature on an Associated Vector Bundle
I am in the midst of trying to solve problem 13 from chapter 5 of Hamilton's Mathematical Gauge theory text, and have come across some terms I am not sure what to do with. I will elaborate below.
Let $$P\rightarrow M$$ be a principal $$G$$ bundle, and $$E=P\times_\rho V\rightarrow M$$ an associated vector bundle. Fix a connection $$A$$ on a $$P$$, the induces the following covariant derivative on $$E$$, letting $$\Phi$$ be a local section of $$E$$:
$$\nabla^A_X\Phi=[s,d\phi(X_x)+\rho_*(A_s(X_x))\phi]$$ where $$s$$ is a local section of $$P$$, $$\phi$$ is a map $$U\subset M\rightarrow V$$, $$X\in \mathfrak{X}(M)$$, and where $$A_s=s^*A$$, can be thought of as a Lie algebra valued one form on the base $$M$$.
In a principal bundle we define curvature as:
$$F=dA+\frac{1}{2}[A,A]$$
The curvature of a covariant derivative in an arbitrary vector bundle is:
$$F^\nabla(X,Y)\Phi=\nabla_X\nabla_Y\Phi-\nabla_Y\nabla_X\Phi-\nabla_{[X,Y]}\Phi$$
I am trying to show that in an associated vector bundle:
$$F^\nabla(X,Y)\Phi=[s,\rho_*(F_s(X,Y))\phi]$$
where $$F_s=s^*F$$. So I calculated the following:
$$\nabla^A_X\nabla^A_Y\Phi=[s,d\left(\rho_*(A_s(X_x)\right)(Y_x)\phi+\rho_*(A_s(X_x))d\phi(Y_x)+\rho_*(A_s(X_x))d\phi(Y_x)+\rho_*(A_s(X_x)A_s(Y_x))]$$ $$\nabla^A_X\nabla^A_Y\Phi=[s,d\left(\rho_*(A_s(Y_x)\right)(X_x)\phi+\rho_*(A_s(Y_x))d\phi(X_x)+\rho_*(A_s(Y_x))d\phi(X_x)+\rho_*(A_s(Y_x)A_s(X_x))]$$ $$\nabla^A_{[X,Y]}\Phi=[s,d\phi([X,Y]_x)+\rho_*(A_s([X,Y]_x)\phi]$$
Putting it all together I obtain that:
$$F^\nabla(X,Y)\Phi=[s,d\left(\rho_*(A_s(Y_x)\right)(X_x)\phi-d\left(\rho_*(A_s(X_x)\right)(Y_x)\phi+\rho_*(A_s(X_x)A_s(Y_x)-A_s(Y_x)A_s(X_x))\phi-d\phi([X,Y]_x)-\rho_*(A_s([X,Y]_x)\phi]$$
I feel like I see all the pieces, but am unsure of how to move them around. Like the third term I'm pretty sure simplifies to $$\frac{1}{2}[A_s,A_s](X_x,Y_x)$$, but I don't know what to do with the exterior derivative of $$\rho_*$$, or what to do with the Lie bracket terms, especially since it seems like I have all the pieces without those terms, so I feel like they should cancel out some how but that doesn't make a ton of sense to me. Any help would be greatly appreciated.
Edit: I suspect I'm messing up something up with the exterior derivative, perhaps $$d(d\phi(X_x)(Y_x)$$ isn't zero? Because you technically contract then apply the exterior derivative again? I am unsure...
• I find this all easier to understand if I start by going backwards. Namely, by starting with a connection on a vector bundle and figuring out how it induces a connection on the appropriate frame bundle. Once you figure that out, you can define what a connection on a frame bundle is and show how it induces a connection on the vector bundle. Finally, that allows you to figure out what a connection on a principal bundle is and how it induces a connection on the vector bundle. Jul 14, 2022 at 12:43
• @Deane I feel like I’m not having issue with the connection, just the actual calculation of curvature using the definition above. I think I may have figured it out using a lie derivative property I had forgotten, will update soon. Jul 14, 2022 at 13:15
First examine the term $$\nabla^A_Y\Phi$$:
$$\nabla^A_Y\Phi=[s,\nabla^A_Y\phi]=[s,d\phi(Y)+\rho_*(A_s(Y))\phi]$$
Note that $$d\phi$$ and $$A_s$$ are vector ($$d\phi$$ takes values in $$V$$, and $$A_s$$ takes values in $$\mathfrak{g}$$) valued one forms on $$M$$, since we are working in a local gauge. This means that the contraction with the vector the vector field $$Y$$ gives us two vector valued zero forms. Since $$d\phi$$ is the exterior derivative of a zero form, we can write the following:
$$\nabla^A_Y\Phi=[s,L_Y\phi+\rho_*(A_s(Y))\phi]$$ Where $$L_Y$$ denote the Lie derivative along $$Y$$. We then see that: $$\nabla^A_X\nabla^A_Y\Phi=[s,d(\nabla^A_Y\phi)(X)+\rho_*(A_s(X))\nabla^A_Y\phi]$$ Examine the exterior derivative term: $$d(\nabla^A_Y\phi)(X)=d(L_Y\phi)(X)+d\rho_*(A_s(Y))(X)\phi+\rho_*(A_s(Y))d\phi(X)$$ Following the same logic as before we can rewrite this as: $$d(\nabla^A_Y\phi)(X)=L_x(L_Y\phi)+\rho_*(L_XA_s(Y))\phi+\rho_*(A_s(Y))d\phi(X)$$ Note that the exterior derivative 'commutes' with $$\rho_*$$ as $$A_s$$ can be written as the tensor product: $$A_s=\sum_{a}^{\text{dim }\mathfrak{g}}\omega^a\otimes T_a$$ where $$T_a$$ are a basis for $$\mathfrak{g}$$, and $$\omega^a$$ are one forms on $$M$$, thus we see that: $$d\rho_*(A_s)=\sum_{a}^{\text{dim }\mathfrak{g}}d(\omega^a)\otimes \rho_*(T_a)$$ as the $$T_a$$'s are constant. Thus we can write: $$d\rho_*(A_s(Y))(X)=\rho_*(dA_s(Y))(X)=\sum_{a}^{\text{dim }\mathfrak{g}}d(\omega^a(Y))(X)\otimes \rho_*(T_a)=\sum_{a}^{\text{dim }\mathfrak{g}}L_X(\omega^a(Y))\otimes \rho_*(T_a)=\rho_*(L_X(A_s(Y)))$$ Looking at the second term in $$\nabla^A_X\nabla^A_Y\Phi$$ we see: $$\rho_*(A_s(X))\nabla^A_y\phi=\rho_*(A_s(X)A_s(Y))\phi+\rho_*(A_s(X))d\phi(Y)$$ By symmetry we easily deduce the form of $$\nabla^A_Y\nabla^A_X\Phi$$, and see that $$\nabla^A_X\nabla^A_Y\Phi-\nabla^A_Y\nabla^A_X\Phi$$ is: $$[s,L_XL_Y\phi-L_YL_X\phi+\rho_*(L_XA_s(Y)-L_YA_s(X))+\rho_*([A_s(X),A_s(Y)])]$$ Finally we see that: $$\nabla_{[X,Y]}\Phi=[s,d\phi([X,Y])+\rho_*(A_s([X,Y]))\phi]$$ $$=[s,L_{[X,Y]}\phi+\rho_*(A_s([X,Y]))\phi]$$ With the identity: $$L_XL_Y f-L_YL_X f=L_{[X,Y]}f$$ we then easily see that the terms of $$F^\nabla\Phi$$ that contain the Lie derivatives of $$\phi$$ cancel, i.e.: $$L_XL_Y \phi-L_YL_X \phi-L_{[X,Y]}\phi=0$$ Hence we see that $$F^\nabla\Phi$$ is: $$F^\nabla\Phi=[s,\rho_*(L_XAs_(Y)-L_YA_s(X)-A_s([X,Y])+[A_s(X),A_s(Y)])\phi]$$ Note that: $$[A_s(X),A_s(Y)]=\frac{1}{2}[A_s,A_s](X,Y)$$ and that for any one form $$\omega$$: $$d\omega(X,Y)=L_X\omega(Y)-L_Y\omega(X)-\omega([X,Y])$$ Therefore we obtain: $$F^\nabla\Phi=[s,\rho_*(dA_s(X,Y)+\frac{1}{2}[A_s,A_s](X,Y))\phi]$$ Using the fact that: $$F_s=dA_s+\frac{1}{2}[A_s,A_s]$$ we finally obtain the desired result:
$$F^\nabla\Phi=[s,\rho_*(F_s(X,Y))\phi]$$ | 2,203 | 5,988 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 73, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-10 | latest | en | 0.750311 |
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