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http://www.readbag.com/ann-jussieu-fr-gostaf-cemracs2009-ff-2009-05-static | 1,566,561,711,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318375.80/warc/CC-MAIN-20190823104239-20190823130239-00230.warc.gz | 294,650,937 | 5,136 | `Stent Design Blood Flow SimulationKirill Pichon GostafEric Bonnetier, Didier Bresch, Vuk Milisic, Amélie Rambaud 15 September 2009 PeopleThis work was done in collaboration with: Eric Bonnetier Prof Didier Bresch DR CNRS Vuk Milisic CR1 CNRS Amélie Rambaud ENS LyonObjective: Asymptotic modeling and numerical simulations of blood flows in arteries with wired multi-layer stents. Mathematical modeling of multi-layered stents inserted in the human cardio-vascular network. Investigation of realistic 3D geometries. Collaboration with an industrial partner: Cardiatis, BelgiumKirill GostafFreefem++ workshop 20092 AneurysmAn aneurysm is an abnormal bulge in the wall of an artery. Often, no symptoms... becomes large, begins to leak blood, ruptures.National Heart, Lung and Blood Institute, USA.Kirill GostafFreefem++ workshop 20093 Stenting Saccular AneurysmThe multilayer stent design reduces flow velocity within the aneurysm, improves laminar flow in the main artery and surrounding branches, pressure and tension reduction within the aneurismal sac, Cardiatis. Frontal flow Tangential flowKirill GostafFreefem++ workshop 20094 Stenting Saccular AneurysmThe multilayer stent design reduces flow velocity within the aneurysm, improves laminar flow in the main artery and surrounding branches, pressure and tension reduction within the aneurismal sac, Cardiatis. Frontal flow Tangential flowKirill GostafFreefem++ workshop 20095 Direct Simulation 2D: Pressure Driven FlowBlood flow impinges directly upon the struts Stokes problemNnodes = 55k-u +p=0 u=0in in on lf ,rt ,strut on in ,outdiv u = 0 u· =0p = pin on in p = pout on out Pressure imposed = Dirichlet velocity Direct finite element simulation to be a reference solution Same number of elements on a stent strutKirill Gostaf Freefem++ workshop 2009 Nnodes = 104k6 Direct Simulation 3D: Two VesselsStokes direct simulation mesh Catia v5 Nnodes = 280K, Ntet =1.5M P2/P1 FE discretization solver FreeFem++ CG symmetric matrix 4.8GbKirill GostafFreefem++ workshop 20097 Direct Simulation 3D: Two VesselsStokes direct simulation mesh Catia v5 Nnodes = 280K, Ntet =1.5M P2/P1 FE discretization solver FreeFem++ CG symmetric matrix 4.8GbEXTREMELY EXPENSIVE!Kirill GostafFreefem++ workshop 20098 Direct Simulation 3D: Two VesselsStokes direct simulation mesh Catia v5 Nnodes = 280K, Ntet =1.5M P2/P1 FE discretization solver FreeFem++ CG symmetric matrix 4.8GbParallel solution EXTREMELY EXPENSIVE!one time step: 120 hoursHomogenization theoryKirill GostafFreefem++ workshop 20099 HomogenizationWe replace struts by the homogenization wall law Bi-domain formulation -u ,j+p ,j = 0in jdiv u ,j = 0 in j R u ,j on [u ,p ] · n =Discontinuous pressure Mono-domain numerical treatment with continuous pressure on velocity Kirill Gostafpressure Freefem++ workshop 2009 10 Domain DecompositionDirichlet-Neumann method [Quarteroni et.al.'88] o Consider an initial guess u on the interface :(1) k k L (u1 , p1 ) = 0 in 1 uk = g1 on 1 \ 1 k k u1 = u on k k L (u2 , p2 ) = 0 in 2 k u2 = g2 on 2 \ k [u ,p ] · n = R u k on 2(2)Correct the solution u until convergence, relaxation parameter :(3)Kirill Gostafk k k u +1 = (1 - ) u + u2Freefem++ workshop 200911 Velocity Profile: Flow Rate0 -0.005 -0.01 Velocity Profile -0.015 -0.02 -0.025 -0.03solid lines = Homogenized symbols = Direct solution6 struts 10 struts 16 struts 20 struts 0.4 0.6 Artery Cross-Section 0.8 1-0.03500.2Kirill GostafFreefem++ workshop 200912 Velocity Profile: Near the Homogenized Border0 Direct -0.005 Homogenized (FEM) Theory Velocity Profile -0.01-0.015-0.02-0.025-0.0300.20.4 0.6 Artery Cross-Section0.81Kirill GostafFreefem++ workshop 200913 Homogenization Error100L2 H1 3/2 10-11Error10-210-310-410-210 -1100Kirill GostafFreefem++ workshop 200914 Aneurysm Side 3D: FreeFem++ codeSolid geometry + mesh created with CATIA v5 CATIA v5 to FreeFem++ convert tool Load mesh, create finite element spacesmesh3 Th("aneurysm_side_h30.mesh"); fespace Xh(Th,P23d); Xh u1,v1, u2,v2, u3,v3; fespace Mh(Th,P13d); Mh p,q;//velocity FE space//pressure FE spaceKirill GostafFreefem++ workshop 200915 Aneurysm Side 3D: FreeFem++ codeVariational form: Stokes equationproblem Stokes([u1,u2,u3,p],[v1,v2,v3,q],solver=UMFPACK) = int3d(Th)( dx(u1)*dx(v1) + dy(u1)*dy(v1) + dz(u1)*dz(v1) + dx(u2)*dx(v2) + dy(u2)*dy(v2) + dz(u2)*dz(v2) + dx(u3)*dx(v3) + dy(u3)*dy(v3) + dz(u3)*dz(v3) - p*dx(v1) - p*dy(v2) - p*dz(v3) - dx(u1)*q - dy(u2)*q - dz(u3)*q - epsPenalty*p*q - int2d(Th,1)( pin*v1) + int2d(Th,2)(pout*v2) //p*div(w) //div(u)*q //penalization //penalization//negative 0X //positive 0Y+ on(1,u2=0,u3=0) //Gamma_in, u2=u3=0 + on(2,u1=0,u3=0) //Gamma_out, u1=u3=0 + on(20,u1=0,u2=0,u3=0);Kirill GostafFreefem++ workshop 200916 Aneurysm Side 3D: Velocity FieldKirill GostafFreefem++ workshop 200917 Aneurysm Side 3D: Pulsatile flowPressure solutionKirill GostafFreefem++ workshop 200918 Aneurysm Side 3D: Pulsatile flowVelocity magnitudeKirill GostafFreefem++ workshop 200919 `
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BETA | 1,755 | 5,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.735161 |
http://www.physicsforums.com/showthread.php?t=690426 | 1,386,763,082,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164035500/warc/CC-MAIN-20131204133355-00079-ip-10-33-133-15.ec2.internal.warc.gz | 483,555,829 | 12,884 | # Inductor opposing change in current.
by Woopydalan
Tags: current, inductor, opposing
P: 704 Hello, I'm wondering, if you have an inductor and resistor that has the switch open, so current is zero. Then you turn on the switch, the inductor is going to resist this change, it wants to be zero, right? Therefore, why doesn't the current indefinitely stay as zero, since the inductor would like to be at that value that it was at initially.
Mentor
P: 37,649
Quote by Woopydalan Hello, I'm wondering, if you have an inductor and resistor that has the switch open, so current is zero. Then you turn on the switch, the inductor is going to resist this change, it wants to be zero, right? Therefore, why doesn't the current indefinitely stay as zero, since the inductor would like to be at that value that it was at initially.
Can you show us the equation that relates current and voltage for an inductor?
P: 704 ε = -L di/dt So if it opposes a change (di/dt = 0) then ε is 0, then no current
Mentor
P: 37,649
## Inductor opposing change in current.
Quote by Woopydalan ε = -L di/dt So if it opposes a change (di/dt = 0) then ε is 0, then no current
There is no negative sign, first of all. And the equation should show the current as a function of time for an applied voltage:
$$V(t) = L \frac{dI(t)}{dt}$$
P: 704 Yeah, so my conceptual understanding of an inductor is that it would like to maintain the same current through it. If that is the case, why does it ever change? I guess unless something can overcome the resistance that the inductor puts forth to resist that change. What is it?
P: 2,366
Quote by Woopydalan Yeah, so my conceptual understanding of an inductor is that it would like to maintain the same current through it. If that is the case, why does it ever change? I guess unless something can overcome the resistance that the inductor puts forth to resist that change. What is it?
Look at the equation berkeman posted again. L is non-zero, V is non-zero, therefore dI/dt must be non-zero. And if dI/dt is non-zero at some time t, then even if I is zero at time t it won't be zero a moment later; the voltage is pushing current through the inductor.
An analogy which is really quite good if you compare the equation with Newton's $$f=ma=m\frac{dv}{dt}$$
Because of inertia, a mass hanging from a rope resists changes in speed unless you apply a force to it. If it's not moving and you push on it, it resists. But it will start swinging, and once it does it will resist any attempt to stop it.
P: 20 Woopydalan, Your equation is correct. It holds true for self-inductance. You are right about the inductor resisting current. It holds true because of Lenz's Law. As for your actual question on why the current doesn't remain zero: it is because, in steady state, flux through the inductor is zero. Hence, as stated by Faraday's Law, induced emf which opposed the current in the first instance eventually disappears, that is, self-inductance (L) becomes zero. Thus, current flows. This is the simplest and only explanation.
P: 1,508
Quote by berkeman There is no negative sign, first of all. And the equation should show the current as a function of time for an applied voltage: $$V(t) = L \frac{dI(t)}{dt}$$
My text books show induced emf = -d∅/dt. = -L.di/dt
The minus sign is taken to be an indication of Lenz's law.
Where is my text book (or my reading of the book!) going wrong
P: 390
Quote by wirefree You are right about the inductor resisting current.
It resists changes in current.
Quote by wirefree As for your actual question on why the current doesn't remain zero: it is because, in steady state, flux through the inductor is zero.
If the current through the inductor is the only source of flux linking its turns, then the flux through the inductor is only zero when the current through it is zero.
If by steady state you mean the rate of change of flux will tend to, or become, zero as t->∞, then you're assuming something which isn't true in general, e.g. it's not true for a constant voltage applied across an ideal inductor.
Quote by wirefree Hence, as stated by Faraday's Law, induced emf which opposed the current in the first instance ...
If the inductor induces an emf, then the rate of change of current through it must be nonzero, i.e. the magnitude of the current changes with time. You have assumed that which makes your conclusion true.
Mentor
P: 11,911
Quote by berkeman There is no negative sign, first of all. And the equation should show the current as a function of time for an applied voltage: $$V(t) = L \frac{dI(t)}{dt}$$
Quote by technician My text books show induced emf = -d∅/dt. = -L.di/dt The minus sign is taken to be an indication of Lenz's law. Where is my text book (or my reading of the book!) going wrong
The negative sign is usually included in intro physics textbooks, and omitted in electrical engineering books. But the two use a different convention for the direction of the current in an inductor, relative to the potential difference. They are equivalent descriptions.
Mentor
P: 11,911
Quote by Woopydalan ε = -L di/dt So if it opposes a change (di/dt = 0) then ε is 0, then no current
If that were true, we would not describe an inductor with that equation. We would simply say ε=0, i=0, and be done with it.
You can think about it this way:
If there is no current in the inductor, then there's no current in the resistor either.
In which case, there is no voltage difference (V=iR) across the resistor.
So then, the inductor must get the full battery voltage ε.
Therefore, di/dt = ε/L is not zero.
P: 704 I guess my question is that if an inductor resists changes to current, if the inductor had 0 current in it, why would it have current in it if the switch is closed? Wouldn't it resist having a nonzero value?
P: 20 When the switch is closed, the current and the resulting magnetic field increases from zero to the maximum value in a short time. This results in a increase in magnetic flux through the inductor and, hence, again induces an electric current in the inductor. The above forms the basis of Faraday's Law which was observed during experiments conducted around 1830.
P: 390
Quote by Woopydalan I guess my question is that if an inductor resists changes to current, if the inductor had 0 current in it, why would it have current in it if the switch is closed? Wouldn't it resist having a nonzero value?
If the inductor didn't oppose changes in current, you would be able to make the current through it "jump" instantaneously from one value to another, like you could for an ideal resistor. At the other end of the spectrum, it would oppose any and all changes to current and the rate of change of current through it would always be zero.
If you apply a voltage V across an ideal inductor with inductance L, you have from Faraday's law:
$$V = L \frac{\mathrm{d}i}{\mathrm{d}t} \Leftrightarrow \frac{\mathrm{d}i}{\mathrm{d}t} = \frac{V}{L}$$
Clearly then, the rate of change of current through the inductor can be nonzero. If you imagine plotting the current through the inductor as a function of time, you'll notice that V/L gives you the slope of this line. Decreasing the inductance L would give you a larger and larger slope, i.e. the change of current through the inductor would occur more rapidly.
You might say the inductance is a measure of "how much" the inductor resists changes in current.
Mentor
P: 11,911
I'll give a short and simple answer.
Quote by Woopydalan I guess my question is that if an inductor resists changes to current, if the inductor had 0 current in it, why would it have current in it if the switch is closed? Wouldn't it resist having a nonzero value?
Resisting change means that the current does not (nearly) instantaneously jump to a nonzero value, as it can in a resistor. Think about what happens in your circuit if there is no inductor.
Sci Advisor Thanks P: 1,751 Perhaps it's easiest to consider the most simple problem of this kind analytically. Take the usual assumptions of quasistationary circuit theory, i.e., the extensions of the whole circuit is much smaller than the wavelength of typical em. waves of the problem, so that the displacement current and retardation can be neglected. Then everything is described by linear ordinary differential equations and the circuit is described by resistance, capacitance and self-inductance. We consider a coil of self-inductance L in series with a resistor, attached to a DC battery. We switch on the circuit at $t=0$. From Faraday's Law we find $$L \dot{i}+R i=U(t),$$ where $$U(t)=U_0 \Theta(t).$$ First we solve the homogeneous equation $$\dot{i}=-\frac{R}{L} i.$$ Obviously the general solution is $$i(t)=A \exp \left (-\frac{R}{L} t \right).$$ Then we need one special solution of the inhomogeneous equation. Obviously this is given by the ansatz $i_0=B =\text{const}$. Plugging this in the equation, we get for $t>0$ that $B=U_0/R.$ The general solution of the full equation thus is for $t>0$ $$i(t)= A \exp \left (-\frac{R}{L} t \right) + \frac{U_0}{R}$$ Since $i(0)=0$ we finally can determine $A$ and then get $$i(t)=\frac{U_0}{R} \left [1-\exp \left (-\frac{R}{L} t \right ) \right ].$$ As you see the current won't flow immediately at full strength but only approaches exponentially the static value $i_{\infty}=U_0/R$.
P: 704 I would like to thank you all for contributing to this thread. I think I am now fully convinced that an inductor can resist change, which doesn't mean that it completely prevents the flow of changing current, much like a resistor doesn't prevent the flow of current entirely.
P: 1,508
Quote by Redbelly98 The negative sign is usually included in intro physics textbooks, and omitted in electrical engineering books. But the two use a different convention for the direction of the current in an inductor, relative to the potential difference. They are equivalent descriptions.
There is something very strange here !!! usually introduced in intro physics textbooks !!!!
Why do some introduce it and others do not, which ones do and which ones don't. The hyperphysics site, often quoted here....what do they do??
I did not know that electrical engineers had a different convention for electric current...how do they cope with kirchoffs laws when communicating with physicists?
An emf is generated across an inductor....not a pd....is there some confusion here about which way current flows.
Your statement does not help anyone wanting to learn physics....especially those using text books.
The minus sign is a recognition of Lenz's law. It is as important in electrical studies as the minus sign is in the analysis of SHM in mechanics. I doubt if anyone would claim that a minus sign is not necessary in SHM and that some intro physics text books decide not to use it.
It is PHYSICS.
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# Help - Probabilty question 3 on test 25
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Help - Probabilty question 3 on test 25 [#permalink]
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06 Apr 2010, 10:08
Question - 4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?
My solution -
Since atleast one professor must be in the committee, I first selected one professor which can be done in 4 ways as there are 4 professor. So I have 4 in the first position.
___4___ ________ _______
Now I have a total of 9 people (10 - 1 professor) and 2 spots in the committee. This can be done in 9C2 ways.
9!
------- = 36
7! *2!
So total ways is
4 * 36 = 144
Please let me know where I went wrong
Tahnks!
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Re: Help - Probabilty question 3 on test 25 [#permalink]
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06 Apr 2010, 10:27
I get the answer 100 ways
we have 4 pr and 6 st.
so, we can form the committee in 3 ways:
2 pr and 1 st = 2C4 * 6= 36
1pr and 2 st = 4 * 2C6 = 60
3 pr 3C4 = 4
Then 36+60+4=100
Am I right?
Last edited by fruit on 09 Apr 2010, 22:50, edited 1 time in total.
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Re: Help - Probabilty question 3 on test 25 [#permalink]
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06 Apr 2010, 10:42
total ways to form a committe with atleast 1 professor = total ways to form committee(both professors and students combined) - total ways to form committe without any professor(only students) = 10c3 - 6c3 = 120 - 20 = 100
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Re: Help - Probabilty question 3 on test 25 [#permalink]
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06 Apr 2010, 12:04
At least an event = All possibilities - No such event occurring
At least a prof = All possibilities - No prof
all possibilities: 10C3 = 10*9*8/6 = 120
No prof involved: 6 students only forming the committee: 6C3 = 20
Therefore, At least a prof = 120 - 20 = 100
Hope this helps
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Re: Help - Probabilty question 3 on test 25 [#permalink] 06 Apr 2010, 12:04
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# Help - Probabilty question 3 on test 25
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,021 | 3,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-47 | latest | en | 0.897524 |
https://math.stackexchange.com/questions/3283393/visualizing-combining-rotational-matrices | 1,653,603,639,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00364.warc.gz | 443,015,719 | 67,371 | # Visualizing Combining Rotational Matrices
I am a bit confused right now in 3D rotations. I have searched this question in many different ways but have not found a resource that could help me mentally/visually understand what is going on.
I'm familiar with matrices and utilizing them quite frequently in graphics, but this particular situation is making me realize I have a flaw in my reasoning at a base level.
I want to take two 3D (3x3 rotational information stored in a 4x4 matrix with last column and row staying as 0, 0, 0, 1 in the following examples) rotational matrices and combine them to perform a rotation in this order: x-axis then y-axis.
so I simply multiply them together to perform the operation. When I apply the matrices individually against a vector of [1, 1, 1, 1] I get the expected results:
Assuming +x toward the screen, +z to the right, +y is up
x-axis matrix (of 90 degrees) produces [1, -1, 1, 1] (a clockwise spin of the x-axis)
y-axis matrix (of 90 degrees) produces [1, 1, -1, 1] (a clockwise spin of the y-axis)
Now when I combine the matrices to rotate x then y and apply it against my vector (both still represent 90 degrees on their respective axis) it produces [1, 1, 1, 1].
I have tested this result against many online calculators that express this result as correct; however, to achieve that result would mean you did a counter-clockwise spin on the x-axis and a counter-clockwise spin on the y-axis.
I do not understand why this is happening. Again as a mental model...mathematically it just works just fine, but it really breaks down some of my perceptions on ordered rotations.
I feel like this has to deal with gimble lock somehow, just not sure how it plays into this especially since I split up the rotations and applied it to my vector one step at a time an lo! The result is more what I expected.
• A 3D rotation matrix is not a 4x4 matrix, it is a 3x3 matrix. Jul 4, 2019 at 21:22
• Yeah I could have expressed it all in 3x3. It's just helpful for me to keep it all in the dimensions I'm using currently to make sure I'm not doing something stupid. I realize, strictly speaking, 3x3 is where the 3d rotational information is stored, but for consistency I assume 4x4 with the last column and row staying as 0, 0, 0, 1. Technically in this case, the vector being 1 does not alter the results anywho. I realize I'm on a math forum though, would it be more proper to be a bit more explicit and less implied here? Jul 4, 2019 at 21:42
• So, it would seem if you use a "left-hand rule" for rotations around an oriented axis, rotating $90^{\circ}$ around the $x$-axis and then rotating $90^{\circ}$ around the $y$-axis does indeed get the point $(1,1,1)$ back to where it started. This is relatively easy to visualize by coloring the vertex of a cube and watching it rotate appropriately. That means the composition of these two rotations must be a rotation around the axis through $(1,1,1)$. Also, if you use a "right-hand rule" to rotate $90^{\circ}$ around the $x$ then $y$-axis, the point $(1,1,1)$ ends up somewhere else. Jul 4, 2019 at 21:54
• @runway44 If you’re working in homogeneous coordinates (usually so that translations can also be represented by matrix multiplication), which is what is going on here, transformation matrices for $\mathbb R^3$ are $4\times4$.
– amd
Jul 6, 2019 at 1:52
• Yes, OP pointed that out after my first comment. Jul 6, 2019 at 4:19
It doesn’t really have to do with anything fancy-sounding like gimbal lock or that the matrices for rotation about the $$y$$-axis look “backwards” because the $$y$$- and $$z$$-axis form a left-handed coordinate system. It’s simply because matrix multiplication is not commutative: the order in which you multiply them matters. In particular, 3-D rotations don’t commute unless they have a common rotation axis. You’re almost certainly multiplying the two primitive rotation matrices together in the wrong order when computing the combined transformation matrix.
You haven’t said whether you’re working with row or column vectors, but either way, when you chain transformations together like this, remember that the input to the next transformation in the chain is the output of the preceding transformation. If you’re working with column vectors so that left-multiplication by a matrix gives you the transformed vector, the result of applying the first rotation is $$\mathbf v'=R_x\mathbf v$$, and the result of applying the second rotation to that is $$R_y\mathbf v' = R_y(R_x\mathbf v) = (R_yR_x)\mathbf v$$, therefore the matrix that represents the combination of the two rotations is $$R_yR_x$$. You likely computed $$R_xR_y$$ instead, which corresponds to first rotating about the $$y$$-axis and then rotating about the $$x$$-axis. An easy way to remember this is that if you left-multiply by a matrix to transform a vector, then the chain of matrices grows to the left.
If you’re instead working with row vectors, then you’d be right-multiplying by the transformation matrices, and the chain would grow to the right. In that case, $$R_xR_y$$ would give you the correct result, since $$\mathbf v(R_xR_y)=(\mathbf vR_x)R_y$$. | 1,296 | 5,157 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-21 | latest | en | 0.937735 |
https://www.bodylanguageacadamy.us/guide-2/protecting-your-hand.html | 1,566,542,596,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318011.89/warc/CC-MAIN-20190823062005-20190823084005-00274.warc.gz | 738,292,009 | 6,009 | You have K-K. A jackal raises before the flop, you then make it t hree bets (reraise), a nd an elephant behind you calls the three bets "cold" (without having any money already invested in the pot). The j ackal then calls the one additional bet.
The flop comes 10-9-2 and the j ackal bets, you raise, and the elephant calls the two bets. The j ackal a lso calls two bets. The turn card comes up a 2, for 10-9-2-2, and now the j ackal b ets out into you. At t his point you should be thinking, "Raise it!" But you're distracted by the conversation going on across the table, and you j ust call the bet. Now, the elephant calls the bet a s weH.
This i s a most costly mistake, since you've now let the elephant c all o nly one bet with his A-9, and the l ast c ard off i s an A, for 10-9-2-2-A. Now the j ackal c hecks and you decide to check as well, because you fear the A may have hit the elephant. Then the elephant bets and the j ackal calls, and you call as weH. The elephant then says, " I have two pair, aces up." You think, " Man, am I unlucky, I cannot believe that he hit an ace on me here!" Wrong! You misplayed this hand! All you had to do was raise after the two came up on fourth street, and the elephant would have been forced to throw his hand away! Your call on the end might also fall i nto the mistake category (even though I've said you should generally not be folding on the end), because the one card you had l egitimate reason to fear, the ace, hit t he board, a nd a bet and a call were already in front of you.
Let's r ewrite the script, then, so you're making the obvious raise on fourth street. A deuce comes off the deck for 10-9-2-2, and the j ackal b ets out i nto you. You don't really t hink the j ackal has a deuce, s o you raise and the elephant reluctantly folds his hand. The j ackal c alls your r aise. The river card is an ace, for 10-9-2-2-A, and the j ackal checks. You conclude t hat the j ackal has a pair of tens, s o you bet out, a nd then the j ackal calls you. You say, " Pocket kings for me" and the j ackal says, " Nice hand." You then pile all t he chips onto your s tack as t he elephant l oudly complains, " Darn it, I would have made aces and nines if I 'd stayed in, b ut I c ouldn't call, b ecause your r aise on fourth street t old me you had me beaten!" You j ust s mile and finish stacking t he chips, thinking, ' ' Looks as though I played that hand perfectly!"
This first example is about "protecting your hand" with a raise on fourth street. If you fail t o do that, you give your opponents a chance to outdraw you for j ust one bet. The next example is another f airly obvious play, b ut i n the other direction—folding!
0 0 | 742 | 2,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.964624 |
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# 4(x2+2x+1)(x2+3x-2)+(x-3)2=(a2+bx+c)2find a2+b2+c2
Grade:10
## 2 Answers
Nirmal Singh.
askIITians Faculty 44 Points
7 years ago
Regards,
Nirmal Singh
Askiitians Faculty
mycroft holmes
272 Points
7 years ago
4(x^2+2x+1)(x^2+3x-2)+(x-3)^2 = 4 (x+1)^2 [(x+1)^2+(x-3)]+(x-2)^2 = 4(x+1)^4 + 4 (x+1)^2 (x-3) + (x-3)^2 = [2(x+1)^2+(x-3)]^2 = [2x^2+5x-1]^2 So a=2, b=5, c=-1
## ASK QUESTION
Get your questions answered by the expert for free | 338 | 834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.72089 |
https://www.deucescracked.com/articles/373-Thoughts-about-winrates-and-variance | 1,527,408,612,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00307.warc.gz | 748,448,681 | 12,523 | # Featured Poker Blog Post by jajvirta
## Thoughts about winrates and variance
Yeah, you heard it first from me: decrease your variance by increasing your winrate.
A couple of disclaimers. First, "variance" is actually not a "thing." I guess you should call the "thing" dispersion. Variance is just a measure of that. "Decreasing variance" would be something like "making kilograms bigger" if you're trying to lose weight or something. Second, what is commonly thought of as variance is the possibility of losing in a given set of hands even if you're a winning player. So it's not like anyone's complaining of the variance of a game where you win either \$10000 or \$20000 even though the variance is huge in such a game. Third, we can't actually decrease the variance, so to speak, without compromising our EV. What most people want is to increase the probability that one's up after a given set of hands.
But anyhow.
I bet you were disappointed when you read the first sentence. It's not like you can increase your winrate on will. The point, however, is that there are ways to decrease your winrate and not doing those will "decrease your variance", so to speak. Again, not anything terribly surprising here.
What prompted me to write about this, then, was that some players advocate the notion that you should increase the amount of hands played to deal with swings. That is, if a winning player can have a 100k breakeven stretch, then the proposed cure is to play so much that 100k hand period so short that you don't have "long lasting" downswings. And it's true too: if you manage to play million hands with the winrate of 1BB/100, it's pretty much absolutely certain that you are up.
But increasing the amount of hands comes with a cost. Presumably you're already playing as much as possible, given the circumstances. You might even think that you should spend more time improving your game instead of playing all the time. (Or, say, hang out with your family. :-) Increasing the amount of tables might not be an option, as you might already have trouble with the four tables you've figured to be feasible.
And some people, *cough coaches cough*, have the chutzpah to claim that you should probably play less tables. "I might have a downswing that lasts a whole year that way," you might yell.
It's not that simple though.
Let's define some notions first. We have a sample of N hands with the winrate of WR which yields the mean expectation, ME. The result of a hand has a standard deviation of X, which we set here to be 1.8BB. Typical values of standard deviation for 100 hands in short-handed limit hold'em is somewhere around 16-18BB/100. (You can find out yours from Poker Tracker's Session tab. There's "More detail" button there or something.) Standard deviation of N hands is standard deviation of a hand times square root of the amount of hands.
Now, we assume that the results are normally distributed. They pretty much are, but it's worth noting that that's what we assume. Next, we plant the normal distribution curve on the spot where our expected result is after N hands. If we play 1000 hands with the winrate of 1BB/100 then the most likely result for us is 10BB. With this method, we can find out what's the probability that we're not losing after a set of hands. Now that we've planted the distribution on the center of the expected value, we just calculate the area of the curve on the left side of the zero result. And since we know the standard deviation of the sample (std dev of a hand * square root of the number of hands) we know the size and the shape of the curve.
Sounds a bit complicated and it is too, in fact. It involves funny looking characters and hairy derivations. But luckily it's all thought out already and we can just enjoy the results. So, let's take a quick example. We have a player with winrate of 1BB/100 and std dev of 1.8 per hand (typical for lower stakes short handed limit hold'em). We ask whether the result is positive after 1000 hands.
So, first, we have the mean value, the expected value,
0.01BB per hand * 1000 = 10BB
Next, we calculate the standard deviation of the sample:
1.8 per hand * sqrt(1000) = 56.92
Then we find the z-zero point, that is, where the zero value is in the curve.
z0 = (0 - mean) / std dev of the sample
or
(0 - 10) / 56.92 = -0.18
In other words, one fifth of a standard devation from the mean. Now we can calculate the area what falls left of this point. To do this, consult a "z-score probability table", for example here: http://www.statsoft.com/textbook/stathome.html?sttable.html&1
this equals to something like 43% (or, 50 - 7). You will see from the table that 0.18 equals to 0.07 (or 7%). This the area from the mean to this point. So the area from this point to the extreme left (infinity) is approximately 43%.
The other way is to use spreadsheet program and use the function NORMSDIST.
So the probability of us being up after 1000 hands with winrate of 1BB/100 and std dev of 1.8 per hand is staggering 57%.
Now, if we expand this idea to playing 5000 hands per month and considering playing a lot more to "decrease the variance", or at least to decrease the probability of being down. Let's say you add a table to your multi-tabling setup and maybe play bit longer sessions. Doubling the amount of hands without adding more time would of course mean doubling the amount of tables you play, which might be infeasible if you already play 4 tables at a time.
Let's further say that your winrate with less tables is 1.5BB/100 and with more tables it drops to 1.0BB/100. The expected value of 5k hand stretch with less tables compared to 10k hands with more is of course a bit lower, 75BB and 100BB respectively. But the funny thing is that the probability that you're on the positive side (that is, winning) is almost the same with both scenarios, 5k and 1.5 winrate and 10k and 1.0 winrate, 71-72%. You can put this calculation in your spreadsheet and compare different scenarios and see for yourself how much the winrate actually affects the variance.
We can of course debate whether playing less tables increases your winrate at all, and if, how much, but I'm personally pretty sure it does, and often pretty significantly so. This little post was intended to show that increasing your winrate does have a significant effect on the aspect of poker which most players find repulsive: losing money while playing better than opponents.
As a final note, what prompted to think about the whole issue was the fact that people tend to worry most about the "poker tracker result". That is, there are sites that take a lot of rake but give some of it back with rakebacks and bonuses. The true winrate is a function of the rakeback, but people want the poker tracker to show green also. But with rakes as high as 4-5BB/100, you are going to have tough going with this goal. If your true non-rake-deducted winrate with this scenario is something like 5BB/100, which is a good winrate I might add, then your "poker tracker winrate" is 1BB/100, which is going to have huge swings and you might expect to have break-even stretches of tens of thousands of hands.
#### Entity
8664 posts
Joined 11/2006
Republished from something written in our forums -- I just thought this was worthy of a little more front-page . Nice work jajvirta.
Rob
#### Squishee
1356 posts
Joined 01/2008
my english or may brain... maybe both.. . so hard to understand for me
#### jajvirta
725 posts
Joined 03/2007
my english or may brain... maybe both.. . so hard to understand for me
Heh, it's not the clearest explanation, now that I've re-read it myself. But it might be partly that it implicitly assumes some knowledge on the basic statistics and if you don't have much, it might not make a lot of sense because of that.
#### DavidC
217 posts
Joined 03/2007
Firstly, from his forum posts, Jaj pretty much rocks, so I don't have anything against him, personally. However, I don't like this post, really. I'm a nit.
---
I'm not sure, but I think when you say "decrease your variance" you mean "decrease your risk" (i.e. reducing var/ev, rather than reducing var).
If you were actually talking about decreasing your variance itself by increasing winrate, I think you'd be discussing stuff like how winners might get put in less marginal spots than losers due to their opponents being more intimidated, or something like that. I feel that situation would reduce actual variance, not just risk, but I'm not totally sure.
Also, you seem to be defining variance as deviation from break-even over a given sample, where you should be defining it as deviation from expectation. If you have a 2/100 winrate and play 10,000 hands, a break-even stretch actually represents a 200bb downswing from your expectation. It's kinda interesting that when you think of people's commonly described massive downswings as being stuff like -500bb (from break-even) over 20,000 hands, that swing size doesn't even include their expectation over that sample, which means it's even bigger than we initially thought!
FWIW, the actual substance of your post seems solid, regardless of the terms used.
---
Finally, I'm curious about how to set my bankroll size to give myself 0.1% ROR over an infinite sample... if anyone knows. Although I suppose determining ROR with a given bankroll size would work too, I could just play with the numbers until I got the right ROR.
#### DJ Sensei
3298 posts
Joined 10/2007
Good post jaj. The toughest part, of course, is figuring out exactly what our winrates are, given our generally small sample sizes, high standard deviations, and the fact that our games are constantly changing. Perhaps your article could be accompanied with a discussion of confidence intervals for winrates over N hands.
DavidC: not exactly what I think you were looking for re: ROR, but its still worth checking out if you haven´t already. http://en.wikipedia.org/wiki/Kelly_criterion
#### DavidC
217 posts
Joined 03/2007
lol I'm actually kinda afraid of the Kelly criterion.
#### jajvirta
725 posts
Joined 03/2007
I'm not sure, but I think when you say "decrease your variance" you mean "decrease your risk" (i.e. reducing var/ev, rather than reducing var).
Yeah, I tried to explain that in the second paragraph, but I guess I hurried over too quickly. I agree with you completely and I think this a common confusion. But after that second pararagraph, I just translated "decrease the risk of being down after a set of hands" to "decrease your variance", of which the latter is somewhat non-sensical as you point out too. But I do think that what most people want is to not lose money and this what was I aiming to talk about in the article.
#### DavidC
217 posts
Joined 03/2007
I'm not sure, but I think when you say "decrease your variance" you mean "decrease your risk" (i.e. reducing var/ev, rather than reducing var).
Yeah, I tried to explain that in the second paragraph, but I guess I hurried over too quickly. I agree with you completely and I think this a common confusion. But after that second pararagraph, I just translated "decrease the risk of being down after a set of hands" to "decrease your variance", of which the latter is somewhat non-sensical as you point out too. But I do think that what most people want is to not lose money and this what was I aiming to talk about in the article.
Yeah, I'm a nit.
nvm.
#### PygmyHero
4246 posts
Joined 08/2007
Hey jajvirta, you know I was a big fan of this when you originally posted it in the Beginner LHE forum, but I just wanted to give you some dap here too - really good work. Thanks for taking the time to put it together.
#### Cblanks
102 posts
Joined 03/2008
How can i find out if the STD DEV in a plo game if based upon my bb/100 on pto?
#### PygmyHero
4246 posts
Joined 08/2007
Cblanks, I'm not sure if I understand your question. The answer might be: Session Notes tab, More Detail button (on the upper right of the screen).
#### DC21
7 posts
Joined 01/2008
thanks pygmy( i know diff name but watching vids with a friend), i also got evpoker and it does it for plo, or atleast tells you how close you are to expectation
#### tylerwest
2 posts
Joined 04/2018
I think this article good describe all win rate chances. It is similar to http://www.worldpressrelease.org/?p=35064 where you can find students chances to cheat with their essays. I hope that you will enjoy this article.
HomePoker BlogsFeatured Posts → Thoughts about winrates and variance | 3,018 | 12,545 | {"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} | 2.640625 | 3 | CC-MAIN-2018-22 | latest | en | 0.980164 |
http://us.metamath.org/mpeuni/pocl.html | 1,638,322,278,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00541.warc.gz | 89,307,732 | 5,480 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > pocl Structured version Visualization version GIF version
Theorem pocl 5002
Description: Properties of partial order relation in class notation. (Contributed by NM, 27-Mar-1997.)
Assertion
Ref Expression
pocl (𝑅 Po 𝐴 → ((𝐵𝐴𝐶𝐴𝐷𝐴) → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝐷) → 𝐵𝑅𝐷))))
Proof of Theorem pocl
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 id 22 . . . . . . 7 (𝑥 = 𝐵𝑥 = 𝐵)
21, 1breq12d 4626 . . . . . 6 (𝑥 = 𝐵 → (𝑥𝑅𝑥𝐵𝑅𝐵))
32notbid 308 . . . . 5 (𝑥 = 𝐵 → (¬ 𝑥𝑅𝑥 ↔ ¬ 𝐵𝑅𝐵))
4 breq1 4616 . . . . . . 7 (𝑥 = 𝐵 → (𝑥𝑅𝑦𝐵𝑅𝑦))
54anbi1d 740 . . . . . 6 (𝑥 = 𝐵 → ((𝑥𝑅𝑦𝑦𝑅𝑧) ↔ (𝐵𝑅𝑦𝑦𝑅𝑧)))
6 breq1 4616 . . . . . 6 (𝑥 = 𝐵 → (𝑥𝑅𝑧𝐵𝑅𝑧))
75, 6imbi12d 334 . . . . 5 (𝑥 = 𝐵 → (((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝐵𝑅𝑦𝑦𝑅𝑧) → 𝐵𝑅𝑧)))
83, 7anbi12d 746 . . . 4 (𝑥 = 𝐵 → ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝑦𝑦𝑅𝑧) → 𝐵𝑅𝑧))))
98imbi2d 330 . . 3 (𝑥 = 𝐵 → ((𝑅 Po 𝐴 → (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))) ↔ (𝑅 Po 𝐴 → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝑦𝑦𝑅𝑧) → 𝐵𝑅𝑧)))))
10 breq2 4617 . . . . . . 7 (𝑦 = 𝐶 → (𝐵𝑅𝑦𝐵𝑅𝐶))
11 breq1 4616 . . . . . . 7 (𝑦 = 𝐶 → (𝑦𝑅𝑧𝐶𝑅𝑧))
1210, 11anbi12d 746 . . . . . 6 (𝑦 = 𝐶 → ((𝐵𝑅𝑦𝑦𝑅𝑧) ↔ (𝐵𝑅𝐶𝐶𝑅𝑧)))
1312imbi1d 331 . . . . 5 (𝑦 = 𝐶 → (((𝐵𝑅𝑦𝑦𝑅𝑧) → 𝐵𝑅𝑧) ↔ ((𝐵𝑅𝐶𝐶𝑅𝑧) → 𝐵𝑅𝑧)))
1413anbi2d 739 . . . 4 (𝑦 = 𝐶 → ((¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝑦𝑦𝑅𝑧) → 𝐵𝑅𝑧)) ↔ (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝑧) → 𝐵𝑅𝑧))))
1514imbi2d 330 . . 3 (𝑦 = 𝐶 → ((𝑅 Po 𝐴 → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝑦𝑦𝑅𝑧) → 𝐵𝑅𝑧))) ↔ (𝑅 Po 𝐴 → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝑧) → 𝐵𝑅𝑧)))))
16 breq2 4617 . . . . . . 7 (𝑧 = 𝐷 → (𝐶𝑅𝑧𝐶𝑅𝐷))
1716anbi2d 739 . . . . . 6 (𝑧 = 𝐷 → ((𝐵𝑅𝐶𝐶𝑅𝑧) ↔ (𝐵𝑅𝐶𝐶𝑅𝐷)))
18 breq2 4617 . . . . . 6 (𝑧 = 𝐷 → (𝐵𝑅𝑧𝐵𝑅𝐷))
1917, 18imbi12d 334 . . . . 5 (𝑧 = 𝐷 → (((𝐵𝑅𝐶𝐶𝑅𝑧) → 𝐵𝑅𝑧) ↔ ((𝐵𝑅𝐶𝐶𝑅𝐷) → 𝐵𝑅𝐷)))
2019anbi2d 739 . . . 4 (𝑧 = 𝐷 → ((¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝑧) → 𝐵𝑅𝑧)) ↔ (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝐷) → 𝐵𝑅𝐷))))
2120imbi2d 330 . . 3 (𝑧 = 𝐷 → ((𝑅 Po 𝐴 → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝑧) → 𝐵𝑅𝑧))) ↔ (𝑅 Po 𝐴 → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝐷) → 𝐵𝑅𝐷)))))
22 df-po 4995 . . . . . . 7 (𝑅 Po 𝐴 ↔ ∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
23 r3al 2935 . . . . . . 7 (∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑥𝑦𝑧((𝑥𝐴𝑦𝐴𝑧𝐴) → (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
2422, 23sylbb 209 . . . . . 6 (𝑅 Po 𝐴 → ∀𝑥𝑦𝑧((𝑥𝐴𝑦𝐴𝑧𝐴) → (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
252419.21bbi 2058 . . . . 5 (𝑅 Po 𝐴 → ∀𝑧((𝑥𝐴𝑦𝐴𝑧𝐴) → (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
262519.21bi 2057 . . . 4 (𝑅 Po 𝐴 → ((𝑥𝐴𝑦𝐴𝑧𝐴) → (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
2726com12 32 . . 3 ((𝑥𝐴𝑦𝐴𝑧𝐴) → (𝑅 Po 𝐴 → (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
289, 15, 21, 27vtocl3ga 3262 . 2 ((𝐵𝐴𝐶𝐴𝐷𝐴) → (𝑅 Po 𝐴 → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝐷) → 𝐵𝑅𝐷))))
2928com12 32 1 (𝑅 Po 𝐴 → ((𝐵𝐴𝐶𝐴𝐷𝐴) → (¬ 𝐵𝑅𝐵 ∧ ((𝐵𝑅𝐶𝐶𝑅𝐷) → 𝐵𝑅𝐷))))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ∧ wa 384 ∧ w3a 1036 ∀wal 1478 = wceq 1480 ∈ wcel 1987 ∀wral 2907 class class class wbr 4613 Po wpo 4993 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-9 1996 ax-10 2016 ax-11 2031 ax-12 2044 ax-13 2245 ax-ext 2601 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1878 df-clab 2608 df-cleq 2614 df-clel 2617 df-nfc 2750 df-ral 2912 df-rab 2916 df-v 3188 df-dif 3558 df-un 3560 df-in 3562 df-ss 3569 df-nul 3892 df-if 4059 df-sn 4149 df-pr 4151 df-op 4155 df-br 4614 df-po 4995 This theorem is referenced by: poirr 5006 potr 5007
Copyright terms: Public domain W3C validator | 2,955 | 3,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-49 | longest | en | 0.268226 |
https://embeddedrelated.com/showthread/lpc2000/5785-1.php | 1,723,633,440,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00860.warc.gz | 174,989,818 | 17,865 | # CAN baudrate setting
Started by February 21, 2005
Dear all, I am using two LPC2129 boards having 12 MHz clock, with CCLK and PCLK set to 60 MHz. They are connected via a CAN link. I am trying to set the bus timing register C1BTR to run the CAN at the baudrate of 1 Mbit/s. The value I tried is : C1BTR = 0x34C005; But it did not work. Anybody can help me to work out the correct setting of this register to let the CAN go for 1 Mbaud. Thanks very much. Mouaaz PS. the correct setting for 125 kbaud was: C1BTR = 0x001C001D;
Dear all, I am using two LPC2129 boards having 12 MHz clock, with CCLK and PCLK set to 60 MHz. They are connected via a CAN link. I am trying to set the bus timing register C1BTR to run the CAN at the baudrate of 1 Mbit/s. The value I tried is : C1BTR = 0x34C005; But it did not work. Anybody can help me to work out the correct setting of this register to let the CAN go for 1 Mbaud. Thanks very much. Mouaaz PS. the correct setting for 125 kbaud was: C1BTR = 0x001C001D;
I've tried making a spreadsheet for this, but I haven't had a chance to actually test the timings. But here are the answers I come up with. These calculations are for SAM = 1 only. Solution Set #1: BRP = 15 SJW = 4 TSeg2 = 4-6 TSeg1 = 10-8 TSeg1 + Tseg2 should be 14 So the values for this set should be 0x0057C00F (Tseg1 = 8) 0x0048C00F (Tseg1 = 9) 0x0039C00F (Tseg1 = 10) Solution Set #2: BRP = 10 SJW = 3-4 Tseg2 = 3-4 Tseg1 = 6-5 Tseg1 + Tseg2 = 9 0x0025800A (SJW = 3, Tseg2=3, Tseg1 = 6) You can probably figure the rest of them out. Solution Set #3 BRP = 6 SJW = 2-4 Tseg2 = 2 Tseg1 = 3 Please let me know if any of these work. If my spreadsheet works, then I might post it, but I won't get a chance to test any of this works for a couple of weeks. Assumptions made, deltaF = 1%, bus length = 10 meters max, propagation time = 6.5m/ns, logic delay and loop delay = 200ns --Peter --- In , "mouaaz" wrote: > > Dear all, > > I am using two LPC2129 boards having 12 MHz clock, with CCLK and PCLK > set to 60 MHz. They are connected via a CAN link. I am trying to set > the bus timing register C1BTR to run the CAN at the baudrate of 1 > Mbit/s. The value I tried is : > > C1BTR = 0x34C005; > > But it did not work. > > Anybody can help me to work out the correct setting of this register > to let the CAN go for 1 Mbaud. > > Thanks very much. > > Mouaaz > > PS. the correct setting for 125 kbaud was: C1BTR = 0x001C001D;
Hi Mouaaz, the Rules of Bit Timing Parameters are not described in the Philips LPC2000 User Manuals. And they are not easy. Please refer to the Bosch Homepage for more Information. But, I give you an Example that works very well in my application: Assumed that you setup the PLL and the cclk = 60 MHz and the VPB Divider = 4 (after RESET) and the pclk = 15 MHz, you can use the following parameters for a first test: // Common CAN bit rates for cclk` MHz, VPBDIV=4: ==> pclk MHz: // CAN Bit Timings in Detail --> // SAM TSEG1 TSEG2 SJW SP% BRP #define CANBit10k_15MHz 0x25C095 // 0+1 5+1 2+1 3+1 70 149+1 #define CANBit25k_15MHz 0x25C03B // 0+1 5+1 2+1 3+1 70 59+1 #define CANBit50k_15MHz 0x25C01D // 0+1 5+1 2+1 3+1 70 29+1 #define CANBit100k_15MHz 0x25C00E // 0+1 5+1 2+1 3+1 70 14+1 #define CANBit125k_15MHz 0x25C00B // 0+1 5+1 2+1 3+1 70 11+1 #define CANBit250k_15MHz 0x25C005 // 0+1 5+1 2+1 3+1 70 5+1 #define CANBit500k_15MHz 0x25C002 // 0+1 5+1 2+1 3+1 70 2+1 #define CANBit1000k_15MHz 0x48C000 // 0+1 8+1 4+1 3+1 67 0+1 The following parameters for VPBDIV = 1 I have calculated on this base: For cclk = 60 MHz and VPBDIV = 1, the following parameters should work: // Common CAN bit rates for cclk` MHz, VPBDIV=1: ==> pclk` MHz: // CAN Bit Timings in Detail --> // SAM TSEG1 TSEG2 SJW SP% BRP #define CANBit10k_60MHz 0x25C257 // 0+1 5+1 2+1 3+1 70 599+1 #define CANBit25k_60MHz 0x25C0EF // 0+1 5+1 2+1 3+1 70 239+1 #define CANBit50k_60MHz 0x25C077 // 0+1 5+1 2+1 3+1 70 119+1 #define CANBit100k_60MHz 0x25C03B // 0+1 5+1 2+1 3+1 70 59+1 #define CANBit125k_60MHz 0x25C02F // 0+1 5+1 2+1 3+1 70 47+1 #define CANBit250k_60MHz 0x25C017 // 0+1 5+1 2+1 3+1 70 23+1 #define CANBit500k_60MHz 0x25C00B // 0+1 5+1 2+1 3+1 70 11+1 #define CANBit1000k_60MHz 0x48C003 // 0+1 8+1 4+1 3+1 67 3+1 Write this Values to the CANBTR Register. The defined CAN Bit Rates works very well in my application (Keil MCB2100 connected to a CAN Analyzer (CANalyzer from vector informatik)). The right Bit Rate Calculation is: Bit rate = cclk / ( VPBDIV * BRP * ( TSEG1+1 + 1 + TSEG2+1)) The Sample Point should be in the Range 50% to 90 %. Please download the Keil Examples for a CAN Driver, that was very helpful for me: http://www.keil.com/download/docs/lpc2_can_examples.zip.asp Greetings from Dietmar. http://www.weiss-electronic.de ----------------------------- --- In , "mouaaz" wrote: > > Dear all, > > I am using two LPC2129 boards having 12 MHz clock, with CCLK and PCLK > set to 60 MHz. They are connected via a CAN link. I am trying to set > the bus timing register C1BTR to run the CAN at the baudrate of 1 > Mbit/s. The value I tried is : > > C1BTR = 0x34C005; > > But it did not work. > > Anybody can help me to work out the correct setting of this register > to let the CAN go for 1 Mbaud. > > Thanks very much. > > Mouaaz > > PS. the correct setting for 125 kbaud was: C1BTR = 0x001C001D;
Dear all: I use LPC2292s CAN , But I cant use its filter properly. 2292 has two cans, I let can1 send data to can2. If I disable the filter,the data is properly received, But if I enable the filter, I see that the CANRFS,CANRID,CANRDA and CANRDB has received the data properly, but the RBS bit is not set to 1. I set the filter ram as the datasheet say, and I have set the SFF_sa,SFF_GRP_sa,EFF_sa,EFF_GRP_sa and EndOfTable, I have set AFMR to 0x00 and filled the unused ram to 0xf7fff7ff. I dont know way. Is there anyother thing I should do? Thanks very much. wangjie click here Terms of Service. | 2,162 | 5,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.926725 |
https://civilseek.com/grading-of-aggregates/ | 1,721,330,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00419.warc.gz | 143,822,824 | 26,719 | # A Detailed Guide on Grading of Aggregates.
## Grading of Aggregates.
Grading of aggregates is determining the average grain size of the aggregates before they are used in construction.
This is applied to both coarse and fine aggregates. The aggregate sample is sieved through a set of sieves and weights retained on each sieve in percentage terms are summed up.
On dividing this sum by 100, The Fineness Modulus of that aggregate is determined. This helps in deciding about the quantity of aggregates of known fineness moduli to be mixed for obtaining a concrete of desired density.
The basis for mixing coarse and fine aggregates of specific fineness modulus is the presence of voids or open spaces when the aggregates are packed together.
In pure coarse aggregates packing may leave 30-40 percent voids, which can be removed only by filling with finer particles.
Similarly, in fine aggregates also, voids are left that have to be filled with still finer particles (of cement).
This is essential to obtain concrete of compact and void-free character.
When some pieces of aggregates of equal size are packed together, voids or open spaces are always left. The percentage of voids may be as high as 45 percent of the total volume of aggregates.
It has been observed that this result (presence of voids) is independent of the size of aggregates used in packing, whether coarse, medium or fine.
Read More: Soundness of Cement & Its Test by Le Chatelier Apparatus.
Only, while packing aggregate should be of equal size, that is of the same grade.
Either all of them should be coarse, or all of them should be medium or fine grade. This implies that even if sand alone is packed, voids to the tune of 40-45 percent are again left.
This principle of formation of voids is the governing principle for the preparation of concrete under this method.
When the coarse aggregates are packed to make the concrete, the voids formed within the mass must be filled by some finer material. Sand is used for this purpose.
But, there will be voids left between the sand grains too. These are filled with the cement particles.
In this way, the resulting concrete mass is a void-less or dense mass. The binding property of cement is made use of to give this dense mass a cohesive stone-like character.
Concrete is, an artificial stone, in a broad sense.
Grading of aggregates is aimed at determining the mean size of the particle in a given batch of aggregates.
This is commonly found by the Method of Fineness Modulus. The method can be used to determine fineness modulus of coarse aggregates, fine aggregates, and all-in aggregates, i.e., mixed aggregates.
In this method, a convenient weight of the sample is taken and sieved through a set of sieves one after another. The number of sieves is five for the coarse aggregates and ten for all-in-aggregates.
It is only six in the case of fine aggregates.
Sieve Size for Grading of Aggregates.
Coarse Aggregates: 80 mm, 40 mm, 20 mm, 10 mm, IS Nos. 480
Fine Aggregates: IS No. 480, 240, 120, 60, 30 and 15.
All in aggregates: 80 mm, 40 mm, 20 mm, 10 mm, Nos: 480, 240, 120, 60, 30 and 15.
Read More: Top 12 Different Types of Cranes used in Construction Works.
It will be noted that each successive sieve has the diameter of mesh reduced to 50 percent.
Calculations involve dividing the cumulative percentage of weights retained on these set of sieves by 100.
The resulting figure gives fineness modulus of the respective aggregate. This is illustrated in the following example.
A weight of 10 kg of Coarse aggregate and 5 kg of fine aggregate has been taken to determine fineness modulus in each case separately.
Let us assume that the weight of aggregate retained in each case and the calculated cumulative percentage of retained weights are as follows:
IS Sieve No.Weight RetainedTotal Weight (kg)Retained Weight (gm)
80 mm000
40 mm000
20 mm2220
10 mm4660
4803.59.595
2400.510100
120100
60100
30100
15100
Total10 Kg6.75 Kg
Cumulative Percentage of weight retained on all the ten sieves – 675 Fineness modulus of coarse aggregate = 675/100 = 6.75 kg.
IS Sieve No.Weight retainedTotal Weight (kg)Retained (gm)
4800.20.24
2400.81.020
1200.51.530
601.22.754
301.54.284
150.85.0100
Total5 kg2.92 kg
Cumulative percentage of weight retained on all the six sieves = 292 Fineness Modulus of Fine Aggregates = 292/100 =2.92 kg.
Now, when it is desired to obtain a combined aggregate of a definite (required) fineness modulus, F, the amount of fine aggregate (x) to be mixed with 1 volume of the coarse aggregate of a given fineness modulus can be easily determined from the relationship:
x=Fc – F / F – Ff
Where “x” is the amount of fine aggregate (by volume)for 1 volume of coarse aggregate.
Fc = Fineness modulus of coarse aggregate.
F = Fineness modulus of desired mixed aggregate.
Ff = Fineness modulus of fine aggregate.
The above relationship is illustrated in the following example.
F = 5.5
Fc = 6.75
Ff = 2.92
To Determine: Volume of fine aggregate (x) in percentage terms of volume of coarse aggregate.
x=Fc – F / F – Ff
=> 6.75 – 5.3 / 5.5-2.92 x 100
=> 48.5 % approximately.
### Limits of Fineness.
Repeated trials with mix designs using different aggregates have shown that following limits of fineness moduli hold good for obtaining concrete mixes of good workability.
Cement Consumption is also reasonable when limits are followed.
Aggregate TypeMaximum Size (mm)Recommended Moduli Maximum.Recommended Moduli Manimum.
Fine2.003.5
Coarse2066.90
406.907.50
807.508.00
1608.008.50
Mixed204.705.10
255.105.5
325.25.7
405.45.9
805.86.3
160657.0
The upper limit of fineness modulus is, therefore, always below 8.5 for coarse and 7.0 for mixed aggregates.
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Read More: Reinforced Cement Concrete | Advantages, Uses, Types, & Purpose.
### 1 thought on “A Detailed Guide on Grading of Aggregates.”
1. Very educative.
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## Circumradius of Icosidodecahedron given surface to volume ratio Solution
STEP 0: Pre-Calculation Summary
Formula Used
circumradius = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(Surface to Volume Ratio*(45+17*sqrt(5))))/2)*(1+sqrt(5))
rc = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(RAV*(45+17*sqrt(5))))/2)*(1+sqrt(5))
This formula uses 1 Functions, 1 Variables
Functions Used
sqrt - Squre root function, sqrt(Number)
Variables Used
Surface to Volume Ratio - Surface to Volume Ratio is fraction of surface to volume. (Measured in Hundred)
STEP 1: Convert Input(s) to Base Unit
Surface to Volume Ratio: 0.5 Hundred --> 0.5 Hundred No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
rc = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(RAV*(45+17*sqrt(5))))/2)*(1+sqrt(5)) --> (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(0.5*(45+17*sqrt(5))))/2)*(1+sqrt(5))
Evaluating ... ...
rc = 21.2242188052286
STEP 3: Convert Result to Output's Unit
21.2242188052286 Meter --> No Conversion Required
(Calculation completed in 00.016 seconds)
## < 5 Circumradius of Icosidodecahedron Calculators
Circumradius of Icosidodecahedron given surface to volume ratio
circumradius = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(Surface to Volume Ratio*(45+17*sqrt(5))))/2)*(1+sqrt(5)) Go
Circumradius of Icosidodecahedron given surface area
### Circumradius of Icosidodecahedron given surface to volume ratio Formula
circumradius = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(Surface to Volume Ratio*(45+17*sqrt(5))))/2)*(1+sqrt(5))
rc = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(RAV*(45+17*sqrt(5))))/2)*(1+sqrt(5))
## What is an icosidodecahedron ?
In geometry, an icosidodecahedron is a polyhedron with twenty (icosi) triangular faces and twelve (dodeca) pentagonal faces. An icosidodecahedron has 30 identical vertices, with two triangles and two pentagons meeting at each, and 60 identical edges, each separating a triangle from a pentagon. As such it is one of the Archimedean solids and more particularly, a quasiregular polyhedron.
## How to Calculate Circumradius of Icosidodecahedron given surface to volume ratio?
Circumradius of Icosidodecahedron given surface to volume ratio calculator uses circumradius = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(Surface to Volume Ratio*(45+17*sqrt(5))))/2)*(1+sqrt(5)) to calculate the Circumradius, Circumradius of Icosidodecahedron given surface to volume ratio formula is defined as a straight line connecting center and any point of circumsphere of truncated cuboctahedron ,Where side= icosidodecahedron edge (a). Circumradius and is denoted by rc symbol.
How to calculate Circumradius of Icosidodecahedron given surface to volume ratio using this online calculator? To use this online calculator for Circumradius of Icosidodecahedron given surface to volume ratio, enter Surface to Volume Ratio (RAV) and hit the calculate button. Here is how the Circumradius of Icosidodecahedron given surface to volume ratio calculation can be explained with given input values -> 21.22422 = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(0.5*(45+17*sqrt(5))))/2)*(1+sqrt(5)).
### FAQ
What is Circumradius of Icosidodecahedron given surface to volume ratio?
Circumradius of Icosidodecahedron given surface to volume ratio formula is defined as a straight line connecting center and any point of circumsphere of truncated cuboctahedron ,Where side= icosidodecahedron edge (a) and is represented as rc = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(RAV*(45+17*sqrt(5))))/2)*(1+sqrt(5)) or circumradius = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(Surface to Volume Ratio*(45+17*sqrt(5))))/2)*(1+sqrt(5)). Surface to Volume Ratio is fraction of surface to volume.
How to calculate Circumradius of Icosidodecahedron given surface to volume ratio?
Circumradius of Icosidodecahedron given surface to volume ratio formula is defined as a straight line connecting center and any point of circumsphere of truncated cuboctahedron ,Where side= icosidodecahedron edge (a) is calculated using circumradius = (((6*(5*sqrt(3)+(3*(sqrt(25)+(10*sqrt(5))))))/(Surface to Volume Ratio*(45+17*sqrt(5))))/2)*(1+sqrt(5)). To calculate Circumradius of Icosidodecahedron given surface to volume ratio, you need Surface to Volume Ratio (RAV). With our tool, you need to enter the respective value for Surface to Volume Ratio and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Circumradius?
In this formula, Circumradius uses Surface to Volume Ratio. We can use 5 other way(s) to calculate the same, which is/are as follows - | 1,472 | 4,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-49 | latest | en | 0.65711 |
http://bbs.magnum.uk.net/?page=001-forum.ssjs&sub=usenet_sciphysm&thread=48912 | 1,719,329,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00881.warc.gz | 2,994,540 | 7,569 | • #### Re: What we know about the gyroscope
From Paul B. Andersen@21:1/5 to All on Mon Dec 4 11:08:24 2023
Den 04.12.2023 06:08, skrev patdolan:
1. All angular momentum has a "rigidity in space", even the so-called minor axis theorem, aka tennis racket theorem.
2. When angular momentum is torqued it is immediately met by an opposing torque of exactly the same magnitude and in exactly the opposite direction.
3. In the event of #2 another, much smaller angular momentum instantly appears which is always perpendicular to the original angular momentum. This is commonly referred to as precession.
4. None of the aforementioned enigmatic behaviors are understood in anyway by any physics, past or present. And they cannot be derived from first principles of any physics, past or present.
Quite.
Physics can't be derived from 'first principles'.
Physics is derived from observations of HOW Nature behaves.
Nobody knows WHY Nature behaves in the enigmatic way she does.
You can't understand WHY Newton's laws of motion are as they are.
But we know that the mathematical model of an aspect of Nature:
Newtonian mechanics, correctly predicts HOW a gyroscope will behave.
Physics doesn't answer or explain WHY, it only predicts HOW.
--
Paul
https://paulba.no/
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• From Tom Roberts@21:1/5 to patdolan on Mon Dec 4 10:05:43 2023
On 12/3/23 11:08 PM, patdolan wrote:
1. All angular momentum has a "rigidity in space", even the
so-called minor axis theorem, aka tennis racket theorem.
Hmmmm. This is poorly stated at best.
For an isolated system in which the Lagrangian is rotationally
invariant, angular momentum is conserved. Note this does not apply to a gyroscope near the surface of the earth and supported against gravity.
2. When angular momentum is torqued it is immediately met by an
opposing torque of exactly the same magnitude and in exactly the
opposite direction.
This is just plain wrong. Invoking MAGIC is useless.
3. In the event of #2 another, much smaller angular momentum
instantly appears which is always perpendicular to the original
angular momentum. This is commonly referred to as precession.
This is also just plain wrong. Invoking MAGIC is useless.
4. None of the aforementioned enigmatic behaviors are understood in
anyway by any physics, past or present. And they cannot be derived
from first principles of any physics, past or present.
This just happens to be right, BECAUSE THEY ARE WRONG.
The precession of a gyroscope is easily derived using Newton's laws
applied to the gyroscope. It is probably easier to use Lagrangian mechanics.
PS--the Big Ben Paradox has again succeeded is raising a lot of hell
How silly, as "the big ben paradox" is merely patdolan's lack of
understanding of very basic physics.
Tom Roberts
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• From Volney@21:1/5 to patdolan on Mon Dec 4 15:21:09 2023
On 12/4/2023 12:08 AM, patdolan wrote:
4. None of the aforementioned enigmatic behaviors are understood in anyway by any physics, past or present. And they cannot be derived from first principles of any physics, past or present.
No, showing the existence of the torque/precession is a second semester freshman physics problem. Best way to treat it is to consider the
rotating mass as many smaller masses instantaneously moving in a
straight line but rigidly attached to the hub and showing how the system
reacts to an instantaneous force. Newton figured this out long ago.
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• From Sylvia Else@21:1/5 to patdolan on Tue Dec 5 13:15:35 2023
On 05-Dec-23 12:54 pm, patdolan wrote:
On Monday, December 4, 2023 at 12:21:12 PM UTC-8, Volney wrote:
On 12/4/2023 12:08 AM, patdolan wrote:
4. None of the aforementioned enigmatic behaviors are understood in anyway by any physics, past or present. And they cannot be derived from first principles of any physics, past or present.
No, showing the existence of the torque/precession is a second semester
freshman physics problem. Best way to treat it is to consider the
rotating mass as many smaller masses instantaneously moving in a
straight line but rigidly attached to the hub and showing how the system
reacts to an instantaneous force. Newton figured this out long ago.
The two most important, most ignored and most unexplained properties of the gyroscope are 1) rigidity, and 2) the instantaneous and perfectly opposing torque which magically appears and exactly balances the applied torque to prevent the g-scope from
tipping over while it precesses.
You've just been told how to analyse a gyroscope, but you ignore that
and go back to magic.
Sylvia.
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• From Volney@21:1/5 to patdolan on Mon Dec 4 22:14:16 2023
On 12/4/2023 8:54 PM, patdolan wrote:
On Monday, December 4, 2023 at 12:21:12 PM UTC-8, Volney wrote:
On 12/4/2023 12:08 AM, patdolan wrote:
4. None of the aforementioned enigmatic behaviors are understood in anyway by any physics, past or present. And they cannot be derived from first principles of any physics, past or present.
No, showing the existence of the torque/precession is a second semester
freshman physics problem. Best way to treat it is to consider the
rotating mass as many smaller masses instantaneously moving in a
straight line but rigidly attached to the hub and showing how the system
reacts to an instantaneous force. Newton figured this out long ago.
The two most important, most ignored and most unexplained properties of the gyroscope are 1) rigidity, and 2) the instantaneous and perfectly opposing torque which magically appears and exactly balances the applied torque to prevent the g-scope from
tipping over while it precesses.
You flunk freshman physics, Semester 2. The properties of gyroscopes is
not "unexplained properties" but are well understood physics as I
stated. Just because you aren't smart enough to understand it doesn't
mean it's wrong or nobody understands it, it just means you're just not
smart enough.
Nor is the gyroscope not tipping over any form of "magic", conservation
of angular momentum is enough to explain "why" a gyroscope doesn't fall
over.
(a common trait of cranks is that they seem to think that if they don't understand something either it's wrong or they project their lack of understanding onto /everyone/, implying /nobody/ understands it)
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• From Mikko@21:1/5 to patdolan on Tue Dec 5 12:45:31 2023
On 05-Dec-23 12:54 pm, patdolan wrote:
The two most important, most ignored and most unexplained properties
of the gyroscope are 1) rigidity,
The approximate rigidity is a consequence of the properties of the
material the gyroscope is made of.
and 2) the instantaneous and perfectly opposing torque which magically appears and exactly balances the applied torque to prevent the g-scope
from tipping over while it precesses.
When you look at the gyroscope you don't see torque, you just see motion.
For the explanation of the motion, see e.g. Goldstein: Classical Mchanics sections 5-6 and 5-7.
Mikko
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• From Mikko@21:1/5 to patdolan on Wed Dec 6 11:28:04 2023
On 2023-12-05 19:07:38 +0000, patdolan said:
On Tuesday, December 5, 2023 at 2:45:36 AM UTC-8, Mikko wrote:
On 05-Dec-23 12:54 pm, patdolan wrote:
The two most important, most ignored and most unexplained properties
of the gyroscope are 1) rigidity,
The approximate rigidity is a consequence of the properties of the
material the gyroscope is made of.
and 2) the instantaneous and perfectly opposing torque which magically
appears and exactly balances the applied torque to prevent the g-scope >>> > from tipping over while it precesses.
When you look at the gyroscope you don't see torque, you just see motion.
For the explanation of the motion, see e.g. Goldstein: Classical Mchanics
sections 5-6 and 5-7.
Look at me, Mikko,
I don't see anything.
and listen very carefully. ALL treatments of gyroscopes, including the
one you reference, start in the middle of the story by starting with precession.
No, Goldstein starts the section 5-7 with equations of kinetic and
potential energy and infers from them that the motion involves
precession and nutation.
Mikko
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• From Mikko@21:1/5 to patdolan on Thu Dec 7 12:17:34 2023
On 2023-12-07 00:55:20 +0000, patdolan said:
On Wednesday, December 6, 2023 at 1:28:09 AM UTC-8, Mikko wrote:
On 2023-12-05 19:07:38 +0000, patdolan said:
On Tuesday, December 5, 2023 at 2:45:36 AM UTC-8, Mikko wrote:
On 05-Dec-23 12:54 pm, patdolan wrote:
The two most important, most ignored and most unexplained
properties of the gyroscope are 1) rigidity,
The approximate rigidity is a consequence of the properties of the
material the gyroscope is made of.
and 2) the instantaneous and perfectly opposing torque which
magically appears and exactly balances the applied torque to
prevent the g-scope from tipping over while it precesses.
When you look at the gyroscope you don't see torque, you just see
motion. For the explanation of the motion, see e.g. Goldstein:
Classical Mchanics sections 5-6 and 5-7.
Look at me, Mikko,
I don't see anything.
and listen very carefully. ALL treatments of gyroscopes, including the
one you reference, start in the middle of the story by starting with
precession.
No, Goldstein starts the section 5-7 with equations of kinetic and>
potential energy and infers from them that the motion involves>
precession and nutation.
Mikko, I have carefully studied Goldstein 5-7 https://www.math.toronto.edu/khesin/biblio/GoldsteinPooleSafkoClassicalMechanics.pdf
and can identify neither an equation(s) or a sentence which "infers" precession and nutation from potential angular energy, kinetic &
potential. Please direct me to the precise eqn(s) or the paragraph.
The equation for procession is (5-50). The equation for nutation is (5-53)
and in another way (5-55). Another equation for precession is (5-66).
The equation for average precession is (5-67).
When a spin axis is first held fixed and then released, keeping ony one
point fixed, the spin axis first starts to fall. This conclusion is
stated in the text before the paragraph that contains the equation (5-58).
The inferences to each of these equations is evertything from the third paragraph to the equation. There is very little other comments in the
text.
Mikko
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* Origin: fsxNet Usenet Gateway (21:1/5) | 2,774 | 10,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.926765 |
http://www.physicsforums.com/showthread.php?t=16210 | 1,368,915,061,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382917/warc/CC-MAIN-20130516092622-00088-ip-10-60-113-184.ec2.internal.warc.gz | 660,868,101 | 9,656 | ## Decoherence & collapse of the wavefunction!
From what I have gathered, whether or not decoherence has solved the measurement problem is still a matter of debate. But to those who say that it does, my question is: how does it solve it? Does it actually cause the collapse of the wavefunction?
These questions are actually pieces of a larger puzzle I am trying to put together: Has the superposition of states of quantum entities which are not being observed by a conscious mind collapsed, or not? E.g. are the photons of light from a light bulb in a superposition of states when noone is looking, or has decoherence already caused the collapse?
If anyone can help me in trying to put the pieces together I'd appreciate it!
P.S. I know that I could gather alot of info on this topic via an engine search, but as a non-scientist I find the accompanying mathematical formulae too hard to understand. I would appreciate if people could keep explanations simple. Thanks!
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Originally posted by SamLuc From what I have gathered, whether or not decoherence has solved the measurement problem is still a matter of debate. But to those who say that it does, my question is: how does it solve it? Does it actually cause the collapse of the wavefunction? These questions are actually pieces of a larger puzzle I am trying to put together: Has the superposition of states of quantum entities which are not being observed by a conscious mind collapsed, or not? E.g. are the photons of light from a light bulb in a superposition of states when noone is looking, or has decoherence already caused the collapse? If anyone can help me in trying to put the pieces together I'd appreciate it! P.S. I know that I could gather alot of info on this topic via an engine search, but as a non-scientist I find the accompanying mathematical formulae too hard to understand. I would appreciate if people could keep explanations simple. Thanks!
Decoherence doesn't actually cause a "collapse" of the wavefunction, although it can if the extra degree of interaction is an act of measurement. Decoherence is an explanation on why we don't see (or very, very seldom see) quantum effects at large scales and long lengths of time.[1] It is why we don't see the weird QM effects on our cats, for example, contrary to the poor creature in Schrodinger's experiment.
It does, however, mean that if we can preserve coherence at larger and larger scales, QM effects should be preserved, and that's what we are beginning to see in those SQUIDs experiments.
Zz.
[1] See, for example, http://www.nature.com/nsu/000120/000120-10.html
Recognitions: Gold Member Science Advisor Staff Emeritus Decoherence is to standard quantum theory what the emergence of species is to population biology: no extra assumptions, postulates etc... are introduced, the standard theory is used fully, one just had to think of it ! What does decoherence in fact say ? It says that quantum systems that are "observed" are necessarily coupled (though an interaction term in the hamiltonian) to a *system with many degrees of freedom*, aka a macroscopic system. This system is nothing else but the observation device in the case of an explicit measurement. It can however, also be the environment in the case of a non-explicitly observed quantum system. It is then mathematically rather simple to deduce that the global state (system + environment) quickly evolves into an entangled state that cannot be written as a product of a state of the system and a state of the environment, but rather as a sum of such terms. If one insists on a 'local system description', then the only possibility left is to consider a statistical ensemble of local states, described by a local density matrix. Moreover, one can show that this density matrix is diagonal in so-called "coherent states", which are the quantum equivalent of macroscopic states, and that the diagonal elements are nothing else but the probabilities you'd calculate naively using just the local description of the quantum system, and the "collapse technique". However, decoherence doesn't explain collapse, or the probabilistic nature of measurements. It *uses* in fact, by introducing the density matrix, the collapse technique in a certain way. What decoherence shows is that you almost never find systems which are coupled to the environment in "non-classical superpositions" which would correspond to non-zero non-diagonal terms in the density matrix. So decoherence explains why we don't observe a Schroedinger cat which is in a superposition of a dead and an alive cat. It doesn't explain the "collapse of the wave function" because it uses the probabilistic interpretation of the wave function by using a density matrix. cheers, Patrick.
## Decoherence & collapse of the wavefunction!
I appreciate the replies guys, although I don't pretend to understand everything contained in them.
Decoherence doesn't actually cause a "collapse" of the wavefunction, although it can if the extra degree of interaction is an act of measurement.
It doesn't explain the "collapse of the wave function" because it uses the probabilistic interpretation of the wave function by using a density matrix.
These responses lead me to believe that the answer to my question (Has the superposition of states of quantum entities which are not being observed by a conscious mind collapsed, or not?) depends on whether or not one believes that observation by a conscious mind is the only thing that actually qualifies as a measurement. Would this belief be correct?
Recognitions:
Gold Member
Staff Emeritus
These responses lead me to believe that the answer to my question (Has the superposition of states of quantum entities which are not being observed by a conscious mind collapsed, or not?)...
As I tried to point out, decoherence answers this question partially: a conscious mind, but also a macroscopic measurement apparatus, will always observe "classically-looking" results, and no "weird superpositions". What one calls "decoherence" is simply the application of orthodox quantum theory (Schroedinger evolution) to a system coupled to an "environment" (= another system with a huge number of degrees of freedom). So the part: why don't I see superpositions of cats and dogs in the street, as naive quantum theory seems to allow for ? is answered by decoherence.
What is NOT answered is whether there is an actual collapse (meaning a non-unitary change in the state of the wave function of the universe) or not. In fact, decoherence seems to indicate that to find this out is an untractable problem because of the almost immediate entanglement with the rest of the universe. Indeed, the only way to find out something about a collapse would be to work with superpositions of "classical" states and see when their interferences disappear. However, decoherence shows us that from the moment we get "macroscopic", these interferences disappear due to entanglement with the environment.
It seems that decoherence has made of the question, of whether there is a collapse or not, a non-scientific question because non-falsifiable by experiment. (but that's my own interpretation !).
cheers,
Patrick.
Ok. It's becoming a little clearer now; I'm beginning to get my head around what you're saying. Thanks again. | 1,569 | 7,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.954772 |
http://www.finderchem.com/is-6-an-composite-number.html | 1,500,656,513,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423787.24/warc/CC-MAIN-20170721162430-20170721182430-00595.warc.gz | 422,988,195 | 7,421 | # Is 6 an composite number?
It's a great way to review the concepts of prime and composite numbers, ... 6 or 9, it divides by 3; numbers that yield an even answer when divided by 2 also divide ... - Read more
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## Is 6 an composite number? resources
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Related Questions
Recent Questions | 1,499 | 6,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-30 | latest | en | 0.861026 |
http://www.transtutors.com/homework-help/corporate-finance/financial-statement-analysis/cash-flow-to-stockholders/ | 1,511,251,720,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806327.92/warc/CC-MAIN-20171121074123-20171121094123-00117.warc.gz | 508,381,498 | 25,418 | ## What is Cash Flow To Stockholders?
What is cash flow? Who is a stockholder?
The net amount of cash and cash-equivalents moving into and out of a business is known as Cash Flow. Stockholder also known as shareholder is an individual, group, or organization that holds one or more shares in a company, and in whose name the share certificate is issued.
The revenues generated by the company is not only for the stockholders. The company has some expenses, reinvestment opportunities and debt obligations which are to be paid. Cash Flow to Stockholders is the cash that company gives to its stockholders after all its obligations are fulfilled. The stockholders are mainly concerned with the cash dividends paid during a reporting period. Stockholders are worried about the future potential of the company to pay dividends as dividends are the main source of income for them.
## Sources of Cash Flows for a Stockholder
What are the two main sources of cash flow for a stockholder?
The main source of cash flow for a stockholder is dividends. Firms also issue new stock periodically and also repurchase their own stock. This transaction is recorded on the balance sheet as treasury stock.
Cash flow for stockholder = Cash dividends – Cash received from stockholders to buy additional shares from the company + Cash paid to investors to repurchase their shares.
For instance, \$50,000 cash is given out as a dividend by the company, the company buys back \$20,000 of shares from stockholders, and sells \$80,000 of stock to stockholders. The outcome is negative cash flow to stockholders of \$10,000 (50,000+20,000-80,000).
## Calculation of cash flow to stockholder
The cash flow to stockholders can be calculated in different ways:
• Common dividends paid less net new common equity raised. Cash flow to stakeholders = Dividends paid - (Ending common stock - Beginning common stock) - (Ending capital surplus - Beginning capital surplus) + (Ending treasury Stock - Beginning treasure stock).
• Cash flow to stockholder = Net income + NCC – Fixed capital expenditure - Change in net working capital + Net borrowing.
• Cash flow to stockholder = CFO - Fixed capital expenditure + Net borrowing.
• Cash flow to stockholder = Cash flow from asset – Interest (1-t) + Net borrowing.
## Understanding and treatment of items of Cash Flow to Stockholder
### Dividends
The dividend is the main source of cash flow to the stockholder. The dividend is a share of the after-tax profit of a company. It is circulated to its stockholders as per the number and class of shares held by them.
Dividends vary with company to company.
The board of directors decide the amount and timing of the dividend. They likewise figure out if it is paid out of current profit or the past income kept as reserve. Larger and public companies pay dividends every quarter whereas smaller companies typically distribute dividends at the end of an accounting year.
Preference shareholders receive the dividend first and the dividends are fixed. A common shareholder is entitled to receive any amount of dividend, based on the level of profit and the company's need for cash for expansion or other purposes. Dividends other than cash paid in the form of additional stock or assets are not considered to be cash flow to stockholders.
### Net New Equity
Net New Equity is subtracted from dividends to arrive at cash flow from the stockholder. It is calculated as:
(Ending common stock - Beginning common stock) - (Ending capital surplus - Beginning capital surplus) + (Ending treasury Stock - Beginning treasure stock).
Net new equity is the increase in owner’s equity from the year beginning to the yearend other than retained earnings.
### Net Income
Net income also known as profit is the company's total earnings after paying for the costs. Net income is shown on the company's income statement and is an important measure of how profitable the company is over a period of time Net income is ascertained by taking revenues and subtracting the expenses of doing business, for example, depreciation, interest, taxes and other expenses.
For detailed information on Net Income
### Capital Expenditure
Fixed Capital Expenditure is a net amount.
Fixed Capital Investment = Capital Expenditures – Proceeds from sales of long-term assets.
If no long-term assets were sold during the year, then capital expenditures will also equal the change in the gross PP&E account from the balance sheet.
FCInv = ending net PP&E – beginning net PP&E + depreciation. It is subtracted from net income while calculating cash flow to the stockholder.
For detailed information on Capital Expenditure
### Noncash Charges
Noncash charges are charges which are not in cash. Noncash charges are added back to net income to arrive at cash flow from stockholders as they represent expenses that actually didn’t result in an outflow of cash. Depreciation is the most significant noncash charge. The items that should be added back to net income:
• Depreciation
• Amortisation of intangible should be added back to net income
• Amortization of a bond discount should be added back to net income
• Accretion of the bond premium should be subtracted from net income
### Net new borrowing
The difference between the long and short-term new debt issues and long and short-term debt repayments is known as the net borrowing. It is added to net income to arrive at the cash flow from the stockholder.
### Change in the net working capital
Change in net working capital is equal to the investment in the net working capital only difference is that it does not include cash, cash equivalents, notes payable, and the current portion of long- term debt. It is subtracted from net income to arrive at cash flow from the stockholder.
### Interest
When we take cash flow from the asset as our starting point then we must deduct the interest expense as it represents a financing cash flow to bondholders that is available to the firm before it makes payments to its capital suppliers. Cash flow to a stockholder is arrived after paying all expenses related to the bondholders. Therefore, it is subtracted from cash flow to asset to arrive at cash flow to the stockholder. The entire portion of interest is not subtracted from the cash flow to stockholder only the after-tax interest cost is subtracted from it because paying interest reduces our tax bill.
### Cash Flow from Assets
Cash Flow from Assets is the total cash flow to creditors and cash flow to stockholders. The Cash Flow from assets measures the cash flows generated by the firm's assets. It is also known as the cash flow of the firm. The calculations are done as following:
Cash Flow to Assets =
• Cash Flow from Operations + Interest (1-t) – Fixed Capital Expenditure.
• Net Income + Noncash Charges + Interest (1-t) – Fixed Capital Expenditure – Working Capital Expenditure.
• EBIT (1-t) + Depreciation – Fixed Capital Expenditure – Working Capital Expenditure.
• EBITDA (1-t) + Depreciation (Tax Rate) – Fixed Capital Expenditure – Working Capital Expenditure.
A healthy firm generally has positive cash flow. The firm which is growing or is young will have a negative cash flow from the firm's assets. Is cash flow to stockholder defined as cash flow to the creditor? No, cash flow to a stockholder is not defined as cash flow to the creditor. Cash flow to creditor + Cash flow to stockholder = Cash flow to assets.
## Example of Cash Flow to Stockholders
1. Find out cash flow from the stockholder given the balance sheet and income statement.
Balance Sheet
Assets 2016 2015 Cash \$30 \$15 Accounts receivable \$90 \$45 Inventory \$120 \$90 Current assets \$240 \$150 Gross PPE \$1,200 \$900 Accumulated depreciation \$570 \$420 Total Assets \$870 \$630 Liabilities 2016 2015 Accounts payable \$60 \$60 Short-term debt \$60 \$30 Current liabilities \$120 \$90 Long term debt \$342 \$300 Common stock \$150 \$150 Retained earnings \$258 \$90 Total liabilities and Equity \$870 \$630
Income Statement
2016 2015 Sales \$900 \$750 Cost of goods sold \$360 \$300 Gross profit \$540 \$450 SG&A \$105 \$90 EBITDA \$435 \$360 Depreciation \$150 \$120 EBIT \$285 \$240 Interest expense \$45 \$30 Pre-tax earnings \$240 \$210 Taxes (30%) \$72 \$63 Net Income \$168 \$147
Calculation of Cash Flow to Stockholder
Net Income 168 Depreciation & Amortization 150 Changes in working capital* 75 Fixed capital expenditure* 300 Net Borrowings* 72 Cash flow to stockholder* 15
Changes in net working capital = Increase in accounts receivable + Increase in inventory – Increase in accounts payable = 45 + 30 – 0 = 75
*Fixed capital expenditure = Gross PP&E 2016 – Gross PP&E
= 1200 – 900
*Net borrowings = Short Term Debt = 60 – 30 = 30
Long Term Debt = 342 – 300 = 42
Total Net Borrowings = 30 + 42 = 72
*Cash flow to stockholder = 168 + 150 – 75 – 300 + 72 = 15
2. Find out cash flow to stockholder
Income Statement 2016 Sales 1600 Cost of Goods Sold 1000 Administrative Expenses 50 Depreciation 60 EBIT 490 Interest Expense 50 PBT 440 Taxes 40 Net Income 400 Dividends 200 Addition to Retained Earnings 200 Owners' Equity 2016 2015 Common Stock (\$1 Par) 100 80 Capital Surplus 200 180 Retained Earnings 300 100 Total Owners' Equity 600 360
Cash Flow to Stockholder = Dividends paid - (Ending common stock - Beginning common stock) - (Ending capital surplus - Beginning capital surplus) + (Ending treasury Stock - Beginning treasure stock).
= \$200 - (\$100 - \$80) - (\$200 - \$180) - (\$0 - \$0) = \$160
Cash Flow to Stockholders
## Related Topics
All Finance Topics
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Página 218 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Página 153 - Compute the interest to the time of the first payment ; if that be one year or more from the time the interest commenced, add it to the principal, and deduct the payment from the sum total. If there be after payments made, compute the interest on the balance due to the next payment, and then deduct the payment as above; and, in like manner, from one payment to another, till all the payments are absorbed ; provided the time between one payment and another be one year or more.
Página 155 - Compute the interest on the principal sum, from the time when the interest commenced to the first time when a payment was made, which exceeds either alone or in conjunction with the preceding payment...
Página 150 - The rule for casting interest, when partial payments have been made, is to apply the payment, in the first place, to the discharge of the interest then due. If the payment exceeds the interest, the surplus goes towards discharging the principal, and the subsequent interest is to be computed on the balance of principal remaining due. If the payment be less than the interest, the surplus of interest must not be taken to augment the principal; but interest continues on the former principal until the...
Página 199 - ... but if the blank fall under the first or second term, the proportion is inverse ; then multiply the third and fourth terms together for a divisor, and the other three for a dividend, and the quotient will be the answer. EXAMPLES.
Página 235 - Operations with Fractions A) To change a mixed number to an improper fraction, simply multiply the whole number by the denominator of the fraction and add the numerator.
Página 7 - Decedents," and to repeal said original sections, -and to repeal sections one (1), two (2), three (3), four (4), five (5), six (6), seven...
Página 241 - Multiply all the numerators together for a new numerator, and all the denominators together for a new denominator.
Página 242 - Then multiply the second and third terms together, and divide the product by the first term: the quotient will be the fourth term, or answer.
Página 155 - In casting interest upon bonds, notes, &.c., upon which partial payments have been made, every payment is to be first applied to keep down the interest ; but the interest is never allowed to form a part of the principal, so as to carry interest, for the effect in such case would be to give compound interest, which the law does not allow. | 688 | 3,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-33 | latest | en | 0.805336 |
https://www.york.ac.uk/students/studying/manage/programmes/module-catalogue/module/MAT00004C/2024-25 | 1,726,142,755,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00286.warc.gz | 1,004,591,493 | 7,523 | Accessibility statement
# Introduction to Probability & Statistics - MAT00004C
« Back to module search
• Department: Mathematics
• Credit value: 20 credits
• Credit level: C
• Academic year of delivery: 2024-25
## Module summary
This module: introduces the basic concepts of probability theory and statistics, illustrated by a full range of examples and applications; introduces an important statistical computing package (R); provides secure and solid foundations for higher level probability and mathematical statistics modules.
## Related modules
• None
### Prohibited combinations
• None
Post-requisite modules:
Statistics stream
## Module will run
Occurrence Teaching period
A Semester 1 2024-25
## Module aims
This module: introduces the basic concepts of probability theory and statistics, illustrated by a full range of examples and applications; introduces an important statistical computing package (R); provides secure and solid foundations for higher level probability and mathematical statistics modules.
## Module learning outcomes
By the end of the module, students will be able to:
1. model simple experiments using probability theory;
2. perform standard probability calculations;
3. work with independent and correlated random variables;
4. correctly apply simple formal statistical techniques and interpret the results;
5. carry out introductory data analysis and simulations using a statistical computing package
## Module content
Axioms of probability
Independence
Bayes Theorem
Random variables and moments
Joint distributions (mainly discrete) and covariance
LLN and CLT
Statistical models
Estimators (including what it means to be unbiased)
Confidence intervals for mean of a normal distribution (variance known/unknown)
## Indicative assessment
Closed/in-person Exam (Centrally scheduled) 70
Coursework - extensions not feasible/practicable 10
Essay/coursework 20
### Special assessment rules
None
If a student has a failing module mark, only failed components need be reassessed.
NB the 10% “written solution” element of the coursework is merged with the exam, so if a student fails the module they resit the exam for 80%. If they have passed the 20% computer-based coursework they may keep that mark, or they may choose to resit it.
Note:
Due to the pedagogical desire to provide speedy feedback in seminars, extensions to the written coursework and computer exercises are not possible. (This is the current practice in this module).
To mitigate for exceptional circumstances, the written coursework grade will be the best 4 out of the 5 assignments. If more than one assignment is affected by exceptional circumstances, an ECA claim must be submitted (with evidence).
Similarly, the computational grade will be the best 4 out of the 5 exercises. If more than one exercise is affected by exceptional circumstances, an ECA claim must be submitted (with evidence).
For extreme exceptional circumstances cases, the 10% coursework component can be discounted, with the exam mark making up 80% of the module
### Indicative reassessment
Closed/in-person Exam (Centrally scheduled) 80
Essay/coursework 20
## Module feedback
Current Department policy on feedback is available in the student handbook. Coursework and examinations will be marked and returned in accordance with this policy. | 669 | 3,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.759117 |
https://thepoorcoder.com/hackerrank-java-primality-test-solution/ | 1,670,259,671,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00061.warc.gz | 602,994,686 | 12,175 | # Hackerrank Java Primality Test Solution
A prime number is a natural number greater than whose only positive divisors are and itself. For example, the first six prime numbers are , , , , , and .
Given a large integer, , use the Java BigInteger class' isProbablePrime method to determine and print whether it's prime or not prime.
Input Format
A single line containing an integer, (the number to be checked).
Constraints
• contains at most digits.
Output Format
If is a prime number, print prime; otherwise, print not prime.
Sample Input
13
Sample Output
prime
Explanation
The only positive divisors of are and , so we print prime.
### Solution in java8
Approach 1.
import java.util.Scanner;
public class Solution
{
public static void main(String[] args)
{
try (Scanner scanner = new Scanner(System.in);)
{
System.out.println(scanner.nextBigInteger().isProbablePrime(100) ? "prime" : "not prime");
}
}
}
Approach 2.
import java.util.Scanner;
import java.math.BigInteger;
public class Solution {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
BigInteger n = scan.nextBigInteger();
scan.close();
System.out.println(n.isProbablePrime(10) ? "prime" : "not prime");
}
}
Approach 3.
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
BigInteger n = in.nextBigInteger();
in.close(); | 351 | 1,499 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-49 | latest | en | 0.422453 |
https://kr.mathworks.com/matlabcentral/answers/877573-filtered-signal-not-matching-with-the-original-signal | 1,632,215,429,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057199.49/warc/CC-MAIN-20210921070944-20210921100944-00702.warc.gz | 398,200,252 | 29,831 | # FIltered signal not matching with the original signal.
조회 수: 5(최근 30일)
indigo fgo 2021년 7월 13일
댓글: Mathieu NOE 2021년 7월 15일
Hello, im new to filter design, as you can probably tell. I have a signal from a wave guage measuring depth with time, The signal has very high low end frequencies so i am trying to apply a short low pass filter to smooth the signal, im not sure if it is my kernal or order choice but when i plot the original signal with the filtered signal, the filtered signal has lost magnitude. Its was originally intended to be a band stop filter so might not be consistant the whole way through. At this stage im completely stuck and any healp would greatly apreciated.
Thank you so much for your time,
clc
srate = 128;%Hz Sample Rate
nyquist = srate/2;
coff = [7];%Cut off or range
transw = .1;% randomly set
shape = [1 1 0 0];
frex = [1 coff(1)-coff(1)*transw coff(1) nyquist]/nyquist;
hz = linspace(0,srate,8321);%8321 is the number of samples in the data
% Loop for best matching order
order1 = (.2*srate/coff(1))/(srate/1000);%randomly set as I was trying different orders
order2 = (.3*srate/coff(1))/(srate/1000);
orders = round(linspace(order1,order2,2));%step count
fkernx = zeros(length(orders),8321);
figure(1)
for k = 1:length(orders);
fkern = firls(orders(k),frex,shape);
n(k) = length(fkern);
fkernx(k,:) = abs(fft(fkern,8321)).^2;
subplot(2,1,1)
hold on
plot((1:n(k))-n(k)/2,fkern+.1*k)
end
subplot(2,2,3)
hold on
plot(hz,fkernx)
plot(frex*nyquist,shape,'ro-')
xlim([0 64])
% % Loop for best matching transwidth
transwidth = linspace(.1,.4,4);
fkerxx = zeros(length(transwidth),8321);
orderl = round(order1);
figure(2)
for t = 1:length(transwidth);
frex2 = [1 coff(1)-coff(1)*transwidth(t) coff coff(2)+coff(2)*transwidth(t) nyquist]/nyquist;
fkern2 = firls(2,frex2,shape);
n(t) = length(fkern2);
fkernxx(t,:) = abs(fft(fkern2,8321).^2);
subplot(2,1,1)
hold on
plot((1:n(t))-n(t)/2,fkern2+0.1*t)
end
subplot(2,1,2)
hold on
plot(hz,fkernxx)
plot(frex2*nyquist,shape,'ro-')
%Final Filter Design
orderx = round((.3*srate/coff(1))-(srate/1000));%Use best matchig order here
transwidthx = 0.1;
frex3 = [1 coff(1)-coff(1)*transw coff nyquist]/nyquist;
filt28 = firls(orderx,frex3,shape);
%Apply filter design to signal (PDg1)
PDg1s = filtfilt(filt28,1,PDg1);
detfftp1 = detrend(PDg1s);
fftp1 = abs(fft(detfftp1));
%Frequency domain after filter
figure(3)
semilogy(hz, fftp1)
figure(4)
hold on
plot(PT5, PDg1,'r');
plot(PT5, PDg1s,'b')%plot smoothed data
legend('Original','Filtered')
xlabel('TIme (s)')
ylabel('Depth (m)')
title('Depth Against Time')
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### 채택된 답변
Mathieu NOE 2021년 7월 13일
hello
I see that you are doing a detrend action on your data. This will remove the dc (mean) value of the signal , so no surprise that the output is shifted.
My suggestion :
T = csvread('ExampleData PDg1 and time.txt');
y = T(:,1);
Fs = 128;
dt = 1/Fs;
samples = length(y);
t = (0:samples-1)*dt;
%%% filter data %%%
N = 250;
y_LP = smoothdata(y, 'gaussian' , N); % low pass filtered data
%%% plot data %%%
figure(1)
plot(t,y,t,y_LP);legend('Raw','low pass filtered');
title(['Data samples at Fs = ' num2str(Fs) ' Hz / filtered with smoothdata' ]);
xlabel('Time (hours) ')
ylabel('Depth (m) ')
##### 댓글 수: 5표시숨기기 이전 댓글 수: 4
Mathieu NOE 2021년 7월 15일
hello
if my answer has helped you, do you consider accepting it ?
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Translated by | 1,215 | 3,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.748389 |
https://slidetodoc.com/the-central-limit-theorem-the-world-is-normal/ | 1,627,482,465,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00403.warc.gz | 546,635,879 | 10,920 | # THE CENTRAL LIMIT THEOREM The World is Normal
• Slides: 25
THE CENTRAL LIMIT THEOREM The World is Normal Theorem
Sampling Distribution of xnormally distributed population Sampling distribution of x: N( , / 10) n=10 / 10 Population distribution: N( , )
Normal Populations Important Fact: If the population is normally distributed, then the sampling distribution of x is normally distributed for any sample size n.
Non-normal Populations �What can we say about the shape of the sampling distribution of x when the population from which the sample is selected is not normal? Population: interarrival times between consecutive customers at an ATM f(x) 0 time
The Central Limit Theorem (for the sample mean x) If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal. (The larger the sample size, the better will be the normal approximation to the sampling distribution of x. )
The Importance of the Central Limit Theorem When we select simple random samples of size n, the sample means we find will vary from sample to sample. We can model the distribution of these sample means with a probability model that is
How Large Should n Be? For the purpose of applying the central limit theorem, we will consider a sample size to be large when n > 30.
Summary Population: mean ; stand dev. ; shape of population dist. is unknown; value of is unknown; select random sample of size n; Sampling distribution of x: mean ; stand. dev. / n; always true! By the Central Limit Theorem: the shape of the sampling distribution is approx normal, that is x ~ N( , / n)
The Central Limit Theorem (for the sample proportion p) If a random sample of n observations is selected from a population (any population), and x “successes” are observed, then when n is sufficiently large, the sampling distribution of the sample proportion p will be approximately a normal distribution.
The Importance of the Central Limit Theorem When we select simple random samples of size n, the sample proportions p that we obtain will vary from sample to sample. We can model the distribution of these sample proportions with a probability model that is
How Large Should n Be? For the purpose of applying the central limit theorem, we will consider a sample size to be large when np > 10 and nq > 10
Population Parameters and Sample Statistics Population parameter Value p proportion of population with a certain characteristic Unknown µ mean value of a population variable Unknown Sample statistic used to estimate The value of a population parameter is a fixed number, it is NOT random; its value is not known. The value of a sample statistic is calculated from sample data The value of a sample statistic will vary from sample to sample (sampling distributions)
Example
Graphically Sampling distribution of x: N( , / n) = N(15, 4/8) n=64 x = 4/ 64 = 4/8 Population distribution: =4 = 15, = 4 = 15
Example (cont. )
Example 2 The probability distribution of 6 -month incomes of account executives has mean \$20, 000 and standard deviation \$5, 000. a) A single executive’s income is \$20, 000. Can it be said that this executive’s income exceeds 50% of all account executive incomes? ANSWER No. P(X<\$20, 000)=? No information given about distribution of X
Example 2(cont. ) b) n=64 account executives are randomly selected. What is the probability that the sample mean exceeds \$20, 500?
Example 3 A sample of size n=16 is drawn from a normally distributed population with mean E(x)=20 and SD(x)=8.
Example 3 (cont. ) c. Do we need the Central Limit Theorem to solve part a or part b? NO. We are given that the population is normal, so the sampling distribution of the mean will also be normal for any sample size n. The CLT is not needed.
Example 4 Battery life X~N(20, 10). Guarantee: avg. battery life in a case of 24 exceeds 16 hrs. Find the probability that a randomly selected case meets the guarantee.
Example 5 Cans of salmon are supposed to have a net weight of 6 oz. The canner says that the net weight is a random variable with mean =6. 05 oz. and stand. dev. =. 18 oz. Suppose you take a random sample of 36 cans and calculate the sample mean weight to be 5. 97 oz. Find the probability that the mean weight of the sample is less than or equal to 5. 97 oz.
Population X: amount of salmon in a can E(x)=6. 05 oz, SD(x) =. 18 oz § X sampling dist: E(x)=6. 05 SD(x)=. 18/6=. 03 § By the CLT, X sampling dist is approx. normal § P(X 5. 97) = P(z [5. 97 -6. 05]/. 03) =P(z -. 08/. 03)=P(z -2. 67)=. 0038 § How could you use this answer?
Suppose you work for a “consumer watchdog” group If you sampled the weights of 36 cans and obtained a sample mean x 5. 97 oz. , what would you think? Since P( x 5. 97) =. 0038, either you observed a “rare” event (recall: 5. 97 oz is 2. 67 stand. dev. below the mean) and the mean fill E(x) is in fact 6. 05 oz. (the value claimed by the canner) the true mean fill is less than 6. 05 oz. , (the canner is lying ).
Example 6 X: weekly income. E(x)=600, SD(x) = 100 n=25; X sampling dist: E(x)=600 SD(x)=100/5=20 P(X 550)=P(z [550 -600]/20) =P(z -50/20)=P(z -2. 50) =. 0062 Suspicious of claim that average is \$600; evidence is that average income is less.
Example 7 § 12% of students at NCSU are left-handed. What is the probability that in a sample of 50 students, the sample proportion that are lefthanded is less than 11%? | 1,386 | 5,476 | {"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} | 4.40625 | 4 | CC-MAIN-2021-31 | latest | en | 0.865389 |
http://biosig-consulting.com/matlab/help/freetb4matlab/statistics/base/cov.html | 1,409,143,832,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829393.78/warc/CC-MAIN-20140820021349-00268-ip-10-180-136-8.ec2.internal.warc.gz | 19,945,998 | 1,808 | Home > freetb4matlab > statistics > base > cov.m
# cov
## PURPOSE
% Compute covariance.
## SYNOPSIS
function c = cov (x, y)
## DESCRIPTION
% -*- texinfo -*-
% @deftypefn {Function File} {} cov (@var{x}, @var{y})
% Compute covariance.
%
% If each row of @var{x} and @var{y} is an observation and each column is
% a variable, the (@var{i}, @var{j})-th entry of
% @code{cov (@var{x}, @var{y})} is the covariance between the @var{i}-th
% variable in @var{x} and the @var{j}-th variable in @var{y}.
% @iftex
% @tex
% $$% \sigma_{ij} = {1 \over N-1} \sum_{i=1}^N (x_i - \bar{x})(y_i - \bar{y}) %$$
% where $\bar{x}$ and $\bar{y}$ are the mean values of $x$ and $y$.
% @end tex
% @end iftex
% If called with one argument, compute @code{cov (@var{x}, @var{x})}.
% @end deftypefn
## CROSS-REFERENCE INFORMATION
This function calls:
• center % If @var{x} is a vector, subtract its mean.
This function is called by:
• cor % Compute correlation.
• corrcoef % Compute correlation.
Generated on Sat 16-May-2009 00:04:49 by m2html © 2003 | 347 | 1,032 | {"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} | 2.640625 | 3 | CC-MAIN-2014-35 | longest | en | 0.592081 |
https://docs.microsoft.com/it-it/dotnet/api/system.math.cos?view=netcore-2.2 | 1,590,550,811,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392057.6/warc/CC-MAIN-20200527013445-20200527043445-00569.warc.gz | 324,813,852 | 12,133 | # Math.Cos(Double) Metodo
## Definizione
Restituisce il coseno dell'angolo specificato.Returns the cosine of the specified angle.
``````public:
static double Cos(double d);``````
``public static double Cos (double d);``
``static member Cos : double -> double``
``Public Shared Function Cos (d As Double) As Double``
d
Double
#### Restituisce
Coseno di `d`.The cosine of `d`. Se `d` è uguale a NaN, NegativeInfinity o PositiveInfinity, questo metodo restituisce NaN.If `d` is equal to NaN, NegativeInfinity, or PositiveInfinity, this method returns NaN.
## Esempio
Nell'esempio seguente viene usato Cos per valutare alcune identità trigonometriche per gli angoli selezionati.The following example uses Cos to evaluate certain trigonometric identities for selected angles.
``````// Example for the trigonometric Math.Sin( double )
// and Math.Cos( double ) methods.
using namespace System;
// Evaluate trigonometric identities with a given angle.
void UseSineCosine( double degrees )
{
double angle = Math::PI * degrees / 180.0;
double sinAngle = Math::Sin( angle );
double cosAngle = Math::Cos( angle );
// Evaluate sin^2(X) + cos^2(X) == 1.
Console::WriteLine( "\n Math::Sin({0} deg) == {1:E16}\n"
" Math::Cos({0} deg) == {2:E16}", degrees, Math::Sin( angle ), Math::Cos( angle ) );
Console::WriteLine( "(Math::Sin({0} deg))^2 + (Math::Cos({0} deg))^2 == {1:E16}", degrees, sinAngle * sinAngle + cosAngle * cosAngle );
// Evaluate sin(2 * X) == 2 * sin(X) * cos(X).
Console::WriteLine( " Math::Sin({0} deg) == {1:E16}", 2.0 * degrees, Math::Sin( 2.0 * angle ) );
Console::WriteLine( " 2 * Math::Sin({0} deg) * Math::Cos({0} deg) == {1:E16}", degrees, 2.0 * sinAngle * cosAngle );
// Evaluate cos(2 * X) == cos^2(X) - sin^2(X).
Console::WriteLine( " Math::Cos({0} deg) == {1:E16}", 2.0 * degrees, Math::Cos( 2.0 * angle ) );
Console::WriteLine( "(Math::Cos({0} deg))^2 - (Math::Sin({0} deg))^2 == {1:E16}", degrees, cosAngle * cosAngle - sinAngle * sinAngle );
}
// Evaluate trigonometric identities that are functions of two angles.
void UseTwoAngles( double degreesX, double degreesY )
{
double angleX = Math::PI * degreesX / 180.0;
double angleY = Math::PI * degreesY / 180.0;
// Evaluate sin(X + Y) == sin(X) * cos(Y) + cos(X) * sin(Y).
Console::WriteLine( "\n Math::Sin({0} deg) * Math::Cos({1} deg) +\n"
" Math::Cos({0} deg) * Math::Sin({1} deg) == {2:E16}", degreesX, degreesY, Math::Sin( angleX ) * Math::Cos( angleY ) + Math::Cos( angleX ) * Math::Sin( angleY ) );
Console::WriteLine( " Math::Sin({0} deg) == {1:E16}", degreesX + degreesY, Math::Sin( angleX + angleY ) );
// Evaluate cos(X + Y) == cos(X) * cos(Y) - sin(X) * sin(Y).
Console::WriteLine( " Math::Cos({0} deg) * Math::Cos({1} deg) -\n"
" Math::Sin({0} deg) * Math::Sin({1} deg) == {2:E16}", degreesX, degreesY, Math::Cos( angleX ) * Math::Cos( angleY ) - Math::Sin( angleX ) * Math::Sin( angleY ) );
Console::WriteLine( " Math::Cos({0} deg) == {1:E16}", degreesX + degreesY, Math::Cos( angleX + angleY ) );
}
int main()
{
Console::WriteLine( "This example of trigonometric "
"Math::Sin( double ) and Math::Cos( double )\n"
"generates the following output.\n" );
Console::WriteLine( "Convert selected values for X to radians \n"
"and evaluate these trigonometric identities:" );
Console::WriteLine( " sin^2(X) + cos^2(X) == 1\n"
" sin(2 * X) == 2 * sin(X) * cos(X)" );
Console::WriteLine( " cos(2 * X) == cos^2(X) - sin^2(X)" );
UseSineCosine( 15.0 );
UseSineCosine( 30.0 );
UseSineCosine( 45.0 );
Console::WriteLine( "\nConvert selected values for X and Y to radians \n"
"and evaluate these trigonometric identities:" );
Console::WriteLine( " sin(X + Y) == sin(X) * cos(Y) + cos(X) * sin(Y)" );
Console::WriteLine( " cos(X + Y) == cos(X) * cos(Y) - sin(X) * sin(Y)" );
UseTwoAngles( 15.0, 30.0 );
UseTwoAngles( 30.0, 45.0 );
}
/*
This example of trigonometric Math::Sin( double ) and Math::Cos( double )
generates the following output.
Convert selected values for X to radians
and evaluate these trigonometric identities:
sin^2(X) + cos^2(X) == 1
sin(2 * X) == 2 * sin(X) * cos(X)
cos(2 * X) == cos^2(X) - sin^2(X)
Math::Sin(15 deg) == 2.5881904510252074E-001
Math::Cos(15 deg) == 9.6592582628906831E-001
(Math::Sin(15 deg))^2 + (Math::Cos(15 deg))^2 == 1.0000000000000000E+000
Math::Sin(30 deg) == 4.9999999999999994E-001
2 * Math::Sin(15 deg) * Math::Cos(15 deg) == 4.9999999999999994E-001
Math::Cos(30 deg) == 8.6602540378443871E-001
(Math::Cos(15 deg))^2 - (Math::Sin(15 deg))^2 == 8.6602540378443871E-001
Math::Sin(30 deg) == 4.9999999999999994E-001
Math::Cos(30 deg) == 8.6602540378443871E-001
(Math::Sin(30 deg))^2 + (Math::Cos(30 deg))^2 == 1.0000000000000000E+000
Math::Sin(60 deg) == 8.6602540378443860E-001
2 * Math::Sin(30 deg) * Math::Cos(30 deg) == 8.6602540378443860E-001
Math::Cos(60 deg) == 5.0000000000000011E-001
(Math::Cos(30 deg))^2 - (Math::Sin(30 deg))^2 == 5.0000000000000022E-001
Math::Sin(45 deg) == 7.0710678118654746E-001
Math::Cos(45 deg) == 7.0710678118654757E-001
(Math::Sin(45 deg))^2 + (Math::Cos(45 deg))^2 == 1.0000000000000000E+000
Math::Sin(90 deg) == 1.0000000000000000E+000
2 * Math::Sin(45 deg) * Math::Cos(45 deg) == 1.0000000000000000E+000
Math::Cos(90 deg) == 6.1230317691118863E-017
(Math::Cos(45 deg))^2 - (Math::Sin(45 deg))^2 == 2.2204460492503131E-016
Convert selected values for X and Y to radians
and evaluate these trigonometric identities:
sin(X + Y) == sin(X) * cos(Y) + cos(X) * sin(Y)
cos(X + Y) == cos(X) * cos(Y) - sin(X) * sin(Y)
Math::Sin(15 deg) * Math::Cos(30 deg) +
Math::Cos(15 deg) * Math::Sin(30 deg) == 7.0710678118654746E-001
Math::Sin(45 deg) == 7.0710678118654746E-001
Math::Cos(15 deg) * Math::Cos(30 deg) -
Math::Sin(15 deg) * Math::Sin(30 deg) == 7.0710678118654757E-001
Math::Cos(45 deg) == 7.0710678118654757E-001
Math::Sin(30 deg) * Math::Cos(45 deg) +
Math::Cos(30 deg) * Math::Sin(45 deg) == 9.6592582628906831E-001
Math::Sin(75 deg) == 9.6592582628906820E-001
Math::Cos(30 deg) * Math::Cos(45 deg) -
Math::Sin(30 deg) * Math::Sin(45 deg) == 2.5881904510252085E-001
Math::Cos(75 deg) == 2.5881904510252096E-001
*/
``````
``````// Example for the trigonometric Math.Sin( double )
// and Math.Cos( double ) methods.
using System;
class SinCos
{
public static void Main()
{
Console.WriteLine(
"This example of trigonometric " +
"Math.Sin( double ) and Math.Cos( double )\n" +
"generates the following output.\n" );
Console.WriteLine(
"Convert selected values for X to radians \n" +
"and evaluate these trigonometric identities:" );
Console.WriteLine( " sin^2(X) + cos^2(X) == 1\n" +
" sin(2 * X) == 2 * sin(X) * cos(X)" );
Console.WriteLine( " cos(2 * X) == cos^2(X) - sin^2(X)" );
UseSineCosine(15.0);
UseSineCosine(30.0);
UseSineCosine(45.0);
Console.WriteLine(
"\nConvert selected values for X and Y to radians \n" +
"and evaluate these trigonometric identities:" );
Console.WriteLine( " sin(X + Y) == sin(X) * cos(Y) + cos(X) * sin(Y)" );
Console.WriteLine( " cos(X + Y) == cos(X) * cos(Y) - sin(X) * sin(Y)" );
UseTwoAngles(15.0, 30.0);
UseTwoAngles(30.0, 45.0);
}
// Evaluate trigonometric identities with a given angle.
static void UseSineCosine(double degrees)
{
double angle = Math.PI * degrees / 180.0;
double sinAngle = Math.Sin(angle);
double cosAngle = Math.Cos(angle);
// Evaluate sin^2(X) + cos^2(X) == 1.
Console.WriteLine(
"\n Math.Sin({0} deg) == {1:E16}\n" +
" Math.Cos({0} deg) == {2:E16}",
degrees, Math.Sin(angle), Math.Cos(angle) );
Console.WriteLine(
"(Math.Sin({0} deg))^2 + (Math.Cos({0} deg))^2 == {1:E16}",
degrees, sinAngle * sinAngle + cosAngle * cosAngle );
// Evaluate sin(2 * X) == 2 * sin(X) * cos(X).
Console.WriteLine(
" Math.Sin({0} deg) == {1:E16}",
2.0 * degrees, Math.Sin(2.0 * angle) );
Console.WriteLine(
" 2 * Math.Sin({0} deg) * Math.Cos({0} deg) == {1:E16}",
degrees, 2.0 * sinAngle * cosAngle );
// Evaluate cos(2 * X) == cos^2(X) - sin^2(X).
Console.WriteLine(
" Math.Cos({0} deg) == {1:E16}",
2.0 * degrees, Math.Cos(2.0 * angle) );
Console.WriteLine(
"(Math.Cos({0} deg))^2 - (Math.Sin({0} deg))^2 == {1:E16}",
degrees, cosAngle * cosAngle - sinAngle * sinAngle );
}
// Evaluate trigonometric identities that are functions of two angles.
static void UseTwoAngles(double degreesX, double degreesY)
{
double angleX = Math.PI * degreesX / 180.0;
double angleY = Math.PI * degreesY / 180.0;
// Evaluate sin(X + Y) == sin(X) * cos(Y) + cos(X) * sin(Y).
Console.WriteLine(
"\n Math.Sin({0} deg) * Math.Cos({1} deg) +\n" +
" Math.Cos({0} deg) * Math.Sin({1} deg) == {2:E16}",
degreesX, degreesY, Math.Sin(angleX) * Math.Cos(angleY) +
Math.Cos(angleX) * Math.Sin(angleY));
Console.WriteLine(
" Math.Sin({0} deg) == {1:E16}",
degreesX + degreesY, Math.Sin(angleX + angleY));
// Evaluate cos(X + Y) == cos(X) * cos(Y) - sin(X) * sin(Y).
Console.WriteLine(
" Math.Cos({0} deg) * Math.Cos({1} deg) -\n" +
" Math.Sin({0} deg) * Math.Sin({1} deg) == {2:E16}",
degreesX, degreesY, Math.Cos(angleX) * Math.Cos(angleY) -
Math.Sin(angleX) * Math.Sin(angleY));
Console.WriteLine(
" Math.Cos({0} deg) == {1:E16}",
degreesX + degreesY, Math.Cos(angleX + angleY));
}
}
/*
This example of trigonometric Math.Sin( double ) and Math.Cos( double )
generates the following output.
Convert selected values for X to radians
and evaluate these trigonometric identities:
sin^2(X) + cos^2(X) == 1
sin(2 * X) == 2 * sin(X) * cos(X)
cos(2 * X) == cos^2(X) - sin^2(X)
Math.Sin(15 deg) == 2.5881904510252074E-001
Math.Cos(15 deg) == 9.6592582628906831E-001
(Math.Sin(15 deg))^2 + (Math.Cos(15 deg))^2 == 1.0000000000000000E+000
Math.Sin(30 deg) == 4.9999999999999994E-001
2 * Math.Sin(15 deg) * Math.Cos(15 deg) == 4.9999999999999994E-001
Math.Cos(30 deg) == 8.6602540378443871E-001
(Math.Cos(15 deg))^2 - (Math.Sin(15 deg))^2 == 8.6602540378443871E-001
Math.Sin(30 deg) == 4.9999999999999994E-001
Math.Cos(30 deg) == 8.6602540378443871E-001
(Math.Sin(30 deg))^2 + (Math.Cos(30 deg))^2 == 1.0000000000000000E+000
Math.Sin(60 deg) == 8.6602540378443860E-001
2 * Math.Sin(30 deg) * Math.Cos(30 deg) == 8.6602540378443860E-001
Math.Cos(60 deg) == 5.0000000000000011E-001
(Math.Cos(30 deg))^2 - (Math.Sin(30 deg))^2 == 5.0000000000000022E-001
Math.Sin(45 deg) == 7.0710678118654746E-001
Math.Cos(45 deg) == 7.0710678118654757E-001
(Math.Sin(45 deg))^2 + (Math.Cos(45 deg))^2 == 1.0000000000000000E+000
Math.Sin(90 deg) == 1.0000000000000000E+000
2 * Math.Sin(45 deg) * Math.Cos(45 deg) == 1.0000000000000000E+000
Math.Cos(90 deg) == 6.1230317691118863E-017
(Math.Cos(45 deg))^2 - (Math.Sin(45 deg))^2 == 2.2204460492503131E-016
Convert selected values for X and Y to radians
and evaluate these trigonometric identities:
sin(X + Y) == sin(X) * cos(Y) + cos(X) * sin(Y)
cos(X + Y) == cos(X) * cos(Y) - sin(X) * sin(Y)
Math.Sin(15 deg) * Math.Cos(30 deg) +
Math.Cos(15 deg) * Math.Sin(30 deg) == 7.0710678118654746E-001
Math.Sin(45 deg) == 7.0710678118654746E-001
Math.Cos(15 deg) * Math.Cos(30 deg) -
Math.Sin(15 deg) * Math.Sin(30 deg) == 7.0710678118654757E-001
Math.Cos(45 deg) == 7.0710678118654757E-001
Math.Sin(30 deg) * Math.Cos(45 deg) +
Math.Cos(30 deg) * Math.Sin(45 deg) == 9.6592582628906831E-001
Math.Sin(75 deg) == 9.6592582628906820E-001
Math.Cos(30 deg) * Math.Cos(45 deg) -
Math.Sin(30 deg) * Math.Sin(45 deg) == 2.5881904510252085E-001
Math.Cos(75 deg) == 2.5881904510252096E-001
*/
``````
``````' Example for the trigonometric Math.Sin( Double ) and Math.Cos( Double ) methods.
Module SinCos
Sub Main()
Console.WriteLine( _
"This example of trigonometric " & _
"Math.Sin( double ) and Math.Cos( double )" & vbCrLf & _
"generates the following output." & vbCrLf)
Console.WriteLine( _
"Convert selected values for X to radians " & vbCrLf & _
"and evaluate these trigonometric identities:")
Console.WriteLine( _
" sin^2(X) + cos^2(X) = 1" & vbCrLf & _
" sin(2 * X) = 2 * sin(X) * cos(X)")
Console.WriteLine(" cos(2 * X) = cos^2(X) - sin^2(X)")
UseSineCosine(15.0)
UseSineCosine(30.0)
UseSineCosine(45.0)
Console.WriteLine( _
vbCrLf & "Convert selected values for X and Y to radians" & _
vbCrLf & "and evaluate these trigonometric identities:")
Console.WriteLine(" sin(X + Y) = sin(X) * cos(Y) + cos(X) * sin(Y)")
Console.WriteLine(" cos(X + Y) = cos(X) * cos(Y) - sin(X) * sin(Y)")
UseTwoAngles(15.0, 30.0)
UseTwoAngles(30.0, 45.0)
End Sub
' Evaluate trigonometric identities with a given angle.
Sub UseSineCosine(degrees As Double)
Dim angle As Double = Math.PI * degrees / 180.0
Dim sinAngle As Double = Math.Sin(angle)
Dim cosAngle As Double = Math.Cos(angle)
' Evaluate sin^2(X) + cos^2(X) = 1.
Console.WriteLine( _
vbCrLf & " Math.Sin({0} deg) = {1:E16}" & _
vbCrLf & " Math.Cos({0} deg) = {2:E16}", _
degrees, Math.Sin(angle), Math.Cos(angle))
Console.WriteLine( _
"(Math.Sin({0} deg))^2 + (Math.Cos({0} deg))^2 = {1:E16}", _
degrees, sinAngle * sinAngle + cosAngle * cosAngle)
' Evaluate sin(2 * X) = 2 * sin(X) * cos(X).
Console.WriteLine( _
" Math.Sin({0} deg) = {1:E16}", _
2.0 * degrees, Math.Sin(2.0 * angle))
Console.WriteLine( _
" 2 * Math.Sin({0} deg) * Math.Cos({0} deg) = {1:E16}", _
degrees, 2.0 * sinAngle * cosAngle)
' Evaluate cos(2 * X) = cos^2(X) - sin^2(X).
Console.WriteLine( _
" Math.Cos({0} deg) = {1:E16}", _
2.0 * degrees, Math.Cos(2.0 * angle))
Console.WriteLine( _
"(Math.Cos({0} deg))^2 - (Math.Sin({0} deg))^2 = {1:E16}", _
degrees, cosAngle * cosAngle - sinAngle * sinAngle)
End Sub
' Evaluate trigonometric identities that are functions of two angles.
Sub UseTwoAngles(degreesX As Double, degreesY As Double)
Dim angleX As Double = Math.PI * degreesX / 180.0
Dim angleY As Double = Math.PI * degreesY / 180.0
' Evaluate sin(X + Y) = sin(X) * cos(Y) + cos(X) * sin(Y).
Console.WriteLine( _
vbCrLf & " Math.Sin({0} deg) * Math.Cos({1} deg) +" & _
vbCrLf & " Math.Cos({0} deg) * Math.Sin({1} deg) = {2:E16}", _
degreesX, degreesY, Math.Sin(angleX) * Math.Cos(angleY) + _
Math.Cos(angleX) * Math.Sin(angleY))
Console.WriteLine( _
" Math.Sin({0} deg) = {1:E16}", _
degreesX + degreesY, Math.Sin(angleX + angleY))
' Evaluate cos(X + Y) = cos(X) * cos(Y) - sin(X) * sin(Y).
Console.WriteLine( _
" Math.Cos({0} deg) * Math.Cos({1} deg) -" & vbCrLf & _
" Math.Sin({0} deg) * Math.Sin({1} deg) = {2:E16}", _
degreesX, degreesY, Math.Cos(angleX) * Math.Cos(angleY) - _
Math.Sin(angleX) * Math.Sin(angleY))
Console.WriteLine( _
" Math.Cos({0} deg) = {1:E16}", _
degreesX + degreesY, Math.Cos(angleX + angleY))
End Sub
End Module 'SinCos
' This example of trigonometric Math.Sin( double ) and Math.Cos( double )
' generates the following output.
'
' Convert selected values for X to radians
' and evaluate these trigonometric identities:
' sin^2(X) + cos^2(X) = 1
' sin(2 * X) = 2 * sin(X) * cos(X)
' cos(2 * X) = cos^2(X) - sin^2(X)
'
' Math.Sin(15 deg) = 2.5881904510252074E-001
' Math.Cos(15 deg) = 9.6592582628906831E-001
' (Math.Sin(15 deg))^2 + (Math.Cos(15 deg))^2 = 1.0000000000000000E+000
' Math.Sin(30 deg) = 4.9999999999999994E-001
' 2 * Math.Sin(15 deg) * Math.Cos(15 deg) = 4.9999999999999994E-001
' Math.Cos(30 deg) = 8.6602540378443871E-001
' (Math.Cos(15 deg))^2 - (Math.Sin(15 deg))^2 = 8.6602540378443871E-001
'
' Math.Sin(30 deg) = 4.9999999999999994E-001
' Math.Cos(30 deg) = 8.6602540378443871E-001
' (Math.Sin(30 deg))^2 + (Math.Cos(30 deg))^2 = 1.0000000000000000E+000
' Math.Sin(60 deg) = 8.6602540378443860E-001
' 2 * Math.Sin(30 deg) * Math.Cos(30 deg) = 8.6602540378443860E-001
' Math.Cos(60 deg) = 5.0000000000000011E-001
' (Math.Cos(30 deg))^2 - (Math.Sin(30 deg))^2 = 5.0000000000000022E-001
'
' Math.Sin(45 deg) = 7.0710678118654746E-001
' Math.Cos(45 deg) = 7.0710678118654757E-001
' (Math.Sin(45 deg))^2 + (Math.Cos(45 deg))^2 = 1.0000000000000000E+000
' Math.Sin(90 deg) = 1.0000000000000000E+000
' 2 * Math.Sin(45 deg) * Math.Cos(45 deg) = 1.0000000000000000E+000
' Math.Cos(90 deg) = 6.1230317691118863E-017
' (Math.Cos(45 deg))^2 - (Math.Sin(45 deg))^2 = 2.2204460492503131E-016
'
' Convert selected values for X and Y to radians
' and evaluate these trigonometric identities:
' sin(X + Y) = sin(X) * cos(Y) + cos(X) * sin(Y)
' cos(X + Y) = cos(X) * cos(Y) - sin(X) * sin(Y)
'
' Math.Sin(15 deg) * Math.Cos(30 deg) +
' Math.Cos(15 deg) * Math.Sin(30 deg) = 7.0710678118654746E-001
' Math.Sin(45 deg) = 7.0710678118654746E-001
' Math.Cos(15 deg) * Math.Cos(30 deg) -
' Math.Sin(15 deg) * Math.Sin(30 deg) = 7.0710678118654757E-001
' Math.Cos(45 deg) = 7.0710678118654757E-001
'
' Math.Sin(30 deg) * Math.Cos(45 deg) +
' Math.Cos(30 deg) * Math.Sin(45 deg) = 9.6592582628906831E-001
' Math.Sin(75 deg) = 9.6592582628906820E-001
' Math.Cos(30 deg) * Math.Cos(45 deg) -
' Math.Sin(30 deg) * Math.Sin(45 deg) = 2.5881904510252085E-001
' Math.Cos(75 deg) = 2.5881904510252096E-001
``````
## Commenti
L'angolo, `d`, deve essere in radianti.The angle, `d`, must be in radians. Moltiplica per Math.PI/180 per convertire i gradi in radianti.Multiply by Math.PI/180 to convert degrees to radians.
I valori accettabili di `d` variano da circa-9223372036854775295 a circa 9223372036854775295.Acceptable values of `d` range from approximately -9223372036854775295 to approximately 9223372036854775295. Per i valori che non rientrano in questo intervallo, il metodo Cos restituisce `d` invariato anziché generare un'eccezione.For values outside this range, the Cos method returns `d` unchanged rather than throwing an exception. | 6,311 | 18,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-24 | latest | en | 0.239852 |
https://graduatednerds.com/2021/09/17/discussion-question-4-statistics-homework-help/ | 1,685,384,620,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00294.warc.gz | 318,476,863 | 12,962 | Home » Discussion question 4 | Statistics homework help
# Discussion question 4 | Statistics homework help
The answer lies in how clearly you articulate the criteria for selecting data sources; (b) your ability to purposefully select cases; and (c) the extent to which those cases are “information-rich… for in-depth study” (Patton, 2015, p. 264) with respect to the purpose of the study.
As you prepare for this week’s Discussion, consider turning your attention to the variety of purposeful sampling strategies you may consider in developing your research plan. Also consider that qualitative researchers seek a threshold or cut-off point for when to stop collecting data. There is no magic number (although there are guidelines). Rather, saturation occurs as an interface between the researcher and the data and (b) between data collection and data analysis to determine when enough is enough.
For this Discussion, you will critique a sampling strategy used in a research article.
To prepare for this Discussion:
• Review the Guest, Bunce, and Johnson article; the Yob and Brewer article; and the Learning Resources related to sampling and saturation for this week.
Prepare a critique of the sampling strategy used by Yob and Brewer (n.d.). Include the following your critique:
• The purpose of the study
• Research questions
• Site selection
• The type of purposeful sampling strategy the researchers applied. (Note: Use Table 4.3 in the Ravitch & Carl text or from Patton’s Chapter 5 to identify and describe the strategy that you think best fits what they described.)
• An alternative sampling strategy that the researchers could have considered. Explain your choice in terms of how the strategy is consistent with their research purpose and criteria for selecting cases.
• Provide a data saturation definition and evaluate the work of the researchers in this article regarding their efforts to achieve data saturation. Note what the researchers could have done differently to convince you that the relevant and important themes emerged.
Be sure to support your main post and response post with reference to the week’s Learning Resources and other scholarly evidence in APA style.
Ravitch, S. M., & Carl, N. M. (2016). Qualitative research: Bridging the conceptual, theoretical, and methodological. Thousand Oaks, CA: Sage Publications.
Chapter 4, “Design and Reflexivity in Data Collection” (pp. 111–144)
Table 4.3, “Purposeful Sampling Strategies” (pp. 129–137)
Rubin, H. J., & Rubin, I. S. (2012). Qualitative interviewing: The art of hearing data (3rd ed.). Thousand Oaks, CA: Sage Publications.
Chapter 3, “Qualitative Data-Gathering Methods and Style” (previously read in Week 3)
Patton, M. Q. (2015). Chapter 5, Module 30: Purposeful sampling and case selection: Overview of strategies and options. In Qualitative research and evaluation methods (4th ed., pp. 264–315). Thousand Oaks, CA: Sage Publications.
Guest, G., Bunce, A., & Johnson, L. (2006). How many interviews are enough? An experiment with data saturation and variability. Field Methods, 18(1), 59–82.
Retrieved from the Walden Library databases.
Mason, M. (2010). Sample size and saturation in PhD studies using qualitative interviews. Forum Qualitative Social Research Sozialforschung, 11(3).
Yob, I., & Brewer, P. (n.d.). Working toward the common good: An online university’s perspectives on social change, 1-25. (previously read in Week 1)
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By sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language. | 1,173 | 5,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-23 | latest | en | 0.902853 |
https://www.physicsforums.com/threads/need-help-with-understanding-this-solution-fluid-pressure-problem.1048134/ | 1,718,725,728,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00084.warc.gz | 830,569,150 | 17,905 | # Need help with understanding this solution (fluid pressure problem)
• bigmike94
In summary, the pressure is due to the oil and water mixture being at different levels, and the measurements are taken at the point where the oil meets the water.
bigmike94
TL;DR Summary
I dont fully understand the solution
So here’s the question (I am only talking about the pressure at point B, the other 2 I can understand.)
And here is the solution
Here is what I am not understanding, why is the pressure due to the water negative and the oil positive, and why are all measurements only made from where the oil meets the water?
i must be missing something really obvious
bigmike94 said:
why are all measurements only made from where the oil meets the water?
Because that's how we find the difference between ##p_{atm}## and ##p_B##
Going down contributes a positive ##\Delta p## and up a negative ##\Delta p##.
One can just as well take the bottom as the turning point: the ##\Delta p## of the extra 1.25 m down is cancelled by ##\Delta p## going up the same 1.25 m
##\ ##
Chestermiller and hutchphd
BvU said:
Because that's how we find the difference between ##p_{atm}## and ##p_B##
Going down contributes a positive ##\Delta p## and up a negative ##\Delta p##.
One can just as well take the bottom as the turning point: the ##\Delta p## of the extra 1.25 m down is cancelled by ##\Delta p## going up the same 1.25 m
##\ ##
Thank you that helped a lot. I attempt the problems quite a bit after I learnt about that method of finding the pressure. Hopefully I’ll remember it
BvU
I think of the problem as moving along a path from the external reference surface to the destination, at point B.
1. Start at the surface with atmospheric pressure. Specify absolute or gauge pressure for the reference.
2. Compute the increasing hydrostatic pressure as you move down through the oil, to the oil-water interface surface.
3. Compute the reducing hydrostatic pressure as you move up through the oil to the destination at point B, on the water-air interface.
bigmike94 said:
Here is what I am not understanding, why is the pressure due to the water negative and the oil positive, and why are all measurements only made from where the oil meets the water?
i must be missing something really obvious
This arrangement can be calculated like a U-shaped manometer, if you can imagine the bend located just underneath the central partition.
https://pressbooks.online.ucf.edu/osuniversityphysics/chapter/14-2-measuring-pressure/
## 1. What is fluid pressure?
Fluid pressure is defined as the force per unit area exerted by a fluid on its container or any object in contact with it. It is caused by the collisions of the fluid particles with the walls of the container or object.
## 2. How is fluid pressure calculated?
Fluid pressure can be calculated using the formula P = F/A, where P is the pressure, F is the force exerted by the fluid, and A is the area over which the force is applied. The SI unit for pressure is Pascal (Pa).
## 3. What is the difference between absolute and gauge pressure?
Absolute pressure is the total pressure exerted by a fluid, including atmospheric pressure. Gauge pressure, on the other hand, is the difference between the absolute pressure and the atmospheric pressure. It is often used to measure pressure changes in a system.
## 4. How does fluid pressure affect objects?
Fluid pressure can affect objects in different ways depending on their shape and orientation. For example, a flat surface will experience a greater force from fluid pressure compared to a curved surface. Additionally, objects will experience a net force in the direction of lower pressure.
## 5. How is fluid pressure used in real-world applications?
Fluid pressure has many practical applications in various industries, such as hydraulics, aerodynamics, and hydrology. It is used to design and operate systems like hydraulic lifts, aircraft wings, and dams. It is also important in understanding weather patterns and predicting natural disasters like hurricanes and floods.
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1K | 1,034 | 4,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.925787 |
https://codeforces.com/blog/entry/54094 | 1,685,815,684,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00569.warc.gz | 209,462,958 | 18,526 | ### skpro19's blog
By skpro19, history, 6 years ago,
The question is this.
The solution is something like this:
dp[1] = 1, dp[2] =3 ;
for(int i =3; i <= n ; i++) dp[i] = dp[i — 1] + dp[ i- 2] + 2;
How did one come with the formulation dp[i] = dp[ i — 1 ] + dp[i — 2] + 2 ?
Any help would be really appreciated.
Peace!
• 0
| Write comment?
» 6 years ago, # | ← Rev. 7 → +2 Its a result of this simple calculation... - we can write dp[i] as all the ways to select such that ith one is selected. So now we have dp[i]=dp[i-1]+dp[i-3]+dp[i-5]....1 using the same rule we can write...2 dp[i-1]=dp[i-2]+dp[i-4]+....3 on adding these two... we see that Result:dp[i]+dp[i-1]=dp[i-1]+dp[i-2]+dp[i-3]+...(all the ters upto 1). now dp[i] in eq 1 can also be written as dp[i]=1+(1+dp[i-2]+dp[i-4]+.....)+dp[i-3]+dp[i-4]... which is dp[i]=2+dp[i-1]+dp[i-2]...from the above result. I guess this proof suffices... you can also derive this by directly taking prefix sum in DP but this i guess is more understandable. The one is added because we are not considering single element in all the previous DP's | 374 | 1,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-23 | latest | en | 0.858611 |
https://mathoverflow.net/questions/69541/almost-direct-product-and-1-formality/75569 | 1,716,296,619,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00685.warc.gz | 329,175,227 | 26,058 | # Almost-direct product and 1-formality
Let $G$ be a finitely presented group. To $G$ is associated in a functorial way a Malcev Lie algebra which can be constructed in several equivalent ways. Roughly speaking, it is the quotient of the completed free Lie algebra on the same number of generators as $G$, by the (formal) logarithm of the defining relations of $G$ (see edit below).
$G$ is called 1-formal if its Malcev Lie algebra is isomorphic as a filtered Lie algebra to the degree completion of a finitely presented Lie algebra with quadratic relations. (If $X$ is a path-connected topological space such that $G=\pi_1(X)$, this is equivalent to say that the Malcev Lie algebra of $G$ is isomorphic to the holonomy Lie algebra of $X$)
It is quite well known that fundamental groups of complementary of hyperplane arrangements in $\mathbb{C}^n$ (e.g. pure braid groups) are 1-formal.
On the one hand, an important fact in the study of these groups is that they are (under some conditions on the underlying arrangement) iterated "almost-direct product" of free groups, and free groups are themselves 1-formal. An almost direct product is a semi-direct product $H\rtimes K$ for which the action of $K$ on the abelianization of $H$ is trivial. On the other hand, according to this interesting survey http://www.arxiv.com/abs/0903.2307 the direct product of two 1-formal groups is again 1-formal, even if no proof is given (it's probably easy..)
So it is temptating to ask:
is an almost-direct product of 1-formal groups again 1-formal ?
Edit: Some details about the construction and its relation with almost direct product. If $G$ is an abelian group, then one can take the tensor product $G \otimes_{\mathbb{Z}} \mathbb{Q}$ leading to a uniquely divisible abelian group. The Malcev construction extends this to any nilpotent group, leading to a uniquely divisible nilpotent group. Let $G^{(0)}=G$, $G^{(n+1)}=[G^{(n)},G]$ be the lower central serie, then the quotients $G/G^{(n)}$ are nilpotent by construction.
The Malcev completion of $G$ is the inverse limit of the $(G/G^{(n)}) \otimes_{\mathbb{Z}} \mathbb{Q}$. It is a pro-unipotent group, hence it has a (pro-nilpotent) Lie algebra $\mathfrak g$. On the other hand, one can define a Lie algebra by $$gr\ G=\mathbb{Q}\otimes_{\mathbb{Z}} \bigoplus G^{(n)}/G^{(n+1)}$$ the bracket being induced by the commutator in $G$. It as a graded Lie algebra, and in fact it is the associated graded of the filtered Lie algebra $\mathfrak g$.
A group is called 1-formal if there is an isomorphism of filtered Lie algebras $gr\ G \cong \mathfrak g$. Now, the point is that almost-direct product behave well with respect to the lower central series. If $G=G_1\rtimes G_2$ is an almost-direct product, then it seems to me that a result of Falk and Randell implies that we have $$G^{(n)}=G_1^{(n)}\rtimes G_2^{(n)}$$ and $$gr\ G = gr\ G_1 \rtimes gr\ G_2$$.
And no, the almost direct product of two 1-formal groups need not be 1-formal. A proof is given in the same paper, Example 8.2. The group in question is a semi-direct product of the form $G=F_4\rtimes F_1$, where $F_n$ is the free group of rank $n$, which is of course 1-formal. The action of $F_1$ on $F_4$ is given by a certain pure braid $\beta \in P_4$, acting via the Artin representation on $F_4$; thus, the action is trivial on $H_1(F_4)$. For this extension, the tangent cone formula" fails: the tangent cone to the characteristic variety $V_2(G)$ is strictly included in the resonance variety $R_2(G)$. In view of Theorem A from the cited paper, the group $G$ is not 1-formal.
It is worth noting that $G$ is the fundamental group of the complement of a certain link of 5 great circles in $S^3$. Alternatively, $G$ can be realized as the fundamental group of the complement of an arrangement of 5 planes in $\mathbb{R}^4$, meeting transversely at the origin (of course, this real arrangement cannot be isotoped to an arrangement of 5 complex lines in $\mathbb{C}^2$). For more details on the construction and properties of such arrangements, see arxiv:math.GT/9712251. In particular, the pure braid $\beta$ is described there in Propositions 4.4, 4.6, and 4.9.
• For instance, it is known that the full McCool group $P\Sigma_n$ (the group of basis-conjugating automorphisms of $F_n$) is 1-formal; see <a href="dx.doi.org/10.1142/S0218216509007257" rel="nofollow">this</a> paper by Berceanu and Papadima. But, as far as I know, it is an open question whether the upper-triangular McCool group $P\Sigma_n^+$ is 1-formal (for $n\ge 4$). Note that this group can be realized as an iterated almost-direct product of free groups, $P\Sigma_n^+=F_{n-1}\rtimes \cdots \rtimes F_2\rtimes F_1$, much the same as the pure braid group $P_n$. Sep 16, 2011 at 18:17
• Careful: the surface pure braid groups $P_{g,n}$ are always 1-formal, except when $g=1$ and $n\ge 3$, when they are not. A proof that $P_{1,n}$ is not 1-formal for $n\ge 3$ is given in arxiv.org/abs/0902.1250, Example 10.1, again using the tangent cone formula". Sep 18, 2011 at 2:08 | 1,479 | 5,053 | {"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} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.905177 |
https://gateoverflow.in/242726/isi2016-mma-4 | 1,579,423,745,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00098.warc.gz | 475,636,181 | 15,703 | 32 views
The $a, b, c$ and $d$ satisfy the equations$$\begin{matrix} a & + & 7b & + & 3c & + & 5d & = &16 \\ 8a & + & 4b & + & 6c & + & 2d & = &-16 \\ 2a & + & 6b & + & 4c & + & 8d & = &16 \\ 5a & + & 7b & + & 3c & + & 5d & = &-16 \end{matrix}$$Then $(a+d)(b+c)$ equals
1. $-4$
2. $0$
3. $16$
4. $-16$
recategorized | 32 views
+1 vote
Let the given equations be 1,2,3,4 in the order they are given.
Adding equation 2 and 3 -
10a + 10b + 10c + 10d = 0; Let this be equation 5
Adding equation 1 and 4 -
6a + 10b + 10c + 6d = 0; Let this be equation 6
Equation 5 minus 6 :
we get : 4a + 4d =0
so (a+d) =0
if (a+d) =0 then (a+d)(b+c)=0
by Active (2.3k points) | 322 | 679 | {"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} | 4.15625 | 4 | CC-MAIN-2020-05 | longest | en | 0.51036 |
http://self.gutenberg.org/articles/eng/G-test | 1,618,068,249,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057142.4/warc/CC-MAIN-20210410134715-20210410164715-00088.warc.gz | 90,318,454 | 20,175 | #jsDisabledContent { display:none; } My Account | Register | Help
# G-test
Article Id: WHEBN0000914820
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Title: G-test Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:
### G-test
In statistics, G-tests are likelihood-ratio or maximum likelihood statistical significance tests that are increasingly being used in situations where chi-squared tests were previously recommended.[1]
The general formula for G is
G = 2\sum_{i} {O_{i} \cdot \ln\left(\frac{O_i}{E_i}\right)},
where Oi is the observed frequency in a cell, Ei is the expected frequency under the null hypothesis, ln denotes the natural logarithm, and the sum is taken over all non-empty cells.
G-tests have been recommended at least since the 1981 edition of the popular statistics textbook by Robert R. Sokal and F. James Rohlf.[2]
## Contents
• Distribution and usage 1
• Relation to the chi-squared test 2
• Relation to Kullback–Leibler divergence 3
• Relation to mutual information 4
• Application 5
• Statistical software 6
• References 7
## Distribution and usage
Given the null hypothesis that the observed frequencies result from random sampling from a distribution with the given expected frequencies, the distribution of G is approximately a chi-squared distribution, with the same number of degrees of freedom as in the corresponding chi-squared test.
For very small samples the multinomial test for goodness of fit, and Fisher's exact test for contingency tables, or even Bayesian hypothesis selection are preferable to the G-test .
## Relation to the chi-squared test
The commonly used chi-squared tests for goodness of fit to a distribution and for independence in contingency tables are in fact approximations of the log-likelihood ratio on which the G-tests are based. The general formula for Pearson's chi-squared test statistic is
\chi^2 = \sum_{i} {\frac{\left(O_i - E_i\right)^2}{E_i}} .
The approximation of G by chi squared is obtained by a second order Taylor expansion of the natural logarithm around 1. This approximation was developed by Karl Pearson because at the time it was unduly laborious to calculate log-likelihood ratios. With the advent of electronic calculators and personal computers, this is no longer a problem. A derivation of how the chi-squared test is related to the G-test and likelihood ratios, including to a full Bayesian solution is provided in Hoey (2012).[3]
For samples of a reasonable size, the G-test and the chi-squared test will lead to the same conclusions. However, the approximation to the theoretical chi-squared distribution for the G-test is better than for the Pearson's chi-squared test.[4] In cases where O_i >2 \cdot E_i for some cell case the G-test is always better than the chi-squared test.
For testing goodness-of-fit the G-test is infinitely more efficient than the chi squared test in the sense of Bahadur, but the two tests are equally efficient in the sense of Pitman or in the sense of Hodge and Lehman.[5][6]
## Relation to Kullback–Leibler divergence
The G-test quantity is proportional to the Kullback–Leibler divergence of the empirical distribution from the theoretical distribution.
## Relation to mutual information
For analysis of contingency tables the value of G can also be expressed in terms of mutual information.
Let
N = \sum_{ij}{O_{ij}} \; , \; \pi_{ij} = \frac{O_{ij}}{N} \; , \; \pi_{i.} = \frac{\sum_j O_{ij}}{N} \; , and \; \pi_{. j} = \frac{\sum_i O_{ij}}{N} \;.
Then G can be expressed in several alternative forms:
G = 2 \cdot N \cdot \sum_{ij}{\pi_{ij} \left( \ln(\pi_{ij})-\ln(\pi_{i.})-\ln(\pi_{.j}) \right)} ,
G = 2 \cdot N \cdot \left[ H(r) + H(c) - H(r,c) \right] ,
G = 2 \cdot N \cdot MI(r,c) \, ,
where the entropy of a discrete random variable X \, is defined as
H(X) = - {\sum_{x \in \text{Supp}(X)} p(x) \log p(x)} \, ,
and where
MI(r,c)= H(r) + H(c) - H(r,c) \,
is the mutual information between the row vector r and the column vector c of the contingency table.
It can also be shown that the inverse document frequency weighting commonly used for text retrieval is an approximation of G applicable when the row sum for the query is much smaller than the row sum for the remainder of the corpus. Similarly, the result of Bayesian inference applied to a choice of single multinomial distribution for all rows of the contingency table taken together versus the more general alternative of a separate multinomial per row produces results very similar to the G statistic.
## Statistical software
• The R programming language has the likelihood.test function in the Deducer package.
• In SAS, one can conduct G-test by applying the /chisq option after the proc freq.[8]
• In Stata, one can conduct a G-test by applying the lr option after the tabulate command.
• Fisher's G-test in the GeneCycle Package of the R programming language (fisher.g.test) does not implement the G-test as described in this article, but rather Fisher's exact test of Gaussian white-noise in a time series.[9]
## References
1. ^ McDonald, J.H. (2014). "G–test of goodness-of-fit". Handbook of Biological Statistics (Third ed.). Baltimore, Maryland: Sparky House Publishing. pp. 53–58.
2. ^ Sokal, R. R.; Rohlf, F. J. (1981). Biometry: The Principles and Practice of Statistics in Biological Research (Second ed.). New York: Freeman.
3. ^ Hoey, J. (2012). "The Two-Way Likelihood Ratio (G) Test and Comparison to Two-Way Chi-Squared Test".
4. ^ Harremoës, P.; Tusnády, G. (2012). "Information divergence is more chi squared distributed than the chi squared statistic". Proceedings ISIT 2012. pp. 538–543.
5. ^ Quine, M. P.; Robinson, J. (1985). "Efficiencies of chi-square and likelihood ratio goodness-of-fit tests".
6. ^ Harremoës, P.; Vajda, I. (2008). "On the Bahadur-efficient testing of uniformity by means of the entropy".
7. ^ Dunning, Ted (1993). "Accurate Methods for the Statistics of Surprise and Coincidence", Computational Linguistics, Volume 19, issue 1 (March, 1993).
8. ^ G-test of independence, G-test for goodness-of-fit in Handbook of Biological Statistics, University of Delaware. (pp. 46–51, 64–69 in: McDonald, J. H. (2009) Handbook of Biological Statistics (2nd ed.). Sparky House Publishing, Baltimore, Maryland.)
9. ^ Fisher, R. A. (1929), "Tests of significance in harmonic analysis", Proceedings of the Royal Society of London: Series A, Volume 125, Issue 796, pp. 54–59. | 1,628 | 6,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-17 | latest | en | 0.869904 |
https://electronics.stackexchange.com/questions/118397/in-a-not-gate-circuit-does-charge-not-flow-through-both-the-transistor-and-outp | 1,717,093,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00716.warc.gz | 192,136,006 | 44,127 | # In a NOT gate circuit, does charge not flow through both the transistor and output line when the transistor is "on"?
I've tried to recreate a circuit from what I think would be a NOT gate, based on some other schematics and breadboard examples that I've seen.
simulate this circuit – Schematic created using CircuitLab
For this to be practical, I think SW1 would go away, and instead be substituted for some other voltage source as an input, but I've inserted SW1 and attached it to the same voltage source as everything else just to simplify the diagram...
Anyway, my assumption is that when there is charge going into the base of the transistor, the charge flows through the collector, out the emitter, and straight to ground (the negative end of the voltage source). So, in this case, does the charge not flow through R2 and through the LED, somehow? It seems to me that this conclusion must be correct, or else the LED would light up, right? But apparently, electricity does not take the "path of least resistance", so to speak.
Conversely, if the base of the transistor does not have any current, I assume the current stops at the collector of the transistor, but since the circuit must be complete, pressure moves the charge through R2 and the LED, then to ground.
I feel that I am missing some fundamental principle involved in this. Any thoughts?
• I assume a resistor is missing between the switch and the base of the transistor? Otherwise you'll force VBE = 9 V in the transistor where usually this should be 0.7 V or so. The circuit as is should let out the magic smoke. Jul 9, 2014 at 1:25
• @swineone, yeah, you're right, thanks for pointing that out. Just a "typo"; edited my question to reflect your correction. Jul 9, 2014 at 1:38
It is conventional to draw circuits so signals flow from left to right, like so:
simulate this circuit – Schematic created using CircuitLab
(You need a base resistor, R3, to limit base current, and a resistor from base to ground to ensure that the transistor does turn off when the switch is open)
With the switch open, the base will be held at ground by R4, so no current will flow through the transistor, so the collector will rise to a voltage determined by R1, R2, and the 2 volts or so voltage drop in the LED - this will allow current to flow through the LED and light it.
With the switch closed, the base will be pulled up, allowing current to flow through the transistor. If the resistors are selected correctly the transistor will be saturated, pulling the collector down to about 0.2 volts. As the LED requires about 2 volts across it, it will not pass any current.
I really dislike the sentence "Current always takes the path of least resistance", as many beginners seem to read it as "Current takes only the path of least resistance". In face, an electric current takes all possible paths, withthe lower resistance paths passing higher currents than the higher resistance paths.
• Thanks for the answer. What do you mean about needing a resistor from base to ground to ensure the transistor doesn't turn off when the switch is open? I thought you would want the transistor to be off (i.e. not allowing current to pass from collector to emitter) when the swith is open. Jul 9, 2014 at 11:13
• Also, what do you mean by "pulling" the collector down to 0.2v? Why does this make it so the LED cannot be forward biased? Jul 9, 2014 at 11:21
• @JoshBeam he says a resistor is needed to ensure the transistor DOES turn off. If you just open the switch the base is left floating, and I'd expected some accumulated charge there might not lead to an immediate turn-off (that should certainly be the case with a MOSFET rather than a BJT). As for your second question, see my answer below about the issue of biasing the LED. Jul 9, 2014 at 12:03
• Accepted this answer because of this: "an electric current takes all possible paths, withthe lower resistance paths passing higher currents than the higher resistance paths." I think that was the key to my whole misunderstanding. Jul 12, 2014 at 20:56
You are correct for the most part, but what you have here is essentially two switches that are working in inverted states.
If you drive the base of the transistor then you will draw about 29mA collector current from V1 and force, assuming saturation, the collector to be about 200mV (Vce sat). Current would flow through the diode if the voltage at that node was sufficient to forward bias the PN junction of the device, but that is not the case. So, what happens here is that when the transistor is ON, the diode switch is reverse biased (off), and when the transistor is OFF, it provides a high-resistance path (an open) and current will flow through the two series resistances and drive the diode.
simulate this circuit – Schematic created using CircuitLab
• Definitely suggest the pull-down base resistor as shown below in Peter's schematic. Jul 9, 2014 at 1:57
• Thanks for answering. When the transistor base has charge pushed through it, why is the voltage at the LED not sufficient to forward bias it? What is the threshold the voltage has to pass in order to forward bias, say, a red LED, and why can the voltage not pass that threshold only when the transistor has charge flowing through it? Jul 9, 2014 at 11:18
• @JoshBeam, That may be fit for an entirely new question. For information as to why the LED needs certain biasing conditions to conduct see here and the section just below it as well. Perhaps for a bit on insight as to why the collector voltage (which is driving your LED in this case) drops, see here. So, in short, the construction of the diode requires an E field to be applied across the device such that the inherent field is negated thereby allowing current flow. Jul 9, 2014 at 12:39
• .. This negation closes what is known as the depletion region between the P and N doped regions. When the diode is reverse biased, this region acts like a very high resistance, and when forward biased the region has some on-state resistance (perhaps a few ohms). In the case of your schematic, if the transistor saturates and Vce(sat) = 200mV, then you are not allowing sufficient biasing voltage for the LED and consequentially it does not conduct (i.e. depletion region is open and as such the path is very high resistance). Jul 9, 2014 at 12:42
• thanks. I think what I'm not understanding is, why does the saturation level of the transistor have anything whatsoever to do with enough voltage going to the LED (thus, disallowing a forward bias in the LED)? Jul 9, 2014 at 13:12
It's simple, really. When the transistor conducts, the current has to choose between going through the collector of Q1 to the emitter and then ground, or going through R2 and D1. The diode only allows current to flow when biased, which in practice (for the range of currents in which the LED is simultaneously visible yet not damaged) means a constant voltage drop in the LED, perhaps 1.5V to 2.5V depending on LED color and other properties. There's also an IR voltage drop in R2; say you must apply at least 1 mA for the LED to be visible, then you're looking at a 0.3 V voltage drop in R2 ($300 \Omega \times 1 \textrm{mA}$).
So, unless you apply somewhere between 1.8 V and 2.8 V in the branch of the circuit containing the LED, it'll basically look like an open circuit (very high resistance). Now the transistor conducts current much more easily, clamping the voltage at a Vce(sat) = 200 mV as mentioned by the other answer. At that voltage, the LED is not biased at all, so it does look like an open circuit, while the transistor looks almost like a perfect short circuit (save for the voltage drop of 200 mV). So indeed the current chooses the past of least resistance, which is the transistor in this case.
When the transistor is not conducting, then no amount of voltage applied to the collector will allow current to pass. On the other hand, when applying 1.8 to 2.8 V to the branch containing the LED, it will bias the LED and let current through. So again the path of least resistance is being chosen: the resistance through a transistor with no current flowing into the base is effectively infinite. | 1,883 | 8,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-22 | latest | en | 0.956274 |
http://www.popflock.com/learn?s=Bending | 1,631,912,418,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00608.warc.gz | 126,487,692 | 38,463 | Bending
Get Bending essential facts below. View Videos or join the Bending discussion. Add Bending to your PopFlock.com topic list for future reference or share this resource on social media.
Bending
Bending of an I-beam
In applied mechanics, bending (also known as flexure) characterizes the behavior of a slender structural element subjected to an external load applied perpendicularly to a longitudinal axis of the element.
The structural element is assumed to be such that at least one of its dimensions is a small fraction, typically 1/10 or less, of the other two.[1] When the length is considerably longer than the width and the thickness, the element is called a beam. For example, a closet rod sagging under the weight of clothes on clothes hangers is an example of a beam experiencing bending. On the other hand, a shell is a structure of any geometric form where the length and the width are of the same order of magnitude but the thickness of the structure (known as the 'wall') is considerably smaller. A large diameter, but thin-walled, short tube supported at its ends and loaded laterally is an example of a shell experiencing bending.
In the absence of a qualifier, the term bending is ambiguous because bending can occur locally in all objects. Therefore, to make the usage of the term more precise, engineers refer to a specific object such as; the bending of rods,[2] the bending of beams,[1] the bending of plates,[3] the bending of shells[2] and so on.
## Quasi-static bending of beams
A beam deforms and stresses develop inside it when a transverse load is applied on it. In the quasi-static case, the amount of bending deflection and the stresses that develop are assumed not to change over time. In a horizontal beam supported at the ends and loaded downwards in the middle, the material at the over-side of the beam is compressed while the material at the underside is stretched. There are two forms of internal stresses caused by lateral loads:
• Shear stress parallel to the lateral loading plus complementary shear stress on planes perpendicular to the load direction;
• Direct compressive stress in the upper region of the beam, and direct tensile stress in the lower region of the beam.
These last two forces form a couple or moment as they are equal in magnitude and opposite in direction. This bending moment resists the sagging deformation characteristic of a beam experiencing bending. The stress distribution in a beam can be predicted quite accurately when some simplifying assumptions are used.[1]
### Euler-Bernoulli bending theory
Element of a bent beam: the fibers form concentric arcs, the top fibers are compressed and bottom fibers stretched.
Bending moments in a beam
In the Euler-Bernoulli theory of slender beams, a major assumption is that 'plane sections remain plane'. In other words, any deformation due to shear across the section is not accounted for (no shear deformation). Also, this linear distribution is only applicable if the maximum stress is less than the yield stress of the material. For stresses that exceed yield, refer to article plastic bending. At yield, the maximum stress experienced in the section (at the furthest points from the neutral axis of the beam) is defined as the flexural strength.
Consider beams where the following are true:
• The beam is originally straight and slender, and any taper is slight
• The material is isotropic (or orthotropic), linear elastic, and homogeneous across any cross section (but not necessarily along its length)
• Only small deflections are considered
In this case, the equation describing beam deflection (${\displaystyle w}$) can be approximated as:
${\displaystyle {\cfrac {\mathrm {d} ^{2}w(x)}{\mathrm {d} x^{2}}}={\frac {M(x)}{E(x)I(x)}}}$
where the second derivative of its deflected shape with respect to ${\displaystyle x}$ is interpreted as its curvature, ${\displaystyle E}$ is the Young's modulus, ${\displaystyle I}$ is the area moment of inertia of the cross-section, and ${\displaystyle M}$ is the internal bending moment in the beam.
If, in addition, the beam is homogeneous along its length as well, and not tapered (i.e. constant cross section), and deflects under an applied transverse load ${\displaystyle q(x)}$, it can be shown that:[1]
${\displaystyle EI~{\cfrac {\mathrm {d} ^{4}w(x)}{\mathrm {d} x^{4}}}=q(x)}$
This is the Euler-Bernoulli equation for beam bending.
After a solution for the displacement of the beam has been obtained, the bending moment (${\displaystyle M}$) and shear force (${\displaystyle Q}$) in the beam can be calculated using the relations
${\displaystyle M(x)=-EI~{\cfrac {\mathrm {d} ^{2}w}{\mathrm {d} x^{2}}}~;~~Q(x)={\cfrac {\mathrm {d} M}{\mathrm {d} x}}.}$
Simple beam bending is often analyzed with the Euler-Bernoulli beam equation. The conditions for using simple bending theory are:[4]
1. The beam is subject to pure bending. This means that the shear force is zero, and that no torsional or axial loads are present.
2. The material is isotropic (or orthotropic) and homogeneous.
3. The material obeys Hooke's law (it is linearly elastic and will not deform plastically).
4. The beam is initially straight with a cross section that is constant throughout the beam length.
5. The beam has an axis of symmetry in the plane of bending.
6. The proportions of the beam are such that it would fail by bending rather than by crushing, wrinkling or sideways buckling.
7. Cross-sections of the beam remain plane during bending.
Deflection of a beam deflected symmetrically and principle of superposition
Compressive and tensile forces develop in the direction of the beam axis under bending loads. These forces induce stresses on the beam. The maximum compressive stress is found at the uppermost edge of the beam while the maximum tensile stress is located at the lower edge of the beam. Since the stresses between these two opposing maxima vary linearly, there therefore exists a point on the linear path between them where there is no bending stress. The locus of these points is the neutral axis. Because of this area with no stress and the adjacent areas with low stress, using uniform cross section beams in bending is not a particularly efficient means of supporting a load as it does not use the full capacity of the beam until it is on the brink of collapse. Wide-flange beams (I-beams) and truss girders effectively address this inefficiency as they minimize the amount of material in this under-stressed region.
The classic formula for determining the bending stress in a beam under simple bending is:[5]
${\displaystyle \sigma _{x}={\frac {M_{z}y}{I_{z}}}={\frac {M_{z}}{W_{z}}}}$
where
• ${\displaystyle {\sigma _{x}}}$ is the bending stress
• ${\displaystyle M_{z}}$ - the moment about the neutral axis
• ${\displaystyle y}$ - the perpendicular distance to the neutral axis
• ${\displaystyle I_{z}}$ - the second moment of area about the neutral axis z.
• ${\displaystyle W_{z}}$ - the Resistance Moment about the neutral axis z. ${\displaystyle W_{z}=I_{z}/y}$
### Extensions of Euler-Bernoulli beam bending theory
#### Plastic bending
The equation ${\displaystyle \sigma ={\tfrac {My}{I_{x}}}}$ is valid only when the stress at the extreme fiber (i.e., the portion of the beam farthest from the neutral axis) is below the yield stress of the material from which it is constructed. At higher loadings the stress distribution becomes non-linear, and ductile materials will eventually enter a plastic hinge state where the magnitude of the stress is equal to the yield stress everywhere in the beam, with a discontinuity at the neutral axis where the stress changes from tensile to compressive. This plastic hinge state is typically used as a limit state in the design of steel structures.
#### Complex or asymmetrical bending
The equation above is only valid if the cross-section is symmetrical. For homogeneous beams with asymmetrical sections, the maximum bending stress in the beam is given by
${\displaystyle \sigma _{x}(y,z)=-{\frac {M_{z}~I_{y}+M_{y}~I_{yz}}{I_{y}~I_{z}-I_{yz}^{2}}}y+{\frac {M_{y}~I_{z}+M_{z}~I_{yz}}{I_{y}~I_{z}-I_{yz}^{2}}}z}$[6]
where ${\displaystyle y,z}$ are the coordinates of a point on the cross section at which the stress is to be determined as shown to the right, ${\displaystyle M_{y}}$ and ${\displaystyle M_{z}}$ are the bending moments about the y and z centroid axes, ${\displaystyle I_{y}}$ and ${\displaystyle I_{z}}$ are the second moments of area (distinct from moments of inertia) about the y and z axes, and ${\displaystyle I_{yz}}$ is the product of moments of area. Using this equation it is possible to calculate the bending stress at any point on the beam cross section regardless of moment orientation or cross-sectional shape. Note that ${\displaystyle M_{y},M_{z},I_{y},I_{z},I_{yz}}$ do not change from one point to another on the cross section.
#### Large bending deformation
For large deformations of the body, the stress in the cross-section is calculated using an extended version of this formula. First the following assumptions must be made:
1. Assumption of flat sections - before and after deformation the considered section of body remains flat (i.e., is not swirled).
2. Shear and normal stresses in this section that are perpendicular to the normal vector of cross section have no influence on normal stresses that are parallel to this section.
Large bending considerations should be implemented when the bending radius ${\displaystyle \rho }$ is smaller than ten section heights h:
${\displaystyle \rho <10h.}$
With those assumptions the stress in large bending is calculated as:
${\displaystyle \sigma ={\frac {F}{A}}+{\frac {M}{\rho A}}+{\frac {M}{{I_{x}}'}}y{\frac {\rho }{\rho +y}}}$
where
${\displaystyle F}$ is the normal force
${\displaystyle A}$ is the section area
${\displaystyle M}$ is the bending moment
${\displaystyle \rho }$ is the local bending radius (the radius of bending at the current section)
${\displaystyle {{I_{x}}'}}$ is the area moment of inertia along the x-axis, at the ${\displaystyle y}$ place (see Steiner's theorem)
${\displaystyle y}$ is the position along y-axis on the section area in which the stress ${\displaystyle \sigma }$ is calculated.
When bending radius ${\displaystyle \rho }$ approaches infinity and ${\displaystyle y\ll \rho }$, the original formula is back:
${\displaystyle \sigma ={F \over A}\pm {\frac {My}{I}}}$.
### Timoshenko bending theory
Deformation of a Timoshenko beam. The normal rotates by an amount ${\displaystyle \theta }$ which is not equal to ${\displaystyle dw/dx}$.
In 1921, Timoshenko improved upon the Euler-Bernoulli theory of beams by adding the effect of shear into the beam equation. The kinematic assumptions of the Timoshenko theory are:
• normals to the axis of the beam remain straight after deformation
• there is no change in beam thickness after deformation
However, normals to the axis are not required to remain perpendicular to the axis after deformation.
The equation for the quasistatic bending of a linear elastic, isotropic, homogeneous beam of constant cross-section beam under these assumptions is[7]
${\displaystyle EI~{\cfrac {\mathrm {d} ^{4}w}{\mathrm {d} x^{4}}}=q(x)-{\cfrac {EI}{kAG}}~{\cfrac {\mathrm {d} ^{2}q}{\mathrm {d} x^{2}}}}$
where ${\displaystyle I}$ is the area moment of inertia of the cross-section, ${\displaystyle A}$ is the cross-sectional area, ${\displaystyle G}$ is the shear modulus, ${\displaystyle k}$ is a shear correction factor, and ${\displaystyle q(x)}$ is an applied transverse load. For materials with Poisson's ratios (${\displaystyle \nu }$) close to 0.3, the shear correction factor for a rectangular cross-section is approximately
${\displaystyle k={\cfrac {5+5\nu }{6+5\nu }}}$
The rotation (${\displaystyle \varphi (x)}$) of the normal is described by the equation
${\displaystyle {\cfrac {\mathrm {d} \varphi }{\mathrm {d} x}}=-{\cfrac {\mathrm {d} ^{2}w}{\mathrm {d} x^{2}}}-{\cfrac {q(x)}{kAG}}}$
The bending moment (${\displaystyle M}$) and the shear force (${\displaystyle Q}$) are given by
${\displaystyle M(x)=-EI~{\cfrac {\mathrm {d} \varphi }{\mathrm {d} x}}~;~~Q(x)=kAG\left({\cfrac {\mathrm {d} w}{\mathrm {d} x}}-\varphi \right)=-EI~{\cfrac {\mathrm {d} ^{2}\varphi }{\mathrm {d} x^{2}}}={\cfrac {\mathrm {d} M}{\mathrm {d} x}}}$
## Beams on elastic foundations
According to Euler-Bernoulli, Timoshenko or other bending theories, the beams on elastic foundations can be explained. In some applications such as rail tracks, foundation of buildings and machines, ships on water, roots of plants etc., the beam subjected to loads is supported on continuous elastic foundations (i.e. the continuous reactions due to external loading is distributed along the length of the beam)[8][9][10][11]
Car crossing a bridge (Beam partially supported on elastic foundation, Bending moment distribution)
## Dynamic bending of beams
The dynamic bending of beams,[12] also known as flexural vibrations of beams, was first investigated by Daniel Bernoulli in the late 18th century. Bernoulli's equation of motion of a vibrating beam tended to overestimate the natural frequencies of beams and was improved marginally by Rayleigh in 1877 by the addition of a mid-plane rotation. In 1921 Stephen Timoshenko improved the theory further by incorporating the effect of shear on the dynamic response of bending beams. This allowed the theory to be used for problems involving high frequencies of vibration where the dynamic Euler-Bernoulli theory is inadequate. The Euler-Bernoulli and Timoshenko theories for the dynamic bending of beams continue to be used widely by engineers.
### Euler-Bernoulli theory
The Euler-Bernoulli equation for the dynamic bending of slender, isotropic, homogeneous beams of constant cross-section under an applied transverse load ${\displaystyle q(x,t)}$ is[7]
${\displaystyle EI~{\cfrac {\partial ^{4}w}{\partial x^{4}}}+m~{\cfrac {\partial ^{2}w}{\partial t^{2}}}=q(x,t)}$
where ${\displaystyle E}$ is the Young's modulus, ${\displaystyle I}$ is the area moment of inertia of the cross-section, ${\displaystyle w(x,t)}$ is the deflection of the neutral axis of the beam, and ${\displaystyle m}$ is mass per unit length of the beam.
#### Free vibrations
For the situation where there is no transverse load on the beam, the bending equation takes the form
${\displaystyle EI~{\cfrac {\partial ^{4}w}{\partial x^{4}}}+m~{\cfrac {\partial ^{2}w}{\partial t^{2}}}=0}$
Free, harmonic vibrations of the beam can then be expressed as
${\displaystyle w(x,t)={\text{Re}}[{\hat {w}}(x)~e^{-i\omega t}]\quad \implies \quad {\cfrac {\partial ^{2}w}{\partial t^{2}}}=-\omega ^{2}~w(x,t)}$
and the bending equation can be written as
${\displaystyle EI~{\cfrac {\mathrm {d} ^{4}{\hat {w}}}{\mathrm {d} x^{4}}}-m\omega ^{2}{\hat {w}}=0}$
The general solution of the above equation is
${\displaystyle {\hat {w}}=A_{1}\cosh(\beta x)+A_{2}\sinh(\beta x)+A_{3}\cos(\beta x)+A_{4}\sin(\beta x)}$
where ${\displaystyle A_{1},A_{2},A_{3},A_{4}}$ are constants and ${\displaystyle \beta :=\left({\cfrac {m}{EI}}~\omega ^{2}\right)^{1/4}}$
The mode shapes of a cantilevered I-beam
1st lateral bending
1st torsional
1st vertical bending
2nd lateral bending
2nd torsional
2nd vertical bending
### Timoshenko-Rayleigh theory
In 1877, Rayleigh proposed an improvement to the dynamic Euler-Bernoulli beam theory by including the effect of rotational inertia of the cross-section of the beam. Timoshenko improved upon that theory in 1922 by adding the effect of shear into the beam equation. Shear deformations of the normal to the mid-surface of the beam are allowed in the Timoshenko-Rayleigh theory.
The equation for the bending of a linear elastic, isotropic, homogeneous beam of constant cross-section under these assumptions is[7][13]
{\displaystyle {\begin{aligned}&EI~{\frac {\partial ^{4}w}{\partial x^{4}}}+m~{\frac {\partial ^{2}w}{\partial t^{2}}}-\left(J+{\frac {EIm}{kAG}}\right){\frac {\partial ^{4}w}{\partial x^{2}~\partial t^{2}}}+{\frac {Jm}{kAG}}~{\frac {\partial ^{4}w}{\partial t^{4}}}\\[6pt]={}&q(x,t)+{\frac {J}{kAG}}~{\frac {\partial ^{2}q}{\partial t^{2}}}-{\frac {EI}{kAG}}~{\frac {\partial ^{2}q}{\partial x^{2}}}\end{aligned}}}
where ${\displaystyle J={\tfrac {mI}{A}}}$ is the polar moment of inertia of the cross-section, ${\displaystyle m=\rho A}$ is the mass per unit length of the beam, ${\displaystyle \rho }$ is the density of the beam, ${\displaystyle A}$ is the cross-sectional area, ${\displaystyle G}$ is the shear modulus, and ${\displaystyle k}$ is a shear correction factor. For materials with Poisson's ratios (${\displaystyle \nu }$) close to 0.3, the shear correction factor are approximately
{\displaystyle {\begin{aligned}k&={\frac {5+5\nu }{6+5\nu }}\quad {\text{rectangular cross-section}}\\[6pt]&={\frac {6+12\nu +6\nu ^{2}}{7+12\nu +4\nu ^{2}}}\quad {\text{circular cross-section}}\end{aligned}}}
#### Free vibrations
${\displaystyle EI~{\cfrac {\mathrm {d} ^{4}{\hat {w}}}{\mathrm {d} x^{4}}}+m\omega ^{2}\left({\cfrac {J}{m}}+{\cfrac {EI}{kAG}}\right){\cfrac {\mathrm {d} ^{2}{\hat {w}}}{\mathrm {d} x^{2}}}+m\omega ^{2}\left({\cfrac {\omega ^{2}J}{kAG}}-1\right)~{\hat {w}}=0}$
This equation can be solved by noting that all the derivatives of ${\displaystyle w}$ must have the same form to cancel out and hence as solution of the form ${\displaystyle e^{kx}}$ may be expected. This observation leads to the characteristic equation
${\displaystyle \alpha ~k^{4}+\beta ~k^{2}+\gamma =0~;~~\alpha :=EI~,~~\beta :=m\omega ^{2}\left({\cfrac {J}{m}}+{\cfrac {EI}{kAG}}\right)~,~~\gamma :=m\omega ^{2}\left({\cfrac {\omega ^{2}J}{kAG}}-1\right)}$
The solutions of this quartic equation are
${\displaystyle k_{1}=+{\sqrt {z_{+}}}~,~~k_{2}=-{\sqrt {z_{+}}}~,~~k_{3}=+{\sqrt {z_{-}}}~,~~k_{4}=-{\sqrt {z_{-}}}}$
where
${\displaystyle z_{+}:={\cfrac {-\beta +{\sqrt {\beta ^{2}-4\alpha \gamma }}}{2\alpha }}~,~~z_{-}:={\cfrac {-\beta -{\sqrt {\beta ^{2}-4\alpha \gamma }}}{2\alpha }}}$
The general solution of the Timoshenko-Rayleigh beam equation for free vibrations can then be written as
${\displaystyle {\hat {w}}=A_{1}~e^{k_{1}x}+A_{2}~e^{-k_{1}x}+A_{3}~e^{k_{3}x}+A_{4}~e^{-k_{3}x}}$
## Quasistatic bending of plates
Deformation of a thin plate highlighting the displacement, the mid-surface (red) and the normal to the mid-surface (blue)
The defining feature of beams is that one of the dimensions is much larger than the other two. A structure is called a plate when it is flat and one of its dimensions is much smaller than the other two. There are several theories that attempt to describe the deformation and stress in a plate under applied loads two of which have been used widely. These are
• the Kirchhoff-Love theory of plates (also called classical plate theory)
• the Mindlin-Reissner plate theory (also called the first-order shear theory of plates)
### Kirchhoff-Love theory of plates
The assumptions of Kirchhoff-Love theory are
• straight lines normal to the mid-surface remain straight after deformation
• straight lines normal to the mid-surface remain normal to the mid-surface after deformation
• the thickness of the plate does not change during a deformation.
These assumptions imply that
{\displaystyle {\begin{aligned}u_{\alpha }(\mathbf {x} )&=-x_{3}~{\frac {\partial w^{0}}{\partial x_{\alpha }}}=-x_{3}~w_{,\alpha }^{0}~;~~\alpha =1,2\\u_{3}(\mathbf {x} )&=w^{0}(x_{1},x_{2})\end{aligned}}}
where ${\displaystyle \mathbf {u} }$ is the displacement of a point in the plate and ${\displaystyle w^{0}}$ is the displacement of the mid-surface.
The strain-displacement relations are
{\displaystyle {\begin{aligned}\varepsilon _{\alpha \beta }&=-x_{3}~w_{,\alpha \beta }^{0}\\\varepsilon _{\alpha 3}&=0\\\varepsilon _{33}&=0\end{aligned}}}
The equilibrium equations are
${\displaystyle M_{\alpha \beta ,\alpha \beta }+q(x)=0~;~~M_{\alpha \beta }:=\int _{-h}^{h}x_{3}~\sigma _{\alpha \beta }~dx_{3}}$
where ${\displaystyle q(x)}$ is an applied load normal to the surface of the plate.
In terms of displacements, the equilibrium equations for an isotropic, linear elastic plate in the absence of external load can be written as
${\displaystyle w_{,1111}^{0}+2~w_{,1212}^{0}+w_{,2222}^{0}=0}$
In direct tensor notation,
${\displaystyle \nabla ^{2}\nabla ^{2}w=0}$
### Mindlin-Reissner theory of plates
The special assumption of this theory is that normals to the mid-surface remain straight and inextensible but not necessarily normal to the mid-surface after deformation. The displacements of the plate are given by
{\displaystyle {\begin{aligned}u_{\alpha }(\mathbf {x} )&=-x_{3}~\varphi _{\alpha }~;~~\alpha =1,2\\u_{3}(\mathbf {x} )&=w^{0}(x_{1},x_{2})\end{aligned}}}
where ${\displaystyle \varphi _{\alpha }}$ are the rotations of the normal.
The strain-displacement relations that result from these assumptions are
{\displaystyle {\begin{aligned}\varepsilon _{\alpha \beta }&=-x_{3}~\varphi _{\alpha ,\beta }\\\varepsilon _{\alpha 3}&={\cfrac {1}{2}}~\kappa \left(w_{,\alpha }^{0}-\varphi _{\alpha }\right)\\\varepsilon _{33}&=0\end{aligned}}}
where ${\displaystyle \kappa }$ is a shear correction factor.
The equilibrium equations are
{\displaystyle {\begin{aligned}&M_{\alpha \beta ,\beta }-Q_{\alpha }=0\\&Q_{\alpha ,\alpha }+q=0\end{aligned}}}
where
${\displaystyle Q_{\alpha }:=\kappa ~\int _{-h}^{h}\sigma _{\alpha 3}~dx_{3}}$
## Dynamic bending of plates
### Dynamics of thin Kirchhoff plates
The dynamic theory of plates determines the propagation of waves in the plates, and the study of standing waves and vibration modes. The equations that govern the dynamic bending of Kirchhoff plates are
${\displaystyle M_{\alpha \beta ,\alpha \beta }-q(x,t)=J_{1}~{\ddot {w}}^{0}-J_{3}~{\ddot {w}}_{,\alpha \alpha }^{0}}$
where, for a plate with density ${\displaystyle \rho =\rho (x)}$,
${\displaystyle J_{1}:=\int _{-h}^{h}\rho ~dx_{3}~;~~J_{3}:=\int _{-h}^{h}x_{3}^{2}~\rho ~dx_{3}}$
and
${\displaystyle {\ddot {w}}^{0}={\frac {\partial ^{2}w^{0}}{\partial t^{2}}}~;~~{\ddot {w}}_{,\alpha \beta }^{0}={\frac {\partial ^{2}{\ddot {w}}^{0}}{\partial x_{\alpha }\,\partial x_{\beta }}}}$
The figures below show some vibrational modes of a circular plate.
## References
1. ^ a b c d Boresi, A. P. and Schmidt, R. J. and Sidebottom, O. M., 1993, Advanced mechanics of materials, John Wiley and Sons, New York.
2. ^ a b Libai, A. and Simmonds, J. G., 1998, The nonlinear theory of elastic shells, Cambridge University Press.
3. ^ Timoshenko, S. and Woinowsky-Krieger, S., 1959, Theory of plates and shells, McGraw-Hill.
4. ^ Shigley J, "Mechanical Engineering Design", p44, International Edition, pub McGraw Hill, 1986, ISBN 0-07-100292-8
5. ^ Gere, J. M. and Timoshenko, S.P., 1997, Mechanics of Materials, PWS Publishing Company.
6. ^ Cook and Young, 1995, Advanced Mechanics of Materials, Macmillan Publishing Company: New York
7. ^ a b c Thomson, W. T., 1981, Theory of Vibration with Applications
8. ^ HETÉNYI, Miklos (1946). Beams on Elastic Foundation. Ann Arbor, University of Michigan Studies, USA.
9. ^ MELERSKI, E., S. (2006). Design Analysis of Beams, Circular Plates and Cylindrical Tanks on Elastic Foundations (2nd ed.). London, UK: Taylor & Francis Group. p. 284. ISBN 978-0-415-38350-9.
10. ^ TSUDIK, E. Analysis of Beams and Frames on Elastic Foundation. USA: Trafford Publishing. p. 248. ISBN 1-4120-7950-0.
11. ^ FRYDRÝ?EK, Karel; Tvrdá, Katarína; Jan?o, Roland; et al. (2013). Handbook of Structures on Elastic Foundation (1st ed.). Ostrava, Czech Republic: VSB - Technical University of Ostrava. pp. 1-1691. ISBN 978-80-248-3238-8.
12. ^ Han, S. M, Benaroya, H. and Wei, T., 1999, "Dynamics of transversely vibrating beams using four engineering theories," Journal of Sound and Vibration, vol. 226, no. 5, pp. 935-988.
13. ^ Rosinger, H. E. and Ritchie, I. G., 1977, On Timoshenko's correction for shear in vibrating isotropic beams, J. Phys. D: Appl. Phys., vol. 10, pp. 1461-1466. | 6,651 | 24,071 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 102, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.924352 |
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[vox-tech] Graphing Analysis
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# [vox-tech] Graphing Analysis
```Hi,
I'm doing some data analysis, stock data analysis to be exact, but I figure
it might as well be anything one could graph. I want to find peaks and
valleys. I want to do with my computer what I do naturally with my eyes.
So for example, some random set like this series:
3, 4, 6, 2, 7, 8, 8, 7, 6, 5, 4, 5, 6, 7, 8, 9, 11, 13,14,13, 9, 8, 9, 10,
11, 12, 15, 20, 15...
I can visually, if I put my face directly up to the paper (i.e look at each
number with regard to only the numbers directly before and after) pick out
peaks and valleys:
peaks: 6, 8, 14, 20
valleys: 3, 2, 4(after the 8), 8 (after 14)
so if I graphed this I'd use: 3,6,2,8,4,14,8,20
Then again if I pull the paper a little away, look at it with regard to 5
numbers before and after my current number, I'd say my series is:
3,8,4,14,8,20
That is, 8 was the highest number, in the 5 points after the initial data
point of 3, also as a whole, the numbers were growing up to the 8, then
decreased to the 4, then grew till the 14, then decreased till the 8, grew
till 20.
If I pull the paper far enough away I have 2 points:
3, 20
What I'm asking is there some math function(s) for analyzing trends and
graphs. Or do I need to write functions for doing what I'm doing with my
eyes?
Thanks
Jay
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```
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Hosting provided by: Sunset Systems Sunset Systems offers preconfigured Linux systems, remote system administration and custom software development. LUGOD: Linux Users' Group of Davis PO Box 2082, Davis, CA 95617 Contact Us LUGOD is a 501(c)7 non-profit organization based in Davis, California and serving the Sacramento area. "Linux" is a trademark of Linus Torvalds. Sponsored in part by: For a significant contribution towards our projector, and a generous donation to allow us to continue meeting at the Davis Library. | 734 | 2,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-15 | longest | en | 0.902278 |
https://physics.stackexchange.com/questions/433643/why-can-we-not-use-the-specific-heat-to-determine-the-work-done-on-steam-in-a-ra | 1,712,965,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00700.warc.gz | 429,636,222 | 38,748 | # Why can we not use the specific heat to determine the work done on steam in a Rankine cycle?
Okay, I am trying to work backwards from an exam which I messed up and I am trying to figure out what the lecturer did based on his answers posted online. Looking back, I realise one mistake I made was to use the specific heat of water given to calculate the enthalpy change in a transition from one pressure to the next.
In a transition within a Rankine cycle: I have P1 = P2 = 200 bar. Moving from 1 to 2, the water vapour turns to two-phase water and the enthalpy at point 2 is given as 2,353 kJ/kg. We also know that the enthalpy at 1 is 145 (worked out from a constant volume work problem and it checks out with the lecturer's answer).
Long story short: if I had tried to use the specific heat of water to determine the change in enthalpy and work from there to the enthalpy, I would have been way off. Instead, students were meant to use the given enthalpy value and then find the correct temperature from a table of saturated water provided.
I understand that the specific heat of water will change with temperature, but looking up the temperatures I am working at (changes from 30 C to 388 C) I can see that even an averaged out specific heat does not allow for a computation of a change in enthalpy from 145 to 2,353 (the second value was given). So I'm trying to figure out what I don't understand; why is a specific heat/temperature change calculation insufficient to calculate the enthalpy change here.
• You forgot to consider the heat of vaporization. Oct 10, 2018 at 12:05
• To make @ChesterMiller's point more explicitly: the Rankine cycle involves change of phase. Oct 10, 2018 at 14:12
I think you mean the water turns to two-phase water, not the water vapor turns to two-phase water.
See figure below.
200 Bar and h = 2353 puts point 2 in the two-phase region.
From point 1 to point 1A you can use specific heat to calculate enthalpy change. From point 1A to 2 you have liquid and vapor at constant temperature. The change in enthalpy (527) is entirely due to the heat of vaporization.
Hope this helps.
• Okay, that makes sense. I'm sorry but my upvotes won't be worth much until I increase my credibility score or whatever.
– Abed
Oct 11, 2018 at 16:42
• @user209194 I believe you can still select the "accept" button (one with checkmark) Oct 11, 2018 at 19:31 | 594 | 2,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.956506 |
https://www.everythingmaths.co.za/read/science/grade-10/motion-in-one-dimension/21-motion-in-one-dimension-02 | 1,545,023,527,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828318.79/warc/CC-MAIN-20181217042727-20181217064727-00309.warc.gz | 863,407,112 | 13,107 | We think you are located in South Africa. Is this correct?
# Reference Frame
## 21.2 Reference frame (ESAGT)
The first thing to focus on when studying motion of an object or person is their position. The word position describes your location (where you are). However, saying that you are here or there is meaningless, you have to use known points (reference points) to help specify your position.
For example, if you were in a classroom and wanted to tell a classmate where you were standing you would first give them a reference point. The reference point might be the classroom door. You would then be able to say that you are $$\text{2}$$ $$\text{m}$$ from the doorway. This still doesn't give your position precisely. We need to provide a reference point and a coordinate system to use to define the location precisely.
Then you can say that you are, for example, $$\text{2}$$ $$\text{m}$$ from the door directly inside the classroom. The classroom door is a reference point and inside/outside is the coordinate system you have chosen. A frame of reference or reference frame is reference point which serves as the origin for a coordinate system. The coordinate system can be up or down, inside or outside, left or right or even forward or backward. These are all examples that define a $$\text{1}$$-dimensional coordinate system. We choose one of the directions as the positive direction.
Frame of reference
A frame of reference is a reference point combined with a set of directions.
A graphical representation of a $$\text{1}$$-dimensional frame of reference:
You can define different frames of reference for the same problem but the outcome, the physical results, will be the same. For example, a boy is standing still inside a train as it pulls out of a station. Both you and the boy define your location as the point of reference and the direction train is moving as a where you are standing as the point of reference and the direction the train is moving in as forward.
You are standing on the platform watching the train move from left to right. To you it looks as if the boy is moving from left to right, because relative to where you are standing (the platform), he is moving. According to the boy, and his frame of reference (the train), he is not moving.
A frame of reference must have an origin (where you are standing on the platform) and at least a positive direction. The train was moving from left to right, making to your right positive and to your left negative. If someone else was looking at the same boy, his frame of reference will be different. For example, if he was standing on the other side of the platform, the boy will be moving from right to left.
For this chapter, we will only use frames of reference in the $$x$$-direction. By doing this we restrict ourselves to one dimensional motion. We can use the sign of the position value (positive or negative) to indicate the direction relative to the origin.
One dimensional motion
An object is constrained to move back and forth along a line.
For example the blue dot in the figure below can only move along the $$x$$-axis.
### Position (ESAGU)
Position
Position is a measurement of a location, with reference to an origin.
Quantity: Position ($$x$$) Unit name: metre Unit symbol: $$\text{m}$$
A position is a measurement of a location within a reference frame. This means that positions can be negative or positive depending on the choice for the reference frame's coordinate system.
Depending on which reference point we choose, we can say that the school is $$\text{300}$$ $$\text{m}$$ from Kosma's house (with Kosma's house as the reference point or origin) or $$\text{500}$$ $$\text{m}$$ from Kevin's house (with Kevin's house as the reference point or origin).
The shop is also $$\text{300}$$ $$\text{m}$$ from Kosma's house, but in the opposite direction as the school. When we choose a reference point, we have a positive direction and a negative direction. If we choose the direction towards the school as negative, then the direction towards the shop is positive. A negative direction is always opposite to the direction chosen as positive.
The origin is at Kosma's house and the position of the school is $$-\text{300}$$ $$\text{m}$$. Positions towards the left are defined as negative and positions towards the right are defined as positive.
Note that we could also choose the positive direction to be towards the school. In this case Kosma's house is still $$\text{300}$$ $$\text{m}$$ away from the school, but it is now in the positive direction.
The origin is at Kosma's house and the position of the school is $$\text{+300}$$ $$\text{m}$$. Positions towards the left are defined as positive and positions towards the right are defined as negative.
## Reference points
Divide into groups of $$\text{5}$$ for this activity. On a straight line, choose a reference point. Since position can have both positive and negative values, discuss the advantages and disadvantages of choosing
1. either end of the line,
2. the middle of the line. (This reference point can also be called “the origin”.)
Stand in a straight line, take turns choosing a different group member as the origin. Let the group member choose which direction along the line is positive. Everyone should then try to define their position. You don't need to measure your position very precisely but can just approximate it. It is important to understand whether your position is positive or negative for each difference origin and choice of coordinate system.
Notice that your position is different every time but that you didn't actually move. How you write an answer might be affected by the choice of a coordinate system but physical processes should never be affected.
## Position
Exercise 21.1
Write down the positions for objects at $$A$$, $$B$$, $$D$$ and $$E$$. Do not forget the units.
Solution not yet available
Write down the positions for objects at $$F$$, $$G$$, $$H$$ and $$J$$. Do not forget the units.
Solution not yet available
There are $$\text{5}$$ houses on Newton Street, $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$. For all cases, assume that positions to the right are positive.
1. Draw a frame of reference with house $$A$$ as the origin and write down the positions of houses $$B$$, $$C$$, $$D$$ and $$E$$.
2. You live in house $$C$$. What is your position relative to house $$E$$?
3. What are the positions of houses $$A$$, $$B$$ and $$D$$, if house $$B$$ is taken as the reference point?
Solution not yet available
### Displacement and distance (ESAGV)
Distance
Distance is the total length of the path taken in going from the initial position, $$\vec{x}_{i}$$, to the final position, $$\vec{x}_{f}$$. Distance is a scalar.
Quantity: Distance ($$D$$) Unit name: metre Unit symbol: $$\text{m}$$
The symbol Δ is read out as delta. Δ is a letter of the Greek alphabet and is used in Mathematics and Science to indicate a change in a certain quantity, or a final value minus an initial value. For example, $$\Delta x$$ means change in $$x$$ while $$\Delta t$$ means change in $$t$$.
In the simple map below you can see the path that winds because of a number of hills from a school to a nearby shop. The path is shown by a dashed line. The initial point, $$\vec{x}_{i}$$, is the school and the final point, $$\vec{x}_{f}$$, is the shop.
Distance is the length of dashed line. It is how far you have to walk along the path from the school to the shop.
Displacement
Displacement is the change in an object's position. It is a vector that points from the initial position ($$\vec{x}_{i}$$) to the final position ($$\vec{x}_{f}$$).
Quantity: Displacement ($$\Delta \vec{x}$$) Unit name: metre Unit symbol: $$\text{m}$$
The displacement of an object is defined as its change in position (final position minus initial position). Displacement has a magnitude and direction and is therefore a vector. For example, if the initial position of a car is $$\vec{x}_{i}$$ and it moves to a final position of $$\vec{x}_{f}$$, then the displacement is:
$\Delta \vec{x} = \vec{x}_{f} - \vec{x}_{i}$
To help visualise what the displacement vector looks like think back to the tail-to-head method. The displacement is the vector you add to the initial position vector to get a vector to the final position.
However, subtracting an initial quantity from a final quantity happens often in Physics, so we use the shortcut Δ to mean final - initial. Therefore, displacement can be written:
$\Delta \vec{x} = \vec{x}_{f} - \vec{x}_{i}$
The following diagram illustrates the concept of displacement:
For example, if you roll a ball $$\text{5}$$ $$\text{m}$$ along a floor, in a straight line, then its displacement is $$\text{5}$$ $$\text{m}$$, taking the direction of motion as positive, and the initial position as $$\text{0}$$ $$\text{m}$$.
The words initial and final will be used very often in Physics. Initial refers to the situation in the beginning of the description/problem and final to the situation at the end. It will often happen that the final value is smaller than the initial value, such that the difference is negative. This is ok!
Displacement does not depend on the path travelled, but only on the initial and final positions. We use the word distance to describe how far an object travels along a particular path.
If we go back to the simple map repeated below you can see the path as before shown by a dashed line.
We will use $$D$$ in this book, but you may see $$d$$ used in other books.
We use the expression 'as the crow flies' to mean a straight line between two points because birds can fly directly over many obstacles.
Distance is the length of dashed line. The displacement is different. Displacement is the straight-line distance from the starting point to the endpoint – from the school to the shop in the figure as shown by the solid arrow.
#### Illustration of distance and displacement
If we use the same situation as earlier we can explore the concepts in more detail. Consider our description of the location of the houses, school and the shop.
Komal walks to meet Kevin at his house before walking to school. What is Komal's displacement and what distance did he cover if he walks to school via Kevin's house?
Komal covers a distance of $$\text{400}$$ $$\text{m}$$ to Kevin's house and another $$\text{500}$$ $$\text{m}$$ from Kevin's house to the school. He covers a total distance of $$\text{900}$$ $$\text{m}$$. His displacement, however, is only $$\text{100}$$ $$\text{m}$$ towards the school. This is because displacement only looks at the starting position (his house) and the end position (the school). It does not depend on the path he travelled.
To calculate his distance and displacement, we need to choose a reference point and a direction. Let's choose Komal's house as the reference point, and towards Kevin's house as the positive direction (which means that towards the school is negative). We would do the calculations as follows:
\begin{align*} \text{Distance } (D) & = \text{path travelled } \\ & = \text{400}\text{ m} + \text{500}\text{ m} \\ & = \text{900}\text{ m} \end{align*}\begin{align*} \text{Displacement } \left(\Delta \vec{x}\right) & = \vec{x}_{f} - \vec{x}_{i} \\ & = -\text{100}\text{ m} + \text{0}\text{ m} \\ & = -\text{100}\text{ m} \\ & = \text{100}\text{ m} \text{ (in the negative } x \text{ direction)} \end{align*}
Very often in calculations you will get a negative answer. For example, Komal's displacement in the example above, is calculated as $$-\text{100}$$ $$\text{m}$$. The minus sign in front of the answer means that his displacement is $$\text{100}$$ $$\text{m}$$ in the opposite direction (opposite to the direction chosen as positive in the beginning of the question). When we start a calculation we choose a frame of reference and a positive direction. In the first example above, the reference point is Komal's house and the positive direction is towards Kevin's house. Therefore Komal's displacement is $$\text{100}$$ $$\text{m}$$ towards the school. Notice that distance has no direction, but displacement has a direction.
Kevin walks to school with Komal and after school walks back home. What is Kevin's displacement and what distance did he cover? For this calculation we use Kevin's house as the reference point. Let's take towards the school as the positive direction.
\begin{align*} \text{Distance } (D) & = \text{ path travelled} \\ & = \text{500}\text{ m} + \text{500}\text{ m} \\ & = \text{1 000}\text{ m} \end{align*}\begin{align*} \text{Displacement } \left(\Delta \vec{x}\right) & = \vec{x}_{f} - \vec{x}_{i} \\ & = \text{0}\text{ m} + \text{0}\text{ m} \\ & = \text{0}\text{ m} \end{align*}
It is possible to have a displacement of $$\text{0}$$ $$\text{m}$$ and a distance that is not $$\text{0}$$ $$\text{m}$$. This happens whenever you end at the same point you started.
#### Differences between distance and displacement
The differences between distance and displacement can be summarised as:
Distance Displacement 1. depends on the path 1. independent of path taken 2. always positive 2. can be positive or negative 3. is a scalar 3. is a vector
## Frames of reference, displacement and distance
Exercise 21.2
Use this figure to answer the following questions.
1. Kogis walks to Kosma's house and then to school, what is her distance and displacement?
2. Kholo walks to Kosma's house and then to school, what is her distance and displacement?
3. Komal walks to the shop and then to school, what is his distance and displacement?
4. What reference point did you use for each of the above questions?
Solution not yet available
You stand at the front door of your house (displacement, $$\Delta \vec{x} = \text{0}\text{ m}$$). The street is $$\text{10}$$ $$\text{m}$$ away from the front door. You walk to the street and back again.
1. What is the distance you have walked?
2. What is your final displacement?
3. Is displacement a vector or a scalar? Give a reason for your answer.
Solution not yet available | 3,382 | 14,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-51 | latest | en | 0.945655 |
http://mathnotations.blogspot.com/2008/10/adding-and-subtracting-mixed-numerals.html | 1,532,339,441,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596204.93/warc/CC-MAIN-20180723090751-20180723110751-00154.warc.gz | 241,056,488 | 37,006 | ## Saturday, October 18, 2008
### Adding and Subtracting Mixed Numerals - A Survey of Methods Taught
Now that our family health crisis has abated (my daughter is doing well), I guess it's time to jump back in with both feet. A math program leader in a district with which I am consulting, asked for my opinion on an important issue of curriculum and instruction.
How much time should middle school teachers spend on the traditional vertical algorithm for adding and subtracting mixed numerals vs. converting to improper fractions immediately?
I assumed that both methods are still commonly taught with about equal time given to each, but I wasn't all that sure about how that was across the country. This is where I need the help of my informed readers.
First, my thoughts. From a practical perspective of those who utilize fractions in their occupation, I would guess that the mixed numeral form is most commonly employed. Whether it's the carpenter taking measurements to see how many board feet of wood must be ordered (or for precise measurement to the nearest sixteenth of an inch) or someone following a recipe in the kitchen, I can't imagine that converting to improper fractions would be their first choice. On the other hand when I personally need to add fractions in a math problem, I usually use improper. I took an informal survey of one of the groups I'm working with and the majority stated they were taught both methods and some preferred working with mixed numerals and others said it's more complicated that way.
Are the number of steps roughly the same?
Mixed Numerals Algorithm for Subtraction:
(1) Convert the proper fractions to common denominator form.
(2) If needed, regroup, i.e., "borrow" 1 from the whole number part of the larger mixed number, convert the 1 into common denominator form and combine this with the other fraction (of course students are shown short-cuts for this which they blithely and mechanically follow without much thought).
(3) Subtract the whole numbers and the proper fractions.
(4) If the resulting fraction is improper convert it and add the whole number part to the previous result.
Improper Fraction Algorithm:
(1) Convert each mixed numeral to an improper fraction by the traditional algorithm (again blithely and mechanically without much thought).
(2) Determine a common denominator (or the lcd) and convert each fraction.
(3) Subtract the fractions.
(4) Convert the answer to mixed numeral form by the traditional division algorithm.
Now I may have combined steps or there are oversights but essentially they appear to be roughly the same number of steps. However, the difficulty or complexity level of the steps
may not be equivalent.
I also feel that the mixed numeral form requires somewhat more conceptual understanding even if the child does it routinely. It may also prepare the youngster for working with algebraic expressions like A + B/C, but that's debatable. Further there seems to me to be a strong connection between the Mixed Numerals Algorithm and adding and subtracting denominate numbers. For example:
Subtract
15 hr 37 min
9 hr 46 min
I doubt that we would encourage students to convert both to minutes first, subtract, then convert back to hrs and min. I could be wrong there!
I feel there are arguments on both sides here. My instinct is that both need to be taught but it's not clear to me how much time should be spent on each method. Certainly some youngsters could handle both with facility while some would struggle mightily with at least one of these methods.
Further, I suspect there are some youngsters who convert mixed numerals to improper fractions procedurally without full conceptual understanding that a mixed numeral is an addition problem!
Maria Miller said...
I'm sorry but I can't tell you any answer to your question.
I just wanted to point out that there are several ways of doing a subtraction in the mixed numeral form, once the fractional parts have the same denominator.
For example:
3 1/5 - 1 3/4
= 3 4/20 - 1 15/20
Here's one way: subtract in parts.
3 4/20 - 1 15/20
= 3 4/20 - 1 - 4/20 - 11/20
= 2 - 11/20 = 1 9/20
Another way, like you said, convert 1 whole into 20/20:
3 4/20 - 1 15/20
= 2 24/20 - 1 15/20
= 1 9/20
And yet another, allowing a negative fraction in between the steps (this is similar to my first one):
3 4/20 - 1 15/20
= (3 - 1) + (4/20 - 15/20)
= 2 + (-11/20) = 1 9/20.
Would this "flexibility" in methods perhaps make this method (I mean where we just convert the fractional parts to like fractions) superior in the sense that more kids will be able to grasp at least one way of doing it?
Then again maybe it just adds to the confusion. But perhaps you could first show them the first step:
3 1/5 - 1 3/4
= 3 4/20 - 1 15/20
and then let the students figure it out by themselves... see what they come up with.
Dave Marain said...
Thank you, Maria. Some interesting alternate methods there that deepen understanding.
I'm hoping to hear from teachers currently following a middle school curriculum. However, I am interested in others' views on the relative merits of these methods. If some teachers feel that improper fraction methods are easier for students, the other methods may be deemphasized. Do you think this is ok?
As students get older they reach for the fraction keys on their calculator or change everything to decimals and convert back to fractions at the end. Of course graphing calculators do not allow entry of mixed numerals.
Alex said...
I've not taught this yet, but one thing that occured to me is you have to manipulate much bigger numbers with the top-heavy fractions method... so weak students will be put off by that. Another reason to go mixed.
David said...
Here's a middle school perspective.
I agree with your comment that a higher conceptual level is required for the mixed numerals method. But most
students don't have that, and I think it is a dangerous method to teach as it leads to so many conceptual errors
later.
Teaching shortcuts only really works when students have sufficient grasp of the basic concept to appreciate what it
is that the shortcut is doing. Otherwise they lose track of the whole idea, their learning becomes
compartmentalized (if I get one looking like that I do it that way, if I get one looking like this I do it this
way......) and they can't see how ideas link together.
I think it is best to teach a straightforward method which can always be applied in any situation. That's the general addition method. Then they already know how to convert between mixed and improper, and vice versa, so it's just combining two ideas they already have - nothing new. That's plenty complicated enough for the vast majority of students.
The smarter ones will start to see shortcuts of their own, and then, maybe, we can teach them the mixed numerals method.
I think you have to be very careful bringing in carpenters and cooks here - i.e. applications for mixed numerals. Many students won't be carpenters, and many won't cook - so what's the point. In addition - most won't have any appreciation for the application, even those who may, eventually, end up as carpenters or cooks (even if only as a hobby), until much, much later. (and the world's going metric anyway!) If you are going to have your fractions lessons out in the carpentry shop or in the kitchen baking muffins, then perhaps yes, it is appropriate, but not generally.
It's very dangerous linking maths to the real world. It is true that it can be done, but it has to be a context which resonates with students sufficiently well for it to stick in their minds. So we need to pick methods that can consistently be relied upon to work, and only later, as their appreciation of the concepts develops, do we show shortcuts and clever little tricks and oh-look-we-can-do-this kind of things.
Sorry about this - I think I've gone off on a bit of a rant (and you should see the bit I saved in a file for later use). Fractions is one of those topics I've spent a LOT of time teaching and thinking about. Often I think that they are one of THE most useful things we teach, they are part of that doorway into advanced mathematical ideas that so many find it hard to get through. I've always been able to DO fractions, but to be honest, I don't think I ever really UNDERSTOOD them until I began teaching them. They still blow me away. They're so simple, yet so complex.
We have to be really, really careful when teaching them, feeling our way through our students thoughts to make sure everything is set up right. If that sounds like someone playing with a ticking bomb then I agree - because if we don't get things set up right we make SUCH a mess of things for later!
I don't think enough people appreciate this.
David said...
Oh, and the 'improper fractions means bigger numbers' thing? Come on! They have to learn to deal with that, that's our job as teachers. I can assure you, any kid who's scared of big numbers is going to run a MILE when faced with any fraction, let alone a mixed one.
Dave Marain said...
Alex and David--
David, I was hoping for the kind of passionate response you supplied from someone who has lived with this issue for some time and reflects deeply on his(her0 profession. Your arguments are swaying me, but I need to hear from others. Actually, it would be so beneficial to students if teachers had the opportunity to regularly meet and discuss issues of content and pedagogy.
I still believe there are some advantages to working with mixed numerals. A sense about the magnitude of these numbers can be lost when converting to improper and the final result may be way off without the student realizing it. Mixed numerals allow for quicker estimating at the beginning of the problem and, perhaps, some exercises in estimation should be assigned for that purpose. Further, while we may not all become carpenters or chefs, we will all have occasion to measure objects (metric is still a ways off here!).
Perhaps, the mixed numerals algorithm for adding/subtracting should be taught and practiced only with relatively simple fractions. The fact that the improper fraction method is then taught and most students might prefer it does not mean the other method was a waste of time. Analogously, in algebra, we commonly teach 3 methods for solving a system of linear equations: graphing, substitution and elimination (linear combinations). Students may prefer one method over another, but I've seen students make choices when needed depending on the form of the given system. Nothing wrong with that. Yes, these students are older and more mature, but most (not all) middle schoolers can handle more than we currently expect IMO, at least here in the US. I suspect, David, you have much more to say on this issue - I welcome that and so will my readers!
mathmom said...
I'm not really seeing much need to add or subtract mixed numbers even in carpentry or cooking. I had occasion recently to want to make 3/4 or a recipe that already contained lots of mixed number measurements (so it would fit in my bread machine) but that was multiplying, and for that I definitely converted to improper fractions.
My intuition is to stick with converting to improper fractions. To me adding and subtracting with mixed numbers is more of a curiosity, similar to adding and subtracting times.
mathmom said...
Oh, and Dave, if your students ever realize when their result is way off, you're doing something right. :) IME "sanity checking" is a very hard practice to drive home.
David said...
I think I would agree that maturity has a lot to do with it. Consider the age at which we deal with these topics - systems of equations come much later than fractions. Consider also the processes involved. Systems of equations involve an understanding of general methods - the fact that each problem is basically a variation of a previous problem. Follow the algorithm and you'll get the right answer.
But fractions, which do involve an understanding of general methods, are taught way, way earlier than systems of equations are, and often at that stage we haven't really started talking about the general nature of solutions and methods. I think perhaps we should. Anything that involves an algorithm is an example of a general solution.
For example, I often throw huge, enormous numbers into problems. The kids freak out, but my response is 'what? what's the fuss? So it's a big number? Who cares, is it REALLY any different from the previous problem?' I'm trying to get them to look at the PROBLEM itself, not the numbers. Once kids start doing that, once they start looking at the STRUCTURE of what's going on, you're winning. Once they start to realise that each problem is just a variation of another, THEN they're learning. Then they're thinking mathematically.
Note, this is NOT the same as getting them to do umpteen variations of the same problem over and over again. That's drill and kill. No, I'm talking about discussion, challenge, forcing them to LOOK at what's happening, to compare and contrast problems - to apply higher order thinking skills to what they are doing rather than just zoning out. It's not easy, for some kids it's hell being required to actually think about this stuff, but it's worth it in the end.
I think as teachers we really need to have some appreciation of why mathematics even exists in the first place. It came from our inbuilt need to find order in the world around us. Now, that's a desire to generalise, it's not a desire to come up with itty bitty separate results all the time. So we need to work with those ideas with students. The numbers are important, but they're the building blocks, it's the structure and generalities that really matter.
Back to fractions. Again, I think it's important to work on general solutions. To develop an understanding of why they work. What exactly are we doing? What on earth is going on here? Once there's more understanding of that, we can start looking at shortcuts and alternative methods.
I do agree that mixed numbers methods could be great for developing estimation skills - I agree with that wholeheartedly. But that's different - I mean, why not just ignore the fractional bit and go with the whole numbers. Then it's just addition or subtraction of whole numbers.
Oh, I just remembered - that was one of my points - if we teach mixed number methods then later they try to apply them to multiplication and division. Now those are ideas which are complicated enough - you want to try working on the distributive property with them as well? (Have you ever really tried to convert a fraction division problem into plain language that kids understand? It's HARD!) Why not just teach them that to work with mixed numbers in maths they need to be converted so they're all one type of number, and then we can work with that. (Think about it - most of the time when we have various quantities involved in a problem, we convert to just the one).
And I agree about the discussion of content and pedagogy. It's professionally and personally enriching, and it makes us think. The more we share ideas, the better teachers we become.
Dave Marain said...
You folks are certainly making me think!
Now I'm wondering where this topic resides in the K-8 Achieve, CA or Mass Math Standards. I don't remember seeing this explicitly stated. If anyone knows how it's worded in their state's standards, I'd be interested. Usually, this level of specificity is missing.
At this point, I'm still not convinced that students should not be exposed to both methods.
Anonymous said...
Dave,
I'm guessing that for word problems mixed numbers are preferred and for "math" without application improper fractions are preferred.
Time would always fall in the former category.
Subtract
15 hr 37 min
9 hr 46 min
?
6 hours with a shortfall of 9 minutes, but that's me.
Doubling back, I actually teach my high school students to do the word problem = mixed, math = improper business. The improper fractions are easier to work with, but you can get some pretty angry stares for asking for five-quarters of a pound of something or other - and you may not even get it.
Jonathan
Burt said...
As a newcomer to these math blogs, first I want to say these blogs are great. Thanks Dave for running this blog. I think it is very important for teachers to have discussions, just as it is important for kids to have discussions in class about concepts and strategies.
David, your posts seem contradictory to me. In the first post you say most kids can’t conceptualize a mixed number solution, so let’s make it easier for them. In a later post you say you don’t want them freaking out over the numbers; you want them to focus on the structure of the problem and solution. I agree strongly with what you say here: “The numbers are important, but they're the building blocks, it's the structure and generalities that really matter.”
My question then is: isn’t a mixed number solution with borrowing the same in structure as subtraction with whole numbers with borrowing? The main difference is that the numbers are different. The rate of decomposing the borrowed whole number is not 1 to 10, unless the fraction is tenths, but the concept is the same. It seems to me that this is an opportunity to help the students better understand what fractions are. If decomposing the whole number and combining it with the fraction is the problem, then that can be addressed separately before doing the subtraction with mixed numbers. In other words, they are not ready for this kind of mixed number problem if they do not understand that 1 + 4/20 = 20/20 + 4/20 = 24/20. If this is not where the difficulty is, then please tell me where the difficulty is.
Dave Marain said...
Burt,
After re-reading all the comments I believe a balanced approach is still the best. You made a critical point in your argument. Identify those parts of the algorithm that seem to create the most difficulty and reinforce those ideas prior to teaching the algorithm. Analyzing and anticipating what steps cause the problems is so fundamental to effective instruction. This comes with experience and reflection.
On the other hand, many youngsters also have trouble with the synthesis aspect. That is, they struggle with multi-step processes and, particularly those with learning disabilities, need the recipe in front of them for some time if not always.
I still feel that students do need to develop better "fraction sense" than they currently have -- I guess that's obvious to our readers here! Expressing 1 as 2/2 or 5/5 or 13/13 is very important. I don't think we can simply deemphasize this without thinking carefully about the ramifications down the road. I also believe many middle schoolers are capable of more sophisticated thinking than we give them credit for. I see how well they can manipulate their parents -- surely they can figure out how to manipulate a mixed numeral!
BTW, I forgot that you were addressing the other 'David'in this thread and I was initially trying to figure out where I contradicted myself!
I also believe David had some excellent arguments and his sentiments are shared by many teachers with whom I've spoken.
Burt, thanks for your thoughtful comment.
Dave | 4,150 | 19,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-30 | latest | en | 0.970396 |
https://behindthesciences.com/towards-5g/channelmodelssuitableforstudyingdsss | 1,642,512,824,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00371.warc.gz | 193,541,133 | 39,690 | # Channel models suitable for studying DSSS
The typical channels that model the multipath behaviour of the wireless link are either a combination of the exponential channel model, or the Rayleigh channel model, together with AWGN [1]. These models present the most suitable conditions for the wireless environment studied in the DSSS systems.
1. AWGN channel
System simulations for this channel correspond to the simplest case: an AWGN channel which was programmed as a random signal in [2]. This signal is characterized by having no statistical correlation between two values in the time domain. As a result, its power spectral density is a constant. Therefore, it is a flat graph, which means that it contains all the frequencies and all of them have the same power. This is the reason why it is named “white”. As its name indicates, its probability density function follows a Gaussian or normal distribution. Hence, both characteristics describe this type of channel. In addition, it is additive because is added to the transmitted signal in the receiver’s input. This channel does not introduce delays or attenuation in the transmitted signal; it only adds the random sequence described above. Its power its described by the following relation:
where
is the noise density, which depends on the energy per bit, Eb.
2. Rayleigh channel and equal power multipath channel
In the multipath Rayleigh channel model, taps are distributed with uniform random phase and Rayleigh distribution for the amplitudes. In this model, the delay is considered smaller than the sample rate and it simulates an environment where fading occurs. An important parameter to characterize this channel is the rms delay spread, which in comparison with the symbol duration, indicates whether the channel introduces flat fading (when it is smaller) or selective fading (when it is greater). The rms can be computed by applying the following expression:
As mentioned, the paths in this channel are equal power, which means that they are independent and identically distributed (i.i.d.) Gaussian variables.
3. Exponentially decaying channel
The exponential model is characterized by independent Gaussian variables which represent taps that have an exponentially decaying average power profile. This model is consistent for narrowband systems. In order to model this channel, we need to describe the following parameters:
• The standard deviation is used to normalize the taps power to 1 and it is defined as:
K is the number of paths introduced by the channel, and according to the IEEE802.11b, it can be truncated in order to have a finite impulse response as:
The variance is expressed as:
Ts is the sampling period, and σk is a normalization factor, defined as:
The expression above has a real and an imaginary part. Therefore, the standard deviation results in:
We can fix τrms to typical values for indoor environments, such as 30 ns.
REFERENCES
[1] S. Halford, K. Halford, and M. Websteri (2000). Evaluating the performance of HRb proposals in the presence of multipath, Intersil Corporation, doc.: Ieee 802.11-00/xxx | 630 | 3,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.941962 |
https://wiki.openoffice.org/wiki/Documentation/How_Tos/Calc:_BETAINV_function | 1,709,303,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00053.warc.gz | 627,821,090 | 7,881 | # BETAINV function
FUNCTIONS
Probability and statistics
Correlation and line fitting
Ranking
## BETAINV
Calculates the inverse of the BETADIST function.
### Syntax:
BETAINV(p; α; β; a; b)
The beta distribution is a family of continuous probability distributions, defined for an interval a to b, where α and β are parameters controlling the shape of the distribution.
BETAINV returns the value n, lying between a and b, such that BETADIST(n; α; β; a; b) is p.
a and b are optional parameters which default (if omitted) to 0 and 1.
### Example:
BETAINV(0.96; 3; 4)
returns approximately 0.75. | 165 | 599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.730465 |
http://threesixty360.wordpress.com/category/uncategorized/ | 1,369,176,386,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700842908/warc/CC-MAIN-20130516104042-00044-ip-10-60-113-184.ec2.internal.warc.gz | 260,352,541 | 24,340 | Archive for the ‘Uncategorized’ Category
Luggage Math Mistake No More!
September 19, 2010
Remember that post on August 1 about the math mistake?
The mistake here is that while individual dimensions were correctly converted to centimeters (by multiplying by 2.54), the Capacity was incorrectly converted, since the 4720 cubic inches were multiplied by 2.54 (and not the cube of 2.54) to get 11988.8 cubic centimeters.
I found this error a month or so before posting it so I know it was around for a while, but I just discovered that the error has been fixed. Behold, the new stats!
In this case, the somehow-determined capacity of 4720 cubic inches is multiplied by (2.54)³, or approximately 16.387 cubic cm per cubic inch, to get 77,346.9 cubic centimeters. But one liter is, conveniently [although not exactly coincidentally], exactly 1000 cubic centimeters, so this translates to 77.3469 liters, which does round to 77.3 liters. The stats for other pieces of luggage were similarly updated.
I’m so happy! I’d like to think that the People In Charge actually read this blog, or at least the email I sent them about it, but alas I have no evidence of this. Still, it’s nice to see the mistake corrected.
Counting with 0
June 10, 2010
We’ve just painted our living room, for the first time since moving in ten years ago, and decided to celebrate the new jazzed up look by getting a new lightswitch for the porch door to replace the old, painted over one. (Question: Why did the previous painters think it was a good idea to paint over every light switch and every electrical outlet?) I eventually found a place that had some pretty neat ones, but I noticed that the two pages of lightswitches [11 items, listed 6 to a page] were ordered in an unusual way:
YES — they started counting at 0. Can you imagine how happy that made me? Although the fact that our living room was newly painted, with 10 years of crayons and scratch marks obliterated in just 2 4 6 days of work might have also had something to do with it.
Thirds
April 14, 2010
As far as fractions go, halves and quarters get a lot of the glory — it’s fairly easy to cut things in half for one, and one could argue that more integers are divisible by two than by any larger integer (although one could also argue the opposite, so that’s probably moot). Lately I’ve been thinking that thirds are really the sleeper fraction.
Take the calendar year: we divide it into four seasons, but they’re not quite equal, at least in upstate NY. Winter spans three of them most years, which can feel a little depressing at times. So about a year ago I started wondering if there was a way to redistrict the year in order to contain winter more thoroughly, and the best configuration I could come up with was:
Fall: August-November
Winter: December-March
Spring-Summer: April-July
It’s not quite perfect, but actually fits our academic schedule pretty well — school starts at the end of August so most of August is spent getting ready for school, and not so much in summer mode. November is probably more like winter, but at least the first half of it is usually pretty fall-like. Then December through March are clearly winter, no two ways about it, but April…ah, April. I love April. It might be my favorite month, except it’s always so busy, but that’s when the daffodils come up and the grass thinks about growing, and we can open the windows more often. Like I said, April tends to be really busy, but it also has that happy Summer’s Coming feel, and the seniors are thinking about graduation, and it’s really a jump into summer. And May is finals and graduation, which lead to June and July. So for me at least, these thirds are a little more natural that the quarterly seasons. [And for schools that start in September, a one-month shift might work, depending on whether school or the weather is the dominant feeling of winter.]
I suspect that thirds might work well for a clock, too, where the AT hours are for work/school, the BT hours are for homework and play, and the CT hours are for sleep, although when I try to put numbers to this it doesn’t work quite as well, since our work day itself tends to be split into three parts (early morning at home, day at work, and then evening at home again, stealing time from the other parts). But still, in theory it makes sense, at least as much as 12-hour divisions.
Goooooo…thirds!
March 14, 2010
If you haven’t been there yet, head over to Google [their image links to a search for Pi Day). If you’ve missed it, it might show up in their archive page here. Happy Pi Day everyone!
February 8, 2010
Here’s what happens when you didn’t get any all of your grading done this weekend (but the Saints won!!!!!!! Yay New Orleans!), and then instead of catching up before class your department gathers informally and starts to talk about Friday’s wacky trillion point grading scheme, and if there’s a class that lends itself to that, and someone mentions that Physics would really be the ideal course for this, because they have to deal with measures of scale so often.
Well, there’s only one place for this conversation to go: the other extreme. What if you had a class that, instead of having 500 points for the semester, or even 1,000,000,000,000, had only 1 point. Total. Exams could be worth 2/10 of a point, each homework assignment might only be worth 1/250 of a point, etc. If someone had an unexcused absence, you’d knock 1/500 off a grade. It might make it easier to defend points deducted, too, because you could say, “Look, I only took off 5 thousandths of a point for that mistake, so it’s hardly worth arguing about.”
Plus, this could be a learning experience because you could use Official Prefixes: “This exam is worth 2 decipoints (not to be confused with decapoints)”. The trouble is, most science students are pretty familiar with deci, centi, and milli so if you really want to make it…memorable….having 1 point for the whole semester is still too much.
The problem is, in looking online, we don’t really seem to have enough prefixes. The prefixes start off simply enough (deci, centi, and milli being 1/10, 1/100, and 1/1000 respectively) but then decrease by factors of 1000: the next smallest amount would be a micropoint, which is actually a thousandth of a millipoint, or 1/1,000,000 of an actual point. Then there are nanopoints (1/1000 of a micropoint) and picopoints (1/1000 of a nanopoint) and while those words are fun to say, we’d need something closer in scale to a micropoint to be able to distinguish amounts, or else it becomes essentially a 1000 point grading scheme with a twist: “This exam is worth 200 yoctopoints out of a total of 1 zeptopoint for the semester.”
I guess maybe this isn’t so practical after all. And that’s truly a shame, because who wouldn’t want to write a grading scheme that used yoctopoints?
What has 8 legs and no room for anything else? A yoctopus!
January 2, 2010
It seems (perhaps only to me) like it ought to factor nicely, because 20 is twice 10. But once you factor out that 10 and that 3 you’re just left with a prime, since 2010 is just 2·3·5·67. (Speaking of four primes, did you know that you can get four prime New York Steaks for \$132.95 on Amazon? I was relieved to see that they were not available for Super Saver Shipping.)
Wolfram Alpha points out that 2010 itself is a factor of 296-1. And Number Gossip adds that it’s untouchable, which means that there aren’t any numbers whose proper divisors add up to 2010.
It can be written as 133122 in Base 4, which is kind of cool, and as 6, 3, 12 in Base 18; my favorite, however, is that it is 5, 10, 15 in Base 19.
Finally, it’s equal to:
669+670+671
400+401+402+403+404
127+128+…+141 and several others
[Hmmm...I can find a string for each of the 7 odd factors, but I'm not sure that exhausts all of the possibilities.]
While getting ready to post this, I noticed that MathNotations has a similar post from yesterday. Whoops!
And by circles we mean….
August 27, 2009
There was an interesting article in ScienceNOW last week about how a study has shown that people walk in circles if they are blindfolded and set loose in a field (unless it’s sunny out; then they stay in a straight line). The article referred to various novels and movies, indicating that as a plot devise walking in circles isn’t unrealistic.
As I read it, I kept thinking, “Circles? Really? But wouldn’t maintaining constant curvature be about as remarkable as walking in a straight line?” Yet no mention was made of this, and when I looked at the accompanying picture it turns out that circle was used in a more literary sense: the circles were really just closed loops occurring within the path. And although that too is interesting because I’d be freaked out by crossing my own path again, my mathematician self was sorry that they weren’t actually circles.
This squiggle picture actually has a lofty pedigree: it appeared on page 1 of La Peau de Chagrin by Honore de Balzac in 1901.
Kitten math
August 1, 2009
We adopted kittens this week, within 24 hours of arriving home. This was the culmination of several months of deciding whether or not it was a good idea given that we already have two adult cats who weren’t actually asking for younger siblings. So far, the kittens are happy, one adult cat is curious and I think will be fine, and the other adult cat has at least progressed to the point where she’ll eat snacks right by the doorway to the room the kittens are primarily staying in.
Anyway, the place we got them from advertises that one adult female can produce 420,000 offspring over 7 years. That seemed a little high, so I decided to check the math.
The first thing to wonder about was how to count offspring. Clearly this was more than just kittens: it must be counting all future generations. But predicting how many offspring males can sire was difficult, because it really only takes a few males to father a LOT of kittens. So I decided to initially look only at the female lines.
In looking around, it seems that cat pregnancies last about two months, and a female can get pregnant again about a month after that. In theory this would mean 4 litters a year, but The Internet implies that most cats have 2-3 litters per year. I decided to go with 3, since I was trying to see if that 420,000 was even resonable (as opposed to typical).
There’s also the question of how old cats have to be until they are able to mate. It turns out to be around 6 months, though for convenience (since I was going by 4 month intervals) I decided to go with 8. Or, looking at it another way, I figured that kittens would have their own litter on their 1st birthday (and every 4 months after that), so that’s assuming mating at about 10 months old.
Finally, there’s the size of the little. I decided to go with 4 kittens per litter, two of which were female.
Then I made a table. It started off like this:
4 months: 1 adult, 2 (female) newborn kittens
8 months: 1 adult, 2 newborns, 2 kid-kittens
12 months: 3 adults, 2 newborns, 2 kids
16 months: 5 adults, 6 newborns, 2 kids
20 months: 7 adults, 10 newborns, 6 kids
By the end of 7 years, there were 35951 adults, 42410 newborns, and 25006 kid-kittens, leading to a grand total of 103,367 cats, including the original one, all along female-descendent lines. If you double it in order to count the males in each litter (but no separate offspring of those males), you get around 206, 732 offspring.
Initially it sounds like the 420,000 is an overestimate, but the wording only is that cats can have that that many youngsters, not that they normally do. If you assume that there are 3 females in each litter (all of whom procreate, etc, since this is just a ballpark estimate) you actually get 898798 offspring along the all-female lines, or over 1.5 million offspring in 7 years. So while I doubt that having half a million offspring within 7 years is the norm, it certainly seems possible in extreme cases.
Math Teachers at Play #9 is up at HB
June 12, 2009
Or, rather, Math Teachers at Play #9 is up at Homeschool Bytes (who have hosted MTaP before as well!) The theme is Game Time, and the post has several games and magic tricks.
Speaking of games, did you know that they have a bunch of math games at coolmath? You can play Mancala (although it seems like there are a million variations to that), the Tower of Hanoi, and lots of other stuff. Watch out for the sound effects, because that can get a little overwhelming at times. (Fifteen minutes later: I just finished a game of Sudoku, helped out by the 8-9 year olds in the living room. Why didn’t I play with this site before? I think Sudoku is one of the best ways to teach Proof by Contradiction, since once you get past the infinitude of the natural numbers, the infinitude of the primes, and the square root of ___ is irrational it’s harder to come up with good examples.
(Incidentally, in courses where I teach that the square root of 2 is irrational I usually make it homework to show that the square root of 3 is irrational. Once they’re comfortable with that, I have them prove that the square root of 4 is irrational, and to find the flaw in the proof. This proves to be really challenging to the sophomore math majors.)
I think I’ll end my stream of consciousness here.
WordPress does the math right
June 3, 2009
This is hardly worthy of a blog post (but, really, if I waited for stuff that was there would be like 3 posts a month), but it still really amused me.
WordPress lets you look at stats for individual posts, which is fun because you can see which posts were the most popular over time (Scoring March Madness by a landslide, thanks to Basketball Guy). There’s also a column for the % change each week, so if 2 let’s say 200 people look at that particular post one week, and 3 300 people look at it the next week, the percentage increase is +50%, from (300-200)/200. But what happens when you first post? The previous week clearly no one looked at the post, so finding the percentage increase causes problems. But not for WordPress!
The percentage increase (in this case of the views of Godzilla makes a hexaflexagon) is . That little attention to detail makes me happy.
DragonFable Math
May 11, 2009
The 8½ year old in our household recently announced, “I have an idea for a blog post! DragonFable Math!” DragonFable is a role-playing game in which you “walk around, go on quests, and slaughter stuff.” Young E insisted that there was math in it, however, explaining:
If you don’t look at the melee (aka damage) [the damage points that you've scored against your opponent] and just look at the hit points, you can figure out how much damage was sent.
A character’s hit points measure their ability to endure damage, and they decrease by the number of damage points in each fight. When your hit points reach zero, you’re toast!
So what Young E was describing is the missing addend model for subtraction. And actually, it turns out that there’s more than just that. By Googling “DragonFable Math” painstaking research, I found several DragonFable Game Formulae. For example, there’s :
and
Total Stats Possible At Any Given Level= 3*(Your Level-1)
complete with examples. Many of these are at a good level for upper elementary school, and sort of real-life (virtual life?). Moreover, the formulas were derived by players,which to my mind suggests opportunities for even more interesting problems of the what’s-the-pattern variety.
So credit to E, for showing me that there’s more to DragonFable than just dragons.
More Big Bills
May 4, 2009
Yesterday I posted about the biggest bills in the US, and that was interesting to me in a historical sense. However, the part of me that likes big numbers really liked seeing what other countries had to offer.
This proved to be difficult. There are a lot of countries, and even individual countries change their currencies from time to time, especially if there is high inflation [which would be tied to really big numbers on the bills]. Here is but a small sample.
There’s a 100,000 rubles bill (2000) from Belarus, worth about US\$36.
There used to be a 2,000,000 złotych (1992) in Poland but in 1995 the złoty was formed by dropping 4 of the zeroes, leaving the largest current note as 200 złoty. It’s worth about US\$61 and has Ignacy Jan Paderewski on the front, who was a Prime Minster of Poland and also a world-class pianist.
From 1990-2005 there was a 20,000,000 note in Turkish Lira. Isn’t it colorful? It’s got Mustafa Kemal Atatürk on it. He was the first president of Turkey, and in 1927 gave a speech that lasted 36 hours.
But the numbers don’t stop in the millions, no siree. About 15 years ago Yugoslavia, right before it split up, had a 500 billion dinar note featuring Serbian poet Jovan Jovanović Zmaj:
From the National Bank of Serbia (NBS Copyright)
And that’s not even the biggest. Zimbabwe is still undergoing hyperinflation, and has a 100 billion dollar bill (2008):
Even with all those zeroes it’s not worth much, less than US\$1, and by the time you read this the value will have decreased even more. Less than four months ago newspapers reported that Zimbabwe would start printing notes in the trillions, up to and including a 100,000,000,000,000 dollar note, worth about US\$30 at the time.
Gum and Math: Is there a connection? REALLY?
April 28, 2009
Last week I was all excited about an article showing a connection between schools with playgrounds and higher math scores. As it happens, at the very same time there was another news report about higher test scores. This one linked them to gum.
The first article I read, from the LA Times, explained:
The study was conducted by researchers at Baylor College of Medicine and was sponsored by the Wrigley Science Institute. The study included 108 students, ages 13 to 16, who were assigned either to chew sugar-free gum during math class, while doing math homework and during math tests, or to refrain from gum-chewing. After 14 weeks, the students took a math test and their grades were assessed.
The result was increased test scores for the gummy kids on the Texas Assessment of Knowledge and Skills achievement test. The increase was small — only 3% — but the National Post says it was statistically significant. Several articles drew the connection that chewing gum can lower stress, and that may have helped with the testing.
The National Post article turned out to be one of the more detailed ones that I ran across (explaining, for example, that the gum-chewers chewed gum 86% of the time in school and 36% of the time while doing homework) and it shared a fact that I didn’t see in many other articles: there were TWO standardized tests given, not just one. The other standardized test was the Woodcock Johnson III Tests of Achievement, and on this test there was no statistically significant difference in the scores of those who chewed gum and those who didn’t.
Furthermore, when I looked further, I found this nutrition site that revealed that the course grades for both groups actually decreased over time, although the gum chewers had less of a decrease. So all in all, it’s kind of interesting but perhaps not worth the hype of headlines like “Chewing Gum Improves Math Scores… No Lie” and “Teenage Brain Power Boosted by Chewing Gum: Wrigley Study”.
Isn’t that a gross photo at the top? That’s Bubblegum Alley in my hometown [photo taken by Samir and posted on wiki under Creative Commons License]. When I was about 10 a friend and I wrote “Sit on It” in gum on the wall of Bubblegum Alley. We felt very sneaky.
Summer Jobs Puzzle
April 9, 2009
I’m in the mood for a puzzle. There are about a million on the web, but I liked this one from Puzzlers Paradise because it talked about summer and because there was a cool grid that you can fill online here. [You click once in a square to X it out, twice to put in the Yeah-I-got-it! circle, and three times thrice to clear it.]. The same site has a whole bunch of these, and a lot more puzzles on their main page.
This summer, five friends each found a different job to earn money. As it happened, they all started their new jobs in the same week; each on a different day and each earning a different amount. Determine the type of job each got, the day of the week each started working, each friend’s mode of transportation to their new job, and how much per hour each friend was earning.
1. Chris didn’t have a newspaper route but he did use the bus. The boy who mowed lawns took the train to work.
2. The boy who used a bicycle started working on Friday. Joe didn’t get a job as a painter.
3. The five friends are represented, in no particular order, by the following: the boy who made \$6.00 per hour, the boy who started work on Thursday, the boy who used a car to get to work, Chris, and the boy who mowed lawns.
4. Peter started work the day before the boy who worked as a cashier but two days after the boy who made \$7.00 per hour. The boy who worked as a cashier made \$0.50 per hour more than Greg made.
5. The boy who started work on Tuesday made \$6.25 per hour. Rob didn’t start work on Monday. Peter made more per hour than the painter but less than the boy who used a car.
6. Rob walked to work. The person who started on Wednesday didn’t make \$6.50 per hour. The golf caddy used a car to get to work.
Good Luck!
The photo of the beach has nothing to do with summer jobs, but it was taken during the summer [by TwoPi, in Oregon].
Tax Math: Mesopotamia Stye
April 6, 2009
We’ve talked about Aztec Taxes and Inca Taxes; what about taxes in Mespotamia?
There are many many tablets that can attest to the fact that they had them. According to Dr. Tonia Sharlach at the University of Pennsylvania, lots and lots of things were taxed year round [Here's a whole article about the salt tax!], with payment of objects instead of money. This apparently led to a lot of smuggling, and Dr. Sharlach’s article quotes a letter from about 3900 years ago written by an office to an employee:
Irra’s son sent smuggled goods to Pushuken but his smuggled goods were intercepted. The Palace then threw Pushuken in jail! The guards are strong…please don’t smuggle anything else!
Apparently one of the worst taxes was the labor tax (sometimes called the “burden”). While it sometimes feels like a certain amount of my job amounts to a labor tax, it’s admittedly quite a bit nicer than being somewhere else to work for a few months.
Upcoming…why is tax day April 15 anyway?
The map was posted under GNU-FDL by Nareklm. | 5,520 | 22,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.960245 |
https://www.eit.edu.au/can-you-calculate-pressure-in-a-tank/ | 1,675,208,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00729.warc.gz | 776,156,330 | 70,444 | and on August 9th, 2012
Dear Colleagues
An interesting little (simple) brain teaser for you – no matter whether what discipline or type of engineering professional you are.
The Challenge
A tank at atmospheric pressure contains 1 kg of air. The tank is then pressurized with an additional 3 kgs of air. What is the resultant gauge pressure (in bars) in the tank after this 3kgs of air has been added? Absolute pressure = gauge pressure + atmospheric pressure
A. 1 bar
B. 3 bar
C. 4 bar
You could use the Universal Gas Law in your deliberations:
PV = nRT
Where V is volume; P is absolute pressure; n is the number of moles (‘molecules’); R is a constant and T is the temperature.
Assume atmospheric pressure is 1 bar (the initial pressure) – naturally this is a number which could change depending on what altitude you are at.
Solution (suggested – you could do it in a myriad of different ways)
There is initially 1kg of air occupying V0 initial volume.
With 3kgs of air added; there is now a total of 4kg of air in the tank (thus the final number of moles or particles is effectively 4 times that of the initial number).
The volume (V0) and temperature (TO) of the tank still stays the same.
Thus:
P0 V0 = n0 R T0 (initial state)
P1 V0 = n1 R T0 (final state where n1 = 4 x n0)
Thus – dividing each side of equation 1 and 2 we get:
P1 V0 / P0 V0 = 4 x n0 R T0 / n0 R T0
P1 V0 / P0 V0 = 4 x n0 R T0 / n0 R T0
Thus P1 = 4 P0
Hence, the final pressure P1 = 4 bar absolute or 3 bar gauge pressure (thus answer B above)
Do you agree?
My humble appreciation to Dr Rodney Jacobs and David Spitzer (the flow guru) for their contributions and critique.
Albert Einstein remarked: I never teach my pupils. I only attempt to provide the conditions in which they can learn.
Yours in engineering learning
Steve
• This field is for validation purposes and should be left unchanged.
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http://newvillagegirlsacademy.org/math/?page_id=4069 | 1,519,061,569,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812758.43/warc/CC-MAIN-20180219171550-20180219191550-00240.warc.gz | 247,893,339 | 15,187 | # 7.3 – Simplifying Rational Expressions
## Review
Reducing Fractions to Simplest Form
• Step 1: To simplify a fraction, factor the its numerator and denominator.
• Step 2: Look for the greatest common factor (GCF) and cancel it.
• Ex. $\frac{6}{24}$ can be canceled by dividing the numerator and denominator by $\frac{6}{6}$
• So, $\frac{6}{24}\div \frac{6}{6}=\frac{1}{4}$
## Notes
Rational Expressions
• The first step in simplifying a rational expression is to factor its numerator and denominator.
• Then cancel those factors by dividing the numerator and denominator by the GCF.
• Example of a rational expression: $\frac{x^{2}-5x-14}{x-7}$
GCF – Greatest Common Factor
• The GCF of a rational expression is the largest factors that can be canceled from both the numerator and denominator.
• Ex. $\frac{3x-15}{x^{2}-x-20}$
• Factor the numerator and denominator: Ex. $\frac{3(x-5)}{(x-5)(x+4)}$
• Since the numerator and denominator both have the factor: (x-5), this is the GCF.
## Examples
Ex 1. Which of the following is equal to the rational expression when x ≠ 1 or -1? $\frac{5(x-1)}{(x+1)(x-1)}$ Note: The denominator cannot equal 0, which is why x ≠ 1 and x ≠ -1 Cancel common factors from the numerator and denominator. Divide the numerator and denominator by $\frac{(x-1)}{(x-1)}$ and they will cancel. Answer: $\frac{5}{x+1}$ Ex 2. Which of the following is equal to the rational expression when x ≠ -6? $\frac{x^{2}-36}{x+6}$ Note: The denominator cannot equal 0, which is why x ≠ -6 First, factor the numerator into: $(x+6)(x-6)$ Divide the numerator and denominator by $\frac{(x+6)}{(x+6)}$ and they will cancel. Answer: $x-6$ Ex 3. Which of the following is equal to the rational expression when x ≠ 5 or x ≠ -3? $\frac{x^{2}+x-6}{x^{2}-2x-15}$ Factor the numerator and denominator and look for factors to cancel. $\frac{(x+3)(x-2)}{(x-5)(x+3)}$ Cancel the (x+3) factors. Answer: $\frac{(x-2)}{(x-5)}$ Ex 4. A rational expression has been simplified below. $\frac{(x+4)(x-2)}{9(x-2)}=\frac{x+4}{9}$ For what values of x are the two expressions equal? First, ask yourself, “What would make the denominator undefined?” Second, cancel any factors that exist in the numerator and denominator. The answer is “All real numbers except the number(s) that make the denominator undefined.” For this problem, $x-2=0$ is in the denominator, so 2 would make the epxression undefined. Answer: All real numbers except 2 Ex 5. What is the simplified form of the rational expression below? $\frac{6x^{2}-54}{5x^{2}+15x}$ Factor the numerator and denominator: $\frac{6(x+3)(x-3)}{5x(x+3)}$ The GCF is (x+3), so cancel it. Answer: $\frac{6(x-3)}{5x}$ | 804 | 2,659 | {"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": 23, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2018-09 | latest | en | 0.784934 |
https://www.civilengineeringhandbook.tk/reactor-engineering/d-t.html | 1,552,940,864,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201672.12/warc/CC-MAIN-20190318191656-20190318213656-00206.warc.gz | 741,001,064 | 8,166 | ## D t
In these equations, n is the molecular flux of momentum and g and F are gravitational acceleration and external body forces, respectively. The physical interpretation of the various terms appearing in these equations again follows similar lines: the first term is the rate of increase in momentum per unit volume; the second term represents change in momentum per unit volume, caused by convection; the fourth and fifth terms represent the gravitational force per unit volume and any other external force, if present, respectively. The third term represents molecular contributions, which include pressure and viscous force per unit volume and is given as where p is pressure and t is the viscous stress tensor. It may be noted that the second term of the right-hand side is not simple divergence because of the tensorial nature of t .
In order to use these general momentum conservation equations to calculate the velocity field, it is necessary to express viscous stress terms in terms of the velocity field. The equations which relate the stress tensor to the motion of the continuous fluid are called constitutive equations or rheological equations of state. Although the governing momentum conservation equations are valid for all fluids, the constitutive equations, in general, vary from one fluid material to another and possibly also from one type of flow to another. Fluids, which follow Newton's law of viscosity (although it is referred to as a law, it is just an empirical proposition) are called Newtonian fluids. For such fluids, the viscous stress at a point is linearly dependent on the rates of strain (deformation) of the fluid. With this assumption, a general deformation law which relates stress tensor and velocity components can be written:
where Sij is the Kronecker delta (Sj = 1 if i = j and Sij = 0 if i = j) function, ^ is the coefficient of viscosity and k is the coefficient of bulk viscosity. The superscript 'T' denotes the transpose of a tensor quantity. In general, it is believed that, except in the study of shock waves and in the absorption and attenuation of acoustic waves, it is convenient to ignore the coefficient of bulk viscosity. Substitution of Eq. (2.7) into Eq. (2.6) and Eq. (2.5) results in the complete momentum conservation equation. A special case of these momentum conservation equations for constant density and constant viscosity fluids is the famous Navier-Stokes equation, which provides the usual starting point for the analysis of flow processes (Bird et al., 1960; Deen, 1998).
Any fluid, which does not obey Newton's law of viscosity, is called a non-Newtonian fluid. This class covers widely different materials/fluids, varying from those exhibiting slight deviation from Newtonian behavior to almost elastic solids. Fluids exhibiting slight deviations from Newtonian behavior, such as power law fluids (which exhibit a power law relationship between stress and strain) or Bingham plastic fluids (which require finite yield stress for flowing), can be readily modeled using the same framework. The molecular viscosity term appearing in Eq. (2.7) is replaced by an effective viscosity term, which may be a function of local stress and strain values. More complex behavior, e.g. viscoelastic behavior, requires a completely different framework to develop satisfactory constitutive equations. The subject of developing suitable constitutive equations for viscoelastic fluids, is extremely complex and outside the scope of this book. As stated earlier, the focus in this book is on simulating turbulent, multiphase and reactive flows. Detailed discussion about the rheology and motion of complex fluids can be found in Tanner (1985), Bird et al. (1987) (constitutive equations, models) and Crochet et al. (1984) (numerical simulation).
2.1.3. Conservation of Energy
Application of the law of conservation of energy can be used to derive transport equations for total energy. In order to derive an equation for internal energy, it is first necessary to derive a transport equation for mechanical energy, which can then be subtracted from the equation for total energy to obtain the governing equation for internal energy (Bird et al., 1960; Bird, 1998). Without going into details, the equation for internal energy written in terms of static enthalpy is given below:
dt (ph) + V-(p Uh) = -V-(q) + Dp - (t : V U) -V ■ ^ hk jk j + Sh (2.8) Here h is an enthalpy, which is defined as
0 0 | 934 | 4,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-13 | latest | en | 0.925135 |
http://www.python-forum.org/viewtopic.php?f=10&t=6861&p=8884 | 1,503,372,280,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00117.warc.gz | 652,942,236 | 6,367 | broken power law script
A forum for general discussion of the Python programming language.
broken power law script
Hey evrybody... I am totally new to python... I found the following script... Please help me understand it...I know about MATLAB...either help me converting it to matlab or help me in understanding it so that i can run it...
Code: Select all
import agpy.mpfit as mpfit
import numpy as np
def powerfit(xax,data,err=None,alphaguess=-2.0,scaleguess=1.0,quiet=True):
"""
Fit a power law (a line in log-space) to data as a function of x
differs from 'plfit' because plfit fits a power law distribution,
this code simply fits a power law
"""
logdata = np.log10(data)
if err is None: err = np.ones(data.shape,dtype='float')
def mpfitfun(data,err):
def f(p,fjac=None): return [0,np.ravel(((np.log10(p[0])+np.log10(xax)*p[1])-data)/err)]
return f
mp = mpfit.mpfit(mpfitfun(logdata,err),xall=[scaleguess,alphaguess],quiet=quiet)
fitp = mp.params
return fitp,mp
def brokenpowerfit(xax, data, err=None, alphaguess1=0.0, alphaguess2=-2.0, scaleguess=1.0,
breakpoint=None, quiet=True):
"""
Fit a broken power law (a line in log-space) to data as a function of x
differs from 'plfit' because plfit fits a power law distribution,
this code simply fits a power law
This is a lot more intricate than the simple power law fit, since it involves
fitting two power laws with different slopes
Parameters:
p[0] - scale
p[1] - breakpoint
p[2] - power 1 (xax < breakpoint)
p[3] - power 2 (xax >= breakpoint)
There are 5 parameters (NOT 4) returned because there are two scales that are *NOT*
independent
returns: scale1,scale2,breakpoint,alpha1,alpha2
"""
logdata = np.log10(data)
if err is None: err = np.ones(data.shape,dtype='float')
def brokenpowerlaw(p):
lowerhalf = (np.log10(p[0]) + np.log10(xax)*p[2]) * (xax < p[1])
# find the location at which both functions must be equal
scale2loc = np.argmin(np.abs(xax - p[1]))
scale2 = np.log10(xax[scale2loc])*(p[2] - p[3]) + np.log10(p[0])
upperhalf = (scale2 + np.log10(xax)*p[3]) * (xax >= p[1])
# DEBUG print "scale1: %15g scale2: %15g xaxind: %5i xaxval: %15g lower: %15g upper: %15g" % (p[0],scale2,scale2loc,np.log10(xax[scale2loc]),lowerhalf[scale2loc-1],upperhalf[scale2loc])
return lowerhalf+upperhalf
def mpfitfun(data,err):
def f(p,fjac=None): return [0,np.ravel((brokenpowerlaw(p)-data)/err)]
return f
if breakpoint is None: breakpoint = np.median(xax)
parinfo = [{}, {'mpminstep':xax.min(),'mpmaxstep':xax.max(),'step':xax.min()}, {}, {}]
mp = mpfit.mpfit(mpfitfun(logdata,err),xall=[scaleguess,breakpoint,alphaguess1,alphaguess2],quiet=quiet,parinfo=parinfo)
fitp = mp.params
scale2loc = np.argmin(np.abs(xax - fitp[1]))
scale2 = 10**( np.log10(xax[scale2loc])*(fitp[2] - fitp[3]) + np.log10(fitp[0]) )
fitp = np.array( [fitp[0],scale2] + fitp[1:].tolist() )
return fitp,mp
I have a file with 3 columns(on which i have to run the script)... TIME MAGNITUDE MAGNITUDE_ERROR
how do we give the 3 columns as input in python...??
Thanks
Last edited by Yoriz on Sat Sep 14, 2013 11:29 pm, edited 1 time in total.
Posts: 1
Joined: Sat Sep 14, 2013 10:54 pm
Re: broken power law script
What is the objective of your analysis? Do you only want to calculate the power fit? In MatLab you can do this via:
Code: Select all
lobf = polyfit( log(x), log(y), 1 )
a = exp( lobf(2) )
b = lobf(1)
% Assuming equation has form: y = a*x^b
As far as inputting data in Python, you first must load the data in some usable way. e.g.
Code: Select all
import numpy as np
data = np.double( np.array( [r.split(delimiter) for r in data] ) )
You now have an r x 3 array and can call the data as needed. e.g.
Code: Select all
# Remember, Python operates on with zero as the first counter, not one like MatLab
x1 = data[0,0]
y1 = data[0,1]
err = data[0,2]
Python: 2.7 via Anaconda
Numpy: 1.7
Pandas: 0.11
OS: Windows 7
IDE: Spyder/IPython
tnknepp
Posts: 153
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In Personal Finance
# What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow ( Full Answer )
In Classical Music
# Who was Felix mendelssohn?
Felix Mendelssohn is regarded by classical music aficionados and critics alike, as one of the most prolific and gifted composers the world has ever known. Even those who could ( Full Answer )
In Acronyms & Abbreviations
# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo ( Full Answer )
In Coins and Paper Money
# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
In Math and Arithmetic
# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
In Volume
# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
In Numerical Analysis and Simulation
# What is the answer for 5c equals -75?
The 'answer' is the number that 'c' must be, if 5c is really the same as -75. In order to find out what number that is, you could use 'algebra'. First, write the equatio ( Full Answer )
In iPhone 5
# How many pixels does the iPhone 5c have?
The iPhone 5c is 640 x 1136 pixels. That is about 326 pixels persquare inch (ppi).
In Temperature
# What is minus 5c in Fahrenheit?
(-5) degrees Celsius = 23 degrees Fahrenheit. Formula: [°F] = [°C] à 9 â 5 + 32
In iPhone 5
# How many inches is a iPhone 5c?
The screen is 4" big. The height is 4.9", width is 2.33" and thedepth is 0.35" | 571 | 1,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-13 | latest | en | 0.910694 |
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# Fractions (Pri-3)
/ Fractions Ahmad Ibrahim Primary School / Homework (2018)
$Q1 What is the product of 4/9 and 3/12?$
Ho
July 6, 2018
0
0
232
CT
$\frac{4}{9}×\frac{3}{12}=\frac{4}{9}×\frac{1}{4}=\frac{1}{9}\phantom{\rule{0ex}{0ex}}\mathrm{Ans}:\frac{1}{9}$
0
Aug 26, 2018 03:55 AM
Correct!! Great working
August 26, 2018
0
August 26, 2018
Wa
$7\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}|{c}_{n}·$
0
August 23, 2018
Wa
$\frac{3}{3,}·1$
0
August 23, 2018
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http://catalog.ysu.edu/courses/math/ | 1,519,295,766,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00702.warc.gz | 60,899,562 | 24,435 | MATH 1501 Elementary Algebraic Models 5 s.h.
Arithmetic of integers and of rational numbers; linear equations and inequalities in one variable; polynomials, factoring, algebraic fractions, radicals and quadratic equations; linear systems in two variables; graphs. Does not count toward a degree.
Prereq.: Level 10 on Math Placement Test.
MATH 1505 Intermediate Algebra with Applications 5 s.h.
This course is intended to prepare STEM students for their college-level mathematics requirement. Topics include linear and nonlinear equations and inequalities; problem solving; relations of function types that include linear, polynomial, radical, rational, exponential, and logarithmic; applications. Math Placement Level 10 or higher. Does not count toward a degree.
MATH 1507 Intermediate Algebra 3 s.h.
Topics include functions of the following: linear, polynomial, rational, exponential, and logarithmic. Emphasis on function relations and graphing by algebraic techniques and technology. Solving linear, nonlinear equations and inequalities. Does not count toward a degree.
Prereq.: MATH 1501 or Level 20 on Math Placement Test.
MATH 1510 College Algebra 4 s.h.
This course is primarily intended to prepare STEM students (along with MATH 1511) for MATH 1570 or 1571 and business students for MATH 1552. Topics include real numbers, equations and inequalities, linear, quadratic, polynomial, exponential, and logarithmic functions, graphing techniques, systems of equations, and applications. The course fulfills the general education requirements for mathematics.
Prereq.: MATH 1505 or MATH 1507 with a "C" or better or Math Placement Level 35 or higher.
Gen Ed: Mathematics.
MATH 1510C Corequisite Support for College Algebra 1-3 s.h.
This course is intended to provide corequisite support for students requiring remediation in mathematics while they are concurrently enrolled in MATH 1510 (College Algebra). Emphasis will be placed on prerequisite skills needed for college algebra as well as just in time review through the use of appropriate technology. Does not count toward a degree.
Prereq.: Concurrent enrollment in MATH 1510.
MATH 1511 Trigonometry 3 s.h.
This course, along with MATH 1510 is primarily intended to prepare STEM students for MATH 1570 or MATH 1571. Topics include algebraic structure and graphs of trigonometric functions and inverse trigonometric functions, angle measurements, similar triangles, trigonometric identities, vectors, complex numbers, polar coordinates and solving trigonometric equations with applications.
Prereq.: MATH 1505 or MATH 1507 or Math Placement Level 35.
Gen Ed: Mathematics.
MATH 1511C Corequisite Support for Trigonometry 1-3 s.h.
This course is intended to provide corequisite support for students requiring remediation in mathematics while they are concurrently enrolled in MATH 1511 (Trigonometry). Emphasis will be placed on prerequisite skills needed for trigonometry as well as just in time review through the use of appropriate technology. Does not count toward a degree.
Prereq.: Concurrent enrollment in MATH 1511.
MATH 1513 Algebra and Transcendental Function 5 s.h.
Function concepts including trigonometric, exponential, and logarithmic functions. Application problems and graphing. Supplemental topics.
Prereq.: Math Placement Level 45 or higher.
Gen Ed: Mathematics.
MATH 1552 Applied Mathematics for Management 4 s.h.
Apply functions, linear systems, linear programming to business including use of technology; mathematics of finance and an introduction to limits, derivatives and integrals with business applications. No credit for students who have completed MATH 1570 or MATH 1571.
Prereq.: MATH 1510 or at least Level 45 on the Mathematics Placement Test.
Gen Ed: Mathematics.
MATH 1552C Corequisite Support for Applied Math for Management 1-3 s.h.
This course is intended to provide corequisite support for students requiring remediation in mathematics while they are concurrently enrolled in MATH 1552 (Applied Math for Management). Emphasis will be placed on prerequisite skills needed for business calculus as well as just in time review through the use of appropriate technology. Does not count toward a degree.
Prereq.: Concurrent enrollment in MATH 1552.
MATH 1564 Foundations of Middle School Mathematics 1 4 s.h.
Conceptual foundations of topics from number theory, operations, functions, algebra, geometry, measurement, probability, and data analysis. Emphasis on multiple approaches and representations, problem solving, and communication of mathematical reasoning. Includes inquiry-based laboratory experiences with manipulatives and computing technology.
Prereq.: At least Level 40 on the Mathematics Placement Test or concurrent registration in MATH 1507 (for both).
MATH 1570 Applied Calculus 1 4 s.h.
The elements of differential and integral calculus, with emphasis on applications. Analytical geometry, differentiation and integration techniques and series representations. Introduction to differential equations, transform calculus, and Fourier analysis. This is a basic methods course particularly adapted for those who require applied topics in mathematics. Not applicable toward the Mathematics major. Credit will not be given for both MATH 1549 and MATH 1570.
Prereq.: MATH 1513, or MATH 1510 and MATH 1511, or at least Level 50 on the Mathematics Placement Test.
Gen Ed: Mathematics.
MATH 1571 Calculus 1 4 s.h.
A sequence of integrated courses in analytic geometry and calculus. A detailed study of limits, derivatives, and integrals of functions of one and several variables with applications.
Prereq.: MATH 1513, minimum grade of "C", or MATH 1510 and MATH 1511, minimum grade of "C" in both courses, or at least Level 70 on the Mathematics Placement Test.
Gen Ed: Mathematics.
MATH 1571C Corequisite Support for Calculus 1 1-3 s.h.
This course is intended to provide corequisite support for students requiring remediation in mathematics while they are concurrently enrolled in MATH 1571 (Calculus I). Emphasis will be placed on prerequisite skills needed for calculus as well as just in time review through the use of appropriate technology. Does not count toward a degree.
Prereq.: Concurrent enrollment in MATH 1571.
MATH 1571H Honors Calculus 1 4 s.h.
A sequence of integrated courses in analytic geometry and calculus. A detailed study of limits, derivatives, and integrals of functions of one and several variables with applications.
Prereq.: MATH 1513, minimum grade of "C", or MATH 1510 and MATH 1511, minimum grade of "C" in both courses, or at least Level 70 on the Mathematics Placement Test.
Gen Ed: Mathematics.
MATH 1572 Calculus 2 4 s.h.
A sequence of integrated courses in analytic geometry and calculus. A detailed study of limits, derivatives, and integrals of functions of one and several variables with applications.
Prereq.: MATH 1571.
Gen Ed: Mathematics.
MATH 1572H Honors Calculus 2 4 s.h.
A sequence of integrated courses in analytic geometry and calculus. A detailed study of limits, derivatives, and integrals of functions of one and several variables with applications.
Prereq.: MATH 1571.
Gen Ed: Mathematics.
MATH 1580H Honors Biomathematics 1 2 s.h.
Counting techniques, probability, matrices and linear systems. Emphasis on the role of mathematical models in explaining and predicting phenomena in life sciences.
MATH 1581H Honors Biomathematics 2 4 s.h.
A study of functions, differential and integral calculus. Emphasis on the role of mathematical models in explaining and predicting phenomena in life sciences. Credit will not be given for both MATH 1581H and MATH 1571.
Gen Ed: Mathematics.
MATH 1585H Accelerated Honors Calculus 1 5 s.h.
A sequence of honors courses in analytical geometry and calculus which cover essentially the same material as MATH 1571, 1572, 2673, in two semesters instead of three. A detailed study of limits, derivatives, and integrals of functions of one and several variables and their applications. This sequence will be offered at most once during each academic year.
Prereq.: ACT math subscore of 32, AP Calculus score of 4 or higher, or at least one unit of high school calculus with a score of 28 or higher on placement exam or instructor permission.
Gen Ed: Mathematics.
MATH 1586H Honors Calculus Laboratory 1 1 s.h.
Introduction to mathematical modeling of topics covered in calculus. Emphasizes the use of technology such as computer algebra systems, technical document processing, and graphics software for solving problems and reporting solutions.
Prereq.: MATH 1571 or concurrent with 1585H.
MATH 2623 Quantitative Reasoning 3 s.h.
Mathematics models emphasizing basic ideas in mathematics and statistics, stressing concept formation rather than manipulative skills. Credit will not be given for both MATH 2623 and STAT 2625.
Prereq.: MATH 1501 or Level 20 on the Mathematics Placement Test.
Gen Ed: Mathematics.
MATH 2623C Corequisite Support for Quantitative Reasoning 1-3 s.h.
This course is intended to provide corequisite support for students requiring remediation in mathematics while they are concurrently enrolled in MATH 2623 (Quantitative Reasoning). Emphasis will be placed on prerequisite skills needed for MATH 2623 as well as just in time review through the use of appropriate technology. Does not count toward a degree.
Prereq.: Concurrent enrollment in MATH 2623.
MATH 2623H Honors Quantitative Reasoning 3 s.h.
Mathematics models emphasizing basic ideas in mathematics and statistics, stressing concept formation rather than manipulative skills. Credit will not be given for both MATH 2623 and STAT 2625.
Prereq.: MATH 1501 or Level 20 on the Mathematics Placement Test.
Gen Ed: Mathematics.
MATH 2651 Mathematics for Early Childhood Teachers 1 3 s.h.
A conceptual development of mathematics topics underlying today's Pre-K-grade 3 curriculum. Emphasis on multiple approaches, problem solving, and communication of mathematics. Incorporates classroom activities, manipulatives, technology, and activities developmentally appropriate for young children.
Prereq.: MATH 1501 or Level 20 on the Mathematics Placement Test.
MATH 2652 Mathematics for Early Childhood Teachers 2 3 s.h.
A conceptual development of mathematics topics underlying today's Pre-K-grade 3 curriculum. Emphasis on multiple approaches, problem solving, and communication of mathematics. Incorporates classroom activities, manipulatives, technology, and activities developmentally appropriate for young children.
Prereq.: MATH 2651.
Gen Ed: Mathematics.
MATH 2665 Foundations of Middle School Mathematics 2 4 s.h.
Conceptual foundations of topics from number theory, operations, functions, algebra, geometry, measurement, probability, and data analysis. Emphasis on multiple approaches and representations, problem solving, and communication of mathematical reasoning. Includes inquiry-based laboratory experiences with manipulatives and computing technology.
Prereq.: At least Level 40 on the Mathematics Placement Test or concurrent registration in MATH 1507 (for both).
Gen Ed: Mathematics.
MATH 2670 Applied Calculus 2 5 s.h.
The elements of differential and integral calculus, with emphasis on applications. Analytical geometry, differentiation and integration techniques and series representations. Introduction to differential equations, transform calculus, and Fourier analysis. This is a basic methods course particularly adapted for those who require applied topics in mathematics. Not applicable toward the Mathematics major. Credit will not be given for both MATH 1549 and MATH 1570.
Prereq.: MATH 1570.
Gen Ed: Mathematics.
MATH 2673 Calculus 3 4 s.h.
A sequence of integrated courses in analytic geometry and calculus. A detailed study of limits, derivatives, and integrals of functions of one and several variables with applications.
Prereq.: MATH 1572.
MATH 2673H Honors Calculus 3 4 s.h.
A sequence of integrated courses in analytic geometry and calculus. A detailed study of limits, derivatives, and integrals of functions of one and several variables with applications.
Prereq.: MATH 1572.
MATH 2686H Accelerated Honors Calculus 2 5 s.h.
A sequence of honors courses in analytical geometry and calculus which cover essentially the same material as MATH 1571, 1572, 2673, in two semesters instead of three. A detailed study of limits, derivatives, and integrals of functions of one and several variables and their applications. This sequence will be offered at most once during each academic year.
Prereq.: MATH 1585H.
Gen Ed: Mathematics.
MATH 2687H Honors Calculus Laboratory 2 1 s.h.
Introduction to mathematical modeling of topics covered in calculus. Emphasizes the use of technology such as computer algebra systems, technical document processing, and graphics software for solving problems and reporting solutions.
Prereq.: MATH 1572 or concurrent with MATH 1572H or 1586H.
MATH 3701 Biomathematics Seminar 1 s.h.
Introduction to interdisciplinary research in biology and mathematics. Topics include current research by faculty and students, cross disciplinary communication, report writing, technical presentations, literature reading, laboratory techniques and safety. May be repeated once. Listed also as BIOL 3701.
Prereq.: MATH 1571 or BIOL 2601 or BIOL 2602.
MATH 3702 Problem Solving Seminar for Secondary Mathematics 3 s.h.
Approaches to and practice with problem solving with examples from a broad spectrum of mathematics. Emphases include problems at the level of the Praxis II examination for mathematics and problems suitable for high school contests such as the American Mathematics Competition 10 and 12.
Prereq.: MATH 1572 or consent of instructor.
MATH 3705 Differential Equations 3 s.h.
Methods and theory of solving differential equations with applications. Existence, uniqueness. First order equations. Higher order linear equations. Introduction to partial differential equations and boundary value problems, including Laplace's equation.
Prereq.: MATH 2673.
MATH 3705H Honors Differential Equations 3 s.h.
Methods and theory of solving differential equations with applications. Existence, uniqueness. First order equations. Higher order linear equations. Introduction to partial differential equations and boundary value problems, including Laplace's equation.
Prereq.: MATH 2673.
MATH 3715 Discrete Mathematics 3 s.h.
A course in discrete mathematical structures to prepare students for advanced courses. Topics include set theory, functions and relations, logic and quantifiers, truth tables and Boolean expressions, induction and other techniques of proof, and graphs. Credit will not be given for both CSCI 3710 and MATH 3715.
Prereq.: MATH 1572.
MATH 3720 Linear Algebra and Matrix Theory 3 s.h.
Matrices; matrix operations; linear transformations; applications.
Prereq.: MATH 1572.
MATH 3721 Abstract Algebra 1 4 s.h.
Introduction to abstract algebra investigating fundamental concepts in group and ring theory. Topics include groups, subgroups, cyclic groups, permutation groups, cosets, direct products, homomorphisms, factor groups, rings, integral domains and polynomial rings.
Prereq.: MATH 3715 and MATH 3720.
MATH 3745 Topics in Mathematical Modeling 3 s.h.
This course exposes students to methods of mathematical modeling through applications. Tools used to develop, refine, test, and present mathematical models will be discussed. Topics covered and projects undertaken may vary with each course offering and are designed to expose students to the types of problems modeled by applied mathematicians working in business, government, industry, or research. Course may be repeated depending on projects or topics presented.
Prereq.: MATH 2673 or permission of the instructor.
MATH 3750 History of Mathematics 3 s.h.
A survey of the historical development of mathematics.
Prereq.: MATH 3715.
MATH 3751 Real Analysis 1 4 s.h.
Introduction to the properties of the real number system and metrics and metric properties, with critical analysis of limits, continuity, differentiability, integration, and other fundamental concepts underlying the calculus.
Prereq.: MATH 2673 and MATH 3715.
MATH 3760 Numerical Analysis 1 3 s.h.
The theory and techniques of numerical computation. The solution of a single equation, interpolation methods, numerical differentiation and integration, direct methods for solving linear systems.
Prereq.: MATH 3720 and CSIS 2610.
MATH 3767 Algebra/Geometry for Middle School Teachers 1 4 s.h.
An integrated, conceptual, and function-centered approach to the foundations of algebra, geometry, and trigonometry for preservice middle childhood mathematics specialists. Emphasis on multiple approaches and representations, problem solving, and communication of mathematical reasoning. Includes inquiry-based laboratory experiences. Not applicable to the mathematics major.
Prereq.: MATH 1564 and either 40 on the Mathematics Placement test or MATH 1507.
MATH 3768 Algebra/Geometry for Middle School Teachers 2 4 s.h.
An integrated, conceptual, and function-centered approach to the foundations of algebra, geometry, and trigonometry for preservice middle childhood mathematics specialists. Emphasis on multiple approaches and representations, problem solving, and communication of mathematical reasoning. Includes inquiry-based laboratory experiences. Not applicable to the mathematics major.
Prereq.: MATH 2665 and either 40 on the Mathematics Placement test or MATH 1507.
MATH 3785 Numerical Methods 3 s.h.
Matrices, matrix operations, and the application of numerical methods. Not applicable to the Mathematics major.
Prereq.: MATH 2670 and ENTC 1505, or equivalent.
MATH 3795 Topics in Mathematics 2-3 s.h.
The study of a mathematical topic or the development of a special area of mathematics. May be repeated once.
Prereq.: MATH 1549 or MATH 1570 or MATH 1571 or MATH 2623 or MATH 2651.
MATH 4822 Abstract Algebra 2 3 s.h.
A continuation of MATH 3721 with special emphasis on fields. Additional topics in pure or applied algebra.
Prereq.: MATH 3721 or equivalent.
MATH 4823 Abstract Algebra 3 3 s.h.
This course introduces advanced topics in field theory. Topics may include principal ideal domains, irreducibility, quotient rings, algebraic extensions, finite fields, splitting fields, and the Galois group.
Prereq.: MATH 4822.
MATH 4830 Foundations of Geometry 3 s.h.
The development of Euclidean and non-Euclidean geometries from postulate systems.
Prereq.: MATH 3715.
MATH 4832 Euclidean Transformations 3 s.h.
General properties of functions and transformations; isometries and transformations of the Euclidean plane; the complex plane, its geometry and subfields; transformational, analytical, and vector approaches to Euclidean geometry; connections to other branches of mathematics and applications.
Prereq.: MATH 3720 and MATH 4830.
MATH 4843 Theory of Probability 3 s.h.
The mathematical foundation of probability theory including the study of discrete and continuous distributions. Other topics selected from limit theorems, generating functions, applications. Credit will not be given for MATH 4843 and STAT 4843.
Prereq.: STAT 3743 and MATH 2673.
MATH 4855 Ordinary Differential Equations 3 s.h.
A second course in differential equations with emphasis on nonlinear problems and qualitative methods or on boundary value problems. Topics are chosen from: proofs of fundamental theorems, phase plane analysis, limit cycles and the Poincare-Bendixon theorem, biological models, stability via Liapunov functions, asymptotic methods, and boundary value problems.
Prereq.: MATH 3705 and MATH 3720.
MATH 4857 Partial Differential Equations 3 s.h.
Introduction to partial differential equations (PDE) including solution techniques and applications. Classifications of the basic types of PDE's (hyperbolic, parabolic and elliptic) and dependence on boundary and initial conditions. Topics include Fourier series, integral transforms (Fourier, Laplace), and applications in vibrations, electricity, heat transfer, fluids or other selected topics.
Prereq.: MATH 3705 and MATH 3720.
MATH 4869 Functions, Calculus, and Applications for Middle School Teachers 3 s.h.
Polynomial and exponential functions, limits, derivatives, integrals, and applications. Interpretation of slope and area in graphs of functions from applied settings. Applications of limits to the derivations of geometric formulas. Relations between tables, graphs, and the symbolic representation of functions.
Prereq.: MATH 3767 or consent of instructor.
MATH 4870 Mathematics Seminar for Middle School Teachers 2 s.h.
Problem solving from a broad spectrum of mathematics topics (Number Sense and Operations; Algebra, Functions, and Calculus; Measurement and Geometry; Statistics, Probability, and Discrete Mathematics) designed to prepare future middle school mathematics teachers to address Common Core Standards. May be repeated 2 times.
Prereq.: MATH 3767, MATH 3768, MATH 4869, and either STAT 2601 or MATH 2625.
MATH 4875 Complex Variables 3 s.h.
Complex numbers and their geometric representation, analytic functions of a complex variable, contour integration, Taylor and Laurent series, residues and poles, conformal mapping.
Prereq.: MATH 3751 or equivalent.
MATH 4880 Introduction to Topology 3 s.h.
An introduction to the basic concepts of general topology: compactness, connectedness, and continuity in topological spaces.
Prereq.: MATH 3721 and MATH 3751.
MATH 4882 Biomathematics Research 1-2 s.h.
Interdisciplinary and individualized study of a topic in biology and mathematics. Student project mentored jointly by faculty in biology and mathematics. May be repeated once. Grading is Traditional/PR. Listed also as BIOL 4882.
Prereq.: MATH 3701, BIOL 3701, senior status and permission of the department chairperson.
MATH 4884 Mathematical Logic 3 s.h.
An introduction to the study of theories in formalized languages and to the theory of models.
Prereq.: MATH 3721 or PHIL 3719.
MATH 4896 Senior Undergraduate Research Project 2 s.h.
Individualized study of a topic in mathematics culminating in a written report and an oral presentation at a national or regional meeting or a local seminar. May be repeated once.
Prereq.: 24 s.h. of mathematics applicable to the mathematics major including either MATH 3721 or MATH 3751 and permission of the department chairperson.
Gen Ed: Capstone.
MATH 4897H Thesis 2 s.h.
Individualized study of a topic in mathematics culminating in a written report and an oral presentation at a national or regional meeting or a local seminar.
Prereq.: 24 s.h. of mathematics applicable to the mathematics major including both MATH 3721 and MATH 3751 and permission of the department chairperson.
MATH 5821 Topics in Abstract Algebra 4 s.h.
A course in abstract algebra aimed at developing a broad understanding of the subject. Credit will not be given for both MATH 3721 and MATH 5821.
Prereq.: MATH 3715 and MATH 3720.
MATH 5825 Advanced Linear Algebra 3 s.h.
A study of abstract vector spaces, linear transformations, duality, canonical forms, the spectral theorem, and inner product spaces.
Prereq.: MATH 3721.
MATH 5828 Number Theory 3 s.h.
A study of congruences, Diophantine equations, quadratic residues, special number theory functions, and selected applications.
Prereq.: MATH 3721.
MATH 5835 Introduction to Combinatorics and Graph Theory 3 s.h.
The pigeonhole principle; permutations, combinations, the binomial theorem; the inclusion-exclusion principle; recurrence relations; graphs and digraphs, paths and cycles, trees, bipartite graphs and matchings.
Prereq.: MATH 3715 and MATH 3720.
MATH 5845 Operations Research 3 s.h.
An introduction to operations research with emphasis on mathematical methods. Topics may include: linear programming, sensitivity analysis, duality theory, transportation problems, assignment problems, transshipment problems, and network problems.
Prereq.: MATH 3715 and MATH 3720.
MATH 5851 Topics in Analysis 4 s.h.
A course in analysis aimed at developing a broad understanding of the subject. Credit will not be given for both MATH 3751 and MATH 5851.
Prereq.: MATH 2673, MATH 3720, and MATH 3715.
MATH 5852 Real Analysis 2 3 s.h.
Uniform convergence of sequences of functions and some consequences; functions on n-space: derivatives in vector spaces, mean value theorem, Taylor's formula, inverse mapping theorem, implicit mapping theorem.
Prereq.: MATH 3720 and MATH 3751 or equivalent.
MATH 5860 Topics in Numerical Analysis 3 s.h.
A course in numerical analysis aimed at developing a broad understanding of the subject. Credit will not be given for both MATH 3760 and MATH 5860.
Prereq.: MATH 3720 and CSIS 2610.
MATH 5861 Numerical Analysis 2 3 s.h.
Numerical methods of initial-value problems, eigenvalue problems, iterative methods for linear and nonlinear systems of equations, and methods involving least squares, orthogonal polynomials, and fast Fourier transforms.
Prereq.: MATH 2673 and MATH 3760 or equivalent.
MATH 5875 Complex Variables 3 s.h.
Complex numbers and their geometric representation, analytic functions of a complex variable, contour integration, Taylor and Laurent series, residues and poles, conformal mapping.
Prereq.: MATH 3751 or equivalent.
MATH 5895 Selected Topics in Mathematics 2-3 s.h.
The study of a standard mathematical topic in depth or the development of a special area of mathematics. May be repeated twice.
Prereq.: 24 s.h. of mathematics applicable to the mathematics major including either MATH 3721 or MATH 3751.
MATH 5895W Selected Topics in Mathematics Topology 2 2-3 s.h.
The study of a standard mathematical topic in depth or the development of a special area of mathematics. May be repeated twice.
Prereq.: 24 s.h. of mathematics applicable to the mathematics major including either MATH 3721 or MATH 3751.
MATH 6905 Teaching Practicum 1 s.h.
Intensive preparation for teaching lower-level mathematics courses, featuring formal instruction and orientation on teaching issues, evaluated presentations, mentored classroom instruction, and weekly teaching seminars. Topics include course design, policies, syllabi, grading; classroom teaching problems; orientation in Mathematics Assistance Center, specific lower-level mathematics courses, online tutorial services. Required of and limited to graduate assistants in the Department of Mathematics and Statistics. To be taken each semester student is a graduate assistant. Grading is S/U. Does not count toward credit in the program.
MATH 6910 Advanced Engineering Mathematics 1 3 s.h.
Theory and solution techniques used in engineering applications. Topics include brief review of ordinary differential equations and linear algebra; vector calculus, integral theorems, complex analysis, series, residue theory, potential theory, special functions, integral transforms, partial differential equations and applications in mathematical modeling.
Prereq.: MATH 3705.
MATH 6911 Advanced Engineering Mathematics 2 3 s.h.
Theory and solution techniques used in engineering applications. Topics include brief review of ordinary differential equations and linear algebra; vector calculus, integral theorems, complex analysis, series, residue theory, potential theory, special functions, integral transforms, partial differential equations and applications in mathematical modeling.
Prereq.: MATH 6910.
MATH 6915 Mathematical Foundations 3 s.h.
Order-theoretic and monadic foundations of mathematics: ordered structures; topologies; powerset operators of a function; applications to continuity, compactness, algebra, logic, and calculus.
Prereq.: MATH 3721 Abstract Algebra I and MATH 3751 Real Analysis I, or permission of graduate coordinator.
MATH 6922 Advanced Topics in Group and Ring Theory 3 s.h.
A continuation of MATH 5821 with special emphasis on groups acting on sets, Sylow's Theorem and its applications, ring homomorphisms, ideals, and polynomial rings. Credit will not be given for MATH 4822 and MATH 6922.
Prereq.: MATH 3721 or MATH 5821.
MATH 6923 Advanced Topics in Field Theory 3 s.h.
This course introduces the major results in advanced field theory. These results include splitting fields, algebraic extensions, finite extensions, cyclotomic polynomials, and finite fields. Credit will not be given for MATH 4823 and MATH 6923.
Prereq.: MATH 4822 or MATH 6922.
MATH 6924 Galois Theory 3 s.h.
An introduction to Galois Theory with special emphasis on the Galois group, the Fundamental Theorem of Galois Theory, and radical extensions.
Prereq.: MATH 4823 or MATH 6923.
MATH 6928 Advanced Number Theory 3 s.h.
Advanced study of number theory: theory and distribution of primes, computational number theory, and additive number theory.
Prereq.: MATH 5828.
MATH 6930 Differential Geometry 3 s.h.
Classical differential geometry of curves and surfaces, differentiable manifolds with tensors.
Prereq.: MATH 5852.
MATH 6942 Advanced Operations Research 3 s.h.
Topics may include integer programming, advanced linear programming, nonlinear programming, dynamic programming, queuing theory, Markov analysis, game theory, and forecasting models.
Prereq.: MATH 5845 and STAT 3743 Probability and Statistics.
MATH 6955 Advanced Differential Equations 3 s.h.
Proofs of existence and uniqueness of nonautonomous, nonlinear equations. Additional topics may include advanced linear systems, partial differential equations, and integral equations.
Prereq.: MATH 5852 and either MATH 3705 Differential Equations or MATH 4855, or permission of graduate coordinator.
MATH 6957 Partial Differential Equations 3 s.h.
An introduction to partial differential equations (PDE) and their applications. The classification of the basic types of linear partial differential equations, development of how boundary and initial conditions affect solutions, exploration, and application of solution techniques for PDEs and explosions in orthogonal functions will be presented.
Prereq.: MATH 3705 and MATH 3720 or equivalent .
MATH 6965 Abstract Analysis 1 3 s.h.
Lebesgue integration and measure on the real line. General measure theory and functional analysis, including the Radon-Nikodym theorem, the Fubini theorem, the Hahn-Banach theorem, the closed graph and open mapping theorems, and weak topology.
Prereq.: MATH 5852 and either MATH 4880 or MATH 6915 or permission of graduate coordinator.
MATH 6975 Complex Analysis 1 3 s.h.
Analytic and meromorphic functions of a complex variable, contour integration, the Cauchy-Goursat theorem, Taylor and Laurent series, residues and poles, conformal mapping. Credit will not be given for both MATH 5875 and MATH 6975.
Prereq.: MATH 3751 Real Analysis I, or permission of graduate coordinator.
MATH 6980 Topology 1 3 s.h.
Basic concepts of topological spaces and mappings between them, including compactness, connectedness, and continuity. Credit will not be given for both MATH 4880 and MATH 6980.
Prereq.: MATH 3721 Abstract Algebra I and MATH 3751 Real Analysis I, or permission of graduate coordinator.
MATH 6981 Topology 2 3 s.h.
Separation, metrization, compactification. Additional topics will be selected from point-set topology, fuzzy topology, algebraic topology, combinatorial topology, topological algebra.
Prereq.: MATH 4880 or MATH 6980, or permission of graduate coordinator.
MATH 6990 Independent Study 1-3 s.h.
Study under the supervision of a staff member. May be repeated.
MATH 6995 Special Topics 1-3 s.h.
Specialized topics selected by the staff. May be repeated up to 12 semester hours.
Prereq.: Permission of graduate coordinator and department chair.
MATH 6996 Mathematical Project 1-3 s.h.
Individual research project culminating in a written report or paper, though not as broad in scope as a thesis. May be repeated once if the second project is in a different area of mathematics.
MATH 6999 Thesis 3 s.h.
A student may register for six semester hours in one semester or for three semester hours in each of two semesters.
MATH 7005 Advanced Topics in Categorical Topology 3 s.h.
Content varies with each offering. Implements ideas from MATH 6915, MATH 6980, MATH 6981, and studies categorical methods in topology and related concrete categories. Emphasis on current literature and open questions. May be repeated with approval of graduate coordinator.
Prereq.: MATH 6915, MATH 6980, MATH 6981, or equivalent, or permission of the graduate coordinator.
MATH 7015 Advanced Topics in Foundations of Topology 3 s.h.
Content varies with each offering, implements ideas from MATH 6915, MATH 6980, MATH 6981, and studies foundations of topology from a variety of viewpoints (algebraic, categorical, logical, order theoretic, powerset theoretic, set theoretic, etc.). Emphasis on current literature and open questions. May be repeated with approval of graduate coordinator.
Prereq.: MATH 6915, MATH 6980, MATH 6981, or equivalent, or permission of graduate coordinator.
MATH 7025 Advanced Topics in General Topology 3 s.h.
Content varies with each offering, implements ideas from MATH 6915, MATH 6980, MATH 6981, and studies various topics in point-set topology. Emphasis on current literature and open questions. May be repeated with approval of graduate coordinator.
Prereq.: MATH 6980, MATH 6981, or equivalent, or permission of graduate coordinator.
MATH 7035 Advanced Topics in Lattice-Valued Topology 3 s.h.
Content varies with each offering. Implements ideas from MATH 6915, MATH 6980, MATH 6981, and studies topology from the standpoint of lattice-valued (fuzzy) subsets. Emphasis on current literature and open questions. May be repeated with approval of graduate coordinator.
Prereq.: MATH 6915, MATH 6980, MATH 6981, or equivalent, or permission of the graduate coordinator.
MATH 7045 Advanced Topics in Topological Analysis 3 s.h.
Content varies with each offering. Implements ideas from MATH 6915, MATH 6965, MATH 6966, MATH 6980, MATH 6981, and studies the overlap between topology and abstract analysis (topological games, topological groups, separate versus joint continuity, etc.). Emphasis on current literature and open questions. May be repeated with approval of graduate coordinator.
Prereq.: MATH 6915, MATH 6965, MATH 6980, MATH 6981, or equivalent, or permission of graduate coordinator.
MATH 7055 Seminar in Topology and Abstract Analysis 3 s.h.
Content varies with each offering. Implements ideas from MATH 6915, MATH 6930, MATH 6965, MATH 6980, MATH 6981, MATH 6984, and focuses on current research activities of seminar participants. Student registrants are expected to make at least one major presentation each month of the term. May be repeated with approval of graduate coordinator. | 8,227 | 35,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.859645 |
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# greatest area
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31 Jan 2009, 04:33
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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31 Jan 2009, 07:50
Try with either issoceles or equilateral triangle.
The one with greater area is the answer.
1: issoceles triangle = 1/2 (1x1) = 1/2
2: equilateral triangle = 1/2 (1xsqrt3/2) = sqrt3/4
1/2 > sqrt3/4.
so B.
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Kudos [?]: 843 [0], given: 19
Re: greatest area [#permalink] 31 Jan 2009, 07:50
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Every Question Helps You Learn
Does this quadrilateral have only one pair of equal opposite angles?
Quadrilaterals belong to the polygon family, and are so named as they have four sides (this comes from Latin, with QUAD meaning 4, and LATERAL meaning sides). See how well you know the properties of quadrilaterals in this GCSE Maths quiz.
Quadrilaterals, like triangles, appear just about everywhere, and so it is important to have a good understanding of their properties. Starting at a very general level, the sum of the interior angles will add up to 360o. Can you remember why this is so? Draw in a diagonal, and it will form 2 triangles. The sum of the angles in a triangle is 180o, so the sum of two triangles will be 360o.
Once you start adding on other properties, such as a pair of parallel sides, opposite sides are equal length, diagonals bisect each other, then you create special quadrilaterals, that each have their own name.
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# Calculating the Maximum Volume of Fermentor that can be used without Additional Cooling
The heat generated by fermentor is 60 kcal/hr L. The cooling capacity of jacket is 10,000 kcal/hr m^2. For a cylindrical fermentor with H/D of 3, the jacket area (m^2) is porportional to fermentor volume (m^3) to the 2/3 power.
Area of Jacket (m^2) = 7*v^(2/3) (m^3)
What is the maximum volume of the fermentor that can be used without additional cooling capacity?
#### Solution Preview
The heat generated by a fermentor with volume v [L] is
Qgen=60*v ...
#### Solution Summary
Using equalites relating heat generated and heat removed, the maximum volume of the fermentor that can be used without additional cooling is calculated. The calculation steps are shown.
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http://www.ams.org/mathimagery/thumbnails.php?album=lastup | 1,490,502,174,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189092.35/warc/CC-MAIN-20170322212949-00160-ip-10-233-31-227.ec2.internal.warc.gz | 409,066,928 | 11,229 | The connection between mathematics and art goes back thousands of years. Mathematics has been used in the design of Gothic cathedrals, Rose windows, oriental rugs, mosaics and tilings. Geometric forms were fundamental to the cubists and many abstract expressionists, and award-winning sculptors have used topology as the basis for their pieces. Dutch artist M.C. Escher represented infinity, Möbius ands, tessellations, deformations, reflections, Platonic solids, spirals, symmetry, and the hyperbolic plane in his works.
Mathematicians and artists continue to create stunning works in all media and to explore the visualization of mathematics--origami, computer-generated landscapes, tesselations, fractals, anamorphic art, and more.
"OSU Triptych No. 2," (view 2) by Robert OrndorffPaper and acrylic, 20 x 46 cm, 2015 Honorable Mention, 2016 Mathematical Art Exhibition My paper folding art is math plus paper. Folded paper is simply tangible math. This is a permanent manifestation of an ephemeral artwork, namely, one solution for a specific one-straight-cut problem. Such problems are usually stated as follows: How must one fold a paper rectangle into a flat figure such that one straight cut through all of the layers will produce a given planar straight-line graph? Here the problem has been solved with paper and then represented in acrylic. To a significant degree the work relies on transmitted and reflected light, and so it never looks the same twice. The figure (the letters "OSU") has been divided into three frames. The crease patterns for the left and right letters are pedestrian but that for the central letter is sublime. --- Robert Orndorff Jun 23, 2016
"A Bird in Flight (2016)," by Hamid Naderi YeganehThis image shows all circles of the form: (x-A(k))^2+(y-B(k))^2=(R(k))^2, for k=-10000, -9999, ... , 9999, 10000, where A(k)=(3k/20000)+sin((π/2)(k/10000)^7)((cos(41πk/10000))^6)+(1/4)((cos(41πk/10000))^16)((cos(πk/20000))^12)sin(6πk/10000), B(k)=-cos((π/2)(k/10000)^7)(1+(3/2)(cos(πk/20000)cos(3πk/20000))^6)((cos(41πk/10000))^6)+(1/2)(cos(3πk/100000)cos(9πk/100000)cos(18πk/100000))^10, R(k)=(1/50)+(1/10)((sin(41πk/10000)sin(9πk/100000))^2)+(1/20)((cos(41πk/10000))^2)((cos(πk/20000))^10). --- Hamid Naderi YeganehMar 23, 2016
"Ring," by Hamid Naderi YeganehThis image shows 5,600 ellipses. For each k=1,2,3,...,5600 the foci of the k-th ellipse are: A(k)+iB(k)+C(k)e^(44πik/5600) and A(k)+iB(k)-C(k)e^(44πik/5600) and the eccentricity of the k-th ellipse is D(k), where A(k)=(cos(28πk/5600))^3, B(k)=sin(28πk/5600)+(1/4)(cos((14πk/5600)-(7π/4)))^18, C(k)=(1/70)+(1/6)+(1/6)sin(28πk/5600), D(k)=(399/400)-(1/6)(sin(28πk/5600))^8. --- Hamid Naderi YeganehMar 23, 2016
"8,000 Ellipses," by Hamid Naderi YeganehThis image shows 8,000 ellipses. For each k=1,2,3,...,8000 the foci of the k-th ellipse are: A(k)+iB(k)+C(k)e^(300πik/8000) and A(k)+iB(k)-C(k)e^(300πik/8000) and the eccentricity of the k-th ellipse is D(k), where A(k)=(3/4)sin(2πk/8000)cos(6πk/8000)+(1/4)sin(28πk/8000), B(k)=(3/4)cos(2πk/8000)cos(8πk/8000)+(1/4)cos(28πk/8000), C(k)=(1/18)+(1/20)cos(24πk/8000), D(k)=(49/50)-(1/7)(sin(10πk/8000))^4. --- Hamid Naderi YeganehMar 23, 2016
"Heart," by Hamid Naderi YeganehThis image shows 2,500 ellipses. For each k=1,2,3,...,2500 the foci of the k-th ellipse are: A(k)+iB(k)+C(k)e^(68πik/2500) and A(k)+iB(k)-C(k)e^(68πik/2500) and the eccentricity of the k-th ellipse is D(k), where A(k)=(-3/2)((sin(2πk/2500))^3)+(3/10)((sin(2πk/2500))^7), B(k)=sin((2πk/1875)+(π/6))+(1/4)(sin((2πk/1875)+(π/6)))^3, C(k)=(2/15)-(1/8)cos(πk/625), D(k)=(49/50)-(1/7)(sin(4πk/2500))^4. --- Hamid Naderi YeganehMar 23, 2016
"Butterfly (3)," by Hamid Naderi YeganehThis image shows 40,000 circles. For k=1,2,3,...,40000 the center of the k-th circle is (X(k), Y(k)) and the radius of the k-th circle is R(k), where X(k)=(3/2)((cos(141πk/40000))^9)(1-(1/2)sin(πk/40000))(1-(1/4)((cos(2πk/40000))^30)(1+(cos(32πk/40000))^20))(1-(1/2)((sin(2πk/40000))^30)((sin(6πk/40000))^10)((1/2)+(1/2)(sin(18πk/40000))^20)), Y(k)=cos(2πk/40000)((cos(141πk/40000))^2)(1+(1/4)((cos(πk/40000))^24)((cos(3πk/40000))^24)(cos(21πk/40000))^24), R(k)=(1/100)+(1/40)(((cos(141πk/40000))^14)+(sin(141πk/40000))^6)(1-((cos(πk/40000))^16)((cos(3πk/40000))^16)(cos(12πk/40000))^16). --- Hamid Naderi YeganehMar 23, 2016
"Butterfly (1)," by Hamid Naderi YeganehThis image shows 40,000 circles. For k=1,2,3,...,40000 the center of the k-th circle is (X(k), Y(k)) and the radius of the k-th circle is R(k), where X(k)=(6/5)((cos(141πk/40000))^9)(1-(1/2)(sin(πk/40000))^3)(1-(1/4)((cos(2πk/40000))^30)(1+(2/3)(cos(30πk/40000))^20)-((sin(2πk/40000))^10)((sin(6πk/40000))^10)((1/5)+(4/5)(cos(24πk/40000))^20)), Y(k)=cos(2πk/40000)((cos(141πk/40000))^2)(1+(1/4)((cos(πk/40000))^24)((cos(3πk/40000))^24)(cos(19πk/40000))^24), R(k)=(1/100)+(1/40)(((cos(2820πk/40000))^6)+(sin(141πk/40000))^2)(1-((cos(πk/40000))^16)((cos(3πk/40000))^16)(cos(12πk/40000))^16). --- Hamid Naderi YeganehMar 23, 2016
"Olive Branch," by Hamid Naderi YeganehThis image shows 4,000 circles. For k=1,2,3,...,4000 the center of the k-th circle is (X(k), Y(k)) and the radius of the k-th circle is R(k), where X(k)=(2k/4000)+(1/28)sin(42πk/4000)+(1/9)((sin(21πk/4000))^8)+(1/4)((sin(21πk/4000))^6)sin((2π/5)(k/4000)^12), Y(k)=(1/4)(k/4000)^2+(1/4)(((sin(21πk/4000))^5)+(1/28)sin(42πk/4000))(cos((π/2)(k/4000)^12)), R(k)=(1/170)+(1/67)((sin(42πk/4000))^2)(1-((cos(21πk/4000))^4)). --- Hamid Naderi YeganehMar 23, 2016
"Snap-together Super-Bottle of Genus 4/σ," by Carlo Séquin (University of California, Berkeley)ABS plastic, printed on an FDM machine, 16 x 20 x 14 cm, 2015 Stimulated by the LEGO-Knot project, I aimed to design a set of modular parts that permits to compose not only various handle-bodies, but also single-sided surfaces of higher genus. The modular parts employed in my sculptures are tubular 3-way junctions, where one of the tubular stubs exposes the opposite side of the surface shown by the other two stubs. Depending on how the parts are connected, the resulting compositions remains orientable or becomes single-sided; in the latter case they correspond to sums of multiple Klein bottles; which I call "Super-Bottles." The two identical parts of which the sculpture is composed can be put together in three different ways. In two cases the resulting surface is single-sided (σ = 1) and in the third case it is double-sided (σ = 2). The genus of the resulting surface is 4/σ. The configuration shown is a non-orientable surface of genus 4, corresponding to the connected sum of two Klein Bottles, with two punctures. The insets show the two individual parts, and an assembly of them resulting in a 2-hole torus of genus 2 (with two punctures). --- Carlo Séquin Mar 09, 2016
"Stereographic projection (grid)," by Henry Segerman (Oklahoma State University, Stillwater)3D printed nylon plastic, lamp, 10 x 9 x 9 cm, 2014 The light rays from the lamp are partly blocked by the shrinking design on the sphere; the resulting shadow is a regular tiling of the plane by squares. This illustrates how stereographic projection transforms the sphere, minus the north pole, into the plane. Note how shapes are slightly distorted near the south pole, and dramatically distorted near the north pole. --- Henry Segerman Mar 09, 2016
"Kokabi Stars," by Reza Sarhangi (Towson University, Towson, MD)Tile, 50 x 50 cm, 2015 I am interested in Persian geometric art and its historical methods of construction. Kokabi Star (the great pentagram) can be constructed using the lines of the 10/3 star polygon. Patterning this star can be achieved using different approaches. Some of the presented stars in this artwork have been made based on the actual tiling on existing buildings. Some others have been constructed based on old treatises and scrolls. Some of the patterns have been created using the traditional compass-straightedge process. Modularity is another approach in this regard. Moreover, the two decorated quasicrystal patterns of Star and Sun (the only two quasicrystal patterns with global five-fold rotational symmetry) and their striking relationships with Kokabi Star have been presented. Is this relationship a theorem? --- Reza Sarhangi Mar 09, 2016
"Levy Dragon Outside Tapestry," by Larry Riddle (Agnes Scott College, Decatur, GA)Back stitch embroidery on 18 point canvas, 31 x 31 cm, 2013 I have been working with needle crafts since graduate school. I have also been interested in fractals and fractal geometry for more than 20 years. The Levy Dragon Outside Tapestry consists of four copies of the Levy Dragon built from the edges of a square. The iteration steps are repeated on each of the four sides of the square with the initial triangle motif placed outside the square. This back stitch design shows the twelfth iteration for this outside construction.The outside tapestry was done in two shades of blue to better show each of the four copies of the Levy dragon. --- Larry Riddle Mar 09, 2016
"Walnut Star," by David Reimann (Albion College, Albion, MI)Walnut veneer and brass fasteners, 38 x 38 x 38 cm, 2015 I am interested in creating patterns that convey messages at multiple levels and scales using a wide variety of mathematical elements and media. This form is based on the small rhombicosidodecahedron, an Archimedean solid with 120 edges. The underlying polyhedral edges have been replaced by 4.75 cm squares made from laser-cut paper-backed walnut veneer and connected at their corners with brass split-pin fasteners. The 62 faces (squares, hexagons, and pentagons) and 60 vertices of the underlying polyhedron are transformed into open negative space. The expansion of linear edges into squares results in a sphere-like shape with 20 knobs. --- David Reimann Mar 09, 2016
"OSU Triptych No. 2," by Robert OrndorffPaper and acrylic, 20 x 46 cm, 2015 Honorable Mention, 2016 Mathematical Art Exhibition My paper folding art is math plus paper. Folded paper is simply tangible math. This is a permanent manifestation of an ephemeral artwork, namely, one solution for a specific one-straight-cut problem. Such problems are usually stated as follows: How must one fold a paper rectangle into a flat figure such that one straight cut through all of the layers will produce a given planar straight-line graph? Here the problem has been solved with paper and then represented in acrylic. To a significant degree the work relies on transmitted and reflected light, and so it never looks the same twice. The figure (the letters "OSU") has been divided into three frames. The crease patterns for the left and right letters are pedestrian but that for the central letter is sublime. --- Robert Orndorff Mar 09, 2016
"Shield 1," by Kerry Mitchell (Phoenix, AZ)Digital print on aluminum panel, 40 x 40 cm, 2014 My work is composed primarily of computer generated, mathematically-inspired, abstract images. I draw from the areas of geometry, fractals and numerical analysis, and combine them with image processing technology. This image was created using a dynamic version of the Chaos Game algorithm. The Chaos Game is a simple example illustrating chaotic motion and strange attractors. It is typically implemented using three anchor points, which become the vertices of a Sierpinski triangle. The Dynamic Chaos Game allows the anchor points to move each iteration. Here, the image comprises seven panels. In each, the three anchor points slid along line segments, with different speeds. The pixels were then colored according how frequently that point was visited during the iteration. --- Kerry Mitchell Mar 09, 2016
574 files on 39 page(s) 1 2 3 4 5 6 7 8 9 - 39
Art & Music, MathArchives Geometry in Art & Architecture, by Paul Calter (Dartmouth College) Harmony and Proportion, by John Boyd-Brent International Society of the Arts, Mathematics and Architecture Journal of Mathematics and the Arts Mathematics and Art, the April 2003 Feature Column by Joe Malkevitch Maths and Art: the whistlestop tour, by Lewis Dartnell Mathematics and Art, (The theme for Mathematics Awareness Month in 2003) MoSAIC - Mathematics of Science, Art, Industry, Culture Viewpoints: Mathematics and Art, by Annalisa Crannell (Franklin & Marshall College) and Marc Frantz (Indiana University) Visual Insight, blog by John Baez | 3,785 | 12,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-13 | latest | en | 0.926283 |
https://analystnotes.com/cfa_question.php?p=DZ8WE5YSF | 1,653,775,992,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00441.warc.gz | 147,674,048 | 5,919 | ### CFA Practice Question
Consider a call option with an exercise price of \$105 on a stock that is worth \$95. Calculate the moneyness of the option.
A. the call is \$10 in the money
B. the call is \$10 out of the money
C. none of these answers.
Explanation: A call option is out of the money when the exercise price exceeds the stock price. In this example, the call is \$10 out of the money.
### User Contributed Comments5
User Comment
jpducros sometimes there is an easy question...just to fill up your ego...
amyjones What about the cost of the option?
chester21 Agree with AmyJones - The cost of the option premium should be include....
nike When calculating moneyness of an option you should simply use formula exercise price - stock price. No option cost is needed (it's a sunk cost - you already have the option).
thebkr7 These questions make me feel smart...
Too bad there's only 1 in 240 | 216 | 902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-21 | latest | en | 0.880359 |
http://www.eulersprint.org/problem/288 | 1,516,201,966,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886946.21/warc/CC-MAIN-20180117142113-20180117162113-00580.warc.gz | 443,318,901 | 2,916 | Problem 288
An enormous factorial
For any prime p the number N(p,q) is defined by N(p,q) = n=0 to q Tn*pn
with Tn generated by the following random number generator:
S0 = 290797
Sn+1 = Sn2 mod 50515093
Tn = Sn mod p
Let Nfac(p,q) be the factorial of N(p,q).
Let NF(p,q) be the number of factors p in Nfac(p,q).
You are given that NF(3,10000) mod 320=624955285.
Find NF(61,107) mod 6110
These problems are part of Project Euler and are licensed under CC BY-NC-SA 2.0 UK | 157 | 474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-05 | longest | en | 0.796586 |
https://www.programcreek.com/python/example/57201/numpy.nanmedian | 1,653,738,182,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00717.warc.gz | 1,106,416,415 | 17,177 | # Python numpy.nanmedian() Examples
The following are 30 code examples for showing how to use numpy.nanmedian(). These examples are extracted from open source projects. You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example.
You may check out the related API usage on the sidebar.
You may also want to check out all available functions/classes of the module , or try the search function .
Example 1
```def test_equ_med_far_vector(self):
"""
Test that median reddening is correct in the far limit, using a vector
of coordinates as input.
"""
l = [d['l']*units.deg for d in self._test_data]
b = [d['b']*units.deg for d in self._test_data]
dist = [1.e3*units.kpc for bb in b]
c = coords.SkyCoord(l, b, distance=dist, frame='galactic')
ebv_data = np.array([np.nanmedian(d['samples'][:,-1]) for d in self._test_data])
ebv_calc = self._bayestar(c, mode='median')
# print 'vector:'
# print r'% residual:'
# for ed,ec in zip(ebv_data, ebv_calc):
# print ' {: >8.3f}'.format((ec - ed) / (0.02 + 0.02 * ed))
np.testing.assert_allclose(ebv_data, ebv_calc, atol=0.001, rtol=0.0001) ```
Example 2
```def test_equ_med_scalar(self):
"""
Test that median reddening is correct in at arbitary distances, using
individual coordinates as input.
"""
for d in self._test_data:
l = d['l']*units.deg
b = d['b']*units.deg
for reps in range(10):
dm = 3. + (25.-3.)*np.random.random()
dist = 10.**(dm/5.-2.)
c = coords.SkyCoord(l, b, distance=dist*units.kpc, frame='galactic')
ebv_samples = self._interp_ebv(d, dist)
ebv_data = np.nanmedian(ebv_samples)
ebv_calc = self._bayestar(c, mode='median')
np.testing.assert_allclose(ebv_data, ebv_calc, atol=0.001, rtol=0.0001) ```
Example 3
```def _nanmedian(a, axis=None, out=None, overwrite_input=False):
"""
Private function that doesn't support extended axis or keepdims.
These methods are extended to this function using _ureduce
See nanmedian for parameter usage
"""
if axis is None or a.ndim == 1:
part = a.ravel()
if out is None:
return _nanmedian1d(part, overwrite_input)
else:
out[...] = _nanmedian1d(part, overwrite_input)
return out
else:
# for small medians use sort + indexing which is still faster than
# apply_along_axis
# benchmarked with shuffled (50, 50, x) containing a few NaN
if a.shape[axis] < 600:
return _nanmedian_small(a, axis, out, overwrite_input)
result = np.apply_along_axis(_nanmedian1d, axis, a, overwrite_input)
if out is not None:
out[...] = result
return result ```
Example 4
```def test_allnans(self):
mat = np.array([np.nan]*9).reshape(3, 3)
for axis in [None, 0, 1]:
with suppress_warnings() as sup:
sup.record(RuntimeWarning)
assert_(np.isnan(np.nanmedian(mat, axis=axis)).all())
if axis is None:
assert_(len(sup.log) == 1)
else:
assert_(len(sup.log) == 3)
# Check scalar
assert_(np.isnan(np.nanmedian(np.nan)))
if axis is None:
assert_(len(sup.log) == 2)
else:
assert_(len(sup.log) == 4) ```
Example 5
```def _nanmedian(a, axis=None, out=None, overwrite_input=False):
"""
Private function that doesn't support extended axis or keepdims.
These methods are extended to this function using _ureduce
See nanmedian for parameter usage
"""
if axis is None or a.ndim == 1:
part = a.ravel()
if out is None:
return _nanmedian1d(part, overwrite_input)
else:
out[...] = _nanmedian1d(part, overwrite_input)
return out
else:
# for small medians use sort + indexing which is still faster than
# apply_along_axis
# benchmarked with shuffled (50, 50, x) containing a few NaN
if a.shape[axis] < 600:
return _nanmedian_small(a, axis, out, overwrite_input)
result = np.apply_along_axis(_nanmedian1d, axis, a, overwrite_input)
if out is not None:
out[...] = result
return result ```
Example 6
```def _nanmedian1d(arr1d, overwrite_input=False):
"""
Private function for rank 1 arrays. Compute the median ignoring NaNs.
See nanmedian for parameter usage
"""
c = np.isnan(arr1d)
s = np.where(c)[0]
if s.size == arr1d.size:
warnings.warn("All-NaN slice encountered", RuntimeWarning, stacklevel=3)
return np.nan
elif s.size == 0:
return np.median(arr1d, overwrite_input=overwrite_input)
else:
if overwrite_input:
x = arr1d
else:
x = arr1d.copy()
# select non-nans at end of array
enonan = arr1d[-s.size:][~c[-s.size:]]
# fill nans in beginning of array with non-nans of end
x[s[:enonan.size]] = enonan
# slice nans away
return np.median(x[:-s.size], overwrite_input=True) ```
Example 7
```def _nanmedian(a, axis=None, out=None, overwrite_input=False):
"""
Private function that doesn't support extended axis or keepdims.
These methods are extended to this function using _ureduce
See nanmedian for parameter usage
"""
if axis is None or a.ndim == 1:
part = a.ravel()
if out is None:
return _nanmedian1d(part, overwrite_input)
else:
out[...] = _nanmedian1d(part, overwrite_input)
return out
else:
# for small medians use sort + indexing which is still faster than
# apply_along_axis
# benchmarked with shuffled (50, 50, x) containing a few NaN
if a.shape[axis] < 600:
return _nanmedian_small(a, axis, out, overwrite_input)
result = np.apply_along_axis(_nanmedian1d, axis, a, overwrite_input)
if out is not None:
out[...] = result
return result ```
Example 8
```def test_out(self):
mat = np.random.rand(3, 3)
nan_mat = np.insert(mat, [0, 2], np.nan, axis=1)
resout = np.zeros(3)
tgt = np.median(mat, axis=1)
res = np.nanmedian(nan_mat, axis=1, out=resout)
assert_almost_equal(res, resout)
assert_almost_equal(res, tgt)
# 0-d output:
resout = np.zeros(())
tgt = np.median(mat, axis=None)
res = np.nanmedian(nan_mat, axis=None, out=resout)
assert_almost_equal(res, resout)
assert_almost_equal(res, tgt)
res = np.nanmedian(nan_mat, axis=(0, 1), out=resout)
assert_almost_equal(res, resout)
assert_almost_equal(res, tgt) ```
Example 9
```def _nanmedian1d(arr1d, overwrite_input=False):
"""
Private function for rank 1 arrays. Compute the median ignoring NaNs.
See nanmedian for parameter usage
"""
c = np.isnan(arr1d)
s = np.where(c)[0]
if s.size == arr1d.size:
warnings.warn("All-NaN slice encountered", RuntimeWarning)
return np.nan
elif s.size == 0:
return np.median(arr1d, overwrite_input=overwrite_input)
else:
if overwrite_input:
x = arr1d
else:
x = arr1d.copy()
# select non-nans at end of array
enonan = arr1d[-s.size:][~c[-s.size:]]
# fill nans in beginning of array with non-nans of end
x[s[:enonan.size]] = enonan
# slice nans away
return np.median(x[:-s.size], overwrite_input=True) ```
Example 10
```def _nanmedian(a, axis=None, out=None, overwrite_input=False):
"""
Private function that doesn't support extended axis or keepdims.
These methods are extended to this function using _ureduce
See nanmedian for parameter usage
"""
if axis is None or a.ndim == 1:
part = a.ravel()
if out is None:
return _nanmedian1d(part, overwrite_input)
else:
out[...] = _nanmedian1d(part, overwrite_input)
return out
else:
# for small medians use sort + indexing which is still faster than
# apply_along_axis
if a.shape[axis] < 400:
return _nanmedian_small(a, axis, out, overwrite_input)
result = np.apply_along_axis(_nanmedian1d, axis, a, overwrite_input)
if out is not None:
out[...] = result
return result ```
Example 11
```def test_allnans(self):
mat = np.array([np.nan]*9).reshape(3, 3)
for axis in [None, 0, 1]:
with warnings.catch_warnings(record=True) as w:
warnings.simplefilter('always')
warnings.simplefilter('ignore', FutureWarning)
assert_(np.isnan(np.nanmedian(mat, axis=axis)).all())
if axis is None:
assert_(len(w) == 1)
else:
assert_(len(w) == 3)
assert_(issubclass(w[0].category, RuntimeWarning))
# Check scalar
assert_(np.isnan(np.nanmedian(np.nan)))
if axis is None:
assert_(len(w) == 2)
else:
assert_(len(w) == 4)
assert_(issubclass(w[0].category, RuntimeWarning)) ```
Example 12
```def compute_mean_metrics(json_folder, compute_averages=True):
files = glob.glob(os.path.join(json_folder, "*.json"))
sdr_inst_list = None
for path in files:
#print(path)
with open(path, "r") as f:
if sdr_inst_list is None:
sdr_inst_list = [list() for _ in range(len(js["targets"]))]
for i in range(len(js["targets"])):
sdr_inst_list[i].extend([np.float(f['metrics']["SDR"]) for f in js["targets"][i]["frames"]])
#return np.array(sdr_acc), np.array(sdr_voc)
sdr_inst_list = [np.array(sdr) for sdr in sdr_inst_list]
if compute_averages:
return [(np.nanmedian(sdr), np.nanmedian(np.abs(sdr - np.nanmedian(sdr))), np.nanmean(sdr), np.nanstd(sdr)) for sdr in sdr_inst_list]
else:
return sdr_inst_list ```
Example 13
```def _nanmedian(a, axis=None, out=None, overwrite_input=False):
"""
Private function that doesn't support extended axis or keepdims.
These methods are extended to this function using _ureduce
See nanmedian for parameter usage
"""
if axis is None or a.ndim == 1:
part = a.ravel()
if out is None:
return _nanmedian1d(part, overwrite_input)
else:
out[...] = _nanmedian1d(part, overwrite_input)
return out
else:
# for small medians use sort + indexing which is still faster than
# apply_along_axis
# benchmarked with shuffled (50, 50, x) containing a few NaN
if a.shape[axis] < 600:
return _nanmedian_small(a, axis, out, overwrite_input)
result = np.apply_along_axis(_nanmedian1d, axis, a, overwrite_input)
if out is not None:
out[...] = result
return result ```
Example 14
```def test_allnans(self):
mat = np.array([np.nan]*9).reshape(3, 3)
for axis in [None, 0, 1]:
with suppress_warnings() as sup:
sup.record(RuntimeWarning)
assert_(np.isnan(np.nanmedian(mat, axis=axis)).all())
if axis is None:
assert_(len(sup.log) == 1)
else:
assert_(len(sup.log) == 3)
# Check scalar
assert_(np.isnan(np.nanmedian(np.nan)))
if axis is None:
assert_(len(sup.log) == 2)
else:
assert_(len(sup.log) == 4) ```
Example 15
```def _nanmedian(a, axis=None, out=None, overwrite_input=False):
"""
Private function that doesn't support extended axis or keepdims.
These methods are extended to this function using _ureduce
See nanmedian for parameter usage
"""
if axis is None or a.ndim == 1:
part = a.ravel()
if out is None:
return _nanmedian1d(part, overwrite_input)
else:
out[...] = _nanmedian1d(part, overwrite_input)
return out
else:
# for small medians use sort + indexing which is still faster than
# apply_along_axis
# benchmarked with shuffled (50, 50, x) containing a few NaN
if a.shape[axis] < 600:
return _nanmedian_small(a, axis, out, overwrite_input)
result = np.apply_along_axis(_nanmedian1d, axis, a, overwrite_input)
if out is not None:
out[...] = result
return result ```
Example 16
```def test_allnans(self):
mat = np.array([np.nan]*9).reshape(3, 3)
for axis in [None, 0, 1]:
with suppress_warnings() as sup:
sup.record(RuntimeWarning)
assert_(np.isnan(np.nanmedian(mat, axis=axis)).all())
if axis is None:
assert_(len(sup.log) == 1)
else:
assert_(len(sup.log) == 3)
# Check scalar
assert_(np.isnan(np.nanmedian(np.nan)))
if axis is None:
assert_(len(sup.log) == 2)
else:
assert_(len(sup.log) == 4) ```
Example 17
```def test_equ_med_far_scalar(self):
"""
Test that median reddening is correct in the far limit, using a single
location on the sky at a time as input.
"""
for d in self._test_data:
c = self._get_gal(d, dist=1.e3*units.kpc)
# print d['samples']
ebv_data = np.nanmedian(d['samples'][:,-1])
ebv_calc = self._bayestar(c, mode='median')
# print 'ebv_data:', ebv_data
# print 'ebv_calc:', ebv_calc
# print ''
# print r'% residual: {:.6f}'.format((ebv_calc - ebv_data) / (0.001 + 0.001 * ebv_data))
np.testing.assert_allclose(ebv_data, ebv_calc, atol=0.001, rtol=0.0001) ```
Example 18
```def test_equ_med_vector(self):
"""
Test that median reddening is correct at arbitary distances, using a
vector of coordinates as input.
"""
for reps in range(10):
l = [d['l']*units.deg for d in self._test_data]
b = [d['b']*units.deg for d in self._test_data]
dm = 3. + (25.-3.)*np.random.random(len(self._test_data))
dist = [d*units.kpc for d in 10.**(dm/5.-2.)]
dist_unitless = [d for d in 10.**(dm/5.-2.)]
c = coords.SkyCoord(l, b, distance=dist, frame='galactic')
ebv_samples = np.array([
self._interp_ebv(datum, d)
for datum,d in zip(self._test_data, dist_unitless)
])
ebv_data = np.nanmedian(ebv_samples, axis=1)
ebv_calc = self._bayestar(c, mode='median')
# print 'vector arbitrary distance:'
# print r'% residual:'
# for ed,ec in zip(ebv_data, ebv_calc):
# print ' {: >8.3f}'.format((ec - ed) / (0.02 + 0.02 * ed))
np.testing.assert_allclose(ebv_data, ebv_calc, atol=0.001, rtol=0.0001) ```
Example 19
```def _rsp_rrv_time(bbi):
diff_bbi = np.diff(bbi)
out = {} # Initialize empty dict
# Mean based
out["RMSSD"] = np.sqrt(np.mean(diff_bbi ** 2))
out["MeanBB"] = np.nanmean(bbi)
out["SDBB"] = np.nanstd(bbi, ddof=1)
out["SDSD"] = np.nanstd(diff_bbi, ddof=1)
out["CVBB"] = out["SDBB"] / out["MeanBB"]
out["CVSD"] = out["RMSSD"] / out["MeanBB"]
# Robust
out["MedianBB"] = np.nanmedian(bbi)
# # Extreme-based
# nn50 = np.sum(np.abs(diff_rri) > 50)
# nn20 = np.sum(np.abs(diff_rri) > 20)
# out["pNN50"] = nn50 / len(rri) * 100
# out["pNN20"] = nn20 / len(rri) * 100
#
# # Geometrical domain
# bar_y, bar_x = np.histogram(rri, bins=range(300, 2000, 8))
# bar_y, bar_x = np.histogram(rri, bins="auto")
# out["TINN"] = np.max(bar_x) - np.min(bar_x) # Triangular Interpolation of the NN Interval Histogram
# out["HTI"] = len(rri) / np.max(bar_y) # HRV Triangular Index
return out ```
Example 20
```def mad(x, constant=1.4826):
"""Median Absolute Deviation: a "robust" version of standard deviation.
Parameters
----------
x : Union[list, np.array, pd.Series]
A vector of values.
constant : float
Scale factor. Use 1.4826 for results similar to default R.
Returns
----------
float
Examples
----------
>>> import neurokit2 as nk
>>> nk.mad([2, 8, 7, 5, 4, 12, 5, 1])
3.7064999999999997
References
-----------
- https://en.wikipedia.org/wiki/Median_absolute_deviation
"""
median = np.nanmedian(np.ma.array(x).compressed())
Example 21
```def _standardize(data, robust=False, window=None, **kwargs):
# Compute standardized on whole data
if window is None:
if robust is False:
z = (data - np.nanmean(data, axis=0)) / np.nanstd(data, axis=0, ddof=1)
else:
z = (data - np.nanmedian(data, axis=0)) / mad(data)
# Rolling standardization on windows
else:
df = pd.DataFrame(data) # Force dataframe
if robust is False:
z = (df - df.rolling(window, min_periods=0, **kwargs).mean()) / df.rolling(
window, min_periods=0, **kwargs
).std(ddof=1)
else:
z = (df - df.rolling(window, min_periods=0, **kwargs).median()) / df.rolling(
window, min_periods=0, **kwargs
# Fill the created nans
z = z.fillna(method="bfill")
# Restore to vector or array
if z.shape[1] == 1:
z = z[0].values
else:
z = z.values
return z ```
Example 22
```def _nanmedian1d(arr1d, overwrite_input=False):
"""
Private function for rank 1 arrays. Compute the median ignoring NaNs.
See nanmedian for parameter usage
"""
arr1d, overwrite_input = _remove_nan_1d(arr1d,
overwrite_input=overwrite_input)
if arr1d.size == 0:
return np.nan
return np.median(arr1d, overwrite_input=overwrite_input) ```
Example 23
```def _nanmedian_small(a, axis=None, out=None, overwrite_input=False):
"""
sort + indexing median, faster for small medians along multiple
dimensions due to the high overhead of apply_along_axis
see nanmedian for parameter usage
"""
m = np.ma.median(a, axis=axis, overwrite_input=overwrite_input)
warnings.warn("All-NaN slice encountered", RuntimeWarning, stacklevel=3)
if out is not None:
out[...] = m.filled(np.nan)
return out
return m.filled(np.nan) ```
Example 24
```def test_mutation(self):
# Check that passed array is not modified.
ndat = _ndat.copy()
np.nanmedian(ndat)
assert_equal(ndat, _ndat) ```
Example 25
```def test_keepdims(self):
mat = np.eye(3)
for axis in [None, 0, 1]:
tgt = np.median(mat, axis=axis, out=None, overwrite_input=False)
res = np.nanmedian(mat, axis=axis, out=None, overwrite_input=False)
assert_(res.ndim == tgt.ndim)
d = np.ones((3, 5, 7, 11))
# Randomly set some elements to NaN:
w = np.random.random((4, 200)) * np.array(d.shape)[:, None]
w = w.astype(np.intp)
d[tuple(w)] = np.nan
with suppress_warnings() as sup:
sup.filter(RuntimeWarning)
res = np.nanmedian(d, axis=None, keepdims=True)
assert_equal(res.shape, (1, 1, 1, 1))
res = np.nanmedian(d, axis=(0, 1), keepdims=True)
assert_equal(res.shape, (1, 1, 7, 11))
res = np.nanmedian(d, axis=(0, 3), keepdims=True)
assert_equal(res.shape, (1, 5, 7, 1))
res = np.nanmedian(d, axis=(1,), keepdims=True)
assert_equal(res.shape, (3, 1, 7, 11))
res = np.nanmedian(d, axis=(0, 1, 2, 3), keepdims=True)
assert_equal(res.shape, (1, 1, 1, 1))
res = np.nanmedian(d, axis=(0, 1, 3), keepdims=True)
assert_equal(res.shape, (1, 1, 7, 1)) ```
Example 26
```def test_small_large(self):
# test the small and large code paths, current cutoff 400 elements
for s in [5, 20, 51, 200, 1000]:
d = np.random.randn(4, s)
# Randomly set some elements to NaN:
w = np.random.randint(0, d.size, size=d.size // 5)
d.ravel()[w] = np.nan
d[:,0] = 1. # ensure at least one good value
# use normal median without nans to compare
tgt = []
for x in d:
nonan = np.compress(~np.isnan(x), x)
tgt.append(np.median(nonan, overwrite_input=True))
assert_array_equal(np.nanmedian(d, axis=-1), tgt) ```
Example 27
```def test_result_values(self):
tgt = [np.median(d) for d in _rdat]
res = np.nanmedian(_ndat, axis=1)
assert_almost_equal(res, tgt) ```
Example 28
```def test_empty(self):
mat = np.zeros((0, 3))
for axis in [0, None]:
with warnings.catch_warnings(record=True) as w:
warnings.simplefilter('always')
assert_(np.isnan(np.nanmedian(mat, axis=axis)).all())
assert_(len(w) == 1)
assert_(issubclass(w[0].category, RuntimeWarning))
for axis in [1]:
with warnings.catch_warnings(record=True) as w:
warnings.simplefilter('always')
assert_equal(np.nanmedian(mat, axis=axis), np.zeros([]))
assert_(len(w) == 0) ```
Example 29
```def test_scalar(self):
```def test_nanmedian(self): | 5,327 | 17,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.59746 |
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# Explosives in the WTC 7 bought it down...I believe now...
page: 3
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posted on Jan, 14 2009 @ 03:40 PM
- What is the "path of most resistance" as it relates to this instance?
- What would be the "path of LEAST resistance"?
Path of MOST resistance would be the building itself, thousands of tons of construction steel bolted and welded together producing a solid structure. The path of LEAST resistance would the open air around it.
All three buildings fell in a vertical path, building 7 was collapsed into it's own footprint, the towers were not so much. The towers were too tall to have the outer walls to fall inwards, so a 'conventional' demo was impossible. I've heard it said there had been talk before 9/11 that the towers would have to be demolished the way they were built, piece by piece.
posted on Jan, 14 2009 @ 03:47 PM
How could this take place if the rest of the building was beneath these top floors?
Have you looked into the tilt of the top of WTC2? It started to fall off the top, angular momentum, and then the building underneath it collapsed faster then the tops momentum, which proves that top section could not have caused the collapse by itself.
So the resistance you refer to is the resistance encountered by the object which slows its gravitational acceleration?
Resistance is anything that stops momentum. Thousand of tons of bolted and welded steel columns, unaffected by the fires or planes impact, would be your resistance in this context.
The only way that resistance would not effect the collapse speed is if that resistance was first removed. This is scientific facts that can be reproduced in a lab. Try reproducing NISTs hypothesis in a lab.
posted on Jan, 14 2009 @ 03:52 PM
I'm sure it has been answered already but your first question is a silly one. What would people think if the building collapsed with the other two when it had hardly ANY damage at all? It HAD to burn for a bit so the excuse of its collapse would be something stupid and absolutely retarded like a new phenom like 'thermal expansion'.
The lies produce lies which produce more. And in the end it is a jumbled mess of garbage that people seeking the truth have to try and wade through. It's a simple tactic of people in power. Create as much garbage as possible and the answers will never come forth.
I knew this thread was a hoax once I spotted the author.
posted on Jan, 14 2009 @ 04:22 PM
posted by UKWO1Phot
Least resistance = Top floors fall to ground and leaves rest of building standing.
How could this take place if the rest of the building was beneath these top floors?
It could not unless the structure below was first destroyed.
posted by UKWO1Phot
Most resistance = Top floors slam straight down through the rest of the building structure.
So the resistance you refer to is the resistance encountered by the object which slows its gravitational acceleration?
But the collapse floors were slowed very little indicating that the most resistence below was non-existent.
But both towers (and WTC7) took the path of most resistence didn't they, down through the supposedly rigid structure below, at one time bolted and welded together into a very strong framework?
How could the towers (and WTC7) fall through the path of most resistence except that strong structure below was already destroyed beforehand ahead of the collapse wave; turning it into the path of least resistance? Both towers (and WTC7) visually fell straight down, although something (explosives?) blew the 8000 pound outer perimeter wall sections away from the towers in every direction up to 600 feet away.
posted on Jan, 14 2009 @ 05:57 PM
OMG! What are you doing showing photos that reveals proof of an explosion?
You are going to make people angry, stop showing the truth it is hurting me, I can’t look anymore (some one help me!) How does 8000 pound outer wall, land 600 feet away?
If, the WTC just collapse then you would see a normal pile of debris in it collapse zone, and you would see large chunks of concrete slabs, broken, and smash everywhere. But you don’t see that why? Why is all the concrete pulverized to a fine powder? Why are steal beams blown in to other buildings?
How could the towers (and WTC7) fall through the path of most resistence except that strong structure below was already destroyed beforehand ahead of the collapse wave; turning it into the path of least resistance? Both towers (and WTC7) visually fell straight down, although something (explosives?) blew the 8000 pound outer perimeter wall sections away from the towers in every direction up to 600 feet away.
posted on Jan, 14 2009 @ 06:34 PM
Originally posted by Swampfox46_1999
A "textbook" demolition, does not damage surrounding buildings. WTC 7's collapse caused so much damage to 30 West Broadway that it was torn down.
But, the question is (and has yet to be satisfactorily answered by ANYONE) how did WTC 7 get wired for demolition and no one noticed?
Quite frankly, anyone that believes the three buildings were controlled demolitions dont know a doggone thing about demolitions.
If the aim was to take 'em down with no one knowing, I wouldn't expect perfection in the results. It's not as if they were looking to the judges for a score in an Olympic competition.
Despite a bit of a sloppy job, the collapse was still "textbook," moving the mass downward into a footprint through the path of greatest resistance.
Aw, geez. I had promised myself I wasn't going to post in any more Official Conspiracy Theorists' threads, and here I am doing so!
posted on Jan, 14 2009 @ 07:15 PM
Originally posted by Gonenuts
How does 8000 pound outer wall, land 600 feet away?
If, the WTC just collapse then you would see a normal pile of debris in it collapse zone, and you would see large chunks of concrete slabs, broken, and smash everywhere.
Remember that I am truther when I say this:
I believe it possible due to moments acting on wall sections. Like taking an uncooked piece of spaghetti and pushing on the ends. It could cause a huge bending moment. I'm just saying, it's not impossible.
posted on Jan, 14 2009 @ 07:26 PM
You should know that Steven Jones has been completely discredited. His university has relieved him of his teaching duties and the civil engineering department has stated that his claims are completely unsupported.
He has no evidence for thermite, thermate or any other such material.
He claimed that the moltem metal pouring out of the tower was steel from thermite and not melted airplane aluminum because molten aluminum is always silver and never any other color. This is absolutely not true.
I worked for ALCOA and USSteel in the past and the color of hot metal is due to its temperature, not its composition. Ask any puddler.
Steven Jones is either the most incompetent physicist on the planet or a pathological liar looking to get in the spotlight and stay there.
posted on Jan, 14 2009 @ 07:34 PM
Originally posted by pteridine
He claimed that the moltem metal pouring out of the tower was steel from thermite and not melted airplane aluminum because molten aluminum is always silver and never any other color. This is absolutely not true.
Care to back this up?
I worked for ALCOA and USSteel in the past and the color of hot metal is due to its temperature, not its composition. Ask any puddler.
Steven Jones is either the most incompetent physicist on the planet or a pathological liar looking to get in the spotlight and stay there.
Again, not that I'm trying to discredit you, but, could you please back up your claims? Thanks.
posted on Jan, 14 2009 @ 07:38 PM
You said:"Both towers (and WTC7) visually fell straight down, although something (explosives?) blew the 8000 pound outer perimeter wall sections away from the towers in every direction up to 600 feet away. "
Maybe this is evidence that not everything fell straight down. Big pieces fell away from the buildings. Maybe not a demolition, at that. After all, part of the inner core and part of a wall were left standing for a while before they collapsed 10-15 seconds later. Where is the demolition?
posted on Jan, 14 2009 @ 09:11 PM
Well a double BS on you. As usual, you failed to read the whole post. Badgerprints claimed it was textbook, my post was a response to him. You, as you normally do, jumped in with your BS. Why I didnt respond to the thread about the Greening paper...I try to stay out of battles of the wits with unarmed opponents.
posted on Jan, 14 2009 @ 09:13 PM
And yet again, no one can explain who the Supermen were that managed to wire a 47 story building (tallest ever) for implosion in less than a week.
posted on Jan, 15 2009 @ 02:11 AM
You should know that Steven Jones has been completely discredited.
civil engineering department has stated that his claims are completely unsupported.
Unsupported by whom? Care to show whom in his engineer department, who disagree with Jones report and who has done a report to support their claim that Jones is wrong in his department? I know for a fact, Jones left on his own terms, to continue in his project in the truth movement. His work is well respected. Jones is no kook only the lying media makes that claim, that’s why I do not watch TV News because, I do not support lairs. Jones work has not been peer review because the government will not let it happen, if by chance, it does get peer review and it is accepted. Then there will be an out rage of anger against this government by We the people. You can guess what that means.
He has no evidence for thermite, thermate or any other such material.
LOL On the contrary, he has proven the goods, do you care to show where he has not?
He claimed that the moltem metal pouring out of the tower was steel from thermite and not melted airplane aluminum because molten aluminum is always silver and never any other color. This is absolutely not true.
I worked for ALCOA and USSteel in the past and the color of hot metal is due to its temperature, not its composition.
I do not care what your opinion is, can you show me with science to back your claim then you will make me a believer. And please show your sources.
Steven Jones is either the most incompetent physicist on the planet or a pathological liar looking to get in the spotlight and stay there.
This is the kind of responds, I get from people, who have not done their research, and parrot the lying media.
However, since you feel Steven Jones is a pathological liar, why don’t you demonstrate to us, where he tell lies over, and over, and over. To make a claim that Steven Jones is incompetent shows you have never read a single report he has written and you have done *no* reasurch in to 911. This ridiculous comments you are making is not new to me, or people who are in search of the truth, we see this all the time, by person who have formed an opinion from watching to much mainstream News. Come on! Do a little research on the subject, and I promise you will not feel the same about 911, that is a fact.
[edit on 15-1-2009 by Gonenuts]
[edit on 15-1-2009 by Gonenuts]
posted on Jan, 15 2009 @ 08:45 AM
if your aluminum at alcoa glowed red it was not pure it must have been ore or an alloy (bauxite is where most aluminum ore is found )
it had to be mixed with cryoltes to get the pure aluminum out threw
then electrolysis process according to temperature recorede that day remind you these where flash teperatures was 1750 so with that fact the only metal that should have been molten was these
www.muggyweld.com...
aluminum(alloys)
brass
bronze
magnesium
Phosphorous
tin and zinc
if the flame stayed 1750 long enough silver would have started to become effected
now just need to find out how much of these metals where present from the flowing metal (laugh) coming from the build
and how it stayed molten (any metal for that matter)
falling threw air 900 feet and staying is the mystery id like to know first
molten metal of anykind no matter its alloy make up needs constant heat to stay molten
and that is a thermodynamic fact
so this should not have been possible its lying on the ground still molte with huge beams and earth around to draw the heat away, so whats keeping it molten ?? a chemical reaction maybe ?
www.investigate911.com...
[edit on 15-1-2009 by lycopersicum]
[edit on 15-1-2009 by lycopersicum]
[edit on 15-1-2009 by lycopersicum]
[edit on 15-1-2009 by lycopersicum]
posted on Jan, 15 2009 @ 11:42 AM
Originally posted by dariousg
I knew this thread was a hoax once I spotted the author.
This is nice. I asked 2 simple questions that no one can seem to answer?
They waited? For what? the right time, I mean, if they were rigged with explosives, bring them all down at once. This would cause massive destruction at one time and secondary explosions would not have been heard or noticed. Those are the theories I am looking for and not links to prisonplanet.
The investigators at NIST were not the tards alot of posters make them out to be. To think some of you have the mental capability to believe Alex Jones over Noble winning scientists is amazing. It took seven years which is a very thorough investigation.
The second question is, where is the evidence. All of the steel was not whisked away in days and some of you believe. I gave a post where there are tons rusting in a field. Why doesn't Alex take a trip and take some, no one would notice. Alex Jones is a hack. I mean, if he predicted the attacks, why did he not stop them???? Trust me, I know who he is.
The prejudice on this board is astronomical. I do not believe your conspiracy theories so this thread is labeled as a hoax. Grow a set and try to answer those 2 simple questions with some thought of your own.
Griff, you starting to think this could have been natural without explosives? I can send you some pics of those big heavy columns that did twist to look like spaghetti.
Also, the terrorist plan was to hit the towers so at least one would fall on the other, just like how they planned the 93 bombing.
posted on Jan, 15 2009 @ 11:54 AM
Simple, a change of plan took some time to come up with answers from the
Illuminati 911 think tank in control of the operation.
Most likely the same answers we got seven years later.
If you don't look for explosives you don't find them.
The whole issue was skirted as the master 911 plan went down with ease.
posted on Jan, 15 2009 @ 11:54 AM
Griff, you starting to think this could have been natural without explosives?
I've never denied the possibility of it. If that's what you mean.
posted on Jan, 15 2009 @ 11:56 AM
Originally posted by ANOK
Originally posted by Gonenuts
Your statement that there are no physical evidences shows you have not done very much research in to 911.
With esdad it's more like he just ignores anything that doesn't fit the 'official story'.
He knows all the 911myths type responses, which he parrots again and again regardless of any counter argument.
I mean he still thinks we believe the buildings fell at free-fall-speed, and he still thinks it's a relevant argument.
Esdad has been around long enough to know the arguments of both sides inside out, and he uses that to throw straw man arguments around. You get nowhere debating him, her, it...
(pretty much goes for most of our resident debunkers who have been around a lot longer than their present accounts would suggest)
Wow, what a nice guy you are. Tell me where the evidence is then Captain Knowledge. That is the problem with conspiracy theorists, they always want someone else to prove their theory for them.To think I do not believe in conspiracy is not true is wrong.
Could our government blow up a building on our soil? Yes
Does the government hide things from the public? Yes
Did they pull of an operation the size of 9/11? No
You see, it is not that I do not want to see the other side. I am asking for evidence. Give me some physical proof of demolition not these same old tired videos. This is no straw man argument i was actually trying to engage in a singular part of my not believing the other side.
If you have known me long enough you would know that I am not trying to troll. I was in another thread getting beat up with conspiracy theory and wanted to see what I could get with a few questions? Maybe some engaging questions or prople trying to convince me. No, just insult. Rock on ANOK.
posted on Jan, 15 2009 @ 12:00 PM
I never heard you deny it so that is why I asked. It has already bothered me that it took so long for the WTC 7 to fall when people theorize that it was bought down with explosives. Why wait?
posted on Jan, 15 2009 @ 12:06 PM
It has already bothered me that it took so long for the WTC 7 to fall when people theorize that it was bought down with explosives. Why wait?
If all three buildings came down together, then wouldn't that pretty much be the definitive moment when people say: "Hey, now that couldn't happen naturally"?
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2 | 3,928 | 17,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.979804 |
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MathsYear 5Autumn Block 5 (Perimeter and Area)01 Measure Perimeter › Free Year 5 Measure Perimeter IWB Perimeter and Area Activity
# Free Year 5 Measure Perimeter IWB Perimeter and Area Activity
Step 1: Free Year 5 Measure Perimeter IWB Perimeter and Area Activity
This Year 5 Measure Perimeter IWB Activity includes three questions designed to check pupils' understanding of measuring the perimeter of rectilinear shapes. Pupils will identify which shape has a given perimeter from a selection, measure lengths with a ruler to find the perimeter and answer a true or false question involving perimeter.
More resources for Autumn Block 5 Step 1.
#### National Curriculum Objectives
Mathematics Year 5: (5M7a) Measure and calculate the perimeter of composite rectilinear shapes in centimetres and metres
This resource is available to play with a Taster subscription. | 212 | 955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.831547 |
https://it.khanacademy.org/video?lang=it&format=lite&v=YEZ_gP0VcMw | 1,498,355,873,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320386.71/warc/CC-MAIN-20170625013851-20170625033851-00218.warc.gz | 777,505,072 | 28,827 | # Dimostrazione: sin(a+b) = (cos a)(sin b) + (sin a)(cos b)
Help us caption and translate this video on Amara.org: http://www.amara.org/en/v/B2v8/ Esercizi: http://www.khanacademy.org/exercise/trig_identities_1 Traduzione in Italiano del video: Proof: sin(a+b) = (cos a)(sin b) + (sin a)(cos b) http://www.youtube.com/watch?feature=player_embedded&v=zw0waJCEc-w http://www.khanacademy.org/video/proof--sin-a-b------cos-a--sin-b-----sin-a--cos-b?playlist=Trigonometry | 158 | 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} | 2.90625 | 3 | CC-MAIN-2017-26 | longest | en | 0.425779 |
https://www.extremeoptimization.com/QuickStart/VisualBasic/BigNumbers.aspx | 1,550,625,532,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494125.62/warc/CC-MAIN-20190220003821-20190220025821-00333.warc.gz | 806,091,017 | 8,669 | Data Analysis Mathematics Linear Algebra Statistics
New Version 7.0!
QuickStart Samples
# BigNumbers QuickStart Sample (Visual Basic)
Illustrates the basic use of the arbitrary precision classes: BigInteger, BigRational, BigFloat in Visual Basic.
```Option Infer On
Imports System.Diagnostics
' Basic generic types live in Extreme.Mathematics.Generics.
Imports Extreme.Mathematics.Generic
' We'll also need the big number types.
Imports Extreme.Mathematics
Namespace Extreme.Numerics.QuickStart.VB
' Illustrates the use of the arbitrary precision number
' classes in the Extreme Optimization Mathematics Library for .NET.
Module BigNumbers
Sub main()
' In this QuickStart sample, we'll use 5 different methods to compute
' the value of pi, the ratio of the circumference to the diameter of
' a circle.
Dim pi As BigFloat
' The number of decimal digits.
Dim digits As Integer = 10000
' The equivalent number of binary digits, to account for round-off error:
Dim binaryDigits As Integer = CInt(8 + digits * Math.Log(10, 2))
' The number of digits in the last correction, if applicable.
Dim correctionDigits As Double
' First, create an AccuracyGoal for the number of digits we want.
' We'll add 5 extra digits to account for round-off error.
Dim goal As AccuracyGoal = AccuracyGoal.Absolute(digits + 5)
Console.WriteLine("Calculating {0} digits of pi:", digits)
' Create a stopwatch so we can time the results.
Dim sw As New Stopwatch()
'
' Method 1: Arctan formula
'
' pi/4 = 88L(172) + 51L(239) + 32L(682) + 44L(5357) + 68L(12943)
' Where L(p) = Arctan(1/p)
' We will use big integer arithmetic for this.
' See the helper function Arctan later in this file.
Console.WriteLine("Arctan formula using integer arithmetic:")
sw.Start()
Dim coefficients As Integer() = {88, 51, 32, 44, 68}
Dim arguments As Integer() = {172, 239, 682, 5357, 12943}
pi = BigFloat.Zero
For k As Integer = 0 To 4
pi = pi + coefficients(k) * Arctan(arguments(k), binaryDigits)
Console.WriteLine("Step {0}: ({1:F3} seconds)", k + 1, sw.Elapsed.TotalSeconds)
Next
' The ScaleByPowerOfTwo is the fastest way to multiply
' or divide by a power of two:
pi = BigFloat.ScaleByPowerOfTwo(pi, 2)
sw.Stop()
Console.WriteLine("Total time: {0:F3} seconds.", sw.Elapsed.TotalSeconds, pi)
Console.WriteLine()
'
' Method 2: Rational approximation
'
' pi/2 = 1 + 1/3 + (1*2)/(3*5) + (1*2*3)/(3*5*7) + ...
' = 1 + 1/3 * (1 + 2/5 * (1 + 3/7 * (1 + ...)))
' We gain 1 bit per iteration, so we know where to start.
Console.WriteLine("Rational approximation using rational arithmetic.")
Console.WriteLine("This is very inefficient, so we only do up to 10000 digits.")
sw.Start()
Dim an = BigRational.Zero
Dim n0 = binaryDigits
If digits <= 10000 Then n0 = CInt(8 + 10000 * Math.Log(10, 2))
For n = n0 To 1 Step -1
an = New BigRational(n, 2 * n + 1) * an + 1
Next
pi = New BigFloat(2 * an, goal, RoundingMode.TowardsNearest)
sw.Stop()
Console.WriteLine("Total time: {0:F3} seconds.", sw.Elapsed.TotalSeconds, pi)
Console.WriteLine()
'
' Method 3: Arithmetic-Geometric mean
'
' By Salamin & Brent, based on discoveries by C.F.Gauss.
' See http://www.cs.miami.edu/~burt/manuscripts/gaussagm/agmagain-hyperref.pdf
Console.WriteLine("Arithmetic-Geometric Mean:")
sw.Reset()
sw.Start()
Dim x1 = BigFloat.Sqrt(2, goal, RoundingMode.TowardsNearest)
Dim x2 = BigFloat.One
Dim S = BigFloat.Zero
Dim c = BigFloat.One
For k = 0 To Integer.MaxValue
S = S + BigFloat.ScaleByPowerOfTwo(c, k - 1)
Dim aMean = BigFloat.ScaleByPowerOfTwo(x1 + x2, -1)
Dim gMean = BigFloat.Sqrt(x1 * x2)
x1 = aMean
x2 = gMean
c = (x1 + x2) * (x1 - x2)
' GetDecimalDigits returns the approximate number of digits in a number.
' A negative return value means the number is less than 1.
correctionDigits = -c.GetDecimalDigits()
Console.WriteLine("Iteration {0}: {1:F1} digits ({2:F3} seconds)", k, correctionDigits, sw.Elapsed.TotalSeconds)
If (correctionDigits >= digits) Then Exit For
Next
pi = x1 * x1 / (1 - S)
sw.Stop()
Console.WriteLine("Total time: {0:F3} seconds.", sw.Elapsed.TotalSeconds, pi)
Console.WriteLine()
'
' Method 4: Borweins' quartic formula
'
' This algorithm quadruples the number of correct digits
' in each iteration.
' See http:'en.wikipedia.org/wiki/Borwein's_algorithm
Console.WriteLine("Quartic formula:")
sw.Reset()
sw.Start()
Dim sqrt2 As BigFloat = BigFloat.Sqrt(2, goal, RoundingMode.TowardsNearest)
Dim y = sqrt2 - BigFloat.One
Dim a = New BigFloat(6, goal) - BigFloat.ScaleByPowerOfTwo(sqrt2, 2)
Dim y2 = y * y
Dim y3 As BigFloat
Dim y4 = y2 * y2
Dim da As BigFloat
correctionDigits = 0
For k As Integer = 0 To Integer.MaxValue
If 4 * correctionDigits > digits Then Exit For
Dim qrt = BigFloat.Root(1 - y4, 4)
y = BigFloat.Subtract(1, qrt, goal, RoundingMode.TowardsNearest) _
' y = BigFloat.Divide(1 - qrt, 1 + qrt, AccuracyGoal.InheritAbsolute, RoundingMode.TowardsNearest)
y2 = y * y
y3 = y * y2
y4 = y2 * y2
da = (BigFloat.ScaleByPowerOfTwo(y + y3, 2) + (6 * y2 + y4)) * a _
- BigFloat.ScaleByPowerOfTwo(y + y2 + y3, 2 * k + 1)
da = da.RestrictPrecision(goal, RoundingMode.TowardsNearest)
a += da
correctionDigits = -da.GetDecimalDigits()
Console.WriteLine("Iteration {0}: {1:F1} digits ({2:F3} seconds)", k, correctionDigits, sw.Elapsed.TotalSeconds)
Next
pi = BigFloat.Inverse(a)
sw.Stop()
Console.WriteLine("Total time: {0:F3} seconds.", sw.Elapsed.TotalSeconds, pi)
Console.WriteLine()
'
' Method 5: The built-in method
'
' The method used to compute pi internally is an order of magnitude
' faster than any of the above.
Console.WriteLine("Built-in function:")
sw.Reset()
sw.Start()
pi = BigFloat.GetPi(goal)
sw.Stop()
Console.WriteLine("Total time: {0:F3} seconds.", sw.Elapsed.TotalSeconds, pi)
' The highest precision value of pi is cached, so
' getting pi to any precision up to that is super fast.
Console.WriteLine("Built-in function (cached):")
sw.Reset()
sw.Start()
pi = BigFloat.GetPi(goal)
sw.Stop()
Console.WriteLine("Total time: {0:F3} seconds.", sw.Elapsed.TotalSeconds, pi)
Console.Write("Press Enter key to exit...")
End Sub
' Helper function to compute Arctan(1/p)
' p: The reciprocal of the argument.
' binaryDigits: The number of binary digits in the result.
' Returns; Arctan(1/p) to "binaryDigits" binary digits.
Function Arctan(ByVal p As Integer, ByVal binaryDigits As Integer) As BigFloat
' We scale the result by a factor of 2^binaryDigits.
' The first term is 1/p.
Dim power = BigInteger.Pow(2, binaryDigits) / p
' We store the sum in result.
Dim result = power
Dim subtract = True
Dim k = 0
Do While (Not power.IsZero)
k = k + 1
' Expressions involving big integers look exactly like any other arithmetic expression:
' The kth term is (-1)^k 1/(2k+1) 1/p^2k.
' So the power is 1/p^2 times the previous power.
power = power / (p * p)
' And we alternately add and subtract
If (subtract) Then
result -= power / (2 * k + 1)
Else
result += power / (2 * k + 1)
End If
subtract = Not subtract
Loop
' Scale the result.
Return BigFloat.ScaleByPowerOfTwo(New BigFloat(result), -binaryDigits)
End Function
End Module
End Namespace``` | 2,052 | 7,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-09 | longest | en | 0.512053 |
https://devforum.roblox.com/t/defining-elements-in-a-table-that-is-within-another-table/994194 | 1,716,370,080,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00375.warc.gz | 166,710,739 | 5,834 | # Defining elements in a table that is within another table?
I’ve been having a hard time with this and I’m not sure why it’s so difficult for me to understand.
Let’s say I have a shopping cart.
``````local cart = {
bananas = 2;
apples = 4;
}
``````
Then let’s add the purse as well.
``````local cart = {
bananas = 2;
apples = 4;
purse = {
cash = 20;
debit = 100;
};
}
``````
When I define the element within purse I get this nil from output, which confuses me?
``````print(table.find(cart.purse,2))
> nil
``````
Or if I wanted to call the element specifically on its own it still calls nil.
``````print(table.find(cart.purse,"cash"))
> nil
``````
What am I doing wrong here?
table.find only works on arrays, not dictionaries. Lua arrays are ordered, so a linear search algorithm is performed whereas this wouldn’t be possible with dictionaries that are unordered.
1 Like
Would that really be necessary tho? If you index for something that doesn’t exist, `nil` is returned.
``````local t = { a = 1 }
print(t.b) --> nil
``````
1 Like
So if you don’t recommend this solution, what’s an alternative I can try?
Look at the above reply by @sjr04. Simply check if `cart.purse` exists and if so, check if `cart.purse.cash` exists.
``````local foundCash = cart.purse and cart.purse.cash
--nil or cash value
``````
1 Like
To define the element within the purse, do
``````cart.purse.cash
``````
If you want to call the element specifically on its own, do
``````_cashInPurse = cart.purse.cash
``````
(This will error if purse is nil, but it won’t if cash is nil, it will just return nil)
1 Like | 447 | 1,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.88534 |
https://sacalecheselectrico.org/how-to-find-increasing-and-decreasing-intervals-on-a-graphing-calculator-references | 1,656,768,667,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00313.warc.gz | 538,489,987 | 15,653 | # How To Find Increasing And Decreasing Intervals On A Graphing Calculator References
How To Find Increasing And Decreasing Intervals On A Graphing Calculator References. How to determine increasing and decreasing intervals on a graph. To find the an increasing or decreasing interval, we need to find out if the first derivative is positive or negative on the given interval.
Graph the function (i used the graphing calculator at desmos.com). Select the correct choice below and fil in any answer boxes in your choi the furpction. Decreasing increasing constant decreasing increasing decreasing when we describe where the function is increasing, decreasing, and
### By Signing Up, You’ll Get.decreasing Intervals Occur When The Values Of Y Are Decreasing.determine The Interval Over Which The Graph Is Constant.determine The Intervals Where The Graph Is Increasing, Decreasing, And Constant.
Select the correct choice below and fil in any answer boxes in your choi the furpction. Graph the function (i used the graphing calculator at desmos.com). From the graph above, we can observe, (a) [1, 5] is the largest interval on which f is increasing.
### If F ′ Is A Quotient, Factor The Numerator And Denominator (Separately).
A function is decreasing when the graph goes down as you travel along it from left to right. Figure 3 shows examples of increasing and decreasing intervals on a function. Choose random value from the interval and check them in the first derivative.
### If The Derivative Is Positive, Then The Function Is Increasing.
The intervals where a function is increasing (or decreasing) correspond to the intervals where its derivative is positive (or negative). Take the square root of both sides of the equation to eliminate the exponent on the left side. Decreasing increasing constant decreasing increasing decreasing when we describe where the function is increasing, decreasing, and
### If Playback Doesn't Begin Shortly, Try Restarting Your Device.
If f' (c) < 0 for all c in (a, b), then f (x) is said to be decreasing in the interval. Put solutions on the number line. The complete solution is the result of both the positive and negative portions of the solution.
### X 2 = 75 3 X 2 = 75 3.
Answer to use a graphing calculator to find the intervals on which the function is increasing or decreasing, and find any relative maxima or minima. So if we want to find the intervals where a function increases or decreases, we take its derivative an analyze it to. How to determine increasing and decreasing intervals on a graph. | 537 | 2,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-27 | latest | en | 0.824408 |
https://www.experts-exchange.com/questions/28579289/Crystal-Reports-Subreport-array.html | 1,490,583,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189377.63/warc/CC-MAIN-20170322212949-00284-ip-10-233-31-227.ec2.internal.warc.gz | 900,583,083 | 27,402 | ## Welcome to Experts Exchange
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# Crystal Reports - Subreport array
Posted on 2014-12-11
181 Views
I have a Subreport that pulls accounts canceled grouped monthly by Status Date. It has two arrays - one for the number of accounts canceled and one for the amount canceled. Here are the formulas:
shared NumberVar Array mthAmtCan;
shared NumberVar Array mthNumCan;
numberVar monthCount:=DateDiff("m",{?Start Date},{Status Date})+1;
mthAmtCan[monthCount]:= Sum ({Balance}, {Status Date}, "monthly");
mthNumCan[monthCount]:= DistinctCount ({Account ID}, {Status Date}, "monthly");
..................................................................................
In my Main Report (grouped monthly by Tran Effective Date) I declare the shared arrays:
shared NumberVar Array mthAmtCan;
shared NumberVar Array mthNumCan;
numberVar ArrayDeclare:=DateDiff("m",{?Start Date},{?End Date})+1;
Redim mthAmtCan[ArrayDeclare];
Redim mthNumCan[ArrayDeclare];
"";
...........................................................
Then display the monthly value on the Group Footer of the Tran Effective Date :
shared NumberVar Array mthNumCan;
numberVar monthCount:=DateDiff("m",{?Start Date},{Tran Effective Date})+1;
mthNumCan[monthCount]
It displays fine unless I have more than 2 years of data, then anything more than 2 years displays as zeros. I can't be past the 1000 limit on the array data because I'm grouping by month. Any idea on why it can't display more than 2 years of data? My Subreport is linked to my main report by the parameters {?Start Date} and {?End Date} and both parameters are used as the criteria for both reports.
0
Question by:altosax
• 3
LVL 100
Expert Comment
ID: 40495374
WHat error do you get?
WHere do the 0's start?
Early months or last months?
Can you upload the report file?
mlmcc
0
LVL 100
Accepted Solution
mlmcc earned 500 total points
ID: 40495386
I assume the main report declaration is in the report header.
mlmcc
0
LVL 100
Expert Comment
ID: 40495388
If you are grouping by month and displaying in the group footer, why the need for the array?
mlmcc
0
LVL 34
Expert Comment
ID: 40495884
It would be best if you could u/l the report file, so that we could see everything for ourselves. In the mean time:
Can you post your actual formulas? {Status Date}, for example, isn't valid in CR, so I know that those aren't your real formulas. We have no way of knowing what might have been "lost in translation".
Exactly where are all of the items that you mentioned (the subreport and the formulas)?
Where is the subreport located in the main report?
I'm guessing that the first formula is in the subreport. Where in the subreport?
Where are the other two formulas located?
In the first formula, you use {Status Date} to set the array index. In the last formula, you use {Tran Effective Date}. Without knowing anything about your data, I can only ask if that really makes sense. Are those two fields related in some way?
James
0
Author Closing Comment
ID: 40513366
I had the main report declaration in the Page Header. When I moved it to the Report Header, it now gives me data for all of the years.
0
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Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaac… | 1,091 | 4,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-13 | longest | en | 0.651712 |
http://www.thestudentroom.co.uk/showthread.php?t=4232010 | 1,477,623,515,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721555.36/warc/CC-MAIN-20161020183841-00094-ip-10-171-6-4.ec2.internal.warc.gz | 728,465,392 | 34,868 | # Proability
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1. 30% of toy adverts are directed at adults and 70% are directed at children. The chance an adult will buy the toy on seeing the advert is 5%, while the chance children will persuade their parents to get the toy bought for them when they see the advert is 1%.
What is the chance a toy will be purchased if someone in the family sees the advert?
What percentage of toy purchases are due to adverts directed at children?
Are the answers just 5% and 1%??
This is 4 marks in a uni exam
2. Forget 5% and 1% (way off)... You're looking for a probability...
https://www.mathsisfun.com/data/probability.html
3. (Original post by Deannnn97)
30% of toy adverts are directed at adults and 70% are directed at children. The chance an adult will buy the toy on seeing the advert is 5%, while the chance children will persuade their parents to get the toy bought for them when they see the advert is 1%.
What is the chance a toy will be purchased if someone in the family sees the advert?
What percentage of toy purchases are due to adverts directed at children?
Are the answers just 5% and 1%??
This is 4 marks in a uni exam
Does it say how many people are in the family? And does it say how many people actually watch the advert?
If it is one child and adult and they both watch the advert, the answer to a is 6%. If it is a two parent two child family the odds double to 12%, assuming they all watch the advert.
(.7/.01) /(.3/.05)= 46.666666666
There is some working out for the second part of the question. i don't think that this is the final answer though. You will really have to do your own working out for this one!
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Contributor II
## Set analysis to get lowest sales amount for top sales person
Hi, I need to get lowest sales amount for top sales person.
Below is my expression: FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}>}LineSalesAmount),SalesPerson,ProductName,OrderID))
However, I will only be able to get the list of lowest sales amount for each of sales person through table.
While if I implement that expression using KPI it will give the lowest sales amount among all sales person..as this is not i wanted.. I've tried to do nested aggr and firstsortedvalue but nothing is working.
I've attached screenshot for the issue for your reference. You will see in the table list its showing the correct lowest sales for each sales person but at the above total row is showing the lowest sales among all sales person.
1 Solution
Accepted Solutions
MVP
Hi @shnqqs
Try like below.
If(Rowno()=0 or isnull(Rowno()), FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}, SalesPerson={"\$(=FirstSortedValue(SalesPerson, Aggr(Sum({<Year={1997}>}LineSalesAmount),ProductName)))"} >}LineSalesAmount),SalesPerson,ProductName,OrderID)),
FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}>}LineSalesAmount),SalesPerson,ProductName,OrderID)))
If its not works, please provide the sample file
Thanks & Regards, Mayil Vahanan R
8 Replies
MVP
HI @shnqqs
For Total column , generally, it will display the values based on the entire chart.
In order to resolve this, you can use different expression for total column like
If(Rowno()=0 or isnull(Rowno()), FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}, SalesPerson={"YourTopSalesPersonlogictobringthevalue"} >}LineSalesAmount),SalesPerson,ProductName,OrderID)),
FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}>}LineSalesAmount),SalesPerson,ProductName,OrderID)))
Thanks & Regards, Mayil Vahanan R
Contributor II
Author
Hi Mayil,
Big thanks to you for helping me really appreciate it!
However, I've tried using you expression its still giving the lowest sales among all of the sales person..while my concern is to find the lowest sales for top sales person.
Anyways, could you help me to understand the function for 'SalesPerson={"YourTopSalesPersonlogictobringthevalue"} ' that you used in the expression?
MVP
HI @shnqqs
SalesPerson={"YourTopSalesPersonlogictobringthevalue"}
In the place of "YourTopSalesPersonlogictobringthevalue", you need to replace the logic of "Top Sales Employee" expression.. so it will occur like SalesPerson = {'Jamet Leverling'}
Thanks & Regards, Mayil Vahanan R
Contributor II
Author
Hi Mahil,
I've replaced it with the logic, however is still giving the lowest sales of all sales person. Below is the amended expression.
If(Rowno()=0 or isnull(Rowno()), FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}, SalesPerson={"FirstSortedValue(SalesPerson, Aggr(Sum({<Year={1997}>}LineSalesAmount),ProductName))"} >}LineSalesAmount),SalesPerson,ProductName,OrderID)),
FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}>}LineSalesAmount),SalesPerson,ProductName,OrderID)))
//FirstSortedValue(SalesPerson, Aggr(Sum({<Year={1997}>}LineSalesAmount),ProductName))
Thanks!
MVP
Hi @shnqqs
Try like below.
If(Rowno()=0 or isnull(Rowno()), FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}, SalesPerson={"\$(=FirstSortedValue(SalesPerson, Aggr(Sum({<Year={1997}>}LineSalesAmount),ProductName)))"} >}LineSalesAmount),SalesPerson,ProductName,OrderID)),
FirstSortedValue(TOTAL LineSalesAmount, Aggr(Sum({<Year={1997}>}LineSalesAmount),SalesPerson,ProductName,OrderID)))
If its not works, please provide the sample file
Thanks & Regards, Mayil Vahanan R
Contributor II
Author
Thanks Mayil, you are a start! It works!
Just one last doubt..would you eleborate what 'Rowno()=0 or isnull(Rowno())' is doing in the function? what is the logic being used using that?
Thanks!
MVP
Hi @shnqqs
Total will occur in Zeroth Row. So, we are replacing the total value with new expression based on our requirement.
In Pivot table, row no is null for total.
Thanks & Regards, Mayil Vahanan R | 1,124 | 4,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.753197 |
http://www.ck12.org/book/Basic-Probability-and-Statistics-A-Full-Course/r1/section/2.4/ | 1,493,035,117,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119356.19/warc/CC-MAIN-20170423031159-00367-ip-10-145-167-34.ec2.internal.warc.gz | 486,613,326 | 30,557 | # 2.4: Review Questions
Difficulty Level: At Grade Created by: CK-12
Answer the following questions and show all work (including diagrams) to create a complete answer.
1. A bag contains 3 red balls and 4 blue balls. Thomas reaches in the bag and picks a ball out at random from the bag. He places it back into the bag. Thomas then reaches in the bag and picks another ball at random.
1. Draw a tree diagram to represent this problem.
2. What is the probability that Thomas picks:
1. 2 red balls
2. a red ball in his second draw
2. A teacher has a prize box on her front desk for when students do exceptional work in math class. Inside the box there are 20 math pencils and 10 very cool erasers. Janet completed a challenge problem for Ms. Cameron, and Ms. Cameron rewarded Janet’s innovative problem-solving approach with a trip to the prize box. Janet reaches into the box and picks out a prize and then drops it back in. Then she reaches in again and picks out a prize a second time.
1. Draw a tree diagram to represent this problem.
2. What is the probability that Janet reaches into the box and picks out an eraser on the second pick?
3. Determine whether the following situations would require calculating a permutation or a combination:
1. Selecting 3 students to attend a conference in Washington, D.C.
2. Selecting a lead and an understudy for a school play
3. Assigning students to their seats on the first day of school
4. Solve for \begin{align*}{_7}P_5\end{align*}.
5. Evaluate \begin{align*}{_4}P_2 \times {_5}P_3\end{align*}.
6. How many different 4-digit numerals can be made from the digits of 56987 if a digit can appear just once in a numeral?
7. In how many ways can the letters of the word REFERENCE be arranged?
8. In how many ways can the letters of the word MISSISSIPPI be arranged?
9. In how many ways can the letters of the word MATHEMATICS be arranged?
10. If there are 4 chocolate chip, 2 oatmeal, and 2 double chocolate cookies in a box, in how many different orders is it possible to eat all of these cookies?
11. A math test is made up of 15 multiple choice questions. 5 questions have the answer A, 4 have the answer B, 3 have the answer C, 2 have the answer D, and 1 has the answer E. How many answer sheets are possible?
12. In how many ways can you select 17 songs from a mix CD of a possible 38 songs?
13. If an ice cream dessert can have 2 toppings, and there are 9 available, how many different selections can you make?
14. If there are 17 randomly placed dots on a circle, how many lines can be formed using any 2 dots?
15. A committee of 4 is to be formed from a group of 13 people. How many different committees can be formed?
16. There are 4 kinds of meat and 10 veggies available to make wraps at the school cafeteria. How many possible wraps have 1 kind of meat and 3 veggies?
17. There are 15 freshmen and 30 seniors in the Senior Math Club. The club is to send 4 representatives to the State Math Championships.
1. How many different ways are there to select a group of 4 students to attend the State Math Championships?
2. If the members of the club decide to send 2 freshmen and 2 seniors, how many different groupings are possible?
18. Students in BDF High School were asked about their preference regarding the new school colors. They were given a choice between green and blue as the primary color and red and yellow as the secondary color. The results of the survey are shown in the tree diagram below. You can see that 75% of the students choose green as the primary color. Of this 75%, 45% chose yellow as the secondary color. What is the probability that a student in BDF High School selected red as the secondary color if he or she chose blue as the primary color?
19. 2 fair dice are rolled. What is the probability that the sum is even given that the first die that is rolled is a 2?
20. 2 fair dice are rolled. What is the probability that the sum is even given that the first die rolled is a 5?
21. 2 fair dice are rolled. What is the probability that the sum is odd given that the first die rolled is a 5?
22. Steve and Scott are playing a game of cards with a standard deck of playing cards. Steve deals Scott a king. What is the probability that Scott’s second card will be a red card?
23. Sandra and Karen are playing a game of cards with a standard deck of playing cards. Sandra deals Karen a seven. What is the probability that Karen’s second card will be a black card?
24. Donna discusses with her parents the idea that she should get an allowance. She says that in her class, 55% of her classmates receive an allowance for doing chores, and 25% get an allowance for doing chores and are good to their parents. Her mom asks Donna what the probability is that a classmate will be good to his or her parents given that he or she receives an allowance for doing chores. What should Donna's answer be?
25. At a local high school, the probability that a student speaks English and French is 15%. The probability that a student speaks French is 45%. What is the probability that a student speaks English, given that the student speaks French?
26. At a local high school, the probability that a student takes statistics and art is 10%. The probability that a student takes art is 60%. What is the probability that a student takes statistics, given that the student takes art?
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# 20092ee132A_1_hwk4 - to the noise power for the following...
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EE132A, Spring 2009 Communication Systems Prof. John Villasenor Handout #9 TA: Pooya Monajemi and Erica Han Homework 4 Assigned: Monday, April 20, 2009 Due: Monday, April 27, 2009 Reading Assignment: Proakis & Salehi, Chapter 8 (8.3), Chapter 9 (9.1, 9.2) 1. P(f) = [sin( 2 π fT )/ 2 π fT ] 2 , for all real f , is the power spectral density function. What is the corresponding autocorrelation function? 2. A communication system transmits binary data at 64kbit/sec. The data are transmitted using PAM with a raised cosine spectrum. Calculate the bandwidth required for transmission for roll off factors 0.3, 0.5, 1.0 in the following cases: (a) Data are transmitted using binary PAM. (b) Data are transmitted using 16-level PAM by grouping each set of 4 consecutive bits into one symbol. 3. Find and sketch the impulse response of the matched filter that maximizes the ratio of the signal power
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Unformatted text preview: to the noise power for the following pulse shapes: (a) ≤ ≤ = otherwise for ) ( T t t t g (b) ≤ ≤ =-otherwise for ) ( T t e t g t (c) ≤ ≤ = otherwise for 2 sin ) ( T t T t t g π For each case, Compute the value of ) ( T p produced by sampling the matched filter output at time T , the output noise power )] ( ˆ [ 2 t n E , and the signal to noise ratio at the output of the matched filter. 4. (a) Show that sinc( t ) is a Nyquist pulse shape for symbol period 1 = s T . (b) Show that sinc 2 ( t ) is a Nyquist pulse shape for symbol period 1 = s T . 5. Problem 7.2 in Proakis and Salehi....
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# 2023-11-16
< 2023-11-15 2023-11-17 >
## Escapements
Last night I was dreaming about escapements and had a few interesting ideas for designs that reduce sliding friction at the pallets:
• Crank escapement: the "escape wheel" has no teeth, but has a crank on it, and the conrod connects the crank horizontally to the top of the (without loss of generality) pendulum, and as the escape wheel spins round and round it pushes the pendulum back and forth.
• Cam escapement: similar to the crank, but the "escape wheel" instead has lots of lobes on it, and there is a spring-loaded cam follower which (via a rocker, if you like) connects it to the top of the pendulum.
• Magnetic escapement: instead of transmitting force by pallets physically touching the escape wheel, we instead use magnets to transmit force. This has the advantage that it is totally frictionless, but the disadvantages that I can't actually imagine how to do it, and that putting magnets inside a clock is generally a bad idea.
• Eddy current brake: we try to rig up some permanent magnets so that the escape wheel is simply allowed to spin, and is slowed down by the eddy currents from the magnets. (Maybe better to put the magnets on the wheel instead, so that the heat is created in the frame instead of in the wheel).
I think all of these would be totally silent as there are no pallet impacts.
I think the crank and cam versions would be very poor timekeepers because they don't allow the pendulum to swing freely. Take the crank for example:
You can see that the amplitude of the pendulum is constant! The escape wheel either has enough torque to drive it to this amplitude (in which case it runs faster than it naturally would, getting worse as torque increases), or it does not (in which case it stalls).
One downside of the crank escapement is that it needs a higher-ratio gear train than a normal escapement, or the cam escapement, because you only get one cycle of the pendulum for each revolution of the escape wheel.
I think the eddy current version be an even worse timekeeper because I think the eddy current brake will not make it run at a constant speed, and also presumably the braking will change based on temperature as the resistance in the escape wheel changes.
The "magnetic escapement" idea is definitely interesting but I can't quite see how it would work. You can imagine some permanent magnets on the escape wheel (of alternating polarity?) and some on the top of the pendulum, and as the pendulum swings back and forth it alternately lets a north and a south pole pass above it? Not quite sure how it would work. Probably it would also be a poor timekeeper without something to actually stop the escape wheel, else it still has the problem that the escape wheel is constantly applying force to the pendulum.
I think I would be interested to analyse the crank escapement to work out how badly it would perform in practice. Maybe a Matter.js simulation is the easiest way to play with it?
## SCAMP binary compression
Devine Lu Linvega put a post on the CollapseOS mailing list about a data compression format for small systems: http://wiki.xxiivv.com/site/ulz_format.html - although it doesn't appear to have an encoding algorithm yet.
But maybe SCAMP would context switch faster if binaries were stored on disk in a compressed form!
Compressing `lib.o` with gzip takes it from 9K words to 3.5K words, so could potentially be useful. Although I've just remembered that disk IO is basically CPU-bound anyway, so making the CPU do more work is probably counter-productive.
## 3d-printed escapement
Anyway, first job is to work out why the behaviour of the impulse pin in the slot does not match the CAD.
One hypothesis is that it is caused by flex in the frame. The only thing constraining the left half of the model to the right half is the flimsy back-plate.
An interesting design might be to have the balance cock "underneath" the top plate. So you'd have a bottom plate spanning the whole model, then the balance cock mounted on one side, and then a top plate above it also spanning the whole model.
Something like this:
Another hypothesis is that something about the assembly of the pallet fork is crooked, but The pallet pins are slightly crooked, but it doesn't look too bad to me, and they're both pointing in the same direction.
Well anyway, whether it is the frame or the pallet fork that is crooked, it pretty much works anyway, and I could modify the slot to have tolerance for it, so I will do that.
So, print the new pallet fork, and then run some isochronism tests with different weights, and assuming it's isochronism is reasonable, I think I'm done with this model, and it becomes the new preferred design for the watch. And maybe I try printing a 1x scale model with the 0.25mm nozzle? It would be good to have a way to scale a model in FreeCAD, to save doing it all by hand.
It's running pretty well now. This time with just the allen keys and no room spray for a weight (which comes to 370g) the impulse pin is sometimes hitting the balance cock! The amplitude does vary throughout the escape wheel rotation, I think there is still some eccentricity, but not as bad as with the previous escape wheel.
So to run a watch like this would require only 10.7 Nm instead of 16.4 from yesterday. Still too much, obviously, but made smaller and with less friction it would be less.
The impulse pin is still rubbing slightly against the top of the slot, but I'm still not totally sure why. I think the pallet pins are pretty straight now, so that's not the problem. I tried twisting the frame by hand but I can't fix it. I tried re-mounting both cocks and that hasn't fixed it.
The crazy part is that in CAD it shows it getting closer to the lower edge of the slot. (And, actually, still rubbing slightly). Oh, well, whatever. Let's do some isochronism tests with different weights.
I was surprised to find it runs with just 1 nut, which is 87 grams, albeit with quite low amplitude. That corresponds to only 2.5 Nm to run the watch (but we'd want more amplitude). With 5 or more nuts the impulse pin hits the balance cock during the "good" part of the escape wheel.
If you ignore the 5 and 6 nut cases (where the impulse pin is hitting the balance cock), then from 1 nut to 4 nuts (factor of 4 torque increase), the tick rate varied within a range of about 0.550 Hz to 0.564 Hz (2.5%). I didn't plot a best fit line because of the obvious outliers at 5 and 6 nuts, and because only 4 different weights is not really a fair comparison to the previous tests. But the previous best was:
In which, over a range of 6 to 14 nuts (factor of ~2.3), tick rate varied within a range of about 1.318 Hz to 1.340 Hz (1.7%).
That escapement didn't even run over a factor of 4 in torque variation, so a really fair comparison can't be made, but over the range of 4 to 14 (all that I tested), the tick rate varied between about 1.30 and 1.34 Hz (3.1%).
Really I'm not interested in minimising the tick rate variation over such a large torque variation, what I really is to minimise the tick rate variation over the actual torque variation that I will have in the watch, variation of which I obviously will want to minimise.
Anyway, I think the experiment has proven that this new captive lever escapement performs broadly at least as well as the old one, with the obvious advantages that it runs over a wider range of torques, and can support larger amplitudes.
## 1x scale escapement
That's mostly the same design as before, but scaled down, and with a slightly different frame.
Shafts/pins required:
• 2x pallet pin: 0.5mm diameter, 3.5mm long
• impulse pin: 0.5mm diameter, 8mm long
• 2x shafts: 1mm diameter shaft with 0.5mm diameter pivots 2mm long, <13mm between pivots
• barrel shaft: 1mm diameter, no pivots, >18mm long
• balance shaft: 1mm diameter shaft with 0.5mm diameter pivots 2mm long, <8mm between pivots
A 2h59m print with 0.25mm nozzle.
The stringing is quite bad, will take a lot of cleaning up with a knife, and probably under the loupe. Also bad elephant's foot, maybe I should be using a raft for these very small parts. The barrel gear is too flimsy for me to easily trim the elephant's foot with a knife.
Great, I printed the base piece without any screw holes. That's a 1h33m print on its own, more than 50% of what I spent so far. Also I forgot to put bosses around the barrel shaft holes. I'll print another barrel while I'm at it, just in case I get better luck with the elephant's foot this time.
## Scotch Yoke Escapement
Here's another weird idea for an escapement: put a Scotch Yoke on the pendulum, and a pin on the "escape wheel":
This will make the pendulum swing faster in one direction than the other, which seems obviously not ideal, and it still has the problem that the system will run faster if it has more drive torque, but might be interesting to try anyway.
I'd obviously put a ball bearing on the pin like on the keyboard switch tester, to reduce sliding friction in the yoke:
You could make the radial position of the pin on the escape wheel adjustable, so that the amplitude can be adjusted. I'm not sure whether adjusting the amplitude or the pendulum length would be a better adjustment, or whether you'd want to adjust them separately for different things. For example, maybe you'd want to adjust the pendulum length so that its free-swinging period is correct, and then adjust the radial position of the Scotch Yoke pin so that the amplitude is high enough that the drive torque can just barely run it, to minimise the deviation from the pendulum's natural period? I don't know, I'm just guessing. I should build one and find out.
If you assume the pin is adjustable in the slot in the disc, then this arrangement can support amplitudes from about 15 degrees up to 47.5 degrees.
I've made a start on it anyway. Still need to design the frame, and there are a lot of parts "implied" that I need to make separately. I'm only modelling the ones I need to 3d print. Also needed are:
• 5x 105ZZ bearings (4 for pivots, 1 for yoke pin) | 2,378 | 10,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.944927 |
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### Topic: A prism - geometrical optics (Read 6119 times)
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#### Hunt
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##### A prism - geometrical optics
« on: May 04, 2006, 07:16:27 PM »
A typical prism is a solid figure, a transparent homogenous and isotropic medium bounded by 2 non-parallel plane surfaces ( each forms a parm ). In geometrical optics, we can view a prism as a triangle characterized by its vertex angle alpha ( suppose 'a' ) and its index of refraction n. A prism can be used to cause dispersion of a beam of white light dividing it into its respective components , each of its own wavelength or frequency.
Suppose a ray of light propogating through air hits one face of an isosceles prism , the emergent ray is deviated towards the base of the prism relative to the incident ray. The angle of deviation is that between the 1st incident ray and 2nd refracted ray , designated ?d.
1-Assuming the 2nd angle of incidence ?i2 ? ?critical, and that Sin ( ? ) ? ? ( in radians ) for small angles, prove that for small incident angles & a small angle alpha 'a' ( of the prism ), ?d = a ( n - 1 ) i.e. independent of ?i2
2-Incase of minimum deviation, find the expression of n in terms of ?m ( minimum angle of deviation ) and alpha 'a'. Deduce, incase of small angles , n = ( ?m + a ) / a
Only basic algebra + trigonometry involved ... happy hunting Xiankai.
« Last Edit: May 04, 2006, 07:18:43 PM by Vant_Hoff » | 431 | 1,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-45 | latest | en | 0.872454 |
https://www.calculatorall.com/bandwidth-calculator | 1,722,649,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00526.warc.gz | 556,307,076 | 10,085 | # Bandwidth Calculator - Calculatorall.com
A bit is smallest unit understood and used by computers and it is also used in the information technology applications. A bit can either be 0 or 1. Eight bits combine to form a byte and one byte can be used for representing values from 0 to 255. The unit of communication bandwidth in IT that's widely used is bit. For instance, if the transfer rate is 1 Mbits that would mean that the communication is taking place at 1 Megabits per second. This is the rate at which information is passed from one machine to the other. The unit for information storage, however, is byte. You can say a hard disk or memory of 8GB would be able to hold information up to 8 Gigabytes of data. But what does this bandwidth calculator do?
#### Bandwidth/Size Converter
bits (b) kilobits (kb) megabits (mb) gigabits (gb) terabits (tb) Bytes (B) Kilobytes (KB) Megabytes (MB) Gigabytes (GB) Terabytes (TB) = ?
File Size
Bytes (B) Kilobytes (KB) Megabytes (MB) Gigabytes (GB) Terabytes (TB)
Select a speed:
bps Kbps Mbps Gbps Tbps
#### Website Bandwidth Calculator
Page Views Per Second Per Minute Per Hour Per Day Per Week Per Month Per Year Average Page Size Bytes (B) Kilobytes (KB) Megabytes (MB) Gigabytes (GB) Terabytes (TB)
#### Hosting Bandwidth Converter
Monthly Usage Bytes (B) Kilobytes (KB) Megabytes (MB) Gigabytes (GB) Terabytes (TB) is equivalent to Bandwidth bit/s Kbit/s Mbit/s Gbit/s Tbit/s
Bandwidth Calculator
Well, this isn't just a single calculator rather there are four different calculators that you can find on this page. Each of these calculators performs a different function and calculates different values. However, each of these values is related to bandwidth. For instance, one calculator is used for measuring website bandwidth while the other is used for hosting bandwidth.
How It Works?
The first bandwidth calculator works as a bandwidth/size converter. You have to enter bandwidth in one of the given units - make sure to select the appropriate unit - and the calculator converts it into different other units that are used for measuring bandwidth and shows you the result in all those units available.
The second calculator on the page can be used for measuring the download or upload time for a specific file with available connection bandwidth. You have to enter the size of the file that you want to download or upload and select an appropriate unit for size. After that you have to enter the bandwidth available for that upload or download with appropriate unit. The calculator will show you the time that will be needed for download/upload.
The next on the list is website bandwidth calculator. This calculator requires you to enter the number of page views per day - you can select other measures as well like per week or per month – and the average size of your website pages. Select appropriate unit for the page size as well. As you hit the calculate button, the bandwidth calculator will tell you how much of bandwidth you will need for your website to perform at optimum levels.
The last calculator is the hosting bandwidth converter. It accepts two values as input, but you have to enter one value and the calculator will convert it into the other and show you the result. For instance, if you provide your monthly usage, the calculator will tell you how much it is in terms of bandwidth and vice versa.
So, use any bandwidth calculator as per your need and calculate the desired results. You will certainly have great insights into your internet usage and will be able to make smarter decisions in terms of what you need or how you should go about using the available bandwidth. | 799 | 3,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.817189 |
http://entrepreneursjunction.net/2021/10/08/excel-finance-class-81-xnpv-function-xirr-function-see-algorithm-that-xnpv-uses/ | 1,675,486,546,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00083.warc.gz | 16,293,047 | 16,334 | # Excel Finance Class 81: XNPV Function XIRR Function – See Algorithm That XNPV uses
Welcome to Finance in
this workbook for chapter 9, click on the link
directly below the video. And scroll all the way down
to the Finance Excel Class section. Hey, we got to talk about the
XNPV and the XIRR function. But mostly we want to
discounted cash flow analysis to value assets,
absolutely have to use this. Now, this function will
solve a lot of problems. It is what you want to use if
you're doing discounted cash flow analysis in most cases. However, in earlier
versions, you may have to add the
Analysis ToolPak. So 2007 and 2010, this
function is already there. In earlier version, you have to
solves a bunch of problems we've had so far, or
some of which we haven't talked about. We actually already did one
video on this last chapter, but we're going to be
much more detailed here.
The problem is is
oftentimes you maybe don't have the exact time, the
equal time between each cash flow. So here you can see four years
out, we get this cash flow. But then we don't get another
cash flow until six years out. And so the NPV function
wouldn't be able to handle that. If when we did NPV in earlier
chapters, we did a RATE comma. And then we highlighted the
values not including zero. But guess what? It just counts. This function counts–
1, 2, 3, 4, 5. So it would assume that this
is 5 when really it's 6. So XNPV will solve that problem. Also we saw with
the NPV function, we always had to go plus
the negative cash flow at time 0, which
is not a big deal. But it's as if someone who
first programmed this function didn't realize that we
needed that time 0 cash flow.
And guess what? XNPV, absolutely no problem. You just highlight from time
0 all the way to the end. So there's a couple
things that just make it totally much easier. Let's just see how this works. And then I'm actually going
to break it down and show you the algorithm of how it
calculates, because it can get a slightly different answer
than the PV function, which a lot of people use when
they get into this situation. It's no problem. PV function or the XNPV will be
a few dollars or pennies off. But we're talking about
estimation here anyway. So estimation, a few pennies or
dollars usually doesn't matter.
Now let's just go ahead and see. XNPV– it needs a rate,
still needs a discount rate, because we're discounting
it at our required return– comma the values,
including time 0– I love that– comma
and then the dates, including the date for time 0. And that's it. Irregular cash flows,
irregular time periods, and it includes time 0. Now let's see how we did it– in this class, we actually did
do it the long way using the PV function in earlier videos. But let's see how
that works, because we want to compare the answer we
get using this to the XNPV. So present value,
that just means we have these future
value amounts, and we need to discount
them back one at a time. So the rate F4 NPER,
that one, comma. There is no PMT. These are all lump sum,
so comma comma and then the lump sum, close parentheses.
Now because we're
going to dump– if you're using present
value of a negative, it gives you a positive. But we don't want
that, because we're going to put these into the– add them up as a cash flow. So we need to actually
put a negative here. And now when I copy this down– and I like that. So we can see that
here this will work, because we didn't
use the Date column. But we very carefully
typed 0, 1, 2, 3, 4, and then put a 6 there.
So it knows, the
PV function knows, this cash flow should be
discounted back six periods. Now I'm going to come over here. And I'm going to add these up. This is net present value
using the PV function– Alt-Equals. And I'm going to
highlight these. So you can see
it's about \$10 off. And actually, I want to
do cash flows for PV. And then I'm going to
wrap text, Home ribbon. And this should be cash flows. I think that's how
you spell algorithm. OK, so why is there a
difference, a \$10 difference? And the reason why
is the algorithm. So I want to show you the
algorithm about how XNPV works. And the reason– it
comes down to it– when it's out here
at this cash flow, it's going to say,
hey, how do I calculate the number of periods? It's going to take the later
date minus the earlier date. And in Excel when you
subtract a later date, you take a later date and
subtract an earlier date, it gives you number of
days between two dates. So if we take the number of
days and divide it by 365, it'll give us an
estimation of years.
Now, in order to see how
the XNPV algorithm works, I'm going to start my
formula right here. Now we're going to do
the longhand formula for discounting a
future lump sum. So I'm going to say equals. Here's the future
value, lump sum, divided by 1 plus our discount rate. And I'm going to have to
lock that with the F4 key, because we're going to
copy this up and down. Now, the only trick is,
this is an annual rate.
And so whatever exponent we
put for our number of periods to discount this over,
it's got to be in years. So we do caret, which
is our exponent. And as we mentioned, we're going
to have to subtract two dates and divide by– so subtracting two dates will
give us the number of days. And we divide by 365– so two open parentheses and then
later date minus earlier date. Now, as we copy this
formula up and down, this one needs to go down. But that always
needs to be locked. So I'm going to hit the
F4 key, close parentheses. That's the number of
days divided by 365. Now, just right
now, we could just do a little
investigating to prove to ourself that this actually
is the number of days between these two dates.
So I'm going to highlight
that and hit the F9– 2,192. So I'm going to take all
the days divided by 365. And notice because we're
using days and an estimation, because some years don't have– actually that's–
this right here, sometimes since we're
subtracting dates, the dates know that
there's leap year. This number does not. So we'll get a little
inaccuracy there. Now I'm going to
Control-Z. All right, let's highlight this whole thing
and prove to ourselves that it's actually the
exponent is in years– sure enough, 6.0054. Control-Z, close parentheses
and Control-Enter. Now we– oops. We need to copy
this up and down. I'm going to check it here. OK, so it's got the later
date minus the earlier date. It worked fine. Let's look up here. And by the way, why is
this working for the 0? Oh, because this date
minus this date is 0. That means the whole
exponent is 0, which means, oh, anything raised to the 0 is
1, so that thing divided by 1. OK, so the trick now
is to add it all up. So we're checking to see the
net present values in XNPV algorithm.
So Alt-Equals. And then we get the
same amount here. And so it's just a matter of– for PV, we were actually
using just whole years. And for the XNPV,
we were actually using day, so a little bit of
a difference in our estimation. But ultimately because this
is so easy to use, we can– and it's all estimation. These future cash flows
are all estimation. This will be just fine. Now let's look at the XIRR. The XIRR is the same as the XNPV
in that it can handle values, including time 0,
irregular amounts comma and then the dates,
dates that are different. Now, we go ahead and if
a guess is only necessary if we get an error. If there is, it's an
iterating function. And so sometimes you can
have trouble in iteration. So you could put in a guess,
maybe something close to that. But we shouldn't
have a problem here. So there is the XIRR. Now we want to look at even– we've already seen
how awesome this is.
XNPV solved the
irregular time period. And it now includes time 0. But watch this. Up here it was all years. We were just separated by years. Down here, look at this. We have a project. It doesn't matter what date,
because the exponent is using days in calculating
an exponent, which is a partial year or a whole year. You can put anything. So this project, at time 0,
we put in spend this amount. And then we have these cash
flows coming in months. These are months. And then next cash flow doesn't
come in for another year and then another year. And then here, whoa, days– three 700s coming in, boom,
boom, boom, days in a row. Wow. And you're probably not going
to have cash flows coming in on days like this. But we want to illustrate that
the XNPV function will handle that, just no problem at all. So the rate– there's a discount
rate, values, including time 0, dates. Doesn't matter
what kind of dates as long as they're real
dates, serial number dates. There we have it. The long way equals,
and we'll take the present value of this
divided by 1 plus our discount rate, F4 caret.
And then we got to take
this date minus the– this is the later date
minus the earlier date. So I'm going to lock the earlier
date, because as we go down, we want the purple box,
the first bigger date, to move relatively. Close parentheses
divided by 365. Come over here,
if I can do this. If I come over here,
oh, look I typed that. This should be NPV. And then I'm going
to add this up. Again, this is– you
wouldn't do this column here if you're out there working. But it does– it is helpful
to learn how this actually is working and why it's a
little bit different than the PV method.
Finally, we could do our XIRR. Got to be kidding me– just the
values, time 0, and the dates. Now that would be hard
to calculate by hand. I'm sure glad that's XIRR. All right, that's a lot
about the amazing function XNPV and XIRR. We'll see you next video.. | 2,521 | 9,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-06 | latest | en | 0.96737 |
https://www.scribd.com/document/366716733/Theory | 1,566,136,863,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313889.29/warc/CC-MAIN-20190818124516-20190818150516-00382.warc.gz | 931,681,838 | 67,424 | You are on page 1of 12
# THEORY
Consider a belt which moves on the pulley with the small angle of lap, . Consider that T is the
tension at point a, and T + dT is the tension at point b. R is the small surface for the pulley rim
and the touching belt (see Figure 1). Figure 2 shows the triangle force, where R can be measured.
Assume that the change on angle is so small, d and the change of the belt tension is dT,
therefore ;
R = T d---------------(1)
The coefficient of friction (COF), also known as a frictional coefficient or friction coefficient,
symbolized by the Greek letter , is a dimensionless scalar value which describes the ratio of the
force of friction between two bodies and the force pressing them together. The coefficient of
friction depends on the materials used; for example, ice on steel has a low coefficient of friction,
while rubber on pavement has a high coefficient of friction. Coefficients of friction range from
near zero to greater than one under good conditions, a tire on concrete may have a coefficient
of friction of 1.7.
When the surfaces are conjoined, Coulomb friction becomes a very poor approximation (for
example, adhesive tape resists sliding even when there is no normal force, or a negative normal
force). In this case, the frictional force may depend strongly on the area of contact. Some drag
racing tires are adhesive in this way. However, despite the complexity of the fundamental physics
behind friction, the relationships are accurate enough to be useful in many applications.
The force of friction is always exerted in a direction that opposes movement (for kinetic friction)
or potential movement (for static friction) between the two surfaces. For example, a curling stone
sliding along the ice experiences a kinetic force slowing it down. For an example of potential
movement, the drive wheels of an accelerating car experience a frictional force pointing forward;
if they did not, the wheels would spin, and the rubber would slide backwards along the
pavement. Note that it is not the direction of movement of the vehicle they oppose, it is the
direction of (potential) sliding between tire and road.
## The coefficient of friction is an empirical measurement it has to be measured experimentally,
and cannot be found through calculations. Rougher surfaces tend to have higher effective values.
Most dry materials in combination have friction coefficient values between 0.3 and 0.6. Values
outside this range are rarer, but Teflon, for example, can have a coefficient as low as 0.04. A
value of zero would mean no friction at all, an elusive property even magnetic levitation
vehicles have drag. Rubber in contact with other surfaces can yield friction coefficients from 1 to
2. Occasionally it is maintained that is always < 1, but this is not true. While in most relevant
applications < 1, a value above 1 merely implies that the force required to slide an object along
the surface is greater than the normal force of the surface on the object. For example, silicone
rubber or acrylic rubber-coated surfaces have a coefficient of friction that can be substantially
larger than 1.
Both static and kinetic coefficients of friction depend on the pair of surfaces in contact; their
values are usually approximately determined experimentally. For a given pair of surfaces, the
coefficient of static friction is usually larger than that of kinetic friction; in some sets the two
coefficients are equal, such as Teflon-on-Teflon.
In the case of kinetic friction, the direction of the friction force may or may not match the
direction of motion: a block sliding atop a table with rectilinear motion is subject to friction
directed along the line of motion; an automobile making a turn is subject to friction acting
perpendicular to the line of motion (in which case it is said to be 'normal' to it). The direction of
the static friction force can be visualized as directly opposed to the force that would otherwise
cause motion, were it not for the static friction preventing motion. In this case, the friction force
exactly cancels the applied force, so the net force given by the vector sum, equals zero. It is
important to note that in all cases, Newton's first law of motion holds.
While it is often stated that the COF is a "material property," it is better categorized as a "system
property." Unlike true material properties (such as conductivity, dielectric constant, yield
strength), the COF for any two materials depends on system variables like temperature, velocity,
atmosphere and also what are now popularly described as aging and deaging times; as well as on
geometric properties of the interface between the materials. For example, a copper pin sliding
against a thick copper plate can have a COF that varies from 0.6 at low speeds (metal sliding
against metal) to below 0.2 at high speeds when the copper surface begins to melt due to
frictional heating. The latter speed, of course, does not determine the COF uniquely; if the pin
diameter is increased so that the frictional heating is removed rapidly, the temperature drops, the
pin remains solid and the COF rises to that of a 'low speed' test.
Known that the coefficient of friction, = friction / action of touching surface, therefore
dT = R --------------(2)
## Replace (1) into (2) ;
dT = (Td) dT/T = d
## From the equation (3), e antilog of log 10 x = a
Antilog of 0.4342
= angle of lap
T1 = m1g
T2 = (m1-m2) g
=m2g
= 90
Log e T1/T2 =
## Coefficient of friction between belt and pulley, = [Log e T1/T2] /
=0.66/ 90 = 0.0073
EQUIPMENT
QUANTITY APPARATUS
wall with ;
## Flat belt groove, large
groove, and fit groove
1 Flat belt
1 V - belt
hook m2
2 Surface plate
500g, and 1kg.
PROCEDURES
## 1. Prepare the apparatus of experiment as shown in figure 3.
2. Fix the flat belt at an angle of 60 degees and through the correct groove.
## 3. Hang the 500g at bottom end of belt.
4. Wind the cord as many as you can on pulley in the opposite direction of flat belt and hang
500g weight hook at the end of cord and state 500g as the first value of m2.
5. Support the weight hook m2 by hand at early stage to avoid from moving downwards due
to the rotation of pulley. Then add load little by little at weight hook m1 until m2 become
stable. State the load as the value of m1.
6. Repeat step 4 and 5 by using the value of m2 as 700g, 900g, 1100g and 1300g. state the
total load of m1 and m2.
7. Then, repeat step 2 until 6 by fixing the belt at different angle of lap, 90 degrees , 120
degrees and 180 degrees. State the total value of load m1 and m2.
## 1. Prepare the apparatus of experiment as shown in figure 4.
2. Fix the V-belt at an angle of 90 degrees and through the incorrect V-groove at (b).
## 3. Hang the 500g at bottom end of belt.
4. Wind the cord as many as you can on pulley in the opposite direction of flat belt and hang
500g weight hook at the end of cord and state 500g as the first value of m2.
5. Support the weight hook m2 by hand at early stage to avoid from moving downwards due
to the rotation of pulley. Then add load little by little at weight hook m1 until m2 become
stable. State the load as the value of m1.
6. Repeat step 4 and 5 by using the value of m2 as 700g, 900g, 1100g and 1300g. state the
total load of m1 and m2.
7. Then, repeat step 2 until 6 by fixing the belt for the V-belt with correct fitting of groove
(c). state the total value of load m1 and m2.
RESULTS
Table 2- Test 1
## Angel of lap of flat belt,
No 90 120 150 180
m1(g) m2(g) m1(g) m2(g) m1(g) m2(g) m1(g) m2(g)
1 1400 500 1200 500 1000 500 900 500
2 2200 700 1700 700 1400 700 1300 700
3 2700 900 2200 900 1900 900 1700 900
4 3300 1100 2700 1100 2400 1100 2000 1100
5 3500 1300 3200 1300 3000 1300 2400 1300
## T1 (N) T2 (N) T1 (N) T2 (N) T1 (N) T2 (N) T1 (N) T2 (N)
1 13.73 8.83 11.77 6.87 9.81 4.91 8.83 3.92
2 21.58 14.72 16.68 9.81 13.78 6.87 12.75 5.89
3 26.49 17.66 21.53 12.75 18.64 9.81 16.68 7.85
4 32.37 21.58 26.49 15.7 23.54 12.75 19.62 8.83
5 34.34 21.58 31.39 18.64 29.43 16.68 23.54 10.79
T1/T2 1.52 1.69 1.87 2.18
loge
[T1/T2] 0.66 0.73 0.81 0.95
Angel of lap of belt, at angle of , = 90
No Flat belt (a) V-belt large groove (b) V-belt with correct / fit groove (c)
m1(g) m2(g) m1(g) m2(g) m1(g) m2(g)
1 1400 500 600 500 600 500
2 2200 700 800 700 900 700
3 2700 900 1100 900 1100 900
4 3300 1100 1300 1100 1400 1100
5 3500 1300 1400 1300 1500 1300
## T1 (N) T2 (N) T1 - T2 (N) T1 (N) T2 (N) T1 - T2 (N) T1 (N) T2 (N) T1 - T2 (N)
1 13.73 8.83 4.9 5.89 0.98 4.91 5.89 0.98 4.91
2 21.58 14.72 6.86 7.85 0.98 6.87 8.83 1.96 6.87
3 26.49 17.66 8.83 10.79 1.96 8.83 10.79 1.96 8.83
4 32.37 21.58 10.79 12.75 1.96 10.79 13.73 2.94 10.79
5 34.34 21.58 12.76 13.73 0.98 12.75 14.72 1.96 12.76
Table 3-Test 2
OBSERVATION
A cord was stuck into a screw when we change the pulley into angle 180 and was make a
rectification to the result.
When we increase the angle of lap, the value of weight that we need to stabilize the m2
hook also decrease.
In the test 2, flat belt need more weight,m1 to stabilize the hook,m2 than the V belt large
groove and v belt with correct or fit groove.
When we increase the angle of lap, the value of tight belt will decrease as we can look at
table 2-test 1.
CALCULATION
Calculation to get T1 :
T1 = m1g
We assume that gravity, g is 9.81 m/s
## Example : we take m1(g) no 1 from angle 90 = 1400g @ 1.4 kg
T1 = 1.4x9.81
= 13.73 N.
Calculation to get T2 :
T2 = (m1-m2)g
## Example : we take m1 and m2 no 1 from angle 90 :
m1 = 1400g @1.4 kg
m2 = 500g @ 0.5kg
T2 = (1.4-0.5)9.81
= 8.829 N
= 128.51N
## T2 = 8.83 + 14.72 + 17.66 + 21.58 + 21.58
= 84.37 N
T1/T2 = 128.51/84.37=1.52
= 0.66
= 90
Log e T1/T2 =
## Coefficient of friction between belt and pulley, = [Log e T1/T2] /
=0.66/ 90 = 0.0073
DISCUSSIONS
1. Based on data T1 , T2 from TEST 1, plot the graph of T1 versus T2 according to the
angle of lap on the same graph paper. Discuss the relation between the changes of driving
force with the angle of lap.
From the graph of T1 versus T2, we can see the relation between the changes of driving
force with the angle of lap. The relation is when the angle of lap increase, so the force
that we need to stabilize the m2 hook also will increase.
2. Based on data from TEST 1, plot the graph of log[T1/T2] versus angle of lap. Then
calculate the value of coefficient of friction of flat belt and review on the result.
= 90
Log e T1/T2 =
## Coefficient of friction between belt and pulley,= [Log e T1/T2] /
=0.66/ 90 = 0.0073
= 120
## As we know from formula (3),
Log e T1/T2 =
Coefficient of friction between belt and pulley, = [Log e T1/T2] / =0.73/ 120 = 0.0061
= 150
Log e T1/T2 =
= 180
## As we know from formula (3),
Log e T1/T2 =
Coefficient of friction between belt and pulley, = [Log e T1/T2] / =0.95/ 180 =
0.0053
As what we can see from a graph log e[T1-T2] versus angle of lap, we can say that when
value angle of the lap increase, so the value of tension, T will decrease. And from the
value of coefficient that we get above , we also can say that, when the value of angle lap
increase, the value of coefficient will be decrease.
3. Based on data from the TEST 2, plot the graph (T1-T2) versus T1 according to the
different type of belt on the same graph. Review the results.
From our observation of the graph (T1-T2) versus T1 , we can say that V-belt large
groove(b) need more tension (T) to stabilize a m2 hook than the V- belt with correct or fit
groove (c) and flat belt. So we can say that , the V-belt large groove posses a less friction
than the V- belt with correct groove and flat belt.
CONCLUSION
In conclusion, from this belt friction experiment, we can learn the relation between tight belt and
angle of lap for flat belt. From the result f this experiment, we can say that the tight of the belt
will be decrease when the angle of lap for flat belt increase. Beside that, during this experiment
student also can look and compare the driving force at certain angle of lap of the flat belt, v belt
with incorrect groove and v belt with correct groove. We also can make a conclusion that the
ideal belt is that of the lowest tension which does not slip in high loads.
REFERENCE
## 1. Engineering laboratory IV (BDA 2721) lab sheet-belt friction, FKMP UTHM
2. http://en.wikipedia.org/wiki/Friction
3. http://en.wikipedia.org/wiki/Belt_(mechanical) | 3,756 | 12,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-35 | latest | en | 0.941067 |
https://justaaa.com/accounting/19934-four-flags-is-a-retail-department-store-on | 1,686,398,587,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00700.warc.gz | 367,135,873 | 11,178 | Question
# Four Flags is a retail department store. On January 1, 2019, Four Flags' accountants used the...
Four Flags is a retail department store. On January 1, 2019, Four Flags' accountants used the following data to develop the master budget for Four Flags for 2019: Cost Fixed Variable (per sales dollar) Cost of Goods Sold \$0 \$0.660 Selling and Promotion Expense \$205,000 \$0.080 Building Occupancy Expense \$180,000 \$0.020 Buying Expense \$145,000 \$0.050 Delivery Expense \$120,000 \$0.010 Credit and Collection Expense \$72,000 \$0.002 Expected unit sales in 2019 were 1,200,000, and 2019 total revenue was expected to be \$12,000,000. Actual 2019 unit sales turned out to be 1,050,000, and total revenue was \$10,500,000. Actual total costs in 2019 were: Cost of Goods Sold \$6,000,000 Selling and Promotion Expense \$1,100,000 Building Occupancy Expense \$320,000 Buying Expense \$520,000 Delivery Expense \$190,000 Credit and Collection Expense \$75,000 Required Compute the flexible budget variances in 2019 for the following two cost items (NOTE: enter favorable variances as positive numbers and unfavorable variances as negative numbers):
Actual results Flexible budget Flexible budget variance Cost of goods sold \$6,000,000 \$6,930,000 (10,500,000*\$0.660) \$930,000 Favorable Selling and promotion expense 1,100,000 1,045,000 (10,500,000*\$0.080+205,000) 55,000 Unfavorable Building occupancy expense 320,000 390,000 (10,500,000*\$0.020+180,000) 70,000 Favorable Buying expense 520,000 670,000 (10,500,000*\$0.050+145,000) 150,000 Favorable Delivery expense 190,000 225,000 (10,500,000*\$0.010+120,000) 35,000 Favorable Credit and collection expense 75,000 93,000 (10,500,000*\$0.002+72,000) 18,000 Favorable
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 518 | 1,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.841953 |
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Sigmoid functions are used to model many natural processes such as population growth of virus. One example of a Sigmoid function $\mathrm{f}$ is given by
$\mathrm{f}: x \mapsto \frac{1}{1+\mathrm{e}^x}, x \in \mathbb{R} .$$(i) Sketch the graph of$y=\mathrm{f}(x)$, indicating clearly the equation(s) of any asymptote(s) and the coordinates of any points where the curve crosses the axes. [2] (ii) Find$\mathrm{f}^{-1}(x)$in similar form. Another function$\mathrm{g}$is given by$\mathrm{g}: x \mapsto 3 x-1, x \in \mathbb{R}, 0 \leq x \leq 2$. [3] (iii) Show that fg exists and find the range of$f g$, expressing your answer in terms of e. [4] (iv) Describe a sequence of transformations which transform the graph of$y=\mathrm{f}(x)$onto the graph of$y=f g(x)\$.
[2] | 257 | 793 | {"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} | 3.03125 | 3 | CC-MAIN-2024-26 | latest | en | 0.668299 |
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# Leading Index for Utah
2015-12: 1.20 Percent
Monthly, Seasonally Adjusted, UTSLIND, Updated: 2016-02-03 12:01 PM CST
1yr | 5yr | 10yr | Max
The leading index for each state predicts the six-month growth rate of the state's coincident index. In addition to the coincident index, the models include other variables that lead the economy: state-level housing permits (1 to 4 units), state initial unemployment insurance claims, delivery times from the Institute for Supply Management (ISM) manufacturing survey, and the interest rate spread between the 10-year Treasury bond and the 3-month Treasury bill.
Release: State Leading Indexes
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Graph updated. | 470 | 1,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-07 | latest | en | 0.815625 |
https://yuqinghu.blog/2015/05/09/real-analysis-review-day-3-relations-part-ii/ | 1,611,490,832,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00105.warc.gz | 1,085,979,663 | 27,468 | #### Mappings
Definition. A mapping $f: X\to Y$ is a relation $R$ from $X$ to $Y$ with the property that for every $x\in X$ there is a unique $y\in Y$ such that $xRy$, in which case we write $y=f(x)$. Mappings are sometimes called maps or functions; and we shall generally reserve the latter name for the case when $Y$ is $\mathbb C$ or some subset thereof.
If $f: X\to Y$ and $g: Y\to Z$ are mappings, we denote by $g\circ f$ their composition:
$g\circ f: X\to Z$, $g\circ f(x)=g(f(x))$.
If $D\subset X$ and $E\subset Y$, we define the image of $D$ and the inverse image of $E$ under the mapping $f: X\to Y$ by:
$f(D)=\{f(x): x\in D\}$, $f^{-1}(E)=\{x: f(x)\in E\}$.
The map $f^{-1}: \mathcal P(Y)\to\mathcal P(X)$ defined by the second formula commutes with union, intersections, and complements:
$f^{-1}(\cup_{\alpha\in A}E_\alpha)=\cup_{\alpha\in A}f^{-1}(E_\alpha)$,
$f^{-1}(\cap_{\alpha\in A}E_\alpha)=\cap_{\alpha\in A}f^{-1}(E_\alpha)$,
$f^{-1}(E^c)=(f^{-1}(E))^c$.
If $f: X\to Y$ is a mapping, $X$ is called the domain of $f$ and $f(X)$ is called the range of $f$. $f$ is said to be injective if $f(x_1)=f(x_2)$ only when $x_1=x_2$surjective if $f(X)=Y$, and bijective if it is both injective and surjective. If $f$ is bijective, then it has an inverse $f^{-1}: Y\to X$ such that $f^{-1}\circ f$ and $f\circ f^{-1}$ are the identity mappings on $X$ and $Y$, respectively. If $A\subset X$, we denote by $f|A$ the restriction of $f$ to $A$:
$(f|A): A\to Y$, $(f|A)(x)=f(x)$ for $x\in A$.
A sequence in a set $X$ is a mapping from $\mathbb N$ into $X$.
A finite sequence is a map from $\{1, 2, ..., n\}$ into $X$ where $n\in \mathbb N$.
If $f:\mathbb N\to X$ is a sequence and $g: \mathbb N\to\mathbb N$ satisfies $g(n) whenever $n, the composition $f\circ g$ is called a subsequence of $f$.
If $\{X_\alpha\}_{\alpha\in A}$ is an indexed family of sets, their Cartesian product $\prod_{\alpha\in A} X_\alpha$ is the set of all maps $f: A\to\cup_{\alpha\in A}X_\alpha$ such that $f(\alpha)\in X_\alpha$ for every $\alpha\in A$.
If $X=\prod_{\alpha\in A}X_\alpha$ and $\alpha\in A$, we define the $\alpha$th projection or coordinate map $\pi_\alpha: X\to X_\alpha$ by $\pi_\alpha(f)=f(\alpha)$. We also frequently write $x$ and $x_\alpha$ instead of $f$ and $f(\alpha)$ and call $x_\alpha$ the $\alpha$th coordinate of $x$.
If the sets $X_\alpha$ are all equal to some fixed set $Y$, we denote $\prod_{\alpha\in A}X_\alpha$ by $Y^A$:
$Y^A=$ the set of all mappings from $A$ to $Y$.
If $A=\{1, 2, ..., n\}$, $Y^A$ is denoted by $Y^n$ and may be identified with the set of ordered $n$-tuples of elements of $Y$.
“To let go does not mean to stop caring, it means I can’t do it for someone else. To let go is not to cut myself off, it’s the realization I can’t control another. To let go is not to enable, but allow learning from natural consequences. To let go is to admit powerlessness, which means the outcome is not in my hands. To let go is not to try to change or blame another, it’s to make the most of myself. To let go is not to care for, but to care about. To let go is not to fix, but to be supportive. To let go is not to judge, but to allow another to be a human being. To let go is not to be in the middle arranging all the outcomes, but to allow others to affect their destinies. To let go is not to be protective, it’s to permit another to face reality. To let go is not to deny, but to accept. To let go is not to nag, scold or argue, but instead to search out my own shortcomings and correct them. To let go is not to adjust everything to my desires, but to take each day as it comes and cherish myself in it. To let go is not to criticize or regulate anybody, but to try to become what I dream I can be. To let go is not to regret the past, but to grow and live for the future. To let go is to fear less and LOVE more. Remember: the time to love is short.” ~ Author Unknown
To be continued tomorrow 🙂
References:
[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 3-4.
[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf | 1,365 | 4,217 | {"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": 86, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-04 | latest | en | 0.795554 |
http://compsci.ca/v3/viewtopic.php?p=269805 | 1,670,420,541,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00294.warc.gz | 8,376,897 | 11,412 | Computer Science Canada Programming C, C++, Java, PHP, Ruby, Turing, VB Username: Password: Wiki Blog Search Turing Chat Room Members
q5 clarify
Author Message
Panphobia
Posted: Wed Feb 20, 2013 5:22 pm Post subject: q5 clarify
I understand the question to the most part, but when it says for the input, that is a maximum length the pattern has to be, is that, is one pattern touching all 9 buttons? so if it was of length 1 would it be a maximum of touching all 9?
Tony
Posted: Wed Feb 20, 2013 6:01 pm Post subject: RE:q5 clarify
I don't understand what you are asking about the 9 buttons. There's a sample case of "size 1" in the question. Does the output to the sample case confirm your understanding of the requirement?
Tony's programming blog. DWITE - a programming contest.
Panphobia
Posted: Wed Feb 20, 2013 6:05 pm Post subject: RE:q5 clarify
I am ok now I think I understand
Panphobia
Posted: Wed Feb 20, 2013 6:30 pm Post subject: RE:q5 clarify
I got pretty bad on this dwite, on the first one did 24 - instead of - 24, but for that last question as I understood you were supposed to figure out the total amount of patterns with a certain length, I made every single button have an integer value on the possible steps from that position, now I understood how you got 56 from 1, you just add up all the moves for each button, but when you do 2, wouldnt you take the add the different 2 length patterns so if you were on button 1 that would have 5 locations and you would go total += location[1]*location[possible Edges]?
Tony
Posted: Wed Feb 20, 2013 6:45 pm Post subject: RE:q5 clarify
AJ would be the best person to talk to about Q5. I think he's planning on some writeups over the next couple of days.
Tony's programming blog. DWITE - a programming contest.
A.J
Posted: Thu Feb 21, 2013 6:15 pm Post subject: RE:q5 clarify
Yeah, just posted the write up: http://compsci.ca/v3/viewtopic.php?p=269811#269811
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https://mathedu.hbcse.tifr.res.in/forums/topic/u-k49s-lotto-and-love-how-theyre-the-same/ | 1,721,456,100,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00309.warc.gz | 342,260,656 | 19,984 | Home Forums Middle School Mathematics Education u K49s Lotto And Love – How They’re The same
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<br> The Rhode Island Lottery doesn’t assure cost of a prize inside any specific period of time. Winners can choose to both select their prize as an annuity, or for considerably much less, a lump-sum cost. Nonetheless, winners can choose to not have their picture taken. This isn’t appropriate however, as you even have precisely the same probability for sure numbers to appear within the lottery draw each time you play. You don’t need to get all the balls appropriate in sequence within the draw to be able to win, as there are several methods of profitable, nonetheless, to win the large jackpot which begins at \$15 million USD, it’s good to get them all right. Which keeps building up your win charges draw after draw and the results are for real! Additionally, you will robotically be entered into the Lotto Plus Raffle Draw. Equivalent to Powerball Lottery, Lotto Max, Lotto 6/49, Mega Millions, Texas Lotto, California Lottery, New York Lotto, plus many different lottery games so we give this winning system a strong 2 thumbs up! In the course of the broadcast, the 5 winning numbers from the set of 50 balls are drawn plus a bonus number. So your syndicate should wheel their numbers.<br> | 966 | 4,603 | {"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} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.951447 |
cshaw3.github.io | 1,695,434,428,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00286.warc.gz | 220,994,407 | 3,438 | ## What is a vector?
Computer graphics best friend, the vector has a long storied history stretching back centuries. Starting as the humble tool of mathematicians, used help explain the world around us. Over time its uses have grown to help fill many niches.
The classical definition of a vector is something that contains both a magnitude (or length) and a direction. This definition is a bit confusing on its face so we won't dwell on it too long. What is important to take from this definition is that a vector is a singular object made up of multiple sub parts. A vector composed of two parts is referred to as a 2D vector, a vector with three parts is referred to as 3D vector, four parts 4D, 5 parts 5D, and so on into infinity. For our needs we won't be going beyond 4D vectors.
The image on the left side shows the 2D vector points that make up the image on the right. Try clicking the center line to see the difference. The points are connected by bezier curves which give the image its shape and form.
## Applications Of Vectors
Now that we have a basic understand of what makes a vector let’s take a look at some examples of vectors used in computer graphics.
An example of a vector used in computer graphics would be position of an object. Now because we live in the future we can represent position in both the second and third dimension.
A 2D position vector could be represented like this: (X,Y). Where X denotes how far left or right on the screen the object is, while Y denotes how far up or down something is. If you have ever played a 2D Mario game Mario’s position on the screen can be represented by a 2D vector.
A 3D position vector is just a 2D vector with an extra part and can be represented like this: (X,Y,Z). Where just like the 2D vector you have X and Y controlling 2 different directions Z is now a third direction we can go.
While mathematically possible to create and work with a 4D position vector it would confuse our poor little 3D minds so instead we will use color as our example of a 4D vector.
A 4D color vector can represented like this: (R,G,B,A). Unlike the last two vectors where the numbers that made up the parts of the vector could theoretically be any number value the numbers that make up the color vector usually go between 0 and 255. This is due to some computer bit byte nonsense that wont be of importance here, its just the standard. The screen your reading this on is made up of tiny little dots called pixels. Every pixel has with it a 4D color vector telling that pixel what color it should be. The R, G, and B components control how red, green, and blue, a pixel is while the A which stands for alpha controls how transparent or see through the pixel is. for example a (255,0,0,255) is a pure red pixel while the color (255,0,255,255) would be half full red and half full blue so they would mix and form a bright pinkish purple color, and the color (0,255,0,128) would be full green but at half alpha meaning it would blend 50% transparent (there is not actually a pixel behind other pixels some fancy math just blends the two colors together)
This demo shows off how vectors can be used to store information. As you can see this square is moving back and forth on the screen and its position is represented by a 2D vector, where the first component of the vector is the X-coordinate (or where it is left and right on the screen) and the second component is the Y-coordinate (or where it is up and down on the screen). Try hovering over the demo and you will see the square follows your cursor, notice how the X and Y coordinates change when you move the square around. | 825 | 3,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.952711 |
http://jiffyclub.github.io/numpy/reference/generated/numpy.random.vonmises.html | 1,685,927,511,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00398.warc.gz | 30,279,318 | 4,064 | #### Previous topic
numpy.random.uniform
#### Next topic
numpy.random.wald
# numpy.random.vonmises¶
numpy.random.vonmises(mu, kappa, size=None)
Draw samples from a von Mises distribution.
Samples are drawn from a von Mises distribution with specified mode (mu) and dispersion (kappa), on the interval [-pi, pi].
The von Mises distribution (also known as the circular normal distribution) is a continuous probability distribution on the unit circle. It may be thought of as the circular analogue of the normal distribution.
Parameters: mu : float or array_like of floats Mode (“center”) of the distribution. kappa : float or array_like of floats Dispersion of the distribution, has to be >=0. size : int or tuple of ints, optional Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if mu and kappa are both scalars. Otherwise, np.broadcast(mu, kappa).size samples are drawn. out : ndarray or scalar Drawn samples from the parameterized von Mises distribution.
See also
scipy.stats.vonmises
probability density function, distribution, or cumulative density function, etc.
Notes
The probability density for the von Mises distribution is
p(x) = \frac{e^{\kappa cos(x-\mu)}}{2\pi I_0(\kappa)},
where \mu is the mode and \kappa the dispersion, and I_0(\kappa) is the modified Bessel function of order 0.
The von Mises is named for Richard Edler von Mises, who was born in Austria-Hungary, in what is now the Ukraine. He fled to the United States in 1939 and became a professor at Harvard. He worked in probability theory, aerodynamics, fluid mechanics, and philosophy of science.
References
[R270] Abramowitz, M. and Stegun, I. A. (Eds.). “Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing,” New York: Dover, 1972.
[R271] von Mises, R., “Mathematical Theory of Probability and Statistics”, New York: Academic Press, 1964.
Examples
Draw samples from the distribution:
>>> mu, kappa = 0.0, 4.0 # mean and dispersion
>>> s = np.random.vonmises(mu, kappa, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> from scipy.special import i0
>>> plt.hist(s, 50, normed=True)
>>> x = np.linspace(-np.pi, np.pi, num=51)
>>> y = np.exp(kappa*np.cos(x-mu))/(2*np.pi*i0(kappa))
>>> plt.plot(x, y, linewidth=2, color='r')
>>> plt.show() | 630 | 2,455 | {"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} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.79672 |
https://www.geeksforgeeks.org/delete-all-nodes-from-the-doubly-linked-list-which-are-divisible-by-k/?ref=rp | 1,713,453,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.54/warc/CC-MAIN-20240418124808-20240418154808-00537.warc.gz | 723,761,815 | 53,252 | # Delete all nodes from the doubly linked list which are divisible by K
Last Updated : 01 Feb, 2023
Given a doubly-linked list containing N nodes, the task is to delete all nodes from the list which are divisible by K.
Examples:
Input: List = 15 <=> 16 <=> 6 <=> 7 <=> 17, K = 2
Output: Final List = 15 <=> 7 <=> 17
Input: List = 5 <=> 3 <=> 4 <=> 2 <=> 9, K = 3
Output: Final List = 5 <=> 4 <=> 2
Approach: The idea is to traverse the nodes of the doubly linked list one by one and get the pointer of the nodes which are divisible by K. Delete those nodes by following the approach used in this post.
Below is the implementation of the above idea:
## C++
`// C++ implementation to delete all` `// from the doubly linked list which` `// are divisible by K` `#include ` `using` `namespace` `std;` `// Node of the doubly linked list` `struct` `Node {` ` ``int` `data;` ` ``Node *prev, *next;` `};` `// function to insert a node at the beginning` `// of the Doubly Linked List` `void` `push(Node** head_ref, ``int` `new_data)` `{` ` ``// allocate node` ` ``Node* new_node = (Node*)``malloc``(``sizeof``(``struct` `Node));` ` ``// put in the data` ` ``new_node->data = new_data;` ` ``// since we are adding at the beginning,` ` ``// prev is always NULL` ` ``new_node->prev = NULL;` ` ``// link the old list of the new node` ` ``new_node->next = (*head_ref);` ` ``// change prev of head node to new node` ` ``if` `((*head_ref) != NULL)` ` ``(*head_ref)->prev = new_node;` ` ``// move the head to point to the new node` ` ``(*head_ref) = new_node;` `}` `// function to delete a node in a Doubly Linked List.` `// head_ref --> pointer to head node pointer.` `// del --> pointer to node to be deleted` `void` `deleteNode(Node** head_ref, Node* del)` `{` ` ``// base case` ` ``if` `(*head_ref == NULL || del == NULL)` ` ``return``;` ` ``// If node to be deleted is head node` ` ``if` `(*head_ref == del)` ` ``*head_ref = del->next;` ` ``// Change next only if node to be` ` ``// deleted is NOT the last node` ` ``if` `(del->next != NULL)` ` ``del->next->prev = del->prev;` ` ``// Change prev only if node to be` ` ``// deleted is NOT the first node` ` ``if` `(del->prev != NULL)` ` ``del->prev->next = del->next;` ` ``// Finally, free the memory occupied by del` ` ``free``(del);` ` ``return``;` `}` `// function to delete all nodes from the` `// doubly linked list divisible by K` `void` `deleteDivisibleNodes(Node** head_ref, ``int` `K)` `{` ` ``Node* ptr = *head_ref;` ` ``Node* next;` ` ``while` `(ptr != NULL) {` ` ``next = ptr->next;` ` ``// if true, delete node 'ptr'` ` ``if` `(ptr->data % K == 0)` ` ``deleteNode(head_ref, ptr);` ` ``ptr = next;` ` ``}` `}` `// function to print nodes in a` `// given doubly linked list` `void` `printList(Node* head)` `{` ` ``while` `(head != NULL) {` ` ``cout << head->data << ``" "``;` ` ``head = head->next;` ` ``}` `}` `// Driver program` `int` `main()` `{` ` ``// start with the empty list` ` ``Node* head = NULL;` ` ``// create the doubly linked list` ` ``// 15 <-> 16 <-> 7 <-> 6 <-> 17` ` ``push(&head, 17);` ` ``push(&head, 6);` ` ``push(&head, 7);` ` ``push(&head, 16);` ` ``push(&head, 15);` ` ``int` `K = 2;` ` ``cout << ``"Original List: "``;` ` ``printList(head);` ` ``deleteDivisibleNodes(&head, K);` ` ``cout << ``"\nModified List: "``;` ` ``printList(head);` `}`
## Java
`// Java implementation to delete all ` `// from the doubly linked list which ` `// are divisible by K ` `class` `GFG` `{` ` ` `// Node of the doubly linked list ` `static` `class` `Node ` `{ ` ` ``int` `data; ` ` ``Node prev, next; ` `}; ` `// function to insert a node at the beginning ` `// of the Doubly Linked List ` `static` `Node push(Node head_ref, ``int` `new_data) ` `{ ` ` ``// allocate node ` ` ``Node new_node = ``new` `Node(); ` ` ``// put in the data ` ` ``new_node.data = new_data; ` ` ``// since we are adding at the beginning, ` ` ``// prev is always null ` ` ``new_node.prev = ``null``; ` ` ``// link the old list of the new node ` ` ``new_node.next = (head_ref); ` ` ``// change prev of head node to new node ` ` ``if` `((head_ref) != ``null``) ` ` ``(head_ref).prev = new_node; ` ` ``// move the head to point to the new node ` ` ``(head_ref) = new_node; ` ` ``return` `head_ref;` `} ` `// function to delete a node in a Doubly Linked List. ` `// head_ref -. pointer to head node pointer. ` `// del -. pointer to node to be deleted ` `static` `Node deleteNode(Node head_ref, Node del) ` `{ ` ` ``// base case ` ` ``if` `(head_ref == ``null` `|| del == ``null``) ` ` ``return` `null``; ` ` ``// If node to be deleted is head node ` ` ``if` `(head_ref == del) ` ` ``head_ref = del.next; ` ` ``// Change next only if node to be ` ` ``// deleted is NOT the last node ` ` ``if` `(del.next != ``null``) ` ` ``del.next.prev = del.prev; ` ` ``// Change prev only if node to be ` ` ``// deleted is NOT the first node ` ` ``if` `(del.prev != ``null``) ` ` ``del.prev.next = del.next; ` ` ``return` `head_ref; ` `} ` `// function to delete all nodes from the ` `// doubly linked list divisible by K ` `static` `Node deleteDivisibleNodes(Node head_ref, ``int` `K) ` `{ ` ` ``Node ptr = head_ref; ` ` ``Node next; ` ` ``while` `(ptr != ``null``) ` ` ``{ ` ` ``next = ptr.next;` ` ` ` ``// if true, delete node 'ptr' ` ` ``if` `(ptr.data % K == ``0``) ` ` ``deleteNode(head_ref, ptr); ` ` ``ptr = next; ` ` ``} ` ` ``return` `head_ref;` `} ` `// function to print nodes in a ` `// given doubly linked list ` `static` `void` `printList(Node head) ` `{ ` ` ``while` `(head != ``null``)` ` ``{ ` ` ``System.out.print( head.data + ``" "``); ` ` ``head = head.next; ` ` ``} ` `} ` `// Driver code ` `public` `static` `void` `main(String args[])` `{ ` ` ``// start with the empty list ` ` ``Node head = ``null``; ` ` ``// create the doubly linked list ` ` ``// 15 <. 16 <. 7 <. 6 <. 17 ` ` ``head = push(head, ``17``); ` ` ``head = push(head, ``6``); ` ` ``head = push(head, ``7``); ` ` ``head = push(head, ``16``); ` ` ``head = push(head, ``15``); ` ` ``int` `K = ``2``; ` ` ``System.out.print( ``"Original List: "``); ` ` ``printList(head); ` ` ``head=deleteDivisibleNodes(head, K); ` ` ``System.out.print( ``"\nModified List: "``); ` ` ``printList(head); ` `} ` `}` `// This code is contributed by Arnab Kundu`
## Python3
` ``# Python3 implementation to delete all` `# from the doubly linked list which` `# are divisible by K` `import` `math` `# Node of the doubly linked list` `class` `Node: ` ` ``def` `__init__(``self``,data): ` ` ``self``.data ``=` `data ` ` ``self``.``next` `=` `None` `# function to insert a node at the ` `# beginning of the Doubly Linked List` `def` `push(head_ref, new_data):` ` ` ` ``# allocate node` ` ``new_node ``=` `Node(new_data)` ` ` ` ``# put in the data` ` ``new_node.data ``=` `new_data` ` ``# since we are adding at the beginning,` ` ``# prev is always None` ` ``new_node.prev ``=` `None` ` ``# link the old list of the new node` ` ``new_node.``next` `=` `head_ref` ` ``# change prev of head node to new node` ` ``if` `(head_ref !``=` `None``):` ` ``head_ref.prev ``=` `new_node` ` ``# move the head to point` ` ``# to the new node` ` ``head_ref ``=` `new_node` ` ``return` `head_ref` `# function to delete a node ` `# in a Doubly Linked List.` `# head_ref --> pointer to head node pointer.` `# del --> pointer to node to be deleted` `def` `deleteNode(head_ref, delete):` ` ` ` ``# base case` ` ``if` `(head_ref ``=``=` `None` `or` `delete ``=``=` `None``):` ` ``return` ` ``# If node to be deleted is head node` ` ``if` `(head_ref ``=``=` `delete):` ` ``head_ref ``=` `delete.``next` ` ``# Change next only if node to be` ` ``# deleted is NOT the last node` ` ``if` `(delete.``next` `!``=` `None``):` ` ``delete.``next``.prev ``=` `delete.prev` ` ``# Change prev only if node to be` ` ``# deleted is NOT the first node` ` ``if` `(delete.prev !``=` `None``):` ` ``delete.prev.``next` `=` `delete.``next` ` ``# Finally, free the memory ` ` ``# occupied by del, free(del)` ` ``return` `# function to delete all nodes from the` `# doubly linked list divisible by K` `def` `deleteDivisibleNodes(head_ref, K):` ` ``ptr ``=` `head_ref` ` ``#next` ` ``while` `(ptr !``=` `None``) :` ` ``next` `=` `ptr.``next` ` ` ` ``# if true, delete node 'ptr'` ` ``if` `(ptr.data ``%` `K ``=``=` `0``):` ` ``deleteNode(head_ref, ptr)` ` ``ptr ``=` `next` `# function to print nodes in a` `# given doubly linked list` `def` `printList(head):` ` ``while` `(head !``=` `None``):` ` ``print``(head.data, end ``=` `" "``)` ` ``head ``=` `head.``next` `# Driver Code` `if` `__name__``=``=``'__main__'``: ` ` ``# start with the empty list` ` ``head ``=` `None` ` ``# create the doubly linked list` ` ``# 15 <- 16 <- 7 <- 6 <- 17` ` ``head ``=` `push(head, ``17``)` ` ``head ``=` `push(head, ``6``)` ` ``head ``=` `push(head, ``7``)` ` ``head ``=` `push(head, ``16``)` ` ``head ``=` `push(head, ``15``)` ` ``K ``=` `2` ` ``print``(``"Original List: "``, end ``=` `"")` ` ``printList(head)` ` ``deleteDivisibleNodes(head, K)` ` ``print``(``"\nModified List: "``, end ``=` `"")` ` ``printList(head)` `# This code is contributed by AbhiThakur`
## C#
`// C# implementation to delete all ` `// from the doubly linked list which ` `// are divisible by K ` `using` `System;` `class` `GFG` `{` ` ` `// Node of the doubly linked list ` `public` `class` `Node ` `{ ` ` ``public` `int` `data; ` ` ``public` `Node prev, next; ` `}; ` `// function to insert a node at the beginning ` `// of the Doubly Linked List ` `static` `Node push(Node head_ref, ``int` `new_data) ` `{ ` ` ``// allocate node ` ` ``Node new_node = ``new` `Node(); ` ` ``// put in the data ` ` ``new_node.data = new_data; ` ` ``// since we are adding at the beginning, ` ` ``// prev is always null ` ` ``new_node.prev = ``null``; ` ` ``// link the old list of the new node ` ` ``new_node.next = (head_ref); ` ` ``// change prev of head node to new node ` ` ``if` `((head_ref) != ``null``) ` ` ``(head_ref).prev = new_node; ` ` ``// move the head to point to the new node ` ` ``(head_ref) = new_node; ` ` ``return` `head_ref;` `} ` `// function to delete a node in a Doubly Linked List. ` `// head_ref -. pointer to head node pointer. ` `// del -. pointer to node to be deleted ` `static` `Node deleteNode(Node head_ref, Node del) ` `{ ` ` ``// base case ` ` ``if` `(head_ref == ``null` `|| del == ``null``) ` ` ``return` `null``; ` ` ``// If node to be deleted is head node ` ` ``if` `(head_ref == del) ` ` ``head_ref = del.next; ` ` ``// Change next only if node to be ` ` ``// deleted is NOT the last node ` ` ``if` `(del.next != ``null``) ` ` ``del.next.prev = del.prev; ` ` ``// Change prev only if node to be ` ` ``// deleted is NOT the first node ` ` ``if` `(del.prev != ``null``) ` ` ``del.prev.next = del.next; ` ` ``return` `head_ref; ` `} ` `// function to delete all nodes from the ` `// doubly linked list divisible by K ` `static` `Node deleteDivisibleNodes(Node head_ref, ``int` `K) ` `{ ` ` ``Node ptr = head_ref; ` ` ``Node next; ` ` ``while` `(ptr != ``null``) ` ` ``{ ` ` ``next = ptr.next;` ` ` ` ``// if true, delete node 'ptr' ` ` ``if` `(ptr.data % K == 0) ` ` ``deleteNode(head_ref, ptr); ` ` ``ptr = next; ` ` ``} ` ` ``return` `head_ref;` `} ` `// function to print nodes in a ` `// given doubly linked list ` `static` `void` `printList(Node head) ` `{ ` ` ``while` `(head != ``null``)` ` ``{ ` ` ``Console.Write( head.data + ``" "``); ` ` ``head = head.next; ` ` ``} ` `} ` `// Driver code ` `public` `static` `void` `Main(String []args)` `{ ` ` ``// start with the empty list ` ` ``Node head = ``null``; ` ` ``// create the doubly linked list ` ` ``// 15 <. 16 <. 7 <. 6 <. 17 ` ` ``head = push(head, 17); ` ` ``head = push(head, 6); ` ` ``head = push(head, 7); ` ` ``head = push(head, 16); ` ` ``head = push(head, 15); ` ` ``int` `K = 2; ` ` ``Console.Write( ``"Original List: "``); ` ` ``printList(head); ` ` ``head = deleteDivisibleNodes(head, K); ` ` ``Console.Write( ``"\nModified List: "``); ` ` ``printList(head); ` `} ` `}` `// This code contributed by Rajput-Ji`
## Javascript
``
Output
```Original List: 15 16 7 6 17
Modified List: 15 7 17 ```
Complexity Analysis:
• Time Complexity: O(N), where N is the total number of nodes.
• Space Complexity: O(1) since using constant variables
Previous
Next | 4,591 | 13,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-18 | latest | en | 0.786913 |
http://230nsc1.phy-astr.gsu.edu/hbase/mom.html | 1,695,387,139,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00839.warc.gz | 373,461 | 1,779 | # Momentum
The momentum of a particle is defined as the product of its mass times its velocity. It is a vector quantity. The momentum of a system is the vector sum of the momenta of the objects which make up the system. If the system is an isolated system, then the momentum of the system is a constant of the motion and subject to the principle of conservation of momentum.
The basic definition of momentum applies even at relativistic velocities but then the mass is taken to be the relativistic mass.
The most common symbol for momentum is p. The SI unit for momentum is kg m/s.
You may insert numbers for any of the quantities. Then click on the text or symbol in the formula above for the quantity you wish to calculate.The numbers will not be forced to be consistent until you click on the desired quantity.
### Momentum p = kg m/s = kg * m/s
Index
Conservation of momentum
HyperPhysics***** Mechanics R Nave
Go Back | 197 | 931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.893853 |
https://www.anagrammer.com/scrabble/regularise | 1,575,829,311,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514475.44/warc/CC-MAIN-20191208174645-20191208202645-00144.warc.gz | 631,891,855 | 11,715 | # Scrabble?! REGULARISE
Is regularise valid for Scrabble? Words With Friends? Lexulous? WordFeud? Other games?
Yes! (61 pts)
Yes! (61 pts)
Yes! (63 pts)
Yes! (61 pts)
Yes! (10 pts)
Yes! (61 pts)
No!
## Definitions of REGULARISE in various dictionaries:
The Benjamin–Bona–Mahony equation (or BBM equation) – also known as the regularized long-wave equation (RLWE) – is the partial differential equation u t + u x + u u x − u x x t = 0. {\displaystyle u_{t}+u_{x}+uu_{x}-u_{xxt}=0.\,} This equation was studied in Benjamin, Bona, and Mahony (1972) as an improvement of the Korteweg–de Vries equation (KdV equation) for modeling long surface gravity waves of small amplitude – propagating uni-directionally in 1+1 dimensions. They show the stability and uniqueness of solutions to the BBM equation. This contrasts with the KdV equation, which is unstable in its high wavenumber components. Further, while the KdV equation has an infinite number of integrals of motion, the BBM equation only has three.Before, in 1966, this equation was introduced by Peregrine, in the study of undular bores.A generalized n-dimensional version is given by u t − ∇ 2 u t + div φ ( u ) = 0. {\displaystyle u_{t}-\nabla ^{2}u_{t}+\operatorname {div} \,\varphi (u)=0.\,} where φ {\displaystyle \varphi } is a sufficiently smooth function from R {\displaystyle \mathbb {R} } to R n {\displaystyle \mathbb {R} ^{n}} . Avrin & Goldstein (1985) proved global existence of a solution in all dimensions.
## WORD SOLVER
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## There are 10 letters in REGULARISE ( A1E1G2I1L1R1S1U1 )
To search all scrabble anagrams of REGULARISE, to go: REGULARISE?
Rearrange the letters in REGULARISE and see some winning combinations
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### 2 letters out of REGULARISE
Anagrammer is a game resource site that has been extremely popular with players of popular games like Scrabble, Lexulous, WordFeud, Letterpress, Ruzzle, Hangman and so forth. We maintain regularly updated dictionaries of almost every game out there. To be successful in these board games you must learn as many valid words as possible, but in order to take your game to the next level you also need to improve your anagramming skills, spelling, counting and probability analysis. Make sure to bookmark every unscrambler we provide on this site. Explore deeper into our site and you will find many educational tools, flash cards and so much more that will make you a much better player. This page covers all aspects of REGULARISE, do not miss the additional links under "More about: REGULARISE" | 703 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-51 | latest | en | 0.85679 |
https://www.proofwiki.org/wiki/Octal_Notation/Examples/371.24/Mistake | 1,660,640,847,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00402.warc.gz | 801,491,269 | 10,942 | # Octal Notation/Examples/371.24/Mistake
octal
## Mistake
... for example, $371.24_8$ represents the number
$\paren {3 \times 8^2} + \paren {7 \times 8^1} + \paren {1 \times 8^0} + \paren {2 \times 8^{-1} } + \paren {4 \times 8^{-2} } = 249.25_{10}$.
## Correction
The actual number in decimal is $249.3125_{10}$.
It appears that the last octal digit was omitted from the calculation. | 140 | 390 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-33 | latest | en | 0.617302 |
http://spaceguard.rm.iasf.cnr.it/tumblingstone/dictionary/orb-elements.htm | 1,495,736,080,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608120.92/warc/CC-MAIN-20170525180025-20170525200025-00426.warc.gz | 339,125,016 | 2,312 | To know the state of motion of a body, it would be necessary to know the values of the six parameters corresponding to its position and its speed. But, during a general motion, these values change constantly and knowing them at every moment is a very difficult task.
I
n the reality, for what concerns constant, stable orbits, the motion of a body can be completely described knowing the values of 6 constant parameters, called orbital elements that characterize the orbit. In this hyothesis, knowing the position and the speed of the object at every instant becomes useless.
First of all, let's consider the orbit of an object of the Solar System (a planet or a minor body). This orbit can be an ellipse (as the first of Kepler's laws states), or an opened orbit such as parabola or hyperbola. Anyhow, this orbit will always be a conic and it will always lie on a plane called the orbital plane. The intersection between this plane and a reference plane (which can be chosen as the plane where the Earth's orbit lies) is called nodal line. This nodal line passes through the ascending and descending node.
The first 2 orbital elements, needed to define size and shape of the orbit on this plane, are the semimajor axis a and the eccentricity e.
These two parameters can be defined on the orbital plane; for an elliptical orbit: - 2a is the length of the major axis (M in the image on the left); - e, the eccentricity, gives an indication of how much the ellipse is elongated: it is 0 for a circle, and tends to 1 for more and more elongated orbits. It can be calculated with the formula (where m and M are defined in the image):
Three more parameters are needed to describe where is the orbit, and how it is oriented: - i the inclination of the orbital plane, the angle that this plane forms with a reference plane; - the longitude of node, the angle between a reference direction and the nodal line, that goes from the Sun to the ascending node; - the argument of perihelion, the angle from the nodal line to the line joining the Sun and the perihelion, the point on the orbit closest to the Sun. Finally, the position on the orbit of the body of interest can be specified by giving the sixth parameter: - T, the time of passage at the perihelion. | 503 | 2,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-22 | longest | en | 0.922494 |
https://shakyradunn.com/slide/microeconomic-theory-eiu-su202h | 1,618,928,341,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00150.warc.gz | 609,771,072 | 49,212 | # Microeconomic Theory - Eiu
Chapter 15 APPLIED COMPETITIVE ANALYSIS MICROECONOMIC THEORY BASIC PRINCIPLES AND EXTENSIONS EIGHTH EDITION WALTER NICHOLSON Copyright 2002 by South-Western, a division of Thomson Learning. All rights reserved. Price Controls Sometimes the government may seek to control prices at below equilibrium levels will lead to a shortage We can look at the changes in producer and consumer surplus from this policy to analyze its impact on welfare Price Controls Price
Initially, the market is in long-run equilibrium at P1, Q1 SS LS P1 Demand increases to D D D Q1 Quantity Price Controls Price In the short run, price rises to P2 SS
P2 LS P3 P1 D Firms would begin to enter the industry The price would end up at P3 D Q1 Quantity Price Controls Price
Suppose that the government imposes a price ceiling at P1 SS LS P3 P1 D There will be a shortage equal to Q2 - Q1 D Q1 Q2 Quantity
Price Controls Price Some buyers will gain because they can purchase the good for a lower price SS LS P3 P1 D This gain in consumer surplus is the shaded rectangle D
Q1 Q2 Quantity Price Controls Price The gain to consumers is also a loss to producers who now receive a lower price SS LS P3 P1 D D
Q1 Q2 The shaded rectangle therefore represents a pure transfer from producers to consumers No welfare loss there Quantity Price Controls Price SS LS P3 P1 This shaded triangle
represents the value of additional consumer surplus that would have been attained without the price control D D Q1 Q2 Quantity Price Controls Price SS LS
P3 P1 This shaded triangle represents the value of additional producer surplus that would have been attained without the price control D D Q1 Q2 Quantity Price Controls Price SS
LS P3 This shaded area represents the total value of mutually beneficial transactions that are prevented by the government P1 D This is a measure of the pure welfare costs of this policy D Q1
Q2 Quantity Disequilibrium Behavior Assuming that observed market outcomes are generated by Q(P1) = min [QD(P1),QS(P1)] suppliers will be content with the outcome but demanders will not This could lead to black markets
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• NamelyHinduism, Buddhism, Jainism, and Sikhism. Dharmic worldview. dharma - harmonious and eternal truth within each human being. Minority Religions in India. Sikhism. Arose in the 15th century . Hearth = Pakistan Punjab state of northwestern India. ... PowerPoint Presentation Last...
• Reduction of corner links to partial order. Four trees (special case of corner links) is sufficient to represent non-degenerate floorplans. Characterization of valid floorplans in the partial order representation. Partial order representation can give absolute coordinates and dimensions of all... | 930 | 4,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-17 | latest | en | 0.899565 |
https://brilliant.org/problems/its-the-returning-journey/ | 1,495,633,101,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607846.35/warc/CC-MAIN-20170524131951-20170524151951-00082.warc.gz | 729,372,565 | 17,857 | # It's the returning journey
An electron with a velocity V1 leaves the surface of a wire of radius a and carrying a current I perpendicular to the surface. Find the maximum distance it will go before turning back.
Details: e=2.718; a=0.3679m V1=1.00m/s I=28.4E(-6) A
× | 80 | 271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-22 | longest | en | 0.810965 |
http://ibmmainframes.com/about37526.html | 1,524,757,025,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00439.warc.gz | 149,512,072 | 11,324 | Portal | Manuals | References | Downloads | Info | Programs | JCLs | Mainframe wiki | Quick Ref
Author Message
Akash Sharma
New User
Joined: 13 Jan 2009
Posts: 36
Location: India
Posted: Thu Jan 22, 2009 6:41 pm Post subject: Problem using arithmetic operators. HI All, I have developed an ISPF Panel. In the ISPF Panel, there is an input field. Whatever is there in the input field, should get stored on to a PDS through a CLIST. The functionality is working fine, but i am facing 1 problem. Whenever i give input as ABC-DEF or any other arithmetic operators, the program fails to store it into PDS. Could you please help here??? I have tried defining many things in VER ####, VER (&STYPE,NB,####,MSG=SRCH020) but cant get through. Please help.
enrico-sorichetti
Global Moderator
Joined: 14 Mar 2007
Posts: 10457
Location: italy
Posted: Thu Jan 22, 2009 6:48 pm Post subject: Reply to: Problem using arithmetic operators.
You did not post anything useful,
post Your rexx ( using the code tags )
both the original source
and a run with TRACE "I" turned on
are You really using a CLIST
in this case You are not likely to get too much help
Akash Sharma
New User
Joined: 13 Jan 2009
Posts: 36
Location: India
Posted: Thu Jan 22, 2009 7:56 pm Post subject: Re: Reply to: Problem using arithmetic operators.
Hi Enrico,
Following is the exact problem im facing.
Code: PROC 0 ALLOCATE FILE(RSTR) DA('EV4.RT.SCH.QUEUE') MOD REU OPENFILE RSTR OUTPUT SET &RSTR = - WRITE LASTCC1 = &LASTCC PUTFILE RSTR WRITE LASTCC2 = &LASTCC CLOSFILE RSTR EXIT CODE(0)
The error response is:
Code: SET &RSTR = WRITE LASTCC1 = 0 THIS STATEMENT HAS AN EXPRESSION WITH OPERATORS OUT OF SEQUENCE ***
This error i believe is coming because of - being the symbol for continuation in next line. How can i handle this?
expat
Global Moderator
Joined: 14 Mar 2007
Posts: 8657
Location: Back in jolly old England
Posted: Thu Jan 22, 2009 8:05 pm Post subject:
Have you tried
Code: SET &RSTR = &STR(-)
Akash Sharma
New User
Joined: 13 Jan 2009
Posts: 36
Location: India
Posted: Thu Jan 22, 2009 8:24 pm Post subject:
expat wrote:
Have you tried
Code: SET &RSTR = &STR(-)
Hi expat,
That almost solved my problems.
But it still fails for &.
ie the code fails if :
Code: PROC 0 ALLOCATE FILE(RSTR) DA('EV4.RT.SCH.QUEUE') MOD REU OPENFILE RSTR OUTPUT SET &RSTR = &STR(&) WRITE LASTCC1 = &LASTCC PUTFILE RSTR WRITE LASTCC2 = &LASTCC CLOSFILE RSTR EXIT CODE(0)
The O\P is:
Code: SET &RSTR = &STR(&) THIS STATEMENT HAS AN INVALID SYMBOLIC VARIABLE ***
expat
Global Moderator
Joined: 14 Mar 2007
Posts: 8657
Location: Back in jolly old England
Posted: Thu Jan 22, 2009 8:33 pm Post subject: Well it amazes me that I even remembered &STR as it's over 20 years since I wrote or even looked at a CLIST. There is a variation on &STR - might be &NSTR or something like that. Take a look in the manuals and see.
Akash Sharma
New User
Joined: 13 Jan 2009
Posts: 36
Location: India
Posted: Thu Jan 22, 2009 8:55 pm Post subject:
expat wrote: Well it amazes me that I even remembered &STR as it's over 20 years since I wrote or even looked at a CLIST. There is a variation on &STR - might be &NSTR or something like that. Take a look in the manuals and see.
Hi,
I looked into the manuals.
U were right, &NRSTR is the command for &
but it works for if there are even nos of &s
like &&, &&&&.
but not for odd....
Akash Sharma
New User
Joined: 13 Jan 2009
Posts: 36
Location: India
Posted: Thu Jan 22, 2009 9:02 pm Post subject: Reply to: Problem using arithmetic operators.
if i have,
case1
Code: SET &ROBTSTR = &NRSTR(&)
o\p is
Code: SET &ROBTSTR = &NRSTR(&) THIS STATEMENT HAS AN INVALID SYMBOLIC VARIABLE
case 2
Code: SET &ROBTSTR = &NRSTR(STRA&STRB)
the o\p written to pds is STRA
case3
Code: SET &ROBTSTR = &NRSTR(STRA&&STRB)
the o\p written is STRA&&STRB
But the o\p should be exactly the same what input is for all the cases.....
Akash Sharma
New User
Joined: 13 Jan 2009
Posts: 36
Location: India
Posted: Fri Jan 23, 2009 2:58 pm Post subject: Reply to: Problem using arithmetic operators. Hi All, Looked extensively into the manuals...but could not figure out the way ahead. Could you please help here?
Pedro
Senior Member
Joined: 01 Sep 2006
Posts: 2104
Location: Silicon Valley
Posted: Sat Jan 24, 2009 2:41 am Post subject: Reply to: Problem using arithmetic operators.
Have you tried the 'no substitution' function?
Code: &SYSNSUB(level,expression)
Akash Sharma
New User
Joined: 13 Jan 2009
Posts: 36
Location: India
Posted: Tue Jan 27, 2009 1:14 pm Post subject: Re: Reply to: Problem using arithmetic operators.
Pedro wrote:
Have you tried the 'no substitution' function?
Code: &SYSNSUB(level,expression)
Thanks a lot.
This worked.
Everything working fine now.
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Job Vacancies | Forum Rules | Bookmarks | Subscriptions | FAQ | Polls | Contact Us | 1,756 | 6,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.593498 |
http://math.stackexchange.com/questions/123/real-world-uses-of-hyperbolic-trigonometric-functions/233 | 1,440,924,501,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064951.43/warc/CC-MAIN-20150827025424-00040-ip-10-171-96-226.ec2.internal.warc.gz | 150,936,591 | 22,294 | Real world uses of hyperbolic trigonometric functions
I covered hyperbolic trigonometric functions in a recent maths course. However I was never presented with any reasons as to why (or even if) they are useful.
Is there any good examples of their uses outside academia?
-
"trigonometric" (not "trigonomic" or "triganomic") – Jason S Jul 20 '10 at 23:04
Outside academia — such as? – kennytm Jul 21 '10 at 12:23
@Kenny anything... – Jacob Jul 21 '10 at 13:26
Lorentz transforms can be understood as hyperbolic rotations. The caternary curve (a dangling string/chain) is really just cosh – crasic Oct 30 '10 at 23:48
If you take a rope, fix the two ends, and let it hang under the force of gravity, it will naturally form a hyperbolic cosine curve.
-
See Catenary for more analysis on the curve that is mentioned in this answer. – Justin L. Jul 20 '10 at 21:22
In real life you use the catenary shape to know how much cable to place between two poles in high power transmission lines. Too much cable and it sags too much making it a hazzard. Too little cable and it breaks due to high tension as it stretches. Quite important to get it right when you have 768kV @ 6000 Amps through the cable. – ja72 Jan 8 '11 at 5:33
On a map using the Mercator projection, the relationship between the latitude L of a point and its y coordinate on the map is given by y = atanh(sin(L)), where atanh is the inverse of the hyperbolic tangent function.
-
The catenary has been mentioned a number of times, but apparently not the corresponding surface of revolution, the catenoid. It and the plane are the only surfaces of revolution that have zero mean curvature (i.e. they are minimal surfaces). This surface is the form a soap bubble (approximately) takes when it is stretched across two rings:
(image from here)
-
I don't know if you consider General Relativity "outside acadamia"(and I don't care to argue the point!) but if you do,
the group of symmetries with respect to the Lorentzian Metric can be written as Matrices containing hyperbolic trig functions as elements.
Note Kenny's comment.
-
(1) That's called Minkowski metric. Or do you mean Lorentz transforms? (2) It is just special relativity because no curved spacetime is involved. – kennytm Jul 24 '10 at 20:08
Thanks Kenny, it's been a while... – BBischof Jul 24 '10 at 20:20
I'll count it. Here's a bit more. Relativity is interested in the space-time interval. If time is the base of a triangle and distance the hypotenuse the interval is the other base. So space-time = d^2 - t^2. So in the d and t plane we have this weird distance metric. Now comes the hand waving. Take the geometric product of d and t = d * t + d ^ t = 0 + d ^ t. Squared it equals 1. Usually it equals -1. So when we expand e^(dt)x = 1 + dtx + x^2 + dtx^3 ... There is no alternating signs anymore. e^(dt)x = cosh(x) + dt*sinh(x). Explained here geocalc.clas.asu.edu/pdf/CompGeom-ch2.pdf – Jonathan Fischoff Jul 24 '10 at 20:28
Velocity addition in (special) relativity is not linear, but becomes linear when expressed in terms of hyperbolic tangent functions.
More precisely, if you add two motions in the same direction, such as a man walking at velocity $v_1$ on a train that moves at $v_2$ relative to the ground, the velocity $v$ of the man relative to ground is not $v_1 + v_2$; velocities don't add (otherwise by adding enough of them you could exceed the speed of light). What does add is the inverse hyperbolic tangent of the velocities (in speed-of-light units, i.e., $v/c$).
$$\tanh^{-1}(v/c)=\tanh^{-1}(v_1/c) + \tanh^{-1}(v_2/c)$$
This is one way of deriving special relativity: assume that a velocity addition formula holds, respecting a maximum speed of light and some other assumptions, and show that it has to be the above.
-
Many kinds of nonlinear PDE have wave solutions explicitly expressed using hyperbolic tangents and secants: shock-wave profiles, solitons, reaction-diffusion fronts, and phase-transition fronts, for starters.
-
An equation for a catenary curve can be given in terms of hyperbolic cosine. Catenary curves appear in many places, such as the Gateway Arch in St. Louis, MO.
-
Apart from occuring naturally in the form of the catenary mentioned, hyperbolic trig functions are also fundamentally important in calculus. I hope you'll agree that calculus has many, many real world applications!
-
Can you elaborate further? I am only aware of the use of their inverses in helping with integration. – Justin L. Jul 24 '10 at 23:59
If A occurs in B, and B in general is very important, it doesn't make A important: most of the meaty applications of B need not involve A. Therefore the appearance of hyperbolic trig functions in calculus along with the importance of calculus does not really show why hyperbolic trig functions are important. – KCd Jan 21 at 0:43
Most curves that look parabolic are actually Catenaries, which is based in the hyperbolic cosine function. A good example of a Catenary would be the Gateway Arch in Saint Louis, Missouri.
The tractrix, which is based in the hyperbolic secant, is also known as the pursuit curve, which models objects like cargo trucks turning corners, or a dog on a porch starting to chase a car in the street.
-
If a pendulum oscillates near its stable equilibrium, the equation of motion is $x''=-\omega^2 x$, and the solution is any linear combination of $\sin\omega t$ and $\cos\omega t$. If the pendulum has a stiff arm (rather than a string), then there is a second, unstable equilibrium, where it's straight up. This is like balancing a pencil on its tip. The equation of motion is $x''=\omega^2 x$, and the solution is any linear combination of $\sinh\omega t$ and $\cosh\omega t$.
- | 1,468 | 5,732 | {"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} | 3.546875 | 4 | CC-MAIN-2015-35 | longest | en | 0.914822 |
https://www.displayr.com/how-to-plot-data-on-a-globe-in-displayr/ | 1,674,890,625,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00602.warc.gz | 727,078,216 | 59,481 | If you have geographic data, you can use a bit of R code to plot it on an interactive globe. This visualization is completely customizable and allows rotation and zoom.
In this earlier post we analysed the location of meteorite impacts from this dataset, including plotting their fall locations on a globe. In this latest post we are going to explain how to create the globe with code. We then go on to describe other options and variations on the same theme.
## Preparing the data
To import the data in to Displayr, click on the New Data Set button on the Home menu, then click URL and paste in this link: https://data.nasa.gov/api/views/gh4g-9sfh/rows.csv?accessType=DOWNLOAD. We then right click on the data under Data Sets on the left to rename the data Meteorites.
Next, we are going to convert the year (which is imported as text string of date and time) into an age in years. Click on Insert > New R > Numeric Variable and paste in the following code.
```current <- as.numeric(format(Sys.Date(), "%Y"))
age <- current - as.numeric(substr(Meteorites\$Questions\$year, 7, 10))
```
## Plotting the data
Now we'll make the map. On the Insert tab, insert an R Output and paste in the following code (which we explain below).
```library(threejs)
library(flipChartBasics)
# Make a data.frame of the required information
x <- data.frame(long = Meteorites\$Questions\$reclong,
lat = Meteorites\$Questions\$reclat,
mass = Meteorites\$Questions\$`mass (g)`,
age = age)
# Set colors on a scale of 1 to 10 by percentile
colors <- as.numeric(cut(x\$age,
breaks = quantile(x\$age,
probs = seq(0, 1, 0.1),
include.lowest = TRUE,
na.rm = TRUE)))
palette <- ChartColors(10, "Reds", reverse = TRUE)
colors <- palette[colors]
# Plot the data on the globe
globejs(lat = x\$lat,
long = x\$long,
val = 2 * log(x\$mass),
color = colors,
pointsize = 0.5,
atmosphere = TRUE)
```
Stepping through the code, we first import required libraries then create a data frame with the variables that we need. We then convert age into a number from 1 to 10 (a.k.a its decile), then map those numbers to shades of red. Finally we plot the data on the globe where val determines the length of each line and point size determines the line thickness.
The output is slightly different from the original in the meteorites post because we have not excluded any data. You can spin the globe around with your mouse and zoom in or out.
## Capital cities
As a second example, we'll plot capital city data available here. Import the data using New Data Set as for previous example except with this URL: https://esa.un.org/unpd/wup/cd-rom/WUP2014_XLS_CD_FILES/WUP2014-F13-Capital_Cities.xls. You can ignore the warning about multiple sheets. Rename the imported data set CapitalCities.
The Population (thousands) variable is imported as text (which you can tell by the letter "a" before the variable name). To convert it to a number, click on it then under Properties on the right of the screen change Structure to Numeric. This creates a new numeric variable in the Data Set called Population (thousands)1.
The code to make the globe is broadly similar to that above. Enter the following code in the Object Inspector on the right after selecting Insert > R Output.
```library(threejs)
library(flipChartBasics)
# Make a data.frame of the required information
x <- data.frame(long = Longitude,
lat = Latitude,
population = `Population (thousands)1`,
city = `Capital City`)
# Set colors according to first letter of city name
first.letters <- sapply(substring(x\$city, 1, 1),
utf8ToInt) - utf8ToInt("A") + 1
palette <- ChartColors(26, "Blues")
colors <- palette[first.letters]
# Plot the data on the globe
earth <- "http://eoimages.gsfc.nasa.gov/images/imagerecords/73000/73909/world.topo.bathy.200412.3x5400x2700.jpg"
globejs(img = earth,
lat = x\$lat,
long = x\$long,
val = 10 * log(x\$population),
color = colors,
pointsize = 5,
atmosphere = FALSE,
bg = "white")
```
A few notable differences are:
• Addition of a more colorful background world image via the img argument.
• Line lengths related to population with colors according to the first letter of the city name (light blue = A, dark = Z).
• Removal of atmosphere, increase of point size and amending the default background (bg) color.
Finally, the globe image can be changed. If you wanted to confuse people by presenting the same information with an arbitrary citrus theme, you could make the following:
## Conclusion
The globes are created with the threejs package. You can follow see and modify the examples in this post by following this link. | 1,147 | 4,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-06 | longest | en | 0.750828 |
https://www.limsforum.com/informatics-educational-institutions-programs/?rdp_we_resource=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FPower_%28physics%29 | 1,627,367,793,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00308.warc.gz | 863,161,878 | 53,347 | Power
Common symbols
P
SI unitwatt (W)
In SI base unitskgm2s−3
Derivations from
other quantities
Dimension
In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity.[1][2][3] Power is a scalar quantity.
The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft. The power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle. In classical mechanics, as quantified from a stationary frame of reference, the motive power of a jet-propelled vehicle is the product of the engine thrust and the velocity of the vehicle (note that by this definition, a propelled vehicle hovering at stationary elevation over a gravitational body, where the upward thrust exactly cancels the downward acceleration of gravity, the motive power is zero). The rate at which a light bulb converts electrical energy into light and heat is measured in watts – the electrical energy used per unit of time.[4][5]
## Definition
Power is the rate with respect to time at which work is done; it is the time derivative of work:
where P is power, W is work, and t is time.
If a constant force F is applied throughout a distance x, the work done is defined as . In this case, power can be written as:
If instead the force is variable over a three-dimensional curve C, then the work is expressed in terms of the line integral:
From the fundamental theorem of calculus, we know that . Hence the formula is valid for any general situation.
## Units
The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second. Other common and traditional measures are horsepower (hp), comparing to the power of a horse; one mechanical horsepower equals about 745.7 watts. Other units of power include ergs per second (erg/s), foot-pounds per minute, dBm, a logarithmic measure relative to a reference of 1 milliwatt, calories per hour, BTU per hour (BTU/h), and tons of refrigeration.
## Average power
As a simple example, burning one kilogram of coal releases much more energy than detonating a kilogram of TNT,[6] but because the TNT reaction releases energy much more quickly, it delivers far more power than the coal. If ΔW is the amount of work performed during a period of time of duration Δt, the average power Pavg over that period is given by the formula:
It is the average amount of work done or energy converted per unit of time. The average power is often simply called "power" when the context makes it clear.
The instantaneous power is then the limiting value of the average power as the time interval Δt approaches zero.
In the case of constant power P, the amount of work performed during a period of duration t is given by:
In the context of energy conversion, it is more customary to use the symbol E rather than W.
## Mechanical power
One metric horsepower is needed to lift 75 kilograms by 1 metre in 1 second.
Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity, or the product of a torque on a shaft and the shaft's angular velocity.
Mechanical power is also described as the time derivative of work. In mechanics, the work done by a force F on an object that travels along a curve C is given by the line integral:
where x defines the path C and v is the velocity along this path.
If the force F is derivable from a potential (conservative), then applying the gradient theorem (and remembering that force is the negative of the gradient of the potential energy) yields:
where A and B are the beginning and end of the path along which the work was done.
The power at any point along the curve C is the time derivative:
In one dimension, this can be simplified to:
In rotational systems, power is the product of the torque τ and angular velocity ω,
where ω measured in radians per second. The represents scalar product.
In fluid power systems such as hydraulic actuators, power is given by
where p is pressure in pascals, or N/m2 and Q is volumetric flow rate in m3/s in SI units.
If a mechanical system has no losses, then the input power must equal the output power. This provides a simple formula for the mechanical advantage of the system.
Let the input power to a device be a force FA acting on a point that moves with velocity vA and the output power be a force FB acts on a point that moves with velocity vB. If there are no losses in the system, then
and the mechanical advantage of the system (output force per input force) is given by
The similar relationship is obtained for rotating systems, where TA and ωA are the torque and angular velocity of the input and TB and ωB are the torque and angular velocity of the output. If there are no losses in the system, then
which yields the mechanical advantage
These relations are important because they define the maximum performance of a device in terms of velocity ratios determined by its physical dimensions. See for example gear ratios.
## Electrical power
Ansel Adams photograph of electrical wires of the Boulder Dam Power Units, 1941–1942
The instantaneous electrical power P delivered to a component is given by
where
is the instantaneous power, measured in watts (joules per second)
is the potential difference (or voltage drop) across the component, measured in volts
is the current through it, measured in amperes
If the component is a resistor with time-invariant voltage to current ratio, then:
where
is the resistance, measured in ohms.
## Peak power and duty cycle
In a train of identical pulses, the instantaneous power is a periodic function of time. The ratio of the pulse duration to the period is equal to the ratio of the average power to the peak power. It is also called the duty cycle (see text for definitions).
In the case of a periodic signal of period , like a train of identical pulses, the instantaneous power is also a periodic function of period . The peak power is simply defined by:
The peak power is not always readily measurable, however, and the measurement of the average power is more commonly performed by an instrument. If one defines the energy per pulse as:
then the average power is:
One may define the pulse length such that so that the ratios
are equal. These ratios are called the duty cycle of the pulse train. | 1,420 | 6,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-31 | latest | en | 0.933455 |
https://www.coursehero.com/file/9092103/09-kilometers-Herea0andb1609-Thistransformationchangestheunitswithout-changingtheoriginadista/ | 1,493,446,758,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123276.44/warc/CC-MAIN-20170423031203-00632-ip-10-145-167-34.ec2.internal.warc.gz | 899,046,093 | 23,493 | Linear_Transformations
# 09 kilometers herea0andb1609
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Unformatted text preview: each quiz X = old grade X* = new grade X* = x + 5 a = 5, b = 1 We changed the origin ( 0 = 5 ) Examples Examples • If a distance x is measured in miles, the same distance in kilometers is: x* = 1.609(x) – That is 10 miles is the same as 16.09 kilometers. – Here a = 0 and b = 1.609. – This transformation changes the units without changing the origin a distance of 0 miles is the same as a distance of 0 kilometers. Example Example • A temperature x measured in degrees Fahrenheit (°F) must be reexpressed in degrees Celsius (°C) to align with every other country in the world. – The transformation is: X* = 5/9( X – 32 ) • What does “a” and “b” equal? • a = 160/9 • b = 5/9 – This changes both the unit size and the origin of the measurements. 0º C is 32º F. Problem #1 Problem #1 • The average salary for an employee at Acme Corporation is \$30,000 per year. This year, management awards the following bonuses to every employee. – A Christmas bonus of \$500 – An incentive bonus equal to 10% of the employee’s salary. • What is the mean bonus received by employees? – (A) \$500 (B) \$3,000 (C) \$3,500 (D) None of the above. (E) There is not enough information to answer this question. Problem #1 Continued Problem #1 Continued • The correct answer is C. To compute the bonus, management applies the following linear transformation to the each employee's salary: – x* = bx + a Y = 0.10 * x + 500 – where x* is the transformed variable (the bonus), x is the original variable (the salary), b is the multiplicative constant 0.10, and a is the additive constant 500. • Since we know that the mean salary is \$30,000, we can compute the mean bonus from the following equation: – Y = bx + a Y = 0.10 * \$30,000 + \$500 = \$3,500 Homework Homework • Textbook pp. 90 – 99 – # 23, 24, 35, 36, 46 • Chapter 3 Project Due Monday, September 12th!! • Unit 1 Test – Thursday, September 15th!...
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## This document was uploaded on 03/30/2014 for the course MATH AP Statist at Richard Montgomery High.
Ask a homework question - tutors are online | 632 | 2,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-17 | longest | en | 0.902405 |
https://fahmyalhafidz.com/44jmhk/ | 1,659,985,231,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00435.warc.gz | 240,184,965 | 11,065 | # A domain error occurs if x is negative and y is not an integral
A domain error occurs if x is negative and y is not an integral value for the function pow(double x, double y).
A. true
B. false
This question was posed to me during a job interview.
Enquiry is from Mathematical functions in division C Library of C
The correct answer is A. true
Explanation: The pow() function computes x raised to the power y. A domain error occurs if x is negative and y is not an integral value. A domain error occurs if the result cannot be represented when x is zero and y is less than or equal to zero. A range error may occur. | 142 | 623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.869459 |
https://www.enotes.com/homework-help/how-far-does-mercury-rise-capillary-when-382503 | 1,670,513,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00079.warc.gz | 808,096,066 | 19,029 | # How far does the mercury rise in the capillary when the temperature changes from 0.0 C to 25.0 C? The density of mercury at these temperatures is 13.596 g/cm^3 and 13.534 g/cm^3 , respectively. Mercury is often used as an expansion medium in a thermometer. The mercury sits in a bulb on the bottom of the thermometer and rises up a thin capillary as the temperature rises. Suppose a mercury thermometer contains 3.350 grams of mercury and has a capillary that is 0.210 mm in diameter. Express your answer with the appropriate units. Thank you!
## Expert Answers
The first thing to do is to get the volume of Mercury for both temperatures. Then from the volume we can get the height of the mercury it has occupied at the temperature given using the formula for cylinder. Finally, we get the difference between the distances calculated.
`density =...
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The first thing to do is to get the volume of Mercury for both temperatures. Then from the volume we can get the height of the mercury it has occupied at the temperature given using the formula for cylinder. Finally, we get the difference between the distances calculated.
`density = (mass)/(volume) = (g)/(cm^(3))`
`volume at speci fic temperature = (mass)/(density at speci fic temperature)`
`volume at 0^(o) C = (3.350 g Hg)/(13.596 (g Hg)/(cm^(3)))`
`volume at 0^(o)C = 0.2464 cm^(3)`
`volume at 25^(o)C = (3.350 gHg)/(13.534 (gHg)/(cm^(3)))`
`volume at 25^(o)C = 0.4275 cm^(3)`
---------------------
The height of mercury for both temperatures:
`volume of cyli nder = volume of Hg = pi*r^(2)*h`
`h = (volume)/(pi*r^(2))`
where, `pi` = 3.1416; h = height and r = radius= diameter/2
radius = 0.210 mm/2 = 0.105mm = 0.0105cm
`h at 0^(o)C = (0.2464 cm^(3))/(3.1416*(0.0105cm)^2)`
`h at 0^(o)C = 711.386 cm`
`h at 25^(o) C = (0.2475cm^3)/(3.1416*(0.0105cm)^2)`
`h at 25^(o) C = 714.64 cm`
By subtracting the distances measured, we can have a value of 3.26 cm which is the final answer.
Approved by eNotes Editorial Team | 635 | 2,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-49 | latest | en | 0.841045 |
https://forum.leasehackr.com/t/lexus-rx-450h-lease-to-buy/561570 | 1,718,507,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00142.warc.gz | 221,473,733 | 9,326 | # Lexus RX 450H+ Lease to Buy
I got an offer for a lease deal on a Lexus RX 450H+ from a dealership. I’m planning on keeping the car for a long time, so I want to buy out the lease immediately. I’m very new to the leasing process and I had two main questions that is preventing me from moving forward.
1. When I buy out my lease, am I paying the Adjusted Capitalized Cost (9c)?
2. I’m not sure what 32D means for the difference between the Adjusted Capitalized Costs and app depreciation and other amortized amounts accrued through the date of early termination calculated in accordance with the constant yield method.
Based on my understanding, OTD pricing on this car will be \$64,416.19 + tax (10.5%) which would put me at 71,179.88 total OTD. I think there are some other fees associated such as transfer & documentation fees (finance manager quoted \$15 + \$80) for these two.
I’m confused because this seems like a really great deal. Every other dealership that I called said that I should just take this but I’m not sure if I’m missing any gotchas. Is my understanding correct?
Any help would be great! Thank you in advance!
Yes or very close to it with some minor adjustments for daily interest.
Some banks have a purchase option fee but sounds like yours doesn’t.
As @max_g indicated, it is very close. Your buyout is computed as follows…
Initial balance = Adj. Cap – Base Payment
= 64416.19 – 943.20
= 63472.99
The 7.688596% is the constant yield rate for your lease contract. It can be calculated using the RATE function in Excel. I’ve referenced this, in some detail, in several of my posts in other threads.
(Adj. Lease Bal. + any purchase option fee) x (1 + tax rate) + applicable fees
Your sales tax rate is 10.25%. Your buyout is calculated assuming you buy the vehicle within 30 days of lease inception. Remember that payments including lease finance charges are due and collected one month in advance.
5 Likes
Go Wildcats
1 Like
1 Like
Can confirm that the buyout is close to the adjusted cap cost. We just did the same process last month for our new RX450h+
1 Like
If you don’t mind me asking how much did you end up paying for the car in total?
Right at \$65K, plus my state taxes. I am still waiting on the final paperwork from Lexus. Depending on how my trade was reported to the state, I am either gonna owe \$4K or right around \$2K LOL
Apologize for the late reply on this topic. Did you pay all your lease payments (the lexus one pay lease), or just make the first month payment and then pay Lexus financial after the dealer transaction? I’m getting so many different answers. arg.
My “first” payment was included in the pricing. So technically, my 2nd payment I did pay, and now my third, because they’re going slow as bananas to process my buyout.
I sent in my \$20K last week, and “should” be getting a call back from them tomorrow letting me know it finally came off my buyout amount so my actual financed amount will be roughly \$43.5K
Sticker was originally almost \$76K, and when my contract was loaded, at the time, my buyout was \$65,0xx so I saved almost \$11K
Ah, so you did it differently than the original post author. You didn’t pay a large amount at the dealer - you paid 20K later. Can I ask how you found out the value to send? (20K). This isn’t the total of the lease payments - it must be something less that was listed on your lease document?
I sent \$20K. Didn’t specify how much to send them
I knew how much to put down based on where I want my finance payments to be at for the buyout.
In my instance \$63.5K originally, sent them \$20K, I am now financing \$43.5K or roughly \$820 a month for 5yrs.
I only paid the first month which was taken out of the rebate (~\$1000). I also just sent in the full amount to purchase the car outright and it ended up being around \$64K pre tax which was right around what the dealership quoted me for the adjusted capital cost.
My taxes are expensive so I’m looking at being around 70K-71K total OTD for the car. | 972 | 4,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.961037 |
http://stackoverflow.com/questions/4196417/can-someone-explain-the-behavior-of-the-functions-mkpp-and-ppval/4196606 | 1,369,436,795,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705195219/warc/CC-MAIN-20130516115315-00026-ip-10-60-113-184.ec2.internal.warc.gz | 250,488,281 | 11,502 | # Can someone explain the behavior of the functions mkpp and ppval?
If I do the following in MATLAB:
``````ppval(mkpp(1:2, [1 0 0 0]),1.5)
ans = 0.12500
``````
This should construct a polynomial `f(x) = x^3` and evaluate it at `x = 1.5`. So why does it give me the result `1.5^3 = .125`? Now, if I change the domain defined in the first argument to `mkpp`, I get this:
``````> ppval(mkpp([1 1.5 2], [[1 0 0 0]; [1 0 0 0]]), 1.5)
ans = 0
``````
So without changing the function, I change the answer. Awesome.
Can anyone explain what's going on here? How does changing the first argument to `mkpp` change the result I get?
-
The function MKPP will shift the polynomial so that `x = 0` will start at the beginning of the corresponding range you give it. In your first example, the polynomial `x^3` is shifted to the range `[1 2]`, so if you want to evaluate the polynomial at an unshifted range of `[0 1]`, you would have to do the following:
``````>> pp = mkpp(1:2,[1 0 0 0]); %# Your polynomial
>> ppval(pp,1.5+pp.breaks(1)) %# Shift evaluation point by the range start
ans =
3.3750 %# The answer you expect
``````
In your second example, you have one polynomial `x^3` shifted to the range `[1 1.5]` and another polynomial `x^3` shifted to the range of `[1.5 2]`. Evaluating the piecewise polynomial at `x = 1.5` gives you a value of zero, occurring at the start of the second polynomial.
It may help to visualize the polynomials you are making as follows:
``````x = linspace(0,3,100); %# A vector of x values
pp1 = mkpp([1 2],[1 0 0 0]); %# Your first piecewise polynomial
pp2 = mkpp([1 1.5 2],[1 0 0 0; 1 0 0 0]); %# Your second piecewise polynomial
subplot(1,2,1); %# Make a subplot
plot(x,ppval(pp1,x)); %# Evaluate and plot pp1 at all x
title('First Example'); %# Add a title
subplot(1,2,2); %# Make another subplot
plot(x,ppval(pp2,x)); %# Evaluate and plot pp2 at all x
axis([0 3 -1 8]) %# Adjust the axes ranges
title('Second Example'); %# Add a title
``````
-
I've been staring at this for so long... Thank you! If I want to evaluate it at some vector of points how would I do it? Is there a better function to use than ppval? – Xodarap Nov 16 '10 at 16:53 @Xodarap: You can pass a vector of points to PPVAL as illustrated in my sample plotting code above. – gnovice Nov 16 '10 at 17:01 | 755 | 2,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.787814 |
http://poker.stackexchange.com/questions/2783/can-you-call-when-the-pot-odds-dont-favour-you?answertab=active | 1,462,458,253,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860127496.74/warc/CC-MAIN-20160428161527-00076-ip-10-239-7-51.ec2.internal.warc.gz | 223,521,487 | 17,878 | # Can you call when the pot odds don't favour you?
Say you're drawing to a made hand in a \$10-\$20 limit hold'em game and you have N outs.
On the turn, one line of thought is that you should only call if the pot odds are in your favour, i.e. if there is X in the pot and
X > (46 - N)/N * 20
However, this seems to ignore the fact that you can probably extract an extra bet if you make your hand. Taking that into account your EV is
EV = N/46 * (X + 20) + (46 - N)/N * (-20)
so you should call if
X + 20 > (46 - N)/N * 20
or
X > [ (46 - N)/N - 1 ] * 20
i.e. you can effectively subtract 1 from the required pot odds to account for the fact that you can extract an extra bet when you make your hand. If you think that that there's only some probability p of extracting that extra bet, you could subtract p from the required pot odds instead of subtracting 1.
This seems to have the largest effect when the pot is small and you have many outs, for example if you have 9 outs (flush draw) you would normally require pot odds of 4.1 : 1 to call, but this suggests that folding with less than 3.1 : 1 is actually throwing away value. If you have 15 outs (both straight and flush draw) you would normally require 2 : 1 but this suggests that if you don't call when you have better than evens, you are throwing money away.
Is my analysis here correct, and if so, why is the advice that you should call only if the pot odds justify it so prevalent? Is there a name for this concept of extracting extra bets from your opponent when you make your hand?
-
This appears in Sklanky's "Theory of Poker" book. He suggest the right way to look at the pot odds is by taking into account future bets too, for that you need to estimate the probability that the rest of the players will call or reraise. – Jubbat Mar 20 '14 at 1:41
Yes but be realistic about what you can get out. Out of position on an obvious draw if you hit you might get a 1/4 pot size bet at best. It is just called EV. It is best if you have disguised your hand - limping from middle says draw. – Paparazzi Jan 4 at 20:17
Yes, you can and you should. The concept you're describing is called implied odds (the estimated profit you'll make if you make your hand).
Notice is a much less concrete value as it is an estimation of whether your opponent will call when the draw comes and the amount he'll be willing to pay. There's also the concept of reverse implied odds which are the odds or you completing your draw and still loosing (for example against a bigger draw or a full house)
-
I only read a part of your question but you can indeed call when odds are not in favor. Implied pot odds is what it's called. Basically you calculate that what you payed too much, you will receive back at the next street(s). An example:
Hero and Villain play a pot. On the turn you have a flush draw (giving you 9 outs). This is more or less 19% chance of hitting it. Villain bets exactly half the pot size (but he puts in half of his stack as well). Knowing this, you might want to call in some cases cause you know that on the river he can't fold to you with only this much chips left (less than half of his original stack).
In numbers: Villain has a stack size of 1200... All the action before the turn costs him 400 chips. You called everything so the pot is now 800 chips. At the turn Villain has 800 chips left and puts 400 in the pot for half the amount of his money and half the pot size. This makes the pot 1200 (giving you 400/1600 odds). Your outs are about 19%, but your odds are 25%, which is EV-... However, knowing that at the river Villain will only be able to commit another 400 chips, giving you maximum 400/2400 odds, you should call.
If you know Villain is really aggressive and pushes a lot on the river, it's even more interesting for you because you can earn ALL his chips, even if he has a large stack...
- | 975 | 3,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-18 | latest | en | 0.959328 |
https://www.datasciencemadesimple.com/round-up-round-down-and-round-off-in-pyspark-ceil-floor/ | 1,709,239,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00340.warc.gz | 723,798,062 | 25,146 | # Round up, Round down and Round off in pyspark – (Ceil & floor pyspark)
Round up or ceil in pyspark uses ceil() function which rounds up the column in pyspark. Round down or floor in pyspark uses floor() function which rounds down the column in pyspark. Round off the column is accomplished by round() function. Let’s see an example of each.
• Round up or Ceil in pyspark using ceil() function
• Round down or floor in pyspark using floor() function
• Round off the column in pyspark using round() function
• Round off to decimal places using round() function.
We will be using dataframe df_states
#### Round up or Ceil in pyspark using ceil() function
Syntax:
ceil(‘colname1’)
colname1 – Column name
ceil() Function takes up the column name as argument and rounds up the column and the resultant values are stored in the separate column as shown below
```## Ceil or round up in pyspark
from pyspark.sql.functions import ceil, col
df_states.select("*", ceil(col('hindex_score'))).show()
```
So the resultant dataframe with ceil of “hindex_score” is shown below
#### Round down or Floor in pyspark using floor() function
Syntax:
floor(‘colname1’)
colname1 – Column name
floor() Function in pyspark takes up the column name as argument and rounds down the column and the resultant values are stored in the separate column as shown below
```## floor or round down in pyspark
from pyspark.sql.functions import floor, col
df_states.select("*", floor(col('hindex_score'))).show()
```
So the resultant dataframe with floor of “hindex_score” is shown below
#### Round off in pyspark using round() function
Syntax:
round(‘colname1’,n)
colname1 – Column name
n – round to n decimal places
round() Function takes up the column name as argument and rounds the column to nearest integers and the resultant values are stored in the separate column as shown below
```######### round off
from pyspark.sql.functions import round, col
df_states.select("*", round(col('hindex_score'))).show()
```
So the resultant dataframe with rounding off of “hindex_score” column is shown below
#### Round off to decimal places using round() function
round() Function takes up the column name and 2 as argument and rounds off the column to nearest two decimal place and the resultant values are stored in the separate column as shown below
```########## round off to decimal places
from pyspark.sql.functions import round, col
df_states.select("*", round(col('hindex_score'),2)).show()
```
So the resultant dataframe with Rounding off of “hindex_score” column to 2 decimal places is shown below
#### Other Related Topics:
## Author
• With close to 10 years on Experience in data science and machine learning Have extensively worked on programming languages like R, Python (Pandas), SAS, Pyspark. | 642 | 2,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-10 | longest | en | 0.640816 |
https://naturalmaths.com.au/blog/complete-these-number-sentences/ | 1,713,071,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00808.warc.gz | 376,916,723 | 21,600 | # Complete these Number Sentences
With the best will and planning, there will still be some odd minutes during the day that could have been better utilized. Careful choice of short burst tasks and challenges can engage students and provide windows into mathematical thinking and understanding. A short burst task is a task that can be:
• given to students on arrival to class while usual morning routines are carried out,
• used during transitions,
• given to students to responsibly work on and think about in odd moments throughout the day,
• quickly debriefed or followed up with short whole-class or small-group focused learning episodes,
• a quick formative assessment task or exit pass.
Here is an example that you can download by clicking on the picture below.
# Formative Assessment Opportunities
This is what we saw and heard when we took this short burst challenge into a year 6-7 classroom recently.
Set 1
All students worked out each answer (a few using fingers) but only a few noticed that the commutative property (4 + 6 and 6 + 4, 4 × 6 and 6 × 4) could have been used, meaning that there were actually only 2 discrete sums to do. In fact a pair of students did notice and were heard to say “6 × 4 is 24 and 4 × 6 is 24 so these are the same” with surprise in their voices as if this was a revelation to them.
Set 2
Clearly an understanding of integers is needed for these questions but as well as revealing understanding about integers they also revealed that some students had been over taught rules for subtraction using the traditional method of subtraction.
For this number sentence, some students actually said “6 from 4 you can’t and so its 6 – 4 equals 2 not minus 2.” For some students though this was not such a problem because they used empty number line thinking and one partnership modelled the following:
This was all it took to prompt others to use an empty number line to show how to work out – 6 + 4 = –2. It was interesting to see how powerful empty number line thinking was for most students.
Set 3
The last two questions were meant to be challenging and we actually suggested that students try drawing or diagramming! We also suggested that one question might be easier than the other question. For some students, this hint fell on deaf ears and they struggled with the first question whereas students who did the second question first were more successful. Pizzas were the main device used, with students sharing the first 4 pizzas, one each, and then cutting the remaining two in half, which meant that each person got 1½ pizzas.
For 4 6 = ⅔ there was a struggle for many students until an unexpected student excitedly called out “Cut them in thirds!”. Some students had difficulty drawing thirds initially and one group drew this diagram:
Actually, we have reproduced it in colour not their lead pencil version, which was harder to follow. So, question? Why do we make kids do maths in grey and black and white? Bring on the gel pens and highlighters.
Have a safe weekend! Best wishes from Johnny and I. | 666 | 3,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | longest | en | 0.97865 |
https://bisohukixoqokuv.3g-architecture.com/static-analysis-in-fea-pdf-18429gy.html | 1,638,923,336,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00266.warc.gz | 215,840,810 | 7,420 | # Static Analysis In Fea Pdf
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Static analysis in fea pdf | 2,243 | 10,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.929822 |
https://www.teachoo.com/13129/495/Ex-2.3--1-iii/category/Ex-2.3/ | 1,723,183,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00842.warc.gz | 783,224,083 | 21,084 | Division Algorithm
Chapter 2 Class 10 Polynomials
Serial order wise
### Transcript
Ex2.3, 1 Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: (iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2 Quotient = – x2 – 2 Remainder = –5x + 10 | 96 | 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} | 3.578125 | 4 | CC-MAIN-2024-33 | latest | en | 0.864306 |
http://doubts.dronstudy.com/question/solve-quicklyak/?order_by=newest | 1,590,969,684,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00512.warc.gz | 36,437,257 | 15,101 | a stationary man observe that the rain strike him at angle of 60 degree to the horizontal when he begin to move with a velocity of 25 m/s then the drop appear to strike him at angle 30 degree from horizontal calculate the velocity of the rain drop | 53 | 247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.911123 |
https://stat.ethz.ch/pipermail/r-help/2011-July/284707.html | 1,670,266,983,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00665.warc.gz | 573,487,018 | 3,074 | # [R] squared "pie chart" - is there such a thing?
Liaw, Andy andy_liaw at merck.com
Mon Jul 25 17:28:45 CEST 2011
Has anyone suggested mosaic displays? That's the closest I can think of as a "square pie chart"...
> -----Original Message-----
> From: r-help-bounces at r-project.org
> [mailto:r-help-bounces at r-project.org] On Behalf Of Naomi Robbins
> Sent: Sunday, July 24, 2011 7:09 AM
> To: Thomas Levine
> Cc: r-help at r-project.org
> Subject: Re: [R] squared "pie chart" - is there such a thing?
>
> I don't usually use stacked bar charts since it is difficult
> to compare
> lengths that don't have
> a common baseline.
>
> Naomi
>
> On 7/23/2011 11:14 PM, Thomas Levine wrote:
> > How about just a stacked bar plot?
> >
> > barplot(matrix(c(3,5,3),3,1),horiz=T,beside=F)
> >
> > Tom
> >
> > On Fri, Jul 22, 2011 at 7:14 AM, Naomi
> Robbins<nbrgraphs at optonline.net> wrote:
> >> Hello!
> >> It's a shoot in the dark, but I'll try. If one has a total of 100
> >> (e.g., %), and three components of the total, e.g.,
> >> mytotal=data.frame(x=50,y=30,z=20), - one could build a
> pie chart with
> >> 3 sectors representing x, y, and z according to their
> proportions in
> >> the total.
> >> I am wondering if it's possible to build something very
> similar, but
> >> not on a circle but in a square - such that the total area of the
> >> square is the sum of the components and the components (x,
> y, and z)
> >> are represented on a square as shapes with right angles (squares,
> >> rectangles, L-shapes, etc.). I realize there are many possible
> >> positions and shapes - even for 3 components. But I don't
> really care
> >> where components are located within the square - as long
> as they are
> >> there.
> >>
> >> Is there a package that could do something like that?
> >> Thanks a lot!
> >>
> >> -----
> >>
> >> I included waffle charts in Creating More Effective Graphs.
> >> The reaction was very negative; many readers let me know
> >> that they didn't like them. To create them I just drew a table
> >> in Word with 10 rows and 10 columns. Then I shaded the
> >> backgrounds of cells so for your example we would shade
> >> 50 cells one color, 30 another, and 20 a third color.
> >>
> >> Naomi
> >>
> >> -------------
> >>
> >>
> >> Naomi B. Robbins
> >> 11 Christine Court
> >> Wayne, NJ 07470
> >> 973-694-6009
> >>
> >> naomi at nbr-graphs.com<mailto:naomi at nbr-graphs.com>
> >>
> >> http://www.nbr-graphs.com
> >>
> >> Author of Creating More Effective Graphs
> >> <http://www.nbr-graphs.com/bookframe.html>
> >>
> >> //
> >>
> >>
> >>
> >> [[alternative HTML version deleted]]
> >>
> >> ______________________________________________
> >> R-help at r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >
>
>
> --
>
> --
>
> Naomi B. Robbins
>
> NBR
>
> 11 Christine Court
>
> Wayne, NJ 07470
>
> Phone: (973) 694-6009
>
> naomi at nbr-graphs.com <mailto:naomi at nbr-graphs.com>
>
> http://www.nbr-graphs.com <http://www.nbr-graphs.com/>
>
>
> Author of /Creating More Effective Graphs
> <http://www.nbr-graphs.com/bookframe.html>/
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help | 1,014 | 3,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.937092 |
http://slideplayer.com/slide/4104628/ | 1,503,386,494,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110485.9/warc/CC-MAIN-20170822065702-20170822085702-00317.warc.gz | 387,655,535 | 27,022 | # COSC 6114 Prof. Andy Mirzaian. TOICS General Facts on Polytopes Algorithms Gift Wrapping Beneath Beyond Divide-&-Conquer Randomized Incremental.
## Presentation on theme: "COSC 6114 Prof. Andy Mirzaian. TOICS General Facts on Polytopes Algorithms Gift Wrapping Beneath Beyond Divide-&-Conquer Randomized Incremental."— Presentation transcript:
COSC 6114 Prof. Andy Mirzaian
TOICS General Facts on Polytopes Algorithms Gift Wrapping Beneath Beyond Divide-&-Conquer Randomized Incremental
References: [M. de Berge et al] chapter 11 [Preparata-Shamos’85] chapter 3 [O’Rourke’98] chapter 4 [Edelsbrunner’87] chapter 8 Branko Grünbaum, "Convex Polytopes," 2nd edition (prepared by Volker Kaibel, Victor Klee, Günter Ziegler), Springer, 2003. Günter M. Ziegler, "Lectures on Polytopes," Graduate Texts in Mathematics 152, Springer 1995. Revised sixth printing 2006. Arne Brondsted, "An Introduction to Convex Polytopes," Springer, 1983. Applications: Clustering, graphics, CAD/CAM, pattern recognition, Operations Research Voronoi Diagrams & Delaunay Triangulations …
General Facts on Polytopes
half-space in d : { p d | 1 p 1 + 2 p 2 + … + d p d d+1 } hyper-plane in d : { p d | 1 p 1 + 2 p 2 + … + d p d = d+1 } convex polyhedron in d : Intersection of finitely many closed half-spaces. d-polytope P: a d dimensional bounded convex polyhedron. supporting hyper-plane of P: a hyper-plane that intersects the boundary but not the interior of the polytope. face of P : intersection of P with a supporting hyper-plane. This is a lower dimensional polytope itself. (We consider P itself and the empty set to be faces of P also.) k-face = a k dimensional face of P, k = -1, 0, 1, 2, …, d d-face : body (polytope P itself) (d-1)-faces : facets (d-2)-faces : sub-facets (or ridges) 1-faces : edges 0-faces : vertices (-1)-face : the empty set d-simplex : CH of d+1 affinely independent points in d. (e.g., 2-simplex is a triangle, 3-simplex is a tetrahedron, … ) simplicial polytope: each facet has d sub-facets.
body facets sub-facets edges vertices empty set If Q and R are two faces of P, then Q R is also a face of P. Face lattice incidence Data Structure
Face lattice incidence 3D Example
FACT: Consider a d-polytope P with n vertices. Define i = # i-faces of P Victor Klee [1966]: d-1 = ( n d/2 ) (# facets) Branko Grünbaum [1967]: d-2 = ( n d/2 ) (# sub-facets) Note 1: These boundes are tight in the worst-case. Note 2: These bounds are O(n) for d=2,3. Bernard Chazelle [1990]: CH of n points in d in worst-case optimal time: O( n log n + n d/2 ).
FACT: The 1-skeleton of a 3-polytope is a planar graph. Proof: Step (1): Central projection: N p q p’ q’ [1-skeleton : vertex-edge incidence graph of the polytope.] Step (2): Stereographic projection:
FACT: [Steinitz 1922] A simple graph is the 1-skeleton of a 3-polytope if and only if it is planar and 3-connected. FACT: [Balinski 1961] The 1-skeleton graph of a d-polytope is d-connected. Terminology: simple graph: no self loops (i.e., edge from a vertex to itself) and no parallel edges (i.e., multiple edges between the same pair of vertices). d-connected: need to remove at least d vertices (and their incident edges) to disconnect the graph. In other words, between each pair of vertices there are at least d paths that are disjoint except at the two ends. k-skeleton: incidence “hypergraph” of all faces of dimension at most k. 1-skeleton: the vertex-edge graph of the polytope.
3D non-convex POLYTOPES Boundary consists of a finite set V (vertices, 0-faces), E (edges, 1-faces), F (facets, 2-faces), such that the boundary’s topology is a closed 2-manifold of genus 0 (i.e., a topological sphere). “link” of a vertex v = 1-skeleton sub-graph induced by the set of vertices adjacent to v. FACT: 1.The intersection of any two facets is either empty, a vertex or an edge. 2. The link of each vertex is a simple closed polygonal chain. (This also implies that each edge is incident to exactly two facets.) 3. The 1-skeleton (V, E) is connected. (1)disallows: (2)Disallows: (3)Disallows disconnected pieces.
3D Regular POLYTOPES In 2D there are infinitely many regular polygons; an n-regular polygon for each n=3,4,5,… In 3D there are only 5 regular 3D polytopes, also called Platonic Solids. (i) p = # vertices of each facet (a regular p-gon) (ii) q = # facets incident to each vertex (i) each angle = (p-2) /p = (1 – 2/p) (ii) q (1 – 2/p) < 2 (p-2)(q-2) < 4 (p 3, q 3) There are only 5 (p,q) pairs that satisfy this condition. Dodecahedron (p=5, q=3) Icosahedron (p=3, q=5) [V = 20, E = 30, F = 12] [V = 12, E = 30, F = 20] Tetrahedron (p=3, q=3) Cube (p=4, q=3)Octahedron (p=3, q=4) [V = 4, E = 6, F = 4] [V = 8, E = 12, F = 6][V = 6, E = 12, F = 8]
d Dim CH Algorithms
The Gift Wrapping Method [Chand-Kapur 1970] Generalization of Jarvis March. Analysis done by Bhattacharya [1982]. FACT: A sub-facet is shared by exactly 2 facets. In a simplicial d-polytope: (a) Each facet has d vertices (is a (d-1)-simplex). (b) Two facets F 1 and F 2 share a sub-facet e if and only if e is determined by a common subset of d-1 vertices of the vertex sets determining F 1 & F 2 (F 1 & F 2 are said to be adjacent on e). Preliminary Assumption: P is in general position, i.e., CH(P) is a simplicial d-polytope.
Input: P = { p 1, p 2, …, p n } d Output: CH(P) 1. L (* output list of facets of CH(P) *) 2. Q (* frontier facets, discovered but not wrapped around *) 3. F find an initial convex hull facet (* see next slide *) 4. T sub-facets of F (* dictionary of un-wrapped sub-facets *) 5. Q F (* push F into Q *) 6. while Q do 7. F Q (* extract front facet from Q *) 8. T F sub-facets of F 9. for each e T F T do (* e is a gift wrapping candidate *) 10. F’ facet sharing sub-facet e with F (* gift wrap *) 11. insert into T all sub-facets of F’ not yet present, 12. and delete all those already present 13. end-for 14. L F (* push facet F into output list *) 15. end-while 16. output L (* list of facets of CH(P) *) end ALGORITHM Gift Wrapping
Initialization: 3. F find an initial convex hull facet (* see below *) 4. T sub-facets of F (* dictionary of un-wrapped sub-facets *) ALGORITHM Gift Wrapping Method: P’ Project the n points of P on the (d-1) dimensional subspace of the d-1 coordinates excluding the 1 st coordinate. P’ d-1. G’ a facet of CH(P’) (found recursively) in d-1. H the hyper-plane in d that contains G’ and is parallel to the 1 st axis. G set of d-1 points in P whose projection is G’ H is a supporting hyper-plane of P and contains G (why?) Rotate H (á la gift wrapping) about G until it hits another point p P. F G {p}. FACT: This initialization takes O(nd 2 ) time, since it follows the recurrence: Q(n,d) = Q(n,d-1) + O(nd), Q(n,1) = O(n).
The Gift Wrapping Method Analysis T : a balanced search tree. Member sub-facets are lexicographically ordered (d-1)-component vector of vertex indices that define the sub-facet. | T | = O( d-2 ). Lines 9,11,12: Each op. search/insert/delete on T takes O(d log d-2 ) time. Line 10: O(nd + d 3 ) time each. THEOREM: Convex hull of a set of n points in d can be constructed by gift-wrapping in time O((nd + d 3 ) d-1 d d-2 log d-2 ) = O( n d/2 +1 ) (assuming dimension d is fixed). Removing “preliminary assumption” of simpliciality is not too complecated.
The beneath-beyond Method [Kallay 1981] On-line algorithm. Add points of P incrementally and update CH(P). THEOREM: Time complexity of beneath-beyond is also O( n d/2 +1 ). P F aff(F) beyond F beneath F P f3f3 f1f1 f2f2 p (1) (2a) (2b) THEOREM: [McMullen-Shepard,1971] Let P be a polytope, p d, P’ = CH(P {p}). Faces of P’ are: (1)A face f of P is also a face of P’ facet F of P s.t. f F & p is beneath F. (2)If f is a face of P, then f’ = CH(f {p}) is a face of P’ (a) or (b) below holds: (a) among facets of P containing f, there is at least one s.t. p is beneath it, and at least one s.t. p is beyond it. (b) p aff(f).
Initialize H 3 tetrahedron (p 1, p 2, p 3, p 4 ) for k 5.. n do for each facet f of H k-1 do Compute volume of tetrahedron determined by f and p k mark f visible iff volume < 0. (* p k is beyond f *) end-for if no facets are visible then discard p k (* it’s inside H k-1 *) else do for each border edge e of H k-1 do construct cone facet determined by e and p k for each visible facet f do delete f update H k end-if end-for return H n end Beneath-Beyond 3D CH Algorithm O(n 2 ) time
Procedure CH(S): Base: if |S| 7 then return trivial answer. Divide: Partition S into two (almost) equal halves L and R around the x-median of S. Conquer: P 1 CH(L) & P 2 CH(R). Merge: P MERGE (P 1, P 2 ) (See next page.) O(1) O(n) 2T(n/2) O(n) T(n) = or + + T(n) = 2 T(n/2) + O(n) = O( n log n). Matches with the lower bound (n log n). Pre-sort p 1, p 2, …, p n lexicographically on (x, y, z). Call CH({p 1, p 2, …, p n }). Divide-&-Conquer 3D CH Algorithm [Preparata-Hong,1977]
Merge two Convex Hulls CH(P 1 P 2 ) P1P1 P2P2 FACT: Each merge-face uses an edge of P 1 or P 2 (called a boundary edge). Q: Do the boundary edges of P i (i=1,2) always form a simple chain? A: No! (See pp 113-114 of [O’Ro-98]
Merge two Convex Hulls DATA STRUCTURE: DCEL, WEDS, or QEDS How to do P CH(P 1 P 2 ) in linear time? P’ i projection of P i on the (x, y) plane, i=1,2 e’ a supporting edge of P’ 1, P’ 2. e original edge whose projection is e’ (e is an edge of P. Why?) “Rotate” a plane through e = (a,b) and “wrap around” P 1 P 2 to obtain a “cylindrical triangulation”: Scan DCEL of P 1 & P 2 for edges around a &b, respectively, to find triangles (a,b, ) & (a,b, ) that form max angle with current triangle (a,b,c). Then, compare (a,b, ) & (a,b, ) & suitably declare the winner as the next triangle. x y z P’ 1 P’ 2 e’ a’ b’ P2P2 P1P1 a b c This step takes time O(# edges incident to a or b that become “beneath”). Total such # edges is O(n). Merge DCEL’s & remove parts that fall beneath. current triangle
A Randomized version of the beneath-beyond method Initialize tetrahedron (p 1, p 2, p 3, p 4 ) Randomly permute {p 5, p 6, …, p n } Let P k = {p 1, p 2, …, p k }, k = 1..n for k 5.. n do update CH(P k-1 ) to CH(P k ) by inserting p k end Randomized Incremental 3D CH Algorithm CH(P k-1 ) pkpk horizon Details on next slides
Bipartite Conflict Graph G After iteration k, we have CH(P k ), P k = {p 1, p 2, …, p k }. For facet f of CH(P k ) and any point p P - P k, (f,p) is a conflict pair if facet f of CH(P k ) is visible from p (i.e., p is beyond f w.r.t. CH(P k )). F conflict (p) = { f | f is a facet of CH in conflict with p }, p P - P k. P conflict (f) = { p | (p,f) is a conflict pair w.r.t. current CH }, facet f of current CH F conflict (p k ) are the facets of CH(P k-1 ) that must be removed when p k is inserted. p f F conflict (p) P conflict (f) points facets conflicts
How to Update the Conflict Graph G Suppose insertion of p k into CH(P k-1 ) creates a new face f with the horizon edge e. pk pk f1 f1 e f2 f2 f e = f 1 f 2 P (e) P conflict (f 1 ) P conflict (f 2 ) P conflict (f) P(e) (a) For each new facet f do: (b) Insert node f in G (c) For each p P(e) s.t. f is visible from p, add conflict edge (p,f) to G (d) Remove nodes corresponding to facets visible from p k (e.g., f 2 above) and their incident edges (e) Remove node p k from G and all its incident edges. Notes: In (d) all facets visible from p k should be removed even those not incident to the horizon. In the figure above, if f is co-planar with f 1, simply extend f 1 to include f. In this case P conflict (f 1 ) remains as before (only point is removed in part (e)). What is P conflict (f)?
Randomized Incremental 3D CH Algorithm Input: P = { p 1, p 2, …, p n } 3 Output: CH(P) 1. Find 4 points p 1, p 2, p 3, p 4 in P that form a tetrahedron (i.e., are not co-planar) 2. C CH({p 1, p 2, p 3, p 4 }) 3. Randomly permute (p 5, p 6, …, p n ) 4. Initialize conflict graph G with all visible pairs (p k,f), where f is a facet of C, k>4 5. for k 5.. n do (* insert p k into C *) 6. if F conflict (p k ) then do 7. delete f from C, f F conflict (p k ) 8. L list of horizon edges of C w.r.t. p k (* traverse f F conflict (p k ) *) 9. for e L do 10. create the new facet f = CH(e, p k ) (* or extend old one if co-planar *) 11. if f is not co-planar with old one then do (* determine conflicts for f *) 12. create a new node f in G 13. Let e = f 1 f 2 14. P (e) P conflict (f 1 ) P conflict (f 2 ) 15. for p P(e) do if f is visible from p then add (p,f) to G 16. end-if 17. end-for 18. Delete p k and all facets F conflict (p k ) and all their incident edges from G 19. end-if 20. end-for 21. return C end
Randomized Incremental 3D CH Algorithm - Analysis LEMMA: Expected # facets created by the algorithm is 6n – 20. Proof: For each vertex p i of CH(P k ) define: deg (p i, CH(P k )) = # edges incident to p i in CH(P k ) = # facets incident to p i in CH(P k ). i deg (p i, CH(P k )) = 2 ( # edges of CH(P k ) ) 2(3k – 6). expected # new facets created in iteration k Backwards analysis: Fix P k P, k 5. Randomly choose p k P k – {p 1, p 2, p 3, p 4 } with probability 1/(k-4) each.
Randomized Incremental 3D CH Algorithm - Analysis THEOREM: Expected Time Complexity O(n log n). Proof: By Lemma, expected # facets created is O(n). Each facet destroyed was previously created, and won’t be re-created. Hence, Creating (and destroying) edges of G dominate the computation time and that, over all iterations, is [Proof of the latter bound is similar to the randomized QuickSort. See Exercise.]
Exercises
1.Modify the stated Gift-Wrapping algorithm so that it outputs the face lattice incidence graph of CH(P). 2.Complete the proof of the O(n log n) expected time bound on the randomized incremental 3D convex hull algorithm. 3.Design & analyze the 2D version of the beneath-beyond convex hull algorithm. 4.Design & analyze the 2D version of the randomized incremental convex hull algorithm. 5.Design & analyze the 3D version of the QuickHull convex hull algorithm. [See Slide 2.] 6.We are given a set P={ p 1,…,p n } of n>3 points and another point q, all in 3. Assume P and q are in general position, i.e., no 4 points are co-planar. Our task is to determine whether or not q is in convex hull of P, and output a certificate in each case. [Note that convex hull of P is NOT given.] Design and analyze an O(n)-time algorithm that (i) if q CH(P), it outputs 3 points p i, p j, p k P, such that their affine hull plane aff(p i, p j, p k ) separates q from P, or (ii) if q CH(P), it outputs 4 points p i, p j, p k, p l P, such that q CH(p i, p j, p k, p l ).
7.Preprocess a given convex 3-polytope P (of size n) for queries of the following type: (a) Given a query point q, is q P? (b) Given a query plane H, determine H P. (c) Given a query ray r, determine the first & lowest dimensional face of P that r intersects. [Hint: aim at O(log n) query time for parts (a) and (c), and O(K H + log n) query time for part (b), where K H is the # faces of P intersected by H. For preprocessing use a hierarchical decomposition of P similar to Kirkpatrick’s triangulation refinement method.] 8.Let P be a non-convex 3-polytope with O(n) faces. Give an algorithm that determines, in O(n) time, whether a given query point q is inside P. 9.Design & analyze a 2D randomized incremental algorithm that computes the intersection of n given half-planes. [Hint: maintain a conflict graph between vertices of the current intersection and the half-planes yet to be inserted.] 10.Design & analyze a 3D randomized incremental algorithm that computes the intersection of n given half-spaces. [Hint: maintain a conflict graph between vertices of the current intersection and the half-spaces yet to be inserted.] 11.Using Geometric duality transform, relate the problems of 3D CH versus intersection of 3D half-spaces.
12.Cauchy’s Rigidity Theorem: If two 3-dimensional convex polytopes P and P’ are combinatorially equivalent (i.e., their face lattice graphs are isomorphic) with corresponding facets being congruent, then also the angles between corresponding pairs of adjacent facets are equal (and thus P is congruent to P’). Prove this theorem.
END
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Similar presentations | 5,613 | 16,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.66384 |
https://scienceoxygen.com/how-is-concentration-expressed-in-chemistry/ | 1,669,837,588,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00055.warc.gz | 529,415,145 | 17,124 | # How is concentration expressed in chemistry?
Concentration is usually expressed in terms of mass per unit volume. However, the solute concentration may also be expressed in moles or units of volume. Instead of volume, concentration may be per unit mass.
## What are 3 ways concentration can be expressed?
Chemists can express concentrations in various ways including: Molarity (M), Parts per million (ppm), % composition, or gram/Liter (g/L).
## How do you show concentration?
It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams …
## What are the methods of expressing concentration?
• Concentration in Parts Per Million (ppm) The parts of a component per million parts (106) of the solution.
• Mass Percentage (w/w):
• Volume Percentage (V/V):
• Mass by Volume Percentage (w/V):
• Molarity (M):
• Molality (m):
• Normality.
• Formality.
## What is the simplest ways to express concentration of solutions?
The simplest way to describe the concentration of solution is mass by mass and mass by volume.
## How do you express the concentration of a solution quantitatively?
1. 1) Mass Percentage (w/w) When we express the concentration of a solution as the percent of one component in the solution, we call it the mass percentage (w/w).
2. 2) Volume percentage (V/V)
3. 3) Mass by Volume Percentage (w/V)
4. 4) Molarity (M)
5. 5) Molality (m)
6. 6) Mole Fraction.
## Which is best method for determining concentration of solution?
Molarity (M) If the weight or volume of the solvent is specified, this means that you’re interested in the amount of solvent before the solute has been added. Molarity is probably the most commonly used way of measuring concentration and is defined as the number of moles of solute per liters of solution.
## What is concentration in chemistry?
In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical description can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration.
## Which is a way to express concentration of a solution quizlet?
The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution.
## What are the 3 types of concentration?
• Percent Composition (by mass)
• Molarity.
• Molality.
• Mole Fraction.
## What is the most common way to expressing the concentration of solution?
Answer. Answer: Most commonly, the concentration of a solution is expressed as mass or volume percentage and molarity or molality, although molarity is temperature dependent, it is used quite commonly used to express the concentration of a solution.
## What is concentration in chemistry examples?
The concentration of the solution tells you how much solute has been dissolved in the solvent. For example, if you add one teaspoon to two cups of water, the concentration could be reported as 1 t salt per 2 c water. The vinegar label will report that the solution is 5% by weight acetic acid.
## What best defines concentration?
The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute.
## Which statement best describes the most concentrated solution?
Which statement best describes the most concentrated solution? It has the most solute for a given volume.
## Which of the following is not a way to describe the concentration of a solution?
Answer and Explanation: The answer is that d, enthalpy, is not a concentration unit. Mass per volume is defined as the ratio of mass of solute vs the volume of the solvent, which is the density. Molarity is defined as the ratio between the number of moles and the molar mass.
## Is a common way to express the concentration of biological solutions?
Most commonly, the concentration of a solution is expressed as mass or volume percentage and molarity or molality, although molarity is temperature dependent, it is used quite commonly used to express the concentration of a solution.
## Which of the following is used to measure concentration?
Molarity. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution.
## Is concentration the same as molarity?
Molarity is not the same as concentration, although they are very similar. Concentration is a measure of how many moles of a substance are dissolved in an amount of liquid, and can have any volume units. Molarity is a type of concentration, specifically moles per liter of solution.
## Which unit can be used to express the concentration of a solution?
Mass Percent: The mass percent is used to express the concentration of a solution when the mass of a solute and the mass of a solution is given: Mass Percent=Mass of SoluteMass of Solution×100%
## Why is concentration important in chemistry?
The concentration of a solute is very important in studying chemical reactions because it determines how often molecules collide in solution and thus indirectly determines the rates of reactions and the conditions at equilibrium (see chemical equilibrium). Concentration may be expressed in a number of ways.
## Why is it important to know the concentration of a solution?
In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions.
## Which factor is more important in concentration of solution?
Answer and Explanation: Volume of the solution in liter.
## Which of the following will decrease the concentration of a solution?
b. The volume of the solvent is inversely proportional to the concentration of the solution. Therefore, adding more solvent decreases the concentration of the solution.
## What are the three ways to measure the concentration of a solution explain?
What are three ways to measure the concentration of a solution? Concentration can be expressed as percent by volume, percent by mass, and molarity.
## What is concentration and how is it measured?
Concentration is a measurement of how much solute there is per volume of solution. In chemistry, concentration is most often expressed in molarity by units of molar (M), which is moles per liter solution. | 1,421 | 6,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-49 | latest | en | 0.894896 |
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